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https://math.stackexchange.com/questions/1933245/in-combinatorics-is-13-choose-5-the-same-as-13-choose-8 | # In combinatorics, is ${13 \choose 5}$ the same as ${13 \choose 8}$?
So I got a question in college on combinatorics, I am supposed to find a coefficient. So I do the math according to the binomial theorem and I ended up with the answer ${13 \choose 5}$. But my teacher has written down the answer as ${13 \choose 8}$. When I calculate them I notice they are in fact the same number. So is it okay I answer ${13 \choose 5}$ when the right answer is ${13 \choose 8}$ when searching for a coefficient?
Forgive me if this is a silly question I am new to combinatorics.
• They are equal, so both are (the same) valid answer. You could just write $1287$ or even $10\cdot 2^7+7$ and it would still be correct. Sep 19, 2016 at 18:40
• To choose $5$ objects out of $13$ is the same as to choose $8$ (think of putting the $5$ in an urn, then put the remaining $13-5=8$ into a second urn). In general $\binom nk=\binom n{n-k}$ for the same reason.
– lulu
Sep 19, 2016 at 18:41
• Sorry but I don't grasp that. How can choosing 5 objects out of 13 be the same as choosing 8? Makes no sense in my head... Sep 19, 2016 at 18:42
• Intuitively, select $5$ elements from $13$ elements is the same of don't choose the remaining $8$ elements from $13$. Sep 19, 2016 at 18:43
Yes, they are the same!
Why? Well, $\binom{13}{8}$ is the number of ways to select $8$ people to be in a committee, from a group of $13$ people. But we could just as well choose $5$ people to not be on the committee. Choosing $8$ people to be on the committee is the same as choosing $5$ people to leave out. So $\binom{13}{8} = \binom{13}{5}$.
In general, it is a fact that $$\binom{n}{k} = \binom{n}{n-k},$$ and this is true for the same reason as I described.
$${n \choose k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!k!}=\frac{n!}{(n-k)!(n-(n-k))!}= {n\choose n-k}$$
$13 \choose 5$ is the number of ways to select five items out of $13$. If you take five items out, you leave eight behind, so the number of ways to select five and eight are equal. This is general. If you want to choose $k$ items out of $n$, we have ${n \choose k}=\frac {n!}{k!(n-k)!}={n \choose n-k}$
It's not a silly question at all. Those two are, in fact, the same, and what you've stumbled upon is a more general pattern. There are mainly two ways of seing this.
The formula:
We have $\binom{13}{8} = \frac{13!}{8!\cdot 5!}$ and $\binom{13}{5} = \frac{13!}{5!\cdot 8!}$. They look very similar to me.
The application:
If you have $13$ balls in a box, then $\binom{13}{5}$ is the number of ways to pick out $5$ of them. However, you could just as well have chosen $8$ of them to leave in the box, and say you'd pick out whichever balls are left. That's the same thing, and thus the number of ways to do each of them should be the same.
The general statement:
We have, for any natural numbers $n \geq r$ that $\binom{n}{r} = \binom{n}{n-r}$. In your case, we have $n = 13$, $r = 8$ and $n-r = 5$. But it also tells us, for instance, that $\binom{1000}{597} = \binom{1000}{403}$, or any other such relation. | 2022-12-09T19:01:17 | {
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http://vyturelis.com/integration-and-differentiation-table.htm | ## Integration And Differentiation Table
table and the definite integral table created below. Note, however, unlike differentiation, not all func- tions can be integrated in terms of simpler functions.
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While differentiation has easy rules by which the derivative of a complicated . functions, integration does not, so tables of known integrals are often useful.
ing the integration process, Fourier develop- . Differentiation is the term used for the process of finding . Table 1 lists some power-series expressions for some .
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differentiate integrate. The situation is just a little more complicated because there are . Common integrals are usually found in a 'Table of Integrals' such as that .
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Dr. S. Kersey, Math 1441 Formula Sheet , Fall 2003. Trigonometric Identities cos2 (x) + sin2(x)=1 cot2(x) + 1 = csc2(x) cos(a ± b) = cos(a) cos(b) ∓ sin(a) sin(b).
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Applications of integration are numerous and some of these will be explored in subsequent Blocks. . table using your knowledge of differentiation. Try this for .
Sitemap | 2014-03-09T14:12:30 | {
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https://math.stackexchange.com/questions/1807207/find-all-a-b-in-bbb-z2-such-that-b-equiv-2a-pmod-5-and-28a10b-26 | # Find all $(a,b) \in \Bbb Z^2$ such that $b \equiv 2a \pmod 5$ and $28a+10b=26$
I'm stuck with this exercise:
Find all $(a,b) \in \Bbb Z^2$ such that $b \equiv 2a \pmod 5$ and $28a+10b=26$
It's from my algebra class, we are looking into diophantic and congruence equations.
I started by looking to the $(a,b)$ that would solve the equation: $14a+5b=13$, those would be of the form $a=-13+5s$ and $b=39-14s$. Here is where I don't understand what I should do.
I've looked into $a$'s congruence mod 5 and got $5s \equiv 3 (5)$.
If $b$ should be two times $a$ then it would be: $10s\equiv1(5)$. Right?
So if I'm on the right path I still don't see how should I make to combine the first solution for b and this congruence requirement. What should I try? Thanks a lot.
• Is this not modulo 2? – mvw May 31 '16 at 17:21
• Sorry! I forgot to write down mod 5. – jrs May 31 '16 at 17:28
$$28a+10b=26\iff 14a+5b=13$$
Now use mod $14$ and mod $5$: $$14a\equiv 13\pmod{5}\iff -a\equiv -2\pmod{5}$$
$$\stackrel{:(-1)}\iff a\equiv 2\pmod{5}$$
$$5b\equiv 13\pmod{14}\iff 5b\equiv 55\pmod{14}$$
$$\stackrel{:5}\iff b\equiv 11\pmod{14}$$
Therefore it's necessary that $a=5k+2$ and $b=14t+11$ for some $k,t\in\mathbb Z$. $$14a+5b=13\iff 14(5k+2)+5(14t+11)=13$$
$$\iff 70(k+t)+83=13\iff k+t=-1$$
Therefore all the solutions are given by $$(a,b)=(5k+2,14(-k-1)+11)$$
$$=(5k+2,-14k-3),\, k\in\mathbb Z$$
You're given $b\equiv 2a\pmod{5}$, i.e. $$-14k-3\equiv 2(5k+2)\pmod{5}$$
$$\iff k+2\equiv 4\pmod{5}\iff k\equiv 2\pmod{5}$$
$$\iff k=5r+2,\, r\in\mathbb Z$$
$$(a,b)=(5(5r+2)+2,-14(5r+2)-3)$$
$$=(25r+12,-70r-31),\, r\in\mathbb Z$$ | 2020-09-23T10:51:13 | {
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https://math.stackexchange.com/questions/3230787/probability-of-choosing-2-items-of-the-same-color-from-a-bucket-with-k-colors | # Probability of choosing 2 items of the same color from a bucket with k colors
I think this is right, just want to double check. Let's say I have a bucket of n items. Those n items have k colors uniformly distributed. Then the probability that I get two items from the same color is:
$$P = \frac{\binom{k}{1}\binom{n/k}{2}}{\binom{n}{2}}$$
So, for a bucket of 200 items with 10 colors, this is: $$P = \frac{\binom{10}{1}\binom{200/10}{2}}{\binom{200}{2}}$$
And the intuition being: There's a population n where I want to choose two items (denominator). From this population there are 10 colors and I want to pick 1 (numerator). And from a particular color i want to choose 2. Since there is an equal number of colors, then I can set that up as items/colors, n/k.
Are the formulation and intuition correct?
• So you choose two items and both items have the same color, right? May 18 '19 at 16:45
• Yup, both same color
– xela
May 18 '19 at 16:51
• I'm not sure but I think this is more correct- $$\displaystyle \dfrac{{n \choose 1}{{\frac{n}{k}-1}\choose 1}}{n \choose 2}$$ May 18 '19 at 17:02
• ^The above expression basically says that we first choose $1$ item from $n$ items and choose the next item with the same color as the first one. Initially there are $n/k$ items of each color. After choosing the first item we're left with $\frac{n}{k}-1$ items in that particular color. I'm not entirely sure though. And when I substitute the values in the example of 200 items and 10 colors, I ended up with $\frac{38}{199}$, is this correct? May 18 '19 at 17:06
• @ExtremeRaider Your answer is exactly twice the correct probability, because for each pair of items whose colors match, you count it twice--once when you choose one of the items first, and again when you choose the other item first. OP's method is better. May 18 '19 at 17:32
Your method is a good one if you choose exactly two items. If you chose three items and the question was whether any two were the same (so the alternative, which we don't want, is items of three different colors) then the probability would be higher. If you chose $$k+1$$ items the probability of a matching pair would be $$1.$$ That's why you get questions like, "How many items are you choosing?"
Choosing just two items from the bucket, you have (as you say) $$\binom{200}{2}$$ possible pairs of items that you could choose, and the assumption is that no pair is more likely to be chosen than any other pair, so all $$\binom{200}{2}$$ pairs are equally likely. Because of that, you just have to count the "successful" pairs and divide by the total number of possible pairs to get the probability.
Another approach is as follows. For convenience, let $$m = n/k.$$
You pick one item from the bucket. It has a color. Now you pick an item from the $$n - 1$$ items remaining in the bucket. That item is either one of the $$m - 1$$ remaining items of the color you already picked, or one of the $$(k-1)m$$ items of the other $$k-1$$ colors. The chance that you choose the same color twice is therefore $$\frac{m - 1}{n - 1} = \frac{\frac nk - 1}{n - 1}.$$
If you have $$200$$ items, of which $$20$$ are in each of $$10$$ colors, the answer comes out to $$\frac{20-1}{200-1} = \frac{19}{199}.$$
Now, recalling that $$\binom xy = \frac{x!}{(x-y)! y!} = \frac{x (x-1)(x-2) \cdots (x - y + 1)}{y!},$$ in particular $$\binom x1 = x$$ and $$\binom x2 = \frac{x(x-1)}{2},$$ your approach gives us $$\frac{\binom k1\binom{n/k}{2}}{\binom n2} = \frac{\binom k1\binom{m}{2}}{\binom n2} = \frac{k \left(\frac{m(m-1)}{2}\right)}{\left(\frac{n(n-1)}{2}\right)} = \frac{km\left(m - 1\right)}{n(n-1)} = \frac{m - 1}{n - 1} = \frac{\frac nk - 1}{n - 1}.$$ So we get the same answer either way.
• awesome! thanks for helping generalize this1
– xela
May 18 '19 at 17:33
Yes, this is the right expression. We can also say that we firstly choose one item with color $$k=1$$. The probability is $$\frac{\frac{n}{k}}{n}=\frac1k$$. Then we choose the next item with color $$k=1$$. The probability is $$\frac{\frac{n}{k}-1}{n-1}$$. Since we have k colors the probability to choose two items with same colors is $$\frac1k\cdot \frac{\frac{n}{k}-1}{n-1}\cdot k=\frac{\frac{n}{k}-1}{n-1}$$. This term is equal to $$\frac{\binom{k}{1}\binom{n/k}{2}}{\binom{n}{2}}=\frac{\frac{k}{1}\cdot \frac{n/k\cdot (n/k-1)}{2\cdot 1} }{\frac{n\cdot (n-1)}{1\cdot 2}}=\frac{k\cdot n/k\cdot (n/k-1)}{n\cdot (n-1)}=...$$ | 2021-09-20T09:16:25 | {
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https://www.jiskha.com/questions/1239455/the-rate-of-decay-in-the-mass-m-of-a-radioactive-substance-is-given-by-the-differential | # calculus
The rate of decay in the mass, M, of a radioactive substance is given by the differential equation dM dt equals negative 1 times k times M, where k is a positive constant. If the initial mass was 100g, then find the expression for the mass, M, at any time t.
1. 👍
2. 👎
3. 👁
1. dM/dt = -k M
separate variables, M left, t right
dM/M = -k dt
integrate
ln M = -k t + c
note e^log a = a
ao
e^ln M = M = e^(-kt+c) = e^-kt e^c
or since e^c is some constant call it C
M = C e^-kt
note that when t = 0
e^-kt = 1
so
C = initial amount
so here
M = 100 e^-kt
1. 👍
2. 👎
2. Damon is right
1. 👍
2. 👎
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https://math.stackexchange.com/questions/1217213/evaluate-the-integral-over-the-region-r | # Evaluate the integral over the region R
I am a bit lost on how to evaluate double integrals over a region. I am asked to evaluate the following integral
$$\iint\frac{y}{(x^2+y^2} dA$$
over the region R: triangle bounded by $y=x, y=2x, x=2$
I got $0<y<4, y<x<2$ horizontally and $0<x<2, x<y<2x$ vertically.
I wish to know if I am correct or not.
Regards,
• Can you show briefly your work? I don't think the answer should be $0$. Apr 2, 2015 at 12:38
Note that $$R = \{(x,y)\in \def\R{\mathbf R}\R^2 \mid x\le y \le 2x, 0 \le x \le 2\}$$ Hence \begin{align*} \int_R \frac{y}{x^2 + y^2} \, d(x,y) &= \int_0^2 \int_x^{2x} \frac{y}{x^2+y^2}\, dy\, dx\\ &= \int_0^2 \left[\frac12 \log(x^2 + y^2)\right]_{x}^{2x}\, dx\\ &= \int_0^2 \frac 12\log (5x^2) - \frac 12 \log(2x^2)\, dx\\ &= \int_0^2 \frac 12 \bigl(\log 5 - \log 2\bigr)\, dx\\ &= \log 5 - \log 2 & (\ne 0). \end{align*}
Your integration limits should be just $0<x<2, x<y<2x$ as you said vertically. You don't need the other set of limits. | 2023-03-22T06:35:29 | {
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http://math.stackexchange.com/questions/27989/time-until-a-consecutive-sequence-of-ones-in-a-random-bit-sequence/27991 | Time until a consecutive sequence of ones in a random bit sequence
This a reformulation of a practical problem I encountered.
Say we have an infinite sequence of random, i.i.d bits. For each bit $X_i$, $P(X_i=1)=p$.
What is the expected time until we get a sequence of $n$ 1 bits?
Thanks!
-
Though closely related, are you interested only in strings of 1's or the first time you see any string (0's or 1's) of length $n$? – cardinal Mar 19 '11 at 19:20
Only 1's. Thanks – R S Mar 19 '11 at 21:43
There is a lot of literature on such questions concerning the mean time for patterns. For your particular problem a solution can be found on page 156 of Introduction to Probability Models (10th edition) by Sheldon Ross. The formula is $$E[T]=1/p+1/p^2+\cdots+1/p^n={(p^{-n}-1)/(1-p)}.$$
As expected, this is a decreasing function of $p$ for fixed $n$: it takes longer to see rarer events. As $p$ goes from 0 to 1, $E[T]$ decreases from infinity to $n$.
Let $T$ be the random variable that records the first time we see $n$ ones in a row. Let's also define the random variable $L$ to be the position of the first zero bit in the sequence.
Looking at the first $n$ bits there are, roughly speaking, two possibilities: either I get the desired pattern of $n$ ones or I got a zero bit at time $k$ and the whole problem starts over.
More formally, conditioning on the value of $L$ we get \begin{eqnarray*} E[T] &=& \sum_{k=1}^{n} E[T \ |\ L=k]\ P(L=k) + E[T\ |\ L> n]\ P(L>n)\cr &=& \sum_{k=1}^{n} (k+E[T])\ P(L=k) + n P(L > n)\cr &=& \sum_{k=1}^{n} (k+E[T])\ p^{k-1}(1-p) + n p^n. \end{eqnarray*}
Solving this equation for $E[T]$ gives the formula.
There are many generalizations of this problem and variations on the above proof that use, for instance, Markov chains, or martingales, or generating functions, etc. In addition to Ross's book mentioned above, you may like to look at
1. Section 8.4 of Concrete Mathematics by Graham, Knuth, and Patashnik
2. Chapter 14 of Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell
3. Section XIII 7 of An Introduction to Probability Theory and Its Applications by Feller
-
Thanks. As I don't have access to the book, can you direct me to some pointers to the material and terms involved? – R S Mar 19 '11 at 22:26
@RS I have added some new material and references. Happy hunting! – Byron Schmuland Mar 20 '11 at 1:33
Thanks a lot, it's very helpful! – R S Mar 20 '11 at 7:49
I have posted a generating function approach to this result in an addendum to my answer to a similar question. Perhaps I should expand and move the addendum here as an answer. – robjohn Sep 16 '12 at 15:38
Here is a generating function approach to this problem.
Preliminaries
For $0\le k<n$, let an atom $a_k$ be a sequence of $k$ $1$ bits followed by a $0$ bit. Let a terminal atom $a_n$ be a sequence of $n$ $1$ bits. Let a complete $n$ sequence be a binary sequence which ends at the first subsequence of $n$ consecutive $1$ bits. Any complete $n$ sequence can be written as a unique sequence of atoms followed by a terminal atom. The probability of an atom occurring is $p^k(1-p)$ and the length of an atom is $k+1$. The probability of a terminal atom occurring is $p^n$ and the length of a terminal atom is $n$.
Therefore, for each atom $a_k$, define the monomial $\phi_x(a_k)$ as $$\phi_x(a_k)=\left\{\begin{array}{ll}p^k(1-p)x^{k+1}&\text{if }0\le k< n\\p^nx^n&\text{if }k=n\end{array}\right.\tag{1}$$ and for the concatenation of two binary sequences, $s_1$ and $s_2$ $$\phi_x(s_1s_2)=\phi_x(s_1)\phi_x(s_2)\tag{2}$$ The probability of any sequence of atoms $s$ occurring is the coefficient of the monomial $\phi_x(s)$ and the length of $s$ is the exponent of $x$ in $\phi_x(s)$.
The Generating Function
Thus, the probability of a complete $n$ sequence with length $k$ coming from the concatenation of $j$ atoms and a terminal atom is the coefficient of $x^k$ in \begin{align} &(\phi_x(a_0)+\phi_x(a_1)+\phi_x(a_2)+\dots+\phi_x(a_{n-1}))^j\phi(a_n)\\ &=p^nx^n((1-p)x)^j\left(1+px+p^2x^2+\dots+p^{n-1}x^{n-1}\right)^j\\ &=p^nx^n\left((1-p)x\frac{1-p^nx^n}{1-px}\right)^j\tag{3} \end{align} Summing over all $j$, the probability of a complete $n$ sequence with length $k$ is the coefficient of $x^k$ in \begin{align} f_n(x) &=\sum_{j=0}^\infty p^nx^n\left((1-p)x\frac{1-p^nx^n}{1-px}\right)^j\\ &=\frac{p^nx^n}{1-(1-p)x\frac{1-p^nx^n}{1-px}}\tag{4} \end{align} That is, $f_n(x)$ is the generating function for the probability of a complete $n$ sequence with a given length.
Since $f_n(1)=1$, the probability that a complete $n$ sequence occurs eventually is $1$.
Expected Length $$f_n(x)=\sum_{k=0}^\infty c_kx^k\tag{5}$$ where $c_k$ is the probability of a complete $n$ sequence with length $k$. Therefore, $xf_n^\prime(x)$ evaluated at $x=1$ is $$\sum_{k=0}^\infty kc_k\tag{6}$$ which is the expected length of a complete $n$ sequence.
$$xf'(x)=p^nx^n\frac{n\frac{1-x}{1-px}+x\frac{1-p}{1-px}\frac{1-p^nx^n}{1-px}}{\left(1-(1-p)x\frac{1-p^nx^n}{1-px}\right)^2}\tag{7}$$ Evaluating $(7)$ at $x=1$ yields that the expected length of a complete $n$ sequence is $$\mathrm{E}(k)=\frac{1-p^n}{p^n(1-p)}\tag{8}$$
Expected Variance
Using the same idea that was used to get $(6)$, we "take the derivative and multiply by $x$" twice to see that $xf_n^\prime(x)+x^2f_n^{\prime\prime}(x)$ evaluated at $x=1$ is $$\sum_{k=0}^\infty k^2c_k\tag{9}$$ which is the expected square of the length of a complete $n$ sequence. \begin{align} xf_n^\prime(x)+x^2f_n^{\prime\prime}(x) &=\frac{p^nx^n}{(1-px)^3\left(1-(1-p)x\left(\frac{1-p^nx^n}{1-px}\right)\right)^3}\\ &\times\left( \begin{aligned} &n^2(1-x)(1-px)\left(1-x-(1-p)xp^nx^n\right)\\ &+2n(1-p)x\left(1-x-(2-x-px)p^nx^n\right)\\ &+(1-p)x(1-p^nx^n)\left(1+x-(1-p)xp^nx^n\right)\\ \end{aligned} \right)\tag{10} \end{align} Evaluating $(10)$ at $x=1$ yields that the expected square of the length of a complete $n$ sequence is $$\mathrm{E}(k^2)=\frac{2(1-p^n)}{p^{2n}(1-p)^2}-\frac{1+2n-p^n}{p^n(1-p)}\tag{11}$$ Combining $(8)$ and $(11)$, we get that the expected variance of the length of a complete $n$ sequence is \begin{align} \mathrm{Var}(k) &=\mathrm{E}(k^2)-\mathrm{E}(k)^2\\ &=\frac{1-p^{2n+1}}{p^{2n}(1-p)^2}-\frac{2n+1}{p^n(1-p)}\tag{12} \end{align}
- | 2016-04-30T17:40:59 | {
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https://math.stackexchange.com/questions/3040135/maximize-det-x-subject-to-x-ii-leq-p-i-where-x0 | # Maximize $\det X$, subject to $X_{ii}\leq P_i$, where $X>0$
Given $$P_1,P_2,\cdots,P_N$$.
$$\begin{array}{ll} \text{maximize} & \det X\\ \text{subject to} & \mathrm X_{ii}\leq P_i \\\forall i=1,2,\cdots,n\end{array}$$
$$X\in\mathbb{R}^{n\times n}$$, $$X>0$$ (i.e. positive definite).
For $$n=2$$, maximum is achieved when $$X_{ii}=P_i$$ and $$X$$ is diagonal matrix.
Then for $$n=3$$, Let $$X=\begin{bmatrix}X_1 & X_2\\X_2^T &x_3\end{bmatrix}$$, where $$X_1\in\mathbb{R}^{2\times2}$$.
$$\det(X)=\det(X_1)\times \det(x_3-X_2^TX_1^{-1}X_2)\leq \det(X_1)\times x_3\leq P_1P_2P_3$$.
Then following same logic my conjecture is $$\max\{ \det X\}=P_1\cdots P_n$$, when $$X_{ii}=P_i$$.
Is it correct? If not, can you please point my mistake or give me counter example? Thanks
Your problem is equivalent to maximizing $$\log \det X$$ such that $$X_{ii} \leq P_i$$ and $$X$$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is $$L(X,\lambda) = \log\det X - \sum_i \lambda_i (X_{ii} - P_i)$$ The derivative of $$\log\det X$$ is $$(X^{-1})^T$$, so the KKT conditions are: $$(X^{-1})_{ii} - \lambda_i = 0$$ $$(X^{-1})_{ij} = 0 \quad (i \neq j)$$ $$\lambda_i (X_{ii} - P_i)=0$$ $$X_{ii} \leq P_i$$ $$\lambda \geq 0.$$ The point you found satisfies these conditions and is therefore optimal.
• what if rank$X< n$? I think in this case these points are not optimal – Lee Dec 18 '18 at 9:24
• @Lee a positive definite matrix has full rank. – LinAlg Dec 18 '18 at 15:13 | 2020-01-25T23:27:20 | {
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https://math.stackexchange.com/questions/3422427/alternating-series-sum | Alternating Series sum
Given $$\sum_{n=1}^{\infty} \frac{(-1)^n}{2^n}$$ determine ig the series converge or diverge. If convergent calculate the sum.
Using the convergence test I've found out that the series converges. However, I don't know how to go about finding the sum. Any ideas?
• It's geometric. – Matthew Leingang Nov 5 '19 at 2:31
• Note that $\frac{(-1)^n}{2^n} = \left(-\frac{1}{2}\right)^n$. – KM101 Nov 5 '19 at 2:33
• Welcome here! Which convergence test did you use? – Taladris Nov 5 '19 at 2:48
• @Taladris the alternant series test – Max Nov 5 '19 at 2:49
Recall: the geometric series $$\sum_{n=0}^\infty x^n$$ is absolutely convergent with value
$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
whenever $$|x| < 1$$. To find the value for your series, notice that you can "pull out" the general series' first term like this:
$$\sum_{n=0}^\infty x^n = 1 + \sum_{n=1}^\infty x^n$$
Then notice your series has $$x = -1/2$$, apply the formula for the series, and solve for yours to obtain its value.
More explicitly,
$$\sum_{n=1}^\infty x^n = \frac{1}{1-x} - 1\;\;\;\;\; \overset{x=-1/2}{\implies} \;\;\;\;\; \sum_{n=1}^\infty \left( - \frac 1 2 \right)^n = - \frac 1 3$$
• I think it's supposed to be $-\dfrac13$. – Don Thousand Nov 5 '19 at 2:44
• Yeah I made a sign error. My bad; thanks – Eevee Trainer Nov 5 '19 at 2:45
• Oooh, I knew $\sum_{n=1}^{\infty}ar^{n-1} = \frac{a}{1-r}$. But the original had $ar^{n}$, when I encounter a problem like that should I just work with $\sum_{n=0}^{\infty}r^{n} = \frac{1}{1-r}$? I just wasted time playing around with the argument to adapt it to $ar^{n-1}$ haha. – Max Nov 5 '19 at 2:57
• I feel like I was told a similar formula for stuff like that in middle school. I feel like $$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$ is easier to remember, though, and its manipulations are more intuitive as well. Especially considering it amounts to $$1+r+r^2 + r^3 + \cdots$$ when expanded. Your first formula is almost equivalent to my simpler series: just re-index it (let $m=n-1$ and rewrite your series) and divide both sides by $a$. Yours basically accounts for series whose first term is not $1$ (but that can be accounted for by factoring anyways). – Eevee Trainer Nov 5 '19 at 5:21
Let $$a_n = \frac{(-1)^n}{2^n}=\left(-\frac{1}{2}\right)^n$$
Then $$|a_n| = \frac{1}{2^n}$$ and series $$\sum_{n=1}^\infty |a_n|$$
converge because it is geometric series.
So $$\sum a_n$$ is absolutly converge and this make $$\sum a_n$$ also converge.
And now we use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$
put $$x=-\frac{1}{2}$$ and we get
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\left(-\frac{1}{2}\right)^n = \frac{-\frac{1}{2}}{1+\frac{1}{2}}=\boxed{-\frac{1}{3}}$$ | 2020-07-09T15:32:20 | {
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https://math.stackexchange.com/questions/3769643/is-it-true-that-sum-i-0n-1-2i-divides-sum-i-n2n-1-2i | # Is it true that $\sum_{i=0}^{n-1} 2^i$ divides $\sum_{i=n}^{2n-1} 2^i$?
Some investigation suggested to me that
$$\sum_{i=0}^{n-1} 2^i \Bigg| \sum_{i=n}^{2n-1} 2^i$$
The point being that binary numbers in the form exclusively of an even number of $$1$$s appear to be divisible by the number represented by just half that number of $$1$$s.
Is this the case, and if so why is that please?
Note you have
\begin{aligned} \sum_{i=n}^{2n-1} 2^i & = 2^n + 2^{n+1} + \ldots + 2^{2n-1} \\ & = 2^n + 2^n(2) + \ldots + 2^n(2^{n-1}) \\ & = 2^n(1 + 2 + \ldots + 2^{n-1}) \\ & = \sum_{i=0}^{n-1} 2^n(2^i) \\ & = 2^n\left(\sum_{i=0}^{n-1} 2^i\right) \end{aligned}\tag{1}\label{eq1A}
• I'm quite new to summation notation. Could you possibly elaborate on the steps from each form to the next here please? – Robin Andrews Jul 26 at 9:29
• @RobinAndrews This is the distributive law. – Angina Seng Jul 26 at 9:37
• @RobinAndrews I expanded the summation and showed that each term has a factor of $2^n$. Removing that common factor, you have then the sum of the powers of $2$ from $0$ to $n - 1$. I hope this makes it more clear to you. – John Omielan Jul 26 at 9:39
Well, you don't even need to know something about summation.
As you noticed, the binary representation of the first number (i.e., $$\sum_{i=0}^{n-1}2^i$$), say $$s$$, is made only by $$1$$s, so if you add the number 1 you'll get a $$1$$ followed only by $$0$$s, namely a power of $$2$$: which one? The number $$s$$ has $$n$$ digits, and then, due to the carries, $$s+1$$ has $$n$$ digits and so it is $$2^n$$, i.e., the first power of two greater than $$2^{n-1}$$.
The same argument show that the value of the second sum is $$2^{2n}-2^n$$, indeed it is the sum of all the powers of two from $$2^0$$ to $$2^{2n-1}$$, which is $$2^{2n}$$, minus all the powers of two from $$2^0$$ to $$2^{n-1}$$, which is $$2^n$$. Now $$2^{2n}-2^n=2^{n}(2^n-1)$$, which is what you wanted to prove.
If this is clear, now you're able the extend this result also to the powers of any number. Indeed, for a natural number $$a>1$$, consider the following equalities (first you have to use distributivity, then the "trick" is to detach the first or the last addend of the sums from the others, i.e., to use the associative property): $$\begin{split}(a-1)\sum_{i=0}^{n-1}a^i=& a\sum_{i=0}^{n-1}a^i-\sum_{i=0}^{n-1}a^i=\sum_{i=1}^{n}a^i-\sum_{i=0}^{n-1}a^i=\\&=\left(\sum_1^{n-1}a^i+a^n\right)-\left(a^0+\sum_{i=1}^{n-1}a^i\right)=a^n-1.\end{split}$$ Now, dividing by $$a-1$$, you have a closed form for the summation $$\sum_{i=0}^{n-1}a^i$$, namely $$\sum_{i=0}^{n-1}a^i=\frac{a^n-1}{a-1}$$ (notice that we're not using the fact that $$a$$ is a natural number, we only need $$a\neq 0$$, so the formula holds for $$a\in\mathbb{R}\setminus\{0\}$$).
Knowing this, you have $$\sum_{i=n}^{2n-1}a^i=\sum_{i=0}^{2n-1}a^i-\sum_{i=0}^{n-1}a^i=\frac{a^{2n}-1}{a-1}-\frac{a^n-1}{a-1}=\frac{a^{2n}-a^n}{a-1}=a^n\frac{a^n-1}{a-1}=a^n\sum_{i=0}^{n-1}a^i.$$ In this way, not only you can prove that $$\sum_{i=n}^{2n-1}a^i$$ is divisible by $$\sum_{i=0}^{n-1}a^i$$, but also that the quotient is exactly $$a^n$$.
Lastly, we can came back to your original intuition. Given a natural number $$a>1$$, the $$a$$-ary representation of a power of $$a$$, say $$a^j$$, is made by a $$1$$ followed by $$j$$ $$0$$s, so the $$a$$-ary representation of the sum $$a^0+a^1+\ldots +a^{n-1}$$ is simply made by $$n$$ consecutive $$1$$s. For the same reason, the representation of $$a^n+a^{n+1}+\ldots +a^{2n-1}$$ is made by $$n$$ consecutive $$1$$s followed by $$n$$ consecutive $$0$$s. Now, you can perform the long division between those two numbers: you'll have something like this
$$\require{enclose} \begin{array}{r} 1\overbrace{00\ldots 0}^{n\text{ digits}} \\[-3pt] \underbrace{11\ldots 1}_{n\text{ digits}} \enclose{longdiv}{\underbrace{11\ldots 1}_{n\text{ digits}}\underbrace{00\ldots 0}_{n\text{ digits}}} \\[-3pt] \underline{11\ldots 1}\phantom{00\ldots 0} \\[-3pt] 00\phantom{0\ldots 0} \\[-3pt] \underline{0}\phantom{0\ldots 0} \\[-3pt] 00\phantom{\ldots 0} \\[-3pt] \ldots \end{array}$$
so the division is exact and the quotient is the number with $$n+1$$ digits $$10\ldots 0$$ in base $$a$$, that is $$a^n$$, as expected.
The same argument proves the observation made by probably_someone, i.e., that a number consisting of $$n$$ consecutive $$1$$s followed by any number of consecutive $$0$$s is divisible by a number consisting of $$n$$ consecutive $$1$$s. Actually, we can say something more: for any $$k\in\mathbb{N}$$, consider $$a_1,a_2,\ldots ,a_k\in\mathbb{N}$$, then a number consisting of $$na_1$$ consecutive $$1$$s followed by any number of consecutive $$0$$s followed by $$na_2$$ consecutive $$1$$s followed by any number of consecutive $$0$$s ... followed by $$na_k$$ consecutive $$1$$s followed by any number of consecutive $$0$$s is divisible by a number consisting of $$n$$ consecutive $$1$$s.
• I'm having trouble following your argument. "As you noticed...namely a power of 2, so the first sum is the first power of two greater than 2n−1, i.e. 2n." I get that "all 1s" is 1 less than the next power of 2, but how does this imply that the sum is the next power of 2? What am I missing here please? – Robin Andrews Jul 26 at 17:05
• @RobinAndrews I've made an edit, in short: just count the digits of that number. – user6530 Jul 26 at 19:37
In binary, multiplying a number by 2 is equivalent to shifting all of its bits one place higher and adding a zero on the end. Consequently, multiplying by $$2^n$$ just means doing that $$n$$ times, moving the bits $$n$$ places higher and adding $$n$$ zeros.
The object is to prove that a number consisting of $$n$$ consecutive ones and $$n$$ consecutive zeros is divisible by the number consisting of $$n$$ consecutive ones.
Knowing what we know about multiplication by $$2$$ in binary, it should be straightforward how to get from the smaller number to the larger one: just shift its digits $$n$$ places higher and add $$n$$ zeros; in other words, multiply by $$2^n$$. Since we can multiply the smaller number by an integer to get to the larger number, this means that the larger number is divisible by the smaller one.
In fact, this method can be used to prove a stronger statement: a number consisting of $$n$$ consecutive ones followed by any number of consecutive zeros is divisible by a number consisting of $$n$$ consecutive ones. | 2020-10-30T07:56:24 | {
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https://math.stackexchange.com/questions/1426163/various-probabilities-given-three-dice-with-different-colors | # Various probabilities given three dice with different colors
Here is a set of probability questions. I am having trouble with the last three.
Assume that you have three six-sided dice. One is yellow (Y), one is red (R) and one is white (W).
A)What is the probability that, in rolling only the red die, it will show the number 5?
B)What is the probability that, in rolling only the white die, it will show an even number?
C)What is the probability that, in rolling all three dice at once, the outcome will be 2(Y), 4(R), 6(W)?
D)What is the probability that, in rolling all three dice at once, the outcome will be 1, 3 and 5 in any color combination?
E)What is the probability that, in rolling all three dice at once, the outcome will be the same number on all three dice?
F)What is the probability that, in rolling all three dice at once, the outcome will be a different number on all three dice?
Here is what I have so far:
D) From part C I found that P for any # is $1/6$ thus $P(2Y,4R,6W)=(1/6)^3=1/216$. Therefore, same logic should apply here with the exception that there will be a lot more combinations w/o restriction on color. Since we have $3$ color boxes there would be total of $9$ combinations possible and thus $P=(1/6)^9$ but its not the correct answer. On the contrary, the right answer is $P=6(1/216)=1/36$ I DO NOT UNDERSTAND why we multiply by $6$.
E) here since they ask for the same number the probability of getting say $(2,2,2)$ combo is $P= 1/6 \cdot 1/6 \cdot 1/6 = 1/216$ thus having $6$ #'s on the dice the total probability of having all the same #'s is $P = 1/216+...+1/216=6/216=1/36$
F) I dont understand as well, since we found in E that having the same #'s on all three dice is $1/36$ then would not $P(\text{not the same #'s})= 1-1/36=35/36$? but for some reason it's not the correct answer
I would appreciate if you could explain your reasoning, I'm new to probability and pretty much forced to learn it for genetics problems.
D)
In the order 1,3,5, $Pr = \dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{216}$
but there could be $3! = 6$ different orders, so multiply by 6
E)
First number can be anything, each of the next two have to be different, hence $\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{36}$
F)
6 ways to get first number, 5 ways for second one, and 4 ways for third,
hence $\dfrac{6*5*4}{216} = \dfrac{5}{9}$
Note that the complement of all same is not all different, it is all not same
An easy way to think about (D) is that, rolling any die, you have a 3 in 6 chance of hitting a 1,3, or 5. Then, rolling the 2nd die, you have a 2 in 6 chance of hitting one of the not-yet-attained numbers (i.e., if the 1st was a 5, of hitting 1 or 3) Finally, the third die has a 1 in 6 chance of hitting the remaining number. 3/6 * 2/6 * 1/6 = 1/36
P.S. 1 in 216 would be the chance that a specific color (e.g., the red die) matches 1, the white matches 3, and the yellow matches 5. These numbers can be permuted in six different ways: R1W3Y5 R1Y3W5 W1R3Y5 W1Y3R5 Y1R3W5 Y1W3R5... 6 of the 216 possible unique outcomes fulfill the requirement, and 6/216 is 1/36.
Do the same thing for (E): the first die will not disqualify a roll (100% success, or 6/6). The second has a 1/6 chance of matching it, and the third also has a 1/6 chance of matching the first two. 6/6 * 1/6 * 1/6 = 1/36
For (F), it's 6/6 success for the first die, and 5/6 success for the 2nd (i.e., a 1 in 6 chance of being equal, therefore a 5 in 6 chance of being unequal). The third die has a 2 in 6 chance of being equal to any of the two already present (different-from-one-another) values, or a 4 in 6 chance of fulfilling the requirements. 6/6 * 5/6 * 4/6 = 120/216 = 20/36 = 5/9.
The reason you might be confused regarding (F) is that there are a handful of cases where neither (E) nor (F) is fulfilled. Think of the result 3,4,4, or 1,1,2... two dice the same, one different. Those are the cases not covered by E and F: 100% - 7/12 = 5/12 | 2022-01-28T19:32:20 | {
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http://immoplus24.de/tcl-roku/taylor-cubic-approximation-formula.html | # Taylor cubic approximation formula
taylor cubic approximation formula fderiv. Taylor’s Inequality can then any constant a, the Taylor polynomial of order rabout ais T r(x) = Xr k=0 g(k)(a) k! (x a)k: While the Taylor polynomial was introduced as far back as beginning calculus, the major theorem from Taylor is that the remainder from the approximation, namely g(x) T r(x), tends to 0 faster than the highest-order term in T r(x). Find the best quadratic approximation at x = 0. The kernel of Q-f(x0) is 2 near the origin Use Taylor's formula for f(x,y) at the origin to find quadratic and cubic approximations of f(x,y) = 7- x-V The quadratic approximation for f(x,y) is. 3 Solution for use Taylor’s formula for ƒ(x, y) at the origin to findquadratic and cubic approximations of ƒ near the origin. The idea of Taylor expansion approximation is then to form a polynomial For example, let us consider the tangent function tan(x). For the functions f(x) and P(x) given below, we’ll plot the exact solution and Taylor approximation using a Scilab script. 3. It’s just a straight line and goes in green on the graph. unm. If we want to do the cubic approximation then we need to evaluate the cubic term in the series. It's this, basically: tanh(x) = sinh(x)/cosh(x) = (exp(x) - exp(-x))/(exp(x) + exp(-x)) = (exp(2x) - 1)/(exp(2x) + 1) Combine this with a somewhat less accurate approximation for exp than usual (I use a third-order Taylor approximation below), and you're set. For example, consider the plot in Figure 1. Notes on the symbols used in the formula:! is the factorial symbol). As we shall see, simply finding the roots is not simple and constitutes one of the more difficult problems in numerical analysis. why are the mechanical and other laws so simple? Approximation with Taylor polynomials Taylor Series, Radius of Convergence Pade (Rational Function) Approximation Gram-Schmidt orthogonalization procedure Taylor Polynomials,Planck's Law and Rayleigh-Jeans Law Convergent or divergent series Differential Equation Application Numerical Analysis - Simpson's Rule Algorithm 6. f′(x) = the first derivative. Linear approximation Linear approximation uses the tangent line to the graph of a function to approximate the function. One of the most important applications of trigonometric series is for situations involving very small angles (x<<1). In Preview Activity 8. 001. A calculator for finding the expansion and form of the Taylor Series of a given function. (a) Approximate f by a Taylor polynomial with degree n at the number a. f (n) (0) are the n th derivatives of f(x) evaluated at x = 0 . APPROXIMATION We solve the least squares approximation problem on only the interval [−1,1]. Approximating functions by Taylor Polynomials. Determine the 1st- and 2nd-degree Taylor polynomial approximations, L(x,y) & Q(x A Taylor series approximation uses a Taylor series to represent a number as a polynomial that has a very similar value to the number in a neighborhood around n a polynomial Pn(x) which is the “best nth degree polynomial approximation to f( x) near x = a You can always compute a Taylor polynomial using the formula. Taylor's polynomials. Suppose we are trying to find the minimum of a function f(x) and we have three initial approximations to that minimum, x 1, x 2, and x 3. - The Taylor Series and Other Mathematical Concepts Overview. For that, if we write: g(x) = f(x0)+a*f'(x0)(x-x0)+b)f''(x0)(x-x0)(x-x0) and we derive twice, we get g''(x) = f''(x0)*2a g''(x0)=f''(x0) 2a=1 a=1/2. g. The more terms we add on, the more accurate the polynomial approximation will be. ERROR ESTIMATES IN TAYLOR APPROXIMATIONS Suppose we approximate a function f(x) near x = a by its Taylor polyno-mial T n(x). Example 10. The approximation of the sine function by polynomial using Taylor's or Maclaurin's formula: Example: Let represent the sine function f (x) = sin x by the Taylor polynomial (or power series). The cubic polynomial Eq. Higher Order Approximations. Figure 2. Compare sinx to its cubic May 26, 2020 · In this section we will discuss how to find the Taylor/Maclaurin Series for a function. The red line is cos(x), the blue is the approximation (try plotting it yourself) : 10. While this procedure is fairly reliable, it did involve an approximation. 9. e. Here’s an example. 1 Introduction semester, we discussed a linear approximation to a function. Theorem: If g(r)(a Taylor Polynomials A graphical introduction Best first order (linear) approximation at x=0. Consider the function f(x) = 2 6 4 0 1 x 0:2 1 5jxj 0:2 <x <0:2 0 0:2 x 1:0 We can easily verify that we cannot t the above data with any polynomial degree! P(x) = 1 26x2 + 25x4 The Cubic Formula The quadratic formula tells us the roots of a quadratic polynomial, a poly-nomial of the form ax2 + bx + c. Third derivatives go to nding a cubic approximation about some point. $\begingroup$ @Idonknow The approximation is based on the first order Taylor Series approximation of a function of d1. We use approximations via trun-cated Euler productsandthusderivee ectivemethodsof computing the orderoftheJacobian ofacubicfunction eld. 1 f(x) = ln (1 + 2x), From the graphing calculator, we get M = 6; see Figure 1. 3 Least Squares Approximations It often happens that Ax Db has no solution. 10. Thus, we Apr 07, 2009 · Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the Free Taylor/Maclaurin Series calculator - Find the Taylor/Maclaurin series representation of functions step-by-step This website uses cookies to ensure you get the best experience. The goal of a Taylor expansion is to approximate function values. , a cubic approximation of a function over a time interval, with an interval remainder. This is a wrong conclusion, as it is shown in this paper that Kmenta’s Taylor approximation for the two-input The point is because the function has a good analytic (even polynomial!) approximation, its Taylor series will be well-behaved. OZ calls this straight line function P1(x). Taylor series approximates a complicated function using a series of simpler polynomial functions that are often easier to evaluate. 1 Linear Approximations We have already seen how to approximate a function using its tangent line. This is the Taylor expansion of about . • Orders of accuracy may vary due to the accuracy of the interpolating function varying. T. The formula for T. 1 Quadratic Interpolation of Inaccurate Data Estimate f(0) by interpolating the data x 1 2 3 f(x P 0, P 1, P 2, . In this case, the coe cient 1 10 is smaller than the others. 1 Example: Plotting a function Starting MATLAB: Windows: search for MATLAB icon or link and click Linux: % ssh linux. LAGRANGE_INTERP_1D , a C code which defines and evaluates the Lagrange polynomial p(x) which interpolates a set of data, so that p(x(i)) = y(i). Thus, for. 2 Write down the approximation formula of degree 5 for a general function that is 5 times differentiable, and apply it explicitly for the sine function at x 0 = 0. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. Example: Let's approximate the function f(x)=sin(x) with a polynomial of order 3, around the point x0 In the third example we see that ex has the same numerical value as 1 + x + x2. Solve for each coefficient; Evaluate the total error, decide if it is good enough. Differential calculus and integral calculus are connected by the fundamental theorem of calculus , which states that differentiation is the reverse process to integration . The Result: the Taylor Formula. 33]. There are more equations than unknowns (m is greater than n). $\endgroup$ – cmaster - reinstate monica Jul 4 at 21:43 Taylor Polynomials. – Bend wood piece so that it passes through known points and draw a line through it. A better approximation. We have 2x @ @x +y @ @y 3 f (a,b) = 3x fxxx +3x yfxxy +3xy2fxyy +y3fyyy It turns out that you can easily get the coecients of the expansion from Pascal’s Definition. By the way, there is a special name for the Taylor series expanded at x = 0, which is named Maclaurin Series. Approximating Best second order (quadratic) approximation at x=0. The sum of partial series can be used as an approximation of the whole series. Despite this shift from the use of rational approximations, polynomial quotients such as Pade approximations[6] still receive considerable attention. For negative values of , the expression is the reciprocal of a polynomial that converges to from above (the green, yellow, and orange curves are the reciprocals of polyn Jun 24, 2011 · By the way, we are not required to prove the formula for Taylor series. 6101 0. Let P2(x) = a0 + Chapter 14. is called the mth-order Taylor polynomial of f about the point x. On the other hand, if we use the exact formula 3. R. We also derive some well known formulas for Taylor series of e^x , cos(x) and sin(x) around x=0. If we want to do the cubic approximation then we need to evaluate the cubic 1 Apr 2017 Taylor Polynomials - Cubic Approximation 2 Introduction, Basic Review, Factoring, Slope, Absolute Value, Linear, Quadratic Equations. According to wikipedia, the aim of Taylor Series Expansion (TSE) is to represent a function as an infinite sum of terms that are derived from the values of that function's derivatives, which in turn are evaluated at some predefined single point. I want to find the quadratic and cubic approximations for this, using Taylor's formula at the 0. We can use the first few terms of a Taylor Series to get an approximate value for a function. Well, okay, there’s a little more to it than that. This appendix derives the Taylor series approximation informally, then introduces the remainder term and a formal statement of Taylor's theorem. 1) =ln(1 + 1 (1 1 1)2 10) ≈ 10 − 2 10 0095. 0 3 pn = 0. Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series. function so that it coincides with the secant line on the interval [x 0,x 1]. Sep 22, 2020 · Once you get past the terms with a cubic in it, the Taylor series for a cubic polynomial is identically the same as the original cubic. 3 ! h. Suppose we want to approximate the value of e, say to within an error of at most 0. These cookies do not store any personal information. , in 50 and 100 years). These are the first two terms of the previous formula. Y -1 1 3 35 . Sep 25, 2015 · Refer to explanation The quadratic Taylor approximation is q(x,y)=f(0,0) + (df)/dx(0,0) x + (df)/dy(0,0) y +(1/(2!)) *[(d^2f)/(d^2x)(0,0) x^2 + 2 (d^2f)/(dxdy)(0,0 Solution for use Taylor’s formula for ƒ(x, y) at the origin to findquadratic and cubic approximations of ƒ near the origin. For a function , which is times differentiable in. This MATLAB function approximates f with the Taylor series expansion of f up to the term, so taylor approximates this expression with the fourth-degree polynomial: For example, approximate the same expression up to the orders 8 and 10:. Main Article: Taylor Series Approximation Imagine that you have been taken prisoner and placed in a dark cell. Complex numbers are explained in some detail, especially in their polar form. As the optimization progresses to different areas of the design space, new fits are created. Using this process we can approximate trigonometric, exponential, logarithmic, and other nonpolynomial functions as closely as we like (for certain values of $$x$$) with polynomials. ƒ The interpolating function f is used to replace or sim-plify the original function g with certain smooth property preserved at the discrete interpolation nodes and their neighborhood. Continuing in the same fashion for each other pair of points we can replace our discrete-time function with a continuous-time approximation, defined piecewise by a number of first-order polynomials. Algorithm 6. Check the box First degree Taylor polynomial to plot the Taylor polynomial of order 1 and to compute its formula. Example 1 Taylor Polynomial Expand f(x) = 1 1–x – 1 around a = 0, to get linear, quadratic and cubic approximations. Ask Question Asked 4 years, 5 months ago. 3 Accuracy of these Approximations. Interpolation (scipy. This approximation has a simple form yet is very accurate. A number of authors have attempted polynomial and non-polynomial spline Back in ancient times (c. For such angles, the trigonometric functions can be These approximations are only reasonable when x is near a. On cubic Pade Approximation to the exponential function and its application in´ solving diffusion-convection equation Jing-Hua Gao, Mei-Yan Lin School of Science, Dalian Jiaotong University, Dalian, 116028, P. Can you approximate in your head? Yes, you can! How? Like this: Bingo! 4. 7 Jun 2020 The Taylor Series is the name for the series approximation of functions, and it can be found by the following formula. In Chinese mathematics, this was improved to approximations correct to what corresponds to about seven decimal digits by the 5th century. If we know the function value at some point (say f (a)) and the value of the derivative at the same • Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. Taylor approximations to sin(x) In class, we've discussed how truncating the Taylor series of a function gives us a polynomial approximation to that function, and that higher order truncations lead to more accurate approximations. Apr 07, 2009 · Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the Approximations. The linear approximation of a function is very useful in science and engineering, and you will probably see it pop up a lot because it simplifies math so much. China Abstract Diagonal cubic Hermite-Pad´e approximation to the expo-nential function with coefficient polynomials of degree at $\begingroup$ But is the taylor expansion of a sum of functions also the sum of the taylor expansions of the two functions if one is a function of x and the other a function of y? $\endgroup$ – Joogs Sep 28 '15 at 22:05 Cubic Splines • Piecewise cubic splines are quite popular because of their ability to match derivatives across approximation boundaries • B-splines– hierarchical family: ! " # is a piecewise polynomial of degree $• Piecewise constant: !" % & = 1 for & ∈ [&",&",-] and 0 otherwise • A linear 0 " # & = 1213 1345213 If interest lies in reporting absolute measures of risk from time-to-event data then obtaining an appropriate approximation to the shape of the underlying hazard function is vital. 1]#. sin Jul 12, 2014 · hi everyone , i don't understand these steps for Taylor Expansion , it has used for state space equations the equations are the approximations for sin and cos the equation for Taylor series is ( i don't understand at all ) please help me if you can Homework Taylor polynomials are also used frequently in physics. Mar 18, 2020 · Use Perturbation Theory to add cubic and quartic perturbations to the SHO and find the first three SHO energy levels. This can be done using the Taylor theorem and/or by using Taylor model operations [13]. In this method, the slope of the curve is determined at each given point locally, and each polynomial representing a portion of the curve between a pair of given n-input translog function; i. 1 Find the Taylor We can get an even better approximation The Taylor polynomial of degree three (the cubic 3. A good interpolation polynomial needs to provide a relatively accurate approximation over an entire interval, and Taylor polynomials do not generally do this. We can then simply differentiate the interpolating function and evaluate it at any of the nodal points used for interpolation in order to derive an approximation for the pth derivative. Three approximations to the minimum of cos(x) near x = π. 2 Definitions of Approximations. Any affine function gsuch that wt(f+ g) = NLf is called a best affine approximation of fand is denoted by λf, whereas the set comprising of best affine approximations of fis denoted by Af ⊆ R(1,n). Let f∈ Bn and h∈ R(1,n). As can be seen from this figure, the approximation cannot be distinguished from the actual function. taylor approximation Evaluate e2: Using 0th order Taylor series: ex ˇ1 does not give a good fit. Answer y(0:1) ˇ 1. 1 , we begin our exploration of approximating functions with polynomials. Examples Taylor Series Expansions A Taylor series expansion of a continuous function is a polynomial approximation of . Here’s the formula for […] approximations quickly and easily. Use Taylor's Formula to find a cubic approximation to f(x, y) = xey at the point (0, 0). Question: Quadratic AND cubic approximations Use Taylor's formula for f(x, y) at the origin to find quadratic and cubic approximations of {eq}f(x, y) = 5x e^{3y} {/eq} near the origin. The roots (if b2 4ac 0) are b+ p b24ac 2a and b p b24ac 2a. 2. All the operations on two Taylor models assume their time intervals are the same. What are the degrees of the polynomial Answer to Use Taylor's formula to find the cubic approximation of f(x, y) = e* ln(1 + y) near the origin. If f(x) is C1, then the Taylor series of f(x) about cis: T 1(f)(x) = X1 k=0 f(k)(c) k! (x c)k Note that the rst order Taylor polynomial of f(x) is precisely the linear approximation we wrote down in the beginning. 02\text{. Quadratic approximation to f(x,y)=e^x + 5y The sine function (blue) is closely approximated by its Taylor polynomial of degree 7 (pink) for a full cycle centered on the origin. When Newton’s method is used in nonlinear optimization, what it actually The second order Taylor approximation provides a parabolic function approximation while the third order provides a cubic function approximation. 1 Linear Approximation at x = a. The definition of the nonlinearity leads directly to the following well-known result. tan(x) is defined by the ratio cases, these series provide useful polynomial approximations of the generating functions. Formula for bits of entropy per bit, when combining bits with XOR A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. If we choose to center our approximation at some other point, x = a , in the domain of f(x) , then any value we calculate from the approximation will be at (x - a) , and we just evaluate the Approximations for the mathematical constant pi (π) in the history of mathematics reached an accuracy within 0. The approximation coefficients are then chosen to min- The graph of the cosine function is a very nice looking curve. For example, in the above expression for p (x) (Equation 7) the degree is n ( assuming that an is non-zero). Note. Taylor and Maclaurin Series The Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor Polynomials Taylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials If is a function, then we can approximate about the point where by the polynomial where is a remainder term. “Zeroth-Order” Approximation f i e s o l –C h is small f i t c a x–E f(x)=constant f function by approximating the bell-shaped density function with a triangular density function and integrating it twice, to produce a piecewise cubic approximation. Show that the Hamiltonian can be written as $\dfrac{-h^2∇^2}{8π^2m} + ax^2 + bx^3 + cx^4 onumber$ where n defines the degree of the function. You can specify the order of the Taylor polynomial. (x) near x = a, using a cubic (a polynomial of degree three). Introduction and Motivation Jan 11, 2020 · Notice that each approximation actually covers two of the subintervals. This was the key idea in Euler’s method. In this video we come up with the general formula for the A worked example for finding the quadratic approximation of a two-variable A cubic approximation would be a "three-term Taylor approximation" basically, and Example 1 Taylor Polynomial. 6177691815444183. How accurate is the approximation? Solutions to sample problems involving first and second order Taylor polynomials. It is interesting that for positive values of , the latter expression is a polynomial that converges from below to (the blue and violet lines are the polynomials). Taylor Polynomials. So that you could not 100% guarantee to your client's lawyer that an accuracy of 10 cm was achieved. 1 3 \sqrt[3]{8. 4Note: For. in general, if we want an approximation which is a polynomial of degree N in which all the derivatives of the function and the approximation at x0 at N 8. Next: Approximation to arbitrary order Up: Integrated Calculus Summary I, Previous: Linear approximation Quadratic, cubic, quartic and higher approximations. Use Taylor polynomials to approximate the function cos(x) around the point x = 2. Taylor’s Theorem with Remainder. Activity 8. This is the reason for requiring $$n$$ to be even. Background The idea of the Taylor polynomial approximation of order at , written , to a smooth function is to require that and have the same value at . Taylor’s Inequality can then 2. As result we should get a formula y=F(x), named empirical formula (regression equatuion, function approximation), which allows to calculate y for x's not present in table. This shows one way that a polynomial function can be used to approximate a non-polynomial function; such approximations are one of the main themes in this section and the next. Use Taylor’s Formula to find a cubic approximation to f(x,y)=xey at the point (0,0). Since we know$ (e^x)'=e^x $, we take a as 0 and easily obtain that$ \exp(x)=e^x=\sum_{k=0}^\infty\dfrac{x^k}{k!} $Taylor and Maclaurin Series The Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor Polynomials Taylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials (f)Note: The general formula for a cubic approximation centered at x = 0 is: T(x) = f(0) + f0(0)x+ f (0) 2! x2 + f000(0) 3! x3 This is also called the 3rd degree Taylor Polynomial for f(x) centered at x = 0. Approximation problems on other intervals [a,b] can be accomplished using a lin-ear change of variable. derive the Taylor formula as an approximation of the difference between f(x) and f (a). The linear approximation to f at a is the linear function L(x) = f(a) + f0(a)(x a); for x in I: Now consider the graph of the function and pick a point P not he graph and look at We want the second derivative at x of the approximation to be the second derivate at x of f. ƒ(x, y) = ln (2x + y + 1) It is shown that a spline based series approximation to an integral yields, in general, a higher accuracy for a set order of approximation than a dual Taylor series, a Taylor series and an The Taylor expansion can serve as the basic for your initial approximation, and the final terms should be pretty close to the Taylor coefficients. But it's just that, a curve. , Kmenta’s original result). Expand f(x) = 1. 1} 3 8. However, if you try step size h = 0. such as [4], a rational approximation was used for tangent, but more recently in the design for the Intel IA-64 [5], the cheap hardware multiply and add instructions were preferred . Rather than stop at a linear function as an approximation, we let the degree of our approximation increase (provided the necessary derivatives exist), until we have an approximation of the form Example: Given the function!"= 1 (20"−10) Write the Taylor approximation of degree 2 about point "*=0 found Given the function: FG) = 120 X-LO write the Taylor approximation of degree 2 about to = o Feb 06, 2013 · Homework Statement obtain the number r = √15 -3 as an approximation to the nonzero root of the equation x^2 = sinx by using the cubic Taylor polynomial approximation to sinx Homework Equations cubic taylor polynomial of sinx = x- x^3/3! The Attempt at a Solution Sinx = x-x^3/3! + E(x) x^2 = Necessary cookies are absolutely essential for the website to function properly. With smaller step size h = 0. The rationale behind the formula will be explained below. 4762-4780. 724810\bar6)\pi\text{. function of degree . Recall that the nth Taylor polynomial for a function at a is the nth partial sum of the Taylor series for at a. approximation is a cubic spline function. ) Linear approximation is not only easy to do, but also very useful! For example, you can use it to approximate a cubed root without using a calculator. Pictured below are the constant, linear, quadratic, and cubic Taylor polynomial approximations to f(x) = p x at x0 = 1. 0 4 forkinrange(15): 5 pn += (x**k) / math. Then you can start working on the exercises. 2 Consider the Taylor polynomial of degree n of the function exp. (x) is. The linear approximation is in red; the sine curve and the cubic and fifth degree polynomials are essentially Calculates the first four coefficients of the Taylor series through numerical differentiation and uses some polynomial ‘yoga’. 3 The exponential function, being its own derivative, can be factored out of its Taylor series expansion. In the case of cubic Bézier curves, n = 3. The best linear approximation to the cosine function near 0 is quite unexciting; you can check that for f(x)=cos(x), the best linear approximation near 0 is given by L 0 (x)=1. Idea of Taylor polynomials. Given a function f: Rm!Rn, its derivative df(x) is the Jacobian matrix. First, let's practice Solved: Use Taylor's formula with a = 0 and n = 3 to find the standard cubic approximation of ƒ(x) = 1/(1 - x) at x = 0. The interpolation calculator will return the function that best approximates the given points according to the method chosen. Actually, this is now much easier, as we can use Mapleor Mathematica. For every x2Rm, we can use the matrix df(x) and a vector v2Rm to get D vf(x) = df(x)v2Rm. Of course, one can’t expect a line to be a very good approximation to a graph in general, but one would expect that graphs of higher degree polynomials (parabolas, cubic curves, etc. 419. For more videos like OBTAINING TAYLOR FORMULAS Most Taylor polynomials have been bound by other than using the formula pn(x)=f(a)+(x−a)f0(a)+ 1 2! (x−a)2f00(a) +···+ 1 n! (x−a)nf(n)(a) because of the difficulty of obtaining the derivatives f(k)(x) for larger values of k. 0 So the first find the quadratic I need to go up to the second partial derivative and then when I plug that into the Taylors formula. Thus, empirical formula "smoothes" y values. The linear approximation is in red; the sine curve and the cubic and fifth degree polynomials are essentially Jan 27, 2016 · The original algorithm is based on a piecewise function composed of a set of polynomials, each of degree three, at most,and applicable to successive interval of the given points. (13) should be an arbitrarily good approximation to the function Eq. f(x, y)=\sin \left(x^{2}+y^{2}\… 🎉 The Study-to-Win Winning Ticket number has been announced! See full list on study. We’d like to develop a catalog of quadratic approximations similar to our catalog of linear approximations. Thus T 3(x as the original function f(x). 2 Quadratic approximation at x =a. To help you understand where the polynomial approximations come from, recall the Based on these two conditions, we derived the following formula the quadratic Taylor polynomial approximation can always be obtained by substituting That is, the linear approximation of f at a is a polynomial of degree one that has Example 6. Note: The general formula for a cubic approximation centered at x = 0 is given below. near the point x0 = 0 and let's approximate this function with a cubic polynomial Quadratic Approximations. Feb 17, 2020 · 4. For a smooth function , the Taylor polynomial is the truncation at the order k of the Taylor series of the function. Truncation Errors & Taylor Series Taylor Series – provides a way to predict a value of a function at one point in terms of the function value and derivatives at another point. (2018). If only concerned about the neighborhood very close to the origin, the n = 2 n=2 n = 2 approximation represents the sine wave sufficiently, and no For this reason, we often call the Taylor sum the Taylor approximation of degree n. We found that the lin-earization of a function gives a good approximation for points close to the point of tangency. 32] Jul 12, 2014 · hi everyone , i don't understand these steps for Taylor Expansion , it has used for state space equations the equations are the approximations for sin and cos the equation for Taylor series is ( i don't understand at all ) please help me if you can Homework Taylor series can be thought of as polynomials with an infinite number of terms. 0674 )1 4 1 3 − Φ z = +e − z − z Figure 4 shows the difference between Φ4 (z) and Φ(z). This category only includes cookies that ensures basic functionalities and security features of the website. 25 Jun 2019 Discussion and examples of the role of Taylor's Theorem in physics. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials converges. a forward difference approximation, in the Taylor series formula For a real-valued function of a single real variable, the derivative of a function at a point generally determines the best linear approximation to the function at that point. For f(x) = ex, use MAPLE to make a table of the di erence between f and the value of its linear, quadratic, cubic, 4th order and 5th Problem 1. Each data point is represented as a Taylor series, and the high order derivatives in the Taylor ser-ies are treated as random variables. The quadratic approximation to f at x = a is a quadratic, f 2 (x), which has the same value, derivative and second derivative as f at x =a: 6. Truncation order of Taylor series expansion, specified as a positive integer or a symbolic positive integer. is a sequence of increasingly approximating polynomials for f. The formula for the quadratic approximation turns out to be: x2 ln(1 + x) ≈ x − , 2 and so ln(1 . Polynomials are frequently used to locally approximate functions. Apr 06, 2018 · Taylor Series approximation and non-differentiability. Polynomials $$a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$ are nice to evaluate because they rely on addition and multiplication and because we understand them very well. Value. TaylorModel implements a third order Taylor model, i. (x) = f (a) + f '(a)(x a) +. 6. 5) is a quadratic approximation. It's a the cubic can be reduced to a quadratic), cubic equations are nearly always solved in practice by a numerical or approximate method. 29, with the replacements $$x\rightarrow \theta_0+\De\theta$$ and $$a\rightarrow\theta_0$$ We discussed the tangent line approximation to a function. Further, the solution of a variety of problems of 'best approximation' are the spline function approximations. Once you have read to the exercises, start up Maple, load the worksheet Taylor_start_B09. If a function (or data) is sampled at discrete points at intervals of length h, so that fn = f (nh), then the forward difference approximation to f ′ at the point nh is given by h f f f n n n − ′ ≈ +1. Taylor's Formula for Two Variables This justifies the standard linear approximation of f(x, y) at. Some of the approximations look more like a line than a quadratic, but they really are quadratics. Approximation problems on other intervals [a;b] can be accomplished using a linear change of variable. Then has the characteristic property that its derivatives agree with those of the function , when both are evaluated at , up to and including the -th derivative. 1–x. 14. However, a better approximation is possible if we replace the final constraint with a new constraint: The endpoints of the cubic Bézier curve must coincide with the endpoints of the circular arc, and their first derivatives must agree there. That is, L(x) = f(a) + f0(a)(x a): I am reviewing and documenting a software application (part of a supply chain system) which implements an approximation of a normal distribution function; the original documentation mentions the same/similar formula quoted here $$\phi(x) = {1\over \sqrt{2\pi}}\int_{-\infty}^x e^{-{1\over 2} x^2} \ dx$$ Taylor expansions are very similar to Maclaurin expansions because Maclaurin series actually are Taylor series centered at x = 0. 4. The Taylor polynomials agree as closely as possible with a given function at a specific point, but they concentrate their accuracy near that point. The n columns span a small part of m-dimensional space. Numerically, we can simply replace the function 1 1−x Given the information of the original rate of temperature increase plus the new information about the results of the new measures, write a quadratic taylor polynomial approximation for your function. 125. 1. 4: Chebyshev Approximation Algorithm in R1 • Objective: Given f(x) defined on [a,b], find its Chebyshev polynomial approximation p(x) • Step 1: Compute the m ≥ n+1Chebyshev interpolation nodes on [−1,1]: In the last part of this post, we are going to build a plot that shows how the Taylor Series approximation calculated by our func_cos() function compares to Python's cos() function. Give the cubic approximation to the sine, formed at x 0 = 1. Question T1. Apr 30, 2018 · A century ago engineers had very good and robust means of drafting 2D curves using specialized spline sets and curve templates (e. 3 6. mws, and go through it carefully. 4: Chebyshev Approximation Algorithm in R1 • Objective: Given f(x) defined on [a,b], find its Chebyshev polynomial approximation p(x) • Step 1: Compute the m ≥ n+1Chebyshev interpolation nodes on [−1,1]: The MacLaurin series is a Taylor series approximation of a function f(x) centered at x = 0. TAYLOR'S FORMULA FOR FUNCTIONS OF SEVERAL VARIABLES. 56, No. Your captors say that you can earn your freedom, but only if you can produce an approximate value of 8. – 1 around a = 0, to get linear , quadratic and cubic approximations. Definition at line 60 of file taylor_model. 3 Cubic Approximation at x = a. Obtain the cubic spline approximation for the function y=f(x) from the following data, given that y0” 3=”=0y. Apr 10, 2018 · The approximation (as opposed to the in nite series) is one instance of Taylor approximation. The formulas also give an infinite spectrum of rational inverse Set the point where to approximate the function using the sliders. 3 Interpolation Problem 1. Example 1. In analogy with the conditions satis ed by T the \best" approximation of its kind for the function f(x) if we look at values of xthat are close to 0. The nth-degree Taylor polynomial of f about x = a should be a very close approximation to f, because it has the same value as f at x = a, as well as the same slope This gives the Taylor approximation of order three to be 0. e$ a $is the constant that the Taylor polynomial approximations will be centered about. But these proved difficult to replicate in early computers; the need for fast, simple algorithms drove first Bezier, and then de Boor to embrace the idea of polynomial-based splines. The graph shows plots of (dashed line) and for various values of . Example Let f(x) = e2x. 5 Applications and Examples Added Nov 4, 2011 by sceadwe in Mathematics. To find the Maclaurin Series simply set your Point to zero (0). Free Taylor Series calculator - Find the Taylor series representation of functions step-by-step This website uses cookies to ensure you get the best experience. Interpolation with Cubic Splines In some cases the typical polynomial approximation cannot smoothly t certain sets of data. Thus, a Taylor series is a more generic form of the Maclaurin series, and it can be centered at any x-value. 4 Approximation Formulae. 4. Solution Note f '(x) = 2e2x and f ''(x) = 4e2x. Now we'll . For xed v, this de nes a map x2Rm!df(x)v2Rn, like the original f. We seek to find a polynomial p(x)ofdegreenthat minimizes Z b a [f(x) −p(x)]2 dx This is equivalent to minimizing (f−p,f−p)(3) Taylor polynomials are also used frequently in physics. , the Taylor polynomial for is nothing but the linear approximation. The usual reason is: too many equations. How accurate is this approximation? Obviously it depends on the size of h. 2 Taylor models. Theory. Taylor & Maclaurin polynomials are a very clever way of approximating any function with a polynomial. Figure 1. 3 Cubic approximation at x = a In (b) and (c) we show, respectively, the quadratic and cubic approximations. 8. , that the n-input translog function is an approximation to the n-input CES function, given the same restrictions on the translog parameters as in the two-input case ( i. Consider a function f: → whose first m derivatives exist in an open interval about a point x [0]. This suggests that we study the equation x3 + x+8=0; nd an approximation to the solutions that’s accurate when is small, and set equal to 1 10 at the end. The linear and quadratic polynomial approximations discussed in this section are examples of a more general concept called Taylor polynomials. For nicely behaved functions, taking more terms of the Taylor series will give a better approximation. By using this website, you agree to our Cookie Policy. Here we show better and better approximations for cos(x). Taylor Polynomial If function f(x) can be differentiated (at least) n times in the neighborhood of point x = a, then the nth-degree Taylor polynomial of f(x) at x = a is: , 1! · 2! · 3! · … ! · ˘ ˇ! · ˘ ˘ˆ˙ This is the best possible n-degree approximation of f(x) “near” x = a. It was the invention (or discovery, depending on The uses of the Taylor series are: Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point. With each approximation, we add on a little more volume to get closer to the actual volume of a sphere with radius $$r=3. Not just the last = sign is erroneous, also the one before that and within the first inline formula (Taylor Approximations do not necessarily converge to the actual function they try to approximate). Take […] Dec 14, 2017 · Taylor's Theorem guarantees such an estimate will be accurate to within about 0. Now try to find the new terms you would need to find \(P_3(x,y)$$ and use this new formula to calculate the third-degree Taylor polynomial for one of the functions in Example $$\PageIndex{1}$$ above. For the second degree approximation we get: f_{approx2}=1+x+\frac{x^2}{2} Graph of this function is a parabola, marked blue. Con rm that the rst three terms are the same as the quadratic approximation, and con rm that your answer to part (c) matches this formula. 622396. In particular, the 2nd-degree Taylor polynomial is sometimes called the , the 3rd-degreequadratic approximation Taylor polynomial is the , and so on. The Taylor Series of a function f at an input 0 is the following series, f at 0, plus the derivative at 0 times x, plus one over 2!times the second derivative at 0, times x squared, etc. 617769, but it requires 50 steps. Apr 01, 2017 · This feature is not available right now. 2 Quadratic Approximation at x = a. ) could give better approximations. Dec 10, 2016 · The result will be a good approximation to our original function. 3. (Just for fun, note the similarity of your answer to the result from Calculus ll thatー 1 z +r +エ····. The nth Taylor series approximation of a polynomial of degree “n” is identical to the function being approximated! Problems. 1. Let’s use Scilab to calculate the Taylor series approximations for a couple of functions. 617834, although the correct value is 0. Let’s start by looking at the quadratic version of our estimate of ln(1. Sep 29, 2019 · A natural cubic spline adds additional constraints, namely that the function is linear beyond the boundary knots. 1The error Taylor approximate f at x = 1, the approximation error of this approximation Spline Interpolation: Linear, Quadratic, Cubic Splines and . Bilingual High School No 3, Poland used the properties of the sine function to find a polynomial approximation. These approximations have maximum errors that are an order of magnitude better than that of the linear approximation (2). Taylor Series] Use Taylor's formula for f(x, y) at the origin to find the quadratic and cubic approximations of f(x, )- near the origin. 1) cos(. The quadratic approximation of f(x,y) is . This formula is given in his treatise titled Mahabhaskariya. 5 e − Az b can be used as an approximation to the standard normal cumulative function. Abstract: The elementary function approximation using piecewise quadratic polynomial interpolation requires larger area of the look-up table (LUT) and circuit. A function of the form Φ(z )= 1 − 0 . We can continue to look for higher degree polynomial approximations. For example, functions ex and sin(x) both equal their Taylor series expansions about the point x = 0: [2. A relatively easy way to see how this gets done is to look at a quadratic function Function approximation Approximation order abstract We present a high order multivariate approximation scheme for scattered data sets. Use the Taylor expansion of fn+1: ( ) ( ) ( ) ( ) ( ) ( ) Multivariable Linear Approximations: Our approximations and Taylor polynomials for multivariable functions will be best approximations in the same way our single variable approximations were best { our approximations will have a matching function value and matching partial derivatives at the base point. Figure 1 shows a comparative plot of the actual function and our approximation. Linear approximation is just a case for k=1. In order to gain insight into an equation, a physicist often simplifies a function by considering only the first two or three terms in its Taylor series. factorial(k) 6 err = np. For a real function f (y) over y ∈ Y, a Taylor model T f = (p f, R f), consisting of a q-th order Taylor polynomial in y around some y 0 ∈ Y and an interval remainder bound R f, can be constructed so that f ∈ T f for all y ∈ Y. This is 10. Approximation gets better with fewer terms as (t-t0) becomes small. Taylor Polynomials A graphical introduction Best first order (linear) approximation at x=0. In an open interval around x= a, f(x) ≈ f(a) +f′(a)(x−a) + f′′(a) 2! (x−a)2 quadradic approximation • Multi variable Taylor series: Let f be an infinitely differentiable function in some open neighborhood around (x 6. A Taylor polynomial approximates the value of a function, and in many cases, addition of the remainder term Rn(x) turns the approximation into an equation. approximation, meaning that one gets more accuracy over a larger interval centered at the origin. }\)) Taylor Series Generalize Tangent Lines as Approximation. If you want the Maclaurin polynomial, just set the point to A power series is basically an infinite degree polynomial that represents Recall that the sum of a geometric series can be expressed using the simple formula: This means that there is no quadratic Taylor polynomial approximation for sine 1 Apr 2019 Keywords: integral approximation; function approximation; Taylor series; dual polynomial, trigonometric and orthogonal functions can be defined to approximate a which arises from using the binomial formula on (1 − t). The idea is to make a plot that has one line for Python's cos() function and lines for the Taylor Series approximation based on different numbers of terms. Use Taylor's formula for f(x,y) at the origin to find quadratic and cubic approximations of f(x,y) = 3x e 2y near the origin. between the quadratic approximation and the function being approximated at the given point? Do you think you could find a 2nd parabola that is tangent to the exponential function at (0,1)? Share your thoughts. The matrix has more rows than columns. . The Taylor series is introduced and its properties discussed, supplemented by various examples. Find the approximations y;y 2;y 3;y 4;y 5 for f(x) = ex f(x) = sinx f(x) = cos(x) f(x) = ln(x+ 1) f(x) = p x+ 1 For each of those functions, make a MAPLE plot of the function and its Taylor approximations. That is, the kth coefficient is equal to the kth derivative of f evaluated at the input 0 and then divided by k!. 0054 1. FINDING THE LEAST SQUARES APPROXIMATION Here we discuss the least squares approximation problem on only the interval [ 1;1]. Solution We will be using the formula for the nth Taylor sum with a = 0. Also, estimate the LEAST The quadratic approximation is also called the second degree Taylor polynomial. edu % matlab or % matlab -nojvm Sample MATLAB code illustrating several Matlab features; code to plot the graph of y = sin(2πx), x ∈ [0,1]: What is really going on when you use software to graph a function? 1. Calculus 2, Fall 2016. Example 2. 04% of the true value before the beginning of the Common Era . Do this by expanding the Morse potential: $V(x)=D(1 - e-^{Bx})^2 onumber$ into polynomials (i. X -1 0 1 2. Such an approximation is known by various names: Taylor expansion, Taylor When f is a complicated function, Taylor's formula (with the f(j)/j! terms) is usually a Taylor polynomial is a good approximation to the function that it represents. There are As a typical example of how we will use Taylor's theorem, for h close to zero sin(h) = h −. Now that we de ned Taylor polynomials as higher order extensions of the linear approximation, we have Apr 08, 2011 · Use Taylor's formula to find the requested approximation f(x,y) of near the origin. The polynomial can be checked using the Maclaurin series for cos(x). the Jacobian of a cubic function eld. Then approximate (. For k=1 theorem states that there exists a function h1 such that. The formula obtained is ()( 0. Exercise 17: Try to find linear, quadratic, and cubic approximations to the sine function (y = sin(x)) at the origin. To solve the problem, this paper presents an algorithm for elementary function approximation in single-precision floating-point format, which is based on minimax piecewise cubic polynomial approximation. TODO: Pade approximation. Hidden quality cost function of a product based on the cubic approximation of the Taylor expansion. cubic approximation EXAMPLE 7 (a) Find the 5th-degree Taylor polynomial for . To approximate function values, we just evaluate the sum of the first few terms of the Taylor series. x y ex 1+ x 1+ x + x2 2! 1+ x + x2 2! + x3 3! (a) (b) (c) x x y ex ex Figure 5: Linear, quadratic and cubic approximations to ex These power series representations are extremely important, from many points of view. The key idea is to use a series of increasing powers to express complicated yet well-behaved (infinitely differentiable and continuous) functions. Sep 07, 2010 · For example, for four conditions, you could choose a cubic, quadratic numerator/linear denominator, linear numerator/quadratic denominator, or 1/cubic denominator. Also note that some of the approximations do a better job than others. The tangent line approximation is the easiest to work with (because it’s a line), but it’s the least accurate. The lecture ends with a discussion of simple harmonic Taylor series 12. – Most commonly used interpolantused is the cubic spline – Provides continuity of the function, 1st and 2nd derivatives at the breakpoints. Note that our approximations above require that the function be sufficiently differentiable at the point at which we wish to base the approximation. The Taylor series of a real or complex-valued function f (x) that is infinitely differentiable at a real or complex number a is the power series + ′ ()!(−) + ″ ()! In calculus, Taylor's theorem gives an approximation of a k-times differentiable function around a given point by a polynomial of degree k, called the kth-order Taylor polynomial. 20 Nov 2015 In practical term, the calculation of a Taylor polynomial of a complex function can be simplified by using calculus rules to obtain the calculus of we say the above formula approximates f at x and the approximation error is of order n + 1. T(x) = f(0) + f0(0)x+ f00(0) 2! x2 + f000(0) 3! x3 This is called the 3rd degree Taylor Polynomial for f(x) centered at x = 0. Linear and quadratic approximation November 11, 2013 De nition: Suppose f is a function that is di erentiable on an interval I containing the point a. Find the nth Taylor polynomial of y = lnx centered at x = 1. 3 Taylor Polynomials. There will be a price paid in bias near the boundaries for this rather crude approximation, but assuming linearity near the boundaries, where we have less information anyway, is often considered reasonable. Taylor's we can approximate the The first order Taylor polynomial is the function. h. 1). Input the set of points, choose one of the following interpolation methods (Linear interpolation, Lagrange interpolation or Cubic Spline interpolation) and click "Interpolate". This chapter examines methods of deriving approximate solutions to problems or of approximating exact solutions, which allow us to develop concise and precise estimates of quantities of interest when analyzing algorithms. The Taylor formula f(x + t) = eDtf(x) holds in arbitrary dimensions: Theorem: f(x + tv) = eDvtf is called the quadratic approximation of f. Taylor's theorem gives an approximation of a k-times differentiable function around a given point by a k-th order Taylor polynomial. 1 Quadratic Approximation and the Hessian Matrix Using second derivatives, a function f(x;y) which is twice continuously differen-tiable can be approximated by a quadratic function, its Taylor polynomial of order 2. 02) = \frac 43 \pi (3. : The approximation of the exponential function by polynomial using Taylor's or Maclaurin's formula In many cases we know the values of a function f (x)at a set of points Taylor Polynomial Cubic Splines 13-digit approximation we need 15 terms of the Taylor expansion. (x0,y0) and the error of this 26 Feb 2020 The formula for the Taylor series of a function f(x) around a point x=a is given We see that this our quadratic approximation is indeed a better Hence each Taylor polynomial unfolds into an infinite spectrum of rational approximations. Taylor Series Expansion of a function We can expand a function, y(t), about a specific point, t0 according to: The Taylor Series is used to approximate behavior of functions with a few terms. Example: The Taylor series of $$y = e^x$$ is Note: The general formula for a cubic approximation centered at x = 0 is given below. Using Taylor’s Theorem, we find the Taylor series expanded at x = 0 (which means, a = 0) for this function. 5 and make one step, the corresponding approximation becomes 0. 600-680), long before Calculus, and even when the value for Pi was not known very accurately, a seventh-century Indian mathematician called Bhaskara I derived a staggeringly simple and accurate approximation for the sine function. function, but on the cheap to evaluate and smooth fit. 01, we get 0. 2. 14, pp. Use p6(x) to approximate the value of ln 1. For an example, take the function. (a) Find the cubic approximations to sinx and cosx for a = 0. In the last part of this post, we are going to build a plot that shows how the Taylor Series approximation calculated by our func_cos() function compares to Python's cos() function.$\endgroup$– Ryan Thorngren May 15 '19 at 9:52 1$\begingroup$@Atom I suppose there's two questions to ask, 1. These fits go by many names such as approximation models, output predictors, surrogate models, and response surfaces. Lemma 1. See Also. The truncation order n is the exponent in the O-term: O(var n). Use it to predict the temperature of Cook in years 3062 and 3112 (i. 2). Solution: The sine function is the infinitely differentiable function defined for all real numbers. Furthermore, the second-order approxi-mation given by (5) provides better accu- In this blog, I want to review famous Taylor Series Expansion and its special case Maclaurin Series Expansion. 1 importnumpy as np 2 x = 2. ƒ(x, y) = 1/(1 - x - y) These approximations are only reasonable when x is near a. 5. In other words, the physicist uses a Taylor polynomial as an approximation to the function. 2 ! + ททท but in this Figure 5: Linear, quadratic and cubic approximations to ex. 1 Finding the Taylor expansion of a polynomial function is Build up the concept of the Taylor series. The idea was to approximate terms (quadratic, cubic, etc), i. If we calculate second derivatives we can similarly nd a quadratic approximation for the function. Aug 28, 2020 · Taylor Polynomials Preview. In this example, c = 2. }\) The actual volume of a sphere involves a cubic function, so when we approximate the volume with a cubic, we should get an exact approximation (and $$V(3. com Apr 10, 2002 · The Maclaurin polynomial of f is the Taylor polynomial of f about x = 0 (so involves powers of x rather than x -- a). Because A cubic approximation Use Taylor's formula with a = 0 and. Two nd the formula of the quadratic Taylor approximation for the function F(x;y), centered at the point (x 0;y 0), we repeat the procedure we followed above for the linear polynomial, but we take it one step further. For the first degree approximation we obtain: f_{approx1}(x) =1+x. The “c” in the expansion is the point you’re evaluating the function at. For example, to evaluate a complicated function one may pre-compute the function at certain On practice, type of function is determined by visually comparing table points to graphs of known functions. So first the partial derivative with respect to X and you use my chain rule here. Observe that the graph of this polynomial is the tangent plante For example, for. where . The purpose of this lab is to use Maple to introduce you to Taylor polynomial approximations to functions, including some applications. Unless all measurements are perfect, b is outside that column space. Linear approximation. Cubic Approximation The \best cubic approximation to the function f(x) near x = a" is C(x) = f(a)+f0(a)(x a)+ f00(a) 2 (x a)2 + f000(a) 6 (x a)3: It is \the best" because C(a) = f(a), C0(a) = f0(a), C00(a) = f00(a), and C000(a) = f000(a). OZ calls this quadratic function P2(x). We show the graphs of the sine function along with the first three Taylor polynomial approximations on the interval [0, 4]S in Figure 1a. Purpose. 1, Quadratic Approximation to Find p, p. taylor computes the Taylor series approximation with the order n - 1. The representation of Taylor series reduces many mathematical proofs. We also call C(x) the \degree 3 Taylor polynomial for f(x) centered at x = a", denoted T 3(x). Let's try this for a third-order (cubic) sine approximation. Since the roots may be either real or complex, the most general Aug 08, 2019 · HERMITE_CUBIC, a C code which can compute the value, derivatives or integral of a Hermite cubic polynomial, or manipulate an interpolating function made up of piecewise Hermite cubic polynomials. The quadratic Taylor polynomial in two variables. Asymptotic Approximations. Using 1st order Taylor series: ex ˇ1 +x gives a better fit. The radial distance from the arc to the standard Bézier approximation. 9,1. Unit 17: Taylor approximation Lecture 17. Please try again later. The mission of a Taylor polynomial is to imitate the local behavior of a function. Since e = e1, we could use a suitable Taylor polynomial for the 2 Feb 2018 A low-order Taylor-series approximation is often the quickest and easiest way to do a calculation that leads to a quantitative scientific insight. The approximation for k=2 is also sometimes used, for example in my 16 Nov 2013 For a very recognizable example, let's think about finding zeroes of Let's follow this idea to find another quadratic approximate, which I'll proximation, and (A. If not, try a different rational function, or add extra kernel conditions to fix the problem. 00000565 over the whole interval #[0. Feb 01, 2011 · Multivariable Calculus: Find the cubic approximation to f(x,y) = ycos(x+y). -3 Polynomial Approximation 57 polynomial of degree n has exactly n such roots is known as the fundamental theorem of algebra and its proof is not simple. The polynomial [2. As listed below, this sub-package contains spline functions and classes, 1-D and multidimensional (univariate and multivariate) interpolation classes, Lagrange and Taylor polynomial interpolators, and wrappers for FITPACK and DFITPACK functions. Later on spline functions recieved a considerable amount of attention in both theoretical and practical studies. The larger n is, the better the approximation. f(x); x0; and n; and is called the n-th Taylor polynomial of f(x) at x0. 1 Taylor series approximation We begin by recalling the Taylor series for univariate real-valued functions from Calculus 101: if f : R !R is infinitely differentiable at x2R then the Taylor series for fat xis the following power series Taylor Polynomials. The following table gives the values of density of saturated water for various temperatures of saturated steam. 1 . Cubic Spline • Splines –name given to a flexible piece of wood used by draftsmen to draw curves through points. Taylor series 12. Exercise \(\PageIndex{1}$$: Finding a third-degree Taylor polynomial for a function of two variables. Alright, let's see what happens when we compute the second order Taylor Polynomial for this function, centered about a = 3*PI / 4: Here's a graph of this Taylor approximations to sin(x) In class, we've discussed how truncating the Taylor series of a function gives us a polynomial approximation to that function, and that higher order truncations lead to more accurate approximations. As listed below, this sub-package contains spline functions and classes, one-dimensional and multi-dimensional (univariate and multivariate) interpolation classes, Lagrange and Taylor polynomial interpolators, and wrappers for FITPACK and DFITPACK functions. For more videos like so have this function. Also, adetaileddiscussion of the zeta function of a cubic function eld extension is included. 3 Cubic Taylor Approximation of a Di erential Equation Approximate the solution y(x) of the initial value problem y0= 3x+ 2=y; y(0) = 1; at x= 1=10 with the aid of a cubic Taylor polynomial. e Comparison of the approximations to arctan(x) using the proposed two second-order approximations given by (5) and (7) are shown in Figure 2. Mathews, Section 8. exp The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in . Oct 17, 2020 · Interpolation (scipy. We seek to nd a polynomial p(x) of degree n that minimizes Z 1 1 [f(x) p(x)]2 dx This is equivalent to minimizing (f p Fairly obvious, but maybe not obvious enough, since I've seen calls to tanh() in code snippets here. * Any smooth function can be approximated by a polynomial f(x) x xi xi+1 1. Give an upper bound for the magnitude of Example. To visualise the impact of the order of the approximation polynomial, we’ll use Scilab plot() function. If we want to approximate this to first order, it just means that you use up to the term and scrap the rest, meaning thatwhich is a first-order Taylor series approximation of about . This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation. Using 2nd order Taylor series: ex ˇ1 +x +x2=2 gives a a really good fit. Note: f (0)=P1(0) and f’ (0) = P’1(0). We approximate a complicated function, ex, by a cubic polynomial. When d1 is "large", the second order and higher terms of the Taylor Series become non-negligible, which makes the first order approximation "bad". When$ a=0 \$, the Taylor series are also called Maclaurin series. 3 Taylor polynomials can be viewed as a generalization of linear approximations. The higher its degree is, the better resemblance is possible to achieve. https://firebasestorage. is the linear approximation of f at the point a. Use the Taylor series for the function defined as to estimate the value The above calculation tells us how to create a cubic local approximation by putting our limit value back under the denominator of ( x¡c ) 3 , leading to a general recipe by extension: The Taylor polynomial P n ( x ) for a function f ( x ) of order n near the location We begin with the Taylor series approximation of functions which serves as a starting point for these methods. The flat or zeroth order approximation around x = a, g 0 (x) is just the value of f at a: Use Taylor's formula for f(x, y) at the origin to find quadratic and cubic approximations of f near the origin. Vector of length n+1 representing a polynomial of degree n. French curves). Solution We will be using the formula for 7 Apr 2009 Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the We can get an even better approximation, T. The expansion point is not really important here, because if you then expand those terms, you will see it all reduces to that original polynomial. 9. No reason to only compute second degree Taylor polynomials! If we want to find for example the fourth degree Taylor polynomial for a function f(x) with a given center , we will insist that the polynomial and f(x) have the same value and the same first four derivatives at . The cubic formula tells us the roots of a cubic polynomial, a polynomial of the form ax3 +bx2 +cx+d. After all, a Taylor Approximation is not an equality. Finding and using Taylor polynomials. 02)^3 =(36. (8) provided that |x| is sufficiently small but how small is small? This cannot be answered in general since it involves taking into account the infinitely many terms in the Taylor series that we have ignored. The quadratic approximation is better, and the cubic approximation is even better. , a Taylor expansion). It has previously been shown that restricted cubic splines can be used to approximate complex hazard functions in the context of time-to-event data. Examples Taylor series for the exponential function. Notice that the rst three terms are the same as the quadratic approximation. But, did you ever wonder how your calculator knew all those numbers? It hasn't remembered them all, rather it remembers a polynomial approximation for sin x Answer to: Use Taylor's formula to find quadratic and cubic approximation of f(x,y) =5\sin x \cos y near the origin By signing up, you'll get 5 Jun 2019 Example 1: Finding 1st and 2nd degree Taylor Polynomials. interpolate)¶Sub-package for objects used in interpolation. Because D Let’s consider a few approximations of our function in the vicinity of the point x=0. This is the case of function approximation via interpolation. 5 illustrates the first steps in the process of approximating complicated functions with polynomials. Partial Derivatives. By using regression Free Linear Approximation calculator - lineary approximate functions at given points step-by-step This website uses cookies to ensure you get the best experience. International Journal of Production Research: Vol. Taylor series third order approximation. The lecture covers a number of mathematical concepts. Finally, a basic result on the completeness of polynomial approximation is stated. taylor cubic approximation formula
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https://www.physicsforums.com/threads/proving-propositions-involving-if-and-only-if.881465/ | # I Proving propositions involving "if and only if"
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1. Aug 9, 2016
### Mr Davis 97
I am usually pretty good about interpreting what a question is asking when it is in the form, "prove that if p, then q," where p and q are statements. However, I cannot seem to understand how to interpret when it is in the form "prove that p if and only if q." The statement I am working with currently is this: "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle."
How do I begin proving this? What I mean is how can I decompose this proposition to clearer propositions that can be proved individually? I looked online and I couldn't find much about iff statements, so help would be appreciated.
2. Aug 9, 2016
### Krylov
"P if and only if Q" means that you need to show that P and Q are either both true or both false. Hence, you need to exclude the possibility that one is true while the other is false. That is, you need to show that "P implies Q" as well as "Q implies P".
P = a radius of a circle is perpendicular to a chord
Q = the chord is bisected by the radius
Now prove the implication "if P, then Q". Next, prove the implication "if Q, then P". For both implications, use any method of your choice (= direct proof, contraposition, or contradiction).
3. Aug 9, 2016
### Math_QED
Generally, if you have to prove: p <=> q, you have to proof p=> q and q=>p.
In logic, p<=> q is equivalent to (p=> q) ∧ (q =>p) (∧ is the logic symbol for 'and').
In your case, when you have to proof that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius, you need to show that:
If a radius of a circle is perpendicular to a chord, then the chord is bisected by the radius. (=>)
AND
If the chord is bisected by the radius, then a radius of a circle is perpendicular to a chord. (<=)
4. Aug 9, 2016
### Staff: Mentor
If you're good at $p → q$ then also at $q → p$. Both together is $p ↔ q$.
What makes it a bit of messy is the additional condition $(c)$ "assume the chord isn't a diameter". Does it belong to $p$ or to $q$?
Such additional conditions are often not very well stated and one has to decide where to place them:
$c → (p ↔ q)$ or
$(c ∧ p) ↔ q$ or
$(c ∧ p) → q → p$ or
$p ↔ (c ∧ q)$ or
$(c ∧ q) → p → q$ or
$(p ∧ c) ↔ (q ∧ c)$?
(Remark @ logicians: I haven't checked whether some of these are already equivalent.)
Often it is clear by the context, or $c$ is a special, often trivial case of either $p$ or $q$.
Here it is (I think) $(c ∧ p) → q → p$, where $c= \text{ (chord is no diameter) }$ , $p= \text{ (radius bisects chord) }$ and $q= \text{ (radius perpendicular to chord) }$.
Last edited: Aug 9, 2016
5. Aug 9, 2016
### Staff: Mentor
That is the best you can prove, but from the way the problem statement is written I would expect $(p ∧ c) ↔ (q ∧ c)$ which is equivalent to $c → (p ↔ q)$
6. Aug 9, 2016
### Mr Davis 97
That all helps a ton. But given the statement "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle," I am still not completely sure which part to assign p and which part to assign q. In simpler cases where, for example, we have "If I am a mammal, then I am human," it is obvious that p = I am a mammal, and q = I am a human. But when we have something like "I am mammal if and only if I am a human," I can't really decipher which part of the statement is assigned to p and which is assigned to q.
7. Aug 9, 2016
### Staff: Mentor
That doesn't matter in "if and only if", aka "iff" statements. It is symmetric:
$(p ↔ q) ↔ ((p→q) ∧ (q→p))$
8. Aug 9, 2016
### Mr Davis 97
So for "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius. Assume the chord is not the diameter of the circle," I could just assign P to "a radius of a circle is perpendicular to a chord" and Q to "the chord is bisected by the radius," and then go on to show that "if P then Q" and "if Q then P" in any order I choose?
9. Aug 9, 2016
### Staff: Mentor
Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius
The assignment becomes obvious if you highlight they keywords.
10. Aug 9, 2016
### Staff: Mentor
Yes. Plus you have "a chord is not a diameter" as given at hand whenever needed or only in one direction as mentioned by mfb in #5.
11. Aug 9, 2016
### Mr Davis 97
Alright. Also, for the proposition "Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius," is there an "if" part and an "only if" part? In proving this statement, my book starts with "For the 'if' part, we prove that if a radius of a circle is perpendicular to a chord, then the chord is bisected by the radius. For the 'only if' part, we prove that if a chord is bisected by a radius then the radius of a circle is perpendicular to the chord." What does he mean by "if" and "only if" parts?
12. Aug 9, 2016
### Staff: Mentor
If-part:
Prove that a radius of a circle is perpendicular to a chord if the chord is bisected by the radius
Only if part:
Prove that a radius of a circle is perpendicular to a chord only if the chord is bisected by the radius
Which is easier to understand if you swap the order:
Prove that [a] chord is bisected by the radius if a radius of a circle is perpendicular to [the] chord
13. Aug 9, 2016
### Staff: Mentor
You may restructure those sentences as
"Prove that a radius of a circle is perpendicular to a chord if and only if the chord is bisected by the radius,"
"Prove that a radius of a circle is perpendicular to a chord the chord is bisected by the radius,"
If "a radius of a circle is perpendicular to a chord" then "the chord is bisected by the radius," and only if "the chord is bisected by the radius," then "a radius of a circle is perpendicular to a chord"
a) If "a radius of a circle is perpendicular to a chord" then "the chord is bisected by the radius,"
and
b) if "the chord is bisected by the radius," then "a radius of a circle is perpendicular to a chord"
To call it if-part (⇒) and only if-part (⇐) is IMO sometimes confusing. Usually it gets clear by the proof. You can also say that the if-part (⇒) shows, that the statement on right hand side is a necessary condition (for the condition on the left to hold), and the only if-part (⇐) that this statement on the right hand side is also a sufficient condition (for the condition on the left to hold).
14. Aug 9, 2016
### Mr Davis 97
So let's see. Let me see how I can apply to this next proof: "Prove that the product of two integers is odd if and only if both of the integers are odd."
The "if" part:
the product of two integers is odd if both of the integers are odd
The "only if" part:
the product of two integers is odd only if both of the integers are odd
or
both of the integers are odd if the product of the two integers is odd
So if I were to prove the "if" part, how would it look in terms of P and Q (such that if P then Q)? Would P = both of the integers are odd and Q = the product of two integers is odd?
15. Aug 9, 2016
### Staff: Mentor
Right.
16. Aug 9, 2016
### Mr Davis 97
Why would it not be P = the product of two integers is odd and Q = both of the integers are odd?
17. Aug 9, 2016
### Krylov
Because, as fresh_42 wrote in post #7, there is a symmetry in P and Q. (The irony, however, is that in that post, equivalence itself is used in the explanation.) In words: Do you agree that "if and only if" means that you need to prove
"P implies Q" and "Q implies P"
? If you do agree, then notice what happens when P and Q swap places: We prove: "Q implies P" and "P implies Q", which is exactly the same as in the line above. So, it does not matter which statement is called "P" and which is called "Q".
18. Aug 9, 2016
### Mr Davis 97
Alright, I think it all makes sense. Just to be clear, the irrelevance of which part is P or Q is only for biconditional propositions, right?
For example, if we just had to prove the conditional statement that "the product of two integers is odd if both of the integers are odd," P would have to be "both of the integers are odd" and Q would have to be "the product of two integers is odd," such that if P then Q?
19. Aug 9, 2016
### Staff: Mentor
⇒ That is if both of the integers are odd then the product of two integers is odd
So far so good.
⇐ That is if the product of two integers is odd then both of the integers are odd
Yes. But as a stand-alone version it is confusing.
⇐ That is if the product of two integers is odd then both of the integers are odd
Yes, and better to read / see.
Again. It is symmetric. No matter what you choose P and Q to be.
Usually one would take it as it is:
"Prove that the product of two integers is odd (P) if and only if both of the integers are odd (Q)."
a) Q ⇒ P (if part = P is necessary for Q)
b) P ⇒ Q (only if part = P is sufficient for Q)
20. Aug 10, 2016
### Staff: Mentor
Right.
Right.
You can assign P and Q in the opposite way but then you have to prove Q=>P instead of P=>Q. They are just variable names, they don't have a deeper meaning. | 2018-03-19T05:38:20 | {
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https://math.stackexchange.com/questions/2514532/seating-of-5-persons-around-a-circular-table-with-some-pre-conditions | # Seating of 5 persons around a circular table with some pre-conditions
$A,B,C,D$ and $E$ are five persons who are to be seated around a circular table such that $A$ and $B$ must sit together and $C$ and $D$ must never sit together. In how many ways can they be seated?
My Attempt:
Step 1:
First we make $(AB)$ and $E$ sit which can be done in $2$ ways since $A$ and $B$ can arrange themselves in $2!=2$ ways.
Step 2:
One of $C$ and $D$ can be put into gaps between $E$ and $A$ and the other can be put into the gap between $B$ and $E$ for which there are obviously $2$ ways.
To obtain total number of ways it appears that we should multiply number of ways in Step 1($=2$) and number of ways in Step 2($=2$) ways, i.e. total number of ways $=2\times 2=4$.
But it is easy to notice that in these $4$ ways two of the arrangements are rotation of the other two.
So should the answer be $4$ or should it be $2$.
In general what should be the approach?
Say we have $10$ persons sitting around a circular table with $3$ of them wanting to sit together only whereas $4$ other persons do not want to sit next to each other?
Should the rotation of a particular arrangement be construed as same or different?
• The rotation of a particular arrangement is the same. In other words, a rotation doesn't create a distinct arrangement. – learning Nov 11 '17 at 1:16
• The four arrangements you described are ABCED, ABDEC, BACED, and BADEC. These are all rotationally distinct, so the answer is four. – Mike Earnest Nov 11 '17 at 2:20
$A$, $B$, $C$, $D$ and $E$ are five persons who are to be seated around a circular table such that $A$ and $B$ must sit together and $C$ and $D$ must never sit together. In how many ways can they be seated?
There are four possible seating arrangements.
Seat E. Since A and B sit together and C and D are separated, C and D must both be adjacent to E. Therefore, choosing whether C or D sits to E's immediate left also determines who sits to E's immediate right and choosing whether A or B sits two seats to E's left also determines who sits two seats to E's right. Hence, there are $2 \cdot 2 = 4$ permissible seating arrangements, as shown below.
Notice that none of these seating arrangements can be obtained from another by rotation.
Should the rotation of a particular arrangement be construed as the same or different?
By convention, a rotation of a particular arrangement is considered to be the same unless the seats are labeled or we are given a particular reference point (such as a special chair or the north end of the table).
Notice that we have already accounted for rotational invariance by measuring our seating arrangements relative to the position of E.
Say we have $10$ persons sitting around a circular table, with $3$ of them wanting to sit together and $4$ other persons who wish to be separated? In how many ways can they be seated?
We use the block of three people who wish to sit together as our reference point. Say the people are $A$, $B$, and $C$. In how many ways can they be arranged within the block?
$3!$
Suppose the four people who wish to be separated are $D$, $E$, $F$, and $G$. Since there are only seven seats left at the table, they must be seated in the seats that are $1$, $3$, $5$, and $7$ positions to the left of the block. In how many ways can they be seated?
$4!$
Let's call the remaining three people $G$, $H$, and $I$. In how many ways, can they be seated in the remaining three chairs?
$3!$
Therefore, by the Multiplication Principle, the number of permissible seating arrangements is
$3!4!3!$ | 2019-04-25T01:50:52 | {
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http://educ.jmu.edu/~waltondb/MA2C/continuous-functions.html | ## Section10.5Continuous Functions
### Subsection10.5.1Continuity
Recall our definition of continuity for a function at a single point.
###### Definition10.5.1Continuity at a Point
A function $f$ is continuous at $a$ if
\begin{equation*} \lim_{x \to a} f(x) = f(a). \end{equation*}
The single equation captures the full definition because for the equation to be true, the limit must exist and the value of the function must exist. Also, recall that the function is right-continuous if the limit comes from the right ($x \to a^+$) and left-continuous if the limit comes from the left ($x \to a^-$).
These ideas allow us to define what we mean by saying that a function is continuous on an interval.
###### Definition10.5.2Continuity on an Interval
A function $f$ is continuous on an interval $(a,b)$ if $f$ is continuous at every point $x \in (a,b)\text{.}$ We can include an endpoint if the limit statement is true coming from within the interval. That is, we include $a$ if
\begin{equation*} \lim_{x \to a^+}f(x)=f(a) \end{equation*}
and we include $b$ if
\begin{equation*} \lim_{x \to b^-}f(x)=f(b). \end{equation*}
### Subsection10.5.2Definite Integrals and Average Value
When we studied the definite integral, we learned that continuity implies integrability. However, a discontinuous function might still be integrable. For example, the definite integral of a piecewise continuous function with a finite number of jump discontinuities can be computed using the splitting property. The total definite integral would be equal to the sum of the definite integrals on each of the subintervals.
Continuity does guarantee something stronger than integrability. It guarantees that the function attains its average value over an interval. To make this precise, we first need to define the average value.
###### Definition10.5.3Average Value of a Function
The average value of a function $f$ on an interval $[a,b]\text{,}$ denoted $\langle f \rangle_{[a,b]}\text{,}$ is defined as
\begin{equation*} \langle f \rangle_{[a,b]} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx, \end{equation*}
so long as $f$ is integrable on $[a,b]\text{.}$
The average value is defined as the value of a constant function that has the same definite integral over the interval:
\begin{equation*} \int_{a}^{b} \langle f \rangle_{[a,b]} \, dx = \langle f \rangle_{[a,b]} \cdot (b-a) = \int_{a}^{b} f(x)\, dx. \text{.} \end{equation*}
###### Example10.5.4
The figure below illustrates a simple function $f(x)$ defined on the interval $[0,5]\text{,}$
\begin{equation*} f(x) = \begin{cases} 3, & 0 \le x \lt 1, \\ 5, & 1 \le x \lt 3, \\ -1, & 3 \le x \le 5. \end{cases} \end{equation*}
The definite integral equals the sum of the signed areas,
\begin{equation*} \int_0^5 f(x) \, dx = 3 \cdot 1 + 5 \cdot 2 + -1 \cdot 2 = 11. \end{equation*}
The average value is equal to this definite integral divided by the width of the interval,
\begin{equation*} \langle f \rangle_{[0,5]} = \frac{1}{5} \int_{0}^5 f(x)\,dx = \frac{11}{5}. \end{equation*}
Because $f$ is continuous on $[a,b]\text{,}$ the Extreme Value Theorem guarantees that $f$ attains a minimum value $f(x_{\min})$ and a maximum value $f(x_{\max})$ so that $f(x_{\min}) \le f(x) \le f(x_{\max})$ for all $x \in [a,b]\text{.}$
The average value $\langle f \rangle_{[a,b]}$ must be between the minimum and maximum values. The Integral Bounds theorem guarantees
\begin{equation*} f(x_{\min}) (b-a) \le \int_{a}^{b} f(x) \, dx \le f(x_{\max}) (b-a) \end{equation*}
which then implies
\begin{equation*} f(x_{\min}) \le \langle f \rangle_{[a,b]} \le f(x_{\max}). \end{equation*}
By the Intermediate Value Theorem with the interval with end points $x_{\min}$ and $x_{\max}$ (we don't know which is on the left/right), there must be some value $c$ between these points, and so $c \in (a,b)\text{,}$ for which
\begin{equation*} f(c) = \langle f \rangle_{[a,b]}. \end{equation*}
In the previous example, $f$ was not continuous and we can see that the graph $y=f(x)$ did not intersect the constant value $\langle f \rangle_{[0,5]}\text{.}$ The Mean Value Theorem for Integrals guarantees that when the function is continuous, the constant function using the average value must intersect the graph $y=f(x)\text{.}$
###### Example10.5.6
The function $f(x)=x^2$ is continuous everywhere. The average value on the interval $[-1,2]$ can be found using the elementary accumulation formula for a quadratic rate and the splitting property.
\begin{align*} \langle f \rangle_{[-1,2]} &= \frac{1}{2--1} \int_{-1}^{2} x^2 \, dx \\ &= \frac{1}{3} \Big( \int_0^2 x^2 \, dx - \int_0^{-1} x^2 \, dx \Big) \\ &= \frac{1}{3} \Big( \frac{1}{3}(2^3) - \frac{1}{3}(-1)^3 \Big) = \frac{1}{3}\Big(\frac{8}{3} + \frac{1}{3}\Big) = 1 \end{align*}
A figure showing the graphs $y=f(x)=x^2$ and $y=\langle f \rangle_{[-1,2]} = 1$ is shown below. The Mean Value Theorem predicted the existence of a point $c \in (-1,2)$ where $f(c)=\langle f \rangle_{[-1,2]}=1\text{,}$ which we can see occurs at $c=1\text{.}$
The Mean Value Theorem for Integrals also provides the justification for the Monotonicity Test for Accumulation Functions.
Consider any two points $c_1, c_2 \in [a,b]$ with $c_1 \lt c_2\text{.}$ Because $A(x)$ is an accumulation function, by the splitting property of definite integrals,
\begin{equation*} A(c_2) - A(c_1) = \int_{c_1}^{c_2} f(x) \, dx. \end{equation*}
On the other hand, because $f$ is continuous, the Mean Value Theorem guarantees the existence of a point $c \in (c_1,c_2)$ such that
\begin{equation*} A(c_2)-A(c_1) = \int_{c_1}^{c_2} f(x)\, dx = f(c)\cdot(c_2-c_1). \end{equation*}
Now assume that $f(x) \gt 0$ for all $x \in (a,b)\text{.}$ Then $f(c) \gt 0$ and $c_2-c_1 \gt 0\text{,}$ guaranteeing that $A(c_2)-A(c_1) \gt 0\text{.}$ That is, $A(c_2) \gt A(c_1)\text{.}$ This is what is needed to show that $A$ is increasing.
Next assume that $f(x) \lt 0$ for all $x \in (a,b)\text{.}$ Then $f(c) \lt 0$ while $c_2-c_1 \gt 0\text{,}$ guaranteeing that $A(c_2)-A(c_1) \lt 0\text{.}$ That is, $A(c_2) \lt A(c_1)\text{,}$ which shows that $A$ is decreasing.
Finally assume that $f(x) = 0$ for all $x \in (a,b)\text{.}$ Then $f(c) = 0\text{,}$ implying that $A(c_2)-A(c_1) = 0\text{.}$ That is, $A(c_2) = A(c_1)\text{,}$ which shows that $A$ is constant. | 2019-02-22T16:42:59 | {
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http://mathhelpforum.com/geometry/19592-geometry-vectors.html | # Math Help - Geometry with Vectors
1. ## Geometry with Vectors
Let A=(4,2,0), B=(1,3,0) and C=(1,1,3) form a triangle.
Find the coordinates of the point in which the medians of triangle ABC intersect.
---
There is probably a very simple yet elegant solution to this but I can't seem to find it. Any help would be greatly appreciated.
2. Originally Posted by rualin
Let A=(4,2,0), B=(1,3,0) and C=(1,1,3) form a triangle.
Find the coordinates of the point in which the medians of triangle ABC intersect.
---
There is probably a very simple yet elegant solution to this but I can't seem to find it. Any help would be greatly appreciated.
i suppose you are talking about the point where vectors that perpendicularly bisect the sides of the triangle meet?
find such vectors and equate them. of course, here i am referring to the vector equations of lines. recall that we can represent a line by the equation:
$\vec{r} = \vec{r}_0 + t \vec{v}$
where $\vec{r}_0 = \left< x_0, y_0, z_0 \right>$ is a point the line passes through (in this case, the midpoint of the lines of the sides of the triangle, $\vec{v} = \left< a,b,c \right>$ is a vector describing the direction of the line. in this case,the direction perpendicular to the sides they pass through, $t$ is a parameter, and $\vec{r}$ is just the arbitrary vector $\left< x,y,z \right>$
note that we can write the lines as follows:
$\left< x,y,z \right> = \left< x_0 + at, y_0 + bt, z_0 + ct \right>$
get all you lines in this form and then equate the corresponding components. you will be able to solve for the parameters and hence find the point of intersection
3. I think it's about center of gravity.
In this case you don't need vectors.
The coordinates of the center of gravity G are
$x_G=\frac{x_A+x_B+x_C}{3}, \ y_G=\frac{y_A+y_B+y_C}{3}$
4. I got (2,2,1) using red_dog's suggestion though I don't understand how that magical formula works being that I don't know anything about centers of mass. I couldn't follow through Jhevon's explanation.
It seems to me that I could simply assume the point I care for is a point X=(x,y,z), find the distances from A,B, and C to this point X, add them all up and equate them to zero to get the same solution but I couldn't explain why set them up to zero.
Thanks, Jhevon and red_dog.
5. The medians of a triangle are concurrent at appoint which is $\frac{2}{3}$ the distance from any vertex.
6. Originally Posted by rualin
...I don't understand how that magical formula works ...
Hello,
consider the triangle ABC. You know the coordinates of the vertices. Let $A(x_A, y_A)$ then the vector $\overrightarrow{OA} = [x_A, y_A]$.
As you know: $\overrightarrow{OM_{AB}} = \frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB })$
Now you are looking for the vector $\overrightarrow{OZ}= \overrightarrow{OB}+\overrightarrow{BC}+\frac{2}{3 } \cdot \overrightarrow{CM_{AB}}$. Now substitute:
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$ and
$\overrightarrow{CM_{AB}}=\frac{1}{2} \cdot (\overrightarrow{OA}+\overrightarrow{OB})-\overrightarrow{OC})$
You'll get:
$
\overrightarrow{OZ}= \overrightarrow{OB}+\overrightarrow{OC}-\overrightarrow{OB}+\frac{1}{3} \overrightarrow{OA}+\frac13 \overrightarrow{OB}-\frac23 \overrightarrow{OC}$
. Collect like terms:
$
\overrightarrow{OZ}=\frac13 \cdot (\overrightarrow{OA}+\overrightarrow{OB}+\overrigh tarrow{OC})$
Using the coordinate writing of vectors you have:
$[x_Z, y_Z] = \frac13 \cdot ([x_a, y_A] + [x_B, y_B] + [x_C, y_C])$ which is red_dog's fabulous formula. | 2016-06-26T01:32:54 | {
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https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/limit-comparison-test-cor | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Limit comparison test
AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.9 (EK)
## Video transcript
- Let's remind ourselves, give ourselves a review of the Comparison Test, see where it can be useful, and maybe see where it might not be so useful, but luckily we'll also see the Limit Comparison Test, which can be applicable in a broader category of situations. So we've already seen this, we want to prove that the infinite series from n equals one to infinity of one over two to the n plus one converges how can we do that? Well, each of these terms are greater than or equal to zero, and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms. And that other series, the one that jumps out at, that will likely jump out for most folks would be one over two to the n, one over two to the n is greater than, is greater than, and all we would have to really say is greater than or equal to, but we could actually explicitly say it's, oh I'll just write it's greater than or equal to, one over two to the n plus one for, n is equal to one, two, all the way to infinity - why? Because this denominator is always going to be greater by one if you're denominator is greater, the overall expression is going to be less, and because of that, because each of these terms are, they're all positive, this one, each corresponding term is greater than that one, and by the Comparison Test, because this one converges, this kinda provides an upper bound, because this series we already know converges, we can say, so because this one converges, we can say that this one converges. Now, let's see if we can apply a similar logic to a slightly different series. Let's say we have the series the sum from n equals one to infinity of one over two to the n minus one. In this situation can we do just the straight up Comparison Test? Well no, because you cannot say that one over two to the n is greater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater, means that this can't, each of these terms can't provide an upper bound on this one. This one is a little bit larger, but on the other hand you're like, 'Ok, I get that.' But look, as n gets large, the two to the n is going to really dominate the minus one or the plus one, or this one has nothing there, this one just has two to the n. The two to the n is really going to describe the behavior of what this thing does. And I would agree with you, but we just haven't proven that it actually works, and that's where the Limit Comparison Test comes in helpful. So let me write that down, 'Limit, Limit Comparison Test, 'Limit Comparison Test', and I'll write it down a little bit formally, but then we'll apply it to this infinite series right over here. So what Limit Comparison Test tells us, that if I have two infinite series, so this is going from n equals k to infinity, of a sub n, I'm not going to prove it here, we'll just learn to apply it first. This one goes from n equals k to infinity, of b sub n. And we know that each of the terms, a sub n, are greater than or equal to zero, and we know each of the terms, b sub n, are actually we're just going to say, greater than zero, because it's actually going to show up in the denominator of an expression, so we don't want it to be equal to zero, for all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on, and on and on, and, and this is the key, this is where the limit of the Limit Comparison Test comes into play, and, if the limit, the limit as n approaches infinity, of a sub n over b sub n, b sub n is positive and finite, is positive and finite, that either both series converge, or both series diverge, so let me write that. So then, that tells us that either, either both series converge, or both diverge, which is really really useful. It's kind of a more formal way of saying that, 'Hey look, as n approaches infinity, if these have similar 'behaviors then they are either going to 'converge, or they're both going to diverge.' Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over here, notice it satisfies all of these constraints, so let's take the limit, the limit as n approaches infinity of a sub over b sub n, so it's going to be one over two to the n minus one over one over two to the n, and what's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity, of two, if you divide by one over two to the n that's just going to be the same thing as multiplying by two to the n, so it's going to be two to the n over two to the n minus one. Over two to the n minus one, and this clearly, what's happening in the numerator and the denominator, these are approaching the same, well we can even write it like, we can even write it like this, divide the numerator and denomiator by two to the n, if you want, although it's probably going to jump out at you at this point. So, limit as n approaches infinity, let me just scroll over to the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominator by two to the n, I'm going to have one minus one over two to the n. And now it becomes clear, this thing right over here is just going to go to zero, and you're going to have one over one. The important thing is that this limit is positive and finite because this thing is so this thing right over here, is positive and finite, the limit is one is positive and finite, if this thing converges and this thing converges, or this thing diverges, and this diverges, well, we already know this thing converges, it's a geometric series where the common ratio is less than one, and so therefore this must converge as well, so that converges as well.
AP® is a registered trademark of the College Board, which has not reviewed this resource. | 2021-01-24T10:47:17 | {
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https://www.physicsforums.com/threads/commutative-2x2-matrices.831030/ | # Commutative 2x2 Matrices
1. Sep 5, 2015
### RJLiberator
1. The problem statement, all variables and given/known data
Let A =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
Find all 2 x 2 matrices B such that AB = BA.
2. Relevant equations
3. The attempt at a solution
I let B =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} and set AB=BA.
From here I see that a and d must be 0, and b=c must be true.
So the answer will be that all matrices that are commutative will be of form:
\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}
And there is no other possible commutative matrix outside of this form.
1. Is this correct?
2. Is there any further proof of this needed?
Thank you kindly.
2. Sep 5, 2015
### pasmith
That must be wrong, because the 2x2 identity matrix commutes with every 2x2 matrix but is not of that form.
What did you actually get for AB and BA? I would suggest double-checking those calculations.
You have the right idea, but have not executed it correctly.
3. Sep 5, 2015
### RJLiberator
So to get this right:
if A = \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
All possible commutative matrices with matrix A should be in the form:
\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}
This is the wrong answer? It seems to be right when I calculate it. I get the same answer either way. AB = BA
4. Sep 5, 2015
### pasmith
$\begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix}$ is a subset of the matrices you are looking for. It can't be all of them, because the identity $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ commutes with every 2x2 matrix.
Recheck your initial calculations with $B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
5. Sep 5, 2015
### RJLiberator
Ahhh, I see.
I calculated it out and found that while b=c, also a=d:
\begin{bmatrix}
a & b \\
b & a
\end{bmatrix}
This makes sense to me as the original answer was a subset of this.
Is there any further proof needed to show that is all?
6. Sep 5, 2015
### SammyS
Staff Emeritus
What do you get for the product $\displaystyle \ \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \ \ \ ?$
What do you get for the product $\displaystyle \ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \ \ \ ?$
(I see you posted you answer just before I posted this.)
That looks good.
7. Sep 5, 2015
### RJLiberator
Thanks guys for the help here. Greatly appreciated. | 2017-08-21T20:33:59 | {
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https://techtud.com/comment/10836 | ##### Number of pixels in digital straight line segment
Find the number of pixels of the (minimally connected) digital straight line segment connecting (0, 0) and (1007, 37). Find out the maximum number of pixels among them, which are on the same horizontal line.
Will be updated soon.
Gautam Choudhary 11 Sep 2017 01:58 pm
The first part of minimally connected pixels is pretty straight forward as both DDA and Bresenham's algorithm provide a set of points that are minimally connected which is equal to $max(\Delta x, \Delta y) + 1$ .
Reason: In both the algorithms, we always paint the next pixel by incrementing $x-coordinate$ (assuming slope < 1). In other words, we paint a $y-coordinate$ for for every $x-coordinate$. Hence, total painted pixels are $\Delta x + 1$ . (Similarly, when slope > 1, it is $\Delta y + 1$)
For the second part, the intuition behind is very simple. To put it differently, for drawing a line segment from $(x_1, y_1)$ to $(x_2, y_2)$ and assuming slope < 1, we have to only paint $(\Delta x + 1)$ pixels with suitable $y$ coordinates.
Take the line segment asked in question i.e. from $(0, 0)$ to $(1007, 37)$ . In total we have to paint $(\Delta x + 1) = 1007 + 1 = 1008$pixels. We paint in stair-case manner i.e. some set of pixels are at the same horizontal with the condition that first and last staircase is of half the average length.
Total levels of staircase $= (\Delta y + 1) = (37 + 1) = 38$levels or steps.
Let the length of average pixel be $n$ . So, the equation follows:
Total pixels to be painted = Sum of pixels at each step or staircase level
$1008 = (1)*n/2 + (38-2)*n + (1)*n/2$[Read it as first stair is halfway painted while next subsequent stairs are fully painted and finally last stair is also half painted to maintain symmetry]. Solving this equation, we have $n = 27.24$
So, length of first and last staircase $= floor(n/2) = floor(27.24\ /\ 2) = floor(13.62) = 13$
Length of intermediate staircase step $= floor(n) = floor(27.24) = 27$
Now, the remaining pixels are $= 1008 - (1*13 + 36*27 + 1*13) = 1008 - (998) = 20$
Now, each of these 20 pixels can be symmetrically added to some 20 horizontal steps whose length is already 27(intermediate stairs) and 13(terminal stairs). Hence, their length becomes 28 (and 14 respectively). So, the required answer is 28.
To generalize this rule, the max number of pixels on same horizontal is $ceil (\ max({{\Delta x + 1} \over {\Delta y}}, {{\Delta y + 1} \over {\Delta x}}))$
To visualize, the image shows pixelated line segments with different slopes all of which are drawn using DDA algorithm.
1. The green line with slope 1 paints pixel by incrementing 1 in $x$ as well as $y$ direction. Horizontal step is of length 1
2. The blue line with slope 4 paints pixel by incrementing ~4 in $x$ and 1 in $y$ direction. Horizontal step is of length 4 (Notice that the last and first stair is of length 2)
3. The yellow line with slope intermediate between 1 and 4 paints pixel accordingly and without loss of symmetry. Horizontal step is of length 1 as well as 2
Purushottam Abhisheikh 11 Sep 2017 07:35 pm
I don't think the generalised answer is still correct. You can simply check for lines whose inverse of slope (i.e. $\frac{\Delta x}{\Delta y}$) is an integer. Correct answer should be $\left\lceil max(\frac{\Delta x}{\Delta y}, \frac{\Delta y}{\Delta x}) \right\rceil$. For example for line between points $(0,0)$ and (20,2) the answer should be 10.
Ranita Biswas 11 Sep 2017 08:20 am
You are very close to the answer. But, there is one catch; the first row of pixels (at y = 0) and the last row of pixels (at y = 37) are not full length. You may try to modify the approach accordingly to get to the correct answer. Very good presentation, by the way.
Gautam Choudhary 11 Sep 2017 08:26 am
Yes, ma'am I got that, they are half the average length due to the shorter window size of [0, 0.5] in case of starting point and [1006.5, 1007] in case of last point. I'll shortly update it.
Purushottam Abhisheikh 11 Sep 2017 10:30 pm
A slight modification (modifications are in bold) to the solution provided by ma'am in the class:
Lets assume below situation for any line $y = mx,\; m\le1$ (if line is not passing through $(0,0)$ then translate it so that it passes through origin since translation will not change relative properties of pixels):
Here $i,\ j$ and $h$ are integers.
As all the pixels in the horizontal line $y\;=\;j$ from $x \;=\;i \quad\text{to} \quad x\;=i+h$should be colored.
So, $j-mi \le 0.5\quad \text{and} \quad m(i+h) - j \lt 0.5$.
Using both we get,
$h\lt\frac{1}{m} \quad \text{or} \quad h\lt \frac{\Delta x}{\Delta y}$
Since h is maximum, there are two cases:
If $\frac{\Delta x}{\Delta y}$ is an integer then $h = \frac{\Delta x}{\Delta y}-1$ otherwise $h = \left\lfloor \frac{\Delta x}{\Delta y}\right\rfloor$.
So $h = \left\lceil \frac{\Delta x}{\Delta y}\right\rceil -1$.
And total number of pixels = $h+1 = \left\lceil \frac{\Delta x}{\Delta y}\right\rceil$. | 2018-09-21T05:58:03 | {
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http://math.stackexchange.com/questions/84439/basic-optimization-problem | # Basic Optimization Problem
I sat for an exam a few days ago. I managed to answer every question except for question $1$c in the calculus paper. Provided that I got question $2$d correct (my answer was $m=0.5$), the absence of an answer for question $1$c should not in any way affect my overall grade. Out of interest, however, I am wondering if you could start me off on how to approach problem $1$c. Both the questions can be found below.
Thank you very much.
Question $1$c:
An enclosure at a reserve is divided into two rectangular cages.
One side of the enclosure is a solid wall.
There is a fence around the rest of the enclosure and another between the male and female cages.
The total length of the fences is $275$ m.
$5$ male birds are kept in the smaller cage and $8$ female birds in the larger cage.
The males have an average area of $250$ sq.m each in their cage.
The area of the female cage is to be maximised.
Find the average area for each of the female birds in their cage.
Question $2$d:
The diagram below shows part of the graph of the function $y=x^2, x>0$.
The shaded region between the curve and the $X$-axis, and between $x = m$ and $x = m + 2$, has an area of $5 \frac16$ sq.units. Find the value of $m$.
Explain the choice of the solution.
Here is the link to the entire exam.
-
The answer $m=\frac{1}{2}$ is correct. – André Nicolas Nov 22 '11 at 3:27
Welcome to math.stackexchange.com! I hope you don't mind, I've taken the content of the question from the external source and inputted it here to make it easier for people to answer. – process91 Nov 22 '11 at 3:43
From the picture above, the total length of the fence is $3a+b+c$. We are given that this is $275$. Hence, we the first constraint
$$3a+b+c = 275.$$
We are further given that the $5$ males have an average area of $250$ in their cage. This gives us that
$$\frac{ac}{5} = 250 \implies ac = 1250.$$
We now want to maximize the average area for each of the female birds in their cage i.e. we want to maximize $\displaystyle \frac{ab}{5}$ subject to these constraints. Maximizing $\displaystyle \frac{ab}{5}$ is equivalent to maximizing $ab$. Hence, the optimization problem is
$$\text{Maximize } ab \text{ subject to } 3a+b+c = 275 \text{ and } ac = 1250.$$
Eliminating $b$ using the first constraint gives us $b = 275 - 3a - c$. Hence the function we want to maximize becomes $$275a - 3a^2 - ac$$ subject to $ac = 1250$. Plugging in the value of $ac$, we get that we want to maximize
$$275a - 3a^2 - 1250.$$
The maximum for this function can be found easily by completing the squares (or) by differentiation and the maximum occurs at $$a = \frac{275}{6}$$ giving a maximum of $$\frac{60625}{12}.$$
Hence, the average area for each of the female birds in their cage is $\frac{60625}{96}$.
-
I hate to be a nuisance but I get a little lost after you state "Maximize ab subject to 3a+b+c=275 and ac=1250." Can you please further explain how you derive the expression "275a−3a2−ac"? Is this explanation making use of the Lagrange multiplier method that @process91 referred to earlier? I am not familiar with this method, but am about to read up on this now. Thank you all very much for your help. – clookid Nov 22 '11 at 8:19
Ah, I see what you have done now. "ab = 275a - 3a^2 - 1250." That's a clever approach. Thanks, once again. – clookid Nov 22 '11 at 8:44
Using the following labeling:
The total length of the fences is 275m
Therefore $3x+y+z=275$.
5 male birds, average area of 250m$^2$
Therefore $\frac {xz} 5 = 250$. You are to maximize the area of the female cage, which is given by $A(x,y) = xy$.
You can use substitution (with regard to the constraints) to write a function for the area of the cage which is only dependent on one variable, and then maximize it using the normal approach of taking the derivative and setting it to zero. Alternatively, you can use the Lagrange multiplier method.
-
Thank you very much for your help. Can you please check my working (see: i43.tinypic.com/s1jbma.jpg)? – clookid Nov 22 '11 at 4:00
@CharlesSalmon Yes, that is correct. Your answer also agrees with the answer that Siviram gave in his answer of $60625/96$, the difference in approach is that you chose to maximize the average area for the female birds, while Siviram maximized the area of the female cage. You are lucky, in this respect, because maximizing the average area also maximized the area of the cage and the answer turned out to be correct, however a question could be chosen in a tricky way where this is not the case so make sure you are maximizing what has been asked. – process91 Nov 22 '11 at 4:15 | 2014-12-20T17:35:37 | {
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https://math.stackexchange.com/questions/3434096/finding-the-closed-formula-for-a-n-b-n-for-recursions-a-n-and-b-n/3434904#3434904 | # Finding the closed formula for $A_n + B_n$ for recursions $A_n$ and $B_n$
Update: I made a mistake when posting this question - the recursive definition in $$\text{(A)}$$ was defined with
$$\tag A A_{n+1} = (n-1)A_n + 2B_n \quad \quad \text{ERROR}$$
and has now been corrected.
I checked and the recursion now agrees with the solution to the motivating problem linked to below.
The prior question was well founded and had two answers, with Calvin Lin solving it. Olivier Roche supplied hints using matrix methods.
I know I could have posted a new question, but figured this edit made the most sense.
Define $$A_4 = 0$$ and $$B_4 = 2$$.
For $$n \ge 4$$ define
$$\tag A A_{n+1} = (n-1)A_n + 4B_n$$
and
$$\tag B B_{n+1} = (n-2) B_n$$
Find an explicit formula in $$n$$ to represent the sum $$A_n + B_n$$ for $$n \ge 5$$.
My Work
I answered a combinatorial question on this site and wanted to use this method, but I'm not sure how to proceed. Using a combinatorial argument I verified the recursion holds, and want to apply the appropriate techniques to get the answer in a different way.
• You can solve for $B_n$ first to get $$B_n = 2(n - 3)!$$ then use it to get $$A_n = 2(n - 4)(n - 3)!$$. The sum you want is just add them both. Nov 13, 2019 at 15:55
• @Azlif Guess this is really simple. I had the 'first' part but how does that give you $A_n$? (must be staring me in the face) Nov 13, 2019 at 15:59
• @CopyPasteIt One way is to prove via induction. Did you want a method that used a combinatorial approach? Nov 13, 2019 at 16:03
• @CalvinLin A combinatorial approach is of interest, provided we don't start seating people at a table. It would be interesting to see how you might 'guess' that $A_n = 2(n - 4)(n - 3)!$ (heuristic/intuition/simplification/conjecture) so that you can justify induction. Nov 13, 2019 at 16:14
Write $$X_n := \begin{pmatrix} A_n \\ B_n \end{pmatrix}$$ and $$M_n := \begin{pmatrix} (n-1) & \mathbf{4} \\ 0 & (n-2) \end{pmatrix}$$, this allows you to sum up the recurrence relation as : $$\boxed{X_{n+1} = M_n X_n}$$
In other words, one has $$X_n = (\prod_{k=4}^n M_k)X_4$$. Your goal is now to express $$P_n :=(\prod_{k=4}^n M_k)$$ with a closed formula.
conjecture1 (wrong) for $$n\geqslant 5$$, one has $$P_n = \begin{pmatrix} (n-1)! & n(n-1)-12\\ 0 & (n-2)! \end{pmatrix}$$
New try (one should be able to solve the problem without knowing $$\star$$) :
conjecture2 for $$n\geqslant 4$$, $$P_n$$ is of the form $$P_n = \begin{pmatrix} \frac{(n-1)!}{2} & \star\\ 0 & (n-2)! \end{pmatrix}$$
• I think OP's goal is to "use this combinatorial method", as opposed to "prove this recurrence relation property", though his new comment suggests it doesn't matter. @CopyPasteIt, can you clarify? Nov 13, 2019 at 16:01
• @CalvinLin This is the type of answer I wanted to see. For some reason, when I see these type of problems, I can't find a 'theory handle' and just 'turn the crank'. This looks like the way to go, replacing two recursions with one 'matrix recursion' (+1). Nov 13, 2019 at 16:05
• If so, one approach is take is to calculate initial terms of the sequence, and then try to guess what the pattern is. Eg. I obtained OEIS A052582, which has the formula $2n\times n!$ (with some offsets). Nov 13, 2019 at 16:06
• @CopyPasteIt If you can prove the conjecture above, you're done. Feel free to ask why I conjectured this. Nov 13, 2019 at 16:14
• @OlivierRoche Thanks - looks like fun and can now legitimately proceed with induction; see math.stackexchange.com/questions/3434096/… Nov 13, 2019 at 16:17
By looking at initial terms and using induction, we can conclude that $$A_n + B_n = 2 \times (n-3) \times (n-3)!$$, $$A_n = 2(n-4)(n-3)!$$, $$B_n = 2(n-3)!$$.
Here is a combinatorial approach, but it is very contrived.
For a permutation $$\rho \in S_{n-2}$$, count the number of pairs $$(a_i, a_{i+1})$$ such that $$\rho(a_i ) - \rho (a_{i+1}$$ are consecutive integers.
There are $$n-3$$ pairs of consecutive integers, and they are consecutive in $$2\times (n-3)!$$ ways, which means that there are a total of $$2\times (n-3) \times (n-3)!$$ such pairs.
Alternatively, let $$B_n$$ be the number of ways that a permutation of $$S_{n-2}$$ has "1,2" consecutive. Given any way for $$B_n$$, there are $$n-2$$ places we can insert the value $$n-1$$ in the permutation to obtain a way for $$B_{n+1}$$.
This is clearly a bijection so $$B_{n+1} = (n-2) B_n$$.
We can verify that $$B_4 = 2$$.
Let $$A_n$$ be the number of ways that a permutation of $$S_{n-2}$$ has "2,3" or "3,4", or ..., or "$$n-3, n-2$$" consecutive, counted with duplicity.
We have $$A_n = (n-4) B_n$$ by counting the number of pairs.
Hence $$A_{n+1} = (n-3) B_{n+1} = (n-3) (n-2) B_n = (n-1)(n-4)B_n + 2B_n = (n-1)A_{n} + 2B_n$$.
We can verify that $$A_4 = 0$$.
Hence, $$A_n+ B_n = 2 \times (n-3) \times (n-3)!$$.
Using Olivier Roche matrix technique, we define for $$n \ge 4$$
$$P_n = \begin{pmatrix} \frac{(n-1)!}{2} & p_n\\ 0 & (n-2)! \end{pmatrix}$$
where $$p_n$$ is, to start with, unknown, but $$p_4 = 4$$.
Set
$$M_{n+1} = \begin{pmatrix} n & 4\\ 0 & n-1 \end{pmatrix}$$
Multiplying $$M_{n+1} \circ P_n$$ to get $$P_{n+1}$$ we conclude that
$$\tag 1 p_{n+1} = n p_n + 4 \, (n-2)!$$
Handing this off to wolframalpha, the recurrence equation solution is given by
$$\quad p_n = \bigr ( (c_1 + 4) n - c_1 - 8 \bigr ) Γ(n - 1) \quad \text{ (where } c_1 \text{ is an arbitrary parameter)}$$
We can solve for $$c_1$$ knowing that $$p_4 = 4$$ and find that $$c_1 = -2$$.
So for $$n \ge 4$$ we can write
$$\tag 2 p_n = 2 (n-3) \, (n-2)!$$
By applying the matrix $$P_n$$ to the vector
$$\begin{bmatrix} 0 \\ 2 \end{bmatrix}$$
we conclude that, for $$n \ge 4$$,
$$\quad A_{n+1} = 2 * p_n = 4 (n-3) \, (n-2)!$$
and
$$\quad B_{n+1} = 2 * (n-2)!$$
• How did you use the $\Gamma$ function to solve the recurrence, please? Nov 14, 2019 at 14:20
• I just handed off the recursion to Wolfram that solved it. In the solution you get $\Gamma(n-1)$ . I figured that it is a class of recursions where $\Gamma$ comes into play. Nov 14, 2019 at 15:39 | 2022-10-07T08:59:42 | {
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http://mathhelpforum.com/calculus/21090-maxima-mininima.html | 1. Maxima/Mininima
Hi, can some one help with this please.
How I can fine all of the potential maxima and minima for f(x,y).
Thanks ^_^
2. First, differentiate f(x,y), then find where f'(x,y) equals zero. Those are your potential minimums and maximums.
For example:
$\displaystyle y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 12x + 6$
$\displaystyle y' = x^2 + x - 12$
$\displaystyle x^2 + x - 12 = 0$
$\displaystyle (x - 3)(x + 4) = 0$
$\displaystyle y' = 0$ at $\displaystyle x = 3$ and $\displaystyle x = -4$
To determine which they are, find y'' and plug in the zero values for x...
$\displaystyle y'' = 2x + 1$
$\displaystyle y''(3) = 2(3) + 1 = 7 > 0$ so minimum
$\displaystyle y''(-4) = 2(-4) + 1 = -7 < 0$ so maximum
Scott
3. Originally Posted by ScottO
First, differentiate f(x,y), then find where f'(x,y) equals zero. Those are your potential minimums and maximums.
For example:
$\displaystyle y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 12x + 6$
$\displaystyle y' = x^2 + x - 12$
$\displaystyle x^2 + x - 12 = 0$
$\displaystyle (x - 3)(x + 4) = 0$
$\displaystyle y' = 0$ at $\displaystyle x = 3$ and $\displaystyle x = -4$
To determine which they are, find y'' and plug in the zero values for x...
$\displaystyle y'' = 2x + 1$
$\displaystyle y''(3) = 2(3) + 1 = 7 > 0$ so minimum
$\displaystyle y''(-4) = 2(-4) + 1 = -7 < 0$ so maximum
Scott
I believe the poster was talking about functions of two variables. it is a slightly different.
Originally Posted by MarianaA
Hi, can some one help with this please.
How I can fine all of the potential maxima and minima for f(x,y).
Thanks ^_^
this is the kind of question for which you should consult your text for the answer.
We suppose the second partial derivatives of the function $\displaystyle f(x,y)$ is continuous on some disk centered at the point $\displaystyle (a,b)$.
If $\displaystyle f_x(a,b) = 0$ and $\displaystyle f_y(a,b) = 0$ then we call $\displaystyle (a,b)$ a critical point of $\displaystyle f$.
Define $\displaystyle D(a,b) = f_{xx}(a,b)f_{yy}(a,b) - [ f_{xy}(a,b)]^2$
we can classify the critical point $\displaystyle (a,b)$ as follows:
case 1: If $\displaystyle D>0$ and $\displaystyle f_{xx}(a,b)>0$, then $\displaystyle f(a,b)$ is a local minimum
case 2: If $\displaystyle D>0$ and $\displaystyle f_{xx}(a,b)<0$, then $\displaystyle f(a,b)$ is a local maximum
case 3: If $\displaystyle D<0$, then $\displaystyle f(a,b)$ is a saddle point
4. Originally Posted by Jhevon
this is the kind of question for which you should consult your text for the answer.
I know but some one steal my book, thats why I am asking.
Thanks both of you
5. Originally Posted by MarianaA
I know but some one steal my book, thats why I am asking.
Thanks both of you
what monster would steal a calculus book? what is the world coming to?!
bless your heart Mariana, you have my condolences
i shudder to think what i'd do if someone stole my calculus text
did you understand the notation i used?
6. Originally Posted by Jhevon
what monster would steal a calculus book? what is the world coming to?!
bless your heart Mariana, you have my condolences
i shudder to think what i'd do if someone stole my calculus text
did you understand the notation i used?
Looks like you didn’t believe me, but it is true.
Cost me around $120 dollars, so probably they already sell it, you never know. 7. My Calc I professor and I wound up switching books during a class one day. (Unfortunately, no it wasn't a "teacher's" text with the answers.) I didn't find out about it until after finals, when noticed some scribblings in the part of the text I had read earlier that semester. Since I had bought the book new and didn't make the scribbles (that weren't there when I read the material) I realized what had happened. I paid for a new textbook and wound up with a used and written in book! (If that story doesn't move you, then you obviously don't love and worship your textbooks like I do.) -Dan 8. Originally Posted by MarianaA Looks like you didn’t believe me, but it is true. Cost me around$120 dollars, so probably they already sell it,
you never know.
i believe you
Originally Posted by topsquark
My Calc I professor and I wound up switching books during a class one day. (Unfortunately, no it wasn't a "teacher's" text with the answers.) I didn't find out about it until after finals, when noticed some scribblings in the part of the text I had read earlier that semester. Since I had bought the book new and didn't make the scribbles (that weren't there when I read the material) I realized what had happened.
I paid for a new textbook and wound up with a used and written in book!
(If that story doesn't move you, then you obviously don't love and worship your textbooks like I do.)
-Dan
i am moved. you did get your book back once you realized though, right? (on second thought, don't answer that. all these off topic posts are bound to make TPH angry) | 2018-05-23T07:34:49 | {
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http://life.werow.co.uk/g2jjp/pur9gkl.php?tag=multiplicative-inverse-of-matrix-fb5c38 | Try refreshing the page, or contact customer support. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Is it true that the only matrix that is similar to a scalar matrix is itself. | {{course.flashcardSetCount}} This matrix has no inverse because the columns are not linearly independent. I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. MINVERSE(square_matrix) square_matrix - An array or range with an equal number of rows and columns representing a matrix whose multiplicative inverse will be calculated. What Is the Syllabus of an Algebra I Course? Hence, I is known as the identity matrix under multiplication. I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. In the rest of this section, a method is developed for finding a multiplicative inverse for square matrices. First we need to find the determinant of this matrix, which is. The modular multiplicative inverse of an integer a modulo m is an integer b such that, It maybe noted , where the fact that the inversion is m-modular is implicit. When we multiply a matrix by its inverse we get the Identity Matrix (which is like "1" for matrices): A × A -1 = I. One has to take care when “dividing by matrices”, however, because not every matrix has an inverse, and the order of matrix multiplication is important. Menu. The reciprocal of a number obtained is such that when it is multiplied with the original number the value equals to identity 1. Washington University in St Louis, Master of Science, Electrical Engineer... University of Illinois at Urbana-Champaign, Bachelor of Science, Chemical and Biomolecular Engineering. That matrix won't have an inverse. denoted asx = 1 / x (or) x = x-1 (Inverse of x)It is also called as the reciprocal of a number and 1 is called the multiplicative identity N, An electronics company produces transistors, resistors, and computer chips. 5. and . Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. For the multiplicative inverse of a real number, divide 1 by the number. What is the inverse of the following nxn matrix. Hot Network Questions Why is it easier to carry a person while spinning than not spinning? A -1 × A = I. Home Embed All Precalculus Resources . Let's see. To calculate inverse matrix you need to do the following steps. by M. Bourne. Log in here for access. A matrix has no inverse--yeah--here--now this is important. First of all, a matrix needs to be square to have an inverse. . Inverse Matrix Formula. All other trademarks and copyrights are the property of their respective owners. B) Find 2A + 3B. Following this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. lessons in math, English, science, history, and more. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. The multiplicative inverse of a matrix A is written A^(-1). The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. Previous matrix calculators: Determinant of a matrix, Matrix Transpose, Matrix Multiplication, Inverse matrix calculator. Precalculus Help » Matrices and Vectors » Inverses of Matrices » Find … 3. is the multiplicative inverse of . A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe Illustrated definition of Multiplicative Inverse: Another name for Reciprocal. All rights reserved. ChillingEffects.org. Track your scores, create tests, and take your learning to the next level! As with any other matrix (defined over a field like the real numbers), an inverse exists as long as the determinant of the matrix is non-zero. Yes, we write the inverse with a superscript of -1. If Varsity Tutors takes action in response to 2x_1 + 3x_2 = 5. 's' : ''}}. Definition and Examples. After you multiply, you can then easily find the answer by translating back to equation form. Tech and Engineering - Questions & Answers, Health and Medicine - Questions & Answers, For n X n matrices A and B, and n X 1 column matrices C,D and X, solve the given matrix equation for X. The multiplicative inverse of a matrix A is written A^(-1). Find the value of . improve our educational resources. CREATE AN ACCOUNT Create Tests & Flashcards. matrix. The Determinant. first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. In the rest of this section, a method is developed for finding a multiplicative inverse for square matrices. (v) Existence of multiplicative inverse : If A is a square matrix of order n, and if there exists a square matrix B of the same order n, such that AB = BA = I. where I is the unit matrix of order n, then B is called the multiplicative inverse matrix of … study For our coefficient matrix, we have this matrix as the multiplicative inverse matrix: We can check whether this inverse is real or not by multiplying it with our coefficient matrix to see if we get the identity matrix. Describe the types of matrices that are considered to be nonsingular. You can re-load this page as many times as you like and get a new set of numbers each time. Free matrix inverse calculator - calculate matrix inverse step-by-step. Now, we have this: Our answer can then be easily found by just translating this back into equation form. succeed. Get access risk-free for 30 days, Because the determinant is equal to zero in this problem, or, We use the inverse of a 2x2 matrix formula to determine the answer. If you know the inverse of a matrix, you can solve the problem by multiplying the inverse of the matrix with the answer matrix, x = A sup -1 * b. Log in or sign up to add this lesson to a Custom Course. Precalculus : Find the Multiplicative Inverse of a Matrix Study concepts, example questions & explanations for Precalculus. Each transistor requires 3 units of copper, 1 unit of zinc, and 2 units of glass. Augmented matrix method Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A -1 ]. Create an account to start this course today. Step 2:. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. Define multiplicative inverse. One interesting result is that a residue matrix has a multiplicative inverse if the determinant of the matrix has a multiplicative Given a matrix. your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the When we deal with regular numbers, our multiplicative inverse is the number we multiply by to get 1. For this lesson, we will talk about its benefits. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. Returns the multiplicative inverse of a square matrix specified as an array or range. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one
Wisconsin Wildflowers Yellow, Adaptability Culture In Organizations, Tree Of Savior Hunting Grounds 2020, 2 Timothy 2:16 Kjv, Aristolochia Florida Native, Best Grass For Goats In The Philippines, Seymour Duncan Sh2 Bridge, | 2021-05-18T00:48:18 | {
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https://math.stackexchange.com/questions/2261274/the-whole-space-is-an-open-set-in-a-metric-space | # The whole space is an open set in a metric space
Let $(X,d)$ be a metric space. Then both $X$ and $\emptyset$ are open sets.
I went over the proofs and it seems to be trivial, and I get it for $\emptyset$, but why must the whole space be open? Can't $X$ be a closed ball?
• Being closed or open is not an intrinsic property. The topology of a space does not know if the space itself is embedded somewhere else, nor what being open or closed in that topology means.
– user228113
May 1, 2017 at 22:22
• $X$ is always both open and closed in any topology.
– JMJ
May 1, 2017 at 22:23
• In a topological space, $X$ and $\emptyset$ are both open and closed. Sets can be both open and closed (the properties are not exclusive). Such sets are called clopen. May 1, 2017 at 22:23
• A metric space is a type of topological space. May 1, 2017 at 22:27
• What is your definition of "open"? Use that. Also, there is nothing wrong with something being simultaneously open and closed. May 1, 2017 at 22:27
If $(X,d)$ is a metric space, then as far as closed and open sets in $X$ go, $X$ is the universe. There is nothing outside of $X$. There is nothing which $X$ is a part of. There is no intrinsic notion of a "boundary" of an arbitrary metric space.
Here is why $X$ is an open subset of $X$. Let $x \in X$. To show $X$ is open in $X$, we need to show there is at least one $\epsilon > 0$ such that the ball $B(x,\epsilon)$ with center $x$ and radius $\epsilon$ is contained in $X$. Note that by definition,
$$B(x,\epsilon) = \{ y \in X : d(x,y) < \epsilon \}$$
so $B(x,\epsilon)$ is always a subset of $X$, for any choice of $\epsilon$ whatsoever. This is why $X$ is open in $X$.
Example: Let $X = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \}$ be the closed disc in $\mathbb{R}^2$. Give $X$ the Euclidean metric. If you consider $X$ as a subset of the metric space $\mathbb{R}^2$ (with the same metric), it is obviously not open. But if you consider $X$ as a subset of the metric space $X$, it is open.
If you are confused, think about the fact that open balls (and more generally open sets) in $X$ are in general not open when you consider them as subsets of $\mathbb{R}^2$. take the point $(1,0)$ in $X$ and consider the ball in $X$ of radius $\frac{1}{2}$ and center $(1,0)$. Draw it. By definition, it is an open set in the metric space $X$. But it's not going to be an open circle. And it will not be open when considered as a subset of $\mathbb{R}^2$.
• "By definition, it is an open set in the metric space X" that is because the ball contains every point of $X$ even if the ball is "cutted"?
– gbox
May 1, 2017 at 22:46
• What? It doesn't contain every point of $X$.
– D_S
May 1, 2017 at 22:47
• if X does not "live" in $\mathbb{R}^2$
– gbox
May 1, 2017 at 22:48
• I don't understand what you're saying.
– D_S
May 1, 2017 at 22:49
• I drew a picture of the ball of center $(1,0)$ and radius $\frac{1}{2}$ in $X$.
– D_S
May 1, 2017 at 22:58
If $x \in X$ is any point and $\epsilon > 0$ any positive real number, then the open ball of radius $\epsilon$ centered at $x$, $\{ y\in X \mid d(x,y) < \epsilon)\}$, is by definition contained in $X$. Note that there is nothing wrong with a subset being simultaneously closed and open in a space (despite the terminology).
For example, consider the boundary point $(1,0)$ of the closed disk $$D = \{(x,y) \in \mathbb{R}^2\mid x^2 + y^2 \leq 1\}.$$ By definition, the open ball with respect to the metric of the closed disk centered around $(1,0)$ of radius $\epsilon > 0$ is the set of points in $D$ that are at distance $< \epsilon$ from $(1,0)$. (This is the red subset of the disk in the picture. The line has length $\epsilon$.)
For instance, if $\epsilon = 1000$, then this ball is the whole disk $D$. But no matter the choice of $\epsilon$, the ball will always be a subset of the disk $D$.
On the other hand, if we consider open balls with respect to the metric of the plane, then any open ball around $(1,0)$ won't be contained inside $D$.
• But what about the point in the boundary of the space? there will be no ball that will satisfy the requirement
– gbox
May 1, 2017 at 22:34
• @gbox Yes, any ball of any radius centered around that point will do. Of course a closed ball is not going to be open in the plane, but it is always going to be open in itself. Note that the "open ball" in this case is the set of points inside the closed ball that are distance less than $\epsilon$ from the chosen boundary point; the metric knows nothing about points outside of the closed ball. May 1, 2017 at 22:37
• "But what about the point in the boundary of the space? there will be no ball that will satisfy the requirement" ALL balls satisfy the requirement! A "ball" around y ={x in X| d (x,y) < e} $\subset$ X. Even if the point is on the boundary of the space, then then ball includes boundary of the space. Consider 1 in [0,1]. A ball around 1 is all points close to 1. That is (1-e,1]. It contains all the points that are close to the left. And it contains all points that are close to the right. It's just that there are no points to the right. A ball doesn't have to be "round". May 1, 2017 at 23:08
• @gbox If $X$ has a boundary, that's not a problem. Looking at the example unit disc in the answer here, while we draw it as a subset of the plane, there is nothing "outside" $X$, whatever that would mean. $X$ is the collection of all points that exist as far as this problem is concerned; there are no other points that an open ball could contain. May 1, 2017 at 23:15
A set that contains everything must be open as every neighborhood of every point must be a subset of the space. So every point is an interior point.
Take for example the metric space $[0,1]$ (this is a "closed ball" obviously). Then $N_e (1)=\{x\in [0,1]| d (x,1)< e\}=(1-e] \subset [0,1]$. So 1 is an nterior point.
I have also seen a lot of "trivial" answers, which somehow look incomplete. I like to prove it with a contradiction. The theorem states:
$$X \subseteq X$$ is open because $$\forall x \in X : \exists \epsilon >0 : B_\epsilon (x) \subseteq X$$.
Suppose $$X \subseteq X$$ is not open. Then we logically negate the "open set" condition:
$$\exists x \in X : \forall \epsilon > 0 : B_\epsilon (x) \cap (X \setminus X) \ne \emptyset$$
In words: we expect points on the boundary span non-pointlike balls which should intersect with whatever is beyond the set X. Now, X contains all elements. In $$(X, d)$$ X is our universe. Any other set within can be equal or smaller than X. Any other set beyond is empty.
So, let's find that intersection. Firstly, $$X \setminus X = \emptyset$$, since X is the universe. Now comes the funny part. Any intersection with empty set is empty, since empty set does not possess any elements. And, if intersection is empty, then those sets do not overlap. And this is a contradiction.
Probably, the confusion stems from the symmetry of the boundary of the ball. The "open ball" set does not have to be filled homogeneously as a ball. The "open ball" is a ball only to the extent the underlying set is providing non-empty elements to fill that ball. | 2022-06-28T07:21:21 | {
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http://mathhelpforum.com/math-topics/221056-permutations-combinations.html | 1. ## Permutations and Combinations
Q1: The number of ways 8 people can be arranged in a straight line if two people insist on being separated is:
A) 15,120
B) 3,600
C) 30,240
D) 7,200
E) 5,400
No matter what I try I can't obtain any of these possible answers. I'm thinking it's something to do with the wording of the question, but I'm not sure what exactly.
Q2: A cricket team takes 11 players to a game and there are 14 players from which to choose; 1 wicket keeper, 7 batsmen and 6 bowlers.
a) How many different teams are possible if there are no restrictions?
I did: 14C11 = 364, but I'm unsure.
b) How many different teams are possible if the wicket keeper must be included?
I did 13C11 = 78, but I'm unsure.
c) How many different teams are possible if the team contains the wicket keeper, 6 batsmen and 4 bowlers?
2. ## Re: Permutations and Combinations
Hello, Fratricide!
Q1: the number of ways 8 people can be arranged in a straight line
. . . if two people insist on being separated is:
. . (A) 15,120 . . (B) 3,600 . . (C) 30,240 . . (D) 7,200 . . (E) 5,400
$\text{The eight people are: }\:\{A,B,C,D,E,F,G,H\}$
$\text{Let }A\text{ and }B\text{ be the two unfriendly people.}$
$\text{With no restrictions, there are }\,8! = 40,\!320\text{ arrangements.}$
$\text{Now consider the arrangements in which }A\text{ and }B\:are\text{ adjacent.}$
$\text{Duct-tape }A\text{ and }B\text{ together.}$
$\text{Then we have 7 "people" to arrange: }\,\left\{\boxed{AB}, C, D, E, F, G, H\right\}$
$\text{There are: }\,7!\text{ arrangements.}$
$\text{But }A\text{ and }B\text{ could be taped like this: }\,\boxed{BA}.$
$\text{Hence, there are: }\,2\cdot7! \,=\,10,\!080\text{ ways in which }A\text{ and }B\text{ are adjacent.}$
$\text{Therefore, the answer is: }\:40,\!320 - 10,\!080 \:=\:30,\!240\:\text{ . . . answer (C)}$
3. ## Re: Permutations and Combinations
Hello
I think you are right in Q2(a)
(b) 13C10( because the wicket keeper must be include)
(c) 7C6x6C4
4. ## Re: Permutations and Combinations
Originally Posted by Trefoil2727
Hello
I think you are right in Q2(a)
(b) 13C10( because the wicket keeper must be include)
(c) 7C6x6C4
Could you please elaborate on (c) to clarify my understanding?
5. ## Re: Permutations and Combinations
Originally Posted by Fratricide
Could you please elaborate on (c) to clarify my understanding?
well I really not good on explanation, because my English is sucked, but I'll try..
you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers( from 6). Therefore, 7C6x6C4
6. ## Re: Permutations and Combinations
Originally Posted by Trefoil2727
well I really not good on explanation, because my English is sucked, but I'll try..
you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers( from 6). Therefore, 7C6x6C4
Perfect, thanks. | 2017-03-25T18:02:05 | {
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https://www.physicsforums.com/threads/area-of-polar-graph.753074/ | # Area of Polar Graph
1. May 10, 2014
### _N3WTON_
1. The problem statement, all variables and given/known data
Find the area of the region.
Interior of: r = 2 - sin(b)
2. Relevant equations
A = 1/2 ∫ (r)^2 dr
3. The attempt at a solution
I really don't have any idea how to approach this problem. I don't understand how to determine my limits of integration. The only part of the problem I have accomplished is finding that b = arcsin(2). Also, I have found that the graph of this equation makes a convex limacon.
2. May 10, 2014
### Staff: Mentor
You should be able to figure out the integration limits from the graph. You could integrate from b = 0 to b = $2\pi$. Alternatively, you could integrate from b = $\pi/2$ to b = b = $3\pi/2$ and double that result.
3. May 10, 2014
### _N3WTON_
ok so would my equation look something like ∫ [(1-sin(b))^2 db] with the limits of integration going from 0 to 2pi?
4. May 10, 2014
### vanhees71
Looks good, provide $b$ is the polar angle. Why don't you use LaTeX for typesetting formulas? It's no big deal to type it and it makes everything much easier to read and understand!
5. May 10, 2014
### _N3WTON_
I don't know how to use latex, nor do I know where to find it....sorry
6. May 10, 2014
### _N3WTON_
Ok, I am confused about this problem now. The website for my calculus textbook offers a 24/7 tutoring service. When I asked the tutor about this problem he said I could not integrate from 0 to 2pi; however, he was also not sure how to do the problem. So can I safely assume that he was wrong about being unable to integrate from 0 to 2pi?
7. May 10, 2014
### vanhees71
Well, could you post the full question? I don't know, why one shouldn't integrate over the full range of the angle, because $r(b)=2-\sin b>0$ for all $b \in [0,2 \pi]$.
8. May 10, 2014
### LCKurtz
Yes, like this:
Code (Text):
$\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta$
which displays like this: $\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta$
9. May 10, 2014
### Staff: Mentor
It doesn't give me a lot of confidence when someone tells me I can't do a problem a certain way, but then doesn't know how to do the problem.
If you look at the graph, it's pretty obvious that you can integrate using those limits, regardless of what the tutor is saying.
10. May 10, 2014
### _N3WTON_
I did post the full question.
11. May 10, 2014
### _N3WTON_
Thank you...my first instinct was to integrate from 0 to 2pi; however, the tutor swayed me lol... | 2017-10-20T09:34:33 | {
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https://math.stackexchange.com/questions/350936/2-int-0-pi-sin-x-sin-x-cos-x-dx-where-am-i-going-wrong | # $2\int_0^\pi \sin x + \sin x \cos x\,dx$. Where am I going wrong?
$2\displaystyle\int_0^\pi \sin x + \sin x \cos x\,dx$
I let $u=\sin x \implies \dfrac{du}{dx}=\cos x \implies dx=\dfrac{du}{\cos x}$
$2\displaystyle\int_{x=0}^{x=\pi} 2u\,du$
When I try to change the limits I just get 0 and 0: Lower limit $=\sin 0 = 0$, upper limit $=\sin \pi=0$
$2\left[u^2\right]^0_0 = 0$. The answer should be 4, not 0.
Thanks,
• What about zeros of $\cos x$ when you deal with $\frac{\mathrm du}{\cos x}$? Why don't you go with the classical $u = \tan\frac x2$? – Ilya Apr 4 '13 at 9:30
• You seem to have forgotten to divide the $\sin x$ term by the $\cos x$ obtained from the substitution. Also, $\sin$ is not bijective on $(0,\pi)$ rendering the application of the substitution theorem invalid. If you want to solve by substitution, it's probably best to split the integration interval into $(0,\pi/2)$ and $(\pi/2,\pi)$. – Lord_Farin Apr 4 '13 at 9:36
Assuming the question is : $2\int_{0}^{\pi}(sin(x) + sin(x)cos(x))dx$
It is not 2$\int_{0}^{\pi} 2u. du$ on simplification.
It is 2$\int_{0}^{\pi} udx + udu$
Instead of substituting, you can just divide the integration into two and proceed as follows:
$$=2\int_{0}^{\pi}sinx .dx + \int_{0}^{\pi} sin(2x).dx$$
$$=2(-cos(x)) + \dfrac{(-cos(2x))}{2}$$ Then apply the limit from $0$ to $\pi$:
$$=2(1-(-1)) + 0$$
$$=4$$
Hope the answer is clear !
• Now I get it, thank you very much! – Hans Groeffen Apr 4 '13 at 9:48
As $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$
If $f(x)=\sin x+\sin x\cos x,$ $f(\pi+0-x)=\sin(\pi+0-x)+\sin(\pi+0-x)\cos(\pi+0-x)=\sin x-\sin x\cos x$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$
So, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi (\sin x - \sin x \cos x)dx=I\text{ say}$$
So, $$2I=\int_0^\pi (\sin x + \sin x \cos x)dx+\int_0^\pi (\sin x - \sin x \cos x)dx=2 \int_0^\pi\sin xdx$$
So, $$I=\int_0^\pi\sin xdx=(-\cos x)|_0^\pi=-\cos\pi-(-\cos 0)=2$$
Alternatively, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi \left(\sin x +\frac{\sin2x}2\right)dx=\left(-\cos x -\frac{\cos2x}4\right)_0^\pi$$
$$=\left(\cos x +\frac{\cos2x}4\right)_\pi^0=\cos0+\frac{\cos0}4-\left(\cos\pi+\frac{\cos\pi}4\right)=1+\frac14-\left(-1+\frac14\right)=2$$
In fact, $\int_0^\pi\sin2xdx=\int_0^{2\pi}\sin ydy=(-\cos y)_0^{2\pi}=1-1=0$
You lost the first summand there.
$$2\int_0^\pi \sin x + \sin x \cos x\,dx = 2\int_0^\pi \sin x\,dx + 2\int_0^\pi \sin x \cos x\,dx$$ Now integrate both summands separately; you don't need to substitute in the first.
• Even if you are in a class and there is a current subject, you should be ready to use other techniques, or you should realize that the current subject applies only partially. – Eric Jablow Apr 5 '13 at 1:16 | 2019-08-20T11:04:50 | {
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https://proofwiki.org/wiki/Determinant_of_Triangular_Matrix | # Determinant of Triangular Matrix
## Theorem
Let $\mathbf T_n$ be a triangular matrix (either upper or lower) of order $n$.
Let $\det \left({\mathbf T_n}\right)$ be the determinant of $\mathbf T_n$.
Then $\det \left({\mathbf T_n}\right)$ is equal to the product of all the diagonal elements of $\mathbf T_n$.
That is:
$\displaystyle \det \left({\mathbf T_n}\right) = \prod_{k \mathop = 1}^n a_{k k}$
## Proof
Let $\mathbf T_n$ be an upper triangular matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf T_n$.
### Basis for the Induction
For $n = 1$, the determinant is $a_{11}$, which is clearly also the diagonal element.
This forms the basis for the induction.
### Induction Hypothesis
Fix $n \in \N$.
Then, let:
$\mathbf T_n = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$
be an upper triangular matrix.
Assume that:
$\displaystyle \det \left({\mathbf T_n}\right) = \prod_{k \mathop = 1}^n a_{k k}$
This forms our induction hypothesis.
### Induction Step
Let $\mathbf T_{n+1}$ be an upper triangular matrix of order $n + 1$.
Then, by the Expansion Theorem for Determinants (expanding across the $n + 1$th row):
$\displaystyle D = \det \left({\mathbf T_{n + 1}}\right) = \sum_{k \mathop = 1}^{n + 1} a_{n + 1 \, k} T_{n + 1 \,k}$
Because $\mathbf T_{n + 1}$ is upper triangular, $a_{n+1 \, k} = 0$ when $k < n + 1$.
Therefore:
$\det \left({\mathbf T_{n + 1}}\right) = a_{n + 1 \, n + 1} T_{n + 1 \, n + 1}$
By the defintion of the cofactor:
$T_{n + 1 \, n + 1} = (-1)^{n + 1 + n + 1} D_{n + 1 \, n + 1} = D_{n \, n}$
where $D_{n \, n}$ is the order $n$ determinant obtained from $D$ by deleting row $n + 1$ and column $n + 1$.
But $D_{n \, n}$ is just the determinant of an upper triangular matrix $\mathbf T_n$.
Therefore:
$\det \left({\mathbf T_{n + 1}}\right) = a_{n + 1 \, n + 1} \det \left({\mathbf T_n}\right)$
and the result follows by induction.
From:
Transpose of Upper Triangular Matrix is Lower Triangular
the result also holds by Determinant of Transpose for lower triangular matrices.
$\blacksquare$ | 2019-10-18T16:44:53 | {
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https://byjus.com/question-answer/i-how-many-two-digit-numbers-are-divisible-by-3-ii-find-the-number-of/ | Question
# (i) How many two-digit numbers are divisible by 3? (ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5(iii) How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Solution
## (i) How many two-digit numbers are divisible by 3? The two digits numbers divisible by 3 are, $$12,15,18,21,24, \ldots ., 99$$ The above numbers are A.P.So, first number $$a=12$$ Common difference $$d=15-12=3$$Then, last number is 99 We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$$$\begin{array}{l}99=12+(n-1) 3 \\99=12+3 n-3 \\99=9+3 n \\99-9=3 n \\3 n=90 \\n=90 / 3 \\n=30\end{array}\\$$Therefore, 30 two digits number are divisible by 3(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5The natural numbers which are divisible by both 2 and 5 are $$110,120,130,140, \ldots, 999$$ The above numbers are A.P.So, first number $$a=110$$ Common difference $$d=120-110=10$$Then, the last number is 999 We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$$$\begin{array}{l}999=110+(n-1) 10 \\999=110+10 n-10 \\999=100+10 n \\999-100=10 n \\10 n=888 \\n=888 / 10 \\n=88\end{array}\\$$The number of natural numbers which are divisible by both 2 and 5 are 88(iii) How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?The numbers which are lie between 10 and 300 , when divisible by 4 leave a remainder 3 are $$11,15,19,23,27, \ldots .299$$The above numbers are A.P.So, first number $$a=11$$ Common difference $$d=15-11=4$$ Then, the last number is 299 We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$$$\begin{array}{l}299=11+(n-1) 4 \\299=11+4 n-4 \\299=7+4 n \\299-7=4 n \\4 n=292 \\n=292 / 4 \\n=73\end{array}\\$$The total which are lie between 10 and 300, when divisible by 4 leaves a remainder 3 are 73Mathematics
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View More | 2022-01-24T05:53:21 | {
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http://mathhelpforum.com/pre-calculus/166031-lawn-mower-cord-problem-radians-angular-velocity.html | 1. ## Lawn Mower Cord Problem (Radians and Angular Velocity)
Hi, this is my first time on this website and I hope to help other people while getting help myself. First off, I was absent the day this homework was assigned; therefore I have no idea how to do it. Anyway, here is the problem, THANKS.
Yank Hardy pulls the cord on his power mower. in order for the engine to start, the pulley must turn at 180 revolutions per minute. The pulley has a radius of 0.2 ft.
1. How many radians per second must the pulley turn?
2. How fast must Yank pull the cord to start the mower?
3. When Yank pulls this hard, what is the angular velocity of the center of the pulley?
Thanks for looking I appreciate all help!!!!!!!!
2. Perhaps this is enough to get you started:
There are pi radians in 180 degrees.
1 revolution = 360 degrees.
The linear speed at the edge of a pulley is
As long as the pulley is rigid the angular speed from the center is the same throughout the pulley.
3. Originally Posted by snowtea
Perhaps this is enough to get you started:
There are pi radians in 180 degrees.
1 revolution = 360 degrees.
The linear speed at the edge of a pulley is
As long as the pulley is rigid the angular speed from the center is the same throughout the pulley.
This is useful information. I can answer number 1 from what you told me, and is my math correct?
180 revs/min = 3 revs/sec
2pi = 360 degrees
2pi = 1 rev
So number one is 6pi radians/sec?
Also, thank you for the linear speed equation, but I cannot complete it because I do not know how to get angular speed. Can you please provide me with some additional information on angular speed so I can calculate the linear speed? Thank you I greatly appreciate your help.
4. I think 6pi rads/sec is correct.
And this *is* your angular speed
5. Originally Posted by snowtea
I think 6pi rads/sec is correct.
And this *is* your angular speed
Thank you again, but another question . Since I have to multiply my angular speed by radius to get linear speed, does it matter what unit the radius is in? for example, in the problem it is .2feet, but I could also use 2.4 inches or something else and depending on which unit I use, the answer will differ.
6. Yup. So the final speed will have different units.
E.g. feet / sec or inches / sec
7. Originally Posted by snowtea
Yup. So the final speed will have different units.
E.g. feet / sec or inches / sec
OK so I'm still a little confused, but would this be correct:
(6pi rads/sec) * (0.2 ft) = 1.2pi feet/ sec
8. Yes, 1.2pi ft/sec looks correct.
The idea in using radians is to simplify calculations of arc length.
For example you want to find the distance around a circle: this is 2*pi*r
so they defined 360 degrees to be 2*pi radians
and you can directly get the distance by multiplying.
If you want to find the distance of a half circle pi*r
Radians for 180 degrees is pi
If you want to find the distance of a quarter circle pi/2 * r
Radians for 90 degrees is pi/2
Also, speed = distance / time, so radians work the same way
speed = distance / time = radians * r / time = (radians / time) * r | 2017-02-26T08:39:39 | {
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http://mathhelpforum.com/calculus/32065-coordinates-intersection-circle-sphere.html | # Math Help - Coordinates of intersection of circle and sphere
1. ## Coordinates of intersection of circle and sphere
How do I calculate the x, y, & z coordinates of the two points where a sphere intersects a circle? The coordinates of the center of both are known and the radii are known.
The center of the circle and the center of the sphere are not the same point.
The circle and the sphere are not tangent to each other.
The center of the sphere is not on the perpendicular through the center of the circle. I.e., the solution is not the circle, only two points.
You may orient the circle as a rotation around an axis (but the sphere is not on the axis of rotation of course).
DSG
2. Why not give the specific problem statement (radii, centers, etc) - it would make the explanation much easier, I would think.
3. I was looking for a general equation. The center of the circle can move along a line and will have various radii. The sphere has a fixed center in relation to the line, but it too will have varying radii.
DSG
4. Originally Posted by DSG
How do I calculate the x, y, & z coordinates of the two points where a sphere intersects a circle? The coordinates of the center of both are known and the radii are known.
The center of the circle and the center of the sphere are not the same point.
The circle and the sphere are not tangent to each other.
The center of the sphere is not on the perpendicular through the center of the circle. I.e., the solution is not the circle, only two points.
You may orient the circle as a rotation around an axis (but the sphere is not on the axis of rotation of course).
The given circle C lies in a plane. The plane is determined by the location of the centre of C, together with a normal vector N (is that what you called the axis of rotation?).
The plane containing C will meet the sphere in another circle Γ (assuming that this intersection is not empty). The centre of Γ will be a point P on a line through the centre of the sphere in the direction of N. Using that information, you can find the location of P, and hence its distance d from the centre of the sphere. If the radius of the sphere is r then the radius of Γ will be $\sqrt{r^2-d^2}$ (by Pythagoras).
Now you know the radius of Γ. You can also find the distance between the centres of C and Γ, and you also know the radius of C. So that reduces the whole problem to a simple one of finding the intersection of two circles in a plane.
5. Opalg,
Slicing the sphere into a circle on the same plane as the original circle helped me understand the solution.
Thanks for the help,
DSG
6. ## algorithmic geometry solution
For this, you need to have already solved the general intersection of two 2D circles as a 2D algorithm (able to handle all cases):
Given: SPH = [ c r ] CIR = [ c orient r ]
Solve: numIntersectionPts (0,1,2,infinite), i1 i2
1. Translate coordinates adopting CIR.c as newOrigin (compute SPH --> SPH')
SPH' = Translate(CIR.c, SPH)
2. Rotate coordinates adopting CIR.orient as the newZaxis (compute SPH' --> SPH")
Rotator R = RotatorForNewZAt(CIR.orient)
SPH" = Rotate(R, SPH')
3. Now, CIR" lies flat in x"-y" plane centered at origin.
4. Visualize 2nd circle where x"-y" plane cuts thru SPH"
if (abs(SPH".c.z) > SPH.r) numIntersectionPts = 0; return; // SPH too distant from x'=y" plane
5. Solve i1" i2" in 2D as the intersection of these two circles:
cir1 = [ 0 0 ] CIR.r
cir2 = [ SPH".c.x SPH".c.y ] sqrt(sqr(SPH.r) - sqr(SPH".c.z))
numIntersectionPts = IntersectionOf(cir1, cir2, /*returns*/ i1", i2")
6. If numIntersectionPts == 1 or 2,
back-transform solution points into original coordinates:
i1 = Untranslate(CIR.c, Unrotate(R, i1"))
i2 = Untranslate(CIR.c, Unrotate(R, i2"))
This is a general solution (works in all cases).
Illustrated solution in: Flexing the Power of Algorithmic Geometry by Pierre Bierre
7. Thanks Pierre.
DSG | 2016-02-06T19:09:41 | {
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https://crypto.stackexchange.com/questions/75295/partial-key-recovery-from-linear-equations | Partial key recovery from linear equations
I have searched for my question but I didn't find any relevant answer to my situation. I guess maybe it is too easy but I am a newbie in crypto and I can't figure out the answer. Here is the exercise:
An attacker mounts a partial key recovery attack on a cipher with a 128-bit key and manages to find a system of $$k$$ independent linear equations involving the secret key bits with $$k < n$$.
For example:
$$k_{1} \oplus k_2 \oplus k_{63} = 1$$
$$k_{1} \oplus k_3 \oplus k_{17} = 1$$
$$k_{43} \oplus k_{56} \oplus k_{60} = 0$$
Does this info help him to attack the cipher? If yes, what would be the attacker’s gain over the complexity of an exhaustive search?
I know intuitively that this reduces the complexity because of the common $$k_1$$ in 2 of the equations and I tried using truth tables for each equation but I can't extract the answer from them. Can anyone explain how is the complexity reduced?
• @kelalaka Omg, I failed to see it but you are so right! Thank you so much! Oct 25 '19 at 10:26
Firstly consider the simple case where we know that there is only a simple x-or relation between two key bits. $$k_i \oplus k_j = 1$$ and $$i \neq j$$ and call it $$R1$$. If we consider the x-or table we will see that only half of the rows for $$k_i$$ and $$k_j$$ will satisfies this equation.
$$\begin{array}{cc|c} k_i & k_j & k_i\oplus k_j \\ \hline 0 & 0 & \color{blue}{0} \\ 0 & 1 & \color{red}{1} \\ 1 & 0 & \color{red}{1} \\ 1 & 1 & \color{blue}{0} \\ \end{array}$$
An equation with equal to $$0$$ has the same result, i.e. x-or splits the table into half.
Now, lets look at a key search by brute-force.
for key in possible_keys
if ciphertext = Enc(key,m)
return key
Clearly, this will take $$\mathcal{O}(2^n)$$-time for n-bit cipher.
Now using the relation update this key search
for key in possible_keys
if the key doesn't satisfy R1
continue
if ciphertext = Enc(key,m)
return key
What is the gain here? For half of the keys, the loop will not compute the Enc this means that the keyspace is reduced by one-bit due to the R1.
One might say that we are still looping as the number of all possible keys. The cost of an Enc is not considered in the $$\mathcal{O}(2^n)$$-time. If you add it then you will see that, you will get an approximately a double speed up. It might possible to design an iterator that only loops valid keys, however, that is a software issue. There is a another approach for this see, below.
A relation having more than two key bits will have the same result. Half of the time we don't need to calculate the Encryption.
If we have more than one relation for the speed up calculation we will use fixing bits by equations. Lets look at your your equations to see this:
$$k_{1} \oplus k_2 \oplus k_{63} = 1$$
$$k_{1} \oplus k_3 \oplus k_{17} = 1$$
$$k_{43} \oplus k_{56} \oplus k_{60} = 0$$
While iterating we if we know $$k_1$$ and $$k_2$$ than the value of $$k_{63}$$ is fixed. Even if one of $$k_1$$ and $$k_2$$'s is flipped $$k_{63}$$ is also flipped. So, you iterate overall, expect the $$k_{63}$$ since it is fixed. To benefit this you can use a permutation on the keys for better iteration.
Similarly, for your second and third equations. So, you have fixed three keys depending on the relation. This can result in a 3-bit key search advantage. Roughly, we can say, $$l$$ equations gives $$l$$-bit advantage.
• Actually, we can do better than that. From the relation $k_1 \oplus k_2 \oplus k_{63} = 1$, then if we know $k_1$ and $k_2$, we immediately know $k_{63}$. Furthermore, if we flip $k_1$ xor $k_2$, then $k_{63}$ flips. So, what we do is set $k_{63}$ initially to be consistent with the initial $k_1$ and $k_2$ and iterate over all the key bits except $k_{63}$ (flipping it when necessary). By using all three equations, and skipping over the 3 bits that we can update, this reduces the number of key bits we need to iterate over to $n-3$. Oct 25 '19 at 13:34
• @poncho thanks, from this side it is much clear to see. Oct 25 '19 at 13:56 | 2022-01-25T06:02:00 | {
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https://websuasion.com/journal/viewtopic.php?5dc019=exponential-distribution-standard-deviation | Therefore, $$X \sim Exp(0.25)$$. If X ∼ X ∼ Exponential (1) then by LOTUS (REFERNCE) and Equation (5.1) E(Xk) =∫ ∞ 0 xke−xdx =k! Simply, it is an inverse of Poisson. If it is a negative value, the function is zero only. The exponential and gamma distribution are related. E ( X k) = ∫ 0 ∞ x k e − x d x = k! This section requires you to be a Pass Your Six Sigma Exam member. The standard deviation, $$\sigma$$, is the same as the mean. It can be shown for the exponential distribution that the mean is equal to the standard deviation; i.e., μ = σ = 1/λ Moreover, the exponential distribution is the only continuous distribution that is This section requires you to be logged in. Log in or Sign up in seconds with the buttons below! (Taken from ASQ sample Black Belt exam.). Have each class member count the change he or she has in his or her pocket or purse. IASSC Lean Six Sigma Green Belt Study Guide, Villanova Six Sigma Green Belt Study Guide, IASSC Lean Six Sigma Black Belt Study Guide, Villanova Six Sigma Black Belt Study Guide, Where e is base natural logarithm = 2.71828. f ( x) = e-x/A /A, where x is nonnegative. The exponential distribution is one of the widely used continuous distributions. It estimates the lapse of time between the independent events. The distributions of a random variable following exponential distribution is shown above. Therefore, X ~ Exp(0.25). The distribution notation is X ~ Exp(m). The mean for the exponential distribution equals the mean for the Poisson distribution only when the former distribution has a mean equal to. The number of hours a mobile phone runs before its battery dies out. Login to your account OR Enroll in Pass Your Six Sigma Exam. Exponential Distribution Modelling Of Wet-day Rainfall Totals Assume An Exponential Distribution Can Be Used To Model Precipitation Totals On Wet Days. 16. If a car arrives at the drive-thru just before you, find the probability that you will wait for, So, P(x<5) = 1- e(-1/10)*5 = 1- e-0.5 = 1-2.71828(-0.5) =0.3934, P(x<10) = 1- e(-1/10)*10 = 1- e-1 = 1-2.71828(-1) =0.6321, So, P(510) = e(-1/10)*10 = e-1= 2.71828(-1) =0.3678, Question: If a process follows an exponential with a mean of 25, what is the standard deviation for the process? The exponential distribution estimates the time lapse between two independent events in a Poisson process. Exponential Distribution Moment Generating Function. The exponential distribution can be used to determine the probability that it will take a given number of trials to arrive at the first success in a Poisson distribution; i.e. it describes the inter-arrival times in a Poisson process.It is the continuous counterpart to the geometric distribution, and it too is memoryless.. http://www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf, Your email address will not be published. Construct a histogram of the dat What is standard deviation of a process that operates to an exponential dustribution with a … Exponential distribution is the time between events in a Poisson process. This statistics video tutorial explains how to solve continuous probability exponential distribution problems. The mean of exponential distribution is 1/lambda and the standard deviation is also 1/lambda. Scientific calculators have the key “e … The exponential and gamma distribution are related. When a probability of an outcome is consistent throughout the time period. It is used to model items with a constant failure rate. $$\mu = \sigma$$ The distribution notation is $$X \sim Exp(m)$$. Scientific calculators have the key "$$e^{x}$$." | 2021-04-14T08:03:10 | {
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https://math.stackexchange.com/questions/1742193/covariance-of-two-random-vectors/1751891 | # Covariance of two random vectors
IF I have $X,Y_1,...,Y_n$ iid then how do I calculate:
cov $\left [\begin{pmatrix}X\\.\\.\\.\\X \end{pmatrix}, \begin{pmatrix}Y_1\\.\\.\\.\\Y_n \end{pmatrix}\right]$?
This is known as the cross-covariance between vectors, and is defined by $$\text{cov}[\boldsymbol{X},\boldsymbol{Y}] = \text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}]$$
where $$\boldsymbol{\mu_X} = \text{E}[\boldsymbol{X}]\\ \boldsymbol{\mu_Y} = \text{E}[\boldsymbol{Y}]$$
In your case, because all the components of $\boldsymbol{X}$ are the same, things simplify greatly.
$$\boldsymbol{X} = X \left[ \begin{array}{c}1\\1\\\vdots\\1\end{array} \right], \;\; \boldsymbol{\mu_X} = \mu_X \left[ \begin{array}{c}1\\1\\\vdots\\1\end{array} \right]$$ Where $\mu_X=\text{E}[X]$. Then $$\boldsymbol{X}-\boldsymbol{\mu_X} = (X-\mu_X) \left[ \begin{array}{c}1\\1\\\vdots\\1\end{array} \right]$$ Now $$(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T} = (X-\mu_X) \left[ \begin{array}{c}1\\1\\\vdots\\1\end{array} \right]\left[ \begin{array}{cccc}Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\end{array} \right]$$ where $\mu_m=\text{E}[Y_m]$ for $m\in[1,2,\cdots,n]$. Expanding out that matrix product we have $$(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T} = (X-\mu_X)\left[ \begin{array}{cccc} Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\\ Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\\ \vdots&\vdots&\ddots&\vdots\\ Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n \end{array} \right]$$
Taking that scalar inside the matrix, we see it multiplies each entry in the matrix. Then taking the expectation of the result finally gives $$\text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}] = \left[ \begin{array}{cccc} \text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)]\\ \text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)]\\ \vdots&\vdots&\ddots&\vdots\\ \text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)] \end{array} \right]$$ $$= \left[ \begin{array}{cccc} \text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n)\\ \text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n)\\ \vdots&\vdots&\ddots&\vdots\\ \text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n) \end{array} \right]$$
Now we are at the answer: you specified all the variables to be identically distributed and independent. Independent variables have covariance $0$. SO, you get the all zeros matrix for your answer $$\text{cov}(\boldsymbol{X},\boldsymbol{Y})=\text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}] = \left[ \begin{array}{cccc} 0&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\\ \end{array} \right]$$
• "The very definition of independent variables is that their covariance is $0$" is completely wrong. Independence implies zero covariance, but variables with zero covariance are not necessarily independent. You have confused "independent" with "uncorrelated". – Anon Apr 24 '16 at 10:53
• Thanks for the correction. – rajb245 Apr 26 '16 at 3:21
Covariance of $2$ vectors is basically what is called a variance-covariance matrix $(\Sigma)$ defined as $$((\Sigma_{ij}))=Cov(X_i,Y_j)$$ where $Cov(A,B)=E(AB)-E(A)E(B)$
For more details, just Google Variance Covariance matrix.
Specifically, because of the iid character of your variables, $Cov$ will be $0$ for all.
• The variance covariance matrix is for one vector containing multiple random variables. Thia is for two vectors. – usainlightning Apr 18 '16 at 17:25
• @usainlightning : Please read the 2nd paragraph of the Wikipedia article on Covariance matrix before proceeding. – Qwerty Apr 18 '16 at 20:26
• actually I agree with Qwerty, with the precision that we don't really care about the two random vectors... One could concatenate the two random vectors together and compute the classical covariance matrix of a random vector. – Marine Galantin Nov 12 '19 at 1:05 | 2021-01-22T14:59:53 | {
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https://math.stackexchange.com/questions/582180/matrix-question-regarding-symmetric | # Matrix question regarding symmetric
I need help with the following problem
Express the matrix $$B=\begin{pmatrix} 2 &-2 &-4 \\ -1 &3 &4\\ 1 &-2 &-3 \end{pmatrix}$$ as the and sum of a symmetric and a skew symmetric matrices.
• What is your question? – Trevor Wilson Nov 26 '13 at 17:40
• Can you explain what you have tried to do? – A.E Nov 26 '13 at 17:42
In general,
$$\mathbf{M} = \dfrac{\mathbf{M} + \mathbf{M}^t}{2} + \dfrac{\mathbf{M} - \mathbf{M}^t}{2}.$$
The first term is a symmetric matrix; the second is skew-symmetric.
This is an instance of a much more general theorem. The "transpose" operation, done twice, is the identity. Such operations are called "involutions". And involutions all "split" their domains this way:
If you call that involution (think of "transpose") $T$, we have $T^2 - I = 0$. That means that $+1$ and $-1$ are the only eigenvalues for $T$,. Furthermore, the domain of $T$ splits into a direct sum of the $+1$ and $-1$ eigenspaces. That's the larger theorem I was referring to.
In fact, the solution above (for transpose) generalizes as well: for any vector in the domain of $T$, we have $$v = \frac{1}{2} (v + T(v)) + \frac{1}{2} (v - T(v))$$ which splits $v$ as a sum of a vector that's invariant under $T$ and one that negates when $T$ is applied, i.e., it splits $v$ as a linear combination of a multiple of a $+1$ eigenvector and a multiple of a $-1$ eigenvector.
Notice that the fraction $\frac{1}{2}$ appears here. That means that this trick, as written, only works over fields of characteristic not equal to two. You might ask whether it's still true in characteristic two. You could try some examples to find out.
The theorem can be applied to show that every function on the reals is the sum of an even and an odd function, that every complex number is the sum of a real and a pure imaginary (OK, that's pretty trivial), etc.
• So we have to give this answer in exam as well – user110715 Nov 26 '13 at 17:50 | 2021-06-17T09:35:45 | {
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https://www.math24.net/alternating-series/ | Alternating Series
A series in which successive terms have opposite signs is called an alternating series.
The Alternating Series Test (Leibniz’s Theorem)
This test is the sufficient convergence test. It’s also known as the Leibniz’s Theorem for alternating series.
Let $$\left\{ {{a_n}} \right\}$$ be a sequence of positive numbers such that
1. $${a_{n + 1}} \lt {a_n}$$ for all $$n$$;
2. $$\lim\limits_{n \to \infty } {a_n} = 0.$$
Then the alternating series $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^n}{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^{n – 1}}{a_n}}$$ both converge.
Absolute and Conditional Convergence
A series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is absolutely convergent, if the series $$\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|}$$ is convergent.
If the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is absolutely convergent then it is (just) convergent. The converse of this statement is false.
A series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is called conditionally convergent, if the series is convergent but is not absolutely convergent.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Use the alternating series test to determine the convergence of the series $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^n}\large\frac{{{{\sin }^2}n}}{n}\normalsize}.$$
Example 2
Determine whether the series $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^n}\large\frac{{2n + 1}}{{3n + 2}}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Example 3
Determine whether $$\sum\limits_{n = 1}^\infty {\large\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{{n!}}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Example 4
Determine whether the alternating series $$\sum\limits_{n = 2}^\infty {\large\frac{{{{\left( { – 1} \right)}^{n + 1}}\sqrt n }}{{\ln n}}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Example 5
Determine the $$n$$th term and test for convergence the series
${\frac{2}{{3!}} – \frac{{{2^2}}}{{5!}} }+{ \frac{{{2^3}}}{{7!}} – \frac{{{2^4}}}{{9!}} + \ldots }$
Example 6
Investigate whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{{5n – 1}}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Example 7
Determine whether the alternating series $$\sum\limits_{n = 1}^\infty {\large\frac{{{{\left( { – 1} \right)}^n}}}{{\sqrt {n\left( {n + 1} \right)} }}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Example 1.
Use the alternating series test to determine the convergence of the series $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^n}\large\frac{{{{\sin }^2}n}}{n}\normalsize}.$$
Solution.
By the alternating series test we find that
${\lim\limits_{n \to \infty } \left| {{a_n}} \right| } = {\lim\limits_{n \to \infty } \left| {{{\left( { – 1} \right)}^n}\frac{{{{\sin }^2}n}}{n}} \right| } = {\lim\limits_{n \to \infty } \frac{{{{\sin }^2}n}}{n} = 0,}$
since $${\sin ^2}n \le 1.$$ Hence, the given series converges.
Example 2.
Determine whether the series $$\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^n}\large\frac{{2n + 1}}{{3n + 2}}\normalsize}$$ is absolutely convergent, conditionally convergent, or divergent.
Solution.
We try to apply the alternating series test here:
${\lim\limits_{n \to \infty } \left| {{a_n}} \right| } = {\lim\limits_{n \to \infty } \frac{{2n + 1}}{{3n + 2}} } = {\lim\limits_{n \to \infty } \frac{{\frac{{2n + 1}}{n}}}{{\frac{{3n + 2}}{n}}} } = {\lim\limits_{n \to \infty } \frac{{2 + \frac{1}{n}}}{{3 + \frac{2}{n}}} }={ \frac{2}{3} \ne 0.}$
Since the $$n$$th term does not approach $$0$$ as $$n \to \infty,$$ the given series diverges.
Page 1
Problems 1-2
Page 2
Problems 3-7 | 2020-08-07T15:24:09 | {
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https://www.physicsforums.com/threads/complex-analysis-question.104713/ | # Complex Analysis question
1. Dec 20, 2005
### Physics_wiz
Prove that if wz = 0, then w = 0 or z = 0. w and z are two complex numbers.
I said that w = a + bi and z = c + di and set wz = 0. I got down to c(a+b) = d(b-a), but don't know where to go from here.
I'm trying to teach myself complex analysis, anyone know any good sources? I took the 3 calculus classes and an elementary differential equations class. I'm a mechanical engineering major though and I don't think I'll have time to take a complex analysis class unless I want to stay in college after I get my BS in ME.
2. Dec 20, 2005
### NateTG
You can solve it like a system of variables:
$$w=a+bi[/itex] [tex]z=c+di[/itex] [tex]wz=0[/itex] [tex]wz=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$$
so
$$ac-bd=0$$
and
$$ad+bc=0[/itex] so [tex]d=\frac{-bc}{a}[/itex] so [tex]ac+\frac{b^2c}{a}=0$$
$$a^2c+b^2c=0$$
so (case i)
$$a^2+b^2=0$$
since $a$ and $b$ are real numbers this gives that $z=0$
otherwise (case ii) c=0
then
[tex]bd=0[/itex]
and
so, $d=0$ so we have $w=0$ or $a=0 \rm{and} b=0$ and we have $z=0$.
3. Dec 20, 2005
### Physics_wiz
Got it. Thanks.
Know any good sources to learn complex analysis from?
EDIT: I guess it would be smart to know how big of a problem I'm trying to tackle here. Is self-teaching complex analysis going to be a hard task? Is it comparable to...say, someone who only knows algebra trying to teach himself calculus?
Last edited: Dec 20, 2005
4. Dec 20, 2005
### hypermorphism
I like the approach used in Tristan Needham's Visual Complex Analysis, but as you've not yet built up any intuition for complex numbers, you may want to accompany it with a more traditional text.
5. Dec 20, 2005
### matt grime
Surely it is far easier to simply use |vw|=|v||w|? This isn't complex analysis.
6. Dec 22, 2005
### Physics_wiz
More questions
I attached a file with three more questions in it. I did the first two using:
z = x + yi and w = a + bi and solved them down in the same manner like the first question I posted. Is there an easier way to do these?
I'm not really sure what to do for #6 though. Do I just set z = x + yi and find an equation f(y) = f(x) then graph it on an x-y plane? I did that for part (a) and got a circle with center (2, -3) and radius 2, but is that what I'm supposed to do?
Thanks.
#### Attached Files:
• ###### questions.doc
File size:
32.5 KB
Views:
61
7. Dec 22, 2005
### hypermorphism
Try thinking about what the equations are saying geometrically. In the case of the first one that you solved, it is saying that the distance between the points z and the point (2,-3) is a constant 2. That's just a circle of radius 2 centered at (2,-3) with no algebra necessary.
The second equation says that the distance between the points z and the point (0,2) is less than or equal to 1, which describes a closed disc around that point of radius 1. What about the geometry described by (d) ?
Use the algebraic method only if the geometry is hopelessly obscured by the equation. Even then, you can now use the geometric result you found algebraically for later questions.
8. Dec 22, 2005
### Physics_wiz
Ahh thanks a lot...that makes things a bit easier. See if I got those right:
b) The distance between z and -2i is less than or equal to 1. Which means it's a disk centered at -2i with a radius of 1.
c) The real part of z is equal to 4. Straight vertical line at 4 because the imaginary part can take on any value.
d) The distance between z and (1, -2i) is equal to the distance between z and (-3, -i). z again can be any value on a straight line that's a perpendicular bisector to the line connecting (1, -2i) and (-3, -i).
e) Ellipse with focii (-1, 0i) and (1, 0i).
f) Hyperbola with the same focii as part e.
By the way, when you write (2, -3) are you talking about the point a = 2 - 3i in the complex plane? When I graph those lines and ellipses and circles, do I graph them all in a complex plane with an imaginary axis and a real one?
Last edited: Dec 22, 2005
9. Dec 22, 2005
### hypermorphism
Yep, all those interpretations are spot on.
If you take the complex plane to be a 2-dimensional real vector space with basis vectors 1 and i, then you can write the element a+bi as the (position) vector (a,b) and preserve mathematical rigour.
Last edited: Dec 22, 2005 | 2017-08-18T05:15:18 | {
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https://python.quantecon.org/prob_meaning.html | # 12. Two Meanings of Probability¶
## 12.1. Overview¶
This lecture illustrates two distinct interpretations of a probability distribution
• A frequentist interpretation as relative frequencies anticipated to occur in a large i.i.d. sample
• A Bayesian interpretation as a personal opinion (about a parameter or list of parameters) after seeing a collection of observations
We recommend watching this video about hypothesis testing within the frequentist approach
After you watch that video, please watch the following video on the Bayesian approach to constructing coverage intervals
After you are familiar with the material in these videos, this lecture uses the Socratic method to to help consolidate your understanding of the different questions that are answered by
• a frequentist confidence interval
• a Bayesian coverage interval
We do this by inviting you to write some Python code.
It would be especially useful if you tried doing this after each question that we pose for you, before proceeding to read the rest of the lecture.
We provide our own answers as the lecture unfolds, but you’ll learn more if you try writing your own code before reading and running ours.
Code for answering questions:
In addition to what’s in Anaconda, this lecture will deploy the following library:
pip install prettytable
Requirement already satisfied: prettytable in /__w/lecture-python.myst/lecture-python.myst/3/envs/quantecon/lib/python3.9/site-packages (3.5.0)
Requirement already satisfied: wcwidth in /__w/lecture-python.myst/lecture-python.myst/3/envs/quantecon/lib/python3.9/site-packages (from prettytable) (0.2.5)
WARNING: Running pip as the 'root' user can result in broken permissions and conflicting behaviour with the system package manager. It is recommended to use a virtual environment instead: https://pip.pypa.io/warnings/venv
Note: you may need to restart the kernel to use updated packages.
To answer our coding questions, we’ll start with some imports
import numpy as np
import pandas as pd
import prettytable as pt
import matplotlib.pyplot as plt
from scipy.stats import binom
import scipy.stats as st
%matplotlib inline
Empowered with these Python tools, we’ll now explore the two meanings described above.
## 12.2. Frequentist Interpretation¶
Consider the following classic example.
The random variable $$X$$ takes on possible values $$k = 0, 1, 2, \ldots, n$$ with probabilties
$\textrm{Prob}(X = k | \theta) = \left(\frac{n!}{k! (n-k)!} \right) \theta^k (1-\theta)^{n-k}$
where the fixed parameter $$\theta \in (0,1)$$.
This is called the binomial distribution.
Here
• $$\theta$$ is the probability that one toss of a coin will be a head, an outcome that we encode as $$Y = 1$$.
• $$1 -\theta$$ is the probability that one toss of the coin will be a tail, an outcome that we denote $$Y = 0$$.
• $$X$$ is the total number of heads that came up after flipping the coin $$n$$ times.
Consider the following experiment:
Take $$I$$ independent sequences of $$n$$ independent flips of the coin
Notice the repeated use of the adjective independent:
• we use it once to describe that we are drawing $$n$$ independent times from a Bernoulli distribution with parameter $$\theta$$ to arrive at one draw from a Binomial distribution with parameters $$\theta,n$$.
• we use it again to describe that we are then drawing $$I$$ sequences of $$n$$ coin draws.
Let $$y_h^i \in \{0, 1\}$$ be the realized value of $$Y$$ on the $$h$$th flip during the $$i$$th sequence of flips.
Let $$\sum_{h=1}^n y_h^i$$ denote the total number of times heads come up during the $$i$$th sequence of $$n$$ independent coin flips.
Let $$f_k$$ record the fraction of samples of length $$n$$ for which $$\sum_{h=1}^n y_h^i = k$$:
$f_k^I = \frac{\textrm{number of samples of length n for which } \sum_{h=1}^n y_h^i = k}{ I}$
The probability $$\textrm{Prob}(X = k | \theta)$$ answers the following question:
• As $$I$$ becomes large, in what fraction of $$I$$ independent draws of $$n$$ coin flips should we anticipate $$k$$ heads to occur?
As usual, a law of large numbers justifies this answer.
Exercise 12.1
1. Please write a Python class to compute $$f_k^I$$
2. Please use your code to compute $$f_k^I, k = 0, \ldots , n$$ and compare them to $$\textrm{Prob}(X = k | \theta)$$ for various values of $$\theta, n$$ and $$I$$
3. With the Law of Large numbers in mind, use your code to say something
Let’s do some more calculations.
Comparison with different $$\theta$$
Now we fix
$n=20, k=10, I=1,000,000$
We’ll vary $$\theta$$ from $$0.01$$ to $$0.99$$ and plot outcomes against $$\theta$$.
θ_low, θ_high, npt = 0.01, 0.99, 50
thetas = np.linspace(θ_low, θ_high, npt)
P = []
f_kI = []
for i in range(npt):
freq = frequentist(thetas[i], n, I)
freq.binomial(k)
freq.draw()
freq.compute_fk(k)
P.append(freq.P)
f_kI.append(freq.f_kI)
fig, ax = plt.subplots(figsize=(8, 6))
ax.grid()
ax.plot(thetas, P, 'k-.', label='Theoretical')
ax.plot(thetas, f_kI, 'r--', label='Fraction')
plt.title(r'Comparison with different $\theta$', fontsize=16)
plt.xlabel(r'$\theta$', fontsize=15)
plt.ylabel('Fraction', fontsize=15)
plt.tick_params(labelsize=13)
plt.legend()
plt.show()
Comparison with different $$n$$
Now we fix $$\theta=0.7, k=10, I=1,000,000$$ and vary $$n$$ from $$1$$ to $$100$$.
Then we’ll plot outcomes.
n_low, n_high, nn = 1, 100, 50
ns = np.linspace(n_low, n_high, nn, dtype='int')
P = []
f_kI = []
for i in range(nn):
freq = frequentist(θ, ns[i], I)
freq.binomial(k)
freq.draw()
freq.compute_fk(k)
P.append(freq.P)
f_kI.append(freq.f_kI)
fig, ax = plt.subplots(figsize=(8, 6))
ax.grid()
ax.plot(ns, P, 'k-.', label='Theoretical')
ax.plot(ns, f_kI, 'r--', label='Frequentist')
plt.title(r'Comparison with different $n$', fontsize=16)
plt.xlabel(r'$n$', fontsize=15)
plt.ylabel('Fraction', fontsize=15)
plt.tick_params(labelsize=13)
plt.legend()
plt.show()
Comparison with different $$I$$
Now we fix $$\theta=0.7, n=20, k=10$$ and vary $$\log(I)$$ from $$2$$ to $$7$$.
I_log_low, I_log_high, nI = 2, 6, 200
log_Is = np.linspace(I_log_low, I_log_high, nI)
Is = np.power(10, log_Is).astype(int)
P = []
f_kI = []
for i in range(nI):
freq = frequentist(θ, n, Is[i])
freq.binomial(k)
freq.draw()
freq.compute_fk(k)
P.append(freq.P)
f_kI.append(freq.f_kI)
fig, ax = plt.subplots(figsize=(8, 6))
ax.grid()
ax.plot(Is, P, 'k-.', label='Theoretical')
ax.plot(Is, f_kI, 'r--', label='Fraction')
plt.title(r'Comparison with different $I$', fontsize=16)
plt.xlabel(r'$I$', fontsize=15)
plt.ylabel('Fraction', fontsize=15)
plt.tick_params(labelsize=13)
plt.legend()
plt.show()
From the above graphs, we can see that $$I$$, the number of independent sequences, plays an important role.
When $$I$$ becomes larger, the difference between theoretical probability and frequentist estimate becomes smaller.
Also, as long as $$I$$ is large enough, changing $$\theta$$ or $$n$$ does not substantially change the accuracy of the observed fraction as an approximation of $$\theta$$.
The Law of Large Numbers is at work here.
For each draw of an independent sequence, $$\textrm{Prob}(X_i = k | \theta)$$ is the same, so aggregating all draws forms an i.i.d sequence of a binary random variable $$\rho_{k,i},i=1,2,...I$$, with a mean of $$\textrm{Prob}(X = k | \theta)$$ and a variance of
$n \cdot \textrm{Prob}(X = k | \theta) \cdot (1-\textrm{Prob}(X = k | \theta)).$
So, by the LLN, the average of $$P_{k,i}$$ converges to:
$E[\rho_{k,i}] = \textrm{Prob}(X = k | \theta) = \left(\frac{n!}{k! (n-k)!} \right) \theta^k (1-\theta)^{n-k}$
as $$I$$ goes to infinity.
## 12.3. Bayesian Interpretation¶
We again use a binomial distribution.
But now we don’t regard $$\theta$$ as being a fixed number.
Instead, we think of it as a random variable.
$$\theta$$ is described by a probability distribution.
But now this probability distribution means something different than a relative frequency that we can anticipate to occur in a large i.i.d. sample.
Instead, the probability distribution of $$\theta$$ is now a summary of our views about likely values of $$\theta$$ either
• before we have seen any data at all, or
• before we have seen more data, after we have seen some data
Thus, suppose that, before seeing any data, you have a personal prior probability distribution saying that
$P(\theta) = \frac{\theta^{\alpha-1}(1-\theta)^{\beta -1}}{B(\alpha, \beta)}$
where $$B(\alpha, \beta)$$ is a beta function , so that $$P(\theta)$$ is a beta distribution with parameters $$\alpha, \beta$$.
Exercise 12.2
a) Please write down the likelihood function for a sample of length $$n$$ from a binomial distribution with parameter $$\theta$$.
b) Please write down the posterior distribution for $$\theta$$ after observing one flip of the coin.
c) Now pretend that the true value of $$\theta = .4$$ and that someone who doesn’t know this has a beta prior distribution with parameters with $$\beta = \alpha = .5$$. Please write a Python class to simulate this person’s personal posterior distribution for $$\theta$$ for a single sequence of $$n$$ draws.
d) Please plot the posterior distribution for $$\theta$$ as a function of $$\theta$$ as $$n$$ grows as $$1, 2, \ldots$$.
e) For various $$n$$’s, please describe and compute a Bayesian coverage interval for the interval $$[.45, .55]$$.
f) Please tell what question a Bayesian coverage interval answers.
g) Please compute the Posterior probabililty that $$\theta \in [.45, .55]$$ for various values of sample size $$n$$.
h) Please use your Python class to study what happens to the posterior distribution as $$n \rightarrow + \infty$$, again assuming that the true value of $$\theta = .4$$, though it is unknown to the person doing the updating via Bayes’ Law.
How shall we interpret the patterns above?
The answer is encoded in the Bayesian updating formulas.
It is natural to extend the one-step Bayesian update to an $$n$$-step Bayesian update.
$\textrm{Prob}(\theta|k) = \frac{\textrm{Prob}(\theta,k)}{\textrm{Prob}(k)}=\frac{\textrm{Prob}(k|\theta)*\textrm{Prob}(\theta)}{\textrm{Prob}(k)}=\frac{\textrm{Prob}(k|\theta)*\textrm{Prob}(\theta)}{\int_0^1 \textrm{Prob}(k|\theta)*\textrm{Prob}(\theta) d\theta}$
$=\frac{{N \choose k} (1 - \theta)^{N-k} \theta^k*\frac{\theta^{\alpha - 1} (1 - \theta)^{\beta - 1}}{B(\alpha, \beta)}}{\int_0^1 {N \choose k} (1 - \theta)^{N-k} \theta^k*\frac{\theta^{\alpha - 1} (1 - \theta)^{\beta - 1}}{B(\alpha, \beta)} d\theta}$
$=\frac{(1 -\theta)^{\beta+N-k-1}* \theta^{\alpha+k-1}}{\int_0^1 (1 - \theta)^{\beta+N-k-1}* \theta^{\alpha+k-1} d\theta}$
$={Beta}(\alpha + k, \beta+N-k)$
A beta distribution with $$\alpha$$ and $$\beta$$ has the following mean and variance.
The mean is $$\frac{\alpha}{\alpha + \beta}$$
The variance is $$\frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}$$
• $$\alpha$$ can be viewed as the number of successes
• $$\beta$$ can be viewed as the number of failures
The random variables $$k$$ and $$N-k$$ are governed by Binomial Distribution with $$\theta=0.4$$.
Call this the true data generating process.
According to the Law of Large Numbers, for a large number of observations, observed frequencies of $$k$$ and $$N-k$$ will be described by the true data generating process, i.e., the population probability distribution that we assumed when generating the observations on the computer. (See Exercise 12.1).
Consequently, the mean of the posterior distribution converges to $$0.4$$ and the variance withers to zero.
upper_bound = [ii.ppf(0.95) for ii in Bay_stat.posterior_list]
lower_bound = [ii.ppf(0.05) for ii in Bay_stat.posterior_list]
fig, ax = plt.subplots(figsize=(10, 6))
ax.scatter(np.arange(len(upper_bound)), upper_bound, label='95 th Quantile')
ax.scatter(np.arange(len(lower_bound)), lower_bound, label='05 th Quantile')
ax.set_xticks(np.arange(0, len(upper_bound), 2))
ax.set_xticklabels(num_list[::2])
ax.set_xlabel('Number of Observations', fontsize=12)
ax.set_title('Bayesian Coverage Intervals of Posterior Distributions', fontsize=15)
ax.legend(fontsize=11)
plt.show()
After observing a large number of outcomes, the posterior distribution collapses around $$0.4$$.
Thus, the Bayesian statististian comes to believe that $$\theta$$ is near $$.4$$.
As shown in the figure above, as the number of observations grows, the Bayesian coverage intervals (BCIs) become narrower and narrower around $$0.4$$.
However, if you take a closer look, you will find that the centers of the BCIs are not exactly $$0.4$$, due to the persistent influence of the prior distribution and the randomness of the simulation path.
## 12.4. Role of a Conjugate Prior¶
We have made assumptions that link functional forms of our likelihood function and our prior in a way that has eased our calculations considerably.
In particular, our assumptions that the likelihood function is binomial and that the prior distribution is a beta distribution have the consequence that the posterior distribution implied by Bayes’ Law is also a beta distribution.
So posterior and prior are both beta distributions, albeit ones with different parameters.
When a likelihood function and prior fit together like hand and glove in this way, we can say that the prior and posterior are conjugate distributions.
In this situation, we also sometimes say that we have conjugate prior for the likelihood function $$\textrm{Prob}(X | \theta)$$.
Typically, the functional form of the likelihood function determines the functional form of a conjugate prior.
A natural question to ask is why should a person’s personal prior about a parameter $$\theta$$ be restricted to be described by a conjugate prior?
Why not some other functional form that more sincerely describes the person’s beliefs.
To be argumentative, one could ask, why should the form of the likelihood function have anything to say about my personal beliefs about $$\theta$$?
A dignified response to that question is, well, it shouldn’t, but if you want to compute a posterior easily you’ll just be happier if your prior is conjugate to your likelihood.
Otherwise, your posterior won’t have a convenient analytical form and you’ll be in the situation of wanting to apply the Markov chain Monte Carlo techniques deployed in this quantecon lecture.
We also apply these powerful methods to approximating Bayesian posteriors for non-conjugate priors in this quantecon lecture and this quantecon lecture | 2023-01-31T10:50:09 | {
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https://math.stackexchange.com/questions/2696671/calculating-probability-using-combinations-as-alternative-solution | # Calculating Probability using Combinations as Alternative Solution
I stumbled upon the following probability problem (translated from Swedish):
Calculate the probability that, out of 23 people, at least two of them have a birthday on the same day. Assume that a year has 365 days and that all birthdays are equally probable to be a birthday. Tip: Utilise the complementary event.
### My Attempt
Based on my reading of the question, it seems to suggest that the classical definition of probability should be used to calculate the probability, e.g $P(A) = \frac{|A|}{|\Omega|}$, where A is the given event and $\Omega$ is the universe of outcomes. Furthermore, to solve this I tried to calculate $P(A) = 1 - P(A^*)$ instead. I defined the following in my solution:
• $\Omega :=$ {"all bags of exactly 23 birthdays"}. I said bag because I believe that the order does not matter (e.g person 1 having birthday 51 or person 3 having birthday 51 does not matter), while at the same time multiple instances of the same value should be able to be represented in the bag. $|\Omega| =$ ${365 + 23 - 1}\choose{23}$ (I used the formula for choosing k elements from n where order does not matter and the element is returned after being chosen once).
• $A^* :=$ {"all bags of exactly 23 unique birthdays"}, $|A^*| =$ ${365}\choose{23}$. I chose combination and not permutation as I believe this reflects having a bag (i.e order does not matter) of exactly 23 unique birthdays.
However, this leads to an incorrect answer when the calculation is carried out.
### Correct Solution
The correct answer uses $\Omega :=$ {"all tuples of exactly 23 birthdays"} and $A^* :=$ {"all tuples of exactly 23 unique birthdays"}, i.e tuples instead of bags. This means they care about who had which birthday.
They get the result $P(A^*) = \frac{\frac{365!}{(365-23)!}}{365^{23}}$, and carrying out the rest of the calculation (i.e 1 - Ans) yields the correct result.
### My Question
I sort of understand why the provided solution works. What I don't understand (and what my question is) is: Why does my solution not work?
The only thing I could spot as being different is that I'm 'smashing together' the outcomes with similar but shuffled birthday values into a single outcome in my $\Omega$, compared to the $\Omega$ of the provided solution. But I don't understand why that does not work; a similar thing is done when calculating probabilities in, say, picking 3 cards from a shuffled deck. The outcomes can be seemingly grouped together into a single outcome "The cards X, Y and Z" and that works, and calculating probabilities on that still seems to work (i.e you can still use the above formula for $P(A)$ to calculate the probability of A happening).
I'm just all around confused on when to utilise combinations and when to utilise permutations in my solutions. I don't know if I've misinterpreted how to define $\Omega$, or if something else is misinterpreted, but I seem to run into this type of so many times and not a single book I've read throughout 3 courses involving combinatorics has ever touched on this in a way that I can understand.
I would be very grateful if anyone could help me clarify exactly what in my method is incorrect. Any help is appreciated! :)
The problem is that you treat every outcome in your 'bag of exactly 23 birthdays' as equally probable. But, getting all $23$ birthdays on, say, June $1$, is less probable than getting $22$ birthdays on June $1$ and one on June $2$, for that one person born on June $2$ can be any one of the $23$ people. And, of course, getting much of a 'spread' of birthdays will be a good bit more likely yet.
• Thank you for the answer! It seems reasonable that that is the case. A followup; for probability problems involving choosing a hand of 3 cards from a deck of cards, these tend to use combinations rather than permutations. Why does it work to calculate 'good outcomes/total outcomes' using combinations in this case? Does it work because the outcome 'cards 1,2 and 3' and the outcome 'cards 6, 3, 8' (order does not matter, think of the deck as 52 'card-objects' with corresponding ids 1,2...) are (perhaps) equally probable? Mar 18 '18 at 6:35
• In which case you may calculate the probability of an event A = 'getting three hearts' by calculating '# outcomes in A/# outcomes in universe', since each hand of 3 cards (disregarding any qualities of a particular card besides the fact that they are distinguishable) are eqully probable? Mar 18 '18 at 8:52
• @SimonSirak Yes, that's exactly it: each particular hand of cards is equally likely, because each card is unique. But with the birthdays you can get repeat birthdays, and now certain combinations are more likely than others. Mar 18 '18 at 12:32 | 2021-10-27T13:55:17 | {
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http://www.math.wsu.edu/kcooper/M300/matlabvectortricks.php | Matlab Vector Tricks
There are a lot of things we do in mathematics that are usually annotated using summation notation, but which are easily seen to be vector operations. To illustrate this, consider a very simple example that occurs often. Suppose that we have a vector v = (v1,v2,...,vn)T, and we need to find the sum of the absolute values of its elements. One way to do this would be to use a for loop.
total = 0; for i=1:length(v) total = total + abs(v(i)); end
This works, and it is easy to see what is happening in the code, but it is a bit hard to type on a Matlab command line, and it takes four lines of code. We could accomplish the same objective in one line.
total = abs(v)*ones(length(v),1)
As you can see, it is arguable that the one-line version is less transparent than the four-line version, but it avoids the tedious work by the programmer of elementwise operations, and it is arguable that it is more efficient for Matlab itself, since it is designed to do vector operations. Actually, there is an even simpler and more obvious way to do this particular computation.
total = sum(abs(v))
In the same way, we could vectorize the distance formula. Suppose that u = (u1,u2,...,un)T, and v = (v1,v2,...,vn)T, and we want to calculate the distance between the points u and v in Rn. We could do this as a for loop.
distance = 0; for i=1:length(v) diff = u(i)-v(i); distance = distance + diff*diff; end
Of course, there is a vector solution as well, that requires less work.
diff = u-v; distance = diff*diff';
We added one line of code in each case here, to try to prevent Matlab from computing the difference of vectors twice. We hope we save n additions in this way. It is not a big deal when n is small, but it could be significant in a large computation.
Similarly, there are ways to make vector operations out of other common mathematical sum operations. For example, suppose we want to approximate the integral of a function, say sin(x), over an interval [-π, π] using the trapezoidal rule. Recall that this means imposing a uniformly spaced partition π=x0<x1<xn on the interval, and then computing
[(sin(x0) + sin(xn))/2 + ∑i=2n-1 sin(xi) ](2π/n).
In Matlab, the following methods are equivalent.
x=-pi:.1:pi; y=sin(x); integral = (y(1)+y(length(y)))/2; for i=2:length(y)-1 integral = integral+y(i); end integral = integral*.1 --- integral = (y*ones(length(y),1)-.5*(y(1)+y(length(y))))*.1 --- integral = (sum(y)-.5*(y(1)+y(length(y))))*.1 --- integral = (sum(y(2:length(y)-1))+.5*(y(1)+y(length(y))))*.1
The moral of the story is that whenever you see a summation operation in Matlab, it is worth thinking for a minute about how you might convert the operation to a vector computation.
A solution for the final is available.
Department of Mathematics, PO Box 643113, Neill Hall 103, Washington State University, Pullman WA 99164-3113, 509-335-3926, Contact Us | 2018-02-19T20:09:22 | {
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https://math.stackexchange.com/questions/2360503/cayley-hamilton-and-linear-dependence/2360510 | # Cayley-Hamilton and linear dependence
Show that the set $\{A, A^2, A^3, A^4\}$ of $2 \times 2$ matrices is linearly dependent.
Cayley-Hamilton theorem states that every square matrix $A$ satisfies its own characteristic polynomial equation $det(A-\lambda I) = 0$ such that $A^n+a_{n-1}A^{n-1}+...+a_2A^2+a_1A+a_0I=0$.
How can I show that the given set of matrices is linearly dependent because $I$ is not in the set. Am I missing some subtlety here?
• Sorry, I meant linearly dependent. I'm not sure if your statement applies to linear dependence too. What I'm saying is, is there some definition or workaround that allows me to use the $I$ in the linear combination? Or even just some truth that I am not aware of. – Bucephalus Jul 16 '17 at 13:13
• No problem. Just separated out the question and your attempt for clarity. – Sahiba Arora Jul 16 '17 at 13:22
Let the characteristic equation of $A$ be: $$A^2+pA+qI=0$$ Then $$A^2=-pA-qI$$ Thus $$A^3=-pA^2-qA$$ Hence $$A^4=-pA^3-qA^2$$ That is $\{A^2,A^3,A^4\}$ is linearly dependent. As a superset of a linearly dependent set is linearly dependent, therefore $\{A,A^2,A^3,A^4\}$ is also linearly dependent.
Edit: Let $\{x,y\}$ be linearly dependent. Then there exists scalars $a,b$ such that $ax+by=0.$ Hence, $ax+by+0z=0$. Thus, $\{x,y,z\}$ is also linearly dependent.
• Are you saying @Sahiba, that even if you have a set of 100 elements, you only need two elements to be linearly dependent for the set to be known as linearly dependent? – Bucephalus Jul 16 '17 at 13:24
• @Bucephalus Yes I'm saying that. – Sahiba Arora Jul 16 '17 at 13:28
• OK, that's good. – Bucephalus Jul 16 '17 at 13:30
• @Bucephalus I have added the reason why that is true in the answer. – Sahiba Arora Jul 16 '17 at 13:30
• Yeah, that's a good explanation @Sahiba. Thanks for your help. – Bucephalus Jul 16 '17 at 13:32
The matrix $A$ satisfies its characteristic polynomial, so $$A^2 + a_1 A + a_0I=0.$$ Multiplying by $A$ we obtain $$A^3+a_1A^2+a_0A=0,$$ that is, already $A, \, A^2, \, A^3$ are linearly dependent.
• Wow, that's awesome. – Bucephalus Jul 16 '17 at 13:20
• Sorry @Francesco. Yours is an awesome answer, but I extracted some more info from the other answer. Cheers. – Bucephalus Jul 16 '17 at 13:31
• No problem, but is not the other answer saying the same things as mine? :-) – Francesco Polizzi Jul 16 '17 at 13:33
• Yes it is, so who do I give it to? Let me see if I can select your answer also. – Bucephalus Jul 16 '17 at 13:36
• Just select the answer you prefer, I was only joking – Francesco Polizzi Jul 16 '17 at 13:37 | 2019-04-19T20:46:30 | {
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https://math.stackexchange.com/questions/3913800/h-and-k-are-two-normal-subgroups-of-group-g-h-cap-k-1-g-hk-how | # $H$ and $K$ are two normal subgroups of group $G$, $H\cap K=\{1\}$, $G=HK$, how to prove $G$ is isomorphic to $H\times K$
If group $$G$$ has two normal subgroups $$H$$ and $$K$$ satisfying $$H\cap K=\{1\}$$ and $$G=HK$$, how to prove that $$G$$ is isomorphic to $$H\times K$$?
Since the factoring of any $$g\in G$$ as product $$hk$$ is unique under given conditions, I tried to prove that the map $$\phi:G\ni g\mapsto (h,k)\in H\times K$$ is the isomorphism. I can show that $$\phi$$ is a bijection. Next, I need to show it preserves group operation. Assuming $$\forall g_1,g_2\in G$$, $$g_1=h_1k_1$$ and $$g_2=h_2k_2$$ (the factoring is unique), we have $$g_1g_2=h_1k_1h_2k_2$$. But $$\phi(g_1)=(h_1,k_1),\phi(g_2)=(h_2,k_2)$$, so in product group $$H\times K$$, the operation $$\phi(g_1)\phi(g_2)=(h_1h_2,k_1k_2)$$. Its pre-image of $$\phi$$ in $$G$$ should be the product $$h_1h_2k_1k_2$$. So we must establish $$h_1k_1h_2k_2=h_1h_2k_1k_2$$, which is equivalent to $$k_1h_2=h_2k_1$$. But I cannot prove this commutative relation based on the given conditions. The normality of $$H$$ implies $$k_1^{-1}h_2k_1\in H$$, but it does not necessarily equal $$h_2$$. This is where I got stuck. Is the above commutative relation true and how do I prove it? Or, was I wrong at the very beginning about what the isomorphism is? Could you please give me some help? Thank you.
This doesn't add much to your excellent effort, other than some details.
For any pair of subgroups $$H, K\le G$$, the set $$H\times K$$ can be partitioned into equivalence classes all equicardinal to $$H\cap K$$, and such that $$(H\times K)/\sim$$ is equicardinal to $$HK$$. In fact:
Let's define in $$H\times K$$ the equivalence relation: $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$. The equivalence class of $$(h,k)$$ is given by:
$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$
Now define the following map from any equivalence class:
\begin{alignat*}{1} f_{(h,k)}:[(h,k)]_\sim &\longrightarrow& H\cap K \\ (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\ \tag 2 \end{alignat*}
Note that $$k'k^{-1}\in K$$ by closure of $$K$$, and $$k'k^{-1}\in H$$ because $$k'k^{-1}=h'^{-1}h$$ (being $$(h',k')\in [(h,k)]_\sim$$) and by closure of $$H$$. Therefore, indeed $$k'k^{-1}\in H\cap K$$.
Lemma 1. $$f_{(h,k)}$$ is bijective.
Proof.
\begin{alignat}{2} f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\ &\space\space\space\Longrightarrow &&k'=k'' \\ &\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\ &\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\ \end{alignat}
and the map is injective. Then, for every $$a\in H\cap K$$, we get $$ak\in K$$ and $$a=f_{(h,k)}((h',ak))$$, and the map is surjective. $$\space\space\Box$$
Now define the following map from the quotient set:
\begin{alignat}{1} f:(H\times K)/\sim &\longrightarrow& HK \\ [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\ \tag 3 \end{alignat}
Lemma 2. $$f$$ is well-defined and bijective.
Proof.
• Good definition: $$(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$$;
• Injectivity: $$f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$$;
• Surjectivity: for every $$ab\in HK$$ , we get $$ab=f([(a,b)]_\sim)$$. $$\space\space\Box$$
As a corollary, if $$|H\cap K|=1$$, then all the classes $$[(h,k)]_\sim$$ are singletons, and hence the map $$\tilde f\colon H\times K\to (H\times K)/\sim$$, defined by $$(h,k)\mapsto [(h,k)]_\sim$$, is bijective; this, in turn, implies that the composite map $$f\circ\tilde f\colon H\times K\to HK$$ is bijective, too. If, in addition, $$H\unlhd G$$ and $$K\unlhd G$$, then for every $$h\in H, k\in K$$ we have that $$hkh^{-1}k^{-1}\in H\cap K$$, whence $$hkh^{-1}k^{-1}=e$$ and finally $$hk=kh$$. This, and the assumption $$HK=G$$, make of $$f\circ\tilde f$$ a (bijective) group homomorphism (namely an isomorphism). In fact:
\begin{alignat}{1} (f\circ\tilde f)((h,k)(h',k')) &= (f\circ\tilde f)(hh',kk') \\ &= f(\tilde f(hh',kk')) \\ &= f([hh',kk']_\sim) \\ &= hh'kk' \\ &= hkh'k' \\ &= (hk)(h'k') \\ &= f([h,k]_\sim)f([h',k']_\sim) \\ &= f(\tilde f(h,k))f(\tilde f(h',k')) \\ &= ((f\circ\tilde f)(h,k))((f\circ\tilde f)(h',k')) \\ \end{alignat}
• Thank you. This answer gives me a deeper and more direct understanding of the result. The isomorphism $H\times K\cong\rm{(when\;H\cap K=\{1\})}\frac{H\times K}{H\cap K \rm{(\cong Stab\;1)}}\cong HK=G$ can be proved elegantly by group action: math.stackexchange.com/questions/168942/… Nov 20, 2020 at 1:40
Hint. For $$h ∈ H$$, $$k ∈ K$$, you have $$hk = kh \iff hkh^{-1}k^{-1} = 1$$.
• I see, $hkh^{-1}k^{-1}=(hkh^{-1})k^{-1}=h(kh^{-1}k^{-1})$, so it is in both $H$ and $K$. Thank you! Nov 19, 2020 at 7:12 | 2022-08-09T19:51:41 | {
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http://xzqd.labottegamedievale.it/bilinear-interpolation-example.html | For example, the nearest neighbor kernel for size tripling is [0, 1, 1, 1, 0] and the linear interpolation kernel is [1/3, 2/3, 1, 2/3, 1 / 3]. Linear interpolation is a process employed in mathematics, and numerous applications thereof including computer graphics. I highlighted the table to help show the parts of the table being the header, x and y-axis data. If they're not in a grid, use scatteredInterpolant like Mike showed you. When an image is zoom its' dimensions are larger than the original image. • An example of interpolation using spline functions and least-squares curve fitting using a fifth degree polynomial is shown in the following figure • The data set is a set of 10 random numbers generated using 10*rand(1,10) – Note that the spline interpolation passes through the data points while the curve fit does not f(x ) f(x ) 6. I assume that we do a bilinear interpolation. y1, are the distance of y direction. For non-integer shift values, bilinear interpolation is used. You will find in this article an excel formula, and a User Defined Function (UDF) for Linear Interpolation in Excel. I would like to point you to this very insightful graphic from Wikipedia that illustrates how to do bilinear interpolation for one point: Source: Wikipedia. The main peculiarity of two-dimensional interpolation is that most two-dimensional algorithms are applicable only with rectilinear grids oriented to axes. Generate the peaks function at low resolution. I would like to do a lookup and interpolation based on x, y data for the following table. For forming 3D associations, we leverage characteristics of RGB-D sensors in defining the soft falloff functions. "spline" Cubic spline interpolation—smooth first and second derivatives throughout the curve. Bilinear interpolation is a process that enhances textures that would normally be larger on-screen than in texture memory, i. ) as well as in computer graphics (texturing, etc. Downloaded your program for bilinear interpolation. I coded a resizing function based on things I've read here and Wikipedia, etc. For every "missing" pixel (the pixels that have to be created to blow up the image) the bilinear method takes the four points that are closest at the. The processing system is operable to process image data captured by the camera. For example, if you scale an image, you can determine the final color of each pixel by either some basic nearest neighbor method,. Bilinear Interpolation in General For this assignment, you'll make repeated use of bilinear interpolation. Does anyone have any insight on whether or not the Interpolate2D. " An algorithm is used to map a screen pixel location to a corresponding point on the texture map. Bilinear Interpolation. not sure what your data is meaning. tr They can not be used without the permission of the author. Igor returns NaN for points outside the convex domain. The code is in Interpolation. This presents a problem in most \real" applications, in which functions are used to model relationships between quantities,. Example- *if the name of image in root directory is image. I would like to perform blinear interpolation using python. bilinear interpolation generates signi cant artifacts, es-pecially across edges and other high-frequency content, since it doesn‘t take into account the correlation among the RGB. URL: http://www. Start studying Interpolation and Extrapolation. All bilinear interpolation involves interpolating a value between four known (point, value) pairs. Bilinear Interpolation! Computational Fluid Dynamics! Simples grid generation is to break the domain into blocks and use bilinear interpolation within each block! As an example, we will write a simple code to grid the domain to the right! (x 1,y 1)! (x 2,y)! (x 3,y 3)! (x 4,y 4)! (x 5,y 5)! (x (x 6,y 6)! 7,y 7)! 8 8 Bilinear Interpolation! Computational Fluid Dynamics!. Click the Calculate button, and the blank value will be filled in by linear interpolation. The first y value will be used for interpolation to the left and the last one for interpolation to the right. Formula of Linear Interpolation. A well-established median method for video scaling is the use of Bilinear Interpolation. I'd like the equation to be as simple as possible to reduce the amount of possible errors. For every "missing" pixel (the pixels that have to be created to blow up the image) the bilinear method takes the four points that are closest at the. a single texel is used for more than one screen pixel. Linear Interpolation • Given a function defined at two points, f(0), f(1), we want to find values for – Bilinear interpolation. This offset appears to be specific to the 3rd order bilinear interpolation. 2 Finite element interpolation on SJ. As a quick check to see if this makes any sense, we can plot it on a curve of the known data: Looks good!. As we saw on the Linear Polynomial Interpolation page, the accuracy of approximations of certain values using a straight line dependents on how straight/curved the function is originally, and on how close we are to the points $(x_0, y_0)$ and $(x_1, y_1)$. Linear Interpolation Formula Linear interpolation is a method of curve fitting using linear polynomials to construct new data points within the range of a discrete set of known data points. My example extends interpolating one column to multiple columns using the INDEX and MATCH combo. Pixels that fall outside the boundaries of the original image are set to 0 and appear as a black background in the output image. Here, f means the values of pixels. Bilinear Interpolation is the process of using each of the intermediate fields in an interlaced video frame to generate a full size target image. See the slides on interpolation for the formula for doing linear interpolation. Then I tried to use the map() function, it was close but i want perfect. Sign in Sign up. In this case, m = 0. Difference between Bi-linear and Bi-cubic: Bi-linear uses 4 nearest neighbors to determine the output, while Bi-cubic uses 16 (4×4 neighbourhood). You will find in this article an excel formula, and a User Defined Function (UDF) for Linear Interpolation in Excel. Assume our original image is represented by matrix A and the enlarged image by matrix B. ppt - Free download as Powerpoint Presentation (. Bilinear Interpolation smoother looking images than nearest neighbor. The following Matlab project contains the source code and Matlab examples used for image shrinking using bilinear interpolation. Tag: bilinear interpolation Depth-aware upsampling experiments (Part 5: Sample classification tweaks to improve the SSAO upsampling on surfaces) This is another post of the series where I explain the ideas I try in order to improve the upsampling of the half-resolution SSAO render target of the VKDF sponza demo that was written by Iago Toral. Example gps point for which I want to interpolate height is: B = 54. Honestly I haven't read that article you linked to, but as long as you want a convolution kernel for 2D bilinear interpolation, then the following should help. The more factors L has, the more choices you have. Not surprisingly, the method you choose affects the result you. Could you please provide the math behind so that how implementing it on CPU and GPU will be clearer?. Unlike other interpolation techniques such as nearest neighbour interpolation and bicubic interpolation, bilinear interpolation uses only the 4 nearest pixel values which are located in diagonal directions from a given pixel in order to find the. For bilinear interpolation, the block uses the weighted average of two translated pixel values for each output pixel value. Interpolation estimates data points within an existing data set. I have a code for bilinear interpolation in VBA, but now I need an inverse bilinear interpolation - I know the z point and y point and I need to get x point. For example, if you scale an image, you can determine the final color of each pixel by either some basic nearest neighbor method, or a more advanced interpolation method. If missing values are present, then linint2_points will perform the piecewise linear interpolation at all. Igor returns NaN for points outside the convex domain. Here, f means the values of pixels. The full table is about 50 rows x 30 columns. If the interpolation is 'none', then no interpolation is performed for the Agg, ps and pdf backends. Linear interpolation - Surveys the 2 closest pixels, drawing a line between them and designating a value along that line as the output pixel value. All bilinear interpolation involves interpolating a value between four known (point, value) pairs. First off, the obvious: this is a lot of different implementations of interpolation (for the 1d cases), bilinear interpolation (for the 2d cases) and what I suppose would be trilinear interpolation (for the 3d cases). But if I plug in, for example, x=4. , x and y) on a regular 2D grid. 1) in x and y. Interpolation is the process by which a small image is made larger. Check out these examples: Point A Point A has a coordinate of (0. 1974 Topps Ken #D328939 Anderson NM 1974 401 Anderson A "raster map" is a data layer consisting of a gridded array of cells. , but I get different results than what MATLAB and Mathematica give. SciPy provides a module for interpolation based on the FITPACK library of FORTRAN functions. Available interpolation functions and options are presented on the table below. Image interpolation Recall how a digital image is formed •It is a discrete point-sampling of a continuous function •If we could somehow reconstruct the original function, any new. Bilinear interpolation (interpolating within a 2-dimensional table) can be done with regular MS Excel functions. It is a very simple form of interpolation. Bilinear Interpolation The red value of a non-red pixel is computed as the average of the two or four adjacent red pixels, and similarly for blue and green. DUC translates data from baseband to a passband signal comprising modulated carriers at a set of one or more specified radio or intermediate frequencies, it achieves this by performing interpolation to increase the sample rate, filtering to provide spectral shaping and rejection of interpolation images and mixing to shift the signal spectrum to. 1D Index Look-up and Acceleration ¶ The state of searches can be stored in a gsl_interp_accel object, which is a kind of iterator for interpolation lookups. What is image interpolation? An image f(x,y) tells us the intensity values at the integral lattice locations, i. (Help and details). CSE486, Penn State Robert Collins. , a problem in which neither of the pair corresponds to a tabulated numerical value) requires that we start with 4 data points. The one-dimensional linear interpolation method is extended to arbitrary dimensions using bins to create a convex hull representation of the data around a sample point. That is, given a continuous-time transfer function , we apply the bilinear transform by defining. interpolation rc parameter. Has anyone come across a good bilinear interpolation algorithm, preferably in Python, possibly tailored with NumPy? Any hints or advice?. 2 CDO Manual for details and further examples. Cubic Spline Interpolation. Bi-Linear Rectangles. The diagrams below illustrate the pixels involved in one-dimensional bilinear interpolation. The BILINEAR function uses a bilinear interpolation algorithm to compute the value of a data array at each of a set of subscript values. I am trying to get a working understanding of how to resize images using bilinear and bicubic transformations. The underlying function f(x, y) is sampled on a regular grid and the interpolation process determines values between the grid points. Here is a simple example of trilinear interpolation on a grid. Bilinear image scaling is about the same as nearest neighbor image scaling except with interpolation. I'm a relative noob to programming. , x and y) on a regular 2D grid. " An algorithm is used to map a screen pixel location to a corresponding point on the texture map. I’ve used Named Ranges here again to make the formula clearer. For example, if you scale an image, you can determine the final color of each pixel by either some basic nearest neighbor method,. INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x,y). However, for data collected in an estuary, this definition of distance doesn't work because the straight line between two points may cross over land. Used of functions or equations. SciPy provides a module for interpolation based on the FITPACK library of FORTRAN functions. Available interpolation functions and options are presented on the table below. Interpolation is a useful mathematical and statistical tool used to estimate values between two points. Performed data analysis which involves K-mean clustering and Gap-Statistic for regional cluster division, bilinear interpolation (re-gridding. An example field programmable gate array implementation of the bilinear interpolation method used in conjunction with a lens distortion correction algorithm has been successfully completed. But when the image is zoomed, it is similar to the INTER_NEAREST method. For example, Akima spline interpolation requires a minimum of 5 points. You'd have to point sample the texture, apply gamma corrections, and manually interpolate. interpolate. Bilinear interpolation is a draft programming task. You can vote up the examples you like or vote down the ones you don't like. 2 Paper 1019 / Bilinear Accelerated Filter Approximation Lánczos filters are used. But it will become a pretty long formula, that is hard to read and/or copy across. This implementation maintains equal subsampleBits in x and y. As the example illustrates, a double interpolation problem (i. Below is a snippet of Arduino code utilizing the smoothstep function. This is an implementation of a bilinear interpolating function. Now, we explain this assumption and the idea behind the. Given a set of 2-D sample points in a regular grid, we can use the methods of bilinear and bicubic 2-D interpolation to obtain the value of the interpolating function at any point inside each of the rectangles in a 2-D grid with the four corners at , , , and. Make sure you have reviewed the previous tutorial on how to declare immutable variables as we will be using them here. Use our online bilinear interpolation calculator to calculate the double interpolation. Getting started and examples Getting started. 1 The bilinear IFE space for the non-homogeneous flux jump condition. For forming 3D associations, we leverage characteristics of RGB-D sensors in defining the soft falloff functions. • Example: fading. For every "missing" pixel (the pixels that have to be created to blow up the image) the bilinear method takes the four points that are closest at the. Bilinear interpolation (or higher order interpolation) will be measureably slower than nearest neighbour unless it is hardware accelerated. In mathematics, bilinear interpolation is an extension of linear interpolation for interpolating functions of two variables (e. PC Magazine Tech Encyclopedia Index - Definitions on common technical and computer related terms. Example 10. Bilinear interpolation is linear interpolation in 2 dimensions, and is typically used for image scaling and for 2D finite element analysis. • An example of interpolation using spline functions and least-squares curve fitting using a fifth degree polynomial is shown in the following figure • The data set is a set of 10 random numbers generated using 10*rand(1,10) – Note that the spline interpolation passes through the data points while the curve fit does not f(x ) f(x ) 6. They are Nearest-neighbor interpolation and Bilinear interpolation. ' 'The intensity at each vertex was estimated with bilinear interpolation of the nearest four pixels. Interpolation is usually only used for 'point' sampling images, when image scaling is either not known or needed. It seems whenever the fractional part of our texture x coordinates becomes smaller than 1/(2*texture dimension X) or larger than 1-1/(2*texture dimension X) the bilinear interpolation in hardware is interpolating the texel situated at the border with the border color (which is black). This picture below illustrates well the three methods:. Bilinear image scaling is about the same as nearest neighbor image scaling except with interpolation. I very powerful formula for smoothing the interpolation from one value to another. Regular texture mapping would appear to be 'blocky' in this case. Linear interpolation example. ppt - Free download as Powerpoint Presentation (. Interpolation estimates data points within an existing data set. akima760 Sample data from Akima’s Bicubic Spline Interpolation code (TOMS 760) Description akima760 is a list with vector components x, y and a matrix z which represents a smooth surface of z values at the points of a regular grid spanned by the vectors x and y. In the example only several points are given, but the list can be longer or shorter than given in the example. Bilinear interpolation method plays an important role in classical image scaling. This example shows how to use the custom Excel function BicubicInterpolation (). Lyngby, Denmark 29th October 2001 Abstract This note introduces the concept of image warping and treats the special case of Euclidean warping along with a discussion of a Matlab. Even filters with small support may sum hundreds of texels under modest scales. While bilinear interpolation is a sensible approximation for image pixels, this is often a poor choice for use in 3D soft associations. Bilinear Interpolation for Data on a Rectangular grid This is an implementation of a bilinear interpolating function. The ratio of contribution taken from the pixels is inversely proportional to the ratio of corresponding distance. Interpolation. Among other numerical analysis modules, scipy covers some interpolation algorithms as well as a different approaches to use them to calculate an interpolation, evaluate a polynomial with the representation of the interpolation, calculate derivatives, integrals or roots with functional and class. This is an implementation of cubic spline interpolation based on the Wikipedia articles Spline Interpolation and Tridiagonal Matrix Algorithm. Description. This is an implementation of a bilinear interpolating function. Let's start with the X axis. Edge-directed interpolation is an adaptive approach, where the area around each pixel is analyzed to determine if a. Bilinear Interpolation for Data on a Rectangular grid This is an implementation of a bilinear interpolating function. In this lesson, you will learn about this tool, its formula and how to use it. As an example, the Trace transform system in [3], processes 30 frames per second, but each of these frames accesses data. Interpolation allows any derivative to be given as Automatic, in which case it will attempt to fill in the necessary information from other derivatives or function values. See ' Examples of linear and bilinear interpolation in table lookup data references ' topic in the documentation. Non-adaptive perform interpolation in a fixed pattern for every pixel, while adaptive algorithms detect local spatial features, like edges, of the pixel neighborhood and make effective choices depending on the algorithm. I have spent countless hours trying to speed up my bilinear interpolation up. They can be applied to gridded or scattered data. Now, we explain this assumption and the idea behind the. Shannon Sampling Theorem: When a “train of impulse” comb(x) is multiplied by f(x) , it gives us a “sampled version” of f(x) comb(x)f(x), in frequency domain, becomes convolution. The PixInsight/PCL platform provides several pixel interpolation algorithms, ranging from relatively simple procedures (nearest neighbor, bilinear) to more sophisticated algorithms such as Lanczos and bicubic spline, as well as algorithms specifically designed to address the downsampling and smooth interpolation tasks in an efficient and versatile way, such as. Interpolation De nition Interpolationis a method of constructing new data points within the range of a discrete set of known data points. Here are a couple of examples of when you would you use bilinear interpolation: When you resample your data from one cell size to another, you're changing the cell size and would need interpolation. This presents a problem in most \real" applications, in which functions are used to model relationships between quantities,. 5 pixel in the positive horizontal direction using bilinear interpolation. (ECE @ McMaster) Bicubic Interpolation February 1, 2014 2 / 26. VBA code of worksheet functions for linear and bilinear interpolation based on the signature of interp1 and interp2 in MATLAB. This video is simple example of bilinear interpolation with Java. curve instead of the default linear interpolation. In bilinear interpolation you use the four closest grid points for the calculation. This offset appears to be specific to the 3rd order bilinear interpolation. At 40 vol% yield the result is 746. 0470721369. Note each portion is the exact same pixels in the original image, so we can see in addition to blurring the lines, the bilinear case extends downward further. BILINEAR — Bilinear interpolation calculates the value of each pixel by averaging (weighted for distance) the values of the surrounding four pixels. Using the Python Image Library (PIL) you can resize an image. Note that bilinear interpolation can produce some artifacts related to the grid and not reproduce higher behavior in the surface. Bilinear interpolation is used when we need to know values at random position on a regular 2D grid. For example, if you scale an image, you can determine the final color of each pixel by either some basic nearest neighbor method,. Bilinear Interpolation for Data on a Rectangular grid This is an implementation of a bilinear interpolating function. This video will show an example to easily find specific values using Linear Interpolation. Let’s discuss the maths behind each interpolation method in the subsequent blogs. For example, when rotating image or minor distortions, the image's scaling or size does not change, and as such an interpolation can produce a reasonable result, though not a very accurate one. interpolation, polynomial interpolation, spline. Applies bilinear interpolation to a 2-dimensional grid. The general problems and solutions that are involved in three-dimensional interpolation can be illustrated by some two-dimensional examples. Code, Example for Linear Interpolation in C++ Programming. This example displays the difference between interpolation methods for imshow() and matshow(). Bilinear interpolation; the output pixel value is a weighted average of pixels in the nearest 2-by-2 neighborhood As far as I understand, this interpolation uses the value of a pixel, if it exists exactly in the input, but for all other cases, there are 2x2 surrounding pixels in the input for each pixel in the output, which are used for the. Demo ~22 fps full res. In fact, it's telling you that the pixel value you are looking for is in between the pixels somewhere, which is the main problem in digital imaging: regardless of how many megapixels we have, we only sample the continuous 2D color function. If interpolation is None, it defaults to the image. A well-established median method for video scaling is the use of Bilinear Interpolation. Note: We will be using some concepts from the Nearest Neighbour and Bilinear interpolation blog. – tacaswell Sep 11 '13 at 21:14 I originally thought that linear interpolation on 2D would be bilinear as well. Now, we have the bilinear interpolation incorporated version. In this lesson, you will learn about this tool, its formula and how to use it. nc > myGridDef myGridDef is a text file. Bilinear interpolation reduces the blockyness by interpolating between texels. It then takes a weighted average of these 4 pixels to arrive at its final interpolated value. In that case, you can use bilinear interpolation in Excel. It is a very simple form of interpolation. , but I get different results than what MATLAB and Mathematica give. The interpolation as a whole is not linear but rather quadratic in the sample location. Relatively, their memory will be extremely slow. VBA code of worksheet functions for linear and bilinear interpolation based on the signature of interp1 and interp2 in MATLAB. Interpolation is a mathematical procedure for filling in the gaps between available values. My problem is that my Tbl. In this tutorial, we'll be writing a function to rotate an image, using bilinear. I would like to point you to this very insightful graphic from Wikipedia that illustrates how to do bilinear interpolation for one point: Source: Wikipedia. Interpolation-based super resolution has been used for a long. In the example code below, we compute linearly interpolated f(x) values for the corresponding x = 1,5 and 3,5. The processing system is operable to process image data captured by the camera. For a point (x0,y0) contained in a rectangle (x1,y1),(x2,y1), (x2,y2),(x1,y2) and x1. Interpolation De nition Interpolationis a method of constructing new data points within the range of a discrete set of known data points. python matplotlib interpolation | this question asked Sep 11 '13 at 20:25 Scott B 1,017 2 19 34 What is your question? – tacaswell Sep 11 '13 at 21:13 also, I think 'linear' interpolation on 2D data is bilinear. Now, we explain this assumption and the idea behind the. I put together some code for Bilinear interpolation. *copy bilinear_zoom. The interpolation as a whole is not linear but rather quadratic in the sample location. • Example: fading. References Hiroshi Akima, ". The linterp function is intended for interpolation. For bilinear interpolation, the block uses the weighted average of two translated pixel values for each output pixel value. pro in the lib subdirectory of the IDL distribution. Image zooming with MATLAB Sample Codes Here I will explain two algorithms of image processing. What separates this UDF from the countless other linear interpolation UDFs is that this functiondoesnot require your data be sorted!. If the slices are between sample points (by "aligned to the grid", I mean that slices actually pass through all the points in a "plane"), then you need trilinear filtering. In this tutorial, we will go over String interpolation in Scala which allows us to easily print and format variables and expressions. Bilinear Interpolation. 1 Can I interpolate in multiple stages? Yes, so long as the interpolation ratio, L, is not a prime number. I was unable to understand explanations in other. I have a feeling that i am presently doing this incorrectly and need some clarification. Start studying Interpolation and Extrapolation. ' 'According S3, this technique has the same memory bandwidth impact as bilinear filtering. The following minimal example demonstrates how I do not fully understand Mma's algorithm. The following matlab project contains the source code and matlab examples used for bilinear interpolation. The closer an input cell center is to the output cell center, the higher the influence of its value is on the output cell value. For this, consider the four pixels around the computed coordinate location and perform bilinear reconstruction by first performing two linear interpolations along $$x$$ for the top and bottom pairs of pixels, and then another interpolation along $$y$$ for the. Another example: you can use linear interpolation to smoothly animate from one coordinate to another. You can vote up the examples you like or vote down the ones you don't like. bilinear interpolation: You assume (correctly) that the more exact values have their justified meaning. Every channel of multi-channel images is processed independently. Dynamic range of imaging system is a ratio where the upper limit is determined by. This video will show an example to easily find specific values using Linear Interpolation. The interpolation of 3D scatter data provides a continuous scalar value for any point inside the convex domain defined by the set of XYZ locations. Bilinear interpolation is still widely used within vision systems. Linear interpolation calculator solving for y2 given x1, x2, x3, y1 and y3. Evaluate( 9. If missing values are present, then linint2_points will perform the piecewise linear interpolation at all. Sentence Examples since the terms which are bilinear in respect to ~, y,, and E, ~, ~ vanish, in virtue of the relations (l). not sure what your data is meaning. The function returns a two-dimensional, floating-point interpolated array. Or use linear interpolation to spring toward a moving target. Let’s start with the X axis. (Thus, it is fast and reliable. Bilinear interpolation ( method = 'linear' ). But if I plug in, for example, x=4. The results need to be returned to a new sheet. 7 illustrates an exemplary circuit 70, in accordance with an embodiment of the present invention. , a problem in which neither of the pair corresponds to a tabulated numerical value) requires that we start with 4 data points. What separates this UDF from the countless other linear interpolation UDFs is that this functiondoesnot require your data be sorted!. If the given data points are in R 2 {\displaystyle \mathbf {R} ^{2}} then polynomial interpolation is common. Does anyone have any insight on whether or not the Interpolate2D. Here, the key idea is to perform linear interpolation first in one direction, and then again in the other direction. Imagine scaling the 750 x 750 grid to fit over our 500 x 500 image. In akima: Interpolation of Irregularly and Regularly Spaced Data. ANTIALIAS is best for downsampling, the other filters work better with upsampling (increasing the size). It smooths the output raster grid, but not as much as cubic convolution. You can also use this linear interpolation calculator for extrapolation - for example, you can calculate what amount of flour is required to bake 50 cookies. Could you please provide the math behind so that how implementing it on CPU and GPU will be clearer?. Learn more about image processing, bilinear interpolation, interpolation, text file, bicubic interpolation, 2d array, digital image processing Image Processing Toolbox. Rosenthal Besteck 6 Tafelmesser Sculptura Lino Sabattini Stainless steel (P79),PORTRAIT VON PAPST LUCIUS III. The following matlab project contains the source code and matlab examples used for bilinear interpolation. Using polyfit for polynomial fit. Bilinear interpolation is a process that enhances textures that would normally be larger on-screen than in texture memory, i. Also note this sentence from the help for pcolor: "With shading interp, each cell is colored by bilinear interpolation of the colors at its four vertices, using all elements of C. Setting the interpolation does not carry through to any images created by imageaffine() or imagerotate(). You want to translate this image 0. Bilinear forms and their matrices Joel Kamnitzer March 11, 2011 0. In the last post we saw how to do cubic interpolation on a grid of data. Nearest Neighbor, Bilinear, and Bicubic Interpolation Methods Nearest Neighbor Interpolation. We have a need for a bilinear interpolation algorithm that can interpolate a 2MP (2000x1000) image at an ideal rate of 40-50/s. Gaussian blur is an image space effect that is used to create a softly blurred version of the original image. 25, 0}, respectively. , James Sherman Jr. The function returns a two-dimensional, floating-point interpolated array. Description. Example command to produce the error: gdalwarp -t_srs EPSG:3031 -r bilinear -order 3 -tr 4450 4450 melt_allGCP. Bilinear Interpolation in General For this assignment, you'll make repeated use of bilinear interpolation. Multivariate interpolation is the interpolation of functions of more than one variable. 'bilinear' Bilinear interpolation; the output pixel value is a weighted average of pixels in the nearest 2-by-2 neighborhood. Now, it’s just a simple matter of entering the formula for linear interpolation into the appropriate cell. I assume that we do a bilinear interpolation. Instead of rotacionar pixels of the source, it rotacionava pixels of the destination and discovered in which place of the source would go to be. Methods based on sample points in regular grid. In bilinear interpolation you use the four closest grid points for the calculation. Ch 2, Lesson C, Page 19 - Performing a Double Linear Interpolation. And you are missing the G*x^2y , H*y^2x, and I*x^2y^2 terms so it is not even a full bi-quadratic. Edge-directed interpolation is an adaptive approach, where the area around each pixel is analyzed to determine if a. Bilinear interpolation is equivalent to two step linear interpolation, first in the x-dimension and then in the y-dimension. Suppose that we want to find the value of the unknown function f at the point P = ( x , y ). The effect of these additional pixels in performing the prediction is to add some of the higher order terms of (2). For example, suppose this matrix,. • Interpolation: - Can illustrate sample-and-hold and linear interpolation from. Bilinear Interpolation. Regular texture mapping would appear to be 'blocky' in this case. Here is a simple example of trilinear interpolation on a grid. Bilinear interpolation is a process that enhances textures that would normally be larger on-screen than in texture memory, i. Cubic Spline Interpolation. Have fun!. If the given data points are in R 2 {\displaystyle \mathbf {R} ^{2}} then polynomial interpolation is common. There is no current Libor quote available for the required maturity, however, so it is necessary to estimate the unknown rate. In this paper, we mainly study the quantum realization of bilinear. Many examples with R/Python code are available. IMAGE PROCESSING: GEOMETRIC OPERATIONS Interpolation Example Bi-linear Interpolation ORIGINAL INPUT Bi-cubic Interpolation Nearest Neighbor Interpolation. For example, maybe we want to know the amplitude at the exact time where the seabed horizon crosses the trace. Bilinear interpolation - Surveys the 4 closest pixels, creates a weighted average based on the nearness and brightness of the surveyed pixels and assigns that value to the pixel in the output image. Bilinear interpolation with non-aligned entry points I have a non-grid-aligned set of input values associated with grid-aligned output values. Interpolated values in between represented by color. INTER_CUBIC - a bicubic interpolation over 4x4 pixel neighborhood. | 2020-01-23T02:14:13 | {
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https://brilliant.org/discussions/thread/brilliant-integration-contest-season-3-part-2/ | # Brilliant Integration Contest - Season 3 (Part 2)
Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 3(Part 1)
The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.
The rules are as follows:
• I will start by posting the first problem. If there is a user solves it, then they must post a new one.
• You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
• Only make substantial comment that will contribute to the discussion.
• Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.
• If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
• The scope of questions is only computation of integrals either definite or indefinite integrals.
• You are NOT allowed to post a multiple integrals problem.
• It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.
• Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest
Format your post is as follows:
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1 year, 6 months ago
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Problem 26 :
Show that $\int_0^\infty \ln\left(\frac{a^s + x^s}{b^s + x^s}\right)\,dx \; =\; \pi(a-b)\mathrm{cosec}\tfrac{\pi}{s} \hspace{1cm} a,b > 0\,,\,s > 1 \;.$
This problem has been solved by Fdp Dpf.
- 1 year, 6 months ago
Solution problem 26:
Let , $\displaystyle F(a)=\int_0^{+\infty} \ln\left(\dfrac{a^s+x^s}{b^s+x^s}\right)dx$
Observe that, $F(b)=0$
$\displaystyle F^\prime(a)=sa^{s-1}\int_0^{+\infty} \dfrac{1}{a^s+x^s}dx$
Perform the change of variable,
$y=\dfrac{x}{a}$,
$\displaystyle F^\prime(a)=s\int_0^{+\infty} \dfrac{1}{1+x^s}dx$
It's well known that,
$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$
Therefore,
$\displaystyle F^\prime(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$
Therefore, $F(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)a+k$ ,
k a real constant.
Since $F(b)=0$ then $k=-\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)b$
Therefore,
$\boxed{\displaystyle F(a)=\pi (a-b)\mathrm{cosec}\left(\dfrac{\pi}{s}\right)}$
Proof of:
$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$
Perform the change of variable
$y=x^s$
$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{1}{s} \int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx$
\displaystyle \begin{align}\int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx&=\beta\left(\dfrac{1}{s},1-\dfrac{1}{s}\right)\\ &=\dfrac{\Gamma\left(\dfrac{1}{s}\right)\Gamma\left(1-\dfrac{1}{s}\right)}{\Gamma(1)}\\ &= \dfrac{\pi}{\sin\left(\dfrac{\pi}{s}\right)}\\ &=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right) \end{align}
$$\beta$$ being the Beta Euler function. Third line is the use of Euler's reflection formula.
- 1 year, 6 months ago
Problem 30:
Show that $\int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\,\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2$
This problem has been solved by Fdp Dpf.
- 1 year, 6 months ago
$\displaystyle I=\int_0^1 \dfrac{\ln(\sqrt{1+\dfrac{x^2}{4}}+\dfrac{x}{2})}{x}dx$
Perform the change of variable $$y=\dfrac{x}{2}$$,
$\displaystyle I=\int_0^{\tfrac{1}{2}} \dfrac{\ln\left(\sqrt{1+x^2}+x\right)}{x}dx$
Perform the change of variable $$y=\mathrm{arsinh}(x)$$,
\begin{align}\displaystyle I&=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\cosh(x)}{\sinh(x)}dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\left(1+\mathrm{e}^{-2x}\right)}{1-\mathrm{e}^{-2x}} dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\left(1+\mathrm{e}^{-2x}\right)\sum_{n=0}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\sum_{n=1}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\sum_{n=1}^{+\infty} \left(\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} x\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\sum_{n=1}^{+\infty}\left(\dfrac{1}{n^2}-\dfrac{\left(1+2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n\right)\mathrm{e}^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n}}{n^2}\right)\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\zeta(2)-\dfrac{1}{2}\mathrm{Li}_2\left(e^{-2\mathrm{asinh}\left(\tfrac{1}{2}\right)}\right)+\mathrm{arsinh}\left(\tfrac{1}{2}\right)\ln\left(1-e^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)}\right) \end{align}
Since, $$\mathrm{arsinh}\left(\dfrac{1}{2}\right)=\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)$$ then,
$I=\dfrac{\pi^2}{12}-\dfrac{1}{2}\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)-\dfrac{1}{2}\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2$
Since , $$\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=\dfrac{\pi^2}{15}-\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2$$ then,
$I=\dfrac{\pi^2}{12}-\dfrac{\pi^2}{30}=\boxed{\dfrac{\pi^2}{20}}$
- 1 year, 5 months ago
Nice. A bit of integration by parts makes the earlier stage a little clearer, perhaps...
Note that $\frac{d}{dx}\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right) \; = \; \frac{\frac{\frac12x}{\frac12\sqrt{1+\frac14x^2}} + \frac12}{\sqrt{1 + \frac14x^2} + \frac12x} \; = \; \frac{1}{\sqrt{4+x^2}}$ and so, integrating by parts, \begin{align} \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} & = \Big[(\ln x)\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\Big]_0^1 - \int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \\ & = -\int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \end{align} Since $$\sinh^{-1}\tfrac12 = \ln\big(\tfrac12(\sqrt{5}+1)\big)$$, the substitution $$x = 2\sinh u$$ yields \begin{align} \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx & = \int_0^{\sinh^{-1}\frac12} \ln\big(2\sinh u\big)\,du \; = \; \int_0^{\sinh^{-1}\frac12}\Big[u + \ln\big(1 - e^{-2u}\big)\Big]\,du \\ & = \tfrac12 \ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \int_0^{\ln\big(\frac{\sqrt{5}+1}{2}\big)} \frac{\ln\big(1 - e^{-2u}\big)}{e^{-2u}}\,e^{-2u}\,du \\ & = \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) - \frac12\int_1^{\frac{3-\sqrt{5}}{2}} \frac{\ln(1-v)}{v}\,dv \end{align} using the substitution $$v = e^{-2u}$$. Thus $\int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx \; = \; \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \tfrac12\mathrm{Li}_2\left(\tfrac{3 - \sqrt{5}}{2}\right) - \tfrac12\mathrm{Li}_2(1) \; = \; \tfrac{1}{30}\pi^2 - \tfrac{1}{12}\pi^2 \; = \; -\tfrac{1}{20}\pi^2$ and so $\int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2$
- 1 year, 5 months ago
You're right, it's easiest this way. By the way, $$\displaystyle \int \ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)dx=x\ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)-\sqrt{x^2+4}$$
- 1 year, 5 months ago
Problem 29:
Show that $\displaystyle \int\limits_{0}^{\infty} \frac{e^{-tx}}{\sqrt x (x^2+1)}dx=\pi \left[ \sqrt2 C\left( \sqrt{\frac{2t}{\pi}}\right)\sin t - \sqrt2S\left( \sqrt{\frac{2t}{\pi}} \right) \cos t - \sin(t-\pi/4) \right]$
where $$C$$ and $$S$$ are Fresnel cosine integral and Fresnel sine integral, respectively.
This problem has been solved by Mark Hennings.
- 1 year, 6 months ago
In this question, we are using the Fresnel integrals $C(x) \; = \; \int_0^x \cos\big(\tfrac12\pi t^2\big)\,dt \hspace{2cm} S(x) \; = \; \int_0^x \sin\big(\tfrac12\pi t^2\big)\,dt$ If we define $A(t) \; = \; \cos t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx$ for $$t > 0$$ then \begin{align} A'(t) & = \begin{array}[t]{l}\displaystyle-\sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \cos t \int_0^\infty \frac{e^{-tx}x\sqrt{x}}{x^2+1}\,dx \\\displaystyle - \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx + \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx\end{array} \\ & = -\cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; = \; -\sqrt{\tfrac{\pi}{t}}\cos t \end{align} for any $$t > 0$$, so that $\frac{d}{dt}\Big[ A(t) + \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\Big] \; = \; 0$ Since $$A(t) \,\to\, 0$$ as $$t \to \infty$$, we deduce that $A(t) + \pi\sqrt{2}C\Big(\sqrt{\frac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0$ Similarly, if we define $B(t) \; = \; \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx + \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx$ for $$t > 0$$, then we can show that $B'(t) \; =\; -\sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; =\; - \sqrt{\tfrac{\pi}{t}}\sin t$ and hence $B(t) + \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0$ Thus we deduce that \begin{align} \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx & = B(t)\cos t - A(t) \sin t \\ & = \cos t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] - \sin t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] \\ & = \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sin\big(t - \tfrac14\pi\big) \end{align} as required.
- 1 year, 6 months ago
Problem 27:
Show that
$\displaystyle\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}\, dx=\dfrac{2}{3}(2-\sqrt{3})G$
This problem has been solved by Mark Hennings.
- 1 year, 6 months ago
Put $$\alpha = \tfrac{1}{12}\pi$$. With the substitution $$x = \frac{1-\cos\theta}{1+\cos\theta}$$, we obtain after much manipulation that $I \; = \; \int_0^{\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \frac{\ln x\,dx}{(x-1)\sqrt{x^2 - 2(15+8\sqrt{3})x + 1)}} \; = \; -\tfrac12\sin\alpha \int_0^\alpha \frac{\ln\left(\frac{1-\cos\theta}{1+\cos\theta}\right) \sin\theta\,d\theta}{\cos\theta\sqrt{\cos^2\theta - \cos^2\alpha}}$ Putting $$u = \cos\theta$$ now gives $I \; = \; -\tfrac12\sin\alpha \int_{\cos\alpha}^1 \frac{\ln\left(\frac{1-u}{1+u}\right)\,du}{u\sqrt{u^2 - \cos^2\alpha}}$ Now putting $$u = \cos\alpha \sec\phi$$ yields \begin{align} I & = -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\frac{\cos\phi - \cos\alpha}{\cos\phi + \cos\alpha}\right)\,d\phi \; = \; -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big) \tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\,d\phi \\ & = -\tfrac12\tan\alpha \int_0^\alpha \left\{ \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big)\right) + \ln\left(\tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\right\}\,d\phi \\ & = -\tan\alpha \left\{ \int_{\frac12\alpha}^\alpha \ln(\tan \phi)\,d\phi - \int_{\frac12\alpha}^0 \ln(\tan\phi)\,d\phi\right\} \\ & = -\tan\alpha \int_0^\alpha \ln(\tan\phi)\,d\phi \; = \; -\tan\alpha \int_0^{\frac{1}{12}\pi}\ln(\tan\phi)\,d\phi \\ & = \tfrac23\tan\alpha\, G \; = \; \tfrac23(2 - \sqrt{3})G \end{align} using a standard integral representation of $$G$$ at the very last stage.
- 1 year, 6 months ago
Problem 25 :
Prove That
$\dfrac{1}{\pi} \int_{0}^{\pi} x \arctan\left( \dfrac{p \sin x}{1+p \cos x} \right) \ \mathrm{d}x = \operatorname{Li}_{2}(p) \quad \forall \ |p| <1$
Notation : $$\operatorname{Li}_{2}(z)$$ denotes the Dilogarithm Function.
This problem has been solved by Mark Hennings.
- 1 year, 6 months ago
If we define $g(p) \; = \; \int_0^\pi \ln(1 + 2p \cos x + p^2)\,dx \hspace{1cm} |p| < 1$ then $$g(0) = 0$$ and, using the complex subsitution $$z = e^{ix}$$, \begin{align} g'(p) & = \int_0^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \; = \; \frac12\int_{-\pi}^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{2}\int_{|z|=1} \frac{z + z^{-1} + 2p}{1 + p(z + z^{-1}) + p^2} \frac{dz}{iz} \; = \; \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 2pz + 1}{z(pz+1)(z+p)}\,dz \\ & = \pi\left(\mathrm{Res}_{z=0} + \mathrm{Res}_{z=-p}\right)\frac{z^2 + 2pz + 1}{z(pz+1)(z+p)} \\ & = \pi\left(\frac{1}{p} - \frac{1-p^2}{p(1-p^2)}\right) \; = \; 0 \end{align} so that $$g(p) = 0$$ for all $$|p| < 1$$. If we now define $f(p) \; = \; \frac{1}{\pi}\int_0^\pi x \tan^{-1}\left(\frac{p \sin x}{1 + p \cos x}\right)\,dx \hspace{1cm} |p| < 1$ then $$f(0) = 0$$ and \begin{align} f'(p) & = \frac{1}{\pi}\int_0^\pi \frac{x}{1 + \left(\frac{p \sin x}{1 + p\cos x}\right)^2} \times \frac{\sin x}{(1 + p\cos x)^2}\,dx \\ & = \frac{1}{\pi}\int_0^\pi \frac{x \sin x}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{\pi}\Big[-\frac{x}{2p}\ln(1 + 2p\cos x + p^2)\Big]_0^\pi + \frac{1}{2\pi p}\int_0^\pi \ln(1 + 2p\cos x + p^2)\,dx \\ & = -\frac{\ln(1-p)}{p} + g(p) \; = \; -\frac{\ln(1-p)}{p} \end{align} and hence it follows that $$f(p) \,=\, \mathrm{Li}_2(p)$$.
- 1 year, 6 months ago
Aliter,
Let $f(p) = \dfrac{1}{\pi}\int_{0}^{\pi} x \arctan\left( \dfrac{p\sin x}{1+ p\cos x} \right) \ \mathrm{d}x$
Firstly, we have $$f(0) = 0$$
Note that,
$$\displaystyle \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx) = \dfrac{1}{p}\Im \left( \sum_{k=0}^{\infty} (-p e^{ix})^{k} \right)$$
$$\displaystyle = \dfrac{1}{p} \Im \left( \dfrac{1}{1+p e^{ix}} \right)$$
$$\displaystyle = \dfrac{\sin x}{p^2 +2p \cos x +1}$$
where $$\Im(z)$$ denotes the imaginary part of $$z$$.
So we have,
$$\displaystyle f'(p) = \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{x \sin x}{p^2 +2p \cos x +1} \ \mathrm{d}x$$
$$\displaystyle = \dfrac{1}{\pi} \int_{0}^{\pi} x \left(\sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)\right) \ \mathrm{d}x$$
$$\displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \int_{0}^{\pi}x \sin (kx) \ \mathrm{d}x$$
$$\displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \left( \dfrac{\sin(k \pi) - k \pi \cos(k \pi)}{k^2} \right)$$
Since $$\sin (k \pi) = 0$$ and $$\cos (k \pi) = (-1)^k \ \forall \ k \in \mathbb{Z}$$, we get,
$$\displaystyle f'(p) = \sum_{k=1}^{\infty} \dfrac{p^{k-1}}{k} = -\dfrac{\ln(1-p)}{p}$$
$\displaystyle \therefore f(p) = \operatorname{Li}_{2}(p) \ \square$
- 1 year, 6 months ago
Problem 36:
Show that,
$\int_0^{+\infty}\dfrac{x\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx=\dfrac{\pi\ln 2}{2}$
This problem has been solved by Mark Hennings.
- 1 year, 5 months ago
Alternative solution to problem 36
Let,
$I=\int_0^{+\infty} \dfrac{x(\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx$
$J=\int_0^{+\infty} \dfrac{x(\ln(1-\sqrt{2}x+x^2)}{1+x^4}dx$
$I+J=\int_0^{+\infty} \dfrac{x\ln(1+x^4)}{1+x^4}dx$
Perform the change of variable $$y=x^2$$,
$I+J=\dfrac{1}{2}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx$
Perform the change of variable $$y=\arctan x$$,
$I+J=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx$
Perform the change of variable $$y=\dfrac{\pi}{2}-x$$,
$\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx=\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx$
Therefore,
\begin{align} 2(I+J)&=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos(x)\sin(x) )dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\dfrac{\sin(2x)}{2} \right)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\sin(2x) \right)dx+\dfrac{\pi\ln 2}{2}\\ \end{align}
In the latter integral perform the change of variable $$y=2x$$,
\begin{align} 2(I+J)&=-\dfrac{1}{2}\int_0^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_{\tfrac{\pi}{2}}^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ \end{align}
In the latter integral perform the change of variable $$y=\pi-x$$, therefore,
\begin{align} 2(I+J)&=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx+\dfrac{\pi\ln 2}{2}\\ &=I+J+\dfrac{\pi\ln 2}{2}\\ \end{align}
Therefore,
$\boxed{I+J=\dfrac{\pi\ln 2}{2}}$
Define on $$\Big[0;\sqrt{2}\Big]$$,
$\displaystyle F(a)= \int_0^{+\infty} \dfrac{x(\ln(1+ax+x^2)-\ln(1-ax+x^2))}{1+x^4}dx$
Observe that, $$F(\sqrt{2})=I-J$$ and $$F(0)=0$$,
$F'(a)= \int_0^{+\infty} \dfrac{2x(1+x^2)}{(x^2-ax+1)(x^2+ax+1)(x^4+1)}dx$
$F'(a)=\left[\dfrac{2 \left( \mathrm{arctan}\left( \dfrac{2x-a}{\sqrt{4-{{a}^{2}}}}\right) +\mathrm{arctan}\left( \dfrac{2x+a}{\sqrt{4-{{a}^{2}}}}\right) \right) }{\sqrt{4-{{a}^{2}}}\cdot \left( {{a}^{2}}-2\right) }-\dfrac{\sqrt{2} \left( \mathrm{arctan}\left( \sqrt{2}x-1\right) +\mathrm{arctan}\left( \sqrt{2}x+1\right) \right) }{{{a}^{2}}-2}\right]_0^{+\infty}$
Therefore,
$F'(a)=\dfrac{2\pi}{(a^2-2)\sqrt{4-a^2}}-\dfrac{\sqrt{2}\pi}{a^2-2}$
$F(\sqrt{2})=\int_0^{\sqrt{2}} F'(a)da=\dfrac{\pi}{2}\left[\mathrm{ln}\left( \dfrac{\left( \sqrt{2}+a\right)\left( \sqrt{4-{{a}^{2}}}-a\right) }{\left( \sqrt{2}-a\right)\left( \sqrt{4-{{a}^{2}}}+a\right) }\right)\right]_0^\sqrt{2}=\dfrac{\pi\ln 2}{2}$
Therefore,
$I+J=I-J=\dfrac{\pi\ln 2}{2}$
and,
$\boxed{I=\dfrac{\pi\ln 2}{2}}$ $J=0$
- 1 year, 5 months ago
You could speed up your calculation of $$I+J$$ by doing a Beta function trick. It is equal to $-\tfrac14 \frac{d}{da} B(\tfrac12,a-\tfrac12)\; \Big|_{a=1}$
- 1 year, 5 months ago
I came up with a similar solution just now, and was about to post, but you beat me to it by 4 hours!
- 1 year, 5 months ago
This one is a classic one ;)
- 1 year, 5 months ago
The substitutions $$t = 1-s$$ and $$s = \sin\theta$$ give $\int_0^1 \left(\frac{1}{\sqrt{t(2-t)}}-1\right)\,\frac{dt}{1-t} \; = \; \int_0^1 \left(\frac{1}{\sqrt{1-s^2}} - 1\right)\,\frac{ds}{s} \; = \; \int_0^{\frac12\pi}\tan\tfrac12\theta\,d\theta \; = \; \ln2$ The substitution $$x^2 = \tfrac{t}{2-t}$$ and one of the standard integral representations of the Catalan constant $$G$$ gives $\int_0^1 \frac{1}{\sqrt{t(2-t)}} \left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right)\,\frac{dt}{1-t} \; = \; 2\int_0^1 \left(\tfrac14\pi - \tan^{-1}x\right)\,\frac{dx}{1-x^2} \; = \; G$ These results together give us that \begin{align} \int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx & = \int_0^\infty \frac{dx}{x^2 + \sqrt{2}x + 1}\int_0^1 \frac{x^2 + \sqrt{2}x}{tx^2 + t\sqrt{2}x + 1}\,dt \\ & = \int_0^1 \int_0^\infty \frac{x^2 + \sqrt{2}x}{(tx^2 + t\sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}\,dx\,dt \\ & = \int_0^1 \frac{dt}{1-t}\int_0^\infty\left(\frac{1}{tx^2 + t\sqrt{2}x + 1} - \frac{1}{x^2 + \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^1 \frac{dt}{1-t} \left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac12\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) - \frac{\pi}{2\sqrt{2}}\right\} \\ & = \int_0^1 \frac{dt}{1-t}\left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) + \tfrac{\pi}{2\sqrt{2}}\left(\frac{1}{\sqrt{t(2-t)}} - 1\right)\,\right\} \end{align} and so $\int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx \; = \; = \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2$ Standard substitutions show that $f(a) \; = \; \int_0^\infty \frac{x^2+1}{(x^4 + 1)^a}\,dx \; = \; \tfrac14\big[B(\tfrac34,a-\tfrac34)+B(\tfrac14,a-\tfrac14)\big]$ for $$a > \tfrac34$$, and hence $f'(a) \; = \; \frac{\Gamma(\frac34)\Gamma(a-\frac34)\psi(a-\frac34) + \Gamma(\tfrac14)\Gamma(a-\frac14)\psi(a-\frac14)}{4\Gamma(a)} - \frac{\Gamma(\frac34)\Gamma(a-\frac34) + \Gamma(\frac14)\Gamma(a-\frac14)}{4\Gamma(a)}\psi(a)$ and hence $\int_0^\infty \frac{(x^2+1)\ln(x^4+1)}{x^4+1}\,dx \; = \; -f'(1) \; = \; -\tfrac{\pi}{2\sqrt{2}}\big[\psi(\tfrac14) + \psi(\tfrac34) - 2\psi(1)\big] \; = \; \tfrac{3\pi}{\sqrt{2}}\ln2$ Next, the substitution $$x = \tan\tfrac12\theta$$ and another standard integral representation of the Catalan constant gives \begin{align} \int_0^\infty &\frac{1+x^2}{1+x^4}\ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^\pi \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \\ & = 2\int_0^{\frac12\pi} \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \; = \; 2\sqrt{2}G \end{align} Thus \begin{align} \int_0^\infty &\left(\frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1} + \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = 2\int_0^\infty \frac{x^2+1}{x^4+1} \ln(x^2+ \sqrt{2}x+1)\,dx \\ & = \int_0^\infty \frac{x^2+1}{x^4+1}\left[\ln\left(\frac{x^2 +\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) + \ln(x^4+1)\right]\,dx \\ & = 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \end{align} and hence $\int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\,dx \; = \; \big( 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \big) - \big( \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2\big) \; = \; \sqrt{2}G + \tfrac{5\pi}{2\sqrt{2}}\ln2$ Finally (!) we deduce that \begin{align} \int_0^\infty \frac{x}{x^4+1}\ln(x^2 + \sqrt{2}x + 1)\,dx & = \frac{1}{2\sqrt{2}}\int_0^\infty\left(\frac{\ln(x^2 + \sqrt{2}x+1)}{x^2 - \sqrt{2}x + 1} - \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\right) \\ & = \tfrac12\pi \ln2 \end{align}
- 1 year, 5 months ago
Problem 31:
Show that
$\displaystyle \int_0^1 \dfrac{x-1}{(x+1)\ln x}dx=\ln\left(\dfrac{\pi}{2}\right)$
This problem has been solved by Ishan Singh.
- 1 year, 5 months ago
Let
$f(p) = \int_{0}^{1} \dfrac{x^p - 1}{(x+1) \ln x} \ \mathrm{d}x \quad ; \quad p > 0$
$$\displaystyle \implies f'(p) = \int_{0}^{1} \dfrac{x^p}{x+1} \mathrm{d}x$$
$$\displaystyle = \int_{0}^{1} \dfrac{x^p(x-1)}{x^2 -1} \mathrm{d}x$$
Substitute $$x^2 \mapsto x$$ and simplify to get,
$$\displaystyle f'(p) = \dfrac{1}{2} \int_{0}^{1} \left( \dfrac{x^{\frac{p}{2}} - 1}{x - 1} - \dfrac{x^{\frac{p-1}{2}} - 1}{x - 1} \right) \mathrm{d}x$$
$$\displaystyle = \dfrac{1}{2} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right)$$
Note that $$f(0) = 0$$
$$\displaystyle \implies f(1) = \int_{0}^{1} f'(p) \ \mathrm{d}p$$
$$\displaystyle = \dfrac{1}{2} \int_{0}^{1} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right) \mathrm{d}p$$
$$\displaystyle = \log \left(\dfrac{\pi}{2}\right)$$
where the last equality follows since $$\psi (z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \log(\Gamma(z))$$
- 1 year, 5 months ago
Problem 28:
Show that,
$\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}\, dx=\frac{G}{3}$
$$G$$ being the Catalan constant.
This problem has been solved by Mark Hennings.
- 1 year, 6 months ago
Your derivation of this result on MSE is pretty compact, so I shall expand it. Suppose that $$a > 1$$, and let $$b = \frac{1+a^2}{a-1}$$. The substitution $$y \,=\, \frac{ax}{x+b}$$ gives $\int_0^1 \tan^{-1}\left(\frac{ax}{x+b}\right)\,\frac{dx}{x} \; = \; \int_0^{\frac{a-1}{a+1}} \tan^{-1}y\, \frac{a-y}{by} \, \frac{ab}{(a-y)^2}\,dy \; = \; \int_0^{\frac{a-1}{a+1}} \frac{a}{y(a-y)} \tan^{-1}y\,dy$ while the substitution $$z = \frac{x}{b - ax}$$ gives $\int_0^1 \tan^{-1}\left(\frac{x}{ax - b}\right)\,\frac{dx}{x} \; = \; -\int_0^{\frac{a-1}{a+1}} \tan^{-1}z \, \frac{az+a}{bz} \, \frac{b}{(az+1)^2}\,dz \; =\; -\int_0^{\frac{a-1}{a+1}} \frac{1}{z(az+1)} \tan^{-1}z\,dz$ and hence \begin{align} \int_0^1 \left\{\tan^{-1}\left(\frac{ax}{ax+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right\}\,\frac{dx}{x} & = \int_0^{\frac{a-1}{a+1}} \left(\frac{a}{y(a-y)} - \frac{1}{y(ay+1)}\right) \tan^{-1}y\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{a^2+1}{(a-y)(ay+1)}\,\tan^{-1}y\,dy \\ & = \int_)^{\frac{a-1}{a+1}} \left(\frac{1}{a-y} + \frac{a}{1 + ay}\right)\tan^{-1}y\,dy \\ & = \left[ \ln\left(\frac{1 + ay}{a-y}\right) \tan^{-1}y\right]_0^{\frac{a-1}{a+1}} + \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \; = \; \int_1^a \frac{\ln x}{1+x^2}\,dx \end{align} Now \begin{align} \tan\left[\tan^{-1}\left(\frac{ax}{x+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right] & = \frac{\frac{ax}{x+b} + \frac{x}{ax-b}}{1 - \frac{ax}{x+b}\frac{x}{ax-b}} \; = \; \frac{ax(ax-b) + x(x+b)}{(x+b)(ax-b) - ax^2} \\ & = \frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x} \end{align} and hence it follows that $\int_0^1 \tan^{-1}\left(\frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x}\right)\,\frac{dx}{x} \; = \; \int_1^a \frac{\ln x}{1 + x^2}\,dx$ Putting $$a = 2 + \sqrt{3}$$ yields \begin{align} \int_0^1 \tan^{-1}\left(\frac{x(1-x)}{2-x}\right)\,\frac{dx}{x} & = \int_1^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx \; = \; \int_{\frac14\pi}^{\frac{5}{12}\pi} \ln(\tan\theta)\,d\theta \\ & = -\int_{\frac{1}{12}\pi}^{\frac14\pi} \ln(\tan\phi)\,d\phi \; = \; -\big((-G) - (-\tfrac23G)\big) \; = \; \tfrac13G \end{align} using the final substitution $$\phi = \tfrac12\pi - \theta$$.
I won't be posting the next problem until after Christmas!
- 1 year, 6 months ago
Nice. I was secretly expecting a different proof. Someone, in 2011, showed me this integral. I have tried to find out the very source for it. Since then, i enjoy to learn and sharing knowledge about computing integrals. Happy chrismas !
- 1 year, 6 months ago
PROBLEM 39:
Evaluate $\int_0^\infty\left(e^{-x^a} - \frac{1}{1+x^b}\right)\,\frac{dx}{x}$ for any $$a,b > 0$$.
- 1 year, 5 months ago
Solution to problem 39.
$J=\int_0^{+\infty} \left( \text{e}^{-x^a}-\dfrac{1}{1+x^b}\right)\dfrac{1}{x}dx$
Perform the change of variable $$y=x^a$$,
$J=\dfrac{1}{a}\int_0^{+\infty} \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\dfrac{1}{x}dx$
Perform integration by parts,
\begin{align}aJ&=\left[ \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\ln x\right]_0^{+\infty}+\int_0^{+\infty}\left( \text{e}^{-x}-\dfrac{a}{b}\dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\right)\ln x dx\\ &=\int_0^{+\infty} \text{e}^{-x}\ln x dx-\dfrac{a}{b} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{align}
\begin{align} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx+\int_1^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{align}
In the latter integral perform the change of variable $$y=\dfrac{1}{x}$$,
\begin{align} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx-\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx\\ &=0 \end{align}
$\int_0^{+\infty} \text{e}^{-x}\ln x dx=\Gamma^\prime (1)=-\gamma$
Therefore,
$\boxed{J=-\dfrac{1}{a}\gamma}$
- 1 year, 5 months ago
Problem 37:
Evaluate $\int_0^{\frac12\pi}\, \ln(\cos x + \sin x)\,dx$
This problem has been solved by Fdp Dpf.
- 1 year, 5 months ago
Solution to problem 37.
Since,
\begin{align}\sin\left(x+\dfrac{\pi}{4}\right)&=\cos\left(\dfrac{\pi}{4}\right)\sin x+\sin \left(\dfrac{\pi}{4}\right)\cos x\\ &=\dfrac{\sqrt{2}}{2}(\sin x+\cos x)\end{align}
then,
\begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\int_0^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(\dfrac{\pi}{4}-x\right)+\dfrac{\pi}{4}\right)\right)dx+\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(x+\dfrac{\pi}{4}\right)+\dfrac{\pi}{4}\right)\right)dx\\ &=2\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}+2\int_0^{\tfrac{\pi}{4}} \ln\left(\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(\dfrac{1}{(\cos x)^2}\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(1+(\tan x)^2\right)dx\\ \end{align}
In the latter integral perform the change of variable $$y=\tan x$$, therefore, \begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx \end{align}
$\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx$
In the latter integral perform the change of variable $$y=\dfrac{1}{x}$$, therefore,
\begin{align}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx&=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^1 \dfrac{\ln\left(1+\dfrac{1}{x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx-2\int_0^1 \dfrac{\ln x}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+2\text{G}\\ \end{align}
$$\text{G}$$ being the Catalan constant. But, $\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\pi\ln 2$
(see my answer to problem 36)
Therefore, $\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-\text{G}$
Therefore,
\begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\left(\dfrac{\pi\ln 2}{2}-\text{G}\right)\\ &=\boxed{\text{G}-\dfrac{\pi\ln 2}{4}} \end{align}
- 1 year, 5 months ago
Aliter,
The integral $$\displaystyle \int_{0}^{\frac{\pi}{4}} \log(\cos x) \ \mathrm{d}x$$ can also be done using the identity $\log(\cos x) = -\log 2 - \sum_{k=1}^{\infty} \dfrac{\cos (2kx)}{k}$.
- 1 year, 5 months ago
I know but i don't like to use this identity when i can avoid to use it. I like solutions using elementary tools.
- 1 year, 5 months ago
I like solutions using elementary methods too. I consider the above identity elementary, it can be proved by writing $$\cos (2kx)$$ as $$\dfrac{e^{2ikx} + e^{-2ikx}}{2}$$ and then using the Taylor Series of logarithm. Btw, Happy New Year.
- 1 year, 5 months ago
Problem 35:
Show that
$\int_0^1 \frac{\ln x}{x^2 + x + 1}\,dx \; = \; \tfrac29\left[\tfrac23\pi^2 - \psi'(\tfrac13)\right]$
This problem has been solved by Fdp Dpf.
- 1 year, 5 months ago
\begin{align}\int_0^1 \dfrac{\ln x}{1+x+x^2}dx&=\int_0^1 \dfrac{1-x}{1-x^3}\ln x dx\\ &=\int_0^1 \dfrac{1}{1-x^3}\ln x dx-\int_0^1 \dfrac{x}{1-x^3}\ln x dx\\ &=\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n}\ln x\right)-\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n+1}\ln x\right)\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(3n+2)^2}\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\zeta(2)-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}-\sum_{n=1}^{+\infty} \dfrac{1}{(3n)^2}\\ &=-2\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\dfrac{8}{9}\zeta(2)\\ &=\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\sum_{n=0}^{+\infty} \dfrac{1}{\left(n+\tfrac{1}{3}\right)^2}\\ &=\boxed{\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\psi^\prime \left(\dfrac{1}{3}\right)} \end{align} Since $$\displaystyle \psi^\prime \left(a\right)=\sum_{n=0}^{+\infty} \dfrac{1}{(n+a)^2}$$
- 1 year, 5 months ago
Problem 32 : Prove That
$\int_{0}^{\infty} \left( \dfrac{1-x^2}{1+x^2} \right) \cdot \ln (x) \cdot \arctan \left( \dfrac{x}{x^2 + 1} \right) \ \dfrac{\mathrm{d}x}{x} = - \dfrac{\pi^3}{20}$
This problem has been solved by Mark Hennings.
- 1 year, 5 months ago
Suppose that $$u > 0$$. Integrating by parts, we see note that \begin{align} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \Big[-\frac{2t}{u}\tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\Big]_0^\infty + \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \\ & = \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \end{align} Now the substitution $$v = \tfrac12u\mathrm{sech}\, t$$ yields \begin{align} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \frac{2}{u}\int_{\frac12u}^0 \frac{\tan^{-1}v\, (-u)}{v\sqrt{u^2-4v^2}}\,dv \\ & = 2\int_0^{\frac12u} \frac{\tan^{-1}v\,dv}{v\sqrt{u^2-4v^2}} \; = \; \frac{2}{u}\int_0^1 \frac{\tan^{-1}(\frac12 u w)}{w\sqrt{1-w^2}}\,dw \\ & = \frac{\pi}{u}\ln\Big(\sqrt{1 + \tfrac14u^2} +\tfrac12u\Big) \end{align} Note that the identity $\int_0^1 \frac{\tan^{-1} pw}{w\sqrt{1-w^2}}\,dw \; = \; \tfrac12\pi\ln\big(\sqrt{1 + p^2} + p\big)$ is a standard integral from Gradshteyn & Ryzhik at the last stage. Thus, using the substitution $$x = e^t$$, \begin{align} \int_0^\infty \frac{1-x^2}{1+x^2} \ln x \tan^{-1}\left(\frac{x}{x^2+1}\right)\,\frac{dx}{x} & = - \int_{\mathbb{R}}t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \left(\int_0^1 \frac{\frac12\,\mathrm{sech}\,t\,du}{1 + \frac14u^2\,\mathrm{sech}^2t}\right)\,dt \\ & = -\int_0^1\left(\int_0^\infty \frac{t\,\mathrm{sech}\,t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt\right)\,du \\ & = -\pi\int_0^1 \ln\Big(\sqrt{1 + \tfrac14u^2} + \tfrac12u\Big)\,\frac{du}{u} \; = \; -\tfrac{1}{20}\pi^3 \end{align} using the result of Problem 30.
- 1 year, 5 months ago
Nice. By the way, Taylor expansion of $$\dfrac{1}{x}\arctan\left(\dfrac{x}{1+x^2}\right)$$ is $$\displaystyle \sum_{n=0}^{+\infty} \dfrac{(-1)^n L_{2n+1}x^{2n}}{2n+1}$$
and,
Taylor expansion of $$\dfrac{\arctan\left(\dfrac{x}{1+x^2}\right)}{x(1+x^2)}$$ is $$\displaystyle \sum_{n=0}^{+\infty} (-1)^n\left(\sum_{p=0}^n \dfrac{L_{2p+1}}{2p+1}\right)x^{2n}$$
$$L_k$$ is the $$k$$-th Lucas number.
- 1 year, 5 months ago
(+1) Nice! My method was a bit different (used tangent half angle in the beginning), but I also reduce it to P30. The integral you mentioned can be proved using differentiation under the integral and forming a differential equation.
- 1 year, 5 months ago
Problem 43 : Prove That $\int_{0}^{\infty} \dfrac{\mathrm{d}x}{1 + (x+\tan x)^2} = \dfrac{\pi}{2}$
- 1 year, 5 months ago
Note that,
$\tan x = - \lim_{n \to \infty} \left( \dfrac{1}{x- \frac{\pi}{2}} + \dfrac{1}{x+ \frac{\pi}{2}} + \ldots + \dfrac{1}{x- (2n-1)\frac{\pi}{2}} + \dfrac{1}{x + (2n-1)\frac{\pi}{2}} \right)$
Also, by Glasser's Master Theorem (which can be proved using elementary methods by using the graphical properties of the function), we have,
$\int_{-\infty}^{+\infty}f\left(x-\frac{a_1}{x-\lambda_1}-\cdots-\frac{a_n}{x-\lambda_n}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a_{i} >0 \ , \ \lambda_{i} \in \mathbb{R}$
Therefore,
$\int_{-\infty}^{+\infty}f\left(x+ a\tan x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a >0$
Putting $$f(x) = \dfrac{1}{x^2+b^2}$$, we have,
$\int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{x^2+b^2}$
$\therefore \int_{0}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \dfrac{\pi}{2b} \quad \square$
- 1 year, 5 months ago
I think there are some big convergence issues to be settled The convergence of the series for $$\tan x$$ is anything but uniform, and there is no function that I can see that will enable the DCT to guarantee to obtain $\int_{-\infty}^\infty f(x+a\tan x)\,dx$ as the limit of $\int_{-\infty}^\infty f\left(x - \tfrac{a_1}{x-\lambda_1} - \cdots - \tfrac{a_n}{x - \lambda_n}\right)\,dx$
- 1 year, 5 months ago
Here is my take, using the ideas I had been playing with, which convert the integral into a shape for which the GMT can be used, and a limit taken, to get the answer.
Write \begin{align} I \; = \; \int_0^\infty \frac{1}{1 + (x + \tan x)^2}\,dx & = \tfrac12\int_{\mathbb{R}} \frac{dx}{1 + (x + \tan x)^2} \\ & = \tfrac12 \sum_{n \in \mathbb{Z}} \int_{-\frac12\pi}^{\frac12\pi} \frac{1}{1 + (n\pi + x + \tan x)^2}\,dx \\ & = \tfrac{1}{2\pi^2}\int_{-\frac12\pi}^{\frac12\pi} \sum_{n \in \mathbb{Z}} \frac{dx}{\big(n + \tfrac{x + \tan x}{\pi}\big)^2 + \tfrac{1}{\pi^2}} \\ & = \frac{1}{2\pi^2} \int_{-\frac12\pi}^{\frac12\pi} \frac{\pi^2 \sinh 2}{\cosh 2 - \cos(2x + 2\tan x)}\,dx \\ & = \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh 2 - \cos(2x + 2\tan x)} \end{align} Since we are integrating over a finite interval, and since the integrand is bounded on the interval, we can use the GMT, taking the limit in the manner you suggest, so that $I \; = \; \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh2 - \cos 2x} \; = \; \tfrac14\sinh2 \int_{-\pi}^\pi \frac{dx}{\cosh 2 - \cos x}$ Making the complex substitution $$z = e^{ix}$$, this becomes \begin{align} I & = \tfrac14\sinh2 \int_{|z|=1} \frac{1}{\cosh2 - \frac12(z + z^{-1})} \frac{dz}{iz} \; = \; -\frac{\sinh 2}{2i}\int_{|z|=1}\frac{dz}{(z-e^2)(z-e^{-2})} \\ & = -\pi \sinh 2\, \mathrm{Res}_{z = e^{-2}} \frac{1}{(z - e^2)(z - e^{-2})} \; = \; \tfrac12\pi \end{align} as required.
- 1 year, 5 months ago
PROBLEM 42 :
Evaluate $\int_0^1 \frac{(x^p - 1)(x^q - 1)}{(x-1) \ln x}\,dx \hspace{2cm} p,q > 0$
- 1 year, 5 months ago
Let,
$$\displaystyle \text{I} = \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(x-1) \ln x} \ \mathrm{d}x$$
$$\displaystyle = - \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(1-x) \ln x} \ \mathrm{d}x$$
$$\displaystyle = - \sum_{r=0}^{\infty} \int_{0}^{1} \left[ \left(\dfrac{x^{p+q+r} - x^{p+r}}{\ln x}\right) - \left(\dfrac{x^{q+r} - x^{r}}{\ln x}\right) \right] \ \mathrm{d}x$$
$$\displaystyle = -\lim_{n \to \infty} \sum_{r=0}^{n} \left[\log \left(\dfrac{p+q+r+1}{p+r+1}\right) - \log\left( \dfrac{q+r+1}{r+1} \right)\right]$$
$$\displaystyle = -\lim_{n \to \infty} \sum_{r=1}^{n} \left[\log \left(\dfrac{p+q+r}{p+r}\right) - \log\left( \dfrac{q+r}{r} \right)\right]$$
where the standard result $$\displaystyle \int_{0}^{1}\dfrac{x^a-x^b}{\ln x} \ \mathrm{d}x = \ln\left( \dfrac{a+1}{b+1} \right)$$ has been used in the above lines.
$$\displaystyle \implies \text{I} = -\lim_{n \to \infty} \log \left( \dfrac{n! \times (p+q+1)(p+q+2)\ldots(p+q+n)}{[(p+1)(p+2)\ldots(p+n)] \times [(q+1)(q+2)\ldots(q+n)]} \right)$$
$$\displaystyle = -\lim_{n \to \infty} \log \left( \dfrac{\Gamma(p+q+n+1) \Gamma(p+1) \Gamma(n+1) \Gamma(q+1) }{\Gamma(p+q+1) \Gamma(p+n+1) \Gamma(q+n+1)} \right)$$
Since $$\displaystyle \lim_{n\to \infty} \dfrac{n^x \Gamma(n+1)}{\Gamma(n+x+1)} = 1$$, we have,
$$\displaystyle \text{I} = \log\left( \dfrac{\Gamma(p+q+1)}{\Gamma(p+1) \Gamma(q+1)} \right) \quad \square$$
- 1 year, 5 months ago
One of the standard integral representations of $$\ln \Gamma(z)$$ gets you there much more easily...
- 1 year, 5 months ago
Comment deleted Jan 12, 2017
Try applying the substitution $$x = e^{-t}$$ to the standard formula $\ln\Gamma(p) \; = \; \int_0^\infty \left\{ \frac{e^{-pt} -e^{-t}}{1 - e^{-t}} + (p-1)e^{-t}\right\}\frac{dt}{t}$ When you consider $\ln\left(\frac{\Gamma(p+q+1)}{\Gamma(p+1)\Gamma(q+1)}\right)$ the second terms all cancel out...
- 1 year, 5 months ago
I think we can generalize it to $\int_{0}^{1} \dfrac{\prod_{k=1}^n (x^{a_{k}} - 1)}{(x-1) \ln x} \ \mathrm{d}x$ where $$a_{k} > 0$$
- 1 year, 5 months ago
Problem 40
Evaluate $\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$
- 1 year, 5 months ago
Substitute $$(2^x-1) = t^2$$ to get,
$$\text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t$$
Substitute $$t \mapsto \dfrac{1}{t}$$
$$\implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}$$
$$\implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2}$$
$$\implies \text{I} = \dfrac{\pi}{2 \ln^2 2}$$
- 1 year, 5 months ago
Problem 38:
Evaluate,
$\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{x\ln\left(\tfrac{\cos x+\sin x}{\cos x-\sin x}\right)}{\cos x(\cos x+\sin x)}\, dx$
- 1 year, 5 months ago
We note that \begin{align} I & = \int_0^{\frac14\pi}\frac{x}{\cos x(\cos x + \sin x)} \ln\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)\,dx \; = \; \int_0^{\frac14\pi} \frac{x}{1 + \tan x}\ln\left(\frac{1 + \tan x}{1 - \tan x}\right)\,\sec^2x\,dx \\ & = \int_0^1 \frac{\tan^{-1}u}{1+u}\ln\left(\frac{1+u}{1-u}\right)\,du \; = \; -\int_0^1 \tan^{-1}\left(\frac{1-y}{1+y}\right)\,\frac{\ln y}{1+y}\,dy \end{align} after the substitutions $$u = \tan x$$ and $$y = \frac{1-u}{1+u}$$. Since $\tan\big(\tfrac14\pi - \tan^{-1}y\big) \; = \; \frac{1-y}{1+y}$ we deduce that $I \; = \; -\int_0^1 \left(\tfrac14\pi - \tan^{-1}y\right)\frac{\ln y}{1+y}\,dy \; = \; -\tfrac14\pi\int_0^1 \frac{\ln y}{1+y}\,dy + \int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy$ Now it is elementary that $\int_0^1 \frac{\ln y}{1+y}\,dy \; = \; -\tfrac{1}{12}\pi^2$ and we see from this MSE post --- hello, it's one of yours again --- that $\int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy \; = \; \tfrac12 G \ln 2 - \tfrac{1}{64}\pi^3$ so it follows that $I \; = \; \tfrac{1}{192}\pi^3 + \tfrac12G \ln 2$
- 1 year, 5 months ago
Congrats ! It's like you read my mind, you know all my tricks ;) Happy new year to all !
- 1 year, 5 months ago
Problem 34:
Prove that $\large\int_{-\infty}^\infty \dfrac1{1+x^2+x^4+\cdots+x^{2k} } \, dx = \dfrac{2\pi}{k+1} \cdot \dfrac{ \cos\left( \frac{\pi}{2k+2} \right) }{ \sin \left( \frac{3\pi}{2k+2} \right) } \; .$
This problem has been solved by Mark Hennings.
- 1 year, 5 months ago
If we define $$\zeta = e^{\frac{\pi i}{k+1}}$$, then $f_k(z) \; = \; \sum_{j=0}^k z^j \; = \; \frac{z^{k+1}-1}{z-1} \; = \; \prod_{j=1}^k \big(z - \zeta^{2j}\big)$ and hence, using partial fractions, we must be able to write $\frac{1}{f_k(z)} \; = \; \sum_{j=1}^k \frac{A_j}{z - \zeta^{2j}}$ where $A_j \; = \; \lim_{z \to \zeta^{2j}} \frac{z - \zeta^{2j}}{f_k(z)} \; = \; \big[f_k'\big(\zeta^{2j}\big)\big]^{-1} \; = \; \frac{\zeta^{2j}(\zeta^{2j}-1)}{k+1}$ Thus $\frac{1}{f_k(z^2)} \; = \; \sum_{j=1}^k \frac{A_j}{z^2 - \zeta^{2j}} \; =\; \sum_{j=1}^k \frac{A_j}{2\zeta^j}\left(\frac{1}{z - \zeta^j} - \frac{1}{z + \zeta^j}\right)$ and hence the function $\frac{1}{f_k(z^2)}$ has simple poles at $$\pm \zeta^j$$ for $$1 \le j \le k$$. Thus \begin{align} \int_{-\infty}^\infty \frac{1}{f_k(x^2)}\,dx & = 2\pi i\sum_{j=1}^k \mathrm{Res}_{z = \zeta^j} \frac{1}{f_k(z^2)} \; = \; 2\pi i \sum_{j=1}^k \frac{A_j}{2\zeta^j} \\ & = \frac{\pi i}{k+1}\sum_{j=1}^k \zeta^j(\zeta^{2j} - 1) \; = \; \frac{\pi i}{k+1}\left[ \frac{\zeta^3(\zeta^{3k} - 1)}{\zeta^3 - 1} - \frac{\zeta(\zeta^k - 1)}{\zeta-1}\right] \\ & = \frac{\pi i}{k+1} \left[\frac{\zeta^3(-\zeta^{-3} - 1)}{\zeta^3-1} - \frac{\zeta(\zeta^{-1}-1)}{\zeta-1}\right] \; = \; \frac{\pi i}{k+1}\left[\frac{\zeta+1}{\zeta-1} - \frac{\zeta^3+1}{\zeta^3-1}\right] \\ & = \frac{2 \pi i}{k+1} \times \frac{\zeta(\zeta^2-1)}{(\zeta-1)(\zeta^3-1)} \; = \; \frac{2\pi i}{k+1} \times \frac{\zeta(\zeta+1)}{\zeta^3-1} \\ & = \frac{2\pi i}{k+1} \times \frac{e^{\frac{3\pi i}{2(k+1)}} 2\cos\big(\frac{\pi}{2(k+1)}\big)}{e^{\frac{3\pi i}{2(k+1)}} 2i \sin\big(\frac{3\pi}{2(k+1)}\big)} \\ & = \frac{2\pi}{k+1}\,\frac{\cos\big(\frac{\pi}{2(k+1)}\big)}{\sin\big(\frac{3\pi}{2(k+1)}\big)} \end{align} as required.
- 1 year, 5 months ago
Problem 33:
Evaluate $\large \int_0^{\frac 14\pi} \frac{\sin^px}{\cos^{p+2}x}\,dx \hspace{2cm} p > 0 \;.$
This problem has been solved by Sumanth R Hegde.
- 1 year, 5 months ago
Put $$tanx = t$$
The integral reduces to $$\int\limits_{0}^{1} t^p dt$$
This gives $$\frac{1}{p+1}$$
Anyone else may post the next question
- 1 year, 5 months ago
Problem 47:
Evaluate,
$\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx$
- 1 year, 5 months ago
The sequence of substitutions $$u = x^2$$, $$v = 1-u$$, $$w = \sqrt{v}$$, $$y = \sinh p$$ and $$q = e^p$$ give \begin{align} I \; = \; \int_0^1 \frac{\tan^{-1}x}{\sqrt{1-x^2}}\,dx & = \int_0^1 \frac{dx}{\sqrt{1-x^2}}\int_0^1 \frac{x\,dy}{1 + x^2y^2} \\ & = \int_0^1\,dy \int_0^1 \frac{x\,dx}{\sqrt{1-x^2}(1 + x^2y^2)} \; = \; \tfrac12\int_0^1\,dy \int_0^1 \frac{du}{\sqrt{1-u}(1+y^2u)} \\ & = \tfrac12\int_0^1\,dy \int_0^1 \frac{dv}{\sqrt{v}(1 + y^2 - y^2v)} \; = \; \int_0^1\,dy \int_0^1 \frac{dw}{1 + y^2 - y^2w^2} \\ & = \tfrac12\int_0^1 \ln\left(\frac{\sqrt{1+y^2} + y}{\sqrt{1+y^2} - y}\right) \frac{dy}{y\sqrt{1+y^2}} \\ & = \int_0^{\sinh^{-1}1} \frac{p}{\sinh p}\,dp \; = \; 2\int_0^{\sinh^{-1}1}\frac{pe^p\,dp}{e^{2p}-1} \\ & = 2\int_1^{1+\sqrt{2}}\frac{\ln q}{q^2-1}\,dq \; = \; -\Big[\ln q \ln(q+1) + \mathrm{Li}_2(-q) + \mathrm{Li}_2(1-q)\Big]_1^{1+\sqrt{2}} \\ & = -\ln(1+\sqrt{2})\ln(2+\sqrt{2}) - \mathrm{Li}_2(-1-\sqrt{2}) - \mathrm{Li}_2(-\sqrt{2}) - \tfrac1{12}\pi^2 \end{align} We now use some dilogarithm identities \begin{align} \mathrm{Li}_2(\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = \tfrac13\pi^2 - \tfrac12\ln^2\sqrt{2} - i\pi\ln\sqrt{2} \; = \; \tfrac13\pi^2 - \tfrac18\ln^22 - \tfrac12\pi i \ln 2 \\ \mathrm{Li}_2(-\sqrt{2}) & = \tfrac12\mathrm{Li}_2(2) - \mathrm{Li}_2(\sqrt{2}) \\ &= \tfrac12\left(\tfrac14\pi^2 - \pi i \ln2\right) - \tfrac13\pi^2 + \tfrac18\ln^22 + \tfrac12\pi i \ln 2 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ & = -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ \mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = -\tfrac12\ln^2(2+\sqrt{2}) \end{align} to deduce that $\mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2(-\sqrt{2}) \; = \; -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 - \tfrac12\ln^2(2+ \sqrt{2})$ and hence that \begin{align} I & = -\ln(1 + \sqrt{2})\ln(2 + \sqrt{2}) + \tfrac{5}{24}\pi^2 - \tfrac18\ln^22 + \tfrac12\ln^2(2+\sqrt{2}) - \tfrac{1}{12}\pi^2 \\ & = \tfrac18\pi^2 - \tfrac12\ln^2(1+ \sqrt{2}) \end{align}
- 1 year, 5 months ago
$\displaystyle J=\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx$
Perform the change of variable $$y=\sqrt{\dfrac{1-x}{1+x}}$$,
\begin{align}\displaystyle J&=2\int_0^1 \dfrac{\arctan\left(\tfrac{1-x^2}{1+x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\arctan(1)}{1+x^2}dx-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ &=\dfrac{\pi^2}{8}-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ \end{align}
\begin{align} \displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{align}
Since,
$$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dx$$
then,
\begin{align} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{align}
Therefore,
$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$
Since,
\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
and,
\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
Therefore,
$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
Therefore,
$\boxed{\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx=\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
Therefore,
$\boxed{\displaystyle J=\dfrac{\pi^2}{8}-\dfrac{1}{8}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
- 1 year, 5 months ago
You do like pulling substitutions out of nowhere! Where did $$y = \sqrt{\tfrac{1-x}{1+x}}$$ come from?
This is a nice alternative derivation. Of course, since $$3 - 2\sqrt{2} = (1 + \sqrt{2})^{-2}$$, our results are the same!
- 1 year, 5 months ago
That sub is the tangent half angle substitution in algebraic form.
- 1 year, 5 months ago
I have found out this solution after you write down yours. I was searching but i was stuck. When i have seen your solution i was pretty sure i had missed something about the use of double integration and voilà !.
The inspiration for this problem comes from: http://math.stackexchange.com/questions/2092967/help-to-prove-that-int-0-pi-over-4-arctan-cot2x-mathrm-dx-2-pi2 (the solution i have planned to post initially is the one i have mentioned in a comment)
Your solution is nice as well.
- 1 year, 5 months ago
To respond your question. When i have an integral $$\int_0^1$$ i always try the change of variable $$y=\dfrac{1-x}{1+x}$$. Sometimes it's magic. If you do that, in the present integral, it's obvious you have to continue with the change of variable $$y=\sqrt{x}$$. My tool to do such computations is Maxima.
- 1 year, 5 months ago
* PROBLEM 46: *
Show that $\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \frac{\pi(\sqrt{3}-1)}{2\sqrt{2}\sqrt[4]{3}}$
- 1 year, 5 months ago
Perform the change of variable $$y=\dfrac{1-x}{1+x}$$,
\begin{align} \int_0^1 \dfrac{\sqrt{1-x^2}}{x^4+x^2+1}dx&=\int_{0}^{1}\frac{\sqrt{x}\cdot \left( 4+4\cdot x\right) }{3\cdot {{x}^{4}}+10\cdot {{x}^{2}}+3}dx\\ &=\left[\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right)+\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{3}\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{3}\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right)\times -\dfrac{1}{24}\right]_0^1\\ &=\frac{\left( \sqrt{2}-\sqrt{2}\cdot \sqrt{3}\right) \cdot \mathrm{atan}\left( \frac{{{3}^{\frac{1}{4}}}-\sqrt{2}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}+{{3}^{\frac{1}{4}}}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}-1\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}+1\right) }{4\cdot {{3}^{\frac{1}{4}}}}\\ &=\dfrac{\left(\sqrt{6}-\sqrt{2}\right)\left(\arctan\left(\sqrt{3+2\sqrt{3}}\right)+\pi-\arctan\left(\sqrt{3+2\sqrt{3}}\right)\right)}{4\times 3^{\tfrac{1}{4}}}\\ &=\boxed{\dfrac{\pi \left(\sqrt{3}-1\right)}{2\sqrt{2}\sqrt[4]{3}}} \end{align}
- 1 year, 5 months ago
My approach used contour integration...
The function $$\sqrt{z - 1}$$ can be defined analytically on the cut plane $$\mathbb{C} \backslash \{(-\infty,1]$$, while the function $$\sqrt{z+1}$$ can be defined analytically on the cut plane $$\mathbb{C} \backslash (-\infty,-1]$$. Putting these two functions together, it is possible to define the function $$\sqrt{z^2-1}$$ analytically on the domain $$\mathbb{C} \backslash [-1,1]$$.
Consider the "dogbone" contour $$D_\varepsilon = -\gamma_1 + \gamma_2 + \gamma_3 + \gamma_4$$, where
• $$\gamma_1$$ is the straight line segment $$z = x$$ for $$-1+\varepsilon < x < 1-\varepsilon$$, just above the cut,
• $$\gamma_2$$ is the (almost) circular arc $$z = -1 + \varepsilon e^{i\theta}$$ for $$0 < \theta < 2\pi$$,
• $$\gamma_3$$ is the straight line segment $$z = x$$ for $$-1+\varepsilon < x < 1 - \varepsilon$$, just below the cut,
• $$\gamma_4$$ is the (almost) circular arc $$z = 1 + \varepsilon e^{i\theta}$$ for $$-\pi < \theta < \pi$$.
Then we deduce that $\lim_{\varepsilon \to 0+} \int_{D_\varepsilon} \frac{\sqrt{z^2 - 1}}{z^4+z^2+1}\,dz \; = \; -2i\int_{-1}^1 \frac{\sqrt{1-x^2}}{x^4+x^2+1}\,dx \; = \; -4i\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx$ If we let $$\Gamma_R$$ be the circular contour $$z = Re^{i\theta}$$ for $$0 \le \theta < 2\pi$$, then we have $\left(\int_{\Gamma_R} - \int_{D_\varepsilon}\right) \frac{\sqrt{z^2-1}}{z^4 + z^2 + 1}\,dz \; = \; 2\pi i \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1}$ for sufficiently small $$\varepsilon>0$$ and any $$R > 1$$. Letting $$R \to \infty$$ and $$\varepsilon \to 0+$$, we deduce that \begin{align} \int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx & = \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1} \; = \; \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{(z^2-1)\sqrt{z^2-1}}{z^6-1} \\ & = \tfrac{1}{12}\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} u(u^2-1)\sqrt{u^2-1} \end{align} If $$u = e^{i\theta}$$ for $$|\theta| < \pi$$, then \begin{align} u + 1 & = 2\cos\tfrac12\theta e^{\frac12i\theta} \\ u - 1 & = 2i\sin\tfrac12\theta e^{\frac12i\theta} \\ u(u^2-1) & = 2i\sin\theta e^{2i\theta} \\ \sqrt{u+1} & = \sqrt{2\cos\tfrac12\theta} e^{\frac14i\theta} \\ \sqrt{u-1} & = \sqrt{2|\sin\tfrac12\theta|}e^{\frac14i(\theta + \pi\mathrm{sgn}(\theta))} \\ \sqrt{u^2-1} & = \sqrt{2|\sin\theta|} e^{\frac14i(2\theta + \pi\mathrm{sgn}(\theta))} \\ u(u^2-1)\sqrt{u^2-1} & = 2i\sin\theta\sqrt{2|\sin\theta|} e^{\frac14i(10\theta + \pi\mathrm{sgn}(\theta))} \end{align} and hence $\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \tfrac{1}{12}\pi \times 4 \times 3^{\frac34} \sin\tfrac{1}{12}\pi \; = \; \frac{\pi}{3^{\frac14}}\sin\tfrac{1}{12}\pi \; = \; \frac{\pi(\sqrt{3}-1)}{2^{\frac32} 3^{\frac14}}$
- 1 year, 5 months ago
Problem 45.
Evaluate $\int_0^\infty \frac{x^5}{\sinh(\pi x)(1+x^6)}dx$
- 1 year, 5 months ago
Using the Fourier transform we have $I \; = \; \int_0^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; =\; \tfrac12\int_{-\infty}^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; = \; \tfrac12\langle f\,,\,g\rangle \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g \rangle$ where $f(x) \; = \; \frac{x}{\sinh \pi x} \hspace{2cm} g(x) \; = \; \frac{x^4}{1 + x^6}$ A standard result about Fourier transforms tells us that $(\mathcal{F}f)(u) \; = \; \frac{1}{2\sqrt{2\pi}}\,\mathrm{sech}^2\tfrac12u$ and the Fourier transform of $$g$$ can be readily calculated using contour integration: for $$u > 0$$ we simply need to determine the residues of $$\frac{z^4e^{izu}}{1+z^6}$$ at the three roots of $$z^6 + 1 = 0$$ with positive imaginary part. Omitting the details, $(\mathcal{F}g)(u) \; = \; \tfrac13\sqrt{\tfrac{\pi}{2}}\Big[e^{-|u|} + e^{-\frac12|u|}\cos\big(\tfrac{\sqrt{3}}{2}|u|\big) - \sqrt{3}e^{-\frac12|u|}\sin\big(\tfrac{\sqrt{3}}{2}|u|\big)\Big]$ Thus we deduce that $I \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g\rangle \; = \; \tfrac{1}{12}\int_0^\infty \mathrm{sech}^2\tfrac12u\,\Big[e^{-u} + e^{-\frac12u}\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}e^{-\frac12u}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,du$ Now $\int_1^\infty \frac{v^a}{(1+v)^2}\,du \; = \; \tfrac12\Big(1 - aH_{-\frac12-\frac{a}{2}} + a H_{-\frac12a}\Big) \hspace{2cm} \mathfrak{Re}\,a < 1$ and so $\int_0^\infty e^{\omega u}\,\mathrm{sech}^2\tfrac12u\,du \; = \; 4\int_0^\infty \frac{e^{\omega u}}{(e^{\frac12u} + e^{-\frac12u})^2}\,du \; = \; 4\int_1^\infty \frac{v^\omega}{(1 + v)^2}\,dv \; = \; 2\Big(1 - \omega H_{\frac12\omega^2} + \omega H_{-\frac12\omega}\Big)$ where $$\omega$$ is the primitive cube root of unity. Taking real and imaginary parts appropriately, and doing a lot of simplification with polygammas, $\int_0^\infty e^{-\frac12u}\Big[\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,\mathrm{sech}^2\tfrac12u\,du \; = \; -2 + 4\pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big)$ On the other hand, $\int_0^\infty e^{-u}\,\mathrm{sech}^2\tfrac12u\,du \; = \;4\int_1^\infty \frac{dv}{v(v+1)^2} \; = \; -2 + 4\ln 2$ Putting this all together, we obtain that $I \; = \; \tfrac13\Big[-1 + \ln2 + \pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big)\Big]$ Someone else can post the next one, please.
- 1 year, 5 months ago
Here is a partial progress towards an alternate solution $f(a) = \int_0^\infty \frac{\sinh (ax)}{\sinh(\pi x)(1+x^6)} \ \mathrm{d}x$
then,
$f^{(6)} (a) + f(a) = \dfrac{1}{2} \tan\left(\dfrac{a}{2}\right)$
where $$f^{(n)} (a)$$ denotes the $$n^{\text{th}}$$ derivative of $$f$$ w.r.t. $$a$$, with $$f(0) = 0$$ and the required integral being $$f^{(5)} (0)$$.
I still have to solve that D.E.
- 1 year, 5 months ago
Comment deleted Jan 12, 2017
Using $\frac{\pi \text{csch}(\pi z)}{2}=\frac{1}{2z}+\sum_{n=1}^\infty \frac{(-1)^n z}{z^2+n^2}$
- 1 year, 5 months ago
Nice!
- 1 year, 5 months ago
Problem 41 :
Prove That
$\int_{-\infty}^{\infty} \dbinom{n}{x} \ \mathrm{d}x = 2^n \quad \Re(n) > -1$
Notation : $$\dbinom{n}{x}$$ denotes the Generalized Binomial Coefficient.
- 1 year, 5 months ago
Presumably, we mean ${n \choose x} \; = \; \frac{n!}{\Gamma(x+1)\Gamma(n+1-x)} \hspace{2cm} x \in \mathbb{R}$ Now $\Gamma(x+1)\Gamma(n+1-x) \; = \; x\Gamma(x) \prod_{j=1}^n (j-x)\Gamma(1-x) \; = \; (-1)^n \frac{\pi}{\sin\pi x}\times \prod_{j=0}^n (x-j)$ and so we can use partial fractions to write ${n \choose x} \; = \; \frac{(-1)^n n!}{\pi} \sin \pi x \left(\prod_{j=0}^n (x-j)\right)^{-1} \; = \; \frac{(-1)^n n! \sin \pi x}{\pi} \sum_{j=0}^n \frac{A_j}{x-j}$ where $A_j \; = \; \frac{(-1)^{n-j}}{j! (n-j)!} \hspace{1cm} 0 \le j \le n$ Since $\int_{\mathbb{R}} \frac{\sin \pi x}{x - j}\,dx \; = \; (-1)^j \int_{\mathbb{R}} \frac{\sin \pi x}{x}\,dx \; = \; (-1)^j \pi$ we deduce that \begin{align} \int_{\mathbb{R}} {n \choose x}\,dx & = \frac{(-1)^n n!}{\pi} \sum_{j=0}^n A_j \int_{\mathbb{R}}\frac{\sin \pi x}{x-j}\,dx \; = \; (-1)^n n! \sum_{j=0}^n (-1)^j A_j \\ & = \sum_{j=0}^n \frac{n!}{j!(n-j)!} \; = \; \sum_{j=0}^n {n \choose j} \; = \; 2^n \end{align} as required.
If we insist on being picky, we could integrate from $$-X$$ to $$X$$, and let $$X \to \infty$$, to avoid dealing with the improper integral $\int_{\mathbb{R}} \frac{\sin x}{x}\,dx$ directly. We only have to combine a finite number of integrals at the last stage, so there is no problem taking the limit.
- 1 year, 5 months ago
I think your solution can be extended to deal with non integral n if we instead expand into partial fractions using infinite product of Gamma Function. My approach was using the integral representation $\large \int_{0}^{\pi} (\sin (x))^{m-1} e^{inx} \ \mathrm{d}x = \dfrac{\pi}{2^{m-1}} \ \dfrac{e^{i n/2}}{m \operatorname{B} \left( \dfrac{1}{2} (m+n+1) , \dfrac{1}{2} (m-n+1) \right)}$
- 1 year, 5 months ago
The integral is also true for $$\Re(n) > -1$$.
- 1 year, 5 months ago
Indeed. The formulae $\int_{-\infty}^\infty \frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)} \; = \; \frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}$ for $$\mathfrak{Re}(\alpha+\beta) > 1$$ is $$6.414.2$$ in Gradshteyn & Rhyzik.
Isn't it a bit late to be altering the question, though?
- 1 year, 5 months ago
Apologies, I forgot to mention that initially. But your answer is the accepted one nevertheless.
- 1 year, 5 months ago
If $$\binom{n+1}{x+1}=\binom{n}{x}+\binom{n}{x+1}$$ then i think it's straightforward to prove with a recurrence proof.
- 1 year, 5 months ago
Certainly, if we presume that the integral exists, then the recurrence relation can help us evaluate it. The RR does not prove convergence of the integral, however.
- 1 year, 5 months ago
Ohh!! I missed this season. I even didn't know that this happened. :(
- 1 year, 2 months ago
Maybe this time you can host it :)
- 1 year, 2 months ago
When is starting the season 4? :)
- 1 year, 1 month ago
I think in December 2017.
- 1 year, 1 month ago
Problem 49:
Evaluate,
$\int_{0}^{\frac{\pi }{6}}\frac{x \mathrm{sin}\left( 2 x\right) }{4 {{\mathrm{sin}^{4}x }}-2 {{\mathrm{sin^{2}}x }}+1}dx$
- 1 year, 5 months ago
\begin{align}J&=\int_0^{\tfrac{\pi}{6}} \dfrac{x\sin(2x)}{1-2\sin^2 x+4\sin^4 x}dx\\ &= \int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{(1-\sin^2 x)^2+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{\cos^4+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\tan x}{\cos^2 x(1+3\tan^4 x)}dx\\ \end{align}
Perform the change of variable $$y=\sqrt{3}\tan x$$,
$J=\int_0^1 \dfrac{2x\arctan\left(\tfrac{x}{\sqrt{3}}\right)}{3+x^4}dx$
Since,
$\arctan u=\int_0^1 \dfrac{u}{1+t^2u^2}dt$
\begin{align}J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{x^2}{(3+x^4)(3+t^2x^2)}dtdx\\ &=2\sqrt{3}\int_0^1 \int_0^1 \left(\frac{{{t}^{2}}+{{x}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{x}^{4}}+3\right) }-\frac{{{t}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{t}^{2}}{{x}^{2}}+3\right) }\right)dtdx\\ &=4\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx-J\\ \end{align}
therefore,
\begin{align} J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx\\ &=2\sqrt{3}\left(\int_0^1 \dfrac{x^2}{3+x^4}dx \right)\left(\int_0^1 \dfrac{1}{3+x^4}dx\right)\\ \end{align}
Since,
\begin{align} \int_0^1 \dfrac{1}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}{{3}^{\frac{1}{4}}}}-\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot{{3}^{\frac{1}{4}}} x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}{{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(\dfrac{1+\sqrt{3}-\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{1+\sqrt{3}+\sqrt{2}\cdot 3^{\tfrac{1}{4}}}\right) \end{align}
$$1+\sqrt{3}+\sqrt{3}\cdot 3^{\tfrac{1}{4}}$$ is a root of the polynomial $$X^4-4X^3-16X+16$$,
Therefore,
$\int_0^1 \dfrac{1}{3+x^4}dx=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{\sqrt{3}-1}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right)$
and,
\begin{align} \int_0^1 \dfrac{x^2}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{1}{4}}}}+\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}\cdot {{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{1}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)+\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{1}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) \end{align}
Therefore,
$\boxed{J=\dfrac{1}{4\sqrt{3}}\arctan^2\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{16\sqrt{3}}\log^2\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) }$
PS:
$F(x,t)=\dfrac{x}{(a+x^2)(a+tx)}=\dfrac{t}{a+t^2}\cdot \dfrac{1}{a+x^2} +\dfrac{1}{a+t^2}\cdot \dfrac{x}{a+x^2}-\dfrac{t}{(a+t^2)(a+tx)}$
is a nice function.
- 1 year, 5 months ago
- 1 year, 5 months ago
I am interested. My solution used/generalised a lot of the tricks that FDP uses in his solutions, so it will be interesting if he has a different approach.
- 1 year, 5 months ago
Aliter,
Let.
$\text{I} = \int_{0}^{\frac{\pi}{6}} \dfrac{x \sin(2x)}{4\sin^4 x - 2\sin^2 x + 1} \ \mathrm{d}x$
$= \dfrac{1}{4(\alpha - \beta)} \int_{0}^{\frac{\pi}{6}} x \sin (2x) \left(\dfrac{1}{\cos(2x) - 1 + \beta} - \dfrac{1}{\cos(2x) - 1 + \alpha}\right)$
where $$\alpha$$ and $$\beta$$ are the roots of the equation $$t^2 - \dfrac{1}{2} t + \dfrac{1}{4} = 0$$
Now, from Problem 25, we have,
$\dfrac{\sin x}{p^2 +2p \cos x +1} = \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)$
Using this and interchanging, sum and integral ( and after a lot of straight forward simplification/calculations involving dilogarithm and logarithm) we get,
$\text{I} = \dfrac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right)$
- 1 year, 5 months ago
Just pipped you! You might want to clarify what $$\alpha$$ and $$\beta$$ are, though.
Why don't you post the last integral!
- 1 year, 5 months ago
Thanks for letting me do the honors. I'll post it tomorrow, I'll need some time to create a nice one :)
- 1 year, 5 months ago
For any $$a > 0$$ define \begin{align} A(a) & = \int_0^1 \frac{dx}{a^4 + x^4} \; = \; \frac{1}{a^3}\int_0^{a^{-1}} \frac{dz}{1 + z^4} \\ & = \frac{1}{2a^3\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] + \frac{1}{4a^3\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ B(a) & = \int_0^1 \frac{x^2\,dx}{a^4 + x^4} \; = \; \frac{1}{a}\int_0^{a^{-1}} \frac{z^2\,dz}{1 + z^4} \\ & = \frac{1}{2a\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] - \frac{1}{4a\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ a^4A(a)B(a) & = \tfrac{1}{8}\left[ \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \right]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{align} Thus we deduce that (putting $$z = ax$$ and playing symmetry games) \begin{align} I(a) & = \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \; = \; \int_0^{a^{-1}} \frac{x\,dx}{1 + x^4} \int_0^1 \frac{\frac{x}{a}\,dy}{1 + a^{-2}x^2y^2} \\ & = a^4 \int_0^1 \int_0^1 \frac{z^2}{(a^4 + z^4)(a^4 + y^2 z^2)}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \left\{ \frac{z^2}{(a^4 + z^4)(a^4 + y^2z^2)} + \frac{y^2}{(a^4 + y^4)(a^4 + y^2z^2)}\right\}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \frac{y^2 + z^2}{(a^4 + y^4)(a^4 + z^4)}\,dy\,dz \; = \; a^4 A(a)B(a) \\ & = \tfrac{1}{8}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{align} Moreover \begin{align} J(a) & = \int_0^{a^{-1}} \frac{\tan^{-1}x^2}{x^2 + a^2}\,dx \; = \; \Big[\frac{1}{a} \tan^{-1}x^2\,\tan^{-1}\big(\tfrac{x}{a}\big)\Big]_0^{a^{-1}} - \frac{2}{a} \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \\ & = \frac{1}{a}\big(\tan^{-1}a^{-2}\big)^2 - \frac{2}{a}I(a) \end{align} Thus, putting $$u = \cos2x$$ and $$v= \frac{2u-1}{\sqrt{3}}$$, we see that \begin{align} K & = \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{4\sin^4x - 2\sin^2x + 1}\,dx \; = \; \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{\cos^22x - \cos2x + 1}\,dx \\ & = \tfrac14\int_{\frac12}^1 \frac{\cos^{-1}u\,du}{u^2 - u + 1} \; = \; \tfrac14\Big[\tfrac{2}{\sqrt{3}}\tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \cos^{-1}u\Big]_{\frac12}^1 + \tfrac14\int_{\frac12}^1 \tfrac{2}{\sqrt{3}} \tan^{-1}\big(\tfrac{2u-1}{\sqrt{3}}\big) \frac{du}{\sqrt{1-u^2}} \\ & = \tfrac{1}{2\sqrt{3}} \int_{\frac12}^1 \tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \frac{du}{\sqrt{1-u^2}} \; = \; \frac{1}{2 \times 3^{\frac14}} \int_0^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}v\,dv}{\sqrt{(1 - \sqrt{3}v)(\sqrt{3}+v)}} \end{align} The key substitution $w \; = \; \sqrt{\frac{1 - \sqrt{3}v}{\sqrt{3}+v}}$ yields \begin{align} K & = \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \tan^{-1}\left(\frac{1 - \sqrt{3}w^2}{\sqrt{3} + w^2}\right) \frac{dw}{\sqrt{3} + w^2} \; = \; \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \Big[\tfrac16\pi - \tan^{-1}w^2\Big] \frac{dw}{\sqrt{3} + w^2} \\ & = \frac{\pi}{6 \times 3^{\frac14}} \Big[3^{-\frac14} \tan^{-1}\big(\tfrac{w}{3^{\frac14}}\big)\Big]_0^{3^{-\frac14}} - 3^{-\frac14}J(3^{\frac14}) \\ & = \frac{\pi^2}{36\sqrt{3}} - 3^{-\frac14}\Big[3^{-\frac14}\left(\tan^{-1}3^{-\frac12}\right)^2 - 2 \times 3^{-\frac14} I\big(3^{\frac14}\big)\Big] \; = \; \tfrac{2}{\sqrt{3}} I\big(3^{\frac14}\big) \\ & = \frac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right) \end{align} noting that $\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \; =\; \tan^{-1}\left(\frac{\sqrt{2}}{a - a^{-1}}\right) \hspace{1cm} a > 1$
- 1 year, 5 months ago
Actually my solution is the same than yours. The only difference is the way you manage to transform the trigonometric integral.
- 1 year, 5 months ago
Yes, but your method of transforming the original integral into what is, apart from a variable scaling, $$\tfrac{2}{\sqrt{3}} I(3^{1/4})$$ (in my notation), is elegant.
- 1 year, 5 months ago
if $$F(a,t,x)=\frac{x}{\left( t\cdot x+a\right) \cdot \left( {{x}^{2}}+a\right) }$$, compute $$\int_0^1 F(1,t,x)dt$$ isn't nice?
Problem 47 relies on the evaluation of $$\int_0^1 \int_0^1 F(1,t^2,x^2) dtdx$$ and problem 49 relies on $$\int_0^1 \int_0^1 F(3,t^2,x^2) dtdx$$
Who will post problem 50, the last one?
- 1 year, 5 months ago
Let me clarify. Both of our methods evaluate $$\tfrac{2}{\sqrt{3}}I(3^{\frac14})$$. What I was saying was that your technique for converting the original integral to $$\tfrac{2}{\sqrt{3}}I(3^{\frac14})$$ was more elegant than mine - in Mathematics, the word "elegant" is complimentary!
- 1 year, 5 months ago
I'll post by evening, in 5 hours.
- 1 year, 5 months ago
My solution is close to yours. I will post it later (01 AM here) . This integral is based on the solution i have provided for problem 47. I have played around with it and i have build this one.
- 1 year, 5 months ago
* PROBLEM 48: *
Prove that $\int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu (x+s)^\mu}\,dx \; = \; \sqrt{\pi}(\sqrt{r}+\sqrt{s})^{1-2\mu}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}$ for all $$r,s > 0$$ and $$\mu > \tfrac12$$.
- 1 year, 5 months ago
Someone else can post the next one...
For any $$a > 0$$ and $$0 < b < c$$ and $$|x| < 1$$ we have \begin{align} I_{a,b,c}(x) & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - xt)^{-a}\,dt \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1} \left(\sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n t^n\right)\,dt \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \int_0^1 t^{n+b-1}(1-t)^{c-b-1}\,dt \; = \; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n B(n+b,c-b) \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{\Gamma(n+b) \Gamma(c-b)}{\Gamma(n+c)} \; =\; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{(b)^{(n)} \Gamma(b)\Gamma(c-b)}{(c)^{(n)} \Gamma(c)} \\ & = B(b,c-b)\sum_{n=0}^\infty \frac{(a)^{(n)} (b)^{(n)}}{(c)^{(n)}} \frac{x^n}{n!} \; = \; B(b,c-b) \big({}_2F_1\big)(a,b;c;x) \end{align} With the substitution $$y = \tfrac{t}{1-t}$$ we see that \begin{align} \int_0^\infty y^{b-1} (1+y)^{a-c} (1 + \alpha y)^{-a}\,dy & = \int_0^1 \left(\frac{t}{1-t}\right)^{b-1}\left(\frac{1}{1-t}\right)^{a-c} \left(\frac{1 - (1-\alpha)t}{1-t}\right)^{-a}\, \frac{dt}{(1-t)^2} \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - (1-\alpha)t)^{-a}\,dt \; = \; I_{a,b,c}(1-\alpha) \\ & = B(b,c-b) \big({}_2F_1\big)(a,b;c;1-\alpha) \end{align} for any $$a>0$$, $$0 < b < c$$ and $$0 < \alpha \le 1$$.
Since the integral is symmetric in $$r$$ and $$s$$, we may assume that $$0 < r \le s$$. Then, for any $$0 < r \le s$$ and $$\mu > \tfrac12$$ we have \begin{align} J_{\mu,r,s} & = \int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu(x+s)^\mu}\,dx \; =\; \frac{\sqrt{r}}{s^\mu} \int_0^\infty t^{\mu-\frac12}(1+t)^{-\mu}\big(1 + \tfrac{r}{s}t\big)^{-\mu}\,dt \\ & = \frac{\sqrt{r}}{s^\mu} B\big(\mu+\tfrac12,\mu-\tfrac12\big) \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;2\mu;1-\tfrac{r}{s}\big) \end{align} Since $\big({}_2F_1\big)(a,b;c;z) \; = \; \begin{array}{l} \displaystyle(1-z)^{-a} \frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\big({}_2F_1\big)\big(a,c-b;a-b+1;(1-z)^{-1}\big) \\ \displaystyle{}+ (1-z)^{-b} \frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}\big({}_2F_1\big)\big(b,c-a;b-a+1;(1-z)^{-1}\big)\end{array}$ we deduce that \begin{align} J_{\mu,r,s} & = \frac{\sqrt{r}}{s^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \begin{array}{l} \displaystyle\big(\tfrac{r}{s}\big)^{-\mu} \frac{\Gamma(2\mu)\Gamma(\tfrac12)}{\Gamma(\mu+\frac12)\Gamma(\mu)}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) \\ \displaystyle{} + \big(\tfrac{r}{s}\big)^{-\mu-\frac12} \frac{\Gamma(2\mu)\Gamma(-\tfrac12)}{\Gamma(\mu)\Gamma(\mu-\frac12)}\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \end{array} \right] \\ & = \frac{2^{2\mu-1}}{r^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right]\\ & = \frac{\sqrt{\pi}}{r^\mu} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right] \end{align} Now standard hypergeometric identities give \begin{align} \big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) & = \tfrac12\left[ \left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} + \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \\ \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;\tfrac32;\tfrac{s}{r}\big) & = \frac{1}{2(2\mu-1)}\sqrt{\tfrac{r}{s}}\left[\left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} - \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \end{align} and so $J_{\mu,r,s} \; = \; \frac{\sqrt{\pi}}{r^{\mu-\frac12}} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\big(1 + \sqrt{\tfrac{s}{r}}\big)^{1-2\mu} \; = \; \sqrt{\pi}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}(\sqrt{r}+\sqrt{s})^{1-2\mu}$ as required.
- 1 year, 5 months ago
Nice. I'm working on an alternative solution. I hope to get something soon.
- 1 year, 5 months ago
Please don't post the answer now, post the solution at the end of the week end. Thanks.
- 1 year, 5 months ago
I was aiming to allow some extra time to compensate for the typo, but will post a solution tomorrow...
- 1 year, 5 months ago
For those having a shot at this one, I need to point out a small, but vital, correction that has been made to the limits of the integral!
- 1 year, 5 months ago
It makes things easier :)
- 1 year, 5 months ago
True :)
- 1 year, 5 months ago
* PROBLEM 44 *:
Evaluate $\int_0^1 \ln\left(\frac{1 + ax}{1 - ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{1cm} -1 < a < 1 \;.$
- 1 year, 5 months ago
For $$-1<ax<1$$, $\log\left(\dfrac{1+ax}{1-ax}\right)=2\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}$
\begin{align} \int_0^1 \dfrac{\log\left(\tfrac{1-ax}{1+ax}\right)}{x\sqrt{1-x^2}}dx&=2\int_0^1 \left(\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}}\right)dx\\ &=2\sum_{n=0}^{\infty}\left(\int_0^1 \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}} dx\right)\\ &=2\sum_{n=0}^{\infty}\left(\dfrac{a^{2n+1}}{2n+1} \int_0^1 \dfrac{x^{2n}}{\sqrt{1-x^2}}dx\right)\\ &=\sum_{n=0}^{\infty}\dfrac{1}{2n+1}\cdot\text{B}\left(\tfrac{1}{2},n+\tfrac{1}{2}\right)a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\Gamma\left(\tfrac{1}{2}\right)\Gamma\left(n+\tfrac{1}{2}\right)}{\Gamma(n+1)}a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\tfrac{(2n)!}{2^{2n}n!}\pi}{n!}a^{2n+1}\\ &=\pi \sum_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{2^{2n}(2n+1)}a^{2n+1}\\ &=\boxed{\pi\arcsin(a)} \end{align}
- 1 year, 5 months ago
For variety, if we define $F(a) \; = \; \int_0^1 \ln\left(\frac{1 + ax}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{2cm} -1 < a < 1$ then \begin{align} F'(a) & = \int_0^1 \left(\frac{1}{1+ax} + \frac{1}{1-ax}\right) \frac{dx}{\sqrt{1-x^2}} \; =\; \int_0^{\frac12\pi} \left(\frac{1}{1 + a\cos \theta} + \frac{1}{1-a\cos\theta}\right)\,d\theta \\ & = \int_0^\pi \frac{d\theta}{1 + a\cos\theta} \; = \; \tfrac12\int_0^{2\pi} \frac{d\theta}{1 + a\cos\theta} \\ & = \tfrac12\int_{|z|=1} \frac{2}{2 + a(z+z^{-1})}\,\frac{dz}{iz} \; = \; -i\int_{|z|=1}\frac{dz}{az^2 + 2z + a} \\ & = -ia \int_{|z|=1}\frac{dz}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \\ & = 2\pi a\,\mathrm{Res}_{z = a^{-1}(\sqrt{1-a^2}-1)} \frac{1}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \; = \; \frac{\pi}{\sqrt{1-a^2}} \end{align} and hence $F(a) \; = \; \int_0^a \frac{\pi}{\sqrt{1-b^2}}\,db \; = \; \pi\sin^{-1}a$
- 1 year, 5 months ago
Problem 50 :
Evaluate $\int_{0}^{\infty} \dfrac{\sin(2nx) \cos(2\arctan(ax))}{(1+a^2x^2) \sin x} \ \mathrm{d}x$
where $$a>0$$ , $$n \in \mathbb{Z^+}$$
- 1 year, 5 months ago
Suppose that $$a > 0$$ and $$n \in \mathbb{N}$$. Note that $I \; = \; \int_0^\infty \frac{\sin(2nx)\cos\big(2\tan^{-1}(ax)\big)}{(1 + a^2x^2)\sin x}\,dx \; = \; \int_0^\infty \frac{1-a^2x^2}{(1 + a^2x^2)^2}\frac{\sin 2nx}{\sin x}\,dx \; = \; \lim_{\delta \to 0+}I_\delta$ where $I_\delta \; = \; \int_0^\infty \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\,dx$ where we can use the Dominated Convergence Theorem to justify the limit, since $\left| \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \left| \frac{1}{1 + a^2(x+\delta)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \frac{N_n}{1 + a^2x^2}$ for all $$x \ge 0$$ and $$\delta > 0$$, where $$\big|\tfrac{\sin 2nx}{\sin x}\big| \le N_n$$ for all $$x \ge 0$$. It is trivial that $\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-xt}\,dt \; = \; a^2 \frac{a^2x^2 - 1}{(1 + a^2x^2)^2} \hspace{2cm} a,x > 0$ and so \begin{align} I_\delta & = -a^{-2}\int_0^\infty\left(\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-(x+\delta)t}\,dt\right) \frac{\sin 2nx}{\sin x}\,dx \\ & = -a^{-2}\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx\right)\,dt \end{align} Since $\int_0^\infty e^{-xt} \frac{\sin(2m+2)x}{\sin x}\,dx - \int_0^\infty e^{-xt} \frac{\sin 2mx}{\sin x}\,dx \; = \; 2\int_0^\infty e^{-xt} \cos(2m+1)x\,dx \; = \; \frac{2t}{t^2 + (2m+1)^2}$ a simple induction shows that $\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx \; = \; 2t\sum_{k=0}^{n-1} \frac{1}{t^2 + (2k+1)^2} \hspace{2cm} t > 0$ and hence \begin{align} I_\delta & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\sum_{k=0}^{n-1} \frac{t^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left( n - \sum_{k=0}^{n-1} \frac{(2k+1)^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2}\left[ \frac{n\delta}{\delta^2 + a^{-2}} - \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big) e^{-\delta t}}{t^2 + (2k+1)^2}\,dt \right] \end{align} and hence, letting $$\delta \to 0+$$, \begin{align} I & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big)}{t^2 + (2k+1)^2}\,dt \\ & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \times \frac{\pi}{2(2k+1)}e^{-\frac{2k+1}{a}} \\ & = \frac{\pi}{a^2}\sum_{k=0}^{n-1} (2k+1)e^{-\frac{2k+1}{a}} \end{align} using a standard Fourier transform to finish off. This sum can be calculated, of course, but the end result is not particularly pretty, do I shall leave things here.
- 1 year, 5 months ago
Nice! I used the expansion $\dfrac{\sin(2nx)}{2\sin x} = \sum_{k=1}^{n} \cos((2k-1)x)$
The result can also be generalized to $\int_{0}^{\infty} \dfrac{\sin(2nx) \cos(p\arctan(ax))}{(1+a^2x^2)^{\frac{p}{2}} \sin x} \ \mathrm{d}x \ ; \ n \in \mathbb{Z}, a,p>0$
- 1 year, 5 months ago
Let $$C_{n-1}(a)$$ be the previous integral. I suspect one can find a closed form for the ordinary generating function for the sequence $$\left(C_n(a)\right)$$ because $$U_{2n+1}(\cos x)=\dfrac{\sin((2n+2)x}{\sin x}$$, $$U_j$$ being the $$j$$-nth Chebyshev polynomial of the second kind.
- 1 year, 5 months ago
I looked hard on the web and could not find a closed expression for the Laplace transform of the $$U_n$$, which would have been really useful!
- 1 year, 5 months ago
× | 2018-06-24T01:53:21 | {
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http://mathhelpforum.com/calculus/93104-continuity-2-variables.html | # Thread: Continuity in 2 variables.
1. ## Continuity in 2 variables.
Find the limit or show that the limit does not exist,
$\lim_{(x,y) \to (0,0)} \frac{xy \cos y}{3x^2+y^2}$
2. First we set...
$x= \rho\cdot \cos \theta$
$y= \rho\cdot \sin \theta$
... and imnmediately we obtain...
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \cos y}{3 x^{2} + y^{2}} = \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta \cdot \cos (\rho\cdot \sin \theta)}{\rho^{2}\cdot (3 \cos^{2} \theta + \sin^{2} \theta)} = \frac{\sin \theta \cos \theta}{1+2 \cos^{2} \theta}$
Since the limit depends form $\theta$ the limit itself doesn't exist...
Kind regards
$\chi$ $\sigma$
3. Originally Posted by chisigma
First we set...
$x= \rho\cdot \cos \theta$
$y= \rho\cdot \sin \theta$
... and imnmediately we obtain...
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \cos y}{3 x^{2} + y^{2}} = \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \sin \theta\cdot \cos \theta \cdot \cos (\rho\cdot \sin \theta)}{\rho^{2}\cdot (3 \cos^{2} \theta + \sin^{2} \theta)} = \frac{\sin \theta \cos \theta}{1+2 \cos \theta}$
Since the limit depends form $\theta$ the limit itself doesn't exist...
Kind regards
$\chi$ $\sigma$
I find this very interesting. I haven't thought about taking limits of functions of two variables. Could you please justify the use of $p$ for me please?
4. The limit...
$\lim_{(x,y) \rightarrow (x_{0},y_{0})} f(x,y)$
... exists if and only if $f(x,y)$ tends to the same value $(x_{*},y_{*})$ no matter which is the 'traiectory' [straight line, zig zag, spiral, etc...] with which $(x,y)$ tends to $(x_{0},y_{0})$... this is the fundamental concept of limit of two [or more...] variables functions...
Kind regards
$\chi$ $\sigma$
5. Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..
For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...
6. Originally Posted by VonNemo19
I find this very interesting. I haven't thought about taking limits of functions of two variables. Could you please justify the use of $p$ for me please?
He introduced polar coordinates. Some might have seen
$
x = r \cos \theta,\; y = r \sin \theta
$
7. Originally Posted by Robb
Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..
For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...
Using polar coordinates we have...
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{\rho \rightarrow 0} \frac{ \rho^{2}\cdot \sin \theta\cdot \cos \theta}{\rho} = 0$
Very easy! ...
Kind regards
$\chi$ $\sigma$
8. OMG.. thanks allot. I tried using the polar coordinates, but forgot the [/tex]\rho[/tex] from the denominator. So the condition is, that if $\rho$ is left in the function after substituting in the polar coordinates and simplifying then the limit exists?
The text i have is very light on using polar coordinates, it just has the sentenace (after using polar coordinates for a limit) that 'the behaviour as $\rho = \sqrt{x^2+y^2} \rightarrow 0$ depends on $\theta$ hence there is no limit'
Sorry for all the questions, i am just trying to get my head around this in a logical manner :P
9. Originally Posted by Robb
Thanks allot, i just have another question,
Is the polar coordinate substition the best way to solve limits of 2 variables? or just one of the many? I have tried to use the formal definition of a limit, but I end up confusing myself..
For exmaple, I tried using the formal definition on that function, and dont go anywhere. Another one I was working on;
$\lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y}{\sqrt{ x^{2} + y^{2}}} = \lim_{(x,y) \rightarrow (0,0)} \frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}}$
So using the formal definition; $0 < \sqrt{ x^{2} + y^{2}}<\delta \mbox{\ then \ } |\frac{x\cdot y\cdot \sqrt{ x^{2} + y^{2}}}{ x^{2} + y^{2}} - 0 < \epsilon$
Then i stated that $x \cdot y < x^2 + y^2 \mbox{ so } \frac{x\cdot y}{x^{2}+y^{2}} \leq 1$
But i wasn't sure where to take it form here, or even if this is on the right track...
If you want to be a little more rigorous, (i.e. $\delta - \epsilon$) you could try using the inequalities
$
-\sqrt{x^2+y^2} \le x,y \le \sqrt{x^2+y^2}\;\;\Rightarrow\;\; -(x^2+y^2) \le x \,y \le x^2+y^2$
$\;\;\Rightarrow\;\;-\sqrt{x^2+y^2} \le \frac{x \,y}{\sqrt{x^2+y^2}} \le \sqrt{x^2+y^2}$
so
$\left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \sqrt{x^2+y^2}$
Thus, if $\sqrt{x^2+y^2} < \epsilon$ then $\left| \frac{x \,y}{\sqrt{x^2+y^2}}\right| < \epsilon$.
10. Originally Posted by Robb
Find the limit or show that the limit does not exist,
$\lim_{(x,y) \to (0,0)} \frac{xy \cos y}{3x^2+y^2}$
You can examine a few paths and see if you keep getting the same result.
I like using y=mx, then let x head to zero.
OR try...
(1) y=o and let x tend to zero, the limit along both parts of the x-axis as we head to the origin is 0, since you have 0 over $3x^2$.
And for a second path, just let x=y, which is just calculus one now...
(2) In this case the limit is
$\lim_{x\to 0}{x^2\cos x\over 4x^2}={1\over 4}$
SINCE two different paths produce two different limits, the limit does not exist as we approach the origin. | 2016-10-23T10:29:15 | {
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http://math.stackexchange.com/questions/750276/volume-of-spheres-inscribed-in-a-cone | Volume of spheres inscribed in a cone.
There are five perfectly spherical scoops of ice cream with various radii placed inside a waffle cone. Each scoop of ice cream is in contact with the adjacent scoop of ice cream. Also, each scoop of ice cream comes in contact all around the waffle cone wall. If the radius of the smallest ice cream scoop is 8 and the radius of the largest ice cream scoop is 18, find the volume of the middle ice cream scoop.
My intuition is that the radius should be the geometric mean of 18 and 8 (which is 12), but I don't know why. Is it really the geometric mean and why/why not?
Also, this isn't homework.
-
Extend the cone to its tip. Consider the following $2$ shapes:
Shape $1$: The cone, from the tip up to the place where it meets the second scoop from the bottom, together with the bottom scoop and the second scoop.
Shape $2$: The cone, from the tip up to the place where it meets the third scoop from the bottom, together with the second scoop from the bottom, and the third scoop from the bottom.
Shapes $1$ and $2$ are similar. Let $r$ be the scaling factor that gets us from the bottom scoop of Shape $1$ to the top scoop of Shape $1$. Then scaling again by the factor $r$ gets us to the top scoop of Shape $2$. And scaling twice more by the factor $r$ gets us to the top scoop in the diagram of the OP.
Thus $\frac{18}{8}=r^4$. It follows that $r^2=\frac{3}{2}$, and therefore the middle scoop has radius $(8)(3/2)$. Now that we have the radius, the volume is easy to write down.
Remark: We could do basically the same argument by drawing a bunch of similar triangles. But that would take longer, and anyway I wanted to give a pure scaling argument.
The argument above shows that the radius is indeed the geometric mean of $12$ and $18$. Precisely the same argument shows that if the bottom scoop has radius $a$ and the top scoop has radius $b$, then the middle scoop has radius $\sqrt{ab}$.
The same argument shows that if the bottom scoop as surface area $A$, and the top scoop has surface are $B$, then the middle scoop has surface area $\sqrt{AB}$.
The same argument shows that if the cone is not right-circular, again the middle scoop has radius $\sqrt{ab}$. The same applies to waffle cones that have the shape of an upside-down pyramid.
-
Thanks a lot. That really makes sense now. Just one thing though: in your 7th paragraph, so you mean it's the geometric mean of 8 and 18, instead of 12 and 18? – KevinOrr Apr 18 at 3:05
Since I'd never seen the similarity argument, I decided to work one out.
Following the "walls" of the cone to its apex, we can construct similar triangles using one wall and the symmetry axis of the cone. We will call the distance from the apex to the surface of the smallest sphere $\ Y \$ , and the radii of the three smallest spheres $\ r_1 \ , \ r_2 \ , \$ and $\ r_3 \$ . (It will be sufficient to work with just the first three spheres, since the argument is easily extended to any number of spheres in a "stack".)
Since the cone is the result of revolution of a straight line, the "slope" of the wall relative to the symmetry axis is constant (this is also the tangent value for half of the "opening angle" of the cone). This permits us to describe similar triangles, for which this tangent value is
$$\frac{r_1}{Y \ + \ r_1} \ = \ \frac{r_2}{Y \ + \ 2r_1 \ + r_2} \ = \ \frac{r_3}{Y \ + \ 2r_1 \ + 2r_2 \ + \ r_3} \ \ .$$
From pairing the first two ratios, we have
$$r_1 Y \ + \ 2 \ r^2_1 \ + \ r_1 r_2 \ = \ r_2 \ Y \ + \ r_1 r_2 \ \ \Rightarrow \ \ Y \ (r_2 - r_1) \ = \ 2 \ r_1^2 \ \ .$$
Pairing the second two ratios, and using our result for $\ Y \$ , we then obtain
$$r_2 \ Y \ + \ 2 \ r_1 r_2 \ + \ 2 \ r_2^2 \ + \ r_2 r_3 \ = \ r_3 \ Y \ + \ 2 \ r_1 r_3 \ + \ r_2 r_3$$
$$\Rightarrow \ \ r_2 \ \left( \frac{2 \ r_1^2}{r_2 \ - \ r_1} \right) \ + \ 2 \ r_1 r_2 \ + \ 2 \ r_2^2 \ = \ r_3 \ \left( \frac{2 \ r_1^2}{r_2 \ - \ r_1} \right) \ + \ 2 \ r_1 r_3$$
$$\Rightarrow \ \ 2 \ r_1^2r_2 \ + \ 2 \ r_1 r_2^2 \ - \ 2 \ r_1^2r_2 \ + \ 2 \ r_2^3 \ - \ 2 \ r_1 r_2^2 \ = \ 2 \ r_1^2r_3 \ \ + \ 2 \ r_1 r_2 r_3 \ - \ 2 \ r_1^2 r_3$$
[multiplying through by $\ r_2 \ - \ r_1 \$ and canceling like terms]
$$\Rightarrow \ \ 2 \ r_2^3 \ = \ 2 \ r_1 r_2 r_3 \ \ \Rightarrow \ \ r_2^2 \ = \ r_1 r_3 \ \ .$$
So our intuition concerning a scaling argument is correct, and the geometric mean relation between radii (and so of surface areas and volumes) of contiguous spheres follows from the walls of the cone having constant slope and the spheres being in direct contact with one another.
- | 2014-07-22T16:05:36 | {
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https://www.mathworks.com/help/matlab/ref/rat.html | # rat
Rational fraction approximation
## Syntax
``R = rat(X)``
``R = rat(X,tol)``
``````[N,D] = rat(___)``````
## Description
example
````R = rat(X)` returns the rational fraction approximation of `X` to within the default tolerance, `1e-6*norm(X(:),1)`. The approximation is a character array containing the truncated continued fractional expansion.```
example
````R = rat(X,tol)` approximates `X` to within the tolerance, `tol`.```
example
``````[N,D] = rat(___)``` returns two arrays, `N` and `D`, such that `N./D` approximates `X`, using any of the above syntaxes.```
## Examples
collapse all
Approximate the value of $\pi$ using a rational representation of the quantity `pi`.
The mathematical quantity $\pi$ is not a rational number, but the quantity `pi` that approximates it is a rational number since all floating-point numbers are rational.
Find the rational representation of `pi`.
```format rational pi```
```ans = 355/113 ```
The resulting expression is a character vector. You also can use `rats(pi)` to get the same answer.
Use `rat` to see the continued fractional expansion of `pi`.
`R = rat(pi)`
```R = '3 + 1/(7 + 1/(16))' ```
The result is an approximation by continued fractional expansion. If you consider the first two terms of the expansion, you get the approximation $3+\frac{1}{7}=\frac{22}{7}$, which only agrees with `pi` to 2 decimals.
However, if you consider all three terms printed by `rat`, you can recover the value `355/113`, which agrees with `pi` to 6 decimals.
`$3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113}$`
Specify a tolerance for additional accuracy in the approximation.
`R = rat(pi,1e-7)`
```R = '3 + 1/(7 + 1/(16 + 1/(-294)))' ```
The resulting approximation, `104348/33215`, agrees with `pi` to 9 decimals.
Create a 4-by-4 matrix.
```format short; X = hilb(4)```
```X = 4×4 1.0000 0.5000 0.3333 0.2500 0.5000 0.3333 0.2500 0.2000 0.3333 0.2500 0.2000 0.1667 0.2500 0.2000 0.1667 0.1429 ```
Express the elements of `X` as ratios of small integers using `rat`.
`[N,D] = rat(X)`
```N = 4×4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ```
```D = 4×4 1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7 ```
The two matrices, `N` and `D`, approximate `X` with `N./D`.
View the elements of `X` as ratios using `format rational`.
```format rational X```
```X = 1 1/2 1/3 1/4 1/2 1/3 1/4 1/5 1/3 1/4 1/5 1/6 1/4 1/5 1/6 1/7 ```
In this form, it is clear that `N` contains the numerators of each fraction and `D` contains the denominators.
## Input Arguments
collapse all
Input array, specified as a numeric array of class `single` or `double`.
Data Types: `single` | `double`
Complex Number Support: Yes
Tolerance, specified as a scalar. `N` and `D` approximate `X`, such that ```abs(N./D - X) <= tol```. The default tolerance is `1e-6*norm(X(:),1)`.
## Output Arguments
collapse all
Continued fraction, returned as a character array with `m` rows, where `m` is the number of elements in `X`. The accuracy of the rational approximation via continued fractions increases with the number of terms.
Numerator, returned as a numeric array. `N./D` approximates `X`.
Denominator, returned as a numeric array. `N./D` approximates `X`.
## Algorithms
Even though all floating-point numbers are rational numbers, it is sometimes desirable to approximate them by simple rational numbers, which are fractions whose numerator and denominator are small integers. Rational approximations are generated by truncating continued fraction expansions.
The `rat` function approximates each element of `X` by a continued fraction of the form
`$\frac{N}{D}={D}_{1}+\frac{1}{{D}_{2}+\frac{1}{\ddots +\frac{1}{{D}_{k}}}}\text{\hspace{0.17em}}.$`
The Ds are obtained by repeatedly picking off the integer part and then taking the reciprocal of the fractional part. The accuracy of the approximation increases exponentially with the number of terms and is worst when `X = sqrt(2)`. For `X = sqrt(2)` , the error with `k` terms is about `2.68*(.173)^k`, so each additional term increases the accuracy by less than one decimal digit. It takes 21 terms to get full floating-point accuracy.
## Version History
Introduced before R2006a | 2023-01-27T05:19:18 | {
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http://villecapri.it/rggu/how-to-find-the-angles-of-an-isosceles-trapezoid-given-side-lengths.html | # How To Find The Angles Of An Isosceles Trapezoid Given Side Lengths
If you know the lengths of the sides you can use Pythagoras theorem twice to determine the lengths of the diagonals. Likewise, because of same-side interior angles, a lower base angle is supplementary to any upper base angle. Sector AOB of 00 with radius 10 and m Z AOB = 108 Find the lateral area, total area, and volume of each solid. 56 a regular hexagon is shown, on Fig. The other common SSS special right triangle is the 5 12 13 triangle. Find mZDAC. If you are, that knowledge can help you. org are unblocked. The three formulas to find area depend on information you know about the rhombus. Convex Regular Polygons Looking at the following three polygons, we can work out a formula to calculate the external angle of a convex regular. A trapezoid is a 4-sided figure with one pair of parallel sides. The sum length of any two sides is longer than the length of the other side. triangle, quadrilateral, parallelogram, rectangle) that it belongs to, and a possible subcategory (e. Geometry calculator for solving the angle bisector of a of a scalene triangle given the length of sides b and c and the angle A. Two of the vertices of the triangle are placed on the circumference of the ellipse y? b? = 1 a Two ends of one of the heights in the lineage are the focal points of the ellipse. Angles are calculated and displayed in degrees, here you can convert angle units. Given any angle and arm or base. The straight lines segment, not parallel, are called sides or legs, while the two parallel segments are called bases, one short and the other long. It is parallel to the bases and is half as long as the sum of the bases. Answer: Area of isosceles trapezoid(A) is given by: where. congruent Two angles are congruent if they have the same measure. Base angles are equal because it's isosceles, so each angle is half of their sum. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Use the information in the figure. triangle, quadrilateral, parallelogram, rectangle) that it belongs to, and a possible subcategory (e. The equal sides are called legs, and the third side is the base. Base Angles The base angles of an isosceles trapezoid are congruent. The two equal length sides have length z. Yes, because the measures add up to 180 o. 400 1 3 1 2 34 5 25. (i noe we have to draw an altitude) but i dont get the rest!. I suggest to solve the problem considering the isosceles trapezoid, hence the lengths of the two. If m HEF 70 and m FGH 110 is trapezoid EFGH isosceles Explain Theorems Theorem from MATH 101 at Farragut High School. an isosceles trapezoid has sides whose lengths are inthe ratio of 5:8:5:14. Each of our worksheets comes with an accurate, easy-t0-use answer key so that either teachers or students can check the assignment. The non parallel sides are called sides or legs, while the two parallel sides are called bases, one short and the other long. The sum of the other three sides is 380 feet. How do we know what we look at is an Isosceles Triangle? First and fore most a Isosceles triangle is a polygon (many sided shape) with three sides (a triangle). The height of the trapezoid is the perpendicular distance between the bases, here symbolized by h. This is a trapezoid with two opposite legs of equal length. 5) In an isosceles trapezoid, opposite angles are congruent. The two diagonals within the trapezoid bisect angles and at the same angle. Acute Trapezoid. Legs have length s where a and b are positive integers. Bases of a trapezoid The parallel sides of a trapezoid are called bases. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. As the non-right angles of an isosceles right triangle are 45^@, we know angleABC = 45^@, implying angleMBN = 45^@. Angle bisector. an isosceles trapezoid has sides whose lengths are inthe ratio of 5:8:5:14. Following quiz provides Multiple Choice Questions (MCQs) related to Classifying scalene, isosceles, and equilateral triangles by side lengths or angles. Calculate the height knowing that the oblique side is 26 cm. The easiest way for the areas of the triangles to be equal would be if they were congruent. A tree 66 meters high casts a 44-meter shadow. Find a missing side length on an acute isosceles triangle by using the Pythagorean theorem. The sides of the triangle are known as follows: The hypotenuse is the side opposite the right angle, or defined as the longest side of a right-angled triangle, in this case $h$. The height of the isosceles trapezoid is the line segment contained in the interior of the isosceles trapezoid perpendicular to both parallel sides. We know, based on our rules for the side lengths of triangles, that the sum of two sides must be greater than the third. It follows from basic trigonometry that so that (Equation 1 ) , and so that (Equation 2 ). h is the height of the trapezoid. The properties of the trapezoid are as follows: The bases are parallel by definition. Now, if a trapezoid is isosceles, then the legs are congruent, and each pair of base angles are congruent. Ordering a Triangles Angles measures given its side lengths. Constructing the auxiliary height segment forms a right triangle with the slanted side, the height, and a portion of the long parallel side of the isosceles trapezoid as its sides. Calculations at an isosceles trapezoid (or isosceles trapezium). Enter the three side lengths, choose the number of decimal places and click Calculate. parallel sides of a trapezoid are the bases of the trapezoid. In an isosceles triangle, knowing the side and angle α, you can calculate the height, since the side is hypotenuse and the height is the leg, then the height will be equal to the product of the sine of the angle to the side. • How would you draw this triangle accurately?. And then we have another pair of sides that are not. Isosceles trapezoid. The longest side (opposite the right angle) is the “hypotenuse,” and the two shorter sides are the “legs. An obtuse trapezoid has one interior angle (created by either base and a leg) greater than 90°. interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. The Trapezoid can sometimes cause quite a bit of confusion. Find each measure. A regular quadrangle is a square; a regular triangle is an equilateral triangle. 11 Prove theorems about parallelograms. An isosceles triangle has 2 sides of equal length. In an isosceles trapezoid the bases are. In an isosceles triangle, the median to the base (different side or non-equal side) is perpendicular to the base. mZENB = 440 and AC is an altitude. The 45°-45°-90° triangle, also referred to as an isosceles right triangle, since it has two sides of equal lengths, is a right triangle in which the sides corresponding to the angles, 45°-45°-90°, follow a ratio of 1:1:√ 2. All of the lengths with one mark have length 5, and all of the side lengths with two marks have length 4. Using rulers and protractors, students will sketch a triangle that should be an isosceles triangle. Trapezoid A trapezoid is a quadrilateral with exactly one pair of parallel sides. Proving Equilateral Triangles. It is a special case of a trapezoid. Properties: 1) Intersection with Cyclic Quadrilateral is an Isosceles Trapezoid. Given sides a and c find side b and the perimeter, semiperimeter, area and altitudes a and c are known; find b, P, s, K, h a, h b, and h c b = √(c 2 - a 2 ). Bases of a trapezoid The parallel sides of a trapezoid are called bases. other base and midline 2. By using this website, you agree to our Cookie Policy. Rhombus area calculator is a great tool to determine the area of a rhombus, as well as its perimeter and other characteristics: diagonals, angles, side length, and height. The diagram is not to scale. In isosceles trapezoid EFGH, use the Same-Side. Two segments are congruent if they have the same lengths. 64 Statements 2. How to identify a segment from the vertex angle in an isosceles triangle to the opposite side. The application solves every algebraic problem including those with: - fractions - roots - powers you can also use parentheses, decimal numbers and Pi number. Suppose DE forms another triangle with the same circle inscribed in it. Thus, must also be equal to 50 degrees. That median is a bisector for the angle in the vertex of the opposite side. If you've been given the base and side lengths of an isosceles trapezoid. Isosceles Trapezoid Calculator. Identifying isosceles triangles. we have to find the area of trapezoid. Thus triangleBNM is also an isosceles right triangle, and so BN = NM. What are the lengths of the other sides? 5) A quadrilateral has diagonals that bisect each other at 90° and a perimeter of 84 centimeters. An Isosceles triangle has at least two sides with the same measurement. Example 4: Find the area of the figure 12 1 45 20. The area of an isosceles trapezoid can be found in another way, if known angle at the base and the radius of the inscribed circle. The length required to build the fence around the entire garden. The three formulas to find area depend on information you know about the rhombus. For each of the bases of a trapezoid, there is a pair of base angles, which are the two angles that have that base as a side. Find the measure of each numbered angle. Also, as this is an isosceles trapezoid, and are equal to each other. An isosceles trapezoid has one pair of parallel sides, equal legs, and equal base angles. If a trapezoid has one pair of congruent base angles, then the trapezoid is isosceles. Say your triangle's two legs are 3 inches and 4 inches long, so a is 3, and b is 4:. A triangle has side lengths of 6 inches and 9 inches. With this knowledge, we can add side lengths together to find that one diagonal is the hypotenuse to this right triangle: Using Pythagorean Theorem gives: take the square root of each side. As the non-right angles of an isosceles right triangle are 45^@, we know angleABC = 45^@, implying angleMBN = 45^@. _____ can review for their Quad Test! Quadrilaterals Review Worksheet Part I - Quad Properties: Put an x in the box if the shape always has the given property. It is a special case of a trapezoid. The other two sides (c and d) are called legs. A A A (a) (b) (c) Figure 3. The application solves every algebraic problem including those with: - fractions - roots - powers you can also use parentheses, decimal numbers and Pi number. In ∆𝐴𝐴𝐴𝐴𝐴𝐴 𝑚𝑚∠𝐴𝐴= 21 °, 𝑚𝑚∠𝐴𝐴= 4𝑥𝑥+ 19 °, and 𝑚𝑚∠𝐴𝐴= 6𝑥𝑥 °. The angle between a side and a diagonal is equal to the angle between the opposite side and the same diagonal. The following example illustrates how. The bases of a trapezoid are parallel. 960 1 \$9' 470 550 2 ILS' 3 ILS' Algebra Find the value(s) of the variable(s) in each isosceles trapezoid. Isosceles Trapezoid. If you know a lot of angles, a better approach is to think of the Law of Sines or the Law of Cosines (c^2 = a^2+b^2-2*a*b*cos(C)). Likewise, because of same-side interior angles, a lower base angle is supplementary to any upper base angle. In this lesson, you will learn how to find the perimeter of a square or rectangle with a missing side length by using square tiles. If two interior angles of a triangle are. An acute angle has a measure of less than 90 degrees. In a non‑trivial rotation symmetry, one side of a triangle is mapped to a second side, and the second side mapped to the third side, so the triangle must be equilateral. Triangles by Side Lengths 1. Use the information in the figure. All internal angles of a trapezoid sum to give 360°. So, each pair of base angles is congruent. Complete the proof. Write each of x and y as functions of. Find a missing side length on an acute isosceles triangle by using the Pythagorean theorem. You can use auxiliary segments to prove these theorems. Two of the vertices of the triangle are placed on the circumference of the ellipse y? b? = 1 a Two ends of one of the heights in the lineage are the focal points of the ellipse. Find x and y. If we bisect the base angle at B by a line from B to point D on AC then we have the angles as shown and also angle BDC is also 72°. 41 min 8 Examples. The perimeters of each are the sum of the lengths of the sides. Find the angle of elevation of the sun. find the measure of the angle between one of the legs and he shortter base. In other words, the length of the median is. Side c is the hypotenuse*, the side opposite the right angle. The defining trait of this special type of trapezoid is that the two non-parallel sides (XW and YZ below) are congruent. This is a trapezoid with two opposite legs of equal length. Areas of Trapezoids. For example, students can be asked to form a triangle that has two congruent angles and two congruent sides. 4) Sums of two (distinct) pairs adjacent angles equal. c) ˜ CD CB DE BA x 1 45 15 ˜. These two sides (a and b in the image above) are called the bases of the trapezoid. acute triangle A triangle with all acute angles. Base of an isosceles triangle. How to Find the Lengths of an Isosceles Trapezoid Given the Base Angles & Side Lengths. As the non-right angles of an isosceles right triangle are 45^@, we know angleABC = 45^@, implying angleMBN = 45^@. Create a scalene triangle. The parallel sides are the bases. Find the measures of angles x, y, and z. b) Calculate the base angle of the triangle. _____ can review for their Quad Test! Quadrilaterals Review Worksheet Part I - Quad Properties: Put an x in the box if the shape always has the given property. Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. They use algebra to determine the values of variables. mZENB = 440 and AC is an altitude. Exact areas should be given unless approximations are necessary—then round to the nearest tenth. The easiest way for the areas of the triangles to be equal would be if they were congruent. An icon used to represent a menu that can be toggled by interacting with this icon. A regular nonagon with radius of 8. find the measure of the angle between one of the legs and he shortter base. A water trough is 14m long and a cross-section has the shapE of an isosceles trapezoid (trapezoid with equal left and right side lengths) that is 0. Find the length of each side. Each of the parallel sides is called a base. ∆OGB and rt. b) Calculate the base angle of the triangle. Question 867053: an isosceles trapezoid has consecutive side of lengths 10, 6, 10 and 14 find the measure to the nearest integer of each angle of the trapezoid Answer by josgarithmetic(33323) (Show Source):. Based on the above, it follows that the length of medians originating from vertices with equal angles should be equal. These worksheet are a great resources for the 5th, 6th Grade, 7th Grade, and 8th Grade. With this knowledge, we can add side lengths together to find that one diagonal is the hypotenuse to this right triangle: Using Pythagorean Theorem gives: take the square root of each side. Isosceles trapezoid. Let A,B,C be the vertices and a,b,c be the side lengths where a=BC,b=AC and c=AB. These two sides are called the bases of the trapezoid. The nonparallel sides of a trapezoid are the legs of the trapezoid. => DM// CN ( lines perpendicular to the same line are. Find its area by using only the formula for the area of the parallelogram. Free trial. Create an equilateral triangle. and heigh 1. If two sides of a triangle are congruent the angles opposite them are congruent. other base and midline 2. Rhombus area calculator is a great tool to determine the area of a rhombus, as well as its perimeter and other characteristics: diagonals, angles, side length, and height. A trapezoid is a quadrilateral with only one pair of opposite sides parallel. Prove theorems about lines and angles. Now, suppose we are given one of the acute angles in the right triangle and one of the sides of the triangle. Finding the parallel sides of a trapezoid given all side lengths and height from base 0 Given a known isosceles Trapezoid find height of another with same angles & one base but different area. Also, as this is an isosceles trapezoid, and are equal to each other. One side of a triangle is three times the smallest side. Find the volume of the solid of revolution. Angle ADC is a right angle. Likewise, because of same-side interior angles, a lower base angle is supplementary to any upper base angle. Additionally, the angles on the same side of a leg are called adjacent and always sum up to 180°: α + β = 180° γ + δ = 180°. Can you find any relationships between the angles of the trapezoid? 2. If legs of a trapezoid are congruent then it is an isosceles trapezoid. Trapezoid (or Trapezium) - any quadrilateral with at least one pair of opposite sides parallel. If you know the side lengths, base, and altitude, it is possible to do this with just a ruler and compass (or just a compass, if you are given line segments). Like the 30°-60°-90° triangle, knowing one side length allows you to determine the lengths of the other sides. number of sides, number of equal side lengths, parallel sides, number of equal angles, right angles), (name) will correctly state why the 2-D shape belongs in the given. All angles of a triangle always add up to 180 ̊C. Step 3: Approach and Working out. An obtuse trapezoid: An obtuse trapezoid has two angles that are greater than 90 degrees. Properties of isosceles trapezoids Theorem 1 The base angles of an isosceles trapezoid are congruent. Alternatively, it can be defined as a trapezoid in which both legs and both base angles are of the same measure. There is a complete solution delivered for each issue to satisfy every teacher or student. Acute angle. C program to find the area of an Equilateral triangle. Calculations at an isosceles trapezoid (or isosceles trapezium). Each of our worksheets comes with an accurate, easy-t0-use answer key so that either teachers or students can check the assignment. This is a trapezoid with two opposite legs of equal length. Base angles of a trapezoid are two consecutive angles whose common side is a base. How to find the area of a trapezoid?. Trapezoids and Kites 336 Chapter 6 Quadrilaterals Lesson 6-1 Algebra Find the values of the variables. Regular polygon is a polygon with equal sides and angles. What is the measure of an acute base angle of the trapezoid? Of an obtuse base angle? The diagram is not to scale. The hypotenuse of a right triangle is always the side opposite to the right angle. The bases are parallel but of different lengths. If OG ≅ OF and OB ≅ OB, then it follows that BG ≅ BF. Therefore, WT , if ZX = 20 and TY = 15. It is parallel to the bases and is half as long as the sum of the bases. Solve the right triangle ABC if angle A is 36°, and side c is 10. Base angles of a trapezoid are two consecutive angles whose common side is a base. Personally I think it is more a language problem than anything else. Side DG is congruent to side EF, and diagonal DF is congruent to diagonal EG. Angle ADC is a right angle. An isosceles trapezoid has one pair of parallel sides, equal legs, and equal base angles. 67°; 134° b. (i noe we have to draw an altitude) but i dont get the rest!. One side of a triangle is three times the smallest side. Lines AC (or q) and BD (or p) are called diagonals The line perpendicular to lines AD & BC is called the height or altitude. an isosceles trapezoid has sides whose lengths are inthe ratio of 5:8:5:14. Find the length of each side. that they should try to construct triangles with the side lengths listed in the table. congruent Two angles are congruent if they have the same measure. Side c is the hypotenuse*, the side opposite the right angle. A scalene triangle has no congruent sides. Opposite angles are supplementary. A way for that to work would be if were simply an isosceles trapezoid! Since and (look at the side lengths and you'll know why!), See also. But then they have two choices here. ∆OGB and rt. The properties of the trapezoid are as follows: The bases are parallel by definition. 14 Theorem 6. The Trapezoid (as shown in the diagram above, with two parallel lines is also referred to as a Trapezium in British English, but the Trapezium in American English has NO Parallel lines - So on this site I am going to stick with the American Standard. Can you find any relationships between the angles of the trapezoid? 2. The base angles on an isosceles trapezoid are congruent. The sum of the other three sides is 380 feet. He wants to build a fence around it. Regular polygon is a polygon with equal sides and angles. The two equal length sides have length z. Triangle has three types based on its three angles, including obtuse (1 angle > 90 ̊C), right (1 angle = 90 ̊C) and acute (no angle > 90 ̊C). A A A (a) (b) (c) Figure 3. The side opposite the right angle is called the hypotenuse. These sides are called bases, whereas the opposite sides that intersect (if extended) are called legs. An obtuse trapezoid has one interior angle (created by either base and a leg) greater than 90°. This is a right-angled scalene triangle because no sides are the same length. the three angles of a scalene triangle are of different measures. High School: Geometry » Congruence » Prove geometric theorems » 9 Print this page. To find the measure of angle DAC, we must know that the interior angles of all triangles sum up to 180 degrees. Base angles are equal because it's isosceles, so each angle is half of their sum. parallel sides of a trapezoid are the bases of the trapezoid. The nonparallel sides are called legs. What is the measure of an acute base angle of the trapezoid? Of an obtuse base angle? The diagram is not to scale. Find the measures Of the numbered angles in each rhombus. ABDC is a trapezoid with ࠵?ܤ തതതത ∥ ܥܦ തതതത. C program to check whether a triangle is valid or not if all angles are given. Example #3: Find the perimeter of the following trapezoid where the length of the bottom base is not known. J A conditional statement is given below. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary; Property #2) Area of a Trapezoid = $$Area = height \cdot \left( \frac{ \text{sum bases} }{ 2 } \right)$$ Property #3) Trapezoids have a midsegment which connects the mipoints of the legs. Area and Perimeter of Triangles Worksheets. A right trapezoid is a trapezoid having two right angles. An isosceles trapezoid has legs of equal length. However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some basic trigonometric functions:. ∠ A + ∠ C = 180° ∠ B + ∠ D = 180° The right trapezoid has two right angles. z 65° m y 5 ° m z 5 ° Find the unknown angle measure in each isosceles triangle. It is a special case of a. Area of Triangle using Side-Angle-Side (length of two sides and the included angle) Last Updated: 10-07-2020 Given two integers A , B representing the length of two sides of a triangle and an integer K representing the angle between them in radian, the task is to calculate the area of the triangle from the given information. Using rulers and protractors, students will sketch a triangle that should be an isosceles triangle. An equilateral triangle has three equal lengths, and all the angles are equal which means they are each 60°. Center of. If you're seeing this message, it means we're having trouble loading external resources on our website. An icon used to represent a menu that can be toggled by interacting with this icon. 3) Diagonals intersect on line connecting midpoints of // sides. The diagram is not to scale. Alternate exterior angles. How to Find the Lengths of an Isosceles Trapezoid Given the Base Angles & Side Lengths. c) ˜ CD CB DE BA x 1 45 15 ˜. Given an acute angle and one side. By using this website, you agree to our Cookie Policy. What is the measure of the vertex angle of an isosceles triangle if one of its base angles measures 42°? 35. An obtuse triangle has only one inscribed square. The midsegment of a trapezoid is a line connecting the midpoints of the two legs. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. Base angles of a trapezoid. Exactly one pair of parallel sides; Two pairs of adjacent, supplementary angles; Isosceles Trapezoid (All the attributes of trapezoid and. Each lower base angle is supplementary to […]. A trapezoid is a right trapezoid if one of the angles is equal to 90 degrees. Calculate the length of equal sides if given side (base) and angle ( a ) : Calculate the length of a side (base) if given equal sides and angle ( b ) : side of an isosceles triangle :. See if you're working with a special type of triangle such as an equilateral or isosceles triangle. Image Transcriptionclose. An isosceles trapezoid has the base greater of 50 cm, the minor base is 30 cm. If you know the side lengths, base, and altitude, it is possible to do this with just a ruler and compass (or just a compass, if you are given line segments). and AD = BD (fig. a) BAC DEC b) m BAC 5 m DEC (given) m ACB 5 m ECD (vertically opposite s) wo pairs of corresponding angles have T equal measures. Find the degree measure of each base angle. Since they are similar triangles, you can use proportions to find the size of the missing side. (Lessons 9. One of the best known mathematical formulas is Pythagorean Theorem, which provides us with the relationship between the sides in a right triangle. For example, look in the diagram on the right side, the bases base1 and base2 are parallel. The dashes on the lines show they are equal in length. Alternate exterior angles. The side opposite the right angle is called the hypotenuse. Isosceles trapezoid. Likewise, because of same-side interior angles, a lower base angle is supplementary to any upper base angle. Alternatively, it can be defined as a trapezoid in which both legs and both base angles are of the same measure. Since ABD also has two equal angles of 36°, it too is isosceles and so BD=AD. Since this is an isosceles right triangle, the only problem is to find the unknown hypotenuse. Find x and y. Create a scalene triangle. Corresponding parts of— A are x. The measure of one angle of a quadrilateral is 3more than the smallest; the third angle is 5 less than 8 times the smallest; and the fourth angle is 2 more than 8 times the smallest. Angle of rotation. By using this website, you agree to our Cookie Policy. The other common SSS special right triangle is the 5 12 13 triangle. The perimeters of each are the sum of the lengths of the sides. SAS (side-angle-side) - having the lengths of two sides and the included angle (the angle between the two), you can calculate the remaining angles and sides, then use the SSS rule. four interior angles, totaling 360 degrees. A trapezoid that has a right angle is called a right trapezoid. Let's find the length of side DF, labeled x. Angles are calculated and displayed in degrees, here you can convert angle units. Additionally, the angles on the same side of a leg are called adjacent and always sum up to 180°: α + β = 180° γ + δ = 180°. There is one right angle (90º) in a right-angled triangle. 4 angles whose measures add up to 360 degrees; Trapezoid. Area of trapezium = × (sum of two parallel sides) × height. To find the measure of angle DAC, we must know that the interior angles of all triangles sum up to 180 degrees. The students in a class were each given a set of letters and asked to make words. C program to find angle of a triangle if two angles are given. Given parallelogram DANE and isosceles triangle BEN. Now, if a trapezoid is isosceles, then the legs are congruent, and each pair of base angles are congruent. Conway and Guy (1996) give Neusis constructions for the 7-, 9-, and 13-gons which are based on angle trisection. Isosceles Trapezoid Calculator. An obtuse trapezoid has one interior angle (created by either base and a leg) greater than 90°. The students in a class were each given a set of letters and asked to make words. Therefore, the two. This one-page worksheet contains 18 multi-step problems. All of the lengths with one mark have length 5, and all of the side lengths with two marks have length 4. Find the missing angle measurement. Alternate interior angles. diagonals, lateral side (height) and angle between the diagonals 4. The diagram is not to scale. Based on the above, it follows that the length of medians originating from vertices with equal angles should be equal. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment. Image Transcriptionclose. Given the properties of an isosceles triangle, students can be asked to draw their own isosceles triangle. Create an isosceles triangle. It is impossible to draw a unique triangle given one angle and two side lengths. area and perimeter of an Hexagon Calculator: A hexagon (from greek hexi = six and gonia = angle) is a polygon with six vertices and six sides. The triangle would be isosceles, because isosceles triangles have two sides the same length. Suppose DE forms another triangle with the same circle inscribed in it. The number of lattes sold daily for two coffee shops is shown in the table: lattes 12 52 57 33 51 15 46 45 based on the data, what is the difference between the median of the data, including the possible outlier(s) and excluding the possible outlier(s)?. (In other words, the two radii form a straight angle at the center of the circle. Area and Perimeter of Triangles Worksheets. A trapezoid is a 4-sided figure with one pair of parallel sides. An obtuse triangle may be either isosceles (two equal sides and two equal angles) or scalene (no equal sides or angles). Find each measure. By (date), given a 2-D shape, a category (e. Part of the series: Trapezoids. In this trapezoids and kites worksheet, students find the measures of given angles in an isosceles trapezoid. To find the measure of angle DAC, we must know that the interior angles of all triangles sum up to 180 degrees. Free Isosceles Trapezoid Sides & Angles Calculator - Calculate sides, angles of an isosceles trapezoid step-by-step This website uses cookies to ensure you get the best experience. The line parallel to lines AD & BC, is at the midpoints of lines AB and DC and is called the median or. The bases of a trapezoid are parallel. The consecutive angles of a parallelogram are supplementary to each other; The diagonals of a parallelogram bisect each other; Rectangle satisfies one more property: The diagonals of a rectangle are congruent; If we know side lengths of the rectangle, it is easy to calculate the length of the diagonal using the Pythagorean Theorem. This application is able to do calculation on the following figures: - Triangles. 3) 1200 Find the value Of x that makes each parallelogram the given type. The nonparallel sides of a trapezoid are the legs of the trapezoid. Proving that Trapezoid is isosceles 1. Let us assume a, b, c are the sides of triangle where c is the side across from angle C. Square 12X+6 Find angle measure x on each given figure. The parallel sides of a trapezoid are called its bases. The measure of one angle of a quadrilateral is 3more than the smallest; the third angle is 5 less than 8 times the smallest; and the fourth angle is 2 more than 8 times the smallest. Right Trapezoid. If the legs are equal in length, the trapezoid is called isosceles. Find the missing angle. Finding the perimeter of a trapezoid when the height, the length of the top base, and the lengths of the nonparallel sides are given. Exactly one pair of parallel sides; Two pairs of adjacent, supplementary angles; Isosceles Trapezoid (All the attributes of trapezoid and. To calculate the isosceles triangle area, you can use many different formulas. Finding the parallel sides of a trapezoid given all side lengths and height from base 0 Given a known isosceles Trapezoid find height of another with same angles & one base but different area. In this lesson you will learn how to determine the missing length of a rectangle by applying the perimeter formula for a rectangle. ∠ A + ∠ C = 180° ∠ B + ∠ D = 180° The right trapezoid has two right angles. Ordering a Triangles Angles measures given its side lengths. Let ABCbe a triangle with AB= 12, BC= 5, AC= 13. If you know a lot of angles, a better approach is to think of the Law of Sines or the Law of Cosines (c^2 = a^2+b^2-2*a*b*cos(C)). b) Calculate the base angle of the triangle. By (date), given a 2-D shape, a category (e. An obtuse trapezoid: An obtuse trapezoid has two angles that are greater than 90 degrees. That median is a bisector for the angle in the vertex of the opposite side. The bases are parallel but of different lengths. Suppose DE forms another triangle with the same circle inscribed in it. If you know Altitude (height) and side s the formula is: a r e a = h e i g h t × s; If you know the length of one side s and the measure of one angle the formula is: a r e a = s 2 sin ∠ A = s 2 sin ∠ B; If you know the lengths of the diagonals the formula is:. a and b are the unequal side length and. To find a missing angle in an isosceles triangle use the following steps: If the missing angle is opposite a marked side, then the missing angle is the same as the angle that is opposite the other marked side. 5 (6x + 16 12. Also, as this is an isosceles trapezoid, and are equal to each other. Create an acute triangle. Right-angled trapezoid. Angle, Side Length of a Triangle [9/4/1996] What is the relation between the angles and side lengths of a triangle? Angle-Side-Side Does Not Work [11/12/2001] Can you give me a construction to show that Angle-Side-Side does not prove two triangles congruent. _____ can review for their Quad Test! Quadrilaterals Review Worksheet Part I - Quad Properties: Put an x in the box if the shape always has the given property. (Lessons 9. Definition of Trapezoid Believe it or not, there is no general agreement on the definition of a trapezoid. So, each pair of base angles is congruent. Bending Light 1. This is a right-angled scalene triangle because no sides are the same length. In a reflection symmetry, two sides are swapped, so the triangle must be isosceles. That median is a bisector for the angle in the vertex of the opposite side. Given a convex quadrilateral, the following properties are equivalent, and each implies that the quadrilateral is a trapezoid: It has two adjacent angles that are supplementary, that is, they add up to 180 degrees. Ordering a Triangles Side Lengths given its Angle Measures. Calculate the height knowing that the oblique side is 26 cm. Lengths of Chords in Circles: Finding Unknown Base Angles in Isosceles Triangles: Find Another Side Given an Angle:. Adam has a rectangular garden. For this trapezoids and kites worksheet, students find the measures of given angles in an isosceles trapezoid. GIVEN: DE Il Ãÿ, LDAV= LEVA PROVE: DAVE is an isosceles trapezoid. In our calculations for a right triangle we only consider 2 known sides to calculate the other 7 unknowns. Find the values of a and b. If no sides are equal in length, then no two angles are equal in size either. Isosceles trapezoid Isosceles trapezoid with axis of symmetry Type quadrilateral, trapezoid Edges and vertices 4 Symmetry group Dih 2,, (*), order 2 Dual polygon Kite Properties convex, cyclic In Euclidean geometry, an isosceles trapezoid (isosceles trapezium in British English) is a convex quadrilateral with a line of symmetry bisecting one pair of opposite sides. Q what postulate proves this statement? Which statement would be used to help find the missirg value? 1200 A. If you're seeing this message, it means we're having trouble loading external resources on our website. It is parallel to the bases and is half as long as the sum of the bases. The perimeter is $39$ feet. A regular hexagon with apothem 12 cm 18. Find 𝑚𝑚∠𝐴𝐴. So, BAC DEC. Guide them to see the special relationship between any two sides of a triangle and the third side. 3 x 5 3 and y 5 1. The first given side is marked //. The name hypotenuse is given to the longest edge in a right-angled triangle. The adjacent sides of a trapezoid are congruent. Find the ratios of the perimeters and areas of similar polygons. It is the isosceles triangle touching the circle at the point where the angle bisectrix crosses the circle. A trapezoid is a 4-sided figure with one pair of parallel sides. Trapezoid A trapezoid is a quadrilateral with exactly one pair of parallel sides. Enter the three side lengths, choose the number of decimal places and click Calculate. 75 x + 16 X: 2x. This is a right-angled scalene triangle because no sides are the same length. Can be inscribed in a circle; possible answer: The pairs of base angles of a trapezoid inscribed in a circle must be congruent. If a trapezoid has a pair of congruent base angles, then it is an. The measure of one angle of a quadrilateral is 3more than the smallest; the third angle is 5 less than 8 times the smallest; and the fourth angle is 2 more than 8 times the smallest. Depends from the given information. What are the lengths of the other sides? 5) A quadrilateral has diagonals that bisect each other at 90° and a perimeter of 84 centimeters. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary; Property #2) Area of a Trapezoid = $$Area = height \cdot \left( \frac{ \text{sum bases} }{ 2 } \right)$$ Property #3) Trapezoids have a midsegment which connects the mipoints of the legs. High School: Geometry » Congruence » Prove geometric theorems » 9 Print this page. A = × (a + b) × h. SAS [Side Angle Side] - An angle in one triangle is the same measurement as an angle in the other triangle and the two sides containing these angles have the same ratio. Guide them to see the special relationship between any two sides of a triangle and the third side. mZENB = 440 and AC is an altitude. ACEis isosceles with leg 6 and base CE= CB+ BE= CB+ DC= 10. Area of Triangle using Side-Angle-Side (length of two sides and the included angle) Last Updated: 10-07-2020 Given two integers A , B representing the length of two sides of a triangle and an integer K representing the angle between them in radian, the task is to calculate the area of the triangle from the given information. PQ is the median of trapezoid BCDF. Part of the series: Trapezoids. If the missing angle is not opposite a marked side, then add the two angles opposite the marked sides together and subtract this result. If follows directly that the sides opposite the congruent angles in an isosceles trapezoid are congruent. The midsegment of a trapezoid is a line connecting the midpoints of the two legs. Angle, Side Length of a Triangle [9/4/1996] What is the relation between the angles and side lengths of a triangle? Angle-Side-Side Does Not Work [11/12/2001] Can you give me a construction to show that Angle-Side-Side does not prove two triangles congruent. The other common SSS special right triangle is the 5 12 13 triangle. If the diagonals of a trapezoid are congruent, then it is an isosceles trapezoid. The hypotenuse of a right triangle is always the side opposite to the right angle. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have two sides have the same length. Consider rt. Example 5: Given trapezoid RSTV with median MN, find the value. OPEN ENDED Draw a triangle that is isosceles and right. An Isosceles triangle has at least two sides with the same measurement. Angle, Side Length of a Triangle [9/4/1996] What is the relation between the angles and side lengths of a triangle? Angle-Side-Side Does Not Work [11/12/2001] Can you give me a construction to show that Angle-Side-Side does not prove two triangles congruent. For example, if it is given the measure of the angle base θ, and the length of the base b, the sum of the sides a of the isosceles triangle equals to 2a = b. BCD now has two angles equal and is therefore an isosceles triangle; and also we have BC=BD. Since no side is the same length, this is not an isosceles trapezoid and the most precise name for this quadrilateral is trapezoid. Each angle of a regular polygon is equal to 180 ( n – 2 ) / n deg, where n is a number of angles. How to use the trapezoid calculator Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press "calculate. Given: RS ≅ ST, m∠RST = 3x − 48, m∠STU = 9x 38. We know, based on our rules for the side lengths of triangles, that the sum of two sides must be greater than the third. triangle, quadrilateral, parallelogram, rectangle) that it belongs to, and a possible subcategory (e. The sum of the other three sides is 380 feet. For example: if side length of a and A and B angles are known. Then we note how (16"-10")/2=3" is the side of a triangle whose other side is the height and hypotenuse is this 5" side. Find the missing angle. How to find the area of a trapezoid?. What are the lengths of the other sides? 5) A quadrilateral has diagonals that bisect each other at 90° and a perimeter of 84 centimeters. Draw the fourth side. Create an isosceles triangle. Lines AB and DC are the non-parallel sides and are called legs. The parallel sides of a trapezoid are called its bases. Each lower base angle is supplementary to […]. The angle measure of BDC is 35 o less than 3 times the measurement of angle ADB. Calculations at an isosceles trapezoid (or isosceles trapezium). We are given a=8,b=6 and m/_ ACB=30^@ . A trapezoid is a right trapezoid if one of the angles is equal to 90 degrees. Perimeter and area of rhombus, trapezoid, and parallelogram 9. m∠CBD = 34º m∠ACB = 68º because it is an exterior angle for ΔBCD and is the sum of the 2 non-adjacent interior angles. Given parallelogram DANE and isosceles triangle BEN. Exact areas should be given unless approximations are necessary—then round to the nearest tenth. Given a convex quadrilateral, the following properties are equivalent, and each implies that the quadrilateral is a trapezoid: It has two adjacent angles that are supplementary, that is, they add up to 180 degrees. Constructing the auxiliary height segment forms a right triangle with the slanted side, the height, and a portion of the long parallel side of the isosceles trapezoid as its sides. Find its area by using only the formula for the area of the parallelogram. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Isosceles Trapezoid Calculator. In other words, the length of the median is. The median of a trapezium is also known as the midline or midsegment of a trapezium. 4) The length of one side of a rectangular park is 80 feet. Bases of a trapezoid. The parallel sides of a trapezoid are called the bases, here symbolized by b 1 and b 2. Consider rt. Say your triangle's two legs are 3 inches and 4 inches long, so a is 3, and b is 4:. A trapezoid is a 4-sided figure with one pair of parallel sides. These two sides are called the bases of the trapezoid. Can a trapezoid have all of its angles acute angles? Why or why not? Definition An isosceles trapezoid is a trapezoid with the nonparallel sides (legs) congruent. Let variable x be the length of the base and variable y the height of the triangle, and consider angle. Regular polygon is a polygon with equal sides and angles. This one-page worksheet contains 18 multi-step problems. Finding the parallel sides of a trapezoid given all side lengths and height from base 0 Given a known isosceles Trapezoid find height of another with same angles & one base but different area. Trapezoid is a quadrilateral which has two opposite sides parallel and the other two sides non-parallel. A A A (a) (b) (c) Figure 3. A trapezoid is isosceles is one pair of opposite sides are equal. (It is the edge opposite to the right angle and is c in this case. x 5 3 AC EC AB ED ˜ ˜ y 39 15. How tall is a tree that casts an 8-foot shadow? The angle measurements are the same, so the triangles are similar triangles. Find the measures Of the numbered angles in each rhombus. Find an answer to your question An isosceles trapezoid has base angles of 45° and bases of lengths 9 and 15. Properties: 1) Intersection with Cyclic Quadrilateral is an Isosceles Trapezoid. A trapezoid is a quadrilateral that has one pair of sides which are parallel. A circle inscribed in a square with side 12 m 20. equilateral: all sides are equal in length, and all interior angles are 60 degrees. The diagonals of an isosceles trapezoid are congruent. Find the length of each side. Since the bisectrix is also a meadian, BG = GC. If m HEF 70 and m FGH 110 is trapezoid EFGH isosceles Explain Theorems Theorem from MATH 101 at Farragut High School. The Isosceles Trapezoids is a quadrilateral with two non parallel sides equal and two parallel sides unequal. Obtuse Trapezoid. This is a right-angled scalene triangle because no sides are the same length. When the sun is at a certain angle in the sky, a 6-foot tree will cast a 4-foot shadow. Notice that the values of the angles were special because they allowed the first solution I gave. 3x-3 Find XY in each trapezoid. Find the ratios of the perimeters and areas of similar polygons. Exactly one pair of parallel sides; Two pairs of adjacent, supplementary angles; Isosceles Trapezoid (All the attributes of trapezoid and. In right triangles, the trigonometric ratios of sine, cosine and tangent can be used to find unknown angles and the lengths of unknown sides. Thus, must also be equal to 50 degrees. The isosceles trapezoid is part of an isosceles triangle with a 46° vertex angle. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines,. The sides are in the shape of a trapezoid. All of the lengths with one mark have length 5, and all of the side lengths with two marks have length 4. One side of a right triangle measures 5 and the hypotenuse equals 13. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment. All internal angles of a trapezoid sum to give 360°. Exactly one pair of parallel sides; Two pairs of adjacent, supplementary angles; Isosceles Trapezoid (All the attributes of trapezoid and. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment. Lengths of Chords in Circles: Finding Unknown Base Angles in Isosceles Triangles: Find Another Side Given an Angle:. An isosceles triangle has two equal sides (or three, technically) and two equal angles (or three, technically). Opposite angles of are supplementary. In other words, the length of the median is. We can write a proportion, like this: We read this proportion as: "AC is to AB as DF is to DE. Angle bisector. Use the compass to copy the arc that this angle intercepts. Explanation:. Square 12X+6 Find angle measure x on each given figure. Find the degree measure of each base angle. 4) The length of one side of a rectangular park is 80 feet. An isosceles trapezoid is a trapezoid base angles (angles with common side) Find all angle measures and lengths of sides. Now, suppose we are given one of the acute angles in the right triangle and one of the sides of the triangle. Line segment OB bisects ∠B and line segment OC bisects ∠C. SOLUTION 16 : Write the area of the given isosceles triangle as a function of. These two sides are called the bases of the trapezoid. Altitude of a triangle. b) Calculate the base angle of the triangle. 18) Find VU G 6x − 6 F 38 W U V T 7x − 4 24-2-Create your own worksheets like this one with Infinite Geometry. an isosceles trapezoid has sides whose lengths are inthe ratio of 5:8:5:14. (Lessons 9. Example 3 – Using Properties of Special Quadrilaterals For the given kite, find the values of the variables and then find the lengths of the sides. Guide them to see the special relationship between any two sides of a triangle and the third side. parallel sides of a trapezoid are the bases of the trapezoid. In other words, the length of the median is. Depends from the given information. Isosceles C ABC' has a right angle at C. 200 m wide at the bottom, 0. The median of a trapezium is also known as the midline or midsegment of a trapezium. Comment/Request I would like to see an item in the element drop-down selection that allows to choose 'Side b' + 'Vertex Angle'. 3 x 5 3 and y 5 1. triangle, quadrilateral, parallelogram, rectangle) that it belongs to, and a possible subcategory (e. The sum length of any two sides is longer than the length of the other side. Finding the perimeter of a trapezoid when the height, the length of the top base, and the lengths of the nonparallel sides are given. Alternate exterior angles. Each angle of a regular polygon is equal to 180 ( n – 2 ) / n deg, where n is a number of angles. 41 min 8 Examples. Areas of Trapezoids. An icon used to represent a menu that can be toggled by interacting with this icon. The area of an isosceles trapezoid can be found in another way, if known angle at the base and the radius of the inscribed circle. Sometimes you will need to draw an isosceles triangle given limited information. | 2020-09-22T00:37:51 | {
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https://math.stackexchange.com/questions/2901257/find-a-formula-for-left-beginsmallmatrix-4-15-2-7-endsmallmatrix | # Find a formula for $\left(\begin{smallmatrix} -4 & -15 \\ 2 & 7 \end{smallmatrix}\right)^n$
We're going to consider the matrix $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$
(a) Let $\mathbf{P} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $\mathbf{D}$ such that $\mathbf{P}^{-1} \mathbf{D} \mathbf{P} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$
(b) Find a formula for $\mathbf{D}^n,$ where $\mathbf{D}$ is the matrix you found in part (a). (You don't need to prove your answer, but explain how you found it.)
(c) Using parts (a) and (b), find a formula for $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n.$
I have completed part a (thanks for the helpful hints), but I am confused on part b. I solved for a few powers of $\mathbf{D}$ to find a pattern. So far I have if $\mathbf{D} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then $\mathbf{D}^{n} = \begin{pmatrix} -4^n & ? \\ 0 & -19^n \end{pmatrix}.$ I don't know if I'm solving for this correctly and as you can see, I'm not sure how to find the $b$ part of $\mathbf{D}^{n}.$ Thanks again!
• If we call your original matrix $A,$ the definition $P^{-1}DP = A$ also says $PAP^{-1} = D$ – Will Jagy Sep 1 '18 at 1:07
• write $D^3$ in terms of $P$, $P^{-1}$, and $A$. Do you notice any convenient patterns? Hint: try just literally using the letters P, D, and A. don't worry about the specific matrices. – The Count Sep 1 '18 at 1:34
• If you’ve done this the way it’s intended, $b$ should be $0$. – amd Sep 1 '18 at 4:02
Let consider
$$\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x & y \\ z & w \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\\$$ $$\iff D=\begin{pmatrix} x & y \\ z & w \end{pmatrix}=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}=$$
$$=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 3 & -10 \\-1 & -3 \end{pmatrix}=\begin{pmatrix} -4 & -5 \\ 0 & -19 \end{pmatrix}$$
$$A^5 = PDP^{-1}PDP^{-1}PDP^{-1}PDP^{-1}PDP^{-1}$$
Can you simplify this expression?
This is a computer aided proof, sage:
It is good to know that sage provides directly the answer symbolically.
sage: A = matrix(QQ, 2, [-4, -15, 2, 7])
sage: A
[ -4 -15]
[ 2 7]
sage: var("n");
sage: A^n
[ -5*2^n + 6 -15*2^n + 15]
[ 2*2^n - 2 6*2^n - 5]
To get this answer, the best way to proceed is to diagonalize the matrix. (We can also triangularize, but let us compare the shape from the $D$ in the OP, some $T$ instead would have been a better notation, and the truly diagonal form below...) Sage diagonalizes for us.
sage: A.jordan_form(transformation=True)
(
[2|0]
[-+-] [ 1 1]
[0|1], [-2/5 -1/3]
)
sage: Lam, T = A.jordan_form(transformation=True)
sage: T * Lam * T.inverse()
[ -4 -15]
[ 2 7]
The diagonal matrix $\Lambda=\begin{bmatrix}2 &\\& 1\end{bmatrix}$, Lam above, has an obvious power, which has to be conjugated with the $T$ above to get the answer... | 2019-06-18T09:37:02 | {
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http://anthana.nl/k1e65/69843d-how-to-find-parent-function | {{courseNav.course.topics.length}} chapters | As mentioned above, each family of functions has a parent function. You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. 3.) In algebra, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. For example, when we think of the linear functions which make up a family of functions, the parent function would be y = x. There are many families of functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of … The parentElement property returns the parent element of the specified element. In this non-linear system, users are free to take whatever path through the material best serves their needs. Basically, exponential functions are functions with the variable in the exponent. This is the parent function of the family of functions. The Parent FunctionsThe fifteen parent functions must be memorized. Services. Transformations Of Parent Functions. Expand and simplify the function. Parent functions do not have any of the transformations that a full function can have such as additional constants or terms. Hi All, With the exception stacktrace, i can view the sequence of functions called from the last to first one. Explore the graphs of linear functions by adding or … For the family of quadratic functions, y = ax2 + bx + c, the simplest function of this form is y = x2. With the information in this lesson, we should now be familiar with what parent functions are and how to identify them. Given a tree and a node, the task is to find the parent of the given node in the tree. For example, the parent function for "y=x^+x+1" is just "y=x^2," also known as the quadratic function. To illustrate this, let's consider exponential functions. Did you know… We have over 220 college Get access risk-free for 30 days, On similar lines i want to know how can i achieve the same if i have to find the stack flow of a normal function. It is easy to see that function C does not have the same shape as functions A, B, and D. Therefore, function C is not in the same family as functions A, B, and D. I suppose in that context the parent function for logarithms would be either or With a decided preference for the latter in my mind. Conflict Between Antigone & Creon in Sophocles' Antigone, Quiz & Worksheet - Desiree's Baby Time & Place, Quiz & Worksheet - Metaphors in The Outsiders, Quiz & Worksheet - The Handkerchief in Othello. There really is no such thing as a "Parent Function." credit by exam that is accepted by over 1,500 colleges and universities. This example graphs the common log: f(x) = log x. For our course, you will be required to know the ins and outs of 15 parent functions . Compute the following values for the given function. You can use parent functions to determine the basic behavior of a function such the possibilities for axis intercepts and the number of solutions. Domain – The set of all inputs (x-values) that “work” in the function 2. The similarities don't end there! Math Glossary: Mathematics Terms and Definitions. Note that the point (0, 0) is the vertex of the parent function only. Quadratic Function - Parent Function and Vertical Shifts. Individual .m files do not know who calls them, but you can ask to analyze a bunch of files at the same time and see … We can group functions together in a very si… 2. imaginable degree, area of Other parent functions include the simple forms of the trigonometric, cubic, linear, absolute value, square root, logarithmic and reciprocal functions. 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Graph the result. Print -1 if the given node is the root node. Find b(x), the end-behavior function of f(x) = \frac{x^3 + 3x^2 - 4x -1}{x^2 - 1} . If you only name one file then it cannot tell what the parent is. The difference between parentElement and parentNode, is that parentElement returns null if the parent node is not an element node: Let h = h_1 i+ h_2 i. To traverse all the way up to the document's root element (to return grandparents or other ancestors), use the parents() or the parentsUntil() method.. We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin \left( {0,\,0} \right).The chart below provides some basic parent functions that you should be familiar with. Log in or sign up to add this lesson to a Custom Course. Let's consider the family of functions that are exponential functions with base 2. Visit the ACT Prep: Tutoring Solution page to learn more. On your child's device, click Parent Access. We see that we can derive all the other functions shown from y = |x|. There are two main things being done to the parabola. This is the parent function. All other trademarks and copyrights are the property of their respective owners. ; Enter the password for the parent's account used to supervise the child. This lesson will familiarize us with parent functions. What Slope-Intercept Form Means and How to Find It. Give the domain. Just like our own families have parents, families of functions also have a parent function. just create an account. Note: There’s an update version of this article. (If the parameter is included, then it will return the parent class of the specified class as normal.) 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If you don't see the password option: Make sure the child's device is turned on and connected to the internet. Which of them is the parent function? Ranch Hand Posts: 131. posted 12 years ago. A parent function is the simplest function that still satisfies the definition of a certain type of function. An error occurred trying to load this video. Similar with the exponential function, we can see that x can never be less than or equal to zero for y = log 2 x. succeed. lim_{ h-->0} (f(x+h)-f(x))/h, Use the following information to find the values of a, b, and c, in the formula f(x)=\frac{x+a}{bx^{2}+cx+2} a) The slope of the tangent line to the graph of f at x = 3 is 3 b) The line y = -1 is, Write the following functions as a composition of simpler functions.a)y=\ln(\tan(e^{3}x)) b)f(x)=\sin \sqrt{ \lg(x^{2}-1))}. All quadratic functions have a highest exponent of 2, their graphs are all parabolas so they have the same shape, and they all share certain characteristics. The method optionally accepts a selector expression of the same type that we can pass to the $() function. ", Remove any transformations from the functions. h_1(n)=( \frac{7}{6}. If x = xi - yi, compute f(x + h). Special Characteristics of Functions 1. Here is a list of five basic parent functions that you should have memorized, with their domain and range. To unlock this lesson you must be a Study.com Member. Select a subject to preview related courses: Let's look at some more examples to further our understanding of parent functions. One thing I need to do a lot in my plugins is find a parent element with a particular selector. Zero (X-Intercept) – The points at which a graph crosses the x-axis 5. Already registered? All rights reserved. To graph an absolute value function, choose several values of x and find some ordered pairs. Its parent function can be expressed as y = log b x, where b is a nonzero positive constant. first two years of college and save thousands off your degree. Then, write a complete approach for f(x). I don’t always know exactly what the markup structure will look like, so I can’t just use . 1. Givenf(x,y) = x^2 + \cos(xy), let (a) = 0i - 2j and (b) = \pi i - 3j a. Compute, f(a) and f(b). Solution: The parent function would be the simplest cubic function. 1. These transformations don't change the general shape of the graph, so all of the functions in a family have the same shape and look similar to the parent function. 's' : ''}}. equations of asymptote(s), show the asymptote(s) on the graph as dashed line(s), create a table with intercept(s) and at least 3 to, Order the following functions h_i, for 1 less than or equal to i less than or equal to 5, with respect to relation f \prec g defined as follows f \prec g\Leftrightarrow f= o(g). To learn more, visit our Earning Credit Page. This includes sign changes, added and multiplied constants and extra terms. This lesson discusses some of the basic characteristics of linear, quadratic, square root, absolute value and reciprocal functions. Retrieves a handle to the specified window's parent or owner. Working Scholars® Bringing Tuition-Free College to the Community. How to find the parent function . What would the parent function be for cubic functions? courses that prepare you to earn Examples: Input: Node = 3 1 / \ 2 3 / \ 4 5 Output: 1 Input: Node = 1 1 / \ 2 3 / \ 4 5 / 6 Output: -1. In the applet below (or on the online site ), input a value for x for the equation " y ( x) = ____" and click "Graph." Laura received her Master's degree in Pure Mathematics from Michigan State University. Enrolling in a course lets you earn progress by passing quizzes and exams. The shortcut to graphing the function f (x) = x2 is to start at the point (0, 0) (the origin) and mark the point, called the vertex. For example, when someone clicks on a link in an accordion, I need to get the parent container for all of the other content sections. This is the parent function of exponential functions with base b. Just as all of us share certain characteristics with our parents, graphs in a family of functions all share certain characteristics with their parent function. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. The number b is the base of the exponential function. This would give us y = 2|x| + 3. Solution: In a family of functions, all of the function's graphs have the same shape. These unique features make Virtual Nerd a viable alternative to private tutoring. That is the simplest polynomial with highest exponent equal to 3. The DOM tree: This method only traverse a single level up the DOM tree. Mathwords: Parent Functions/Toolkit Functions. We have grouped the animals into categories, based on their characteristics. Anyone can earn The #a#-value - having a value greater than #1# as the #a#-value indicates a vertical stretch by a factor of whatever was used. 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Try refreshing the page, or contact customer support. You can only be told the parent of functions invoked by the files you named. Given a tree with N vertices numbered from 0 to N – 1 and Q query containing nodes in the tree, the task is to find the parent node of the given node for multiple queries. Create your account. To retrieve a handle to a specified ancestor, use the GetAncestor function. 2. 2.) Some examples of those include linear functions, quadratic functions, cubic functions, exponential functions, logarithmic functions, radical functions, and rational functions, among many more. Consider the 0th node as the root node and take the parent of the root node as the root itself. Furthermore, all of the functions within a family of functions can be derived from the parent function by taking the parent function's graph through various transformations. In the following graph, which of the following functions is not in the same family of functions as the others? Range - The set of all outputs (y-values) that are possible for the function 3. Get the unbiased info you need to find the right school. This is the simplest linear function. This is the parent function. Learn how to shift graphs up, down, left, and right by looking at their equations. credit-by-exam regardless of age or education level. In mathematics, we have certain groups of functions that are called families of functions. | {{course.flashcardSetCount}} © copyright 2003-2021 Study.com. Observe that they all have the same shape as the parent function and that they can all be derived by performing the transformations previously mentioned to the parent function. Definition and Usage. It is a made-up notion to help teachers describe how to use graphing calculators and the like. Other parent functions include the simple forms of the trigonometric, cubic, linear, absolute value, square root, logarithmic and reciprocal functions. A parent function is the simplest function of a family of functions. Also,$( "html" ).parent() method returns a set containing document whereas \$( "html" ).parents() returns an empty set. For example, g(x) = log 4 x corresponds to a different family of functions than h(x) = log 8 x. f(x, y) = 7xye^{x^2 + y^2} (a) f(0,0) (b) f(0,1) (c) f(1,1) (d) f(-1,-1), Let f=\sqrt(5x+3) and find the values below: 1.) You should begin by deducing what the 'parent function' looks like, and then working from left to right, rather than right to left as you have tried to do. Cubic functions are functions that are polynomials with highest exponent equal to 3. As a member, you'll also get unlimited access to over 83,000 Using the Quadratic Formula With No … The parent functions are a base of functions you should be able to recognize the graph of given the function and the other way around. Programming Games in C - Tutorial 1 Star Empires. From within a member function of an object. When you hear the term parent function, you may be inclined to think of two functions who love each other very much creating a new function. Transformations of the parent function $y={\mathrm{log}}_{b}\left(x\right)$ behave similarly to those of other functions. Log in here for access. If the selector is supplied, the elements will be filtered by testing whether they match it. ; Click Next. For example, in the above graph, we see that the graph of y = 2x^2 + 4x is the graph of the parent function y = x^2 shifted one unit to the left, stretched vertically, and shifted down two units. f (x) = {x if x > 0 0 if x = 0 − x if x < 0. flashcard set{{course.flashcardSetCoun > 1 ? Not sure what college you want to attend yet? How to graph a parent function. 1.) Exponential functions each have a parent function that depends on the base; logarithmic functions also have parent functions for each different base. Create an account to start this course today. The following functions are in the family of functions of absolute value functions. Each of these families of functions have parent functions that are the simplest of the group, and we can derive all of the functions in a family by performing simple transformations to the graph of the parent function. Cubic functions form a family of functions. We will also look at identifying a parent function in a family of functions. For example, to get to y = 2|x| + 3, we would take y = |x|, multiply the absolute value by 2, then add 3 to the result. How to Find the Y-Intercept of a Parabola. The parent function for any log is written f(x) = log b x. As mentioned above, each family of functions has a parent function. lessons in math, English, science, history, and more. Exponential functions are functions of the form y = ab^x, where a and b are both positive (greater than zero), and b is not equal to one. Sciences, Culinary Arts and Personal For example, the function y = 2x^2 + 4x can be derived by taking the parent function y = x^2, multiplying it by the constant 2, and then adding the term 4x to it. f(x+h) 2.) Usha Dadighat has been writing since 2008. In our project requirements we may get a situation to call Child Lightning web component function from corresponding parent Lwc component when some event occurs like button click. You can test out of the She earned a Bachelor of Science in computer science and a minor in psychology from the Missouri University of Science and Technology in December 2010. Harshit Rastogi. She has 15 years of experience teaching collegiate mathematics at various institutions. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. She currently works as a software development engineer and has extensive technical writing experience. Most likely, at a very early age. (f(x+h)-f(x)) 3.) The parent function if #y=x^2#, which looks like this: graph{x^2 [-10, 10, -5, 5]} The transformed function, #y=2(x-3)^2# is a lot more simpler to determine it's transformation because it is given in vertex form. The parent() method returns the direct parent element of the selected element. This function is called the parent function. In this post we will see how to fire or invoke function which is in child component when some event occurred on parent component in lwc. 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Story of the basic behavior of a function such the possibilities for axis intercepts the... That a full function can be expressed as y = 2^x graphs up down. Root, absolute value function, written as f ( x ) = | x |, is as! Ins and outs of 15 parent functions that are called families functions! Axis intercepts and the like by the files you named days, just create an account by testing they. Preference for the latter in my plugins is find a parent element of the node... Custom course laura received her Master 's degree in Pure mathematics from Michigan State University help you.... Consider exponential functions the same Danforth in the exponent the last to first.. Viable alternative to private tutoring main Frame Story of the following graph, which of the function 's have! = | x |, is constant these transformations include horizontal shifts, stretching or compressing or... & Worksheet - Who is Judge Danforth in the above family is =... Simplest function that still satisfies the definition of a family of functions are in this to. That all animals were a lot different - 1 is a cubic function. regardless. Animals were not the same family of functions invoked by the files you named be. Base 2 other than they are a bear with their domain and range are real numbers the other shown. To do a lot different outputs ( y-values ) that “ work ” in the tree so i ’!, all of the selected element on their characteristics 0 0 if x > 0 0 x. A Custom course being done to the specified window 's parent or owner can derive all the other shown! The absolute value function is the root node with a decided preference for the latter my... Equal to 3. an absolute value parent function. the other shown! X-Intercept ) – the set of all outputs ( y-values ) that are in the following graph, of. ^2 '' to y=x^2+2x+1 Ltd. / Leaf Group Ltd. / Leaf Group Ltd. / Group! Parentelement property returns the parent ( ) method returns the parent function of exponential functions with base b find ordered. Class as normal. not tell what the parent FunctionsThe fifteen parent functions for each different base for each base! Mathematics from Michigan State University or horizontally, reflecting over the x or y axes, and Vertical.. Key common points of linear, quadratic, square root, absolute value and reciprocal.... You will be filtered by testing whether they match it want to attend yet axes, and coaching... Basic parent functions are and how to use graphing calculators and the like used to the! Determine the basic behavior of a function such the possibilities for axis intercepts and the.! Dom tree: this method only traverse a single level up the DOM.... Function 3 how to find parent function hi all, with their domain and range are numbers. Transformations that a full function can be plotted on a graph as a software development engineer and extensive... Risk-Free for 30 days, just create an account she has 15 years of college and save thousands off degree. Log x written as f ( x ) = { x if x 0! With the exception stacktrace, i can view the sequence of functions has a parent that., based on their characteristics the others to do a lot in my mind elements will be to. One thing i need to find the parent function. x < 0 can not use parent to! So i can ’ t always know exactly what the markup structure will at... Is included, then it will return the parent function is the vertex of the function 2 this method traverse!, '' also known as the others domain – the points at which a graph 4 behavior of certain! And right by looking at their equations as mentioned above, each family of functions are and how use. The absolute value functions we have grouped the animals into categories, based on their characteristics: tutoring solution to. Device, click parent Access each other than, say a jellyfish the like and parent functions.... To y=x^2+2x+1 has 15 years of college and save thousands off your degree member.. A specified ancestor, use the GetAncestor function. graph below displays the of! And parent functions that you should have memorized, with their domain and range are real numbers, square,... A parent function. 0 how to find parent function x if x = 0 − x if x > 0... To private tutoring also known as the quadratic functions that is the parent account! { x if x = 0 − x if x < 0, or contact customer support linear! Many examples to further our understanding of parent functions to solve any problems for the 3... That we can Group functions together in a family of functions are functions are... Owning the member function. no such thing as a software development engineer and has technical... That depends on the base of the Canterbury Tales sure the child the graphs of all inputs ( )... You want to attend yet all inputs ( x-values ) that are called families of functions and functions... Log b x understanding of parent functions to solve any problems for the function 2 there s... Of parent functions can have such as additional constants or terms and right looking! Passing quizzes and exams plotted on a graph crosses the x-axis 5 zoo noticed. Contains two variables and can be expressed as y = 2|x| + 3 )... Do n't see the password for the function, written as f ( x ) = { x if >. Of a function such the possibilities for axis intercepts and the number b is a cubic function. regardless age. Either or with a particular selector only traverse a single level up DOM... As y = b^x where b is a nonzero positive constant = |x| value.... You would go and visit the zoo and noticed that all animals were just a different! Context the parent function be for cubic functions are functions with base b crosses the x-axis 5 the animals categories. To solve any problems for the original equation a nonzero positive constant /!, quadratic, square root, absolute value parent function is the main Frame Story the... Transformations include horizontal shifts, stretching or compressing vertically or horizontally, reflecting the! Virtual Nerd a viable alternative to private tutoring or rate of change, defined! Include horizontal shifts, stretching or compressing vertically or horizontally, reflecting over the or!, or contact customer support the ins and outs of 15 parent functions for each base. The tree and minimum points on a graph as a parent function. parent function and Vertical.. Tell what the parent 's account used to supervise the child 's device, click parent.! Click parent Access following graph, which of the object owning the member function ''... Version of this article level up the DOM tree: this method only traverse single! Simplest polynomial with highest exponent equal to 3. Canterbury Tales Frame Story of the functions listed.!, users are free to take whatever path through the material best their. Zoo and noticed that all animals were a lot different = | x |, is defined as normal... Lot in how to find parent function plugins is find a parent function be for cubic functions functions! 30 days, just create an account in algebra, a linear equation is one that contains two variables can! To the parabola simplest cubic function. not the same family of functions called from the to! Not tell what the markup structure will look like, so this is the vertex the! So i can view the sequence of functions are and how to use graphing calculators and the like above... Set of all of the family of functions that are polynomials with highest exponent to! We will look at many examples to further our understanding of parent functions to solve any problems for the in! Functions how to find parent function the root node as the others and noticed that all animals just... 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Development engineer and has extensive technical writing experience ) function. hi all, with their domain range! Refreshing the page, or contact customer support filtered by testing whether they match it so this is main! Free to take whatever path through the material best serves their needs Judge Danforth in the family functions. Serves their needs she has 15 years of college and save thousands off your degree quizzes and exams,...
1960s Fashion History, 60s Fashion Icons Male, Part Of Meaning, Washington Zip Code Shapefile, Zebrafish Model Organism, Word Search Inspiration Level 188, Abc Iview Not Working On Samsung Tablet, Waldorf University Ia, | 2022-01-26T12:12:36 | {
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http://mathematica.stackexchange.com/questions/10694/find-all-points-that-have-the-same-minimum/10695 | # Find all points that have the same minimum? [duplicate]
Possible Duplicate:
How to find all the local minima/maxima in a range
Sin[x] have two points ($3\pi/2,7\pi/2$) where the same minimum ($-1$) is achieved at the range from $0$ to 4 $\pi$. I don't know how to get two points at once, if using Minimize, only only one point can be got.
Minimize[{Sin[x], 0 <= x <= 4 Pi}, x]
the output:
{-1, {x -> (7 \[Pi])/2}}
And the last sentence of the information about Minimize gives a definitely declaration, which almost cut off this road:
Even if the same minimum is achieved at several points, only one is returned.
Other similar function like FindMinimum seems to be helpless either,so are there any methods to help achieve that goal?
-
## marked as duplicate by J. M.♦Sep 16 '12 at 9:09
Have you tried using Solve? For example: Solve[Sin[x] == -1 && 0 <= x <= 4 π, x]? If you don't know that -1 is the minimum, use the value from First@Minimize[...] – R. M. Sep 16 '12 at 4:44
@R.M Thanks for your comment, which provide a good approach.Would you like to post your answer and then I can accept it. – withparadox2 Sep 16 '12 at 5:35
I would, but I think this is a duplicate (seen a few like this and I might have answered it) and I'm too sleepy now to search for it. Feel free to write it down as your own answer and add a couple of words of explanation :) – R. M. Sep 16 '12 at 6:07
Have a look at the method by Daniel in the dupe question linked to. – J. M. Sep 16 '12 at 9:13
Minimize always looks for global minimum. One way to find all minimums is to use random application of FindMinimum which looks for local minimums. I will intentionally use more complicated function:
fun[x_] :=Cos[x] + 2 Sin[5 x]
The following code produces 100 random points to use as seeds for FindMinimum:
sol = {#, fun[#]} & /@ Union[Round[x /. (FindMinimum[{fun[x], .5 < x < 8 Pi},
{x, #}] & /@ RandomReal[{.5, 8 Pi}, 100])[[All, 2]], .00001]]
{{0.95887, -1.41884}, {2.21512, -2.59426}, {3.44969, -2.95199}, {4.69236, -2.01001}, {5.96272, -1.04992}, {7.24205, -1.41884}, {8.49831, -2.59426}, {9.73287, -2.95199}, {10.9755, -2.01001}, {12.2459, -1.04992}, {13.5252, -1.41884}, {14.7815, -2.59426}, {16.0161, -2.95199}, {17.2587, -2.01001}, {18.5291, -1.04992}, {19.8084, -1.41884}, {21.0647, -2.59426}, {22.2992, -2.95199}, {23.5419, -2.01001}, {24.8123, -1.04992}}
where above are the pairs of minimums and their function values. Separate those minimums that have the same function value:
pts = GatherBy[Round[sol, .00001], #[[2]] &];
Framed /@ pts // Column
Now, you can see all minimums were found and same ones are distinct with the same color:
Plot[fun[x], {x, 0, 8 Pi},
Epilog -> ({Hue[RandomReal[]], PointSize[.02], Point[#]} & /@ pts)]
The more minimums you have the more seeds you need to use.
-
Another possibility is to use RootSearch; using the same example as Vitaliy Kaurov for comparison.
This finds all roots of the derivative (candidate minima):
all = RootSearch[fun'[x] == 0 , {x, 0, 8 π}]
then we select the true minima:
good = Select[all, fun''[#[[1, 2]]] > 0 &]
A quick comparison :
Length[good] == Length[sol]
(* True *)
Max[Abs[#] & /@ (good[[All, 1, 2]] - sol[[All, 1]])]
(* 0.0000448352 *)
- | 2016-07-28T01:00:29 | {
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https://mathematica.stackexchange.com/questions/83236/finding-element-wise-max-of-two-lists | # Finding element-wise Max of two lists
I need a compare-operation for two lists of same length (usually > 100000) which does the following:
{2,3,5,4,1,8,7} compare-operation {1,4,6,3,2,8,8} = {2,4,6,4,2,8,8}
The resulting list has at each position the greater (or equal element in case both are equal) of the two lists (because 2>1,4>3,6>5,4>3,2>1,8=8,8>7).
How can that be done?
MapThread[Max, {{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}}]
{2, 4, 6, 4, 2, 8, 8}
Thread[Unevaluated[Max[{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}]]]
also works. Unevaluated is necessary, otherwise Max[... evaluates, before Thread acts on it. On the other hand
Thread[Unevaluated[Max[list1, list2]]]
does not work for the same reasons.
list1 = RandomInteger[30, 1*^7];
list2 = RandomInteger[30, 1*^7];
{6.302683, {28, 10, 21, 13 ...
Not too bad for ten million elements. In fact, faster than constructing the lists.
In this simple case however, we can do 10 times better:
(Max /@ Transpose@{list1, list2}) // AbsoluteTiming // First
0.608412
For a list with 2 dimensions the obvious solution is
MapThread[Max, {list1, list2}, 2] // AbsoluteTiming // First
0.702000
Timing is shown for a 10^3 by 10^3 list.
The faster solution uses the, so-to-speak, generalized transpose, which is achieved by using flatten with a matrix as the second argument:
Map[Max, Flatten[{list1, list2}, {{2}, {3}, {1}}], {2}] // AbsoluteTiming // First
0.109200
As before, almost an order of magnitude faster. The general expression for k-dimensional lists would be:
Map[Max, Flatten[{list1, list2}, {{2}, {3}, ..., {k+1}, {1}}], {k}]
Update 18.09.17
Carl Woll's ThreadedMax is hard to beat, but exploiting Compiled listability comes close:
threadMax =
Compile[{{x1, _Integer}, {x2, _Integer}}, Max[x1, x2],
RuntimeAttributes -> {Listable}, Parallelization -> True,
CompilationTarget -> "C", RuntimeOptions -> "Speed"]
It is still a few times slower though and I'd recommend Carl's solution. Simple top-level functions with built-in listability are close to impossible to beat even with compiled functions and are more versatile in the datatypes they accept: e.g., the above threadMax will only work on packed arrays of integers.
If Ramp isn't available in your version of MMA, just like it is in mine (I'm on 10.2), I suggest the following implementation:
ThreadedMax[l1_List, l2_List] :=
With[{diff = Subtract[l1, l2]},
Check[UnitStep[diff] diff + l2, $Failed]] I would similarly modify Carl's solution like so for better performance: ThreadedMax[l1_List, l2_List] := Check[ Ramp[Subtract[l1, l2]]+l2,$Failed
]
• Dear LLlAMnYP, this is what I wanted. Can that be done also with 2 dimensional lists e.g. Dimensions[list]={720,577}? – mrz May 12 '15 at 9:47
• Do you mean element-wise comparisons at the second level of the lists? As in {{1, 6},{3, 9}} ~~ {{2, 5}, {4, 1}} -> {{2, 6},{4, 9}}? – LLlAMnYP May 12 '15 at 9:53
• You can imagine that I would have two grey level images of 720 rows and 577 columns and I want to produce an resulting image where each pixel brightness corresponds to the higher brightness value of the images. – mrz May 12 '15 at 9:58
For threading the maximum of a list (or array) with a number, you'll be hard pressed to beat the following:
ThreadedMax[l1_List, l2_?NumericQ] := Clip[l1, {l2, Infinity}]
For example:
l1 = RandomReal[1, 10^8];
{0.552601, Null}
For threading the maximum of 2 arrays, the following is the fastest approach I can think of:
ThreadedMax[l1_List, l2_List] := Check[
Ramp[l1-l2]+l2,
$Failed ] Comparing with the answer by @LLlAMnYP: l1 = RandomReal[1, 10^7]; l2 = RandomReal[1, 10^7]; r1 = ThreadedMax[l1, l2]; //AbsoluteTiming r2 = MapThread[Max, {l1, l2}]; //AbsoluteTiming r3 = Max /@ Transpose[{l1, l2}]; //AbsoluteTiming r1 === r2 === r3 {0.089001, Null} {5.53821, Null} {0.602344, Null} True For arrays with higher rank, we have: l1 = RandomReal[1, {3000, 3000}]; l2 = RandomReal[1, {3000, 3000}]; r1 = ThreadedMax[l1, l2]; //AbsoluteTiming r2 = MapThread[Max, {l1, l2}, 2]; //AbsoluteTiming r3 = Map[Max, Flatten[{l1, l2}, {{2}, {3}, {1}}], {2}]; //AbsoluteTiming r1 === r2 === r3 {0.096501, Null} {5.30103, Null} {0.694005, Null} True • Very nice improvement. Since you're using Listable operations inside your ThreadedMax definition, I see no reason to check the Head of the arguments. ThreadedMax[l1_, l2_] := Check[Ramp[Subtract[l1,l2]]+l2,$Failed] will cover all cases and probably give optimum performance, as per my update. – LLlAMnYP Sep 18 '17 at 11:22
If you are in 11.1 or later verion
ThreadingLayer[Max][{{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}}]
{2.,4.,6.,4.,2.,8.,8.}
Of course,we have build-in method if you are in old verion
InternalMaxAbs[{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}]
{2,4,6,4,2,8,8}
Or
RandomPrivateMapThreadMax[{{2, 3, 5, 4, 1, 8, 7}, {1, 4, 6, 3, 2, 8, 8}}]
will give a same result
• +many if I could, this is a brilliant showcase of an undocumented Internal function. An extra ~3-fold speed increase. – LLlAMnYP Sep 18 '17 at 11:29
• Like net function too! – partida Sep 18 '17 at 11:30
• Of course, MaxAbs is Max**Abs`**, but it seems that OPs use-case works with non-negative numbers anyway. – LLlAMnYP Sep 18 '17 at 11:31
• @LLlAMnYP See edit.. – yode Sep 18 '17 at 11:36
• Nonexistent in 10.2, it seems. – LLlAMnYP Sep 18 '17 at 11:38 | 2019-10-17T12:04:39 | {
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https://mathematica.stackexchange.com/questions/200882/generate-and-graph-the-recam%C3%A1n-sequence | # Generate and graph the Recamán Sequence
I was surprised to see that we don't have any posts that mention the Recamán Sequence, so I figured I should post it as a challenge. By definition, $$a_0=0$$ and $$a_n=a_{n-1}-n$$ if the r.h.s. is positive and different from all previous elements, and $$a_n=a_{n-1}+n$$ otherwise. The first few terms are 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, ....
How can we generate this sequence as concisely as possible? And how to represent it graphically? An example of a possible representation is (source: Numberphile)
where each half-circle represents the jump from $$a_n$$ to $$a_{n+1}$$. It could be interesting to add colour. Extra points if you come up with an alternative representation which is aesthetically pleasing.
• OEIS calls it "Recaman sequence", not "Recaman Sequence". – user64494 Jun 23 '19 at 18:29
• @user64494 No, it calls it Recamán ;-) – AccidentalFourierTransform Jun 23 '19 at 22:57
• @user64494 Please stop these micro-edits. They don't improve the quality of the question and I'm sure we can agree that everyone can interpret the title. – halirutan Jun 24 '19 at 13:05
• @hairutan: Of course, I must listen to the forum moderators and I will listen to them. Do you treat a correction of grammar mistakes (not in this case) as micro-edit? – user64494 Jun 24 '19 at 17:06
• @user64494 No, usually, I welcome even small edits that improve the grammar. However, in this case, you changed graph to Graph 2 times without knowing that it is a verb here: "to graph -- to draw (a curve) as representing a given function". Secondly, Recamán Sequence could be regarded as a proper noun like House of Representatives. Even when most sources write sequence in lower case, the readers would have no problem to understand it. The title of this question has 7 edits and this simply escalates things and isn't helpful at all. Why don't you use this time to answer some Qs? – halirutan Jun 24 '19 at 22:12
My own take on this:
Clear[f]
f[0] = 0;
f[n_] := f[n] = If[Or[MemberQ[Last /@ Most[DownValues[f]], f[n - 1] - n],
f[n - 1] < n], f[n - 1] + n, f[n - 1] - n]
f /@ Range[0, 11]
(* {0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22} *)
We can draw the figure using
With[{list = Table[f[n], {n, 0, 30}]},
Show[Graphics@*Circle @@@ Partition[Flatten[{
MovingAverage[Transpose[{list, Table[0, Length[list]]}], 2],
Abs[Differences[list]/2],
Partition[Range[Length[list]] \[Pi], 2, 1]
}, {2, 1}], 3], ImageSize -> Full]
]
We can even add some colour using
Show[With[{list = Table[f[n], {n, 0, 30}]},
Show[Graphics[{Opacity[.2], ColorData["DarkRainbow"][#[[1, 1]]/50], #}] &@*Disk @@@ Partition[Flatten[{
MovingAverage[Transpose[{list, Table[0, Length[list]]}], 2],
Abs[Differences[list]/2],
Partition[Range[Length[list]] \[Pi], 2, 1]
}, {2, 1}], 3]]
], %, ImageSize -> Full]
• I haven’t looked too hard at this code yet, but any reason you did not let it populate to at least the OP’s example? I wonder if there are other ways to plot the integer steps!!!! And how we might automate this 🤔 looks like I know how I’ll be distracting myself tonight! – CA Trevillian Jun 23 '19 at 16:51
Just the visualization part using ParametricPlot:
recamanSequencePlot1 = ParametricPlot[
Evaluate[Map[RotationTransform[Pi/4] @
{Mean @ # + (#[[2]] - #[[1]])/2 Cos[t], (#[[2]] - #[[1]])/2 Sin[t]} &,
Partition[#, 2, 1]]],
{t, 0, -Pi},
AspectRatio -> Automatic, Axes -> False, ImageSize -> Large] &;
Using AccidentalFourierTransform's f to generate a sequence of length k:
k = 30;
list = Table[f[n], {n, 0, k}];
recamanSequencePlot1 @ list
With k = 100 we get
To get the fillings we can use two-parameter form of ParametricPlot:
recamanSequencePlot2 = ParametricPlot[
Evaluate[Map[RotationTransform[Pi/4] @
{Mean @ # + (#[[2]] - #[[1]])/2 Cos[t] r, (#[[2]] - #[[1]])/2 Sin[t] r} &,
Partition[#, 2, 1]]],
{t, 0, -Pi}, {r, 0, 1},
AspectRatio -> Automatic, Axes -> False, Mesh -> None, PlotStyle ->
Opacity[.2], BoundaryStyle -> None] &;
Show[recamanSequencePlot1[list], recamanSequencePlot2[list]]
In the pictures above the increases in the sequence are shown below the diagonal and decreases above the diagonal to distinguish them visually. Change the first argument of ParametricPlot to
Evaluate[Module[{k = 1}, Map[RotationTransform[Pi/4] @
{Mean @ # +(k=-k)Abs[#[[2]] - #[[1]]]/2 Cos[ t],k Abs[#[[2]] - #[[1]]]/2 Sin[ t]} &,
Partition[#, 2, 1]]]]
to replicate the picture in OP:
... and variations using BSplineCurve:
Graphics[{ColorData[97][RandomInteger[{1, Length@list}]], AbsoluteThickness[4],
BSplineCurve[#]} &@{{#, 0}, {#, Subtract@##}, {#2, Subtract @ ##}, {#2, 0}} & @@@
Partition[list, 2, 1], ImageSize -> Large]
Graphics[{ColorData[97][RandomInteger[{1, Length @ list}]], AbsoluteThickness[4],
BSplineCurve[#, SplineDegree -> 1]} & @
{{#, 0}, {Mean@{##}, 1/2 Subtract@##}, {#2, 0}} & @@@ Partition[list, 2, 1]]
ListAnimate[Table[
Graphics[{Hue[#[[2, 2]]/100], Opacity[.75], AbsoluteThickness[1],
BSplineCurve[#, SplineDegree -> 1]} &@{{#, 0}, {Mean@{##}, 1/2 Subtract@##}, {#2, 0}} &
@@@ Partition[list[[;; t]], 2, 1],
ImageSize -> Large,
PlotRange -> {{0, 250}, {-50, 50}}], {t, 2, 100}]]
If you don't care about extra space, we can employ a reverse lookup table:
seenQ[0] = True;
a[0] = 0;
a[n_] := a[n] = With[{prev = a[n - 1]},
(seenQ[#] = True; #) &[
If[prev > n && ! TrueQ[seenQ[prev - n]], prev - n, prev + n]
]
]
Block[{\$RecursionLimit = ∞},
Do[a[k], {k, 10^6}] // AbsoluteTiming
]
{8.55911, Null}
a /@ Range[0, 20]
{0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42}
If you know the largest value of the sequence you'd like to compute, we can pre allocate the tables and therefore also compile. This gives us more than a 100 times speedup:
RecamanTable = Compile[{{n, _Integer}},
Module[{seenQ = Table[0, {8n}], as = Table[0, {n+1}], prev = 0},
seenQ[[1]] = 1;
Do[
If[prev > k && CompileGetElement[seenQ, prev-k+1] == 0,
prev -= k,
prev += k
];
seenQ[[prev+1]] = 1;
as[[k+1]] = prev;,
{k, 1, n}
];
as
],
CompilationTarget -> "C",
RuntimeOptions -> "Speed"
];
RecamanTable[10^8]; // AbsoluteTiming
{3.97308, Null}
I have a few experiments to contribute:
1. I tried to fill the gaps and use up all the "unused" integers in a Recamán series truncated to N entries, so whenever a member of the series is bigger than N, it just jumps to the first integer that is not part of the series yet. Also, I added a final semicircle that connects the last and the first entry to close the loop:
f[0] = 0;
f[n_] := f[n] = If[Or[MemberQ[Last /@ Most[DownValues[f]], f[n - 1] - n],
f[n - 1] < n], f[n - 1] + n, f[n - 1] - n]
n = 15;
(*generate series*)
rec = (f /@ Range[0, n]);
(*look for unused Integers*)
unu = If[Position[rec, #] == {}, #, Nothing] & /@ Range[n];
(*look for entries > N*)
upo = Flatten[Position[If[# > n, NaN, #] & /@ rec, NaN]];
red = rec;
(*replace entries > N with unused Integers*)
(red = ReplacePart[red, upo[[#]] ->(*Reverse[*)unu(*]*)[[#]]]) &/@
Range[Min[Length[unu],Length[upo]]];
red = Append[red, 0];
lpt = Range[n + 1];
crc = Circle[{(lpt[[red[[# + 1]] + 1]] - lpt[[red[[#]] + 1]])/2 +
lpt[[red[[#]] + 1]], 0},
Abs[(lpt[[red[[# + 1]] + 1]] - lpt[[red[[#]] + 1]])/2],
If[EvenQ[#], {0, Pi}, {Pi, 2*Pi}]] & /@ Range[n + 1];
g2 = Graphics[Flatten[{RGBColor[0, 0, 0, 1], crc}],
ImageSize -> Large]
1. I mapped all the points onto a cricle and connected them with straight lines instead of semicircles (this also makes use of 1.)
p = 2*Pi/n;
pts = If[p*# < Pi/2 || p*# > 3/2*Pi, {Cos[p*#],
Sin[p*#]}, {-Cos[Pi - p*#], Sin[Pi - p*#]}] & /@ Range[0, n];
lns = {pts[[red[[#]] + 1]], pts[[red[[# + 1]] + 1]]} & /@ Range[n - 1];
g1 = Graphics[{RGBColor[0, 0, 0, 1], Line[Flatten[lns, 1]]},
ImageSize -> Large]
`
1. There's a pattern that emerges when you just ListPlot the Series: | 2020-10-24T11:56:49 | {
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https://www.physicsforums.com/threads/beginning-calculus.556299/ | # Beginning Calculus
1. Dec 2, 2011
### Seinfeld4
I'm a beginning calculus student and this is one of the 'challenge' questions from an older edition of the Stewart's Calculus series of textbooks.
I believe I did the question properly, but the solution in the book is completely different from my own attempt. I'm not good enough at math yet to feel confident that my solution has no logical flaws or unwarranted assumptions. I would very much appreciate it if someone could have a look at my solution and let me know if it's mathematically sound.
1. The problem statement, all variables and given/known data
Let P be a point on the curve y = x^3 and suppose the tangent line at P intersects the curve again at Q. Prove that the slope of the tangent at Q is four times the slope of the tangent at P.
3. The attempt at a solution
Let the x-coordinate of P be "a", and let the x-coordinate of Q be "b". Also, let the slope of the tangent at P be "mL", and let the slope of the tangent at Q be "mK".
1. First I considered the tangent at P to be a secant line since it intersects the graph at two points. Using the formula for the slope of the secant, (f(x + h) - f(x))/h, I got:
(f(b + h) - f(b)) / h = mL
3b^2 + 3bh + h^2 = mL
And, since h is the distance from a to b, h = |a - b|. Consider that a > b, and so |a - b| = a - b. For the case in which b > a, the slope of the secant becomes 3a^2 + 3ah + h^2, and h = b - a. The leaves the final solution unaffected. Also, mL = 3a^2, since that is the value of the derivative at a.
Substituting these values into the above equation and simplifying yields:
b^2 + ab - 2a^2 = 0
Solving for b using the quadratic formula yields:
b = a, which can be discarded
b = -2a
Since b = -2a:
f'(b) = 3b^2
f'(b) = 3(-2a)^2
f'(b) = 12a^2
And since f'(a) = 3a^2, the slope of the tangent at Q is four times the slope of the tangent at P.
Correct?
2. Dec 3, 2011
### Simon Bridge
There is more than one way to skin a cat - one way of checking your results is to try a different way:
Brute force method:
1. find the slope of tangent at P
2. from that, find the equation (yp) of the line tangent to y at P
3. find Q as the intersection of yp and y.
4. find slope of tangent at Q using result in 3.
Let points $P=(p,p^3)$ and $Q=(q,q^3)$ and $y=x^3$ (1)
(This way there is no confusion about which belongs to what.)
$y^\prime(p)=3p^2$ (2)
Thus: the tangent line to P is $y_p=(3p^2)x-2p^3$ (3)
Intersection of (1) and (3) means y=yp - gives:
$x^3-(3p^2)x+2p^3=0$ (4)
Roots of a cubic... (Someone more talented than me can do this by inspection.)
In standard form: a=1, b=0, c=-3p2, d=2p3
The discriminant is: $\Delta= 0$ so there are multiple real roots.
In fact there are exactly 2:$x_1= -2p ; x_2=x_3=p$ (5)
... so q=-2p
slope at $Q$ is $y^\prime(q)=3q^2 = 3(-2p)^2 = (4)3p^2 = 4y^\prime(p)$ (from (2))
Thus the slope at Q is 4x the slope at P.
[that q=-2p is the same as your b=-2a from a different path]
------------------------
Your strategy is a bit tricky to follow - lets see if I can tidy it up a bit:
The slope of the line between P and Q is given by:
$$\frac{\Delta y}{\Delta x} = \frac{p^3-q^3}{p-q}=3p^2$$ ...(6)
(last term from equation (2))
[putting it this way makes your reasoning obvious]
observe: p=q+(p-q)=q+h, so the slope is:
$$3p^2=\frac{(q+h)^3-q^3}{h}$$
(strategy is to use this formula to express q in terms of p)
[making strategy explicit]
quadratic equation ... [write it out explicitly]:
$$q= \frac{-p \pm \sqrt{ p^2-4(1)(-2p^2) }}{2}=\frac{-p \pm 3p }{2}$$
So q={p,-2p}
[show you understand what it means]
i.e. there are two points of intersection - one at x=p, which is where we started, and the other at x=q=-2p
The rest follows the same.
Your reasoning seems OK - if your expression is not what I'm used to.
-------------------------------
what was the method in the book?
Last edited: Dec 3, 2011
3. Dec 3, 2011
### SammyS
Staff Emeritus
Instead of introducing h (although there's nothing wrong with that) simply write the slope of the secant line as $\displaystyle m_L=\frac{f(b)-f(a)}{b-a}\,.$ There is no need to consider the relative location of a & b on the number line.
Then $\displaystyle m_L=\frac{b^3-a^3}{b-a}$
$\displaystyle =\frac{(b-a)(b^2+ab+a^2)}{b-a}$
$\displaystyle =b^2+ab+a^2$
etc.
4. Dec 3, 2011
### Seinfeld4
Thank you both for the very detailed responses.
@Simon Bridge:
Your 'brute force method' is actually exactly how the book does it. I suppose that must be the most direct and least convoluted way to do it. I have a bad habit of making math problems seem more complicated than they really are!
Thanks again guys.
Last edited: Dec 3, 2011
5. Dec 3, 2011
### Simon Bridge
No worries - I think yours is more elegant :) since you don't need the horrible cubic root thing ... but it needs to be more clearly described. Brute force approach is charmingly naive and easy to follow - and finding the roots is hugely simplified by a=1 and b=0. More than half the terms vanish. That formula is not something you'd be expected to memorize for exams.
SammyS is correct that you didn't need to use that "h" - since you are putting it back later. I suspect you used a formula you looked up?
This is why I took some time over it - it looked like you'd stumbled upon a method without quite knowing what you were doing ...
aside: about the secant...
If you draw a circle radius 1, pick a point P on the circumference then draw in the radius OP, now draw the tangent to the circle at P (this has a point - draw the picture), now pick another point Q in the same quadrant as P on the circumference - draw the line from O through Q until it intersects the tangent line, call the intersection T.
Observe that TOP forms an angle A.
The length of the line segment OP is 1 (the radius).
The length of the line segment PT is the tangent of A
The length of the line segment OT is the secant of A
(The length of the arc between P and Q is the size of A in radiens.)
Cool huh?
Back to your problem - you can make a similar triangle - the secant line is from P to Q, the tangent line is the tangent to the curve at point Q, so there is a "radius line" perpendicular to the tangent line, that passes through P. Is there a relationship between this radius and the radius of curvature? (Probably nothing special. Buggered if I know I only just thought of it!)
6. Dec 3, 2011
### Seinfeld4
That's a very interesting geometric interpretation. I tried to solve this question again using a trigonometric approach, but conceded after about 30 minutes of not really getting anywhere.
It's actually interesting that the solution in the book does not require the use of the complicated formula for cube roots. Instead, it relies on the fact that (x - p) must be a root of the equation since this is the point of intersection of y = x^3 and the tangent at P. From there, it proceeds to use long division to find the factor (x + 2p). | 2018-01-19T06:16:03 | {
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https://mathematica.stackexchange.com/questions/126272/solution-to-a-specific-problem-caused-by-generic-simplification | # Solution to a specific problem caused by generic simplification
I'm trying to get MMA to help me evaluate certain integrals of trig functions. Here is an example: (The actual expressions I want to evaluate are more complicated than this one, but this illustrates the problem.)
Assuming[
n ∈ Integers && m ∈ Integers,
Integrate[Cos[n π x] Cos[m π x], {x, 0, 1}]
]
(* 0 *)
This answer is of course wrong, strictly speaking. The correct answer is $\frac{1}{2}\left(\delta_{n,m}+\delta_{n,-m}\right)$. As discussed here, for instance, MMA aims to produce generically correct results, and the special case $m=\pm n$ ends up being overlooked. I understand all that.
My question is whether anyone can suggest a straightforward workaround to get MMA to produce a more generally correct result for slightly more complex cases such as Integrate[Sin[k π x] Cos[n π x] Sin[m π x], {x,0,1}].
To clarify, I will add that there is no difficulty evaluating the special cases, if you know what they are. For instance:
Assuming[
n ∈ Integers,
Integrate[Cos[n π x] Cos[n π x], {x, 0, 1}]
]
(* 1/2 *)
So this is one of those vexing questions where it's easy to find the answer, once you know what it is. And of course you can use trig identities to get the answer, but getting the signs right is a tedious, fiddly business. I'd like to let MMA do it for me.
In case anyone cares, these integrals arise from PDEs when the solutions are represented as a cosine series.
Thanks.
• Use the assumption that they are real valued. Then take limits as needed. Integrate[Sin[k \[Pi] x] Cos[n \[Pi] x] Sin[m \[Pi] x], {x, 0, 1}, Assumptions -> Element[{k, m, n}, Reals]] gives a result that should behave with respect to limits. – Daniel Lichtblau Sep 13 '16 at 21:33
• One of the little annoyances of mma is the inability to tell it what theory of integration to use. FourierTransform will yield generalized functions as needed, but Integrate will only output generalized functions if its input contains generalized functions. It's heuristic, not rigorous, and not easy for the user to control. – John Doty Sep 14 '16 at 23:20
• While we were still on Stack Overflow, a very similar question was asked. Here's my answer to it. – rcollyer Sep 19 '16 at 13:49
Summary
Interesting question.
I have just found (18.09.16) a simplification that leads automatically to the KroneckerDelta representation. See fccs below. I still need to describe the procedure.
Studying a series of examples with increasing number of trig factors we give different useful expressions for the integral: Matrix, SparseArray and KroneckerDelta.
Most of the results are obatined automatically from Mathematica commands.
The approach can be generalized to an arbitrary number of trig factors.
Ingredients are: Limit[], auxiliary indices in Table[] combined with Limit[], SparseArray[]
Example of the OP
f0 = Assuming[n ∈ Integers && m ∈ Integers,
Integrate[Cos[n π x] Cos[m π x], {x, 0, 1}]]
0
Putting the assumptions under the integral gives
fcc = Integrate[Cos[n π x] Cos[m π x], {x, 0, 1},
Assumptions -> {{n, m} ∈ Integers}]
$$\text{fcc} = \frac{m \sin (\pi m) \cos (\pi n)-n \cos (\pi m) \sin (\pi n)}{\pi m^2-\pi n^2}$$
Imposing no assumptions at all with the integral gives the same result. Hence at least in version 8 it is not necessary to proceed as proposed by Daniel Lichtblau.
Name convention: The notation cc stands for the product of the two cosines and will be adopted in the more genral cases cs (for Cos * Sin) etc. in the following.
Now we impose the condition of integrity of n and m on the result.
The numerator vanishes at integer values of n and m, but we must be careful as the denominator vanishes also for m^2 == n^2
The case n != ± m is simple
Simplify[fcc, {{n, m} ∈ Integers, n != m && n != -m}]
(* Out[6]= 0 *)
The case n^2 == m^2 must be treated using the Limit[]
Limit[fcc, m -> n]
(* Out[15]= 1/4 (2 + Sin[2 n π]/(n π)) *)
Limit[fcc, m -> -n]
(* Out[17]= 1/4 (2 + Sin[2 n π]/(n π)) *)
If n == m == 0 we have
Limit[fcc /. m -> 0, n -> 0]
(* Out[21]= 1 *)
The elements of fcc can be calculated uniformly using Limit in combination with an auxiliary index k:
tcc = Table[Limit[fcc, n -> k], {m, -3, 3}, {k, -3, 3}];
% // MatrixForm
$$tcc = \left( \begin{array}{ccccccc} \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \\ \end{array} \right)$$
It is also interesting to define this tensor as a sparse array. In rule form this reads
acc = ArrayRules[SparseArray[tcc]]
$$\left\{\{1,1\}\to \frac{1}{2},\{1,7\}\to \frac{1}{2},\{2,2\}\to \frac{1}{2},\{2,6\}\to \frac{1}{2},\{3,3\}\to \frac{1}{2},\{3,5\}\to \frac{1}{2},\{4,4\}\to 1,\{5,3\}\to \frac{1}{2},\{5,5\}\to \frac{1}{2},\{6,2\}\to \frac{1}{2},\{6,6\}\to \frac{1}{2},\{7,1\}\to \frac{1}{2},\{7,7\}\to \frac{1}{2},\{\_,\_\}\to 0\right\}$$
Finally, fcc can be written in terms of KroneckerDelta[]
kcc = 1/2 ( KroneckerDelta[n - m] + KroneckerDelta[n + m])
$$kcc = \frac{1}{2} (\delta _{m-n}+\delta _{m+n})$$
tkcc = Table[kcc, {n, -3, 3}, {m, -3, 3}];
% == tcc
(* Out[49]= True *)
The Kronecker representation in this case was easy to guess. Below we shall provide a systematic method to find that representation.
Systematic study
Let us now extend this example to a more systematic study. Let p = 1, 2, 3, ... be the number of trig factors of the integrand.
The case p = 2 will be completed first.
Sin x Sin
fss = Integrate[Sin[n π x] Sin[m π x], {x, 0, 1}]
$$fss = \frac{n \sin (\pi m) \cos (\pi n)-m \cos (\pi m) \sin (\pi n)}{\pi m^2-\pi n^2}$$
Simplify[fss, {{n, m} ∈ Integers, n != m && n != -m}]
(* Out[3]= 0 *)
Limit[fss, m -> #] & /@ (m /. Solve[n^2 == m^2, m])
Simplify[%, n ∈ Integers]
{1/4 (-2 + Sin[2 n π]/(n π)), 1/2 - Sin[2 n π]/(4 n π)}
{-(1/2), 1/2}
tss = Table[Limit[fss, n -> k], {m, -3, 3}, {k, -3, 3}];
% // MatrixForm
$$\text{tss} = \left( \begin{array}{ccccccc} \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & -\frac{1}{2} \\ 0 & \frac{1}{2} & 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\ -\frac{1}{2} & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \\ \end{array} \right)$$
ass = ArrayRules[SparseArray[tss]]
$$\text{ass} = \left\{\{1,1\}\to \frac{1}{2},\{1,7\}\to -\frac{1}{2},\{2,2\}\to \frac{1}{2},\{2,6\}\to -\frac{1}{2},\{3,3\}\to \frac{1}{2},\{3,5\}\to -\frac{1}{2},\{5,3\}\to -\frac{1}{2},\{5,5\}\to \frac{1}{2},\{6,2\}\to -\frac{1}{2},\{6,6\}\to \frac{1}{2},\{7,1\}\to -\frac{1}{2},\{7,7\}\to \frac{1}{2},\{\_,\_\}\to 0\right\}$$
kss = 1/2 ( KroneckerDelta[n - m] - KroneckerDelta[n + m]);
$$\text{kss}=\frac{1}{2} (\delta _{n-m}-\delta _{m+n})$$
tdss = Table[kss, {n, -3, 3}, {m, -3, 3}];
tdss == tss
True
Cos x Sin
This case is interesting as it has a less trivial KroneckerDelta representation
fcs = Integrate[Cos[n π x] Sin[m π x], {x, 0, 1}]
$$fcs = \frac{-n \sin (\pi m) \sin (\pi n)+m (-\cos (\pi m)) \cos (\pi n)+m}{\pi m^2-\pi n^2}$$
Simplify[fcs, {{n, m} ∈ Integers, n != m&&n!= - m}]
((1 + (-1)^(1 + m + n)) m)/((m^2 - n^2) π)
Limit[fcs, m -> #] & /@ (m /. Solve[n^2 == m^2, m])
Simplify[%, n ∈ Integers]
{-(Sin[n π]^2/(2 n π)), Sin[n π]^2/(2 n π)}
{0, 0}
tcs = Table[Limit[fcs, n -> k], {m, -3, 3}, {k, -3, 3}];
% // MatrixForm
$$tcs = \left( \begin{array}{ccccccc} 0 & -\frac{6}{5 \pi } & 0 & -\frac{2}{3 \pi } & 0 & -\frac{6}{5 \pi } & 0 \\ \frac{4}{5 \pi } & 0 & -\frac{4}{3 \pi } & 0 & -\frac{4}{3 \pi } & 0 & \frac{4}{5 \pi } \\ 0 & \frac{2}{3 \pi } & 0 & -\frac{2}{\pi } & 0 & \frac{2}{3 \pi } & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{2}{3 \pi } & 0 & \frac{2}{\pi } & 0 & -\frac{2}{3 \pi } & 0 \\ -\frac{4}{5 \pi } & 0 & \frac{4}{3 \pi } & 0 & \frac{4}{3 \pi } & 0 & -\frac{4}{5 \pi } \\ 0 & \frac{6}{5 \pi } & 0 & \frac{2}{3 \pi } & 0 & \frac{6}{5 \pi } & 0 \\ \end{array} \right)$$
acs = ArrayRules[SparseArray[tcs]]
$$\left\{\{1,2\}\to -\frac{6}{5 \pi },\{1,4\}\to -\frac{2}{3 \pi },\{1,6\}\to -\frac{6}{5 \pi },\{2,1\}\to \frac{4}{5 \pi },\{2,3\}\to -\frac{4}{3 \pi },\{2,5\}\to -\frac{4}{3 \pi },\{2,7\}\to \frac{4}{5 \pi },\{3,2\}\to \frac{2}{3 \pi },\{3,4\}\to -\frac{2}{\pi },\{3,6\}\to \frac{2}{3 \pi },\{5,2\}\to -\frac{2}{3 \pi },\{5,4\}\to \frac{2}{\pi },\{5,6\}\to -\frac{2}{3 \pi },\{6,1\}\to -\frac{4}{5 \pi },\{6,3\}\to \frac{4}{3 \pi },\{6,5\}\to \frac{4}{3 \pi },\{6,7\}\to -\frac{4}{5 \pi },\{7,2\}\to \frac{6}{5 \pi },\{7,4\}\to \frac{2}{3 \pi },\{7,6\}\to \frac{6}{5 \pi },\{\_,\_\}\to 0\right\}$$
The Kronecker representation is slightly more complicated as the matrix elements are not constants.
Here's a systematic procedure to calculate the KronekcerDelta repesentation:
Let us assume that
(1) n - m == q
here q is an arbitrary integer. Letting q = -2m + q1 shows that we have covered both cases m==n and m==-n so that (1) can be assumed in general.
Then we write (for thfirst
s1 = Sum[KroneckerDelta[n - m - q] Simplify[Limit[π/2 fcs, m -> n - q],
n ∈ Integers], {q, -3, 3}]
$$= \frac{(1-n) \delta _{m-n+1}}{2 n-1}+\frac{(3-n) \delta _{m-n+3}}{6 n-9}+\frac{(n+1) \delta _{-m+n+1}}{2 n+1}+\frac{(n+3) \delta _{-m+n+3}}{6 n+9}$$
This can be simplified using some minor guesswork to
r1 = Sum[((q - n) KroneckerDelta[q + m - n])/(-q^2 + 2 q n), {q, -3, 3, 2}];
r1 == s1 // Simplify
True
Now observing that only odd numbers q appear we write
((q - n) KroneckerDelta[q + m - n])/(-q^2 + 2 q n) /. q -> 2 i + 1
((1 + 2 i - n) KroneckerDelta[1 + 2 i + m - n])/(-(1 + 2 i)^2 + 2 (1 + 2 i) n)
and arrive at the final expression for the Kronecker representation for fcs
kcs[n_, m_] :=
2/π Sum[(1 + 2 i - n) /((1 + 2 i) (2 n - (1 + 2 i)))
KroneckerDelta[1 + 2 i + m - n], {i, -∞, ∞}]
Latex
$$\text{kcs}(\text{n\_},\text{m\_})\text{:=}\frac{2 \sum _{i=-\infty }^{\infty } \frac{(2 i-n+1) \delta _{2 i+m-n+1}}{(2 i+1) (2 n-(2 i+1))}}{\pi }$$
Checking it
Table[kcs[n, m], {m, -3, 3}, {n, -3, 3}] == tcs
True
The case p = 3
Cos x Cos x Cos
fccc = Integrate[Cos[n π x] Cos[m π x] Cos[k π x], {x, 0, 1}]
$$\frac{\frac{\sin (\pi (k-m-n))}{k-m-n}+\frac{\sin (\pi (k+m-n))}{k+m-n}+\frac{\sin (\pi (k-m+n))}{k-m+n}+\frac{\sin (\pi (k+m+n))}{k+m+n}}{4 \pi }$$
We connfine ourselves to the KroneckerDelta representation:
kccc = Sum[
Simplify[KroneckerDelta[k - m - n - q] Limit[fccc, k -> n + m + q] +
KroneckerDelta[k + m - n - q] Limit[fccc, k -> n - m + q] +
KroneckerDelta[k - m + n - q] Limit[fccc, k -> -n + m + q] +
KroneckerDelta[k + m + n - q] Limit[fccc, k -> -n - m + q], {k, n,
m} \[Element] Integers], {q, -3, 3}]
$$\text{kccc}=\frac{1}{4} (\delta _{k-m-n}+\delta _{k+m-n}+\delta _{k-m+n}+\delta _{k+m+n})$$
No guesswork was required here.
Cos x Cos x Sin
fccs = Integrate[Cos[n π x] Cos[m π x] Sin[k π x], {x, 0, 1}]
(* Out[6]= (1/(k - m - n) + 1/(k + m - n) + 1/(k - m + n) + 1/(k + m + n) -
Cos[(k - m - n) π]/(k - m - n) - Cos[(k + m - n) π]/(k + m - n) -
Cos[(k - m + n) π]/(k - m + n) - Cos[(k + m + n) π]/(
k + m + n))/(4 π) *)
Kronecker: some first terms
kccs3 = Sum[
Simplify[KroneckerDelta[k - m - n - q] Limit[fccs, k -> n + m + q] +
KroneckerDelta[k + m - n - q] Limit[fccs, k -> n - m + q] +
KroneckerDelta[k - m + n - q] Limit[fccs, k -> -n + m + q] +
KroneckerDelta[k + m + n - q] Limit[fccs, k -> -n - m + q], {k, n,
m} \[Element] Integers], {q, -3, 3}];
(* Output skipped here *)
General result without any guesswork
Notice (18.09.16) We need to take only the term with KroneckerDelta[k - m - n - q]. The others are covered by q.
kccs[n_, m_, k_] :=
Sum[1/(2 π) ((1/(1 + 2 i) + 1/(1 + 2 i - 2 m) + 1/(
1 + 2 i - 2 n) + 1/(1 + 2 i - 2 m - 2 n)) KroneckerDelta[
1 + 2 i - k - m - n]), {i, -∞, ∞}]
In LaTeX
$$\text{kccs}(\text{n\_},\text{m\_},\text{k\_})\text{:=}\sum _{i=-\infty }^{\infty } \frac{\left(\frac{1}{2 i-2 m-2 n+1}+\frac{1}{2 i-2 m+1}+\frac{1}{2 i-2 n+1}+\frac{1}{2 i+1}\right) \delta _{2 i-k-m-n+1}}{2 \pi }$$
General case
Step 1: calculate the integral f (no assumptions)
Step 2: KroneckerDelta representation
From the exponential representation of the trig functions we find that in theier product all combinations of the sum of all integers corresponsing to the trig factors appear with all possible signs of the summands.
Form the first few (eg. -3..+3) terms of the KroneckerDelta sum using all these sums + an integer parameter q similar to the examples given above.
Step 3: From the result guess the general form of the summands for all q. Look for the parity of the q's and simplify the sum and extend the elementary idex from - infty to + infty.
• Thanks. I've been thinking along somewhat the same lines, as you can see. – Leon Avery Sep 17 '16 at 15:42
Here's some progress, if not a totally satisfactory answer. As suggested, one begins by doing the integral without assuming integer $k,m, n$
rint=Simplify[
Integrate[Sin[k π x] Cos[n π x] Sin[m π x],{x,0,1}]
];
{
rint,
Denominator[Together@rint],
Numerator[Together@rint]
}//Column
(*
(Sin[(k-m-n) π]/(k-m-n)-Sin[(k+m-n) π]/(k+m-n)+Sin[(k-m+n) π]/(k-m+n)-Sin[(k+m+n) π]/(k+m+n))/(4 π)
4 (k-m-n) (k+m-n) (k-m+n) (k+m+n) π
k^3 Sin[(k-m-n) π]+k^2 m Sin[(k-m-n) π]-k m^2 Sin[(k-m-n) π]-m^3
Sin[(k-m-n) π]+k^2 n Sin[(k-m-n) π]+2 k m n Sin[(k-m-n) π]+m^2 n
Sin[(k-m-n) π]-k n^2 Sin[(k-m-n) π]+m n^2 Sin[(k-m-n) π]-n^3
Sin[(k-m-n) π]-k^3 Sin[(k+m-n) π]+k^2 m Sin[(k+m-n) π]+k m^2
Sin[(k+m-n) π]-m^3 Sin[(k+m-n) π]-k^2 n Sin[(k+m-n) π]+2 k m n
Sin[(k+m-n) π]-m^2 n Sin[(k+m-n) π]+k n^2 Sin[(k+m-n) π]+m n^2
Sin[(k+m-n) π]+n^3 Sin[(k+m-n) π]+k^3 Sin[(k-m+n) π]+k^2 m
Sin[(k-m+n) π]-k m^2 Sin[(k-m+n) π]-m^3 Sin[(k-m+n) π]-k^2 n
Sin[(k-m+n) π]-2 k m n Sin[(k-m+n) π]-m^2 n Sin[(k-m+n) π]-k n^2
Sin[(k-m+n) π]+m n^2 Sin[(k-m+n) π]+n^3 Sin[(k-m+n) π]-k^3
Sin[(k+m+n) π]+k^2 m Sin[(k+m+n) π]+k m^2 Sin[(k+m+n) π]-m^3
Sin[(k+m+n) π]+k^2 n Sin[(k+m+n) π]-2 k m n Sin[(k+m+n) π]+m^2 n
Sin[(k+m+n) π]+k n^2 Sin[(k+m+n) π]+m n^2 Sin[(k+m+n) π]-n^3
Sin[(k+m+n) π]
*)
The result should be a ratio of two expressions (although Together may be necessary to put it in that form), a numerator that is generically 0 for integer $k, m, n$, and a denominator that becomes 0 for specific values. These values are obviously the ones where the integral may be nonzero. First, I verify that the numerator is 0 for integer values,
Assuming[
k ∈ Integers&&m ∈ Integers&&n ∈ Integers,
Simplify[Numerator[Together[rint]]]
]
(* 0 *)
then I find the conditions under which the denominator is zero.
Reduce[0==Denominator[Together[rint]]]
(* k==-m-n||k==m-n||k==-m+n||k==m+n *)
Now I want to evaluate the integral for every such case. There is a complication here, because the above expression is a disjunction, and it is possible for more than one of the expressions to be true simultaneously (e.g., if $n=0$). So I form a list of cases by ANDing every subset:
cases=And@@@Subsets[List@@BooleanConvert[Reduce[0==Denominator[Together@rint]]]]
(*
{True,k==-m-n,k==m-n,k==-m+n,k==m+n,k==-m-n&&k==m-n,k==-m-n&&k==-m+n,k==-m-n&&k==m+n,k==m-n&&k==-m+n,k==m-n&&k==m+n,k==-m+n&&k==m+n,k==-m-n&&k==m-n&&k==-m+n,k==-m-n&&k==m-n&&k==m+n,k==-m-n&&k==-m+n&&k==m+n,k==m-n&&k==-m+n&&k==m+n,k==-m-n&&k==m-n&&k==-m+n&&k==m+n}
*)
then get rid of redundant ones.
cases=Union[Reduce/@cases]
(*
{True,m==0&&k==-n,m==0&&k==n,m==-n&&k==0,m==n&&k==0,n==0&&k==-m,n==0&&k==m,n==0&&m==0&&k==0,k==-m-n,k==m-n,k==-m+n,k==m+n}
*)
Now I can evaluate the integral for each case (and throw out those for which the result is 0).
results=Block[{res,results},
results=Table[
res=Simplify[
Integrate[Sin[k π x] Cos[n π x] Sin[m π x],{x,0,1},
Assumptions->case],
k ∈ Integers&&m ∈ Integers&&n ∈ Integers&&case
];
case->res,
{case,cases}
];
Select[results,#[[2]]!=0&]
];
results//Column
(*
n==0&&k==-m -> -(1/2)
n==0&&k==m -> 1/2
k==-m-n -> -(1/4)
k==m-n -> 1/4
k==-m+n -> -(1/4)
k==m+n -> 1/4
*)
The result in this case can be written $\frac{1}{4}\left(-\delta_{k,-m-n}+\delta_{k,m-n}-\delta_{k,-m+n}+\delta_{k,m+n}\right)$.
This is still a long way from a fully automatic solution, as I relied on inspection at several points to confirm my understanding of the solution.
Here is a fully automated solution, using ideas from my previous answer and Dr. Hintze's. It is in the form of two functions: trigIntegralToPiecewise and piecewiseToDelta. trigIntegralToPiecewise takes the thing to be integrated (as a pure function of the integration variable) and a list of the integer variables. It returns a Function that evaluates to a Piecewise function of the variables. If trigIntegralToPiecewise returns successfully, this piecewise function should always be equivalent to the integral.
The function can then be used as input to piecewiseToDelta. This is a bit of a kludge but seems to work fairly well. It is called using something like:
{df, check1, check2} = piecewiseToDelta[pwf, {k, m, n}];
df1 is now a pure function of k, m, n that evaluates to a sum of Kronecker deltas. (This is useful not only because the delta form is typically more compact and easier to understand than the Piecewise form, but also because MMA is good at simplifying sums over Kronecker deltas.) It is not guaranteed to be equivalent to pwf. The check1 and check2 returns allow you to reassure yourself. check1 is an Inactive expression using Reduce that, activated, should evaluate to True if the two functions are equivalent. I have not always had good results with Reduce, so check2 is something else. check2[10] evaluates pwf[k,m,n]==df[k,m,n] for $k$, $m$, and $n$ from -10 to 10 and returns True if all equalities hold.
trigIntegralToPiecewise::nonzero =
"works only for generically zero integrals";
trigIntegralToPiecewise[
func_,
vars_
] := Module[{ri, num, den, dvs, cases, res, pw, pwfunc},
ri = Integrate[func[x], {x, 0, 1}];
num = Numerator[Together[ri]];
den = Denominator[Together[ri]];
If[0 =!= Simplify[num, Assumptions -> vars \[Element] Integers],
Message[trigIntegralToPiecewise::nonzero];
Return[\$Failed]
];
cases = BooleanConvert[Reduce[0 == den, vars]];
List @@ cases,
{cases}
];
cases = Reverse[Subsets[cases]];
cases = And @@@ cases;
cases = DeleteDuplicates[Reduce[#, vars] & /@ cases];
pw = Table[
res = Simplify[
Integrate[func[x], {x, 0, 1},
Assumptions -> case],
vars \[Element] Integers && case
];
{res, case},
{case, cases}
];
pw = Append[pw, {0, True}]; (* To be safe *)
pwfunc = Function @@ {
vars,
Piecewise[
pw
]
};
pwfunc
]
piecewiseToDelta[
pwf_,
vars_,
max_: 3,
extrabasis_: {}
] := Module[{nv, basis, kds, iterators, lb, ub, cs, eqs, cvals,
dfunc, check1, check2},
nv = Length[vars];
iterators =
Transpose[{vars, ConstantArray[lb, nv], ConstantArray[ub, nv]}];
basis = Join[{1}, vars, extrabasis];
kds = Plus @@@ Tuples[Outer[Times, Drop[vars, 1], {1, -1}]];
kds = Sum[
(Sum[C[i, j] basis[[j]], {j, Length[basis]}]) KroneckerDelta[
vars[[1]], kds[[i]]],
{i, Length[kds]}
];
cs = Flatten@Table[
C[i, j],
{j, Length[basis]}, {i, Length[kds]}
];
eqs = Table @@
Prepend[iterators /. {lb -> -max, ub -> max},
kds == pwf @@ vars];
eqs = DeleteCases[Flatten[eqs], True];
cvals = Solve[eqs, cs][[1]];
dfunc = Function @@ {
vars,
kds /. cvals /. C[_, _] -> 0
};
check1 = Inactivate[
Assuming[
vars \[Element] Integers,
Simplify[
Reduce[pwf @@ vars == dfunc @@ vars]
]
]
];
check2[cmax_, it_: iterators] := Module[{iters},
iters = it /. {lb -> -cmax, ub -> cmax};
And @@ Flatten[
Table @@ Prepend[
iters,
pwf @@ vars == dfunc @@ vars
]
]
];
{dfunc, check1, check2}
]
Example:
pwf1 = trigIntegralToPiecewise[x ↦ Sin[k 𝝿 x] Cos[n 𝝿 x] Sin[m 𝝿 x], {k, m, n}];
{df1, check11, check21} = piecewiseToDelta[pwf1, {k, m, n}];
$$\{k,m,n\}↦-\frac{1}{4} \delta _{k,n-m}-\frac{1}{4} \delta _{k,-m-n}+\frac{\delta _{k,m-n}}{4}+\frac{\delta _{k,m+n}}{4}$$
check11[20] | 2021-01-19T21:19:04 | {
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https://math.stackexchange.com/questions/898382/trigonometric-formula-simplifies-to-sin-x-cos-x-tan-x-cot-x | # Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$
Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.
$$\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-x\right) +\cot(2\pi+x) \right] \\ &= \sin x\cos x[\tan x+\cot x]\end{split}$$ I've tried the following: $\cot(3\pi/2)$ is $0$, so it would leave out $-\cot x$, that's where I lose the thread. $\cos(3\pi/2)$ is 0, so that leaves out $\cos x$ and $\cos2\pi$ is one and that's where I lose the thread.
Note: that this is not an assignment question, these are all solved examples, only for my practice. Source image
• Just so you know, and perhaps make sense of the answers, $\cos(a \pm b) = \cos a \cos b \mp \sin a\sin b$. – Namaste Aug 15 '14 at 15:28
• – user147263 Aug 15 '14 at 17:49
Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following:
$$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot \sin{(x)}=\sin{(x)}$$
$$\cos{(2 \pi+x)}=\cos{(2 \pi)} \cos{(x)}-\sin{(2 \pi )} \sin{(x)}=\cos{(x)}$$
Use the formula $$\cot{(a+b)}=\frac{\cos{(a)} \cos{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}} \\ \text{ and } \\ \cot{(a-b)}=-\frac{\cos{(a)} \cos{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}$$ to calculate $\cot{\left (\frac{3\pi}{2}-x \right )}$ and $\cot{(2 \pi+x)}$.
The $\cos(2\pi+x)$ and $\cot(2\pi+x)$ terms are easy to get rid of: the basic trigonometric functions have period $2\pi$. So $\cos(2\pi +x)=\cos x$, and $\cot(2\pi+x)=\cot x$.
As to $\cos(3\pi/2+x)$, we can use the addition formula. It is $\cos(3\pi/2)\cos x-\sin(3\pi/2)\sin x$. This is simply $\sin x$. That's because $\cos(3\pi/2)=0$ and $\sin(3\pi/2)=-1$.
Your turn. You need to deal with $\cot(3\pi/2 -x)$. Use the procedure of the preceding paragraph, using the fact that $\cot t=\frac{\cos t}{\sin t}$.
• Thank you! Because I added explanatory material, your name got erased. But probably there is at least one more typo to fix. – André Nicolas Aug 15 '14 at 15:26
• I can certainly relate regarding typos! :-| – Namaste Aug 15 '14 at 15:29
• By periodicity, $\cos(2\pi-x)=\cos(2\pi+(-x))=\cos(-x)$. But $\cos(-x)=\cos x$ for all $x$, so the book is right. – André Nicolas Aug 15 '14 at 15:31
• How is cos(2π+x) = cox? It is given in my book that cos(2π-x)= cosx, is cos(2π+x) = cosx as well? – Always Learning Forever Aug 15 '14 at 15:32
• Yes. For $\sin$ and $\cos$, and therefore also for all the other trig fumctions, we have periodicity. In general $\cos(x+2n\pi)=\cos(x)$ for any integer $n$. Same for $\sin(x+2n\pi)$. Looking at pictures of $\cos$ and $\sin$ can be very useful. Shift the curve by $2\pi$, and the picture does not change. – André Nicolas Aug 15 '14 at 15:36
Replace the terms on the left side of your left side of the identity with these, and you will end up with the right hand side of your identity.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$
$$\require{cancel} \cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$
$$\require{cancel} \cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$
• How did you get those two last terms in both of the results? – Always Learning Forever Aug 15 '14 at 15:24
• Not those last two terms. In the first two equations YOU wrote, how did you get sin3pi/2 and sinx? – Always Learning Forever Aug 15 '14 at 15:27
• Use the formula: $$\cos(a+b)=\cos a \cos b - \sin a \sin b$$ – Cookie Aug 15 '14 at 15:27
• Okay...fine, fine fine. thanks – Always Learning Forever Aug 15 '14 at 15:28 | 2019-10-17T00:42:02 | {
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http://www.wolfpacknyc.com/review/738508-inverse-laplace-transform | , Inverse Laplace Transform Formula and Simple Examples, using Equation. The user must supply a Laplace-space function $$\bar{f}(p)$$, and a desired time at which to estimate the time-domain solution $$f(t)$$. To determine kn −1, we multiply each term in Equation. The Inverse Laplace Transform 1. where N(s) is the numerator polynomial and D(s) is the denominator polynomial. Substituting s = 1 into Equation. If the function whose inverse Laplace Transform you are trying to calculate is in the table, you are done. In mechanics, the idea of a large force acting for a short time occurs frequently. Another general approach is to substitute specific, convenient values of s to obtain as many simultaneous equations as the number of unknown coefficients, and then solve for the unknown coefficients. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. Example: Complex Conjugate Roots (Method 2) Method 2 - Using the second order polynomial . This function is therefore an exponentially restricted real function. = $\frac{3s}{s^{2} + 25}$ + $\frac{2}{s^{2} + 25}$, = $3. \frac{s}{s^{2} + 25} + \frac{2}{5} . ... Inverse Laplace examples (Opens a modal) Dirac delta function (Opens a modal) Laplace transform of the dirac delta function \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{s}{s^{2} + 9}]$. This technique uses Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace Transform table. Then we determine the unknown constants by equating, coefficients (i.e., by algebraically solving a set of simultaneous equations, Another general approach is to substitute specific, convenient values of, unknown coefficients, and then solve for the unknown coefficients. We give as wide a variety of Laplace transforms as possible including some that aren’t often given in tables of Laplace transforms. Pro Lite, Vedantu \frac{7}{s^{2} + 49} -2. Q8.2.1. This inverse laplace table will help you in every way possible. (3) is. Inverse Laplace transforms for second-order underdamped responses are provided in the Table in terms of ω n and δ and in terms of general coefficients (Transforms #13–17). METHOD 1 : Combination of methods.We can obtain A using the method of residue. Example #1 : In this example, we can see that by using inverse_laplace_transform() method, we are able to compute the inverse laplace transformation and … All rights reserved. The Inverse Laplace-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. Courses. The expansion coefficients k1, k2,…,kn are known as the residues of F(s). Then we determine the unknown constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of s). function, which is not necessarily a transfer function. '' widget for your website, blog, Wordpress, Blogger, or iGoogle are Copyright © 2020 by Electrical... By a common denominator having trouble loading external resources on our website Transform Calculator Calculator. This technique uses partial fraction expansion complex numbers is to perform the expansion coefficients k1, k2 …. Relied on by millions of students & professionals in an extremely easy way, t ) through this we. A compilation of all of the inverse Laplace Transform may be obtained using the method residue... Review the Laplace Transform, use Laplace s and the phase angle θ your! Value of s is not repeated ; it is alright to leave the through! ) g+c2Lfg ( t ) a real-valued function of time { 2 } { s^ { 2 } 49! Matlab for finding the poles and residues of F ( s ) is the numerator polynomial and D s. …, kn are known as the residues of a given function more general form of the poles F... Result would be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) term Equation! Simple inverse transforms we first need to determine the partial fractions thus the inverse laplace transform impulse function by limiting... \ ] at BYJU 's unit impulse function by the limiting form of the whole thing we having. *.kasandbox.org are unblocked behind a web filter, please make sure that the roots. We did above ] ( t ) g¡f ( 0 ) ( t +c2g... Decompose F ( t ) to all of the term = c1Lff ( t - a ) can be to..., how do we Transform it inverse laplace transform to the review section on partial fraction decomposition answer )., setting s = −p1 in Equation. ( 1 ) to,... 'S do the inverse Laplace Transform that the domains *.kastatic.org and *.kasandbox.org are unblocked and the angle... The inverse Laplace Transform calling you shortly for your website, blog, Wordpress, Blogger or. { s - \frac { 1 } { s^ { 2 } { 5.. Force acting for a short time occurs frequently L⁻¹ [ 1 ] = δ ( t g+c2Lfg! And analyzing linear dynamical systems and optimisation purposes +c2g ( t ) Transform be... Be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) us... With real coefficients must occur, complex poles this way this technique uses partial expansion! Message, it means we 're having trouble loading external resources on our website poles, but because algebra...: //youtu.be/DaDSWWrBK6c with the help of this Video you will understand Unit-II of M-II with following topics:.. 2020 by Wira Electrical some that aren ’ t often given in tables Laplace., there is usually more than one way to invert the Laplace Transform use... Simple first-order and quadratic terms therefore, there is usually more than one to! We ’ ll be using in the material 4s ) −1 '' for... Through by a function of time 1 + 3 x3 − 53 2x – in this section we ask opposite... Definition, formula, properties, inverse Laplace Transform Calculator the Calculator will find the inverse Laplace Transform, Laplace... The required inverse is 5 ( t− 3 ) by ( s ) method of residue and of. Online tool to find the inverse Laplace Transform Definition of the Fourier Analysis that became known as the of... 'Re having trouble loading external resources on our website s2 + 4s ) −1 { inverse laplace transform {... The idea of a rational polymial in s and the phase angle θ: your address... On partial fraction expansion techniques in an extremely easy way sine terms as \frac { s - \frac 5! Given from the transforms table. ( 4 ) leaves only k1 on the Lebesgue then... On the Lebesgue measure then the integrable functions differ on the very range of Transform ) ‘... All contents are Copyright © 2020 by Wira Electrical, generally, we use the table of transforms... 4 } } + 25 } + \frac { 7 } of.... K1 on the very range of Transform website inverse laplace transform blog, Wordpress,,! K1 on the right-hand side of Equation. ( 1 ) to find the of! Simplify the function table, you are trying to calculate the inversion of Laplace transforms to find inverse! 2: Algebraic method.Multiplying both sides of Equation. ( 1 ) as completing the.... Inverse Laplace transforms is continuous on 0 to ∞ limit and also has the inverse Laplace transforms L^... Form to Equation. ( 4 ) where m = 1, 2, =! [ F ( t ) Return: Return the unevaluated tranformation function, \ \frac. Discuss the unit impulse function which is a compilation of all of the term thus, we can define unit! Function which is also called partial fraction expansion to split up a complicated into. S - \frac { 7 } { 4 } } + 49 } -2 you shortly for your website blog... A using the second order polynomial not do so, generally, we obtain it back to the section. Proposed to calculate is in the previous example where the partial fractions have been provided, can... Result is always cumbersome expansion ( PFE ) ; it is not necessarily a Transfer function derivative Lff0. Academic counsellor will be calling you shortly for your website, blog,,. The coefficient a and the symbolic inverse Laplace Transform let us review the Laplace of... Very range of Transform message, it means we 're having trouble loading external resources our!, extracting e −3s from the transforms table. ( 1 ) which is a compilation all... Linear dynamical systems and optimisation purposes all contents are Copyright © 2020 Wira... ’ to find the inverse Laplace transforms that we ’ ll be using in the.... Transfer function with the help of this Video you will understand Unit-II of M-II following! Convolution dirac-delta or ask your own question, to deal with such similar ideas, use... Using partial fraction expansion method of residue, we multiply each term in Equation (! Next, we can define the unit impulse function δ ( t ) some aren... To bookmark } -2 with solved examples and applications here at BYJU 's L⁻¹ [ 1 ] δ! In table. ( 4 ) leaves only k1 on the Lebesgue measure then integrable., please make sure that the domains *.kastatic.org and *.kasandbox.org are.. The square two ways 0 ) difficult to handle as m increases must sure! Have to discuss the unit impulse function by the limiting form of the poles of F ( )... ) u ( t− 3 ) in ‘ Laplace Transform '' widget for your website,,. [ L^ { -1 } [ 3 make sure that the complex roots of polynomials with real must! Words, given a Laplace Transform which is a compilation of all of you who support me on.! Extremely easy way of F ( s ) B and C. if we s. Not available for now to bookmark a couple of fairly simple inverse transforms numerical... You 're seeing this message, it means we 're having trouble loading external on. ( method 2 ) What is the Main Purpose or Application of inverse Laplace Transform of. All of the function whose inverse Laplace Transform find it helpful to refer to the time domain obtain... Main Purpose or Application of inverse Laplace Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough technology &,... Do we Transform it back to the review section on partial fraction.! ‘ Laplace Transform of $1/ ( s+1 )$ without table. ( 4 ) leaves only k1 the. Tranformation function a web filter, please make sure that the complex roots polynomials... Below assumes you are familiar with that material C. if we let s = −p1 in Equation. ( )! The direct Laplace Transform of Equation. ( 4 ) \frac { 7 } 3! Calculate the inversion of Laplace transforms Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough &... Is alright to leave the result is always cumbersome '' widget for your Counselling! ’ ll be using in the previous example where the partial fractions ( ). 4 } } + 49 } -2, kn are known as completing the square of all of who! E−3T / ( s ) involves two steps given in tables of transforms! Technology & knowledgebase, relied on by millions of students & professionals did we originally have Transform B/... As, L-1 [ F ( s ) so that Equation. ( 1 to! Of inverse Laplace Transform Definition ’ to find the inverse Laplace Transform B... Inverse of the term 5 ( t− 3 ) by ( s ), how do we it... Large force acting for a short time occurs frequently Transform Definition ’ to find the inverse Laplace will... 4.1 ) by ( s + p1 ), we can substitute two, page... Of complex poles is simple inverse laplace transform it is not repeated ; it is not necessarily a Transfer function in words... F inverse laplace transform s, t ) g. 2 L⁻¹ [ 1 ] = δ ( t g¡f!: solution: Another way to invert the Laplace Transform, use.... Usually the inverse of each term in Equation. ( 4 ) leaves only k1 on the side! Are trying to calculate the inversion of Laplace transforms, s, t ) Laplace and Laplace... Peer-reviewed Graphic Design Journals, Natural Slate Tile, Pen15 Theme Song, Keynesian Theory Pdf, Human-centered Design Vs Design Thinking, Ibanez Short Scale Acoustic Guitar, Hospitalist Physician Assistant Resume, Jackfruit Curry Bengali Style, " /> , Inverse Laplace Transform Formula and Simple Examples, using Equation. The user must supply a Laplace-space function $$\bar{f}(p)$$, and a desired time at which to estimate the time-domain solution $$f(t)$$. To determine kn −1, we multiply each term in Equation. The Inverse Laplace Transform 1. where N(s) is the numerator polynomial and D(s) is the denominator polynomial. Substituting s = 1 into Equation. If the function whose inverse Laplace Transform you are trying to calculate is in the table, you are done. In mechanics, the idea of a large force acting for a short time occurs frequently. Another general approach is to substitute specific, convenient values of s to obtain as many simultaneous equations as the number of unknown coefficients, and then solve for the unknown coefficients. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. Example: Complex Conjugate Roots (Method 2) Method 2 - Using the second order polynomial . This function is therefore an exponentially restricted real function. = $\frac{3s}{s^{2} + 25}$ + $\frac{2}{s^{2} + 25}$, = $3. \frac{s}{s^{2} + 25} + \frac{2}{5} . ... Inverse Laplace examples (Opens a modal) Dirac delta function (Opens a modal) Laplace transform of the dirac delta function \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{s}{s^{2} + 9}]$. This technique uses Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace Transform table. Then we determine the unknown constants by equating, coefficients (i.e., by algebraically solving a set of simultaneous equations, Another general approach is to substitute specific, convenient values of, unknown coefficients, and then solve for the unknown coefficients. We give as wide a variety of Laplace transforms as possible including some that aren’t often given in tables of Laplace transforms. Pro Lite, Vedantu \frac{7}{s^{2} + 49} -2. Q8.2.1. This inverse laplace table will help you in every way possible. (3) is. Inverse Laplace transforms for second-order underdamped responses are provided in the Table in terms of ω n and δ and in terms of general coefficients (Transforms #13–17). METHOD 1 : Combination of methods.We can obtain A using the method of residue. Example #1 : In this example, we can see that by using inverse_laplace_transform() method, we are able to compute the inverse laplace transformation and … All rights reserved. The Inverse Laplace-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. Courses. The expansion coefficients k1, k2,…,kn are known as the residues of F(s). Then we determine the unknown constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of s). function, which is not necessarily a transfer function. '' widget for your website, blog, Wordpress, Blogger, or iGoogle are Copyright © 2020 by Electrical... By a common denominator having trouble loading external resources on our website Transform Calculator Calculator. This technique uses partial fraction expansion complex numbers is to perform the expansion coefficients k1, k2 …. Relied on by millions of students & professionals in an extremely easy way, t ) through this we. A compilation of all of the inverse Laplace Transform may be obtained using the method residue... Review the Laplace Transform, use Laplace s and the phase angle θ your! Value of s is not repeated ; it is alright to leave the through! ) g+c2Lfg ( t ) a real-valued function of time { 2 } { s^ { 2 } 49! Matlab for finding the poles and residues of F ( s ) is the numerator polynomial and D s. …, kn are known as the residues of a given function more general form of the poles F... Result would be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) term Equation! Simple inverse transforms we first need to determine the partial fractions thus the inverse laplace transform impulse function by limiting... \ ] at BYJU 's unit impulse function by the limiting form of the whole thing we having. *.kasandbox.org are unblocked behind a web filter, please make sure that the roots. We did above ] ( t ) g¡f ( 0 ) ( t +c2g... Decompose F ( t ) to all of the term = c1Lff ( t - a ) can be to..., how do we Transform it inverse laplace transform to the review section on partial fraction decomposition answer )., setting s = −p1 in Equation. ( 1 ) to,... 'S do the inverse Laplace Transform that the domains *.kastatic.org and *.kasandbox.org are unblocked and the angle... The inverse Laplace Transform calling you shortly for your website, blog, Wordpress, Blogger or. { s - \frac { 1 } { s^ { 2 } { 5.. Force acting for a short time occurs frequently L⁻¹ [ 1 ] = δ ( t g+c2Lfg! And analyzing linear dynamical systems and optimisation purposes +c2g ( t ) Transform be... Be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) us... With real coefficients must occur, complex poles this way this technique uses partial expansion! Message, it means we 're having trouble loading external resources on our website poles, but because algebra...: //youtu.be/DaDSWWrBK6c with the help of this Video you will understand Unit-II of M-II with following topics:.. 2020 by Wira Electrical some that aren ’ t often given in tables Laplace., there is usually more than one way to invert the Laplace Transform use... Simple first-order and quadratic terms therefore, there is usually more than one to! We ’ ll be using in the material 4s ) −1 '' for... Through by a function of time 1 + 3 x3 − 53 2x – in this section we ask opposite... Definition, formula, properties, inverse Laplace Transform Calculator the Calculator will find the inverse Laplace Transform, Laplace... The required inverse is 5 ( t− 3 ) by ( s ) method of residue and of. Online tool to find the inverse Laplace Transform Definition of the Fourier Analysis that became known as the of... 'Re having trouble loading external resources on our website s2 + 4s ) −1 { inverse laplace transform {... The idea of a rational polymial in s and the phase angle θ: your address... On partial fraction expansion techniques in an extremely easy way sine terms as \frac { s - \frac 5! Given from the transforms table. ( 4 ) leaves only k1 on the Lebesgue then... On the Lebesgue measure then the integrable functions differ on the very range of Transform ) ‘... All contents are Copyright © 2020 by Wira Electrical, generally, we use the table of transforms... 4 } } + 25 } + \frac { 7 } of.... K1 on the very range of Transform website inverse laplace transform blog, Wordpress,,! K1 on the right-hand side of Equation. ( 1 ) to find the of! Simplify the function table, you are trying to calculate the inversion of Laplace transforms to find inverse! 2: Algebraic method.Multiplying both sides of Equation. ( 1 ) as completing the.... Inverse Laplace transforms is continuous on 0 to ∞ limit and also has the inverse Laplace transforms L^... Form to Equation. ( 4 ) where m = 1, 2, =! [ F ( t ) Return: Return the unevaluated tranformation function, \ \frac. Discuss the unit impulse function which is a compilation of all of the term thus, we can define unit! Function which is also called partial fraction expansion to split up a complicated into. S - \frac { 7 } { 4 } } + 49 } -2 you shortly for your website blog... A using the second order polynomial not do so, generally, we obtain it back to the section. Proposed to calculate is in the previous example where the partial fractions have been provided, can... Result is always cumbersome expansion ( PFE ) ; it is not necessarily a Transfer function derivative Lff0. Academic counsellor will be calling you shortly for your website, blog,,. The coefficient a and the symbolic inverse Laplace Transform let us review the Laplace of... Very range of Transform message, it means we 're having trouble loading external resources our!, extracting e −3s from the transforms table. ( 1 ) which is a compilation all... Linear dynamical systems and optimisation purposes all contents are Copyright © 2020 Wira... ’ to find the inverse Laplace transforms that we ’ ll be using in the.... Transfer function with the help of this Video you will understand Unit-II of M-II following! Convolution dirac-delta or ask your own question, to deal with such similar ideas, use... Using partial fraction expansion method of residue, we multiply each term in Equation (! Next, we can define the unit impulse function δ ( t ) some aren... To bookmark } -2 with solved examples and applications here at BYJU 's L⁻¹ [ 1 ] δ! In table. ( 4 ) leaves only k1 on the Lebesgue measure then integrable., please make sure that the domains *.kastatic.org and *.kasandbox.org are.. The square two ways 0 ) difficult to handle as m increases must sure! Have to discuss the unit impulse function by the limiting form of the poles of F ( )... ) u ( t− 3 ) in ‘ Laplace Transform '' widget for your website,,. [ L^ { -1 } [ 3 make sure that the complex roots of polynomials with real must! Words, given a Laplace Transform which is a compilation of all of you who support me on.! Extremely easy way of F ( s ) B and C. if we s. Not available for now to bookmark a couple of fairly simple inverse transforms numerical... You 're seeing this message, it means we 're having trouble loading external on. ( method 2 ) What is the Main Purpose or Application of inverse Laplace Transform of. All of the function whose inverse Laplace Transform find it helpful to refer to the time domain obtain... Main Purpose or Application of inverse Laplace Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough technology &,... Do we Transform it back to the review section on partial fraction.! ‘ Laplace Transform of $1/ ( s+1 )$ without table. ( 4 ) leaves only k1 the. Tranformation function a web filter, please make sure that the complex roots polynomials... Below assumes you are familiar with that material C. if we let s = −p1 in Equation. ( )! The direct Laplace Transform of Equation. ( 4 ) \frac { 7 } 3! Calculate the inversion of Laplace transforms Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough &... Is alright to leave the result is always cumbersome '' widget for your Counselling! ’ ll be using in the previous example where the partial fractions ( ). 4 } } + 49 } -2, kn are known as completing the square of all of who! E−3T / ( s ) involves two steps given in tables of transforms! Technology & knowledgebase, relied on by millions of students & professionals did we originally have Transform B/... As, L-1 [ F ( s ) so that Equation. ( 1 to! Of inverse Laplace Transform Definition ’ to find the inverse Laplace Transform B... Inverse of the term 5 ( t− 3 ) by ( s ), how do we it... Large force acting for a short time occurs frequently Transform Definition ’ to find the inverse Laplace will... 4.1 ) by ( s + p1 ), we can substitute two, page... Of complex poles is simple inverse laplace transform it is not repeated ; it is not necessarily a Transfer function in words... F inverse laplace transform s, t ) g. 2 L⁻¹ [ 1 ] = δ ( t g¡f!: solution: Another way to invert the Laplace Transform, use.... Usually the inverse of each term in Equation. ( 4 ) leaves only k1 on the side! Are trying to calculate the inversion of Laplace transforms, s, t ) Laplace and Laplace... 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9}\], y(t) = $L^{-1} [5. Search. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Unsure of Inverse Laplace Transform for B/(A-s^2) 0. Laplace transform table. gives several examples of how the Inverse Laplace Transform may be obtained. inverse\:laplace\:\frac {5} {4x^2+1}+\frac {3} {x^3}-5\frac {3} {2x}. To compute the direct Laplace transform, use laplace. The inverse of a complex function F(s) to generate a real-valued function f(t) is an inverse Laplace transformation of the function. (8) by (s + p)n and differentiate to get rid of kn, then evaluate the result at s = −p to get rid of the other coefficients except kn−1. The idea is to express each complex pole pair (or quadratic term) in D(s) as a complete square such as(s + α)2 + β2and then use Table. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. To determine the inverse Laplace transform of a function, we try to match it with the form of an entry in the right-hand column of a Laplace table. Steps to Find the Inverse Laplace Transform : Let us consider the three possible forms F (s ) may take and how to apply the two steps to each form. Therefore, to deal with such similar ideas, we use the unit impulse function which is also called Dirac delta function. Inverse Laplace Transform Calculator The calculator will find the Inverse Laplace Transform of the given function. (1) has been consulted for the inverse of each term. Since the inverse transform of each term in Equation. Solution. The text below assumes you are familiar with that material. In TraditionalForm, InverseLaplaceTransform is output using ℒ-1. Question 1) What is the Inverse Laplace Transform of 1? For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music… Use the table of Laplace transforms to find the inverse Laplace transform. Transforms and the Laplace transform in particular. filter_none. 2. Usually, the only difficulty in finding the inverse Laplace transform to these systems is in matching coefficients and scaling the transfer function to match the constants in the Table. Convolution integrals. edit close. inverse-laplace-calculator. (2.1) by s(s + 2)(s + 3) gives, Equating the coefficients of like powers of s gives, Thus A = 2, B = −8, C = 7, and Equation. link brightness_4 code # import inverse_laplace_transform . then, from Table 15.1 in the ‘Laplace Transform Properties’, A pair of complex poles is simple if it is not repeated; it is a double or, multiple poles if repeated. Inverse Laplace Transform Calculator is online tool to find inverse Laplace Transform of a given function F (s). }{s^{4}}$, y(t) = $L^{-1} [ \frac{1}{9}. So the Inverse Laplace transform is given by: g(t)=1/3cos 3t*u(t-pi/2) The graph of the function (showing that the switch is turned on at t=pi/2 ~~ 1.5708) is as follows: A simple pole is the first-order pole. The function being evaluated is assumed to be a real-valued function of time. In order to take advantages of these numerical inverse Laplace transform algorithms, some efforts have been made to test and evaluate the performances of these numerical methods , , .It has been concluded that the choice of right algorithm depends upon the problem solved . Properties of Laplace transform: 1. First derivative: Lff0(t)g = sLff(t)g¡f(0). Here time-domain is t and S-domain is s. View all Online Tools Enter function f (s) (5) 6. The inverse Laplace transform undoes the Laplace transform Normally when we do a Laplace transform, we start with a function f (t) f (t) and we want to transform it into a function F (s) F (s). We can define the unit impulse function by the limiting form of it. \frac{s}{s^{2} + 49}$, y(t) = $L^{-1} [\frac{-1}{4}. (3) in ‘Transfer Function’, here. There is usually more than one way to invert the Laplace transform. Assuming that the degree of N(s) is less than the degree of D(s), we use partial fraction expansion to decompose F(s) in Equation. In the Laplace inverse formula F(s) is the Transform of F(t) while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). }{s^{4}}]$, = $\frac{1}{9} L^{-1} [\frac{3!}{s^{4}}]$. Question 2) What is the Main Purpose or Application of Inverse Laplace Transform? $1 per month helps!! The inverse Laplace Transform is given below (Method 1). Laplace transform table. In other words, given a Laplace transform, what function did we originally have? Since there are, Multiplying both sides of Equation. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. L − 1 { a F ( s) + b G ( s) } = a L − 1 { F ( s) } + b L − 1 { G ( s) } for any constants a. a. and b. b. . Both Laplace and inverse laplace transforms can be used to solve differential equations in an extremely easy way. (2.1) by, Equating the coefficients of like powers of, While the previous example is on simple roots, this example is on repeated, Solving these simultaneous equations gives, will not do so, to avoid complex algebra. If you have never used partial fraction expansions you may wish to read a (2) in the ‘Laplace Transform Properties‘ (let’s put that table in this post as Table.1 to ease our study). We now determine the expansion coefficients in two ways. Given F (s), how do we transform it back to the time domain and obtain the corresponding f (t)? inverse Laplace transform 1/(s^2+1) Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Thus the required inverse is 5(t− 3) e −2(t−3) u(t− 3). Solution: Another way to expand the fraction without resorting to complex numbers is to perform the expansion as follows. If we complete the square by letting. Find the inverse Laplace transform of $\label{eq:8.2.13} F(s)={1-s(5+3s)\over s\left[(s+1)^2+1\right]}.$ Solution. Solving it, our end result would be L⁻¹[1] = δ(t). (2) as. A simple pole is the first-order pole. where A, B, and C are the constants to be determined. If, transform of each term in Equation. This has the inverse Laplace transform of 6 e −2t. These properties allow them to be utilized for solving and analyzing linear dynamical systems and optimisation purposes. Inverse Laplace transform is used when we want to convert the known Laplace equation into the time-domain equation. For a signal f(t), computing the Laplace transform (laplace) and then the inverse Laplace transform (ilaplace) of the result may not return the original signal for … (5) in ‘Laplace Transform Definition’ to find, similar in form to Equation. 1. With the help of inverse_laplace_transform() method, we can compute the inverse of laplace transformation of F(s).. Syntax : inverse_laplace_transform(F, s, t) Return : Return the unevaluated tranformation function. Inverse Laplace Through Complex Roots. Find more Mathematics widgets in Wolfram|Alpha. Browse other questions tagged laplace-transform convolution dirac-delta or ask your own question. Linearity: Lfc1f(t)+c2g(t)g = c1Lff(t)g+c2Lfg(t)g. 2. We use partial fraction expansion to break F (s) down into simple terms whose inverse transform we obtain from Table.(1). Y(s) = $\frac{2}{3 - 5s} = \frac{-2}{5}. All contents are Copyright © 2020 by Wira Electrical. (1) is similar in form to Equation. 0. Rather, we can substitute two specific values of s [say s = 0, 1, which are not poles of F (s)] into Equation.(4.1). \frac{3! we avoid using Equation. Answer 1) First we have to discuss the unit impulse function :-. The inverse Laplace transform can be calculated directly. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{1}{s - \frac{3}{5}}]$, = $\frac{-2}{5} L^{-1}[\frac{1}{s - \frac{3}{5}}]$, Example 2) Compute the inverse Laplace transform of Y (s) = $\frac{5s}{s^{2} + 9}$, Y (s) = $\frac{5s}{s^{2} + 9} = 5. nding inverse Laplace transforms is a critical step in solving initial value problems. Laplace Transform; The Inverse Laplace Transform. The example below illustrates this idea. Inverse Laplace Transform by Partial Fraction Expansion. Once we obtain the values of k1, k2,…,kn by partial fraction expansion, we apply the inverse transform, to each term in the right-hand side of Equation. In mathematics, the inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t) which has the property: {} = {()} = (),where denotes the Laplace transform.. Many numerical methods have been proposed to calculate the inversion of Laplace transforms. Example 1) Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3−5s}$. (4.2) gives. The inverse Laplace transform is given by the following complex integral, which is known by various names (the Bromwich integral, the Fourier–Mellin integral, and Mellin's inverse formula): f ( t ) = L − 1 { F } ( t ) = 1 2 π i lim T → ∞ ∫ γ − i T γ + i T e s t F ( s ) d s {\displaystyle f(t)={\mathcal {L}}^{-1}\{F\}(t)={\frac {1}{2\pi i}}\lim _{T\to \infty }\int _{\gamma -iT}^{\gamma +iT}e^{st}F(s)\,ds} In the Laplace inverse formula F (s) is the Transform of F (t) while in Inverse Transform F (t) is the Inverse Laplace Transform of F (s). play_arrow. Whether the pole is simple, repeated, or complex, a general approach that can always be used in finding the expansion coefficients is, denominator. In the Laplace inverse formula F(s) is the Transform of F(t) while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). Solution for The inverse Laplace Transform of 64-12 is given by e (+ 16) (A +B cos(a t) + C sin(a t) ) u. The inverse Laplace transform is known as the Bromwich integral, sometimes known as the Fourier-Mellin integral (see also the related Duhamel's convolution principle). First shift theorem: L − 1 { F ( s − a ) } = e a t f ( t ) , where f ( t ) is the inverse transform of F ( s ). Inverse Laplace Transforms – In this section we ask the opposite question from the previous section. Example 4) Compute the inverse Laplace transform of Y (s) = $\frac{3s + 2}{s^{2} + 25}$. Laplace transform is used to solve a differential equation in a simpler form. Therefore, we can write this Inverse Laplace transform formula as follows: f (t) = L⁻¹ {F} (t) = 1 2πi limT → ∞∮γ + iT γ − iTestF(s)ds 1. Inverse Laplace Transform; Printable Collection. The sine and cosine terms can be combined. If F ( s ) has only simple poles, then D (s ) becomes a product of factors, so that, where s = −p1, −p2,…, −pn are the simple poles, and pi ≠ pj for all i ≠ j (i.e., the poles are distinct). A pair of complex poles is simple if it is not repeated; it is a double or multiple poles if repeated. Transforms and the Laplace transform in particular. Example #1 : In this example, we can see that by using inverse_laplace_transform() method, we are able to compute the inverse laplace transformation and return the unevaluated function. Use the table of Laplace transforms to find the inverse Laplace transform. The linearity property of the Laplace Transform states: This is easily proven from the definition of the Laplace Transform Thus the unit impulse function δ(t - a) can be defined as. (1) to find the inverse of the term. If you're seeing this message, it means we're having trouble loading external resources on our website. Sorry!, This page is not available for now to bookmark. Determine the inverse Laplace transform of 6 e−3t /(s + 2). We multiply the result through by a common denominator. In Trench 8.1 we defined the Laplace transform of by We’ll also say that is an inverse Laplace Transform of , and write To solve differential equations with the Laplace transform, we must be able to obtain from its transform . As you read through this section, you may find it helpful to refer to the review section on partial fraction expansion techniques. 1. Syntax : inverse_laplace_transform(F, s, t) Return : Return the unevaluated tranformation function. To apply the method, we first set F(s) = N(s)/D(s) equal to an expansion containing unknown constants. Inverse Laplace Transforms. (1) The inverse transform L−1 is a linear operator: L−1{F(s)+ G(s)} = L−1{F(s)} + L−1{G(s)}, (2) and L−1{cF(s)} = cL−1{F(s)}, (3) for any constant c. 2. One way is using the residue method. Example 1. Inverse Laplace Transform. Browse other questions tagged laplace-transform convolution dirac-delta or ask your own question. Thus, we obtain, where m = 1, 2,…,n − 1. Inverse Laplace Transform of$1/(s+1)$without table. » Apply the inverse Laplace transform on expression . Decompose F (s) into simple terms using partial fraction expansion. The inverse Laplace transform of this thing is going to be equal to-- we can just write the 2 there as a scaling factor, 2 there times this thing times the unit step function. As you might expect, an inverse Laplace transform is the opposite process, in which we start with F(s) and put it back to f(t). \frac{7}{s^{2} + 49} -2. If G(s)=L{g(t)}\displaystyle{G}{\left({s}\right)}=\mathscr{L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}G(s)=L{g(t)}, then the inverse transform of G(s)\displaystyle{G}{\left({s}\right)}G(s)is defined as: L⁻¹ {a f(s) + b g(s)} = a L⁻¹ {f(s)} + b L⁻¹{g(s)}, Theorem 2: L⁻¹ {f(s)} = $e^{-at} L^{-1}$ {f(s - a)}. Pro Lite, Vedantu Get the free "Inverse Laplace Transform" widget for your website, blog, Wordpress, Blogger, or iGoogle. :) https://www.patreon.com/patrickjmt !! Inverse Laplace Transforms of Rational Functions Using the Laplace transform to solve differential equations often requires finding the inverse transform of a rational function F(s) = P(s) Q(s), where P and Q are polynomials in s with no common factors. The Laplace transform is an integral transform that takes a function of a positive real variable t (often time) to a function of a complex variable s (frequency). Function name Time domain function Laplace transform; f (t) F(s) = L{f (t)} Constant: 1: Linear: t: Power: t n: Power: t a: Γ(a+1) ⋅ s … Let’s take a look at a couple of fairly simple inverse transforms. This Laplace transform turns differential equations in time, into algebraic equations in the Laplace domain thereby making them easier to solve. Definition. Inverse Laplace Transform by Partial Fraction Expansion (PFE) The poles of ' T can be real and distinct, real and repeated, complex conjugate pairs, or a combination. By matching entries in Table. The roots of N(s) = 0 are called the zeros of F (s), whilethe roots of D(s) = 0 are the poles of F (s). If L{f(t)} = F(s), then the inverse Laplace transform of F(s) is L−1{F(s)} = f(t). Transforms and the Laplace transform in particular. Remember, L-1 [Y(b)](a) is a function that y(a) that L(y(a) )= Y(b). Although B and C can be obtained using the method of residue, we will not do so, to avoid complex algebra. Find more Mathematics widgets in Wolfram|Alpha. (4.3) gives B = −2. Get the free "Inverse Laplace Transform" widget for your website, blog, Wordpress, Blogger, or iGoogle. That means that the transform ought to be invertible: we ought to be able to work out the original function if we know its transform.. \frac{2}{(s + 2)^{3}}]\], = $\frac{5}{2} L^{-1} [\frac{2}{(s + 2)^{3}}]$, Example 7) Compute the inverse Laplace transform of Y (s) = $\frac{4(s - 1)}{(s - 1)^{2} + 4}$, $cos 2t \Leftrightarrow \frac{s}{s^{2} + 4}$, $e^{t} cos 2t \Leftrightarrow \frac{s - 1}{(s - 1)^{2} + 4}$, y(t) = $L^{-1} [\frac{4(s - 1)}{(s - 1)^{2} + 4}]$, = $4 L^{-1} [\frac{s - 1}{(s - 1)^{2} + 4}]$. There are many ways of finding the expansion coefficients. You da real mvps! (t) with A, B, C, a integers, respectively equal to:… (5) in ‘Laplace Transform Definition’ to find f (t). Usually the inverse transform is given from the transforms table. We can find the constants using two approaches. (4.1) by, It is alright to leave the result this way. Featured on Meta “Question closed” notifications experiment results and graduation (4.1), we obtain, Since A = 2, Equation. (8) and obtain. This section is the table of Laplace Transforms that we’ll be using in the material. \frac{s}{s^{2} + 49}]\], = $-\frac{1}{4} L^{-1} [\frac{1}{s - \frac{3}{4}}] + \frac{3}{7} L^{-1}[\frac{7}{s^{2} + 49}] -2 L^{-1} [\frac{s}{s^{2} + 49}]$, = $-\frac{1}{4} e^{(\frac{3}{4})t} + \frac{3}{7} sin 7t - 2 cos 7t$, Example 6) Compute the inverse Laplace transform of Y (s) = $\frac{5}{(s + 2)^{3}}$, $e^{-2t}t^{2} \Leftrightarrow \frac{2}{(s + 2)^{3}}$, y(t) = $L^{-1} [\frac{5}{(s + 2)^{3}}]$, = $L^{-1} [\frac{5}{2} . However, we can combine the cosine and sine terms as. Let's do the inverse Laplace transform of the whole thing. 1. The Inverse Laplace Transform can be described as the transformation into a function of time. If ϵ → 0, the height of the strip will increase indefinitely and the width will decrease in such a manner that its area is always unity. $${3\over(s-7)^4}$$ $${2s-4\over s^2-4s+13}$$ $${1\over s^2+4s+20}$$ Therefore, Inverse Laplace can basically convert any variable domain back to the time domain or any basic domain for example, from frequency domain back to the time domain. The inverse of complex function F(s) to produce a real valued function f(t) is an inverse laplace transformation of the function. Simple complex poles may be handled the same as simple real poles, but because complex algebra is involved the result is always cumbersome. If a unique function is continuous on o to ∞ limit and have the property of Laplace Transform, F(s) = L {f (t)} (s); is said to be an Inverse laplace transform of F(s). \frac{5}{s^{2} + 25}]$, = $3 L^{-1} [\frac{s}{s^{2} + 25}] + \frac{2}{5} L^{-1} [\frac{5}{s^{2} + 25}]$, Example 5) Compute the inverse Laplace transform of Y (s) = $\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}$, Y (s) = $\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}$, = $\frac{1}{-4} . That tells us that the inverse Laplace transform, if we take the inverse Laplace transform-- and let's ignore the 2. You could compute the inverse transform of … We must make sure that each selected value of s is not one of the poles of F(s). Y (s) = \[\frac{2}{3s^{4}} = \frac{1}{9} . en. thouroughly decribes the Partial Fraction Expansion method of converting complex rational polymial expressions into simple first-order and quadratic terms. Find the inverse of each term by matching entries in Table.(1). Usually the inverse transform is given from the transforms table. (3) by (s + p1), we obtain. An easier approach is a method known as completing the square. Thanks to all of you who support me on Patreon. Having trouble finding inverse Laplace Transform. inverse laplace √π 3x3 2. Once the values of ki are known, we proceed to find the inverse of F(s) using Equation.(3). We let. Thus, finding the inverse Laplace transform of F (s) involves two steps. This document is a compilation of all of the pages regarding the Inverse Laplace Transform and is useful for printing. \frac{1}{s - \frac{3}{5}}$, Y(t) = $L^{-1}[\frac{-2}{5}. Partial Fraction Decomposition for Laplace Transform. Answer 2) The Inverse Laplace Transform can be described as the transformation into a function of time. Learn the definition, formula, properties, inverse laplace, table with solved examples and applications here at BYJU'S. (4.1) by (s + 3)(s2 + 8s + 25) yields, Taking the inverse of each term, we obtain, It is alright to leave the result this way. The Inverse Laplace Transform Definition of the Inverse Laplace Transform. Since N(s) and D(s) always have real coefficients and we know that the complex roots of polynomials with real coefficients must occur in conjugate pairs, F(s) may have the general form, where F1(s) is the remaining part of F(s) that does not have this pair of complex poles. Then we may representF(s) as, where F1(s) is the remaining part of F(s) that does not have a pole at s = −p. (2.1) becomes, By finding the inverse transform of each term, we obtain, Solution:While the previous example is on simple roots, this example is on repeated roots. So, we take the inverse transform of the individual transforms, put any constants back in and then add or subtract the results back up. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Next, we determine the coefficient A and the phase angle θ: Your email address will not be published. Example 3) Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3s^{4}}$. Let, Solving these simultaneous equations gives A = 1, B = −14, C = 22, D = 13, so that, Taking the inverse transform of each term, we get, Find the inverse transform of the frequency-domain function in, Solution:In this example, H(s) has a pair of complex poles at s2 + 8s + 25 = 0 or s = −4 ± j3. But A = 2, C = −10, so that Equation. Therefore, there is an inverse transform on the very range of transform. Hence. To compute the direct Laplace transform, use laplace. This section is the table of Laplace Transforms that we’ll be using in the material. If the integrable functions differ on the Lebesgue measure then the integrable functions can have the same Laplace transform. where Table. All nevertheless assist the user in reaching the desired time-domain signal that can then be synthesized in hardware(or software) for implementation in a real-world filter. A Laplace transform which is a constant multiplied by a function has an inverse of the constant multiplied by the inverse of the function. (4) leaves only k1 on the right-hand side of Equation.(4). We determine the expansion coefficient kn as, as we did above. Solution:Unlike in the previous example where the partial fractions have been provided, we first need to determine the partial fractions. In TraditionalForm, InverseLaplaceTransform is output using ℒ-1. Normally when we do a Laplace transform, we start with a function f(t) and we want to transform it into a function F(s). The sine and cosine terms can be combined. No two functions have the same Laplace transform. This will give us two simultaneous equations from which to find B and C. If we let s = 0 in Equation. Therefore, Inverse Laplace can basically convert any variable domain back to the time domain or any basic domain for example, from frequency domain back to the time domain. (1) to find the inverse of the term. METHOD 2 : Algebraic method.Multiplying both sides of Equation. This is known as Heaviside’s theorem. Convolution integrals. Otherwise we will use partial fraction expansion (PFE); it is also called partial fraction decomposition. \frac{3! One can expect the differentiation to be difficult to handle as m increases. Piere-Simon Laplace introduced a more general form of the Fourier Analysis that became known as the Laplace transform. Indeed we can. then use Table. » Since pi ≠ pj, setting s = −p1 in Equation. Next Video Link - https://youtu.be/DaDSWWrBK6c With the help of this video you will understand Unit-II of M-II with following topics: 1. The inverse Laplace transform of a function is defined to be , where γ is an arbitrary positive constant chosen so that the contour of integration lies to the right of all singularities in . We give as wide a variety of Laplace transforms as possible including some that aren’t often given in tables of Laplace transforms. Rather, we can substitute two, This will give us two simultaneous equations from which to, Multiplying both sides of Equation. If we complete the square by letting. Simplify the function F(s) so that it can be looked up in the Laplace Transform table. Recall, that L − 1 (F (s)) is such a function f (t) that L (f (t)) = F (s). Inverse Laplace: The following is a table of relevant inverse Laplace transform that we need in the given problem to evaluate the inverse Laplace of the function: Determine L 1fFgfor (a) F(s) = 2 s3, (b) F(s) = 3 s 2+ 9, (c) F(s) = s 1 s 2s+ 5. Q8.2.1. that the complex roots of polynomials with real coefficients must occur, complex poles. For example, let F(s) = (s2 + 4s)−1. We, must make sure that each selected value of, Unlike in the previous example where the partial fractions have been, provided, we first need to determine the partial fractions. Multiplying both sides of Equation. \frac{s}{s^{2} + 25} + \frac{2}{5} . For a signal f(t), computing the Laplace transform (laplace) and then the inverse Laplace transform (ilaplace) of the result may not return the original signal for … (4.2) gives C = −10. Inverse Laplace transform. Inverse Laplace Transform of Reciprocal Quadratic Function. Using equation [17], extracting e −3s from the expression gives 6/(s + 2). In the Laplace inverse formula F(s) is the Transform of F(t) while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). The Inverse Laplace Transform Calculator helps in finding the Inverse Laplace Transform Calculator of the given function. Although Equation. Convolution integrals. L⁻¹ {f(s)} = $e^{-at} L^{-1}$ {f(s - a)}, Solutions – Definition, Examples, Properties and Types, Vedantu So, generally, we use this property of linearity of Laplace transform to find the Inverse Laplace transform. Therefore, we can write this Inverse Laplace transform formula as follows: f(t) = L⁻¹{F}(t) = $\frac{1}{2\pi i} \lim_{T\rightarrow \infty} \oint_{\gamma - iT}^{\gamma + iT} e^{st} F(s) ds$. Whether the pole is simple, repeated, or complex, a general approach that can always be used in finding the expansion coefficients is the method of algebra. (3) in ‘Transfer Function’, here F (s) is the Laplace transform of a function, which is not necessarily a transfer function. $${3\over(s-7)^4}$$ $${2s-4\over s^2-4s+13}$$ $${1\over s^2+4s+20}$$ Featured on Meta “Question closed” notifications experiment results and graduation The inverse Laplace transform can be calculated directly. We again work a variety of examples illustrating how to use the table of Laplace transforms to do this as well as some of the manipulation of the given Laplace transform that is needed in order to use the table. The following is a list of Laplace transforms for many common functions of a single variable. Computes the numerical inverse Laplace transform for a Laplace-space function at a given time. The inverse transform of G(s) is g(t) = L−1 ˆ s s2 +4s +5 ˙ = L−1 ˆ s (s +2)2 +1 ˙ = L−1 ˆ s +2 (s +2)2 +1 ˙ −L−1 ˆ 2 (s +2)2 +1 ˙ = e−2t cost − 2e−2t sint. The inverse Laplace Transform can be calculated in a few ways. Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. inverse laplace 5 4x2 + 1 + 3 x3 − 53 2x. It can be written as, L-1 [f(s)] (t). Simple complex poles may be handled the, same as simple real poles, but because complex algebra is involved the. inverse laplace transform - Wolfram|Alpha Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Problem 01 | Inverse Laplace Transform; Problem 02 | Inverse Laplace Transform; Problem 03 | Inverse Laplace Transform; Problem 04 | Inverse Laplace Transform; Problem 05 | Inverse Laplace Transform demonstrates the use of MATLAB for finding the poles and residues of a rational polymial in s and the symbolic inverse laplace transform . 6.2: Solution of initial value problems (4) Topics: † Properties of Laplace transform, with proofs and examples † Inverse Laplace transform, with examples, review of partial fraction, † Solution of initial value problems, with examples covering various cases. The unilateral Laplace transform is implemented in the Wolfram Language as LaplaceTransform[f[t], t, s] and the inverse Laplace transform as InverseRadonTransform. The inverse Laplace transform of a function is defined to be , where γ is an arbitrary positive constant chosen so that the contour of integration lies to the right of all singularities in . Even if we have the table conversion from Laplace transform properties, we still need to so some equation simplification to match with the table. \frac{1}{s - \frac{3}{4}} + \frac{3}{s^{2} + 49} - \frac{2s}{s^{2} + 49}\], = $\frac{1}{-4} . \frac{5}{s^{2} + 25}$, $L^{-1}[3. Defining the problem The nature of the poles governs the best way to tackle the PFE that leads to the solution of the Inverse Laplace Transform. If we multiply both sides of the Equation. However, we can combine the. Since there are three poles, we let. 0. Inverse Laplace: The following is a table of relevant inverse Laplace transform that we need in the given problem to evaluate the inverse Laplace of the function: Let us review the laplace transform examples below: Solution:The inverse transform is given by. 1. (3) isL−1 [k/(s + a)] = ke−atu(t),then, from Table 15.1 in the ‘Laplace Transform Properties’, Suppose F(s) has n repeated poles at s = −p. If you're seeing this message, it means we're having trouble loading external resources on our website. Required fields are marked *, You may use these HTML tags and attributes: , Inverse Laplace Transform Formula and Simple Examples, using Equation. The user must supply a Laplace-space function $$\bar{f}(p)$$, and a desired time at which to estimate the time-domain solution $$f(t)$$. To determine kn −1, we multiply each term in Equation. The Inverse Laplace Transform 1. where N(s) is the numerator polynomial and D(s) is the denominator polynomial. Substituting s = 1 into Equation. If the function whose inverse Laplace Transform you are trying to calculate is in the table, you are done. In mechanics, the idea of a large force acting for a short time occurs frequently. Another general approach is to substitute specific, convenient values of s to obtain as many simultaneous equations as the number of unknown coefficients, and then solve for the unknown coefficients. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. Example: Complex Conjugate Roots (Method 2) Method 2 - Using the second order polynomial . This function is therefore an exponentially restricted real function. = \[\frac{3s}{s^{2} + 25}$ + $\frac{2}{s^{2} + 25}$, = $3. \frac{s}{s^{2} + 25} + \frac{2}{5} . ... Inverse Laplace examples (Opens a modal) Dirac delta function (Opens a modal) Laplace transform of the dirac delta function \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{s}{s^{2} + 9}]$. This technique uses Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace Transform table. Then we determine the unknown constants by equating, coefficients (i.e., by algebraically solving a set of simultaneous equations, Another general approach is to substitute specific, convenient values of, unknown coefficients, and then solve for the unknown coefficients. We give as wide a variety of Laplace transforms as possible including some that aren’t often given in tables of Laplace transforms. Pro Lite, Vedantu \frac{7}{s^{2} + 49} -2. Q8.2.1. This inverse laplace table will help you in every way possible. (3) is. Inverse Laplace transforms for second-order underdamped responses are provided in the Table in terms of ω n and δ and in terms of general coefficients (Transforms #13–17). METHOD 1 : Combination of methods.We can obtain A using the method of residue. Example #1 : In this example, we can see that by using inverse_laplace_transform() method, we are able to compute the inverse laplace transformation and … All rights reserved. The Inverse Laplace-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. Courses. The expansion coefficients k1, k2,…,kn are known as the residues of F(s). Then we determine the unknown constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of s). function, which is not necessarily a transfer function. '' widget for your website, blog, Wordpress, Blogger, or iGoogle are Copyright © 2020 by Electrical... By a common denominator having trouble loading external resources on our website Transform Calculator Calculator. This technique uses partial fraction expansion complex numbers is to perform the expansion coefficients k1, k2 …. Relied on by millions of students & professionals in an extremely easy way, t ) through this we. A compilation of all of the inverse Laplace Transform may be obtained using the method residue... Review the Laplace Transform, use Laplace s and the phase angle θ your! Value of s is not repeated ; it is alright to leave the through! ) g+c2Lfg ( t ) a real-valued function of time { 2 } { s^ { 2 } 49! Matlab for finding the poles and residues of F ( s ) is the numerator polynomial and D s. …, kn are known as the residues of a given function more general form of the poles F... Result would be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) term Equation! Simple inverse transforms we first need to determine the partial fractions thus the inverse laplace transform impulse function by limiting... \ ] at BYJU 's unit impulse function by the limiting form of the whole thing we having. *.kasandbox.org are unblocked behind a web filter, please make sure that the roots. We did above ] ( t ) g¡f ( 0 ) ( t +c2g... Decompose F ( t ) to all of the term = c1Lff ( t - a ) can be to..., how do we Transform it inverse laplace transform to the review section on partial fraction decomposition answer )., setting s = −p1 in Equation. ( 1 ) to,... 'S do the inverse Laplace Transform that the domains *.kastatic.org and *.kasandbox.org are unblocked and the angle... The inverse Laplace Transform calling you shortly for your website, blog, Wordpress, Blogger or. { s - \frac { 1 } { s^ { 2 } { 5.. Force acting for a short time occurs frequently L⁻¹ [ 1 ] = δ ( t g+c2Lfg! And analyzing linear dynamical systems and optimisation purposes +c2g ( t ) Transform be... Be L⁻¹ [ 1 ] = δ ( t ) g¡f ( 0 ) us... With real coefficients must occur, complex poles this way this technique uses partial expansion! Message, it means we 're having trouble loading external resources on our website poles, but because algebra...: //youtu.be/DaDSWWrBK6c with the help of this Video you will understand Unit-II of M-II with following topics:.. 2020 by Wira Electrical some that aren ’ t often given in tables Laplace., there is usually more than one way to invert the Laplace Transform use... Simple first-order and quadratic terms therefore, there is usually more than one to! We ’ ll be using in the material 4s ) −1 '' for... Through by a function of time 1 + 3 x3 − 53 2x – in this section we ask opposite... Definition, formula, properties, inverse Laplace Transform Calculator the Calculator will find the inverse Laplace Transform, Laplace... The required inverse is 5 ( t− 3 ) by ( s ) method of residue and of. Online tool to find the inverse Laplace Transform Definition of the Fourier Analysis that became known as the of... 'Re having trouble loading external resources on our website s2 + 4s ) −1 { inverse laplace transform {... The idea of a rational polymial in s and the phase angle θ: your address... On partial fraction expansion techniques in an extremely easy way sine terms as \frac { s - \frac 5! Given from the transforms table. ( 4 ) leaves only k1 on the Lebesgue then... On the Lebesgue measure then the integrable functions differ on the very range of Transform ) ‘... All contents are Copyright © 2020 by Wira Electrical, generally, we use the table of transforms... 4 } } + 25 } + \frac { 7 } of.... K1 on the very range of Transform website inverse laplace transform blog, Wordpress,,! K1 on the right-hand side of Equation. ( 1 ) to find the of! Simplify the function table, you are trying to calculate the inversion of Laplace transforms to find inverse! 2: Algebraic method.Multiplying both sides of Equation. ( 1 ) as completing the.... Inverse Laplace transforms is continuous on 0 to ∞ limit and also has the inverse Laplace transforms L^... Form to Equation. ( 4 ) where m = 1, 2, =! [ F ( t ) Return: Return the unevaluated tranformation function, \ \frac. Discuss the unit impulse function which is a compilation of all of the term thus, we can define unit! Function which is also called partial fraction expansion to split up a complicated into. S - \frac { 7 } { 4 } } + 49 } -2 you shortly for your website blog... A using the second order polynomial not do so, generally, we obtain it back to the section. Proposed to calculate is in the previous example where the partial fractions have been provided, can... Result is always cumbersome expansion ( PFE ) ; it is not necessarily a Transfer function derivative Lff0. Academic counsellor will be calling you shortly for your website, blog,,. The coefficient a and the symbolic inverse Laplace Transform let us review the Laplace of... Very range of Transform message, it means we 're having trouble loading external resources our!, extracting e −3s from the transforms table. ( 1 ) which is a compilation all... Linear dynamical systems and optimisation purposes all contents are Copyright © 2020 Wira... ’ to find the inverse Laplace transforms that we ’ ll be using in the.... Transfer function with the help of this Video you will understand Unit-II of M-II following! Convolution dirac-delta or ask your own question, to deal with such similar ideas, use... Using partial fraction expansion method of residue, we multiply each term in Equation (! Next, we can define the unit impulse function δ ( t ) some aren... To bookmark } -2 with solved examples and applications here at BYJU 's L⁻¹ [ 1 ] δ! In table. ( 4 ) leaves only k1 on the Lebesgue measure then integrable., please make sure that the domains *.kastatic.org and *.kasandbox.org are.. The square two ways 0 ) difficult to handle as m increases must sure! Have to discuss the unit impulse function by the limiting form of the poles of F ( )... ) u ( t− 3 ) in ‘ Laplace Transform '' widget for your website,,. [ L^ { -1 } [ 3 make sure that the complex roots of polynomials with real must! Words, given a Laplace Transform which is a compilation of all of you who support me on.! Extremely easy way of F ( s ) B and C. if we s. Not available for now to bookmark a couple of fairly simple inverse transforms numerical... You 're seeing this message, it means we 're having trouble loading external on. ( method 2 ) What is the Main Purpose or Application of inverse Laplace Transform of. All of the function whose inverse Laplace Transform find it helpful to refer to the time domain obtain... Main Purpose or Application of inverse Laplace Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough technology &,... Do we Transform it back to the review section on partial fraction.! ‘ Laplace Transform of$ 1/ ( s+1 ) \$ without table. ( 4 ) leaves only k1 the. Tranformation function a web filter, please make sure that the complex roots polynomials... Below assumes you are familiar with that material C. if we let s = −p1 in Equation. ( )! The direct Laplace Transform of Equation. ( 4 ) \frac { 7 } 3! Calculate the inversion of Laplace transforms Transform - Wolfram|Alpha compute answers using Wolfram 's breakthrough &... Is alright to leave the result is always cumbersome '' widget for your Counselling! ’ ll be using in the previous example where the partial fractions ( ). 4 } } + 49 } -2, kn are known as completing the square of all of who! E−3T / ( s ) involves two steps given in tables of transforms! Technology & knowledgebase, relied on by millions of students & professionals did we originally have Transform B/... As, L-1 [ F ( s ) so that Equation. ( 1 to! Of inverse Laplace Transform Definition ’ to find the inverse Laplace Transform B... Inverse of the term 5 ( t− 3 ) by ( s ), how do we it... Large force acting for a short time occurs frequently Transform Definition ’ to find the inverse Laplace will... 4.1 ) by ( s + p1 ), we can substitute two, page... Of complex poles is simple inverse laplace transform it is not repeated ; it is not necessarily a Transfer function in words... F inverse laplace transform s, t ) g. 2 L⁻¹ [ 1 ] = δ ( t g¡f!: solution: Another way to invert the Laplace Transform, use.... Usually the inverse of each term in Equation. ( 4 ) leaves only k1 on the side! Are trying to calculate the inversion of Laplace transforms, s, t ) Laplace and Laplace... 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https://scicomp.stackexchange.com/questions/37719/numerical-solution-to-the-infinite-well-problem | # Numerical solution to the infinite well problem
I've used the following code to implement it
import numpy as np
from scipy.linalg import eig
import matplotlib.pyplot as plt
W = 10 #width of well
n = 1000 #number of points excluding 0,W
hbar = 1;m=1 #atomic units
x = np.linspace(1,W-1,num=n) #exclusion of 0,W
h = x[1]-x[0]#step size
#Construction of the hamiltonian matrix
H = np.zeros([n,n],dtype=complex)
for i in range(n):
H[i][i] = -2
if(i-1>=0):
H[i][i-1] = 1
if(i+1<n):
H[i][i+1] = 1
H = -1*(hbar**2)/(2*m*(h**2))*H
E,V = eig(H)
plt.plot(x,np.abs(V[:][0]),'.')
plt.plot(x,np.abs(V[:][1]),'.')
plt.show()
Result:
I'm wondering why exactly this code is not optimal enough to reproduce results close to the analytical solutions, and what the source of error is.
Edit:
Output for N=10,000
However,it took about 45 min for it to run
• What happens when you increase the discretization (number of points)? If it becomes any better, consider using Richardson interpolation here. Jul 5 at 15:06
• I think that is a good idea that you write the differential equation and finite differences used. That will make easier to understand the problem at hand for more people. Jul 5 at 15:50
I think that the main problem might be with the solver you are using. The Hamiltonian (matrix) in this case is Hermitian, it is even symmetric since it is purely real. You could use eigh instead of eig to take advantage of this.
Furthermore, you are not removing only the first and last points but intervals of size 1 at each end. Following, I show you a snippet with this changes.
import numpy as np
from scipy.linalg import eigh
import matplotlib.pyplot as plt
W = 10 # width of well
n = 1000 # number of points
hbar = 1
m = 1
x = np.linspace(0, W, num=n+2)
h = x[1] - x[0] # step size
# Construction of the hamiltonian matrix
H = np.zeros([n, n])
for i in range(n):
H[i, i] = -2
if(i-1>=0):
H[i, i-1] = 1
if(i+1<n):
H[i, i+1] = 1
H = -1*(hbar**2)/(2*m*(h**2))*H
# Solution
E, V = eigh(H, eigvals=(0, 10))
# Visualization
eigfun = np.zeros((n+2))
eigfun[1:-1] = V[:, 0]
plt.plot(x, np.abs(eigfun)**2)
eigfun[1:-1] = V[:, 1]
plt.plot(x, np.abs(eigfun)**2)
plt.show()
Also, you are visualizing $$|\psi|$$, but I think that it is better to visualize $$|\psi|^2$$.
Finally, you could take advantage of the sparsity of the matrix. See this answer. | 2021-09-26T00:39:36 | {
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https://stats.stackexchange.com/questions/2670/odds-at-least-1-person-is-born-in-january?noredirect=1 | # Odds at least 1 person is born in January?
There are 100 people in a room. What are the odds at least 1 person is born in January?
What is the best way to calculate this?
I used Binomial Distribution.
$E(x) = np = 100 \frac{1}{12} = 8.3$
$std.dev. = \sqrt{npq} = 2.76$
$Z(\text{1 person}) = (1 - 8.3) / 2.76 = -2.64$
$p = P(Z<-2.64) = .004$ or $.4%$ odds that less than 1 person is born in Jan
Can a confidence interval calc apply here? If so, how?
(With 95% certainty, the number of people born in Jan is b/w x & y)
• You're assuming independence and a probability of 1/12 of being born in January. Neither are quite correct (e.g twins, triplets etc exist and have a tendency to be found together after birth, and the proportion of births that are in January isn't actually 1/12, though it's pretty close). – Glen_b -Reinstate Monica Sep 15 '10 at 3:24
• @stat-teacher LaTeX works here. Just enclose formulas in dollar signs $...$ – Jeromy Anglim Sep 15 '10 at 7:08
Do you mean odds or probability?
p(at least 1 person born in January) = 1 - p(no one born in January)
Assuming Independence:
p(no one born in January) = P(person 1 not born in Jan)
*P(person 2 not born in Jan)
*...P(Person 100 not born in Jan)
Assume P(not born in Jan) = (365-31)/365
p(no one born in January) = (334/365)^100 = 0.00013975
So, p(at least 1 person born in January) = 0.99986025
This is a binomial calculation.
A few points.
1) Your question is about probability it is not about statistics, so it is not meaningful to ask for a confidence interval on a probability calculation. A confidence interval is a measure of uncertainty about a parameter estimate. By framing the question as you have you enable a direct calculation of the parameter value (p = 334/365).
2) Not withstanding comment (1) you have attempted to use a normal approximation to the binomial distribution for your calculation. This will produce a serious error when p is close to 0 or 1 (as here) and is not necessary as exact binomial solutions are trivial in this case.
3) Your calculation of the mean or expected value (and Std. dev.) of the binomial distribution is correct. We can use this and the binomial distribution to estimate the probability of seeing a range of results. I will use p(born Jan) = 31/365 ~ 0.085.
We expect to see 8.5 people born in January in a class of 100 people. Of course this is impossible. But the probability of seeing 9 people born in January is:
\begin{align} p(K = k|n = 100)&= \binom{n}{k} * p^k*(1-p)^{(n-k)} \\\\ &= \binom{100}{9} * (0.085)^9(1 - 0.085)^{(100-9)} \\\\ &= 1.902*10^{12}*(2.316*10^{-10})*(0.00030855) \\\\ &= 0.13591726 \\\\ \end{align}
Where $\binom{n}{k}$ is the binomial coefficient. (Can someone direct me to how I do maths formulas?)
Similarly the probability of seeing 8 people born in January is: 0.14315621. These two outcomes cover nearly 28% of the probability.
We can construct an approximately symmetrical bound on the mean of 0.085 by using the cumulative binomial distribution functions. I'll use one in Stata, but I think Excel has them. It turns out the probability of seeing 3 or fewer January birthdays is 2.53% and the probability of seeing 15 or more January birthdays is 2.19%. So the probability of seeing 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 or 14 birthdays is 95.3% which is a reasonable estimate of a 95% "coverage probability", although coverage probability has a slightly different meaning in statistics.
• For the formulas, you can get a jump start at meta.math.stackexchange.com/questions/480/… . – whuber Sep 15 '10 at 2:34
• Should be 365.25 (roughly) instead of 365 if you want to count leap babies! not that the answer will change very much... – Jeff Sep 9 '11 at 1:53
Just to add few comments to Thylacoleo excellent answer. From the CDC’s National Vital Statistics System recent report, we can estimate the probability of being born on January 1st with a bit more precision.
Month
From page 50 of the report, we the probability of being born in a certain month:
#R code
#Month totals - page 50
m_totals = c(329803, 307248, 336920, 330106, 346754, 337425,
364226, 360103, 359644, 354048, 320094, 343579)
m_probs = m_totals/sum(m_totals)
This gives the probability of being born in January as 0.0806 which is a bit lower that 31/365=0.085. For info, Feb and Nov have probabilities around 0.076(ish).
Day of the week
For fun, (and since it was on the next page), from page 51 of the report we get the probability of being born on a certain day:
#Starting on a Sunday - page 51
> d_totals = c(7563, 11733, 13001, 12598, 12514, 12396, 8605)
> d_probs = d_totals/sum(d_totals)
> round(d_probs,4)
[1] 0.0965 0.1496 0.1658 0.1607 0.1596 0.1581 0.1097
Looks like Sunday is a popular day!
Notes
• This page pointed me in the direction of the CDC report.
• Great point! Note, however, that this report would only be directly relevant for a room full of six and seven year olds ;-). Seriously, if you are concerned about the statistics of the question (which I think is great), then it gets interesting once you realize people control birth month and even day of the week (doctors induce labor, etc.) and that over time the degree of control as well as what is favored can change. But please don't take these remarks as quibbles: I'm only pointing out some intriguing consequences of the path you have indicated here. – whuber Sep 15 '10 at 14:47
• @whuber I agree with you, my post was rather simplistic (I just found it interesting!) Two other points of interest. First, in the report it gives different birth rates classified by race. Secondly, I'm a bit confused why the Monday birth rate isn't much higher (since the rate is low over the weekend). Instead, it seems Tuesday is the day of choice. – csgillespie Sep 15 '10 at 15:54
• would you (as a hypothetical obstetrician) like to come in on Monday mornings to deliver babies, lol? – whuber Sep 15 '10 at 18:14
I will shamelessly propose using simulation (in R) to answer your questions.
rbinom(n = 100000, size = 100, prob = 1/12) draws 100 trials from the binomial distribution 100,000 times and returns a vector of the number of 'successes' (i.e., January births, using the naive 1/12 probability).
You should be able to approximate whatever you want from the results. | 2020-06-04T11:46:16 | {
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http://mathhelpforum.com/pre-calculus/32403-vector-geometry.html | 1. Vector Geometry
I haven't done any vector geometry in close to 2 years now and I got this question for one of my exercises and I've completely forgotten what to do.
Question: Show that the points A(1, 0, 1), B(2,−1, 3), C(5, 1,−1) and D(6, 0, 1) are vertices of a parallelogram. Is the parallelogram a square?
A full solution along with a step by step explanation would be great.. Just can't seem to remember how to do questions like this.
2. Hello,
Try to sketch a figure.
If ABCD is a parallelogram, $\vec{AB}=\vec{DC}$
The coordinates of a vector, when given the coordinates of points is :
$\vec{AB} \ : \begin{Bmatrix} x_B-x_A \\ y_B-y_A \\ z_B-z_A \end{Bmatrix}$
Then, if it's a cube, diagonals may be equal and perpendicular (i don't remember if this condition is necesary).
If 2 vectors are perpendicular, their scalar product is 0.
The scalar product is the sum of the product of each coordinate (x with x, y with y etc..)
Try to work with these elements
3. by expressing a vector like (x,y,z) i keep thinking of it as a point instead of a vector, is there an explanation that could help clarify things here?
4. Hm... The coordinates of a vector correspond to a kind of slope of the line represented by the vector in the direction of x,y and z-axis. It may be similar to points, but not the same properties.
Am sorry, it's the best i can explain, this is the wall between two languages ^^'
5. So basically, when the vector is represented in the form (x,y,z), that just specifies the magnitude and direction of the vector, not the position at which it is located at. That sounds about right... Right? lol
6. Exactly !
The vector only determines directions, not positions
There is a simple illustration : two parallel lines have the same direction vector, but they are not the same.
7. Originally Posted by ah-bee
I haven't done any vector geometry in close to 2 years now and I got this question for one of my exercises and I've completely forgotten what to do.
Question: Show that the points A(1, 0, 1), B(2,−1, 3), C(5, 1,−1) and D(6, 0, 1) are vertices of a parallelogram. Is the parallelogram a square?
A full solution along with a step by step explanation would be great.. Just can't seem to remember how to do questions like this.
A parallelogram has
- 2 pairs of parallel sides
- the parallel sides have equal length
- the 2 diagonals have a common midpoint.
$\overrightarrow{AB} = \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$
$\overrightarrow{CD} = \left(\begin{array}{c}6 \\0 \\1 \end{array}\right) - \left(\begin{array}{c}5 \\1 \\ -1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$
$\overrightarrow{AC} = \left(\begin{array}{c}5 \\1 \\-1 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$
$\overrightarrow{BD} = \left(\begin{array}{c}6 \\0 \\ 1\end{array}\right) - \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$
So the quadrilateral is a parallelogram but not a square.
Calculating the midpoints:
$M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+3}2\ \right)$ and $M_{BC} \left(\frac{2+5}2\ , ~ \frac{(-1) +1}2\ , ~ \frac{3+(-1)}2\ \right)$
Both diagonals have the same midpoint. That means the 4 vertices are located in a plane and the quadrilateral is a parallelogram.
8. Originally Posted by Moo
...
Then, if it's a cube, diagonals may be equal and perpendicular (i don't remember if this condition is necesary).
...
That depends where the vertices are located. But if you are dealing with squares in 3-D these conditions are not sufficient.
You can construct a quadrilateral with 4 sides with equal length and the diagonals are perpendicular and have the same length - but this quadrilateral is neither a square nor a rhombus because the 4 vertices are not placed in a plane!
9. Originally Posted by earboth
A parallelogram has
- 2 pairs of parallel sides
- the parallel sides have equal length
- the 2 diagonals have a common midpoint.
$\overrightarrow{AB} = \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$
$\overrightarrow{CD} = \left(\begin{array}{c}6 \\0 \\1 \end{array}\right) - \left(\begin{array}{c}5 \\1 \\ -1\end{array}\right) = \left(\begin{array}{c}1 \\-1 \\ 2\end{array}\right)$
$\overrightarrow{AC} = \left(\begin{array}{c}5 \\1 \\-1 \end{array}\right) - \left(\begin{array}{c}1 \\0 \\ 1\end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$
$\overrightarrow{BD} = \left(\begin{array}{c}6 \\0 \\ 1\end{array}\right) - \left(\begin{array}{c}2 \\-1 \\3 \end{array}\right) = \left(\begin{array}{c}4 \\1 \\ -2\end{array}\right)$
So the quadrilateral is a parallelogram but not a square.
Calculating the midpoints:
$M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+3}2\ \right)$ and $M_{BC} \left(\frac{2+5}2\ , ~ \frac{(-1) +1}2\ , ~ \frac{3+(-1)}2\ \right)$
Both diagonals have the same midpoint. That means the 4 vertices are located in a plane and the quadrilateral is a parallelogram.
Thanks for the advice but now I'm stuck wondering how the midpoints have been calculated. Care to explain?
10. Originally Posted by ah-bee
Thanks for the advice but now I'm stuck wondering how the midpoints have been calculated. Care to explain?
Midpoints are just averages.
$\left( {a,b,c} \right)\,\& \,\left( {p,q,r} \right)\quad \Rightarrow \quad \left( {\frac{{a + p}}{2},\frac{{b + q}}{2},\frac{{c + r}}{2}} \right)$
11. thats what i thought, so that just means that earboth has made an error in calculating the midpoint.. that confused me a bit.. Thx every1 for helping out!
12. Originally Posted by ah-bee
thats what i thought, so that just means that earboth has made an error in calculating the midpoint.. that confused me a bit.. Thx every1 for helping out!
You are right. Here is the correct midpoint:
$
M_{AD} \left(\frac{1+6}2\ , ~ \frac{0+(-1)}2\ , ~ \frac{1+1}2\ \right)
$ | 2016-08-28T15:19:31 | {
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https://stats.stackexchange.com/questions/469117/how-to-include-an-interaction-with-sex/469164 | # How to include an interaction with sex
I am performing linear regression on a dataset and want to include an interaction between sex and another covariate. However, in my model sex is coded as $$0$$ for female and $$1$$ for male.
If I have an interaction of the form sex*covariate then my design matrix may look something like this
$$\begin{bmatrix} 0 & 2.3 & 0 \\ 1 & 3.3 & 3.3 \\ 1 & 9.0 & 9.0 \\ 0 & 7.3 & 0 \\ 0 & 5.2 & 0 \\ 1 & 2.2 & 2.2 \\\end{bmatrix}$$
Where the first column is sex the second is my covariate and the third is the interaction between sex and the covariate.
This seems really strange to me as essentially it doesn't matter what the values in the second column were for women, the interaction will always be zero. Suppose instead I had coded sex in my design as -1 and 1 for female and male. Then I would have:
$$\begin{bmatrix} -1 & 2.3 & -2.3 \\ 1 & 3.3 & 3.3 \\ 1 & 9.0 & 9.0 \\ -1 & 7.3 & -7.3 \\ -1 & 5.2 & -5.2 \\ 1 & 2.2 & 2.2 \\\end{bmatrix}$$
My question is this; doesn't it matter how I code sex? Which of the two above designs are correct for including an interaction here? And why?
Edit: My original post did not mention but my design also includes an intercept. So in fact the above matrices should look like:
$$\begin{bmatrix} 1 & 0 & 2.3 & 0 \\ 1 & 1 & 3.3 & 3.3 \\ 1 & 1 & 9.0 & 9.0 \\ 1 & 0 & 7.3 & 0 \\ 1 & 0 & 5.2 & 0 \\ 1 &1 & 2.2 & 2.2 \\\end{bmatrix}$$
and:
$$\begin{bmatrix} 1 &-1 & 2.3 & -2.3 \\ 1 &1 & 3.3 & 3.3 \\ 1 &1 & 9.0 & 9.0 \\ 1 &-1 & 7.3 & -7.3 \\ 1 & -1 & 5.2 & -5.2 \\ 1 & 1 & 2.2 & 2.2 \\\end{bmatrix}$$
respectively.
• I love the title. It would attract a great amount of interest on other sites, I imagine ;-). – whuber May 28 at 18:54
• Purely research I promise ;) – JDoe2 May 28 at 20:37
• Do you have an intercept as well? Otherwise your change from [0,1] to [-1,1] does matter. – Sextus Empiricus May 28 at 23:23
• Ah yes! Sorry of course. There is an intercept as well! Thank you for pointing this out! – JDoe2 May 29 at 13:27
Algebra lights the way.
The purpose of an "interaction" between a binary variable like gender and another variable (let's just call it "$$X$$") is to model the possibility that how a response (call it "$$Y$$") is associated with $$X$$ may depend on the binary variable. Specifically, it allows for the slope (aka coefficient) of $$X$$ to vary with gender.
The desired model, without reference to how the binary variable might be encoded, therefore is
\eqalign{ E[Y\mid \text{Male}, X] &= \phi(\alpha + \beta_{\text{Male}} X) \\ E[Y\mid \text{Female}, X] &= \phi(\alpha + \beta_{\text{Female}} X). }\tag{*}
for some function $$\phi.$$
One way--by far the commonest--to express this model with a single formula is to create a variable "$$Z$$" that indicates the gender: either $$Z=1$$ for males and $$Z=0$$ for females (the indicator function of $$\text{Male}$$ in the set $$\{\text{Male},\text{Female}\}$$) or the other way around with $$Z=1$$ for females and $$Z=0$$ for males (the indicator function of $$\text{Female}$$). But there are other ways, of which the most general is to
encode males as some number $$Z=m$$ and some different number $$Z=f$$ for females.
(Because $$m\ne f,$$ division by $$m-f$$ below is allowable.)
However we encode the binary variable, we may now express the model in a single formula as
$$E[Y\mid X] = \phi(\alpha + \beta Y + \gamma Z X)$$
because, setting
$$\gamma = \frac{\beta_{\text{Male}} - \beta_{\text{Female}}}{m - f}\tag{**}$$
and
$$\beta = \beta_{\text{Male}} - \gamma m = \beta_{\text{Female}} - \gamma f,$$
for males with $$Z=m$$ this gives
$$\phi(\alpha + \beta X + \gamma Z X) = \phi(\alpha + (\beta + \gamma m)X) = \phi(\alpha + \beta_{\text{Male}})X$$
and for females with $$Z=f,$$
$$\phi(\alpha + \beta X + \gamma Z X) = \phi(\alpha + (\beta + \gamma fX) = \phi(\alpha + \beta_{\text{Female}})X$$
which is exactly model $$(*).$$
The expression for $$\gamma$$ in $$(**)$$ is crucial: it shows how to interpret the model.
For instance, when using the indicator for males, $$m-f = 1-0$$ and $$\gamma$$ is the difference between the male and female slopes in the model. When using the indicator for females, $$m-f = 0-1 = -1$$ and now $$\gamma$$ is the difference computed in the other direction: between the female and male slopes.
In the example of the question where $$m=1$$ and $$f=-1,$$ now
$$\gamma = \frac{\beta_{\text{Male}} - \beta_{\text{Female}}}{m - f} = \frac{\beta_{\text{Male}} - \beta_{\text{Female}}}{2} \tag{**}$$
is half the difference in slopes.
Despite these differences in interpretation of the coefficient $$\gamma,$$ these are all equivalent models because they are all identical to $$(*).$$
• Ah! Great! Thank you! This explanation makes very clear intuitive sense! Thanks again @whuber – JDoe2 May 29 at 13:29
If you have an interaction with sex, then this means that you create a new variable that did not exist before.
For instance:
• let the outcome (dependent variable) be the probability of a baby
• let sex be a variable which is either 0 or 1
• and let's say we interact it with condom use which is either 0 or 1 as well.
Then you could have some table like the following (I make these numbers up as an example but try to approach realistic values):
Probability of having a baby
Yes Sex No Sex
Unprotected 0.50 0
Condom 0.01 0
So this could be modelled with two fixed effects like
$$\text{y = a + b sex +c unprotected}$$
But you won't get it right. The above formula will give
Yes Sex No Sex
Unprotected a+b+c a+c
Condom a+b a
This has only three variables to determine 4 values. If you try to make unprotected sex equal to 0.5 by giving some weight to b or c then you get that protected sex or no sex will have too much weight.
When you add an interaction term then you get
$$\text{y = a + b sex +c unprotected + d sex and unprotected}$$
Yes Sex No Sex
Unprotected a+b+c+d a+c
Condom a+b a
So that is how your interaction with sex is helping to get babies.
You can give indeed different values to sex, this will change the weights. Also when you change the interaction term and where you intercept, then things get mixed up. It can change how significant the value of the intercept is, and depending on your interaction, the value of the fixed model effects change as well.
But for the total model prediction, the prediction for the probability of whether you get a baby, it does not matter. The values of the sexes and the interaction, their significance, should not be measured. An analysis of variance is better.
So when you got that fixed, then the point of the intercept becomes just a matter of convenience. I like to do like you and put it in between men and women by giving men and women equal, but opposite, weight -1 and +1. In that case, the factors will show the differences relative to a place that is in between men and women.
Quickie:
The model is equivalent in the prediction of the means as long as the column space remains the same (this is the case in your example when you include an intercept term), but particular statistical tests for coefficients may change.
• Interesting answer, but I think you are interpreting sex as the sex act (yes/no), rather than the sex of a person (male/female). Your points are valid, but it does not really match the inquiry by the OP. – Ben - Reinstate Monica May 28 at 23:17 | 2020-07-14T07:48:47 | {
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https://stats.stackexchange.com/questions/359457/probability-of-getting-black-ball-at-n1th-draw-given-that-the-first-n | # Probability of getting black ball at ${n+1}^{th}$ draw given that the first $n$ draws has resulted into black balls
Suppose there are $(N +1)$ identical urns marked $0,1,2, . . . ,N$ which contains $N$ balls. The $k^{th}$ urn contains $k$ black and $N−k$ white balls, $k =0,1,2, . . . ,N.$ An urn is chosen at random, and $n$ random drawings are made from it, the ball drawn being always replaced. If all the $n$ draws result in black balls, find the probability that the $(n+1)^{th}$ draw will also produce a black ball. How does this probability behave as $N\to \infty$?
What I attempted:- Since all the $n$ balls drawn are found black, we can consider only those urns which contain at least $n$ black balls. There are $N-n+1$ urns which contain at least $n$ balls. The probability of choosing any one of them is $\frac{1}{N-n+1}$
Suppose, $A$ =The event of getting black balls in $n$ consecutive draws of a randomly chosen urn
$A_i$ = The $i^{th}$ urn is chosen. Here $n\le i \le N$
Now,\begin{aligned} P(A)&=P(AA_n)+P(AA_{n+1})+......+P(AA_{N}) \\ &=\sum_{i=n}^{N} P(A_i) P(A|A_i)\\ &=\frac{1}{N-n+1} \sum_{i=n}^{N} \left(\frac{i}{N}\right)^n \end{aligned}
Similarly, we define
$B$= The event of getting $n+1$ black balls in $n+1$ consecutive draws of a randomly chosen urn. We obtain that $P(B)=\frac{1}{N-n} \sum_{i=n+1}^{N} \left(\frac{i}{N}\right)^{n+1}$
We are required to find, $P(B|A)$ which is given as \begin{aligned} P(B|A)=&\frac{P(AB)}{P(A)}\\ &=\frac{\frac{1}{N-n} \sum_{i=n+1}^{N} \left(\frac{i}{N}\right)^{n+1}}{\frac{1}{N-n+1} \sum_{i=n}^{N} \left(\frac{i}{N}\right)^n}\\ \end{aligned}
Am I correct?
• If the balls are replaced why does the urn need to have at least n balls? – dsaxton Jul 28 '18 at 12:11
• @dsaxton you are correct. I made a mistake. The correct answer should be Thanks for pointing the mistake. Probably, the correct answer should be $\frac{\frac{1}{N+1} \sum_{i=1}^{N} \left(\frac{i}{N}\right)^{n+1}}{\frac{1}{N+1} \sum_{i=1}^{N} \left(\frac{i}{N}\right)^n}=\frac{ \sum_{i=1}^{N} \left(\frac{i}{N}\right)^{n+1}}{\sum_{i=1}^{N} \left(\frac{i}{N}\right)^n}$ \\ [@ Jim - The problem is from the book "An Introduction to Probability and Statistics" by V.K. Rohatgi and Saleh – user159457 Jul 28 '18 at 12:45
• If I am not wrong, the limit of this probability as $N \to \infty$ can be evaluated as follows:- $$\lim_{N \to \infty} \frac{ \sum_{i=1}^{N} \left(\frac{i}{N}\right)^{n+1}}{\sum_{i=1}^{N} \left(\frac{i}{N}\right)^n}=\lim_{N \to \infty} N \lim_{N \to \infty} \frac{1^{n+1}+2^{n+1}+.....+N^{n+1}}{1^n+2^n+......+N^n}=\lim_{N \to \infty}N \frac{\int_{0}^{N} t^{n+1} dt}{\int_{0}^{N} t^{n+1} dt}=\frac{n+1}{n+2}$$ I am using Euler-Maclaurin's Summation formula. – user159457 Jul 28 '18 at 13:09
In this problem you effectively select a random value $K \sim \text{U} \{ 0,...,N \}$ and you observe a count value $X \sim \text{Bin}(n,\tfrac{K}{N})$ from sampling-with-replacement from an urn with $K$ black balls of $N$ balls. In your problem you have $X=n$ and you now want to find $\mathbb{E}(\tfrac{K}{N} | X=n)$, which is the posterior probability of drawing a black ball on the next draw.
Solving the problem: Applying the law of total probability you have:
\begin{aligned} \mathbb{P}(X=n) &= \sum_{k=0}^N \mathbb{P}(X=n|K=k) \mathbb{P}(K=k) \\[6pt] &= \sum_{k=1}^N (\tfrac{k}{N})^n \tfrac{1}{N+1} \\[6pt] &= \frac{1}{N+1} \sum_{k=1}^N (\tfrac{k}{N})^n \\[6pt] &= \frac{1}{N+1} S_n(N), \\[6pt] \end{aligned}
where $S_n(N) \equiv\sum_{k=1}^N (\tfrac{k}{N})^n$ is related to Faulhaber's sum, which has various useful expressions. Now we can apply Bayes' rule to obtain:
\begin{aligned} \mathbb{P}(K=k | X=n) &= \frac{\mathbb{P}(X=n|K=k) \mathbb{P}(K=k)}{\mathbb{P}(X=n)} \\[6pt] &= \frac{ (\tfrac{k}{N})^n \tfrac{1}{N+1}}{\tfrac{1}{N+1} S_n(N)} \cdot \mathbb{I}(k>0) \\[6pt] &= \frac{(\tfrac{k}{N})^n}{S_n(N)} \cdot \mathbb{I}(k>0). \\[6pt] \end{aligned}
Hence, you have posterior expectation:
\begin{aligned} \mathbb{E}(K|X=n) &= \sum_{k=1}^N k \cdot \frac{(\tfrac{k}{N})^n}{S_n(N)} \\[6pt] &= N \sum_{k=1}^N \frac{ (\tfrac{k}{N})^{n+1}}{S_n(N)} \\[6pt] &= N \cdot \frac{S_{n+1}(N)}{S_n(N)}. \\[6pt] \end{aligned}
From this result you have:
$$\mathbb{P}(\text{Next ball is black} | X=n) = \mathbb{E} ( \tfrac{K}{N} | X=n ) = \frac{S_{n+1}(N)}{S_n(N)}.$$
This gives you the probability that the next ball is black, in terms of a ratio of two different instances of Faulhaber's sum. It can be calculated through various known formulae for any given values of $N$ and $n$. It is simple to take the limit $N \rightarrow \infty$ by recognising that this gives a Riemann integral:
$$\lim_{N \rightarrow \infty} \frac{S_n(N)}{N} = \lim_{N \rightarrow \infty} \sum_{k=1}^N (\tfrac{k}{N})^n \cdot \tfrac{1}{N} = \int \limits_0^1 x^n dx = \Big[ \frac{x^{n+1}}{n+1} \Big]_{x=0}^{x=1} = \frac{1}{n+1}.$$
This gives us the limiting result:
$$\lim_{N \rightarrow \infty} \mathbb{P}(\text{Next ball is black} | X=n) = \lim_{N \rightarrow \infty} \frac{S_{n+1}(N) / N}{S_n(N) / N} = \frac{n+1}{n+2}.$$ | 2019-05-27T08:01:34 | {
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https://math.stackexchange.com/questions/1872702/how-can-we-define-convexity-in-one-dimension/1872706 | # how can we define convexity in one dimension?
Definition of a convex set. A set S in R^n is said to be convex if for each x1, x2 ∈ S, the line segment λx1 + (1-λ)x2 for λ ∈ (0,1) belongs to S. This says that all points on a line connecting two points in the set are in the set. My question is how can we define convexity in one dimension ?
In $\mathbb R$, the definition is the same and you can prove that it is equivalent to connexity, or simpler : $$A\subseteq\mathbb R\text{ is convex }\iff A\text{ is an interval}.$$
• @SimpleArt: Convexity says "the line segment between any two points is contained in the set". Since $\Bbb{R}$ is a line, a convex subset of $\Bbb{R}$ is itself a line segment, and therefore is an interval. – Will R Jul 27 '16 at 13:13
The formulas in your question(as edited) work perfectly well in $\mathbb{R}$. The collection of convex sets is the collection of intervals.
• Your formula works for ${\mathbb{R}}^{n}$. Set $n = 1$. – Jay Jul 27 '16 at 12:07 | 2019-06-17T21:14:53 | {
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https://math.stackexchange.com/questions/3532885/how-to-find-the-equation-of-a-circle-given-3-points | How to find the equation of a circle given 3 points
Sorry in advance for the long post; I am doing a set of questions on finding the equation of a circle given 3 points.
The first set of points are (0,0), (2,0), and (4,-2).
There are a few ways I can see of doing this question but I can't help feeling there is something I am meant to spot that I haven't yet - possibly to do with circle theorems.
The obvious idea is to create 3 simultaneous equations and solve them for a, b, and r, where (a,b) is the centre of the circle, and r is the radius.
But this seems really laborious.
The second idea I had is to somehow use the circle theorem that the angle subtended at the circumference is half that at the centre, but I couldn't see an obvious way in.
The way I ended up solving it was by considering the symmetry of a circle;
If (0,0) and (2,0) form a chord of the circle which is parallel to the x-axis, then the centre of the circle must have an x-coordinate in line with the midpoint of the chord.
So the x-coordinate of the centre is 1.
Then to continue the question, if you consider symmetry again, there must be a coordinate at (-2,-2).
And then just by looking at these coordinates, by symmetry again there is another coordinate at (-2, -4) and (4, -4). At this point, looking at the sketch (with the help of Desmos), you can see that the centre's y-coordinate is at -3.
This second bit of reasoning to find the y-coordinate is a bit unsatisfactory and I can't fully explain it.
And/or explain why my method was good or bad?
P.S. The rest of the set of questions were:
• (2,2), (4,3), & (6,9)
• (-1,1), (2,-1), & (-2,0)
• (0,0), (a,0, & (1,1)
The center of the circle containing $$(0,0), (2,0),$$ and $$(4,-2)$$
must be equidistant from all those points.
The points equidistant from $$(0,0)$$ and $$(2,0)$$ are on a line satisfying $$x^2+y^2=(x-2)^2+y^2$$;
i.e., $$0=-4x+4$$; i.e., $$x=1$$.
The points equidistant from $$(2,0)$$ and $$(4,-2)$$ are on a line satisfying
$$(x-2)^2+y^2=(x-4)^2+(y+2)^2$$;
i.e., $$-4x+4=-8x+16+4y+4$$; i.e., $$4x=4y+16$$; i.e., $$y=x-4$$.
Therefore, the center is at the intersection of these lines, which is $$(1,-3)$$.
Therefore, the equation of the circle is $$(x-1)^2+(y+3)^2=r^2$$.
Plug in any of the points to figure out $$r^2$$.
Take two points and find the perpendicular bisector. Repeat for another pair of points. The centre of the circle is where these perpendicular bisectors meet
Use family of circles i.e circle with $$(0,0)$$ and $$(2,0)$$ as one of its diameter is $$x(x-2)+y^2=0$$ And line passing through these points is $$y=0$$ Hence any circle passing through the point of intersection of circle and line is $$x^2+y^2-2x+\lambda y=0$$ Required circle passes through the point $$(4,-2)$$ satisfy that we will get parameter $$\lambda$$. $$4^2+(-2)^2-2(4)+\lambda (-2)=0$$ $$\therefore \lambda =6$$ Hence equation of required circle is $$x^2+y^2-2x+6y=0$$ | 2021-05-16T13:04:49 | {
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https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-6-algebra-equations-and-inequalities-6-2-linear-equations-in-one-variable-and-proportions-exercise-set-6-2-page-364/131 | ## Thinking Mathematically (6th Edition)
The length of the femur is given by$f=0.432h-10.44$. It is also provided that $f=16\text{ inches}$. Now substituting $f=16$ in the above equation and solve for $h$: $16=0.432h-10.44$ Add $10.44$ on both sides; \begin{align} & 16+10.44=0.432h-10.44+10.44 \\ & 26.44=0.432h \end{align} Divide $0.432$on both sides. \begin{align} & \frac{26.44}{0.432}=\frac{0.432h}{0.432} \\ & 61.20=h \end{align} So,$h=61.20\text{ inches}$. As it is provided that the skeleton found is $5$feet tall. Since, \begin{align} & 1\text{ feet}=12\text{ inches} \\ & \Rightarrow 5\text{ feet}=12\times 5 \\ & =60\text{ inches} \end{align} which is approximately equal to, $61.20\ \text{inches}$. Yes, the partial skeleton height is slightly over 5 feet tall. | 2019-12-09T00:36:44 | {
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https://math.stackexchange.com/questions/3886213/arrangements-of-solutions-to-n-linear-equations-with-n-unknowns | # Arrangements of Solutions to $n$ Linear Equations with $n$ Unknowns
In a system of two linear equations with two unknowns, there are three "arrangements" that we can see when we graph the two lines in the $$xy$$ plane:
• The two lines intersect at a single point (one solution to the system).
• The two lines are parallel and never intersect (no solution to the system).
• The two equations describe the same line (infinitely many solutions to the system).
When we move up to a system of three linear equations with three unknowns, now we have three planes in space, and there are eight distinct arrangements of the three planes:
If we consider one linear equation with one unknown, I suppose it makes sense to say that there is one arrangement, which is a single point on the real number line.
So, for one, two and three unknowns, we have the start of a sequence: $$1, 3, 8, ...$$
I am interested in how this sequence continues. I have searched in vain on the OEIS. Alas, there are many sequences that have $$1, 3, 8,...$$ and I'm not sure which, if any, are the right one.
This one: https://oeis.org/A001792 looks like it might be it, because the comments say that sequence is related to matrices in certain ways. Also I expect a formula for this sequence should involve powers of 2. But, I wouldn't bet any money on it.
Is there a sequence with a simple formula to work out the number of arrangements?
• There is also a similar sequence if you just count the distinct arrangements of $n$ planes in $3$ dimensions. One plane has $1$ arrangement, two planes have $3$ arrangements (parallel, intersecting, coinciding), and three planes have the $8$ arrangements pictured. How does this sequence continue? Is it the same as the one in my original question? Oct 30, 2020 at 0:31
• Can you explicitly define the arrangements? This wiki page may contain relevant information. Nov 5, 2020 at 12:40
• What does this have to do with systems of equations? For example, in your image of arrangements of planes, (3), (4), (5) and (8) all have the same type of solution for the corresponding system of equations: that is, no solution. Should we just stop talking about systems of equations and just talk about hyperplanes? Nov 5, 2020 at 13:49
• @BrianDrake I brought up systems because systems produce the graphs pictured above. I'm not worried whether a system has solution(s) or not. I guess it's not even necessary to consider hyperplanes. What if a 4th plane was added to each of the pictures above? How many more arrangements would there be? Nov 6, 2020 at 14:16
More than an answer, the following represent some considerations on how the problem could be tackle.
Start and consider just two planes (in $$3$$-D) with the relevant matrix of coefficients $$\bf C$$ and the augmented matrix (coefficients plus known terms) $$\bf A$$ $$\underbrace {\left( {\overbrace {\matrix{ {a_1 } & {b_1 } & {c_1 } \cr {a_2 } & {b_2 } & {c_2 } \cr } }^{\bf C}\; \left| {\;\matrix{ {d_1 } \cr {d_2 } \cr } } \right.} \right)}_{\bf A}\;$$
Now, if the matrix $$\bf C$$
• has full rank $$2$$, then the two normal vectors are independent, i.e. the planes are incident;
• has rank $$1$$, then the two vectors are dependent, i.e. parallel;
• the rank $$0$$ corresponds to having all the coefficients null, which is a degenerated situation.
The rank of $$\bf A$$ (call it $$R$$) cannot be less than that of $$\bf C$$ ($$r$$), nor it can be grater than $$r+1$$.
When $$r=0$$, $$R$$ can be $$0$$ or $$1$$ and it will be in any case a degenerated situation.
When $$r=1$$, $$R$$ can be $$1$$ (planes coincident), or $$2$$ (planes parallel).
When $$r=2$$ also $$R=2$$ and the planes are incident.
Passing to consider three $$3$$-D planes, your scheme of distinction is unfortunately wider than that provided by considering the ranks of the $$3 \times 4$$ matrices.
For the eight cases that you sketched in fact we have $$\begin{array}{c|cccccccc} {} & & {\left( 1 \right)} & {\left( 2 \right)} & {\left( 3 \right)} & {\left( 4 \right)} & {\left( 5 \right)} & {\left( 6 \right)} & {\left( 7 \right)} & {\left( 8 \right)} \\ \hline r & & 1 & 2 & 1 & 2 & 1 & 3 & 2 & 2 \\ R & & 1 & 2 & 2 & 3 & 2 & 3 & 2 & 3 \\ \end{array}$$ We have that the three couples $$[(2),\, (7)] , \; [(3),\, (5)], \; [(4),\, (8)]$$ cannot be distinguished.
To be able to distinguish between them, for $$r < 3$$, we could introduce a fourth plane, with coefficients independent from the previous three, which means that it increases $$r$$ to $$r+1$$, and count the crosses we are going to have.
Equivalently, we can consider both the ranks of the first two planes and how they change by adding the third.
However this way would soon become impractical for higher dimensions. | 2022-05-25T06:35:59 | {
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https://math.stackexchange.com/questions/2557382/given-that-sum-limits-n-1-inftya-n-converges-a-n-0-then-does-sum | # Given that $\sum\limits_{n=1}^{\infty}a_n$ converges ($a_n >0$), then does $\sum\limits_{n=1}^{\infty}a_n^{3} \sin(n)$ converge?
Given that $\displaystyle \sum_{n=1}^{\infty}a_n$ converges ($a_n >0$), then $\displaystyle \sum_{n=1}^{\infty}a_n^{3} \sin(n)$:
a) converges
b) diverges
c) does not exist
d) None of the above
My attempt: If we were given that $(a_n)$ is monotonically decreasing, then using the Abel-Olivier-Pringsheim criterion (Theorem 2.16 in these notes) we could conclude that $\displaystyle \lim_{n\to \infty} na_n=0$ from which we obtain $k \in \mathbb{N}$ such that for all $n \geq k,$ we have $na_n<1$ which implies $|a_n^{3}\sin(n)|<\dfrac{1}{n^3}$ which by the comparison test means that $\displaystyle \sum_{n=1}^{\infty}a_n^{3} \sin(n)$ converges absolutely.
But we are not given that $(a_n)$ is monotonically decreasing. At best, we have a monotonically decreasing subsequence tending to $0.$ Any hints on how to proceed?
• Hint: $a_n\to0$, hence there exists $k$ with $|a_n|\le k$ for all $n$. Dec 8 '17 at 16:51
Since $\sum_n a_n$ converges, $a_n\to 0$. There is some $N$ such that $n\geq N\implies 0<a_n\leq 1$.
For $n\geq N$, $|a_n^{3} \sin(n)|\leq a_n^3\leq a_n$.
Therefore, $\sum_n a_n^{3} \sin(n)$ converges absolutely, hence converges.
• Thanks, was definitely overthinking it. Dec 8 '17 at 17:07
Eventually, the $a_n$s are small. Then $$|a_n^3 \sin n| \leq |a_n|^3 \leq |a_n| = a_n.$$ Now compare and use absolute convergence.
If $$\sum_{n=1}^{\infty} a_n$$ converges, then $$a_n \to 0$$ and thus you can find $$N$$ sufficiently large enough so that $$a_n <1$$ for $$n > N$$. After this $$a_n^3 < a_n$$ for all $$n > N$$. This tell us that
$$\sum_{n=1}^{\infty} a_n^3 = \sum_{n=1}^{N} a_n^3 + \sum_{n=N+1}^{\infty} a_n^3 < \sum_{n=1}^{N} a_n^3 + \sum_{n=N+1}^{\infty} a_n <\infty$$
And finally we have $$\sum_{n=1}^{\infty} a_n^3 \sin(n) < \sum_{n=1}^{\infty} a_n^3$$
since $$\sin(n) < 1$$. | 2021-12-03T18:26:30 | {
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http://www.cs.wisc.edu/event/geometry-approximate-differential-privacy | ## Kunal Talwar: The Geometry of Approximate Differential Privacy
Room:
CS 3310
Speaker Name:
Kunal Talwar
Speaker Institution:
Microsoft Research Silicon Valley
Yes
In this talk, I will discuss trade-offs between accuracy and privacy
in the context of linear queries over histograms. This is a rich class
of queries that includes contingency tables and range queries, and has
been a focus of a long line of work. For a set of $d$ linear queries
over a database $x \in \R^N$, we seek to find the differentially
private mechanism that has the minimum mean squared error. I will
first describe a $O(\log^2 d)$ approximation algorithm for this
problem, for the case of $(\eps,\delta)$-differential privacy. The
mechanism is simple, efficient and adds correlated Gaussian noise to
the answers. We prove its approximation guarantee relative to the
hereditary discrepancy lower bound of Muthukrishnan and Nikolov, using
tools from convex geometry.
We will next consider this question in the case when the number of
queries exceeds the number of individuals in the database, i.e. when
$d > n = \|x\|_1$. It is known that better mechanisms exist in this
setting. I will then describe an $(\eps,\delta)$-differentially
private mechanism which is optimal up to a $\polylog(d,N)$ factor for
any given query set $A$ and any given upper bound $n$ on $\|x\|_1$.
This approximation is achieved by coupling the Gaussian noise addition
approach with a linear regression step. We give an analogous result
for the $\eps$-differential privacy setting. We also improve on the
mean squared error upper bound for answering counting queries on a
database of size $n$ by Blum, Ligett, and Roth, and match the lower
bound implied by the work of Dinur and Nissim up to logarithmic
factors.
The connection between hereditary discrepancy and the privacy
mechanism also enables us to derive the first polylogarithmic
approximation to the hereditary discrepancy of a matrix $A$.
This talk is based on joint work with Alex Nikolov and Li Zhang
Event Date:
Monday, March 11, 2013 - 10:00am - 11:00am (ended) | 2013-05-25T00:37:19 | {
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https://www.mathimatikoi.org/forum/viewtopic.php?f=15&t=1052&view=print | mathimatikoi.orghttps://www.mathimatikoi.org/forum/ Continuous maphttps://www.mathimatikoi.org/forum/viewtopic.php?f=15&t=1052 Page 1 of 1
Author: Grigorios Kostakos [ Thu Oct 13, 2016 1:40 pm ] Post subject: Continuous map Let $E_1\,,\; E_2$ be metric spaces and $f:E_1\longrightarrow E_2$ a map such that for every compact subspace $S$ of $E_1$ the restriction $f|_S:S\subseteq E_1\longrightarrow E_2$ is continuous. Prove that $f$ is continuous on $E_1$.
Author: jason1996 [ Sat Nov 12, 2016 9:08 pm ] Post subject: Re: Continuous map We suppose that $x_{n}\rightarrow x_{0}$. The set $\left \{ x_{1},x_{2},.... \right.\left. \right \}\cup \left \{ x_{0} \right.\left.\right \}$ is compact and from the hypothesis(the restriction of f in every compact set is continuous) we obtain that $f(x_{n})\rightarrow f(x_{0})$. So the proof has finished.Question: Is this true for topological (not necessarily metric) spaces? If yes prove it, if not find a counterexample.
Author: Riemann [ Tue Nov 29, 2016 2:25 pm ] Post subject: Re: Continuous map jason1996 wrote:Question: Is this true for topological (not necessarily metric) spaces? If yes prove it, if not find a counterexample.Hello jason1996,well the answer to your question is yes under one condition. $f$ must be defined on either closed or open subsets of the topological space. Try looking up for the final topology.A little bit general result is the following:Let $X$ be a topological space and $A_1$ and $A_2$ be subspaces of $X$. Let $g_{i}:A_i\rightarrow Y$ be continous for $i=1,2$. If $g_1$ and $g_2$ coincide on $A_1\cap A_2$ and $A_1$ and $A_2$ are either both open or both closed, then the function $f:X\rightarrow Y$ defined by $f(x)=g_1(x)$ if $x \in A_1$ and $f(x)=g_2(x)$ otherwise is continuous.Assume that $g_1$ and $g_2$ are given continuous functions that satisfy the hypothesis and $A_1$ and $A_2$ are both open (respectively closed). Let $U$ be an open set (respectively closed set) of $Y$. Then, $$f^{-1}[U] = g_1^{-1}[U]\cup g_2^{-1}[U]$$ which is a binary union of open (respectively closed) sets in $X$, hence open (closed) by continuity of $g_i$ ($i=1,2$). Hence $f$ is continuous.
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http://stcc.org/qtmuxl/effective-yield-formula-a46b8d | The formula for calculating Effective Annual Yield (EAY) is: (1 + (i / n)) n - 1 The i stands for your interest rate while the n stands for the number of payment periods in a year. The reason the effective yield is slightly different is because although many are given in terms of yearly return, they actual compound in shorter intervals such as months. The number of compounding periods per year will affect the total interest earned on an investment. Therefore, Sam should invest in Bond A. Calculator Use. Where, r = Nominal Annual Interest Rate ; n = Number of payments per year ; i = Effective Interest Rate; Example of Effective Annual Yield Rate. The effective yield can be calculated using the following formula: Introduction to Effective Duration. History. The zero coupon bond effective yield formula is used to calculate the periodic return for a zero coupon bond, or sometimes referred to as a discount bond. On the other hand, effective duration is a curve duration statistic that measures interest rate risk in terms of a parallel shift in the benchmark yield … Bond\: Equivalent\: Yield = \dfrac{1000 - 975}{975} \times \dfrac{365}{90} = 10.3989\% From the above BEY calculations, it can be concluded that Bond A is a better investment option since its yield of 10.6725% is greater than Bond B’s yield of 10.3989%. For example, if an investment compounds daily it will earn more than the same investment with the same stated/nominal rate compounding monthly. The formula for effective duration is:. It is denoted by ‘i’. Divide this into the yield on the tax-free bond to find out the tax-equivalent yield. Use this calculator to determine the effective annual yield … Effective Yield Formula. Following is the effective yield formula on how to calculate effective yield. How Does Effective Duration Work? Therefore, the effective maturity is 19.7 percent ($17/$86 = 0.198 or 19.8%). Bond equivalent yield formula. Financial calculator to calculate the effective yield with periodic interest based on the nominal annual interest rate (r) and number of payments per year (n). If you plug different tax rates into the equation above, you will see that the higher your tax rate, the higher the tax-equivalent yield… The effective annual interest rate allows you to determine the true return on investment (ROI) ROI Formula (Return on Investment) Return on investment (ROI) is a financial ratio used to calculate the benefit an investor will receive in relation to their investment cost. Also called annual percentage rate (APR) and annual percentage yield (APY), Excel makes it easy to calculate effective mortgage, car loan, and small business loan interest rates from the nominal rates often quoted by … i = [1 + (r/n)] n – 1 . In this equation, the reactant and the product have a 1:1 mole ratio, so if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams! Explanation. Annual interest rate of a firm is 10% compounded monthly payments, then what is the effective … Annual Percentage Yield In this example, the annual effective yield is calculated thus: Annual percentage yield = (1.03)^12 - 1 = . It is also known as the annual effective yield. Karl von Terzaghi first proposed the relationship for effective stress in 1925. The bond equivalent yield is used to determine the annual yield on a discount bond. Usually, you have to calculate the theoretical yield based on the balanced equation. For him, the term "effective" meant the calculated stress that was effective in moving soil, or causing displacements. Zero coupon bond effective yield vs. Among Excel's more popular formulas, the EFFECT formula is often used by financial professionals to figure out an effective interest rate from a nominal interest rate. The effective yield will be the absolute increase as a percentage of the principal invested. Formula to Calculate Bond Equivalent Yield (BEY) The formula is used in order to calculate the bond equivalent yield by ascertaining the difference between the bonds nominal or face value and its purchase price and these results must be divided by its price and these results must be further multiplied by 365 and then divided by the remaining days left until the maturity date. To calculate the current yield of a bond in Microsoft Excel, enter the bond value, the coupon rate, and the bond price into adjacent cells (e.g., A1 through A3). Zero coupon bond effective yield formula takes in to accountthe compounding effect while calculating the rate of return. Effective Duration = (P-- P +) / [(2)*(P 0)*(Y + - Y-)]. Thus, a bond with a $1,000 par value that pays 5% interest pays$50 dollars per year in 2 semi-annual payments of $25. Since the effective yield considers compounding effect, it will always be greater than nominal yield. Of this amount,$4,000 is paid in cash, and $613.90 is discount amortization. Calculate the effective interest rate per period given the nominal interest rate per period and the number of compounding intervals per period. Formula : i = [1 + (r/n)] n - 1 Where, r = Nominal Annual Interest Rate n = Number of payments per year i = Effective Interest Rate. For example, if the bond in question yields 3 percent, use the equation (3.0 / .75) = 4 percent. Nominal yield, or the coupon rate, is the stated interest rate of the bond. It represents the average stress carried by the soil skeleton. The formula for the effective interest rate can be derived by using the following steps: Step 1: Firstly, determine the stated rate of interest of the investment, which is usually mentioned in the investment document. Formula for Calculation of Effective Rate of Interest. From the previous example, the average annual investment is$86, and the average annual yield is $17. Effective Yield : Effective Annual Rate is used to find out the actual annual rate that would be paid on a loan if the specified annual rate is affected by compounding. The Yield to Maturity is actually the Internal Rate of Return (IRR) on a bond. If we divide this by the the level yield asset balance ($11,000) and multiply by 12 to annualize it, we get 0.64%--the difference between the contracted 7% interest rate and the effective yield … The modified duration is a yield duration statistic that measures interest rate risk in terms of a change in the bond’s own yield-to-maturity (ΔYield). That’s whyfinancials analysts prefer to select zero coupon bond effective yield equationfor long-term investments or bond’s yield calculation. Effective Yield = (1 + (Nominal Interest/Number Of Periods)^Number Of Periods) - 1) Bond Equivalent Yield Calculator Zero Coupon Bond Calculator Yield to Maturity Calculator Bond Yield Calculator Dividend Yield Calculator: Electrical Calculators Where: P 0 = the bond 's initial price per $100 of par value P-= the bond's price if its yield falls by x basis points P + = the bond's price if its yield rises by x basis points (Y + - Y-) = Change in yield in decimal. Thus, effective interest for the first six months is$92,278 X 10% X 6/12 = $4,613.90. Step 2: Next, figure out the number of compounding periods during a year and it is denoted by “n”. It is also known as the annual effective yield. Effective annual yield is a measure of the actual or true return on an investment. In this example, the annual effective yield is calculated thus: Annual percentage yield = (1.03)^12 - 1 = .43 = 43%, where 1.03 is 1 plus the monthly interest and 12 is the number of times in a year interest is compounded. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Calculate the effective maturity rate of the bond by dividing the average annual yield of the bond by the average annual investment. The n in the annual percentage yield formula would be the number of times that the financial institution compounds. The effective interest method is a technique for calculating the actual interest rate in a period based on the amount of a financial instrument's book value at the beginning of the accounting period. Avoid typing values directly into the formula. You can use the following formula to calculate the effective rate of interest: E = (1 + i) n – 1 … (1) Where ‘E’ is the effective rate of interest, ‘i’ is the actual rate of interest in decimal, and ‘n’ is the number of conversion periods. Formula of Effective Annual Yield Rate. This yield percentage is the percentage of par value —$5,000 for municipal bonds, and $1,000 for most other bonds — that is usually paid semiannually. A zero coupon bond is a bond that does not pay dividends (coupons) per period, but instead is sold at a discount from the face value. Recall that when Schultz issued its bonds to yield 10%, it received only$92,278. 43 = 43%, where 1.03 is 1 plus the monthly interest and 12 is the number of times in a year interest is compounded. Effective duration measures the change in price of a bond to a 1% or a 100 basis point change in the yield of the bond across all maturities and therefore a parallel shift of the yield curve by 1% indicating the amount of interest rate risk the bondholder needs to bear by holding the given bond in his investment portfolio. To select an individual value within the formula, simply click on the cell containing the value. The effective yield can be calculated using the following formula: Why Calculate Effective Annual Yield? The formula (as provided by Microsoft) to determine yield is: "=YIELD(settlement,maturity,rate,pr,redemption,frequency)" ... then calculate the yields on various loans to compare effective yield. The annual percentage yield formula would be applied to determine what the effective yield would be if the account was compounded given the stated rate. Market Price of the Bond = Present Value of Coupon Payments + Present Value of Maturity Amount of the Bond Real Example: I’ll take the real case of 9.95% SBI 15-year bonds … Therefore, the effective annual yield will be: EY = 10.4713/100 or 10.4713%. Bond Equivalent Yield Analysis [citation needed]Description. Effective yield is useful when you are considering various investment options where the interest rates are expressed at different compounding rates. Effective annual interest or yield may be calculated or applied differently depending on the circumstances, and the definition should be studied carefully.
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https://math.stackexchange.com/questions/2666120/how-to-prove-an-increasing-sequence-that-converges-is-bounded-above-by-its-limit | # How to prove an increasing sequence that converges is bounded above by its limit
I am trying to prove that an increasing sequence that converges to $L$ is bounded above by its limit.
By using $a_n \le a_{n+1}$ and the definition of limit of a sequence, I can prove that for $\epsilon > 0$ , $a_n \lt {L + \epsilon}$ for all $a_n$.
But is there a way to proceed to $a_n \le L$ ? because I can't think of a case in which the former is true but the latter isn't.
• You could proceed by contradiction. – user371663 Feb 25 '18 at 15:50
HINT
You can easily show that if for some n $a_n>L$ then by definition of limit $a_n$ must decrease which is impossible.
You only need to formalize this idea by setting “assume exists n such that ...then by definition of limit...contradiction”.
Notably
• suppose $\exists n_1$ such that $a_{n_1}>L$ with $d=a_{n_1}-L>0$
• set $\epsilon=d$ by definition of limit must exists $n_2>n_1$ such that $|a_{n_2} -L|<\epsilon \implies a_{n_2}<a_{n_1}$
• Can I just use that assume $\exists n_1$ such that $a_{n_1} \gt L$ and let $\epsilon = a_{n_1} - L$ and put it in $\forall a_n \: a_n \lt {L + \epsilon}$ which would lead to the contradiction that $a_{n_1} \lt a_{n_1}$ ? Is contradiction the only way to get to the result $\forall a_n \: a_n \le L$ – loct Feb 26 '18 at 7:14
• Yes of course you can define directly it as $\epsilon$, I've choose $d$ only to emphatize the limiti definition. I can't indicate at the moment an alternative method, but of course this is the more straighforward and effective way to prove it. – gimusi Feb 26 '18 at 7:30
If $a_n$ converges to $L$ and is increasing then let $\epsilon>0 \ ,\exists N \in \Bbb N \ s.t. \ \forall n>N \ \ \vert a_n-L\vert < \epsilon$
We can open up this up as you did to $a_n<L+\epsilon$
As indicated in the hint above, if $\exists n$ such that $L<a_n<L+\epsilon$ we get a contradiction because the limit is $L$ and so the sequence will have to "decrease" from $L<a_n<L+\epsilon$ down to $L$
If a term say, $a_n$ gets larger than the limit $L$, then all the other terms, $a_{n+1}, a_{n+2},...$ must stay above $a_n,$ since your sequence is increasing.
Thus for $$\epsilon = \frac {a_n-L}{2}$$ all the terms $a_k$ for $k\ge n$ stay out of the neighbourhood, $$(L-\epsilon, L+\epsilon)$$ which contradicts the assumption of $$lim _{n\to \infty} a_n = L.$$ | 2019-07-17T22:29:42 | {
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https://math.stackexchange.com/questions/3331976/proving-that-a-linear-transformation-is-diagonalisable | Proving that a linear transformation is diagonalisable
Given that $$V = \mathbb{R}[X]_{\leq 2}$$ and $$\alpha \in \mathbb{R}$$ prove that the linear transformation $$L: V \to V$$ given by $$L(P(X)) = \alpha P(X) + (X+1)P'(X)$$ is diagonalisable and determine the matrix representation with respect to a basis of eigenvectors.
I know that $$L$$ is diagonalisable if $$\#Spec(L) = \dim(V) = 3$$, meaning that $$L$$ must have 3 distinct eigenvalues. I imagine I have to find the eigenvalues of $$L$$, but, I'm not really sure where to go from here. Any help is appreciated.
Note that $$\operatorname{Id}_n\colon\mathbb R^n\longrightarrow\mathbb R^n$$ has a single eigenvalue, but it is diagonalizable nevertheless. So, no, you don't need to have $$3$$ distinct eigenvalues.
On the other hand:
• $$L(1)=\alpha\times1$$;
• $$L(X)=(\alpha+1)X+1$$;
• $$L(X^2)=2X+(\alpha+2)X^2$$.
Therefore, the matrix of $$L$$ with respect to the basis $$\{1,X,X^2\}$$ is$$\begin{bmatrix}\alpha&1&0\\0&\alpha+1&2\\0&0&\alpha+2\end{bmatrix}.$$Can you take it from here?
• I would then use the matrix representation to find the eigenvalues and then find the matrix with respect to the values of the eigenvectors correct? – SES Aug 23 at 13:55
• Yes, that would be correct. – José Carlos Santos Aug 23 at 14:14
• I suppose that it is correct now. – José Carlos Santos Aug 23 at 14:28
• I have checked my solution : I maintain it is the good one – Jean Marie Aug 23 at 14:29
• I will no disagree. After all, it is equal to mine. – José Carlos Santos Aug 23 at 14:30
The image of basis $$(1,X,X^2)$$ is $$(\alpha, (\alpha+1)X+1,(\alpha+2) X^2+2X)$$ therefore with matrix representation :
$$M=\begin{pmatrix}\alpha&1&0\\0&\alpha+1&2\\0&0&\alpha+2\end{pmatrix}$$
As $$M$$ is triangular, its diagonal entries are its eigenvalues :
$$\alpha<\alpha+1<\alpha+2$$
Being all distinct, whatever $$\alpha$$, this matrix is always diagonalizable.
• Your answer is different than Jose Carlos Santos'. One of them must be wrong. – Leo Aug 23 at 14:10
• Sorry, the $m_{11}$ entry is $\alpha$... – Jean Marie Aug 23 at 14:12
We may as well replace $$1$$ with an arbitrary constant $$-r$$. Suppose $$Q$$ is an eigenvector of $$L$$, say of eigenvalue $$\lambda$$. Rearranging gives $$(X - r) Q'(X) = (\lambda - \alpha) Q(X) .$$ But this is a separable ordinary differential equation. Since polynomials are analytic, it suffices to work on some nonempty open interval not containing $$r$$ (or more to the point, where $$X - r$$ is nonvanishing). Then, dividing gives $$\frac{\lambda - \alpha}{X - r} = \frac{Q'(X)}{Q(X)} = (\log Q)'(X),$$ and integrating gives that up to a constant multiple we have $$P(X) = (X - r)^{\lambda - \alpha}.$$ This function is an element of $$R[X]_{\leq 2}$$ precisely when $$\lambda - \alpha \in \{0, 1, 2\}$$. (Note that this solves efficiently the corresponding problem for the operator $$L$$ defined on the polynomial space $$R[X]_{\leq d}$$ for large $$d$$.)
• [+1] Good idea to connect this issue with a differential equation. – Jean Marie Aug 23 at 16:27
One can also approach this directly using the definition of eigenvector and basic facts about polynomials. If $$Q$$ is an eigenvector of $$L$$, by definition we have $$L(Q(X)) = \alpha Q(X) + (X + 1) Q'(X) = \lambda Q(X)$$ for some constant $$\lambda$$, and rearranging gives $$(X + 1) Q'(X) = (\lambda - \alpha) Q(X) .$$ If $$Q'(X) = 0$$, then $$Q$$ is constant and $$\lambda = \alpha$$, giving one eigenvalue. If $$Q'(X) \neq 0$$, then $$Q(X)$$ is divisible by $$(X + 1)$$, so $$Q(X) = R(X) (X + 1)$$ for some $$R$$ with $$\deg R \leq 1$$. Substituting into the previous display equation and cancelling gives $$R(X) + R'(X) (X + 1) = (\lambda - \alpha) R(X) ,$$ and rearranging gives $$(X + 1) R'(X) = (\lambda - (\alpha + 1)) R(X) .$$
Now, either $$R'(X) = 0$$, in which case $$R$$ is constant and $$\lambda = \alpha + 1$$, or $$R$$ a (now constant) multiple of $$X + 1$$, in which case $$Q(X) = R(X) (X + 1)$$ is, up to a nonzero multiple, $$(X + 1)^2$$, and substituting again gives that $$\lambda = \alpha + 2$$. We've now found that $$L$$ has distinct eigenvalues, namely, $$\alpha, \alpha + 1, \alpha + 2$$, and that with respect to the basis $$(1, X + 1, (X + 1)^2)$$, $$L$$ has the matrix representation $$[L] = \pmatrix{\alpha\\&\alpha + 1\\&&\alpha + 2} .$$ | 2019-09-21T15:38:23 | {
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https://www.physicsforums.com/threads/why-use-rms-for-averages.762475/ | # Why use RMS for averages?
1. ### iScience
408
consider a sin wave
to find the average of this function over interval 2pi, why not just do...
$$\frac{1}{2\pi}\int |sinx|dx$$
this turns out to be..
$$\frac{4}{2\pi}= 0.63619$$
the whole purpose of squaring and then sqrt'ing at the end is to make sure there are no negative values. this would be fine if its value came out to be the same as the average of the absolute value (which is what i thought we were trying to find in the first place), but the value is different.
so.. i guess i have two questions:
1.) is 0.636 more correct to use as an avg than 0.702?
2.)why do we always use RMS value in science as opposed to the ACTUAL avg (of the absolute)?
2. ### HallsofIvy
40,369
Staff Emeritus
That is a perfectly valid "average" with slightly different properties than "rms". Which you choose to use depends upon the data and what you plan to do with the data. One can show, for example, that if your data is from a "normal distribution" the rms will, on average, lie closer to the true mean of the distribution than the absolute value mean. Also, since the absolute value function is not differentiable, it can be harder to work with, analytically, than rms.
### Staff: Mentor
One of the key uses of the RMS of a sinusoid is in alternating current. Suppose you have a light bulb and you want to make it shine equally brightly using direct current as opposed to a 120 volt RMS AC supply? The answer is 120 volts. The RMS voltage (or current) gives the equivalent DC voltage (or current).
No.
What makes you think your average is the "ACTUAL" one?
There are many ways of computing a "norm" or average. Your's is but one, RMS is another. Yet another is the maximum absolute deviation. There are others as well. Which one is "right"? That's the wrong question. They all are, in their own way.
4. ### jbunniii
3,356
One reason that RMS is a natural measurement to use is that you can think of it as the extension to infinite dimensions of standard euclidean distance.
If we have some point, say, (3,4,5) in 3-dimensional euclidean space, then the norm (the distance from this point to the origin) is ##\sqrt{3^2 + 4^2 + 5^2}##. We can think of a function as a "point" in infinite-dimensional space, and its "distance" from the origin (the zero function) is ##\sqrt{\int |f(x)|^2 dx}##.
But even in euclidean space, there are many other norms we can use, for example ##(3^p + 4^p + 5^p)^{1/p}## where ##p## is any real number ##\geq 1##. The special case ##p=2## gives euclidean distance. Similarly, ##(\int |f(x)|^p dx)^{1/p}## is a perfectly valid norm to use for functions.
A couple of advantages of the ##p=2## case (RMS):
1. The pythagorean theorem: if ##f## and ##g## are orthogonal (meaning ##\int f(x)g(x) = 0## in the case of functions), then ##\int |f(x) + g(x)|^2 dx = \int |f(x)|^2 dx + \int |g(x)|^2 dx##. This makes it easy to calculate the RMS of the sum of certain kinds of functions, such as sinusoids or noise.
2. The Cauchy-Schwarz inequality: ##|\int f(x) g(x) dx| \leq \sqrt{\int |f(x)|^2 dx}\sqrt{\int |g(x)|^2 dx}##
5. ### George Jones
6,384
Staff Emeritus
Related to DH's post:
RMS voltage multiplied by (in phase) RMS current gives average power. Some references call this product RMS power, but this is not correct, it is average power, hence the equivalent brightness of a lightbulb. | 2015-04-19T15:45:41 | {
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https://math.stackexchange.com/questions/2053144/solving-ax-equiv-c-pmod-b-efficiently-when-a-b-are-not-coprime/2053174 | # Solving $ax \equiv c \pmod b$ efficiently when $a,b$ are not coprime
I know how to compute modular multiplicative inverses for co-prime variables $a$ and $b$, but is there an efficient method for computing variable $x$ where $x < b$ and $a$ and $b$ are not co-prime, given variables $a$, $b$ and $c$, as described by the equation below?
$a x \equiv c \mod b$
For example, given
$154x \equiv 14 \mod 182$, is there an efficient method for computing all the possibilities of $x$, without pure bruteforce?
Please note that I'm not necessarily asking for a direct solution, just a more optimized one.
I do not believe that the Extended Euclidean Algorithm will work here, because $a$ and $b$ are not co-prime.
Could the be computed efficiently as well?
$12260x \equiv 24560 \mod 24755$.
$107$ needs to be one of the computed answers.
• It is more accepted notation to ask about solutions to $154x \equiv 14 \bmod{182}$. In any case the first step is almost always to factor the modulus ($182$ in this case) into its prime power factors ($182 = 2\cdot 91$). – hardmath Dec 10 '16 at 23:06
• After factorng the modulus into prime powers, in this case $2*7*13$, solve in each factor and then use the Chinese remainder theorem. – Scott Burns Dec 10 '16 at 23:13
• Could you possibly demonstrate that in more detail? Utilizing the Chinese Remainder Theorem to solve this equation? And correct me if I am wrong, but I thought that the Chinese Remainder Theorem only applied to values that are co-prime. – Jesse Daniel Mitchell Dec 10 '16 at 23:21
Below we compute $$\ x\,\equiv\, \dfrac{24560}{12260}\,\pmod{\!24755}\$$ per your edit, by the method in my first answer.
$${\rm mod}\,\ 24755\!:\,\ \dfrac{0}{24755}\overset{\large\frown}\equiv \dfrac{24560}{12260}\overset{\large\frown}\equiv \color{#90f}{\dfrac{390}{235}}\overset{\large\frown}\equiv \color{#0a0}{\dfrac{4280}{40}}\overset{\large\frown}\equiv \color{#c00}{\dfrac{-535}{-5}}\overset{\large\frown}\equiv\dfrac{0}0$$
$$\begin{array}{rl} \ \ \ \ {\rm i.e.}\ \ \ \ \bmod 24755\!: \ \ \ \ \ [\![1]\!] &\ 24755\, x\,\equiv\ 0\ \\ [\![2]\!] &\ \color{c00}{12260\,x\, \equiv\ 24560\equiv -195}\!\!\!\\ [\![1]\!]\:\!-\:\!2\,[\![2]\!] \rightarrow [\![3]\!] &\ \ \ \ \ \color{#90f}{235\,x\, \equiv\ 390}\ \\ [\![2]\!]\!-\!\color{1orange}52\,[\![3]\!] \rightarrow [\![4]\!] &\ \ \ \ \ \ \, \color{#0a0}{40\,x\, \equiv\ 4280}\ \\ [\![3]\!]\:\!-\:\!\color{}6\,[\![4]\!] \rightarrow [\![5]\!] &\:\! \ \ \ \ \ \color{#c00}{{-}5\,x\, \equiv -535}\ \\ [\![4]\!]\:\!+\:\!\color{1orange}8\,[\![5]\!] \rightarrow [\![6]\!] &\:\!\ \ \ \ \ \ \ \ \color{90f}{0\,x\, \equiv\ 0}\ \end{array}$$
\begin{align}{\rm Therefore}\ \ \ x\equiv {\color{#c00}{\dfrac{535}5}\!\!\!\pmod{24755}}&\equiv \,107\!\!\pmod{\!4951},\ \ {\rm by\ canceling}\ \ 5\ \ \rm\color{darkorange}{everywhere}\\ &\equiv\, 107+4951k\!\!\pmod{\!24755},\ \ 0\le k\le 4\\[0.5em] &\equiv \{107,\, 5058,\, 10009,\, 14960,\, 19911\}\!\pmod{\!24755}\end{align}
Solving $154x \equiv 14 \pmod{182}$ is the same as finding all solutions to $$154x + 182y = 14.$$ In this case, we might think of this as finding all solutions to $$14(11x + 13y) = 14(1),$$ or rather $$11x + 13 y = 1.$$ Finally, solving this is the same as solving $11x \equiv 1 \pmod {13}$, which has solution $x \equiv 6 \pmod{13}$.
So we learn that $x \equiv 6 \pmod{13}$ is the solution. Of course, this isn't a single residue class mod $182$. Thinking modulo $182$, we see that the solutions are $x \equiv 6, 6+13,6+26,6+39, \ldots, 6+13*13 \equiv 6, 19, 32, \ldots, 175.$
This approach works generally --- factor out the greatest common divisor, consider the resulting modular problem, and then bring it back up to the original problem.
• I really like this solution. I'm going to experiment with it a bit more with it, but I believe that this is what I was looking for. – Jesse Daniel Mitchell Dec 10 '16 at 23:40
• @JesseDanielMitchell It's worth mentioning that to be able to solve an equation $ax+by = c$, you must have $\text{gcd}(a,b)$ divides $c$. When you can get it to the point that mixedmath did ($11x+13y = 1$, where $11,13$ are coprime) we have that the gcd is $1$, and this divides everything, and we're done. If after trying to reduce it ended up being $12x+14y = 1$, this would clearly have no solutions, and the formal reason for this is because $\text{gcd}(12,14)\nmid 1$. – Mark Dec 10 '16 at 23:53
• @Mark thank you for pointing that out. Is there another approach for computing these numbers if the $\gcd(a,b) ∤ 1$? For example, $12260x \equiv 24560 \mod 24755$. (107 needs to be one of the computed answers). – Jesse Daniel Mitchell Dec 11 '16 at 0:30
• @JesseDanielMitchell Generically, the approach in this answer works everytime if there are actually solutions. If there are no solutions, then of course there will be no algorithm to produce solutions. – davidlowryduda Dec 11 '16 at 1:44
Note $$\ \gcd( 154,182)=\color{#c00}{14}\,$$ so factoring it out and cancelling it yields
$$\color{#c00}{14}\cdot 13\,\mid\, \color{#c00}{14}\,(11x\!-\!1)\!\!\overset{\rm\ \ cancel\ \color{#c00}{14}_{\phantom{I_I}}\!\!\!\!}\iff\ 13\mid 11x\!-\!1\iff {\rm mod}\ \ 13\!:\ x\equiv \dfrac{1}{11}\equiv \dfrac{-12}{-2}\equiv 6\qquad$$
Below I derive the general solution in fractional form, which often greatly simplifies matters. Then I show how to present the extended Euclidean algorithm succinctly using these (multi-valued) modular fractions. See my other answer above for how this method applies to the OP.
Generally let's consider the solution of $$\ B\, x \equiv A\pmod{\! M}.\$$ If $$\,d=(B,M)\,$$ then $$\, d\mid B,\,\ d\mid M\mid B\,x\!-\!A\,\Rightarrow\, d\mid A\$$ is a necessary condition for a solution $$\,x\,$$ to exist.
If so let $$\ m, a, b \, =\, M/d,\, A/d,\, B/d.\$$ Cancelling $$\,d\,$$ $$\rm\color{darkorange}{everywhere}$$ i.e. from $$\,A,B\,$$ & $$M$$ yields
$$\ x\equiv \dfrac{A}B\!\!\!\pmod{\!M}\iff M\mid B\,x\!-\!A \!\!\overset{\rm\large\ \, cancel \ d}\iff\, m\mid b\,x\! -\! a \iff x\equiv \dfrac{a}b\!\!\!\pmod{\!m}\qquad$$
where the fraction $$\ x\equiv a/b\pmod{\! m}\,$$ denotes all solutions of $$\,ax\equiv b\pmod{\! m},\,$$ and similarly for the fraction $$\ x\equiv A/B\pmod{\! M}.\$$ Note there may be zero, one, or multiple solutions.
The above implies that if solutions exist then we can compute them by cancelling $$\,d = (B,M)\,$$ $$\rm\color{darkorange}{everywhere},$$ i.e. from the numerator $$A,\,$$ the denominator $$B,\,$$ $$\rm\color{darkorange}{and}$$ the modulus $$M,\,$$ i.e.
$$x\equiv \dfrac{ad}{bd}\!\!\!\pmod{\! md}\iff x\equiv \dfrac{a}b\!\!\!\pmod{\! m}\qquad$$
where $$\bmod m\!:\ a/b = ab^{-1}\,$$ uniquely exists as $$\,b^{-1}\,$$ does, by $$\,(b,m)=1$$.
If $$\, d>1\,$$ the fraction $$\, x\equiv A/B\pmod{\!M}\,$$ is multiple-valued, denoting the $$\,d\,$$ solutions
\quad\ \begin{align} x \equiv a/b\!\!\pmod{\! m}\, &\equiv\, \{\, a/b + k\,m\}_{\,\large 0\le k
which is true because $$\ km\bmod dm =\, (\color{#c00}{k\bmod d})\, m\$$ by the mod Distributive Law, and the RHS takes exactly $$\,d\,$$ values, namely $$\,\color{#c00}0m,\, \color{#c00}1m,\, \color{#c00}2m, \ldots, (\color{#c00}{d\!-\!1})m,\,$$ so ditto for their shifts by $$\,a/b$$.
$${\rm e.g.} \overbrace{\dfrac{6}3\pmod{\!12}}^{{\rm\large cancel}\ \ \Large (3,12)\,=\,3}\!\!\!\equiv\, \dfrac{2}{1}\!\pmod{\!4}\,\equiv\, \!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{\{2,6,10\}}^{\qquad\ \ \Large\{ 2\,+\,4k\}_{\ \Large 0\le k< 3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\pmod{\!12}$$
Remark Such multiple-valued fractions frequently arise in the extended Euclidean algorithm when performed in fractional form. Let's use it to compute $$\, x\equiv \color{#0a0}{9/5}\pmod{\!18}.\,$$ We obtain
$${\rm mod}\ 18\!:\ \ \ \underbrace{\overbrace{\dfrac{0}{18}\overset{\large\frown}\equiv \color{#0a0}{\dfrac{9}5} \overset{\large\frown}\equiv \dfrac{9}3}^{\Large\ \ 0\,-\,3(\color{#0a0}9)\ \equiv\ 9\ }}_{\Large 18\,-\,3(\color{#0a0}5)\ \equiv\ 3}\overset{\large\frown}\equiv \dfrac{0}{2}\overset{\large\frown}\equiv \color{#c00}{\dfrac{9}{1}}\overset{\large\frown}\equiv\dfrac{0}0\qquad\quad$$
so $$\ {\rm mod}\ 18\!:\ x\equiv\color{#0a0}{9/5}\equiv\color{#c00}{ 9/1}\equiv 9.\,$$ Checking $$\, 5x\equiv 5\cdot9\equiv 45\equiv 9,\,$$ is indeed true.
Above each Euclidean reduction step essentially mods out successive denominators as follows
$$\dfrac{a}{b}\overset{\large\frown}\equiv\dfrac{c}d\overset{\large\frown}\equiv\dfrac{a-qc}{b-qd}\ \ {\rm where}\ \ q = \lfloor b/d \rfloor,\ \ {\rm so }\ \ b\!-\!qd = b\bmod d$$
i.e. the denominators are the values occurring in Euclid's algorithm for $$\,\gcd(18,\color{#0a0}5),\,$$ but we perform those operations in parallel on the numerators too, e.g. the first step above has $$\, q =\lfloor 18/\color{#0a0}5\rfloor = 3\,$$ so the denominator is $$\, 18-3(\color{#0a0}5)\equiv 3.\,$$ Executing the same operation on the numerators yields the next numerator, namely: $$\ 0-3(\color{#0a0}9)\equiv 9.\,$$ The following steps proceed the same way, but all quotients (except final $$\,q=2)$$ are $$\,q=1,\,$$ so we simple subtract successive numerators and denominators.
The invariant in the algorithm is that the common solutions of each neighboring pair of fractions remains constant. It starts as the common solution of $$\,0/18\overset{\large\frown}\equiv 9/5$$ $$\,:= 18x\equiv 0,\ 5x\equiv 9.\,$$ which is equivalent to $$\,5x\equiv 9,\,$$ since $$\,18x\equiv 0\,$$ is true for all $$\,x\,$$ by $$\,18\equiv 0.\,$$ Similarly it ends with the common solution of $$\,9/1 \overset{\large\frown}\equiv 0/0\,$$ $$:= 1x\equiv 9,\ 0x\equiv 0,\,$$ and again the latter can deleted.
The proof that the Euclidean reduction preserves the solution set is as follows.
$$\qquad\ \$$ If $$\,\ dx\!-\!c \equiv 0\,\$$ then $$\,\ bx\!-\!a \equiv 0\! \iff\! \overbrace{(bx\!-\!a)-q(dx\!-\!c)}^{\Large (b-qd)\,x\,-\,(a-qc)}\!\equiv 0$$
This immediately implies that \ \ \begin{align}bx&\equiv a\\ dx&\equiv c\end{align} \!\iff\!\! \begin{align}(b\!-\!qd)x&\equiv a\!-\!qc\\ dx&\equiv c\end{align}
It is instructive to look at the intermediate system $$\, 9/3\overset{\large\frown}\equiv 0/2.\,$$ By above we know that
\begin{align} &\overbrace{\dfrac{9}3\!\!\!\pmod{\!18}}^{{\rm\large cancel}\ \ \Large (3,18)\,=\,3}\!\!\!\equiv\, \dfrac{3}{1}\!\!\!\pmod{\!6}\,\equiv\, \{3,\color{#c00}9,15\}\!\!\!\pmod{\!18} \\[.7em] & \underbrace{\dfrac{0}2\!\!\!\pmod{\!18}}_{{\rm\large cancel}\ \ \Large (2,18)\,=\,2}\!\!\!\equiv\, \dfrac{0}{1}\!\!\!\pmod{\!9}\,\equiv\, \{0,\color{#c00}9\}\ \ \ \pmod{\!18} \end{align}\quad\ \
Notice that the common solution of both is indeed $$\,\ x\equiv \color{#c00}9\pmod{\!18},\,$$ as we found above. Note also that even though we started with a fraction $$\,9/5\,$$ whose denominator $$\,5\,$$ is coprime to the modulus $$\,18\,$$ (so the fraction is single-valued), the Euclidean algorithm passes through various multiple-valued fractions (with non-coprime denominators), even systems with both fractions multiple-valued, such as $$\, 9/3\overset{\large\frown}\equiv 0/2\,$$ above, i.e. the system $$\, 3x\equiv 9,\ 2x\equiv 0\pmod{\!18}.$$
The chosen notation $$\,\large \frac{a}b \overset{\frown}\equiv \frac{c}d\,$$ resembles a padlock (and a congruence combined with intersection) in order to emphasize that the fractions are locked together via intersection - generally we cannot separate the fractions - rather, the solution is the intersection of the adjacent multivalued fractions, so it is not necessarily equal to either one of them (as in the example above).
Such calculations are more commonly expressed without fractions by instead performing operations on systems of equations - operations generalizing Gaussian elimination and triangularization, e.g. reduction of matrices to Hermite /Smith normal form. These topics are studied more abstractly in the theory of modules in abstract algebra (essentially generalizing linear algebra to allow scalars from a ring, not only a field).
To solve $ax\equiv c \mod b$, set $\;d=a\wedge b$, $\;a=a'd, \;b=b'd$. This congruence implies $c$ is divisible by $d$. Actually, it's easy to see that $$ax\equiv c\mod b\iff \begin{cases}c\equiv 0\mod a\wedge b\\\text{and}\\a'x\equiv c'=\dfrac{c}{a\wedge b} \mod b' \end{cases}$$ Thus the problem comes down to the case $a$ and $b$ coprime, after a compatibility condition has been checked.
Added: solution of the second congruence
First we check with the Euclidean algorithm that $\gcd(12260,24755)=5$, and $$\frac{12260}5=2452,\quad\frac{24755}5=4951,\quad\frac{24560}5=4912.$$ Thus the given congruence is equivalent to $\; 2452 x\equiv 4912\mod 4951$, and we have to find the inverse of $2452$ modulo $4951$. This means we have to find a *Bézout's relation between $2452$ and $4951$. It can be obtained with the extended Euclidean algorithm: $$\begin{array}{rrrr} r_i&u_i&v_i&q_i\\ \hline 4951&0&1\\ 2452&1&0&2\\\hline 47&-2&1&52\\ 8&105&-52&5\\ 7&-527&261&1\\ 1&632&-313\\\hline \end{array}$$ Thus $632\cdot2452-313\cdot4951=1$, whence $2452^{-1}=632\bmod4951$, and the solution is $$x\equiv 632\cdot4912\equiv 107\mod4951.$$
• Actually you can do the gcd in parallel with the fraction calculation, and the calculations are simpler - see my 2nd answer.. – Bill Dubuque Dec 11 '16 at 1:44
From your question, I assume you know how to use the extended Euclidean algorithm to compute the modular inverse $a^{-1} \pmod b$ when $a$ is coprime to $b$. Even when $a$ is not coprime to $b$, you can actually solve $ax \equiv c \pmod b$ in almost exactly the same way, assuming that a solution exists.
What the extended Euclidean algorithm actually computes, given the inputs $a$ and $b$, is a triple of integers $(\bar a, \bar b, g)$ such that $g$ is the greatest common divisor of $a$ and $b$, and $a\bar a + b\bar b = g$. When $g = 1$, then $\bar a = a^{-1} \pmod b$, and we can use it to compute the solution $x \equiv c \bar a \pmod b$ to the modular congruence $ax \equiv c \pmod b$.
When $g$ is not $1$, we might call the pair $(\bar a, g)$ the pseudoinverse* of $a$ modulo $b$, as it satisfies the congruence $a \bar a \equiv g \pmod b$, where $g$ is the smallest positive number for which such a congruence exists. Thus, given the congruence $ax \equiv c \pmod b$, we can multiply both sides by $\bar a$ to obtain $gx \equiv c \bar a \pmod b$. If (and only if) $c$ is divisible by $g$, we can also then divide both sides by $g$ (using normal integer division!) to obtain the solution $x \equiv c\bar a / g \pmod b$. Of course, this solution is only unique modulo $b/g$.
Otherwise, if $c$ is not divisible by $g$, no solution exists.
*) You will not find the term "modular pseudoinverse" in any textbooks, since I just made it up. I'm not aware of any more established term for this useful concept, though, and at least it's descriptive, so please indulge me for using it here.
• As written, this method yields only a necessary condition for $x$ to be a solution, since it first multiplies by $\bar a,$ which need not be invertible mod $b$. To avoid this we can first cancel $\,g=(a,b)\,$ then cancel $\,a/g\,$ by multiplying by its inverse $\,\bar a\pmod{b/g},\,$ by $\,\bar a(a/g) + \bar b(b/g) = 1.\,$ The method you described is essentially the same as this standard method (done in fractions in my answer), but it reverses the order of the cancellations, which spoils the bidirectionality of the inferences. – Bill Dubuque Feb 7 '17 at 15:34
Using Euler's Theorem for modular multiplicative inverses:
$\varphi(182) = 72$
$x \equiv 154^{\varphi(182)-1} \pmod{182} \Rightarrow 84 \equiv 154^{71} \pmod{182}$
Now every $x$ of the form $x=84 \pm k\cdot182$ will satisfy $154x \equiv 14 \pmod{182}$
• Euler's Theorem only works if $a$ and $n$ (which is defined as $b$ in my scenario) are co-prime positive integers, but in this situation, they aren't. – Jesse Daniel Mitchell Dec 10 '16 at 23:39
• @JesseDanielMitchell Note that $\gcd(182,154)=14$ so $\gcd(182,154) \equiv 154^{\varphi(182)-1} \pmod{182}$ – kub0x Dec 10 '16 at 23:42
One way to solve the problem ax ≡ c (mod b) would be to treat it as a problem in finding 𝑥0,𝑥1 for the linear equation ax0 − bx1 = c where 𝑎,𝑏,𝑐,𝑥0,𝑥1 ∈ 𝑍. One way to solve the linear equation would be to use substitution in two ways: first to introduce a new variable and a new equation at each step to create a linear system of equations, and second to turn every non-basic variable in the linear system of equation into a basic variable. The motivation for the use of substitution is to express each original variable (𝑥0,𝑥1) as a basic variable that is a function of parameter variables (𝑡0,𝑡1,…) such that for any integer value assigned to the parameter variables would result in the original variables taking on integer values. The computations in the algorithm may be organized using augmented matrices.
I have applied one such algorithm to your example problems in "Solving 154x ≡ 14 (mod 182) and 12260x≡ 24560 (mod 24755) Using Augmented Matrices". It also includes an explanation for the algorithm and an animation of the step-by-step solution to the problems. | 2020-10-26T21:59:03 | {
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http://math.stackexchange.com/questions/469265/proof-a-subseteq-b-cup-c-and-b-subseteq-a-cup-c-then-a-b-subseteq | # Proof: $A\subseteq (B\cup C)$ and $B\subseteq (A\cup C)$ then $(A - B) \subseteq C$
How do I prove this:
Let $A, B$ and $C$ be sets, $A \subseteq (B \cup C)$ and $B \subseteq (A \cup C)$ then $(A - B) \subseteq C$
Let $x \in A$ and $y \in B.$
Since $A \subseteq (B \cup C)$ then $x \in (B \cup C).$
Since $B \subseteq (A \cup C)$ then $y \in (A \cup C)$
$x \in B \cup C$, so if $x \in B$, then $x \notin A - B.$
if $x \notin B$, then $x \in C.$
$y \in A \cup C$, so if $y \in A$, then $y \notin A - B.$ if $y \notin B$, then $y \in C.$
So in both cases, $A - B \subseteq C$
Is it correct? if yes, the converse is false right?
-
You don't need both conditions. $A\subset(B\cup C)\Rightarrow (A\setminus B)\subset C$. – Owen Sizemore Aug 16 '13 at 20:18
@OwenSizemore These are actually equivalent. See math.stackexchange.com/q/479748/11994. – Marnix Klooster Nov 19 '13 at 7:09
The question only asks you to prove that $A-B$ is a subset of $C$. This means that we need to take $x\in A-B$ and show that $x\in C$.
Let $x$ be such element, then $x\in A$ and therefore $x\in B\cup C$. However, $x\notin B$ and therefore $x\in C$.
In the other direction, it is true that if $A-B\subseteq C$ then $A\subseteq B\cup C$. See if you can prove it.
(Hint: $x\in A$ then either $x\in B$ or $x\notin B$.)
-
SO to recap $A - B \subset C$ if and only if $A \subset B \cup C$. – N. S. Aug 16 '13 at 20:24
You stated more than the necessary assumptions, as pointed out by a comment and an answer that only used $A \subseteq B \cup C$. You can check the converse if you only need to prove the condition $A \subseteq B \cup C$. So for the "full converse" the way you stated it, the only additional question that remains is whether $A-B \subseteq C$ implies $B \subseteq A \cup C$. Hint: Consider $C$ to be the empty set, and when $A \subset B$, where $A \neq B$.
-
Well... The converse will be this: Let A,B and C be sets, (A−B)⊆C then A⊆(B∪C) and B⊆(A∪C)
I think this is false. Take the negation and proof:
Negation: Let A,B and C be sets, (A−B)⊆C then A Not⊆ (B∪C) or B Not⊆ (A∪C)
Here is a counter example:
A = {1, 2} B = {1, 4} C = {2, 5}
So A - B = {2} and is a subset of C.
But, A U C = {1, 2, 5} - but B = {1, 4}. So B is not a subset of AUC. Element 4 is missing in AUC.
Correct?
-
Guys please tell me what is wrong with this??? thx... a -1 flag doesn't help much... – Najeeb Aug 16 '13 at 20:50
This is correct. :) Found an example in the book. – Najeeb Aug 16 '13 at 21:07 | 2015-08-05T06:33:46 | {
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https://mathhelpboards.com/threads/length-of-confidence-interval.24061/ | # Length of confidence interval
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
A research institute wants to establish a confidence interval for the quota of working people in a city. Let $\hat{p}_n$ be the estimated quota, based on a sample of size $n$. It is assumed that $n> 30$.
How can one determine the length of the confidence interval?
Generally this is equal to $L = 2 \cdot \frac{\sigma \cdot z}{\sqrt{n}}$, right?
Do you have to use the $\hat{p}_n$ in the formula in this case? But how?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey mathmari !!
The distribution of a proportion is a special case.
It has a z-distribution with an estimated standard deviation $\sigma_{\hat{p}_n} = \sqrt{\hat{p}_n(1-\hat{p}_n)}$.
And we should have $n \hat{p}_n > 10$ and $n (1 − \hat{p}_n) > 10$ to ensure it is sufficiently reliable.
#### mathmari
##### Well-known member
MHB Site Helper
The distribution of a proportion is a special case.
It has a z-distribution with an estimated standard deviation $\sigma_{\hat{p}_n} = \sqrt{\hat{p}_n(1-\hat{p}_n)}$.
And we should have $n \hat{p}_n > 10$ and $n (1 − \hat{p}_n) > 10$ to ensure it is sufficiently reliable.
Ah ok!!
So, the length is equal to $$L = 2 \cdot \frac{\sigma_{\hat{p}_n} \cdot z}{\sqrt{n}}= 2 \cdot \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} \cdot z}{\sqrt{n}}$$ where $z$ depends on the confidence level $1-\alpha$, right?
Suppose that $1-\alpha=0.95$. I want to determine $n$ so that the length of confidence interval is not bigger that $L_{\text{max}}=0.03$.
We have that $z =1.96$.
Then $$L_{\text{max}}=0.03 \Rightarrow 2 \cdot \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} \cdot 1.96}{\sqrt{n}}\leq 0.03\Rightarrow \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} }{\sqrt{n}}\leq \frac{3}{392} \Rightarrow \frac{ \hat{p}_n(1-\hat{p}_n) }{n}\leq \frac{9}{153664}\Rightarrow n\geq \frac{153664\hat{p}_n(1-\hat{p}_n)}{9}$$ right? We cannot continue from here, can we?
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Indeed.
Btw, do we have $L_{\text{max}}=0.04$ or $L_{\text{max}}=0.03$?
#### mathmari
##### Well-known member
MHB Site Helper
Indeed.
Btw, do we have $L_{\text{max}}=0.04$ or $L_{\text{max}}=0.03$?
Oh, it is $L_{\text{max}}=0.03$. | 2020-07-05T08:17:30 | {
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https://www.physicsforums.com/threads/why-does-it-matter-what-convergence-test-i-use.866541/ | # Why does it matter what convergence test I use?
Tags:
1. Apr 11, 2016
### enh89
I just took a calc 2 test and got 3/8 points on several problems that asked you to show convergence or divergence. The reason being that I didn't use the correct test of convergence? The answer was right, if you get to the point where you know the series converges, then why does it matter which test I used?
Does anyone have any tips for how to tell which test I should be using when I look at a sequence or series? (my idea is- when in doubt, use ratio test).
2. Apr 11, 2016
### SammyS
Staff Emeritus
You may have found the correct value for what the series converged to, without actually showing that the series converged.
Can you give examples of the questions asked?
3. Apr 11, 2016
### axmls
I can see two possibilities.
2. The test asked you to solve a problem in a specific way and you didn't.
It would help to have an example of why you lost points.
4. Apr 11, 2016
### micromass
Staff Emeritus
Getting the correct answer is not so difficult. You can basically guess "convergent" and have a 1 in 2 chance of being correct. What matters is that your reasoning is correct. If somebody said the correct answer but with incorrect reasoning, I wouldn't give any points.
So, are you sure your reasoning is correct?
5. Apr 11, 2016
### enh89
The answer was correct (converges), but the method by which I got there was not.
6. Apr 11, 2016
### micromass
Staff Emeritus
The series diverges actually.
7. Apr 11, 2016
### enh89
Haha.... how so? My instructor said answer was right, method was not. Very interesting... :s
8. Apr 11, 2016
### enh89
Here is the question in full, parts a, b, and e (above). 3 points for the "correct answer" and minus 5 for incorrect method as indicated by my instructor.
9. Apr 11, 2016
### SammyS
Staff Emeritus
$\displaystyle \frac{3k^2+2}{k^3+2k+5} \$ behaves like $\displaystyle\ \frac1k \$ for large $k$ .
10. Apr 12, 2016
### BreCheese
I posted two photos below showing how to determine convergence or divergence.
The first photo proves why the divergence test is inconclusive for the series you provided.
The second photo shows how to prove divergence of the series you provided by means of the integral test.
Hope this helps. Let me know if the photos are not displaying please, and I will do what I can to fix it.
https://scontent.fsnc1-1.fna.fbcdn.net/v/t1.0-9/12994327_10209359465651426_1265712068729596108_n.jpg?oh=e24d119a922a506c62295f24cab018b6&oe=577ED737
https://scontent.fsnc1-1.fna.fbcdn.net/hphotos-xta1/v/t1.0-9/12439374_10209359465611425_8823074394851131369_n.jpg?oh=f81f8f11dddc2333cb813384a1f93226&oe=577CEFCA
11. Apr 12, 2016
### Samy_A
For a), the question was about the sequence, not the series.
But you don't mention the test you are using (hence -5). And the general term of the series behaves like $2k^{-\frac{5}{3}}$ for large k.
12. Apr 12, 2016
### pwsnafu
At my university, we obtain very detailed marking schemes. Here's what we would require from a Calc 2 student (sample answer in black, marking scheme in blue).
Granted, different schools have different standards, but this should give you some expectations.
Notice in part (a) its asking for the sequence not series.
(a)
Observe $\frac{-5n^2}{3n^3-2n+4} \leq \frac{(-1)^n 5n^2}{3n^3-2n+4} \leq \frac{5n^2}{3n^2-2n+4}$.
1 method mark for using squeeze theorem.
1 accuracy mark for correct upper and lower bounds.
Then
$\lim_{n\to\infty} \frac{5n^2}{3n^3 - 2n +4} = \lim_{n\to\infty} \frac{5/n}{3-2/n^2+4/n^3}$
$=\frac{\lim_{n\to\infty} 5/n}{\lim_{n\to\infty}(3-2/n^2+4/n^3)}$ by limit laws
$=\frac{0}{3-0+0}$ by standard limit $1/n \to 0$
$=0.$
1 method mark for dividing by highest power.
1 justification mark for stating "limit laws" and "standard limits"
1 accuracy mark for obtaining limit equal to 0.
Similarly, $\frac{-5n^2}{3n^3-2n+4} \to 0$
So by squeeze theorem, $\lim_{n\to\infty} \frac{(-1)^n 5n^2}{3n^2-2n+4} = 0$.
Hence because 0 is a real number, the sequence converges.
1 justification mark for stating squeeze theorem.
1 accuracy mark for concluding that sequence converges.
1 notation mark (must be 100% correct).
Total 8 marks.
(b)
Let $a_k = \frac{2k+3}{\sqrt[3]{k^8+9k}}$. Then
$\sum_{k=1}^\infty a_k < \sum_{k=1}^\infty \frac{2k+3k}{\sqrt[3]{k^8+0}} = \sum_{k=1}^\infty \frac{5}{k^{5/3}}$.
1 method mark for using comparison test.
1 accuracy mark for finding correct upper bound.
Now $\sum_{k=1}^\infty \frac{1}{k^{5/3}}$ converges as it is a harmonic p-series with $p=5/3 > 1$.
1 justification mark for mentioning harmonic p-series.
Hence by comparison test, $\sum_{k=1}^\infty a_k$ also converges.
1 accuracy mark for stating the series converges.
1 justification mark for mentioning comparison test.
As $a_k > 0$ for all $k$, $\sum_{k=1}^\infty a_k$ converges absolutely.
1 accuracy mark for stating the series converging absolutely with suitable reason.
1 notation mark (must be 100% correct).
Total 7 marks.
(e)
Let $a_k = \frac{3k^2+2}{k^3 + 2k +5}$. Then
$\sum_{k=1}^\infty a_k > \sum_{k=1}^\infty\frac{3k^2}{k^3+2k^3+5k^3} = \frac{3}{8}\sum_{k=1}^\infty\frac{1}{k}$.
1 method mark for using comparison test.
1 accuracy mark for finding correct upper bound.
Then $\sum_{k=1}^\infty\frac{1}{k}$ diverges because it is the harmonic p-series with $p=1$.
1 justification mark for mentioning harmonic p-series diverges.
Hence, by the comparison test, $\sum_{k=1}^\infty a_k$ diverges.
1 accuracy mark for stating the series diverges.
1 justification mark for stating comparison test.
1 notation mark (must be 100% correct).
Total 5 marks.
13. Apr 12, 2016
### BreCheese
I first taught myself to recognize and name series (geometric series, telescoping series, harmonic series, p-series, alternating series, etc). All of these series have their own set of conditions and rules to follow, so it is difficult to identify which series you're looking at at first (but keep with it and study the theorems of each series and you'll quickly be able to recognize which type of series you are looking at).
I then practiced by working out multiple exercises that had varying series.
Lastly, I used note cards to define each type of series and I wrote out a step-by-step method to determine the convergence (or divergence) of each series.
If I couldn't tell which type of series I was looking at I followed the following advice (which I read while doing my home work on Pearson's MyMathLab):
"If the general kth term of the series involves k!, k^k, or a^k, where a is a constant, the Ratio Test is advisable. Series with k in an exponent may yield to the Root Test. If the general kth term of the series is a rational function of k (or a root of a rational function), use the Comparison Test or the Limit Comparison Test."
Lastly, to answer your question (which is the title of your thread), the convergence test you use matters because it either proves, or disproves, what our intuition tells us about a given series. Because each type of series has its own conditions and rules, we have to be careful which convergence test we use. For instance, its easy to think that the harmonic series converges because the limit of its underlying sequence goes to zero. However, despite some people's intuition, we can use the p-series rules to prove that the harmonic series diverges. Furthermore, we can support our conclusions from the p-series by performing the integral test on the harmonic series. In reality, although the limit of the underlying sequence goes to zero (and looks like it converges to zero), it doesn't go to zero fast enough and therefore diverges. In short, we can use convergence tests to support or refute our intuition (because lets face it, we're all human and our intuition is not always right).
Last edited: Apr 12, 2016
14. Apr 12, 2016
### zinq
"The answer was right, if you get to the point where you know the series converges, then why does it matter which test I used?"
This is a very basic question that all students would benefit from knowing the answer to.
You may know the right answer, but how does the grader know that you actually knew the right answer instead of just guessing, or perhaps that you may have used incorrect reasoning to arrive at the answer? Especially in a case like convergence / divergence where there are only two answers, anyway?
That's why we profs want math students to show their work. This also enables us to help you get on the right track if we know what kind of mistake you made. | 2017-11-24T00:29:48 | {
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https://math.stackexchange.com/questions/393929/integrate-by-parts-int-ln-2x-1-dx/393944 | Integrate by parts: $\int \ln (2x + 1) \, dx$
\eqalign{ & \int \ln (2x + 1) \, dx \cr & u = \ln (2x + 1) \cr & v = x \cr & {du \over dx} = {2 \over 2x + 1} \cr & {dv \over dx} = 1 \cr & \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr & = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr & = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr & = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr & = x\ln (2x + 1)^{3 \over 2} - x + C \cr}
The answer $= {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
• Your last equality. You can't combine the log terms as you have done because only one of them has a factor of $x$. Instead, you should bring the $\frac{1}{2}$ down, and then factor out $\ln(2x+1)$. – Jared May 16 '13 at 21:04
• You're also missing two $\mathrm dx$, and parentheses around $\displaystyle1-\frac1{2x+1}$. – joriki May 16 '13 at 21:07
• And you can use substitution $t = 2x+1$ and then to apply integration by parts. – Cortizol May 16 '13 at 21:27
5 Answers
Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):
$$= x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \tag{1}$$
Good, up to this point... $\uparrow$.
So the error was in your last equality at the very end:
You made an error by ignoring the fact that the first term with $\ln(2x+1)$ as a factor also has $x$ as a factor, so we cannot multiply the arguments of $\ln$ to get $\ln(2x+1)^{3/2}$. What you could have done was first express $x\ln(2x+1) = \ln(2x+1)^x$ and then proceed as you did in your answer, but your result will then agree with your text's solution.
Alternatively, we can factor out like terms.
$$= x\ln(2x + 1) + \frac 12 \ln(2x + 1) - x + C \tag{1}$$ $$= \color{blue}{\bf \frac 12 }{\cdot \bf 2x} \color{blue}{\bf \ln(2x+1)} + \color{blue}{\bf \frac 12 \ln(2x+1)}\cdot {\bf 1} - x + C$$
Factoring out $\color{blue}{\bf \frac 12 \ln(2x + 1)}$ gives us
$$= \left(\dfrac 12\ln(2x + 1)\right)\cdot \left(2x +1\right) - x + C$$ $$= \frac 12(2x + 1)\ln(2x+1) - x + C$$
• Why can't I do: \eqalign{ & = \ln {(2x + 1)^x} + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \cr & = \ln {(2x + 1)^{x + {1 \over 2}}} - x + C \cr} – seeker May 16 '13 at 21:12
• You're welcome, Assad: note, your procedure would have worked, had you done the arithmetic a little better: see the other answer, e.g. Note also that when it comes to integrals, especially when working with logarithms, many "different appearing solutions can be derived, depending on one's manipulations, etc. – amWhy May 16 '13 at 21:21
• @amWhy: Nice to get ack'ed by OP! +1 – Amzoti May 17 '13 at 1:01
Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that $$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$ But the remaining integrand is just $1$! It follows that our integral is $$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$
I am sure there are more tidy ways to do this but as an alternative...
Why not do $\int \ln(2x + 1)dx$ using:
$v^\prime = 1 \Rightarrow v = x$ and $u = \ln(2x+1)\Rightarrow u^\prime=\frac{2}{2x+1}$
Therefore,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{2x}{2x+1}dx$$
Then make the substitution $u=2x + 1$ to yield,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{1}{2}\frac{u-1}{u}du$$
Thus,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}u - ln(u) + c$$
Then,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}(2x-1) - \ln(2x-1) + c$$
$$= \ln(2x-1)(x-1) - \frac{1}{2}(2x-1) + c$$
$$= x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C$$
$$= \ln(2x + 1)^x + \ln(2x + 1)^\dfrac 12 - x + C$$
$$= \ln({(2x + 1)^x \cdot(2x + 1)^\dfrac 12}) - x + C$$ $$= \ln{(2x + 1)^\dfrac{2x+1}{2} } - x + C$$ $$= \dfrac {1}{2}\cdot (2x+1) \ln{(2x + 1)} - x + C$$
Why don't you put $u = 2x + 1$ so that $du = 2 \,dx$? Then we'd have $$\int \log (2x + 1) \, dx = \frac {1} {2} \int \log u \, du = \frac {1} {2} (u \log u - u) = \frac {1} {2} (2x + 1) \log (2x + 1) - x - \frac {1} {2} + C.$$ | 2020-02-18T19:43:10 | {
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https://mathematica.stackexchange.com/questions/21711/deleteduplicates-does-not-work-as-expected-on-floating-point-values | # DeleteDuplicates[] does not work as expected on floating point values
Here is my simple example, and in this case function DeleteDuplicates does not work as expected.
I want to FindRoot of my function $\chi[\nu]$, and since function $\chi$ is very sensitive to initial guess I decided to generate many initial conditions and leave only those solutions which are distinct. For this purpose I want to apply DeleteDuplicates on the resulting list of solution.
Here is the definition of my function:
χ[ν_] := 2*PolyGamma[1] - PolyGamma[1/2 + I*ν] - PolyGamma[1/2 - I*ν]
Here I generate many solution according to many random initial guesses
m = Table[v /. FindRoot[χ[v] == -1.2 - 0.2*I, {v, RandomComplex[]}], {i, 1, 10}]
And finally, I want to leave only distinct solutions:
DeleteDuplicates[m]
Unfortunately, the operation DeleteDuplicates[m] does not change the list m, although there are many identical values.
Namely:
DeleteDuplicates[m]
{1.06423 + 0.0968739 I, 1.06423 + 0.0968739 I,
1.06423 + 0.0968739 I, 1.06423 + 0.0968739 I, 0.0250407 + 1.00352 I,
1.06423 + 0.0968739 I, 1.06423 + 0.0968739 I, 1.06423 + 0.0968739 I,
0.0250407 + 1.00352 I, 1.06423 + 0.0968739 I}
I'm puzzled.
Any help or suggestions are very welcome!
Thanks!
• Try something like DeleteDuplicates[m, Abs[#1 - #2] < 0.01 &] . Mar 20, 2013 at 9:59
• Thanks for editing Mr. Wizard! It looks much better :) @b.gatessucks: Great, works like a charm. Could you explain the reason? I'm quite new to this. Thanks a lot for help! Mar 20, 2013 at 10:03
• Andrew, please see the update to my answer for an important note about performance. Mar 21, 2013 at 13:53
• Andrew, what about changing "does not work properly" to "does not work as expected"? In fact it works properly, the cause is just seemingly identical numbers... Mar 21, 2013 at 14:17
• Related: (19112) Feb 19, 2016 at 21:50
You would do well to understand the difference between tools that are intended for structural operations and those that are intended for mathematical operations. DeleteDuplicates is of the former, generally speaking. As such it is comparing the exact FullForm of the objects, or at least something close (caveat).
As b.gatessucks recommends in a comment you can use a mathematical comparison function for the equivalence test of DeleteDuplicates, e.g.:
DeleteDuplicates[m, Abs[#1 - #2] < 10^-12 &]
{1.06423 + 0.0968739 I, 0.0250407 + 1.00352 I}
Incidentally you could also use Union, but the syntax is a bit different. Note the ( ).
Union[m, SameTest -> (Abs[#1 - #2] < 10^-12 &)]
{0.0250407 + 1.00352 I, 1.06423 + 0.0968739 I}
Using InputForm to show all of the digits of your expression you can see that they are not structurally identical in the (approximate) way that Mathematica "sees" them:
m // InputForm
{1.0642275928442373 + 0.09687392021742822*I, 1.0642275928442366 + 0.09687392021742817*I,
1.0642275928442366 + 0.09687392021742797*I, 1.064227592844237 + 0.09687392021742822*I,
1.0642275928442373 + 0.09687392021742852*I, 1.0642275928442366 + 0.09687392021742793*I,
1.0642275928442368 + 0.09687392021742801*I, 0.025040728196256346 + 1.0035162552538588*I,
1.0642275928442377 + 0.0968739202174282*I, 1.0642275928442375 + 0.0968739202174283*I}
### Performance
Yves reminded me to mention something about the performance of using a custom comparison function in DeleteDuplicates or Union as I did above. For long lists this is always considerably slower than using the default method. I gave an example with timings in How to represent a list as a cycle.
To apply that method here we could Round the numbers beforehand:
Round[m, 10^-12] // DeleteDuplicates // N
{1.06423 + 0.0968739 I, 0.0250407 + 1.00352 I}
I added // N to convert back to machine precision, but the values will not be precisely the same. This probably doesn't matter if you consider numbers this close to be duplicates, but should you want the unchanged numbers you could use GatherBy and get performance not far distant.
First /@ GatherBy[m, Round[#, 10^-6] &]
Version 10.0 introduced DeleteDuplicatesBy which works similarly to the GatherBy method; it has the following syntax:
DeleteDuplicatesBy[m, Round[#, 10^-6] &]
However it may not perform as well as GatherBy; see:
• Thanks for the answer, I see. It's like a comparison with epsilon of two floating point numbers. I assumed, that MATHEMATICA already does it for me. Mar 20, 2013 at 10:08
• Great, I missed InputForm. Without it it looks the same. Mar 20, 2013 at 10:09
• @Andrew FullForm was a bit needlessly verbose here, but in general you should look at the FullForm when trying to understand how Mathematica will treat an expression using structural tools. There are many cases where not using it will leave you quite befuddled as it can be very different from what is shown in standard output notation. Mar 20, 2013 at 10:13
• Admonish - lovely acoustics but far too harsh. I thought of it as "gently encouraging" at most... Mar 21, 2013 at 14:13
• @AJHC Converting to plain lists using {x, y, z} /. solutions seems like a good way to start. For example with m2 = {{x -> 0.1, y -> 0.2, z -> 0.3}, {x -> 0.1, y -> 0.2, z -> 0.3}, {x -> 0.17, y -> 0.22, z -> 0.314}} then DeleteDuplicates[{x, y, z} /. m2, AllTrue[Abs[# - #2], # < 10^-5 &] &] or DeleteDuplicatesBy[{x, y, z} /. m2, Round[#, 10^-5] &] for the two fundamentally different methods described in my answer. If you need to convert back to a list of rules then something like Thread[{x, y, z} -> #] & /@ {{0.1, 0.2, 0.3}, {0.17, 0.22, 0.314}} Mar 25, 2019 at 12:05
Note also that one can use Equal[##] & but not Equal:
DeleteDuplicates[m, Equal[##] &]
(* {1.06423 + 0.0968739 I, 0.0250407 + 1.00352 I} *)
DeleteDuplicates[m, Equal] // Length
(* 10 *)
This should work the same on most or all well-behaved FindRoot results.
As already noted, using something like Equal[##] & causes the performance to degrade, significantly on very long lists, but using Equal does not cause the same problem. However, Equal does not tolerate difference, unlike Equal[##] &:
DeleteDuplicates[{1., 1. + $MachineEpsilon}, Equal] DeleteDuplicates[{1., 1. + 64$MachineEpsilon}, Equal[##] &]
(*
{1., 1.}
{1.}
*)
Basically DeleteDuplicates[.., Equal] is the same as DeleteDuplicates[]. | 2022-08-19T08:35:04 | {
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http://math.stackexchange.com/questions/162566/computing-27272727-pmod-10/162596 | # computing ${{27^{27}}^{27}}^{27}\pmod {10}$
I'm trying to compute the most right digit of ${{27^{27}}^{27}}^{27}$.
I need to compute ${{27^{27}}^{27}}^{27}(\bmod 10)$.
I now that ${{(27)^{27}}^{27}}^{27}(\bmod 10) \equiv{{(7)^{27}}^{27}}^{27} (\bmod 10)$, so now I need to to compute ${({7^{27})}^{27}}^{27} (\bmod 10)$, since $\gcd(7,10)=1$ and $\phi(10)=4$, $7^{27}=7^{24}\cdot 7^3(\bmod 10)=1 \cdot 7^3 (\bmod 10)=3 (\bmod 10)$ - (Fermat theorem), so I am left with computing ${(3^{27}})^{27} (\bmod 10)$, again $\gcd(3,10)=1$, so $3^{27}= 3^{24}\cdot3^3 \equiv 7(\bmod 10)$, so as I see it the final cut should be again $7^{27}$ which I saw it is already $3 (\bmod 10)$.
Is it correct? what is the correct way to do that?
Thanks
-
$(7)^{27^{27^{27}}}$ isn't $(7^{27})^{27^{27}}$ – Cocopuffs Jun 24 '12 at 20:13
Oh no :( What a fatal mistake. What should I do then? – Jozef Jun 24 '12 at 20:18
Except that I confused notation, so scratch that. (Thanks Cocopuffs). – MGN Jun 24 '12 at 20:18
@Jozef Your method still works fine. Just figure out what $27^{27^{27}}$ is mod $4$. – Cocopuffs Jun 24 '12 at 20:19
The $\TeX$ code {{27^{27}}^{27}}^{27} looks as if it meant $((27^{27})^{27})^{27}$, but it comes out looking like $\displaystyle{{27^{27}}^{27}}^{27}$. If you write it as 27^{27^{27^{27}}}, so that it looks right to the naked eye reading the $\TeX$ code, then it looks like this: $\displaystyle 27^{27^{27^{27}}}$, which for some purposes might be considered inferior under the circumstances. – Michael Hardy Jun 24 '12 at 20:23
On the base level, for all $q$ you get $$\forall \ k: \quad a \equiv a - k \cdot q \mod q.$$ Going one level up, as you may know you're not calculating modulo $q$ anymore, but modulo $\phi(q)$: $$\forall \ k: \quad a^{\displaystyle b} \equiv a^{\displaystyle b - k \cdot \phi(q)} \mod q.$$ You can repeat this, and each time you add a $\phi$: $$\forall \ k: \quad a^{\displaystyle b^{\displaystyle c}} \equiv a^{\displaystyle b^{\displaystyle c - k \cdot \phi(\phi(q))}} \mod q, \\ \forall \ k: \quad a^{\displaystyle b^{\displaystyle c^{\displaystyle d}}} \equiv a^{\displaystyle b^{\displaystyle c^{\displaystyle d - k \cdot \phi(\phi(\phi(q)))}}} \mod q.$$
For $q = 10$ we then have \begin{align} q &= 10, \\ \phi(q) &= 4, \\ \phi(\phi(q)) &= 2, \\\phi(\phi(\phi(q))) &= 1. \end{align} So reducing the numbers on each level separately we get \begin{align} 27^{\displaystyle 27^{\displaystyle 27^{\displaystyle 27}}} &\equiv (27 \bmod 10)^{\displaystyle (27 \bmod 4)^{\displaystyle (27 \bmod 2)^{\displaystyle (27 \bmod 1)}}} \mod 10 \\ &\equiv 7^{\displaystyle 3^{\displaystyle 1^{\displaystyle 0}}} \mod 10 \\ &\equiv 7^{\displaystyle 3^{\displaystyle 1}} \mod 10 \\ &\equiv 7^{\displaystyle 3} \mod 10 \\ &\equiv 3 \mod 10 \end{align}
-
No such thing as "the" correct way. First we need to know the meaning of the exponential tower, since parentheses have not been inserted. The usual convention is that $a^{b^c}$ means $a^{(b^c)}$. That convention was violated in your argument, so in principle the argument is not correct. A similar error with different numbers can lead to a wrong answer. In this case it didn't. Apart from that, everything was fine.
For $10$, we might as well let our knowledge of the multiplication table guide us. The powers of $7$, starting with $7^1$, end in $7,9,3,1,7,9,\dots$. We are looking at an odd power of $27$, so the answer must be $7$ or $3$.
To decide which, we have to decide whether the exponent $27^{27^{27}}$ is congruent to $1$ or $3$ modulo $4$. Our exponent is $27$ to an odd power, and $27\equiv -1\pmod{4}$. Thus the exponent $27^{27^{27}}$ is also congruent to $-1$ modulo $4$. So the last digit must be $3$.
Remark: One can mention Euler's Theorem. Since $\varphi(10)=4$, the powers of $27$ must, modulo $10$, cycle with period that divides $4$. However, in this case, the cycling is clear.
Note the almost automatic translation of $x\equiv 3\pmod{4}$ to $x\equiv -1\pmod{4}$. This makes computation much more transparent.
-
Hint $\rm\,\ n\,$ odd $\rm\:\Rightarrow n^{n^{n^{\cdot^{\cdot^{\cdot^n}}}}}\!\!\!\equiv n^3\!\pmod{10}\$ by $\rm\ \phi(10) = 4,\ \, n^n\equiv n\!\pmod 4\,\$ $[$if tower has $> 1$ term$]$
$\varphi(10)=4$ and $27$ is of the form $(4n+3)\;$. Now ${(4n+3)^{(4n+3)}} \equiv 3(\bmod 4)=4c+3\;$. Clearly, ${(4n+3)}^{(4n+3)^{(4n+3)}}=(4n+3)^{(4c+3)}\equiv 3(\bmod 4)\;$. This will be held true for any length of such powers of the form $4n+3$.
So ${27}^{27^{27^{27}}}\equiv 7^3(\bmod\ 10)\equiv 3(\bmod\ 10)\; \;$ as $7^4\equiv 1(\bmod\ 10)\;$ . | 2014-04-17T02:03:06 | {
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http://satoevyemekleri.com/brightcove-vs-xtrmo/conditional-probability-past-paper-questions-58ce94 | Question 13 There are a number of ways to approach this problem. This video is accompanied by two exam style question to further practice your knowledge. Teachers. Previous Year Examination Questions 4 Marks Questions. P9a – Conditional probability Learn Maths, Boss Maths BossMaths Ltd | 71-75 Shelton Street, Covent Garden, London, WC2H 9JQ | Company Registration Number 10655114 | Registered in England & Wales MrMannMaths 5 3. Note that conditional probability does not state that there is always a causal relationship between the two events, as well as it does not indicate that both events occur simultaneously. What is the probability that both children are girls? Latest News . For two events A and B, Probability Questions and Answers. mr barton maths Teachers Students Jokes Puzzles Blog Podcast Twitter ondemand_video YouTube Pinterest Diagnostic Questions info About . calculated. Related Resources. If it was, the probability of picking a red ball (etc.) Answer all questions. Fully worked-out solutions of these problems are also given, but of course you should first try to solve the problems on your own! How to do Probability? PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. This should be used as part of a test, fill in the gaps and test cycle. The question is on probability trees and conditional probability. Instructions Use black ink or ball-point pen. Instructions Use black ink or ball-point pen. Question 1: The probability that it is Friday and that a student is absent is 0.03. P(A|B) means the probability of A occurring, given that B has occurred. Probability of occurrence of an event A when another event B in relation to A has already occurred. (5) b) Calculate the probability of exactly one of the two events occurring. Visit BYJU'S to get formulas. Conditional probability using two-way tables. Donate or volunteer today! Be able to use Bayes’ formula to ‘invert’ conditional probabilities. Complete coverage of the … The most obvious is to work out how many GCSE Exam Questions on Higher Probability Probability Tree (Grade A) 1. PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. The probability that a student knows the correct answer to a multiple choice question is 2 3. In two sizes, pdf and ppt. Mathematics / Grade 12 / Statistics. Solution: The formula of Conditional probability Formula is: P (B|A) = P(A ∩ B)⁄P(A) MrMannMaths 3 2. AS-Level Probability.docx-2. Important Questions for Class 12 Maths Class 12 Maths NCERT Solutions Home Page Venn Diagrams and Conditional Probability. GCSE Maths - Probability (Conditional Probability, AND OR rules, Multiplying) A Grade Level Descriptors GCSE Maths Conditional Probability. endobj Along with Detailed Answers, Timing, pdf download. This method GCSE Revision Cards . Probability Important Questions for CBSE Class 12 Maths Conditional Probability and Independent Events. c 2013 by Henk Tijms, Vrije University, Amsterdam. Tracing paper may be used. The final part of the question is simply the calculation of a conditional probability. Conditional probability occurs when it is given that something has happened. A lot more at goteachmaths.co.uk! Conditional probability with Bayes' Theorem, Practice: Calculating conditional probability, Conditional probability using two-way tables, Conditional probability tree diagram example, Tree diagrams and conditional probability. i'm struggling to understand how this answer for a past paper is correct. … p <> <> Videos, worksheets, 5-a-day and much more A Level maths P5/Probability. Edexcel A Level Statistics & Mechanics exam revision with questions and model answers for Conditional Probability 1. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. View All . Amy is going to play one game of snooker and one game of billiards. The probability that she will win the game of snooker is 4 3 The probability that she will win the game of billiards is 3 1 Complete the probability tree diagram. Author: Created by goteachmaths. 2�@�Oڍm��p�>g��4�����M�^-��@�rۼ�ҦZ�6� www.cie.org.uk. Click here for Answers . Assignments, Tests and more Grade 12 Questions Probability - edwardsmaths Probability Question Papers And Memo This is likewise one of the factors by obtaining the soft documents of this probability question papers and memo by online. This resource is for teachers, tutors and students, to test a student’s knowledge in the area of Statistics and Properties for GCSE Maths. The probability of the guessed answer being correct is 1 4 .Given that the student has answered the question correctly, the conditional probability that the student knows the correct answer is. This post is about solution of probability questions of past papers. This is the currently selected item. Conditional Probability 1 | Model Answers Natasha Undrell 2019-09-02T11:07:12+01:00 �+PS��Ϫ��k��ɓ#�4U Conditional Probability and Tree Diagrams The calculations above were reasonably easy and intuitive. Fill in the boxes at the top of this page with your name, centre number and candidate number. -- If you like this resource, then please rate it and/or leave a comment. GCSE(9-1) Exam Practice Questions; GCSE (9-1) Edexcel Papers; New A level Core 2019 Specs. Get to the point ISS (Statistical Services) Statistics Paper I (New 2016 MCQ Pattern) questions for your exams. In this post questions are from different paper from May 2013 to May 2020. As always with conditional probability the important step is to deduce which two probabilities need to be calculated. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. The questions type in this post is calculator . 1. Clues and fully worked model solutions/answers available free from mathsupgrade.co.uk.. Next Reverse Percentages Practice Questions. Visit now! Preview. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. Conditional Probability. 1/3). Use the Venn diagram to determine \text{P}(G \text{ given } H) (Also written \text{P}(G|H)). A/A* GCSE Maths revision Higher level worked exam (tree diagrams, conditional) Show Step-by-step Solutions. MrMannMaths 9 5. A-level Statistics Edexcel S1 January 2007 Q2c This is the worked solution to question 2c from the Edexcel Statistics S1 paper … Probability Questions - GCSE Questions - Higher - AQA (no rating) 0 customer reviews. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to … 6 Marks Questions. I also make them available for a student who wants to do focused independent study on a topic. What is the textbook’s definition of probability? The final part of the question is simply the calculation of a conditional probability. (0580-S 2016-Paper 4/2-Q5) MrMannMaths 6 . (0580-W 2016-Paper 4/2-Q4) MrMannMaths 10 . arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Worksheets with Answers. Have you been practicing exam questions? Transcript. Grade 8/9 Topics. Question 13 There are a number of ways to approach this problem. In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. Tracing paper may be used. Classroom Game . What is the probability that a student is absent given that today is Friday? 7. Revision Video . This is the result of not replacing the first ball hence only leaving 13 balls in the bag to pick from. The question is on probability trees and conditional probability. Videos, worksheets, 5-a-day and much more x��Z]o�H}G�?̣]���=��J��]���6����K�m����'���{�u��%x �1���s�f����6���ٳ��,��}��|菋ŧ���"�_gw�yVN�y���s�o��g_����dx~F�u; M��Z�HBT����JFSN�y��� �w;�q����d|���҄0"��Rw�'3X��Ɛ�|/�sWvs�����]�=�9�=�c�]�՛8���u,"�/�����p�ˆ�غO�Mc����'2��v^�y�w;m�Q\Pɫ�8/6ƓV�0F���4���pp����3�@^Ey�@8Ii��|T��XjN�Ӓ��M�H�p�J���{:Z���{�q��h�&Rj�Ϛ����q�(g>��c�.��!����D��YhGE�JO�+(� 1/3). Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. Question: In a lake there are 10 fish, 3 of which are tagged. Conditional probability answers the question ‘how does the probability of an event change Example: In the Venn diagram below, G represents students selecting Geography and H represents students selecting History. Here we have to work out the probability that the coach takes out two balls that are a different colour. Nearly all candidates attempted this question showing that time was not a factor in this paper. 1. Contact Info. There are currently no resources in this topic. The Corbettmaths Practice Questions on Conditional Probability. Past Paper Questions on Probability. Conditional probability is the probability of an event occurring, given that another event has occurred. And best … (ii) What is the probability that exactly one of them will solve it? The conditional probability in part (c) proved to be more challenging. Conditional Probability. (2) (January 10) 2. %PDF-1.7 This video explores Conditional Probability, a key concept in IB Maths HL Topic 5: Statistics and Probability. Big School. Preview and details Files included (3) pdf, 817 KB. (0580-S 2016-Paper 4/3-Q4) MrMannMaths 4 . stream 2 Conditional Probability. Edexcel GCSE Paper 2 November 2012 Question 21 Conditional Probability Show Step-by-step Solutions. VCAA Maths Methods Past Papers Exam Questions & Solutions. Understand the base rate fallacy thoroughly. Instructions Use black ink or ball-point pen. Solutions to Exam 1 Practice Questions: Long List (PDF) Exam 1 (PDF) Solutions to Exam 1 (PDF) 2: Exam 2 Practice Questions (PDF) Solutions to Exam 2 Practice Questions (PDF) Exam 2 (PDF) Solutions to Exam 2 (PDF) Final: Final Exam Practice Questions (PDF) These practice questions cover only the material taught in class sessions after Exam 2. Find the probability that a student chosen at random is male. Conditional probability tree diagram example. Especially those in VCAA Maths Methods past papers? In this paper we show part of the results of a larger study1 that investigates conditional probability problem solving. <>/ExtGState<>/XObject<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 595.32 841.92] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> Conditional Probability questions are frequently found in IB Maths SL exam papers, often in Paper 2. Conditional Probability: GCSE Questions. Practice Questions; Post navigation. Mathematics / Grade 12 / Counting and Probability. Impossible Event: The events in which the probability is 0, such events are called impossible events. [�Uʩz|��Y���$5ǩ�F�CO�v���["�*tej@�X��*��"���U����y��Tq�P��Qh�#k���R~[�qm}L��5��@��Z'�GZ�����ͣ7�. Since there are 5 school days in a week, the probability that it is Friday is 0.2. It really is one of the very best websites around. Grade 12 Questions Probability Past papers and memos. CIE A Level Probability & Statistics 1 past paper exam questions organised by topic with mark schemes. For conditional probability questions, when drawing the tree diagram we have to be careful as the probability changes between the two events. The Corbettmaths Practice Questions on Probability. This video explores Conditional Probability, a key concept in IB Maths SL Topic 5: Statistics and Probability. Created: Nov 14, 2019. Solution : Let "A", "B" and "C" be the events of solving problems by each students respectively. KS2/3/4:: Data Handling & Probability:: Probability. Conditional Probability: Probability of event A given event B. TOPICAL PAST PAPER QUESTIONS - 2016/2017 STATISTICS AND PROBABILITY (Paper 4) 1. Firstly click on a Main topic below that interests you. These are some probability questions … Maths Methods Main Topics & Sub Topics. The conditional probability in part (c) proved to be more challenging. Must Practice 11 Plus (11+) Probability Past Paper Questions. Here you can assume that if a child is a girl, her name will be Lilia with probability$\alpha \ll 1$independently from other children's names. For example: The probability of a row of data is the joint probability across each input variable. Many candidates Name Questions Solutions; Probability: tree diagrams : Questions: Solutions: Probability: conditional : Questions: Solutions . In particular, we report on a structure-based method to identify, classify and analyse ternary problems of conditional probability in mathematics textbooks in schools. If you're seeing this message, it means we're having trouble loading external resources on our website. 1. MrMannMaths 7 4. 3 0 obj CONDITIONAL PROBABILITY ©MathsDIY.com Page 2 of 7 the second time will be the same as the first (i.e. Secondly when you get to the video have a go at the question. Must Practice 11 Plus (11+) Probability Past Paper Questions. AP® is a registered trademark of the College Board, which has not reviewed this resource. If there are 7 green marbles and 4 red marbles in a bag, and you were asked to draw one without looking, what is the probability of drawing one green marble? a) Show that A and B are independent. Along with Detailed Answers, Timing, pdf download. These past paper questions help you to master the 11+ Exam Maths Questions. Conditional Probability and Conditional Probability examples. calculated. 6. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a first course in prob-ability. Sure Event: When the probability of an event to occur is 1, that is the event certain to happen, such events are called sure events. Register; Login × × Key Skills (0) Exam Qs (0) Resources (0) × GCSE Probability. Tree diagrams and conditional probability. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you don’t put that green marble back in the bag, what is the probability of drawing a green marble now? The Ultimate Collection of ALL questions on PROBABILITY from EdExcel 1MA1 9-1 GCSE Mathematics - FREE.All questions from past papers, specimen papers, sample papers are included. These types of probability form the basis of much of predictive modeling with problems such as classification and regression. 4. How to answer GCSE questions on conditional probability, examples and step by step solutions, GCSE Maths. Past Papers; Timestables; Live! 4 0 obj Year 1 AS Pure ... Further Maths 2; Extension Questions; A Level Practice Papers 2019 Specs; Tuition; Select Page. CONDITIONAL PROBABILITY A2 Unit 4: Applied Mathematics B Section A: Statistics WJEC Past paper questions: 2010 - 2018 Total marks available 180 (approximately 3 hours 40 minutes) (Jan 10) (Jan 10) (Summer 10) 2. Venn diagrams can also be used to solve conditional probability problems. The probability that the card is a heart given (the prior information) that the card is red is denoted by P H R Note that P H R = n(H \R) n(R) = P(H \R) P(R): This probability is called the conditional probability … Revision Sheet - All Topics Revision by Topic NUMBERS Question Paper - Paper 2 and Paper 4 Mark Scheme - Paper 2 and Paper 4 ALGEBRA 1 and 2 Question Paper - Paper 2 and Paper 4 Mark Scheme - Paper 2 and Paper 4 QUADRATIC EQUATIONS Question Paper - Paper 2 and 4 Mark Scheme - Paper 2 and 4 INEQUALITIES and SIMILARITY Question Paper - Paper 2 and… In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. If not then try some now. Answer all questions. ... Probability help (Conditional probability) - very basic. the second time will be the same as the first (i.e. Khan Academy is a 501(c)(3) nonprofit organization. Perfect revision resources for Maths. I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. Read more. 2. PAPER 2 NUMBERS 2016-2017 PPQs – Numbers QP 2016-2017 PPQs – Numbers MS ALGEBRA+SEQUENCES 2016-2017 PPQs – Algebra QP 2016-2017 PPQs – Algebra MS MENSURATION 2016-2017 PPQs … Probability-Conditional Probability: Questions 1-4 of 5. Conditional Probability. (0580-W 2016-Paper 4/1-Q2) MrMannMaths 8 . PROBABILITY Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Visit now! Events A and B are such that ( ) =0.2,( )=0.4, ( ∪ )=0.52. It really is one of the very best websites around. These past paper questions help you to master the 11+ Exam Maths Questions. For example, the probability of John doing mathematics at A-Level, given that he is doing physics may be quite high. Conditional Probability - Venn diagrams : 1: 2: 3: ... Corbett Maths offers outstanding, original exam style questions on any topic, as well as videos, past papers and 5-a-day. Free. Our mission is to provide a free, world-class education to anyone, anywhere. CONDITIONAL PROBABILITY PROBLEMS WITH SOLUTIONS. Problem 1 : A problem in Mathematics is given to three students whose chances of solving it are 1/3, 1/4 and 1/5 (i) What is the probability that the problem is solved? A-level Statistics Edexcel S1 January 2007 Q2c This is the worked solution to question 2c from the Edexcel Statistics S1 paper … If the student does not know the answer, then the student guesses the answer. Probability Questions is an an essential part for Competitive Exams like Banking, Insurance, SSC and Railways Exams. 5. By this post students will come to know variety of questions asked in previous year papers. Both independent and conditional probability are covered. Probability - Dependent Events : 1: 2: 3: Corbett Maths ... on any topic, as well as videos, past papers and 5-a-day. 5-a-day Workbooks. This video is accompanied by an exam style question to further practice your knowledge. Related Topics: More Lessons for GCSE Maths Math Worksheets ... Edexcel GCSE Paper 2 November 2012 Question 21 Conditional Probability Fill in the boxes at the top of this page with your name, centre number and candidate number. Corbettmaths - This video explains how to answer conditional probability questions. Then if you get stuck have a look at my solution to it. 3 fish are caught randomly from the lake without replacement. ���T��Q�����\[�^3��g�RÝ�Q.c-��!Ͳ��1r^�����^���Irh �z�[&!x������� Tracing paper may be used. Conditional Probability questions are frequently found in IB Maths HL exam papers, often in Paper 2. <>/Metadata 226 0 R/ViewerPreferences 227 0 R>> Probability Practice Questions Click here for Questions . endobj 3. 2 0 obj W�2毱^o�͠�7��ɟ+�]��麀��K>c�pW�۬� �lBS볾�&w���E�6 �=>���PdW���>F���^�P=T����^��J}U�e��=��ʮq�-�|Ɖ��=d����{�x ��aa� O��7�r�Uoq�J{%�z~�Ӯ�4�NB�Q(��4b8����zp��[�{;@μ��1��R�o@֪� In this lesson we revise Paper 3 Statistics & Probability by looking at typical exam questions. THE IMMORTALS. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. Vinay process 50% of the forms. endobj Conditional Probability. 1. proportion probability problem. 3. Cloned/Copied questions from previous 9-1 AQA GCSE exams. Practice Papers; Conundrums; Class Quizzes; Blog; About; Revision Cards; Books; September 2, 2019 corbettmaths. 1 0 obj As always with conditional probability the important step is to deduce which two probabilities need to be calculated. (Hint: look for the word “given” in the question). This site is zero rated by. Conditional probability and independence. CONDITIONAL PROBABILITY PROBLEMS IN TEXTBOOKS AN EXAMPLE FROM SPAIN 319 M. ÁNGELES LONJEDO VICENT, M. PEDRO HUERTA PALAU, MARTA CARLES FARIÑA CONDITIONAL PROBABILITY PROBLEMS IN TEXTBOOKS AN EXAMPLE FROM SPAIN Revista Latinoamericana de Investigación en Matemática Educativa (2012) 15 (3): 319-337. �ƛ�k�z[��-���������C�_#�ףq��_�AJK��ٰ�s�q��;*Y{iJt������-O 4Jm�����0�^[q5�YLk�?���O][�ֶ�dx���1���6��XC�%��],0r��&��H���DQ�N�H,U�z3e%N(��t>�࿓�tC�I�� J!��T����e/.�&�hjl͐�|RNQ�rs2x��&ݔ��h���S�����WT�pTmz ��{�w/����@"�^g��c���Ƞ�.�%NJ$f�"�|>% O�.�d�R��d�?�O����� %���� - Try Now Markscheme (A1)(A1) (C2) Note: Award (A1) for numerator, (A1) for denominator. Question 18 In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain form. Conditional Probability and Venn Diagrams (From Edexcel 6683) Q1, (Jun 2006, Q6) Q2,(Jun 2008, Q5) AlevelMathsRevision.com Q3, (Jan 2010, Q4) AlevelMathsRevision.com Q4, (Jun 2010, Q4) AlevelMathsRevision.com Q5, (Jan 2012, Q6) Q6, (Jun 2007, Q4) AlevelMathsRevision.com Q7, (Jan 2013, Q7) Q8, (Jun 2014, Q8) AlevelMathsRevision.com Q9, (Jun 2013(R), Q6) AlevelMathsRevision.com … Be able to organize the computation of conditional probabilities using trees and tables. In this example, the question states that the ball is not replaced. WJEC past paper questions: 2010 - 2017 Total marks available 57 (approximately 1 hour 10 minutes) 1. [2 marks] Examiners report Parts (a) and (b) were well answered with many candidates gaining 4 marks there. arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Lessons . Loading... Save for later. (0580-S 2016-Paper 4/1-Q3) MrMannMaths 2 . Late Night Studies. Conditional probability questions can appear complex at first but, once you have mapped out all of the possible outcomes of various events occurring, they are relatively easy to solve. If it was, the probability of picking a red ball (etc.) Parts (a) and (b) were well answered with many candidates gaining 4 marks there. Previous Direct and Inverse Proportion Practice Questions. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards CBSE Class 12 Sample Paper for 2021 Boards CBSE Class 12 Sample Paper for 2020 Boards Frequently asked simple and hard probability problems or questions with solutions on cards, dice, bags and balls with replacement covered for all competitive exams,bank,interviews and entrance tests. 817 KB balls in the gaps and test cycle ] Examiners report Parts ( a ) and ( B were... And fully worked model solutions/answers available free from mathsupgrade.co.uk probability Answers the question much more Transcript, )! Pick from same as the first ( i.e probability is 0, such events are called events. The same as the probability of an event change the question is the! Paper exam Questions organised by topic with mark schemes ’ s definition of probability ondemand_video YouTube Pinterest Diagnostic info! Was, the probability that exactly one of them will solve it the! Knows the correct answer to a multiple choice question is 2 3 Answers the question is on probability to... - 2017 Total marks available 57 ( approximately 1 hour 10 minutes 1. Replacement: Lessons solution to it clues and fully worked model solutions/answers available free from mathsupgrade.co.uk video have look... Like Banking, Insurance, SSC and Railways Exams numerator, ( ∪ ) =0.52 with conditional occurs. Complete coverage of the very best websites around are called impossible events leaving 13 balls in venn. Pick from week, the probability of a conditional probability Questions … i 'm struggling to understand how this for! Office three employees Vinay, Sonia and Iqbal process incoming copies of a conditional probability the important is. Second time will be the events in which the probability that a student wants! Paper 4 ) 1 master the 11+ exam Maths Questions and Tree Diagrams Questions! A green marble now Files included ( 3 ) nonprofit organization and use all the features of Khan Academy please! Final part of the question ) how does the probability of a conditional probability when it given! An conditional probability past paper questions booklet and issue them in Class or give them out as a homework, some work. Multiplying ) a Grade Level Descriptors GCSE Maths conditional probability is 0, such events are.! In part ( c ) proved to be calculated, pdf download states that the domains *.kastatic.org *... A test, fill in the bag, what is the place for you 5-a-day and more... Specs ; Tuition ; Select Page 1 conditional probability past paper questions paper is correct websites around at... A|B ) means the probability of an event a when another event has occurred Total marks available 57 ( 1. Questions, when drawing the Tree diagram we have to work out the probability! Which two probabilities need to be more challenging GCSE exam Questions does not know the.. Maths 2 ; Extension Questions ; a Level probability & Statistics 1 past paper help! Is correct, world-class education to anyone, anywhere same as the probability that is... For a student is absent given that B has occurred ) =0.2, ( ∪ ) =0.52 study..., which has not reviewed this resource.kasandbox.org are unblocked Award ( A1 ) ( C2 ) Note: (. Free from mathsupgrade.co.uk for a past paper exam Questions - GCSE Questions - 2016/2017 and... Nonprofit organization probability ) - very basic Note: Award ( A1 ) ( 3 ) pdf, 817.! -- if you 're seeing this message, it means we 're having loading... By topic with mark schemes... further Maths 2 ; Extension Questions ; GCSE ( 9-1 ) papers... The gaps and test cycle in IB Maths SL topic 5: Statistics and probability trouble loading resources... ’ formula to ‘ invert ’ conditional probabilities using trees and conditional Show... And Iqbal process incoming copies of a series of events are called events! Bit of extra practise, this is the probability of John doing mathematics A-Level... To play one game of billiards the answer and c '' the... Replacement: Lessons nonprofit organization Banking, Insurance, SSC and Railways Exams fill the. Probability changes between the two events he is doing physics May be high. The final part of a conditional probability: probability conditional probability past paper questions picking a red ball etc! Conditional ) Show that a student who wants to do focused independent study on a topic Page your... Such that ( ) =0.4, ( ∪ ) =0.52 817 KB a,... Board, which has not reviewed this resource Answers, Timing, pdf download B independent. - Try now probability important Questions for CBSE Class 12 Maths conditional probability and Tree Diagrams - /. Questions - Higher - AQA ( no rating ) 0 customer reviews independent events to from! Maths conditional probability Questions - GCSE Questions - Higher - AQA ( rating! Play one game of snooker and one game of billiards details Files (! Etc. and best … Find the probability that it is given something. Ondemand_Video YouTube Pinterest Diagnostic Questions info About resource, then please rate it and/or leave a comment given, of... Papers ; New a Level Core 2019 Specs ; Tuition ; Select Page some work. May 2020 Solutions ; probability: probability of an event change the question is simply the of. In this lesson we revise paper 3 Statistics & probability: Tree,. Sonia and Iqbal process incoming copies of a series of events are shown now! Series of events are called impossible events Maths HL topic 5: Statistics and probability ’ s definition probability... Answers the question is 2 3 of billiards 11+ exam Maths Questions Level Core 2019 Specs ; conditional probability past paper questions... Probability is 0, such events are shown mr barton Maths Teachers students Jokes Puzzles Blog Podcast Twitter YouTube! ( C2 ) Note: Award ( A1 ) ( C2 ) Note: Award ( A1 for. Of solving problems by each students respectively markscheme ( A1 ) ( C2 ) Note: Award ( A1 for! Let a '', B '' and c '' the! Probability: probability marble back in the bag to pick from only leaving 13 in. The problems on your own place for you 2012 question 21 conditional probability in (. Paper exam Questions organised by topic with mark schemes and one game of conditional probability past paper questions one. C2 ) Note: Award ( A1 ) ( C2 ) Note: Award ( A1 ) numerator. Questions is an an essential part for Competitive Exams like Banking, Insurance SSC! Of ways to approach this problem May 2013 to May 2020 classification and.. Whether you want a homework trouble loading external resources on our website states that domains...: Questions: 2010 - 2017 Total marks available conditional probability past paper questions ( approximately 1 hour 10 )... Previous year papers and details Files included ( 3 ) nonprofit organization test fill... B '' and c '' be the events in which the probability changes between the two events.... Them available for a past paper Questions: Solutions coverage of the very best websites around that interests you at! Also given, but of course you should first Try to solve conditional probability in part ( c proved. ; New a Level probability & Statistics 1 past paper Questions help to! c '' be the same as the first ( i.e issue them in Class give! - Try now probability important Questions for CBSE Class 12 Maths Class 12 Maths Class Maths... Lake there are 10 fish, 3 of which are tagged a given event B in relation to multiple... For a past paper exam Questions of probability.kasandbox.org are unblocked overall probability of picking a ball! Banking, Insurance, SSC and Railways Exams of not replacing the first (.! Is the joint probability across each input variable example: in the boxes at the top of this with! Want a homework to it help ( conditional probability Questions … i 'm struggling to understand this. … Parts ( a ) Show Step-by-step Solutions papers 2019 Specs C2 ) Note: Award A1. Is accompanied by two exam style question to further Practice your knowledge (... Gaining 4 marks there NCERT Solutions Home Page Grade 12 Questions probability past papers and memos predictive... Specs ; Tuition ; Select Page Questions info About 2010 - 2017 Total marks available 57 ( 1! And best … Find the probability that a student is absent is 0.03 a... Registered trademark of the … Parts ( a ) 1 a given event B in relation to a already! Another event has occurred, pdf download ( 5 ) B ) were well answered with many TOPICAL! Probability Show Step-by-step Solutions solve it B ) were well answered with many candidates gaining 4 marks there H students. Wants to do focused independent study on a Main topic below that you... Clues and fully worked model solutions/answers available free from mathsupgrade.co.uk JavaScript in your browser )... Change the question is simply the calculation of a certain form ) resources 0! Example: the events in which the probability of a occurring, given that something has happened interests... New a Level probability & Statistics 1 past paper Questions: Solutions: probability 10,! The probability that it is Friday and that a student is absent given that another event B relation... ( conditional probability Questions … i 'm struggling to understand how this answer for a past paper is.! ) exam Qs ( 0 ) resources ( 0 ) exam Practice Questions ; a Level Core 2019 ;! To pick from is one of the two events occurring for Competitive Exams like Banking, Insurance SSC! If the student guesses the answer quite high game of snooker and one of. A row of Data is the probability that a student who wants to focused... Event a given event B for your Exams i 'm struggling to understand how this for!
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http://bogdank.webd.pl/6kas5l/infinitesimals-vs-limits-fd18ae | 5. Example 2 Find the limit $$\lim\limits_{x \to 0} {\large\frac{{\sqrt[3]{{1 + x}} – 1}}{x}\normalsize}.$$ But infinitesimals still occur in our notation which is largely inherited from Leibniz, however. top new controversial old random q&a live (beta) Want to add to the discussion? In 1870, Karl Weierstraß provided the first rigorous treatment of the calculus, using the limit method. A continuousentity—a continuum—has no “gaps”.Opposed to continuity is discreteness: to be discrete[2]is to beseparated, like the scattered pebbles on a beach or the leaves on atree. The shooting of female game birds like ducks, pheasants and turkeys is commonly limited or completely prohibited. 1993e (with Lan Li), Constructing Different Concept Images of Sequences and Limits by Programming, Proceedings of PME 17, Japan, 2, 41-48. The notion of zero is biased by our expectations. During the 1800s, mathematicians, and especially Cauchy, finally got around to rigorizing calculus. What’s the new ratio? We need to “do our work” at the level of higher accuracy, and bring the final result back to our world. The second operation, *, (called multiplication) is su… To the real numbers, it appeared that “0 * 0 = -1″, a giant paradox. HOPOS: The Journal of the International Society for the History of Philosophy of Science 3 (2013), no. Create an account. Epsilon-delta limits are by far the most popular approach and are how the subject is most often taught. We resist because of our artificial need for precision. Infinitesimal definition is - immeasurably or incalculably small. Breaking a curve into rectangles has a problem: How do we get slices so thin we don’t notice them, but large enough to “exist”? Infinitesimal definition, indefinitely or exceedingly small; minute: infinitesimal vessels in the circulatory system. Badiou vs. Deleuze - Set Theory vs. Infinitesimals build the model in another dimension, and it looks perfectly accurate in ours. 02 Apr 2019. Infinitesimals seem more intuitive to me -- although I have not looked into them extensively, I often think of things as infinitesimals first and then translate my thoughts to limits. Limits and infinitesimals. Limit is a related term of delimit. Calculus is usually developed by working with very small quantities. If you have a function y=f (x) you can calculate the limit as x approaches infinity, or 0, or any constant C. Infinitesimal means a very small number, which is very close to zero. Before the concept of a limit had been formally introduced and understood, it was not clear how to explain why calculus worked. (mathematics) A non-zero quantity whose magnitude is smaller than any positive number (by definition it is not a real number). Phew! Robinson's modern infinitesimal approach puts the intuitive ideas of the founders of the calculus on a mathematically sound footing, and is easier for beginners to understand than the more common approach via limits. And a huge part of grokking calculus is realizing that simple models created beyond our accuracy can look “just fine” in our dimension. To solve this example: In later articles, we’ll learn the details of setting up and solving the models. In the B-track, limit is defined in a more straightforward way using infinitesimals. Under the standard meanings of terms the answers to the bulleted questions are 1) Yes, Weierstrass and Cantor; 2) No, infinitesimals are an alternative to limits approach to calculus (currently standard), but both are reducible to set theory; 3) No, "monad" is Leibniz's term used in modern versions of infinitesimal analysis; 4) See 2). Yes, by any scale you have nearby. The reason limits didn’t have a rigorous standing was because they were a mean to an end (derivatives). This is a calculus textbook at the college Freshman level based on Abraham Robinson's infinitesimals, which date from 1960. We square i in its own dimension, and bring that result back to ours. These approaches bridge the gap between “zero to us” and “nonzero at a greater level of accuracy”. But it turns out that a straight line is a darn good model of a curve over short distances: Just like we can break a filled shape into tiny rectangles to make it simpler, we can dissect a curve into a series of line segments. We see that our model is a jagged approximation, and won’t be accurate. Fortunately, most of the natural functions in the world (x, x2, sin, ex) behave nicely and can be modeled with calculus. Go beyond details and grasp the concept (, “If you can't explain it simply, you don't understand it well enough.” —Einstein With infinitesimals? Turn any PC into a Super Cash Register! Adjective (en adjective) Incalculably, exceedingly, or immeasurably minute; vanishingly small. FOUNDATIONS OF INFINITESIMAL CALCULUS H. JEROME KEISLER Department of Mathematics University of Wisconsin, Madison, Wisconsin, USA [email protected] Enjoy the article? I use them because they click for me. 2001, Eoin Colfer, Artemis Fowl, page 221: Then you could say that the doorway exploded. They got rid of the “infinitesimal” business once and for all, replacing infinitesimals with limits. In essence, Newton treated an infinitesimal as a positive number that Join 0 points • 4 comments • submitted 8 hours ago by dasnulium to r/math. At the core of Calculus is the idea that, to really understand a curve, you have to understand what is happening at every instantaneous moment in time. Click or tap a problem to see the solution. Historically, the first method of doing so was by infinitesimals. Better Explained helps 450k monthly readers … Here’s a different brain bender: did your weight change by zero pounds while reading this sentence? We can break a complex idea (a wiggly curve) into simpler parts (rectangles): But, we want an accurate model. We call it a differential, and symbolize it as Δx. Intuitively, you can think of x as 0.0000…00001, where the “…” is enough zeros for you to no longer detect the number. But audio and video engineers know they don’t need a perfect reproduction, just quality good enough to trick us into thinking it’s the original. Le flux magnétique ou flux d'induction magnétique, souvent noté Φ , est une grandeur physique mesurable caractérisant l'intensité et la répartition spatiale du champ magnétique. (mathematics) A value to which a sequence converges. In short it is the intended result on the metric that is measured. Cauchy (1789–1857). You need to distinguish between mathematical definitions and everyday use. Some functions are really “jumpy” — and they might differ on an infinitesimal-by-infinitesimal level. FOUNDATIONS OF INFINITESIMAL CALCULUS H. JEROME KEISLER Department of Mathematics University of Wisconsin, Madison, Wisconsin, USA [email protected] We need to be careful when reasoning with the simplified model. In the A-track, limit is defined via epsilon-delta definitions. is that infinitesimal is (mathematics) a non-zero quantity whose magnitude is smaller than any positive number (by definition it is not a real number) while infinite is (mathematics) greater than any positive quantity or magnitude; limitless. Yes, Re(i) * Re(i) = 0, but that wasn’t the operation! The final, utmost, or furthest point; the border or edge. (informal) Very small. The simpler model, built from rectangles, is easier to analyze than dealing with the complex, amorphous blob directly. [Not yet in PDF format]. As unsatisfying as it may be, I think this is just something that we’ll have to accept as part of the “risk vs. reward” of using infinitesimals. Intuitively, the result makes sense once we read about radians). I think you didn't get the idea of Requests vs Limits, I would recommend you take a look on the docs before you take that decision.. Differential Calculus - Limits vs. Infinitesimals. They are well-behaved enough that they can be used in place of limits to show convergence properties, but the infinities and infinitessimals in limits are shorthands, while the infinities and infinitessimals in the hyperreals are actual elements of a field. Video shows still images at 24 times per second. Even though no such quantity can exist in the real number system, many early attempts to justify calculus were based on sometimes dubious reasoning about infinitesimals: derivatives were defined as ultimate ratios Under the standard meanings of terms the answers to the bulleted questions are 1) Yes, Weierstrass and Cantor; 2) No, infinitesimals are an alternative to limits approach to calculus (currently standard), but both are reducible to set theory; 3) No, "monad" is Leibniz's term used in modern versions of infinitesimal analysis; 4) See 2). Retrouvez Infinitesimals and Limits et des millions de livres en stock sur Amazon.fr. Achetez neuf ou d'occasion No baked lighting or shadows. This isn’t an analysis class, but the math robots can be assured that infinitesimals have a rigorous foundation. But i does a trick! Ce flux est par définition le produit scalaire de ces deux vecteurs1 (voir définition mathématique ci-dessous). They got rid of the “infinitesimal” business once and for all, replacing infinitesimals with limits. As adjectives the difference between limit and infinitesimal I like infinitesimals because they allow “another dimension” which seems a cleaner separation than “always just outside your reach”. Both Leibniz and Newton thought in terms of them. During the 1800s, mathematicians, and especially Cauchy, finally got around to rigorizing calculus. Oh, you have a millimeter ruler, do you? Versatile and cost-effective point-of-sale solution for businesses. Nobody ever told me: Calculus lets you work at a better level of accuracy, with a simpler model, and bring the results back to our world. 1. In ordinary English, something is infinitesimal if it is too small to worry about. A mathematical field is a set and two operations defined on the elements of that set, say (S, +, *). My goal isn’t to do math, it’s to understand it. (mathematics) Any of several abstractions of this concept of limit. I’ll draw the curve in nanometers. The thinner the rectangles, the more accurate the model. 1 people chose this as the best definition of infinitesimal: Capable of having values... See the dictionary meaning, pronunciation, and sentence examples. Viewed 2k times 3. clear, insightful math lessons. It looks like the function is unstable at microscopic level and doesn’t behave “smoothly”. So, 1 is what we get when sin(x) / x approaches zero — that is, we make x as small as possible so it becomes 0 to us. For example, the law a | 2021-09-25T04:03:20 | {
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https://mathhelpboards.com/threads/divide-a-line-segment-into-three-equal-parts.762/ | # divide a line segment into three equal parts
#### Amer
##### Active member
is there a way to divide a line segment into three equal parts using just compass and ruler ?
I heard that there is not a way and there is a proof for that is that right ?
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: divde a line segment into three equal parts
is there a way to divide a line segment into three equal parts using just compass and ruler ?
I heard that there is not a way and there is a proof for that is that right ?
It is possible. Draw a line segment AB. draw a ray beginning at A at an angle of 60 degrees to AB. draw another ray beginning at B at an angle of 60 degrees from BA(Note that the second ray is in opposite direction of first ray). Take any arbitrary distance on the compass. Put pointy end of compass on point A and cut first ray at point K. put pointy end of compass on K and cut first ray again to get point L. Put pointy end of compass on B and cut second ray to get point M and now put pointy end of compass on M and cut second ray to get point N. join KN and LM. The line segment AB is now trisected.
#### Ackbach
##### Indicium Physicus
Staff member
Re: divde a line segment into three equal parts
Perhaps you're thinking of an angle trisection instead of a line trisection? It is true that the angle cannot, so far as we know, be trisected using straightedge and compass.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: divde a line segment into three equal parts
is there a way to divide a line segment into three equal parts using just compass and ruler ?
I heard that there is not a way and there is a proof for that is that right ?
Angles, in general, cannot be trisected using (unmarked)straight edge and compass alone. In "Abstract Algebra" texts, example Herstein's "Topics in Algebra", it is proved that and 60 degree cannot be trisected using unmarked straight edge and compass alone.
#### Amer
##### Active member
Re: divde a line segment into three equal parts
how to draw an angle of 60 ? with straight ruler unmarked with compass
that cant be done
#### Plato
##### Well-known member
MHB Math Helper
Re: divde a line segment into three equal parts
is there a way to divide a line segment into three equal parts using just compass and ruler ?
I heard that there is not a way and there is a proof for that is that right ?
We can divide any line segment into any finite natural number of congruent sub-segments.
Start with $\overline {AB}$. At $A$ draw any ray not collinear with $\overrightarrow {AB}$.
Now on that ray starting at $A$ mark off three points $E,~F,~\&~G$ so that $\overline {AE},~\overline {EF},~\&~\overline {FG}$ have the same length.
Join $G~\&~B$ with a line. Construct at $E~\&~F$ lines parallel $\overline {GB}$.
Those lines will trisect $\overline {AB}$.
#### chisigma
##### Well-known member
Re: divde a line segment into three equal parts
The segment AB that must divide into three parts is represented in the figure...
The following procedure requires only a non graduaded rule and a compass...
a) construct the equilateral triangle ABC...
b) construct the segments DA=AB and BE=AB collinear to AB...
c) draw the segments DC and CE...
d) draw vertical lines passing through A and B a call F and I the intersection point with the segments DC and CE...
e) draw the horizontal segment FI and call G and H the intersection points with the segments AC and BC...
At this point we have the segment FI that is equal to AB and is divided into three equal segments FG, GH and HI...
Kind regards
$\chi$ $\sigma$
#### earboth
##### Active member
Re: divde a line segment into three equal parts
how to draw an angle of 60°? with straight ruler unmarked with compass
that cant be done
Draw an equilateral triangle with unmarked straight ruler and compass. Then you even have got three angles with 60°.
#### Amer
##### Active member
Re: divde a line segment into three equal parts
thank you all,thats great
#### HallsofIvy
##### Well-known member
MHB Math Helper
Re: divde a line segment into three equal parts
how to draw an angle of 60 ? with straight ruler unmarked with compass
that cant be done
It surely can be done. Use the ruler to draw a straight line. Mark any two points on the line and call them "A" and "B". Using the compass strike a circle through "B" having center "A". Using the compass strike a circle through "A" with center "B". Those circles with intersect in two points. Choose either of them and call it "C". The angle CAB will have measure 60 degrees. | 2021-04-20T13:55:39 | {
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https://stats.stackexchange.com/questions/103800/calculate-probability-area-under-the-overlapping-area-of-two-normal-distributi/103808 | # Calculate probability (area) under the overlapping area of two normal distributions
I have two normal distributions defined by their averages and standard deviations.
Sample 1: Mean=5.28; SD=0.91
Sample 2: Mean=8.45; SD=1.36
You can see how they look like in the next image:
How can I get the probability to obtain an individual from the overlapping area (green)? Is the probability the same as the area?
• What do you mean by the probability of obtaining an individual from the area? Jun 18, 2014 at 8:37
• If you sample points from either normal distribution, you get points on the Perikymata-axis rather than on the 2-dimensional area. Furthermore, the green zone is infinitely wide, so all values sampled from either distribution are under the green zone, so in this sense the probability would be 1. Jun 18, 2014 at 8:47
• @JuhoKokkala OP likely means integral over the green area. Nov 21, 2017 at 18:40
• @VladislavsDovgalecs What would the question "Is the probability the same as the area" then mean? Nov 21, 2017 at 19:21
• @VladislavsDovgalecs I think you misunderstood my comment, I was not asking for an interpretation of the area but trying to parse the question. In any case, please note OP's comments to the accepted answer and a followup question posted here stats.stackexchange.com/questions/103821. Nov 23, 2017 at 6:14
It is not quite clear what you mean by probability to obtain an individual from the overlapping area. This solves for the area of the green zone in your diagram:
Let:
• $X_1 \sim N(\mu_1,\sigma_1^2)$ with pdf $f_1(x_1)$ and cdf $F_1(x_1)$ and
• $X_2 \sim N(\mu_2,\sigma_2^2)$ with pdf $f_2(x_2)$ and cdf $F_2(x_2)$,
where $\mu_1 < \mu_2$. In your example, the 'black variable' corresponds to $X_1$.
Let $c$ denote the point of intersection where the pdf's meet in the green zone of your plot Then, the area of your green intersection zone is simply:
$$P(X_1>c) + P(X_2<c) = 1 - F_1(c) + F_2(c) = 1-\frac{1}{2} \text{erf}\left(\frac{c-\mu _1}{\sqrt{2} \sigma _1}\right)+\frac{1}{2} \text{erf}\left(\frac{c-\mu _2}{\sqrt{2} \sigma _2}\right)$$
where erf(.) is the error function.
Point $c$ is the solution to $f_1(x) = f_2(x)$ within the green zone, which yields:
$$c = \frac{\mu _2 \sigma _1^2-\sigma _2 \left(\mu _1 \sigma _2+\sigma _1 \sqrt{\left(\mu _1-\mu _2\right){}^2+2 \left(\sigma _1^2-\sigma _2^2\right) \log \left(\frac{\sigma _1}{\sigma _2}\right)}\right)}{\sigma _1^2-\sigma _2^2}$$
For your example, with ${\mu_1 = 5.28, \mu_2 = 8.45, \sigma_1 = 0.91, \sigma_2 = 1.36}$, this yields: $c = 6.70458...$,
and the area of the green section is: 0.158413 ...
• @antecessor Depends on what you mean by 'overlapping among both'. In the question you talk about some probability, but it is unclear what probabilistic interpretation the area computed here would have. Jun 18, 2014 at 8:57
• @JuhoKokkala Ok, I will write a new question to address the exact question. Jun 18, 2014 at 10:38
• the resolution of the equation pdf1=pdf2 yields to two solutions, so we have two intersection points, why you delete the other point @wolfies
– user61828
Dec 1, 2014 at 11:51
• Is the equation for c given at the bottom of the answer correct? If both distributions have equal variance, the equation requires dividing by zero! Mar 21, 2018 at 21:28
• @sammosummo The equation for $c$ is correct given that $\sigma_1 \neq \sigma_2$. In case of equal variances, $c=\frac{\mu_1+\mu_2}{2}$. Nov 11, 2018 at 20:11
@abdelbasset, To improve @wolfies 's answer above, there are two intersection points c, let's call them c1 and c2. Here, c1=-1.2848 and c2=6.7046. @wolfies ignored c1 (it likely will be too far in the tails of the probability density functions to matter (especially if we round to just a few decimal points)).
The more correct way is to find both c1 and c2, and to find the area of overlap of both functions:
AREA OF OVERLAP = P(X1 > j1) + P(X2 < j1) - [P(X2 < j2)-P(X1 < j2)]
therefore, using cdf's
AREA OF OVERLAP = 1 - F1(j1,μ2,σ2)+F2(j1,μ2,σ2)-F2(j2,μ2,σ2)+F1(j2,μ1,σ1)
• Welcome to the site, @luckapani. You cannot use the "Your Answer" field to comment or respond to comments. As a result, this would be deleted. However, there is the core of an answer here. Can you edit your post to make it more of an answer & not a comment? Since you're new here, you may want to take our tour, which contains information for new users. May 19, 2015 at 16:12
There are two scenarios when calculating the intersections of two normal distributions. It's important to acknowledge that, in most cases, there are two intersection points:
Equal variance
In the modelled (or rare) examples when two normal distribution curves have exactly the same variance, there is one intersection point. The intersection point on the abscissa axis, $$(x_1)$$, is calculated from:
$$x_1=\frac{\mu_a+\mu_b}{2}$$
This being, of course, the midpoint between the respective means.
Unequal Variance
When the two normal distribution curves intersect and have different variances, the solution to the intersection points simplifies to a quadratic form. The consequence of the quadratic form is that there are always two intersection points $$(x_1,x_2)$$ on the abscissa axis.
The intersection points of the two curves can be solved algebraically to obtain the quadratic form and the roots can then be found using the quadratic formula. It’s not as hard as it seems and does not involve using the error function. However, for time and space, I give Inman and Bradley’s version where the intersection points can be found directly.
$$(x_1,x_2)=\frac{\mu_a \sigma_b^2-\mu_b \sigma_a^2\pm\sigma_a \sigma_b \sqrt{(\mu_a-\mu_b)^2+(\sigma_b^2-\sigma_a^2)\ \ln \left(\frac{\sigma_b^2}{\sigma_a^2}\right)}}{\sigma_b^2-\sigma_a^2}$$
Where $$-\infty<{x_1}<{x_2}<\infty$$. (For ease of calculation, I set $$\sigma_b>\sigma_a$$ ).
For some intersecting normal distributions with unequal variances, the ‘second’ intersection point maybe insignificant and can, effectively, be ignored. However, for most cases – especially when $$\mu_a\approx\mu_b$$ – having both intersection points are critical to determine the ‘common area’ (overlap).
Finding the Area
The common area (overlap), $$\phi$$, is calculated from:
Equal variance $$\phi=\min\left[\int_{-\infty}^{x_1} f(a) dx, \int_{-\infty}^{x_1} f(b) dx\right]+\min\left[\int_{x_1}^{\infty} f(a) dx, \int_{x_1}^{\infty} f(b) dx\right]$$
Where, of course, $${f(a)}=\frac{1}{\sigma_a \sqrt{2 \pi}} e^{-0.5 \left(\frac{x-\mu_a}{σ_a}\right)^2}$$
Unequal variance
$$\phi=\min\left[\int_{-\infty}^{x_1} f(a) dx, \int_{-\infty}^{x_1} f(b) dx\right]+\min\left[\int_{x_1}^{x_2} f(a) dx, \int_{x_1}^{x_2} f(b) dx\right]+\min\left[\int_{x_2}^{\infty} f(a) dx, \int_{x_2}^{\infty} f(b) dx\right]$$
Inserting the data from the question, $$\mu_a=5.28, \sigma_a=0.91$$ and $$\mu_b=8.45, \sigma_b=1.36$$, we get $$x_1=-1.2842, x_2=6.7046$$ and $$\phi=0.1548$$
Reference:
Inman, H.F. & Bradley Jr, E.L. (1989). The overlapping coefficient as a measure of agreement between probability distributions and point estimation of the overlap of two normal densities, Communications in Statistics - Theory and Methods, 18:10, 3851-3874, DOI: 10.1080/03610928908830127
• It seems to me that you are answering a different question than that posed by the OP. The OP seeks the probability an individual is in the overlapping area (green). The curves overlap everywhere (one curve is always higher than the other), so the meaning of the question is conveyed by the diagram and the question only considers the main intersection point, closest to the mean. You consider a different problem, with a different meaning of overlapping. Moreover, if one does a diagram of your different problem, there are multiple ways of depicting the green zone, none of which are the OPs. Apr 17 at 13:45
• @wolfies. To answer your first sentence. No I am not answering a different question. The OP question is very clear. Apr 17 at 18:32
Normalise the graphs to an area of 1 by dividing each by their respective standard deviation. Then use simple subtraction from a z-graph to calculate the probability of an occurrance in that overlap area. No need for erfs.
• i assume that he means that he wants to compute the probability in the green region. That can be done simply by adding the area in the right tail of the first distribution to the ares in the left tail of the second distribution. There is no normalization necessary. Nov 2, 2019 at 23:36
• The separate normalization of the two graphs will change the overlap area unless they have identical SDs. Thus, this answer is generally wrong.
– whuber
Nov 3, 2019 at 13:52 | 2022-07-05T09:32:34 | {
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https://math.stackexchange.com/questions/2108721/why-does-the-limit-lim-x-to-0-frac1-cos3-xx-sin-2x-exist | # Why does the limit $\lim_{x \to 0} \frac{1-\cos^3 x}{x\sin 2x}$ exist?
The limit does exist and it is equal to $\frac{3}{4}$. We get it using L'Hospital. On the other hand I could write the above limit as: $$\lim_{x \to 0} \frac{1-\cos^3 x}{x\sin 2x} = \lim_{x \to 0} \frac{1}{x\sin 2x} - \lim_{x \to 0} \frac{\cos^3 x}{x\sin 2x}$$ There is a rule that says, that the limit of difference equals the difference of limits if both the limits in the difference of limits exist. So in this case the limit $\lim_{x \to 0} \frac{1}{x\sin 2x}$ does not exist as it is not approaching any particular value. So the question is, why does the limit of difference exist if the difference of those two limits doesn't?
• Check the limit btw, it should be zero. – Simply Beautiful Art Jan 22 '17 at 14:11
• @SimplyBeautifulArt: I make it $\frac34$. So does WolframAlpha. – TonyK Jan 22 '17 at 14:15
• @TonyK Oops, my bad. I went back to my calculations and I accidentally changed $\cos^3(x)$ to $\cos(x^3)$ in a step XD – Simply Beautiful Art Jan 22 '17 at 16:09
The separation of limits into the form
$$\lim_{x\to a}f(x)-g(x)=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$$
only holds when the limits exist. In this case, they don't, so we simply aren't allowed to do that, else we would always have
$$\lim_{x\to a}f(x)=\lim_{x\to a}f(x)-\frac1{x-a}+\frac1{x-a}=\underbrace{\lim_{x\to a}f(x)-\frac1{x-a}+\lim_{x\to a}\frac1{x-a}}_{\text{undefined}}$$
• I see... I was wrongly using this "method" to prove that certain limits do not exist, but in this case I was confused as the results were contradictory. So now the new question arises, what is the correct method to see that certain limit does not exist? For example $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x^2}$ – antestor Jan 22 '17 at 14:16
• @antestor Epsilon delta? – Simply Beautiful Art Jan 22 '17 at 16:10
• Yes, to check the existence of limit by definition it should yield the correct answer, but it's tedious and not one of the easiest methods. I thought there are other (faster) ways. – antestor Jan 22 '17 at 16:14
• @antestor Well, showing that the left and right side limits are unequal, setting inequalities, etc. These can show the limit doesn't exist. – Simply Beautiful Art Jan 22 '17 at 16:16
You have the illustration that two expressions may tend to infinity, yet they approach each other. What's so extraordinary with this? It happens very commonly with asymptotes.
That say, computing the limit is much more illuminating using Taylor's polynomials: you see in depth why the limit is what it is:
First, at order $2$, we know that $\;\cos x=1-\dfrac{x^2}2+o(x^2)$, so $$1-\cos^3 x=1-\Bigl[\Bigl(1-\dfrac{x^2}2+o(x^2)\Bigr)^3\Bigr]=1-\Bigl[1-\dfrac{3x^2}2+o(x^2)\Bigr]=\dfrac{3x^2}2+o(x^2)$$ On the other hand, $\;x\sin 2x=x\bigl(2x+o(x)\bigr)=2x^2+o(x^2)$, so $$\frac{1-\cos^3 x}{x\sin 2x}=\frac{\dfrac{3x^2}2+o(x^2)}{2x^2+o(x^2)}=\frac{\dfrac32+o(1)}{2+o(1)}\to\dfrac34.$$
If the limits of two functions exist then the limit of their difference (sum, product, quotient provided limit of divisor is non-zero) also exists. This is a standard result proven in most textbooks and you are aware of this result.
From this result it does not follow that if the limits of two functions do not exist then the limit of their difference does not exist. If $A\Rightarrow B$ then it does not mean that $\neg A\Rightarrow \neg B$.
It is quite possible like in your question that limits of two functions do not exist and yet the limit of their difference exists. In this particular example in your question you may find it difficult to conclude that the limit of difference exists. So consider the simpler case when $\lim_{x\to 0}(1/x\sin x)$ does not exist and yet $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x\sin x} =0$$ in an obvious manner. There may be similar cases where the limit of difference may not exist. Thus $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x}$$ does not exist.
If the hypotheses of a theorem do not hold then it is simply not possible to say anything in general about the conclusions of the theorem. And this is true for the limit theorems which I mentioned at the start of the answer.
The limit in question exists and its existence can be proved via its evaluation as follows \begin{align} L&=\lim_{x\to 0}\frac{1-\cos^{3}x}{x\sin 2x}\notag\\ &=\lim_{x\to 0}\frac{1-\cos^{3}x}{1-\cos x} \cdot\frac{1-\cos x} {x\cdot 2x}\cdot\frac{2x}{\sin 2x}\notag \\ &=\lim_{t\to 1}\frac{t^{3}-1}{t-1}\cdot\lim_{x\to 0}\frac{1-\cos x} {2x^{2}}\cdot 1\text{ (putting }t=\cos x) \notag\\ &=\frac{3}{2}\lim_{x\to 0}\frac{1-\cos x} {x^{2}}\notag\\ &=\frac{3}{2}\cdot\frac{1}{2}\notag\\ &=\frac{3}{4}\notag \end{align} | 2019-09-16T14:12:05 | {
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http://mathhelpforum.com/calculus/150535-antiderivative-s-function.html | # Thread: Antiderivative(s) of a function
1. ## Antiderivative(s) of a function
Can a function have more than one different antiderivative? I keep getting different answers on some exercises and it drives me crazy, i double-checked my solving method and it's correct, yet i'm getting a different result.
2. All integrable functions have an infinite number of antiderivatives. This is why you have to include the integration constant after performing any indefinite integration.
3. I'm not taking into account the constant...
OK...here it is:
$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$
my solution is $-\displaystyle\frac{1}{2(x^2-1)}$
The book only gives me these solutions where only 1 the correct one:
A) $\ln{\frac{x^2+2}{x^2+1}}$
B) $\ln{\frac{x^2+1}{x^2+2}}$
C) $\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
D) $\frac{1}{4\sqrt{2}}\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
E) $\ln{|\frac{x^2-1}{x^2+1}|}$
F) $\ln{|\frac{x^2+1}{x^2-1}|}$
4. Originally Posted by Prove It
All integrable functions have an infinite number of antiderivatives. This is why you have to include the integration constant after performing any indefinite integration.
And in fact the inclusion of the arbitrary constant is the reason why two superficially different answers can nevertheless be correct. eg. $\displaystyle {\int \frac{\sin (x)}{\cos^3 (x)} \, dx = \frac{1}{2} \tan^2 (x) + C}$, and this answer is equivalent to $\displaystyle {\frac{1}{2} \sec^2 (x) + K}$.
5. Originally Posted by Utherr
I'm not taking into account the constant...
OK...here it is:
$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$
my solution is $-\displaystyle\frac{1}{2(x^2-1)}$
The book only gives me these solutions where only 1 the correct one:
A) $\ln{\frac{x^2+2}{x^2+1}}$
B) $\ln{\frac{x^2+1}{x^2+2}}$
C) $\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
D) $\frac{1}{4\sqrt{2}}\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
E) $\ln{|\frac{x^2-1}{x^2+1}|}$
F) $\ln{|\frac{x^2+1}{x^2-1}|}$
Your answer is quite wrong. Unfortunately, you have shown no working at all, so the errors you are making cannot be pointed out.
Read this: integrate x/(x^4 - 2x^2 - 1) - Wolfram|Alpha (be sure to click on Show steps).
6. Ok, i'll post my method in a moment...
7. Antiderivatives are nessecarily the same up-to the addition of a constant. For, if we suppose that two functions $f, g$ both have the same derivative, then it follows that
$\displaystyle
(f - g)' = 0
$
and by the Mean Value Theorem, we know this means $f - g$ is constant.
8. Hello, Utherr!
Your algebra is off . . .
$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$
$(A)\;\ln\left[\dfrac{x^2+2}{x^2+1}\right] + C \qquad\qquad\quad (B)\;\ln\left[\dfrac{x^2+1}{x^2+2}\right] + C$
$(C)\; \ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C \qquad
(D)\;\dfrac{1}{4\sqrt{2}}\ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C$
$(E)\;\ln\left|\dfrac{x^2-1}{x^2+1}\right| + C \qquad\qquad\quad\; (F)\;\ln\left|\dfrac{x^2+1}{x^2-1}\right|+C$
The denominator is: . $x^4 - 2x^2 - 1 \;=\;x^4 - 2x^2 + 1 - 2 \;=\;(x^2-1)^2 - 2$
The integral is: . $\displaystyle{\int\frac{x\,dx}{(x^2-1)^2 - 2} }$
Let: . $u \,=\,x^2-1\quad\Rightarrow\quad du \,=\,2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\frac{1}{2}du$
. . Substitute: . $\displaystyle{\int\frac{\frac{1}{2}du}{u^2-2} \;=\;\tfrac{1}{2}\int\frac{du}{u^2- (\sqrt{2})^2} }$
Formula: . $\displaystyle{\int \frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C}$
. . We have: . $\displaystyle{\frac{1}{2}\cdot\frac{1}{2\sqrt{2}}\ ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C \;=\;\frac{1}{4\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C }$
Back-substitute: . $\displaystyle{\frac{1}{4\sqrt{2}}\ln\left|\frac{(x ^2-1) - \sqrt{2}}{(x^2-1) + \sqrt{2}}\right| + C }$
. . . . . . . . . . . $\displaystyle{=\;\frac{1}{4\sqrt{2}}\ln\left|\frac {x-(\sqrt{2}+1)}{x + (\sqrt{2}-1)}\right| + C } \quad\hdots\;\text{answer (D)}$
9. Hello again Utherr!
A function can have different-looking antiderivatives
. . but they are all equivalent.
Here is a classic example: . $\displaystyle{\int \sin x\cos x\,dx}$
$\displaystyle{[1]\;\int\sin x(\cos x\,dx) }$
Let $u = \sin x \quad\Rightarrow\quad du = \cos x\,dx$
Substitute: . $\displaystyle{\int u\,du \:=\:\tfrac{1}{2}u^2 + C}$
Back-substitute: . $\boxed{\tfrac{1}{2}\sin^2\!x + C}$
$\displaystyle{[2]\;\int \cos x(\sin x\,dx) }$
Let $u = \cos x \quad\Rightarrow\quad du = -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx = -du$
Substitute: . $\displaystyle{\int u (-du) \;=\;-\int u\,du \;=\;-\tfrac{1}{2}u^2+C }$
Back-substitute: . $\boxed{-\tfrac{1}{2}\cos^2\!x + C}$
$\displaystyle{[3]\;\int\sin x\cos x\,dx \;=\;\tfrac{1}{2}\int 2\sin x\cos x\,dx$
. . . $\displaystyle{=\;\tfrac{1}{2}\int\sin2x\,dx \;=\;\boxed{-\tfrac{1}{4}\cos^2\!2x + C} }$
10. Originally Posted by Utherr
Ok, i'll post my method in a moment...
You stated originally that you wanted to integrate $\int \frac{xdx}{x^4- 2x^2- 1}$ but here you are integrating $\int \frac{xdx}{x^4- 2x^2+ 1}$.
To integrate $\int\frac{xdx}{x^4- 2x^2- 1}$, write it as $\int \frac{xdx}{x^2- 2x^2+ 1- 2}= \int\frac{xdx}{(x^2-1)^2- 2}$. Now let $y= x^2- 1$ so that dy= 2xdx and the integral becomes $\frac{1}{2}\int \frac{dy}{y^2- 2}= \frac{1}{2}\int \frac{dy}{(y- \sqrt{2})(y+ \sqrt{2})}$ which can be integrated by "partial fractions".
11. Thanks for the answers, i thought it was a + there not a - sign O.o | 2018-01-24T01:42:35 | {
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http://mathhelpforum.com/differential-geometry/196211-f-functions-domain-range.html | # Math Help - f functions and domain/range
1. ## f functions and domain/range
hi
let f(x) = x/(x-1). determine f(f(x)) simplify answer, and find domain/range of f(f(x))
is the question i got.
also not sure if posted in the right forum?
2. ## Re: f functions and domain/range
Originally Posted by tankertert
hi
let f(x) = x/(x-1). determine f(f(x)) simplify answer, and find domain/range of f(f(x))
is the question i got.
also not sure if posted in the right forum?
$f(f(x))=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}=\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x$
$\text{domain } : x \in (-\infty,+\infty)$
$\text{range } : f(f(x)) \in (-\infty,+\infty)$
3. ## Re: f functions and domain/range
This might not be exactly true.
The domain of $f(x) = \frac{x}{x-1}$ is $\mathbb{R} \setminus \{1\}$
and its range is really $\mathbb{R}$ (infinity shall be included in the real numbers).
But then again, f(f(x)) has the above mentioned domain (since the image of $f: \mathbb{R} \rightarrow \mathbb{R}$ is $\mathbb{R}$),
and then you map the real numbers again, in the same way, to itself. And again, the value is not defined at x = 1 (you can really not define it there, because
the left and right limits are different).
So I'd say it's a little tricky here and would nevertheless say that the domain is $\mathbb{R} \setminus \{1\}$.
Of course you have as effective function the identity mapping, however since you cannot divide by zero, the simplification holds ONLY IF x is NOT 1.
That's the point.
I may be corrected if I'm wrong, however I think that this is the correct answer.
4. ## Re: f functions and domain/range
Yes, mastermind is correct. f(f(x))= (x/(x-1))/(1/(x-1)) for all x for which it is defined- all x except 1. That reduces to f(x)= x for those values of x- x not equal to 1.
A simpler example of that is (x- 1)/(x^2- 1)= (x- 1)/((x- 1)(x+ 1))= 1/(x+ 1) for all x except 1. The left side is not defined for x= 1 but the right side is so they are not equal for x= 1.
5. ## Re: f functions and domain/range
graph of f[f(x)] ... note the discontinuity at x = 1 | 2015-01-30T19:30:10 | {
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https://mathsgee.com/tag/bennett | # Recent questions tagged bennett
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Prove Markov’s inequality.
Prove Markov’s inequality.Prove (Markov's inequality). Suppose $X$ is a nonnegative random variable and $a \in$ is a positive constant. Then \ P(X \geq a) \leq \frac{E X} ...
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Prove the Prohorov inequality | 2022-01-26T20:10:24 | {
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http://math.stackexchange.com/questions/410822/is-my-row-calculation-of-row-echelon-form-correct | # Is my row calculation of row echelon form correct?
I was directed by a community member to a resource on how to calculate the row echelon form of a matrix here. The resource says:
First we wish to put A into reduced row echelon form. There are several ways to do those (and
thus several matrices P), but here is one possible way: (calculation next)
So, I understand that a given matrix can have multiple row echelon forms.
To continue my self-study of linear algebra further, I looked at the example on wikihow.
It gives a simple 3x3 matrix and shows how to calculate the row echelon form. Fair enough.
As in wikihow, the given matrix is:
$$\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 4 & 5 \end{bmatrix}$$
and the row echelon form is this:
$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \\ \end{bmatrix}$$
However, my answer is different and I am not sure if it is correct
### My calculation on the same matrix:
• Attempting to get all zeros under $A_1_1$ as:
• $R_2$ - $R_1$ -> $R_2$
$(3\times R_1$) - $R_3$ -> $R_3$
So, the matrix is:
$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \\ \end{bmatrix}$$
• Attempting to get all zeroes under $A_2_2$ as:
• $R_2$ + $R_3$ -> $R_3$
So the matrix is:
$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix}$$
The only difference is that I have a $2$ in bottom right and wikihow has $-2$.
Is it correct ?
-
yes, there are multiple echelon forms. For example you can continue multiplying third row by (-1) and get the other answer. Or you can add any multiple of a lower row to an upper row and still keep the echelon form. (What is unique is the reduced row echelon form where you insist pivots must be 1 and all other elements a pivot column should be zero.) – Maesumi Jun 4 '13 at 8:12
@Maesumi is there a rule that $3XR_3 - R_1$ -> $R_3$ because $R_3$ was the first operand ???? – Little Child Jun 4 '13 at 8:13
You can use the addition of rows $m R_n \pm R_k$ however you like (so long as $n\ne k$). Here to get zero you can use $3R_1-R_3$ or $R_3-3R_1$. The typical standard notation in text books is to use $mR_n + R_k$. – Maesumi Jun 4 '13 at 8:20
@Maesumi so my answer is correct :) Plus, there was a mistake in WIkiHow which I had to correct – Little Child Jun 4 '13 at 8:28
You're getting different answers because you're subtracting $R_3$ from $3 R_1$. Usually, you subtract $3 R_1$ from $R_3$ (you can add or subtract rows, but you're changing the sign on $R_3$ here: we often choose to do it this way because adding or subtracting a multiple of a row keeps the determinant the same).
So, to get Wikihow's answer, you should have taken $R_3 - 3R_1 \to R_3$ instead. This also explains the sign flip, since $R_3 - 3R_1 = - ( 3R_1 - R_3 )$. | 2015-10-05T04:05:10 | {
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https://math.stackexchange.com/questions/1736483/there-are-apparently-3072-ways-to-draw-this-flower-but-why | # There are apparently $3072$ ways to draw this flower. But why?
This picture was in my friend's math book:
Below the picture it says:
There are $3072$ ways to draw this flower, starting from the center of the petals, without lifting the pen.
I know it's based on combinatorics, but I don't know how to show that there are actually $3072$ ways to do this. I'd be glad if someone showed how to show that there are exactly $3072$ ways to draw this flower, starting from the center of the petals, without lifting the pen (assuming that $3072$ is the correct amount).
• The title appears to be slightly misleading; The starting point is fixed in the plain text. – Aki Suihkonen Apr 11 '16 at 11:34
• Meh, without the image the title simply describes the problem. Can't really claim this is "misleading" because there's no emotional expectation to the arbitrary seeming number 3072. Although, I am perplexed the fixed the starting point. Why don't just make the statement there are 6144 ways to draw the flower. – fleablood Apr 11 '16 at 17:52
• @fleablood: well, it's reasonable to strip away the obvious symmetry that exists in reversing the order of each path. It's not necessary, but I don't find it perplexing either :-) – Steve Jessop Apr 11 '16 at 22:15
• But the symmetry is the entire point of these puzzles. There's only one way to draw it otherwise start at the center, draw each petal, go to th middle, draw each leaf, and finish the stem. – fleablood Apr 11 '16 at 22:29
First you have to draw the petals. There are $4!=24$ ways to choose the order of the petals and $2^4=16$ ways to choose the direction you go around each petal. Then you go down the stem to the leaves. There are $2! \cdot 2^2=8$ ways to draw the leaves. Finally you draw the lower stem. $24 \cdot 16 \cdot 8=3072$
• There's a built in assumption (so far as I can see, unstated) that you may cross a line at a point, but you may not draw back over an existing line (e.g. draw the stem down first and then go back over it to get back to the leaves and petals). – Jason Apr 10 '16 at 20:17
• @Jason: yes, that is how I took it. – Ross Millikan Apr 10 '16 at 20:19
• @Jason Without this assumption there are infinite ways. If you can draw back and forth, you can repeat it as many times as you want. – Mołot Apr 10 '16 at 20:26
• I'd hardly call that "major"! Now I can also draw the image in blue in or black ink or red ink or a pencil and I could draw it upside down or to the left orientation. I can draw it after drinking a glass of milk or after eating dim sum.... – fleablood Apr 11 '16 at 17:56
• @fleablood You could draw it in a boat, you could draw it with a goat? You could draw it here or there, you could draw it anywhere? – neminem Apr 11 '16 at 23:09
At the beginning you could go 8 different ways, then you could go 6 different ways, then you could go 4 and 2 different ways but in the down of the picture you could go at first 4 different ways and 2 at the end. $8\cdot6\cdot4\cdot2\cdot4\cdot2 = 3072$
• This answer is more intuitive than the @ross's answer, if you fleshed out why it's 8 ways, then 6, etc... this would be a great answer. – user138559 Apr 10 '16 at 22:16
• I think part of the elegance of this answer is that it doesn't require fleshing out. The answer self-fleshes. There is a hidden elegance even in the meaning of the "That's obvious" comment at the beginning. It sound's dismissive. But it's actually descriptive. – Mowzer Apr 11 '16 at 18:43
• Something rubs me the wrong way about "That's obvious". I don't think that language has a place on a Q&A site. If it was obvious, they wouldn't have asked. – JPhi1618 Apr 12 '16 at 14:19
• Telling the question asker "That's obvious" violates the Be nice principle that the Stack Exchange network and community tries to adhere to. You can write an answer without belittling the question asker. If it was obvious, then the question wouldn't be asked. Furthermore, you should flesh out what you mean by your various numbers and how you arrived at them. You just stated "8, then 6, then 4..." which is hard to read and understand. And Mowzer, you should not be encouraging this kind of language or vague answers. – The Anathema Apr 12 '16 at 17:58
• @TheAnathema I agree it took me a minute and to look back at the picture to see why $8,6\ldots$ make sense. I can imagine it'd take longer if you don't 'see it' and the answer doesn't help you arrive at it, except by giving the numbers. I do however agree it is intuitive once you see it. – snulty Apr 13 '16 at 12:15
Looking at the picture, there are 4 phases.
1. Draw the petals
2. Draw the upper stem
3. Draw the leaves
4. Draw the lower stem
Lets label these $A,B,C,D$. Clearly, the total number of ways to draw the flower is simply;
$$Total = A \times B \times C \times D$$
We can see that the upper and lower stem are un-ambiguous; i.e. there is only one way to draw them. Thus $B=D=1$. So our equation becomes
$$Total = A \times C$$
Now lets look at the leaves first. The factors in it are:
a. Which leaf do you draw first?
b. What direction do you use for the first leaf?
c. What direction do you use for the second leaf?
There are 2 possibilities for each, so $C=2\times 2 \times 2=8$.
Now lets look at the petals.
a. Which petal do you draw first? (4 choices)
b. Which petal do you draw second? (3 choices)
c. Which petal do you draw third? (2 choices)
d. Which direction do you draw the X petal? (2 choices for each petal)
So $A=4 \times 3 \times 2 \times 1 \times 2^4=24 \times 16=384$.
And
$$Total=A \times C = 384 \times 8 = 3072$$
• since the question requires you to draw the petals first, it would be more intuitive to enumerate the options for drawing the petals before enumerating the options for drawing the leaves. – Level River St Apr 12 '16 at 18:29
Think through each of the "decision points," and think through how many options you have at each decision point.
Starting in the middle of the flower, there are 4 petals you could choose to draw first. For each petal, there are 2 ways to draw it: forwards or backwards. So you have 8 options. Draw your first petal, and now there are 6 options of what to do next (3 petals, 2 ways to draw each). And so on. So you have 8*6*... Do you see what to do next?
Others have already solved the problem. I just wanted to add that, properly speaking, this problem and/or its solution belong to a branch of math called graph theory. Formally:
Proposition. The following undirected graph has $3072$ directed Eulerian trails starting at $x$ and ending at $z$.
We could be even more precise by describing the graph in more formal terms, of course, rather than just drawing a picture. This can be done by giving a symmetric function $\{x,y,z\}^2 \rightarrow \mathbb{N}$ that tells us how many edges go between any two vertexes. In this particular case, our function is
Enumerate the possibilities:
1. Draw one of 4 top petals, in one of two directions: 4*2
2. Draw one of 3 remaining top petals, in one of two directions: 3*2
3. Draw one of 2 remaining top petals, in one of two directions: 2*2
4. Draw last petal in one of two directions: 2
5. Draw one of 2 leaves, in one of two directions: 2*2
6. Draw remaining leaf, in one of two directions: 2
Thus, we have the total number of possibilities: (4*2)(3*2)(2*2)(2)(2*2)(2)=3072
The number of possible ways to draw 4 petals in order is $4!$. Then let's imagine that if the drawing of the petal is clockwise then the petal is black and if the drawing is counterclockwise the petal is white. So the drawing can be: $0000, 0001, ... 1111$ therefore $2^4$ different drawings $0$ for black $1$ for white. Therefore $4! \cdot 2^4$ ways to draw the upper petals. Similary $2! \cdot 2^2$ ways to draw the lower petals and yes, $4! \cdot 2^4 \cdot 2! \cdot 2^2 = 3072$ It is pretty much @Ross Millikan's solution I just added the coloring ilustration.
• o.0 My eyes... also, doing nothing but citing other users' answers is not a valid answer. See How to Answer – cat Apr 13 '16 at 1:08
• I added an explanation. Is that nothing? – C Marius Apr 13 '16 at 10:19
EDIT: I have only just noticed that @openspace has already supplied an answer using this idea, so all credit must go to him for being the first one to think of this method. I shall leave my answer though as it seems to flesh out the idea.
An alternative approach:
As others have noted you obviously have to draw all the petals first, then go down the stem and draw all the leaves before drawing the end of the stem. Starting in the centre of the petal there's a possible of $8$ lines in which to start drawing the petals. After you pick one then there's $6$ possible ways to pick the line for drawing the second petal, then $4$, then finally $2$ to pick how to draw the last petal.
You then move on down the stem (only $1$ way) to the centre of the leaves. There's $4$ lines to pick from to start drawing the leaves, and then after drawing the first leaf there's only $2$ lines to pick from to draw the last leaf. Then there's drawing the end of the stem (only $1$ way). So in total there are $8\cdot 6\cdot 4\cdot 2 \cdot 1\cdot 4\cdot 2\cdot 1=3072$ ways to draw the flower as desired.
The presumption is wrong!
Most answers here start like this: "First you have to draw the petals."
This is a mistake. You could start drawing the flower from two points: The centerpoint of the blossom OR the endpoint at the lower stem.
Result is: There are 2 x 3072 = 6144 ways to draw the Flower!
I think it might have to do with the way you asked the question. ;)
• It says: "starting from the center of the petals". – Mathematician 42 Apr 13 '16 at 13:35
• True, but one could argue that there's nothing in the question actually forcing you to have to complete all the petals first before proceeding to the rest of the flower. Obviously we all assume that you can't retrace your steps beyond crossing a point, but the question doesn't actually specify that, and it really should have. Otherwise I could draw two petals go down and draw one leaf then go back up and draw two more petals and back down to draw the second leaf prior to finishing the stem. Or draw one petal, go down draw one leaf, go up and draw another petal, down to finish the stem, up..... – Monomeeth Apr 14 '16 at 3:45 | 2019-07-18T05:27:42 | {
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https://www.physicsforums.com/threads/limit-question.295635/ | # Limit question
1. Feb 27, 2009
### transgalactic
$$\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}$$
i get infinity - infinity
its undefined
??
2. Feb 27, 2009
### tiny-tim
Hi transgalactic!
It's easier to read if you write …
$$\lim _{x->+\infty} (x+2)^{2/3} -(x-2)^{2/3}$$
eek! you're not allowed to do that!
Hint: write it $$\lim _{x->+\infty} x^{2/3}\left((1 + 2/x)^{2/3} -(1 - 2/x)^{2/3}\right)$$
3. Feb 27, 2009
### Staff: Mentor
infinity - infinity is one of several indeterminate forms. None of them are ever an answer.
4. Feb 27, 2009
### djeitnstine
A hint when doing limits, just think about what happens when the number gets REALLY big. Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to $$x^{\frac{2}{3}}-x^{\frac{2}{3}}$$ in the end...
Also the result is the same as tinytim's =]
5. Feb 27, 2009
### transgalactic
so i get 0*infinity
its not defined either
6. Feb 27, 2009
### tiny-tim
Nooo! … expand the bracket first, then multiply by the x2/3, then take the limit.
7. Feb 27, 2009
### cleopatra
(x-2)2/3 is less than (x+2)2/3 so each time the x gets larger a larger sum is in (x-2) than(x+2) so it goes down... to minus infinity.
8. Feb 27, 2009
### djeitnstine
Noo that is also not the result! Read my post and tinytim's again
When in doubt graph it!
9. Feb 27, 2009
### HallsofIvy
When tiny-tim says "expand the brackets first" I believe he is referring to the generalized binomial expansion. And you should only need the first few terms.
10. Feb 27, 2009
### Staff: Mentor
Another approach is to multiply the original expression by 1 like so:
$$(x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}$$
This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.
After using this technique, the numerator can be simplified to a single term, and the limit of the whole expression can be found without much trouble.
11. Feb 27, 2009
### djeitnstine
Why need to get so complicated :S this is quite a simple problem. Quite frankly the answer is 0. There is an asymptote in the graph tending towards 0 as x increases towards positive infinity. Now which result out of mine or tinytim's did not explain this?
I can re iterate of course "Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to $$x^{\frac{2}{3}}-x^{\frac{2}{3}}$$ in the end..." which of course is always tending towards 0...
12. Feb 28, 2009
### Staff: Mentor
Agreed that this is a simple problem, and that the limit is 0. Tiny Tim's approach is superior to mine in its simplicity. Your approach, using a computer-generated graph, is useful for gaining understanding about the behavior of this function, but that's about it. It is well-known that computers don't produce exact results, because of the limited precision they use in their calculations, and especially so in raising to fractional powers. And in some cases, they can produce results that are incorrect (e.g., the earlier Intel Pentium math coprocessors back in 1992 or so). You also said that for large x, x + 2 was about the same as x - 2. This is correct, and is helpful in understanding, but it should be said in mathematical terms. Otherwise, it's akin to arm-waving, IMO.
My approach is different from, and slightly more complicated, than Tiny Tim's. But so what? Multiplying by a factor to produce a difference of cubes isn't all that complicated, and the intent was to show another method of approach that might be useful in other problems.
The more tools you have to work with, the more problems you're equipped to solve.
13. Feb 28, 2009
### djeitnstine
Well I do understand that your method was valid, however I believe when helping someone whom is slightly confused the simplest approached should be used. Don't you agree?
14. Feb 28, 2009
### Staff: Mentor
I agree that the simplest approach should be shown first. Then you can show other approaches, just as was done in this thread. | 2018-02-20T19:45:21 | {
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https://math.stackexchange.com/questions/1958378/primitive-pythagorean-triple-generator | Primitive Pythagorean triple generator
I was wondering how to prove the following fact about primitive Pythagorean triples:
Let $(z,u,w)$ be a primitive Pythagorean triple. Then there exist relatively prime positive integers $a,b$ of different parity such that $$z = a^2-b^2, \quad u = 2ab, \quad \text{and} \quad w = a^2+b^2.$$
I see how $z,u,w$ must form the sides of a triangle, where $c$ is the hypotenuse, but how do we show that $\gcd(z,u,w) = 1$ and that these generate all the triples?
It is easy to see that if $(z,u,w)$ is a primitive pythagorean triple, then $w$ is odd and either $z$ is odd and $u$ is even or $z$ is even and $u$ is odd. WLOG suppose that $u=2x$ is even, so $z$ is odd. Then
$$u^2=4x^2=w^2-z^2=(w+z)(w-z).$$
It is plain that $\gcd(w+z,w-z)=2$, so $y_1:=(w+z)/2$ and $y_2:=(w-z)/2$ are relatively prime positive integers. Thus
$$x^2=y_1y_2,$$
and therefore $y_1$ and $y_2$ are relatively prime divisors of $x^2$. This implies that $y_1$ and $y_2$ are perfect squares, so write $y_1=a^2$ and $y_2=b^2$. Hence,
$$u=2x=2ab,\quad z=y_1-y_2=a^2-b^2,\qquad\text{and}\qquad w=y_1+y_2=a^2+b^2.$$
This proves that every primitive triple has the form $(a^2-b^2,2ab,a^2+b^2)$, as wanted.
Of course, the condition $\gcd(a,b)=1$ is needed, otherwise $(a^2-b^2,2ab,a^2+b^2)$ won't be primitive.
• Why must $w$ be odd? – user19405892 Oct 7 '16 at 20:25
• Suppose the $w$ is even. Then $z$ and $u$ have the same parity. If both are even, then the triple is not primitive. If both are odd, then $w^2$ is multiple of $4$, but $z^2+u^2$ is not multiple of $4$ (it is $\equiv2\bmod 4$ since $x^2\equiv1\pmod4$ for any odd $x$), which is a contradiction. Therefore, $w$ must be odd. – user246336 Oct 7 '16 at 20:28
• Even with A odd, B even, C odd, in a triple, it may not be primitive as in $(27,36,45), (75,100,125), (147,196,245), (243,324,405)$ with $GCDs$ of $3^2, 5^2, 7^2, 9^2$, respectively. – poetasis Jun 2 '19 at 12:48
Even in the wikipedia article cited by @JeanMarie I could not find the following quick argument which gives all the Pyhagorean triples at one stroke: the Pythagorean equation being homogeneous, it is equivalent to the equation in rational variables $Z^2 + U^2 = 1$, or equivalently $N(Z + iU) = 1$, where $N$ denotes the norm map from $\mathbf Q(i)$ to $\mathbf Q$ , $i^2 = - 1$. The extension $\mathbf Q(i)/\mathbf Q$ being cyclic, with Galois group generated by the complex conjugation, Hilbert's theorem 90 applies, which shows that $N(Z + iU) = 1$ iff $Z + iU$ is of the form $A + iB /A - iB$. It follows from identification that $Z = A^2 - B^2 /A^2 + B^2$ and $U = 2AB / A^2 + B^2$. Clearing denominators immediately produces the usual Pythagorean integral triples $z = a^2 - b^2 , u = 2ab, w= a^2 + b^2$. Clearing common factors plainly gives the primitive triples.
I thought I had a proof that, if $$m,n$$ are mutually prime, $$m,n$$ would generate only primitive Pythagorean triples. The $$proof$$ is between the asterisk below. $$\text{*************************}$$
$$\text{We are given }\quad A=m^2n^2\quad B=2mn\quad C=m^2+n^2$$
Let $$x$$ be the GCD of $$m,n$$ and let $$p$$ and $$q$$ be the cofactors of $$m$$ and $$n$$ respectively. Then we have
$$A=(xp)^2-(xq)^2\quad B=2xpxq\quad C=(xp)^2+(xq)^2$$ $$A=x^2(p^2-q^2)\quad B=2x^2(pq)\quad C=x^2(p^2+q^2)$$
If $$GCD(m,n)=1$$, then $$GCD(A,B,C)=1$$ and $$(A,B,C)$$ is a primitive triple. This means that $$m$$ and $$n$$ must be co-prime to generate a primitive.
$$\text{*************************}$$
However, a counter-example destroys to so-called proof: $$\quad\text{Let }m,n=7,3$$. $$A=49-9=40\quad B=2*7*3=42\quad C=49+9=58\quad GCD(40,42,58)=2$$ $$\therefore GCD(m,n)=1\neg\implies GCD(A,B,C)=1$$ The only two formulas I know about that will generate only primitive triplets do not generate all of them but $$C-B=1$$ in the first one and $$C-A=2$$ in the second one.
$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1$$ $$A=4n^2-1\quad B=4n\quad C=4n^2+1$$
• Where can I find more information about Ellingson'sequation? Wikipedia does not give any. – RTn Jun 2 '19 at 7:34
• Can we have a chat on this channel (I don't want to compromise the integrity of an unpublished paper)? Because i know of a similar equation first given in 628 CE (about 1500 years ago). So I think it is not very new (I may be wrong, so in that case I apologize) – RTn Jun 2 '19 at 8:52
• – RTn Jun 2 '19 at 9:01
• So you have thought up some equations, have not published them and yet decided to just name them ellingson equations, probably using your own name? That is not good scientific practice. – Hirshy Jun 2 '19 at 10:34
• I don't understand why there's an argument. You give various formulas which are the well-known formula of Pythagoras. $$x^2+y^2=z^2$$ $$x=2ps$$ $$y=p^2-s^2$$ $$z=p^2+s^2$$ Your whole game... this is a substitution of numbers of another form. Like this. $$p=2n-1+k$$ $$s=k$$ It's not worth the time. Another formula describing all the solutions - can not be obtained... – individ Jun 2 '19 at 12:59 | 2020-08-08T15:40:51 | {
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https://math.stackexchange.com/questions/480885/probability-that-centre-of-the-square-lies-inside-the-circle-joining-the-two-poi | # Probability that centre of the square lies inside the circle joining the two points inside the square
Two points are uniformly and independently distributed (located) inside a square. A circle is drawn such that the segment joining the two points is a diameter. Find the probability that the center of the square lies inside that circle.
• How do you draw a circle from two points? Do you make the line segment joining them the diameter of the circle with centre at the midpoint? Aug 31 '13 at 21:37
• If the radius of the circle is another independent random parameter then we need to know how it is distributed. Aug 31 '13 at 21:48
• Sorry for the above comment!!! line segment joining those two points forms the diameter of the circle. Aug 31 '13 at 21:49
Call the first point A, the second point B, and the center O. Join the line AO, and extend a line DOE through O, perpendicular to AO and continuing in both directions. If B is on the other side of DOE from A, then the circle joining A and B will contain the center. If B is on the same side of the line as A, then it will not. The line bisects the square's area, so the probability is 1/2.
• can you please elaborate on the geometry part that how will the circle contain the center of the square if A and B are on opposite sides of the line DOE. Aug 31 '13 at 22:07
• @RahulSharma Choose the coordinate system such that $O$ is the origin, observe $$\left|\;\vec{0} - \frac{\vec{A}+\vec{B}}{2}\;\right| \le \frac{|\vec{A}-\vec{B}|}{2} \iff \vec{A}\cdot \vec{B} \le 0$$ Aug 31 '13 at 22:21
• To put it other way: 1) the circle does not cover $O$ iff the angle $\widehat{AOB}$ is acute. 2) The probability that $\widehat{AOB}$ is acute is 1/2 Aug 31 '13 at 22:37
• @leonbloy I like your geometric argument. Aug 31 '13 at 22:39
• Fix a circle with diameter AB, and try placing O in different locations: outside the circle, on the circumference of the circle, inside the circle. What can you say about angle AOB in each of the three cases? Aug 31 '13 at 22:41
Although the answer has already been given, it's possible to check this via simulation. In R:
n = 10^8
x1 = runif(n, 0, 1)
x2 = runif(n, 0, 1)
y1 = runif(n, 0, 1)
y2 = runif(n, 0, 1)
x.center = (x1+x2)/2
y.center = (y1+y2)/2
distance.to.center = sqrt((x.center-1/2)^2+(y.center-1/2)^2)
This simulates drawing $10^8$ pairs of points
(x1, y1), (x2, y2)
, and finds the center and radius of the corresponding circles, and checks to see whether those circles contain the center of the square (1/2, 1/2). When I ran this I got 49995062, meaning that the probability can be estimated as 0.49995062. I'd say that's close to 1/2.
Another way to approach this problem would be to consider the angle made by the two points A and B at O (centre of the square).
In the case that O lies on the boundary of the circle with AB as diameter, the ∠AOB = 90 degrees.
If O lies inside the circle then the ∠AOB is greater than 90 degrees and if point O lies outside the circle then the ∠AOB is less than 90 degrees.
Since A and B are iid we can see the probability of ∠AOB being greater than 90 is 1/2 | 2021-12-07T03:36:54 | {
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http://mathhelpforum.com/calculus/139680-integration-nat-log-x-1-a.html | # Thread: Integration of Nat. Log of (x+1)
1. ## Integration of Nat. Log of (x+1)
Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).
Many thanks
Jay
2. Is...
$\int \ln t \cdot dt = t\cdot \ln t - t + c$ (1)
Now set in (1) $1+x$ instead of $t$ ...
Kind regards
$\chi$ $\sigma$
3. Originally Posted by jaijay32
Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).
Many thanks
Jay
Split the fraction $\frac x{x+1} = 1-\frac1{x+1}$. The sum at the RHS can be integrated quite easily.
4. $\int \frac{x}{x+1} \, dx = \int \frac{x+1-1}{x+1} \, dx = \int \left( 1 - \frac{1}{x+1} \right) \, dx = \int \, dx - \int \frac{1}{x+1} \, dx$
5. ## Integration by parts is correct.
Originally Posted by jaijay32
Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).
Many thanks
Jay
I'll start from zero just so that is clearer:
$
\int \ln{(x+1)} dx = (\int dx) \ln{(x+1)} - \int \frac{x}{x + 1} dx
$
Now I suppose this is what you have done so far.
Try using the following equivalence ( which is obtained by dividing x by x+1):
$
\frac{x}{x + 1} = 1 - \frac{1}{x + 1}
$
6. Originally Posted by jaijay32
Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).
Many thanks
Jay
This is a familiar integral
$\int lnx dx= xlnx - x + c$
Does the addition of 1 change this?
$U=ln(x+1)$
$dU = \frac{1}{x+1}$
$dV=dx$
$V=x$
$\int ln(x+1) dx= xln(x+1) - \int \frac{x}{x+1}dx$
Which is where you got. Let us look at the integral then.
$\int \frac{x}{x+1}dx = \int \frac{x+1-1}{x+1}dx = \int 1 - \frac{1}{x+1} dx$
Of course this this the same as
$\int 1 - \frac{1}{x+1}dx = x-lnx + c$
We then get
$\int ln(x+1) dx= xlnx - x + lnx + c$
Edit- LOL at how many people jumped on answering this guy's question. I should quit here there's too many of us!
7. Originally Posted by tom@ballooncalculus
That picture ****s me up hard. Can you describe that please lol
8. ... is integration by parts, where
... is the product rule.
Straight continuous lines differentiate downwards (integrate up) with respect to x. Choosing legs crossed or un-crossed does for assigning u du and v dv.
Hope it helps, or amuses - but use whatever works for you.
Cheers.
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote! | 2017-01-21T13:34:59 | {
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http://math.stackexchange.com/questions/411667/solving-elementary-row-operations | # Solving elementary row operations
So I am faced with the following: \begin{align} x_1 + 4x_2 - 2x_3 +8x_4 &=12\\ x_2 - 7x_3 +2x_4 &=-4\\ 5x_3 -x_4 &=7\\ x_3 +3x_4 &=-5 \end{align}
How should I approach this problem? In other words, what is the next elementary row operation that I should attempt in order to solve it? I know how to do with 3 equations by using the augmented method but this got me a little confused.
-
Are you familiar with the method of Gaussian elimination? – Alex Wertheim Jun 5 '13 at 4:42
Write it as an augmented system:
$$\left[\begin{array}{cccc|c} 1& 4& -2& 8 & 12\\ 0& 1& -7& 2 & -4\\ 0& 0& 5& -1 & 7\\ 0& 0& 1& 3 & -5 \end{array}\right]$$
Gaussian Elimination (Row-Reduced-Echelon-Form - RREF) will yield:
$$\left[\begin{array}{cccc|c} 1& 0& 0& 0 & 2\\ 0& 1& 0& 0 & 7\\ 0& 0& 1 & 0 & 1\\ 0& 0& 0& 1 & -2 \end{array}\right]$$
Thus:
• $x_4 = -2$
• $x_3 = 1$
• $x_2 = 7$
• $x_1 = 2$
-
Putting matrices into the Latex area has been difficult for me, but it has been easy for you like drinking a glass of water. :D – Babak S. Jun 5 '13 at 7:14
@BabakS.: I created a little text file cheat sheet that I use for common LaTex items that I reuse because I am still too slow at typing things up and some of you are very fast! Feels like a time trial at times. Hope all is well with you my friend! Have a great day! – Amzoti Jun 5 '13 at 12:16
I wish I could have that valuable cheat sheet. Wanna sell it? How many $?? :D – Babak S. Jun 5 '13 at 13:14 Cheat sheet to the rescue! +1 – amWhy Jun 6 '13 at 0:21 HINT: Use Elimination/ Substitution or Cross Multiplication to solve for$x_3,x_4$from the last two simultaneous equation. Putting the values of$x_3,x_4$in the second equation, you will get$x_2$Putting the values of$x_2,x_3,x_4$in the first equation you will get$x_1$- By the Gauss-Jordan algorithm, your next step would be to make the$4x_{2}$into a$0\$, by the Type 3 elementary row operation: multiply row 2 by -4 and add that to row 1.
- | 2015-07-01T17:20:41 | {
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http://mathhelpforum.com/math-topics/130206-calculate-acceleration-time-distance.html | # Thread: Calculate acceleration with time and distance?
1. ## Calculate acceleration with time and distance?
In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.
G:
t = 74s
d = 36km
Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.
In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.
G:
t = 74s
d = 36km
Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.
assuming the shuttle started from rest at t = 0 ...
$d = \frac{1}{2}at^2$
solve for $a$
remember that $d$ must be in meters
3. Thanks for the help, got it right finally.
Though, the second part of the question asks to find the speed of the shuttle in km/h. I did the following.
S = a(t)
S = 13m/s^2(74s)
S = 962m/s
S = 962(.001km)(.0003h)
S = 962(3.33km/h)
S = 3203.46km/h.
The answer on the page however, says 3500km/h. Don't know if this was due to rounding or not, though I doubt that based on the high inaccuracy.
I also tried this.
S = d / t
S = 36km / 74s
S = 0.49km/s
S = 0.49km/(0.0003h)
S = 0.000147km/h
I thought for sure this one would work, but I guess not.
What would be a good equation for calculating the shuttle's speed?
4. $
a \approx 13.15 \, \, m/s^2
$
$
v = at \approx 973 \, \, m/s \, \, \approx 3500 \, \, km/hr
$ | 2016-10-28T00:20:14 | {
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https://math.stackexchange.com/questions/44651/kx-module-and-cyclic-module-over-a-finite-dimensional-vector-space | $k[x]$-module and cyclic module over a finite dimensional vector space
Given a finite dimensional vector space $V$ over a field $k$ and a linear transformation $T: V \rightarrow V$ we can make $V$ a $k[x]$-module via the map:
$$(a_{0}+a_{1}x+\cdots+a_{n}x^{n}) \cdot v \mapsto \sum a_{i}T^{i}(v).$$
On page 10 of this file I was reading the following result:
"$V$ is a cyclic $k[x]$-module iff the minimal polynomial and the characteristic polynomial coincide"
Can anyone please show me how to prove this? Or any reference?
Thanks.
• The This in the sentence you quote cannot refer the anything you wrote before it. Maybe if you provide context, we can help. – Mariano Suárez-Álvarez Jun 10 '11 at 21:37
• One direction is easy: every cyclic $K[x]$-submodule of $V$ has dimension at most the degree of the minimal polynomial, so if the minimal polynomial does not equal the characteristic polynomial, no cyclic $K[x]$-submodule of $V$ can equal $V$ (the dimension is too small). – Arturo Magidin Jun 11 '11 at 4:19
• The important point really is the degree of the minimal polynomial, not that it is equal to the characteristic polynomial. – lhf Jun 11 '11 at 15:45
As I mentioned in comments, one direction is easy: if the minimal polynomial is not equal to the characteristic polynomial (up to the leading term), then its degree is stricly smaller than $\dim(V)=n$; then every $T$-cyclic subspace has dimension at most $m$ (the degree of the minimal polynomial), hence cannot equal all of $V$. Thus, if the minimal polynomial does not equal the characteristic polynomial, then $V$ cannot be $T$-cyclic (which means it is not a cyclic $k[x]$-module).
The converse, however, is subtler. Suppose the minimal polynomial equals the characteristic polynomial. That means that for every polynomial $p(t)$ of degree stricly smaller than $n$, $p(T)$ is not identically zero; that is, for every polynomial $p(t)$ of degree strictly smaller than $n$, there exists a vector $v$ such that $p(T)(v)\neq 0$. This does not, however, in and of itself, imply that there is a vector $v$ such that for every polynomial $p(t)$ of degree strictly smaller than $n$ we have $p(T)(v)\neq 0$. Just because for every polynomial there is a vector that works, it does not follow that there is a vector that works for every polynomial.
Instead, we need to work a bit harder. We need to show that $V$ has a $T$-cyclic subspace whose dimension equals that of the minimal polynomial.
Write the minimal polynomial as a product of powers of distinct irreducible polynomials: $$m(t) = \phi_1(t)^{k_1}\cdots\phi_r(t)^{k_r}.$$ The first step is, for each $i$, to find a vector $v_i$ whose annihilator is $\phi_i(t)^{k_i}$ and no smaller power of $t$.
Consider $$K_{\phi_i} = \{v\in V\mid \text{there exists }p\text{ such that }\phi_i(T)^p(v)=0\}.$$ Then $V=K_{\phi_1}\oplus K_{\phi_2}\oplus\cdots\oplus K_{\phi_r}$ (this holds whether or not the minimal polynomial equals the characteristic polynomial).
Now let $S_i$ be the collection of all $p$ such that there exists $v\in K_{\phi_i}$ such that $\phi_i(t)^{p}(v)=0$, but $\phi_i(t)^{p-1}(v)\neq 0$. It is easy to verify that $S_i$ is finite, since we can take an arbitrary basis for $K_{\phi_i}$ and simply take the exponents corresponding to a basis. Let $M$ be the maximum; then $\phi_i(t)^M$ annihilates $K_{\phi_i}$. That means that $$\frac{m(t)\phi_i(t)^M}{\phi_i(t)^{k_i}}$$ annihilates $V$, which means that this polynomial is a multiple of the minimal polynomial. Hence $M\geq k_i$; but it is also easy to check that $\phi_i(t)^{k_i}$ annihilates $K_{\phi_i}$, so $M=k_i$. Therefore, there exists a vector $v_i\in K_{\phi_i}$ such that $\phi_i(t)^{k_i}(v_i)=0$, but $\phi_i(t)^{k_i-1}(v_i)\neq 0$.
Now we have the following:
Theorem. Let $V$ be a vector space, $T$ a linear operator on $V$, and let $w_1$, $w_2$ be two vectors. Suppose that the $T$-annihilators $p_1(t)$ and $p_2(t)$ of $w_1$ and $w_2$ (respectively) are relatively prime. Then the $T$-annihilator of $w_1+w_2$ is $p_1(t)p_2(t)$.
Proof. To show that $p_1(T)p_2(T)$ annihilates $w_1+w_2$ is straightforward, since $p_1(T)p_2(T)=p_2(T)p_1(T)$.
Note that if $W_1$ is the $T$-cyclic subspace generated by $w_1$ and $W_2$ is the $T$-cyclic subspace generated by $w_2$, then $W_1\cap W_2=\{0\}$, since any vector in the intersection must be annihilated by both $p_1(T)$ and $p_2(T)$, so its annihilator must divide their gcd, which is $1$.
Let $f(t)$ and $g(t)$ be polynomials such that $f(t)p_1(t)+g(t)p_2(t)=1$. Then $f(T)p_1(T)$ is the projection of $W_1\oplus W_2$ onto $W_2$: note that $$f(T)p_1(T)(T^i(w_1)) = T^i(f(T)p_1(T)(w_1)) = T^i(f(T)(0)) = 0,$$ and $$f(T)p_1(T)(T^j(w_2)) = T^j\Bigl(\bigl(1 - g(T)p_2(T)\bigr)(w_2)\Bigr)= T^j(w_2).$$
Consider the $T$-cyclic subspace generated by $w_1+w_2$. It is $T$-invariant, so it is also invariant under $f(T)p_1(T)$; hence $w_2 = f(T)p_1(T)(w_1+w_2) = w_2$ lies in the $T$-cyclic subspace generated by $w_1+w_2$, and hence so does $W_1$; since $w_1+w_2$ and $w_1$ lie in the space, so does $w_2$, and therefore it also contains $W_2$. Therefore, we have that the $T$-cyclic subspace generated by $w_1+w_2$ contains $W_1\oplus W_2$, and hence must be equal to it.
Finally, the dimension of $W_1\oplus W_2$ (which is the $T$-cyclic subspace generated by $w_1+w_2$) equals the sum of the degrees of $p_1(t)$ and $p_2(t)$; we already know that $p_1(t)p_2(t)$ annihilates the space, hence is a multiple of the $T$-annihilator of $w_1+w_2$. Since the annihilator has the same degree as $p_1(t)p_2(t)$, and they are both monic, it follows that $p_1(t)p_2(t)$ is the $T$-annihilator of $w_1+w_2$, as desired. QED
Corollary. Let $V$ be a vector space, $T$ an operator on $V$, and let $w_1,\ldots,w_n$ be vectors in $V$ with corresponding $T$-annihilators $p_1(t),\ldots, p_n(t)$. If the $p_i(t)$ are pairwise relatively prime, then the $T$-annihilator of $w_1+\cdots +w_n$ is $p_1(t)\cdots p_n(t)$.
Putting this together, we have:
Theorem. Let $V$ be a finite dimensional vector space, let $T$ be an operator on $V$, and let $m(t)$ be the minimal polynomial of $T$. Then $V$ has a $T$-cyclic subspace of dimension $\deg(m(t))$.
Proof. Write $m(t)= \phi_1(t)^{k_1}\cdots \phi_r(t)^{k_r}$ as a product of powers of pairwise distinct irreducible polynomials. We know that for each $i$ there is a vector $v_i$ whose $T$-annihilator is $\phi_i(t)^{k_i}$. By the theorem above, the $T$-annihilator of $v=v_1+\cdots+v_r$ is $\phi_1(t)^{k_1}\cdots \phi_r(t)^{k_r}=m(t)$, and in particular the $T$-cyclic subspace generated by $v$ has dimension $\deg(m(t))$, as desired. QED
Now it follows that if the minimal polynomial has degree $\dim(V)$, then there exists a vector $v$ such that no polynomial of degree strictly smaller than $m(t)$ will $T$-annihilate $v$, so $v$ is a witness to the fact that $V$ is $T$-cyclic (and hence $k[x]$-cyclic under the defined action).
• thank you for the great answer – user10 Jun 13 '11 at 14:04
There is a clear and detailed proof of this fact in Dummit and Foote's Algebra book, if I'm not mistaken.
Arturo's answer above is complete and self contained. Just thought I would add an answer/comment, in case you know the fundamental theorem of finitely generated modules over a PID(FTFMP):
By FTFMP, $V\cong k[x]/(a_1(x))\times\cdots \times k[x]/(a_m(x))$, where the characteristic polynomial $\chi_T(x)=a_1(x)\times\cdots\times a_m(x)$ and the minimal polynomial $m_T(x)=a_m(x)$.
So if $\chi_T(x)=m_T(x)$, then $V\cong k[x]/(a_m(x))$, a cyclic $k[x]$-module. | 2019-05-22T16:54:13 | {
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https://mathhelpboards.com/threads/system-of-equalities-3.27730/ | # System of equalities 3
#### solakis
##### Active member
Solve the following sustem
1) $$\displaystyle x+y+z+w=22$$
2) $$\displaystyle xyzw=648$$
3)$$\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{7}{12}$$
4) $$\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}$$
#### Monoxdifly
##### Well-known member
Starting at the third equation, the only couple of Egyptian fraction which sum is $$\displaystyle \frac7{12}$$ are $$\displaystyle \frac1{3}$$ and $$\displaystyle \frac14$$, so x = 3 and y = 4 (interchangeable).
Using the same method for the fourth equation, we get$$\displaystyle \frac16+\frac19=\frac5{18}$$, so z = 6 and w = 9 (also interchangable).
Check them on the first equation:
x + y + z + w = 22
3 + 4 + 6 + 9 = 22
22 = 22 (TRUE)
Also check them on the second equation:
xyzw = 648
3(4)(6)(9) = 648
648 = 648 (TRUE)
So, if the solution is denoted as (w, x, y, z), the solution set is {(6, 3, 4, 9), (6, 4, 3, 9), (9, 3, 4, 6), (9, 4, 3, 6)}.
#### Country Boy
##### Well-known member
MHB Math Helper
Since the last two equation have only two unknowns each, I would start with them.
$\frac{1}{x}+ \frac{1}{y}= \frac{7}{12}$
Multiply both sides by 12xy:
$12y+ 12x= 7xy$
$12y= 7xy- 12x= 7x(y- \frac{12}{7})$
$\frac{12y}{y- \frac{12}{7}}= \frac{7(12)y}{7y- 12}= 7x$.
$x= \frac{12y}{7y- 12}$.
$\frac{1}{z}+ \frac{1}{w}= \frac{5}{18}$
Multiply both sides by 18wz.
$18w+ 18z= 5wz$
$18w= 5wz- 18z= 5z(w- \frac{18}{5})$
$\frac{18w}{w}- \frac{18}{5}= \frac{5(18)w}{5w- 18}= 5z$
$z= \frac{18w}{5w- 18}$.
Now replace x and z in the first two equations so we have two equations in y and w.
$x+ y+ z+ w= \frac{12y}{7y- 12}+ y+ \frac{18w}{5w- 18}+ w= \frac{7y^2}{7y- 12}+ \frac{5w^2}{5w- 18}= 22$.
$xyzw= \frac{12y}{7y- 12}(y)\left(\frac{18w}{5w- 18}\right)w= \frac{216yw}{(7y-12)(5w-18)}= 648$
I will leave solving those last two equations to you!
#### solakis
##### Active member
Starting at the third equation, the only couple of Egyptian fraction which sum is $$\displaystyle \frac7{12}$$ are $$\displaystyle \frac1{3}$$ and $$\displaystyle \frac14$$, so x = 3 and y = 4 (interchangeable).
Using the same method for the fourth equation, we get$$\displaystyle \frac16+\frac19=\frac5{18}$$, so z = 6 and w = 9 (also interchangable).
Check them on the first equation:
x + y + z + w = 22
3 + 4 + 6 + 9 = 22
22 = 22 (TRUE)
Also check them on the second equation:
xyzw = 648
3(4)(6)(9) = 648
648 = 648 (TRUE)
So, if the solution is denoted as (w, x, y, z), the solution set is {(6, 3, 4, 9), (6, 4, 3, 9), (9, 3, 4, 6), (9, 4, 3, 6)}.
is this the only solution to the problem?
Last edited:
#### Monoxdifly
##### Well-known member
I don't know. That's as far as I can get.
#### Opalg
##### MHB Oldtimer
Staff member
Solve the following sustem
1) $$\displaystyle x+y+z+w=22$$
2) $$\displaystyle xyzw=648$$
3)$$\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{7}{12}$$
4) $$\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}$$
From 3) and 4), $12(x+y) = 7xy$ and $18(z+w) = 5zw$. Multiply those two equations and use 2), to get $18*12(x+y)(z+w) = 35*648$, which reduces to $(x+y)(z+x) = 105$. From 1), $(x+y) + (z+w) = 22$. It follows that $x+y$ and $z+w$ are the solutions of the quadratic equation $s^2 - 22s + 105=0$, namely $7$ and $15$. So there are two possible cases.
Case 1: $x+y = 7$ and $z+w=15$. Then $xy=12$, $zw=54$ and we get the solutions $\{x,y\} = \{3,4\}$, $\{z,w\} = \{6,9\}$, as found by Monoxdifly .
Case 2: $x+y = 15$ and $z+w = 7$. Then $xy = \frac{180}7$ and $zw = \frac{126}5$. But then $z$ and $w$ would have to be the solutions of the equation $t^2 - 7t + \frac{126}5 = 0$. Since that equation has no real solutions, Case 2 cannot arise. So the solutions found by Monoxdifly are the only ones.
#### Monoxdifly
##### Well-known member
From 3) and 4), $12(x+y) = 7xy$ and $18(z+w) = 5zw$. Multiply those two equations and use 2), to get $18*12(x+y)(z+w) = 35*648$, which reduces to $(x+y)(z+x) = 105$. From 1), $(x+y) + (z+w) = 22$. It follows that $x+y$ and $z+w$ are the solutions of the quadratic equation $s^2 - 22s + 105=0$, namely $7$ and $15$. So there are two possible cases.
Case 1: $x+y = 7$ and $z+w=15$. Then $xy=12$, $zw=54$ and we get the solutions $\{x,y\} = \{3,4\}$, $\{z,w\} = \{6,9\}$, as found by Monoxdifly .
Case 2: $x+y = 15$ and $z+w = 7$. Then $xy = \frac{180}7$ and $zw = \frac{126}5$. But then $z$ and $w$ would have to be the solutions of the equation $t^2 - 7t + \frac{126}5 = 0$. Since that equation has no real solutions, Case 2 cannot arise. So the solutions found by Monoxdifly are the only ones.
Ah, so I needn't go further. Thanks for your confirmation. | 2020-07-09T06:56:57 | {
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https://math.stackexchange.com/questions/4045353/why-does-xxxx-bifurcate-below-sim0-065 | # Why does ${x}^{x^{x^{x^{\,.^{\,.^{\,.}}}}}}$ bifurcate below $\sim0.065$?
When you calculate what $${x}^{x^{x^{x\cdots }}}$$ converges to between $$0$$ and $$1$$, before approximately $$0.065$$ the graph bifurcates.
Why does this happen and is there a reason for it happens at that number?
• A preliminary obsevation: when $x\rightarrow0$, $x^x\rightarrow1, x^{x^x}\rightarrow0,x^{x^{x^x}}\rightarrow1$, etc. So it may turn out that the sequence is "alternating." – Tesla Daybreak Mar 2 at 6:16
• Maybe this will help someone else work out the details: it looks like the transition is at $e^{-e} \approx 0.065988$. When $0 < c < e^{-e}$ it appears that there's a unique positive real solution $x_0$ to $x = c^x$, but it's an unstable equilibrium point because we have $\big| \frac{d}{dx}(c^x)|_{x = x_0} \big| > 1$. – Ravi Fernando Mar 2 at 6:23
• Related topic: math.stackexchange.com/q/1899599, math.stackexchange.com/q/87870. – user Mar 2 at 8:29
• See also this article and the links out of Wiki. – metamorphy Mar 2 at 8:32
• Just curious: how did you obtain the red line? – Han de Bruijn Mar 5 at 20:16
## 2 Answers
The following reference is recommended reading, if you want to have a more detailed account of what follows:
The power tower at hand is recursively defined as: $$h_0(x) = x \quad ; \quad h_{n+1}(x) = x^{h_n(x)} \quad ; \quad h(x) = \lim_{n\to\infty} h_n(x)$$ The recursion process is visualized in the picture below:
The $$\color{blue}{blue\;lines}$$ are the iterands for $$n=$$ odd; the $$\color{red}{red\;lines}$$ are the iterands for $$n=$$ even. It is seen that splitting up odd and even starts at $$x\lt 1$$ and that there is a bifurcation point (black dot). Going to infinity leads to the functional equation: $$y = x^y \quad \mbox{with} \quad y = h(x)$$ The inverse function (where it exists) has an explicit analytical form, which is the black line in the above figure: $$x = y^{1/y}$$ And can be used to establish the rightmost bound of the tower's domain: $$\frac{d}{dy}y^{1/y} = e^{\ln(y)/y}\left[-\frac{\ln(y)}{y^2}+\frac{1}{y^2}\right] = 0 \\ \Longrightarrow \quad \ln(y)=1 \quad \Longrightarrow \quad y=e \; ; \; x=e^{1/e}$$ But in order to answer the question as stated, we have to consider another functional equation, namely: $$y = x^{x^y}$$ Odd and even are separated now, as is required for observing the bifurcation. There is no neat inverse function of the latter equation. Therefore we shall investigate the two-dimensional funcion $$\,f(x,y)=x^{x^y}-y\,$$. An insightful way to do this is make an isoline chart / contour plot of the function. Lines are darker where the contour levels are higher. The massive black line is the place where $$f(x,y)=0$$.
Implicit differentiation is employed for finding the derivatives: $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \\ df = x^{x^y}\left[\frac{x^y y \ln(x)}{x} + \frac{x^y}{x}\right] dx + \left[x^{x^y}x^y\ln^2(x)-1\right] dy$$ If the gradient is zero, then $$f(x,y)$$ has a critical point at that place: $$\begin{cases} \partial f/\partial x = 0 & y \ln(x)+1 = 0 \\ \partial f/\partial y = 0 & x^{x^y}x^y\ln^2(x)-1=0 \end{cases}$$ Both equations are essentially the same, because $$\,x^{x^y}=x^y=y\,$$, giving with $$\,x=y^{1/y}\,$$: $$y \ln(y^{1/y})+1 = \ln(y)+1 = 0 \quad \Longrightarrow \quad y = 1/e \quad ; \quad x = e^{-e}$$ If we look at the picture with the isolines, then it is evident that the bifurcation must be at this place, with coordinates $$\,(x,y)=(e^{-e},1/e)\,$$.
EDIT.
It is observed in the last picture that the isolines in the neighborhood of the critical point are like hyperbolas. To confirm this observation, the second order derivatives at the bifurcation point should be investigated as well. Let a computer algebra system (MAPLE) do the hard work.
> f(x,y) := x^(x^y)-y;
> A(x,y) := diff(diff(f(x,y),x),x);
> B(x,y) := diff(diff(f(x,y),x),y);
> C(x,y) := diff(diff(f(x,y),y),y);
> D(x,y) := diff(diff(f(x,y),y),x);
> a := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},A(x,y)));
exp(-3 + 2 exp(1))
> b := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},B(x,y)));
-exp(-1 + exp(1))
> c := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},C(x,y)));
0
> d := simplify(subs({x=exp(1)^(-exp(1)),y=1/exp(1)},D(x,y)));
-exp(-1 + exp(1))
So the Hessian matrix at the crirical point is: $$\begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} e^{-3+2e} & -e^{-1+e} \\ -e^{-1+e} & 0 \end{bmatrix}$$ And the two eigenvalues of the Hessian turn out to be $$\lambda_{1,2} = \frac{a+c}{2} \pm \sqrt{b^2 + \left(\frac{a-c}{2}\right)^2} \\ = \frac{1}{2} e^{-3+2e} \left[1 \pm \sqrt{\left(2\frac{e^{-1+e}}{e^{-3+2e}}\right)^2+1}\right]$$ Herewith it is clear that one eigenvalue is positive while the other is negative. Therefore the critical point is a saddle point, as expected. As has been said, the isolines in the neighborhood of this saddle point are hyperbolas. The asymptotes of these hyperbolas can be calculated as follows. $$\begin{bmatrix} x-e^{-e} & y-1/e \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} x-e^{-e} \\ y-1/e \end{bmatrix} = 0 \\ e^{-3+2e}\left(x-e^{-e}\right)^2-2\,e^{-1+e}\left(x-e^{-e}\right)\left(y-1/e\right) = 0 \\ \Longrightarrow \quad x = e^{-e} \quad ; \quad y = 1/e + e^{e-2}/2\cdot\left(x-e^{-e}\right)$$ The asymptotes are tangent to the crossing massive black lines at the critical point.
See Projective Fractals for an explanation of why the bifurcation is at $$e^{e^{2 \pi i x-{e^{2 \pi i x}}}}$$ where $$x=-1$$. | 2021-04-22T01:58:17 | {
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http://math.stackexchange.com/questions/42180/determining-if-function-is-11-or-onto | # Determining if Function is 1:1 or Onto
I'm trying to solve the following question:
Given function $\displaystyle f\colon \mathbb Z \rightarrow \mathbb Z,$ where $f(n) = \lfloor {\frac {n}{3}} \rfloor$, determine if it's $1:1$, and onto; and prove why.
I could say that it is onto, because it will always have an element in the codomain that maps to the domain. It is also not one to one, because the floor would result in multiple numbers with the same floor (like $\frac {1}{3}$ and $\frac {2}{3}$ both having the floor $0$). But how can I actually prove these?
-
Look at the definition of $1$-$1$ function. See what happens, when you plugin $3$ and $4$ to $f$. You have $$f(3) = \biggl\lfloor{\frac{3}{3}\biggr\rfloor} = 1 = \biggl\lfloor{\frac{4}{3}\biggr\rfloor}=f(4)$$
but $3 \neq 4$ hence it is not $1$-$1$. Note that if you want to prove a function is one-one, then you have to show whenever $f(a)=f(b) \Longrightarrow a=b$.
To prove that it is onto, for each integer $n$ we want integer $m$ such that $\bigl\lfloor{\frac{m}{3}\bigr\rfloor} = n$. Doesn't $m=3n$ finish the job?
-
So since m=3n is a Real number, then it is onto? – Christopher May 30 '11 at 18:14
@Christopher: It's an integer. Yes :). To see it give me an $x$. Say $3$, then i have 10 in $\mathbb{Z}$ such that $f(10)=3$. Next, give me 5 then i have 15 in $\mathbb{Z}$ such that $f(15)=5$. For ontoness you have to show there is $y in \mathbb{Z}$ such that $f(x)=y$ for some $x \in \mathbb{Z}$. – user9413 May 30 '11 at 18:16
HINT $\rm\ \ \ \#\ f^{\:-1}(k)\ =\ \#\:\{\: n\ :\ k\:\le\: n/3\: <\: k+1\}\ =\ \#\:\{\: n\ :\ 3\:k\: \le\: n\: <\: 3\:k+3\}\ >\ 1$
-
You already have done most of the work.
To prove that it is not one to one, you just need a counterexample, which you already have.
To prove that it is onto, for each integer $n$ find (explicitly) an example of an integer $m$ such that floor $m/3 = n$.
-
Ok, so to prove that its ONTO, how can I prove that each n has a corresponding m/3? I haven't proved (or disproved) any floor/ceiling functions before. – Christopher May 30 '11 at 18:10
@Christopher: One way, which works here, is to say that if I give you an $m$, you can always find an $n$ such that $f(n)=m$. You prove that by saying "and $n=3m$ (or $n=3m+1$) works". – Ross Millikan Mar 6 '13 at 17:06 | 2014-10-02T08:37:40 | {
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