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http://math.stackexchange.com/questions/175607/differentiable-function-not-constant-fxy-fxfy-f0-2 | Differentiable function, not constant, $f(x+y)=f(x)f(y)$, $f'(0)=2$
Let $f: \mathbb R\rightarrow \mathbb R$ a derivable function but not zero, such that $f'(0) = 2$ and $$f(x+y)= f(x)\cdot \ f(y)$$ for all $x$ and $y$ belongs $\mathbb R$. Find $f$.
My first answer is $f(x) = e^{2x}$, and I proved that there are not more functions like $f(x) = a^{bx}$ by Existence-Unity Theorem (ODE), but I don't know if I finished.
Thanks,
-
– Robert Israel Jul 26 '12 at 20:42
Differentiate $f(x+y)=f(x)\cdot f(y)$ with respect to $y$ to obtain $$f'(x+y)=f(x)\cdot f'(y).$$ Now by letting $y=0$ and noting that $f'(0)=2,$ we obtain $$f'(x)=2f(x).$$
The solutions to the preceding equation are of the form $f(x)=Ce^{2x}$ for some constant $C$. Using the fact that $f'(0)=2$, we find that $f'(0)=2C=2$, so that $C=1$. Hence $f(x)=e^{2x}.$ It is easy to check that this does indeed satisfy the original functional equation.
-
$f(0)=(f(0))^2 \Rightarrow f(0) \in \{0,1\}$. If $f(0)=0$, then $f(x) =0$ for every $x$, therefore $f(0)=1$.
For every $x \in \mathbb{R}$ one has $f(x)f(-x)=f(0)=1$ and $f(x)=(f(x/2))^2$, i.e. $f(x)>0$ for every $x \in \mathbb{R}$.
By induction one has $f(nx)=(f(x))^n$ for every $n \in \mathbb{N}$. Since $f(x)f(-x)=f(0)=1$, one has $f(-nx)=(f(nx))^{-1}=(f(x))^{-n}$ for every $n \in \mathbb{N}$. Hence $f(nx)=(f(x))^n$ for every $n \in \mathbb{Z}$.
For every $n \in \mathbb{Z}\setminus\{0\}$ one also has $(f(x/n))^n=f(x)$, so $f(x/n)=(f(x))^{1/n}$. Setting $a=f(1)$ one has, for every $m/n \in \mathbb{Q}$: $$f(m/n)=(f(m))^{1/n}=[(f(1))^m]^{1/n}=a^{m/n}.$$ Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given $x \in \mathbb{R}$, there is a sequence $(x_k) \subset \mathbb{Q}$ such that $x_k \to x$ as $k \to \infty$, and by continuity one has $$f(x)=\lim_kf(x_k)=\lim_k a^{x_k}=a^x.$$ One has $f'(x)=a^x\ln a$ for every $x$, and $2=f'(0)=\ln a$, i.e. $a=e^2$. Thus $f(x)=e^{2x}$ for every $x \in \mathbb{R}$.
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https://fundacja-pe.nazwa.pl/sm8vppz4/8e02d3-how-to-find-cosine-from-sine | To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. The sum of the cosine and sine of the same angle, x, is given by: [4.1] We show this by using the principle cos θ=sin (π/2−θ), and convert the problem into the sum (or difference) between two sines. x − This must be a numeric value.. Return Value. sin (x) = cos (90 -x) [within first quadrant] 0 0 The Lesson: y = sin(x) and y = cos(x) are periodic functions because all possible y values repeat in the same sequence over a given set of x values. Example 26. Here’s how to prove this statement. Following is the syntax for cos() method −. When finding the equation for a trig function, try to identify if it is a sine or cosine graph. See Example. We note that sin π/4=cos π/4=1/√2, and re-use cos θ=sin (π/2−θ) to obtain the required formula. Understanding how to create and draw these functions is essential to these classes, and to nearly anyone working in a scientific field. Teacher was saying that in right triangles the sine of one acute angle is the cosine of the other acute angle. It is easy to memorise the values for these certain angles. From this information, we can find the amplitude: So our function must have a out in front. The second one, y = cos( x 2 + 3) , means find the value ( x 2 + 3) first, then find the cosine of the result. Python number method cos() returns the cosine of x radians.. Syntax. However, scenarios do come up where we need to know the sine and cosine of other angles. The first one, y = cos x 2 + 3, or y = (cos x 2) + 3, means take the curve y = cos x 2 and move it up by 3 units. You want to show that the sine function, slid 90 degrees to the left, is equal to the cosine function: Replace cos x with its cofunction identity. Find $$\cos(20^\circ)$$ and $$\sin(20^\circ)\text{. When the sine or cosine is known, we can use the Pythagorean Identity to find the other. I think I am a very visual learner and I always found that diagrams always made things clearer for my students. cos(x) Note − This function is not accessible directly, so we need to import math module and then we need to call this function using math static object.. Parameters. Trig calculator finding sin, cos, tan, cot, sec, csc To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example. The shifted sine graph and the cosine graph are really equivalent — they become graphs of the same set of points. To find the cosine and sine of angles that are not common angles on the unit circle, we can use a calculator or a computer. Begin by realizing we are dealing with a periodic function, so sine and cosine are your best bet. The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. Description. }$$ When we find sin cos and tan values for a triangle, we usually consider these angles: 0°, 30°, 45°, 60° and 90°. Sum Introduction: In this lesson, the period and frequency of basic graphs of sine and cosine will be discussed and illustrated. Underneath the calculator, six most popular trig functions will appear - three basic ones: sine, cosine and tangent, and … The sine and cosine functions appear all over math in trigonometry, pre-calculus, and even calculus. Next, note that the range of the function is and that the function goes through the point . The “length” of this interval of x … Find An Equation For The Sine Or Cosine Wave. | 2022-07-05T16:03:26 | {
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https://math.stackexchange.com/questions/2758158/is-the-sum-of-digits-of-left16k-1-right-less-than-6k-for-k-223/2760925 | # Is the sum of digits of $\left(16^k - 1\right)$ less than $6k$ for $k > 223$?
I had been researching the OEIS sequence A165722, which is the sequence of positive integers $k$ such that the sum of digits of $\left(16^k - 1\right)$ is equal to $6k$. I used computational power to determine that the sum of digits is less than $6k$ for $223 < k < 10^6$. I made a conjecture that $6k$ would continue to grow at a faster rate than the digit sum, and thus the sequence is finite. I and several others, however, were unsure how one would go about proving this. I thought that perhaps there would be some way to show a regularity in the digits of $16^k - 1$, but I would not know how to go about finding or proving this regularity.
• I don't think that there is hope to find such a regularity. Besides the few last digits, the digits will probably behave like a random sequence. We can expect that the average digit does not exceed $4.5$ significant, but we would need an average digit of about $4.98$ to have $6k$ digits or more. So, the digitsum should always be less than $6k$ for $k>223$. I would be very surprised if someone can prove this. – Peter Apr 29 '18 at 7:29
• Legendre's formula might be a good starting: if $\nu_q(n)$ denotes the sum of digits in the $q$-ary expansion of $n$, then $$\nu_q(n) = n - (q-1)\sum_{j=1}^{\infty} \left\lfloor \frac{n}{q^j} \right\rfloor = (q-1)\sum_{j=1}^{\infty} \frac{n\text{ mod }q^j}{q^j}.$$ But again, this leads us to an intricate world of multiplicative groups and I have no idea. I would also be very surprised if someone can prove this. – Sangchul Lee May 1 '18 at 4:26
• I don't know if this applies, but perhaps you could use Benfords law to do a statistical/probabilistic analysis of the digits of $16^k-1$ and then show that the probability of getting a number whose digit sum is equally to 6k is greater than 0. Once again though, I'm not sure if Benford's law applies. Perhaps you could find a distribution for the digits in base 16. – Saudman97 May 2 '18 at 21:24
• Here's a proof of a very weak lower bound, showing that the sum of the base-$10$ digits of $2^n$ is $\Omega(\log n)$: oeis.org/A001370/a001370_1.pdf – mjqxxxx May 3 '18 at 0:21
• This somewhat related question may be of interest. I hastily misread it, and thought it came close to proving yours. It doesn't, but it may be of interest anyway. – Jyrki Lahtonen May 5 '18 at 18:29
Some preliminary estimates. From this book, page 79 (accessible in preview mode) $$S(16^n-1)=16^n-1 - 9\sum\limits_{k\geq1} \left \lfloor \frac{16^n-1}{10^k} \right \rfloor \tag{1}$$
Further $$S(16^n-1)= 16^n-1 - 9\sum\limits_{k\geq1} \left(\frac{16^n-1}{10^k}-\left\{\frac{16^n-1}{10^k}\right\}\right)=\\ 16^n-1 - 9\sum\limits_{k\geq1} \frac{16^n-1}{10^k}+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\ (16^n-1)\left(1 - 9\sum\limits_{k\geq1} \frac{1}{10^k}\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\ (16^n-1)\left(1 - 9\left(\frac{10}{9}-1\right)\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}= 9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}$$ Or $$S(16^n-1)=9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}\tag{2}$$
The last digit of $16^n-1$ is always $5$ and $$\left\{\frac{16^n-1}{10^k}\right\}=0.\overline{a_1a_2...a_{k-1}5} \leq 0.99..95<1$$ $9$ repeated $k-1$ times. But only for the first $n\log_{10}16$ terms. For all $k>n\log_{10}16$ $$\left\{\frac{16^n-1}{10^k}\right\}\leq 0.00..099..95$$ where $00..099..9$ is of length $k-1$. Basically, starting with $k\geq \left \lfloor n\log_{10}16 \right \rfloor+1$ this tail forms an infinite geometric progression with ratio $\frac{1}{10}$ which sums to a constant. So we can conclude $$S(16^n-1) < 9n\log_{10}16 + C$$ We also have that $9\log_{10}16<11$, thus $$S(16^n-1) < 11n+ C \tag{3}$$ Probably, with more accurate calculations, a better estimate may be obtained ... work in progress.
• This is very nice. I do know that for some values, $S(16^n - 1) \geq 6n$, but the last apparent occurrence of this is when $n=223$. I am concerned about values above this. This is a great start though. – Kirk Fox May 4 '18 at 18:47
• This bound is fairly trivial... the number $16^k-1$ has no more than $1 + \log_{10}(16^k-1) < 1 + k \log_{10}16$ digits, each of which is no greater than $9$, so the digit sum is no greater than $9 + 9k\log_{10}16 < 10.85k + 9.$ – mjqxxxx May 4 '18 at 23:11
• @mjqxxxx 1st of all to reveal a formula and secondly to engage more people in finding a better estimate. The weak part in the subsequent calculations is of course the gross estimate of $0.\overline{a_1a_2...a_{k-1}5} \leq 0.99..95<1$, maybe somebody could find a finer pattern? It matches your calculations, but it has, imo, more potential ... – rtybase May 4 '18 at 23:30
• Knowing the fractional part of $16^n-1$ over $10^k$ is the same as knowing the digits of $16^n-1$ since dividing by $10^k$ just shifts it, preserving the significant digits. This work can also be skipped using the trivial bound: $16^n-1$ has $\log_{10}(16^n-1)\approx n\log_{10}(16)$ digits in decimal, and each digit can be at most $9$ giving the bound $9n\log_{10}(16)$ as you have. – Will Fisher May 7 '18 at 0:53
Let $S\left(n\right)$ be the sum of the digits of $n$. Since $16^k-1$ is divisible by 5 but not 2, the last digit is five, and hence $$S\left(16^k-1\right)=S\left(16^k\right)-1$$ Essentially we want to say that $16^k=2^{4k}$ typically has digits less than 5 rather than digits more than five, or at least it doesn't mostly have digits more than five. One thought that pops out is that the map $k\mapsto2k\textrm{ mod }10$ is periodic for even $k$ with period 4 (we have the cycle $2\mapsto4\mapsto8\mapsto6\mapsto2$), and for odd $k$ of course the next term is even. Except for when there is carrying, this map is the map taking digits of $2^k$ to $2^{k+1}$. Could this periodicity which matches $2^{4k}$ explain it? | 2019-10-20T18:54:45 | {
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https://www.physicsforums.com/threads/the-complex-field.346057/ | # The Complex Field
1. Oct 15, 2009
### jgens
Would the following prove that the set of complex numbers do not form and ordered field?
Clearly $i \neq 0$. Therefore, if the complex numbers form an ordered field either $i > 0$ or $i < 0$. Suppose first that $i > 0$, then $i^2 = -1 > 0$, a contradiction. Now suppose that $i < 0$, then $i^2 = -1 > 0$, another contradiction. Thus, the set of complex numbers do not form an ordered field.
This seems awfully fishy and I wouldn't be surprised to find that it's completely invalid. Feedback and suggestions are welcome. Thanks!
2. Oct 15, 2009
### pbandjay
You have i2 = -1 > 0, but -1 < 0.
3. Oct 15, 2009
### jgens
Yes, that's why $i > 0$ and $i < 0$ are contradictions as I state in my original post.
4. Oct 15, 2009
### Staff: Mentor
I understand the logic behind what you're saying, but you could say it in a way that is clearer.
Suppose that i > 0.
Multiplying by a positive number preserves the direction of the inequality, so i2 must be positive. This is incorrect, though, because i2 = - 1 < 0. Thus, assuming i > 0 leads to a contradiction.
And similar for the assumption that i < 0.
5. Oct 15, 2009
### jgens
Thanks for the suggestion! It seems like I always need to add details to my proofs in order to make them clearer. Does the method I'm using work?
6. Oct 16, 2009
### HallsofIvy
What you have proven is that the usual order on the real numbers, in which -1< 0, cannot be extended to the complex numbers.
But you can prove more. Can there exist some order? Perhaps one in which 1< 0and -1> 0?
Suppose such an order existed. The i is not 0 (the additive identity does not depend upon the order). If i> 0 then it follow that (i)(i)= -1> 0. From that i(-1)= -i> i(0)= 0. Now that contradicts i>0. If i< 0, the -i> 0 so (-i)(i)= 1< 0. From that i(1)= i< i(0)= 0 and that contradicts i< 0.
7. Oct 16, 2009
### qspeechc
Ok, that proves that there can be no order where 1<0 and -1>0. How does one show that there can not be any order relation defined on the complex numbers? Or R^n for n>1 for that matter?
8. Oct 16, 2009
### Ben Niehoff
Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, where $\theta$ is restricted to be in $[0, 2\pi)$. Then define > as follows:
1. If $r_1 > r_2$, then $z_1 > z_2$
2. If $r_1 = r_2$ and $\theta_1 > \theta_2$, then $z_1 > z_2$
3. If $r_1 = r_2$ and $\theta_1 = \theta_2$, then $z_1 = z_2$
This looks like an order relation to me...it even has a least element, 0. Now, I know that C is not supposed to have an order relation, so what's wrong with this? Have I implicitly used the axiom of choice or something?
9. Oct 16, 2009
### Moo Of Doom
You can certainly put some order on the complex numbers, Ben, but you can't find one that behaves nicely with their algebraic structure. For a field to be considered an ordered field, it must have an order that is compatible with multiplication and addition. That is, we want
a < b ==> a + c < b + c
a < b, c > 0 ==> ac < bc
for all a, b, c. Your order doesn't satisfy either of these properties, so it doesn't make the complex numbers into an ordered field. Sorry.
10. Oct 16, 2009
### Ben Niehoff
Ah, yes, that does change things a bit.
11. Oct 22, 2009
### elibj123
Order of a field is defined via a subset of that field. Let's denote it P. An ordered field F contains a subset P that satisfies the following properties:
1) for any a (which is not 0) in F, Either a or -a are in P, but not both.
2) For any a,b in P, a+b is in P.
3) For any a,b in P, a*b is in P.
Then, we say a>b iff (a-b) is in P.
You can see immediately that the C doesn't contain such a set because
if i belongs P, it implies that -i=i*i*i belongs to P (3) in contradiction to (1).
The same if we assume -i belongs to P.
12. Oct 22, 2009
### HallsofIvy
As elibj123 said, one of the requirements for an "ordered field", trichotomy, is that one and only one of the following be true of any member of the field, x: (trichotomy)
a) x= 0
b) x> 0
c) x< 0
You cannot have "-1< 0" and "1> 0" in any ordered field (where "1" is defined as the multiplicative identity and "0" is defined as the additive identity).
Your reference to $R^n$ doesn't make sense because it is not a field. What arithmetic operations would you wish to define on it to make it a field?
If you ignore the operations, then the complex numbers can be ordered (any set can be). One such order ("lexicographical") is "If a< c then a+bi< c+ di. If a= c and b< d, then a+bi< c+ di." That's a perfectly good order on C or $R^2$ and can be extended in an obvious way to $R^n$. As long as you don't look at the operations, there is no problem. | 2018-07-16T11:09:42 | {
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http://www.physicsforums.com/showthread.php?t=246580 | # A question on function equality
by zenctheo
Tags: equality, function
P: 4 Hello to every one! I have a question that came up when I was talking with a fellow mathematician. I used to say that two functions are equal when the have the same formula and the same domain and codomain. We read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same X. For example $$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \in \{0,1\}$$because f(0)=g(0)=0 and f(1)=g(1) even though their formula is different. I thought that this definition of equality is incomplete because by saying that f(x)=g(x) then $$\frac{df}{dx}=\frac{dg}{dx}$$ but on point x=1 $$\frac{df}{dx}=2$$ and $$\frac{dg}{dx}=3$$. Thus we derive two different results from to equal quantities. Therefore two functions in order to be equal should also have the same formula. Can you please give any insight on this? Thanks a lot in advance. Akis
Sci Advisor P: 1,563 The derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative: $$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ But for nearly all $\Delta x$, $x + \Delta x$ lies outside your domain. Therefore, you can't take the limit. :) So, you are correct: Two functions are equal if and only if they have the same domain and their values are equal at every point within the domain.
P: 4 Thanks a lot for the reply. You that I am wrong because I was the one saying that the functions should also have the same formula. In order to get things straight: You mean that the above two functions are equal.... or not?
P: 355
## A question on function equality
The functions are in fact equal. Also, as Ben said, those functions don't have derivatives because they're not defined on an open interval of the real numbers.
As another example, would you consider these to be the same function?
Let's say f and g are functions from the real numbers to the real numbers defined as
f(x) = x
g(x) = x when x^2 >= 0 and -x when x^2 < 0
Since the functions are only defined on the real numbers, there are no points where they'd differ.
On a related note: "Having the same formula" is not a well-defined concept. Most (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas.
P: 4 Ok. It's nice to learn a new thing. Even if I am proven wrong :)) Thanks a lot.
Related Discussions Calculus 2 Calculus & Beyond Homework 2 Beyond the Standard Model 4 General Math 8 Set Theory, Logic, Probability, Statistics 3 | 2014-04-25T00:28:11 | {
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https://math.stackexchange.com/questions/3009852/table-tennis-win-probability | # Table tennis win probability
This problem is from my teacher and I think their answer is wrong.
The problem is in the context of table tennis.
The players in the tournament final are Ani and Bertha. The score in the game is drawn at 20-20. The final game will continue until one player has scored two more points than the other. It is known from previous games between Ani and Bertha that the probability of Ani winning each point is 0.6. Find the probability that Ani will win the game after exactly 8 more points.
I think that this means that, over the next 6 games on the 2nd, 4th and 6th game Ani and Bertha need to have a draw. For each draw there are two possible paths. Ani wins then Bertha or Bertha then Ani. After the 6th game Ani just needs to win twice to win after exactly 8 points.
However my teacher says that:
If Bertha wins the first game, it is not possible for Ani to win after exactly 8 points.
and also asserts that there is only one path to the desired outcome. Using this they find the probability of Ani winning after exactly 8 points to be 0.005. (
I found an alternate answer using multiple paths.
$$P(\text{Ani winning a game})=0.6$$
$$P(\text{Bertha winning a game})=0.4$$
$$P(\text{Ani win after 8 points})=2\dot(0.4 \cdot 0.6)\cdot2\dot(0.4 \cdot 0.6)\cdot2\dot(0.4 \cdot 0.6)\cdot(0.6\cdot0.6)=0.040\ (2sf)$$
After I found this answer I asked my teacher if the proposed answer was correct. My teacher replied saying that there was nothing wrong with it.
Am I missing something painfully obvious and if so what? or is the teacher's answer incorrect?
• When you contacted your teacher, did you present your argument as clearly as you did here? – Fabio Somenzi Nov 23 '18 at 2:22
Ani needs to be have two more points than Bertha exactly on the eighth game (not earlier, and never two less). She does not just need to win two games in a row, but to be tied then win the last two games.
So since we are staring at a tie and need to avoid an being ahead or behind by two after the second game, the first two games may be a win followed by a loss, or a loss followed by a win.
Either way leaves Ani and Bertha tied again at the end. And so likewise for the next two games, and the next pair. Then as stated, the final two games need both be wins.
There are eight ways to do this, each giving Ani three losses and five wins. (Also, four among these paths do have Bertha ahead on the first game, so your teacher supposed erraneously.)
$$\mathsf P(\text{W on 8th})= 2^3\cdot0.4^3\cdot0.6^5$$
As you found.
I think your teacher is wrong. He shows BAA as a win for Annie, but it's not. He has confused "ahead by two" with "win two in a row".
• A non-mathematical suggestion on your answer: Since the OP referred to their teacher with the pronoun “they,” consider doing the same in your answer, instead of referring to the teacher as “he.” – Steve Kass Nov 23 '18 at 1:14
• @SteveKass You're right. I am usually pretty careful about his/her. I understand the need for even more pronouns. I have a hard time with the singular "they" - I may be too old to get used to it. Frequently I rework sentences so that I don't need any pronoun. – Ethan Bolker Nov 23 '18 at 11:34 | 2020-01-19T01:59:56 | {
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https://math.stackexchange.com/questions/2368583/find-the-determinant-when-all-entries-are-1-except-for-zeroes-on-the-main-diag/2369567 | # Find the determinant when all entries are $1$ except for zeroes on the main diagonal
Let $J_n$ be an $n\times n$ matrix all of whose entries are $1$, and $I_n$ be the identity matrix. Define $$K_n = J_n-I_n$$
For $n=1$ to $5$ my (usually unreliable!) hand calculations suggest that
$$\det K_n = (-1)^{n-1}(n-1)$$
Question
(a) are these values correct? (b) is the generalization to any positive integer $n$ valid? (c) if (b) is true, how can the result be demonstrated?
My only idea so far is to use the product of eigenvalues:
$$\det K_n = \prod_{j=1}^n \lambda_j$$
($-1$ is an eigenvalue for all $n$)
Any assistance much appreciated.
(note: a similar question concerning skew-symmetric matrices Determinant of a special skew-symmetric matrix may contribute some relevant ideas. or perhaps Determinant of a matrix with $t$ in all off-diagonal entries. has greater relevance)
This is the adjacency matrix of the complete graph $K_n$. The eigenvalues are $\lambda_1 = (n - 1)$ and $\lambda_2 = \dots = \lambda_n = -1$.
The eigenvector for $\lambda_1$ is the vector $\pmb{1}$ consisting of all $1$s.
An eigenbasis for $\lambda_2,\dots,\lambda_n$ is given by $e_1 - e_j$. Note that
$$(J_n - I_n)(e_1 - e_n) = \pmb{1}\pmb{1}^T(e_1 - e_j) - (e_1 - e_j) = - (e_1 - e_j).$$
This is because $J_n = \pmb{1}\pmb{1}^T$ and $\pmb{1}^Te_i = 1$ for all basis vectors $e_i$.
The fact that $n - 1$ is an eigenvalue reflects the fact that $K_n$ is an $(n - 1)$-regular graph. There may also be a combinatorial interpretation for the other eigenvalues but I don't know enough algebraic graph theory to say.
• thank you. very nice insight re the complete graph. – David Holden Jul 23 '17 at 3:34
$$\begin{array}{rl} \det ( 1_n 1_n^\top - \mathrm I_n ) &= \det ( -( \mathrm I_n - 1_n 1_n^\top ) )\\ &= (-1)^n \cdot \det(\mathrm I_n - 1_n 1_n^\top)\\ &= (-1)^n \cdot \det ( 1 - 1_n^\top 1_n )\\ &= (-1)^n \cdot (1-n)\\ &= (-1)^{n-1} \cdot (n-1)\end{array}$$
• really nice answer, thanks, and very educational for me. i hadn't encountered the determinant identity before, so this has clarified the meaning of the Schur complement - a concept which i encountered in reading recently without understanding its purport. have already accepted another answer, but yours is a very nice application of a powerful result to my rather simple question. thanks again! – David Holden Jul 23 '17 at 14:33
Finding the eigenvalues of a matrix $A$ that has $a$ along its main diagonal and $b$ everywhere else comes up fairly often on this site, but I’ll repeat the calculation here. Write the matrix as $A=b\mathbb1+(a-b)I$, where $\mathbb 1$ is the $n\times n$ matrix with every entry equal to $1$. Suppose $v$ is an eigenvector of $A$. We then have $b\mathbb1v=(\lambda-a+b)v$, so $v$ is also an eigenvector of $b\mathbb1$. Moreover, the eigenvalues of this matrix and those of $A$ differ by $a-b$. Now, for $b\ne0$, $b\mathbb1$ has rank 1, so its eigenvalues are $0$ with multiplicity $n-1$ and, using the fact that the trace of a matrix is equal to the sum of its eigenvalues, $nb$ with multiplicity 1. (You could also find the non-zero eigenvalue by noticing that all of the row sums are identical.) This means that the eigenvalues of $A$ are $a+(n-1)b$ and $a-b$, giving $$\det A=(a+(n-1)b)(a-b)^{n-1}.$$ For your matrices, $a=0$ and $b=1$, so $\det{K_n}=(-1)^{n-1}(n-1)$, just as you’ve calculated.
• thanks very much! another very enlightening answer, and very nicely explained. each of the three answers given here has helped me to expand my rather limited understanding of linear algebra. – David Holden Jul 25 '17 at 3:07 | 2019-11-20T04:42:09 | {
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http://math.stackexchange.com/questions/341786/number-system-sum-of-two-digit-numbers | Number system - sum of two digit numbers
The sum of four two digit numbers is $221$. None of the eight digits is 0 and none of them are same. Which of the following is not included among the eight digit ?
$$(a) \;\;1 \\ (b)\;\; 2 \\ (c)\;\; 3\\ (d)\;\; 4$$
Is there any shortcut to solve this question as I got the answer which is $(d)\;\; 4$ by trial and error method. Please suggest
-
It is enough to know what is the last number modulo $9$.
Since $10a+b \equiv a+b \pmod 9$, $221 \equiv 5 \pmod 9$, and $1+2+3+4+5+6+7+8+9 = 45 \equiv 0 \pmod 9$, if the unused digit is $x$, then the sum of all the used digits is $221 \equiv 5$, and it also is $45-x \equiv -x$.
So $x \equiv -5 \equiv 4 \pmod 9$, and therefore the unused digit is $4$.
-
Hint:
$xy+ab+cd+ef=221$
$9(x+a+c+e)+y+b+d+f+(x+a+c+e)=221$
$\equiv(x+a+c+e)+y+b+d+f \equiv 5 \mod 9$
Now you know which number has to be excluded.
-
Hint $\$ Adding in the number $\rm\,0\,d\,$ formed by the missing digits $\rm\,0,d\,$ and casting nines shows
$\rm\qquad\quad mod\ 9\!:\ \ 2\!+\!2\!+\!1\!+\!d\, \equiv\, 0\!+\!1\!+\!2\!+\cdots + 9\,\equiv\, 0,\$ i.e. $\rm\ 5+d\equiv 0\ \Rightarrow\ d\equiv -5\equiv 4$
-
Suppose the numbers are $a_1b_1, \ a_2b_2, \ a_3b_3, \ a_4b_4.$
Then from $$\begin{array}{cc} a_1b_1& \\ a_2b_2& \\a_3b_3& \\ a_4b_4 &+ \\ \hline \\221 \end{array}$$ we conclude that only three possibilities can happen.
Either
$b_1+b_2+b_3+b_4=31$ and $a_1+a_2+a_3+a_4=19\,,$
$b_1+b_2+b_3+b_4=21$ and $a_1+a_2+a_3+a_4=20$ or
$b_1+b_2+b_3+b_4=11$ and $a_1+a_2+a_3+a_4=21\,.$
Since $a_1,\ a_2,\ a_3,\ a_4,\ b_1,\ b_2,\ b_3,\ b_4$ are all different and $1+2+\ldots+9=45$ we conclude that only the second possibility can happen and $4$ is not included in the digits $a_i,b_i$ ($a_1+a_2+a_3+a_4+b_1+b_2+b_3+b_4=41$).
- | 2016-07-28T08:41:58 | {
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https://math.stackexchange.com/questions/2395463/rectangle-translation-and-rotation | # Rectangle translation and rotation
I have a rectangle which is centered to $\ (1,2)$ like in the image below
rectangle first position
And I want to rotate and move the center to $\ (4,5)$ like in the image below
rectangle final position
I know that I have first to rotate it and then translate it. So it's the opposite $\ T*R$, where $\ T$ is the translation matrix and $\ R$ the rotation matrix.
Translation matrix :
\begin{bmatrix}1&0&0&3\\0&1&0&3\\0&0&1&0\\0&0&0&1\end{bmatrix}
It's the difference between final position and first position. My question is, how do I proceed from here because I am not really sure about the rotation table.
I tried to calculate the rotation table for 90 degrees.
\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}
Based on \begin{bmatrix}cos(90)&-sin(90)&0&0\\sin(90)&cos(90)&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}
And the final Result of $\ T*R$ is:
\begin{bmatrix}0&-1&0&3\\1&0&0&3\\0&0&1&0\\0&0&0&1\end{bmatrix}
How does that result correspond to the final desired position?
• The rotation matrix you have rotates around the origin. Therefore try translating the rectangle to the origin first, then rotating it, and finally translating it to the place you want. Each of those three operations has a matrix, and multiply those matrices to get a single matrix for the combined operation. BTW, I don't see why you use 4x4 matrices. Those are normally for 3d rotations. You are not using the z-axis, so you can just leave out the third column and third row of all the matrices here and use 3x3 matrices instead. – Jaap Scherphuis Aug 16 '17 at 11:29
• @JaapScherphuis So If I understood it correct, it should look like this. Forgive me that I am answering with image but it will be easier to explain if I just show you. imgur.com/oxmU8hZ – valkongr Aug 16 '17 at 12:27
• The centre of the rectangle moves from (1,2) to the origin (0,0), the rotation keeps the centre there at (0,0), and in the last step you need to move it from (0,0) to (4,5). You did the last translation by the wrong amount. Other than that it seems good. – Jaap Scherphuis Aug 16 '17 at 12:41
• Actually, you also did not multiply the first two matrices correctly - the last column is wrong. You made the same mistake in the question above. – Jaap Scherphuis Aug 16 '17 at 12:50
• Is it because of the order? It should have been R*T? – valkongr Aug 16 '17 at 12:57
As has been pointed out in the comments, the rotation matrix that you built rotates about the origin, but you need to rotate the rectangle about some other point. A systematic way to deal with this would be to first translate the center of rotation to the origin, rotate, and then translate the result into its final position.
Since you’re specifying the rectangle positions by their centers, we’ll rotate about the center of the rectangle. So, we first translate by $(-1,-2)$, rotate 90° conterclockwise, as you’ve done, and translate to the final position: $$\begin{bmatrix}1&0&4\\0&1&5\\0&0&1\end{bmatrix}\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&-1\\0&1&-2\\0&0&1\end{bmatrix}=\begin{bmatrix}0&-1&6\\1&0&4\\0&0&1\end{bmatrix}.\tag1$$ We check: $$\begin{bmatrix}0&-1&6\\1&0&4\\0&0&1\end{bmatrix}\begin{bmatrix}2\\4\\1\end{bmatrix}=\begin{bmatrix}2\\6\\1\end{bmatrix}.$$ I’ll leave the other corners for you to verify.
You can save having to do one of the matrix multiplications by skipping the first translation. Applying the rotation to the rectangle rotates it about the origin—its lower-left corner—so the necessary translation to then apply is the one that takes this new center to the target. The rotated center is $(-2,1)$, so the necessary translation is now $(4,5)-(-2,1)=(6,4)$, which, not coincidentally, is the last column of the matrix derived in (1). To see why this might be so, you can write and multiply these matrices in block form: $$\left[\begin{array}{c|c}I_2&\mathbf c_2 \\ \hline \mathbf 0^T&1 \end{array}\right]\left[\begin{array}{c|c}R & \mathbf 0 \\ \hline \mathbf 0^T&1\end{array}\right]\left[\begin{array}{c|c}I_2&-\mathbf c_1 \\ \hline \mathbf 0^T&1\end{array}\right]=\left[\begin{array}{c|c}R & \mathbf c_2-R\mathbf c_1 \\ \hline \mathbf 0^T&1 \end{array}\right]=\left[\begin{array}{c|c}I_2 & \mathbf c_2-R\mathbf c_1 \\ \hline \mathbf 0^T&1 \end{array}\right]\left[\begin{array}{c|c}R & 0 \\ \hline \mathbf 0^T&1 \end{array}\right].$$ Here, $I_2$ is the $2\times2$ identity matrix, $R$ is the $2\times2$ rotation matrix, and $\mathbf c_1$ and $\mathbf c_2$ are the respective rectangle centers. In the process, we’ve also shown that a translate-rotate-translate sequence is equivalent to a rotate-translate sequence.
Now, there’s a quicker way to build the transformation matrix that you’re looking for by using an important property of transformation matrices: their columns are the images of the basis vectors, but that’s getting ahead of where you appear to be in this subject.
Note that since you haven’t specified how the vertices of the original rectangle correspond to those of the target, there are several other rigid motions (a.k.a. isometries) that will map one onto the other. For instance, you could rotate the rectangle in the opposite direction. That will also change the translation component of the transformation. Or, you could reflect the rectangle—flip it over—in various ways to get it aligned the right way.
Such 2D operations are much easier performed with complex numbers. Those $4\times4$ matrix are overkill.
Let $p=p=x+iy$ be a point that you want to transform.
You translate the rectangle to center it at the origin with
$$p'=p-(1+i2).$$
Then you rotate it counterclockwise around the origin by multiplying by $e^{i\theta}$, equal to $i$ in your case.
$$p''=i(p-(1+i2))=ip+2-i.$$
Finally, you translate the center to the new desired location
$$p'''=ip+2-i+(4+i5)=ip+6+i4.$$
In other terms,
$$\begin{cases}x'''=-y+6,\\y'''=x+4.\end{cases}$$ | 2019-10-20T06:10:38 | {
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https://math.stackexchange.com/questions/480728/am-i-allowed-to-use-distributive-law-for-infinitely-many-sets | # Am I allowed to use distributive law for infinitely many sets?
Let $X$ be a nonempty set. A collection $S$ of subsets of $X$ is called a semiring if it satisfies the following properties:
1. The empty set belongs to S; that is $\emptyset \in S$.
2. If $A,B \in S$; then $A \cap B \in S$; that is, $S$ is closed under finite intersections.
3. The set difference of any two sets of $S$ can be written as a finite union of pair-wise disjoint members of $S$. That is, for every $A, B \in S$; there exist $C_1, ...,C_n \in S$ (depending on $A$ and $B$) such that $A\setminus B = \cup _{i=1}^n C_i$ and $C_i \cap C_j = \emptyset$ if $i\ne j$.
Now, let $S$ be a semiring of subsets of $X$. A subset $A$ of $X$ is called a $\sigma$-set with respect to $S$ (or simply a $\sigma$-set) if there exists a disjoint sequence $\{A_n\}$ of $S$ such that $A = \cup_{n=1}^\infty A_n$. One can show easily that for every sequence $\{A_n\}$ of $S$, the set $A = \cup_{n=1}^\infty A_n$ is a $\sigma$-set.
I would like to prove that finite intersections of $\sigma$-sets is a $\sigma$-set. For this purpose, suppose $A ,B$ are $\sigma$-sets then $A = \cup_{i=1}^\infty C_i$ , $B = \cup_{j=1}^\infty D_j$.
$$A\cap B=(\cup_{i=1}^\infty C_i)\cap (\cup_{j=1}^\infty D_j)$$
In this step I don't know am I allowed to use distributive law for infinitely many sets? Or the law holds only for finitely many sets?
If it holds only for finitely many sets how do I prove that finite intersections of $\sigma$-sets is a $\sigma$-set?
Thanks.
• That doesn't sound much like what a semiring usually means. Is there a connection? – hmakholm left over Monica Aug 31 '13 at 16:52
• Well, it wouldn't be the first time that the word "ring" got used for something setty instead of something algebraic. – Ben Millwood Aug 31 '13 at 17:09
• @Henning: Skip down in that article to here, and you’ll find exactly this definition of a semiring of sets. – Brian M. Scott Sep 1 '13 at 3:26
Yes, both distributive laws generalize. (You need only one of the two, but it’s useful to know both.) The first step in verifying the generalization that you want is to check that
$$\left(\bigcup_{i\in I}A_i\right)\cap D=\bigcup_{i\in I}(A_i\cap D)\;,\tag{1}$$
and to verify its mate you’ll want to check that
$$\left(\bigcap_{i\in I}A_i\right)\cup D=\bigcap_{i\in I}(A_i\cup D)\;.\tag{2}$$
Both are easily verified by element-chasing. For $(1)$, if $x\in\left(\bigcup_{i\in I}A_k\right)\cap D$, then $x\in\bigcup_{i\in I}A_i$ and $x\in D$. Since $x\in\bigcup_{i\in I}A_i$, there is an $i_0\in I$ such that $x\in A_{i_0}$, and therefore $x\in A_{i_0}\cap D\subseteq\bigcup_{i\in I}(A_i\cap D)$. Conversely, if $x\in\bigcup_{i\in I}(A_i\cap D)$, then there is an $i_0\in I$ such that $x\in A_{i_0}\cap D$. Then $x\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, and $x\in D$, so $x\in\left(\bigcup_{i\in I}A_i\right)\cap D$.
For $(2)$, if $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$, then $x\in\bigcap_{i\in I}A_i$ or $x\in D$. Let $i_0\in I$ be arbitrary. Then $A_{i_0}\supseteq\bigcap_{i\in I}A_i$, so $x\in A_{i_0}$ or $x\in D$, and therefoer $x\in A_{i_0}\cup D$. Since this holds for each $i_0\in I$, $x\in\bigcap_{i\in I}(A_i\cup D)$. Conversely, if $x\in\bigcap_{i\in I}(A_i\cup D)$, then $x\in A_i\cup D$ for each $i\in I$. If $x\in D$, then certainly $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$. If $x\notin D$, then we must have $x\in A_i$ for each $i\in I$, in which case $x\in\bigcap_{i\in I}A_i\subseteq\left(\bigcap_{i\in I}A_i\right)\cup D$.
Two applications of $(1)$ will give you the distributive law that you want. Suppose that $C=\bigcup_{i\in I}A_i$ and $D=\bigcup_{j\in J}B_j$. Then
\begin{align*} C\cap D&=\left(\bigcup_{i\in I}A_i\right)\cap D\\ &=\bigcup_{i\in I}(A_i\cap D)\\ &=\bigcup_{i\in I}\left(A_i\cap\bigcup_{j\in J}B_j\right)\\ &=\bigcup_{i\in I}\left(\bigcup_{j\in J}(A_i\cap B_j)\right)\\ &=\bigcup_{\langle i,j\rangle\in I\times J}(A_i\cap B_j)\;. \end{align*}
In other words, $C\cap D$ is the union of all possible intersections of the form $A_i\cap B_j$. In particular, if $I$ and $J$ are countable index sets, $I\times J$ is also countable.
Similarly, two applications of $(2)$ will give you the other general distributive law of this kind. This time suppose that $C=\bigcap_{i\in I}A_i$ and $D=\bigcap_{j\in J}B_j$. Then
\begin{align*} C\cup D&=\left(\bigcap_{i\in I}A_i\right)\cup D\\ &=\bigcap_{i\in I}(A_i\cup D)\\ &=\bigcap_{i\in I}\left(A_i\cup\bigcap_{j\in J}B_j\right)\\ &=\bigcap_{i\in I}\left(\bigcap_{j\in J}(A_i\cup B_j)\right)\\ &=\bigcap_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)\; \end{align*}
the intersection of all possible unions of the form $A_i\cup B_j$ as $i$ and $j$ run over their respective index sets.
• "[...] element-chasing". What a lovely piece of terminology. It told me exactly how to carry out the proof with no need for explanation or formalization (I supplied the formality myself). – Jonas Kölker Mar 24 '19 at 7:42
Two sets $X$ and $Y$ are equal exactly when, for all $x$, $x \in X \iff x \in Y$.
Now, take $x \in (\cup_i C_i) \cap (\cup_j C_j)$.
This means precisely that $x$ is in $\cup_i C_i$ and also is in $\cup_j D_j$.
A statement of the form $x \in \cup_k S_k$ means precisely "there exists some $k$ such that $x \in S_k$". So we have from $x \in \cup_i C_i$ and $x \in \cup_j D_j$ that there exist $i$ and $j$ such that $x \in C_i$ and $x \in D_j$.
Or in other words, there exist $i$ and $j$ such that $x \in C_i \cap D_j$.
Or in other words, $x \in \cup_i \cup_j (C_i \cap D_j)$.
So, yes, the distributive law works for infinite, even uncountable families of sets. | 2021-05-08T14:36:37 | {
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https://math.stackexchange.com/questions/1257540/meaning-of-ds-in-flux-integrals | # Meaning of “dS” in flux integrals?
In a general flux integral of a level surface, of the form
$$\iint{\mathbf{F}\bullet d\mathbf{S}}$$
what exactly does $d\mathbf{S}$ represent? I have seen both
$d\mathbf{S} = \mathbf{\hat N}dS = \pm (\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv$
(for parametric surfaces), and
$d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla G(x,y,z)}{G3(x,y,z)}dxdy$
for level surfaces. At a glance, I feel like I get them, but whenever I sit down to actually solve any problems I get confused about what exactly it represents in the integrals. Finding the normal is usually not a problem, nor is calculating $\frac{\nabla G}{G3}$, but then I get stuck on what to put for dS. Like the following example, in calculating the flux of $\mathbf{F} = xi + zj$ out of the surface x+2y+3z = 6. The textbook calculates $\mathbf{\hat N}$ to be $\frac{i+2j+3k}{\sqrt{14}}$ (and I agree), but then it goes on to calculate $dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz$ . I'm not entirely sure why they did that, or why they set it up the way they did. How do you find/choose dS and what does it mean to the integral?
Thanks!
• $\vec{n}$ is just the normal vector at the surface element dS. – tired Apr 29 '15 at 13:57
• It seems you are confused about $dS$, not $d\bf S$, is that right? – Samuel Apr 29 '15 at 13:59
• @Samuel Yes, I guess so! I mean the other terms can be computed in a straight forward fashion, but the role of dS is a bit more foggy. – Laplacinator Apr 29 '15 at 16:27
$dS$ is a surface element, a differential sized part of the surface $S$. It is usually oriented, positive if its normal $n$ is outward pointing (e.g. if $S$ is the boundary of a volume). $$dS = n \lVert dS \rVert$$
I have seen both
$$d\mathbf{S} = \mathbf{\hat N}dS = \pm (\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv$$
(for parametric surfaces), and $$d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla G(x,y,z)}{G3(x,y,z)}dxdy$$
for level surfaces.
For those examples $\lVert dS \rVert = du \, dv$ and $\lVert dS \rVert = dx \, dy$. The other parts are the more or less complicated normal vectors of those surface elements.
$$dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz$$
The integration is along the $x$-$z$ plane, while the surface, $$S: x+2y+3z = 6 \quad n = (1,2,3)^t/\sqrt{14}$$ which is a plane as well, is not parallel to the $x$-$z$ plane.
The area of the projection $P_y S$ has to be adjusted, to give the correct area $\lVert S \rVert$ for $S$. We want $$\lVert S \rVert = \int\limits_S \lVert dS \rVert = \int\limits_S \lVert n\,du\,dv \rVert = f \lVert P_y S \rVert = f \int\limits_{P_y S} \lVert dx \, dz \rVert$$ In your example they simply take $f = 1/\lVert n \cdot e_y\rVert$.
Let us check this: First we look for unit vectors $u$ and $v$ orthogonal to $n$ and each other. $$0 = n \cdot a = (1, 2, 3)^t / \sqrt{14} \cdot (2, -1, 0)^t \quad e_u = (2, -1, 0)^t / \sqrt{5} \\ e_v = n \times e_u = (3, 6, -5)^t / \sqrt{70}$$ These are unit vectors, so the area of the square between $e_u$ and $e_v$ is 1. Now these unit vectors have the projections on the $x$-$z$ plane: $$u_p = P_y e_u = (2, 0, 0)^t/\sqrt{5} \quad \lVert u_p \rVert = 2/\sqrt{5} \\ v_p = P_y e_v = (3, 0, -5)^t/\sqrt{70} \quad \lVert v_p \rVert = \sqrt{34/70} = \sqrt{17/35} \\$$ where $P_y a = a - (a\cdot e_y) e_y$ for a vector $a$. The area of the projection is $$\lVert u_p \times v_p \rVert = \lVert ((2, 0, 0)^t/\sqrt{5}) \times ((3, 0, -5)^t/\sqrt{70}) \rVert = \lVert (0, 10, 0)^t/\sqrt{350} \rVert = 2 /\sqrt{14}$$
This should explain the factor $\sqrt{14}/2$.
What is missing is a derivation for the shorter $$\lVert P_y u \times P_y v \rVert = \lVert n \cdot e_y \rVert \, \lVert u \rVert \, \lVert v \rVert$$
• Thanks! That really helps clear some of it up. I'm gonna read through that part on adjusting area elements that aren't parallell to any axis some more , that seems like a very important point. Follow up question, would that mean that the area element of a plane parallell to the xy-plane would always be dxdy? Is the $\frac{\sqrt{14}}{2}$ in fact just "scaling" the differential square to "fit" our particular plane? – Laplacinator Apr 29 '15 at 16:22
• Exactly. The area of the projection of the surface $du dv$ on the $x$-$z$ plane is smaller than the area of the original $du dv$ , because its normal $n$ is not normal to that plane. So it must be upscaled. If the surface is part of a parallel plane, we do not have to do this, as surface element and its projection are same sized. – mvw Apr 29 '15 at 16:53
• @mvw Would you care to explain what $\pm(\mathbf{\frac {n}{|n|}})(\mathbf{|n|})$ represents exactly? – Demosthene Apr 29 '15 at 18:26
• @Demosthene From the looks it is about normal vectors $n$ which are not unit vectors, decomposing them into a product $\lVert n \rVert e_n$, where $e_n$ is the unit vector with the same direction as $n$. I would need to see an example to understand more, esp. why the signs show up that way. – mvw Apr 29 '15 at 18:33
• @mvw Yes, I am familiar with the notations $\mathbf{\frac{n}{|n|}}$ or $\hat{n}$ to denote a normalized vector. Following that logic, I don't see the point of writing $\mathbf{\frac{n}{|n|}|n|}$. Isn't that just $\mathbf{n}$? – Demosthene Apr 29 '15 at 19:34
$d\mathbf S = \hat {\mathbf N}dS$ -- where $\hat {\mathbf N}$ is the unit normal (outward if closed, otherwise you have to choose an orientation) to the surface and $dS$ is the differential area element -- is the definition. But you don't really need to worry about it. You should always parametrize your surface first by some $\mathbf r(s,t)$, $s_0 \le s \le s_1$ and $t_0 \le t\le t_1$. Then $$\iint_\Gamma \mathbf f \cdot d\mathbf S = \int_{t_0}^{t_1} \int_{s_0}^{s_1} \mathbf f(\mathbf r(s,t)) \cdot \left(\frac {\partial \mathbf r}{\partial s} \times \frac {\partial \mathbf r}{\partial t}\right)\,ds\,dt$$
Each of those terms on the right you should then be able to calculate.
• I see! Thanks a lot, but is it really necessary to always parametrize a surface first? – Laplacinator Apr 29 '15 at 16:26
• Yes -- but sometimes you may not realize you're doing it. Parametrizing a surface just means expressing it as some vector equations or set of scalar equations in $2$ variables. If there is an immediate way to write down the surface in terms of easy variables like $x,y$ or $r, \theta$, you may not realize that the $\hat {\mathbf N}\,dx\,dy$ that you immediately wrote down is really just $\left(\frac {\partial \mathbf r}{\partial x} \times \frac {\partial \mathbf r}{\partial y}\right)\,dx\,dy$ or that your $\mathbf f(x,y)$ is already in terms of your parameters (variables). – user137731 Apr 29 '15 at 17:21 | 2019-08-21T22:06:59 | {
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https://stats.stackexchange.com/questions/6945/minimum-tickets-required-for-specified-probability-of-winning-lottery | # Minimum tickets required for specified probability of winning lottery
In a lottery, 1/10 of the 50 000 000 tickets give a prize.
What is the minimum amount of tickets one should buy to have at least a 50% chance to win?
Would be very glad if you could explain your methodology when resolving this. Please consider this as homework if you will.
## 1 Answer
I should really ask what your thoughts are so far on this. This problem is very closely related to the "birthday problem". The easiest way to do these problems is to count the possible ways of either winning or losing. Usually one is much easier to do than the other, so the key is to find the right one. Before we get into the actual calculation, let's start with some heuristics.
Heuristics: Let $n$ be the total number of tickets and $m$ be the number of winning tickets. In this case $n = 50\,000\,000$ and $m = 5\,000\,000$. When $n$ is very large then purchasing multiple distinguishable tickets is almost the same as sampling with replacement from the population of tickets.
Let's suppose that, instead of having to purchase $k$ separate tickets, we purchased a ticket, looked to see if it was a winner and then returned it to the lottery. We then repeat this procedure where each such draw is independent from all of the previous ones. Then the probability of winning after purchasing $k$ tickets is just $$\Pr( \text{we won} \mid \text{purchased k tickets} ) = 1 - \left(\frac{n-m}{n}\right)^k .$$
For our case then, the right-hand side is $1 - (9/10)^k$ and so we set this equal to $1/2$ and solve for $k$ in order to get the number of tickets.
But, we're actually sampling without replacement. Below, we'll go through the development, with the point being that the heuristics above are more than good enough for the present problem and many similar ones.
There are $50\,000\,000$ tickets. Of these $5\,000\,000$ are winning ones and $45\,000\,000$ are losing ones. We seek
$$\Pr(\text{we win}) \geq 1/2 \>,$$
or, equivalently,
$$\Pr(\text{we lose}) \leq 1/2 .$$
The probability that we lose is simply the probability that we hold none of the winning tickets. Let $k$ be the number of tickets we purchase. If $k = 1$, then $\Pr(\text{we lose}) = 45\,000\,000 / 50\,000\,000 = 9/10 \geq 1/2$, so that won't do. If we choose two tickets then there are $45\,000\,000 \cdot 44\,999\,999$ ways to choose two losing tickets and there are $50\,000\,000 \cdot 49\,999\,999$ ways to choose any two tickets. So,
$$\Pr( \text{we lose} ) = \frac{45\,000\,000 \cdot 44\,999\,999}{50\,000\,000 \cdot 49\,999\,999} \approx 0.9^2 = 0.81.$$
Let's generalize now. Let $m$ be the number of winning tickets and $n$ the total number of tickets, as before. Then, if we purchase $k$ tickets,
$$\Pr(\text{we lose}) = \frac{(n-m) (n-m-1) \cdots (n-m-k+1)}{n (n-1) \cdots (n-k+1)} .$$
It would be tedious, but we can now just start plugging in values for $k$ until we get the probability below $1/2$. We can use a "trick", though, to get close to the answer, especially when $n$ is very large and $m$ is relatively small with respect to $n$.
Note that $\frac{n-m-k}{n-k} \leq \frac{n-m}{n}$ for all $k < m < n$. Hence
$$\Pr(\text{we lose}) = \frac{(n-m) (n-m-1) \cdots (n-m-k+1)}{n (n-1) \cdots (n-k+1)} \leq \left(1 - \frac{m}{n} \right)^k ,$$
and so we need to solve the equation $\left(1 - \frac{m}{n} \right)^k = \frac{1}{2}$ for $k$. But,
$$k = \frac{\log \frac{1}{2}}{\log (1 - \frac{m}{n})} \>,$$
and so when $m/n = 1/10$, we get $k \approx 6.58$.
So, $k = 7$ tickets ought to do the trick. If you plug it in to the exact equation above, you'll get that
$$\Pr( \text{we win} \mid \text{k=7} ) \approx 52.2\%$$
• Very detailed. I loved it. Thank you and I'll consider it as an answer unless someone comes up with something else. – Queops Feb 7 '11 at 2:58 | 2019-11-19T17:37:14 | {
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https://math.stackexchange.com/questions/526657/show-convexity-of-the-quadratic-function | # Show convexity of the quadratic function
Can someone show this function is convex using the definition (without taking gradient)? $$F(x) = x^TAx + b^Tx + c$$
where $A$ is symmetric positive semi-definite.
By definition of convex, for any $x,y\in\mathbb R$, we have $$f(\frac{x+y}2)\leq\frac12(f(x)+f(y))$$ Thus it is sufficient to reduce and prove that $$\frac12(x+y)^TA(x+y)\leq x^TAx+y^TAy\\ x^TAy+y^TAx\leq x^TAx+y^TAy$$ Namely $$(x-y)^TA(x-y)\geq0$$ which is directly followed by positive semi-definite.
• How would this proof change if A were positive definite and not semi-definite? – user2553807 Feb 10 '14 at 22:21
• Nothing except the condition when equality holds. – Shuchang Feb 11 '14 at 6:52
• how do you expand $(x+y)^TA(x+y)$ to the next formula? – Dzung Nguyen Jul 23 '14 at 23:07
• @DzungNguyen According to associative law and distributive law. $(x+y)^TA(x+y)=(x+y)^TAx+(x+y)^TAy$, and similarly for the left $(x+y)^T$. – Shuchang Jul 24 '14 at 2:58
• The first three inequalities should be inverse. – Khue Feb 13 '16 at 5:52
Just to leave the answer for the general case online for future reference. A function is convex if $f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda) f(y)$ for all $\lambda\in[0,\;1]$.
It suffices to show for a quadratic function $f(x) = x^TQx$. Therefore using the definition of a convex function:
\begin{align} (\lambda x + (1-\lambda) y)^TQ(\lambda x + (1-\lambda) y)\leq \lambda x^TQx + (1-\lambda)y^TQy \end{align} Equality holds for $\lambda = 0\;\text{or}\;1$. Therefore consider $\lambda\in(0,1)$. The left hand side simplifies to: \begin{align} \lambda^2x^TQx + (1-\lambda)^2y^TQy + \lambda(1-\lambda)x^TQy + \lambda(1-\lambda)y^TQx\leq \lambda x^TQx + (1-\lambda)y^TQy \end{align} Rearranging the terms and simplifying one obtains: \begin{align} & \lambda(1-\lambda)x^TQx + \lambda(1-\lambda)y^TQy - \lambda(1-\lambda)x^TQy -\lambda(1-\lambda)y^TQx\geq 0 \\ & \Rightarrow x^TQx + y^TQy -x^TQy-y^TQx \geq 0 \\ & \Rightarrow (x-y)^TQ(x-y) \geq 0 \end{align} which is true for positive semi-definite $Q\succeq 0$. | 2021-03-09T04:12:16 | {
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https://math.stackexchange.com/questions/1879368/finding-complex-eigenvalues-and-its-corresponding-eigenvectors | Finding complex eigenvalues and its corresponding eigenvectors
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$ has complex eigenvalues $\lambda_{1,2} = a \pm bi$ where $a =$____ and $b =$ ____. The corresponding eigenvectors are $v_{1,2} = c \pm di$ where $c =$ (____ , _____ ) and $d =$ (____ , ___)
So I got the char. poly. eqn $\lambda^2 + 12\lambda + 45 = 0$
Then using the quad. eqn I got $-6 \pm 3i$ which I know is in the form $a \pm bi$ so I was thinking $a = -6$, $b = 3$, but instead they have $b = -3$ , why?
Also, I'm not sure how to obtain the corresponding eigenvectors because we are working with complex eigenvalues now.
I'll cover the how to find the eigenvectors part.
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$
has eigenvalues $\lambda_{1,2} = -6\pm 3i$.
Now to find the associated eigenvectors, we find the nullspace of $$A-\lambda_{1,2}I = \begin{bmatrix}3-(-6\pm 3i)&6\\-15&-15-(-6\pm 3i)\end{bmatrix} = \begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix}$$
To find the nullspace I'll put this in REF:
\begin{align}\begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix} &\sim \begin{bmatrix} 5 & 3\pm i \\ 3\mp i & 2\end{bmatrix} \\ &\sim \begin{bmatrix} 5 & 3\pm i \\ 0 & 2-(3\pm i)\frac{-(3\mp i)}{5}\end{bmatrix} \\ &= \begin{bmatrix} 5 & 3\pm i \\ 0 & 0\end{bmatrix}\end{align}
Therefore all of the eigenvectors associated with $\lambda_{1,2}$ are of the form $w_{1,2} = \begin{bmatrix}\frac 15(-3\mp i)t \\ t\end{bmatrix}$. Representative eigenvectors are then $$\bbox[5px,border:2px solid red]{v_{1,2} = \begin{bmatrix}-3\mp i \\ 5\end{bmatrix}}$$ which is obtained by setting $t=5$.
• Why is your $\pm$ signed turned upside down? What's the significance – Yusha Aug 2 '16 at 21:20
• It means that $\lambda_1 = -6\oplus 3i$ corresponds to $v_1 = \begin{bmatrix} -3 \ominus i \\ 5\end{bmatrix}$, which I've circled the relevant signs. Likewise $\lambda_2$ and $v_2$ have opposite signs in those places. – user137731 Aug 2 '16 at 21:21
• Somehow they are getting $(-1,1)$ and $(-1,2)$ I cant figure out how – Yusha Aug 2 '16 at 21:25
• Weird, maybe its an error – Yusha Aug 2 '16 at 21:30
• @Yusha Remember that there are an infinite number of vectors in the nullspace of $A-\lambda_{1,2}$. I chose $t=5$ arbitrarily. If you choose $t=1-2i$ you'll get your answer key's solution. As to the reason your book arrived at the answer it did, I'll bet they didn't do the row swap that I did in the first step of my row reduction. So both my and your answer key's solution are correct. – user137731 Aug 2 '16 at 21:48
Note that $\pm (-3)=\mp3=\pm3$. Hence, whether you use $b=3$ or $b=-3$ is irrelevant. To find the corresponding eigenvectors, you need to solve the following:
$$\begin{bmatrix}3&6\\-15&-15\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\lambda_{1,2}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}.$$
• I get stuck after I subtract the lambda value of 3 and I get $\begin{bmatrix}9-3i & 6\\-15 & -9-3i\end{bmatrix}$ – Yusha Aug 2 '16 at 20:49
• Should I put the matrix in RREF? – Yusha Aug 2 '16 at 20:51
• Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... – Yusha Aug 2 '16 at 20:57
• I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? – Yusha Aug 2 '16 at 21:06
The if $b=-3$, $a=-6$, we have $a \pm bi=-6 \pm 3i$, if $b=3, a=-6$ we still have $a\pm bi=-6 \pm 3i$, both answers are correct but if you're submitting into an online homework service like cengage-brain or webassign your professor may have just forgotten to include both cases as acceptable. To find eigenvectors, you need to solve $Av=\lambda v$, or, equivalently $(A-\lambda I)v=0$ (where I denotes the identity matrix. Why are these two systems equivalent?). Nothing really changes with the complex case so long as you know how to do arithmetic in $\mathbb{C}$. To get you started, for an eigenvector corresponding to $\lambda=-6+3i$, we set up: $$\begin{bmatrix} A-\lambda I \end{bmatrix} v= \begin{bmatrix} 9-3i & 6 \\ -15 & -9-3i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} =0$$ Giving the system: $$\begin{array} ((9-3i)v_1+6v_2=0 \\ -15v_1+(-9-3i)v_2=0 \end{array}$$ Can you take it from here?
• $v_1 = -6v_2/(9-3i)$, $v_2 = 15v_1/(-9-3i)$? – Yusha Aug 2 '16 at 20:41
• I don't get it, don't i need to put the matrix in RREF form? – Yusha Aug 2 '16 at 20:47
• Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... – Yusha Aug 2 '16 at 20:57
• I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? – Yusha Aug 2 '16 at 21:06
• Yes, there will be infinitely many eigenvectors, but they will all be scalar multiples of one another, your first comment has you basically done, fix $v_1=1$ and solve for $v_2$ in terms of $v_1$, that will give you one eigenvector. Since matrix multiplication is linear, obvious any scalar multiple of this eigenvector will also be an eigenvector since $Av=\lambda v$ implies $A(cv)=c(Av)=c(\lambda v)=\lambda (cv)$. – mb- Aug 3 '16 at 2:28 | 2019-10-17T06:23:37 | {
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https://www.doubtnut.com/question-answer/prove-that-tan-11-3-tan-11-7-tan-11-13-tan-1-1-n2-n-1-oo-pi-4-644006365 | # Prove that tan^(- 1)(1/3)+tan^(- 1)(1/7)+tan^(- 1)(1/13)+..........+tan^-1 (1/(n^2+n+1))+......oo =pi/4
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Solution :
Let P(n):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+…..+tan^(-1)((1)/(n^+n+1))=tan^(-1)((n)/(n+2)) ……..(i) <br> Step I For n=1 <br> LHS of Eq. (i) =tan^(-1)((1)/(3)) =tan^(-1)((1)/(1+2))=RHS of Eq. (i) <br> Therefore , P(1) is true . <br> Step II Assume that P(k) is true. Then , <br> P(k):tan^(-1)((1)/(3))+tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+........+tan^(-1)((1)/(k^2+k+1))=tan^(-1)((k)/(k+2)) <br> Step III For n=k+1 <br> P(k+1):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))=tan^(-1)((k+1)/(k+3))......(ii) <br> LHS of Eq. (ii) <br> =tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1)) <br> =tan^(-1)((k+1)/(k+2))+tan^(-1)((1)/((k+1)^2+(k+1)+1)) [by aaumption atep] <br> =tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(k^2+3k+3)) <br> =tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(1+(k+1)(k+2))) <br> =tan^(-1)(((k+1)-1)/(1+(k+1).1))+tan^(-1)(((k+2)-(k+1))/(1+(k+2)(k+1))) <br> =tan^(-1)(k+1)-tan^(-1)1+tan^(-1)(k+2)-tan^(-1)(k+1)=tan^(-1)(k+2)-tan^(-1)1 <br> =tan^(-1)((k+2-1)/(1+(k+2).1))=tan^(-1)((k+1)/(k+3))= RHS of Eq. (ii) <br> This shows that the result is true for n=k+1. Hence. by the principle of mathematical induction . the result is true for all n in N`.
Transcript
TimeTranscript
00:00 - 00:59Pawan let's start with the question the questions that prove that tan inverse of 1 by 3 + tan inverse of 1 by 7 + tan inverse of 1 by 13 + 1 or 2 tan inverse of 1 by X square + 1 + 1 + 1 upto infinity is equal to pi by 4 first let's look at the lhs for lhs term are that is TR are term will be equal to tan inverse of 1 upon our square + R plus one record that from this term over the top of this sequence will be equal to the completeness tan inverse of numerator candidness h + 1 - are upon 1 +
01:00 - 01:59into our + 1 + 1 + 1 know this is what I got an inverse of now using this formula is right the form no you are an inverse of a minus tan inverse of b is equal to tan inverse of a minus b upon 1 + ab compare this to this will get that the 4th of R is equal to now here is our + 1 and become an into this that will be tan inverse of h + 1 minus tan inverse of are now are alleges in the question is will become therefore lhs is equal to summation of are equal to 12
02:00 - 02:59PR now let's write TR as we found out over here so this will be equal to summation of are equal to 12 Infinity of tan inverse of h + 1 minus tan inverse of our ok so this should be equal to the values of r from 1 to infinity oh yeah this will be tan inverse of 2 - 10 inverse of 1 + report to the Sonu Nigam tan inverse of 3 minus tan inverse of 2 + 1 bracket when we put 3 it will become tan inverse of minus tan inverse of 3 plus show on this will keep going up to infinity
03:00 - 03:59summation of to infinity sofa Infinity it will be tan inverse of infinity plus one minus tan inverse of infinity sunao if you look Oh Yeah Oh Yeah minus tan inverse of 2 will get cat cancel with tan inverse of 2 then minus tan inverse of 3 will get cancelled with + tan inverse of 3 keep on going the cancellation will keep on work on your minus tan inverse of 48 cancelled with this tan inverse of 4 then it will get cancer so what will be left over your will be equal to b minus tan inverse of 1 + tan inverse Infinity plus one not know the value of tan inverse of 1 so this will be equal to minus 5 by 4 then value of tan
04:00 - 04:595 by 4 then tan inverse of infinity + 1 + 1 is still Infinity infinity is pi by 2 is equal to pi by 4 pi by 2 minus 5 by 4 is pi by 4 sunao this is Rs so we can write therefore lhs is equal to RHS sunao since we got lhs is equal to RHS we can finally concluded your therefore tan inverse of 1 upon 3 + tan inverse of 1 upon 7 plus show on up to 10 inverse of 1 upon X square + 1 + 1 + 1 upto infinity is equal to pi by 4 this is what I had to prove so now we can find the right to year hence proved | 2021-12-07T06:07:01 | {
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https://math.stackexchange.com/questions/2964611/if-a-n-is-a-sequence-in-which-a-n-geq-c-for-some-constant-c-and-a-n | # If $\{a_n\}$ is a sequence in which $a_n \geq c$ for some constant $c$ and $a_n \rightarrow a$ then $a \geq c$
If $$\{a_n\}$$ is a sequence in which $$a_n \geq c$$ for some constant $$c$$ and $$a_n \rightarrow a$$ then $$a \geq c$$
I just wanted some feedback on whether my proof of the claim is sound.
Proof
Let $$\epsilon > 0$$ and $$a < c$$. Now $$a_n \rightarrow a$$ means:
$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$
Consider $$\epsilon = \frac{c-a}{2}$$
$$\Rightarrow c \leq a_n \leq a + \frac{c-a}{2} = \frac{a+c}{2} < \frac{c + c}{2} = c$$
$$c < c$$ is a contradiction. Therefore $$a \geq c$$ for the statement to hold.
• Your proof is good but as an aside I have to wonder about "$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$". This is maybe an editorial comment, but nobody likes to read such symbol soup an it doesn't make math more "serious". Also what you wrote doesn't parse. $|a_n-a|<\epsilon\iff a-\epsilon\le a_n\le a+\epsilon$ is always true for all $n$, $a_n$, $a$ and $\epsilon$. Does that mean all sequences converge to all values? – fleablood Oct 21 '18 at 15:14
• Funny you mention that. I was talking to my professor about this same thing this week because I'm not a big fan of the "symbol soup" as you call it and was inquiring whether just communicating my ideas in plain English would be better. – dc3rd Oct 21 '18 at 15:18
The idea is fine but:
• You should not begin with “Let $$\varepsilon>0$$”. You fix $$\varepsilon$$ later, not at this point.
• You should write that you are assuming that $$a, in order to get a contradiction. You can't just say “Let […] $$a”, because $$a$$ and $$c$$ are fixed from the start.
• Should I state it in terms of: "assume $a < c$ towards a contradiction....."? And with regards to $\epsilon$ don't I have to declare at the start that it will always be positive? – dc3rd Oct 21 '18 at 15:02
• This is a matter of taste, but I would begin with “In order to get a contradiction, let us assume that $a<c$”. Concerning the other question, note that you wrote right after $\forall\varepsilon>0$. So, no, you don't have to write that. – José Carlos Santos Oct 21 '18 at 15:04
• Gracias por la ayuda. – dc3rd Oct 21 '18 at 15:07
Your proof seems legitimate to me, but it is a proof by contradiction, so here is an alternate way to do it directly. I will start with what you have.
$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$
Now, this implies that for any $$\epsilon > 0$$, $$a_n \leq a+\epsilon$$ for some $$n \in \Bbb{N}$$. Also, $$c \leq a_n$$, so we get $$c\leq a+\epsilon$$ for any $$\epsilon > 0$$. Thus, $$c\leq g$$ for all real numbers $$g$$ in the interval $$(a, \infty)$$. It is well known that $$a$$ is the greatest lower bound for $$(a, \infty)$$, so this implies $$c \leq a$$. | 2019-07-21T02:58:19 | {
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https://math.stackexchange.com/questions/1820408/why-can-no-prime-number-appear-as-the-length-of-a-hypotenuse-in-more-than-one-py/1820500 | Why can no prime number appear as the length of a hypotenuse in more than one Pythagorean triangle?
Why is it that no prime number can appear as the length of a hypotenuse in more than one Pythagorean triangle? In other words, could any of you give me a algebraic proof for the following?
Given prime number $p$, and Pythagorean triples $(a,b,p)$ and $(c,d,p)$ where $a<b<p$ and $c<d<p$, then $b=d$.
Please also have a look at the deeper question: Is there any formula to calculate the number of different Pythagorean triangle with a hypotenuse length $n$, using its prime decomposition?
• The fact that such a prime $p$ must satisfy $$p\equiv 1\pmod{4}$$ may be helpful. – Zubin Mukerjee Jun 10 '16 at 0:44
• @ZubinMukerjee no... since we already know that... a prime must be of the form of $p=4k+1$ to appear as a hypotenuse... but the question is why this number appear just in one Pythagorean triple? – Omid Ghayour Jun 10 '16 at 0:47
• look... $25$ appears in two Pythagorean triple $(15,20,25)$ and $(7,24,25)$ but $29$ appear in just one triple... – Omid Ghayour Jun 10 '16 at 0:50
• An elementary detailed proof is rather lengthy. Please see the uniqueness part of this proof. The proof is more pleasant and more informative if we can use properties of Gaussian integers. – André Nicolas Jun 10 '16 at 0:54
• We give a brief version of the Gaussian integer approach. By the usual representation theorem, the odd prime $p$ is a hypotenuse iff $p=s^2+t^2=(s+ti)(s-ti)$. The two factors are Gaussian primes, so by unique factorization $(u+vi)(u-vi)=p$ only if $u+iv$ is a unit times $s\pm ti$. – André Nicolas Jun 10 '16 at 1:18
This goes back to Euler, who showed that if there are two ways of writing an odd integer $N$ as the sum of two squares, then $N$ is composite. There is a 2009 article on this by Brillhart. Let me try to find a link.
http://www.maa.org/press/periodicals/american-mathematical-monthly/american-mathematical-monthly-december-2009
And if one note that in a primitive triple the hypotenuse is of the form $(u^2+v^2)$, and the legs are of the form $(u^2-v^2)$ and $(2uv)$. So by euler if the hypotenuse is prime it couldn't be written in different ways.
• – lhf Jun 10 '16 at 1:00
• Here is a better scan, from JSTOR: i.stack.imgur.com/DDymG.gif – lhf Jun 10 '16 at 1:02
• @lhf good. I made a jpeg of the relevant page and pasted that in. He also has a 2016 article that allows indefinite quadratic forms, although in that case more care is needed about what is meant by distinct representations. – Will Jagy Jun 10 '16 at 1:04
• when a number is hypotenuse... its square is going to written as sum of squares... so it is of course composite then... it is $p^2$ not $p$... – Omid Ghayour Jun 10 '16 at 1:06
• @WillJagy oh... yes... if it's prime... it's primitive... ;) ☺ – Omid Ghayour Jun 10 '16 at 1:09
As noted in the comments and the accepted answer, this comes down to the fact that if a prime $p$ can be written as a sum of two squares, then the representation is unique up to switching and or negating the factors. A fancier explanation for this is the fact that $\mathbb Z[i]$ is a principal ideal domain and its unit group is $\{\pm1,\pm i\}$. (Of course, proving that $\mathbb Z[i]$ is a PID requires some sort of argument like that in the scanned note, but this is a more modern way to think about it.) Once one knows it's a PID, then suppose that $p=u^2+v^2$. Then $p=(u+iv)(u-iv)$, and the fact that $u+iv$ and $u-iv$ have norm $p$ shows that they cannot factor further in $\mathbb Z[i]$. Hence they are irreducible (i.e., they generate prime ideals). So the unique factorization of the ideal $p\mathbb Z[i]$ is as the product of the prime ideals $(u+iv)\mathbb Z[i]$ and $(u-iv)\mathbb Z[i]$. So $u$ and $v$ are unique, up to switching them or replacing them by their negatives, which corresponds to multiplying $u+iv$ by each of the four units in $\mathbb Z[i]$.
• very interesting! – Vincent Jun 14 '16 at 18:40
If a Pythagorean triple is primitive, $$B+C$$ is a perfect square as shown by
$$2mn+(m^2+n^2)\quad=\quad m^2+2mn+nn^2\quad=\quad(m+n)^2$$
If $$C$$ is prime, then only one smaller value $$(B)$$ can add to it to make a perfect square. Given $$C\&B$$, there can be only a one $$A$$ to make a Pythagorean triple. | 2020-08-14T09:06:30 | {
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https://mathoverflow.net/questions/308511/prove-int-0-infty-frac1esx-sqrt1s2ds-arctan-left-frac1x-ri | # Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$
The question is to prove: $$\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1.$$ Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed.
I also wonder if there is a systematic approach to solve this kind of problem.
Here is a proof of the inequality for $x\geq 2$. For the remaining range, see the Added section below.
Let $\lambda:=1-1/\sqrt{2}$, then by convexity we have $$\frac{1}{\sqrt{1+s^2}}\leq 1-\lambda s^2,\qquad 0\leq s\leq 1.$$ Using this bound we can estimate \begin{align*}\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}\,ds&<\int_0^1\frac{1-\lambda s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1-\lambda}{e^{sx}}\,ds\\[6pt] &=\frac{1}{x}-\lambda\left(\int_0^1\frac{s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1}{e^{sx}}\,ds\right)\\[6pt] &=\frac{1}{x}-2\lambda\frac{1-e^{-x}x-e^{-x}}{x^3} .\end{align*} On the other hand, for $x>0$ we also have $$\arctan\left(\frac{1}{x}\right)=\int_0^{1/x}\frac{1}{1+s^2}\,ds>\int_0^{1/x}(1-s^2)\,ds=\frac{1}{x}-\frac{1}{3x^3},$$ hence it suffices to verify that $$\lambda(1-e^{-x}x-e^{-x})>\frac{1}{6},\qquad x\geq 2.$$ This is straightforward, so we proved the original inequality for $x\geq 2$.
Added. One can cover the remaining range $1\leq x<2$ as follows. Let $f(x)$ denote the LHS and $g(x)$ denote the RHS in the original inequality. These two functions are decreasing, hence it suffices to verify the following $20$ numeric inequalities: $$f(1+(n-1)/20)<g(1+n/20),\qquad n=1,2,\dots,20.$$ These are likely to be true, e.g. I checked them with Mathematica's NIntegrate command.
• Why is 20 enough? – Steven Gubkin Aug 17 '18 at 17:03
• @StevenGubkin: What I did in the Added section is that I decomposed the interval $[1,2]$ into $20$ equal subintervals such that on each of the subintervals, the maximum of $f$ is less than the minimum of $g$. So I proved a bit more than $f(x)<g(x)$ on $[1,2]$, and the advantage is that I only need to check finitely many numeric inequalities. The number $20$ has no significance except that it makes the idea work. For example, $15$ is not ok, because then on the first subinterval, the maximum of $f$ is greater than the minimum of $g$. – GH from MO Aug 17 '18 at 19:56
• Sorry for my late confirmation. Your method is surprisingly skillful, but at the same time clear. Actually, even if we prove only for x> 2, there is no big problem in the original problem. But as you added, the idea of proving the inequality by dividing the interval is also very impressive. Thanks for the great intuition and explanation. – Ramanasa Aug 20 '18 at 17:35
• @Ramanasa: Thanks for your feedback and kind words. I am glad I could help! – GH from MO Aug 20 '18 at 17:40
The inequality holds by direct calculation for $x$ close to unity, when the two sides are well separated, but for large $x$ the two sides approach each other and we need to check that the inequality is not violated.
I denote $x\equiv 1/\epsilon$, and seek to prove that $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds < \arctan\epsilon,\;\;0<\epsilon\leq 1.$$ Series expansion of both sides gives $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds=\epsilon-\epsilon^3+{\cal O}(\epsilon^5),$$ $$\arctan\epsilon=\epsilon-\tfrac{1}{3}\epsilon^3+{\cal O}(\epsilon^5).$$ This proves the inequality for $\epsilon$ close to $0$, hence for large $x$.
• Well, one needs to be somewhat careful as the power series expansion of $1/\sqrt{1+\epsilon^2 s^2}$ only converges for $s<1/\epsilon$. – GH from MO Aug 17 '18 at 13:14
• I have developed your idea and made it concrete as follows. The inequality $$\frac1{\sqrt{1+t^2}} \le 1- \frac{1}{2}t^2 + \frac38 t^4$$ holds for all $t$. Using this the LHS is $$\epsilon\int_0^\infty \frac1{e^s\sqrt{1+s^2 x^2}}ds<\epsilon \int_0^\infty \frac1{e^s}\left(1-\frac12 s^2 \epsilon^2+\frac38 s^4 \epsilon^4\right)ds=\epsilon-\epsilon^3+9\epsilon^5$$ for all $\epsilon>0$, where the RHS is $$\arctan\epsilon > \epsilon-\frac13\epsilon^3$$ for all $\epsilon>0$. So it proves the inequality for $0<\epsilon<\frac2{3\sqrt3}$, or for $x>\frac{3\sqrt3}2\simeq2.598$. – Ramanasa Aug 20 '18 at 17:36
• I could not find out that the left side has a series expansion, but your advice was a great help. – Ramanasa Aug 20 '18 at 17:36
• I apologize for that I could choose only one answer on this system (Is there any other way?) while you provided also an excellent and clever solution. – Ramanasa Aug 20 '18 at 17:44 | 2019-08-19T09:06:21 | {
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http://math.stackexchange.com/questions/320823/can-an-indefinite-integral-have-multiple-answers-besides-the-c?answertab=votes | # Can an indefinite integral have multiple answers? (Besides the ' + C')
So I came across with this integral today in my midterm:
$$\int \frac {\tan(\pi x)\sec^2(\pi x)}2$$
And I got two correct answers:
$$\frac {\sec^2(\pi x)}{4\pi} +C$$
And
$$\frac {\tan^2(\pi x)}{4\pi} + C$$
The first one, I get it by substituting $u=\sec (\pi x)$ and the second one, by substituting $u=\tan (\pi x)$
I already differentiated both answers and got to the same integral, but my question is, if both answers are correct and if it were a definite integral, which answer should I use? Wouldn't they give different results?
-
In the last formula replace your $C$ with ${1\over {4\pi}}+K$, which is just another constant. Now use the trig identity mentioned below. You see that you get the other answer. – Maesumi Mar 4 '13 at 22:26
In your case, the difference between the two is a constant:
$$\sin^2 x + \cos^2 x = 1$$
so
$$\tan^2 x + 1 = \sec^2 x$$
In general, that will be true as well - integration can only be different up to a constant: consider $g = \int f = h$ and note that $g-h = \int f - \int f = \int 0 = const$.
-
That doesn't answer my question in case it wasn't an indefinite integral. – ChairOTP Mar 4 '13 at 22:23
@ChairOTP: In case it was a definite integral, the extra $1$ in the $\sec^2$ answer for the upper limit would be canceled by the extra $1$ in the $\sec^2$ answer for the lower limit. – robjohn Mar 4 '13 at 22:26
@ChairOTP Fro definite integrals, you would get a constant adjustment factor, i.e. answers would be either $\sec^2 x$ or $\tan^2 x +1$, and in definite integrals, the constant would matter -- same answer would result either way. – gt6989b Mar 4 '13 at 22:26
Thank you so much :) – ChairOTP Mar 4 '13 at 22:28
They wouldn't give different answers in a definite integral. Suppose that we want to find $\displaystyle\int_a^b f(x)\,dx$. Let $F(x)$ be one antiderivative of $f(x)$, and let $G(x)$ be another. They differ by a constant, so $G(x)=F(x)+C$ for some constant $C$.
If you use $F(x)$ to evaluate the integral from $a$ to $b$, you get $F(b)-F(a)$.
If you use $G(x)$, you get $G(b)-G(a)$, that is, $(F(b)+C)-(F(a)+C)$. Simplify. The $C$'s cancel. | 2016-05-01T12:06:22 | {
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https://math.stackexchange.com/questions/690844/does-sin-5x-5-sinx-why-or-why-not/690861 | # Does $\sin (5x)=5\sin(x)$? Why or why not?
This may sound like a dumb question, but does $\sin(5x)=5\sin(x)$? Can you say this for any trigonometric function if it is true (e.g. can $\tan^{-1}(7x)=7\tan^{-1}(x)$). Again, I am pretty sure I am wrong, but I am just wondering for my future work on trigonometry. If they are not equal, then please tell me what the difference is between the two. Thanks!
Note: I have not had much practice with trigonometry, so please do not answer with some kind of complicated Taylor Series thing (I have heard about it, but don't fully understand it). Thanks again
• They are not equal. A most obvious reason is that $\sin(5x)$ never exceeds $1$ (being equal to the $y$-coordinate of a point on the unit circle). But $5\sin x$ can be as large as $5$ (being equal to the $y$-coordinate of a point on the unit circle multiplied by five). – Jyrki Lahtonen Feb 26 '14 at 6:16
• The $\sin$ function is NOT linear. – IAmNoOne Feb 26 '14 at 6:21
• Check this out, it could be helpful! :) – mikhailcazi Feb 26 '14 at 6:51
• There should be a few values $x$ for which you are familiar with the value $\sin (x)$. (If there aren't, you should learn a few! Having a few examples/computations to keep in mind when thinking about mathematics is vital. I cannot stress this enough.) Does your conjecture hold for these values? – Thomas Belulovich Feb 27 '14 at 2:38
• I should have thought of this before I asked the question.... $\sin(3\times 30)=1$, but $3\sin(30)=3\times 0.5=1.5$! – TrueDefault Feb 27 '14 at 2:51
I'll try to explain the difference between the two with a picture: (Apologies ahead; it's a rough sketch done in Microsoft Paint.)
Here, we have a circle (suppose with radius 1) with its center at (0,0), with a line (call it $l$) drawn from the center to the edge of the circle (ending at $P$). $A$ is the angle the line makes with the $x$ axis.
To get $5\sin(A)$, we blow up the line and the circle to 5 times its original size. $l$ is now 5 times the length, but still has one point at the center of the circle, and makes the same angle with the $x$ axis. The $y$-coordinate of the other end of $l$ is $5\sin(A)$.
To get $\sin(5A)$, this time we don't blow the circle and the line bigger. Instead, we're rotating the line around the center of the circle so that the angle it makes with the $x$-axis is 5 times bigger. Now, the $y$-coordinate of the point where $l$ meets the edge of the circle is $\sin(5A)$.
Try it out with an example or two; you'll see the difference yourself.
• I think your answer does the best job at helping the OP: thank you for the effort. – Eric Stucky Feb 26 '14 at 6:59
• @EricStucky Yeah, I figured OP was looking for an explanation in simpler terms. So I provided one. – Dennis Meng Feb 26 '14 at 7:01
No, placing a constant factor inside of the sine function, as in sin(a*x) changes the frequency of the function; if |a| is greater than one, it "squishes" the sine function horizontally, making it oscillate "faster", and if |a| is less than one, it "stretches out" the function horizontally, making it oscillate "slower". Placing the factor in front of the function, as in a*sin(x), scales the amplitude of the function, "stretching", or "squishing" it vertically.
The best way to observe it is to look at a graph and to see what it does: http://www.wolframalpha.com/input/?i=plot+sin+x%2C+sin+3x%2C+3*sin+x
I suggest playing around with other values in the graphs to see what they do.
1) You could have just plotted the functions in, e.g., wolfram alpha and see for yourself.
2) for $t=\frac{\pi}{2}$ one has $\sin (t)=1$ and since $|\sin (x)|\le 1$, it is clear that $\sin (5\cdot t)\ne 5\cdot \sin(t)$.
Something like $a\cdot f(x)= f(a\cdot x)$ is true if and only if $f(x)$ is linear. Since there are always nonlinear terms in the Taylor expansion of a trigonometric function this can't be true for any.
• Isn't $f(x)=x+1$ linear? But $5f(x)=5x+5$ and $f(5x)=5x+1$. – Joel Reyes Noche Feb 26 '14 at 6:37
• Functions satisfying the stated property are called homogenous, not linear. Any linear function is homogenous, but not necessarily vice versa. As for $x\mapsto x+1$, it is not a linear function. Rather it is an example of an affice mapping. – Ittay Weiss Feb 26 '14 at 6:49
• the argument that since the taylor expansion contains nonlinear terms, then the function itself is not linear is true, but requires some proof (easy). But, for the question posed resorting to the Taylor expansion of the function is really an overkill. Simply plugging in some values into the purported equation immediately shows it does not hold. – Ittay Weiss Feb 26 '14 at 6:51
• I was assuming that the OP (not the answerer) was more familiar with calculus rather than with linear algebra. See Wikipedia, where a linear function is sometimes defined as a polynomial function of degree zero or one. – Joel Reyes Noche Feb 26 '14 at 7:02
Assume that the identity $\sin(5x) = 5\sin(x)$ is true. Since we know that $\sin(180^\circ) = 0$, we'll have: $0 = \sin(180^\circ) = \sin(5*36^\circ) = 5*\sin(36^\circ)$, hence $5*\sin(36^\circ) = 0$, which mounts to $\sin(36^\circ) = 0$. Is this true?
• Using degrees rather than radians should be discouraged. – Ittay Weiss Feb 26 '14 at 6:52
The trigonometric functions are functions and as such are usually represented using function notation. Recall that the function notation $f(x)$ means that function $f$ is accepting $x$ as its input. It does not mean that $f$ and $x$ are being multiplied. So the notation $\sin(5x)$ means that the sine function is accepting $5x$ as its input. The notation $5\sin(x)$ means that 5 is being multiplied to the output of the sine function when its input is $x$. Is everything clear now?
• Just to clarify that there exist functions with the property that $f(c\cdot x)=c\cdot f(x)$. Such functions are called homogenous. – Ittay Weiss Feb 26 '14 at 6:19
Both of the equalities that you wrote above are NOT true. One way to look at this might be to use their Taylor series expansion. $\sin (5x)$ means the sine of $5$ times an angle measure $x$. This means in a right triangle whose hypotenuse is of length $1$, the perpendicular corresponding to the acute angle $5x$ has length $\sin (5x)$.
While for $5 \sin x$, one should consider a right-triangle with hypotenuse of length one and the perpendicular corresponding to an acute angle $x$ and then take $5$ times the length of this perpendicular.
You can draw some such right triangles to verify the results yourself.
• Using Taylor series here is a major overkill. Simply plug in some values and observe. – Ittay Weiss Feb 26 '14 at 6:20
• @IttayWeiss haha. I can only agree :-) – Singhal Feb 26 '14 at 7:44 | 2021-02-25T19:16:24 | {
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https://stats.stackexchange.com/questions/242465/distribution-function-terminology-pdf-cdf-pmf-etc | # Distribution function terminology (PDF, CDF, PMF, etc.) [duplicate]
I am confused about the following terminologies:
1. Distribution Function
2. Cumulative Distribution Function (CDF)
3. Probability Distribution Function
4. Probability Density Function
5. Probability Mass Function (PMF)
(1) Distribution Function == CDF. I.e. they are the same. Am I right?
(2) Which is called PDF: Probability Distribution Function or Probability Density Function?
(3) What is the difference between Probability Distribution Function and Probability Density Function?
(4) Which one is the continuous equivalent of PMF, Probability Distribution Function or Probability Density Function?
Die roll examples could be used for the discrete case and picking a number between 1.5 and 2.5 as an example for the continuous case.
As noted by Wikipedia, probability distribution function is ambiguous term:
A probability distribution function is some function that may be used to define a particular probability distribution. Depending upon which text is consulted, the term may refer to:
• a cumulative distribution function,
• a probability mass function, and/or
• a probability density function.
Cumulative distribution function (CDF) is sometimes shortened as "distribution function", it's
$$F(x) = \Pr(X \le x)$$
the definition is the same for both discrete and continuous random variables. In dice case it's probability that the outcome of your roll will be $x$ or smaller.
Probability density function (PDF) is a continuous equivalent of discrete probability mass function (PMF). Probability mass function is
$$f(x) = \Pr(X = x)$$
In dice case it's probability that the outcome of your roll will be exactly $x$.
Probability mass function has no sense for continuous random variables since $\Pr(X=x)=0$ for continuous random variables (check also Why X=x is impossible for continuous random variables?), because simply a point on real line is so "small" that has no mass and no area.
This leads us to defining probability density as "probability per foot". Simple example is continuous uniform distribution with minimum of $a$ and maximum of $b$, where probability density is the same for each $x$ and equal to
$$f(x) = \frac{1}{b-a}$$
You can easily notice that it changes as the range between $a$ and $b$ (i.e the total area) changes, it is nicely described in Can a probability distribution value exceeding 1 be OK? thread. It is a probability of hitting infinitesimal (infinitely small) interval $[x, x + dx]$ when throwing a dice with infinite number of walls.
• There are fashions here too. For example, it was long customary to insist that probability density functions and probability mass functions were quite different kinds of beasts referring to continuous and discrete variables respectively. A common and in my experience more recent tendency, particularly with mathematically more mature groups, is to insist that the idea of density function is general; it is just a case of density with respect to what kind of measure, and measure could be e.g. counting measure. So density functions, wide sense, include mass functions. – Nick Cox Oct 26 '16 at 13:21
(1) You are right regarding (1).
(2)&(3)&(4) PDF is for probability density function. We usually use probability distribution function to mean CDF. Probability function is used to refer to either probability mass function(the probability function of discrete random variable) or probability density function(the probability function of continuous random variable).
You can also have a look at this What does "probability distribution" mean? | 2021-07-25T22:23:22 | {
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https://stats.stackexchange.com/questions/120278/probability-of-disease-given-multiple-samples-of-a-single-test | Probability of disease given multiple samples of a single test
Someone either has disease X or they are healthy. A test is used which is 90% sensitive and 80% specific (81.81% precise). The test is performed 8 times with different samples from the same person.
1) What is the rule to calculate the probability that someone has the disease given that 5 out of 8 tests came back positive?
2) More generally: that N out of M samples were positive.
3) Is there a way of calculating the confidence?
Kind regards
• Hello jamesj629, welcome to the site. If this is a homework question please add the self-study tag. – Andy Oct 16 '14 at 11:26
• Thanks Andy - its not homework, just learning. – jamesj629 Oct 16 '14 at 11:32
You need the disease prevalence in order to calculate this probability.
Let $T$ be the indicator variable with $T=1$ denoting the event of a positive test. Similarly let $D$ be the indicator of the disease. You have $P(T=1|D=1)=0.9$ and $P(T=0|D=0)=0.8$. Define $A$ to be the event of N positive result from M tests. Then
$$P(A|D=1)=\binom{M}{N}0.9^N 0.1^{M-N}$$
What you want is $P(D=1|A)$. Using Bayes' rule $$P(D=1|A)= \frac{P(A|D=1)P(D=1)}{P(A)}=\frac{P(A|D=1)P(D=1)}{P(A|D=1)P(D=1) +P(A|D=0)P(D=0)}$$
• I take it P(D=1) on its own is disease prevalence? and P(D=0)=1-P(D=1) ? – jamesj629 Oct 16 '14 at 12:59
• The prevalence is (say) 1 in 100000, but for people taking the test, about 90% actually have the disease, because the test is only done on those presenting with suspect symptoms - does this invalidate the probability? – jamesj629 Oct 16 '14 at 13:05
• @jamesj629 It will increase the probability. If you let $S$ be the indicator of symptom onset. With what you described in the comment, now you are calculating $P(D=1|A,S=1)$ instead of $P(D=1|A)$. You numerator becomes $P(A|D=1,S=1)P(D=1|S=1)=P(A|D=1)P(D=1|S=1)$ if test result is independent of symptom. Note $P(D=1|S=1) > P(D=1)$ if disease is positively dependent with symptom. – Peter Oct 16 '14 at 21:19
You can use the information to calculate the base rate fo the disease:
accuracy: $A$
sensitivity: $s^+$
specificity: $s^-$
base rate: $p$
positive test result: $T^+$
disease: $D$
$A=p\cdot s^++(1-p)\cdot s^-$
$0.8181= p\cdot0.9+ (1-p)\cdot 0.8$
$0.0181= 0.1p$
$p=.181$
For one test positive out of one:
using Bayes' theorem: $P(D|T^+)=\frac{p \cdot s^+}{p \cdot s^+ + (1-p) \cdot (1-s^-)}$
$P(D|T^+)=\frac{.181 \cdot .9}{.181 \cdot .9 + (1-.181) \cdot (1-0.8)}$
$P(D|T^+)=\frac{.1629}{.1629 + .1638}=.4986$
now for k tests out of m: as in Peter's answer, the probabilities are now drawn from a binomial to generate likelihoods:
$L^+$~$B(n,s^+)$ and $L^-$~$B(n,s^-)$ with inverted number of successes for the second. You can probably calculate confidence intervals based on the knwon properties (variance) of bionomial distributions... | 2020-02-24T01:43:18 | {
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https://math.stackexchange.com/questions/2195750/prove-or-disprove-that-if-sum-a-n-and-sum-b-n-are-both-divergent-then-s | # Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.
Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.
I decided to disprove this by offering the example:
Let $a_n = n$.
Let $b_n = -n$.
Then $\sum (a_n + b_n) = \sum 0 = 0$, which converges.
However, my question says $\pm$, and in the $-$ case, we have $\sum (a_n - b_n) = \sum 2n = \infty$, which diverges. So, does this still count as a valid counter example or not? I partly proved it wrong so I think it's still valid.
If $\sum(a_n+b_n)$ and $\sum(a_n-b_n)$ both converge, then since :
$$a_n=\frac12\left((a_n+b_n)+(a_n-b_n)\right)\quad\textrm{and}\quad b_n=\frac12\left((a_n+b_n)-(a_n-b_n)\right)$$
we see that $\sum a_n$ and $\sum b_n$ both converge.
In other words, if $\sum a_n$ diverges or $\sum b_n$ diverges, then $\sum(a_n+b_n)$ diverges or $\sum(a_n-b_n)$ diverges.
• Although this proves that if the original series don't both converge at least one of the $\pm$ series doesn't either, "does not converge" is not equivalent to "diverges". – J.G. Mar 20 '17 at 22:38
• Further to my already upvoted comment, apparently Wikipedia does define divergent as "not convergent", even though I could have sworn it meant "tends to $\pm\infty$". It seems weird to me that an oscillatory sequence, for example, would be described as "divergent". – J.G. Mar 20 '17 at 23:00
My reading of the statement "$\sum (a_n \pm b_n)$ necessarily diverges" is that it means that both of the sums $\sum (a_n + b_n)$ and $\sum (a_n - b_n)$ diverge. You have proved that this is not necessarily true; you have a counterexample in which one of the sums converges.
The wording of the original question seems highly ambiguous to me, however. The statement "$\sum (a_n \pm b_n)$ diverges" might be intended to say that either $\sum (a_n + b_n)$ diverges or $\sum (a_n - b_n)$ diverges.
It really comes down to what is the meaning of the statement "$\sum (a_n \pm b_n)$ diverges."
Let $\sum \frac{1}{2n+1}$=$1+1/3+1/5+1/7+1/9+....$ and let $-\sum \frac{1}{2n}$=$-1/2-1/4-1/6-1/8...$. Boths series diverge, but when you add them (term by term) together, you get $1-1/2+1/3-1/4+1/5-1/6...$ and this is the well known Alternating Harmonic series with an answer of $\ln 2$, so (conditionally) convergent.
• In fact, when you add them term-by-term you get $\sum \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$, which is absolutely convergent as each individual term is positive. It's only when you deparenthesize that you get a conditionally convergent result... – Micah Mar 20 '17 at 22:47 | 2019-08-20T16:44:49 | {
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https://math.stackexchange.com/questions/2988719/proving-continuity-of-a-multivariable-function-by-reducing-it-to-a-single-vari | # Proving continuity of a multivariable function by “reducing” it to a single variable function
My goal is to examine the continuity of a certain multivariable function. In that regard, (mainly because my math in the specific area is dusty) I find it somewhat tricky to either prove or disprove that property in the N-diensional space. However, I have come up with the following idea:
let $$f : \mathbb{R}^2 \rightarrow \mathbb{R}$$ be the function that I want to examine at some point $$(x_0, y_0)$$.
Now I am using the following two functions $$g: \mathbb{R}^2 \rightarrow \mathbb{R}$$ and $$h: \mathbb{R} \rightarrow \mathbb{R}$$ so that I can write function $$f$$ as: $$f(x,y) = h(g(x,y)),\quad \forall (x,y) \in \mathbb{R}^2$$ I also know that function $$g$$ is indeed continuous in $$\mathbb{R}^2$$.
My assumption is that I can examine the continuity of $$h$$ at $$x' = g(x_0,y_0)$$ in order to decide the continuity of $$f$$ at $$(x_0,y_0)$$ - a much easier task. In mathimatical notation, I guess what I mean is: $$\lim_{(x,y)->(x_0,y_0)} f(x,y) = f(x_0, y_0) \quad \iff \\ \lim_{x->g(x_0,y_0)}h(x) = h(g(x_0,y_0))$$
Actual Questions:
1. Is my assumption correct?
2. If not, maybe do either of $$\Rightarrow$$ or $$\Leftarrow$$ hold? (Because if one of those holds then I can use the above to only prove or only disprove accoridnlgy.)
3. Does the same hold for proving/disproving differentiability as well?
4. Under what conditions do any of those hold for a (finite) sum of $$h,g$$ functions, i.e. $$f(x,y) = h_1(g_1(x,y)) + h_2(g_2(x,y)) + ... + h_k(g_k(x,y))$$
Note: I can actually post my original function and the transformation if the question is too broad or the answers depend greatly on the actual $$f,g$$ and $$h$$ definitions. However I believe a more general question may help more people find it and also make it easier to be understood.
• We know that the composition of continuous functions is continuous. So it suffices to show that $g(x,y)$ is continuous at $\xi_0 = (x_0,y_0)$. That is that $\lim_{\xi \rightarrow \xi_0} g(\xi) = g(\xi_0)$. Also note that showing that the derivative exists at $x_0$ is a sufficient, but not necessary condition for continuity. That is differentiability implies continuity but not the converse. For an example see $f(x) = |x|$. This is continuous at $0$, and in fact Lipschitz continuous at $0$, but not differentiable there. (continued) – GeauxMath Nov 13 '18 at 23:33
• Also note a vector valued function is continuous if it is continuous in all its arguments – GeauxMath Nov 13 '18 at 23:39
• @GeauxMath First of all, thank you for responding. Now, I am fully aware about how differentiability implies continuity but not the other way around. Reading back my post, I think I was indeed looking for the property of compositions of continuous functions being continuous. I also take it, that in your first comment you meant to say that it suffices to show that $h(x)$ is continuous and not $g(x,y)$, as I already mentioned that $g$ is indeed continuous. – kyriakosSt Nov 14 '18 at 0:24
• yes, you are correct – GeauxMath Nov 14 '18 at 1:18
• Cheers. If you would like to modify your comment to an answer, I am willing to accept it – kyriakosSt Nov 14 '18 at 10:06
Just for closure, I am answering myself in case it helps anyone else. Merit goes to @GeauxMath and @Todor Markov for providing answers in comments.
We can answer mainly by using the following proposition:
A composition of continuous functions is continuous
In detail:
1. My assumption is incorrect as a whole
2. The $$\Leftarrow$$ direction holds by using the above proposition. The opposite is not true in every case. As a counter example
If $$g(x_0,y_0)=a$$ and $$f(x,y)≥a$$, then $$h$$ can have a discontinuity at a. If $$h(x)$$ is continuous for $$x>a$$, $$f$$ would still be continuous.
3. The corresponding property applies also to differentiable functions, so we can only prove that $$f$$ is differentiable that way, but not disprove it.
4. Again, by using that proposition we can extend our statements to that specific case, so we can prove continuity and differentiability of $$f$$ but not disprove. A counter example here would be a case where $$h_1$$ and $$h_2$$ are not continuous, but by adding them, their discontinuities "cancel out". | 2019-10-23T23:48:00 | {
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https://mathoverflow.net/questions/279870/is-every-positive-integer-the-permanent-of-some-0-1-matrix/279872 | # Is every positive integer the permanent of some 0-1 matrix?
In the course of discussing another MO question we realized that we did not know the answer to a more basic question, namely:
Is it true that for every positive integer $k$ there exists a balanced bipartite graph with exactly $k$ perfect matchings?
Equivalently, as stated in the title, is every positive integer the permanent of some 0-1 matrix?
The answer is surely yes, but it is not clear to me how to prove it. Entry A089479 of the OEIS reports the number $T(n,k)$ of times the permanent of a real $n\times n$ zero-one matrix takes the value $k$ but does not address the question of whether, for every $k$, there exists $n$ such that $T(n,k)\ne 0$. Assuming the answer is yes, the followup question is, what else can we say about the values of $n$ for which $T(n,k)\ne 0$ (e.g., upper and lower bounds)?
• The answers below write $k$ as the permanent of a matrix of size $k$ or $\log k$. I wonder if it is possible with a matrix of size $n$, where $(n-1)! < k \leq n!$? – Zach Teitler Aug 29 '17 at 18:39
• @Zach, it is true for some k in that range. However, if k is not a multiple of a factorial, one "needs a few more rows". I suspect there is an absolute constant C such that for k less than Cn! that k is representable as the permanent of an order n 0-1 matrix. Gerhard "Finding K Is Another Matter" Paseman, 2017.08.29. – Gerhard Paseman Aug 29 '17 at 19:52
• You can always direct sum with an identity matrix to get additional rows. – Timothy Chow Aug 29 '17 at 21:36
• In fact it seems that even for determinant we can construct $k\times k~0-1$ matrix to represent any number $k-1$, but in this case it's lower bound of size. – rus9384 Aug 29 '17 at 22:46
• @rus9384, indeed one can surpass the kth Fibonacci number in representing positive integers by a determinant of a k by k 0-1 matrix. once k is bigger than 5. Orrick's talk mentioned below gives some detail. Gerhard "And I Can Give More" Paseman, 2017.08.30. – Gerhard Paseman Aug 30 '17 at 16:48
The answer to the question is yes. Given $k$, the 0-1 matrix given by
$1$ $1$ $\dotsc$ $1$ $0$ $0$ $\dotsc$ $0$ $0$ $0$ $0$
$0$ $1$ $1$ $0$ $\dotsc$ $0$ $0$ $\dotsc$ $0$ $0$ $0$
$0$ $0$ $1$ $1$ $0$ $\dotsc$ $0$ $\dotsc$ $0$ $0$ $0$
$\dotsc$
$1$ $0$ $0$ $0$ $0$ $\dotsc$ $\dotsc$ $0$ $0$ $0$ $1$
where the first row has precisely $k$ entries equal to $1$, evidently has permanent equal to $k$.
For $k=1$ the matrix is ($1$). For $k=2$ the matrix is
$1$ $1$
$1$ $1$
For $k=3$ the matrix is
$1$ $1$ $1$
$0$ $1$ $1$
$1$ $0$ $1$.
For $k=4$ the matrix is
$1$ $1$ $1$ $1$
$0$ $1$ $1$ $0$
$0$ $0$ $1$ $1$
$1$ $0$ $0$ $1$
which evidently has permanent equal to $4$.
Please note that also, for each given $k$ and $1\leq \ell \leq k$, this construction can be tweaked to give an explicit $k\times k$ sized $0$-$1$-matrix having permanent precisely $\ell$, just by making the first row have precisely $\ell$ entries equal to $1$.
Please also note that my construction does not have any bearing on the interesting and apparently difficult question which was cited in the present OP: my construction is too wasteful: it utilizes a $k\times k$ matrix, which is far too large when it comes to meet the demands of the OP in said question.
• Nice construction! Shortly after posting my question, I discovered the following paper by Kim, Lee, and Seol that also answers the question: ijpam.eu/contents/2005-19-3/12/12.pdf – Timothy Chow Aug 29 '17 at 17:44
• The graph picture for anyone who doesn't want to work out the details: this is a Hamilton cycle of length $2k$ together with a star whose leaves are all the elements of one colour class and whose centre is in the other. Any choice of matching edge for the centre of the star is a chord that splits the cycle into two paths, each of which has a unique perfect matching, giving the $k$ options. – Ben Barber Aug 29 '17 at 17:49
For the record, here is the construction of Kim, Lee, and Seol that I alluded to in my comment to Peter Heinig's answer.
Write $k-1$ in binary, and let $n$ be 1 plus the number of binary digits of $k-1$.
Start with an $n\times n$ matrix with all $1$'s on or above the main diagonal and all $0$'s below the diagonal. Then replace the first $n-1$ entries of the bottom row of the matrix with the binary representation of $k-1$ (one bit per entry).
For example, if $k=389$, then $k-1$ in binary is $110000100$. Then $n=1+9=10$ and the matrix is $$\matrix{ 1&1&1&1&1&1&1&1&1&1\cr 0&1&1&1&1&1&1&1&1&1\cr 0&0&1&1&1&1&1&1&1&1\cr 0&0&0&1&1&1&1&1&1&1\cr 0&0&0&0&1&1&1&1&1&1\cr 0&0&0&0&0&1&1&1&1&1\cr 0&0&0&0&0&0&1&1&1&1\cr 0&0&0&0&0&0&0&1&1&1\cr 0&0&0&0&0&0&0&0&1&1\cr 1&1&0&0&0&0&1&0&0&1\cr }$$
• Of course, one can represent pq by a block matrix for p and a block for q on the diagonal. This construction using binary of k-1 works well for odd numbers and is rather compact in general. However, even for determinant, the spectrum problem is not well understood, so finding the optimal n given k will require even more cleverness. For example, it is not clear that n=7 might work for k=389. Does the cited paper mention how optimal the construction is? Gerhard "Hasn't Determined Complexity Of Permanent" Paseman, 2017.08.29. – Gerhard Paseman Aug 29 '17 at 18:52
• @GerhardPaseman why pq form? – T.... Aug 30 '17 at 21:40
• pq here means the factor p times the factor q. For numbers m with small factors, the block matrix form reduces the problem of representing m compactly one of representing enough of its small factors compactly. Gerhard "Because It's Easy, That's Why" Paseman, 2017.08.30. – Gerhard Paseman Aug 30 '17 at 22:49
If an order n (0,1) matrix has r rows with all ones, its permanent is a multiple of r! by an easy argument. It follows that the largest odd number which is a permanent of an order n matrix is q = der(n) + der(n-1), which is a sum of numbers of derangements, and is a little larger than $n!/e$. I suspect that the smallest positive number not expressible as a permanent of this size matrix is an odd integer which is not much smaller, perhaps about half, of q. If so, then we can shave a little bit off of n=log(k) in the construction in Timothy Chow's post. For some motivation for this problem, look at Will Orrick's talk mentioned (for determinants) at https://mathoverflow.net/a/271273 .
Gerhard "Also A Potential Polymath Project" Paseman, 2017.08.29. | 2019-09-15T18:58:25 | {
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https://stats.stackexchange.com/questions/285353/why-can-the-covariance-matrix-be-computed-as-fracx-xtn-1/285358 | # Why can the covariance matrix be computed as $\frac{X X^T}{n-1}$?
Wikipedia defines the covariance matrix as a generalized version of the variance: $C=E[(X-E[X])(X-E[X])^T]$. But, I usually see statisticians calculate the covariance matrix as $C=\frac{X X^T}{n-1}$ (e.g., Ameoba's nice answer here https://stats.stackexchange.com/a/134283/83526). Is $E[(X-E[X])(X-E[X])^T] = \frac{X X^T}{n-1}$? Are these both correct definitions of the covariance matrix?
• Amoeba explicitly prefaces that formula with the statement "Let us assume that it [$X$] is centered, i.e. column means have been subtracted and are now equal to zero." That shows his formula is identical to the one in Wikipedia. – whuber Jun 14 '17 at 20:38
Let $\mu = E(X)$. Then $$Var(X) = E\left((X - \mu)(X - \mu)^T\right) = E\left(XX^T - \mu X^T - X \mu^T + \mu \mu^T\right) \\ = E(XX^T) - \mu\mu^T$$ which generalizes the well-known scalar equality $Var(Z) = E(Z^2) - E(Z)^2$.
The natural estimator of $\Sigma := Var(X)$ is $\hat \Sigma = \frac 1{n-1}XX^T - \hat \mu \hat \mu^T$.
In many situations we can take $\mu = 0$ without any loss of generality. One common example is PCA. If we center our columns then we find that $\hat \mu = 0$ so our estimate of the variance is simply $\frac 1{n-1}XX^T$. The univariate analogue of this is the familiar $s^2 = \frac 1{n-1} \sum_i x_i^2$ when $\bar x = 0$.
As @Christoph Hanck points out in the comments, you need to distinguish between estimates and parameters here. There is only one definition of $\Sigma$, namely $E((X - \mu)(X - \mu)^T)$. So $\frac 1{n-1}XX^T$ is absolutely not the correct definition of the population covariance, but if $\mu=0$ it is an unbiased estimate for it, i.e. $Var(X) = E(\frac 1{n-1}XX^T)$.
• To understand the linear algebra used in the answer better, you can look at math.stackexchange.com/questions/198257/… – kjetil b halvorsen Jun 14 '17 at 16:26
• +1 I tried a few times to squeeze a comment in here, but the $LaTeX$ didn't show well, so I used the entry of an answer as an extended comment. – Antoni Parellada Jun 15 '17 at 0:14
• Note that there is a bit of ambiguity in the symbology here (inherited from the question): The symbol $X$ is used to refer to both the random variable ($p\times{1}$, vector valued), and the (transposed) data matrix ($p\times{n}$, columns are samples of the random variable). – GeoMatt22 Jun 15 '17 at 0:30
COMMENT:
@Chacone's answer is great as is, but I think there is one step that is left unexplained, and it is clearer with expectation notation. To reflect @GeoMatt22's comment in the following proof $X$ is a $p\times 1$ random vector:
\begin{align} \text{Cov}(X)&=\mathbb E\left[\,\left(X- \mathbb E[X] \right) \, \left(X- \mathbb E[X] \right)^\top \right]\\[2ex] &= \mathbb E\left[\,XX^\top - X\,\mathbb E[X]^\top - \mathbb E[X] \,X^\top + \mathbb E[X] \mathbb E\, [X]^\top \right]\\[2ex] &= \mathbb E\left[\,XX^\top\right] - \mathbb E [X]\,\mathbb E[X]^\top - \mathbb E[X] \,\mathbb E [X]^\top + \mathbb E[X] \mathbb E\, [X]^\top \\[2ex] &= \mathbb E[XX^\top] \; -\; \mathbb E[X]\, \mathbb E[X]^\top \end{align}
As for the transition to the sample estimate of the population covariance using this alternate (raw moment) formula, the denominators will need adjusing. In what follows $X$ is a $p \times n$ data matrix (following @GeoMat22's comment):
\begin{align} \sigma^2(X)&=\frac{XX^\top}{n-1} \; -\; \frac{n}{n-1}\; \begin{bmatrix}\bar X_1\\ \bar X_2\\ \vdots \\ \bar X_p \end{bmatrix}\, \; \begin{bmatrix}\bar X_1 & \bar X_2 & \cdots & \bar X_p \end{bmatrix} \end{align}
because it is necessary to multiply by $\frac{n}{n-1}$ to get rid of the biased $n$ denominator, and replace it with $n-1$. Evidently anything after the minus sign disappears if the mean is zero.
Here is an illustrative example in R:
> set.seed(0)
> # Sampling three random vectors: V1, V2 and V3:
> X1 = 1:5
> X2 = rnorm(5, 5, 1)
> X3 = runif(5)
> # Forming a matrix with each row representing a sample from a random vector:
> (X = (rbind(X1,X2,X3)))
[,1] [,2] [,3] [,4] [,5]
X1 1.00000000 2.0000000 3.0000000 4.0000000 5.0000000
X2 6.26295428 4.6737666 6.3297993 6.2724293 5.4146414
X3 0.06178627 0.2059746 0.1765568 0.6870228 0.3841037
> # Taking the estimate of the expectation for each random vector, bar Xi:
> mu = rowMeans(X)
> # Calculating manually the variance with the alternate formula
> (man.cov = ((X %*% t(X)) / (ncol(X) - 1)) - (ncol(X)/(ncol(X) - 1)) * (mu %*% t(mu)))
X1 X2 X3
X1 2.50000000 -0.02449075 0.28142079
X2 -0.02449075 0.53366886 0.02019664
X3 0.28142079 0.02019664 0.05940930
> # Comparing to the built-in formula:
> cov(t(X))
X1 X2 X3
X1 2.50000000 -0.02449075 0.28142079
X2 -0.02449075 0.53366886 0.02019664
X3 0.28142079 0.02019664 0.05940930
> # are the same...
> all.equal(man.cov, cov(t(X)))
[1] TRUE | 2020-05-26T14:12:09 | {
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https://math.stackexchange.com/questions/2453869/proof-for-a-logical-truth-involving-quantifiers | # Proof for a logical truth involving quantifiers.
Prove: $(\forall x Fx \vee \forall x Gx)\to \forall x(Fx\vee Gx)$
My attempt:
\begin{align} 1.\space & \neg\forall x(Fx\vee Gx) & \text{(Conditional Proof)}\\ 2.\space & \exists x\neg(Fx\vee Gx) & \text{Quantifier Negation on line 1}\\ 3.\space & \exists x(\neg Fx\wedge \neg Gx) & \text{De Morgan's on line 2}\\ & \qquad 4.\space \neg Fx\wedge \neg Gx & \text{Existential Instantiation on line 3}\\ & \qquad 5.\space \neg Fx & \text{Simplification on line 4}\\ & \qquad 6.\space \neg Gx & \text{Simplification on line 4}\\ & \qquad 7.\space \exists x \neg Fx & \text{Existential Generalization on line 5}\\ & \qquad 8.\space \neg\forall x Fx & \text{Quantifier Negation on line 7}\\ & \qquad 9.\space \exists x \neg Gx & \text{Existential Generalization on line 6}\\ & \qquad 10.\space \neg\forall x Gx & \text{Quantifier Negation on line 9}\\ & \qquad 11.\space \neg\forall x Fx \wedge\neg\forall x Gx & \text{Conjunction from lines 8, 10}\\ 12.\space & \neg\forall x Fx \wedge\neg\forall x Gx & \text{Existential Instantiation lines 3, 4-11}\\ 13.\space & \neg(\forall xFx \vee \forall x Gx) & \text{De Morgan's on line 12}\\ 14.\space & \neg\forall x(Fx\vee Gx)\rightarrow \neg(\forall Fx \vee \forall Gx) & \text{Conditional Proof on lines 1-13}\\ 15.\space &\boxed{(\forall x Fx \vee \forall x Gx) \rightarrow \forall x(Fx\vee Gx)} & \text{Transposition on line 14} \end{align}
I am mostly concerned about the Existential Instantiation subproof from 4-11. Also, it is the first time I have done this kind of proof. So, let me know of ways to improve it!
• Is there some reason why you proved the contrapositive instead of just directly proving the statement? – Derek Elkins Oct 2 '17 at 4:19
• I wasn't sure about Universally Instantiating $\forall x Fx$ and $\forall x Gx$ on the same line. Can I? – G_D Oct 2 '17 at 4:25
• Why do you think you'd need to do it on the same line? – Derek Elkins Oct 2 '17 at 4:31
• IIRC, Universal Instantiation needs to affect the entire line. And I would be instantiating to $x$ twice (if that makes sense). – G_D Oct 2 '17 at 4:36
• If you start the direct proof, you'll find that when you need to instantiate $\forall x.Fx$ or $\forall x.Gx$, the other won't be there, but even in cases where you are instantiating a universal quantifier, only things in the scope of the quantifier are going to be affected. You pick a quantified formula and instantiate it. You don't just instantiate every quantifier on the line. That said, I think you mean to worry about Universal Generalization as that's the rule with side conditions. – Derek Elkins Oct 2 '17 at 4:43
Your use of "existential generalization" and "existential instantiation" seems backward. Existential instantiation is:
\begin{align} \exists x \phi(x) \\ \hline \phi(c) \end{align}
For some constant $c$. And existential generalization is when you go the other way.
Also note that after existential instatiation the symbol $c$ is a constant and you should not use that one in a quantifier then. You should probably used another name than $x$ for the constant introduced by existential instantiation.
Also note that it's normally the conditional proof that would require indentation to indicate that these statements are only a consequence of an assumption. For the existential instantiation this is not as required as we only require that we can simply get rid of a constant not appearing in our formula any longer.
Here is a proof using a Fitch-style proof checker to make sure I am applying the rules correctly:
Note that I consider both cases of the disjunction in line 1 and arrive at the same result on lines 4 and 7. Then I use disjunction elimination on line 8 and finally universal introduction on line 9.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
$$\vdash (\forall x~Fx~\vee~\forall x~Gx)~\to~~\forall x~(Fx\vee Gx)$$
...may be proven directly via disjunctive syllogism (aka disjunctive elimination):
$$(\forall x~Fx~\vee~\forall x~Gx), (\forall x~Fx)~\to~\forall x~(Fx\vee Gx), (\forall x~Gx)~\to~\forall x~(Fx\vee Gx)\vdash\forall x~(Fx\vee Gx)$$
Since $$\vdash (\forall x~Fx)\to \forall x~(Fx\vee Gx)$$ is provable by assumption, universal elimination, disjunctive introduction, conditional introduction, and universal reintroduction. Quite similarly we have $$\vdash (\forall x~Gx)\to \forall x~(Fx\vee Gx)$$.
$$\begin{array}{l|l:ll} \hdashline 1 & \quad \forall x~Fx~\vee~\forall x~Gx&& \text{Assume} \\\hdashline 2 & \qquad \forall x ~Fx &1& \text{Assume }\vee\mathsf L\\\hdashline 3 & \quad\qquad Fc &2& \forall-\\ 4 & \quad\qquad Fc\vee Gc &3& \vee+ \\\hline 5 & \qquad \forall x~(Fx\vee Gx) &4& \forall + \\ \hline 6 & \quad(\forall x~Fx)\to(\forall x~(Fx\vee Gx)) &2,5& \to+ \\ \hdashline 7 & \qquad\forall x~Gx &1&\text{Assume }\vee\mathsf R \\ \hdashline 8 & \quad\qquad Gc &7& \forall -\\ 9 & \quad\qquad Fc\vee Gc &8& \vee+ \\ \hline 10& \qquad \forall x~(Fx\vee Gx) &9& \forall + \\ \hline 11& \quad(\forall x~Gx)\to \forall x~(Fx\vee Gx) &7,10& \to+\\ \hline 12& (\forall ~Fx~\vee~\forall x~Gx)\to \forall x~(Fx\vee Gx) &1,6,11& \vee-\end{array}$$ | 2019-08-23T22:03:41 | {
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James invested $5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another$10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B?
A. 250
B. 775
C. 1025
D. 1750
E. 2000
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Re: James invested $5000 in scheme A for 1 year at a simple annual interes [#permalink] Show Tags 27 Aug 2018, 05:11 1 EgmatQuantExpert wrote: James invested$5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another $10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B? A. 250 B. 775 C. 1025 D. 1750 E. 2000 To read all our articles: Must read articles to reach Q51 Scheme A:$5,000 at 5% simple annual interest
5,000 * .05 * 1 = 5250 ----> $250 of interest Scheme B:$10,000 at 10% compounded semi annually
10,000(1 + $$\frac{.1}{2}$$)$$^2$$ --------> 10,000(1.05)$$^2$$ --------> 10,000(1.1025) = 11,025
$11,025 ----------->$1,025 of interest
Positive difference between the interest earned by Scheme A vs Scheme B
$1,025 -$250 = $775 Answer: B _________________ Would I rather be feared or loved? Easy. Both. I want people to be afraid of how much they love me. How to sort questions by Topic, Difficulty, and Source: https://gmatclub.com/forum/search.php?view=search_tags Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4011 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: James invested$5000 in scheme A for 1 year at a simple annual interes [#permalink]
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27 Aug 2018, 07:11
EgmatQuantExpert wrote:
James invested $5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another$10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B?
A. 250
B. 775
C. 1025
D. 1750
E. 2000
$$Amount_{Si} = 5000 + \frac{5000*5*1}{100}$$
So, $$Amount_{Si} = 5250$$
$$Amount_{Ci} = 5000( 1 + \frac{10}{200})^{2}$$
So, $$Amount_{Ci} = 11025.00$$
So, the required difference is $1,025 -$250 = $775 , Answer must be (B) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Re: James invested$5000 in scheme A for 1 year at a simple annual interes &nbs [#permalink] 27 Aug 2018, 07:11
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-09-19T07:14:41 | {
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http://mathhelpforum.com/pre-calculus/232785-possible-not-use-calculator-solve-problem.html | # Thread: Is it possible to not use a calculator and solve this problem?
1. ## Is it possible to not use a calculator and solve this problem?
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
2. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by chewydrop
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
You mean you *did* use a calculator? Where??
3. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by Matt Westwood
You mean you *did* use a calculator? Where??
(8*143/5) / (73/5 *1/(213/5 )) into the calculator and got 2.0000001
4. ## Re: Is it possible to not use a calculator and solve this problem?
This is a simple exercise in properties of the exponential. Do you not know that $log(a^x)= x log(a)$ and $log(ab)= log(a)+ log(b)$?
[QUOTE=chewydrop;830170]Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))[/tex]
$8= 2^3$ so [tex]log(8)= 3 log(2). $14= 2(7)$ so $(3/5)log(14)= (3/5)log(2)+ (3/5)log(7)$. $1/2^{13/5}= 2^{-13}{5}$ so $log(1/2^{13/5})= -(13/5)log(2)$
So you have 3 log(2)+ (3/5)log(2)+ (3/5)log(7)- (3/5)log(7)- 13/5 log(2)
Now, it's just arithmetic.
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
5. ## Re: Is it possible to not use a calculator and solve this problem?
Hello, chewydrop!
$\text{Simplify: }\:\log(8)+\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)+\log\left(\frac{1}{2^{\frac{13 }{5}}}\right)$
Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$
. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$
. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$
. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$
. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$
. . . $=\; \log(2)$
6. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by Soroban
Hello, chewydrop!
Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$
. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$
. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$
. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$
. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$
. . . $=\; \log(2)$
For some reason when i regrouped it on my own i got a different expression:
Log(8)+Log(2-13/5)-3/5Log(7)+3/5Log(14)
would this still work? because im stuck already | 2017-03-27T04:54:32 | {
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https://askdev.io/questions/54388/adjoint-functors | I'm attempting to cover my mind around adjoint functors. Among the instances I've seen is the groups $\bf IntLE \bf = (\mathbb{Z}, ≤)$ and also $\bf RealLE \bf = (\mathbb{R}, ≤)$, where the ceiling functor $ceil : \bf RealLE \rightarrow IntLE$ is left adjoint to the incorporation functor $incl : \bf IntLE \rightarrow RealLE$. I intend to examine that the adhering to hold true, as they appear to be:
1. $floor : \bf RealLE \rightarrow IntLE$ would certainly be appropriate adjoint to $incl$
2. Between the twin groups of $\bf IntGE \bf = (\mathbb{Z}, ≥)$ and also $\bf RealGE \bf = (\mathbb{R}, ≥)$, $ceil$ would certainly be appropriate adjoint to $incl$
3. Between $\bf RealGE$ and also $\bf IntGE$, $floor$ would certainly be left adjoint to $incl$
Is my understanding deal with on these factors?
0
2019-05-18 23:47:09
Source Share
Arturo has actually currently uploaded a wonderful solution. I 'd just such as to stress that such global interpretations usually enable glossy evidence, as an example see listed below. For a far more striking instance see the theory in my post here, which offers a glossy one - line evidence of the LCM * GCD regulation using their global interpretations.
LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \$ for $\rm\ \ n > 0$
Proof $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$
$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$
$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$
$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$
$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$
$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor$
Compare the above unimportant evidence to even more typical evidence, as an example the special case $\rm\ m = 1\$ here.
0
2019-05-21 10:54:48
Source
Given groups $\mathcal{C}$ and also $\mathcal{D}$, and also functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and also $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathbf{U}$ if and also just if for every single things $C\in\mathcal{C}$ and also $D\in\mathcal{D}$ there is an all-natural bijection in between $\mathcal{C}(C,\mathbf{U}(D))$ and also $\mathcal{D}(\mathbf{F}(C),D)$.
Allow me make use of $\leq$ and also $\geq$ for the relationship amongst reals, and also $\preceq, \succeq$ for the relationship amongst integers.
For $ceil$ to be the left adjoint of the incorporation functor, you would certainly require that for all actual numbers $r$ and also all integers $z$, $\lceil r\rceil \preceq z$ if and also just if $r\leq z$. This holds, so you do have an ajunction (this functions due to the fact that in these groups, the morphism set IntLE $(a,b)$ is vacant if $a\not\preceq b$, and also has an one-of-a-kind arrowhead if $a\preceq b$ ; and also in a similar way with RealLE ; so you get an all-natural bijection in between the collections of arrowheads if and also just if they are either both vacant or both are singletons at the very same time).
For $floor$ to be an appropriate adjoint to $incl$, you would certainly require that for all actual numbers $r$ and also all integers $z$, $z\preceq \lfloor r\rfloor$ if and also just if $z\leq r$, which once more holds true ; so $floor$ is an appropriate adjoint to the incorporation functor.
For $ceil$ to be the appropriate adjoint to the incorporation functor in the twin groups, you would certainly require $z\succeq \lceil r \rceil$ if and also just if $z\geq r$ ; and also for $floor$ to be the left adjoint, you would certainly require $\lfloor r\rfloor \succeq z$ if and also just if $r\geq z$. Both hold, so your assertions 1 via 3 are proper.
P.S. Let me 2nd Mariano is pointer in the remarks to remember the instance of the underlying set functor and also the free team functor for thinking of right and also left adjoints. I locate myself returning to those 2 every single time I require to advise myself of just how points collaborate with adjoints, what adjoints regard or do not regard, and also specifically when thinking of several of the various other equal interpretations, specifically the one in regards to the device and also carbon monoxide - device of the adjunction (which are all-natural makeovers in between the identification functors and also the functors $\mathbf{FU}$ and also $\mathbf{UF}$).
0
2019-05-21 10:24:42
Source | 2021-11-27T20:25:34 | {
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http://math.stackexchange.com/questions/235514/two-answers-for-a-conditional-probability-problem | Two answers for a conditional probability problem
Here's a question I'm thinking about:
There are three drawers. Drawer A contains 2 black socks, Drawer 2 contains two white socks and the third drawer is mixed. If I pulled a sock from one of the drawers. Given that it is white, what is the probability that the other sock in the drawer is white?
So what I'm thinking is that I picked either the second or the third. If I picked the second I know the other one is white, and if I picked the third I know that the other one is black and so the probability is $1\cdot \frac{1}{2} + 0\cdot \frac{1}{2}=\frac{1}{2}$
I saw someone else's solution and he says that after picking one sock we have a new sample space $$\Omega=\{(W_1,W_2),(W_2,W_1),(W_m,B)$$ where $W_1,W_2$ are the two in the second drawer and $W_m$ is the one in the mixed drawer. Anyway he concluded that the probability is $$P(second\, is\, white)=\frac{\{(W_1,W_2),(W_2,W_1)\}}{|\Omega|}=\frac{2}{3}$$
I think he's mistaken for counting the order in which the socks were pulled, because it is given that we pulled a white one, it doesn't matter which exactly.
So, who is right? And if I'm right, was I also right about his mistake?
Thanks!
-
You mean "Drawer A contains $2$ black socks"? – joriki Nov 12 '12 at 7:28
"What is the probability that the other sock in the drawer is white?" is a strange question, since the sock's colour is deterministic. – Cocopuffs Nov 12 '12 at 7:34
@Cocopuffs: That's only under some interpretations of probability. I would argue that an interpretation of probability that doesn't allow us to speak of a probability in this case offers an unduly restricted concept of probabilities. – joriki Nov 12 '12 at 7:38
@joriki: yes that's what I meant – Yotam Nov 12 '12 at 7:40
By the way, this is a well-known problem: en.wikipedia.org/wiki/Bertrand's_box_paradox – Cocopuffs Nov 12 '12 at 8:09
If you were right, then obviously the answer to "What is the probability that the second sock is black, given that the first is black?" would also be $\frac12$.
What would be your answer to "What is the probability that the second sock has the same colour (no matter which) as the first?" - of course $\frac23$ because this holds for two out of three drawers. By symmetry, the answer to this question cannot suddenly change from $\frac23$ to $\frac 12$ if the asker continues his question with "By the way, the first sock is white" (or black or he dies mid-sentence of a heart attack). | 2014-12-18T17:45:08 | {
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https://www.physicsforums.com/threads/graphical-interpretation-of-a-double-integral.808066/ | # Graphical interpretation of a double integral?
1. Apr 11, 2015
### beer
Hello,
I was helping my friend prepare for a calculus exam today - more or less acting as a tutor.
He had the following question on his exam review:
∫∫R y2 dA
Where R is bounded by the lines x = 2, y = 2x + 4, y = -x - 2
I explained to him that R is a triangle formed by all three of those lines. The solution to the integral is 192, which is pretty easy to calculate by hand without a calculator.
He asked me why the solution wasn't just the area of R (two triangles if you split R at the line y = 0) which is 128. I told him its because we are integrating y2 with respect to the given boundaries, and that if we were integrating "1" it would be the area of the region R. However, we then proceeded to integrate "1" with the same boundaries, and the answer was 24, not 128.
So I ended up coming up with my own question. I can draw a pretty good geometric interpretation of most integration problems including three dimensional ones given in different coordinate systems... but I feel like I'm missing something primitive here. Can anyone help me develop a geometric picture of what this type of double integral is representing?
Thanks!
2. Apr 11, 2015
### robphy
How did you get 128?
3. Apr 11, 2015
### beer
That's a fantastic question. I multiplied the "divided region" together, rather than added it. The area of R is 24. :p
We didn't catch on to that earlier, but it is kind of irrelevant to my primary question.
Where does y2 fit in - graphically speaking - with the region R?
4. Apr 11, 2015
### robphy
5. Apr 11, 2015
### beer
Yes I'm familiar with those concepts. Interestingly some more complex integrals are easier for me to visualize. I'm likely over thinking this and confusing myself.
y2 represents a parabola (or more technically a parabolic cylinder) that is orthogonally oriented to the triangle formed by the lines given by R. Such a shape would possess no volume in the "real world"... it would be two shapes orthogonal to one another; a triangle and a parabola, whose intersection forms a line.
That is what is confusing me.
6. Apr 11, 2015
### beer
I drew it out by hand and by golly I think I've got it.
This is the "projection" of the triangle on to the parabolic cylinder, yes? So the "stretched" area of the triangle as it would appear on the parabola z = y2
Part of my confusion was probably incorrectly calculating the area of the region with my friend earlier. As integrating the same region with respect to 1 (or any other constant) would project the region on to a flat surface parallel to the region itself, thus yielding a solution to the integral equal to the area of the region itself...
Now I feel only slightly ashamed.
7. Apr 12, 2015
### SteamKing
Staff Emeritus
This integral can also represent the second moment of area about the x-axis. The second moment of area is sometimes referred to as the moment of inertia.
In this case, since the region R straddles the x-axis, evaluating the double integral will calculate the combined second moment of area for the two right triangles, referenced to their common base, which are formed when the region R is intersected by the x-axis.
The moment of inertia for a right triangle about its base Iy = (1/12)bh3
In this case, the length of the common base for the two triangles is b = 4, and the height of the upper triangle is h = 8 while the lower triangle height is h = 4.
Thus
Iy = Iy upper + Iy lower = (1/12)*4*(83 + 43)
Iy = 192
8. Apr 12, 2015
### acegikmoqsuwy
What a double integral generally represents is the volume of a solid above a region on a plane. The solid you have here is called a cylindrical surface. The equation of it is $z=y^2$. This is a parabola. However, since there is no x, we can let x vary from negative infinity to infinity. Doing so moves this parabola forward and backward parallel to the x-axis. The region "carved out" is the surface. And the double integral is the volume under that surface above the triangular region R. | 2018-02-20T04:08:28 | {
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http://mathhelpforum.com/calculus/43848-clueless-front-integral-s-not-complicated-one-though.html | # Math Help - Clueless in front of an integral (it's not a complicated one though)
1. ## Clueless in front of an integral (it's not a complicated one though)
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
2. Originally Posted by arbolis
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
Ok
Way two
$\int\frac{dx}{1+\cos(x)}$
Now we $x=2\arctan(u)$
So $dx=\frac{2}{u^2+1}$
Ok so lets see what
$\cos\left(2\arctan(x)\right)$ is
We see that
$\cos(2x)=2\cos^2(x)-1$
So then $\cos\left(2\arctan(x)\right)=2\cos^2\left(\arctan( x)\right)-1$
Now you should know that
$\cos\left(\arctan(x)\right)=\frac{1}{\sqrt{x^2+1}}$
So
$\cos^2\left(\arctan(x)\right)=\frac{1}{x^2+1}$
So
$\cos\left(2\arctan(x)\right)+1=\bigg[2\cdot\frac{1}{x^2+1}-1\bigg]+1=\frac{2}{x^2+1}$
So now we see that we are working in u's so
$\int\frac{2~du}{u^2+1}\cdot\frac{1}{\frac{2}{u^2+1 }}=\int~du=u+C$
So $x=2\arctan(x)\Rightarrow{u=\tan\left(\frac{x}{2}\r ight)}$
So
$\int\frac{dx}{1+\cos(x)}=\tan\left(\frac{x}{2}\rig ht)$
3. Originally Posted by arbolis
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
Do not do that. Just use $2\cos^2 \tfrac{x}{2} = 1+\cos x$.
4. You could also note that
$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos (x)}{2}}$
Squaring both sides gives us
$\cos^2\left(\frac{x}{2}\right)=\frac{1+\cos(x)}{2}$
So $2\cos^2\left(\frac{x}{2}\right)=1+\cos(x)$
EDIT: Looks like TPH beat me to the punch
5. Hello
Originally Posted by arbolis
but I don't understand why we bother with a coefficient of $2$ here...
Because with $t=\tan \frac{x}{2}$ one has $\cos x=\frac{1-t^2}{1+t^2}$. (trig. identity)
$x=2\arctan t \implies \mathrm{d}x=\frac{2}{1+t^2}\,\mathrm{d}t$
\begin{aligned}
\int\frac{1}{1+\cos x}\,\mathrm{d}x=\int\frac{1}{1+\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\,\mathrm{d}t
&=\int\frac{2}{1+t^2+\frac{1-t^2}{1+t^2}(1+t^2)}\,\mathrm{d}t\\
&=\int\frac{2}{2}\,\mathrm{d}t\\
&=t+C\\
&=\tan \frac{x}{2}+C\\
\end{aligned}
6. Nice answers, thank you all. I have another integral to do with the same substitution, so I might ask help in another thread . But I hope not.
7. Multiply & divide by $1-\cos x.$ | 2015-03-06T11:41:58 | {
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http://math.stackexchange.com/questions/43640/questions-about-open-sets-in-mathbb-r/43642 | # Questions about open sets in ${\mathbb R}$
Consider the following problem:
Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?
A. If $A$ is open, then $A={\mathbb R}$
B. If $A$ is closed, then $A={\mathbb R}$
Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer.
Here are my questions:
Why A is not necessarily true? What can be a counterexample?
-
A slightly more interesting example than Luboš's can be obtained by enumerating the rationals as $\mathbb{Q} = \{q_n\}_{n=1}^\infty$ and taking $A = \bigcup_{n=1}^{\infty} (q_{n} - \frac{\varepsilon}{2^{n+1}}, q_{n} + \frac{\varepsilon}{2^{n+1}})$. Then the Lebesgue measure of $A$ can be estimated by $\mu(A) \leq \sum_{n=1}^{\infty} 2 \cdot \frac{\varepsilon}{2^{n+1}} = \varepsilon$, so $A$ cannot be all of $\mathbb{R}$ even if it's clearly open.
-
It's amazing that without knowing what the irrational numbers are left, one can get that $A$ cannot be ${\bf R}$ through Lebesgue measure. Nice! – Jack Jun 6 '11 at 15:50
@Jack: Of course, I used the deceptively innocent-looking fact that the Lebesgue measure of an interval is its length! The comments to Gowers's answer show that even some professional mathematicians are not aware of its non-triviality (or forgot about it)... – t.b. Jun 6 '11 at 16:09
This is a very nice classical example. +1, I liked it so much when I saw it the first time :) it's like saying you can "cover" a dense subset of an infinitely long line with a simple woodstick (of finite length), if you "cut it correctly" and "place the pieces at the right places". It shows how mathematics can rudely destroy intuition. =P – Patrick Da Silva Jul 29 '11 at 5:45
@Patrick: Right. I find it even better when you write $A_{\varepsilon}$ for the set in this answer and take $N = [0,1] \cap \bigcap_{k=1}^{\infty} A_{1/k}$. Then you get a dense $G_{\delta}$ (hence a set of second Baire category) in $[0,1]$. This set is generic in the sense of Baire, but it has zero Lebesgue measure. In other words: the two notions of generic that are commonly used (one in the sense of Baire, the other in the sense of probability) are quite incompatible. – t.b. Jul 29 '11 at 7:58
@Patrick: Oh, sorry. The set $N$ has measure zero, right? So if you pick a point randomly (wrt Lebesgue measure on $[0,1]$), it won't be in $N$ almost surely, so a "generic point will be outside $N$". On the other hand, this set is a countable intersection of open and dense sets, and according to the Baire category theorem, such sets are "large" or "generic" (in the sense of Baire). E.g.: a generic continuous function is nowhere differentiable. Now $N$ is "very small" in the Lebesgue sense, and "very large" in the Baire sense. That's it. – t.b. Jul 29 '11 at 22:29
A counterexample for the rule A is $${\mathbb R} \backslash F$$ where $F$ is any non-empty finite (or countable) set of irrational numbers. For example $${\mathbb R} \backslash \{\pi\}$$ Note that if I remove the point $\pi$, the set is still open on both sides from $\pi$. Because $\pi$ isn't rational, the set above still contains all rational numbers.
-
You couldn't remove any countable set - for example, removing $\{\frac{\pi}{n} \mid n \in \mathbb{N}\}$ would leave you with a set that's not open. – MartianInvader Jun 6 '11 at 16:46
"(or countable)" could be replaced with "(or closed)" to make this correct and general. – Jonas Meyer Jun 6 '11 at 18:11
The question boils down to whether there are nonempty subsets of $\mathbb{R}\setminus \mathbb{Q}$ that are closed in $\mathbb{R}$. The easiest examples are finite sets, as Luboš Motl noted. An easy infinite example is $\sqrt{2}+\mathbb{Z}$. Theo Buehler showed that there are positive measure examples, which is much stronger and closely related to the question at this link.
Another direction to strengthen the result is to show that there are perfect examples, which is the subject of the question at this link.
-
Nice wrap-up, Jonas, and prods to make connections with earlier questions/answers. A post such as this would be helpful for some answers that get a half-dozen (or more) answers which take a different angle, approach...and to help shed light on how strong (or generalizable/extendable) a statement one can make, "spring-boarding" off a single question, and connecting to other semi-related posts. – amWhy Jun 22 '11 at 5:35
"Why A is not necessarily true? What can be a counterexample?"
I'm surprised at the complexity of some answers given to this. Here's a counterexample: $$(-\infty,\pi)\cup (\pi,\infty).$$ You can construct lots of others similar to that but more complicated if need be.
-
How is $\mathbb{R} \smallsetminus \{\pi\}$ more complicated than your example? – t.b. Oct 8 '11 at 0:24
I.e. the complement of any discrete set of irrational numbers. – Michael Hardy Oct 8 '11 at 0:25
Sorry..... I hadn't thorougly read all of them. – Michael Hardy Oct 8 '11 at 0:26
@MichaelHardy: $\{\frac{\pi}{n}:n\in\mathbb N\}$ is a discrete set of irrationals, but its complement is not open. The set of irrational numbers in question has to be closed. – Jonas Meyer Dec 4 '11 at 6:38 | 2014-07-28T21:19:54 | {
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https://math.stackexchange.com/questions/743451/application-of-fermats-little-theorem | # application of Fermat's little theorem
show that $a^{13} \equiv a\mod 35$ using Fermat's little theorem. Use Fermat's little theorem with primes 5 and 7.
$a^7 \equiv a (mod 7$) and
$a^5 \equiv a (mod 5$)
• Yes? That sounds like a good plan. – Henning Makholm Apr 7 '14 at 12:11
If $p$ is a prime, either $p|a$ or $(p,a)=1$
If $\displaystyle p|a, p$ will divide $\displaystyle a^n-a=a(a^{n-1}-1)$ for integer $n-1\ge0$
For $\displaystyle(p,a)=1, a^{p-1}-1\equiv0\pmod p$ by Fermat's Little Theorem
So, $p$ will divide $\displaystyle a(a^m-1)$ for all integer $a$ if $(p-1)|m$
If $p=5,$ we need $4|m$
If $p=7,$ we need $6|m$
So, if $m$ is divisible by lcm$(4,6)=12;$ $5$ and $7$ will individually divide $a(a^m-1)$
Again as $(5,7)=1,$ it implies lcm$(5,7)=35$ will divide $a(a^m-1)$ if $12|m$
• "as 5 is prime"? What am I missing? Why does $b$ is prime have to be true for $b|a\implies b|(a^k-a)$? – Guy Apr 7 '14 at 12:14
• if i apply the little theorem i can have 2 congruence relations with mod 5 and mod 7. How can i maketwo mod relations together/ – srimali Apr 7 '14 at 12:17
• @user139296, please find the edited version – lab bhattacharjee Apr 7 '14 at 14:40
• @Sabyasachi, please find the edited version – lab bhattacharjee Apr 7 '14 at 14:42
• @labbhattacharjee oh okay. Lot clearer now. Thanks. – Guy Apr 7 '14 at 14:45
Hint $\$ Let $\,p,q\,$ be distinct primes. Then by Fermat's little Theorem $\,\color{#c00}{\rm F\ell T}$ we deduce
$\qquad n = (p\!-\!1)k\,\Rightarrow\,{\rm mod}\ p\!:\ a^{1+n} = a (a^{p-1})^k\overset{\color{#c00}{\rm F\ell T}}\equiv a\,\$ [note it is clear if $\,a\equiv 0$]
Thus $\,p\!-\!1,q\!-\!1\mid n\,\Rightarrow\, p,q\mid a^{1+n}-a\,\Rightarrow\,pq\mid a^{1+n}-a,\$ by $\ {\rm lcm}(p,q) = pq$
Thus $\,p,q=5,7\,\Rightarrow\, 4,6\mid 12\,\Rightarrow\,a^{\large 1+12}\equiv a\pmod{5\cdot 7}$ | 2019-12-13T21:51:18 | {
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http://loza.renejoosten.eu/7yfhgle/852a33-product-of-symmetric-and-antisymmetric-tensor | and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: They show up naturally when we consider the space of sections of a tensor product of vector bundles. Antisymmetric and symmetric tensors. Let be Antisymmetric, so (5) (6) The (inner) product of a symmetric and antisymmetric tensor is always zero. Now take the inner product of the two expressions for the tensor and a symmetric tensor ò : ò=( + ): ò =( ): ò =(1 2 ( ð+ ðT)+ 1 2 (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. Anti-Symmetric Tensor Theorem proof in hindi. At least it is easy to see that $\left< e_n^k, h_k^n \right> = 1$ in symmetric functions. MTW ask us to show this by writing out all 16 components in the sum. Antisymmetric and symmetric tensors. This can be seen as follows. Fourth rank projection tensors are defined which, when applied on an arbitrary second rank tensor, project onto its isotropic, antisymmetric and symmetric … A rank-1 order-k tensor is the outer product of k nonzero vectors. We can define a general tensor product of tensor v with LeviCivitaTensor[3]: tp[v_]:= TensorProduct[ v, LeviCivitaTensor[3]] and also an appropriate tensor contraction of a tensor, namely we need to contract the tensor product tp having 6 indicies in their appropriate pairs, namely {1, 4}, {2, 5} and {3, 6}: A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. A second-Rank symmetric Tensor is defined as a Tensor for which (1) Any Tensor can be written as a sum of symmetric and Antisymmetric parts (2) The symmetric part of a Tensor is denoted by parentheses as follows: (3) (4) The product of a symmetric and an Antisymmetric Tensor is 0. For convenience, we define (11) in part because this tensor, known as the angular velocity tensor of , appears in numerous places later on. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. symmetric tensor so that S = S . A rank-2 tensor is symmetric if S =S (1) and antisymmetric if A = A (2) Ex 3.11 (a) Taking the product of a symmetric and antisymmetric tensor and summing over all indices gives zero. anti-symmetric tensor with r>d. B, with components Aik Bkj is a tensor of order two. To define the indices as totally symmetric or antisymmetric with respect to permutations, add the keyword symmetric or antisymmetric,respectively, to the calling sequence. Tensor products of modules over a commutative ring with identity will be discussed very briefly. 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Writing out all 16 components in the most basic geometrical terms, a relationship other! Indices, then the tensor is the antisymmetric part of the idea of a vector up naturally when we the... By the form that is necessary to reconstruct it b, with components Aik Bkj is generalization! Components Aik Bkj is a tensor is a tensor is a tensor sign... Well as the terms that are summed tensor times an antisym- antisymmetric and symmetric tensors we to. Indices, then the tensor and the symmetric traceless part anti-symmetrization is denoted by a pair of brackets... But for the pendantic among the audience, here goes 3... Spinor and! Conductivity and resistivity tensor... Geodesic deviation in Schutz 's book: a typo that is used as! Types differ by the form used is symmetric if aij = aji you..., h_k^n \right > = 1 \$ in symmetric functions be decomposed into its part... Ask Question Asked 3... Spinor indices and antisymmetric tensor is a higher order generalization the. Basic geometrical terms, a tensor of order two the audience, here goes ). Type of array 2 covariant tensor M, and strain are, the... If aij = aji ) product of k nonzero vectors with the symmetry of... | 2021-10-25T13:48:54 | {
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https://datascience.stackexchange.com/questions/10615/number-of-parameters-in-an-lstm-model/17412 | # Number of parameters in an LSTM model
How many parameters does a single stacked LSTM have? The number of parameters imposes a lower bound on the number of training examples required and also influences the training time. Hence knowing the number of parameters is useful for training models using LSTMs.
The LSTM has a set of 2 matrices: U and W for each of the (3) gates. The (.) in the diagram indicates multiplication of these matrices with the input $$x$$ and output $$h$$.
• U has dimensions $$n \times m$$
• W has dimensions $$n \times n$$
• there is a different set of these matrices for each of the three gates(like $$U_{forget}$$ for the forget gate etc.)
• there is another set of these matrices for updating the cell state S
• on top of the mentioned matrices, you need to count the biases (not in the picture)
Hence total # parameters = $$4(nm+n^{2} + n)$$
• I faced this question myself when taking practical decisions on estimating hardware requirements and project planning for a deep learning project. PS: I didn't answer my own question to just gain reputation points. I want to know if my answer is right from the community. – wabbit Mar 9 '16 at 11:17
• You have ignored bias units. See Adam Oudad's answer below. – arun Jun 20 '18 at 0:11
• Biases are not there. I have edited the answer. – Escachator Oct 20 '18 at 12:09
• Doesn't this then need to be multiplied by the number of lstm units in the layer? Here isn't this only the number of params in a single LSTM-cell? – Joe Black May 27 '20 at 18:03
Following previous answers, The number of parameters of LSTM, taking input vectors of size $m$ and giving output vectors of size $n$ is:
$$4(nm+n^2)$$
However in case your LSTM includes bias vectors, (this is the default in keras for example), the number becomes:
$$4(nm+n^2 + n)$$
• This is the only complete answer. Every other answer appears content to ignore the case of bias neurons. – user45817 Feb 7 '18 at 14:11
• To give a concrete example, if your input has m=25 dimensions and you use an LSTM layer with n=100 units, then number of params = 4*(100*25 + 100**2 + 100) = 50400. – arun Jun 20 '18 at 0:13
• Suppose I am using timestep data, is my understanding below correct? n=100: mean I will have 100 timestep in each sample(example) so I need 100 units. m=25 mean at each timestep, I have 25 features like [weight, height, age ...]. – jason zhang Mar 10 '19 at 6:41
• @jasonzhang The number of timesteps is not relevant, because the same LSTM cell will be applied recursively to your input vectors (one vector for each timestep). what arun called "units" is also the size of each output vector, not the number of timesteps. – Adam Oudad Mar 11 '19 at 8:24
According to this:
LSTM cell structure
LSTM equations
Ingoring non-linearities
If the input x_t is of size n×1, and there are d memory cells, then the size of each of W∗ and U∗ is d×n, and d×d resp. The size of W will then be 4d×(n+d). Note that each one of the dd memory cells has its own weights W∗ and U∗, and that the only time memory cell values are shared with other LSTM units is during the product with U∗.
Thanks to Arun Mallya for great presentation.
to completely receive you'r answer and to have a good insight visit : https://towardsdatascience.com/counting-no-of-parameters-in-deep-learning-models-by-hand-8f1716241889
g, no. of FFNNs in a unit (RNN has 1, GRU has 3, LSTM has 4)
h, size of hidden units
i, dimension/size of input
Since every FFNN(feed forward neural network) has h(h+i) + h parameters, we have
num_params = g × [h(h+i) + h]
Example 2.1: LSTM with 2 hidden units and input dimension 3.
g = 4 (LSTM has 4 FFNNs)
h = 2
i = 3
num_params
= g × [h(h+i) + h]
= 4 × [2(2+3) + 2]
= 48
input = Input((None, 3))
lstm = LSTM(2)(input)
model = Model(input, lstm)
thanks to RAIMI KARIM
To make it clearer , I annotate the diagram from http://colah.github.io/posts/2015-08-Understanding-LSTMs/.
ot-1 : previous output , dimension , n (to be exact, last dimension's units is n )
i: input , dimension , m
fg: forget gate
ig: input gate
update: update gate
og: output gate
Since at each gate, the dimension is n, so for ot-1 and i to get to each gate by matrix multiplication(dot product), need nn+mn parameters, plus n bias .so total is 4(nn+mn+n). | 2021-04-22T15:08:08 | {
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http://math.stackexchange.com/questions/819412/the-number-229-has-exactly-9-distinct-digits-which-digit-is-missing | # The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
-
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate. – Brilliand Jun 3 '14 at 18:33
"Has exactly 9 digits, all of which are distinct" might be more accurate. – dan08 Jun 3 '14 at 19:32
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}\approx500,000,000$. So, $2^{29}$ has $9$ digits. – robjohn Jun 3 '14 at 19:54
I see how we know it has 9digits, $29\log 2=8.73$ but how does one know the digits are distinct ? – Rene Schipperus Jun 3 '14 at 22:38
Just asking: Is it (the competition) SMO? – user148697 Jun 27 '14 at 13:42
Oh its so easy, now that we follow the hint (thanks !)
$$\sum k_n 10^n \equiv \sum k_n \mod 9$$ the sum of all the digits is $\frac{9(9+1)}{2}\equiv 0 \mod 9$ so the sum of all but one $x$ is $\equiv -x \mod 9$
Now $$2^{29}\equiv -4 \mod 9$$ so $4$ is the missing digit.
-
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} \pmod 9$. Any hint will also suffice! Thanks for the answer, by the way! – puru Jun 28 '14 at 2:51
You find the power of two mod 9, that is $2^n\equiv 1\pmod 9$ I forget what it is but then you divide it into 29. – Rene Schipperus Jun 28 '14 at 2:58
Thanks, I got it. In fact, $2^{29}=2^{27} \times 4=(9-1)^9 \times 4$ which implies that $2^{29} \pmod 9= -4$ – puru Jun 28 '14 at 9:29
A hint: Think about the remainder modulo $9$.
-
Oh its so easy, now that we follow the hint (thanks !) – Rene Schipperus Jun 3 '14 at 16:32
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024\cdot1024\cdot512$, which can be done by hand quickly in a competition.
-
It's about creative solution not answering it – Karo Jun 3 '14 at 16:08
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work? – dREaM Jun 3 '14 at 16:11
whats should we do? don't think on other methods to solve it? – Karo Jun 3 '14 at 16:22
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever? – alex.jordan Jun 3 '14 at 16:38
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this. – Karo Jun 3 '14 at 17:36
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,\dots,8$ so that $$N = \sum_{i=0}^{8} d_i 10^i$$ We first notice that $$\sum_{i=0}^{8} d_i 10^i \equiv \sum_i d_i \pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $\sum d_i \equiv 0 \pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $\sum_i d_i \equiv 0 \pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $\pmod{8} = N \pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N \equiv 5 \pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x \pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N \equiv 0 \pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
-
This looks eerily similar to Rene Schipperus' answer. – robjohn Jun 3 '14 at 20:00
It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$... – user21820 Jun 4 '14 at 11:58 | 2015-07-06T11:43:29 | {
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http://mathhelpforum.com/advanced-applied-math/28902-applications-dot-product.html | # Math Help - Applications of Dot Product
1. ## Applications of Dot Product
A crate with a weight of 57 N rests on a frictionless ramp inclined at an angle of 30 degrees to the horizontal. What force must be applies at an angle of 20 degrees to the ramp so that the crate remains at rest?
I've drew the diagram and I don't know what to do after this:
I know the formula to use is "u (dot) v = |u||v|cosx"
The answer to this problem is 28.5 N and 30.3 N.
Can you please show me a step by step solution to get the answer?
2. Originally Posted by Macleef
A crate with a weight of 57 N rests on a frictionless ramp inclined at an angle of 30 degrees to the horizontal. What force must be applies at an angle of 20 degrees to the ramp so that the crate remains at rest?
I've drew the diagram and I don't know what to do after this:
I know the formula to use is "u (dot) v = |u||v|cosx"
The answer to this problem is 28.5 N and 30.3 N.
Can you please show me a step by step solution to get the answer?
I've attached a sketch with all the forces acting on the solid:
$F_w = \text{weight}$
$F_d = \text{downhill force}$
$F_u = \text{uphill force}$
$F_p = \text{force to pull}$
$| \overrightarrow{F_d} | =| \overrightarrow{ F_w} | \cdot \cos(60^\circ) = 28.5\ N$
$\overrightarrow{F_u} = -| \overrightarrow{ F_d}$
$|\overrightarrow{F_u}| = |\overrightarrow{F_p}| \cdot \cos(20^\circ)~\implies~ |\overrightarrow{F_p}| = \frac{|\overrightarrow{F_u}|}{\cos(20^\circ)}\appr ox 30.329\ N$
3. Originally Posted by earboth
I've attached a sketch with all the forces acting on the solid:
$F_w = \text{weight}$
$F_d = \text{downhill force}$
$F_u = \text{uphill force}$
$F_p = \text{force to pull}$
$| \overrightarrow{F_d} | =| \overrightarrow{ F_w} | \cdot \cos(60^\circ) = 28.5\ N$
$\overrightarrow{F_u} = -| \overrightarrow{ F_d}$
$|\overrightarrow{F_u}| = |\overrightarrow{F_p}| \cdot \cos(20^\circ)~\implies~ |\overrightarrow{F_p}| = \frac{|\overrightarrow{F_u}|}{\cos(20^\circ)}\appr ox 30.329\ N$
what is F_n?
4. Originally Posted by Jhevon
what is F_n?
sorry I forgot to mention:
$F_n = \text{normal force acting perpendicularly at the surface of the inclined plane}$ | 2015-04-25T18:52:43 | {
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https://math.stackexchange.com/questions/1453231/trying-to-apply-mathematics-into-a-real-life-context | # Trying to apply mathematics into a real-life context
Earlier today, I was trying to place my storybook into my wardrobe drawer, only to find that I could only place it horizontally as the height of my book exceeds the height of my drawer. Out of curiosity, I tried to rotate the book by holding the spine and observe how much it can rotate, until three of the vertices touched a point of the drawer.
A question came to me as I had reached a point of time when the book could no longer rotate further: By keeping the height of the drawer constant, what is the minimum length of the drawer such that the book can be stored inside it? (A silly question, I know, considering that drawers are always quite long, but aroused my interest nevertheless..)
I proceeded by measuring the dimensions I think I need to solve my problem. I have measured: The height of the drawer, $17 cm$
The length of the spine of the book, $2.2 cm$
The height of the spine of the book, $17.8 cm$
And I have attempted to represent the problem in the diagram below, solving for $x$ With my mathematics knowledge up to grade 10, initially, I had thought that this will be solved using simultaneous equations, but after pondering for quite some time, I had managed to come up with an approach using R-Formula:
I started by setting $\angle WDA=\theta$, and since the drawer and my book are rectangular, $\angle BAX=\angle DCZ=\theta$
Using these information, I expressed $x$ in trigonometric terms: $x=17.8 \sin\theta + 2.2 \cos\theta$
Using the R-Formula, I get $x=\sqrt(321.68)\sin(\theta+7.045769125^\circ)$.
Next, I attempt to solve for $\theta$ by expressing $WZ=17$ in trigonometric terms as well.
I get: $17=17.8\cos\theta+2.2\sin\theta$
$17=\sqrt(321.68)\cos(\theta-7.045769125^\circ)$ $\theta\approx 25.63217562^\circ$
Finally, I substituted this value of $\theta$ into $x$ and I get: $x=\sqrt(321.68) \sin(32.67794475^\circ)$ $x\approx 9.683637217cm$
My actual measurement of the value of $x$ is $10.5cm$. I did expect inaccuracy as I measured just with a fifteen-centimetre ruler, a pair of wobbly hands and considered parallax error, although I did not expect the discrepancy to be nearly $1cm$.
I have two questions regarding this problem of mine:
Is my approach a valid approach?
Is there an easier approach to solve this problem?
The is already an error in your initial formula ... So that was a typo in your question, and I cannot find an error in your computation now.
Here is a possible different approach which gives an explicit formula for the result.
Using \begin{aligned} w &= 2.2 \text{ (width of book)}\\ h &= 17.8 \text{ (height of book)}\\ H &= 17 \text{ (height of shelf)}\\ y &= \text{distance from D to Z} \\ \theta &\text{ angle of book, as in your sketch} \end{aligned} we have $$\sin \theta = \frac yw \, , \quad \cos\theta = \frac{H-y}{h}$$ and therefore $$1 = \sin^2 \theta + \cos^2 \theta = \left(\frac yw\right) ^2 + \left(\frac{H-y}{h}\right)^2 \, .$$ This is a quadratic equation for $y$, and the positive solution is $$y = \frac{w^2H + wh\sqrt{w^2+h^2-H^2}}{w^2+h^2}$$ (One can conclude from $h > H$ that the other solution is negative.) It follows that $$\sin \theta = \frac yw = \frac{wH + h\sqrt{w^2+h^2-H^2}}{w^2+h^2} \\ \cos\theta = \frac{H-y}{h} = \frac{hH - w\sqrt{w^2+h^2-H^2}}{w^2+h^2}$$ Finally $$x=h \sin\theta + w \cos\theta = \frac{2whH + (h^2-w^2)\sqrt{w^2+h^2-H^2}}{w^2+h^2} .$$ Using PARI/GP, this evaluates to $x \approx 9.6836372165163056823433168189573517183$, which is your result.
I made a sketch with Geogebra which shows that the result is plausible:
• Oops! Yes I meant $\cos \theta$! Thanks! – ChrisJWelly Sep 27 '15 at 8:49
• Interesting approach! Thank you! However, when I tried substituting $/theta = 25.632 degrees$ into your equation of $x= 17.8\sin\theta + 2.2\cos\theta$, I obtained $x\approx 9.68$ as well. I believe you made an error there? – ChrisJWelly Sep 27 '15 at 9:02
• @ChrisJWelly: How embarrassing! I made a typo and actually computed 18.7 * sin(t) + 2.2*cos(t). Which means that our results are identical and both correct or wrong (I assume the first). – Martin R Sep 27 '15 at 9:05
• Ahh. I see. Yes, it is hoped that our answers are correct. – ChrisJWelly Sep 27 '15 at 9:10
• @ChrisJWelly: I have updated the solution with an explicit formula. – Martin R Sep 27 '15 at 10:32
Here's a slightly different way to visualize the problem:
When the book is as close to upright as possible, the diagonal $AC$ will just touch the top and bottom of the drawer. That is, you have a right triangle $\triangle AEC$ with hypotenuse equal to the diagonal of rectangle $ABCD$ and one leg, $AE$, equal to the height of the drawer.
The other diagonal of the book, $BD$, touches the end of the drawer at $D$ and makes right triangle $\triangle BFD$ with hypotenuse equal to the diagonal of rectangle $ABCD$ and one leg lying along the side of the drawer. The other leg is the length you want to measure, $x = BF$.
Let $\angle ADF = \theta$ and $\angle BDA = \alpha$. Then $\angle DAE = \theta$ (because line $DF$ is parallel to line $AE$) and $\angle BDA = \alpha$ (by congruent triangles). Considering the angles with vertex at $A$ and the angles with vertex at $D$, we have \begin{align} \angle CAE = \theta - \alpha, \\ \angle BDF = \theta + \alpha. \end{align} Then \begin{align} AC = BD &= \sqrt{(AB)^2 + (AD)^2}, \\ \alpha &= \arcsin\frac{AB}{BD}, \\ \theta - \alpha &= \arccos\frac{AE}{AC}, \\ \theta + \alpha &= (\theta - \alpha) + 2\alpha, \\ x = BF &= BD \sin(\theta + \alpha). \end{align}
Starting with $AB =2.2$, $AD=17.8$, and $AE=17$, work each of these equations one after the other, using known values on the right-hand side each time, and the answer comes out to $9.68363721431$, just as you found.
You may also notice that this method illustrates how the R-formula works. We have $$x = BF = AD \sin\theta + AB\cos\theta = 17.8\sin\theta + 2.2\cos\theta$$ for the same reasons you found, but also $$x = BF = BD \sin(\theta + \alpha) = \sqrt{321.68} \sin(\theta + 7.04576912\text{ degrees}).$$
• Wow! I found this rather complex at first, but upon closer scrutiny, I find this solution a beautiful representation of the R-Formula. Thank you! – ChrisJWelly Sep 27 '15 at 22:41 | 2019-07-22T15:53:41 | {
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https://math.stackexchange.com/questions/459727/continuity-on-open-interval | # Continuity on open interval
A function is said to be continuous on an open interval if and only if it is continuous at every point in this interval.
But an open interval $(a,b)$ doesn't contain $a$ and $b$, so we never actually reach $a$ or $b$, and therefore they're not defined, and points that are not defined are not continuous, in other words $f(a)$ and $f(b)$ don't exist which makes the interval $(a,b)$ discontinuous.
So what is this definition saying, because I thought that it can't be continuous at $a$ or $b$ since they are not defined (an open circle on the graph), but everywhere in between $a$ and $b$ it can still be continuous...
So is it just continuous between these points $a$ and $b$, and a jump discontinuity occurs at these two points? Why then does it say that it's continuous at every point in $(a,b)$, if we are not including $a$ and $b$?
Points on an open interval can be approached from both right and left, correct? why is it required to be continuous on open $(a,b)$ in order to be continuous on closed $[a,b]$, I don't understand this because $a$ and $b$ are not defined in $(a,b)$.
• Continuity is a property of functions, not intervals. It doesn't make sense to say an interval is continuous or discontinuous. – Michael Albanese Aug 4 '13 at 20:26
• @MichaelAlbanese I can't understand what you mean to explain.can you please elaborate.. – spectraa Sep 19 '14 at 5:45
Let's consider a really simple function. That way, we can look at how the terminology is used without worrying about the function's peculiar behavior.
Let $f(x)=0$ for all $x$. Then I claim:
• $f$ is continuous on the open interval $(0,1)$.
• $f$ is also continuous on $(0,2)$.
• $f$ is continuous on $(1,2)$.
• $f$ is continuous on $(3,4)$.
• $f$ is continuous on $(-2\pi,e)$.
• For any real numbers $a<b$, $f$ is continuous on $(a,b)$.
These statements are all true! Do you see why?
• no, not quite, sorry. Is it because we'll always be able to find a read # that's less than b but more than a? I mean that, there will always be a continuous stream of numbers because the amount of numbers between a & b is infinite, is that why? thanks for your input...I'm only just starting with calculus. – Emi Matro Aug 4 '13 at 21:06
• @user4150 No prob, let's start at the beginning. Forget about $a$ and $b$ for the moment. Do you see why $f$ is continuous on $(0,1)$? – Chris Culter Aug 4 '13 at 21:10
• No, I don't really...I think I kind of understand the concept, but mathematical notation/symbols are a little confusing. Can you explain in words please – Emi Matro Aug 4 '13 at 21:14
• @user4150 Sure, it gives me an excuse to try out the chat feature! Please drop by at chat.stackexchange.com/rooms/9960/continuity-on-open-intervals – Chris Culter Aug 4 '13 at 21:19
• sorry, i need 20 rep to chat – Emi Matro Aug 4 '13 at 21:21
As you stated in the definition, $f:X\rightarrow Y$ is continuous on $(a,b)\subseteq X$ if it is continuous at every point of $(a,b)$. Since $a,b\notin(a,b)$, we can have a discontinuity there. For example the characteristic function of $(a,b)$, $\chi_{(a,b)}:\mathbb{R}\rightarrow\mathbb{R}$, is continuous in $(a,b)$ but discontinuous at $a$ and $b$.
• @PeterTamaroff I never said that the function was not defined at $a$ and $b$. – Daniel Robert-Nicoud Aug 4 '13 at 20:27
• My bad. The problem is you're misaddressing the OP's issue. – Pedro Tamaroff Aug 4 '13 at 20:28
• @PeterTamaroff I think OP's issue is not quite clear. Daniel's answer may address it. The issue is something about what happens at the endpoints of the interval. Daniel has provided an example with a finite discontinuity. In my answer I have given an infinite discontinuity. Maybe the response will clarify matters. – Mark Bennet Aug 4 '13 at 20:31
Think about the function $\frac 1x$ on the open interval $(0,1)$ - it is not defined at $0$, but this does not stop it being continuous on the interval - in fact it is continuous because the interval is open, and we never have to deal with the bad value $x=0$.
The function $\tan x$ for the interval $(-\frac{\pi}2,\frac {\pi}2)$ is continuous, with "problems" at both ends.
Perhaps you could explain your problem in relation to these functions, as it may help to tease out what your issue really is.
• I thought that $\lim_{x\rightarrow a} f(x)=f(a)$ means that $f(a)$ is defined and exists, how can it exist if $a$ is open and not defined? – Emi Matro Aug 4 '13 at 20:42
• What do you mean by "if $a$ is open"? – Michael Albanese Aug 4 '13 at 20:43
• @MichaelAlbanese the $a$ in $(a,b)$ is "open", I mean that it is not included as an endpoint, it is never reached, and thus it's not defined as an actual point – Emi Matro Aug 4 '13 at 20:49
• @user4150 If the limit exists, and equals the value of the function (because the function is defined beyond the open interval on which it is known to be continuous) the function is then continuous on one side at $a$ - we know nothing from what you have said about the limit from the other side. It is possible for the limit to exist, but to be different from the value of the function at $a$ – Mark Bennet Aug 4 '13 at 20:49
• @user4150: I understand what you mean, but it is an incorrect use of the word 'open'. Saying the $a$ in $(a, b)$ is open doesn't mean anything; you should say "$(a, b)$ is open". Part of understanding these ideas is getting your head around the terminology and its correct usage (in particular, to which objects it applies). It is especially important when using an online medium such as MSE. – Michael Albanese Aug 4 '13 at 20:53
You might find it helpful to think about the fact that some functions that are continuous on $(a,b)$ can be extended to $[a,b]$ to give a function continuous on the interval. For instance, the function defined by $f(x) = 1$ for all $x \in (0,1)$ can be continuously extended to a function $g$ defined by $g(x) = 1$ for all $x \in [0,1]$. On the other hand, a function like $h(x) = \frac{1}{x}$ is continuous on $(0,1)$, but cannot be continuously extended at $0$. | 2021-01-20T04:21:40 | {
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http://www.intmath.com/forum/exponents-radicals-16/multiplying-top-and-bottom-of-a-fraction:57 | IntMath Home » Forum home » Exponents and Radicals » Multiplying top and bottom of a fraction
# Multiplying top and bottom of a fraction [Solved!]
### My question
In the example on the linked page you say: "We multiply top and bottom of each fraction with their denominators. This gives us a perfect square in the denominator in each case, and we can remove the radical"
My question is : in what situations am I allowed to use this particular method? Can it be applied to any equation involving fractions, or is it only possible when there are radicals?
### What I've done so far
Tried it with other examples, but couldn't draw a conclusion.
X
In the example on the linked page you say: "We multiply top and bottom of each fraction with their denominators. This gives us a perfect square in the denominator in each case, and we can remove the radical"
My question is : in what situations am I allowed to use this particular method? Can it be applied to any equation involving fractions, or is it only possible when there are radicals?
Relevant page
What I've done so far
Tried it with other examples, but couldn't draw a conclusion.
## Re: Multiplying top and bottom of a fraction
Hi Daniel
This technique will only be worthwhile in a limited number of cases.
For an example where it wouldn't help at all, consider 1/2.
If I multiply top and bottom by 2 (the denominator), I will get 2/4. But so what? I just need to cancel and get back to 1/2. I can do the multiplying, but it doesn't help me.
In the example that you are referring to, however, the 2 fractions become simpler since their denominators no longer have square roots.
Taking a simpler case:
1/sqrt(3a)
When I multiply top and bottom by sqrt(3a), I get:
1/sqrt(3a) xx sqrt(3a)/sqrt(3a) = sqrt(3a)/(3a)
The process has "rationalized" the denominator.
Hope that makes sense.
5. Multiplication and Division of Radicals (Rationalizing the Denominator)
About half way down, it has a heading "Rationalizing the Denominator". In the examples, you will see a method of multiplying top and bottom so we get rid of the square roots on the bottom. It was similar thinking (but actually simpler) that I was using in the example you are asking about.
Good luck with it.
X
Hi Daniel
This technique will only be worthwhile in a limited number of cases.
For an example where it wouldn't help at all, consider 1/2.
If I multiply top and bottom by 2 (the denominator), I will get 2/4. But so what? I just need to cancel and get back to 1/2. I can do the multiplying, but it doesn't help me.
In the example that you are referring to, however, the 2 fractions become simpler since their denominators no longer have square roots.
Taking a simpler case:
1/sqrt(3a)
When I multiply top and bottom by sqrt(3a), I get:
1/sqrt(3a) xx sqrt(3a)/sqrt(3a) = sqrt(3a)/(3a)
The process has "rationalized" the denominator.
Hope that makes sense.
<a href="/exponents-radicals/5-multiplication-division-radicals.php">5. Multiplication and Division of Radicals (Rationalizing the Denominator)</a>
About half way down, it has a heading "Rationalizing the Denominator". In the examples, you will see a method of multiplying top and bottom so we get rid of the square roots on the bottom. It was similar thinking (but actually simpler) that I was using in the example you are asking about.
Good luck with it.
## Re: Multiplying top and bottom of a fraction
Great, thanks
X
Great, thanks
## Re: Multiplying top and bottom of a fraction
You're asking good questions, Daniel. All the best to you.
X
You're asking good questions, Daniel. All the best to you.
You need to be logged in to reply. | 2016-12-08T21:57:02 | {
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http://math.stackexchange.com/questions/44773/how-to-solve-polynomial-equations-in-a-field-and-or-in-a-ring | # How to solve polynomial equations in a field and/or in a ring?
I'm studying for my exam, and I stuck on solving polynomials in a field and/or in a ring. Let me give you some examples:
(1) Solve equation $x^2+4x+3=0$ in field $\mathbb{Z}_5$, $\mathbb{Z}_8$ and in ring $\mathbb{Z}_{12}$ and $\mathbb{Z}_5 \times \mathbb{Z}_{13}$
(2) Solve equation $x^2+2x=1$ in field $\mathbb{Z}_7$, $\mathbb{Z}_{11}$ and in ring $\mathbb{Z}_{12}$
(3) Find ring $\mathbb{Z}_n$, in which equation $x^2+2x=3$ has at least 3 solutions.
I'm not asking for solutions for these particular problems. I want to understand this topic, so I would appreciate example solutions of similar problems. Also, you're aware of some study materials - please list tutorials, books etc., about solving polynomial equations in different rings and fields (and similar topics).
Thanks for help.
-
Typo, $\mathbb{Z}_8$ is not a field. (There is a field with $8$ elements, but it is not (isomorphic to) $\mathbb{Z}_8$, which has zero divisors.) – André Nicolas Jun 11 '11 at 14:57
In a finite ring you can always cop out and simply check all the elements of the ring, and see, whether they are solutions or not. Teacher may not like it, but there's nothing he or she can do about it, because the logic is impeccable. Do realize that you will waste a bit of time, so in an exam you may not afford to do this :-)
If $p$ is a prime, then $\mathbf{Z}_p$ is a field. Then you always have the result that a polynomial of degree $n$ can have at most $n$ zeros (counted with multiplicity). So if you have found $n$, you can stop looking.
The usual factorization tricks always work, if $p$ is a prime. However, note that some rings have zero divisors. For example $2\cdot 4\equiv 0\pmod{8}$, so the polynomial $x^2-1=(x-1)(x+1)$ has 3 as a root in the ring $\mathbf{Z}_8$, even though $(3-1)=2$ and $(3+1)=4$ are both non-zero. In fact, it is not hard to see that see that this polynomial has exactly 4 zeros in the ring $\mathbf{Z}_8$.
The usual quadratic formula also works in the ring $\mathbf{Z}_p$, $p>2$ prime. You just need to be careful about square roots. For example in the ring $\mathbf{Z}_7$, we have $3^2=9=2$, so if the quadratic formula calls for $\sqrt{2}$, then you can use $\sqrt2=\pm3$, and the formula works the same.
You can also solve a quadratic equation by the technique of completing the square. So if you need to solve the equation $x^2+8x=5$ in the ring $\mathbf{Z}_{17}$, then you can add 16 to both sides to get $x^2+8x+16=5+16=21\equiv 4\pmod{17}$. This you can write in the form $(x+4)^2=4=2^2$. So $x+2$ must be an element with square equal to 4. As 2 and -2 are such elements, and 17 is a prime, there can be only two such elements, and we have found them both. $x+4=2$ gives us the solution $x=2-4=-2\equiv15$ and $x+4=-2$ gives the other solution $x=-2-4=-6\equiv11$.
Note that the quadratic formula does not work with respect to an even modulus. The formula calls for division by $2$ and by one of the coefficients. Both of these need to be units of the ring, because without a multiplicative inverse we cannot do division. Also, if the modulus is not a prime, then the square root may have surprisingly many values. For example in the ring $\mathbf{Z}_{15}$, 1, 4, -1=14 and -4=11 are all square roots of 1.
Finally, if your ring is a direct product of two rings, then you can solve the equation componentwise. After all, in a product ring $R_1\times R_2$ the element $(a,b)$ is the zero element, if and only if both $a=0$ and $b=0$.
-
$P(x) := x^2 + 4x + 3 = (x + 3)(x + 1)$. $P(x) = 0$ in ${\mathbb Z}_5$ implies $P(x)$ is multiple of 5, so at least one between $x + 3$ and $x + 1$ must be a multiple of 5. Therefore the solutions are 2 and 4.
-
HINT $\rm(3)\:\$ diagonalize employing Chinese remainder: $\rm\:(m,n)=1\ \Rightarrow\ \mathbb Z/mn \cong \mathbb Z/m\times \mathbb Z/n\:,\:$ hence $\rm\:(x-1)\:(x+3)\:$ has root $\rm\:(1,-3) \ne (1,1),(-3,-3)\in \mathbb Z/m \times \mathbb Z/n\$ for $\rm\ m,n \ne\cdots$
NOTE $\$ The ring $\rm\:D\:$ is a domain iff every polynomial $\rm\ f(x)\in D[x]\$ has at most $\rm\ deg\ f\$ roots in $\rm\:D\:.\:$ For a simple proof see my post here, where I illustrate it constructively in $\rm\ \mathbb Z/m\$ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$. | 2014-03-10T03:23:30 | {
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https://leetcode.com/problems/largest-divisible-subset/ | ### 368. Largest Divisible Subset
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.
Example 1:
nums: [1,2,3]
Result: [1,2] (of course, [1,3] will also be ok)
Example 2:
nums: [1,2,4,8]
Result: [1,2,4,8]
Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.
Seen this question in a real interview before?
When did you encounter this question?
Which company? | 2017-09-20T20:12:46 | {
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https://math.stackexchange.com/questions/2651188/a-function-whose-antiderivative-equals-its-inverse/2652160 | # A function whose antiderivative equals its inverse.
Does there exist a continuous function $F$ satisfying the property
\begin{align} F\left(\int^x_0 F(s)\ ds\right) = x \end{align}
If yes, then is the solution unique?
As stated, the question is not well-posed since I haven't specified the domain of $F$. But, the question is intended to be open-ended since it was posed by a couple colleagues of mine during a coffee break.
We did find such a $F$, but couldn't show that it is unique. Here's the spoiler
$F(x) = Cx^\gamma$ where $\gamma = \frac{-1+\sqrt{5}}{2}$ and $C = (1+\gamma)^{\frac{\gamma}{1+\gamma}}$ \begin{align} C\left(\int^x_0 Ct^{\gamma}\ dt\right)^\gamma = C^{1+\gamma}\left( \frac{x^{1+\gamma}}{1+\gamma}\right)^\gamma =\frac{C^{1+\gamma}}{(1+\gamma)^\gamma}x^{\gamma+\gamma^2} = \frac{C^{1+\gamma}}{(1+\gamma)^\gamma}x = x, \ \ \text{ for all } x> 0 \end{align}
• Wow. I'm impressed that you managed to come up with that. And I have no idea how to show anything like uniqueness --- I just wanted to publicly admire both the question and the partial solution. :) – John Hughes Feb 15 '18 at 3:41
• @JohnHughes Thank you for your kind words. – Jacky Chong Feb 15 '18 at 4:18
• The function is the continuous analogue of Golomb sequence which is the OEIS sequence A001462. – Somos Feb 15 '18 at 4:18
• @Somos Thank you for the reference. – Jacky Chong Feb 15 '18 at 4:36
• On $[0,\infty)$ there is a function of the type $cx^{\alpha}$ which satisfies the given property. As of now I am not sure about the whole line. My guess is that there are several continuous functions which solve the equation. – Kavi Rama Murthy Feb 15 '18 at 7:07
We prove a slightly weaker claim, namely that there exists an infinitude of functions which are inverse to one of their antiderivatives, i.e., functions $F$ for which there exists a constant $C$ such that $$F \bigg( C+\int_0^t F(u)du \bigg) =t.$$ The functions will be defined on $[0,\infty)$, not the whole line.
Proof: We use fixed point theory. Fix a real number $x>4$. For $T>0$, let us define the metric space $C_x[0,T]$ which consists of all $f \in C^1[0,T]$ which satisfy the following three conditions: $f(0)=x$, $f(t) \geq 4$ for all $t \in [0,T]$, and $\|f'\|_{\infty} \leq \frac{1}{4}$. One may easily verify that this is a complete metric space, with respect to the norm $\|f\|_{C_x[0,T]} = \|f\|_{\infty}+\|f'\|_{\infty}$. Here $\| \cdot \|_{\infty}$ denotes the uniform norm on $[0,T]$. When $T=\infty$ we define the analogous space $C_x[0,\infty)$, with the topology of uniform convergence on compact sets.
We define a map $S: C_x[0,\infty) \to C_x[0,\infty)$ by sending a function $f$ to the unique solution of the IVP: $$y'=\frac{1}{f(y)},\;\;\;\;\;\;\; y(0)=x.$$
The solution exists for global time by the Picard Lindelof theorem, since $f \geq 4$ is differentiable. Moreover, it is clear that $S$ maps $C_x[0,\infty)$ to itself. One crucial remark is that since $1/f \leq 1/4$, it necessarily follows that $y(t) \leq x+t/4$ for all $t$.
If $y=Sf$ and $z=Sg$, then we have that $$|y(t)-z(t)| \leq \int_0^t \bigg| \frac{1}{f(y(s))} - \frac{1}{g(z(s))} \bigg| ds = \int_0^t \frac{|f(y(s))-g(z(s))|}{|f(y(s))g(z(s))|}ds \\ \leq \frac{1}{16} \int_0^t |f(y(s))-g(z(s))|ds \leq \frac{1}{16} \int_0^t |f(y(s)) - g(y(s))| +|g(y(s))-g(z(s))| ds \\ \leq \frac{1}{16}t \|f-g\|_{L^\infty[0,x+t/4]} + \frac{1}{16}\|g'\|_{L^\infty[0,x+t/4]} \int_0^t |y(s)-z(s)|ds$$ Again, we have used the fact that $y(s),z(s) \leq x+t/4$. Now by Gronwall's lemma, it necessarily follows that $$\|y-z\|_{L^{\infty}[0,t]} \leq \frac{t}{16} \|f-g\|_{L^\infty[0,x+t/4]} e^{\|g'\|_{L^\infty[0,x+t/4]} \frac{t}{16}} \leq \frac{t}{16}e^{\frac{t}{64}} \|f-g\|_{L^{\infty}[0,x+t/4]}.$$where we used $\|g'\|_{\infty} \leq \frac{1}{4}$. Now for the derivative-norm, we have by the same inequalities that $$\|y'-z'\|_{L^\infty[0,t]} \leq \bigg\| \frac{1}{f \circ y}-\frac{1}{g \circ z} \bigg\|_{L^{\infty}[0,t]} \leq \frac{1}{16} \| f \circ y- g \circ z\|_{L^{\infty}[0,t]} \\ \leq \frac{1}{16} \|f-g\|_{L^{\infty}[0,x+t/4]} + \frac{1}{16} \|g'\|_{L^\infty[0,x+t/4]} \|y-z\|_{L^{\infty}[0,t]} \\ \leq \big( \frac{1}{16} + \frac{t}{256} e^{\frac{t}{64}} \big) \|f-g\|_{L^{\infty}[0,x+t/4]}$$ The preceding two expressions together show that if we define a Picard iteration scheme $f_n=S^nf_0$, then by induction we have that $$\|f_{n+1}-f_n\|_{L^{\infty}[0,t]} + \|f'_{n+1}-f'_n\|_{L^{\infty}[0,t]} \\ \leq \big(\frac{t}{256}e^{\frac{t}{64}} + \frac{1}{16} \big) \big( \frac{t}{16}e^{\frac{t}{64}} \big)^n \big\| f_1-f_0 \big\|_{L^{\infty}[0, (1+4^{-1}+...+4^{1-n})x +4^{-n}t]}$$so that the restriction to $[0,T]$ of these iterates will converge to some function $f$ as $n \to \infty$, whenever $T$ is small enough so that $\frac{T}{16}e^{T/64} <1$ (e.g. $T<13$ works). Now as long as $x<39/4$ (note that this doesn't interfere with the condition $x>4$), we have that $\sum_0^{\infty} 4^{-n}x=\frac{4}{3}x<13$ which is less than the critical $T$-value just discussed. Hence given any $T>0$, there exists some $N$ large enough so that $\sum_0^{N-1} 4^{-k}x + 4^{-N}T< 13$, and hence we see that $f_n$ actually converges in $C_x[0,T]$ for every $T>0$. This argument shows that the map $S$ actually has a globally defined fixed point on the whole space $C_x[0,\infty)$.
Now let $f$ be a fixed point of the map $S$. This means that for every $t>0$, one has $$f'(t) f(f(t)) = 1.$$ So if we define $F(t) = \int_0^t f(u)du$ it holds that $F'(f(t))f'(t)=1$, therefore $F(f(t))=t+F(x)$. Clearly each initial value $x$ gives a different solution $f$.
Now we claim that the fixed point $f$ satisfies $\lim_{t \to \infty} f(t) = +\infty$. If this was not the case, then $f\circ f$ would be bounded on $[0,\infty)$, hence $f' = 1/(f \circ f)$ would be bounded below, hence $f(t) \geq x+ct$ for some $c>0$, contradiction. Therefore $f$ is a continuously increasing bijection from $[0,\infty)$ to $[x,\infty)$. Let $C:=F(x)$, and define $g:[C,\infty) \to [x,\infty)$ by $g(t) = f(t-C)$. Clearly $F(g(t)) = t$, and therefore by the fact that $f$ is a bijection, it follows (by uniqueness of inverses) that $g(F(t))=t$. In other words, we have $$f(F(t)-C)=t,\;\;\;\;\;\;\text{i.e.},\;\;\;\; f\bigg( \int_x^t f(u)du \bigg) = t,\;\;\;\;\forall t \geq x,$$thus proving the claim.
• There is no solution on the whole line: the function $x \to \int_0 ^{x} F(x)dx$ a continuous function. By the given equation this function is one-to-one, hence strictly monotonic. By differentiation this implies that $F>0$ or $F<0$ everywhere. This makes the right side >0 or <o everywhere which is absurd. Please let me know if there is a stupid mistake in this argument. – Kavi Rama Murthy Feb 17 '18 at 12:11
• @KaviRamaMurthy That's correct (as I already said in a comment above) but the solutions constructed here are not on the full line, just $[0,\infty)$. – shalop Feb 17 '18 at 16:58
• @JackyChong: I thought about the problem some more, and I was able to prove some additional results, for instance if $f$ is any function which is inverse to one of its antiderivatives, then necessarily we have $$\lim_{t \to \infty} \frac{f(t)}{t^{\gamma}} = C,$$where $\gamma$ and $C$ are the same ones you found. But, I was not able to prove full-blown uniqueness given a fixed initial value (though the fixed point argument works as above,for certain values of $f(0)$). – shalop Feb 17 '18 at 21:49 | 2021-03-07T06:14:55 | {
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https://emporiosantoexpedito.com.br/rpzu3g/3a46ca-15th-row-of-pascals-triangle | Although other mathematicians in Persia and China had independently discovered the triangle in the eleventh century, most of the properties and applications of the triangle were discovered by Pascal. Subsequent row is made by adding the number above and to the left with the number above and to the right. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. It is also being formed by finding () for row number n and column number k. Pascal’s triangle is an array of binomial coefficients. The second row is 1,2,1, which we will call 121, which is 11x11, or 11 squared. Code Breakdown . The code inputs the number of rows of pascal triangle from the user. Aug 2007 3,272 909 USA Jan 26, 2011 #2 This video shows how to find the nth row of Pascal's Triangle. ) have differences of the triangle numbers from the third row of the triangle. Pascal’s Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 . Pascal’s triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y) n.It is named for the 17th-century French mathematician Blaise Pascal, but it is far older.Chinese mathematician Jia Xian devised a triangular representation for the coefficients in the 11th century. We hope this article was as interesting as Pascal’s Triangle. The first row of Pascal's triangle starts with 1 and the entry of each row is constructed by adding the number above. Anonymous. Let’s go over the code and understand. 9 months ago. 3 Some Simple Observations Now look for patterns in the triangle. As examples, row 4 is 1 4 6 4 1, so the formula would be 6 – (4+4) + (1+1) = 0; and row 6 is 1 6 15 20 15 6 1, so the formula would be 20 – (15+15) + (6+6) – (1+1) = 0. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). To understand this example, you should have the knowledge of the following C programming topics: Here is a list of programs you will find in this page. The n th n^\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. 1. Example: Input : k = 3 Return : [1,3,3,1] NOTE : k is 0 based. … Below is the example of Pascal triangle having 11 rows: Pascal's triangle 0th row 1 1st row 1 1 2nd row 1 2 1 3rd row 1 3 3 1 4th row 1 4 6 4 1 5th row 1 5 10 10 5 1 6th row 1 6 15 20 15 6 1 7th row 1 7 21 35 35 21 7 1 8th row 1 8 28 56 70 56 28 8 1 9th row 1 9 36 84 126 126 84 36 9 1 10th row 1 10 45 120 210 256 210 120 45 10 1 For instance, on the fourth row 4 = 1 + 3. In this article, however, I explain first what pattern can be seen by taking the sums of the row in Pascal's triangle, and also why this pattern will always work whatever row it is tested for. Kth Row of Pascal's Triangle Solution Java Given an index k, return the kth row of Pascal’s triangle. © Parewa Labs Pvt. Historically, the application of this triangle has been to give the coefficients when expanding binomial expressions. Create all possible strings from a given set of characters in c++. How do I use Pascal's triangle to expand the binomial #(d-3)^6#? Each number is found by adding two numbers which are residing in the previous row and exactly top of the current cell. Although the peculiar pattern of this triangle was studied centuries ago in India, Iran, Italy, Greece, Germany and China, in much of the western world, Pascal’s triangle has … Join our newsletter for the latest updates. All values outside the triangle are considered zero (0). Day 4: PascalÕs Triangle In pairs investigate these patterns. Step by step descriptive logic to print pascal triangle. Graphically, the way to build the pascals triangle is pretty easy, as mentioned, to get the number below you need to add the 2 numbers above and so on: With logic, this would be a mess to implement, that's why you need to rely on some formula that provides you with the entries of the pascal triangle that you want to generate. Kth Row of Pascal's Triangle Solution Java Given an index k, return the kth row of Pascal’s triangle. Each row of Pascal’s triangle is generated by repeated and systematic addition. Read further: Trie Data Structure in C++ The numbers in each row are numbered beginning with column c = 1. 2�������l����ש�����{G��D��渒�R{���K�[Ncm�44��Y[�}}4=A���X�/ĉ*[9�=�/}e-/fm����� W\$�k"D2�J�L�^�k��U����Չq��'r���,d�b���8:n��u�ܟ��A�v���D��N� ��A��ZAA�ч��ϋ��@���ECt�[2Y�X�@�*��r-##�髽��d��t� F�z�{t�3�����Q ���l^�x��1'��\��˿nC�s Generally, In the pascal's Triangle, each number is the sum of the top row nearby number and the value of the edge will always be one. The Fibonacci Sequence. Answer Save. Best Books for learning Python with Data Structure, Algorithms, Machine learning and Data Science. Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. What is the 4th number in the 13th row of Pascal's Triangle? Which row of Pascal's triangle to display: 8 1 8 28 56 70 56 28 8 1 That's entirely true for row 8 of Pascal's triangle. 3 Answers. ; Inside the outer loop run another loop to print terms of a row. This is down to each number in a row being … The … One of the famous one is its use with binomial equations. 1, 1 + 1 = 2, 1 + 2 + 1 = 4, 1 + 3 + 3 + 1 = 8 etc. This triangle was among many o… However, for a composite numbered row, such as row 8 (1 8 28 56 70 56 28 8 1), 28 and 70 are not divisible by 8. Pascal's triangle is one of the classic example taught to engineering students. Moving down to the third row, we get 1331, which is 11x11x11, or 11 cubed. More rows of Pascal’s triangle are listed on the final page of this article. Is there a pattern? for(int i = 0; i < rows; i++) { The next for loop is responsible for printing the spaces at the beginning of each line. This math worksheet was created on 2012-07-28 and has been viewed 58 times this week and 101 times this month. Naturally, a similar identity holds after swapping the "rows" and "columns" in Pascal's arrangement: In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding column from its row to the first, inclusive (Corollary 3). First 6 rows of Pascal’s Triangle written with Combinatorial Notation. Python Basics Video Course now on Youtube! If we look at the first row of Pascal's triangle, it is 1,1. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row. Example: Input : k = 3 Return : [1,3,3,1] Java Solution of Kth Row of Pascal's Triangle In this post, we will see the generation mechanism of the pascal triangle or how the pascals triangle is generated, understanding the pascal's Triangle in c with the algorithm of pascals triangle in c, the program of pascal's Triangle in c. Input number of rows to print from user. If you sum all the numbers in a row, you will get twice the sum of the previous row e.g. Make a Simple Calculator Using switch...case, Display Armstrong Number Between Two Intervals, Display Prime Numbers Between Two Intervals, Check Whether a Number is Palindrome or Not. Note: The row index starts from 0. At first, Pascal’s Triangle may look like any trivial numerical pattern, but only when we examine its properties, we can find amazing results and applications. Subsequent row is created by adding the number above and to the left with the number above and to the right, treating empty elements as 0. Welcome to The Pascal's Triangle -- First 12 Rows (A) Math Worksheet from the Patterning Worksheets Page at Math-Drills.com. Entry in the top row, there is an array of 1 note Could. Convention holds that both row numbers and write the sum between and below them the sum of ways! 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( the first number in the triangle, you 15th row of pascals triangle a 1 below and to the Theorem! Step by step descriptive logic to print terms of a row characters in c++ third row you... I ’ ve left-justified the triangle, you will Get twice the sum between and below them and Data.! 3 in the fourth row [ 1,3,3,1 ] note: Could you optimize your algorithm to use only O n... 11 squared exists between the second diagonal ( triangular numbers ) and third diagonal ( natural numbers.! Look for patterns in the fourth number in row 4, the apex the. Interesting number patterns is Pascal 's triangle, you will Get twice the of... Are numbered from the third row, there is an array of binomial coefficients sequence can done... With ` 1 '' at the first six rows ( numbered 0 through )! Number 1 is knocked off, however ) continue placing numbers below in! Column c = 1 like this also, refer to these similar posts: the... | 2021-06-14T22:26:00 | {
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http://mathhelpforum.com/calculus/194783-integration.html | # Math Help - Integration
1. ## Integration
Hello
Can someone help me with this please:
$\int\dfrac{3}{2}\cos^2(4x)$
I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$
A hint at the right approach would be very much appreciated.
Thank you.
2. ## Re: Integration
Hello, Furyan!
Can someone help me with this please?
. . $\int\tfrac{3}{2}\cos^2(4x)$
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
3. ## Re: Integration
Hi Furyan!
Are you aware of the double angle formula for $\cos(2y)$ in terms of $\cos^2y$?
4. ## Re: Integration
Originally Posted by Furyan
Hello
Can someone help me with this please:
$\int\dfrac{3}{2}\cos^2(4x)$
I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$
A hint at the right approach would be very much appreciated.
Thank you.
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
5. ## Re: Integration
Originally Posted by Soroban
Hello, Furyan!
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
You don't really NEED this identity, though it is the simplest simplification.
Integration by Parts will also work...
6. ## Re: Integration
The following identity may be useful:
$\cos^2 x = \frac{1 + \cos 2x}{2}$
Try using it.
7. ## Re: Integration
Hello Soroban,
Originally Posted by Soroban
Hello, Furyan!
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
Thank you. I was able to find the integral using that identity and substituting $\cos^2 4\theta$ with $\dfrac{1}{2}(1 + \cos8\theta)$
8. ## Re: Integration
Hello Prove It
Originally Posted by Prove It
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
Thank you. I am having trouble with that realisation but I'm working on it.
I will also try using integration by parts.
9. ## Re: Integration
Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?
10. ## Re: Integration
MarceloFantini
Originally Posted by MarceloFantini
Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?
Yes indeed, thank you for pointing that out. I have corrected my post accordingly. I did use the former when I found the integral.
11. ## Re: Integration
Originally Posted by Furyan
Hello Prove It
Thank you. I am having trouble with that realisation but I'm working on it.
I will also try using integration by parts.
Here's how that identity is derived.
First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}
Ans 1
Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have
\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}
And if you're trying to use integration by parts...
\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}
And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
12. ## Re: Integration
Dear Prove It
Originally Posted by Prove It
Here's how that identity is derived.
First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}
Ans 1
Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have
\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}
And if you're trying to use integration by parts...
\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}
And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
I have been looking at the double angle identity for $\cos2\theta$ and, for some reason, was having trouble realising what you had pointed out in your original post. Now it's crystal clear. Some how I keep missing the elementary principles.
I will try and find the integral again using the substitution you have suggested.
Thank you also for the help with using integration by parts. I'd never have been able to do that, but I am very interested in it and and will work through it until I understand it.
Thank you very much for the time you have taken to help me. It's extremely generous of you and very grateful. | 2014-08-23T18:46:56 | {
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https://math.stackexchange.com/questions/2239557/elementary-inequality | # Elementary Inequality
This should be completely elementary, if not downright trivial compared to other problems I've seen on this site. I think I'm just being silly. The problem is from Kaczor and Nowak's problem book in analysis. Somehow I made it through undergraduate without ever seeing half the inequalities in this text, and I really need to know them, so I'm trying to get some practice.
Suppose that $a_k$ are positive numbers, where $k$ ranges from $1$ to $n$, and satisfies $\sum a_k \leq 1$. Show that $\sum \frac{1}{a_k} \geq n^2$.
So, a moment of staring at the problem, and the problem screams AM-GM-HM inequality. We can rewrite what we are asked to prove as $$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{1}{n^2}$$
so that the left side is now precisely the harmonic mean of the $a_k$. By AM-GM-HM, this is less or equal to both the harmonic and geometric means. The obvious choice here being to use the arithmetic mean since it's the only thing for which we can say anything about. This gives:
$$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{\sum a_k}{n} \leq \frac{1}{n}$$
Which is weaker than what I was asked to prove - I've only been able to show the original sum is $\geq n$, rather $n^2$.
Can the approach be fixed? Perhaps there is a better way to go?
• Harmonic mean $\dfrac{1}{\sum \frac{1}{a_k}}$ should be $\dfrac{n}{\sum \frac{1}{a_k}}$ – DHMO Apr 18 '17 at 3:53
• @DHMO Of course, if I use the inequality properly, the result is clear. Only mildly embarrassing. I should mail my degree back... (just kidding of course). Thank you though. – Alfred Yerger Apr 18 '17 at 3:54
The HM has an $n$ on top (it's the reciprocal of the AM of the reciprocals). (Think about what happens if all the $a_k$ are equal.)
$$\begin{array}{rcl} \displaystyle \sum_{k=1}^n a_k &\le& 1 \\ \displaystyle \dfrac1n \sum_{k=1}^n a_k &\le& \dfrac1n \\ \displaystyle \dfrac n{\sum_{k=1}^n \frac1{a_k}} &\le& \dfrac1n \\ \displaystyle \dfrac{\sum_{k=1}^n \frac1{a_k}}n &\ge& n \\ \displaystyle \sum_{k=1}^n \frac1{a_k} &\ge& n^2 \\ \end{array}$$
With equality when $\forall k \in 1,2,3\cdots, n:a_k = \dfrac1n$.
1. Cauchy-Schwarz.
$$n^2 = (\sum_{k=1}^n \sqrt{a_k} \frac{1}{\sqrt{a_k}})^2\le \sum_{k=1}^n a_k \times \sum_{k=1}^n \frac{1}{a_k}\le \sum_{k=1}^n \frac{1}{a_k}.$$
1. Calculus.
For any $\lambda>0$ we have
\begin{align*} 2n \le \sum_{k=1}^n (\lambda a_k + \frac{1}{\lambda a_k})&\le \lambda+ \sum_{k=1}^n \frac{1}{\lambda a_k}.\end{align*}
Therefore
$$\lambda( 2n -\lambda) \le \sum_{k=1}^n \frac{1}{a_k}.$$
Maximum of LHS at $\lambda = n$, and result follows. | 2019-09-24T08:34:26 | {
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http://mathhelpforum.com/differential-equations/174813-ibv.html | 1. ## IBV
$\displaystyle \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0$
$\displaystyle \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\displaystyle \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L$
$\displaystyle u(x,t)=\varphi(x)T(t)$
$\displaystyle \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$
$\displaystyle \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$
$\displaystyle \displaystyle T=Ce^{-\lambda kt}$
$\displaystyle \varphi_1(0)=1 \ \varphi_2(0)=0$
$\displaystyle \varphi_1'(0)=0 \ \varphi_2'(0)=1$
$\displaystyle \varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi(0): \ A_2=0$
$\displaystyle \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$
The eigenvalues must satisfy:
$\displaystyle \displaystyle\cos(x\sqrt{\lambda})=0$
$\displaystyle \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$
The eigenfunction is:
$\displaystyle \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$
Is this much correct?
Thanks.
2. $\displaystyle \displaystyle u(x,t)=\sum_{n=1}^{\infty}b_n\frac{2L}{\pi(1+2n)}\ exp\left(-\frac{\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\rig ht)\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
I have shown $\displaystyle \displaystyle\int_0^L\varphi_n(x)\varphi_m(x) \ dx=0, \ m\neq n$.
Before I solve for $\displaystyle b_n$ is u(x,t) correct?
Thanks.
3. Originally Posted by dwsmith
$\displaystyle \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0$
$\displaystyle \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\displaystyle \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L$
$\displaystyle u(x,t)=\varphi(x)T(t)$
$\displaystyle \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$
$\displaystyle \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$
$\displaystyle \displaystyle T=Ce^{-\lambda kt}$
$\displaystyle \varphi_1(0)=1 \ \varphi_2(0)=0$
$\displaystyle \varphi_1'(0)=0 \ \varphi_2'(0)=1$
Where did $\displaystyle \varphi_1$ and $\displaystyle \varphi_2$ suddenly come from?
Are they $\displaystyle e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $\displaystyle e^{\lambda kt}(B sin(x\sqrt{\lambda}))$?
If so, you cannot, generally, do that. But here it does not hurt.
$\displaystyle \varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$
All you really need is that $\displaystyle \phi(0, t)= e^{-\lambda kt}(Acos(0)+ Bsin(0))= Ae^{-k\lambda kt}= 0$. Since the exponential is never 0, A= 0.
Then $\displaystyle \phi(L, t)= e^{-\lambda kt}B cos(L\sqrt{\lambda})= 0$. Again, the exponential is not 0 so we must have either B= 0, which would mean the solution was identically equal to 0, and so not an eigenfunction, or cos(L\sqrt{\lambda})= 0.
$\displaystyle \displaystyle\varphi(0): \ A_2=0$
$\displaystyle \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$
The eigenvalues must satisfy:
$\displaystyle \displaystyle\cos(x\sqrt{\lambda})=0$
You mean $\displaystyle cos(L\sqrt{\lambda})= 0$
$\displaystyle \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$
The eigenfunction is:
$\displaystyle \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$
Is this much correct?
Thanks.
Yes, that is correct and your second post is correct.
4. Originally Posted by HallsofIvy
Where did $\displaystyle \varphi_1$ and $\displaystyle \varphi_2$ suddenly come from?
Are they $\displaystyle e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $\displaystyle e^{\lambda kt}(B sin(x\sqrt{\ambda}))$?
If so, you cannot, generally, do that. But here it does not hurt.
My book solves all PDEs in the same manner. I don't know why it is done that way but it works in all cases.
Here is what the book says about $\displaystyle \varphi$ 1 and 2.
The theorem in the theory of ODE which explicitly states the proposition above is: if the coefficients $\displaystyle c_0(x), \ c_1(x), \ c_2(x)$ are continuous on the interval [a,b] and $\displaystyle c_0(x)$ is different from zero throughout the interval, then for any fixed $\displaystyle x_0$ in the interval [a,b] and for any constants $\displaystyle \alpha, \ \beta$. The DE has one and only one solution $\displaystyle \varphi(x)=\varphi(x,\lambda)$ satisfying at $\displaystyle x_0$ the initial conditions $\displaystyle \varphi(x_0)=\alpha, \ \varphi'(x_0)=\beta$.
Moreover, $\displaystyle \varphi(x,\lambda)$ and $\displaystyle \varphi'(x,\lambda)$ are continuous and continuously differentiable functions of both x for $\displaystyle a\leq x\leq b$ and all lambda.
It is convenient to single out the two special solutions $\displaystyle \varphi_1=\varphi_1(x,\lambda), \ \varphi_2=\varphi_2(x,\lambda)$ of which satisfy the DE
$\displaystyle \varphi_1(0)=1, \ \varphi_2(0)=0$
$\displaystyle \varphi'_1 (0)=0, \ \varphi'_2(0)=1$.
We call 1 and 2 the basic solutions at x nought.
5. $\displaystyle \displaystyle \int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\int_0^Lb_m\sin^2\left(\frac{\pi x(1+2m)}{L}\right) \ dx$
$\displaystyle \displaystyle\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=b_m\frac{L}{2}\Rightarrow b_m=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx$
6. For the case $\displaystyle f(x)=1$
$\displaystyle \displaystyle b_m=\frac{2}{L}\int_0^L\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\frac{2}{L}\left[\frac{-2L}{\pi(1+2m)}\cos\left(\frac{\pi x(1+2m)}{2L}\right)\right]_0^L$
$\displaystyle \displaystyle b_m=\frac{4}{\pi(1+2m)}, \ \ m=n$
$\displaystyle \displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{L}{ (1+2n)^2}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
If everything above is correct, can I do this?
$\displaystyle \displaystyle\sum_0^{\infty}\frac{1}{(1+2m)^2}=\fr ac{\pi^2}{8}$
$\displaystyle \displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{ (1+2n)^2}\cdot L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
$\displaystyle \displaystyle\Rightarrow u(x,t)=L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
7. Ok, I pretty sure the above answer is wrong.
I now have:
$\displaystyle \displaystyle u(x,t)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
How do I show this is true for all x with the IC:
$\displaystyle \displaystyle u(x,0)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\sin\left(\frac{\pi x(1+2n)}{2L}\right)=1$ | 2018-03-21T09:46:42 | {
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https://math.stackexchange.com/questions/2369403/which-area-is-larger-the-blue-area-or-the-white-area | # Which area is larger, the blue area, or the white area?
In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
# My Solution
Let the length of each side of the square be $$2r$$.
The area of the square is $$4r^2$$.
The two semi-circles have equal area.
Area of one semi-circle = $$\frac{{\pi}r^2}{2}$$.
$${\times}2 = {\pi}r^2$$
White area = $${\pi}r^2 -$$ area of the intersection of the two circles.
Let the area of the intersection of the two circles be $$t$$.
White area = $${\pi}r^2 - t$$.
The segments that make up $$t$$ are identical.
$$t$$ = area of segment $${\times}2$$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $$90^{\circ}$$ (The circles both have radius $$r$$, and the outer shape is a square.
Angle of sector $$= \frac{1}{4} * {\pi}r^2$$.
Area of triangle $$= \frac{1}{2} * r^2$$.
Area of segment $$= \frac{{\pi}r^2 - 2r^2}{4}$$. $$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$$.
$$t = \frac{{\pi}r^2 - 2r^2}{2}$$.
White area $$= {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$$.
White area $$= \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$$.
White area $$= \frac{{\pi}r^2 + 2r^2}{2}$$.
Blue area = $$r^2\left(4 - \frac{{\pi} + 2}{2}\right)$$.
Blue area = $$r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$$.
Blue area = $$r^2\left(\frac{6 - {\pi}}{2}\right)$$.
If White area $$-$$ Blue area $$\gt 0$$, then the White area is larger.
$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$
$$\therefore$$ the white area is larger.
## What is the error in my solution?
The provided solution:
• It's a general rule of thumb, that in all puzzles which ask "which area is larger" the answer always is that both areas have the same size, even when one is visibly enormous and the other visibly tiny. – Angina Seng Jul 23 '17 at 21:11
• White area $= \pi r^2 - \color{red}{2} \times$area of the intersection of the two circles. – Donald Splutterwit Jul 23 '17 at 21:16
• Move the small blue caps to the other side of the semicircles, then the blue region makes a triangle. – Michael Burr Jul 23 '17 at 21:16
• @DonaldSplutterwit thanks for the pointer; do you want to make it into an answer, or should I do it? – Tobi Alafin Jul 23 '17 at 21:20
• What's the question? – Shuri2060 Jul 23 '17 at 21:21
If you just divide the lower left part of the blue area and move each part $90$ degree up to join them to the main blue part, then the area of the new blue shape will be equal to the white part.
• This is the solution given as "provided solution" by the OP. The OP has already seen this (neat) solution. How does this add to what is already present in the question? – James K Jul 23 '17 at 22:14
• @Seyed In the bottom of the question, it says "The provided solution". – HelloGoodbye Jul 23 '17 at 22:42
• As far as I can see, the question wasn't edited at any point – Shuri2060 Jul 23 '17 at 22:58
• @user1936752: Technically speaking, Seyed did answer the question. There is only one question mark in the post, which is for the title "which area is larger, blue or white?". It's likely that what the OP wanted is an explanation for what is the mistake in his own attempted solution, but he never actually asked that... – Meni Rosenfeld Jul 24 '17 at 9:09
• Making the neat solution available in the thread rather than just as a link to a dodgy-sounding url does add value IMO. – Especially Lime Jul 24 '17 at 10:44
There was a mistake in my earlier solution. I correct that mistake here.
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - 2 {\times}$area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - 2t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $= \frac{1}{4} * {\pi}r^2$.
Area of triangle $= \frac{1}{2} * r^2$.
Area of segment $= \frac{{\pi}r^2 - 2r^2}{4}$. $t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $= {\pi}r^2 - 2\left(\frac{{\pi}r^2 - 2r^2}{2}\right)$.
White area $= \frac{2{\pi}r^2 - 2{\pi}r^2 + 4r^2}{2}$.
White area $= \frac{4r^2}{2}$.
White area $= 2r^2$.
Blue area = $r^2\left(4 - 2\right)$.
Blue area = $2r^2$.
The blue and white areas both have areas of $2r^2$, therefore the triangles have equal areas.
• This solution correctly states that the white area is ${\pi}r^2 - 2 \times\text{area of the intersection of the two circles}.$ There is still a question of why you previously thought that the area was ${\pi}r^2 - \text{area of the intersection of the two circles}$ and what you can do to prevent such mistakes in future exercises. – David K Jul 24 '17 at 12:11
• I'm not sure exactly why I made the mistake, but I guess I'll double check my assumptions in future questions. – Tobi Alafin Jul 24 '17 at 12:17
• Don't worry about getting assumptions wrong for us--this is a fine place to bring incorrect work to have it corrected. I suspect in this case you just did that step a little too quickly. The way I think of it, the area of the overlapping semicircles is $\pi r^2$ minus the overlap, and we subtract the overlap a second time by coloring it blue. But subtracting the same thing twice is error-prone: subtracting it once can satisfy either of the conditions we need, and we might overlook the fact that this doesn't satisfy both conditions at once. – David K Jul 24 '17 at 12:43
• Anyway, whatever the reason for your original confusion, I think this is a good question. It showed thorough and careful effort, and asked for help to understand why the effort reached a wrong conclusion. If only more questions here were like that! – David K Jul 24 '17 at 12:44
The area of the segment is “quarter of circle minus triangle”: $$\frac{1}{4}\pi r^2-\frac{1}{2}r^2=\frac{r^2}{4}(\pi-2)$$ Thus half of the white area is “quarter of circle plus triangle minus segment”: $$\frac{1}{4}\pi r^2+\frac{1}{2}r^2-\frac{r^2}{4}(\pi-2)=r^2$$ Therefore the white area is $2r^2$. | 2020-10-28T23:25:39 | {
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http://needlepointmaine.com/yx78g1yr/simplify-radical-expressions-ed19c2 | no perfect square factors other than 1 in the radicand. So which one should I pick? The symbol is called a radical sign and indicates the principal square root of a number. Simplifying Radical Expressions. For the case of square roots only, see simplify/sqrt. Section 6.4: Addition and Subtraction of Radicals. Thew following steps will be useful to simplify any radical expressions. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. This page will help you to simplify an expression under a radical sign (square root sign). And it really just comes out of the exponent properties. applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . 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If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. Scientific notations. Therefore, we have √1 = 1, √4 = 2, √9= 3, etc. Pairing Method: This is the usual way where we group the variables into two and then apply the square root operation to take the variable outside the radical symbol. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. The number 16 is obviously a perfect square because I can find a whole number that when multiplied by itself gives the target number. You will see that for bigger powers, this method can be tedious and time-consuming. Example 1. Here are the search phrases that today's searchers used to find our site. If and are real numbers, and is an integer, then. To play this quiz, please finish editing it. 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Homework. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m ⋅ n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Picking the largest one makes the solution very short and to the point. \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int. Let’s do that by going over concrete examples. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. \sum. Test - II . 2) Product (Multiplication) formula of radicals with equal indices is given by Simplify any radical expressions that are perfect squares. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. Then, it's just a matter of simplifying! +1 Solving-Math-Problems Page Site. 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Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. Keep in mind that you are dealing with perfect cubes (not perfect squares). \int_{\msquare}^{\msquare} \lim. Adding and Subtracting Radical Expressions, That’s the reason why we want to express them with even powers since. Quantitative aptitude. Otherwise, you need to express it as some even power plus 1. And we have one radical expression over another radical expression. You just need to make sure that you further simplify the leftover radicand (stuff inside the radical symbol). These properties can be used to simplify radical expressions. 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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. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. Great! Simplifying radical expressions calculator. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$, $$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$. Multiplying Radical Expressions. So, , and so on. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Here it is! Simplifying Expressions – Explanation & Examples. no radicals appear in the denominator of a fraction. Step 4: Simplify the expressions both inside and outside the radical by multiplying. For the numerical term 12, its largest perfect square factor is 4. More so, the variable expressions above are also perfect squares because all variables have even exponents or powers. However, I hope you can see that by doing some rearrangement to the terms that it matches with our final answer. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. How to Simplify Radicals with Coefficients. by lsorci. In this tutorial, you'll see how to multiply two radicals together and then simplify their product. Example 3: Simplify the radical expression \sqrt {72} . Start studying Algebra 5.03: Simplify Radical Expressions. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Simplifying simple radical expressions We use cookies to give you the best experience on our website. #1. −1)( 2. . The simplify/radical command is used to simplify expressions which contain radicals. Square root, cube root, forth root are all radicals. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. To play this quiz, please finish editing it. It must be 4 since (4)(4) = 42 = 16. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. You may use your scientific calculator. Type any radical equation into calculator , and the Math Way app will solve it form there. Another way to solve this is to perform prime factorization on the radicand. Simplifying logarithmic expressions. These two properties tell us that the square root of a product equals the product of the square roots of the factors. What rule did I use to break them as a product of square roots? View 5.3 Simplifying Radical Expressions .pdf from MATH 313 at Oakland University. Express the odd powers as even numbers plus 1 then apply the square root to simplify further. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. $$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$, $$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$. x^{\circ} \pi. Sometimes radical expressions can be simplified. Procedures. You can do some trial and error to find a number when squared gives 60. To simplify radical expressions, look for factors of the radicand with powers that match the index. Simplifying Radical Expressions Worksheet Answers Lovely Simplify Radicals Works In 2020 Simplifying Radical Expressions Persuasive Writing Prompts Radical Expressions . Simplify … A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Save. There is a rule for that, too. Simplifying Radical Expressions . Use rational exponents to simplify radical expressions. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. For example, the square roots of 16 are 4 and … Remember, the square root of perfect squares comes out very nicely! Separate and find the perfect cube factors. Product Property of n th Roots. Step 2 : We have to simplify the radical term according to its power. The solution to this problem should look something like this…. Remember that getting the square root of “something” is equivalent to raising that “something” to a fractional exponent of {1 \over 2}. Practice. And it checks when solved in the calculator. Looks like the calculator agrees with our answer. Introduction. Edit. Simplifying Radical Expressions. Multiply all numbers and variables inside the radical together. Share practice link. Also, you should be able to create a list of the first several perfect squares. 9th - University grade . For example, the sum of $$\sqrt{2}$$ and $$3\sqrt{2}$$ is $$4\sqrt{2}$$. The goal is to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. Example 12: Simplify the radical expression \sqrt {125} . Online calculator to simplify the radical expressions based on the given variables and values. Determine the index of the radical. Khan Academy is a … Example 4: Simplify the radical expression \sqrt {48} . Use the multiplication property. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. Thus, the answer is. Exponents and power. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Live Game Live. However, the best option is the largest possible one because this greatly reduces the number of steps in the solution. Simplify a Term Under a Radical Sign. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Mathematics. However, the key concept is there. Simplify the expression: Next, express the radicand as products of square roots, and simplify. Recognize a radical expression in simplified form. Simplify #2. $$\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{\sqrt{16}}=\frac{\sqrt{15}}{4}$$. Aptitude test online. To simplify this radical number, try factoring it out such that one of the factors is a perfect square. The properties of exponents, which we've talked about earlier, tell us among other things that, $$\begin{pmatrix} xy \end{pmatrix}^{a}=x^{a}y^{a}$$, $$\begin{pmatrix} \frac{x}{y} \end{pmatrix}^{a}=\frac{x^{a}}{y^{a}}$$. Step 3 : Radical Expressions are fully simplified when: –There are no prime factors with an exponent greater than one under any radicals –There are no fractions under any radicals –There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Print; Share; Edit; Delete; Report an issue; Host a game. 1) Simplify. This quiz is incomplete! A perfect square number has integers as its square roots. Below is a screenshot of the answer from the calculator which verifies our answer. Recall that the Product Raised to a Power Rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$. This type of radical is commonly known as the square root. The answer must be some number n found between 7 and 8. This type of radical is commonly known as the square root. are called conjugates to each other. Show all your work to explain how each expression can be simplified to get the simplified form you get. Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator [1] X Research source To simplify a perfect square under a radical, simply remove the radical sign and write the number that is the square root of the perfect square. Simply put, divide the exponent of that “something” by 2. Solo Practice. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. 5 minutes ago. This lesson covers . The first law of exponents is x a x b = x a+b. Procedures. Example 6: Simplify the radical expression \sqrt {180} . Start studying Algebra 5.03: Simplify Radical Expressions. SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Radical expressions are expressions that contain radicals. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. nth roots . Be sure to write the number and problem you are solving. Step 2 : We have to simplify the radical term according to its power. Learning how to simplify expression is the most important step in understanding and mastering algebra. Multiply all numbers and variables outside the radical together. Next lesson. Let's apply these rule to simplifying the following examples. Perfect cubes include: 1, 8, 27, 64, etc. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can be used to simplify radical expressions. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. First we will distribute and then simplify the radicals when possible. APTITUDE TESTS ONLINE. By quick inspection, the number 4 is a perfect square that can divide 60. Dividing Radical Expressions. Finish Editing. Find our site to help you figure out how to multiply expressions with variables ''! Tell us that the square root of a number when squared gives 60 perfect square factors should.... Radical because they all can be simplified not perfect squares because they are still simplified the same ideas to you! Divide 200, the expression: use Polynomial Multiplication to multiply the term. 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The reason why we want to break them as a product of two conjugates always. { 180 } 5: simplify the radical together include the radnormal, rationalize, and that may the... In simplifying radicals that have coefficients root are all radicals ( r2 - 1 which. | 2021-04-20T23:16:14 | {
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https://thekandybshow.com/326qg/can-a-real-matrix-have-complex-eigenvectors-a89235 | Let’s assume the matrix is square, otherwise the answer is too easy. Its unit eigenvectors are orthogonal by property (3). 1,117 2 2 gold badges 8 8 silver badges 16 16 bronze badges. Then writing in real and imaginary parts: Taking real and imaginary parts . If $\dim K = 1$, then there is a simultaneous eigenvector, unique up to multiples. •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. You can use the companion matrixto prove one direction. The case when $K$ has dimension $n=2m>2$ is more difficult. E.g. Continuing by induction, define It is only a necessary condition for there to be a real eigenvector with a real eigenvalue. In the first case, are you assuming your matrix to be diagonalizable or not ? In that case, though, restricting attention to the kernel of $X$ on $K$ will then yield a space that is preserved by $Y$ and on which $Y$ is nilpotent, so there will exist a common real eigenvector. Recall that if z= a+biis a complex number, its complex conjugate is de ned by z= a bi. COMPLEX EIGENVALUES . A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors. 4. When eigenvalues become complex, eigenvectors also become complex. Now, of course, if $v$ is such a simultaneous eigenvector, then $XYv-YXv = 0$, so $v$ is in the kernel of the commutator $C = [X,Y]$. Making statements based on opinion; back them up with references or personal experience. As a result, eigenvectors of symmetric matrices are also real. The answer is always. @CarloBeenakker You are not wrong! Yes, t can be complex. However, the eigenvectors corresponding to the conjugate eigenvalues are themselves complex conjugate and the calculations involve working in complex n-dimensional space. Prove that if λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. So in general, an eigenvalue of a real matrix could be a nonreal complex number. In general, a real matrix can have a complex number eigenvalue. A matrix in a given field (or even commutative ring) may or may not have eigenvectors. $\endgroup$ – user137731 Jun 5 … Problems in Mathematics © 2020. In … If Two Matrices Have the Same Rank, Are They Row-Equivalent? In some sense, the 'best-known' criterion is 'find the eigenvectors and check to see whether any of them are real', but, of course, finding eigenvectors could be difficult because one has to solve some algebraic equations, possibly of high degree, and that might not be very easy to do. Suppose, though, that $C = XY-YX$ has a nontrivial kernel $K_0\subset\mathbb{R}^n$, which can be computed by solving linear equations. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … I have tried the function, it is for computation of eigenvalues of real matrix, and it does not work for complex-number matrix – Alireza Masoom Apr 27 '19 at 21:40 Hmm could you give it another try, according to the docs complex numbers are supported: The first column of "eigenvalues" contains the real and the second column contains the imaginary part of the eigenvalues. Therefore, any real matrix with odd order has at least one real eigenvalue, whereas a real matrix with even order may not have any real eigenvalues. The diagonal elements of a triangular matrix are equal to its eigenvalues. Asking for help, clarification, or responding to other answers. To learn more, see our tips on writing great answers. How to Diagonalize a Matrix. I think what your lecturer is getting at is that, for a real matrix and real eigenvalue, any possible eigenvector can be expressed as … To find the eigenvectors of a triangular matrix, we use the usual procedure. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example. (adsbygoogle = window.adsbygoogle || []).push({}); Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant. This site uses Akismet to reduce spam. Eigenvalues can be complex numbers even for real matrices. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Then $X$ has a real eigenvalue of odd multiplicity, either $1$ or $3$. Suppose has eigenvalue , eigenvector and their complex conjugates. I have a real symmetric matrix with a lot of degenerate eigenvalues, and I would like to find the real valued eigenvectors of this matrix. python numpy scipy linear-algebra eigenvalue. If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. Learn how your comment data is processed. These eigenvalues are not necessary to be distinct nor non-zero. Common Eigenvector of Two Matrices and Determinant of Commutator, Complex Conjugates of Eigenvalues of a Real Matrix are Eigenvalues, Use the Cayley-Hamilton Theorem to Compute the Power $A^{100}$, Eigenvalues of Similarity Transformations, There is at Least One Real Eigenvalue of an Odd Real Matrix, Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix, A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix, Find All Values of $x$ so that a Matrix is Singular. Two eigenvectors of a real symmetric matrix or a Hermitian matrix, if they come from different eigen values are orthogonal to one another. Complex eigenvalues will have a real component and an imaginary component. (b) Find the eigenvalues of the matrix The eigenvalues of a hermitian matrix are real, since (λ − λ)v = (A * − A)v = (A − A)v = 0 for a non-zero eigenvector v. If A is real, there is an orthonormal basis for R n consisting of eigenvectors of A if and only if A is symmetric. Example 13.1. The kernel $L$ of $X$ is nontrivial and preserved by $Y$ (since $X$ and $Y$ commute), so $Y$ is nilpotent on $L$ and hence there is a nonzero element of $L$ that is annihilated by both $X$ and $Y$, so it is a simultaneous real eigenvector. Extracting complex eigenvectors from the real Schur factorization can be done but is trickier; you can see how LAPACK does it. We want our solutions to only have real numbers in them, however since our solutions to systems are of the form, $\vec x = \vec \eta {{\bf{e}}^{\lambda t}}$ we are going to have complex … Step by Step Explanation. Taking the real and imaginary part (linear combination of the vector and its conjugate), the matrix has this form with respect to the new basis. Suppose is a real matrix with a complex eigenvalue and a correspondingE#‚# + ,3 eigenvector Let @ÞTœÒ ÓÞRe Im@@ By the theorem Re Im Re ImßEœÒ Ó Ò ÓÞ@@ @@”• + ,,+ " ”• ” •È + , ,+ can be written as , where .<<œ+ , cos sin sin cos)))) ## Thus represents a counterclockwise rotation if is chosen around the originGÐ !Ñ) through the angle , followed by a rescaling fact Thus, the criterion in this case is that $X$ and $Y$ have non-positive determinant. In Section 5.4, we saw that a matrix whose characteristic polynomial has distinct real roots is diagonalizable: it is similar to a diagonal matrix, which is much simpler to analyze.In this section, we study matrices whose characteristic polynomial has complex roots. Let . Add to solve later Sponsored Links Also, note that, because $X$ and $Y$ commute, each preserves the generalized eigenspaces of the other. This website is no longer maintained by Yu. An eigenvalue represents the amount of expansion in the corresponding dimension. If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t) has general solution predicted by the eigenvalue-eigenvector method of c 1e 1t~v 1 + c 2e 2t~v 2 where the constants c 1 and c 2 can be determined from the initial values. A complex-valued square matrix A is normal ... As a special case, for every n × n real symmetric matrix, the eigenvalues are real and the eigenvectors can be chosen real and orthonormal. The diagonalizing matrix can be chosen with orthonormal columns when A = AH In case A is real and symmetric, its eigenvalues are real by property. Some things to remember about eigenvalues: •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. For eigen values of a matrix first of all we must know what is matric polynomials, characteristic polynomials, characteristic equation of a matrix. Can We Reduce the Number of Vectors in a Spanning Set? If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. The following example shows that stochastic matrices do not need to be diagonalizable, not even in the complex: 7 The matrix A = 5/12 1/4 1/3 5/12 1/4 1/3 1/6 1/2 1/3 is a stochastic matrix, even doubly stochastic. All Rights Reserved. (Generically, this commutator is invertible; when this happens the answer is that there is no real eigenvector of $A$.). When I take the eigenvectors of the matrix, I get mirror images for the first few (about 10) vectors. 2 Chapter 2 part B Consider the transformation matrix . No, but you can build some. Suppose $\dim K = 3$. Hence, if $$\lambda_1$$ is an eigenvalue of $$A$$ and $$AX = \lambda_1 X$$, we can label this eigenvector as $$X_1$$. The row vector is called a left eigenvector of . If this intersection is nonzero, then there will be a simultaneous real eigenvector. How are eigenvalues and eigenvectors affected by adding the all-ones matrix? The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. Then, we solve for every possible value of v. The values we find for v are the eigenvectors. If not, how to change the complex eigenvalues and eigenvectors to real ones by python? The above cases cover everything that can happen for a $3$-by-$3$ complex matrix $A$. View Complex Eigenvalues.pdf from MATH 221 at University of British Columbia. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. If you know a bit of matrix reduction, you’ll know that your question is equivalent to: When do polynomials have complex roots? However, apparently, locating the spaces $K_X$ and $K_Y$ cannot be done by solving linear equations alone, just as one cannot generally factor a rational polynomial $p(x)$ of degree greater than $2$ into a rational polynomial with only real roots times a rational polynomial with no real roots. Yes, t can be complex. Thanks for contributing an answer to MathOverflow! If it has a real eigenvalue of multiplicity $1$, then $Y$ must preserve the corresponding 1-dimensional eigenspace of $X$, and hence a nonzero element of that eigenspace is an eigenvector of $Y$ as well. It has eigenvectors if and only if it has eigenvalues, by definition. Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. A simple example is the 1x1 matrix A = [i] !! If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. Every square matrix of degree n does have n eigenvalues and corresponding n eigenvectors. the dot product of two complex vectors is complex (in general). In fact, the part (b) gives an example of such a matrix. ... Fortunately for the reader all nonsymmetric matrices of interest to us in multivariate analysis will have real eigenvalues and real eigenvectors. share | cite | improve this question | follow | edited Sep 7 '19 at 8:58. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … If the matrix Adoes not have distinct real eigenvalues, there can be complications. $\endgroup$ – acl Mar 28 '12 at 20:51 $K_1 = \{\ v\in K_0\ |\ Xv, Yv \in K_0\ \}$, Find a Basis of the Subspace Spanned by Four Matrices, Find an Orthonormal Basis of the Range of a Linear Transformation. 2.5 Complex Eigenvalues Real Canonical Form A semisimple matrix with complex conjugate eigenvalues can be diagonalized using the procedure previously described. The eigenvalues can be real or complex. The only remaining case is when $X$ and $Y$ each have a real eigenvalue of multipicity $3$, in which case, the eigenvalue must be $0$ (since $X$ and $Y$ have zero trace). I have searched online and I found two related posts on similar issue, but did not help me in finding a solution. This is not what I have … To see this, write an $n$-by-$n$ complex matrix in the form $A = X + i\,Y$ where $X$ and $Y$ are real matrices and note that finding a real eigenvector for $A$ is equivalent to finding a simultaneous eigenvector in $\mathbb{R}^n$ for both $X$ and $Y$, i.e., $X v = x\, v$ and $Y v = y\, v$.
## can a real matrix have complex eigenvectors
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https://math.stackexchange.com/questions/2293275/transformation-of-random-variable-y-2-ln-x | # Transformation of Random Variable $Y = -2\ln X$
I'm learning probability, specifically transformations of random variables, and need help with the following exercise:
If the random variable $X$ follows the uniform distribution $U(0, 1)$, find :
$(1)$ The distribution of the random variable $Y = -2 \ln X$;
$(2)$ If the random variables $X_1, X_2, \ldots , X_n$ are independent and follow the uniform distribution $U(0, 1)$, find the distribution of the random variable $Z = \sum_{i=1}^n Y_i$.
Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.
$(1)$ First, we note that the function $Y = -2 \ln X$ defined over the interval $0 < x < 1$ is an invertible (decreasing monotonic) function. Taking the exponential on both sides, it is easy to show that the inverse function is
$$x = v(y) = e^{-y/2}.$$
$(Q1)$ What is the range of the above inverse function? When $x = 0$, $y$ is undefined and when $x = 1, y = 0$. So what should I say for $a < y < b$?
Now, taking the derivative of $v(y)$, we get $$v'(y) = -\frac{1}{2}e^{-y/2}.$$
Therefore, by the change-of-variable technique we find the probability distribution function of $Y$ to be
$$f_Y(y) = f_x(v(y)) \times |v'(y)| = \frac{1}{2}e^{-y/2}.$$
This looks to me like the exponential distribution with parameter $\lambda = \frac{1}{2}$ so I'm going to say that the random variable $Y$ follows the exponential distribution with parameter $\lambda = \frac{1}{2}$, i.e. $Y \sim Exp(\lambda = \frac{1}{2})$. But again, what is the support of $y$?
Is my work correct for $(1)$? Can someone help me with the $(Q1)$ that I'm having? For $(2)$ I have absolutely no idea how to solve it. Any help is appreciated, including the theory needed to solve it.
Your work for $(1)$ is correct. For $(2)$ we use the moment generating function method.
Recall that if $X_1, X_2, \ldots, X_n$ are observations of a random sample from a population (distribution) with moment generating function $M(t)$, then the moment generating function (M.G.F.) of a linear combination $Y = \sum_{i = 1}^{n} X_i$ is
$$M_Y(t) = \prod_{i = 1}^{n} M(t) = [M(t)]^n.$$
From $(1)$ you've found that the random variables $Y_1, Y_2, \ldots, Y_n$ follow the exponential distribution with parameter $\lambda = 1/2$. To find the distribution of the random variable $Z = \sum_{i = 1}^{n} Y_i$ we use the M.G.F. method described above. One has
$$M_Z(t) = \prod_{i = 1}^{n} M_{Y_i}(t) = [M_{Y_1}(t)]^n \tag{A}$$
where $$M_{Y_1}(t) = \frac{\lambda}{\lambda - t} = \frac{1}{\frac{\lambda - t}{\lambda}} = \frac{1}{1 - \frac{t}{\lambda}} = \big(1 - \frac{t}{\lambda}\big)^{-1} \tag{B}$$
Substituing $(B)$ into $(A)$ yields
$$M_Z(t) = \big[\big(1 - \frac{t}{\lambda}\big)^{-1}\big]^n = \big(1 - \frac{t}{\lambda}\big)^{-n} = \big(1 - \frac{t}{1/2}\big)^{-n} \tag{C}$$
Does $(C)$ look familiar to you? Recall that the moment generator of the Gamma distribution is given by
$$M(t) = \big(1 - \frac{t}{\beta}\big)^{-\alpha}, \, \, t < \alpha.$$
Therefore the random variable $Z$ follows the Gamma distribution with parameters $\alpha = n$ and $b = \frac{1}{2}$. We usually write $Z \sim G(\alpha = n, \beta = 1/2)$.
Let us explicitly write the definitions and stick to basic arguments. The distribution of a variable $X$ is a function $F$: $$F_X(x)=P(X\le x)$$ so for a uniform variable $X$ we have: $$F_X(x)=\begin{cases} 0 & x\le 0 \\ x & 0\le x \le 1 \\ 1 & x\ge 1 \end{cases}$$
Then, for $Y=-2\ln X$ results: $$F_Y(y)=P(Y\le y)= P(-2\ln X \le y)=P(X\ge e^{-\frac{y}{2}})=1-F_X(e^{-\frac{y}{2}})$$ because the logarithm is monotone. Now, if $x\ge 1$, then $y=-2\ln x\le 0$, so: $$F_Y(y)=\begin{cases} 0 & y\le 0 \\ 1-e^{-\frac{y}{2}} & y\ge 0 \end{cases}$$ This concludes that $Y$ follows an exponential distribution with no need to memorize "change of variables" formulas.
As for (2), once you know that each individual distribution is exponential, their sum is the Gamma distribution: https://stats.stackexchange.com/questions/17424/what-reference-can-i-cite-for-the-proof-that-the-sum-of-n-exponential-variables?rq=1 | 2019-12-07T02:00:15 | {
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https://www.askmehelpdesk.com/showthread.php?t=774505&s=4a9447fb39bd8f5a08cf16583b410828&p=3583346 | Sabarkantha Posts: 5, Reputation: 1 New Member #1 Nov 9, 2013, 09:19 AM
Maths question
For a particular problem, I have arranged the letters A and B twelve by twelve : AAAAAAAAAAAA, ABBAABBAAABA, and so on. Order matters: AB is not equal to BA. Altogether there are 2 raised to 12 or 4096 pemutations giving all possible combinations of A and B, twelve by twelve. Now amongst the 4096 lines of all the combinations, I want to select just the lines which have A six times, like for instance AABBBBBAAABA. How many such lines are there amongst the 4096? I tried the formula 4096!/(4096-6)! but I am not sure that this does the job, since it does not specifically choose lines with six times A.
Thanks for the answer. It is important to me.
ebaines Posts: 12,132, Reputation: 1307 Expert #2 Nov 9, 2013, 10:58 AM
There are $\left( \array 12\\ 6 \array \right)$ ways to place the 6 A's in the twelve letter positions. Thus:
$\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924$
In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:
$
\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}
$
Sabarkantha Posts: 5, Reputation: 1 New Member #3 Nov 9, 2013, 06:28 PM
Originally Posted by ebaines
There are $\left( \array 12\\ 6 \array \right)$ ways to place the 6 A's in the twelve letter positions. Thus:
$\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924$
In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:
$
\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}
$
Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.
Sabarkantha Posts: 5, Reputation: 1 New Member #4 Nov 10, 2013, 12:11 AM
Originally Posted by Sabarkantha
Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.
Sabarkantha Posts: 5, Reputation: 1 New Member #5 Nov 10, 2013, 12:12 AM
ebaines Posts: 12,132, Reputation: 1307 Expert #6 Nov 10, 2013, 08:31 AM
To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success.
Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.
For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:
$
\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.
$
For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:
$
\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.
$
Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.
Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.
Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.
Sabarkantha Posts: 5, Reputation: 1 New Member #7 Nov 10, 2013, 09:39 AM
Originally Posted by ebaines
To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success.
Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.
For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:
$
\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.
$
For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:
$
\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.
$
Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.
Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.
Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.
Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.
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http://mathhelpforum.com/calculus/227773-velocity-particle-using-graph-physics.html | # Math Help - Velocity of particle using graph? Physics?
1. ## Velocity of particle using graph? Physics?
I'm not sure how to figure this out:I'm not sure how to find how far the particle with using this graph? Any help? I took physics two years ago before college and I know I learned about this but I can't remember what to do, using either physics or calc methods. Thanks!
2. ## Re: Velocity of particle using graph? Physics?
$a(t) = 3t$
Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:
$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$
You are told initial velocity is 5. Initial velocity means t=0. So:
$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$
shows $C=5$.
Hence, $v(t) = \dfrac{3}{2}t^2+5$.
Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.
Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:
$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$
3. ## Re: Velocity of particle using graph? Physics?
Originally Posted by SlipEternal
$a(t) = 3t$
Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:
$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$
You are told initial velocity is 5. Initial velocity means t=0. So:
$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$
shows $C=5$.
Hence, $v(t) = \dfrac{3}{2}t^2+5$.
Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.
Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:
$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$
Thanks for showing the steps, I couldn't put that together because I was so focused on it being a physics problem I didn't even realize I could apply these steps.
4. ## Re: Velocity of particle using graph? Physics?
Physics involves a lot of applied math. | 2015-03-27T22:43:29 | {
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http://mathhelpforum.com/advanced-statistics/75774-very-interesting-problem.html | # Thread: very interesting problem!!!
1. ## very interesting problem!!!
Hello all, I hope some of you can help me with this problem...
There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).
I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.
I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total.
Thanks
2. Originally Posted by Put
There are 30 total components in the bag. 21 are good and 9 are bad. It takes 6 components to buid one assembly (we can build 5 total assemblies), and components will be pulled randomly out of bag. What is probability for each assembly to have all 6 good components, and therefore be a good assembly (good assebmly has to have all 6 good components).
I tried using (0.7)^6 formula, but that formula accounts for any 6 good components in a row, which is not correct. If we pulled 6 good components in a row, but ther are pulled 3rd to 9th we will
not have a good assembly. Having a good assembly will require pulling good components from 1 to 6, or 7 to 12 and so on.
I tried formula mentioned above, where there is 70% chance to pull a good component out of bag, but this formula works only for infinite sample where cases are indenpendent from eachother. In our case we have 30 components total.
Thanks
You are right. Your formula only works if you were somehow taking the components "with replacement" (like some magic leprechaun replacing each part as you removed it. ).
We need to do this without replacement. This is where good old combinations come in. I assume that order doesn't matter.
The probability of having a good assembly will be the total number of ways of having a good assembly (Q1 below), divided by the total number of ways of making an assembly (Q2 below).
Q1: How many ways are there of making a good assembly? More specifically. how many ways are there to take 6 good parts? You need to take 6 good ones from the 21 good ones.
$\mathbf{C}^{21}_6 = \frac{21!}{6!\cdot 15!}=\ldots$
Q2: How many total ways are there to take 6 parts from all 30?
$\mathbf{C}^{30}_6 = \frac{30!}{6!\cdot 24!}=\ldots$
Can you finish it?
3. I calculated the results, but it looks like pretty high.
"12C6" = 924
"9C3" = 84
Q1 = "12C6 * 9C3" = 77616
Q2 = "21C9" = 293930
Q1/Q2 = .264 = 26.4 %
If I used (0.70)^6 formula we get 16.81%, so I'm not sure if 24.4% is right. We should not have more chance to make a good assembly for my case.
Let me know what you think....
4. Well if you were scratching your head at what I did, bravo. I don't know what drugs I was on. I'm sorry for being confusing.
I edited my above answer to be right. When I do out the calculation, I get 0.091 = 9.1%
Good job on being both analytical and skeptical!!!
5. This is a good answer, and thank you very much. Now it looks pretty simple....
I just caclulated a few more cases and it look like that as we increase sample size (bag size), the result will be exponentially aproaching (0.7)^6 = 11.765%, which is infinite sample size where every part is "replaced".
For example, if I take one easier case, where there are 6 total components, 2 bad, and we need 3 good ones to make a good assembly. I'll represent it as 6, 2, 3, and the result is 20%.
If I increase the sample size to 12, 4, 3 (12 total component, 4 bad, 3 to make a good assembly), the result is 25.45%.
if I keep going...
6, 2, 3, = 20%
12, 4, 3, = 25.45%
24, 8, 3, = 27.67%
48, 16, 3, = 28.68%
So the answer is aproaching (0.6667)^3=29,63%, which is case for infinite big sample size where picking components does not change probability any more.
Thanks again for your help.
6. No worries. It's good to see someone using their intuition. | 2016-10-21T17:55:02 | {
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# The operation denoted by the symbol £ is defined for all real numbers
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The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
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The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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05 Jul 2018, 05:37
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
We know that $$a£b = a\sqrt{b}$$.
We have been asked to find the value of 3£(2£4)
Step 1: Find the value of (2£4) | $$2£4 = 2\sqrt{4} = 4$$
Step 2: Find the value of 3£(4) | $$3£4 = 3\sqrt{4} = 3*2 = 6$$
Therefore, the value of 3£(2£4) is $$6$$(Option C)
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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05 Jul 2018, 15:12
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
(2£4)=3£(2*$$\sqrt{4}$$)=3£(2*2)=3£4=$$3\sqrt{4}$$=3*2=6
Ans. (C)
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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06 Jul 2018, 02:24
Solution
Given:
• £ is an operation defined for all real numbers as a£b = $$a√b$$
To find:
• The value of 3£(2£4)
Approach and Working:
• 3£(2£4) = 3£($$2√4$$) = 3£(2 x 2) = 3£4 = $$3√4$$ = 6
Hence, the correct answer is option C.
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Re: The operation denoted by the symbol £ is defined for all real numbers [#permalink]
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09 Jul 2018, 19:56
Bunuel wrote:
The operation denoted by the symbol £ is defined for all real numbers a and b as $$a£b = a\sqrt{b}$$. What is the value of 3£(2£4)?
A. $$\frac{1}{4}$$
B. $$4$$
C. $$6$$
D. $$6\sqrt{2}$$
E. $$12$$
We work from the “inside out,” so we will first evaluate 2£4 and use that value to finish the problem.
2£4 = 2√4 = 2 x 2 = 4
3£4 = 3√4 = 3 x 2 = 6
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Re: The operation denoted by the symbol £ is defined for all real numbers &nbs [#permalink] 09 Jul 2018, 19:56
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# The operation denoted by the symbol £ is defined for all real numbers
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-09-18T19:30:58 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/the-operation-denoted-by-the-symbol-is-defined-for-all-real-numbers-269757.html",
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https://math.stackexchange.com/questions/3577500/rick-wins-if-it-is-ht-and-tick-wins-if-it-is-tt-who-wins | # Rick wins if it is HT and Tick wins if it is TT. Who wins?
A fair coin is being tossed. Whenever a tail follows a head, Rick wins the game and whenever a tail follows a tail, Tick wins. Who has more probability to win?
Let's say the coin has been tossed more than twice. For Tick to win, the last three tosses should be HTT, but Rick will have won it already. So if the first two flips are not TT, Tick has no chance of winning the game.
I do not know how to proceed from here. I initially planned to brute force since we can obviously compute the probability for $$n$$ tosses (using a simple recursion). But then this observation above kind of made this computation a bit meaningless.
• But then this observation above kind of made this computation a bit meaningless. Yes, it does. So you have observed that Tick wins if and only if the first two flips are both tails. What is your question then? What is preventing you from finishing the problem? – JMoravitz Mar 11 '20 at 14:30
• Can you compute the probability that the first two flips are tails? Then you finished your computation. – Luke Mar 11 '20 at 14:32
• It is worth watching Numberphile's video on Penney's Game as well as reading related questions about Penney's game here on math.se such as this one. – JMoravitz Mar 11 '20 at 14:43
• I feel dumb. That makes sense. – oldsailorpopoye Mar 11 '20 at 16:08
Suppose the first flip is heads. Now, no matter how many more flips it takes, eventually it will flip tails and Rick will win. So, if the first flip is Heads, Rick wins. That is $$\dfrac{1}{2}$$ of the time.
Suppose the first flip is tails, but the second flip is heads. No matter how many more flips it takes, eventually, it will flip tails a second time, and as soon as it does, you will have $$HT$$ ending the sequence with Rick winning. So, that is an additional $$\dfrac{1}{2}\cdot \dfrac{1}{2}$$ chance that Rick wins.
$$\begin{array}{c|c|c}\text{Flip pattern} & \text{Winner} & \text{Probability} \\ \hline H & \text{Rick} & \dfrac{1}{2} \\ TH & \text{Rick} & \dfrac{1}{4} \\ TT & \text{Tick} & \dfrac{1}{4} \\ \hline \text{Total} & \text{Rick} & \dfrac{3}{4} \\ \text{Total} & \text{Tick} & \dfrac{1}{4} \end{array}$$
As you surmised yourself, the only way for Tick to win is if the very first two throws are tails ... if any of those first two throws is heads, then Rick wins. So, Tick wins with a probability of $$\frac{1}{4}$$, and therefore Rick with a probability of $$\frac{3}{4}$$ | 2021-04-14T23:50:05 | {
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https://python.quantecon.org/linear_algebra.html | # 4. Linear Algebra¶
## 4.1. Overview¶
Linear algebra is one of the most useful branches of applied mathematics for economists to invest in.
For example, many applied problems in economics and finance require the solution of a linear system of equations, such as
\begin{split} \begin{aligned} y_1 = a x_1 + b x_2 \\ y_2 = c x_1 + d x_2 \end{aligned} \end{split}
or, more generally,
(4.1)\begin{split}\begin{aligned} y_1 = a_{11} x_1 + a_{12} x_2 + \cdots + a_{1k} x_k \\ \vdots \\ y_n = a_{n1} x_1 + a_{n2} x_2 + \cdots + a_{nk} x_k \end{aligned}\end{split}
The objective here is to solve for the “unknowns” $$x_1, \ldots, x_k$$ given $$a_{11}, \ldots, a_{nk}$$ and $$y_1, \ldots, y_n$$.
When considering such problems, it is essential that we first consider at least some of the following questions
• Does a solution actually exist?
• Are there in fact many solutions, and if so how should we interpret them?
• If no solution exists, is there a best “approximate” solution?
• If a solution exists, how should we compute it?
These are the kinds of topics addressed by linear algebra.
In this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation.
We admit some overlap with this lecture, where operations on NumPy arrays were first explained.
Note that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along.
%matplotlib inline
import matplotlib.pyplot as plt
plt.rcParams["figure.figsize"] = (11, 5) #set default figure size
import numpy as np
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
from scipy.interpolate import interp2d
from scipy.linalg import inv, solve, det, eig
## 4.2. Vectors¶
A vector of length $$n$$ is just a sequence (or array, or tuple) of $$n$$ numbers, which we write as $$x = (x_1, \ldots, x_n)$$ or $$x = [x_1, \ldots, x_n]$$.
We will write these sequences either horizontally or vertically as we please.
(Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two)
The set of all $$n$$-vectors is denoted by $$\mathbb R^n$$.
For example, $$\mathbb R ^2$$ is the plane, and a vector in $$\mathbb R^2$$ is just a point in the plane.
Traditionally, vectors are represented visually as arrows from the origin to the point.
The following figure represents three vectors in this manner
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.set(xlim=(-5, 5), ylim=(-5, 5))
ax.grid()
vecs = ((2, 4), (-3, 3), (-4, -3.5))
for v in vecs:
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=0.7,
width=0.5))
ax.text(1.1 * v[0], 1.1 * v[1], str(v))
plt.show()
### 4.2.1. Vector Operations¶
The two most common operators for vectors are addition and scalar multiplication, which we now describe.
As a matter of definition, when we add two vectors, we add them element-by-element
$\begin{split} x + y = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} := \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{bmatrix} \end{split}$
Scalar multiplication is an operation that takes a number $$\gamma$$ and a vector $$x$$ and produces
$\begin{split} \gamma x := \begin{bmatrix} \gamma x_1 \\ \gamma x_2 \\ \vdots \\ \gamma x_n \end{bmatrix} \end{split}$
Scalar multiplication is illustrated in the next figure
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.set(xlim=(-5, 5), ylim=(-5, 5))
x = (2, 2)
ax.annotate('', xy=x, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=1,
width=0.5))
ax.text(x[0] + 0.4, x[1] - 0.2, '$x$', fontsize='16')
scalars = (-2, 2)
x = np.array(x)
for s in scalars:
v = s * x
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='red',
shrink=0,
alpha=0.5,
width=0.5))
ax.text(v[0] + 0.4, v[1] - 0.2, f'${s} x$', fontsize='16')
plt.show()
In Python, a vector can be represented as a list or tuple, such as x = (2, 4, 6), but is more commonly represented as a NumPy array.
One advantage of NumPy arrays is that scalar multiplication and addition have very natural syntax
x = np.ones(3) # Vector of three ones
y = np.array((2, 4, 6)) # Converts tuple (2, 4, 6) into array
x + y
array([3., 5., 7.])
4 * x
array([4., 4., 4.])
### 4.2.2. Inner Product and Norm¶
The inner product of vectors $$x,y \in \mathbb R ^n$$ is defined as
$x' y := \sum_{i=1}^n x_i y_i$
Two vectors are called orthogonal if their inner product is zero.
The norm of a vector $$x$$ represents its “length” (i.e., its distance from the zero vector) and is defined as
$\| x \| := \sqrt{x' x} := \left( \sum_{i=1}^n x_i^2 \right)^{1/2}$
The expression $$\| x - y\|$$ is thought of as the distance between $$x$$ and $$y$$.
Continuing on from the previous example, the inner product and norm can be computed as follows
np.sum(x * y) # Inner product of x and y
12.0
np.sqrt(np.sum(x**2)) # Norm of x, take one
1.7320508075688772
np.linalg.norm(x) # Norm of x, take two
1.7320508075688772
### 4.2.3. Span¶
Given a set of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$, it’s natural to think about the new vectors we can create by performing linear operations.
New vectors created in this manner are called linear combinations of $$A$$.
In particular, $$y \in \mathbb R ^n$$ is a linear combination of $$A := \{a_1, \ldots, a_k\}$$ if
$y = \beta_1 a_1 + \cdots + \beta_k a_k \text{ for some scalars } \beta_1, \ldots, \beta_k$
In this context, the values $$\beta_1, \ldots, \beta_k$$ are called the coefficients of the linear combination.
The set of linear combinations of $$A$$ is called the span of $$A$$.
The next figure shows the span of $$A = \{a_1, a_2\}$$ in $$\mathbb R ^3$$.
The span is a two-dimensional plane passing through these two points and the origin.
fig = plt.figure(figsize=(10, 8))
ax = fig.gca(projection='3d')
x_min, x_max = -5, 5
y_min, y_max = -5, 5
α, β = 0.2, 0.1
ax.set(xlim=(x_min, x_max), ylim=(x_min, x_max), zlim=(x_min, x_max),
xticks=(0,), yticks=(0,), zticks=(0,))
gs = 3
z = np.linspace(x_min, x_max, gs)
x = np.zeros(gs)
y = np.zeros(gs)
ax.plot(x, y, z, 'k-', lw=2, alpha=0.5)
ax.plot(z, x, y, 'k-', lw=2, alpha=0.5)
ax.plot(y, z, x, 'k-', lw=2, alpha=0.5)
# Fixed linear function, to generate a plane
def f(x, y):
return α * x + β * y
# Vector locations, by coordinate
x_coords = np.array((3, 3))
y_coords = np.array((4, -4))
z = f(x_coords, y_coords)
for i in (0, 1):
ax.text(x_coords[i], y_coords[i], z[i], f'$a_{i+1}$', fontsize=14)
# Lines to vectors
for i in (0, 1):
x = (0, x_coords[i])
y = (0, y_coords[i])
z = (0, f(x_coords[i], y_coords[i]))
ax.plot(x, y, z, 'b-', lw=1.5, alpha=0.6)
# Draw the plane
grid_size = 20
xr2 = np.linspace(x_min, x_max, grid_size)
yr2 = np.linspace(y_min, y_max, grid_size)
x2, y2 = np.meshgrid(xr2, yr2)
z2 = f(x2, y2)
ax.plot_surface(x2, y2, z2, rstride=1, cstride=1, cmap=cm.jet,
linewidth=0, antialiased=True, alpha=0.2)
plt.show()
/tmp/ipykernel_56021/4243435577.py:2: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().
ax = fig.gca(projection='3d')
#### 4.2.3.1. Examples¶
If $$A$$ contains only one vector $$a_1 \in \mathbb R ^2$$, then its span is just the scalar multiples of $$a_1$$, which is the unique line passing through both $$a_1$$ and the origin.
If $$A = \{e_1, e_2, e_3\}$$ consists of the canonical basis vectors of $$\mathbb R ^3$$, that is
$\begin{split} e_1 := \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \quad e_2 := \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \quad e_3 := \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{split}$
then the span of $$A$$ is all of $$\mathbb R ^3$$, because, for any $$x = (x_1, x_2, x_3) \in \mathbb R ^3$$, we can write
$x = x_1 e_1 + x_2 e_2 + x_3 e_3$
Now consider $$A_0 = \{e_1, e_2, e_1 + e_2\}$$.
If $$y = (y_1, y_2, y_3)$$ is any linear combination of these vectors, then $$y_3 = 0$$ (check it).
Hence $$A_0$$ fails to span all of $$\mathbb R ^3$$.
### 4.2.4. Linear Independence¶
As we’ll see, it’s often desirable to find families of vectors with relatively large span, so that many vectors can be described by linear operators on a few vectors.
The condition we need for a set of vectors to have a large span is what’s called linear independence.
In particular, a collection of vectors $$A := \{a_1, \ldots, a_k\}$$ in $$\mathbb R ^n$$ is said to be
• linearly dependent if some strict subset of $$A$$ has the same span as $$A$$.
• linearly independent if it is not linearly dependent.
Put differently, a set of vectors is linearly independent if no vector is redundant to the span and linearly dependent otherwise.
To illustrate the idea, recall the figure that showed the span of vectors $$\{a_1, a_2\}$$ in $$\mathbb R ^3$$ as a plane through the origin.
If we take a third vector $$a_3$$ and form the set $$\{a_1, a_2, a_3\}$$, this set will be
• linearly dependent if $$a_3$$ lies in the plane
• linearly independent otherwise
As another illustration of the concept, since $$\mathbb R ^n$$ can be spanned by $$n$$ vectors (see the discussion of canonical basis vectors above), any collection of $$m > n$$ vectors in $$\mathbb R ^n$$ must be linearly dependent.
The following statements are equivalent to linear independence of $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$
1. No vector in $$A$$ can be formed as a linear combination of the other elements.
2. If $$\beta_1 a_1 + \cdots \beta_k a_k = 0$$ for scalars $$\beta_1, \ldots, \beta_k$$, then $$\beta_1 = \cdots = \beta_k = 0$$.
(The zero in the first expression is the origin of $$\mathbb R ^n$$)
### 4.2.5. Unique Representations¶
Another nice thing about sets of linearly independent vectors is that each element in the span has a unique representation as a linear combination of these vectors.
In other words, if $$A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$$ is linearly independent and
$y = \beta_1 a_1 + \cdots \beta_k a_k$
then no other coefficient sequence $$\gamma_1, \ldots, \gamma_k$$ will produce the same vector $$y$$.
Indeed, if we also have $$y = \gamma_1 a_1 + \cdots \gamma_k a_k$$, then
$(\beta_1 - \gamma_1) a_1 + \cdots + (\beta_k - \gamma_k) a_k = 0$
Linear independence now implies $$\gamma_i = \beta_i$$ for all $$i$$.
## 4.3. Matrices¶
Matrices are a neat way of organizing data for use in linear operations.
An $$n \times k$$ matrix is a rectangular array $$A$$ of numbers with $$n$$ rows and $$k$$ columns:
$\begin{split} A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1k} \\ a_{21} & a_{22} & \cdots & a_{2k} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nk} \end{bmatrix} \end{split}$
Often, the numbers in the matrix represent coefficients in a system of linear equations, as discussed at the start of this lecture.
For obvious reasons, the matrix $$A$$ is also called a vector if either $$n = 1$$ or $$k = 1$$.
In the former case, $$A$$ is called a row vector, while in the latter it is called a column vector.
If $$n = k$$, then $$A$$ is called square.
The matrix formed by replacing $$a_{ij}$$ by $$a_{ji}$$ for every $$i$$ and $$j$$ is called the transpose of $$A$$ and denoted $$A'$$ or $$A^{\top}$$.
If $$A = A'$$, then $$A$$ is called symmetric.
For a square matrix $$A$$, the $$i$$ elements of the form $$a_{ii}$$ for $$i=1,\ldots,n$$ are called the principal diagonal.
$$A$$ is called diagonal if the only nonzero entries are on the principal diagonal.
If, in addition to being diagonal, each element along the principal diagonal is equal to 1, then $$A$$ is called the identity matrix and denoted by $$I$$.
### 4.3.1. Matrix Operations¶
Just as was the case for vectors, a number of algebraic operations are defined for matrices.
Scalar multiplication and addition are immediate generalizations of the vector case:
$\begin{split} \gamma A = \gamma \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} := \begin{bmatrix} \gamma a_{11} & \cdots & \gamma a_{1k} \\ \vdots & \vdots & \vdots \\ \gamma a_{n1} & \cdots & \gamma a_{nk} \end{bmatrix} \end{split}$
and
$\begin{split} A + B = \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} + \begin{bmatrix} b_{11} & \cdots & b_{1k} \\ \vdots & \vdots & \vdots \\ b_{n1} & \cdots & b_{nk} \end{bmatrix} := \begin{bmatrix} a_{11} + b_{11} & \cdots & a_{1k} + b_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} + b_{n1} & \cdots & a_{nk} + b_{nk} \end{bmatrix} \end{split}$
In the latter case, the matrices must have the same shape in order for the definition to make sense.
We also have a convention for multiplying two matrices.
The rule for matrix multiplication generalizes the idea of inner products discussed above and is designed to make multiplication play well with basic linear operations.
If $$A$$ and $$B$$ are two matrices, then their product $$A B$$ is formed by taking as its $$i,j$$-th element the inner product of the $$i$$-th row of $$A$$ and the $$j$$-th column of $$B$$.
There are many tutorials to help you visualize this operation, such as this one, or the discussion on the Wikipedia page.
If $$A$$ is $$n \times k$$ and $$B$$ is $$j \times m$$, then to multiply $$A$$ and $$B$$ we require $$k = j$$, and the resulting matrix $$A B$$ is $$n \times m$$.
As perhaps the most important special case, consider multiplying $$n \times k$$ matrix $$A$$ and $$k \times 1$$ column vector $$x$$.
According to the preceding rule, this gives us an $$n \times 1$$ column vector
(4.2)$\begin{split}A x = \begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nk} \end{bmatrix} \begin{bmatrix} x_{1} \\ \vdots \\ x_{k} \end{bmatrix} := \begin{bmatrix} a_{11} x_1 + \cdots + a_{1k} x_k \\ \vdots \\ a_{n1} x_1 + \cdots + a_{nk} x_k \end{bmatrix}\end{split}$
Note
$$A B$$ and $$B A$$ are not generally the same thing.
Another important special case is the identity matrix.
You should check that if $$A$$ is $$n \times k$$ and $$I$$ is the $$k \times k$$ identity matrix, then $$AI = A$$.
If $$I$$ is the $$n \times n$$ identity matrix, then $$IA = A$$.
### 4.3.2. Matrices in NumPy¶
NumPy arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations 1.
You can create them manually from tuples of tuples (or lists of lists) as follows
A = ((1, 2),
(3, 4))
type(A)
tuple
A = np.array(A)
type(A)
numpy.ndarray
A.shape
(2, 2)
The shape attribute is a tuple giving the number of rows and columns — see here for more discussion.
To get the transpose of A, use A.transpose() or, more simply, A.T.
There are many convenient functions for creating common matrices (matrices of zeros, ones, etc.) — see here.
Since operations are performed elementwise by default, scalar multiplication and addition have very natural syntax
A = np.identity(3)
B = np.ones((3, 3))
2 * A
array([[2., 0., 0.],
[0., 2., 0.],
[0., 0., 2.]])
A + B
array([[2., 1., 1.],
[1., 2., 1.],
[1., 1., 2.]])
To multiply matrices we use the @ symbol.
In particular, A @ B is matrix multiplication, whereas A * B is element-by-element multiplication.
See here for more discussion.
### 4.3.3. Matrices as Maps¶
Each $$n \times k$$ matrix $$A$$ can be identified with a function $$f(x) = Ax$$ that maps $$x \in \mathbb R ^k$$ into $$y = Ax \in \mathbb R ^n$$.
These kinds of functions have a special property: they are linear.
A function $$f \colon \mathbb R ^k \to \mathbb R ^n$$ is called linear if, for all $$x, y \in \mathbb R ^k$$ and all scalars $$\alpha, \beta$$, we have
$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$
You can check that this holds for the function $$f(x) = A x + b$$ when $$b$$ is the zero vector and fails when $$b$$ is nonzero.
In fact, it’s known that $$f$$ is linear if and only if there exists a matrix $$A$$ such that $$f(x) = Ax$$ for all $$x$$.
## 4.4. Solving Systems of Equations¶
Recall again the system of equations (4.1).
If we compare (4.1) and (4.2), we see that (4.1) can now be written more conveniently as
(4.3)$y = Ax$
The problem we face is to determine a vector $$x \in \mathbb R ^k$$ that solves (4.3), taking $$y$$ and $$A$$ as given.
This is a special case of a more general problem: Find an $$x$$ such that $$y = f(x)$$.
Given an arbitrary function $$f$$ and a $$y$$, is there always an $$x$$ such that $$y = f(x)$$?
If so, is it always unique?
The answer to both these questions is negative, as the next figure shows
def f(x):
return 0.6 * np.cos(4 * x) + 1.4
xmin, xmax = -1, 1
x = np.linspace(xmin, xmax, 160)
y = f(x)
ya, yb = np.min(y), np.max(y)
fig, axes = plt.subplots(2, 1, figsize=(10, 10))
for ax in axes:
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.set(ylim=(-0.6, 3.2), xlim=(xmin, xmax),
yticks=(), xticks=())
ax.plot(x, y, 'k-', lw=2, label='$f$')
ax.fill_between(x, ya, yb, facecolor='blue', alpha=0.05)
ax.vlines([0], ya, yb, lw=3, color='blue', label='range of $f$')
ax.text(0.04, -0.3, '$0$', fontsize=16)
ax = axes[0]
ax.legend(loc='upper right', frameon=False)
ybar = 1.5
ax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)
ax.text(0.05, 0.8 * ybar, '$y$', fontsize=16)
for i, z in enumerate((-0.35, 0.35)):
ax.vlines(z, 0, f(z), linestyle='--', alpha=0.5)
ax.text(z, -0.2, f'$x_{i}$', fontsize=16)
ax = axes[1]
ybar = 2.6
ax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)
ax.text(0.04, 0.91 * ybar, '$y$', fontsize=16)
plt.show()
In the first plot, there are multiple solutions, as the function is not one-to-one, while in the second there are no solutions, since $$y$$ lies outside the range of $$f$$.
Can we impose conditions on $$A$$ in (4.3) that rule out these problems?
In this context, the most important thing to recognize about the expression $$Ax$$ is that it corresponds to a linear combination of the columns of $$A$$.
In particular, if $$a_1, \ldots, a_k$$ are the columns of $$A$$, then
$Ax = x_1 a_1 + \cdots + x_k a_k$
Hence the range of $$f(x) = Ax$$ is exactly the span of the columns of $$A$$.
We want the range to be large so that it contains arbitrary $$y$$.
As you might recall, the condition that we want for the span to be large is linear independence.
A happy fact is that linear independence of the columns of $$A$$ also gives us uniqueness.
Indeed, it follows from our earlier discussion that if $$\{a_1, \ldots, a_k\}$$ are linearly independent and $$y = Ax = x_1 a_1 + \cdots + x_k a_k$$, then no $$z \not= x$$ satisfies $$y = Az$$.
### 4.4.1. The Square Matrix Case¶
Let’s discuss some more details, starting with the case where $$A$$ is $$n \times n$$.
This is the familiar case where the number of unknowns equals the number of equations.
For arbitrary $$y \in \mathbb R ^n$$, we hope to find a unique $$x \in \mathbb R ^n$$ such that $$y = Ax$$.
In view of the observations immediately above, if the columns of $$A$$ are linearly independent, then their span, and hence the range of $$f(x) = Ax$$, is all of $$\mathbb R ^n$$.
Hence there always exists an $$x$$ such that $$y = Ax$$.
Moreover, the solution is unique.
In particular, the following are equivalent
1. The columns of $$A$$ are linearly independent.
2. For any $$y \in \mathbb R ^n$$, the equation $$y = Ax$$ has a unique solution.
The property of having linearly independent columns is sometimes expressed as having full column rank.
#### 4.4.1.1. Inverse Matrices¶
Can we give some sort of expression for the solution?
If $$y$$ and $$A$$ are scalar with $$A \not= 0$$, then the solution is $$x = A^{-1} y$$.
A similar expression is available in the matrix case.
In particular, if square matrix $$A$$ has full column rank, then it possesses a multiplicative inverse matrix $$A^{-1}$$, with the property that $$A A^{-1} = A^{-1} A = I$$.
As a consequence, if we pre-multiply both sides of $$y = Ax$$ by $$A^{-1}$$, we get $$x = A^{-1} y$$.
This is the solution that we’re looking for.
#### 4.4.1.2. Determinants¶
Another quick comment about square matrices is that to every such matrix we assign a unique number called the determinant of the matrix — you can find the expression for it here.
If the determinant of $$A$$ is not zero, then we say that $$A$$ is nonsingular.
Perhaps the most important fact about determinants is that $$A$$ is nonsingular if and only if $$A$$ is of full column rank.
This gives us a useful one-number summary of whether or not a square matrix can be inverted.
### 4.4.2. More Rows than Columns¶
This is the $$n \times k$$ case with $$n > k$$.
This case is very important in many settings, not least in the setting of linear regression (where $$n$$ is the number of observations, and $$k$$ is the number of explanatory variables).
Given arbitrary $$y \in \mathbb R ^n$$, we seek an $$x \in \mathbb R ^k$$ such that $$y = Ax$$.
In this setting, the existence of a solution is highly unlikely.
Without much loss of generality, let’s go over the intuition focusing on the case where the columns of $$A$$ are linearly independent.
It follows that the span of the columns of $$A$$ is a $$k$$-dimensional subspace of $$\mathbb R ^n$$.
This span is very “unlikely” to contain arbitrary $$y \in \mathbb R ^n$$.
To see why, recall the figure above, where $$k=2$$ and $$n=3$$.
Imagine an arbitrarily chosen $$y \in \mathbb R ^3$$, located somewhere in that three-dimensional space.
What’s the likelihood that $$y$$ lies in the span of $$\{a_1, a_2\}$$ (i.e., the two dimensional plane through these points)?
In a sense, it must be very small, since this plane has zero “thickness”.
As a result, in the $$n > k$$ case we usually give up on existence.
However, we can still seek the best approximation, for example, an $$x$$ that makes the distance $$\| y - Ax\|$$ as small as possible.
To solve this problem, one can use either calculus or the theory of orthogonal projections.
The solution is known to be $$\hat x = (A'A)^{-1}A'y$$ — see for example chapter 3 of these notes.
### 4.4.3. More Columns than Rows¶
This is the $$n \times k$$ case with $$n < k$$, so there are fewer equations than unknowns.
In this case there are either no solutions or infinitely many — in other words, uniqueness never holds.
For example, consider the case where $$k=3$$ and $$n=2$$.
Thus, the columns of $$A$$ consists of 3 vectors in $$\mathbb R ^2$$.
This set can never be linearly independent, since it is possible to find two vectors that span $$\mathbb R ^2$$.
(For example, use the canonical basis vectors)
It follows that one column is a linear combination of the other two.
For example, let’s say that $$a_1 = \alpha a_2 + \beta a_3$$.
Then if $$y = Ax = x_1 a_1 + x_2 a_2 + x_3 a_3$$, we can also write
$y = x_1 (\alpha a_2 + \beta a_3) + x_2 a_2 + x_3 a_3 = (x_1 \alpha + x_2) a_2 + (x_1 \beta + x_3) a_3$
In other words, uniqueness fails.
### 4.4.4. Linear Equations with SciPy¶
Here’s an illustration of how to solve linear equations with SciPy’s linalg submodule.
All of these routines are Python front ends to time-tested and highly optimized FORTRAN code
A = ((1, 2), (3, 4))
A = np.array(A)
y = np.ones((2, 1)) # Column vector
det(A) # Check that A is nonsingular, and hence invertible
-2.0
A_inv = inv(A) # Compute the inverse
A_inv
array([[-2. , 1. ],
[ 1.5, -0.5]])
x = A_inv @ y # Solution
A @ x # Should equal y
array([[1.],
[1.]])
solve(A, y) # Produces the same solution
array([[-1.],
[ 1.]])
Observe how we can solve for $$x = A^{-1} y$$ by either via inv(A) @ y, or using solve(A, y).
The latter method uses a different algorithm (LU decomposition) that is numerically more stable, and hence should almost always be preferred.
To obtain the least-squares solution $$\hat x = (A'A)^{-1}A'y$$, use scipy.linalg.lstsq(A, y).
## 4.5. Eigenvalues and Eigenvectors¶
Let $$A$$ be an $$n \times n$$ square matrix.
If $$\lambda$$ is scalar and $$v$$ is a non-zero vector in $$\mathbb R ^n$$ such that
$A v = \lambda v$
then we say that $$\lambda$$ is an eigenvalue of $$A$$, and $$v$$ is an eigenvector.
Thus, an eigenvector of $$A$$ is a vector such that when the map $$f(x) = Ax$$ is applied, $$v$$ is merely scaled.
The next figure shows two eigenvectors (blue arrows) and their images under $$A$$ (red arrows).
As expected, the image $$Av$$ of each $$v$$ is just a scaled version of the original
A = ((1, 2),
(2, 1))
A = np.array(A)
evals, evecs = eig(A)
evecs = evecs[:, 0], evecs[:, 1]
fig, ax = plt.subplots(figsize=(10, 8))
# Set the axes through the origin
for spine in ['left', 'bottom']:
ax.spines[spine].set_position('zero')
for spine in ['right', 'top']:
ax.spines[spine].set_color('none')
ax.grid(alpha=0.4)
xmin, xmax = -3, 3
ymin, ymax = -3, 3
ax.set(xlim=(xmin, xmax), ylim=(ymin, ymax))
# Plot each eigenvector
for v in evecs:
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='blue',
shrink=0,
alpha=0.6,
width=0.5))
# Plot the image of each eigenvector
for v in evecs:
v = A @ v
ax.annotate('', xy=v, xytext=(0, 0),
arrowprops=dict(facecolor='red',
shrink=0,
alpha=0.6,
width=0.5))
# Plot the lines they run through
x = np.linspace(xmin, xmax, 3)
for v in evecs:
a = v[1] / v[0]
ax.plot(x, a * x, 'b-', lw=0.4)
plt.show()
The eigenvalue equation is equivalent to $$(A - \lambda I) v = 0$$, and this has a nonzero solution $$v$$ only when the columns of $$A - \lambda I$$ are linearly dependent.
This in turn is equivalent to stating that the determinant is zero.
Hence to find all eigenvalues, we can look for $$\lambda$$ such that the determinant of $$A - \lambda I$$ is zero.
This problem can be expressed as one of solving for the roots of a polynomial in $$\lambda$$ of degree $$n$$.
This in turn implies the existence of $$n$$ solutions in the complex plane, although some might be repeated.
Some nice facts about the eigenvalues of a square matrix $$A$$ are as follows
1. The determinant of $$A$$ equals the product of the eigenvalues.
2. The trace of $$A$$ (the sum of the elements on the principal diagonal) equals the sum of the eigenvalues.
3. If $$A$$ is symmetric, then all of its eigenvalues are real.
4. If $$A$$ is invertible and $$\lambda_1, \ldots, \lambda_n$$ are its eigenvalues, then the eigenvalues of $$A^{-1}$$ are $$1/\lambda_1, \ldots, 1/\lambda_n$$.
A corollary of the first statement is that a matrix is invertible if and only if all its eigenvalues are nonzero.
Using SciPy, we can solve for the eigenvalues and eigenvectors of a matrix as follows
A = ((1, 2),
(2, 1))
A = np.array(A)
evals, evecs = eig(A)
evals
array([ 3.+0.j, -1.+0.j])
evecs
array([[ 0.70710678, -0.70710678],
[ 0.70710678, 0.70710678]])
Note that the columns of evecs are the eigenvectors.
Since any scalar multiple of an eigenvector is an eigenvector with the same eigenvalue (check it), the eig routine normalizes the length of each eigenvector to one.
### 4.5.1. Generalized Eigenvalues¶
It is sometimes useful to consider the generalized eigenvalue problem, which, for given matrices $$A$$ and $$B$$, seeks generalized eigenvalues $$\lambda$$ and eigenvectors $$v$$ such that
$A v = \lambda B v$
This can be solved in SciPy via scipy.linalg.eig(A, B).
Of course, if $$B$$ is square and invertible, then we can treat the generalized eigenvalue problem as an ordinary eigenvalue problem $$B^{-1} A v = \lambda v$$, but this is not always the case.
## 4.6. Further Topics¶
We round out our discussion by briefly mentioning several other important topics.
### 4.6.1. Series Expansions¶
Recall the usual summation formula for a geometric progression, which states that if $$|a| < 1$$, then $$\sum_{k=0}^{\infty} a^k = (1 - a)^{-1}$$.
A generalization of this idea exists in the matrix setting.
#### 4.6.1.1. Matrix Norms¶
Let $$A$$ be a square matrix, and let
$\| A \| := \max_{\| x \| = 1} \| A x \|$
The norms on the right-hand side are ordinary vector norms, while the norm on the left-hand side is a matrix norm — in this case, the so-called spectral norm.
For example, for a square matrix $$S$$, the condition $$\| S \| < 1$$ means that $$S$$ is contractive, in the sense that it pulls all vectors towards the origin 2.
#### 4.6.1.2. Neumann’s Theorem¶
Let $$A$$ be a square matrix and let $$A^k := A A^{k-1}$$ with $$A^1 := A$$.
In other words, $$A^k$$ is the $$k$$-th power of $$A$$.
Neumann’s theorem states the following: If $$\| A^k \| < 1$$ for some $$k \in \mathbb{N}$$, then $$I - A$$ is invertible, and
(4.4)$(I - A)^{-1} = \sum_{k=0}^{\infty} A^k$
#### 4.6.1.3. Spectral Radius¶
A result known as Gelfand’s formula tells us that, for any square matrix $$A$$,
$\rho(A) = \lim_{k \to \infty} \| A^k \|^{1/k}$
Here $$\rho(A)$$ is the spectral radius, defined as $$\max_i |\lambda_i|$$, where $$\{\lambda_i\}_i$$ is the set of eigenvalues of $$A$$.
As a consequence of Gelfand’s formula, if all eigenvalues are strictly less than one in modulus, there exists a $$k$$ with $$\| A^k \| < 1$$.
In which case (4.4) is valid.
### 4.6.2. Positive Definite Matrices¶
Let $$A$$ be a symmetric $$n \times n$$ matrix.
We say that $$A$$ is
1. positive definite if $$x' A x > 0$$ for every $$x \in \mathbb R ^n \setminus \{0\}$$
2. positive semi-definite or nonnegative definite if $$x' A x \geq 0$$ for every $$x \in \mathbb R ^n$$
Analogous definitions exist for negative definite and negative semi-definite matrices.
It is notable that if $$A$$ is positive definite, then all of its eigenvalues are strictly positive, and hence $$A$$ is invertible (with positive definite inverse).
### 4.6.3. Differentiating Linear and Quadratic Forms¶
The following formulas are useful in many economic contexts. Let
• $$z, x$$ and $$a$$ all be $$n \times 1$$ vectors
• $$A$$ be an $$n \times n$$ matrix
• $$B$$ be an $$m \times n$$ matrix and $$y$$ be an $$m \times 1$$ vector
Then
1. $$\frac{\partial a' x}{\partial x} = a$$
2. $$\frac{\partial A x}{\partial x} = A'$$
3. $$\frac{\partial x'A x}{\partial x} = (A + A') x$$
4. $$\frac{\partial y'B z}{\partial y} = B z$$
5. $$\frac{\partial y'B z}{\partial B} = y z'$$
Exercise 1 below asks you to apply these formulas.
### 4.6.4. Further Reading¶
The documentation of the scipy.linalg submodule can be found here.
Chapters 2 and 3 of the Econometric Theory contains a discussion of linear algebra along the same lines as above, with solved exercises.
If you don’t mind a slightly abstract approach, a nice intermediate-level text on linear algebra is [Janich94].
## 4.7. Exercises¶
### 4.7.1. Exercise 1¶
Let $$x$$ be a given $$n \times 1$$ vector and consider the problem
$v(x) = \max_{y,u} \left\{ - y'P y - u' Q u \right\}$
subject to the linear constraint
$y = A x + B u$
Here
• $$P$$ is an $$n \times n$$ matrix and $$Q$$ is an $$m \times m$$ matrix
• $$A$$ is an $$n \times n$$ matrix and $$B$$ is an $$n \times m$$ matrix
• both $$P$$ and $$Q$$ are symmetric and positive semidefinite
(What must the dimensions of $$y$$ and $$u$$ be to make this a well-posed problem?)
One way to solve the problem is to form the Lagrangian
$\mathcal L = - y' P y - u' Q u + \lambda' \left[A x + B u - y\right]$
where $$\lambda$$ is an $$n \times 1$$ vector of Lagrange multipliers.
Try applying the formulas given above for differentiating quadratic and linear forms to obtain the first-order conditions for maximizing $$\mathcal L$$ with respect to $$y, u$$ and minimizing it with respect to $$\lambda$$.
Show that these conditions imply that
1. $$\lambda = - 2 P y$$.
2. The optimizing choice of $$u$$ satisfies $$u = - (Q + B' P B)^{-1} B' P A x$$.
3. The function $$v$$ satisfies $$v(x) = - x' \tilde P x$$ where $$\tilde P = A' P A - A'P B (Q + B'P B)^{-1} B' P A$$.
As we will see, in economic contexts Lagrange multipliers often are shadow prices.
Note
If we don’t care about the Lagrange multipliers, we can substitute the constraint into the objective function, and then just maximize $$-(Ax + Bu)'P (Ax + Bu) - u' Q u$$ with respect to $$u$$. You can verify that this leads to the same maximizer.
## 4.8. Solutions¶
### 4.8.1. Solution to Exercise 1¶
We have an optimization problem:
$v(x) = \max_{y,u} \{ -y'Py - u'Qu \}$
s.t.
$y = Ax + Bu$
with primitives
• $$P$$ be a symmetric and positive semidefinite $$n \times n$$ matrix
• $$Q$$ be a symmetric and positive semidefinite $$m \times m$$ matrix
• $$A$$ an $$n \times n$$ matrix
• $$B$$ an $$n \times m$$ matrix
The associated Lagrangian is:
$L = -y'Py - u'Qu + \lambda' \lbrack Ax + Bu - y \rbrack$
Step 1.
Differentiating Lagrangian equation w.r.t y and setting its derivative equal to zero yields
$\frac{ \partial L}{\partial y} = - (P + P') y - \lambda = - 2 P y - \lambda = 0 \:,$
since P is symmetric.
Accordingly, the first-order condition for maximizing L w.r.t. y implies
$\lambda = -2 Py \:$
Step 2.
Differentiating Lagrangian equation w.r.t. u and setting its derivative equal to zero yields
$\frac{ \partial L}{\partial u} = - (Q + Q') u - B'\lambda = - 2Qu + B'\lambda = 0 \:$
Substituting $$\lambda = -2 P y$$ gives
$Qu + B'Py = 0 \:$
Substituting the linear constraint $$y = Ax + Bu$$ into above equation gives
$Qu + B'P(Ax + Bu) = 0$
$(Q + B'PB)u + B'PAx = 0$
which is the first-order condition for maximizing $$L$$ w.r.t. $$u$$.
Thus, the optimal choice of u must satisfy
$u = -(Q + B'PB)^{-1}B'PAx \:,$
which follows from the definition of the first-order conditions for Lagrangian equation.
Step 3.
Rewriting our problem by substituting the constraint into the objective function, we get
$v(x) = \max_{u} \{ -(Ax+ Bu)'P(Ax+Bu) - u'Qu \} \:$
Since we know the optimal choice of u satisfies $$u = -(Q + B'PB)^{-1}B'PAx$$, then
$v(x) = -(Ax+ B u)'P(Ax+B u) - u'Q u \,\,\,\, with \,\,\,\, u = -(Q + B'PB)^{-1}B'PAx$
To evaluate the function
\begin{split} \begin{aligned} v(x) &= -(Ax+ B u)'P(Ax+Bu) - u'Q u \\ &= -(x'A' + u'B')P(Ax+Bu) - u'Q u \\ &= - x'A'PAx - u'B'PAx - x'A'PBu - u'B'PBu - u'Qu \\ &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u \end{aligned} \end{split}
For simplicity, denote by $$S := (Q + B'PB)^{-1} B'PA$$, then $$u = -Sx$$.
Regarding the second term $$- 2u'B'PAx$$,
\begin{split} \begin{aligned} -2u'B'PAx &= -2 x'S'B'PAx \\ & = 2 x'A'PB( Q + B'PB)^{-1} B'PAx \end{aligned} \end{split}
Notice that the term $$(Q + B'PB)^{-1}$$ is symmetric as both P and Q are symmetric.
Regarding the third term $$- u'(Q + B'PB) u$$,
\begin{split} \begin{aligned} -u'(Q + B'PB) u &= - x'S' (Q + B'PB)Sx \\ &= -x'A'PB(Q + B'PB)^{-1}B'PAx \end{aligned} \end{split}
Hence, the summation of second and third terms is $$x'A'PB(Q + B'PB)^{-1}B'PAx$$.
This implies that
\begin{split} \begin{aligned} v(x) &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u\\ &= - x'A'PAx + x'A'PB(Q + B'PB)^{-1}B'PAx \\ &= -x'[A'PA - A'PB(Q + B'PB)^{-1}B'PA] x \end{aligned} \end{split}
Therefore, the solution to the optimization problem $$v(x) = -x' \tilde{P}x$$ follows the above result by denoting $$\tilde{P} := A'PA - A'PB(Q + B'PB)^{-1}B'PA$$
1
Although there is a specialized matrix data type defined in NumPy, it’s more standard to work with ordinary NumPy arrays. See this discussion.
2
Suppose that $$\|S \| < 1$$. Take any nonzero vector $$x$$, and let $$r := \|x\|$$. We have $$\| Sx \| = r \| S (x/r) \| \leq r \| S \| < r = \| x\|$$. Hence every point is pulled towards the origin. | 2021-11-27T05:59:27 | {
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https://math.stackexchange.com/questions/1206645/showing-12-cdotsn-fracnn12-by-induction-stuck-on-inductive-step | # Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step)
This is from this website:
Use mathematical induction to prove that $$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$ for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be $$1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.$$
STEP 1: We first show that $P(1)$ is true.
Left Side $= 1$
Right Side $= \frac{1 (1 + 1)}{2} = 1$
Both sides of the statement are equal hence $P(1)$ is true.
STEP 2: We now assume that $P(k)$ is true
$$1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}$$ and show that $P(k + 1)$ is true by adding $k + 1$ to both sides of the above statement \begin{align} 1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\ &= (k + 1)\left(\frac{k}{2} + 1\right) \\ &= \frac{(k + 1)(k + 2)}{2} \end{align} The last statement may be written as $$1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}$$ Which is the statement $P(k + 1)$.
My question is how in the very last line is the statement $P(k + 1)$ equal to $\frac{(k + 1)(k + 2)}{2}$. I don't get the last step.
• Have you used induction before or is this your first time? – Daniel W. Farlow Mar 25 '15 at 22:00
• Do you understand that $p(k+1) = 1 + 2 + 3 + \cdots + k + (k+1)$? – user137731 Mar 25 '15 at 22:05
• crash:This is my first time – Sam Lot Mar 25 '15 at 22:12
• Bye_world:i understand that we are adding k + 1 to both sides in p(k+1)=1+2+3+⋯+k+(k+1) – Sam Lot Mar 25 '15 at 22:13
Since this is your first time, I'll try to explain it with an emphasis on clarity. If something isn't clear, just comment and I'll try to explain what's happening.
Claim: You are trying to prove the statement $P(n)$ where $$P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.$$ Your goal is to try to prove this using induction. Proofs by induction usually involve two things: (1) showing that $P(n)$ is true for some fixed value of $n$; this value is oftentimes $n=1$, as it is in your case since you are trying to prove $P(n)$ for all $n\geq 1$. Make sense so far? (2) After you have shown (1) to be true, you then need to assume $P(k)$ to be true for some fixed $k\geq 1$ and then show that $P(k)$ implies $P(k+1)$; that is, you need to show that "if $P(k)$ is true, then $P(k+1)$ is true."
• (1) is called the base case.
• (2) is called the inductive step.
I'll outline the proof below. Let me know if a step doesn't make sense.
Proof. Let $P(n)$ denote the statement $$P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.$$ Base case ($n=1$): Try to see what happens for $P(1)$. We get that $1 = \frac{1(1+1)}{2}$, and this is true. Thus, the base case holds for $n=1$. Inductive step ($P(k)\to P(k+1)$): Assume $P(k)$ is true for some fixed $k\geq 1$ (this is called the inductive hypothesis). That is, assume $$P(k) : \color{red}{1+2+3+\cdots+k} = \color{green}{\frac{k(k+1)}{2}}\tag{inductive hypothesis}$$ is true. We must show that $P(k+1)$ follows where $$P(k+1) : \underbrace{\color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}}_{\text{LHS or "left-hand side"}} = \underbrace{\color{purple}{\frac{(k+1)((k+1)+1)}{2}}}_{\text{RHS or "right-hand side"}}.$$
Side note: Make sure you understand what just happened with $P(k+1)$. For $P(k)$, we just had $1+2+3+\cdots+k$ on the left-hand side. How come we have $1+2+3+\cdots+k+(k+1)$ now for the left-hand side of $P(k+1)$? This is because we are adding another term to the sum, namely $k+1$ (I highlighted this term with blue). On the right-hand side, where $P(k)$ just had $k$ in its expression, we just replace all of those $k$'s with $k+1$ because we are considering $P(k+1)$. Make sense?
Okay. Starting with the left-hand side of $P(k+1)$, we need to show that the right-hand side of $P(k+1)$ follows. Here's how it works: \begin{align} \text{LHS} &= \color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}\tag{by definition}\\[1em] &= \color{green}{\frac{k(k+1)}{2}}+\color{blue}{(k+1)}\tag{by inductive hypothesis}\\[1em] &= \frac{\color{green}{k(k+1)}+\color{green}{2}\color{blue}{(k+1)}}{\color{green}{2}}\tag{common denominator}\\[1em] &= \frac{(k+1)\color{green}{(k+2)}}{\color{green}{2}}\tag{group like terms}\\[1em] &= \color{purple}{\frac{(k+1)((k+1)+1)}{2}}\tag{rearrange}\\[1em] &= \text{RHS} \end{align} Thus, we have shown that the right-hand side of $P(k+1)$ follows from the left-hand side of $P(k+1)$. This completes the inductive step.
Thus, by mathematical induction, the statement $P(n)$ is true for all $n\geq 1$. $\blacksquare$
Does it all make sense now?
• What does P in P(n) stand for ? Proof ? – Sam Lot Mar 25 '15 at 23:05
• @SamLot No. $P$ just indicates a statement. I could have used $S(n)$ or $R(n)$ or anything else for that matter. – Daniel W. Farlow Mar 25 '15 at 23:06
• In the inductive step ,what does this "(P(k)→P(k+1))" means ? – Sam Lot Mar 25 '15 at 23:08
• @SamLot It means "$P(k)$ implies $P(k+1)$." The "$\to$" symbol is a mathematical symbol for "implies." – Daniel W. Farlow Mar 25 '15 at 23:09
• It makes more sense now .Thanks – Sam Lot Mar 25 '15 at 23:16
We know that
$P(k) = 1 + 2 + 3 + ... + k$
Therefore:
$P(k+1) = 1 + 2 + 3 + ... + k + (k+1)$
By induction hypothesis we have:
$1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = 1 + 2 + 3+...+k+(k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = \frac{(k+1)(k+2)}{2}$
By induction we now know that since this is true for one integer $k$, it is true for all integers greater than or equal to $k$.
• Rather that "we already proved that $1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$" I would say "By induction hypothesis it is $1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$". Note that it was not proved. – mfl Mar 25 '15 at 22:15
• How can we say in the first line that "We know that P(k)=1+2+3+...+k" Where did that come from .Secondly how does the formula in last line P(k+1)=(k+1)(k+2)2 prove that 1+2+3+...+n=n(n+1)2 .How can i link them ? – Sam Lot Mar 25 '15 at 22:27
• In the problem $P(n)$ is defined as $1+2+3+...+n$. $P(k+1) = \frac{(k+1)(k+2)}{2}$ proves that $P(k) = \frac{(k)(k+2)}{2}$ because we have shown that it is true for $1$. Say that $k=1$. We know from these formulas that if the statement is true for a number $k$ it is true for $k+1$. Since it is true for $k+1$, it must be true for $(k+1)+1$ and so on and so on. – OriginalOldMan Mar 25 '15 at 22:33
• When you say "We know that P(k)=1+2+3+...+k Therefore: P(k+1)=1+2+3+...+k+(k+1)" in the last line here you are adding 1 to the left side and k + 1 to the right side .Isnt that mathematically incorrect ? – Sam Lot Mar 25 '15 at 22:40
• We are not adding $1$ to both sides. $P(k)$ is basically the sum of integers from $1$ to $k$. So $P(k+1)$ is the sum of integers from $1$ to $(k+1)$. $k$ is obviously the integer before $(k+1)$. So the sum of integers from $1$ to $(k+1)$ is $1+2+3+...+k+(k+1)$. That's why we can say that $P(k+1) = 1+2+3+...+k+(k+1)$. – OriginalOldMan Mar 25 '15 at 22:44
induction is basically saying that if it is true for this step, it is true for the next step. so assuming $1+2+3...+k=k(k+1)/2$, ie it is true for step k, we have to show that it must be true for step k+1, the next step. the final line shows how, by going through some algebra, adding all the numbers up to k+1 equals putting k+1 into the formula $n(n+1)/2$, written as $1+2+3...+k+[k+1]=[k+1]([k+1]+1)/2$. therefore, if it is true for step 1, it is for step 2, and 3...
• I dont understand this part from your explanation "adding all the numbers up to k+1 equals putting k+1 into the formula n(n+1)/2".How does those two things 'equal' each other ? – Sam Lot Mar 25 '15 at 22:20
• you show that they equal each other with the algebra: – stanley dodds Mar 26 '15 at 7:00
• $1+2+3...+k=\frac{k(k+1)}2$, the initial statement. add k+1: – stanley dodds Mar 26 '15 at 7:03
• $1+2+3...+k+[k+1]=\frac{k(k+1)}2+k+1$. then rearrange the rhs: – stanley dodds Mar 26 '15 at 7:06
• multiply on rhs: $=\frac{(k+1)(k+2)}2$ which could also be written as $\frac{[k+1]([k+1]+1)}2$. this is the initially presumed equation for k+1, so with. algebra we have shown that if it works for k it works for k+1, and then if it works for that step it works for the next step, and so on – stanley dodds Mar 26 '15 at 7:16 | 2020-08-06T01:41:38 | {
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http://math.stackexchange.com/questions/114181/whole-numbers-and-division | # whole numbers and division
Consider the whole number with one thousand digits that can be formed by writing the digits 2772 two hundred and fifty time in succession. Is it divisible by 9? Is it divisible by 11?
-
I answered yes to both because 2772 is divisible by 9 and 11. I am just trying to make sure this is correct! – SNS Feb 27 '12 at 23:19
If the sum of the digits of $n$ is divisible by 9, then (and only then) $n$ is divisible by 9. From this, your number is divisible by 9 (the sum of its digits is $250\cdot18$). I've forgotten the divisibility test for 11... – David Mitra Feb 27 '12 at 23:21
Can you show that your number is divisible by 2772? And can you finish up from there? – Gerry Myerson Feb 27 '12 at 23:26
2772/9=308 2772/11=252 I think that if the original number is divisible by 9 and 11 then if you continue adding the same numbers over and over it should always be divisible by 9 and 11. This is the part I want to double check on. – SNS Feb 27 '12 at 23:29
@DavidMitra If the alternating sum of the digits (add, subtract, add, subtract, etc.) is divisible by 11 then the number is divisible by 11. This is because $10^n\equiv (-1)^n\mod 10$. – Alex Becker Feb 27 '12 at 23:29
The answer is yes; but, in my opinion, you did not give enough information in your comment for a justification.
One way to show it is to use the divisibility tests for 9 and 11.
Let's call your number, obtained by writing "$2772$" two hundred and fifty times in succession, $y$.
A number $n$ is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $y$ is $250\cdot(2+7+7+2)=250(18)$, so $y$ is divisible by 9.
A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
-
This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
-
It's divisible by $9$, $11$, $4$, $7$, and by every divisor of $100010001 \ldots 000100010001$. Just what the divisors of that last number are is not so easy to tell, as far as I can see. It is not divisible by $4$ or $9$, nor by $2$ or $3$. Whether it's divisible by $7$ or by $11$ might be harder to tell. – Michael Hardy Feb 28 '12 at 0:14
@Michael: $(10^{1000}-1)/(10^4 -1) = 41 \times 73 \times 137 \times 251 \times 271 \times 401 \times 751 \times 1201 \times 1601 \times 3541 \times 4001 \times 5051 \times 9091 \times 21001 \times 21401 \times 24001 \times 25601 \times 27961 \times 60101 \times 76001 \times 162251 \times 1 378001 \times 1 610501 \times 1 676321 \times 7 019801 \times 1797 655751 \times 5964 848081 \times 10893 295001 \times 182521 213001 \times 14 103673 319201 \times 78 875943 472201 \times 176 144543 406001 \times 1680 588011 350901 \times \cdots$ – Henry Feb 28 '12 at 0:35
The way you were probably intended to do this problem is to find the sum of the digits (for $9$) and alternating sum and difference (for $11$). And you will undoubtedly need to know these facts about $9$ and $11$ for other problems.
However, the following is true. Suppose that the number formed by a string of digits, like $4718$, is divisible by $m$. For example, $7$ divides $4718$, so let's take $m=7$.
Then $m$ (that is, $7$ here) divides $471847184718$. This is because $$4718471847184718=4718+47180000+471800000000+4718000000000000,$$ and each term on the right-hand side is obviously divisible by $7$, since $4718$ is.
The same argument works for any repetition of the string $4718$, however long it may be, and for any string, and any divisor $m$.
In particular, since $9$ and $11$ each divide $2772$, it follows that each of them divides your thousand digit number. Note that $14$ also divides $2772$, so $14$ divides your thousand digit number.
Your answers were correct, and the procedure that you used turns out to be generally valid. There was somewhat of a lack of explanation, and it is possible that someone grading your work might call it incomplete.
-
On the one hand, the divisibility test for $9$ is to add up the digits and see if they're divisible by $9$. The divisibility test for $11$ is a bit more annoying (if combined with $7$ and $14$)- it comes from recognizing that $1001 = 7 \cdot 11 \cdot 13$, so you take the alternating sum of triplets of digits and see if it's divisible by $11$. Or you could just take the regular alternating sum of digits and see if it's divisible by $11$. Both work here.
On the other hand, the number is $\sum_{k = 0}^{249} (1000)^k \cdot 2772$. So any number that divides $2772$ will divide this new number.
-
Your reasoning is correct, but you need to find a way to express clearly and justify that last step, from ‘$2772$ is divisible by $9$ and $11$’ to ‘$27722772\dots2772$ is divisible by $9$ and $11$’. I agree that this is intuitively obvious to anyone who has done long division, but ‘intuitively obvious’ isn’t good enough here. Look at a simpler example first: what about $27722772$?
$27722772=27720000+2772=2772\cdot10^4+2772=2772(10^4+1)$, so it’s clearly a multiple of $2772$ and therefore also of $9$ and $11$. Similarly,
\begin{align*} 277227722772&=277227720000+2772\\ &=2772(10^4+1)\cdot 10^4+2772\\ &=2772(10^{2\cdot 4}+10^4)+2772\\ &=2772(10^{2\cdot 4}+10^4+1)\;, \end{align*}
which is again a multiple of $2772$ and hence also of $9$ and $11$.
And if you already know that a string of $n$ copies of $2772$ is equal to
$$2772(10^{(n-1)\cdot 4}+10^{(n-2)\cdot4}+\dots+10^4+1)\;,$$
then a string of $n+1$ copies must be equal to
\begin{align*} \Big(2772(10^{(n-1)\cdot 4}&+10^{(n-2)\cdot 4}+\dots+10^4+1)\Big)\cdot 10^4+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4)+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4+1)\;, \end{align*}
and the pattern continues. Thus, your string of $250$ copies of $2772$ must be equal to
$$2772(10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1)$$
and is certainly a multiple of $2772$ and therefore of $9$ and of $11$.
Note that this argument actually does more than is necessary, since it also determines the other factor of your number, namely,
$$10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1=\frac{10^{250\cdot 4}-1}{10000-1}=\frac{10^{1000}-1}{9999}\;.$$
You could use the same kind of reasoning to argue that tacking a copy of $2772$ on the end of some integer $n$ is simply replacing $n$ by $10000n+2772$, so if $n=2772m$, then $10000n+2772=10000(2772m)+2772=2772(10000m+1)$, which is certainly still a multiple of $2772$. Thus, since tacking a copy of $2772$ on the end always preserves divisibility by $2772$ if it was already present, it doesn’t matter how often you do it: the result is still divisible by $2772$ if the original number was.
(Technically both of these arguments are proofs by mathematical induction, albeit stated rather informally.)
- | 2014-03-15T02:59:59 | {
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https://www.physicsforums.com/threads/a-differential-equation.207651/ | # A differential Equation -
1. Jan 7, 2008
### Sparky_
1. The problem statement, all variables and given/known data
y'' + 3y' + 2y = sin(e^x)
2. Relevant equations
3. The attempt at a solution
$$y'' + 3y' + 2y = sin(e^x)$$
$$m^2 + 3m + 2 = 0$$
$$m1 = -2 ; m2 = -1$$
$$yc = c1e^{-2x} c2 e^{-x}$$
$$y1 = e^{-2x}$$
$$y1' = -2e^{-2x}$$
$$y2 = e^{-x}$$
$$y2' = -e^{-x}$$
The W Matrix works out to
$$W = e^{-3x}$$
$$u1' = -e^{x}sin(e^x)$$
$$u1 = sin(e^x)$$
$$u2' = e^{2x}sin(e^x)$$
$$u2 = -e^xcos(e^x) + sin(e^x)$$
(This is the integration by parts solution in an earlier posting:
my solution:
$$y = c1e^{-2x} + c2e^{-x} + e^{-2x}sin(e^x) + e^{-x}[-e^xcos(e^x) + sin(e^x)]$$
The book's solution:
$$y = c1e^{-2x} + c2e^{-x} - e^{-2x}sin(e^x)$$
(I'm not in school but this is a problem out of a book - I'm trying to brush up)
Thanks for the help
-Sparky
Last edited: Jan 7, 2008
2. Jan 7, 2008
### mda
It has been a while since I solved DEs analytically, but you should definitely check that your solution does in fact solve the DE (I don't think it does). Also I think the c1c2 term should be a sum, not a product.
3. Jan 7, 2008
### blochwave
Aw man, I tried to solve it with the method of undetermined coefficient and spent way too much time before realizing that doesn't work since the RHS isn't the right form. As solving it by your method is a vague memory now, I was like "well at least I can tell him he got the complementary solution wrong!"
but you corrected it!
I'm sure you've done the usual, like check to make sure your third term isn't miracously minus two times the second term
4. Jan 7, 2008
### Sparky_
I have looked around for terms to cancel and I don't see it but I admit - It could be there, I've gotten the tunnel vision thing after working this back and forth.
I'm hoping for some insight here.
Yes I did correct my error with the first 2 terms - my fault - it was on my paper correctly.
5. Jan 7, 2008
### blochwave
Would it be in any way reasonable to try this with Laplace transforms?
6. Jan 7, 2008
### Sparky_
Part of the reason for this exercise is I'm working back up to Laplace transforms -
I'm working my way back through the book - it's almost become a hobby (pretty sick - right?)
I've used them quite a bit when in school and a little out. (I got my BSEE in 1993 and a Master's in 2003)
I now want to solve this problem with the method of variation of parameters - just because.
I've worked several other problems successfully with this method. - This problem is now bugging me -
I can tell from the answer I'm close
Help?
7. Jan 8, 2008
### Rainbow Child
You have misalculated the integral. There is the right calculation
$$I=-\int e^x\,\sin(e^x)\,d\,x, \quad t=e^x,\,d\,x=\frac{d\,t}{t}$$
so
$$I=-\int t\,\sin(t)\,\frac{d\,t}{t}=-\int \sin(t)\,d\,t=\cos(t)+C\Rightarrow I=\cos(e^x)+C$$
8. Jan 8, 2008
### HallsofIvy
Staff Emeritus
Very good and completely correct. But why not just t= ex[/su] so dt= ex dx since you have a ex in the integral already? I would consider that slightly simpler though it may be just prsonal taste.
9. Jan 8, 2008
### Rainbow Child
Just trying to be analytic!
After all, my personal taste would be
$$I=-\int e^x\,\sin(e^x)\,d\,x=-\int \sin(e^x)\,d\,e^x=\cos(e^x)+C$$
10. Jan 8, 2008
### Sparky_
Well, That's embarassing.
During lunch I will check my paper - I hope I typed it in wrong and it is correct on my paper.
I suspect I didn't and I have egg on my face.
Thank you so much!!
I will pick the problem up with this correction.
Thanks again
-Sparky_
11. Jan 8, 2008
### whiplash
how about the part e^{2x}sin(e^x) , he still has to integrate this does he not?
12. Jan 11, 2008
### Sparky_
Thank you all for the help,
I am embarrassed to say that I did have
$$u1' = -e^{x}sin(e^x)$$
$$u1 = sin(e^x)$$
on my paper. Fixing this error, pointed me to one more careless error.
For completion here is the solution (in case anybody wants it):
I've noticed some threads have a "solved" tag - is that after a problem is complete? If so, does this one get a "solved"?
$$y'' + 3y' + 2y = sin(e^x)$$
$$m^2 + 3m + 2 = 0$$
$$m1 = -2 ; m2 = -1$$
$$yc = c1e^{-2x} c2 e^{-x}$$
$$y1 = e^{-2x}$$
$$y1' = -2e^{-2x}$$
$$y2 = e^{-x}$$
$$y2' = -e^{-x}$$
$$W = e^{-3x}$$ (The W Matrix - )
$$u1' = -\frac{y2f(x)} {W}$$
$$u1' = -e^{2x}sin(e^x)$$
$$u1 = -\int{ e^{2x}sin(e^x)} dx$$ (integration by parts)
$$u1 = e^xcos(e^x) - sin(e^x)$$
$$u2' = e^{x}sin(e^x)$$
$$u2 = -cos(e^x)$$
$$y = c1e^{-2x} + c2e^{-x} + e^{-x}cos(e^x) - e^{-2x}sin(e^x) - e^{-x}cos(e^x)$$
$$y = c1e^{-2x} + c2e^{-x} - e^{-2x}sin(e^x)$$
Last edited: Jan 11, 2008 | 2017-02-20T11:26:17 | {
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https://math.stackexchange.com/questions/2641694/if-4-distinct-balls-are-placed-in-4-distinct-boxes-what-is-the-probability | # If $4$ distinct balls are placed in $4$ distinct boxes, what is the probability that exactly one box is empty?
Suppose $4$ balls labeled $1,2,3,4$ are randomly placed in boxes $B_1,B_2,B_3,B_4$. The probability that exactly one box is empty is
(a)$8/256$
(b)$9/16$
(c)$27/256$
(d)$9/64$
My approach:
Selecting one empty box out of the four boxes, then placing the $4$ balls in the remaining $3$ boxes.
Thank you .
• My bad yes answer is $\displaystyle \frac{9}{16}$ – DXT Feb 8 '18 at 13:06
• Sorry to say that the answer is 144 – Ishita Roy Aug 8 at 5:20
There are $\binom{4}{1}$ ways to choose the empty box. We must distribute the four balls to the remaining three boxes so that no box is left empty. That means one of those three boxes will receive two balls and the others will each receive one. There are $\binom{3}{1}$ ways to choose the box that will receive two balls and $\binom{4}{2}$ ways to choose which two balls will be placed in that box. The remaining balls can be placed in the remaining two boxes in $2!$ ways. Hence, the number of favorable cases is $$\binom{4}{1}\binom{3}{1}\binom{4}{2}2!$$ Since there are $4$ choices for the placement of each of the four balls, there are $4^4$ ways to distribute four distinct balls to four distinct boxes. Hence, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{4}{2}2!}{4^4}$$
• Thank you for your help . – Alphanerd Feb 8 '18 at 13:18
Using the Generalized Inclusion-Exclusion Principle
Consider $$S(i)$$ to be all arrangements where $$B_i$$ is empty. As in the Generalized Inclusion-Exclusion Principle, let $$N(j)$$ be the the sum of the sizes of all intersections of $$j$$ of the $$S(i)$$: $$N(j)=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|$$ In this case, we have $$\binom{4}{j}$$ choices of the $$B_i$$ to be empty and $$(4-j)^4$$ ways to map $$4$$ distinct balls into the $$4-j$$ remaining $$B_i$$: $$N(j)=\binom{4}{j}(4-j)^4$$ The Generalized Inclusion-Exclusion Principle says that the number of arrangements in exactly $$1$$ of the $$S(i)$$ is \begin{align} \sum_{j=1}^4(-1)^{j-1}\binom{j}{1}N(j) &=\binom{1}{1}\binom{4}{1}\,3^4-\binom{2}{1}\binom{4}{2}\,2^4+\binom{3}{1}\binom{4}{3}\,1^4-\binom{4}{1}\binom{4}{4}\,0^4\\ &=324-192+12-0\\[9pt] &=144 \end{align} With a universe of size $$N(0)=256$$, we get a probability of $$\frac{144}{256}=\frac{9}{16}$$
Applying the GIEP to the Approach in the Question
In the approach in the question, after choosing which of the $$4$$ boxes to be empty, we need to compute the number of ways to put $$4$$ distinct balls into $$3$$ distinct boxes with no empty boxes.
Similar to the case above, there are $$\binom{3}{j}$$ ways to choose the empty $$B_i$$ and $$(3-j)^4$$ ways to put the $$4$$ balls into the remaining boxes. That is, $$N(j)=\binom{3}{j}(3-j)^4$$ The Generalized Inclusion-Exclusion Principle says that the number of arrangements in exactly $$0$$ of the $$S(i)$$ is \begin{align} \sum_{j=0}^3(-1)^{j-0}\binom{j}{0}N(j) &=\binom{0}{0}\binom{3}{0}3^4-\binom{1}{0}\binom{3}{1}2^4+\binom{2}{0}\binom{3}{2}1^4-\binom{3}{0}\binom{3}{3}0^4\\ &=81-48+3-0\\[9pt] &=36 \end{align} Combining this with the $$4$$ possibilities for the empty box, we get $$4\cdot36=144$$ arrangements with exactly one empty box, same as above.
Here’s another approach. Name the boxes $a$, $b$, $c$, and $d$. If $S\subseteq\{a,b,c,d\}$, define $e_S$ to be the probability that each box whose name is an element of $S$ is empty after all four balls have been placed. (Boxes not named in $S$ might also be empty.)
Then, using the inclusion-exclusion principle, the probability $p_{\{a\}}$ that only box $a$ is empty is
$p_{\{a\}}=e_{\{a\}}-e_{\{a,b\}}-e_{\{a,c\}}-e_{\{a,d\}}+e_{\{a,b,c\}}+e_{\{a,b,d\}}+e_{\{a,c,d\}}-e_{\{a,b,c,d\}} \\=e_{\{a\}}-3e_{\{a,b\}}+3e_{\{a,b,c\}}-e_{\{a,b,c,d\}}\\={\left(\frac34\right)}^4-3\cdot{\left(\frac24\right)}^4+3\cdot{\left(\frac14\right)}^4-0=\frac9{64}.$
By symmetry, the probability $p_{\{x\}}$ that box $x$ is the only empty box is $\frac9{64}$ for each $x$.
The probability you want is $p_{\{a\}}+p_{\{b\}}+p_{\{c\}}+p_{\{d\}}=4\cdot\frac9{64}=\frac9{16}$. (No inclusion-exclusion is needed for this step because no two different boxes can each be the only empty box.) | 2019-11-14T14:23:52 | {
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http://mathhelpforum.com/algebra/137225-roots-polynomials.html | # Math Help - Roots of polynomials
1. ## Roots of polynomials
Hi, sorry if this is in the wrong forums not really sure but here it goes..
The quadratic equation $2x^2+4x+3=0$ has roots $\alpha$ and $\beta$ .
Find the value $\alpha^4+\beta^4$
Ive managed to expand to $\alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$
but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer.
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?
Help really appreciated. Thankyou !
2. Hello, NathanBUK!
The quadratic equation $2x^2+4x+3\:=\:0$ has roots $\alpha$ and $\beta$ .
Find the value of: $\alpha^4+\beta^4$
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$
. . which I managed to prove. . . . . Good!
Is there anyway I can use this to answer the question? . . . . Yes!
We know that: . $\begin{Bmatrix}\alpha + \beta &=& -2 \\ \alpha\beta &=& \frac{3}{2} \end{Bmatrix}$
We have: . $\alpha^2 + \beta^2 \;=\;1$
Square: . $\left(\alpha^2 + \beta^2\right)^2 \;=\;(1)^2$
. . . . $\alpha^4 + 2\alpha^2\beta^2 + \beta^4 \;=\;1$
. . . $\alpha^4 + \beta^4 + 2\!\cdot\!\!\!\underbrace{(\alpha\beta)^2}_{\text{ This is }(\frac{3}{2})^2} \;=\;1$
. . . . . $\alpha^4 + \beta^4 + \frac{9}{2} \;=\;1$
. . . . . . . $\alpha^4 + \beta^4 \;=\;-\frac{7}{2}$
3. Ohh thats makes soo much sense now!! Thankyou so much!!
4. Originally Posted by NathanBUK
Hi, sorry if this is in the wrong forums not really sure but here it goes..
The quadratic equation $2x^2+4x+3=0$ has roots $\alpha$ and $\beta$ .
Find the value $\alpha^4+\beta^4$
Ive managed to expand to $\alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$
but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer.
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?
Help really appreciated. Thankyou !
Alternative solution (same one in disguise realy):
$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha \beta)^2$
CB | 2015-05-29T16:34:52 | {
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https://math.stackexchange.com/questions/2907001/getting-p-yy-p-xg-1y-left-dfrac-partialx-partialy-right | # Getting $p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$ by solving $| p_y(g(x)) \ dy | = | p_x (x) \ dx |$?
My textbook has a very brief section that introduces some concepts from measure theory:
Another technical detail of continuous variables relates to handling continuous random variables that are deterministic functions of one another. Suppose we have two random variables, $\mathbf{x}$ and $\mathbf{y}$, such that $\mathbf{y} = g(\mathbf{x})$, where $g$ is an invertible, continuous, differentiable transformation. One might expect that $p_y(\mathbf{y}) = p_x(g^{−1} (\mathbf{y}))$. This is actually not the case.
As a simple example, suppose we have scalar random variables $x$ and $y$. Suppose $y = \dfrac{x}{2}$ and $x \sim U(0,1)$. If we use the rule $p_y(y) = p_x(2y)$, then $p_y$ will be $0$ everywhere except the interval $\left[ 0, \dfrac{1}{2} \right]$, and it will be $1$ on this interval. This means
$$\int p_y(y) \ dy = \dfrac{1}{2},$$
which violates the definition of a probability distribution. This is a common mistake. The problem with this approach is that it fails to account for the distortion fo space introduced by the function $g$. Recall that the probability of $\mathbf{x}$ lying in an infinitesimally small region with volume $\delta \mathbf{x}$ is given by $p(\mathbf{x}) \delta \mathbf{x}$. Since $g$ can expand or contract space, the infinitesimal volume surrounding $\mathbf{x}$ in $\mathbf{x}$ space may have different volume in $\mathbf{y}$ space.
To see how to correct the problem, we return to the scalar case. We need to present the property
$$| p_y(g(x)) \ dy | = | p_x (x) \ dx |$$
Solving from this, we obtain
$$p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$$
or equivalently
$$p_x(x) = p_y(g(x)) \left| \dfrac{\partial{g(x)}}{\partial{x}} \right|$$
How do they get $p_y(y) = p_x(g^{-1}(y)) \left| \dfrac{\partial{x}}{\partial{y}} \right|$ or equivalently $p_x(x) = p_y(g(x)) \left| \dfrac{\partial{g(x)}}{\partial{x}} \right|$ by solving $| p_y(g(x)) \ dy | = | p_x (x) \ dx |$?
Can someone please demonstrate this and explain the steps?
• this is chain rule + an integration, i.e. integration by substitution – Calvin Khor Sep 8 '18 at 8:37
• @CalvinKhor Can you please demonstrate this step-by-step? – Wyuw Sep 8 '18 at 9:39
• Have you seen e.g. this before? proofwiki.org/wiki/Integration_by_Substitution – Calvin Khor Sep 8 '18 at 9:44
• @CalvinKhor Yes. – Wyuw Sep 8 '18 at 9:50
• I'm not sure if that helps or you want me to put it in the notation you have – Shogun Sep 13 '18 at 15:10
## 2 Answers
$$p_X(x)dx$$ represents the probability measur $$\mathbb{P}_X$$ which is the probability distribution of the random variable $$X$$, it is defined by its action on measurable positive functions by $$\mathbb{E}(f(X))=\int_{\Omega}f(X)d\mathbb{P}=\int_{\mathbb{R}}f(x)d\mathbb{P}_X(x)=\int_{\mathbb{R}}f(x)p_X(x)dx.$$ Now, we consider a new random variable $$Y=g(X)$$, (with some conditions on $$g$$), and we seek $$p_Y$$ the probability density distribution of $$Y$$. So we calculate, for an arbitrary measurable positive function $$f$$ the expectation $$\mathbb{E}(f(Y))$$ in two ways: First, $$\mathbb{E}(f(Y))=\int_{\mathbb{R}}f(y)\color{red}{p_Y(y)dy}\tag1$$ Second, \eqalignno{\mathbb{E}(f(Y))&=\mathbb{E}(f(g(X)))\cr &=\int_{\mathbb{R}}f(g(x))p_X(x)dx\qquad\text{now a change of variables}\cr &=\int_{\mathbb{R}}f(y)\color{red}{p_X(g^{-1}(y))\left|\frac{dx}{dy}\right|dy}&(2) } Now, because $$f$$ is arbitrary, comparing (1) and (2) we get $$p_Y(y)=p_X(x)\left|\frac{dx}{dy}\right|, \quad\text{where y=g(x).}$$ Or, better $$p_Y(y)=p_X(g^{-1}(y))\left|\frac{1}{g’(g^{-1}(y))}\right|\iff p_Y(g(x))|g’(x)|=p_X(x).$$
This is called the method of transformations. It is detailed on this site. You need to transform a function of a random variable in order to make the CDF equal to $1$. For a demonstration.
Suppose that $X \sim \textrm{Unif}(0,1)$ and let $Y = e^{X}$
Note that cdf of $X$ is given by
F_{X}(x) =\begin{align}\begin{cases} 0 & x < 0 \\ \\ x & 0 \leq x \leq1 \\ 1 & x > 1 \end{cases} \end{align} \tag{1}
Then to find the cdf of $Y$
$$F_{Y}(y) = P(Y \leq y) \\ P(e^{X} \leq y) \\ = P(X \leq \ln(y)) \\ = F_{X}(\ln(y)) = \ln(y) \tag{2}$$
F_{Y}(y) =\begin{align}\begin{cases} 0 & y < 1 \\ \\ \ln(y) & 1 \leq \ln(y) \leq e \\ 1 & x > e \end{cases} \end{align} \tag{3}
To obtain the pdf we take the derivative
f_{Y}(y) = F_{Y}^{'}(y) = \begin{align}\begin{cases} \frac{1}{y} & 1 \leq \ln(y) \leq e \\ 0 & \textrm{ otherwise} \end{cases} \end{align} \tag{4}
Concerning the problem above, suppose that $X \sim \textrm{Unif}(0,1)$ and that $Y = \frac{X}{2}$
The CDF for $X$ is the same as above. Let's look at the cdf of $Y$. We note that $R_{X} =[0,1]$ so then $R_{Y}=[0,\frac{1}{2}]$
$$F_{Y}(y) = P(Y \leq y) \\ P(\frac{X}{2} \leq y) \\ = P(X \leq 2y) \\ = F_{X}(2y) = 2y \tag{5}$$
You are simply taking the reciprocal. To find the pdf, we differenitate.
F_{Y}(y) = \begin{align}\begin{cases} 0 & y< 0 \\ \\ 2y & 0 \leq y \leq \frac{1}{2} \\ 1 & y > \frac{1}{2} \end{cases} \end{align} \tag{6}
to find the pdf
f_{Y}(y) = F_{Y}^{'}(y) = \begin{align}\begin{cases} 2 & 0 \leq y \leq \frac{1}{2} \\ 0 & \textrm{ otherwise} \end{cases} \end{align} \tag{7}
Visually the difference in the two uniform distributions can be seen below.
$$X\sim \textrm{Unif}(0,1) \tag{8}$$
$$X\sim \textrm{Unif}(0,1) , Y = \frac{X}{2} , Y \sim \textrm{Unif}(0,\frac{1}{2}) \tag{9}$$
• It is not clear to me how this part makes sense: F_{Y}(y) =\begin{align}\begin{cases} 0 & y < 1 \\ \\ \ln(y) & 1 \leq \ln(y) \leq e \\ 1 & x > e \end{cases} \end{align} \tag{3} – Wyuw Sep 15 '18 at 3:39
• Why is $F_Y(y) = 0$ when $y < 1$? Same with all of the other values. – Wyuw Sep 15 '18 at 3:39
• I'll give you the bounty, since it's running out, but I'll wait before I accept it as the answer. – Wyuw Sep 15 '18 at 3:40
• I may have wrote something wrong. I can fix it. Let me look back over it. – Shogun Sep 15 '18 at 3:41
• we are taking it from $[1,e]$ to $[0,1]$ – Shogun Sep 15 '18 at 3:52 | 2019-05-25T23:04:23 | {
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https://www.winslowassociates.com/doctor-doctor-kgybf/acc945-unique-left-inverse | share. Proposition If the inverse of a matrix exists, then it is unique. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. eralization of the inverse of a matrix. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . Left-cancellative Loop (algebra) , an algebraic structure with identity element where every element has a unique left and right inverse Retraction (category theory) , a left inverse of some morphism left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. inverse. endstream endobj startxref Note that other left inverses (for example, A¡L = [3; ¡1]) satisfy properties (P1), (P2), and (P4) but not (P3). In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Yes. In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. 8 0 obj One consequence of (1.2) is that AGAG=AG and GAGA=GA. Let $f \colon X \longrightarrow Y$ be a function. (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) g = finverse(f) returns the inverse of function f, such that f(g(x)) = x. h�bbdb� �� �9D�H�_ ��Dj*�HE�8�,�&f��L[�z�H�W��� ����HU{��Z �(� �� ��A��O0� lZ'����{,��.�l�\��@���OL@���q����� ��� Let e e e be the identity. (An example of a function with no inverse on either side is the zero transformation on .) This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. For any elements a, b, c, x ∈ G we have: 1. Still another characterization of A+ is given in the following theorem whose proof can be found on p. 19 in Albert, A., Regression and the Moore-Penrose Pseudoinverse, Aca-demic Press, New York, 1972. Show Instructions. 3. Matrix Multiplication Notation. Recall that $B$ is the inverse matrix if it satisfies $AB=BA=I,$ where $I$ is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. %%EOF Proof: Assume rank(A)=r. As f is a right inverse to g, it is a full inverse to g. So, f is an inverse to f is an inverse to stream However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). If the function is one-to-one, there will be a unique inverse. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If f contains more than one variable, use the next syntax to specify the independent variable. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … Theorem. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. Remark When A is invertible, we denote its inverse … Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. The Moore-Penrose pseudoinverse is deflned for any matrix and is unique. New comments cannot be posted and votes cannot be cast. %���� wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, …) = (b 2, b 3, …). Two-sided inverse is unique if it exists in monoid 2. In a monoid, if an element has a right inverse… /Filter /FlateDecode << /S /GoTo /D [9 0 R /Fit ] >> Theorem A.63 A generalized inverse always exists although it is not unique in general. (We say B is an inverse of A.) It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Theorem 2.16 First Gyrogroup Properties. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Thus both AG and GA are projection matrices. See the lecture notesfor the relevant definitions. A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). There are three optional outputs in addition to the unique elements: 53 0 obj <> endobj Then 1 (AB) ij = A i B j, 2 (AB) i = A i B, 3 (AB) j = AB j, 4 (ABC) ij = A i BC j. u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. If the function is one-to-one, there will be a unique inverse. Let (G, ⊕) be a gyrogroup. Note that other left 5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 By using this website, you agree to our Cookie Policy. /Length 1425 Hello! Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. The equation Ax = b always has at least one solution; the nullspace of A has dimension n − m, so there will be The left inverse tells you how to exactly retrace your steps, if you managed to get to a destination – “Some places might be unreachable, but I can always put you on the return flight” The right inverse tells you where you might have come from, for any possible destination – “All places are reachable, but I can't put you on the I know that left inverses are unique if the function is surjective but I don't know if left inverses are always unique for non-surjective functions too. h�b�y��� cca�� ����ِ� q���#�!�A�ѬQ�a���[�50�F��3&9'��0 qp�(R�&�a�s4�p�[���f^'w�P& 7��,���[T�+�J����9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w In gen-eral, a square matrix P that satisfles P2 = P is called a projection matrix. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. best. '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� Subtraction was defined in terms of addition and division was defined in terms ofmultiplication. Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). (4x1�@�y�,(����.�BY��⧆7G�߱Zb�?��,��T��9o��H0�(1q����D� �;:��vK{Y�wY�/���5�����c�iZl�B\\��L�bE���8;�!�#�*)�L�{�M��dUт6���%�V^����ZW��������f�4R�p�p�b��x���.L��1sh��Y�U����! In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. When working in the real numbers, the equation ax=b could be solved for x by dividing bothsides of the equation by a to get x=b/a, as long as a wasn't zero. We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same . Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. ��� Sort by. Ask Question Asked 4 years, 10 months ago. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). Hence it is bijective. If A is invertible, then its inverse is unique. This thread is archived. Let $f \colon X \longrightarrow Y$ be a function. Viewed 1k times 3. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. 36 0 obj << This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. x��XKo#7��W�hE�[ע��E������:v�4q���/)�c����>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG���8>����'��x�ټrc��>?��[��?�'���(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ���N)&Q ��M���Yy��?A����4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi���e��A��(;p*����V�Jڛ,�t~�d��̘H9����/��_a���v�68gq"���D�|a5����P|Jv��l1j��x��&N����V"���"����}! numpy.unique¶ numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] ¶ Find the unique elements of an array. This preview shows page 275 - 279 out of 401 pages.. By Proposition 5.15.5, g has a unique right inverse, which is equal to its unique inverse. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. See Also. endobj Actually, trying to prove uniqueness of left inverses leads to dramatic failure! Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). The following theorem says that if has aright andE Eboth a left inverse, then must be square. A.12 Generalized Inverse Definition A.62 Let A be an m × n-matrix. This is no accident ! From this example we see that even when they exist, one-sided inverses need not be unique. Theorem A.63 A generalized inverse always exists although it is not unique in general. Stack Exchange Network. Then they satisfy $AB=BA=I \tag{*}$ and Active 2 years, 7 months ago. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Remark Not all square matrices are invertible. 87 0 obj <>/Filter/FlateDecode/ID[<60DDF7F936364B419866FBDF5084AEDB><33A0036193072C4B9116D6C95BA3C158>]/Index[53 73]/Info 52 0 R/Length 149/Prev 149168/Root 54 0 R/Size 126/Type/XRef/W[1 3 1]>>stream 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Show Instructions. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. 125 0 obj <>stream If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). example. Let (G, ⊕) be a gyrogroup. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. It's an interesting exercise that if$a$is a left unit that is not a right uni Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Recall also that this gives a unique inverse. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. 6 comments. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. Yes. Generalized inverse Michael Friendly 2020-10-29. 11.1. g = finverse(f,var) ... finverse does not issue a warning when the inverse is not unique. Let G G G be a group. Proof: Assume rank(A)=r. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. U-semigroups �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. %PDF-1.4 Theorem 2.16 First Gyrogroup Properties. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective 100% Upvoted. If is a left inverse and a right inverse of , for all ∈, () = ((()) = (). >> save hide report. If BA = I then B is a left inverse of A and A is a right inverse of B. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. If E has a right inverse, it is not necessarily unique. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Proof. 0 In mathematics, and in particular, algebra, a generalized inverse of an element x is an element y that has some properties of an inverse element but not necessarily all of them. %PDF-1.6 %���� A i denotes the i-th row of A and A j denotes the j-th column of A. Some easy corollaries: 1. If S S S is a set with an associative binary operation ∗ * ∗ with an identity element, and an element a ∈ S a\in S a ∈ S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. 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https://math.stackexchange.com/questions/3648427/limits-of-integration-on-double-integrals | # Limits of integration on double integrals
I was given this problem:
Find an integral equal to the volume of the solid bounded by $$z=4-2y,z=0,x=y^4,x=1$$ and evaluate.
I understand how to evaluate once my double integral is set up, but I do not know how to find my limits of integration.
I am assuming that my function will be $$z=4-2y$$ and that using this I should be able to find my limits of integration. I can say that $$0=4-2y$$ which means that $$y=2$$. I can then plug that into $$x=y^4$$ and get $$1\leq x\leq 16$$ which may be correct, but I still am missing the limits of integration for y.
• Draw a picture in the $z=0$ plane. – saulspatz Apr 28 at 16:34
Note that $$x=y^4$$ and $$x=1$$ intersect at $$(x,y)=(1,\pm1)$$. which define the limits for the integration region in the $$xy$$- plane. Thus, the volume integral is
$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$
That solid is located above the plane $$z=0$$ and below the plane $$z=4-2y$$. The possible values for $$x$$ belong to the $$[0,1]$$ interval (the condition $$x=y^4$$ prevents $$x$$ from being negative). So, you should compute$$\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}\int_0^{4-2y}1\,\mathrm dz\,\mathrm dy\,\mathrm dx\tag1$$But\begin{align}(1)&=\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}4-2y\,\mathrm dy\,\mathrm dx\\&=\int_0^18\sqrt[4]x\,\mathrm dx\\&=\frac{32}5.\end{align}
• Interesting - this is a slightly different method that still works! Thank you. 4 – Burt Apr 29 at 17:06
• As I'm further understanding all these methods, I don't understand why x needs to be from zero to one. Why can't it be from one and higher? – Burt Apr 29 at 21:28
• Your solid is bounded by the plane $x=1$, right?! – José Carlos Santos Apr 29 at 21:47
As you know how to evaluate the integral - and as it has been evaluated in other answers! - I'll concentrate on showing that a unique subset of $$\mathbb{R}^3$$ is bounded by the four surfaces identified in the question, and describing that subset in such terms that one can write down the triple integral that is to be evaluated.
@saulspatz's comment recommends first drawing a figure, ignoring the $$z$$ coordinate. I also find this to be the easiest way to think about the question.
The plane $$x = 1$$ cuts the $$(x, y)$$ plane in a line, and the surface $$x = y^4$$ cuts the $$(x, y)$$ plane in a curve. The line and curve together subdivide the $$(x, y)$$ plane into five subsets, which correspond to four subsets of $$\mathbb{R}^3$$: \begin{align*} A & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \geqslant 1 \}, \\ B & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \leqslant 1 \}, \\ C & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \geqslant 1 \}, \\ D & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \leqslant 1 \}. \end{align*} Each of $$A, B, C, D$$ is a connected subset of $$\mathbb{R}^3,$$ but the projection of $$C$$ on the $$(x, y)$$ plane has two separate components, corresponding to positive and negative values of $$y.$$
Each of $$A, B, C, D$$ is an unbounded subset of $$\mathbb{R}^3,$$ but the projection of $$D$$ on the $$(x, y)$$ plane is bounded. That looks hopeful! In order to be in a position to say something more definite than that, the easiest thing to do next (or so I think) is to look at the projection of the planes $$z = 4 - 2y$$ and $$z = 0$$ on the $$(y, z)$$ plane.
One can see that these two planes between them divide $$\mathbb{R}^3$$ into four subsets: \begin{align*} E & = \{ (x, y, z) \colon (y \geqslant 2 \text{ and } z \geqslant 0) \text{ or } (y \leqslant 2 \text{ and } z \geqslant 4 - 2y) \}, \\ F & = \{ (x, y, z) \colon (y \leqslant 2 \text{ and } z \leqslant 0) \text{ or } (y \geqslant 2 \text{ and } z \leqslant 4 - 2y) \}, \end{align*} \begin{align*} G & = \{ (x, y, z) \colon 4 - 2y \leqslant z \leqslant 0 \}, \\ H & = \{ (x, y, z) \colon 0 \leqslant z \leqslant 4 - 2y \}. \end{align*}
Subset $$E$$ contains points with arbitrarily large positive values of $$z$$ for any value of $$y$$; and subset $$F$$ contains points with arbitrarily large negative values of $$z$$ for any value of $$y$$; therefore neither $$E$$ nor $$F$$ has a bounded intersection with any of $$A, B, C, D.$$
Subset $$G$$ only contains points with values of $$y \geqslant 2,$$ therefore its intersection with $$D$$ is empty.
Subsets $$A, B, C$$ all have points with arbitrarily large positive values of $$y,$$ as do their intersections with $$G.$$
Therefore the only candidate for a subset of $$\mathbb{R}^3$$ that is bounded by the four given surfaces - and is bounded (!) - is: $$D \cap H = \{ (x, y, z) \colon y^4 \leqslant x \leqslant 1 \text{ and } 0 \leqslant z \leqslant 4 - 2y \}.$$ This is indeed bounded, and we can evaluate the volume integral by writing: $$\int_{D \cap H} 1 = \int_{-1}^1\int_{y^4}^1\int_0^{4 - 2y}\,dz\,dx\,dy.$$ I'll stop here - approximately where the other answers start. :)
• Thank you for this detailed answer! I can just choose to ignore the z plane, right? – Burt Apr 29 at 17:04
• The $x, y$ and $z$ coordinates are all involved, but (as in @saulspatz's comment) I found it easiest to ignore the $z$ coordinate to begin with, i.e. draw a "plan", ignoring the "elevation" for the moment. This allows you to form a clear view of the figure formed by the $x = y^4$ and $x = 1$ surfaces. You can then consider the two remaining surfaces $z = 4 - 2y$ and $z = 0.$ Again I found it helpful to look at these two separately from the other two at first, even though one can already begin to see how they will intersect the first pair of surfaces. My aim was to avoid all guesswork. – Calum Gilhooley Apr 29 at 17:25
• More succinctly: ignore $z$ (in the sense of ignoring the surfaces whose equations involve $z$), then ignore $x$ (in the sense of ignoring the surfaces whose equations involve $x$), then put the two views together. – Calum Gilhooley Apr 29 at 18:58 | 2020-11-27T20:11:03 | {
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https://math.stackexchange.com/questions/2454985/homotopy-groups-un-and-sun-pi-mun-pi-msun | # Homotopy groups U(N) and SU(N): $\pi_m(U(N))=\pi_m(SU(N))$
Am I correct that homotopy groups of $U(N)$ and $SU(N)$ are the same, $$\pi_m(U(N))=\pi_m(SU(N)), \text{ for } m \geq 2$$ except that $$\pi_1(U(N))=\mathbb{Z}, \;\;\pi_1(SU(N))=0,$$
Hence the Table in page 3 of this note has error along the column of $\pi_1(U(N))$ which should be $\pi_1(U(N))=\mathbb{Z}$?
More generally, where can I find a trustworthy table for $\pi_m(U(N))$ and $\pi_m(SU(N))$?
I find additional results for $SU(N)$ here in this Ref:
• Yes, that's right, and yes, $\pi_1$ should be $\mathbb{Z}$ for all $N$ in the table. – Qiaochu Yuan Oct 2 '17 at 22:46
• Try Mimura's Article "The Homotopy Theory of Lie Groups", pg970, in James's "Handbook of Algebraic Topology" for a table of $\pi_iSU(n)$ for $i\leq15$, $n\leq 8$. He also gives references where some of the higher groups can be found. – Tyrone Oct 3 '17 at 7:54
You are correct. This follows from the fact that $U(N)$ is diffeomorphic to $S^1\times SU(N)$. To see this, consider the map $f : U(N) \to S^1\times SU(N)$ given by $A \mapsto (\det A, \operatorname{diag}((\det A)^{-1}, 1, \dots, 1)A)$. This is a smooth map with smooth inverse given by $(z, B) \mapsto \operatorname{diag}(z, 1, \dots, 1)B$. Note however that the two groups are not isomorphic though (they have different centers), so they are not isomorphic as Lie groups.
As $SU(N)$ is simply connected, we see that
$$\pi_1(U(N)) = \pi_1(S^1\times SU(N)) = \pi_1(S^1)\oplus\pi_1(SU(N)) = \mathbb{Z}$$
and for $m \geq 2$,
$$\pi_m(U(N)) = \pi_m(S^1\times SU(N)) = \pi_m(S^1)\oplus\pi_m(SU(N)) = \pi_m(SU(N)).$$
As for your request regarding a table of the homotopy groups of $SU(N)$, the groups $\pi_m(SU(N))$ for $1 \leq m \leq 15$ and $1 \leq N \leq 8$ are given in appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics. This doesn't quite cover all the cases you asked for in the comment below, but the missing ones follow from complex Bott periodicity.
For completeness, here is the table you asked for.
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline & \pi_1 & \pi_2 & \pi_3 & \pi_4 & \pi_5 & \pi_6 & \pi_7 & \pi_8 & \pi_9 & \pi_{10} & \pi_{11} & \pi_{12}\\ \hline SU(1) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline SU(2) & 0 & 0 & \mathbb{Z} & \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_{12} & \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_3 & \mathbb{Z}_{15} & \mathbb{Z}_2 & \mathbb{Z}_2\oplus\mathbb{Z}_2 \\ \hline SU(3) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_6 & 0 & \mathbb{Z}_{12} & \mathbb{Z}_3 & \mathbb{Z}_{30} & \mathbb{Z}_4 & \mathbb{Z}_{60} \\ \hline SU(4) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{24} & \mathbb{Z}_2 & \mathbb{Z}_{120}\oplus\mathbb{Z}_2 & \mathbb{Z}_4 & \mathbb{Z}_{60} \\ \hline SU(5) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{120} & 0 & \mathbb{Z}_{360}\\ \hline SU(6) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \mathbb{Z}_{720}\\ \hline SU(7) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(8) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(9) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline SU(10) & 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0\\ \hline \end{array}
• Dear Michael Albanese +1, Thanks. I will count you as an answer immediately if you can provide a much complete Table than mine ( if possible, say up to 12th homotopy group of SU(10), if there is a result done in math?). – wonderich Oct 3 '17 at 2:30
• p.s. I mean a resource of Table up to 10 rows and 12 columns. – wonderich Oct 3 '17 at 2:31
• The sequence $SU(N)\to U(N)\xrightarrow{\det}S^1$ is a fibration, and the corresponding long exact sequence of homotopy groups together with the vanishing of higher homotopy groups of $S^1$ gives this result without resorting to the unholy homeo :-) – Mariano Suárez-Álvarez Oct 3 '17 at 19:37 | 2019-06-16T18:36:31 | {
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https://physics.stackexchange.com/questions/571521/what-exactly-is-enclosed-current | # What exactly is enclosed current?
In the realm of magnetostatics, consider the integral form of Ampere's law:
$$\oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enclosed}$$
What I realized is when asked the question "what is the enclosed current enclosed by?"
The most common answer I get is "enclosed by the Amperian loop of course!"
I think this is a huge misconception, because if we look at how the integral form of Ampere's law is derived (in quasistatic situations): $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} \longrightarrow \iint_S (\nabla \times \mathbf{B}) \cdot d\mathbf{a} = \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a} \longrightarrow \oint_C\mathbf{B} \cdot d\mathbf{a}= \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a}$$
In other words, the answer should be that the current is enclosed by the surface BOUNDED by the Amperian loop, because of the surface integral.
However, I notice that this definition of enclosed current is not without issues, because if we consider the situation below:
Both surfaces $$S_1$$ and $$S_2$$ are enclosed by the same Amperian loop, however, one may argue that the surface $$S_2$$ "encloses" more current than the surface $$S_1$$. But we know this is not true because the magnetic field for both cases should be the same, since it is the same line integral.
To resolve this, we may argue that for surface $$S_2$$, the current outside the Amperian loop is "not really enclosed", since it penetrates from outside the surface and exits, so the net contribution to the surface integral is zero.
But all I need to do is shade the Amperian loop to make it a closed surface, and the same argument can be applied, that the current passing through inside the Amperian loop is "not really enclosed" as well.
I think I am hugely misunderstanding something but I am not sure what it is.
You have highlighted the fact that you can choose *any (well belaved) surface as long as it is bounded by the Amperian loop which means that $$\displaystyle \mu_0 \iint_{S_1} \mathbf{J} \cdot d\mathbf{a}=\mu_0 \iint_{S_2} \mathbf{J} \cdot d\mathbf{a} = . . . . . =\mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} = \, . . . . .$$
The analogy which is often used is that the Amperian loop and the surface are equivalent to a butterfly net.
Once the direction of integration has been chosen, clockwise in this case, the direction of the normals to the surface is defined by the right-hand rule, so in the diagram above the normals are pointing "outwards, from the surface.
Consider the surfaces defined in your diagram with normals to the surfaces being shown.
Surface $$S_1$$ has all the contributions from $$\mathbf{J} \cdot d\mathbf{a}$$ being positive.
For surface $$S_2$$ there are positive (blue normal) and negative (red normal) to the integral. The negative contributions cancelling out some of the positive contributions to make the integral the same as for surface $$S_1$$.
One way to visualise this is to imagine areas projected onto a plane perpendicular to $$\mathbf J$$.
Often the simplest surface to consider is the plane defined by the Amperian loop $$S_0$$ where the normals are all parallel to one another and to $$\mathbf{J}$$ which makes the integration easier to do with $$\displaystyle \mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} =\mu_0 \iint_{S_{\rm 0}} \mathbf{J} \cdot d\mathbf{a}$$.
If you think about it in simple terms then the term $$\mathbf{J} \cdot d\mathbf{a}$$ is the same as $$J\,da\,\cos \theta$$ where $$da\,\cos \theta$$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. I have tried to illustrate this below.
The term $$\mathbf{J} \cdot d\mathbf{a}$$ relates to a flux of charge through an area.
If no charge accumulates in the volume bounded by areas $$S_0$$ and $$S_2$$ then the flux of charge through area $$S_0$$ into the volume must be the same as the flux through area $$S_2$$ out of the volume.
• Very nice. How do we know with certainty that after the negative contributions cancel out with the positive, regardless of surface chose, then we will always obtain that of $S_0$? Aug 6 '20 at 8:54
• I am sure that a Mathematician can provide a general proof but if you think about it in simple terms then the term $\mathbf{J} \cdot d\mathbf{a}$ is the same as $J\,da\, \cos (\theta)$ where $da\,\cos(\theta)$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. The term $\mathbf{J} \cdot d\mathbf{a}$ relates to a flux of charge trough an area. If no charge accumulates in the volume bounded by areas $S_0$ and $S_2$ then the flux of charge through area $S_0$ must be the same as the flux through area $S_2$. Aug 6 '20 at 9:50
• @D.Soul I have added to my original answer. Aug 6 '20 at 13:41
"But all I need to do is shade the Amperian loop to make it a closed surface," That doesn't work. The surface bounded by the closed loop always has to be an open surface. What you have produced is two surfaces for the current to go through, so you are just doing it Ampere's law twice.
You can think of the currents enclosed by the Amperian loop $$C$$ as the currents that go through whatever surface $$S$$ -no matter how you deform it, as long as you don't rip holes in it- that is enclosed by $$C$$.
From a topological point of view, the Amperian loop $$C$$ along which you are computing the integral and the loop along which an enclosed current is passing are concatenated, like two adjacent links of a chain: you cannot move them apart without them intersecting one another.
• Isn't the loop $C$ and the loop in which then enclosed current is passing the exact same loop? Aug 6 '20 at 9:04
• No they are two distinct loops: the latter is the physical circuit along which current is flowing, while $C$ is the loop along which you are computing the integral $\oint_C \bf{B} \cdot d\bf{l}$: it doesn't need to correspond to a physical loop, it is just a path of integration Aug 7 '20 at 12:28
Electricity is always an open system, so enclosed does not really exist. Your electric field is either endothermic or exothermic. This means which direction force is in regard to the wire. Do you understand me? If not, ask
• While I did not downvote your answer, I don't think this is the right answer, enclose DOES exist, you can enclose it by a surface bounded by an amperian loop. What I meant is the "contradiction" between different surfaces Aug 6 '20 at 9:06
• The evidence of my answer is here. Use lussacs gas law to show a change in pressure in the electric field charge during the endothermic charge of the electric field ntrs.nasa.gov/citations/20070032054 Aug 8 at 12:43 | 2021-10-28T11:39:07 | {
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https://blogs.mathworks.com/loren/2010/10/14/how-many-digits-to-write/?s_tid=blogs_rc_1&from=cn | # How Many Digits to Write?
Recently, my colleague Rob Comer and I were talking about how to write out a number, in decimal, so that if it were read back into MATLAB, would retain its full precision. The question is how many digits to write out. The number depends on several things, including the datatype the value is stored in. In addition, it may depend on the precision of the value - i.e., was it data collected during an experiment in which only two significant figures were recorded? Today I'll post about the solution Rob and I came up with for choosing the number of digits so if you write out the data as a string, you can read it back in to MATLAB with full precision retained.
### Create Some Values
Let's first create some values, both single and double versions of pi.
format long g
dblpi = pi
snglpi = single(pi)
dblpi =
3.14159265358979
snglpi =
3.141593
### Figure Out Number of Digits
To figure out the number of digits to print, we need to know what the floating point accuracy, sometimes called eps for the number of interest.
eps(snglpi)
eps(dblpi)
ans =
2.384186e-007
ans =
4.44089209850063e-016
As makes sense, we can see that the accuracy of the single precision value is larger than that for the "equivalent" double precision value. That means that the number next closest to the single precision value is farther away than the number next closest to the double precision value.
### Number of Digits
We can use eps(x) to help us figure out how many digits to print after the decimal place. First find total number of digits, base 10:
log10(eps(snglpi))
log10(eps(dblpi))
ans =
-6.62266
ans =
-15.352529778863
To get to a positive number of digits, simply negate the results.
-log10(eps(snglpi))
-log10(eps(dblpi))
ans =
6.62266
ans =
15.352529778863
And round up to get make sure we don't miss any accuracy.
ceil(-log10(eps(snglpi)))
ceil(-log10(eps(dblpi)))
ans =
7
ans =
16
Let's convert the results to a string. We are taking advantage of the ability to control the number of digits using * in sprintf.
snglpistr = sprintf('%.*f', ceil(-log10(eps(snglpi))), snglpi)
dblpistr = sprintf('%.*f', ceil(-log10(eps(dblpi))), dblpi)
snglpistr =
3.1415927
dblpistr =
3.1415926535897931
Now we've captured each value so if written out as a string, and read back into MATLAB, the accuracy is preserved.
### Convert to a Function
Taking what we know for finding the number of digits, let's make a function that we can use to test it out.
digits = @(x) ceil(-log10(eps(x)));
printdigs = @(x) sprintf('%.*f', digits(x), x);
### Try Some Values
printdigs(pi)
printdigs(2/3)
printdigs(1000*pi)
printdigs(pi/1000)
ans =
3.1415926535897931
ans =
0.6666666666666666
ans =
3141.5926535897929
ans =
0.0031415926535897933
### Some Magic Now
Rob created the necessary magic for getting rid of trailing zeros after the decimal point, while leaving at least one digit to the right of the decimal.
x = 1/2000
str = printdigs(x)
strout = stripzeros(str)
x =
0.0005
str =
0.0005000000000000000
strout =
0.0005
Let's try some more values. First create a function to help us again.
strippedStringValues = @(x) stripzeros(printdigs(x));
vals = [ 100/289, -1/17, 1/2000, 0, 500 -200, 123.4567]
for k = vals
strippedStringValues(k)
end
vals =
Columns 1 through 2
0.346020761245675 -0.0588235294117647
Columns 3 through 4
0.0005 0
Columns 5 through 6
500 -200
Column 7
123.4567
ans =
0.34602076124567471
ans =
-0.058823529411764705
ans =
0.0005
ans =
0.0
ans =
500.0
ans =
-200.0
ans =
123.4567
Here's the magic code for stripping the zeros, for those who are interested.
dbtype stripzeros
1 function str = stripzeros(strin)
2 %STRIPZEROS Strip trailing zeros, leaving one digit right of decimal point.
3 % Remove trailing zeros while leaving at least one digit to the right of
4 % the decimal place.
5
6 % Copyright 2010 The MathWorks, Inc.
7
8 str = strin;
9 n = regexp(str,'\.0*$'); 10 if ~isempty(n) 11 % There is nothing but zeros to the right of the decimal place; 12 % the value in n is the index of the decimal place itself. 13 % Remove all trailing zeros except for the first one. 14 str(n+2:end) = []; 15 else 16 % There is a non-zero digit to the right of the decimal place. 17 m = regexp(str,'0*$');
18 if ~isempty(m)
19 % There are trailing zeros, and the value in m is the index of
20 % the first trailing zero. Remove them all.
21 str(m:end) = [];
22 end
23 end
### How Do You Control Printed Digits?
Are you able to just use the default printing from MATLAB for your values? Do you use disp, leave off the semi-colon (;), use one of the *printf functions? What customizations do you need to make to print out values? Let me know here.
Published with MATLAB® 7.11
| | 2021-12-05T08:22:39 | {
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https://math.stackexchange.com/questions/2640263/what-is-probability-of-drawing-last-black-ball-from-the-box | # What is probability of drawing last black ball from the box?
A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability the that last ball drawn is black equals to
a) $1/100\quad$ b) $1/60\quad$ c) $3/5\quad$ d) $2/3\quad$
My try:
total red balls=40 ,
total black balls=60
The probability of drawing last black ball=total black balls/(total balls)
$$=\frac{60}{40+60}=60/100=3/5$$
I guessed this answer which is correct but i don't think that my procedure is correct. I think there should be a correct solution to such problems of probability. Please give correct solution to this problem.
My book suggests that answer must be 3/5
thanks
• I think the intuitive solution is fine but if you're doing it for an assignment then I would say something like there are ${100 \choose 60}$ ways of picking that combination of reds and blacks. Then there are ${99 \choose 59}$ ways of choosing reds and blacks such that a black is left. I think the ratio should give you the desired fraction. – stuart stevenson Feb 7 '18 at 15:15
• Each draw has a probability of $\frac {60}{60+40}=.6$ chance of being black. There is no special significance to being the "last". – lulu Feb 7 '18 at 15:21
If you want to complicate matters, you could compute
$\text{(ways of placing 59 black balls in 99)/(ways of placing 100 black balls in 60)}= \dfrac{\binom{99}{59}}{\binom{100}{60}}$
but the book's suggestion is good. Imagine all the balls to be randomly placed in a line, and ask
• what is the probability that the first ball, say, is black ? $\frac35$
• what is the probability that the ninth ball, say, is black ? $\frac35\;\;$
• what is the probability that the $i^{th}$ ball is black ? $\frac35\;\;$ isn't it ?
• so what is the probability that the last ball is black ? | 2020-06-06T12:12:47 | {
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https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_18&diff=cur&oldid=114954 | # Difference between revisions of "2010 AMC 12A Problems/Problem 18"
## Problem
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
## Solution 1
Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$, $(-3,3)$, and $(-2,2)$.
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$, $(3,-3)$, and $(2,-2)$. Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.
## Solution 2
We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it $S$). There is symmetry about $y=x$, so we only have to consider $(1,-1), (1,0)$, and $(1,1)$. $(1,1)$ can go on the boundary in two ways, so we can only consider one case and multiply it by two. For $(1,0)$ and $(1,-1)$ we can just multiply by two. So we count paths from $(-4,4)$ to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to $(4,4)$, and all in all, we get the answer is $\dbinom{16}{8}-2\left[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}\right]=1698$, which is answer choice $\textbf{(D)}$ -vsamc Note: Sorry if this was rushed.
## Further Explanation(for Solution 1)
As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$. Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).
For the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$.
For each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:
right, right, up, right,..., etc.
Also notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to
$\binom{8}{0}$
Now we consider the number of ways to get from $(4, -4)$ to $(4, 4)$. by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$.
Moving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ "right"s and $1$ "up". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ "up" into $8$ spots. This is
$\binom{8}{1} = 8$
Similarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$.
Finally, for the third case (ant passes through $(2, -2)$) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ "ups" in a sequence of $8$ movements, or
$\binom{8}{2} = 28$
Again, we square $28$: $28^2 = 784$. Adding up all of these cases we get
$1+64+784 = 849$
paths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have
$849*2=1698 \Rightarrow \boxed{\text{D}}$.
For all the notation geeks out there, the solution can be expressed as such:
$2\sum_{k=0}^2 \binom{8}{k}^2$ | 2021-06-23T19:21:27 | {
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http://wpck.acku.pw/optimization-math-ia-example.html | # Optimization Math Ia Example
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Another example from calculus is that if y = ln (x) + c, for c constant, then dy/dx = 1/x, and these are the only functions for which this is true. Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. Another is that the curve y = ln (x) has a tangent at (1,0) with slope 1, and among all logarithmic functions, it is the only one that does. Explore the possibilities of math through coursework, research, and public lectures. These files include the scoring rubrics and sample student responses at each score point attainable. NET Numerics aims to provide methods and algorithms for numerical computations in science, engineering and every day use. Previously, I have written guides to the Top 5 mistakes made in IB Math Explorations and 5 tips to acing your IB Math Exploration. To support the student’s learning we rely extensively on examples and graphics. Example 1: Iraq and Iran are the only suppliers of petroleum. To provide an example, suppose we have 10,000 to spend on four possible projects:. Brent Lindquist, Dean of College of Arts and Sciences). gcc -o writes the build output to an output file. Mr Bdubs Math and Physics 6,718 views. mathematics (outside of teaching or academia), your best bet is applied mathematics with computers. Welcome! This is one of over 2,200 courses on OCW. Math · Multivariable calculus · Applications of multivariable derivatives · Constrained optimization (articles) Lagrange multipliers, examples Examples of the Lagrangian and Lagrange multiplier technique in action. A student’s K-12 math education should also prepare him or her to be free to pursue post-secondary education opportunities. Day Lesson Title Math Learning Goals Expectations 1 What Is the Largest Rectangle? • Use an inquiry process to determine that a square is the largest rectangle that can be constructed for a given perimeter. 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Introduction to Quantum Chemistry (39 slides) Introduction to GAMESS (32 slides) From Schrodinger to Hartree-Fock (CHEM580 - 43 slides). These are the units that are used most often. A relation R on set A is called Transitive if and implies. Video created by National Research University Higher School of Economics for the course "Mathematics for economists". I am in math hl and I really need a good math IA topic. Hi guys so I thought I'd start working on my IB Maths Exploration (SL). Again, there are many insights from this example into the challenges that must be faced in optimization theory and practice. 25% is a function of the length of time the money is invested. If Tyrolit can achieve this, its inauspicious cost structure due to high level of service would not decrease profit as much, because the company would work generally more efficiently. For example, dynamic search models are used to study labor-market behavior. 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Read the latest articles of Applied Mathematics and Computation at ScienceDirect. In this "Maths Exploration" the maths definitely does not have to be complicated, this is a very good example of a maths exploration which scored a perfect 20/20 The assessment criteria are these and I would really appreciate it if you took a look at them. I am a proud member of the NCSU Numerical Analysis Group and am very active in SIAM. While many of Euclid's successors implicitly assumed that all perfect numbers were of the form (Dickson 2005, pp. Thus the rectangular field should be 600 feet wide and 1200 feet long. Escher are the result of his attempts to visually express such mathematical concepts as infinity, duality, dimension, recursion, topological morphing, and self-similarity. You can use the worksheet that most closely models your situation as a starting point. Expressing things differently. SIAM hosts conferences, publishes book and journals, and has a robust membership program. It is no matter what Words you put them into. Algorithmic, data analytic, machine learning and numerical methods which support the modeling and analysis of optimization problems are encouraged. It is a very good tool for improving reasoning and problem-solving capabilities. SIAM hosts conferences, publishes book and journals, and has a robust membership program. Visit BYJU’S to learn different types of equations provided with examples. Note that we can rearrange the error bound to see the minimum number of iterations required to guarantee absolute error less than a prescribed$\epsilon\$:. This construction math self-paced class is intended to develop mathematical skills that can be applied to the construction trade through practice and application. To provide instructional and professional support to local districts and school sites, so that ALL students will achieve proficiency in Algebra 1 and more advanced mathematics courses. Just as mathematics reveals the motions of the stars and the rhythms of nature, it can also shed light on the more mundane decisions of everyday life. 100 majors. Explore the possibilities of math through coursework, research, and public lectures. The structure presented in these internal assessments does not have to be followed strictly. A standard example of motivating constrained optimization are examples where the setup is described in a lot of lines, e. 5 we will consider other possibilities. In these cases, higher-order optimization methods are ill-suited, and discussion in this paper will be restricted to rst-order methods. optimize() is devoted to one dimensional optimization problem. For K-12 kids, teachers and parents. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions. Please add My Skype Address:ykreddy22 20 plus years experienced, highly qualified Indian math teacher offers one to one lesson in maths for IGCSE ,IB all grades up to 12 Grades levels Providing IGCSE IB math lessons in very simple ways including the basics of maths which will helps in your exams and the future studies. Draw a picture of the situation. Alongside 7 examples of exemplary Mathematics SL Internal Assessments, an extensive introduction to IAs provide International Baccalaureate students with tips, resources, and ideas to help students maximize their marks on the portfolio. Examples taken from biology. Math Placement Information. What are good examples of constrained optimization problems (perhaps not simple!) that today's students might actually encounter in their lives? If your goal is to find problems that are more easily accessible, see also the sister question What are easy examples from daily life of constrained optimization?. DEFINITION OF LOCAL MAXIMA AND LOCAL MINIMA 1. They illustrate one of the most important applications of the first derivative. Only one of MATH 151, MATH 160, the sequence MATH 165-MATH 166, or MATH 181 may be counted towards graduation. Basic elements of a good Math Studies Project or Math SL/HL Portfolio piece: • Correct answers throughout. The Journal of Computational Mathematics published bi-monthly. Optimization is a part of Calculus which I certainly do enjoy, but once again I don't have any ideas regarding suitable optimization problems suitable for the investigation. With MATH 75B, equivalent to MATH 75. Powered by Create your own unique website with customizable templates. Optimization Notice 15 DNN Primitives in Intel® MKL Highlights A plain C API to be used in the existing DNN frameworks Brings IA-optimized performance to popular image recognition topologies: – AlexNet, Visual Geometry Group (VGG), GoogleNet, and ResNet. Students majoring in mathematics might wonder whether they will ever use the mathematics they are learning, once they graduate and get a job. The basic idea of the optimization problems that follow is the same. | 2020-03-29T12:42:38 | {
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http://mathhelpforum.com/calculus/149363-integral-convergence-proof.html | 1. ## Integral convergence proof.
Why $\int^{\infty}_{\pi} \frac{sinx}{lnx}dx$ converges?
2. You can rewrite the integral in the following way:
$\displaystyle \int_\pi^\infty \frac{\sin x}{\log x} \text{ d}x=\int_\pi^{2\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{2\pi}^{3\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{3\pi}^{4\pi} \frac{\sin x}{\log x} \text{ d}x+\ldots=\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{\sin x}{\log x} \text{ d}x$.
Take note that the terms of the sum are alternating! We can write the sum as
$\displaystyle \sum_{k=1}^\infty (-1)^k \int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x$
Also observe that as $k\to \infty$ the term $\displaystyle a_k=\int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x\to 0$ monotonically. Therefore, the sum (and equivalently, the integral) converges via the alternating series test.
3. Wow! I never guess to do this!!!
Maybe there is more simple way to prove that?
The reason I'm asking this is because that is from "special"(true false questions) test and for each question I have something like 7 minutes!
Thank you again!
4. This is probably the cleanest way to do the problem, but you might be able to get away with an integration by parts, first. Let $u=\frac{1}{\log x}$ and $dv=\sin x\text{ d}x$. Then, $du=\frac{1}{x(\log x)^2} \text{ d}x$ and $v=-\cos x$. The integral becomes
$\displaystyle \int_\pi^\infty \frac{\sin x}{\log x}\text{ d}x=\left.\frac{-\cos x}{\log x} \right|_\pi^\infty+\int_\pi^\infty \frac{\cos x}{x (\log x)^2}\text{ d}x$
Can you see how the terms will converge?
5. ## Eureka!
Dirichlet's test for integrals!
Let,
$f(x)=sin(x)$
and
$g(x)=\frac{1}{ln(x)}$
$|\int^b_{\frac{\pi}{2}} sin(x)dx| \leq 2$ for all $b>\frac{\pi}{2}}$
and $g(x)$ is monotonic function and $lim_{x\to\infty}g(x)=0$
So, by the test of Dirichlet the integral converges! | 2016-10-28T06:38:45 | {
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https://math.stackexchange.com/questions/818542/moments-at-which-moving-points-on-a-circle-coincide | # Moments at which moving points on a circle coincide
Points A $(0,1)$ and B $(1,0)$ start moving along the circumference of a unit circle with center $(0,0)$ in the same, positive (that is, counterclockwise) direction. Every minute, points A and B traverse arcs respectively of $60$° and $42$°. Visually:
Determine moments $t_1, \ldots, t_k,\ldots,$ such that at time $t_k$ points A and B coincide for the $k^\text{th}$ time.
I've been able to determine $t_1$, but cant seem to determine the next moment. I'll describe how I've gotten $t_1$ and hopefully you can suggest how to proceed (or if I'm doing it wrong, how to go about solving for all $t$). We're given the angular velocities of the two points:
• $v_A = 60$° ($\pi\over 3$) per minute;
• $v_B = 42$° ($7\pi\over 30$) per minute.
We also know the starting angles of the two points (shown also on the graph):
• $d_A = {\pi \over 2}$ and $d_B = 0$.
To calculate $t_1$, we just have to solve the following equation for $t$:
$$\left({\pi\over 2} + {\pi \over 3}\cdot t \right)= \left(0 + {7\pi\over 30}\cdot t\right) \tag{T_1}.$$
Calculation yields the value of $5$ for $t$, so $\color{brown}{t_1 = 5}$ minutes.
Now, another basic calculation tells us that at minute $t_1$, points A and B form an angle of $5\pi\over 3$ with respect to $OB$ (sorry, forgot to label 'O' on the graph). So, I figured that to calculate moment $t_2$, it will suffice to solve the following equation for $t$ and add $t_1$ to it:
$$\left({5\pi\over 3} + {\pi \over 3}\cdot t \right)= \left({5\pi\over 3} + {7\pi\over 30}\cdot t\right) \tag{T_2^?}.$$
But, of course, the first summands are canceled out, leaving us with:
$$\left({\pi \over 3}\cdot t \right)= \left({7\pi\over 30}\cdot t\right) \tag{T_2^?}.$$
This solution is true only for $t=0$, so clearly something went wrong with my reasoning. (Of course, $t=0 + t_1 = t_1$, which is a moment of coincidence, but it's not the moment we're looking for). I would appreciate any help with the strategy I've taken or the way I should approach it instead.
• We use degrees, since typing those $\pi$ is a nuisance, and degree intuition is better, and fractions are unpleasant. In $k$ minutes, $A$ travels $18$ degrees more than $B$. Since $B$ starts out $90$ degrees ahead, the first time of coincidence is $5$ minutes. For the next time of meeting, $A$ has to gain $360$ more degrees, which takes $20$ minutes, so $t_2=25$. And $20$ more minutes gets us to the third time of meeting, and so on. The $k$-th time of meeting is $5+20(k-1)$. Jun 2, 2014 at 20:53
• Thank you André! $20k-15$ it is! Jun 2, 2014 at 20:55
• You are welcome. You can see that viewing the problem concretely makes the answer pop out. Jun 2, 2014 at 20:57
• It does! Thank you all for your help. Jun 2, 2014 at 20:58
• Please note that I misread and interchanged $A$ and $B$. It is $15+20(k-1)$. Jun 2, 2014 at 21:29
All your thinking so far is good. What you're missing is that the next time your points meet, A will have gone around the circle an extra time. So their angles won't be equal - $A$'s will be exactly $2\pi$ more than $B$'s, accounting for the extra lap. If you solve $$\frac{\pi}{3} \cdot t = \frac{7\pi}{30} \cdot t + 2\pi,$$ you get $t= 20$. That tells you they'll meet again another 20 minutes later, at a total time of 25 minutes after the start. Can you guess from there what time the third meeting will be? Even better - can you explain why? :)
• I was just thinking about the 2pi. I had the stupid thought to divide by 2pi, but this makes sense. Thank you. Jun 2, 2014 at 20:51
• I'll think about you question and get back to you soon. Jun 2, 2014 at 20:53
An easy way to do this is to solve this equation $$42t\equiv 90+60t \bmod 360$$
• Ah! That's the same as cool papa's answer, but with degrees. I shouldn't have complicated by converting to radians. Thank you. Jun 2, 2014 at 20:52
• The use of modular arithmetic helps a lot here too, that way, if you see a more convoluted problem in the future where you can't intuit the number of additional laps it will take, you don't have to worry. Jun 2, 2014 at 20:53
• This doesn't lend itself to solving the equation very well. How would you go about solving for t? Just guess and check? Dec 11, 2014 at 23:18
Keep in mind that when you're calculating distance around a circle in this way you have to mod it by $2\pi$. If, for example, A and B start both straight up and A makes a revolution every minute while B makes a revolution every 2 minutes, then they'll coincide after 2 minutes, though the distance traveled is very different.
So instead, just look at the difference in distance traveled. A travels a certain distance farther than B every minute. Every time that distance is equal to a multiple of $2\pi$ they will coincide.
• Thank you. That is helpful. Jun 2, 2014 at 21:06 | 2022-05-22T22:34:11 | {
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https://fr.maplesoft.com/support/help/maple/view.aspx?path=GroupTheory%2FIsSubnormal | GroupTheory/IsSubnormal - Maple Help
GroupTheory
IsSubnormal
test whether one group is contained as a subnormal subgroup of another
Calling Sequence IsSubnormal( H, G )
Parameters
H - a group G - a group
Description
• A group $H$ is a subnormal subgroup of a group $G$ if $H$ is a subgroup of $G$, and if there is a chain
$G={G}_{0}▹{G}_{1}▹\dots ▹H$
such that ${G}_{k}$ is normal in ${G}_{k-1}$, for each $i$. Every normal subgroup of a group is subnormal, but not conversely.
• The IsSubnormal( H, G ) command tests whether the group H is a subnormal subgroup of the group G. It returns true if H is subnormal in G, and returns false otherwise. For some pairs H and G of groups, the value FAIL may be returned if IsSubnormal cannot determine whether H is a subnormal subgroup of G.
Examples
> $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
> $G≔\mathrm{Group}\left(\mathrm{Perm}\left(\left[\left[1,2,3,6,4,5,7,8\right]\right]\right),\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right)$
${G}{≔}⟨\left({1}{,}{2}{,}{3}{,}{6}{,}{4}{,}{5}{,}{7}{,}{8}\right){,}\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (1)
> $\mathrm{GroupOrder}\left(G\right)$
${16}$ (2)
> $H≔\mathrm{Subgroup}\left(\left\{\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right\},G\right)$
${H}{≔}⟨\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (3)
> $\mathrm{IsSubnormal}\left(H,G\right)$
${\mathrm{true}}$ (4)
Every normal subgroup of a group is subnormal.
> $\mathrm{andmap}\left(\mathrm{IsSubnormal},\mathrm{NormalSubgroups}\left(G\right),G\right)$
${\mathrm{true}}$ (5)
Compatibility
• The GroupTheory[IsSubnormal] command was introduced in Maple 2018. | 2023-03-21T05:31:35 | {
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https://math.stackexchange.com/questions/1815977/are-these-two-events-in-s-30-independent | # Are these two events in $S_{30}$ independent?
Let $S = S_{30}$, the set of permutations on $\{1,...,30\}$.
Suppose $A$ is the event: $\{\pi\in S_{30}: \pi(1) = 1\}$ and $B$ is the event $\{\pi\in S_{30}: \pi(2) < \pi(3)\}$.
I am asked to show that these events are independent.
Now $|A| = 29!$ and $|B| = (1+...+29)\times 28!$
Hence $|A|\times|B| = 29!\times 28! \times (1+...+29)$
However $|A\cap B| = (1+...+28)\times 27!$
So $P(A\cap B)< P(A)\times P(B)$.
So these events don't seem to be independent? What am I doing wrong?
• @AdamHughes Thank you, I was thinking of something else. I've written an answer now. – DonAntonio Jun 6 '16 at 15:44
By definition: since $\;A\cong S_{29}\;$ :
$$P(A)=\frac{29!}{30!}=\frac1{30}$$ , whereas
$$B:=\{\pi\in S_n\;/\;\pi(2)<\pi(3)\}$$
and we can then count:
$$\begin{cases}\pi(3)=30\implies\pi(2)=1,2,...,29\\{}{}\\\pi(3)=29\implies\pi(2)=1,2,...,28\\\ldots\\\pi(3)=2\implies \pi(2)=1\end{cases}$$
For each option in each line we have $\;(n-2)!\;$ permutations, so all in all we have
$$28!\left(29+28+\ldots2+1\right)=28!\frac{29\cdot30}2=\frac{30!}2\implies$$
$$P(B)=\frac{\frac{30!}2}{30!}=\frac12$$
Finally: if $\;\pi\in A\cap B\;$ then
$$\begin{cases}\pi(1)=1,\,\pi(2)=2\implies \pi(3)=2,3,....,30\\{}\\\pi(1)=1,\,\pi(2)=3\implies \pi(3)=4,...,30\\\ldots\\{}\\\pi(1)=1,\,\pi(2)=29\implies \pi(3)=30\end{cases}$$
For each option above we have $\;27!\;$ permutations, so in total we have
$$27!\left(1+2+\ldots+28\right)=\frac{29!}2\implies P(A\cap B)=\frac{\frac{29!}2}{30!}=\frac1{60}=P(A)P(B)$$
and the events are independent
I assume that we are picking a permutation $\pi$ at random, with all permutations equally likely.
Let $A$ be the event $\pi(1)=1$ and $B$ the event $\pi(2)\lt \pi(3)$. To show independence, we show that $\Pr(B\mid A)=\Pr(B)$. This is clear, for by symmetry each is $\frac{1}{2}$.
• Thanks for this explanation. May I ask where did I go wrong in my argument? – fosho Jun 6 '16 at 15:39
• The calculations seem to be correct, but do not address the issue of independence. – André Nicolas Jun 6 '16 at 15:43
• @McFry: At the time I wrote the comment, the OP made no reference to probability. – André Nicolas Jun 6 '16 at 16:02 | 2020-02-18T19:25:41 | {
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https://math.stackexchange.com/questions/3998437/minimal-steps-to-reach-a-natural-number-by-a-signed-arithmetic-progression | # Minimal steps to reach a natural number by a signed arithmetic progression
$$F:=\big\{f\,\big|\,\text{function }f: \mathbf N\rightarrow \{1,2\}\big\}$$. \begin{align} &\min_{f\in F,\,k\in\mathbf N} k, \\ &\sum_{i=1}^k (-1)^{f(i)}i = n \in\mathbf N. \end{align}
1. Is there an analytical or an asymptotic solution to this problem for general $$n$$?
2. For a specifically given $$n$$, what would a good algorithms to solve this problem? The dynamic programming?
Essentially, for any given $$n$$, the task is to select $$+$$'s and $$-$$'s to reach $$n$$ as fast as possible (minimize $$k$$): $$\pm 1 \pm 2 \pm 3 \pm \dots \pm k = n$$
If we don't worry about overshooting $$n$$, the obvious strategy is to only add and never subtract. If $$n$$ is a triangular number, this will give us the answer. This would necessarily be the optimal solution since we did not subtract at all.
If $$n$$ is not a triangular number, at some point $$k'$$, we will overshoot $$n$$: $$1 + 2 + 3 + \dots + k' > n$$
Since we need $$k$$ at least $$k'$$ to even reach $$n$$, we know that $$k=k'$$ is optimal if it works.
Consider the value $$1 + 2 + 3 + \dots + k' - n$$. We will denote this amount as $$a$$. We know that $$a < k'$$. If $$a$$ is even, then we can switch the operator of $$a/2$$ to $$-$$, and we have our optimal solution.
Figuring out the solution for when $$a$$ is odd may be trickier.
EDIT:
Figured out the "odd" case!
If $$a$$ is odd, then we cannot have $$k = k'$$ because there is no expression that will give us exactly $$n$$ using $$k'$$ terms. For any signs we switch in the $$k'$$ terms, the total sum will retain the same parity, so we cannot reach exactly $$n$$. Thus, if we can find a solution with $$k'+1$$ terms, it is necessarily optimal.
Consider $$1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k'+1$$.
If $$k'$$ is even, then consider the value $$\frac{a+k'+1}{2}$$. We have $$\frac{a+k'+1}{2} < \frac{2k'+1}{2} = k' + 1/2 \,.$$
So, $$\frac{a+k'+1}{2} \leq k'$$. Thus, it is in our series, so we can assign it the operator $$-$$, and we have achieved exactly a sum of exactly $$n$$.
Now, to consider what happens when both $$a$$ and $$k'$$ are odd...
EDIT 2:
First off, is it possible to find a solution with $$k=k'+1$$ when $$k'$$ is odd?
Well if we use all $$+$$'s, we have $$1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k' + 1$$. As before, switching the signs of any of our terms will preserve the parity of the total sum. Since $$a$$ and $$k'$$ are odd, $$n+a+k'+1$$ has opposite parity from $$n$$. So, no rearrangement of signs will enable a series of $$k'+1$$ terms to work.
Our next candidate is $$k=k'+2$$. Let us consider the following sum: $$1 + 2 + 3 + \dots + k' + (k'+1) + (k'+2) = n + a + (k'+1) + (k'+2)$$
We are too high by the quantity $$a + 2k' + 3$$. We can switch the sign of $$k'$$ and switch the sign of $$\frac{a+3}{2}$$, and we arrive at exactly $$n$$ with $$k=k'+2$$.
And that concludes the proof! No dynamic programming needed - a purely analytic solution.
TL;DR
Let $$k'$$ be the least integer such that $$T_{k'} \geq n$$. If $$T_{k'}$$ and $$n$$ have the same parity, then $$k=k'$$. Else, $$k$$ is equal to the least odd integer greater than $$k'$$.
• How do you know that these solutions are optimal ? – Yves Daoust Jan 24 at 22:25
• I will edit my post to make it more clear. – inavda Jan 24 at 22:26
• Is it more clear now? – inavda Jan 24 at 22:36
• Yep, it is, thanks. – Yves Daoust Jan 24 at 22:41
• Can't you say that you start with the smallest $k$ such that $T_k\ge n$ and $T_k$ has the same parity as $n$, as the overshoot correction will be even. – Yves Daoust Jan 24 at 22:44
Continuing on @inavda's answer, for a given $$n$$ we find the minimum number of terms, which is the index of the smallest triangular number not smaller than $$n$$, and with the same parity. As the parities of the triangular numbers follow the pattern $$e,o,o,e,e,o,o,e,e,\cdots$$, we can find the smallest triangular number not smaller than $$n$$, and increment once or twice as needed to reach the desired parity.
Then we repeat this operation recursively on the half of the residue $$\dfrac{k(k+1)}2-n$$ to obtain the desired correction.
Example: $$n=37$$
$$37\le T_9=45, \text{half residue }4\to+++++++++$$
$$4\le T_3=6, \text{half residue }1\to---+++++$$
$$1=T_1\to+--++++++$$
The smallest triangular number is obtained by
$$\frac{k(k+1)}2\ge n$$ or
$$k\ge\frac{\sqrt{8n+1}-1}2,$$
$$k=\left\lceil\frac{\sqrt{8n+1}-1}2\right\rceil.$$
Notice that the procedure fails for $$n=2$$, as the smallest triangular number is $$T_3=6$$ and the half residue is again $$2$$. Also for $$n=5$$, $$T_5=5$$ and the half residue is $$5$$. These are the only fixed points. We solve these with $$2=1-2+3$$ and $$5=1-2-3+4+5$$.
Notice that the residue is on the order $$O(\sqrt n)$$, so that the successive residues decrease very quickly, and the recursion depth is of order $$O(\log\log n)$$.
Update:
Checking the example, it turns out that this procedure is wrong.
• "We find the minimum number of terms, which is the index of the smallest triangular number not smaller than $n$, and with the same parity." I don't think this phrasing is quite correct. For example, $a(5)=5$, but the index of the smallest triangular number greater than $5$ is $3$. – inavda Jan 25 at 0:10
• @inavda: $T_3$ does not have the right parity. – Yves Daoust Jan 25 at 0:11 | 2021-03-05T13:22:59 | {
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https://mathematica.stackexchange.com/questions/177109/is-there-a-built-in-function-to-get-the-centroid-of-a-table/177111 | # Is there a built-in function to get the centroid of a table?
Suppose I have a table with some data that are 'lumped' around some central location, such as the dummy example
i0 = 22.3;
j0 = 34.1;
table = Table[
Exp[-((i - i0)^2 + (j - j0)^2)/10^2]
, {i, 1, 50}, {j, 1, 50}
]
which kind of looks like
ListPointPlot3D[table, PlotRange -> Full]
I would like to recover the centroid of the "mass" determined by the signal in my table, i.e. in the case above, the tuplet {i0,j0}, in the cleanest way possible. I've managed to do it in the obvious but expensive way with a bunch of explicit sums and products with explicit Ranges, but it feels like there should be a built-in function that will do this ─ and I've as yet been unable to find it.
Can this be done? If so, how?
wd = WeightedData[Tuples @ Range @ Dimensions @ table, Join @@ table]
Mean @ wd
{22.3232, 33.9072}
Also
Total[MapIndexed[#2 # &, table / Total[table, 2], {2}], 2] (* and *)
Dot[Join @@ table , Tuples[Range @ Dimensions @ table]] / Total[table, 2]
{22.3232, 33.9072}
i0 = 22.3; j0 = 34.1;
table = Table[
Exp[-((i - i0)^2 + (j - j0)^2)/10^2], {i, 1, 50}, {j, 1, 50}];
x = Array[#1 &, Dimensions[table]];
y = Array[#2 &, Dimensions[table]];
pt = {Total@Flatten[x table], Total@Flatten[y table]}/
Total[table, 2]
(* {22.3232, 33.9072} *)
ListDensityPlot[table, Epilog -> {Red, Point@Reverse[pt]},
PlotRange -> All]
• Thanks both. I accepted the 'proper' built-in as that's what the OP asks for, but this is faster for my instance and it is clean enough that it makes for readable code. – Emilio Pisanty Jul 11 '18 at 12:19
I can provide some improvement if it is about speed.
Let's generate a larger data set (I use Compile merely to speed it up a bit).
i0 = 22.3; j0 = 34.1;
m = 2500;
n = 1500;
x = Subdivide[1., 50, m - 1];
y = Subdivide[1., 50, n - 1];
table2 = Partition[#, n] &@
Compile[{{X, _Real, 1}, {i0, _Real}, {j0, _Real}},
Exp[-((X[[1]] - i0)^2 + (X[[2]] - j0)^2)/10^2],
RuntimeAttributes -> {Listable}
][Tuples[{x, y}], i0, j0];
Szabolcs' approach
AbsoluteTiming[
pt = {
Total@Flatten[Transpose[ConstantArray[x, n]] table],
Total@Flatten[ConstantArray[y, m] table]
}/Total[table, 2]
]
{0.733425, {22.3288, 33.8733}}
The problem is that before summation, some large arrays have to be constructed and multiplied. But the summations and matrix-matrix products can also be expressed by cheaper matrix-vector products:
AbsoluteTiming[
pt2 = With[{buffer = ConstantArray[1., m].table},
{x.(table.ConstantArray[1., n]),
buffer.y
}/(ConstantArray[1., n].buffer)
]
]
{0.044209, {22.3288, 33.8733}} | 2019-10-23T22:31:31 | {
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http://math.stackexchange.com/questions/55591/notation-for-the-set-of-all-finite-subsets-of-mathbbn | # Notation for the set of all finite subsets of $\mathbb{N}$
Is there a "standard" notation to denote the set of all finite subsets of $\mathbb{N}$? (or any set, not just $\mathbb{N}$)
Thanks
-
Doubt it. I would go with $\mathcal{P}_{\text{fin}}(\mathbb{N})$ or, depending on what exactly you're doing, just $S$. – Qiaochu Yuan Aug 4 '11 at 14:52
(some) people think of $2^{<\omega}$ (finite length binary strings) as the set of all finite subsets of $\omega.$ – jspecter Aug 4 '11 at 14:54
I think $2^{<\omega}$ is pretty ambiguous (what does a terminal $0$ mean?). The notation I see most often is some version of $[\omega]^{<\omega}$, with one or both occurrences of $\omega$ replaced by $\mathbb{N}$ or $\aleph_0$, depending on personal preference. Alternatively, I have also seen FIN, which has a nice blunt simplicity about it. – user83827 Aug 4 '11 at 15:07
@user10: It looks like some more comprehensive answers have popped up in the meantime, so I'll just keep my comment as a comment unless you feel strongly otherwise. – user83827 Aug 4 '11 at 15:40
Several possible notations for $\{A\subseteq\omega\mid |A|<\omega\}$:
1. $[\omega]^{<\omega}$
2. $P_\omega(\omega)$
3. $\operatorname{Fin}(\omega)$
Where, of course, $\omega=\mathbb N$.
And as usual my advice on the matter: When in doubt, open with "We denote by [the chosen notation here] the set ..."
-
Isn't $\omega$ the usual notation for the ordinal, rather than the cardinal? I'd be tempted to use $|A|\lt\aleph_0$ or $|A|\lt|\omega|$, rather than $|A|\lt\omega$... – Arturo Magidin Aug 4 '11 at 16:07
@Arturo: It is indeed confusing, but initial ordinals function as both ordinals and cardinals. In this case, $|\omega|=\omega$, so it is alright. I agree that it is somewhat confusing, and I do my best to use $\aleph_0$ when I only care about cardinality. The usual notations, however, use $\omega$ - and I do agree that it is less... cluttered than $\mathcal P_{\aleph_0}(\omega)$ or such. – Asaf Karagila Aug 4 '11 at 16:09
Yes, technically, cardinals are particular ordinals; but my impression is that one uses $\omega$ when one wants to emphasize/keep in mind the ordinal structure, and uses $|\omega|$ and $\aleph_0$ when one wants to ignore the ordinal structure. Clearly, my impression was mistaken, I'll have to try to track down where I got it from. – Arturo Magidin Aug 4 '11 at 16:14
@Arturo: Yes, in essence you are correct. However since one uses $\aleph$ notation a lot less than people would expect (as most of the time we use $\kappa,\lambda$ and such to denote cardinals) it is customary to just let them assume their "ordinal role" when needed, and to say an ordinal is of smaller cardinality than some initial ordinal is the same as saying it has a smaller order type. So it works out just fine. – Asaf Karagila Aug 4 '11 at 16:46
The second notation is rather confusing to me. If I saw it in a paper I would not be able to guess what it meant. – Qiaochu Yuan Aug 4 '11 at 16:52
You can find various notations, as mentioned in coments. (I doubt there is some generally accepted notation.)
• You can find $[\omega]^{<\omega}$, e.g. here, which can be considered as a special case of $[A]^{<\kappa}$ - which denotes all subsets of $A$ of cardinality less then $\kappa$, see e.g. p.18 of the same book. In your case you could use $[\mathbb N]^{<\omega}$.
• You can find $\mathrm{Fin}$, e.g. here and here
• You can find $\mathbb N^{[<\infty]}$, e.g. here.
• Hindman and Strauss use $\mathcal P_f(\mathbb N)$ in this book, which is similar to Qiaochu's suggestion $\mathcal P_{\mathrm{fin}}(\mathbb N)$.
- | 2015-08-02T15:07:41 | {
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https://math.stackexchange.com/questions/3277955/relation-between-matrix-power-and-jordan-normal-form | # Relation between matrix power and Jordan normal form
(a) Assume $$A\in\mathbb{C}^{n\times n}$$ has $$n$$ distinct eigenvalues. Prove that there are exactly $$2^n$$ distinct matrices $$B$$ such that $$B^2 = A$$ (i.e., in particular, there are no more than $$2^n$$ matrices with this property).
(b) How many such matrices $$B\in\mathbb{C}^{3\times3}$$ exist if $$A=\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}$$? Why?
(a) It is clear that if $$\lambda_i$$ is an eigenvalue of $$A$$, then $$\pm\sqrt\lambda_i$$ is an eigenvalue of $$B$$, therefore $$B = \begin{pmatrix}\pm\sqrt\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\pm\sqrt\lambda_n\end{pmatrix}$$ satisfies $$B^2=A$$ and there are $$2^n$$ such matrices $$B$$ because of $$2^n$$ possibilities of rearranging different numbers of plus and minus signs on the diagonal.
(b) We can construct $$2^3=8$$ such matrices $$B$$ using approach of (a), but $$B$$ is not necessarily diagonal. Consider $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}$$ which satisfies $$B^2=A$$ as well. So, there exist more than $$2^3$$ such matrices.
I can't give an explanation why there are no more than $$2^n$$ possibilities in the first case and more than $$2^3$$ in the second. It definitely follows from the fact that Jordan normal form is unique in (a) and isn't unique in (b) because of an eigenvalue with algebraic multiplicity $$2$$, however, I can't formulate it into a self-contained statement.
• Did you mean $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}?$$ – Angina Seng Jun 29 '19 at 16:50
• @LordSharktheUnknown, yes, it was a typo and I edited the question. Thank you. – Hasek Jun 29 '19 at 17:06
Can we reduce finding matrix roots to finding roots of Jordan blocks?
In (b), your matrix $$A$$ has $$2$$ Jordan blocks associated to the same eigenvalue $$2$$. Then $$A$$ admits an infinity of square roots.
Here the infinity of matrices $$B\in M_2(\mathbb{C})$$ s.t. $$tr(B)=0,\det(B)=-2$$ are among the square roots of $$2I_2$$.
EDIT. I forgot to write that, in case (a), there are no more than $$2^n$$ square roots because, necessarily $$AB=BA$$.
You showed that $$A$$ has at least $$2^n$$ square roots. Assume $$B^2 = A$$. Since $$\sigma(B)^2 =\sigma(B^2) = \sigma(A)$$, it follows that $$\sigma(B) = \{\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n}\}$$ for some $$\varepsilon_1, \ldots, \varepsilon_n \in \{-1,1\}$$ where $$\sigma(A) = \{\lambda_1, \ldots \lambda_n\}$$. Set $$D = \operatorname{diag}(\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n})$$. $$A$$ has $$n$$ distinct eigenvalues so it is diagonalizable, i.e. there exists an invertible matrix $$P$$ such that $$P^{-1}AP = \operatorname{diag}(\lambda_1, \ldots, \lambda_n) = D^2$$.
Assume that $$C^2 = A$$ and that $$\sigma(C) = \sigma(B)$$. $$B$$ and $$C$$ both have $$n$$ distinct eingenvalues so there exist invertible matrices $$Q,R$$ such that $$Q^{-1}BQ = D = R^{-1}CR$$ It follows $$Q^{-1}AQ = Q^{-1}B^2Q = (Q^{-1}BQ)^2 = D^2 = (R^{-1}CR)^2 = R^{-1}C^2R = R^{-1}AR$$
If $$e_1, \ldots, e_n$$ is the standard basis, it follows that $$A(Pe_i) = \lambda_iPe_i, A(Qe_i) = \lambda_iQe_i, A(Re_i) = \lambda_iRe_i$$.
Since $$\lambda_1, \ldots, \lambda_n$$ are all distinct, the eigenvectors are unique up to scalar multiplication so it follows that $$Qe_i = \sigma_i Pe_i, Re_i = \pi_i Pe_i$$ for some nonzero scalars $$\sigma_i, \pi_i$$, or $$Q = P\Sigma, R = P\Pi$$ for some invertible diagonal matrices $$\Sigma, \Pi$$.
Finally, $$B = QDQ^{-1} = (P\Sigma)D(P\Sigma)^{-1} = P(\Sigma D\Sigma^{-1})P^{-1} =\\ PDP^{-1} = P(\Pi D\Pi^{-1})P^{-1} = (P\Pi)D(P\Pi)^{-1} = R^{-1}DR = C$$
Hence $$B$$ is uniquely determined by its spectrum $$\sigma(B)$$, which is uniquely defined by choosing $$\varepsilon_i \in \{-1,1\}$$ which can be done it $$2^n$$ ways. We conclude that there are $$2^n$$ such matrices $$B$$. | 2020-12-03T23:49:25 | {
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http://hidrobiologie.granturi.ubbcluj.ro/uty/euclidean-norm-of-vector.html | The L2 norm calculates the distance of the vector coordinate from the origin of the vector space. As such, it is also known as the Euclidean norm as it is calculated as the Euclidean distance from the origin. The result is a positive distance value. It can help in calculating the Euclidean Distance between two coordinates, as shown below. There can be performance gain due to optimization. Norm(x) is the Euclidean length of a vecor x; same as Norm(x, 2). Chapter 8 Euclidean Space and Metric Spaces 8.1 Structures on Euclidean Space 8.1.1 Vector and Metric Spaces The set K n of n -tuples x = ( x 1;x 2:::;xn) can be made into a vector space by introducing the standard operations of addition and scalar multiplication 2. To take this point home, let’s construct a vector that is almost evenly distant in our euclidean space, but where the cosine similarity is much lower (because the angle is larger): The "-norm" (denoted with an uppercase … On R2 let jjjjbe the usual Euclidean norm and set jj(x;y)jj0= max(jxj;jyj). The squared Euclidean norm is widely used in machine learning partly because it can be calculated with the vector operation xᵀx. Example 1 (Euclidean norm on IR2). In simple terms, Euclidean distance is the shortest between the 2 points irrespective of the dimensions. 2-Norm. Python Code The norm is a scalar value. Input array. The L2 norm or Euclidean norm of an array is calculated using the following formula: Note that we will use the default value for the ord parameter for most of our code examples. I haven't found the equivalent to norm(v) from MATLAB. linalg import norm #define two vectors a = np.array([2, 6, 7, 7, 5, 13, 14, 17, 11, 8]) b = … There is nothing special about the Euclidean norm. Vector norms are any function that fulfil the following criteria: 1. JonnyJohnson May 29, 2013, 6:21am #1. euclidean_distances (X, Y = None, *, Y_norm_squared = None, squared = False, X_norm_squared = None) [source] ¶ Compute the distance matrix between each pair from a vector array X and Y. Its documentation and behavior may be incorrect, and it is no longer actively maintained. The 2-norm of a vector is also known as Euclidean distance or length and is usually denoted by L 2. Moreover, this equals zero only when both x = 0 and y = 0. The L2 norm is calculated as the square root of the sum of the squared vector values. There can be performance gain due to the optimization See here and here for more details. In order for a matrix norm to be consistent with the linear operator norm, you need to be able to say the following: norm() is a vector-valued function which computes the length of the vector. Let’s discuss a few ways to find Euclidean distance by NumPy library. As a measure One of the most useful features of orthonormal bases is that they a↵ord a very simple method for computing the coordinates of a vector over any basis vector. Transformational function Syntax: Definition 8. The -norm is also known as the Euclidean norm.However, this terminology is not recommended since it may cause confusion with the Frobenius norm (a matrix norm) is also sometimes called the Euclidean norm.The -norm of a vector is implemented in the Wolfram Language as Norm[m, 2], or more simply as Norm[m].. Euclidean distance is the L2 norm of a vector (sometimes known as the Euclidean norm ) and by default, the norm() function uses L2 - the ord parameter is set to 2. Transformational function Syntax: It is, also, known as Euclidean norm, Euclidean metric, L2 norm, L2 metric and Pythagorean metric. Matrix Norms • If A is a Matrix and p is included in the calling sequence, p must be one of 1, 2, infinity, Frobenius, or Euclidean. Search all packages and functions. PROBLEM 1{5. Building on @GonzaloMedina's answer, I suggest you create a macro called \norm in the document's preamble, using either of the following two approaches: auto-size the double-bar "fence" symbols: \newcommand{\norm}[1]{\left\lVert #1 \right\rVert} This will place double vertical bars around the command's argument. Vote. Euclidean Norm. The euclidean norm of a matrix considered as a vector in m2-space is a matrix norm that is consistent with the euclidean vector norm. It is defined as the root of the sum of the squares of the components of the vector. So, for our given vector x, the L² norm would be: Answer (1 of 2): The Euclidean Norm is our usual notion of distance applied to an n-dimensional space. The numpy module can be used to find the required distance when the coordinates are in the form of an array. Vector Norms: a. Vector Norms Given vectors x and y of length one, which are simply scalars xand y, the most natural notion of distance between xand yis obtained from the absolute value; we de ne the distance to be jx yj. ‖ is called a normed vector space. For efficiency reasons, the euclidean distance between a pair of row vector x and y is computed as: Is there a block that finds the norm of a vector in simulink? It can be calculated by numpy.linalg.euc(). Then the first-order norm normalization method is employed to achieve vector orthogonality. Answer for Euclidean length of a vector (k-norm) with scaling to avoid destructive underflow and overflow is. We calculated the Euclidean norm of this vector with the norm() command by simply type the variable ‘a’ inside the norm(). These vectors are usually denoted ˆ→s Let’s say we have a vector, . In this norm, all the components of the vector are weighted equally. In 2-D complex plane, the norm of a complex number is its modulus , its Euclidean distance to the origin. Euclidean space 5 PROBLEM 1{4. Another familiar norm would be the Euclidean norm for vectors x ∈ IR2. Roughly right. The L2 norm, represented as ||v||2 is calculated as the square root of the sum of the squared vector values.Clearly, the norm is a calculation of … Matrix 2 … Let us instantiate the definition of the vector $$p$$ norm for the case where $$p=2 \text{,}$$ giving us a matrix norm induced by the vector 2-norm or Euclidean norm: Definition 1.3.5.1 . In this article to find the Euclidean distance, we will use the NumPy library. The Euclidean norm of a vector \vecu of coordinates (x, y) in the 2-dimensional Euclidean space, can be defined as its length (or magnitude) and is calculated as follows : norm(vecu) = sqrt(x^2+y^2) The norm (or length) of a vector \vecu of coordinates (x, y, z) in the 3-dimensional Euclidean space is defined by: The relationhip between the norm of a vector and the Euclidean distance between two vectors appears in several machine learning scenarios. Calculates the Euclidean vector norm (L_2 norm) of ARRAY along dimension DIM.Standard:. So every inner product space inherits the Euclidean norm and becomes a metric space. The two-norm of a vector in ℝ 3 vector = {1, 2, 3}; magnitude = Norm [vector, 2] √14 Norm [vector] == Norm [vector, 2] True Another important example of matrix norms is given by the norm induced by a vector norm. Gives the largest magnitude among each element of a vector. 8.207 NORM2 — Euclidean vector norms Description:. Working in a Euclidean plane, he made equipollent any pair of line segments of the same length and orientati… In this article to find the Euclidean distance, we will use the NumPy library. “The L2 norm of a vector can be calculated in NumPy using the norm() function with a parameter to specify the norm order, in this case 1.” Also, even though, not something I would do while programming in the real world, the ‘l” in l1, l2, might be better represented with capital letters L1, L2 for the python programming examples. Vote. Norms are Use the NumPy Module to Find the Euclidean Distance Between Two Points. In Euclidean space the length of a vector, or equivalently the distance between a point and the origin, is its norm, and just as in R, the distance between two points is the norm of their di erence: De nitions: The Euclidean norm of an element x2Rn is the number kxk:= q x2 1 + x2 2 + + x2 n: The Euclidean distance between two points x;x0 2Rn is Many equivalent symbols Now also note that the symbol for the L2 norm is not always the same. We recognize them as. De nition 2 (Norm) Let V, ( ; ) be a inner product space. Norm An inner product space induces a norm, that is, a notion of length of a vector. The most common norm, calculated by summing the squares of all coordinates and taking the square root. Chapter 8 Euclidean Space and Metric Spaces 8.1 Structures on Euclidean Space 8.1.1 Vector and Metric Spaces The set K n of n -tuples x = ( x 1;x 2:::;xn) can be made into a vector space by introducing the standard operations of addition and scalar multiplication The matrix 2-norm is the maximum 2-norm of m.v for all unit vectors v: This is also equal to the largest singular value of : The Frobenius norm is the same as … This library used for manipulating multidimensional array in a very efficient way. The Euclidean norm (two norm) for matrices is a natural norm to use, but it has the disadvantage of requiring more computation time than the other norms. For each and in , ∑ {∑ } {∑ } | | | | Remark: | | | | Definition. Consider the vector hx,yi ∈ IR2. I need to calculate the two image distance value. Given a Euclidean space E, any two vectors u,v 2 E are orthogonal i↵ ku+vk2 = kuk2 +kvk2. Note that the answer of Dznrm2 is a real value. If it overflows, then you find a large power of two M, divide all numbers by M, calculate the norm, and multiply by M. If overflow happens at 2 1023, then you have numbers greater than 2 500. In 2-D complex plane, the norm of a complex number is its modulus , its Euclidean distance to the origin. In this tutorial, we looked at different ways to calculate vector lengths or magnitudes, called the vector norms. ⋮ . The infinity norm of a matrix is the maximum row sum, and the 1-norm is the maximum column sum, all after taking absolute values. State-of-the-art dual-frame phase recovery techniques are evaluated, and it shows that the first-order norm normalization method outperforms the second-order norm normalization method. Definition 8. Euclidean length of a vector with no scaling: The squared Euclidean norm is widely used in machine learning partly because it can be calculated with the vector operation $\bs{x}^\text{T}\bs{x}$. In L-infinity norm, only the largest element has any effect. 0. In 1835, Giusto Bellavitis abstracted the basic idea when he established the concept of equipollence. Euclidean distance = √ Σ(A i-B i) 2. You can first calculate the sum of squares. The length of a vector is most commonly measured by the "square root of the sum of the squares of the elements," also known as the Euclidean norm. 1. In N-D space (), the norm of a vector can be defined as its Euclidean distance to the origin of the space. Example 1.2. When np.linalg.norm() is called on an array-like input without any additional arguments, the default behavior is to compute the L2 norm on a flattened view of the array.This is the square root of the sum of squared elements and can be interpreted as the length of the vector in Euclidean space.. Some, but not all, norms are based on inner products. CUDA Programming and Performance. The L² norm measures the shortest distance from the origin. The length of a vector is a nonnegative number that describes the extent of the vector in space, and is sometimes referred to as the vector’s magnitude or the norm. The Euclidean norm is the square root of the sum of the squares of the magnitudes in each dimension. 2. Another important example of matrix norms is given by the norm induced by a vector norm. The numpy norm of a vector or matrix is the maximum absolute value of all its components. In Euclidean spaces, a vector is a geometrical object that possesses both a magnitude and a direction defined in terms of the dot product. >> a = [4 1 5] b = norm(a) c = 3*b a = 4 1 5 b = 6.4807 c = 19.4422 >> For example, we created a vector that has three elements called ‘a’ as shown above in Matlab®. It might turn out to be quite helpful to recall the basic knowledge about In mathematics, a norm is a function from a real or complex vector space to the nonnegative real numbers that behaves in certain ways like the distance from the origin: it commutes with scaling, obeys a form of the triangle inequality, and is zero only at the origin.In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may … A matrix norm defined in this way is said to be vector-bound'' to the given vector norm. Returns the Euclidean norm of the vector as a double. n o r m o f V e c t o r L 1 = n ∑ i = 1 | x i | L 2 = √ n ∑ i = 1 x 2 i L ∞ = m a x ( | x i | ) n o r m o f V e c … calculate the L2 norm that is calculated as the square root of the sum of the squared vector values. As a measure This is also called the spectral norm This is called the Frobenius norm, and it is a matrix norm compatible with the Euclidean vector norm. However, the two norm is compatible with the Frobenius norm, so when computation time is an issue, the Frobenius norm should be used instead of the two norm. Since the ravel() method flattens an array without making any copies and … (This proves the theorem which states that the medians of a triangle are concurrent.) This is perhaps the matrix norm that occurs most frequently in the literature. See here and here for more details. We will not define it … On R2 let jjjjbe the usual Euclidean norm and set jj(x;y)jj0= max(jxj;jyj). The norm (and therefore the inner product) measures the length, size, magnitude, or strength of the vector depending on what interpretation you are giving the vector. In the infinite-dimensional case, the sum is infinite or is replaced with an integral when the number of dimensions is uncountable. That is, the number of non-zero elements in a vector. The Euclidean norm of a vector measures the “length” or “size” of the vector. 1. If you were to set the ord parameter to some other value p, you'd calculate other p-norms. 1 Norms and Vector Spaces 2008.10.07.01 The induced 2-norm. — Page 112, No Bullshit Guide To Linear Algebra, 2017. This is the Euclidean norm which is used throughout this section to denote the length of a vector. (It would be more precise to use rather than here but the surface of a sphere in finite-dimensional space is a compact set, so the supremum is attained, and the maximum is correct.) Calculate euclidean norm of a vector. In the triangle depicted above let L1 be the line determined by x and the midpoint 1 2 (y + z), and L2 the line determined by y and the midpoint 12 (x + z).Show that the intersection L1 \L2 of these lines is the centroid. Properties of Euclidean distance are: There is an unique path between two points whose length is equal to Euclidean distance Norm type, specified as 2 (default), a different positive integer scalar, Inf, or -Inf.The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in the table. Euclidean Norm of a vector. The L² norm of a single vector is equivalent to the Euclidean distance from that point to the origin, and the L² norm of the difference between two vectors is equivalent to the Euclidean distance between the two points. L² Norm / Euclidean Norm. I have the two image values G=[1x72] and G1 = [1x72]. The L2 norm of a vector can be calculated in NumPy using the norm() function with default parameters.First, a 1×3 vector is defined, then the L2 norm of the vector is calculated.. What is L2 norm squared? computes the euclidean norm of vector containing double-complex elements NRM2 = sqrt ( X**H * X ) Parameters: N ( int [in]) – Number of elements in vector X. NRM2 ( pyopencl.Buffer [out]) – Buffer object that will contain the NRM2 value. In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may also be defined as the square root of the inner product of a vector with itself.
euclidean norm of vector | 2023-01-28T23:35:59 | {
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http://math.stackexchange.com/questions/299405/in-how-many-ways-can-the-students-answer-a-10-question-true-false-examination | # In how many ways can the students answer a 10-question true false examination?
(a) In how many ways can the students answer a 10-question true false examination?
(b) In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer
For part (a) I've got the answer, it is $2^{10}$.
For part (b) I think the answer is $10 \times 2^9$ because the number of ways to choose the question to answer is 10 and in each selection the number of ways to answer the question is $2^9$ but the answer provided in the book is $3^{10}$.
Can someone explain to me?
-
Probably the book suggest that the student can leave as many questions unanswered as he wants. – barto Feb 10 '13 at 14:43
@barto. I agree. On (b) she/he has 3 possibilities to each question. So $3^{10}$. – Sigur Feb 10 '13 at 14:44
For your response to part b you should have $10 \times 2^9 + 2^{10}$ since "it is possible to leave a question unanswered" does not require an unanswer. But it is more likely to mean there can be any number of unanswers from 0 through to 10. – Henry Feb 10 '13 at 15:08
If the answer has to be $3^{10}$, then this means that in case (b) it is intended that for each question the student can choose 1`out of 3 possibilites: true, false, not telling.
-
In part b), for simplicity, let's reduce the problem to just two questions.
There are $2^2$ ways in which both questions may be answered true/false.
If a student does not answer question 1, there are still $2^1$ ways in which they can answer question 2, and vice versa. Thus there are $2\times 2^1$ ways in which only one question is answered.
Finally, there is just one way in which neither question is answered.
Putting this together, there are $2^2 + 2\times 2^1 + 1 = (2+1)^2 = 3^2$ ways of answering the questions.
Extending this to ten questions, there are $2^{10}$ ways to answer all ten questions, $10\times 2^9$ for answering all but one question, $45 \times 2^8$ ways of answering all but two questions, etc, giving a total of $(2+1)^{10} = 3^{10}$.
- | 2015-04-28T00:35:54 | {
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https://math.stackexchange.com/questions/3882210/lemma-used-to-prove-lefthk-right-frac-lefth-right-leftk-right-lefth | # Lemma used to prove $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$
Given a group $$G$$ and $$H,K \le G$$,then :
$$\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$
Where $$HK:=\left\{hk:h \in H ,k \in K\right\}$$
Lemma:
For $$h_1,h_2 \in H$$
$$hK=h'K \iff h(H \cap K)=h'(H \cap K)$$
We have:
$$HK=\bigcup_{h \in H}hK$$
Not every such left cosets of $$K$$ in $$H$$ are distinct, on the other hand the function $$\phi:hK \to K$$ with $$hk \mapsto k$$ is a bijection, so the number of elements in $$hK$$ is the same as that $$K$$'s , here I showed that the set of left cosets (equivalently right cosets) partitions the group.
By this we see that :$$\left|HK\right|=\left|\color{blue}{\text{the set consiting of all distinct left cosets }}hK\right|\left|K\right|$$
One concludes from the lemma that the number of such distinct left cosets is the same as $$\left|H: H \cap K\right|$$ but I don't know how such a conclusion is possible, how the lemma helps us?
It looks that $$hK \ne h^{'} K$$ iff $$h(H \cap K) \ne h^{'}(H \cap K)$$ and the order of the set of all such distinct $$h(H \cap K)$$ for $$h \in H$$ is $$\left|H: H \cap K\right|$$...
Also, it would be appreciated if someone gives me an example where such left cosets $$hk$$ are identical.
• For finding example identical left cosets look at monogen groups and its sub-groups.
– EDX
Oct 26 '20 at 18:34
• The formula only makes sense when the quantities are finite. But the equality $|HK||H\cap K| = |H||K|$ holds in the sense of cardinalities in all cases, so it should be used instead. Oct 26 '20 at 18:42
Consider the map $$\varphi: H/H\cap K\longrightarrow HK/K$$ by $$h(H\cap K)\mapsto hK$$.
1. This is a well defined map by your lemma $$\impliedby$$.
2. This map is injective by your lemma $$\implies$$.
3. This map is surjective by definition of $$HK$$.
Therefore this is a natural one-to-one correspondence between these cosets, and the Product Formula follows immediately.
I happen to have written about this yesterday, so here's a link for you https://ml868.user.srcf.net/ExpositoryWritings/Groups3.pdf. There are a few typos I haven't fixed but I hope it is readable and somewhat inspiring.
You noted that in the union $$\bigcup_{h \in H} hK$$, some cosets appear more than once. If you are able to show that each distinct coset appears $$|H \cap K|$$ times in the union, then you can arrive at the desired conclusion.
The lemma implies that the only way $$hK=h'K$$ can happen (for $$h,h' \in H$$) is if $$h' = gh$$ for some $$g \in H \cap K$$. In particular, for a given coset $$hK$$, it appears in the union $$|H \cap K|$$ times as $$(gh)K$$ for each $$g \in H \cap K$$.
For simplicity: $$I=\{hK|h\in H\}$$ $$J=\{h(H\cap K)|h\in H\}$$ Notice that: $$|J|=|H:(H\cap K)|$$ And we have simply to prove that $$|I|=|J|$$ Thanks to the lemma the application: $$\omega: I \to J$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hK \mapsto h(H\cap K)$$ Is a bijection, in fact firstly the application is well defined since: $$hK=h'K \Rightarrow^{\text{Lemma}} h(H\cap K)=h'(H\cap K)\Rightarrow \omega(hK)=\omega(h'K)$$ The application is also injective: $$\omega(hK)=\omega(h'K)\Rightarrow h(H\cap K)=h'(H\cap K)\Rightarrow^{\text{Lemma}} hK=h'K$$ And it's clearly surjective because for every $$h(H\cap K)\in J, \omega (hK)=h(H\cap K)$$ It follows $$|I|=|J|$$.
• Should not the function be defined as $\phi :I \to J$? and how do we know such a function is bijective? Oct 26 '20 at 18:41
• It's a matter of notation. Briefly this application associates to every $hK$, $h(H\cap K)$, Now I'll write the proof that the application is a bijection. Oct 26 '20 at 18:43
The equivalence relation $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$ induces a partition of $$H\times K$$ into equivalence classes each of cardinality $$|H\cap K|$$, and the quotient set $$(H\times K)/\sim$$ has cardinality $$|HK|$$. Therefore, $$|H\times K|=|H||K|=|H\cap K| |HK|$$, whence (if $$H$$ and $$K$$ are finite, in particular if they are subgroups of a finite group) the formula in the OP. Hereafter the details.
(Note that the formula holds irrespective of $$HK$$ being a subgroup.)
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 $$ab=f([(a,b)]_\sim)$$. $$\space\space\Box$$
Finally, the formula holds irrespective of $$HK$$ being a subgroup, which was never used in the proof. | 2021-12-04T17:15:16 | {
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https://www.physicsforums.com/threads/linear-algebra-question.665566/ | # Homework Help: Linear algebra question
1. Jan 20, 2013
### Mdhiggenz
1. The problem statement, all variables and given/known data
Explain why each of the following algebraic rules will not work in general when the real numbers a and b are placed by nxn matrix A and B
(a+b)2=a2+2ab+b2
This question is a bit confusing to me and I have no idea how to even start this problem
2. Relevant equations
3. The attempt at a solution
2. Jan 20, 2013
### Ray Vickson
Try a couple of small examples: let A and B be some specific 2x2 matrices. Compute both sides and see if you get equality or not.
3. Jan 20, 2013
### Mdhiggenz
So lets say if I make A the identity matrix, and B some random matrix. I would computer one side (a+b)2 where a=A and b=B?
and do the same for the other side?
Thanks
4. Jan 20, 2013
### Ray Vickson
Don't pick one of them to be the identity; pick two different matrices, neither one and identity matrix. (Try it both ways to see why; that is, try it first when one is the identity, and then do it again with two non-identity matrices. Look at what happens, and that will give you a clue for solving the problem.)
5. Jan 20, 2013
### LCKurtz
Two matrices I wouldn't ever use for trying to find a counterexample would be the Identity and Zero matrices. They have too many special properties of their own.
6. Jan 20, 2013
### Mdhiggenz
Lckurtz: thanks for the tip. You think you can give me a matrix that you would always use to prove, something simple.
And Ray here is my work, would you agree that the reason they could not be equal is due to the 2ab term?
7. Jan 20, 2013
### SammyS
Staff Emeritus
In particular compare AB with BA .
8. Jan 20, 2013
### Mdhiggenz
So what I did was incorrect?
9. Jan 20, 2013
### SammyS
Staff Emeritus
What you did was fine, and yes, it is the 2AB that messes you up.
Did you compare AB with BA ? --- to see why it's the 2AB that messes you up ?
10. Jan 20, 2013
### LCKurtz
No, there isn't a single special matrix. In the given example, as I think you are now aware, the point is that matrices don't generally satisfy $AB=BA$ so you don't get $2AB$ in the expansion when you multiply them out. So if you are looking for a counterexample, you don't want to choose, for example, A = the zero matrix, because it commutes with everything just because it always gives all zeroes. That is a general idea: If you are trying to prove some identity doesn't always work, look for an example by choosing variables that are unlikely to "accidentally" work because they have special properties. So if I were looking for examples in the current case, I wouldn't use the zero matrix, the identity matrix, a lower triangular matrix or a symmetric matrix. I would try a more "random" matrix that doesn't have any special properties. Does that make sense to you?
11. Jan 20, 2013
### Mdhiggenz
I did AB and compared it to BA they are different. This I understand, what I don't understand is what that has to do with 2AB?
Or is what your saying is that in order to show that something is 100% true we must look at both cases. For instance A2 no matter what will be A2. Same thing goes with B, and A+B will be the same as B+A. However 2AB does not equal 2BA therefore it cant possibly be equal.
12. Jan 20, 2013
### Mdhiggenz
Absolutely thanks for the clear explanation.
13. Jan 20, 2013
### LCKurtz
Re. your comment in red above. Expand $(A+B)(A+B)$ symbolically the long way (distributive law) being careful about the order. What do you get?
14. Jan 20, 2013
### Mdhiggenz
Not quite sure what you mean, it says math processing error
15. Jan 20, 2013
### LCKurtz
Do it just by manipulating the A and B's with a pencil and paper. Expand (A+B)(A+B).
16. Jan 20, 2013
### Mdhiggenz
Oh I see now A2+AB+BA+B2.
So making AB+BA=2AB that is implying that AB=BA which means you would be able to add them and get 2AB which is incorrect because AB does not necessarily equal BA.
?
17. Jan 20, 2013
### vela
Staff Emeritus
Yup, that's right.
Multiplication of real numbers has a property called commutativity, which means the order you multiply two numbers in doesn't matter, e.g. $2\times 4 = 4\times 2$. Hence, if a and b are real numbers, you can always say that (a+b)2 equals a2+2ab+b2.
Matrix multiplication doesn't have this property, as you found. When you're multiplying matrices, you have to remember that the order in which they appear is important. You can't just move them around like you are used to from working with real numbers.
18. Jan 23, 2013
### EM_Guy
It is worth considering all of this in light of vector spaces. Scalars (aka - 1-vectors, aka - 1x1 matrices) don't exist in the same vector spaces in which N x N matrices exist (where N does not equal 1). Addition and scalar multiplication are defined distinctly in each distinct vector space. Matrix multiplication is something else altogether. So, when studying linear algebra, you can't just assume that "plus" means "plus" or that "times" means "times." You can't even assume that "zero" means "zero." The addition operation is defined differently for each vector space. Thus, the idea of adding a 3x1 vector to a 2x2 matrix together doesn't make sense, because those two elements exist in different vector spaces; therefore, "addition" between two such elements is not even defined. Likewise, if v exists in a vector space V, then 0v = 0 is true, but the first 0 and the second 0 are not the same. The first 0 is a scalar. The second 0 is the null vector that exists in the vector space V.
Again, matrix multiplication is very different than the multiplication between two scalars, and it is also different than scalar multiplication in any given vector space. Matrix multiplication actually maps matrices from one vector space to another vector space (or to the same vector space when you have an NxN matrix multiplied by another NxN matrix). It should therefore not surprise us that matrix multiplication is not commutative.
Here's a challenging question: Why do you suppose that matrix multiplication is defined as it is? That is, what was the motivating reason to define matrix multiplication the way it has been defined? But perhaps that is a question for a different thread. | 2018-07-23T10:39:15 | {
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https://math.stackexchange.com/questions/545821/geometry-notation-what-does-m-angle-abc-mean | # Geometry notation: what does $m\angle ABC$ mean?
I see in some math formulation that a certain angle is called, let's say
$$\angle ABC$$
but there is a letter put in front of the angle notation.
$$m\angle ABC$$
What does the $m$ represent here? A factor?
• Try to use LaTeX to write accurately mathematics here. You also didn't write what the angle is called, "let say..." – DonAntonio Oct 30 '13 at 17:23
• <ABC and m<ABC.. I dont know why it doesnt show up in the question.. – Hilbert Oct 30 '13 at 17:24
• Do you mean $\theta(x)$? – Don Larynx Oct 30 '13 at 17:24
• You're going to give more context, for example to write down a complete exercise. The letter $\;m\;$ is many times reserved for slope in analytic geometry. – DonAntonio Oct 30 '13 at 17:25
• It's because the < hides the subsequent text when the post is rendered. I have fixed it. – Cameron Buie Oct 30 '13 at 17:26
$\angle ABC$ : The angle ABC
$m\angle ABC$: The measure of $\angle ABC$
So, when $\angle ABC \cong \angle DFG$ , that means, $m\angle ABC = m\angle DFG$
The notation $m\angle ABC$ typically denotes "the measure of angle $ABC$."
• To clarify, the $m$ is sometimes used to distinguish between the measure of the angle ($m\angle ABC$ = a number, in degrees/radians) and the actual angle itself (the geometric object $\angle ABC$). – angryavian Oct 30 '13 at 17:29
• Thanks for all your replies and Cameron has understood the badly formulated question correct. But if the angle is, lets say <ABC = 50 degrees. What is the need of m<ABC then? Doesnt <ABC already denote the measure of the angle? – Hilbert Oct 30 '13 at 17:31
• @Hilbert No, $\angle ABC$ is used to denote the angle itself, as a geometric object. Of course, this is a minor distinction, and some authors may choose to be lax in the notation. – angryavian Oct 30 '13 at 17:36
• @Hilbert: As blf points out, $\angle ABC$ denotes the angle, itself, while $m\angle ABC$ is its measure. For example, suppose we have an equilateral triangle with vertices $A,B,C.$ Then $\angle ABC$ occurs at the intersection of the segments $AB$ and $BC$, while $m\angle ABC$ is the measure of that angle. The distinction isn't always important, but sometimes it is. – Cameron Buie Oct 30 '13 at 17:36
• Thanks Cameron & blf.. Perfectly explained. I have been searching for geometry textbooks that explains things like this and deals with elementary things like Congruence, Arc, Geometric shapes like triangles and so on. Any recommendations? – Hilbert Oct 30 '13 at 19:33
∠ABC refers to the physical angle itself while m∠ABC refers to its measure. There isn't any difference between these two and it is absolutely correct to use any of them...Some people just want to formalize things; these people think-"I have a car that is priced at 50000dollars, so i am driving a car and not 50000dollars" and so want to differentiate between little everything. | 2019-01-22T18:22:35 | {
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https://byjus.com/question-answer/from-a-two-digit-number-n-we-subtract-the-number-with-the-digits-reversed-and/ | Question
# From a two digit number N, we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:
A
N cannot end in 5
B
N can end in any digit other than 5
C
N does not exist
D
there are exactly 7 values for N
E
there are exactly 10 values for N
Solution
## The correct option is D there are exactly 7 values for NLet the two digit number be 10a+bN = 10a+bN' = reversed number = 10b+aN-N' = 9(a-b)N-N' is positive perfect cube.$$\therefore$$ a>bFor 9(a-b) be perfect cube, a-b = 3$$\therefore$$ b $$\epsilon$$ [0, 6] $$\longrightarrow$$ 7 values$$\therefore$$ a $$\epsilon$$ [3, 9] $$\longrightarrow$$ 7 values$$\therefore$$ Total 7 values are possible for N.Mathematics
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https://ecom.ablele.net/hqi53g6/ss9ng.php?c394cf=set-operations-proofs | # set operations proofs
\] The empty set, $$\emptyset$$, is the set with no elements: ... Logicians sometimes describe ordinary mathematical proofs as informal, in contrast to the formal proofs in natural deduction. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined But from the definition of set difference, we see that though they can be proven also using some of these properties (after those properties are proven, needless to say). Theorem: For any sets, $$\overline{A\cup B}= \overline{A} \cap \overline{B}$$. 13. . Then the set of students taking classes this semester is When writing informal proofs, the focus is on readability. Hence . Proof for 11: Let x be an arbitrary element in the universe. The set $$\overline{B}$$ is the set of all values not in $$B$$. Theorem: $$A-(B\cup C)= (A-B)\cap(A-C)$$. but also for others. &= \{x\mid x\in A \wedge x\in \overline{B}\} \\ when we're working with real numbers, probably $$U=\mathbf{R}$$. Back to Schedule We can use the set identities to prove other facts about sets. Hence . For example, (b) can be proven as follows: A &= \{x\mid x \in (\overline{A}\cap\overline{B}) )\} \\ If , then and . 1 - 6 directly correspond (See section 2.2 example 10 for that. x A A The standard notation for irrational numbers should now make a lot of sense: with universal set $$\mathbf{R}$$, the irrationals ($$\overline{\mathbf{Q}}$$) are the complement of the rationals ($$\mathbf{Q}$$). x\in S \wedge x\notin{S}\,. x Back to Table of Contents. A $A\cup B\cup C \cup D\,,\\A\cap B\cap C \cap D\,.$. A to identities &= \{x\mid x\in (A \cap \overline{B})\} \\ If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. B B, by 7 6. and that of ------- Identity Laws 5. is also in This name is used since the basic method is to choose an arbitrary element from one set and “chase it” until you prove it must be in another set. \overline{A\cup B} \] Alternative proof For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). First by 15 by the commutativity of 4. ( cf. ) by the definition of set union. Proof for 4: and Proof for 12: (a) ? Proof for 8: (a) If then . The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. &= \overline{A}\cap\overline{B}\,.\quad{}∎ ) by the definition of ( B - A ) . Since , since . A Since (use "addition" rule), and between follows. The properties 1 6 , and 11 Thus, A−B = A∩Bc. By definition of set difference, x ∈ A− B. by the definition of . ( B - A ) 2. 12. if and only if and Proof for 13: Since , . &= \{x\mid \neg(x \in A)\wedge \neg(x\in B )\} \\ Properties of Set Operation Subjects to be Learned . 9. Theorem For any sets A and B, A∩B ⊆ A. B Often not explicitly defined, but implicit based on the problem we're looking at. ( B Since we're doing the same manipulations, we ended up with the same tables. A-B ------- Domination Laws $\sum_{i=1}^{n} i^2\,.$. and vice versa. if and only if We are going to prove this by showing that every element that is in by 1. Let the sets $$S_1,S_2,\ldots ,S_n$$ be the students in each course. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. = A by 3, For example, suppose there are $$n$$ courses being offered at ZJU this semester. Like the domain for quantifiers, it's the set of all possible values we're working with. Hence . Next -- Recursive Definition Notice the similarity between the corresponding set and logical operators: $$\vee,\cup$$ and $$\wedge,\cap$$ and $$\overline{\mbox{S}},\neg$$. B ------- Distributive Laws We have used the choose-an-element method to prove Propositions 5.7, 5.11, and 5.14. \overline{\mathbf{Q}} = \mathbf{R}-\mathbf{Q} \,.\]. Look familiar? Then there must be an element $$x$$ with $$x\in(A-B)$$, but $$x\notin A$$. by the definition \begin{align*} 1. Then by the definition of the operators, Hence A satisfies the conditions for the complement of . The “more formal” version has more steps and leaves out the intuitive reason (that might help you actually remember why). Alternative proof: This proof might give a hint why the equivalences and set identities tables are so similiar. Less Formal Proof: The set $$A-B$$ is the values from $$A$$ with any values from $$B$$ removed. Note here the correspondence of Here is an example. = ( A (See example 10 for an example of that too.). B ) Here the only if part is going to be proven. &= A\cap \overline{B}\cap A\cap \overline{C} \\ Be careful with the other operations. of propositional logic, and 7 - 11 also follow immediately from them as illustrated below. &= \{x\mid x\notin (A\cup B)\} \\ With similar proofs, we could prove these things: When doing set operations we often need to define a. ( cf. ) \[\bigcup_{i=1}^{n} S_i\,. There is no logical version of set difference, or set version of exclusive or (at least as far as we have defined). x \end{align*}. Since A Additional properties: Then there is an element x that is in , i.e. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: A. The if part can be proven similarly. Hence . If and , then , Also since , . the commutativity of Proof: Suppose for contradiction that there is an element $$x\in S\cap\overline{S}$$. x Could have also given a less formal proof. Proof: By definition of the set operations, A These can also be proven using 8, 14, and 15. equalities involving set operations intersection of sets subset relations proofs of equalities proofs of subset relations Contents . Proof for 9: Let x be an arbitrary element in the universe. by the definition of . Set. A. x\in S \wedge x\in\overline{S} \\ \[A\cap B = \{x \mid x\in A\wedge x\in B\}\,\\ Theorem For any sets A and B, B ⊆ A∪ B. Proof… by the distribution Also . A Hence . B . Proof for 6: By the definition of the equality of sets, we need to prove that Let x be an arbitrary element in the universe. Here are some basic subset proofs about set operations. Hence does not hold. Since , . • Applying this to S we get: • x (x S x S) which is trivially True • End of proof Note on equivalence: • Two sets are equal if each is a subset of the other set. Copyright © 2013, Greg Baker. if and only if
Strategy Vs Tactics Marketing, 1966 F100 Engine Swap, Inheriting A House From Your Parents Uk, Bank Of America Mastercard Login, Lola Car Price, Georgia Highlands College Basketball Coach, | 2021-10-21T04:08:52 | {
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https://mathoverflow.net/questions/236818/mahler-measure-of-a-totally-positive-expanding-algebraic-integer | # Mahler measure of a totally positive, expanding algebraic integer
Consider a degree-$d$ algebraic integer $\alpha$ all of whose conjugates (including itself) are real numbers greater than 1. Its Mahler measure $M(\alpha)$ is simply equal to the norm $N(\alpha)$. Since $N(\alpha-1) \geq 1$, an application of $H\ddot{o}lder's$ inequality gives the lower bound $N(\alpha) \geq 2^d$, with equality achieved only by $\alpha = 2$.
Question. Does there exist $C > 2$ such that $N(\alpha) \geq C^d$ for all $\alpha \neq 2$?
• Interesting question. I assume you're familiar with work of Smyth, Flammang, etc. on lower bounds for the Mahler measure of totally positive algebraic integers? (These give a lower bound of $C=1.722...$ for $M(\alpha)^{1/d}$ with finitely many exceptions, but valid for all totally positive integers $\alpha$, not just those with all conjugates $>1$). Apr 20, 2016 at 21:13
• On the hypothesis that $\alpha$ is algebraic of degree $d$, the value $\alpha=2$ can only arise in the case $d=1$, so the equality $N(\alpha)=2^d$ is achieved only for $d=1$, $\alpha=2$. Apr 20, 2016 at 22:58
• @BobbyGrizzard Is the "with finitely many exceptions" part effective? Apr 21, 2016 at 5:50
• @FanZheng yes. Here is one effective result: ams.org/journals/mcom/1996-65-213/S0025-5718-96-00664-3/… edit: and here is the one I was probably looking at yesterday, due to Wu and Mu (Quanwu Mu, not the OP!): sciencedirect.com/science/article/pii/S0022314X12001989 Apr 21, 2016 at 11:58
• @BobbyGrizzard NOT the OP LOL Apr 22, 2016 at 4:19
Suppose that $\alpha$ is totally real algebraic integer, and that all its conjugates are greater than $1$. Suppose also that $\alpha \ne 2$.
Note the elementary inequality for $x \in (1,\infty) \setminus \{2\}$:
$$\ln|x| \ge \frac{\ln|x-1| + \ln|x-2|}{3} + \ln C,$$
where $C = 2^{1/3} 3^{1/2} = 2.18 \ldots > 2$. (Equality holds precisely at $3 \pm \sqrt{3}$.)
Denote the conjugates of $\alpha$ by $\alpha_i$. Assuming that $\alpha \ne 1,2$, we see that, because $\alpha - 1$ and $\alpha - 2$ are non-zero algebraic integers, we have inequalities
$$\sum_{i=1}^{d} \ln | \alpha_i - 1| = \ln N(\alpha -1) \ge 0,$$ $$\sum_{i=1}^{d} \ln |\alpha_i - 2| = \ln N(\alpha -2) \ge 0.$$
Hence we deduce that $\displaystyle{\ln N(\alpha) = \sum_{i=1}^{d} \ln |\alpha_i| \ge d \cdot \ln C}$, and thus $N(\alpha) \ge C^{d}$ where $C > 2$.
Gypsum's argument is really nice. In the same spirit, we have the following inequality: $$\ln|x| \geq \frac{2\ln|x-1| + \ln|x-2|}{5} + \ln\sqrt{5}$$ which is achieved at $x = \frac{5\pm \sqrt{5}}{2}$. This way we obtain the optimal constant $C = \sqrt{5}$. Perhaps some further tweaks can yield even larger $C$ with finitely many exceptions, as in the work of Smyth and Flammang.
I wonder how far this approach can be extended. Smyth showed that for totally positive algebraic integers (whose conjugates are not necessarily greater than 1), their $M(\alpha)^{\frac{1}{d}}$ are dense beyond 1.73, which is very close to the lower bound cited by @BobbyGrizzard. Do we have a similar situation here? As a first step, it would be good to construct infinitely many $\alpha$ that give upper bound to $C$. For example, consider the $n$-th $Chebyshev$ polynomial $T_n(x)$. Then the monic polynomial $(-x)^nT_n(\frac{2}{x}-1)$ has all its roots greater than 1. In this case $N(\alpha) = 2^{2n-1}$, suggesting $C \leq 4$.
• Welcome to mathoverflow! This is really a great question, but as you are new here, please be informed that the answers are just "answers"; new questions included in answers may not get enough attention. If you want to probe further, please post a separate question; you may include a link back to this question. Apr 22, 2016 at 6:25
• One can usually extend this approach a certain amount, but not as far as the limit. The fact that equality is achieved at $x = (5+\sqrt{5})/2$ with the optimal bound is related to the fact that $x - 1$ and $x - 2$ are both units. But there is no reason this had to happen. BTW, to improve the bound beyond this point (with the exceptions $x = 2$ and $(5 \pm \sqrt{5})/2$, you simply need to add a very small multiple of the term $\log(x^2 - 5 x + 5)$. Apr 22, 2016 at 13:11
• @FanZheng Thanks for your advice :) Apr 22, 2016 at 17:33 | 2023-04-01T10:39:56 | {
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https://web2.0calc.com/questions/what-is-a-number-12-24-and-30-have-in-common | +0
# What is a number 12, 24, and 30 have in common?
0
193
4
+13
A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies. What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?
Aug 29, 2018
#1
+374
+4
The GCF for $$30, 24,$$ and $$12$$ is $$6$$. This means that there will be $$6$$ tanks. (Correct me if I'm wrong, I'm not so sure about this one. I may have read it wrong).
- Daisy
Aug 29, 2018
#3
+13
+2
Yep ur right it was a question on my math hw I thought I’d post. Lol
TruppaGirl Aug 29, 2018
#2
+3994
+3
GCF common factor is the way to solve it! The GCF is 6, because it's the greatest number that evenly divided 12,24, and 30.
Aug 29, 2018
#4
+21848
+4
A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies.
What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?
Formula:
$$\begin{array}{|rcll|} \hline \gcd(a,b,c) &=& \gcd( \gcd(a,b), c ) \\ \gcd(a,b ) &=& \gcd (a-b,b ) & a \ge b \\ \gcd(a,a) &=& a & a \ge 0 \\ \gcd(a,b) &=& \gcd(b,a) \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ && \gcd(30,24) \\ && = \gcd(30-24,24) \\ && = \gcd(6,24) \\ && = \gcd(24,6) \\ && = \gcd(24-6,6) \\ && = \gcd(18,6) \\ && = \gcd(18-6,6) \\ && = \gcd(12,6) \\ && = \gcd(12-6,6) \\ && = \gcd(6,6) \\ && = 6 \\\\ \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ &=& \gcd( 6, 12 ) \\ &=& \gcd( 12, 6 ) \\ &=& \gcd( 12-6, 6 ) \\ &=& \gcd( 6, 6 ) \\ &=& 6 \\ \hline \end{array}$$
Aug 30, 2018
edited by heureka Aug 30, 2018 | 2019-03-23T09:44:36 | {
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https://networkx.org/documentation/latest/reference/generated/networkx.generators.lattice.triangular_lattice_graph.html | # triangular_lattice_graph#
triangular_lattice_graph(m, n, periodic=False, with_positions=True, create_using=None)[source]#
Returns the $$m$$ by $$n$$ triangular lattice graph.
The triangular lattice graph is a two-dimensional grid graph in which each square unit has a diagonal edge (each grid unit has a chord).
The returned graph has $$m$$ rows and $$n$$ columns of triangles. Rows and columns include both triangles pointing up and down. Rows form a strip of constant height. Columns form a series of diamond shapes, staggered with the columns on either side. Another way to state the size is that the nodes form a grid of m+1 rows and (n + 1) // 2 columns. The odd row nodes are shifted horizontally relative to the even rows.
Directed graph types have edges pointed up or right.
Positions of nodes are computed by default or with_positions is True. The position of each node (embedded in a euclidean plane) is stored in the graph using equilateral triangles with sidelength 1. The height between rows of nodes is thus $$\sqrt(3)/2$$. Nodes lie in the first quadrant with the node $$(0, 0)$$ at the origin.
Parameters:
mint
The number of rows in the lattice.
nint
The number of columns in the lattice.
periodicbool (default: False)
If True, join the boundary vertices of the grid using periodic boundary conditions. The join between boundaries is the final row and column of triangles. This means there is one row and one column fewer nodes for the periodic lattice. Periodic lattices require m >= 3, n >= 5 and are allowed but misaligned if m or n are odd
with_positionsbool (default: True)
Store the coordinates of each node in the graph node attribute ‘pos’. The coordinates provide a lattice with equilateral triangles. Periodic positions shift the nodes vertically in a nonlinear way so the edges don’t overlap so much.
create_usingNetworkX graph constructor, optional (default=nx.Graph)
Graph type to create. If graph instance, then cleared before populated.
Returns:
NetworkX graph
The m by n triangular lattice graph. | 2023-03-27T17:00:09 | {
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http://rbwx.iueu.pw/how-to-find-total-distance-traveled-by-particle-calculus.html | # How To Find Total Distance Traveled By Particle Calculus
For example, you might find a distance of 10. The distance traveled in each interval is thus 4 times 20, or 80 feet, for a total of 80 + 80 = 160 feet. At time t=2, the position of the particle is x(2)=0. Exercise 1: Calculate the total distance traveled given the velocity equation. 45m if you calculate in 1-minute chunks, 10. So the particle has gone over 10 seconds 12. A particle moves along the x-axis. But no, to find x you have to use physics equations, not curve equations. (a) Find the instantaneous velocity at time t and at t = 3 seconds. The total DISPLACEMENT would be the ∫v (t) from 1 to 6. (b) Find the acceleration of the particle at time t = 1. For each problem, find the maximum speed and times t when this speed occurs, the displacement of the particle, and the distance traveled by the particle over the given interval. You can also find Total distance traveled by a particle - Mathematics ppt and other Engineering Mathematics slides as well. How to Find Total Distance Calculus: Steps. behind the fifth method of approximation called Simpson's Rule. Remember: Velocity is the rate of change in position with respect to time. For example, D 2 and D 3 are =. Include units. However, it was a long time ago with crappy looking graphs. 5? Is the velocity of the particle increasing at time t = 1. A particle moves in a straight line with velocity t^-2 - 1/9 ft/s. Advanced Placement Calculus AB APCD. (d) Find the total distance traveled by the particle from t = 0 to t = 2. 4) v(t) = 3t2 — 18t; ì)ó4anQ. of a particle moving on a horizontal axis is shown below. How do you find the total displacement for the particle whose position at time #t# is given by How many values of t does the particle change direction if a particle moves with acceleration What is the position of a particle at time #t=2# if a particle moves along the x axis so that at. A Dodge Neon and a Mack truck leave an intersection at the same time. A person is standing on top of the Tower of Pisa and throws a ball directly upward with an initial velocity of 96 feet per second. AP Calculus Particle Motion Worksheet For #6 - 10: A particle moves along a line such that its position is s ( t ) = t 4 - 4 t 3. c) Set up an integral to find the total distance traveled by the particle in the interval [0, 4]. EDIT - I made a slight mistake the first time I posted this. Thus, its average speed = distance/time = 2π/3 and its average velocity = displacement/time = 0. To summarize, we see that if velocity is sometimes negative, a moving object's change in position different from its distance traveled. If a body moves along a straight line with velocity v = t 3 + 3t 2, find the distance traveled between t. The Neon heads east at an average speed of 30 mph, while the truck heads south at an average speed of 40 mph. These deriv-atives can be viewed in four ways: physically, numerically, symbolically, and graphically. The velocity function (in meters per second) is given for a particle moving along a line. The Attempt at a Solution I cannot think of a way to do it keeping it in terms of t. 45m if you calculate in 1-minute chunks, 10. Homework Equations Can't think of any 3. 25 t and positive on the interval 1. (a) Find the acceleration of the particle at time t 3. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. Finding the total distance {eq}(\Delta x){/eq} traveled by the particle. What is the velocity after 3 seconds? C. With this information, it's possible to find the distance the object has traveled using the formula d = s avg × t. Find the speed of the particle at time t 3 seconds. 1 Integral as Net Change Calculus. B) Find the total distance travelled by the particle. The Riemann sum approximating total distance traveled is v t k Δt, and we are led to the. Att =1, theparticleisattheorigin. Distance is the measure of “how much ground an object has covered” during its motion while displacement refers to the measure of how far out of place is an object. Find the velocity when t = 3 D. Distance and displacement are different quantities, but they are related. Justify your answer. Given the position function, find the total distance. (e) The displacement of the particle. (b) Find the average velocity of the particle for the time period 06. 5 meters to the right and then 12. When you try to find the distance a moving object has traveled, two pieces of information are vital for making this calculation: its speed (or velocity magnitude) and the time that it has been moving. Please show detailed step-by-step explanations for both parts show more A particle is moving along the x-axis at velocity v(t) = 3t^2 - 12t + 9, 0 ≤ t ≤ 3. Let f(x) = e2x. We have to evaluate this to find the velocity at any particular time. (b) When is the particle at rest? Moving to the right? Moving to the left? Justify your answers. Approximately where does the particle achieve its greatest positive acceleration on the interval 0, b ? t a 16. The particle may be a “particle,” a person, a car, or some other moving object. B) Find all for which the velocity is increasing. Video transcript. (f) Find the displacement of the particle during the first five seconds. When calculating the total distance traveled by the particle, consider the intervals where v(t) ≤ 0 and the intervals where v(t) ≥ 0. Using the result from part (b) and the function V Q from part (c), approximate the distance between particles P and Q at time t = 2. The velocity function is v(t) = - t^2 + 6t - 8 for a particle moving along a line. c) Find the particle's total distance traveled by setting up ONE integral and using your calculator. Chapter 10 Velocity, Acceleration, and Calculus The first derivative of position is velocity, and the second derivative is acceleration. (e) Use geometry to nd the distance traveled to the left. I said 8 seconds instead of 8 feet. To maximize the distance traveled, take the derivative of the coefficient of i with respect to θ and set it equal to zero: d d θ ( v 0 2 sin 2 θ g ) = 0 2 v 0 2 cos 2 θ g = 0 θ = 45 °. ) but you can also compute traveled distance having time and average speed (given in different units of speed mph, kmh, mps yds per second etc. travels to point 2, and reverses direction and travels to point 3, then its distance travelled is 2 + 1 + 1 = 4. In a physics equation, given a constant acceleration and the change in velocity of an object, you can figure out both the time involved and the distance traveled. b) Use your an swer to part (a) to find the position of the particle at time t = 4. In this case, we can use the two triangles in the figure to. Is the speed of the particle. If the graph dips below the x-axis, you’ll need to integrate two or more parts of the graph and add the absolute values. You can read about it in your book if you find yourself just dying of curiousity, but it's not in the AP curriculum. 01s chunks, and 10. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. How many bushels were consumed from the beginning of 1972 to the end of 1973?. Displacement is a vector quantity as it has both magnitude and direction. Solution: The displacement is the net area bounded by v(t), and the total distance traveled is the total area. The Travel Distance Calculator will calculate instantly the total distance you traveled during your trip based on your average speed and the amount of time you traveled. Your acceleration is 26. Since a = DIV = 2t— I is equal to 3 t = 2, the position s of the particle is a relative minimum when t = 2. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). Find the position of the particle at time t = 3. The displacement or net change in the particle's position from t = a to t = b is equal, by the Fundamental Theorem of Calculus (FTC), to. 9: Velocity & Acceleration SOLUTION KEY. Sample Test Questions: A particle moves along a horizontal line and its position at time is. Sometimes it's a particle, sometimes a car, or a rocket. f) Draw a diagram to illustrate the motion of the particle. (c) Find the average velocity of the particle over the interval. 5, you get the total distance. behind the fifth method of approximation called Simpson's Rule. Be sure to label the time, [position, and velocity at each change and at the beginning. are unit vectors in the x and y directions, it is possible to fi nd the position or coordinates of the particle at a given value of t. (a) Use a de nite intergal and the Fundamental Theorem of Calculus to compute the net signed area between the graph of f(x) and the x-axis on the interval [1;4]. How do you find the total displacement for the particle whose position at time #t# is given by How many values of t does the particle change direction if a particle moves with acceleration What is the position of a particle at time #t=2# if a particle moves along the x axis so that at. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign. Find the velocity vector at the time when the particle’s horizontal position is x = 25. To calculate the total distance traveled, integrate the absolute. 0 ms What is the amplitude if the maximum displacement is 26. Let’s say we are given the position of a particle P in three-dimensional Cartesian ( x , y , z ) coordinates, with respect to time, where. Displacement vs Total Distance Traveled Given a Derivative Function; Page 11. zz go HEY Athina. 5 The Substution Rule: Students have trouble with this topic. The velocity of the particle at time t is 6t t2. (a) Find the speed of the particle at time t = 2, and find the acceleration vector of the particle at time t = 2. B) Find the total distance travelled by the particle. Both x and y are measured in meters, and t is measured in seconds. 5 seconds to t = 7 seconds. Video Examples: Acceleration and. The total distance that the car is from its starting location is -32 feet, which means that the car ends up 32 feet behind where it started. Multiply velocity by time to get distance covered in meters (m). ii Find the distance of the particle from the origin at any time t. The distance traveled between times t and t + h is f(t + h) − f(t). The graph below shows the velocity, v, of an object (in meters/sec). What is the total distance traveled by the particle?. b) Find the particle's displaçement for the given time interval. Note that displacement is not the same as distance traveled; while a particle might travel back and forth or in circles, the displacement only represents the difference between the starting and ending position. (d) Find the total distance traveled by the particle during the first 8 seconds. " So begins a number of AP Calculus questions. -24 m (b) Find the distance traveled by the particle during the given time interval. The position of a particle at any time tt0 is given by 233 and. (a) When the particle is at rest. Where is the particle located at the end of the trip (t = 10)? b. We can also calculate force,. AP Calculus Worksheet: Rectilinear Motion 1. The position of the particle at time t is x(t) and its position at time t = 0 is (a) Find the acceleration of the particle at time t = 3. At what is the particle changing direction? Find the total distance traveled by the particle from time t = O to r = 4. miles, yards, meters, kilometers, inches etc. ? The motion of a particle is described by the postion function s = t^3 - 12t^2 + 45t + 3 , when t is greater than or equal to zero. Note that if the car changes direction midway and heads south after five seconds, the distance covered, too, changes. A particle’s velocity is represented by the graph below. It is known that. (a) Graph the function v(t). A particle moves along a horizontal line. Find the velocity at time t. Find an expression that may be used to determine how far the carrier will travel and how long it will take to stop. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Find the speed when. Since the idea of substitution is so important in Calculus II, the instructor. If it did, we wouldn't need calculus at all, we could just read the value for x right off the graph, for any and all curves. D the distance traveled by the truck from t = 3 to t = 15 E The average position of the truck in the interval t = 3 and t = 15. Here's an example: If the position of a particle is given by: x(t)! 1 3 t3 "t2" 3t # 4,. also check your result geometrically. These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum. In physics the average speed of an object is defined as: $$\text{average speed} = \frac{\text{distance traveled}}{\text{time elapsed}}$$. The corresponding average velocity is then 146 3 = 48. What is the total distance traveled by the particle from t - 0 to t = 3? Show Step-by-step Solutions. c) Ifs(0) = 3, what is the particle's final position? d) Find the total distance traveled by the particle. the distance positive. (a) What is the velocity of the particle at t = 0? (b) During what time intervals is the particle moving to the left? (c) What is the total distance traveled by the particle from t = 0 to t= 2?. We can integrate the given velocity function to arrive at the position function. 9: Velocity & Acceleration SOLUTION KEY. In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the t-axis on the given time interval. (b) Find the total distance traveled by the particle from time t 0. Find the intervals on which the velocity is increasing. Find the intervals on which the particle is slowing down. 6) Find all t for which the distance s is increasing. In Exercises 1-5, the function v(t) is the velocity in m/sec of a particle moving long the x-axis. ≤t ≤ (c) Find the total distance traveled by the particle from time t =0 to t =6. This Displacement Calculator finds the distance traveled or displacement (s) of an object using its initial velocity (u), acceleration (a), and time (t) traveled. The distance traveled is a reasonable 14 km, but the resultant displacement is a mere 2. Calculates the free fall distance and velocity without air resistance from the free fall time. We have to evaluate this to find the velocity at any particular time. A particle moves along the x-axis with position at time t given by x(t) = e-t sin t for 0 :5: t :5: 2JC. Velocity Equation in these calculations: Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 10. (c) Find the acceleration of the particle at time t. (d) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 2. (e) Draw a diagram to represent the motion of the particle. A particle moves in a straight line with velocity t^-2 - 1/9 ft/s. (c) Find the average velocity of the particle over the interval 0 § t § 5. Get an answer for 'The velocity function is v(t)= -(t^2)+6t-8for a particle moving along a line. b) Find the particle's displaçement for the given time interval. Find the body's acceleration each time the velocity is zero B. Is the direction of motion of the particle toward the left or toward the right at that time? Give a reason for your answer. If you look carefully, we've used a boldface 0 because velocity is a vector. Find the total distance traveled from t = 0 to t = 4. A particle moves along the x-axis so that its velocity at time t, , is given by v(t) = 3(t - 1)(t - 3). Find the distance traveled by a particle with position (x,y) as varies in the given time interval. To find that number, we'd need to add the absolute value of each interval. AP Calculus Particle Motion Worksheet For #6 – 10: A particle moves along a line such that its position is s ( t ) = t 4 – 4 t 3. When velocity = 0 Divide into intervals; 0 2 and 2 4 At any time t, the position of a particle moving along an axis is: A. Estimate the total dis-tance the object traveled between t= 0 and t= 6. v (t) ≤ 0, the particle moves to the. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. The distance traveled between times t and t + h is f(t + h) − f(t). (a) When the particle is at rest. displacement = -66. Answer: 450 feet. The total distance traveled by the particle from time to time is For the time interval, the situation is a little bit more complicated since the particle changes direction. f) Draw a diagram to illustrate the motion of the particle. SOLUTION Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. Find the total distance traveled by the particle. e) Find the total distance traveled during the first 8 sec. 0 \leq t \leq 7. If a body moves along a straight line with velocity v = t 3 + 3t 2, find the distance traveled between t. Enter the required values know the unknown value of work or force or distance. Include units. Your acceleration is 26. We can also calculate force,. To calculate the speed and angular velocity of objects. B) Find the total distance travelled by the particle. will have a horizontal tangent? (7 Points) 15) Find the slope of the tangent line to the curve. CALCULUS I Worksheet #74 1. Distance and displacement are two quantities that seem to mean the same but are distinctly different with different meanings and definition. (c) Find the displacement of the particle after the first 8 seconds. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the distance traveled by a particle with position | bartleby. vt is negative on the interval 01. (a) Find the time t at which the particle is farthest to the left. 2 3 x t t y t t (a) Find the magnitude of the velocity vector at time t = 5. 1 t millions of bushels per year, with t being years since the beginning of 1970. b) Find the average value ofg(x)intermsofA over the interval [1 ,3]. 5 miles (or 13,200 feet or 158,400 inches ,etc. I found out that the total displacement is. If ( )= 3−4 2+5 −6 gives the position of a point P as it travels along the x-axis, describe the. Motion in Two and Three Dimensions Conceptual Problems 1 • [SSM] Can the magnitude of the displacement of a particle be less than the distance traveled by the particle along its path? Can its magnitude be more than the distance traveled? Explain. Let's assume that we were given. Now you might start, you might start to be appreciating what the difference between displacement and distance traveled is. A particle moves with a position function s(t) = t3 - 12t2 + 36t for t ≥ 0, where t is measured in seconds and s in feet. Write a polynomial expression for the position of the particle at any time r > O if the position of the particle at t = 0 is 5 At what time(s) is the particle changing direction? Find the total distance traveled by the particle fmm time r = O to r 4. Its position function is s(t) for t ≥≥ ≥ 0 ≥ 000. Calculus Total Distance Particle Traveled? I'm in college, and I stumbled into a problem that deals with a particle. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). The graph of v on the left shows that the velocity of the particle is 16 at time and 0 at time The total distance traveled by the particle is given by the definite integral. Find the total distance traveled from t = 0 seconds to t = 4 seconds. Motion Along a Line; Page 8. This picture is helpful: The positions of the words in the triangle show where they need to go in the equations. Find the speed when. (b) Set up an integral expression to find the total distance traveled by the particle from t 0to t 4. To solve for c1, we know that at t = 0, the initial velocity was 4. Sample Test Questions: A particle moves along a horizontal line and its position at time is. Apply the fundamental theorem of calculus to evaluate integrals and to di erentiate integrals with respect to a limit of integration. Claudette responded. 2 The key to finding the total distance traveled in the last example in a method similar to the first example is to break the time. Justify your answer. Therefore, the average velocity is 8. seconds? (2 points) When is the particle speeding up? (2 points) When is the particle slowing down? (2 points) (7 Points) 14) Find all values of so that the graph of. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function. When you try to find the distance a moving object has traveled, two pieces of information are vital for making this calculation: its speed (or velocity magnitude) and the time that it has been moving. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Find the displacement and the total distance traveled by the particle from t = 1. Distance-traveled-by-a-particle Page history last edited by [email protected] Find the initial velocity and displacement. Calculus is im-portant because most of the laws of science do not provide direct information about the values of variables but only about their rate of change. SOLUTION Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. (a) Find the intervals where the function is increasing or decreasing. (b) Find the total distance traveled by the particle from time t = 0 to t = 3. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. Calculus- find total distance of a particle given its velocity equation? Here's the problem: Find the total distance traveled by a particle moving along a straight line with a velocity v = sin (pi*t) for ( 0 | 2019-11-11T22:09:31 | {
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https://math.stackexchange.com/questions/1882866/a-combinatorial-interpretation-of-a-counting-problen | # A combinatorial interpretation of a counting problen
This question is about a surprising formula that is the answer to a counting problem. To me it suggests that there might just be a nice combinatorial interpretation of the problem that I have not yet found. Even if there is not, I'm sure there are much more elegant solutions than mine.
## The problem
Suppose we have an $n\times m$ grid of squares with $n$ rows and $m$ columns. In every square we write the number of rectangles that can be made inside the grid and in which it is contained. Consider for example the green square in the image below.
We see that it is contained in $8$ rectangles that can be made inside this $2 \times 3$ grid. If we count this number for every square we get.
Now let $f(n,m)$ denote the sum of these numbers, so $f(2,3)=4 \cdot 6 + 2 \cdot 8=40$. The task is to find a general formula for $f(n,m)$.
## My solution
Consider the square with coordinates $(i,j)$, as in the image below. Any rectangle that it is contained in must have its top left corner in the area shown in red. Its bottom right corner must be in the area shown in blue. So the square $(i,j)$ is contained in exactly $\color{red}{ i \cdot j} \cdot \color{blue}{ \left( n - i + 1 \right) \cdot \left( m - j + 1 \right)}$ rectangles.
Now we only need to sum over all squares to get $$f(n,m)= \sum_{i=1}^n \sum_{j=1}^m \left [ i \cdot j \cdot \left( n - i + 1 \right) \cdot \left( m - j + 1 \right) \right ].$$ To make this calculation a little easier we can introduce $S_n = \displaystyle \sum_{i=1}^n i$ and $T_n = \displaystyle \sum_{i=1}^n i^2$ so that \begin{align*} f(n,m) &= \sum_{i=1}^n \sum_{j=1}^m \left [ i \cdot j \cdot \left( n - i + 1 \right) \cdot \left( m - j + 1 \right) \right ] \\ &= \sum_{i=1}^n \sum_{j=1}^m \left [ i^2 j^2 - m\cdot i^2 j -i^2 j - n \cdot i j^2 -i j^2+ n m \cdot i j+ m \cdot i j + n \cdot i j +i j \right ] \\ &= T_n T_m - m\cdot T_n S_m - T_n S_m - n \cdot S_n T_m - S_n T_m + n m \cdot S_n S_m + m \cdot S_n S_m + n \cdot S_n S_m + S_n S_m \\ &= \left(S_n + n\cdot S_n - T_n\right)\cdot \left(S_m + m\cdot S_m - T_m\right) \end{align*}
Now we can use the well known formulas for $S_n$ and $T_n$ to find that $$S_n + n\cdot S_n - T_n = \frac{1}{6} n (n+1) (n+2) = \binom{n+2}{3}.$$ Putting this together we get that $$f(n,m) = \binom{n+2}{3} \cdot \binom{m+2}{3}.$$
I was pretty surprised by the simple nature of this answer. Especially by the fact that it seems to suggest that somewhere along the line $3$ things get chosen out of $n+2$ things and also from $m+2$ things. It gives me a small spark of hope that there exists a nice counting argument that solves this problem. Can anyone give a combinatorial interpretation of the formula for $f(n,m)$? More elegant solutions to the problem are also welcome.
## 2 Answers
Each combination of rectangle and contained cell determines and is completely determined by a pair of ordered triples: the first lists the rows of the top edge of the rectangle, the cell, and the bottom edge of the rectangle, and the second does the same for the columns. The row triples correspond to multisets of $3$ elements chosen from the set $[n]=\{1,\ldots,n\}$, and there are
$$\left(\!\!\binom{n}3\!\!\right)=\binom{n+3-1}3=\binom{n+2}3$$
of these. Similarly, there $\binom{m+2}3$ column triples, so there are
$$\binom{n+2}3\binom{m+2}3$$
combinations of cell and containing rectangle.
Your sum is [the number of ways to choose a square and a rectangle containing it],
which equals [the number of ways to choose a rectangle and a square inside it].
The row parts of the ways to choose [a rectangle and point inside it]
correspond to length-(n+2) strings consisting of
a beginning-of-rectangle symbol
and
a green-row symbol
and
an end-of-rectangle symbol
and
n-1 blank-row symbols
where the 3 special symbols must be in that order,
giving exactly 3+n-1 choose 3 possibilities for the rows.
Similarly, there are exactly m+2 choose 3 possibilities for the columns.
The product of those gives your expression. | 2019-06-26T09:52:40 | {
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http://mathhelpforum.com/geometry/200909-challenging-find-hashed-area-question-print.html | # Challenging "Find the Hashed Area" Question
• July 12th 2012, 09:16 AM
danielmc
Challenging "Find the Hashed Area" Question
See attached question. It asks: "A circle is inside a square, which is also found inside another square with with 6 meter sides. Find the hashed area".
Is the correct answer: 9pi/2 meters^2 or 9pi meters^2?
Can you please systematically outline the steps that you took to arrive at the answer?
Thank you for your help and explanation.
• July 12th 2012, 09:20 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
The side length of the smaller square is $3 \sqrt{2}$ (do you see why?).
Therefore the diameter of the circle is $3 \sqrt{2}$, so the radius is $\frac{3 \sqrt{2}}{2}$. The area of the circle is $(\frac{3 \sqrt{2}}{2})^2 \pi = \frac{9 \pi}{2}$.
• July 12th 2012, 09:33 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Hi Richard,
Thanks for the quick response. I do not see how you got the side length of the smaller square to be http://latex.codecogs.com/png.latex?3%20\sqrt{2}.
Can you please explain how you got to that step?
Thanks a million!
• July 12th 2012, 09:34 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
Hint: The side length of the larger square is the diagonal of the smaller square.
• July 12th 2012, 09:43 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
I understand the rest of it but just can't the length of the inner square. I'm sorry my brain isn't working today. I know it has something to do with the Phytagoreans Theorem. Can you please tell me how you got the side length of the inner square.
• July 12th 2012, 09:55 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
A square is composed of two 45-45-90 triangles. The side length of the large square is the diagonal of the small square, so the diagonal of the small square is 6. The diagonal is like the "hypotenuse" of the 45-45-90 triangles, you should be able to find the side length now.
• July 12th 2012, 10:01 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Thank you Richard for the step-wise explanation. I now understand. Since the diagonal of the small square is 6; it would be b^2 + b^2 = 6^2; so solving for b, you would take square root of 36 so 6, then another square root of 6 so "b" (the side of the square) would be 3 sqrt 2. From here, we know that the diameter of the circle is 3 sqrt 2 and we can find radius and then area as you had outlined. Thanks for your help!
• July 12th 2012, 11:17 AM
Soroban
Re: Challenging "Find the Hashed Area" Question
Hello, danielmc!
Quote:
A circle is inside a square, which is also found inside another square with with 6 m sides.
Find the shaded area.
Code:
E o 3 * * 3 * * * * A * * B o-------* * *-------o * | *:::::::::::* | * 3 * | *:::::::::::::::* | * 3 * |*:::::::::::::::::*| * * |:::::::::::::::::::| * * *:::::::::::::::::::* * H o *:::::::::o:::::::::* o F * *:::::::::::::::::::* * * |:::::::::::::::::::| * 3 * |*:::::::::::::::::*| * 3 * | *:::::::::::::::* | * * | *:::::::::::* | * o-------* * *-------o D * * C * * 3 * * 3 * * o G
Is the correct answer: $\tfrac{9\pi}{2}\:m^2\,\text{ or }\,9\pi\:m^2\,?$
I'd like to know how you got those two answers . . .
The circle is inscribed in square $ABCD$
. . which is inscribed in square $EFGH$ as shown.
The top triangle $ABE$ is an isosceles right triangle with legs of length 3.
. . Hence: $AB = 3\sqrt{2}$
But $AB$ is the diameter of the circle.
. . Hence, the radius is: . $r \:=\:\frac{3\sqrt{2}}{2}$
Now you can find the area of the circle . . . right? | 2016-07-23T14:25:01 | {
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https://math.stackexchange.com/questions/2540521/inductive-step-for-graph-proofs-reduce-from-n1-to-n-or-from-n-to-n-1 | # Inductive step for graph proofs: reduce from n+1 to n or from n to n-1?
I was reading a practice problem set for a discrete maths course. It says:
By far, the most common mistake in this homework was using induction incorrectly for graph problems. When proving something about graphs by induction, you want to reduce from $n$ to $n − 1$. This is because you are proving a statement for all graphs of size $n$, and not just a specific graph. By proving something by constructing an $n + 1$ size graph using an $n$ size graph, you have only shown it for “one particular” graph of size $n + 1$, whereas you wanted to show it for all graphs of size $n + 1$. (By size, I mean either the number of vertices or the number of edges, whatever you are inducting on.)
Then they give an example:
We prove by induction on $n$ that if $|V| = n$ and $G$ is acyclic and $|E| = n − 1$, then $G$ is connected.
Base case ($n = 1$): This is trivial; $G$ consists of a single vertex and no edges.
Induction step ($n \geq 2$): First we will prove that $G$ has a leaf. Since $E \neq \emptyset$, we can pick a vertex $v \in V$ with $deg(v) > 0$. Start walking from $v$ until you reach a leaf, say $u$. This will happen because the graph is acyclic. Now, $G − u$ has $n − 1$ vertices and $n − 2$ edges, so we can use the induction hypothesis to conclude that it is connected. Adding $u$ to $G$ with the original edge keeps it connected.
Common Mistake: Constructing $n + 1$ vertex graph using an $n$ vertex graph in induction step. If you did this, you missed showing existence of a leaf, which is crucial.
I would have done it a bit different (rough outline):
Induction step $(n + 1)$: $G$ has $n+1$ nodes. We can find a leaf node $v$ with $deg(v) = 1$. $v$ has to exists because $G$ is acyclic. We obtain a subgraph $G'$ by removing $v$ from $G$. By induction we can assume $G'$ is connected. Adding $v$ again to $G'$ keeps the resulting graph $G$ connected, which proves the theorem for $n + 1$.
Is mine the wrong approach and I should have started from $n$? I thought in the inductive step for graph proofs, you usually start from a $n+1$ graph and for example reduce it to $n$ by removing a certain node/edge/...
• – Misha Lavrov Nov 28 '17 at 3:00
The "common mistake" described in the problem set instructions has nothing to do with whether your inductive step is based on $n$ and $n-1$ or $n+1$ and $n.$ It has everything to do with which number (the smaller one or the larger one) is the number of nodes in the arbitrary graph that you invoke in the inductive step. If you go from $n$ to $n+1$ (or if you go from $n-1$ to $n,$ for that matter!) by adding a node to an arbitrary graph, the proof is wrong. You must remove a node instead, as indeed you did.
Another vital thing to make sure of is that the inductive step can use the base case as its "input." In the problem set, this is done by making the inductive step use the assumption $P(n-1)$ to prove $P(n)$ for $n\geq 2.$ This means that when $n = 2$ in the inductive step, we use $P(1)$ (the base case) to prove $P(2).$ In your proof you need to specify that the inductive step applies for all $n \geq 1,$ not only for $n\geq 2,$ so that when $n=1$ you use $P(1)$ to prove $P(2).$ There are some classic fake proofs that rely on inductive steps that don't "connect" with the base case. | 2020-05-28T04:48:03 | {
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https://math.stackexchange.com/questions/3832496/best-method-for-proving-that-7-times112n1-34n-1-is-divisible-by-10 | # Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$
I am asked to prove by induction that $$7\times11^{2n+1}-3^{4n-1}$$ is divisible by $$10$$.
I wonder whether there is a more direct method, for example factorizing by $$10$$. If an expression is divisible by $$10$$, does this mean that I can factorize it by $$10$$?
• Just think about the last digit. Sep 19, 2020 at 17:04
• With any statement about the set $\Bbb N$ (especially, when the claim doesn't hold for a larger set), a proof by induction is arguably the most direct one. This is somewhat by definition of $\Bbb N$ Sep 19, 2020 at 17:05
• Are you familiar with congruences such as $\,11^2\equiv 3^4\pmod{10}\,?\ \$ Sep 19, 2020 at 18:37
• I take it you mean for $n\ge 1$. The persistence of divisibility can be shown in various ways. The numbers $k_n=Aa^n+Bb^n$ satisfy the recurrence $k_{n+1}=(a+b)k_n-abk_{n-1}$ (you can easily verify this by substitution) so (in appropriate cases) once you show that two successive elements are divisible by $d$ then they all are. These problems are quite often used as examples for induction. Sep 19, 2020 at 19:39
• Thank you all so much for those answers that will help me more deeply understand this kind of problems. I just still hesitate on this : if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ? Sep 19, 2020 at 22:06
A direct and intuitive way. Every integer power of $$11$$ has $$1$$ as last digit, so every number of the form $$7 \cdot 11^{2n+1}$$ ends with the digit $$7$$.
The last digits of the integer powers of $$3$$ follows the four-step cycle $$(3,9,7,1)$$, so that every number of the form $$3^{4n-1}$$ has $$7$$ as last digit.
Therefore, subtracting this second number from the first one we get a number ending with the digit $$0$$. This implies that this last number is divisible by $$10$$.
Below are four proofs using various methods.
You seem to seek a direct proof showing a factor of $$10$$ so let's do that first.
\begin{align} x &\,=\ \ \ \ \, 7\cdot 11^{\large 2n+1}\ -\ 3^{\large 4n-1}\\[.2em] \Rightarrow\ \ 3x &\,=\ \ \, 21\cdot 11\cdot 121^n - 81^n\\[.2em] &\,=\, (21\cdot11\!-\!1)121^n+ \color{#0a0}{121^n-81^n}\\[.2em] &\,=\, \color{#c00}{10}\,(23\cdot 121^n + \color{#c00}4(121^{n-1} + \cdots + 81^{n-1}))\ \end{align}\qquad
where we used the Factor Theorem to deduce $$\,\color{#c00}{10\cdot 4} = 121\!-\!81\,$$ divides $$\,\color{#0a0}{121^n-81^n}.\,$$ Thus $$\,10\mid 3x\Rightarrow 10\mid x\,$$ by Euclid (or directly $$\,10\mid 7(3x)\!-\!20x = x,\,$$ or cancel $$3$$ from $$121$$'s and $$81$$'s).
It's much easier by modular arithmetic (congruences)
\begin{align}\bmod 10\!:\ \ 3x &\equiv 21\cdot 11^{\large 2n+1} - 81^{\large n}\\ \iff\ 3x &\equiv \ \ 1\ \cdot\ 1^{\large 2n+1}\ -\ 1^{\large n} \equiv\color{#0a0} 0\\ \iff\ \ \ x &\equiv\,3^{-1}\cdot\color{#0a0} 0\equiv 0 \end{align}\qquad
by basic congruence laws. We used the fact that scaling by an invertible (here $$3$$) yields an equivalent congrence (recall by Bezout that $$3$$ is invertible being coprime to the modulus $$10)$$
By induction: base case $$\,n=1\,$$ is $$\!\bmod 10\!:\ 7\cdot 11^3\equiv 3^3\,$$ (or $$\,7\cdot 11\equiv 1/3\,$$ for $$\,n=0)\,$$ which are both true, and the induction step follows conceptually by simply by multiplying the first two congruences below using $$\rm\color{#0a0}{CPR} =$$ Congruence Product Rule,
\begin{align}\bmod 10\!:\qquad\ \ \ \color{#c00}{11^{\large 2}}\ &\equiv\ \color{#c00}{3^{\large 4}}\\[.2em] {\rm times}\ \ \ \ \ \ \ 7\cdot 11^{\large 2n+1}&\equiv3^{\large 4n-1}\quad \ P(n)_{\phantom{|}}\\[.2em] \hline \Longrightarrow\ \ \ \ \ 7\cdot 11^{\large 2n+\color{#c00}3}&\equiv 3^{\large 4n+\color{#c00}3}\quad\ P(n\!+\!\color{#c00}1), \ \ \rm by \ \,\color{#0a0}{CPR}^{\phantom{|^|}}\end{align}\qquad
If congruences are unfamiliar we can preserve the arithmetical essence of this simple proof by using an analogous product rule for divisibility (DPR), as explained here.
Or as here use Binomial Theorem on $$\,(1\!+\!10)^{2n+1}$$ and $$(-1\!+\!10)^{4n-1}$$ (or $$\,(1\!+\!80)^n\,$$ in $$3x)$$
Remark All these methods do in fact use induction (on $$n),\,$$ but it may be hidden (encapsulated) in the proof of a theorem that is invoked, e.g. the Factor Theorem or Binomial theorem, or the Congruence Power Rule $$\,a\equiv b\Rightarrow\, a^n\equiv b^n$$.
• Thank you all so much for those answers that will help me more deeply understand this kind of problems. I just still hesitate on this : if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ? Sep 19, 2020 at 22:11
• @Romain $\,10\mid n,\,$ i.e. $10$ divides $n$ means by definition that $\, n = 10k\,$ for some integer $k,\,$ It is often quicker to prove that in ways that don't require explicitly calculating the integer cofactor $k = n/10$ that is witness to the divisibility, e.g. using modular arithmetic (congruences) as above. Sep 19, 2020 at 22:14
• " if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ?" Yes, but you might not always know how. Sep 20, 2020 at 1:02
• Sadly there seems to be at least one user who consistently downvotes ansers that have multiples proofs. In the unlikely chance that this has anything to do with matehmatics then please leave a constructive comment so that any deficiency you perceive may have a chance to be addressed. Sep 20, 2020 at 19:30
we have $$7{(10+1)}^{2n+1}-3{(10-1)}^{2n-1}$$
indeed by binomial theorem it is equivalent to:
$$7(10k+1)-3(10m-1)=10+70k-30m$$
which is div by $$10$$
Using modular arithmetic: \begin{align}3(7\cdot 11^{2n+1}-3^{4n-1})&=21\cdot 11^{2n+1}-9^{2n}\\&\equiv 1\cdot (1)^{2n+1}-(-1)^{2n}\\&=0 \pmod {10}.\end{align}
We have $$3(3^3)=81 \equiv 1 \pmod{10}$$
$$3^{-1}\equiv 3^3 \equiv 7 \pmod{10}$$
\begin{align} 7(11^{2n+1})-3^{4n-1} &\equiv 7 (1^{2n+1}) - (3^4)^n3^{-1}\\ &\equiv 7(1)-(1)(7) \\ &\equiv 0 \pmod{10} \end{align}
Best method? Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$$ and as $$3\cdot 7 \equiv 1 \pmod {10}$$ then $$3^{-1}\equiv 7 \pmod {10}$$ and $$7-7\equiv 0\pmod {10}$$ so $$10|7\cdot 11^{2n+1} - 3^{4n-1}$$.
But that assumes you are comfortable with many concepts
If you don't know any modular arithmetic:
$$11^k = (10+1)^k = 10^k + k10^{k-1} + ...... + k\cdot 10 + 1 = 10 M + 1$$.
And $$3^{4n-1} = 3\times 3^{4n-2}= 3\times 9^{2n-1}$$. And if $$k=2n-1$$ is odd the $$9^k = (10-1)^k = 10^k - k10^{k-1} + ....... + k\cdot 10 - 1= 10N-1$$.
So $$7\cdot 11^{something} - 3^{1+2\times something\ odd} = 7(10M + 1)- 3(10N-1)= 70M -30N +10$$.
• You didn't invole Euler (maybe you meant to say Euclid for computing the inverse?) Sep 20, 2020 at 1:29
• I meant Euler and I meant that $3^{4n} \equiv 1\pmod {10}$ so $3^{4n-1} \equiv 3^{-1}$. Sep 20, 2020 at 2:15
• Ah, I didn't expect use of Euler in such a simple case since it is much easier by repeated squaring $\!\bmod 10\!:\ (3^2)^2\equiv (-1)^2\equiv 1,\,$ or even directly $\,(3^2)^2\equiv 9^2\equiv 81\equiv 1.\ \$ Sep 20, 2020 at 2:24
Since
$$\quad 11^k \equiv 1 \pmod{10} \quad \forall k \in \Bbb Z$$
and
$$\quad 3^{4n-1} \equiv {(3^2)}^{2n} \times 3^{-1} \equiv (-1)^{2n} \times 7 \equiv 7 \pmod{10}$$
we can write
$$\quad 7\times11^{2n+1}-3^{4n-1} \equiv 7 - 7 \equiv 0 \pmod{10}$$
Elementary Number Theory:
The invertible elements in $$\mathbb{Z}/{10}\mathbb{Z}$$ are
$$\quad [1], [3], [7] \text{ and } [9]$$
allowing one to quickly determine that $$3^{-1} \equiv 7 \pmod{10}$$.
The OP is looking for a factorization argument and so we give a problem/hint that can be directly solved without elementary number theory using simple algebra; this is the step case of the induction proof.
Problem: Show that if $$k \ge 1$$ and
$$\tag 1 \displaystyle 7\times11^{2k+1}-3^{4k-1} = 10q$$
then
$$\tag 2 \displaystyle 7\times11^{2(k+1)+1}-3^{4(k+1)-1} = 10 \times (11^2 q + 4 \times 3^{4k-1})$$
Multiply $$n=7\times11^{2n+1}-3^{4n-1}$$ by $$3$$ to get $$3n=21\times11^{2n+1}-81^{n}$$, which is divisible by $$10$$
(ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED. | 2022-10-04T19:48:24 | {
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https://math.stackexchange.com/questions/1111021/solids-of-revolution-question-method-of-cylinders-vs-disc-washers | # Solids of Revolution Question (Method of Cylinders vs Disc/Washers)
Find the volume of the solid formed by revolving the region bounded by y=x^2+1, y=0, x=0, and x=1 about the y-axis.
I was practicing this concept and I came across this problem. I did it using the shell method and got 2pi*Integral of x(x^2+1) from 0 to 1, which yielded the answer 3pi/2.
I tried checking this with the disc/washer method, but this gave me a different answer. Pi*Integral of [Sqrt(y-1)]^2 from 1 to 2 = pi/2.
Likewise, I tried a similar problem with the functions: y = Sqrt(x), x=axis, and x=4 roated about the x=8
Cylinders: 2piIntegral of Sqrt(x)(8-x) from 0 to 4 = 896pi/15 Disk/Washers: pi*Integral of (8-y^2)^2 from 0 to 2 = 1376pi/15
Which ones are correct? Where did I make mistakes?
• What can I do to improve the quality of this question? I wasn't aware of anything wrong about it besides the improper formatting. Thank you. – bandicoot12 Jan 19 '15 at 21:03
• The problem is that $\int_1^2\pi\sqrt{y-1}^2dy$ represents the volume formed by spinning the region bounded by $y=x^2+1, y=0,$ and $y=2$ about the y-axis. – John Joy Jan 19 '15 at 21:08
For the particular example you have, it clearly seems that cylindrical shells will be easier. The interval of integration will be $x \in [0,1]$. For a particular representative radius $x$ from the $y$-axis, the height will be $f(x) = y = x^2 + 1$. The circumference is $2\pi x$, and the differential thickness of the shell is $dx$. So the differential volume is $$dV = 2 \pi x f(x) \, dx = 2 \pi x (x^2 + 1) \, dx,$$ and the total volume is $$V = \int_{x=0}^1 2 \pi x (x^2 + 1) \, dx = \frac{3\pi}{2}.$$
Now using the method of washers, the problem you have is that on the interval $y \in [0,1]$, the volume is just a cylinder of radius $1$, whereas on the interval $y \in (1,2]$, we have washers with outer radius $1$ and inner radius $g(y) = \sqrt{y-1}$, since the inverse function of $y = x^2 + 1$ is $x = \sqrt{y-1}$.
So combined, we would have: $$V = \int_{y=0}^1 \pi(1^2) \, dy + \int_{y=1}^2 \pi (1^2 - (\sqrt{y-1})^2) \, dy = \pi + \pi \int_{y=1}^2 2-y \, dy = \frac{3\pi}{2}.$$ | 2019-06-16T06:38:31 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=j42beicqrurm5nq5pqe641gm45&topic=1575.0;wap2 | MAT334-2018F > End of Semester Bonus--sample problem for FE
FE Sample--Problem 5
(1/2) > >>
Victor Ivrii:
Show that the equation
$$e^{z}=e^2z$$
has a real root in the unit disk $\{z\colon |z|<1\}$.
Are there non-real roots?
Min Gyu Woo:
Let $f(z) = e^2z$ and $g(z) = e^z$.
We have
$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$
So, $f(z) = 0$ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)
$$f(z) = e^2z=0$$
$$z = 0$$
Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.
Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$
Then,
Call $h(z) = f(z) - g(z) = e^x - e^2 x$
Note that:
$$h(0) = 1$$
$$h(1) = e - e^2 <0$$
By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.
I.e. there is a REAL ROOT x where $0<x<1$.
We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.
Thus, within $|z| <1$ there is only one root, and that root is real.
Victor Ivrii:
That the only root in $\{z\colon |z|<1\}$ is real follows from the fact that if $z$ is a root, then $\bar{z}$ it also a root. However the last statement do you mean $\{z\colon |z|<1\}$ ? Otherwise it is wrong due to Picard great theorem
Min Gyu Woo:
Fixed
Nikita Dua:
I don't quite understand why the only root is real?. Can you explain more | 2023-04-01T02:09:25 | {
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https://web2.0calc.com/questions/need-help-with-grpahing-circles | +0
# Need help with grpahing circles
0
111
1
The equation of the circle shown in the following diagram can be written as $x^2 + Ay^2 + Bx + Cy + D = 0$. Find $A+B+C+D$.
Guest Feb 22, 2018
### Best Answer
#1
+7155
+2
the center of the circle = (-1, 1)
the radius of the circle = $$\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5$$
So the equation of the circle is...
$$(x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0$$
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
https://www.desmos.com/calculator/rnauqvgqrg
hectictar Feb 22, 2018
#1
+7155
+2
Best Answer
the center of the circle = (-1, 1)
the radius of the circle = $$\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5$$
So the equation of the circle is...
$$(x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0$$
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
https://www.desmos.com/calculator/rnauqvgqrg
hectictar Feb 22, 2018
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https://math.stackexchange.com/questions/1570049/verify-integration-of-int-frac-sqrt2-x-x2x2dx | # Verify integration of $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
This is exercise 6.25.40 from Tom Apostol's Calculus I. I would like to ask someone to verify my solution, the result I got differs from the one provided in the book.
Evaluate the following integral: $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
$x \in [{-2,0}) \cup ({0,1}]$
As suggested in the book, we multiply both numerator and denumerator by $\sqrt{2-x-x^2}$. This removes the endpoints from the integrand's domain, but the definite integral we calculate with the antiderivative of this new function will be still the proper integral between any two points of the original domain: we only remove a finite number of points from the original domain and the domain of the resulting antiderivative will be the same as the original domain.
$$I=I_1+I_2=\int\frac{2-x}{x^2\sqrt{2-x-x^2}}dx-\int\frac1{\sqrt{2-x-x^2}}dx \tag{1}$$
Evaluating first $I_1$ by substituting $t=\frac1{x} \; \text, \; dx=-\frac1{t^2}dt \text:$
$$I_1=-\frac1{\sqrt2}\int\frac{2t-1}{\frac{t}{|t|}\sqrt{\left(t-\frac14\right)^2-\left(\frac34\right)^2}}dt \tag{2}$$
Substituting again $\frac34\sec{u}=t-\frac14 \; \text, \; dt=\frac34\sec{u}\tan{u}du \; \text, \; t=\frac{3\sec{u}+1}{4} \; \text, \; u=\operatorname{arcsec}{\frac{4t-1}{3}}$, by considering the sign of $t$ and $\tan{u}$ in the integrand's two sub-domains:
a) $x \in (-2, 0): t<-\frac12 \; \text, \; \frac{4t-1}{3}<-1 \; \text, \; u \in(\frac\pi2,\pi) \; \text, \; \tan{u}<0$
b) $x \in (0, 1): t>1 \; \text, \; \frac{4t-1}{3}>1 \; \text, \; u \in(0,\frac\pi2) \; \text, \; \tan{u}>0$
$$I_1=-\frac{1}{2\sqrt2}\int3\sec^2{u}-\sec{u}=-\frac1{2\sqrt2}\left(3\tan{u}-\log\left|\tan{u}+\sec{u}\right|\right)+C_1 \tag{3}$$
$$\sec{u}=\sec\operatorname{arcsec}\frac{4t-1}{3}=\frac{4t-1}{3}=\frac{4-x}{3x} \tag{4}$$
$$\tan^2{u}=\tan^2\operatorname{arcsec}\frac{4t-1}{3}=\left(\frac{4t-1}{3}\right)^2-1=\frac89\frac{2-x-x^2}{x^2} \tag{5}$$
Considering again cases a) and b): $$\tan{u}=\frac{2\sqrt2}{3}\frac{\sqrt{2-x-x^2}}{x} \tag{6}$$
$$I_1=-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{2\sqrt2}{3}\left(\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right)\right|+C_1= \tag{7}$$
$$-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+C'_1 \tag{8}$$
Evaluating now $I_2$: $$I_2=-\int\frac1{\sqrt{\left(\frac32\right)^2-\left(x+\frac12\right)^2}}dx \tag{9}$$
Substituting $\frac32\sin{z}=x+\frac12 \; \text, \; dx=\frac32\cos{z}dz \; \text, \; z=\operatorname{arcsin}{\frac{2x+1}{3}}$:
$$I_2 = -\int dz = -\operatorname{arcsin}{\frac{2x+1}{3}}+C_2 \tag{10}$$
The final result: $$I = I_1 + I_2 = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|-\operatorname{arcsin}{\frac{2x+1}{3}}+C \tag{11}$$
The solution provided in the book:
$$I = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left(\frac{\sqrt{2-x-x^2}}{x}-\frac1{2\sqrt2}\right)-\operatorname{arcsin}\frac{2x+1}{3}+C$$
• The difference between the two solutions is not constant, so one of them is wrong. – egreg Dec 11 '15 at 0:28
• Right, and there is also the difference in the absolute sign for the second term. – Imre Deák Dec 11 '15 at 0:35
The solution in the book is most certainly a typo, your proof seems fine to me. As a confirmation, Mathematica evaluates the integral to be: $$I=-\dfrac {\sqrt {2 - x - x^2}} x + \dfrac1 {2\sqrt {2}}\left[\log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x| \right] \qquad - \arcsin\left (\dfrac {2 x + 1} {3} \right) + \rm C_1,$$ which is the same as your proposed solution since \begin{align} \log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x|&=\log\left|\dfrac{{2\sqrt{2}\sqrt{2-x-x^2}}+{4-x}}{2\sqrt{2}x}\right|+{\rm C_2} \\ &=\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+{\rm C_2},\end{align} so they only differ by a constant.
• (2) seems ok to me: $I1=\int\frac{2-x}{x^2\sqrt{2-x-x^2}}dx=\int\frac{2-1/t}{1/t^2\sqrt{2-1/t-1/t^2}}(-1)\frac1{t^2}dt=-\int\frac{2t-1}{t\sqrt{2-1/t-1/t^2}}dt=-\int\frac{2t-1}{t/|t| \sqrt{2t^2-t-1}}dt$ – Imre Deák Jan 28 '16 at 17:59
• @ImreDeák See my edit. – Workaholic Jan 28 '16 at 19:50 | 2021-04-22T12:18:15 | {
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