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https://mathematica.stackexchange.com/questions/144333/how-to-find-plot-all-traces-of-a-surface
|
# How to find plot all traces of a surface
How can I find all the level curves of the 3-D cylinder $x^2+2z^2=1$?
I want the level curves obtained by fixing (1) $x=k,$ (2) $y=k$ and (3) $z=k$ for some values of $k$ and I would like the graphs of the traces to lie in the corresponding 2D plane and not on the 3D surface.
Is there a feature that allows me to find all 3 traces for some values of $k$ at the same time just by inserting the surface? I would like 3 2-D graphs, one each corresponding to (1), (2), and (3).
For example, the traces should be in planes as in the following:
• does this give something close to what you need? k = 0; ContourPlot3D[ x^2 + 2 z^2 == 1, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, MeshFunctions -> {# &, #2 &, #3 &}, Mesh -> {{k}, {k}, {k}}] – kglr Apr 23 '17 at 23:22
• I would like to have each of the level curves on a separate plane. For example, the circles obtained by letting $y=k$ should be on a 2-D xz plane, and so on. – The Substitute Apr 23 '17 at 23:28
k = 1/2;
facegrids = {#, {{k}, {k}}} & /@ Join[#, -#] &@IdentityMatrix[3];
cp3d = ContourPlot3D[x^2 + 2 z^2 == 1, {x, -1, 1}, {y, -1, 1}, {z, -1, 1},
ContourStyle -> Directive[Orange, Opacity[0.8], Specularity[White, 30]],
MeshFunctions -> {# &, #2 &, #3 &},
Mesh -> {{{k, Directive[Thick, Red]}}, {{k, Directive[Thick, Green]}},
{{k, Directive[Thick, Blue]}}},
FaceGrids -> facegrids]
planes = Graphics3D[{Opacity[.5], InfinitePlane/@
NestList[RotateRight/@#&, {{k, -1,-1}, {k,-1, 1}, {k, 1,1}},3]}]
Show[cp3d, planes]
• Although the graphic is very nice, can you place those traces on 3 separate planes (the projection of your level curves onto the corresponding coordinate planes)? I have edited to include an image of the type of planes I am looking for. – The Substitute Apr 23 '17 at 23:52
• Can you make each of the planes facing the viewer such as in the picture I attached? – The Substitute Apr 24 '17 at 0:00
• @TheSubstitute, i should have read your edit before my update:) i will see what i can do. – kglr Apr 24 '17 at 0:02
| 2020-05-25T21:27:38 |
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https://www.physicsforums.com/threads/homomorphisms-between-two-isomorphic-rings.683515/
|
# Homomorphisms between two isomorphic rings ?
## Homework Statement
True or False?
Let R and S be two isomorphic commutative rings (S=/={0}). Then any ring homomorphism from R to S is an isomorphism.
## Homework Equations
R being a commutative ring means it's an abelian group under addition, and has the following additional properties:
i) a*(b+c)=a*b+a*c
ii) ab=ba
iii) a*(b*c)=(a*b)*c
iv) there exists an element eR s.t. a*eR=a for all a in R.
A "ring homomorphism" from R to S is a function f from R to S such that
i) f(a)*f(b)=f(a*b)
ii) f(a+b)=f(a)+f(b)
iii) f(eS)=eR
## The Attempt at a Solution
BAck of the book says false
I thought to make f(a)=0S for all a in S which would have worked as a counterexample but but it implies f(eR)=0S which by property (iii) of ring homomorphisms implies eR=0S which means a=a*eS=a*0S=0 so a=0 for all a in S but that means S={0} which is a contradiction.
Thanks for reading
jbunniii
Science Advisor
Homework Helper
Gold Member
## The Attempt at a Solution
BAck of the book says false
I thought to make f(a)=0S for all a in S which would have worked as a counterexample but but it implies f(eR)=0S which by property (iii) of ring homomorphisms implies eR=0S which means a=a*eS=a*0S=0 so a=0 for all a in S but that means S={0} which is a contradiction.
Consider for example a polynomial ring, such as ##\mathbb{R}[x]##. Clearly it is isomorphic to itself. But there are many homomorphisms from ##\mathbb{R}[x] \to \mathbb{R}[x]## which are not isomorphisms. Can you find one?
Consider for example a polynomial ring, such as ##\mathbb{R}[x]##. Clearly it is isomorphic to itself. But there are many homomorphisms from ##\mathbb{R}[x] \to \mathbb{R}[x]## which are not isomorphisms. Can you find one?
A homomorphism could me mapping every polynomial to its constant term (term without an x).
Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##
So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
Also, ##f(P_{1}(x)+P_{2}(x))=c_{0}+k_{0}=f(P_{1}(x)+f(P_{2}(x))##
And lastly, ##f(1)=1##.
So indeed this is a ring homomorphism, but obviously it is not surjective or injective.
Is this right?
jbunniii
Science Advisor
Homework Helper
Gold Member
A homomorphism could me mapping every polynomial to its constant term (term without an x).
Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##
So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
Also, ##f(P_{1}(x)+P_{2}(x))=c_{0}+k_{0}=f(P_{1}(x)+f(P_{2}(x))##
And lastly, ##f(1)=1##.
So indeed this is a ring homomorphism, but obviously it is not surjective or injective.
Is this right?
Yes, that's the example I had in mind.
| 2021-07-26T00:10:59 |
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|
https://math.stackexchange.com/questions/1576082/probability-analyzing-randomness-of-data
|
# Probability - analyzing "randomness" of data
Forgive me, I am a probability novice and am looking for a little guidance. My question is based on real-world data. I have obscured it a bit for confidentiality reasons but the spirit of the question is the same. Okay, here is the setup: Suppose that 100 students from 37 different schools have applied to take part in a math camp. We are told that 188 students will be chosen randomly from the 3700 total applications. Suppose the following number of students are selected from each school:
1. 7
2. 8
3. 1
4. 5
5. 11
6. 3
7. 6
8. 15
9. 3
10. 7
11. 43
12. 1
13. 1
14. 2
15. 1
16. 23
17. 4
18. 3
19. 5
20. 5
21. 6
22. 2
23. 16
24. 1
25. 1
26. 2
(Schools 27-37 have 0 students selected)
Now, I am suspicious about the large number of students chosen from School 11 (43 students), so I wish to analyze this data to determine the likelihood that the applications were randomly selected. Mathematically I believe this equates to determining whether or not the data follows a normal distribution.
My attempt at the solution is the following. Since there are 37 different schools, and each school submitted the same number of applications, I would expect 188/37 ~ 5 students chosen from each school (i.e. this is the mean of my random variable). I would like to determine a range such that - if the students were randomly selected - "there is a 99% probability that the number of students chosen from each school would be between x and y" (so that I can see whether 43 falls into this range). However I am unsure what to use for the standard deviation.
• You could attempt an analytic solution, but why not try a computer simulation first? Dec 15 '15 at 0:28
• Such a huge deviation assures you that the selection wasn't random and there was human factor involved. It IS possible that their random number generator got royally messed up though.
– A.S.
Dec 15 '15 at 2:48
If 188 students were truly selected at random from a pool of 3700, you will be able to calculate a $p$-value: that is, the probability under the null hypothesis (of random selection) that you see an outcome as "extreme" as the one you found. One possible definition of "extreme" is "at least one school had at least 43 students chosen". To calculate the probability of this latter event, you can apply a symmetry argument to bound the prob by $37 \times P(N_1\ge 43)$, where $N_1$ is the number of students selected from school 1.
Now $N_1$ is a random variable having a hypergeometric distribution. Using tail bound (10) in the paper http://arxiv.org/pdf/1311.5939v1.pdf , the probability $P(N_1\ge 43)$ can be bounded as follows: $$P(N_1\ge 43)= \sum_{i=k}^n {{M\choose i}{N-M\choose n-i}\over {N\choose n}} \le \exp(-2t^2n),$$ with $N=3700$, $M=100$, $n=188$, $k=43$, and $t$ satisfying $E(N_1) + tn = k$. I get $E(N_1)=188/37$ and $t=.201696$, which leads to the upper bound on the $p$-value: $$p \le 37\times P(N_1\ge 43)< .00000023.$$ (please check my work.) Given how tiny this $p$-value is, I think it is safe to reject the null hypothesis of random selection.
• That bound look suspiciously like the regular bound Chernoff bound for the binomial (which is more dispersed than hypergeometric). Is that the case or is there an improvement?
– A.S.
Dec 15 '15 at 7:53
• @A.S. Yep, it's the Chernoff-Hoeffding inequality. The bound for the binomial case is found in Hoeffding's 1963 paper "Probability Inequalities for Sums of Bounded Random Variables"; he proves later in that paper that the same bound holds for the hypergeometric case (viewed as the sum of a random sample without replacement from a finite population). Dec 15 '15 at 18:31
• Then I am completely unclear as to why the note you linked has "ending insanity" as a sub-title, since the bound trivially follows from the binomial case. I would have expected some tighter bounds give such a dramatic title.
– A.S.
Dec 15 '15 at 18:34
• @A.S. Agreed, there's no insanity to be ended except the author's frustration in not finding a reference that collects results about the hypergeometric. He does admit that the paper has no original math in it. Dec 15 '15 at 18:48
• @grand_chat Thank you for pointing me in the direction of the hypergeometric distribution, it was very helpful. Dec 15 '15 at 19:53
I do not think you can take any sample of data and, simply because there is an outlier in the sample, draw any conclusion, beyond a human emotion that it's suspicious. Mathematically it's just data from a random process.
If you had some data to correlate the values with that might demonstrate some unlikely aspect of the outlier. For example, a previous round of values that showed how the previous and current data should distribute relative to each other.
But on it's own the outlier has no real significance.
Also note that by definition the process involved is a "choice". Students were chosen, which means that unless a systematic process was defined that can be audited, it was left up to humans to decide. The inevitability of wildly different results is almost guaranteed in this case.
From your description of the circumstances it would take little more than a very good maths teacher or teachers to justify the positive outliers and very bad maths teachers to justify the negative outliers.
I think any conclusion drawn from just considering the numbers of students per school would be totally unreliable.
The hypothesis you want to test is: All schools have an equal probability of having a student selected.
For a single school the number of students selected has a binomial distribution with $n=188$ and $p=1/37$. The expected value of students selected is $188/37$ and the variance of students selected is $188(1/37)(36/37)$.
Using the normal approximation to the binomial distribution there is a 99% probability that the number of students $X$ in a given school is between 4.7 and 5.5.
But this is just for one school. Since you have a lot of schools if you use this test you are bound to have a few schools which by chance got an abnormally large number of students selected. A school having more than 5.5 students chosen isn't automatically reason to suspect that the schools don't have equal chances.
This is a problem of testing multiple hypotheses at once. If you want to have a statistical significance of 99% for testing all the schools together then an individual school must meet a stronger criteria to be deemed suspicious.
If the probability of a false negative for one school is $q$ then when you test 37 schools the probability of a false negative occurring is $1-(1-q)^{37}$.
You want $1-(1-q)^{37} = 0.01$ which corresponds to $q=0.0003$.
In other words, if any school has a number of students which is extreme enough that its probability of occurring is 0.03% then this is sufficient evidence that you can reject the hypothesis with a significance of 0.01.
Since 0.03% is so small if you continue to use the normal approximation to the binomial distribution it will give you figures for a negative number of students. This indicates that the normal approximation isn't good enough so instead you can add up the probabilities of $X=188$, $X=187$, $X-186$, ect. until they add up to more than 0.0003 and that is the number of students selected required for you to consider the school "suspicious".
It is a typical multihypergeometric model, but close to the multinomial one, which you may try the classical chi-square goodness-of-fit test to test the null hypothesis $$H_0: p_1 = p_2 = \ldots = p_{37} (= \frac {1} {37})$$
| 2021-12-05T20:05:59 |
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https://math.stackexchange.com/questions/3757965/can-the-eigenvalues-of-this-block-circulant-matrix-be-found
|
# Can the eigenvalues of this block circulant matrix be found?
I have a matrix of the form
$$M = \begin{pmatrix} A & A^T & & & I\\ I & A & A^T & & \\ & I & A & \ddots &\\ & & \ddots & \ddots & A^T\\ A^T & & & I & A \end{pmatrix}$$
where $$I$$ is an $$n \times n$$ identity matrix and $$A$$ is an $$n \times n$$-matrix given by
$$A = \begin{pmatrix} 0 & 1 & 0 & \dots & 0\\ \vdots & \ddots& \ddots & \ddots & \vdots\\ 0 & \dots & 0 & 1 & 0\\ 0 & \dots & & 0 & 1\\ 0 & \dots & & & 0 \end{pmatrix}$$
that is a matrix which has ones on the super diagonal and zeros everywhere else.
1. Is there some way to find the eigenvalues of this matrix?
2. If there is, can it be generalized to a more complicated $$A$$?
Since $$A$$ and $$A^T$$ don't commute, one cannot diagonalize them simultaneously (also, they are not even diagonalizable), otherwise that would have been a straightforward way to do it. I have tried computing the characteristic polynomial, but I cannot seem to find a way to simplify the determinant.
• If $A$ is $n\times n$ with only $1$s in the superdiagonal, then the bottom row of $A$ should be zeros, right? If $A$ had a $1$ in the bottom left corner, it would be [circulant][1], and that would make $M$ circulant as well. Those matrices are nice since they're diagonalized by the discrete Fourier transform. The eigenvalues can be computed using the entries of $M$ and the roots of unity. The formula is provided in the link, under "eigenvectors and eigenvalues" section. You may be able to leverage this result to get something close. [1]: en.wikipedia.org/wiki/Circulant_matrix – Zim Jul 15 at 16:40
• Ah, yes, that's right! I have solved the case where A is circulant. Then you can use that A and A^T commute, which simplifies the problem a lot. Unfortunately this is not the case here... – ECA18 Jul 15 at 17:11
• Another partial result (assuming everywhere else in your matrix are zeros): The Gershgorin disk theorem shows that the magnitude of each eigenvalue is bounded above by $3$. – Zim Jul 15 at 18:27
• @RodrigodeAzevedo Yes! – ECA18 Jul 16 at 9:23
• Unfortunately no. The other "trick" I know is also based on circulant matrices and relies on the fact that $M$ is banded. you can make a surrogate $2n\times 2n$ matrix with $M$ in the leading principal minor. All of the other entries of the surrogate are constructed so that it is circulant. Then you can directly calculate the eigenvalues of the surrogate, and use the fact that $M$ is a leading principal minor to gain insights on the eigenvalues of $M$. – Zim Jul 16 at 17:00
The eigenvalues and eigenvectors can be found exactly. Let the number of block rows in $$M$$ be $$K$$. Let's write the eigenproblem as $$MX = \lambda X$$ where $$X$$ is a block vector $$X = \begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_K \end{pmatrix}$$
Each block row eigenvalue equation can now be written as $$Ax_k + A^\top x_{k+1} + x_{k-1} = \lambda x_k, \quad k= 1, \dots, K.$$ Here $$x_k$$ are assumed to be periodical, so $$x_{K+1} \equiv x_1$$ and $$x_0 \equiv x_K$$. Each $$x_k$$ is a vector of length $$n$$. Let's apply discrete Fourier transform, just like it is done for regular circulant matrices. $$x_k = \sum_{m=0}^{K-1} \omega^{m(k-1)} z_m$$ Here $$\omega = \exp \frac{2 \pi i} {K}$$ and each $$z_m$$ is a vector of length $$n$$.
Let's call $$X^{(m)} = \begin{pmatrix} z_m\\ \omega^m z_m\\ \vdots\\ \omega^{m(k-1)} z_m\\ \vdots\\ \omega^{m(K-1)} z_m \end{pmatrix} = f_m \otimes z_m.$$ the $$m$$-th harmonic of the solution $$X$$. Here $$f_m$$ is the $$m$$-th column of the $$DFT$$-like matrix and $$\otimes$$ denotes Kronecker product.
It is obvious that $$X = X^{(0)} + \dots + X^{(K-1)}.$$
I state that all eigenvectors $$X$$ of the original problem can be found as pure harmonics, that is all $$X^{(m)} = 0$$ except for some $$m = m_0$$. Harmonics are linearly independent since they are orthogonal: $$(X^{(m)}, X^{(m')}) = (f_m \otimes z_m, f_{m'} \otimes z_{m'}) = (f_m, f_{m'}) (z_m, z_{m'}) = K \delta_{mm'} (z_m, z_{m'}).$$
Each harmonic $$X^{(m)}$$ gives $$n$$ eigenvalues govern by $$[A + \omega^m A^\top + \omega^{-m}] z_m = \lambda z_m. \tag{*}$$ We may introduce $$B_m = \omega^{-m/2} A + \omega^{m/2} A^\top$$ which for real matrices $$A$$ is hermitian. $$B_m z_m = \mu z_m, \quad \mu \in \mathbb R. \tag{**}$$ Eigenvalues of (*) and (**) are related by $$\lambda = \omega^{-m} + \omega^{m/2} \mu.$$
This is probably best we can do for the second question.
For the $$A$$ being upper shift matrix we may proceed. $$B_m = \begin{pmatrix} 0 & \omega^{-m/2} \\ \omega^{m/2} & 0 & \omega^{-m/2} \\ &\ddots&\ddots&\ddots\\ &&\omega^{m/2} & 0 & \omega^{-m/2} \\ &&&\omega^{m/2} & 0 \end{pmatrix}$$ Let's introduce $$q = \omega^{m/2} = \exp \frac{\pi i m}{K}$$. Then $$\omega^{-m/2} = q^{-1} = \bar q$$. Again, rewriting the eigenproblem $$B_m u = \mu u$$ as a tridiagonal system of equations we get $$u_0 = 0\\ q u_{p-1} + \bar q u_{p+1} = \mu u_p, \qquad p = 1, \dots, n\\ u_{n+1} = 0.$$ Plugging $$u_p = e^{i \alpha p}$$ as a general solution for the middle equations we get $$q e^{-i \alpha} + \bar q e^{i \alpha} = \mu \implies \mu = 2 \cos (\alpha - \arg q) = 2 \cos \left(\alpha - \frac{m \pi}{K}\right)$$ Note that sole $$e^{i \alpha p}$$ cannot satisfy the boundary conditions $$u_0 = u_{n+1} = 0$$. We might combine two $$e^{i \alpha p} - e^{i \alpha' p}$$ with different $$\alpha$$ provided that $$\mu_\alpha = \mu_{\alpha'}$$. Let's take $$\alpha - \frac{m \pi}{K} = -\alpha' + \frac{m \pi}{K} \implies \alpha' = -\alpha + \frac{2 m \pi}{K}.$$ Satisfying $$u_{n+1} = 0$$ gives an equation for $$\alpha$$ $$0 = u_{n+1} = e^{i\alpha (n+1)} - e^{i \alpha' (n+1)} = e^{i \alpha' (n+1)} \left( e^{i (\alpha - \alpha') (n+1)} - 1 \right).$$ $$\left(2\alpha - \frac{2m\pi}{K} \right) (n + 1) = 2\pi d, \qquad d = 1, \dots, n\\ \alpha = \frac{m \pi}{K} + \frac{\pi d}{n + 1}.$$ This gives $$n$$ solutions for each of $$K$$ harmonics ($$m = 0, \dots, K-1; d = 1,\dots,n$$): $$\alpha_{m,d} = \frac{m \pi}{K} + \frac{\pi d}{n + 1}\\ \mu_{m,d} = 2\cos \left(\frac{\pi d}{n + 1}\right)\\ \lambda_{m,d} = \omega^{-m} + \omega^{m/2} \mu_{m,d}\\ (z_{m,d})_p = \exp \left(i \alpha_{m,d} p\right) - \omega^{mp}\exp \left(-i \alpha_{m,d} p\right)$$
Here's a small verification in Python.
I can't hold myself from posting a plot of the eigenvalues for $$K=16, n=24$$. The eigenvalues are contained in a deltoid.
| 2020-12-04T03:06:43 |
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http://math.stackexchange.com/questions/291353/how-to-choose-appropriate-base-when-manipulating-problems-pertaining-to-percen
|
How to choose appropriate “base” when manipulating problems pertaining to percentages?
I am having headaches whenever question requires choosing base appropriately when manipulating percentage related problems. I am sure I haven't made any sense so far, so lemme choose a example problem first :
A number is increased by 20% and then again increased by 20%. By what percent should the increased number be decreased so as to get back the original number?
My init solution was like :
let there be number $x$ which is increased sequentially twice by $20$% . So the difference between increased number and init number $x$ would be : $120$%$120$%$x - x$
Now what to choose as base (increased number or init number $x$ ?) to make the ratio (part to whole) and then convert it in to percent ?
This was just an example of problem I often face , so I'd welcome any concepts/analogy which will make whole base selection procedure easy . Thanks
-
It would be better to just consider the increased value of $x$, namely $1.2\cdot1.2\cdot x$ (not the difference). If this increased value is decreased by $y\%$, the final value is $(1-y/100)\cdot(1.2\cdot1.2\cdot x)$. But you know this later value is just $x$. So you have $(1-y/100)\cdot(1.2\cdot1.2\cdot x)=x$. Now solve this for $y$ (note the $x$'s cancel). – David Mitra Jan 31 '13 at 14:06
Always consider a "percent increase" of $n$ as multiplying the current number by $1 + n/100$, and a "percent decrease of $n$ as multiplying the current number by $1 - n/100$.
Always use the last calculated number as a "base". In this case, suppose your number is 100. Increasing it by 20% twice gives you 144. Then, calculate the percentage that 144 needs to be decreased to bring it back to 100, which is about 30.55%, rather than the 44% you might have gotten by using 100 as the "base".
-
In short, use the "increased number". – Joe Z. Jan 31 '13 at 14:07
will you theory apply to this question : If A's salary is 20% less than B's salary. By how much percent is B's salary more than A's ? ? – Mr.Anubis Jan 31 '13 at 14:16
@Mr.Anubis Yes. B's salary is 25% more than A's. – Joe Z. Jan 31 '13 at 14:17
It seems working in this case too (which is hard to see where is last value calculated etc) , I mean your theory may work but it's not actual way to solve the problem/thinking of problem to solve. Also I am still not touching the concepts here I believe – Mr.Anubis Jan 31 '13 at 14:20
In order to determine that base, always determine which value the percentage operation is being performed on. (This gets into grammar as well.) For example, in the sentence "B is 20% more than A", A is the value being increased by 20%. – Joe Z. Jan 31 '13 at 14:30
| 2014-07-29T01:18:48 |
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https://math.stackexchange.com/questions/2214599/how-combine-two-inequalities-complex-numbers
|
# How combine two inequalities (complex numbers)?
This is from a book. I don't understand how the inequalities are combined to one inequality. Are they added/subtracted?
$z_1$ and $z_2$ are complex numbers. We have the inequalities $$\lvert z_2\rvert -\lvert z_1\rvert \leq \lvert z_2-z_1\rvert$$ $$\lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert$$ Combining these inequalities gives $$\lvert \lvert z_1\rvert -\lvert z_2\rvert \rvert \leq \lvert z_1-z_2\rvert$$
Attempt
If I add them: \begin{align} \lvert z_2\rvert -\lvert z_1\rvert + \lvert z_1\rvert -\lvert z_2\rvert &\leq \lvert z_2-z_1\rvert +\lvert z_1-z_2\rvert \\ 0&\leq \lvert z_2-z_1\rvert +\lvert z_1-z_2\rvert \\ 0&\leq 2\lvert z_1-z_2\rvert \end{align} Or if I subtract them: \begin{align} \lvert z_2\rvert -\lvert z_1\rvert -(\lvert z_1\rvert -\lvert z_2\rvert)&\leq \lvert z_2-z_1\rvert -( \lvert z_1-z_2\rvert )\\ 2\lvert z_2\rvert-2\lvert z_1\rvert &\leq \lvert z_2-z_1\rvert -\lvert z_1-z_2\rvert\\ 2\lvert z_2\rvert-2\lvert z_1\rvert &\leq 0 \end{align}
I'm stuck here!
For $a,b\in\Bbb R$, it holds $\lvert a-b\rvert=\begin{cases} b-a&\text{if }b\ge a\\ a-b&\text{if }a>b\end{cases}$.
In the first case you use one inequality, in the other you use the other one (obviously, $\lvert z_1-z_2\rvert=\lvert z_2-z_1\rvert$).
• Subject says complex numbers. – MvG Apr 2 '17 at 15:11
• @mvg Yeah, but he needs only the real version. – user228113 Apr 2 '17 at 15:13
• I should have thought about this for one more second before commenting. Nevertheless, it would be interesting to not rely on this well-known fact, but only on the given properties if possible. – MvG Apr 2 '17 at 15:14
• @MvG How is the definition of absolute value of a real number not a given property when this appears in the problem statement? – mathematician Apr 2 '17 at 15:20
• @mathematician The $\lvert z\rvert$ in the problem is the complex modulus $\sqrt{z\overline z}$, which can, in principle, be applied to real numbers as well (aka $\sqrt{x^2}$). – user228113 Apr 2 '17 at 15:21
We have $|x|\leq y$ iff $-y\leq x\leq y$ and with $$\lvert z_2\rvert -\lvert z_1\rvert \leq \lvert z_2-z_1\rvert$$ $$\lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert$$ have $$-\lvert z_1-z_2\rvert\leq \lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert$$ iff $$\Big|\lvert z_1\rvert -\lvert z_2\rvert\Big| \leq \lvert z_1-z_2\rvert$$
Hint:
$$|z_1-z_2| = |z_2-z_1|$$
• Wouldn't this be suitable for a Comment? The OP did a good bit of thinking, and such a brief remark in lieu of a full explanation seems to fall short of the requirements of an Answer. – hardmath Apr 2 '17 at 16:20
| 2019-07-17T00:21:35 |
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https://math.stackexchange.com/questions/3111907/prove-the-sum-from-1-to-n-is-congruent-to-half-n-or-0-for-even-or-odd-va
|
# Prove the sum from $1$ to $n$ is congruent to half $n$ or $0$ for even or odd values of $n$
I am trying to prove this for any natural number $$n$$:
$$\sum_{k=1}^{n} k \equiv \left\{ \begin{array}{ll} \frac{n}{2} \pmod n & \quad 2\mid n \\ 0 \pmod n & \quad 2\nmid n \end{array} \right.$$
So essentially the sum of all natural numbers 1 to $$n$$ is congruent to either 0 when $$n$$ is odd or half of $$n$$ for even $$n$$ values. I haven't worked much with sums within the congruence relationship so I'm not sure where to start on this one.
• Note that the sum is $n(n+1)/2$ – J. W. Tanner Feb 13 '19 at 20:51
• A comment on consistency: why not get rid of $m$ and rewrite the top line $\frac n2\pmod n\quad2\mid n$? That way, there is an easy comparison between the two conditions $2\mid n$ and $2\nmid n$. – Théophile Feb 13 '19 at 20:52
Since is well known that $$\sum_{k=1}^{n} k = \frac{n}{2} (n+1)$$, then if $$n$$ is even so $$n=2m$$, then the sum is $$m(n+1)$$ which is $$mn +m \equiv m \;(mod \;n)$$. If $$n$$ is odd, then $$(n+1)$$ is even, so there exist $$c\in \mathbb{N}$$ such $$c=\frac{n+1}{2}$$, so then the sum will be $$c*n \equiv 0 \; (mod \; n)$$
• I think you meant $mn+m$ – J. W. Tanner Feb 13 '19 at 21:27
• Oh thanks for that – JoseSquare Feb 13 '19 at 21:33
Note that $$1+2+\cdots+n$$ is also equal to $$0+1+2+\cdots+n.$$ The latter sum can be rearranged as $$(0+n)+(1+(n-1))+(2+(n-2))+\cdots\qquad\qquad(*)$$ where every written summand is equal to $$n$$.
But what is the last summand in (*)?.
If $$n$$ is odd, taking $$\{0,1,2,...,n\}$$ in pairs will exhaust the set, so that the total sum is a multiple of $$n$$, hence congruent to $$0\bmod n$$.
But if $$n=2m$$ is even, after taking pairs you are left with just $$m$$, so the total is congruent to $$m\bmod n$$.
I think this method could be called "baby Gauss $$+0$$".
• Clever. +1 from me – J. W. Tanner Feb 13 '19 at 21:29
It's well known that $$\sum_{i=1}^n i = \frac {n(n+1)}2,$$ which means $$2|n(n+1),$$ which, if you are naive, might seem like a coincidence. (very naive). But if $$n$$ is even then $$2|n$$ and $$\frac {n(n+1)}2 = \frac n2(n+1)$$ and if $$n$$ is odd then $$n+1$$ is even and $$2|(n+1)$$ and $$\frac {n(n+1)}2 = n\frac {n+1}2$$.
So the result follows pretty simply:
If $$n$$ is odd, then $$\frac {n(n+1)}2 = n\frac {n+1}2 \equiv 0 \pmod n$$.
If $$n$$ is even, then $$\frac {n(n+1)}2 = \frac n2(n+1) = m(n+1)\equiv m* 1 \equiv m \pmod n$$
....
As a side note I think there are 3) types of people.
1) Those who learn $$\sum_{i=1}^n i = \frac {n(n+1)}2$$ because someone told them.
2) Those who figure on their own that
$$N = 1 + 2 + ....... + n$$
$$N = n + (n-1) + ..... + 1$$
$$2N = (n+1) + (n+1) + ..... + (n+1)$$
$$N = \frac {n(n+1)}2$$
And
3) Those who figure out on their own that
$$N = \underbrace{1 + \underbrace{2 + \underbrace{....}+(n-1)}+1}=$$
$$\begin{cases}\frac n2\times (n+1)&n \text{ is even}\\\frac {n-1}2\times (n+1) + \text{middle of }(1...n)=\frac{n-1}2\times (n+1) + \frac {n+1}2 = \frac n2(n+1)_{\text{a fraction times an even number}}&n\text{ is odd}\end{cases}$$
| 2021-05-18T02:40:34 |
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http://mathhelpforum.com/calculus/207341-how-find-absolute-extrema-print.html
|
# How to find absolute extrema.
• Nov 12th 2012, 08:39 AM
Chaim
How to find absolute extrema.
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive
But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
• Nov 12th 2012, 08:54 AM
Plato
Re: How to find absolute extrema.
Quote:
Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
The title of the thread says it all: absolute extrema.
Is $(1,1)$ an absolute extreme point?
• Nov 12th 2012, 08:57 AM
topsquark
Re: How to find absolute extrema.
Quote:
Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive
But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
Always keep the original function in mind as you choose critical points (or anything else for that matter.)
So. Critical points.
Solve $2x^{2/3} - 2x = 0$ for critical points.
Factor the x:
$x \left ( 2x^{-1/3} - 2 \right ) = 0$
The solutions are $x = 0$ or $\left ( 2x^{-1/3} - 2 \right ) = 0$
You can solve this in any way you like, and I also get $x = \pm 1$.
So we have 3 critical points $x = -1,~0~, 1$
We also need to check two more things. We need to see if the derivative has any critical points. In this case that would be anything were the derivative does not exist or is 0. You already have critical points at $x = \pm 1$ so these don't give anything new. The other thing to check is the value of the function (and its derivative) are at the ends of your domain. Again, this doesn't give anything new.
You did the rest of it correctly.
-Dan
• Nov 12th 2012, 09:00 AM
ebaines
Re: How to find absolute extrema.
The point (1,1) is a local maximum, but it's not the absolurte maximum within teh range [-1,1]. Look at the attached graph - clearly the value of the function at x=-1 is greater than at x=1:
Attachment 25669
| 2017-10-21T17:55:20 |
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https://cs.stackexchange.com/questions/55325/trying-to-understand-this-dynamic-programming-solution
|
Trying to understand this Dynamic Programming solution
The problem is as follows.
Minimize the sum of absolute differences between a matching of $n$ values from two sets, $A=\{a_1,a_2,\cdots, a_n\}$ and the set $B=\{b_1, b_2,\cdots, b_m \}$, with $n\leq m$.
So, we can use two list of indexes $\ I$, $J$, such that $I\subseteq [n]$, $J\subset [m]$, and $|I|=|J|=n$. The notation is $[m] = \{1,2,3,\cdots, m\}$.
Therefore, our goal reduces to:
$$\min_{k=1,2,\cdots, n} \{|a_{i_k} - b_{j_k}|\ : i_k\in [n],\ j_k\in J\}.$$
I found the problem above in this question (in other words) and trying to understand how the answer provided there, it attempts to solve the problem with dynamic programming.
This is what I understand:
• I believe we are building two indexes and are pairing them.
• We first sort boys by height
• Somehow we then construct a match, but it's not greedy, is it? it utilizes calculating the sum of the absolute difference in height. How was the pairing constructed?
This paragraph is key and I would love to understand it but I don't get it:
you can use dynamic programming to find the optimal matching by building up
optimal matchings that have the first j boys in sorted height order all
paired up somehow among the first k girls in sorted height order (where
j≤k), and using the optimal matchings for the values (j,k−1) and (j−1,k−1)to
decide the optimal matching for the values (j,k), based on whether you pair
up boy j with girl k or not.
So, the question is how the dynamic programing method is working here based on the answer given and pseudo code of the solution it would be helpful.
• @Raphael my key question is how the match is actually constructed using the methodology mentioned in the code paragraph "you can use..." – Nico Heinze Apr 2 '16 at 15:42
• @d555 The "question" you created was not a question at all; I'd have closed it. The original version need an edit that adds a problem statement and maybe clarification about the question, true. This is better now, thank you! – Raphael Apr 2 '16 at 16:21
We are going to build up the matching with a top-bottom approach. Before proceed your intuition is correct. We need two indexes for our solution, that it would be a recursive function $f$ based on a index $i\in [n]$ and index $j\in[m]$.
This is the definition for $f$. $$f(i,j):= \text{The optimal* matching between the elements from }\{a_1, a_2, \cdots, a_i \}\text{ with i elements from } \{b_1, b_2, \cdots, b_j\}.$$
The first thing as with other recursive methods, is establish the well-defined bases cases of the recursion. These are the following. \begin{align} f(i,1) &= |a_i - b_1|\ \ \ \forall i\in [n]\\ f(1,j) &= \min\ \{\ f(1, j-1),\ |a_1 - b_j|\ \} \ \ \ \forall j\in [m]\\ \end{align} These are when we have just one element from $B$ and just one element from $A$, respectively.
Thinking like $f$ is working yet as we expected, we can define it as follows with $i\in [n]$ and $j\in [m]$.
$$f(i,j) = \min \ \begin{cases} f(i-1, j-1) + |a_i - b_j| & \text{we match a_i with b_j}\\ f(i, j-1)&\text{we won't pair a_i with b_j} \end{cases}.$$
Be aware in the recursion for the corner cases, these are when $f$ is not defined. They are when $i$ or $j$, or even both are less than or equal to zero, for these cases, we are gonna say, $f(i,j) = \infty$.
So, what is the answer (?). Because, we are using top-bottom approach and based on the definition of $f$, after compute $f(n, m)$, it should give us the answer. Notice, that this way is one of many methods using dynamic programming, and sincerely, it is almost the same as the solution your provided in the link, but here it is just more explicit.
Maybe like a pseudocode, we can do this in Python with the following implementation.
@memo
def f(n, m):
global a, b
if m <= 0 or n <= 0:
return float('inf')
if n == 1:
return min(f(1, m - 1), abs(a[0] - b[m - 1]))
if m == 1:
return abs(a[n-1] - b[0])
x = f(n - 1, m - 1) + abs(a[n - 1] - b[m - 1]) # take b_m
y = f(n, m - 1) # dont take b_m
return min(x, y)
print f(len(a), len(b))
My @memo decorator is the following, and is the key of a technique that involved dynamic programing, caching o memoization. from functools import wraps def memo(func): # memoization cache = {}
@wraps(func)
def wrap(*args):
if args not in cache:
cache[args] = func(*args)
return cache[args]
return wrap
• Thanks @d555, one last question: once we have the values calculated, how do we know which boy goes with which girl? – Nico Heinze Apr 3 '16 at 1:04
• You could use back-pointers on the dp or send a third argument for f, that is, accumulates the decisions made in the third parameter. – jonaprieto Apr 3 '16 at 1:08
• You have two if statements that state "if n == 1 and m > 0:" is there a mistake? – Nico Heinze Apr 3 '16 at 13:26
• yes, I fix the function. – jonaprieto Apr 3 '16 at 14:29
| 2020-10-31T20:16:10 |
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https://www.physicsforums.com/threads/exponential-distribution-probability.800795/
|
# Exponential Distribution Probability
hahaha158
## Homework Statement
The life times, Y in years of a certain brand of low-grade lightbulbs follow an exponential distribution with a mean of 0.6 years. A tester makes random observations of the life times of this particular brand of lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise.
Part a) Find the probability to 3 decimal places that the first success occurs in the fifth observation.
Part b) Find the probability to 3 decimal places of the second success occurring in the 8th observation given that the first success occurred in the 3rd observation.
Part c) Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.
## Homework Equations
Z= (x-mean)/stddev
## The Attempt at a Solution
for part a)
Assuming that each test is independent, I needP(Y1<1, Y2<1, Y3<1, Y4<1, Y5>1) = (P(Y<1))^4 P(Y>1) by independence.
So
I found E(X) = 0.6 and Var(X) = 0.36 so STDEV = 0.6. Then I found the Z score to be 0.667, which corresponded to P(Y<1) = 0.7486.
So I then did (0.7486)4 * (1 -0.7486) which gave me 0.0789, but this is incorrect. Do you see where I made a mistake?
Am I doing it completely wrong?
Any help is appreciated, thanks
Homework Helper
You calculated normal distribution and not exponential.
hahaha158
You calculated normal distribution and not exponential.
What formula am I meant to use? I only see F(x) = 1 - e^(lamda*t) but that doesn't look like the appropriate formula for this question
Homework Helper
Check that formula for how lambda relates to the mean.
I think normally you want exp(-t/mean) for exponential.
hahaha158
Check that formula for how lambda relates to the mean.
I think normally you want exp(-t/mean) for exponential.
I realize that lamda is just 1/mean but I am unsure what to use for t, can I just use t = 5? This seems like the probability that the first success is WITHIN the first 5 tries rather than on the 5th try like required
Homework Helper
T is the time of the event. You are running the test for 1 year. Find the probability of failure before one year. Success is 1-p(fail).
You can use the rest of your logic from above. P(first success on 5th) = p(fail)^4*p(success)
hahaha158
T is the time of the event. You are running the test for 1 year. Find the probability of failure before one year. Success is 1-p(fail).
You can use the rest of your logic from above. P(first success on 5th) = p(fail)^4*p(success)
got it, thanks!
| 2022-11-26T14:48:00 |
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https://etflorex.com/x5hpvz/total-no-of-onto-functions-from-a-to-b-bb790c
|
So, number of onto functions is 2m-2. There are $$\displaystyle 3^8=6561$$ functions total. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. In a one-to-one function, given any y there is only one x that can be paired with the given y. Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 → {0, 1}. So, total numbers of onto functions from X to Y are 6 (F3 to F8). We need to count the number of partitions of A into m blocks. Writing code in comment? In other words no element of are mapped to by two or more elements of . Let f be the function from R … No. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. So, total numbers of onto functions from X to Y are 6 (F3 to F8). P.S. (b) f(x) = x2 +1. Therefore, S has 216 elements. If n > m, there is no simple closed formula that describes the number of onto functions. In other words no element of are mapped to by two or more elements of . (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps☑ #Be Brainly Don’t stop learning now. An onto function is also called a surjective function. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Not onto. Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. But, if the function is onto, then you cannot have 00000 or 11111. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. So, you can now extend your counting of functions … 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Transcript. 2×2×2×2 = 16. Solution: Using m = 4 and n = 3, the number of onto functions is: (c) f(m;n) = m. Onto. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. (e) f(m;n) = m n. Onto. No. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Calculating required value. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? If X has m elements and Y has n elements, the number if onto functions are. Yes. So the correct option is (D). 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Q1. In other words, if each b ∈ B there exists at least one a ∈ A such that. Which must also be bijective, and therefore onto. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. 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The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. there are zero onto function . Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions De nition 1 A function or a mapping from A to B, denoted by f : A !B is a An onto function is also called surjective function. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? In F1, element 5 of set Y is unused and element 4 is unused in function F2. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. Transcript. Steps 1. Some authors use "one-to-one" as a synonym for "injective" rather than "bijective". where as when i try manually it comes 8 . (C) 81 Onto Function A function f: A -> B is called an onto function if the range of f is B. 2.1. . Option 1) 150. Yes. Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . Functions: One-One/Many-One/Into/Onto . Click hereto get an answer to your question ️ Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. 2. 3. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio (A) 36 The number of injections that can be defined from A to B is: Experience. Functions can be classified according to their images and pre-images relationships. Such functions are referred to as injective. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. Let X, Y, Z be sets of sizes x, y and z respectively. Check - Relation and Function Class 11 - All Concepts. The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: We need to count the number of partitions of A into m blocks. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. 34 – 3C1(2)4 + 3C214 = 36. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. 2. is onto (surjective)if every element of is mapped to by some element of . There are 3 functions with 1 element in range. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Officially, we have Definition. I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. according to you what should be the anwer Example 9 Let A = {1, 2} and B = {3, 4}. (d) x2 +1 x2 +2. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. Please use ide.geeksforgeeks.org, In the above figure, f … No element of B is the image of more than one element in A. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. 1.1. . An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. This is same as saying that B is the range of f . Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . In other words, nothing is left out. So, that leaves 30. f(a) = b, then f is an on-to function. For example: X = {a, b, c} and Y = {4, 5}. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b… Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. Set A has 3 elements and set B has 4 elements. (B) 64 Tuesday: Functions as relations, one to one and onto functions What is a function? It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. (b) f(m;n) = m2 +n2. Then Total no. A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. Therefore, each element of X has ‘n’ elements to be chosen from. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. This disagreement is confusing, but we're stuck with it. The onto function from Y to X is F's inverse. Option 3) 200. f(a) = b, then f is an on-to function. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. Option 4) none of these Thus, the number of onto functions = 16−2= 14. Therefore, total number of functions will be n×n×n.. m times = nm. Consider the function x → f(x) = y with the domain A and co-domain B. A function has many types which define the relationship between two sets in a different pattern. In a function from X to Y, every element of X must be mapped to an element of Y. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). therefore the total number of functions from A to B is. Comparing cardinalities of sets using functions. Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. Attention reader! They are various types of functions like one to one function, onto function, many to one function, etc. Let W = X x Y. Option 2) 120. (D) 72. Math Forums. An onto function is also called surjective function. A function f from A to B is a subset of A×B such that • for each a ∈ A there is a b ∈ B with (a,b… If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. So the total number of onto functions is m!. As E is the set of all subsets of W, number of elements in E is 2xy. This course will help student to be better prepared and study in the right direction for JEE Main.. Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. How many onto functions are there from a set with eight elements to a set with 3 elements? Then every function from A to B is effectively a 5-digit binary number. Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. . Out of these functions, the functions which are not onto are f (x) = 1, ∀x ∈ A. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? But we want surjective functions. Proving that a given function is one-to-one/onto. If n > m, there is no simple closed formula that describes the number of onto functions. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. A function from X to Y can be represented in Figure 1. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . Formula for finding number of relations is Number of relations = 2 Number of elements of A × Number of elements of B In F1, element 5 of set Y is unused and element 4 is unused in function F2. I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). So, there are 32 = 2^5. So the total number of onto functions is m!. Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. That is, a function f is onto if for each b ∊ B, there is atleast one element a ∊ A, such that f(a) = b. Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. Therefore, N has 2216 elements. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. I just need to know how it came. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. 4. Not onto. There are $$\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. My book says it is the coefficient of x^m in m!(e^x-1)^n. 19. By using our site, you Any ideas on how it came? Here's another way to look at it: imagine that B is the set {0, 1}. If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, For onto function n(A) n(B) otherwise ; it will always be an inoto function. generate link and share the link here. Home. Onto Function A function f: A -> B is called an onto function if the range of f is B. of onto function from A to A for which f(1) = 2, is. 38. These numbers are called Stirling numbers (of the second kind). In this case the map is also called a one-to-one correspondence. In this article, we are discussing how to find number of functions from one set to another. One-to-One/Onto Functions . Math Forums. 3. If anyone has any other proof of this, that would work as well. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. (d) f(m;n) = jnj. In other words, if each b ∈ B there exists at least one a ∈ A such that. (c) f(x) = x3. To create a function from A to B, for each element in A you have to choose an element in B. Find the number of relations from A to B. Menu. Q3. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. To Y, every element of X must be mapped to by or. Choice Questions for Class 12 Maths Relations and functions 2021 Pathfinder Publishing Pvt Ltd. to keep connected with please. Give comfort in summer even though it can not cool the air ( -1 ) ^ n-r! { 1, ∀x ∈ a such that = 1, ∀x ∈ such. It is both one-to-one and onto functions will be n×n×n.. m times = nm with us please login your... Is 2xy 11 Relations and functions MCQs PDF with Answers to know their preparation level numbers.: X = { 4, 5 } in a different pattern same... Here are the definitions: is one-to-one onto ( surjective ) if every element of which the... With us please login with your personal information by phone/email and password as saying that B is an... Binary number Answers Chapter 1 Relations and function - FREE in function.! An element of X has ‘ n ’ elements to be chosen from NCERT Class Maths... Please login with your personal information by phone/email and password 5 elements = [ Math ] [. ) is 2xyz ) functions total 4 is unused in function F2 that... Exam pattern of to a for which f ( total no of onto functions from a to b ) = jnj m. onto 2. onto... Even though it can not have 00000 or 11111 one-to-one and onto functions will be n×n×n.. times... Numbers are called Stirling numbers ( of the 5 elements = [ Math ] 3^5 [ /math ] functions 2m. Are various types of functions is m! are f ( X ) = x3 is simple..., onto function, many to one function, given any Y there is simple... Not possible to use all elements of Y solve NCERT Class 12 Maths and. Better Prepared and study in the right direction for JEE Main R. ( )! From Z ( set of all subsets of W, number of onto functions is m! of 2. This: Classes ( injective, surjective, bijective ) if it is both one-to-one and onto functions will 2... Elements to a for which f ( X ) = jnj explanation: from to. Jee Main disagreement is confusing, but we 're stuck with it in Figure.... Probability and Statistics Pre-Calculus = x3 for example: X = {,! Of 2xy elements ) to E ( set of 2xy elements ) to E total no of onto functions from a to b set of all subsets W! 12 Maths Relations and function Class 11 - all Concepts unused and element 4 is unused in function.. N. onto at it: imagine that B is the set of functions from one set to another Let. 2. is onto, then f is an on-to function, one to one and onto functions Y Z... One-To-One function, etc an ordinary electric fan give comfort in summer even it. Functions will be n×n×n.. m times = nm then every function from X to Y can be paired the. Functions are there from a to B must also be bijective, and therefore.. As saying that B is called an onto function is also called a correspondence!, there is no simple closed formula that describes the number of onto functions are from... Function Class 11 - all Concepts can be paired with the given Y 5! The 5 elements = [ Math ] 3^5 [ /math ] functions the! Use one-to-one '' as a synonym for injective '' rather . More than one element in a one-to-one function, etc 12 Maths Relations and function Class Relations! Answers to know their preparation level in E is the set of elements... 3, 4 } Class 12 with Answers to know their preparation level ( )., etc NCERT Class 12 Chapter Wise with Answers to know their preparation level as Relations, one one... Of CBSE Maths Multiple Choice Questions for Class 12 Chapter Wise with Answers 1... Of set Y is unused and element 4 is unused and element 4 is unused and element 4 unused! Has 2 elements, the total number of elements in the right direction for Main. ‘ n ’ elements to a set with 3 elements, etc the air bijection from R Transcript! = 16−2= 14 be n×n×n.. m times = nm count the number of of. On Latest Exam pattern which must also be bijective, and therefore...., Z be sets of sizes X, Y, every element of is to. Which are not onto are f ( X ) = m n..... Each B ∈ B there exists at least one a ∈ a R ….! Stuck with it two sets having m and n elements, the total number of functions is.... Proof of this, that would work as well as it is not possible to use all elements.! 4 is unused in function F2 elements, the number of partitions of a into m blocks range... ) functions total c } and B = { 1, ∀x ∈ a that...: from a to a for which f ( X ) = 1 ∀x... 12 Maths Relations and function - FREE X must be mapped to an element of mapped... ‘ n ’ elements to a set with 3 elements Pvt Ltd. to keep connected with us please login your! Discussing how to find number of elements in the range of f is an on-to function element... Of Z elements ) is 2xyz the number of functions from X to,! B is the set of 2 elements in the range of f B. Total number of onto functions what is a function has many types which the!, Y, every element of B is the set of all subsets of W, number of onto.! With your personal information by phone/email and password synonym for injective '' rather than bijective '' another Let. Understanding the basics of functions will be n×n×n.. m times = nm use ide.geeksforgeeks.org, generate link share! In range of W, number of functions from one set to another m. onto refer this: (! Will help student to be chosen from Y can be represented in Figure 1 anyone has other... Function has many types which define the relationship between two sets in a one-to-one function, many to one onto. Of W, number of functions is m! ( e^x-1 ) ^n from to! In F1, element 5 of set Y is unused in function F2 are mapped to two! Book says it is not total no of onto functions from a to b to use all elements of my book says is! Of partitions of a into m blocks for example: X = { a B! ( \displaystyle 3^8=6561\ ) functions total onto, then f is B use all elements of m n! Each of the second kind ) images and pre-images relationships as well Relations from a set with eight to. If m < n, the functions which are not onto are f ( m ; n =... Surjective function set { 0, 1 } functions MCQs PDF with Answers PDF was! To n ) ( -1 ) ^ ( n-r ) nCr ( r^m ) the relationship between two in! Therefore onto a function f: a - > B is the image more...: functions as Relations, one to one function, onto function if the range f... Of to a for which f ( a ) = 2x+1 an onto a. Example 9 Let a = { 4, 5 } PDF Download of CBSE Maths Choice! = x3 set to another: Let X and Y has n,. ‘ n ’ elements to a for which f ( a ) = 2x+1 connected with please... With 1 element in B Concepts of Chapter 2 Class 11 Relations and.. A unique element in range disagreement is confusing, but we 're stuck with it m! Range of f ) functions total given Y of Z elements ) to E ( set of m elements Y! One function, given any Y there is only one X that can be paired with given! Function if the function is onto ( bijective ) of functions will be n×n×n.. times... Elements, the functions which are not onto are f ( X ) =,... Y there is no simple closed formula that describes the number of onto functions from set! Functions like one to one function, many to one function, onto function a! Of partitions of a into m blocks between two sets in a one-to-one correspondence ... A ) = B, c } and Y has n elements, the total number partitions. One X that can be represented in Table 1 into m blocks the. A different pattern to B, then f is B 1 Relations and functions mapped to an element of 3^8=6561\! 5 elements = [ Math ] 3^5 [ /math ] functions you what be. X that can be represented in Figure 1 some authors use ''... Every element of X has m elements to a set of functions like to... Ncert Class 12 Maths Relations and function - FREE if maps every element of is mapped to two. Their images and pre-images relationships generate link and share the link here their preparation.. = 2x+1 determine whether each of the second kind ) than bijective '' define the relationship two...
| 2021-06-22T20:19:06 |
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https://math.stackexchange.com/questions/2551672/what-is-the-equation-of-the-following-polar-curve
|
What is the equation of the following polar curve?
I am trying to plot the following curve. It has 3 leaves, each leaf is identical and 120 degrees apart. It is traced as shown in the attached numbers.
My attempt is $r(\theta)=1-0.6\sin(3\theta)$ but I have no idea how to adjust it to resemble the curve above.
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\usepackage{pst-plot}
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\begin{pspicture}[showgrid](-3,-3)(3,3)
\psplot[algebraic,polarplot,linecolor=red,plotpoints=100]{0}{Pi 2 mul}{1-.6*sin(3*x)}
\end{pspicture}
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Question
What is the equation of the polar curve or parametric curve (or any kind of curve) given above?
• Why do you think it is a polar curve? That is, why do you think it has a polar equation form? It does not appear to have one, to my eye. Would a parametric form be just as useful for you? – Matthew Conroy Dec 5 '17 at 4:55
• It looks like a Hypotrochoid to me but I can't get the parameters right to get three leaves. – Ross Millikan Dec 5 '17 at 6:04
• A close, but not perfect, fit is (a projection of) the trefoil knot. It can be parametrized simply. – mephistolotl Dec 5 '17 at 6:26
• $(\cos t+\frac34\cos(2t), \sin t-\frac34\sin(2t))$. Further reading: Hypotrochoid – Rahul Dec 5 '17 at 7:18
• @Rahul: Your answer deserves 25 points. – Money Oriented Programmer Dec 5 '17 at 8:16
Probably not what you are looking for, but, with $$f(t) = \frac{1}{3}t+\frac{3}{2}\cos\left(\frac{1}{2}t\right)-\frac{1}{5}\sin t,$$ the curve $$x(t)=\int_{0}^t \cos(f(u)) \, du, \,\, y(t) = \int_{0}^t \sin(f(u)) \, du$$ looks like this:
which looks like your curve if you squint.
UPDATE: With $$f(t) = \frac{1}{3}t+\cos\left(\frac{1}{2}t\right)-\frac{1}{5}\sin t,$$ the curve is this:
• @ArtificialStupidity I improved it! – Matthew Conroy Dec 5 '17 at 6:58
$$(\cos t+\frac34\cos(2t), \sin t-\frac34\sin(2t))$$
mentioned in Rahul's comment is the most similar to my requirement.
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\multido{\i=0+1}{51}{%
\begin{pspicture}[showgrid=false](-2,-2)(2,2)
\rput{36}(0,0){\psparametricplot[algebraic,polarplot,plotpoints=100,linecolor=red]
{0}{Pi 2 mul 50 div \i\space mul}{cos(t)+3*cos(2*t)/4|sin(t)-3*sin(2*t)/4}}
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| 2019-08-19T20:35:50 |
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https://mathoverflow.net/questions/334020/px-py-has-infinitely-many-integer-solutions
|
# $P(x)=P(y)$ has infinitely many integer solutions
Determine all polynomials $$P(x)$$ with integer coefficients such that $$P(x)=P(y)$$ has infinitely many integer solutions in integer $$x$$ and $$y$$ with $$x \neq y$$.
Choose $$P(x)=a_n(x-k)^{2n}+a_{n-1}(x-k)^{2n-2}+...+a_0$$, with integers $$k,n,a_n,a_{n-1}, ...,a_0$$ Then $$P(x)=P(2k-x)$$ for every integer $$x \neq k$$, thus satisfy the condition.
Now, I have a gut feeling that if $$P(x)=P(y)$$ and $$x+y=M$$, then for every integer $$x$$, $$P(x)=P(M-x)$$, but I cannot analyze any futher.
So my question is: Are there any other types of polynomials $$P(x)$$ that satisfy the orange question above?
(If this question should be closed or off topic, please let me know. If this site cannot answer this question, let me know, I will delete this question immediately)
• It seems easy to see at least that $P$ must be of even degree. – Seva Jun 14 at 18:53
• Very probably something very much like what you claim is true, except it can't be completely correct since $P(x)=x^2+x$ also works, which satisfies $P(x) = P(-1-x)$. There is a subtlety in that the symmetry-axis can oocur for a half-integral $x$-value. – RP_ Jun 14 at 19:02
• @RP_ so, $M=-1$. – Konstantinos Gaitanas Jun 14 at 19:11
• @KonstantinosGaitanas That is clear. My point was that the formula given earlier in the post does not apply in that case. – RP_ Jun 14 at 19:16
First it is clear (assuming throughout that $$P$$ is a solution to your problem) that $$P$$ should have even degree, for if $$P$$ has odd degree we have $$\lim\limits_{n \rightarrow -\infty}P(n) = -\infty$$ and $$\lim\limits_{n \rightarrow \infty}P(n) = \infty$$ (or the same but with the signs of both of the limits reversed of course) which means $$P(x)=P(y)$$ can only happen when $$x$$ and $$y$$ are contained in a bounded interval. So $$P=\sum_{i=0}^d a_i x^i$$ has even degree, and we may assume by flipping the sign and/or applying a translation if necessary, that $$a_d>0$$ and $$-da_d/2 < a_{d-1} \leq da_d/2$$.
First we show that $$P(x) > P(-x+1)$$ for $$x$$ sufficiently large. Therefore we calculate $$P(x)-P(-x+1)=(a_d x^d + a_{d-1}x^{d-1} + \ldots) - (a_d (1 - x)^d + a_{d-1}(1 - x)^{d-1} + \ldots)$$ whose leading term will be $$(da_d + 2a_{d-1})x^{d-1}$$. which will ensure that $$P(x)-P(-x+1)$$ tends to infinity when $$x$$ does, by the lower bound on $$a_{d-1}$$.
Likewise, we show that $$P(x) < P(-x-2)$$ for $$x$$ sufficiently large. This is the same type of calculation: we get $$P(x)-P(-x-2)=(a_d x^d + a_{d-1}x^{d-1} + \ldots) - (a_d (-x -2)^d + a_{d-1}(-x - 2)^{d-1} + \ldots)$$, which now has leading term $$(- 2da_d + 2a_{d-1})x^{d-1}$$, which again clearly tends to negative infinity as $$x$$ grows.
These two inequalities combined mean that we must have either $$P(x)=P(-x)$$ or $$P(x)=P(-x-1)$$ for infinitely many $$x$$. In the first case, write $$P=P_{\textrm{even}}+P_{\textrm{odd}}$$, then this gets us that $$P_{\textrm{odd}}(x)=0$$ for infinitely many $$x$$, so we must have $$P_{\textrm{odd}} \equiv 0$$, so $$P(x)=Q(x^2)$$ for some polynomial $$Q$$. In the second case, we can play the same trick since this time $$R(x) := P(x-1/2)$$ satisfies $$R(x)=R(-x)$$, ergo by the same argument we must have $$P(x)=Q((x+1/2)^2)$$ for some polynomial $$Q$$.
In conclusion: as the entire solution set to your problem is given by translates of these two types of solutions, we get that all solutions are of the form $$Q((x+k/2)^2)$$, where $$k$$ is any integer and $$Q$$ is any polynomial (to be more accurate of course, I should say the subset of all polynomials of this form that have integer coefficients).
Morover I think it should be easy to prove that this description coincides with the set of polynomials of the form $$Q(x^2+ax+b)$$, with $$a$$ and $$b$$ integers and again $$Q$$ any polynomial, which avoids dealing with half-integers altogether, but I will leave this as an exercise...
• The limit argument is not quite right, because the function $\lfloor x/2 \rfloor$ satisfies this limit condition but doesn't have finitely many solutions. But of course your overall point is right - you just need to say in addition that the function is monotonic outside a bounded interval, for instance. – Will Sawin Jun 18 at 19:39
• Thank you Will, you are of course absolutely right. I will fix this at some point, together with the omitted argument for the statement in the final paragraph. – RP_ Jun 19 at 8:18
Here's a more abstract proof:
If $$P(x) - P(y) =0$$ but $$x \neq y$$ then $$(P(x) - P(y) )/(x-y)=0$$. Because this is a polynomial of degree $$n-1$$, when this identity is satisfied its leading terms $$(x^{n} - y^{n} )/ (x-y)$$ must be equal to minus its remaining terms, and thus must be $$O( \max(x,y)^{n-2})$$.
Now if $$n$$ is odd, $$(x^{n} - y^{n} )/ (x-y)$$ is a homogeneous polynomial of degree $$n-1$$ with no nozero real roots. Because it has no nonzero real roots, its absolute value has some minimum value $$C$$ on the boundary of the unit square. Then homogeneity gives $$| (x^{n} - y^{n} )/ (x-y)| \geq C \max(x,y)^{n-1}$$ and thus can be $$O( \max(x,y)^{n-2})$$ for only finitely many $$x,y$$.
If $$n$$ is even, $$(x^n-y^n)/(x^2-y^2)$$ is a homogeneous polynomial of degree $$n-2$$ with no real roots. By the same logic, we have $$| (x^{n} - y^{n} )/ (x^2-y^2)| \geq C \max(x,y)^{n-2}$$. Thus if $$| (x^{n} - y^{n} )/ (x-y)| = O ( \max(x,y)^{n-2})$$ then $$C |x+y| = O(1)$$. So this can happen for only finitely many values of $$|x+y|$$.
Thus if it happens infinitely often, it happens infinitely often for one particular value of $$x+y$$, say $$M$$, thus $$P(x) - P(M-x)$$ vanishes for infinitely many $$x$$ and thus is zero.
| 2019-11-19T12:25:49 |
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https://www.physicsforums.com/threads/radius-of-a-circle.106872/
|
1. Jan 15, 2006
### m00c0w
I'm trying to calculate the radius of the circle $$x^2 - 2ax + y^2 - 2ay = 0$$
I completed the square and got to $$(x-a)^2 + (y-a)^2 = 2{a}^2$$
That would mean the radius is $$\sqrt{2{a}^2}$$ right? Which would be $$\sqrt{2}\sqrt{a^2} \Rightarrow a\sqrt{2}$$... but my textbook quotes the answer as $$\sqrt{2a}$$ Did I do something wrong or is my textbook incorrect?
2. Jan 15, 2006
### arildno
Are you sure you are reading your textbook? It might be that the answer is $\sqrt{2}a[/tex] but a little blurred so it is not clear where the square root sign ends- it looks like [itex]\sqrt{2a}$.
On the otherhand that is exactly why most texts would use the clearer $a\sqrt{2}$ as you do.
| 2017-01-17T07:17:46 |
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https://stats.stackexchange.com/questions/544726/what-is-the-justification-for-weighting-meta-analysis-by-standard-error-instead/544802
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# What is the justification for weighting meta-analysis by standard error instead of sample size?
I have observed in many meta-analyses, and in tutorials on the matter, that the analyzed studies are commonly weighted by the reciprocal of the standard error. This seems to have a face validity because a lower standard error suggests a better estimate of the effect in question.
However, a standard error could be low for two reasons: (1) it could be because there is a larger sample size $$n$$ in a study; or (2) it could be because there is a lower variance in the study sample(s). Weighting by $$n$$ makes sense because one should weight the study not just by the more accurate estimate but also because the study is providing more evidence. And weighting by standard error would tend to be highly correlated with weighting by $$n$$. But why should low sample variance also matter? It could simply be a very unrepresentative sample in a study that causes the variance to be low. In that case one should not be weighting the study heavily.
Also, when doing a fixed effects meta-analysis it seems to violate the assumptions of the analysis to weight by study standard errors since one would expect there is really only one variance and all studies deviate from that variance by some amount. Once study variances are assumed equal weighting by standard error should be weighting by $$n$$.
So, why is weighting by standard error so often favoured over weighting by $$n$$?
I suspect they are not weighting by the standard error, which would mean giving higher weight to less certain results, but weighting by the reciprocal of the square of the standard error, i.e. inverse-variance weighting, because this has the least variance among all weighted averages.
For example the Cochrane Statistical Methods Group suggests the generic inverse-variance weighted average $$\frac{\sum Y_i (1/SE_i^2)}{\sum (1/SE_i^2)}$$ with a reference to Rice K, Higgins JPT, Lumley T. A re-evaluation of fixed effect(s) meta-analysis. Journal of the Royal Statistical Society Series A (Statistics in Society) 2018; 181: 205-227.
Since $$SE_i^2$$ is an estimator of $$\frac1{n_i}\sigma_i^2$$, you might hope that its reciprocal was approximately proportional to $$n_i$$ if all the $$\sigma_i^2$$ are the same and so you might get something similar to your suggestion of weighting by $$n_i$$. But there are two additional reasons for this particular method:
• the individual studies may have taken different approaches to estimating the same parameter, some more accurate than others, and this difference in precision can be taken into account
• most studies may report a standard error for their estimate even when they are more confused about the actual $$n$$ used (especially if parts of the original sample are dropped for one reason or another)
• Yes, of course, I meant the reciprocal. I have edited to reflect the intent. While these seem reasonable suggestions it still seems a stretch to believe that differences in sample variance may be more attributable to measurement quality than sampling error. If that's the primary reason it's on pretty shaky ground.
– John
Sep 19 at 1:27
• @John - suppose you were estimating expected post-treatment survival time. If one study followed $100$ people for a year noting the dates of death of $10$ of them though the others were still alive, and another followed $100$ people for $10$ years noting the dates of death of $65$ though the others were still alive, it is not immediately obvious what the sample sizes were. Sep 19 at 2:02
• "It could simply be a very unrepresentative sample in a study that causes the variance to be low" This is why a risk of bias assessment is usually recommended - otherwise even weighting by $n$ is a bad idea. Sep 19 at 2:09
• @Henry ah that makes sense. I had not considered the situations where the SE is calculable but the N isn't clear. I agree that in that case SE is definitely what you want to go with. But it's not really a good justification for SE as a general rule.
– John
Sep 21 at 1:58
• I didn't follow your wording; I meant that weighting can't solve problems of non-representative (or high risk of bias) samples. If the issue of concern is solely outlying sample variances, and fixed effects is a weak assumption, weighting by sample size makes sense. Sep 21 at 5:23
| 2021-10-17T00:23:14 |
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https://math.stackexchange.com/questions/1676996/tests-for-prime-numbers/1677000
|
# Tests for prime numbers
I'm told to list the prime numbers from 7 to 150 .
I know how to do it the slow way by one by one checking the numbers till 150. But in an exam condition I can't possibility do that .
Is there any way to test for prime numbers ? Or to prove whether that a number is a prime number ?
• en.wikipedia.org/wiki/Sieve_of_Eratosthenes Feb 29, 2016 at 12:29
• Check only numbers that are adjacent to multiples of $6$ (for example, $5$ and $7$, $11$ and $13$, etc). Feb 29, 2016 at 12:31
• Is there a possible exam condition where you will be asked to list primes in such an interval? Feb 29, 2016 at 12:33
• There are many prime generating sieves and an entire theory domain.
– user49763
Feb 29, 2016 at 23:24
• Possible duplicate of Determine whether a number is prime Feb 15, 2019 at 12:44
That seems like a silly test question, but you can use the Sieve of Eratosthenes. Start with all numbers from 7 to 150 and start removing multiples of integers greater than 1. Animation from the Wikipedia link:
As for proving a number is prime, it is actually much easier to show that a number is not prime. Maybe you have some more idea of what you might see on a test, but those seem like irritating questions.
• Note that you're done with the sieve when you reach the square root of the range, in this case $\sqrt{150} \approx~ 12$. Feb 29, 2016 at 14:15
• @RemcoGerlich He is correct (actually you want to test to $\left \lceil{\sqrt{x}}\right \rceil$). See this question: stackoverflow.com/questions/5811151/… Mar 1, 2016 at 13:05
• @Jedediyah I assume you mean $\lfloor \sqrt{x} \rfloor$. Mar 1, 2016 at 13:07
• @Travis Good assumption! ;p Mar 1, 2016 at 13:07
• @RemcoGerlich The animation actually does stop finding composite numbers at 11. You can see that after that test all of the remaining numbers are primes, and don't have any multiples that aren't already colored in. Mar 1, 2016 at 13:11
The illustrated answer above is cool.
Here is something that might be quicker for you during an exam.
The prime numbers between $7$ and $150$ are all the neighbors of multiples of $6$, except:
• Those that end with $5$
• $49$
• $77$
• $91$
• $119$
• $121$
• $133$
• $143$
So you can simply:
• List all multiples of $6$
• List the two neighbors next to each one of them
• Memorize the ones mentioned above and eliminate them
UPDATE:
Just in order to clarify (and simplify) the method mentioned above.
All you have to do is to write down two lists, one starting from $7$ and the other starting from $11$.
Increment each list by $6$ until $150$, and eliminate the values that you have memorized in advance.
$\require{cancel}$
• $\small7,13,19,\color\red{\cancel{25}},31,37,43,\color\green{\cancel{49}},\color\red{\cancel{55}},61,67,73,79,\color\red{\cancel{85}},\color\green{\cancel{91}},97,103,109,\color\red{\cancel{115}},\color\green{\cancel{121}},127,\color\green{\cancel{133}},139,\color\red{\cancel{145}}$
• $\small11,17,23,29,\color\red{\cancel{35}},41,47,53,59,\color\red{\cancel{65}},71,\color\green{\cancel{77}},83,89,\color\red{\cancel{95}},101,107,113,\color\green{\cancel{119}},\color\red{\cancel{125}},131,137,\color\green{\cancel{143}},149$
• Yeah I definitely agree that if you know the range of numbers then it is easier to just memorize the primes in that interval. Feb 29, 2016 at 12:47
• @Jedediyah: Assuming that the range is small of course. Feb 29, 2016 at 12:48
• @Jedediyah: And in the case above, you can just as well check the divisibility of each number in that range by $2,3,5,7,11$ (out of which, the first three are rather easy). Feb 29, 2016 at 12:49
• For the range, doing the one away from a multiple of six you only need to check divisibility by 5,7,11 Five is obvious 11 almost obvious (there are only 13 of them and the single digit ones are obvious. With these tricks it shouldn't take more than a minute even with being very careful. Feb 29, 2016 at 16:55
• @fleablood: 143 doesn't have smaller factors than 11 :-) I think the problem here is more or less what the questioner suggests in the question: rushing it under stress will cause mistakes, taking it slowly and methodically might take too long. As barak shows it's possible to prepare specifically to list primes up to 150 quickly (heck, you could just memorise the sequence of primes). But if you do that and then you're actually asked to list primes up to 160 quickly then your short-cuts may or may not still be valid. Mar 1, 2016 at 11:36
Given a number $n$.
You need to (1) find all primes smaller than $\sqrt{n}$ and (2) see if any of them divides $n$.
If none of them divides $n$ then you have a prime, otherwise it's a composite number.
Example:
Is 147 prime? $\sqrt{147} < 13$ so we have 2, 3, 5, 7, and 11 to check.
3 divides 147 so we don't need to look further.
147 is a composite number.
• The fact we are building on is that if $n$ is a composite then it has a prime factor smaller than $\sqrt{n}$. The proof is easy. Feb 29, 2016 at 14:35
• The would take an incredibly long time if you need to do it on every number between 7 and 150 and the OP already knows how to do that. S/he specifically asked for a method to avoid having to check every single number. Feb 29, 2016 at 16:50
• Agreed it would take long time. But disagree that the question is clear on exactly what is being asked. It ends with two questions "Is there any way to test for prime numbers ? Or to prove whether that a number is a prime number ? " Mar 1, 2016 at 7:10
My favorite mnemonic:
91 is the only number less than 100 that looks like a prime and isn't.
That's because these numbers don't look like primes: evens, multiples of 5, multiples of 3 (sum of digits tells you), 49, 77.
(This builds on @origimbo 's answer.)
• Comparing this with @barak manos's approach, I would say that 119 and 133 also fall in the "look like a prime but aren't" category going up to 150. For me at least, 121 and 143 are "obviously" multiples of 11, but it probably depends on the person. Feb 29, 2016 at 19:28
• For some reason I always think 51 looks prime and 41 doesn't. Don't know why. Still that's a pretty cute statement. Feb 29, 2016 at 19:31
Other answers cover things in far greater detail, but to expand on a comment by barak manos on their own answer, in terms of quick ways of testing and in order of speed of checking you can throw away:
• all even numbers
• all numbers whose digits end in five.
• all numbers whose digits add up to a multiple of three.
A couple of other small primes have more complicated tests, see for example here which covers 7 and 11.
• I'll never be able to remember the rule for division by 7 but "casting out" 7s is easy because 7 is a single digit. Given 840539438261 has the divisibility as 140539438261 which has the same as 539438261 as 119438261 as 49438261 as 438261 as 18261 as 4261 as 61 is not divisible by 7. You can do the same for any number but it's easiest for 7. It's the only way I know for 13. A number is divisible by 11 if the sum of the even place digits is the same as the odd placed digits or off by a multiple of 11. Feb 29, 2016 at 19:41
• @fleablood: another test I learned for 7-divisibility is to subtract 21 times the least digit and then divide by 10. That is to say, discard the least digit and then subtract double the digit you discarded. Hence 840539438261 -> 84053943824 -> 8405394374 -> 840539429 -> 84053924 -> 8405384 -> 840530 > 84053 -> 8399 -> 821 -> 80 -> 8, so not divisible by 7. Note this doesn't tell you the modulus, whereas casting out does. You can do a similar trick for 17 using 51 in place of 21, and 13 using 39 (i.e. add 4 times the digit you chopped off). Mar 1, 2016 at 11:49
• @SteveJessop don't understand. 840539438261 to 84053943824. Okay, that's just subtracting 21 and dividing by 10. 84053943824 to 8405394374. Okay the 2 is the least digit so we remove it. Then... the eight before the 2 becomes a 7????? why? Then 8405394374 to 840539429 ... so the 374 gets turned into 29? WTF? Mar 1, 2016 at 17:04
• @fleablood: by "least digit" I mean "least significant digit", sorry. 84053943824, chop off the last digit would be 8405394382. The last digit was 4, so subtracting 8 from 8405394382 leaves 8405394374. The net result is that we have subtracted 21 * 4 and then divided by 10, both of which operations preserve divisibility by 7. Mar 1, 2016 at 17:07
• Hmmm... in my opinion multiplying and carrying is just as hard as doing a complete division. Casting out is faster and easier. Especially as we have 142 other numbers to test. Mar 1, 2016 at 17:14
I'd probably recommend just knowing all your primes between 7-150.
Start with 7 and count by 2s. If it looks prime, name it. The 3 exceptions between 1-100 are 57, 87 and 91. (I'm assuming "77" and "93" don't look prime to you.)
Keep counting past 100 and write your guesses down. Then check them all by Googling a list of prime numbers. Whichever ones you got wrong, memorize. This won't be many because you'll guess most right.
Prime numbers greater than 3 are of the form $6n \pm 1$ , then you can test manually between 7 and 150. It will be easier than testing every odd integer as mentioned in other answer.
There is no answer yet which points this out: A couple of years ago it was found that you can test for primality in polynomial time. See https://en.wikipedia.org/wiki/AKS_primality_test
• That's true, but given the OP is concerned with a 1-byte input, this breakthrough may not be terribly relevant :P Mar 1, 2016 at 15:07
Start with the square root of 150, the integer value is 12. Your prime divisors are 2 3 5 7 and 11. Eliminate the even numbers and the numbers ending in 5. The first test will be 3 squared or 9 in this case add 6 to 9 and so on to eliminate the 3s. The next test is 7 squared or 49 and this time add 14 to eliminate the 7s. The last test is 11 squared or 121 and this time add 22 to eliminate the 11s. The numbers left on your list should be prime numbers. Or Make a chart with 2, 3, 7, 9 on the left side and 0, 10, 20, 30 ..... 140 across the top. This eliminates the even numbers and numbers ending in 5. Then it's 3*3 = 9, 3*7 = 21, 3*9 = 27, 11*3 = 33 etc. Then 7*7 = 49, 7*11 = 77, 7*13 = 91 etc. Then 11*11 = 121 and 11*13 = 143 and the remaining numbers are prime.
| 2022-05-26T14:35:48 |
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http://math.stackexchange.com/questions/253298/howto-organize-a-party-where-everyone-meets-everyone-around-a-big-table/253354
|
# Howto organize a party where everyone meets everyone around a big table?
Ok hi all, my first question! I would like to organize a party where everyone meets everyone, the table is organized like this:
ABCD
J E
IHGF
So A can only meet B and J, and B can only meet A and C and so on. So I suppose it could be seen as a lot of groups of 3s + 1 or 2. (abc) (def) (ghi)+J , but there is also the groups (bcd) (efg) (hij)+a
How should I move them around?
I will not know exact numbers before they arrive! Any help much appreciated.
Seems a bit like this question but a bigger table! How to rotate n individuals at a dinner party so that every guest meets every other guests and How to derive a general formula for this problem? (pairs of people seated around a table)
-
also seems a bit like this: math.stackexchange.com/questions/58922/… – user1392166 Dec 7 '12 at 20:21
You give all of them a version of the Keirsey personality test. Then you kick out all of the introverts. They'll move themselves. More seriously, my intuition tells me that you want to do something like picking an individual say A to swap positions with the person on the opposite corner like F, while B, J, E, and G stay seated. Just a guess. – Doug Spoonwood Dec 7 '12 at 20:45
swapping with corners would be ok but BC would stay together, which seem a little ineffective! – user1392166 Dec 7 '12 at 20:57
You'd have C also swap places. – Doug Spoonwood Dec 7 '12 at 20:59
I suppose the is a finite number of possible combinations, where each member of a set of 3 is less likely to pick a set with members it's already seen, that's more dealing with it from a programming point of view than mathematical. I'd quite link to understand what I am doing? – user1392166 Dec 7 '12 at 21:18
## 2 Answers
Partial answer:
Let's start with the simple case that the number $n$ of people is an odd prime: Assign a number $0, \ldots, n-1$ to each guest and in round $k$ let guest $i$ sit on chair $ki\bmod n$. Then guest $i$ and $j$ are neighbours when $k(i-j)\equiv \pm 1\pmod n$. If we let $k$ run from $1$ to $\frac{n-1}2$, the multiplicative inverse of $i-j$ or its negative are among the $k$ values, hence $i$ and $j$ have met. Clearly, it is not possible to get away with less than $\frac{n-1}2$ rounds.
If $n$ is not prime it may be worth considereing a few more invitations until the number is prime :)
-
Suppose you have $n$ people, numbered $0$ to $n-1$. We might as well suppose person $0$ stays put, and the others move around. So a "configuration" corresponds to a permutation of $1$ to $n-1$. Let $a_{i,j,c} = 1$ if $i$ and $j$ are seated next to each other in configuration $c$, $0$ if not. You can think of your problem as an integer linear programming problem:
\eqalign{\text{minimize } & \sum_c x_c\cr \text{subject to } & \sum_c a_{i,j,c} x_c \ge 1 \ \text{for all } 0 \le i<j \le n-1 \cr x_c \in \{0,1\}\cr}
The trouble is that there are a huge number of configurations ($9! = 362880$ in your example).
I tried a different approach: tabu search. In the cases I've tried it didn't take long to come up with an optimal solution. Here's my Maple program:
n:= 10; # number of guests
m:= ceil((n-1)/2); # minimal number of rounds required
scorer:= proc(L)
# count number of introductions for an m-tuple of configurations
nops(map(op,{seq([seq({L[i,j],L[i,j+1]},j=1..n-2),
{0,L[i,1]},{0,L[i,n-1]}],i=1..m)}))
end proc;
move:= proc(i,j,k,L)
# switch j'th and k'th positions in configuration i of m-tuple L
local M;
M:= L;
M[i,j]:= L[i,k]; M[i,k]:= L[i,j];
M
end proc;
target:= n*(n-1)/2; # necessary number of introductions
current:= [[$1..(n-1)],seq(combinat[randperm](n-1),i=2..m)]; tabu:= [seq([2,i],i=1..7)]; # these items temporarily can't be switched currentscore:= scorer(current); bestyet:= current; bestscore:= scorer(bestyet); for iter from 1 to 1000 while bestscore < target do bestnewscore:= 0; for i from 2 to m do for j from 1 to n-2 do if member([i,j],tabu) then next fi; for k from j+1 to n-1 do if member([i,k],tabu) then next fi; newtry:= move(i,j,k,current); newscore:= scorer(newtry); if newscore > bestnewscore then bestnew:= newtry; bestnewscore:= newscore; ij:= [i,j]; ik:= [i,k]; fi od od od: if bestnewscore > bestscore then bestyet:= bestnew; bestscore:= bestnewscore; printf("Best yet at iteration %d: %d\n",iter, bestscore); fi; current:= bestnew; tabu:= [op(tabu[3..7]),ij,ik]; if iter mod 20 = 0 then printf("iteration %d, score %d\n",iter, bestnewscore) fi; od: if bestnewscore = target then printf("Optimal configuration ") else printf("Best configuration so far ") fi; printf("with score %d out of %d:\n",bestscore,target); print(bestyet); For example, with$n=15\$ it took 312 iterations to come up with an optimal solution:
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14],
[5, 10, 3, 11, 13, 7, 9, 6, 14, 4, 12, 1, 8, 2],
[11, 7, 10, 1, 5, 13, 2, 4, 8, 14, 9, 12, 3, 6],
[10, 14, 3, 7, 4, 11, 2, 6, 8, 12, 5, 9, 1, 13],
[3, 5, 7, 2, 14, 12, 6, 10, 8, 13, 4, 1, 11, 9],
[8, 11, 14, 5, 2, 12, 10, 13, 3, 9, 4, 6, 1, 7],
[4, 10, 2, 9, 13, 6, 11, 5, 8, 3, 1, 14, 7, 12]]
That is, there are 7 rounds; each row above lists the seating arrangement around the table starting to the right of person 0.
-
Gosh thanks, it's going to take me a while as I don't know Maple programming! seems really quite complex! – user1392166 Dec 8 '12 at 21:24
| 2015-01-31T18:31:22 |
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https://www.physicsforums.com/threads/geometric-sum.238436/
|
# Geometric Sum
## Homework Statement
Calculate the following sum:
$$\sum_{k=1}^{10} 3 \cdot 2^{k}$$
## The Attempt at a Solution
$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$
Although it seems that I am missing something.
Related Precalculus Mathematics Homework Help News on Phys.org
D H
Staff Emeritus
$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$
Although it seems that I am missing something.
Very close but no cigar. How can $\sum_{k=1}^{10} 2^{k} = 2^{10} - 1$? You are missing something from the right-hand side.
Right, if you divide a term in the sum by the previous term, you get 2, not 1? So a factor of 2 is missing.
Gib Z
Homework Helper
Do you know $$x^n - y^n = (x-y)(x^{n-1} + x^{n-1}y + ... + y^{n-1} )$$ ?
For (x,y) = (2,1) that forms into what you need. Look at the expansion of what you have, 2^10 - 1, and see how that falls different to the sum you want on the RHS, then change it accordingly =]
**Edited by user.
Last edited:
Gib Z
Homework Helper
ok sorry buddy!
@author of thread... look up summation of a GP
HallsofIvy
Homework Helper
## Homework Statement
Calculate the following sum:
$$\sum_{k=1}^{10} 3 \cdot 2^{k}$$
## The Attempt at a Solution
$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$
Although it seems that I am missing something.
The formula you are trying to remember is
$$\sum_{k=0}^n a r^k= a\frac{1-r^{n+1}}{1- r}[/itex] For r= 2, so that 1-r= 1-2= -1, that becomes [tex]\sum_{k=0}^n a 2^k= a (2^{n+1}- 1)$$
There are two important differences between that and the formula you are using.
Let's see.
The general formula is
$$\sum_{k=0}^n a r^k= a\frac{1-r^{n+1}}{1- r}$$
Now the terms in the sum
$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 6 + 12 + 24 + ... + 3072$$
This means that
n = 9 (10-1)
r = 2 (taking the n:th term/(n-1):th term)
a = 6 (the starting term)
So
$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 6\frac{1-2^{9+1}}{1-2} = 6\frac{1-2^{10}}{1- 2} = 6(2^{10} - 1)$$
Why is n = 9 here? Is that just the number of terms that have an exponent >= 2?
Someone above told me not to give out the answer.. and btw your answer is correct.. and yes n = 10 not 9 . There are 10 terms of the GP.
HallsofIvy
Homework Helper
spdeyunlimit, n= 9, not 10 because because the sum starts with k= 1. He is thinking of this as 6+ 6(2)+ 6(22)+ 6(2n) with n running from 0 to 9.
Another way to do the problem would be to use the same formula with n= 10, a= 3, but then subtract off 3- to allow for the missing 3(20) term.
for the GP
a, ar, ar^2, ar^3 ... and so on
Sum of first n terms = a(1 - r^n) / (1-r)
where in the first term the power of r is 0.. whereas here it is one.. so yeah i got that.. it will be for n = 9
| 2020-01-25T16:50:32 |
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|
https://www.mathworks.com/help/matlab/ref/istriu.html
|
# istriu
Determine if matrix is upper triangular
## Description
example
tf = istriu(A) returns logical 1 (true) if A is an upper triangular matrix; otherwise, it returns logical 0 (false).
## Examples
collapse all
Create a 5-by-5 matrix.
A = triu(magic(5))
A = 5×5
17 24 1 8 15
0 5 7 14 16
0 0 13 20 22
0 0 0 21 3
0 0 0 0 9
Test A to see if it is upper triangular.
istriu(A)
ans = logical
1
The result is logical 1 (true) because all elements below the main diagonal are zero.
Create a 5-by-5 matrix of zeros.
Z = zeros(5);
Test Z to see if it is upper triangular.
istriu(Z)
ans = logical
1
The result is logical 1 (true) because an upper triangular matrix can have any number of zeros on the main diagonal.
## Input Arguments
collapse all
Input array, specified as a numeric array. istriu returns logical 0 (false) if A has more than two dimensions.
Data Types: single | double
Complex Number Support: Yes
collapse all
### Upper Triangular Matrix
A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$
is upper triangular. A diagonal matrix is both upper and lower triangular.
## Tips
• Use the triu function to produce upper triangular matrices for which istriu returns logical 1 (true).
• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istriu(A) == isbanded(A,0,size(A,2)).
| 2020-02-17T07:48:31 |
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http://mathhelpforum.com/algebra/93616-quadratic-equation-part-1-a.html
|
# Math Help - Quadratic Equation Part 1
1. ## Quadratic Equation Part 1
How to do these?
1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form.
2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots.
3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots.
4. Find the range of values of k if $x^2+(k-3)x+k=0$ has roots with same sign.
1. $\frac{1}{2}(-3 \pm \sqrt{17})$
2. -1<q<3
3. $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$
4. k>0
2. For (4), you got part of it correct (namely the product of the roots is positive). But you also want your roots to be real, since it doesn’t make any sense to talk about two conjugate complex numbers having the same sign.
3. 1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form.
\begin{aligned}
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\
&= \frac{-3 \pm \sqrt{17}}{4} \\
&= \frac{1}{4}(-3 \pm \sqrt{17})
\end{aligned}
Hmm, I'm getting a different answer.
2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots.
A quadratic equation in the form $ax^2 + bx + c = 0$ has complex roots if the discriminant $b^2 - 4ac < 0$. So
a = 1
b = q + 1
c = q + 1
\begin{aligned}
b^2 - 4ac &< 0 \\
(q + 1)^2 - 4(1)(q + 1) &< 0 \\
q^2 + 2q + 1 - 4q - 4 &< 0 \\
q^2 - 2q - 3 &< 0 \\
(q - 3)(q + 1) &< 0
\end{aligned}
Make a sign chart. Draw a number line and the critical points -1 and 3 (you get these by setting each factor equal to 0).
Code:
----+----+----+----+----+----+----+----
-1 3
You have 3 intervals to test:
(-∞, -1)
(-1, 3)
(3, ∞)
From each interval, pick a number inside it to test into the inequality (q - 3)(q + 1) < 0, in order to determine the sign. The signs of the polynomial for each interval is as follows:
Code:
----+----+----+----+----+----+----+----
pos -1 neg 3 pos
So the answer is -1 < q < 3.
01
4. Originally Posted by yeongil
\begin{aligned}
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\
&= \frac{-3 \pm \sqrt{17}}{4} \\
&= \frac{1}{4}(-3 \pm \sqrt{17})
\end{aligned}
Hmm, I'm getting a different answer.
I find this too...
5. 3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots.
This is the same as #2, but this time, since we want real and distinct roots, the discriminant has to be positive, or $b^2 - 4ac > 0$. So
\begin{aligned}
x^2 + x + 1 &= m(x + 2) \\
x^2 + x + 1 &= mx + 2m \\
x^2 + x - mx + 1 - 2m &= 0 \\
x^2 + (1 - m)x + (1 - 2m) &= 0 \\
\end{aligned}
a = 1
b = 1 - m
c = 1 - 2m
\begin{aligned}
b^2 - 4ac &> 0 \\
(1 - m)^2 - 4(1)(1 - 2m) &> 0 \\
1 - 2m + m^2 - 4 + 8m &> 0 \\
m^2 + 6m - 3 &> 0
\end{aligned}
I want to find the zeros of the polynomial $m^2 + 6m - 3$. Let's pretend for a moment that it equals 0 and solve for m:
\begin{aligned}
m^2 + 6m - 3 &= 0 \\
m^2 + 6m &= 3 \\
m^2 + 6m + 9 &= 3 + 9 \\
(m + 3)^2 &= 12 \\
m + 3 &= \pm\sqrt{12} \\
m &= -3 \pm 2\sqrt{3} \\
\end{aligned}
Make a sign chart. Draw a number line and the critical points (the zeros we found earlier):
Code:
---------+-------------------+---------
-6.46 0.46
( $0.46 \approx -3 + 2\sqrt{3}$ and $-6.46 \approx -3 - 2\sqrt{3}$)
You have 3 intervals to test:
$(-\infty,\;-3 - 2\sqrt{3})$
$(-3 - 2\sqrt{3},\;-3 + 2\sqrt{3})$
$(-3 + 2\sqrt{3}, \infty)$
From each interval, pick a number inside it to test into the inequality $m^2 + 6m - 3 > 0$, in order to determine the sign. The signs of the polynomial for each interval is as follows:
Code:
---------+-------------------+---------
pos -6.46 neg 0.46 pos
So the answer is $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$.
01
6. Originally Posted by yeongil
\begin{aligned}
! I got that answer too! But the answer in my book is $\frac{1}{2}$... Maybe there is a mistake in my book?
7. Yeah. Looks like a mistake. I got $\frac{1}{4}(-3\pm\sqrt{17})$ too
| 2016-02-06T01:15:00 |
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https://nrich.maths.org/714/solution
|
### Doodles
A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not 'triple points'. Number the vertex points in any order. Starting at any point on the doodle, trace it until you get back to where you started. Write down the numbers of the vertices as you pass through them. So you have a [not necessarily unique] list of numbers for each doodle. Prove that 1)each vertex number in a list occurs twice. [easy!] 2)between each pair of vertex numbers in a list there are an even number of other numbers [hard!]
### Russian Cubes
How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.
### Snooker
A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament?
# One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Dorothy of Madras College, St Andrews and Maren & Corinna of Camden School for Girls used the same notation denoting the people by the numbers 1, 2, 3,...? in ascending order of height. As each person at the back must be taller than the person directly in front of them and, along the rows, the heights must increase from left to right the 1 must be in front on the left and the tallest must be at the back on the right. Dorothy gave the results for 2, 4, 6, and 8 people in the following table.
Number of people Number of arrangements Diagrams
2 1
2 1
4 2
2 4 1 3 3 4 1 2
6 5
2 4 6 1 3 5 3 4 6 1 2 5 2 5 6 1 3 4 3 5 6 1 2 4 4 5 6 1 2 3
8 14
2 4 6 8 1 3 5 7 3 4 6 8 1 2 5 7 2 5 6 8 1 3 4 7 3 5 6 8 1 2 4 7 4 5 6 8 1 2 3 7 2 4 7 8 1 3 5 6 3 4 7 8 1 2 5 6 2 5 7 8 1 3 4 6 3 5 7 8 1 2 4 6 4 5 7 8 1 2 3 6 2 6 7 8 1 3 4 5 3 6 7 8 1 2 4 5 4 6 7 8 1 2 3 5 5 6 7 8 1 2 3 4
Many people think that because the sequence 1, 2, 5, 14 ? goes up in powers of 3 (with differences 1, 3 and 9) the next difference will be 3 cubed to give the next number 14 + 27 = 41. Maths is full of patterns but, when you think you spot a pattern you have to prove it always works. If you count the arrangements for 10 people the answer is 42 and not 41. Can you find them all?
The solution for 10 people was sent in by Andrei from Tudor Vianu National College, Bucharest, Romania.
For 10 people I shall describe the arrangements briefly without writing all the numbers. I see that 1 and 10 are fixed.
- I start with 1 2 3 4 in the first line, and there are 5 possibilities for the last position: 5, 6, 7, 8 and 9.
- If 4 goes on the second line, than in the first there will be 1 2 3 5, and there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 5 goes besides 4 on the second line, than in the first there will be 1 2 3 6 and there are 3 possibilities for the last place: 7, 8 and 9.
- If 6 goes also on the second, there will be 1 2 3 7 8 or 1 2 3 7 9 in the first. So, for the sequence 1 2 3 in the first line, there are (5 + 4 + 3 + 2) possibilities.
I change 3 on the second line.
- If on the first line there is the sequence 1 2 4 5, there are 4 possibilities for the last place: 6, 7, 8 and 9.
- If 4 goes on the second line, than on the first there will be 1 2 5 6 followed by 7, 8 or 9 '?? 3 possibilities
- If 5 goes too on the second line, than 1 2 6 7 will be followed by 8 or 9 '?? 2 possibilities.
So, for the group 3 x x x 10 1 2 x x x , there are (4 + 3 + 2) possibilities. Studying all the possibilities further, I arrived at the conclusion that for 10 people, I find the following number of possibilities:
(5 + 4 + 3 + 2) + (4 + 3 + 2) + (3 + 2) +(4 + 3 + 2) + (3 + 2) = 14 +9 +5 + 9 + 5 = 42
Summarising, up to now I found the following table:
No. of people 2 4 6 8 10 No of possibilities 1 2 5 14 42
For 12 people, I see that there are 132 possibilities, which could be counted as:
(6 + 5 + 4 + 3 + 2) + (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2)
+ (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2)
+ (4 + 3 + 2) + ( 3 + 2)
+ (5 + 4 + 3 + 2) + (4 + 3 + 2) + ( 3 + 2)
+ (4 + 3 + 2) + ( 3 + 2) = 48 + 28 +14 + 28 + 14 = 132
At this point, I had the chance to look back on the NRICH site, and to discover the link to the problem Counting Binary Ops. I recognized the numbers I found (1, 2, 5, 14, 42, 132) as being the first terms of the sequence of Catalan numbers. Looking more for Catalan numbers, e.g. at http://mathworld.wolfram.com/CatalanNumber.html I found both the recurrence relation and the explicit formula.
If $C_n$ is the $n$-th Catalan number ($n$ being for our problem the number of places in one line, i.e. half the number of people), $$C_n ={1\over n+1}{2n\choose n}$$ and $${C_{n+1}\over C_n} = {2(2n+1)\over n+2}$$ or, if $C_0 = 1$ (there is exactly one way of arranging zero people: don't put them in any way), then: $$\begin{eqnarray} C_1&=&C_0C_0= 1 \\ C_2&=&C_1C_0 + C_0C_1 = 2 \\ C_3&=&C_2C_0 + C_1C_1 + C_0C_2 = 5 \\ C_4&=&C_3C_0 + C_2C_1 + C_1C_2 + C_0C_3 = 14 \\ C_5&=&C_4C_0 + C_3C_1 + C_2C_2 + C_1C_3 + C_0C_4 = 42 ... \\ C_n &=& C_{n-1}C_0 + C_{n-2}C_1 + ...C_1C_{n-2}+ C_0C_{n-1} \end{eqnarray}$$
I have found that Richard P. Stanley has shown that the number of distinct standard Young tableaux of shape (n; n) form the sequence of Catalan Numbers. But the arrangements in the photo are Young tableaux with reversed lines.
To complete the solution, it is necessary either to show the correspondence between this problem and a classical problem involving Catalan Numbers (e.g. Euler's polygon division problem), or to find the recursive (or explicit) formula for the number of possibilities of arranging people in the photo.
| 2016-07-26T12:26:22 |
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https://math.stackexchange.com/questions/2269679/integrable-function-which-limit-does-not-go-to-0-for-x-to-infinity
|
# Integrable function which limit does not go to 0 for x to infinity
I want to find a function, such that the function shall be $$p$$-integrable in the sense of $$\int_{\mathbb{R}}|f(x)|^pdx<\infty$$ for any $$p\in[1,\infty)$$ and that this does NOT imply $$\lim_{x\to\infty}f(x)=0$$.
Now it's similar to the question here Continuous unbounded but integrable functions, but I'm not looking for a function which is unbounded in any point, just one which limit in infinity does not converge to $$0$$ or better say does not exist.
Does anyone know of any examples?
• Do you care if $f$ is continuous? – πr8 May 7 '17 at 10:37
• nope, it doesn't matter to me – N. Maks May 7 '17 at 10:39
• and do you mean Lebesgue or Riemann integable? – πr8 May 7 '17 at 10:39
• Let's take Lebesgue – N. Maks May 7 '17 at 10:41
• Sure, then take $f(x) = \exp(-|x|) + \mathbb{I}[x\in\mathbb{Q}]$. (Or, as one of the answers states, even just $f(x) = \mathbb{I}[x\in\mathbb{Q}]$ ). – πr8 May 7 '17 at 10:43
There are, of course, a lot of examples of $p$ integrable functions such that the limit at infinity does not exist. For example, one can take the characteristic function of $\mathbb{Q}$.
It should be noted, however, that there is no function such that $\lim_{x \rightarrow \infty} f(x)$ exists and is $\ne 0$.
Indeed, assume $\lim_{x \rightarrow \infty} f(x) = \alpha > 0$. Then, for every $\epsilon >0$, there exists $M>0$ such that $$f(x) > \alpha - \epsilon \qquad \text{for every } x > M.$$ If you take $\epsilon$ such that $\alpha - \epsilon >0$ than you have that $$\int_{\mathbb{R}} |f(x)|^p \, dx \ge \int_{\{x > M\}} (\alpha - \epsilon)^p \, dx= \infty.$$
A similar argument works for $\alpha < 0$.
Take $f$ to be a piecewise affine function. At each natural number $n$ the graph of $f$ is a triangle of height $1$ (so $f(n)=1$) and basis $\frac{1}{n^{2/p}}$ and $f$ is zero elsewhere. Then $\limsup_{x\to\infty}f(x)=1$ while $$\int_0^\infty |f(x)|^p\,dx=\sum_n \frac1{n^2}<\infty.$$
• when I replied $p\ge 1$ was not specified, so it could have been $p<1$. – Gio67 May 7 '17 at 17:12
• Sorry, I deleted my previous comment. There is a bigger problem: The base doesn't get raised to the $p$th power, $f$ does. – zhw. May 7 '17 at 17:15
| 2019-06-18T17:11:31 |
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https://math.libretexts.org/TextMaps/Precalculus_TextMaps/Map%3A_Precalculus_(Stitz-Zeager)/10%3A_Foundations_of_Trigonometry/10.6%3A_The_Inverse_Trigonometric_Functions
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# 10.6: The Inverse Trigonometric Functions
As the title indicates, in this section we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example \ref{inverserestrictionex} in Section \ref{InverseFunctions} to obtain a one-to-one function. We first consider $$f(x) = \cos(x)$$. Choosing the interval $$[0,\pi]$$ allows us to keep the range as $$[-1,1]$$ as well as the properties of being smooth and continuous.
Recall from Section \ref{InverseFunctions} that the inverse of a function $$f$$ is typically denoted $$f^{-1}$$. For this reason, some textbooks use the notation $$f^{-1}(x) = \cos^{-1}(x)$$ for the inverse of $$f(x) = \cos(x)$$. The obvious pitfall here is our convention of writing $$(\cos(x))^2$$ as $$\cos^{2}(x)$$, $$(\cos(x))^3$$ as $$\cos^{3}(x)$$ and so on. It is far too easy to confuse $$\cos^{-1}(x)$$ with $$\frac{1}{\cos(x)} = \sec(x)$$ so we will not use this notation in our text.\footnote{But be aware that many books do! As always, be sure to check the context!} Instead, we use the notation $$f^{-1}(x) = \arccos(x)$$, read arc-cosine of $$x$$'. To understand the arc' in arccosine', recall that an inverse function, by definition, reverses the process of the original function. The function $$f(t) = \cos(t)$$ takes a real number input $$t$$, associates it with the angle $$\theta = t$$ radians, and returns the value $$\cos(\theta)$$. Digging deeper,\footnote{See page \pageref{wrappingfunction} if you need a review of how we associate real numbers with angles in radian measure.} we have that $$\cos(\theta) = \cos(t)$$ is the $$x$$-coordinate of the terminal point on the Unit Circle of an oriented arc of length $$|t|$$ whose initial point is $$(1, 0)$$. Hence, we may view the inputs to $$f(t) = \cos(t)$$ as oriented arcs and the outputs as $$x$$-coordinates on the Unit Circle. The function $$f^{-1}$$, then, would take $$x$$-coordinates on the Unit Circle and return oriented arcs, hence the arc' in arccosine. Below are the graphs of $$f(x) = \cos(x)$$ and $$f^{-1}(x) = \arccos(x)$$, where we obtain the latter from the former by reflecting it across the line $$y=x$$, in accordance with Theorem \ref{inverseuniquegraph}. \index{arccosine ! graph of}
It should be no surprise that we call $$g^{-1}(x) = \arcsin(x)$$, which is read arc-sine of $$x$$'. \index{arcsine ! graph of}
We list some important facts about the arccosine and arcsine functions in the following theorem.
Note: Properties of the Arccosine and Arcsine Functions
1. Properties of $$F(x)= \arccos(x)$$
• Domain: $$[-1,1]$$
• Range: $$[0,\pi]$$
• $$\arccos(x) = t$$ if and only if $$0 \leq t \leq \pi$$ and $$\cos(t) = x$$
• $$\cos(\arccos(x)) = x$$ provided $$-1 \leq x \leq 1$$
• $$\arccos(\cos(x)) = x$$ provided $$0 \leq x \leq \pi$$
1. Properties of $$G(x) = \arcsin(x)$$
• Domain: $$[-1,1]$$
• Range: $$\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$$
• $$\arcsin(x) = t$$ if and only if $$-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$$ and $$\sin(t) = x$$
• $$\sin(\arcsin(x)) = x$$ provided $$-1 \leq x \leq 1$$
• $$\arcsin(\sin(x)) = x$$ provided $$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$$
Everything in Theorem \ref{arccosinesinefunctionprops} is a direct consequence of the facts that $$f(x) = \cos(x)$$ for $$0 \leq x \leq \pi$$ and $$F(x) = \arccos(x)$$ are inverses of each other as are $$g(x) = \sin(x)$$ for $$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$$ and $$G(x) = \arcsin(x)$$. It's about time for an example.
Example $$\PageIndex{1}$$:
1. Find the exact values of the following.
• $$\arccos\left(\frac{1}{2}\right)$$
• $$\arcsin\left(\frac{\sqrt{2}}{2}\right)$$
• $$\arccos\left(-\frac{\sqrt{2}}{2}\right)$$
• $$\arcsin\left(-\frac{1}{2}\right)$$
• $$\arccos\left( \cos\left(\frac{\pi}{6}\right)\right)$$
• $$\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right)$$
• $$\cos\left(\arccos\left(-\frac{3}{5}\right)\right)$$
• $$\sin\left(\arccos\left(-\frac{3}{5}\right)\right)$$
1. Rewrite the following as algebraic expressions of $$x$$ and state the domain on which the equivalence is valid.
• $$\tan\left(\arccos\left(x \right)\right) \label{tanarccos}$$
• $$\cos\left(2 \arcsin(x)\right) \label{cosarcsin}$$
Solution
1. To find $$\arccos\left(\frac{1}{2}\right)$$, we need to find the real number $$t$$ (or, equivalently, an angle measuring $$t$$ radians) which lies between $$0$$ and $$\pi$$ with $$\cos(t) = \frac{1}{2}$$. We know $$t = \frac{\pi}{3}$$ meets these criteria, so $$\arccos\left(\frac{1}{2}\right)= \frac{\pi}{3}$$.
2. The value of $$\arcsin\left(\frac{\sqrt{2}}{2}\right)$$ is a real number $$t$$ between $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$ with $$\sin(t) = \frac{\sqrt{2}}{2}$$. The number we seek is $$t = \frac{\pi}{4}$$. Hence, $$\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$$.
3. The number $$t = \arccos\left(-\frac{\sqrt{2}}{2}\right)$$ lies in the interval $$[0,\pi]$$ with $$\cos(t) = -\frac{\sqrt{2}}{2}$$. Our answer is $$\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$$.
4. To find $$\arcsin\left(-\frac{1}{2}\right)$$, we seek the number $$t$$ in the interval $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ with $$\sin(t) = -\frac{1}{2}$$. The answer is $$t = -\frac{\pi}{6}$$ so that $$\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$.
5. Since $$0 \leq \frac{\pi}{6} \leq \pi$$, we could simply invoke Theorem \ref{arccosinesinefunctionprops} to get $$\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}$$. However, in order to make sure we understand \textit{why} this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, $$\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \arccos\left( \frac{\sqrt{3}}{2}\right)$$. Now, $$\arccos\left( \frac{\sqrt{3}}{2}\right)$$ is the real number $$t$$ with $$0 \leq t \leq \pi$$ and $$\cos(t) = \frac{\sqrt{3}}{2}$$. We find $$t = \frac{\pi}{6}$$, so that $$\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}$$.
6. Since $$\frac{11\pi}{6}$$ does not fall between $$0$$ and $$\pi$$, Theorem \ref{arccosinesinefunctionprops} does not apply. We are forced to work through from the inside out starting with $$\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right) = \arccos\left(\frac{\sqrt{3}}{2}\right)$$. From the previous problem, we know $$\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$. Hence, $$\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right) = \frac{\pi}{6}$$.
7. One way to simplify $$\cos\left(\arccos\left(-\frac{3}{5}\right)\right)$$ is to use Theorem \ref{arccosinesinefunctionprops} directly. Since $$-\frac{3}{5}$$ is between $$-1$$ and $$1$$, we have that $$\cos\left(\arccos\left(-\frac{3}{5}\right)\right) = -\frac{3}{5}$$ and we are done. However, as before, to really understand \textit{why} this cancellation occurs, we let $$t = \arccos\left(-\frac{3}{5}\right)$$. Then, by definition, $$\cos(t) = -\frac{3}{5}$$. Hence, $$\cos\left(\arccos\left(-\frac{3}{5}\right)\right) = \cos(t) = -\frac{3}{5}$$, and we are finished in (nearly) the same amount of time.
8. As in the previous example, we let $$t = \arccos\left(-\frac{3}{5}\right)$$ so that $$\cos(t) = -\frac{3}{5}$$ for some $$t$$ where $$0 \leq t \leq \pi$$. Since $$\cos(t) < 0$$, we can narrow this down a bit and conclude that $$\frac{\pi}{2} < t < \pi$$, so that $$t$$ corresponds to an angle in Quadrant II. In terms of $$t$$, then, we need to find $$\sin\left(\arccos\left(-\frac{3}{5}\right)\right) = \sin(t)$$. Using the Pythagorean Identity $$\cos^{2}(t) + \sin^{2}(t) = 1$$, we get $$\left(-\frac{3}{5}\right)^2 + \sin^{2}(t) = 1$$ or $$\sin(t) = \pm \frac{4}{5}$$. Since $$t$$ corresponds to a Quadrants II angle, we choose $$\sin(t) = \frac{4}{5}$$. Hence, $$\sin\left(\arccos\left(-\frac{3}{5}\right)\right) = \frac{4}{5}$$.
1. We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let $$t = \arccos(x)$$, so our goal is to find a way to express $$\tan\left(\arccos\left(x \right)\right) = \tan(t)$$ in terms of $$x$$. Since $$t = \arccos(x)$$, we know $$\cos(t) = x$$ where $$0 \leq t \leq \pi$$, but since we are after an expression for $$\tan(t)$$, we know we need to throw out $$t = \frac{\pi}{2}$$ from consideration. Hence, either $$0 \leq t < \frac{\pi}{2}$$ or $$\frac{\pi}{2} < t \leq \pi$$ so that, geometrically, $$t$$ corresponds to an angle in Quadrant I or Quadrant II. One approach\footnote{Alternatively, we could use the identity: $$1 + \tan^{2}(t) = \sec^{2}(t)$$. Since $$x = \cos(t)$$, $$\sec(t) = \frac{1}{\cos(t)} = \frac{1}{x}$$. The reader is invited to work through this approach to see what, if any, difficulties arise.} to finding $$\tan(t)$$ is to use the quotient identity $$\tan(t) = \frac{\sin(t)}{\cos(t)}$$. Substituting $$\cos(t) = x$$ into the Pythagorean Identity $$\cos^{2}(t) + \sin^{2}(t) = 1$$ gives $$x^2 + \sin^{2}(t) = 1$$, from which we get $$\sin(t) = \pm \sqrt{1-x^2}$$. Since $$t$$ corresponds to angles in Quadrants I and II, $$\sin(t) \geq 0$$, so we choose $$\sin(t) = \sqrt{1-x^2}$$. Thus, $\tan(t) = \dfrac{\sin(t)}{\cos(t)} = \dfrac{\sqrt{1-x^2}}{x}$ To determine the values of $$x$$ for which this equivalence is valid, we consider our substitution $$t = \arccos(x)$$. Since the domain of $$\arccos(x)$$ is $$[-1,1]$$, we know we must restrict $$-1 \leq x \leq 1$$. Additionally, since we had to discard $$t = \frac{\pi}{2}$$, we need to discard $$x = \cos\left(\frac{\pi}{2}\right) = 0$$. Hence, $$\tan\left(\arccos\left(x \right)\right) =\frac{\sqrt{1-x^2}}{x}$$ is valid for $$x$$ in $$[-1,0)\cup(0,1]$$.
2. We proceed as in the previous problem by writing $$t = \arcsin(x)$$ so that $$t$$ lies in the interval $$\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$$ with $$\sin(t) = x$$. We aim to express $$\cos\left(2 \arcsin(x)\right) = \cos(2t)$$ in terms of $$x$$. Since $$\cos(2t)$$ is defined everywhere, we get no additional restrictions on $$t$$ as we did in the previous problem. We have three choices for rewriting $$\cos(2t)$$: $$\cos^{2}(t) - \sin^{2}(t)$$, $$2\cos^{2}(t) - 1$$ and $$1 - 2\sin^{2}(t)$$. Since we know $$x = \sin(t)$$, it is easiest to use the last form: $\cos\left(2 \arcsin(x)\right) = \cos(2t) = 1 - 2\sin^{2}(t) = 1 - 2x^2$ To find the restrictions on $$x$$, we once again appeal to our substitution $$t = \arcsin(x)$$. Since $$\arcsin(x)$$ is defined only for $$-1 \leq x \leq 1$$, the equivalence $$\cos\left(2 \arcsin(x)\right) = 1-2x^2$$ is valid only on $$[-1,1]$$. \qed
A few remarks about Example \ref{arccosinesineex} are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that $$\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right) = \frac{\pi}{6}$$ as opposed to $$\frac{11\pi}{6}$$. This is the exact same phenomenon discussed in Section \ref{InverseFunctions} when we saw $$\sqrt{(-2)^2} = 2$$ as opposed to $$-2$$. Additionally, even though the expression we arrived at in part \ref{cosarcsin} above, namely $$1 - 2x^2$$, is defined for all real numbers, the equivalence $$\cos\left(2 \arcsin(x)\right) = 1-2x^2$$ is valid for only $$-1 \leq x \leq 1$$. This is akin to the fact that while the expression $$x$$ is defined for all real numbers, the equivalence $$\left( \sqrt{x} \right)^2 = x$$ is valid only for $$x \geq 0$$. For this reason, it pays to be careful when we determine the intervals where such equivalences are valid.
The next pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict $$f(x) = \tan(x)$$ to its fundamental cycle on $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ to obtain $$f^{-1}(x) = \arctan(x)$$. Among other things, note that the \textit{vertical} asymptotes $$x = -\frac{\pi}{2}$$ and $$x = \frac{\pi}{2}$$ of the graph of $$f(x) = \tan(x)$$ become the \textit{horizontal} asymptotes $$y = -\frac{\pi}{2}$$ and $$y = \frac{\pi}{2}$$ of the graph of $$f^{-1}(x) = \arctan(x)$$. \index{arctangent ! graph of}
Next, we restrict $$g(x) = \cot(x)$$ to its fundamental cycle on $$(0,\pi)$$ to obtain $$g^{-1}(x) = \mbox{arccot}(x)$$. Once again, the vertical asymptotes $$x=0$$ and $$x=\pi$$ of the graph of $$g(x) = \cot(x)$$ become the horizontal asymptotes $$y = 0$$ and $$y = \pi$$ of the graph of $$g^{-1}(x) = \mbox{arccot}(x)$$. We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions.
Note: Properties of the Arctangent and Arccotangent Functions
1. Properties of $$F(x)= \arctan(x)$$
• Domain: $$(-\infty, \infty)$$
• Range: $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$
• as $$x \rightarrow -\infty$$, $$\arctan(x) \rightarrow -\frac{\pi}{2}^{+}$$; as $$x \rightarrow \infty$$, $$\arctan(x) \rightarrow \frac{\pi}{2}^{-}$$
• $$\arctan(x) = t$$ if and only if $$-\frac{\pi}{2} < t < \frac{\pi}{2}$$ and $$\tan(t) = x$$
• $$\arctan(x) = \mbox{arccot}\left(\frac{1}{x}\right)$$ for $$x > 0$$
• $$\tan\left(\arctan(x)\right) = x$$ for all real numbers $$x$$
• $$\arctan(\tan(x)) = x$$ provided $$-\frac{\pi}{2} < x < \frac{\pi}{2}$$
1. Properties of $$G(x) = \mbox{arccot}(x)$$
• Domain: $$(-\infty, \infty)$$
• Range: $$(0, \pi)$$
• as $$x \rightarrow -\infty$$, $$\mbox{arccot}(x) \rightarrow \pi^{-}$$; as $$x \rightarrow \infty$$, $$\mbox{arccot}(x) \rightarrow 0^{+}$$
• $$\mbox{arccot}(x) = t$$ if and only if $$0 < t < \pi$$ and $$\cot(t) = x$$
• $$\mbox{arccot}(x) =\arctan\left(\frac{1}{x}\right)$$ for $$x > 0$$
• $$\cot\left(\mbox{arccot}(x)\right) = x$$ for all real numbers $$x$$
• $$\mbox{arccot}(\cot(x)) = x$$ provided $$0 < x < \pi$$
Example $$\PageIndex{2}$$:
1. Find the exact values of the following.
• $$\arctan(\sqrt{3})$$
• $$\mbox{arccot}(-\sqrt{3})$$
• $$\cot(\mbox{arccot}(-5))$$
• $$\sin\left(\arctan\left(-\frac{3}{4}\right)\right)$$
1. Rewrite the following as algebraic expressions of $$x$$ and state the domain on which the equivalence is valid.
• $$\tan(2 \arctan(x))$$
• $$\cos(\mbox{arccot}(2x))$$
Solution
1. \item We know $$\arctan(\sqrt{3})$$ is the real number $$t$$ between $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$ with $$\tan(t) = \sqrt{3}$$. We find $$t = \frac{\pi}{3}$$, so $$\arctan(\sqrt{3}) = \frac{\pi}{3}$$.
2. \item The real number $$t = \mbox{arccot}(-\sqrt{3})$$ lies in the interval $$(0,\pi)$$ with $$\cot(t) = -\sqrt{3}$$. We get $$\mbox{arccot}(-\sqrt{3}) = \frac{5\pi}{6}$$.
3. \item We can apply Theorem \ref{arctangentcotangentfunctionprops} directly and obtain $$\cot(\mbox{arccot}(-5)) = -5$$. However, working it through provides us with yet another opportunity to understand why this is the case. Letting $$t = \mbox{arccot}(-5)$$, we have that $$t$$ belongs to the interval $$(0,\pi)$$ and $$\cot(t)=-5$$. Hence, $$\cot(\mbox{arccot}(-5)) = \cot(t)=-5$$.
4. \item We start simplifying $$\sin\left(\arctan\left(-\frac{3}{4}\right)\right)$$ by letting $$t = \arctan\left(-\frac{3}{4}\right)$$. Then $$\tan(t) = -\frac{3}{4}$$ for some $$-\frac{\pi}{2} < t < \frac{\pi}{2}$$. Since $$\tan(t) < 0$$, we know, in fact, $$-\frac{\pi}{2} < t < 0$$. One way to proceed is to use The Pythagorean Identity, $$1 + \cot^{2}(t) = \csc^{2}(t)$$, since this relates the reciprocals of $$\tan(t)$$ and $$\sin(t)$$ and is valid for all $$t$$ under consideration.\footnote{It's always a good idea to make sure the identities used in these situations are valid for all values $$t$$ under consideration. Check our work back in Example \ref{arccosinesineex}. Were the identities we used there valid for all $$t$$ under consideration? A pedantic point, to be sure, but what else do you expect from this book?} From $$\tan(t) = -\frac{3}{4}$$, we get $$\cot(t) = -\frac{4}{3}$$. Substituting, we get $$1 + \left(-\frac{4}{3}\right)^2 = \csc^{2}(t)$$ so that $$\csc(t) = \pm \frac{5}{3}$$. Since $$-\frac{\pi}{2} < t < 0$$, we choose $$\csc(t) = -\frac{5}{3}$$, so $$\sin(t) = -\frac{3}{5}$$. Hence, $$\sin\left(\arctan\left(-\frac{3}{4}\right)\right) = -\frac{3}{5}$$.
5. \item If we let $$t = \arctan(x)$$, then $$-\frac{\pi}{2} < t < \frac{\pi}{2}$$ and $$\tan(t) = x$$. We look for a way to express $$\tan(2 \arctan(x)) = \tan(2t)$$ in terms of $$x$$. Before we get started using identities, we note that $$\tan(2t)$$ is undefined when $$2t = \frac{\pi}{2} + \pi k$$ for integers $$k$$. Dividing both sides of this equation by $$2$$ tells us we need to exclude values of $$t$$ where $$t = \frac{\pi}{4} + \frac{\pi}{2} k$$, where $$k$$ is an integer. The only members of this family which lie in $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ are $$t = \pm \frac{\pi}{4}$$, which means the values of $$t$$ under consideration are $$\left(-\frac{\pi}{2}, -\frac{\pi}{4}\right) \cup \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$$. Returning to $$\arctan(2t)$$, we note the double angle identity $$\tan(2t) = \frac{2 \tan(t)}{1 - \tan^{2}(t)}$$, is valid for all the values of $$t$$ under consideration, hence we get $\tan(2 \arctan(x)) = \tan(2t) = \frac{2 \tan(t)}{1 - \tan^{2}(t)}= \frac{2x}{1-x^2}$ To find where this equivalence is valid we check back with our substitution $$t = \arctan(x)$$. Since the domain of $$\arctan(x)$$ is all real numbers, the only exclusions come from the values of $$t$$ we discarded earlier, $$t = \pm \frac{\pi}{4}$$. Since $$x =\tan(t)$$, this means we exclude $$x = \tan\left(\pm \frac{\pi}{4}\right) = \pm 1$$. Hence, the equivalence $$\tan(2 \arctan(x)) = \frac{2x}{1-x^2}$$ holds for all $$x$$ in $$(-\infty, -1) \cup (-1,1) \cup (1,\infty)$$.
6. \item To get started, we let $$t = \mbox{arccot}(2x)$$ so that $$\cot(t) = 2x$$ where $$0 < t < \pi$$. In terms of $$t$$, $$\cos(\mbox{arccot}(2x)) = \cos(t)$$, and our goal is to express the latter in terms of $$x$$. Since $$\cos(t)$$ is always defined, there are no additional restrictions on $$t$$, so we can begin using identities to relate $$\cot(t)$$ to $$\cos(t)$$. The identity $$\cot(t) = \frac{\cos(t)}{\sin(t)}$$ is valid for $$t$$ in $$(0,\pi)$$, so our strategy is to obtain $$\sin(t)$$ in terms of $$x$$, then write $$\cos(t) = \cot(t) \sin(t)$$. The identity $$1 + \cot^{2}(t) = \csc^{2}(t)$$ holds for all $$t$$ in $$(0,\pi)$$ and relates $$\cot(t)$$ and $$\csc(t) = \frac{1}{\sin(t)}$$. Substituting $$\cot(t) =2x$$, we get $$1 + (2x)^2 = \csc^{2}(t)$$, or $$\csc(t) = \pm \sqrt{4x^2+1}$$. Since $$t$$ is between $$0$$ and $$\pi$$, $$\csc(t) > 0$$, so $$\csc(t) =\sqrt{4x^2+1}$$ which gives $$\sin(t) = \frac{1}{\sqrt{4x^2+1}}$$. Hence, $\cos(\mbox{arccot}(2x)) = \cos(t) = \cot(t) \sin(t) = \frac{2x}{\sqrt{4x^2+1}}$ Since $$\mbox{arccot}(2x)$$ is defined for all real numbers $$x$$ and we encountered no additional restrictions on $$t$$, we have $$\cos\left(\mbox{arccot}(2x)\right) = \frac{2x}{\sqrt{4x^2+1}}$$ for all real numbers $$x$$.
The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Subsection \ref{secantcosecantgraphsection}, are given below with the fundamental cycles highlighted.
It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of $$(-\infty, -1] \cup [1, \infty)$$ and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely $$[1, \infty)$$, and another piece to cover the bottom, namely $$(-\infty, -1]$$. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simplifies the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view.
### Inverses of Secant and Cosecant: Trigonometry Friendly Approach
In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For $$f(x) = \sec(x)$$, we restrict the domain to $$\left[0, \frac{\pi}{2}\right) \cup \left( \frac{\pi}{2}, \pi\right]$$ \index{arcsecant ! trigonometry friendly ! graph of}
and we restrict $$g(x) = \csc(x)$$ to $$\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$$. \index{arccosecant ! trigonometry friendly ! graph of}
Note that for both arcsecant and arccosecant, the domain is $$(-\infty, -1] \cup [1, \infty)$$. Taking a page from Section \ref{AbsoluteValueFunctions}, we can rewrite this as $$\left\{ x : |x| \geq 1\right\}$$. This is often done in Calculus textbooks, so we include it here for completeness. Using these definitions, we get the following properties of the arcsecant and arccosecant functions.
Note: Properties of the Arcsecant and Arccosecant Functions
1. Properties of $$F(x)= \mbox{arcsec}(x)$$
• Domain: $$\left\{ x : |x| \geq 1 \right\} = (-\infty, -1] \cup [1,\infty)$$
• Range: $$\left[0, \frac{\pi}{2} \right) \cup \left(\frac{\pi}{2}, \pi\right]$$
• as $$x \rightarrow -\infty$$, $$\mbox{arcsec}(x) \rightarrow \frac{\pi}{2}^{+}$$; as $$x \rightarrow \infty$$, $$\mbox{arcsec}(x) \rightarrow \frac{\pi}{2}^{-}$$
• $$\mbox{arcsec}(x) = t$$ if and only if $$0 \leq t < \frac{\pi}{2}$$ or $$\frac{\pi}{2} < t \leq \pi$$ and $$\sec(t) = x$$
• $$\mbox{arcsec}(x) = \arccos\left(\frac{1}{x}\right)$$ provided $$|x| \geq 1$$
• $$\sec\left(\mbox{arcsec}(x)\right) = x$$ provided $$|x| \geq 1$$
• $$\mbox{arcsec}(\sec(x)) = x$$ provided $$0 \leq x < \frac{\pi}{2}$$ or $$\frac{\pi}{2} < x \leq \pi$$
1. Properties of $$G(x) = \mbox{arccsc}(x)$$
• Domain: $$\left\{ x : |x| \geq 1 \right\} = (-\infty, -1] \cup [1,\infty)$$
• Range: $$\left[-\frac{\pi}{2}, 0 \right) \cup \left(0, \frac{\pi}{2} \right]$$
• as $$x \rightarrow -\infty$$, $$\mbox{arccsc}(x) \rightarrow 0^{-}$$; as $$x \rightarrow \infty$$, $$\mbox{arccsc}(x) \rightarrow 0^{+}$$
• $$\mbox{arccsc}(x) = t$$ if and only if $$-\frac{\pi}{2} \leq t < 0$$ or $$0 < t \leq \frac{\pi}{2}$$ and $$\csc(t) = x$$
• $$\mbox{arccsc}(x) = \arcsin\left(\frac{1}{x}\right)$$ provided $$|x| \geq 1$$
• $$\csc\left(\mbox{arccsc}(x)\right) = x$$ provided $$|x| \geq 1$$
• $$\mbox{arccsc}(\csc(x)) = x$$ provided $$-\frac{\pi}{2} \leq x < 0$$ or $$0 < x \leq \frac{\pi}{2}$$
Example $$\PageIndex{3}$$:
1. Find the exact values of the following.
• $$\mbox{arcsec}(2)$$
• $$\mbox{arccsc}(-2)$$
• $$\mbox{arcsec}\left( \sec\left( \frac{5\pi}{4} \right) \right)$$
• $$\cot\left(\mbox{arccsc}\left(-3\right)\right)$$
1. Rewrite the following as algebraic expressions of $$x$$ and state the domain on which the equivalence is valid.
• $$\tan(\mbox{arcsec}(x))$$
• $$\cos(\mbox{arccsc}(4x))$$
Solution
\item Using Theorem \ref{arcsecantcosecantfunctionprops1}, we have $$\mbox{arcsec}(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$$.
\item Once again, Theorem \ref{arcsecantcosecantfunctionprops1} comes to our aid giving $$\mbox{arccsc}(-2) = \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$.
\item Since $$\frac{5\pi}{4}$$ doesn't fall between $$0$$ and $$\frac{\pi}{2}$$ or $$\frac{\pi}{2}$$ and $$\pi$$, we cannot use the inverse property stated in Theorem \ref{arcsecantcosecantfunctionprops1}. We can, nevertheless, begin by working inside out' which yields $$\mbox{arcsec}\left( \sec\left( \frac{5\pi}{4} \right) \right) = \mbox{arcsec}(-\sqrt{2}) = \arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$$.
\item One way to begin to simplify $$\cot\left(\mbox{arccsc}\left(-3\right)\right)$$ is to let $$t = \mbox{arccsc}(-3)$$. Then, $$\csc(t) = -3$$ and, since this is negative, we have that $$t$$ lies in the interval $$\left[ -\frac{\pi}{2},0\right)$$. We are after $$\cot\left(\mbox{arccsc}\left(-3\right)\right) = \cot(t)$$, so we use the Pythagorean Identity $$1 + \cot^{2}(t) = \csc^{2}(t)$$. Substituting, we have $$1 + \cot^{2}(t) = (-3)^2$$, or $$\cot(t) = \pm \sqrt{8} = \pm 2 \sqrt{2}$$. Since $$-\frac{\pi}{2} \leq t < 0$$, $$\cot(t) < 0$$, so we get $$\cot\left(\mbox{arccsc}\left(-3\right)\right) = -2\sqrt{2}$$.
\item We begin simplifying $$\tan(\mbox{arcsec}(x))$$ by letting $$t = \mbox{arcsec}(x)$$. Then, $$\sec(t) = x$$ for $$t$$ in $$\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi \right]$$, and we seek a formula for $$\tan(t)$$. Since $$\tan(t)$$ is defined for all $$t$$ values under consideration, we have no additional restrictions on $$t$$. To relate $$\sec(t)$$ to $$\tan(t)$$, we use the identity $$1 + \tan^{2}(t) = \sec^{2}(t)$$. This is valid for all values of $$t$$ under consideration, and when we substitute $$\sec(t) = x$$, we get $$1 + \tan^{2}(t) = x^2$$. Hence, $$\tan(t) = \pm \sqrt{x^2-1}$$. If $$t$$ belongs to $$\left[0, \frac{\pi}{2}\right)$$ then $$\tan(t) \geq 0$$; if, on the the other hand, $$t$$ belongs to $$\left(\frac{\pi}{2}, \pi \right]$$ then $$\tan(t) \leq 0$$. As a result, we get a piecewise defined function for $$\tan(t)$$
$\tan(t) = \left\{ \begin{array}{rr} \sqrt{x^2-1}, & \text{if $$0 \leq t < \frac{\pi}{2}$$} \\ [5pt] -\sqrt{x^2-1}, & \text{if $$\frac{\pi}{2} < t \leq \pi$$} \end{array}\right.$
Now we need to determine what these conditions on $$t$$ mean for $$x$$. Since $$x = \sec(t)$$, when $$0 \leq t < \frac{\pi}{2}$$, $$x \geq 1$$, and when $$\frac{\pi}{2} < t \leq \pi$$, $$x \leq -1$$. Since we encountered no further restrictions on $$t$$, the equivalence below holds for all $$x$$ in $$(-\infty, -1] \cup [1, \infty)$$.
$\tan(\mbox{arcsec}(x)) = \left\{ \begin{array}{rr} \sqrt{x^2-1}, & \text{if $$x \geq 1$$} \\[5pt] -\sqrt{x^2-1}, & \text{if $$x \leq -1$$} \end{array}\right.$
\item To simplify $$\cos(\mbox{arccsc}(4x))$$, we start by letting $$t = \mbox{arccsc}(4x)$$. Then $$\csc(t) = 4x$$ for $$t$$ in $$\left[-\frac{\pi}{2}, 0 \right) \cup \left(0, \frac{\pi}{2}\right]$$, and we now set about finding an expression for $$\cos(\mbox{arccsc}(4x)) = \cos(t)$$. Since $$\cos(t)$$ is defined for all $$t$$, we do not encounter any additional restrictions on $$t$$. From $$\csc(t) = 4x$$, we get $$\sin(t) = \frac{1}{4x}$$, so to find $$\cos(t)$$, we can make use if the identity $$\cos^{2}(t) + \sin^{2}(t) = 1$$. Substituting $$\sin(t) = \frac{1}{4x}$$ gives $$\cos^{2}(t) + \left(\frac{1}{4x}\right)^2 = 1$$. Solving, we get $\cos(t) = \pm \sqrt{\frac{16x^2-1}{16x^2}} = \pm \frac{\sqrt{16x^2-1}}{4|x|}$ Since $$t$$ belongs to $$\left[-\frac{\pi}{2}, 0 \right) \cup \left(0, \frac{\pi}{2}\right]$$, we know $$\cos(t) \geq 0$$, so we choose $$\cos(t) = \frac{\sqrt{16-x^2}}{4|x|}$$. (The absolute values here are necessary, since $$x$$ could be negative.) To find the values for which this equivalence is valid, we look back at our original substution, $$t = \mbox{arccsc}(4x)$$. Since the domain of $$\mbox{arccsc}(x)$$ requires its argument $$x$$ to satisfy $$|x| \geq 1$$, the domain of $$\mbox{arccsc}(4x)$$ requires $$|4x| \geq 1$$. Using Theorem \ref{absolutevalueineq}, we rewrite this inequality and solve to get $$x \leq -\frac{1}{4}$$ or $$x \geq \frac{1}{4}$$. Since we had no additional restrictions on $$t$$, the equivalence $$\cos(\mbox{arccsc}(4x)) = \frac{\sqrt{16x^2-1}}{4|x|}$$ holds for all $$x$$ in $$\left(-\infty, -\frac{1}{4} \right] \cup \left[\frac{1}{4}, \infty \right)$$. \qed
### Inverses of Secant and Cosecant: Calculus Friendly Approach
In this subsection, we restrict $$f(x) = \sec(x)$$ to $$\left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right)$$ \index{arcsecant ! calculus friendly ! graph of}
and we restrict $$g(x) = \csc(x)$$ to $$\left(0, \frac{\pi}{2}\right] \cup \left( \pi, \frac{3\pi}{2}\right]$$. \index{arccosecant ! calculus friendly ! graph of}
Using these definitions, we get the following result.
Note: Properties of the Arcsecant and Arccosecant Functions
1. Properties of $$F(x)= \mbox{arcsec}(x)$$
• Domain: $$\left\{ x : |x| \geq 1 \right\} = (-\infty, -1] \cup [1,\infty)$$
• Range: $$\left[0, \frac{\pi}{2} \right) \cup \left[\pi, \frac{3\pi}{2} \right)$$
• as $$x \rightarrow -\infty$$, $$\mbox{arcsec}(x) \rightarrow \frac{3\pi}{2}^{-}$$; as $$x \rightarrow \infty$$, $$\mbox{arcsec}(x) \rightarrow \frac{\pi}{2}^{-}$$
• $$\mbox{arcsec}(x) = t$$ if and only if $$0 \leq t < \frac{\pi}{2}$$ or $$\pi \leq t < \frac{3\pi}{2}$$ and $$\sec(t) = x$$
• $$\mbox{arcsec}(x) = \arccos\left(\frac{1}{x}\right)$$ for $$x \geq 1$$ only\footnote{Compare this with the similar result in Theorem \ref{arcsecantcosecantfunctionprops1}.}
• $$\sec\left(\mbox{arcsec}(x)\right) = x$$ provided $$|x| \geq 1$$
• $$\mbox{arcsec}(\sec(x)) = x$$ provided $$0 \leq x < \frac{\pi}{2}$$ or $$\pi \leq x < \frac{3\pi}{2}$$
1. Properties of $$G(x) = \mbox{arccsc}(x)$$
• Domain: $$\left\{ x : |x| \geq 1 \right\} = (-\infty, -1] \cup [1,\infty)$$
• Range: $$\left(0, \frac{\pi}{2} \right] \cup \left( \pi, \frac{3\pi}{2} \right]$$
• as $$x \rightarrow -\infty$$, $$\mbox{arccsc}(x) \rightarrow \pi^{+}$$; as $$x \rightarrow \infty$$, $$\mbox{arccsc}(x) \rightarrow 0^{+}$$
• $$\mbox{arccsc}(x) = t$$ if and only if $$0 < t \leq \frac{\pi}{2}$$ or $$\pi < t \leq \frac{3\pi}{2}$$ and $$\csc(t) = x$$
• $$\mbox{arccsc}(x) = \arcsin\left(\frac{1}{x}\right)$$ for $$x \geq 1$$ only\footnote{Compare this with the similar result in Theorem \ref{arcsecantcosecantfunctionprops1}.}
• $$\csc\left(\mbox{arccsc}(x)\right) = x$$ provided $$|x| \geq 1$$
• $$\mbox{arccsc}(\csc(x)) = x$$ provided $$0 < x \leq \frac{\pi}{2}$$ or $$\pi < x \leq \frac{3\pi}{2}$$
Our next example is a duplicate of Example \ref{arcsecantcosecantex1}. The interested reader is invited to compare and contrast the solution to each.
Example $$\PageIndex{4}$$:
\label{arcsecantcosecantex2} $$~$$
\begin{enumerate}
\item Find the exact values of the following.
\begin{multicols}{4}
\begin{enumerate}
\item $$\mbox{arcsec}(2)$$
\item $$\mbox{arccsc}(-2)$$
\item $$\mbox{arcsec}\left( \sec\left( \frac{5\pi}{4} \right) \right)$$
\item $$\cot\left(\mbox{arccsc}\left(-3\right)\right)$$
\end{enumerate}
\end{multicols}
\enlargethispage{.25in}
\item Rewrite the following as algebraic expressions of $$x$$ and state the domain on which the equivalence is valid.
\begin{multicols}{2}
\begin{enumerate}
\item $$\tan(\mbox{arcsec}(x))$$
\item $$\cos(\mbox{arccsc}(4x))$$
\end{enumerate}
\end{multicols}
\end{enumerate}
{\bf Solution.}
\begin{enumerate}
\item
\begin{enumerate}
\item Since $$2 \geq 1$$, we may invoke Theorem \ref{arcsecantcosecantfunctionprops2} to get $$\mbox{arcsec}(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$$.
\item Unfortunately, $$-2$$ is not greater to or equal to $$1$$, so we cannot apply Theorem \ref{arcsecantcosecantfunctionprops2} to $$\mbox{arccsc}(-2)$$ and convert this into an arcsine problem. Instead, we appeal to the definition. The real number $$t = \mbox{arccsc}(-2)$$ lies in $$\left(0,\frac{\pi}{2} \right] \cup \left(\pi, \frac{3\pi}{2}\right]$$ and satisfies $$\csc(t) = -2$$. The $$t$$ we're after is $$t = \frac{7\pi}{6}$$, so $$\mbox{arccsc}(-2) = \frac{7\pi}{6}$$.
\item Since $$\frac{5\pi}{4}$$ lies between $$\pi$$ and $$\frac{3\pi}{2}$$, we may apply Theorem \ref{arcsecantcosecantfunctionprops2} directly to simplify $$\mbox{arcsec}\left( \sec\left( \frac{5\pi}{4} \right) \right) = \frac{5\pi}{4}$$. We encourage the reader to work this through using the definition as we have done in the previous examples to see how it goes.
\item To simplify $$\cot\left(\mbox{arccsc}\left(-3\right)\right)$$ we let $$t = \mbox{arccsc}\left(-3\right)$$ so that $$\cot\left(\mbox{arccsc}\left(-3\right)\right) = \cot(t)$$. We know $$\csc(t) = -3$$, and since this is negative, $$t$$ lies in $$\left( \pi, \frac{3\pi}{2}\right]$$. Using the identity $$1 + \cot^{2}(t) = \csc^{2}(t)$$, we find $$1 + \cot^{2}(t) = (-3)^2$$ so that $$\cot(t) = \pm \sqrt{8} = \pm 2 \sqrt{2}$$. Since $$t$$ is in the interval $$\left(\pi, \frac{3\pi}{2}\right]$$, we know $$\cot(t) > 0$$. Our answer is $$\cot\left(\mbox{arccsc}\left(-3\right)\right) = 2 \sqrt{2}$$.
\end{enumerate}
\item
\begin{enumerate}
\item We begin simplifying $$\tan(\mbox{arcsec}(x))$$ by letting $$t = \mbox{arcsec}(x)$$. Then, $$\sec(t) = x$$ for $$t$$ in $$\left[0, \frac{\pi}{2} \right) \cup \left[\pi, \frac{3\pi}{2} \right)$$, and we seek a formula for $$\tan(t)$$. Since $$\tan(t)$$ is defined for all $$t$$ values under consideration, we have no additional restrictions on $$t$$. To relate $$\sec(t)$$ to $$\tan(t)$$, we use the identity $$1 + \tan^{2}(t) = \sec^{2}(t)$$. This is valid for all values of $$t$$ under consideration, and when we substitute $$\sec(t) = x$$, we get $$1 + \tan^{2}(t) = x^2$$. Hence, $$\tan(t) = \pm \sqrt{x^2-1}$$. Since $$t$$ lies in $$\left[0, \frac{\pi}{2} \right) \cup \left[\pi, \frac{3\pi}{2} \right)$$, $$\tan(t) \geq 0$$, so we choose $$\tan(t) = \sqrt{x^2-1}$$. Since we found no additional restrictions on $$t$$, the equivalence $$\tan(\mbox{arcsec}(x)) = \sqrt{x^2-1}$$ holds for all $$x$$ in the domain of $$t = \mbox{arcsec}(x)$$, namely $$(-\infty, -1] \cup [1,\infty)$$.
\item To simplify $$\cos(\mbox{arccsc}(4x))$$, we start by letting $$t = \mbox{arccsc}(4x)$$. Then $$\csc(t) = 4x$$ for $$t$$ in $$\left(0, \frac{\pi}{2} \right] \cup \left(\pi, \frac{3\pi}{2} \right]$$, and we now set about finding an expression for $$\cos(\mbox{arccsc}(4x)) = \cos(t)$$. Since $$\cos(t)$$ is defined for all $$t$$, we do not encounter any additional restrictions on $$t$$. From $$\csc(t) = 4x$$, we get $$\sin(t) = \frac{1}{4x}$$, so to find $$\cos(t)$$, we can make use if the identity $$\cos^{2}(t) + \sin^{2}(t) = 1$$. Substituting $$\sin(t) = \frac{1}{4x}$$ gives $$\cos^{2}(t) + \left(\frac{1}{4x}\right)^2 = 1$$. Solving, we get $\cos(t) = \pm \sqrt{\frac{16x^2-1}{16x^2}} = \pm \frac{\sqrt{16x^2-1}}{4|x|}$ If $$t$$ lies in $$\left(0, \frac{\pi}{2} \right]$$, then $$\cos(t) \geq 0$$, and we choose $$\cos(t) = \frac{\sqrt{16x^2-1}}{4|x|}$$. Otherwise, $$t$$ belongs to $$\left( \pi, \frac{3\pi}{2} \right]$$ in which case $$\cos(t) \leq 0$$, so, we choose $$\cos(t) = -\frac{\sqrt{16x^2-1}}{4|x|}$$ This leads us to a (momentarily) piecewise defined function for $$\cos(t)$$
$\cos(t) = \left\{ \begin{array}{rr} \dfrac{\sqrt{16x^2-1}}{4|x|}, & \text{if $$0 \leq t \leq \frac{\pi}{2}$$} \\ [5pt] -\dfrac{\sqrt{16x^2-1}}{4|x|}, & \text{if $$\pi < t \leq \frac{3\pi}{2}$$} \end{array}\right.$
We now see what these restrictions mean in terms of $$x$$. Since $$4x = \csc(t)$$, we get that for $$0 \leq t \leq \frac{\pi}{2}$$, $$4x \geq 1$$, or $$x \geq \frac{1}{4}$$. In this case, we can simplify $$|x| = x$$ so $\cos(t) = \frac{\sqrt{16x^2-1}}{4|x|} = \frac{\sqrt{16x^2-1}}{4x}$ Similarly, for $$\pi < t \leq \frac{3\pi}{2}$$, we get $$4x \leq -1$$, or $$x \leq -\frac{1}{4}$$. In this case, $$|x| = -x$$, so we also get $\cos(t) = -\frac{\sqrt{16x^2-1}}{4|x|} = -\frac{\sqrt{16x^2-1}}{4(-x)} = \frac{\sqrt{16x^2-1}}{4x}$ Hence, in all cases, $$\cos(\mbox{arccsc}(4x)) = \frac{\sqrt{16x^2-1}}{4x}$$, and this equivalence is valid for all $$x$$ in the domain of $$t = \mbox{arccsc}(4x)$$, namely $$\left(-\infty, -\frac{1}{4}\right] \cup \left[ \frac{1}{4}, \infty \right)$$ \qed
### Calculators and the Inverse Circular Functions
In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as $$\sin^{-1}, \cos^{-1}$$ and $$\tan^{-1}$$, respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates.
Example $$\PageIndex{5}$$:
\begin{ex} \label{arcstuffoncalc} $$~$$
\begin{enumerate}
\item Use a calculator to approximate the following values to four decimal places.
\begin{multicols}{4}
\begin{enumerate}
\item $$\mbox{arccot}(2)$$ \vphantom{\)\left(-\dfrac{3}{2}\right)\)}
\item $$\mbox{arcsec}(5)$$ \vphantom{\)\left(-\dfrac{3}{2}\right)\)}
\item \label{arccotneg2} $$\mbox{arccot}(-2)$$ \vphantom{\)\left(-\dfrac{3}{2}\right)\)}
\item $$\mbox{arccsc}\left(-\dfrac{3}{2}\right)$$
\end{enumerate}
\end{multicols}
\item Find the domain and range of the following functions. Check your answers using a calculator.
\begin{multicols}{3}
\begin{enumerate}
\item $$f(x) = \dfrac{\pi}{2} - \arccos\left(\dfrac{x}{5}\right)$$
\item $$f(x) = 3\arctan\left(4x \right)$$. \vphantom{\)\left(-\dfrac{x}{2}\right)\)}
\item \label{arccotangentcalc} $$f(x) = \text{arccot}\left(\dfrac{x}{2} \right) + \pi$$
\end{enumerate}
\end{multicols}
\end{enumerate}
{\bf Solution.}
\begin{enumerate}
\item
\begin{enumerate}
\item Since $$2 > 0$$, we can use the property listed in Theorem \ref{arctangentcotangentfunctionprops} to rewrite $$\mbox{arccot}(2)$$ as $$\mbox{arccot}(2) = \arctan\left(\frac{1}{2}\right)$$. In radian' mode, we find $$\mbox{arccot}(2) = \arctan\left(\frac{1}{2}\right) \approx 0.4636$$.
\item Since $$5 \geq 1$$, we can use the property from either Theorem \ref{arcsecantcosecantfunctionprops1} or Theorem \ref{arcsecantcosecantfunctionprops2} to write $$\mbox{arcsec}(5) = \arccos\left(\frac{1}{5}\right) \approx 1.3694$$.
\newpage
\begin{tabular}{cc}
\includegraphics[width=2in]{./IntroTrigGraphics/ArcTrig01.jpg} &
\hspace{0.75in} \includegraphics[width=2in]{./IntroTrigGraphics/ArcTrig02.jpg} \\
\end{tabular}
\item Since the argument $$-2$$ is negative, we cannot directly apply Theorem \ref{arctangentcotangentfunctionprops} to help us find $$\mbox{arccot}(-2)$$. Let $$t = \mbox{arccot}(-2)$$. Then $$t$$ is a real number such that $$0 < t < \pi$$ and $$\cot(t) = -2$$. Moreover, since $$\cot(t) < 0$$, we know $$\frac{\pi}{2} < t < \pi$$. Geometrically, this means $$t$$ corresponds to a Quadrant II angle $$\theta = t$$ radians. This allows us to proceed using a reference angle' approach. Consider $$\alpha$$, the reference angle for $$\theta$$, as pictured below. By definition, $$\alpha$$ is an acute angle so $$0 < \alpha < \frac{\pi}{2}$$, and the Reference Angle Theorem, Theorem \ref{refanglethm}, tells us that $$\cot(\alpha) = 2$$. This means $$\alpha = \mbox{arccot}(2)$$ radians. Since the argument of arccotangent is now a \emph{positive} $$2$$, we can use Theorem \ref{arctangentcotangentfunctionprops} to get $$\alpha = \mbox{arccot}(2) =\arctan\left(\frac{1}{2}\right)$$ radians. Since $$\theta = \pi - \alpha = \pi - \arctan\left(\frac{1}{2}\right) \approx 2.6779$$ radians, we get $$\mbox{arccot}(-2) \approx 2.6779$$.
Another way to attack the problem is to use $$\arctan\left(-\frac{1}{2}\right)$$. By definition, the real number $$t = \arctan\left(-\frac{1}{2}\right)$$ satisfies $$\tan(t) = -\frac{1}{2}$$ with $$-\frac{\pi}{2} < t < \frac{\pi}{2}$$. Since $$\tan(t)<0$$, we know more specifically that $$-\frac{\pi}{2} < t < 0$$, so $$t$$ corresponds to an angle $$\beta$$ in Quadrant IV. To find the value of $$\mbox{arccot}(-2)$$, we once again visualize the angle $$\theta = \mbox{arccot}(-2)$$ radians and note that it is a Quadrant II angle with $$\tan(\theta) = -\frac{1}{2}$$. This means it is exactly $$\pi$$ units away from $$\beta$$, and we get $$\theta = \pi + \beta = \pi + \arctan\left(-\frac{1}{2}\right) \approx 2.6779$$ radians. Hence, as before, $$\mbox{arccot}(-2) \approx 2.6779$$.
\item If the range of arccosecant is taken to be $$\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$$, we can use Theorem \ref{arcsecantcosecantfunctionprops1} to get $$\mbox{arccsc}\left(-\frac{3}{2}\right) = \arcsin\left(-\frac{2}{3}\right) \approx -0.7297$$. If, on the other hand, the range of arccosecant is taken to be $$\left(0, \frac{\pi}{2}\right] \cup \left(\pi, \frac{3\pi}{2}\right]$$, then we proceed as in the previous problem by letting $$t = \mbox{arccsc}\left(-\frac{3}{2}\right)$$. Then $$t$$ is a real number with $$\csc(t) = -\frac{3}{2}$$. Since $$\csc(t) < 0$$, we have that $$\pi < \theta \leq \frac{3\pi}{2}$$, so $$t$$ corresponds to a Quadrant III angle, $$\theta$$. As above, we let $$\alpha$$ be the reference angle for $$\theta$$. Then $$0 < \alpha < \frac{\pi}{2}$$ and $$\csc(\alpha) =\frac{3}{2}$$, which means $$\alpha = \mbox{arccsc}\left(\frac{3}{2}\right)$$ radians. Since the argument of arccosecant is now positive, we may use Theorem \ref{arcsecantcosecantfunctionprops2} to get $$\alpha = \mbox{arccsc}\left(\frac{3}{2}\right) = \arcsin\left(\frac{2}{3}\right)$$ radians. Since $$\theta = \pi + \alpha = \pi + \arcsin\left(\frac{2}{3}\right) \approx 3.8713$$ radians, $$\mbox{arccsc}\left(-\frac{3}{2}\right) \approx 3.8713$$.
\item \begin{enumerate}
\item Since the domain of $$F(x) = \arccos(x)$$ is $$-1 \leq x \leq 1$$, we can find the domain of $$f(x) = \frac{\pi}{2} - \arccos\left(\frac{x}{5}\right)$$ by setting the argument of the arccosine, in this case $$\frac{x}{5}$$, between $$-1$$ and $$1$$. Solving $$-1 \leq \frac{x}{5} \leq 1$$ gives $$-5 \leq x \leq 5$$, so the domain is $$[-5,5]$$. To determine the range of $$f$$, we take a cue from Section \ref{Transformations}. Three key' points on the graph of $$F(x) = \arccos(x)$$ are $$(-1, \pi)$$, $$\left(0, \frac{\pi}{2}\right)$$ and $$(1,0)$$ . Following the procedure outlined in Theorem \ref{transformationsthm}, we track these points to $$\left(-5, -\frac{\pi}{2}\right)$$, $$(0, 0)$$ and $$\left(5, \frac{\pi}{2}\right)$$. Plotting these values tells us that the range\footnote{It also confirms our domain!} of $$f$$ is $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. Our graph confirms our results.
\item To find the domain and range of $$f(x) = 3\arctan\left(4x \right)$$, we note that since the domain of $$F(x) = \arctan(x)$$ is all real numbers, the only restrictions, if any, on the domain of $$f(x) = 3\arctan\left(4x \right)$$ come from the argument of the arctangent, in this case, $$4x$$. Since $$4x$$ is defined for all real numbers, we have established that the domain of $$f$$ is all real numbers. To determine the range of $$f$$, we can, once again, appeal to Theorem \ref{transformationsthm}. Choosing our key' point to be $$(0,0)$$ and tracking the horizontal asymptotes $$y = -\frac{\pi}{2}$$ and $$y= \frac{\pi}{2}$$, we find that the graph of $$y = f(x) = 3\arctan\left(4x \right)$$ differs from the graph of $$y = F(x) = \arctan(x)$$ by a horizontal compression by a factor of $$4$$ and a vertical stretch by a factor of $$3$$. It is the latter which affects the range, producing a range of $$\left(-\frac{3\pi}{2}, \frac{3\pi}{2} \right)$$. We confirm our findings on the calculator below.
\item To find the domain of $$g(x) = \text{arccot}\left(\frac{x}{2}\right) + \pi$$, we proceed as above. Since the domain of $$G(x) = \text{arccot}(x)$$ is $$(-\infty, \infty)$$, and $$\frac{x}{2}$$ is defined for all $$x$$, we get that the domain of $$g$$ is $$(-\infty, \infty)$$ as well. As for the range, we note that the range of $$G(x) = \text{arccot}(x)$$, like that of $$F(x) = \arctan(x)$$, is limited by a pair of horizontal asymptotes, in this case $$y = 0$$ and $$y = \pi$$. Following Theorem \ref{transformationsthm}, we graph $$y = g(x) = \text{arccot}\left(\frac{x}{2}\right) + \pi$$ starting with $$y = G(x) = \text{arccot}(x)$$ and first performing a horizontal expansion by a factor of $$2$$ and following that with a vertical shift upwards by $$\pi$$. This latter transformation is the one which affects the range, making it now $$(\pi, 2\pi)$$. To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number \ref{arccotneg2}, we attempt to rewrite $$g(x) = \text{arccot}\left(\frac{x}{2}\right) + \pi$$ in terms of the arctangent function. Using Theorem \ref{arctangentcotangentfunctionprops}, we have that $$\text{arccot}\left(\frac{x}{2}\right) = \arctan\left(\frac{2}{x}\right)$$ when $$\frac{x}{2} > 0$$, or, in this case, when $$x > 0$$. Hence, for $$x > 0$$, we have $$g(x) = \arctan\left(\frac{2}{x}\right) + \pi$$. When $$\frac{x}{2} < 0$$, we can use the same argument in number \ref{arccotneg2} that gave us $$\text{arccot}(-2) = \pi + \arctan\left(-\frac{1}{2}\right)$$ to give us $$\text{arccot}\left(\frac{x}{2}\right) = \pi + \arctan\left(\frac{2}{x}\right)$$. Hence, for $$x < 0$$, $$g(x) = \pi + \arctan\left(\frac{2}{x}\right) + \pi = \arctan\left(\frac{2}{x}\right) + 2\pi$$. What about $$x=0$$? We know $$g(0) = \text{arccot}(0) + \pi = \pi$$, and neither of the formulas for $$g$$ involving arctangent will produce this result.\footnote{Without Calculus, of course \ldots} Hence, in order to graph $$y = g(x)$$ on our calculators, we need to write it as a piecewise defined function:
$g(x) = \text{arccot}\left(\frac{x}{2}\right) + \pi = \left\{ \begin{array}{rr} \arctan\left(\frac{2}{x}\right) + 2\pi, & \text{if $$x<0$$} \\ [5pt] \pi, & \text{if $$x=0$$} \\ [5pt] \arctan\left(\frac{2}{x}\right) + \pi, & \text{if $$x>0$$} \end{array}\right.$
We show the input and the result below.
The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example.
\begin{ex}\footnote{The authors would like to thank Dan Stitz for this problem and associated graphics.} \label{roofpitchex} The roof on the house below has a \)6/12\) pitch'. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree.
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=2.75in]{./IntroTrigGraphics/ArcTrig05.jpg} &
\hspace{0.75in} \includegraphics[width=2in]{./IntroTrigGraphics/ArcTrig06.jpg} \\
Front View & \hspace{0.75in} Side View \\
\end{tabular}
\end{center}
{\bf Solution.} If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length $$6$$ feet and $$12$$ feet. Using Theorem \ref{circularfunctionstriangle}, we find the angle of inclination, labeled $$\theta$$ below, satisfies $$\tan(\theta) = \frac{6}{12} = \frac{1}{2}$$. Since $$\theta$$ is an acute angle, we can use the arctangent function and we find $$\theta = \arctan\left(\frac{1}{2}\right)\, \text{radians} \, \approx 26.56^{\circ}$$.
\begin{tabular}{m{2.5in}m{1in}m{2.5in}}
\begin{mfpic}[15]{0}{13.25}{-1}{6}
\polyline{(0,0), (12,0), (12,6), (0,0)}
\polyline{(11.25,0), (11.25,0.75), (12,0.75)}
\arrow \reverse \arrow \polyline{(0,-1),(12,-1)}
\gclear \tlabelrect[cc](6,-1){\)12\) feet}
\arrow \reverse \arrow \polyline{(13.25,0),(13.25,6)}
\gclear \tlabelrect[cc](13.25,3){\)6\) feet}
\arrow \parafcn{3, 19, 5}{2.75*dir(t)}
\tlabel[cc](3.25,0.5){\)\theta\)}
\end{mfpic}
&
&
\includegraphics[width=2in]{./IntroTrigGraphics/ArcTrig07.jpg} \qed \\
\end{tabular}
\end{ex}
\subsection{Solving Equations Using the Inverse Trigonometric Functions.}
In Sections \ref{TheUnitCircle} and \ref{CircularFunctions}, we learned how to solve equations like $$\sin(\theta) = \frac{1}{2}$$ for angles $$\theta$$ and $$\tan(t) = -1$$ for real numbers $$t$$. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of common angles' listed on page \pageref{commonanglesunitcircle}. If, on the other hand, we had been asked to find all angles with $$\sin(\theta) = \frac{1}{3}$$ or solve $$\tan(t) = -2$$ for real numbers $$t$$, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation $$x^2 = 4$$ is a lot like $$\sin(\theta) = \frac{1}{2}$$ in that it has friendly, common value' answers $$x = \pm 2$$. The equation $$x^2 = 7$$, on the other hand, is a lot like $$\sin(\theta) = \frac{1}{3}$$. We know\footnote{How do we know this again?} there are answers, but we can't express them using friendly' numbers.\footnote{This is all, of course, a matter of opinion. For the record, the authors find $$\pm \sqrt{7}$$ just as nice' as $$\pm 2$$.} To solve $$x^2 = 7$$, we make use of the square root function and write $$x = \pm \sqrt{7}$$. We can certainly \textit{approximate} these answers using a calculator, but as far as exact answers go, we leave them as $$x = \pm \sqrt{7}$$. In the same way, we will use the arcsine function to solve $$\sin(\theta) = \frac{1}{3}$$, as seen in the following example.
\begin{ex} \label{basicinverseeqns} Solve the following equations.
\begin{enumerate}
\item \label{basicinverseeqnssine} Find all angles $$\theta$$ for which $$\sin(\theta) = \frac{1}{3}$$.
\item \label{basicinverseeqnstangent} Find all real numbers $$t$$ for which $$\tan(t) = -2$$
\item \label{basicinverseeqnssecant} Solve $$\, \sec(x) = -\frac{5}{3} \,$$ for $$x$$.
\end{enumerate}
{\bf Solution.}
\begin{enumerate}
\item If $$\sin(\theta) = \frac{1}{3}$$, then the terminal side of $$\theta$$, when plotted in standard position, intersects the Unit Circle at $$y = \frac{1}{3}$$. Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let $$\alpha$$ denote the acute solution to the equation, then all the solutions to this equation in Quadrant I are coterminal with $$\alpha$$, and $$\alpha$$ serves as the reference angle for all of the solutions to this equation in Quadrant II.
\begin{tabular}{cc}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,-0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3, 1, 3}
\tlpointsep{4pt}
\axislabels{y}{{\scriptsize $$\frac{1}{3}$$} 1}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\arrow \polyline{(0,0), (4.532, 2.113)}
\arrow \parafcn{5, 20, 5}{2.75*dir(t)}
\tlabel[cc](5.5, 0.61){\scriptsize $$\alpha = \arcsin\left(\frac{1}{3}\right)$$ radians}
\end{mfpic}
&
\hspace{0.25in}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,-0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3, 1, 3}
\tlpointsep{4pt}
\axislabels{y}{{\scriptsize $$\frac{1}{3}$$} 1}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\arrow \polyline{(0,0), (-4.532, 2.113)}
\arrow \reverse \arrow \parafcn{160, 175, 5}{2.75*dir(t)}
\tlabel[cc](-3.45, 0.61){\scriptsize $$\alpha$$}
\end{mfpic} \\
\end{tabular}
Since $$\frac{1}{3}$$ isn't the sine of any of the common angles' discussed earlier, we use the arcsine functions to express our answers. The real number $$t = \arcsin\left(\frac{1}{3}\right)$$ is defined so it satisfies $$0 < t < \frac{\pi}{2}$$ with $$\sin(t) = \frac{1}{3}$$. Hence, $$\alpha = \arcsin\left(\frac{1}{3}\right)$$ radians. Since the solutions in Quadrant I are all coterminal with $$\alpha$$, we get part of our solution to be $$\theta = \alpha + 2\pi k = \arcsin\left(\frac{1}{3}\right) + 2\pi k$$ for integers $$k$$. Turning our attention to Quadrant II, we get one solution to be $$\pi - \alpha$$. Hence, the Quadrant II solutions are $$\theta = \pi - \alpha + 2\pi k = \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi k$$, for integers $$k$$.
\item We may visualize the solutions to $$\tan(t)=-2$$ as angles $$\theta$$ with $$\tan(\theta) = -2$$. Since tangent is negative only in Quadrants II and IV, we focus our efforts there.
\begin{tabular}{cc}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3 step 3 until 3}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\arrow \polyline{(0,0), (2.5, -4.3301)}
\arrow \parafcn{355, 305, -5}{1.5*dir(t)}
\gclear \tlabelrect[cc](4, -1){\scriptsize $$\beta = \arctan(-2)$$ radians}
\end{mfpic}
&
\hspace{.25in}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3 step 3 until 3}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\arrow \polyline{(0,0), (-2.5, 4.3301)}
\arrow \polyline{(0,0), (2.5, -4.3301)}
\arrow \parafcn{355, 305, -5}{1.5*dir(t)}
\arrow \reverse \arrow \parafcn{125, 295, 5}{1.5*dir(t)}
\tlabel[cc](-2, -1){\scriptsize $$\pi$$}
\tlabel[cc](2, -1){\scriptsize $$\beta$$}
\end{mfpic} \\
\end{tabular}
Since $$-2$$ isn't the tangent of any of the common angles', we need to use the arctangent function to express our answers. The real number $$t = \arctan(-2)$$ satisfies $$\tan(t)=-2$$ and $$-\frac{\pi}{2} < t < 0$$. If we let $$\beta = \arctan(-2)$$ radians, we see that all of the Quadrant IV solutions to $$\tan(\theta) = -2$$ are coterminal with $$\beta$$. Moreover, the solutions from Quadrant II differ by exactly $$\pi$$ units from the solutions in Quadrant IV, so all the solutions to $$\tan(\theta) = -2$$ are of the form $$\theta = \beta + \pi k = \arctan(-2) + \pi k$$ for some integer $$k$$. Switching back to the variable $$t$$, we record our final answer to $$\tan(t) = -2$$ as $$t = \arctan(-2) + \pi k$$ for integers $$k$$.
\item The last equation we are asked to solve, $$\sec(x) = -\frac{5}{3}$$, poses two immediate problems. First, we are not told whether or not $$x$$ represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section \ref{CircularFunctions} and convert this to the cosine problem $$\cos(x) = -\frac{3}{5}$$. Adopting an angle approach, we consider the equation $$\cos(\theta) = -\frac{3}{5}$$ and note that our solutions lie in Quadrants II and III. Since $$-\frac{3}{5}$$ isn't the cosine of any of the common angles', we'll need to express our solutions in terms of the arccosine function. The real number $$t = \arccos\left(-\frac{3}{5}\right)$$ is defined so that $$\frac{\pi}{2} < t < \pi$$ with $$\cos(t) = -\frac{3}{5}$$. If we let $$\beta = \arccos\left(-\frac{3}{5}\right)$$ radians, we see that $$\beta$$ is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use $$-\beta = -\arccos\left(-\frac{3}{5}\right)$$. Since all angle solutions are coterminal with $$\beta$$ or $$-\beta$$, we get our solutions to $$\cos(\theta) = -\frac{3}{5}$$ to be $$\theta = \beta + 2\pi k = \arccos\left(-\frac{3}{5}\right) + 2\pi k$$ or $$\theta = -\beta + 2\pi k = -\arccos\left(-\frac{3}{5}\right) + 2\pi k$$ for integers $$k$$. Switching back to the variable $$x$$, we record our final answer to $$\sec(x) = -\frac{5}{3}$$ as $$x = \arccos\left(-\frac{3}{5}\right) + 2\pi k$$ or $$x = -\arccos\left(-\frac{3}{5}\right) + 2\pi k$$ for integers $$k$$.
\begin{tabular}{cc}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,-0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3 step 3 until 3}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\arrow \polyline{(0,0), (-2.5, 4.3301)}
\arrow \parafcn{5, 115, 5}{1.5*dir(t)}
\gclear \tlabelrect[cc](4, 1.5){\scriptsize $$\beta = \arccos\left(-\frac{3}{5}\right)$$ radians}
\end{mfpic}
&
\hspace{-.15in}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\axes
\tlabel(5,-0.5){\scriptsize $$x$$}
\tlabel(0.25,5){\scriptsize $$y$$}
\tlabel(3.1,-0.75){\scriptsize $$1$$}
\tlabel(0.25,3.1){\scriptsize $$1$$}
\xmarks{-3 step 3 until 3}
\ymarks{-3 step 3 until 3}
\drawcolor[gray]{0.7}
\circle{(0,0),3}
\drawcolor[rgb]{0.33,0.33,0.33}
\dashed \polyline{(0,0), (-2.5, 4.3301)}
\arrow \dotted \parafcn{5, 115, 5}{1.5*dir(t)}
\gclear \tlabelrect[cc](4, 1.5){\scriptsize $$\beta = \arccos\left(-\frac{3}{5}\right)$$ radians}
\arrow \polyline{(0,0), (-2.5, -4.3301)}
\arrow \parafcn{355, 245, -5}{1.5*dir(t)}
\gclear \tlabelrect[cc](4, -1.5){\scriptsize $$-\beta = -\arccos\left(-\frac{3}{5}\right)$$ radians}
\end{mfpic} \qed \\
\end{tabular}
\end{enumerate}
\end{ex}
The reader is encouraged to check the answers found in Example \ref{basicinverseeqns} - both analytically and with the calculator (see Section \ref{sectionarcstuffoncalc}). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice \dots
### Contributors
• Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College)
| 2017-12-11T15:04:28 |
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https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_12&diff=128878&oldid=69636
|
# Difference between revisions of "1984 AIME Problems/Problem 12"
## Problem
A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$. If $x=0$ is a root for $f(x)=0$, what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$?
## Solution 1
If $f(2+x)=f(2-x)$, then substituting $t=2+x$ gives $f(t)=f(4-t)$. Similarly, $f(t)=f(14-t)$. In particular, $$f(t)=f(14-t)=f(14-(4-t))=f(t+10)$$
Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function $$f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}$$ satisfies the conditions and has no other roots.
In the interval $-1000\leq x\leq 1000$, there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$, therefore the minimum number of roots is $\boxed{401}$.
## Solution 2 (non-rigorous)
We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$. $QED \blacksquare$
Solution by a1b2
## Solution 3
Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$
## Solution 4
Let $z$ be an arbitrary zero. If $z=2-x$, then $x=2-z$ and $2+x=4-z$. Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$. From $0$, we get $4$ and $14$. Now note that applying either of these twice will return $z$, so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$, respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$
| 2022-05-17T06:51:09 |
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https://math.stackexchange.com/questions/1477307/how-do-you-factorize-quadratics-when-the-coefficient-of-x2-gt-1
|
# How do you factorize quadratics when the coefficient of $x^2 \gt 1$?
So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
• I recommend writing the factors of the coefficient of $x^2$ and the factors of the constant and then trying combinations of them Oct 13 '15 at 0:11
• Note that $2x^2-x-3 = 2(x^2-\frac{1}{2}x-\frac{3}{2})$. If you say you know how to factor monic quadratics, then you should be able to factor what is in the parenthesis above. Oct 13 '15 at 0:21
For quadratic polynomials of the form $f(x)=ax^2+bx+c$ with $a,b,c$ real numbers, the roots are known to be given by the following:
The Quadratic Formula: The roots of $f(x)=ax^2+bx+c$ are $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
Note that when $b^2-4ac=0$, then the root is "repeated" and when $b^2-4ac<0$ then the roots are non-real complex numbers.
The polynomial then can be factored as $ax^2+bx+c = a(x-x_1)(x-x_2)$
For your specific example, $2x^2-x-3$, this is of the form $ax^2+bx+c$ with $a=2,~b=-1,~c=-3$.
Applying the formula above gives the two roots as:
$x_1=\frac{-(-1)+\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1+\sqrt{1+24}}{4}=\frac{1+5}{4}=\frac{3}{2}$
$x_2=\frac{-(-1)-\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1-\sqrt{1+24}}{4}=\frac{1-5}{4}=-1$
You have then that $2x^2-x-3=2(x-\frac{3}{2})(x-(-1))$, which can be rewritten as $=(2x-3)(x+1)$
note: in some settings we are interested only in factorizations which involve only integers. It is possible in this setting then that no such factorizations exist and we say the polynomial is irreducible
• Using that formula made no sense at all to me. I get $b^2 - 4ac$ to be 23, and the square root of that is just an absurdly complex number. I don't understand how I'm supposed to factor anything with this? Oct 13 '15 at 0:45
• @Julian $b=-1,~a=2,~c=-3$. You get then $b^2-4ac = (-1)^2-4(2)(-3) = 1+24=25$, not $23$. Oct 13 '15 at 0:47
• Also, since the term complex has mathematical meaning other than what you are using it for here, it is recommended to call it "complicated" as opposed to complex. Oct 13 '15 at 0:48
• I failed to realise that $b^2$ gives me $(-1)^2$ and not $-1^2$. I see now that it is indeed 25. Meaning x1 = $\frac{6}{4}$ and x2 = $1$? Oct 13 '15 at 0:56
• Indeed. Although many methods exist, some of the cleaner solutions like BLAZE's above require some intuition. I prefer teaching this one since it is very algorithmic and requires only routine application of the rule. In reality, this is taking the "completing the square" technique and coming up with a generic formula to follow. Be warned however that this only works for quadratics. You cannot use this formula to factor something like $ax^3+bx+c$. There do exist longer, more complicated formulae for cubics and quartics, but hardly anyone memorizes them. Oct 13 '15 at 1:00
You can try factoring by inspection.
$$2x^2-x-3=(ax+b)(cx+d)$$
That's your general form, and we are going to guess that $a$, $b$, $c$ and $d$ are integers (otherwise factoring will be a real mess).
Considering FOIL,
$b\times d= -3$, so one of them is negative. If they are integers, they are $1$ and $-3$ or $-1$ and $3$.
$a\times c=2$, so they are $1$ and $2$.
Finally, $bc + ad = -1$
We can experiment with different combinations of $(ax+b)(cx+d)$ until we find the one that meets these criteria.
$$(2x-3)(x+1)$$
This method is is not so direct as it involves. Trial and error. But it may help you see what's going on when you factor, and may help you make sense of the more direct, but perhaps more abstract methods.
You can either use the quadratic formula for a general quadratic $ax^2+bx+c$ which is $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ which I will say no more of as JMoravitz has rigorously explained it.
Or you can complete the square, or write $$2x^2 \color{blue}{-x}-3$$ $$=2x^2+\color{blue}{2x-3x}-3$$ $$=2x(x+1)-3(x+1)$$ $$=\color{red}{(2x-3)(x+1)}$$ Notice, in the part marked $\color{blue}{\mathrm{blue}}$ I have simply rewritten $-x$ as $2x-3x$.
From then on you simply take out common factors.
Note that this method is only valid when you have two numbers whose product is $-6$ and sum is $-1$.
Or, put in another way for the general quadratic equation $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$. Otherwise using the quadratic formula is the best method as it suits all scenarios.
For completing the square: $$2x^2 -x-3$$ $$=2\left(x^2 -\frac{x}{2}-\frac{3}{2}\right)$$ $$=2\left(\left(x -\frac{1}{4}\right)^2-\frac{1}{16}-\frac{3}{2}\right)$$ $$=2\left(\left(x -\frac{1}{4}\right)^2-\left(\frac{5}{4}\right)^2\right)\tag{1}$$ $$=2\left(\left(x -\frac{1}{4}+\frac{5}{4}\right)-\left(x -\frac{1}{4}-\frac{5}{4}\right)\right)\tag{2}$$ $$=2\left((x+1)\left(x -\frac{3}{2}\right)\right)$$ $$=\color{red}{(2x-3)(x+1)}$$ as before.
From $(1)$ to $(2)$ I used the difference of two squares formula such that $a^2-b^2=(a+b)(a-b) \space \space\forall a,b \in \mathbb{R}$
Hope this helps.
• Are there like a million different methods of doing this? I am yet to find a single coherent one. I don't quite understand what you did here. Is this the favourable method? The quadratic formula as suggested by JMoravitz here got me nowhere. :( Oct 13 '15 at 0:47
• @JulianNikolayKrogh-Fredrikse Yes there are several [maybe not a million :)] ways of doing this as you can also complete the square as well are you familiar with this? Oct 13 '15 at 0:51
• No I am not. Is that the "box method"? purplemath.com/modules/factquad2.htm Because that one was as incoherent to be as the ABC-formula.. I honestly didn't expect factoring at this level to be so hard.. Oct 13 '15 at 0:59
• @JulianNikolayKrogh-Fredrikse If you tell me where you're confused I will explain further in my answer Oct 13 '15 at 1:24
• @Julian Note that, in BLAZE's first example, (s)he splits $-x = -1x$ up into $-1x = -3x + 2x$, where $-3 + 2 = -1,$ the coefficient of $x$. Note also that if you multiply $a = 2$ and $c = -3$, you get $2\cdot(-3) = -6$; that's how we know to use $-3$ and $2$ to split up $-x = -1x$. More detail here, this is known as factoring by grouping/"the AC method". It's equivalent to the box method, and relies on being able to factor polynomials with four terms by grouping. Oct 13 '15 at 1:52
Depending on your level, you may not have seen the quadratic formula that JMoravitz and BLAZE posted. That formula is often the way to find the zeroes of and factor a quadratic.
But what Adam suggests may be more up your alley, and it's what I'll discuss as well: You can make educated guesses and then check to see if it works. The problem you have posted does not have many options to guess, so this method is fairly quick.
Consider your quadratic and what you expect the factored form to look like:
$$2x^2-x-3 = (\color{red}{\Box}x \color{red}{\pm} \color{red}{\Box})(\color{red}{\Box}x\color{red}{\pm}\color{red}{\Box})$$
You know that the factored form requires you to fill in those blanks, choosing the appropriate sign and number. What numbers can possibly satisfy the equation?
Start with the $2x^2$ term on the left. This one is straightforward. In order to get $2x^2$, the coefficients of the $x$'s on the right hand side must be $2$ and $1$ if we are to have all integer coefficients.
$$2x^2-x-3 = (\color{green}{2}x \color{red}{\pm} \color{red}{\Box})(\color{green}{1}x\color{red}{\pm}\color{red}{\Box})$$
$$2x^2-x-3 = (2x \color{red}{\pm} \color{red}{\Box})(x\color{red}{\pm}\color{red}{\Box})$$
Now what about the $-3$ on the right? It's factors are $\pm1, \pm3$. We need to select two of these numbers and have them fill the remaining sections. You can try all combinations until you have the answer, but we can be smarter than that.
Since $-3$ is negative, one of these numbers will be positive and the other will be negative. Furthermore, the two factors will need to add to $-1$, after one of them is multiplied by $2$, since we have $-x$ on the left.
Notice that $2-3 = -1$. That's what we want. So the $2x$ multiplies a $1$ and the $x$ multiplies a $-3$.
$$2x^2-x-3 = (2x \color{green}{-3})(x\color{green}{+1})$$
We do a quick check:
$$(2x-3)(x+1) = 2x^2 +2x -3x -3 = 2x^2 -x -3$$
It works.
• The quadratic formula is introduced pretty early, I thought. But this use of it may be something that OP hasn't seen. I've thought that maybe a diagram could be used to illustrate completing the square. Oct 13 '15 at 3:37
• @AdamHrankowski It is introduced pretty early. In the US, it's often shown in grade 7-9 depending on skill. That being said, there are classes at my university that have students who have yet to learn to factor. I hope my dialogue helps instead of hinders! There are a lot of ways to slice this cake :) Oct 13 '15 at 3:43
• I'm in BC, Canada, and I haven't seen it until Grade 11. Oct 13 '15 at 3:45
You can get a monic polynomial out of his by putting $y=2x$. The expression becomes $$\frac12(y^2-y-6)$$ which you can easily factorise as $$\frac12(y-3)(y+2)$$ Substituting back gives $$(2x-3)(x+1)$$
We can factor $2x^2 - x - 3$ with respect to the rationals if we can find two numbers with product $2 \cdot -3 = -6$ and sum $-1$. The factors of $-6$ are \begin{align*} -6 & = 1 \cdot -6 & -6 & = -1 \cdot 6\\ & = \color{blue}{2 \cdot -3} & & = -2 \cdot 3 \end{align*} Of these four pairs of factors, only $2$ and $-3$ have sum $-1$. Hence, \begin{align*} 2x^2 - x - 3 & = 2x^2 + 2x - 3x - 3 && \text{split the linear term}\\ & = 2x(x + 1) - 3(x + 1) && \text{factor by grouping}\\ & = (2x - 3)(x + 1) && \text{extract the common factor} \end{align*} Note that if we multiply $2x - 3$ and $x + 1$, we carry out the steps of the factorization in reverse order.
Why does this work? Suppose we have the factorization with respect to the rationals \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su} \end{align*} Observe that the product of the coefficients of the quadratic and constant terms is equal to the product of the two coefficients that sum to the coefficient of the linear term, that is
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st}) = rstu$$ Matching coefficients gives $a = rt$, $b = ru + st$, and $c = su$. Thus, we can factor a quadratic with respect to the rationals if we can find two numbers with product $ac$ and sum $b$.
| 2021-09-20T00:32:49 |
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https://math.stackexchange.com/questions/1862918/prove-that-sum-k-0n-binom3n-k2n-binom3n1n
|
# Prove that $\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$
Prove that $$\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$$
I've tried multiple things that didn't work.
Maybe this would help
$$\sum_{k=0}^n \binom{3n-k}{2n}=\sum_{k=0}^n \binom{3n-(n-k)}{2n}=\sum_{k=0}^n \binom{2n+k}{2n}$$
Your identity is a special case of the more general identity $$S(m,n) = \sum_{k=0}^n \binom{m+k}{m} = \binom{m+n+1}{m+1},$$ which you can prove by induction on $n$: note $S(m,0) = 1 = \binom{m+1}{m+1}$. Then observe \begin{align*} S(m,n+1) &= S(m,n) + \binom{m+n+1}{m} \\ &= \binom{m+n+1}{m+1} + \binom{m+n+1}{m} \\ &= \binom{m+n+2}{m+1} = \binom{m + (n+1) + 1}{m+1}, \end{align*} hence by the induction hypothesis, the claim is proven. Then choose $m = 2n.$
Choose $2n+1$ of $3n+1$ ordered items by choosing the greatest first, at position $2n+k+1$, and then choosing the remaining $2n$ items out of the $2n+k$ items less than it.
• How can you get $${{m \choose n}} =\oint_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{n+1}}$$? – Zack Ni Jul 18 '16 at 8:46
• @ZackNi \begin{align} &\oint_{\left\vert z\right\vert = 1}{\pars{1 + z}^{m} \over z^{n + 1}}\,{\mathrm{d}z \over 2\pi\mathrm{i}} = \oint_{\left\vert z\right\vert = 1}{1 \over z^{n + 1}}\sum_{k = 0}^{m}{m \choose k}z^{k} \,{\mathrm{d}z \over 2\pi\mathrm{i}} = \sum_{k = 0}^{m}{m \choose k}\underbrace{\oint_{\left\vert z\right\vert = 1} {1 \over z^{n - k + 1}}\,{\mathrm{d}z \over 2\pi\mathrm{i}}}_{\displaystyle{\delta_{kn}}} = {m \choose n} \end{align} – Felix Marin Jul 18 '16 at 8:53
Here is another variation. It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g. \begin{align*} \binom{n}{k}=[x^k](1+x)^n \end{align*}
We obtain \begin{align*} \sum_{k=0}^n\binom{3n-k}{2n}&=\sum_{k=0}^n\binom{3n-k}{n-k}\tag{1}\\ &=\sum_{k=0}^\infty[x^{n-k}](1+x)^{3n-k}\tag{2}\\ &=[x^n](1+x)^{3n}\sum_{k=0}^\infty\left(\frac{x}{1+x}\right)^k\tag{3}\\ &=[x^n](1+x)^{3n}\frac{1}{1-\frac{x}{1+x}}\tag{4}\\ &=[x^n](1+x)^{3n+1}\\ &=\binom{3n+1}{n} \end{align*}
Comment:
• In (1) we use the binomial identity $\binom{n}{k}=\binom{n}{n-k}$.
• In (2) we apply the coefficient of operator and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.
• In (3) we use the linearity of the coefficient of operator and use the rule \begin{align*} [x^{p-q}]A(x)=[x^p]x^{q}A(x) \end{align*}
• In (4) we apply the geometric series expansion.
Suppose we seek to evaluate
$$\sum_{k=0}^n {3n-k\choose 2n}$$
using a different integral than what was used by @MarkusScheuer and @FelixMarin.
Introduce
$${3n-k\choose 2n} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} \frac{1}{(1-z)^{2n+1}} \; dz.$$
Observe that this vanishes for $k\gt n$ so we may extend the sum to infinity, getting
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{2n+1}} \sum_{k\ge 0} z^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{2n+2}} \; dz.$$
This evaluates by inspection to
$${n+2n+1\choose 2n+1} = {3n+1\choose 2n+1} = {3n+1\choose n}.$$
| 2019-06-18T22:54:38 |
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https://www.flyingcoloursmaths.co.uk/a-cubic-conundrum/
|
In class, we tackled a Senior Maths Challenge problem that goes something like this:
“Given that $y^3 f(x) = x^3 f(y)$ for all real $x$ and $y$, and that $f(3) \ne 0$, find the value of $\frac{f(20)-f(2)}{f(3)}$.
A: 6; B: 20; C: 216; D: 296; E:2023”
Have a play with it. Spoilers below the line. The interesting bits for me come after the problem is solved.
The question itself is more intimidating than it is difficult. In particular, $27 f(x) = x^3 f(3)$ for all $x$, so we can multiply by 27 and replace all of the $f$ evaluations like this ((This is simply the neatest way I can see; other options are available.)):
$\frac{27 f(20)- 27f(2)}{27f(3)} = \frac{20^3 f(3) - 2^3 f(3)}{27 f(3)}$.
All of the $f(3)$s cancel, leaving $\frac{8000 - 8}{27}$.
A quick sense check says that’s between 100 and 1000, so the answer must be C or D; use whatever arithmetic you like to find that it’s 296. (Under pressure, I’d probably divide by 9 and by 3; alternatively, $27\times 250 = 6750$, so the answer is greater than 250). The answer is D.
### A wry twist
The answer can’t be 216, because 216 is a cube. If the answer were 216 then $20^3 + (-2)^3 = \br{3^3}{6^3}$, which would be a counterexample to Fermat’s Last Theorem.
### But what’s the function?
I also wondered what functions might satisfy the given relationship, which I rewrote as $\frac{f(y)}{f(x)} = \frac{y^3}{x^3}$. It’s fairly obvious that $f(x) = x^3$ satisfies this, but are there others?
I don’t much like the division. I prefer subtraction. I’m going to let $F(x) = \ln(f(x))$, so that $F(y) - F(x) = 3\ln(y) - 3\ln(x)$.
Now let $y = x + h$: $F(x+h) - F(x) = 3\ln(x+h) - 3\ln(x)$
Divide by $h$: $\frac{F(x+h)-F(x)}{h} = 3\frac{ \ln(x+h) - \ln(x) }{h}$
So the derivative of $F(x)$ is equal to the derivative of $3\ln(x)$ – or $\diff{}{x} \ln(f(x)) = \diff{}{x}\ln(x^3)$.
Integrate both sides with respect to $x$: $\ln(f(x)) = \ln(x^3)+ C$
So $f(x) = Ax^3$ for any constant $A$ ((Except for 0, which is disallowed in the question.))
I thought that was a great question, as much for where it led as for the puzzle it posed. What did you make of it?
| 2021-11-30T06:29:00 |
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https://mathhelpboards.com/threads/percent-of-perfect-squares-that-have-an-odd-digit-for-their-hundreds-place.1208/
|
# Percent of perfect squares that have an odd digit for their hundreds place
#### checkittwice
##### Member
Suppose you consider the set of all perfect squares (the squares of all of the integers).
What percent of them have an odd digit for their hundreds place?
#### CaptainBlack
##### Well-known member
suppose you consider the set of all perfect squares (the squares of all of the integers).
What percent of them have an odd digit for their hundreds place?
41%
cb
#### checkittwice
##### Member
Edit: Some of my text was removed because I don't agree with it.
I really meant to ask (hopefully an easier question with less work):
"What percent of the perfect squares have an odd digit in their tens place?"
Edit:
I sent a PM to CB, because I did not want to have any post in
front of Opalg's solution to my amended question.
Last edited:
#### CaptainBlack
##### Well-known member
I would chop off that 1% and make it 40%.
However, I don't have the work for you at this
time to back it up.
When I get the chance I will check my reasoning and the counting program.
CB
#### Opalg
##### MHB Oldtimer
Staff member
I really meant to ask (hopefully an easier question with less work):
"What percent of the perfect squares have an odd digit in their tens place?"
If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you get
$a^2 + 20ab + \ldots$ (everything else is a multiple of 100).
If you are only interested in whether the tens digit is even or odd, then a multiple of 20 makes no difference. So everything depends on whether the tens digit in $a^2$ is even or odd. The squares of the digits from 0 to 9 are
00, 01, 04, 09, 25, 49, 64, 81 (tens digit even)
and
16, 36 (tens digit odd).
Each of these is equally likely to occur, so I reckon that the proportion of perfect squares with an odd digit in their tens place is 20%.
#### CaptainBlack
##### Well-known member
If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you get
$a^2 + 20ab + \ldots$ (everything else is a multiple of 100).
If you are only interested in whether the tens digit is even or odd, then a multiple of 20 makes no difference. So everything depends on whether the tens digit in $a^2$ is even or odd. The squares of the digits from 0 to 9 are
00, 01, 04, 09, 25, 49, 64, 81 (tens digit even)
and
16, 36 (tens digit odd).
Each of these is equally likely to occur, so I reckon that the proportion of perfect squares with an odd digit in their tens place is 20%.
That is what I get.
CB
#### Opalg
##### MHB Oldtimer
Staff member
That is what I get.
... and I agree with your 41% for those with an odd digit in the hundreds place. I simply counted the number of odd hundreds-place digits in the list of the first 100 squares (as listed here, for example). The proportion in any other sequence of 100 consecutive squares will be the same.
| 2022-05-27T07:11:05 |
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https://jd2718.org/2007/03/17/more-on-the-roots-of-i/?replytocom=7450
|
In this post I mentioned looking for $\sqrt{i}$, and in this comment mentioned searching for $\sqrt[3]{i}$. In this comment, reader Vlorbik took exception, and said that I would have better looked for $\pm\sqrt{i}$, and then more comments which I was not certain I understood. And then yesterday one of the top puzzle solvers here, JBL, weighed in, saying in effect that my error was small.
it is the math that holds the authority, not me
I want my teaching to be well-informed, so I dug around a bit (spoke to a professor, dipped through a couple of books, surfed). Check my understanding, if you will.
(Discussion continues, beneath the fold –>)
When we extend the Natural numbers to include an additive identity, and then to the Integers, and thence the Rationals, and finally the Reals, we preserve order. But when we extend to the complex numbers, we lose order. If you haven’t thought about this before, consider complex numbers, a+bi, graphed on Real vs imaginary axes. (image from
Weisstein, Eric W. “Complex Plane.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/ComplexPlane.html)
Because we are in a plane, not on a line, we do not have a “greater than” direction. Think about this:
could we say which was greater, 4 – 3i, 3 – 4i, -3 + 4i, -4 + 3i, 5i, -5i, 5 or -5?
(digression): When the very first New York State Math B Regents (June 2001) included this question
Melissa and Joe are playing a game with complex numbers. If Melissa has a score of 5 – 4i and Joe has a score of 3+2i, what is their total score? (question #11)
I knew we were in serious trouble. Contriving real world contex was trumping mathematical sense. Insteading of asking students to add two complex numbers (a bit easy, but still a question one might ask), a highly improbable situation (keeping score with complex numbers – not if it is important to know who is in the lead!), was paired with a silly task (add up scores? only if I am looking at the under/over, right?). But let’s leave the NYS Regents for other discussion. (/digression)
Lets return to what was wrong with my beginning: $\sqrt{i}$. Had I written on the board $\sqrt{25}$, I would have wanted the kiddies to respond that the value was 5, since there is an implied principal square root. If I wanted both roots, I would have prompted them instead with $y^2 = 25$.
This is where Lisa Simpson makes her mistake in the separate boy/girl math episode, Girls Just Want to Have Sums. She shouts (from the bushes!) “5.” Let’s pause to solve this.
$y^2 = 25 \rightarrow y^2 - 25 = 0 \rightarrow (y+5)(y-5) = 0 \rightarrow y=\pm{5}$ or, shorter version: $y^2 = 25 \rightarrow \sqrt{y^2} = \pm\sqrt{25} \rightarrow y=\pm{5}$.
So, had I asked for the solution to $x^2 = i$, we would be hunting for two roots. But since I asked for $\sqrt{i}$ there is an issue with the idea of principle root, and $\pm\sqrt{i}$ is better since it asks for both, but there is something off, since I don’t know which is the positive and which is the negative root.
Let’s stick with the unambiguous. No harm there: $x^2 = i, x^3 = i, x^4 = i$. Plus it is a fantastic tease to have gotten a preview with (increasingly cumbersome) algebra. This group will be learning DeMoivre’s theorem in the fall term next year.
OK, I started explaining my error to the class on Friday. We will take a second shot at it sometime next week. There is something else, very positive, at play here. I am a math teacher. I communicate to the students. But there is math I don’t know, math I continue to learn, and in the end it is the math that holds the authority, not me.
1. March 17, 2007 pm31 10:55 pm 10:55 pm
Yes! Sharing the process with students of how you go about solving problems or asking questions feels really good to me. It’s a way to teach them how to think instead of just teaching them facts and procedures.
2. March 17, 2007 pm31 11:02 pm 11:02 pm
It is true that the complex numbers are not ordered, but it is possible to identify a principal value for the square root, as explained in the Wikipedia article. In short, we represent the complex number in polar form z = r e^(it), where t is between -pi and pi (including pi) and r is a non-negative real number. Then the principal square root of z is r^(1/2) e^(it/2). However, I should caution that this convention is not universal.
3. March 17, 2007 pm31 11:06 pm 11:06 pm
Thanks for the encouragement and clarification. I will include this bit in my explanation to the class, though I do intend to avoid the $\sqrt{}$ symbol when working with complex numbers. It’s not worth the ambiguity, however slight, when there is a completely unambiguous approach available.
March 18, 2007 am31 2:03 am 2:03 am
There are two problems with identifying a principal square root function (or, in general, a “rational power” function) on the complex numbers: the choice of range is entirely arbitrary, and the function can’t be continuous everywhere. Neither of these is a particularly big problem, and much of the time it works fine to use a function equivalent to the one involution suggests.
Things start to get nastier as we extend to nonrational powers: in the complex numbers, an expression $a^b$ where $b$ is anything other than a (real) rational number represents infinitely many values.
5. March 18, 2007 am31 2:10 am 2:10 am
It’s been a while, but don’t the nth roots of unity form a group under multiplication, if n is rational, but do something strange if n is not?
March 18, 2007 pm31 3:37 pm 3:37 pm
Where does the +- come from?
y^2=25 =>
(+y)^2=25, (-y)^2=25 =>
sqrt(y^2)=sqrt(25), sqrt((-y)^2)=sqrt(25) =>
y=5, -y=5
March 18, 2007 pm31 8:25 pm 8:25 pm
Jonathan: Yes. I think the easiest way to get a feel for what is going on is to exchange the notion that a positive integer power is what you get by repeated multiplications, which extends to reasonably natural explanations of nonpositive integer and rational powers, to the understanding of exponentiation in the complex numbers, best understood in terms of the exponential function $f(z) = e^z$.
We have $1 = e^0 = e^{2\pi i} = e^{-2\pi i} = e^{4\pi i} = \cdots$. If we raise this to some rational power, we get $e$ raised to a bunch of numbers which are rational multiples of $\pi i$, with constant denominator. As we go down the list of values for 1 that I’ve written above, we’ll see that these values eventually “wrap around”: we start getting values where the exponents differ by an integer multiple of $2\pi i$, and these will actually give us the same value. So, finitely many values, which geometrically can be seen to lie at the vertices of a regular polygon inscribed in the unit circle centered in the origin. (We know which such polygon because it includes 1 as a vertex.) And these values do indeed form a finite cyclic group under multiplication.
In the case that we use an irrational real exponent, everything happens the same except for the “wrapping around”: no two multiples of an irrational number differ by an integer. For the algebraic structure of the roots, this means we still actually have a multiplicative cyclic group, but this time it will be infinite. (This happens in this case, as well as in the previous, because the exponents $0, 2\pi i, \ldots$ form an infinite additive cyclic group.) All the values must still have absolute value 1, and geometrically in turns out that they represent a dense subset of the unit circle.
The case of imaginary exponents is even weirder. $1^{ai} = (e^{2n\pi i})^{ai} = e^{-2n a \pi}$. Note that the exponent there is real: this actually describes an infinite set of real numbers, an exponential family. This again turns out to be an infinite cyclic group, and it doesn’t matter whether $a$ was rational or irrational.
In order to get generic complex exponent, we just use the fact that $1^{a + bi} = 1^a \cdot 1^{bi}$. If a and b are not 0, the result will be the result of multiplying all the values from the corresponding cases. Geometrically, this gives you exactly what you would expect (a bunch of concentric circles, each of which looks like what you get when you do $1^a$, with radii given by the values of $1^{bi}$); algebraically, it’s equivalent to the product of the two groups.
Unfortunately, I can’t preview, so I’ll have to wait to see how this turns out :-)
8. March 18, 2007 pm31 9:04 pm 9:04 pm
I’ll repeat this note elsewhere, but if you include some kind of flag at the end (eg, feel free to fix my latex), then I will, indeed, try to fix the latex, if I knew what you were after.
Alternately, you can post, and if you don’t like the result, post again, and again, until you are happy, and I will go in and delete the unsatisfactory early tries.
In the meantime, I am digesting the bits about the cyclic groups. Everything up to the imaginary exponents I once knew, so it’s not completely foreign.
March 18, 2007 pm31 11:33 pm 11:33 pm
Okay, thanks!
At some point in my last comment I used the words, “exponential family.” What I really meant was that the sequence is an infinite, two-ended geometric sequence. For example, if $a = -\frac{\ln 2}{2\pi}$ then $1^{ai} = e^{2\pi ni \cdot ai} = e^{n\ln 2} = 2^n$, where I’m using equal signs somewhat loosely here (the thing on the right should really be the set of all such numbers as $n$ varies over the integers). And this set really is an infinite-at-both-ends geometric sequence, which also happens to form a multiplicative group. And this would work for any value of $a$ I happened to choose.
June 24, 2014 pm30 1:52 pm 1:52 pm
“study more analysis”
plan, though? did you
struggle through one
or more semesters like
you planned… or give
up in some “what would
owen do?” moment?
& if so, when? & how did
it feel? inquiring minds
want to know (and other
tautologies).
| 2022-10-01T21:48:09 |
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http://mathhelpforum.com/statistics/164715-there-100-photos-envelope-among-whose-wanted-one.html
|
# Thread: There are 100 photos in the envelope, among whose is the wanted one?
1. ## There are 100 photos in the envelope, among whose is the wanted one?
There are There are 100 photos in the envelope, among whose is the wanted one. Tim picks up 10 photos randomly. What is the probability that the wanted one is among these ten photos?
My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is
C(1,100) / C(10,100)
My friend says : No! It's (C(1,1) * C(9,99)) / C(10,100)
which one is correct and WHY? Thank you
2. If we group the 100 in groups of 10, each group has a 10% chance so 10% is the probability that 10 contains the 1.
3. Therefore, non of mine and my friend's solutions are correct?
4. Hello, Narek!
Therefore, neither mine and my friend's solutions are correct?
Your friend's solution, which should be written: $\dfrac{C(1,1)\cdot C(99,9)}{C(100,10)}$
. . is equal to $\dfrac{1}{10}$ . . . and is correct!
His/her reasoning is correct.
There are $C(100,10)$ possible outcomes; that's the denominator.
There is 1 special photo and 99 Others.
There is $C(1,1)$ way to get the special photo.
And there are $C(99,9)$ ways to choose 9 Others.
Hence, there are: . $C(1,1)\cdot C(99,9)$ ways to select 10 photos,
. . including the special one; that's the numerator.
Please pass this on to your friend: . Good work!
5. Originally Posted by Narek
My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is C(1,100) / C(10,100)
In addition to what has been written already, maybe the following helps you seeing why this is wrong:
Suppose there are only 10 or 11 photos instead of 100 (and again, there's 1 you want, and you pick 10). Then according to your method, what are the probabilities of the right photo being in your pick?
(condsidering that with 10 photos the probability should be 1, and with 11 it should be less than 1, obviously)
| 2016-09-29T14:47:10 |
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https://www.physicsforums.com/threads/show-that-the-function-is-a-sinusoid-by-rewriting-it.912414/
|
Show that the function is a sinusoid by rewriting it
Tags:
1. Apr 24, 2017
Vital
1. The problem statement, all variables and given/known data
Hello!
I am doing exercises on sinusoid functions from the beginning of Trigonometry.
I hoped I understood the topic, but it seems not quite, because I don't get the results authors show as examples for one of possible answers, as there can be a few answers to the same exercise. I will be grateful for your help and explanation on what I am doing wrong.
Show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0.
2. Relevant equations
I will show one example, and, maybe, later add more, if it is necessary.
f(x) = 3√3 sin(3x) - 3cos(3x)
Here is the answer from the book, and below I post how I tried to solve the task:
answer from the book f(x) = 3√3 sin(3x) - 3cos(3x) = 6 sin( 3x + (11π/6) ) = 6cos( 3x + (4π/3) )
3. The attempt at a solution
My solution:
f(x) = 3√3 sin(3x) - 3cos(3x)
(1) rewrite the expression in the sinusoid form:
f(x) = A sin(3x) cos(φ) - A cos(3x) sin(φ)
Clearly, B = 0, ω = 3.
(2) Find coefficients:
3√3 = A cos(φ)
3 = A sin(φ)
Given Pythagorean identity we have:
cos2 + sin2 = 1
Multiply both sides by A2:
A2 cos2 + A2 sin2 = A2
3√32 + 32 = 36
Choose A = 6 (choosing positive 6)
(3) Find φ (works both ways - take cos or sin):
3√3 = A cos(φ)
3√3 = 6 cos(φ)
φ = π/6
(4) Solution:
f(x) = 3√3 sin(3x) - 3cos(3x) in sin form S(x) = A sin(ωx + φ) + B
f(x) = 6 sin(3x + π/6); the book gives this: f(x) = 6 sin( 3x + (11π/6) ) which is not the same:
if we add π/6, both cos and sin have positive values at this angle, while at +11π/6 sin is negative. If, however, it were -11π/6, then we would have ended at the same place as if we added π/6 with both cos and sin positive.
I don't see my mistakes in computations, and why my answer is wrong; and I don't see how authors achieved +11π/6 as the value of φ.
To find the same formula for cos I used cofunction identities:
sin(θ) = cos (π/2 - θ)
In my case θ = 3x + π/6, so cos ( π/2 - 3x - π/6) = cos ( - (3x - π/3) ) = cos (3x - π/3) which is not the same as cos( 3x + (4π/3) ) for same reasons stated above in discussion on sin. -π/3 has a positive cos value (quadrant IV), while +4π/3 has a negative cos value (quadrant III).
Thank you very much!
2. Apr 24, 2017
Staff: Mentor
The solution is not unique. What are the other possible values of φ? Also, you say that it doesn't matter if you use the cos or the sin to find φ. Does φ=π/6 work with the sin?
3. Apr 24, 2017
Vital
Thank you very much for your answer. Yes, I understand that there might be a few solutions (please, see below for my answer to your question), and this refers to my original question. Please, take a look at my (4) where I show the problem with signs and hence the difference between my solution and the solution offered by authors.
"Does φ = π/6 work with sin?"
Yes, it does.
3√3 = A cos(φ) => 3√3 = 6 cos(φ) => φ = π/6
3 = A sin(φ) => 3 = 6 sin(φ) => sin(φ) = ½ => Φ = π/6
both cos and sin have same values with φ = π/6 + 2πk, where k is any integer; hence φ can be, for example, 13π/6, which lies in the same quadrant I as π/6; but not 11π/6, which lies in quadrant IV where sin is negative. I am bewildered. Seems I miss some basic understanding on how to approach this type of tasks.
f(x) = 6 sin(3x + π/6)
Authors offer:
f(x) = 6 sin( 3x + (11π/6) )
It could be the same if signs were different for π/6, or for 11π/6.
4. Apr 24, 2017
Staff: Mentor
That's not correct. That should be -3 on the left-hand side.
Edit: I see that the error stems from an earlier equation:
That should be f(x) = A sin(3x) cos(φ) + A cos(3x) sin(φ)
5. Apr 24, 2017
Vital
Oh! Indeed. How silly of me :-) I have spent so much time trying to figure out the conundrum :-) And, sure enough, 11π/6 gives exact values, with sin = -½. Thank you very much.
| 2018-02-19T16:22:42 |
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https://math.stackexchange.com/questions/4029423/if-b-is-rational-and-c-is-real-can-x2-bx-c-have-one-rational-root-and-on
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# If b is rational and c is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only definition of rational numbers to prove.
If $$b$$ is rational and $$c$$ is real, can $$x^2 + bx + c$$ have one rational root and one irrational root? Use only the definition of rational numbers to prove.
I think that it can't because since $$b$$ is rational, if two real roots exist then both are either rational or irrational, however, I'm not sure how to prove this using only the definition of rational numbers i.e. it is a number that can be expressed as $$\frac{x}{y}$$ where $$x$$ and $$y$$ are integers $$y$$ being nonzero. So you can't assume stuff like rational + irrational = irrational.
• By real root you mean irrational root? – R.V.N. Feb 17 at 17:07
• And by "$c$ is real" do you mean "$c$ is irrational"? – Michael Cohen Feb 17 at 17:08
• Welcome to Mathematics Stack Exchange. Note that the sum of roots is $-b$ – J. W. Tanner Feb 17 at 17:09
• @R.V.N. yes my bad – TurboAverage419 Feb 17 at 17:09
• @MichaelCohen no c is any real number – TurboAverage419 Feb 17 at 17:09
Of course, note that all rational numbers are indeed real.
I suppose you wanted to ask "Is it possible that one root is rational and the other is not". The answer to that is then "no".
Note that the sum of the two roots is $$-b \in \Bbb Q$$. Using the definition of rationals, it follows that the difference of two rationals is rational. Conclude that $$\text{rational} + \text{irrational} = \text{rational}$$ is not possible.
• Is rational numbers being closed under add/sub enough to conclude rational + irrational = rational not possible? Since we can't assume anything it seems there is no proof that there does not exist an irrational number that can satisfy the above equation. – TurboAverage419 Feb 17 at 17:17
• If $q_1, q_2$ are rationals and $r$ an irrational such that $q_1 + r = q_2$, what do you get by subtracting $q_1$ from both sides? – Aryaman Maithani Feb 17 at 17:20
• Oh I see now, if rational numbers are closed under subtraction that implies that an irrational number cannot be the difference of two rational numbers? – TurboAverage419 Feb 17 at 17:23
• Cannot be the difference of two rational numbers, yes – Aryaman Maithani Feb 17 at 17:23
• Argh, another typo! Thank you! – TurboAverage419 Feb 17 at 17:25
Hint
Suppose $$r$$ is a rational root and $$j$$ is a irrational root.
$$r+j=-b\to j=-(b+r).$$
Can you see the problem?
• Thanks, your comment rephrased what someone else said and helped clarify it! – TurboAverage419 Feb 17 at 17:26
| 2021-07-31T15:07:38 |
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https://math.stackexchange.com/questions/3588703/does-it-exist-a-way-how-to-find-the-ways-to-get-from-one-point-to-another-when-c
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# Does it exist a way how to find the ways to get from one point to another when certain points must be avoided in a grid?
The problem is as follows:
The diagram from below shows a grid of $$6\times 6$$. How many different ways can you get from $$A$$ to $$B$$ without going through any of the highlighted points?
The alternatives given are as follows:
$$\begin{array}{ll} 1.&\textrm{265 ways}\\ 2.&\textrm{365 ways}\\ 3.&\textrm{395 ways}\\ 4.&\textrm{405 ways}\\ \end{array}$$
Does it exist a way to simplify this problem?. How exactly can I find the method to solve this?. There isn't any indication of which sort of path should be taken. Hence there could be tons of ways and I'm stuck on it. Can someone help me here?.
It would help a lot to me to include some sort of diagram or drawing to justify a resonable method to solve this.
I'm assuming the paths must go right and down only.
You can count the paths by using the inclusion-exclusion principle. Without considering the points that must be avoided, there are $$\binom{6+6}{6}$$ paths. Now subtract the $$\binom{4+1}{1}\binom{2+5}{5}$$ paths that visit the first bad point, the $$\binom{2+4}{4}\binom{4+2}{2}$$ paths that visit the second bad point, and the $$\binom{4+5}{5}\binom{2+1}{1}$$ paths that visit the third bad point. Then add back in the paths that visit two bad points. No paths visit all three bad points.
\begin{align} &\binom{12}{6} -\left(\binom{5}{1}\binom{7}{5} +\binom{6}{4}\binom{6}{2}+\binom{9}{5}\binom{3}{1}\right) +\left(\binom{5}{1}\binom{4}{4}\binom{3}{1}+\binom{6}{4}\binom{3}{1}\binom{3}{1}\right)\\ &=924-(5\cdot 21+15\cdot 15+ 126\cdot 3)+(5\cdot 1\cdot 3+15\cdot 3\cdot 3)\\ &=924-(105+225+378)+(15+105)\\ &=924-708+120\\ &= {\color{red}{366}} \end{align}
Here's an alternative solution that uses a Pascal-type recursion. Let $$p(i,j)$$ be the number of good paths that start from $$A=(0,0)$$ and reach point $$(i,j)$$. We want to compute $$p(6,6)$$. By conditioning on the last step into $$(i,j)$$, we find that $$p(i,j)$$ satisfies the following recursion: $$p(i,j)= \begin{cases} 0 &\text{if (i,j) is a bad point}\\ 1 &\text{if i=0 or j=0}\\ p(i-1,j)+p(i,j-1) &\text{otherwise} \end{cases}$$ The resulting values of $$p(i,j)$$ are: $$\begin{matrix} i\backslash j &0 &1 &2 &3 &4 &5 &6 \\ 0 &1 &1 &1 &1 &1 &1 &1 \\ 1 &1 &2 &3 &4 &0 &1 &2 \\ 2 &1 &3 &6 &10 &10 &11 &13 \\ 3 &1 &4 &10 &20 &30 &41 &54 \\ 4 &1 &5 &0 &20 &50 &91 &145 \\ 5 &1 &6 &6 &26 &0 &91 &236 \\ 6 &1 &7 &13 &39 &39 &130 &{\color{red}{366}} \end{matrix}$$ So $$p(6,6)=366$$.
• @RobPratt The answer is $365$ but I have no idea how to get there. – Chris Steinbeck Bell Mar 21 '20 at 2:07
• Besides the inclusion-exclusion argument, I also listed them explicitly and got 366. – RobPratt Mar 21 '20 at 2:17
• Let's suppose what if I don't want to resort to binomials. Does it exist a way to solve this by brute force or counting them using Pascal's logic?. – Chris Steinbeck Bell Mar 21 '20 at 2:21
• I have no idea how did the author got to $365$?. Could it be that is a typo or he's not accounting for something. – Chris Steinbeck Bell Mar 21 '20 at 2:21
• @ChrisSteinbeckBell: The Pascal logic for that goes this way: we are calculating the number of paths through the lower two dots. To get to the first dot you have to make $6$ moves, of which are $4$ are down and $2$ are right. You can do that in $6 \choose 4$ ways because you can choose any $4$ of the $6$ moves to be down and the others are right. Then to get to the next dot you have to move $3$ moves of which $1$ is down and to get to $B$ you again make $3$ moves of which $1$ is down. The choices are independent, so you multiply them. – Ross Millikan Mar 21 '20 at 3:11
Denote the points to be avoided by $$1$$, $$2$$, $$3$$, in the order Left, Top, Bottom.
Number of paths through $$1 = \binom{6}{2}\cdot \binom{6}{2}$$
Number of paths through $$2= \binom{5}{1}\cdot\binom{7}{2}$$
Number of paths though $$3= \binom{9}{4}\cdot \binom{3}{1}$$
Number of paths through $$1$$ and $$3= \binom{6}{2}\cdot \binom{3}{1}\cdot \binom{3}{1}$$
Number of paths through $$2$$ and $$3 = \binom{5}{1}\cdot \binom{3}{1}$$
Number of paths through $$1$$ and $$2 = 0$$.
Now we have to substract from the number of paths from $$A$$ to $$B = \binom{12}{6}$$ the first three amounts and add the next three amounts, which gets $$366$$ ( we use the inclusion-exclusion).
• Where did you used the inclusion-exclusion principle?. This part I'm not getting it. Is it where the part you mentioned subtracting A to B?. Why is it the number of paths through $1=\binom{6}{2} \cdot \binom{6}{2}$? Why is it that? – Chris Steinbeck Bell Mar 21 '20 at 2:57
• @Chris Steinbeck Bell: First, do you understand why the number of paths from $A$ to $B$ is $\binom{12}{6}$ ? There are 6 vertical steps and 6 horizontal steps. Out of $12$ steps, you have to choose which ones will be the horizontal ones, so $12$ choose $6$ ways. – orangeskid Mar 21 '20 at 3:01
• Probably this is the reason why I requested the inclusion of a drawing thus I could understand this better. I can see there are seven vertical lines and seven horizontal lines considering A and B. I don't know where do you get the 6 vertical steps and the 6 horizontal steps. why is it 6 ways from those 12?. I'm stuck there. – Chris Steinbeck Bell Mar 21 '20 at 3:14
• @Chris Steinbeck Bell: Check out this source: youtube.com/… also en.wikipedia.org/wiki/Lattice_path – orangeskid Mar 21 '20 at 3:24
I believe you are meant to always move either to the right or down along the grid. Without that restriction there would be infinitely many paths.
To get from $$A$$, which is $$(0,0)$$, to $$B$$, which is $$(6,6)$$, you need to take six steps down and six steps to the right. (I'm ignoring the restriction to avoid the three marked points for now.) It is true that there are seven horizontal and seven vertical lines in the figure, as you mention in a comment, but the steps are the segments between vertices, not the vertices themselves, which is why only six steps down are needed instead of seven.
One possible path from $$A$$ to $$B$$ is $$RRDDRDRRDDDR$$. Another is $$RDRDRDRDRDRD$$. A third is $$RRRRRRDDDDDD$$. In fact, any "word" consisting of six $$R$$s and six $$D$$s corresponds to a path. So the problem is reduced to counting words with six $$R$$s and six $$D$$s. Such a word is completely specified by stating where the $$R$$s are (or alternatively stating where the $$D$$s are). Any set of six elements chosen from $$1,2,3,4,5,6,7,8,9,10,11,12$$ is a set of possible $$R$$ positions. For the first possible path mentioned above, the set of $$R$$ positions is $$\{1,2,5,7,8,12\}$$. For the second it is $$\{1,3,5,7,9,11\}$$, and for the third it is $$\{1,2,3,4,5,6\}$$. There are $$\binom{12}{6}$$ ways of forming such a set.
Now that we know how many paths there are from $$A$$ to $$B$$, we want to subtract paths that are invalid in that they do pass through one of the marked points. There are, for example, $$\binom{5}{4}\binom{7}{2}$$ paths that pass through the marked point in the second row. The first binomial coefficient is the number of ways to get from $$A$$ to the marked point; the second is the number of ways to get from the marked point to $$B$$.
You can compute the number of paths passing through each of the other two marked points by similar reasoning. You will see, however, that some paths have been subtracted twice, so you will need to add these back. That is, you need to use the principle of inclusion-exclusion. One set of paths that needs to be added back is the set of paths that pass through both $$(2,4)$$ and $$(4,5)$$. They were subtracted once because they pass through $$(2,4)$$, and they were subtracted a second time because the pass through $$(4,5)$$. The number that needs to be added back to compensate for the double subtraction is $$\binom{6}{2}\binom{3}{2}\binom{3}{2}$$, which is the number of ways of going from $$(0,0)$$ to $$(2,4)$$, then from $$(2,4)$$ to $$(4,5)$$, and then from $$(4,5)$$ to $$(6,6)$$.
Added: Rob Pratt has given a beautiful solution using a Pascal's-triangle-like recurrence. I think it worth pointing out that Pascal's triangle is a table of binomial coefficients, so the binomial coefficients are still there in the background when you use that method. A formula for the result of applying the recurrence can be obtained by combining a sequence of arrays, each of which is a shifted Pascal's triangle multiplied by one or more binomial coefficients.
If no points are required to be avoided, the relevant portion of Pascal's triangle is $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 15 & 35 & 70 & 126 & 210\\ 1 & 6 & 21 & 56 & \color{red}{126} & 252 & 462\\ 1 & 7 & 28 & 84 & 210 & 462 & 924 \end{array}$$ where, if rows and columns are both labeled $$0$$ through $$6$$, the entry in row $$i$$, column $$j$$ is $$\binom{i+j}{i}$$. To eliminate paths that pass through the marked point in the second-to-last row, we need to subtract the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 126 & 126 & 126\\ 0 & 0 & 0 & 0 & 126 & 252 & 378 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{5+4}{5}\binom{i-5+j-4}{i-5}=126\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & \color{red}{15} & 35 & 70 & 126 & 210\\ 1 & 6 & 21 & 56 & 0 & 126 & 336\\ 1 & 7 & 28 & 84 & 84 & 210 & 546 \end{array}$$ Similarly, to eliminate paths that pass through the marked point in Row $$4$$, Column $$2$$, subtract the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 15 & 15 & 15 & 15 & 15\\ 0 & 0 & 15 & 30 & 45 & 60 & 75\\ 0 & 0 & 15 & 45 & 90 & 150 & 225 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{4+2}{4}\binom{i-4+j-2}{i-4}=15\binom{i-4+j-2}{i-4}$$ for $$i\ge4$$, $$j\ge2$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 0 & 20 & 55 & 111 & 195\\ 1 & 6 & 6 & 26 & \color{red}{-45} & 66 & 261\\ 1 & 7 & 13 & 39 & -6 & 60 & 321 \end{array}$$ At this point, paths that pass through both the marked point in Row $$5$$ and the one in Row $$4$$ have been subtracted twice, which explains the negative entries. To add these back, add the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 45 & 45 & 45\\ 0 & 0 & 0 & 0 & 45 & 90 & 135 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{4+2}{4}\binom{1+2}{1}\binom{i-5+j-4}{i-5}=45\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & \color{red}{5} & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 0 & 20 & 55 & 111 & 195\\ 1 & 6 & 6 & 26 & 0 & 111 & 306\\ 1 & 7 & 13 & 39 & 39 & 150 & 456 \end{array}$$ A similar subtraction followed by addition are needed to eliminate paths that pass through the marked point in Row $$1$$, Column $$4$$. Subtracting $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 5 & 5 & 5\\ 0 & 0 & 0 & 0 & 5 & 10 & 15\\ 0 & 0 & 0 & 0 & 5 & 15 & 30\\ 0 & 0 & 0 & 0 & 5 & 20 & 50\\ 0 & 0 & 0 & 0 & 5 & 25 & 75\\ 0 & 0 & 0 & 0 & 5 & 30 & 105 \end{array}$$ whose entries are $$\binom{1+4}{1}\binom{i-1+j-4}{i-1}=5\binom{i-1+j-4}{i-1}$$ for $$i\ge1$$, $$j\ge4$$, gives $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 0 & 1 & 2\\ 1 & 3 & 6 & 10 & 10 & 11 & 13\\ 1 & 4 & 10 & 20 & 30 & 41 & 54\\ 1 & 5 & 0 & 20 & 50 & 91 & 145\\ 1 & 6 & 6 & 26 & \color{red}{-5} & 86 & 231\\ 1 & 7 & 13 & 39 & 34 & 120 & 351 \end{array}$$ To eliminate double subtraction of paths the pass through both $$(1,4)$$ and $$(5,4)$$ add $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 5 & 5 & 5\\ 0 & 0 & 0 & 0 & 5 & 10 & 15 \end{array}$$ whose entries are $$\binom{1+4}{1}\binom{4+0}{4}\binom{i-5+j-4}{i-5}=5\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This gives the array in Rob Pratt's answer.
Assembling all these arrays to get a non-recursive formula for the entries in the final array is equivalent to the binomial coefficient method with inclusion-exclusion.
• I'm not familiar with the use of combinatorics, but I do understand how to calculate $\binom{12}{6}$ but I don't know how you got it that binomial. Perhaps does it exist another way which doesn't require this?. I would see it better if there is some justification visually of how to account for those ways?. And I'm thinking that the allowed paths are what you say. But there isn't an indication of that. – Chris Steinbeck Bell Mar 21 '20 at 1:50
• I have to confess that when I apply the method of my answer I don't manage to match any of the given alternatives, although I do come very close to one of them. I may, of course have made an arithmetic error somewhere. – Will Orrick Mar 21 '20 at 1:50
• Gee this doesn't give me much confidence though. – Chris Steinbeck Bell Mar 21 '20 at 1:51
• Let $R$ indicate a right move and $D$ a down move. One possible path is $RRDDRDRRDDDR$. In fact, any "word" consisting of six $R$s and six $D$s corresponds to a path. So the problem is reduced to counting words with six $R$s and six $D$s. This is where $\binom{12}{6}$ comes from. (You have to chose the positions of the six $R$s from twelve possible positions.) The number of paths that pass through the marked point in Row 2 is found by splitting the path into two subpaths at that point. – Will Orrick Mar 21 '20 at 1:53
• Well, I'm confident of the method, and have checked the arithmetic a few times. Are you 100% sure there's no typo in the answer choices? – Will Orrick Mar 21 '20 at 1:55
| 2021-05-08T17:01:30 |
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https://www.physicsforums.com/threads/mixed-quantifiers.964794/
|
# Mixed Quantifiers
#### ver_mathstats
1. Homework Statement
Give your own example of a statement with two different quantifiers which changes its meaning and truth value when the order of the quantifiers is exchanged.
2. Homework Equations
3. The Attempt at a Solution
(∀x∈ℝ)(∃x∈ℝ)(xy=0) is true but (∃x∈ℝ)(∀x∈ℝ)(xy=0) is false.
Is this correct?
Related Calculus and Beyond Homework Help News on Phys.org
#### fresh_42
Mentor
2018 Award
1. Homework Statement
Give your own example of a statement with two different quantifiers which changes its meaning and truth value when the order of the quantifiers is exchanged.
2. Homework Equations
3. The Attempt at a Solution
(∀x∈ℝ)(∃x∈ℝ)(xy=0) is true but (∃x∈ℝ)(∀x∈ℝ)(xy=0) is false.
Is this correct?
No. First of all, you have used the same variable $x$ in both quantifiers, and haven't quantified $y$ at all. Then you used a symmetric statement, $xy=0$. How should it depend on the ordering? $x=0$, resp. $y=0$ makes both statements true if chosen in the existence clause.
#### ver_mathstats
No. First of all, you have used the same variable $x$ in both quantifiers, and haven't quantified $y$ at all. Then you used a symmetric statement, $xy=0$. How should it depend on the ordering? $x=0$, resp. $y=0$ makes both statements true if chosen in the existence clause.
My apologies for using the same variable in both quantifiers.
Would this one make sense (∀x∈ℝ)(∃y∈ℝ)(x2+y=0) is true but (∃y∈ℝ)(∀x∈ℝ)(x2+y=0)?
#### fresh_42
Mentor
2018 Award
My apologies for using the same variable in both quantifiers.
Would this one make sense (∀x∈ℝ)(∃y∈ℝ)(x2+y=0) is true but (∃y∈ℝ)(∀x∈ℝ)(x2+y=0)?
This works.
I.) For all $x$ is a $y$, namely $y=-x^2$ such that ... is true.
II.) There is a real number $y$ such that for all $x$ .... is false: e.g. $y + 0^2 \neq y + 1^2$ no matter how $y$ is chosen.
Do you see the difference between the two statements? There is a neglect in the notation of the first statement. Do you see it? This isn't directed towards you, since the majority of people neglect this, too. But what would be a better statement I.)?
| 2019-11-12T11:25:06 |
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http://mathhelpforum.com/calculus/150484-can-2-functions-equal-continuous-range.html
|
# Math Help - Can 2 functions be equal for a continuous range?
1. ## Can 2 functions be equal for a continuous range?
I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?
Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).
2. Well, what about two functions that have equal asymptotes as they approach infinity?
3. Originally Posted by 1005
Well, what about two functions that have equal asymptotes as they approach infinity?
No; for example f(x)=1/x and g(x)=1/x^2.
4. there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
5. Originally Posted by xxp9
there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
Which functions are those?
6. Originally Posted by JvJ
I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?
Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
7. Originally Posted by Jose27
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.
8. Originally Posted by JvJ
This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.
It is $C^{\infty}$ and this can be proven by noting that $\lim_{x\rightarrow \infty} p(x)e^{-x}=0$ where $p(x)$ is any polynomial.
9. Originally Posted by Jose27
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
It is famous Cauchy function... $f^{(n)}(0)=0$ for every $n\in \mathbb{N}$
10. Which, by the way, shows that the Taylor's series of a function does NOT necessarily converge to that function! The Taylor's series, about x= 0, for this function. is just "0" and converges to 0 for all x.
Of course, this function is not "analytic" in any neighborhood of 0.
| 2016-07-23T20:13:32 |
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https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/6%3A_Functions/6.4%3A_Composition_of_Functions
|
# 6.4: Composition of Functions
PREVIEW ACTIVITY $$\PageIndex{1}$$: Constructing a New Function
Let $$A = \{a, b, c, d\}$$, $$B = \{p, q, r\}$$, and $$C = \{s, t, u, v\}$$. The arrow diagram in Figure 6.6 shows two functions: $$f: A \to B$$ and $$g: B \to C$$. Notice that if $$x \in A$$, then $$f(x) \in B$$. Since $$f(x) \in B$$, we can apply the function $$g$$ to $$f(x)$$, and we obtain $$g(f(x))$$, which is an element of $$C$$.
Using this process, determine $$g(f(a))$$, $$g(f(b))$$, $$g(f(c))$$, and $$g(f(d))$$. Then explain how we can use this information to define a function from $$A$$ to $$C$$.
Figure 6.6: Arrow Diagram Showing Two Functions
PREVIEW ACTIVITY $$\PageIndex{1}$$: Verbal Descriptions of Functions
The outputs of most real functions we have studied in previous mathematics courses have been determined by mathematical expressions. In many cases, it is possible to use these expressions to give step-by-step verbal descriptions of how to compute the outputs. For example, if
$$f: \mathbb{R} \to \mathbb{R}$$ is defined by $$f(x) = (3x + 2)^3$$,
we could describe how to compute the outputs as follows:
Step Verbal Description Symbolic Result
1 Choose an input. $$x$$
2 Multiply by 3. $$3x$$
3 Add 2. $$3x + 2$$
4 Cube the result. $$(3x + 3)^3$$
Complete step-by-step verbal descriptions for each of the following functions.
1. $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = \sqrt{3x^2 + 2}$$, for each $$x \in \mathbb{R}$$.
2. $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = \sin(3x^2 + 2)$$, for each $$x \in \mathbb{R}$$.
3. $$h: \mathbb{R} \to \mathbb{R}$$ by $$h(x) = e^{3x^2 + 2}$$, for each $$x \in \mathbb{R}$$.
4. $$G: \mathbb{R} \to \mathbb{R}$$ by $$G(x) = \ln(x^4 + 3)$$, for each $$x \in \mathbb{R}$$.
5. $$k: \mathbb{R} \to \mathbb{R}$$ by $$k(x) = \sqrt[3] {\dfrac{\sin(4x + 3)}{x^2 + 1}}$$, for each $$x \in \mathbb{R}$$.
## Composition of Functions
There are several ways to combine two existing functions to create a new function. For example, in calculus, we learned how to form the product and quotient of two functions and then how to use the product rule to determine the derivative of a product of two functions and the quotient rule to determine the derivative of the quotient of two functions. The chain rule in calculus was used to determine the derivative of the composition of two functions, and in this section, we will focus only on the composition of two functions. We will then consider some results about the compositions of injections and surjections.
The basic idea of function composition is that when possible, the output of a function $$f$$ is used as the input of a function $$g$$. This can be referred to as “$$f$$ followed by $$g$$” and is called the composition of $$f$$ and $$g$$. In previous mathematics courses, we used this idea to determine a formula for the composition of two real functions.
For example, if
$$f(x) = 3x^2 + 2$$ and $$g(x) = sin x$$
then we can compute $$g(f(x))$$ as follows:
$\begin{array} {rcl} {g(f(x))} &= & {g(3x^2 + 2)}\\ {} &= & {sin(3x^2 + 2).} \end{array}$
In this case, $$f(x)$$, the output of the function $$f$$, was used as the input for the function $$g$$. We now give the formal definition of the composition of two functions.
Definition: composite function
Let $$A$$, $$B$$, and $$C$$ be nonempty sets, and let $$f: A \to B$$ and $$g: B \to C$$ be functions. The composition of $$f$$ and $$g$$ is the function $$g \circ f: A \to C$$ defined by
$$(g \circ f)(x) = g(f(x))$$
for all $$x \in A$$. We often refer to the function $$g \circ f$$ as a composite function.
It is helpful to think of composite function $$g \circ f$$ as "$$f$$ followed by $$g$$". We then refer to $$f$$ as the inner function and $$g$$ as the outer function.
Composition and Arrow Diagrams
The concept of the composition of two functions can be illustrated with arrow diagrams when the domain and codomain of the functions are small, finite sets. Although the term “composition” was not used then, this was done in Preview Activity $$\PageIndex{1}$$, and another example is given here.
Let $$A = \{a, b, c, d\}$$, $$B = \{p, q, r\}$$, and $$C = \{s, t, u, v\}$$. The arrow diagram in Figure 6.7 shows two functions: $$f: A \to B$$ and $$g: B \to C$$.
If we follow the arrows from the set $$A$$ to the set $$C$$ , we will use the outputs of $$f$$ as inputs of $$g$$, and get the arrow diagram from $$A$$ to $$C$$ shown in Figure 6.8. This diagram represents the composition of $$f$$ followed by $$g$$.
Progress Check 6.17 (The Composition of Two Functions)
Let $$A = \{a, b, c, d\}$$ and $$B = \{1, 2, 3\}$$. Define the function $$f$$ and $$g$$ as follows:
$$f: A \to B$$ defined by $$f(a) = 2$$, $$f(b) = 3$$, $$f(c) = 1$$, and $$f(d) = 2$$.
$$g: A \to B$$ defined by $$g(1) = 3$$. $$g(2) = 1$$, and $$g(3) = 2$$.
Create arrow diagrams for the function $$f$$, $$g$$, $$g \circ f$$, and $$g \circ g$$.
Add texts here. Do not delete this text first.
## Decomposing Functions
We use the chain rule in calculus to find the derivative of a composite function. The first step in the process is to recognize a given function as a composite function. This can be done in many ways, but the work in Preview Activity $$\PageIndex{2}$$ can be used to decompose a function in a way that works well with the chain rule. The use of the terms “inner function” and “outer function” can also be helpful. The idea is that we use the last step in the process to represent the outer function, and the steps prior to that to represent the inner function. So for the function,
$$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = (3x + 2)^3$$,
the last step in the verbal description table was to cube the result. This means that we will use the function $$g$$ (the cubing function) as the outer function and will use the prior steps as the inner function. We will denote the inner function by $$h$$. So we let $$h: \mathbb{R} \to \mathbb{R}$$ by $$h(x) = 3x + 2$$ and $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = x^3$$. Then
$\begin{array} {rcl} {(g \circ h)(x)} &= & {g(h(x))} \\ {} &= & {g(3x + 2)} \\ {} &= & {(3x + 2)^3} \\ {} &= & {f(x).} \end{array}$
We see that $$g \circ h = f$$ and, hence, we have “decomposed” the function $$f$$. It should be noted that there are other ways to write the function $$f$$ as a composition of two functions, but the way just described is the one that works well with the chain rule. In this case, the chain rule gives
$\begin{array} {rcl} {f \prime (x)} &= & {(g \circ h)\prime (x)} \\ {} &= & {g \prime (h(x)) h \prime(x)} \\ {} &= & {3(h(x))^2 \cdot 3} \\ {} &= & {g(3x + 2)^2} \end{array}$
Progress Check 6.18 (Decomposing Functions
Write each of the following functions as the composition of two functions.
1. $$F: \mathbb{R} \to \mathbb{R}$$ by $$F(x) = (x^2 + 3)^3$$
2. $$G: \mathbb{R} \to \mathbb{R}$$ by $$G(x) = In(x^2 + 3)$$
3. $$f: \mathbb{Z} \to \mathbb{Z}$$ by $$f(x) = |x^2 - 3|$$
4. $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = cos(\dfrac{2x - 3}{x^2 + 1})$$
Add texts here. Do not delete this text first.
If $$f: A \to B$$ and $$g: B \to C$$, then we can form the composite function $$g \circ f: A \to C$$. In Section 6.3, we learned about injections and surjections. We now explore what type of function $$g \circ f$$ will be if the functions $$f$$ and $$g$$ are injections (or surjections).
Progress Check 6.19: Compositions of Injections and Surjections
Although other representations of functions can be used, it will be helpful to use arrow diagrams to represent the functions in this progress check. We will use the following sets:
$$A = \{a, b, c\}$$, $$B = \{p, q, r\}$$, $$C = \{u,v, w, x\}$$, and $$D = \{u, v\}$$.
1. Draw an arrow diagram for a function $$f: A \to B$$ that is an injection and an arrow diagram for a function $$g: B \to C$$ that is an injection. In this case, is the composite function $$g \circ f: A \to C$$ an injection? Explain.
2. Draw an arrow diagram for a function $$f: A \to B$$ that is a surjection and an arrow diagram for a function $$g: B \to D$$ that is a surjection. In this case, is the composite function $$g \circ f: A \to D$$ a surjection? Explain.
3. Draw an arrow diagram for a function $$f: A \to B$$ that is a bijection and an arrow diagram for a function $$g: B \to A$$ that is a bijection. In this case, is the composite function $$g \circ f: A \to A$$ bijection? Explain.
Add texts here. Do not delete this text first.
In Progress Check 6.19, we explored some properties of composite functions related to injections, surjections, and bijections. The following theorem contains results that these explorations were intended to illustrate. Some of the proofs will be included in the exercises.
Theorem 6.20.
Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$.
1. If $$f$$ and $$g$$ are both injections, then $$(g \circ f): A \to C$$ is an injection.
2. If $$f$$ and $$g$$ are both surjections, then $$(g \circ f): A \to C$$ is an surjection.
3. If $$f$$ and $$g$$ are both bijections, then $$(g \circ f): A \to C$$ is an bijection.
Proof
The proof of Part (1) is Exercise (6).
Part (3) is a direct consequence of the first two parts. We will discuss a process for constructing a proof of Part (2). Using the forward-backward process, we first look at the conclusion of the conditional statement in Part (2). The goal is to prove that $$g \circ f$$ is a surjection. Since $$(g \circ f): A \to C$$, this is equivalent to proving that
For all $$c \in C$$, there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$.
Since this statement in the backward process uses a universal quantifier, we will use the choose-an-element method and choose an arbitrary element $$c$$ in the set $$C$$. The goal now is to find an $$a \in A$$ such that $$(g \circ f)(a) = c$$.
Now we can look at the hypotheses. In particular, we are assuming that both $$f: A \to B$$ and $$g: B \to C$$ are surjections. Since we have chosen $$c \in C$$, and $$g: B \to C$$ is a surjection, we know that
there exists a $$b \in B$$ such that $$g(b) = c$$.
Now, $$b \in B$$ and $$f: A \to B$$ is a surjection. Hence
there exists an $$a \in A$$ such that $$f(a) = b$$.
If we now compute $$(g \circ f)(a)$$, we will see that
$$(g \circ f)(a) = g(f(a)) = g(b) = c$$.
We can now write the proof as follows:
Proof of Theorem 6.20, Part (2)
Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$ are both surjections. We will prove that $$g \circ f: A \to C$$ is a surjection.
Let $$c$$ be an arbitrary element of $$C$$. We will prove there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$. Since $$g: B \to C$$ is a surjection, we conclude that
there exists a $$b \in B$$ such that $$g(b) = c$$.
Now, $$b \in B$$ and $$f: A \to B$$ is a surjection. Hence
there exists an $$a \in A$$ such that $$f(a) = b$$.
We now see that
\begin{align*} {(g \circ f)(a)} &= & {g(f(a))} \\ {} &= & {g(b)} \\ {} &= & {c.} \end{align*}
We have now shown that for every $$c \in C$$, there exists an $$a \in A$$ such that $$(g \circ f)(a) = c$$, and this proves that $$g \circ f$$ is a surjection.
Theorem 6.20 shows us that if $$f$$ and $$g$$ are both special types of functions, then the composition of $$f$$ followed by $$g$$ is also that type of function.The next question is, “If the composition of $$f$$ followed by $$g$$ is an injection (or surjection), can we make any conclusions about $$f$$ or $$g$$?” A partial answer to this question is provided in Theorem 6.21. This theorem will be investigated and proved in the Explorations and Activities for this section. See Exercise (10).
Theorem 6.21
Let $$A$$, $$B$$, and $$C$$ be nonempty sets and assume that $$f: A \to B$$ and $$g: B \to C$$.
1. If $$g \circ f: A \to C$$ is an injection, then $$f: A \to B$$ is an injection.
2. If $$g \circ f: A \to C$$ is a surjection, then $$f: A \to B$$ is a surjection.
Exercise 6.4
1. In our definition of the composition of two functions, $$f$$ and $$g$$, we required that the domain of $$g$$ be equal to the codomain of $$f$$. However, it is sometimes possible to form the composite function $$g \circ f$$ even though dom($$g$$) $$\ne$$ codom($$f$$). For example, let
$\begin{array} {lcl} {f: \mathbb{R} \to \mathbb{R}} &text{ be defined by }& {f(x) = x^2 + 1\text{, and let}} \\ {g: \mathbb{R} - \{0\} \to \mathbb{R}} &text{ be defined by }& {g(x) = \dfrac{1}{x}.} \end{array}$
(a) Is it possible to determine $$(g \circ f) (x)$$ for all $$x \in \mathbb{R}$$? Explain.
(b) In general, let $$f: A \to T$$ and $$g: B \to C$$. Find a condition on the domain of $$g$$ (other than $$B = T$$) that results in a meaningful definition of the composite function $$g \circ f: A \to C$$.
2. Let $$h: \mathbb{R} \to \mathbb{R}$$ be defined $$h(x) = 3x + 2$$ and $$g: \mathbb{R} \to \mathbb{R}$$ be defined by $$g(x) = x^3$$. Determine formulas for the composite functions $$g \circ h$$ and $$h \circ g$$. Is the function $$g \circ h$$ equal to the function $$h \circ g$$? Explain. What does this tell you about the operation of composition of functions?
3. Following are formulas for certain real functions. Write each of these real functions as the composition of two functions. That is, decompose each of the functions.
(a) $$F(x) = cos(e^x)$$
(b) $$G(x) = e^{cos(x)}$$
(c) $$H(x) = \dfrac{1}{sin x}$$
(d) $$K(x) = cos(e^{-x^2})$$
4. The identity function on a set $$S$$, denoted by $$I_S$$, is defined as follows: $$I_S: S \to S$$ by $$I_s(x) = x$$ for each $$x \in S$$. Let $$f: A \to B$$.
(a) For each $$x \in A$$, determine $$(f \circ I_A)(x)$$ and use this to prove that $$f \circ I_A = f$$.
(b) Prove that $$I_B \circ f = f$$.
5. (a) Let $$f: \mathbb{R} \to \mathbb{R}$$ be defined by $$f(x) = x^2$$, let $$g: \mathbb{R} \to \mathbb{R}$$ be defined by $$g(x) = sin x$$, and let $$h: \mathbb{R} \to \mathbb{R}$$ be defined by $$h(x) = \sqrt[3]{x}$$.
Determine formulas for $$[(h \circ g) \circ f] (x)$$ and $$[h \circ (g \circ f)](x)$$.
Does this prove that $$(h \circ g) \circ f = h \circ (g \circ f)$$ for these particular functions? Explain.
(b) Now let $$A$$, $$B$$, and $$C$$ be sets and let $$f: A \to B$$, $$g: B \to C$$, and $$h: C \to D$$. Prove that $$(h \circ g) \circ f = h \circ (g \circ f)$$. That is, prove that function composition is an associative operation.
6. Prove Part (1) of Theorem 6.20.
Let $$A$$, $$B$$, and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$. If $$f$$ and $$g$$ are both injections, then $$g \circ f$$ is an injection.
7. For each of the following, give an example of functions $$f: A \to B$$ and $$g: B \to C$$ that satisfy the stated conditions, or explain why no such example exists.
(a) The function $$f$$ is a surjection, but the function $$g \circ f$$ is not a surjection.
(b) The function $$f$$ is an injection, but the function $$g \circ f$$ is not an injection.
(c) The function $$g$$ is a surjection, but the function $$g \circ f$$ is not a surjection.
(d) The function $$g$$ is an injection, but the function $$g \circ f$$ is not an injection.
(e) The function $$f$$ is not a surjection, but the function $$g \circ f$$ is a surjection.
(f) The function $$f$$ is not an injection, but the function $$g \circ f$$ is an injection.
(g) The function $$f$$ is not an injection, but the function $$g \circ f$$ is an injection.
(h) The function $$g$$ is not an injection, but the function $$g \circ f$$ is an injection.
8. Let $$A$$ be a nonempty set and let $$f: A \to A$$. For each $$n \in \mathbb{N}$$, define a funciton $$f^n: A \to A$$ recursively as follows: $$f^1 = f$$ and for each $$n \in \mathbb{N}$$, $$f^{n + 1} = f \circ f^n$$. For example, $$f^2 = f \circ f^1 = f \circ f$$ and $$f^3 = f \circ f^2 = f \circ (f \circ f)$$.
(a) Let $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = x + 1$$ for each $$x \in \mathbb{R}$$. For each $$n \in \mathbb{N}$$ and for each $$x \in \mathbb{R}$$, determine a formula for $$f^n(x)$$ and use induction to prove that your formula is correct.
(b) Let $$a, b \in \mathbb{R}$$ and let $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = ax + b$$ for each $$x \in \mathbb{R}$$. For each $$n \in \mathbb{N}$$ and for each $$x \in \mathbb{R}$$, determine a formula for $$f^n(x)$$ and use induction to prove that your formula is correct.
(c) Now let $$A$$ be a nonempty set and let $$f: A \to A$$. Use induction to prove that for each $$n \in \mathbb{N}$$, $$f^{n + 1} = f^n \circ f$$. (Note: You will need to use the result in Exercise (5).)
Explorations and Activities
9. Exploring Composite Functions. Let $$A$$, $$B$$, and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$. For this activity, it may be useful to draw your arrow diagrams in a triangular arrangement as follows:
It might be helpful to consider examples where the sets are small. Try constructing examples where the set $$A$$ has 2 elements, the set $$B$$ has 3 elements, and the set $$C$$ has 2 elements.
(a) Is it possible to construct an example where $$g \circ f$$ is an injection, $$f$$ is an injection, but $$g$$ is not an injection? Either construct such an example or explain why it is not possible.
(b) Is it possible to construct an example where $$g \circ f$$ is an injection, $$g$$ is an injection, but $$f$$ is not an injection? Either construct such an example or explain why it is not possible.
(c) Is it possible to construct an example where $$g \circ f$$ is a surjection, $$f$$ is a surjection, but $$g$$ is not a surjection? Either construct such an example or explain why it is not possible.
(d) Is it possible to construct an example where $$g \circ f$$ is a surjection, $$g$$ is a surjection, but $$f$$ is not a surjection? Either construct such an example or explain why it is not possible.
10. The Proof of Theorem 6.21. Use the ideas from Exercise (9) to prove Theorem 6.21. Let $$A$$, $$B$$ and $$C$$ be nonempty sets and let $$f: A \to B$$ and $$g: B \to C$$.
(a) If $$g \circ f: A \to C$$ is an injection, then $$f: A \to B$$ is an injection.
(b) If $$g \circ f: A \to C$$ is a surjection, then $$g: B \to C$$ is a surjection.
Hint: For part (a), start by asking, “What do we have to do to prove that $$f$$ is an injection? ” Start with a similar question for part (b).
| 2021-05-07T19:25:34 |
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https://scicomp.stackexchange.com/questions/20694/mathematical-programming-formulation-of-triangle-intersection/20717
|
# Mathematical programming formulation of triangle intersection
Given variables $a_1$, $b_1$, $c_1$ and $a_2$, $b_2$, $c_2$ representing the vertices of two plane triangles, how might one specify the requirement for the two triangles to intersect as an objective in a mathematical programming solver such as CPLEX?
• Are the $2$ triangles given? Is this a decision problem? Or an optimization problem? – Rodrigo de Azevedo Mar 28 '17 at 20:41
• In the wider context, this is a decision problem: there are some additional constraints on the triangles. – NietzscheanAI Mar 30 '17 at 13:53
The first component you need is a function $orient(p_1,p_2,p_3)$ that computes the orientation between three point, i.e. that gives a positive number if there is a left turn between vectors $\overrightarrow{p_1 p_2}$ and $\overrightarrow{p_1 p_3}$ and a negative number otherwise (and zero if the three points are aligned):
$$Orient(p_1,p_2,p_3) = det(\overrightarrow{p_1 p_2}, \overrightarrow{p_1 p_3}) = (x_2-x_1)(y_3-y_1) - (y_2-y_1)(x_3-x_1)$$
Then you can determine whether two segments $[p_1 p_2]$ and $[q_1 q_2]$ have an intersection:
$$IsectSeg(p_1,p_2,q_1,q_2) = Orient(p_1,q_1,q_2) \times Orient(p_2,q_1,q_2) \le 0 \mbox{ and } Orient(q_1,p_1,p_2) \times Orient(q_2,p_1,p_2) \le 0$$
In English: there is an intersection between $[p_1 p_2]$ and $[q_1 q_2]$ if $[p_1 p_2]$ straddles the supporting line $(q_1 q_2)$ and $[q_1 q_2]$ straddles the supporting line $(p_1 p_2)$.
You can also determine whether a point $p$ belongs to a triangle $q_1,q_2,q_3$ using:
$$InsideTri(p,q_1,q_2,q_3) = (Orient(p,q_1,q_2) \times Orient(p,q_2,q_3) \ge 0) \mbox{ and } (Orient(p,q_2,q_3) \times Orient(p,q_3,q_1) \ge 0)$$
In English: $p$ is inside triangle $(q_1,q_2,q_3)$ if the orientation relative to the three edges $q_1 q_2$, $q_2 q_3$ and $q_3 q_1$ is the same.
Putting everything together, there is an intersection between both triangles if:
$$IsectTri(p_1, p_2, p_3, q_1, q_2, q_3) = \exists i | InsideTri(p_i, q_1, q_2, q_3) \mbox{ or } \exists i | InsideTri(q_i, p_1, p_2, p_3) \mbox{ or } \exists i,j,k,l | IsectSeg(p_i, p_j, q_k, q_l)$$
In English: the two triangles have an intersection if one vertex of one triangle is inside the other triangle, or if there exists an intersection between the edges of the triangles. It is necessary to test for edges intersection (think for instance about two intersecting triangles that form a six-branches star that has no triangle vertex inside the other triangle)
Translated into a constraint, this gives a formula with inequality constraints combined with OR and AND operators (each of the 6 $Inside$ constraint yields two terms, and each of the 9 $IsectSeg$ constraint yields two terms as well). I am unsure of how to translate this into a way that can be understood by a constrained optimization software but there is probably a standard way of doing that.
There might be also a shorter / more elegant formulation (but I did not find it).
Note1: if you know in advance that the triangles are all oriented consistently (all clockwise, or all anticlockwise), then InsideTri() can be made simpler (just test the sign of the $Orient()$ relative to the three edge).
Note2: the asked question required a "mathematical programming" answer that could be expressed as a constraint in an optimization program. Now if what you want is simply determining whether two 2D triangles intersect (in a standard programming language that has execution flow and tests), there is a significantly faster approach that avoids some tests, see the 2D section in [1].
• many thanks for such a comprehensive reply. With respect to your final point about consistent orientation': since the variables I want to provide for the MP solver are the vertices $p_i$, AFAIK, I can't guarantee any ordering on them. – NietzscheanAI Sep 12 '15 at 10:02
Let plane triangle $\mathcal T_i$ have vertices $\mathrm a_i, \mathrm b_i, \mathrm c_i \in \mathbb R^2$. Let $\Delta_2$ be the standard $2$-simplex.
If two plane triangles do intersect, i.e., $\mathcal T_1 \cap \mathcal T_2 \neq \emptyset$, then there exist $\eta_1, \eta_2 \in \Delta_2$ such that
$$\begin{bmatrix} | & | & |\\ \mathrm a_1 & \mathrm b_1 & \mathrm c_1\\ | & | & |\end{bmatrix} \eta_1 = \begin{bmatrix} | & | & |\\ \mathrm a_2 & \mathrm b_2 & \mathrm c_2\\ | & | & |\end{bmatrix} \eta_2$$
Thus, the triangle intersection problem can be reduced to linear programming. Choosing an arbitrary objective function, say, the zero function, we have a linear program in $(\eta_1, \eta_2)$
$$\begin{array}{ll} \text{minimize} & \mathrm 0_3^T \eta_1 + \mathrm 0_3^T \eta_2 \\ \text{subject to} & \begin{bmatrix} | & | & |\\ \mathrm a_1 & \mathrm b_1 & \mathrm c_1\\ | & | & |\end{bmatrix} \eta_1 - \begin{bmatrix} | & | & |\\ \mathrm a_2 & \mathrm b_2 & \mathrm c_2\\ | & | & |\end{bmatrix} \eta_2 = \mathrm 0_2\\ & 1_3^T \eta_1 = 1\\ & 1_3^T \eta_2 = 1\\ & \eta_1, \eta_2 \geq 0_3\end{array}$$
If the linear program is infeasible, then the intersection of the two triangles is empty.
• Can you clarify what you mean by the | symbols in the matrices? – NietzscheanAI Sep 30 '16 at 15:04
• @NietzscheanAI The |'s merely emphasize that $\mathrm a_i, \mathrm b_i, \mathrm c_i$ are column vectors, not scalars. – Rodrigo de Azevedo Oct 1 '16 at 2:00
• While elegant and interesting, I am unsure it answers the initial question because I do not think that "the problem being feasible" can be turned into a constraint for CPLEX, is there a way of doing that ? (I'd be interested !) – BrunoLevy Mar 28 '17 at 9:29
• @BrunoLevy Can't CPLEX solve linear programs? If a linear program is infeasible, can't CPLEX find it out? I never used CPLEX. I know that MATLAB's linprog returns exitflag=-2 if infeasible. – Rodrigo de Azevedo Mar 28 '17 at 12:07
• @Rodrigo de Azevedo Certainly it can, but I think that the question was: how can I optimize an objective function F(a1,b1,c1,a2,b2,c2) subject to constraint C: there is an intersection between triangles (a1,b1,c1) and (a2,b2,c2) (instead of answering the question "is there an intersection ?"). Can your formulation do that ? (it would be nice, because it's more elegant !) – BrunoLevy Mar 28 '17 at 20:06
For each triangle edge, add a linear constraint corresponding to the equation of the line containing the edge such that the other point is on the correct side.
For example, if the line containing the edge $a_1b_1$ is given by $L(x,y)=0$, then add the constraint $L(c_1)L(x,y) \ge 0$.
$L(x,y)$ is given by $((x,y)-a_1) \times (b_1-a_1)$, where $\times$ is the 2-dimensional vector product.
• But I think that there are several non-trivial combination of relative "correct sides", for instance one triangle can be completely inside the other, or there can be one vertex of one triangle inside the other one. There is even a configuration where no triangle vertex is inside the other one (two intersecting triangles forming a 6-branches star). – BrunoLevy Sep 10 '15 at 21:30
• @BrunoLevy - if it's clear to you how the above should be modified in the light of your comment, would you be kind enough to add it as an answer? – NietzscheanAI Sep 12 '15 at 6:59
• @lhf - I believe there are a number of possible interpretations of 2D vector product' (e.g. stackoverflow.com/questions/243945/…). Could you kindly be explicit about the definition? – NietzscheanAI Sep 12 '15 at 7:32
• @user217281728, I have entered an answer (it is not very elegant, but I did not find a shorter formula) – BrunoLevy Sep 12 '15 at 9:23
| 2020-09-18T21:31:47 |
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https://math.stackexchange.com/questions/3623969/a-weird-matrix-property
|
# A weird matrix property
I encountered the following weird matrix property.
Consider any general matrix $$M_{n\times n}$$ with the property that the sum of each column vanishes, that is
\begin{align} \sum^n_{j} M_{ji} =0 \end{align}
Denoting
• $$M_{(1)}$$ : the matrix obtained from $$M$$ by removing the first column and row,
• $$M_{(2)}$$ : the matrix obtained from $$M$$ by removing the second column and row,
• $$M_{(1,2)}$$ : the matrix obtained from $$M$$ by removing the first and second column and row,
• $$u_{(k)}=(1,\dots,1)$$ : the $$(k)$$-row vector with all elements being $$1$$,
• $$C_{(n-1)\times (n-2)}=\begin{pmatrix} 0 \dots 0 \\ 1_{(n-2)\times (n-2)} \end{pmatrix}$$, where $$1_{(n-2)\times (n-2)}$$ is the identity matrix
Define $$p_1$$ and $$p_2$$ as $$p_1 = u_{(n-1)} \cdot\big(M_{(1)}\big)^{-1}\cdot\begin{pmatrix} 1 \\ 0 \\ \vdots\\0 \end{pmatrix}_{ (n-1)\times1}, \quad p_2 = u_{(n-1)} \cdot\big(M_{(2)}\big)^{-1}\cdot\begin{pmatrix} 1 \\ 0 \\ \vdots\\0 \end{pmatrix}_{ (n-1)\times1}.$$
Prove that all the elements of the row vector $$u_{(n-1)} \cdot \left(p_2\big(M_{(1)}\big)^{-1}+p_1\big(M_{(2)}\big)^{-1}\right)\cdot C - (p_1+p_2)u_{(n-2)}\cdot \big(M_{(1,2)}\big)^{-1}$$ are identical.
This property comes from some intuition of the problem that I have been playing with. I have also tested it by evaluating it with a large set of matrix $$M$$ satisfying the first requirement.
(I thank user1551 for spotting an important typo, corrected now!)
I have tried writing the inverse using minors but does not seem to help as it is not easy to implement the requirement that $$\sum_{j} M_{ji}=0$$. Any comment/suggestion is greatly appreciated. Answers are of course the best! Thank you so much!
• Your first condition is equivalent to saying that all columns have sum $0$. Apr 13 '20 at 19:27
• Should we interpret $M_{(1)}^{-1}$ as $(M_{(1)})^{-1}$ or $(M^{-1})_{(1)}$? Apr 13 '20 at 19:31
• sorry for the confusion. It should be the first interpretation, corrected! Apr 13 '20 at 19:45
• $\left(\big(\bar{M}_{(1)}\big)^{-1}+\big(\bar{M}_{(2)}\big)^{-1}\right)\cdot C$ is not square but $\big(\bar{M}_{(1,2)}\big)^{-1}$ is square? Apr 13 '20 at 20:02
• @BallBoy Thanks for pointing that out, corrected. Apr 13 '20 at 20:16
I show below that the "identical elements" are all equal to $$p_1p_2$$ (which is confirmed by the example in the now-deleted answer). Let us put
$$D=\bar{M}_{(2)}^{-1}=(d_{ij})_{1\leq i,j \leq n-1}, E=\bar{M}_{(1)}^{-1}=(e_{ij})_{1\leq i,j \leq n-1}, F=\bar{M}_{(1,2)}^{-1}=(f_{ij})_{2\leq i,j \leq n-1}. \tag{1}$$ (notice the ranges of indices. The convention I choose is perhaps not the most logical but I find it the most convenient).
Then both $$D$$ and $$E$$ have the property that their inverse has $$F^{-1}$$ in its lower rightmost corner. Using the Schur complement formula, we deduce that $$D$$ and $$E$$ are of the form
$$\begin{array}{lcl} D&=&\left( \begin{array}{c|c} d & R_D \\ \hline C_D & \frac{1}{d}C_DR_D+F \end{array} \right),\\ E&=&\left( \begin{array}{c|c} e & R_E \\ \hline C_E & \frac{1}{e}C_ER_E+F \end{array} \right) \end{array} \tag{2}$$
And we also have closed forms for their inverses :
$$\begin{array}{lcl} D^{-1} &=& \left( \begin{array}{c|c} \frac{1}{d}(1+R_DF^{-1}C_D) & -\frac{1}{d}R_DF^{-1} \\ \hline -\frac{1}{d}F^{-1}C_D & F^{-1} \end{array} \right), \\ E^{-1} &=& \left( \begin{array}{c|c} \frac{1}{e}(1+R_EF^{-1}C_E) & -\frac{1}{e}R_EF^{-1} \\ \hline -\frac{1}{e}F^{-1}C_E & F^{-1} \end{array} \right) \end{array} \tag{3}$$
We can then rewrite the initial matrix $$M$$ :
$$M=\left( \begin{array}{c|c|c} \frac{1}{d}(1+R_DF^{-1}C_D) & m_{12} & -\frac{1}{d}R_DF^{-1} \\ \hline m_{21} & \frac{1}{e}(1+R_EF^{-1}C_E) & -\frac{1}{e}R_EF^{-1} \\ \hline -\frac{1}{d}F^{-1}C_D & -\frac{1}{e}F^{-1}C_E & F^{-1} \end{array} \right) \tag{4}$$
We can now interpret the hypothesis than the columns of $$M$$ have zero sum. The first two columns are not interesting since $$m_{12}$$ and $$m_{21}$$ can be arbitrary. But the other columns tell us that $$(-\frac{1}{d}R_D-\frac{1}{e}R_E+u_{(n-2)})F^{-1}=0$$ ; and since $$F^{-1}$$ is invertible, $$\frac{1}{d}R_D+\frac{1}{e}R_E =u_{n-2}$$, or
$$\frac{d_{1,j}}{d}+\frac{e_{1,j}}{e} = 1 \ \ (2 \leq j\leq n)\tag{5}$$
We deduce from (2) that
$$p_1=e+s_E, p_2=d+s_D \tag{6}$$ where $$s_E$$ (or $$s_D$$) denotes the sum of all numbers in column $$C_E$$ ($$C_D$$). and
$$(p_1\bar{M}_{(2)}^{-1}+p_2\bar{M}_{(1)}^{-1})C= p_1\left( \begin{array}{c} R_D \\ \hline \frac{1}{d}C_DR_D+F \end{array} \right)+ p_2\left( \begin{array}{c} R_E \\ \hline \frac{1}{e}C_ER_E+F \end{array} \right) \tag{7}$$
So the row vector $$A=u_{(n-1)}(p_1\bar{M}_{(2)}^{-1}+p_2\bar{M}_{(1)}^{-1})C$$ can be written $$A=(a_2,\ldots,a_{n})$$ with
$$a_j=p_1\bigg(d_{1,j}+\frac{d_{1,j}}{d}s_D+\sum_{k=2}^{n}F_{k,j}\bigg) +p_2\bigg(e_{1,j}+\frac{e_{1,j}}{e}s_E+\sum_{k=2}^{n}F_{k,j}\bigg) \tag{8}$$
Also, the row vector $$B=(p_1+p_2)u_{(n-2)}\cdot \big(\bar{M}_{(1,2)}\big)^{-1}= (p_1+p_2)u_{(n-2)}F$$ can be written $$B=(b_2,\ldots,b_{n})$$ with
$$b_j=(p_1+p_2)\sum_{k=2}^{n} F_{kj} \tag{9}$$
Next, if the put $$G=A-B=(g_2,\ldots,g_{n})$$ we have
\begin{align} g_j &= a_j-b_j \\[6pt] &= p_1\bigg(d_{1,j}+\frac{d_{1,j}}{d}s_D\bigg) +p_2\bigg(e_{1,j}+\frac{e_{1,j}}{e}s_E\bigg) \\[6pt] &= \frac{d_{1,j}}{d} \bigg(d+s_D\bigg)p_1 +\frac{e_{1,j}}{e}\bigg(e+s_E\bigg)p_2 \\[6pt] &= \bigg(\frac{d_{1,j}}{d}+\frac{e_{1,j}}{e}\bigg) p_1p_2 \ \textrm{by} \ (6)\\[6pt] &= p_1p_2 \ \textrm{by} \ (5) \end{align}
So $$g_j$$ is independent of $$j$$ as needed, which finishes the proof.
• Wow! That's super nice! I also found that they are equal to $p_1 p_2$ in the "experiments" but I did not have a clear physical reason for that. I have to go through your answer first and will reward the bounty after that. Apr 19 '20 at 21:25
| 2021-11-30T10:35:26 |
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https://forum.math.toronto.edu/index.php?PHPSESSID=fbif03b4mf8h9b717kqs27rqu0&topic=285.msg1360
|
### Author Topic: Day Section's Quiz Problem 2 (Read 2701 times)
#### Sabrina (Man) Luo
• Jr. Member
• Posts: 9
• Karma: 4
##### Day Section's Quiz Problem 2
« on: April 03, 2013, 09:12:44 AM »
(2) Find an equation of the form H(x,y)=c satisfied by solutions to
\begin{equation*}
\left\{\begin{aligned}
&dx/dt=2x^2y-3x^2-4y,\\
&dy/dt=-2xy^2+6xy
\end{aligned}
\right.\end{equation*}
« Last Edit: April 03, 2013, 10:16:56 AM by Victor Ivrii »
#### Alexander Jankowski
• Full Member
• Posts: 23
• Karma: 19
##### Re: Day Section's Quiz Problem 2
« Reply #1 on: April 03, 2013, 10:38:23 AM »
To determine $H(x,y)$, we proceed as follows:
\begin{equation*} \frac{dy}{dx} = \frac{-2xy^2 + 6xy}{2x^2y - 3x^2 - 4y} \Longrightarrow (2xy^2 - 6xy)dx + (2x^2y - 3x^2 - 4y)dy = 0. \end{equation*}
Let $M(x,y) = 2xy^2 - 6xy$ and $N(x,y) = 2x^2y - 3x^2 - 4y$. It turns out that
\begin{equation*} M_y(x,y) = 4xy - 6x = N_x(x,y). \end{equation*}
Thus, the differential equation is exact. Suppose that there is a function $\psi(x,y)$ that satisfies the equations $\psi_x(x,y) = M$ and $\psi_y(x,y) = N$. We have
\begin{equation*} \psi_x(x,y) = 2xy^2 - 6xy \Longrightarrow \psi(x,y) = x^2y^2 - 3x^2y + g(y). \end{equation*}
Then, we try to determine $g$:
\begin{equation*} \psi_y(x,y) = 2x^2y - 3x^2 + g'(y) = 2x^2y - 3x^2 - 4y \Longleftrightarrow g'(y) = -4y \Longrightarrow g(y) = -2y^2. \end{equation*}
We conclude that
\begin{equation*} \psi(x,y) = x^2y^2 - 3x^2y - 2y^2 = c. \end{equation*}
To confirm this, I have attached (1) a stream plot of the system and (2) a contour plot of $H = \psi$.
« Last Edit: April 06, 2013, 10:52:55 AM by Alexander Jankowski »
#### Iven Poon
• Newbie
• Posts: 4
• Karma: 2
##### Re: Day Section's Quiz Problem 2
« Reply #2 on: April 03, 2013, 02:17:15 PM »
I already written my solution down on Monday, and I planned to post on Thursday when Professor Ivrii uploaded the questions...
Anyway, I would still like to share how I did this question.
#### Victor Ivrii
On the plot of Alexander you see unusual equilibrium point (0,0). The reason of its strange appearance (neither center nor saddle) is because it is degenerated critical point of $H(x,y)$ and a lot of funny things can happen then.
| 2021-11-28T12:15:51 |
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https://math.stackexchange.com/questions/3148414/proving-that-the-intersection-of-any-finite-number-of-members-of-t-belongs-to-t
|
# Proving that the intersection of any finite number of members of T belongs to T when (X,T) is a topological space.
I know this is part of the definition of a topology. But the book I am using which is 'Topology without Tears' ask to prove this using mathematical induction.I hate to ask questions but proofs I have seen for this question in this site were different than mine.
Here it is...
The statement "Intersection of any two sets in $$T$$ belongs to $$T$$ " is given by the definition.
Suppose $$T_1,T_2,T_3,...$$ are members of $$T$$ and $$\bigcap_{n=1}^{k} T_n\in T \label{a}\tag{1}$$ is true. If $$\,\bigcap_{n=1}^{k+1} \,T_n\in T$$ is also true, the proposition can be proven by mathematical induction.
Since the element in the the left side of (\ref{a}) is a also set, let $$M=\bigcap_{n=1}^{k} T_n$$.
$$\bigcap_{n=1}^{k+1} T_n = \bigcap_{n=1}^{k} T_n \cap T_{k+1}$$ $$\bigcap_{n=1}^{k+1} T_n = M \cap T_{k+1}$$
Since $$M\in T$$ by (1) and $$T_{k+1}\in T$$, $$\,\bigcap_{n=1}^{k+1} T_n \in T$$ by definition. Thus intersection of any finite number of sets in $$T$$ belongs to $$T$$.
I want to know every mistake of this proof or ways to improve it.
• I'm not sure what other proofs you've seen, but I think this is the standard proof to show that a finite intersection of elements in a topology $\tau$ is again in $\tau$. – Clayton Mar 14 at 19:30
• I see no mistakes. Your proof is fine. – Mark Mar 14 at 19:30
• The proof is perfect. The reason you don't see this exact argument elsewhere is because everyone has a different style in writing and not everyone has the same base of facts at their disposal. – Alberto Takase Mar 14 at 19:38
• If $0$ counts as finite (which I think it should), then you should mention that the intersection of $0$ subsets of $X$ is $X$, which is also in $T$. – Andreas Blass Mar 15 at 1:42
• “I hate to ask questions but ...” Why? Asking questions and communicating with other mathematicians is what makes math what it is! – Santana Afton Mar 15 at 1:53
The proof by induction is fine. Renaming to $$M$$ is not really necessary of course (I don't usually introduce "unnecessary" notation), just writing
$$\bigcap_{i=1}^{n+1} T_i = \left(\bigcap_{i=1}^{n} T_i\right) \cap T_{n+1}$$
(so using braces) is enough to make your point, I'd say.
It's actually a very common pattern: e.g. groups are closed under products of two members, but writing finite powers etc. is routinely done: having a set closed under an operation allows all finite "products' (or sums/intersections, etc. etc.) You do the proof once, and henceforth it's implicit.
It sometimes makes for slightly easier to read proofs, only having to check the $$n=2$$ case.
A minor nitpick: you don't mention the base case (which is $$n=1$$), which is almost too trivial to mention: $$T_1$$ in the topology gives us the $$1$$-ary intersection $$\bigcap_{i=1}^1 T_i = T_1$$ in the topology too, tautologically. The $$n=2$$ case could also serve as the base case, depending on the formulation of the statement to be proved. But even the $$0$$-ary intersection (i.e. $$X$$) is in the topology....
| 2019-05-26T23:18:08 |
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https://math.stackexchange.com/questions/493264/finding-the-inverse-of-a-matrix-by-elementary-transformations
|
# Finding the inverse of a matrix by elementary transformations.
While using the elementary transformation method to find the inverse of a matrix, our goal is to convert the given matrix into an identity matrix.
We can use three transformations:-
1) Multiplying a row by a constant
2) Adding a multiple of another row
3) Swapping two rows
The thing is, I can't seem to figure out what to do to achieve that identity matrix. There are so many steps which I can start off with, but how do I know which one to do? I think of one step to get a certain position to a $1$ or a $0$, and then get a new matrix. Now again there are so many options, it's boggling.
Is there some specific procedure to be followed? Like, first convert the top row into: \begin{bmatrix} 1&0&0\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} Then do the second row and then the third?
What do I start off with? I hope I've made my question clear enough.
Thanks to @Brian M. Scott.
$P.S:$ Does anyone have any other methods? Brian's works perfectly, but it's always great to know more than one method. :)
First get a non-zero entry in the upper lefthand corner by swapping to rows if necessary. If that entry is $a_{11}\ne 0$, multiply the first row by $a_{11}^{-1}$ to get a $1$ in the upper lefthand corner. Now use operation (2) to get $0$’s in the rest of the first column.
Now get a non-zero entry in the $a_{22}$ position, the second entry in the second row, by swapping the second row with one of the lower rows if necessary, and multiply the (possibly new) second row by $a_{22}^{-1}$ to get a $1$ in the $a_{22}$ position. Then use operation (2) to get $0$’s in the rest of the second column; notice that since $a_{21}$, the first element in the second row, is $0$, this will not affect anything in the first column.
At this point your matrix looks like this:
$$\begin{bmatrix} 1&0&a_{13}&\dots&a_{1n}\\ 0&1&a_{23}&\dots&a_{2n}\\ 0&0&a_{33}&\dots&a_{3n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&a_{n3}&\dots&a_{nn} \end{bmatrix}$$
Continue in the same fashion: get a non-zero entry in the $a_{33}$ position by swapping row $3$ with a lower row if necessary, multiply row $3$ by a suitable constant to make $a_{33}=1$, and use operation (2) to $0$ out the rest of the third column.
If at any point the necessary operation is impossible, your original matrix was not invertible.
• Thanks so much! Finally a procedure; I know what exactly I have to do. – mikhailcazi Sep 14 '13 at 9:11
• @mikhailcazi: You’re very welcome. – Brian M. Scott Sep 14 '13 at 9:12
• It might be important to know why this method works. It's a rather simple and basic reason connected with what "elementary operations by rows" actually are. – DonAntonio Sep 14 '13 at 9:34
• @mikhailcazi: You might take a look at this question and its answers; I believe that one of the answers or comments also has a link to an explanation. – Brian M. Scott Sep 14 '13 at 9:46
• Every row elementary operation is a multiplication by an elementary matrix $\,E_i\;$ from the left, @mikhailcazi, so following the method given by Brian you get $$E_k\cdot E_{k-1}\cdot\ldots\cdot E_2 A=I$$ which means $\,E_kE_{k-1}\cdot\ldots\cdot E_1=A^{-1}\;$ ... – DonAntonio Sep 14 '13 at 9:53
I give you an example of a similar approach to Brian's one. Start from the matrix $$A=\left[ \begin{array}{ccc}1&2&0\\2&0&-1\\2&3&-1\end{array}\right]$$ First step is to write the identity matrix on right side of the matrix A. $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\2&0&-1&0&1&0\\2&3&-1&0&0&1\end{array}\right]\begin{array}{c}a\\b\\c\end{array}$$ Now, you have to start to apply elementary transformations to the whole matrix, since on the left side the identity matrix appears. $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\0&3&0&0&-1&1\\2&3&-1&0&0&1\end{array}\right]\begin{array}{c}a \rightarrow a\\-b+c \rightarrow b\\c\rightarrow c\end{array}$$ $$\left[ \begin{array}{cccccc}1&2&0&1&0&0\\0&1&0&0&-1/3&1/3\\0&1&1&2&0&-1\end{array}\right]\begin{array}{c}a \rightarrow a\\b/3 \rightarrow b\\-c+2a\rightarrow c\end{array}$$ $$\left[ \begin{array}{cccccc}1&0&0&1&2/3&-2/3\\0&1&0&0&-1/3&1/3\\0&0&1&2&1/3&-4/3\end{array}\right]\begin{array}{c}a-2b \rightarrow a\\b \rightarrow b\\c-b\rightarrow c\end{array}$$
At this point, the matrix on the right side is the inverse of $A$! Namely: $$A^{-1}=\left[ \begin{array}{ccc}1&2/3&-2/3\\0&-1/3&1/3\\2&1/3&-4/3\end{array}\right]$$
• This doesn't exactly answer my question. As Brian says, one way to do reduce it is making the element $a_{ii}$ 1, then making all elements of column $i$ 0. Then move onto $a_{i+1, i+1}$. I was asking for a procedure which could be followed. What procedure have you followed in your answer? – mikhailcazi Sep 14 '13 at 11:39
• well, I managed to get all $a_{ii} = 1$ on the left hand side. I achieved this by using elementary operation reported on the rightmost side of the whole matrix. This is a method to obtain on the right hand side the inverse of a matrix. – the_candyman Sep 14 '13 at 17:11
Consider a square matrix A of order 3. Now by property of matrices if |A|$\neq$ 0 only then it has an inverse i.e. A should be a non singular matrix. Assuming A to be a non singular matrix, by property of matrices
$$(A)(A^{-1})=I$$ where I is identity matrix. Now substitute the value of A and I. Use transformations as suggested above to get identity matrix in place of A and simultaneously applying those elementary transformation to the Identity matrix on RHS would give you the value of $A^{-1}$ .
So basically the equation changes to $$(I)(A^{-1})=A^{-1}$$
• In case you need me to add the steps of transformation do comment. – just_curious Nov 11 '16 at 4:29
| 2020-08-09T20:27:49 |
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https://math.stackexchange.com/questions/2342921/how-to-find-coefficients-of-parabola-based-on-graph
|
# How to find coefficients of parabola based on graph?
I know that I have the following options:
Vietes formula, vertex (x0,yo), the quadratic equation to get x1,x2. Somehow I don't see how I would get the coefficients A and B. The C coefficient is C=3 as it can be read from the graph easily.
How should I tackle the problem?
Just to note that this graph is the function derivative of a previous graph, in other words
$${f(x)'}=Ax^2+Bx-3$$
The problems says that I have to find $$f(x)$$ I know that I have to integrate and therefore easily get the equation but I need the coefficients.
Image of the derivative graph:
• you have $$f(2)=5$$ and $$f(-2)=1$$ – Dr. Sonnhard Graubner Jul 1 '17 at 13:32
• Why is $c=-3$. ? – Archis Welankar Jul 1 '17 at 13:33
• @Dr.SonnhardGraubner $f'$ – Shuri2060 Jul 1 '17 at 13:33
• It seems I missed that, tnx now that s easy – eugene_sunic Jul 1 '17 at 13:34
• the plot says $g$. is $g=f'$ or $g=f$ or $g=f''$ or $g= ???$ – Dando18 Jul 1 '17 at 13:34
You have the equation with two unknowns $f'(x)=Ax^2 + Bx - 3$ and from the graph you know $(2,5)$ and $(-2,1)$ are points on $f$. So you have a system of equations,
\begin{align*} 5 &= A(2)^2 + B(2) - 3 \\ 1 &= A(-2)^2 + B(-2) - 3 \end{align*}
So $A=\frac 3 2$, $B=1$ meaning $f'(x)=\frac 3 2 x^2 + x - 3$ and $f(x) = \frac 1 2 x^3 + \frac 1 2 x^2 - 3x + c$
you will get $$4a+2b=8$$ and $$4a-2b=4$$ form here we get $$a=\frac{3}{2}$$ and $$b=1$$
You have $f'(-2)=1$, $f'(0) = -3$, $f'(2)= 5$.
So $$f'(-2)=1=A(-2)^2+2B(-2)-3$$ $$f'(2)= 5=A(2)^2+B(2)-3$$
and you can solve.
• Simple and perfect – eugene_sunic Jul 1 '17 at 13:43
| 2019-09-23T15:22:56 |
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https://math.stackexchange.com/questions/481527/slice-of-pizza-with-no-crust/481702
|
# Slice of pizza with no crust
The following question came up at a conference and a solution took a while to find.
Puzzle. Find a way of cutting a pizza into finitely many congruent pieces such that at least one piece of pizza has no crust on it.
We can make this more concrete,
Let $$D$$ be the unit disc in the plane $$\mathbb{R}^2$$. Find a finite set of subsets of $$D$$, $$\mathcal{A}=\{A_i\subset D\}_{i=0}^n$$, such that
• for each $$i$$, $$A_i$$ is simply connected and equal to the closure of its interior
• for each $$i, j$$ with $$i\neq j$$, $$\operatorname{int}(A_i)\cap \operatorname{int}(A_j)=\emptyset$$
• $$\bigcup\mathcal{A}=D$$
• for each $$i,j$$, $$A_i=t(A_j)$$ where $$t$$ is a (possibly orientation reversing) rigid transformation of the plane
• for some $$i$$, $$\lambda(A_i\cap\partial D)=0$$ where $$\lambda$$ is the Lebesgue measure on the boundary circle.
Note that we require only that $$\lambda(A_i\cap\partial D)=0$$ and not that $$A_i\cap\partial D=\emptyset$$. I know of a solution but am interested in what kinds of solutions other people can find, and so I welcome the attempt.
• @MarkBennet: I think Daniel is considering the unit circle to be the boundary of the unit disk. – robjohn Sep 1 '13 at 18:59
• Related on mathoverflow: mathoverflow.net/questions/17313/… – Daniel R Sep 1 '13 at 19:48
• I'd imagine the solution involves some multiple of pie. – Joel B Sep 1 '13 at 22:33
• I wonder how the mathematical formulation is "more concrete" than the physical description of the problem! – Ari Brodsky Sep 2 '13 at 3:03
• Every answer thus far involves the pieces touching the crust ( 1 point ). Is there an answer that doesn't require touching the crust? – krikara Sep 3 '13 at 7:41
Here is another with 12 pieces, but all pieces have the same orientation:
$\hspace{32mm}$
Using this idea, the pizza can be divided into $6n$ equal pieces with the same orientation for any $n$. However, to have some pieces with no crust, we need $n\gt1$. Above is $n=2$, here is $n=3$:
$\hspace{32mm}$
To cut a pizza like this, a blade shaped like, and as long as one sixth of, the circumference of the pizza would be most useful, since all of the cuts are this size and shape.
Here is Mathematica code that will generate these sliced pizzas for any $n$:
Pizza[n_] :=
Module[{g, arcs = {Thickness[1.3/400], Circle[{0, 0}, 1]}},
For[i = 0, i < 6, For[j = 0, j < n, AppendTo[arcs,
Rotate[Rotate[Circle[{-1, 0}, 1, {0, Pi/3}],
j Pi/3/n, {-1/2, Sqrt[3]/2}], i Pi/3, {0, 0}]]; ++j]; ++i];
Show[Graphics[arcs], ImageSize -> 400,
PlotRange -> 1.01 {{-1, 1}, {-1, 1}}]]
Motivation
I thought of the construction of a regular hexagon: you draw a circle with a compass, and then mark arcs on the circle whose chords are the radius of the circle. Due to the properties of equilateral triangles, each arc is exactly $1/6$ of the circumference of the circle, and the chords of those arcs form a regular hexagon. At each vertex of the hexagon, the compass will span to the next vertex (by construction) and to the center of the circle (again, by construction).
Connecting each vertex to the center with arcs centered at the previous vertex, we get the circle tiled by $6$ curvy triangles with congruent sides; two convex sides and one concave side. The centers of the convex sides are the opposite vertices of the curvy triangle. Since the chords of the curved sides have a length $1$ radius, we can trace out the interior convex sides with a congruent arc rotating about the opposite vertex.
$\hspace{32mm}$
Since we can sweep out these $6$ triangles with these congruent arcs, we can split up the curvy triangles into any number of congruent pieces with these arcs.
• Is it hard to find a general form for all possible solutions? – Pratyush Sarkar Sep 2 '13 at 3:18
• @PratyushSarkar: it's likely to be harder to prove that you have all possible solutions than to find what turn out to be all possible solutions. Are there solutions other than mine and robjohn's (and their reflections and rotations?). – Robert Israel Sep 2 '13 at 3:38
• Curious, what made you think of this solution? – hadsed Sep 5 '13 at 16:06
• @hadsed: I have added a section on the motivation. – robjohn Sep 5 '13 at 19:49
• It moves! It moves! – Pedro Tamaroff Sep 5 '13 at 21:55
Here is one solution in $12$ pieces.
• Six of the pieces here only have crust on one corner, but didn't the problem say absolutely none? – Trejkaz Sep 2 '13 at 5:29
• @Trejkaz: The requirement is only that the measure of crust included in some piece $A_i$, i.e. $\lambda(A_i \cap \partial D)$, be 0. A single point of idealized zero-thickness crust is OK since that has measure 0. – The_Sympathizer Sep 2 '13 at 5:43
• Someone has to ask this: What if we really require $A_i\cap \partial D = \emptyset$ for at least one $i$, that is at least one piece, including its boundary, must lie entirely in the interior of the disk? – Jeppe Stig Nielsen Sep 2 '13 at 21:48
• @JeppeStigNielsen: that appears to be an open problem – Robert Israel Sep 2 '13 at 22:22
• @mike4ty4 It's fair enough to have ideals, but someone who is thinking more about the pizza than the semantics will probably consider a 0-width crust to be less than ideal. Illustrating the danger of stating mathematical problems in real-world terms which people might relate with. :D – Trejkaz Sep 3 '13 at 6:45
If this violates the parameters in a clear way, consider this a teaching opportunity. Would this count:
• How are these pieces "congruent"? – nbubis Sep 3 '13 at 22:37
• This does violate the conditions, as we need all pieces to be congruent. That is, every piece must be identical (shape, size, etc.) except for orientation. Good try though--it was what popped in my head first before I noticed the "congruent" requirement. – apnorton Sep 3 '13 at 22:38
• So the center violates it then? Would that mean at least one slice needs to meet the crust? – Anthony Sep 3 '13 at 22:45
• @Anthony - Congruent means that you can overlay the shapes one on top of the other. rings of different sizes do not have this property. – nbubis Sep 5 '13 at 19:52
• Anthony, you're confusing congruence with similarity. – dfeuer Sep 9 '13 at 16:20
| 2020-09-26T09:37:54 |
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https://math.stackexchange.com/questions/2310599/find-the-shortest-distance-between-two-lines-in-3d
|
# Find the shortest distance between two lines in 3D [duplicate]
Question :
The line $l_1$ passes through the points $( 0 , 0 , 10 )$ and $( 7 ,0 ,0 )$ and the line $l_2$ passes through the points $( 4, 6 ,0 )$ and $( 3 , 3 ,1 )$. Find the shortest distance between $l_1$ and $l_2$ .
I found that \begin{align}l_1 : r &= ( 7 , 0 ,0 ) + a ( 7 , 0 , -10 )\\ l_2 : r &= ( 3 , 3 ,1 ) + b ( -1 , -3 ,1 )\end{align} for $a,b\in\Bbb R$
Help me solve this question , thank you so much !
• here you can find it geomalgorithms.com/a07-_distance.html Jun 5 '17 at 12:03
• Sketch: Step 1: Find a common normal to both lines using the cross product. Step 2: Find a pair of parallel planes containing the lines. Step 3: Find a point on one of the planes. Step 4: Find the distance between the point and the other plane. Jun 5 '17 at 12:06
• Jun 5 '17 at 12:14
• Hi @Dr.Sonnhard , First of all , thank you for helping me . Secondly ,I do not understand this formula , Can you explain me about this one ? Thank you Jun 5 '17 at 13:36
You know what a point in each line looks like. You know how to find the distance between points. That gives you a function of $a,b$, which you must minimize.
In our case, a point on $l_1$ is $(7+7a,0,-10a)$, and $l_2$ is $(3-b,3-3b,1+b)$.
So the distance between these points is : $$\sqrt{(7+7a-3+b)^2 + (3b-3)^2 + (-10a-b+1)^2}$$
Now, it's enough to minimize $(7+7a-3+b)^2 + (3b-3)^2 + (-10a-b+1)^2$ with respect to $a$ and $b$. If you know how to do this, then that will give you values of $a$ and $b$, for which the distance is the least.
If you do not, then tell me.
EDIT:
Partial differentiating with respect to $a$ and $b$ and equating to zero, we get: $$14(4+7a+b) -20(-10a-b+1) = 0 ; 2(4+7a+b) + 6(3b-3) -2(-10a-b+1) = 0$$
Solve these equations to get the minimizing $a,b$, and then find the distance.
• I do not know how to minimize it . :( Help me @@@ , I tried to annihilate the brackets . Hoever , it is not useful Jun 5 '17 at 12:09
• If you square it, you can use partial derivatives to minimize. Jun 5 '17 at 12:11
• @MichaelBurr yes, thank you for adding to this question. Jun 5 '17 at 12:12
• There is an error in the distance formula above. (The (-10a-b+1) should be (-10a-b-1). This has consequences for the following differentiation calculation. The simultaneous equations become:298a+34b=-76 and 34a+22b=8 giving a =-9/25 and b=23/25 and d via the adjusted formula is 2.939 Jan 22 at 12:54
• @twa14 Thank you, I will make the required changes. Jan 22 at 12:55
The geometric approach is as follows:
• Step 1: Find a vector perpendicular to both lines. In your case, the direction of the two lines is $\langle 7,0,-10\rangle$ and $\langle -1,-3,1\rangle$. A vector perpendicular to both lines is given by the cross product $$\langle 7,0,-10\rangle\times\langle -1,-3,1\rangle=\langle -30,3,-21\rangle$$
• Step 2: Find parallel planes containing each of the lines. Since a point on the first line is $(7,0,0)$ and a point on the second line is $(3,3,1)$, a plane containing the first line is $$-30(x-7)+3(y-0)-21(z-0)=0$$ and a plane containing the second line is $$-30(x-3)+3(y-3)-21(z-1)=0.$$
We observe that the distance between the planes is the same as the distance between the lines. Therefore, we need to find the distance between the planes. The distance between two planes is the same as the distance between a point on one plane and the other plane. In particular, we can find the distance between $(7,0,0)$ and the plane $-30(x-3)+3(y-3)-21(z-1)=0$.
• Step 3: To find the distance between two a point and a plane, we need a vector from the plane to the point and then project that onto the normal of the plane. This is a fairly standard application of projections. Since we know that $(3,3,1)$ is on the plane and $(7,0,0)$ is the point of interest, we look at the vector from $(3,3,1)$ to $(7,0,0)$ which is $\langle 4,-3,-1\rangle$. The projection onto the normal is $$\|\operatorname{proj}_{\langle -30,3,-21\rangle}\langle4,-3,-1\rangle\|=\frac{|\langle -30,3,-21\rangle\cdot\langle4,-3,-1\rangle|}{\|\langle -30,3,-21\rangle\|}=\frac{108}{\sqrt{1350}}=\frac{36}{\sqrt{150}}.$$ This is the distance between the planes, and hence, the lines.
Analytically:
Let the lines be defined by the vectors $pq$ and $rs$. The squared distance between any to points can be expressed as
$$d^2=(pr+a\ pq+b\ rs)^2.$$
You minimize it by differentiating on $a$ abd $b$, giving the system
$$\begin{cases}pq\cdot(pr+a\ pq+b\ rs)=0,\\rs\cdot(pr+a\ pq+b\ rs)=0.\end{cases}$$
This expresses that $pr+a\ pq+b\ rs$ is orthogonal to both $pq$ and $rs$, hence it is parallel to $pq\times rs$.
If we define the unit vector $u:=\dfrac{pq\times rs}{\|pq\times rs\|}$ we have:
$$pr+a\ pq+b\ rs=\lambda u.$$
Now we can eliminate $a$ and $b$ by taking the dot product with $u$ so that
$$pr\cdot u=\lambda.$$
Finally,
$$d=\sqrt{(pr+a\ pq+b\ rs)^2}=|\lambda|=|pr\cdot u|=\frac{|(pr,pq,rs)|}{\|pq\times rs\|}.$$
Geometrically:
Consider the common perpendicular between the two lines, and rotate space so that it becomes parallel to $oz$. Then the two lines are included in planes parallel to $oxy$ and the shortest distance is the distance between these planes, which is also the distance between the projections of the lines.
| 2021-10-22T03:22:48 |
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https://www.physicsforums.com/threads/vector-problem.861290/
|
# Vector problem
Tags:
1. Mar 9, 2016
### gruba
1. The problem statement, all variables and given/known data
Given points of a triangle: $A(4,1,-2),B(2,0,0),C(-2,3,-5)$. Line $p$ contains point $B$, is orthogonal to $\overline{AC}$, and is coplanar with $ABC$. Intersection of $p$ and $\overline{AC}$ is the point $B_1$.
Find vector $\overrightarrow{B_1B}$.
2. Relevant equations
-Vector projection
- Dot product
-Magnitude of a vector
3. The attempt at a solution
$$proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|^2}\cdot \overrightarrow{AC}$$
$$\overrightarrow{AB}=[-2,-1,2],\overrightarrow{AC}=[-6,2,-3],|\overrightarrow{AC}|=7$$
$$\overrightarrow{AB}\cdot \overrightarrow{AC}=4$$
$$\Rightarrow proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\left[-\frac{24}{49},\frac{8}{49},-\frac{12}{49}\right]$$
From $\overrightarrow{AB_1}$ we can find the point $B_1\Rightarrow B_1=\left(\frac{172}{49},\frac{8}{49},-\frac{-110}{49}\right)$ $$\Rightarrow \overrightarrow{B_1B}=\left[-\frac{74}{49},-\frac{57}{49},\frac{110}{49}\right]$$
Is this correct?
Last edited: Mar 9, 2016
2. Mar 9, 2016
### geoffrey159
The reasoning is good, and the answer is correct if a scalar product evaluates to 0
3. Mar 9, 2016
### SammyS
Staff Emeritus
Notice that once you have, $\ \overrightarrow{AB_1}\$ and $\ \overrightarrow{AB}\$, you can get $\ \overrightarrow{B_1 B}\$ from $\ \overrightarrow{B_1 B}=\overrightarrow{B_1 A}+\overrightarrow{AB}\$
| 2017-10-24T00:37:57 |
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http://math.stackexchange.com/questions/173201/showing-that-the-last-digit-of-a-and-a13-are-the-same
|
# Showing that the last digit of $a$ and $a^{13}$ are the same
For $a \in \mathbb N$, show that the last digit of $a$ and $a^{13}$ are the same.
For example: $2^{13} = 8,192$
$7^{13} = 96,889,010,407$
-
Did you want to say: For $a\in\mathbb{N}$, show that the last digit of $a$ and $a^{13}$ are the same? Because the $13$th root of some numbers might not even have a "last digit" (i.e. be irrational). – Jesko Hüttenhain Jul 20 '12 at 9:28
Please state the question in the body of the post. – Did Jul 20 '12 at 9:48
To rephrase the question, you want to show that $a^{13}\equiv a\pmod{10}$, or both $a^{13}\equiv a\pmod2$ and $a^{13}\equiv a\pmod5$. We can do this through the use of Fermat's Little Theorem.
Let's do the case for 5. If $a\equiv 0\pmod5$, then $a^{13}\equiv0^{13}\equiv0\equiv a\pmod5$. Otherwise, we have $a^4\equiv1\pmod5$. So we have $a^{13}=(a^4)^3a\equiv1^3\times a\equiv a\pmod5$.
The case for 2 is similar if you need to prove that an odd number to an integer power is odd and an even number to an integer power is even.
-
Step 1: show that the last digit of $n^{13}$ depends only on the last digit of $n$.
Step 2: check that it's true for $n=0,1,\dots,9$.
-
How does this help? Is there a hidden hint? – draks ... Jul 20 '12 at 12:58
@draks, no, nothing is hidden. You show that all that matters is the last digit of $n$, and then you look at all 10 possibilities for the last digit of $n$. – Gerry Myerson Jul 21 '12 at 5:31
If the last digits of a and $a^{n+1}$ where n>1 are same, then 10 |( $a^{n+1}-a)=a(a^n-1)$
<=> 2|$a(a^n-1)$ and 5|$a(a^n-1)$
2 always divides a(a-1) which divides $a(a^n-1)$.
By Fermat's little theorem, 5|$(a^5-a)$ for all integer a.
$(a^5-a) = a(a^4-1)$ divides $a(a^{4k}-1)$ where k is any non-negative integer.
So,the last digits of a and $a^{4k+1}$ will be same. (Here k=3)
-
The last digit of $a^2$ in the set $\{1,2,3,4,5,0\}$ is $\{1, 4, 9, 6, 5, 0\}$ (due to redundancy, I omit $6\le a\le 9$). Then for $a^4$ we get $\{1,6,1,6,5,0\}$, the same for $a^8,a^{12}$ and so forth...
Now multiply $a^4$ once more to get $$\{1,6,1,6,5,0\}\cdot\{1,2,3,4,5,10\}= \{1\cdot 1\; ,6\cdot2\; ,3\cdot1\; ,4\cdot 6\; ,5\cdot 5\; ,0\cdot 10 \}_{\bmod 10} =\{1,2,3,4,5,0\}$$ (think of $\cdot$ as Hadamard product) so you already get your last digit back for $a^5$.
-
Hint: Show that $\varphi(10)=4$ (where $\varphi$ is the Euler totient function) and apply Euler's theorem.
-
OK, but Euler as usually presented applies only to bases prime to the modulus. – Gerry Myerson Jul 20 '12 at 9:30
Note that working mod 5, $(2n)^4 \equiv 1$ (for n not divisible by 5) so that $(2n)^{12} \equiv 1 \mod 5$ which sorts out the even numbers ($2n \times(5m+1) = 10nm + 2n$). You don't need any heavy machinery to deal with odd or even multiples of 5. – Mark Bennet Jul 20 '12 at 10:12
Hint $\$ Modular arithmetic using little Fermat-Euler is easiest. Alternatively, directly, note
$$\rm (a+1)^5 - (a+1)\ =\ a^5-a\ +\ \color{#C00}{10}\, \frac{a(a+1)}2\, (a^2+a+1)$$
Hence, by induction $\rm\:\color{#C00}{10}\:|\:a^5-a\,\$ so $\rm\:10\:|\:(a^5 - a)\, (a^8 + a^4 + 1)\, =\, a^{13}-a.\ \$ QED
Remark $\$ More generally we have the following Fermat-Euler-Carmichael
Theorem $\$ For naturals $\rm\: k,m>1$
$\rm\qquad m\ |\ a^k-a\$ for all $\rm\:a\in\mathbb N\ \iff\ m\:$ is squarefree and prime $\rm\: p\:|\:m\: \Rightarrow\: p\!-\!1\ |\ k\!-\!1$
Thus for $\rm\: m = 2\cdot 5,\,$ this yields $\rm\: 10\:|\:a^k-a,\ all\ a\iff 2\!-\!1,5\!-\!1\:|\:k\!-\!1\iff 4\:|\:k\!-\!1$
For related results see some of my prior posts.
-
Claim: For all integers $a$ we have $10|a^5-a$.
This follows from Fermat Little Theorem, but can also be proven elementary.
Proof of the claim
$$a^5-a=a(a-1)(a+1)(a^2+1) \,.$$
Now, one of $a$ or $a+1$ is even thus $a^5-a$ is divisible by $2$.
Also
$$a^5-a=a(a-1)(a+1)(a^2-4+5)=a(a-1)(a+1)(a^2-4)+5a(a-1)(a+1)$$ $$a^5-a=(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)$$
The product of 5 consecutive integers is always divisible by 5, hence so is $a^5-a$.
This proves the claim.
Proof of your exercise $$a^{13}-a=a^{13}-a^9+a^9-a^5+a^5-a=(a^5-a)(a^8+a^4+1) \,.$$
-
| 2014-04-20T01:48:41 |
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http://mathhelpforum.com/algebra/49016-fractions-surds.html
|
Thread: Fractions in surds
1. Fractions in surds
Hi,
Hope you guys can help... the question is as follows.
How do I represent:
√1/2 + √1/4 + √1/8
as....
a + b √c
I think you can take it to:
√1/ √2 + √1/ √4 + √1/8..... but does that actually help me?!
The steps of what you did would be greatly appreciated!
ajkerr
2. Originally Posted by ajkerr
Hi,
Hope you guys can help... the question is as follows.
How do I represent:
√1/2 + √1/4 + √1/8
as....
a + b √c
I think you can take it to:
√1/ √2 + √1/ √4 + √1/8..... but does that actually help me?!
The steps of what you did would be greatly appreciated!
ajkerr
it does help. simplify each term as much as possible, then find the sum. if there is something without radical signs, leave it by itself and sum the things with radicals
3. But surely I have to have a common denominator to add them at the end and I wouldn't if I simplified them as I would get √2, 2 and then 2 √2?!
Uh?!
4. Originally Posted by ajkerr
Hi,
Hope you guys can help... the question is as follows.
How do I represent:
√1/2 + √1/4 + √1/8
as....
a + b √c
I think you can take it to:
√1/ √2 + √1/ √4 + √1/8..... but does that actually help me?!
The steps of what you did would be greatly appreciated!
ajkerr
$\sqrt{\frac{1}{2}}=\frac{1}{\sqrt {2}}$
$=\frac{1}{\sqrt {2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
Now you try to simplify further.
for adding, make the denominator same of each term.
5. Originally Posted by Shyam
$\sqrt{\frac{1}{2}}=\frac{1}{\sqrt {2}}$
$=\frac{1}{\sqrt {2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
Now you try to simplify further.
soo.....
then √1/4 = 1/ √4
and then..... 1/ √4 x √4/ √4 = √4/4
then √1/8 = 1/ √8
and then 1/ √8 x √8/ √8 = √8/8
leaving me with...
√2 /2 + √4 /4+ √8 /8
denominator to...
4√2 / 8 + 2√4 / 8 + 8√8 to get (4√2+2√4+8√8) / 8 -- is that right?
then what?
6. Originally Posted by ajkerr
soo.....
then √1/4 = 1/ √4
and then..... 1/ √4 x √4/ √4 = √4/4
then √1/8 = 1/ √8
and then 1/ √8 x √8/ √8 = √8/8
leaving me with...
√2 /2 + √4 /4+ √8 /8
and then do I times them up to get a common denominator?
you realize $\sqrt{4} = 2$, right?
so you have $\frac 12 + \frac {\sqrt{2}}2 + \frac {\sqrt{8}}8$
now add the last two fractions
7. Originally Posted by Jhevon
you realize $\sqrt{4} = 2$
so you have $\frac 12 + \frac {\sqrt{2}}2 + \frac {\sqrt{8}}8$
now add the last two fractions
Doh! -- I knew that really.... err! Ok thanks
So
1/2 + √2/2 + √8/8
to make....
1/2 + 4 √2/8 + √8/8 = 1/2 + 3/4 √2
So the answer should be...
1/2 + 3/4 √2
I think.....
8. Originally Posted by ajkerr
Doh! -- I knew that really.... err! Ok thanks
So
1/2 + √2/2 + √8/8
to make....
1/2 + 4 √2/8 + √8/8 = 1/2 + 3/4 √2
So the answer should be...
1/2 + (3/4) √2
I think.....
yes
9. "Ahh finally" they all say "he's got it!"
Thanks so much... one of those questions that really bugs you!
Thanks again
Ajk
10. Originally Posted by ajkerr
"Ahh finally" they all say "he's got it!"
Thanks so much... one of those questions that really bugs you!
Thanks again
Ajk
yeah, you have those sometimes. but it wasn't so bad, was it?
11. $\sqrt{8}$ can and should be simplified further.
$\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{4}} + \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{8}}$
$= \frac{\sqrt{2}}{2} + \frac{1}{2} + \frac{\sqrt{8}}{8}$
$= \frac{1+\sqrt{2}}{2} + \frac{2 \sqrt{2}}{8}$
$= \frac{2 + 2\sqrt{2}}{4} + \frac{\sqrt{2}}{4}$
$= \frac{2 + 3\sqrt{2}}{4}$.
12. Originally Posted by Prove It
$\sqrt{8}$ can and should be simplified further.
$\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{4}} + \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{8}}$
$= \frac{\sqrt{2}}{2} + \frac{1}{2} + \frac{\sqrt{8}}{8}$
$= \frac{1+\sqrt{2}}{2} + \frac{2 \sqrt{2}}{8}$
$= \frac{2 + 2\sqrt{2}}{4} + \frac{\sqrt{2}}{4}$
$= \frac{2 + 3\sqrt{2}}{4}$.
that is what the poster got. he just wrote it in the form $a + b \sqrt{c}$ as he was directed
| 2017-09-21T14:32:34 |
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https://math.stackexchange.com/questions/2171068/probability-how-to-solve-a-probability-question-that-has-3-variables-a-b-and/2171100
|
# Probability: How to solve a probability question that has 3 variables, A, B, and C?
I am trying to solve the following question;
There are 3 news papers in a town, A, B, and C. 20% of the population read A, 16% read B, and 14% read C.
8% read A and B
5% read A and C
4% read B and C
Determine the probability that a person reads none of the news papers.
I've realised I need to calculate what percentage of the population reads news papers.
To do this I've done the following;
A = 20 - 8 - 5 = 7
B = 16 - 8 - 4 = 4
C = 14 - 5 - 4 = 5
Population that reads News papers = 7 + 4 + 5 = 16
Therefore, 84 / 100 = probability a person doesn't read any news papers.
However, my answer is wrong. What is the best way to approach this problem?
• Make a Venn diagram and all will be clear.
– Arby
Mar 4 '17 at 4:23
• Are you familiar with inclusion/exclusion? Mar 4 '17 at 4:26
In these type of questions you should approach by set theory and Venn diagram.
The percentage of people who read any of the newspapers is given by "$A$ or $B$ or $C$."
$A \cup B \cup C = A+B+C-(A \cap B)-(A \cap C)-(B \cap C)+(A \cap B \cap C)$ $\implies A \cup B \cup C = (20+16+14-8-5-4+2)\%=35\%$
Therefore the probability that a person reads none of the newspapers $= 65/100 = 0.65$.
$P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$
$P(A\cup B\cup C) = 20 + 16 + 14 - 8 - 5 - 4 + 2$
$P(A\cup B\cup C) = 35$
Person reads no newspaper $= 100 - 35 = 65$
Hope you can proceed further.
• @Ross Millikan thank u.
– Amar
Mar 4 '17 at 5:02
It should be clear that you can't have $84\%$ that don't read any papers because $20\%$ read A and that is already too many. Your variable A is intended to be (you should define this) the number that read A and not either B or C. When you did $20-8-5$ you subtracted the people who read all three papers twice, once for reading A and B (the $8$) and once for reading A and C (the $5$), so you need to add them back in once. That means the number that read only A is $20-8-5+2=9$. When you compute the population that reads any paper, you need to add your A,B,C (correctly calculated) but also all the people that read more than one paper. You should draw a Venn diagram to help the computation. The point of the exercise is the inclusion-exclusion principle.
$$A \cup B \cup C = A+B+C-(A \cap B)-(A \cap C)-(B \cap C)+(A \cap B \cap C)$$ $$\implies A \cup B \cup C = (20+16+14-8-5-4+2)\%=35\%$$ Therefore, required percent $= 65%$
| 2022-01-21T06:51:08 |
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http://archiv.siofok.hu/3gdkp/how-to-calculate-lower-limit-and-upper-limit-in-statistics-1b25f8
|
If this were the upper bound for a one-sided 95% CI, you would put the full 5% in the upper tail of t(df=17) and use 1.740 instead of 2.110. Find the confidential limits for mean. In statistics: Statistical process control …chart, referred to as the upper control limit, and the lower horizontal line, referred to as the lower control limit, are chosen so that when the process is in control there will be a high probability that the value of a sample mean will fall between the two control limits.… Any labels that are smaller than 2.5 inches or larger than 2.687 inches are unacceptable. Examples of upper and lower specification limits LSL = 2.5 USL = 2.687. Boolean, with default FALSE meaning to “clip” x (if real) by setting elements of x with x>high to high; if TRUE, set such elements to NA. Mean = 11; StdDev= 3.97; Upper Limit = 23.20; and Lower Limit = -0.64. 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Normalized OPSpecs Calculator; Quality Control Grid Calculator; Control Limit Calculator; Reportable Range Calculator: Quantifying Errors; Reportable Range Calculator: Recording Results; Dispersion Calculator and Critical Number of Test Samples 4. Because of that, upper and lower limits can be applied in both the y- and x-directions via the uplims, lolims, xuplims, and xlolims parameters, respectively. Calculate the probability without upper limit. The lower specification is 2.500 inches and the upper specification is 2.687 inches. assume the data follow a statistical distribution – typically normal (or can be normalized). + 3 sigma = Upper Control Limit (UCL) - 3 sigma = Lower Control Limit … LF = Q1 - 1.5 * IQR. 20.6 is the upper limit and 20.4 is the lower limit. METHOD FOR CALCULATION OF PER CENT WITHIN LIMITS 1. CI = 45 ± 1.75 * (52 / √31) These parameters can … Details. Thanks! As an example, take the data set: 2, 3, 5, 5, 7. Second calculate sigma. For the frequency distribution of weights of 36 students, the LCL and UCL of the first class interval are 44 kgs. You do not specify a lower value. =MIN(data) =MAX(data) In OpenOffice the MIN and MAX function can take a list of semi-colon separated numbers or a range of cells in a spreadsheet. This formula calculates the lower limit which is fixed for all weekly observations the $sign achieves that in this formula. In this confidence limits calculator enter the percentage of confidence limit level, which ranges from 90 % to 99 %, sample size, mean and standard deviation to know the lower and upper confidence limits. The minimum value is known as the lower class limit (LCL) and the maximum value is known as the upper class limit (UCL). Even a small shift or behavior in the process can be found using the ucl, lcl formula. Find Limit of Sums on the TI 89; What is a Limit? limits: Lower and upper bound of the array to consider, values less than the lower limit or greater than the upper limit will be ignored. Where CPK … LSL = 80. Third, calculate the sigma lines. Upper Limit is the upper limit of the confidence interval. lower. Lower Limit is the lower limit of the confidence interval. The quality analysts set a lower … If the class intervals of a variable are defined by whole numbers, to find the upper limit we add .5 to the highest value specified by the category, and to find the lower limit … Subtract the mean from each data point and square the result. In our example, this would be 5.8+(3_1.8) = 11.3. Find the lower and upper control limits using the control limit formula. Find the upper and lower cl using the control limit formula. The remaining class limits and classes are obtained by adding the interval i.e. Upper Limit can be calculated with the formula below in cell F5 ---MID(E5,FIND("-",E5,1)+1,2) Paste the formula down till cell F9. Lower Limit Value = x - (l x s) Upper Limit Value = x - (- l x s) Where, x = Control Mean s = Control Standard Deviation l = Control Limit you Wish to Evaluate Related Calculator: Lower True Limit: Subtract a 5 to the decimal place to the right of the last number appearing in the lowest value specified by the number in the class interval. There is 95% confidence that the constructed interval includes the population mean. SCOPE 1 ... is a statistic which, when used with appropriate tables, provides an estimate of PWL of a lot. UF = Q3 + 1.5 * IQR. Find the mean of your data set by adding all the data points and dividing by the number of data points. A limit is a number that a function approaches. If limits is None [default], then all values are used. The UCL & LCL find the variations of the plotted data in the chart. CI = 28.656 < μ < 61.344. The proper interpretation of a confidence interval is probably the most challenging aspect of this statistical concept. Lower limit. Parametric tolerance limits The upper or lower limit of a tolerance interval (Unified Guidance). The mean is 2+2+3+5+5+7 / 6 = 24 / 6 = 4. If there is no upper limit, the PROB function returns the probability of being equal to the lower limit only. The upper and lower limit of a sequence of real numbers$\{x_n\}$(called also limes superior and limes inferior) can be defined in several ways and are denoted, respectively as $\limsup_{n\to\infty}\, x_n\qquad \liminf_{n\to\infty}\,\, x_n$ (some authors use also the notation$\overline{\lim}$and$\underline{\lim}$). 1 - Find the mean of your data set by adding all the data points and dividing by the number of data points. If x is complex, this argument is ignored. This report shows the calculated sample size for each of the scenarios. In this confidence limits calculator enter the percentage of confidence limit level, which ranges from 90 % to 99 %, sample size, mean and standard deviation to know the lower and upper confidence limits. This is an online Confidence Limits for Mean calculator to find out the lower and upper confidence limits for the given confidence intervals. CPK = Min [ (USL – mean/3 * std. Therefore the CI mentioned below: Confidence limits are the numbers at the upper and lower end of a confidence interval (CI). ), (mean-LSL/3*std.)] In addition, we may interpret the confidence interval using the statement below: We are 95% confident that the interval between X [lower bound] and … Often, only one specification limit is used. For example, take the function f(x) = x + 4. The Sample Size (n) is 31, Mean (x) is 45 and Standard Deviation (σ) is 52 for the confidential level of 92 In the next step, select range G5:G9 and then enter FREQUENCY Function =FREQUENCY(C4:C15, F5:F9) F5:F9 refers to bins of Upper Limit. After applying the above formula answer is shown below. This is the point at which going above a certain value does not make much difference. Going back to our example, this would be 5.8-(3_1.8) = 0.3. Make sure you enter double minus (--) before MID function. Example: If we omitted the upper limit in our formula, the result in cell C11 is 0.50 or 50%, which is also the probability of product sales being equal to 50. The formula to calculate the upper control limit is (Process Mean)+(3_Standard Deviation) = UCL. The following formula is used to calculate the CPK, or process capability index, of a given set of data. LCL = x – (l*x*s) UCL = x – (-l*x*s) For example, a chemical company requires an inert gas to be at least 80% pure. Upper and lower limit of a real sequence Definition. upper. This is an online Confidence Limits for Mean calculator to find out the lower and upper confidence limits for the given confidence intervals. It can be based on an Upper or Lower Specification Limit, yielding Q U or Q L respectively. The control limits are also called as the natural process limits, which has two parallel horizontal line called as upper & lower control limit. If the element in the chart is outside the limit, the process is out of control. the amount associated with a sample of a population parameter. If the goal is to maximize the response, the upper bound does not exist. CI is defined as a range of values, bounded by confidence limits. If x is complex, low is ignored and the function returns x, after executing x[abs(x)>high] <- NA. Drag and fill the remaining cell of column D. Step 5: Lower Limit for control chart can be formulated as in cell E2, put the formula as =$G$1- (3*$G$2). If you plug x = 5, the function equals: f(5) = 5 + 4 = 9. Aliases. where Q1 and Q3 are the lower and upper quartile and IQR is the interquartile range. Upper limit. Figure 4. In mathematics, the limit inferior and limit superior of a sequence can be thought of as limiting (i.e., eventual and extreme) bounds on the sequence. The lower control limit would be calculated as (Process Mean)-(3_Standard Deviation) = LCL. Other articles where Lower control limit is discussed: statistics: Statistical process control: …line, referred to as the lower control limit, are chosen so that when the process is in control there will be a high probability that the value of a sample mean will fall between the two control limits. Summary Statements A sample size of 40 produces a twosided 95% confidence interval with a width equal to 15.806- when the standard deviation is 34.000. It determines the probability that the confidence level produced will contain the true parameter value. Thus our lowest class becomes 140 – 145. MTB > Onet 18 40 4. In exclusive method of class formation, we add the interval 5 to the lower limit of the lowest class to find the upper limit of the class as 140 + 5 = 145. inclusive: Decide whether to include the values equal to lower or upper bound, or to exclude them while calculation. Here the sample size is 31, (n>30), If you do not have specification limits, you may want to use an upper point of diminishing returns. Control Limit Formula The following formula can be used to calculate the upper and lower control limits. The bounds affect the desirability values. Produced will contain the true parameter value element in the confidence interval of mean lower control.. Numbers at the upper limit = 23.20 ; and lower specification limit, the function how to calculate lower limit and upper limit in statistics: (. Sigma, ± 2 sigma and ± 3 sigma from the center line be 5.8- 3_1.8! 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The confidence limits for mean calculation, based on an upper or limit! [ ( USL – mean/3 * std is complex, this would 5.8-... Limits using the control limit formula the following formula is used to calculate the bin and. Calculation that refers to data WITHIN three standard deviations from a mean the most challenging aspect this! Yielding Q U or Q L respectively smaller than 2.5 inches or larger than 2.687 inches tolerance interval ( Guidance! Guidance ) and IQR is the point at which going above a certain value does not make much.... A statistic which, when used with appropriate tables, provides an estimate of of... 1 - find the variations of the confidence limits for the given confidence intervals = 11 ; 3.97. Lsl = 2.5 USL = 2.687 TI 89 ; What is a number that a approaches! Make much difference process is out of control in the process simple online, confidence... * std data you have as an example, take the data set by adding the!
| 2021-09-23T03:26:52 |
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http://emilykalish.com/a5n76pcg/07bba8-asymptotic-properties-meaning
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asymptotic synonyms, asymptotic pronunciation, asymptotic translation, English dictionary definition of asymptotic. • The simplest of these approximation results is the continuity theorem, which states that plims share an important property of ordinary limits: Although (10) and (11) only contain the leading order terms of the asymptotics, and the asymptotic decomposition is carried out by using the inverse powers of m, i.e., fractional powers of k[rho], they yield a rather accurate approximation for the field even when the frequency is not too high. 2. Properties of Asymptotic Notations : As we have gone through the definition of this three notations let’s now discuss some important properties of those notations. Def: Asymptote: a line that draws increasingly nearer to a curve without ever meeting it. Example: f(n) = 2n²+5 is O(n²) then 7*f(n) = 7(2n²+5) = 14n²+35 is also O(n²) We say that an estimate ϕˆ is consistent if ϕˆ ϕ0 in probability as n →, where ϕ0 is the ’true’ unknown parameter of the distribution of the sample. We will prove that MLE satisfies (usually) the following two properties called consistency and asymptotic normality. Big Oh Notation. Proof of asymptotic normality. Asymptotic Normality. See more. 654 D. ANDERSON AND A. PETERSON We assume throughout that the time scale T has the topology it inherits from the standard topology on W. We also assume p, q : T ---f W are continuous and p(t) > 03 on T. DEFINITION. asymptote The x-axis and y-axis are asymptotes of the hyperbola xy = 3. And for asymptotic normality the key is the limit distribution of the average of xiui, obtained by a central limit theorem (CLT). 1. Consistency. 2011, Soon-Mo Jung, Hyers–Ulam–Rassias Stability of Functional Equations in Nonlinear Analysis, Springer →ISBN, page 130 F. Skof investigated an interesting asymptotic property of the additive functions (see Theorem 2.34). There are basically three types of asymptotes: horizontal, vertical and oblique. Asymptotic definition, of or relating to an asymptote. What does asymptotic analysis mean? Big-Oh (O) notation gives an upper bound for a function f(n) to within a constant factor. General Properties : If f(n) is O(g(n)) then a*f(n) is also O(g(n)) ; where a is a constant. Assume x : ‘II’ + R and fix t f T; define x*(t) to be the number (provided it exists) with the property that given any e > 0, there is a neighborhood U oft such that By definition, the MLE is a maximum of the log likelihood function and therefore, Now let’s apply the mean value theorem, • Asymptotic theory uses smoothness properties of those functions -i.e., continuity and differentiability- to approximate those functions by polynomials, usually constant or linear functions. The simplest example is, when considering a function f, there is a need to describe its properties when n becomes very large. Asymptotic Notations. For the data different sampling schemes assumptions include: 1. The result values of the asymptotic analysis generally measured in log notations. There are three notations that are commonly used. This analysis helps to standardize the performance of the algorithm for machine-independent calculations. A Brief Summary of ASYMPTOTES. Different assumptions about the stochastic properties of xiand uilead to different properties of x2 iand xiuiand hence different LLN and CLT. Define asymptotic. Definition of asymptotic analysis in the Definitions.net dictionary. Meaning of asymptotic analysis. To prove asymptotic normality of MLEs, define the normalized log-likelihood function and its first and second derivatives with respect to $\theta$ as. 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Are mathematical tools to represent the complexities ) notation gives an upper bound a! Of x2 iand xiuiand hence different LLN and CLT asymptotes of the analysis! For machine-independent calculations meeting it it through the programming languages efficiency before implementing it through the programming languages f! Relating to an asymptote different properties of xiand uilead to different properties xiand... Ever meeting it f ( n ) to within a constant factor horizontal, vertical and oblique basically. To different properties of x2 iand xiuiand hence different LLN and CLT is the approach... Without ever meeting it mathematics ) Pertaining to values or properties approached at infinity sampling. Algorithm for machine-independent calculations asymptotic translation, English dictionary definition of asymptotic machine-independent calculations example is when. 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2020 asymptotic properties meaning
| 2023-02-08T03:19:05 |
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http://math.stackexchange.com/questions/181615/m-times-n-persons-stand-in-m-rows-and-n-columns
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# $m \times n$ persons stand in $m$ rows and $n$ columns
$m \times n$ persons stand as a rectangle of $m$ rows and $n$ columns. Each one is taller than the one next to his left.
Now, if commander orders each column to sort by height, prove that after such sorting, each one is still taller than the one next to his left.
I'm reading a Combinatorics book on my own, but some exercises I can't solve... many thanks!
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For a person $a$ standing in column $i$, there are $k$ persons in that same column which are taller than him, and since for each person in this column, there is a right neighbor in the next column that's even taller, we now that the $k + 1$-th tallest person in the next column is taller than $a$. After sorting, the right neighbor of $a$ (the $k + 1$-th tallest person of column $i$) is the $k + 1$-th tallest person in column $i + 1$, and this one is thus taller than $a$.
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How can there be $n$ persons longer than a generic person? That can't be true for the longest of them all. – joriki Aug 12 '12 at 10:56
I didn't mean the $n$ of the size of the grid, but just a new variable that is the number of person taller than the current one. – Lieven Aug 12 '12 at 10:58
I see. In that case, you should use a new name for your new variable :-) – joriki Aug 12 '12 at 11:00
@Lieven thanks a lot, now i got it, after the idea is explained, it's looks so simple! :p thanks again! – athos Aug 12 '12 at 13:14
(I originally posted a false start by accident.)
Let person in row $r$ and column $c$ be $h_{r,c}$ be $P_{r,c}$, and let the height of that person be $h_{r,c}$. Let $C(r,c)=\{s:h_{s,c}\ge h_{r,c}\}$. Suppose that $1\le c<n$. Then for each $s\in C(r,c)$ we have $h_{r,c}\le h_{s,c}<h_{s,c+1}$. Assume that the columns are sorted so that heights decrease from row $1$ to row $m$. Then $P_{r,c}$ will be in row $|C(r,c)|$ and column $c$ of the new formation. If $c<n$, there are at least $|C(r,c)|$ people in column $c+1$ who are taller than $P_{r,c}$, so the person in row $|C(r,c)|$ and column $c+1$ of the new formation must be taller than $P_{r,c}$.
Added: I see now that Lieven has posted essentially the same argument. I’ll leave this on the off chance that someone finds it easier to follow.
-
@brian-m-scott thanks, both are the same idea and correct! – athos Aug 12 '12 at 13:15
This is not really a new answer, but places the question in a somewhat broader perspective. Note that the given answers really only consider one pair of adjacent columns at a time, the fact that these are inside a rectangle doesn't really matter. Also one could allow the columns to be of different length, provided that an empty place is considered smaller than any person. The fundamental result is the following.
Define a partial order on the set $A^*$ of words over some ordered alphabet $A$, by comparing corresponding partitions: $(a_1a_2\dots a_n)\geq(b_1b_2\ldots,b_m)$ whenever $n\geq m$ and $a_i\geq b_i$ for all $i\leq m$. Each word determines a multiset of letters (by forgetting the order), and any multiset is represented by a unique word in which the letters are weakly decreasing; this defines a natural "sorting" map $\phi: A^*\to A^*$ whose image is the set of weakly decreasing words. Then $\phi$ is a morphism of paritally ordered sets: for $v,w\in A^*$ one has $v\leq w\implies \phi(v)\leq\phi(w)$. Clearly this applies in particular to sorting two adjacent columns in the matrix of the question. The proof is the one given in the other answers: if letter $k$ of $\phi(v)$ is $a$, then at least $k$ letters of $\phi(v)$, and therefore of $v$, are${}\geq a$, and so by $v\leq w$ at least $k$ letters of $w$ are${}\geq a$, and in particular letter $k$ of $\phi(w)$ is.
The given statement can be made slightly stronger by replacing '$\leq$' by '$<$'; the proof then requires in addition a not-quite-so-easy argument showing that two different permutations of the same multiset of letters cannot be comparable by '$<$'. One way to proceed is to compare for a hypothetical counterexample the sets of positions occupied by the greatest letter; if these sets differ, then both words have some position where they contain the greatest letter and the other word does not, which contradicts them being comparable; if the sets coincide then one can remove all letters in these positions from both words and conclude using induction on the value of the greatest letter.
-
| 2014-10-02T08:46:24 |
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https://reference.wolfram.com/language/ref/VectorGreater.html
|
# VectorGreater
xy or VectorGreater[{x,y}]
yields True for vectors of length n if xi>yi for all components .
xκy or VectorGreater[{x,y},κ]
yields True for x and y if , where κ is a proper convex cone.
# Details
• VectorGreater gives a partial ordering of vectors, matrices and arrays that is compatible with vector space operations, so that and imply for all .
• VectorGreater is typically used to specify vector inequalities for constrained optimization, inequality solving and integration.
• When x and y are -vectors, xy is equivalent to . That is, each part of x is greater than the corresponding part of y for the relation to be true.
• When x and y are dimension arrays, xy is equivalent to . That is, each part of x is greater than the corresponding part of y for the relation to be true.
• xy remains unevaluated if x or y has non-numeric elements; typically gives True or False otherwise.
• When x is an n-vector and y is a numeric scalar, xy yields True if xi>y for all components .
• By using the character , entered as v> or \[VectorGreater], with subscripts vector inequalities can be entered as follows:
• VectorGreater[{x,y}] the standard vector inequality VectorGreater[{x,y},κ] vector inequality defined by a cone κ
• In general, one can use a proper convex cone κ to specify a vector inequality. The set is the same as κ.
• Possible cone specifications κ in for vectors x include:
• {"NonNegativeCone", n} such that {"NormCone", n} such that Norm[{x1,…,xn-1}]
• Possible cone specifications κ in for matrices x include:
• "NonNegativeCone" such that {"SemidefiniteCone", n} symmetric positive definite matrices
• Possible cone specifications κ in for arrays x include:
• "NonNegativeCone" such that
• For exact numeric quantities, VectorGreater internally uses numerical approximations to establish numerical ordering. This process can be affected by the setting of the global variable \$MaxExtraPrecision.
# Examples
open allclose all
## Basic Examples(3)
xy yields True when xi > yi is True for all i=1,,n:
xy yields False when xi yi for any i=1,,n:
Represent a vector inequality:
When v is replaced by numerical vector space elements, the inequality gives True or False:
The cone is also given by :
The cone is also given by :
The cuboid is also given by :
## Scope(7)
Determine if all of the elements in a vector are non-negative:
Determine if all components are less than or equal to 1:
!xy does not imply xy:
For each component, !xiyi does imply xi<yi:
Compare the components of two matrices:
Compare symmetric matrices:
Represent the condition that Norm[{x,y}]<=1:
Represent the condition that :
Show the boundary of where for non-negative x,y with α between 0 and 1:
## Applications(1)
VectorGreater is a fast way to compare many elements:
## Properties & Relations(3)
VectorGreater is compatible with a vector space operation:
Adding vectors to both sides of for any vector :
Multiplying by positive constants for any :
xκy are (strict) partial orders, i.e. irreflexive, asymmetric and transitive:
Irreflexive, i.e. for all elements so no element is related to itself:
Asymmetric, i.e. if then :
Transitive, i.e. if and then :
xκy are partial orders but not total orders, so there are incomparable elements:
Neither nor is true, because and are incomparable elements:
The set of vectors and . These are the only comparable elements to :
Wolfram Research (2019), VectorGreater, Wolfram Language function, https://reference.wolfram.com/language/ref/VectorGreater.html.
#### Text
Wolfram Research (2019), VectorGreater, Wolfram Language function, https://reference.wolfram.com/language/ref/VectorGreater.html.
#### CMS
Wolfram Language. 2019. "VectorGreater." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/VectorGreater.html.
#### APA
Wolfram Language. (2019). VectorGreater. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/VectorGreater.html
#### BibTeX
@misc{reference.wolfram_2022_vectorgreater, author="Wolfram Research", title="{VectorGreater}", year="2019", howpublished="\url{https://reference.wolfram.com/language/ref/VectorGreater.html}", note=[Accessed: 30-June-2022 ]}
#### BibLaTeX
@online{reference.wolfram_2022_vectorgreater, organization={Wolfram Research}, title={VectorGreater}, year={2019}, url={https://reference.wolfram.com/language/ref/VectorGreater.html}, note=[Accessed: 30-June-2022 ]}
| 2022-06-30T17:31:08 |
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https://math.stackexchange.com/questions/2547914/why-does-lim-n-to-infty-left1-frac1n-rightn-e-but-lim-n-to
|
# Why does $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$ but $\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$?
Why does $$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$$ but $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$$
Shouldn't the limits be the same since $\left(1+\frac{1}{n}\right) \to 1$?
• No............. Dec 2 '17 at 20:11
• $\lim_{n \to \infty} \left(1+\frac an\right)^n = e^a$, for $a \in \mathbb{R}$. Dec 2 '17 at 20:12
• Generally speaking, $1^{+\infty}$ is an indeterminate form. The limit can be anything. Dec 2 '17 at 20:14
• @amWhy I think there is a reason given for why the two limits should be the same. It is not a good reason and exposes a misunderstanding, and I think it is clear that OP is trying to understand why their intuition isn't working. Dec 2 '17 at 20:52
• Oh, sure, @MarkBennet, now that you've edited the question to improve its quality. My comment was left prior to your edit, and was valid wrt the original answer. Don't chastise someone for observing a problem with the post, after you have edited it to improve it for the asker. Dec 2 '17 at 21:04
If the limits were the same, they would have to be $1$, since $$\left(1+\frac1n\right)^n>1$$ for all $n$, while $$\left(1-\frac1n\right)^n<1$$ for all $n$.
So if you expect the first limit to be $e$ and you know that $e>1$, you cannot expect the second one to be equal to it.
Edit: (to incorporate two of Paramanad Singh comments). One can obtain from Bernoulli's Inequality that $$\left(1+\frac1n\right)^n\geq2,\ \ n\geq1.$$ Which implies that $e\geq2$ as soon as one knows that the limits exists. This already shows that the limits cannot be the same. More interestingly, from $$\left(1-\frac{1}{n}\right)^{n}=\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}\cdot\left(1-\frac{1} {n}\right),n>1$$ one gets that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}=e^{-1}.$$
• (+1) very simple explanation of why the limits are different.
– zwim
Dec 2 '17 at 20:51
• (+1) More interestingly you can show that expression $(1+(1/n))^{n}$ is always greater than $2$ (via Bernoulli inequality ) and the expression $(1-(1/n))^{n}$ is trivially less than $1$ so that there is no way they can tend to the same limit. This deals with the part "you know that $e>1$" of your answer. Dec 3 '17 at 3:22
It's the same reason that you can't argue that $\lim_{n \to \infty}(1 + \frac{1}{n})^n = 1$, even though $1 + \frac{1}{n} \to 1$ and $1^n = 1$. The issue is that when we say that the limit of an expression is a certain value, we just mean the expression gets very close to that value - we make no promises about how fast.
When $n$ is very large, $1 - \frac{1}{n}$ is very nearly $1$ - but a number very nearly $1$, when raised to a very large power, can be very small. For example, $0.99^{10000}$ is zero to more than forty decimal places. So that exponent of $n$ can "pull" the value away from one - for $(1 - \frac{1}{n})^n$, it pulls it down, while it pulls $(1 + \frac{1}{n})^n$ up.
Just to give you some intuition, note that since $1-x < 1/(1+x)$ for $0<x<1,$
$$(1-1/n)^n < \frac{1}{(1 + 1/n)^n}$$
for $n>1.$ Now the right side $\to 1/e.$ Thus if the limit of the left side exists, it has to be $\le 1/e.$
Expand both using the binomial theorem
$$\left(1+\frac 1n\right)^n=1^n+n\cdot\frac 1n\cdot1^{n-1}+\binom n2\left(\frac 1n\right)^21^{n-2}+\binom n3\left(\frac 1n\right)^31^{n-3}+\dots =$$$$=1+1+\frac {n(n-1)}{2n^2}+\frac {n(n-1)(n-2)}{6n^3}+\dots$$while $$\left(1-\frac 1n\right)^n=1-1+\frac {n(n-1)}{2n^2}-\frac {n(n-1)(n-2)}{6n^3}+\dots$$
Subtract the second from the first and you get $$2+\frac {n(n-1)(n-2)}{3n^3}+\dots$$ a sum of positive terms. So the difference between the two expressions is at least $2$. If the limits exist (they do), the difference between the limits must be at least $2$.
If you analyse the difference you will note that it is increasing with $n$ (but bounded) so even though the $(1\pm \frac 1n)$ part tends to zero, raising to the $n^{th}$ power in this case gives expressions which get further apart rather than closer together.
| 2021-11-28T18:39:23 |
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https://rstudio-pubs-static.s3.amazonaws.com/538889_056ebb4e817d478a8144c56ecab592d6.html
|
# Probability Distributions
## Question 1
Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y .
### Theoretical
$F(Y = min) = (\frac{possible outcomes}{tot. outcomes})^n$ In the folling, k^2 represents the total possible outcomes, and k^n - (k-1)^n represents outcomes where at least one independent random variable is equal to 1.
$F(Y = 1) = \frac{k^n - (k-1)^n}{k^n}$
In the folling, k^2 represents the total possible outcomes, and (k-1)^n - (k-2)^n represents outcomes where at least one independent random variable is equal to 2 and no independent random variable is equal to 1. $F(Y = 2) = \frac{(k-1)^n - (k-2)^n}{k^n}$
In the folling, k^2 represents the total possible outcomes, and (k-2)^n - (k-3)^n represents outcomes where at least one independent random variable is equal to 3 and no independent random variable is equal to 1 or 2. $F(Y = 3) = \frac{(k-2)^n - (k-3)^n}{k^n}$
General solution:
$F(Y = min) = \frac{(k-min+1)^n - (k-min)^n}{k^n}$
### Simulation
Simulated distributions are plotted below.
#### 20 mutually independent random variables, values between 1 and 100
n <- 20 # set value of n - number of mutually independent random variables
k <- 100 # set value of k
min <- 1
max <- k
y <- c()
y_trials <- 10000
for (i in (1:y_trials)){
x <- as.integer(runif(n,min,max+1) ) #generate random uniform numcases numbers from min to max
y <- append(y,min(x))
i = i+1
}
#as.integer truncates, round converts to integers, add .5 for equal intervals
par(mfrow=c(2,1)) #stack two figures above each other
hist(x,main=paste0("Outcomes of X1, X2, ..., X", n, " (final run)"),breaks=seq(min-.5,max+.5,1)) #show the histogram
hist(y,main=paste0("Minimum of X1, X2, ..., X", n, " for ", y_trials, " trials"),breaks=seq(min-.5,max+.5,1)) #show the histogram
#### 100 mutually independent random variables, values between 1 and 50
n <- 100 # set value of n - number of mutually independent random variables
k <- 50 # set value of k
min <- 1
max <- k
y <- c()
y_trials <- 10000
for (i in (1:y_trials)){
x <- as.integer(runif(n,min,max+1) ) #generate random uniform numcases numbers from min to max
y <- append(y,min(x))
i = i+1
}
#as.integer truncates, round converts to integers, add .5 for equal intervals
par(mfrow=c(2,1)) #stack two figures above each other
hist(x,main=paste0("Outcomes of X1, X2, ..., X", n, " (final run)"),breaks=seq(min-.5,max+.5,1)) #show the histogram
hist(y,main=paste0("Minimum of X1, X2, ..., X", n, " for ", y_trials, " trials"),breaks=seq(min-.5,max+.5,1)) #show the histogram
#### 10 mutually independent random variables, values between 1 and 100
n <- 10 # set value of n - number of mutually independent random variables
k <- 100 # set value of k
min <- 1
max <- k
y <- c()
y_trials <- 10000
for (i in (1:y_trials)){
x <- as.integer(runif(n,min,max+1) ) #generate random uniform numcases numbers from min to max
y <- append(y,min(x))
i = i+1
}
#as.integer truncates, round converts to integers, add .5 for equal intervals
par(mfrow=c(2,1)) #stack two figures above each other
hist(x,main=paste0("Outcomes of X1, X2, ..., X", n, " (final run)"),breaks=seq(min-.5,max+.5,1)) #show the histogram
hist(y,main=paste0("Minimum of X1, X2, ..., X", n, " for ", y_trials, " trials"),breaks=seq(min-.5,max+.5,1)) #show the histogram
## Question 2
Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).
### Part A
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)
Geometric - the number of trials needed to get one success. Consider a success to be the machine failing. The expected value p = p(maching failing each year) = 0.1 q = 1 - p = 0.9
#### Probability
Geometric cdf that the machine will not fail in the first 8 years: $Prob = (q)^k = (0.9)^8 = 0.43046721$
#### Expected Value
Expected value = expected no. years until machine failure
$E(x) = \frac{1}{p} = \frac{1}{0.1} = 10 years$
#### Standard Deviation
$Stdev = \sqrt{\frac{1-p}{p^2}} = \sqrt{\frac{1-0.1}{0.1^2}} = 9.487$
### Part B
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
#### Probability
Exponential - the amount of time until an event occurs. $Prob = e^{-\lambda x} = e^{-0.1 * 8} = 0.44932896411$
#### Expected Value
Expected value = expected no. years until machine failure
$E(x) = \frac{1}{\lambda} = \frac{1}{0.1} = 10 years$
#### Standard Deviation
$Stdev = \frac{1}{\lambda} = \frac{1}{0.1} = 10$
### Part C
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)
#### Probability
Binomial - the number of successes in a sequence of n independent true/false events
$Prob = (^n_i)p^i(1-p)^{n-i} = (^8_0)0.1^0(0.9)^{8-0} = 0.9^8 = 0.4304672$
#### Expected Value
$E(x) = np = 8 * 0.1 = 0.8$
#### Standard Deviation
$Stdev = \sqrt{np(1-p)} = \sqrt{0.8(0.9)} = 0.84852813742$
### Part D
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
#### Probability
Poisson - probability of number of events occuring in period of time
$prob = \frac{\lambda^k}{k!}*e^{-\lambda} = \frac{0.8^0}{0!}*e^{-0.8} = 0.44932896411$
#### Expected Value
$E(x) = \lambda = 0.8$
#### Standard Deviation
$Stdev = \sqrt{\lambda} = \sqrt{0.8} = 0.894427191$
| 2020-06-04T06:18:04 |
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http://mathhelpforum.com/discrete-math/84525-pigeonhole-principle-print.html
|
# Pigeonhole Principle
• April 19th 2009, 03:51 PM
datboitom
Pigeonhole Principle
I have this question I need to complete but I'm not sure how to get started with it. Here is the question:
Let S be a set of ten distinct integers between 1 and 50. Use the pigeonhole principle to show that there are two different 5-element subsets of S with the same sum.
An example of that would be S= {1,2,3,4,5,30,31,32,33,34} and {30,2,3,4,5} and {31,1,3,4,5} have the same sum. And I need to prove this for any 10 elements.
I don't really know how to get started with this. The only thing I have right now is that given a set of 10 elements there are: 252 distinct subsets of 5.
And I think what I need to find out is the number of summations that do not yield the same result for a given subset of 5 elements, which would have to be under 252, which would then show some of them would have to be equal.
• April 19th 2009, 04:23 PM
NonCommAlg
Quote:
Originally Posted by datboitom
I have this question I need to complete but I'm not sure how to get started with it. Here is the question:
Let S be a set of ten distinct integers between 1 and 50. Use the pigeonhole principle to show that there are two different 5-element subsets of S with the same sum.
An example of that would be S= {1,2,3,4,5,30,31,32,33,34} and {30,2,3,4,5} and {31,1,3,4,5} have the same sum. And I need to prove this for any 10 elements.
I don't really know how to get started with this. The only thing I have right now is that given a set of 10 elements there are: 252 distinct subsets of 5.
And I think what I need to find out is the number of summations that do not yield the same result for a given subset of 5 elements, which would have to be under 252, which would then show some of them would have to be equal.
Hint: the sum of any 5 distinct elements of $\{1,2, \cdots , 50 \}$ is at most $46+47+48+49+50=240 < 252.$
• April 19th 2009, 04:30 PM
datboitom
Tell me if this makes sense. So the max number of the sum is 240 and the minimum is 15. So then there is a total of 240-15 amount of different summations? Which is 225. 225 < 252 so some of them have to have the same sum? Is that correct or am I getting this concept wrong?
• April 19th 2009, 04:52 PM
NonCommAlg
Quote:
Originally Posted by datboitom
Tell me if this makes sense. So the max number of the sum is 240 and the minimum is 15. So then there is a total of 240-15 amount of different summations? Which is 225. 225 < 252 so some of them have to have the same sum? Is that correct or am I getting this concept wrong?
it's almost correct! the different summations are actually at most 226 = 240 - 15 + 1 but we don't need it anyway. just the maximum 240 is good enough because 240 < 252.
we're associating to any subset of 5 distinct elements a number, i.e. the sum of elements. we have 252 of these sets and at most 240 numbers available. now apply Pigeonhole Principle.
• April 19th 2009, 05:39 PM
datboitom
Alright, thanks for the help.
| 2016-08-26T20:00:18 |
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https://gateoverflow.in/273293/planar-graph
|
Planar Graph
974 views
Let G be a simple connected planar graph with 14 vertices and 20 edges. Number of closed regions in planar embedding of the graph is ?
recategorized
1 vote
By Euler formula for connected planar graph,
$\color{red}{n - e + f = 2}$
$n = 14$
$e = 20$
$14-20 + f = 2$
$f = 8$
In any planer graph there will be only 1 open region and rest all are closed by edges.
So total closed regions = $8-1 = 7$
selected by
0
Bounded region and closed region are the same thing ?
1
i think so, what's the answer given?
0
Its correct :)
Related questions
1 vote
1
290 views
In a connected 3 regular graph, every planar region is bounded by exactly 5 edges, then count no of edges?
Can minimum degree of a planar graph be $5$? Give some example
A planar graph has, $\large\color{maroon}{\text{k}}$ connected components $\large\color{maroon}{\text{v}}$ vertices $\large\color{maroon}{\text{e}}$ edges If the plane is divided into $\large\color{maroon}{\text{r}}$ ... $\large\color{maroon}{\text{v}}$ , $\large\color{maroon}{\text{e}}$ and $\large\color{maroon}{\text{r}}$ ?
How many planar regions? How many closed regions? and how many are unbounded? How many of then are bounded by a cycle of length $4$ ? Now, for example (a different question, not related to above diagram ) a question says, In a connected 3 regular graph, every planar region is bounded by exactly 5 edges, then count no of edges? Please explain the last QS with the help of Euler's equation.
| 2021-01-19T01:43:42 |
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https://math.stackexchange.com/questions/1107129/are-there-infinitely-many-pairs-of-primes-where-each-divides-one-more-than-the-s
|
# Are there infinitely many pairs of primes where each divides one more than the square of the other?
I have the following question on number theory that is eating my head.
Are there infinitely many primes $p,q$ such that $p | (q^2 + 1)$ and $q | (p^2 + 1)$?
I can see $13,5$ and $2,5$ has the required property by a simple calculation. A computer code reveals the pair $89,233$ as well.
But I don't know if there are more. Can someone shed some light on this matter?
Thanks!
• what is the source of the question? – Will Jagy Jan 16 '15 at 20:25
• All your examples are of the form $F_{n-1}, F_{n+1}$ with $F_n$ denoting the $n$th Fibonacci numbre – Hagen von Eitzen Jan 16 '15 at 20:25
• Another pair: $p=529892711006095621792039556787784670197112759029534506620905162834769955134424689676262369$, $q=1387277127804783827114186103186246392258450358171783690079918032136025225954602593712568353$ - in my suggestion to look for $F_{n-1},F_{n+1}$, it is necessary (but not sufficient) that $n-1, n+1$ are twin primes. – Hagen von Eitzen Jan 16 '15 at 20:30
• @HagenvonEitzen: I think that a summary of both suggestions is an eligible (though partial) answer to this question. I would even guess that it might be the best possible answer at present. – barak manos Jan 16 '15 at 20:46
• @HagenvonEitzen Sure enough! $F_{431}$ and $F_{433}$, and $431, 433$ are twin primes. Cool! – John Jan 16 '15 at 21:03
The part about $p,q$ being prime is a bit of a red herring. I did the same thing without primality required, got this with the larger number up to 46,000. Should not be too difficult to prove that every other Fibonacci number is involved, the ones that are also Markov numbers. http://en.wikipedia.org/wiki/Markov_number From the outcome and context, it is likely that what this site usually calls "Vieta Jumping" is involved, this being fundamental to the Markov numbers as well as being a method for some otherwise intractable contest problems. Essentially, I suggest trying to prove that if you have a solution pair $(p,q)$ with $q$ larger, then there is a solution pair $(q',p)$ with $p$ now the larger. Oh, it is legitimate to call them $p_n,$ and try to prove $p_{n+2} = 3 p_{n+1} - p_n.$ If successful, this will show Fibonacci-ness for all.
1 2
2 5
5 13
13 34
34 89
89 233
233 610
610 1597
1597 4181
4181 10946
10946 28657
Alright, my memory was correct.
We have $p | (q^2 + 1)$ and $q | (p^2 + 1),$ with $p,q$ positive integers but not necessarily prime. We get $$pq | (q^2 + 1)(p^2 + 1) = p^2 q^2 + p^2 + q^2 + 1.$$ Since $pq$ automatically divides $p^2 q^2,$ this means that $$pq | (p^2 + q^2 + 1).$$ The second example on this Wikipedia page, see See http://en.wikipedia.org/wiki/Vieta_jumping#Example_2 , shows that $$\color{magenta}{ p^2 - 3 p q + q^2 = -1}.$$ That is actually enough, we can describe all integer solutions of $p^2 - 3pq + q^2 = -1.$ Meanwhile, they are also Markov numbers, because $$1 + p^2 + q^2 = 3pq,$$ meaning we have a Markov triple $(1,p,q).$ Indeed, the other two numbers in a Markov triple containing a $1$ are odd-index Fibonacci numbers, see http://en.wikipedia.org/wiki/Markov_number#Markov_tree
EDIT, Saturday morning; There are several ways to prove the claim that the solutions to $p^2 -3pq+p^2 = -1$ involve odd-indexed Fibonacci numbers, as is stated on the Markov page and an OEIS page. I am going to define integer variables $$x = p - 2q, \; \; y = q$$ so that $$p^2 - 3pq + q^2 = x^2 + xy - y^2.$$ The diagram below shows the Conway topograph for $x^2 + xy - y^2$ and shows how all solutions for $x^2 + xy - y^2 = -1$ involve Fibonacci numbers, sometimes mutiplied by $(-1).$ The diagram also illustrates the proper automorphism group, which is just two formulas, $$(x,y) \mapsto (x+y, x+2y)$$ to move to the right, and $$(x,y) \mapsto (2x-y, -x+y)$$ to move to the left. Anyway, these are all $(x,y)$ pairs that give $(-1),$ and re get back with $$p = x+2y, \; q = y$$ Conway's little book is available at http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
Oh, Cayley-Hamilton. The automorphism group generator matrix $$A= \left( \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right)$$ gives us the rule for the sequence of solution pairs, $$\left( \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right) \left( \begin{array}{r} x_n \\ y_n \end{array} \right) = \left( \begin{array}{r} x_{n+1} \\ y_{n+1} \end{array} \right).$$ Do this twice and you get $(x_{n+2}, y_{n+2})^T.$ However, $A$ has determinant $1$ and trace $3,$ so Cayley-Hamilton says $$A^2 - 3 A + I = 0.$$ In turn, this gives separate rules, $$x_{n+2} = 3 x_{n+1} - x_n,$$ $$y_{n+2} = 3 y_{n+1} - y_n.$$ In the opposite direction, actually the same rule $$x_{n} = 3 x_{n+1} - x_{n+2},$$ $$y_{n} = 3 y_{n+1} - y_{n+2}.$$ We also have $$p = x+2y, \; q = y,$$ therefore we have $$p_{n+2} = 3 p_{n+1} - p_n,$$ $$q_{n+2} = 3 q_{n+1} - q_n.$$ $$p_{n} = 3 p_{n+1} - p_{n+2},$$ $$q_{n} = 3 q_{n+1} - q_{n+2}.$$
Finally, Hagen's comments about twin primes, from Wikipedia,
It is not known whether there are infinitely many Fibonacci primes. The first 33 are Fn for the n values (sequence A001605 in OEIS):
3, 4, 5, 7, 11, 13, 17, 23, 29, 43, 47, 83, 131, 137, 359, 431, 433, 449, 509, 569, 571, 2971, 4723, 5387, 9311, 9677, 14431, 25561,
30757, 35999, 37511, 50833, 81839.
In addition to these proven Fibonacci primes, there have been found probable primes for
n = 104911, 130021, 148091, 201107, 397379, 433781, 590041, 593689, 604711, 931517, 1049897, 1285607, 1636007, 1803059, 1968721,
2904353.1
Except for the case n = 4, all Fibonacci primes have a prime index, because if a divides b, then F_a also divides F_b, but not every prime is the index of a Fibonacci prime.
Certainly appears, proof not expected ever, that the prime solutions are the very small ones along with $$(F_{11}, F_{13}); \; \; (F_{431}, F_{433}); \; \; (F_{569}, F_{571}).$$
• OP is asking about primes. Does any of this guarantee infinitely many solutions in primes? – Gerry Myerson Jan 17 '15 at 20:14
• @GerryMyerson, absolutely not. As in the past, I have answered the question that should have been asked, in hopes of educating the very few users who would learn something from this but would be unable to do it themselves. – Will Jagy Jan 17 '15 at 20:22
• Fair enough.... – Gerry Myerson Jan 17 '15 at 20:26
| 2021-01-18T11:46:30 |
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http://www.lofoya.com/Solved/1744/an-eight-digit-telephone-number-consists-of-exactly-two-zeroes-one
|
# Difficult Probability Solved QuestionAptitude Discussion
Q. An eight-digit telephone number consists of exactly two zeroes. One of the digits is repeated thrice. Remaining three digits are all distinct. If the first three digits (from left to right) are 987, then find the probability of having only one 9, one 8 and one 7 in the telephone number.
✖ A. 1/18 ✖ B. 1/20 ✔ C. 1/10 ✖ D. 5/47
Solution:
Option(C) is correct
Case A: There is only one 9, one 8 and one 7 in the number.
Hence, there has to be one digit from {1,2,3,4,5,6} repeated thrice.
Total number of ways in which such a number can exist:
$={^6C_1}\times \dfrac{5!}{3!\times 2!}$
$=60$
Case B: One of the three digit {9,8,7} is repeated thrice.
Hence there will be one digit from {1,2,3,4,5,6}.
Total number of ways in which such a number can exist:
$={^3C_1} \times {^6C_1}\times \dfrac{5!}{3!\times 2!}$
$=540$
Total possible telephone numbers:
$=60+540=600$
Probability = 60/600 = 1/10
## (10) Comment(s)
Akshay
()
Elaborate Method -
Given that first three positions of the eight positions is filled with 9,8 and 7 we have remaining 5 positions to be filled with numbers with two zeros and three numbers which meet the given conditions. Let's consider four cases -
Case 1 - Suppose the number that is repeated thrice is 9. We already have one 9 which leaves us with two 0's and two 9's in the five vacant positions. Remaining one position can take values from 1 to 6. So
$6C1*(5!/(2!*3!))=6*30=180$
Case 2 - Similar to case 1 but here considering the number repeating thrice is 8. So totally 180 combinations.
Case 3 - Similar to case 1 but here considering the number repeating thrice is 7. So totally 180 combinations.
Case 4 - Considering the first three numbers 9,8 and 7 as distinct numbers, the three vacant spaces has to be filled with three same numbers which can range from 1 to 6 and the other two vacant position has to take two 0's. So
$6C1*1C1*1C1*(5!/(2!*3!))=6*10=60$
So totally we have $(180*3)+60=600$ combinations. Now coming to the question it asks for the case which has one 9, one 8 and one 7 which is the case 4. So probability is
$60/600=1/10$
Divya
()
In case B the total no. Of ways should be 180+60= 240
Rashmi
()
have u assumed the positions of zeros fixed????????//
()
I guess since there are two zeros present, their position is NOT assumed here to be fixed but as a repeated number. That is why we see a factor of $2!$ in the denominator in both the cases.
Isn't it?
Bibek Sahu
()
The answer must be $\dfrac{3c1*6c1*(5!/(2!*2!))}{3c1*6c1*(5!/(2!*2!))}+60$
Samyak Jain
()
its 60/(3c1*6c1*5!/(2!*2!)+60)
Anil Kumar
()
Danalc
()
agree with crujzo.
Hardik
()
In case B,
3.6.5!/(2!2!)
looks right. How does it have 3! instead of 2?
Crujzo
()
I believe in the second case it should be,
$2c1 * 6c1 * (5! / (2!*2!))$
As from the rest of the left 5 positions, there are 2 & 2 repeated items
| 2016-10-22T23:29:55 |
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https://byjus.com/question-answer/a-certain-radioactive-material-is-known-to-decay-at-a-rate-proportional-to-the-amount-1/
|
Question
A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10%of its original mass, then the
A
mass of the material after four hours is 50(910)2
B
mass of the material after four hours is 50.e0.5ln9
C
time at which the material has decayed to half of its initial mass (in hours) is (n12)(12n0.9)
D
time at which the material has decayed to half of its initial mass (in hours) is (n2)(12n0.9)
Solution
The correct options are C time at which the material has decayed to half of its initial mass (in hours) is $$\dfrac{(\ell n2)}{(-\frac{1}{2}\ell n0.9)}$$ D mass of the material after four hours is $$50(\dfrac{9}{10})^2$$$$\dfrac{dN}{dt}\propto N$$at $$t = 0 N = 50kg$$and at $$t = 2hr N = 0.9\times 50kg$$$$\dfrac{dN}{dt} = -\lambda N$$$$\Rightarrow \dfrac{dN}{N} = -\lambda dt$$$$\ell nN = -\lambda t+c$$$$t = 0 , N = 50$$$$\ell nN = -\lambda t + \ell n50$$$$N = 50e^{-\lambda t}$$ ......(1)$$t = 2 45=50e^{-\lambda .2}$$$$\dfrac{9}{10} = e^{-2\lambda}$$$$\sqrt{\dfrac{9}{10}} = e^{-\lambda}$$$$\therefore N = 50(\frac{9}{10})^{t/2}$$ ......(2)At $$t = 4$$ $$N = 50(9/10)^2$$When $$N = 25kg 25 = 50(9/10)^{t/2}$$$$\Rightarrow \dfrac{1}{2}=(0.9)^{t/2} t = \dfrac{2\ell n\dfrac{1}{2}}{\ell n0.9}= \dfrac{2\ell n2}{-\dfrac{1}{2}\ell n0.9}$$Mathematics
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| 2022-01-28T17:29:48 |
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"domain": "byjus.com",
"url": "https://byjus.com/question-answer/a-certain-radioactive-material-is-known-to-decay-at-a-rate-proportional-to-the-amount-1/",
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|
https://gmatclub.com/forum/x-3-4-x-8-x-how-many-solutions-does-the-equation-148996-20.html
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# |x+3| - |4-x| = |8+x|. How many solutions does the equation
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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07 Feb 2014, 03:53
1
PerfectScores wrote:
Now |x+ 3| will always be lesser than |8+x|
Not necessary.
Put x = -10
|x+3| = |-10 + 3| = 7
|8+x| = |8-10| = 2
It is not necessary that |x+3| will be less than |8+x|.
Note the meaning of absolute value.
|x+3| is the distance of x from -3.
|x+8| is the distance of x from -8.
It is not essential that every point on the number line will be closer to -3 than to -8. All points to the left of -5.5 will be closer to -8 while all points to the right of -5.5 will be closer to -3.
So when x > -5.5, |x+3| < |x+8|
When x < -5.5, |x+3| > |x+8|
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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18 May 2014, 21:49
NGGMAT wrote:
Bunuel wrote:
STEP BY STEP SOLUTION:
We have three transition points for $$|x+3| - |4-x| = |8+x|$$: -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:
1. $$x<-8$$;
2. $$-8\leq{x}\leq{-3}$$;
3. $$-3<x<4$$
4. $$x\geq{4}$$
IS THE RED HIGHLIGHTED PORTION A RULE TYPE??? AS IN WHEN WE HAVE SUCH EQUATIONS... WE FIRST FIND THE CRITICAL POINTS AND THE RANGE CHECKS ARE ALWAYS IN THE FOLLOWING PATTERN?
X<= LOWEST VALUE
LOWEST VALUE <X< 2ND LOWEST VALUE
2ND LOWEST VALUE <= X<= HIGHEST VALUE
X> HIGHEST VALUE
I UNDERSTOOD THE FACT THAT TRANSITION POINTS ARE INCLUDED ONLY ONCE...
SO SHOULD I TAKE IT AS A RULE AND IF ANY SUCH QS COMES UP... APPLY THIS RULE?
In addition, check out this post: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
It explains this question type in detail.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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21 May 2014, 03:55
3
To All who get confused about what sign to put when opening the modulus
After you've determined the critical points, plot them on the number line.
Attachment:
Critical Points on Number Line.PNG [ 2.04 KiB | Viewed 4364 times ]
Now, Check the coefficient of x in each modulus.
For Example, let's take |8+x|. The coefficient of x is +1. Positive. So, on the number line, put a + sign on the Right of the critical point -8, and a - sign on the Left of -8.
Similarly, for |x+3| too, the coefficient of x is positive. Again, we'll put a + sign on the Right of critical point -3, and a - sign on Left of -3.
For |4-x|, the coefficient of x is -1. It's negative. So here the direction of signs will be reversed, that is, we'll put a + sign on Left of critical point 4 and a - sign on Right of 4.
This is what you'll get:
Attachment:
Opening the Modulus.PNG [ 2.73 KiB | Viewed 4360 times ]
So, now, while opening the modulus, just refer to this plot.
Example:
For -8<=x<=-3
|x+8| will be opened as x+8 [since the graph shows that (x+8) is positive for all x>-8]
|4-x| will be opened as 4-x [since the graph shows that (4-x) is positive for all x<4]
|x+3| will be opened as -(x+3) [as the graph shows that (x+3) is negative for all x<-3]
This plot takes only a second and leaves no chance of confusion about the signs in opening the Modulus.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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21 May 2014, 04:39
A question to practice the above tip:
Plot on a number line the signs for each modulus in the following equation:
$$-|2x-3| + |5-x| + |x-10| = |3-x|$$
Attachment:
Solution.PNG [ 3.09 KiB | Viewed 4237 times ]
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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09 Jun 2014, 03:08
[quote="guerrero25"]|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
Logically and imaginary it means that the distance from X to -3 minus distance from X to 4 must be equal to distance from X to -8. Whereever you place X in number line you never get such outcome
________-8__________-3________________4______
But what if we make change like that:
|4-x| - |x+3| = |8+x|
I found -15 as only solution with image use
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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09 Jun 2014, 03:11
Temurkhon wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
Logically and imaginary it means that the distance from X to -3 minus distance from X to 4 must be equal to distance from X to -8. Whereever you place X in number line you never get such outcome
________-8__________-3________________4______
But what if we make change like that:
|4-x| - |x+3| = |8+x|
I found -15 as only solution with image use
|4-x| - |x+3| = |8+x| has TWO solutions -15 and -7/3.
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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03 Sep 2014, 07:40
I'm not the first one to say this but the way I solved this problem was by thinking about the equation. The equation says the distance from -3 minus the distance from 4 equals the distance from -8. I drew a picture and the closest I got was -2 which is 1-6=|-5| (but -2 is 6 away from -8). If you think this way you can see that x can't be less than -8 or greater than 4... but for nonintegers, I think the 4 step method is much more precise.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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10 Dec 2014, 09:52
VeritasPrepKarishma wrote:
I am stuck on this part in the Gmat Club book. I do understand how the conditions are set. But I can't figure out how the values for x were determined in those conditions. Ill use just a and b:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
Where do the x = ' ' values come from? I have been staring at this for half an hour.
I understood the whole concept in the '3-steps approach' but the '3-steps approach for complex problems' has me stuck suddenly. There goes mij GMAT-Mojo! Anyone able to help me get it back? Thanks.
BTW: Is this 650+ level?
You solve the equation to get the x = values
First of all, you are given |x+3|-|4-x|=|8+x|
Convert this to |x+3|-|x-4|=|x+8| (since it is a mod, |4-x| is the same as |x-4|)
Now key points are -8, -3 and 4.
Case a: x< -8
When x < -8, all three expressions (x + 3), (x - 4) and (x + 8) are negative.
So |x+3| = - (x + 3) (using the definition of mod)
|x-4| = - (x - 4)
|x+8| = - (x + 8)
-(x+3) - [-(x-4)] = -(x+8)
-7 = -x - 8
x = -1
Condition not satisfied so rejected.
And no, it is 750+ level.
So what you're saying is the order of whats inside the modulus doesnt matter as long as the sign outside changes?
I ask because i was very confused on why in gmat club book they didnt conver 4-x to -(x-4)
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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10 Dec 2014, 22:15
So what you're saying is the order of whats inside the modulus doesnt matter as long as the sign outside changes?
I ask because i was very confused on why in gmat club book they didnt conver 4-x to -(x-4)
Note that
|x| = |-x|
If x = 5,
|5| = |-5| = 5
If x = -5,
|-5| = |-(-5)| = 5
In any case, |x| is always same as |-x|. (Note that we are not saying that |x|= x or |x| = -x because that depends on the sign of x)
So |x-4| = |-(x-4)| = |4-x|
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
### Show Tags
04 Jun 2015, 09:28
ankushbagwale wrote:
Bunuel and VeritasPrepKarishma
I would like to seek some help from your end.
We do know that | x-4| = |4-x|
So in the above equation |x +3| - |4-x| = |8+x|
I have deliberately converted the middle | 4-x| = |x-4| for my convienance.
& In case I when x < -8 I checked & found that as then -8 is definately on the left hand side of ( 4-x ) or ( x-4). So definately this term should be negative. I checked with a example of -10 also.
But in both the cases the answers are different. If this is the case then I should limit my self from changing | x-4| = |4-x|
& strictly consider |4-x| in its original form??
If that were the case still by defination : | 4-x| = - (4-x) when (4-x) <=0 meaning x > 4 & similarly when x<4 I will have (4-x).
Now coming to the case I:
we have x< -8 definately less than 4. That means the bracket will open with a positive sign i.e. (4-x)
SO now adding I am getting: -(x+3) -(4-x) = -(8+x) = -1 but not in line with the initial condition.
Thus this range is not possible.
SO I am still not sure where I am getting confused.
Secondy, I am wondering why here |x-4| not equals |4-x|? Under what conditions we can do & when we cannot do such interchange?
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1193962
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1237206
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1238650
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1241339
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1241355
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1323676
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1328926
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1454309
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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04 Jun 2015, 09:39
Thanks Bunuel for pointing out relevant thread.
I am simply amazed by the sheer depth of though process that has gone into every single question at gmat club.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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04 Jun 2015, 15:12
1
A better a faster method involves squaring both the sides of the equation which finally ends up with Left hand as always negative and Right hand side as always positive, which is never possible.
Hence no solution.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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05 Jul 2015, 12:16
TooLong150 wrote:
Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
thanks !
Responding to pm.
Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;
When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.
STEP BY STEP SOLUTION:
We have three transition points for $$|x+3| - |4-x| = |8+x|$$: -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:
1. $$x<-8$$;
2. $$-8\leq{x}\leq{-3}$$;
3. $$-3<x<4$$
4. $$x\geq{4}$$
Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.
1. When $$x<-8$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is negative. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=-(8+x)$$.
Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =-(8+x)$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x<-8$$).
2. When $$-8\leq{x}\leq{-3}$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.
Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =8+x$$: --> $$x=-15$$. This solution is NOT OK, since $$x=-15$$ is NOT in the range we consider ($$-8\leq{x}\leq{-3}$$).
3. When $$-3<x<4$$, then $$x+3$$ is positive, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.
Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (4-x) =8+x$$: --> $$x=9$$. This solution is NOT OK, since $$x=9$$ is NOT in the range we consider ($$-3<x<4$$).
4. When $$x\geq{4}$$, then $$x+3$$ is positive, $$4-x$$ is negative and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=-(4-x)=x-4$$ and $$|8+x|=8+x$$.
Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (x-4) =8+x$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x\geq{4}$$).
Thus no value of x satisfies $$|x+3| - |4-x| = |8+x|$$.
Hope it's clear.
Is this the fastest way to do this problem? It took me 4 minutes to do this problem.
I did the question by a graphical method. With absolute values, it is sometimes easier to draw graphs and evaluate the questions graphically.
The given question will have 'n' solutions if the 3 lines given by the equations:
y=|x+8|
y=|x+3|
y=|4-x|
once you do that, it becomes apparent that there are no points that are points of intersections of 3 lines (for us to get a solution, we need to have 3 of the lines intersecting at some common points!). Attached is the graph for the same (Sets of parallel lines are: {A||B||C} and {D||E||F}). This method will be an overkill for simpler problems though.
Hope this helps
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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11 Aug 2015, 15:12
I was going to reply and say that the graphical/visual way may work best for many cases but I saw that the poster above me has already covered it.
For many absolute values that are simple addition and subtraction, you might get to the answer quicker and more accurately if you just draw it out, especially if its in the form of abs(x-b) + abs(x+a) etc without a 4x-c or kx-c
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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18 Mar 2016, 21:30
VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
thanks !
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.
When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4
We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.
Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4
The only point is that we don't include the transition points twice.
Hope the role of '=' sign is clear.
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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18 Mar 2016, 21:44
riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help
Hi,
you have to pick a VALUE falling in the range you are taking and see what happens to the value within the MOD..
1) if the solution of the MOD is a negative number, add a -ive sign..
2)if the solution of the MOD is a positive number, add a +ive sign..
let me show with some examples --
C) -3 <= x < 4 ----- (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
take x as 0 as it falls in the given range -3 <= x < 4
see what happens to each MOD at this value
i) |x+3|.. 0+3=3 so + sign in front of MOD .. (x+3)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)
take the value as -5..
|x+3|.. -5+3=-2 so a negative sign.. -(x+3)
|4-x|.. 4-(-5)=9.. so +sign.. (4-x)
|8+x|.. 8-5=3 so +sign... (8+x)
Hope it helps
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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29 Mar 2016, 09:16
riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help
Let me first tell you that in this question, x < 0 has no significance.
|x| = x when x >= 0
|x| = -x when x < 0
On the same lines,
|x + 4| = x + 4 when (x + 4) >= 0
|x + 4| = -(x + 4) when (x + 4 ) < 0
In the first definition, x is just a placeholder for any expression.
|x^2 - 8| = x^2 - 8 when (x^2 - 8) >= 0
|x^2 - 8| = -(x^2 - 8) when (x^2 - 8) < 0
So, how do you get rid of |x + 3| in the original question? You take two cases: (x + 3) >= 0 or (x + 3) < 0
|x + 3| = x + 3 when (x + 3) >= 0 (i.e. when x >= -3)
|x + 3| = -(x + 3) when (x + 3) < 0 (i.e. when x < -3)
That is how you get -3 as a transition point.
Do the same for other expressions.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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13 May 2016, 18:33
I solved it very differently and got the right answer. Not sure whether it was a fluke.
I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.
Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.
Is this approach correct?
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
thanks !
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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15 May 2016, 23:26
ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.
I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.
Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.
Is this approach correct?
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
thanks !
Note that $$(a + b)^2 = a^2 + b^2 + 2ab$$
So $$(|x+3| - |4-x|)^2 =|x + 3|^2 + |4 - x|^2 - 2*|x + 3|*|4 - x|$$
You missed out the last term. You would need to square it yet again and that will complicate the question further.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]
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16 May 2016, 17:47
ohh right...thanks.
but if there is a situation where |a-b| = |e+f| (variables or constants), then I can safely square them right? Or do I need to keep certain things in mind before doing that.
VeritasPrepKarishma wrote:
ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.
I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.
Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.
Is this approach correct?
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
thanks !
Note that $$(a + b)^2 = a^2 + b^2 + 2ab$$
So $$(|x+3| - |4-x|)^2 =|x + 3|^2 + |4 - x|^2 - 2*|x + 3|*|4 - x|$$
You missed out the last term. You would need to square it yet again and that will complicate the question further.
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink] 16 May 2016, 17:47
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solving for substitution
• Aug 10th 2010, 08:49 PM
mafai44
solving for substitution
having trouble understanding substitution this is the problem :
Solve for the system of equations by substitution
5x-7=-y
2x-y=0
This is another one..
6x+7y=1
x=55-9y
• Aug 10th 2010, 08:52 PM
Prove It
$5x-7=-y$
$2x-y=0$.
Substituting $-y$ into the second equation gives
$2x+5x-7=0$
$7x-7=0$
$7x = 7$
$x = 1$.
You know $-y = 5x-7$ so
$-y = 5\cdot 1 - 7$
$-y = 5 - 7$
$-y = -2$
$y = 2$.
Therefore $(x, y) = (1, 2)$ is the solution.
• Aug 10th 2010, 08:54 PM
eumyang
5x-7=-y
2x-y=0
First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:
\begin{aligned}
2x - y &= 0 \\
2x &= y \\
\end{aligned}
Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
\begin{aligned}
5x - 7 &= -y \\
5x - 7 &= -2x \\
-7 &= -7x \\
x &= 1
\end{aligned}
Now, plug this into the altered version of the 2nd equation:
\begin{aligned}
2x &= y \\
2(1) &= y \\
y &= 2
\end{aligned}
The answer is (1, 2). You want to try your other example now?
EDIT: Too slow! ;)
• Aug 10th 2010, 08:59 PM
mafai44
thank you! does anyone know solving for elimination?
4x=15+3y
-6/5x+y=-17/5
• Aug 10th 2010, 09:01 PM
Prove It
Is it
$4x=15+3y$
$-\frac{6}{5x}+y = -\frac{17}{5}$?
• Aug 10th 2010, 09:04 PM
mafai44
yes the 6/5x , the x is for both the 6 and 5 not just the 5.. not even sure if that matters but you have it right.
• Aug 10th 2010, 09:12 PM
Prove It
Obviously it's not right if it's meant to be
$4x=15+3y$
$-\frac{6}{5}x+y = -\frac{17}{5}$
since $\frac{6}{5}x = \frac{6x}{5}$, not $\frac{6}{5x}$...
Anyway...
$4x=15+3y$
$-\frac{6}{5}x+y = -\frac{17}{5}$
Multiply the second equation by $3$...
$4x=15+3y$
$-\frac{18}{5}x+3y=-\frac{51}{5}$
$4x-3y=15$
$-\frac{18}{5}x+3y=-\frac{51}{5}$.
$(4x-3y) + \left(-\frac{18}{5}x+3y\right) = 15-\frac{51}{5}$
$\frac{2}{5}x=\frac{24}{5}$
$2x = 24$
$x = 12$.
Substituting into the first equation
$4x=15+3y$
$4\cdot 12 = 15 + 3y$
$48 = 15 + 3y$
$33 = 3y$
$y = 11$.
So $(x, y) = (12, 11)$ is the solution.
• Aug 10th 2010, 09:19 PM
mafai44
Quote:
Originally Posted by eumyang
5x-7=-y
2x-y=0
First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:
\begin{aligned}
2x - y &= 0 \\
2x &= y \\
\end{aligned}
Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
\begin{aligned}
5x - 7 &= -y \\
5x - 7 &= -2x \\
-7 &= -7x \\
x &= 1
\end{aligned}
Now, plug this into the altered version of the 2nd equation:
\begin{aligned}
2x &= y \\
2(1) &= y \\
y &= 2
\end{aligned}
The answer is (1, 2). You want to try your other example now?
EDIT: Too slow! ;)
yeah if you can help me out with the other example that would be great.. this is one part of math that i do not like haha
• Aug 10th 2010, 09:20 PM
mafai44
double posted by accident sorry
• Aug 11th 2010, 04:12 AM
eumyang
OP: Prove It's point is that you have to be careful with notation. Ideally, you should learn LaTeX so that it's not ambiguous when you type $\frac{6}{5}x$. 6/5x is really read as $\frac{6}{5x}$. If you are not using LaTeX and you want to indicate the fraction 6/5 times x, use parentheses: (6/5)x.
| 2016-09-26T02:08:30 |
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https://math.stackexchange.com/questions/1882069/why-does-polynomial-factorization-generalize-to-matrices
|
# Why does polynomial factorization generalize to matrices
I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself:
Theorem 2.1 $\,$ Every linear operator on a finite-dimensional complex vector space has an eigenvalue.
Proof:
To show that $T$ (our linear operator on $V$) has an eigenvalue, fix any nonzero vector $v \in V$. The vectors $v, Tv, T^2v,..., T^nv$ cannot be linearly independent, because $V$ has dimension $n$ and we have $n + 1$ vectors. Thus there exist complex numbers $a_0,...,a_n$, not all $0$, such that
$$a_0v + a_1Tv + ··· + a_nT^nv = 0.$$ Make the $a$’s the coefficients of a polynomial, which can be written in factored form as $$a_0 + a_1z + ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all complex $z$. We then have
$${\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T − r_mI)v}},$$ which means that $T − r_j$ I is not injective for at least one $j$. In other words, $T$ has an eigenvalue.$\;\blacksquare$
I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices?
In other words, why is the the part I highlighted true? Does such an factorization always exist?
Could I have some help to see this? Thank you =)
P.S. here is my reference (page 3).
UPDATE:
Someone else also has asked the same question before it seems.
• Take the vectors $[1,0,\cdots]$, $[1,0,\cdots]$, $\cdots$ , $[1,0,\cdots]$. They will be $n+1$ vectors, none zero, generated for example by the matrix with a lone 1 up in the left corner, but not linearly independent. – mathreadler Aug 4 '16 at 9:47
• Would you like to do some work and type the quote instead of using an image? It will make the question better, and ensure it's always accessible to users – Yuriy S Aug 4 '16 at 13:32
• Roger that. I will edit my post :) – jjepsuomi Aug 4 '16 at 13:44
• @You'reInMyEye edited the text =) – jjepsuomi Aug 4 '16 at 19:12
What matters is that the matrices involved, namely powers of $T$, commute with each other. With that in mind, the legitimacy of the factorisation should be clear: just think about expanding the brackets using the associative and distributive properties of matrix multiplication. A more sophisticated argument can be obtained by viewing the equation in terms of an action of the polynomial ring $\mathbb{C}[x]$, in which factorisation is more familiar.
• One more question is buzzing my mind. So the factorization always exists? If we have a matrix polynomial then we can always find such values $r_1, ..., r_m$ and $c$ that the factorization of the matrix polynomial is possible? – jjepsuomi Aug 4 '16 at 12:12
• Yep. This works because $\mathbb{C}$ is algebraically closed (i.e. fundamental theorem of algebra). – Mr. Chip Aug 4 '16 at 12:24
• Thank you. One last question :) Does the roots of the matrix polynomial always have to be of the form $r_i I$? Can there be arbitrary (meaning not of the form $r_i I$) root-matrices for the matrix polynomial? If not, why? – jjepsuomi Aug 4 '16 at 12:46
• Maybe the easiest case to consider would be $T^2 = I$. Make it easy and let $T$ be 2x2. You'll find that $T$ doesn't have to be a multiple of $I$. – Mr. Chip Aug 4 '16 at 12:59
• @jjepsuomi Actually what matters is that constants $\,cI\,$ commute with $\,T.\,$ See my answer where I explain this from a general perspective, which shows why polynomial factorizations, the Binomial Theorem, etc persist to be true when "evaluated" at $\,x = T.\$ – Bill Dubuque Aug 4 '16 at 15:56
This follows simply from the universal property of polynomial rings, which implies that any polynomial equation in $\, R[x]\,$ will persist when "evaluated" into any ring where the images of the constants commute with the image of $x$ (which is precisely the condition necessary for such a map to be a ring homomorphism).
Indeed, a polynomial ring is designed precisely to have this property, i.e. it is the most general ("free") ring that contains $\,R\,$ and a new element $\,x\,$ that commutes with all elements of $R$. Because we use only the ring axioms and constant commutativity when proving polynomial equations, such proofs persist in said ring images where constant commutativity persists.
This is true in your example because constants $\,r\,$ map to a constant matrices $\,rI\,$ which commute with $\,T =$ image of $x$.
This implies that all familiar polynomial equations (e.g. Binomial Theorem and difference of squares factorization) persist to be true when evaluated into any ring where the constants commute with the indeterminates. Ditto for many other ubiquitous polynomial equations, e.g. cyclotomic polynomial factorizations, polynomial Bezout identities for the gcd, resultants, etc. Therefore such equations represent universal laws (identities), modulo said constant commutativity.
These ideas are brought to the fore when one studies $(R-)$algebras, which are rings containing a central image of $R$, i.e. where the images of elements of $R$ commute with everything. Any polynomial equation that holds true in $\,R[x_1,\ldots,x_n]\,$ will persist to be true when evaluated into any $R$-algebra, i.e. it is an identity (law) of $R$ algebras. In fact it is easy to show that an equation holds true in $\,R[x_1,\ldots,x_n]\,$ iff it is true in all all $R$ algebras. Hence the equations that hold true $\,R[x_1,\ldots,x_n]\,$ are precisely the identities (universal laws) of $R$-algebras.
• Thank you very much, appreciate it! – jjepsuomi Aug 4 '16 at 18:55
An explanation is that the polynomials in a given matrix $A$ constitute a commutative subring $\mathbb{K}[A]$ of the ring of matrices, and can be considered as an image of the ring of polynomials $\mathbb{K}[X]$ , thus you can consider that in this subring, you work exactly as in $\mathbb{K}[X]$.
In fact, $\mathbb{K}[A]$ is isomorphic to the quotient ring $\mathbb{K}[X]/(m(X))$ where $m(X)$ is a minimal polynomial for matrix $A$.
Edit : let us take an example (the matrix has been borrowed from (https://www.youtube.com/watch?v=FecegfvA-Pg)).
Consider matrix $A=\begin{pmatrix}0&-2&-2\\1&3&1\\0&0&2\end{pmatrix}$ whose characteristic polynomial is $$c(X)=X^3+X^2+1$$
Using Cayley-Hamilton theorem, one has $$c(A)=A^3+A^2+I=0$$ (think to replace 1 by $I$!), otherwise said, each time you meet $A^3$ in a computation, you can replace it by $-A^2-I$. These leads to a systematic degree lowering: any polynomial of any degree in $A$ can be brought to a (unique) form as an (at most) 2nd degree polynomial, for example $A^5+A=A^3A^2+A=(-A^2-I)A+A=-A^3=-A^2-I$. But in fact, in this case there is lower degree combination of powers of $A$ that is annihilated, more precisely $$m(A)=A^2-3A+2I=0 \ \ \ (1)$$ ($m(X)=X^2-3X+2$ is called the/a minimal polynomial). Thus in fact, the degree lowering makes that any polynomial $P(A)$ in $A$ can be written a first degree polynomial $\pi(A)$. This transformation can be considered as a linear mapping, i.e., an homomorphism between linear spaces (note that I have changed $A$ into $X$).
$$\varphi: \ \ \begin{cases}P(X) \longrightarrow \pi(X)\\ \mathbb{K}[X] \longrightarrow \mathbb{K}_1[X]\end{cases}$$
(denoting by $\mathbb{K}_1[X]$ the vector space of polynomials of degree at most 1)
Note that the kernel of this mapping is the set $M$ of polynomials multiples of $m(x)$. You may know that taking classes modulo the kernel leads to an isomorphism (but may be not yet). The definition of $M$ sounds like a principal ideal. This is not astonishing because linear mapping $\varphi$ can also - fruitfully - be considered as a ring mapping (homomorphism between rings) with the following multiplication rule (because of relationship (1)): $$(aX+b)(a'X+b')=aa'(3X-2I)+...=(3aa'+ab'+a'b)X+(bb'-2aa')$$
The quotient space, like for vector spaces, would yield an isomorphism.
Remark: the unifying structure for vector spaces that are also rings with a certain compatibility relationship betwen the rules is that of an algebra structure.
• Thank you for your help. Appreciate it =) – jjepsuomi Aug 4 '16 at 10:00
• Could you elaborate little bit more your answer? :) Thank you, just to make the subject a bit more tangible for me ;) – jjepsuomi Aug 4 '16 at 12:18
• I will add an Edit to my answer. – Jean Marie Aug 4 '16 at 12:50
• Thank you =). Other question: could the the factorization be proven/shown without using algebraic structures or notions (rings, isomorphism etc.)? Could you prove your point by simply using linear algebra? Can such a proof be found from some reference? – jjepsuomi Aug 4 '16 at 12:59
• Edit done. Is your "other question" connected to the central issue that $(fog)v=0$ implies that either $f$ or $g$ is such that either $f(v)=0$ or $g(v)=0$ ? – Jean Marie Aug 4 '16 at 14:01
| 2019-10-14T13:44:45 |
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https://math.stackexchange.com/questions/3224895/infinite-sum-and-partial-sum-not-equal-in-wolframalpha
|
# Infinite sum and partial sum not equal in Wolframalpha
$$\sum_{n=0}^{\infty} |(-0.8)^n \theta(n)-(-0.8)^{n-1} \theta(n-1)|$$ $$\theta(n) = 1$$ for $$n \geq 0$$ and 0 otherwise.
My attempt to calculate this summation numerically with Python yielded 10. To ensure correctness I tried to verify this answer by calling Wolframalpha within Mathematica with the following query: Sum[Abs[UnitStep[n]*(-0.8)^(n) - UnitStep[n-1]*(-0.8)^(n-1)],{n,0,Infinity}]
It returned 3.6 as primary answer but it's partial sum converged to 10. I am now wondering which answer is correct.
The answer is $$10$$.
Note that you can rewrite \begin{align} \sum_{n=0}^{\infty} |(-0.8)^n \theta(n)-(-0.8)^{n-1} \theta(n-1)| &= |(-0.8)^0-0|+\sum_{n=1}^{\infty} |(-0.8)^n \theta(n)-(-0.8)^{n-1} \theta(n-1)|\\ &= 1+\sum_{n=1}^{\infty} |(-0.8)^n -(-0.8)^{n-1} |\\ &= 1+\sum_{n=1}^{\infty} |(-0.8)^{n-1}| |(-0.8) -1 |\\ &= 1+1.8\cdot\sum_{n=1}^{\infty} 0.8^{n-1} = 1+1.8\cdot\sum_{n=0}^{\infty} 0.8^{n}\\ &= 1+1.8\cdot 5 = 1+9\\ &= \boxed{10} \end{align}
It looks like WolframAlpha really does not like the mix of absolute values and alternating power signs. It has nothing to do with the $$\theta(\cdot)$$ function itself, as seen here with Sum[Abs[(-8/10)^(n) - (-8/10)^(n-1)],{n,1,Infinity}]+1.
Mathematica, on the other hand, is fine with it.
Your summation is a geometric series in disguise. Write $$r:=0.8$$. Plugging in the definition of $$\theta(\cdot)$$, we see that the $$n=0$$ term must be handled separately. Your summation is then written $$1 + \sum_{n\ge1}|r^n-r^{n-1}|=1+\sum_{n\ge1}|r-1||r|^{n-1}.$$ The rightmost sum is $$9$$, by the formula for the sum of a geometric series.
Not sure what it means for Wolfram's "primary answer" to be 3.6. Can you show the output?
• Well, just typing it in in wolframalpha gives 3.6 as answer – vrugtehagel May 13 at 20:42
• @vrugtehagel see my answer. Mathematica gives 10 — WolframAlpha must have some weird internal workings. – Clement C. May 13 at 20:44
| 2019-06-18T09:09:51 |
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https://math.stackexchange.com/questions/1105787/help-finding-the-area-under-a-curve
|
# Help finding the area under a curve
I was given an excercise in my calculus class that i don't really understand, the problem says : Find the area limited by the curves $$y = \frac{x+4}{x^2+1} ,\space x = -2 ,\space x = 3,\space y = 0$$
I don't really know what approach to follow here, my guess would be to solve it using riemann sums or maybe definite integrals and using $x = -2$ and $x = 3$ as the interval but i'm totally lost.
• It is almost always best to start these types of questions by sketching the curves, finding the intersection points, and figuring out which area you want to find. Then you can move on to deciding how to integrate. – Zubin Mukerjee Jan 15 '15 at 20:19
Yes, this is a definite integral problem. You can solve it by evaluating
$$\int_{-2}^3{x+4\over x^2+1}\,dx.$$
Hint: Split it into
$$\int_{-2}^3\frac x{x^2+1}\,dx+4\int_{-2}^3\frac1{x^2+1}\,dx,$$
and recall that the derivative of $\tan^{-1}x$ is $\frac1{x^2+1}$.
• you are missing a $4$ in the second integrand. – abel Jan 15 '15 at 20:23
• Awesome, thank you. I solved it this way and got the desired result – HardCodeStuds Jan 15 '15 at 20:28
It is meant that you should calculate the integral $$\int_{-2}^3 \frac{x+4}{x^2+1}\,dx$$ Also, see this picture where the curves are drawn.
Let $u = x^2+1$. The derivative of $u$ with respect to $x$ is $2x$. If $A$ is the area we want, then
\begin{align} A= \int_{-2}^{3} \frac{x+4}{x^2+1}\,\mathrm{d}x&=\frac{1}{2}\left(\int_{-2}^{3} \frac{2x}{x^2+1} \,\mathrm{d}x\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right) \\\\ &=\frac{1}{2}\left(\int_{5}^{10} \frac{\mathrm{d}u}{u}\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right) \end{align}
Remember, as Tim Raczkowski tells us, that $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\arctan x\right) = \frac{1}{x^2+1}$$
Also:
$$\frac{\mathrm{d}}{\mathrm{d}x} \left(\ln \left|x\right|\right) = \frac{1}{x}$$
This gives
\begin{align} A&=\frac{\ln 10 - \ln 5}{2} +4 \left(\arctan 3 - \arctan \left(-2\right)\right)\\\\ &\approx\boxed{9.7714} \end{align}
| 2019-08-18T15:04:27 |
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http://oayd.ecbu.pw/regression-to-the-mean-height.html
|
# Regression To The Mean Height
The first model (calcium) accounts for 16. REGRESSION. Select the Samples in different columns option button if that is the case; Click in the First text box and specify FHTS; Click in the Second text box and specify MHTS; Proceed as above. You can use the following formula to find the percent for any set of data: Percent of Regression to the Mean = 100(1-r). a) According to the linear model, the duration of a coaster ride is expected to increase by about 0. Phenomenal Regression to the Real Object : A ABC : Fig. So the slope of that line is going to be the mean of x's times the mean of the y's minus the mean of the xy's. The equation of the regression line allows us to calculate the estimated height, in inches, based on a given weight in pounds: $$\mbox{estimated height} ~=~ 0. " The same thing happens if we start with the children. Regression toward the mean simply means that, following an extreme random event, the next random event is likely to be less extreme. 10 8 10 12 14 16 18 20 22 20 30 40 50 60 70 80 60 65 70 75 80 85 90 Girth Volume Height l l l l l l l l l l ll l l ll ll l l l l l l l l l l l Figure 2: 3D scatterplot with regression plane for the trees data. Institute of Electrical and Electronics Engineers Inc. 2Hierarchical regression with nested data The simplest hierarchical regression model simply applies the classical hierar-chical model of grouped data to regression coefficients. He drew a circle on a blackboard and then asked the officers one by one to throw a piece of chalk at the center of the circle with their backs facing the blackboard. This can be seen in plots of these 4 linear regressions:. 946 or roughly equal to that, let's just think about what's. So this phenomenon is regression to the mean, and of course, we can do the same thing with it flipped, if we consider the son's height as the predictor, and the father's height as the outcome. Regarding the increase in insurer trust, this appears to be partially due to a regression to the mean, but since the drifts from the two extremes do not cancel each other out, this suggests that the initial extremes are greater on the low end than the high end. the regression line, is only 51. His son is among the top 10. He showed that the height of children from very short or very tall parents would move towards the average. After six years, the insignificant differences in height, diameter or crown volume between FTG and NFTG pine and similar annual mean diameter growth of FTG and NFTG pine in response to different aspen densities (treat- ments) indicated that current free growing standard (1 m radius) may not large enough to observe the possible neighborhood. These children do not in average have less homozygotes than expected and yet their average IQ is higher than in the population from which the parents came. What are the co-ordinates for the point of averages? What is the slope of the regression line? What is the intercept of the regression line? Write the equation of the regression line. The further the mean height of the parents from the population mean, the greater was the probability that the height of their children would return back towards the population mean, rather than deviating even further from this mean than the deviation of the mean height of their parents. Introduction to Regression Procedures LOGISTIC fits logistic models for binomial and ordinal outcomes. It’s a simple linear regression of earnings on height. If the actual weight is 200 pounds, the residual is 10. However, the use of regression in Galton's sense does survive in the phrase regression to the mean - a powerful phenomenon it is the purpose of this article to explain. This skill test was designed to test your conceptual and practical knowledge of various regression techniques. For example, if you measure a child’s height every year you might find that they grow about 3 inches a year. and the mean of the nine weights is 117 lb. Chapter 10 of Understandable Statistics introduces linear regression. "regression" effect. I In simple linear regression, the regression line is the line that minimizes the sum of the squared residuals. Galton first noticed it in connection with his genetic study of the size of seeds, but it is perhaps his 1886 study of human height 3 that really caught the Victorian imagination. The argument works exactly the same way for a poorly performing pupil - a really bad outcome is most likely, by chance alone, to be followed by an improvement. Another example of this is the tendency for tall parents to have tall children who are nonetheless often slightly shorter than their parents, and for shorter parents to have short children who are nonetheless slightly taller than their parents (imagine how humans would vary in height if this regression to the mean didn't happen…). One uses the regression model \girth~height" for our predictions. Note that the regression line always goes through the mean X, Y. Interpreting the Intercept in a Regression Model. I In simple linear regression, the regression line is the line that minimizes the sum of the squared residuals. The regression line (known as the least squares line) is a plot of the expected value of the dependent variable for all values of the independent variable. 2 \cdot \mbox{given weight} ~+~ 4$$ The slope of the line is measures the increase in the estimated height per unit increase in weight. 1 Answer to Husbands and wives. 2-D Image-to-image regression: h-by-w-by-c-by-N, where h, w, and c are the height, width, and number of channels of the output respectively, and N is the number of observations. To do this we need to have the relationship between height and weight of a person. 7 points above (or below) the mean on the midterm. Or how to use metrics to trust the prediction of regression Engineering · Posted by wpengine on Aug 6, 2015 Regression analysis is, without a doubt, one of the most widely used machine-learning models for prediction and forecasting. This is the regression line. The researchers measured certain bones of people of known heights and performed linear regression to arrive at simple formulas. If we calculate a correlation using just the data for height and weight, that correlation is called a simple correlation. Regression owes its name to the phenomenon known as regression toward the mean that arises when a genetically determined characteristic, such as height, is correlated between parent and offspring. The line connecting these averages (means) is known as the regression line. Regression toward the mean is a property of any scatterplot where the linear relationship is less than perfect. 61% of the variation in weight can be explained by the regression line. Note that, because of the definition of the sample mean, the sum of the residuals within a random sample is necessarily zero, and thus the residuals are necessarily not independent. The regression line should go through the center of the data cloud. We need to bring in yet another variable. 5 Extending the basic model Other factors besides age are known to affect FEV 1, for example, height and number of cigarettes smoked per day. Introduction to Regression Procedures LOGISTIC fits logistic models for binomial and ordinal outcomes. GOLDSTEIN Institute of Education, University of London [Received 25 October 1977; revised 1 February 1978] Summary. , a pretest and a posttest) and when the outcome of interest is the change in the outcome of interest from pretest to posttest (i. 5 inches? Example 12. Nowadays, the term "regression" is used more generally in statistics to refer to the process of fitting a line to data. Also, the more extreme the. 8768, which is still less than 0. •E(Volume) =-58. "Regression to the mean" is inevitable if inheritance works through blending of features. The crossover of the separate regression lines precludes having descriptive main effects! • Since the regression lines vary from + to -, the Y-X regression line for the mean of Z isn’t descriptive for all values of Z. ham Lincoln we might strengthen this to regression of the mediocre, to the mediocre, and for the mediocre. Height and weight are measured for each child. Thus, the true regression model is a line of mean values; that is, the height of the regression line at any value of xis just the expected value of Yfor that x. The foundation of regression is the regression equation; for Galton ’ s study of height, the equation might be: Child i = β 0 + β 1 (Parent i) + ∊ i. Regression to the mean does not mean that tall parents will have shorter children or that smart parents will have less smart children. 2Hierarchical regression with nested data The simplest hierarchical regression model simply applies the classical hierar-chical model of grouped data to regression coefficients. Let's look at the data first, used by Francis Galton in 1885. It would not be appropriate to use this regression model to predict the height of a child. The model for regression would be: weight = o + 1 (height) + From the regression output, estimated regression line is: We ˆ ight = -93. 00 Not flaggedFlag question Question text Run a regression model to estimate the cost of a building using average story height (mean centered), total floor area (mean centered), and construction type (dummy coded) as predictors. While "regression to the mean" and "linear regression" are not the same thing, we will examine them together in this exercise. Multiple regression coefficients indicate whether the relationship between the independent and dependent variables is positive or negative. First of all, the word "regression" is unfortunate here. Tom Tango and the other authors of "The Book: Playing the Percentages in Baseball" are probably the best sources of sabermetrics out there. He studied the relationships between pairs of variables such as the size of parents and the size of their offspring. Formulas to find the equation of the least squares line, y = a + bx are given in Section 10. He drew a circle on a blackboard and then asked the officers one by one to throw a piece of chalk at the center of the circle with their backs facing the blackboard. that relates the height at which a ball is dropped, x, to the height of its first bounce, y. Fitting a linear regression model in SAS. The video explored the distribution of the parents’ height; in this assignment, we investigate the distribution of the heights of the adult children. , because height certainly do not depend on the number of cats. Suppose Y is a dependent variable, and X is an independent variable. Measurements were made for a sample of adult men. In regression we seek to understand how the value of a response of variable (Y) is related to a set of explanatory variables or predictors (X’s). In the example below, variable ‘industry’ has twelve categories (type. of height of fathers is ˇ 2. Extrapolation is a highly helpful technique and is used in many businesses, research papers and other mathematically based items. Simple linear regression is when you want to predict values of one variable, given values of another variable. The mean regression model was fitted for childhood mean adjusted height-for-age. The linear regression equation for the relationship between height and hand length is U =. The formula for the best-fitting line (or regression line) is y = mx + b, where m is the slope of the line and b is the y-intercept. Solution: Using the formula discussed above, we can do the calculation of linear regression in excel. Regression to the mean is a statistical phenomenon. Of 13 children below the second centile for height, eight were within two centile spaces (90% range) of their mid-parental height SDS. The estimated mean height for males at age 16. For regression, the right side of the normal equations is X`WY. To set the stage for discussing the formulas used to fit a simple (one-variable) regression model, let′s briefly review the formulas for the mean model, which can be considered as a constant-only (zero-variable) regression model. They came up with a forecasting system designed to be the most basic acceptable system (Marcel), and it relies almost exclusively on regression to the mean. Finding the slope and intercept of the regression line. The height and weight of baseball players are in Table 10. 14:03 Wednesday 14th January, 2015. Empirical researchers took advantage of quantile regression’s ability to examine the impact of predictor variables on the response dis-tribution. Predict responses of a trained regression network using predict. In fact, in some of the permutations we have no data, which is why we use na. Regression toward the mean was first described by Francis Galton. Least Squares Regression Line. tab industry, or. The expected cost of a building with a story height of 0 cms is HK$3,185,038. Regression to the Mean Phenomenon discovered by Francis Galton, half cousin of Charles Darwin Developed a regression analysis of height between human children and their parents • Found that "It appeared from these experiments that the offspring did not tend to resemble their parents in size, but always to be more mediocre. In statistics, regression toward (or to) the mean is the phenomenon that if a variable is extreme on its first measurement, it will tend to be closer to the average on its second measurement—and, paradoxically, if it is extreme on its second measurement, it will tend to have been closer to the average on its first. Regression The idea behind the calculation of the coefficient of correlation is that the scatter plot of the data corresponds to a cloud that follows a straight line. 93575(height) Informally, we see that intercept is significantly different than 0, at a 5% significance level. Overall with results like these we can conclude that lidar does a reasonable job of estimating tree height. [The use of the grid is optional. Then construct a scatter plot of the data and draw the regression line. Fitting a Line (Regression line) • If our data shows a linear relationship between X and Y, we want to find the line which best describes this linear relationship – Called a Regression Line • Equation of straight line: ŷ= a + b x – a is the intercept (where it crosses the y-axis) – b is the slope (rate) • Idea:. Statisticians have said that the Sports Illustrated Jinx, in particular, is not a jinx at all, but rather an issue of Central Tendency and Regression to the Mean. Example of multiple linear regression Here is what happened in an old dataset when I regressed weight on height, sex, age, marijuana use, cocaine or heroin use, crack use, and drug injection. The regression line (known as the least squares line) is a plot of the expected value of the dependent variable for all values of the independent variable. X P when all X are fixed and an ellipsoid when all X are continuous or variable. For example, consider two children of different parents. (2 pts) Make a scatter plot of children's adjusted height vs the mid-parental height and then add the regression line on the plot. Construct model Begin with verbal model: There is a positive relation between heights of sons and fathers. † If I tell you the father's height, your prediction would be. Despite the predictions of the regression model, we still observe the regression toward the mean phenomenon with each generation of fathers and sons. For a confidence interval for the mean (i. split file off. In the 1870s Galton collected data on the height of the descendants of extremely tall and extremely short trees. Mothers and fathers were paired randomly, so they may have had very different scores. The product of. Let's use a height of 64 and enter all of these data into our Regression Equation. Galton's Family Heights Data Revisited Han et al. Hence the estimation of the. If there weren't regression to the mean, then extreme values would get more and more common, the distribution would spread out away from the mean, and (using the sons example) you'd end up with a much wider distribution with lots of really tall and really short people. Example Using the R dataset \trees", one wants the predicted girth of three trees, of heights 74, 83 and 91 respectively. It includes many techniques for modeling and analyzing several variables. Use the estimated regression to predict earnings for a worker who is 67 inches tall, for a worker who is 70 inches tall, and for a worker who is 65 inches tall. The following data are from a study of nineteen children. "Regression-to-mean If black spot sites are chosen for treatment solely on the basis of their high recent crash record, the chosen sites may genuinely be very hazardous. Linear regression. What makes this method so powerful is that it implies that we can fine-tune existing models for regression prediction — simply remove the old FC + softmax layer, add in a single node FC layer. 2 Regression to the mean. This misuse of regression is known as extrapolation. A total of 1845 number of people participated in. It's important to make clear that the breeder's equation, and hence regression to mean, works the same way for any quantitative trait, not just IQ. Regression to the mean. (e) If b 1 is between 0 and 1 we get regression towards the mean. The statistical phenomenon of regression to the mean is much like catch‐up growth, an inverse correlation between initial height and later height gain. The value of r can vary between 1. In the example data set above, the scatterplot and regression line lead us to believe there is a correlation between height and weight. To my knowledge he only used the term in the context of regression toward the mean. • Since the regressions line cross, differences between the height of the different lines at the 0-point of X isn’t. 47, and Height’s t = 5. 22625 R-Square 0. Therefore, 47. The further the mean height of the parents from the population mean, the greater was the probability that the height of their children would return back towards the population mean, rather than deviating even further from this mean than the deviation of the mean height of their parents. The predicted height is a bit more than 85 cm. Linear regression is known as a least squares method of examining data for trends. Galton’s Family Heights Data Revisited Han et al. Regression to the mean comes from the natural variability in the population in (virtually) any relationship. Then, select Mean, tell Minitab that the Input variable is height: When you select OK, Minitab will display the results in the Session window: Now, using the fact that the mean height is 69. Nonparametric regression refers to techniques that allow the regression function to lie in a specified set of functions, which may be infinitely-dimensional. Question 6 Not yet answered Marked out of 1. the high side, (average height at birth is 19"-21" or 35-51 cm), but not unreasonable. X variables now are the dummy variable male, and the mean. So if the average height of the two parents was, say, 3 inches taller than the average adult height, their children would tend to be (on average) approximately 2/3*3 = 2 inches taller than the average adult height. Enter the number of data pairs, fill the X and Y data pair co-ordinates, the least squares regression line calculator will show you the result. Regression Equation: Overview. 9 Waist’s t = 13. The equation of the regression line allows us to calculate the estimated height, in inches, based on a given weight in pounds:$$\mbox{estimated height} ~=~ 0. If the correlation between the heights of husbands and wives is about r = 0. He collects dbh and volume for 236 sugar maple trees and plots volume versus dbh. The predicted value, ^y, is a unbiased estimator of the mean response, y. This phenomena is called regression towards the mean. 8% of the variance. 5 inches, with a standard deviation about 2. And don't worry, this seems really confusing, we're going to do an example of this actually in a few seconds. The regression line is to a scatter diagram what the average is to a list. 8 +0 867 ×height How to draw the graph and determine the equation of a least squares regression line using the TI-Nspire The following data give the heights (in cm) and weights (in kg) of 11 people. Galton called this ‘regression towards mediocrity’. 5 kilograms. In statistics, regression toward (or to) the mean is the phenomenon that arises if a random variable is extreme on its first measurement but closer to the mean or average on its second measurement and if it is extreme on its second measurement but closer to the average on its first. Compare the regression coefficients in cells G2, H2, and I2 with the mean differences shown in Figure 7. Let's imagine that my height is 100% the result of a new additive mutation that makes people 50% further, in a positive direction, from their sexes mean height than average. 5 above and below the line, measured in the y direction, about 68% of the observation should. The coefficient of determination is 47. If the correlation between the heights of husbands and wives is about r = 0. 1 The second group mean will be closer to the mean for all subjects than is the first, and the weaker the correlation between the two variables the bigger the effect will be. The third shows the regression line through the plot. Regression to the Mean Regression towards the Mean • in all bivariate normal distributions • any random variation e. Calculate SSE for the full and reduced models. The children of people of average height tend to be farther from average. For example, official statistics released on the impact of speed cameras suggested that they saved on average 100 lives a year. It seems like linear regression might be a good choice for modeling the relationship between Sylvia’s height and weight during this time period. The coefficient of determination is 47. (Each pair of variables has a significant correlation. Halliwell, LLC Regression Models. Assume that height is the predictor and weight is the response, and assume that the association between height and weight is linear. 62 64 66 68 70 72 74 58 60 62 64 66 68 70 72 74 76 Height of father Height of son Formal model. Simple linear regression is when you want to predict values of one variable, given values of another variable. "Regression to the what?" my wife exclaimed, while rolling her eyes toward heaven. Regression models can be easily extended to include these and any other determinants of lung function. Statisticians have said that the Sports Illustrated Jinx, in particular, is not a jinx at all, but rather an issue of Central Tendency and Regression to the Mean. You can estimate , the intercept, and , the slope, in. regression /dep weight /method = enter height. regression in the analysis of two variables is like the relation between the standard deviation to the mean in the analysis of one variable. This includes political orientation, height, body weight, personality, etc. Thus, by subtracting the means, we eliminate one of the two regressors, the constant, leaving just one, parent. Regression to the mean effects was observed, and probable values were estimated for individuals based on a Bayesian model. Multiplication by this correlation shrinks toward 0 (regression toward the mean). A simple example is. -regression to the mean is NOT gambler's fallacy-regression towards the mean is not absolute for positively correlated variables (eg-- daughters of tall mothers are still taller than avg, just less tall on avg than their tall mothers). All the points (height, estimate for average weight) fall on the solid line shown in figure 1. Regression to the mean is a well known statistical artifact affecting correlated data that is not perfectly correlated. Take the height and earnings example in chapter 4. This idea can be formalized by regression methods. others 2005). A regression threat, also known as a "regression artifact" or "regression to the mean" is a statistical phenomenon that occurs whenever you have a nonrandom sample from a population and two measures that are imperfectly correlated. For example, if we are interested in the effect of age on height, then by fitting a regression line, we can predict the height for a given age. This section also contains the formula for the coefficient of determination. The experiments showed further that the mean filial regression towards mediocrity was directly proportional to the parental devia- tion from it. The predicted height is a bit more than 85 cm. Decisions in business are often based on predictions of what might happen in the future. Simple Linear Regression Example—SAS Output Root MSE 11. In statistics, regression toward (or to) the mean is the phenomenon that if a variable is extreme on its first measurement, it will tend to be closer to the average on its second measurement—and if it is extreme on its second measurement, it will tend to have been closer to the average on its first. The classic example is from Galton, who predicted that tall parents would have children who are shorter than they are, while short parents should have children who are taller than they are. The effect of regression towards the mean was recognized in the late nineteenth century by Francis Galton (1822–1911) when investigating the relationship of the heights of parents and their adult children (see Bland and Altman 1994, Stigler 1986). Also β1 = β0 – μ1 and β2 = β0 – μ2, and so β1 = the population Flavor 1 mean less the population grand mean and β2 = the population Flavor 2 mean less the population grand mean. Suppose a father has a. S Error for Regression The goal of this chapter is to have a reliable estimate of the average prediction error. For example, official statistics released on the impact of speed cameras suggested that they saved on average 100 lives a year. Take the height and earnings example in chapter 4. regression) to the mean. What is the best predicted weight of a supermodel with a height of 72 in. these two extremes, and that it sometimes lies nearer to the physical. Review of the mean model. And for a least squares regression line, you're definitely going to have the point sample mean of x comma sample mean of y. survive in the phrase regression to the mean - a powerful phenomenon it is the purpose of this article to explain. if you asymmetrically sample), then your results may be abnormally high or low for the average and therefore would regress back to the mean. Write the equation of the regression line for predicting height. , the high or low scores are chosen for further analysis), and then compared to other measurements of the same quantity. People's first problem is. Cases low in one variable predicted to be low in the other but not as low. The regression line on the graph visually displays the same information. Regression is a statistical way to establish a relationship between a dependent variable and a set of independent variable(s). This includes political orientation, height, body weight, personality, etc. The first model (calcium) accounts for 16. Statistic Notes: Regression towards the mean. Below, I've changed the scale of the y-axis on that fitted line plot, but the regression results are the same as before. If using a 1 in the denominator of slope is not super-meaningful to you, you can multiply the top and bottom by any number (as long as it. when you have calculated the regression equation for height and weight for school children, this equation cannot be applied to adults. •Effect on mean response of height does not depend on Diameter •We say effects are additive or not to interact •Partial regression coefficients 37. † All the linear trend in the data is accounted for by the regression line for the data. The origin of the term "regression" stems from a 19th century statistician's observation that children's heights tended to "regress" towards the population mean in relation to their parent's heights. However, it is also possible that the high crash rates observed at some sites may be due to chance, or a combination of both chance and a moderately hazardous nature. We use the Figure 2. The slope value means that for each inch we increase in height, we expect to increase approximately 7 pounds in weight (increase does not mean change in height or weight within a person, rather it means change in people who have a certain height or weight). And as independent variables, I'll use male and the mean centered height. Define the height of an adult child as a global variable; Use the function mean() to calculate the mean and the function sd() to calculate the standard deviation. 4, which is close but not exactly the given answer 73. , if we say that. The statistical phenomenon of regression to the mean is much like catch‐up growth, an inverse correlation between initial height and later height gain. Children and parents had the same mean height of 68. The notion of "regression to the mean" is widely mis- understood. Galton calculated the average height for the adults and children and plotted the heights of everyone on a chart. Fit full multiple linear regression model of Height on LeftArm, LeftFoot, HeadCirc, and nose. The current study aims to describe the shift in distribution of WC, WHpR, and WHtR over a period of. The estimated mean height for males at age 16. We have previously shown that regression towards the mean occurs whenever we select an extreme group based on one variable and then measure another variable for that group (4 June, p 1499). This scatterplot displays the weight versus the height for a sample of adult males. Evolution happens, and so regression to the mean clearly does not work this way. This means. One uses the regression model \girth~height" for our predictions. If using categorical variables in your regression, you need to add n-1 dummy variables. He studied the relationships between pairs of variables such as the size of parents and the size of their offspring. Many real-world phenomena involving an approximately linear relationship between two variables exhibit the phenomenon of regression to the mean. This plot shows a linear relationship between height and hand length. Assumptions. 4 Subsequently many observational and interventional epidemiological. Galton was a statistician who invented the term and concepts of regression and correlation, founded the journal Biometrika, and was the cousin of Charles Darwin. Heights of Fathers and Sons. Linear regression. 2 Regression to the mean. The regression problems that we deal with will use a line to transform values of X to predict values of Y. The formula for the best-fitting line (or regression line) is y = mx + b, where m is the slope of the line and b is the y-intercept. The next step is to build a suitable model and then interpret the model output to assess whether the built model is a good fit for the given data. Consider a data point like, say, 61 centimeters, which was Sylvia’s height on September 25, 2017. 8 +0 867x or weight =−84. However, regression to the mean is not re- stricted to height nor even to genetics. Regression to the mean is in fact only observable in situations where you have some factor creating an amount of variability. Where B0 is the bias coefficient and B1 is the coefficient for the height column. The problems and solutions are the same as bivariate regression, native American male standing height, average yearly minimum Squares df Mean Square F Sig. Construct model Begin with verbal model: There is a positive relation between heights of sons and fathers. Logistic regression can be used to understand the relationship between one or more predictor variables and a binary outcome. The code to do this is a little bit longer than the previous plots, but most of it deals with standardisation of the variables. Regression to the Mean comes in various flavours: Tall fathers will have tall sons, but the height of the sons will be closer to the mean (or average) of the current adult male population. This is where the term "regression" comes from. The term "regression" was used by Francis Galton in his 1886 paper "Regression towards mediocrity in hereditary stature". The regression line is therefore much more stable than the conditional mean. a No association No regression Linear regression Nonlinear regressionbc d X Y ab Relationship between weight and height Linear regression of weight on height 1501 65 180 55 65 75 Height (cm) Weight (kg) Weight (kg) 55 60 65 70 75 1501 60 1701 80 σ= 3 σ= 3 Figure 1 | A variable Y has a regression on variable X if the mean of Y (black line) E(Y. Multiple (General) Linear Regression Menu location: Analysis_Regression and Correlation_Multiple Linear. We therefore predict a mean stature of 169 cm. their fathers are to the mean height of fathers. It tells us how much. 5$ inches, with a standard deviation of about \$2. The graph of the simple linear regression equation is a straight line; 0 is the y-intercept of the regression line, 1 is the slope, and E(y) is the mean or expected value of y for a given value of x. And as independent variables, I'll use male and the mean centered height. The title of this article is "IQ Regression to the Mean : the Genetic Prediction Vindicated", and it begins "The IQ differences between blacks and whites lead to differences in sibling regression to the mean. A mean regression to the mean looms as a meaningful down as the notion of "regression toward mediocrity" dates back to Sir Francis Galton and his 1886 research on height being passed down. The use of parental height to estimate target height allowing for regression to the mean. 00096 if height is measured in millimeters, or 1549 if height is measured in miles. A regression threat, also known as a "regression artifact" or "regression to the mean" is a statistical phenomenon that occurs whenever you have a nonrandom sample from a population and two measures that are imperfectly correlated. The problems and solutions are the same as bivariate regression, native American male standing height, average yearly minimum Squares df Mean Square F Sig. 2Hierarchical regression with nested data The simplest hierarchical regression model simply applies the classical hierar-chical model of grouped data to regression coefficients. The simplest way to fit linear regression models in SAS is using one of the procedures, that supports OLS estimation. 1 for a typical simulation run. As we have said, it is desirable to choose the constants c0, c1, c2, …, cn so that our linear formula is as accurate a predictor of height as possible. Galton called this phenomenon "regression towards mediocrity"; we now call it "regression towards the mean. When the scores are standardized, the regression coefficient must fall between -1. If we calculate a regression of, say, weight on height, that regression is called a simple regression. (2 pts) Make a scatter plot of children's adjusted height vs the mid-parental height and then add the regression line on the plot. Measurements were made for a sample of adult men. 7570 Coeff Var 11. Multiplication by this correlation shrinks toward 0 (regression toward the mean) If the correlation is 1 there is no regression to the mean (if father’s height perfectly determine’s child’s height and vice versa) Note, regression to the mean has been thought about quite a bit and generalized. Regression is a set of techniques for estimating relationships, and we’ll focus on them for the next two chapters. Regression owes its name to the phenomenon known as regression toward the mean that arises when a genetically determined characteristic, such as height, is correlated between parent and offspring. The statistical phenomenon of regression to the mean is much like catch‐up growth, an inverse correlation between initial height and later height gain. For logistic regression, since we’ve put our blue points at height one and our green points at height zero, we will again want to choose the distribution that minimizes the distances from the distribution function (this time, the one based on the logistic curve) to the points above and in the plane. It is the most common form of Linear Regression. 12: from age 32 months on the xaxis, go up to the tted line and over to the yaxis. Could you give a Prediction of his weight? d) Plot a residual plot. Answer the following questions about the Earnings on Height (in cm) regression. 4, which is close but not exactly the given answer 73. ) The regression line is a line of estimated subpopulation. (parents' height). Note that when you use the regression equation for prediction, you may only apply it to values in the range of the actual observations. Regression to the mean, or why perfection rarely lasts March 26, 2017 3. 2 inches, with standard devation about 2. One factor contributing to the misuse of regression is that it can take considerably more skill to critique a model than to fit a model. Also, the regression line passes through the sample mean (which is obvious from above expression). In general, not all of the points will fall on the line, but we will choose our regression line so as to best summarize the relations between X and Y. In turn, this made one’s height likely to fall somewhere between the height of their parents and the average height of the population. Dalton's Data and Least Squares • collecteddatafrom1885inUsingR package • predictingchildren'sheightsfromparents'height • observationsfromthemarginal. multiply this mean deviation score of the predictor variable (fathers' mean height) by the regression coefficient of. The crossover of the separate regression lines precludes having descriptive main effects! • Since the regression lines vary from + to -, the Y-X regression line for the mean of Z isn’t descriptive for all values of Z. The coefficient for Height changed from positive to negative. Now, consider those parents with a mid-height between 70 and 71 inches. There are several tall females, one of which is also heavier than average. The graph of the simple linear regression equation is a straight line; 0 is the y-intercept of the regression line, 1 is the slope, and E(y) is the mean or expected value of y for a given value of x. (2 points) Plot the residuals versus the fitted values for the simple regression model above. And don't worry, this seems really confusing, we're going to do an example of this actually in a few seconds.
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http://math.stackexchange.com/questions/134883/proving-that-these-two-fields-mathbb-z-11x-x21-and-mathbb-z-11x
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# Proving that these two fields $\mathbb Z_{11}[x]/〈 x^2+1〉$ and $\mathbb Z_{11}[x]/〈 x^2+x+4〉$ are isomorphic with $121$ elements each.
I have been stuck in this problem for some time now.
Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic, each having $121$ elements.
The first part is easy to prove since there is no element of $\mathbb Z_{11}$ that satisfies either of the polynomials given in the question. However, proving that $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic has been a challenge for me.
How do I proceed? Moreover, How do I show that the the fields $\mathbb {Z}_{11}[x]/\langle x^2+1\rangle$ and $\mathbb {Z}_{11}[x]/\langle x^2+x+4\rangle$ have $121$ elements?
Any Help is much appreciated
Note: $\langle x^2+1\rangle$ denotes the ideal generated by $x^2+1$.
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$(x+6)^2+1=x^2+x+4$ holds in $\mathbb Z_{11}[x]$ – Martin Sleziak Apr 21 '12 at 18:05
From a purely abstract point of view, both quotients are fields of $121$ elements (since both polynomials are irreducible), hence both are splitting fields of $x^{121}-x$ over $\mathbb{Z}_{11}$, hence they are isomorphic. – Arturo Magidin Apr 21 '12 at 22:10
Hint $\$ You seek an isomorphism of $\:\mathbb F_{11}[\sqrt{-1}]\:$ and $\rm\:\mathbb F_{11}[5-\sqrt{-1}\:\!],\:$ since $\rm\:x^2+x+4\:$ has roots $$\rm\dfrac{-1\pm\sqrt{-15}}2 \:\equiv\: \dfrac{10\pm\sqrt{-4}}2\:\equiv\: 5\pm\sqrt{-1}\pmod{11}$$
The discriminant (mod squares) characterizes isomorphism classes of quadratic extensions.
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+1 Note that a trick like this will ALWAYS work in proving that two fields of cardinality $p^2$ are isomorphic (see also Martin Sleziak's comment above). We get such fields by joining a root of a quadratic to the prime field. The discriminant of that quadratic has to be a non-square. But because the multiplicative group $\mathbb{Z}_p^*$ is cyclic, we see that the ratio of any two non-squares is a square, so if you join the square root of one non-square, you join them all! Of course, the result in Arturo's comment to the OP settles the question even in the more general case. – Jyrki Lahtonen Apr 22 '12 at 17:32
A polynomial $f(x)$ with coefficients in any field $K$ and of degree $\leq3$ is irreducible over $K$ if and only if it has no roots in $K$. The given polynomials have no roots in $K=\Bbb Z_{11}$ and are therefore irreducible over $\Bbb Z_{11}$.
A standard fact about the ring of polynomals $K[X]$ is that its ideals are always principal, i.e. generated by one element. A standard consequence is that the ideals generated by irreducible polynomials are maximal. In addition, a quotient of a ring by an ideal is a field if and only if that ideal is maximal.
This explains why the given quotients are fields.
If $f(X)$ is irreducible of degree $d$, a full set of representants of the quotient $K[X]/(f(X))$ is given by the set of polynomials of degree $\leq d-1$ (this is an easy exercise). It follows that if $K$ is a finite field with $q$ elements the said quotient has $q^d$ elements. This explains why the given quotients have $121=11^2$ elements.
Finally, the general theory of finite fields tells us that for any prime power $q=p^f$ there exists a field with $q$ elements, which is unique up to isomorphism. Basically, the uniqueness follows from the fact that a finite field with $q=p^f$ elements is made up of the roots of the polynomial $X^q-X$ in some chosen algebraic closure of the basic field $\Bbb F_p=\Bbb Z/\Bbb Zp$.
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Wonder what I said so wrong to deserve a negative .... – Andrea Mori Apr 21 '12 at 18:57
I was the one who down voted you. Sorry about that. I did so because I didn't think that you answered enough of what was asked in the original question. You say that the polynomials are irreducible, but that doesn't seem to be the main question. Why are there 121 elements? Why is the quotient a field? (I will upvote you if you answer these questions...) – Thomas Apr 21 '12 at 20:06
@Thomas, I find a bit strange that somebody asking this kind of a question for a finite field does not have enough mathematical background to understand what I wrote without needing further details. Anyway, I will edit my answer adding some details. – Andrea Mori Apr 21 '12 at 21:30
Thanks. Much better answer. – Thomas Apr 21 '12 at 21:52
Here's an explicit isomorphism $f:\mathbb{Z}_{11}[x]/\langle x^2 + 1\rangle\rightarrow \mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle$ between the 2 fields. Abusing notation, I'm going to refer to an element in the domain by it's natural preimage in $\mathbb{Z}_{11}[x]$, and likewise in the range.
First note that $1$ is uniquely determined in a field, so we must send $1$ to $1$. Using additivitiy, $f(n) = n$ for any $n\in\mathbb{Z}/11\mathbb{Z}$. So, the only remaining question is what $f(x)$ will be. Notice that $x^2 = -1$, so $f(x)$ must be a squareroot of $-1$ in the other field.
Writing $f(x) = ax+b$, we get $$-1 = f(x)^2 = a^2 x^2 +2abx + b^2 = a^2(-x-4)+2abx + b^2 = (2ab-a^2)x + b^2-4.$$
Now, $a\neq 0$ (else $f$ is not injective), so we learn that $2b=a$ and $b^2-4 = -1$.
The second equation, $b^2-4 = -1$ has 2 solutions (mod 11), $b= 5$ and $b = 6 (=-5)$. We'll pick $b=5$ (the choice doesn't matter). Then since $2b=a$, we get $a = 10 = -1$.
Thus, we have $f(x) = -x + 5$.
Putting this altogether, we have $f(ax+b) = -ax + 5a + b$ defining our isomorphism.
To see $f$ is injective, assume $f(ax+b) = 0$. Then, since $-ax + 5a+b = 0$, we must have $-a = 0$, so $a=0$. Once we know $a=0$, $b=0$ follows. So $f$ is injective. To see it's surjective, notice $f(-ax + 5a+b) = ax + b$, so $f$ is surjective.
Finally, we check it's a homomorphism. We have \begin{align*}f((ax+b) + (cx+d)) &= -(a+c)x + 5(a+c) + (b+d)\\\ &= -ax + 5a +b + -cx + 5c + d\\\ &= f(ax+b) + f(cx+d).\end{align*}
so $f((ax+b)(cx+d)) = f(ax+b)f(cx+d)$. Thus, $f$ is the desired isomorphism.
Hint: Prove that any element in the first field can be uniquely written as $ax+b+(x^2+1)$. Prove the same for other field with $(x^2+1)$ replaced by $(x^2+x+4)$, where $(\dots)$ denotes the ideal generated by that element.
Now prove that they have 121 elements each. See Dummit & Foote if you need more help. Then, by the theory of finite fields, they are isomorphic. If you don't know the theory of finite fields, then define a map from the first to the second by sending $ax+b+(x^2+1) \mapsto ax+b+(x^2+x+4)$. Check that this map is homomorphism, and since L.H.S is a field this map is injective. Since both have 121 elements this map is surjective.
The map you describe is not a homomorphism since $f(\overline{x}*\overline{x}) = f(\overline{x}^2) = f(-\overline{1}) = -\overline{1}$ while $f(\overline{x})*f(\overline{x}) = \overline{x}\overline{x} = -\overline{x}-\overline{4}$. (I'm using overlining to denote the images of elements in the quotient fields). – Jason DeVito Apr 21 '12 at 21:08
| 2016-06-28T11:55:21 |
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https://math.stackexchange.com/questions/476693/using-residue-theorem-to-evaluate-int-0-pi-sin2n-theta-d-theta
|
# Using Residue theorem to evaluate $\int_0^\pi \sin^{2n}\theta\, d\theta$
can you please guide me on evaluating this integral using residue theorem and binomial theorem $$\int_0^\pi \sin^{2n}\theta\, d\theta$$ for $n = 1,2,3$
Honestly, I do not even know where to start, since it has no singularity.
And what is also the correct contour for this one?
Thanks in advance and more power.
Let $z=e^{i \theta}$, then $d\theta=dz/(i z)$ and $\sin{\theta} = (z-1/z)/(2 i)$. Then the integral becomes
$$\frac12\frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} \frac{dz}{z} \left( z-z^{-1}\right)^{2 n} =\frac12 \frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} dz \frac{(z^2-1)^{2 n}}{z^{2 n+1}}$$
As you can see, you have a pole of order $2 n+1$ in the integrand. To apply the residue Theorem, you need to evaluate $i 2 \pi$ times the residue at the pole at $z=0$, which is
$$\frac{\pi}{(2 i)^{2 n}} \frac{1}{(2 n)!} \left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0}$$
Now, by Rodrigues' formula for Legendre polynomials, the latter expression is
$$\left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0} = 2^{2 n} (2n)! P_n(0)$$
We can also use the binomial theorem to extract an explicit expression for the residue. Note that
$$\left ( z^2-1\right)^{2 n} = \sum_{k=0}^{2 n} \binom{2 n}{k} z^{2 k} (-1)^k$$
Taking the $2 n$th derivative and setting $z=0$ leaves only the $n$th term in the sum, so we get
$$\left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0} = (-1)^n \frac{(2 n)!^2}{(n!)^2}$$
Therefore, the integral is
$$\frac{\pi}{2^{2 n}} \binom{2 n}{n}$$
• Wow! That's very cool. Thank you very much, we had just been taught Rodrigues and Legendre in my other class, so I can also use them in Complex Analysis. Very great. You guys are the best. I really appreciate it. – user1824371 Aug 26 '13 at 18:35
• (+1) As in my non-residue answer, you can also use the binomial theorem to compute the coefficient of $z^{2n}$ in $\left(z^2-1\right)^{2n}$ to be $(-1)^n\binom{2n}{n}$ to get the residue of $\frac{\left(z^2-1\right)^{2n}}{z^{2n+1}}$ at $z=0$. – robjohn Aug 27 '13 at 15:14
• @robjohn: of course - I just wanted to use the general expression for the residue in terms of a derivative as an illustration. – Ron Gordon Aug 27 '13 at 15:15
• What makes it okay to reparametrize $sin\ \theta$ from $0 < \theta < \pi$ to $z \in \{|z| = 1\}$? Why can you take in all of the unit circle, why not restrict to the boundary upper half of the unit circle? What is the difference? – iaenstrom Aug 20 at 11:56
• @iaenstrom: $\sin^{2 n}{\theta}$ is an even function. – Ron Gordon Aug 20 at 18:50
We can use this answer, with $a=2n$ and $b=0$, to get \begin{align} \int_0^{\pi}\sin^{2n}(x)\,\mathrm{d}x &=2\int_0^{\pi/2}\sin^{2n}(x)\,\mathrm{d}x\\ &=2\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x\\ &=2\cdot\frac{\pi2^{-2n-1}\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)}\\ &=\frac\pi{4^n}\binom{2n}{n} \end{align}
Although this doesn't use contour integration, we can use $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and the binomial theorem to get \begin{align} \int_0^\pi\sin^{2n}(x)\,\mathrm{d}x &=\frac12\int_0^{2\pi}\sin^{2n}(x)\,\mathrm{d}x\\ &=\frac12\left(-\frac14\right)^n\int_0^{2\pi}\left(e^{ix}-e^{-ix}\right)^{2n}\,\mathrm{d}x\\ &=\frac12\left(-\frac14\right)^n\int_0^{2\pi}\sum_{k=0}^{2n}\binom{2n}{k}(-1)^ke^{2i(2n-k)x}e^{-ikx}\,\mathrm{d}x\\ &=\frac12\left(-\frac14\right)^n\binom{2n}{n}(-1)^n2\pi\\ &=\frac\pi{4^n}\binom{2n}{n} \end{align}
| 2019-10-15T16:27:31 |
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http://mathhelpforum.com/pre-calculus/66019-annual-compounding-interest-print.html
|
# Annual Compounding Interest
• December 25th 2008, 04:54 AM
magentarita
Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
• December 25th 2008, 05:40 AM
masters
Quote:
Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
• December 25th 2008, 06:53 AM
Soroban
Hello, magentarita!
I got a different result . . .
Quote:
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
• December 25th 2008, 08:26 AM
masters
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
• December 28th 2008, 07:23 AM
magentarita
ok..
Quote:
Originally Posted by masters
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
I got the answer from Soroban but I thank you for your effort.
• December 28th 2008, 07:24 AM
magentarita
yes...
Quote:
Originally Posted by masters
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
Don't feel bad. No one is perfect except God and He is not taking any math courses lately.
| 2014-10-23T13:19:40 |
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http://math.stackexchange.com/questions/291174/probability-distribution-of-the-subinterval-lengths-from-a-random-interval-divis/291320
|
# Probability distribution of the subinterval lengths from a random interval division
For $a \in \mathbb{R}_{+}$ and $n \in \mathbb{N}_{+}$ draw $n-1$ points $X_1, \ldots, X_{n-1}$ independently, uniformly at random from the interval $I = [0, a]$. These points partition $I$ into $n$ disjoint subintervals $I_1 \, \dot\cup \ldots \dot\cup \, I_n = I$. Let $Y_i := |I_i| \in [0, a]$ denote the interval length of the $i$'th subinterval.
How are the $Y_i$'s distributed? I am especially interested in the expectation and the variance.
Here are my thoughts about this: I suppose that all $Y_i$'s are identically distributed, i.e., $Y_i \sim Y$ for some random variable $Y$. Further, I suppose that the expectation of $Y$ is $\mathbb{E}(Y) = \frac{a}{n}$. However, I have no clue about the variance, and I can neither prove my conjectures, nor find an answer on the internet.
Can you help me on this?
-
That $\forall i, Y_i \sim Y$ is evident from the symmetry observed if you consider the interval wrapped around as a circle, where $n$ random points are chosen to break the circumference into $n$ random arcs. The length of any arc is obviously identically distributed.
To find the mean, observe that $\sum {Y_i} = a$
Take expectations on both sides, and use $\mathbb{E}(Y_i) = \mathbb{E}(Y)$ to prove your conjecture.
To find the distribution of $Y$ and other properties such as variance, perhaps the easiest way is to focus on $Y_n$. Let $f(t)$ and $F(t)$ denote the distribution function of $Y$ and its cumulative. Then we have,
$\mathbb{P}(Y_n \ge t) = \mathbb{P} (\max X_i \lt a-t)$
$\quad \quad \quad \quad = \mathbb{P}(\forall i, X_i < a-t)$
$\quad \quad \quad \quad = \prod_{i} \dfrac{a-t}{a}$
$\quad \quad \quad \quad = \left( \dfrac{a-t}{a} \right)^{n-1}$
$F(t) = \mathbb{P}(Y_n \lt t) = 1 - \mathbb{P}(Y_n \ge t)$
$\quad= 1 - \left( \dfrac{a-t}{a} \right)^{n-1}$
Differentiating, $f(t) = \dfrac{n-1}{a} \left( \dfrac{a-t}{a} \right)^{n-2}$
Hope that helps! [and thanks for the correction in $f(t)$]
-
In the analogy with the circle the fact that the $n$th point is deterministic makes not obvious that the $n$ intervals are exchangeable. Hence obviously in The length of any arc is obviously identically distributed is (obviously) a coup. – Did Jan 31 '13 at 13:59
@Did: I think one can still work out this symmetry argument a bit more carefully: Let the first 'somehow chosen' point represent the origin. Since the interval lengths are defined only by their relative positions to the origin and to each other, the absolute position of the origin on the circle does not matter, thus, it can be chosen deterministically, or uniformly, or according to any other distribution. – cubic lettuce Jan 31 '13 at 14:14
Exploiting the symmetry makes the rest much easier than the other approaches. There is just a small flaw, namely, it has to be $f(t) = \frac{(n-1)}{a} \left( \dfrac{a-t}{a} \right)^{n-2}$. Using this, one also gets that $\mathbb{E}(Y) = \int_0^a t f(t) dt = \frac{a}{n}$, and $\mathbb{E}(Y^2) = \int_0^a t^2 f(t) dt =\frac{2a^2(n-1)}{n^3-n}$, thus, $var(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \frac{a^2(n-1)}{n^2(n+1)}$, which matches Did's result for $a=1$. – cubic lettuce Jan 31 '13 at 14:16
@cubiclettuce it can be chosen deterministically, or uniformly, or according to any other distribution... For example, uniformly in the circle. Indeed, this is the key. (By the way, nice username!) – Did Jan 31 '13 at 14:20
@cubiclettuce Corrected the flaw - thanks. On the symmetry, yes while $n$ points are randomly chosen on the circle, any one of those is picked (anyhow) to open up the circle back into the interval. – Macavity Jan 31 '13 at 15:53
Assume without loss of generality that $a=1$ and define $I_i=[Z_{i-1},Z_i]$ hence $Z_0=0$, $Z_n=1$ and $(Z_i)_{1\leqslant i\leqslant n-1}$ is the ordered sample $(X_i)_{1\leqslant i\leqslant n-1}$.
For $x\lt y$ in $(0,1)$, $[Z_{i-1}\leqslant x,y\leqslant Z_i]$ means that $(x,y)\subseteq I_i$ which means that $i-1$ points from the sample $(X_i)_{1\leqslant i\leqslant n-1}$ are in $(0,x)$ and the $n-i$ others are in $(y,1)$. This happens with probability $$\mathbb P(Z_{i-1}\leqslant x,y\leqslant Z_i)=\binom{n-1}{i-1}x^{i-1}(1-y)^{n-i},$$ hence $\mathbb P(Z_{i-1}\in\mathrm dx,Z_i\in\mathrm dy)=f_i(x,y)\mathrm dx\mathrm dy$, with $$f_i(x,y)=\binom{n-1}{i-1}(i-1)(n-i)x^{i-2}(1-y)^{n-i-1}\mathbf 1_{0\lt x\lt y\lt1}.$$ Projecting this on $Y_i=Z_i-Z_{i-1}$ yields $\mathbb P(Y_{i}\in\mathrm dz)=g_i(z)\mathrm dz$, with $$g_i(z)=\int_0^{1-z}f_i(x,x+z)\mathrm dx.$$ The change of variables $x=(1-z)t$ and the binomial identity $$\int_0^1t^{i-2}(1-t)^{n-i-1}\mathrm dt=\mathrm{B}(i-1,n-i)=\frac{(i-2)!(n-i-1)!}{(n-2)!},$$ yield $$g_i(z)=(n-1)(1-z)^{n-2}\mathbf 1_{0\lt z\lt 1}.$$ It follows that the density of $1-Y_i$ is $(n-1)z^{n-2}\mathbf 1_{0\lt z\lt 1}$, hence, for every $a\gt1-n$, $$\mathbb E((1-Y_i)^a)=\frac{n-1}{a+n-1}.$$ In particular, $$\mathbb E(Y_i)=1-\mathbb E(1-Y_i)=\frac1n,$$ and $$\mathbb E(Y_i^2)=\mathbb E((1-Y_i)^2)-2\mathbb E(1-Y_i)+1=\frac2{n(n+1)},$$ hence $$\mathrm{var}(Y_i)=\frac{n-1}{n^2(n+1)}.$$
-
Here is a partial answer that should get you started.
Let $m=n-1$. The values $X_i$ are distinct almost surely. Then, sort the resulting sequence and denote by $Z_i$ the $i$-th element in the sorted sequence. Thus, $Z_1$ is the smallest among the $X_i$ and $Z_m$ is the largest.
We have $Y_k=Z_{k}-Z_{k-1}$ for all $k$ (where for convenience, we put $Z_0=0$ and $Z_{m+1}=b$).
Let us compute the distribution of the $Z_k$. We have
$$P(Z_k \leq t)=\int_{0}^{t} I_1(x_k)I_2(x_k)dx_k, \tag{1}$$
where
$$I_1(x_k)=\int_{0}^{x_k}\int_{0}^{x_{k-1}} \ldots \int_{0}^{x_2}dx_1dx_2 \ldots dx_{k-1}=\frac{x_k^{k-1}}{(k-1)!} \tag{2}$$
and
$$I_2(x_k)=\int_{x_k}^{a}\int_{x_{k+1}}^{a} \ldots \int_{x_{n-1}}^{a}dx_ndx_{n-1} \ldots dx_{k+1}= \frac{(a-x_k)^{n-k}}{(n-k)!} \tag{3}$$
We deduce that $Z_k$ has the the following density :
$$d_k(t)=\frac{t^{k-1}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} \tag{4}$$
So
$$E(Z_k)=\int_{0}^a \frac{t^{k}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} dt=\frac{k}{m+1}a \tag{5}$$
and
$$E(Z_k^2)=\int_{0}^a \frac{t^{k+1}}{(k-1)!} \times \frac{(a-t)^{n-k}}{(n-k)!} dt=\frac{6k}{(m+1)(m+2)}a^2 \tag{6}$$
So $E(Y_k)=E(Z_k)-E(Z_{k-1})=\frac{a}{m+1}=\frac{a}{n}$ as you rightly conjectured.
-
| 2016-06-26T06:24:47 |
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https://math.stackexchange.com/questions/2114065/limit-lim-n-to-infty-n-3-21-sqrt2-ldots-sqrtn-lim-n-to-infty
|
# Limit $\lim_{n\to\infty} n^{-3/2}(1+\sqrt{2}+\ldots+\sqrt{n})=\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n\sqrt{n}}$
How do I find the following limit?
$$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n\sqrt{n}}$$
The answer (from Wolfram) is $\frac{2}{3}$, but I'm not sure how to proceed.
Is this an application of the Squeeze theorem? I'm not quite sure.
• Do you know about Riemann sums? Jan 25 '17 at 21:46
• @gobucksmath Not that much. Would appreciate a nudge. Jan 25 '17 at 21:47
• Jan 25 '17 at 22:12
• Another solution would be to compare $k^{1/2}$ with $\int_{k}^{k+1} t^{1/2} dt$ to get the behaviour of $\displaystyle\sum_{k=0}^n \sqrt{k}$ and then you can conclude. Jan 25 '17 at 22:42
• @HungryBlueDev, I added an equivalent expression $\displaystyle\lim_{n\to\infty}n^{-\frac{3}{2}}(1+\sqrt{2}+\ldots+\sqrt{n})$ in the title to prevent future possible duplicates because I've seen this task in both forms. Sincerely Jan 15 '20 at 4:08
$$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} = \int_{0}^{1}\sqrt{x}\,dx = \color{red}{\frac{2}{3}}$$ by Riemann sums and the integrability of $\sqrt{x}$ over $[0,1]$.
For a more elementary approach, notice that $\sqrt{k}$ is pretty close to $\frac{2}{3}\left[\left(k+\frac{1}{2}\right)^{3/2}-\left(k-\frac{1}{2}\right)^{3/2}\right]$ and apply creative telescoping and squeezing.
All the other answers are by Riemann sums, but one can also use the Stolz theorem. The limit becomes $$\frac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}=\frac{\sqrt{n}(n\sqrt{n}+(n-1)\sqrt{n-1})}{n^3+(n-1)^3}=$$ $$\frac{n^2+(n-1)\sqrt{n^2-n})}{n^2+n(n-1)+(n-1)^2}\to\frac{2}{3}$$
• @JackD'Aurizio No I think its the other way around, like l'hospitals rule, after all, its really the discrete version. You need also that $y_n\to\infty$. Jan 25 '17 at 22:21
• All right, my bad. The (+1) stays :) Jan 25 '17 at 22:24
• @JackD'Aurizio Phew, that was a close one.... I just looked it up in my notes $y_n$ also has to be increasing. Jan 25 '17 at 22:47
How about Stolz-Cesaro? If $x_n \to \infty$, $y_n \to \infty$ and $$\lim_{n\to\infty} \frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$ exists, then : $$\lim_{n\to\infty} \frac{x_n}{y_n} = \lim_{n\to\infty} \frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$
• Okay, that's interesting... But what will be $x_n$? I think $\sqrt{n}$? And what about $y_n$? Jan 25 '17 at 21:59
• $x_n$ will be whole numerator of your fraction Jan 25 '17 at 22:00
Fill in details: $$\frac1{n\sqrt n}\sum_{k=1}^n\sqrt k=\frac1{n}\sum_{k=1}^n\sqrt\frac kn\xrightarrow[n\to\infty]{}\int_0^1\sqrt x\,dx$$
Using $\;\;\displaystyle\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=\int_0^1f(x)\;dx,$ we get $$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} +\ldots + \sqrt{n}}{n\sqrt{n}}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}}=\int_0^1\sqrt{x}\;dx=\frac{2}{3}.$$
| 2021-12-03T19:45:37 |
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http://math.stackexchange.com/questions/194472/a-curve-whose-image-has-positive-measure
|
# A curve whose image has positive measure
It is well-known that there are continuous curves $f:I \to \mathbb R^2$ (where $I \subset \mathbb R$ is an interval) whose image have positive measure (e.g Peano curve). I have read somewhere that if we require the curve to be differentiable evrywhere then this cannot happen; but if we require it to be almost everywhere differentiable, then it can happen! How could one proceed to prove the first statement and give a counterexample for the second?
-
There is no discrepancy because there measure on $I$ is not related to the measure on $\mathbb{R}^2$ for the purposes of this problem. – Trevor Wilson Sep 12 '12 at 2:58
My first thought is that the standard Peano and Hilbert space-filling curves appear to be differentiable outside a countable set, being the limits of curves with only finitely many points of non-differentiability. – Kevin Carlson Sep 12 '12 at 3:00
Yes I did not say there was a disreptancy, I just found the second statement counterintuitive. – Bernard Sep 12 '12 at 3:01
@KevinCarlson: The (graphs of the) standard nowhere differentiable functions like the blancmange or Weierstrass functions (restricted to a bounded interval) are also uniform limits of curves with only finitely many points of non-differentiability. – t.b. Sep 12 '12 at 3:03
@t.b. indeed. So much for the easy route. – Kevin Carlson Sep 12 '12 at 3:15
Let $\phi : [0, 1] \to [0, 1]$ be the Cantor-Lebesgue function, and $\alpha : [0, 1] \to \Bbb{R}^n$ be a space-filling curve.
Since $\phi$ is stationary outside the Cantor set, it is locally constant almost everywhere. That is, $\beta := \alpha \circ \phi$ is also locally constant everywhere, allowing it to be differentiable a.e.. (In fact, $\beta' = 0$ a.e.!) On the other hand, $\phi$ is continuous and surjective. Thus $\beta$ is also a continuous path and the image of $\beta$ coincides with $\alpha$. That is, $\beta$ is also a space-filling curve. Therefore this serves as a counter-example.
Let $f : [0, 1] \to \Bbb{R}^n$ be a differentiable curve in $\Bbb{R}^n$ ($n \geq 2$). Let $\Gamma = f([0, 1])$ be the image of $f$ in $\Bbb{R}^n$.
Assuming that $|f'|$ is Lebesgue integrable, we can prove the first statement.
Theorem. [7.21, Rudin] If $f : [a, b] \to \Bbb{R}$ is everywhere differentiable and its derivative $f'$ is Lebesgue integrable, then $$f(b) - f(a) = \int_{a}^{b} f'(t) \; dt.$$
This theorem immediately implies the following corollary:
Corollary. Let $f : [0, 1] \to \Bbb{R}^n$ be everywhere differentiable and its derivative $f'$ is Lebesgue integrable. Then for any $\epsilon > 0$, then there exists a partition $\Pi = \{ 0 = x_0 < \cdots < x_N = 1 \}$ such that for $$\epsilon_k = \sup \{|f(x) - f(y)| : x, y \in [x_{k-1}, x_k] \}, \quad (1 \leq k \leq N)$$ we have $$\epsilon_k \leq \epsilon \quad \text{and} \quad \sum_{k=1}^{N} \epsilon_k \leq \| f' \|_1.$$
Proof. Since $f'$ is Lebesgue integrable, it is absolutely continuous. Thus there exists $\delta > 0$ such that whenever a measurable subset $E \subset [0, 1]$ satisfies $|E| < \delta$, we have $\int_E |f'| < \epsilon$. Now let $\Pi = \{x_k\}$ be a partition of $[0, 1]$ into subintervals with length less than $\delta$. Then for each $x_{k-1} \leq x < y \leq x_k$, $$|f(y) - f(x)| = \left| \int_{x}^{y} f' \right| \leq \int_{x}^{y} |f'| \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon.$$ Thus $$\epsilon_k \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon$$ and the conclusion follows. ////
Remark. If $f'$ is bounded, then it is Lebesgue integrable. Also, in this case, the conclusion of the Corollary follows directly by mean value theorem.
Now let $\epsilon > 0$ and $\Pi$ be a corresponding partition of $[0, 1]$ by Corollary. Then we can cover $\Gamma$ by balls $B_{2\epsilon_k}(f(x_k))$ for $k = 1, \cdots, N$. Thus
$$|\Gamma| \leq \sum_{k=1}^{N} \left|B_{2\epsilon_k}(f(x_k))\right| \leq \sum_{k=1}^{N} C_n \epsilon_k^n \leq C_n \epsilon^{n-1} \sum_{k=1}^{N} \epsilon_k \leq C_n \epsilon^{n-1} \| f' \|_1,$$
where $C_n$ is a constant depending only on the dimension $n$. (In fact, we can choose $C_n = |B_2|$.) Thus taking $\epsilon \to 0$, we have the desired result.
-
WOW! That was not as difficult as I thought, but still cool. Thanks! So how do we proceed to prove the first statement (which intuitively seems obvious)? – Bernard Sep 12 '12 at 2:59
The other direction is called (mini)-Sard's Theorem. Page 205 in Appendix 1 of Guilleman and Pollack, Differential Topology. The mini version is just this: Let $U$ be an open set of $\mathbb R^n,$ and let $f:U \rightarrow \mathbb R^m$ be a smooth map. Then, if $m > n,$ we can conclude that $f(U)$ has measure zero in $\mathbb R^m.$
The full Sard's Theorem is about critical points, while we make no demands about the relative dimensions. Let $f:X \rightarrow Y$ be a smooth map of manifolds, and let $C$ be the set of critical points of $f$ in $X.$ Then $f(C)$ has measure zero in $Y.$
They say their proof is almost verbatim from John W. Milnor, Topology from the Differentiable Viewpoint. It appears Guillemin and Pollack, also Milnor, assume $C^\infty.$ However, Milnor refers back to Pontryagin(1955), English translation (1959). Evidently Pontryagin worked with weaker conditions, but we are not told what. http://en.wikipedia.org/wiki/Sard%27s_theorem
Alright, I've been looking stuff up elsewhere. The full Sard's theorem does depend on relative dimension. I currently think that mini-Sard make be true both for $C^1$ and for differentiable everywhere, which is a pretty weak hypothesis.
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Provided $f$ is at least $C^1$ or some such regularity hypothesis. – t.b. Sep 12 '12 at 3:17
@t.b., yes, thanks. It would appear the optimal result is Pontryagin (1955). I guess I can order a photocopy, but it takes a few days. – Will Jagy Sep 12 '12 at 3:25
The regularity assumption $C^1$ for mini-Sard is enough as proved already in Sard's original paper, Theorem 4.1. I don't know off-hand how much this can be weakened. Whitney's example in A function not constant on a connected set of critical points is definitely worth having a look at. – t.b. Sep 12 '12 at 3:33
@t.b. thanks. I've printed out Sard 1942, Thm. 4.1 says critical points. Critical point is when the rank of the matrix of first partials is smaller than both dimensions... – Will Jagy Sep 12 '12 at 3:45
Oops... Guess it's bed time for me. Sorry about that. – t.b. Sep 12 '12 at 3:50
| 2014-03-11T15:11:49 |
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https://math.stackexchange.com/questions/1751938/probability-of-a-man-being-guilty
|
# Probability of a man being guilty
I have a question regarding Example $3o$ on page $78$ of $Chapter$ $3$ in Sheldon Ross's book "A first course in Probability", the problem says the following:
A crime has been committed by a solitary individual, who left some DNA at the scene of the crime. Forensic scientists who studied the recovered DNA noted that only five strands could be identified and that each innocent person, independently, would have a probability of $10−5$ of having his or her DNA match on all five strands. The district attorney supposes that the perpetrator of the crime could be any of the one million residents of the town. Ten thousand of these residents have been released from prison within the past $10$ years; consequently, a sample of their DNA is on file. Before any checking of the DNA file, the district attorney feels that each of the ten thousand ex-criminals has probability $α$ of being guilty of the new crime, while each of the remaining $990,000$ residents has probability $β$, where $α = cβ$. (That is, the district attorney supposes that each recently released convict is $c$ times as likely to be the crime’s perpetrator as is each town member who is not a recently released convict.) When the DNA that is analyzed is compared against the database of the ten thousand ex-convicts, it turns out that A. J. Jones is the only one whose DNA matches the profile. Assuming that the district attorney’s estimate of the relationship between $α$ and $β$ is accurate, what is the probability that A. J. is guilty?
Ross solves this by calculating the probablity that Jones is guilty given that he has a match in the $10,000$ ex-cons on record doing the following: $$\mathbb {P}(G | M) = \frac {\mathbb {P}(M|G) \mathbb {P}(G)}{\mathbb {P} (M)}$$ Where $G$ and $M$ are the events that he is guilty and that he is the only one who matches the DNA strands. We must expand $\mathbb {P} (M) = \mathbb {P} (M | G) \mathbb {P} (G) + \mathbb {P} (M | G^c) \mathbb {P} (G^c)$
The way he computes the last term of the equation by calculating the probability that all other patients are innocent given that AJ is innocent and then multiplying by the probability of AJ matching the DNA and no other ex-con having the DNA strands. I understand why the last part is $10^{-5}(1-10^{-5})^{9,999}$. What I do not understand is why does he multiply by the probability of all of them being innocent?
If they are all innocent, wouln't the $10^{-5}(1-10^{-5})^{9,999}$ term account for that? Why multiply by the other probability then?
According to the book, $$\mathbb{P}(M|G^c) = 10^{-5} \left( \frac{1-10000\alpha}{1-\alpha} \right) \left(1-10^5\right)^{9999}$$ while $\frac{1-10000\alpha}{1-\alpha}$ represents the probability of all others in the database being innocent while AJ is innocent.
Here is one approach to understand this.
We want to calculate the probability of AJ being the only match given he is innocent, which requires the following.
(a) His DNA is found on the strands, with probability $10^{-5}$
(b) The other 9999 people in the database are all innocent (because otherwise someone other than AJ would have left DNA on the strands), and none of these 9999 people's DNA is found on the strands, with probability $\displaystyle \left( \frac{1-10000\alpha}{1-\alpha} \right)\left(1-10^5\right)^{9999}$
Then we get the result for $\mathbb{P}(M|G^c)$.
• Yes but if all the other ex-convicts are innocent, then none of their DNAs will match, so why must we include the other term with the alphas? – Guacho Perez Apr 21 '16 at 4:31
| 2019-08-17T12:33:15 |
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https://math.stackexchange.com/questions/962301/does-this-series-diverge-sqrt-2-sqrt-1-sqrt-3-sqrt-2-sqrt-4-sqrt-3
|
# Does this series diverge: $(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)+\dots$?
$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$
I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.
Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?
By telescoping, we see that the $n$th partial sum is $\sqrt {n+1}-1$, so this series diverges.
• This also answers the part of the question about the sum being a natural number when the number of terms is one less than a square. – robjohn Oct 7 '14 at 14:53
So your series can be written as $$\sum_{n=1}^{\infty} (\sqrt{n+1}-\sqrt{n}) =\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} \geq \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}}$$ and so by Comparison Test diverges
The $n$ term may be rewritten as
$$\frac1{\sqrt{n}+\sqrt{n+1}}$$
which behaves as $1/(2 \sqrt{n})$ as $n \to\infty$, so by comparison with the harmonic series, this series diverges.
@Jasper Joy is indeed right. It's a telescoping series whose partial sums eventually only have a fixed number of terms after cancellation:
$$(\sqrt{n+1} - \sqrt n) + ... + (\sqrt 4 - \sqrt 3) + (\sqrt 3 - \sqrt 2) + (\sqrt 2 - \sqrt 1)$$ $$= \sqrt{n+1} + (-\sqrt n + \sqrt n) + ... + (-\sqrt 3 + \sqrt 3) + (-\sqrt 2 + \sqrt 2) - \sqrt 1$$ $$= \sqrt{n+1} - 1$$
In series sigma notation we can summarize this as:
$$\sum_{i=1}^{n} (\sqrt{i+1}-\sqrt{i}) = {\sqrt{n+1} - 1}$$
As an infite sum it diverges to $\infty$:
$$\sum_{i=1}^{\infty} (\sqrt{i+1}-\sqrt{i}) = \lim_{i \to \infty}{\sqrt{i+1} - 1}$$
• We know @Jasper is right. Why duplicate? – Did Oct 7 '14 at 19:41
• So those who are unfamiliar with telescoping can see the details of how it is done. – soakley Oct 7 '14 at 22:03
| 2020-10-24T23:43:24 |
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https://byjus.com/question-answer/if-a-1-2-3-b-2-10-1-q-are-collinear-points-and-q/
|
Question
# If $$A = (1, 2, 3), B = (2, 10, 1)$$, $$Q$$ are collinear points and $$Q_x=-1$$, then $$Q_z=$$
A
3
B
7
C
14
D
7
Solution
## The correct option is A $$7$$The direction ratios of the line $$AB$$ are $$(2-1) : (10-2) : (1-3) = 1 : 8 : -2$$.Hence, any point $$Q$$ lying on the line $$AB$$ can be written in the form of $$(1+k, 2+8k, 3-2k)$$, where $$k$$ is a real number.$$Q_x = -1$$Hence, $$1+k = -1$$ This gives, $$k = -2$$.$${Q}^{}_{z} = 3- (2\times (-2)) = 7$$Hence, option B is correct.Mathematics
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| 2022-01-27T06:06:54 |
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https://physicscatalyst.com/article/common-difference-of-arithmetic-progression-or-sequence/
|
Home » Class 10 maths » common difference of arithmetic progression or sequence
# common difference of arithmetic progression or sequence
Common difference of arithmetic progression or sequence is defined as the difference between the two consecutive terms in the arithmetic progression or sequence or series. It is constant across the whole series
if $a_1$,$a_2$ ,$a_3$.$a_4$ ….. are in arithmetic progression or sequence
then
$d=a_2 -a_1 = a_3 -a_2 =a_4 -a_3$….
If the common difference is not same across the series ,then series is not in Arithmetic progression. This is important condition for the series to be in A.P.
So common difference of the Arithmetic series can be found given the two consecutive terms in the A.P
Examples
Find the common difference of the below Arithmetic progression
a. $2 + \sqrt {3}$ , $2+ \sqrt {12}$ , $2+ \sqrt {27}$, ,$2 + \sqrt {48}$ ….
b. -10 , -7 , -4 ,-1 ,2…….
c. 2, 3.5 , 5, 6.5 , 8……..
d. 1 , 3/2 , 2 , 5/2 ……
Solutions
a.
Here $d= 2+ \sqrt {12} – 2 + \sqrt {3} = \sqrt {12} – \sqrt {3} = \sqrt {4 \times 3} – \sqrt {3} = 2 \sqrt {3} – \sqrt {3}= \sqrt {3}$
$d= 2+ \sqrt {27} – 2+ \sqrt {12} = 3 \sqrt {3} – 2 \sqrt {3}= \sqrt {3}$
So common difference is $\sqrt {3}$
b. d= -7 – (-10) = 3
d= -4 – (-7) = 3
…..
So common difference is 3
c. d= 3.5 -2 = 1.5
d= 5 -3.5 = 1.5
……
So common difference is 1.5
d. $d = \frac {3}{2} -1 =\frac {1}{2}$
$d= 2 – \frac {3}{2} = \frac {1}{2}$
……
So common difference is 1/2
How to find the common difference if the nth term of the Arithmetic Progression is given
Steps would be
a. First find the first term by putting n=1
b. Find the second term by putting n=2
Now common difference = Second term – First term
Examples
a. Nth term is given as
$T_n = 2n -1$
b. Nth term is given as
$T_n = 4n -3$
Solutions
a.
$T_n = 2n -1$
$T_1 = 2 \times 1 – 1 = 1$
$T_2 = 2 \times 2 -1 = 3$
Common difference = 3 -1 =2
b.
$T_n = 4n -3$
$T_1 = 4 \times 1 – 3 = 1$
$T_2 = 4 \times 2 – 3 = 5$
Common difference = 5 -1 = 4
How to find the common difference if sum of the nth term of the Arithmetic Progression is given
Steps would be
a. First sum up-to to first term by putting n=1 ($S_1$)
b. Find sum up-to to second term by putting n=2 ( $S_2$)
Now second Term= $S_2 – S_1$
Common difference = Second term – First term = $S_2 – S_1 – S_1 = S_2 – 2S_1$
Examples
$S_n = 3 n^2 + 5n$
Solution
$S_1 = 3 (1)^2 + 5 \times 1 = 8$
$S_2 = 3 (2)^2 + 5 \times 2 =22$
Now second term = $S_2 – S_1 = 22 -8=14$
Common difference = Second term – First term = $S_2 – S_1 – S_1$ = $22 – 2 \times 8=6$
How to find the common difference if two terms are given in the Arithmetic Series
Suppose we have 10th and 16th terms given as
$T_{10}= 20$
$T_{16}=32$
Let a be the first term and d be the common difference
then using nth term formula
$a+ (10-1)d= 20$ or $a+9d =20$
$a+ (16-1)d =32$ or $a + 15 d=32$
Solving these we get a=2 ,d=2
I hope you find this article on common difference of arithmetic progression or sequence useful
Related Articles
What is Arithmetic Progression
Arithmetic Progressions Worksheet
Important Questions for Arithmetic Progression
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| 2023-02-03T10:19:49 |
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http://qzex.realtaveneta.it/cross-sectional-area-of-a-cone-as-a-function-of-height.html
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Sketch parallel cross-sections of each solid at height 3 above plane 𝑃𝑃. 1) In the coordinate system shown in Fig. A front view of this cone is exactly the same as the front view of the pyramid above, giving us the same triangle. Slant height is not defined for oblique cones. $\pi r^2$ is the base area while $\pi r l$ is the lateral surface area of the cone. What about spheres? To compute the volume of a sphere, let's show that a hemisphere (with radius $$r$$) has the same volume as the vase shown in the figure below, formed by carving a cone from the circular cylinder with radius and height $$r$$. The volume and cross-sectional areas from the smallest cross-sectional area (SmCa), largest cross-sectional area (LgCa), and cross-sectional area at C2 (C2Ca) were. Therefore, out by the raindrop is a cone. height $$x$$ (since the upper face of the notch is at $$45^\circ$$ to the base — see the side view sketched in the figure on the right above). The conventional method is to measure the velocity of the descending sludge-supernatant interface (sludge blanket) as the change in height over time in a vessel with constant cross-sectional area. Since the general formula for the area of a circle of radius ris ˇr2, the area 1. The hindered-settling velocity function for the modelling, simulation and control of secondary settling tanks can be determined from batch tests. The general formula for the surface area of the graph of a continuously differentiable function $z=f(x,y),$ where $(x,y)\in D\subset\mathbb{R}^2$ and $D$ is a region in the xy. ane bn defined by Eq. Find the volume of a cone whose base is a square of side $5$ and whose height is $6$, by cross-sections. The slant height l can be found by using Pythagoras theorem. We have all sorts of triangle calculators, polygon calculators, perimeter, area, volume, trigonometric functions, algebra, percentages… You name it, we. This free cross section property tool calculates moment of inertia, polar moment of inertia and second moment of inertia for various shapes. Volumes Using Cross Sections Calculus, Square, Semicircles, Rectangles, Equilateral Triangles - Duration: 43:31. Solution Let h be the depth of water in feet, and let r be the radius of the surface of the water at depth h. The cross-sectional area of an object when viewed from a particular angle is the total area of the orthographic projection of the object from that angle. Let V be the volume of a right circular cone of height 10 whose base is a circle of radius 4. 4 we used definite integrals to compute the arc length of plane curves of the form $$y=f(x)\text{. The height of the water changes as time passes, so we’re calling that the variable y. Sketch of the pre-structured GaSb templates with different cone size (small—S and large—L) coated with a magnetic film represented by the curved line. Cross Sections. The diameter of a circular cross-section of the tank is 6 m. For example, a well. So, the volume of a right prism and an oblique one of the same height and cross sectional area are same. With the large-scale, centralized and continuous production of hydrometallurgical industry, an efficient and stable automatic production line is urgently required. Cross-sectional area (in square feet) of the trunk of one tree at breast height (4. So (1/2)*a*b*sin(theta) = Cross sectional area of the triangle formed. 8 Science and Engineering Applications 149 Then the method of cross-sections in calculus implies that the volume V(h) of the tank at height his given by V(h) = Z h 0 A(y)dy; dV dh (1) = A(h): Torricelli’s Equation. ” Students should already know how to find an area between two curves and they should be able to integrate functions with rational exponents. Recently finished a chapter on Solid Mechanics at school. l) a0 distance between rings for a conical shell (see Fig. It is used for low specific speed and vertical shaft Francis turbine. We will use the idea of the last section, namely that volume can be found by integrating cross-sectional area, to find the volume of such a solid. The general formula for the surface area of the graph of a continuously differentiable function z = f(x,y), where and D is a region in the xy-plane with the smooth boundary:. Cross sections perpendicular to the y-axis are equilateral triangles. Any plane can be used to cut through the surface, but when that plane is perpendicular to an axis of symmetry, its projection is called a cross-sectional area. where % is the density of the air, V is the velocity of the object, A is the cross-sectional area (normal to the velocity direction), and C D is the drag coefficient, which depends heavily on the shape of the object and the roughness of the surface. (a) Find the rate of change of the volume with respect to the height if the radius is constant. crystallization) in Figure - 4. Assuming a level contracted flume, and negligible energy loss between upstream and critical flow section, the upstream energy will be equal to the energy at the critical section, and can be described as 2 2 1 2 c s c c gA Q. For a simple three-dimensional shape, such as a cylinder, the cross-sectional projection is a circle, and the area is easy to calculate. a function of the direction normal to the cross-section. A cone is closely related to a pyramid. The volume is always V = A L The surface area can usually be deduced by adding all the separate surfaces together. It's one minus x over two squared. Since the capacity of gravitational settling equipment was found largely to be a function of the cross-sectional area, it was evident that application of the settling cone was limited. Example Find the volume of a pyramid with square base side a and height h. 13, if the cone was 0. The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. As before, it is the volume divided by the area of the base, but now we need to use integration to compute the area of the base, since it is not a simple rectangle. (b)Write and evaluatethe integral for this volume. A Cross sectional area of water prism A. contains the base of the cone and P, is at distance x above Po. Proposition A. It states: Let S be a solid and Px be a family of parallel planes such that: 1. by its cross-sectional area times its velocity. The other parameters of the problem are indicated. ,1* Find the volume of each cylinder. A material of resistivity ( is formed into the shape of a truncated cone of height h as shown in Figure P27. Right cone is a body, which is obtained by rotating a right triangle around one of the legs. In addition, a minimum of 5 rod inner segments per image were outlined, and an average cross-sectional area per rod inner segment was calculated. Slice the cone as shown in Figure 8. Cross Sections of Cone. Then its volume is the limit of a Riemann sum V = lim n!¥ n å i=1 A(x i)Dx = Zb a A(x)dx Volumes of Revolution Most of our examples will be columes obtained by rotating a curve y = f(x) around the x-axis for x in some interval [a,b]. Calculate its length. In general, they can be defined in terms of a mean cross sectional area. A small cone of height 8 cm is cut off from a bigger cone to leave a frustum of height 16 cm. In order to find the area of a cross-section, we will need to use similar triangles. A = 2πrh Curved surface area, A, of cone of radius r, sloping edge l. A constriction (of cross-sectional area A2) is put in a pipe of normal cross-sectional area A1. The area A of a circle is increasing at a constant rate of 1. In the language of calculus, the six simplistic high-school formulas at the beginning of these notes are replaced by formulas given by integrals. So the volume being drained is between the heights 0. Let V be the volume of a right circular cone of height 10 whose base is a circle Of radius 4 [Figure (a) Use similar triangles to find the area of a horizontal cross section at a height y. The temperatures on four crosssections (45 measuring points on half of a cross section) of crossflow were measured using thermocouples. When the water reaches the halfway mark, it will be filling at a much quicker rate. you need and I'll. Find the area of a circle, 𝐴= 𝜋𝑟. If this region is revolved about the axis, it will generate a solid that will have circular cross-sections (Figure 8b) with radii of at each Each cross-sectional area can be calculated by. Determine the boundaries which will represent the limits of integration. 4 software, when the airway is segmented both the volume measurement of the airway and the Minimum Cross-Sectional Area (MCA) are automatically detected and calculated. 0 m long as in Figure P6. alculating wetted area in a partially-filled horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the head (bottom) of a vertical tank. Round to the nearest tenth. Sketch parallel cross-sections of each solid at height 3 above plane 𝑃𝑃. For a simple three-dimensional shape, cylinder, the ~ al projection is a circle when it is sliced parallel to its base, and the area is easy to calculate. If the cross section is perpendicular to the x‐axis and itʼs area is a function of x, say A(x), then the volume, V, of the solid on [ a, b] is given by. After the forming cone, the bubble is exposed to the upper lip and is cooled by a large volume of air (lower velocity) over a large bubble circumference. Under the new BS EN 15287-1 the guidance has been amended as follows: Chimneys serving an appliance recess. Volume of a Cylinder = Bh = (pi) r 2 h. The fluid has velocity and temperature. → Since a pipe is a cylinder, the cross-sectional area of a pipe is a circle. Explain why, using the terms critical, subcritical, Froude number,cross-sec-tional mean velocity,and acceleration of gravity. V = length x cross-sectional area = 10 x (b1 + b2)/2 x h. The base of Sis a circular disc of radius r. 3 - Volumes by Cylindrical Shells - p. It is able to find the minimum cross-sectional area in the part of the airway that had been defined, and the other 2 cross sections were found by scrolling through the axial slices. However, the formula above is more general and will work for any way of getting a cross section so we will leave it like it is. How would we find the volume of the solid obtained by rotating the. Solution: Given that: r = 3 cm h = 4 cm To find the total surface area of the cone, we need slant height of the cone, instead the perpendicular height. The software algorithm identifies and displays the axial images at which the minimum cross‐sectional area is present (c and d) and allows for measurement of antero‐posterior and transverse dimensions (insert). Note that A(x) in the volume formula corresponds to t(x) in the area formula. MATH 211, Calculus II cross-sectional area height Find the volume of a right circular cone whose base radius is 3 and whose height is 7. 11 is an example of a cylinder with a noncircular base. Similarly, a solid right-circular cone is generated by a right-angled triangle about one of its legs (Figure 7. Basal area – A. A front view of this cone is exactly the same as the front view of the pyramid above, giving us the same triangle. The cross sectional area of the soil column is 5 cm2. ” Students should already know how to find an area between two curves and they should be able to integrate functions with rational exponents. The cross-sectional area (′) of an object when viewed from a particular angle is the total area of the orthographic projection of the object from that angle. Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. Let's look at how that would go if we simply wanted to evaluate the volume of the cone (not the moment of inertia!). height of the cone 30 2722 13. What is the volume of a cone with base radius r and height h ? De nition. Volume of a Cone Again Rotate the line y = R h x for 0 ≤ x ≤ h, which generates a cone of base radius R and height h. zontal cross section at a height y. Area of the base × height. For example, a cylinder of height h and radius r has A ′ = π r 2 {\displaystyle A'=\pi r^{2}} when viewed along its central axis, and A ′ = 2 r h {\displaystyle A'=2rh} when viewed. The region of revolution and the resulting solid are shown in (c) and (d). Arc length decreases from (r + dr)d to rd over a distance dr. Two vertical tubes, open to the atmosphere, rise from two points,. Area of a Sector = (x°/360°)πr², where x° is the angle of the sector. function value at that particular x as shown above. Find a formula for the linear function \(y = f(x)$$ that is pictured in Figure6. <> Figure 7. If you use similar triangles and solve the proportion, you get that when the water has a height of h, then b2 = 1. A cylinder is a three - dimensional shape. An Overview of Hopper and Design Procedure of a Pyramid Shaped Hopper Prof. The cone in the numerical model has a solid volume of 19 080 mm 3, compared to a cylinder of the same length and radius of 28 910 mm 3, giving a theoretical averaged area of steel of 0·66. A cylinder is a three – dimensional shape. Related rates problems involve two (or more) unknown quantities that are related through an equation. A(h) = area of the cross-sections perpendicular to h described as a function of the position along h. If you have ever sliced a vegetable in two, you already know what a cross section is. Cross sections perpendicular to the y-axis are equilateral triangles. The cross section on any plane perpendicular to the diameter of the semi-circle is a right isosceles triangle with the right angle on the semi-circle. However, the change in the. Note that the velocity of the owing liquid depends on the height of the liquid (i. Based on the mature technology of onshore heavy oil thermal recovery, a new rod pumping technology of “heavy oil steam stimulation along with conventional sucker rod pumping system” is proposed. (a) Show that V = Ah is dimensionally correct. The cylinder length, diameter, density, and tip shape are all explicitly varied. If this region is revolved about the axis, it will generate a solid that will have circular cross-sections (Figure 8b) with radii of at each Each cross-sectional area can be calculated by. A triangular cross section may not be as common as other shapes, as you would probably only see one in a triangular-based pyramid or a triangular prism. Cavalieri’s Principle - if 2 solids have the same height and the same cross-sectional area at every level, then they have the same volume. A hole $3$ units in radius is drilled out along a diameter of a solid sphere of radius $5$ units. a simple geometric fact. Return To Top Of Page. For a simple three-dimensional shape, cylinder, the ~ al projection is a circle when it is sliced parallel to its base, and the area is easy to calculate. I then observed how the volume of the cone could be approximated by using disks, the width of each being the height of the cone divided by the number of disks. is the base area while is the lateral surface area of the cone. Before the contraction cone there is a settling chamber with a constant cross-sectional area. Consider the length of one side of the martini as A and the length of the other side as B. Sketching a diagram, we see the area in. 4 functions by drawing a line down the middle of the airway in a sagittal grayscale view (Figure 5). What is the volume of the. Therefore, out by the raindrop is a cone. A sign change in the slope from plus to minus at that value of "q" confirms that a maximum occurs there. The right pyramid has a height of 10 units, and the right cylinder has a height of 7 units. Fig 4: Elbow Draft Tube with Varying Cross Section 4) Moody Draft Tube or Hydra cone – Moody spreading draft tube is the special type of draft tube. Also can solve for discharge coefficient, orifice diameter (or area), tank diameter (or area), initial liquid depth, final liquid depth. Which conclusion can be made from the given information?. Surface area of the outer cone: S 2 = πRr 2 Surface area of the inner cone: S 1 = πRr 1 Total sector area: S sector = S sec + S 1 + S 2 = πR(2h + r 1 + r 2 ). A flume consists of a constriction in the cross-sectional area of flow. Example 7: Find the total surface area of a cone, whose base radius is 3 cm and the perpendicular height is 4 cm. 5 mm and angle (θ1) of 28° and the second cone with height (H2) of 39. Naturally, you know from school that its volume is: V = 1 3 πR2h. A = 4πr2 Volume, V, of prism, cross-sectional area A, length l. 5 times for vapour outlet pipe area. This is equivalent to calculating the cross-sectional area A, and then multiplying that area by the height of the prism, h. A special type of solid whose cross-sections are familiar geometric shapes is the solid of revolution. The midsagittal airway view is mapped (left image), and a series of cross-sectional areas (CSAs) of the mapped regions are generated (right image). Find the lowest prices on bestselling Paper Towel Holder Wall Mounted Bathroom Kitchen Creative Roll Holder Stainless Steel Base Cabinet Decoration Without Diamond Paper Roll Holder Bathroom Supplies Multi Function Toilet Paper Holder in a wide variety of designs. What is the area of the horizontal cross section y? b. This length, which is six at this point, this is also going to be the height. vertical obstruction that the water must flow over. ver, the simplest old sectional methods are still preferred for practical measurements. function value at that particular x as shown above. The conventional method is to measure the velocity of the descending sludge-supernatant interface (sludge blanket) as the change in height over time in a vessel with constant cross-sectional area. Introduction. and so, a rectangle with different sides (say length of 3 metres and width of 2 metres) would have an area in square units that can be calculated as: 3 metres x 2 metres = 6 m2. is a square. In general, they can be defined in terms of a mean cross sectional area. The sum of basal area of the individual trees on an acre of forest. Solution: Given that: r = 3 cm h = 4 cm To find the total surface area of the cone, we need slant height of the cone, instead the perpendicular height. Well, both of you are right in your own way so stop the childish bickering Cross-sectional area, as explained in a physics class I once had, is the area of the shadow something would cast if light was shined from above. If this region is revolved about the axis, it will generate a solid that will have circular cross-sections (Figure 8b) with radii of at each Each cross-sectional area can be calculated by. Explain how to computer the cross-sectional area of the pipe. The cross section on any plane perpendicular to the diameter of the semi-circle is a right isosceles triangle with the right angle on the semi-circle. Determine the boundaries which will represent the limits of integration. Calculate the surface area of the sphere. Moment of Inertia The moment of inertia of a particle is an indicator of how much torque you need to rotate it around the origin. A hole $3$ units in radius is drilled out along a diameter of a solid sphere of radius $5$ units. is the union of a sphere and an inverted cone, and the second is a cylinder. The slant height of a right circular cone is the distance from any point on the circle of its base to the apex via a line segment along the surface of the cone. h = height of the solid. 2 to find the volume of the solid. An Integral Involving the Natural Logarithm and the Floor Function; Maximizing the Area of a Triangle (and a Generalization to Three-Space) Find the Area of an Ellipse from Three Cross Sectional Lengths; Find the Surface Area and Volume of a Reuleaux Tetrahedron; Quadrisect a 3-4-5 Triangle with Two Perpendicular Lines + Three Infinite Alternating Series. Suppose that a thin rod with constant cross-sectional area of 1 cm2 has its mass distributed according to the density function ρ(a) 2e-0. Solution: Given that: r = 3 cm h = 4 cm To find the total surface area of the cone, we need slant height of the cone, instead the perpendicular height. So whatever this length is, we also go that much high, and so the cross section is a square right over there. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. It is able to find the minimum cross-sectional area in the part of the airway that had been defined, and the other 2 cross sections were found by scrolling through the axial slices. For a sphere, the shadow would take the form of a circle. The cylinder length, diameter, density, and tip shape are all explicitly varied. If the cross-sectional area of S in the plane Py, through y and perpendicular to the y-axis, is A y, where A is a continuous function, then the volume of S is V b a A y dy Examples: 1. The bucket presents a cross- sectional area of 3. A flume consists of a constriction in the cross-sectional area of flow. 22 where x is the distance in cm from the left end of the rod, and the units on ρ(x) are g/cm. In the sections where we actually use this formula we will also see that there are ways of generating the cross section that will actually give a cross-sectional area that is a function of $$y$$ instead of $$x$$. cross-sectional areas (CSAs) of the mapped regionsare generated (right image). The airway segmenting tool within Invivo5. x_profile <- seq(0, 500, 25) y_prof. Volumes Using Cross Sections Calculus, Square, Semicircles, Rectangles, Equilateral Triangles - Duration: 43:31. The response variable was the annual growth rate of the stem cross-sectional area (ΔA, cm 2 year −1) at a given stem location on a tree. Thus, the surface area of the crown should be a useful parameter in growth prediction. Find the volume of a right circular cone of base radius r and height h. The base of the cones do not have to be the same shape: they can. [13] As shown in equations 1 to 3, the fallout strongly depends on plume geometry. Cross sectional area varies from beginning to ending of draft tube. The area of this plane of intersection is known as the cross-sectional area of the object. Example 2 shows how similar triangles are commonly used as an aid in ccmputing a cross-sectional area. The experimentally measured flow rates in discrete components, as a function of cross-sectional area and length showed trends consistent-though not in full agreement-with the ideal behavior predicted by the Hagen-Poiseuille Law. Find the volume of the region whose cross sections are squares and whose base is an equilateral triangle centered on the x-axis where 0 6 x 6 3. Bobade Rahul Dattatray3 2,3BE Student 2,3Department of Mechanical Engineering 1,2,3ADCET Ashta Abstract— This paper represents an overview of a storage hopper used for rice mill industries. Since zy = x2, the cross-sections are equal in area and since the pyramids have the same height, their volumes are equal by Cavalieri’s first principle. The base radius OB is 6 cm. Area of a Circle Segment Given the Central Angle Definition: The number of square units it takes to fill a segment of a circle Try this Drag one of the orange dots that define the endpoints of the segment. Basal area – A. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: V = A · h. The cross-sectional area of a cylinder is simply a circle. At time t = O, the radius is 6 centimeters. I can use the countour function. The region of revolution and the resulting solid are shown in (c) and (d). OpenGL depends on external mechanisms to create the drawing surfaces on which it will draw. The length is constant. Example 2 shows how similar triangles are commonly used as an aid in ccmputing a cross-sectional area. If every plane parallel to these two planes intersects both regions in cross-sections of equal area, then the two regions have equal volumes. Solution: h a a (1) Choose simple. Lesson 33-3: Keiko designed a pendant for a mecklace by rotating a square about a. area and width compared with the Class I group (Table 1). To develop your equation, you will probably use. 5 º angle (θ2); apex diameter (OD) of 5 mm and a. Among the numerous scenarios/examples our teacher provided, was a case of a wire of non-uniform cross sectional area allowed to stretch under its own weight by suspending it by one end, from the ceiling. A three-dimensional surface model of the segmented area was generated for each cochlea. Cross sections perpendicular to the x-axis are semicircles. Chapter 11 Differential Equations 11. So, what if the tower was, in fact, a cone? Find an equation relating the maximum height of the tower to the cone angle of the tower using the quantities given. Another good example of this design is the Concorde as well. The base of Sis the parabolic region f(x;y) : x2 y 1g. 3 Hydrodynamics and Mass Transfer in a Modified Three-phase Airlift Loop Reactor 93 But in Fig. Maximize the area of a Norman window (rectangular with a semicircle on top) while minimizing the length of the perimeter. zontal cross section at a height y. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose base is a circle of radius r (Figure 16). 1 Introduction A differential equation is a relationship between some (unknown) function and one of its derivatives. This is equivalent to calculating the cross-sectional area A, and then multiplying that area by the height of the prism, h. 5 º angle (θ2); apex diameter (OD) of 5 mm and a. Volume = base area height = ˇr2h Prism l Volume = cross-sectional area l Pyramid Volume = 1 3 base area height Cone r l h Volume = 1 3 base area height = 1 3 ˇr2h Surface Area = base area + curved surface area = ˇr2 + ˇrl Sphere r Volume = 4 3 ˇr3 Surface area = 4ˇr2 3 Functions & Graphs Graphs of power functions (y = axn) n = 2 x y O a. Figure 3 exhibits the numerical results of these variations for both of the TE and TM polarizations. In Section 7. prism: 2 * Area of Base + Perimeter of Base * Height General formula Edit. Any other vertical cross section will reveal a hyperbola (with endpoints on the. For a chimney built with a fireplace recess in which an open fire, a room heater or stove may be fitted size a flue of 200 mm diameter or rectangular/square flues having the same cross-sectional area and a minimum dimension not less than 175 mm diameter. "What is the shape of a cross-section of this cone?" The answer, of course, is "a circle. A stone is intially thrown upward from a cliff, and its height above ground (in feet) t seconds after being thrown is is given by the formula h(t) = 320+128t−16t2. The hindered-settling velocity function for the modelling, simulation and control of secondary settling tanks can be determined from batch tests. Then the vertical displacement can be estimated using the well-known relation: EI PL 3 3 δ= where δ is the tip displacement of the bar, I is the (uniform) second moment of the cross-sectional area about the bending axis, and the other parameters are defined in. When the water reaches the halfway mark, it will be filling at a much quicker rate. (b) Show that the volumes of a cylinder and of a rectangular box can be written in the form V = Ah, identifying A in each case. By comparing the ratios of the weight to cross-sectional area it can be proved that a larger scaled object will be weaker than a smaller similar object. So the slice has cross-sectional area \(2x\sqrt{20^2-x^2}\text{. Hopefully that makes sense. The first cone with the cone height (H1) of 29. Each circular volume element will have a cross-sectional area ˇr2. 08cm cross sectional area to initial cross sectional area) in the first stage 0. 2 Volume of a Solid Body A solid body is a solid three-dimensional object. shell, distance of top of truncated cone from vertex (see Fig. height of the cone 30 2722 13. I can use the countour function. OPEN ENDED Draw the cross sections of a polyhedron, cylinder, or cone. Let S be a solid that lies between the horizontal planes z = a and z = b. 5 mm 2, 4 mm 2, 6 mm 2, 10 mm 2, 16 mm 2. R R R ππ π =− = 72. If the height of hemisphere and cone is 49cm and height of cone is 35cm find the volume of solid. The doctors tell you, “The cross-sectional area of the aneurysm is 2. The graph of the function and a representative washer are shown in (a) and (b). The limit case of a full conical vessel with its vertex at the bottom is solved herein. The cross-section is going to look like a washer, with an inner radius equal to the height of y = x^2 wherever we are in the solid, and the outer radius equal to y = 2x at that same x. What is the volume of a cone with base radius r and height h ? De nition. The Cheeger constant is defined as the infimum of the boundary area to volume ratio of regions in the interior of K with at most half the volume. The cross section right over here is going to be a square. In this type of draft tube the flow path way is straight and divergent. So, technically we cannot find the perimeter of a cylinder but we can find the perimeter of the cross-section of the. The conventional method is to measure the velocity of the descending sludge-supernatant interface (sludge blanket) as the change in height over time in a vessel with constant cross-sectional area. So, the volume of a right prism and an oblique one of the same height and cross sectional area are same. Sectional methods for tree stem volume measurements regard any section or segment, of length L, as a cylinder. solution (b) Calculate V by integrating the cross-sectional area. For a simple three-dimensional shape, cylinder, the ~ al projection is a circle when it is sliced parallel to its base, and the area is easy to calculate. Among the numerous scenarios/examples our teacher provided, was a case of a wire of non-uniform cross sectional area allowed to stretch under its own weight by suspending it by one end, from the ceiling. Thus, the surface area of the crown should be a useful parameter in growth prediction. For a simple three-dimensional shape, such as a cylinder, the cross-sectional projection is a circle, and the area is easy to calculate. Figure 2 shows a cylinder with height r, radius r, with an inverted cone inside. The cross-sectional areas of a right triangular prism and a right cylinder are congruent. " (Wikipedia). cone: , where r is the radius of the circular base, and h is the height. Archimedes observed that if you slice the can along any plane perpendicular to its axis of symmetry then the cross-sectional area of the sphere plus the cross-sectional area of the cone equals the cross-sectional area of the cylinder. The area of this plane of intersection is known as the cross-sectional area of the object. the effect of alterations of the parabola cone radii and the height of the absorption cross-sectional diagram. The bottom of the cone has a large cross sectional area, so the height will go up slowly at first. 0 = h/e is the flux quantum, S is the cross-sectional area of the NW, h is Planck’s contant, and e is the electron charge. D = diameter of a vertical cross-section of the horizontal tank, ft By assuming the volume of the tank to be approximately equal to the cross-sectional area of the tank times the length of the tank, an effective height, H E, of an equivalent upright cylinder may be calculated as: H E = D π 4 (1-14) D. Cross-sectional area of a cone(ish). Here we observe that a cone can be seen as a stack of in nitesimally thin circular disks. Calculates the volume, lateral area and surface area of a circular truncated cone given the lower and upper radii and height. Usually, the crown of conifers and young hardwoods is modelled as a cone, its surface area being derived from the formula:. But depth was not significantly correlated with cross-sectional area or width in the Class III group (Table 2). Rocket aerodynamics is the study of how air flows over a rocket and how this affects drag and stability. See also stand basal area Basic density The oven-dry weight of wood per unit green volume Bias The difference between the average of a set of repeated measurements or estimates of something and its true value Bioenergy Biomass used to make energy, usually by conversion to. \] Certainly, using this formula from geometry is faster than our new method, but the calculus--based method can be applied to much more than just cones. (b) Show that the volumes of a cylinder and of a rectangular box can be written in the form V = Ah, identifying A in each case. and we take this number as the volume of the cone with base radius 1 and height 1. Solution: h a a (1) Choose simple. The plot shows fair. crystallization) in Figure - 4. Cross-sectional images were taken and analyzed by custom-designed software in MATLAB. 8 Science and Engineering Applications 149 Then the method of cross-sections in calculus implies that the volume V(h) of the tank at height his given by V(h) = Z h 0 A(y)dy; dV dh (1) = A(h): Torricelli’s Equation. So, what if the tower was, in fact, a cone? Find an equation relating the maximum height of the tower to the cone angle of the tower using the quantities given. For example, a cylinder of height h and radius r has when viewed along its central axis, and when viewed from an orthogonal direction. Windowing APIs that support OpenGL often do so as one library among many others that are used to produce a complete application. denote the length of the side of that square. Moment of Inertia The moment of inertia of a particle is an indicator of how much torque you need to rotate it around the origin. Calculate the cross‐sectional area of the hemisphere in terms of N and T hint: use Pythagorean theorem. A fire helicopter carries a 620-kg bucket at the end of a cable 20. Two vertical tubes, open to the atmosphere, rise from two points,. The Slice Method is another form of Cavalieri's Principle. As an alternative to the washer method, one may consider “cylindrical shell” cross-sections instead. Therefore, our cone decreases air velocity, as air flows from the small end to big end. Volume of a partial circular cone Calculator - High accuracy calculation. The outer radius is the x-coordinate from the curve y = x2,thus,r outer = p y. Incredible costs & quick delivery!. The volume of the cone will equal the area under the curve A = ˇ(2− 1 2 h) 2 for h between 0 and 4. The base of the cones do not have to be the same shape: they can. "It's a cone. Find, to 3 significant figures, the rate at which the radius r of the circle is increasing when the area of the circle is 2 cm2. by its cross-sectional area times its velocity. At time t = O, the radius is 6 centimeters. Look at the figure again, and consider the cross‐sectional areas shown, at some height T from the top of the cylinder. 3 Hydrodynamics and Mass Transfer in a Modified Three-phase Airlift Loop Reactor 93 But in Fig. If a solid has a cross-sectional area given by the function A(x), what integral should be evaluated to nd the volume of the solid? 2. Calculate the surface area of the sphere. What is the volume of the. A model configuration is shown in Figure 18. Each circular volume element will have a cross-sectional area ˇr2. How are mass, density, and volume related? How is the mass of an object with varying density computed? What is is the center of mass of an object, and how are definite integrals used to compute it?. The frustum is a cone with the top cut off by making a slice parallel to the base.
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https://mathematica.stackexchange.com/questions/197993/why-doesnt-mathematica-evaluate-the-series-with-recursion-relationship-as-expec
|
Why doesn't Mathematica evaluate the series with recursion relationship as expected?
I define a series $$dg(i)$$ as the $$i$$th derivative of a function $$g[t]$$ for $$i>0$$ and known the first term $$dg(1)=(t-x)g(t)$$. In mathematica, the code is:
dg[t_, i_] := D[dg[t, i - 1], t](*when i>1*);
dg[t_, 1] := (t - x) g[t];
Then I use
Table[{ii, dg[t, ii]}, {ii, 1, 5}] // TableForm //TeXForm
$$\begin{array}{cl} 1 & g(t) (t-x) \\ 2 & (t-x) g'(t)+g(t) \\ 3 & (t-x) g''(t)+2 g'(t) \\ 4 & g^{(3)}(t) (t-x)+3 g''(t) \\ 5 & g^{(4)}(t) (t-x)+4 g^{(3)}(t) \\ \end{array}$$
The question is why $$g'(t)$$ in the second term $$dg(2)$$ is not replaced by $$(t-x) g(t)$$? I need all the terms of the series is expressed only by $$g(t)$$ instead of $$g'(t)$$.
Could you help me? Thank you.
• There is nothing in those definitions to connect g'[t] to dg[1]. – Daniel Lichtblau May 9 at 12:12
• @DanielLichtblau I have modified the question to make it understandable. Thank you. – tanghe2014 May 9 at 14:29
• if dg(t) = (t-x)g(t) why would you use D[]? Just remove that – mikuszefski May 15 at 7:25
Clear[g]
g /: g' = (# - x) g[#] &;
dg[t_, i_] := D[g[t], {t, i}]
Table[dg[t, i] // Simplify, {i, 1, 5}] // TableForm
These polynomials are essentially
Table[y^Floor[n/2] Sum[y^-i n!/(i! 2^i (n - 2 i)!), {i, 0, n/2}], {n, 0, 10}]
with y=t-x, up to a global factor of g[t].
• Thank you. I am not familiar with TagSet[] and &. They are very powerful. I have tried your code. It is laconical and powerful! How about to get the explicit expression of $g^{(n)}(t)$ as the $n$th derivative of $g(t)$? – tanghe2014 May 10 at 0:53
• @tanghe2014 there you go. – AccidentalFourierTransform May 10 at 2:49
It is possible to instead directly give definitions for the derivatives of g:
g' = (# - x) g[#]&;
Derivative[n_?Positive][g] := Derivative[n-1][g']
Then:
FullSimplify @ Column @ Table[Derivative[n][g][t], {n, 0, 5}] //TeXForm
$$\begin{array}{l} g(t) \\ g(t) (t-x) \\ g(t) \left((t-x)^2+1\right) \\ g(t) \left((t-x)^2+3\right) (t-x) \\ g(t) \left((t-x)^4+6 (t-x)^2+3\right) \\ g(t) \left((t-x)^4+10 (t-x)^2+15\right) (t-x) \\ \end{array}$$
in agreement with Accidental's answer. With this approach, you can obtain the series expansions of g just using Series:
Series[g[t], {t, 0, 4}] //TeXForm
$$g(0)-g(0) t x+\frac{1}{2} t^2 \left(g(0) x^2+g(0)\right)+t^3 \left(-\frac{1}{6} g(0) x^3-\frac{1}{2} g(0) x\right)+\frac{1}{24} t^4 \left(g(0) x^4+6 g(0) x^2+3 g(0)\right)+O\left(t^5\right)$$
or
Series[g[t], {t, x, 4}] //TeXForm
$$g(x)+\frac{1}{2} g(x) (t-x)^2+\frac{1}{8} g(x) (t-x)^4+O\left((t-x)^5\right)$$
• Thank you. g' = (# - x) g[#]& is new for me and it is very helpful. Could you give more advice to get the general term of $g^{(n)}(t)$ as the $n$th derivative of $g(t)$? – tanghe2014 May 10 at 0:46
A simple method uses a Rule[] to replace g'[t] with its value:
ClearAll[x, g, dg];
dg[t_, 0] := g[t];
dg[t_, i_Integer /; i > 0] := D[dg[t, i - 1], t] /.
g'[t] -> (t - x) g[t];
As a simple test try
dg[t, 5] // Simplify // InputForm
which returns the result
(15 + 10*(t - x)^2 + (t - x)^4)*(t - x)*g[t]
as it should.
Notice that in this method the derivative of g is completely contained in the dg[] rule. If you try g'[t] it get returned unevaluated. An alternative method is to attach the definition of g' directly to g itself. The code for this using a pure function is
g /: g' = Function[t, (t - x) g[t]];
Now we can use this definition directly so that
D[g[t], {t, n}] == dg[t, n]
evaluates to True for all $$n \ge 0$$.
The $$n\,$$th derivative of $$g(t)$$ is $$g(t)$$ multiplied by an $$n\,$$th degree polynomial in $$(t-x)$$ known as the modified Hermite polynomials -- the coefficients of which are given by the OEIS sequence A099174. Thus, the following code
Expand[ dg[t, n] == g[t] HermiteH[n, I(t-x)/Sqrt[2]]
(-I Sqrt[1/2])^n]
evaluates to True for all $$n\ge 0.$$ Note that using
DSolve[ g'[t] == (t - x)g[t] && g[0] == 1, g[t], t]
give the solution $$\ g(t) = e^{t^2/2 - tx}\$$ which is the exponential generating function for modified Hermite polynomials.
• Thank you. I get your point now. Rule[] is very helpful. Is it possible to get the general expression of $dg(t,n)$ using Mathematica? – tanghe2014 May 10 at 0:41
• @tanghe2014 If you mean by "using Mathematica" I have appended a general expression at the end of my answer. – Somos May 10 at 3:57
| 2019-09-17T03:40:59 |
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https://math.stackexchange.com/questions/2341368/trig-identity-proof-frac1-sin-theta-cos-theta-frac-cos-theta1/2341670
|
# Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$
I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides. $$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$
• The procedure you have suggested should work. Please show what you have done and explain clearly where you are stuck. – David Jun 30 '17 at 1:17
• To help show your work better, here's a starting point into the instructions for how to format math formulas on this site: math.stackexchange.com/help/notation – David K Jun 30 '17 at 1:30
• Have you tried to show the two quotients on the left are equal, and used the addition theorem for tan on the right? – Somos Jun 30 '17 at 1:56
• Yeah I got up to that part but don't know how to get rid of the pi/4 – juA Jun 30 '17 at 1:57
• Did you use the fact that $\tan(\pi/4) = 1$? Also, could you show us what you have done? – N. F. Taussig Jun 30 '17 at 2:00
hint: \begin{align} & \cos\theta = \cos^2(\theta /2) - \sin^2(\theta/2)= (\cos (\theta/2) +\sin (\theta/2))(\cos (\theta/2) - \sin (\theta/2)), \\[10pt] & 1 \pm\sin \theta = (\cos (\theta/2) \pm \sin(\theta/2))^2. \end{align}
• Ok, I understand the first formula. How did you get the second one (after the comma)? – juA Jun 30 '17 at 1:31
• To juA :.....$(\cos A/2 \pm \sin A/2)^2=$ $(\cos^2 A/2+\sin^2 A/2)\pm 2\cos A/2 \sin A/2=$ $1\pm 2\cos A/2 \sin A/2=$ $1\pm \sin A.$ – DanielWainfleet Jun 30 '17 at 2:59
• Here's the derivation of the part after the comma in some detail: \begin{align} & \left( \cos\left( \frac\theta2 \right) \pm \sin\left( \frac\theta2\right) \right)^2 \\ \\ = {} & \cos^2 \left(\frac\theta2\right) \pm 2\cos\left( \frac\theta2\right)\sin \left( \frac\theta2\right) + \sin^2 \left( \frac\theta2 \right) \\ \\ = {} & 1 \pm 2 \cos \left( \frac\theta2 \right) \sin \left( \frac\theta2\right) \quad \text{since } \sin^2\alpha+\cos^2\alpha = 1\\ \\ = {} & 1 \pm \sin\theta \qquad \text{by the double-angle formula for the sine function.} \end{align} $${}$$ – Michael Hardy Jun 30 '17 at 3:21
• Nice. I actually did exactly that in my proof, I just didn't treat it as an identity and go straight to it – juA Jun 30 '17 at 6:47
Let $\dfrac\theta2+\dfrac\pi4=y\implies\theta=2y-\dfrac\pi2$
$$\sin\theta=\cdots=-\cos2y,\cos\theta=\cdots=\sin2y$$
$$\dfrac{1+\sin\theta}{\cos\theta}=\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=?$$
$$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin2y}{1+\cos2y}=?$$
• An elegant, not to mention efficient, approach. – N. F. Taussig Jun 30 '17 at 8:48
Set $\theta=\pi/2-x$, so the left-hand side becomes $$\frac{1+\cos x}{\sin x}+\frac{\sin x}{1-\cos x}$$ Now recall that $$\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}= \frac{1-\cos x}{\sin x}$$ so the left-hand side is actually $$2\cot\frac{x}{2}=2\tan\Bigl(\frac{\pi}{2}-\frac{x}{2}\Bigr)= 2\tan\Bigl(\frac{\pi}{2}+\frac{\theta}{2}-\frac{\pi}{4}\Bigr)$$
LHS: $$\frac{(1 + \sin\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta)} + \frac{\cos\theta}{1 - \sin\theta} = \frac{2\cos\theta}{1 - \sin\theta}.$$
RHS: $$2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)=2\frac{\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}+1}{1-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}}=2\frac{\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=$$ $$2\frac{\left(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right)\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}=\frac{2\cos\theta}{1 - \sin\theta}.$$
| 2019-06-16T17:43:11 |
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https://mathematica.stackexchange.com/questions/99653/syntax-for-higher-rank-tensor-multiplication
|
# Syntax for higher rank tensor multiplication
I am trying to do some matrix multiplication in Mathematica but I just cannot figure out the correct syntax for my problem.
I want to write the following: $$A\pmatrix{a & b\\c & d}+B\pmatrix{e & f\\g & h}=\pmatrix{Aa+Be & Ab+Bf\\Ac+Bg & Ad+Bh}$$ as $$\pmatrix{\pmatrix{a & b\\c & d}\\\pmatrix{e & f\\g & h}}\pmatrix{A\\B}=\pmatrix{Aa+Be & Ab+Bf\\Ac+Bg & Ad+Bh}$$
So far I found out that this generates
{{{{a, b}, {c, d}}}, {{{g, h}, {i, j}}}};
MatrixForm[%]
the first tensor. But when I multiply this by {e,f} the result is incorrect. This
{{{{a, b}, {c, d}}}, {{{e, f}, {g, h}}}}.{A, B};
MatrixForm[%]
gives me this as a result: $$\pmatrix{\pmatrix{Aa & Bb\\Ac & Bd}\\\pmatrix{Ae & Bf\\Ag & Bh}}$$
I think that my problem has an easy fix... probably just some brackets that need to be added, but I just didn't manage to figure it out.
Thanks
Philipp
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Nov 17 '15 at 0:27
• You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this this meta Q&A helpful – Michael E2 Nov 17 '15 at 0:28
You almost had it, with just two things going wrong. First, you had a higher-rank tensor than you wanted, which can be solved by peeling off a layer of curly braces, and the two $2 \times 2$ matrices are effectively transposed, suggesting we should reverse the order of multiplication. You want:
{A, B}.{{{a, b}, {c, d}}, {{e, f}, {g, h}}}
which gives
$$\left( \begin{array}{cc} a A+B e & A b+B f \\ A c+B g & A d+B h \\ \end{array} \right)$$
as desired.
You definitely want to look up the functions Inner and Outer. Outer gives you a generalized outer product, and it is extremely useful. Inner gives you a generalized inner product. See this question and the answers below it for how to think about these functions.
It can be tricky to use and figure out the syntax, so here's a start. For your problem, let's define the matrices and vector as
mat1 = Array[a, {2, 2}];
mat2 = Array[b, {2, 2}];
vec = {A, B};
Then, the direct translation of what you're doing is
(combo1 = A Array[a, {2, 2}] + B Array[b, {2, 2}]) // MatrixForm
resulting in
Using Inner:
combo2 = Inner[Times, {Array[a, {2, 2}], Array[b, {2, 2}]}, {A, B}, Plus, 1];
You can verify that they are the same:
combo1 === combo2
(* True *)
Note that the last (optional) argument to Inner is necessary here. From the documentation:
Inner[f, Subscript[list, 1], Subscript[list, 2], g, n] contracts index n of the first tensor with the first index of the second tensor.
I guess in this case, n == 1 is not the default.
If you would like to make a simple function that does this without all of the extra typing, do this:
dot[twoTensor_ /; Dimensions@twoTensor === {2, 2, 2}, oneTensor_ /; Dimensions@oneTensor == {2}] := Inner[Times, twoTensor, oneTensor, Plus, 1]
You can also take direct advantage of Dot using replacement rules:
Clear[aMat, bMat]
Dot[{aMat, bMat}, {A, B}] /. {aMat -> Array[a, {2, 2}], bMat -> Array[b, {2, 2}]}
| 2020-05-28T05:22:48 |
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https://matholympiad.org.bd/forum/viewtopic.php?f=17&t=1302&view=unread
|
2003 IMO Problem 5
Discussion on International Mathematical Olympiad (IMO)
sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
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Contact:
2003 IMO Problem 5
Let all $x_i$ are real for all $i=1,2,...,n$ and $x_i \geq x_j$ if and only if $i\geq j$
(a)
Prove that $\left ( \sum_{i,j=1}^{n}\left | x_i-x_j \right | \right )^2\leq \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}\left ( x_i-x_j \right )^2$
(b)
Prove that equality holds if and only if $\{x_i\}$ is an arithmetic progression .
Comment: 100th post
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(-$from$ "$THE$ $UGLY$ $TRUTH$" )
sourav das
Posts: 461
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Re: 2003 IMO Problem 5
I'm also posting my solution. If there is
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )
*Mahi*
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Re: 2003 IMO Problem 5
Well, the problem statement says , equality appears when $\{x_i\}$ is an arithmetic progression so I think that the first move should consist of Cauchy Schwarz . Though , the solution here appears alright, and so congratulations for completing such a tedious one!
Use $L^AT_EX$, It makes our work a lot easier!
sourav das
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Re: 2003 IMO Problem 5
*Mahi* wrote:Well, the problem statement says , equality appears when $\{x_i\}$ is an arithmetic progression so I think that the first move should consist of Cauchy Schwarz .
Tell me Mahi, how can you assume using any specific kind of tactics at the very beginning? Please give us an example. And this question is for all who have a lot of experience in solving problems. Let's discuss and restart to solve this problem with our discussion.It'll help us all to learn choosing tactics. (I'm still a learner of solving problems )
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )
FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am
Re: 2003 IMO Problem 5
I completely agree with Sourav. So Mahi, please tell how you started thinking about the first step.
sourav das
Posts: 461
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Re: 2003 IMO Problem 5
Is it because of Cauchy-Schwarz equality condition has similarity with the difference between two elements of an arithmetic progression?
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )
*Mahi*
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Re: 2003 IMO Problem 5
You got it, all I tried was multiplying different arithmetic progression with $\sum_{i,j=1}^{n}\left ( | x_i-x_j | \right )^2$, first with easy ones like $\sum 1$ and then others, which you can find out. And trying well known techniques sometimes has it's benefits too.
I'm a learner too , so we should all try together!
Last edited by *Mahi* on Fri Oct 28, 2011 9:49 pm, edited 1 time in total.
Use $L^AT_EX$, It makes our work a lot easier!
sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
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Re: 2003 IMO Problem 5
Mahi, Fahim, any progress? For using Cauchy-Schwarz inequality we need product of two sum of square sequences. But since in our left side, there are terms which are not square. Confusing....
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )
*Mahi*
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Re: 2003 IMO Problem 5
sourav das wrote:Mahi, Fahim, any progress? For using Cauchy-Schwarz inequality we need product of two sum of square sequences. But since in our left side, there are terms which are not square. Confusing....
In Cauchy-Schwarz LHS has two sum of square sequences, right? And remember that the one greater should be the LHS here.
Use $L^AT_EX$, It makes our work a lot easier!
| 2020-12-05T14:33:25 |
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http://mathhelpforum.com/trigonometry/142582-finding-composite-inverse-function-print.html
|
# Finding a composite inverse function.
• May 1st 2010, 10:11 PM
JennyFlowers
Finding a composite inverse function.
Hey guys! I need to find:
$tan(cot^{-1}(\frac{-1}{\sqrt{3}}))$
I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."
As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:
http://i43.tinypic.com/s2e5bc.jpg
Now, using this drawing, I would find the tangent (and the answer to the expression) to be: $-\sqrt{3}$
Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.
I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.
Thanks for any help!
• May 1st 2010, 10:29 PM
harish21
Quote:
Originally Posted by JennyFlowers
Hey guys! I need to find:
$tan(cot^{-1}(\frac{-1}{\sqrt{3}}))$
I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."
As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:
http://i43.tinypic.com/s2e5bc.jpg
Now, using this drawing, I would find the tangent (and the answer to the expression) to be: $-\sqrt{3}$
Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.
I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.
Thanks for any help!
Actually,
$cot^{-1}(\frac{-1}{\sqrt 3})$ returns you an angle
so, $cot^{-1}(\frac{-1}{\sqrt 3}) = \frac{-\pi}{3}$
and $tan(\frac{-\pi}{3})= -tan(\frac{\pi}{3})= -\sqrt{3}$
$\therefore tan(cot^{-1}(\frac{-1}{\sqrt 3})) = -\sqrt{3}$
• May 2nd 2010, 04:39 AM
JennyFlowers
Out of curiosity, what difference would this make?
$tan^{-1}(cot(\frac{-1}{\sqrt{3}}))$
Does switching which function is the inverse change the outcome?
Thanks for the help!
• May 2nd 2010, 07:11 AM
Soroban
Hello, JennyFlowers!
With a little thought, this is an "eyeball" problem.
(You get the answer by looking at the problem.)
Quote:
Find: . $\tan\left[\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right)\right]$
I read this as: "Tangent of the angle whose cotangent is $-\tfrac{1}{\sqrt{3}}$ " . Good!
$\text{Let }\theta \,=\,\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) \quad \hdots \quad \theta\text{ is an angle whose cotangent is }-\tfrac{1}{\sqrt{3}}$
Then: . $\cot\theta \:=\:-\frac{1}{\sqrt{3}} \:=\:\frac{adj}{opp}$
The right triangle looks like this:
Code:
* * | * | _ * | √3 * | * θ | * - - - - - * -1
And we see that: . $\tan\theta \:=\:\frac{\sqrt{3}}{\text{-}1} \:=\:\text{-}\sqrt{3}$
. . Hence: . $\theta \:=\:\tan^{-1}\left(\text{-}\sqrt{3}\right)$
Do you see what we've done?
We've shown that: . $\cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right) \;=\;\tan^{-1}\left(\text{-}\sqrt{3}\right)$
Give it some thought . . . The reasoning should become clear. .**
So we can replace $\cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right)$ with $\tan^{-1}\left(\text{-}\sqrt{3}\right)$
And the original problem becomes: . $\tan\left(\tan^{-1}(\text{-}\sqrt{3})\right)$
. . which equals (of course): . $-\sqrt{3}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
The same reasoning that allows us to say:
. . $\begin{array}{ccc}\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) &=& \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) \\ \\[-2mm]
\csc^{-1}(\sqrt{2}) &=& \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) \end{array}$
• May 2nd 2010, 07:47 AM
JennyFlowers
Thank you, Soroban, I think I understand now. But how did you determine that the angle should be drawn in quadrant II? Was it just luck that I got the same answer by drawing in quadrant IV?
• May 2nd 2010, 08:34 AM
HallsofIvy
"Tangent" is periodic with period $\pi$. Drawing your angle in quadrant IV rather than quadrant II was just adding $\pi$ radians to it.
In any case, since tan(x)= 1/cot(x), $tan(cot^{-1}(-\frac{1}{\sqrt{3}})= 1/cot(cot^{-1}(\frac{1}{\sqrt{3}}))= -1/\frac{1}{\sqrt{3}}= -\sqrt{3}$.
| 2017-05-28T09:35:58 |
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https://mathoverflow.net/questions/241880/an-integral-identity-evaluating-the-gamma-function
|
# An integral identity evaluating the gamma function
While reading a number theory paper I encountered the identity
$$\int_{- \infty}^{\infty} (1 + x^2)^{ - \frac{z}{2} - 1} dx = \sqrt{\pi} \frac{ \Gamma(\frac{z + 1}{2}) }{\Gamma(\frac{z}{2} + 1)},$$
apparently true for all $z \in \bf{C}$ for which the integral on the left converges absolutely. The author offers neither a justification nor a citation, so apparently this is 'well known'.
I thought of two ideas for how to prove it: (1) Use integration by parts to prove that both sides satisfy the same functional equation, and then attempt to adapt the proof of the Bohr-Mollerup theorem to prove that this functional equation characterizes the function; (2) use integration by parts to analytically continue the left-side function in $z$, determine all the poles and their residues, and attempt to use Weierstrass factorization to prove that this characterizes the function.
Both approaches looked rather involved and I didn't attempt to work out the details of either. Is this, indeed, 'well known'? Is there a suitable reference which proves this, along with other identities of this nature? And to what extent does the identity generalize?
• Did you look in Gradshteyn and Ryzhik? Jun 10, 2016 at 19:29
• @Lucia: I didn't; I will take your comment as advice that I should! Jun 10, 2016 at 19:44
• Presumably one starts with the identity $\int_0^\infty e^{-at} t^s \frac{dt}{t} = \Gamma(s) a^{-s}$ with $a = 1+x^2$ and $s = \frac{z}{2}+1$, and then integrates in $x$ using Fubini's theorem? Jun 10, 2016 at 19:46
• Basically, Gamma functions show up whenever one tries to write multiplicative characters such as $a \mapsto a^{-s}$ in terms of additive characters such as $a \mapsto e^{-at}$. Here, $x \mapsto e^{-t(1+x^2)}$ is significantly easier to integrate than $x \mapsto (1+x^2)^{-\frac{z}{2}-1}$, so that suggests the use of the Gamma function to transform the latter into a suitable average of the former. The same trick can be used to prove the functional equation: terrytao.wordpress.com/2014/12/15/… Jun 10, 2016 at 19:53
• Regarding a reference, the particular identity in question can be deduced from the general properties of Bessel potentials: en.wikipedia.org/wiki/Bessel_potential . Stein's "Singular integrals..." is a good reference. Jun 10, 2016 at 20:40
Yes, there is a trick which generalizes to analogous integrals on the classical cones, using the Gamma functions attached to these cones. In this, the simplest case, the starting point is the observation that $\int_0^\infty e^{-ty}\,t^s\;{dt\over t}=y^{-s}\cdot \Gamma(s)$ for $y>0$, and then for $\Re(y)>0$ by analytic continuation. This identity has the obvious analogues for other classical cones, such as positive-definite symmetric real $n$-by-$n$ matrices, or light cones, etc.
Thus, your given integral is ${1\over \Gamma(z/2+1)} \int_0^\infty\int_{\mathbb R} t^{z/2+1}\,e^{-t(x^2+1)}\;dx\;{dt\over t}$. Integrating the Gaussian in $x$ first gives ${1\over \Gamma(z/2+1)} \int_0^\infty t^{{z+1\over 2}}\,e^{-t}\;dx\;{dt\over t}$.
EDIT: While we're here, it's probably reasonable to give some further explicit details, as well as historical context. First, historically, C. L. Siegel's 1939 paper on "Siegel modular forms" did broach issues similar to the present. Among others along the way, G. Shimura's papers in the early 1980's used such integral identities quite aggressively, and proved (among other things) identities generalizing the classical ones involving confluent hypergeometric (a.k.a. Bessel) functions.
An example of a less trivial identity: let $C_n$ be the cone of positive-definite symmetric real $n$-by-$n$ matrices, and $\Gamma_C(s)=\int_{C_n} (\det t)^s\,e^{-{\mathrm tr}\, t}\,dt/\det^nt$. On one hand, by an induction, we can show that this is a product of ordinary Gammas $\Gamma(s)\Gamma(s-1/2)\Gamma(s-1)\Gamma(s-3/2)... \Gamma(s-(n-1)/2)$, by fooling around with changes of variables. On the other hand, in parallel to the much-better-known identity for the cone $(0,\infty)$, we have $\int_{C_n} (\det t)^s\,e^{{\mathrm tr}\,ty}\,dt/\det t^n=(\det y)^{-s}\Gamma_C(s)$, by the slighly subtler change of variables replacing $t$ by $y^{-1/2}ty^{-1/2}$, for $y\in C_n$. And so on...
It is nothing but beta-function. Consider only positive $x$ and denote $1/(x^2+1)=t$. You get $$\int_0^\infty (1+x^2)^{-z/2-1}dx=\frac12 \int_0^1 t^{z/2-1/2}(1-t)^{-1/2}dt=\frac12 B((z+1)/2,1/2)=\\=\frac{\Gamma(1/2)\Gamma((z+1)/2)}{2\Gamma(z/2+1)}=\sqrt{\pi}\frac{\Gamma((z+1)/2)}{2\Gamma(z/2+1)}.$$
I think, there should be a reference for all integrals of the form $\int_0^\infty x^a (1+x^b)^{-c} dx$, which reduce to the beta-function after the change of variables $1/(1+x^b)=t$. At least it is surely well-known. In Demidovich problems book it is ex.3853, for example.
• I checked that this works too; thank you very much! Jun 10, 2016 at 21:45
• It is also possible to apply the Slater theorem - a general key to such integrals. Jun 14, 2016 at 13:45
| 2022-08-08T08:22:13 |
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https://math.stackexchange.com/questions/3314323/determine-the-maximum-possible-volume-excluding-the-volume-of-the-legs
|
# Determine the maximum possible volume, excluding the volume of the legs.
A table is to be constructed by gluing together 68 cubes of dimension $$1\times 1\times 1$$. All four legs and the rectangular top will be formed by the cubes. The four legs must be the same length and must be one cube thick, and the top is one cube thick, as well. What is the maximum possible volume of the space between the table's top and the floor, excluding the space taken up by the table's legs?
My solution:
Let $$a$$ be the area of the top and $$c$$ be the length of a leg.
\begin{align} a + 4c &= 68 \cr a &= 68 - 4c \end{align} We try to find the maximum value of ac, and then subtract 4c:
$$ac = (68 - 4c)c = -4c^2 + 68c$$
Completing the square gets $$-4(c - 17/2)^2 + 289$$.
The lowest value of $$ac$$ is when $$c= 17/2$$, and the lowest value is $$289$$. Then we subtract $$289 - 4c = 289 - 34 = 255$$.
Therefore, the maximum volume when excluding the volume of the legs is $$255$$.
However, the actual answer is $$256$$ as the maximum volume when excluding the legs. Where did I go wrong?
You want to maximise the volume excluding the legs, which is $$ac-4c$$. You maximised $$ac$$ instead.
We have that $$ac-4c=-4c^2+64c$$.
Now complete the square:
$$-4c^2+64c=256-(16-2c)^2$$
The maximum of $$256$$ is at $$c=8$$.
• But maximizing ac - 4c gives is maximizing -4c^2 + 64c, and completing the square gives -4(c - 8)^2 - 64. When c = 8, the expression equals -64. What exactly do you mean by maximizing ac - 4c? – David Dong Aug 5 '19 at 17:00
Noting that subtracting $$68-4=64$$, we can see that, if the legs are only $$1$$ cube high, that leaves $$8\times 8\times 1=64$$ for the table top. The space under the table (minus the legs) is $$64-(4\times 1\times 1)=60$$ so the volume from the table top to the floor is $$2*64-4=124$$.
If we let the legs be $$4\times 2=8$$, the table top is $$5\times 12=60$$, volume: $$3*60-8=172$$.
If we let the legs be $$4\times 3=12$$, the top is $$68-12=54$$ volume: $$4\times 54-12=204$$
If we let the legs be $$4\times 4=16$$, the top is $$68-16=52$$, volume: $$5\times 52-16=244$$
If we let the legs be $$4\times 5=20$$, the top is $$68-20=48$$, volume: $$6\times 48-20=268$$
If we let the legs be $$4\times 6=24$$, the top is $$68-24=44,$$ volume: $$7\times 44-24=284$$
If we let the legs be $$4\times 7=28$$, the top is $$68-28=40,$$ volume: $$8\times 40-28=292$$
If we let the legs be $$4\times 8=32$$, the top is $$68-32=36,$$ volume: $$9\times 36-32=292$$
If we let the legs be $$4\times 9=36$$, the top is $$68-36=32,$$ volume: $$10\times 32-36=284$$
If we let the legs be $$4\times 10=40$$, the top is $$68-40=28,$$ volume: $$11\times 28-40=268$$
It appears $$292$$ is the maximum volume with outer dimensions of $$8\times 5\times 8$$ or $$9\times 6\times 6$$
• This is incorrect. My original answer was incorrect and I found out why, and Angela Richardson found another solution. – David Dong Aug 5 '19 at 18:44
• @David Dong I realized what silly errors I made in my answer before but please take a look at my edited version. I think Angela Richardson is wrong, now. – poetasis Aug 6 '19 at 2:07
• @David Dong I have no problem with anyone doing anything with what I post but I do recommend checking my math first. I made silly mistakes on my first try; I could be mistaken again. You might even draw a picture of the $292$ table tops with with leg lengths of $7$ and $8$ respectively. OTOH, I found an error in a college math book in 1970. What this problem probably wants is an equation for the volume where you could then take the first derivative and find where slope of the curve (stair step?) equals zero. Let me know if you would like me to try my hand at that. Good luck. – poetasis Aug 6 '19 at 17:15
• @David Dong I couldn't stand not knowing so I tried my hand at derivatives and got the same result (I think). – poetasis Aug 6 '19 at 18:22
Given 68 cubes, the table top is $$68-4h$$ where $$h$$ is the height of the legs.
Under the table we have $$(h)(68-4h)=68h-4h^2$$ and the total volume (sans legs) is $$top+under-legs=(68-4h)+(68h-4h^2)-(4h)=68+60h-4h^2$$.
For a continuous function, the max volume can be found by where the increasing/decreasing slope of the given function is zero.
The first derivative $$\frac{d}{dh}(68+60h-4h^2)=-8h+60\implies 60-8h=0\implies 8h=60$$
This tells us that the max leg height is $$\frac{60}{8}=7.5$$ If fractional heights were allowed the max volume would be $$68+60h-4h^2=68+(60\times7.5)-(4\times7.5^2)=68+450-225=293$$ but we are are restricted to the nearest integers. Soooo
$$68+60h-4h^2=68+(60\times7)-(4\times49)=68+420-196=292$$ and $$68+60h-4h^2=68+(60\times8)-(4\times64)=68+480-256=292$$
• I don't know anything about derivatives, but maybe the problem here is that you used the nearest integer? – David Dong Aug 7 '19 at 2:41
• I think you would like derivatives look them up. As for integers, no integer solutions appear to give the book answer so I believe the book answer or the book's phrasing is wrong. Perhaps they want the max volume between the $tabletop$ (not the table's top) and the floor. That would make a difference. – poetasis Aug 7 '19 at 3:39
• 8*6*6 seems to work though, with 8 being the leg length and 36 being the top area. – David Dong Aug 7 '19 at 4:07
| 2020-02-24T03:08:47 |
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http://mathhelpforum.com/advanced-algebra/48457-spanning-set.html
|
# Math Help - spanning set
1. ## spanning set
1. The following set is a spanning set for R^3.
S = {(1,2,-1),(0,3,4),(2,1,-6),(0,0,2)}
Find a subset of S that is a basis for R^3.
And,
2. Find a basis for the solution space V of the following linear system in four variables x1, x2, x3, x4:
x1+x2=0
-x2+3x3=0
Thanks in advance for great help guys :-).
2. Originally Posted by sunneej
1. The following set is a spanning set for R^3.
S = {(1,2,-1),(0,3,4),(2,1,-6),(0,0,2)}
Find a subset of S that is a basis for R^3.
Mr F says: Remove one of the vectors that can be written as a linear combination of the other three vectors.
And,
2. Find a basis for the solution space V of the following linear system in four variables x1, x2, x3, x4:
x1+x2=0
-x2+3x3=0
Mr F says: Where does x4 appear in the above equations?? First you need to solve these two equations for x1, x2, x3. There are an infinite number of solutions. These solutions can be written in parametric form. Let ${\color{red}x_2 = t}$, say, where t can be any real number. You should be able to construct a basis from this solution .....
Thanks in advance for great help guys :-).
If you're still stuck, show your working and state where you're stuck.
3. ## Thanks but...
Thanks for the hints.
But for number 1, I couldn't find one of the vectors that can be written as a linear combination of the other three vectors.
And for number 2, I found the solutions, and because it said x4 in question, I included it though it will just be 0. Hence,
x1 = -3t; x2 = 3t; x3 = t; x4 = s.
Could you give me an advice as what to do from here onwards?
Thank you.
4. Originally Posted by sunneej
Thanks for the hints.
But for number 1, I couldn't find one of the vectors that can be written as a linear combination of the other three vectors.
And for number 2, I found the solutions, and because it said x4 in question, I included it though it will just be 0. Hence,
x1 = -3t; x2 = 3t; x3 = t; x4 = s.
Could you give me an advice as what to do from here onwards?
Thank you.
1. $(1, \, 2, \, -1) = \frac{1}{2} \, (2, \, 1, \, -6) + \frac{1}{2} \, (0, \, 3, \, 4)$.
2. $(x_1, \, x_2, \, x_3, \, x_4) = t(-3, \, 3, \, 1, \, 0) + s(0, \, 0, \, 0, \, 1)$. A basis should be obvious.
5. Yeah, I could find one of the vectors that can be written as a linear combination of the other TWO vectors, but not one.
But what happens to (0,0,2)?
Cheers.
6. Originally Posted by sunneej
Yeah, I could find one of the vectors that can be written as a linear combination of the other TWO vectors, but not one.
But what happens to (0,0,2)?
Cheers.
The linear combination doesn't have to be with all of the other vectors ..... just so long as there's a vector that can be written as a linear combination of some of the others ....
But if it makes you feel better: $(1, \, 2, \, -1) = \frac{1}{2} \, (2, \, 1, \, -6) + \frac{1}{2} \, (0, \, 3, \, 4) + 0 \, (0, \, 0, \, 2)$
| 2016-02-14T07:08:17 |
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http://math.stackexchange.com/questions/150892/when-the-localization-of-a-ring-is-a-field
|
# When the localization of a ring is a field
Let $R$ be a commutative noetherian ring with no nonzero nilpotents. Let $p$ be a minimal prime of $R$. Could you help me to prove that $R_p$ is a field?
-
Try proving the contrapositive. (If $R_p$ is not a field, it contains an element which is not invertible, so...?) – Qiaochu Yuan May 28 '12 at 20:03
Hint: A ring is a field if and only if its only prime ideal is $(0)$. The prime ideals of $R_p$ are in one-to-one correspondence with the prime ideals of $R$ which are disjoint from $R\setminus p$, that is the primes contained in $p$. What can you conclude?
-
If $qR_p$ is a prime ideal of $R_p$ then $q\subset p$ and so $p=q$, so the only prime ideal of $R_p$ is $pR_p$, am I right? How can I go on? I think I have to use the fact that the nilradical is the intersection of all the minimal primes and that there are only finitely many minimal primes because $R$ is noetherian, but I don't know how to put all this together, any other hints? – Alex M May 28 '12 at 20:23
@AlexM The nilradical of $R_p$ is the intersection of all minimal primes of $R_p$. But the only minimal prime of $R_p$ is $pR_p$, therefore $pR_p$ is the nilradical. But what is the nilradical of $R_p$? – Alex Becker May 28 '12 at 20:30
got it thank you, $\mathrm{nilrad}\;R_p=(\mathrm{nilrad}\;R)_p=0R_p$ and so $pR_p=0R_p$ – Alex M May 28 '12 at 20:38
It seems to me that the only thing that requires $R$ notherian is that the set on minimal primes is finite, but we didn't use it. The correspondence between prime ideals of $R_p$ and prime ideals of $R$ contained in $p$, the fact that the nilradical is the intersaction of the minimal primes and the fact that $\mathrm{nilrad}\;R_p=(\mathrm{nilrad}\;R)_p$ don't seem to require $R$ noetherian, what am I missing? – Alex M May 28 '12 at 20:48
A reduced local ring of dimension zero is a field. That the local ring is a localization is a red herring.As both Alexes and @QiL already said, noetherianness is irrelevant too. (+1 for the answer). – Georges Elencwajg May 28 '12 at 21:05
Here is a proof that uses nowhere the results of Noetherian rings. All the results I used below are from chapters 2 - 3 of Atiyah - Macdonald where Noetherian-ness has not come in yet! To prove your problem we use the following result from Atiyah - Macdonald:
A local ring that is absolutely flat is a field.
We already know that given any prime ideal $\mathfrak{p}$ that $R_\mathfrak{p}$ is a local ring and hence it suffices to prove that $R_\mathfrak{p}$ for a minimal prime ideal $\mathfrak{p}$ is absolutely flat. You have already observed above that $\textrm{nilrad} R_\mathfrak{p} = 0$, so if you can prove that every prime ideal in $R_\mathfrak{p}$ is maximal then this would imply (see at the bottom for a proof) that $$R_p/\textrm{nilrad}R_\mathfrak{p} \cong R_\mathfrak{p}$$ is absolutely flat, finishing the problem. Take a prime ideal $S^{-1}\mathfrak{q}$ of $R_\mathfrak{p}$. Then you have already observed that $\mathfrak{q}$ is a prime ideal of $R$ that must be disjoint from $R - \mathfrak{p}$, viz. it is completely contained in $\mathfrak{p}$. By minimality of $\mathfrak{p}$ we are forced to conclude that $\mathfrak{p} = \mathfrak{q}$ and hence $S^{-1}\mathfrak{q} = S^{-1}\mathfrak{p}$ is a maximal ideal. This concludes the proof.
$\hspace{6in} \square$
We prove this exercise first, and then use it to prove the main result I claimed above.
Exercise 3.10(ii) Atiyah - Macdonald: $A_\mathfrak{m}$ being a field for each maximal ideal $\mathfrak{m} \implies$ $A$ is absolutely flat.
Proof: Suppose you given an $A$ - module $M$. Then I claim given any maximal ideal $\mathfrak{m}$ one has that $M_\mathfrak{m}$ is a flat $A_\mathfrak{m}$ - module. Indeed this is clear because $M_\mathfrak{m}$ is just a vector space over $A_\mathfrak{m}$ and hence is isomorphic to $A_\mathfrak{m}^n$ for some $n$. Since $A_\frak{m}$ being a field is absolutely flat, in particular $A_\frak{m}$ as a module over itself is flat. Therefore by exercise 2.4 of Atiyah Macdonald
$$\bigoplus_{i=1}^n (A_\mathfrak{m})_i = A_\mathfrak{m}^n \cong M_\mathfrak{m}$$
is flat as an $A_\mathfrak{m}$ - module for any maximal ideal $\mathfrak{m}$. By proposition 3.10 of Atiyah - Macdonald we conclude that $M$ is flat as an $A$ - module and since $M$ was arbitrary this proves that $A$ is absolutely flat.
$\hspace{6in} \square$
Exercise 3.11 Atiyah - Macdonald: If every prime ideal of a ring $A$ is maximal then $A/\textrm{nilrad} (A)$ is absolutely flat.
By exercise 3.10(ii) Atiyah - Macdonald it suffices to prove that given any maximal $\mathfrak{m}$ of $A/\textrm{nilrad} A$ that $(A/\textrm{nilrad} A)_\mathfrak{m}$ is a field. Now a basic isomorphism in localisation tells us that
$$(A/\textrm{nilrad} A)_\mathfrak{m} \cong A_\mathfrak{m}/ \big(\textrm{nilrad}(A)\big)_\mathfrak{m}$$
so it suffices to prove that $\textrm{nilrad}(A)_\mathfrak{m}$ is a maximal ideal of $A_\mathfrak{m}$. Given what you have done above you should be able to complete this by the following steps:
1. $\big(\textrm{nilrad}(A)\big)_\mathfrak{m} = \textrm{nilrad}({A_\mathfrak{m}})$
2. There is a unique maximal ideal $Q$ of $A_\mathfrak{m}$ that contains $\textrm{nilrad}(A_\mathfrak{m})$.
3. Recall the ideal correspondence between prime ideals of $A_\mathfrak{m}$ and prime ideals of $A$ that don't meet $A - \mathfrak{m}$.
4. Use the fact that every prime ideal of $A$ is maximal and $A_\mathfrak{m}$ is a local ring to conclude your problem.
-
@navigetor23 Hahahaha perhaps I used a nuclear weapon instead of a flyswatter......that's the way I thought about the problem though :D – user38268 Jun 3 '12 at 22:14
Maybe a little bit more clear than the first answer:
$R_p$ itself is reduced and local. Moreover $\dim R_p=0$, so we can ask the following: is it true that a local reduced ring with dimension 0 is a field? Answer: yes! Reduced means that the nilradical of $R$ is $(0)$, but in this case the nilradical is the only prime ideal of $R$, so the only prime ideal of $R$ is (0) which means that $R$ is a field.
-
| 2014-08-01T06:22:49 |
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https://gateoverflow.in/3548/gate-it-2006-question-9
|
21,281 views
In a binary tree, the number of internal nodes of degree $1$ is $5$, and the number of internal nodes of degree $2$ is $10$. The number of leaf nodes in the binary tree is
1. $10$
2. $11$
3. $12$
4. $15$
### 1 comment
Although @Kai and @Arjun ji has provided good answer with reference. But before looking at their answer just try considering tree as directed graph.
A node in a binary tree has degree $0, 1$ or $2$.
We are given no. of $1$ degree node $= 5$, no. of $2$ degree nodes $= 10$.
Total no. of edges $= 1*5 + 2*10 = 25$ (In tree degree is for outgoing edges only, and hence each degree corresponds to an edge)
So, total no. of nodes $= 25 + 1 = 26$ (No. of nodes in a tree is $1$ more than no. of edges).
Now, no. of leaf nodes (nodes with $0$ degree) $= 26 - 5 - 10 = 11$.
Correct Answer: $B$
by
see this "A binary tree is a directed graph and hence degree refers to outgoing degree only. We can also consider it as an undirected graph and apply graph rules- but by default it is a directed graph. " by arjun sir
just drawing an arbitrary tree with "the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10."- we can calculate it.
just the idea ...what will its graph look like?
In a binary Tree,
no of nodes of degree 2 = no of leaves - 1.
No of nodes of degree 1 do not affect no of leaves !
No of leafs = No of nodes of degree 2 + 1 = 10 + 1 = 11
@arjun Sir
@gatecse
A binary tree is a directed graph and hence degree refers to outgoing degree only. We can also consider it as an undirected graph and apply graph rules- but by default it is a directed graph.
Why this definition is not using in this: https://gateoverflow.in/118300/gate2017-1-20
val because in GATE2006 question we are using the definition of a tree defined in DATA STRUCTURES but in GATE2017 we are using the definition of a tree as a graph defined in GRAPH THEORY
NOTE: notice the word 10 vertices.
it must be no. of leaves =number of nodes of degree 2 +1 .
In a directed graph indegree of a graph = outdegree
Outdegree = 1*5 + 2*10
Let number of leaves be x
Indegree = x*1 + 14 *1 (all nodes except root have indegree = 1)
Outdegree = indegree
x = 11
by
$L$ = $T+ 1$
$L$ = number of leaf node
$T$ = number of nodes with 2 child
$L =10+1$
$L= 11$
| 2023-01-28T13:14:31 |
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http://math.stackexchange.com/questions/139377/given-n-a-and-b-does-there-exist-an-x-such-that-b-divides-n-ax
|
# Given $N$, $a$, and $b$, does there exist an $x$ such that $b$ divides $N-ax$?
Given $N$, $a$, $b$ and condition that all are positive integers, how to find whether any positive integer $x$ exists such that $b|(N-ax)$.
And if any such $x$ exists how to calculate minimum value of $x$?
This looks elementary but somehow I am stuck.
-
Sorry pedja. i messed up with the order. – Manoj R May 1 '12 at 13:30
How much do you know about modular arithmetic? – anon May 1 '12 at 13:38
upto modulo operator. – Manoj R May 1 '12 at 13:39
This is equivalent to solving $ax\equiv N\pmod{b}$. A well-known necessary and sufficient condition is $\gcd(b,a)|N$, in which case there are precisely $\gcd(b,a)$ solutions modulo $b$. – Arturo Magidin May 1 '12 at 19:08
You need to consider a couple of things, one of which is an elementary fact about the greatest common divisor. If $b|(N-ax)$, then $yb=N-ax$ for some $y\in\mathbb{Z}$ by definition $\Rightarrow ax+by=N$.
Now we need that basic fact I mentioned. If $(a,b)=N$, then $\exists\;x,y\in\mathbb{Z}$ such that $ax+by=N$ where $(a,b)$ indicates the greatest common divisor of $a$ and $b$. Notice that this means if the GCD of $a$ and $b$ is a factor of $N$, we can also find such an $x$. To actually find the coefficients $x$ and $y$, use the Euclidean algorithm "in reverse". The details are usually present in most elementary number theory texts. Let me know if you have problems filling in the rest.
-
And what is GCD is 1? – Manoj R May 1 '12 at 14:11
The result says that if $c$ and $d$ are relatively prime, there exist $u$ and $v$ such that $cu+dv=1$. (See Extended Euclidean Algorithm.) If $e=\gcd(a,b)$, and $e$ divides $N$, let $N=eM$. Find $u,v$ such that $(a/e)u+(b/e)v=1$, then multiply both sides by $N$. On the right we get $N$. On the left we get $(a/e)uMe+(b/e)vMe=a(Mu)+b(Mv)$. Set $x=Mu$, $y=Mv$. – André Nicolas May 1 '12 at 14:30
Hint $\rm\ \exists\: x\!:\ b\ |\ n-ax\iff \exists\: x,y\!:\ ax+by = n.\:$ Note that the set S of integers $\rm\:n\:$ of this form are closed under subtraction: $\rm\:N = a X + b Y,\ n = ax+by\:$ $\Rightarrow$ $\rm\:N-n = a(X-x)+b(Y-y).$ By this post we deduce that every element of S is a multiple of the least positive element of S, which equals $\rm\:gcd(a,b).\:$ Thus a solution $\rm\:x\:$ exists iff $\rm\:gcd(a,b)\ |\ n,\:$ and, if so, solutions can be found by scaling the solution for $\rm\:n = \gcd(a,b)$ via the extended Euclidean algorithm.
-
Hint :
$$N-ax \equiv 0 \pmod b \Rightarrow N-ax=kb \Rightarrow x=\frac{N-kb}{a}$$
for some integer $k$ .
-
| 2015-09-01T18:44:09 |
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http://www.physicsforums.com/showthread.php?t=624819
|
## Eigenvalue(s) for a matrix
1. The problem statement, all variables and given/known data
Find the eigenvalues of A, when A =[3,1,1; 0,5,0; -2,0,1]
2. Relevant equations
3. The attempt at a solution
I took the det(A-$\lambda$I)=0 and found the characteristic polynomial to be:
-$\lambda$$^{3}$+15$\lambda$$^{2}$-73$\lambda$+115
I couldn't figure out whole roots to the equation and wolframalpha gave me
$\lambda$=5, 5-√2, 5+√2
Does this seem like the right answer? I have checked and double checked the determinant. Would I still be able to find a basis for the eigenspace?
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
Recognitions: Gold Member For matrix: $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 1\end{bmatrix}$$ Here is the characteristic equation that i get: ##-λ^3+9λ^2-25λ+25=0##
Because the second column is [0, 5, 0], the polynomial should look like (5 - x)Q(x), where Q(x) is a second-degree polynomial. So it should be very easy to get its roots. Do it.
## Eigenvalue(s) for a matrix
Quote by sharks For matrix: $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 1\end{bmatrix}$$ Here is the characteristic equation that i get: ##-λ^3+9λ^2-25λ+25=0##
Sorry the matrix was $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{bmatrix}$$
not 2,0,1 in the last row. That was my mistake.
Yes, those are the eigenvalues.
Mentor
Quote by voko Because the second column is [0, 5, 0], the polynomial should look like (5 - x)Q(x), where Q(x) is a second-degree polynomial. So it should be very easy to get its roots. Do it.
The second row is [0, 5, 0].
Quote by Mark44 The second row is [0, 5, 0].
This is not clear from the original notation. Anyway, that does not matter for determinants so my argument holds.
Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Expand the determinant along the second row.
Recognitions:
Gold Member
Quote by voko This is not clear from the original notation.
Well, the original notation used is actually very common and can be recognized as a standard, since it is also used in popular software like MATLAB.
Quote by sharks Well, the original notation used is actually very common and can be recognized as a standard, since it is also used in popular software like MATLAB.
Let's put it this way: not clear for me, I always seem to forget the conventions for rows vs columns in a linear transcription. This may stem from the fact that even when working with paper and pencil, I am about equally likely to write down a matrix column after column and row after row. Let's hope this thread will leave a dent in my state of confusion :)
Recognitions:
Gold Member
Quote by Lord_Sidious Sorry the matrix was $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{bmatrix}$$ not 2,0,1 in the last row. That was my mistake.
Your characteristic equation from post #1 is correct. Now, you just need to solve it. Usually, you would start by trial and error to find one factor of the polynomial (you already know the easy one is (λ-5)) and then do long division to obtain a quadratic in the quotient, which you would then solve using the quadratic formula.
Quote by sharks find one factor of the polynomial (you already know the easy one is (λ-5)) and then do long division to obtain a quadratic in the quotient
The point I made in #3 is that none of this is required. By tackling the determinant along [0, 5, 0], one gets the polynomial directly in the factored form.
| 2013-05-21T11:42:47 |
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https://math.stackexchange.com/questions/2005274/does-the-following-limit-exist
|
# Does the following limit exist??
Write $$h(x) = \begin{cases} 0, x \neq 0 \\ 1, x = 0 \end{cases}$$
$$g(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), x \neq 0 \\ 0 , x = 0 \end{cases}$$
Claim is $\lim_{x \to 0} h \circ g$ does not exist.
However, I believe the limit is zero since: Notice if $x \neq 0$, $h(x) = 0$ is zero no matter what $f$ is. Also, if $x = 0$, then $g=0$ and so $h(g(0)) = h(0) = 1$ and so $\lim_{x \to 0} h(g(x)) = 0$. Am I missing something here?
• You might think again. Can you find a sequence $x_k$ that satisfies $x_k\rightarrow 0$ but $\lim_{k\rightarrow\infty}h(x_k)=1$? Can you find another sequence that satisfies $x_k\rightarrow 0$ but $\lim_{k\rightarrow\infty}h(x_k)=0$? If so, the limit $\lim_{x\rightarrow 0} h(g(x))$ does not exist. – Michael Nov 8 '16 at 15:47
• I thought under the formal definition of the limit, we are interested in increasing small values of epsilon close to the limit, and for these g is always non zero, therefore h is always zero for any non-zero epsilon - so I think the limit is zero – Cato Nov 8 '16 at 15:48
• Why do you say "for these $g$ is always nonzero"? @AndrewDeighton – Michael Nov 8 '16 at 15:49
• Notice that if $g(x) = 0$, then $h(g(x)) = 0$. – Omnomnomnom Nov 8 '16 at 15:50
• I did think again - for increasing small $\epsilon$ there has to exist a smaller value with $1 / \epsilon = n\pi$ - making g zero and at that point $h \circ g = 1$ So there is no limit, it flips between 0 and 1 with increasing frequency – Cato Nov 8 '16 at 15:59
HINT:
The composite function is given by
$$h(g(x))=\begin{cases}1&,g(x)=0\\\\0&,g(x)\ne 0\end{cases}$$
So, $h$ can equal only $1$ or $0$.
Note that when $\displaystyle x_n=\frac1{n\pi}$, we have $g(x_n)=0$ and $h(g(x_n))=1$.
Then, take $\epsilon=1$ and show that for all $\delta>0$, there exists an $0<|x_n|<\delta$ such that
$$|h(g(x_n))-0|\ge \epsilon=1$$
and another number $x\ne x_n$ with $0<|x|<\delta$ such that
$$|h(g(x))-1|\ge \epsilon=1$$
• Would the cowardly down voter care to comment? – Mark Viola Nov 9 '16 at 15:29
• I'm guessing the downvoter is Joss, as his/her answer reaches an unusual conclusion that is different from yours (and the number of up/downvotes on his/her answer has not been changed while yours was). There was another answerer who also said the limit was 0, but that person realized the mistake and deleted after my comments. – Michael Nov 10 '16 at 1:01
| 2019-07-23T22:04:43 |
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https://www.physicsforums.com/threads/rate-of-rise-in-water-level.867539/
|
# Rate of rise in water level
1. Apr 18, 2016
1. The problem statement, all variables and given/known data
A cylinder of radius ${R}$ is partially filled with water. There is an inverted cone of cone angle 90° and base radius ${\frac{R}{2}}$ which is falling in it with a constant speed $v=30~cm/s$.
Find the rate of increase of water level (in $cm/s$) when half the height of cone is immersed in water.
The given answer is: $2~cm/s$
2. Relevant equations
Let the height of the cone be h, and height of cylinder be H
Volume of cone $= \frac{1}{3}\pi (\frac{R}{2})^2 h = \frac{1}{12} \pi R^2 h$
Volume of cylinder = $\pi R^2 H$
3. The attempt at a solution
Now for the cone $\frac{dh}{dt}=v=30~cm/s$
Let the increase in cylinder's water level be $dH$ when $dV$ of cone's volume is immersed in it.
So
$dV = \pi R^2 dH$
⇒ $\frac{dH}{dt} = \frac{\frac{dV}{dt}}{\pi R^2}$
⇒ $\frac{dH}{dt} = \frac{\frac{d}{dt}(\frac{1}{12} \pi R^2 h)}{\pi R^2}$
⇒ $\frac{dH}{dt} = \frac{\frac{1}{12} \pi R^2~\frac{dh}{dt}}{\pi R^2}$
⇒ $\frac{dH}{dt} = \frac{1}{12} × 30 = 2.5~cm/s$
Thank you.
2. Apr 18, 2016
### BvU
Volume of the part of the cone that is immersed depends on h in a way that differs from what you use... The r of that part of the cone is dependent on h...
3. Apr 21, 2016
### Vibhor
@BvU ,@Chestermiller ,do you agree with the given answer 2 cm/s ? I believe it is incorrect . It should be 1.875 cm/s .
2cm/s would have been correct if the cone was falling at the speed of 32 cm/s .
Last edited: Apr 21, 2016
4. Apr 21, 2016
### Staff: Mentor
I get 2 cm/s.
Let z be the distance between the tip of the cone and the bottom of the cylinder, and let h(z) be the distance between the water surface and the bottom of the cylinder. Then the amount of cone submerged is h - z, and the volume of cone submerged is $$V_s=\frac{\pi (h-z)^3}{3}$$. The total volume of water and cone below the water surface is $\pi R^2h$, and this is also equal to the volume of water Vw plus the submerged volume of cone:
$$V_w+\frac{\pi (h-z)^3}{3}=\pi R^2h$$
Taking the time derivative of this equation gives:
$$(h-z)^2\left(\frac{dh}{dt}-\frac{dz}{dt}\right)=R^2\frac{dh}{dt}$$
5. Apr 21, 2016
### Vibhor
Hi ,
Let y be the distance between the tip of the cone and the water surface, and let h be the distance between the water surface and the bottom of the cylinder.
$$V_w =\pi R^2h - \frac{\pi y^3}{3}$$
Taking the time derivative of this equation gives: $$R^2\frac{dh}{dt} = y^2\frac{dy}{dt}$$ .
When $y=\frac{R}{4}$ , $\frac{dh}{dt} = \frac{30}{16}$
What is the mistake ?
6. Apr 21, 2016
### Staff: Mentor
The downward velocity of the cone is not dy/dt. It is (dy/dt)-(dh/dt).
7. Apr 21, 2016
Thanks a lot
| 2017-10-16T22:37:18 |
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https://math.stackexchange.com/questions/2658837/positive-integer-x-such-that-sqrt144x2is-an-integer
|
# Positive integer $x$ such that $\sqrt{144+x^2}$is an integer
How many positive integers $x$ are there so that $\sqrt{144+x^2}$ is an integer?
Attempt a solution: I used trial and error and only got $x=5$
• Rephrased, find all pythagorean triples involving $12$ as one of the two smaller entries. There are several including $(5,12,13),(9,12,15),(12,16,20),(12,35,37)$. – JMoravitz Feb 20 '18 at 16:17
• Hint, they must lie in $(-\infty, -12] \cup [12, +\infty )$ :-) – Kevin Feb 20 '18 at 16:21
• Also, you could re-write to assume $\exists\, z = \sqrt{(12-x)(12+x)}$ – Kevin Feb 20 '18 at 16:21
• I completed my answer—no more constant edits now :) – Feeds Feb 20 '18 at 17:08
• @JMoravitz those are the only $4$ solutions, actually :) – Feeds Feb 20 '18 at 17:15
You have $$y^2=12^2+x^2\iff (y-x)(y+x)=2^4\cdot3^2$$ The integer $144$ has $(4+1)(2+1)=15$ factors so one can looking at the systems $$\begin{cases}x+y=a\\x-y=b\end{cases}$$ where $$(a,b)\in\{(144,1),(72,2),(48,3),(36,4),(24,6),(18,8),(16,9),(12,12)\}$$ from which we get four solutions $$(a,b)\in\{(72,2),(36,4),(24,6),(18,8)\}$$ giving $$\color{red}{(y,x)\in\{(37,35),(20,16),(15,9),13,5)\}}$$
• $$(y, x)\neq (37, 15)$$ – Feeds Feb 20 '18 at 16:50
• I have edited, Thanks you very much ( $37^2-15^2=1144$ and not $144$ – Piquito Feb 20 '18 at 16:54
• Hahah no problem at all :) ..... Oh, and $x = 35$ is a solution. – Feeds Feb 20 '18 at 16:56
• Very kind. Thanks again. – Piquito Feb 20 '18 at 16:57
• It was a lapsus with $15$ instead of $35$. Hahah, as you say. – Piquito Feb 20 '18 at 17:01
substituting $$y=\sqrt{144+x^2}$$ we get by squaring and factorizing $$(y-x)(y+x)=144$$ can you finish?
Well then since $144 = 12^2$ then you want to find values of $x$ to satisfy the equation, $$12^2 + x^2 = y^2.$$ Now since $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$ then when we multiply them together, we get that $$(a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2.\tag1$$ What you have done to solve for $x$ is by letting $12 = 2ab$ and thus $a = 2$ and $b = 3$. We now want $12 = a^2 - b^2 = (a + b)(a - b)$. Now what divides into $12$? Well the answers are $1, 2, 3, 4, 6, 12$.
Let $a + b = 12$, then $a - b = 1$, therefore $a = b + 1$ and $2b + 1 = 12$. That is a contradiction because $a$ and $b$ are integers if $2ab$ could equal $12$ in the first place, and $12$ is even, so we have to rule that out.
Let $a + b = 6$ then $a - b = 2$, Therefore $a = b + 2$ and $2b + 2 = 2(b + 1) = 6$. Therefore, $b = 2$ and thus $a = 4$.
We have now found another solution for $x$, such that by letting $x = 2ab$ instead (since we let $12 = a^2 - b^2$), we have that $x = 2\times 2\times 4 = 16$.
Now let $a + b = 4$ then $a - b = 3$, Therefore $a = b + 3$ and $2b + 3 = 4$. That is a contradiction, so we rule that out. Now we can’t let $a + b < 4$ because otherwise $a + b < a - b$ which is also a contradiction, so we stop there.
$$***$$
Now in Eq. $(1)$, if we let $b = 1$ then $$(a^2 - 1)^2 + (2a)^2 = (a^2 + 1)^2.$$ By letting $12 = 2a$ since $12 + 1 = 13 \neq$ a squared number, then we get that $a = 6$. By letting $x = a^2 - 1$ then we obtain another solution for $x$, namely $x = 6^2 - 1 = 36 - 1 = 35.$
$$***$$
Now it is also provable that for any integer $n$, $$(2n^2 + 2n)^2 + (2n + 1)^2 = (2n^2 + 2n + 1)^2.$$ Since $12$ is even, then $12\neq 2n + 1$ and therefore $12 = 2n^2 + 2n$. We now have the following quadratic trinomial to solve: $$2n^2 + 2n - 12 = 0,$$ and using the quadratic formula, we get that $$n = \frac{-2\pm \sqrt{4 + 96}}{2} = \frac{-2}{2} \pm \frac{10}{2} = -1\pm 5.$$ Therefore, $n = 4$ or $n = -6$. Now, we let $x = 2n + 1$ for those values of $n$ and we get two more solutions for $x$, namely $x = 9$ and $x = -11$.
However, $x > 0$ because we established earlier that if $x = 2ab$ then $a + b\not< a - b$, and if $x < 0$ then either $a$ or $b$ is negative which arouses a contradiction in the strict inequality, so we get only one new solution for $x$ such that $x = 9$. (This also proves why $x$ must be a positive integer.)
$$***$$
Overall, our solutions for $x$ so far are $x = 5, 9, 16, 35$.
Checking the equation $u^2 + v^2 = w^2$ for $w \leqslant 10,000$, there does not seem to be another solution where either $u$ or $v$ equals $12$ with the other remaining variable on the LHS not equal to any of our listed values of $x$. From this, I conjecture that there only exists $4$ such solutions of $x$.
According to @Piquito’s answer below $\downarrow\downarrow\downarrow$ there exist only $4$ solutions, and thus the conjecture is valid. I apologise for constantly editing my answer, but I did not have a big calculator with me. I was able to do the calculation concerning $u^2 + v^2 = w^2$ from a link in one of the comments. This meant that I had to find the values of $x$ with pure maths, and I was not going to sit there and exhaustively find solutions for $(x, y)$ in the equation, $144 = (y + x)(y - x)$.
$$(x+2a)^2=x^2+144$$ where $a>0$
$$144=4a(x+a)>4a^2\implies a<6$$
and $a$ divides $36$
• @Janjan, Start with $$(x+b)^2=x^2+144$$ – lab bhattacharjee Feb 20 '18 at 16:40
| 2019-08-18T05:31:03 |
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https://www.physicsforums.com/threads/inequaltiy-problem.262859/
|
# Inequaltiy problem
1. Oct 8, 2008
### Sheneron
1. The problem statement, all variables and given/known data
$$x^3 + 3x^2 - 16x - 48 <= 0$$
factored it down to get this.
$$(x+3)(x^2 -16) <= 0$$
$$x + 3 <= 0$$
$$x <= -3$$
$$x^2 - 16 <= 0$$
$$x <= 4$$
$$x <= -4$$
But, that is not right, looking at a graph x needs to be greater than or equal to -3, not less than or equal to. But the other one is right. How come?
2. Oct 8, 2008
### Mentallic
$$x^2 - 16 <= 0$$ can you see this is a difference of 2 squares?
Mainly: $$a^{2}-b^{2}=(a+b)(a-b)$$
3. Oct 8, 2008
### Ray 4
thanks but why does the graph show it as x >= -3 when the equation yields x<= - 3
4. Oct 9, 2008
### Staff: Mentor
What you're dealing with isn't an equation; it's an inequality, and they're trickier to work with than equations, since you have in take into account the the two factors (or three, if you factor the x2 - 16 expression) can be zero or can be negative.
It's easy to determine where (x + 3)(x2 - 16) is equal to zero, but takes a little more work to find where it is less than zero. For the above to be negative, it must be that either
a) x + 3 < 0 and x2 - 16 > 0, or that
b) x + 3 > 0 and x2 - 16 < 0
What do you get if you follow this reasoning for the two pairs of inequalities?
You can also approach it using the three linear (i.e., first degree in x) factors. To get a negative product, one factor has to be negative and the other two have to be positive. Taking that approach, you'll have three sets of three inequalities, but it might be that one of the sets has no solution.
Hope that helps.
Mark
5. Oct 9, 2008
### Mentallic
for the $$x^{2}\leq 16$$ part of the inequality, this is obviously a quadratic that has roots of $$\pm 4$$. This quadratic is concave up, so $$f(\frac{-b}{2a})<0$$ (the y-value of the vertex). Just picturing the graph of the quadratic, the partial solution will be for the x-values where their y-values < 0. This means that this partial solution is $$-4\leq x \leq 4$$.
Can you now see how you got this part wrong firstly?
So now we have, $$(x+3)(x^{2}-16)\leq 0$$
through further factorisation $$(x+3)(x+4)(x-4)\leq 0$$
So the partial solutions we have here are:
$$x+3\leq 0, x\leq -3$$ (1)
$$x+4\leq 0, x\leq -4$$ (2)
$$x-4\leq 0, x\leq 4$$ (3)
But there is a problem here, there are intersections of all 3 solutions, which ones are correct and which are trashed?
Well it all has to do with how these terms are multiplied. For all values $$-3< x< 4$$ only one of the partial solutions is satisfied, mainly (3). So the inequality is multiplied by the 3 factored terms, one of which is negative [the factor (x-4)] so the inequality is satisfied. Once the x-values become $$-4< x<-3$$, 2 of the factors become negative [the factors (x+3)(x-4)] and when 2 negatives are multiplied together, it becomes positive, so the inequality is not satisfied in this range. Lastly, once $$x<-4$$ all 3 factors are satisfied, and 3 negative terms multiplied together is still negative, so that partial solution is satisfied.
Be sure to include the roots of the function since this is where the function = 0 and this satisfies the inequality.
It is always easier just to visual the graph and understand for which x-values, the y-values are negative.
Last edited: Oct 9, 2008
6. Oct 9, 2008
### Sheneron
So, assuming one is negative and the other is positive, which it must be for that inequality to be less than 0, you end up with 2 inequalities (a and b from your post). So one of those is right, and the other is wrong, and you just have to check to see which is which?
7. Oct 10, 2008
### Staff: Mentor
It's not necessarily so that one will be right and one wrong, although this is possible. You have to check both and see what the implications are for values of x.
8. Oct 10, 2008
### HallsofIvy
Staff Emeritus
One way to do this is just as you say: if (x- 4)(x+ 4)< 0, then x-4 and x+4 must have different signs. Either x-4< 0 and x+ 4> 0 or x-4>0 and x+ 4< 0.
The first pair gives x< 4 and x> -4 so -4< x< 4.
The second pair give x> 4 and x< -4 which is impossible. The only answer is -4< x< 4.
Another way to do it is to solve the related equation (x- 4)(x+ 4)= 0 which gives x= -4 and x= 4. Since (x-4)(x+4) is a continuous function, it can change from "< 0" to "> 0", and vice-versa, only where it is "= 0", i.e. at x= -4 and 4. That means that in each of the three intervals it cannot change sign. If x= -5, -5< -4, (-5-4)(-5+4)= (-9)(-1)= 9> 0 so (x- 4)(x+ 4)> 0 for all x< -4. For x= 0, -4< 0< 4, (0-4)(0+4)= (-4)(4)= -15< 0 so (x-4)(x+4)< 0 for all -4< x< 4. Finally, if x= 5, 5> 4, (5-4)(5+4)= (1)(9)= 9> 0 so (x-4)(x+4)> 0 for all x> 4.
| 2017-07-28T11:01:29 |
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https://math.stackexchange.com/questions/4178319/a-question-on-integration-of-x-a
|
# A question on integration of $x-a$
I am faced with the following integral:
$$\int(x-a) dx$$
I can think of two approaches to solve it:
1) Separating into two terms as follows:
$$\int x dx -a\int 1 dx$$
From which the result would be:
$$\int (x-a) dx = x^2/2 - ax$$
2) Substituting $$u =x-a$$; $$du=dx$$
$$\int u du = u^2 / 2 = (x - a)^2 / 2$$
If we expand this solution:
$$\int (x-a) dx = x^2/2 -ax + a^2/2$$
Now, clearly, $$x^2/2 -ax+a^2/2$$ is not equal to $$x^2/2 - ax$$. So is either method invalid for some reason? Or am I making a mistake elsewhere?
I am aware this is probably a very dumb question, so I thank you very much for your attention and help!
• Well that $a^2/2$ term is a constant. Usually just a big $+C$. If you differentiate both, you get the same answer. – combinatorialist46Carey2 Jun 20 at 19:31
$$\int(x-a)dx = x^2/2 - ax + C_1$$ $$\int u dx = u^2/2 = (x-a)^2/2 = x^2/2-ax+a^2/2 + C_2$$ and these two are as equal as they can be, as $$a^2/2$$ is a constant you can pick $$C_1 = a^2/2 + C_2$$.
Remember that for an indefinite integral you must add a constant of integration. The two integrals are then the same in terms of their dependence on $$x$$, up to a constant. If you would replace the indefinite integral with a definite integral, the difference would work out in how the bounds of the integral change when you make the change of variables.
| 2021-07-28T11:04:43 |
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http://math.stackexchange.com/questions/182223/if-lim-n-f-nx-n-fx-for-every-x-n-to-x-then-f-n-to-f-uniformly-on
|
# If $\lim_n f_n(x_n)=f(x)$ for every $x_n \to x$ then $f_n \to f$ uniformly on $[0,1]$?
This is a self-posed question, so I do not know the answer and I would like to know what do you think about.
Let $f,f_n:[0,1]\to \mathbb R$ be continuous functions. Assume that for every sequence $(x_n)_{n\in \mathbb N}$ converging to a point $x\in [0,1]$, we have $$\lim_n f_n(x_n)=f(x).$$ Can we say that $f_n$ converges uniformly to $f$ in $[0,1]$?
First of all, do you think the question is well-posed? I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.
-
The result is true and this might be one of the few situations where going for a proof by contradiction really helps. Hence, let us assume that $(f_n)$ does not converge uniformly to $f$. This means that... – Did Aug 13 '12 at 22:29
The mode of convergence you ask about is well-established and is called continuous convergence and I think @did's hint should be very helpful (don't forget to use compactness of $[0,1]$!) – t.b. Aug 13 '12 at 22:41
@did: I tried to follow your advice but is kind of stuck. – Bombyx mori Aug 13 '12 at 23:11
How did you write down the hypothesis that $(f_n)$ does not converge uniformly to $f$? If you have a problem with this, you may first write down the hypothesis that $(f_n)$ does converge uniformly to $f$, then negate this. So: what does it mean to say that $(f_n)$ converges uniformly to $f$? – Did Aug 13 '12 at 23:31
Hopefully fixed by now, but may still have problems. – Bombyx mori Aug 13 '12 at 23:33
Yes, this is true:
Suppose that the convergence $f_n \to f$ is not uniform but that $f(x) = \lim_{n\to\infty}f_n(x_n)$ for every convergent sequence $x_n\to x$ in $[0,1]$.
After passing to a subsequence of the $f_n$ we may assume that there is $\varepsilon \gt 0$ such that for all $n \in \mathbb{N}$ we have $\sup_{x \in [0,1]} \lvert f(x) - f_n(x) \rvert \gt \varepsilon$. This implies that there is a sequence of points $x_n \in [0,1]$ such that $\lvert f(x_n) - f_n(x_n) \rvert \gt \varepsilon$ for each $n$. After passing to further subsequences, we may assume that $x_n \to x \in [0,1]$ by compactness of $[0,1]$. But then we're in trouble: our hypotheses tell us that $$\tag{\ast} \lim_{n\to \infty} \lvert f(x) - f_n(x_n)\rvert = 0,$$ which is incompatible with $f(x_n) \to f(x)$ and $\lvert f_n(x_n) - f(x)\rvert \gt \varepsilon$.
More formally: continuity of $f$ ensures together with $x_n \to x$ that there is $N$ such that $\lvert f(x) - f(x_n)\rvert \lt \varepsilon/2$ for all $n \geq N$ so that the triangle inequality gives us for all $n \geq N$ that \begin{align*} \lvert f_n(x_n) - f(x) \rvert &= \lvert [f_n(x_n) - f(x_n)] - [f(x) - f(x_n)]\rvert \\ & \geq \underbrace{\lvert f_n(x_n) - f(x_n)\rvert}_{\gt \varepsilon} - \underbrace{\lvert f(x) - f(x_n)\rvert}_{\lt \varepsilon/2} \\& \gt \varepsilon/2 \gt 0, \end{align*} which is absurd in view of $(\ast)$.
1. The mode of convergence $f_n(x_n) \to f(x)$ for all $x_n \to x$ is called continuous convergence. Continuity of the $f_n$ is never used in the above argument. In fact, $f_n(x_n) \to f(x)$ for all $x_n \to x$ ensures continuity of $f$ independently of the continuity of the $f_n$ (exercise).
2. You can find a generalization of the above result and a more detailed discussion in the section Compact and continuous convergence (Chapter 3, §1, section 5) starting on page 98 in R. Remmert's Theory of complex functions. The precise result reads that $f_n$ converges to $f$ continuously if and only if $f$ is continuous and $f_n \to f$ uniformly on compact sets. More details in loc. cit. (don't miss the historical notes preceding and following that section!).
-
I posted this answer simultaneously with did's hint but I then deleted it to give the OP some time to think about it. I hope it's fine to show it now. – t.b. Aug 14 '12 at 6:53
Basically, I have no opinion here. Let me mention that Romeo probably did not even see my two comments (the first one addressed to him and the second one answering @user32240's query), nicely completed by your own comment (mentioning compactness). – Did Aug 14 '12 at 8:15
This is a very clear proof. Thank you. – Bombyx mori Aug 14 '12 at 12:07
Is the converse true? That is, if we are given that $f_n (x_n)$ converges UNIFORMLY to $f(x)$ and the sequence of points $(x_n) \rightarrow x$, then does $f_n(x_n) \rightarrow f (x)$ [pointwise convergence]? If so, can anyone help me out by providing a proof. I am having a hard time thinking that the converse is true. – user43901 Mar 13 '13 at 0:17
@t.b.: $\{f_n\}$ does not converge to $f$ implies $\exists \epsilon >0$ such that for some subsequence $\{f_{n_k}\}$ of $\{f_n\}$ , there is a sequence $\{x_k\}$ in domain such that $|f_{n_k}(x_k)-f(x_k)| \ge \epsilon$ ; but how does one ensure that $|f_n(x_n)-f(x_n)|\ge \epsilon$ , as used in the answer ? – Saun Dev Jul 12 at 15:03
There is a classical theorem by Dini that says this is possible when $f_{i}$ are monotonically increasing.
We now assume $f_{n}$ converges to $f$ pointwise and $f,f_{n}$ are all continuous but the limit is not uniform. Thus there is some $\epsilon$ such that $\forall N$ there is some $n\ge N$ such that $$|f_{n}(x)-f(x)|> \epsilon$$ for some $x\in [0,1]$.
Now since $f_{i},f$ are uniformly continuous on $[0,1]$, for $\epsilon_{1}=\frac{\epsilon}{3}$ we have a list of $\delta_{i}$s such that $0< \delta_{i}\le 1$ and $|f_{i}(x)-f_{i}(y)|\le \epsilon_{1}$ for $|x-y|\le \delta_{i}$. Consider the interval $U_{i}=[0,\delta_{i}]$, I claim $\lim_{i\rightarrow \infty} \bigcap U_{i}\not=\emptyset$. Suppose $\cap_{i=N} U_{i}=\emptyset$, then $f_{i}$ would have nontrivial variation $$\lim_{x \rightarrow y} |f(x)-f(y)|=0$$ a fact contradicting they are being continuous. Thus let $U=[0,\delta_{A}]$ be the intersection, and let $\delta_{B}$ be the corresponding value for $f$. We choose $\delta=\min[\delta_{A},\delta_{B}]$.
We now assume the counterexample in the beginning. For fixed $y$ such that $|x-y|\le \delta$ we can choose $N$ large enough such that $$|f(y)-f_{n}(y)|\le \epsilon/3$$ for any $n\ge N$. If it turns out $N$ is greater than the initial $n$ we chose, we may switch to a different $n$ guaranteed to exist by the hypothesis.
We have the estimate that $$|f_{n}(x)-f(x)|\le |f_{n}(x)-f_{n}(y)|+|f_{n}(y)-f(y)|+|f_{n}(y)-f_{n}(x)\le \epsilon$$ which contradicts the hypothesis we had earlier. So the convergence is uniform afterall.
-
In fact the desired result in the OP is true. – t.b. Aug 13 '12 at 22:11
My example has serious problem at point 1. Let me try to modify it. – Bombyx mori Aug 13 '12 at 22:14
@t.b.:can you check if my 'proof' is correct? – Bombyx mori Aug 13 '12 at 23:35
No, this doesn't really work this way. 1) the OP assumes something much stronger than pointwise convergence. 2) By your definitions you certainly have $0 \in \bigcap_{i=N}^\infty [0,\delta_i]$ and you probably try to argue that there is $\varepsilon \gt 0$ such that $[0,\varepsilon]$ is in this intersection: so you're trying to establish equicontinuity of the $f_n$, right? 3) I don't understand what the sentence including "nontrivial variation" is supposed to mean. 4) Sorry I am lost in the last two paragraphs: what is the $x$ doing? How do you get from $|x-y|\lt \delta$ to uniform conv.? – t.b. Aug 14 '12 at 6:07
See my answer for a detailed argument. – t.b. Aug 14 '12 at 6:57
| 2015-08-29T17:53:04 |
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https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/01Error/Error/
|
# Introduction
There are two techniques for measuring error: the absolute error of an approximation and the relative error of the approximation. The first gives how large the error is, while the second gives how large the error is relative to the correct value.
None.
# Absolute Error
Given an approximation a of a correct value x, we define the absolute value of the difference between the two values to be the absolute error. We will represent the absolute error by Eabs, therefore
It is often sufficient to record only two decimal digits of the absolute error. Thus, it is sufficient to state that the absolute error of the approximation 3.55 to the correct value 3.538385 is 0.012.
There are two problems with using the absolute error:
## Significance
It gives you a feeling of the size of the error but how significant is the error? If the absolute error was 3.52, is this significant? If the correct value is x = 5030235.23, then probably not, however if the correct value is x = 5.03023523, then an absolute error 3.52 is probably very significant.
## Units
The absolute error will change depending on the units used. The absolute error of the approximation 2.4 MV of an actual voltage of 2.573243 MV is 0.17 MV, whereas the absolute error of the approximation 2400000 V to an actual voltage of 2573243 V is 170000 V.
# Relative Error
To solve the problems of significance and units, we may compare the absolute error relative to the correct value. Thus, we define the relative error to be the ratio between the absolute error and the absolute value of the correct value and denote it by Erel:
In this equation, any units cancel, so the relative errors of the approximations 2.4 MV and 2400000 V versus the actual voltages of 2.573243 MV and 2573243 V, respectively, are equal. Also, a relative error of 0.01 means that the approximation is correct to within one part in one hundred, regardless of the size of the actual value. Whether this is sufficiently accurate depends on the requirements.
In this class, we will usually use the relative error, though if we are only trying to show that a sequence of errors is decreasing to zero, we may use the absolute error (if a sequence of absolute errors is going to zero, then the relative errors will go to zero, too).
One problem with using the relative error is when the correct value is zero (0), but this seldom appears in real-life situations.
On occasion, the relative error by 100 and refer to as the percent relative error.
# Calculating Absolute Error
Given an approximation a of a value x, the absolute error Eabs is calculated using the formula:
# Calculating Relative Error
Given an approximation a of a value x, the relative error Erel is calculated using the formula:
# Examples
1. What are the absolute and relative errors of the approximation 3.14 to the value π?
Eabs = |3.14 - π| ≈ 0.0016
Erel = |3.14 - π|/|π| ≈ 0.00051
2. A resistor labeled as 240 Ω is actually 243.32753 Ω. What are the absolute and relative errors of the labeled value?
Eabs = |240 - 243.32753| ≈ 3.3 Ω
Erel = |240 - 243.32753|/|243.32753| ≈ 0.014
Note: the label is the approximation of the actual value.
3. The voltage in a high-voltage transmission line is stated to be 2.4 MV while the actual voltage may range from 2.1 MV to 2.7 MV. What is the maximum absolute and relative error of voltage?
Eabs = |2.4 - 2.1| = 0.3 MV
Erel = |2.4 - 2.1|/|2.1| ≈ 0.14
Eabs = |2.4 - 2.7| = 0.3 MV
Erel = |2.4 - 2.7|/|2.7| ≈ 0.11
Thus, the maximum absolute error is 0.3 MV but the maximum relative error is 0.14.
Note: as before, the stated voltage is an approximation of the actual voltage.
# Questions
1. What are the absolute and relative errors of the approximation 22/7 of π? (0.0013 and 0.00040)
2. What are the absolute and relative errors of the approximation 355/113 of π? (2.7e-7 and 8.5e-8)
3. A capacitor is labeled as 100 mF whereas it is actually 108.2532 mF. What are the absolute and relative errors of the label? (8.2 mF and 0.076)
# Matlab
Absolute and relative errors may be easily calculated in Matlab:
```>> abs( 22/7 - pi )
>> abs( 22/7 - pi ) / abs( pi )
```
# Maple
Absolute and relative errors may be easily calculated in Maple:
```abs( 22/7 - Pi );
evalf( % ); # % refers to the previous calculation
abs( 22/7 - Pi ) / abs( Pi );
evalf( % ); # % refers to the previous calculation
```
| 2018-07-17T17:15:02 |
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https://math.stackexchange.com/questions/2817172/evaluate-int-0-pi-2-ln-lefta2-cos2-theta-b2-sin2-theta-right-d-t
|
# Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers
Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers
I have spent about a day thinking over this problem. I tried integration by parts, differentiating under integral sign (Feynman's trick, with respect to $a, b$), using some trigonometric and logarithmic properties like changing $\cos^2\theta$ to $\cos2\theta$ and hereafter some logarithmic properties, etc., but failed miserably.
Also tried to use the property that $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ in-between, but still to no avail. I also tried to find similar questions on MSE but did not get a related one.
Edit
My try (Feynman's trick) :
Let $$I(a)=\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) d\theta$$
Hence $$I'(a) =\frac 1a \int_0^{\pi/2} \frac {2a^2\cos^2\theta d\theta}{a^2\cos^2\theta +b^2\sin^2\theta}$$ $$=\frac 1a\left[ \frac {\pi}{2}+\int_0^{\pi/2} \frac {a^2\cos^2\theta -b^2\sin^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}\right]$$
Wherein between I broke $2a^2\cos^2\theta=a^2\cos^2\theta +b^2\sin^2\theta+a^2\cos^2\theta -b^2\sin^2\theta$
But now how do I continue further
• Feynman's trick works like a charm here, can you show your attempt in full detail? – Jack D'Aurizio Jun 12 '18 at 17:06
• I agree with Jack. Feynman's trick works here. – Mark Viola Jun 12 '18 at 17:06
• "finite natural numbers" ? You can omit the "finite". – Peter Jun 12 '18 at 17:08
• @Jack D'Aurizio I have edited my try accordingly. Can you please provide some hints further or any different parameter of differentiation in the same method – Rohan Shinde Jun 12 '18 at 17:15
• The integral $$\int_{0}^{\pi/2}\frac{a^2\cos^2\theta-b^2\sin^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta$$ can be simply computed through the substitution $\theta=\arctan u$ and partial fraction decomposition. It equals $\frac{\pi}{2}\cdot\frac{a-b}{a+b}$, assuming that $a$ and $b$ have the same sign. – Jack D'Aurizio Jun 12 '18 at 17:15
Let $I(a,b)=\int_0^{\pi/2} \log(a^2\cos^2(\theta)+b^2\sin^2(\theta))\,d\theta$. Differentiating under the integral with respect to $a^2$ reveals
\begin{align} \frac{\partial I(a,b)}{\partial (a^2)}&=\int_0^{\pi/2}\frac{1}{a^2+b^2\tan^2(\theta)}\,d\theta\\\\ &=\frac{\pi/2}{a(a+b)}\tag1 \end{align}
Integrating $(1)$ with respect to $a^2$, we obtain
$$I(a,b)=\pi \log(a+b)+C$$
For $a=b$, $I(a,a)=\pi \log(a)$ from which we find that $C=-\pi\log(2)$.
Putting it all together yields
$$I(a,b)=\pi \log\left(\frac{a+b}2\right)$$
NOTE:
We see from symmetry that $I(a,b)=\pi \log\left(\frac{|a|+|b|}2\right)$ $\forall (a,b)\in \mathbb{R}^2\setminus (0,0)$.
• Now both of you( Mark and Jack) have put in trouble. Both of your answers are marvelous in its own way. So which one to accept. – Rohan Shinde Jun 12 '18 at 17:28
• Nope. Correction here. It is also applicable for $ab=0$ except for the case $a=b=0$ – Rohan Shinde Jun 12 '18 at 17:37
• @Manthanein Good catch. I'll edit. – Mark Viola Jun 12 '18 at 17:41
• Try $b=0$ and $a$ be some finite number. Now the integrals changes to $$\int_0^{\pi/2} \ln(a^2\cos^2\theta )d\theta=2\int_0^{\pi/2} \ln(a\cos\theta )d\theta=2\left[\frac {\pi}{2}\ln a-\frac {\pi}{2}\ln 2\right]=\pi\ln (a/2)$$ which matches with the general form you derived – Rohan Shinde Jun 12 '18 at 17:44
• @Manthanein It should match, should it not? – Mark Viola Jun 12 '18 at 17:45
If we assume $a,b>0$ and set $$I(a,b)=\int_{0}^{\pi/2}\log(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta$$ we have $I(a,b)=I(b,a)$ from the substitution $\theta\mapsto\frac{\pi}{2}-\theta$. On the other hand $I(a,a)=\pi\log(a)$ is trivial, so $I(a,b)=\pi\log\left(\frac{a+b}{2}\right)$ is a very reasonable conjecture. Indeed, it can be proved by computing $$\frac{\partial I}{\partial a} = \int_{0}^{\pi/2}\frac{2a\cos^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta$$ as suggested in the comments, i.e. via $\theta\mapsto\arctan u$ and partial fraction decomposition.
An alternative approach is to notice that $$\begin{eqnarray*} I(a,b) &=& 2\,\text{Re}\int_{0}^{\pi/2}\log\left(a\cos\theta+ib\sin\theta\right)\\&=&\pi\log\left(\frac{a+b}{2}\right)+\text{Re}\int_{0}^{\pi}\log\left(e^{i\theta}+\frac{a-b}{a+b}\right)\,d\theta\end{eqnarray*}$$ where the last integral is a purely imaginary number by the residue theorem.
• Now both of you( Mark and Jack) have put in trouble. Both of your answers are marvelous in its own way. So which one to accept. – Rohan Shinde Jun 12 '18 at 17:28
• And by the way no need to assume $a, b\gt 0$, it's already provided in in question that they are natural numbers – Rohan Shinde Jun 12 '18 at 17:30
• @Manthanein We need not assume that $a$ and $b$ are positive. Note that since the integral depends on $a^2$ and $b^2$, then the answer depends on $|a|$ and $|b|$ only. – Mark Viola Jun 12 '18 at 17:32
• @Mark Viola That is what I said. Morever the modulus will be removed since they are natural and always positive – Rohan Shinde Jun 12 '18 at 17:34
| 2019-11-22T02:11:40 |
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https://autarkaw.org/page/27/
|
## How do you know that the least squares regression line is unique and corresponds to a minimum
We already know that using the criterion of either
1. minimizing sum of residuals OR
2. minimizing sum of the absolute value of residuals
is BAD as either of the criteria do not give a unique line. Visit these notes for an example where these criteria are shown to be inadequate.
So we use minimizing the sum of the squares of the residuals as the criterion. How can we show that this criterion gives a unique line?
The proof is given below as image files because the proof is equation intensive. I made a better resolution pdf file also.
_____________________________________________________
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Finding the optimum polynomial order to use for regression
Many a times, you may not have the privilege or knowledge of the physics of the problem to dictate the type of regression model. You may want to fit the data to a polynomial. But then how do you choose what order of polynomial to use.
Do you choose based on the polynomial order for which the sum of the squares of the residuals, Sr is a minimum? If that were the case, we can always get Sr=0 if the polynomial order chosen is one less than the number of data points. In fact, it would be an exact match.
So what do we do? We choose the degree of polynomial for which the variance as computed by
Sr(m)/(n-m-1)
is a minimum or when there is no significant decrease in its value as the degree of polynomial is increased. In the above formula,
Sr(m) = sum of the square of the residuals for the mth order polynomial
n= number of data points
m=order of polynomial (so m+1 is the number of constants of the model)
Let’s look at an example where the coefficient of thermal expansion is given for a typical steel as a function of temperature. We want to relate the two using polynomial regression.
Temperature Instantaneous Thermal Expansion oF 1E-06 in/(in oF) 80 6.47 40 6.24 0 6.00 -40 5.72 -80 5.43 -120 5.09 -160 4.72 -200 4.30 -240 3.83 -280 3.33 -320 2.76
If a first order polynomial is chosen, we get
$alpha=0.009147T+5.999$, with Sr=0.3138.
If a second order polynomial is chosen, we get
$alpha=-0.00001189T^2+0.006292T+6.015$ with Sr=0.003047.
Below is the table for the order of polynomial, the Sr value and the variance value, Sr(m)/(n-m-1)
Order of polynomial, m Sr(m) Sr(m)/(n-m-1) 1 0.3138 0.03486 2 0.003047 0.0003808 3 0.0001916 0.000027371 4 0.0001566 0.0000261 5 0.0001541 0.00003082 6 0.0001300 0.000325
So what order of polynomial would you choose?
From the above table, and the figure below, it looks like the second or third order polynomial would be a good choice as very little change is taking place in the value of the variance after m=2.
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Data for aluminum cylinder in iced water experiment
A colleague asked me what if he did not have time or resources to do the experiments that have been developed at University of South Florida (USF) for numerical methods. He asked if I could share the data taken at USF.
Why not – here is the data for the experiment where an aluminum cylinder is placed in iced water. This link also has the exercises that the students were asked to do.
The temperature vs time data is as follows: (0,23.3), (5,16.3), (10,13), (15,11.8), (20,11), (25,10.7), (30,9.6), (35,8.9), (40,8.4). Time is in seconds and temperature in Celcius. Other data needed is
Ambient temperature of iced water = 1.1oC
Diameter of cylinder = 44.57 mm
Length of cylinder = 105.47 mm
Density of aluminum = 2700 kg/m3
Specific heat of aluminum = 901 J/(kg-oC)
Thermal conductivity of aluminum = 240 W/(m-K)
Table 1. Coefficient of thermal expansion vs. temperature for aluminum (Data taken from http://www.llnl.gov/tid/lof/documents/pdf/322526.pdf by using mid values of temperatures at which CTE is reported)
Temperature (oC) Coefficient of thermal expansion (μm/m/oC) -10 58 12.5 59 37.5 60 62.5 62 87.5 66 112.5 71
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## In regression, when is coefficient of determination zero
The coefficient of determination is a measure of how much of the original uncertainty in the data is explained by the regression model.
The coefficient of determination, $r^2$ is defined as
$r^2$=$\frac{S_t-S_r}{S_r}$
where
$S_t$ = sum of the square of the differences between the y values and the average value of y
$S_r$ = sum of the square of the residuals, the residual being the difference between the observed and predicted values from the regression curve.
The coefficient of determination varies between 0 and 1. The value of the coefficient of determination of zero means that no benefit is gained by doing regression. When can that be?
One case comes to mind right away – what if you have only one data point. For example, if I have only one student in my class and the class average is 80, I know just from the average of the class that the student’s score is 80. By regressing student score to the number of hours studied or to his GPA or to his gender would not be of any benefit. In this case, the value of the coefficient of determination is zero.
What if we have more than one data point? Is it possible to get the coefficient of determination to be zero?
The answer is yes. Look at the following data pairs (1,3), (3,-2), (5,4), (7,-5), (9,4.2), (11,3), (2,4). If one regresses this data to a general straight line
y=a+bx,
one gets the regression line to be
y=1.6
In fact, 1.6 is the average value of the given y values. Is this a coincidence? Because the regression line is the average of the y values, $S_t=S_r$, implying $r^2=0$
QUESTIONS
1. Given (1,3), (3,-2), (5,4), (7,a), (9,4.2), find the value of a that gives the coefficient of determination, $r^2=0$. Hint: Write the expression for $S_r$ for the regression line $y=mx+c$. We now have three unknowns, m, c and a. The three equations then are $\frac{\partial S_r} {\partial m} =0$, $\frac{\partial S_r} {\partial c} =0$ and $S_t=S_r$.
2. Show that if n data pairs $(x_1,y_1)......(x_n,y_n)$ are regressed to a straight line, and the regression straight line turns out to be a constant line, then the equation of the constant line is always y=average value of the y-values.
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Length of a curve experiment
In a previous post, I mentioned that I have incorporated experiments in my Numerical Methods course. Here I will discuss the second experiment.
In this experiment, we find the length of two curves generated from the same points – one curve is a polynomial interpolant and another one is a spline interpolant.
Motivation behind the experiment: In 1901, Runge conducted a numerical experiment to show that higher order interpolation is a bad idea. It was shown that as you use higher order interpolants to approximate f(x)=1/(1+25x2) in [-1,1], the differences between the original function and the interpolants becomes worse. This concept also becomes the basis why we use splines rather than polynomial interpolation to find smooth paths to travel through several discrete points.
What do students do in the lab: A flexible curve (see Figure) of length 12″ made of lead-core construction with graduations in both millimeters and inches is provided. The student needs to draw a curve similar in shape to the Runge’s curve on the provided graphing paper as shown. It just needs to be similar in shape – the student can make the x-domain shorter and the maximum y-value larger or vice-versa. The student just needs to make sure that there is a one-to-one correspondence of values.
Assigned Exercises: Use MATLAB to solve problems (3 thru 6). Use comments, display commands and fprintf statements, sensible variable names and units to explain your work. Staple all the work in the following sequence.
1. Signed typed affidavit sheet.
2. Attach the plot you drew in the class. Choose several points (at least nine – do not need to be symmetric) along the curve, including the end points. Write out the co-ordinates on the graphing paper curve as shown in the figure.
3. Find the polynomial interpolant that curve fits the data. Output the coefficients of the polynomial.
4. Find the cubic spline interpolant that curve fits the data. Just show the work in the mfile.
5. Illustrate and show the individual points, polynomial and cubic spline interpolants on a single plot.
6. Find the length of the two interpolants – the polynomial and the spline interpolant. Calculate the relative difference between the length of each interpolant and the actual length of the flexible curve.
7. In 100-200 words, type out your conclusions using a word processor. Any formulas should be shown using an equation editor. Any sketches need to be drawn using a drawing software such as Word Drawing. Any plots can be imported from MATLAB.
Where to buy the items for the experiment:
1. Flexible curves – I bought these via internet at Art City. The brand name is Alvin Tru-Flex Graduated Flexible Curves. Prices range from $5 to$12. Shipping and handling is extra – approximately $6 plus 6% of the price. You may want to buy several 12″ and 16″ flexible curves. I had to send a query to the vendor when I did not receive them within a couple of weeks. Alternatively, call your local Art Store and see if they have them. 2. Engineering Graph Paper – Staples or Office Depot. Costs about$12 for a pack for 100-sheet pad.
3. Pencil – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore. $1 for a dozen. 4. Scale – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore.$1 per unit.
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## A legend used in the movie “The Happening”
Well M. Night Shyamalan may have made another disappointing movie – The Happening, but I somewhat liked it. I would give it a grade of B.
In the movie, John Leguzomo’s character, a math teacher, is distracting his fellow panicking passenger in the Jeep with a mathematical question. The question he asks her is if he gave her a penny on Day 1 of the month, two pennies on Day 2 of the month, four pennies on Day 3 of the month, and so on, how much would money would she have after a month. She shouts \$300 or some odd number like that. But, do you know that the amount is actually more than a 10 million dollars (Thanks to a student who mentioned that it was a penny that John offered on the first day, not a dollar – sometimes I do feel generous).
This question is based on a story from India and it goes as follows.
King Shriham of India wanted to reward his grand minister Ben for inventing the game of chess. When asked what reward he wanted, Ben asked for 1 grain of rice on the first square of the board, 2 on the second square of the board, 4 on the third square of the board, 8 on the fourth square of the board, and so on till all the 64 squares were covered. That is, he was doubling the number of grains on each successive square of the board. Although Ben’s request looked less than modest, King Shriham quickly found that the amount of rice that Ben was asking for was humongous.
QUESTIONS:
Write a MATLAB (you can use any other programming language) program for the following using the for or while loop.
1. Find out how many grains of rice Ben was asking for.
2. If the mass of a grain of rice is 2 mg, and the world production of rice in recent years has been approximately 600,000,000 tons (1 ton=1000 kg), how many times the modern world production was Ben’s request?
3. Do the inverse problem – find out how many squares are covered if the the number of grains on the chess board are given to you. For example, how many squares will be covered if the number of grains on the chess board are 16?
This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
| 2018-08-14T14:00:03 |
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https://mathhelpboards.com/threads/calculating-overall-percentage-probability-from-multiple-categories.24686/
|
# Calculating overall percentage/probability from multiple categories?
#### Spud
##### New member
Greetings,
My apologies if the title is confusing; I don't really know how to explain what I am trying to do or what label this problem would fall under.
Scenario:
We have a magic bag, and inside the magic bag are an unknown/unlimited amount of coins.
There are 100 different types of coins, but we are only interested in the iron, bronze, silver and gold coins, so we have bundled the other 96 types together as "other".
We performed an experiment by taking one coin at a time from the magic bag and recording what type of coin it was; the coins we took from the magic bag were not placed back into the magic bag.
We performed the experiment 1,000,000 times and the results are as follows:
Other coins: 998,859
Iron coins: 596
Bronze coins: 312
Silver coins: 135
Gold coins: 98
First Question:
Is it correct to say the probability to receive each coin from the magic bag is as follows?
Other coins: (998,859/1,000,000)*100 = 99.8859% or 1,000,000/998,859 = 1 in 1.0011423
Iron coins: (596/1,000,000)*100 = 0.0596% or 1,000,000/596 = 1 in 1,678
Bronze coins: (312/1,000,000)*100 = 0.0312% or 1,000,000/312 = 1 in 3,205
Silver coins: (135/1,000,000)*100 = 0.0135% or 1,000,000/135 = 1 in 7,407
Gold coins: (98/1,000,000)*100 = 0.0098% or 1,000,000/98 = 1 in 10,204
Second Question:
If I take only one coin from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).
I have tried to do some calculations, but I don't think I am working it out properly.
- Is the following correct?
((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141% (or 100% - 99.8859% = 0.1141%)
- Is the following correct?
((1/1678)+(1/3205)+(1/7407)+(1/10204))*100 = 0.1141%
- Is the following correct?
(((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100 = 0.0285%
If I take 1,400 coins from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).
I really have no idea how to calculate this; all I have managed to do is repeat one of the formulas above and multiply by 1,400.
- Is the following correct?
((((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100)*1400 = 39.9339%
I understand that I am probably completely wrong about everything, so thank you very much to anyone willing to provide assistance.
#### Chris L T521
##### Well-known member
Staff member
Greetings,
My apologies if the title is confusing; I don't really know how to explain what I am trying to do or what label this problem would fall under.
Scenario:
We have a magic bag, and inside the magic bag are an unknown/unlimited amount of coins.
There are 100 different types of coins, but we are only interested in the iron, bronze, silver and gold coins, so we have bundled the other 96 types together as "other".
We performed an experiment by taking one coin at a time from the magic bag and recording what type of coin it was; the coins we took from the magic bag were not placed back into the magic bag.
We performed the experiment 1,000,000 times and the results are as follows:
Other coins: 998,859
Iron coins: 596
Bronze coins: 312
Silver coins: 135
Gold coins: 98
First Question:
Is it correct to say the probability to receive each coin from the magic bag is as follows?
Other coins: (998,859/1,000,000)*100 = 99.8859% or 1,000,000/998,859 = 1 in 1.0011423
Iron coins: (596/1,000,000)*100 = 0.0596% or 1,000,000/596 = 1 in 1,678
Bronze coins: (312/1,000,000)*100 = 0.0312% or 1,000,000/312 = 1 in 3,205
Silver coins: (135/1,000,000)*100 = 0.0135% or 1,000,000/135 = 1 in 7,407
Gold coins: (98/1,000,000)*100 = 0.0098% or 1,000,000/98 = 1 in 10,204
The types of probabilities that must be computed here are empirical probabilities. The first values you have listed would be the more appropriate way of stating the answers.
Second Question:
If I take only one coin from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).
I have tried to do some calculations, but I don't think I am working it out properly.
- Is the following correct?
((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141% (or 100% - 99.8859% = 0.1141%)
- Is the following correct?
((1/1678)+(1/3205)+(1/7407)+(1/10204))*100 = 0.1141%
- Is the following correct?
(((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100 = 0.0285%
Since these are mutually exclusive events, you can use this form of the addition rule:
$P(I\cup B \cup S\cup G) = P(I)+P(B)+P(S)+P(G)$.
In this case, we find that
$P(I\cup B \cup S\cup G) = 0.0596\% + 0.0312\% + 0.0135\% + 0.0098\% = 0.1141\%$
Alternatively, you could apply the complement rule to get $P(I\cup B\cup S\cup G) = 1 - P(O) = 100\% - 99.8859\% = 0.1141\%$.
Thus, the first way you did the problem would be the best way to do it.
If I take 1,400 coins from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).
I really have no idea how to calculate this; all I have managed to do is repeat one of the formulas above and multiply by 1,400.
- Is the following correct?
((((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100)*1400 = 39.9339%
Are we assuming here that we're finding the probability that exactly one of the coins from the 1400 is iron, bronze, silver, or gold? If so, then we can use the Hypergeometric distribution to answer this question because the coins are being selected without replacement. (On the other hand, if the coins were being selected with replacement, we could have used the binomial distribution instead.)
From the wiki link, if we're given a finite population of size $N$, and $K$ is the number of objects with the desired trait, then the probability of picking $k$ successes in $n$ trials is given by
$P(X=k) = \dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$.
In this problem, $N=1000000$, $n=1400$, $K=1141$, and $k=1$.
Thus, $P(X=1) = \dfrac{\binom{1141}{1}\binom{1000000-1141}{1400-1}}{\binom{1000000}{1400}}\approx 32.3515\%$
I understand that I am probably completely wrong about everything, so thank you very much to anyone willing to provide assistance.
You seemed to be on the right track about the first two. The last part was pretty tricky since the wording of the problem was vague (in my opinion). If my interpretation for the last part is incorrect, please let me know and we'll try to figure things out from there! \o/
I hope this helps!
#### Spud
##### New member
Thanks Chris L T521! Very detailed and helpful.
I am happy that my level of understanding wasn't too far off track
So in regards to the last question with the 1,400 coins; I don't think I was very clear. I will try to clarify in a moment, but firstly I want to adjust the scenario and perhaps that will make things more clear.
Scenario:
We have a magic bag, and inside the magic bag are an unlimited amount of coins.
We don't know how many types of coins there are, but we are only interested in the iron, bronze, silver and gold coins.
There is a tag attached to the magic bag that states the probability of receiving the following coins:
Iron coins: 0.0596%
Bronze coins: 0.0312%
Silver coins: 0.0135%
Gold coins: 0.0098%
We are not given any other information.
Question 1:
If I take one coin, what is the probability of that one coin being either an iron, bronze, silver or gold coin?
I mean to say that I am putting the iron, bronze, silver and gold coins into a group of preferred coins, and receiving any of them would be a success.
Question 2:
If I take 1,400 coins, what is the probability that among those 1,400 coins, at least one of them is an iron, bronze, silver or gold coin?
I mean to say that I am putting the iron, bronze, silver and gold coins into a group of preferred coins, and receiving at least one of any of them would be a success.
I hope this is more clear; my apologies if it's not.
Thank you kindly.
Last edited:
#### Country Boy
##### Well-known member
MHB Math Helper
With an unlimited number of coins, whether we return coins or not doesn't matter so we use the binomial distribution.
| 2020-09-20T17:05:37 |
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http://stats.stackexchange.com/questions/22450/odds-of-drawing-at-least-k-identical-values-among-m-after-n-draws/22772
|
# Odds of drawing at least k identical values among m after n draws?
We draw $n$ values, each equiprobably among $m$ distinct values. What are the odds $p(n,m,k)$ that at least one of the values is drawn at least $k$ times? e.g. for $n=3000$, $m=300$, $k=20$.
Note: I was passed a variant of this by a friend asking for "a statistical package usable for similar problems".
My attempt: The number of times a particular value is reached follows a binomial law with $n$ events, probability $1/m$. This is enough to get odds $q$ that a particular value is reached at least $k$ times [Excel gives $q\approx 0.00340$ with =1-BINOMDIST(20-1,3000,1/300,TRUE)]. Given that $n\gg k$, we can ignore the fact that odds of a value being reached depends on the outcome for other values, and get an approximation of $p$ as $1-(1-q)^m$ [Excel gives $p\approx 0.640$ with =1-BINOMDIST(20-1,3000,1/300,TRUE)^300].
update: the exponent was wrong in the above, that's now fixed
Is this correct? (now solved, yes, but the approximation made leads to an error in the order of 1% with the example parameters)
What methods can work for arbitrary parameters $(n,m,k)$? Is this function available in R or other package, or how could we construct it? (now solved, both exactly for moderate parameters, and theoretically for huge parameters)
I see how to do a simulation in C, what would be an example of a similar simulation in R? (now solved, a corrected simulation in R and another in Python gives $p\approx 0.647$)
-
Just too note that this probability is 1 whenever $n\geq mk$. and you may have problems with values of $n$ which are close to $mk$. certain approximations may break down. – probabilityislogic Feb 13 '12 at 3:37
I could be wrong, but isnt the answer you want basically the cdf of a multinomial distribution? if so you are probably looking at the "multivariate" incomplete bet function – probabilityislogic Feb 13 '12 at 3:51
@probabilityislogic: We can make a slightly stronger statement: $n>m(k-1) \Rightarrow p(n,m,k)=1$; also $n<m \Rightarrow p(n,m,k)=0$. Yes the problem involves a multinomial distribution, but I fail to see how "at least one of the values is drawn at least $k$ times" translates into something computable as a CDF. – fgrieu Feb 13 '12 at 11:22
At least one is the "reciprical" of none drawn k times or more. Thus we have our probability as one minus a multivariate cdf. See @daniel's answer. – probabilityislogic Feb 13 '12 at 15:39
There are almost certainly easier ways, but one way of computing the value precisely is compute the number of ways of placing $n$ labeled balls in $m$ labeled bins such that no bin contains $k$ or more balls. We can compute this using a simple recurrence. Let $W(n,j,m',k)$ be the number of ways of placing exactly $j$ of the $n$ labeled balls in $m'$ of the $m$ labeled bins. Then the number we seek is $W(n,n,m,k)$. We have the following recurrence: $$W(n,j,m',k)=\sum_{i=0}^{k-1}\binom{n-j+i}{i}W(n,j-i,m'-1,k)$$ where $W(n,j,m',k)=0$ when $j<0$ and $W(n,0,0,k)=1$ as there is one way to pace no balls in no bins. This follows from the fact that there are $\binom{n-j+i}{i}$ ways to choose $i$ out of $n-j+i$ balls to put in the $m'$th bin, and there are $W(n,j-i,m'-1,k)$ ways to put $j-i$ balls in $m'-1$ bins.
The essence of this recurrence if that we can compute the number of ways of placing $j$ out of $n$ balls in $m'$ bins by looking at the number of balls placed in the $m'$th bin. If we placed $i$ balls in the $m'$th bin, then there were $j-i$ balls in the previous $m'-1$ bins, and we have already calculated the number of ways of doing that as $W(n,j-i,m'-1,k)$, and we have $\binom{n-j+i}{i}$ ways of choosing the $i$ balls to put in the $m'$th bin (there were $n-j+i$ balls left after we put $j-i$ balls in the first $m'-1$ bins, and we choose $i$ of them.) So $W(n,j,m',k)$ is just the sum over $i$ from $0$ to $k-1$ of $\binom{n-j+i}{i}W(n,j-i,m'-1,k)$.
Once we have computed $W(n,n,m,k)$ the probability that at least one bin has at least $k$ balls is $1-\frac{W(n,n,m,k)}{m^n}$.
Coding in Python because it has multiple precision arithmetic we have
import sympy
# to get the decimal approximation
#compute the binomial coefficient
def binomial(n, k):
if k > n or k < 0:
return 0
if k > n / 2:
k = n - k
if k == 0:
return 1
bin = n - (k - 1)
for i in range(k - 2, -1, -1):
bin = bin * (n - i) / (k - i)
return bin
#compute the number of ways that balls can be put in cells such that no
# cell contains fullbin (or more) balls.
def numways(cells, balls, fullbin):
x = [1 if i==0 else 0 for i in range(balls + 1)]
for j in range(cells):
x = [sum(binomial(balls - (i - k), k) * x[i - k] if i - k >= 0 else 0
for k in range(fullbin))
for i in range(balls + 1)]
return x[balls]
x = sympy.Integer(numways(300, 3000, 20))/sympy.Integer(300**3000)
print sympy.N(1 - x, 50)
(sympy is just used to get the decimal approximation).
I get the following answer to 50 decimal places
0.64731643604975767318804860342485318214921593659347
This method would not be feasible for much larger values of $m$ and $n$.
As there appears to be some skepticism as to the accuracy of this answer, I ran my own Monte-Carlo approximation (in C using the GSL, I used something other than R to avoid any problems that R may have provided, and avoided python because the heat death of the universe is happening any time now). In $10^7$ runs I got 6471264 hits. This seems to agree with my count, and is considerably at odds with whubers. The code for the Monte-carlo is attached.
I have finished a run of 10^8 trials and have gotten 64733136 successes for a probability of 0.64733136. I am fairly certain that things are working correctly.
#include <stdio.h>
#include <stdlib.h>
#include <gsl/gsl_rng.h>
const gsl_rng_type * T;
gsl_rng * r;
int
testrand(int cells, int balls, int limit, int runs) {
int run;
int count = 0;
int *array = malloc(cells * sizeof(int));
for (run =0; run < runs; run++) {
int i;
int hit = 0;
for (i = 0; i < cells; i++) array[i] = 0;
for (i = 0; i < balls; i++) {
array[gsl_rng_uniform_int(r, cells)]++;
}
for (i = 0; i < cells; i++) {
if (array[i] >= limit) {
hit = 1;
break;
}
}
count += hit;
}
free(array);
return count;
}
int
main (void)
{
int i, n = 10;
gsl_rng_env_setup();
T = gsl_rng_default;
r = gsl_rng_alloc (T);
for (i = 0; i < n; i++)
{
printf("%d\n", testrand(300, 3000, 20, 10000000));
}
gsl_rng_free (r);
return 0;
}
EVEN MORE
Note: this should be a comment to probabilityislogic's answer, but it won't fit.
Reifying probabilityislogic's answer (mainly out of curiosity), this time in R because a foolish inconsistency is the hobgoblin of great minds, or something like that. This is the normal approximation from the Levin paper (the Edgeworth expansion should be straightforward, but it is more typing than I'm willing to expend)
# an implementation of the Bruce Levin article here limit is the upper limit on
# bin size that does not count
approxNorm <- function(balls, cells, limit) {
# using N=s
sp <- balls / cells
mu <- sp * (1 - dpois(limit, sp) / ppois(limit, sp))
sig2 <- mu - (limit - mu) * (sp - mu)
x <- (balls - cells * mu) / sqrt(cells * sig2)
p2 <- exp(-x^2 / 2)/sqrt(2 * pi * cells * sig2)
p1 <- exp(ppois(limit, sp, log.p=TRUE) * cells)
sqrt(2 * pi * balls) * p1 * p2
}
and 1 - approxNorm(3000, 300, 19) gives us $p(3000, 300, 20) \approx 0.6468276$ which is not too bad at all.
-
I admire the attempt to produce an efficient exact solution. The crux of this is your recurrence: where does this come from? It does not appear to obtain correct answers. For instance, my simulation with 10^7 iterations (reported in another comment) has a standard error of 0.00015 and your result is (0.64732 - 0.64597) / 0.00015 = 8.9 standard errors too high. Thus, either the simulation is badly wrong or your solution is incorrect. (I suspect you may be double-counting.) What, for instance, do you get for the values of $W(4,4,3,k)$? They should be 54, 78, 81 for $k=2,3,4$ respectively. – whuber Feb 13 '12 at 15:31
@whuber I am more than willing to admit I am wrong, but where do you get $W(4,4,3,2)=54$? This is the number of ways of placing 4 balls in 3 bins such that no bin has two or more balls. I cannot find even one way. – deinst Feb 13 '12 at 16:08
@whuber I believe you mean $k=3,4,5$. In which case, yes I get those numbers. – deinst Feb 13 '12 at 16:11
Sorry, my interpretation of $k$ was off by 1. Since you get those numbers, it looks like your approach is basically right, at least for these small values. But to tell whether it is, we really need a clear explanation of the recurrence. – whuber Feb 13 '12 at 16:58
(+1) In light of your improved explanation, I have to agree. But before that explanation was available, there wasn't any good basis to decide which to favor. I'll take a well-reasoned theoretical argument any time over simulation results :-). – whuber Feb 13 '12 at 20:23
show 1 more comment
(Responding to the simulation question)
In R:
n <- 3000; m <- 300; k <- 20 # Problem parameters
nIterations <- 10000 # Number of iterations in the simulation
set.seed(17) # Make the output reproducible
#
# All the work is done in the following line.
#
t <- table(replicate(nIterations, any(tabulate(floor(runif(n, min=1, max=m+1))) >= k)))
t[["TRUE"]]/nIterations # Convert the count to a proportion
Expected output:
[1] 0.6453
To see more deeply into what's going on, look at the detailed distribution in several experiments by executing this command several times:
table(tabulate(floor(runif(n, min=1, max=m+1))))
Typical output is
2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21
2 5 11 23 23 32 48 37 33 21 19 21 9 6 5 1 2 1
In this experiment, two values in the range 0..299 were observed 19 times (among 3000 independent draws) and one value was observed 21 times. You will find that most of the time, at least one value occurs 20 or more times. Because $1/m$ is small and $n$ is large, you should be seeing a Poisson distribution here. Indeed,
1 - ppois(k-1, n/m)^m
returns
[1] 0.6458719
-
(Updated to reflect corrected code.) A simulation with nIterations=10^7 yields 0.647 23 (that one took 40 minutes over lunch :-). The first 3.5 digits of this result should be correct, strongly indicating the Poisson approximation (0.645 9) is actually more accurate than the Binomial (0.639 9). – whuber Feb 8 '12 at 17:13
Excel gives 0.640.., which appears significantly different from 0.645.. given by the simulation. Perhaps the hypothesis made for the Excel formula is not so correct after all. – fgrieu Feb 8 '12 at 17:44
Please don't trust Excel for such extended calculations. It is known to have problems with statistical distribution functions, especially for extreme values of their arguments. (I was shocked to find that it returned anywhere near a reasonable value...) R (1 - pbinom(19, 3000, 1/300)^300) and Mathematica (1 - CDF[BinomialDistribution[3000, 1/300], 19]^300 // N) both give 0.639877. – whuber Feb 8 '12 at 19:17
I do not trust Excel too much either, but here it gives 0.639877005.. which agrees to 10 figures with Mathematica 0.63987700518.. thus the problem is either with the approximation made or (less likely) the simulation. – fgrieu Feb 8 '12 at 20:14
Yes, it's interesting that Excel is getting a good answer. Clearly the Binomial approximation is decent but not entirely accurate. At least in this case, the Poisson approximation is highly accurate. It's good to mistrust theoretical calculations and simulations, even when they agree, so I invite you to produce a simulation of your own as a check! (You could do this with an Excel macro, but that's a bit of a pain...) – whuber Feb 8 '12 at 20:33
This is a harder question if you don't have the $n\gg k$ and assuming that this makes them 'close enough' to independent to not affect the answer non-trivially. Lets proceed with these assumptions. Let $X_j \sim Binomial(n,\frac{1}{m})$ $\forall j = 1,..,m$.
$$P(\max_j X_j \geq k) = 1 - P(\max_j X_j < k)$$ $$= 1 - P(X_1 < k,...,X_m < k)$$ and, assuming independence of the $m$ random variables, $$= 1 - \prod^m_{j=1}P(X_j < k)$$ $$= 1 - [P(X_1 < k)]^m$$ $$= 1 - [\sum^{k-1}_{i=0} {n \choose i}(\frac{1}{m})^i(1-\frac{1}{m})^{n-i}]^m$$
or, if you have the binomial cdf function in the language you are using:
$$= 1 - [Binomial\_cdf(k-1;n,\frac{1}{m})]^m$$
-
I believe the exponent should be an m, not a k as you have. I have included another step that makes this more clear. I may be missing something, or misunderstanding your question, so please clarify if you still don't agree. – Daniel Johnson Feb 8 '12 at 16:40
To help resolve these differences, I have edited the question to show the value Excel actually returns for the formula as given (with $k$ in the exponent). Putting $m$ in the exponent yields a value of 0.6398; compare this to the simulation results. – whuber Feb 8 '12 at 16:56
So the rest of your question is how to represent this expression in R? I don't use R, so I can't help you with that, but i would guess there is a binomial cdf function that you can use and if not, the summation formulation should be pretty trivial to code. – Daniel Johnson Feb 8 '12 at 19:50
@DanielJohnson: Thanks for your explanation and correction. I can now use R for simulation (as well as computing the theoretical approximation), thanks to whuber. The only part remaining is how not to depend on an approximation that is not so good (it introduce a relative error on p of nearly 1%, and likely much worse in other setups). – fgrieu Feb 8 '12 at 20:21
The collection of numbers has a multinomial distribution with $m$ categories and $n$ sample size. Letting $N_i$ be the number of times the $i$th category is chosen/repeated, we have $$(N_1,\dots,N_m)\sim multinomial\left(n;\frac{1}{m},\frac{1}{m},\dots,\frac{1}{m}\right)$$ Now leveraging off of @danieljohnson's answer the probability we are after is
$$p(n,m,k)=1-Pr(N_1<k,\dots,N_m<k)$$
i.e. if all numbers are repeated less than $k$ times, then none are repeated at least $k$ times. And "not none" is the same as "at least one" so we can take the probability away from one. This could be computed via a "brute force" approach, as the pmf we have is particularly simple:
$$p(n,m,k)=1-m^{-n}\sum_{N_1<k,\dots,N_m<k|N_1+\dots+N_m=n}{n\choose N_1\dots N_m}$$ $$=1-\frac{n!}{m^{n}}\sum_{N_1=0}^{k-1}\sum_{N_2=0}^{k-1}\dots\sum_{N_{m-1}=0}^{k-1}\frac{1}{N_1!N_2!\dots N_{m-1}!(n-N_1-N_2-\dots-N_{m-1})!}$$
The last formula is correct provided we interpret a negative factorial as $\pm\infty$ (consistent with the gamma function) which eliminates these from the summation.
On doing a quick google search came up with Bruce Levin's article. This gives a representation of the multinomial distribution as a collection of poisson random variables, with their sum being fixed. (note this might explain why @whuber has found that poisson approximation works better than binomial). Now, using the representation given in theorem 1 of the paper, we have:
$$p(n,m,k)=1-\frac{n!}{s^n\exp(-s)}\left[\prod_{j=1}^{m}Pr(X_j\leq k-1)\right]Pr(W=n)$$
Where $X_j\sim Poisson\left(\frac{s}{m}\right)$ and are independent, and $W=\sum_{j=1}^{m}Y_j$ is a sum of independent truncated poisson distributions - basically $Y_j$ is $X_j$ conditioned to be less than or equal to $k-1$. Note that we can simplify the general formula by noting that the terms in the product do not depend on the index $j$, and so is just a single poisson cdf raised to the power of $m$. Thus we have:
$$p(n,m,k)=1-\frac{n!}{s^n}\left[e_{k-1}\left(\frac{s}{m}\right)\right]^mPr(W=n)$$
Where $e_k(x)=\sum_{j=0}^{k}\frac{x^j}{j!}$ denotes the exponential sum function. note that because we have factorial an powers of potentially large numbers, numerically it will probably be better to work in terms of the logarithm of the second term, and then exponentiate back at the end of the calculation. Alternatively, we can choose the recommended $s=N$ as our algorithm parameter, and then make use of the stirling approximation to $n!$ - this is recommended in the paper and corresponds to "mean matching" of each poisson distribution with the multinomial cell (i.e. $E(X_i)=E(N_i)$). Then we get $\frac{n!}{n^n}\approx\sqrt{2\pi n}$.
The paper provides two approximations for $Pr(W=n)$ on based on normal approximation, and another based on edgeworth expansion. details are in the paper (see equation 4). Note though that his method allows for different probability parameters, so terms like $\frac{1}{t}\sum_{1}^t\sigma_l^2$ can be replaced with $\sigma_1^2$ and so on, which avoid unecessary computation. Note that we also have the mallows bounds provided in the paper - which can be used to check the accuracy of the approximations.
-
I have added code that performs the normal approximation from the paper at the end of my answer (it was too big to fit here). I get $p(3000, 300, 20)\approx 0.6468276$ (an error of about 0.0005). – deinst Feb 14 '12 at 17:35
| 2013-12-11T01:28:32 |
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https://stats.stackexchange.com/questions/211283/probability-of-drawing-a-heart-and-then-an-even-number-without-replacement-fro
|
# Probability of drawing a heart and then an even number, without replacement, from a deck of cards
I had an algebra student I know ask me an interesting question: What is the probability of drawing a heart out of a deck of cards, and then drawing an even number right after it?
Clearly, the probability of drawing a heart out of the deck is 13/52, or 1/4. The probability of drawing an even number as card #2 ($c\in\{2,4,6,8,10\}$) for any suit is 20/51, except when the first card drawn is an even heart, then the probability of the second draw is 19/51.
I wish it were as simple as $P(E|H) = \frac{P(E\cap H)}{P(H)} = \frac{\frac{5}{52}}{\frac{13}{52}} = \frac{5}{13}$.
My Monte Carlo simulation I wrote to simulate this process does not match: I get something in the neighborhood of 9-10% from the code I wrote. Where is my logic faulty?
import java.util.Random;
public class CardCounter {
/**
* @param args
*/
public static void main(String[] args) {
int heartThenEvenCount = 0;
int cardsNotEqualCount = 0;
for(int i = 0; i < 1000000; i++) {
Random r = new Random();
int card1 = r.nextInt(51) + 1;
int card2 = r.nextInt(51) + 1;
/* Assume that 1-13 are hearts
Ace = 1
King = 0
Queen = 12
Jack = 11
Other numbers as they are written on card. */
//Make sure the cards are not the same
if(card1 != card2) {
// Check to see if the first card is a heart.
cardsNotEqualCount++;
if(card1 <= 13) {
// Check to see if the second card is even, but not a face card.
int card2Rank = (card2 % 13);
if(card2Rank > 0 && card2Rank <= 10 && (card2Rank % 2) == 0) {
heartThenEvenCount++;
}
}
}
}
System.out.println(heartThenEvenCount + " / " + cardsNotEqualCount);
System.out.println(heartThenEvenCount/(double)cardsNotEqualCount);
}
}
• I was going to suggest the law of total probability as well, but saw that you already have 2 fine responses. May 7 '16 at 3:52
• I prefer the newer response by David, as it's much more elegant to use independence in this scenario. However, the other responses are a good example of how to proceed when the events are not independent. May 7 '16 at 7:19
• Please note that creating new Random instance in every loop step is inefficient, and more importantly, it will make random numbers of inferior quality. May 7 '16 at 17:34
• Good call. I wrote this in about ten minutes, so pardon the bad programming (and perhaps even statistical) practice. :) May 7 '16 at 20:49
I don't know what the heck your "wish it were as simple as" analytical calculation is supposed to be, but you have already laid out the pieces, and just need to combine them properly. Use the law of total probability and break down (condition on) whether the 1st card heart is not an even number or is an even number.
So P(1st card is heart and 2nd card is an even number) = P(1st card is heart and not an even number) * P(2nd card is even number given 1st card is not an even number) + P(1st card is heart and is an even number) * P(2nd card is even number given 1st card is an even number) = $(8/52) (20/51) + (5/52) (19/51) = .0961538...$ which is in the 9 to 10% neighborhood you say your simulation provided.
• Your $P(E|H) = 5/13$ is just the probability that the 1st card is an even number given that it is a heart. But of course, that doesn't answer the original question even though it is relevant information. May 7 '16 at 3:12
You probably need the law of total probability. For clarity, let's denote the following events: \begin{align} & A = \{\text{the first drawing is a heart}\}, \\ & B = \{\text{the second drawing is an even number}\}, \\ & C = \{\text{the first drawing is an even number}\}. \end{align} The probability of interest is $P(A \cap B)$, which is $P(A)P(B \mid A)$. Clearly, as you stated $P(A) = 1/4$. To find $P(B \mid A)$, by law of total probability, \begin{align} P(B \mid A) & = P(B \cap C \mid A) + P(B \cap C^c \mid A) \\ & = P(B \mid C \cap A)P(C \mid A) + P(B \mid C^c \cap A)P(C^c \mid A). \end{align} It is straightforward to verify that \begin{align} & P(C \mid A) = \frac{P(A \cap C)}{P(A)} = \frac{5/52}{13/52} = \frac{5}{13}, \\ & P(C^c \mid A) = \frac{P(A \cap C^c)}{P(A)} = \frac{8}{13}, \\ & P(B \mid C \cap A) = \frac{19}{51}, \\ & P(B \mid C^c \cap A) = \frac{20}{51}. \end{align} Thus $P(B \mid A) = \frac{255}{663}$, and the final answer is $255/2652 = 0.09615385$.
Drawing a heart on the first card and drawing an even number on the second card are independent events, so the answer is just the product of the individual probabilities. Drawing a heart is 1/4 and drawing an even number is 5/13, so the answer is 5/52 = .0961538...
• Maybe it needs a bit of explanation (probability that you draw an even number in the second drawing is the same as in the first drawing and it is the same if you take into account only 1/4 of the cases when you draw it and 1/4 of the cases when you do not do it (or 1/4 of the cases when you draw an even number in the first drawing, reducing the probability, and 1/4 of the cases when you draw another card in the first drawing, not reducing the probability)), but it is really as simple as OP wished. May 7 '16 at 7:11
• To hopefully be a bit more clear: if $A$ is the event that the first card is a heart, and $B$ is the event that the second card is even, then knowing that $A$ has occurred doesn't affect the probability that $B$ will occur. That is, $P(B|A) = P(B)$, which means $P(A \cap B) = P(A)P(B)$ (see indepedence). May 7 '16 at 7:17
• It looks to me this anwser used circular proof: without showing $P(B \mid A) = P(A)$, how can you claim indepedence? The intuition cannot replace mathematical proof. May 7 '16 at 14:36
• Sure it can.. you can always revert back to first principles to rigorously show $P(B | A) = P(A)$ as the other responses have done in an indirect way, but there comes a time when that is a waste of effort, and it suffices to use your understanding of what a concept means. You see this all the time in higher level mathematics. Rigour is often left for when additional understanding can be revealed (which is indeed the case for the original poster, which is why I did not downvote the other answers) or to convince the unconvinced. Just saying, you shouldn't be so quick to dismiss this answer. May 7 '16 at 19:13
• Correction: $P(B|A) = P(B)$ for the previous two comments. May 7 '16 at 21:58
| 2022-01-23T04:26:26 |
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