Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Visual representation of the domain and range of a function? An excerpt in the book "College Algebra by Michael Sullivan is that:
When the graph of a function is given, its domain may be viewed as the shadow created by the graph on the x- axis by vertical beams of light. Its range can be viewed as the shadow created by the graph on the y-axis by horizontal beams of light.
I did't get it and i also sought in google to get some idea but i can't find one.
Can anyone help me to understand it maybe with some graphics.
Thanks.
| Consider a particular point on the function. If you cast a vertical light beam through that point it would cast a shadow on the x-axis according to the x value of that point. Since the domain is the set of all possible x values of a function, imagine repeating the vertical light beams on every point on the function. Then the resulting shadow on the x-axis represents the domain. It is a set of values on the x-axis. By a similar process you can represent the range on the y-axis.
| {
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Epsilon delta proof for $\lim_\limits{x\to a} \sqrt{x} = \sqrt{a}$ where $a>0$ $f(x) = \sqrt{x}$. For right limit, $x-a > 0$. Let $x-a < \delta$ then $(\sqrt{x}-\sqrt{a}) \lt \frac{\delta}{\sqrt{x}+\sqrt{a}}$. Then chose we must choose $\epsilon $ so that $\epsilon > \frac{\delta}{\sqrt{x}+\sqrt{a}}$. Is this correct? How would I conclude the proof?
Edit - Follow up
From answers, I try a proof for the same limit but this time Left side:
For left side limit, $x-a < 0$ so $a-x > 0$. Let $a-x < \delta$ then $\sqrt{a}-\sqrt{x} \lt \frac{\delta}{\sqrt{a}+\sqrt{x}} \lt \frac{\delta}{\sqrt{a}}$.
So let me begin the proof by saying to choose $\delta = \epsilon \sqrt a$ then $a-x \lt \delta $ so $(\sqrt a - \sqrt x )(\sqrt a + \sqrt x) \lt \epsilon \sqrt a$ so $\sqrt a - \sqrt x \lt \epsilon$
| For $x >0$, $a>0$:
$|√x-√a| =\dfrac {|x-a|}{√x+√a}\lt$
$\dfrac{|x-a|}{√a}.$
Let $\epsilon >0$ be given.
Choose $\delta \le \epsilon √a.$
Then
$|x-a| \lt \delta$ implies
$|√x-√a| \lt \dfrac{|x-a|}{√a} \lt $
$\dfrac{\delta}{√a} \le\epsilon.$
| {
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Understanding Galois theory, swapping roots I am trying to understand the rudiments of galois theory but I have a hard time answering the following question: Suppose we have $f(X)=x^5-2$ over $\mathbb{Q}$. What is $Gal(E_f: \mathbb{Q})$?
After seeing similar questions here I am more confused: Isn't it true that since $f$ is irreducible, every possible permutation of its roots can be extended linearly to a endomorphism $E_f \rightarrow E_f$ while holding $\mathbb{Q}$ constant? If that is so then shouldn't the Galois Group be $S_5$?
I understand that this isn't the case, but can't really understand why...
| First note that the splitting field of $f$ is given by $L = \mathbb{Q}(\sqrt[5]{2}, \zeta_5)$, where $\zeta_5$ is the primitive root of unity.
Now we know that any automorphism on $L$ is uniquely determined by the action on the adjoined elements $\sqrt[5]{2}$ and $\zeta_{5}$. Additionally we know that if $\alpha_1$ and $\alpha_2$ are roots of the same irreducible polynomial over $\mathbb{Q}$ then the identity automorphism on $\mathbb{Q}$ can be extended to a automorphism $\sigma$ of the splitting field s.t. $\sigma(\alpha_1) = \alpha_2$. In particular in our case we can have automorphism of $L$ s.t. $\sigma(\sqrt[5]{2}) = \zeta_5^2\sqrt[5]{2}$. Now we can do the same for $\zeta_5$ and for example get an automorphism $\tau$ on $L$, s.t. $\tau(\zeta_5) = \zeta_5^3$ and it is identity on $\mathbb{Q}$.
However note that this doesn't give you every possible permutation of roots, as it's not possible to permute the first three roots, while to keep the other two fixed. So to summarize there is an element of the Galois group of $f$ (assuming $f$ is irreducible) s.t. that it send any root to any other root of $f$, but it's not necessarily keeping the other roots fixed or permuting them in arbitrary manner.
| {
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Add together numbers to make final number? I have this problem where I want to add some numbers together to make a final number. I ask how many ways are there to make this final number using exactly k numbers? Here is the condition:
(1) The numbers are in a consecutive sequence but with a number missing like 1,3,4 or 2,3,4,5.
(2) I want to have at least one of each number.
(3) I can use any number as many times like the permutation like 1,1,3,4,4.
I have this examples:
I want to make final number 12 with 1,3,4 in exactly 5 numbers. I can do only 1,1,3,3,4. And can do it 5!/(2!2!)=30 ways.
I want to make final number 21 with 1,2,3,4,6 in exactly 8 numbers. I can do only 1,1,2,2,2,3,4,6 or 1,1,1,2,3,3,4,6. I can do both in 8!/(2!3!)=3360 ways.
I think it involves partitions or combinations but I can't find a nice or any formula to do it. Is it possible to do this?
Thank you in advance. Sorry for the format I am new to maths exchange.
| Your question is similar to this one if not the sameFinding all possible combinations of numbers to reach a given sum
i wanted to comment but couldnt so i had to put it as an answer
| {
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Hypothesis testing - Experiment with replacement Exercise :
A box contains 4 balls out of which $\theta$ are white and $4-\theta$ are black. Suppose that you want to check the null hypothesis $H_0 : \theta =2$ against the alternative $H_1 : \theta \neq 2$. You draw 2 balls with replacement from the box and if they are both of the same color you reject the null hypothesis.
i) Calculate the confidence level $a$ of the test above.
ii) If the box contains 3 white balls, calculate the probability of the type two error for the test above.
Question :
First of all doea the last phrase means that you draw one ball first and then replace it and draw a second one too? Can someone please clarify me the test mentioned?
Generally, I know that :
$$a = \mathbb{P}(reject H_0 | H_0 true)$$
How would I express this probability though ? Obviously due to replacement he events are independent and then the probability of drawing 2 of the same color is :
$$\mathbb{P}(2black) = \mathbb{P}(black)\cdot \mathbb{P}(black) = \frac{(4- \theta)^2}{16}$$
$$\mathbb{P}(2white) = \mathbb{P}(white) \cdot \mathbb{P}(white) = \frac{\theta^2}{16}$$
Also the type II error is :
$$P(accept H_0 | H_1 true)$$
Can someone help me formulate the expression for $a$ ? I would really appreciate a thorough explanation.
| Assuming this is a test of drawing two balls with replacement......
A type I error is rejecting the null when you shouldn't. The only scenario where this can occur is when there are two white and two black and two of the same color are chosen in the test. The probability of that is $0.5$. The first choice can be either color, the probability is limited to picking a second ball the same color as the first. This is $2/4 = 0.5$. Hence the confidence level, which is the same as the probability of making a type I error, is $0.5$
A type II error is the probability of not rejecting the null when you should. If there are 3 white and one black, this is limited to the probability of selecting one of each color. If the first is a white $3/4$, the probability of the second being a black is $1/4$ hence $P1 = 3/16$. If the first is a black $1/4$, the probability of the second being white is $3/4$ hence $P2 = 3/16$. The total probability of making a type II error is therefore $P = 3/16 + 3/16 = 3/8$.
| {
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Permutation and characteristic polynomial of a matrix I have a matrix $A\in \mathbb{R}^{n\times n}$, after some manipulations I find:
$$\tilde{A}=\begin{pmatrix} 0 & I_n&0&\cdots&0\\0&0&I_n&\cdots&0\\ \vdots&\vdots& \vdots&\ddots &\vdots\\0&0&0&\cdots&I_n\\ A & 0 &0 &\cdots &0 \end{pmatrix};\; \mathcal{A}=\begin{pmatrix} I_n&0&\cdots&0&0\\0&I_n&\cdots&0&0\\ \vdots& \vdots&\ddots &\vdots&\vdots\\0&0&\cdots&I_n&0\\ 0 & 0 &\cdots &0&A \end{pmatrix} ; \tilde{A},\mathcal{A}\in\mathbb{R}^{kn\times kn}$$
by simple permutation $\tilde{A}$ becomes $\mathcal{A}$.
Q: my objective is to find if there exist, a relation between eigenvalues of $A$ and $\tilde{A}$ (so $\mathcal{A}$) in term of characteristic polynomial?
Is there any relation between $P_\mathcal{A}(\lambda)$ and $P_\tilde{A}(\lambda)$, where $\mathcal{A}$ is a simple rotation of $\tilde{A}$? or just a relation between their eigenvalues?
Thanks a lot,
| I see no easy way to relate the eigenvalues of $\tilde A$ and $\mathcal A$ directly. However, both of these have eigenvalues that are neatly related to those of $A$.
Characteristic polynomial of $\mathcal A$: Note that the vectors of the form
$$
v = \pmatrix{x_1\\\vdots \\ x_{k-1} \\ 0}, \quad v = \pmatrix{0\\ \vdots\\ 0 \\ w}
$$
where $x_i \in \Bbb C^n$ are arbitrary and $w$ is an eigenvector of $A$ can be used to form an eigenbasis of $\mathcal A$. Conclude that $\mathcal A$ has characteristic polynomial
$$
p_{\mathcal A}(\lambda) = (\lambda - 1)^{n(k-1)}p_{A}(\lambda)
$$
Although we assumed that $A$ is diagonalizable, this carries over to the case where $A$ fails to be diagonalizable by the continuous dependence of the characteristic polynomial on matrix entries. We could also make an argument appealing to the Jordan form of $A$.
Characteristic polynomial of $\tilde A$: a vector $v = (x_1,\dots,x_k)$ will be an eigenvector of $\tilde A$ if and only if we have
$$
\pmatrix{x_2\\ \vdots \\ x_k\\ Ax_1} = \lambda\pmatrix{x_1\\ \vdots \\ x_{k-1} \\ x_k}
$$
That is, the vectors $x_i$ must satisfy
$$
x_i = \lambda x_{i-1} \qquad i=2,\dots,k\\
Ax_1 = \lambda x_k
$$
By substituting, we see that this amounts to
$$
Ax_1 = \lambda^k x_1
$$
So, $\lambda^k$ will be an eigenvalue of $\tilde A$ whenever $\lambda$ is an eigenvalue of $A$. We can use this to conclude that
$$
p_{\tilde A}(\lambda) = p_A(\lambda^k)
$$
| {
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Presentation of the holomorph of $\mathbb Z/5 \mathbb Z$ When I look up the presentation of the holomorph of $\mathbb Z/5 \mathbb Z$ it reads like the following:
$\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^2\right\rangle$
See https://groupprops.subwiki.org/wiki/General_affine_group:GA(1,5)
The automorphisms of $N := \mathbb Z/5 \mathbb Z$ are as follows $Aut(N) = \{\iota, \psi, \psi^2, \psi^3\}$
where $\psi: \mathbb Z/5 \mathbb Z \to \mathbb Z/5 \mathbb Z$ via $x \mapsto 3\cdot x$ and $\iota: \mathbb Z/5 \mathbb Z \to \mathbb Z/5 \mathbb Z$ via $x \mapsto x$. This means in detail:
*
*$\psi(x) = 3\cdot x$
*$\psi^2(x) = -x$
*$\psi^3(x) = 2\cdot x$
*$\psi^4(x) = x$
Define the holomorph of $N$ as $G := N \rtimes Aut(N)$ via $(g_1, \eta_1)\star (g_2, \eta_2) := (g_1 + \eta_1(g_2), \eta_1\circ\eta_2)$.
Note $(0,\iota)$ is the identity element in $G$.
Define (possible) generators $x := (1,\iota)$ and $y := (0, \psi)$ of $G$.
*
*$x^2 = (1,\iota)\star (1,\iota) = (1 + \iota(1),\iota\circ\iota) = (1 + 1,\iota\circ\iota) = (2,\iota)$
*$x^3 = (1+2,\iota) = (3,\iota)$
*$x^4 = (4,\iota)$
*$x^5 = (0,\iota)$
*$y^2 = (0, \psi) \star (0, \psi) = (0 + \psi(0), \psi\circ\psi) = (0+0,\psi^2) = (0,\psi^2)$
*$y^3 = (0, \psi^3)$
*$y^4 = (0, \psi^4) = (0,\iota)$
*$y\star x \star y^{-1} = (0, \psi)\star (1,\iota) \star (0, \psi^3) = (0+\psi(1),\psi) \star (0, \psi^3) = (3,\psi) \star (0, \psi^3) = (3+\psi(0),\psi^4) = (3,\iota)= x^3$
So, everything works out except for $y\star x \star y^{-1} =x^3$ instead of $y\star x \star y^{-1} = x^2$ as in the presentation given above.
Questions:
*
*Did I make an error? (I apologize if it is obvious to you.)
*Are the presentations $\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^2\right\rangle$ and $\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^3\right\rangle$ equivalent? If so, is there an easy argument?
Thank you for your thoughts!
| Let let $x := (1,\iota)$ and $y := (0, \psi^3)$ as motivated by the comment of Tobias Kildetoft above.
*
*$y^2 = (0, \psi^3) \star (0, \psi^3) = (0 + \psi^3(0), \psi^3\circ\psi^3) = (0+0,\psi^2) = (0,\psi^2)$
*$y^3 = (0, \psi^5) = (0, \psi)$
*$y^4 = (0, \psi^4) = (0,\iota)$
*$y\star x \star y^{-1} = (0, \psi^3)\star (1,\iota) \star (0, \psi) = (0+\psi^3(1),\psi^3) \star (0, \psi) = (2,\psi^3) \star (0, \psi) = (2+\psi^3(0),\psi^4) = (2,\iota)= x^2$
| {
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Is this complex exponential inequality true? Is it true that $|f(z)|\leq M$ implies that $|e^{f(z)}|\leq e^{M}$. If not, what is a simple counterexample?
| For $f=u+iv$ we have $u(z)\le |f(z)|\le M$ and
$$
|e^{f(z)}|=|e^{u(z)}e^{iv(z)}|=e^{u(z)}\le e^M
$$
since $e$ is increasing.
| {
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Finding $\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}$.
Find an expression for
$$\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}.$$
This was given in the chapter for APs. However, I do not see how this relates to them. I tried using telescopic sums, but I am not proficient in them and was thererfore unable to solve this question.
I tried calculating the sum for a few consecutive values, and tried constructing a polynomial that covered them all, but it proved to be difficult.
Any hints on how to solve this question?
Please help.
| Let $a_k = k/(1+k^2+k^4)$.
Then:
$$a_k =
\underbrace{\frac{k(1-k)}{2(1-k+k^2)}}_{b_k}
+
\underbrace{\frac{k(1+k)}{2(1+k+k^2)}}_{c_k}.$$
Now observe that
$$b_{k+1} = \frac{(k+1)(-k)}{2(-k+(k+1)^2)}
=
\frac{-k(1+k)}{2(1+k+k^2)} = -c_k.$$
Can you conclude?
| {
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Find the following limit: $\lim \limits_{x,y \to 0,0} \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}$ Recently I came upon a limit which confused me. The reason is that when I try to solve the following limit using polar coordinates I get a constant which I do not know if it gives me information.
Let : $$\lim_{x,y \to 0} \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}$$
Using polar coordnates I get this:
$$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{\sin\left(r\sin\left(\theta\right)\right)+\log\left(1+r\cos\left(\theta\right)\right)}$$
Which is equal to:
$$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{r\sin\left(\theta\right)+r\cos\left(\theta\right)}=1$$
I already know this limit does not exist. Actually it was quite difficult to find a path for which I get a different limit...
My question is: If I use polar coordinates and the result is not something that depends on $r,\theta$ then what I get is basically useless information? (I know that if that limit goes to infinity the limit of the function does not exist)
| Note that
*
*$x=0,\, y=t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}=\frac{t-\frac{1}{2}t^2}{\sin\left(t\right)}\to 1$
*$x=-t+\frac12t^2,\, y=t,\, t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}
=\frac{-t+\frac12t^2+t-\frac12t^2}{\sin\left(t\right)+\log\left(1-t+\frac12t^2\right)}=0$
| {
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Why is $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}}>0$ for all $({{x}_{1}},{{x}_{2}},{{x}_{3}})\ne (0,0,0)$ if and only if $|\alpha |<1$? Why is $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}}>0$ for all $({{x}_{1}},{{x}_{2}},{{x}_{3}})\ne (0,0,0)$ if and only if $|\alpha |<1$? I tried solving the inequality and just ended up with $\alpha >-\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{2{{x}_{1}}{{x}_{2}}}$ (if ${{x}_{1}}{{x}_{2}}>0$ ) and $\alpha <-\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{2{{x}_{1}}{{x}_{2}}}$(if ${{x}_{1}}{{x}_{2}}<0$).
| There are probably easier ways to do it. But whatever, consider the matrix representing this quadratic form: $$A = \begin{pmatrix} 1 & \alpha & 0 \\ \alpha & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}.$$Your condition is equivalent to the matrix $A$ being positive-definite, which by Sylvester's Criterion happens if and only if all principal minors are positive. We compute $$\begin{align} \Delta_1 &= 1, \\ \Delta_2 &= 1-\alpha^2, \\ \Delta_3 &= 1-\alpha^2. \end{align}$$These guys are all positive if and only if $1-\alpha^2 > 0$, which is the same as $|\alpha|<1$.
| {
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Annihilation and direct sums On
http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf
page 87, ex. 3.8, we read:
Given $V=M\oplus N$, prove that $V^*= M^0 \oplus N^0$, being $V^*$ the dual space of $V$ and $M^0$ the annihilator of M (same for $N$).
If $V$ is finite, I can see that $N$ is isomorphic to $M^0$ and vice-versa since $d(V)= d(M)+d(M^0)=d(N)+d(N^0)$ (but this is not exactly the proof, nor $V$ is assumed finite).
Any hint?
| Just use the definitions. If one desires to check that $V^* = M^0 \oplus N^0$, there are two things to be done.
1) $M^0 \cap N^0 = 0$. Let $f \in M^0 \cap N^0$. Then $f|_M = 0$ and $f|_N = 0$ gives $f|_{M \cup N} = 0$. But $f$ is linear, so it follows that $f|_{V}= f|_{M\oplus N} = f|_{{\rm span}(M\cup N)} = 0$.
2) $V^* = M^0 + N^0$. Let $f \in V^*$. We want to write $f = f_1+f_2$ with $f_1 \in M^0$ and $f_2 \in N^0$. Let's see what $f_1$ and $f_2$ must be. If an element decomposes as $v = m+n$, then we must have $$f(v) = f_1(n) + f_2(m).$$This motivates us to define $f_1 = f\circ \pi_N$ and $f_2 = f\circ \pi_M$, where $\pi_M$ and $\pi_N$ are the projections of $V$ onto $M$ and $N$, respectively. Clearly $f_1$ and $f_2$ are linear, with $f_1 \in M^0$ and $f_2 \in N^0$. And $$f_1(v) + f_2(v) = f(\pi_N(v)) + f(\pi_M(v)) = f(n)+f(m) = f(n+m) = f(v)$$as wanted.
| {
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How do I use the given vector equation to resolve vector $p$ into a parallel and perpendicular component? I am working on the following problem:
Here's what I've done so far:
I know that dotting the first component with q should equal one to show that it is parallel and dotting the second component with q should equal to 0 to show that it is equal to zero to show that it is perpendicular. I haven't been getting those two results. Please help!
|
I know that dotting the first component with q should equal one to
show that it is parallel
Not in general:
$$
p_\parallel \cdot q
= \lVert p_\parallel \rVert \lVert q \rVert \cos \angle(p_\parallel, q)
= \lVert p_\parallel \rVert
$$
and dotting the second component with q
should equal to 0 to show that it is equal to zero to show that it is
perpendicular.
Check your calculation:
$$
(p \times q)_1 = -6/\sqrt{17} - (-1 \cdot -2/\sqrt{17}) = -8 / \sqrt{17}
$$
etc.
| {
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Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$ I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$
My attempt: Letting
$$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$
$$
\begin{aligned}
I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta)
\end{aligned}
$$
$$
I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha
$$
But I can't go further.
| Taking integration by parts,
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$
Now
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx
= \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$
and the first integral is easily computed by integration by parts:
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx
= -\log 2 + \frac{\pi}{2}. $$
The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion
\begin{align*}
\log \sec\theta
= -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right|
&= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\
&= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta),
\end{align*}
we have
\begin{align*}
\int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx
&= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\
&= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,}
\end{align*}
where $G$ is Catalan's constant. Combining altogether, we obtain
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$
Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The unit group of a finite dimensional associative algebra is a Lie group? I am reading Serre's "Lie algebras and Lie groups" p.103.
Let $k$ be a complete valued field(for example $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{Q}_p$) and $R$ be a finite dimensional associative $k$-algebra.
Surely $R$ is an additive Lie group.
The book asserts that the unit group $G_m(R)$ is a multiplicative Lie group
and also contains the proof, but I cannot understand it.
I copy the text here.
"We contend that $G_m(R)$ is an analytic group which is open as a subset of $R$. To show that $G_m(R)$ is open in $R$ it suffices to show that there is a neighborhood of $1$ contained in $G_m(R)$. Now, there exists an open neighborhood $U$ of $0$ in $R$ such that for $x \in U$ the series $\sum x^n$ converges. It follows $V=\{1-x:x \in U\} \subset G_m(R)$ and $V$ is a neighborhood of $1$. To show that $G_m(R)$ is an analytic group it remains to show that multiplication is a morphism. This follows since multiplication in $R$ is bilinear."
I cannot understand the first step and the final step:
*
*Why does there exist an open set $U$ which satisfies $\sum x^n$ converges?
*Why is multiplication a manifold morphism? (Also, It seems that we need $x\mapsto x^{-1}$ is a morphism.)
From googling, I've found (ex1) of http://www.math.cornell.edu/~sjamaar/classes/6520/problems/2016-10-26.pdf
, but still I cannot solve it.
| *
*For $x$ of norm $<1$, $(\displaystyle\sum_{k=0}^n x^k)_n$ is a Cauchy sequence (by the triangle inequality) , thus by completeness of $k$ and finite dimensionality of $R$, it converges in $R$. The open neighbourhood is $||x||<1$.
*Multiplication $R\times R\to R$ is bilinear so it must be smooth, thus so is its restriction to $G_m(R)$.
It remains the question of why $x\mapsto x^{-1}$ is smooth. First of all note that for $y=1-x\in V$, $y^{-1}= \sum_n x^n$, so for $y\in V$, $y^{-1}=\sum_n (1-y)^n$, thus the inversion is smooth on a neighbourhood of $1$.
If $x\in G_m(R)$, and $y\in xV$, then $y=xv$ for some $v\in V$ and $y^{-1}= v^{-1}x^{-1}= (\sum_n (1-v)^n)x^n$, so $y^{-1}= (\sum_n (1-x^{-1}y)^n)x^{-1}$ so it is smooth as well
| {
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"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Does this matrix equality hold? How to prove it? $B \bullet A=B\bullet U\Lambda U^T=U^TBU\bullet A$? Suppose $A,B$ are all symmetric matrices, How tp prove that $B \bullet A=B\bullet U\Lambda U^T=U^TBU\bullet A$, in which $\bullet$ is the inner product of a matrix defined as $A\bullet B=\sum_{i,j=1}A_{ij}B_{ij}$. $\Lambda $ is a diagonal matrix, $U$ is a orthognal matrix.
| Presumably, we're given that $A = U\Lambda U^T$.
Note that $A \bullet B = \operatorname{Tr}(A^TB)$. With that in mind, we note that
$$
B \bullet A = \operatorname{Tr}(B^TA) = \operatorname{Tr}([B^TU\Lambda] U^T) =
\operatorname{Tr}(U^T[B^TU \Lambda]) =
\operatorname{Tr}([U^TBU]^T\Lambda) = [U^TBU] \bullet \Lambda
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. As the question states:
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related question, and understand the steps to be:
*
*find the first few terms of the Taylor polynomial.
*Generalize the terms by making use of an infinite sum to represent the function as the Taylor series.
*use the infinite sum in the ratio test to find the radius of convergence.
Progress so far:
*
*The first 6 terms (n = 0 to n = 5) of the Taylor polynomial I have calculated to be:
$x^3 \cdot \ln{(\sqrt{x})} + \frac{1}{2}(x-a) + \frac{5}{4}(x-a)^2 + \frac{11}{12}(x-a)^3 + \frac{1}{8}(x-a)^4 - \frac{1}{40}(x-a)^5$
It is at this point however that I fall over. It is not intuitive to me how I can write the c-terms as a function without utilizing some sort of online maths engine for fitting the data to a curve.
Is there some sort of first-year-student-friendly technique for modelling these data points systematically? Alternatively, does someone have an intuition they would be willing to share for solving this problem?
| $$f(x+1)=(x+1)^3\ln(\sqrt{x+1})=\frac{(x+1)^3}{2}\ln(x+1)$$
Now write
$$\ln(x+1) = \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} $$
And multiply the infinite sum by the polynomial.
Then you can switch back to $f(x)$. Your approach is good in some cases, but often it's just guessing.
Easiest way to derive the Taylor series for $\ln(1+x)$ is, i think, as follows.
We know that $$\int\limits_0^x\frac{1}{1-t}dt = -\ln(1-x) $$
But we also know the Taylor series for $\frac{1}{1-t}$.
$$ \frac{1}{1-t}=\sum\limits_{n=0}^\infty t^n $$
We can integrate the sum, and we can do it term by term, and we get:
$$ -\ln(1-x)=\sum\limits_{n=1}^\infty \frac{x^n}{n} $$
So that
$$ \ln(1+x)=\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} $$
You can also try your guessing method to derive the Taylor series for $\ln(x+1)$, that will be more straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Intersection of surfaces is a curve if intersection of tangent planes is line I think I need a hint in solving the following exercise:
Let $S_1$, $S_2 \subseteq \mathbb R^3$ be surfaces (i.e. 2-dimensional submanifolds), with nonempty intersection $\Gamma := S_1 \cap S_2.$ Show that, if $V :=T_pS_1 \cap T_pS_2$ is a line for each $p \in \Gamma$, then $\Gamma$ is a curve.
Now this is intuitively clear, but the problem is how to prove it formally.
Thoughts:
Fix $p\in S_1$. Since $S_1$ is a surface we can find an open neighbourhood of $p$, $U\subseteq \mathbb R^3 $ and a differentiable map $f: U \to \mathbb R$ with $S_1\cap U = f^{-1}(\{0\})$ and $\nabla f(q) \neq 0$ for each $q\in S_1 \cap U$.
Now $T_pS_1$ = $\ker (D_p f)$ and therefore $D_p f(V) = \{0\}$. Is this going in the right direction? Another idea was to modify $f$ to get a map $\tilde f:W \to \mathbb R$ for some open subset $W \subseteq \mathbb R^2$ which then describes the curve, i.e. $\Gamma = f^{-1}(\{0\})$. Any hints appreciated!
| Let $p \in S_1\cap S_2$.
Show that $T_pS_1 \cap T_pS_2$ $($and here I'm guessing the tangent planes are thought of as embedded in $\mathbb R^3)$ is a line if and only if $S_1$ and $S_2$ cross (rather than touch) at $p$. Can you conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Limit of a function with two variables existance I have the function $f(x,y)=\frac{x^3}{y^2}e^{\frac{-x^2}{y}}$, and I have to show whether the following limit exists: $\lim\limits_{(x,y)\to(0,0)}f(x,y)$. If I set $y=x$, then $\lim\limits_{(x,x)\to(0,0)}f(x,x)=\lim\limits_{x \to 0}f(x,x)=0$; and if I set $y=x^2$ the limit does not exist. Is this enough to say that the $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ does not exist?
| Yes it is correct since
*
*$x=0 \implies \frac{x^3}{y^2}e^{\frac{-x^2}{y}}=0$
*$y=x^2 \quad x\to 0^+ \implies \frac{x^3}{y^2}e^{\frac{-x^2}{y}}=\frac{1}{xe}\to +\infty$
the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the series representation of $\int_{0}^{t} \exp(-x^n)\, dx$ where $n$ is a positive integer? I have searched in the web to get series representation of $$\int_{0}^{t} e^{-x^n}\,dx$$ where $n$ is a positive integer, but I haven’t succeeded; really I want its representation as hypergeometric series.
| Integrating the Power Series
We have the power series
$$
e^{-x^n}=\sum_{k=0}^\infty\frac{(-1)^kx^{kn}}{k!}\tag1
$$
which leads to the alternating power series
$$
\begin{align}
\int_0^t e^{-x^n}\,\mathrm{d}x
&=\sum_{k=0}^\infty\frac{(-1)^kt^{kn+1}}{(kn+1)k!}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,{}_1F_1\!\left(\tfrac1n;1+\tfrac1n;-t^n\right)}\tag2
\end{align}
$$
Mathematica verification:
Simplify[D[t Hypergeometric1F1[1/n,1+1/n,-t^n],t]]
Unilateral Power Series
The problem with $(2)$ is that for large $t$, the terms get huge and oscillate. This means that a lot of significance can get lost in cancellation. It is better to have a unilateral series where nothing is lost to cancellation.
Setting
$$
f(t)=e^{t^n}\int_0^te^{-x^n}\,\mathrm{d}x\tag3
$$
we have
$$
f'(t)=nt^{n-1}f(t)+1\tag4
$$
which, since $f(0)=0$, says not only that
$$
f(t)=t+O\!\left(t^{n+1}\right)\tag5
$$
but also gives the coefficient recurrence
$$
a_{k+n}=\frac{n}{k+n}\,a_k\tag6
$$
From $(5)$ and $(6)$, we get
$$
a_{kn+1}=\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\tag7
$$
and $a_k=0$ for $k\not\equiv1\pmod n$. This gives the unilateral series:
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=e^{-t^n}\sum_{k=0}^\infty\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\,t^{kn+1}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,e^{-t^n}{}_1F_1\!\left(1;1+\tfrac1n;t^n\right)}\tag8
\end{align}
$$
Note that the terms in the sum in $(8)$ peak when $k\sim t^n$.
Mathematica verification:
Simplify[D[t Exp[-t^n]Hypergeometric1F1[1,1+1/n,t^n],t]]
Asymptotic Expansion
$$
\begin{align}
\int_t^\infty e^{-x^n}\,\mathrm{d}x
&=\frac1n\int_{t^n}^\infty e^{-x}x^{\frac1n-1}\,\mathrm{d}x\tag9\\
&=\frac{e^{-t^n}}n\int_0^\infty e^{-x}\left(t^n+x\right)^{\frac1n-1}\mathrm{d}x\tag{10}\\
&=\frac{e^{-t^n}t^{1-n}}n\int_0^\infty e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x\tag{11}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\int_0^{t^n/2}e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x+O\!\left(e^{-t^n/2}\right)\right]\tag{12}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\int_0^{t^n/2} e^{-x}\frac{\Gamma\left(k+1-\frac1n\right)}{k!\,\Gamma\left(1-\frac1n\right)}\frac{x^k}{t^{nk}}\mathrm{d}x+O\!\left(\frac1{t^{nm}}\right)\tag{13}\right]\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\tag{14}
\end{align}
$$
Explanation:
$\phantom{0}(9)$: substitute $x\mapsto x^{1/n}$
$(10)$: substitute $x\mapsto t^n+x$
$(11)$: substitute $x\mapsto x/t^n$
$(12)$: $\int_{t^n/2}^\infty e^{-x}\,\mathrm{d}x=e^{-t^n/2}$
$(13)$: Binomial Theorem
$(14)$: $\int_{t^n/2}^\infty e^{-x}x^k\,\mathrm{d}t=O\!\left(e^{-t^n/3}\right)$
Therefore,
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\\
&=\bbox[5px,border:2px solid #C0A000]{\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\,{}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)}\tag{15}
\end{align}
$$
where ${}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)$ is not convergent, but is to be taken in the asymptotic sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The $p^{th}, q^{th},r^{th}$ term of an A.P are in G.P . Find the ratio of the $p^{th}$ and the $r^{th}$ term. I took the $p^{th}, q^{th},r^{th}$ term as $t_p,t_q,t_r$ respectively. Then $t_r = t_p * c^2$ (If c is the common ratio). and the common difference of the A.P is $\frac{t_r -t_p}{r-p}$. But I think I need to relate the common difference and the common ratio.
| Hint:
If $a,d(\ne0)$ be the first term & the common difference of the AP
$$\{a+(p-1)d\}\{a+(r-1)d\}=\{a+(q-1)d\}^2$$
$a^2+ad(p+r-2)+(r-1)(p-1)d^2=a^2+2ad(q-1)+(q-1)^2d^2$
As $d\ne0$
$$a(p+r-2)+(r-1)(p-1)d=2a(q-1)+(q-1)^2d$$
Express $a$ in terms of $d$
Now we need $$\dfrac{a+(p-1)d}{a+(r-1)d}$$
Replace the value of $a$ in terms of $d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802713",
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"source": "stackexchange",
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Show that a compact operator is bounded
Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence
has a cauchy subsequence $\iff $ totally bounded).
Prove: Let $T: V \to W$ be a compact linear operator. Show that $T$ is
bounded.
My attempt:
Suppose $T$ is not bounded. Then,
$$\forall M > 0: \exists v_M \in V: \Vert T v_M \Vert > M \Vert v_M \Vert$$
I then tried to construct a sequence without cauchy sequence in $\{Tv \mid \Vert v \Vert \leq 1\}$
So, let $p > q$. Then, $$\left\Vert T \frac{v_p}{\Vert v_p \Vert} - T \frac{v_q}{\Vert v_q \Vert}\right\Vert \geq \left|\frac{\Vert Tv_p \Vert}{\Vert v_p \Vert} - \frac{\Vert Tv_q \Vert}{\Vert v_q \Vert}\right|$$
but was unable to conlude something because off the minus sign.
Any ideas?
EDIT: This is not a duplicate, as other posts use other definitions of compact operators.
| It follows directly from the definitions. Since $T$ is compact, the image $T(V_1)\subset W$ of the closed unit ball of $V$ is precompact. Consider the cover $W\subset \bigcup_n W_n$, where $W_n$ is the ball of radius $n$. As $\overline{T(V_1)}$ is compact, the cover has a finite subcover, which means that there exists $m$ with $\overline{T(V_1)}\subset W_m$. In particular, $\|Tv\|\leq m$ for all $v\in V_1$, which leads to $$ \|Tv\|\leq m\|v\|,\ \ \ v\in V.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difference between $x\in[0,1]$ and $x\in \{0,1\}$? If I write
$$
x\in [0,1] \tag 1
$$
does it mean $x$ could be ANY number between $0$ and $1$?
Is it correct to call $[0,1]$ a set? Or should I instead write $\{[0,1]\}$?
Q2:
If I instead have
$$
x\in \{0,1\} \tag 2
$$
does it mean $x$ could be only $0$ OR $1$?
| $[0,1]$ is (defined as) the set $\{ x \in \mathbb{R} : 0 \leq x \leq 1 \}$, i.e. it is a set that contains every real number between $0$ and $1$ (inclusive). It contains an uncountable number of elements.
$\{0,1\}$ is a set containing 2 elements: $0$, and $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Can two different models of arithmetic have non-comparable views of peano arithmetic? For a given model of arithmetic $M$, we say that models view of peano arithmetic, $V(M)$, is $\{\phi : M \models (PA \vdash \phi) \}$.
For example the view of the standard model is $\{\phi : PA \vdash \phi \}$. On the other hand, for any model $X$ of $PA + \lnot Con(PA)$, $V(X)$ is the set of all statements in arithmetic. For a model $Y$, of $PA + Con(PA) + \lnot Con(ZFC)$, $(ZFC \vdash 0=1) \in V(Y)$ but $0=1 \notin V(Y)$. So $V(\text{standard model}) \subset V(Y) \subset V(X)$.
Can we have two model, $M$ and $M'$, of arithmetic (which are models of peano arithmetic) such that $V(M) \nsubseteq V(M')$ and $V(M') \nsubseteq V(M)$.
| First of all, I believe you would be interested in a paper by Kikuchi and Kurahashi, "Illusory Models of Peano Arithmetic". They explore related questions in detail in that paper. In their notation, what you call "V(M)" they call $\mathrm{Thm}_{\mathsf{PA}}(M)$.
Let's first note that if $M \models \lnot \mathrm{Con}(\mathsf{PA})$, then $\mathrm{Thm}_{\mathsf{PA}}(M)$ will contain all sentences in the language. They refer to such models as "insane", and models of $\mathrm{Con}(\mathsf{PA})$ as "sane". So the real question here is whether, given sane models $M$ and $N$, the sets $\mathrm{Thm}_{\mathsf{PA}}(M)$ and $\mathrm{Thm}_{\mathsf{PA}}(N)$ are necessarily linearly ordered. The answer is no; in fact the main result regarding this in their paper actually shows that the family $\mathcal{T} = \{ \mathrm{Thm}_{\mathsf{PA}}(M) : M$ is sane $ \}$ has cardinality $2^{\aleph_0}$.
The main crux behind getting these independence results is that $\mathrm{Thm}_{\mathsf{PA}}(M)$ is determined entirely by the $\Sigma_1$ theory of $M$. Further, given any recursively enumerable theory $T$, there is a $\Sigma_1$ statement $\phi$ such that $T + \phi$ and $T + \lnot \phi$ are both consistent (this is a usual incompleteness argument, though they use a stronger version in order to get that $|\mathcal{T}| = 2^{\aleph_0}$). so let $T$ be the theory $\mathsf{PA} + \mathrm{Con}(\mathsf{PA})$, and then you can find two models of PA which have different $\Sigma_1$ theories, and therefore have incompatible "theorems of PA".
I strongly recommend reading that paper as they ask and answer many related questions that you might be interested in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The nth prime calculator How to prove the correctness of the following algorithm :
Input: $n$ : the ordinal number
Output: $x$ : the nth prime number
$b_1=6$ , $b_2=6$ , $b_3=6$
If $n==1$ , then $x=2$ , else $x=3$
$k=4$ , $m=3$
While $m \leq n$ do:
$\phantom{5}$ $b_4=b_1+ \operatorname{lcm}(k-2,b_1)$
$\phantom{5}$ $a=b_4/b_1-1$
$\phantom{5}$ $k=k+1$
$\phantom{5}$ $b_1=b_2$ , $b_2=b_3$ , $b_3=b_4$
$\phantom{5}$ If $x<a$ then $x=a$ , $m=m+1$
Return $x$
You can run this code here .
GUI application that implements this algorithm can be found here .
Fast command line program that implements this algorithm can be found here .
I can confirm that algorithm produces correct results for all $n$ up to $10000$ .
| Updated 11.06.18
The proposed calculator enumerates the prime numbers on the pass.
Let us construct the similar algorithm based on the ideas of Eratosthenes sieve.
m = 1;
x = 2;
b = 2;
while (m < n)
{
for(k = 3; ; k = k+2)
{
if(gcd(b,k)==1)
{
x = k;
m = m + 1;
b = b * x;
}
}
}
return x;
In the both of the algorithms, a sequence of positive integers is tested for the presence of a common factor with a high composite number $b.$ If it equals to 1, then the integer is prime.
The algorithm above shows that it's sufficiently to use $b$ which equals the primorial $x\#$.
The OP calculator algorithm forms the greater value of $b,$ which contains any prime $p$ in the degree $d,$ wherein
$$p^d\le x < p^{d+1},$$
but uses its delayed value.
This means that the OP calculator can be improved.
At this time, analysis of delaying influence shows that OP calculator is correct.
| {
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"url": "https://math.stackexchange.com/questions/2803156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.
| I have obtained a very interesting form of fraction which will follow all the conditions. Unfortunately I am not seasoned in number theory and Hence my solution is lengthy, confusing and has a large no. Of variable, and has a few assumptions. It would be helpful if someone could suggest a better and shorter solution. The form is (210k+2)/(420k+2) .
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with central limit theorem Let ($\xi_{k}$)$_{k \in \mathbb{N}}$ be a sequence of i.i.d random variables with expectation $\mu$ and variance $\sigma^2$ $\in (0, \infty)$.
We define $X_k$ = $\xi_k$ - $3\xi_{k+1}$ + $\xi_{k+2}$, $S_n$ = $\sum_{k=1}^n X_k$, $k,n \in \mathbb{N}$
Compute for $x \in \mathbb{R}$ the limit
$\lim_{n \to \infty}$ $\mathbb{P}$$(\frac{S_n - n\mathbb{E}[X_1]}{\sqrt{n\mathbb{Var}[X_1]}}$ $\le$ $x$).
My thoughts on this:
First note that $\mathbb{E}[X_1]$ = -$\mu$,$\mathbb{Var}$[$X_1$] = $11\sigma^2$
Then with the central limit theorem, it holds
$\frac{S_{n}-n\mu}{\sigma\sqrt{n}}$ $\overset d \longrightarrow$ ${N}(0,1)$ $\Rightarrow$ $S_{n} - n\mu$ $\overset d \longrightarrow$ $N(0,n\sigma^2)$
$\Rightarrow$ $S_n$ $\overset d \longrightarrow$ $N(n\mu,n\sigma^2)$
It then follows that $S_n +n\mu$ $\overset d \longrightarrow$ $N(2n\mu,n\sigma^2)$ $\Rightarrow$ $\frac{S_n +n\mu}{n}$ $\overset d \longrightarrow$ $N(2\mu,\frac{\sigma^2}{n})$
I would now like to reach $\frac{S_n+n\mu}{\sqrt{11n\sigma^2}}$ $\overset d \longrightarrow$ $N(2\mu,\frac{1}{11})$, as I feel this is the correct result, but I am unable to get to the correct steps.
I am grateful for any tip and suggestion
| Your intuition, in this case, is off the target. But that's a very interesting mistake you made (in a good sense!).
The problem is that the version of the CLT you want to use is valid for sequences of independent random variables. And the $X_k$ are not independent. For instance, if $\xi_3$ is large and positive, then $X_1$ and $X_3$ will be large and positive, but $X_2$ will be very large and negative. Then, when you sum them, a compensation occurs.
On average, the $X_k$ have expectation $-\mu$, so $S_n$ grow like $-\mu n$. However, these compensations make the variance of the sum smaller; that is, $Var(S_n) < 11 \sigma^2 n$.
If you want to get the correct result, try to express $\sum_{k=1}^n X_k$ using $\sum_{k=1}^n \xi_k$. There are some boundary effects in the sum, but the asymptotic behaviour should be clearly visible.
In addition, be more cautious in the way you divide by $n$ or $\sqrt{n}$. The CLT tell you that $(S_n-n\mu)/(\sigma \sqrt{n}) \to N(0,1)$, but the expression $S_n \to N(n \mu, n \sigma^2)$ has no meaning. It's as if I were saying that the sequence $(n+\sqrt{n})$ converge to $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}$ I did experiments with a Pari/GP program that suggest that the numerical series
$$\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}\tag{1}$$
is convergent, where for an integer $n>1$ $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p \text{ prime}}}p$$
is the product of distinct prime factors dividing $n$, with the definition $\operatorname{rad}(1)=1$. You can see the Wikpedia Radical of an integer to see the properties of such arithmetic function (
it is the famous function of the abc conjecture). We know the size/asymptotic of the least common multiple of the first $n$ integers $\operatorname{lcm}(1,2,\ldots,n)$ and its relationship to the so-called second Chebyshev function, see for example this MathWorld's article Chebyshev Functions.
Question. Provide help, details or hints to prove that our series $$\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}$$
is convergent. Many thanks.
Remarks (About that seems that previous Question isn't obvious using comparisons).
1) Notice that the suare-free kernel $\operatorname{rad}(n)$ is an arithmetic function satisfying that $\operatorname{rad}(n)\leq n$, but the ratio test implies that $\sum_{k=A}^\infty k!e^{-k}$ is divergent, for each fixed positive integer $A$.
2) We know also Legendre's formula to calculate to evaluate the $p$-adic valuation of $n$, I mean the symbol $\nu_p(n!)$.
Computational fact. Using a Pari/GP program our series $(1)$ is about $\approx 6.26851$.
| $\sum_{k=1}^\infty\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}
$
From its definition,
$rad(k!)
=\prod_{p \le k} p
$
so
$\ln rad(k!)
=\sum_{p \le k} \ln p
=\theta(k)
$
and
$\ln(\operatorname{lcm}(1,2,\ldots,k))
=\psi(k)
$,
where
$\theta$ and $\psi$
are the two
Chebychev
prime counting functions
(https://en.wikipedia.org/wiki/Chebyshev_function).
Since
$\psi(x)
=\sum_{n=1}^{\log_2x}\theta(x^{1/n})
$
so
$\psi(x)-\theta(x)
=\sum_{n=2}^{\log_2x}\theta(x^{1/n})
\gt\theta(x^{1/2})
\sim x^{1/2}
$.
Therefore
$\frac{\operatorname{rad}(k!)}{\operatorname{lcm}(1,2,\ldots,k)}
=\exp(\theta(k)-\psi(k))
\lt \exp(-k^{1/2})
$
and this sum converges
since,
for any $a>0$,
$k^{1/2}
\gt a\ln(k)
$
for large enough $k$.
Choosing $a=2$
gives
$\exp(-k^{1/2})
\lt 1/k^2
$.
Therefore
the sum converges.
| {
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An interesting series converging to a constant Let $K>0$ be a constant. Suppose $\{z_n\}_{n=1}^\infty$ is a non-decreasing positive sequence. Then the series
$$\sum_{n=1}^\infty\frac{z_n}{(K+z_1)(K+z_2)\cdots(K+z_n)}K^n=K$$
This is a quite interesting result as the series is convergent and the limit doesn't depend on the choice of $\{z_n\}_{n=1}^\infty$, as long as it is non-decreasing and positive
I have run computer simulations and this result seems to hold. However, I am not sure how to prove it.
| The series can be transformed into a telescoping series in the following way:
$$
\sum_{n=1}^{\infty}\frac{z_n}{\prod_{j=1}^n(K+z_j)}K^n=\sum_{n=1}^{\infty}\left(\frac{K^n}{\prod_{j=1}^{n-1}(K+z_j)}-\frac{K^{n+1}}{\prod_{j=1}^n(K+z_j)}\right)=K
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Circumference inscribed in a square What is the area of the hatched region, knowing that the arc AC is 1/4 of the circumference with a center in D?
I've tried using algebra to solve this but it seemed insufficient, I thought of using integrals to find the area, by finding the analytic geometry equations for the circles but it went nowhere.
|
Sector 1 Angle $= 2\cdot \sin^{-1}(\sqrt(\frac{31}{128})\cdot 2)= 2.78617$ rad
Sector 2 Angle $= 2\cdot \sin^{-1}(\sqrt(\frac{31}{128})) = 1.02906$ rad
Shaded Area = Area Chord Segment 1 – Area Chord Segment 2
Area Seg 1 = Area Sector 1 - Area Triangle 1
$\frac{2.78617}{2π}\cdot π(\frac{a}{2})^2 - (\sqrt(\frac{31}{128})a)( \sqrt(\frac{97}{128})a - \sqrt(2)\frac{a}{2})$
$= .26785 a^2$
A Seg 2 = Area Sector 2 - Area Triangle 2
$\frac{1.02906}{2π}\cdot πa^2 – (\sqrt(\frac{31}{128})a(\sqrt(\frac{97}{128})a)$
$= .08612 a^2$
Shaded Area $= (.26785 - .08612) a^2 = .18173 a^2$
| {
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Why is $\sqrt{x^2}=|x|$ instead of $\pm x$? $\sqrt{a}=\pm a$ for any $a$. $x^2$ always removes the negative, meaning that it will result in a positive number for $a$, but that doesn’t change the ambiguity of the square root operation. Thus I would think that $\sqrt{x^2}=\pm x$, where you select which output to use based on the problem in question, or leave it ambiguous if there is not enough information.
Yet in usage, it seems like the positive root is assumed if $a$ can be written as the square of something (ie $y=\sqrt{x^2+4x+4}=\sqrt{(x+2)^2}$ is assumed to have range $y\ge 0$). This would seem to contradict the definition, since ANY $a$, be it an equation or a number, can be considered the square of something and thus any use of the square root could be considered to be acting on a square.
What is going on here?
| If $x\in\mathbb{R};x\ge 0$, then $\sqrt{x}$ is defined as "the non-negative real number which, when squared, equals to $x$.
Saying this is correct:
$\sqrt{x^2}=|x|=\begin{cases}x, \text{if }x\ge 0\\-x, \text{if }x\le 0\end{cases}$
I will give you another example first before considering your question:
$x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$
For this case, $x=1$ and $x=-1$ are both true for the equation.
However:
$\sqrt{x^2}=|x|=\pm x$
$\sqrt{x^2}=|x|$ is true, however $|x|=\pm x$ is not true, because $|x|=x$ is only true if $x\ge 0$ and $|x|=-x$ is only true if $x\le 0$, they cannot both be correct at the same time (except for $x=0$, but saying $|0|=\pm{0}=0$ is quite pointless).
| {
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Which integers can be written as $x^2+2y^2-3z^2\ $?
For which integers $n$ has the diophantine equation $$x^2+2y^2-3z^2=n$$ solutions ?
These theorems
https://en.wikipedia.org/wiki/15_and_290_theorems
do not apply because the given quadratic form is not positive (or negative) definite. It seems that the quadratic form is universal (for every integer $n$ a solution exists) , but I have no idea how this can be proven.
| We claim that any $n\in\mathbb{Z}$ can be written as $x^2+2y^2-3z^2$.
Note that any such integer $n$ is $0$ or it is equal to a $4^a\cdot 2^b\cdot d$ where $a$ is a non negative integer, $b\in\{0,1\}$ and $d$ is a signed odd number. Then we consider the following cases.
0) If $n=0$ then let $x=0$, $y=0$ and $z=0$.
1) If $n=d=2k+1$ then let $x=k+1$, $y=k$ and $z=k$:
$$x^2+2y^2-3z^2=(k+1)^2-k^2=2k+1=n.$$
2) If $n=2d=2(2k+1)$ then let $x=k$, $y=k+1$ and $z=k$:
$$x^2+2y^2-3z^2=2(k+1)^2-2k^2=4k+2=n.$$
3) If $n=4^ak$ and $k$ can be written as $x_k^2+2y_k^2-3z_k^2$ then
let $x=2^ax_k$, $y=2^ay_k$ and $z=2^az_k$:
$$x^2+2y^2-3z^2=4^a(x_k^2+2y_k^2-3z_k^2)=4^ak=n.$$
| {
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Linear Transformation $T\colon V\rightarrow V$ defined by $T(y)=y+e^{-2t}\frac{dy}{dt}$
Consider the set of functions $B=\left\{ {1,e^{2t},e^{4t}}\right\}$. It is given that B is a linearly independent set. Write $V=\text{span}(B)$ and let $T\colon V\rightarrow V$ be the linear transfomration defined by $$T(y)=y+e^{-2t}\frac{dy}{dt}$$
Using the matrix M of T with respect to the ordered basis $B$, decide whether the equation $$y+e^{-2t}\frac{dy}{dt}=1+2e^{2t}+3e^{4t}$$
has no solution, one solution or more than one solution $y$ in $V$. Find the solutions (if any).
I managed to determine the matrix $M=\left( {\begin{array}{cc}
1 & 2 & 0 \\
0 & 1 & 4 \\
0 & 0 & 1 \\
\end{array} } \right)$ as
\begin{align*}
T(1)&=1\\
T(e^{2t})&=2+e^{2t} \\
T(e^{4t})&=4e^{2t}+e^{4t} \\
\end{align*}
But how do you determine where the equation in the question has a solution or not? Intuition tells me it has 3 solutions, but I have no valid reasoning to approach this hypothesis.
| Since teh detreminant of $M$ is 1 there can be only one solution. Equate $T(a+be^{2t}+ce^{4t})$ to $1+2e^{2}+3e^{4t}$ and you will get $a=13, b=-10,c=3$.
| {
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If $f(x) \le M$ and $\lim_{x \to a} f(x) = A$, prove that $A \le M$. I am an adult software developer who is trying to do a math reboot. I am working through the exercises in the following book.
Ayres, Frank , Jr. and Elliott Mendelson. 2013. Schaum's Outlines Calculus Sixth Edition (1,105 fully solved problems, 30 problem-solving videos online). New York: McGraw Hill. ISBN 978-0-07-179553-1.
Writing proofs is not yet intuitive for me and I would like to know if my proof for the following problem is complete or if I skipped steps. I have reviewed the following document about writing proofs.
Cheng, Eugenia. 2004. "How to write proofs: a quick guide." University of Chicago. Last modified October. http://cheng.staff.shef.ac.uk/proofguide/proofguide.pdf.
Chapter 7 Limits, problem 26.
Prove: If $f(x) \le M$ for all $x$ in an open interval containing $a$ and if $\lim_{x \to a} f(x) = A$, then $A \le M$.
(Hint: Assume $A>M$. Choose $\epsilon = \frac{1}{2}(A-M)$ and derive a contradiction.)
My understanding.
The key to this proof is whether or not $A>M$ or $A \le M$. I do not see how the factor of $\frac{1}{2}$ in the hint helps.
My proof.
Let $f(x) \le M$ for $x \in (a-Q, a+Q)$ where $0 < Q$.
Let $\lim_{x \to a} f(x) = A$. For any $0 < \epsilon$ there exists $0<\delta$ such that $|x-a| < \delta$.
Suppose $\epsilon = A - M$, which implies $M < A$ because $0<\delta$.
$$
|f(x) - A| < \epsilon \\
-\epsilon < f(x) - A < \epsilon \\
-(A - M) < f(x) - A < A - M \\
M - A < f(x) - A < A - M \\
M < f(x) < 2A - M \ \ {\Large \#} \\
$$
By definition $f(x) \le M$, so $\epsilon$ can not possibly have a value that implies $M < A$. Therefore, $A \le M$. $\square$
| When presenting this proof we usually use the factor $\frac{1}{2}$ to make it very clear that $f(x) \geq M + \varepsilon / 2 > M$, and not have to bother with strict inequalities. For instance sometimes in defining the continuity of $f$ at point $a$ you could end your sentence by using a large inequality, those two definitons would be equivalent. It's actually a good exercice to proove it. But then if you use a large inequality to define the continuity at point $a$ you have to use $\varepsilon = (A - M)/2$ in this proof, otherwise your last inequality would be $ M \leq f(x)$ which is not in contradiction with $f(x) \leq M$. I don't know if I made my point clear, but that's reason behind this hint in your book. ;)
Another remark, you should not say $0<Q$ but rather $Q>0$ (easier to read). On this point it might be just a personnal preference though...
| {
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Clear Definition of Two Graphs Being Isomorphic I am wanting to deal with canonical graph homorphisms and other graph homorphisms much more deeply, but I need a definition for isomorphism between graphs to start. I know the definition of two groups being isomorphic, but I do not know how to tell if two graphs are isomorphic to each other. I watched a video which explains how to tell if two graphs are isomorphic. Now, the video states the following definition of isomorphic graphs: $G_1\cong G_2$ iff $\exists f : V(G_1)\xrightarrow[]{\text{bijection}} V(G_2)$ ST $f(u)f(v)\in G_2 \leftrightarrow uv\in G_1$. Personally, I find this definition hard to work with for two reasons:
*
*Reason 1: How can I draw a table similar to that of groups in abstract math?
*Reason 2: How is there even an operation defined in this way? It seems as though the operation is not closed (it also seems to be more of a directed graph than an undirected graph in the way that it is not a set), and it does not resemble that of the abstract algebra definition which would be in this case by the functions codomain the following: $G_1\cong G_2$ iff $\exists f : V(G_1)\xrightarrow[]{\text{bijection}} V(G_2)$ ST $\forall u, v \in V(G_1)$, [$f(u\cdot v)=f(u)\diamond f(v)$].
*
*QUESTION:
Is there a good definition of isomorphism between graphs that I should be using by you graph theory/abstract algebra/other experts? Any help would be greatly appreciated!
Above are two graphs which I represented as a table like that of an adjacency matrix if this helps answer the question by giving an example to talk about. So, $G_1:=(V(G_1), E(G_1))$ (shown on the left) and $G_2:=(V(G_2), E(G_2))$ (shown on the right).
So, in this context, $f=\begin{pmatrix}
v_1 & v_2 & v_3 \\
v_4 & v_5 & v_6
\end{pmatrix}$.
\begin{array}{c||c||c||c}
Adjacency(G_1) & v_1 & v_2 & v_3 \\
\hline
v_1 & 0 & 2 & 0 \\
\hline
v_2 & 2 & 0 & 1 \\
\hline
v_3 & 0 & 1 & 0
\end{array}
\begin{array}{c||c||c||c}
Adjacency(G_2) & v_4 & v_5 & v_6 \\
\hline
v_4 & 0 & 2 & 0 \\
\hline
v_5 & 2 & 0 & 1 \\
\hline
v_6 & 0 & 1 & 0
\end{array}
| Consider the following definition of group isomorphism: it is a bijection $f:G_1\to G_2$ such that
$$uvw^{-1}=e_{G_1} \iff f(u)f(v)f(w)^{-1}=e_{G_2}\quad \forall u,v,w\in G_1.$$
And now compare graphs:
$$uv\in E(G_1) \iff f(u)f(v)\in E(G_2)\quad \forall u,v\in V(G_1).$$
Not much difference really.
| {
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What is the best way to guess a number in a limited number of guesses? A random integer is picked from 0 to 100. You can make 5 guesses at what the number is, and after each guess, you are told if your guess was too high or too low. What strategy maximizes your probability of guessing the number?
| You can work your way back from the end of the game to optimize the strategy. At each stage, you know that the number is in some interval. On the last guess, you'll just guess any number in the interval, and your chance to win is one over the length of the interval. On the penultimate guess, you have some interval $[i,j[$ (closed at the beginning and open at the end to simplify the length calculations), and you can choose some integer $k\in[i,j[$ to maximize the remaining winning probability. You have probability $\frac1{j-i}$ to win right away by guessing right. You have probability $\frac{k-i}{j-i}$ to guess too high, and then you have probability $\frac1{k-i}$ to guess right on the last guess; and you have probability $\frac{j-k-1}{j-i}$ too guess too low, and then you have probability $\frac1{j-k-1}$ to guess right on the last guess; so the total winning probability as function of $k$ is
$$
\frac1{j-i}+\frac{k-i}{j-i}\cdot\frac1{k-i}+\frac{j-k-1}{j-i}\cdot\frac1{j-k-1}=\frac3{j-i}\;,
$$
which doesn't in fact depend on $k$. Each of the three possibilities contributes $\frac1{j-i}$ to the winning probability, independent of $k$.
But that means we don't have to do much more work for the previous stages – they work essentially like the penultimate one, except the cases of guessing too high or too low contribute larger winning probabilities, whereas the case of guessing right still just contributes the reciprocal of the interval length. So on the third guess, your winning probability is $3+1+3=7$ over the interval length, on the second guess it's $7+1+7=15$ over the interval length, and on the first guess it's $15+1+15$ over the interval length, that is, $\frac{31}{101}$ – irrespective of which numbers you choose to guess.
Well, not completely irrespective, that can't be, since you obviously only have a $\frac5{101}$ winning probability if you guess $0$, $1$, $2$, $3$, $4$. The argument is only valid as long as you always leave enough numbers on either side to have all three possibilities – if you don't, you lose the contribution from that possibility. That means that you have to leave at least $2^{5-n}-1$ numbers on either side on the $n$-th guess, so for instance your first guess must be between $15$ and $85$.
| {
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The area between the graphs $f_t(x)=e^{x-t}-t$ and $g_t(x)=\log(x+t)+t$ is irrational only if $t$ is irrational.
Find the smallest $t \in \mathbb{R}$ such that the graphs of $f_t:x\mapsto \exp(x-t)-t$ and $g_t:x\mapsto\log(x+t)+t$ intersect. Note that $\log$ is meant as the natural logarithm.
Hint: I guess it's $\frac{1}{2}$.
Now the real problem (it's hard):
Prove or disprove: whenever the area $A$ between $f_t$ and $g_t$ exists and $A \in \mathbb{R} \setminus \mathbb{Q}$ then $t\in \mathbb{R} \setminus \mathbb{Q}$.
| Assume it to be true, then
If $t \in \mathbb{Q} \Rightarrow $ Area between $f$ and $g$ $\in \mathbb{Q}$
Then specially for $t = 0$
$$
\int_{0}^{\infty} e^{x}- \log(x) dx = \left[x+ e^x + x (-\log (x)) \right]_{0}^{\infty} = \\
\lim_{x \to \infty} x + e^x + x (-\log (x)) - \lim_{x \ \to 0 } x+ e^x + x (-\log (x)) = \infty
$$
| {
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Are there intractable problems? Obviously, there is no problem in NP which can be shown to be intractable (by which I mean: not in P).
Is there a problem (outside of NP) which can be shown to be intractable (not lying in P)?
| What you want is the Time Hierarchy Theorem, which broadly speaking states that for any "reasonable" function $f$ there are problems that take about $f(n)$ time to decide.
Concretely, each of the known EXPTIME-complete problems would be an example of something that has been proved not to lie in P.
Here is a related question on compsci.
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The inverse of a continuous one-to-one function that is defined on a connected set is not always continuous I am trying to find a function $f:B \subset \Bbb R^n \rightarrow \Bbb R^m$ for $B$ a connected set that is continuous, one-to-one where $f^{-1} = f(B) \rightarrow B$ is discontinuous. The hint I have been given in my textbook is to choose $m>1$. I know that the image will be a connected set. The only idea I had is to send an angle to the unit circle $\theta \rightarrow (\cos(\theta),\sin(\theta)), \theta \in [0, 2\pi)$ and then take the inverse to be $f^{-1}:(x,y)\rightarrow (\arctan(y/x))$ but I feel that this function is continuous and not appropriate. Any hints appreciated.
I have seen a similar question here:
Inverse function that takes connected set to non-connected set
| I was working on this, managed to prove the opposite (If wrong, please point out the mistake)
I claim that $f^{-1}$ is continuous. Let's call it $g$ for convenience of notation, i.e. $g\circ f (x) = x$ $\forall x \in B$
Enough to show that $g^{-1}(U)$ is open in $f(B)$ if $U$ is open in $B$
Now, $g^{-1}(U) = \{ x : g(x) \in U , x \in f(B)\}$ and
$g^{-1}(U)$ is open $\iff$ $f(U)$ is open, as both are the same
Let $U' = f(U)$. Since $f$ is continuous, $U'$ is open $\iff$ $f^{-1}(U')$ is open
$f^{-1}(U') = \{ y : f(y) \in U' , y \in B\} = \{ y : f(y) \in f(U) , y \in B\}$. Now, as $f$ is one-one, $f(y) \in f(U) \iff y \in U$.
So, $f^{-1}(U') = U$.
Thus, $U' = f(U) = g^{-1}(U)$ is open $\iff$ $U$ is open.
P.S : Apologies if this proof is wrong.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Other methods for solving homogeneous differential equation I want to find a general solution of the following homogeneous equation
$$\frac{dy}{dt}=\frac{3t+12y}{t+14y}$$
I have tried using the substitution $z = y/t$ to make the equation separable, but then it gets a little bit tedious to solve.
$$\frac{dy}{dt}=\frac{3+12z}{1+14z}=z+\frac{dz}{dt}t$$
$$\int \frac{1+14z}{-14z^2+11z+3} dz = \ln(t)+c_1$$
So I'm wondering if there's another way to solve it?
| My teacher in ODEs show me this "rarely unknown"substitution method, it Works with homogeneous and almost homogeneous ODEs.
1° Let $y=p^S$ and $t=q^R$, respectively $dy=Sp^{S-1}dS$, $dt=Rq^{R-1}dR$; R and S are constants. Then substitute in the ODE:
$$\frac{dy}{dx}=\frac{3t+12y}{t+14y}\rightarrow\frac{Sp^{S-1}dS}{Rq^{R-1}dR}=\frac{3q^R+12p^S}{q^R+14p^S}$$
$$\frac{S\ dS}{R\ dR}=\frac{q^{R-1}(3q^R+12p^S)}{p^{S-1}(q^R+14p^S)}$$
$$\frac{S\ dS}{R\ dR}=\frac{3q^{2R-1}+12p^Sq^{R-1}}{p^{S-1}q^R+14p^{2S-1}}$$
Match the all the upper exponents with all the lowers
$$2R-1+S+R-1+=S-1+R+2S-1\\3R+S=3S+R\\R=S$$
The only restriction is $(R,S)>0\quad$ I'll do it easier with $R=1$ we obtain:
$$\frac{dS}{dR}=\frac{3q+12p}{q+14p}$$ we got the same initial ODE, so we're right.
$\star\quad $ 2° Method we want $F(\lambda^\alpha t,\lambda^\beta y)=\lambda^{\beta-\alpha}F(t,y)\ $ to be homogeneous, and the substitution must be $y=\sigma t^{\frac{\alpha}{\beta}}$
$$\frac{3\lambda t+12\lambda y}{\lambda t+14\lambda y}\\\lambda^0\left(\frac{3t+12y}{t+14y}\right)$$ it works with bigger exponents but in this case $\alpha=\beta=1$ then $y=\sigma t$ the easiest substitution .
$\star\quad $ 3° Another method could be; from $dy/dt=3(t-4y)/(t+14y)\ $Let $u=t+4y\ $ then $du=dt+4dy$
$$\frac{\frac14(du-dt)}{dt}=\frac{3u}{t+\frac72(u-t)}\\\frac14\frac{du}{dt}-\frac14=\frac{6u}{7u-5t}\\\frac{du}{dt}=\frac{31u-5t}{7u-5t}$$ but its unnecessary cause its the same traditional method, in another form.Also you can do it letting $\omega=t+14y$
All these methods are long and again unnecessary, your integral is easier enough, the first method its for different exponents for a ODE like this: $y'=(x^2y+yx^3)/(y+y^2x)$
| {
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Geometric construction to divide a segment Given a segment AB, I would like to construct using only straightedge and compass, a point C on the segment AB such that $\frac{AC}{CB}$ is equal to $\frac{\phi}{2}$, where $\phi$ is the golden ratio, $\phi = 1.61803..$ .
The Wikipedia article on the golden ratio offers a construction for dividing a segment in the ratio $\phi$, but I cannot figure out how to do it to divide in $\phi/2.$ Any ideas?
| Render
$\frac{1+\phi /2}{\phi /2}=\sqrt{5}$
So the entire length divided by the shorter piece is $\sqrt{5}$.
Start with the given segment $AB$. Bisect this segment to identify the midpoint $M$ and construct a circle $Z$ centered at $M$, passing through $A$ and thus also through $B$. Construct the perpendicular to $AB$ through $A$, and mark off a point $C$ on this perpendicar, such that $AC$ is congruent to $AM$. Draw the hypoteneuse $BC$ which intersects circle $Z$ at point $X$ (distinct from $B$). Triangle $XBA$ is similar to triangle $ABC$ so $AB$ is $\sqrt{5}$ times as long as $AX$. Construct $Y$ on $AB$ with $AY$ congruent to $AX$ and render $Y$ as the required dividing point.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of improper integral $\to\pi$ $$\int_0^\pi\frac{\sin^2x}{\pi^2-x^2}\mathrm dx$$
$\sin$ is converging, but the rest has infinite limit in this segment, while $ \to\pi$, so Abel doesn't work. Dirichlet is not working either. I tried to bound this integral, but unsuccessfully. Not sure what else to use. Any throughts?
| Split the integral in half and apply $x\mapsto\pi-x$ to the $x>\pi/2$ integral. You'll see each integral ends up with an integrand whose numerator and denominator are respectively $O(x^2)$ and $O(x^{0\,\text{or}\,1})$ for small $x$, and the integrals converge.
| {
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Let $ A\in M_n(\mathbb{C}) $ be an invertible and non-diagonalizable matrix. Prove that for all $k\ge 1 \Rightarrow A^k$ is not diagonalizable.
Let $ A\in M_n(\mathbb{C}) $ be an invertible and non-diagonalizable matrix. Prove that for all $k\ge 1 \Rightarrow A^k$ is not diagonalizable.
Hi all. Since $A$ is over $\mathbb{C}$ then $A$ must have a Jordan normal form which is not a diagonal matrix. Therefor at least one of its basic Jordan blocks is of the form $\begin{pmatrix}
\lambda & 1 & 0 & \dots & 0 \\
0 & \lambda & 1 & \ddots & \vdots \\
0 & 0 & \ddots & \ddots & 0 \\
0 & 0 & \ddots & \ddots & 1 \\
0 & 0 & \dots & 0& \lambda \\
\end{pmatrix} = J_n(\lambda) $
($\lambda \neq 0$, for $A$ is invertible.)
Since $A$ is similar to some $J_A$ Jordan matrix then $P^{-1}AP=J_A \implies P^{-1}A^kP=(J_A)^k$ then $A^k$ is similar to $(J_A)^k$ which is a block-diagonal matrix with one of its blocks being $(J_n(\lambda))^k$.
I had already proven that
$$(J_n(\lambda))^k = \begin{pmatrix}
\lambda^k & \binom{k}{1}\lambda^{k-1} & \binom{k}{2}\lambda^{k-2} & \cdots & \cdots & \binom{k}{j-1}\lambda^{k-j+1} \\
& \lambda^k & \binom{k}{1}\lambda^{k-1} & \cdots & \cdots & \binom{k}{j-2}\lambda^{k-j+2} \\
& & \ddots & \ddots & \vdots & \vdots\\
& & & \ddots & \ddots & \vdots\\
& & & & \lambda^k & \binom{k}{1}\lambda^{k-1}\\
& & & & & \lambda^k
\end{pmatrix}$$
and we notice that $$(J_n(\lambda))^k=f(J_n(\lambda))=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$ where $f$ is the polynomial $f(t)=t^k$
I wanted to prove this statement using this, with the differentiation operator $D$ (for example, I know that $f\in \ker(D-\lambda I) \iff D^k(e^{-\lambda t}f)=0 $, and I thought of using that- but I am not sure how).
Please notice:
I know how to prove this using the minimal polynomial, but I would like to prove this using the Jordan-normal form since this is the direction our prof' wants us to go.
I would love to hear your thoughts, thank you :)
| We can assume that $A$ is in the Jordan form.
$A$ is not diagonalizable so $\exists\lambda \in \sigma(A)$ such that the $A$ has a block $J_j(\lambda)$ with $j \ge 2$, which is equivalent to $\dim\ker (A - \lambda I) < \dim\ker (A - \lambda I)^2$.
We have
$$J_j(\lambda) = \begin{pmatrix}
\lambda & 1 & 0 & \dots & 0 \\
0 & \lambda & 1 & \ddots & \vdots \\
0 & 0 & \ddots & \ddots & 0 \\
0 & 0 & \ddots & \ddots & 1 \\
0 & 0 & \dots & 0& \lambda \\
\end{pmatrix} \implies
J_j(\lambda)^k = \begin{pmatrix}
\lambda^k & k\lambda^{k-1} & * & \dots & * \\
0 & \lambda^k & k\lambda^{k-1} & \ddots & \vdots \\
0 & 0 & \ddots & \ddots & * \\
0 & 0 & \ddots & \ddots & k\lambda^{k-1} \\
0 & 0 & \dots & 0& \lambda^k \\
\end{pmatrix}$$
so $$(J_j(\lambda)^k - \lambda^kI)^2 = J_j(\lambda)^k = \begin{pmatrix}
0 & 0 & k^2\lambda^{2k-2} & \dots & * \\
0 & 0 & 0 & \ddots & \vdots \\
0 & 0 & \ddots & \ddots & k^2\lambda^{2k-2} \\
0 & 0 & \ddots & \ddots & 0 \\
0 & 0 & \dots & 0& 0 \\
\end{pmatrix}$$
having nonzero terms $k^2\lambda^{2k-2}$ on the diagonal two places above the main diagonal. So $$\dim\ker (J_j(\lambda)^k - \lambda^k I)^2 = 2 > 1 =\dim\ker (J_j(\lambda)^k - \lambda^k I)$$
A block-diagonal matrix retains its block-diagonal structure when taking powers, so $\dim\ker (A^k - \mu^k I)$ is simply the sum of dimensions of kernels of its blocks (which are of the same sizes as those for $A$). Same holds for $\dim\ker (A^k - \mu^k I)^2$.
Hence $\dim\ker (A^k - \lambda^k I)^2 > \dim\ker (A^k - \lambda^k I)$ which implies that $A^k$ has a Jordan block of size $\ge 2$ associated to $\lambda^k$.
We conclude that $A^k$ is not diagonalizable.
| {
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Visualisation of dot/cross product If this question is a duplicate / the answer already exists I am very sorry.
I recently started working with vectors in 3 dimensions. My visualisation of the dot product is the following:
Lets say we have the vector v(0,0,1) with the origin O. The dot product between v and the vector with the origin in O and end in P is: positive if P is situated anywhere "above" O and negative otherwise. It's like O would be the centre of a sphere and P would be situated in the upper/lower hemisphere. The radius is infinite. It surely sounds pretty stupid to smarter people but this is how I see it.
Could you point me so something similar regarding the cross product?
Thank you very much!
| Cross product between two vectors
$$\vec A=(x_1,x_2,x_3)\;,\;\;\;\vec B=(y_1,y_2,y_3)$$
is defined as
$$\vec C=\vec A\times \vec B:=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2\,,\,x_3y_1-x_1y_3\,,\,x_1y_2-x_2y_1)$$
and it turns out to be a vector $\vec C$ orthogonal to the plane spanned by $\vec A$ and $\vec B$.
The direction of $\vec C$ is given by the right hand rule.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Connected, compact, orientable smooth manifold with finite fundamental group The problem is prove that a connected, compact, orientable smooth manifold $M$ with finite fundamental group has $H^{1}_{dR}(M)=0$. The hint is to apply
Suppose $\tilde M$ and $M$ are smooth $n$-manifolds and $\pi:\tilde M\to M$
is a smooth $k$-sheeted covering map. If $\tilde M$ and $M$ are oriented and $\pi$ is orientation-preserving, show that $\int_{\tilde M}\pi^*\omega=k\int_{M}\omega$ for any compactly supported $n$-form $\omega$ on $M$.
but i don't know how this proves the result!
Any tips?
| I honestly don't know what I was thinking when I wrote that hint -- it doesn't seem to be useful at all.
A more appropriate hint would have been something like this: "If $\omega$ is a closed $1$-form on $M$, let $\widetilde\omega$ be the pullback of $\omega$ to the universal cover of $M$, and let $\tilde f$ be a potential function for $\omega$. Define $f\colon M\to \mathbb R$ by letting $f(x)$ be the average value of $\tilde f$ over the preimages of $x$, and show that $f$ is a smooth potential function for $\omega$."
This answer by @TedShifrin gives a little bit more detail about how to carry out that argument.
There's a different proof of this result in the second edition of my Introduction to Smooth Manifolds -- see Corollary 17.18.
| {
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"timestamp": "2023-03-29T00:00:00",
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Doubt related to the $\epsilon-\delta$ definition of limit $$\text{If}\quad\lim\limits_{x\to a}{f(x)}=L\quad\text{then:}$$
$$\text{If}\quad\forall (\epsilon > 0)\;\exists (\delta>0\;\text{and}\;\forall x\quad((x\neq a\;\text{and}\;|x-a|<\delta)\quad\Rightarrow\quad|f(x)-L|< \epsilon)).$$
I have understood the intiution behind this definition as: it says that for every $x$ closer and closer to $a$, if we have $f(x)$ closer and closer to $l$, here closeness is in terms of $\delta$ and $\epsilon $ then we say that limit of function equals $l$ as $x$ approaches $a$.
So my doubt is that why definition is not defined by putting less than equal to in place of less than?
I mean what is wrong if we take equality sign, I mean why distance cannot be taken as $\delta$ or $\epsilon$
$$\text{If}\quad\lim\limits_{x\to a}{f(x)}=L\quad\text{then:}$$
$$\text{If}\quad\forall (\epsilon > 0)\;\exists (\delta > 0\;\text{and}\;\forall x\quad((x\neq a\;\text{and}\;|x-a|\leq\delta)\quad\Rightarrow\quad|f(x)-L|\leq\epsilon)).$$
| There is nothing wrong if you put $\leq$ instead of $<$, its just that conventionally a topology is described by its open sets, even if it can also be defined in terms of the closed sets. Here, behind the idea of taking less than, the topological aspect plays a big role so that the definition can easily be extended in a more general setting.
| {
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Can the integers $n$ expressible as $2a^2-3b^2$ be classified? Let $S$ be the set of integers $n$ that are expressible with $$n=2a^2-3b^2$$ with integers $a,b$
Can $S$ be classified by an iff-condition that allows to decide whether $n\in S$ , when the factorization of $n$ is known ?
I found out several partial results, but I could not find a final structure.
What I found out :
*
*The residues $1$ and $3$ modulo $8$ are impossible as well as the residue $1$ modulo $3$
*A prime $p\ge 5$ can only be in $S$, when $p\equiv 5\mod 24$ or $p\equiv 23\mod 24$
*$n\in S$ if and only if the pell-like equation $$c^2-6b^2=2n$$ has an integer solution
*A non-zero perfect square (including $1$) cannot be in $S$
Looking at the impossible residues modulo $2^k$, it seems that for $k\ge 3$ there are $2^{k-1}-2$ impossible residues, so with every additional factor $2$, $2$ new impossible residues occur. Is this true, and if, why ? Also, the residue $3^k$ modulo $3^{k+1}$ seems to be impossible. Again, is this true, and if , why ?
| The integer solutions of $2 x^2 - 3 y^2 = n$ are invariant under the mapping
$(x,y) \to (5x+6y, 4x+5y)$. Thus if there is an integer solution, there must be one with $\sqrt{n/2} \le x \le 5 \sqrt{n/2}$ (if $n > 0$), or $\sqrt{-n/3} \le y \le 6 \sqrt{-n/3}$ (if $n < 0$).
| {
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Determine the arithmetic and geometric sequence given the relationships betweeen the first four terms No matter what I try, I can't solve this problem. I'm almost done but I need to get just one more thing to be able to finish it.
Problem
The first, second and fourth term of the arithmetic and of the geometric sequence are equal, respectively. The third term of the arithmetic sequence is by $18$ greater than the third term of the geometric sequence. Determine both sequences.
My attempt
I first wrote down the relationships between the terms.
$$a_1=b_1$$
$$a_2=b_2$$
$$a_3=b_3+18$$
$$a_4=b_4$$
We can rewrite the last three relationships as
$$a_1+d=b_1r$$
$$a_1+2d=b_1r^2+18$$
$$a_1+3d=b_1r^3$$
Then we can square $a_2=b_2$
$$a^2_2=(a_1+d)^2=b^2_2=(b_1r)^2$$
Rewrite $(b_1r)^2$
$$(b_1r)^2=b^2_1r^2=b_1b_1r^2=b_1b_3$$
So, we now know that $a^2_2=b_1b_3$. Based on that, we can write
$$(a_1+d)^2=b_1b_3$$
And since $b_1=a_1$
$$(a_1+d)^2=a_1b_3$$
We also know that $a_3=b_3+18 \leftrightarrow b_3=a_3-18$
$$(a_1+d)^2=a_1(a_3-18)$$
Which gives
$$(a_1+d)^2=a_1(a_1+2d-18)$$
Now, let's rewrite that as
$$a^2_1+d^2+2a_1d=a^2_1+2a_1d-18a_1$$
From that we have
$$d^2+18a_1=0$$
Now I just need to get another equation to form a system of equations. From that, I can get $d$ or $a_1$, and then I can use the variable that I've got to get the second variable (i.e. the one I did not got). When I get $a_1$, I also get $b_1$ since $a_1=b_1$. Then I'll have $d$, $a_1$ and $b_1$. If I know these variables, I can easily get $r$. And then I can finally write out both sequences.
Of course, I also have the solution. It is
Arithmetic sequence: $$\langle-2,4,10,16,...\rangle$$
Geometric sequence: $$\langle-2,4,-8,16,...\rangle$$
| Here we know that $r\neq1$ otherwise all terms would be equal and it would not be possible to have the third term of the AP being greater than the third term of the GP.
Let $T_i$ be the $i-$th term of the AP.
$$\begin{align}
\frac {T_4-T_2}{T_2-T_1}&=\frac {4-2}{2-1}\\
\frac {r^3-r}{r-1}&=2
&&\scriptsize (T_i=ar^{i-1}\text{ for } i=1,2,4)\\
r^2+r-2&=0
&&\scriptsize (\text{as }r\neq 1)\\
(r-1)(r+2)&=0\\
r&=-2
&&\scriptsize (\text{as } r\neq 1)\\ \\
T_2-T_1=d&=a(r-1)\\
d&=-3a\\ \\
T_3=a+2d&=18+ar^2\\
a-6a&=18+4a\\
a&=-2\\\\
\Longrightarrow d&=6
\end{align}$$
Hence
*
*$\color{red}{\text{AP}: \qquad -2,4,10,16}$
*$\color{red}{\text{GP}:\qquad -2,4,-8, 16}$
| {
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How to show the double sum identity $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $ I am not sure how to go about showing this: $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $
It is a bit like the formula for $\sum ^{n}_{i=1}i^{2}$
and this $\sum_{i=1}^n \sum_{j=1}^i \frac{i-j}{nm} + \sum_{i=1}^{n} \sum_{j=i}^m \frac{j-i}{nm} = \frac{2 n^2 - 3 n m + 3 m^2 - 2}{6m}$
As I really do not know how to proceed.
| I would suggest an inductive proof to that identity. Assume the induction hipotesis
$$
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j= \dfrac{1}{6}n\left( n-1\right) \left( n+1\right).
$$
It's easy to verify identity for $n=1$, $n=2$ and $n=3$. Consider the scheme.
$$
\begin{array}{rl}
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j=&0
\\
+&0+1
\\
+&0+1+2
\\
+&0+1+2+3
\\
+&0+1+2+3+4
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\, \vdots \;\;\;\, \ddots
\\
+&0+1+2+3+4+\cdots +i
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\,\vdots \;\;\;\, \quad \;\;\;\, \vdots \;\;\;\,\ddots
\\
+&0+1+2+3+4+\cdots +i+\cdots+(n-1)
\end{array}
$$
Note by scheme that
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\sum^{n}_{i=1}\sum^{i}_{j=1}i-j\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
By induction hipotesis we have
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\dfrac{1}{6}\left( n-1\right)n\left( n+1\right)\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
And since we know that $\dfrac{1}{2}n(n+1)$ is the result of the sum $1+2+3+\ldots +n$ we have
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \dfrac{1}{6}( n-1)n(n+1)+ \dfrac{1}{2}n(n+1)
\\
&= \dfrac{1}{6}(n-1)n(n+1)+ \dfrac{3}{6}n(n+1)
\\
&= \Big[(n-1)+ 3\Big]\dfrac{1}{6}n( n+1)
\\
&= \dfrac{1}{6}n\cdot (n+1)(n+2)
\\
&= \dfrac{1}{6}[\color{red}{(n+1)}-1]\cdot [\color{red}{(n+1)}][\color{red}{(n+1)}+1]
\end{align}
| {
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Expanding $\prod_{m=1}^{n-1}(1+a_m+b_m)$. Quite a complicated product in the form below occured in my research. I'm having trouble getting started in evaluating it.
Let $n\in\Bbb N\setminus \{1\}$, $(a_m)_{m\in\overline{1, n-1}}\in \Bbb R^{n-1}$, and $(b_m)_{m\in\overline{1, n-1}}\in \Bbb R^{n-1}$. Consider the product $$P:=\prod_{m=1}^{n-1}(1+a_m+b_m).$$ Expand $P$.
Thoughts:
I expect binomial coefficients to show up.
If we let $c_m:= a_m+b_m$, then $P$ becomes
$$\begin{align}
\prod_{m=1}^{n-1}(1+c_m)&= (1+c_1)\prod_{m=2}^{n-1}(1+c_m) \\
&=\prod_{m=2}^{n-1}(1+c_m)+c_1\prod_{m=2}^{n-1}(1+c_m),
\end{align}$$
which suggests that induction might work if I can guess what the expansion would look like.
I think I should be able to do this myself but I've been stuck for longer than I care to mention.
Please help :)
NB: Here $0\notin \Bbb N$.
| $$\begin{align}\prod_{m=1}^{n-1}(1+c_m)&=(1+c_1)(1+c_2)\cdots(1+c_{n-1})\\&=1+\sum_{\text{cyc}}c_1+\sum_{\text{cyc}}c_1c_2+\cdots+\prod_{i=1}^{n-1}c_i\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Basel Problem - Area of $\frac 16$ of Circle with Radius $\sqrt{\pi}$. There are several proofs to the solution of the well-known Basel Problem, i.e.
$$\sum_{n=1}^\infty \frac 1{n^2}=\frac {\pi^2}6$$
Is is possible to create a geometrical interpretation of this identity in the form of the area of $\frac 16$ of a circle with radius $\sqrt{\pi}$?
|
TOPIC: Area of $\frac{1}{6}$ of Circle with Radius $\sqrt{\pi}$
Last time I show how to get an easy approximation for radius via Pythagorean theorem:
$$ (\sqrt{\pi})^2\geq1.7^2+0.5^2
$$
and have give a suggestion to do different- let me show for $\pi\leq\frac{22}{7}$ (by using intercept theorem for $\frac{\sqrt{7}}{7}$) and fully geometrical solving this time:
| {
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Cone of tangents of $S=\{(x,y):y\ge-x^3\}$ Find the cone of tangents for the set $S$ at the point $\overline x=(0\ 0)^t$.
$$S=\{(x,y):y\ge-x^3\}$$
The cone of tangents of $S$ at $\overline x$ is the set of directions $d$ such that $d=\lim_{k\to\infty}\lambda_k(x_k-\overline x),\ \lambda_k>0,x_k\in S$ for all $k$ and $x_k\to\overline x$.
In this particular case, $d=\lim_{k\to\infty}\lambda_k(x,y)$
what should I do next?
I've searched online but I only find 'solved' exercises that just write the answer and does not says how the cone was found with details.
I greatly appreciate any assistance you may provide.
| There are several first-order cones that appear in context of KKT conditions and constraint qualifications (like feasible directions, attainable directions etc), but all those cones, including the tangent cone, are between two standard cones for the linearized problem that are easy to calculate (here $I$ is the set of indices for active constraints):
\begin{align}
G_0&=\{d\colon \nabla g_i(\bar x)^Td<0,\,i\in I\},\\
G'&=\{d\colon \nabla g_i(\bar x)^Td\le 0,\,i\in I\}.
\end{align}
In our case, $g(x,y)=-x^3-y\le 0$ is active at the origin, so
\begin{align}
G_0&=\{d\colon (0\ \ {-}1)\,d<0\}=\{d\colon d_2>0\},\\
G'&=\{d\colon (0\ \ {-}1)\,d\le 0\}=\{d\colon d_2\ge 0\}.
\end{align}
Since
$$
\text{closure}\,G_0\subset T\subset G'
$$
and $\text{closure}\,G_0=G'$ (the closed upper half-plane), the tangent cone $T$ is also the closed upper half-plane.
P.S. The condition $\text{closure}\,G_0=G'$ is known as Cottle's constraint qualification (which is equivalent to the Mangasarian-Fromovitz constraint qualification in case of no equality constraint).
| {
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Find rectangular box which has biggest volume Exercise: Find rectangular box with surface area 10 that has minimum volume
My solution: We know that surface area is $$S= 2(wl +hl+hw)$$and volume $$V=whl$$ ,where w-width,h-heigh, l=length.
We will use lagrange function$$L = whl - \lambda(2(wl +hl+hw)-10)$$.
Now we will find partial derivatives w.r.t $w,h,l,\lambda$ respectively
$$1)hl-2h\lambda-2l\lambda \\ 2)wl-2l\lambda - 2wl\lambda \\ 3)wh-2w\lambda-2h\lambda \\ 4) 2(wl +hl+hw) - 10$$
By setting this equations to 0 we will get that $l=h=w$. Then inset into Surface equation and we will get that the volume will be the biggest when $w=h=l= \sqrt{\frac{5}{3}}$.
Did i solve it correctly?
| As an alternative and to check let use AM-GM inequality
$$\frac {S} 6= \frac{wl +hl+hw}{3}\ge\sqrt[3]{w^2l^2h^2}\implies V^{\frac23}\le\frac {5}3\implies V\le\sqrt{\left(\frac{5}3\right)^3}$$
with equality for $$wl=hl=hw \implies w=l=h=\sqrt{\frac{5}3}$$
| {
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Can you find the result of a linear transformation by projecting a given vector onto known eigenvectors? I have a practice exercise I'm struggling with. Essentially, it gives you the eigenvalues and eigenvector for a given linear transformation and then asks you to find the result of the transformation on some other vector.
$$
A\begin{bmatrix}-1\\-1\\-2\end{bmatrix} = 1\begin{bmatrix}-1\\-1\\-2\end{bmatrix}
$$
$$
A\begin{bmatrix}1\\1\\1\end{bmatrix} = 0\begin{bmatrix}1\\1\\1\end{bmatrix}
$$
$$
A\begin{bmatrix}-1\\-4\\-3\end{bmatrix} = 2\begin{bmatrix}-1\\-4\\-3\end{bmatrix}
$$
Find $A\begin{bmatrix}3\\-4\\3\end{bmatrix}$.
My first thought is to split this vector into components along the known eigenvector, calculate the result, and then add the vectors back together, but I don't know if this is valid. If it isn't, how would one go about this problem?
Thanks for any and all help.
| Yes, this is the right approach; the idea is to appeal to linearity. First decompose the given vector into a linear combination of the eigenvectors, then:
$$\begin{align}
A\left(c_1\begin{bmatrix}-1\\-1\\-2\end{bmatrix} + c_2\begin{bmatrix}1\\1\\1\end{bmatrix} + c_3\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right) &= c_1\left(A\begin{bmatrix}-1\\-1\\-2\end{bmatrix}\right) + c_2\left(A\begin{bmatrix}1\\1\\1\end{bmatrix}\right) + c_3\left(A\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right)\\
&=c_1\left(1\begin{bmatrix}-1\\-1\\-2\end{bmatrix}\right) + c_2\left(0\begin{bmatrix}1\\1\\1\end{bmatrix}\right) + c_3\left(2\begin{bmatrix}-1\\-4\\-3\end{bmatrix}\right)\\
&=c_1\begin{bmatrix}-1\\-1\\-2\end{bmatrix} + 2c_3\begin{bmatrix}-1\\-4\\-3\end{bmatrix}
\end{align}
$$
| {
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Show that the Euler characteristic of a chain complex is equal to the Euler characteristic of its homology Let $C_*$ be a chain complex such that each $C_i$ is a torsion-free, finite-range abelian group with $C_i=0$ for all $i<0$. Suppose that $C_i=0$ for all $i$ is large enough. The Euler characteristic of $C_*$ chain complex is defined as $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(C_i)$$ Prove that $$\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$$
I have to prove that $\sum_{i\geq 0}(-1)^iRank(C_i)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))$, I think that one way to do this is by showing that $Rank(C_i)=Rank(H_i(C_*))$ but I do not know if this is true in general, in a nutshell, I want to show that the cardinality of the base of any $C_i$ is the same as the cardinality of the basis of the corresponding homology, how can I do this? Thank you
| Your chain complex looks like
$$0\to C_n\to C_{n-1}\to\cdots\to C_0\to0$$
where the $C_i$ are nonzero outside this range. Proceed by induction
on $n$. Call this complex $\mathbf C$. Let $\mathbf{C}'$ be the subcomplex
$$0\to0\to C_{n-1}\to\cdots\to C_0\to0.$$
Then there is a short exact sequence of complexes
$$0\to\mathbf{C}'\to\mathbf{C}\to\mathbf{C}''\to0$$
where $C''$ consists just of $C_n$ in dimension $n$. This gives a long
exact sequence of homology. This starts
$$0\to H_n(\mathbf{C})\to C_n\to H_{n-1}(\mathbf{C'})\to H_{n-1}(\mathbf{C})\to0$$
so that
$$\textrm{rank}(H_n(\mathbf{C}))
-\textrm{rank}(C_n)+\textrm{rank}(H_{n-1}(\mathbf{C'}))-\textrm{rank}(H_{n-1}(\mathbf{C}))=0.\tag{1}$$
For $k<n-1$ another piece of the long exact sequence is
$$0\to H_{k}(\mathbf{C'})\to H_{k}(\mathbf{C})\to0$$
so that
$$\textrm{rank}(H_k(\mathbf{C'}))=\textrm{rank}(H_k(\mathbf{C})).
\tag{2}$$
From (1) and (2) one gets the relation between Euler characteristics:
$$\chi(\mathbf{C})=\chi(\mathbf{C'})+(-1)^n\textrm{rank}(C_n)$$
which gives the inductive step.
| {
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Maxima & Minima Word Problem Problem: Given the following profit-versus-production function for a certain commodity: $P=200000-x-(\frac{1.1}{1+x})^8$. Where P is the profit and x is the unit of production. Determine the maximum profit.
Solution: Taking its first derivative, $\frac{dP}{dx} = -1-8(\frac{1.1}{1+x}^7) * (\frac{-1.1}{(1+x)^2})$, then equate to $0$, the value of x would be equal to $0.371$. Then substituting it to the original equation would result to $199,999.46$ which is the maximum profit.
Question:
*
*How to solve if instead, the problem asked for the minimum profit?
*In some problems, the minimum is the value of x (example: the 0.371 in the problem above) after differentiating the given equation and equating it to 0. But in some problem the minimum is the value after substituting that x, so in some problem, that 199,999.46 is the minimum instead. So how can I know which is which?
Any help or tip would be appreciated.
| By Second Derivative test we can decide function is maximum or minimum at point x. http://mathworld.wolfram.com/SecondDerivativeTest.html
http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-maxmin-2009-1.pdf
| {
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Two variants of proof for Baire's category theorem One of the variants of the so called Baire's Category Theorem (BCT) says that
Given a (possibly nonempty) complete metric space $(X,d)$ and a system of open dense subsets $(O_n)$ of $X,$ then $\bigcap_{n \in \mathbb{N}} O_n$ is dense in $X.$
Most proofs of this theorem goes like this: to show that the intersection of $O_n$'s is dense, it is enough to show that for any $y \in X,$ $\bigcap_{n \in \mathbb{N}} O_n \cap B_{r}(y)$ with $B_{r}(y)$ an open ball of radius $r$ and centered at $y,$ is nonempty. Since $O_1$ is dense in $X,$ the intersection $O_1 \cap B_r(y)$ is nonempty and we can choose $x_1$ and $r_1 < r/2$ such that $B_{r_1}(x_1) \subset O_1 \cap B_r(y).$ In fact, by shrinking $r_1,$ we can arrange things in such a way that $\overline{B_{r_1}(x_1)} \subset O_1 \cap B_r(y).$ Then, since $O_2$ is dense, we can find $x_2 \in B_{r_1}(x_1) \cap O_2$ such that $\overline{B_{r_2}(x_2)} \subset O_2 \cap B_{r_1}(x_1)$ with $r_2 < r/4.$ Going this way, we recursively define sequences $(x_n)$ and $(r_n)$ such that $\overline{B_{r_{n+1}}(x_{n+1})} \subset O_{n+1} \cap B_{r_n}(x_n)$ with $r_{n+1} < r/2^{n+1},$ showing that $(x_n)$ is Cauchy with a limit $x.$ Since $\overline{B_{r_n}(x_n)}$ is closed and $x_i$ lies in $\overline{B_{r_n}(x_n)}$ for every $i>n$ with $n \in \mathbb{N}$ arbitrary, it follows that $x$ lies in $\overline{B_{r_n}(x_n)}$ for every $n,$ hence it lies in every $O_n$ and it lies also in $B_r(y),$ which completes the proof.
A slightly more straightforwardly looking variant of this proof is given for example here, the only difference is that it gives the sequences $(x_n)$ and $(r_n)$ in such a way that we have only $B_{r_{n+1}}(x_{n+1}) \subset O_{n+1} \cap B_{r_n}(x_n)$ (without ever taking the closure of the balls). The author then goes to the conclusion that, due to its construction, the limiting point $x$ (which no doubt exists, since $X$ is complete), lies in $\bigcap_{n \in \mathbb{N}} O_n \cap B_r(y).$ Is this conclusion correct? If so, why?
My understanding is that essentially the same argument with the closures must be somehow implicitly involved, or is there another way that bypasses the use of closures (like for example contradiction with cauchyness of the sequence involved)?
| I think his conclusion at that point is wrong or incomplete.
Let's consider the example
$$
X=[0,1], d(x,y)=|x-y|,
O_n = (0,1).
$$
Then $O_n$ is open and dense for every $n\in\mathbb N$.
However, we can assume that the construction of $x_n$ and $r_n$ yields
$$
x_n= 2^{-n},
r_n = 2^{-n}.
$$
Then $B_{r_{n+1}}(x_{n+1}) \subset O_n \cap B_{r_n}(x_n)$ is true,
(because $B_{r_{n+1}}(x_{n+1})\subset (0,1)$)
as well as $r_n<\frac1n$ as required in the proof that you linked.
However, the limit is $x=\lim x_n = 0$, which is not in the intersection of $\cap_{n\in\mathbb N} O_n = (0,1)$.
| {
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Formula for the floor function I found the following formula for the floor function:
$$\lfloor x \rfloor = -\frac12+x+\frac{\arctan(\cot\pi x)}{\pi}$$
for all $x$ not an integer.
My question is where I can find the proof of this formula.
| Consider
$$f(x) = \frac{1}{\pi}\int_{0}^{\cot(\pi x)}\frac{dy}{1+y^2}=\frac{\arctan(\cot(\pi x))}{\pi},$$
with the latter equation obtained by substituting $y=\tan u.$ Since $x\to \cot(\pi x)$ is manifestly periodic of period $1,$ and $f$ integrates to $0$ over one period (since $\cot$ is an odd function), $f$ also is periodic.
When $x$ is not an integer $f$ is differentiable at $x$ because both $\cot(\pi x)$ and the integral (qua function of its upper limit) are. The Chain Rule and Fundamental Theorem of Calculus together imply
$$f^\prime(x) = \frac{1}{\pi}\left(\frac{1}{1 + (\cot(\pi x))^2}\right) \frac{d}{dx}\left(\cot(\pi x)\right)=\frac{-\pi\csc(\pi x)^2}{\pi\csc(\pi x)^2}=-1.$$
This is the key insight, because it shows $f$ has the basic properties needed to construct functions that are periodic and linear between their points of discontinuity. The rest is just algebra.
Since $f(1/2)$ is an integral from $0$ to $0=\cot(\pi/2),$
$$f(1/2) = 0.$$
This information completely determines $f.$ To summarize, at nonintegral values $f$ falls linearly with slope $-1,$ equals $0$ at $1/2,$ and repeats this pattern between each successive pair of integers. Consequently the function
$$\frac{1}{2} - f(x)$$
must rise linearly from $0$ at $x=0$ up to a limit of $1$ as $x\to 1.$ Because it is periodic, it drops back to $0$ when $x=1$ and repeats this pattern ad infinitum in both directions. Obviously that describes the remainder ("fractional part") function. That is,
$$x - \lfloor x \rfloor = \frac{1}{2} - f(x).$$
Solving for the floor,
$$\lfloor x \rfloor = x - \left( \frac{1}{2} - f(x)\right) = -\frac{1}{2} + x + \frac{\arctan(\cot(\pi x))}{\pi}.$$
| {
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Question about Spivaks Inverse function theorem proof In the opening lines of Spivaks "Calculus on Manifolds" proof of the inverse function theorem he writes the equation below. let $\lambda = Df(a)$
$$D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$$
I don't see how $ D(\lambda^{-1})(f(a)) = \lambda^{-1}$ in the second equality
and in the statment of the theorem makes the standard assumptions of f being $c^1$ and $f'(a)$ being nonsingular.
| It's simply because $\lambda$ is a linear map. So, its inverse is also a linear map and the derivative of a linear map at every point of its domain is again that linear map.
| {
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Making a substitution in a PDE - how did the author get to this result?
How did the author get from $k_t = k^2 k_{\theta \theta} + k^3$ to that form I outlined? I've tried computing $k_{t}$ and $k_{\theta \theta}$ using $k(\theta, t)^2 = A(\theta) + B(t)$ and substituting that in the PDE, but it was all a mess and not close to what the author got. Also, why is it that $A''(\theta) + 4A(\theta)$ is a constant?
link to where I got this from. page 40
| Differentiating $k(\theta,t)^2=A(\theta)+B(t)$ twice with respect to $\theta$ and once with respect to $t$ yields
\begin{eqnarray*}
2kk_\theta&=&A'\;,\\
2k_\theta^2+2kk_{\theta\theta}&=&A''\;,\\
2kk_t&=&B'\;.
\end{eqnarray*}
Solve the first equation for $k_\theta$, substitute the result into the second equation, solve that for $k_{\theta\theta}$, solve the third equation for $k_t$ and substitute everything into the differential equation, multiply through by $k$, replace $k^2$ by $A+B$ everywhere, and you (should) get the desired result.
$A''+4A$ must be constant (with respect to $\theta$) because $(A''+4A)B$ is the only term that mixes $\theta$ and $t$. Consider the differential equation at fixed $\theta$ for two different times $t_1$ and $t_2$, and subtract the two resulting equations. The terms that depend only on $\theta$ cancel, so the only remaining term that depends on $\theta$ is $(A''(\theta)+4A(\theta))(B(t_2)-B(t_1))$. It follows that unless $B$ is constant (in which case the second factor would always be zero), the first factor cannot depend on $\theta$.
| {
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The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$. The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain complex values
Yet the reasoning for the implication itself is
\begin{align}
x+\frac {1}{x}&=1 \\
\color {green}{x^2}&=\color {green}{x-1 }\\
x^3&=\color {green}{x^2}-x \\
&=(\color {green}{x-1})-x \\
&=-1.
\end{align}
Hence $x^6=1$, which implies $x^7=x$.
I had just started reading this book when it was presented, asking you to prove the implication without finding the roots first.
So my question: is this implication really justified? Should the author have said to only prove it for an implicit (complex) specific value of $x$? If I didn't know any better, I'd say he presented it like it was true for some range of values...he couldn't of...
| You shouldn't be trying to find where $y_1=y_2$, but whether a root of $y_1$ is necessarily a root of $y_2$.
That is,
$$\forall x\Big(y_1(x)=0\implies y_2(x)=0\Big)$$
or equivalently,
$$\forall x\Big(y_1(x)\neq0\lor y_2(x)=0\Big)$$
| {
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Closed form / computationally efficient method for binomial + factorial sum I'm trying to evaluate the sum given below
$$
S(x,y) = \frac{1}{k!}\sum_{j=0}^k \binom{k}{j} (-x)^j y^{k-j} j!
$$
where $x,y > 0$. Is there any clever closed form way of writing this sum? If not, maybe there is a clever way of writing it as a product? I was considering products of $(y-jx)$ factors perhaps $\pm (y-x)^k$ but could not get this work. Any ideas?
| Since ${k \choose j} = {k \choose k-j} = \frac{k!}{j!(k-j)!}$, upon making the substitution $j \leftrightarrow k-j$ you can write your sum as
$$ S(x,y) = (-x)^k \sum_{j=0}^k \frac{1}{j!} \left( - \frac{y}{x} \right)^j = (-x)^k S(1, -y/x) \; .$$
Now we recognize $S(u) := S(1, u) = \sum_{j=0}^k \frac{u^j}{j!}$ as the $k$-th partial sum of the MacLaurin expansion of $e^{u}$, which admits a closed form in terms of the (incomplete) Gamma function, as expressed in this MSE post. This gives some perspective on the other possible 'closed forms' one might expects.
| {
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Convex sum order If I have a strictly convex function $f(x)$ with $f''(x)>0$
and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have
$$a+b+c = x+y+z$$
$$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$
can I conclude that at least one of the following must be true about the order?
$$ a \le x \le y \le b \le c \le z$$
$$ x \le a \le b \le y \le z \le c$$
Perhaps an equivalent question is to consider coincident centroids of two triangles with vertices on the convex curve, looking something like this diagram with pairs of vertices on the inside of the order
though I realise that I cannot push this analogy too far as it would not be necessarily true if say the curve was a circle and the triangles equilateral
| In geometry a triangle $(ABC)$ on the plane $xOy$ with coordinates $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ has its gravity center (baricenter) the point
$$
G=\left(\frac{a_1+b_1+c_1}{3},\frac{a_2+b_2+c_2}{3}\right).
$$
Hence what you ask is:
Given a curve (function) $c_f:y=f(x)$ or $M=(x,f(x))$, with positive curvature, find all triangles $(ABC)$ inscribed in $c_f$ such that they have the same gravity center.
In geometry, the gravity center of (ABC) is defined as the point of intersection of all the dimedians i.e.
In $A$ edge the dimedian is the sigment $AM_1$ with $M_1$ being the median of $BC$ side. In $B$ edge the coresponding dimedian is $BM_2$, where $M_2$ is the median of $AC$ and in $C$ edge the sigment $CM_3$.
For a given baricenter there are finite number of triangles inscribed in $c_f$ with the same baricenter $G=\left(\frac{a+b+c}{3},\frac{f(a)+f(b)+f(c)}{3}\right)$.
A theorem of Leibniz says
Theorem.
If $M$ is an arbitrary point of plane of the triangle $(ABC)$ and $G$ is the baricenter, then
$$
MA^2+MB^2+MC^2=3MG^2+\frac{1}{3}(AB^2+BC^2+CD^2)
$$
Hence your broblem reduces to find all points $A=(a,f(a))$, $B=(b,f(b))$, $C=(c,f(c))$ of curve $y=f(x)$ such
$$
GA^2+GB^2+GC^2=\frac{1}{3}(AB^2+BC^2+CA^2)
$$
or equivalently for given $f$ find all $a,b,c$ such
$$
(g_1-a)^2+(g_2-f(a))^2+(g_1-b)^2+(g_2-f(b))^2+(g_1-c)^2+(g_2-f(c))^2=\frac{1}{3}\left((a-b)^2+(f(a)-f(b))^2+(b-c)^2+(f(b)-f(c))^2+(c-a)^2+(f(c)-f(a))^2\right),\tag 1
$$
where
$$
g_1=\frac{a+b+c}{3}\textrm{ and }g_2=\frac{f(a)+f(b)+f(c)}{3}.\tag 2
$$
Hence we have three equations with three unknowns $a,b,c$. Hence the number of points will be finite.
For example assume the parabola $y=x^2$. Then equations $(2)$ become
$$
a=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
or
$$
a=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
The value of $c$ is given from
$$
\frac{4 a^3}{3}-\frac{2 a^2 b^2}{3}-\frac{2 a^2 c^2}{3}-\frac{2 a b}{3}-\frac{2 a c}{3}+\frac{4 b^3}{3}-\frac{2 b^2 c^2}{3}-\frac{2 b c}{3}+\frac{4 c^3}{3}=0
$$
| {
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Find the limit of $(\tan(x) + \sec (x))^{1/\sin(x)}$ $$ \lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = e $$
how to arrive at e, according to wolfram alpha, that this is the answer?
| Note that:
$$\begin{align}\lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = &\lim_{x \rightarrow 0} \left[\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right]^{1/\sin x} =\\ &\lim_{x \rightarrow 0} \frac{[1+\sin x]^{1/\sin x}}{[\cos x]^{1/\sin x}} =\\
&\frac{\lim_{x \rightarrow 0}[1+\sin x]^{1/\sin x}}{\lim_{x \rightarrow 0}[\cos x]^{1/\sin x}} =\\
&\frac{e}{1}=e.\end{align}$$
| {
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The number of functions $ f : \left \{ 1, 2, . . . , 10 \right \} \rightarrow \left \{ 1, 2, . . . , 10 \right \}$ such that $f(x) \neq x$ for all $x$ Question
The number of functions $ f : \left \{ 1, 2, . . . , 10 \right \} \rightarrow \left \{ 1, 2, . . . , 10 \right \}$ such that $f(x) \neq x$ for all $x$ is
Approach
Total number of function possible $=10^{10}$
But with the restriction ,$f(x) \neq x$ for all $x$,
Eg-:$f(1)\neq 1 $ each element in the domain will have 1 less available option in the range .
So total number of function possible$=9^{10}$
Am I correct?
please help.
| Case 1: If functions are not bijective.
It's a simple case for every $i^{th}$ element ,$1\le i\le 10$, you have 9 possibilities to choose from for the output.
Hence total number of functions=$ 9^{10}$
Case 2: If functions are bijective
On close observation you might notice that what you need is exactly the number of Derangements that could be done with $10$ elements. That means you need to find $$!10=\left\lfloor \frac {9!}{e}+\frac 12\right\rfloor =362 880 $$
| {
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Uniform convergence of $\sum_{n=1}^\infty \frac{1}{n^3+x}$ How do is show that for $-1<x<1$ the series $\sum_{n=1}^\infty \frac{1}{n^3+x}$ converges uniformly.
For $x\geq0$ I can make a convergent Majorant series, choosing $M_n = 1/n^3$, but I can't seem to determine a convergent Majorant series for $-1<x<0$.
| Applying Weierstrass. For every $r \in (-1,1)$ you can consider $A_r = (r,1)$. Now apply M-test using that $\forall x \in A_r$
$$
\frac{1}{n^3+x} \leq \frac{1}{n^3+r}
$$
And obviously
$$
\sum_{n=1}^\infty \frac{1}{n^3+r} < \infty
$$
You can check it by comparation with $\sum_n 1/n^3$. So you prove uniform convergence in $A_r$. Now take limit $r\to -1$.
| {
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Find a short expression for the long sum First of all, I'm quite new here, so sorry if this is not asked in the correct place.
The sum is $$X={{100}\choose{1}}+3\cdot{{100}\choose{3}}+5\cdot{{100}\choose{5}}+...+97\cdot{{100}\choose{97}}+99\cdot{{100}\choose{99}}$$ I have noticed that I can get a much simpler sum:
$$X={{100}\choose{1}}+99\cdot{{100}\choose{99}}+3\cdot{{100}\choose{3}}+97\cdot{{100}\choose{97}}+5\cdot{{100}\choose{5}}+95\cdot{{100}\choose{95}}+...+49\cdot{{100}\choose{49}}+51\cdot{{100}\choose{51}}$$.
Since $${{100}\choose{k}}={{100}\choose{100-k}}$$, it follows that the sum equals to:
$$X=100\cdot{{100}\choose{1}}+100\cdot{{100}\choose{3}}+...+100\cdot{{100}\choose{49}}$$
or:
$$X=100\cdot\sum_{n=0}^{24}{{100}\choose{2n+1}}$$
I'm not sure how to continue. I couldn't find a short term for the last sum.
|
We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^{24}\binom{100}{2n+1}}&=\frac{1}{2}\sum_{n=0}^{49}\binom{100}{2n+1}\tag{1}\\
&=\frac{1}{2}\left(1\cdot\sum_{n=0}^{49}\binom{100}{2n+1}+0\cdot \sum_{n=1}^{50}\binom{100}{2n}\right)\tag{2}\\
&=\frac{1}{2}\sum_{n=0}^{100}\frac{1-(-1)^n}{2}\binom{100}{n}\tag{3}\\
&=\frac{1}{4}\sum_{n=0}^{100}\binom{100}{n}-\frac{1}{4}\sum_{n=0}^{100}(-1)^n\binom{100}{n}\tag{4}\\
&=\frac{1}{4}\cdot 2^{100}-\frac{1}{4}\left(1+(-1)\right)^{100}\tag{5}\\
&\,\,\color{blue}{=2^{98}}\tag{6}
\end{align*}
Comment:
*
*In (1) we extend the region to all odd $n$ between $0$ and $100$ using the symmetry $\binom{n}{k}=\binom{n}{n-k}$.
*In (2) we add zero times all the even summands between $0$ and $100$.
*In (3) we collect the sums.
*In (4) we split the sum (somewhat differently to (2)).
*In (5) we apply the binomial theorem to both sums.
*In (6) we do a final simplification.
| {
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Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof (if you need it):
$0 \le (a - b)^2 = a^2 + b^2 -2ab \implies 2ab \le a^2 + b^2$
$(a + b)^2 = a^2 + b^2 + 2ab$ which by previous $ \le 2(a^2 + b^2) $.
Application:
If $f, g$ are positive functions then $(f + g)^2$ is integrable $\iff$ $f^2, g^2$ are integrable since $f^2 + g^2 \le (f + g)^2 \le 2(f^2 + g^2) $ pointwise.
| This inequality can as weel be seen as a particular case of the equivalence between $p$-norms in $\Bbb R^n$. Indeed for $1<p\leq q<\infty$, it holds
$$\|x\|_q \leq \|x\|_p \leq n^{1/p-1/q}\|x\|_q$$
In the particular case $n=2$, $p=1$ and $q=2$ we get
$$(|a|+|b|)=\|(a,b)\|_1\leq 2^{1-1/2}\|(a,b)\|_2 =\sqrt{2(a^2+b^2)},$$
which is even slightly tighter as $|a+b| \leq |a|+|b|$.
I should nevertheless point out that the equivalence between $p$-norms is proved using the Cauchy-Schwarz (or more generally the Hölder) inequality mentioned by @Ihf
| {
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Fourier transform of $\lvert x^2-1\rvert $ One of the exercises of my assignment was to determine the Fourier transform of function
$$f(x)=\lvert x^2-1\rvert$$
The domain wasn't specified. First I was puzzled since $f$ isn't a $L^1$ function.
If I were to calculate
$$\mathcal{F}(f)(\xi)=\int_\mathbb{R} \lvert x^2-1\rvert e^{-2\pi i x \xi}\,dx$$ I would calculate it separately on $\langle-\infty, -1], \langle-1,1\rangle$ and $[1, +\infty \rangle$ but in case of real domain, it doesn't converge (first and third integral).
In case of complex domain, separate parts converge depending on the imaginary part of $\xi$, but not at the same time so the whole integral diverges.
Am I missing something here? I would have said at first that the Fourier transform of this function isn't defined, but why would it be an exercise then...
| The Fourier transform of a distribution is defined as
$$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]).$$
That is, the action of $\mathcal F[f]$ on $\phi$ is given by the rhs of this identity, where we know that $\mathcal F[\phi]$ is well-defined and is again a valid test function.
We can find the transforms of $1$ and of $x^2$ by finding the transform of $\delta^{(n)}$ and also can find the transform of $|x^2 - 1| - (x^2 - 1)$ directly to get
$$(|x^2 - 1|, e^{-2 \pi i \xi x}) =
-\frac {\delta''(\xi)} {4 \pi^2} - \delta(\xi) +
\frac {\sin 2 \pi \xi - 2 \pi \xi \cos 2 \pi \xi} {\pi^3 \xi^3}.$$
| {
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Courant–Fischer Theorem Proof [Meyer]. Intersection of two subspaces with same dimension. Suppose we have 2 subspaces $\mathcal{A} \subseteq \mathbf{R}^n$ and $\mathcal{B}\subseteq \mathbf{R}^n$ that have the same dimension, say dim $\mathcal{A}=$ dim $\mathcal{B}=l>0$. Is it true that they have a non null intersection?
Ignore this part and focus on the top question answered correctly by @paf
I'm reading a section of Meyer's book, Matrix Analysis and Applied Linear Algebra, on the Courant–Fischer Theorem. I'm following the proof of pages 550-551, but I don't seem to understand why $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$ . . . This two subspaces have the same dimension so I figure it might be a general property that it's being used. Is this the case? If not, how would your show that $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$, following the reasoning of Meyer?
| I'm not sure of your question, hence I'll give two answers.
1. Two vector subspaces of $\Bbb R^n$ always have non-empty ($\ne \varnothing$) intersection. Indeed, by definition, every vector subspace must contain the zero vector.
2. But their intersection can be exactly $\{0\}$ (thus having dimension $0$). Simply take two distinct vector subspaces of dimension 1 in $\Bbb R^2$ (i.e. 2 lines through 0)! To have an example valid for all $n$, if $V = \{a_1x_1 + \dots + a_nx_n = 0\}\subset \Bbb R^n$, then the line $W$ generated by $(a_1,\dots,a_n)$ intersects $V$ only at the zero vector.
So don't confuse $\varnothing$ and $\{0\}$...
| {
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Predicate formulas on the natural numbers using only $\le$. We want to find predicate formulas about the natural numbers using only the $\le$ predicate and no constants. For instance, the following predicate formula defines equality:
$[x=y] ::= [x \le y \; \land y \le x].$
And then, using that we can define $[x>0]$:
$[x>0]::= \exists y. x\neq y \; \land y \le x.$
$[x =0]$ is then:
$[x =0]::= [\forall y \ x \le y].$
Now, then, how would I define $[x = y+1]$?
| $x=y+1$ iff both $x>y$ and there is no $z$ with $x>z$ and $z>y$.
| {
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Show that infinite Cartesian product is uncountable let $A_i=\{0,1\} \forall i \in \mathbb Z^+$ and $A^{\omega}= \prod _\limits{i \in \mathbb Z^+}A_i$ thus $A^\omega=\{(a_i)|a_i=1,2 \forall i\in \mathbb Z^+\}$
now here is my problem
let $(\underline x_n)_{n \in \mathbb Z^+}$ where $\underline x_n=(x_{nk})_{k \in \mathbb Z^+}$ be a sequence in $A^\omega$. Define $\underline y=(y_n)_{n \in \mathbb Z^+}$ by $y_n=1-x_{nn}$ for all $n \in \mathbb Z^+$
a) Show that $\underline y \in A^\omega $ and $\underline y \neq \underline x_n$ for all $n\in \mathbb Z^+$
b) $\mathbf{using}$ $\mathbf{a)}$ show that $A^\omega$ is not countable
I did the a) part. And I can prove the part b) $A^\omega$ is uncountable by using diagonal argument. but here i should use the part a). my attempt was
first, I suppose $ A^\omega$ is countable.
then there exists a bijection from $f:\mathbb Z^+\to A^\omega$.
then I can list out all elements in $A^\omega$.
then since I should use a) I have to find a connection between this list and $\underline x_n$s. I am stuck here.
| Given any function $f \colon \mathbb{Z}^+ \to A^\omega$, let $\underline{x}_n = f(n)$. Then use (a) to show that $f$ can't be surjective.
| {
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prove that $S_{\tau \land n} \to S_{\tau}$ in $L^1$ for a random walk with $E\tau^{1/2} < \infty$
Let $X_i$ be a sequence of iid $L^2$ RVs with $EX_i = 0$ and define the martingale $S_n = \sum_1^n X_i$. I want to show that if $\tau$ is a stopping time and $E\tau^{1/2} < \infty$, then $ES_{\tau} =0$.
I have been given the hint that if $Y_n$ is an $L^2$ martingale, then $E(\sup_n | Y_n|) \leq 3E(\sqrt{A_{\infty}})$, where $A = \langle Y \rangle$ is the quadratic variation process of $Y$, i.e. $A_0 =0$, $$A_n = \sum_{k=1}^nE(Y_k^2 \mid \mathcal{F}_{k-1}) - E(Y_{k} \mid \mathcal{F}_{k-1})^2$$ and $\lim_n A_n =A_{\infty}$.
My work:
Following the idea in Doob's OST, clearly have that $S_{\tau \land n}$ is a martingale and therefore $E[S_{\tau \land n}] = E[S_0] = 0$. Since $\tau< \infty$ a.s, I need to show $\lim_n E[S_{\tau \land n}] = E[S_{\tau}]$ (that is, $S_{\tau \land n} \to S_{\tau}$ in $L^1$).
I have tried to follow this hint, and noted that since $X_i$ are $L^2$, $S_n = \sum_{i=1}^n X_i \in L^2$. Also,
$$\begin{aligned}
A_n &= \langle S \rangle_n =\sum_{k=1}^nE(S_k^2 \mid \mathcal{F}_{k-1}) -0\\
& = E \left(\sum_{k=1}^nX_k^2 + 2\sum_{i<j \leq n} X_iX_j \mid \mathcal{F}_{k-1} \right)\\
& = E(X_n^2) + \sum_{k=1}^{n-1}X_k^2 + 2 \sum_{i<j<n}X_iX_j \\
&= E(X_n^2)+S_{n-1}^2
\end{aligned}$$
I don't know that this has a limit. If I use the hint on the $X_i$s, I get that $$\langle X\rangle_n = E(X_n^2) \implies E(\sup_n|X_n|) \leq 3(X_{\infty}^2)$$ which I don't think is helpful.
I'm not sure exactly which route I should be going down to show convergence, there are so many theorems on it with slightly different conditions.
I'm so lost, any ideas would be much appreciated, thanks!
| First of all: Your calculation of $A_n$ is not correct; note that
$$A_n = \sum_{k=1}^n \mathbb{E}(S_k^2 \color{red}{-S_{k-1}^2} \mid \mathcal{F}_{k-1})$$
and then you will end up with $$A_n = \sum_{k=1}^n \mathbb{E}(X_k^2) = n \mathbb{E}(X_1^2). \tag{1}$$
If we consider the stopped process $Y_n := S_{n \wedge T}$, then it follows from $(1)$ that its compensator is given by $A_n := \mathbb{E}(X_1^2) \min\{n,T\}$. Applying the inequality which you were given in the hint, we thus find
$$\mathbb{E} \left( \sup_{n \geq 1} |S_{n \wedge T}| \right) \leq 3 \mathbb{E}(\sqrt{T}) < \infty.$$
Now it follows easily from the dominated convergence theorem that $S_{T \wedge n} \to S_T$ in $L^1$.
| {
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Denote Riemann integral as sum of infinite series One of the standard definitions of Riemann Integral is as follows:
Let $f$ be bounded on $[a, b]$. For any partition $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ of $[a, b]$ and any choice of points $t_{k} \in [x_{k - 1}, x_{k}]$ the sum $$S(P, f) = \sum_{k = 1}^{n}f(t_{k})(x_{k} - x_{k - 1})$$ is called a Riemann sum for $f$ over $P$ and tags $t_{k}$. The norm of $P$ denoted by $||P||$ is defined as $||P|| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})$. A number $I$ is said to be Riemann integral of $f$ over $[a, b]$ if for any arbitrary $\epsilon > 0$ there exists a $\delta > 0$ such that $$|S(P, f) - I| < \epsilon$$ for all Riemann sums $f$ over any partition $P$ with $||P|| < \delta$. When such a number $I$ exists we say that $f$ is Riemann integrable over $[a, b]$ and we write $$I = \int_{a}^{b}f(x)\,dx$$
Suppose that if $f$ is in $R[0, 1]$. Does it necessarily follow that $$\lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^{n}f\left(\frac{k}{n}\right) = \int_{0}^{1}f(x)\,dx?$$
My attempt : Since $f$ is in $R[0,1]$, for any $\epsilon>0$, exist $\pi_0$ s.t for every $\pi$ which is refinement of $\pi_0$, $U(f,\pi)-L(f,\pi)<\epsilon$. If there exist $k$ s.t $\pi' = \{0,\frac{1}{k}, ..., 1\}$ is refinement of $\pi_0$, then it is done. But when if $\pi_0$ contains irrational point, then any $\pi_0$ cannot be refinement of $\pi_0$, Is there any other method to prove this?
| YES.
Set $$
P_n=\left\{t_0=0,t_1=\frac{1}{n},t_2=\frac{2}{n},\ldots,t_n=1\right\}.
$$
Then
$$
\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=S(P_n,f), \quad \text{with $t_k=\frac{k}{n}$}.
$$
If we choose an arbitrary $\varepsilon>0$, then, according to the definition in the OP, there exists a $\delta>0$, such that if $\|P\|<\delta$, then $|S(P,f)-I|<\varepsilon$.
Hence if $n_0\in\mathbb N$, such that $\delta>\frac{1}{n_0}$, then
$$
\left|\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)-I\right|<\varepsilon,
$$
for all $n\ge n_0$.
| {
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Solve $\log_3(x^2+2x+1)=\log_2(x^2+2x)$ Solve $\log_3(x^2+2x+1)=\log_2(x^2+2x)$
I have tried to do to as followed:
$\log_3(x^2+2x+1)=\frac{\log_3(x^2+2x)}{\log_3(2)}$
$\iff\log_3(x^2+2x+1).\log_3(2)=\log_3(x^2+2x)$
Is it possible to proceed this way? Or should one approach this differently?
| If $\log_3(x^2+2x+1)=\log_2(x^2+2x)=y$
$f(y)=3^y-2^y=1$
Now $f(y)$ is an increasing function in $[0,\infty)$ and decreasing in $(-\infty,0]$
| {
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Inequality $x^2\leq y$ I have a question about how to handle this inequality:
$$x^2\leqslant y \to x\leqslant \pm\sqrt{y}$$ or it should be $$\sqrt{x^2}\leq \sqrt{y}\Rightarrow$$
either $$x\leq\sqrt{y}$$
or
$$-x\leq\sqrt{y}\Rightarrow x\geq-\sqrt{y}$$ so$$-\sqrt{y}\leq x\leq\sqrt{y}$$
Is my way of thinking is correct?
| I assume you work in $\mathbb{R}$. If $y$ is negative, then no $x$ satisfies the inequality. If $y$ is non-negative, then $-\sqrt{y} \leq x \leq \sqrt{y}$.
| {
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Determining the point from which the most area in a polygon is visible I am wondering about the following problem: Given a polygon and the set of points $S$ inside it, what are the point(s) in $S$ from which the most area in $S$ is visible? Furthermore, what is the maximum visible area?
Here, I define $q$ to be visible from $p$ if the line segment between $p$ and $q$ is contained in $S$. This intends to capture the intuitive idea of what points in a room are visible when standing somewhere in the room. For example, in the figure below, the dark blue area is visible from point P, at the center of the top left quarter. The light blue area is not.
While the answer for any star-shaped domain is clear, finding the answer for arbitrary polygons seems difficult.
Question: How can we find the solution to the problem for a given polygon?
For example, the problem is not so easy for the polygon below...
| This is not an answer to your question, but an answer to a related question
that you might find interesting. The paper below computes
"the maximum clique in the visibility graph $G$ of a simple polygon $P$ in time
$O(n^2 e)$,
where $n$ and $e$ are number of vertices and edges of $G$ respectively."
Ghosh, Subir Kumar, Thomas Caton Shermer, Binay Kumar Bhattacharya, and Partha Pratim Goswami. "Computing the maximum clique in the visibility graph of a simple polygon." Journal of Discrete Algorithms 5, no. 3 (2007): 524-532.
(Elsevier link to HTML.)
| {
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Independence of shifted squares mod p Given an odd prime $p$, let $S\subseteq \mathbb F_p$ be the set of quadratic residues modulo $p$. Given $a,b\in \mathbb F_p$ we write $aS+b$ for the set $$aS+b:=\{t\in\mathbb F_p:\ t=ax^2+b \text{ for some $x\in\mathbb F_p$}\}.$$
What is the cardinality of $S\cap ((S+1)\cup (2-S))$?
I would say that they are $\frac 3 8 p+ o(p)$, because of the following heuristic:
*
*the residues of the form $x^2$ are $\sim \frac p 2$ out of $p$;
*there's small correlation between a residue mod p being of the form $x^2$ or $1+y^2$;
*so $\# S\cap (S+1)$ should be $\sim p/4$.
This in fact is quickly provable in a number of ways.
*
*similar reasoning for the equation $\# S\cap (2-S)$.
Now, one would guess that, given random $x,y,z$, the equalities $x^2=1+y^2$ and $x^2=2-z^2$ are achieved for $1/4$ of the $x^2$s. So by Inclusion-Exclusion we get the heuristic.
I expect this to be provable in finite time by standard computations and the error term would be related to Weil's bound, aka Riemann Hypothesis on finite fields.
But how is it possible to make the above heuristic reasoning rigorous up to (1+o(1)), in a simple, direct and general way?
In other words, how to make oneself safe to assume the approximate independence of equations as above when doing this sort of ballpark estimates?
Let's restrict to linear equations between quadratic residues or to the specific example above, in this question. But of course the same question is interesting for other sets of residues that should "distribute randomly" modulo p, for example sets related to sums of characters.
Possible approaches so far:
*
*Estimate $\# S\cap S\cap(S+1)\cap(2-S)$ via Inclusion-Exclusion after counting precisely enough the solutions of the equation $(x^2-1-y^2)(x^2-2+z^2)=0$.
*Compute first effectively the expected Weil error bound, then compute the cardinality numerically for a large enough prime.
*Transform the problem into character sums as in Jyrki Lahtonen comment below.
| $S \cap (S+1)$ is approximately $\frac 14 \operatorname{Card}\{(x,y) \in \Bbb F_p^2 \mid x^2 = y^2+1 \}$.
This set is an algebraic curve of dimension $1$ (and of genus $0$) so it has $p+O(1)$ points. And so $\operatorname{Card}(S \cap (S+1)) = p/4+O(1)$
Similarly, $S \cap (2-S)$ will be related to the curve $x^2 = 2-y^2$ and $(S+1) \cap (2-S)$ to $x^2+1 = 2-y^2$. They are all curves of genus $0$ so those intersections also have $p/4 + O(1)$ elements.
Finally, $S \cap (S+1) \cap (2-S)$ corresponds to a spatial curve, the intersection of $x^2 = y^2+1$ and $x^2 = 2-z^2$. At a glance, this is an elliptic curve, so the curve has $p + 2\sqrt p u(p) + O(1)$ points, where $|u(p)| \le 1$, and then the intersection has $p/8 + \sqrt p u(p)/4 + O(1)$ elements.
Gathering up everything, you get :
$\operatorname{Card}(S \cap ((S+1) \cup (2-S))) = \\
\operatorname{Card}(S \cap (S+1)) + \operatorname{Card}(S \cap (2-S)) - \operatorname{Card}(S \cap (S+1) \cap (2-S)) = \\
3p/8 - \sqrt p u(p)/4 + O(1)$
where $|u(p)| \le 1$
Note that this doesn't require $p$ to be prime, this is valid over any finite field $\Bbb F_q$
| {
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Show that this stochastic process is a.s. strictly positive. Consider the one-dimensional SDE
$$
dX_t = f(X_t)dt + \sigma(X_t)dW_t,\quad X_0 = 1,
$$
where $W_t$ is a Brownian motion under the measure $\mathbb{P}$, in its natural filtration. Suppose that $f(x)-\sigma^2(x)/2x=0$ and $\sigma^2(x)\le x^2$. Show that $X_t>0$ almost surely.
My initial attempt was to consider the log-process $\log(X_t)$. Using Ito's lemma one finds
$$
d\log(X_t) = \frac{1}{X_t}\left(f(X_t)
- \frac{\sigma^2(X_t)}{2X_t}\right)dt +
\frac{\sigma(X_t)}{X_t}dW_t = \frac{\sigma(X_t)}{X_t}dW_t.
$$
Integrating now leads to
$$
\log(X_t)=\int_0^t\frac{\sigma(X_s)}{X_s}dW_s.
$$
By the Ito isometry, the variance of the above process is bounded by $t$:
$$
\text{Var}\big[\log(X_t)\big] = \mathbb{E}\left[\int_0^t\frac{\sigma^2(X_s)}{X_s^2}ds\right] \le t.
$$
I feel like this is heading in the right direction, but I'm struggling to see what the next step is. Any suggestions would be very welcome.
| Define stopping times by
$$T_k := \inf\{t \geq 0; X_t \leq k^{-1}\}, \qquad k \in \mathbb{N}$$
and
$$T := \inf\{t \geq 0; X_t \leq 0\}.$$
Following the reasoning in your question we find that
$$\mathbb{E} \left( \left| \log(X_{t \wedge T_k}) \right|^2 \right) \leq t.$$
By the continuity of the sample paths of $(X_t)_{t \geq 0}$, we have $X_{T_k}(\omega) = k^{-1}$ for $\omega \in \{T_k<\infty\}$, and therefore we get that
$$\mathbb{E} \bigg( 1_{\{T_k \leq t\}} \underbrace{|\log(X_{T_k})|^2}_{=|\log(k^{-1})|^2} \bigg) \leq t.$$
Since $\{T \leq t\} \subseteq \{T_k \leq t\}$ this shows, in particular, that
$$\mathbb{E}\left(1_{\{T \leq t\}} |\log(k^{-1})|^2 \right) \leq t,$$
i.e.
$$\mathbb{P}(T \leq t) \leq \frac{t}{|\log(k^{-1})|^2}.$$
As $|\log(k^{-1})|^2 \to \infty$ as $k \to \infty$ we conclude that $\mathbb{P}(T \leq t)=0$. As $t>0$ is arbritrary, this proves the assertion.
| {
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A student must pick 5 classes from 12 courses, if he must have at least one WH class or USH class, how many different choices does he have? The World History course and United States history course is already part of the 12 courses he must choose from. It is required to have at least one of these classes in his new class schedule. Sorry if my question didn't make sense earlier, this question was from a practice exam I'm just recalling the question from memory. I was short on time so I just solved C(12,5) and I know its wrong because I forgot about the at least. This has been bugging me all day and I'd appreciate it if someone can explain.
| We will take "At least one USH or WH" to mean that you could take all five courses from this set. This is odd, but it is the best semantic match for the question wording. If it is supposed to mean one or both, this approach wont work.
In this case, the "stars and bars" technique applies. Ken Ribet is a better teacher than I. If you don't have time for a sixteen minute video, here's a summary:
Your question can be reframed by asking how many ways can you toss k balls into n buckets. Or, how many ways can you insert n - 1 separators into a list of k potential courses. The spaces in between separators represent different buckets (potential courses). There are one fewer separators than buckets because extremal areas count as buckets so the formula is:
${n+k-1\choose k}$
Your question is further complicated by the initial condition of at least one course of two potential types. Now we can count the possibilities by separating into ways of choosing courses from the first two types:
1 History: ${1+2-1\choose1} * {4+10-1\choose4}$ +
2 History: ${2+2-1\choose2} * {3+10-1\choose3}$ +
3 History: ${3+2-1\choose3} * {2+10-1\choose2}$ +
4 History: ${4+2-1\choose4} * {1+10-1\choose1}$ +
5 History: ${5+2-1\choose5}$
| {
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General question about Self-adjoint operators on Hilbert Spaces so if I have a self-adjoint operator $K \in B(H,H)$. Then I found a theorem that said that
$A$ is invertible if and only if $(A^*Ax,x) \geq c||x||$ and $(AA^*x,x) \geq c||x||$ for all $x$.
Then does that mean that if I have a self adjoint operator like $K$.
$(K^*Kx,x) = (Kx,Kx) \geq 0$, but then as long $Kx \not = 0$ for all $x$ i.e. $K = 0$, then $K$ is invertible because I have $(Kx,Kx)> 0$. But that is weird to me because that means every self-adjoing bounded linear operator is invertible.
| In an infinite-dimensional vector space, there's a big difference between $(Ax,Ax)\geq c\|x\|^{2}$ for all $x\in H$ and $(Ax,Ax)>0$ for all nonzero $x\in H$. For an example, consider $\ell^{2}(\mathbb{N})$. Then if $A(e_{i})=2^{-i}e_{i}$ for all $i\geq 0,$ we can see that the inverse must be $A^{-1}(e_{i})=2^{i}e_{i}$, but this is not a bounded linear operator.
We see that $(Ax,Ax)=\sum_{i=0}^{\infty}2^{-2i}x_{i}^{2}>0$ whenever $x\neq0$, but for any $c>0,$ if $x\in\mathrm{span}(\{e_{i}\}_{i\geq I}),$ where $I$ is large enough that $2^{-2I}<c$, then $(Ax,Ax)\leq 2^{-2I}\sum_{i=0}^{\infty}x_{i}^{2}<c\|x\|^{2}$. Therefore there is no $c>0$ for which this bound holds.
EDIT: Consider the optimization problem $$\min_{x\neq0}\frac{(Ax,Ax)}{\|x\|^{2}},$$ which is equivalent to $$\min_{\|x\|=1}(Ax,Ax).$$ The unit sphere $\{x\in H:\|x\|=1\}$ is compact if and only if $H$ is finite-dimensional, so in this case, there is some $x$ which achieves this minimum $c$, which gives $(Ax,Ax)\geq c\|x\|^{2}$ for all $x\in H$. When this minimization is not over a compact set, there is no such guarantee (just consider the example I gave above).
| {
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Spivak's Calculus: Proofs concerning Pascal's Triangle Problem 3 of Chapter 2 in Spivak's Calculus poses 5 problems associated with Pascal's Triangle. The first of these asks you to prove that $\binom{n+1}{k}$ = $\binom{n}{k-1} + \binom{n}{k}$ which was fairly straightforward.
The second task was to prove by induction that $\binom{n}{k}$ is always a natural number. For this, I took the recursion approach: given the demonstration of the equality above, any $\binom{n}{k}$ can be decomposed into $\binom{n-1}{k-1} + \binom{n-1}{k}$ and so on, with each value decomposing until every binomial reaches either an $n$ or $k$ value of $0$, at which point they evaluate to 1 and can be summed. $1$ is a natural number, so the sum of any number of $1$s must therefore be a natural number. This proof seems to me as though it lacks formality, but I think the logic is appropriate.
The third task, however, I am simply unsure what to make of. It reads: "Give another proof that $\binom{n}{k}$ is a natural number by showing that $\binom{n}{k}$ is the number of sets of exactly $k$ integers each chosen from $1, \cdots,n$." The word induction is not used in the prompt, and I can see that this appears to be true across the first few rows of the triangle, but how one would rigorously prove something like this, I do not know. Additionally, having proven that this is true, is the link from that to $\binom{n}{k}$ always being a natural number just the fact that you cannot have partial or negative sets, and therefore it must be a natural number?
| For convenience, let $S_n = \{x_1, x_2, x_3, \dots, x_n\}$ represent a set of $n$ distinct objects.
Lets use $_nC_k$ to represent the number of ways of selecting $k$ objects from $S_n$.
So what does $_{n+1}C_k$ look like?
An element of such a selection will either include the object $x_{n+1}$ or it will not. We already know that there are $_nC_k$ ways to choose $k$ objects from the objects $x_1, x_2, \dots, x_n$. So how do we count the number of ways to select $k$ objects that contain $x_{n+1}$. Easy; we select $k-1$ objects from $S_n$ {there are $_nC_{k-1}$ such ways} and then add $x_{n+1}$ to the collection.
It follows that $_nC_{k-1} + _nC_k = _{n+1}C_k$.
| {
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Is $Q + A$ postive semidefinte? Let $Q$ be a Matrix and $V \subseteq \Bbb R^n$ be a vector subspace on which $Q$ is positive semidefinite i.e, $$\langle x , Q x \rangle \ge 0 ~~~~~ \quad \forall x \in V $$
Prove or provide a counter example
There exist a matrix $A$ such that $Q + A$ is positive semidefinite (on whole $\Bbb R^n$) and
$$\langle x , Q x \rangle = \langle x , (Q + A) x \rangle \quad \quad x \in V.$$
| If $P$ is a projection onto $V$, then for any $x \in \mathbb{R}^n$ we have
$$\langle x, P^T QPx \rangle = \langle Px, QPx \rangle \ge 0$$
since $Px \in V$, so $P^T QP$ is positive semidefinite. Moreover if $x \in V$ then $Px = x$, so
$$\langle x, P^T QPx \rangle = \langle Px, QPx \rangle = \langle x, Qx \rangle.$$
So pick $A = P^T QP - Q$ and we're done.
| {
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Maximal ideals in a group ring when the ring is a field Is there any simple characterization of the maximal ideals in a group ring $R[G]$ when $R$ is a field, perhaps in terms of maximal subgroups of $G$?
| The maximal ideals of $R[G]$ are in canonical bijection with the isomorphism classes of simple $R[G]$-modules. In one direction, if $I$ is a maximal ideal, then $R[G]/I$ is a simple ring, hence has a unique simple module up to isomorphism, which you can lift to a simple module of $R[G]$. In the other direction, given a simple $R[G]$-module $M$, take $I$ to be its annihilator.
Whether or not this characterisation helps you in any way depends on further properties of $G$ and $R$. For example if $G$ is finite, then there is a very rich theory of simple $R[G]$-modules, equivalently of irreducible representations of $G$ over $R$, namely character theory if the characteristic of $R$ does not divide $\#G$, and the theory of Brauer characters in the other, the modular case. But in the generality that you have posed the question in, I am not sure whether anything more enlightening can be said.
| {
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Well definedness of factor group multiplication I have been reading on this site and the internet, and I do not quite understand the comments which are being made with respect to the problem of proving that factor group multiplication is well defined. According to John Fraleigh, A First Course in Abstract Algebra, 7th ed (page 137), if we make the following multiplication rule for factor groups,
$$
(xH)(yH) = (xy)H,
$$
we must check whether it is well defined. I have searched on this website and the internet to try and find out what the concern is, and it seems that the problem is that we must make sure that if $xH = x'H$ and $yH = y'H$, then $(xy)H = (x'y')H$.
First question: Is this Fraleigh's concern?
If the answer to this question is indeed yes. How can it be a problem, since if $xH = x'H$ and $yH = y'H$, then $$(x'y')H = (x'H)(y'H)=(xH)(yH) = (xy)H$$ is automatic? This is my second question.
|
(x′y′)H=(x′H)(y′H)=(xH)(yH)=(xy)H
This is known as "proof by notation". It's a trap you have to be careful about falling into. Simply because you declare some expression to represent a value, doesn't mean that the value exists or is well-defined.
For instance, suppose we define f(p) to be the area of triangles with perimeter p. We can "prove" that given a perimeter, all triangles with that perimeter have the same area with the following argument:
Given triangle T1 and T2 with the same perimeter p, the area of T1 is f(p), and the area of T2 is also f(p). Thus, their areas are the same.
| {
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Proof that if $f:S^{n} \to \mathbb{R} $ is continuous, then is not injective I have to prove that if $f:S^{n} \to \mathbb{R} $ (where $S^{n}= \{(x_1, ..., x_{n+1})\in\mathbb{R}^{n+1} | x_1^2+...+x_{n+1}^2=1\} $ is continuous, then is not injective. If possible, I would like it to be proven by using connectivity arguments.
My attempt: suppose $f$ is injective. Let $p\in f(S^n)$ and, since $f$ is injective, there exists only one $q\in S^n$ such that $f(q)=p$. We consider now $f_{|S^n \setminus\{q\}}:{S^n \setminus\{q\}} \to f(S^n\setminus\{q\})=f(S^n)\setminus\{p\} $, which is continuous and bijective (because $f$ was continuous and injective). Then, since ${S^n \setminus\{q\}}$ is path-connected and $f$ is continuous and exhaustive, we have that $f(S^n)\setminus\{p\}$ is path-connected as well. Here I would like to use that $\mathbb{R}\setminus \{p\}$ is not path-connected and arrive at a contradiction, but this is wrong.
| Since $S^n$ is compact, if $f$ was injective, it would be a homeomorphism onto $f(S^n)$. But $S^n$ is compact and connected and the only subsets of $\mathbb R$ which are compact and connected are the the intervals $[a,b]$. However, if you remove one point from the middle of this interval, it becomes disconnected. No point of $S^n$ has that property.
| {
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How many trees over ${1,2,3,...n}$ with conditions I’m stuck on this question in graph theory.
The question is:
How many labeled trees are there over $V={0,1,2,...n}$
with which vertices 1,2,3 are leaves, and distance between any two of these leaves is 3 or more.
I tried using Cayley theorem but I don’t know how to apply it in this specific question.
| Like every other question about counting trees, this can be answered using Prüfer codes.
Each tree with vertex set $\{1,2,\dots,n\}$ corresponds to a unique Prüfer code, which is a sequence of $n-2$ elements of $\{1,2,\dots,n\}$. Moreover, a vertex of degree $k$ in the tree appears $k-1$ times in the Prüfer code.
So to count the trees which have $1$, $2$, and $3$ as leaves, it suffices to count Prüfer codes which don't include the elements $1$, $2$, and $3$. There are $(n-3)^{n-2}$ of these.
To deal with the condition that these leaves are distance $3$ apart, it's easiest to use inclusion-exclusion. Starting with the quantity $(n-3)^{n-2}$,
*
*subtract the number of trees that have $1$, $2$, and $3$ as leaves, with $1$ and $2$ only distance $2$ apart.
*subtract the number of trees that have $1$, $2$, and $3$ as leaves, with $1$ and $3$ only distance $2$ apart.
*subtract the number of trees that have $1$, $2$, and $3$ as leaves, with $2$ and $3$ only distance $2$ apart.
*add back, twice, the number of trees that have $1$, $2$, and $3$ as leaves, with all three of them only distance $2$ apart.
We can compute these by observing that any tree in which vertices $1$, $2$, and $3$ are leaves, and $1$ and $2$ are distance $2$ apart, can be built by starting with a tree on vertex set $\{1,3,\dots,n\}$ in which $1$ and $3$ are leaves, and adding the vertex $2$ to the unique neighbor of $1$. So there are $(n-3)^{n-3}$ such trees. The other two cases are similar (except in the last case, we add two vertices), so we get a final answer of
$$
(n-3)^{n-2} - 3(n-3)^{n-3} + 2(n-3)^{n-4}.
$$
We can also reason more directly, though this requires making use of more details of the Prüfer code. From the algorithm to convert a tree into a Prüfer code (see the Wikipedia article for details) it is clear that when vertices $1$, $2$, and $3$ are all leaves, the first number in the code is the parent of vertex $1$, the second number is the parent of vertex $2$, and the third number is the parent of vertex $3$.
All three of these numbers must be different to ensure that the three vertices are not too close together. Therefore the number of ways to choose a Prüfer code for such a tree is the product of:
*
*$n-3$ ways to choose the first number out of $\{4,5,\dots,n\}$;
*$n-4$ ways to choose the second number out of $\{4,5,\dots,n\}$, different from the first;
*$n-5$ ways to choose the third number out of $\{4,5,\dots,n\}$, different from the first and second;
*$(n-3)^{n-5}$ ways to choose the remaining $n-5$ numbers out of $\{4,5,\dots,n\}$.
The product of these is $(n-3)^{n-4}(n-4)(n-5)$, which is equivalent to the previous formula.
| {
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gradient of least squares loss function derivation I am trying to derive the derivative of the loss function from least squares. If I have this (I am using ' to denote the transpose as in matlab)
(y-Xw)'(y-Xw)
and I expand it
=(y'- w'X')(y-Xw)
=y'y -y'Xw -w'X'y + w'X'Xw
=y'y -y'Xw -y'Xw + w'X'Xw
=y'y -2y'Xw + w'X'Xw
Now I get the gradient
=-2y'Xw + X'(Xw) + X(w'X')
=-2y'Xw + X'(Xw) + X'(Xw)
=-2y'Xw + 2X'(Xw)
And that is the intended result. Now, I saw in this post Vector derivation of $x^Tx$
That the gradient of x'x=2x, So I am trying to get the same result applying that, and the chain rule to get the gradient of
=(y-Xw)'(y-Xw)
So I think this might be
=2(y-Xw)(-X)
=-2yX + 2XwX
The result is similar but the transpositions are missing so it would not work... What am I doing wrong? My mathematical background has almost disappeared and I just started to begin the recovery, so please be patient if I did something terribly wrong...
| Here is a piece of background information that we must be clear on at the beginning. If $F:\mathbb R^p \to \mathbb R^q$ is differentiable at a point $z$, then $F'(z)$ is a $q \times p$ matrix.
I assume $X$ is a real $m \times n$ matrix and $y$ is an $m \times 1$ column vector. Let $g:\mathbb R^m \to \mathbb R$ be defined by $g(u) = u^T u$.
Note carefully that for any $u \in \mathbb R^m$,
$g'(u) = 2 u^T$ is a $1 \times m$ matrix.
Define $h:\mathbb R^n \to \mathbb R^m$ by $h(w) = y - X w$, and note that
$h'(w) = - X$.
Now let $f:\mathbb R^n \to \mathbb R$ be defined by
$$
f(w) = g(h(w)) = (y - X w)^T(y - X w).
$$
The chain rule tells us that
$$
\underbrace{f'(w)}_{1 \times n} = \underbrace{g'(h(w))}_{1 \times m} \underbrace{h'(w)}_{m \times n}.
$$
With our particular choices of $g$ and $h$, we have
$$
f'(w) = 2 (y - X w)^T(-X).
$$
If we use the convention that the gradient is a column vector, then we have
$$
\nabla f(w) = f'(w)^T = 2 X^T (X w - y).
$$
| {
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Prove or show a counter example for: $\forall a, b, c \in \mathbb Z$, if $ab\mid c$ then $a\mid c$ and $b\mid c$ I'm working on my Discrete Mathematics homework and they are asking me this:
Prove or show a counterexample for: $\forall a, b, c \in \mathbb Z$, if $ab\mid c$ then $a\mid c$ and $b\mid c$
I'm not completely sure how to prove it. I was thinking to find a counterexample like zero but I'm not sure if that it's going to work. Or maybe try to approach it backwards?
I'm a little bit lost
| Hint: If $ab\mid c$, then, since $a\mid ab$, …
| {
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prove there exists a unique polynomial $p\in P_{m}(\mathbb{F})$ such that $p(z_j) = w_j$ Suppose $m$ is a nonnegative integer, $z_1,\cdots, z_{m+1}$ are distinct elements of $\mathbb{F}$, and $w_1, \cdots, w_{m+1} \in \mathbb{F}$. Prove there exists a unique polynomial $p\in P_{m}(\mathbb{F})$ such that $p(z_j) = w_j$ for $j = 1, \cdots, m+1$.
When I attempted this I really had no idea where to start, then I saw a proof online that did this:
Define $T: P_{m}(\mathbb{F}) \rightarrow \mathbb{F^{m+1}}$ by: $$Tp = (p(z_{1}),\cdots,p(z_{m+1}))$$
From here we have to show that the map is linear and bijective. I have no issue with that portion. My question is the IDEA of coming up with a linear map to prove this problem. Where did the motivation to attempt something like that come from? It has me perplexed.
Note: I know there are other forms of proving this, but I am focusing on the linear algebra form.
| The question is about to find a certain polynomial of degree $\le m$, and these form a vector space.
The standard basis of $P_m(\Bbb F) $ is $1,x, x^2,\dots, x^m$, which makes coordinates from the coefficients.
The determinant of (the matrix written in the standard basis of) the given transformation is called Vandermonde determinant and it's nonzero if $z_i$'s are different:
$$\left\vert\matrix{1&1&\dots&1\\z_1&z_2&&z_{m+1}\\
z_1^2&z_2^2&&z_{m+1}^2\\
&\vdots&\vdots&\\
z_1^m&z_2^m&\dots&z_{m+1}^m} \right\vert$$
| {
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Where does the following formula come from? Where does the following formula come from?
For a Laurent polynomial $f(z)=\sum a_j z^j$ and a positive integer $n$ we have $$\sum_{k\equiv \alpha\pmod n} a_k=\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega).$$
I hope someone can answer this question or give some refferences about it ! Thanks a lot!
| This formula comes from Thomas Simpson's Series Multisection Theory. Speak in the concrete，
a multisection of the series of an analytic function
$$f(z) = \sum_{n=0}^\infty a_n\cdot z^n.$$
has a closed-form expression in terms of the function $f(x)$:
$$\sum_{m=0}^\infty a_{qm+p}\cdot z^{qm+p} = \frac{1}{q}\cdot \sum_{k=0}^{q-1} \omega^{-kp}\cdot f(\omega^k\cdot z),$$
where $\omega = e^{\frac{2\pi i}{q}}$ is a Primitive nth root of unity primitive ''q''-th root of unity . This solution was first discovered by Thomas Simpson.
| {
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Simplify $\arccos{\sqrt{ 2\over 3}}−\arccos{\frac{\sqrt6+1}{2\sqrt3}}$
Prove $$\arccos{\sqrt{ 2\over 3}}−\arccos{\dfrac{\sqrt6+1}{2\sqrt3}} = \dfrac\pi6$$
How to proceed with this question? I have tried changing them to $\arctan$ and applying $\arctan a- \arctan b$ but ended up getting some numbers which cant be simplied further.
| Let $$\alpha = \arccos{\sqrt2\over \sqrt3}\;\;\;\;{\rm and}\;\;\;\;\beta =\arccos{\sqrt6+1\over 2\sqrt3}$$
so $$ \cos \alpha = \sqrt{2\over 3}\;\;\;\;{\rm and}\;\;\;\;\cos \beta = {\sqrt{6}+1\over 2\sqrt{3}}$$
and
$$ \sin \alpha = \sqrt{1-{2\over 3}} ={1\over \sqrt{3}}\;\;\;\;{\rm and}\;\;\;\;\sin \beta = \sqrt{1-{7+2\sqrt{6}\over 12}} = \sqrt{{5-2\sqrt{6}\over 12}}$$
so $$\cos (\alpha -\beta) = \sqrt{2\over 3} {\sqrt{6}+1\over 2\sqrt{3}}+{1\over \sqrt{3}}\sqrt{{5-2\sqrt{6}\over 12}}=$$
$$ = {\sqrt{12}+\sqrt{2}+\sqrt{5-2\sqrt{6}}\over 6} $$
$$ = {2\sqrt{3}+\sqrt{2}+\sqrt{(\sqrt{3}-\sqrt{2})^2}\over 6} $$
$$ = {\sqrt{3}\over 2} $$
so $\alpha -\beta = \pi/6$
| {
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Find $f(3)$ if $f(f(x))=3+2x$ A function $f\colon \mathbb{R} \to \mathbb{R}$ is defined as $f(f(x))=3+2x$
Find $f(3)$ if $f(0)=3$
My try:
Method $1.$ Put $x=0$ we get $f(f(0))=3$ $\implies$ $f(3)=3$
Method $2.$ Replace $x$ with $f(x)$ we get
$$f(f(f(x)))=3+2f(x)$$ $\implies$
$$f(3+2x)=3+2f(x)$$
Put $x=0$
$$f(3)=9$$
I feel Method $2.$ is Correct since $f(f(x))=3+2x$ is Injective which means $f(x)$ should be Injective.
| Not only is $f(f(x))=3+2x$ injective, it is bijective, and $f(f(x))\not=x$ except when $x=-3$
so you must have $f(x)\not =x$ and thus must have have $f(f(x)) \not =f(x)$, except for the case $x=-3$ in which case you must have $f(-3)=-3$
in particular you cannot have $f(0)=f(f(0))=3$
In fact you can have $f(0)$ with any values apart from $-3$ or those in $\left\{\ldots, -\frac{45}{16},-\frac{21}{8},-\frac{9}{4},-\frac{3}{2},0,3,9,21,45,\ldots\right\}$, and you will then have $f(3)=3+2f(0)$
For a general solution to the functional equation,
*
*you partition the reals into equivalence classes with the equivalence relation $y = 3+2x \implies y R x$
*then pair these equivalence classes up, apart from $\{-3\}$ as a special case paired with itself, (there are some easy ways of doing this, and many more difficult ones)
*so $f(x)=z$ takes you from one equivalence class to its pair
*and $f(z)=f(f(x))=3+2x$ takes you back but one step up
One example solution to the functional equation is $f(x)=-6-x$ when $x\ge -3$ and $f(x)=-9-2x$ when $x\le -3$, which has $f(0)=-6$ and $f(3)=-9$
| {
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Function with positive Fourier coefficients decaying as $\frac{1}{n}$ Is there an example (I'm not looking for a sufficient or necessary condition but just for an example) of a bounded Rieman-integrable function $f\colon [-\pi,\pi]\rightarrow\mathcal{R}$ with Fourier coefficients
$c_n(f) = \int_{-\pi}^\pi{f(t)e^{-int}\,dt}$ that are positive and decay as $\frac{1}{n}$ to zero (so that the Fourier coefficients are not summable)?
From this discussion a sufficient condition would be to find a function which has Fourier coefficients decaying as $n^{-1/2}$ and I could then take the convolution of this function with itself to obtain what I want.
| No, there is no such function.
Suppose $f(t)\sim\sum_nc_ne^{int}$ where $c_n\ge0$ and $\sum c_n=\infty$. Then $f$ is not bounded (hence not Riemann integrable).
Not-quite proof: Let $t=0$: it follows that $f(0)=+\infty$.
Of course that's not quite a proof, since there's no reason a priori that the series should converge to $f(0)$ for $t=0$. It does make it clear that the answer must be no, and if you have any feeling for real analysis it seems clear that the condition $c_n\ge0$ means that it can't be hard to fix the argument.
Hint for an actual proof: Show that if $f$ is bounded then the Fejer means (or the Abel means) of the Fourier series must be uniformly bounded.
Edit: It appears that the hint was not sufficient. The argument is very simple. First, if $f$ is bounded then the Fejer means are uniformly bounded, as hinted: $\sigma_n=f*K_n$, so $$||\sigma_n||_\infty\le||f||_\infty||K_n||_1=||f||_\infty.$$
On the other hand, it's obvious that if $c_n\ge0$ and $\sum c_n=\infty$ then the Fejer means are not uniformly bounded:
$$\sigma_n(0)=\sum_j(1-|j|/n)^+c_j\ge\frac12\sum_{|j|<n/2}c_j.$$
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Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of $z$. Unfortunately, the magnitude doesn't seem to be a nice number. I feel like this might be a trigonometry question in disguise, but converting $e^{i\pi/7}=\cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right)$ hasn't been very fruitful.
| Let
$$z=\frac1{1-e^{it}}$$
where $t$ is real and $e^{it}\ne1$.
Then
$$z+\overline z=\frac1{1-e^{it}}+\frac1{1-e^{-it}}
=\frac1{1-e^{it}}+\frac{e^{it}}{e^{it}-1}=
\frac{1-e^{it}}{1-e^{it}}=1.$$
Therefore the real part of $z$ equals $1/2$.
| {
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Minimize $ \mbox{tr} ( X^T A X ) + \lambda \mbox{tr} ( X^T B ) $ subject to $ X^T X = I $ - Linear Matrix Function with Norm Equality Constraint We have the following optimization problem in tall matrix $X \in\mathbb R^{n \times k}$
$$\begin{array}{ll} \text{minimize} & \mbox{tr}(X^T A X) + \lambda \,\mbox{tr}(X^T B)\\ \text{subject to} & X^T X = I_k\end{array}$$
where $A \in \mathbb R^{n \times n}$ is symmetric and positive semidefinite, $B \in \mathbb R^{n \times k}$ and $n>k$.
What is the solution? Is there a closed-form solution?
| EDIT 1. We assume that $\lambda=1$ (change $B$ with $\lambda B$) and $A$ is only symmetric (the non-negativity has nothing to do here). $M_{n,k}$ denotes the real $n\times k$ matrices.
Let $f:X\in M_{n,k}\rightarrow tr(X^TAX)+ tr(X^TB)$. Since $Z=\{X;X^TX=I_k\}$ is compact, then the minimum of $f$ is reached in a point $X$ s.t. its derivative $Df_X(H)$ is $0$ for every $H$ in the tangent space $T_XZ$ of $Z$.
$\textbf{Proposition}.$ The $X$ that realize the minimum of $f$ are among the finite (generically) set of the solutions of the system (S):
$X^TX=I_k,X^TB=B^TX,(I_n-XX^T)(2AX+B)=0$.
$\textbf{Proof}$.
$Df_X:H\in T_XZ\rightarrow tr(2H^TAX+ H^TB)$.
Now $H\in T_XZ$ iff $H^TX=K$, a skew-symmetric matrix. Since $rank(X)=k$, $X^+=X^T$ and $H^T=KX^T+U(I_n-XX^T)$ where $U\in M_{k,n}$ is arbitrary.
Finally, the condition is: for every skew-symmetric $K$ and for every $U\in M_{k,n}$:
$tr((KX^T+U(I-XX^T))(2AX+ B))=0$. It suffices to consider the following 2 cases.
Case 1. $U=0$. then $X^T(2AX+ B)$ is symmetric, that is $X^TB$ is symmetric (generically, we obtain $k(k-1)/2$ relations).
Case 2. $K=0$. Then $(I-XX^T)(2AX+ B)=0$. $\square$
Of course, there is no closed-form solution.
EDIT 2. ** In order to gain 70% in computing time, we can diagonalize $A$.
Inded $A=PDP^T$ where $P\in O(n)$ and $D$ is diagonal. We put $Y=P^TX$; note that $Y^TY=X^TPP^TX=I_k$. On the other hand, $f(X)=g(Y)=tr(Y^TDY+Y^TP^TB)$.
Thus we may assume that $A$ is diagonal (change $B$ with $P^TB$).
** An example. When $k=3,n=5$, the number of complex solutions $X$ of the system (S) is (in the generic case) $248$, that is (S) admits, on average, $\approx 16$ real solutions that remain to be tested.
EDIT 3. Answer to @Royi . I reduce the problem to solving the matricial algebraic system (S). I tested the method using the Grobner basis theory (under Maple); that works until $n=6$. For larger $n$, you (it's your business) must use numerical methods in order to obtain -at least locally- the solutions. About that, there is a problem:
We fix a large $n$. If $k=1$, then $(S)$ admits (generically) $2n$ complex solutions that is, in average, $\sqrt{2n}$ real solutions. Then, with a good software, we can find and test all these solutions.
Unfortunately, when $k$ increases , the number of solutions of $(S)$ increases dramatically until $k=n-1$; indeed, when $k=n-1$, the number of complex solutions is $2^{2n-2}$, that is, in average, $2^{n-1}$ real solutions, what is associated to an exponential complexity. Therefore, in a first step, we must localize a region (not too large) where the function $f$ reaches its minimum.
| {
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Convergence or not of infinite series: $\sum^{\infty}_{n=1}\frac{n}{1+n^2}$ How can we prove that the series $\displaystyle \sum^{\infty}_{n=1}\frac{n}{1+n^2}$ is convergent or divergent?
Solution I try:
$$\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{1+n^2}<\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{n^2}$$
Did not know how I can solve that problem from that point.
| Another method is the integral test: $$\int_1^\infty\frac{x}{1+x^2}dx=\frac{1}{2}[\ln (1+x^2)]_1^\infty =\infty.$$(Note that $\frac{x}{1+x^2}=\frac{1}{x+1/x}$ is maximised at $x=1$, by the AM-GM inequality.)
| {
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How to solve this multiple-absolute-value equation using regions-in a number line method? How to solve this multiple-absolute-value equation using three-region number line?
I can solve it with combination of giving each absolute value a negative sign and leaving it as it is. There are four combinations. The method using regions in number line only requires three combinations instead of four. But it fails. So please help me to solve it using region-in-a number line method. (The answer should be {-6, -2/3})
| Case 1: Let $x<-2$ therefore $$|2x+4|-|3-x|=-2x-4-(3-x)=-1$$which yields to valid answer $x=-6$
Case 2: Let $-2\le x\le3$ therefore $$|2x+4|-|3-x|=2x+4-(3-x)=-1$$which yields to valid answer $x=-\dfrac{2}{3}$
Case 3: Let $x>3$ therefore $$|2x+4|-|3-x|=2x+4+(3-x)=-1$$which yields to invalid answer $x=-8$
| {
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Minimizing a sum of decaying exponentials subject to constraints Consider the following problem
\begin{align*}
\min_{x}\quad\sum_i\sum_j&\exp(-a_{ij}x_{ij})\\
\text{subject to}\quad\sum_i\sum_j&x_{ij}=n\\
&x_{ij}\ge0\,\,\forall i,j
\end{align*}
where $x\in\mathbb{R}^{I\times J}$ is the optimization variable and each $a_{ij}>0$, $i=1,\ldots,I$, $j=1,\ldots,J$, and $n\in\mathbb{N}$ are given. The optimal solution is claimed to be (from a paper) $x_{ij}^* = n\frac{1/a_{ij}}{\sum_i\sum_j1/a_{ij}}$ -- I am trying to prove this.
My attempt:
Since the problem is convex, the KKT conditions are necessary and sufficient. First write the problem as
\begin{align*}
\min_{x}\quad\sum_i\sum_j&\exp(-a_{ij}x_{ij})\\
\text{subject to}\quad\sum_i\sum_j&x_{ij}-n=0\\
&-x_{ij}\le0\,\,\forall i,j
\end{align*}
and associate dual variable $\lambda$ with the first constraint and $\mu_{ij}$ with each of the next $IJ$ constraints. Following the definition, the KKT conditions for the above problem are:
*
*Stationarity:
\begin{align*}
a_{ij}\exp(-a_{ij}x_{ij}^*) = \lambda-\mu_{ij}\,\,\forall i,j
\end{align*}
(as pointed out by David M. in the comments/chat, since the gradients of $-x_{ij}\le0$ form a basis, the stationarity condition takes this form)
*Primal feasibility:
\begin{align*}
\sum_i\sum_jx_{ij}^*-n&=0\\
-x_{ij}^*&\le0\,\,\forall i,j
\end{align*}
*Dual feasibility:
\begin{align*}
\mu_{ij}\ge0\,\,\forall i,j
\end{align*}
*Complementary slackness:
\begin{align*}
\mu_{ij}x_{ij}^*=0\,\,\forall i,j
\end{align*}
Using the stationarity condition, I obtain
\begin{align*}
x_{ij}^* = -\frac{1}{a_{ij}}\log\left(\frac{\lambda - \mu_{ij}}{a_{ij}}\right)
\end{align*}
To use the primal feasibility condition, sum
\begin{align*}
\sum_i\sum_jx_{ij}^* &= -\sum_i\sum_j\frac{1}{a_{ij}}\log\left(\frac{\lambda - \mu_{ij}}{a_{ij}}\right)\\
&= -\sum_i\sum_j\frac{1}{a_{ij}}[\log(\lambda - \mu_{ij})-\log(a_{ij})]\\
&= -\sum_i\sum_j\frac{\log(\lambda - \mu_{ij})}{a_{ij}}+\sum_i\sum_j\frac{\log(a_{ij})}{a_{ij}}
\end{align*}
so primal feasibility states that
\begin{align*}
-\sum_i\sum_j\frac{\log(\lambda - \mu_{ij})}{a_{ij}}+\sum_i\sum_j\frac{\log(a_{ij})}{a_{ij}} = n
\end{align*}
If we can show that all $x_{ij}^*$ are nonzero (is this true?) then $\mu_{ij}=0$ by complementary slackness and the above equations can be solved to give
\begin{align*}
x_{ij}^* = \frac{1}{a_{ij}}\left(\log(a_{ij})+\frac{n-\sum_k\sum_l\frac{\log(a_{kl})}{a_{kl}}}{\sum_k\sum_l\frac{1}{a_{kl}}}\right)
\end{align*}
This seems to satisfy the KKT conditions, so it appears the original claim of the optimal solution is incorrect.
Question/Update:
Thanks to Alex, the proposed solution was shown to be incorrect. I've derived what I think is that optimal (but I'm questioning if $x_{ij}^*$ is indeed positive). Thanks to all for the help.
| Note, that since you are not asked to find the optimal solution, but only prove that the given one is optimal, you don't need to solve the KKT system. You only need to substitute the given solution and show that there exist $\lambda, \mu$ such that the the KKT system holds. From convexity and Slater any point is optimal if and only if this point satisfies KKT.
First, note that primal feasibility indeed holds - all the given $x_{ij}^*$ sum to $n$, and they are all non-negative.
From complementarity slackness, since all $x_{ij}^*$ are nonzero, you get that $\mu_{ij} = 0$.
Substituting $\mu_{ij} = 0$ into stationarity, we get
$$
a_{ij} exp(-a_{ij} x_{ij}^*) = \lambda
$$
Substituting the given $x_{ij}^*$ we get
$$
a_{ij} exp \left (-\frac{n}{\sum_i \sum_j \frac{1}{a_{ij}}} \right) = \lambda
$$
And whoops - we get a contradiction! The equation above can be true only if all $a_{ij}$ are equal! Otherwise, it cannot be true!
So unless I made some algebraic mistake here, the point you think is optimal cannot be optimal in general.
| {
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What is the relationship between different theorems all called Hilbert's Nullstellensatz? The following statements are all named the Hilbert's Nullstellensatz, but they appear at first to be completely unrelated to each other. What is the relationship between them exactly?
*
*(Theorem 1.3A on page 4 of Hartshorne's Algebraic Geometry) Let $k$ be an algebraically closed field, $a$ be an ideal in $A = k[x_1, ... x_n]$, and $f \in A$ be a polynomial which vanishes on all points of $Z(a)$, then $f^r \in a$ for some integer $r > 0$
*("weak Nullstellensatz" listed on Wikipedia) An ideal $I \subseteq k[x_1, ... x_n]$ contain 1 iff the polynomials in $I$ do not contain any common zeroes in $k^n$
*(Theorem 3.2.4 on page 107 of Vakil's notes) The only maximal ideals in the ring $k[x_1, ... x_n]$ are of the form $(x_1 - a_1, ... x_n - a_n)$, for $(a_1, ... a_n) \in k^n$
*(Theorem 3.2.5 on page 107 of Vakil's notes) If $k$ is any field, every maximal ideal of $k[x_1, ... x_n]$ has residue field a finite extension of $k$. Any field extension of $k$ that is finitely generated as a ring is also finitely generated as a module
*(Theorem 3.7.1 on page 128 of Vakil's notes) Let $A$ be a commutative ring with identity, then $V(.)$ and $I(.)$ give an inclusion reversing bijection between closed subsets of $Spec(A)$ and radical ideals of $A$
| Number 4. is sometimes called Zariski's lemma. The relation to Nullstellensatz is that if $k$ is algebraically closed, then the residue field is a finite algebraic extension of $k$, which must be $k$ itself. From this follows 3.; it is not hard to see that $k[x_1, \dots, x_n]/ \mathfrak{m} = k$ implies that $\mathfrak{m}$ is of the form $(x_1 -a_1, \dots, x_n - a_n)$. Therefore if $k$ is closed, then $k[x_1, \dots, x_n]/\mathfrak{m}= k$ for any maximal ideal (by 4), and hence all maximal ideals are of this form.
Item 2. is an immediate corrollary. You know that $V(\cdot)$ and $I(\cdot)$ are order reversing, and every ideal is contained in a maximal ideal. Clearly $V(\mathfrak{m})$ when $\mathfrak{m} = (x_1 -a_1, \dots, x_n - a_n)$ is maximal contains a point, and so $V(\mathfrak{A})$ for any other ideal contains at least a point. Thus if $V(I)$ is empty, then $I$ contains $(1)$. The reverse is obvious. Items 2,3,4 are variously/collectively called the weak Nullstellensatz by different authors.
The first one is proved from 2, 3, 4 by algebraic trickery here and is typically called the strong Nullstellensatz.
| {
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