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H: Banach space, Normed vector space
Help me please with this question.
Let's $Y$ be Banach space, $Z$ - Normed vector space and $(T_{n})_{\mathbb{N}}$ - the sequence in $B(Y,Z)$ so that all sequence $(y_{n})_{\mathbb{N}}$ in Y holds:
if $\left \| y_n \right \|_{n \to \infty }\rightarrow 0$ then $\left \| T_ny_n \right \|_{n \to \infty }\rightarrow 0$.
Prove that: $ \underset{\mathbb{N}}{\sup}\left \| T_n \right \|< \infty $
Thanks!
AI: Suppose not: then we can find a subsequence $\{T_{n_k}\}$ such that $\lVert T_{n_k}\rVert\geq 2k$. Pick $x_{n_k}$ of norm $1$ such that $\lVert T_{n_k}(x_{n_k})\rVert\geq \frac{\lVert T_{n_k}\rVert}2$. Then the sequence $\{y_{n_k}:=\frac 1{\lVert T_{n_k}\rVert}x_{n_k}\}$ converges to $0$. But $$\lVert Ty_{n_k}\rVert =\lVert T_{n_k}\left(\frac 1{\lVert T_{n_k}\rVert}x_{n_k}\right)\rVert=\frac 1{\lVert T_{n_k}\rVert}\lVert T_{n_k}(x_{n_k})\rVert\geq\frac 12,$$
a contradiction.
Note that we don't need $Y$ to be a Banach space. |
H: Continuity of Multivariable function
While doing revision I stumbled upon this problem:
Is this function continuous at the origin?
$$
f_3(x,y) =
\begin{cases}
\frac{x^3-y^3}{x^2+y^2}, & \text{if }(x,y)\not= (0,0) \\
0, & \text{if } (x,y)=(0,0) \end{cases}
$$
The answer is yes, but how do I prove it?
Sincere thanks for help.
AI: For every $u=(x,y) \ne 0=(0,0)$ we have
$$
|f_3(0)-f_3(u)|=|f_3(u)| \le \frac{|x|^3+|y|^3}{|u|_2^2}
\le \frac{2|u|_2^3}{|u|_2^2}=2|u|_2.
$$
For every $\epsilon>0$, if $|u|_2<\epsilon/2$ then $|f_3(0)-f_3(u)|<\epsilon$, i.e.
$f_3(u) \to f_3(0)=0$ as $u \to 0$. |
H: a question on distributions.
Suppose $L\in \mathcal{S}$ is a tempered distribution, and for each $\phi \in \mathcal{S} :\phi \geq 0 \implies L(\phi) \geq 0$.
Prove that there exists a borel measure $\mu$ with polynomial growth such that $L=L_{\mu}$, where $L_{\mu}(\phi) = \int \phi \:d\mu$.
I think I need to use here Riesz Representation Theorem, but not sure how.
Any hints?
Thanks.
AI: 1. Proof of the existence of Borel measure. Consider restriction $\tilde{L}$ of $L$ on the space of smooth compactly supported functions $C_c^\infty(\mathbb{R}^p)$. This is a positive linear functional, hence using ideas of this answer one can show that $\tilde{L}$ is continuous. Since $C_c^\infty(\mathbb{R}^p)$ is dense in the space of continuous compactly supported functions $C_c(\mathbb{R}^p)$ with respect to the $\sup$-norm, then there exist unique continous linear extension $\hat{L}$ of $\tilde{L}$ to the space $C_c(\mathbb{R}^p)$.
Now we want to prove that $\hat{L}$ is positive. Take arbitrary non-negative function $f\in C_c(\mathbb{R}^p)$, and consider its convolutions with molifiers $\omega_{1/n}$, i.e. consider functions $f_n=f*\omega_{1/n}$. It is known that they are smooth, and also they are compactly supported becase $f$ is compactly supported. Since $f$ is non-negatie, then does $f_n$. Since $f$ is compactly supported and continuous, then $f_n$ converges uniformly to $f$, i.e. in $\sup$-norm. Since $\hat{L}$ is the continuous extension of $\tilde{L}$, then
$$
\hat{L}(f)=\hat{L}\left(\lim\limits_{n\to\infty} f_n\right)=\lim\limits_{n\to\infty} \hat{L}(f_n)=\lim\limits_{n\to\infty} \tilde{L}(f_n)\geq 0
$$
Since $f\in C_c(\mathbb{R}^p)$ is arbitrary, $\hat{L}$ is positive. Then by Reisz representation theorem there exist Borel regular measure $\mu$ such that
$$
\hat{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in C_c(\mathbb{R}^p)
$$
In particular
$$
\tilde{L}(f)=\hat{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in C_c^\infty(\mathbb{R}^p)
$$
Since $\tilde{L}$ is continuous with respect to $\sup$-norm of $C_c^\infty(\mathbb{R}^p)$, then it is continuous with respect to the locally convex topology in $C_c^\infty(\mathbb{R}^p)$ coming from $S(\mathbb{R}^p)\supset C_c^\infty(\mathbb{R}^p)$.
Note that locally convex space $C_c^\infty(\mathbb{R}^p)$ is dense in $S(\mathbb{R}^p)$, so you can uniquely extend $\tilde{L}$ to the continuous linear functional $\overline{L}$ on the whole space $S(\mathbb{R}^p)$. Now we want to prove that $\overline{L}$ acts on the functions as integral. Consider arbitrary $f\in S(\mathbb{R}^p)$. Since $C_c(\mathbb{R}^p)$ is dense in $S(\mathbb{R}^p)$ there exist sequence $\{f_n:n\in\mathbb{N}\}$ of compactly supported functions that uniformly converges to $f$, i.e. converges in $\sup$-norm. Since $\hat{L}$ is the continuous extension of $\tilde{L}$, then using dominated convergence theorem we get
$$
\overline{L}(f)=\overline{L}(\lim\limits_{n\to\infty} f_n)=
\lim\limits_{n\to\infty}\overline{L}(f_n)=
\lim\limits_{n\to\infty}\tilde{L}(f_n)=
$$
$$
\lim\limits_{n\to\infty}\int\limits_{\mathbb{R}^p} f_n(x)d\mu(x)=
\int\limits_{\mathbb{R}^p} \lim\limits_{n\to\infty}f_n(x)d\mu(x)=
\int\limits_{\mathbb{R}^p} f(x)d\mu(x)
$$
Thus,
$$
\overline{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
By definition $\tilde{L}=L|_{C_c(\mathbb{R}^p)}$ and by construction $\tilde{L}=\overline{L}|_{C_c(\mathbb{R}^p)}$, then from the uniqueness we conclude that $L=\overline{L}$, i.e.
$$
L(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
2. Proof of polynomial gowth property. Now we proceed to the proof that $\mu$ is of polynomial growth. Since $L$ is a continuous functional on $S$, then it is continuous with respect to the semi-norm of Schwartz space
$$
p_{\alpha,\beta}(f)=\sup\limits_{|\alpha|\leq k}\sup\limits_{x\in\mathbb{R}^p}|x^\alpha(\partial^\beta f)(x)|\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
for some $k\in\mathbb{Z}_+$. Hence for some $C>0$ we have
$$
|L(f)|\leq C \max\limits_{i=1,\ldots,m} p_{\alpha_i,\beta_i}(f)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
Consider sequence of smooth functions $\{\eta_n:n\in\mathbb{N}\}\subset S(\mathbb{R}^n)$ such that
for all $n\in\mathbb{N}$ function $\eta_n$ is compactly supported
for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ we have $0\leq\eta_n(x)\leq 1$.
for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ such that $|x|<n$ we have $\eta_n(x)=1$.
for all $n\in\mathbb{N}$ we have $\max\limits_{i=1,\ldots,m} p_{\alpha_i,\beta_i}((1+|x|^2)^{-\alpha}\eta_n)\leq 1$
Speaking informally $\eta_n$ is a smooth analogue of the characteritic functions of the ball $B(0,n)$. Paragraph 4 means that derivatives of $\omega_n$ varies
very slowly. The proof of the existence is very messy, but if you need it I'll write it later.
Then define $g_n=(1+|x|^2)^{-\alpha}\eta_n$. From paragraph 1 we see that $\{g_n:n\in\mathbb{N}\}\subset C_c(\mathbb{R}^p)\subset S(\mathbb{R^p})$. From paragraph 4 for all $n\in\mathbb{N}$ we obtain
$$
\left|\int\limits_{\mathbb{R}^p} \eta_n(x)\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|=|L(g_n)|\leq C p_{\alpha,\beta}(g_n)=C p_{\alpha,\beta}((1+|x|^2)^{-\alpha}\eta_n)\leq C
$$
Paragraph 2 and inequality above allows us to apply dominated convergence theorem. Then we get
$$
\left|\int\limits_{\mathbb{R}^p}\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|=
\left|\int\limits_{\mathbb{R}^p}\lim\limits_{n\to\infty} \eta_n(x)\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|
$$
$$
\lim\limits_{n\to\infty}\left|\int\limits_{\mathbb{R}^p} \eta_n(x)\frac{d\mu(x)}
{(1+|x|^2)^\alpha}\right|\leq C
$$
Thus measure $\mu$ is of polynomial growth. |
H: Characteristic equation for 2-nd order ODE
Given a differential equation $\dot x = Ax$, $x \in \mathbb{R}^n$ we define its characteristic equation as $\chi(\lambda) = \det (\lambda I - A)$. Consider now the second order ODE
$$
\ddot x + A x + B \dot x = 0, \;\;\; x \in \mathbb{R}^n
$$
With substitution $u = x$, $v = \dot x$ we can rewrite this ODE as a system
$$
\left\{\begin{array}{rcl}
\dot u & = & v \\
\dot v & = & -Au - Bv.
\end{array}\right.
$$
This is an ODE with matrix
$$
\begin{bmatrix}
0 & I \\
-A & -B
\end{bmatrix}
$$
and hence with characteristic equation
$$
\chi(\lambda) = \det \begin{bmatrix} \lambda I & - I \\ A & \lambda
I+B\end{bmatrix}
.
$$
I know that there is a representation of such determinant as a determinant of $n \times n$ matrix with $a_{ii} + \lambda^2$ on the diagonal. How to obtain such representation? How to find off-diagonal items explicitly?
AI: We have for block matrices, with $A$ invertible, that
$$\pmatrix{A&B\\ C&D}=\pmatrix{A&0\\ C&I}\cdot\pmatrix{I&A^{-1}B\\0&D-CA^{-1}B}.$$
In our case, when $\lambda\neq 0$, we get
$$\chi(\lambda)=\lambda^n\det\left(\lambda I+B-A\frac 1{\lambda}I(-I)\right)=\lambda^n\det(\lambda I+B+\frac 1{\lambda} A)=\det(\lambda^2I+\lambda B+A).$$ |
H: How can an ordinary function be a distribution?
I think distributions are linear and continuous functionals on the set of testfunctions. In a textbook I saw this question:
Let $f$ be a $2\pi$-periodic function with $f(t) = \frac{\pi}{4}|t|$
for $t \in [-\pi, \pi]$, show that it is a distribution!
How could this be, its not even a functional?
AI: Locally integrable functions can be indentified to distributions in the sense that if $f$ is locally integrable one associates to it the linear functional $L_f$ defined by $L_f(\phi)=\int f\phi$ for every test function $\phi$. |
H: Slope of curve in $\mathbb{R}^3$
While doing revision, I came across this problem:
The surface given by $z=x^2-y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. Find the slope of this curve at the point $(1,3,-8)$.
I tried substituting $y=3x$ into $z=x^2-y^2$, yielding $z=-8x^2$. Then, $\frac{dz}{dx}=-16x=-16$.
However the answer is $-8\sqrt{\frac{2}{5}}$.
Thank you very much for any help.
AI: This is a very badly posed question, and does not have an answer. (Read the comments.)
The following is a solution to a rephrased question which can be answered, however.
Question:
The surface given by $z=x^2−y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane.
Treating the intersection as a curve in the said plane with vertical axis along $(0,0,1)$ and horizontal axis along $(1,3,0)/\sqrt{10}$, find the slope of this curve at the point (1,3,−8).
Solution:
Any point on the plane has Cartesian coordinates in the form $$\frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix}.$$ Substituting this into $z=x^2−y^2$, we get $b = -4a^2/5$.
So the "slope" at a point on this intersection, with $a$ and $b$ given, is $$\frac{db}{da} = -\frac{8a}{5}.$$
Setting $$\begin{pmatrix}1 \\3\\-8\end{pmatrix} = \frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix},$$ we get $a = \sqrt{10}$ and so the "slope" at this point is $-8\sqrt{\frac{2}{5}}$.
Solution using grad:
Let $f:=x^2-y^2-z$ and $g:=y-3x$. At the point $(1,3,-8)$, $\nabla f=(2,-6,-1)$ and $\nabla g=(-3,1,0)$. Their cross product, $(1,3,-16)$, is along the tangent direction of the intersecting curve produced by the surface and the plane, at the point $(1,3,-8)$.
Denote the angle between $(1,3,-16)$ and $(1,3,0)$ (i.e. the "horizontal") by $\theta$. Then, using the dot product, $\cos\theta = \sqrt{\frac{5}{133}}$. The "slope" is $$\tan \theta = - \sqrt{\frac{1}{\cos^2 \theta}-1} = -8\sqrt{\frac{2}{5}},$$ where the negative square root is taken because the "vertical" is along $(0,0,1)$. |
H: Does this sequence converge to $\pi$?
I have a problem with the following sequence
$$ \lim_{n \to \infty} g_n \stackrel{?}{=} \pi $$
where
$$g_n = \sum_{k=1}^{n-1} \frac{\sqrt{\frac{2n}{k}-1}}{n-k}
+ \sum_{k=n+1}^{2n-1}\frac{\sqrt{\frac{2n}{k}-1}}{n-k}.$$
Does it converge to $\pi$? I tested experimentally that it
does, but I was unable to prove it by hand. Could anybody help,
or offer some methods of approach?
AI: Setting $i=n-k$ and $j=k-n$ we have
\begin{eqnarray}
g_n&=&
\sum_{i=1}^{n-1}\frac{1}{i}\sqrt{\frac{2n-n+i}{n-i}}-\sum_{j=1}^{n-1}\frac{1}{j}\sqrt{\frac{2n-n-j}{n+j}}\cr
&=&\sum_{i=1}^{n-1}\frac{1}{i}\sqrt{\frac{n+i}{n-i}}-\sum_{j=1}^{n-1}\frac{1}{j}\sqrt{\frac{n-j}{n+j}}\cr
&=&\sum_{k=1}^{n-1}\frac{1}{k}\left(\sqrt{\frac{n+k}{n-k}}-\sqrt{\frac{n-k}{n+k}}\right)\cr
&=&\sum_{k=1}^{n-1}\frac{1}{k}\frac{(n+k)-(n-k)}{\sqrt{n^2-k^2}}\cr
&=&2\sum_{k=1}^{n-1}\frac{1}{\sqrt{n^2-k^2}}=\frac{2}{n}\sum_{k=1}^{n-1}\frac{1}{\sqrt{1-(k/n)^2}}=-\frac{2}{n}+\frac{2}{n}\sum_{k=0}^{n-1}f(k/n),
\end{eqnarray}
with $f(x)=1/\sqrt{1-x^2}$.
Therefore
$$
\lim_{n\to \infty}g_n=2\int_0^1f(x)dx=2\int_0^{\pi/2}\frac{\cos t}{\sqrt{1-\sin^2t}}dt=2\int_0^{\pi/2}dt=\pi.
$$ |
H: Solving $\sin (5\phi)-\sin \phi=\sin (2\phi)$
The question is :
Find the general solution of this equation
$$
\sin (5\phi)-\sin \phi=\sin (2\phi)
$$
I tried to expand $\sin (5\phi)$ and $\sin (2\phi)$,
so the equation only contains $\cos\phi$ and $\sin \phi$.
But I can't make it into a form like $\sin(\phi+a)=n$
to get the general solution.
There's another question like this :
Find the general solution of this equation
$$
\sin 2x+\sin 4x=\cos 2x+\cos 4x
$$
AI: Just for your request from avtar:
$\sin(n\theta)=\binom{n}{1}\cos^{n-1}(\theta)\sin(\theta)-\binom{n}{3}\cos^{n-3}(\theta)\sin^3(\theta)+...$
Now put $n=5$. |
H: Accumulation points / Cluster points / Closed sets
In a topological space $X$, call $x\in X$ an accumulation point if $\forall$ open set $U\ni x$, $U \cap A \neq \emptyset$, and $y\in X$ a cluster point if $\forall$ open set $U\ni y$, $U\cap A\setminus \{y\} \neq \emptyset$. (These are the terminologies used by my lecturer. I'm aware that different ones exist.)
Call a set $A\subseteq X$ closed if its complement is open.
My lecturer gave us a proof that $A$ is closed iff $A$ contains all of its accumulation points (see below). However, I managed to modify it to show that $A$ is closed iff $A$ contains all of its cluster points (see below, marked with []). What went wrong here? If the latter is false in general, in what special cases is it true (I heard it's true in metric spaces)?
The proof:
($\Rightarrow$): Suppose $A$ is closed and $x_0 \in X \setminus A$. Take $U:= X\setminus A$, an open set containing $x_0$. Now $U\cap A =\emptyset$, so $x_0$ is not an accumulation point.
[$x_0$ is not an accumulation point and so it is not a cluster point either.]
($\Leftarrow$): Suppose $A$ is not closed, then $X\setminus A$ is not open. $\exists x_0 \in X\setminus A$ such that no open set $U\ni x_0$ is contained in $X\setminus A$, i.e. any open set $U\ni x_0$ satisfies $U\cap A \neq \emptyset$. So $x_0$ is an accumulation point of $A$ but not in $A$.
[For this $x_0$, note that $x_0 \notin U\cap A$ because $x_0 \notin A$. So any open set $U\ni x_0$ satisfies $U\cap A \setminus \{x_0\} \neq \emptyset$, i.e. $x_0$ is a cluster point of $A$ but not in $A$.]
AI: Your result is correct, as is your argument. You can even prove directly that if $A$ contains all of its cluster points, then it contains all of its accumulation points. Suppose that a set $A$ contains all of its cluster points but fails to contain its accumulation point $x$. Then $x$ is not a cluster point, so $x$ has an open nbhd $U$ such that $U\cap A\subseteq\{x\}$. But $x\notin A$, so $U\cap A=\varnothing$, contradicting the assumption that $x$ was an accumulation point of $A$.
Added: Your lecturer could have proved a stronger result. Let $\operatorname{cl}A$ be the set of accumulation points of $A$; then $A$ is closed iff $A=\operatorname{cl}A$. Suppose first that $A$ is closed. You’ve already proved that $A\supseteq\operatorname{cl}A$, and it’s clear that every point of $A$ is an accumulation point of $A$, so $A=\operatorname{cl}A$. Conversely, if $A$ is not closed, you already know that it fails to contain some accumulation point, so $A\ne\operatorname{cl}A$.
This stronger result fails for cluster points. Let $X$ be any $T_1$-space with at least two points, and let $x\in X$. Then $\{x\}$ is closed, but it has no cluster points, so it can’t be equal to the set of its cluster points. |
H: Prove that $K$ is a field
Let be $K$ the set of real numbers that can be written as $a+b\sqrt2$, with $a$ and $b$ rational numbers. Prove that $K$ is a field.
I have already proved that $0$ and $1$ $\in K$, and that sum and product of two elements $\in K$. I have also already proved that the opposite $\in K$. I don't know how to prove that the reciprocal $\in K$. Can you help me?
AI: Let $a+b\sqrt{2}\in K$ and $a+b\sqrt{2}\neq 0$. If $b=0$, then $a+b\sqrt{2}=a\neq 0$ and $1/a$ is its multiplicative inverse. Therefore, we assume $b\neq 0$, which implies that
$a^2-2b^2\neq 0$. Otherwise, if $a^2-2b^2=0$, then $\sqrt{2}=\frac{a}{b}$ which is rational since $a, b$ are rational, which contradicts to the fact that $\sqrt{2}$ is irrational.
Define
$$\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}.$$
First note that it is well-defined because $a^2-2b^2\neq 0$. Moreover, it belongs to $K$ because $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are rational which follows from the fact that $a,b$ are rational numbers. Finally, we have
$$(a+b\sqrt{2})\cdot\left(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\right)=\frac{(a+b\sqrt{2})(a-b\sqrt{2})}{a^2-2b^2}=\frac{a^2-2b^2}{a^2-2b^2}=1.$$
That is to say, $\displaystyle\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$ is the multiplicative inverse of $a+b\sqrt{2}$. |
H: Logic question proving something about compactness
Let $\Sigma$ be a set of formulas. There's a finite set $\Lambda \subseteq \Sigma$.
I'm asked to prove or disprove that $\Sigma$ has a model if and only if $\Lambda$ has a model.
It seems to me it's just true using compactness, but that sounds too easy.
AI: It is not enough that there is a finite $\Lambda$ that has a model. In order for compactness to be applicable (or for the claim to be true at all), you need to assume that for every possible finite $\Lambda$ it has a model.
On the other hand, if you rewrite the assumption in that way, what you're being asked to prove is just compactness theorem. And then you shouldn't be using that theorem itself in the proof.
On the other hand, it looks like the exercise also allows for the solution to be a disproof -- so if you're sure you have reproduced the claim correctly, you should be looking for a counterexample instead. (If so, hint: The empty set is finite, and the empty theory has a model). |
H: Discontinuous optimizer but continuous optimal
Consider a locally-bounded, continuous, positive-semidefinite function $f: X \times Y \rightarrow \mathbb{R}_{\geq 0}$, where $X \subset \mathbb{R}^n$ is compact, $Y \subseteq \mathbb{R}^m$.
For each $y \in Y$, define:
$$ f^*(y) := \min_{x \in X} f(x,y) $$
$$ x^*(y) := \arg\min_{x \in X} f(x,y) $$
Assume that the optimal & optimizer always exist and are finite.
It is known that, under these conditions, $f^*$ is a continuous function.
1) Provide an example in which $x^*$ is not continuous (while $f^*$ necessarily does).
2) Provide an example in which $x^*$ is not continuous outside the origin.
AI: It depend on what you mean by "the $\arg\min$ exists".
When $x^*$ is not unique, it can indeed be discontinuous: see $f(x,y)=1+xy$ with $X=Y=[-1,1]$, then $x^*(y)=-\operatorname{sgn} y$ when $y\ne 0$.
But when $x^*$ is unique, it is necessarily continuous. If $x^*(y)$ does not have a limit as $y\to y_0$, by compacity of $X$ we can find two points $x_1\ne x_2$ such that $x^*(y_i^{(n)})\to x_i$ and $y_i^{(n)}\to y_0$. Since $f^*(y_1^{(n)})=f\left(x^*(y_1^{(n)}),y_1^{(n)}\right)$, by continuity of $f$ and $f^*$, $f^*(y_0)=f(x_1,y_0)$. Likewise $f^*(y_0)=f(x_2,y_0)$. This means the $\arg\min$ can be any of $x_1$ or $x_2$, contradicting uniqueness of $x^*$. |
H: Intuition for the following change of index of summation
I am working through Concrete Mathematics. I came across the following change in index of summation while going through the number theory chapter.
$$\sum_{m|n}^{ } \sum_{k|m}{ } a_{k,m} = \sum_{k|n}^{ } \sum_{l|(n/k)} a_{k,kl}$$
If I list out the complete summation mechanically I can verify that both the sides are the same. The Left hand side is (I think) for every divisor $m$ of $n$, you have $k$ running through the divisors of $n$ less or equal to $m$. But I can't find some interpretation of the RHS. So I want to know how the right hand side is rearranging the terms.
AI: On both sides you’re taking a sum of $a_{k,m}$ over all pairs $\langle k,m\rangle$ such that $k\mid m$ and $m\mid n$. The first summation chooses the middle element first and then sums over its divisors; the second chooses the smallest element first and then sums over the possible intermediate elements. For a fixed $k$ dividing $n$, as $l$ runs over divisors of $\frac{n}k$ in the inner summation on the righthand side, $kl$ runs over all $m$ such that $k\mid m$ and $m\mid n$. Both are thus equal to
$$\sum_{k\mid m\text{ and }m\mid n}a_{k,m}\;.$$ |
H: Closed form expression for the k-th term of a sequence.
Let
$$
x_{n+1} = \left\{ \begin{array}{c} x_{n}^2 & \mbox{if $b_{n+1}=1$} \\ \alpha x_n & \mbox{if $b_{n+1}=0$}\end{array} \right. ,
$$
for $n\ge 0$ and $\alpha > 1$. Can we write $x_k$ in terms of $x_0$ and $P$ where
$$
P = \sum_{i=1}^{k} b_i.
$$
For $\alpha=1$,
$$
x_k = x_0^{\left( 2^P \right)}.
$$
I am unable to figure out the expression for any $\alpha>1$.
AI: No. Let $\alpha > 1$ and $x_0 \ne 0$. For $b_1 = 0, b_2 = 1$ you have $x_2 = \alpha^2 x_0^2$ and for $b_1 = 1, b_2 = 0$ you have $x_2 = \alpha x_0^2$. These aren't equal, so $x_2$ doesn't depend on $x_0$ and $P = b_1 + b_2$ alone. |
H: how to rotate a Gaussian?
Lets suppose that we have a 2D Gaussian with zero mean and one covariance and the equation looks as follows
$$f(x,y) = e^{-(x^2+y^2)}$$
If we want to rotate in by an angle $\theta$, does it mean that we rotate the values $x$ and $y$ and then see how the Gaussian is rotated or do we actually rotate the graph of the function.
How this rotation actually be computed analytically and how the graph would look like. Is there any intuitive way of understanding.
How do we explain rotating a general function analytically and geometrically?
Thanks a lot
AI: If the covariance is the $2\times2$ identity matrix, then the density is
$$e^{−(x^2+y^2)/2}$$
multiplied by a suitable normalizing constant. If $\begin{bmatrix} X \\ Y \end{bmatrix}$ is a random vector with this distribution, then you rotate that random vector by multiplying on the left by a typical $2\times 2$ orthogonal matrix:
$$
G \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} X \\ Y \end{bmatrix}.
$$
If the question is how to "rotate" the probability distribution, then asnwer is that it's invariant under rotations about the origin since it depends on $x$ and $y$ only through the distance $\sqrt{x^2+y^2}$ from the origin to $(x,y)$.
If you multiply on the left by a $k\times2$ matrix $G$, you have
$$
\mathbb{E}\left(G\begin{bmatrix} X \\ Y \end{bmatrix}\right) = G\mathbb{E}\begin{bmatrix} X \\ Y \end{bmatrix}
$$
and
$$
\operatorname{var}\left( G \begin{bmatrix} X \\ Y \end{bmatrix} \right) = G\left(\operatorname{var}\begin{bmatrix} X \\ Y \end{bmatrix}\right)G^T,
$$
a $k\times k$ matrix. If the variance in the middle is the $2\times2$ identity matrix and $G$ is the $2\times 2$ orthogonal matrix given above, then it's easy to see that the variance is
$$
GG^T
$$
and that is just the $2\times 2$ identity matrix. The only fact you need after that is that if you multiply a multivariate normal random vector by a matrix, what you get is still multivariate normal. I'll leave the proof of that as an exercise. |
H: If a function has a finite limit at infinity, does that imply its derivative goes to zero?
I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L < \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?
It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:
$$
\begin{align}
\lim_{x \to +\infty} f'(x) &= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\
&= \lim_{h \to 0} \frac1{h}(L-L) \\
&= \lim_{h \to 0} \frac{0}{h} \\
&= 0
\end{align}
$$
In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is a similar question already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.
AI: The answer is: No. Consider $f(x)=x^{-1}\sin(x^3)$ on $x\gt0$. The derivative $f'(x)$ oscillates between roughly $+3x$ and $-3x$ hence $\liminf\limits_{x\to+\infty}\,f'(x)=-\infty$ and $\limsup\limits_{x\to+\infty}\,f'(x)=+\infty$. |
H: Vector Identity and the Scalar Product
Ok, I have a question that probably has a very simple answer but for some reason I can't see it. Let $a$ and $r$ be two vectors of nonzero length with a common origin and let $\theta$ be the nonzero angle between them. Then, by definition of the cosine funtion,
$$
\cos \theta = \frac{|a|}{|r|}
$$
where $|\cdot|$ denotes the norm. On the other hand, the scalar product is given by
$$
\langle a, r \rangle = |a|\cdot |r| \cos \theta.
$$
Putting these facts together we have
$$
\langle a, r \rangle = |a|\cdot |r| \cdot \frac{|a|}{|r|} = |a|\cdot |a| = |a|^2 = \langle a, a \rangle
$$
which is a result that is independent of $r$ and thus makes no sense. What is my error?
AI: Your error is that
$$\cos(\theta)=\frac{|a|}{|r|}$$
is not correct except when $a$ and $r$ are a leg and the hypotenuse of a right triangle, respectively, which is not the case in general. |
H: About Kirby Diagrams
I'm reading R.E. Gompf and A.I. Stipsicz, 4-Manifolds and Kirby Calculus. There is something I don't understand on page 116 (Google Books link to page 116; alternatively, here are images of page 115 and page 116)
Now, we consider compact 4-manifolds, and they have handle decomposition.
0-handle and $m$ 1-handles form $X_{1}$, which is diffeomorphic to $\natural m S^1 \times D^3$. Let $X_{2}$ be $X_{1}\cup $ 2-handles. I understand those facts, but I can't understand
$$\partial X_{2}=\partial(\natural m S^1 \times D^3).$$
If you can help me, then please teach me the reason for this.
AI: The idea is to use duality to notice that the union of $3$ and $4$ handles must be isomorphic to some handlebody $\natural nS^1\times D^3$. (Here $m$ doesn't have to equal $n$. If you look at the book carefully, it uses both $n$ and $m$.) Recall that a $k$-handle can be thought of as an $(n-k)$-handle if you read the decomposition backward. (This is the dual decomposition.) Thus the $3$ and $4$ handles represent $0$ and $1$-handles in the dual decomposition. The union of the rest of the handles must attach to the boundary of this, so their boundary must equal $\partial(\natural nS^1\times D^3)$. |
H: If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $
For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
AI: Either $\frac{1}{2a}\leq \frac{1}{2b}$, in which case
$$\frac{1}{b}=\frac{1}{2b}+\frac{1}{2b}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2a}+\frac{1}{2a}=\frac{1}{a},$$
or $\frac{1}{2b}\leq \frac{1}{2a}$, in which case
$$\frac{1}{a}=\frac{1}{2a}+\frac{1}{2a}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2b}+\frac{1}{2b}=\frac{1}{b}.$$
Now take reciprocals. |
H: Product of varieties
If we have two rational varieties (i.e varieties which are birational to some projective space) is their product also a rational variety? would this rely on the fact that the Zariski topology is finer than the product topology and that $\mathbb{P}^{r} \times \mathbb{P}^{s}$ is birational to $\mathbb{P}^{r+s}$?
AI: What you write is correct but it's even simpler than that: projective space needn't be invoked, nor the product topology.
To say that $X$ (resp. $X'$) is a rational variety means that some non-empty open subset $U\subset X$ (resp. $U'\subset X'$) is isomorphic to some open subset $V\subset \mathbb A^n$ (resp. $V'\subset \mathbb A^{n'}$).
But then the open subset $U\times U'\subset X\times X'$ is isomorphic to the open subset $V\times V'\subset \mathbb A^{n}\times \mathbb A^{n'}=\mathbb A^{n+n'}$, proving that $X\times X'$ is rational. |
H: Infinite intersection of kernels
Let $X$ be an infinite dimensional Banach space. And let $X^{\mathrm{*}}$ be the space of linear continuous functionals( $f : X \rightarrow \mathbb{R}$ linear and continuous). Assume $X^{\mathrm{*}}$ is separable, and take $(x^{\mathrm{*}}_n)$for $n \in \mathbb{N}$ be a dense subset of $X^{\mathrm{*}}$. Is it true that $\bigcap _{i=0}^{\infty} \mathrm{ker}(x^{\mathrm{*}}_n)$ is trivial?
AI: Consider $x\in \bigcap\limits_{n=0}^\infty\mathrm{Ker} (x_n^*)$, so $x_i^*(x)=0$ for all $n\in\mathbb{Z}_+$. Consider arbitrary functional $x^*\in X^*$. Since $\{x_n^*:n\in\mathbb{Z}_+\}$ is dense in $X^*$, then there exist some sequence $\{x_{n_k}^*:k\in\mathbb{Z}_+\}$ that is norm convergent to $x^*$, i.e. $\lim\limits_{k\to\infty}\Vert x^*-x_{n_k}^*\Vert=0$. In particular
$$
\Vert x^*(x)\Vert=\lim\limits_{k\to\infty}\Vert x^*(x)-x_{n_k}^*(x)\Vert\leq
\left(\lim\limits_{k\to\infty}\Vert x^*-x_{n_k}^*\Vert\right)\Vert x\Vert=0
$$
Thus for all $x^* \in X^*$ we have $x^*(x)=0$. By corollary of Hahn-Banach theorem $x=0$. Since $x\in \bigcap\limits_{i=0}^\infty\mathrm{Ker} (x_n^*)$ is arbitrary we conclude
$$
\bigcap\limits_{n=0}^\infty\mathrm{Ker} (x_n^*)=\{0\}
$$ |
H: Computing roots of high degree polynomial numerically.
Here is my problem ; for my research, I believe that the complex numbers I am looking at are precisely the (very large) set of roots of some high degree polynomial, of degree $\sim 2^n$ where $1 \le n \le 10 \sim 15$. Mathematica has been running for the whole day on my computer just for $2^{10}$ even though $2^9$ took half an hour, and I wondered if any other program out there would be faster than Mathematica so that I could compute more of those roots. If I had more examples to compute it would REALLY help. The thing I need the program to be able to do is simple : I give you a polynomial of very high degree, and I want to numerically plot its complex roots. I don't care about multiplicity.
Thanks in advance,
EDIT: Marvis, here is my code for computing the nested polynomial $p^m(x) - x$, where $p^2(x) = p(p(x))$.
function r = polycomp(p,q);
r = p(1);
for k = 2:length(p);
r = conv(r,q);
r(end) = r(end) + p(k);
end
All I do afterwards is a loop with
r = [1 0]
for i = 1:n
r = polycomp(r,p)
end
where $n$ is my loop length and $p$ is my polynomial.
AI: The MATLAB command roots() seems to do a fine job and at a decent speed. I tried using roots for polynomials of degree from $2^7$ to $2^{14}$ and below are the timings. The time for computing the roots using roots() seem to scale as $N^{\log_2(5)}$. |
H: Evaluating $\int{\frac{x}{1+x^4}\ dx}$
I have this integral:
$$\int{\frac{x}{1+x^4}\ dx}$$
The solution should be:
$$\frac{1}{2} \arctan{x^2}+C$$
But I have only seen how to integrate when in denominator I have an expression with real roots. Here, with $1+x^4$ I dont have any real roots and I don't know how I can integrate it.
AI: Substitute $t = x^2$. This gives us $xdx = \dfrac{dt}2$. Then we get that $$I = \int \dfrac{x}{1+x^4} dx = \dfrac12 \int \dfrac{dt}{1+t^2}$$
Can you take it from here?
Move your mouse over the gray area for the answer.
We get that $\displaystyle I = \dfrac12 \int \dfrac{dt}{1+t^2}$. Now substitute $\theta = \arctan(t)$. This gives us that $t = \tan(\theta)$ and $dt = \sec^2(\theta) d \theta$. Also, note that $1 + \tan^2(\theta) = \sec^2(\theta)$. Hence, $$I = \dfrac12 \int \dfrac{dt}{1+t^2} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{1+\tan^2(\theta)} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{\sec^2(\theta)} \\= \dfrac12 \int d \theta = \dfrac{\theta}2 + C = \dfrac{\arctan(t)}2 + C = \dfrac{\arctan(x^2)}2 + C$$ |
H: Why do irrationality proofs of $\sqrt x$ not apply when $x$ is a perfect square?
When trying to prove that a particular root (say $\sqrt{2}$ or $\sqrt{10}$) cannot be rational, I always see a particular indirect proof that goes something like this:
Suppose $\sqrt{x}$ were rational; then, there would be two integers $a$ and $b$ such that $a/b$ was $\sqrt{x}$. We can also assume that $a$ and $b$ have no common factors, because we can simplify $a/b$ as much as we want before we begin.
Then, $(a/b)^2 = x$, or $a^2/b^2=x$, or $a^2 = x * b^2$. But, if $a * a$ is a multiple of $x$ then $a$ must also be a multiple of $x$, so we can rewrite $a$ as $a = cx$ and substitute:
$(c * x)^2 = x * b^2$, or $x^2 * c^2 = x * b^2$. Divide by $x$ and we get that $b^2 = x * c^2$.
But we've now shown that both $a$ and $b$ have a factor of x, contradicting our original assumption; this means that $\sqrt{x}$ cannot be rational.
I see why this proof works, what I don't see is why you can't plug in a number with an actual rational root, like 16, and not form the same proof that $\sqrt{16}$ cannot be rational.
AI: The proof will work for any integer $\rm\,x\,$ (or rational) that is not a perfect square, since it will have a prime factor occurring to an odd power, which is precisely what's needed for the proof to succeed. From $\rm\:a^2 = x b^2\:$ we deduce, by unique prime factorization, that the power of every prime $\rm\:p\:$ dividing $\rm\:x\:$ must be even, being the difference of two even integers (the power of $\rm\:p\:$ in $\rm\:a^2\:$ minus the power in $\rm\:b^2.\,$
Similarly for $\,\rm n$'th roots: from $\rm \,a^n = x b^n\,$ we deduce that all primes occur to powers divisible by $\rm\,n,\,$ so $\,\rm x\,$ is an $\rm\,n$'th power.
Generally it is not true that $\rm\:x\:|\:a^2\:\Rightarrow\:x\:|\:a\ $ (e.g. let $\rm\:x = a^2 > 1)$. In fact this property is true iff $\rm\:x\:$ is squarefree, which is why the proof works for $\rm\:x\:$ prime (or a product of distinct primes) which, having at least one prime to the power one, certainly does have a prime occurring to an odd power. The general case reduces to this case by pulling the square part of $\rm\:x\:$ out of $\rm\:\sqrt{x},\:$ i.e. if $\rm\:x = n^2 y,\:$ with $\rm\:y\:$ squarefree then we have $\rm\:\sqrt{x} = \sqrt{n^2 y} = n\sqrt{y}\in \mathbb Q\iff\sqrt{y}\in\mathbb Q.$ Therefore your method of proof will work if you first reduce this way to the case where $\rm\:x\:$ is squarefree. |
H: PEMDAS:How to solve this exercise?
I have the following problem:
$$100*\left\{24+100*[1001-4*(25*6+25*4)]\right\}$$
I'm very frustrated that I can't solve exercises like this. I have read about PEMDAS and followed the steps but somehow I am making a mistake because the result I get is not correct.(looked at answers in the book).
Can someone provide the steps in order to solve correctly this problem?
AI: First, do the innermost products, $25*6$ and $25*4$. Then add them. Then multiply the result by $4$. Then subtract this from $1001$. Then multiply the answer by $100$. Then add $24$. Then multiply the result of that by $100$.
Or:
You will need to multiply $100$ by the result of what is in curly brackets;
What is in curly brackets requires you to add $24$ to the result of multiplying $100$ by the answer you get from inside the square brackets;
-What is inside the square brackets requires you to start with $1001$, and subtract the result of multiplying $4$ by the result of what you have inside the round parentheses.
To compute what is inside the round parentheses, you first do the products ($25$ times $6$ and $25$ times $4$), and add them.
Now "unfold" that to get the answer. |
H: A question on modules over Noetherian ring
If $G$ is a module over the non-trivial commutative Noetherian ring $R$ then is it possible that for all maximal ideal $M$ of $R$ we have $MG=G$ ?
I guess the answer is no.
AI: Actually the answer is yes:
$R=\mathbb Z, G=\mathbb Q$ . |
H: Velocity Question & Acceleration
Below I have a question that I tried to solve on a exam. I am curious as to the actual way to approach the question. What I did was set the equation equal to $0$ and get $t = -3$ then I plugged in $3$ into the derivative of the equation and I got $128$ ft/sec. But apparently I approached it wrong. What is the right way to approach the question and solve it for the questions below?
If a rock is thrown upward on Earth, with an initial velocity of 32
ft/sec from the top of a 48 ft building, we can model the height of
the rock at time t by $s(t) = −16t^2+ 32t + 48$. After throwing
the rock up, it will at some point be level again with the top of the
48 ft building.
What will the velocity be at this time?
Find the acceleration, why is it constant?
AI: Recall that for constant vertical acceleration only, the position of the object is given by:
$$s\left(t\right) = s_0 + v_ot + \frac {1}{2} a t^2,$$
with constant vertical acceleration $a$, initial velocity and position $v_0$ and $s_0$, respectively, and current vertical position $s$.
Since our initial height and final height are both $48$ feet, we have:
$$48 = 48 + 32t - 16t^2 \tag{2}$$
Solving for the time to where the rock is at the same height it started at leads us to:
$$0= -16t^2 + 32t \implies t = 0, 2 \space \text{sec} \tag{3} $$
Obviously the rock is at the same height at $t = 0$, so at $t = 2$ sec is where it makes a parabola-like movement and comes back down to the original height of the $48$ foot building.
Taking the derivative of our position function, $s(t)$ and plugging in $t = 2$ sec leaves us:
$$s'(t) = -32t + 32 \implies v = -32 \space \text{feet/sec} \tag{4} $$
Taking the derivative again leads us to:
$$s''(t) = -32 \space \text{feet per second per second} \tag{5}$$
As expected, $s''(t)=a$ is indeed constant, since the ball is only accelerating due to gravity which is roughly $-32$ feet per second per second. Also, you could note that since acceleration is the second derivative of a position function and our position function was a quadratic, it follows that the acceleration function is constant since taking the derivative each time lowers the power. Further, as noted in equation (1), the formula only holds for constant acceleration, so it is a bit circular to see that since your given equation fits the model formula (1), the object undergoes constant acceleration. |
H: Techniques of Integration w/ absolute value
I cannot solve this integral.
$$
\int_{-3}^3 \frac{x}{1+|x|} ~dx
$$
I have tried rewriting it as:
$$
\int_{-3}^3 1 - \frac{1}{x+1}~dx,
$$
From which I obtain:
$$
x - \ln|x+1|\;\;\bigg\vert_{-3}^{\;3}
$$
My book (Stewart's Calculus 7e) has the answer as 0, and I can intuitively see this from a graph of the function, it being symmetric about $y = x$, but I cannot analytically solve this.
AI: Your rewriting is incorrect; when $-3\leq x\leq 0$, the original function takes the values
$$\frac{x}{1+|x|} = \frac{x}{1-x} \neq 1 - \frac{1}{x+1}$$
(your rewriting is only valid when $x\geq 0$)
To solve the problem, either notice that your function is odd:
$$f(-x) = \frac{-x}{1+|-x|} = - \frac{x}{1+|x|} = -f(x)$$
and conclude that the answer is $0$ because your interval is symmetric about the origin; or
divide the integral into two:
$$\int_{-3}^3\frac{x}{1+|x|}\,dx = \int_{-3}^0\frac{x}{1+|x|}\,dx + \int_{0}^3\frac{x}{1+|x|}\,dx.$$
In the first integral, you know $x$ is negative so $|x|=-x$; in the second, you have $|x|=x$. Now you can solve each part separately. The second integral can be solved with your decomposition:
$$\frac{x}{1+x} = \frac{-1+1+x}{1+x} = -\frac{1}{1+x} + 1$$
while the first one can be done similarly, with
$$\frac{x}{1-x} = -\frac{-x}{1-x} = -\left(\frac{-1+1-x}{1-x}\right) = \frac{1}{1-x} -1.$$ |
H: Is this proof that $\sqrt 2$ is irrational correct?
Suppose $\sqrt 2$ were rational. Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.
Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.
Squaring both sides,
$2 = \frac 21 = \frac{a^2}{b^2}$.
Lowest terms representations of rational numbers are unique, so we have $a^2 = 2$ and $b^2=1$.
But there is no such integer $a$, and therefore we have a contradiction and $\sqrt 2$ is irrational.
I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound.
AI: Your proof is correct. Quicker, from $\rm\:\color{#0a0}{(a,b)\!=\!1},\ 2\:\!b^2 = a^2\:$ we deduce by Euclid's Lemma ($\color{#c00}{\rm EL}$) that $\rm\:\color{#0a0}{b\:|\:a}\:\!a\:\Rightarrow\:b\:|\:a\,\ $ so $\rm\:\sqrt{2} = a/b\in \mathbb Z.\:$
In your proof, the only "uniqueness" result we need on reduced fractions is the following:
$\rm{\bf Lemma}\ \ (a,b)\!=\!1,\ \dfrac{a}b = \dfrac{A}B\:\Rightarrow\:b\:|\:B\ \ $ Proof $\rm\ \ Ab=aB\:\Rightarrow\:b\:|\:aB\:\Rightarrow\:b\:|\:B\:$ by $\rm\color{#c00}{EL}.\ \small\bf QED$
$\rm Hence\quad\ \, (a,b)\!=\!1,\ \dfrac{a}b = \dfrac{2b}{a}\ \Rightarrow\:b\:|\:a\ \Rightarrow\ \sqrt{2}=\dfrac{a}b\in\mathbb Z.\ $ So you don't need $\rm\:(a^2,b^2)=1.$
This method of proof generalizes to arbitrary square roots, e.g. see here.
The converse of the Lemma is also true, i.e. Euclid's Lemma is equivalent to this uniqueness property of reduced fractions. For more on this topic see my posts on unique fractionization. |
H: Compute the limit of $\frac1{\sqrt{n}}\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)^{1/n^2}$
Compute the following limit:
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}} $$
I'm interested in almost any approaching way for this limit. Thanks.
AI: Let's begin
$$
\lim\limits_{n\to\infty}\frac{\left(\prod\limits_{k=1}^n k^k\right)^{\frac{1}{n^2}}}{\sqrt{n}}=
\lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log k - \frac{1}{2}\log n\right)=
$$
$$
\lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log\left(\frac{k}{n}\right)+\frac{1}{n^2}\sum\limits_{k=1}^n k\log n - \frac{1}{2}\log n\right)=
$$
$$
\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\log n\left(\frac{n^2+n}{n^2}-1\right)\right)=
$$
$$
\exp\left(\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\lim\limits_{n\to\infty}\frac{\log n}{n}\right)=
$$
$$
\exp\left(\int\limits_{0}^1 x\log x dx\right)=\exp\left(-1/4\right)
$$
And now we are done! |
H: Fundamental Group of the complement of $\mathbb{S}^1$ union with $z$ axis in $\mathbb{R}^3$
Can you help me to compute the fundamental group of $\mathbb{R}^3\setminus X$, where $$X = \mathbb{S}^1\cup \{(0,0,z)\in\mathbb{R}^3\mid z\in\mathbb{R}\}\ ?$$
I'm sorry if I repeated the question and I'm sorry if the question it's not well written and ... thanks in advance ! I'd say that the fundamental group is isomorphic to $F_2$, but I'm not even sure if the intuition is right or not.
AI: Thanks to S.V.K., you can add a point at infinity to obtain $S^3 - X$ where X are your now linked circles without changing the fundamental group. Then in a way similar to deformation retracting $\mathbb{R}^3 - S^1$ to obtain $S^2 \vee S^1$, you can deformation retract your new space to obtain $S^2 \vee T^2$, so that again by S.V.K. the fundamental group becomes $\mathbb{Z} \times \mathbb{Z}$. |
H: Method of Frobenius
Can someone please explain the green text to me?
Maybe I am not reading it right, but that sentence makes no sense to me. "that (2) is an equation For which $xp(x)$ and $x^2q(x)$ " what does this part mean? The next part that follows sys something about being constants. Are they trying to say we should replace equation (2) with $xp(x)$ and $x^2q(x)$ like
$$y''(x) + xp(x)y'(x) + x^2q(x)y(x) = 0 $$ So that
$$y''(x) + p_0y'(x) + q_0y(x) = 0$$
Either case it makes no sense to me at all, especially why they started writing out the $xp(x)$ as a series.
AI: Right below Eq. (2) it is said that $p(x)=p_0/x, \ q(x)=q_0/x^2$ which means $xp(x)$ and $x^2q(x)$ are constants. Later on he assumes that these functions are not constants! |
H: proof of inequality-using a convex function
Let $a_{1},a_{2},\ldots,a_{n},b_{1},b_{2},\ldots,b_{n}$ be positive numbers. We need to prove that:
$$(a_{1}+b_{1})^{\frac{1}{n}}(a_{2}+b_{2})^{\frac{1}{n}}\cdots(a_{n}+b_{n})^{\frac{1}{n}}\geq a_{1}^{\frac{1}{n}}a_{2}^{\frac{1}{n}} a_{n}^{\frac{1}{n}}+b_{1}^{\frac{1}{n}}b_{2}^{\frac{1}{n}}\cdots b_{n}^{\frac{1}{n}}$$
I came up with a proof for this problem by simply using Arithmetic-Geometric Mean Inequality as follows:
$$
\frac{a_{1}^{\frac{1}{n}}a_{2}^{\frac{2}{n}}\cdots a_{n}^{\frac{1}{n}}+b_{1}^{\frac{1}{n}}b_{2}^{\frac{1}{n}}\cdots b_{n}^{\frac{1}{n}}
}{(a_{1}+b_{1})^{\frac{1}{n}}(a_{2}+b_{2})^{\frac{1}{n}}\cdots (a_{n}+b_{n})^{\frac{1}{n}}}=(\frac{a_{1}}{a_{1}+b_{1}})^{\frac{1}{n}}(\frac{a_{2}}{a_{2}+b_{2}})^{\frac{1}{n}}\cdots(\frac{a_{n}}{a_{n}+b_{n}})^{\frac{1}{n}}+(\frac{b_{1}}{a_{1}+b_{1}})^{\frac{1}{n}}(\frac{b_{2}}{a_{2}+b_{2}})^{\frac{1}{n}}\cdots(\frac{b_{n}}{a_{n}+b_{n}})^{\frac{1}{n}}=\sqrt[n]{(\frac{a_{1}}{a_{1}+b_{1}})(\frac{a_{2}}{a_{2}+b_{2}})\cdots(\frac{a_{n}}{a_{n}+b_{n}})}+\sqrt[n]{(\frac{b_{1}}{a_{1}+b_{1}})(\frac{b_{2}}{a_{2}+b_{2}})\cdots(\frac{b_{n}}{a_{n}+b_{n}})}\leq \frac{1}{n}\left [ (\frac{a_{1}}{a_{1}+b_{1}})+(\frac{a_{2}}{a_{2}+b_{2}})+\cdots+(\frac{a_{n}}{a_{n}+b_{n}}) \right ]+\frac{1}{n}\left [(\frac{b_{1}}{a_{1}+b_{1}})+ (\frac{b_{2}}{a_{2}+b_{2}})+\cdots+(\frac{b_{n}}{a_{n}+b_{n}}) \right ]=1$$
where last inequality follows from applying the Arithmetic-Geometric Mean inequality.
The professor said there is a short way to solve the problem by proving that the function (I don't know which function he was talking about) is convex, and then the inequality follows immediately. I am really interested to know this method, so I appreciate if someone shares it with me.
AI: The proof by AM-GM inequality is indeed elegant, you can
see it again here.
In Exercise 6.5 of J. Michael Steele's The Cauchy-Schwarz Master Class, you are asked to find an alternative proof
using Jensen's inequality. Here is Steele's solution:
To build a proof with Jensen's inequality, we first divide by $(a_1a_2\cdots a_n)^{1/n}$ and write
$c_k$ for $b_k/a_k$, so the target inequality takes the form
$$1+(c_1c_2\cdots c_n)^{1/n}\leq \left\{(1+c_1)(1+c_2)\cdots(1+c_n)\right\}^{1/n}. $$
Now if we take logs and write $c_j$ as $\exp(d_j)$, we find it takes the form
$$\log(1+\exp(\bar d))\leq {1\over n}\sum_{j=1}^n \log(1+\exp(d_j)), $$
where $\bar d=(d_1+d_2+\cdots +d_n)/n$. Finally, the last inequality is simply
Jensen's inequality for the convex function $x\mapsto \log(1+e^x)$, so the solution is complete. |
H: Help Verifying Trigonometric Identity
I could really use help, hint or otherwise, in proving a trigonometric identity:
We are only allowed to work on one side of the equation.
$$\dfrac{2\sin^2(x)-5\sin(x)+2}{\sin(x)-2} = 2\sin(x)-1$$
AI: HINT: Factorize the numerator and cancel terms arguing why the terms you are canceling are not zero.
Move your mouse over the gray area for the answer.
$$\dfrac{2 \sin^2(x) - 5 \sin(x) + 2}{\sin(x) - 2} = \dfrac{2 \sin^2(x) - 4 \sin(x) - \sin(x) + 2}{\sin(x) - 2}\\ = \dfrac{2 \sin(x) (\sin(x) - 2) - ( \sin(x) - 2)}{\sin(x) - 2}\\= \dfrac{(\sin(x) - 2) (2 \sin(x) - 1)}{\sin(x) - 2} = 2 \sin(x) - 1$$ We are allowed to cancel $\sin(x) - 2$ since $\sin(x) \neq 2$, $\forall x \in \mathbb{R}$. |
H: Bounds for $\zeta$ function on the $1$-line
I was going over my notes from a class on analytical number theory and we use a bound for the $\zeta$ function on the $1$ line as $\vert \zeta(1+it) \vert \leq \log(\vert t \vert) + \mathcal{O}(1)$ for $t$ bounded away from $0$, say $\vert t \vert \geq 1$. I don't seem to have a proof for this in my notes.
How does one prove the following bound for $\zeta(s)$ on the one line?
$$\zeta(1+it) \leq \log(|t|) + \mathcal{O}(1) \, \, \forall t \geq 1$$
AI: Here is a long answer. The main ingredient to get your bound is to find the analytic continuation to a region which includes the $1$ line.
You can skip the first part if you know how to continue $\zeta(s)$ analytically to the left of $\text{Re}(s) =1$.
First note that for $\text{Re}(s) > 1$, we have $$\zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$$
As stated earlier, to have a bound on the $1$ line, we need to have $\zeta(s)$ to be defined in a half-plane containing the $1$ line i.e. we need to extend the above definition from $\text{Re}(s) > 1$ to a region $\text{Re}(s) > a$ where $a < 1$.
Analytic continuation of $\zeta(s)$ to the left of $1$ line:
One way to do it is as follows. We know that for $\text{Re}(s) > 1$, $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$. We will extend this analytically to the region $\text{Re}(s) > 0$. In some sense, this is all we need not just for this problem but also for most of the problems on $\zeta$, since most often we are only interested in the behavior of the $\zeta(s)$ inside the critical strip i.e. $\{s \in \mathbb{C}: \text{Re(s)} \in (0,1)\}$. We make use of the powerful Euler–Maclaurin formula to make this analytic extension.
$$\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s} = \sum_{n=1}^{N} \dfrac1{n^s} + \sum_{n=N+1}^{\infty} \dfrac1{n^s}$$
The reason why we split this into two parts will become clear as well proceed. But to give a quick answer, the reason for this splitting into a sum $\leq N$ and a sum $>N$, is to allow us to play around with $N$ and optimize the error term accordingly.
We will now focus our attention on $\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s}$. Note that $$\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s} = \int_{N^+}^{\infty} \dfrac{d \lfloor t\rfloor}{t^s} = \left. \dfrac{\lfloor t \rfloor}{t^s} \right \vert_{N^+}^{\infty} + s \int_{N^+}^{\infty} \dfrac{\lfloor t \rfloor}{t^{s+1}} dt$$
So far, we have $\text{Re}(s) > 1$. Hence, $\left. \dfrac{\lfloor t \rfloor}{t^s} \right \vert_{N^+}^{\infty} = - \dfrac{N}{N^s} = - N^{1-s}$. Further, $$\int_{N^+}^{\infty} \dfrac{\lfloor t \rfloor}{t^{s+1}} dt = \int_{N^+}^{\infty} \dfrac{t - \{ t \}}{t^{s+1}} dt = \int_{N^+}^{\infty} \dfrac{dt}{t^s} - \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$ Again, since $\text{Re}(s) > 1$, we have $$\int_{N^+}^{\infty} \dfrac{dt}{t^s} = - \dfrac{N^{1-s}}{1-s}$$
Hence, putting all these together, we now have that $$\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s} = - N^{1-s} - \dfrac{s}{1-s} N^{1-s} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt = \dfrac{ N^{1-s}}{s-1} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$
Hence, $$\zeta(s) = \sum_{n=1}^{N} \dfrac1{n^s} + \dfrac{N^{1-s}}{s-1} - s \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$$
However, note the little miracle here. The expression on the right now makes sense for $\text{Re}(s) > 0$, since $\displaystyle \int_{N^+}^{\infty} \dfrac{\{ t \}}{t^{s+1}} dt$ converges for $\text{Re}(s) >0$, and is analytic, and it matches with $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^s}$ for $\text{Re}(s) > 1$. This means that this is the analytic extension we were after.
Back to the problem
Now setting $s = 1 +it$, we get that
\begin{align*}
\zeta(1+it) & = \sum_{n \leq N} \frac1{n^{1+it}} - \frac{N^{1-(1+it)}}{1-(1+it)} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy\\
& = \sum_{n \leq N} \frac1{n^{1+it}} + \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy
\end{align*}
We will now get a bound for $E(N,t) = \displaystyle \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy$.
\begin{align*}
\left \lvert E(N,t) \right \rvert & \leq \left \lvert \frac{N^{-it}}{it} \right \rvert + \displaystyle \left \lvert 1+it \right \rvert \left \lvert \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy \right \rvert\\
& \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \left \lvert \int_{N^+}^{\infty} \frac1{y^{2+it}} dy \right \rvert\\
& \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \int_{N^+}^{\infty} \left \lvert \frac1{y^{2+it}} \right \rvert dy\\
& \leq \frac1{\lvert t \rvert} + \displaystyle \left \lvert 1+it \right \rvert \int_{N^+}^{\infty} \frac1{y^{2}} dy\\
& = \frac1{\lvert t \rvert} + \frac{\left \lvert 1+it \right \rvert}{N}
\end{align*}
Now we need to optimally choose $N$ to get a good error bound. This is why we split the original summation into two parts.
Note that since $\lvert t \rvert \geq 1$, we get that $\left \lvert 1 + it \right \rvert \leq \lvert t \rvert \sqrt{2}$.
If we choose $N = \lfloor \lvert t \rvert \rfloor$, we get that $$\left \lvert E(N,t) \right \rvert \leq \frac1{\lvert t \rvert} + \frac{\lvert t \rvert \sqrt{2}}{\lfloor \lvert t \rvert \rfloor}.$$
Since, $\displaystyle |t| \geq 1$, we get that $\displaystyle \frac1{\lvert t \rvert} \leq 1$ and also that $\displaystyle \frac{\lvert t \rvert}{\lfloor\lvert t \rvert \rfloor} \leq 2$. Hence, we get that $$\displaystyle \left \lvert E(N,t) \right \rvert \leq 1 + 2 \sqrt{2}$$ i.e. $E(N,t) = \mathcal{O}(1)$. Hence, we have that $$\zeta(1+it) = \sum_{n \leq N} \frac1{n^{1+it}} + \frac{N^{-it}}{it} - (1+it) \int_{N^+}^{\infty} \frac{\{y\}}{y^{2+it}} dy\\ = \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + E(N,t) = \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + \mathcal{O}(1).$$
Finishing steps
Now we are almost done. Taking the absolute value we get that
\begin{align*}
\lvert \zeta(1+it) \rvert & = \left \lvert \sum_{n \leq \lvert t \rvert} \frac1{n^{1+it}} + E(N,t) \right \rvert\\
& \leq \sum_{n \leq \lvert t \rvert} \left \lvert \frac1{n^{1+it}} \right \rvert + \left \lvert E(N,t) \right \rvert\\
& = \sum_{n \leq \lvert t \rvert} \frac1{n} + \left \lvert E(N,t) \right \rvert\\
& \leq \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1) + \mathcal{O}(1) \text{ (Since $\displaystyle \sum_{n \leq |t|} \dfrac1n$ goes as $\log(|t|) + \gamma + \mathcal{O}(1/|t|)$)}\\
& = \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1).
\end{align*}
Hence, we get what we want, viz, $$\lvert \zeta(1+it) \rvert \leq \log \left( \left \lvert t \right \rvert \right) + \mathcal{O}(1).$$
You should also look here where another asymptotic for $\zeta(s)$ on the left of the $0$-line is obtained by different argument. You could also follow a similar method as described in the other question i.e. use the functional equation described in that to obtain a bound for $\zeta(s)$ on the $1$-line. You will need bounds for $\zeta(it)$, if you want to make use of the functional equation. But either way you cannot escape some long arguments. And I think what I have written is a bit more straight-forward than trying to make use of the functional equation. |
H: Does the curvature determine the metric?
Here I asked the question whether the curvature deterined the metric. Since I am unfortunately completely new to Riemannian geometry, I wanted to ask, if somebody could give and explain a concrete example to me, as far as the following is concerned:
At the MO page (as cited above) I got the following answer to the question
Given a compact Riemannian manifold M, are there two metrics g1 and g2, which are not everywhere flat, such that they are not isometric to one another, but that there is a diffeomorphism which preserves the curvature?
If the answer is yes:
Can we chose M to be a compact 2-manifold?
On the positive side, if $M$ is compact of dimension $\ge 3$ and has nowhere constant sectional curvature, then combination of results of Kulkarni and Yau show that
a diffeomorphism preserving sectional curvature is necessarily an isometry.
Concerning 2-dimensional counter-examples: First of all, every surface which admits an open subset where curvature is (nonzero) constant would obviously yield a counter-example. Thus, I will assume now that curvature is nowhere constant. Kulkarni refers to Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164, for a counter-example attributed to Stackel and Wangerin. You probably can get the book through interlibrary loan if you are in the US.
I looked up the example in Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164:
If we rotate the curve $x_3=\log x_1$ about the $x_3$-axis in space,
we obtain the surface of revolution $X(u_1,u_2)=(u_2\cos(u_1), u_2\sin(u_1),\log(u_2))$, $u_2>0$.
This is diffeomorphic to the helicoid $X(u_1,u_2) =(u_2\cos(u_1),u_2\sin(u_1),u_1)$.
I think, these manifolds are not compact (but I assumed compactness of the manifold in my question on MO).
I don't understand, how to manipulate this example in order to get a compact manifold.
Thank you for your help.
AI: Here's another example.
First, imagine a short cylinder $S^1\times [0,1]$ and a long cylinder $S^1\times [0,10^{10}]$, but with the same radii. Smoothly cap off both ends of the cylinders in the same way using spaces homeomorphic to discs.
The resulting manifolds are both homeomorphic to $S^2$, are not isometric (since one has a much larger diameter than the other), but there is a curvature preserving diffeomorphism between them.
To see this, just convince yourself there is a diffeomorphism $f:S^1\times[0,1]\rightarrow S^1\times [0,10^{10}]$ with the property that $f$ is an isometry when restricted to $[0,\frac{1}{4}]$ and $[\frac{3}{4},1]$. This condition allows you to extend $f$ to a curvature preserving diffeo of both compact manifolds. |
H: Does the series converge or diverge?
Let
$$
a_n=\frac{\sqrt{3n+1}}{n^2}
$$
I cannot find a suitable $b_n$ to use for the comparison test, and when I try to use the ratio test it really becomes a mess and I cannot find the limit as a result. I am going to guess there is something simple I am missing.
AI: I am assuming that you are interested in the convergence of the series $\sum_n a_n$. Either way the same argument will work.
HINT
First note that $\forall n \geq 1$, we have $0 \leq 3n+1 \leq 4n$. Now can you bound the terms and argue?
Move your mouse over the gray area for the answer.
Note that $\forall n \geq 1$, we have $0 \leq 3n+1 \leq 4n$. Hence, we have that $$0 \leq \sum_n \dfrac{\sqrt{3n+1}}{n^2} \leq \sum_n \dfrac{\sqrt{4n}}{n^2} = \sum_n \dfrac2{n^{3/2}}$$ Recall that $\displaystyle \sum_n \dfrac1{n^p}$ converges for all $p > 1$. Hence, we have that $\sum_n a_n$ converges. You may look here for proofs of $\displaystyle \sum_n \dfrac1{n^p}$ converges for all $p > 1$. |
H: Proof an inequality
I'm trying to prove that
$$ \frac{3-2\sqrt{1-15 m^2}}{1+12 m^2}\geq 1+3 m^2$$
I have obtained in a CAS software the Taylor expansion in $m=0$
One posibility to prove the inequality is showing coeficients in Taylor expansion are non-negative, by I don't find how.
Really I want only to obtain inequality. Some idea?
EDIT
$m$ must be between $0<m<\frac{1}{\sqrt 15}$
AI: $$ \frac{3-2\sqrt{1-15m^2}}{1+12m^2} \geq 1+3m^2 \iff $$
$$ 3-2\sqrt{1-15m^2} \geq (1+3m^2)(1+12m^2) \iff $$
$$ 2\sqrt{1-15m^2} \leq 3-(1+3m^2)(1+12m^2) \iff $$
$$ \sqrt{1-15m^2} \leq \frac{3-(1+3m^2)(1+12m^2)}{2} \iff $$
$$ \sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} $$
Note that on the interval you're concerned about, the right hand side is always positive. Proof: it's obviously decreasing on $\left(0,\frac{1}{\sqrt{15}}\right)$, and is equal to $\frac{21}{50}$ at the right endpoint. Therefore squaring both sides is legal here with an $\iff$ statement.
$$ \sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} \iff $$
$$ 1-15m^2 \leq \left(\frac{2-15m^2-36m^4}{2}\right)^2 \iff $$
$$ 1-15m^2 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4 - 15m^2 + 1 \iff $$
$$ 0 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4 $$
This last statement is clearly true. |
H: Spanning tree of a strongly connected directed graph.
Given a strongly connected directed graph $G=(V,E)$, and a node $r \in V$. Let $T_r$ be the set of spanning trees of $G$ with $r$ as root and all edges pointing to $r$. Is is it possible that there is an edge $e$ such that for all $t \in T_r$, we have $e \in E(Tr)$? In other words, is it possible that there is a directed edge of $G$ belongs to all possible spanning trees with $r$ as root?
BTW : My guess is that it is impossible.
Edit : I meant to say "directed graph"
AI: Unless I misunderstand something, of course it is possible. There will always be such an edge if the graph can be divided into two parts with only two edges (one in each way) between them.
E.g. the full simple graph on two vertices. There is only one spanning tree starting at each vertex and it has one edge. |
H: Every Hilbert space operator is a combination of projections
I am reading a paper on Hilbert space operators, in which the authors used a surprising result
Every $X\in\mathcal{B}(\mathcal{H})$ is a finite linear combination of orthogonal projections.
The author referred to a 1967 paper by Fillmore, Sums of operators of square zero. However, this paper is not online.
I wonder whether someone has a hint on how this could be true since there are all kinds of operators while projections have such a regular and restricted form.
Thanks!
AI: I believe the answers you're looking for can be found in a paper by Pearcy and Topping, Sums of small numbers of idempotents., which is openly accessible. |
H: Integral of $\int{\frac{dx}{(\arcsin{x})\sqrt{1-x^2}}}$
I am having a problem solving an integral. I am stuck in an infinite loop. Integral is:
$$\int{\frac{dx}{\sqrt{1-x^2}\arcsin{x}}}$$
I have separated it in dv and u on this way:
$$u = \frac{dx}{\sqrt{1-x^2}}$$
$$dv = \frac{1}{\arcsin{x}}$$
And the using:
$$u v - \int{v \, du}$$
I get again:
$$\int{\frac{dx}{\sqrt{1-x^2}\arcsin{x}}}$$
I dont know, but probably, I am doing something wrong. I am new at solving Integrals so I am learning :) According to my book the result should be:
$$\ln({\arcsin{x}})-C$$
And it will be true if I didn't had $$\sqrt{1-x^2}$$ but on this way I have no idea.
AI: Do not use Integration by Parts. Use $u$-substitution. Let $u=\sin^{-1}(x)$. Then $du=dx/\sqrt{1-x^2}$
So now your integral is $$\int \frac{du}{u}$$ |
H: Linear algebra: Inverse of a matrix?
How does $$\mathbf P^{-1}\mathbf A \cdot \mathbf I \cdot \mathbf P = \mathbf A\mathbf P^{-1}\cdot \mathbf P = \mathbf A\cdot \mathbf I\ ?$$
I thought $\mathbf P^{-1}$ couldn't be "moved around" within an equation.
It's from a theorem that similar matrices have the same eigenvalues.
Let A and B be similar. Then B = $\mathbf P^{-1}$ AP, for some nonsingular matrix P. We prove that A and B have the same characteristic polynomials, pA(λ) and pB(λ), respectively. We have:
pB(λ) = det(λIn - B) = det(λIn - $\mathbf P^{-1}$ AP) = det($\mathbf P^{-1}$ λInP - $\mathbf P^{-1}$ AP).
AI: If $A = P^{-1}BP$, then
$\begin{align*}
p_A(x) &= \det(xI - A) = \det(xI - P^{-1}BP) = \det (P^{-1}xIP - P^{-1}BP) = \det \big(P^{-1}(xI - B)P\big) = \\ &= \det P^{-1} \cdot \det(xI - B) \cdot \det P = \det(xI - B) \cdot \det P^{-1} \cdot \det P= \\ &= \det(xI - B) (\det P)^{-1} \cdot \det P = \det(xI - B)= p_B(x)
\end{align*}
$
Note that we don't really move $P^{-1}$ around; we apply the product property of determinants: $\det(AB) = \det A \cdot \det B$, and since the determinant is a real number, you can move it around because the real numbers are associative and commutative with respect to multiplication.
Now, why does $xI = P^{-1}xIP$?
We call scalar matrices all matrices of the form $kI$, where $k$ is a scalar (note that these matrices have zeroes everywhere except on the diagonal, which has all its entries equal to $k$). It turns out that scalar matrices commute with every matrix, that is, for any matrix $A$, $(kI)A = A(kI)$. (Actually, there are no other matrices that commute with all the matrices.)
Let's prove this equality entrywise; $n$ stands for the appropriate dimension of the matrix. The important thing to note in both cases is that the matrix $kI$ has all zeroes except in the diagonal, so the sums reduce to only one term:
$\displaystyle \big((kI)A\big)_{ij} = \sum_{l=1}^n (kI)_{il}A_{lj} = kA_{ij}$ (if $l \neq i$, then $(kI)_{il} = 0$; the only remaining term is with $l = i$)
$\displaystyle \big(A(kI)\big)_{ij} = \sum_{l=1}^n A_{il}(kI)_{lj} = A_{ij}k$ (if $l \neq j$, then $(kI)_{lj} = 0$; the only remaining term is with $l = j$)
Then we have $\big((kI)A\big)_{ij} = \big(A(kI)\big)_{ij}$ for all $i$, $j$, so $(kI)A = A(kI)$. And in our particular case, since $xI$ is a scalar matrix, $P^{-1}(xI)P = (xI)P^{-1}P = (xI)I = xI$. |
H: How to compute $\int\frac{x^2}{x^2+4}\,dx$ and $\int\frac{1}{\sqrt{4-x^2}}\,dx$?
My calculus teacher gave us a list of integrals to solve through the substitution method. I've been trying everything for hours, and I just can't find out how to compute these:
$$∫\frac{x^2}{x^2+4}\,dx\qquad\text{and}\qquad \int\frac{1}{\sqrt{4-x^2}}\,dx.$$
Is it even possible to solve them through substitution?
AI: The first one can be solved by rewriting it first
$$\int\frac{x^2}{x^2+4}\,dx = \int\frac{x^2+4-4}{x^2+4}\,dx = \int\,dx - 4\int\frac{dx}{x^2+4}.$$
Then we can solve the second one of these integrals (first one is easy) by factoring out $4$ from the denominator and then using the substitution $u=\frac{x}{2}$:
$$\begin{align*}
\int\frac{dx}{x^2+4} & = \frac{1}{4}\int\frac{dx}{\left(\frac{x}{2}\right)^2+1}\\
&= \frac{1}{2}\int\frac{du}{u^2+1}.\end{align*}$$
The last integral has a direct antiderivative.
For the second integral, try $x=2u$. Then $dx = 2du$, so
$$\int\frac{1}{\sqrt{4-x^2}}\,dx = \int\frac{2du}{\sqrt{4-4u^2}} = \int\frac{du}{\sqrt{1-u^2}}.$$
Does the last one look familiar?
Alternatively, do the same trick: factor out $4$ in the denominator inside the square root:
$$\begin{align*}
\int\frac{1}{\sqrt{4-x^2}}\,dx &= \int\frac{1}{\sqrt{4(1 - (\frac{x}{2})^2}}\,dx\\
&= \int\frac{1}{2\sqrt{1 - (\frac{x}{2})^2}}\end{align*}$$
which may suggest the substitution $w=\frac{x}{2}$.
Added. In light of the comments...
My first impulse with the second integral would have been to use a trigonometric substitution. That's because integrals that involve $\sqrt{a^2-x^2}$ but which do not have an $x\,dx$ factor are the traditional proving grounds for the trigonometric substitution. Here, $a=2$, so we would try the substitution $x=2\sin\theta$, with $0\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, $dx = 2\cos\theta\,d\theta$. This wold turn the $\sqrt{4-x^2}$ factor into
$$\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|=2\cos\theta$$
with the last equality because $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, so $\cos\theta\geq 0$. Then the integral would simplify to
$$\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{\cos\theta\,d\theta}{2\cos\theta} = \frac{1}{2}\int\,d\theta,$$
which is trivial to do; finally, we would return to $x$ using $\theta=\frac{1}{2}\arcsin(x)$. |
H: Regarding direction of gradient of a function
I was reading this book related to optimizing a function f(x,y) with constraints g(x,y) = 0. The book says that let us suppose g(x,y) define a surface, then gradient of the g(x,y) will be orthogonal to the surface. But I am a bit confused how is it possible. I mean gradient always points in the direction of maximum increase. It is not necessary for it to be orthogonal.
For eg lets take the function
f(x) = x^2
If I take the gradient of the function at x=1, then its value is 2i with i giving the direction. This direction is not perpendicular to the function or its tangent at the point (1,f(1)).
I am a bit confused. So any clarifications?
AI: The function that you give is a function of $1$ variable only. So there is only $1$ way that the function can increase at $x=1$.
Suppose we had, instead, $f(x,y)=x^2-y$. This is the actual graph you are describing: $y=x^2$. In particular, we are interested with the level curve $f(x,y)=0$.
This is now a function of two variables. Now take the gradient $\nabla f=\langle2x,-1\rangle$, and at $(1,1)$, this becomes $\nabla f=\langle2,-1\rangle$. Look where this graph is pointing! It is perpendicular to the tangent line of the level curve $f(x,y)=0$. |
H: Trying to rederive an exponential approximation
So I was reading a paper where the following approximation was made. Note that $p$ is small, $L$ is large, and $pL$ is $O(1)$:
$$\left[1-e^{-4p(1+p)L}\right]^{L/2}=\textrm{exp}\left[\frac{2(pL)^2}{e^{4pL}-1}\right](1-e^{-4pL})^{L/2}+O(p^3L^2)$$
I just don't see where this approximation came from, despite playing around with it for a while. Any insight would be much appreciated.
AI: If $p$ is small, then $1+p \sim 1$, so let's pull that out:
$$ e^{-4p(1+p)L} = e^{-4pL} e^{-4p^2L} = e^{-4pL} (1 + O(p^2 L))$$
Easy enough. $e^{-4pL} \in O(1)$ so we can simplify a bit:
$$ 1 - e^{-4p(1+p)L} = (1 - e^{-4pL}) (1 + O(p^2 L)) $$
Oh, but this approximation isn't good enough to compute $(1 + O(p^2 L))^{L/2}$. So let's do better:
$$ e^{-4p(1+p)L} = e^{-4pL} (1 - 4p^2L + O(p^4 L^2))$$
now, we have
$$ (1 - e^{-4p(1+p)L}) = 1 - e^{-4pL} + e^{-4pL} 4 p^2 L + O(p^4 L^2) $$
a little messier to factor:
$$ \cdots = (1 - e^{-4pL}) \left(1 + \frac{e^{-4pL} 4 p^2 L}{1 - e^{-4pL}} + O(p^4 L^2)\right) $$
It's still going to be a pain to raise that to the $L/2$ power. Fortunately, we can rewrite the second factor in a convenient way:
$$ \cdots = (1 - e^{-4pL}) \exp\left( \frac{e^{-4pL} 4 p^2 L}{1 - e^{-4pL}} + O(p^4 L^2) \right) $$
Great! I don't usually use this trick, but that's probably because I usually forget about it! You could do without it (binomial theorem). Anyways, raising to the $L/2$ power...
$$ (1 - e^{-4p(1+p)L})^{L/2} = (1 - e^{-4pL})^{L/2} \exp\left( \frac{e^{-4pL} 4 p^2 L^2}{2(1 - e^{-4pL})} + O(p^4 L^3) \right) $$
Now let's move that $O$ out of the exponent:
$$ \cdots = (1 - e^{-4pL})^{L/2} \exp\left( \frac{e^{-4pL} 2 p^2 L^2}{(1 - e^{-4pL})} \right) (1 + O(p^4 L^3))$$
We can simplify that exponential a bit more:
$$ \cdots = (1 - e^{-4pL})^{L/2} \exp\left( \frac{2 p^2 L^2}{e^{4pL} - 1} \right) (1 + O(p^4 L^3))$$
Also, since the first factor has magnitude less than 1 and the second is $O(1)$, we can easily distribute the last factor:
$$ \cdots = (1 - e^{-4pL})^{L/2} \exp\left( \frac{2 p^2 L^2}{e^{4pL} - 1} \right) + O(p^4 L^3)$$
So the approximation is actually better than what the paper claims. Honestly, I didn't expect to get the same approximation when I started this calculation; I figured I'd get something similar, but show enough of the ideas that you could work the rest of the details on your own to get the exact form you wanted.
(EDIT) Warning: I had assumed $pL \in \Theta(1)$ in my derivation. If you are interested in the case of $pL \in o(1)$, you may (or may not) need to tweak some things. One thing to be careful of is that $(1 - e^{-4pL})^{-1}$ would no longer be $O(1)$. |
H: Maximum and Minimum Perimeter of a Triangle
I cant figure out the following question:
A triangle has sides with lengths of 9,14 and h. if h is an integer what is the difference between the maximum and minimum possible perimeter of the rectangle ? (Ans=17.5)
Any suggestions on how this should be solved ??
AI: Note that the third side must be less than the sum of the other two, so it must be less than $23$. But the third side is an integer, so its largest possible value is $22$. It is easy to see that $22$ is achievable. Just put a hinge where the sides $9$ and $14$ meet, and open up the hinge so that the sides $9$ and $14$ almost form a straight line (that is, make an angle that is almost $180^\circ$. If you wish, you could compute the suitable angle by using the Cosine Law.
As to the smallest value of the third side, we need to make sure that whatever it is, it plus $9$ is bigger than $14$. The smallest integer that works is $6$.
The difference between the largest possible perimeter and the smallest possible perimeter is therefore $22-6$, since the other two sides are the same for each triangle.
If $h$ must be an integer, then the perimeter must be an integer, and the answer of $17.5$ is not possible.
Remark: The Triangle Inequality (the sum of any two sides must be greater than the third side) is an important fact about distances. |
H: Linear algebra: diagonalization
Let $L: P_2 \longrightarrow P_2$ be the linear operator defined by $L(p(t)) = p'(t)$ for $p(t) \in P_2$, the space of real polynomials of degree at most $2$. Is $L$ diagonalizable? If it is, find a basis $S$ for $P_2$ with respect to which $L$ is represented by a diagonal matrix.
Answer: $L$ is not diagonalizable. The eigenvalues of $L$ are $\lambda_1 = \lambda_2 = \lambda_3 = 0$. The set of associated eigenvectors does not form a basis for $P_2$.
I don't how you solve this problem. How do you transform the polynomial into a matrix so that I can find the eigenvalues? I can solve these questions if they state the matrix but this one concerns polynomials. Can someone please help me?
AI: Hints:
Pick a basis of $P_2$. What is the standard basis?
Write the matrix of $L$ with respect to that basis (you'll have to apply $L$ to each element of the basis of $P_2$, and calculate its coordinates in that basis; that'll give you the columns of the matrix).
Work with the matrix now: is it diagonalizable?
1.
We can choose the basis $B = \{1, x, x^2\}$.
2.
$L(1) = 1' = 0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \\ L(x) = x' = 1 = 1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \\ L(x^2) = (x^2)' = 2x = 0 \cdot 1 + 2 \cdot x + 0 \cdot x^2 \\ \text{Then} \\ \left[L\right]_B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 &0 \end{pmatrix}$
3.
Find the roots of $\det(xI - \left[L\right]_B)$, which will be the eigenvalues, and the corresponding eigenvectors. Do you end up with a basis of $\mathbb{R}^3$? |
H: Evaluating a double integral with change of coordinates. $\int\!\!\int_{R} (x+y)\,dA$
It looks simple, but I'm having a bit of difficulty with trying to obtain the proper change of coordinates.
Evaluate the following integral
$$\displaystyle\int\!\!\!\int_{R} (x+y)\,dA$$
with region R in the first quadrant bounded by $-1\le y-x\le 1$ and $1\le x^2+y^2 \le 4$
I've tried $u=x^2+y^2$ and $v=y-x$, but I can't seem to express $x+y$ in terms of $u$ and $v$. Is it possible to express $x+y$ in terms of $u$ and $v$ with the above substitutions? Or do we need another change of variables?
What other change of coordinates is possible to approach this problem?
Based on what region $R$ looks like, I don't think polar coordinates would work for this even though there is an $x^2+y^2$ term (though I could be mistaken).
AI: Below is the region (both the blue and red) over which you are integrating.
$$I = \int_{\Omega} (x+y) dA = \int_{\text{Red region}} (x+y) dA + \int_{\text{Blue region}} (x+y) dA$$
Denoting $\arctan((4-\sqrt{7})/3)$ as $\theta_1$ and $\arctan((4-\sqrt{7})/3)$ as $\theta_2$, we get that
In the red region, you have the $$\int_{\text{Red region}} = \int_{\theta = \theta_1}^{\theta = \theta_2} \int_{r=1}^{r=2} (\cdot) dr d \theta$$
In the blue region, you have the $$\int_{\text{Blue region}} = \underbrace{\int_{\theta = 0}^{\theta = \theta_1} \int_{r=1}^{r=f_1(\theta)} (\cdot) dr d \theta}_{\text{Blue region below}} + \underbrace{\int_{\theta = \theta_2}^{\theta = \pi/2} \int_{r=1}^{r=f_2(\theta)} (\cdot) dr d \theta}_{\text{Blue region above}}$$
Now all you need to find are the functions $f_1,f_2$ and perform the integral. The functions $f_1,f_2$ are easy to find since you know the equation of the line below is $x-y = 1$ and the equation of the line above is $y-x=1$. Hence, $$f_1(\theta) = \dfrac1{\cos(\theta) - \sin(\theta)} \text{ and } f_2(\theta) = \dfrac1{\sin(\theta) - \cos(\theta)}$$ |
H: Is $z=x^2+y^2$ a bijection?
I am learning basic set theory and was doing some exercises where we are to determine whether a given relation is a function, an injection, a surjection and if it is a bijection.
The question in this case was: Determine whether the relation from $\mathbb{R}^2$ to $\mathbb{R}$ defined by $(x,y)Rz$ if and only if $z=x^2+y^2$ is $(1)$ a function, $(2)$ an injection, $(3)$ a surjection and $(4)$ a bijection.
$(1)$ is clearly true. $(2)$ It is not an injection as $(x,y)$ and $(y,x)$ both map to the same point in $\mathbb{R}$. $(3)$ Neither is it a surjection as $x^2+y^2\ge0, \forall x,y\in\mathbb{R}$. $(4)$ This leads me to believe that it is therefore not a bijection. However the answers says it is. This is strange as the book defines a bijection to be a function which is both an injection and a surjection. Clearly the given relation does not satisfy either condition so cannot be a bijection.
I think the answers might have some typos though since it is missing the answer to the next exercise (A relation from $\mathbb{Z}\times\mathbb{Z}$ to $\mathbb{Z}\times\mathbb{Z}$ where $(a,b)R(x,y)$ if an only if $y=a$ and $x=b$; which I found to be a bijection) and so might have mixed up the answer to this exercise with the previous. However, I am new to set theory and so was wondering whether I just made a silly mistake.
AI: Recall the definitions:
$F\colon A\to B$ is a bijection if two things are true:
For every $x,y\in A$ if $x\neq y$ then $F(x)\neq F(y)$ (injectivity); and
For every $b\in B$ there is some $a\in A$ such that $F(a)=b$ (surjectivity).
Now we identify this in the problem, $A=\mathbb R^2$ and $B=\mathbb R$, and $F(\langle x,y\rangle)=x^2+y^2$.
Is this $F$ injective (do the ordered pairs $\langle a,b\rangle$ and $\langle b,a\rangle$ have different images)?
Is this $F$ surjective (can the sum of two non-negative numbers be $-1$)?
Even if only one of these is true the definition of bijection no longer holds, in our case - both are true. |
H: Line integral with only one unknown
Let $\Gamma$ be a circumference with center $3i$ and radius $5$
oriented counter-clockwise. Calculate: $$\displaystyle\oint_{\Gamma} \frac{z}{(z^2 -2z)(z^2 - 4z + 13)} dz$$
At a first glance I thought it was a line integral but then I realized the function $f(z)$ has only one unknown ($z$), while it should have 2 such that I can parameterized it. Any hint is appreciated.
Thanks,
rubik
P.S. Sorry for the english: the problem is a translation from another language and I don't know whether I used the correct terms.
AI: You need to use Cauchy Residue Theorem. In your case, the integrand has poles at $z = 0,2,2 \pm 3i$. The poles $0,2,2+3i$ (except $2-3i$) are all inside the circle over which you are integrating. Hence you need to compute the residue at all these poles and add them up and multiply by $2 \pi i$ to get the value of the integral. |
H: Generating the Sorgenfrey topology by mappings into $\{0,1\}$, and on continuous images of the Sorgenfrey line
Show that the topology of the Sorgenfrey line can be generated be a family of mappings into a two-point discrete space.
Verify that the Sorgenfrey line can be mapped onto $D(\aleph_0)$ but cannot be mapped onto $D(\mathfrak{c})$ (where $D(\kappa)$ is the discrete space of cardinality $\kappa$).
AI: For each real $a$ define a function $f_a : \mathbb{R} \to \{ 0,1 \}$ by $$f_a (x) = \begin{cases}0, &\text{if } x < a \\ 1, &\text{if }x \geq a.\end{cases}$$ Show that the topology on $\mathbb{R}$ generated by this family of functions is the Sorgenfrey line.
(a) Consider $\mathbb{Z}$ with the discrete topology. Define $f : \mathbb{R} \to \mathbb{Z}$ by $f (x) = \lfloor x \rfloor$ (where $\lfloor x \rfloor$, the floor of $x$, denotes the greatest integer not (strictly) greater than $x$). Show that this is continuous (and onto).
(b) Note that the Sorgenfrey line is separable (consider $\mathbb{Q}$, the set of rationals), and recall that any continuous image of a separable space is separable. Is $D(\mathfrak{c})$ separable? |
H: Compute: $\lim_{n\to\infty} ({x_{2}}^n+{x_{3}}^n)$
Let be $ x_{1}, x_{2}, x_{3}$ the roots of $x^3-x^2-1=0$. If $x_{1}$ is the real root of the equation, then calculate:
$$\lim_{n\to\infty} ({x_{2}}^n+{x_{3}}^n)$$
I wonder if this limit can be computed without being necessary to know the exact values of $x_{2}$ and $x_{3}$.
AI: We will prove that the magnitude of the complex roots is less than $1$. Let $f(x) = x^3-x^2-1$. Then $f(1) = -1 <0$ and $f(2) = 3>0$. Hence, there is a real root, $x_1$, in the interval $(1,2)$.
You can prove that this is the only real root and the other two have to be complex. This can be done by looking at the derivative in the interval $(1,2)$ which will turn out to be positive and a local maximum occurs at $x=0$ and a local minimum occurs at $x= \dfrac23$. These guarantee that the remaining two roots are complex.
The complex roots $x_2$ and $x_3$ can be written as $x_2 = r e^{i \theta}$ and $x_3 = r e^{- i \theta}$ since the complex roots occur in conjugate pairs.
But the product of the roots is $1$ i.e. $r^2 x_1 = 1 \implies r^2 = \dfrac1{x_1} < 1$.
The complex roots must have magnitude less than $1$. Hence, $$x_2^n + x_3^n = 2r^n \cos(n \theta)$$
Hence, $$\lim_{n \to \infty}\left( x_2^n + x_3^n \right) = \lim_{n \to \infty} 2r^n \cos(n \theta) =0 \text{ (Since $0<r < 1$)}$$ |
H: Finding all integer solutions for $x^2 - 2y^2 =2 $
I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2 =1$ $y=3, x=4$, so form Pell I get that $\alpha= (4+3\sqrt{2})^n$ for every integer $n$, and a private solution for $-\frac{1}{2}x^2+ y^2 =-1$ is $y=1, x=2$, so the total solution is $\alpha= (1+\sqrt{2}) \cdot (+/- (4+3\sqrt{2})^n)$. Are all these steps correct? and if not- how should I solve this one?
Thank you!
AI: The standard way is to find one solution to $x^2-2y^2=2$, e.g., $x=2$, $y=1$, and find the fundamental solution to $x^2-2y^2=1$, which is $x=3$, $y=2$, and then go $(2+\sqrt2)(3+2\sqrt2)^n$, etc., etc. |
H: Eigenvectors of a matrix and its diagonalization
I'm trying to understand the relation (if any) between the eigenvectors of similar matrices and in particular of a matrix and its diagonalization.
Given $A,D\in M^F_{n\times n}$ and invertible $P$ such that $P^{-1}AP=D$ then $AP=PD$ and the eigenvectors of A are the columns of $P$ because $AP_i=\lambda_iP_i$ and $P$ is a change of basis matrix from whatever basis $A$ is in to whatever basis $PD$ is in. $D$ itself is obviously diagonalizable, and its eigenvectors are the columns of $I$ which won't equal $P$ unless $A=D$. And that's as far as I can get at the moment.
EDIT
As Marvis noted, the heart of the question is what is the relationship ( if any ) between the eigenvectors of two similar matrices.
AI: If $A$ has eigenvector $v$ with eigenvalue $a$, and $A$ is similar to $B$, say $B=C^{-1}AC$, then let $w=C^{-1}v$; then $$Bw=C^{-1}ACC^{-1}v=C^{-1}Av=C^{-1}av=aC^{-1}v=aw$$ so $w$ is an eigenvector of $B$. |
H: What is a formal definition of series?
Is there a formal definition for series?
For example, cardinal sum has a formal definition such that $\sum a_i$ = $\bigcup a_i$.
Is there any clear definition for series of real or complex number?
The definition on my book is the sum of $a_0 + ... + a_n$. This seems very intuitive to me so i don't like it..
I tried to define series such that $\gamma(0) = a_0$ and $\gamma(n+1) = f_n(\gamma(n))$ where $f_n(x) = x+ a_n$
Since $f_n ≠ f_{n+1}$, i can't apply finite recurssion theorem. However it seems obvious that $\gamma$ is a function and unique. How do i prove the existence and uniqueness of $\gamma$?
I have proved that "If A is a set, 'c' a fixed point in A and $f_n : A →A$ a function for every $n\in \mathbb{N}$, then there exists a unique function $\gamma : \mathbb{N} →A$ such that $\gamma(0) = c$ and $\gamma(n+1) = f_n(\gamma(n))$. This is a bit generalized form of original finite recursion theorem.
Let $\alpha$ be a sequence. Let $f_n(x)=x+\alpha(n+1) : F→F$. (F denotes an arbitrary field here) Then by above theorem, we can construct $\gamma$. It can be easily checked that $\gamma(n)$ is a summation of $a_0,...,a_n$.
By the way, i've never said series is a 'finite summation'. Of course, series is a sequence..
AI: Take any sequence $(a_0,a_0,\ldots)$ of real numbers. Define a function $r$ on finite sequences of real numbers such that if $s$ has length $n-1>0$, $r(s)$ is the last value of $s$ plus $a_n$. If the length of $s$ is $0$ (which we allow), then $r(s)=a_0$. The (countable) recursion theorem gives you a unique sequence $(b_0,b_1\ldots)$ such that $b_{n+1}=b_n+a_{n+1}$ and that defines your series. Note that the recursion is allowed to depend on all finite sequence of values.
There are two things you might think are not rigorous here: The "length" and the "last value". A finite sequence is a function with domain being a finite ordinal. This ordinal is the length, which is therefore well-defined. the elements of an ordinal are again ordinals and the value at the largest such ordinal in the domain is the "last value". |
H: A problem on joint distribution
Suppose that the joint distribution of $X$ and $Y$ is uniform over the region in the $xy$-plane bounded by $x=-1,x=1,y=x+1, \text{ and }y=x-1$.
What is $\mathbb{P}(XY>0)$?
What is the conditional p.d.f. of $Y$ given that $X=x$?
AI: The region in the $XY$-plane is as shown below.
HINT for the first part. Identify the regions where $XY > 0$. And integrate over the region to get $\mathbb{P}(XY > 0)$.
HINT for the second part. Recall that $f_{Y|X=x} = \dfrac{f_{XY}}{f_X}$, where $f_X = \displaystyle \int_y f_{XY} dy$. |
H: Preimage of a point by a non-constant harmonic function on $\mathbb{R}$ is unbounded
Let $u$ be a non-constant harmonic function on $\mathbb{R}$. Show that $u^{-1}(c)$ is unbounded.
I am not getting what theorem or result to apply. Could anyone help me?
AI: Let $u(x)=x$, for all $x \in \mathbb{R}$. Then $u''(x)=0$ for all $x$. But $u^{-1}(\{c\}) = \{c\}$. I think somebody is cheating you :-) |
H: Construction a sequence of real numbers
Can we construct a sequence $\{a_{i}\}$, where $0<a_{i}\leq 1$ for all $i$, such that $\sum_{i=1}^{n}a_{i}=\frac{B_{n}}{2^{n}}$ such that $B_{n}\to b$ as $n\to \infty$ for some $1\leq b <\infty$?
Edit: What about $\sum_{i=1}^{n}a_{i}=\frac{n}{n+1}$, or $\sum_{i=1}^{n}a_{i}=B_{n}$ with $B_{n}\to b$, could we find such $a_{i}$?
AI: Whether $(B_n)$ converges or not does not really matter. That such a sequence exists is equivalent to $$0<\underbrace{\frac{B_{n+1}}{2^{n+1}}-\frac{B_n}{2^n}}_{\text{ this would be }a_{n+1}}\leq 1,$$ for $n\geq 1$, since this uniquely determines the sequence $(a_n)$ as above with $a_1=B_1/2$. |
H: Counting number of ways of splitting a card deck
Suppose that we have a $52$-card deck. We are interested to find how many different combinations there could be if we divide this $52$-card deck in two parts, so that in each part there are $2$ aces.
What I am thinking is that we have $4$ aces, we can choose two of them in $\dfrac{4!}{2!}=12$ ways.
Also $52$ cards, we can divide in two parts in $\dfrac{52!}{2!}$ ways, so I have to multiply $12$ by $\dfrac{52!}{2!}$? It is a huge number, so am I missing something?
AI: There are 3 ways to split the aces: SH vs DC, SD vs HC, and SC vs HD (that's Spades, Hearts, Diamonds, Clubs).
Having split the aces, each of the other 48 cards has 2 possible destinations. So the answer would be $$3\times2^{48}$$ |
H: System of Parameters.
Let $R=k[X,Y,U,V]/(XV-YU)$, where $k$ is field of characteristic $0$. Consider $S=R_m$, where $m$ is the maximal ideal $(X,Y,U,V)/(XV-YU)$. How can we find a system of parameters for $S$ and what are they? We know that there are 3 elements in any system of parameters, as $S$ is a 3-dimensional domain.
AI: Alternatively, $\{x_1, \ldots, x_s\} \subset S$ is a SOP if the ideal $(x_1, \ldots, x_s)$ they generate is $m$-primary, in other words, if the ring $S/(x_1,\ldots,x_s)$ is $0$-dimensional. Then, we see that choosing $\{\overline x,\overline v,\overline y-\overline u\}$ yields $S/(\overline x,\overline v,\overline y-\overline u) \cong K[\overline y]/(\overline y^2),$ which is $0$-dimensional, by the commutativity of localization with quotients. Thus, $\{\overline x,\overline v,\overline y-\overline u\}$ is indeed a SOP for S.
(This definition of SOP is in Matsumura's Commutative Ring Theory, by the way.) |
H: Polynomial of same degree
Let $p(z)$ and $q(z)$ are two polynomial of same degree and zeroes of $p(z)$ and $q(z)$ are inside open unit disc, $|p(z)|=|q(z)|$ on unit circle then show that $p(z)=\lambda q(z)$ where $|\lambda|=1$.
Please just give a hint not the whole solution. Thank you.
AI: Let $d$ the common degree of $P$ and $Q$. Define $P_1(z):=z^dP(1/z)$ and $P_2(z):=z^dQ(1/z)$. These polynomial doesn't vanish in the unit disk. We apply maximum modulus principle to $P_1/P_2$ and $P_2/P_1$. This gives that $P_1/P_2$ is constant (otherwise we get a contradiction because $P_1/P_2$ would be a holomorphic function with a constant modulus on a connected open set, hence would be itself constant).
Note that the fact that $P$ and $Q$ have the same degree is necessary. Indeed, $P(z)=z$ and $Q(z)=z^2$ have their root in the open unit disk, the same modulus on the unit sphere but of course are not equal up to a constant. |
H: Unique isomorphism between fields generated by a domain.
Suppose $F$ and $K$ are fields both generated by a common subring $D$, which is a domain. My question is, why is there a unique isomorphism between $F$ and $K$ which is the identity on $D$?
Wouldn't the field generated by $D$ be unique? So $F=K$, and then any isomorphism is determined by its images on a generating set, that is, $D$, so such an isomorphism is unique? If this is so, how can we be sure that there exists an isomorphism which restricts to the identity on $D$?
Second thoughts: Let $F$ be the field of fractions of $D$. Then I let $\iota\colon D\to F_1$ be the identity embedding, which is known to have a unique extension to a monomorphism of $F$ into $F_1$. Since $F_1$ is generated by $D$, it is isomorphic to $F$, so this unique monomorphism is an isomorphism? Likewise, there is a unique isomorphism between $F$ and $F_2$ which is the identity on $D$, so there is a unique isomorphism between $F_1$ and $F_2$ which is the isomorphism on $D$. Is this right?
AI: Yes you are right: every field generated by a given domain $D$ is canonically isomorphic to the field of fractions of $D$, by the argument you given under "Second thoughts". And this isomorphism is the unique one that extends the embedding of $D$. So as long as your two fields have the same domain $D$ in common, there is a unique way to exend this identification on $D$ to one on all of the fields.
Should your fields instead just have isomorphic generating subdomains, then you must deal with the possibility that there are multiple isomorphisms between those subdomains, each of which will extend differently to isomorphisms between the fields. But that does not seem to be the situation given in the question. |
H: Modular arithmetic for negative numbers
If I have the congruence
$$m^2 \equiv -1 \pmod {2k+1}$$
how do I solve for the solutions to this congruence (given that I know $k$)?
AI: As others have pointed out, when dealing with congruences the concept of a negative number is meaningless (as is the concept of a positive number). Judging from the comments you are, in addition to this, asking about the solvability of the congruence $m^2\equiv -1\pmod q$, where $q$ is some odd integer. Here the theory says that this congruence has solutions, if and only if all the prime divisors of $q$ are congruent to $1\pmod 4$.
For example, when $q=7$ there are no solutions, because $7\not\equiv1\pmod4$, but when $q=13$, there are solutions $m\equiv\pm5$. When $q$ is a prime, $m=\pm((q-1)/2)!$ are the only non-congruent solutions.
The general theory goes via the usual route: solve it for prime number moduli => solve for prime power moduli (by "lifting") => solve for general moduli with the aid of Chinese remainder theorem. The number of non-congruent solutions depends on the number of prime factors of $q$.
[Edit: In response to Thomas' comment.] If $m^2\equiv-1\pmod q$, then this can be lifted to a solution modulo $q^k$ for all positivie integers $k$. Lifting means that if we have, for some positive integer $k$, a solution $m_k$ such that $m_k\equiv m\pmod q$ and $m_k^2\equiv -1\pmod{q^k}$, then we can find an integer $m_{k+1}$ such that $m_{k+1}=m_k+aq^k$ and $m_{k+1}^2\equiv -1\pmod{q^{k+1}}$. This is not difficult, because we know that $m_k^2=-1+bq^k$ for some integer $b$, and can use this to solve $a$ from the
congruence
$$
m_{k+1}^2=m_k^2+2am_kq^k+a^2q^{2k}\equiv-1+(b+2am_k)q^k\pmod{q^{k+1}},
$$
because this is equivalent to the linear congruence
$$
b+2am_k\equiv0\pmod q.
$$
Here $\gcd(2m_k,q)=1$ so the solutions $a$ of this congruence form a unique residue class modulo $q$.
As an example, let us lift the solution $m=m_1=5$ of the congruence $m^2\equiv -1\pmod{13}$
to a square root of $-1$ modulo $13^2$. Here $m_1^2=25=-1+2\cdot13$, so $b=2$. We want to solve the linear congruence
$$
b+2am_1=2+2\cdot5a\equiv0\pmod{13}.
$$
The usual method for solving a linear congruence yields $a\equiv5\pmod{13}$ as the solution. Therefore $m_2=5+13a=5+5\cdot13=70$ should be a solution. Indeed,
$$
m_2^2=4900=-1+4901=-1+13\cdot377=-1+13^2\cdot29\equiv-1\pmod{13^2}.
$$
[/Edit]
[Edit^2: An example on using the CRT]
Assume that we are given the task to solve the equation
$$
m^2\equiv-1\pmod{2873}.\tag{1}
$$
The process begins by factoring $2873$. If you know this an advance, you are lucky, because otherwise this could be tricky. We simply observe that
$2873=13^2\cdot17$. It is our lucky day, because all the prime factors are $\equiv 1\pmod 4$ (of course I reverse engineered this example a bit). The idea is that we next solve the
congruences
$$
m^2\equiv-1\pmod{13^2}\tag{2}
$$
and
$$
m^2\equiv -1\pmod{17}\tag{3}$$
separately. In the previous example I showed how to lift the "guessable" solution $5^2\equiv1\pmod{13}$ to a solution $m=70$ of equation $(2)$, so let's reuse that.
We observe that $-70\equiv99\pmod{13^2}$ is then the other solution of $(2)$.
Finding the solutions of $(3)$ is also easy, because $17$ is a relatively small number.
There are several methods. Either we simply observe that $m=\pm4$ are solutions. Or, as above, we can calculate that $(17-1)/2=8$ and $8!=40320\equiv 13\equiv-4\pmod{17}$. We can also use the quicker (for primes a bit larger than $17$) but non-deterministic method that for any integer $a, 0<a<17$, the power $x=a^{(p-1)/4}=a^4$ is a solution of either the equation $x^2\equiv1$ or $x^2\equiv-1$. The non-deterministic part is that we don't know which it is. Let's try. With $a=2$ we get $x=16\equiv-1$, which is not a solution of $(3)$.
But with $a=3$ we get $x=81\equiv13\equiv-4$, which we already knew to be a solution. Anyway, we now know that $(3)$ holds, if and only if $m\equiv\pm4\pmod{17}.$
Because $17$ and $13^2$ are coprime (no common prime factors), the Chinese remainder theorem tells us that $m$ satisfies congruence $(1)$, if and only if it satisfies both congruences $(2)$ and $(3)$. We can also conclude from CRT that there is a single residue class modulo $2873$ that satisfies e.g. both congruences $m\equiv70\pmod{13^2}$ and $m\equiv 4\pmod{17}$. The extended Euclidean algorithm gives us a method for finding integers $u$ and $v$ such that $$u\cdot 169+v\cdot 17=1,$$ for example $u=-1,v=10$. What this means is that the number $e_1=v\cdot17=170$ is congruent to $1$ modulo $13^2$ to $0$ modulo $17$, and OTOH the number $e_2=u\cdot169=-169$ is congruent to $0$ modulo $13^2$ and to $1$ modulo $17$.
Therfore
$$m_1=70\cdot e_1+4\cdot e_2=70\cdot170-4\cdot169=11224\equiv2605\pmod{2873}$$ is congruent to $70$ modulo $13^2$ and to $4$ modulo 17. So it should be a solution to $(1)$. With the aid of a calculator we can check
$$
2605^2=6786025\equiv-1\pmod{2873},
$$ so this is, indeed a solution. Using other solution to congruences $(2)$ and $(3)$ we get a total of of four non-congruent solutions:
$$
m_2=99\cdot e_1+4\cdot e_2=16154\equiv1789\pmod{2873}$$
is congruent to $99$ modulo $13^2$ and to $4$ modulo $17$. Using the residue class $-4$ modulo $17$ gives us two more solutions
$$
m_3=70\cdot e_1-4\cdot e_2=12576\equiv 1084\pmod{2873},$$
and
$$m_4=99\cdot e_2-4\cdot e_2=17506\equiv 268\pmod{2873}.$$ |
H: $|G|=12$ and no elements of order $2$ in $Z(G)$
I am thinking on the following problem:
If $|G|=12$ and there is no element of order $2$ in its center then $3$-Sylow subgroup of $G$ cannot be normal in $G$.
I was told that to assume the $3$-Sylow subgroup of $G$, say $P$, is normal in the group and go to reach a contradiction.
My attept: $|P|=3$ so it is cyclic, $P=\langle x\rangle=\{1,x,x^2\}$. As above hint, I would have for all $g\in G$ two possibilities: $gxg^{-1}=x$ or $gxg^{-1}=x^2$. Cannot to go any further.
I am in doubt if the hint leads me to a contradiction properly and if so, it does with a long proof. Any help or any other way are welcome to me. Thanks for your time.
AI: Hint #1: Show that all Sylow 2-subgroups are abelian. So if an element of a Sylow 2-subgroup commutes with the elements of $P$, then it belongs to the center.
Hint #2: At this point the questions you should ask from yourself are: 1) What possibilities (up to isomorphism) are there for the Sylow 2-subgroup H? 2) Given that $P$ was assumed to be normal, what possibilities are there for the conjugation action for the non-trivial elements of $H$. Remember that conjugation action is a homomorphism from $H$ to $Aut(P)$. |
H: How to show the dimensionality of a tangent space defined via equivalence classes of curves?
Lets have smooth $n-$ manifold $M\subset R_N$ and define its tangent space at $p\in M$ to be set of equivalence classes of smooth curves with $\gamma: I\to M$, $\gamma(0)=p$ with relation $\gamma_1\sim\gamma_2$ if $\frac{d\gamma_1(0)}{dt}=\frac{d\gamma_2(0)}{dt}$. How to prove that $T_p(M)$ is $n$ dimensional vector space?
AI: Let $(U,\varphi)$ a chart in $p$. Let $\theta_\varphi:T_pM\rightarrow\mathbb R^n$ defined by $\theta_\varphi([\gamma])=(\varphi\circ\gamma)'(0)$.
Then $\theta_\varphi$ is clearly injective (by the definition of your equivalence relation). We check that it's surjective : let $v\in\mathbb R^n$, let $\gamma:t\mapsto\varphi^{-1}(tv+\varphi(p))$ then $v=\theta_\varphi([\gamma])$.
So, $\theta_\varphi$ is a bijection and we can use it to transfer the vector space structure from $\mathbb R^n$ to $T_pM$ :
If $\xi,\eta\in T_pM$, we define $\xi+\eta=\theta_\varphi^{-1}(\theta_\varphi(\xi)+\theta_\varphi(\eta))$ and if $\lambda\in\mathbb R$, we define $\lambda\xi=\theta_\varphi^{-1}(\lambda\theta_\varphi(\xi))$.
You can easily check that you get a space vector and that $\theta_\varphi$ is a linear isomorphism.
This construction is independant of the choice of $(U,\varphi)$ : if $(V,\psi)$ is another chart containing $p$, then $\theta_\varphi\circ\theta_\psi^{-1}(v)=d_{\psi(m)}(\varphi\circ\psi^{-1})(v)$. |
H: Normal subgroups of p-groups
Let $G$ be a group of order $p^\alpha$, where $p$ is prime. If $H\lhd G$, then can we find a normal subgroup of $G/H$ that has order $p$?
AI: Theorem: a group $\,G\,$ of order $\,p^n\,,\,p\,$ a prime, $\,n\in\mathbb N\,$ , always has normal subgroup of order $\,p^m\,\,,\,\,\forall\, m\leq n\,\,,\,m\in \mathbb N$
Proof: Exercise, using that always $|Z(G)|>1\,$ and induction on $\,n\,$
So the comments by Gerry and Geoff close the matter. |
H: Solving $\sin 7\phi+\cos 3\phi=0$
The question is find the general solution of this equation:$$\sin(7\phi)+\cos(3\phi)=0$$
I tried to use the "Sum-to-Product" formula, but found it only suitable for $\sin(a)\pm \sin(b)$ or $\cos(a)\pm \cos(b)$. So I tried to expand $\sin 7\phi$ and $\cos 3\phi$, but the equation became much more complicated..
I'm self studying BUT There's nothing about how to solve this type of equations on my textbook..
reeeaaaaally confused now..
AI: Hint: Use sum to product!
$$\sin 7\phi+\sin \left(\frac{\pi}{2}-3\phi \right)=0$$ |
H: Finding a proper sequence
The following function was given to me $$f(x)=\lfloor x\rfloor+\lfloor-x\rfloor$$
wherein $\lfloor x\rfloor$ is the floor function of $x$. I was asked to select a proper sequence for showing that this function has no limit at $\infty$.
Honestly, my knowledge about analysis is weak. Thank you
AI: Take $$f(n)=[n]+[-n]=n-n=0, \quad n\in\mathbb{N},$$ and then take $$f\left(\frac{2n+1}{2}\right)=n+(-n-1)=-1, \quad n\in\mathbb N$$
Then you have two different sequences with different limits when $\,n\to\infty\,$ and thus the limit at $\,\infty\,$ doesn't exist. |
H: How to denote a "boolean" (on/off, 1/0) variable in mathematics?
Is there any conventional notation for variables that can only take the value 0 or 1? (I'm looking for something of the nature of an overbar, a caret, etc.)
AI: I don't know of such notation. You can always define that $\dot x$ means that $x$ is a Boolean variable with values in $\{0,1\}$.
Of course the dot can be replaced by other symbol. Be forewarned, though, that there are many many different contexts in which these symbols already have meaning. If you specify what you are going to use this notation for (logic, comp. sci., etc.) maybe some better suggestion will be given.
Until then, I think my first suggestion should probably fit. |
H: The congruence $f(x)=x^3+3x+9 \equiv 0 (\bmod 5^n)$ has only one solutions for every $n \geq 2$
I need to prove that the congruence $f(x)=x^3+3x+9 \equiv 0 (\bmod 5^n)$ has only one solutions for every $n \geq 2$.
I checked with Hensel theorem that for $n=2$ there is one solution indeed.
I want to use induction and to use Hensel theorem again, so I assumed that for $n$ there is one solution,r, but for using Hensel Lemma for $n+1$ how do i know that $f'(r)\not\equiv 0 (\bmod 5) $?
Thanks!
AI: Here $f'(x)=3x^2+3=3(x^2+1)$ vanishes modulo $5$, iff $x\equiv\pm2\pmod 5$. Modulo $25$ considerations dictate $r\equiv19\pmod{25}$, so you always have $r\equiv-1\pmod5$, and the assumptions of Hensel's lemma are satisfied. |
H: how to change polar coordinate into cartesian coordinate using transformation matrix
I would like to change $(3,4,12)$ in $xyz$ coordinate to spherical coordinate using the following relation
It is from the this link. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. Also please check my previous question building transformation matrix from spherical to cartesian coordinate system. Please I need your insight on building my concept.
Thank you.
EDIT::
I understand that $ \left [ A_x \sin \theta\cos \phi \hspace{5 mm} A_y \sin \theta\sin\phi \hspace{5 mm} A_z\cos\theta\right ]$ gives $A_r$ but how is other coordinates $ (A_\theta, A_\phi)$ equal to their respective respective rows from Matrix multiplication?
AI: The transformation from Cartesian to polar coordinates is not a linear function, so it cannot be achieved by means of a matrix multiplication. |
H: Finding the set of solutions
Kindly asking to find the set of possible solutions if they exist of the equation $$\lfloor x+\lfloor x+\lfloor x\rfloor\rfloor\rfloor+3\lfloor x\rfloor=18$$ Of course I have 4 intervals as choices:
$[\frac{5}{2}, \frac{7}{2})$
$[3, 5)\smallsetminus \{4\}$
$[3, 4)$
$\emptyset$
$\lfloor x\rfloor$ is largest integer not greater than $x$. Thank you very much.
AI: Hint: show that if $n$ is an integer, and $r$ is any real number, then
$$\lfloor r+n\rfloor=\lfloor r\rfloor+n.$$
Apply this to the expression $\lfloor x+\lfloor x+\lfloor x\rfloor\rfloor\rfloor$. |
H: Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite?
Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite?
AI: Yes. For every $r\in\mathbb R$ choose a sequence of rational numbers $\{r_n\in\mathbb Q\mid n\in\mathbb N\}$ which converges monotonically to $r$, this sequence is of course a subset of $\mathbb Q$ - a countable set.
If $r\neq s$ are two real numbers then the sequence we chose for them must intersect at a finite subset, otherwise we had a subsequence of the two which would converge to two different limit points.
Since $\mathbb R$ is uncountable (and in fact has cardinality as $\mathcal P(\mathbb Q)$) we have indeed uncountably many subsets of $\mathbb Q$ with the wanted property. |
H: Definition and meaning of "Proof Schema", "Class Sign"
I'm a newbie in advanced mathematics, and I'm trying to understand Godel's theorem. I came across these two words which I couldn't understand clearly.
"Proof Schema" and "Class-Sign"
Can anybody provide me definition of these, and describe what these term mean in simple words?
It's from Kurt Godel's book (translated into English) "On Formally Undecidable Propositions of Principia Mathematica and Related Systems".. by Dover publications.. I'll just quote lines where it's introduced (Page No 39)-
"It can be shown that "formula", "proof-schema", and "provable formala" are definable in the system of PM.
Class-Sign:
"A formula of PM with just one free variable, and that of the type of natural numbers(class of classes), we shall designate a 'class-sign'..
Thank you
AI: By "proof schema" Gödel just means a formal proof. You can see this quite clearly from the context: a formula is a finite sequence of natural numbers, each of which codes one of the symbols of the formula, while a proof schema is a finite sequence of finite sequences of natural numbers, in other words a finite sequence of codes of formulae.
The basic idea of Gödel coding is very simple: take the symbols of a formal language (the alphabet) and represent each symbol by a natural number in such a way that given any string of such symbols we can produce a finite sequence of symbols which represent them (encoding); and given any natural number there is an effective procedure by which we can calculate which symbol it represents (decoding).
A "class sign" is what we would normally in English call a predicate: a formula with one free variable $\varphi(x)$ whose extension is the class of objects $\left\{ x : \varphi(x) \right\}$. Since we're only considering natural numbers here the extension in question will of course be a set, that is, the set of natural numbers $n$ such that $\varphi(n)$. (Being fussy, we might say that a class sign is a unary predicate since it has only one free variable; one can have predicates of any finite arity.)
Defining these things in a precise manner is key to the method of arithmetisation, whereby logical notions are shown to be expressible in the language of arithmetic, and thus sentences in that language can be constructed which refer to themselves. This is called the method of diagonalisation. |
H: How can I calculate this limit: $\lim\limits_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$?
I was given this limit to solve, without using L'Hospital rule. It's killing me !! Can I have the solution please ?
$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$
AI: $$ \lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} $$
$$ = \lim_{x\rightarrow 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} \times \frac{2+\sqrt{2+x}}{2+\sqrt{2+x}} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3} }{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}$$
$$ = \lim_{x\rightarrow 2} \frac{2 - x}{-(2-x)} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}{2 + \sqrt{2+x}}$$
$$= -\frac{3 (2) ^{2/3}}{4} = -\frac{3}{2(2)^{1/3}}$$ |
H: How is $\Vert x \Vert_3$ constrained by $\Vert x \Vert_2$ and $\Vert x \Vert_4$ for $x\in \ell^2$?
It seems we can find some $x\in \ell^2$ with $\Vert x \Vert_2=1$ that has $\Vert x \Vert_4=a$ for any $0<a\le 1$.
But can we find an $x$ with $\Vert x \Vert_2=1,\Vert x \Vert_3=b,\Vert x \Vert_4=a$ for every choice of $0<a<b<1$?
(This was inspired by this longstanding MO question that made me curious about the flexibility of the three norms.)
AI: This is extended version of D. Thomine's comment. It is known that for a given real numbers $\{\theta_k:k\in\{1,\ldots,n\}\}$$\subset(0,1)$, such that $\sum_{k=1}^n \theta_k=1$ and real numbers $\{p_k:k\in\{1,\ldots,n\}\}\subset\mathbb{R}_+$ we have the following generalized Hölder inequality:
$$
\Vert x\Vert_{p}\leq\prod\limits_{k=1}^n\Vert x\Vert_{p_k}^{\theta_k}
$$
where $p^{-1}=\sum_{k=1}^n\theta_k p_k^{-1}$.
In your particular case we take $n=2$, $\theta_1=1/3$, $\theta_2=2/3$, $p_1=2$ and $p_2=4$. Then we get $p=1/3$ and
$$
\Vert x\Vert_3\leq\Vert x\Vert_2^{1/3}\Vert x\Vert_4^{2/3}
$$
which is equivalent to $b\leq a^{2/3}$. Thus in general we can't find $x\in\ell_2$ satisfying all conditions mentioned above. |
H: Show $|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$ for complex $a$, $b$, $c$
How to prove for any complex numbers $a$, $b$, $c$, the inequality $$|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$$ is correct?
AI: Both sides are non-negative, so it suffices to show that the square of the left-hand-side is at least the square of the right-hand-side. That is, we wish to show:
$$
|a|^2+|b|^2+|c|^2+|a+b+c|^2+2|ab|+2|bc|+2|ac+2(|a|+|b|+|c|)|a+b+c|
\geq\\
|a+b|^2+|b+c|^2+|a+c|^2+2(|a(a+b+c)+bc|+|b(a+b+c)+ac|+|c(a+b+c)+bc|)
$$
The square terms cancel:
$$
|a|^2+|b|^2+|c|^2+|a+b+c|^2 = 2|a|^2+2|b|^2+2|c|^2+2\operatorname{Re}(ab+bc+ac)=|a+b|^2+|b+c|^2+|a+c|^2
$$
and by the triangle inequality we have $|a(a+b+c)|+|bc|\geq |a(a+b+c)+bc|$ and cyclic permutations. |
H: Does $\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}$ converge?
Does the following series converge or diverge?
$$
\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}
$$
The methods I have at my disposal are geometric and harmonic series, comparison test, limit comparison test, and the ratio test.
AI: It is not hard to see that
$$\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^\infty(\sqrt{n+1}-\sqrt{n})$$
As you know this series is divergent. |
H: How to integrate $\int{\frac{dx}{\sqrt{16-9x^2}}}$
Me again, probably someone is going to blame me. I have this:
$$\int{\frac{dx}{\sqrt{16-9x^2}}}$$
I have asked an old teacher of mine an he said me I should let $x=\frac{4}{3}\sin{\big(\frac{3}{4}x\big)}$ but I don't how he realise that. I have tried $u-substitution$ and Integration by Parts but the square root is making the whole thing much harder.
I tried multiplying both terms by $\sqrt{16-9x^2}$ but nothing. Also tried to factor to eliminate square root but nothing.
I don't want that anyone solve my integral, just give me a hint about how do it, and how I can realise what to do in this cases :)
AI: In general, computing an integral is a guessing process. We usually rely on the fundamental theorem of calculus to provide us the initial guess i.e. if you want to compute $\displaystyle \int f(x) dx$, we need to first guess $F(x)$ such that $F'(x) = f(x)$. Then we make use of the fundamental theorem of calculus to conclude that $\displaystyle \int f(x) dx = F(x) + c$.
If you have an integral of the form $\displaystyle \int \dfrac{dx}{\sqrt{b^2 - a^2x^2}}$, recall that you would seen something similar before. You would have learnt that the derivative of $\arcsin(x)$ is $\dfrac1{\sqrt{1-x^2}}$. But now you the $a$ and $b$ hanging around. The goal now is to convert $\sqrt{b^2 - a^2x^2}$ into something like $\sqrt{1-u^2}$, so that we can then recognize that to be the derivative of $\arcsin(u)$.
Consider $\sqrt{b^2 - a^2x^2}$. The first thing is to pull out the $b$. This gives us $$\sqrt{b^2 - a^2x^2} = b \sqrt{1 - \dfrac{a^2}{b^2}x^2}$$ Now it looks very similar to a scaled version of $\sqrt{1-u^2}$ except for the coefficient in front of $x^2$. This gives us the motivation to make the substitution. $u^2 = \dfrac{a^2}{b^2}x^2$ i.e. $u = \dfrac{a}b x$.
In our case $a = 3$ and $b=4$. Hence, substitute $u = \dfrac34x$. We then get $du = \dfrac34 dx$. Plug this into the integral and recall that the derivative of $\arcsin(u)$ is $\dfrac1{\sqrt{1-u^2}}$ to get the integral.
Move you mouse over the gray area for the answer.
$$ I = \int \dfrac{dx}{\sqrt{16-9x^2}} = \int \dfrac43 \dfrac{du}{\sqrt{16 - 9 \times \dfrac{16}9 u^2}} = \int \dfrac43 \dfrac{du}{4\sqrt{1-u^2}} = \dfrac13 \arcsin(u) + c = \dfrac13 \arcsin \left( \dfrac34x\right) + c$$ |
H: cardinality of the set of countable partitions of $\mathbb{R}$
What is the cardinality of the set
$$A=\{ P| P\ \text{is a countable partition of the reals} \}$$ ?
I am searching on this for a while. I think the cardinality is $2^w$ where $w$ is the cardinality of $\mathbb{N}$. Clearly $|A|\geq |\mathbb{R}|=2^w$, but I cannot prove the other direction. I tried (but did not succeed)to construct an injection $A \rightarrow C$ for some set $C$ with cardinality equal to $2^w$.
AI: I think it's not very hard. A countable partition of reals can be seen as just a function from the reals into natural numbers, up to a permutation of the naturals. So we have
$$\lvert \mathbf N^{\mathbf R}\rvert=\aleph_0^{2^{\aleph_0}}=2^{2^{\aleph_0}}$$
(because $\aleph_0\leq 2^{\aleph_0}$ we can do the last substitution)
On the other hand, there are only $\aleph_0^{\aleph_0}=\mathfrak c<2^{\mathfrak c}$ permutations of natural numbers, so there are $2^{2^{\aleph_0}}$ equivalence classes of functions from reals to naturals under the relation of equivalence up to a permutation of naturals (because each has at most $\mathfrak c$ elements), which are precisely correspondent to the countable partitions of reals. |
H: Multiplying by invertible matrices maintains similarity
I know that given any two similar matrices $A,B\in M^F_{n\times n}$ and any two invertible matrices $P,Q$ of the same order then $P^{-1}AP$ is similar to $Q^{-1}BQ$.
However I don't completely understand why. Could someone explain it?
AI: We use the fact that if $A_1$ is similar to $A_2$ and $A_2$ to $A_3$ then $A_1$ is similar to $A_3$. To see that, write $A_2=P^{-1}A_1P$ and $A_3=Q^{-1}A_2Q$. Then
$$A_3=Q^{-1}P^{-1}A_1PQ=(PQ)^{-1}A_1PQ$$
and $PQ$ is invertible.
Now, to solve the problem, note that $P^{-1}AP$ is similar to $A$ and $Q^{-1}BQ$ is similar to $B$. |
H: For prime $p$, $p\equiv 5 (\bmod 8)$, and $a$ such that $\left(\frac{a}{p}\right)=1$: $a^{\frac{p-1}{4}}$ is either $1$ or $-1$
I'd really love your help with the following problem: For prime $p$, $p= 5 (\bmod 8)$, and $a$ such that $(\frac{a}{p})=1$ (Legendre symbol): I need to show that $a^{(p-1)/4}$is either $1$ or $-1$ and that if it's 1, so $x=a^{(p+3)/8}$ is a solution for $x^2 \equiv a(\bmod p)$.
For the first part, Can I look on $x^4 \equiv 1(p)$, and tell that if $a^{(p-1)/4}$ is not 1 or -1 I get contradiction to Fermat theorem, but is it enough? This equation can have up to 4 solutions, what else should I use? I don't see how the other information helps.
From the information given I know that $x^2 \equiv 2 (\bmod p)$ is not solvable (Does it help?)
For the second part of the question, Isn't it just that $(a^{(p+3)/8})^2\equiv a^{(p-1)/4}a \equiv a \equiv a(\bmod p) $?
Thanks a lot.
AI: Using Euler's criterion we know that:
$a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right) \bmod p$.
So:
$a^{\frac{p-1}{2}} \equiv 1 \bmod p$.
Now using difference of two squares we see that:
$(a^{\frac{p-1}{4}} + 1)(a^{\frac{p-1}{4}} - 1) \equiv 0 \bmod p$.
The fact that $p$ is prime now tells us that $a^{\frac{p-1}{4}} \equiv \pm 1 \bmod p$.
For the second part notice that we needed to know that $\left(\frac{a}{p}\right) = 1$ for there to actually exist a solution to $x^2 \equiv a \bmod p$. You have checked that if this is the case then $a^{\frac{p+3}{8}}$ would be a solution.
However, is $\frac{p+3}{8}$ necessarily an integer? In this case yes because $p \equiv 5 \bmod 8$. So this was an important condition too. |
H: On the Dirichlet beta function sum $\sum_{k=2}^\infty\Big[1-\beta(k) \Big]$
Given the Dirichlet beta function,
$$\beta(k) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^k}$$
(The cases k = 2 is Catalan's constant.) It seems,
$$\sum_{k=2}^\infty\Big[1-\beta(k) \Big] = \frac{1}{4}\big(\pi+\log(4)-4\big)=0.131971\dots$$
or, in general, for some constant p > 0,
$$\sum_{k=2}^\infty\left[1-\sum_{n=0}^\infty\frac{(-1)^n}{(pn+1)^k} \right] = \sum_{m=1}^\infty\frac{1}{2p^2m^2+3pm+1}$$
Anyone knows how to prove the general proposed equality? (This is similar to the question on the zeta sum here.)
AI: Here is a way to derive a slightly different looking result:
Notice that $$\sum_{k=2}^{\infty}\left[1-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(pn+1)^{k}}\right]=\sum_{k=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{k}}$$
$$=\sum_{n=1}^{\infty}(-1)^{n-1}\sum_{k=2}^{\infty}\frac{1}{(pn+1)^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{2}}\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}.$$ Now, since $$\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}=\frac{1}{1-\frac{1}{pn+1}}=\frac{pn+1}{pn},$$ our series is
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}.$$
Plugging in the case $p=2$ seems to agree with your first identity.
Remark: Using partial fractions, we can go a bit further. Notice that $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{pn}-\frac{1}{pn+1}\right)=\frac{\log 2}{p}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn+1} $$
Suppose $p$ is an integer, and let $\zeta_{p}$ be a $p^{th}$ root of unity. Then consider $$\frac{\log\left(1+z\right)}{z}+\frac{\log\left(1+\zeta_{p}z\right)}{\zeta_{p}z}+\cdots+\frac{\log\left(1+\zeta_{p}^{p-1}z\right)}{\zeta_{p}^{p-1}z}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}z^{n-1}\sum_{k=0}^{p-1}\zeta_{p}^{k(n-1)} $$
$$=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}z^{pn}.$$ Letting $z=1,$ we have the identity $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}=\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z},$$ so our original series is $$\frac{1}{p}\log2+\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z}.$$ |
H: Is $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$?
Why does my calculator return false when I input $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$ but W|A returns true?
I'm thinking my calculator is wrong because I know that $\displaystyle \log_{5}{(-3)} = \frac{\log(-3)}{\log(5)}$ and $\displaystyle \log(-x) = \log(x) + \pi i$ so that means that $\displaystyle \log_{5}{(-3)} =\frac{\log(3) + \pi i}{\log(5)}$.
So why does my calculator return false?
AI: Complex logarithms are multivalued, since the exponential is periodic with period $2 \pi i$. Thus, properly speaking,
$$ \log_5(-3) = \frac{\log(3) + \pi i + 2 n \pi i}{\log(5) + 2 k \pi i} n,k \in \mathbb{Z} $$
So, your calculator may be using different values of $n$ and $k$ from your $n=k=0$. In the vocabulary of complex analysis, we might say that your calculator is returning a value from a different branch than you expect.
Which calculator are you working with? |
H: What does it mean to say an integral exists 'in the distributional sense'?
What exactly does it mean to say that an integral exists 'just in the distributional sense'? For example, the Fourier transform of $x^2 e^{-\lambda x}$ or of $H(R-|x|)$ where $R > 0$ and $H$ is the Heaviside-Function, are often said to exist only in that sense.
AI: While the derivative in the distributional sense always exist, the same is not true for the Fourier transform or arbitrary (convolution) integrals. The Fourier transform in the distributional sense can be defined for tempered distributions. The space of tempered distributions $\mathcal{S}'$ is the dual space of the Schwartz space $\mathcal{S}$ (which is the function space of functions all of whose derivatives are rapidly decreasing). The Fourier transform is an automorphism on the Schwartz space, which enables to define the Fourier transform for its dual space by transposition, i.e. the Fourier transform of the tempered distribution $T\in \mathcal{S}'$ is defined via $$<\mathcal{F}(T),\phi>:=<T,\mathcal{F}(\phi)> \forall \phi\in\mathcal{S}$$
Regarding your question, $H(R-|x|)$ is a tempered distribution (or can be identified with a ...), but $x^2e^{-\lambda x}$ is not a tempered distribution. Perhaps you meant $x^2e^{-i\lambda x}$, which would be a tempered distribution. I propose this because your question might be motivated by the inverse Laplace transformation, where the $i$ is omitted.
Things get more tricky when it comes to multiplication (or convolution) of tempered distributions. These operations are not always defined. Even so it's easy to decide for two given tempered distributions whether their product (or convolution) is defined, writing down a set of sufficient conditions general enough to cover all cases which can occur in practice is nearly impossible. |
H: How to evaluate $\displaystyle{\int{\frac{1}{\sqrt{e^{2x}-4}}}\,dx}$
Please, could someone correct my exercise? I got it by myself, but the result that I found is differend by the WolframAlpha result... Could you tell me, please, if I'm wrong an where?
Thanks in advance!
AI: There is an error in the beginning. Setting $u^2 - 4 = s^2 \implies u^2 = s^2 +4$ and not $s^2-4$ as you have written. Because of this the entire integration is incorrect. |
H: a question on one one function on an open subset of $\mathbb{R}^n$
$1$)Let $A\subseteq\mathbb{R}^n$ be an open set and $f:A\rightarrow \mathbb{R}^n$ a continuously differentiable $1-1$ function such that $det f'(x)\neq 0$
for all $x$. Show that $f(A)$ is an open set and $f^{-1}: f(A )\rightarrow A$ is differ
entiable. Show also that $f(B)$ is open for any open set $B\subseteq A$.
$2$) Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is $\mathbb{C}^{\infty}$, show that $f$ can not ne $1-1$
Could any one tell me whether my solutions are correct or not?
$1$)For every $y\in f(A)$,there is an $x\in A$, By Inverse function theorem, there is an open set $U\subsetneq A$ and open set $V\subsetneq \mathbb{R}^n$ such that $x\in U$ and $f(U)=V$, since clearly $y\in V$ $f(A)$ is open(Can I say this?) Furthermore $f^{-1}:V\rightarrow U$ is differentiable( by IFT), this implies $f^{-1}$ is differentiable at $y$ and as $y$ was chosen arbitrarily from $f(A)$ so $f^{-1}:f(A)\rightarrow A$ is differentiable.
$2$)I show the result is true even if $f$ is only defined in a non-empty open subset of$\mathbb{R}^2$, we know that $f$ is not constant in any open set,So, suppose we have $D_1f(x_0,y_0)\neq 0$ so there is a nbd $U$ of $(x_0,y_0$) where $D_1f(x,y)\neq 0\forall (x,y)\in U$ The function $g:U\rightarrow \mathbb{R}^2$ defined by $g(x,y)=(f(x,y),y)$ satisfies $det g'(x,y)\neq 0\forall (x,y)\in U$ , here $f$ and $g$ are one-to-one and We can apply the result of problem $1$, The inverse function is clearly of the form $(h(x,y),y)$ and hence $(f(h(x,y),y),y)=(x,y)\forall (x,y)\in V=\{(f(x,y),y):(x,y)\in U\}$, Now $V$ is open but each horizontal line intersects $U$ atmost once since $f$ is one-one, This is a contradiction since $U$ is non-empty and open.
AI: 1) The statement "$f^{-1}\colon f(A)\to A$ is differentiable" is false. For example, take $n=2$ and $f(x,y)=(e^x\cos y, e^x \sin y)$. This map satisfies all the assumptions with $A=\mathbb R^2$, and we have $f(A)=\mathbb R^2$. Yet, there is no inverse $f^{-1}\colon f(A)\to A$.
What is true is that for sufficiently small neighborhoods there is an inverse, and this inverse is differentiable. Your proof is fine.
2) is fine. |
H: Is it assumable that $2^{1/12}$ is irrational because $2^{1/2}$ is?
I need to prove that $2^{1/12}$ is irrational but I need to connect this to $2^{1/2}$ being irrational.
I know how to prove that $2^{1/2}$ is irrational, but can I assume that $2^{1/12}$ is irrational because $2^{1/2}$ is and why?
Cheers
AI: If $2^{1/12}$ were rational, then $(2^{1/12})^6$ would also be rational. |
H: Formula for the sequence repeating twice each power of $2$
I am working on some project that needs to calculate what $a_n$ element of the set of numbers $$1, 1, 2, 2, 4, 4, 8, 8, 16, 16 \ldots$$ will be.
$n$ can be quite big number so for performance issues I have to calculate formula for this set of numbers (Note: first element may be different). What I've done so far is that I managed to find out how to calculate next element using element before. My formula for this is:
$$a_n = a_{n-1}\cdot 2^{\frac{1 + (-1)^{n-1}}2}$$
Now from this I want to calculate $a_n$ element using $a_1$ element. with this I am stuck.
AI: $a_n$ is termed as a "sequence" and not a series. Getting back to the question,
$$a_n = 2^{\lfloor n/2 \rfloor}$$ should do the job, where $\lfloor x \rfloor$ is the greatest integer $\leq x$, where $n$ goes from $0$. Usually the function to do this, is available through the command floor() in most languages.
If your first element is $a$ and not $1$, your $$a_n = a \times 2^{\lfloor n/2 \rfloor}$$ |
H: Small categories
Why $R$-Mod is a small category? There is a way to recognize small categories?
For example Grp (i.e. category of all groups) is large because every set can be equiped with a group structure.
AI: For any ring $R$, the category $R$-Mod isn't a small category: for any set $S$ one can form the $R$-module $R^{S}=\{f:S\to R\}$, and $R^S\neq R^T$ for any two distinct sets $S\neq T$ (though of course they may be isomorphic), so there are "at least as many $R$-modules as sets", and so the collection of all $R$-modules is a proper class.
In my experience with categories so far, I've never come across a situation where it wasn't clear from the outset (i.e., using what we already know about whatever the objects of our category are) whether a category was small, whether it was large, or whether it made no difference to the discussion at hand. |
H: Integral of $\frac{1}{\sin z}$ along a path
Suppose $\gamma$ is a simple, closed path, with $0$ in its interior and $\{\pi n:n\in\mathbb{Z}\setminus\{0\}\}\subset\mathbb{C}\setminus|\gamma|$. Find
$$
\int_{\gamma} \frac{1}{\sin z} dz
$$
Perhaps it's a simple question of Cauchy's formula or Cauchy's theorem.
Thanks.
AI: You can write $$\frac{1}{\sin z} = \frac{g(z)}{z}$$ where $g(z) := \frac{z}{\sin z}$ is a holomorphic function inside of $\gamma$ if we set $g(0)$ to be $1$. (L'Hopital's formula, etc.).
Cauchy's formula gives $$\int_{\gamma} \frac{dz}{\sin z} = \int_{\gamma} \frac{g(z)}{z} = 2\pi i g(0) = 2\pi i$$ |
H: Some properties of Yosida-Moreau transform
Let $f(x)$ be a continuous function on $\mathbb{R}^n$, $f(x) \geqslant 0$ for any $x$. Define
$$
f_{\alpha}(x) = \inf\limits_{y}\left( f(y) +\frac{|x-y|^2}{2\alpha} \right)
$$
where $\alpha > 0$. How to show that $f_{\alpha}(x)$ is locally Lipschitz continuous and $f_{\alpha} \to f$ when $\alpha \to 0+0$ uniformly on any compact $K$ in $\mathbb{R}^n$?
AI: Let us restrict attention to $|x|\le R$. Let $M=\sup_{|x|\le R}f$. The main observation is that the infimum can be taken only over $y$ such that $|x-y|\le \sqrt{2\alpha M}$. Indeed, when $|x-y|>\sqrt{2\alpha M}$, the second term makes the sum greater than $M$, hence greater than $f(x)$.
1) Let $R'=R+\sqrt{2\alpha M}$. For any fixed $y$ such that $|y|\le R'$, the function $x\mapsto f(y)+|x-y|^2/(2\alpha)$ is Lipschitz with a constant $L=L(R',\alpha)$. It is easy to see that the infimum of any family of $L$-Lipschitz functions is $L$-Lipschitz (or $\equiv -\infty$, which is what we rule out with a lower bound for $f$).
2) is now easy: if $\alpha$ is small, $\sqrt{2\alpha M}$ is small, so the infimum will not be far away from $f(x)$ by the uniform continuity of $f$ on $\{|x|\le R'\}$. |
H: Integral - using Euler Substitution
I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot get this result. Below is my solution of the problem. I've checked it many times, must be something very obvious that I'm missing:
(original image)
Euler Substituion
$\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}$
Let $\sqrt{x^2+x+1}=t-x$.
$x^2+x+1=t^2-2xt+x^2$
$x(1+2t)=t^2-1\implies x=\dfrac{t^2-1}{1+2t}$
$dx=\left(\dfrac{t^2-1}{1+2t}\right)'dt=\dfrac{2t(1+2t)-(t^2-1)2}{(1+2t)^2}=\dfrac{2t+4t^2-2t^2+2}{(1+2t)^2}=\dfrac{2(t^2+t+1)}{(1+2t)^2}$
$\sqrt{x^2+x+1}=t-x=t-\dfrac{t^2-1}{1+2t}=\dfrac{t^2+t+1}{1+2t}$
$\implies\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}=2\int\frac{\frac{t^2+t+1}{(1+2t)^2}\;dt}{\frac{t^2-1}{1+2t}\cdot\frac{t^2+t+1}{1+2t}}=2\int \frac{1}{t^2-1}\,dt$
$\dfrac{1}{t^2-1}=\dfrac{1}{(t+1)(t-1)}=\dfrac{A}{t+1}+\dfrac{B}{t-1}\implies \begin{eqnarray}&&At-A+Bt+B=1\\&&A+B=0\implies A=-B\\ &&B-A=1\implies B=\frac{1}{2},A=-\frac{1}{2}\end{eqnarray}$
$\implies \displaystyle 2\int \frac{1}{2}\frac{1}{2t-1}\,dt-2\int\frac{1}{2}\frac{1}{t+1}\,dt=\int\frac{1}{t-1}\,dt-\int\frac{1}{t+1}\,dt=$
$=\ln|t-1|-\ln|t+1|=\ln\left|\dfrac{t-1}{t+1}\right|$
$t-x=\sqrt{x^2+x+1}\implies t=\sqrt{x^2+x+1}+x$
$\implies \ln\left|\dfrac{t-1}{t+1}\right|=\ln\left|\dfrac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|$
I'll appreciate any help.
Thanks in advance!
AI: What you have done is correct. All you need to do is to rewrite it a different form.
$$\begin{align}
\ln \left( \sqrt{x^2+x+1} + x - 1\right) & = \ln \left( \dfrac{\left(\sqrt{x^2+x+1} + x - 1 \right) \left(\sqrt{x^2+x+1} - x + 1 \right)}{\left(\sqrt{x^2+x+1} - x + 1 \right)}\right)\\
& = \ln \left(\left(x^2 + x + 1 - (x^2 - 2x + 1) \right) \right) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right)\\
&= \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right)
\end{align}
$$
Hence, $$\begin{align}
\ln \left( \dfrac{\sqrt{x^2+x+1} + x - 1}{\sqrt{x^2+x+1} + x + 1}\right) & = \ln \left( \sqrt{x^2+x+1} + x - 1\right) - \ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\ & = \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right) -\ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\
& = \ln(x) - \left(\ln \left(\sqrt{x^2+x+1} - x + 1 \right) +\ln \left(\sqrt{x^2+x+1} + x + 1 \right) \right)\\
& = \ln(x) - \ln \left( \left(\sqrt{x^2+x+1}+1 \right)^2 - x^2\right)\\
& = \ln(x) - \ln \left( x^2 + x + 1 + 1 +2 \sqrt{x^2+x+1} - x^2\right)\\
& = \ln(x) - \ln \left( 2 \sqrt{x^2+x+1} + x + 2\right)
\end{align}$$ |
H: Linear algebra: power of diagonal matrix?
Let A = $\begin{pmatrix} 3 & -5 \\ 1 & -3 \end{pmatrix}$. Compute $A^{9}$. (Hint: Find a matrix P such that $P^{-1}AP$ is a diagonal matrix D and show that $A^{9}$= $PD^{9}P^{-1}$
Answer: $\begin{pmatrix} 768 & -1280 \\ 256 & -768 \end{pmatrix}$
I keep getting $\begin{pmatrix} -768 & 1280 \\ -256 & 768 \end{pmatrix}$ but could it be still right? I have D=$\begin{pmatrix} -2 & 0\\ 0& 2\end{pmatrix}$ and P =$\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ with $P^{-1}$= $\begin{pmatrix} \frac{1}{4} & \frac{1}{-4} \\ \frac{1}{-4} & \frac{5}{4} \end{pmatrix}$
AI: Your answer is not correct. Please note that the eigenvectors should be corresponding to the eigenvalues. So, if you choose $$D=\left(
\begin{array}{cc}
-2 & 0 \\
0& 2
\end{array}
\right),$$ then your $P$ should be $$P=\left(
\begin{array}{cc}
1 & 5 \\
1& 1
\end{array}
\right),$$
because $(1,1)$ is the eigenvector corresponding to $-2$. |
H: Which theories and concepts exist where one calculates with sets?
Recently I thought about concepts for calculating with sets instead of numbers. There you might have axioms like:
For every $a\in\mathbb{R}$ (or $a\in\mathbb{C}$) we identify the term $a$ with $\{a\}$.
For any operator $\circ$ we define $A\circ B := \{a\circ b : a\in A\land b\in B\}$.
For any function $f$ we define $f(A) := \{ f(a) : a\in A \}$. (More general: For a function $f(x_1,\ldots, x_n)$ we define $f(A_1,\ldots, A_n):= \{f(a_1,\ldots, a_n): a_1\in A_1 \land \dots \land a_n\in A_n \}$).
One has to find a good definition for $f^{-1}(A)$ which might be the inverse image of $A$.
((3.) is just the normal definition of the image and (2.) is a special case of (3.))
Now I am interested to learn about theories and concepts where one actually calculates with sets (similar to the above axioms).
After a while I found interval arithmetic. What theories or approaches do you know?
Because there will not be just one answer to my question, I will accept the answer with the most upvotes.
Update: The theories do not have to follow the above axioms. It's okay when they make there own definitions how a function shall act on sets. It is just important that you calculate with sets in the theory, concept or approach.
AI: I like Minkowski addition, aka vector addition. It is a basic operation in the geometry of convex sets. See: zonotopes & zonoids, Brunn-Minkowski inequality, polar sets... and here's a neat inequality for an arbitrary convex set $A\subset\mathbb R^n$:
$$
\mathrm{volume}\,(A-A)\le \binom{2n}{n}\mathrm{volume}\,(A)
$$
with equality when $A$ is a simplex. (Due to Rogers and Shepard, see here)
The case $n=1$ isn't nearly as exciting. |
H: Showing that the product and metric topology on $\mathbb{R}^n$ are equivalent
I'm new to topology, and can't figure out why the metric and product topologies over $\mathbb{R}^n$ are equivalent. Could someone please show me how to prove this?
AI: The product topology is induced by this norm.
$$\|x\|_{\rm prod} = \max\{|x_k|, 1\le k \le n\}$$
Let us use $\|\cdot\|$ for the Euclidean norm. Then
$$\|x\| = \left(\sum_{k=1}^n x_k^2\right)^{1/2}\le \left(\sum_{k=1}^n \|x\|_{\rm prod}^2\right)^{1/2} = \|x\|_{\rm prod}\sqrt{n}.$$
Now for a reverse inequality.
We have
$$|x_k |\le \left(\sum_{k=1}^n x_k^2\right)^{1/2}, \qquad 1\le k \le n,$$
so $$\|x\|_{\rm prod} \le \|x\|.$$
The norms are equivalent. |
H: Generalisation of a polynomial factorisation
I know of a theorem in algebra, that every polynomial $p\left(X\right)=a_{0}+a_{1}X+\ldots+a_{n}X^{n}$
that is nonconstant and has real coefficients, admits a factorisation
of the form
$$
p\left(X\right)=c\left(X-\lambda_{1}\right)\ldots\left(X-\lambda_{m}\right)\left(X^{2}+\alpha_{1}X+\beta_{1}\right)\ldots\left(X^{2}+\alpha_{M}X+\beta_{M}\right),
$$
where $m+M\geqslant1$, $c,\lambda_{1},\ldots,\lambda_{m}\in\mathbb{R}$
and $\left(\alpha_{j},\beta_{j}\right)\in\mathbb{R}^{2}$ with $\alpha_{j}<4\beta_{j}$
for $j\in\left\{ 1,\ldots,M\right\} $. My questions are:
1) Is there a name for this theorem ?
2) a) Is there an analogue of this theorem for " polynomials consisting
of infinite sums", i.e. for Laurent polynomials ?
b) Is there an analogue of this theorem for convergent sums ? (So that they be expressed as a convergent product)
AI: 1) It's a special case of the Fundamental Theorem of Algebra.
2) I think the closest analogue is the Weierstrass Factorization Theorem |
H: Is Frenet-Serret frame valid for non-natural parametrized curves when one normalizes the tangent vector?
When one have a curve $\beta(s)$ which is parametrized by arc length (has natural parametrization) one is able to obtain the tangent, normal and binormal vectors by using Frenet-Serret frame equations:
$T = \beta'(s)$, $N=\frac{T'(s)}{|T'(s)|}$, $B = T \times N$
But are those formulas valid for non-regular parametrizations when one normalizes the tangent vector?
$T=\frac{\beta'(s)}{|\beta'(s)|}$, $N$ and $B$ are calculated as above.
AI: Yes, as the Wikipedia article says at the end. |
H: What kind of analytical function could resemble that plot?
I am in the middle of making a model, and I am looking for an analytical expression which could resemble this evolution for one of the parameters:
In short: a sudden increase from 0 to a global maximum, followed by slower decrease. Once I have an idea of a functional form that presents the right characteristics, I can tweak it (peak height, peak position) to my liking.
The only expression I've been able to come up with so far involves rather high powers (12 and 6) of $1/x$:
(In case you wonder how I came to think of this one, it's the Boltzmann factor of a Lennard-Jones potential…)
AI: So is this supposed to be for $r > 0$? You might try more generally $y = k e^{a/r^b - c/r^d}$, where $b < d$. The maximum is at $r = \left( \frac{cd}{ab}\right)^{1/(d-b)}$, and
the limit as $r \to \infty$ is $k$. |
H: Riemann zeta sums and harmonic numbers
Given the nth harmonic number of order s,
$$H_n(s) =\sum_{m=1}^n \frac{1}{m^s}$$
It can be empirically observed that, for $s > 2$, then,
$$\sum_{n=1}^\infty\Big[\zeta(s)-H_n(s)\Big] = \zeta(s-1)-\zeta(s)$$
Can anyone prove this is true?
AI: $$\sum_{n=1}^{\infty} (\zeta(s) - H_n(s)) = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^s}$$ $$= \sum_{k=2}^{\infty} \frac{k-1}{k^s} = \sum_{k=2}^{\infty} k^{1-s} - \sum_{k=2}^{\infty} k^{-s} $$$$=\zeta(s-1) - 1 - (\zeta(s) - 1) = \zeta(s-1) - \zeta(s)$$
since each term $\frac{1}{k^s}$ appears in exactly $k-1$ of the sums $\sum_{m=n+1}^{\infty} \frac{1}{m^2}$ (namely, for $n=1,..,k-1$). |
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