问:
Asking for reference for challenging problems for Functional Analysis 1st course
I am a masters 1st year student and my functional analysis course ended 1 week ago. The instructor followed  Irwin Krieszig's Book " Introductory Functional Analysis with Applications" for teaching and although he didn't gave any assignments or exercise problems, I solved problems from exercises given in Krieszig's Book. I was able to understand the concepts given in the book clearly and then do exercises of Krieszig without much difficulty. But I noticed that the problems of the book are  much easier for a masters student but as I was busy in some extra - curricular activities also , so I didn't tried other problems or asked for reference books or materials here.

But Now I think that I must solve good number of Challenging problems in Functional Analysis as I will apply for phd in pure mathematics soon and I got my December break time to it.

So, Can you please tell any book in functional analysis course whose exercises are challenging as compared to Krieszig's  / any assignments which any professor of a university gave which are available on web?

Also, I am adding my course syllabus in the following images so that you come to know exactly which topics should be  available( I absolutely no problem If more topics are covered). All these topics were covered which are given in the images and none except them were covered.

Edit: I am good in measure theory , so if any functional analysis book you are recommending assumes knowledge of measure theory  then I am fine with it.

Looking for your help.

答:
You may follow the following references:

$1.~~$ "Theorems and Problems in Functional Analysis" by A. A. Kirillov, A. D. Gvishiani

$2.~~$ "Problems in Real and Functional Analysis" by Alberto Torchinsky

$3.~~$ "Theorems and Problems in Functional Analysis" by J. D. Knowles

$4.~~$ "A Glimpse at Hilbert Space Operators" by Paul R. Halmos

These books contains many standard problems. Most of the problems are hard but very good once it is done. I think you may get benefit from here so just try it.


问:
Conceptual question regarding absolute value
When taking cases for let's say $|x|$ we say :

It is $x$ for all $x \geq 0$
And

It is $-x$ for all $x<0$

I don't understand why we have "or equal to" in the first case

Why can't we have $\leq$ for the second case ?

Does this imply zero is positive?

This is something which I have blindly learnt without understanding why, hopefully I can change that.

Thanks in advance guys.

答:
Because $0 = -0$, the following ways of writing $|x|$ are all equivalent:
\begin{align}
|x| &= \begin{cases}
-x, &x<0\\
x, & x\ge 0
\end{cases} \tag{1}\\
|x| &= \begin{cases}
-x, &x\le 0\\
x, &x \ge 0
\end{cases}\tag{2}\\
|x| &= \begin{cases}
-x, &x \le 0\\
x, &x > 0
\end{cases}\tag{3}
\end{align}

Some authors/instructors may prefer their domains of definition to be disjoint, and so prefer either $(1)$ or $(3)$, but that's just a matter of preference.

ContentLicense: CC BY-SA 4.0


问:
Rational fraction constructions with origami
I'm writing a paper on origami in maths and currently looking at approximating rational fractions using various methods: crossing diagonals, Fujimoto, Haga, Noma methods.

Reading Origami and Geometric Constructions by Robert J. Lang page 14:

I would be very grateful if someone could explain why the bottom edge is divided into fractions $y=w/(w+x)$ and $z=x/(w+x)$

(assuming the $w$ on the right side of the pic is meant to be an $x$)

答:
By similar triangles, and assuming $y+z=1$, we have $y=\dfrac{h}{x}$ and $z=\dfrac{h}{w}$. This means that
$$\dfrac{h}{x}+\dfrac{h}{w}=1$$
and rearranging gives
$$h=\frac{wx}{w+x}$$


问:
The Limit of a Wallis-like Integral
Let

$\displaystyle I_n=\int^{\pi}_{0} e^x \sin^n x dx$,

then, what number does

$\displaystyle \sqrt{n} I_n$

tend to?

According to wolfram alpha it tends to 12.05... but I want to know what is this number.

I posted similarly titled question about two months ago: The limit and Asymptotic Behavior Wallis-like Integral but the theme here is not that.

答:
A key idea here is that the kernel

$$K_n(x) = \sqrt{n}\sin^n (x), \qquad 0 \leq x \leq \pi$$

is an approximation-to-the-identity (up to constant multiple). More precisely, for each continuous function $\varphi$ on $[0, \pi]$, the following is true:

$$ \lim_{n\to\infty} \int_{0}^{\pi} \varphi(x) K_n(x) \, \mathrm{d}x = \sqrt{2\pi} \phi\Bigl(\frac{\pi}{2}\Bigr) $$

Here is a proof:

\begin{align*}
\int_{0}^{\pi} \varphi(x) K_n(x) \, \mathrm{d}x
&= \int_{-\frac{\pi}{2}\sqrt{n}}^{\frac{\pi}{2}\sqrt{n}} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \, \mathrm{d}u \tag{$\textstyle x = \frac{\pi}{2} + \frac{u}{\sqrt{n}}$}.
\end{align*}

Now by noting that $ \log(\cos t) \leq - \frac{t^2}{2} $ holds for all $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we have

$$ \cos^n(u/\sqrt{n}) \leq e^{-u^2/2}. $$

Moreover, using the asymptotic formula $\log(\cos t) = -\frac{t^2}{2}(1 + o(1))$ as $t \to 0$, it follows that for each fixed $u$ in the domain of integration,

$$ \cos^n(u/\sqrt{n}) \xrightarrow[n\to\infty]{} e^{-u^2/2}. $$

So by the dominated convergence theorem,

\begin{align*}
&\lim_{n\to\infty} \int_{-\frac{\pi}{2}\sqrt{n}}^{\frac{\pi}{2}\sqrt{n}} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \lim_{n\to\infty} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \mathbf{1}_{\{ -\frac{\pi}{2}\sqrt{n} < u < \frac{\pi}{2}\sqrt{n}\}} \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \phi\Bigl( \frac{\pi}{2} \Bigr) e^{-u^2/2} \, \mathrm{d}u \\
&= \sqrt{2\pi} \phi\Bigl( \frac{\pi}{2} \Bigr).
\end{align*}

As a corollary,

\begin{align*}
\lim_{n\to\infty} \int_{0}^{\pi} e^x K_n(x) \, \mathrm{d}x
= \sqrt{2\pi} e^{\pi/2}
\approx 12.05807862.
\end{align*}

Title: The Limit of a Wallis-like Integral


问:
Expressing a number in the least possible space
I need to express a list of integers in a book, all falling between 1 and about 50,000, but I have limited space to print them. For e.g. "300009" takes up 6 characters. Put simply, I need to save paper. I'd like to either find a way to reduce the number of characters needed to express the numbers or replace the numbers with something that takes less horizontal or vertical space, but still represents the information reasonably well.

If the numbers are above 100, their exact value is unimportant. I only need to display a ballpark figure, i.e., to say that the value is somewhere between 100 and 199, 200 and 299, ... 1000 and 1999, 2000 and 2999, ... 10000 and 19999, ... 100000 and 199999, etc. Unfortunately, I find this does not seem to help me, as rounding the number up or down doesn't reduce the number of characters used and displaying the range for "300009" takes up 13 characters (300000-399999), which is considerably longer.

I have tried looking at using other bases, but these seem to make the numbers unreadable. The numbers should not require too much effort to understand.

I explored various alternative numeral systems from other cultures (past and present), as these are not so difficult to learn, but could not find anything more condensed way to express "300009" or a ballpark range for this.

Does math notation provide any solution? Is there a short-hand notation which can express these kinds of ranges with fewer than 6 characters to express "300009"? Are there some other numeral systems I should consider?

答:
If the numbers are close together, you can use xxyyyy, where xx is set in the leader columns, and the tabular entries give yyyy.  This is fairly common on old books of tables.

Railway enthusiest use a scheme where only the last changed digit are shown, eg the number list 81003/8/14/04 means that engines 810003, 810008, 81014 and 81004 are hooked up in that order.

If you want to display digits, where the known digits are important, use X for the missing digits, eg 31XXXX.  You can use other letters if the XXXX has several meanings, eg 31XYYY works.

ContentLicense: CC BY-SA 3.0


问:
Tangent of Ellipse
Given the equation of an ellipse $x^2+4y^2=36$, I am asked to find the equation of the 2 tangent lines that intersect at the point $(12,3)$. I am not sure of what to do after doing the implicit differentiation, I tried forming an equation for the gradient, $(y_1-3)/(x_1-12)$ then equate it to $\mathrm{d}y/\mathrm{d}x$. I'm not sure if I'm approaching it the right way but I can't find an answer.

答:
Since you are interested in just the equations of the two tangents, this approach can sometimes prove useful. Consider a line passing through $(12, 3)$ with a slope $m$. It will have the equation $y = mx + (3 - 12m)$. If this line intersects the ellipse at a point $(u,v)$, then we will have the relation
\begin{align*}
u^2 + 4(mu + (3 - 12m))^2 & = 36 \\
\implies (1 + 4m^2)u^2 + 8m(3 - 12m)u + 4(3 - 12m)^2 - 36 & = 0.
\end{align*}
Moreover, since the line is a tangent, the above quadratic in $u$ should have two repeated roots. Therefore, the discriminant should be zero.  On imposing this condition, you will get a quadratic in $m$ whose two solutions would be the slopes of the two required tangents.

Title: Tangent of Ellipse


答:
When you define a function piecewise, so to say $$f(x)=\begin{cases} g_1(x)&\text{if }\mathcal A(x)\\ g_2(x)&\text{if }\mathcal B(x)\\ g_3(x)&\text{if }\mathcal C(x)\end{cases},$$

it is good practice to divide the cases in such a way that conditions $\mathcal A(x)$, $\mathcal B(x)$ and $\mathcal C(x)$ are mutually exclusive. In principle, there is nothing wrong with having them not be mutuallly exclusive, but then one would need to check that, for instance, $g_1(x)=g_2(x)$ for all $x$ such that $\mathcal A(x)\land \mathcal B(x)$. Otherwise we are not defining a function. This necessity makes the exposition unnecessarily messy, other than being a cause of errors for people whose strong suit is the neglect of details.

In the instance of $\lvert x\rvert$, there is no reason to put the case $x=0$ with $x>0$ rather than with the other: for all that matters, you could even put it by itself and say that $$\lvert x\rvert=\begin{cases}x&\text{if }x>0\\ \arcsin x&\text{if }x=0\\ -x&\text{if }x<0\end{cases},$$ though it's better to have fewer and simpler cases.

Title: Conceptual question regarding absolute value

Body:
When taking cases for let's say $|x|$ we say :

It is $x$ for all $x \geq 0$
And

It is $-x$ for all $x<0$

I don't understand why we have "or equal to" in the first case

Why can't we have $\leq$ for the second case ?

Does this imply zero is positive?

This is something which I have blindly learnt without understanding why, hopefully I can change that.

Thanks in advance guys.


问:
Show that $\angle PAB = \angle CAQ$
Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.

Prove that $\angle PAB = \angle CAQ $

I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$... wait this looks like an apolonian circle.

Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.

Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$

答:
Let $R$ be the intersection of $PQ$ with $AC$ and let $O$ be the circumcenter of $ABC$. Clearly, $O$ lies on the line $PQR$.

Clearly $\angle PBA = \angle PBO + \angle OBA$ and $\pi - \angle ACQ = \angle CQR + \angle QRC$. Since $\angle OBA = \frac \pi 2 - \angle ACB = \angle QRC$ and $\angle PBA + \angle ACQ = \pi$ (by assumption), it follows that $\angle PBO = \angle CQR = \angle OQB$. This proves that $\triangle OPB \sim \triangle OBQ$. This yields $OB^2=OP\cdot OQ$. As $O$ is the circumcenter of $ABC$, we have $OB=OA$ and therefore $OA^2=OP\cdot OQ$. This implies that $\triangle OPA \sim \triangle OAQ$, hence $\angle OAP = \angle OQA$. Finally,
$$\angle PAB = \angle OAB - \angle OAP = \angle QRC - \angle OQA = \angle QAC.$$

Title: Show that $\angle PAB = \angle CAQ$


问:
Prove that $e^{i2a} - 1 = 2ie^{ia}\sin a$
Here is my solution and I think I got a bit close but I literally have no idea how to continue from here

a is a real number. Just a note.

$2i(\cos(a) + i\sin(a))\sin(a)$

$2i(\cos(a)\sin(a) + i\sin^{2}(a))$

By using $\cos$ and $\sin$ identities:

$2i(\tfrac{\sin(2a)}{2} + i\sin^{2}(a))$

$i\sin(2a) -2\sin^{2}(a)$

Now...what? I don't have $\cos(2a)$ so I can't get "$e^{2ai}$" out and I have no idea how to get "-1" out too.

Any ideas?

答:
You are closer than you think. Recall the identity:

$$\cos (2a) = 1-2\sin^2 a$$


问:
How a set whose existence is guarantied by one formula can be substituted into another formula?
I try to combine the Power Set Axiom and the Axiom Schema of Specification to prove that for any given set there exist another set that contains the given one as its only element. Moreover, I would like to do it as formal / strict as possible.

So, I start from the following two statements:

$\forall X : \exists P : \forall p : [(p \in P) \Leftrightarrow (p \subseteq X)]$

$\forall X : \forall B : \exists Y : \forall y : [(y \in Y) \Leftrightarrow ((y \in B) \wedge (y = X))]$

One can think of these two statements as "functions". In the first case we take a set X and for this set we can "generate" set P. In the second case we take sets X and B and "generate" set Y. Now I want to combine these "functions". In more details, in the second formula for all X I would like to use its P (as defined by the first formula) instead of B. Or, in other words, in the second formula for any X, I can use any B and I decided to use P as B.

So, the first part of my question is: How can I substitute B by P formally? In other words, how can I write P instead of B in the second formula such that P comes in there together with its definition provided by the first formula?

The second part of my question (assuming that we can write a formula described by the first part of the question) is if we can make such step-by-step transformations of this formula to get the following formula:

$\forall X : \exists Y : \forall y : [(y \in Y) \Leftrightarrow (y = X)]$

答:
This is accomplished via existential instantiation, with universal instantiation also playing a key role in the proof.

Specifically, we reason as follows:

Introduce a symbol $u$ standing for an arbitrary set; we want to show, using the $\mathsf{ZFC}$ axioms, that $\exists Z\forall z(z\in Z\leftrightarrow z=u)$. If we can do this we'll apply universal generalization to conclude $\forall X\exists Z\forall z(z\in Z\leftrightarrow z=X)$ as desired.

By powerset and universal instantiation, we get $\exists P\forall p(p\in P\leftrightarrow p\subseteq u)$.

Now we apply existential instantiation: this lets us introduce a new symbol $v$ and the formula $\forall p(p\in v\leftrightarrow p\subseteq u)$.

And this $v$ we can plug directly into an instance of the separation scheme as desired, via - again - universal instantiation.

Of course there are many different proof systems out there (e.g. compare natural deduction and Hilbert systems), but the above sketch will go through with only minor elaboration in any of them that I'm aware of.


答:
Your calculation is correct.
$i\sin 2a-2\sin^2 a$ = $i\sin 2a -(1-\cos 2a)$
= $i\sin 2a +\cos 2a -1$ = $e^{i2a}-1$, as desired.

Title: Prove that $e^{i2a} - 1 = 2ie^{ia}\sin a$

Body:
Here is my solution and I think I got a bit close but I literally have no idea how to continue from here

a is a real number. Just a note.

$2i(\cos(a) + i\sin(a))\sin(a)$

$2i(\cos(a)\sin(a) + i\sin^{2}(a))$

By using $\cos$ and $\sin$ identities:

$2i(\tfrac{\sin(2a)}{2} + i\sin^{2}(a))$

$i\sin(2a) -2\sin^{2}(a)$

Now...what? I don't have $\cos(2a)$ so I can't get "$e^{2ai}$" out and I have no idea how to get "-1" out too.

Any ideas?


问:
"Why do I always get 1 when I keep hitting the square root button on my calculator?"
I asked myself this question when I was a young boy playing around with the calculator.

Today, I think I know the answer, but I'm not sure whether I'd be able to explain it to a child or layman playing around with a calculator.

Hence I'm interested in answers suitable for a person that, say, knows what the square root of a square number is, but doesn't know about sequences, functions, convergence and the like.

答:
If $x>1$ then, since
$x-1=(\sqrt{x}-1)(\sqrt{x}+1)$ and
$\sqrt{x}>1$, so
$(\sqrt{x}-1)/(x-1)
=1/(\sqrt{x}+1)
\lt 1/2$,
so $\sqrt{x}$ is closer to $1$
than $x$ by a factor of at least $2$.

A similar argument holds when
$0 < x < 1$.

ContentLicense: CC BY-SA 2.5


问:
continiouty of maping from set back into itself.
Let  $f: [a,b] \to [a,b]$ be continuous. Show that the equation $f(x) = x$ has at least one solution in $[a,b]$.
Firstly im going to assume $x \in [a,b]$ thus a is the min and b is that max or vice versa assume the first. thus x >a and $x <b$ so we can divide the interval up into $(A=[a,x) ) \cup (B=[x,b])$ then i am going to assume that f maps the interval back onto itself but doesn't not include $f(x)=x$. thus $f(A) \cap f_{closure}(B) = \phi $ and $f(B) \cap f_{closure}(A) = \phi $ thus f is not connected so clearly $f(x)=x$

im fairly certain im supposed to use the intermediate value theorem but i don't remember it. this work?

答: If there are two points $x,y$ s.t. $f(x)\geq x$ and $f(y)\leq y$, then the intermediate value theorem will give a solution. Therefore, assume that $f(x)>x$ for all $x\in (a,b)$. Now think about what happens when $x\mapsto b$.</p>


问:
Why can the absolute value of the derivative be used in the 1-dimensional change of variable formula?
In Spivak's 'Calculus on Manifolds', page 67, the following statement is made.

Suppose $g: [a,b]\rightarrow\mathbb{R}$ is continuously differentiable and $f: \mathbb{R}\rightarrow\mathbb{R}$ is continuous. Then,
\begin{equation}\int_{g(a)}^{g(b)} f=\int_{a}^{b}(f \circ g) \cdot g^{\prime}\end{equation}
It is left to the reader to show that if $g$ is 1-1, then the above formula can be written as
\begin{equation}\int_{g((a,b))} f=\int_{(a,b)} f \circ g \cdot\left|g^{\prime}\right|\end{equation}
Now, I have tried to show this and came up with the following case:
Let $f(x)=\sin(x)$ and $g(x)=-\pi x$, then
\begin{equation}\int_{g(0)}^{g(1)} f = \int_{0}^{1}(f \circ g) \cdot g^{\prime}
= \int_{0}^{1}\sin(-\pi x)\cdot -\pi = \pi\int_{0}^{1}\sin(\pi x) = 2 \end{equation}
but
\begin{equation}\int_{g((0,1))} f=\int_{(0,1)} f \circ g \cdot\left|g^{\prime}\right|
= \int_{(0,1)} \sin(-\pi x) \cdot\left|-\pi\right| = -\pi\int_{(0,1)} \sin(\pi x) = -2 \end{equation}
Clearly, the two formulas give different results. The only way I can reconcile the two is if $\int_{(0,1)}$ actually means $\int_{1}^{0}$, but I don't see why that would be the case.

Can someone explain why my result disagrees with Spivak's statement?

答:
You aren't contradicting the formulas.  The first of Spivak's equations says, $$\int_0^{-\pi}\sin( x)=\int_0^1\sin(-\pi x)(-\pi)$$  Then second says,
$$\int_{-\pi}^0\sin(x)=\int_0^1\sin(\pi x)\pi$$

These are obviously equivalent.

If you look at the first expression in each of your formulas, you are trying to show that $$\int_{g(a)}^{g(b)}f=\int_{(g(a),g(b))}f$$ which is not necessarily true, as you yourself have pointed out.


问:
Two bijections between set of integers
I have the following interesting question:

Let $b$ be a $\mathbb{Z} \rightarrow \mathbb{Z}$ bijection, where $\mathbb{Z}$ denotes the set of integers. Is it possible that there exist a bijection between $\mathbb{Z}$ and $k+b(k)$ (where $k\in \mathbb{Z}$)?

That is: is it possible that $k+b(k)$ are all distinct integers and for every integer $S$ there exist (exactly one) an integer $n$ for which $S=n+b(n)$?

答:
We construct a sequence of maps $b_n\colon A_n\to \Bbb Z$ such that

$A_n\subsetneq A_{n+1}$

$b_{n+1}|_{A_n}=b_n$

$b_n$ is injective

$c_n:=b_n+\operatorname{id}$ is injective

if $n>6m$ then $m\in A_n$, $m\in b_n[A_n]$, $m\in c_n[A_n]$

Then $b(n):=b_{6n}(n)$ will have the desired properties.

We start with $A_0=\{0\}$ and $b_0(0)=0$.
To obtain $A_{n+1}$, we pick suitable $x\in\Bbb Z\setminus A_n$, $y\in \Bbb Z\setminus b_n[A_n]$, $z\in \Bbb Z\setminus c_n[A_n]$ and set $A_{n+1}=A_n\cup\{x\}$ and $b_{n+1}(x)=y$. We just need to take care that $x+y=z$.

If we first pick a (valid) $x$, the conditions on $y$ and $z$ exclude only finitely many values, hence we can certainly find a valid $y$ (and then $z$)

If we first pick a (valid) $y$, the conditions on $x$ and $z$ exclude only finitely many values, hence we can certainly find a valid $x$ (and then $z$)

If we first pick a (valid) $z$, the conditions on $x$ and $y$ exclude only finitely many values, hence we can certainly find a valid $x$ (and then $y$)

So by in turns picking the absolutely smallest "missing" $x$ or $y$ or $z$, we mange to comply with the last bullet point above.

Title: Two bijections between set of integers


问:
Expectation of an Ito integral involving arbitrary function in integrand
Suppose, we have a have stochastic integral $\int ^t_0 I(t) dB(t)$ and $\int ^t_0 \frac{S(t)}{I(t)} dB(t)$, where $I(t)$ and $\frac{S(t)}{I(t)}$ are continuous bounded functions and they are actually Susceptible and infected populations of an epidemic model. I have to prove that the expectation of both the integrals are zero.

My thought: I write $\mathbb{E}\int ^t_0 I(t) dB(t)= \mathbb{E} ~\Sigma_{j} I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}= \Sigma_{j} \mathbb{E}\bigg\{I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}\bigg\}=\Sigma_{j} \mathbb{E}\big\{I(t_j)\big\} \mathbb{E}\big\{B_{t_{i+1}}-B_{t_{i}}\big\}$. Now since Brownian motion has mean zero, so $\mathbb{E}\int ^t_0 I(t) dB(t)=0$. Is it right? Please help to prove it more elaborately.

答:
The integral $I = \int_a^bf(s)dB_s$ is the is the mean square limit of the partial sums $S_n = \sum_{i=0}^{n - 1}f(t_i)(B_{t_{i+1}} - B_{t_i}).$ Then
$$E[S_n] = \sum_{i=0}^{n - 1}E[f(t_i)]E[(B_{t_{i+1}} - B_{t_i})]$$

We used the fact that $f(t_i)$ is independent of $B_{t_{i+1}} - B_{t_i}$ to get that expectation. Now we use the fact that the increments of Brownian motion are normally distributed with mean zero and variance $t_{i+1} - t_i$,
$$B_{t_{i+1}} - B_t \sim N(0, t_{i+1} - t_i).$$

This gives

$$E[S_n] = \sum_{i=0}^{n - 1}f(t_i)E[(B_{t_{i+1}} - B_{t_i})] = 0$$

and since $I = \int_a^bf(s)dB_s = \lim_{n \to \infty} S_n$. We have that
$$E[I] = \lim_{n \to \infty} E[S_n] = 0$$


问:
Converse of Dirichlet test: a result about existence
I am wondering whether the following converse to the Dirichlet's test of convergence is true.

Let $\lambda_n$ be a sequence of complex numbers such that $\sum_{k=1}^n \lambda_k$ is unbounded as $n \to \infty$. Then there exists a sequence $x_n \to 0$ such that $\sum \lambda_n x_n$ diverges.

There is a delicate balance between $\lambda$ and $x$, which makes it really hard to prove the above result or find a counterexample.

Is the statement above true?

In fact, if a counterexample exists, then there might be some possibilities of strengthening the Dirichlet test to a stronger result.

答:
First step : Let's begin with the case where $\lambda_k \in \mathbb{R}_+$ for all $k$. In that case, the sequence $(S_n)_{n \geq 1}$ defined for all $n \geq 1$ by $$S_n = \sum_{k=1}^n \lambda_k$$ is increasing and diverges to $+\infty$. Considering $$x_n = \frac{1}{S_n}$$

You have for all $N, p \in \mathbb{N}^*$,
$$\sum_{k=N}^{N+p} \lambda_k x_k = \sum_{k=N}^{N+p} \frac{\lambda_k}{S_k} \geq \sum_{k=N}^{N+p} \frac{\lambda_k}{S_{N+p}} = \frac{S_{N+p}-S_N}{S_{N+p}} = 1 - \frac{S_N}{S_{N+p}}$$

which tends to $1$ as $p$ tends to $+\infty$, so you see that the sequence $\sum \lambda_k x_k$ is not Cauchy, so it does not converge.

Second step : Now we come back to the general case, where $\lambda_k \in \mathbb{C}$. We can suppose without loss of generality that for every $k \geq 1$, one has $\lambda_k \neq 0$ (just removing the zero terms from the sums).

First, notice that
$$\sum_{k=1}^n |\lambda_k| \geq \left|\sum_{k=1}^n \lambda_k \right|$$

so because $\sum \lambda_k$ is supposed to be unbounded, then $\sum |\lambda_k|$ diverges to $+\infty$. By the first case applied to $|\lambda_k|$, there exists a sequence $(x_n)$ which tends to $0$, such that $\sum |\lambda_k|x_k$ diverges. Define $(y_n)$ by
$$y_n = \frac{\overline{\lambda_n}}{|\lambda_n|}x_n$$

Of course $|y_n|=|x_n|$ so $(y_n)$ tends to $0$, and $$\sum \lambda_k y_k = \sum \lambda_k \frac{\overline{\lambda_k}}{|\lambda_k|}x_k = \sum |\lambda_k|x_k$$

which diverges by definition of the sequence $(x_n)$.


问:
Marker Exercise 2.5.10: universal part of a theory and supermodel
I'm trying to solve Exercise 2.5.10 in Marker's Model Theory: An Introduction.

It goes:

Let T be an $\mathcal L$-theory and $T_\forall$ be all of the universal
  sentences  $\phi$ such that $T \models \phi$. Show that $\mathcal A \models T_\forall$ if and only if there is
  $\mathcal M \models T$ with $\mathcal A \subseteq \mathcal M$.

WLOG we may assume that $T$ is consistent and that $T$ is closed under $\models$. $T_\forall$ is then just the set of universal (aka $\forall_1$ or $\Pi_1$) sentences in $T$.  Now "only if" part is easy.  I try to show the "if" part by resorting to Diagram Lemma: i.e., by showing that the union of the diagram of $\mathcal A$ and $T$ is consistent.  Assume not and let a quantifier-free formula $\phi(\vec x)$ be such that $T \models \neg \phi(\vec a)$ and $\mathcal A \models \phi(\vec a)$, where $\vec a \in \mathcal A$. ...

This is where I'm stuck.  I'm trying to replace the use of parameter with a quantifier but with no luck.  What can I do?  Is it possible to provide me with a clue?

答:
(I'll use Marker's notation)

We want to show that $Diag(\mathcal{A}) \cup T$ is satisfiable. By compactness, it's enough to show that $T \cup \{\psi_1(\bar a), \ldots, \psi_n(\bar a)\}$ is satisfiable for any finite subset $\{\psi_1(\bar a), \ldots, \psi_n(\bar a)\} \subset Diag(\mathcal{A})$. Note that $\bar a$ here is a finite tuple of symbols representing elements of $A$ (in the language $\mathcal{L}_A$).

Suppose not. Then

$T \models \neg \bigwedge_i \psi_i(\bar a)$ in the language $L_A$

Because this is true for any interpretation of the symbols $\bar a$,

$T \models \forall \bar v \:\neg \bigwedge_i \psi_i(\bar v)$

Let $\phi$ be the sentence $\forall \bar v \:\neg \bigwedge_i \psi_i(\bar v)$. It is universal (since the $\psi_i$ don't have quantifiers) and it follows from $T$. Therefore, $\phi \in T_{\forall}$, and so $\mathcal{A} \models \phi$. But $\phi$ is clearly false in $\mathcal{A}$, a contradiction.


问:
Inequality using AM GM
How could you prove using only the AM GM inequality:

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$

Where $abc = 1$ and $a,b,c$ are positive real numbers?

答:
Because $abc=1$ we can consider some real numbers $x,y,z$ such that $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. Computing this into your equation, we get:

$$\frac{\frac{x}{y}}{\frac{y}{z}}+\frac{\frac{y}{z}}{\frac{z}{x}}+\frac{\frac{z}{x}}{\frac{x}{y}}\geq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ which is equivalent to $$\frac{xz}{y^2}+\frac{yx}{z^2}+\frac{zy}{x^2}\geq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ and multiplying by $(xyz)^2$ both sides, we get $$(xy)^3+(yz)^3+(zx)^3\geq x^3z^2y+y^3x^2z+z^3y^2x$$

Now write $xy=m$, $yz=n$, $zx=p$ and we get $$m^3+n^3+p^3\geq p^2m+n^2m+m^2p$$ which is a known inequality.

Title: Inequality using AM GM


问:
General equation of an ellipse in 3D space with respect to cylindrical coordinate systems
The regular ellipse formula in 2D is $x^2/a^2 + y^2/b^2 = 1$ but how can it be transformed into a 3D formula including the parameter of $r, \theta$ and $z$?

答:
To translate to cylindrical coordinates you use the following.

$$x = r\cos\theta$$

$$y = r\sin\theta$$

This translates $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ into $$\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2} = 1$$

Notice that $z$ does not enter into the transformation. That is because in three dimensions the equation for an ellipse describes an elliptical cylinder.


Title: Inequality using AM GM

Body:
How could you prove using only the AM GM inequality:

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$

Where $abc = 1$ and $a,b,c$ are positive real numbers?

Body:
Hint: $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2b}{b^2c}}=3\sqrt[3]{a^3}=3a$$
$$\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\ge 3\sqrt[3]{\frac{b^2c}{c^2a}}=3\sqrt[3]{b^3}=3b$$
$$\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\ge 3\sqrt[3]{\frac{c^2a}{a^2b}}=3\sqrt[3]{c^3}=3c$$

Add all: what do you get?


问:
Question about a proof of Gödel's 1st incompleteness theorem
I am confused about one of the standard proofs of this theorem that I've seen. I believe it's a small thing that I missing as authors seem to skip straight through it. The proof I am referring to is the one that:

(1) encodes formulas of a logical system (such as RA) using Gödel numbering, having agreed on numerical labels for the symbols of the language. Then encodes sequences using the encoding for formulas.

(2) is able to encode '$p$ is a proof of $q$'

(3) is able to encode 'there exists a proof of $p$'

(4) creates an 'I am not provable' sentence

It's the nature of (2) and (3) that I am iffy about:

are we talking about merely encoding those statements, or is it about whether the system able to 'check' whether $p$ is a proof of $q$, in some sort of computable way? I can see how it would do so, it decodes the list of formulas presented by $p$ using arithmetic, and checks whether they follow the rules of logic established in advance by our system, also using arithmetic.

if the latter on my previous question, how does one get (3) from (2)? It seems authors go from checking $p$ proves $q$ to 'given $q$, we can establish whether it has a proof or not in the system' (or the system can, viewing all of this has code/arithmetic), but it's not clear to me how exactly?

I would appreciate any help!

答:
For simplicity, throughout I'll conflate a natural number $k$ with the corresponding numeral $\underline{k}$. Also I'll talk about $\mathsf{PA}$ instead of $\mathsf{RA}$, since I don't know what the latter is and the argument is the same for all appropriate theories.

I'm not sure I quite understand your first question, but I believe the following will clear things up - the key point being the difference between definability and provability, and the subsequent notion of representability which connects the two. We have two different objects at play here: the structure $\mathfrak{N}=(\mathbb{N};+,\times)$, that is the standard model of arithmetic, and the theory $\mathsf{PA}$. "Talking about provability in $\mathsf{PA}$" amounts to two observations, which I've separated below:

First, we have the purely model-theoretic fact.

Fixing an appropriate Godel numbering system $\ulcorner\cdot\urcorner$, we whip up a formula $\varphi(x,y)$ in the language of arithmetic such that for each finite string of symbols $\alpha$ and each finite string of finite strings of symbols $\pi$ we have $$\mathfrak{N}\models\varphi(\ulcorner \pi\urcorner, \ulcorner\alpha\urcorner)\quad\iff\quad\mbox{$\alpha$ is a sentence and $\pi$ is a $\mathsf{PA}$-proof of $\alpha$}.$$ Note that the theory is not involved yet. $\mathsf{PA}$-provability then corresponds to the formula $$\psi(y)\equiv \exists x\varphi(x,y).$$ What this all says is that the "is-a-$\mathsf{PA}$-proof-of" relation and the "is-$\mathsf{PA}$-provable" predicate are each definable in the structure $\mathfrak{N}$ - after "pushing through" the Godel numbering system, that is.

Remark: When I say "we whip up a formula $\varphi$ such that [stuff]," I'm glossing over some serious work. This is one of the three parts of this proof of the first incompleteness theorem - the second is proving the diagonal lemma, and the third is combining the first two to get the desired result - and is the only one of the three which is actually long: the third is quite short and sweet, and the second is short (if really slippery at first). To see how to construct $\varphi$, you should read the details of the proof carefully.

That said, once we have $\varphi$ we quickly get $\psi$: $\psi$ is just $\varphi$ with an existential quantifier slapped on.

Now we show that $\mathsf{PA}$ "understands" the formula $\varphi$ we've produced above - or at least, understands "basic facts" about $\mathsf{PA}$. Specifically, we claim that for every $m,n\in\mathbb{N}$ we have $$\mathfrak{N}\models\varphi(m,n)\quad\iff\quad \mathsf{PA}\vdash\varphi(m,n)$$ and $$\mathfrak{N}\models\neg\varphi(m,n)\quad\iff\quad\mathsf{PA}\vdash\neg\varphi(m,n).$$ In jargon, the relation $\varphi$ is representable in $\mathsf{PA}$.

With this in hand, we'll be able to prove the first incompleteness theorem: we prove the diagonal lemma and apply it to the formula $\neg\psi$, and then a quick analysis gives $\mathsf{PA}\not\vdash\theta$ and $\mathsf{PA}\not\vdash\neg\theta$. These last two steps are quite short. By far the longest part of the proof is constructing the desired $\varphi$.

A couple final comments:

First, Rosser's improvement uses the same idea, except it looks at something more compicated than $\psi$. First we whip up a formula $\upsilon(x,y)$ which defines in $\mathfrak{N}$ the relation "$y$ is the Godel number of the sentence with Godel number $x$" and which is $\mathsf{PA}$-representable. This takes some work, but not nearly as much as whipping up $\varphi$. Then we apply the diagonal lemma, not to $\neg\psi(y)$, but to the more complicated formula $$\chi(y)\equiv\forall x(\varphi(x,y)\rightarrow\exists z,w(z<x\wedge\upsilon(y,w)\wedge \varphi(z, w))$$ ("for every proof of me there is a shorter disproof of me"). A minor modification of Godel's original "quick analysis" mentioned above then shows that the resulting fixed point is $\mathsf{PA}$-undecidable using only the hypothesis that $\mathsf{PA}$ is consistent (as opposed to $\omega$-consistent).

Second, it's worth noting that when a formula $\gamma(x,y)$ is representable in $\mathsf{PA}$, the formula $\exists x\gamma(x,y)$ will not in general be representable - but it will be "half-representable." Specifically, we'll have for each $m\in\mathbb{N}$ that $$\mathfrak{N}\models\exists x\gamma(x,m)\quad\implies\quad\mathsf{PA}\vdash\exists x\gamma(x,m).$$ In particular, each $\mathsf{PA}$-provable sentence is $\mathsf{PA}$-provably $\mathsf{PA}$-provable; in the modal logic of provability this is expressed by the axiom $\Box p\rightarrow \Box\Box p.$


答: Another approach: since $e^{\pm ia}=\cos a\pm i\sin a$, $\sin a=\frac{e^{ia}-e^{-ia}}{2i}$.


问:
For a symmetric matrix, the geometric and algebraic multiplicities are equal
Could anyone tell me how to prove the following proposition?

Let $\lambda$ be an eigenvalue of a symmetric matrix. Then, its geometric multiplicity equals its algebraic multiplicity.

答: Every symmetric matrix is diagonalizable (this can be proved by small perturbation argument), that is: it has a full set of orthogonal eigenvectors and is conjugate to a diagonal matrix. So, you only need to prove the statement for diagonal matrix. Symmetric matrices have no Jordan block in their spectral decomposition, that cause discrepancy in the geometric and algebraic multiplicities of eigenvalues. </p>


问:
Is this function convex? Is my proof correct?
I want to prove (or disprove) the convexity of a function $f:\mathbb{R}^n \times \mathbb{R}^n \rightarrow\mathbb{R}$ defined as
$$f(x,y) = \frac{1}{2}\|x-y\|_2^2.$$
My attempt: We know that
$$f \text{ is convex} \iff \text{Hessian matrix is positive semi-definite}.$$
Hence, we want to show
$$ H =
\begin{bmatrix}
\frac{\partial^2f}{\partial x_1^2} & \frac{\partial^2f}{\partial x_1\partial x_2} & \dots & \frac{\partial^2f}{\partial x_1\partial x_n} & \frac{\partial^2f}{\partial x_1\partial y_1} & \dots & \frac{\partial^2f}{\partial x_1\partial y_n}\\
\frac{\partial^2f}{\partial x_2 \partial x_1} & \frac{\partial^2f}{\partial x_2^2} & \dots & \frac{\partial^2f}{\partial x_2\partial x_n} & \frac{\partial^2f}{\partial x_2\partial y_1} & \dots & \frac{\partial^2f}{\partial x_2\partial y_n}\\
\vdots & \vdots & \ddots\\
\frac{\partial^2f}{\partial x_n\partial x_1} & \frac{\partial^2f}{\partial x_n\partial x_2} && \frac{\partial^2f}{\partial x_n^2}\\
\frac{\partial^2f}{\partial y_1\partial x_1} & \frac{\partial^2f}{\partial y_1\partial x_2} &&& \frac{\partial^2f}{\partial y_1^2}\\
\vdots & \vdots &&&& \ddots\\
\frac{\partial^2f}{\partial y_n\partial x_1} & \frac{\partial^2f}{\partial y_n\partial x_2} &&&&& \frac{\partial^2f}{\partial y_n^2}
\end{bmatrix}
$$
is positive semi-definite. Here, I am constructing $H$ by treating the function as a multivariate function in $\mathbb{R}$ like so:
$$f(x_1, \dots, x_n, y_1, \dots, y_n) = \frac{1}{2}\sum_{i=1}^{n}(x_i-y_i)^2.$$
Then, I look at each possible element of the Hessian matrix:
$$\frac{\partial^2 f}{\partial x_i \partial x_j} = \delta_{ij}$$
$$\frac{\partial^2 f}{\partial x_i \partial y_j} = \frac{\partial^2 f}{\partial y_i \partial x_j} = -\delta_{ij}$$
$$\frac{\partial^2 f}{\partial y_i \partial y_j} = \delta_{ij}$$
where $\delta_{ij} := \begin{cases}1,\ \ i=j\\0,\ \ i\ne j\end{cases}$.

This means our Hessian will be
$$
H =
\begin{bmatrix}
I^{n\times n} & -I^{n \times n}\\
-I^{n \times n} & I^{n\times n}
\end{bmatrix}
$$
where $I^{n\times n}$ is the $n\times n$ identity matrix.
Now, all I need to do is determine if $H$ is positive semi-definite. That is, if
$$x^THx \ge 0, \ \ \forall x\in \mathbb{R}^n.$$

By playing around with small examples of $H$ I believe the general formula for $x^THx$ is
$$x^THx = \sum_{i=1}^{n} x_i^2 - 2\sum_{j=1}^{n/2} x_ix_{i+n/2}.$$
(although I don't know how to show this).
An example here.

From here, I use the fact that $x_i^2 + x_j^2 > 2x_ix_j$ to claim
$$x^THx \ge 0$$
and hence $H$ is positive semidefinite. Thus, $f$ is convex.

Is this reasoning correct? How might I prove the determinant is what I claim?

答:
Looks good.

In the verification, you could have used the stacked vector $\pmatrix{x\\y}$, then
$$
\pmatrix{x\\y}^TH\pmatrix{x\\y} = x^Tx - 2x^Ty + y^Ty = \|x-y\|_2^2 \ge0.
$$

This task gets easier, if one realizes that $f$ is the composition of a linear map $\pmatrix{I\\-I}$ with the convex function $\frac12\|\cdot\|_2^2$. Such a composition is always convex.


问:
Maximum module principle
The question is

Let $f $ be analytical and not constant in $\{z\in\mathbb C:|z|\leq 1\}$. If there is some $z_0\in \{|z|\leq 1\}$ such that $$|f(z_0)|=\max_{|z|\leq 1} |f(z)|\,,$$ then there must be $|z_0|=1 $ and $f’(z_0)\neq 0$.

The first part is just the maximum principle, but for the derivative, I have no idea how solve it. Looking forward to some hints or answers, thank you.

答:
The hypothesis means that $f$ has an analytic extension to a neighborhood of $z_0$ (actually it means that $f$ is analytic on a slightly larger open disc, but we need only the weaker assumption above)

$f$ nonconstant means that $f(z_0) \ne 0$ and $|f(re^{it}z_0)| < |f(z_0)|, 0 \le r <1$. To simplify the proof, by a rotation and homothety let wlog $f(z_0)=1$ (take $g(z)=Af(\alpha z), |\alpha| =1, A>0$ as that doesn't change the problem).

Assuming $f'(z_0)=0$ we have a power series $$f(z)=1+a(z-z_0)^n+O(|z-z_0)|^{n+1}, a \ne 0, n \ge 2,$$ for $|z-z_0|$ small enough.

But now if $\arg a =\theta$, since $n \ge 2, n\arg (z-z_0)$ covers more that an arc of angle length $\pi$ when $z$ is close to $z_0$ and $|z|<1$ so we can find such for which $\theta+n\arg (z-z_0) \in (-\pi/2, \pi/2)$ modulo $2\pi$ which means that $\Re a(z-z_0)^n >0$.

Letting $z \to z_0$ radially we can insure that $\Re (a(z-z_0)^n+O(|z-z_0)|^{n+1}) >0$ as the remainder decays faster, so one can find $|w|<1, \Re (a(w-z_0)^n+O(|w-z_0)|^{n+1}) >0$ which means that $|f(w)| \ge \Re f(w) >1=|f(z_0)|$ and that is a contradiction!

Title: Maximum module principle


问:
Getting specified correlation matrix from i.i.d. zero mean Gaussian random variables
Say there are two zero mean, unit variance Gaussian random variables, $X_1$ and $X_2$. The covariance matrix of the vector $\bf{X} = [X_1, X_2]^T$ is then
\begin{align*}
C_X =
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}.
\end{align*}
Now I would like to get a new vector $\bf{Y}$ from $\bf{X}$ using linear transformations, such that the covariance matrix of $\bf{Y}$ is
$$
C_Y =
\begin{bmatrix}
1 & \rho \\
\rho & 1
\end{bmatrix}.
$$
I tried the following linear transformation
$$
\bf{Y} = \begin{bmatrix}
1 & \rho \\
\rho & 1
\end{bmatrix} \bf{X}.
$$
This gives me the covariance matrix
$$
C_Y =
\begin{bmatrix}
1+\rho^2 & 2\rho \\
2\rho & 1+\rho^2
\end{bmatrix}.
$$
I can reduce the variance of $X_1$ and $X_2$ to $\frac{1}{2}$ so that I can have $\rho$ in the off diagonal entries. Under this adjustment $Y_1$ and $Y_2$ would each have variance $\frac{1+\rho^2}{2}$, which I can normalize to get $1$ in the diagonal entries.

So now I have two questions

Is this the best way to get the required covariance matrix?

If the dimensionality of $\bf{X}$ is $n$, I would need to divide $\bf{X}$ by $\sqrt{n}$. Is that right? How would I normalize the diagonal entries of $\bf{Y}$ to be $1$? In this case I would require the covariance matrix to have $\rho$ in all the off-diagonal entries and $1$ along diagonal entries.

Any help is appreciated.

Body:
Choose $Y = C_Y^{\frac{1}{2}}X$,

$$ \begin{align*}
 \mathbb{E}[YY^T] &= \mathbb{E}[C_Y^\frac{1}{2} X X^T C_Y^\frac{1}{2}] \\
 &= C_Y^\frac{1}{2} \mathbb{E}[X X^T] C_Y^\frac{1}{2} \\
&= C_Y^\frac{1}{2} C_X C_Y^\frac{1}{2} \\
&= C_Y
\end{align*}$$


问:
Counter Example About $L^{\infty}$ function
Let $(\Omega,\Sigma,\mu)$ be a measure space.

Let $1<p<\infty$ and $q={p\over p-1}.$

Let $f\in L^p(\Omega,\Sigma,\mu).$

Then by the Riesz Representation Theorem, we know that, $$||f||_p=\sup_{||g||_q\le1} |\int fg\,d\mu|. $$

My text says that, this equality does not always hold if $p=\infty.$ Intuitively I think this is true, because the dual space of $L^\infty(\Omega,\Sigma,\mu)$ is not $L^1(\Omega,\Sigma,\mu).$ But I cannot come up with a concrete counter example.

My question is, can any one help me with coming up such a counter example? Thanks.

答:
I think the claim is true if the measure space has the following property:  every $A\in \Sigma$ with $\mu(A)>0$ contains a subset of finite, positive measure. Define $A:=\{x:\ |f| \ge \|f\|_\infty-\epsilon\}$. Take $B\subset A$ with $0<\mu(B)<+\infty$, define $g:=\mu(B)^{-1}\chi_B\cdot sign(f)$.

Hence, a counter-example has to use some exotic measure space:
Take the measure space $\Omega =\{0\}$, $\mu(\Omega)=+\infty$. Then $L^1(\mu)$ contains only the null function. So this supremum is zero. $L^\infty(\mu)$ contains all functions from $\Omega$ to $\mathbb R$.


问:
Find the general solution for tan x = tan 4x
I am studying maths as a hobby and have come to the following problem:

Find the general solution for $\tan x = \tan 4x$

My book says the general form of such an answer is:

$\tan x = \tan \alpha \implies x = n\pi +\alpha$

When I look at the answer at the back of the book it says it is

$\frac{n\pi}{3}$, which is not in the same format.

In any case, I cannot see how this was worked out. I have tried using double angle formulae but it gets very messy and I get confused.

答: $$\tan (x) = \tan (4x)\xrightarrow{{n \in \mathbb{Z}}}4x = n\pi  + x \to 3x = n\pi  \to x = \frac{{n\pi }}{3}$$


问:
There is something wrong in this contour integral
I am trying to prove that $$I=\int ^{\infty}_{-\infty}{\frac {dy} {1+y^2} \frac {s+\cosh {(\pi y)}} {s^2+2s\cosh {(\pi y)} +1}} =\frac {\pi} {s-1}-\frac {\pi} {s\ln {s}}$$ One easily can set any real value to $s$ (providing $s>1$) and calculate this integral at any online integral calculator (Wolfram for example), compare it with the right side of equation value and convince himself that the statement above is true. But when I carry out the integration, I find that above the real line there are three poles of integrand: $+i$, $-\frac{\ln{s}}{\pi}+i$, and $\frac{\ln{s}}{\pi}+i$. So I can write $$\oint {\frac {dz} {1+z^2} \frac {s+\cosh {(\pi z)}} {s^2+2s\cosh {(\pi z)} +1}}=I+\lim_{R \rightarrow \infty}\int ^{\pi}_{0}\frac{Re^{i\theta}id\theta}{1+R^2e^{2i\theta}}\frac{s+\cosh{(\pi Re^{i\theta})}}{s^2+2s\cosh{(\pi Re^{i\theta})}+1}$$
The integral in $\theta$ vanishes as $R$ goes to infinity. So, $I$ is the sum of residues at the three poles. Now the trouble appears: the first pole, $i$, generates the $\pi/(s-1)$; and the two other poles generate the term $$-\frac{\pi}{s\ln{s}}\frac{4\pi^2}{(\ln{s})^2+4\pi^2} $$ I must be making some mistake, because that term produces the wrong value, but I can't figure out what is it. If anyone finds it, would be great! Thank you.

答:
The integral along the upper semicircular arc does not necessarily vanish as $R \to \infty$, since the contour will pass through infinitely many poles as $R$ grows.

Indeed, assuming $s > 1$, note that the integrand

$$ f(z) = \frac{1}{z^2 + 1} \frac{s + \cosh(\pi z)}{s^2 + 2s \cosh(\pi z) + 1} $$

has infinitely many simple poles in the upper-half plane, which can be classified into three types:

$z = i$

$z = z^{+}_k := \frac{1}{\pi} \log s + (2k+1) i$ for $k = 0, 1, 2, \dots$

$z = z^{-}_k := -\frac{1}{\pi} \log s + (2k+1) i$ for $k = 0, 1, 2, \dots$

Now let $K$ be a positive integer and $R$ be a sufficiently large positive real, and consider the contour integral of $f$ along the boundary of the rectangle with corners $\pm R$ and $\pm R + 2 K i$. If $[z_0, z_1]$ denotes the directed line segment from $z_0$ to $z_1$, the residue theorem tells that

\begin{align*}
\newcommand{\Res}{\mathop{\operatorname{Res}}}
\int_{[-R,R]} f(z) \, \mathrm{d}z
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^-_k} f(z) \biggr) \\
&\quad - \int_{[R,R+2Ki]} f(z) \, \mathrm{d}z - \int_{[R+2Ki,-R+2Ki]} f(z) \, \mathrm{d}z - \int_{[-R+2Ki,-R]} f(z) \, \mathrm{d}z.
\end{align*}

Letting $R \to \infty$, the integrals along $[R,R+2Ki]$ and $[-R+2Ki,-R]$ vanish, hence

\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^-_k} f(z) \biggr) + \int_{-\infty}^{\infty} f(x + 2Ki) \, \mathrm{d}x
\end{align*}

Then it is not hard to check that the last integral vanishes as $K \to \infty$, and so,

\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{\infty} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{\infty} \Res_{z=z^-_k} f(z) \biggr)
\end{align*}

In order to compute this sum, note that

\begin{align*}
\Res_{z=z^+_k} f(z)
&= \frac{1}{1 + (z^+_k)^2} \frac{s + \cosh(\pi z^+_k)}{2 \pi s \sinh(\pi z^+_k)} \\
&= \frac{1}{(z^+_k + i)(z^+_k - i)} \frac{s - \bigl( \frac{1 + s^2}{2s} \bigr)}{2\pi s \bigl( \frac{1 - s^2}{2s} \bigr)} \\
&= -\frac{1}{4\pi i s} \biggl( \frac{1}{z^+_k - i} - \frac{1}{z^+_k + i} \biggr) \\
&= -\frac{1}{4\pi i s} \biggl( \frac{\pi}{\log s + 2k \pi i} - \frac{\pi}{\log s + 2(k+1)\pi i} \biggr)
\end{align*}

and

\begin{align*}
\Res_{z=z^-_k} f(z)
&= -\frac{1}{4\pi i s} \biggl( \frac{\pi}{\log s - 2k \pi i} - \frac{\pi}{\log s - 2(k+1)\pi i} \biggr).
\end{align*}

So the sums are telescoping and hence yield

\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= \frac{\pi}{s-1} - \frac{\pi}{s \log s}.
\end{align*}


问:
If $f$,$g$ (not necessarily differentiable) are convex, non-negative, and increasing on $(a,b)$, then $f \cdot g$ is also convex on $(a,b)$.
The question is in the title. Show that $f\cdot g$ is convex on $(a,b)$, provided that $f, g$ are non-negative, increasing, and convex on $(a,b)$. I know that if $f,g$ are differentiable, then I can simply show that $(f\cdot g)'' >0$ on the interval, but in this case differentiability is not specified. I tried to answer it by simply multiplying the inequalities that result from the definition
$$
h(tx+(1-t)y)\leq th(x) + (1-t)h(y)
$$
for $h=f$, $h=g$ and $t \in [0,1]$, but I couldn't really get anywhere with it. I know there is a question on here about this before, but all of the answers there relating to this specific case assumed differentiability, and in my case the question does not say that the functions are differentiable. I would appreciate any guidance on this problem.

答:
Here is a partial answer that does not use differentiability but also does not use the non-negativity of $f$ and $g$. Let $x,y\in(a,b)$ and assume wlog that $x\leq y$ and let also $t\in[0,1]$. Then,
\begin{align*} h(tx+(1-t)y)&=f(tx+(1-t)y)\cdot g(tx+(1-t)y) \\ &\leq[tf(x)+(1-t)f(y)]\cdot[tg(x)+(1-t)g(y)] \\ &=t^{2}(fg)(x)+t(1-t)f(x)g(y)+t(1-t)f(y)g(x)+(1-t)^{2}(fg)(y) \\ &\leq t^{2}(fg)(x)+t(1-t)(fg)(y)+t(1-t)(fg)(y) +(1-t)^{2}(fg)(y) \\ &= t^{2}(fg)(x)+(fg)(y)[(1-t)^{2}+2t(1-t)] \\ &=t^{2}(fg)(x)+(1-t^{2})(fg)(y)=t^{2}h(x)+(1-t^{2})h(y) \end{align*}
where the first inequality uses the convexity of $f$ and $g$, and where the second inequality uses the fact that both $f$ and $g$ are increasing.


问:
“$B$ does not follow from $A$” seems different from $\lnot(A\to B)$
I am a beginner in logic. I have a basic question about implication. If "$B$ follows from $A$", we can write $A\rightarrow B$; but what about "$B$ does not follow from $A$"? Someone might suggest $\neg(A\rightarrow B)$, but that's not what "$B$ does not follow from $A$" means. Because $\neg(A\rightarrow B)$ is logically equal to $A\wedge\neg B$. But saying that "$B$ does not follow from $A$", it might be the case that $B$ is undecidable given $A$. So, I wonder how this sentence can be analyzed using logical tools. Thanks in advance!

答:
$A\; \implies \; B $ says that if $ A $ is true, we are sure and certain that $ B $ is true and cannot be false.

$ A $ does not imply $ B $ says that even if $ A $ is true, we can say Nothing about $ B $.

For example,
$(p\vee q)\wedge q $ does not imply $ p$.

$ p $ could true and false as well.

The best way to understand the "implication" begins by the notion of "Argument".


问:
To prove that a function is linear or two series are convergent

Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function and let
$\{a_n\}_n$ and $\{b_n\}_n$ be sequences of real numbers such that:

$$\lim_{n\to\infty}\int_0^1|f(x)-a_nx-b_n|\text{dx}=0$$

Prove that:

a$)$ The sequences $\{a_n\}_n$ and $\{b_n\}_n$ are convergent

b$)$ The function $f$ is linear

In this particular problem  which was given in an entrance examination of $10+2$ level,my approach was to do the epsilon delta definition of limit with integration ,but of no avail.
I cannot start and arrive at the result... Kindly help anyone.

答: Hints only. First, note that $N$ defined on $E=\mathbb{R}^2$ by $N(u,v)=\int_0^1|ux+v|dx$ is a norm. As all the norms on $E$ are equivalent (and ${\rm Max}\{|u|,|v|\}$ is another norm on $E$) there exists a $c>0$ such that $c{\rm Max}\{|u|,|v|\} \leq \int_0^1|ux+v|dx$ for all $(u,v)$. Now put $w_n=\int_{0}^{1}|f(x)-a_nx-b_n|dx$. We have that $|(a_m-a_n)x+b_m-b_n|\leq |f(x)-a_nx-b_n|+|f(x)-a_mx-b_m|$, hence $c{\rm Max}\{|a_m-a_n|,|b_m-b_n|\}\leq w_n+w_m$. Can you take it from here ?


问:
Would the negation of Russell's axiom of infinity serve this purpose?
For any positive integer n, we can construct a sentence Ψ_n in FOL such that an interpretation        I ⊨ Ψ_n iff the domain of I has exactly n elements. Suppose we wanted to construct a sentence Φ  such that an interpretation I ⊨  Φ iff the domain of I is finite. Would the negation of Russell's axiom of infinity serve this purpose?</p>

答: As I mention in my comment, not only will the negation of the axiom of infinity not do the trick, nothing will. If $\Phi$ is satisfied by arbitrarily large finite $I$ (and you certainly want there to be arbitrarily large such $I$ if $I\vDash\Phi$ is to mean "$I$ is finite") then the compactness theorem tells us that $\Phi$ will be satisfied by some infinite structure.


问:
Basis of $\mathbb R$ seen as a $\mathbb Q$-vector space
$\mathbb R$ is clearly a $\mathbb Q$-vector space since all axioms needed to be a vector space are verified. Its basis is infinite because it would need to have all roots, $\pi$, $e$...

Is there any quick and "relatively easy" but solid proof that this statement is true ? I don't know how to formally write "the basis needs to contain all (actually not all but still a lot) irrationals and |$\mathbb R$ \ $\mathbb Q$| would be bigger than some uncountable set.

Can we write the basis in a simple form ? What I mean is that, the basis of polynomials $\mathbb R$[t] is infinite but can be written as {$t^k$ , k $\in$ $\mathbb N$}

答:
By definition, the powers of $\pi$ are linearly independent over $\mathbb Q$ since $\pi$ is transcendental. Therefore, $\mathbb Q[\pi]$ is a infinite-dimensional subspace of $\mathbb R$.

No description of a basis for $\mathbb R$ over $\mathbb Q$ is known. In most if not all definitions, a description is a string in a language and so there are only countably many descriptions, not enough to describe an uncountable set.


问:
$0\in \mathfrak h^*$ is a weight of $V(\lambda)$ if and only if $\lambda \in \Lambda^+$ is a sum of roots.
This question appears in Humphrey's book on Lie algebras (Introduction to lie algebras and representation theory) page 116. Let $L$ be a semisimple Lie algebra with root system $\Phi$ and base $\Delta = \{\alpha_1,\cdots,\alpha_\ell\}$. Denote by $\Lambda^+$ the set of dominant weights.

If $\lambda \in \Lambda^+$, then $0$ occurs as a weight of the irreducible highest weight $L$-module $V(\lambda)$ if and only if $\lambda$ is a sum of roots.

The direction $\Rightarrow$ is simple: since $V(\lambda)$ is generated by a maximal vector $v$ of weight $\lambda$, its weights $\mu$ all satisfy $\mu\leq \lambda$, so $\lambda - \mu$ is a sum of positive roots. In particular,  $\mu = 0$ shows that $\lambda$ is a sum of (positive) roots.

I'm actually struggling to prove the direction $\Leftarrow$ . We may write $\lambda = \sum \langle \lambda, \alpha_i\rangle \omega_i$ in terms of the fundamental weights. Since the fundamental weights are a linear comb. of simple roots with rational positive coefficients and $\lambda \in \Lambda^+$, then $\lambda$ is a fortiori a sum of positive roots. Let then $\lambda = \sum k_i \alpha_i$ with $k_i\in \mathbb Z_{\geq 0}.$ Now the vectors $y_{\alpha_1}^{i_1}\cdots y_{\alpha_\ell}^{i_\ell}v \in V(\lambda)$ have weight $\lambda - \sum i_j \alpha_j$, so I must show that $y_{\alpha_1}^{i_1}\cdots y_{\alpha_m}^{i_m}v \neq0$ for some $\{i_1,\cdots, i_\ell\} = \{k_1,\cdots, k_\ell\}$. How can I prove that? I already know that $y_{\alpha_j}^{m_j+1}v = 0$ for $m_i = \langle \lambda,\alpha_j \rangle$, so why we cannot have $k_j>m_i$? If this situation can happen, how can I cancel out the factor $k_j\alpha_j$ of $\lambda$?

Any hint or help is very much appreciated.

答:
Denote by $\Pi(\lambda)$ the set of weights of $V(\lambda)$. By proposition 21.3, it is saturated. We prove that if the set $\{\mu \in \Pi(\lambda): \mu \mbox{ is a sum of positive roots}\}\neq \emptyset$, then $0 \in \Pi(\lambda)$. In particular,  $\mu = \lambda$ solves the exercise.

Proceed by induction on $\mbox{ht}\,\mu$, the case $\mbox{ht}\,\mu = 0$ being clear, since $\mu = 0 $ if and only if $\mbox{ht}\,\mu =0$ and $\mu \in \Pi(\lambda)$ by hyphotesis.

The inductive hyphotesis states that if $\{\nu \in \Pi(\lambda): \nu \mbox{ is a um of positive roots and } \mbox{ht}\,\nu < \mbox{ht}\,\mu\}\neq \emptyset$, then $0 \in \Pi(\lambda)$.

Now, if $\mu \neq 0,$ let us write $\mu = \sum k_i\alpha_i, k_i\in \mathbb Z_{\geq 0}$.
Then, $0<(\mu,\mu) = \sum k_i(\mu,\alpha_i)$ implies that $(\mu,\alpha_i)>0$ for some index $i$ with $k_i\geq 1$, so it follows that $\langle \mu, \alpha_i \rangle\geq 1.$ Since $\Pi(\lambda)$ is saturated, we have at least that $\mu - \alpha_i \in \Pi(\lambda)$, and $k_i\geq 1$ assures that $\mu-\alpha_i$ is still a sum of positive roots. The induction hyphotesis now gives us that $0\in \Pi(\lambda)$.


问:
Ordinary least squares give $P = A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}}$, prove ${P^2} = P$ and ${P^{\text{T}}} = P$
Ordinary least squares give a solution matrix $P = A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}}$

Multiplying with $P$ is projecting the vectors orthogonal to $A \in {\mathbb{R}^{m \times n}}$ subspace.

Now, I need to prove ${P^2} = P$ and ${P^{\text{T}}} = P$. I don't know where to even start, and after some research I'm not familiar with matrix image and kernel concepts.

答:
For the first part, note that:
$$\begin{gathered}
  P = A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}} \to  \hfill \\
  {P^2} = PP = A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}}A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}} \hfill \\
  \xrightarrow{{M = {A^{\text{T}}}A}}{P^2} = A{M^{ - 1}}M{M^{ - 1}}{A^{\text{T}}} = A{M^{ - 1}}{A^{\text{T}}} \hfill \\
   \to {P^2} = A{\left( {{A^{\text{T}}}A} \right)^{ - 1}}{A^{\text{T}}} \to {P^2} = P \hfill \\
\end{gathered}$$
and for the second part:
$$\begin{gathered}
  {P^{\text{T}}} = {\left( {A{{\left( {{A^{\text{T}}}A} \right)}^{ - 1}}{A^{\text{T}}}} \right)^{\text{T}}} = {\left( {A{M^{ - 1}}{A^{\text{T}}}} \right)^{\text{T}}} \hfill \\
  \left. \begin{gathered}
   \to {P^{\text{T}}} = A{\left( {{M^{\text{T}}}} \right)^{ - 1}}{A^{\text{T}}} \hfill \\
  {M^{\text{T}}} = {\left( {{A^{\text{T}}}A} \right)^{\text{T}}} = {A^{\text{T}}}A = M \hfill \\
\end{gathered}  \right\} \to {P^{\text{T}}} = A{M^{ - 1}}{A^{\text{T}}} = P \hfill \\
\end{gathered}$$
P.S. Note that:
$${\left( {AB} \right)^{\text{T}}} = {B^{\text{T}}}{A^{\text{T}}}$$
and also:
$${\left( {{A^{\text{T}}}} \right)^{\text{T}}} = A$$


问:
Continuous extension
Is there any bounded continuous map f:A to R (A is open) which can not be extended on whole R?

This is a question posed by myself.
My attempt: Let A=(1,2) then we can extend it. If A is finitely many intervals it can be extended. If A is countable many intervals then it can also be extended.

But the last claim is based on A is countable union of disjoint open interval.

Am i right or wrong?

答: $\sin(1/x)$ is continuous and bounded on $(0,1)$, but if you extend it, what is the limit at $x =0$?

Title: Continuous extension


答:
If $A$ is open and neither empty nor all of $\Bbb R$, then we can pick $a\in A$, $b\notin A$.
Wlog $a<b$. Let $s=\inf([a,b]\setminus A)$. Then $a<s\le b$ and  $s\notin A$ and we can define $f\colon A\to \Bbb R$ as $f(x)=\sin\frac1{x-s}$. There is no continuous extension of $f$ to $A\cup\{s\}$.

Title: Continuous extension

Body:
Is there any bounded continuous map f:A to R (A is open) which can not be extended on whole R?

This is a question posed by myself.
My attempt: Let A=(1,2) then we can extend it. If A is finitely many intervals it can be extended. If A is countable many intervals then it can also be extended.

But the last claim is based on A is countable union of disjoint open interval.

Am i right or wrong?


问:
Legendre symbol proof
Show that if $a$ is a quadratic residue of the odd prime $p$, then the solutions of $x^2\equiv{a}\pmod {p}$ are $x\equiv{\pm a^{n+1}}\pmod {p}$  if
$p=4n+3$. </p>

<p>Proof
Assume $p=4n+3$ and $x^2\equiv{a}\pmod {p}$.  $p=4n+3 \Longleftrightarrow n=(p-3)/4$.
To see if $\pm{a}\pmod{p}$ Plugging it into $x^2\equiv{a}\pmod {p}$ we get
${(\pm a^{n+1})}^2\equiv {a}\pmod{p}$ We plug $n=(p-3)/4$. We get ${(\pm a^{\frac{P-3}{4}+1})}^2\equiv {a}\pmod{p}$ which is  ${(\pm a^{\frac{P+1}{4}})}^2\equiv {a}\pmod{p}$
which is ${( a^{\frac{P+1}{2}})}\equiv {a}\pmod{p}$ But $\frac{p-1}{2}$ is odd since $p=4n+3$ is an odd prime. Let $\frac{p-1}{2}=2k+1$ where $k$ is a positive integer. Now we get
${(a^{{2k+1}})}\equiv {a}\pmod{p}$ Which is $(a^{k})^2*a\equiv {a}\pmod{p}$ Divide both sides by $a$ but since $(a,p)=1$ the modulus stays the same.  Since $(a^k)^2\equiv {1}\pmod{p}$ since it is of the form  $x^2\equiv{a}\pmod {p}$ it follows $x\equiv{\pm a^{n+1}}\pmod {p}$
Not sure if I have this right or that $(a,p)=1$. </p>

答:
All congruences are done modulo $p$.  Assume that $x = \pm a^{n+1}$.  Then,

\begin{align*}
x^2 \equiv a^{2n + 2} = a^{\frac{4n+4}{2}} = a^{\frac{p+1}{2}}.
\end{align*}

We need to show that $a^{\frac{p+1}{2}} \equiv a$ for all $p = 4n + 3$. Note that $x^2 \equiv b^2 \implies x \equiv \pm b$ since the integers modulo $p$ forms a field.  Since $\left( a^{\frac{p+1}{2}}\right)^2 \equiv a^2$, by Fermat's little theorem, we really need to show that $a^{\frac{p+1}{2}} \not\equiv -a$.  However, since $p = 4n +3$, and since $a$ is a quadratic residue, then $-a$ is not.  Above we showed that $a^{\frac{p+1}{2}}$ is a quadratic residue.  Hence $-a$ and $a^{\frac{p+1}{2}}$ cannot be equal, and thus we arrive at $a^{\frac{p+1}{2}} \equiv a$ as desired.

I should note that my "proof" is not as self-contained as I would like.  In particular, it may or may not be clear why it is true that if $a$ is a quadratic residue $\pmod p$, then $-a$ is a quadratic nonresidue.  It may also not be clear that the integers modulo $p$ form a field.  Because of this lack of self-containment, perhaps there is a more elementary answer.

Title: Legendre symbol proof


问:
Proving that the set of continuous linear maps is complete w.r.t the operator norm
I am trying to prove that the set of continuous (i.e. bounded) linear maps $B(V,V)$ is complete with respect to the operator norm, where $V$ is a complete normed vector space.

I have got this far:

Let $(l_n)_n$ be a Cauchy sequence, where $l_n \in B(V,V)$. Define $f_{ni}= l_n e_i$ where $(e_i)_i$ is a standard basis of $V$.

Then $\forall \epsilon > 0$, $\exists N$ such that $n,m > N \implies \|l_n - l_m\|_{op} < \varepsilon$.

So, using implicit summation, $\sup_{\|v\|=1} \|l_n v - l_m v\| = \sup_{\|v\|=1} |v_i|\|f_{ni} - f_{mi}\| \le \sum_i \| f_{ni} - f_{mi}\|< \varepsilon$.

Hence $\|f_{ni}-f_{mi}\|<\varepsilon$ so $(f_{ni})_n$ is a Cauchy sequence for each $i$.

$V$ is complete so $(f_{ni})_n$ converges. Define $f_i = \lim_{n \to \infty} f_{ni}$.

Now define a linear map $l$, which sends $e_i \mapsto f_i$. It remains to show that $l_i \to l$ under the operator norm and that $l \in B(V,V)$ i.e. that $l$ is bounded. The latter is simple to do once convergence has been shown.

Let $\varepsilon > 0$. I need to show that there exists $N$ such that if $n > N$ then $\sup_{\| v \| = 1} \| (l_n - l) v \| < \varepsilon$

We have $$\| (l_n - l) v \| = \| (l_n - l_m) v \| + \| (l_m - l) v \|$$

For all $v$ with $\| v \| = 1$, the first term on the RHS can be made arbitrarily small because $(l_n)_n$ is a Cauchy sequence by assumption. I just need to show that for a fixed $v$ with $\| v \| = 1$ that $(l_m v)_m \to lv$, which boils down to showing that $ \sum_i \| f_{mi} - f_i\|< \varepsilon$ for $m$ sufficiently large. I am not sure how to prove this though... any hints would be appreciated.

答: Let $(f_n)_n $ be a Cauchy sequence in $B(V, V) $. So $||f_n-f_m||<\epsilon $ for $n, m>N$ for some $N$. So $$||(f_n-f_m) (x) ||< \epsilon ||x|| $$ for all $x \in V$. So from this you can show that for each $x$ in $V$ that $(f_n(x)) _n $ is a Cauchy sequence in $V$. Now you can use that $V$ complete.


问:
How to prove this inequality $\frac{a}{b}\leq \frac{a+c}{b+d}\leq \frac{c}{d}$
How can I show that for positive reals

$$
\frac{a}{b}\leq \frac{c}{d},
$$
that
$$
\frac{a}{b}\leq\frac{a+c}{b+d}\leq\frac{c}{d}.
$$

Thanks in advance.

答:
Hint:

$$\dfrac{a+c}{b+d}-\dfrac ab=\dfrac{ab+bc-(ab+ad)}{b(b+d)}=\dfrac{bd\left(\dfrac cd-\dfrac ab\right)}{b(b+d)}$$ which will be $\ge0$ if $\dfrac d{b+d}\ge0$


问:
Power series representation of an integral, centered at 1?
We let $$\int_{1}^{x} \frac{\ln(t)}{t-1} \,dt $$
I want to represent this integral as a power series centered at $1$. My initial strategy is to take a Taylor series of $\ln(t)$ and divide that by $(t-1)$. Then interchange summation and integral.  Is this strategy valid or am I making a mistake?

So    $\ln(t) = (t-1) - \frac{1}{2}(t-1)^2 +\frac{1}{3}(t-1)^3+\ .... $ = $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(t-1)^n$

Dividing by (t-1) gave me $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(t-1)^{n-1}$

Body:
First, Let's try to solve the integral
$$\int^x_1 \frac{\ln(x)}{x-1}$$
Let $y=x-1$($dx=dy$).
$$\int^{x-1}_0 \frac{\ln(y+1)}{y}dy =\sum^\infty_{n=1}\frac{(-1)^n}{n}\int_0^{x-1}\frac{y^{n}}{y}=\sum^{\infty}_{n=1} \frac{(-1)^n}{n}(0-(-\frac{(x-1)^{n}}{n}))=\sum^\infty_{n=1}\frac{(-1)^{n+1}(x-1)^{n}}{n^2}$$


问:
Positive genus and Abelian-ness
So if a Seifert surface of a knot K has positive genus, what can we say about the its complement's first fundamental group, is it Abelian because the only knot with Seifert surface of positive knot is the unknot?</p>

答: The only knot whose complement has an abelian fundamental group is the unknot. The fundamental group of a knot complement abelianizes to $\mathbb Z$ by Alexander Duality. So in fact the only possible abelian fundamental group for a knot complement is $\mathbb Z$. It is known that the unknot is the only knot where $\pi_1$ of the complement is $\mathbb Z$.</p>

Title: Positive genus and Abelian-ness


问:
Why does base $b$ have digits from $0$ to $b-1$?
Base system with $b \in \mathbb N$ consists of $b$ digits $d_0,d_1,d_2\dots d_{b-1}$. A number $a$ is expressed by some weighted sum of (integer) powers of $b$, where the digits $d_0,d_1,d_2\dots d_{b-1}$ are the weights.

My question is, why do we, if $b$ is a natural number greater than $1$, select the weights or the digits to be whole numbers and why do we need $b$ of them?

In the usual base $2$ system, with weights $0$ and $1$, number $15$ in base $10$, for example, would be expressed as $1111$.

However, if we say that the base two system has weights $0$ and $\frac{1}{3}$, which we name $a$, $15_{10}$ would be $a0aa0a$, ie. $\frac{1}{3} 2^5+\frac{1}{3} 2^3+\frac{1}{3} 2^2+\frac{1}{3} 2^0$.

It is a tad longer than with the usual weights. Is that the only reason to use $0$ and $1$, instead of $0$ and $1/3$?

How about base $3$ with weights $0$ and $\frac{2}{5}$, denoted by $b$? For example $15_{10}$ would be $bb0b.\overline {bbb}$ or $\frac{2}{5}3^3 + \frac{2}{5}3^2 + \sum_{n=0}^{\infty} \frac{2}{5}3^{-n}$.

Can I choose the weights or base digits in any way I want and there would always be a way to write any number in any base using them? Or do some other complications arise?

答:
All that is needed is that there a digit equal to each number 0 to b-1, modulo b.  That is, you can add or subtract b from various d where d is in 0 to b-1.

Base 3 works with digits $0,1,2$ or $0,1,-1$ or $3,1,-1$ or $3,1,2$.  Except for 3,1,-1, the bases are correctly spelt: ie sorted spelling is sorted size.

On the other hand, one can quite easily have negative bases, like $-3$, and even complex bases like $\sqrt{-3}$.  Both of these rely on the digits $-1, 0, 1$ in their usual form, the latter is implemented by carrying and borrowing over two columns.

The following is a multiplication of $7+4\sqrt{3}$ by $7-4\sqrt{3}$, in base $\sqrt{3}$, which is ordinary base 3, with carry and borrow over two columns (ie 3 = 1 0 0).

```<code>     7+4 sqrt(3)     1  1  m  1  1   = 1 0 m 0 1 + 1 0 1
     7-4 sqrt(3)     1  m  m  m  1   = 1 0 m 0 1 - 1 0 1
                     -------------
                     1  1  m  1  1
                  m  m  1  m  m
              m   m  1  m  m
           m  m   1  m  m
        1  1  m   1  1
        ---------------------------
        1  0 -3   0  1  0 -3  0  1    carryless sum
        0  0  0   0  0  0  0  0  1    carry the -3 as -1 0 0.
</code>```

Body:
All that is needed is that there a digit equal to each number 0 to b-1, modulo b.  That is, you can add or subtract b from various d where d is in 0 to b-1.

Base 3 works with digits $0,1,2$ or $0,1,-1$ or $3,1,-1$ or $3,1,2$.  Except for 3,1,-1, the bases are correctly spelt: ie sorted spelling is sorted size.

On the other hand, one can quite easily have negative bases, like $-3$, and even complex bases like $\sqrt{-3}$.  Both of these rely on the digits $-1, 0, 1$ in their usual form, the latter is implemented by carrying and borrowing over two columns.

The following is a multiplication of $7+4\sqrt{3}$ by $7-4\sqrt{3}$, in base $\sqrt{3}$, which is ordinary base 3, with carry and borrow over two columns (ie 3 = 1 0 0).

     7+4 sqrt(3)     1  1  m  1  1   = 1 0 m 0 1 + 1 0 1
     7-4 sqrt(3)     1  m  m  m  1   = 1 0 m 0 1 - 1 0 1
                     -------------
                     1  1  m  1  1
                  m  m  1  m  m
              m   m  1  m  m
           m  m   1  m  m
        1  1  m   1  1
        ---------------------------
        1  0 -3   0  1  0 -3  0  1    carryless sum
        0  0  0   0  0  0  0  0  1    carry the -3 as -1 0 0.


问:
Test if population standard deviation is less than 7.2.
With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon.

With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers.

Id like to test the hypothesis that the waiting times for a single line are longer.
For this problem, I get Chi critical value of 36.41502850. Which is much larger than noted in the solution. If Chi squared test statistic value is larger than the critical value, then the null hypothesis ( that the waiting times for singe line has a more variation could be rejected? ) What if one would like to determine if one has to wait longer than 7.2 minutes? Then I guess the appropriate F score would be 1-$/alpha$ quantile?

答:
First you have to assume Normality...

Then your Hypothesis' System to be verified is the following

$$\begin{cases}
\mathcal{H}_0:  & \sigma^2=51.84 \\
\mathcal{H}_1:  & \sigma^2<51.84
\end{cases}$$

thus your critical chi value at 5% is $13.8484$

As your Observed Statistic is $\frac{24\times3.5^2}{7.2^2}\approx5.67$ this means that you are at the left of  your critical value: you reject $\mathcal{H}_0$ that is: there are enough evidences against the null hypothesis to agree with the claim that "a single line causes lower variation among waiting times for customers".


Title: Power series representation of an integral, centered at 1?

Body:
We let $$\int_{1}^{x} \frac{\ln(t)}{t-1} \,dt $$
I want to represent this integral as a power series centered at $1$. My initial strategy is to take a Taylor series of $\ln(t)$ and divide that by $(t-1)$. Then interchange summation and integral.  Is this strategy valid or am I making a mistake?

So    $\ln(t) = (t-1) - \frac{1}{2}(t-1)^2 +\frac{1}{3}(t-1)^3+\ .... $ = $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(t-1)^n$

Dividing by (t-1) gave me $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(t-1)^{n-1}$

Body:
$$f(x)=\int_1^x \frac{\log (t)}{t-1} \, dt$$

$$\log t=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(t-1)^n$$

$$\frac{\log t}{t-1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(t-1)^{n-1}$$ $$\int_1^x \frac{(-1)^{n+1} (t-1)^{n-1}}{n} \, dx=\frac{(-1)^{n+1} (x-1)^n}{n^2}$$

$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1} (x-1)^n}{n^2}$$


问:
Expectation of an Ito integral involving arbitrary function in integrand
Suppose, we have a have stochastic integral $\int ^t_0 I(t) dB(t)$ and $\int ^t_0 \frac{S(t)}{I(t)} dB(t)$, where $I(t)$ and $\frac{S(t)}{I(t)}$ are continuous bounded functions and they are actually Susceptible and infected populations of an epidemic model. I have to prove that the expectation of both the integrals are zero.

My thought: I write $\mathbb{E}\int ^t_0 I(t) dB(t)= \mathbb{E} ~\Sigma_{j} I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}= \Sigma_{j} \mathbb{E}\bigg\{I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}\bigg\}=\Sigma_{j} \mathbb{E}\big\{I(t_j)\big\} \mathbb{E}\big\{B_{t_{i+1}}-B_{t_{i}}\big\}$. Now since Brownian motion has mean zero, so $\mathbb{E}\int ^t_0 I(t) dB(t)=0$. Is it right? Please help to prove it more elaborately.

答:
I was thinking in the way @user1 and thanks for his answer. But now I am thinking in the following way:

Since the integrated $I(t)$ is a non-anticipative continuous function and also
$\mathbb{E} \bigg\{\int_0^ t |I(t)|^2 dt\bigg\} < \infty $, so we can write
$$\int_0^t I(t)~ dB(t)= \lim _{\Delta t \rightarrow 0} \Sigma_{j=1}^N~ I(t_{j-1})~(B(t_j)-B(t_{j-1})).$$ Now $\mathbb{E} \bigg\{\int_0^t I(t)~ dB(t)\bigg\}=\lim _{\Delta t \rightarrow 0} \Sigma_{j=1}^N~ \mathbb{E}(I(t_{j-1}))~\mathbb{E}((B(t_j)-B(t_{j-1})))$. Now as  increments of Brownian motion are normally distributed with mean zero, variance $t_j−t_{j-1}$ and the function is non-anticipative, so we can say that $\mathbb{E} \bigg\{\int_0^tI(t)~dB(t)\bigg\}=0.$

Can I do it this way? Please give some advice. Thank you.


问:
Express an integral in terms of another integral.
Question:

Let $$I=\int_0^{\pi/6}\tan^{10}(2x)\sec(2x)\,dx.$$ Express the value of $$\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx$$ in terms of $I$.

Attempts:

$$\begin{align}\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx&=\int_0^{\pi/6}\tan^{10}(2x)(\sec^2(2x)-1)\sec(2x)\,dx\\&=\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx-I\end{align}$$

From here I can’t continue. How should I express $\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx $? Thanks a lot.

答:
Integrals of powers of secant ant tangent is a bit common calculus problem. Here I will summarize the techniques that we can used to solve this problem. First thing is the Pythagoras theorem $$1+\tan^2\theta=\sec^2\theta,$$ which we can used to interchange between them. Next we have few elementary integrals related to these two functions $$\displaystyle\int\sec^2\theta\,d\theta=\tan\theta,\qquad\qquad \displaystyle\int\sec\theta\tan\theta\,d\theta=\sec\theta,$$ and $$\displaystyle\int\tan\theta\,d\theta=\ln|\sec\theta|,\qquad\qquad \displaystyle\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|. $$

Now we are ready to attack the general question $\displaystyle\int\sec^m\theta\tan^n\theta\,d\theta.$

We can solve most of them using substitutions:

$$\begin{array}{|l|c|r |}
m & n & \text{substitution} \\
\hline
\text{even} & \text{odd} & x=\tan\theta,\,x=\sec\theta \\
\text{even} & \text{even} & x=\tan\theta \\
\text{odd} & \text{odd} &  x=\sec\theta \\
\text{odd} & \text{even} &  \text{None}
\end{array}
$$

The odd case "neither does secant have an even power nor does tangent have an odd power" type problems can be handle via integration by parts followed by applying Pythagoras theorem. Since this last case is very interesting, its second simplest problem even has a Wikipedia page.


问:
Find the extrema values of a multivariable function with one inequality constraint
Question: Find the absolute maximum and minimum values of $f(x, y) = xy^2$ on the set $D=\{{(x, y), x^2+y^2 <= 3}\}$.

My attempted Solution: First I computed the following gradients of the functions:

Gradient of $f(x,y)=<y^2, 2xy>$

Gradient of $g(x,y)=<2x, 2y>$

Critical point 1: Since $<y^2, 2xy> = <0, 0>$, $(x, y) = (0, 0)$

By using the Lagrange multipliers, I obtained the following three functions:

$$y^2=2x\lambda$$
$$2xy=2y\lambda$$
$$x^2+y^2=3$$

However, I'm not sure how to continue on from this point to compute the critical points. Since $x$ and $y$ can be zero in both these cases, I'd assume that we can't divide either from both sides of the equation, thus making it hard to solve.

I'd be much appreciated if someone could show me how this is done. Thanks in advance!

答:
An alternative approach with change of variables.
Let $x=r\cos{\theta}$ and $y=r\sin{\theta}$.

Then we have as our inequality $r^2\leq 3$ and we seek to find the extreme values for $f(r,\theta)=r^3 \cos{\theta} \sin^2{\theta}$.

From here we can use single variable calculus to find max and min values of $f$. Max and min for $r^3$ are 27 and 0, respectively.

Let $ g(\theta)=\cos{\theta}\sin^2{\theta}=\cos{\theta}-\cos^3{\theta}$.

$g'(\theta)=-\sin{\theta}+3\sin{\theta}\cos^2{\theta}$. It's zeros occur when theta is an integer multiple of $\pi$ or $\cos^2{\theta}=1/3$. $g$ is zero at integer multiples of pi and takes on extremes of $\frac{2}{3\sqrt{3}}$ and $\frac{-2}{3\sqrt{3}}$. $g$ flips signs by adding $\pi$ to the argument.

Altogether, $f$ has the extreme values $6\sqrt{3}$ and $-6\sqrt{3}$.


问:
Fixed point iteration problem of $f(u)=u^3-u-1$
I was thinking about the following problem:

Let $f(u)=u^3-u-1$. Then I have to verify whether the following statements are true/false?

1.Starting with the initial guess $u^{(0)}=1.5,$ the fixed point iterates of the equation $u=g(u)$,where $g(u)=u^3-1$ converges

2.If $u^{\star}$ is a root of the equation $f(u)=0$ and $u^{\star} >1,$ then $u^{\star} $ is a stable fixed point of the equation $u=g(u)$

3.$f(u)=0$ has a root between $1$ and $2$

4.Staring with the initial guess $u^{(0)}=1.5,$ the fixed point iterates of the equation $u=\tilde{g}(u),\text{where} \space \tilde{g}(u)=\sqrt{1+u^3} $ converge.

Here,option 3 is true as we see that $f(1)<0  \space \text{whereas} f(2)>0$ .But I am not sure about the other options .How can I check other options?

Can someone point me in the right direction?

Thanks in advance for your time.

答:
Hints

There is a root at: $u = 1.3247179572447460260$ and the other two roots are imaginary.

Part 1.

Here are the iterates. What do you notice is happening from iterate to iterate?

$u_1 = 2.375$

$u_2 = 12.396484375$

$u_3 = 1904.00277223$

$u_4 = 6902441412.84$

$u_5 = 3.28857830393E+29$

$u_6 = 54608.3924665$

$\ldots$

$u_{40} = 5153.0115226$

So, what can you say about convergence?

Part 2

You just need to think about this one a bit (go back and look at the FP theory.

Part 3

Correct. We could have also plotted the function to see:

Part 4

$u_1 = 2.09165006634$

$u_2 = 3.18605854313$

$u_3 = 5.77421697131$

$u_4 = 13.9112014761$

$u_5 = 51.8952497366$

$u_6 = 373.846193564$

$\ldots$

$u_{40} = 1.5659769437E+15$

So, what can you say about convergence?


问:
For $1 \le p \lt q \le \infty$ show that neither $L^p(\Bbb{R},\lambda)$ or $L^q(\Bbb{R},\lambda)$ contain the other
Let $1 \le p \lt q \le \infty$, and $\lambda$ denote the Lebesgue measure. Show that $L^p(\Bbb{R},\lambda) \not\subset L^q(\Bbb{R},\lambda)$ and $L^q(\Bbb{R},\lambda) \not\subset L^p(\Bbb{R},\lambda)$ by constructing functions where each belongs to one but not the other.

Hint: You will likely want to handle $q = \infty$ separately.

My idea:

After some searching online, I came across this suggestion:

For the first non-inclusion, consider $f(x) = \begin{cases} x^{\frac{1}{p}} &\text{if } 0 \le x \le 1 \\ 0 &\text{o.w.} \end{cases}$

Here is my attempt to use this function:

Then $\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^p = \int_0^1xd\lambda(x) \lt \infty$

And$\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^q = \int_0^1x^{\frac{q}{p}}d\lambda(x)$ $?= \infty$

I'm not quite sure how to show this. If anyone can help that would be great!

答:
My professor in grad school would say this is about spikes and tails. A spike is something like
$$f(x) = \frac1{x^r} \chi_{(0,1)},$$
where $\chi_A$ is the characteristic function of $A$, and a tail is something like
$$g(x) = \frac1{x^r} \chi_{[1,\infty)}.$$
For what values of $r$ is $f$ integrable? For what values of $r$ is $g$ integrable? You should be able to use this to answer your question.

While you're at it, what can you say about the containment

between $L^p((0,1),\lambda)$ and $L^q((0,1),\lambda)$, when $p<q$?

between $L^p([0,\infty),\lambda)$ and $L^q([0,\infty),\lambda)$, when $p<q$?


问:
Gaussian convolution
$X$ and $Y$ are standard normal independent random variables, let $Z=XY$. Is the distribution of $Z$ equal to $N(0,\frac{1}{2})$?

I tried using convolution to get the distribution of $Z$. Is it correct?

答:
There is a similar formula for the distribution of the product $Z=XY$ of two nicely behaved random variables: $$ f_Z(z)=\int_{-\infty}^\infty f_X(t)f_Y(z/t)\frac{1}{|t|}dt.$$ So in this case with two standard normals, that's $$ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}(t^2+\frac{z^2}{t^2})}\frac{dt}{|t|}.$$ There's no way to integrate this in terms of elementary functions, but we can get it into a well-known integral representation of a well-known special function:

If we let $u= t/\sqrt{|z|},$ this gives $$ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-|z|\frac{1}{2}(u^2+1/u^2)}\frac{du}{|u|}$$ and letting $w=2\ln(u)$ gives $$ f_Z(z) = \frac{1}{\pi}\int_{0}^\infty e^{-|z|\cosh(w)}dw = \frac{1}{\pi}K_0(|z|)$$ where $K_0$ is the modified Bessel function. (The relevant integral formula is both on the wikipedia page and here.)

You mentioned in the comments going the route of characteristic equations, and that works too. We have $$ \phi_Z(t)=E(e^{itZ}) = E(e^{itXY}) =\frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty e^{itxy-x^2/2-y^2/2}dxdy.$$ You can complete the square to $$ \phi_Z(t)=\frac{1}{2\pi}\int_{-\infty}^\infty dy e^{-y^2(1+t^2)/2}\int_{-\infty}^\infty e^{-\frac{1}{2}(x-ity)^2}dx.$$ The $x$ integral is just a shifted standard Gaussian (shifted in an imaginary direction, but that doesn't wind up mattering), so is just $\sqrt{2\pi},$ so we get $$ \phi_Z(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{1}{2}(1+t^2) dy} = \frac{1}{\sqrt{1+t^2}}\int_{-\infty}^\infty e^{-\frac{1}{2}u^2} \frac{du}{\sqrt{2\pi}} = \frac{1}{\sqrt{1+t^2}}.$$

But then, if you want the PDF, there's the work of doing the inverse Fourier transform to compute it from the characteristic function. Using this answer, we obtain $$ F_Z(z) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-izt}\frac{1}{\sqrt{1+t^2}}dt = \frac{1}{2\pi} 2K_0(|z|) = \frac{1}{\pi}K_0(|z|),$$ exactly as above.

Title: Gaussian convolution


问:
Prove $\frac{d}{dx} e^x = e^x$ using $\frac{d}{dx} a^x = xa^{x-1}$
The hint in the textbook says,

You can determine $\frac{d}{dx} e^x = e^x$ by first determining that $\frac{d}{dx}$ ln $x = \frac{1}{x}$ using inverse functions. Then consider $x^p =e^{p\ln x}$.

I understand intuitively what this means, but I'm having trouble showing it.

We need to prove $\frac{d}{dx} e^x = e^x$

Let $f(x)=e^x$. Equivalently, $\ln y = x$. Thus, $f^{-1}(y)=\ln y$.

Now recall that $(f^{-1})'[f(c)]=\frac{1}{f'(c)}$

Since $f^{-1}(y)=\ln y$, by the rule stated above, $$\frac{d}{dx}\ln y = \frac{1}{y'}$$

$$y'= \frac{1}{\frac{d}{dx}\ln y}$$

But $y'$  [used interchangeably with $f'(x)$] is what we're trying to show. Where am I going wrong here?

答:
What is creating confusion to you is the notation. Use the prime notation, I suggest.

Instead of using $\frac{d}{dx}(\cdot)$ in the rule, use the prime notation $(\cdot)'$ as
$$(\ln y)'=\frac{1}{y'}$$
$$\frac{1}{y}=\frac{1}{y'}$$
$$y'=y$$

Or if you wish to go your way, then change $\frac{d}{dx}$ to $\frac{d}{dy}$ in the last step.
$$\therefore y'=\frac{1}{\frac{d}{dy}\ln y}=y$$

In brief, your confusion was created due to the confusion in the variable of differentiation.

Edit

We wish to prove that
$$\frac{de^x}{dx}=e^x$$

We shall use the fact that $\frac{dx^p}{dx}=px^{p-1}$ and $x^p=e^{p\ln x}$.

To clear things a bit, we let $u=p\ln x$. So, $x^p=e^u$. By Chain Rule,
$$\frac{de^u}{dx}=\frac{de^u}{du}\cdot\frac{du}{dx}$$
$$\frac{dx^p}{dx}=\frac{de^u}{du}\cdot\frac px$$
Use the Power Rule.
$$px^{p-1}=\frac{de^u}{du}\cdot\frac px$$
$$x^p=\frac{de^u}{du}$$
$$\frac{de^u}{du}=e^u$$
as desired.

Hope it helps. Ask if anything is not clear :)


问:
Periodicity and non-periodicity of compositions
Prove $\sin(\sqrt x)$ is not periodic.

My wrong assumption was non-periodicity is due to the injectivity of the square root, but, then this would work for any positive real number, but it doesn't, and $\sin(x^2)$ isn't periodic either, so injectivity fails.

I find this example more interesting because it is defined on the interval $[0,+\infty)$ so it can't be periodic on the interval $(-\infty,0).$

On the other hand, $\;\sin\left(\sqrt{x}\right)\;=\sin\left(\sqrt{x+\tau}\right)$ seems hopeless.

How does composition behave in general when talking about periodicity?

Our assistant said the answer lays in irrationality in examples of this kind:
$\sin\left(\sqrt{2}x\right),$

but it seems clumsy, incomplete and too intuitive to claim there is no common multiple halfway the solution...

At some point I get some expression I don't know what to do with.

Also, what does almost periodic in English mean?

Are constants, signum, Dirichlet's and Moebius function almost periodic? as far as I'm concerned, they don't have the least period? Or does it refer to floor and ceiling?

答:
Let $f$ be periodic with period $T$. Then for $f(g(u))$ to be periodic with period $U$, we need

$$g(u+U)=g(u)+kT$$ for some integer constant $k$.

Now if

$$h(u):=g(u)-kT\frac uU,$$

$$h(u+U)=g(u+U)-kT\frac{u+U}U=g(u)+kT-kT\frac{u+U}U=h(u)$$ and $h$ is periodic.

Finally, $g$ must be of the form

$$g(u)=kT\frac uU+h(u)$$ where $h$ is periodic of period $U$.

E.g. let $f(x)=\{x\}$, which is periodic with period $1$ and let us choose $h(x)=\sin(x)$ (period $2\pi$). We also choose $k=2$ and get

$$g(x)=\frac x\pi+\sin(x).$$

This gives us a nice psychedelic function.


问:
Some question on localization of polynomial ring
Let $S=A[x_1,\dots, x_r](r \geq 2)$ be a polynomial ring where $A$ is a commutative ring.
Then is it true that $S=\bigcap_{i=1}^r S_{x_i}$? If $S$ is $A$-algebra and $x_i$ are not zero divisors, then is it true?</p>

答:
Let's assume $r=2$ and prove $S=S_{x_1}\cap S_{x_2}$. (As Hartshorne mentions the intersection takes place in $S_{s_1x_2}$.) Pick $w\in S_{x_1}\cap S_{x_2}$. Then $w=f/x_1^m=g/x_2^n$ with $f,g\in S$. If $m=0$ or $n=0$ we are done. Suppose $m,n\ge 1$. Write $f=a_0(x_1)+a_1(x_1)x_2+\cdots+a_s(x_1)x_2^s$ and $g=b_0(x_1)+b_1(x_1)x_2+\cdots+b_t(x_1)x_2^t$ with $a_s\neq 0$, $b_t\neq 0$. Since $x_2^nf=x_1^mg$ we get $$x_2^n[a_0(x_1)+a_1(x_1)x_2+\cdots+a_s(x_1)x_2^s]=x_1^m[b_0(x_1)+b_1(x_1)x_2+\cdots+b_s(x_1)x_2^t].$$ Now consider both sides as polynomials in $x_2$ with coefficients in $A[x_1]$ and identify the coefficients. One obtains $t=n+s$ and $x_1^mb_0(x_1)=\cdots=x_1^mb_{n-1}(x_1)=0$. But $x_1$ is a non-zerodivisor in $A[x_1]$, so $b_0(x_1)=\cdots=b_{n-1}(x_1)=0$. This shows that $x_2^n\mid g$, and thus $w\in S$.

For $r\ge 2$ the proof is the same with some minor changes.

If $S$ is an $A$-algebra and $x_i$ are non-zerodivisors the property can fail: consider $S=K[x,y]$ with $x^2=y^3$. This is an integral domain and $\frac{1}{x^2}\in S_x\cap S_y$, but $\frac{1}{x^2}\notin S$.


问:
Approaching the limit of an inverse matrix
So I was trying to find the inverse of the $3\times 3$ matrix of ones. I found that its adjoint is the zero matrix and its determinant is $0$ so the entries of the inverse matrix are in an indeterminate form. Suppose that there was a function $f: \mathbb{R}^{6\times1} \rightarrow \mathbb{R}^{3\times3}$. f would change the entries of the first two rows of $3\times3$ matrix of ones and then calculate its inverse. Does there exist a path for which we can take the limit along as f approaches $(1,1,1,1,1,1)$? I already know that the limit doesn't exist in general so now I'm trying to see along what paths does the limit exists and if all such paths yield the same limit.

EDIT: Sorry this was my mistake. The function $f$ is defined as $$f = \frac{adj(A)}{|{A}|}$$
where $adj(A)$ is the Adjoint of A and $|A|$ is determinant of A.

答:
First of all I think the function you are defining is this:
$\mathbf{f}:\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)\in D\subset\mathbb{R}^6\to \begin{bmatrix}
    x_1 & x_2 & x_3 \\
    x_4 & x_5 & x_6 \\
    1 & 1 & 1 \\
    \end{bmatrix}^{-1} \in \mathbb{R}^{3\times 3}  $

Where $D=\left\{  \mathbf{x} \in \mathbb{R}^6:\exists\begin{bmatrix}
    x_1 & x_2 & x_3 \\
    x_4 & x_5 & x_6 \\
    1 & 1 & 1 \\
    \end{bmatrix}^{-1}\right\}$

Now you are asking about:
$$\lim_{\mathbf{x}\to (1,1,1,1,1,1)} \mathbf{f}(\mathbf{x})$$
(Where $\mathbb{R}^6$ is equipped with euclidean metric and $\mathbb{R}^{3\times 3}$ with Frobenius metric)
First of all I would like to make you notice that two metrics above are both norm-induced metric so in both spaces I can use the respective norms and:
$$d(x,y)=||x-y||$$
(For simplicity I'll use the same symbol for the two norms/metrics even if they are different).
A theorem(that is pretty advanced!I just had the luck to read it during a numerical calculus course, if you want to know something about it it's a consequence of Eckart-Young theorem)tells us that if $A$ is a non-singular matrix of order $n$(let $\Sigma$ be the set of all the singular matrices of order $n$):
$$\min_{S \in \Sigma}(||A-S||)=\frac{1}{||A^{-1}||}$$
Clearly this implies in our particular space that:
$$\left\|A-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|\geq \frac{1}{||A^{-1}||} \ \ \ \  \forall A : \exists A^{-1}$$
So:
$$\left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|\geq \frac{1}{||\mathbf{f}(\mathbf{x})||} \ \ \ \  \forall \mathbf{x} \in D$$
So:
$$||\mathbf{f}(\mathbf{x})|| \geq \left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|^{-1} \forall \mathbf{x} \in D$$
Notice that if our limit exists finite along a path $\mathcal{P}\subseteq D$(such that $(1,1,1,1,1,1)$ is of accumulation for $\mathcal{P}$), then:
$$\forall \varepsilon>0 \exists \delta >0 :\forall \mathbf{x} \in \mathcal{P} \ \  \ (||\mathbf{x}-(1,1,1,1,1,1)||<\delta \Rightarrow ||\mathbf{f}(\mathbf{x})-L||<\varepsilon)$$

So if $\mathbf{x}$ is in a sufficiently small ball(let's call it $B$) of center $(1,1,1,1,1,1)$ ,then $||\mathbf{f}(\mathbf{x})||$ is bounded in $B \cap \mathcal{P}$(you can easily prove it using norm subtraction property on  $||\mathbf{f}(\mathbf{x})-L||<\varepsilon$).
But:
$$||\mathbf{f}(\mathbf{x})|| \geq \left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|^{-1} \forall \mathbf{x} \in \mathcal{P}\subseteq D$$

So the boundness of  $||\mathbf{f}(\mathbf{x})||$ , with $\mathbf{x} \in B \cap \mathcal{P}$ requires $ \left\|\mathbf{f}(x)^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|^{-1}$ with $\mathbf{x} \in B\cap \mathcal{P}$ to have an upper bound. That is to say that $ \left\|\mathbf{f}(\mathbf{\mathbf{x}})^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|$ with $\mathbf{\mathbf{x}} \in B \cap \mathcal{P}$ must have a non-zero lower bound.
But:
$$\left\|\mathbf{f}(\mathbf{\mathbf{x}})^{-1}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|=\left\|\begin{bmatrix}
    x_1 & x_2 & x_3 \\
    x_4 & x_5 & x_6 \\
    1 & 1 & 1 \\
    \end{bmatrix}-\begin{bmatrix}
    1 & 1 & 1 \\
   1 & 1 & 1 \\
    1 & 1 & 1 \\
    \end{bmatrix}\right\|=\sqrt{\sum_{i=1}^6 (x_i-1)^2}=$$
$$=||\mathbf{x}-(1,1,1,1,1,1)||$$
But clearly:
$$\inf_{\mathbf{x} \in B \cap \mathcal{P}}(||\mathbf{x}-(1,1,1,1,1,1)||)=0$$
Because $(1,1,1,1,1,1)$ is an accumulation point of $\mathcal{P}$(and also of $B \cap \mathcal{P}$, since $B$ is a neighborhood of $(1,1,1,1,1,1)$). So there is not non-zero lower bound and by consequence the limit cannot exist finite along any path.
This concludes the proof.


答:
Here is how you wish to prove it:

$$f(x) = e^{x}$$

$$f^{-1}(x) = \ln(x)$$

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{x}$$

$$\frac{1}{f'(\ln(x))} = \frac{1}{x}$$

$$f'(\ln(x)) = x$$

Because $e^{x}$ is the only function with domain $\mathbb{R}$ satisfying this property:

$$\boxed{f'(x) = e^{x}}$$

Sidenote: Here is a proof which I like better.

$$\begin{align}\frac{d}{dx}e^{x} &= \lim_{h\to 0}\frac{e^{x + h} - e^{x}}{h}\\ &=\lim_{h\to 0}\frac{e^{x}(e^{h} - 1)}{h}\\ &=e^{x}\lim_{h\to 0}\frac{e^{h} - 1}{h}\end{align}$$

This proof requires $\displaystyle\lim_{n\to-\infty}\bigg(1 + \frac{1}{n}\bigg)^{n} = e$, which is similar to the definition of $e$ but requires justification. We have:

$$\lim_{n\to-\infty}\bigg(1 + \frac{1}{n}\bigg)^{n} = \lim_{n\to\infty}\bigg(\frac{1}{1 - \frac{1}{n}}\bigg)^{n} = \lim_{n\to\infty}\bigg(\frac{n}{n-1}\bigg)^{n} = \lim_{n\to\infty}\bigg(1 + \frac{1}{n-1}\bigg)^{n}$$
$$ = \lim_{n\to\infty}\bigg(1 + \frac{1}{n-1}\bigg)\lim_{n\to\infty}\bigg(1 + \frac{1}{n-1}\bigg)^{n-1} = 1\cdot e = e$$

Then, noting that $h\to 0\Longleftrightarrow\frac{1}{h}\to\pm\infty$:

$$\begin{align}&=e^{x}\lim_{h\to 0}\frac{\bigg(\big(1 + h\big)^{\frac{1}{h}}\bigg)^{h} - 1}{h}\\ &=e^{x}\lim_{h\to 0}\frac{1 + h - 1}{h}\\ &=e^{x}\lim_{h\to 0}1\\&=\boxed{e^{x}}\end{align}$$
You could make the part where the limit definition of $e$ is substituted in more rigorous by taking the one sided limits at $0$ and proving they both equal $1$.

Title: Prove $\frac{d}{dx} e^x = e^x$ using $\frac{d}{dx} a^x = xa^{x-1}$

Body:
The hint in the textbook says,

You can determine $\frac{d}{dx} e^x = e^x$ by first determining that $\frac{d}{dx}$ ln $x = \frac{1}{x}$ using inverse functions. Then consider $x^p =e^{p\ln x}$.

I understand intuitively what this means, but I'm having trouble showing it.

We need to prove $\frac{d}{dx} e^x = e^x$

Let $f(x)=e^x$. Equivalently, $\ln y = x$. Thus, $f^{-1}(y)=\ln y$.

Now recall that $(f^{-1})'[f(c)]=\frac{1}{f'(c)}$

Since $f^{-1}(y)=\ln y$, by the rule stated above, $$\frac{d}{dx}\ln y = \frac{1}{y'}$$

$$y'= \frac{1}{\frac{d}{dx}\ln y}$$

But $y'$  [used interchangeably with $f'(x)$] is what we're trying to show. Where am I going wrong here?


问:
derivative complex number
I have a problem in a demonstration where I have to do a derivative $\frac{df}{dc}$ of

$f=(y-cx)(y-cx)^{*}$

where * denotes the complex conjugate.

In my demonstration the derivative $\frac{df}{dc}$is equal to 0. So I tried to solve it by applying the chain rule:

$x(y-cx)^{*}+x^{*}(y-cx)=0$

obtaining the result:

$c=\frac{xy^{*}+yx^{*}}{2xx^{*}}$

Unfortunately I should have obtained

$c=\frac{yx^{*}}{xx^{*}}$

Can you help me with this derivative of complex number? What am I doing wrongly?

答: I'm confused; $c \mapsto \overline{c}$ is not analytic and $f=|y|+|cx|-\overline c \overline x y - cx\overline y$, so why should $f$ be complex-differentiable?

Title: derivative complex number


问:
Question from Section -5 Subsection -Splitting Fields Of Hungerford Algebra
I am trying some questions from Textbook hungerford algebra and was unable to solve this particular problem. I am self studying so can't ask anyone else.

If $f \in K[x]$ has degree n and F is a spitting field of f over K, then [F : K] divides
n!.

I let degree of f =1 then it splits in K and hence assertion is true. but induction can't be used(clearly).

Can you please tell how to approach this question? I am at loss of ideas on why should always this happen.

Thank You!!

答:
We can use strong induction on $\mathrm{deg}(f)$.

Assume that we know that for all $k \leq n$, we have that the splitting field of a degree $k$ polynomial divides $k!$.

Let $\mathrm{deg}(f)=n+1$. If $f$ is not irreducible, we may factor $f=gh$ where $\mathrm{deg}(g)=a,\mathrm{deg}(h)=b$ satisfy $a,b \leq n$, so by the induction hypothesis gives that if $F'$ is splitting field of $g$ over $K$ and $F$ is the splitting field of $h$ over $F'$, then $F$ is also the splitting field of $f$ over $K$. We have by the induction hypothesis $[F:K]=[F:F'][F':K] \mid a!b! \mid (a+b)!=(n+1)!$. Here we used that the binomial coefficient $\binom{a+b}{a}=\frac{(a+b)!}{a!b!}$ is an integer.

So suppose that $f$ is irreducible over $K$. Let $\alpha$ be a root of $K$ in $F$. Then we can write $f=(x-\alpha)g$ in $K[\alpha][x]$ where $g$ has degree $n$. As $f$ is irreducible, we have $[K[\alpha]:K]=n+1$. $F$ is the splitting field of $g$ over $K[\alpha]$, so by the induction hypothesis, we have $[F:K[\alpha]] \mid n!$, thus $[F:K]=[F:K[\alpha]][K[\alpha]:K] \mid n!(n+1)=(n+1)!$


问:
Proving A=RU where R is nonnegative self-adjoint mapping and U is unitary
I'm trying to following proof of this.

$A$ which is linear mapping of complex Euclidean space into itself.

Then $A$ can be factorized by

$A=RU$ where $R$ is nonnegative self-adjoint mapping, and $U$ is unitary.

I already did when $A$ is invertible, but I'm stuck with not invertible case.

If $A$ is not invertible,

Scalar product $(AA^*x,x)=(A^*x,A^*x) \geq 0$. Thus, $AA^*$ is nonnegative.

By Theorem, $AA^*$ has unique nonnegative self-adjoint square root. Denote it $R$.

$\parallel Rx \parallel^2=(Rx,Rx)=(R^2x,x)=(AA^*x,x)=(A^*x,A^*x)=\parallel A^*x\parallel^2$

Suppose $Rx=Ry$.

Then $\parallel R(x-y)\parallel=0$. Thus, $\parallel A^*(x-y)\parallel=0$. Therefore, $A^*x=A^*y$

---Until this, I can easily follow. But here, I cannot follow.---

This shows that for any $u$ in the range of $R$, $u=Rx$, we can define $Vu$ as $A^*x$

According to upper equation, which is

$\parallel Rx \parallel^2=(Rx,Rx)=(R^2x,x)=(AA^*x,x)=(A^*x,A^*x)=\parallel A^*x\parallel^2$,

$V$ is an isometry.

---Here, I can't understand why $V$ is an isometry from that equation.

Can someone give me some help?

答:
$V$ is an isometry because it maps $Rx$ to $A^*x$, and by your "upper equation", $\|Rx\|=\|A^*x\|$. By the way, this only gives $V$ as an isometry from $\operatorname{ran}R$ to $\operatorname{ran}A^*$. You now have to extend it to an isometry on the whole space.

You're looking at the polar decomposition: https://en.wikipedia.org/wiki/Operator_theory#Polar_decomposition

If you have access to Axler's book ("Linear Algebra Done Right"), he has a good exposition in 7.D (p.234).