问:
Conjectural closed-form representations of sums, products or integrals
What are some examples of infinite sums, products or definite integrals that have conjectural closed form representations that were confirmed by numerical calculations up to whatever maximum precision anybody tried but still remain unproved?

I am also interested in values of special functions at certain points that have conjectural representations in terms of simpler functions (e.g. special values of hypergeometric functions, Meijer G-function or Fox H-function that representable in terms of elementary functions and well-known constants like $\pi$, $e$, Catalan, Euler–Mascheroni, Glaisher–Kinkelin or Khinchin).

To give some examples:

Gourevitch conjecture mentioned at MathOverflow:
$$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7\stackrel{?}{=}\frac{32}{\pi^3}.$$

Riemann hypothesis (in an unusual form, also mentioned at MathOverflow)
$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt\stackrel{?}{=}\frac{\pi(3-\gamma)}{32}.$$

Another conjecture from MathOverflow attributed to J. M. Borwein, D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century:
$$\frac{\displaystyle\int_{\pi/3}^{\pi/2}\log\left|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right|dt}{\displaystyle\sum_{n=1}^\infty\left(\frac n7\right)\frac{1}{n^2}}\stackrel{?}{=}\frac{7\sqrt{7}}{24},$$
where $(\frac{n}{7})$ denotes the Legendre symbol.

答:
You might want to have a look at the work of Zhi-Wei Sun and especially his 181 Conjectural Series for $\pi$ and Other Constants. Some examples:

$$\sum_{k=1}^\infty\frac{(10k-3)8^k}{k^3\binom{2k}{k}^2\binom{3k}{k}}=\frac{\pi^2}{2},$$

$$\sum_{k=1}^\infty\frac{(28k^2-18k+3)(-64)^k}{k^5\binom{2k}{k}^4\binom{3k}{k}}=-14\zeta(3),$$

$$\sum_{n=1}^\infty\frac{16n+5}{12^n}\sum_{k=1}^\infty\binom{2k}{k}\binom{2(n-k)}{n-k}\binom{-1/4}{k}\binom{-3/4}{n-k}=\frac{8}{\pi}.$$


问:
Solve integrals $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$ and $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$?
Can the solution to the integrals

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$$

and

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$$

be computed analytically - or approximated? I have only figured out the trivial solution for $\beta=1$ where I can compute the second integral by

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-v}}~\mathrm{d}v = \left[e^{-e^{-v}}\right]_{\alpha}^{\infty} = 1 - e^{-e^{-a}}.$$

答:
$$I=\int_\alpha^\infty e^{-v}e^{-e^{-\beta v}}dv$$
$$x=e^{-v}\Rightarrow dv=-e^vdx$$
$$I=-\int_{e^{-\alpha}}^0e^{-x^\beta}dx=\int_0^{e^{-\alpha}}e^{-x^\beta}dx$$
then you could use the series expansion of $e^z$ and interchange the summation and integral

ContentLicense: CC BY-SA 4.0


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: Since by the chain & power rules $\exp^\prime(p\ln x)=\frac{(x^p)^\prime}{p/x}=x^p=\exp(p\ln x)$, in general $\exp^\prime y=\exp y$; or, if we relabel variales, $\exp^\prime x=\exp x$, as required.


问:
Convergence of sequence in a finite dimensional vector space
$\mathbf {The \ Problem \ is}:$ $V$ is a finite dimensional vector space of dimension $d$ and $\alpha$,  $\beta$ : $V \to \mathbb R^d$ are two vector space isomorphisms .

If $v_n$ is a sequence in $V$ , then $\alpha(v_n)$ converges to $\alpha(v)$ iff $\beta(v_n)$ converges to $\beta(v)$.

$\mathbf {My \ approach}:$  Actually , I was trying to link the matrices of $\alpha$ and  $\beta$ (matrix under some fixed ordered basis of $V$), but I couldn't do anything .

A small hint is greatly appreciated .

答: Hint: The linear map $\beta \circ \alpha^{-1}: \Bbb R^d \to \Bbb R^d$ is linear and therefore continuous.


问:
Homologically-trivial codimension-2 submanifolds must bound codimension-1 submanifolds
I'm not sure about the exact statement, but I've heard that a homologically-trivial codimension-2 submanifolds must bound codimension-1 submanifolds. Is there a reference a proof for this statement? Thanks in advance.</p>

答:
In the setting of smooth manifolds (judging by the tags this is the case you are interested in), see Theorem 3 on page 50 of

Kirby, Robion C., The topology of 4-manifolds, Lecture Notes in Mathematics, 1374. Berlin etc.: Springer-Verlag. vi, 108 p. DM 25.00 (1989). ZBL0668.57001.

I am not sure about the topological category (PL should work the same). I remember, Mike Miller wrote a more detailed account of this proof (and mentioned it in one of the MSE questions), I just forgot where it was. You may want to ask Mike directly, he is at  Columbia U.

Edit. In the answer I was assuming that you are using homology with integer coefficients and your codimension 2 submanifold $M\subset N^n$ is closed, connected and oriented, and $N$ is also oriented (not sure if this assumption is essential here, but it is used in the proof). Then the fundamental class $[M]$ of $M$ is well-defined and the condition that $[M]=0\in H_{n-2}(N)$ is well-posed.

Here is the proof taken from Kirby's book:


问:
Constructing projective resolution of a chain complex
I am trying to construct the projective resolution in the category of chain complexes of

$\dots \to 0 \to M \to 0 \to \dots$

It seems like it should be possible to do this in terms of the projective resolution of $M$ but I am completely stuck.

I know a projective chain complex is split exact and formed by projectives, so if we think of the resolution as a half plane double complex, the column with $M$ must be a projective resolution of $M$.

I was trying to use the trick of $0 \to P \to P \to 0$ is a projective complex whenever $P$ is projective, but if I put that on top of our complex we don't necessarily get exactness.

答:
In this case, you are in the category of bounded above complexes, where a $\textit{projective resolution}$ of a complex (in this case $\bar{M}:\cdots\rightarrow 0\rightarrow M\rightarrow0\rightarrow\cdots$) means a bounded-above complex of projectives $P$ with a quasi-isomorphism $P\rightarrow \bar{M}$. So, if you take the usual projective resolution of $M$ as a module, $$\cdots\rightarrow P^{-n}\rightarrow P^{-n+1}\rightarrow\cdots\rightarrow P^{-1}\rightarrow P^{0}\rightarrow M\rightarrow0\rightarrow\cdots$$
we can construct the projective resolution of $\bar{M}$ as follows
$\require{AMScd}$
\begin{CD}
\cdots @>>>P^{-1} @>>>P^{0} @>>> 0  @>>>\cdots\\
@V{f^{-2}}VV @V{f^{-1}}VV @V{f^{0}}VV @V{f^{1}}VV @V{f^{1}}VV\\
\cdots@>>>0@>>>M @>>> 0 @>>> \cdots
\end{CD}
where the arrow $f:\bar{P}\rightarrow \bar{M}$ is obviously a quasi-isomorphism.

In the homotopic category $K(\mathscr{A})$ (where $\mathscr{A}$ is an abelian category such as the category of modules over a ring ) you can generalize this and talk about $K$-projective resolutions, complexes $X$ in $K(\mathscr{A})$ which verify that $Hom(X,Z)=0\ ,\ \forall Z\in\mathscr{Z}=\lbrace Z\in K(\mathscr{A})\ \text{such that}\ H^{n}(Z)=0\ \forall \ n\in\ \mathbb{N} \rbrace $.

The good thing is that if $P$ is a bounded above complex of projectives, then is $K$-projective.


问:
Monoton function proof of discontinuity
Let  $f:[a,b] \to \mathbb {R}$ be nondecreasing.

(a) Recall from Thm. 2.1.9 (c) that  $f(x-)$ and $f(x+)$ exist at every $x\in [a,b]$,
and conclude that $f$ is discontinuous at $x\in S $ iff $f(x-) < f(x+)$.
Im assuming this true because the if the $f(x-)$ < the x+ then we would have a constant function not an increasing funtion. the second part is more obvious since f is nondecreasing then $f(x+) \geq f(x-)$ or it would violate the original statement. how would i try and lay this out with $ e$ and $ \delta$'s?

(b) Give an upper bound for the number of points $x$ at which the `jump' $(f(x+)- f(x-))$ is greater than
a given $r>0$.

one point?

(c) Prove that the set $S = \{ x\in [a,b]: f \text{ is discontinuous at }x \}$ is at most countable.

Think i got this one figured out, suppose that S is uncountable  then S must contain a Limit Point, but f is discontinuous at x so there exists a B(r,x) r>0 s.t x is isolated for all x in S if this wasn't the case then f(x) would be continuous thus S is countable.

答: You have that $S=\bigcup_{n\in \mathbb{N}} \{x\in [a,b]\mid f(x+)-f(x-)<1/n\}$. What can you say about the sets in the union?</p>

ContentLicense: CC BY-SA 3.0

Title: Monoton function proof of discontinuity


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:
I'll recommend Exercises in Analysis, Part I by Gasiński, and Papageorgiou. It's a brilliant source of thoughtful and challenging exercises in mathematical analysis; starting from basics of metrics spaces, then topology, measure theory and functional analysis. Also, it comes with full solutions.

For more advanced topics there's also a second part too by the same authors.


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:
For 'follows from', you should not use the material implication $\to$, but rather the logical implication, for which you can use $\Rightarrow$ or $\vDash$

The difference is that $\to$ is the truth-functional operator you are familiar with, but the $\Rightarrow$ is a meta-logical symbol that claims some logical relationship between two statements.

$P \to Q$ is true in some world as long as it is not true in that world that $P$ is true and $Q$ is false.

$P \Rightarrow Q$ is true is it impossible for $P$ to be true and $Q$ to be false, i.e. that there is no world in which $P$ is true and $Q$ is false.

It is the latter that we typically mean by 'follows from': no matter what the circumstances are (i.e. no matter what world we're dealing with), if $P$ is true, then $Q$ will be true.

So, to say that $Q$ does not follow from $P$, we can write $P \not \Rightarrow Q$ ... but we can not write $\neg (P \Rightarrow Q)$, because then we are mixing up logic with meta-logic. We should simply say that 'it is not the case that $P \Rightarrow Q$.

And indeed, just because it is not the case that $P \Rightarrow Q$, does not mean anything about the truth-value of $P$ or $Q$ when evaluated in some particular world. Notably, it does not mean that $P$ is true and $Q$ is false. Moreover, if in some specific world $P$ is true, $Q$ can still be true or false. So these are all things you were looking for.

In sum, the $\Rightarrow$ (or $\vDash$) is what you should use when thinking about logical implication, rather than the $\to$.


问:
I found $\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$ to be -infinity
Here is what I tried to calculate:

$\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$

When I calculated for 0-(I mean negative zero) I found it to be -infinity but the solution found it to be 1

Can anyone verify it for me please?

My solution:

I found the equation as it is to have an intermediate form(1 - 0/0) so I tried to remove that by simplifying it to

$1 - \sqrt{\frac{1}{x^2} + 1} + \frac{1}{x}$

So I calculated limit to 0-(Negative zero) and I found it to be -infinity

答:
Imho easiest way:
$$\frac{1}{x}\left(\sqrt{1+x^2}-1\right)=\frac{x}{\sqrt{1+x^2}+1}\to0$$

Now about your solution: you have little mistake, it should be
$$\frac{1}{x}\left(\sqrt{1+x^2}-1\right)=\text{sign(x)}\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}$$
so you obtain not $-\infty$ but $\infty-\infty$ which is undefined.


问:
Partial Converse of Holder's Theorem
Holder's Theorem is the following: Let $E\subset \mathbb{R}$ be a measurable set. Suppose $p\ge 1$ and let $q$ be the Holder conjugate of $p$ - that is, $q=\frac{p}{p-1}.$ If $f\in L^p(E)$ and $g\in L^q(E),$ then $$\int_E \vert fg\vert\le \Vert f\Vert _p\cdot\Vert g \Vert_q$$

I am trying to prove the following partial converse is true. Suppose $g$ is integrable, $p>1$ and $$\int_E \vert fg\vert\le M\Vert f\Vert_p$$ when $f\in L^p(E)$ is bounded, for some $M\ge0$. I am trying to prove that this implies that $g\in L^q(E)$, where $q$ is the conjugate of $p$. I am interested in knowing if my proof is correct or if it has any potential.

Attempt: Assuming this is true, I want to show $\int_E \vert g\vert^q<\infty$. Note that $g^{q-1}\in L^p(E)$ since $\int_E \vert g\vert^{pq-p}=\int_E \vert g \vert ^q<\infty$ (I think this is true since $g$ is integrable - I know it holds when $q$ is a natural number). Take $f=g^{q-1}$ and the hypothesis tells us that $$\int_E \vert g \vert^q\le M\Vert g^{q-1}\Vert_P\\<\infty \,\,(\text{since }g^{q-1}\in L^p)$$so $g$ is integrable.

答:
An alternative proof.

Let $T_g(f):=\int_E f g\,,\, \forall f\in L^p.$

It's clear that $T_g$ is linear.

By the condition $\int_E \vert fg\vert\le M\Vert f\Vert_p$, we see $T_g$ is bounded.

So, $T_g\in (L^p)^*.$

By the Riesz Representation Theorem for the Dual of $L^p$(c.f. p. 160, Real Analysis, 4th Edition, Royden), we know there exists an unique $L^q$ function $\tilde{g}$ s.t. $T_g(f)=\int_E f \tilde{g}\,,\, \forall f\in L^p.$  By the uniqueness, we conclude that $g=\tilde{g}\in L^q.$

$\tag*{$\blacksquare$}$

The following is the theorem I quoted in my answer.

$\tag*{$\blacksquare$}$

Title: Partial Converse of Holder's Theorem


答:
Your method is incorrect as it is of the $-\infty+\infty$ form and nothing can be said about it.Also $\frac{\sqrt{1+x^2}}{x}=\sqrt{1+\frac{1}{x^2}}$ need not be true (what happens when $x$ is negative?)

let $x=\tan t$ we have to calculate $$\lim_{t\to 0}1-\frac{\sqrt{1+\tan^2t}-1}{\tan t}=\lim_{t \to 0}(1-\tan (t/2))=1$$

Title: I found $\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$ to be -infinity

Body:
Here is what I tried to calculate:

$\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$

When I calculated for 0-(I mean negative zero) I found it to be -infinity but the solution found it to be 1

Can anyone verify it for me please?

My solution:

I found the equation as it is to have an intermediate form(1 - 0/0) so I tried to remove that by simplifying it to

$1 - \sqrt{\frac{1}{x^2} + 1} + \frac{1}{x}$

So I calculated limit to 0-(Negative zero) and I found it to be -infinity


答:
It sounds like the question is, "Given $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and the Power Rule $\frac{d}{dx}(x^p) = px^{p-1}$ ($p$ any real number), show $\frac{d}{dx}(e^x) = e^x$." As the other answers show, only the first fact is needed to derive the conclusion, but this approach might be more in the spirit of the question (if I've read it correctly).

We know that $x^p = e^{p\ln x}$. Write $\operatorname{exp}(x) = e^x$. Taking the derivative of the left-hand side gives $px^{p-1}$, while taking the derivative of the right-hand side (and using Chain Rule) gives $\operatorname{exp}'(p\ln x) \cdot \frac{p}{x}$. Thus,
$$ px^{p-1} = \operatorname{exp}'(p\ln x) \cdot \frac{p}{x}.$$
Multiplying by $x/p$ on both sides shows that $x^p = \exp'(p\ln x)$, hence
$$\operatorname{exp}(p \ln x) = e^{p \ln x} = x^p = \exp'(p\ln x).$$
From this, we may deduce that $\operatorname{exp} = \operatorname{exp}'$ since the range of the function $x \mapsto p\ln x$ is the whole real line whenever $p \neq 0$ (so just pick any $p \neq 0$).

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?


答:
$A \to B$ does not mean "$B$ follows from $A$". It means "either $A$ is false or $B$ is true in this particular situation". It does not make sense to say such a thing as "$A \to B$ is true" without specifying in which situation/model it is to have that truth value. To say that a $A \to B$ is false in a given situation is indeed logically equivalent to $A \land \neg B$ being true in that situation.

If what you want to say is "$B$ logically follows from $A$", then you're not talking about truth in a specific situation, but preservance of truth  from the premises to the conclusion across any possible situation. In that case, what you really want is logical consequence, $\vDash$:

$A \vDash B$

= "In every situation in which $A$ is true, $B$ is true as well."

This permits for situations in which $B$ is false, as long as in those situations $A$ is false as well. It also permits for situations in which $B$ is true despite $A$ being false. What it rules out is the existence of situations in which $A$ is true but $B$ is false.

The negation of this is "$B$ does not follow from $A$":

$A \not \vDash B$

= "Not in every situation in which $A$ is true, $B$ is true as well."

This does not rule out the existence of situations in which $B$ is true or  $A$ false or both of that, but it does assert the existence of at least one situation in which $A$ is true but $B$ is not. It also permits for, but does not entail, the possibility that $B$ is false not only in some but all situation in which $A$ is true.

If you want to say that $B$ is undecidable from $A$, i.e. that there exist situations in which $A$ is true and $B$ false but there also exist situations in which $A$ is true and $B$ true, you have to make two claims:

$A \not \vDash B$ and $A \not \vDash \neg B$

meaning that from $A$ you can neither definitely conclude $B$ nor $\neg B$.

We do have the following connection between the notions of material conditional and logical consequence (called the deduction theorem):

$$A \vDash B \text{ if and only if } \vDash A \to B$$

i.e. $B$ logically follows from $A$ iff the material conditional is $A \to B$ is tautological.

The crucial point here is that logical consequence commits the material conditional to being true in all situations; when the consequence is denied, $\not \vDash A \to B$, this merely negates its universalness, leading to the conditional being false in some situations -- which should be what you want. This is different from $\vDash \neg (A \to B)$, which would be a stronger claim meaning that $A \to B$ is false and thereby $A \land \neg B$ true in all situations, which, as you figured out, is not the intended state of affairs.

Title: “$B$ does not follow from $A$” seems different from $\lnot(A\to B)$

Body: 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!


问:
Newton's Method for initial approximation
In order to find an approximation to the root of the equation $-x^3-\cos x=0$ by using Newton's method, which of the following initial approximations can be chosen?
$x_0=1 , x_0=0, x_0=2 , x_0=1$ or $x_0=0$ and $x_0=1$ Which one?

答: If you plot the function, you will see that the only solution is $x\approx-0.865474$. Try all of these initial values in your code and you will see, what happens. In one case it converges however only if the damped Newton is used.


问:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$
I would like to prove that for every $x \in \mathbb{R}, n$ natural, we have $$\!\left|\sin(x)+\frac{\sin(2x)}{2}+\ldots+\frac{\sin{(nx)}}{n}\right| \le \int_{0}^{\pi}\frac{\sin x}{x}\,dx$$

We can of course restrict our attention for $x \in [0, 2\pi]$.
So far I tried a few things: letting $S_n(x)$ denote the quantity inside the absolute value, we have$$S_n'(x) = \sum_{k=1}^{n}\cos(kx) = -\frac1{2}+\frac{\sin(n+\frac12)x}{2\sin\frac{x}{2}}$$

and hence $S_n(x)$ has critical points at $x = \frac{2k\pi}{n}$ or $x = \frac{(2k+1)\pi}{n+1}$ where $k$ is an integer. Moreover:$$S_n(x) = -\frac{x}{2}+\int_{0}^{x}\frac{\sin(n+\frac12)t}{2\sin\frac{t}{2}}\,dt$$
I also know that the function $F(x) = \int_{0}^{x}\frac{\sin(t)}{t}\,dt$ has an absolute maximum for $x = \pi$. I struggle to bound the integral in the expression for $S_n$. Can you help me? Or maybe there's a different approach to solve this?

答:
This is the partial sum of the Fourier series for the sawtooth wave where $S(x) = \sum_{n=1}^\infty \frac{\sin nx}{n} = \frac{\pi-x}{2}$ on $(0,2\pi)$.  The waveform has a discontinuity at $x=0$ and $S(x) \to \frac{\pi}{2}$ as $x \to 0+$.

Note that

$$\int_0^\pi \frac{\sin x}{x} \, dx  = 1.85194\ldots > \frac{\pi}{2}$$

However, the partial sums overshoot the waveform (Gibbs phenomena) and it is proved here that

$$\limsup_{n \to \infty}\sup_{x \in [0,\pi]} S_n(x) \geqslant  \int_0^\pi \frac{\sin x}{x} \, dx $$

We can prove that $\displaystyle|S_n(x)| \leqslant \int_0^\pi \frac{\sin x}{x} \, dx$ for all $n \in \mathbb{N}$ and $x \in [0,2\pi]$.

By periodicity and anti-symmetry it is enough to consider the interval $[0,\pi]$. Here the relative maxima of $S_n (x)$ occur at $x_{n,m} = \frac{2m+1}{n+1}\pi$ for $m = 0,1, \ldots, \lfloor\frac{n-1}{2} \rfloor$.

It can be shown that $S_n(x_{n,m})  > S_n(x_{n,m+1})$ so that there is an absolute maximum of $S_n(x)$ on $[0,\pi]$ at $x_{n,0} = \frac{\pi}{n+1}$. It can also be shown that $S_n(x_{n,0})$ is an increasing sequence, such that

$$S_n(x_{n,0}) \nearrow \lim_{n \to \infty}\sum_{k=1}^n \frac{\sin k x_{n,0}}{k} = \lim_{n \to \infty}\frac{\pi}{n+1}\sum_{k=1}^n \frac{\sin \frac{k\pi}{n+1}}{\frac{k\pi}{n+1}} \\ = \int_0^\pi \frac{\sin x}{x} \, dx$$


问:
Show that $H(U)$ the space of holomorphic functions on $U$ is not normable.
I'm trying to show the following result:

Let $U \subset \mathbb{C}$ be an open set. Show that the vector space $H(U)=\{f:U \to \mathbb{C} \; \text{analytic}\}$ is not normable.

I know that $H(U)$ is a Frechet space. If we take $A_{k}:=\{z \in U\ : |z| \leq k \; , \; \text{dist}(\partial U,z) \geq \frac{1}{k} \}$ and $p_{k}(f)=\sup_{z \in A_{k}}|f(z)|$ for each $k \in \mathbb{N}$, we have
$\cup_{k}A_{k}=U$ and metric

$$d(f,g)=\sum_{k=1}^{\infty}\frac{1}{2^{k}} \frac{p_{k}(f-g)}{1+p_{k}(f-g)}.$$

I'm not sure how to work with this. Any help would be greatly appreciated.

答:
It is easy to prove this by using Montel theorem. It states that if $f_n \in H(U)$ is a bounded sequence (i.e. it is bounded by means of seminorms $p_k$) then there is a convergent subsequence $f_{n_m}$. Thus, if $H(U)$ is normable then its unit ball is a bounded subset and, therefore, by Montel theorem it is compact. Now we observe that a normable space with compact unit ball is finite dimensional. Therefore, $H(U)$ is not normable.

There is an abstract notion of Montel space: a barrelled locally convex space in which all closed bounded subsets are compact. $H(U)$ is a Montel space (a simple corollary of Montel theorem). The foregoing argument shows that a normable space is Montel iff it is finite dimensional. By the same reason the space $C^\infty(U)$ is not normable (with the topology of uniform convergence of all derivatives on compact subsets of $U$).


问:
If A is idempotent and $A^5=3A^3-2A$
Stuck at this question for a while.

Question : If A is idempotent and $A^5=3A^3-2A$ , Prove $A=0$

My try :

By $A^2-A=0$ , we know that the minimal polynomial devides $\Rightarrow x(x-1)$

So the options for the minimal polynomial is only : $x , x-1 , x(x-1)$

I tried assuming by contradiction that the minimal polynomial is $x$ and then we get $A=0$ and we are finished.

if $x-1$ is the minimal then $A-I=0 \Rightarrow A=I$ , but we have an eigenvalue that is $0$ so $|A|=0$ and the matrix can't be $I$.

but I'm stuck with $x(x-1)$.

any hints? Thank you

答:
The claim is false. Just take any nonzero idempotent matrix.

Indeed, $A^2=A$ implies $A^n=A$ for all $n\ge 1$. Therefore $A^5-3A^3+2A=A-3A+2A=0$ and so $A^5=3A^3-2A$ gives no new information.

Title: If A is idempotent and $A^5=3A^3-2A$


答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Your stated goal is to avoid summations, but it also aids comprehension (and reduces algebraic errors) to avoid sprawling component-based calculations.

Use a colon to denote the trace/Frobenius product
$$A:B \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; {\rm Tr}(A^TB)$$
This product can also be applied to vectors by treating them as rectangular matrices (set $\,n={\tt1}$) in which case it corresponds to the usual dot product.  The cyclic property of the trace allows the terms in such products to be rearranged in many equivalent ways, e.g.
$$\eqalign{
A:BC &= B^TA:C = AC^T:B \\
B:C &= B^T:C^T = C:B\\
}$$
Note that terms on the RHS and LHS of the colon must have identical dimensions.

For typing convenience, define the vector
$$\eqalign{
 y &= AWx-b \\
}$$
Write the function in terms of this new variable.
Then calculate its differential.
$$\eqalign{
 f &= \tfrac 12y:y \\
df &= y:dy \\
 &= y:(A\,dW\,x + AW\,dx) \\
 &= A^Tyx^T:dW + W^TA^Ty:dx \\
}$$
Holding $x$ constant (i.e. setting $dx=0$) yields
the gradient with respect to $W$
$$\eqalign{
\p{f}{W} &= A^Tyx^T = A^T(AWx-b)x^T \\
}$$
while holding $W$ constant yields the gradient with respect to $x$
$$\eqalign{
\p{f}{x} &= W^TA^Ty = W^TA^T(AWx-b) \\
}$$

Title: Vectorize a long term

Body:
I have the matrices $A \begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{pmatrix}, W = \begin{pmatrix}w_{11} & w_{12} & w_{13} \\ w_{21} & w_{22} & w_{23} \\ w_{31} & w_{32} & w_{33} \end{pmatrix}, x = \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}, b = \begin{pmatrix}b_1 \\ b_2\end{pmatrix}$

I'm interested in the derivatives of the function $f(W,x) = \frac{1}{2}|AWx-b|^2$

Writing it explicitly, I found that

$f(W,x) = \frac{1}{2}(a_{11}w_{11}x_1 + a_{11}w_{12}x_2 + a_{11}w_{13}x_3 + a_{12}w_{21}x_1 + a_{12}w_{22}x_2 + a_{12}w_{23}x_3 +a_{13}w_{31}x_1 + a_{13}w_{32}x_2 + a_{13}w_{33}x_3-b_1)^2 + \frac{1}{2}(a_{21}w_{11}x_1 + a_{21}w_{12}x_2 + a_{21}w_{13}x_3 + a_{22}w_{21}x_1 + a_{22}w_{22}x_2 + a_{22}w_{23}x_3 +a_{23}w_{31}x_1 + a_{23}w_{32}x_2 + a_{23}w_{33}x_3-b_2)^2$

I want to find the derivatives of this function with respect to all variables, and vectorize the result (as in, write it in terms of matrix product rather than summations).

For instance:

$\frac{\partial f}{\partial x_1} = (a_{11}w_{11}x_1 + a_{11}w_{12}x_2 + a_{11}w_{13}x_3 + a_{12}w_{21}x_1 + a_{12}w_{22}x_2 + a_{12}w_{23}x_3 +a_{13}w_{31}x_1 + a_{13}w_{32}x_2 + a_{13}w_{33}x_3-b_1)(a_{11}w_{11}+a_{12}w_{21} + a_{13}w_{31}) + (a_{21}w_{11}x_1 + a_{21}w_{12}x_2 + a_{21}w_{13}x_3 + a_{22}w_{21}x_1 + a_{22}w_{22}x_2 + a_{22}w_{23}x_3 +a_{23}w_{31}x_1 + a_{23}w_{32}x_2 + a_{23}w_{33}x_3-b_2)(a_{21}w_{11}+a_{22}w_{21}+a_{23}w_{31})$

This is clearly a scalar product of $\begin{pmatrix}a_{11}w_{11}x_1 + a_{11}w_{12}x_2 + a_{11}w_{13}x_3 + a_{12}w_{21}x_1 + a_{12}w_{22}x_2 + a_{12}w_{23}x_3 +a_{13}w_{31}x_1 + a_{13}w_{32}x_2 + a_{13}w_{33}x_3-b_1 \\ a_{21}w_{11}x_1 + a_{21}w_{12}x_2 + a_{21}w_{13}x_3 + a_{22}w_{21}x_1 + a_{22}w_{22}x_2 + a_{22}w_{23}x_3 +a_{23}w_{31}x_1 + a_{23}w_{32}x_2 + a_{23}w_{33}x_3-b_2\end{pmatrix}$

With the vector $\begin{pmatrix}a_{11}w_{11}+a_{12}w_{21} + a_{13}w_{31} \\ a_{21}w_{11}+a_{22}w_{21}+a_{23}w_{31}\end{pmatrix}$

the first vector, the long one, is $AWx-b$, so overall this derivative is

$\begin{pmatrix}a_{11}w_{11}+a_{12}w_{21} + a_{13}w_{31} & a_{21}w_{11}+a_{22}w_{21}+a_{23}w_{31}\end{pmatrix}(AWx-b)$.

This is just for $x_1$. If we do the same for $x_2$ and $x_3$, we should reach that

$\nabla _xf(W,x) = \begin{pmatrix}a_{11}w_{11}+a_{12}w_{21} + a_{13}w_{31} & a_{21}w_{11}+a_{22}w_{21}+a_{23}w_{31} \\ a_{11}w_{12}+a_{12}w_{22} + a_{13}w_{32} & a_{21}w_{12}+a_{22}w_{22}+a_{23}w_{32} \\ a_{11}w_{13}+a_{12}w_{23} + a_{13}w_{33} & a_{21}w_{13}+a_{22}w_{23}+a_{23}w_{33}\end{pmatrix}(AWx-b)$

My question is how to vectorize the first term, and how to repeat this process for $\nabla _W f(W,x)$

Edit:

$\begin{pmatrix}a_{11}w_{11}+a_{12}w_{21} + a_{13}w_{31} & a_{21}w_{11}+a_{22}w_{21}+a_{23}w_{31} \\ a_{11}w_{12}+a_{12}w_{22} + a_{13}w_{32} & a_{21}w_{12}+a_{22}w_{22}+a_{23}w_{32} \\ a_{11}w_{13}+a_{12}w_{23} + a_{13}w_{33} & a_{21}w_{13}+a_{22}w_{23}+a_{23}w_{33}\end{pmatrix} = W^TA^T$.

Which means $\nabla _x f(W, x) = W^TA^T(AWx-b)$.

Is this result correct? Is it correct for larger matrices as well?

Just need to repeat this process for $\nabla _W f(W,x)$.

Edit 2:

In general, $\frac{\partial f}{\partial w_{ij}} = \begin{pmatrix}a_{1i}x_j & a_{2i}x_j\end{pmatrix}(AWx-b)$

So

$\nabla W f(W,x) = \begin{pmatrix}\begin{pmatrix}a_{11}x_1 & a_{21}x_1\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{11}x_2 & a_{21}x_2\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{11}x_3 & a_{21}x_3\end{pmatrix}(AWx-b) \\ \begin{pmatrix}a_{12}x_1 & a_{22}x_1\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{12}x_2 & a_{22}x_2\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{12}x_3 & a_{22}x_3\end{pmatrix}(AWx-b) \\ \begin{pmatrix}a_{13}x_1 & a_{23}x_1\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{13}x_2 & a_{23}x_2\end{pmatrix}(AWx-b) & \begin{pmatrix}a_{13}x_3 & a_{23}x_3\end{pmatrix}(AWx-b)\end{pmatrix}$

How do I represent this as a product of $A, W, x, b$?


问:
finding a value for $\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$
I have been trying to find the sum:
$$S=\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$$
And I can't find if anyone else has posted about this. Interestingly wolfram alpha claims it converges for $n=0,1$ as well which cannot be correct. I think maybe there is some way of solving this by changing the order of summation:
$$S=\sum_{n=2}^\infty\frac 1{n!}\sum_{k=1}^\infty\frac{1}{k^n}$$
but I am struggling to justify this move, also trying to find:
$$\sum_{n=2}^\infty\frac{1}{k^n}$$
seems to be a geometric series so can I just do it like so, or will the bounds of the summation have changed? Thanks

From the advice of answers I have:
$$\sum_{n=2}^\infty\frac{(1/k)^n}{n!}=\sum_{n=0}^\infty\frac{(1/k)^n}{n!}-\frac1k-1=e^{1/k}-\frac1k-1$$
so we now have:
$$S=\sum_{k=1}^\infty e^{1/k}-\left(1+\frac1k\right)$$

答:
It's a positive-term series, so rearrangements are fine (Baby Rudin, Ch.8, #3).

However, you won't get the $\frac1{n!}$ out of a series in $n$, so you'll have $\sum_{n=2}^\infty \frac{(1/k)^n}{n!}$. Not a geometric series, but still one that you should recognize ...


问:
Intuition for the invariance lemma
The invariance lemma, as stated below, is commonly used to prove Lie’s theorem.

Lemma (Invariance lemma). Let $\mathbb{k}$ be a field of characteristic zero.
Let $\mathfrak{g}$ be a $\mathbb{k}$-Lie algebra, let $M$ be a finite-dimensional representation of $\mathfrak{g}$ and let $I$ be an ideal of $\mathfrak{g}$.
Let $\lambda$ be an element of the dual space $I^*$ and let
$$
  M_\lambda
  :=
  \{
    m \in M
    \mid
    \text{$x m = \lambda(x) m$ for all $x \in I$}
  \}
$$
be the corresponding $I$-weight space of $M$.
This weight space is already a $\mathfrak{g}$-subrepresentation of $M$.

Every time I’m reading through my old notes on representation theory of Lie algebras I’m wondering where this lemma comes from.
I have no intuition for why it should be true, and if I hadn’t seen a proof of it I would probably even suspect it to be false.

What is the intuition behind the invariance lemma?

答:
This is a slight extension of the idea that for commuting endomorphisms $S,T$ of a vector space $V$, the $\lambda$-eigenspace $V_\lambda$ of $T$ is stabilized by $S$. Maybe good to recall the one-line proof: for $v\in V_\lambda$,
$$
T(S v) \;=\; (TS)v \;=\; (ST)v \;=\; S(Tv)\;=\; S(\lambda v) \;=\; \lambda Sv
$$
so $Sv$ is in $V_\lambda$.

In your case, $\lambda:I\to k$, and $\mathfrak g$ does not quite commute with $I$, but stabilizes it. But the proof is essentially the same.

Title: Intuition for the invariance lemma


问:
find a function $f$ such that $\| f \|_p$ is finite if $p\in(1,2)$ and infinite for $p=1$ or $p\in[2,\infty]$
I'm stuck in this problem and really I would appreciate a hint, I'm looking a Borel measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $\|f \|_p=\infty$ if $p=1$ or $p\in[2,\infty]$ and $\|f \|_p<\infty$ if $p\in(1,2)$

Thank you in advance.

Body:
Using the fact that $\|x^{-r} \chi_{(0,1)}(x)\|_1 = \infty$ if $r > 1$ and $\|x^{-r} \chi(1, \infty)(x)\|_1 = \infty$ if $r \in (0,1)$, you can use
$$f(x) = x^{-1/2} \chi_{(0,1)}(x) + x^{-1} \chi_{(1, \infty)}(x)$$


问:
Help showing that a function is surjective
Consider $G_n=\{ \overline{x}\in\mathbb{Z}/n\mathbb{Z}; \gcd(x,n)=1 \}$, and $q$ an odd prime number.

For all $x\in \mathbb{Z}$, $\overline{x}$ represents, in this question, the class of $x$ modulo $q^\alpha$, and let's use the notation $\dot{x}$ for the class of $x$ modulo $q$ and consider the morphism of groups
\begin{align*}
        \Psi : G_{q^\alpha} &\rightarrow G_q \\
        \overline{x}&\mapsto \dot{x}
    \end{align*}
Prove that $|\operatorname{Ker}\Psi|=q^{\alpha-1}$ (with $|\operatorname{Ker}\Psi|$ the notation for order). This is a step on a bigger exercise to show that $G_{q^\alpha}$ is cyclic.

Someone gave me a hint to show that $\Psi$ is surjective, but how do I show that?

答:
If for some $x \in \Bbb Z$, we have $\gcd(x,q)=1$, then $q$ does not divide $x$, so $x$ and $q^\alpha$ don't have any prime factor in common, thus $\gcd(x,q^\alpha)=1$, this implies that $\Psi$ is surjective.

Since you also tagged this with Galois theory, one can note that using the canonical isomorphisms $\mathrm{Gal}(\Bbb Q(\zeta_{q^\alpha})/\Bbb Q) \cong G_{q^{\alpha}}$ and $\mathrm{Gal}(\Bbb Q(\zeta_q)/\Bbb Q) \cong G_q$, one can identify $\Psi$ with the restriction map $\mathrm{Gal}(\Bbb Q(\zeta_{q^\alpha})/\Bbb Q) \to \mathrm{Gal}(\Bbb Q(\zeta_q)/\Bbb Q)$ which is surjective by general theory.

You don't really need the surjectivity of $\Psi$ to show that $|\ker \Psi|=q^{\alpha-1}$, though. You can just show that $1+qa$, where $a=0,1, \dots, q^{\alpha-1}-1$ is a system of representatives for $\ker \Psi$.


问:
Zero to the zero power – is $0^0=1$?
Could someone provide me with a good explanation of why $0^0=1$?

My train of thought:

$x>0$

$0^x=0^{x-0}=0^x/0^0$, so

$0^0=0^x/0^x=\,?$

Possible answers:

$0^0\cdot0^x=1\cdot0^0$, so $0^0=1$

$0^0=0^x/0^x=0/0$, which is undefined

PS. I've read the explanation on mathforum.org, but it isn't clear to me.

答:
It's pretty straight forward to show that multiplying something by $x$ zero times leaves the number unchanged, regardless of the value of $x$, and thus $x^0$ is the identity element for all $x$, and thus equal to one.

For the same reason, the sum of any empty list is zero, and the product is one.  This is when a product or sum of an empty list is applied to a number, it leaves it unchanged.  Thus if the product $\Pi()$ = 1, then we immediately see why $0! = 0^0 = 1$.

Without this property, one could prove that $2=3$, by the ruse that there are zero zeros in the product on the left (zero is after all, a legitimate count), and thus $2*0^0$, and since $0^0$ as indeterminate, could be 1.5, and thus $2=3$.  I think not.

The approaches to $0^0$ by looking at $x^y$ from different directions, fails to realise that for even lines close to $x=0$, the line sharply sweeps up to 1 as it approaches $y=0$, and that the case for $x=0$, it may just be a case of not seeing it sweep up.  On the other hand, looking from the other side, even in a diagonal line (ie $(ax)^x$), all do rapidly rise to 1, as x approaches 0.  It's only when one approaches it from $0^x$ that you can't see it rising.  So the evidence from the graph of $x^y$ is that $0^0$ is definitely 1, except when approached from $y=0$, when it appears to be zero.

Title: Zero to the zero power – is $0^0=1$?


答:
Using the standard limit properties:
$$\begin{align}
\lim_{x\to 0} (1 - \frac{1}{x}(\sqrt{x^2+1}-1)) &= \lim_{x\to 0}1 - \lim_{x\to 0} \frac{\sqrt{x^2+1} - 1}{x} \\
&= 1 - \lim_{x\to 0}\left( \frac{\sqrt{x^2+1} - 1}{x}\cdot \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}\right) \\
&= 1 - \lim_{x\to 0}\frac{x^2+1-1}{x(\sqrt{x^2+1}+1)} \\ &= 1 - \lim_{x\to 0}\frac{x}{\sqrt{x^2+1}+1} \\ &= 1 - \frac{\lim_{x\to 0}x}{\lim_{x\to 0}(\sqrt{x^2+1}+1)} \\ &= 1 - \frac{0}{\sqrt{0^2+1}+1} = 1 - 0 = 1 \end{align}$$

$\underline{\text{EDIT}}$: In response to a question in the comments, here's how to find the limit of the same function at infinity: $$\begin{align}
\lim_{x\to \infty}\left( 1-\frac{1}{x}(\sqrt{x^2+1}-1)\right) & = \lim_{x\to\infty}1 - \lim_{x\to\infty}\left( \frac{\sqrt{x^2+1}}{x}-\frac{1}{x}\right) \\ &= 1 - \left( \lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x} - \lim_{x\to\infty}\frac{1}{x} \right) \\ &= 1 - \left( \lim_{x\to\infty} \frac{\sqrt{x^2+1}}{x}-0 \right) \\ &= 1 - \lim_{x\to\infty} \frac{\sqrt{x^2\left( 1+\frac{1}{x^2}\right) }}{x} \\ &= 1 - \lim_{x\to\infty} \frac{x\sqrt{1+\frac{1}{x^2}}}{x} \\ &= 1 - \lim_{x\to\infty} \sqrt{1+\frac{1}{x^2}} = 1-1=0
\end{align}$$


问:
Multivariable Calculus Changing the order of Integration
enter image description here

$\int_0^2 \int_0^{4-x^2} \int_0^{4-y} \, dz \, dy \, dx$

Change the limits of integration to match the specified order below.

$\int_a^b \int_c^d \int_e^f \, dx \, dy \, dz$

I did the work and got an answer of

$a=0, b=4, c=0, d = 4-z, e= -\sqrt{4-y}, f = \sqrt{4-y}$.

However my $e$ is incorrect, wonder what I did wrong / what's the right answer.

答:
As you can see in the first integral that $0 \leq x \leq 2$ so lower bound of $x$ in your second integral is incorrect.

Your integral should be $\displaystyle \int_0^4 \int_0^{4-z} \int_0^{\sqrt{4-y}} \, dx \, dy \, dz$.


问:
How to find the SVD of a matrix with one zero eigenvalue?
I am trying to find the SVD of the matrix \begin{pmatrix}1&2\\3&6\end{pmatrix}

$A^TA=\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}10&20\\20&40\end{pmatrix}$

$\lambda_1=50$

$\lambda_2=0$

$\lambda_1>\lambda_2>=0$

$\Sigma=\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}$

When $\lambda_1=50$:

$v_1=\begin{pmatrix}1\\2\end{pmatrix}$

Normalize:

$u_1=\begin{pmatrix}\frac{1}{\sqrt{5}}\\\frac{2}{\sqrt{5}}\end{pmatrix}$

When $\lambda_1=0$:

$v_2=\begin{pmatrix}-2\\1\end{pmatrix}$

Normalize:

$u_2=\begin{pmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{pmatrix}$

$V=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{-2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$V^T=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$A=U\Sigma V^T$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}u_1&u_2\\u_3&u_4\end{pmatrix}\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}5\sqrt{2}u_1&0\\5\sqrt{2}u_3&0\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$

$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}\sqrt{10}u_1&2\sqrt{10}u_1\\\sqrt{10}u_3&2\sqrt{10}u_3\end{pmatrix}$

$u_1=\frac{1}{\sqrt{10}}$

$u_3=\frac{3}{\sqrt{10}}$

How can I find $u_2$ and $u_4$?

答:
Note that the given matrix is rank-$1$ and that in $\Bbb R^2$ it is easy to find orthogonal vectors. Hence,

$$\begin{aligned} {\rm A} := \begin{bmatrix} 1 & 2 \\ 3 & 6\end{bmatrix} = \begin{bmatrix} 1 \\ 3\end{bmatrix} \begin{bmatrix} 1 \\ 2\end{bmatrix}^\top &= \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix}^\top \\ &= \left( \frac{1}{\sqrt{10}} \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \right) \begin{bmatrix} \sqrt{50} & 0\\ 0 & 0\end{bmatrix} \left( \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix} \right)^\top = {\rm U} \Sigma {\rm V}^\top\end{aligned}$$

Lastly, do note that the nonzero singular value is $\sqrt{\mbox{tr} \left( {\rm A}^\top {\rm A} \right) } = \| {\rm A} \|_{\text{F}} = \sqrt{50} = \| {\rm A} \|_2$.

matrices svd singular-values


问:
Definition of $\omega$-inconsistency
I am currently studying Gödel's Incompleteness Theorems, and I have a question about $\omega$-inconsistency.

Some say that a theory $T$ is $ω$-inconsistent if $T$ proves $P(0)$, $P(1)$, $P(2)$,... and also proves that there is a number $N$ for which $P(N)$ fails.

Others instead say that $T$ is $\omega$-inconsistent if it proves that there is a number for which $P(N)$ holds, but proves that $P(0)$ fails, $P(1)$ fails,...

Which one is wrong? Or (as I think) are they equivalent?

答:
Turning Rob Arthan's initial comment into an answer to kick this off the "unanswered" queue, and CW-ifying to avoid reputation gain for someone else's work:

The two definitions are equivalent: given one, we get the other by swapping $P$ and $\neg P$. (This reveals an omitted clause in the definitions: we need to begin with "for some $P$" There's no single $P$ which "tests for $\omega$-inconsistency.")

Title: Definition of $\omega$-inconsistency


问:
Is $\int_{0}^{x}{|\sin t|dt}$ asymptotically equivalent to $2x$?
Or to what? For $x\to+\infty$.

In the previous question, I wrote

$lim_{x->+\infty}{\int_{0}^{x}|\sin t|dt=\sum_{1}^{+\infty}{2}=lim_{n->+\infty}{2n}=+\infty}$.

Is this correct?

答:
As I said in the comment, if you call $\begin{cases}n=\lfloor \frac x{\pi}\rfloor & n\in\mathbb Z\\r=x-n\pi & r\in[0,\pi)\end{cases}$

$$I(x)=\int_0^x|\sin(t)|dt=\sum\limits_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}|\sin(t)|dt+\int_{n\pi}^x|\sin(t)|dt$$

The integral on $[k\pi,(k+1)\pi]$ is constant of value $2$.

The integral on $[n\pi,x]$ is $\displaystyle \int_{n\pi}^{n\pi+r}|\sin(t)|dt=\int_0^r|\sin(t)|dt=1-\cos(r)$

Therefore $$I(x)=2n+1-\cos(r)$$

Now you can reintroduce $x$ to get $I(x)=2\dfrac{x-r}{\pi}+1-\cos(r)=\dfrac{2x}{\pi}+\underbrace{\bigg(1-\dfrac{2r}{\pi}-\cos(r)\bigg)}_\text{bounded}$

The last part is bounded because $r\in[0,\pi)$ and $|\cos(r)|\le 1$, thus $I(x)\sim \dfrac{2x}{\pi}$

Note:

it is not mandatory to get the exact expression for $I(x)$, you can say that $\int_0^r|\sin(t)|dt\le\int_0^\pi|\sin(t)|dt\le 2$ for the same conclusion (bounded)


问:
Vector space of $M_{n\times n}(\mathbb{R})$.
Let $E=\{A \in M_{n\times n}(\mathbb{R}): A^T = A\}$ and $F = \{A \in M_{n\times n}(\mathbb{R}): A^T = -A\}$. How can you prove that $E$ and $F$ are vector subspaces of $M_{n\times n}(\mathbb{R})$?. Besides, what would its dimension even be, for $E$, $F$ and $M_{n\times n}(\mathbb{R})$? Maybe $M_{n\times n}(\mathbb{R})$ is direct sum of $E$ and $F$. I'm really stuck with this questions. How can matrices form a vector space? And, provided that, how can you actually compute the dimension of all those three?

答:
You seem to be thinking of the elementary definition of vector space as being a Euclidean space $\mathbb{R}^n$, or some subspace of it. There is a more general definition you should read about here: https://en.wikipedia.org/wiki/Vector_space

Since you are given that $M_{n \times n}(\mathbb{R}$ is a vector space, all you have to do to show that $E$ or $F$ is a subspace is to verify the closure axioms: closed under addition and closed under scaling.

Closed under addition: Suppose $B$ and $C$ are elements of $E$. Now can you show that $B+C$ is an element of $E$? Something is in $E$ if and only if it equals its own transpose. To see if $B+C$ is in $E$, you should take its transpose and see if you get the original thing, i.e., verify the equation $(B+C)^T=B+C$. In order to do this, you will need the fact that $B$ and $C$ were taken from $E$, i.e., that $B^T=B$ and $C^T=C$.

Closed under scaling: Suppose $B$ is an elements of $E$ and that $\lambda$ is a scalar. Now can you show that $\lambda B$ is an element of $E$? Play the same game ...


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:
There is a condition you must check to verify that the $g(x)$ you've chosen converge:

$$|g'(u)| \leq \lambda <1$$

As for the suggested $g(x)$, apparently both of them diverge.


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:
Your text had a HUGE typo.

It should be obvious that $\frac {da^x}{dx} \ne xa^{x-1}$ (spoiler alert: $\frac {da^x}{dx} = \ln a\cdot a^x$).

They must mean $\frac {dx^a}{dx} = ax^{a-1}$ that fits in with the clue that $x^p = e^{p\ln x}$.

Now just do it.

$$\frac {dx^p}{dx} = px^{p-1}$$

But by the chain rule,

$$\frac {dx^p}{dx}=\frac {de^{p\ln x}}{dx}=\frac {de^{p\ln x}}{d(p\ln x)}\frac {d(p\ln x)}{dx}=\frac {de^{p\ln x}}{d(p\ln x)}\cdot \frac px$$

$$\frac {de^{p\ln x}}{dp\ln x}p\cdot \frac 1x=  px^{p-1}$$

$$\frac {de^{p\ln x}}{d(p\ln x})p\cdot \frac 1x\cdot \frac xp = px^{p-1}\cdot \frac xp$$

$$\frac {de^{p\ln x}}{d(p\ln x)}= x^p$$

But $x^p = e^{p\ln x}$  so

$$\frac {de^{p\ln x}}{d(p\ln x)}= e^{p\ln x}$$
and if we just swap out variables and let $u = p\ln x$ then

$$\frac {de^u}{du} = e^u$$
which proves the claim.


问:
understanding the solution to a particular differential equation?
so I ran across this problem:

$(3x^2+1)y' -2xy = 6x$

I solved it by integrating factor $ e^{\int\frac{-2x}{3x^2+1} dx}$

.

.

.

$\int D[(3x^2+1)^{\frac{-1}{3}}y] = \int 6x(3x^2+1)^{\frac{-4}{3}} dx$

.

.

.

$(3x^2+1)^{\frac{-1}{3}}y = -3(3x^2+1)^{\frac{-1}{3}} + C$

$y = -3 + C(3x^2+1)^{\frac{1}{3}}$

my question is interpreting the solution $y=-3 + C(3x^2+1)^{\frac{1}{3}}$. So I understand why the integrating factor method works in solving the problem. Why did the answer come out so simple after doing so much work? Could I have solved it 'faster?'

答:
You could transform the equation simply as
$$
(3x^2+1)y'=2x(y+3)
\\~\\
y'=\frac{2x}{3x^2+1}(y+3)
$$
This is a separable equation and the case of the constant solution $y=-3$ appears directly as the one to exclude for performing the separation of variables. Then continue
$$
\int\frac{dy}{y+3}=\int\frac{2x\,dx}{3x^2+1}.
$$


问:
Finding double integrals using polar coordinates.
So Im new to using polar coordinates to work out double integrals and was wondering if anyone could help me.

So I am looking at the question:

$I_{6}=\int\int_{Q}x/\pi(x^{2}+y^{2}) dxdy$ where $Q=\{(x,y)|(x-2)^{2}+y^{2}\leq 4, y\geq0\}$

Looking at the first equation:

$(x-2)^{2}+y^{2} \leq4$   which gives
$x^{2} + y^{2} = 4x$

and in polar form:

$r^{2}cos^{2}\theta +r^{2}sin^{2}\theta =4rcos\theta$

which simplifies to :

$r^{2}=4rcos\theta$ which equals $r=4cos\theta$

So my first integral limits will be between $0$ and $4cos\theta$

So the first part of the equation looks like:

$\int_{0}^{4cos\theta} (rcos\theta)/\pi(r^2cos^2\theta+r^2sin^2\theta) rdr$

Which simplifies to:

$\int_{0}^{4cos\theta}(cos\theta/\pi) dr$

which gives me: $\large[(cos\theta/\pi)r\large]_{0}^{4cos\theta}$

and gives:

$4cos^2\theta/\pi$

but im unsure on what my next set of limits will be. I know from a previous question that ive asked that I need to draw a diagram with the centre of the circle at (2,0) but I dont undestand how to utilise this to get my second limit.

Any help would be greately appreciated.

答:
I rather would use polar coordinates adapted to the given disk centered in $(2,0)$, so we have
$$
\begin{aligned}
x &= 2+r\cos t\ ,\\
y &= 0+r\sin t\ ,\\
dx &= \cos t\; dr - r\sin t\; dt\ ,\\
dy &= \sin t\; dr + r\cos t\; dt\ ,\\
dx\wedge dy &= (\cos t\; dr - r\sin t\; dt)\wedge(\sin t\; dr + r\cos t\; dt)\\
&=r\cos^2 t\; dr\wedge dt - r\sin^2 t\; dt\wedge dr\\
&=r\cos^2 t\; dr\wedge dt + r\sin^2 t\; dr\wedge dt\\
&=r\; dr\wedge dt\ ,\qquad\text{ so the Jacobian is $|r|=r$}\\
dx\; dy &=r\; dr\; dt\\[3mm]
%
I_6 &=
\iint_D\frac x{\pi(x^2+y^2)}\; dx\; dy\\
&=
\frac 1\pi
\iint_{\substack{r\in[0,2]\\t\in[0,\pi]}}
\frac {2+r\cos t}{(2+r\cos t)^2+(r\sin t)^2}\; r\; dr\; dt\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\int_0^\pi
\frac {2+r\cos t}{r^2+ 4r\cos t + 4}\;  dt\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac 14 t
+
\frac 12\arctan\frac{(2-r)\sin t}{(2+r)(1+\cos t)}
\right]_0^\pi
\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac 14 t
+
\frac 12\arctan\left(\frac{2-r}{2+r}\tan \frac t2\right)\right]_0^\pi
\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac \pi4
+
\frac \pi4
\right]
\\
&=1\ .
\end{aligned}
$$

CAS check. Using polar coordinates as above. The computer algebra system (CAS) used is sage:

```<code>sage: var('r,t');
sage: INT = integral
sage: X, Y = 2 + r*cos(t), r*sin(t)
sage: I6 = 1/pi * INT( INT( X/(X^2+Y^2) * r, r, 0, 2), t, 0, pi)
sage: I6
1
</code>```

CAS check. Using Fubini. sage again:

```<code>sage: var('x,y');
sage: integral( integral( x/(x^2+y^2), y, 0, sqrt(x*(4-x)) ), x, 0, 4) / pi
1
</code>```

Human check. (Using Fubini again.)
$$
\begin{aligned}
I_6
&=
\frac 1\pi
\int_0^4 dx
\int_0^{\sqrt{4-(x-2)^2}}
\frac {x\; dy}{x^2+y^2}
\\
&=
\frac 1\pi
\int_0^4 dx
\left[
\arctan\frac yx
\right]_0^{\sqrt{4-(x-2)^2}}
\\
&=
\frac 1\pi
\int_0^4
\arctan\frac {\sqrt{4-(x-2)^2}}x
\; dx
\\
&\qquad\text{(Substitution: $x=2+2\cos u$, $u\in[0,\pi]$)}
\\
&=
\frac 1\pi
\int_0^\pi
\arctan\frac {2\sin u}{2(1+\cos u)}\; 2\sin u\; du
\\
&=
\frac 1\pi
\int_0^\pi
\arctan\frac {4\sin \frac u2\cos \frac u2}{2\cos^2 \frac u2)}\; 2\sin u\; du
\\
&=
\frac 1\pi
\int_0^\pi
\frac u2\; 2\sin u\; du
\\
&= 1\ .
\end{aligned}
$$

Body:
I rather would use polar coordinates adapted to the given disk centered in $(2,0)$, so we have
$$
\begin{aligned}
x &= 2+r\cos t\ ,\\
y &= 0+r\sin t\ ,\\
dx &= \cos t\; dr - r\sin t\; dt\ ,\\
dy &= \sin t\; dr + r\cos t\; dt\ ,\\
dx\wedge dy &= (\cos t\; dr - r\sin t\; dt)\wedge(\sin t\; dr + r\cos t\; dt)\\
&=r\cos^2 t\; dr\wedge dt - r\sin^2 t\; dt\wedge dr\\
&=r\cos^2 t\; dr\wedge dt + r\sin^2 t\; dr\wedge dt\\
&=r\; dr\wedge dt\ ,\qquad\text{ so the Jacobian is $|r|=r$}\\
dx\; dy &=r\; dr\; dt\\[3mm]
%
I_6 &=
\iint_D\frac x{\pi(x^2+y^2)}\; dx\; dy\\
&=
\frac 1\pi
\iint_{\substack{r\in[0,2]\\t\in[0,\pi]}}
\frac {2+r\cos t}{(2+r\cos t)^2+(r\sin t)^2}\; r\; dr\; dt\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\int_0^\pi
\frac {2+r\cos t}{r^2+ 4r\cos t + 4}\;  dt\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac 14 t
+
\frac 12\arctan\frac{(2-r)\sin t}{(2+r)(1+\cos t)}
\right]_0^\pi
\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac 14 t
+
\frac 12\arctan\left(\frac{2-r}{2+r}\tan \frac t2\right)\right]_0^\pi
\\
&=
\frac 1\pi
\int_0^2
r\; dr\;
\left[
\frac \pi4
+
\frac \pi4
\right]
\\
&=1\ .
\end{aligned}
$$

CAS check. Using polar coordinates as above. The computer algebra system (CAS) used is sage:

sage: var('r,t');
sage: INT = integral
sage: X, Y = 2 + r*cos(t), r*sin(t)
sage: I6 = 1/pi * INT( INT( X/(X^2+Y^2) * r, r, 0, 2), t, 0, pi)
sage: I6
1

CAS check. Using Fubini. sage again:

sage: var('x,y');
sage: integral( integral( x/(x^2+y^2), y, 0, sqrt(x*(4-x)) ), x, 0, 4) / pi
1

Human check. (Using Fubini again.)
$$
\begin{aligned}
I_6
&=
\frac 1\pi
\int_0^4 dx
\int_0^{\sqrt{4-(x-2)^2}}
\frac {x\; dy}{x^2+y^2}
\\
&=
\frac 1\pi
\int_0^4 dx
\left[
\arctan\frac yx
\right]_0^{\sqrt{4-(x-2)^2}}
\\
&=
\frac 1\pi
\int_0^4
\arctan\frac {\sqrt{4-(x-2)^2}}x
\; dx
\\
&\qquad\text{(Substitution: $x=2+2\cos u$, $u\in[0,\pi]$)}
\\
&=
\frac 1\pi
\int_0^\pi
\arctan\frac {2\sin u}{2(1+\cos u)}\; 2\sin u\; du
\\
&=
\frac 1\pi
\int_0^\pi
\arctan\frac {4\sin \frac u2\cos \frac u2}{2\cos^2 \frac u2)}\; 2\sin u\; du
\\
&=
\frac 1\pi
\int_0^\pi
\frac u2\; 2\sin u\; du
\\
&= 1\ .
\end{aligned}
$$


问:
Ways to construct set models of $\mathsf{ZFC}$ from existing ones
So I'm trying to solve this exercise, but I am having trouble solving it without any further assumptions. This is the exercise:

Assume $M$ is a transitive model of $\mathsf{ZFC}$ such that if $\alpha = M\cap \mbox{Ord}$, then $|\alpha| = \aleph_1$. Show that there exists a transitive model $M'$ of $\mathsf{ZFC}$ such that $M'\cap \mbox{Ord} = \alpha$ and $M'\neq M$.

Now if $M\neq L_\alpha$, we can take $M' = L_\alpha$. Otherwise if $M = L_\alpha$ and $(\omega_1)^M < \omega_1$, using the Rasiowa-Sikorski lemma we can force over $M$ with the Cohen forcing and take $M' = M[G]$. But if $(\omega_1)^M = \omega_1$, I don't see a clear way of using forcing since we don't have $\mathsf{MA} + 2^{\aleph_0} > \omega_1$ or any other helpful hypothesis. I'm stuck here.

One other semi-idea I had, was to take some $X\in V_\alpha - M$ and consider some sort of constructible closure of $M \cup \{X\}$, but that didn't work out either.

What are some other ways of constructing models of set theory without much assumptions? (Aside from Lowenheim-Skolem theroem which may give ill-founded models.)

答:
This was a question I asked many years ago on MathOverflow. Namely, can there be only one transitive model of an uncountable height.

Françios Dorais gave a fantastic answer:

As you point out, we may assume $M=L_\alpha$.

First option, $\alpha=\omega_1$, in that case every $x\in M$ is countable in $V$, so the Cohen forcing over $M$ has only countably many dense open sets. Therefore we can find a generic over it in the usual sense.

Well, the first option failed. So $\omega_1\in M$. In that case, $\operatorname{Add}(\omega_1,1)^L\in M$, and it is $\sigma$-closed, and since $|M|=|L_\alpha|=|\alpha|=|\aleph_1|$, there are only $\aleph_1$ dense open sets in $M$ itself. So we can diagonalise over them to produce a generic.

This is the same usual Rasiowa–Sikorski lemma, where at limit steps you use the $\sigma$-closure.


问:
Is there a way to solve $\arcsin\left(\frac{d}{r}\cdot \sin(x)\right)=\phi-x+\frac{\theta}{2}$ for $x$?
While solving a geometry problem, I derived the following equation:

$$\arcsin\left(\frac{d}{r}\cdot \sin(x)\right)=\phi-x+\frac{\theta}{2}$$

And, in order to solve such problem, I need to solve for $x$, but I see no way to do that.

答:
$$\arcsin\left(\frac{d}{r}\cdot \sin(x)\right)=\phi-x+\frac{\theta}{2}$$

$$\frac{d}{r}\cdot \sin(x)=\sin \left(\phi-x+\frac{\theta}{2} \right)$$

$$\frac{d}{r}\cdot \sin(x)=\sin \left(\phi+\frac{\theta}{2} \right)\cos \left(x\right)-\cos \left(\phi+\frac{\theta}{2} \right)\sin \left(x \right)$$

$$\sin(x) \left(\frac{d}{r}+ \cos \left(\phi+\frac{\theta}{2} \right)\right)=\sin \left(\phi+\frac{\theta}{2} \right)\cos \left(x\right)$$

$$\tan(x)=\frac{\sin \left(\phi+\frac{\theta}{2} \right)}{\frac{d}{r}+ \cos \left(\phi+\frac{\theta}{2} \right)} $$


问:
Proof regarding unitary self-adjoin linear operators
I'm suck on how to do the following Linear Algebra proof:

Let $T$ be a self-adjoint operator on a finite-dimensional inner product space $V$.

Prove that for all $x \in V$, $||T(x)\pm ix||^2=||T(x)||^2+||x||^2.$

Deduce that $T-iI$ is invertible and that $[(T-iI)^{-1}]^{*}=(T+iI)^{-1}.$

Furthermore, show that $(T+iI)(T-iI)^{-1}$ is unitary.

My attempt at a solution (to the first part):

$||T(x)\pm ix||^2=\left< T(x)\pm ix, T(x)\pm ix\right>$

$=\left< T(x), T(x) \pm ix\right>\pm \left<ix,  T(x)\pm ix \right>$

...

$=\left<T(x), T(x) \right>+ \left<x,x \right>$

$=||T(x)||^2+||x||^2$

The ... is the part I'm stuck on (I know, it's the bulk of the first part).
I have yet to consider the next parts since I'm still stuck on this one.

Any help would be appreciated! Thanks.

答:
I think this is should be a correct sketch of the proof...I will leave the details to you

1). $||T(x) \pm ix||^2 = <T(x) \pm ix,T(x) \pm ix>$
by definition

2).$<T(x) \pm ix,T(x) \pm ix> = <T(x), T(x) > + <ix, ix> \pm <T(x), ix> \pm <ix, T(x)>$

3). But.. we know that $<T(x) , ix> = -<ix,T(x)> $ since $T = T^*$  and $ix = (-ix)^*$ which means the last two terms of the RHS of step two cancel each other out leaving you with the result. $<T(x), T(x)> + <ix, ix> = ||T(x)||^2 + ||x||^2$ as desired.


问:
Limit of complex function help.
What is this limit:

$$\lim_{z \to 2i}\, \left[\frac{(3z^2 + 2)}{(z-1)(z+2i)}\right]$$

when I plug in I get:

$$= \frac{-10}{(2i-1)(4i)} = \frac{-10}{-8 - 4i}$$

but the answer in my book appears to be:

$$= 1 - \frac{1}{2}i$$

Where did I go wrong? $3(2i)^2 = 3(4i^2) = 3(-4) = -12$ right?

答:
Your answer is correct, you just need to simplify it:

$$ \dfrac{-10}{-8-4i} = \dfrac{10}{8+4i} \cdot \dfrac{8-4i}{8-4i} = \dfrac{80-40i}{64+16} = 1-\frac{1}{2}i $$

Title: Limit of complex function help.


问:
Minimizer of a convex function.
I found this result on some lectures about optimization:
.

Let $f : X \rightarrow \mathbb{R}$ a differentiable convex function. Then $x$ is a minimizer of $f$ if and only if

$$\langle x'-x,\nabla f(x)\rangle \ge 0 \qquad \forall x'$$

Shouldn't be $\langle x'-x,\nabla f(x)\rangle = 0 \quad \forall x'$, if $x$ is a minimizer of $f$?

答:
Let $f : X \rightarrow \mathbb{R}$ a differentiable convex function. Then $x$ is a minimizer of $f$ if and only if

$$\langle x'-x,\nabla f(x)\rangle \ge 0 \qquad \forall x'$$

Note that this result holds for a general convex set $X$. A proof can be found in this answer.

Shouldn't be $\langle x'-x,\nabla f(x)\rangle = 0 \quad \forall x'$,
if $x$ is a minimizer of $f$?

You were probably thinking about the common optimality condition for unconstrained problem (e.g. when $X=\mathbb{R}^n$), in which case $\nabla f(x) = 0$. It turns out that this result even holds for a general open convex set $X$.

Indeed, if $X$ is open, then for $\epsilon > 0$ small enough, both $x+\epsilon\nabla f(x)$ and $x-\epsilon\nabla f(x)$ belongs to $X$. Replacing $x'$ in $\langle x'-x,\nabla f(x)\rangle \ge 0$ with these two vectors in turn, we obtain $\epsilon \nabla f(x) \ge 0 \ge \epsilon \nabla f(x)$, which means $\nabla f(x) = 0$.

Title: Minimizer of a convex function.


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:
Let $A$ and $B$ be propositions, the claim that $B$ doesn't follow from $A$ can be alternatively described as: either (i) $A$ implies $\neg B$ or (ii) $B$ is irrelevant to $A$ (i.e., given $A$, whether $B$ is true or not is not determined).

(i) can be easily expressed as (using your notation): $A\rightarrow\neg B$.

(ii) can hardly be expressed in the object language of formal logic. In order to express it, first define $p$ as a mapping from a proposition to a scale of probability ranging from 0 to 1, both ends included:
$$p:\textsf{Prop}\rightarrow[0,1].$$
So, the probability of $A$ and $B$ shall be expressed as $p(A)$ and $p(B)$ respectively. Now, consider (ii), it amounts to say that the probability of $B$ given $A$ is equal to the probability of $B$ itself but is irrelevant to the probability of $A$:
$$P(B|A)=P(B).$$
This might seem a little bit complicated, but I think it's the easiest way to capture (ii).


问:
Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$
I need to find max and min of $f(x,y)=x^3 + y^3 -3x -3y$ with the following restriction: $x + 2y = 3$.

I used the multiplier's Lagrange theorem and found $(1,1)$ is the minima of $f$. Apparently, the maxima is $(-13/7, 17/7)$ but I could not find it via Lagrange's theorem.

Here's what I did:

I put up the linear system:

$\nabla f(x,y) = \lambda \, \nabla g(x,y)$

$g(x,y) = 0$

then,

$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$

$x + 2y -3 = 0$

Solving for $\lambda$, I got $\lambda = 0$, which gave me $x = 1$ and $y = 1$.

How can I find the maxima if lambda only gives one value which is $0$?

答:
In this case, it is really not a great idea to use Lagrange multipliers. We can write $x$ in terms of $y$ (or vice-versa) using the restriction and reduce this question to a one variable optimisation problem. Substituting $x = 3 - 2y$, the problem reduces to finding the extrema of $f(y)=-7 y^3+36 y^2-51 y+18$. The critical points are $y = \frac{17}{7}$ and $y = 1$ and, considering that $f''(1)>0$ and $f''(\frac{17}{7})<0$, they are a local minimum and maximum, respectively.

It is also worth noticing that, since $\displaystyle \lim_{y \to \pm \infty} f(y) = \mp \infty$, there will not be any global extrema.


问:
How to prove that the mean follows this expression
Let $X\sim N_p(\mu,\Sigma)$, where $N_p$ denotes the mulvariate normal distribution of dimension $p$ and let $\alpha\in\mathbb{R}^p$. Prove that:
$$
E[(\alpha'(X-\mu))^k] = \left\{\begin{array}{lc}
\frac{(2m)!}{2^mm!}(\alpha'\Sigma\alpha)^m, & \text{ si } k=2m\\
0, & \text{ si } k=2m-1
\end{array}\right.
$$
where $\alpha'$ is the transpose of $\alpha$.

In order to prove this, I have found that $X\sim N_p(0,\Sigma)$, then
$$
E[X^k] = \left\{\begin{array}{lc}
\sigma^k(k-1)!!, & \text{ si } k \text{ even}\\
0, & \text{ si } k \text{ odd}
\end{array}\right.
$$
where $!!$ denotes the double factorial, defined by the product of all integers between 1 and $n$ with the same parity.

I think that result should help a lot with the proof, but I do not know where to use it. I mean, I have tried to expand the term $E[(\alpha'(X-\mu))^k]$ to find, at some point, the term $E[X^k]$ and use that result.

Any suggestion?

答:
Hint: show that $\alpha'(X-\mu) \sim N(0, \alpha' \Sigma \alpha)$.
Then with $\sigma^2 := \alpha' \Sigma \alpha$, the result you found regarding moments of univariate normal random variables implies $$E[(\alpha'(X-\mu))^k] = \begin{cases}(\alpha' \Sigma \alpha)^{k/2} (k-1)!! & \text{$k$ even} \\ 0 & \text{$k$ odd}\end{cases}$$
which the desired result.


问:
Boolean Expression Simplification $A'B'CD + A'B'C'D + ABC'D' + A'BC'D + AB'C'D'$
I want to simplify the below equation but I can't get the same result with an online calculator (the result for the calculator is the correct - I check it with Logisim).

$Y = A'B'CD + A'B'C'D + ABC'D' + A'BC'D + AB'C'D'$

$Y = A'B'(C'D + CD) + A'BC'D + C'D'(AB' + AB)$

--> From Redundancy Laws: $AB + AB' = A$

$Y = A'B'D + A'BC'D + C'D'A$

--> I can simplify until here

The complete simplification is the below, but I can't find a boolean's law to use to remove the $B$.

Does anyone knows which law I have to use in order to remove the variable?

$Y = A'B'D + A'C'D + C'D'A$

Body:
Here is something you can do:

$A'B'D + A'BC'D + C'D'A =$

$A'D(B' + BC') + C'D'A =$

$A'D(B' + B)(B'+C') + C'D'A =$

$A'D1(B'+C') + C'D'A =$

$A'D(B'+C') + C'D'A =$

$A'B'D + A'C'D + C'D'A$


问:
Showing that X such that $X^TQX=Q$ is orthogonal transformation with respect to the $\langle u,v \rangle = (Qu,v)$ scalar product
I am given the following scalar product : $\langle u,v \rangle  = (Qu,v)$ where $(\cdot,\cdot)$ is the usual scalar product of $\mathbb{R}^n$. Now consider a matrix $X$ such that $X^TQX = Q$. Is this matrix $X$ orthogonal with respect to the scalar product $\langle\cdot,\cdot \rangle$? I think I have to show that $\langle Xu,Xv \rangle=\langle u,v \rangle$. But starting from the left side $\langle Xu,Xv \rangle=(QXu,Xv)$. I am not sure how to continue (note that maybe I have to add some additional assumptions that the matrix is orthogonal or something)

答: Hint: For any matrix $A$ and vectors $x,y,$ we have $(x,Ay) = (A^Tx,y)$.


问:
Show that a complex, compact, abelian linear group is finite (using Liouville's theorem)
I'm trying to solve the following problem from the second chapter of Rossmann's book on Lie groups:

This seems like a very interesting result (and a cool application of Liouville's theorem), but I'm not sure on how to approach this.

I thought about maybe looking at the matrix norm/determinant as analytic functions from the matrices to $\mathbb C$ (which should be bounded since the group is compact), but the version I know of Liouville's theorem requires an analytic function who's domain is all of $\mathbb C$ (and here I'm defining it on a group of matrices). So I'm not at all sure what function I should use, and what set to define it on.

Does anyone have an idea on how one should approach this problem? Where should the finiteness come from?

Thanks in advance.

答: If complex linear group means a complex submanifold of $GL_n(\Bbb{C})$, then the maximum modulus principle is more natural: let $M$ be a compact complex submanifold of $\Bbb{C}^k$ then  $\phi:M\to \Bbb{C},\phi(z)=z_j$ is analytic and $|\phi|$ attains its maximum at some $z\in M$ thus $\phi$ is constant near $z$, doing so with each connected component shows that $\phi$ is everywhere locally constant so $M$ is discrete thus finite.


问:
finding a value for $\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$
I have been trying to find the sum:
$$S=\sum_{n=2}^\infty\frac{\zeta(n)}{n!}$$
And I can't find if anyone else has posted about this. Interestingly wolfram alpha claims it converges for $n=0,1$ as well which cannot be correct. I think maybe there is some way of solving this by changing the order of summation:
$$S=\sum_{n=2}^\infty\frac 1{n!}\sum_{k=1}^\infty\frac{1}{k^n}$$
but I am struggling to justify this move, also trying to find:
$$\sum_{n=2}^\infty\frac{1}{k^n}$$
seems to be a geometric series so can I just do it like so, or will the bounds of the summation have changed? Thanks

From the advice of answers I have:
$$\sum_{n=2}^\infty\frac{(1/k)^n}{n!}=\sum_{n=0}^\infty\frac{(1/k)^n}{n!}-\frac1k-1=e^{1/k}-\frac1k-1$$
so we now have:
$$S=\sum_{k=1}^\infty e^{1/k}-\left(1+\frac1k\right)$$

答:
Well, here is an approximate value of $S$.

Observe that as $n$ gets bigger $\zeta(n)$ tends to $1$ moreover $\zeta(n)$ is decreasing sequence.  So for large $n$ we can consider $\zeta(n)$ to be approximately equal to $1$.

So
$$
S \approx \sum_{n=2}^{N} \frac{\zeta(n)}{n!} + \sum_{n = N+1}^{\infty} \frac{1}{n!} \approx e - 2 + \sum_{n = 2}^{N} \frac{\zeta(n) - 1}{n!}
$$

Actually $\zeta(n)$ quickly converge to $1$ as $n$ becomes bigger. For example $\zeta(6) = \pi^6/945 \approx  1.017... $ and  $\zeta(8) = \pi^8/9450 \approx $1.004...

So depending on the need of accuracy you need to choose $N$ and can calculate the value of $S$.

Bonus thing : We have an exact expression for zeta function at even positive integers given by
$$
\zeta(2n) = (-1)^{n-1}\frac{(2\pi)^{2n}B_{2n}}{2(2n)!}
$$
where $B_n$ are Bernoulli's number. I leave it to you use this if you need more exact expression even though we do not know have any closed expression at odd integers.


问:
If A is idempotent and $A^5=3A^3-2A$
Stuck at this question for a while.

Question : If A is idempotent and $A^5=3A^3-2A$ , Prove $A=0$

My try :

By $A^2-A=0$ , we know that the minimal polynomial devides $\Rightarrow x(x-1)$

So the options for the minimal polynomial is only : $x , x-1 , x(x-1)$

I tried assuming by contradiction that the minimal polynomial is $x$ and then we get $A=0$ and we are finished.

if $x-1$ is the minimal then $A-I=0 \Rightarrow A=I$ , but we have an eigenvalue that is $0$ so $|A|=0$ and the matrix can't be $I$.

but I'm stuck with $x(x-1)$.

any hints? Thank you

答:
I agree with the other answer that the claim is false, and won't repeat that argument.

However, assuming that there is a typo in the problem, and it should've been e.g. $A^5=3A^3+2A^2$... Here is a procedure that helps in problems like this one:

We know that $p(A)=0$ and $q(A)=0$, where $p(X)=X^5-3X^3-2X^2$ and $q(X)=X^2-X$. Let's do Euclidean division of $p$ by $q$ (with a remainder):

$$X^5-3X^3-2X^2 = (X^2-X)(X^3+X^2-2X-4)-4X$$

So $p(X)=q(X)s(X)+r(X)$ where $s(X)=X^3+X^2-2X-4$ and $r(x)=-4X$.

This immediately implies that $r(A)=p(A)-q(A)s(A)=0-0\cdot s(A)=0$, i.e. $-4A=0$. I hope you are working in the field $\mathbb R$ (or another field of characteristic $\ne 2$), in which $-4\ne 0$ and so $A=0$.

(Note in a field of characteristic $2$ the claim is just as false as with the original problem, as in characteristic $2$ we have not really fixed the typo - the "fixed" problem is the same as the original one. This makes me think that, perhaps, there is a different typo in the original problem.)

Title: If A is idempotent and $A^5=3A^3-2A$


问:
A continuous bijection that is not homeomorphism
I am trying to solve a question from Pugh's Real Mathematical Analysis 2nd ed. The question seemed a little bit odd to me, and is as follows;

Construct a subset A ⊂ $\mathbb{R}$ and a continuous bijection $f: A → A$ that is not a homeomorphism.

Since the question does not imply anything about using different metrics, I suppose A is equipped with the usual metric of $\mathbb{R}$. In addition, I know that A must be noncompact. But I cannot go further to construct such a subset.

答:
Let $A = \mathbb N \cup \{0\} \cup \{\frac{1}{n}  \mid n \in \mathbb N \}$.  Note that $0$ is the only non-isolated point of $A$.

Define
$$f : A \to A, f(x) = \begin{cases} 0 & x= 0 \\ \frac{1}{2n} & x = \frac{1}{n}, n \in \mathbb N \\ m & x = 2m, m> 0 \\ \frac{1}{2m+1} & x = 2m+1,m > 0 \end{cases}$$
It is easy to verify that this map is a bijection. It is continuous in all $x  \ne 0$ since these points are isolated. Let $\epsilon > 0$. If $\lvert x - 0 \rvert = x < \min(\epsilon,1)$, then either $x = 0$ or $x = \frac{1}{n}$ for some $n$. In both cases $f(x) =  \frac{x}{2}$ and $\lvert f(x) - f(0) \rvert = \frac{x}{2} < \epsilon$. Therefore $f$ is continuous in $0$. If it were a homeomorphism, the closed subset $\mathbb N \subset A$ would be a closed image in $A$. But this is not true because $f(\mathbb N)$ has $0$ as cluster point.


答:
You might like to look at modal logic, which adds a new operator: $\square p$ means “$p$ is necessarily true”.  The idea is that we can imagine alternate worlds which are plausible, but different from the world in which we find ourselves.  For example, we find ourselves in a universe in which I am wearing a green shirt.  Is the necessarily the case?  No, of course not, we could easily have been in a universe in which I was wearing a blue shirt.  But could we have found ourselves in a universe in which $2+2=5$?  No, that is impossible; $2+2=4$ in every possible universe.  “I am wearing a green shirt” is only locally true, but $2+2=4$ is globally true.

The exact behavior and semantics of $\square$ is a philosophical question, and there are different ways to define it, just like the way it is a philosophical question  whether $p\to q$ is identical with $\lnot(p\land \lnot q)$.  (Classical logic says yes, intuitionistic logic says they are different.) One typical axiom is $\square p\to p$, which says that if $p$ is true in every possible universe, it is true in the particular universe in which we live. Another is $$\square(p\to q) \to (\square p\to \square q).$$

Here $\square(p\to q)$ means that $q$ necessarily follows from $p$:  in every possible universe where $p$ is true, $q$ is true in that universe also.

Suppose we agree that $\square p\to p$, so by contrapositive $\lnot p\to\lnot \square p$.  This says that if $p$ is false (in our universe), then it is not necessarily true.  But it might still be true in some other universe. We cannot conclude from $\lnot\square p$ ($p$ is not necessarily true) that $\square\lnot p$ ($p$ is necessarily false).

Now what you asked for is available: in modal logic, $\square(a\to b)$ means that $b$ necessarily follows from $a$, and $\lnot\square(a\to b)$ means that $b$ does not necessarily follow from $a$—but it might follow anyway, locally if not globally. Modal logic usually has a second operator $\diamond p$ which means that $p$ is possibly true; usually $\diamond p$ is taking to be an abbreviation for $\lnot\square\lnot p$, so that $\lnot\diamond p$ and $\square\lnot p$ are two ways of writing that $p$ is impossible.  So consider:

$$\square(p\to q)\\\\
p\to q\\\\
\diamond(p\to q)\\\\
$$

The first one says that from $p$ we can deduce $q$ in every possible universe.  The second only says that from $p$ we can deduce $q$ in the contingent universe we are actually in.  The last one only says that there is some universe, not necessarily this one, in which we can deduce $q$ from $p$.  Applying some simple transformations to the last one we can turn it into $$\lnot\square(p\land\lnot q)$$ which says that it is not necessarily the case that $p$ is true and $q$ is false.

Stanford Encyclopedia of Philosophy article on modal logic

Wikipedia


问:
Computing a triple integral with cylindrical coordinates

Compute $$\int_{-2}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{2}xz \ dz \ dx \ dy$$

Trying to compute this in cylindrical coordinates, I have that

$$\int_{0}^{2\pi}\int_{0}^{2}\int_{r}^{2}r^2\cos(\theta)z \ dz \ dr \ d\theta$$ which evaluates to $0$. What might I be missing here, this shouldn't evaluate to zero right?

答: The integral is in fact zero. You can guess the value looking at the symmetry of the function over the region of integration.</p>


问:
Convergence/Divergence of $\sum \frac{-a_n}{\log a_n}$
Assume that $0<a_n<1$ for all $n\in\Bbb N$ is a sequence in $c_0\setminus\ell^1$, that is, such that $\sum a_n$ is divergent but $a_n\to0$.

What can be said about the convergence/divergence of $\displaystyle \sum \frac{-a_n}{\log a_n}$?

If one thinks on $a_n=\frac{1}{n}$ (probably the classical example of sequence in $c_0\setminus\ell^1$), then $\frac{-a_n}{\log a_n}=\frac{1}{n\log n}$ wich os also divergent.

My intuition says that the answer yo my question is DIVERGENT for all such sequences $(a_n)_n$, but I cannot find a way to prove/disprove this conjecture.

If this were true, I think I can give a complete answer to this question about the Dirichlet Test.

答: With $a_n = \frac1{nlogn}$ we get $- \frac{a_n}{log(a_n)} \sim \frac1{n (logn)^2}$ wich is the general term of a convergent serie.


问:
Polar is equal to the radical axis, special case.

In the above image, $C_0$ contains $C$ where $C$ has $OP$ as diameter and $O$ is the center of $C_0$. Also, $e$ is the polar of $P$.How can I show that $e$ is also the radical axis of both circumferences?

When $P$ is in $C_0$ or out it's easy to proof, but I'm struggling with this case.

答:
Clearly line $e$ is perpendicular to line $OP$ ( polar ) and if we define $Y$ as midpoint of $OP$ ie. centre of $C$, we also get $e \perp OY$.

Define, $X = OP \cap e$ ,and $A$ and $B$ as the tangency point from $X$ to $C_0$.

Now, it's well know that radical axis and line joining the centres are perpendicular.

And we have $OY\perp e$ at $X$ . So it's enough to show $\Bbb{P}(X,C)=\Bbb{P}(X,C_0)$ or enough to show that $XP\cdot XO= XA^2$ , which is true by similar triangles .


答:
As others have said, you don't necessarily need to use Lagrange multipliers. But since you've set up the system, we can see what happens:

$$
3x^2-3=\lambda\\
3y^2-3=2\lambda\\
x+2y-3=0
$$

From the first two equations, we have $3y^2-3=2(3x^2-3)$, which simplifies to $y^2-1=2(x^2-1)$. Rearranging the linear constraint, we have $x=3-2y$. Putting this information together leads to
$$7y^2-24y+17=0.$$

You could solve this using the quadratic formula, but it is quicker to observe that $-24 = -7-17$:

$$7y^2-7y-17y+17=0$$

and so $(7y-17)(y-1)=0$.

This will give you the two local extrema.

Title: Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$

Body:
I need to find max and min of $f(x,y)=x^3 + y^3 -3x -3y$ with the following restriction: $x + 2y = 3$.

I used the multiplier's Lagrange theorem and found $(1,1)$ is the minima of $f$. Apparently, the maxima is $(-13/7, 17/7)$ but I could not find it via Lagrange's theorem.

Here's what I did:

I put up the linear system:

$\nabla f(x,y) = \lambda \, \nabla g(x,y)$

$g(x,y) = 0$

then,

$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$

$x + 2y -3 = 0$

Solving for $\lambda$, I got $\lambda = 0$, which gave me $x = 1$ and $y = 1$.

How can I find the maxima if lambda only gives one value which is $0$?


问:
Dependent or independent samples?
For the scenario given below, I am confused about if the samples are dependent or independent since the scenario does not mention anything about the samples being paired/related or vice versa.

I am aware if terms such as paired, repeated measurements, within-subject effects, matched pairs, and pretest/posttest are instructed in scenarios then it indicates that the samples are dependent and the opposite applies to independent samples, but I am clueless for the given scenario. Any help would be appreciated.

Alice and Bob work evening shifts in a supermarket. Alice has complained to
the manager that she works, on average, much more than Bob. The manager claims that on
average they both work the same amount of time, i.e. the competing claim is that the average
working hours are different. After a short discussion between the manager and Alice, the manager
randomly selected 50 evenings when Alice and Bob both worked.

答:
After a short discussion between the manager and Alice, the manager randomly selected 50 evenings when Alice and Bob both worked.

Under what circumstances would the manager's approach be invalid?  Suppose that Alice normally works long hours, but on nights that Bob works, she is (hypothetically) so repulsed by Bob that she normally goes home early.  She compensates by working even more hours on the nights when Bob isn't there.

This would make the manager's analysis invalid, since the average number of hours that Alice works in an evening is not accurately reflected by the average number of hours that Alice works when Bob is also working.

Of course, the symmetrical argument is that Bob tries to work longer hours when Alice is also working because he likes being in the same building with Alice.

The fact that the manager chose the approach that he did pre-supposes that none of the above considerations apply.  This means that the manager is assuming that the average number of hours that Alice works is independent of whether Bob is also present.

The manager is also assuming that the average number of hours that Bob works is independent of whether Alice is also present.

Addendum

It occurs to me (well after the fact) that there is a subtle flaw in my conclusion.  It is (theoretically) possible for the strategy chosen by the manager to be valid, even when the average number of hours that Alice works is dependent on whether Bob is in the building.  Similarly, the strategy chosen by the manager can be valid, even when the average number of hours that Bob works is dependent on whether Alice is in the building.

Suppose (for example) that Alice and Bob find each other equally repulsive, and this causes each of them to shorten their typical work hours in equal measure, on the nights that they both work together.  In this scenario, the average number of hours that both Alice and Bob work are both dependent on whether the other person is in the building.

At the same time, under this artificial scenario, the strategy chosen by the manager should still provide an accurate assessment.  For example, suppose that on the nights that Alice and Bob both work, Alice works more hours.  Since it is assumed that Bob and Alice are each (typically) cutting back their hours when they are both present, to the same degree, then you can conclude that on the nights that only one of them is present, Alice's average work hours will still exceed Bob's average work hours.

Unfortunately, this is only the start of the analysis of when the manager's strategy may still be valid despite the assumption that the average number of hours that Alice and/or Bob work is dependent on the presence of the other person.

For example, suppose that Bob is indifferent to Alice's presence.  Further suppose that one of the days that Alice works with Bob is Saturday, when she goes home 1 hour early, and Bob does not.  Further suppose that Alice works (without Bob) on Thursday, typically working an extra hour.

Assuming that Alice and Bob each work 5 days a week, the dependence of Alice's work hours on Bob's presence will lower her average work hours per day by 0.2 hours (i.e. cause a variance of -0.2).  Following the manager's strategy, if the comparison between Alice's average work hours and Bob's average work hours is greater than the variance, then the manager will automatically still come to the right conclusion.

In a typical work setting, the strategy chosen by the manager seems typical (i.e. quick and easy).  No need for random sampling of work days of each of Alice and Bob (more work for the manager).  Instead, just use a quick and easy side by side comparison.

So, what the manager is really assuming is that whatever the net effect is on Alice's variance vs Bob's variance (if any) that is caused by Alice and Bob both working in the building on the same evening will either be

less than the side by side comparison of their respective work hours

or in the same direction as the side by side comparison of their work hours (i.e. if Alice works an hour more than Bob on the nights that they work together and if Alice's global average of work hours per day exceeds Bob's global average by only 0.2 hours).

Title: Dependent or independent samples?


问:
How to prove that $1/n!$ is less than $1/n^2$?
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

Body:
$2(n-1)>n$

$3(n-2)>n$

$4(n-3)>n$

$\cdots$

$\cdots$

$(k+1)(n-k)>n$

$\cdots$

$\cdots$

$(n-1)2>n$

Multiplying all, $~$ $(n-1)^{2}\times (n-2)^{2}\times\cdots\times 2^{2}\times 1^{2}>n^{n-2}$

$\therefore$ $n^{2}\times (n-1)^{2}\times \cdots 2^{2}\times 1^{2}>n^{n}$

$\therefore$ $n!>n^{\frac{n}{2}}$ $~$ $(n\geq 3)$

$\therefore$ $n!>n^{\frac{n}{2}}\geq n^{2}$ $~$ $(n\geq 4)$


答:
This is from your working -

$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$

$3x^2 - 3 = \lambda, 3y^2-3 = 2\lambda$

Equating $\lambda$ from both equations,

$6x^2-6 = 3y^2-3 \implies 2x^2 - y^2 = 1$

Substitute $x$ from $x+2y = 3$

$2(3-2y)^2 - y^2 = 1$

$\implies 7y^2 - 24y + 17 = 0 \, $ or $(7y-17)(y-1) = 0$

Can you take it from here and find possible points for extrema?


问:
What happens if the Euler-Lagrange function is always zero?
As titled, I was considering the mimimization problem where $y(x)$ has two endpoints fixed. That is, minimizing $$\int_a^b L(x,y(x),y'(x)) \, dx$$

where $$\ y(a)=m, y(b)=n $$for all $y(x)$.

if the Euler-Lagrange equation is always zero for all functions $y(x)$, I think the original integral $$\int_a^b L(x,y(x),y'(x)) \, dx$$ is constant for all $y(x)$.

Which means if
$$\frac {d}{dx} \frac {\partial L}{\partial y'} = \frac {\partial L}{\partial y}$$ for all $y(x)$, then
$$\int_a^b L(x,y(x),y'(x)) \, dx = C$$ for some constant $C$.

But I don't know how to prove it. I start by using integration by parts, writing $$\int_a^b L(x,y(x),y'(x)) \, dx$$ as $$\left.\ x*L(x,y(x),y'(x))\right|_a^b - \int_a^b x*\frac {d}{dx}L(x,y(x),y'(x)) \, dx$$
But I don't know how to continue from here, or maybe I shouldn't use IBP here.

Any help will be appreciated.

答:
More generally, one may show that

if Euler-Lagrange (EL) equations are always satisfied, and

if the $x$- and $y$-spaces are contractible spaces,

then the Lagrangian density is a total divergence, i.e. the action functional is a boundary term, cf. e.g. Refs. 1-3.

If furthermore the boundary is fixed by boundary conditions, then the action functional is a constant, as OP already suspected.

See also this related Phys.SE post.

References:

P.J. Olver, Applications of Lie Groups to Differential Equations, 1993.

I. Anderson, Introduction to variational bicomplex, Contemp. Math. 132 (1992) 51.

G. Barnich, F. Brandt & M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.


问:
How do I find the point between two point at a specific distance (in x,y coordinates)
I am working on a programming project and have run into a problem.

I need to find a point that is on a line. For example if my line is AC and I have the coordinates for point A and C, how can I get the coordinates for B on the same line between A and C.  B will be a set distance from A, for example 60.

So if the coords for point A are (20,45) and the coords for point C are (120, 240), how do I get the x,y coords for point B that is a distance of 60 from point A.

I hope that made sense.

It has been a long time since math class, so please keep it as simple as possible.

This is not homework, it is just a problem that has been bugging me for hours.

答:
Let $A=(a,b)$ and $C=(c,d)$. The points on the line $AC$ have coordinates $(x_t,y_t)$ where
$$x_t=(1-t)a+tc,\qquad y_t=(1-t)b+td,$$
where $t$ ranges over the reals. (For the line segment $[A,B]$, the parameter $t$ ranges over the interval $[0,1]$.)

for positive $t$, the distance from  $A$ to $(x_t,y_t)$ is $t$ times the distance from $A$ to $C$. So for your problem, calculate the distance $d$ from $A$ to $C$, and let $t=\frac{60}{d}$.

In your example, $d=\sqrt{(120-20)^2+(240-45)^2}\approx  219.146$.


问:
Non cyclic Galois extension of degree 12
Are there examples of Galois extensions of degree 12 which are not cyclic(i.e the corresponding Galois group is non cyclic)?
If yes, please give me an hint on how to come up with such an extension?</p>

答:
Hint 1: Consider the polynomial $(x^2+1)(x^2-3)(x^3-2)$.

Hint 2: Show that the Galois group has two distinct elements of order $2$.


问:
What is the difference between the Comprehension scheme an Replacemnt scheme?
I am currently studying ZFC axioms and having trouble understanding the difference between the comprehension scheme and the replacement scheme. Here's what I understand about the two schemes so far.
The comprehension scheme allows us to create a new set whose elements are members of a given set obeying some property and the replacement scheme let us create a new set which is the image of an existing set under a function.
So both the scheme just creates a new set with some elements from already existing set according to some property P.</p>

答: Comprehension requires that the elements already be part of a set.  You are making a subset from the elements that have the property of interest.  Replacement allows the elements of the new set to come from anywhere as the range of the function may not be known to be a set.  If I apply comprehension to $\Bbb N$ I can get any subset of the naturals that I can describe.  If I apply replacement to $\Bbb N$ I can get any countable set that I can describe.


问:
Show that $\int_0^1\sqrt{1+x^4}dx\geq \frac{\sqrt{10}}{3}.$
Show that $$\int_0^1\sqrt{1+x^4}dx\geq \frac{\sqrt{10}}{3}.$$
Actually, I am trying to make Cauchy-Schwarz work for it, but it doesn't work. Could someone give some idea to fix it?

答:
(Fill in the gaps as needed. If you're stuck, show your work.)

From the expansion of $ \sqrt{ 1 + x^4}$ or otherwise, conclude that for $0 \leq x \leq 1,$

$$ \sqrt{ 1 + x^4 } \geq 1 + \frac{x^4}{2} - \frac{x^8}{8}. \quad \quad (1)$$

Hence, show that

$$ \int_0^1 \sqrt{ 1+x^4} \, dx \geq \int_0^1 1 + \frac{x^4}{2} - \frac{x^8}{8} \, dx = \frac{391}{360} > \frac{ \sqrt{10} } { 3} . $$

Notes

In fact, inequality $(1)$ holds for all $x$. This can be shown by

A) RHS is negative when $ x^4 \geq 8$ and

B) Squaring both sides to show it's true for $ x^4 < 8$.


答:
Write the equation of a line
Say A=(a,b) and C=(c,d) then equation of line is (y-d)(c-a)=(d-b)(x-c) now you know the distance of B from A and equation of line AC.
let B=(k,l)so you have two equations
1. Point B lies on line AC i.e. it satisfies equation of line AC
2.and also we know that the distance between AB=squart[(a^2-k^2)+(b^2-l^2)]

so two equations and two unknowns
you get coordinates of point B

Title: How do I find the point between two point at a specific distance (in x,y coordinates)

Body:
I am working on a programming project and have run into a problem.

I need to find a point that is on a line. For example if my line is AC and I have the coordinates for point A and C, how can I get the coordinates for B on the same line between A and C.  B will be a set distance from A, for example 60.

So if the coords for point A are (20,45) and the coords for point C are (120, 240), how do I get the x,y coords for point B that is a distance of 60 from point A.

I hope that made sense.

It has been a long time since math class, so please keep it as simple as possible.

This is not homework, it is just a problem that has been bugging me for hours.


问:
Solve the recurrence $\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n) $
Solve the following recurrence:

$$\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n)$$

I tried to just unroll it to see if I can understand, but it is too complicated .

$\ T(n) = n^{\frac{1}{2}} \cdot T(n^{\frac{1}{2}}) +n\cdot \log(n)\\
= n^{\frac{1}{2}} \cdot \left( n^{\frac{1}{4}} \cdot T(n^{\frac{1}{4}} ) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n \cdot \log(n) \\
= n^{\frac{1}{2}} \left( n^{\frac{1}{4}} \cdot \left( n^{\frac{1}{8}} \cdot T(n^{\frac{1}{8}}) +  \frac{n}{4}\cdot \log(\frac{n}{4})  \right) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n\cdot \log(n) $

Not really clear how to proceed here? I also tried recurrence tree, but couldn't see a clear pattern. I've found a similar question here but I could not understand the answer

答: The Ansatz $T(n)\sim An^p\log^qn$ gives$$An^p\log^qn=\frac{A}{2^q}n^{(p+1)/2}\log^qn+n\log n\implies A=2,\,p=q=1.$$


问:
How to find the remainder of big number divisions using congruences?
I´m asked to find the remainder of dividing $\sum_{i=0}^{1080}i^5$ by $14$. How can I do this using only basic results from modular arithmetic? Only one thing comes to my mind, here's my idea:
we know that each number has a representative $r$ in the class of $\pmod{14}$, that satisfies $0\leq r<14$, so in order to simplify things, it is only needed to find this representative for $n^5$ for the integers $n$ between $0$ and $13$, because, for example, if I wanted to find the residue of $\sum_{i=0}^{27}i^5$ divided by $14$, then assuming I know that $k_n$ is the representative of $n$ for each $n\in\{ 0,\dots,13 \}$, then $14\equiv0\pmod{14}$ implies $14^5\equiv0^5\equiv k_0\pmod{14}$, $15\equiv1\pmod{14}$ implies $15^5\equiv1^5\equiv k_1\pmod{14}$, and so on until $27^5\equiv13^5\equiv k_{13}\pmod{14}$. In this way, it is now possible to know that, given that
\begin{equation*}\sum_{i=0}^{27}i^5=\sum_{i=0}^{13}i^5+(i+14)^5
\end{equation*}, then for $i\in\{ 0,\dots,13 \}$, $i^5\equiv (i+14)^5\pmod{14}\Rightarrow i^5+(i+14)^5\equiv2i^5\equiv2k_i\pmod{14}$, which implies that:
\begin{equation*}\sum_{i=0}^{13}i^5+(i+14)^5\equiv\sum_{i=0}^{13}2k_i\equiv R_k\pmod{14}
\end{equation*}
where $R_k$ is the representative of $\sum_{i=0}^{13}2k_i$ in $\mathbb{Z}_{14}$. That's the residue we're looking for.

Following the pattern, I'd need to find out how many times a number which is congruent to a number between $0$ and $13$ in $\mathbb{Z}_{14}$ appears between $0$ and $1080$, and then rewrite the original sum in terms of how many times the "repeated" (by this I mean in the sense that they are equivalent in $\mathbb{Z}_{14}$) numbers appear.

(I already did this, but as I said, it's a very long and tedious process).

Another idea is to use the formula for the sum of the first $n$ fifth powers.

Any other idea of an easier process, or a check to mine would be really appreciated. Thanks in advance.

答:
We have

$\tag 1 \displaystyle \sum_{i=0}^{13}i^5 \equiv 0 + + 7^5  +\sum_{i=1}^{6}\bigr(i^5+(-i)^5\bigr)  \equiv 7^5 \equiv 7 \pmod{14} $

Also, since

$\quad 1081=77\times14+3$ we can write

$ \tag 2 \displaystyle \sum_{i=0}^{1080}i^5 \equiv 77\cdot7 + 0^5 + 1^5 + 2^5 \equiv 7 + 0 + 1 + 4 \equiv 12\pmod{14} $


问:
Product and sum of iid random variables converges a.s.
The set of independent and identically distributed r.v. $Y_{1},Y_{2},...,Y_{n}$ has a distribution $Pr(Y_i=1)=Pr(Y_i=1/2)=1/2\ \forall i>0$ . If $Z_n=\prod_{i=1}^{n}Y_i$ and $S_n=\sum_{i=1}^{n}Z_i \ \forall n>0$. Does $Sn$ converge a.s. to some limit r.v.? If so, find the mean and variance of the limit r.v.

I used the properties of iid r.v. to rewrite Sn as following: $$S_n=\sum_{i=1}^{n}\prod_{i=1}^{n}Y_i=\sum_{i=1}^{n}(Y_1)^{n}=(Y_1)+(Y_1)^{2}+...+(Y_1)^{n}.$$

According to the distribution, Yi=1 or Yi=1/2 but I am not sure how to implement that fact here. Moreover, the Law of Large Numbers and a.s. convergence involve the sum of r.v., so I don't know if they can apply here as well.

答: Note that $S_n - S_{n-1} = Z_n$. Use this to show that $S_n$ is a Cauchy sequence in $L^2$, thus $S_n$ has an $L^2$ limit $S_{\infty}$. Then $S_n$ must converge to $S_{\infty}$ almost surely along some subsequence. But since $S_n$ is a monotonic sequence, convergence along a subsequence implies convergence for the entire sequence.


问:
$F\cup G \le V$ if and only if $F \subseteq G$ or $G\subseteq F$
In an exercise I'm asked to prove the following result:

Let $V$ be a vector space and $F,G \le V$. Prove that $F\cup G \le V$ if and only if $F \subseteq G$ or $G\subseteq F$.

I was able to prove that $F \subseteq G \vee G\subseteq F \to F\cup G \le V$ But I'm not being able to prove the opposite.

How can I do so?

答:
If there are elements $x\in F\setminus G$ and $y\in G\setminus F$ then $x+y$ is a sum of elements in $F\cup G$, and since by assumption this is a subspace we have $x+y\in F\cup G$. So $x+y\in F$ or $x+y\in G$. Let's assume $x+y\in F$. Then, since $F$ is a subspace:

$y=(x+y)-x\in F$

But this in a contradiction, because $y\in G\setminus F$. Similarly, if $x+y\in G$ then:

$x=(x+y)-y\in G$

Which is again a contradiction, because $x\in F\setminus G$.