答:
This is a long comment. I wish I could give a definitive answer, but what WA does is unclear.

That generating function implies a polynomial $n$th term. As with most choices of a sequence's first five terms, the lowest-order polynomial expression for the $n$th term that's correct for all data is quartic. You can see this by verifying these terms' third differences are two different values, $2,\,0$. It's likely WA determined the quartic first, then obtained its generating function (which will be a polynomial divided by $(z-1)^5$), so it's conceptually simplest to get the quartic directly.

The first differences are $7,\,4,\,3,\,2$; the second are $-3,\,-1,\,1$; the third are $2,\,0$; the only fourth difference is $-2$. If fourth differences are constant and $25$ is the $0$th term, the $n$th term is $25+7n-3\binom{n}{2}+2\binom{n}{3}-2\binom{n}{4}$. This sequence has generating function $\frac{25}{1-z}+\frac{7z}{(1-z)^2}-\frac{3z^2}{(1-z)^3}+\frac{2z^3}{(1-z)^4}-\frac{2z^4}{(1-z)^5}$.

I doubt this matches what you got, because the highest power of $(1-z)^{-1}$ is different. My comment on the OP noted WA gave me a different generating function anyway (which, by the same logic, probably doesn't match the above either). Further, this $k$th-difference technique predicts the next term is $40$, not $42$ (although WA gave me $42$ as the next term as well).

Title: Finding Nth term using a generating function

Body:
Good day everyone

I am in need of help, I was using an online program to solve a sequence, the name of the program is WolframAlpha and using it I was able to get the actual nth term that I was looking for several times. But in order to get that nth term the program used the following generating function

$$\frac{ z(-34z  ^ { 2  }  +77z-41)  }{ (z-1)  ^ { 3  }    }  +25$$

Note my sequence was 25,32,36,39,41 and gave me nth term 42 which is what I actually needed.
Now I need to find out how did the program get to find the nth using this generating function, I don't understand how -34z and 77z came to be.

Thank you while I wait for the answer


问:
How do I calculate a line integral of $ x^i dx^j - x^j dx^i $ with polar coordinates?
I've come across this general integral $$S^{ij} = \oint_{\text{loop on the surface}}  x^i dx^j - x^j dx^i \approx \text{Area Enclosed By The Loop.}$$

which, in $(x,y)$ coordinates is

$$S^{xy} = \oint_{\text{loop on the surface}}  x dy - y dx \approx \text{Area Enclosed By The Loop.}$$

and this is easily seen to be $2\pi r^2$ when $x$ and $y$ are parameterized and integrated around a circle.

My question is this.

Should I also be able to calculate that as

$$S^{r\theta} = \oint^{\theta=2\pi} _{\theta=0}  r d\theta - \theta dr = r \theta-0=2\pi r$$

Why doesn't this also work?

(I have transformed the rank 2 tensor to polar coordinates and, unless I messed up, it does give this answer...which isn't an "area." So the meaning of the tensor seems to have changed and that doesn't seem right or acceptable.)

答:
You cannot replace one system of coordinates by another without invovling some "correction" terms: here you cannot just use the substitution $(x,y) \to (r,\theta)$ by just saying $x = r$ and $y=\theta$ in the integral.

Here is a proper way: say $x = r \cos \theta$ and $y = r \sin \theta$. Hence,
\begin{align}
\mathrm{d}x &= \cos \theta \mathrm{d}r - r \sin \theta \mathrm{d}\theta,\\
\mathrm{d}y &= \sin \theta \mathrm{d}r + r \cos\theta \mathrm{d}\theta.
\end{align}
From this, we deduce
\begin{align}
x \mathrm{d}y - y \mathrm{d}x &= r\cos\theta\left(\sin \theta \mathrm{d}r + r \cos\theta \mathrm{d}\theta \right) - r\sin\theta\left(\cos \theta \mathrm{d}r - r \sin \theta \mathrm{d}\theta \right) \\
&= r^2 \mathrm{d}\theta.
\end{align}
And from now, you should be able to recover the right answer.

Comment. A quick way to show the result is by defining $\alpha = x\mathrm{d}y - y\mathrm{d}x$, then $\mathrm{d}\alpha = 2\mathrm{d}x\wedge \mathrm{d}y$, which is twice the standard volume form in coordinates. By Stokes formula, it follows that the integral is twice the area enclosed, that is, $2 \times \pi r^2$ if $r$ is constant.

ContentLicense: CC BY-SA 4.0


问:
On eigenvalues of complex-orthogonal matrices
Suppose that $A$ is a complex matrix satisfying $A^TA = I$ (so $A$ is the entrywise transpose, not the conjugate transpose).  What can be said about the eigenvalues of $A$, if $A$ is "complex-orthogonal" in this sense?

Of course, for any eigenpair $(\lambda,x)$, we have
$$
x^Tx = x^TA^TAx = (Ax)^TAx = \lambda^2 (x^Tx)
$$
which allows us to conclude that $\lambda^2 = 1$... so long as $x^Tx \neq 0$.  Can anything else be said?  Does the case in which $A$ has real entries allow us to conclude that $|\lambda| = 1$?

答:
This has been thoroughly studied in the paper "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices" by Horn and Merino (1999) and also in Olga Ruff's master thesis "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices: characterisation and examples" (2007). In particular, theorem 1.2.3 (pp. 31-32) of Ruff's thesis states that

An $n\times n$ complex matrix is similar to a complex orthogonal matrix if and only if its Jordan Canonical Form can be expressed as a direct sum of matrices of only the following three types:

(a) $J_k(\lambda)\oplus J_k(\lambda^{-1})$ for $\lambda\in\mathbb C\setminus\{0\}$ and any $k$,

(b) $J_k(1)$ for any odd $k$ and

(c) $J_k(-1)$ for any odd $k$.

In particular, every nonzero complex number is an eigenvalue of some complex orthogonal matrix, and for each complex orthogonal matrix, all eigenvalues $\ne\pm1$ must occur in reciprocal pairs.

ContentLicense: CC BY-SA 3.0


问:
Prove $|f(x)-f(y)|<\varepsilon \implies \sup f - \inf f \leq \varepsilon$
$f$ is continuous on $[a,b]$ and $|f(x)-f(y)|<\varepsilon$. Show that $\sup f - \inf f \leq \varepsilon$

Suppose that $\sup f - \inf f > \varepsilon$.

$|f(x)|-|f(y)| \leq |f(x)-f(y)| < \varepsilon < \sup f - \inf f \Rightarrow |f(x)|-|f(y)| < \sup f - \inf f$

It follows that $\begin{cases}|f(x)|<\sup f\\-|f(y)|<-\inf f \end{cases} \Rightarrow \begin{cases}|f(x)|<\sup f\\|f(y)|>\inf f \end{cases}$

In particular $f(x)<\sup f$ for all $x \in [a,b]$. But $f(x)$ has a maximum on $[a,b]$ since it's continuous and $\max(f)<\sup f$, hence $\sup f$ is not a least upper bound. I've derived a contradiction, so $\sup f - \inf f \leq \varepsilon$.

Not sure if this proof is correct. Is there a better one?

答:
Just another approach.

Since $f$ is continuous, there are two sequences $(x_n),(y_n)\subset[a,b]$, such that $f(x_n)\to\inf f$ and $f(y_n)\to\sup f$.

But $|f(x_n)-f(y_n)|<\varepsilon$, now letting $n\to\infty$, you get $$\sup f-\inf f=|\inf f-\sup f|\le\varepsilon.$$


问:
How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$?
I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.

答:
Rewrite it as $5\sin^2(x)+8\sin x \cos x-3(\sin^2 x + \cos^2 x)$, which gives $2 \sin^2 x + 8 \sin x \cos x - 3 \cos^2 x$, a quadratic in $\sin x$.

Using the quadratic formula, the roots of $2u^2 + 8u - 3$ are $-2 - \sqrt{11/2}$ and $-2 + \sqrt{11/2}$. Hence this is equivalent to $(u + 2 + \sqrt{11/2})(u + 2 - \sqrt{11/2}) = 0$, so the original factorises as $(\sin x + (2 + \sqrt{11/2})\cos x)(\sin x + (2 - \sqrt{11/2})\cos x) = 0$.


问:
Proving that $\int u^2 dx < Ce^{-at}$ where $u$ solves a linear pde
Let the unit sphere in $\mathbb{R^n}$ be $B_1(0)$ and let $u$ be the smooth solution of
$$ \begin{cases} u_{tt} + a^2(x) u_t - \Delta u = 0 & B_1(0) \times (0,\infty)\\
u(x,t) = 0 & \partial B_1(0) \times (0,\infty) \\ u(x,0) = g, \; u_t(x,0) = h & B_1(0) \times \{0\} \end{cases}$$

Here, $g,h$ and $a(x)$ are smooth functions and $g,h$ vanish on the boundary of $B_1.$  Prove that $\int_{B_1(0)} u^2(x,t) dx \leq C e^{-At}$, where $A$ is the minimum of $a^2(x)$ and $C$ is a positive real number.

Usually the methods I have seen for problems like these has been to use some sort of an energy, so here is my twice updated attempt, now using Poincare's inequality in the first step:

Let
\begin{align}
\frac{d}{dt} e(t) &:= \frac{d}{dt}\frac{1}{2} \int_{\Omega} u^2 \,dx \\
&\leq \frac{d}{dt}\frac{C}{2} \int_{\Omega} u_t^2 + |\nabla u|^2 dx\\
&= C\int_{\Omega} u_tu_{tt} + \nabla u_t \cdot \nabla u \; dx\\
&= C\int_{\Omega} u_t( u_{tt} - \Delta u) \; dx + \underbrace{\int_{\partial\Omega} u_t\nabla u \cdot n \, dS}_{\text{0 Due to BCs}}\\
&= -C\int_{\Omega} a^2(x)u_t^2 \; dx \leq 0
\end{align}

So we have $\frac{d}{dt}e(t) \leq 0,$ and therefore we have $e(t) \leq e(0) = \int_{B_1(0)} g^2 \; dx.$  Still, this does not show the desired inequality.

EDIT: Another idea that comes to mind, although I realize this is a finite domain, is to look at the Fourier transform, and then show that $||{\hat u(\xi,t)}||_{L_2}^2 \leq C e^{-At},$ since the $L_2$ norm of the transform is more or less equivalent to the $L_2$ norm in the spatial domain.  The finite domain aspect seems to ruin this though.

答:
You are trying to show that there is damping.  What is the equation satisfied by $v(x,t)=u(x,t)\exp(\lambda t)$? We find
$$
v(x,0)=g,\quad v_t(x,0)=h+\lambda g,\quad v=0\text{ on }\partial B,
$$
so no noticeable changes (just a factor). When it comes to the pde, it becomes
$$
v_tt + v_t(a^2-2\lambda) + v\lambda (-\lambda +a^2) -\Delta v=0
$$
If $a^2-2\lambda\geq0$, your analysis above carries through without a change. So you have proved
$$
\int_\Omega  v^2 \text{d} x \leq \int_{B^1} g^2  \text{d} x,
$$
for  $\lambda =\frac{1}{2}\min a^2$. The conclusion follows by writing this inequality in terms of $u$.


Title: How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$?

Body: I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.

Body:
$$2(5\sin^2(x)+8\sin x \cos x-3)=5(1-\cos2x)+8\sin2x-6=-5\cos 2x+8\sin 2x-1=0.$$

This is a classical linear trigonometric equation.

$$8s=5c+1$$
$$64(1-c^2)=(5c+1)^2$$
$$c=\frac{-5\pm16\sqrt{22}}{89},s=\frac{5c+1}8.$$


问:
How to find the Laurent expansion for $1/\cos(z)$
How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$</p>

Body:
Note $\frac{1}{\cos z}=-\frac{1}{\sin (z-\frac{\pi}{2})}$.

Let $t:=z-\frac{\pi}{2}$. Then $0<|t|<1$, $\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots$, so
$$
\begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
                &=\frac{1}{t}\frac{1}{1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)}\\
                &=\frac{1}{t}\left[1+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)^2+\dots\right]\\
                &=t^{-1}+\frac{1}{3!}t+\left[\left(\frac{1}{3!}\right)^2-\frac{1}{5!}\right]t^3+\dots
\end{align}
$$
It seems no closed forms for higher terms.


问:
Showing that a group has infinite order
Let $G$ be an abelian group such that for all $g \in G$ and for all positive integers $k$, there exists some $h \in G$ such that $h^k = g$.

Now, I would like to show that the group is either the trivial group, or of infinite size. The trivial case is straightforward, but I am having trouble proving the infinite cardinality.

I start by supposing that $G$ has finite cardinality with $|G| = m>1$ where $m$ is some integer. I now pick an abitrary $g$ and $k=m+1$, and the corresponding $h$ so that we have $h^k=g$.

If I am on the right track, I'm lost as to where to go to from here. If I am not on the right track, well then I am quite lost!

Thank you for any help!

答: Suppose $G$ is finite. Let $g \in G$ be arbitrary. Then there exists some $h \in G$ such that $h^{\lvert G \rvert} = g$. But $h^{\lvert G \rvert} = e$ by Lagrange's theorem! Since $g$ was arbitary, we conclude that $G = \{e\}$ is trivial.

Title: Showing that a group has infinite order


答: If $m:=|G|$ is finite, then $h^m=e$ for all $h\in G$. So with $k=m$ (not $m+1$), we conclude from the existence of $h$ with $h^k=g$ that $g=e$.

Title: Showing that a group has infinite order

Body:
Let $G$ be an abelian group such that for all $g \in G$ and for all positive integers $k$, there exists some $h \in G$ such that $h^k = g$.

Now, I would like to show that the group is either the trivial group, or of infinite size. The trivial case is straightforward, but I am having trouble proving the infinite cardinality.

I start by supposing that $G$ has finite cardinality with $|G| = m>1$ where $m$ is some integer. I now pick an abitrary $g$ and $k=m+1$, and the corresponding $h$ so that we have $h^k=g$.

If I am on the right track, I'm lost as to where to go to from here. If I am not on the right track, well then I am quite lost!

Thank you for any help!


问:
Constructing Very "Flat" Functions
Consider the function $f(x)=\tan(\sin(x))-\sin(\tan(x))$. It has (what I think is) a remarkable property. The Taylor series expansion of $f(x)$ around $x=0$ is
$$\frac{1}{30}x^7+\frac{29}{756}x^9+\frac{1913}{75600}x^{11}+o(x^{13}).$$

Since $\tan(x)$ and $\sin(x)$ are odd functions, $f(x)$ is also an odd function. But its first nonzero term is $x^7$. This tells us that that $f(x)$ is very "flat" near the origin, which it is indeed.

I haven't calculated an explicit closed form for the series coefficients but looking at the wild oscillations around $x=\pm \pi/2$ (and confirming it by looking at it in the complex plane), the radius of convergence is $\pi/2$. This is simply because of the tangent function.

I have the following questions.

Is this just a coincidence or is there something deeper going on here? Is there any way to build/find more such examples? How can we find two real functions $g_1(x)$ and $g_2(x)$ such that

they are both distinct,

they are both elementary and can be explicitly written in closed form,

and $g_1(g_2(x))-g_2(g_1(x))$ is $o(x^n)$ for some $n>1$ with a strictly positive radius of convergence.

Neither should be a polynomial because this is trivial to do with polynomials. So the Taylor expansion of $g_1(x)$, $g_2(x)$, and $g_1(g_2(x))-g_2(g_1(x))$ should be infinite.

Addressing Calvin Khor's comment

Is it really trivial with polynomials? The first few terms of the Taylor series only depend on the first few derivatives, so it seems to me that understanding how to do it for polynomials should help understand what's happening in your example

Yes, it is trivial with polynomials. Consider
$$g_1(x)=x^{11}+x^{13}$$
$$g_2(x)=x^{15}+x^{17}$$
then
$$g_2(g_1(x))-g_1(g_2(x))=4 x^{167}+50x^{169}+290x^{171}+\cdots$$
which is indeed a finite series with an infinite radius of convergence, because it is only a composition of polynomials. So it is trivial to build such a (finite series) example for any desired $o(x^n)$.

I was looking for a general construction for any given $n$, so that the series is infinite and can be written in explicit closed form using elementary functions, or a proof that it is not possible.

答:
This occurs because the two functions are very similar to  $x$ for small $x$, so that the given composition cancels them efficiently.

Consider

$$\sin x\approx\arctan x\approx x,$$

then

$$\sin(\arctan x)-\arctan(\sin x)$$
is flat. The Taylor expansion is also of order $7$.

https://www.wolframalpha.com/input/?i=Taylor+expansion+of+sin%28arctan+x%29-arctan%28sin+x%29

For a nonic case, combine

$$\frac x{1-x^2}$$ and $$x-x^3+2x^5.$$

Title: Constructing Very "Flat" Functions


答:
Since you have $|f(x)-f(y)| \geq f(x)-f(y)$ for all $x,y$. Take a supremum over all $x,y$ from both sides, you will get that $\sup(|f(x)-f(y)|)\geq \sup f-\inf f$. Because of $\varepsilon \geq \sup(|f(x)-f(y)|)$ you have the needy inequality.
You proof has a small mistake because $\sup f-\inf f>|f(x)|-|f(y)|$ does not imply both strict inequalties you have written, just one strict and one nonstrict. But since $|f| $ is still continuos it is all okey.

Title: Prove $|f(x)-f(y)|<\varepsilon \implies \sup f - \inf f \leq \varepsilon$

Body:
$f$ is continuous on $[a,b]$ and $|f(x)-f(y)|<\varepsilon$. Show that $\sup f - \inf f \leq \varepsilon$

Suppose that $\sup f - \inf f > \varepsilon$.

$|f(x)|-|f(y)| \leq |f(x)-f(y)| < \varepsilon < \sup f - \inf f \Rightarrow |f(x)|-|f(y)| < \sup f - \inf f$

It follows that $\begin{cases}|f(x)|<\sup f\\-|f(y)|<-\inf f \end{cases} \Rightarrow \begin{cases}|f(x)|<\sup f\\|f(y)|>\inf f \end{cases}$

In particular $f(x)<\sup f$ for all $x \in [a,b]$. But $f(x)$ has a maximum on $[a,b]$ since it's continuous and $\max(f)<\sup f$, hence $\sup f$ is not a least upper bound. I've derived a contradiction, so $\sup f - \inf f \leq \varepsilon$.

Not sure if this proof is correct. Is there a better one?


问:
Sobolev space with negative index
The Dirac delta function is in the Sobolev space $H^{-1/2-\epsilon}(\mathbb{R})=W^{-1/2-\epsilon,2}(\mathbb{R})$ for $\epsilon>0$, but it is a distribution as opposed to a function in the traditional sense. On the other hand we have $H^{0}(\mathbb{R})=L^{2}(\mathbb{R})$, so I presume $H^{s}(\mathbb{R})$ only has traditional functions for $s\geq0$. Now what can I say for $-1/2\leq s<0$? We can naturally expect something larger than an $L^2$ function but I don't know if there are any non-traditional functions there. I tried like $\delta^{1/2}(x)$ but it doesn't really seem to be a distribution.

答:
The following is based on what LL sketched in the comments using the Cantor function, which can be modified to find elements in $H^s$ for each $s<0$ that are not represented by locally integrable functions. My answer may seem a bit unsatisfying as it relies on some technical machinery and is not entirely explicit, but hopefully I can convince you that it is a fairly natural apporach. I will use results from the following reference

Mattila, Pertti, Geometry of sets and measures in Euclidean spaces. Fractals and rectifiability, Cambridge Studies in Advanced Mathematics. 44. Cambridge: Univ. Press. xii, 343 p. (1995). ZBL0819.28004.

Motivation and technical results

I will start with some motivation for the construction, which will also serve as an opportunity for me to state some technical results I will need later. When looking for distributions that are not locally integrable, a natural starting point is to look at Radon measures (zeroth order distributions) which are compactly supported. For such measures $\mu$ the structure of their Fourier-Stieltjes transform gives a close connection to the fractal dimension of its support, which is (sort of) seen in the following result.

Lemma (12.12 from Mattila): Let $\mu$ be a compactly supported Radon measure on $\Bbb R.$ Then for $0<s<1$ there is a constant $c(s)>0$ such that
$$ I_s(\mu) = c(s)\int_{\Bbb R} |\xi|^{s-1} |\hat\mu(\xi)|^2 \,\mathrm{d}\xi,$$
where $I_s$ is the $s$-energy defined as
$$ I_s(\mu) = \int_{\Bbb R}\int_{\Bbb R} |x-y|^{-s} \,\mathrm{d}\mu(x) \,\mathrm{d}\mu(y). $$

Since $s-1<0,$ we have $(1+|\xi|^2)^{\frac{s-1}2} \leq |\xi|^{s-1}$ and so we evidently have $\mu \in H^{\frac{s-1}2}(\Bbb R)$ provided $I_s(\mu)<\infty.$ This recasts our problem into finding subsets with finite $s$-energy, but what exactly is this quantity? (And what does this have to do with fractal dimension?)

These $s$-energies can be used to define (see 8.4 in Mattila) the Riesz $s$-capacity of a set $A \subset \Bbb R$ as
$$ C_s(A) = \sup\left\{ \frac1{I_s(\mu)} : \mu \in \mathcal M(\Bbb R),\, \operatorname{supp} \mu \subset A,\, \mu(\Bbb R) = 1 \right\}.$$
That is, we consider $1/I_s(\mu)$ over all Radon probability measures supported on $A.$ Whether this quantity is non-zero or not is closely related to the fractal dimension of the associated set $A,$ and a consequence of the so-called Frostman's lemma we have the following.

Theorem (8.9 in Mattila): Let $A \subset \Bbb R$ be a Borel set and $s>0.$ If $C_s(A) = 0,$ then $\mathcal H^t(A) = 0$ for all $t>s.$

And thus we have reduced the problem of finding a $H^s$ function to finding a set of suitable fractal dimension. The idea is that if $A$ has Hausdorff dimension $s$ which is close but strictly less that $1,$ then it will lie in a small but negative $H^{-t}$ space but will not be represented by a locally integrable function. Using these results the construction follows fairly easily, and we also see a close connection between these two seemingly unrelated topics.

Main construction

I will show that for all $s \in (0,1/2),$ there exists a Radon measure $\mu \in H^{-s}(\Bbb R)$ which is compactly supported and singular with respect to the Lebesgue measure; that is it is not represented by a locally integrable function.

The construction follows by the two results, and is a matter of putting everything together and by exhibiting sets with the correct fractal dimension. This is where we need the Cantor-type construction, whose construction I will sketch from Section 4.10 in Mattila.

Fix $\lambda \in (0,1/2),$ and let $I_{0,1} = [0,1].$ We inductively define closed intervals $I_{k,1},\dots,I_{k,2^k}$ for each $k\geq 0$ by deleting the middle $(1-2\lambda)$ of each $I_{k-1,j}$ to obtain $I_{k,2j-1}, I_{k,2j}.$ To clarify the construction the first few terms are
\begin{align*}
  &I_{0,1} = [0,1] \\
  &I_{1,1} = [0,\lambda], \ \ I_{1,2} = [1-\lambda,1] \\
  &I_{2,1} = [0,\lambda^2], \ \ I_{2,2} = [\lambda(1-\lambda),\lambda], \ \ I_{2,3} = [(1-\lambda),\lambda(1-\lambda)], \ \ I_{2,4} = [(1-\lambda)^2,1],
\end{align*}
from which you can hopefully see the general pattern emerging. We then define $C^{\lambda}_k = \bigcup_{j=1}^{2^k} I_{k,j}$ and define
$$ C^{\lambda} = \bigcap_{k \geq 0} C^{\lambda}_k. $$

Evidently this is a closed subset of $[0,1].$ We wish to show this has Hausdorff dimension $s_{\lambda} = \frac{\log 2}{\log(1/\lambda)},$ for which note that $\operatorname{diam}(I_{k,j}) = \lambda^k$ for each $k,j$ so
$$ \mathcal H^{s_{\lambda}}(C^{\lambda}) \leq \lim_{k \to \infty}\mathcal H^{s_{\lambda}}_{\lambda^k}(C^{\lambda}_k) = \lim_{k \to \infty} 2^k\lambda^{ks_{\lambda}} = 1, $$
by choice of ${s_{\lambda}}.$ It is further shown in the text that $\mathcal H^{s_{\lambda}}(C^{\lambda}) \geq \frac14;$ it actually equals one but we just need that it is non-zero.

Hence as $\mathcal{H}^{s_{\lambda}}(C^{\lambda})>0,$ we have $C_s(C^{\lambda})) > 0$ for all $s > s_{\lambda}.$ That is, there exists a Radon probability measure $\mu$ supported on $C^{\lambda}$ such that $I_s(\mu) < \infty$ and so $\mu \in H^{\frac{s-1}2}(\Bbb R).$ Note that $s_{\lambda} \to 1$ as $\lambda \to \frac12,$ so we can find such measures in $H^{-t}$ for any $t>0.$ Note also that since $\mu$ is supported on $C^{\lambda}$ which is Lebesgue-null, it is necessarily singular with respect to the Lebesgue measure.

Remark: I believe it is possible, as suggested in the comments, to take $\mu(A) = \mathcal H^{s_{\lambda}}(A \cap C^{\lambda})$ (the uniform measure on $C^{\lambda}$). This lends itself to a more explicit representation, as one can take an analogous staircase function $f_{\lambda},$ whose derivative gives a singular measure coinciding with $\mu$ above. Checking this works is quite technical, but this would provide alternative ways to construct suitable examples.

Title: Sobolev space with negative index


问:
I need help finding the roots of a characteristic polynomial
I am trying to find the roots of characteristic polynomials.

This can be difficult to do by hand, especially when I have a characteristic polynomial of the 3rd or 4th degree.

I always try to factorise the polynomial, but it doesn't always work.

I start by guessing simple roots like 1,2,-1,-2...

I want help in finding the roots in a shorter period of time.

答:
You have the rational root  test.

If you have a polynomial of this type:

${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0}$
with integer coefficients ${\displaystyle a_{i}\in \mathbb {Z} }$.

The test tells you that each rational solution $x = \frac{p}{q}$, written in lowest terms so that p and q are relatively prime, satisfies:

p is an integer factor of the constant term $a_0$, and
q is an integer factor of the leading coefficient $a_n$.

All the rational (racionals and integers) roots are of this form.

In other words, if there is a rational root in can be written  like $x = \frac{p}{q}$ such that p divides $a_0$ and q divides $a_n$.

If you try all of these combinations of numerators and denominators and none of them is a root, then there are no roots in $\mathbb{Q}$.


问:
Complex-Orthogonal Similarity
First, a quick definition: I say that a matrix $M$ with complex entries is complex-orthogonal if $MM^T = I$.  Note that $T$ here refers to the transpose, not the conjugate transpose.  That is, we are taking the adjoint with respect to a bilinear (as opposed to sesquilinear) dot-product over $\Bbb C$.  A complex-orthogonal matrix need not be unitary, and vice versa (unless of its entries are in $\Bbb R$).

Consider the following claim:

Claim: For any $A \in \Bbb C^{n \times n}$, there exists a complex-orthogonal $U$ such that $UAU^T$ has equal diagonal entries.

My Question: Does this claim hold?

It may be helpful to take a look at this paper which proves that for any $A$, there is an $S$ such that $SAS^{-1}$ has equal diagonal entries.  It goes on to show that such an $S$ may be taken to be a unitary matrix.  Another proof of the same fact is presented in p. 77 of R.A. Horn and C.R. Johnson’s Matrix Analysis (1985/7, Cambridge Univ. Press).

Notably, both of these proofs exploit facts for which an analog fails to exist in my case.  For Horn and Johnson's proof, we need the fact that the unitary matrices form a compact subset of $\Bbb C^{n \times n}$.  For the proof linked, we need the numerical range to have certain "nice properties", which I believe fail in my case.

Any insight here is appreciated.

答:
An easier justification of ComptonScattering's counterexample: since $A=\pmatrix{i&1\\ 1&-i}$ is nilpotent and complex symmetric, if $UAU^T$ has equal diagonal entries, we must have
$$
UAU^T=a\pmatrix{0&1\\ 1&0}
$$
for some $a\in\mathbb C$. But then $0=U(A^2)U^T=(UAU^T)^2=a^2I$. In turn, we get $a=0$ and $A=0$, which is a contradiction.

Title: Complex-Orthogonal Similarity


答:
If $m:=|G|$ is finite, then $h^m=e$ for all $h\in G$. So with $k=m$ (not $m+1$), we conclude from the existence of $h$ with $h^k=g$ that $g=e$ (and so $m=1$).

Alternative approach:
Suppose $G$ is not trivial and let $1\ne a\in G$. If $a$ has inifnite order, $G$ is infintei and we are done. So assume $a^m=1$ for some $m>1$. Then $x\mapsto x^m$ is not injective, but by assumption is surjective. This is possible only for infinite sets.

Title: Showing that a group has infinite order


问:
Find the marginal PDF
I am having a hard time trying to find the marginal of the following joint pdf.

$$f_{U,V}(u,v) =2 \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1}$$
where $ u<v<\frac{u+\theta+1}{2}$ and $\theta < u <\theta+1$.

This was my attempt

$\begin{align*}
f_V (v) &= \int_{\theta}^{\theta + 1} \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
&=  \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\int_{\theta}^{\theta + 1}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
\end{align*}$

答:
The given joint density  looks like some joint density of some order statistic but is is clearly not a nice density.

To realize that $f_{UV}$ is wrong just set $n=2$ and $\theta=1$ getting

$$f_{UV}(u,v)=2\cdot\mathbb{1}_{(1;2)}(u)\cdot\mathbb{1}_{(u;1+u/2)}(v)$$

Where the integral over all the support is $2\cdot 0.25=0.5\ne 1$

Title: Find the marginal PDF


答:
Since $$lim_{z \rightarrow \frac{\pi}{2}} (z-\frac{\pi}{2})(\frac{1}{\cos(z)})=lim_{z \rightarrow \frac{\pi}{2}} \frac{1}{-\sin(z)} = -1$$
we know that $\frac{\pi}{2}$ is a pole of order 1 of our function $\frac{1}{cos(z)}$ with residue -1. Therefore we know that our Laurent series looks like
$$\frac{-1}{(z-\frac{\pi}{2})} + \sum_{i=0}^{+\infty}a_0(z-\frac{\pi}{2})^i$$
Since $\frac{1}{\cos(z)}\cos{z}=1$, we can determine the remaining a's out of the laurent series of $\cos(z)$.
$$\left(\frac{-1}{(z-\frac{\pi}{2})} + \sum_{i=0}^{+\infty}a_0(z-\frac{\pi}{2})^i\right)\left(-(z-\frac{\pi}{2})+\frac{1}{6}(z-\frac{\pi}{2})^3-\frac{1}{120}(z-\frac{\pi}{2})^5 + ...\right)=1$$
We find for example that $-a_0(z-\frac{\pi}{2})=0(z-\frac{\pi}{2})$. So $a_0=0$.
You can find the complete series here.</p>

Title: How to find the Laurent expansion for $1/\cos(z)$

Body: How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$</p>


问:
Let $f\in C([a,b])$ and $\int_{a}^{x}f=\int_{x}^{b}f \ \forall x\in[a,b]$. Show that $f=0$ on $[a,b]$.

Let $f\in C([a,b])$ and $\int_{a}^{x}f=\int_{x}^{b}f \ \forall x\in[a,b]$. Show that $f=0$ on $[a,b]$. I would like to know if my proof holds, please. My attempt is to pass by primitive of $f$.

As $f\in C([a,b])$, then it has a primitive such that $F'(x)=f$. Moreover, as $$\int_{a}^{x}f=\int_{x}^{b}f \ \forall x\in[a,b]$$ we have the following:

$$F(x)-F(a)=F(b)-F(x) \iff F(x)=\frac{F(b)+F(a)}{2}.$$

The previous equality holds $\forall x \in[a,b]$. Therefore $F$ is constant $\implies$ $f=0$ (as $F'(x)=f(x)$).

答:
Your proof is correct. It works even in the case that $f$ is “only” Lebesgue integrable on $[a, b]$, in that case the conclusion is that $f(x) = 0$ almost everywhere on $[a, b]$.

As an alternative one can observe that the integral vanishes over all subintervals $[c,d]\subset [a, b]$:
$$
\int_c^d f(x) \, dx = \int_a^d f(x) \, dx - \int_a^c f(x) \, dx \\
= \int_d^b f(x) \, dx - \int_c^b f(x) \, dx = -\int_c^d f(x) \, dx
$$
implies
$$
\int_c^d f(x) \, dx = 0 \, .
$$
If $f$ is continuous and not identically zero then it is strictly positive or strictly negative on some subinterval, and that gives a contradiction.


问:
Partial derivatives of $|Ax - b|^2$?
I'm trying to work out the partial derivatives of a function $L$ in terms of $x_i$:

$$ A \in \mathbb{R}^{m x n} \quad b \in \mathbb{R}^m \quad x \in \mathbb{R}^n $$

$$\begin{aligned} L(x) &= \left\|{Ax - b }\right\|^2 \\&= (Ax-b)^T(Ax-b) \\ &= x^TA^TAx - b^TAx - x^TA^Tb + b^Tb\end{aligned}$$

All four of these terms are scalars so I think I can transpose the third term to get:

$$\begin{aligned} &= x^TA^TAx - 2b^TAx + b^Tb\end{aligned}$$

I'm a little stuck on how to transform $x^TA^TAx$ further.

Calculating the partial derivatives, the final term $b^Tb$ is constant so goes to zero, the second term has coefficient vector $2b^TA$ so we just drop the $x$, its again the first term I'm stuck on:

$$\frac{\partial{L(x)}}{\partial x} = ??? -2b^TA$$

What's the derivative of $x^TA^TAx$ in terms of $x$?  How you work it out?

Update:

After looking at the potential duplicate, I think I'm mostly covered - however it's not immediately obvious to me why:

$$\frac {\partial(x^TMx)} {\partial x}=(M+M^T)x$$

How is this rule derived?

答:
We have $\frac{\partial}{\partial x} = \begin{bmatrix}
\frac{\partial}{\partial x_1} \\
\frac{\partial}{\partial x_2}\\
\ldots \\
\frac{\partial}{\partial x_n} \\
\end{bmatrix}$,

Let $M$ be a $n\times n$ matrix, then $\frac{\partial}{\partial x}(x^TMx)$ is a $n\times1$ vector. Consider the $k^{th}$ element of this vector

$\frac{\partial}{\partial x_k}(x^TMx)=\frac{\partial}{\partial x_k}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}x_im_{ij}x_j=\sum\limits_{i=1}^{n}x_im_{ik}+\sum\limits_{j=1}^{n}x_jm_{kj}$

$\implies \frac{\partial}{\partial x}(x^TMx) = Mx+M^Tx=(M+M^T)x$

Title: Partial derivatives of $|Ax - b|^2$?


问:
concentration inequality and how to bound densities of a sequence of random variables
Let $X_n$ be a sequence of random Variables such that $E(X_n)=0$ and $Var(X_n)=1$
I need to bound the following:

$$
P(|X_n|<\delta_n)\lesssim \delta_n
$$

for a positiv sequence $\delta_n$ tending to zero.
My question is under which asumptions this inequality holds.
Lets first assume that there is a density $f_{X_n}$ according to the lebesgue measure for each $X_n$. Then it follows
$$
P(|X_n|<\delta_n)=\int^{\delta_n}_{-\delta_n}f_{X_n}.
$$
Do i need additional asumptions to bound $f_{X_n}<c$ for all $n$ in order to bound the Integral by $2c\delta_n$ or is it somehow enough to assume that $Var(X_N)=1$?

答:
If i assume that $X_n$ is integrable, then i get by Chernoff (https://en.wikipedia.org/wiki/Concentration_inequality):
$$
P(|X_n|<\delta_n)\leq \delta_n E(|X_n|^{-1})
$$
and by the jensen's inequality i get for all $n$
$$
\delta_n E(|X_n|^{-1})\leq\delta_n E(X_n^{2})^{-\frac{1}{2}}
$$
and by $Var(X_n)=1$:
$$
P(|X_n|<\delta_n)\leq \delta_n
$$
Where did i make a mistake?


问:
Squarefree totient sum
Does anybody have a reference/proof for the asymptotic growth rate of
$$A(x) = \!\!\!\!\!\!\sum_{\substack{n \leqslant x \\ n \ \text{squarefree}}} \!\!\!\!\!\! \varphi(n)$$
as $x \to \infty$? Here $\varphi(n)$ denotes Euler's totient function. Numerical evidence suggests it grows like $cx^2$ where $c \approx 0.21$.

I tried using the identity
$$ \varphi(n) = n \sum_{d \mid n} \frac{\mu(d)}{d}$$
along with the fact that $\mu(n)^2 = 1$ iff $n$ is squarefree, and I got the expression
$$A(x) = \sum_{d \le x} \mu(d) \sum_{m \le x/d} m \mu(md)^2.$$
I realize that $md$ is squarefree iff $m$ and $d$ are both squarefree and coprime, but it feels like that just makes the problem harder. Any help would be appreciated!

Title: Squarefree totient sum

Body:
$$\sum_{n\ge 1} \mu(n)^2 \phi(n) n^{-s}=\prod_p (1+(p-1)p^{-s})=
\prod_p \frac{1-p^{-s}+(p-p^2) p^{-2s}}{1-p^{1-s}}
$$ $$=(\sum_{d\ge 1} d^{1-s})(\sum_{m\ge 1} a(m) m^{-s})$$ where $\sum_{m\ge 1} \frac{a(m)}{m^s}$ converges absolutely for $s> 3/2$, whence $\sum_{m\le x} |a(m)|=O(x^{5/3})$ and $$\sum_{n\le x} \mu(n)^2 \phi(n)=\sum_{n\le x} \sum_{dm=n} a(m) d=\sum_{m\le x} a(m) \sum_{d\le x/m} d$$ $$=\frac12\sum_{m\le x} a(m) (\lfloor x/m\rfloor^2+\lfloor x/m\rfloor)$$

$$=\frac12\sum_{m\le x/\log x} a(m) \frac{x^2}{m^2} (1+o(1))+
\sum_{x/\log x<m\le x} a(m) O(\log^2 x)$$
$$ = \frac{x^2}2\sum_{m\ge 1} \frac{a_m}{m^2}\ \ + o(x^2)$$

$\sum_{m\ge 1} \frac{a(m)}{m^2}=\prod_p (1-2p^{-2}+p^{-3})$ doesn't have a closed form.


问:
Calculus - Showing that Min does not exists
I have the set $A = \{f(x) : x > -4\}$, while $f(x)= \frac{x^2+12x+32}{x+5}$ for every $x>-4$. I want to show that $\min(A)$ does not exists.

My attempt:

I showed that: $\lim\limits_{x \to -4} f(x)=0$, so $A$ is bounded from below by $0$, and $0 \notin A$.

Now I want to show that $\inf(A)=0$, I tried using the epsilon property, and tried assuming towards contradiction, with no luck.

Any help will be amazing!

Thanks!

答:
$f(x)=\frac {(x+4)(x+8)} {x+5}$ so $f(x) >0$ for all $x >-4$ and $f(x) \to 0$ as $x \to -4$. This implies that the infimum of $f$ is $0$ and this infimum is not attained.

If you want to prove that the infimum  is $0$ using definition of infimum  you can do the following: Since $f(x) >0$ for all $x$ it follows that $0$ is a lower  bound. Let us show that no positive number is a lower bound. Let $0<\epsilon < 1$. Take $x=-4+\frac  {\epsilon } 5$. Then $f(x) < \frac {(\frac  {\epsilon } 5)(5)} {1}=\epsilon$ Hence, $\epsilon $ is not  a lower bound.  Obviously, and $\epsilon \geq 1$ also cannot be a lower bound.


问:
Polar integration confusion
The integral
$$
\int_{\phi = 0}^{2\pi}\int_{\theta = 0}^{\pi} (\vec{a}\cdot\hat{r})(\vec{b}\cdot\hat{r})\sin\theta\,d\theta\,d\phi
$$
$\vec{a}$ and $\vec{b}$ being constant vectors, equals $\frac{4\pi}{3}\vec{a}\cdot\vec{b}$ as proven here, but if we write $\vec{a} = a_r\hat{r} + a_\theta\hat{\theta} + a_\phi\hat{\phi} $ and $\vec{b}$ similarily, we can simply calculate it to be $4\pi a_rb_r$. Am I doing anything wrong or $\vec{a}\cdot\vec{b} = 3 a_rb_r$?

答:
Take the vector $(1,0)$ in 2D polar for simplicity's sake. At any point along the positive $x$ axis, this vector field is $+\hat{r}$. But on the negative $x$ axis, this would be $-\hat{r}$. Another example would be on the positive $y$ axis, where it would be $-\hat{\phi}$. Putting this information together, we have that

$$(1,0) = \hat{x} = \cos\phi \hat{r} -\sin\phi \hat{\phi}$$

which is no longer constant. You would have to integrate whatever function that may be.

Title: Polar integration confusion


问:
What is the optimal dimension $k$ for which $S^2\times S^2$ into $\mathbb{R}^k$?
Clearly $k\geq4$ since $S^2\times S^2$ is a 4-manifold and $k\leq6$ since $S^2$ embeds in $\mathbb{R}^3$.
Also, since $S^2\times S^2$ is compact, "invariance of domain" argument also show that $k$ cannot be 4.
Knowing the answer for the lower-dimensional case, (i.e., $S^1\times S^1\hookrightarrow \mathbb{R}^3$ is optimal) I tried to imitate the proof  for $S^2\times S^2\hookrightarrow\mathbb{R}^5$, but that didn't go very well.

My proof for $S^1\times S^1\hookrightarrow \mathbb{R}^3$: embed $S^1$ in $\mathbb{R}^3$ and take a tubular neighborhood, which is trivial since the orientable Euclidean rank 2 vector bundle over $S^1$ is unique: $[S^1,BSO(2)]=\pi_1(BSO(2))=\pi_2(SO(2))=\pi_2(S^1)=0$. The boundary of the unit disk bundle gives a desired embedded $S^1\times S^1$.

Problem with imitating the proof for $S^2\times S^2$: the isomorphism classes of orientable Euclidean rank 3 vector bundles over $S^2$ is not unique: we consider the bundle as the result of gluing two trivial vector bundles over two hemispheres $S^2_+$ and $S^2_-$.
The isomorphism class depends on "gluing" $S^1\to SO(3)$, and there are $2$ ways: $[S^1,SO(3)]=\pi_1(SO(3))=\mathbb{Z}_2$.
Thus, the tubular neighborhood might not be an honest product.

Is there a clever way to show that this tubular neighborhood is in fact trivial (or not)?
Or, is there a totally different way of finding the optimal dimension $k$ ($5$ or $6$)?

答:
Just by "brute-forcing" it: The usual embedding $S^1\times S^1\to\mathbb{R}^3$ is obtained by starting with $S^1=(x_1^2+x_2^2=R^2)$, and using polar coordinates $(\rho,\theta)$ on the other $S^1$ ($\rho=r$), with $R>r$ to avoid self-intersection to get $(\rho-r)^2+x_2^2=R^2$ on $\mathbb{R}^3$, $\rho^2=y_1^2+y_2^2$.  So it suggests you look at $(\rho-r)^2+x_2^2+x_3^3=R^2$ in $\mathbb{R}^5$ with coordinates $(y_1,y_2,y_3,x_2,x_3)$, where $\rho^2=y_1^2+y_2^2+y_3^2$ and again $R>r$.

In other words, the normal bundle to $S^2\hookrightarrow\mathbb{R}^3$ is trivial rank 1.  So $S^2\times\mathbb{R}\hookrightarrow\mathbb{R}^3$ and we have a trivial rank-3 bundle $S^2\times\mathbb{R}^3\hookrightarrow\mathbb{R}^5$.  Now pick a $S^2\subset\mathbb{R}^3$ on that second factor.

The upshot is a product of round spheres of any dimension can be (smoothly) realised as a hypersurface in Euclidean space.


问:
Why are we setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$ in exercise 1.9.a in Concrete Mathematics?
In the book Concrete Mathematics, exercise 1.9, one is given:

$P(n): x_1...x_n \leq (\frac{x_1 + ... + x_n}{n})^n$ if $x_1, ..., x_n \geq 0$.

and is asked the following question:

By setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$, prove that $P(n)$ implies $P(n-1)$ whenever $n > 1$.

The solution to this is given here: Backwards induction to show that $x_1\cdots x_n \leq ((x_1+\cdots+x_n)/n )^n$, but my question is:

Why are we setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$? This equation isn't true for all $x_i$ and all $n$, since taking $x_1 = 0$, $x_2 = 1$ and $n = 2$, the above equation results in $x_2 = x_1$. Aren't we trying to prove $P(n)$ in general, so for all all $x_i$ and all $n$?

EDIT: to be clear, my question is not on exercise 1.9.b, nor on 1.9.c, but as stated, on 1.9.a. More precisely, I don't understand how one is allowed to conclude the general statement of $P(n)$ implies $P(n-1)$ by choosing a particular value of $x_n$.

答:
You have left out an important part. The point of the exercise is that it is being shown that $P(n)\implies P(2n)$. This shows $P(2^n)$ for $n\in\mathbb{N}$, but misses $P(k)$ for $2^{n-1}\lt k\lt2^n$. Showing $P(k)\implies P(k-1)$ fills this gap.

From the Comments

We are given $P(n)$. This works for all non-negative values of $x_1,x_2,x_3,\dots,x_n$. We are choosing $x_n=(x_1+x_2+x_3+\dots+x_{n-1})/(n-1)$. Apply $P(n)$ which works for all values of $x_1,x_2,x_3,\dots,x_n$. Then we see that this implies $P(n-1)$.


问:
What is the difference in accuracy in answers from Definite Integration and Simpson's rule?
I have worked out the same equation with the Definite Integration and Simpson's Rule. I now need to comment on the accuracy of these answers. I know Simpson's rule is an approximation, but does that make the answer from the Definite Integration a $100$% accurate? or is it just a more accurate estimation?

答: If you have been able to find the definite integral, then that is the exact area required. You can then find the error in your Simpson's Rule approximation by working out the difference between your two values. The relative error is given by dividing the error by the true value; it can be expressed as a percentage or as a decimal.</p>


问:
How do you prove $37^{100} - 37^{20}$ is a multiple of $10$ using induction?
I've tried making it $37^{5n} - 37^n$ is a multiple of $10.$ Then I made the base case be $n = 0,$ so $1 - 1 = 0$ which is a multiple of $10.$ I assumed $37^{5n} - 37^n$ is a multiple of $10$ for all $n \geqslant0.$ The inductive step is really hard however. I did $37^{5(n + 1)} - 37^{n+1}$ which I simplified to $37^{5n + 5} - 37^{n+1}.$ I tried factoring and made it $37^5\cdot37^{5n} - 37\cdot(37^n).$ I don't know where to go from here though. </p>

答:
= 37^100 − 37^20    ---------> (i)

As 37 and 100 are co-prime numbers so by Euler's theorem Euler's theorem Wikipedia 37^100 ≡ 1

and 37 and 20 are co-prime, so similarly 37^20 ≡ 1.

So equ.(i) can be written as:

≡  1 - 1

≡  0

which is multiple of 10.


答:
You already know that $$\forall x>-4:\frac{x^2+12x+32}{x+5}>0$$ so that $0$ is a lower bound.

Now take $\epsilon>0$ and solve

$$\frac{x^2+12x+32}{x+5}=\epsilon$$

or, with $t:=x+4$,

$$t\frac{t+4}{t+1}=\epsilon.$$

It seems clear that, as the fraction is continuously growing from $1$, the equation always has a solution and $\epsilon$ cannot be the minimum.

Title: Calculus - Showing that Min does not exists

Body:
I have the set $A = \{f(x) : x > -4\}$, while $f(x)= \frac{x^2+12x+32}{x+5}$ for every $x>-4$. I want to show that $\min(A)$ does not exists.

My attempt:

I showed that: $\lim\limits_{x \to -4} f(x)=0$, so $A$ is bounded from below by $0$, and $0 \notin A$.

Now I want to show that $\inf(A)=0$, I tried using the epsilon property, and tried assuming towards contradiction, with no luck.

Any help will be amazing!

Thanks!


问:
Squeeze theorem 2 bounded functions
$$\lim_{(x,y) \to (0,0)} \frac{x^2 \sin (y)}{x^2+y^2} $$

Squeeze Theorem:

$$g(x) \leq f(x) \leq h(x)$$

$$-1 \leq \sin(y) \leq 1$$

$$- \frac{x^2}{x^2+y^2}\leq \frac{x^2 \sin (y)}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} $$
Also
$$ 0 \leq \frac{x^2}{x^2+y^2} \leq 1 $$
What should be my next step
How does this comes out to be zero

答:
$-1 \leq \sin y \leq 1$ is not helpful in this case.

Use the fact that $|\sin y | \leq |y|$ so $-|y| \leq \sin y \leq |y|$. Now you get the bounds $\pm \frac  {x^{2}|y|} {x^{2}+y^{2}}$. If you note that $\frac  {x^{2}} {x^{2}+y^{2}} \leq 1$ you get $-|y| \leq f(x,y) \leq |y|$ and you can now apply Squeeze Theorem.

Title: Squeeze theorem 2 bounded functions


问:
Center of a nilpotent group
I'm trying to solve an exercise from Humphreys' Linear algebraic groups. It asks to show that center Z of a nilpotent algebraic group G has positive dimension if G has positive dimension.

I'm trying to consider the lower central series of G, say $C_{0}G=G \supset C_{1}G \supset ....C_{n-1}G\supset C_n{G}=\{e\} $. By definition of being nilpotent, we get that $C_{n-1}G\subset Z$ and $C_{n-1}G\neq \{e\}$. I'm stuck on how to proceed further.

There are 2 things which might be useful: $C_{n-1}G$ is closed and normal in G and if G is connected then the center Z has positive dimension. Any help is appreciated.

答: Choose the biggest $i$, s.t. $C^{i}G$ is not discrete, then $C^{i}(G)^{0}$ has positive dimension, and $C^{i+1}(G)$ is discrete. Now take any element $g\in G$, consider the map $C^{i}(G)^{0}\longrightarrow C^{i+1}(G)$, $x\longrightarrow [x,g]$.

Title: Center of a nilpotent group


问:
Direct/Lagrange to show max/min
The domain Is defined by the half unit circle: $\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \}$

for the function $f:D\rightarrow \mathbb{R}$. The function is $f(x,y)=2xy^2-2x^2$. I have shown that a min/max must exists since the function is continuous and the domain is both bounded and closed. I do not know how to proceed to find these points where the function is equal to its min/max. I tried with the Lagrange multiplier but it is just too messy for me to do. How would one do it directly? I found that the gradient vanishes at $(0,0)$

答:
We may start by saying something about the function  $ \ f(x,y) \ = \ 2xy^2 \ - \ 2x^2 \ = \ 2x · (y^2 - x) \ . $  Because the coordinate variable $ \ y \ $ appears in only one term with an even exponent, the value of the function is unaltered by a sign-change of $ \  y \ $; in other words, the function has symmetry about the $ \ x-$ axis.  There is no symmetry for the variable $ \ x \ $ since it appears with both even and odd exponents.  Further, we can say that $ \ f(0,y) \ = \ 0 \ $ and $ \ f(y^2,y) \ = \ 0 \ . $  This divides the plane into four regions, with $ f(x,y) \ < \ 0  \ $ for $ \ x \ < \ 0 \ $ and $ \ x \ > \ y^2 \ $ , and $ f(x,y) \ > \ 0  \ $ otherwise.  The graph below presents the situation, with the unit circle marked in red; this will be useful in the discussion to come.

You found from the first partial derivative equations $ \ f_x \  = \ 2y^2 \ - \ 4x \ = \ 0 \ $ and $ \ f_y \ = \ 4xy \ = \ 0 \ $ that there is a single critical point for the function at the origin.  We see, however, that it sits at the junction of alternating regions of positive and negative function values, so $ \ (0,0) \ $ is a saddle point.

You presumably found the Lagrange equations

$$ 2y^2 \ - \ 4x \ \ = \ \ \lambda·2x \ \ \Rightarrow \ \ y^2 \ = \ (\lambda \ + \ 2)·x \ \ \ , \ \ \ 4xy \ \ = \ \ \lambda·2y \ \ \Rightarrow \ \ y·(2x \ - \ \lambda) \ \ = \ \ 0 \ \ .  $$

It is best to deal with the second equation first, which produces the two cases $ \ y \ = \ 0 \ $ and $ \ \lambda \ = \ 2x \ \ : $

$ \mathbf{y = 0 :  } \quad  x^2 \ + \ 0^2 \ = \ 1 \ \ \Rightarrow \ \ x \  =  \ \pm 1 \ \ \Rightarrow \ \ f(\pm 1 \ , \ 0) \ = \ 2·(\pm 1)·(0^2  -  [\pm 1] \ ) \ = \ -2 \ \ , $

which are the absolute minima on the unit disk (for the condition $ \ x \ \ge \ 0 \ $ , we can discard the result for $ \ x  =  -1 \ $ ) ;

$ \mathbf{\lambda = 2x :  } \quad  y^2 \ = \ (2x \ + \ 2)·x \ = \ 2x^2 \ + \ 2x \ \ , $ which upon insertion into the constraint equation produces

$ x^2 \ + \ (2x^2 \ + \ 2x) \ = \ 3x^2 \ + \ 2x \ - \ 1 \ = \ (3x \ - \ 1)·(x \ + \ 1) \ = \ 0 \ \ . $  The root $ \ x \ = \ -1 \ $ tells us nothing new; the other root yields

$$ x \ = \ \frac13 \ \ \Rightarrow \ \ \left( \frac13 \right)^2  + \ y^2 \ = \ 1 \ \ \Rightarrow \ \ y^2 \ = \ \frac89 \ \ \Rightarrow \ \ y \ = \ \pm \frac{2 \sqrt{2}}{3} $$

$$ \Rightarrow \ \ f \left(\frac13 \ , \ \pm \frac{2 \sqrt{2}}{3} \right) \ = \ 2·\frac13  ·(\frac89  -  \frac13  ) \ = \ 2·\frac13  ·\frac59 \ = \ \frac{10}{27} \ \ , $$

which are the absolute maxima. [The parabola $ \ y^2 \ = \ x \ $ intersects the unit circle at the points found from $ \ x^2 \ + \ x \ = \ 1 \ \Rightarrow   \ x \ = \ \frac{-1 \ + \ \sqrt{1^2 \ - \ 4·1·(-1)}}{2} \ = \frac{-1 \ + \ \sqrt{5}}{2} \ \approx \ 0.618 \ , $ confirming that the maximal points we obtained are in the positive-value regions of $ \ f(x,y) \ \ . $  ]

Title: Direct/Lagrange to show max/min


问:
How the Sobolev embedding used in this case
Assume $\{u_k\}\subset H^1(S)$ and
$$u_k \rightharpoonup u_0 \quad in \quad H^1(S), $$ where “$\rightharpoonup$” means “converges weakly”, $S$ be the unit circle parameterized by angle $\theta$ and $H^1(S)$ means Hilbert space equipped with norm $$\|u\|=\left\{\int_S[(u^{\prime})^2+u^2]d\theta\right\}^{\frac{1}{2}}.$$
Then by Sobolev imbedding, we have
$$u_k\rightarrow u_0 \quad in \quad C^{\beta}, \forall \beta<\frac{1}{2}.$$
My question is how the Sobolev Embedding Theorem used in this step?

The Sobolev Imbedding Theorem is equal to Sobolev Embedding Theorem ?(This answer is in the comment)

Why the $\beta <\frac{1}{2}$ ? Thanks in advance!

答:
If $\{u_n\}$ converges weakly to $u$ in the $H^1-$norm, then it converges strongly in the $H^s-$norm, for every $s<1$, $s\in\mathbb R$.

So, if $\beta<1/2$, then there exists an $s<1$, such that $ H^s(S)\subset C^\beta(S)$, i.e.
$$
s=\beta+½.
$$


问:
prove for bell number using induction on n
hi guys I have to prove this equality $$B_n=e^{-1}\sum_{k=0}^{\infty}\frac{k^n}{k!},$$ that is called bell equality only using induction on $n$ . How can i do this? I have tried by  substituting the recursive formula  $\sum\limits_{k=0}^{n} \binom{n}{k} B_{k}$ in the first one. But I am completely lost with indexes.
I have already tried to search some informations on the internet but nothing using induction

答:
Using the Bell´s formula we have that $$B(n)=\sum\limits_{k=0}^nS(n,k).$$
This is the total number of ways to put $n$ balls in an arbitrary number of
boxes (no empty boxes remaining). To count them we look at the number of balls (at this parameter
we will call it $a$, and it can be from $0$ to $(n - 1)$) that accompany any ball (for example ball
number $1$) on your box. To do this:

First we choose the $a$ companions of ball $1$ from among the $(n - 1)$ possible (which can be done in $\binom{n-1}{a}$ different ways).

Then, for each way of selecting the companions, we count how many ways they can be entered
the remaining $(n − 1 − a)$ balls in an arbitrary number of boxes (we have called this $B(n−1−a)$)

$$\boxed{B(n)}=\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}}\sum\limits_{k=1}^{n-1-a}{S(n-1-a)}=\boxed{\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}B(n-1-a)}}$$

Now, if we define the $DOB$ function this way: $$DOB(n)=\sum\limits_{k=1}^{\infty}\dfrac{k^n}{n!},$$ we can see that

\begin{align*}
DOB(0)&=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots\\
DOB(1)&=\phantom{OO}\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}+\dfrac{4}{4!}+\cdots\\
DOB(2)&=\phantom{OO}\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+\cdots\\
DOB(3)&=\phantom{OO}\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots
\end{align*}

By combining these numbers we get that

\begin{gather*}
\color{red}{1}DOB(0)+\color{red}{2}DOB(1)+\color{red}{1}DOB(2)=1+\dfrac{4}{1!}+\dfrac{9}{2!}+\dfrac{27}{3!}\cdots=\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots=DOB(3)\\
\color{red}{1}DOB(0)+\color{red}{3}DOB(1)+\color{red}{3}DOB(2)+\color{red}{1}DOB(3)=1+\dfrac{8}{1!}+\dfrac{27}{2!}+\dfrac{64}{3!}\cdots=\dfrac{1^4}{1!}+\dfrac{2^4}{2!}+\dfrac{3^4}{3!}+\dfrac{4^4}{4!}+\cdots=DOB(4)
\end{gather*}

What Dobinsky wanted to prove is that both sequences fulfill the same recurrence, and that have the following property:

$$DOB(n)=\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB}(j)$$

To prove this just use the binomial Theorem:

\begin{align*}
 \boxed{\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB(j)}}=&\ e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}DOB(j)}=e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{n!}}}\\
 =&\ e+\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}\underbrace{\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}k^j}}_{=(k+1)^{n-1}-1}=\ e-\underbrace{\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}}_{=e-1}+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k}}\\
 =&1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k!}=1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^n}{(k+1)!}}=\sum\limits_{k=1}^{\infty}{\dfrac{j^n}{j!}}}=\boxed{DOB(j)}
\end{align*}

This tells us that:
\begin{equation}\label{Formula_Dobinsky}
eB(n)=DOB(n) \rightarrow \boxed{B(n)=\dfrac{1}{e}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{k!}}}\quad \text{for each }n=1,2,3\ldots
\end{equation}

This formula is way more friendly and computerwise more stable. To see that just check that

\begin{equation*}
B(10)=115975;\qquad\dfrac{1}{e}\sum\limits_{k=1}^{15}{\dfrac{k^{10}}{k!}}=115974.978
\end{equation*}


答:
I would use polar coordinates and equivalence of functions:
$$\frac{x^2 \sin (y)}{x^2+y^2}=\frac{r^2\cos^2\theta\sin(r\sin\theta)}{r^2}\sim_{r\to 0}r\sin\theta\cos^2\theta.$$

Title: Squeeze theorem 2 bounded functions

Body:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 \sin (y)}{x^2+y^2} $$

Squeeze Theorem:

$$g(x) \leq f(x) \leq h(x)$$

$$-1 \leq \sin(y) \leq 1$$

$$- \frac{x^2}{x^2+y^2}\leq \frac{x^2 \sin (y)}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} $$
Also
$$ 0 \leq \frac{x^2}{x^2+y^2} \leq 1 $$
What should be my next step
How does this comes out to be zero


问:
Can someone explain how to get the minimal polynomial for this matrix?
This is example 8.45 from Sheldon Axler's Linear Algebra Done Right. $T$
is the operator on $\mathbb{C}^5$ whose matrix with respect to the standard basis is:\begin{matrix}
0 & 0 & 0 & 0 & -3\\
1 & 0 & 0 & 0 & 6\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0
\end{matrix}
Find the minimal polynomial of $T$.
The solution is supposed to be $z^5-6z+3$ but I don't understand the steps to get to this.

答:
Without loss of specificity (i.e., more generally) assume the final column has entries $c_0,\ldots,c_4$; here is how one can show that the minimal polynomial is $X^5-c_4X^4-c_3X^3-c_2X^2-c_1X-c_0$. Since the first $4$ columns of the matrix are equal to the last $4$ columns of the identity matrix $I_5$, in other words column $j$ equals the standard basis vector $e_{j+1}$ for $j=1,2,3,4$, one has (by induction if you will) $T^j(e_1)=e_{j+1}$ for those $j$ (and also for $j=0$), and the final column gives $T^5(e_1)=T(e_5)=c_0e_1+c_2e_2++c_2e_3+c_3e_4+c_4e_5$. [This is a nice point to remark that the habit of numbering things, like standard basis vectors and columns, from $1$ rather than $0$ is unfortunate.]

Now consider annihilating polynomials $P$ of $T$, namely those for which $P[T]=0$. A necessary condition for this (i.e., a weaker condition) is that $P[T](e_1)=0$. If $P$ were a nonzero polynomal of degree${}<5$, then this gives a nontrivial relation between $T^j(e_1)$ for $j=0,1,2,3,4$, but those are the standard basis vectors $e_1,e_2,e_3,e_4,e_5$ which are linearly independent; this cannot happen. On the other hand we do have (by the computation of $T^5(e_1)$) that $P[T](e_1)=0$ for $P=X^5-c_4X^4-c_3X^3-c_2X^2-c_1X-c_0$, and for no other monic polynomial of degree$~5$. So we know that the minimal polynomial of $T$ is certainly not of degree${}<5$, and if it is of degree$~5$ then it must be the mentioned polynomial.

It remains to show that $P[T]=0$ for that polynomial, in other words that $P[T](v)=0$ for any $v\in\Bbb C^5$, not just for $v=e_1$. Since the vectors $T^j(e_1)$ for $j=0,1,2,3,4$ span $\Bbb C^5$, there is a (unique) polynomial $Q$ of degree${}<5$ such that $v=Q[T](e_1)$. Then
$$P[T](v)=P[T](Q[T](e_1)) =(PQ)[T](e_1) =Q[T](P[T](e_1)) =Q[T](0)=0,$$ since $Q[T]$ is a linear map. QED.


答:
Show that $f'(x)\gt 0 \;\;\forall x\gt -4$. It follows that $f(x)\gt 0$ for all $x\gt -4$. Hence, $0$ is a lower bound for $A$.

Claim: $0$ is $\inf A$.

Suppose not and let $t\gt 0$ be a lower bound for $A$. It follows that $t\le f(x)\;\forall x\gt -4$.

Note that $\lim_{x\to -4^+} f(x)=0$ and therefore, for every $\epsilon \gt 0, \exists \delta_\epsilon\gt 0: -4\lt x\lt -4+\delta_\epsilon \implies f(x)\lt \epsilon$.

Choose $\epsilon=t$ and get a contradiction. Therefore, $0$ is $\inf A$.

To prove that $0$ is not attained, solve $f(x)=0$ and show that there is no $x\in \mathbb R$ that satisfies the equation.

PS: $f'(x)\gt 0$ is not really required here. Note that $N^r$ and denominator of $f$ are always positive for $x\gt -4\implies f(x)\gt 0 \;\; \forall x\gt -4$.


问:
In a random sequence of N letters, what is the expected number of occurrences of a specific M-letter sequence?
In a random sequence of N letters, what is the expected number of occurrences of a specific M-letter sequence?

Edit:

As a specific example, in a single-stranded DNA having N letters, what is the expected number of occurrences of the letter "GG"?

答:
If $X$ is your random variable (number of occurrences), then you can write it as:

$$X=X_1+X_2+\ldots+X_{N-M+1}$$

where each $X_i$ is the indicator:

$$X_i=\begin{cases}1&\text{there is a desired sequence starting at index }i\\0&\text{there isn't}\end{cases}$$

Now, for all $i=1,2,\ldots,N-M+1$:

$$E(X_i)=\mathcal P(\text{there is a desired sequence starting at index }i)=p$$

where $p$ depends on the exact probabilities of different symbols. If you have $k$ symbols, equally probable, then $p=\frac{1}{k^M}$. In your DNA example, $p=\frac{1}{4^2}=\frac{1}{16}$.

By linearity of expectation:

$$E(X)=\sum_{i=1}^{N-M+1} E(X_i)=(N-M+1)p$$

which is a curious result, as one would intuitively expect that it would hold if the individual random variables were independent (and we know that $X_i$ are definitely not independent). However, linearity of expectation is satisfied even for random variables that are not independent.

Answer: Applying the above formula for $p=\frac{1}{16}$ and $M=2$, we get that the expected number of DNA pairs GG in sequence of length $N$ is $\frac{1}{16}(N-1)$.

Note: For avoidance of confusion, just one more remark on counting. Let's take the DNA example with, say $N=4$, and let the sequence be GGGA. That sequence has two occurrences of GG, at positions $1$ and $2$, respectively, and that is how it went into the above calculation. If we were counting something else (e.g. sequences of G's of length $\ge 2$), then the calculation would be completely different.


问:
What is the combinatorial interpretation for this expression$ \sum_{i=0}^{n} (-1)^{i} \binom{n}{i} = 0$
What is the combinatorial interpretation for this expression  $$\sum_{i=0}^{n} (-1)^{i} \binom{n}{i} = 0\:?$$

答:
This says that there are the same number of ways to choose 1) an even number of things, or 2) an odd number of things, from a set of $n$ distinct objects.

Case I) This is easy to see, by symmetry, if $n$ is odd.

Case II) If $n$ is even, can you see an argument for it? Spoiler below.

 Fix one arbitrary object ("the first object"), and take two cases. Case 1: Choose the first object. Now, an odd number of $n-1$ objects remain, so follow Case I). Case 2: Do not choose the first object. Then follow Case I) again.


问:
How to calculate the number of decimal digits for a binary number?
I was going to ask this on Stack Overflow, but finally decided this was more math than programming. I may still turn out to be wrong about that, but...

Given a number represented in binary, it's fairly easy to derive a decimal representation using integer division and remainders to extract digits in reverse order. You can then simply count how many digits you extracted.

However, I wondered about calculating an exact number of decimal digits more efficiently by avoiding calculating the actual digits.

At first sight this is easy. First, count the binary digits needed (ie determine the position of the highest set bit). Many processors have an operation to do this directly, and there are bit-fiddling algorithms to do it in O(log n) time where n varies with the width of a machine register. In an arbitrary position integer, you normally know how many machine words you have, so you can usually jump directly to the most significant word.

Once you know the size in binary digits, you can scale by $\log 2 \over \log 10$ to get the number of decimal digits. Sort of.

The problem is that this scale factor is (I think) an irrational number. If you have a maximum number of digits you need to worry about, you can use a rational approximation and you only need to worry about getting the rounding right.

The question is, therefore - is there a way to determine this number of decimal digits efficiently and precisely for numbers of any size? Is there (for example) a completely different approach to the calculation that I haven't thought of?

答:
With no better answer, I'll just add a little to what I said in the question to get this closed.

There are series calculations for logarithms (http://en.wikipedia.org/wiki/Logarithm) so I could in principle calculate the base 10 log directly. This would be awkward and slow, though, as for arbitrary precision integers a machine float/double isn't precise enough. I'd probably need a rational library, and the rationals involved wouldn't be simple.

It may be possible to use an imprecise base-converting factor. With an underestimate factor, you'd get an underestimate log value. Divide through by 10^x for that log value, then repeat as often as necessary to top up your underestimate.

The problem is that dividing through by 10^x isn't trivial unless you happen to be working in base 10.

Probably best to stick with repeated division by 10 - or powers of 10, at least. Constructing a list of powers (10^1, 10^2, 10^4, 10^8 and so on) by squaring wouldn't be too hard. Stop when you exceed the target number, then do the divisions largest to smallest. Each power should only need to be considered once.

Even then, it may be better to not build powers of 10 that won't fit in a machine word, and just do repeated division by (small powers of) 10.

Also, this sounds now more like programming than math, so sorry about that.

ContentLicense: CC BY-SA 2.5


问:
How to prove the following O-notation?
$f(n)=O(g(n))$ and $f(n),g(n)>0$ for $n ≥ 1$, then there is a
constant c such that $0 ≤ f(n) ≤ c·g(n)$ for $n ≥ 1$

答: $f(n)=O(g(n))$ means that $(\frac{f(n)}{g(n)})$ is a bounded sequence.

Title: How to prove the following O-notation?


问:
End extension example for complete theory
I elaborate my previous question about example of complete theory,  that has two models, $\mathfrak{A}$ and $\mathfrak{B}$, that $\mathfrak{A} \le_{end} \mathfrak{B}$, but $\mathfrak{A} \npreceq \mathfrak{B}$.

Could you prompt me,  now,  such example with predicate symbol in its signature?  Or,  maybe with function. Cause I need to do some task(I will do it myself:),  about checking of keeping truth over end extensions for non sigma - formulas,  so I need some theories,  where I would check it for formulas with $\forall$ quantifiers, in way,  it would be more convenient.

Update: And, if it possible, can it be theories on some intervals, like (0,3) or something like that? Initially I wanted to take theory of dense linear order without endpoints over field of rationals with only multiplication and without addition, but I ain't sure, that it will be right.

答:
The answer to your original question ($\mathfrak{B}=$ $\mathbb{N}$ with the reversed order as an end extension of $\mathfrak{A}_m=\mathbb{N}_{>m}$ with the reversed order, for your favorite $m\in\mathbb{N}$) can be modified in various ways. Specifically, say that a $k$-ary relation $R$ on $\mathbb{N}$ is $m$-shift invariant iff we have $R(a_1+m,...,a_k+m)\iff R(a_1,...,a_k)$, and say that a $k$-ary function $f$ on $\mathbb{N}$ is $m$-shift invariant iff we have $f(a_1+m,..., a_k+m)=f(a_1,...,a_k)+m$. Given any $m$-shift invariant relations and functions, we have that the expansion of $\mathfrak{B}$ by those relations and functions is an elementary equivalent end extension of the expansion of $\mathfrak{A}_m$ by (the restrictions of) those same relations and functions, while of course this extension is not elementary. (Shift invariance comes in when we argue for elementary equivalence - in fact, the relevant structures are isomorphic, via the "shift-by-$m$" map.)

For example, the unary relation naming the even numbers is $m$-shift invariant for every even $m$, while the function $a*b=\max\{a,b\}+\vert b-a\vert$ is $m$-shift invariant for every $m$ whatsoever.

I suspect this sort of minor tweak won't help you too much, but without knowing more about your particular goal it's not clear where to look for a better example. (If you're familiar with for


问:
Calculating mixed strategy of $3 \times 3$ game

The Question

We consider the following zero-sum strategic game in matrix form

\begin{array}{c|lcr}
 & \text{A} & \text{B} & \text{C} \\
\hline
A & 0 & +\epsilon & -\delta \\
B & -\epsilon & 0 & 0 \\
C & \delta & 0 & 0
\end{array}

where $\epsilon$ and $\delta$ are nonnegative real numbers.

(a) Find all equilibria for this game (show your working) assuming $\epsilon \gt 0$ and $\delta \gt 0$.

(b) Assume some probabilities for playing $A$ or playing $B$, both for focal player and opponent. Calculate the expected payoffs for each strategy.

My Understanding

I am completely stuck as I seem to calculate that the strategy $(C,C)$ is a saddle point but then if we reduce it to the $2 \times 2$ game $(A,B)$ then $(A,A)$ is a saddle point therefore there can't be a mixed strategy correct?

答:
The game is symmetric (i.e. the payoff matrix is skew-symmetric) so you know its value must be $\ 0\ $. Therefore any optimal mixed strategy $\ \big(p_1,p_2,p_3\big)\ $ for the second player must guarantee that the expected payoff to the first player be non-positive. It must therefore satisfy the inequalities
\begin{align}
&\epsilon p_2-\delta p_3&\le0\\
-\epsilon p_1&&\le 0\\
\delta p_1&&\le0\\
&p_i\ge0&\text{for }\ i=1,2,3,
\end{align}
and the equation
$$
p_1+p_2+p_3=1\ .
$$
The second and third inequalities imply that $\ p_1=0\ $, while the equation and the first inequality give
$$
0\ge\epsilon p_2-\delta p_3=(\epsilon+\delta)p_2-\delta\ \ \text{, or}\\
0\le p_2\le\frac{\delta}{\epsilon+\delta}\ .
$$
Conversely, if $\ p_2\ $ satisfies this final pair of inequalities, $\ p_3=1-p_2\ $, and $\ p_1=0\ $, then all six inequalities and the equation are satisfied, so $\ \big(0,p_2,p_3\big)\ $ is an optimal strategy.

Therefore, a mixed strategy $\ \big(p_1,p_2,p_3\big)\ $  is optimal if and only if:

$\ p_1=0\ $,

$\ 0\le p_2\le\frac{\delta}{\epsilon+\delta}\ $, and

$\ p_3=1-p_2\ $.


答:
I assume you mean $f,g:\Bbb N\to\Bbb R$. Consider $N$ and $c'$ coming from the definition of $f\in O(g)$.

$c'$ covers all the values $n\ge N$. For $1\le n<N$, call $c_n=\frac{f(n)+1}{g(n)}$. Then, $c=\max\{c',c_1,c_2,\cdots, c_{N-1}\}$ works for all $n\ge1$.

On the other hand, for $f\ge0$ and $g>0$ your condition clearly implies $f\in O(g)$.

Title: How to prove the following O-notation?

Body:
$f(n)=O(g(n))$ and $f(n),g(n)>0$ for $n ≥ 1$, then there is a
constant c such that $0 ≤ f(n) ≤ c·g(n)$ for $n ≥ 1$


问:
Link between deformation and velocity gradient: $L_v = \dot{M}_t \cdot M_t^{-1}$
I'm studying fluid dynamics and I'm really struggling with notations and passages in some proofs that simply vanish.

In this question I need some guide-lines to show how to link the defomation map of a fluid and the velocity gradient: Let $B_0$ be a body that after a certain amount of time $t$ takes a different shape $B_t$. Supposing the change of shape is smooth enought into time, I used to call $M_t$ the deformation map which takes $B_0 \to B_t$.

Now, a part the clarifications about eulerian and lagrangian descripition of motion that is almost clear to me, I struggling to prove a theorem that states:
$$L_v = \dot{M}_t \cdot M_t^{-1}$$
where $L_v$ is used ad a notation of the gradient of velocity (which one: the eulero's or lagrange's one?) and the dot operator is the convective/material derivative.

Can anyone help me to dissolve my doubts and give me a sketch of proof (in the case the theorem I stated is true)?

答:
$M_t=\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}$, so
$$
\begin{split}
\dot{M}_t M_t^{-1} = \left[\frac{D}{\mathrm{d}t}\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]\left[\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]^{-1}\\
=\left[\frac{\partial\mathbf{v}(t)}{\partial\mathbf{x}(0)}\right]\cdot\left[\frac{\partial\mathbf{x}(0)}{\partial\mathbf{x}(t)}\right]=\frac{\partial\mathbf{v}(t)}{\partial\mathbf{x}(t)}=L_\mathbf{v}
\end{split}
$$
This is an Eulerian quantity because we are taking derivative with respect to the current $\mathbf{x}(t)$ not a fixed reference $\mathbf{x}(0)$.


问:
ODE Euler approximation converging to exact solution
I have this ODE: $dy/dt = -y$, $y(0)=10$.
How can I prove that its Euler numerical solution converges to the exact solution?

This is what I have attempted:
The Euler's scheme for this would be: $y_n = y_{n-1}(1 - \Delta_t) = y_0(1 - \Delta_t)^n$.
As $\Delta_t$ goes to $0$, $y_n$ would go to $y_0$, which is 10.

The exact solution would be $y(t)=10e^{-t} = 10e^{-n\Delta_t}$. As $\Delta_t$ goes to $0$, t goes to $0$ (I don't know if this is valid), and thus the solution converges to 10 as well.

I am really unsure about what I have done. Can someone help me with this? Thank you!

答:
You need to increase $n$ so that $nΔt=T$ remains constant, as you want to compare the numerical appriximations at the same time point. Then you get the more familiar limit
$$
\lim_{n\to\infty}\left(1-\frac{T}{n}\right)^n.
$$


问:
Finding zeros of a function
I need to find when $f(x)=0$, where:
\begin{equation}
f(x)=1-\dfrac{k}{x}-\dfrac{k}{3}\dfrac{e^{-ax}}{x}+\dfrac{4k}{3}\dfrac{e^{-bx}}{x}
\end{equation}
Here, $k$, $a$ and $b$ are positive constants. I think this problem analytically is impossible, but I really don't know. I've tried to use Mathematica but it doesn't give any answer and I'm pretty desperate.

Thank you so much.

答:
I prefer to add another answer for better notations.

Let $$b=\lambda\, a \qquad \text{and} \qquad t=2\,\frac{ a\, k \,(4 \lambda -1)-3}{a^2 \,k \,\left(4 \lambda ^2-1\right)}$$
Now
$$x=\sum_{n=1}^\infty \Bigg[\frac a {4 \lambda ^2-1}\Bigg]^{n-1}\, \frac{P_{n}(\lambda)}{c_n}\,\, t^n$$ where the first $c_n$ are
$$\{1,3,36,270,6480,45360,2721600,\cdots \}$$ and the $P_n$ are polynomials of degree $3(n-1)$ in $\lambda$. The very first are

$$\left(
\begin{array}{cc}
n & P_n(\lambda) \\
 1 & 1 \\
 2 & 4 \lambda ^3-1 \\
 3 & 80 \lambda ^6+12 \lambda ^4-64 \lambda ^3+12 \lambda ^2+5 \\
 4 & 1088 \lambda ^9+456 \lambda ^7-1800 \lambda ^6+618 \lambda ^5-222 \lambda
   ^4+450 \lambda ^3-114 \lambda ^2-17
\end{array}
\right)$$

Using this notations, it will be quite simple to generate as many terms as required.

Edit

After a comment about the case $(k=1,a=5,b=2)$, I realise that the first condition is not $g'(0) <0$ but $g(0) \leq 0$. If we face the case where $g'(0)=0$, the problem is a bit different.
$$g'(0)=1+\frac{a k}{3}-\frac{4 a \lambda  k}{3}\implies k=\frac{3}{a (4 \lambda -1)}$$ In such a case, the series expansion of $g(x)$ around $x=0$ starts with a term in $x^2$.

$$g(x)=\frac{x^2} {4\lambda-1}\,\sum_{n=0}^\infty (-1)^n \frac{a^{n+1} \left(4 \lambda ^{n+2}-1\right)}{(n+2)!}\, x^n$$

The inversion of the sum does not make any problem but leads to other polynomials in $$t=\frac{3 \left(4 \lambda ^2-1\right)}{a \left(4 \lambda ^3-1\right)}$$

Title: Finding zeros of a function


问:
Fast way to put boundaries of integration on the 3-dimensional integral

Calculate the triple integral
$$
\iiint_{G}(x+z) \,d x d y d z, \text { domain } G\text { is bounded by planes }\\ x+y=1, x-y=1
,x+z=1, z=0, x=0
$$

My question is, what is the easiest and most efficient way to put boundaries on such integral? For integral, that I provided, I built a 3D representation, but from it it very difficult to understand, how $x,y,z$ are changed. And especially it hard to do with only 2D representation.

答:
If you visualize the domain and also from the equations of planes that $x-y=1, x+ y = 1, x+z = 1$, they all meet at point $(1,0,0)$.Using given equations to find the limits,

First for the limit of $z$, note that $z \geq 0$ and $z \leq 1-x$

Limits of $y$ is between $-(1-x) \leq y \leq (1-x)$

$0 \leq x \leq 1$

The integral is

$\displaystyle \int_0^1 \int_{-(1-x)}^{1-x}\int_0^{1-x} (x+z) \ dz \ dy \ dx$


问:
Constructing Very "Flat" Functions
Consider the function $f(x)=\tan(\sin(x))-\sin(\tan(x))$. It has (what I think is) a remarkable property. The Taylor series expansion of $f(x)$ around $x=0$ is
$$\frac{1}{30}x^7+\frac{29}{756}x^9+\frac{1913}{75600}x^{11}+o(x^{13}).$$

Since $\tan(x)$ and $\sin(x)$ are odd functions, $f(x)$ is also an odd function. But its first nonzero term is $x^7$. This tells us that that $f(x)$ is very "flat" near the origin, which it is indeed.

I haven't calculated an explicit closed form for the series coefficients but looking at the wild oscillations around $x=\pm \pi/2$ (and confirming it by looking at it in the complex plane), the radius of convergence is $\pi/2$. This is simply because of the tangent function.

I have the following questions.

Is this just a coincidence or is there something deeper going on here? Is there any way to build/find more such examples? How can we find two real functions $g_1(x)$ and $g_2(x)$ such that

they are both distinct,

they are both elementary and can be explicitly written in closed form,

and $g_1(g_2(x))-g_2(g_1(x))$ is $o(x^n)$ for some $n>1$ with a strictly positive radius of convergence.

Neither should be a polynomial because this is trivial to do with polynomials. So the Taylor expansion of $g_1(x)$, $g_2(x)$, and $g_1(g_2(x))-g_2(g_1(x))$ should be infinite.

Addressing Calvin Khor's comment

Is it really trivial with polynomials? The first few terms of the Taylor series only depend on the first few derivatives, so it seems to me that understanding how to do it for polynomials should help understand what's happening in your example

Yes, it is trivial with polynomials. Consider
$$g_1(x)=x^{11}+x^{13}$$
$$g_2(x)=x^{15}+x^{17}$$
then
$$g_2(g_1(x))-g_1(g_2(x))=4 x^{167}+50x^{169}+290x^{171}+\cdots$$
which is indeed a finite series with an infinite radius of convergence, because it is only a composition of polynomials. So it is trivial to build such a (finite series) example for any desired $o(x^n)$.

I was looking for a general construction for any given $n$, so that the series is infinite and can be written in explicit closed form using elementary functions, or a proof that it is not possible.

Title: Constructing Very "Flat" Functions

Body:
For any two functions $f(x)=x+a_1x^3+a_2x^5+a_3x^7+O(x^9)$, $g(x)=x+b_1x^3+b_2x^5+b_3x^7+O(x^9)$, we have
$$\begin{align}
f(g(x))=x&+(a_1+b_1)x^3+(3a_1b_1+a_2+b_2)x^3\\
&+(3a_1b_1^2+3a_1b_2+5a_2b_1+a_3+b_3)x^7+O(x^9),\\
g(f(x))=x&+(a_1+b_1)x^3+(3a_1b_1+a_2+b_2)x^3\\
&+(3a_1^2b_1+5a_1b_2+3a_2b_1+a_3+b_3)x^7+O(x^9).
\end{align}$$


问:
Using layer cake representation in proving Hardy-Littlewood-Sobolev inequality

This is Theorem 4.3 page 106 in "Analysis" of Lieb and Loss.

There is a step in the proof that I cannot understand:
(I am writing in a less elegant way for the sake of clarity)
$$\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\chi_{\{x:f(x)>a\}}{(x)}\chi_{\{x:h(x)>b\}}{(y)}\chi_{\{x:|x|<c\}}{(x-y)}\,dxdy$$
$$\leq
\frac{\int_{\mathbb{R}^{n}}\chi_{\{x:f(x)>a\}}{(x)}dx\,\int_{\mathbb{R}^{n}}\chi_{\{x:h(x)>b\}}{(x)}\,dx\,
C_n\, c^n}{\max\{\int_{\mathbb{R}^{n}}\chi_{\{x:f(x)>a\}}{(x)}dx,\,\int_{\mathbb{R}^{n}}\chi_{\{x:h(x)>b\}}{(x)}\,dx,\,
C_n\, c^n\}}
$$
It says in the book, the integrals in $x,y$ can be estimated from above by replacing one of the characteristic functions by 1.

My question is: why can we bound

$$\int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} f(x)g(y) \chi_{\{x:|x|<R\}}{(x-y)}\, dx\,dy=
\int_{|x-y|<R}f(x)g(y) \, dx\,dy$$
by
$$
\frac{\int_{\mathbb{R}^{n}} f(x)dx\,\int_{\mathbb{R}^{n}} g(x)dx\; |B_{R}(o)|}{\max\{\int_{\mathbb{R}^{n}} f(x)dx,\,\int_{\mathbb{R}^{n}} g(x)dx,\; |B_{R}(o)|\}
}$$

答:
You have to use $0 \le f(x),g(y) \le 1$.

We always have $\int_{|x-y|<R} f(x)g(y)dxdy \le \int_{(\mathbb{R}^n)^2} f(x)g(y)dxdy = \int_{\mathbb{R}^n} f(x)dx\int_{\mathbb{R}^n} g(y)dy$, so we're done if the maximum is $|B_R(0)|$. We also have $\int_{|x-y| < R} f(x)g(y)dxdy \le \int_{|x-y|<R} g(y)dxdy = |B_R(0)|\int_{\mathbb{R}^n}g(y)dy$. Same thing with $f$.


答:
For the sake of illustration, the plot shows the functions

$$10\left(1-\frac tn\right)^n$$ for $n=1,2,3,4$. The dark green curve is the exponential solution.

Title: ODE Euler approximation converging to exact solution

Body:
I have this ODE: $dy/dt = -y$, $y(0)=10$.
How can I prove that its Euler numerical solution converges to the exact solution?

This is what I have attempted:
The Euler's scheme for this would be: $y_n = y_{n-1}(1 - \Delta_t) = y_0(1 - \Delta_t)^n$.
As $\Delta_t$ goes to $0$, $y_n$ would go to $y_0$, which is 10.

The exact solution would be $y(t)=10e^{-t} = 10e^{-n\Delta_t}$. As $\Delta_t$ goes to $0$, t goes to $0$ (I don't know if this is valid), and thus the solution converges to 10 as well.

I am really unsure about what I have done. Can someone help me with this? Thank you!


问:
Show that the annihilator space is a closed linear subspace of the dual space.
The question of interchanging inner product and limit has been posted here several times, especially in the case of Hilbert space. There are also several posts about the proof of the closeness of annihilator space.  However, I have a more specific issue.

Let $(V, \|\ \cdot\ \|)$ be a Banach space on $\mathbb{C}$ and let $V^{*}$ be its dual, endowed with the dual norm $$\|f\|_{V^{*}}=\|f\|_{op}=\sup_{\|x\|_{V}\leq 1}|\langle x,f \rangle_{V, V^{*}}|.$$ For $M\subset V$, we define the annihilator of $M$ as $$M^{\bot}:=\{f\in V^{*}:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ x\in M\}\subset V^{*}.$$

I want to show:

$M^{\bot}$ is a closed linear subspace of $V$.

I have proved that $M^{\bot}$ is a linear subspace but I got stuck when I tried to show the closeness. There are several online notes that have proved this, but not in the language of inner product which is the one that brings the issue to us.

So, let $f\in \overline{M^{\bot}}$, then there exists a sequence of bounded linear functional, $(f_{n})_{n=1}^{\infty}\subset M^{\bot}$ such that $\lim_{n\rightarrow\infty}f_{n}=f.$ It then follows that, for any $x\in M,$ we have $$\langle x, f\rangle_{V,V^{*}}=\langle x, \lim_{n\rightarrow\infty}f_{n}\rangle_{V, V^{*}}=0,$$ since $\langle x, f_{n}\rangle_{V, V^{*}}=0$ for every $n$.

Is this correct? It looks really weird to me in the sense that: what does it mean by $\lim_{n\rightarrow\infty}f_{n}=f$ here? pointwise? in the norm?

If in the norm, do we actually have $$\langle f-f_{n}\rangle_{V^{*}}=\sup_{\|x\|_{V}\leq 1}|\langle x, f-f_{n}\rangle_{V,V^{*}}|\longrightarrow 0\ \ \text{as}\ \ n\rightarrow\infty?$$ If this is correct, then I do not know what to do next...

If pointwise, does it actually mean $$\lim_{n\rightarrow\infty}\langle x, f_{n}\rangle_{V,V^{*}}=\langle x, f\rangle_{V, V^{*}}?$$

Well then I do not even need to consider the interchange of limit and inner product.

I am really confused the sense of $\lim_{n\rightarrow\infty}f_{n}=f$ here...

答:
You have to start with: let $(f_n)$ be a convergent sequence in $M^{\perp}$ with limit $f \in V^*$.

This means: $||f_n-f||_{op} \to 0.$

We have to show that $f \in M^{\perp}.$

Since

$|f_n(x)-f(x)| \le ||f_n-f||_{op} \cdot ||x||_V$ for all $n$ and all $x \in V,$ we get

$$f_n(x) \to f(x)$$

for all $x \in V$.

Since $f_n(x)=0$ for all $n$ and all $x \in M$, we conclude that $f(x)=0$ for all $x \in M.$

This gives  $f \in M^{\perp},$ thus  $M^{\perp}$ is closed.


问:
Discrete arithmetic subgroup
Let $F=\mathbb{Q}(\sqrt{d})$ for some $d>0$ be a real quadratic field and $G=\rm{GL}_2$, say over $\mathbb{Z}$. For any embedding $F\hookrightarrow \mathbb{R}$, the group $G(\mathcal{O}_d)$ is not discrete in $G(\mathbb{R})$. But for the pair of the two embeddings of $F$ into $\mathbb{R}$, the group $G(\mathcal{O}_d)$ becomes discrete in $G(\mathbb{R})\times G(\mathbb{R})$. This is confusing me. Is $G(\mathcal{O}_d)$ now a discrete group in its own or not? Why is it discrete in one group, but not in the other. Are there different topologies or what is the matter?

答:
You need to know the notion of induced topology.

A priori, there is no topology on the set $G(\mathcal O_d)$. But we have two embeddings $\iota_{1, 2}: G(\mathcal O_d) \hookrightarrow G(\Bbb R)$, given by the two real embeddings of $F$ into $\Bbb R$. They together give another embedding $\iota: G(\mathcal O_d) \hookrightarrow G(\Bbb R) \times G(\Bbb R)$ such that $\iota(x) = (\iota_1(x), \iota_2(x))$.

On the other hand, we have canonical topologies on $\Bbb G(\Bbb R)$ and on $G(\Bbb R) \times G(\Bbb R)$ (product topology). The three embeddings $\iota_1, \iota_2, \iota$ then induce three different topologies on $G(\mathcal O_d)$.

It turns out that, under the first two induced topologies, $G(\mathcal O_d)$ is not discrete; but under the third induced topology (via $\iota$), $G(\mathcal O_d)$ is discrete.

Is that clear now?

Title: Discrete arithmetic subgroup


问:
Probabilistic Polynomial Time Algorithm
Let $C$ be a language and $M$ be a probabilistic polynomial-time algorithm where $w \notin C$ implies $\mathbb{P}\{M\; accepts\; w\} \leqslant 1/8$, and  $w \in C$ implies $\mathbb{P}\{M\; accepts\; w\} \geqslant 1/4$.

How do I go about proving that there is another probabilistic polynomial-time algorithm $M'$ that on input $x \in C$, accepts with probability $\geqslant 9/16$ and if $x \notin C$ then it accepts with probability $\leqslant 7/16$.

Then, also prove $M''$ with accepts $\geqslant 2/3$ when $x \in C$, and if not accepts with probability $\leqslant 1/3$

Not really sure how to approach the problem. Any help would be appreciated.

Thanks

答:
Assuming that the "probabilistic" qualifier means that the algorithm returns results that satisfy the probability inequalities independently on successive runs, then let $M'(w)$ be the algorithm that checks $M(w)$ three times and accepts if any run of $M$ would have accepted.

Then $M'$ satisfies the condition and determines membership in $C$ correctly with probability $>1/2$. So we can get an algorithm that determines membership in $C$ correctly with any desired probability $<1$ by repeating $M'$ a finite number of times.

Title: Probabilistic Polynomial Time Algorithm


问:
Let $f\in C([a,b])$, $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ for all continuous and differentiable functions $\phi$. Show $f$ is constant

Let $f\in C([a,b])$ and $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ for all continuous and differentiable functions $\phi$ on $[a,b]$ with $\phi(a)=\phi(b)=0$. Show that $f$ is constant on $[a,b]$.

I would like to know if my proof holds, please. My attempt is to pass by primitive of $f$.

First of all, as $f\in C([a,b])$, $f$ has a primitive $F\in C^1([a,b])$ such that $F'(x)=f(x) \ \forall x \in [a,b]$. Then, as $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ holds for all continuous and differentiable functions $\phi$ on $[a,b]$, we can choose $\phi=F$. So we have to show that $\int_{a}^{b}f(x)F'(x) \ dx=\int_{a}^{b}f^2(x)=0 \implies f$ constant.

Let $G$ be a primitive of $f^2$. We have that $G'(x)=f^2(x)\ge 0 \ \forall x \in [a,b]$. So, $G$ is increasing on $[a,b]$. Moreover, $G(b)-G(a)=0 \iff G(a)=G(b)$. So, we conclude that $G$ is constant as $a\le x\le b\iff G(a)\le G(x)\le G(b)=G(a)$. Thus,

$G'=f^2=0 \Rightarrow f=0$. So we showed that $f$ is constant.

I'm curious as well if there is a maneer to prove the statement using upper and lower Darboux sum considering by absurd that $f\neq 0$ so there exists $c\in[a,b]: f(c)\neq0$. I tried to prove the statement using it, but I feel lost in the inequalities. I tried to pass by integrability of th functions implies integrability of absolute value of functions and by Upper Darboux sum.

答:
Of course you can drop the requirement $\phi(a) = \phi(b) = 0$ to make your proof work. However, it turns out that you only need to tweak your approach a little bit to prove the given statement:

Define the mean $\overline{f} = \frac{1}{b-a} \int_a^b f(x) \, \mathrm{d} x$ of $f$. Instead of $\phi = F$ choose
$$\phi(x) = F(x) - F(a) - (x-a) \overline{f}$$
for $x \in [a,b]$. You can check that $\phi$ satisfies all hypotheses and $\phi' = f - \overline{f}$. Therefore,
$$ 0 = \int \limits_a^b f(x) \phi'(x) \, \mathrm{d} x - \bar{f} \int \limits_a^b \phi'(x) \, \mathrm{d} x = \int \limits_a^b \left[f(x) - \overline{f}\right] \phi'(x) \, \mathrm{d} x = \int \limits_a^b \left[f(x) - \overline{f}\right]^2 \, \mathrm{d} x \, .$$
You have already shown that this implies $f - \overline{f} = 0$, so $f$ must be constant.


答:
This is not an unswer, just a hint and too long for a comment: The integral above may be transformed with  $\theta = \frac{\arccos(y)}{2}$

$$2\, \left(1+2\, r^2\right)^{1/3} \int_{-1}^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dy$$
Next we can split the integral in two parts with $p=-y$ in the first one:

$$2\,\left(1+2\, r^2\right)^{1/3} \left(\int_0^1 \left(1+r^2 p\right)^{-1/3} \left(1-p^2\right)^{-\frac{1}{2}} \, dp +  \int_0^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dp \right) $$

and evaluate them in form of hypergeometric functions. For this step e.g. we can express the integrand in form of MeijerG - Function. The first integral will be transformed with $q = p^2$  apply the formular Mathematical Functions Site and simplify:

$$I_1 = \frac{\left(1+2\, r^2\right)^{1/3}}{\gamma\left[\frac{1}{3}\right]} \int_0^1 (1-q)^{-\frac{1}{2}} q^{-\frac{1}{2}} \,G_{1,1}^{1,1}\left( \sqrt{q}\, r^2\left\vert
\begin{array}{c}
\frac{2}{3}\\
0%
\end{array}%
\right. \right)\,
\, dq$$
$$ = \left(1+2 \,r^2\right)^{1/3} \left(\pi  \,_{2}F_{1}\left[\frac{1}{6},\frac{2}{3},1,r^4\right]-\frac{2}{3} r^2 \,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\
\left\{\frac{3}{2},\frac{3}{2}\right\},r^4\right]\right)$$

The second integral $I_2$ is performed in the same way. The limit $r \to 1$ Mathematical Functions Site and Mathematical Functions Site exists e.g. for $I_1$
$$\lim_{r \to 1}\,I_1=\frac{\frac{3 \pi\,  \gamma\
\left[\frac{1}{6}\right]}{\gamma\left[\frac{1}{3}\right]\, \gamma\left[\frac{5}{6}\right]}-2\,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\left\{\frac{3}{2},\frac{3}{2}\right\},1\right]}{3^{2/3}}$$
and for  $I_2$ in the same way. The sum of $I = I_1  + I_2$
delivers just the half of your answer:

$$I= \frac{2^{1/3}\, 3^{5/6}\, \gamma\left[\frac{1}{3}\right]^2}{\gamma\left[\frac{2}{3}\right]} $$.

Therefore, inspite of a calculation error of factor 2, @user1055 is right, the limit may be also calculated inside the integral.

Title: Find $ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta$

Body:
Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta  $$
Question- Can I take the limit inside the integral?

My try-

$$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left(\frac{1+2r^2}{1-r^2\cos2\theta} \right)^{1/3}\, d\theta  $$
$$ I= \int_{-\pi}^{\pi}   \lim_{r \to 1^{-}}  (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta  $$
$$ I=3^{1/3} \int_{-\pi}^{\pi}  \frac{1}{(1-\cos2\theta)^{1/3}}d\theta  $$
$$ I= 3^{1/3}2\int_{0}^{\pi}    \frac{1}{(1-\cos2\theta)^{1/3}}   d\theta       $$
$$ I= 3^{1/3}4\int_{0}^{\pi/2}    \frac{1}{(1-\cos2\theta)^{1/3}}   d\theta       $$
$$ I= (3/2)^{1/3}4\int_0^{\pi/2}
\sin^{-2/3}\theta  \  d\theta       $$
$$I= 4(3/2)^{1/3} \frac{\Gamma(1/6)\Gamma(1/2) }{\Gamma(2/3)}. $$


问:
$ \int |y-x| d \mu (y) \leq \int |y-x| d \nu (y)$ then for $\phi$ convex $ \int \phi(y) d \mu (y) \leq \int \phi(y) d \nu (y)$
How to prove that for two measures with finite moments,

$$ \forall x \in \mathbb R,  \quad \int |y-x| d \mu (y) \leq  \int |y-x| d \nu (y)$$
$ \qquad \qquad \implies \forall \phi, \text{ convex function, we have that }$
$$\int \phi(y) d \mu (y) \leq  \int \phi(y) d \nu (y) $$

Perhaps there should also be the condition: $$ \int d \mu = \int d \nu $$

but for this I am not sure. That is the reason why I would like to find the proof.

My attempt.

I tried to approximate any convex function with functions like $(y-x)^+$ but it is hard to get jumps with continuous function like this one. I was thinking of multiplying by a constant $(y-x)^+$  to get a steeper slope, and combining $(y-x_0)^+$  with $(y-x_1)^+$  in order to get a constant function over the interval $[x_0, x_1]$, but it seems to me it is not exactly the right approach.

Source.

It is from the book Peacocks and Associated Martingales, with Explicit Constructions, exercise 1.7. The topics related are ordering of measures, convex order of measures...

答:
After multiple thoughts, I think I have settle down this problem. Here is my approach:

We work under the measure $\mu$.

First we prove that we can  approximate any characteristic function by a sequence of linear combination of Call function.

We prove the result for real valued functions, since transferring the problem to multi-dimensional spaces is straight-forward. We only prove the statement for sets of the form : $[a, \infty[$ and $a \in \mathbb  R$.  This proves for any set of the Borel set of $\mathbb R$ using the $\pi-\lambda$ theorem.

\vspace{0.2cm}

The proof goes as follows: first we fix $a, c \in \mathbb R$ and we are searching for a sequence of function $\phi_n$, constructed with Call functions such that $\phi_n$ approximates $c \cdot 1_{ [a,\infty[}$.

We set $\forall n\in \mathbb N, \phi_n(x) = c \cdot n \left [ ( x - (a - \frac 1 n))^+ - (x - a)^+ \right ]$.

Then, the error of the approximation is given by:

\begin{align*}
\int_{\mathbb R^d} | c \cdot 1{ [a,\infty[} - \phi_n | d \mu &=
\int_{a - \frac 1 n}^{a} | c - c \cdot n  [  x - a + \frac 1 n ] | d \mu \\
&= \int_{a - \frac 1 n}^{a} | c \cdot n ( x - a ) | d \mu \\
&=  \frac c {2n} .
\end{align*}

Then, $\phi_n$ converges to $1_{ [a,\infty[}$ in $L^1 (d  \mu)$. Now, since any positive measurable function can be approximated by a sequence of linear combination of simple functions, we can approximate any positive measurable function by a linear combination of Call functions.

Then, it is easy to prove the statement with dominated convergence :)


问:
What is the optimal dimension $k$ for which $S^2\times S^2$ into $\mathbb{R}^k$?
Clearly $k\geq4$ since $S^2\times S^2$ is a 4-manifold and $k\leq6$ since $S^2$ embeds in $\mathbb{R}^3$.
Also, since $S^2\times S^2$ is compact, "invariance of domain" argument also show that $k$ cannot be 4.
Knowing the answer for the lower-dimensional case, (i.e., $S^1\times S^1\hookrightarrow \mathbb{R}^3$ is optimal) I tried to imitate the proof  for $S^2\times S^2\hookrightarrow\mathbb{R}^5$, but that didn't go very well.

My proof for $S^1\times S^1\hookrightarrow \mathbb{R}^3$: embed $S^1$ in $\mathbb{R}^3$ and take a tubular neighborhood, which is trivial since the orientable Euclidean rank 2 vector bundle over $S^1$ is unique: $[S^1,BSO(2)]=\pi_1(BSO(2))=\pi_2(SO(2))=\pi_2(S^1)=0$. The boundary of the unit disk bundle gives a desired embedded $S^1\times S^1$.

Problem with imitating the proof for $S^2\times S^2$: the isomorphism classes of orientable Euclidean rank 3 vector bundles over $S^2$ is not unique: we consider the bundle as the result of gluing two trivial vector bundles over two hemispheres $S^2_+$ and $S^2_-$.
The isomorphism class depends on "gluing" $S^1\to SO(3)$, and there are $2$ ways: $[S^1,SO(3)]=\pi_1(SO(3))=\mathbb{Z}_2$.
Thus, the tubular neighborhood might not be an honest product.

Is there a clever way to show that this tubular neighborhood is in fact trivial (or not)?
Or, is there a totally different way of finding the optimal dimension $k$ ($5$ or $6$)?

答:
Your question is essentially a duplicate of An explicit embedding of $S^m \times S^n$ into $\mathbb R^{m + n + 1}$.  It shows that $k = 5$ in your case.

The proof of the this theorem is very easy. I didn't close your question as a duplicate because you present some nice general ideas towards a solution.


问:
Splitting of conjugacy class in alternating group
Browsing the web I came across this:

The conjugacy class of an element $g\in A_{n}$:

splits if the cycle decomposition of $g\in A_{n}$ comprises cycles of distinct odd length. Note that the fixed points are here treated as cycles of length $1$, so it cannot have more than one fixed point; and

does not split if the cycle decomposition of $g$ contains an even cycle or contains two cycles of the same length.

Anybody with a proof?

答:
Note the following: (1) The conjugacy class in $S_n$ of an element $\sigma \in A_n$ splits, iff there is no element $\tau \in S_n\setminus A_n$ commuting with $\sigma$. For if there is one, for each $\tau' \in S_n \setminus A_n$ we have
$$ \tau'\sigma{\tau'}^{-1} = \tau'\sigma\tau\tau^{-1}\tau'{}^{-1}
    = (\tau'\tau)\sigma(\tau'\tau)^{-1}
$$
and $\tau\tau' \in A_n$. On the other hand, if $\tau\sigma\tau^{-1}$ and $\sigma$ with $\tau \in S_n\setminus A_n$ are conjugate in $A_n$, then for some $\tau' \in A_n$, we have $\tau\sigma\tau^{-1} = \tau'\sigma\tau'^{-1}$, giving
$$ \tau'{}^{-1}\tau \sigma = \sigma\tau'{}^{-1}\tau $$
and hence $\tau'{}^{-1}\tau \in S_n\setminus A_n$ commutes with $\sigma$.

Now suppose, $\sigma$ has a cycle $c_i$ of even length. A cycle of even length is an element of $S_n \setminus A_n$, and as $\sigma$ commutes with its cycles, we are done by the above. If $\sigma$ has two cycles $(a_1\ldots a_\ell)$ and $(b_1 \ldots b_\ell)$ of the same odd length $\ell$, then $(a_1b_1) \ldots (a_\ell b_\ell)$ is a product of $\ell$ transpositions (hence odd, so an element of $S_n \setminus A_n$) commuting with $\sigma$.

Now suppose $\sigma = c_1 \cdots c_s$ is a product of odd cycles $c_i$ of distinct length $d_i$. Let $\tau \in S_n$ be a permutation commuting with $\sigma$. Then $\tau$ must fix each of the $c_i$, that is, $\tau$ must be of the form $\tau = c_1^{a_1} \cdots c_s^{a_s}$ for some $a_i \in \mathbb Z$. But as the $c_i$ are even permutations (as cycles of odd length), we have $\tau \in A_n$. So no $\tau \in S_n \setminus A_n$ commutes with $\sigma$ and we are done.


问:
Difference between homotopy function and homotopic space
I'm having some troubles understanding the concept of homotopy and homotopic spaces.
I understood that given two function $f,g:X\to Y$ we can say that there's an homotopy from $f$ to $g$ if $\exists$ a continuous $H:X\times [0,1]\to Y$ such as $H(x,0)=f(x)$ and $H(x,1)=g(x)$, and that two spaces $X,Y$ are homotopic equivalent if we can find two function $h:X\to Y$ and $k:Y\to X$ such as $h\circ k\simeq\operatorname{id}_{Y}$ and $k\circ h\simeq\operatorname{id}_{X}$.

My problem is that I can't understand how the fact that I have and homotopy between two function interact with the space used by the function.

To clarify I'll give you an example: let's say that I have a function $F:D^2\to X$ which is homotopic to cost$:D^2\to X$, the map that $\forall x\in D^2$ cost(x)=c with $c\in\ X$. What properties does $D^2$ get from the existence of those two homotopic function?

答:
The main idea is that a homotopy encompasses the notion of a 'continuous' deformation. The fact that two maps are homotopic means that one map can be 'deformed' into the other. For example, consider two maps $f,g: I\to \mathbb{R}^2$ with the same endpoints. Intuitively, these maps will look like two strings with the same endpoints, and you can imagine that you can deform either one into the other. Since this process is continuous, we say that $f$ and $g$ are homotopic to one another.

On the other hand, two spaces $X$ and $Y$ are homotopy equivalent if one can be 'continuously' deformed into the other, and viceversa. This can be expressed by saying that a map $f: X\to Y$ is a homotopy equivalence, which means there exists a homotopy inverse $g: Y\to X$ to $f$, i.e. the composition of these maps is homotopic to the identity.

For example, the cylinder and the circle are homotopy equivalent. Why? Consider the inclusion of $S^1$ into the cylinder, and consider a map from $S^1\times I$ to $S^1$ given by 'squashing' the cylinder into the bottom circle. The composition of these maps, in either order, amounts to doing nothing in each of the spaces, and thus it is homotopic to the identity map.


问:
What is the set of stochastic processes that have a deterministic quadratic variation?
It is know that the quadratic variation of Brownian motion $B_t$ is deterministic, $[B]_t=t$.

What other stochastic processes $X_t$ have a deterministic quadratic variation $f_X(t)$ ? Can we characterise this set ?

This set must include at least all process of the form $B_{g(t)}$ for any continuous $g: \mathbb R^+ \rightarrow \mathbb R^+$. What other elements are in this set ?

答:
I hope the following is an answer in the way you wanted it to be. If there are any mistakes in it, please correct me.

As far as I know, the quadratic variation is only defiened for semimartingales, so we just consider this type of process. As $[X]_0=X_0^2$ we need $|X_0|$ to be deterministic. For simplicity let's focus first on continous processes. Since semimartingales are sums of local martingales and FV-processes, and since the continuous part of the quadratic variation is zero for FV-processes, we can always add a continuous FV-process to any stochastic process with a deterministic quadratic variation without destroying this property. As all contiuous local martingales are time-changed Brownian motions
and since their quadratic variation is the change-of-time, the set contains all time-changed Brownian motions with a deterministic change-of-time (it might sound a bit silly, but this seems to be the answer). Now consider the discontinuous part. As the jumps of the quadratic variation are exactly the squared jumps of the process $\Delta [X]_t=(\Delta X_t)^2$ we need to have deterministic jumps. In summary the set (Hint: This is not unique decomposition.) consists of all processes
$$X=M+A+J$$
(with a deterministic value $|X_0|$), where $M$ is a time-changed Brownian motions with a deterministic change-of-time, $A$ is a continuous FV-process and $J$ is a deterministic pure jump process. $\big($I'm not sure about this: As a non-constant deterministic pure jump process can't be a local martingale, the process $J$ can be included in the (noncontinuous) FV-part, thus $X$ is the sum of a a time-changed Brownian motions with a deterministic change-of-time and a FV-process with deterministic jumps.$\big)$

Another way to look at this is to consider purely discontinuous semimartingales. These are sums of FV-processes and purely discontinuous local martingales, which means that continuous part of the quadratic variation ist zero. As every semimartingale $X$ decomposes uniquely as $$X=X^c+X^d,$$
where $X^c$ is a continuous local martingale with ${X^c_0=0}$ and ${X^d}$ is a purely discontinuous semimartingale, the desired set consists of processes $X$ (with a deterministic value $|X_0|$) which are the sum of a purely discontinuous semimartingale with deterministic jumps and a time-changed Brownian motion with a deterministic change-of-time.