问:
modular automorphism group of faithful weight

Theorem 2.11 is from Takeski's book (Vol 2. Chapter VIII, Section 2).

It mentions that we can get $\sigma^{\psi}_t(x)=(he_n)^{it}\sigma^{\varphi}_t(x)(he_n)^{-it}$ by Lemma 2.10.
I have a question: $x\in M_{e_n}$, it is also in $M$, by Lemma 2.10, we have $\sigma^{\psi}_t(x)=(he_n)^{it}\sigma^{\varphi}_t(x)(he_n)^{-it}$$\sigma^{\psi}_t(x)=(h)^{it}\sigma^{\varphi}_t(x)(h)^{-it}$. Why should we replace $h$ by $he_n$.

Another question: how to derive $(he_n)^{it}=(h)^{it}$?

In the proof of Theorem of

答:
The fact that $h$ is affiliated with $\mathcal M_\varphi$ implies that $he_n\in \mathcal M_\varphi$. So, for $x\in e_n\mathcal M e_n$
\begin{align}
\sigma_t^\psi(x)
&=
(he_n)^{it}\sigma_t^\varphi(x)(he_n)^{it}=h^{it}e_n\sigma_t^\varphi(x)e_nh^{it}\\[0,3cm]
&=h^{it}\sigma_t^\varphi(e_n)\sigma_t^\varphi(x)\varphi(e_n)h^{it}=h^{it}\sigma_t^\varphi(e_nxe_n)h^{it}\\[0.3cm]
&=h^{it}\sigma_t^\varphi(x)h^{it}.
\end{align}
The first equality is due to Lemma 2.10 applied to $he_n\in e_n\mathcal Me_n$.


问:
Finding $a$ given 2 linear equations with 2 other constants b and c
I am trying to find the value of $a$ in the given the equations:

$x_1 + x_2 = b$ and $2x_1 + ax_2 = c$ considering there are also 2 extra constants $b , c$ that belong to $\Bbb{R}$.

I fall short when I start to compute for a, giving $a = 1 - (b - c)x_2$

But I can't get rid of $x_2$ since I have to find $a$. Any help?

答:
$$\begin{cases}
x_1 + x_2 = b\\
2x_1 + ax_2 = c\\
\end{cases}
$$
The matrix of the coefficients is
$$A=\left(
\begin{array}{cc}
 1 & 1 \\
 2 & a \\
\end{array}
\right)$$
According to Cramer's rule, the system has one and only one solution if $\det A\ne 0$

$\det A=a-2\ne 0\to a\ne 2$

Thus the system has one unique solution if $a\ne 2$.

ContentLicense: CC BY-SA 4.0


问:
Prove that $f(x) = \sum_{n=1}^{\infty} 2^{-n} |x-1/n|$ is differentiable on $x_0 \notin \{1, 1/2, 1/3, ... \}$
Prove that $f(x) = \sum_{n=1}^{\infty} 2^{-n} |x-1/n|$ is differentiable on $x_0 \notin \{1, 1/2, 1/3, ...\}$

I think the relevant theorem is the term-by-term differentiation theorem which states:

Let $f_n$ be differentiable functions on A and assume $\sum_{n=1}^{\infty} f_n'(x)$ converges uniformly and that $\exists x_0 \in A$ where $\sum_{n=1}^{\infty} f_n(x)$ converges. Then
$f'(x) =\sum_{n=1}^{\infty} f_n'(x)$ on A.

Here I know that $\sum_{n=1}^{\infty} f_n(x)$ converges for all R. So what I have left to show is that

Each $f_n$ is differentiable on $\{1, 1/2, 1/3, ...\}^c$

$\sum_{n=1}^{\infty} f_n'(x)$ converges uniformly on $\{1, 1/2, 1/3, ...\}^c$

But I'm stuck on showing 1. Once I write the limit definition I'm stuck.

答:
Take $n\in\Bbb N$. Then $f_n(x)=2^{-n}\left(x-\frac1n\right)$ when $x>\frac1n$, and therefore $f_n'(x)=2^{-n}$ then. And $f_n(x)=-2^{-n}\left(x-\frac1n\right)$ when $x<\frac1n$, and therefore $f_n'(x)=-2^{-n}$ then.

This also proves that $(\forall n\in\Bbb N)\left(\forall x\in\Bbb R\setminus\left\{1,\frac12,\frac13,\ldots\right\}\right):|f_n'(x)|=2^{-n}$. Therefore, the series $\sum_{n=1}^\infty f_n'(x)$ is uniformly convergent.


问:
Two random variable with the same variance and mean
Let $Y\in L^{2}(\Omega,\Sigma,P)$ and let $E[Y^2|X]=X^2$ and $E[Y|X]=X$. Could we prove that $Y=X$ almost surely.

My partial answer:

By the definition of conditional expectation we have $E[Y^2]=\int_{\Omega} E[Y^2|X] dP=\int_{\Omega}X^2 dP=E[X^2]$ and $E[Y]=\int_{\Omega} E[Y|X] dP=\int_{\Omega}X dP=E[X]$. Hence, Var$(X)=$Var$(Y)$.

答:
We have that
$$E((X-Y)^2) = E(X^2) + E(Y^2) - 2 E(XY) = 2E(X^2) -2 E(XY)$$
the last equality being for the things you already proved.
On the other hand
$$ E(XY) = E ( E(XY|X) ) = E (X E(Y|X) ) = E(X \cdot X) = E(X^2) $$
The second equality is due a property of conditional expectations (that is essential!). Finally the variables are equal in $L^2$ which implies that they are equal a.s.</p>

ContentLicense: CC BY-SA 3.0


问:
Why almost sure convergence holds if $X_n = n$ w.p. $1/n$?
I've encountered these two examples (used to show how a.s. convergence doesn't imply convergence in $R$th mean and visa versa). In one case we have a random variable $X_n = n$ with probability $\frac{1}{n}$ and zero otherwise (so with probability $1-\frac{1}{n}$). In another case same deal with only difference being $X_n=1$, not $n$ with probability $\frac{1}{n}$. Assume $X_n$'s are independent in both.

It is clear to me that in the second case, where $X_n = 1$ w.p.  $\frac{1}{n}$, the summation is $\sum_{n=1}^\infty P(X_n=1) =\sum_{n=1}^\infty \frac{1}{n} =\infty $, so by Borel Cantelli Lemma 1 we have that "$X_n=1$" occurs infinitely often and hence $X_n$ doesn't converge almost surely to zero.

In the first case however, where $X_n = n$ w.p. $\frac{1}{n}$, the argument goes that $X_n$ converges to zero almost surely. Why is that? Isn't it the same, only here we are summing $P(X_n \ne 0)$ for $n=1$ through $\infty$ and these probabilities are again resulting in a diverging harmonic series?

答:
Sure there is a typo in the textbook. The first case is, for example, $X_n=\frac{1}{n}$ w.p. $\frac{1}{n}$ and zero elsewhere.

This sequence converges


问:
3 subspaces $A, B, C$ of linear space $V$ and $A + B = A + C \ iff B = C$ and $dim (B) = dim(C) \iff dim (A \cap B) = dim (A \cap C)$
There are $3$ subspaces $A, B, C$ of linear space $V$ and $A + B = A + C$.

Then it means that: $B = C$ and that $dim (B) = dim(C) \iff dim (A \cap B) = dim (A \cap C)$. Am I thinking right?

I am trying to get better understanding of that subject.

答:
Let $V=\mathbb{R}^2$, $A=Z(x)$, $B=Z(x-y)$, and $C=Z(2x-y)$ where $Z(f)$ means zeros of $f \in \mathbb{R}[x,y]$ in $\mathbb{R}^2$.

Check that these spaces satisfy the conditions you wrote but $B \neq C$. Besides, the dimension related parts are true.


问:
Can the graph of $x^x$ have a real-valued plot below zero?
The function $f(x) = x^x$ gives a complex number only if x has an even denominator. I'm not sure about irrational numbers. Why, then, is the best graph I can find of that function that of Wolfram Alpha, which plots adjacent sinusoidal waves for the real and imaginary parts, even at well-defined values such as $x = -\frac{1}{3}$? (And that's the plot that includes imaginary numbers; the real-valued plot shows nothing below zero.) I suspect this might be because the results are computed in a way that requires imaginary numbers.

The number of values for which the function is defined is infinite, but the values for which it is not are also infinite. So how should the graph look below zero?

(Note that in order to avoid another point where the function may be undefined I'm assuming that $0^0 = 1$.)

答:
Although it doesn't deal directly with $x^x$, here's a WolframAlpha blog post that details how real and complex roots are currently treated by WolframAlpha.

As far as $x^x$ goes, the plot does indeed show the real and imaginary parts of the principal value of $x^x$ on the same axis.  Another approach is to plot the complex point itself in a plane that's perpendicular to the $x$-axis at the corresponding point.  This leads to a spectacular image called the $x^x$-spindle, which was described in a great paper in Mathematics Magazine back in 1996.  This looks like so:

Using the fact that the complex logaritm is multi-valued, this can be generalized to obtain more threads on the spindle:

It sounds like you've seen the claim that $(p/q)^{p/q}$ is defined for $p$ negative and $q$ odd and positive.  Thus the graph might look something like so.

From the complex perspective, the dots arise as spots where one of the spiral threads punctures the $x$-$z$ plane.

Note that the Mathematica code for these images is all provided in this answer over on mathematica.SE.


问:
Polynomial Diophantine equation under modulo
This problem appered in my textbook and i cant solve it, it's in the chapter of the chinese reminder theorom so that's probably related. A solution or hint would be great.

答:
$\require{AMScd}$Hint:By CRT we have a commutative diagram
\begin{CD}
\frac{\Bbb Z}{m_1m_2\Bbb Z}@>\sim>>\frac{\Bbb Z}{m_1\Bbb Z}\times\frac{\Bbb Z}{m_2\Bbb Z}\\
@VVV@VVV\\
\frac{\Bbb Z}{m_1m_2\Bbb Z}@>>\sim>\frac{\Bbb Z}{m_1\Bbb Z}\times\frac{\Bbb Z}{m_2\Bbb Z}\\
\end{CD}
which maps:
\begin{CD}
x+m_1m_2\Bbb Z@>>>(x+m_1\Bbb Z,x+m_2\Bbb Z)\\
@VVV@VVV\\
f(x)+m_1m_2\Bbb Z@>>>(f(x)+m_1\Bbb Z,f(x)+m_2\Bbb Z)\\
\end{CD}


问:
Why it is important to write a function as sum of even and odd functions?
For the function $f(x)$ we can write it as sum of even and odd functions:

$$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{\text{Even}}+\underbrace{\frac{f(x)-f(-x)}{2}}_{\text{Odd}}$$

My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?

答:
One really neat application for this decomposition (which I saw on the YouTube channel "Flammable Maths") is evaluating integrals of the form $$\int_{-a}^a\Bigg(\frac{E(x)}{1+t^{O(x)}}\Bigg)dx$$ where $t,a>0$ are constants, $E(x)$ is a (continuous) even function, and $O(x)$ is a (continuous) odd function. If you set $f(x)=\frac{E(x)}{1+t^{O(x)}}$ and write $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$ you can say that $$\int_{-a}^a\Bigg(\frac{E(x)}{1+t^{O(x)}}\Bigg)dx=\int_{-a}^a\Bigg(\frac{f(x)+f(-x)}{2}\Bigg)dx+\int_{-a}^a\Bigg(\frac{f(x)-f(-x)}{2}\Bigg)dx$$ The last integral on the RHS vanishes since we're integrating an odd function on a symmetric domain. With a little algebra $\frac{f(x)+f(-x)}{2}=\frac{1}{2}E(x)$ giving us the awesome result $$\int_{-a}^a\frac{E(x)}{1+t^{O(x)}}dx=\int_{0}^aE(x)dx$$ which is really cool! This means we can say something like $$\int_{-1}^1\Bigg(\frac{x^4-x^2+1}{1+3^{\sin^2(x)\tan(x)+x^5+x}}\Bigg)dx=\int_0^1\big(x^4-x^2+1\big)dx=\frac{13}{15}$$ This can also be used to calculate some pretty nasty double integrals! $$\int_0^1 \int_{-x^2}^{x^2}\Bigg(\frac{xy^2+x^3}{1+3^{x\tan^{11}(y)+e^x\sin^7(y)}}\Bigg)dydx=\int_0^1 \int_0^{x^2}(xy^2+y^3)dydx=\frac{5}{24}$$ Love it.

Edit: This integration technique actually generalizes to integrals of the form $$\int_{-a}^a\Bigg(\frac{E_1(x)}{1+\big(E_2(x)\big)^{O(x)}}\Bigg)dx$$ where $E_1(x),E_2(x)$ are arbitrary (continuous) even functions while $O(x)$ is an arbitrary (continuous) odd function. Using the exact same procedure delineated above we can say $$\int_{-a}^a\Bigg(\frac{E_1(x)}{1+\big(E_2(x)\big)^{O(x)}}\Bigg)dx=\int_{0}^aE_1(x)dx$$ which means $$\int_{-1}^1\Bigg(\frac{x^4+x^2+1}{1+\big(x^2e^{-x^4}+\cos(x)\sin(x^2)\big)^{x^2\tan(x^3)+x}}\Bigg)dx=\int_0^1(x^4+x^2+1)dx=\frac{23}{15}$$


问:
Hermitian Matrix true or false?
I am trying to prove or disprove the statement:

If $A,B$ are n by n complex matrix, then if $A+B$ is Hermitian, then both $A$ and $B$ are Hermitian.

Is this statement true or false?

答:
A real matrix is Hermitian if it is symmetric. Use

$\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
1 & 0
\end{bmatrix}+\begin{bmatrix}
0 & 0\\
-1 & 1
\end{bmatrix}$

LHS is hermitian but the matrices in RHS are not Hermitian.

Title: Hermitian Matrix true or false?


Title: Hermitian Matrix true or false?

Body:
I am trying to prove or disprove the statement:

If $A,B$ are n by n complex matrix, then if $A+B$ is Hermitian, then both $A$ and $B$ are Hermitian.

Is this statement true or false?

Body: $$\begin{bmatrix}0&i\\-i&0\end{bmatrix}=\begin{bmatrix}0&0\\-i&0\end{bmatrix}+\begin{bmatrix}0&i\\0&0\end{bmatrix}$$


问:
Prove that nearly all positive integers are equal to $a + b + c$ where $a | b$ and $b | c$, $a \lt b \lt c$

If a positive integer $n$ is equal to $a + b + c$ where $a | b$, $b | c$ and $a \lt b \lt c$, let it be called "faithful". Prove that nearly all numbers are faithful and list the non-faithful numbers.

Here's where I am:

Let the number be $n$, $b = ap$ and $c = bq = apq$ where $p ,q \in Z_+$.
Let the set of faithful numbers be $F$.

$$n = a + b + c$$
$$\therefore n = a + ap + apq$$

Since $1 \lt p \lt q$ and $p, q \in Z_+$, $n \ge 7$. Hence $1, 2 \ldots 7 \notin F$.

Now what?

答:
Let $a=2^k$ for some $k\ge 0$, and let $p=2$. Then $a+ap+apq=2^k(3+2q)$, where we must have $q\ge 2$. $\{3+2q:q\ge 2\}$ is the set of odd numbers greater than $5$, so every number of the form $2^k(2m+1)$ with $m\ge 3$ is faithful. The potential exceptions are therefore the numbers of the form $2^k,2^k\cdot3$, and $2^k\cdot 5$ for $k\ge 0$.

Suppose that one of these numbers can be written in the form $a+ap+apq$ for some $a\ge 1$ and $p,q\ge 2$. Then it can be written as $a(1+p+pq)$, where $1+p+pq\ge 7$.

If $k\ge 4$, $2^k=2^{k-4}\cdot16=2^{k-4}(1+3+3\cdot4)$, so we may take $a=2^{k-4}$, $p=3$, and $q=4$. Thus, $2^k$ is faithful for $k\ge 4$, and we need consider only $2,4$, and $8$ individually; all are easily seen to be unfaithful.

Similarly, $2^k\cdot3=3\cdot2^{k-4}(1+3+3\cdot4)$, so we may take $a=3\cdot2^{k-4}$, $p=3$, and $q=4$ and catch all numbers of the form $2^k\cdot3$ except $12$ and $24$, and $2^k\cdot5=5\cdot2^{k-4}(1+3+3\cdot4)$, so we may take $a=5\cdot2^{k-4}$, $p=3$, and $q=4$ and catch all numbers of the form $2^k\cdot 5$ except $10,20$, and $40$.

We’ve now shown that all positive integers are faithful except $1,2,3,4,5,8$, and possibly $10,12,20,24$, and $40$. $10=1+3+6$ is faithful. It’s not hard to check that $12$ is not faithful: its only factor larger than $6$ is itself, so $a$ would have to be $1$, and $11=p(1+q)$ would have to be composite. $20=2\cdot20=2+6+12$ is faithful, and similarly $40$ is faithful. That leaves only $24$ to be checked, and I’ll leave it to you.


问:
If $\lambda$ is an eigenvalue of $A$, then $\overline{\lambda}$ must be an eigenvalue of $A$?
Is this true or false?

If $A$ is an n by n Real matrix, then if $\lambda$ is an eigenvalue of A, then $\overline{\lambda}$ must an eigenvalue of $A$

There is a theorem say that, if $A$ has a Non-real eigenvalue $\lambda$, then $\overline{\lambda}$ is also an eigenvalue of $A$,
so this statement is true??  since all conjugate of real is just it self?

答: $Ax=\lambda x \to \overline{Ax}=\overline{\lambda x} \to A \overline{x} = \overline{\lambda} \overline{x} \to Ay = \overline{\lambda} y$


问:
How to prove a PDE preserves mass?
My general question is: suppose you are given a PDE (possibly with boundary conditions). What does it mean to say the PDE "conserves mass"?

Specifically, if you are given the PDE
$$- \nabla \cdot (a(x) \nabla u(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u = 0 \quad \text{ on } \partial \Omega,$$
how would you determine whether "mass is conserved"?

My guess would be that the mass is the integral $\int_\Omega u(x) \, dx$. But if this is the case, I don't know how to prove whether it is constant or not.

Edit: In this specific problem, there is no time involved, so you can't actually use a time derivative. However, it is the starting problem of a homogenization procedure, so actually it reads
$$- \nabla \cdot (a_\epsilon(x) \nabla u_\epsilon(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u_\epsilon = 0 \quad \text{ on } \partial \Omega.$$

So the mass conservation might be related to the $\epsilon$, but I don't think so (because from the context I gather that determining whether the problem conserves mass is supposed to be done for a fixed $\epsilon$).

Any help is appreciated!

答:
Conservation laws imply some notion of 'time' in which a quantity $M(t)$ like the mass $$\int_\Omega u(x,t) \,\mathrm d x$$ is conserved.

This is equivalent to proving that
$$\frac {\mathrm d}{\mathrm d t}\int_\Omega u(x,t) \,\mathrm d x = \int_\Omega \frac {\partial}{\partial t} u(x,t) \,\mathrm d x = 0$$

For example, an equation of the form
$$\frac {\partial u}{\partial t} = \nabla \cdot (F(u,\nabla u,x,t)) \quad \text{and}\quad G(u,\nabla u,x,t)|_{\partial \Omega} = 0$$
has the property that
$$\int_\Omega \frac {\partial}{\partial t} u(x,t) \,\mathrm d x = \int_\Omega \nabla \cdot F \,\mathrm d x = \int_{\partial\Omega} F \cdot \mathrm d S $$
so that mass is conserved if e.g. $G=0 \implies F=0$.

So if $F=u^2 \nabla u$ and $G=u$ then this last integral is zero as the boundary conditions imply $F$ vanishes on the boundary.

Title: How to prove a PDE preserves mass?


答: Let $A$ be a matrix, which is not hermitian and $B=-A$, then $A+B$ is hermitian.

Title: Hermitian Matrix true or false?


问:
Interesting solutions for cyclic infinite nested square roots of 2
I have derived cosine values for following cyclic infinite nested square roots of 2 ( Hereafter simply referred as $cin\sqrt2$)

$cin\sqrt2[1-]$ represents $\sqrt{2-\sqrt{2-...}}$

$cin\sqrt2[1-1+]$ represents $\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}$

$cin\sqrt2[1-1+]$ = $2\cos\frac{2\pi}{5}$

$cin\sqrt2[2-2+]$ = $2\cos\frac{4\pi}{15}$

$cin\sqrt2[3-3+]$ = $2\cos\frac{24\pi}{65}$

$cin\sqrt2[4-4+]$ = $2\cos\frac{16\pi}{51}$

$cin\sqrt2[5-5+]$ = $2\cos\frac{352\pi}{1025}$

$cin\sqrt2[6-6+]$ = $2\cos\frac{64\pi}{195}$

$cin\sqrt2[7-7+]$ = $2\cos\frac{5504\pi}{16385}$

$cin\sqrt2[8-8+]$ = $2\cos\frac{256\pi}{771}$

$cin\sqrt2[9-9+]$ = $2\cos\frac{87552\pi}{262145}$

$cin\sqrt2[10-10+]$ = $2\cos\frac{1024\pi}{3075}$

.

.

.

We need pattern for $cin\sqrt2[n-n+]$ ($n\in N$)

At present I have figured out the pattern as follows

If n is even $$cin\sqrt2[n-n+] = 2\cos(\frac{2^n\pi}{3\times(2^n+1)})$$

If n is odd $$cin\sqrt2[n-n+] = 2\cos(\frac{\frac{2^n+1}{3}\times 2^n\times\pi}{2^{2n}+1})$$

In both the cases, it is quite obvious that
$$\lim_{n\to\infty} cin\sqrt2[n-n+]=2\cos\frac{\pi}{3}=1$$

My question is how to combine both the equations into a single one?

答:
We can multiply the fraction in the even-$n$ case by $2^n-1$ on top and bottom, yielding
$$\operatorname{cin}\sqrt2[n-n+]=2\cos\frac{2^n\pi(2^n-(-1)^n)}{3(2^{2n}-(-1)^n)}$$


问:
What is the Sobolev Lemma?
In the paper I am reading the authors state that $|\nabla u|_\infty$ can be replaced by $|u|_3$ using the Sobolev Lemma. I am trying to find this lemma but its turned out to be very difficult.

The context is the following:

a smooth bounded domain $\Omega \subset \mathbb{R}^3$

$|\cdot|_s$ denotes the Sobolev norm of the space $W^{s,2}(\Omega)=H^2(\Omega)$ and $|\cdot|_\infty$ the norm in $L^{\infty}(\Omega)$

$u$ is a vector valued function (the velocity of a fluid)

This has to be one of the many imbedding theorems which should give
$$|\nabla u|_\infty \leq C \: |u|_3$$
where $C$ is a constant depending on $\Omega$ alone I suppose.

I'd appreciate if you can give me a reference as well. ThX!

答:
I'm assuming that your function $u$ belongs to $W^{3,2}(\Omega)$. We have the following result that can be found in any good book of Sobolev spaces (for example in Leoni's book or even in Brezi's book, but in the later you have to iterate the estimates that he find only to $W^{1,p}(\Omega)$).

If $\Omega$ is a bounded regular domain, $p\geq 1$, $k>\frac{n}{p}$ then $$W^{k,p}(\Omega)\hookrightarrow C^m(\overline{\Omega}),\ \forall\ 0\leq m<k-\frac{n}{p}$$

In your case: $p=2$, $n=3$ and $k=3$, then $m\in[0,3/2]$, which implies your result.

Title: What is the Sobolev Lemma?


问:
Equality condition for convolution's $L^p$ norm.
Suppose that $1< p< \infty$, $f\in L^1(R)$, and $g\in L^p(R)$ and that $\|f*g\|_p=\|f\|_1\|g\|_p$. Show that then either $f=0$ a.e. or $g=0$ a.e.
I have solved for $g=0$ a.e. if $\|f\|_1>0$ using the equality condition for Hölder ; $\alpha f^p=\beta g^q$ for some $\alpha, \beta \ge 0$. Is this still useful for supposing $\|g\|_p>0$ and deriving $f=0$ a.e?

答:
I am 6 years late but I might as well answer. What you have shown is enough. You have shown that if $||f||_1 > 0$ then $g = 0$ a.e.

Now assume $||g||_p > 0$. This means "$g=0$ a.e" doesn't hold. So the contrapositive of what you have shown means $||f||_1 = 0$, so $f = 0$ a.e.


问:
From an infinite summation to an integral
It is a well known formula that, given a suitably well behaved function, we have

$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=1}^{n} f(k/n) = \int_0^1 f(x)dx.
$$

I am interested in an infinite version of this. Is it true that

$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-\infty}^{\infty} f(k/n) = \int_{-\infty}^\infty f(x)dx?
$$

Edit: We can assume that the function has sufficient smoothness and decays sufficiently fast for the summations and integrations to make sense.

答:
In my opinion it would not work.

You should do something like this:
$$\lim_{n\to\infty} \sum_{h=-\infty}^{+\infty}\frac{1}{n}\sum_{k=1}^{n}f\bigg(\frac{k}{n}+h\bigg)$$


问:
Find X such that 22 divides $11^{2n−1}+X^{3n+1}$.
Need help on inductions, having a few issues.</p>

答:
I take this to mean, find all positive integers $A$ such that $$A^{3n+1}+13^{2n-1}\equiv0\pmod{11}$$ for every positive integer $n$.  This is the same as $$A^{3n+1}\equiv-2^{2n-1}$$ for every positive inter $n$. Suppose this equation is true for some $n$.  When $n$ increases by $1$, the left-hand side is multiplied by $A^3$ and the right-hand side is multiplied by $4$, so we must have $$A^3\equiv4\pmod{11}$$  We find by trial that the only solution mod $11$ is $A\equiv5$.  We have to check $n=1$, and we find $5^4\equiv9\equiv-2\pmod{11}$, so the answer is all integers $$A\equiv5\pmod{11}$$

EDIT

There is one point in the argument that relies on the fact that $11$ is prime.  Do you see where I failed to mention that?


问:
Show the following continuous function has a solution $f(c) = 0$
Let $S = [0,1] \times [0,1]$, and let $f: S \rightarrow \mathbb{R}$ be continuous. Assume for each $a \in S$, there exists a $b \in S$ such that $|f(b)| \leq \frac{1}{2}|f(a)|$. Show that there exists a point $c \in S$ such that $f(c) = 0$.

My idea is to use the intermediate value theorem. So let $a \in S$. Then, there exists $b \in S$ such that $|f(b)| \leq \frac{1}{2}|f(a)|$.

If $f(a) = 0$ or $f(b) = 0$, then we're done. So assume $f(a) \neq 0$ and $f(b) \neq 0$.

Then, from $|f(b)| \leq \frac{1}{2}|f(a)|$, we have $-\frac{1}{2}|f(a)| \leq f(b) \leq \frac{1}{2}|f(a)|$.

But I don't see how I can apply IVT from here since we are not guaranteed that $\pm \frac{1}{2}|f(a)|$ is in the range of $f$. Perhaps $f$ maps $S$ to only positive reals.

答:
Hint

Consider $a_1\in S.$ (If $f(a_1)=0$ we are done. So assume $f(a_1)\ne 0.$) There exists $a_2\in S$ such that $|f(a_2)|\le \dfrac {|f(a_1)|}{2}.$ And there exists $a_3\in S$ such that $|f(a_3)|\le \dfrac {|f(a_2)|}{2}=\dfrac {|f(a_1)|}{2^2}.$

Use that $S$ is compact and $f$ is continuous.


答:
Fix ANY $a_0\in S$.

If $f(a_0)=0$, you are done.

Assume $|f(a_0)|>0$. for $a_0\in S$, there exists $a_1\in S$ such that $|f(a_1)|\le \frac{1}{2}|a_0|$.

For $a_1\in S$, there exissts $a_2\in S$ such that $|f(a_2)|\le \frac{1}{2}|f(a_1)|\le \frac{1}{2^2}|f(a_0)$.

So, by induction, you construct a sequence $(a_n)_{n\ge 0}$ such that $|f(a_n)|\le\frac{1}{2^n}|f(a_0)|$. In particular, $f(a_n)\to 0$.

Since $S$ is compact, there is a subsequence $(a_{n_k})_k$ of $(a_n)$ converging to some point $a\in S$.

Now, since $f$ is continuous, $f(a)=\lim_k f(a_{n_k})=0$.

Title: Show the following continuous function has a solution $f(c) = 0$

Body:
Let $S = [0,1] \times [0,1]$, and let $f: S \rightarrow \mathbb{R}$ be continuous. Assume for each $a \in S$, there exists a $b \in S$ such that $|f(b)| \leq \frac{1}{2}|f(a)|$. Show that there exists a point $c \in S$ such that $f(c) = 0$.

My idea is to use the intermediate value theorem. So let $a \in S$. Then, there exists $b \in S$ such that $|f(b)| \leq \frac{1}{2}|f(a)|$.

If $f(a) = 0$ or $f(b) = 0$, then we're done. So assume $f(a) \neq 0$ and $f(b) \neq 0$.

Then, from $|f(b)| \leq \frac{1}{2}|f(a)|$, we have $-\frac{1}{2}|f(a)| \leq f(b) \leq \frac{1}{2}|f(a)|$.

But I don't see how I can apply IVT from here since we are not guaranteed that $\pm \frac{1}{2}|f(a)|$ is in the range of $f$. Perhaps $f$ maps $S$ to only positive reals.


问:
power series radius of convergence
How can I prove that if the coefficients $\{a_k\}$ of the power series $\sum_{0}^{\infty} \{a_k\}x^k$ form a bounded sequence, then the radius of convergence is at least 1?</p>

答: Use the comparison test on $\sum |a_kx^k|$.</p>

Title: power series radius of convergence


问:
Two random variable with the same variance and mean
Let $Y\in L^{2}(\Omega,\Sigma,P)$ and let $E[Y^2|X]=X^2$ and $E[Y|X]=X$. Could we prove that $Y=X$ almost surely.

My partial answer:

By the definition of conditional expectation we have $E[Y^2]=\int_{\Omega} E[Y^2|X] dP=\int_{\Omega}X^2 dP=E[X^2]$ and $E[Y]=\int_{\Omega} E[Y|X] dP=\int_{\Omega}X dP=E[X]$. Hence, Var$(X)=$Var$(Y)$.

Body: Yes, try $var(X-Y)=E(E[(X-Y)^2|X])$


答:
In general, if you want to prove a statement of the form $A\implies (B\ \text{or}\ C)$, it is logically equivalent to show $(A\ \text{and}\ (\text{not}\ B))\implies C$. The other answer essentially shows how this works in this particular case, but also you could convince yourself of this by making a truth table with columns

$A$,

$B$,

$C$,

$A\implies (B\ \text{or}\ C)$,

$(A\ \text{and}\ (\text{not}\ B))\implies C$,

and then checking that no matter what values of $0$ or $1$ ($\text{False}$ or $\text{True}$) you assign to the first three statements $A, B, C$, both of the last two statements will agree.

Title: Equality condition for convolution's $L^p$ norm.

Body:
Suppose that $1< p< \infty$, $f\in L^1(R)$, and $g\in L^p(R)$ and that $\|f*g\|_p=\|f\|_1\|g\|_p$. Show that then either $f=0$ a.e. or $g=0$ a.e.
I have solved for $g=0$ a.e. if $\|f\|_1>0$ using the equality condition for Hölder ; $\alpha f^p=\beta g^q$ for some $\alpha, \beta \ge 0$. Is this still useful for supposing $\|g\|_p>0$ and deriving $f=0$ a.e?


问:
First Fundamental Theorem of Calculus question regarding F'(x)
The first fundamental theorem is defined:

$F(x) = \int^x_c f(t) dt$ where $c$ is constant.

The corollary:

$\int^b_a f(t) dt = F(b) - F(a)$

My question is how can the integral $[c, x]$ in the first theorem be calculated without the corollary?  Otherwise would $c$ be assumed to be $0$?

As an example:

$f(t) = t^2$,
$F(t) = \int^x_c f(t) dt = \int^x_c t^2 dt$

Antiderivative:
$F(t) = \frac {t^3} 3$

However, my understanding is the antiderivative is the area under the curve from $t = 0$.

Appreciate any guidance.

答:
You are skipping a step. The corollary can be stated:

For a function $F'(x)=f(x)$ (such a function is called an antiderivative)

$$\int_{a}^{b}f(t)dt=F(b)-F(a)$$

Hence it tells you that in order to calculate the definite integral you need to find an antiderivative $F$.


问:
How do you scale along a non canonical direction?
Silly question but I am not figuring it out on my own.

In 3D you can scale a point / shape / field along a canonical direction quite easily:

$p = (x_0, x_1, x_2)$ becomes $p = (ax_0, x_1, x_2)$ for example if we wanted to scale along the first axis. This map has the property that distances in any plane parallel to the plane $(x_1, x_2)$ remain unaffected.

Now given an arbitrary vector $N$ I want to induce the same transformation along the direction of $N$.

One possible solution, of course, is to find a rotation that transforms $N$ into one of the canonical directions, scale that direction, then invert the transformation. i.e. $RSR^{-1}$.

However this is inelegant, I am trying to do it in a way that involves only $N$ and requires no change of basis. i.e. I want to scale points along $N$ without changing the coordinate space, and perhaps better, in a coordinate free way.

答:
The way to construct a matrix that does this is as follows:

The component of $P$ parallel to $N$ is $N\cdot P N = (N^TP)N = NN^TP$

Thus $-NN^T P + P = (-NN^T + I)P$ is P minus its component orthogonal to $N$. Thus $(-NN^T + I)P + sNN^TP$ is the scaling of $P$ along the direction $N$ as described in Chris Eagle's answer.

We get:

$$(-NN^T + I)P + sNN^TP = (-NN^T + I + sNN^T)P$$
$$ = (NN^T(-I + sI) + I)P$$
$$ = (NN^T(s - 1) + I)P$$

Thus the matrix $((s - 1)NN^T + I)$ Represents a linear scaling along the direction $N$ for any $P$ in a coordinate free way.

Notice that matrix must be symmetric btw.


问:
Figuring out shape of four points scattered in a plane(Quadrilateral)

Suppose you're given coordinates of four points $x_1,x_2,x_3,x_4$ , given the fact that the points lie in a plane and polygon is convex. Figure out the shape formed by these four.

So, in a 'brute force' way I have two methods. One is sketching the points in space and the second is to figure out the angles and lengths to classify if it is a parallelogram, square etc.

Are there any other faster algorithms / alternate methods to quickly figure out the shape formed by these points?

Allowed shapes:

Parellogram

Rhombus

Rectangle

Square

Trapezoid

答:
If you know that the points occur along the perimeter of the quadrilateral in the sequence $x_1,$ $x_2,$ $x_3,$ $x_4$
(clockwise or counterclockwise does not matter), then define the vectors

\begin{align}
\mathbf a &= x_2 - x_1, \\
\mathbf b &= x_3 - x_2, \\
\mathbf c &= x_4 - x_3, \\
\mathbf d &= x_1 - x_4.
\end{align}

Suppose you have written each vector in two-dimensional Cartesian coordinates.
Let's denote these coordinates by subscripts $x$ and $y$,
that is, $u_x$ and $u_y$ are the first and second coordinates of the vector
$\mathbf u.$
Define the "dot" product of two vectors $\mathbf u$ and $\mathbf v$ by

$$\mathbf u \cdot \mathbf v = u_xv_x + u_yv_y $$

and the "cross" product of two vectors $\mathbf u$ and $\mathbf v$ by

$$\mathbf u \times \mathbf v = u_xv_y - u_yv_x. $$

(The "cross" product is the dot product of $\mathbf u$ with a vector of the same length as $\mathbf v$ in a direction $90$ degrees from the direction of $\mathbf v$.)

Then $\mathbf a,$ $\mathbf b,$ $\mathbf c,$ and $\mathbf d$
give the distance and direction you travel along each edge of the quadrilateral as you make one full circuit around the perimeter.

If $\mathbf a \times \mathbf c = 0$ then the two opposite sides represented by $\mathbf a$ and $\mathbf c$ are parallel.
If $\mathbf b \times \mathbf d = 0$ then the two opposite sides represented by $\mathbf b$ and $\mathbf d$ are parallel.

If neither pair is parallel, you have an unclassified quadrilateral.

If one pair is parallel and the other is not, you have a trapezoid.

If both pairs are parallel you have some variety of parallelogram
(using the definition of "parallelogram" in which rectangles and rhombi are parallelograms),
which you can further classify using the following criteria:

If $\mathbf a \cdot \mathbf b = 0$ then the two adjacent sides represented by
$\mathbf a$ and $\mathbf b$ are perpendicular. Otherwise they are not perpendicular.

If $\mathbf a \cdot \mathbf a = \mathbf b \cdot \mathbf b$
then the two sides are equal in length. Otherwise they are unequal in length.

Using those facts, you can decide whether the shape is a square,
a non-square rhombus, a non-square rectangle,
or a parallelogram that is neither a rectangle nor a rhombus.

If you have two-dimensional Cartesian coordinates for all points but
do not know the sequence in which the given points appear around the quadrilateral, then make a guess that they have been presented in some cyclic sequence (either clockwise or counterclockwise without skipping).
Define the vectors $\mathbf a,$ $\mathbf b,$ $\mathbf c,$ and $\mathbf d$
as at the start of this answer.

Compute $\mathbf a \times \mathbf b,$ $\mathbf b \times \mathbf c,$
and $\mathbf c \times \mathbf d.$
If the answers all have the same sign, the points are in cyclic order
and you can simply continue as described above.
Otherwise you will have to swap two points.

If $\mathbf a \times \mathbf b$ and $\mathbf b \times \mathbf c$
have opposite signs, you can swap the labels of the two endpoints of $\mathbf a$
or $\mathbf c$.
If $\mathbf b \times \mathbf c$ and $\mathbf c \times \mathbf d$
have opposite signs, you can swap the labels of the two endpoints of $\mathbf b$
or $\mathbf d$.
(But don't swap two pairs of points!)

If you are not guaranteed that the four points are vertices of a non-trivial convex quadrilateral then you need some additional tests to ensure that no point is on or inside the triangle formed by the other three.


问:
Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$

Let $T \in \text {SO} (3) : = \left \{T \in M_3 (\Bbb R)\ |\ TT^t = T^tT = I,\ \det (T) = 1 \right \}.$ Prove that there exists $v \in \Bbb R^3 \setminus \{0\}$ such that $T(v) = v.$ Hence conclude that the elements of $\text {SO} (3)$ are rotations by some angle about some uniquely determined axis.

The first part is easier. To prove that $T(v) = v,$ for some $0 \neq v \in \Bbb R^3$ it is enough to show that $\det (T - I) = 0.$ Now $$\begin{align*} \det (T - I) & = 1 \cdot \det (T - I) \\ & = \det (T) \det (T - I) \\ & = \det (T^t) \det (T - I) \\ & = \det (T^t T - T^t) \\ & = \det (I - T^t) \\ & = \det (I^t - T^t) \\ & = \det ((I - T)^t) \\ & = \det (I - T) \\ & = (-1)^3 \det (T - I) \\ & = - \det (T - I) \end{align*}$$ This shows that $2 \det (T - I) = 0 \implies \det (T - I) = 0,$ as required.

How to prove the second conclusion? Any help in this regard will be appreciated.

Thanks in advance.

答: Take $v\ne0$ such that $T(v)=v$. Let $w\in\Bbb R^3$ be a vector orthogonal to $v$. Then\begin{align}\langle T(w),v\rangle&=\langle T(w),T(v)\rangle\\&=\langle w,T^tT(v)\rangle\\&=\langle w,v\rangle\\&=0.\end{align}Now let $u=v\times w$. Then $u$ and $v$ are orthogonal and therefore $T(u)$ and $v$ are orthogonal too. And, since $w$ and $u$ are orthogonal, then so are $T(w)$ and $T(u)$. The restriction of $T$ to $\operatorname{span}(\{w,u\})$ is a rotation (since it is an isometry and its determinant is $1$). So, $T$ is a rotation around $v$.


问:
How to Tell That this Relation is AntiSymmetric
We have $A = \{1,2,3,4 \}$ and the relation $R = \{(2,1), (3,1), (3,2), (4,1), (4,2), (4,3) \}$.

My book is saying that this is antisymmetric, but I cannot figure out why. I know the definition is $\forall a \forall b(((a,b)\in R \land (b,a)\in R) \rightarrow (a=b)).$

I used the definition here and I am not getting anywhere. Does someone else have a recommendation on how to look at this problem?

答: Are there elements $a,b\in A$ such that both $(a,b)$ and $(b,a)$ belong to $R$? No, there aren't. So, the assertion$$(a,b)\in R\wedge(b,a)\in R\implies a=b$$is vacuously true.


问:
Generalize a sum given first elements
Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as

$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$

generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)

$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$

I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.

答:
Try $f(n)=(n+1)+(n)\rho+(n-1)\rho^2+\cdot\cdot\cdot+(1)\cdot\rho^n=\sum\limits_{i=1}^{n+1} i\rho^{n+1-i} = \sum\limits_{i=0}^{n} (i+1)\rho^{n-i}$

$f(n)$ has $n+1$ terms in its sum where the sum of the coefficient of each term and the power of $\rho$ is $n+1$.

Title: Generalize a sum given first elements


答:
One possible generalized $f$  is $$f(n)=\sum_{i=0}^{n} (i+1){\rho}^{n-i}$$.
There can be many $f$. But here it fits.

Title: Generalize a sum given first elements

Body:
Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as

$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$

generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)

$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$

I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.


答: I'm not sure what you're trying to do with your example, but one way of expressing $f(n)$ with a general formula is $$f(n) = \sum_{i=0}^{n} (n-i+1)\rho^i.$$

Title: Generalize a sum given first elements


问:
Eigenvector Proof $(I+A)^{-1}$.
Show that the eigenvectors of the $n \times n$ matrix A are also eigenvectors of the matrix $$M = (I+A)^{-1} $$ Where I is the $n \times n$ unit matrix. Determine the eigenvalues.

My Work:

$$Mx=(I+A)^{-1}x = ???$$

答:
Hint: if you have
$$
Ax=\lambda x
$$
then what do you get when you add $x$ to both sides?
$$
(I+A)x = (1+\lambda)x
$$
Now do a bit of matrix algebra to get $(I+A)^{-1}$ on the right and $1+\lambda$ on the left.

(Alternative way to think of it - if $(I+A)^{-1}x = \Lambda x$, then how can you rearrange this to get $Ax$ on its own?)

Title: Eigenvector Proof $(I+A)^{-1}$.


问:
Is every triangle with 3 rational angles (in degrees) similar to one with 3 integer-length sides?
The same way right triangles have pythagorean triples, does every non-right class of similar triangles also have some integer triples, if we say that all the angles have to be rational numbers in degrees?

Are there other interesting results about integer triples if we relax/tighten the restrictions on the angle values?

答: No. Consider a triangle with angles 30, 60, 90.</p>


答:
Let $x_n=A^{3n+1} + 13^{2n−1}$. Then $x_{n+2}= a x_{n+1} + b x_n$, where $a,b \in \mathbb Z$ whose values are irrelevant; the important point is that they are integers.

Therefore, if $11$ divides $x_1$ and $x_2$, then $11$ divides $x_n$ for all $n$. Now,

$11$ divides $x_1=A^4+13$ iff $A \equiv \pm 5 \bmod 11$.

$11$ divides $x_2=A^7+13^3$ iff $A \equiv 5 \bmod 11$.

Bottom line: $A \equiv 5 \bmod 11$.

Title: Find X such that 22 divides $11^{2n−1}+X^{3n+1}$.

Body: Need help on inductions, having a few issues.</p>


问:
example of subspaces involving inequalities
$A+B=\left \{ a+b : a\in A, b\in B \right \}$

Let $a_1,\ldots,a_n$ be elements of $A$ and let $b_1,\ldots,b_k$ be elements of $B$ be two families of vectors in $C$. Let $A=\operatorname{Span}(a_1,\ldots,a_n)$ and $b=\operatorname{Span}(b_1,\ldots,b_k)$

problem
I have shown that $A+B$ is a subspace of $C$ but I am unsure on some inequalities using dimension of the spaces.

Find subspaces $A_1, B_1$ of $C=\mathbb{R}^3$ of dimension $n=1$ and  $k=2$ such that $\dim(A_1+B_1) <n+k=3$ and an example of a subspace $A_2, B_2$ of $C=\mathbb{R}^3$ of same dimension $n=1$ and $k=2$ but this time so that $\dim(A_2+B_2)=n+k=3$

Body:
If $A_1=\{(x,0,0)\mid x\in\Bbb R\}$ and $B_1=\{(x,y,0)\mid x,y\in\Bbb R\}$, then $\dim A_1=1$, $\dim B_1=2$, and, since $A_1+B_1=B_1$, $\dim(A_1+B_1)=2$.

But if $A_2=\{(x,0,0)\mid x\in\Bbb R\}$ and $B_2=\{(0,x,y)\mid x,y\in\Bbb R\}$, then $\dim A_2=1$, $\dim B_2=2$, and, since $A_2+B_2=\Bbb R^3$, $\dim(A_2+B_2)=3$.


答:
$f$ can be represented as a reccurence relation by defining $f(0)=1$.

$$f(r)=\rho f(r-1)+(r+1)\forall r\in\mathbb{Z^{+}}$$

Title: Generalize a sum given first elements


问:
Let $X_i$ be i.i.d discrete random variables and $M = max_i(X_i)$. Find $\mathbb{P}(M = m)$
A known approach for a similar problem, $\mathbb{P}(M \le m)$, can be formulated as

\begin{eqnarray}
\mathbb{P}(M \le m) = \prod_{i}\mathbb{P}(X_i \le m)
\end{eqnarray}

However, by dealing with a specific value, there can be multiple ways of achieving the desired outcome. For instance, let $i = \{{ 1, 2\}}$:

\begin{eqnarray}
\mathbb{P}(M = m) = \mathbb{P}(X_1 = m;\; X_2 < m) \; \cup \;\mathbb{P}(X_1 < m;\; X_2 = m)\; \cup \;\mathbb{P}(X_1 = m;\; X_2 = m)
\end{eqnarray}

In order to overcome the necessity of counting for all the possible combinations, I thought about considering the 'range approach', as

\begin{eqnarray}
\mathbb{P}(M = m) &=& 1 - \mathbb{P}(M < m) - \mathbb{P}(M > m) \\
&=& 1 - \prod_{i}\mathbb{P}(X_i < m) - \prod_{i}\mathbb{P}(X_i > m)
\end{eqnarray}

I was wondering if there is any mistake in this approach to tackle the original problem. If not, what would be other options?

Thanks in advance!

答:
Your first equality is correct. However:

\begin{eqnarray}
\mathbb{P}(M = m) &=& 1 - \mathbb{P}(M < m) - \mathbb{P}(M > m) \\
&=& 1 - \prod_{i}\mathbb{P}(X_i < m) - (1-\prod_{i}\mathbb{P}(X_i \leq m))\\
&=&\prod_{i}\mathbb{P}(X_i \leq m))-\prod_{i}\mathbb{P}(X_i < m).
\end{eqnarray}

Otherwise you can use this approach.


问:
How to avoid overflow when evaluating the exponential smoothing function?
The exponential smoothing function is $f:\Bbb R^n\to \Bbb R$ defined as

$$f(x):= \log\left(\sum_{i=1}^{n}e^{x_i}\right).$$ Obviously, when $x_i$ is large for some $i$, the term inside the logarithmic function goes to infinity. How should we implement the exponential smoothing function in a computer so as to avoid overflow?

The only thing I can think of is to use the identity

$$f(x):= \log\left(\sum_{i=1}^{n}e^{(x_i- L)}\right) + L$$ for any $L$, and choose $L$ large enough. Are there other practical(better) ways to do this?

答:
Let $L=x_n=\max_ix_i$ wlog. The function may then be written as

$$f=x_n+\log\left(1+\sum_{i=1}^{n-1}\exp(x_i-x_n)\right)$$

which does not incur any overflow for any individual terms, which are all less than $1$. It is worth noting here that for small $x$ one has

$$\log(1+x)\sim x$$

so that if the exponential terms in the sum underflow it should not be an issue, since

$$x_n+\log(1+\epsilon)\sim x_n+\epsilon\sim x_n$$

cannot be differentiated in floating point arithmetic unless $x_n=0$.

Aside from over/underflow, this form has the advantage of one less exponential term to compute and it is more accurate to compute $\operatorname{Lnp1}(x)=\log(1+x)$ directly for small $x$ instead of adding $1$ and then taking the logarithm. See here.


问:
Homology of $S^1 \times (S^1 \vee S^1)$
I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology.

Question: Calculate the homology groups of $X=S^1 \times (S^1 \vee S^1)$.

My attempt: I try to use the Mayer-Vietoris sequence.

Let $A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1$ and $B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1$. (Unsure how to express this, but hopefully this is clear enough.)

Then $A\cap B=S^1\times \{\text{point}\}=S^1$ and $A\cup B=X$.

The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows:

$$0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$

Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps?

答:
If your coefficients are in a field, then you can use the Künneth formula, which says that if $X=Y \times Z$
$$ H^i(X) = \bigoplus_{j+l=i} H^j(Y) \otimes H^{l}(Z). $$

So, for example to calculate $H^1(X)$, there are two terms in sequence, namely when $(j,l)=(0,1)$ and when $(j,l)=(1,0)$. So
$$ H^1(X) = H^1(Y) \otimes H^0(Z) \bigoplus H^0(Y) \otimes H^1(Z).$$

In your case, $Y=S^1$ and $Z=S^1 \lor S^1$, so that
$$ H^1(X) = H^1(S^1) \otimes H^0(S^1 \lor S^1) \bigoplus H^0(S^1) \otimes H^1(S^1 \lor S^1)$$
$$ = k \otimes k \bigoplus k \otimes(k^2) = k \oplus k^2 = k^3.$$

Title: Homology of $S^1 \times (S^1 \vee S^1)$


问:
Relation between isomorphism and bases
Let $V$ and $W$ be finite dimensional vector spaces over $\Bbb{F}$ and $T : V → W$ is a linear map. Prove that $T$ is an isomorphism if and only if ${T(v_1 ),T(v_2 ),...,T(v_n)}$ is a basis for $W$ whenever ${v_1 ,v_2 ,...,v_n}$ is a basis for $V$.

答:
Let $T:V \rightarrow W$ be an isomorphism then $Ker(T)=\{0\}$ and $v_1,...,v_n$ be a basis

$a_1T(v_1)+...+a_nT(v_n)=0$ $\Rightarrow$ $T(a_1v_1+...+a_nv_n)=0$

$T(a_1v_1+...+a_nv_n)=0$   $\Rightarrow$ $a_1v_1+...+a_nv_n=0$

$a_1v_1+...+a_nv_n=0$ $\Rightarrow$ $a_1=...=a_n=0$

which means $T(v_1),...,T(v_n)$  forms a basis

Let ${T(v_1 ),T(v_2 ),...,T(v_n)}$ be a basis  and $\{v_1,...,v_n\}$ be a basis for $V$

$T(a_1v_1+...+a_nv_n)=T(b_1v_1+...+b_nv_n)=0$   $\Rightarrow$ $T((a_1-b_1)v_1+...+(a_n-b_n)v_n)=0$

$T((a_1-b_1)v_1+...+(a_n-b_n)v_n)=(a_1-b_1)T(v_1)+...+(a_n-b_n)T(v_n)=0$ $\Rightarrow$ $a_1=b_1 ,...,a_n=b_n$

which means all elements of $Ker(T)$ are equal each other so $Ker(T)=\{0\}$ that implies $T$ is an isomorphism

Title: Relation between isomorphism and bases


问:
Prove two graphs are isomorphic from geometric duals.
Prove that two graphs $G_1$ and $G_2$ are isomorphic if and only if their geometric duals $G_1^*$ and $G_2^*$ are also isomorphic.

答:
This is false. Consider the following graphs:

Here, clearly $G \cong H$ but $G^* \ncong H^*$ since $G^*$ has a vertex of degree $5$ whereas $H^*$ does not. Note that the colors of the vertices are chosen so that they represent the color of the faces of the original graphs. White is used for the exterior face.


问:
How can one calculate the geodesic curvature of arbitrary arcs on a sphere?
Consider that one regular hexagon and one square cut a sphere of radius $r$ as shown in the attached figure. A circular arc of length $s_1$ and a circular arc of length $s_2$ are formed (blue arcs). The arc $AB$ is a part of a small circle centered at a given point $O_1$, while the arc $AC$ is an arbitrary arc of a circle centered at the point $O_2$.

Assuming that the coordinates of the points $A, B, C$ with respect to a common coordinate system are known, how can we compute the geodesic curvature of these circular arcs?

Any tip/help would be useful.

My approach: Given the formula presented in Geodesic curvature of sphere parallels, the small circle geodesic curvature is given by $\kappa_g =
\frac{\sqrt{r^2 - r_1^2}}{rr_1}$, where $r_1$ denotes the radius of the small circle centered at $O_1$.

Similarly, how can we compute $\kappa_g$ for the other arbitrary circular arc?

答: That seems an exceedingly cumbersome formula and I haven't checked it. It is a standard formula (which can be checked with a tiny bit of trig, or by using the formula $\kappa_g = \pm\sqrt{\kappa^2-\kappa_n^2}$) that a geodesic circle of radius $R$ on a sphere of radius $r$ has $\kappa_g = \dfrac1r\cot R$. Does this help you? (Your circular arcs are arcs of geodesic circles centered at $O_1$ and $O_2$.)


问:
Defining "distributive category" by requiring the existence of an isomorphism instead of requiring that the canonical morphism is an iso
A category $\mathfrak C$ is distributive if it has binary products $\times$ and binary coproducts $\sqcup$, and the canonical morphism
$$X\times Y\sqcup X\times Z\longrightarrow X\times(Y\sqcup Z)$$
is an iso.

Why does one require that this particular morphism is an isomorphism? Why doesn't one require that there exists an isomorphism between $X\times Y\sqcup X\times Z$ and $X\times(Y\sqcup Z)$?

In particular: Is a category distributive if and only if there's a natural isomorphism between $X\times Y\sqcup X\times Z$ and $X\times(Y\sqcup Z)$ (natural in $X,Y,Z$)?

答:
Distributivity can be phrased abstractly as follows. There are 2-monads $\mathrm{Cart}$ and $\mathrm{Cocart}$ for cartesian categories and cocartesian categories respectively, between which there exists a distributive law establishing $\mathrm{Cocart} \circ \mathrm{Cart}$ as a composite 2-monad. The algebras for this 2-monad are exactly distributive categories. An algebra for a distributive law is an algebra for each of the two monads forming the composite, together with the requirement that the latter structure is preserved by the former structure: this is precisely the condition Qiaochu Yuan mentions in the comments about the finite product functors preserving finite coproducts. This shows that the distributivity condition is very natural, arising from distributive laws of 2-monads.

However, it happens (surprisingly) that the distributivity condition can be replaced by the requirement for any such natural transformation to be invertible. This is proven in Lack's Non-canonical isomorphisms. One should not expect this in general, but under certain circumstances holds more generally: for this, see Nunes's Pseudoalgebras and non-canonical isomorphisms.


问:
What is the Direction of a Zero (Null) Vector?
To be more precise, I am interested in knowing if the intuition that a Euclidean zero vector does not have a particular direction is actually correct, and if there is a rigorous formulation that would back it up.

Wikipedia's entries on the zero vector seems to agree with that intuition, but as always, one shouldn't blindly trust Wikipedia: in one place, it is stated that a zero vector "is orthogonal to all other vectors with the same number of components," while in another, it is stated that "two vectors can be considered orthogonal if and only if their dot product is zero, and they have non-null length." Correct me if I am wrong, but these two statements contradict each other.

This question popped up in my head when I heard someone arguing that since a Euclidean vector is defined as a geometric entity that has both a magnitude and direction, and since a zero vector is a vector with length 0, then it is only fitting for a zero vector to have a "direction 0." I am personally inclined to say that a zero vector does not have a particular direction, but as I said, I would like to know if there is a rigorous formulation that would lead to this conclusion.

And to put the question in a less "discussion-inducing" form, is there an agreement on the direction of a Euclidean zero vector?

答: when we take cross product of two vectors which are perpendicular, we get a vector having magnitude zero. But we know the direction of this vector by using right hand thumb rule or screw rule. so we know the direction of a null vector...! what do you say about this?</p>

ContentLicense: CC BY-SA 2.5


问:
Why squaring two absolute value functions never gets answers that don't satisfy the domain?

$|x-1| = |3x-1|+1$

Where $f(x)=|x−1|$ then $f(x)=x−1$ in a certain domain, $x≥1$.
And $f(x)=−(x−1)$ in a certain domain, $x≤1$.

And $g(x)=|3x-1|+1$ then $g(x)=3x-1+1$ in a certain domain, $x≥1/3$ and $g(x)= -(3x-1)+1$ in a certain domain, $x≤1/3$.

I know this can be solved by squaring both sides and I'll get no solutions. And drawing them on a graph paper supports that.

$(x-1)^2 = (|3x-1|+1)^2$  →$8x^2-4x+2|3x-1|+1=0$ , $8x^2-4x+1$ can never be negative and so is $2|3x-1|$. So, there's no solution.

But doesn't we mean by squaring both sides that $x-1$ can be either negative or positive.

So it's like we are saying:

$-(x-1)$ can be equal to $-(3x-1)+1$, which has an answer $x = 0.5$.

However, this answer doesn't satisfy the domain of $g(x)$ when its equal to $-(3x-1)+1$, $x≤1/3$ so it's not correct.

Or

$-(x-1)$ can be equal to $3x-1+1$, which also has an answer $x=0.25$.

However this answer doesn't satisfy the domain of $g(x)$ when its equal to $(3x-1)+1$, $x≥1/3$ so isn't correct.

Etc.

But these answers since they are out of the domains of both functions aren't right but how can squaring guess that, why are the answers we get out of squaring always in the domain (if both functions are absolute value, I'm not talking about a case like this when only one function has an absolute value sign: $|x-1|=3x-1+1$) how can squaring know where the domain ends and starts, it seems like magic which surely can't be the case.

I tried to ask my teacher but he couldn't get my point, I hope someone will.

答: I don't know what you mean by "why are the answers we get out of squaring always in the domain".  For example, take the equation $|x| = -1$.  Obviously this equation has no solutions. If we square both sides, we get $x^2 = 1$ whose solutions are $x = \pm 1$, but neither value is a solution of the original equation. Is this what you are referring to?


问:
From an infinite summation to an integral
It is a well known formula that, given a suitably well behaved function, we have

$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=1}^{n} f(k/n) = \int_0^1 f(x)dx.
$$

I am interested in an infinite version of this. Is it true that

$$
\lim_{n\to\infty} \dfrac{1}{n}\sum_{k=-\infty}^{\infty} f(k/n) = \int_{-\infty}^\infty f(x)dx?
$$

Edit: We can assume that the function has sufficient smoothness and decays sufficiently fast for the summations and integrations to make sense.

答:
Let $\varphi_n(t)=f\left(\frac{\lfloor nt\rfloor}{n}\right)$, then
$$ \int_{-\infty}^{+\infty}\varphi_n(t)dt=\sum_{k=-\infty}^{+\infty}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\varphi_n(t)dt=\frac{1}{n}\sum_{k=-\infty}^{+\infty}f\left(\frac{k}{n}\right) $$
However for all $t\in\mathbb{R}$, $\lim\limits_{n\rightarrow +\infty}\varphi_n(t)=f(t)$ (I suppose that $f$ is continuous). If we get an upper bound $|\varphi_n|\leqslant g$ where $g\in L^1(\mathbb{R})$ then we can use the dominated convergence theorem, therefore
$$ \lim\limits_{n\rightarrow+\infty}\int_{-\infty}^{+\infty}\varphi_n(t)dt=\int_{-\infty}^{+\infty}f(t)dt $$
and you get the desired equality. In order to have such an upper bound, we can suppose that $f$ is monotone, thus we have $|\varphi_n(t)|\leqslant |f(t)|$ if $f$ is non-decreasing and $|\varphi_n(t)|\leqslant |f(t-1)|$ if $f$ is decreasing. We can also suppose that there exists $\delta>0$ such that $f(t)=\underset{t\rightarrow\pm\infty}{\mathcal{O}}\left(\frac{1}{t^{1+\delta}}\right)$ because there would exist $M>0$ such that $|f(t)|\leqslant\frac{M}{(1+|t|)^{1+\delta}}$ for all $t\in\mathbb{R}$ and therefore $|\varphi_n(t)|\leqslant\frac{C}{(1+|t|)^{1+\delta}}$ where $C>0$.


问:
Does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ where $A = A_0 + O_p(n^{-1})$?
Let $A$ be a random matrix that converges in probability to a constant matrix $A_0$ at the rate $O_p(n^{-1})$:
$$
A = A_0 + O_p(n^{-1}).
$$

I want to know if $A^{-1} = A_0^{-1} + O_p(n^{-1})$?

Let $I$ be the identity matrix. Here is my attempt at proving it, which I'm not sure is valid mainly because I don't know if we are allowed to define a series expansion in the form of $O_p$ variables.

$$
\begin{align}
A^{-1} &= (A_0 + O_p(n^{-1}))^{-1} \\
&= (A_0(I + A_0^{-1}O_p(n^{-1}))^{-1} \\
&= (A_0(I + O_p(n^{-1}))^{-1} \\
&= (I + O_p(n^{-1}))^{-1}A_0^{-1}.
\end{align}
$$
Now we perform a Neumann series expansion which gives
$$
\begin{align}
(I + O_p(n^{-1}))^{-1}
&= \sum_{k=0}^\infty (-1)^{k} (O_p(n^{-1}))^{k} \\
&= I - O_p(n^{-1}) \\
&= I + O_p(n^{-1}).
\end{align}
$$
I have assumed here that $(O_p(n^{-1}))^0$, is this true?

Then we substitute this into $(I + O_p(n^{-1}))^{-1}A_0^{-1}$ to get

$$
\begin{align}
A^{-1} &= (I + O_p(n^{-1}))A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1})A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1}).
\end{align}
$$

In the deterministic case, if we wanted to invert the matrix $(I+K)$ we need to have $||K|| < 1$ to ensure convergence. So does this mean that in the probabilistic case, to invert the matrix $(I + O_p(n^{-1}))$ we need to have $||O_p(n^{-1})|| < 1$? So the norm of an $O_p(n^{-1})$ term is required to be less than one..does this phrase even make sense?

It seems an $O_p(n^{-1})$ term will be increasingly likely to have a value less than $1$ as $n \to $\infty$, since the probability is becoming more concentrated around zero, but I don't think it can ever be fully guaranteed to be below one?

So does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ really hold and if it does is my proof valid? If not, what is a valid proof?

答:
Set $A_n$ be a random matrix which converges to some constant matrix $A_0$ in probability at rate $\mathcal{O}(n^{-1})$, as $n\rightarrow \infty$. That is, for all $\epsilon>0$, there exists a constants $M, N>0$, with the property that
$$ \mathbb{P}\left( \left\|A_N - A_0\right\| > \frac{M}{n} \right)< \epsilon \qquad , n\geq N.
$$
Under the assumption that $A_n$ and $A_0$ are invertible, we write $B_n = A_n - A_0$ and observe that by the Neumann series expansion gives
\begin{equation} A^{-1}_n = A^{-1}_0 \left( I + B_n A^{-1}_0\right)^{-1} = A^{-1}_0\left( I + \sum_{k=1}^{\infty} \left(-B_n A^{-1}_0\right)^k \right),
\end{equation}
at least whenever $\left\|B_n A^{-1}_0\right\|< 1$, which may possibly depend on $n$ and points in underlying sample space. Anyway, to remedy this effect, we can simply use the conditional probability to write
$$ \mathbb{P}\left( \left\|A^{-1}_n - A^{-1}_0\right\| > \frac{M}{n} \right) \leq \mathbb{P}\left( \left\|A^{-1}_n - A^{-1}_0\right\| > \frac{M}{n} , \left\|A_n - A_0\right\|< \frac{M_1}{n} \right) + \mathbb{P}\left(\left\|A_n - A_0\right\|\geq \frac{M_1}{n} \right),
$$
Where $n > M_1 ,M$, for appropriately chosen $M,M_1>0$ to be determined at the end.
In order to treat the first term, we primarily notice that the previously derived Neumann series expansion implies
$$ \left\|A^{-1}_n - A^{-1}_0\right\| \leq \frac{\left\|A^{-1}_0\right\|\cdot \left\|B_n A^{-1}_0\right\|}{1-\left\|B_n A^{-1}_0\right\|},
$$
whenever $\left\|B_n A^{-1}_0\right\| < 1$, which in particular holds as long as
$\left\|A_n - A_0\right\| < \frac{M_{1}}{n}$, for sufficiently large $n>1$. With this at hand we get
$$ \mathbb{P}\left( \left\|A^{-1}_n - A^{-1}_0\right\| > \frac{M}{n} , \left\|A_n - A_0\right\|< \frac{M_1}{n} \right) \leq \mathbb{P} \left(\frac{\left\|A^{-1}_0\right\|\left\|B_n A^{-1}_0\right\|}{1-\left\|B_n A^{-1}_0\right\|} > \frac{M}{n} \right)= \mathbb{P} \left( \left\|B_n A^{-1}_0\right\|> \frac{M/n}{M/n+ \left\|A^{-1}_0\right\|}\right)
$$
$$ \leq \mathbb{P}\left(\left\|A_n- A_0\right\| > \frac{1}{n}\frac{M \left\| A^{-1}_0\right\|^{-1}}{1+ \left\|A^{-1}_0\right\|} \right).
$$
Adding up and choosing $M_1, M$ accordingly, answers your question in the affirmative.


问:
Why well-defined function sometimes means a group homomorphism function?
If we have the map $\phi_s$ that takes $\overline{1}\in\mathbb{Z}/p^a$ to $\overline{p^{s+b}/d}\in\mathbb{Z}/p^b$, where $d=\gcd(p^a,p^b)$. Why in order to see that $\phi_s$ is a well-defined map, we need to show that $\overline{p^a}\phi_s(\overline{1})=\overline{0}\in\mathbb{Z}/p^b$?

EDIT:

Is the meaning of well defined if we have $\mathbb Z_{n}'s$ as domain and target is just to send the identity to the identity?If so, for me, that definition of well defined is very different from the one ordinary used (if $x=y$ then $f(x) = f(y)$)

Is there any connection between both definitions?

答:
It seems that the question implicitly requires the map to be an abelian group homomorphism, i.e. $\phi_s(x+y)=\phi_s(x)+\phi_s(y)$ for all $x,y$ in the domain. In particular, if we write $n\cdot x$ for $x+\ldots+x(n$ times$)$ (which is essentially viewing abelian groups as $\def\Z{\mathbb Z}\Z$-modules), then $\phi_s(n\cdot x)=n\cdot\phi_s(x)$. In this case it is the same as $\bar n\phi_s(x)$.

For the map to be well-defined, the problem is $\phi_s(\bar0)$. In $\Z/p^a\Z$, $\bar0=\bar{p^a}=p^a\cdot\bar1$. Hence $\bar0=\phi_s(\bar0)=\phi_s(p^a\cdot\bar1)=p^a\cdot\phi_s(\bar 1)=\bar p^a\phi_s(\bar 1)$, which is what you want to check.


答:
Let $\lambda$ be the eigenvalue for the eivenvector $x$.

Your work, with a minor addition:

$$(I+A)^{-1}x=\lambda x\implies x=(I+A)\lambda x $$

By expanding and reorganizing, you will see $x$ is an eigenvector for $A$, but the eigenvalue that appears might be something you didn't guess.

Can you take it from here?

Title: Eigenvector Proof $(I+A)^{-1}$.

Body:
Show that the eigenvectors of the $n \times n$ matrix A are also eigenvectors of the matrix $$M = (I+A)^{-1} $$ Where I is the $n \times n$ unit matrix. Determine the eigenvalues.

My Work:

$$Mx=(I+A)^{-1}x = ???$$


问:
How to use slopes (3 points are given) to prove that they form a right triangle?
Question: Use slopes to show that $A(-3, -1)$, $B(3, 3)$ and $C(-9, 8)$ are vertices of a right triangle.

My try at the problem: I know that we can find the slopes of $AB$, $BC$ and $CA$ and then prove that the slope of one line is the negative reciprocal of another and hence, two lines are perpendicular. However, it can be the case that two of the lines are perpendicular, but the other one is not perpendicular/parallel to either of them as follows:

In these case, we do see two lines perpendicular to each other, but no right triangle. How to we actually solve the problem them?

答:
There are actually two process where we can show its a right triangle

By showing that any two of those line's slope are perpendicular.
Actually Its a right triangle so basically multiplying the slope of the hypotenuse wont get the result with either of the lines. We have to find out all the slopes of those lines and multiply all of them simultaneously.

Given, A(−3,−1), B(3,3) and C(−9,8)
Slope=(difference of y/difference of x)
Slope of AB=2/3,
Slope of AC=-3/2,
Slope of BC=-5/12
Now we have to multiply any two of them two find out whether they are perpendicular or not
By using my intuitive, I could spot it and it should be AB x AC=-1
If you are still unable to reckon that you can try all of them
You can do all of them AB x BC, BC x AC.

Another method is calculating the distance between those points and find out the length then use the Pythagorean theorem
Thanks for asking


问:
asymptotic statement and one example
how I can logically understand this is false ($c$ is constant):

$ \frac{n!}{c} \leq 2^n$

why it was false for large value from asymptotic notation concept? ( i see this is true for some $n < n_0$ but not hold for large value).

答:
For each value of $c$ you can find out exactly where the inequality switches over by checking numerically. For example, for $c=3$ the change is between $n=4$ and $n=5$. There is no formula.

The argument that for any particular $c$ it happens eventually does not depend on knowing just where. It usually does not matter just where - that's when knowing what happens "asymptotically" is all you care about.

Edit in response to comment.

You have to show that for any particular $c$, eventually
$$
x_n = \frac{n!}{2^n} > c.
$$
When $n > 5$, show that each $x_n$ is more than twice its predecessor. That should be enough to finish the proof.

Title: asymptotic statement and one example


问:
Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$

Let $T \in \text {SO} (3) : = \left \{T \in M_3 (\Bbb R)\ |\ TT^t = T^tT = I,\ \det (T) = 1 \right \}.$ Prove that there exists $v \in \Bbb R^3 \setminus \{0\}$ such that $T(v) = v.$ Hence conclude that the elements of $\text {SO} (3)$ are rotations by some angle about some uniquely determined axis.

The first part is easier. To prove that $T(v) = v,$ for some $0 \neq v \in \Bbb R^3$ it is enough to show that $\det (T - I) = 0.$ Now $$\begin{align*} \det (T - I) & = 1 \cdot \det (T - I) \\ & = \det (T) \det (T - I) \\ & = \det (T^t) \det (T - I) \\ & = \det (T^t T - T^t) \\ & = \det (I - T^t) \\ & = \det (I^t - T^t) \\ & = \det ((I - T)^t) \\ & = \det (I - T) \\ & = (-1)^3 \det (T - I) \\ & = - \det (T - I) \end{align*}$$ This shows that $2 \det (T - I) = 0 \implies \det (T - I) = 0,$ as required.

How to prove the second conclusion? Any help in this regard will be appreciated.

Thanks in advance.

答:
As you showed that there is $v_1$ such that $Tv_1=v_1$. Let $P$ be the plane whose normal vector is $v_1$. Firstly we should show $Tx \in P$ for all $x \in P$. This equivalent to show that $Tx$ and $v_1$ are orthogonal.

$<Tx,v_1>=<Tx,Tv_1>=x^TT^TTv_1=x^Tv_1=0 \hspace{0.2cm} \text{since $x$ and $v_1$ are orthogonal.}$

Secondly, we must show that the induced map $T|_P$ is rotation. Let this be denoted by $G$. So $G$ can be represented by $2 \times 2$ matrix with eigenvalues $e,f$.

Actually $e$ and $f$ are eigenvalues of $T$, too. So they are $e=e^{i \theta}$ and $f=e^{-i \theta}$.

Since the eigenvalues are distinct, we can directly say that $G$ is similar to the matrix $\begin{bmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta
\end{bmatrix}$. We know that similar matrices represent same linear transformation, we can directly to say that $G$ is a rotation.


问:
Result on determinant of matrix
Let $A$ and $B$ be two $2×2$ real matrix. Then which of the following is true:

$det(A+B)+det(A-B)=det(A)+det(B)$

$det(A+B)+det(A-B)=2\times det(A)-2\times det(B)$

$det(A+B)+det(A-B)=2\times det(A)+2\times det(B)$

$det(A+B)-det(A-B)=2\times det(A)+2\times det(B)$

My try: Take $A=I$ and $ B=-I$ then clearly option 1,2 and 4 are discarded. But I cannot prove option 3. So please help me. Thank you in advance

答:
We have
$$
\det(A+B)+\det(A-B)=2a_1a_4 - 2a_2a_3 + 2b_1b_4 - 2b_2b_3=2\det(A)+2\det(B)
$$
by a direct computation. Here $A$ has the entries $a_i$ and $B$ the entries $b_i$.

Title: Result on determinant of matrix


问:
Second order differential equation question.
Consider the equation

$\dfrac{d^2y}{d^2t}+k\cdot\dfrac{y}{m}=0$

a) let $y(t)= \cos (at)$. Under what conditions on $a$ is $y(t)$ a solution?

b) What initial condition in the yv-plane corresponds to this solution?

c) In term of m and n , what is the period of this solution?

d) sketch the solution curve associate to this point.

I have no idea how to start. I really want to understand this so I wonder if you can explain this problem step by step for me please.

答:
Let's write this equation in the following way:

$y''=-\dfrac{k}{m}y$. and the function $y(t)=\cos(at)$ is a solution if it solves this equation. Let's plug it in:

$ (\cos(at))''=-\dfrac{k}{m}\cos(at) \Leftrightarrow (-a \sin(at))'=-\dfrac{k}{m}\cos(at) \Leftrightarrow a^2\cos(at)=-\dfrac{k}{m}\cos(at) $

Let's note that $\cos(at)$ isn't 0 for all real t, and therefore, we can divide by $\cos(at)$, in order to get that $ a^2=-\dfrac{k}{m} \Rightarrow a=\pm\sqrt{-\dfrac{k}{m}}$.


问:
Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$
I have to solve the following statements, finding the tangent planes to $f$ and its points of tangency:

that contains the line in $\mathbb{R}^3$ that passes through the points $(3,0,3)$ and $(0,-3,3)$

and this other statement:

that contains the point $(0,0,2)$

For the first statement, I tried first making the parametric equation and I found that $L(t) = (3-3t, -3t, 3)$, and I also know that the equation of the plane is given by $$p(x,y) = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) = -2x_0 (x-x_0) -2y_0 (y-y_0) + (1 -x_0^2 - y_0^2)$$ but after that, I don´t see what else to do. Could you give me any hint for both statements?

答:
The partial derivatives are:$$f'_x = -2x, \quad f'_y = -2y$$
Thus, we have
$$\begin{align}z &= \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) \\
&= -2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1\tag{1}\end{align}$$
Since $z(0,0) = 2$, we have
$$x_0^2 +y_0^2 + 1 = 2 \implies x^2_0 + y_0^2 = 1 \tag{2}$$
From this, $(1)$ simplifies to
$$\begin{align}z&=-2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1 \\&= -2x_0x-2y_0y+x_0^2+y_0^2+1 \\&= -2x_0x-2y_0y + 2 \tag{3}\end{align} $$

For all points $(x_0,y_0)$ satisfying $(2)$, the equation $(3)$ gives a tangent plane that passes through $(0,0,2)$: For example, when $x_0 = \sqrt{\frac{2}{3}}$ and $y_0 = \sqrt{\frac{1}{3}}$ you get the following picture:

One more example with another random choice of a tangency point satisfying $(2)$:


答:
By the multilinearity of the determinant function,
\begin{aligned}
&\det(u+x,v+y)+\det(u-x,v-y)\\
=&\big(\det(u,v)+\det(u,y)+\det(x,v)+\det(x,y)\big)
+\big(\det(u,v)-\det(u,y)-\det(x,v)+\det(x,y)\big)\\
=&2\det(u,v)+2\det(x,y).
\end{aligned}

Title: Result on determinant of matrix

Body:
Let $A$ and $B$ be two $2×2$ real matrix. Then which of the following is true:

$det(A+B)+det(A-B)=det(A)+det(B)$

$det(A+B)+det(A-B)=2\times det(A)-2\times det(B)$

$det(A+B)+det(A-B)=2\times det(A)+2\times det(B)$

$det(A+B)-det(A-B)=2\times det(A)+2\times det(B)$

My try: Take $A=I$ and $ B=-I$ then clearly option 1,2 and 4 are discarded. But I cannot prove option 3. So please help me. Thank you in advance


问:
Prove or disapprove $\pi = 3 + \sum_{n=1}^{\infty} \frac{(-1)^{n}(x_1^n + x_2^n)}{n+2}$ (hint needed)
Prove or disapprove $\pi = 3 + \sum_{n=1}^{\infty} \frac{(-1)^{n}(x_1^n + x_2^n)}{n+2}$,

where $x_1$ and $x_2$ are the roots of $2x^2+2x+1=0$.

I've tried rewriting the series but it didn't work. There is definitely a trick but I don't see it. I will appreciate any help. Thanks!

答: Borrowing @MatthewPilling's calculation, the series (not including the $3$) is $\sum_\pm\sum_{n\ge1}\frac{z_\pm^n}{n+2}$ with $z_\pm:=\tfrac{1}{\sqrt{2}}\exp\frac{\pm\pi i}{4}$, i.e.$$-\sum_\pm\left(\tfrac{z_\pm+\ln(1-z_\pm)}{z_\pm^2}+\tfrac12\right)=-\sum_\pm\left(\sqrt{2}\exp\tfrac{\mp\pi i}{4}-\tfrac{\pi}{2}\pm i\ln2+\tfrac12\right)=\pi-3.$$Numerical analysis agrees.


问:
Eigenvector Proof $(I+A)^{-1}$.
Show that the eigenvectors of the $n \times n$ matrix A are also eigenvectors of the matrix $$M = (I+A)^{-1} $$ Where I is the $n \times n$ unit matrix. Determine the eigenvalues.

My Work:

$$Mx=(I+A)^{-1}x = ???$$

答: If $Ax=\lambda x$, then $(A+I)x = (\lambda + 1)x$. If $A+I$ is invertible, then one can multiply with $(A+I)^{-1}$ on both sides to get $(A+I)^{-1}x = \frac{1}{\lambda + 1}x$</p>

Title: Eigenvector Proof $(I+A)^{-1}$.


问:
Understanding partial expectation.
I am new to probability and statistics, I am reading an article on survival analysis and when defining the expectation of a parametric model, log-normal distribution, the author used the partial expectation. I did a research and found nothing significant, or rather nothing that provide me with a good understanding, so what is that partial expectation, definition, properties. Thankful</p>

答:
I would consider the Wikipedia article section on partial expectation of the lognormal distribution to be a superior reference, since it:

gives a clear mathematical definition

is not paywalled

fits with the original question's context in regard to survival models and lognormal distributions

is presumably more stable as a link for future reference than a publisher's link to an abstract.

For the sake of completeness, the partial expectation of a continuous random variable $X$ is given by $$\operatorname{E}[X \mid X > k]\Pr[X > k] = \int_{x=k}^\infty x f_X(x) \, dx.$$

Title: Understanding partial expectation.


问:
A Partial order can be extended to a linear ordering
This question has already been asked over here.

I was following this answer, but I'm having difficulties to prove that (with the notation given in the linked answer) for the maximal element $\langle M,\preceq_M\rangle$ given by Zorn's Lemma, we have $M=P$.

I tried to do it by contradiction. If $M\subsetneq P$, there must be some $m\in P-M$. If no element of $P-\{m\}$ is comparable to $m$ with respect to $\leq$, then one can show that $\langle M,\preceq_M\rangle$ isn't maximal (just take the sum $\langle N,\preceq_N\rangle$ of $\langle M,\preceq_M\rangle$ and $\langle \{m\},\{(m,m)\}\rangle$, for which one has $\langle M,\preceq_M\rangle\sqsubseteq \langle N,\preceq_N\rangle$).

How can one proceed if some element of $P-\{m\}$ is comparable to $m$ with respect to $\leq$?

答:
The trick is to insert $m$ into an appropriate Dedekind cut in $M$.

Let $L=\{x\in M:x<m\}$ and $R=\{x\in M:m<x\}$. For each $x\in L$ and $y\in R$ we have $x<m<y$, so $x<y$, and therefore $x\prec_M y$. For $x\in M$ let $x\!\!\downarrow=\{y\in M:y\preceq_M x\}$, and let $L^+=\{m\}\cup\bigcup_{x\in L}x\!\!\downarrow$. Let $M^+=M\cup\{m\}$ and

$$\preceq_{M^+}=\preceq_M\cup\{\langle x,m\rangle:x\in L^+\}\cup\{\langle m,x\rangle:x\in M\setminus L^+\}\,;$$

then $\langle M^+,\preceq_{M^+}\rangle\in\mathscr{L}$, and $\langle M,\preceq_M\rangle\sqsubset\langle M^+,\preceq_{M^+}\rangle$, contradicting the maximality of $\langle M,\preceq_M\rangle$.


答:
Often times, when solving equations, you use steps that only work in one direction (from left to right, $\implies\!$). At the end, when you have solution candidates, you shall always check that they satisfy the original equation.

Here's an easy example. Consider the equation $|x|=-5$. Squaring both sides gives $x^2=25$, so $x=5$ or $x=-5$. These are our solution candidates. However, plugging these back into our original equation solves that none of them works. Hence, there is no solution to our equation (which, in this case, was obvious).

If, after squaring and solving, you obtain only complex numbers, and you are only interested in real values, you don't even have any solution candidates. In this case, you can immediately conclude that your equation has no solutions.

Title: Why squaring two absolute value functions never gets answers that don't satisfy the domain?

Body:
$|x-1| = |3x-1|+1$

Where $f(x)=|x−1|$ then $f(x)=x−1$ in a certain domain, $x≥1$.
And $f(x)=−(x−1)$ in a certain domain, $x≤1$.

And $g(x)=|3x-1|+1$ then $g(x)=3x-1+1$ in a certain domain, $x≥1/3$ and $g(x)= -(3x-1)+1$ in a certain domain, $x≤1/3$.

I know this can be solved by squaring both sides and I'll get no solutions. And drawing them on a graph paper supports that.

$(x-1)^2 = (|3x-1|+1)^2$  →$8x^2-4x+2|3x-1|+1=0$ , $8x^2-4x+1$ can never be negative and so is $2|3x-1|$. So, there's no solution.

But doesn't we mean by squaring both sides that $x-1$ can be either negative or positive.

So it's like we are saying:

$-(x-1)$ can be equal to $-(3x-1)+1$, which has an answer $x = 0.5$.

However, this answer doesn't satisfy the domain of $g(x)$ when its equal to $-(3x-1)+1$, $x≤1/3$ so it's not correct.

Or

$-(x-1)$ can be equal to $3x-1+1$, which also has an answer $x=0.25$.

However this answer doesn't satisfy the domain of $g(x)$ when its equal to $(3x-1)+1$, $x≥1/3$ so isn't correct.

Etc.

But these answers since they are out of the domains of both functions aren't right but how can squaring guess that, why are the answers we get out of squaring always in the domain (if both functions are absolute value, I'm not talking about a case like this when only one function has an absolute value sign: $|x-1|=3x-1+1$) how can squaring know where the domain ends and starts, it seems like magic which surely can't be the case.

I tried to ask my teacher but he couldn't get my point, I hope someone will.


问:
Proving Jordan Curve theorem doesn't hold for Torus and Real Projective plane.
I want to prove Jordan Curve Theorem doesnt hold for Torus ($T$) and Real Projective plane ($P$) ie I want to find embedding $f:S^1 \rightarrow S^1\times S^1 $ and $h:S^1 \rightarrow P$ such that $S^1\times S^1-f(S^1), P-h(S^1)$ are connencted.

Intuitivly I can see that. But I'm having trouble with rigourus proof.
For instance this is what I have tried with the torus case:

I defined $f:S^1 \rightarrow S^1\times S^1 $ as $f(z) = (1,z)$ which is obiviously an embedding. But Im not sure how to show $ S^1\times S^1 - f(S^1) $ is connected. All I know is that I should get a space homeomorphic to cylinder.

Any advice?

答:
In the case of $S^1 \times S^1$ and using $f$ with the formula given, you are right, the space $S^1 \times S^1 - f(S^1)$ is indeed homeomorphic to a cylinder. So your task, of course, is to prove this is so, by constructing a homeomorphism
$$h : S^1 \times (0,1) \to (S^1 \times S^1) - f(S^1)
$$
Perhaps if you write out the formulas for the domain and range, the formula for the required homeomorphism $h$ will be obvious. So, the domain is
$$S^1 \times (0,1) = \{(z,t) \mid z \in S^1, t \in (0,1)\}
$$
and
$$(S^1 \times S^1) - f(S^1) = \{(z,w) \mid z \in S^1, w \in S^1 - \{1\}\}
$$
Can you write down the formula now?


问:
Indeterminate form : $ 0 \cdot \infty$
I have a question about the limit of :
$$\lim_{t \to \infty}t\cdot e^{-t}$$

Why is that equal to $0$ ? Shoudln't $ 0 \cdot \infty$ be an inderminate form ?
It's true that with the graphic we can determine that this limit is equal to  $0$ :

But all I know is that  : $ 0 \cdot \infty$ is supposed to be an inderminate form .

答:
$$\lim_{t\to \infty} te^{-t}=0$$
as your graph clearly shows. But: $$\int_0^\infty xe^{-x} dx =\lim_{t\to\infty}\int_0^t xe^{-x}dx=\lim_{t\to\infty}\bigg[1-e^{-t}(1+t)\bigg]=1$$

Your graph clearly shows the integral (area under the curve) is non-zero.

The answer to the broader question you pose about whether $0\cdot \infty$ is indeterminate is "Not necessarily".

Suppose $f(x)=x^2, g(x)=x$. Then $f(x)\to\infty, \frac{1}{g(x)}\to 0$, but $\frac{f(x)}{g(x)}=x\to\infty$

Swap the two functions around and we see $\frac{g(x)}{f(x)}=\frac 1x \to 0$

What the function converges to just depends on the rate of convergence of each side.

Title: Indeterminate form : $ 0 \cdot \infty$


答:
$ 0.\infty $ is of course an indeterminate form but in this particular case, it is zero.

Put $ t=\ln(u)$.

then

$$te^{-t}=\ln(u)e^{\ln(\frac 1u)}=\frac{\ln(u)}{u}$$

when $ t\to +\infty $, $ u\to +\infty$, and, as you know, $\frac{\ln(u)}{u}\to 0$

because, for $ u $ great enough, it is easy to prove that $ \ln(u)<\sqrt{u}.$

Title: Indeterminate form : $ 0 \cdot \infty$

Body:
I have a question about the limit of :
$$\lim_{t \to \infty}t\cdot e^{-t}$$

Why is that equal to $0$ ? Shoudln't $ 0 \cdot \infty$ be an inderminate form ?
It's true that with the graphic we can determine that this limit is equal to  $0$ :

But all I know is that  : $ 0 \cdot \infty$ is supposed to be an inderminate form .


问:
Show that if $f'$ is greater than $0$, then $f(t_2) \geq f(t_1)$ if $t_1 < t_2$
$f: [a,b] \to R, C^0$ on $[a,b]$, and Differentiable $(a,b)$

If $f'(x) \geq 0$ on $(a,b)$ then $f(t_2) \geq f(t_1)$ if $a \leq t_1 < t_2 \leq b$

I can clearly see that it is true. However, I'm honestly not sure how I would show this. All I know is that $f'(x) \geq 0$. I don't know anything about what $f(x)$ would send $t_2$ or $t_1$ to. So I'm not sure how to get the result from these assumptions. Any help is appreciated!

答:
Well yes,restrict the domain to $[t_{1} , t_{2}]$ then as $f$ is differentiable by Lagrange mean value theorem there exist $c\in (t_{1} ,t_{2})$ such that $f^{'}(c)=\frac{f(t_{2})-f(t_{1})}{t_{2}-t_{1}} \geq 0$ so $f(t_{2})\geq f(t_{1})$ as $t_{2}>t_{1}$.
Hope this helps.


问:
Generalize a sum given first elements
Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as

$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$

generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)

$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$

I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.

答:
I would like to add one more approach.
Consider
$$
f(n) = (n+1) + n \rho + (n-1) \rho^2 + \cdots + 1 \rho^n = \sum_{k=1}^{n+1} k \rho^{n+1-k}.
$$
Obviously $f(0)= 1$, $f(1) = 2 + \rho$ and so on.
Now make this function a bit more complicated
$$
f(n,x) = (n+1) x^n + n x^{n-1} \rho + (n-1) x^{n-2} \rho^2 + \cdots + 1 \rho^n.
$$
You may see that $f(n,1) = f(n)$.
Let's represent our new function as a derivative and do few transformations
$$
\begin{align}
f(n,x)
&= \frac{\partial}{\partial x} \left(x^{n+1} + x^{n} \rho + x^{n-1} \rho^2 + \cdots + x \rho^n\right) =\\
&= \frac{\partial}{\partial x} \left( x \left[x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n \right] \right) =\\
&= \frac{\partial}{\partial x} \frac{x (x - \rho) (x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n)}{x - \rho} =\\
&= \frac{\partial}{\partial x} \frac{x (x^{n+1} - \rho^{n+1})}{x - \rho}
\end{align}
$$
Doing all the boring stuff with derivatives and setting $x=1$, at the end of the day we get
$$
f(n) = \frac{(n+1) - \rho(n+2) + \rho^{n+2}}{(\rho - 1)^2}
$$
One may check that
$$
f(0) = \frac{1 - 2\rho + \rho^2}{(\rho - 1)^2} = 1,
$$
$$
\begin{align}
f(1) &= \frac{2 - 3\rho + \rho^3}{(\rho - 1)^2}
      = \frac{2 - 4\rho + 2\rho^2 + \rho^3 - 2\rho^2 + \rho}{(\rho - 1)^2} =\\
     &= \frac{2(1-\rho)^2+ \rho(\rho - 1)^2}{(\rho - 1)^2}
      = 2 + \rho
\end{align}
$$
and so on.

Title: Generalize a sum given first elements


答:
You have already proven that there exists $u$ such that $T(u) = u$. Complete $u$ to an orthogonal basis $u, v, w$. The matrix of $T$ in this basis is

$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & c & d
\end{pmatrix}
$$

where the form of the first column follows from $T(u) = u$ and the form of the first row follows from the fact that rows and columns of an orthogonal matrix have unit 2-norm. Applying this property to the second column and second row, we also see that

$$
a^2 + b^2 = 1 \\
a^2 + c^2 = 1
$$

so $|b| = |c|$. Similarly, $|a| = |d|$. Therefore we can write

$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & \pm b & \pm a
\end{pmatrix}
$$

where so far all four possibilities for the signs in the bottom row are possible. However, since the columns are orthogonal to each other we see that either

$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & b & -a
\end{pmatrix}
$$

or

$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & -b & a
\end{pmatrix}\tag1.
$$

However, the first case is ruled out because then $\det(M(T)) = -a^2 - b^2 = -1$.

Now, notice that $a \in [-1, 1]$, so there exists an angle $\beta \in [0, \pi]$ such that $a = \cos\beta$. Also, comparing $a^2 + b^2 = 1$ to $\cos^2\beta + \sin^2\beta = 1$ we see that $b = \sin\beta$ or $b = -\sin\beta$. In the letter case, we can absorb the minus sign in the angle by defining $\theta = 2\pi - \beta$. In the first case, we set $\theta = \beta$. Substituting this into $(1)$, we get

$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{pmatrix}
$$

which we recognize as the rotation matrix around the $u$ axis by angle $\theta$.

Title: Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$

Body:
Let $T \in \text {SO} (3) : = \left \{T \in M_3 (\Bbb R)\ |\ TT^t = T^tT = I,\ \det (T) = 1 \right \}.$ Prove that there exists $v \in \Bbb R^3 \setminus \{0\}$ such that $T(v) = v.$ Hence conclude that the elements of $\text {SO} (3)$ are rotations by some angle about some uniquely determined axis.

The first part is easier. To prove that $T(v) = v,$ for some $0 \neq v \in \Bbb R^3$ it is enough to show that $\det (T - I) = 0.$ Now $$\begin{align*} \det (T - I) & = 1 \cdot \det (T - I) \\ & = \det (T) \det (T - I) \\ & = \det (T^t) \det (T - I) \\ & = \det (T^t T - T^t) \\ & = \det (I - T^t) \\ & = \det (I^t - T^t) \\ & = \det ((I - T)^t) \\ & = \det (I - T) \\ & = (-1)^3 \det (T - I) \\ & = - \det (T - I) \end{align*}$$ This shows that $2 \det (T - I) = 0 \implies \det (T - I) = 0,$ as required.

How to prove the second conclusion? Any help in this regard will be appreciated.

Thanks in advance.


问:
Complicated derivative question from a calc 2 student
Question from a Calc 2 student.

To solve the following derivative
$$\frac{d}{dn}\left(\frac{1}{2^{2n+1}}\frac{\left(\frac{1}{2}\right)^{2n+2}}{\left(2n+1\right)\left(2n+2\right)}\right)$$

can a dummy variable be used for $2n+2$ or $2n+1$ to make this more manageable?

答:
If you let, say, u= 2n+ 1 and v= 2n+ 2 then your function becomes $f(u,v)= \frac{1}{2^u}\frac{\left(\frac{1}{2}\right)^v}{uv}= \frac{2^{-u}2^{-v}}{uv}$.

Now you need to use the "chain rule": $\frac{df}{dn}= \frac{\partial f}{\partial u}\frac{du}{dn}+ \frac{\partial f}{\partial v}\frac{dv}{dn}$.

(I am surprisingly uncomfortable using "n" as a continuous variable!)


问:
Can the radius of this semicircle be calculated from the given data?
The semicircle highlighted by the red ellipse has an unspecified radius. Can it be calculated given available data? If not, what additional data is needed?

Note: As stated, all features are symmetrical about the dashed diagonal.

答:
Converting Nick Pavlov's comment into an answer:

Two things are problematic: there's no way to work out the thickness of those back "wings" and also, the lines around the semicircle don't look exactly parallel (so it isn't really a semi-circle) and nothing allows you to figure out the angle between them.


问:
Number of possible bit sequences of length m with at least n consecutive 1's in them
I have seen similar questions to this but they each seem to be special cases of this general question. Answering this would be beneficial to my research, but I am not a combinatorics expert, and this seemingly simple question eludes me. Is there a simple formula to calculate this? Everything I have seen online has been centered around things like "either 2 consecutive 1's or 0's" or "contains no ..".

If it helps, I know that for $m = 8$ bits and say the sequence is denoted $S(m,n)$
$$
S(m = 8, n = 1) = 255 \\
S(8,2) = 201  \\
S(8,3) = 107  \\
S(8,4) = 48   \\
S(8,5) = 20   \\
S(8,6) = 8   \\
S(8,7) = 3   \\
S(8,8) = 1
$$

Interestingly I'm finding that $S(8,4)=S(9,5)=S(10,6)=S(11,7)=48$
I haven't tested $S(12,8)$ because I don't want my computer to melt but I'm seeing a pattern... However this does not seem to work for $m<8$.

答:
If the string is $m$ bits long and you demand a run of exactly $n\ 1$s we can find a formula for $n \ge \frac m2$.  Let us call this $T(m,n)$.  If the run is at one end of the string ($2$ choices) you need a $0$ at the end of the run and have $2^{m-n-1}$ choices for the other bits.  If the run is not at the end of the string, there are $m-n-1$ places it can start and you have $2^{m-n-2}$ choices to complete the string.  If $m-n-2$ is negative there are no other bits to fill in. So
$$T(m,n)=\begin {cases} 1&n=m\\2&n+1=m\\2^{m-n}+(m-n-1)2^{m-n-2}&n+2 \le m \end {cases} $$
and the fact that it only depends on $m-n$ is clear.  Then
$$S(m,n)=\sum_{i=n}^mT(m,i)$$
I repeat that this only works for $n \ge \frac m2$.  The reason it only depends on $m-n$ is because if you take a string of the type $(m,n)$ you can find a unique string of type $(m+1,n+1)$ by extending the run by one more bit.


问:
power series radius of convergence
How can I prove that if the coefficients $\{a_k\}$ of the power series $\sum_{0}^{\infty} \{a_k\}x^k$ form a bounded sequence, then the radius of convergence is at least 1?</p>

答:
If $|x| \lt 1$ then $\frac{1}{1-|x|} = \sum_{n = 0}^{\infty} |x|^{n}$ by the summation formula for the geometric series. Now use this, the triangle inequality and the assumption that $|a_{n}| \leq C$ for all $n$ to estimate your given series from above, hence it series converges absolutely for $|x| \lt 1$.

Alternatively, you can easily show that the radius of convergence $\rho^{-1} = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$ satisfies $\rho^{-1} \leq 1$, since $\sqrt[n]{C} \; \xrightarrow{n \to \infty}\; 1$ for all $C \gt 0$. If you look at the proof of this formula for the radius of convergence (usually called the Cauchy-Hadamard theorem), you'll see that this essentially comes down to the same as the first paragraph: a comparison with a geometric series which is known to converge.

Title: power series radius of convergence


问:
meaning of some lines in a book
Would you please anyone translate these three lines into a simpler meaning? it's written in very advanced level.

答:
The prediction error has a zero mean: some predicted values will be too high, and some will be too low. The mean of these errors will be 0.

The prediction error does not co-vary with any linear function of the inputs: It doesn't matter if some predictor variable has a very low or a a very high value, the error in the outcome variable will not be systematically affected. It also doesn't matter if some combination of predictor variables (for example, x_1 * x_2) has a very low or a very high value, the error will not be affected.

These are basic assumptions in much statistical modelling. If a high value of a predictor gives you bigger prediction error, you have a problem with your model (usually). So these lines are basically just saying "we did our modelling properly".

OwnerDisplayName: Mio Unio

Title: meaning of some lines in a book

OwnerDisplayName: legatrix


答:
The mathematics itself is a "red herring". Regardless of the details of the mathematical argument, "means ..." in this mathematical context is used in the sense of "implies that ...", "shows that" or "leads to the logical conclusion that ..."

mean = to have a particular result

Cambridge dictionary

mean = to serve or intend to convey, show, or indicate

Merriam Webster

In response to comment and query I add that the full meaning cannot be reliably given, although there are hints of meaning everywhere in the quotation.

The quote is clearly about the linear regression (y = w0 + w1 x) of a set of n values (xi, yi). J is a function of these values and of the two parameters w0 (intercept) and w1 (slope).

A linear combination of two variables is merely the sum of them when each is numerically multiplied in some way (perhaps with the addition of a constant number, or not). If they are A and B, for example, a linear combination would be (3A +2.89B). (2A + 5B + AB) is not linear because A and B are multiplied together. Nor is (A squared).

If something co-varies with something else, it means that if you change one, the other changes. If this is not the case, they do not covary. The two partial differential equations ( ... =0) show that J does not change if you change w0 or w1. (This likely follows from the maths, because w0 and w1 are chosen by calculating from the n values of (xi, yi)). Hence J does not covary with w0 or w1. It is not clear how this argument finishes.

A full and reliable meaning cannot be given without further information because the mathematical part of the quotation is incomplete in three respects: the function J is not defined other than by its partial derivatives; the significance of the * is not given; w is not defined. All other symbols (including the x with a bar over it, representing the mean value of x) are conventional and understandable.

I hope this helps understand the English that is used in such technical ways. I could do better but that would lead in to a mathematical explanation rather than discussing the use of English.

Title: meaning of some lines in a book

Body: Would you please anyone translate these three lines into a simpler meaning? it's written in very advanced level.

OwnerDisplayName: Anton


问:
Prove theorem of Bolzano-Weierstrass by least upper bound property
Let be $(a_n)_{n\in\mathbb{N}}$ a bounded sequence of real numbers and we want to show that it contains a convergent subsequence.

My approach:

We know, by the least upper bound property, that the set $M:=\{a_n\mid n\in\mathbb{N}\}$ has a supremum $S$. As $S$ is the least upper bound of $M$, for each $\epsilon>0$ there exists a $a_n$ such that $S-\epsilon<a_n<S+\epsilon$.

Let be $\epsilon>0$ then there exists a $k_0\in\mathbb{N}$ such that $\frac{1}{k_0}<\epsilon$. We define a subsequence $\left( a_{n_k}\right)_{k\in\mathbb{N}}$ as follows:

There exists a $a_n$ which we set $a_{n_{k_0}}:=a_n$ such that

$$
S-\frac{1}{k_0}<a_{n_{k_0}}<S+\frac{1}{k_0}.
$$
For all $k>k_0$ it follows that
$$
S-\frac{1}{k_0}<S-\frac{1}{k}<a_{n_k}<S+\frac{1}{k}<S+\frac{1}{k_0}\\
\implies |a_{n_k}-S|<\frac{1}{k}<\epsilon.
$$

Hence, $(a_n)_{n\in\mathbb{N}}$ is convergent.

My tutor said that it is flawed!? I have spent hours revising it and couldn't find any mistakes?

答:
Other comments and answers have told you where your errors are.

The existential problem with this question is the following: When you have seen just a finite number of the $a_n$'s you have no idea where a point of convergence could be. This means that you have to encompass the full sequence $(a_n)_{n\geq1}$ at each step of the process, in order to find a point of convergence.

We may assume $0\leq a_n\leq1$ for all $n$. Let $S$ be the set of all $x\in{\mathbb R}$ such that for infinitely many $n$ we have $a_n\geq x$. Then $0\in S$, and $S$ is bounded above by $1$. The set $S$ therefore has a supremum $\sigma\in{\mathbb R}$. I claim that there is a subsequence $k\mapsto a_{n_k}$ with $\lim_{k\to\infty} a_{n_k}=\sigma$.

Proof. Put $n_1:=1$, and assume that for a $k\geq1$ the "selection" $n_k$ has been constructed such that $|a_{n_k}-\sigma|\leq{1\over k}$. We then  shall find an $n_{k+1}>n_k$ such that $$|a_{n_{k+1}}-\sigma|\leq{1\over k+1}\ .\tag{1}$$

By definition of the sup there is an $s\in S$ with $s>\sigma-{1\over k+1}$, and there are infinitely many $a_n\geq s$. On the other hand there are only finitely many $a_n>\sigma+{1\over k+1}$. It follows that there is an $n_{k+1}>n_k$ satisfying $(1)$.


问:
Compactness of a metric space
If a metric space $(X,d)$ is compact then for every equivalent metric $\sigma$, $(X,\sigma)$ is complete. This is because, for any Cauchy sequence in $(X,\sigma)$ has a convergent subsequence due to fact $(X,\sigma)$ is a compact metric space, hence original sequence is convergent. My question is, does the converse also hold ?

In other words, let $(X,d)$ be a metric space such that for every equivalent
metric $\sigma$ on $X$ is complete. Does this imply $(X,d)$ is
compact?

答:
Here is yet another proof.

Suppose $(X,d)$ is not compact. Since it is a metric space, it follows that there is a sequence $\bar x = (x_n)_{n\in \mathbf N}$ with no convergent subsequence. We will find an equivalent metric $d'$ which makes a subsequence of $\bar x$ a Cauchy sequence, so $d'$ is necessarily incomplete.

We may assume without loss of generality that $\bar x = (x_n)_n$ contains no $d$-Cauchy subsequence (otherwise, $d=d'$ is incomplete). It follows that there is some $\varepsilon>0$ such that for all $n$, there is some $m>n$ such that $d(x_n,x_m)>\varepsilon$. We may assume without loss of generality that $d(x_n,x_m)>\varepsilon$ for all $n,m$ (passing to a subsequence if necessary).

Now, for each $n$, let $B_n$ be the ball around $x_n$ of radius $\varepsilon/2$, and write $X'$ for $X\setminus \bigcup_n B_n$. Now, define a metric $d'$ by the following formula:
$$
d'(x,y) = \begin{cases}
d(x,y) & x,y\in X' \\
d(x_n,y)+\varepsilon/n & x\in B_n, y\in X'\\
d(x,x_n)+\varepsilon/n & x\in X', y\in B_n\\
\varepsilon/\min(m,n) & x\in B_n, y\in B_m, n\neq m \\
d(x,y)/n & x,y\in B_n
\end{cases}
$$
It remains to check that:

$d'$ is a metric (the only nontrivial axiom is the triangle inequality, but it is easily checked by considering all possible cases),

$d'$ is equivalent to $d$ (this is very easy),

$(x_n)_n$ is Cauchy in $d'$ (this is trivial).

Title: Compactness of a metric space


问:
Computing the derivative of a matrix-valued function
Suppose $\mathbf{A} :\mathbb{R}^m\to\mathbb{R}^{m\times m}$ such that $\mathbf{A}(\theta)$ is a symmetric matrix. Let $\mathbf{A}(\theta)$ have the eigen-decomposition $\mathbf{A}(\theta) = \mathbf{U}(\theta)~\mathrm{diag}(\lambda_1(\theta),\ldots,\lambda_m(\theta))~\mathbf{U}(\theta)^\top$. If I have a function $f:\mathbb{R}\to\mathbb{R}$ and I define
\begin{align}
f(\mathbf{A}(\theta)) = \mathbf{U}(\theta) ~\mathrm{diag}(f(\lambda_1(\theta)),\ldots,f(\lambda_m(\theta)))~\mathbf{U}(\theta)^\top
\end{align}
How does one compute $\frac{\partial}{\partial\theta_i} f(\mathbf{A}(\theta))$?

答:
For the real function $$F(\theta) \doteq f(A(\theta))$$
the most practical method for calculating the derivative is
the Complex Step Approximation
$$\eqalign{
\frac{dF}{d\theta} &= {\rm{Im}}\!\left(\frac{F(\theta+i\eta)}{\eta}\right),
\qquad \eta &\approx 10^{-90} \\
}$$
unless your function has some special structure which can be exploited.

Update

Invoking the Daleckii-Krein theorem leads to a closed-form expression
$$\eqalign{
\def\BR#1{\Big(#1\Big)}
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\Diag#1{\op{Diag}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\l{\lambda}
\def\b{\lambda}
\def\p{\partial}
\def\grad#1#2{\frac{\p #1}{\p #2}}
A &= UL\,U^T \qiq L=\Diag{\b_k} \\
F &= f(A) \\
dF &= U\,\BR{R\odot\LR{U^TdA\:U}\!}\,U^T \\
\\
}$$
$$\eqalign{
R_{jk}
   =  \begin{cases}
  {\Large\frac{f(\b_j)\,-\,f(\b_k)}{\b_j\,-\,\b_k}}\qquad{\rm if}\;\b_j\ne\b_k \\
\\
  \quad{f'(\b_k)}\qquad\qquad{\rm otherwise} \\
\end{cases} \\
}$$
where $(\odot)$ denotes the elementwise/Hadamard product and $dA$ can be interpreted as either the differential or
as the derivative with respect to $\theta_i$


问:
Complete the square for $f(x) = 2x^2 + 4x - 6$
I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

答:
Rewrite your expression as
$$
2x^2+4x+b-6-b
$$
for any $b>0$. To close the square you need the expression in the form
$$
(\alpha x +\beta)^2-(6+b)=\alpha^2x^2+2 \alpha \beta x+\beta^2-(6+b)
$$
You get immediately $\alpha=|\sqrt{2}|,\beta=|\sqrt{b}|$, hence your expression is
$$
(|\sqrt{2}|x+|\sqrt{b}|)^2-(6+b)
$$
If you want you can set $b=2$ to get
$$
2(x+1)^2-(6+2)
$$</p>


问:
Why squaring two absolute value functions never gets answers that don't satisfy the domain?

$|x-1| = |3x-1|+1$

Where $f(x)=|x−1|$ then $f(x)=x−1$ in a certain domain, $x≥1$.
And $f(x)=−(x−1)$ in a certain domain, $x≤1$.

And $g(x)=|3x-1|+1$ then $g(x)=3x-1+1$ in a certain domain, $x≥1/3$ and $g(x)= -(3x-1)+1$ in a certain domain, $x≤1/3$.

I know this can be solved by squaring both sides and I'll get no solutions. And drawing them on a graph paper supports that.

$(x-1)^2 = (|3x-1|+1)^2$  →$8x^2-4x+2|3x-1|+1=0$ , $8x^2-4x+1$ can never be negative and so is $2|3x-1|$. So, there's no solution.

But doesn't we mean by squaring both sides that $x-1$ can be either negative or positive.

So it's like we are saying:

$-(x-1)$ can be equal to $-(3x-1)+1$, which has an answer $x = 0.5$.

However, this answer doesn't satisfy the domain of $g(x)$ when its equal to $-(3x-1)+1$, $x≤1/3$ so it's not correct.

Or

$-(x-1)$ can be equal to $3x-1+1$, which also has an answer $x=0.25$.

However this answer doesn't satisfy the domain of $g(x)$ when its equal to $(3x-1)+1$, $x≥1/3$ so isn't correct.

Etc.

But these answers since they are out of the domains of both functions aren't right but how can squaring guess that, why are the answers we get out of squaring always in the domain (if both functions are absolute value, I'm not talking about a case like this when only one function has an absolute value sign: $|x-1|=3x-1+1$) how can squaring know where the domain ends and starts, it seems like magic which surely can't be the case.

I tried to ask my teacher but he couldn't get my point, I hope someone will.

答:
First of all, if you want to communicate your doubts to your teacher you need a common language: if you improperly use maths words, you cannot expect him to understand what your difficulties are.

By reading between the lines, I guess you do not understand why the equation
$$|f(x)| = |g(x)| \tag{1}\label{1}$$
is equivalent to (has the same solutions as)
$$f^2(x) = g^2(x).\tag2 \label{2}$$
Also you seem to assume that this has something to do with the values of $x$ for which the arguments of the absolute values are positive or negative (this is probably what you erroneously call domain). But, as a matter of fact, these are irrelevant, as far as the equivalence is concerned. Note in fact that the last equation is in turn equivalent to
$$[f(x)-g(x)][f(x) + g(x)]=0.\tag3 \label{3}$$

To show equivalence between \eqref{1} and \eqref{2}, you can show that any solution to \eqref{1} is a solution to \eqref{2}, and viceversa.

Suppose $x_0$ is a solution to \eqref{1}. This means that
$$|f(x_0)| = |g(x_0)|.\tag4 \label{4}$$
If $f(x_0)$ and $g(x_0)$ have the same sign, \eqref{4} implies
$$f(x_0) = g(x_0).$$
If they have opposite sign, \eqref{4} implies
$$f(x_0) = -g(x_0).$$
This shows that in any case $x_0$ is a solution to \eqref{3} and hence to \eqref{2}.

On the other way around, suppose $x_1$ solves \eqref{2}, and thus \eqref{3}. This means that either
$$f(x_1) = g(x_1)$$
or
$$f(x_1) = -g(x_1).$$
And this gives
$$|f(x_1)| = |g(x_1)|,$$
that is $x_1$ solves also \eqref{1}.