问:
Probability of Error: Binary Communication Channel?
I have a binary communication channel shown below:

with the following parameters:

$\pi_0 = 0.3$

$\pi_1 = 0.7$

$\epsilon_0 = 0.2$

$\epsilon_1 = 0.1$

Goal: compute the probability of error and probability of false alarm for the decision rule: always choose 0

Starting out:
$$P(Error) = P(A_0)P(B_1 | A_0) + P(A_1)P(B_0 | A_1)$$
and from the diagram,

$P(B_0|A_0) = 1 - .2 = 0.8$

$P(B_0|A_1) = 0.1$

$P(B_1 | A_0) = 0.2$

$P(B_1 | A_1) = 0.9$

so I can compute the following quantities:

$P(B_0) = P(B_0|A_0)P(A_0) + P(B_0|A_1)P(A_1) = (.8)(.3) + (.1)(.7) = .31$

$P(B_1) = P(B_1|A_1)P(A_1) + P(B_1|A_0)P(A_0) = (.9)(.7) + (.2)(.3) = .69$

$P(A_0|B_0) = \frac{P(B_0|A_0)P(A_0)}{P(B_0)} = \frac{(.8)(.3)}{.31} = .774$

$P(A_1|B_1) = \frac{P(B_1|A_1)P(A_1)}{P(B_1)} = \frac{(.9)(.7)}{.69} = .913$

Question:
Where does the decision rule: always choose $0$ come into play? Is the probability of error $.69$ since that's the value of $P(B_1)$? What if the decision rule was different: say keep the observation?

Thanks!

答: Obviously the optimum decision rule is: choose 1 if you receive 1, and choose 0 if you receive 0, because $P(A_1|B_1)>P(A_0|B_1)$, and $P(A_0|B_0)>P(A_1|B_0)$. The question is: what happens if you disregard what you have received and always choose 0. In that case, the probability of error would be $P(B_0|A_1)P(A_1)$, because the error only occurs when 1 is sent.


问:
Identifying a sequence of numbers from an optimization problem in $L^1$

Question

Does there exist general closed form solutions (or some sort of recurrence relation) to the system of equations:
$$\begin{align}
x_0 &= -1\\
x_{k+1} &= 1\\
\sum_{j = 0}^k (-1)^j (x_{j+1} - x_{j}) & = 0 \\
\sum_{j = 0}^k (-1)^j (x_{j+1}^2 - x_{j}^2) &= 0 \\
\vdots \\
\sum_{j = 0}^k (-1)^j (x_{j+1}^k - x_{j}^k) &= 0
\end{align}$$
with $-1 \leq x_1 \leq x_2 \leq \ldots \leq x_k \leq 1$?

For $k$ small the solutions can be computed by hand to be

$k = 1$: $\{0\}$

$k = 2$: $\{\pm \frac12\}$

$k = 3$: $\{0, \pm\frac12\}$

$k = 4$: $\{(\pm 1\pm\sqrt{5})/4\}$

$k = 5$: $\{0, \pm\frac12, \pm\frac{\sqrt{3}}2\}$

The non-trivial part of the system can be summarised as
$$  \begin{pmatrix}
x_0 & x_1 & x_2 & \ldots & x_k & x_{k+1} \\
x_0^2 & x_1^2 & x_2^2 & \ldots & x_k^2 & x_{k+1}^2 \\
x_0^3 & x_1^3 & x_2^3 & \ldots & x_k^3 & x_{k+1}^3 \\
& \vdots \\
x_0^k & x_1^k & x_2^k & \ldots & x_k^k & x_{k+1}^k \end{pmatrix}
\begin{pmatrix}
-1 \\
2\\
-2 \\
2 \\
-2 \\
\vdots \\
(-1)^{k-1} 2\\
(-1)^k\end{pmatrix} = 0 $$
if it helps.

Motivation

This question asked for the minimizer among linear functions of the functional $\int_0^1 |e^x - \ell(x)|dx$. The solution given by anonymous user surprised me slightly: it turns out that there is a universal construction that is independent of the target function (in this case, $e^x$). Namely, the user has actually proven that

Theorem Let $f\in C^2([-1,1])$ such that $f''$ is strictly signed. Then the minimizer of the functional $\int_{-1}^1 |f(x) - \ell(x)|dx$ among linear functions $\ell(x)$ is attained when
$$ \ell(x) = (x + 1/2) \left( f(1/2) - f(-1/2)\right) + f(-1/2) $$
in other words, it is the linear interpolation function between $(-1/2, f(-1/2))$ and $(1/2, f(1/2))$.

I gave another write-up of the proof here on my blog.

It turns out that the theorem extends to the case of "higher order convexity", in the following sense:

Theorem There exists a list of numbers $\{x_1, \ldots, x_k\}\subset [-1,1]$ such that for every $f\in C^k([-1,1])$ such that $f^{(k)}$ is strictly signed, the minimizer of the functional $\int_{-1}^1 |f(x) - p(x)|dx$ among polynomials of degree at most $k-1$ is attained by the $(k-1)$-order polynomial interpolation based on the points $(x_j, f(x_j))$.

I wrote up a proof again on my blog.

The question above is asking for general closed form expressions of the numbers $\{x_1, \ldots, x_k\}$ in terms of $k$.

(If someone happens to know a literature reference for this result, that would also be very nice.)

答:
First off, your answer for $k=3$ is wrong -- the correct values are $\{0, \pm \frac{1}{\sqrt{2}} \}$.

Second, the answer is $x_{k+1-m} = \cos{\frac{m\pi}{k+1}}, \, m=0\ldots k+1$.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base.
While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about the Taylor expansion , and try to prove this in the expanded form only, how to will we prove it ?

It's already clearly  visible for $ x > 0$ as well as for $ -1<x<0 $ . But how to prove it for $x<-1$ ? I have tried much but not able to build up a strong convincing proof.

Any help would be highly appreciated. thanks!

P.S : prove it 'WITHOUT' using $e^{x}$ function.

答:
To elaborate on some of the hints in the comments, let $x,y \in \mathbb{R}$. Then:

\begin{align*}
e^{x+y} &= \sum_{n=0}^{\infty}\frac{(x+y)^n}{n!} \\
&= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{x^ky^{n-k}}{n!} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\
&= 1 + \left(x+y\right) + \left(\frac{x^2}{2!}+xy + \frac{y^2}{2!}\right) + \left(\frac{x^3}{3!} + \frac{x^2y}{2!} + \frac{xy^2}{2!} + \frac{y^3}{3!}\right) + \ldots \\
&= \left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) + y\left(1+x+\frac{x^2}{2!}  + \ldots\right) + \frac{y^2}{2!}\left(1+x + \ldots\right) + \ldots \\
&= \left(1+y+\frac{y^2}{2!} + \frac{y^3}{3!} + \ldots\right)\left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) \\
&= e^{x}e^{y}
\end{align*}

This factoring trick is what people in the comments are calling the Cauchy Product. I thought that seeing the algebra done by hand (sort of) might convince you it's not a crazy sophisticated trick. It follows easily from this identity that $e^{-x} = 1/e^{x} > 0$.


问:
Show $\cos\frac\gamma2=\sqrt{\frac{p(p-c)}{ab}}$

Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.

I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$

Is this a well known formula for angle bisectors?

Can we use it somehow?

Thank you in advance!

答:
As for the last part (how we can use the bisector).. The property that the ratio of sides is the ratio in which opposite side $c$ is intersected

$$ e=\dfrac{AC}{ CB}=\dfrac{  AL}{  LB},$$

a constant that defines the Apollonius circle of constant $e$. Bisection can be done either for the internal or for external angle.


问:
The proof of comparing the fractions.
Hi I came up with this method but couldn't find any proof can you provide one thanks

To compare $\frac{a}{b}$ with $\frac{c}{d}$ , compare the cross products as follows:

If ad > bc, then:
$\frac{a}{b}$>$\frac{c}{d}$

If ad < bc, then:
$\frac{a}{b}$>$\frac{c}{d}$

If ad = bc, then:
$\frac{a}{b}$ = $\frac{c}{d}$

I mean how can we conclude the bigger or smaller ones.

答:
Simply note that $\dfrac ab$ is greater than $\dfrac cd$ if $$\dfrac ab -\dfrac cd >0 \implies \dfrac{ad-bc}{bd} >0$$

Now, your method only works if $bd>0$. Similarly reverse the inequality for the other case.

Title: The proof of comparing the fractions.


问:
Show that if $4 | (n + 1)$, then there is a prime number $p$ with $p|n$ and $4|(p + 1)$.

Show that if $4 | (n + 1)$, then there is a prime number $p$ with $p|n$ and $4|(p + 1)$.

答:
Proof by induction.

We can manually check a few numbers.

$n = 3,7,11,15$

For each $n$ there as a corresponding $p = 3,7,11,3$ respectively.

Suppose our proposition is true for all $n\equiv -1 \pmod 4$ and $n\le k$

Is our proposition true for the smallest $n >k$ such than $n\equiv -1 \pmod 4$?

If $n$ is prime, we are done.

If $n$ is composite, $n = ab$ such that $a\equiv 1\pmod 4$ and $b\equiv -1\pmod 4$

By the inductive hypothesis $b$ has a prime factor $p$ such that $4|(p+1)$


问:
Show that $\alpha(a) = h$ if and only if $a = gk$ for some $k \in \text{ker} (\alpha)$

Let $\alpha : G → H$ be a homomorphism, and suppose that $\alpha (g) = h$. For any
$a \in G$, show that $\alpha(a) = h$ if and only if $a = gk$ for some $k \in \text{ker} (\alpha)$.

My attempt: Suppose that $a=gk, k \in \text{ker} (\alpha)$. Then

$\alpha(a) =\alpha(gk) = \alpha(g) \alpha(k) =\alpha(g) e=\alpha(g) =h$

Conversely, suppose that $\alpha$ is a homomorphism. If $\alpha(a) =\alpha(g) $, then $\alpha(a)(\alpha(g)) ^{-1} =e$. So

$h=\alpha(a) =\alpha(g)\alpha(a)( \alpha(g)) ^{-1}$

Let $k=ag^{-1}$ then

$h=\alpha(a) =\alpha(gk) $.

Thus $a=gk, k\in  \text{ker} (\alpha)$.

Am I right?

答:
Your proof is fine.

You could improve it by making it a little clearer how you conclude that $k\in\ker(\alpha)$.


答:
It is easy applying the basic rule  for use of inequalities:

You can multiply an inequality by any number $C$ without reversing the inequality if $C>0$, reversing it if $C<0$.

So here, starting from $\frac ab<\frac cd$, if you multiply by $bd$,

If $b,d$ have the same sign, $bd>0$ and therefore
$$\frac a{\not b} \cdot \not bd <\frac c{\not d}d\cdot b\not d\iff ad<bc.$$

If $b,d$ have different signs, $bd<$ so
$$\frac a{\not b} \cdot \not bd >\frac c{\not d}d\cdot b\not d\iff ad>bc.$$

Title: The proof of comparing the fractions.

Body:
Hi I came up with this method but couldn't find any proof can you provide one thanks

To compare $\frac{a}{b}$ with $\frac{c}{d}$ , compare the cross products as follows:

If ad > bc, then:
$\frac{a}{b}$>$\frac{c}{d}$

If ad < bc, then:
$\frac{a}{b}$>$\frac{c}{d}$

If ad = bc, then:
$\frac{a}{b}$ = $\frac{c}{d}$

I mean how can we conclude the bigger or smaller ones.


问:
What is the proper substitution for this integration?

Evaluate the integral $$\int_{D} \frac{x}{y} d x d y$$ $D=\left\{(x, y): x^{2}+y^{2}<4, x y>\sqrt{3}, x>0\right\}$

$$\int_{D} \frac{x}{y} d x d y = \int_{0}^{4} \frac{1}{y}d y\int_{0}^\sqrt{4-y^2}x dx-\int_0^2 \frac{1}{y}dy\int_0^{\frac{\sqrt 4}{y}}x dx$$

I can't find the correct substitution: both direct integration and polar coordinates lead to "$y=\log0$".

答:
Your limits for $y$ is incorrect. As $xy \gt \sqrt3$, $y$ cannot be zero. See the shaded region in the sketch.

Intersection of the circle $x^2+y^2=4$ and curve $xy = \sqrt3$ will be given by

$x^2 + \frac{3}{x^2} = 4 \implies x = 1, y = \sqrt3, x = \sqrt3, y = 1$ (as $x\gt0$, we have only considered positive values of $x$).

So the integral should be $\displaystyle \int_1^\sqrt3 \int_{\sqrt3/y}^\sqrt{4-y^2} \frac{x}{y} \ dx \ dy$


问:
Compact formula for the $n$th derivative of $f\left(\sqrt{x+1}\right)$?
I'm interested in a general formula for

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big].$$

In particular, Fàa di Bruno's formula gives

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]=\sum_{k=1}^n f^{(k)}\left(\sqrt{x+1}\right)B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big),$$

where $g(x)=\sqrt{x+1}$ and $B_{n,k}$ denote Bell polynomials.

I suspect we can do better than this, using the identities

$$\frac{d^k}{dx^k}\Big[\sqrt{x+1}\Big]=\left(\frac{1}{2}\right)^\underline{k}x^{1/2-k}=-\frac{(2k-3)!!}{(-2)^k}x^{1/2-k}=\frac{\sqrt{\pi}}{2\Gamma(3/2-k)}x^{1/2-k}.$$

In particular, every term in $B_{n,k}$ for fixed $n$ and $k$ will be some numerical factor times $(x+1)^{k/2-n}$ (see notes below), so the result is always of the form

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=1}^n a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}}$$

for some constants $a_{n,k}.$ Is there an explicit expression for the $a_{n,k}$ above? For example, it seems likely that there is some expression using Stirling numbers, factorials, binomial coefficients, etc., given the combinatorial nature of the problem.

Current progress:

Using the first identity above for derivatives of $\sqrt{x+1}$ [where $(\cdot)^\underline{k}$ denotes the $k$th falling factorial], we have

$$B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big)
=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1}x^{1/2-1},\left(\frac{1}{2}\right)^\underline{2}x^{1/2-2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}x^{1/2-(n-k+1)}\right].$$

The definition of the Bell polynomials guarantees that the power of $x$ in every term of $B_{n,k}(\cdots)$ comes out to $x^{k/2-n}$, with pre-factor given by

$$a_{n,k}=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1},\left(\frac{1}{2}\right)^\underline{2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}\right],$$

so the remaining task is to simplify the above expression, if possible.

答:
A formula to calculate the $n$-th derivative of composite functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule

Let $D_g$ represent differentiation with respect to $g$ and $g=g(x)$. Hence $D^n_x f(g)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid for $n\geq 1$:
\begin{align*}
D_x^n f(g)=\sum_{k=1}^nD_g^kf(g)\frac{(-1)^k}{k!}\sum_{j=1}^k(-1)^j\binom{k}{j}g^{k-j}D_x^ng^j\tag{1}
\end{align*}

Here we have $g(x)=\sqrt{1+x}$. The right-most part of (1) is
\begin{align*}
g(x)^{k-j}D_x^ng(x)^j&=(1+x)^{\frac{k-j}{2}}D_x^n(1+x)^{\frac{j}{2}}\\
&=(1+x)^{\frac{k-j}{2}}\left(\frac{j}{2}\right)^{\underline{n}}(1+x)^{\frac{j}{2}-n}\\
&=(1+x)^{\frac{k}{2}-n}\left(\frac{j}{2}\right)^{\underline{n}}\tag{2}\\
\end{align*}
We obtain with $g(x)=\sqrt{1+x}$ from (1) and (2):
\begin{align*}
\color{blue}{D_x^n f\left(\sqrt{1+x}\right)
=\sum_{k=1}^n(1+x)^{\frac{k}{2}-n}D_g^kf(g)\frac{(-1)^k}{k!}\sum_{j=1}^k(-1)^j\binom{k}{j}\left(\frac{j}{2}\right)^{\underline{n}}}
\end{align*}


问:
If $M$ is a compact Riemannian manifold and $g$ and $\tilde{g}$ are metrics on $M$, then $\frac{1}{C} g \leq \tilde{g} \leq C g$ for $C > 1$
I am reading Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities by Emmanuel Hebey and he stated on page $22$:

Let $M$ be a compact manifold endowed with two Riemannian metrics $g$ and $\tilde{g}$. As one can easily check, there exists $C > 1$ such that
$$\frac{1}{C} g \leq \tilde{g} \leq C g$$
on $M$, where such inequalities have to be understood in the sense of the bilinear forms.

I would like to help to prove this, because I can not give a satisfactory proof with my attempt, but I put it below to show my effort. I also would like to apologize if my proof is very detailed, but I would like to see if I understood very well the argument and what hypothesis are used and how they are used.

It is sufficient to prove that $\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i$ on $M$ for some constant $C > 1$. Suppose that $\tilde{g}$ is a Riemannian metric which is geodesic normal coordinates at $p$ without loss of generality because if the inequalities above are proved, then the inequalities are true for the metric $\tilde{g}$ which is not geodesic normal coordinates at $p$ only changing $C$ by $\frac{C}{A}$, where $A$ denotes the Jacobian of the change of the coordinates. Now, consider $M$ connected (the author assumes in the beginning of the book that manifolds are connected, I think this is used here to define the next metric on $M$) and endowed with the metric $d(p,q) := \inf \left\{ l(\alpha) \ ; \ \alpha \ \text{is a piecewise differentiable curve joining} \ p \ \text{to} \ q \right\}$. Recall that the Riemannian metric $\tilde{g}$ is smooth in the sense that the map

\begin{align*}
\tilde{g}: (M,d) &\longrightarrow (\mathscr{L}^2(T_pM \times T_pM, \mathbb{R}),||\cdot||_{op})\\
p &\longmapsto \tilde{g}(p)
\end{align*}

is smooth ($||\cdot||_{op}$ denotes the operator norm over $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$), in particular, the map above is a continuous map defined over a compact metric space, then it is uniformly continuous. This part I am stuck, but I want to  define a norm $||\cdot||$ over the image of the Riemannian metric $\tilde{g}$ in order to, for every $\varepsilon > 0$, there exists $\delta(\tilde{g}) > 0$ such that

$$q \in B_{\delta(\tilde{g})}(p) \Longrightarrow |\tilde{g}_{ij}(q) - \tilde{g}_{ij}(p)| \leq = ||\tilde{g}(q) - \tilde{g}(p)|| < \varepsilon$$

Choosing $C > 1$ and $\varepsilon := \frac{1}{2} \left( C - \frac{1}{C} \right)$, we have

$$\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i \ (1)$$

on $B_{\delta(\tilde{g})}(p)$ for each $p \in M$.

I do not sure how to do this, once that $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$ and the coordinate fields vary with $p$, therefore I think I can not take simply the operator norm of this space to be $||\cdot||$, but if I can overcome this difficult, then we can do an analogous reasoning for $g$ to obtain

$$\frac{1}{C} \delta_j^i \leq g_{ij} \leq C \delta_j^i \ (2)$$

on $B_{\delta(g)}(p)$ for each $p \in M$.

Defining $\delta := \min \{ \delta(\tilde{g}), \delta(g) \}$, $(1)$ and $(2)$ hold on $B_{\delta}(p)$ for each $p \in M$. Combining $(1)$ and $(2)$ and observing that $\{ B_{\delta}(p) \ ; \ p \in M \}$ is an cover for $M$, we proved the inequalities desired.

$\textbf{EDIT:}$

We know that

$$\frac{1}{A} g_p(v,v) \leq \tilde{g}_p(v,v) \leq A g_p(v,v) \ (\star)$$

for all $v \in T_pM$ based on what DIdier_ proved. Analogously,

$$\frac{1}{B} \tilde{g}_p(v,v) \leq g_p(v,v) \leq B \tilde{g}_p(v,v) \ (\star \star)$$

for all $v \in T_pM$.

I will try to prove that

$$\frac{1}{C} g_p(u,v) \leq \tilde{g}_p(u,v) \leq C g_p(u,v)$$

for all $u,v \in T_pM$.

Let $q_{g_p}(v) := g_p(v,v)$ and $q_{\tilde{g}_p}(v) := \tilde{g}_p(v,v)$ be the quadratic forms associated to the $g_p$ and $\tilde{g}_p$ respectively, then

$$g_p(u,v) = \frac{q_{g_p}(u+v) - q_{g_p}(u) - q_{g_p}(v)}{2} \ \text{and} \ \tilde{g}_p(u,v) = \frac{q_{\tilde{g}_p}(u+v) - q_{\tilde{g}_p}(u) - q_{\tilde{g}_p}(v)}{2}.$$

This, $(\star)$ and $(\star \star)$ imply that

$$\tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$

and

$$g_p(u,v) \leq \left( B - \frac{1}{B} \right) \tilde{g}_p(u,v)$$

for all $u,v \in T_pM$, therefore

$$\frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$

for all $u,v \in T_pM$.

Choosing $C > 1$ sufficiently large such that

$$\frac{1}{C} g_p(u,v) \leq \frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v) \leq C g_p(u,v)$$

for all $u,v \in T_pM$ gives the result.

答:
The point of this answer is to explain what the question is; the other answer has a perfect proof.
As stated in your first quote, $\frac{1}{C} g \leq \tilde{g} \leq C g$ is to be understood in the sense of quadratic forms.
This means that for all $x\in M$ and $v\in T_xM$ we have
$$
\frac{1}{C} g_x(v,v) \leq \tilde{g}_x(v,v) \leq C g_x(v,v)
$$
or equivalently in local coordinates
$$
\frac{1}{C} \sum_{i,j}g_{ij}(x)v^iv^j \leq \sum_{i,j}\tilde{g}_{ij}(x)v^iv^j \leq C \sum_{i,j}g_{ij}(x)v^iv^j.
$$
This means that the two norms on every $T_xM$ are bi-Lipschitz equivalent and the constant is independent of $x$.

Even when we write $\frac{1}{C} g_{ij} \leq \tilde{g}_{ij} \leq C g_{ij}$, it can be a shorthand for the inequalities in the sense of quadratic forms.
That is indeed a much more likely interpretation than a componentwise result.

To underline the importance of working with quadratic forms rather than individual components, let me define three (partial) orders for symmetric square matrices:

In the sense of quadratic forms: $A\leq_{qf}B$ means that $v^TAv\leq v^TBv$ for all $v$.

Componentwise: $A\leq_{cw}B$ means that $A_{ij}\leq B_{ij}$ for all indices.

For all pairs: $A\leq_{p}B$ means that $u^TAv\leq u^TBv$ for all $u$ and $v$.

Now take
$$
A
=
\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}
$$
and
$$
B
=
\begin{pmatrix}
1&10\\
10&1
\end{pmatrix}.
$$
Clearly $A\leq_{cw}B$, but for $v=(1,-1)$ we have
$$
2
=
v^TAv
>
v^TBv
=
-18.
$$
Thus $A\leq_{cw}B$ does not imply $A\leq_{qf}B$.

In the case of Riemannian metrics proving $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is insufficient, and in general it does not even hold.
For example, if $\tilde g$ is the Euclidean metric (the identity matrix) and $g$ is a Riemannian metric with non-zero (perhaps both positive and negative) off-diagonal entries at some point, the componentwise version is false but the version with quadratic forms is still valid.

In general, $A\leq_{p}B$ implies both $A\leq_{qf}B$ (use the same vector twice) and $A\leq_{cw}B$ (choose two basis vectors).
While the order given by pairs of vectors implies the correct one, it often fails because because the componentwise one does even though the desired estimate holds true.

What you need is $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$, not $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ or $\frac1Cg\leq_{p}\tilde g\leq_{p}Cg$.
Unfortunately your proof that $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$ implies $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is invalid.


问:
Would this proof by induction for the sequence of the product of odd numbers be correct? Is there something I can do better with it?
I'm new to proofs by induction, and I was challenged by my teacher to learn this when I came up with a form to represent this sequence. Would this be correct?

$$1\cdot3\cdot\ldots\cdot(2n-1)=\frac{(2n)!}{2^nn!}$$
$$\text{If $n=1$, then }1=\frac{(2\cdot1)!}{2^1\cdot1!}$$
$$1=1$$
$$\text{Suppose the proposition holds for $n=k$ so that }1\cdot3\cdot\ldots\cdot(2k-1)=\frac{(2k)!}{2^kk!}$$
$$\text{We show that the proposition holds for $n=k+1$ holds, i.e. }1\cdot3\cdot\ldots\cdot(2(k+1)-1)=\frac{(2(k+1))!}{2^{k+1}(k+1)!}$$
$$1\cdot3\cdot\ldots\cdot(2k+1)=1\cdot3\cdot\ldots\cdot(2k-1)\cdot(2k+1)=\frac{(2k)!}{2^kk!}\cdot(2k+1)$$
$$=\frac{(2k+1)!}{2^kk!}=\frac{(2k+1)!}{2^kk!}\cdot\frac{(2k+2)}{(2k+2)}=\frac{(2k+2)!}{2^kk!(2k+2)}=\frac{(2k+2)!}{2^kk!\cdot2\cdot(k+1)}$$
$$=\frac{(2k+2)!}{2^{k+1}(k+1)!}=\frac{(2(k+1))!}{2^{k+1}(k+1)!}$$
If so, is there anything I could improve on it format-wise?

答:
"Assume $n=k$ holds" is a bit of an odd phrasing. It's rather "Suppose the proposition holds for $n=k$." Similarly, "Show that $n=k+1$ holds so that blah" is really "We show that the proposition holds for $n=k+1$, i.e. blah".

Otherwise it looks fine to me, except that of course you have omitted to prove a base case. You have provided no way to deduce the result for $n=1$.


问:
A Verification using Schwarz inequality
I'm having difficulty following this reasoning:

How is equality being used in the Schwarz inequality? I'm getting lost between using it between the usual dot product and the inner product between $L^2$ spaces ($\nabla u_0 \in L^2$).

How are they concluding that $u_0(r \xi) = r u_0(\xi)$ (i.e. $u_0$ is 1-homogeneous).

答:
The author is applying the Schwarz inequality to the functions $f(r)=1$ and $g(r)=\nabla u_0(r\xi)\cdot\xi$ on the interval $[0,t]$. We have
$$\left(\int_0^t f(r)g(r)\,dt\right)^2 \le \int_0^t f(r)^2\,dr \int_0^t g(r)^2\,dr,$$
and equality holds iff $g$ is a scalar multiple of $f$, i.e., $g$ is a constant function of $r$.

Title: A Verification using Schwarz inequality


问:
Infinite prime numbers from a sum of powers
I am not sure if it's possible to get infinite prime numbers from this sum:
$$p=k^j+j^k$$
with $j\in\mathbb{N}, k\in\mathbb{N}$
I tried for $j=1,2,...9,k=1,2,...9$ and I get only eleven prime numbers.
If I consider the matrix:
$$A(k,j)=k^j+j^k$$
in which the components $A(j,k)=1$ iff $p$ is prime
this is a sparse matrix in which the prime numbers are mostly in the first four rows. Can someone give me some hint to prove $A$ contains infinite prime numbers in the limit $j\to \infty$, $k\to\infty$</p>

答: Hint You get infinitely many primes when $k=1$ or $j=1$.


问:
Cardinality of set of functions
Let $\mathbb{N}=\{0,1,2,\ldots\}$ and we consider the following set:

$$\{f:\mathbb{N}\longrightarrow\mathbb{N} \mid\forall n\in\mathbb{N} \ , f(n)=f(n+1)+f(n+2)\}$$

I can tell that this set has at least one element $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)=0$ but can we say more about its cardinal? and what would happen in the case of decreasing functions

答:
Let $f$ be a function in the given set. Then for some fixed $n$:
$$
\begin{aligned}
f(0) &= f(1)+f(2)\ ,\\
f(1) &= f(2)+f(3)\ ,\\
f(2) &= f(3)+f(4)\ ,\\
f(3) &= f(4)+f(5)\ ,\\
f(4) &= f(5)+f(6)\ ,\\
&\qquad\text{ and so on up to}\\
f(n) &= f(n+1)+f(n+2)\ .
\end{aligned}
$$
We add and get:
$$
f(0) = f(2)+f(3)+f(4)+\dots + f(n) + \text{ some more terms}\ .
$$
The value $f(0)$ is finite, on the R.H.S we have integer terms $\ge 0$, so only finitely many are $\ne 0$. So starting with some index, all "terms" are zero. Now we proceed inductively from this index backwards to see that all values of $f$ are zero.

Title: Cardinality of set of functions


问:
Prove $\|x_1 + \cdots + x_m\|_p \leq m^{\text{max}\{0, 1/p - 1\}}(\|x_1\|_p + \cdots + \|x_m\|_p)$ for $p \in (0, \infty), x_i \in \mathbb{C}^n$.
This is a problem or fact presented in some compressed sensing literature as the $\ell^p$ quasinorm $(p \in (0,1])$ is of particular interest for sparse signal reconstruction. To address the problem, I suspect the previously introduced results will be needed or useful:

(1). For $0 < p < 1$, $\|x_1 + x_2\|_p^p \leq \|x_1\|_p^p + \|x_2\|_p^p$, $x \in \mathbb{C}^n$

(2). For $0 < q < p \leq \infty$, $\|x\|_p \leq \|x\|_q \leq n^{1/q - 1/p}\|x\|_p$, $x \in \mathbb{C}^n$.

First off, for $p \geq 1$, $\text{max}\{0, 1/p -1\} = 0$, and so the result just becomes the standard triangle inequality of a norm. For the more interesting part, suppose $p \in (0,1)$. If I look at fact (2), noticing that $0 < p < 1$, then

$$\|x\|_p \leq n^{1/p-1}\|x\|_1$$

which looks encouraging. I see another useful fact in that, for an appropriate choice of vector, the sum $\sum_{i=1}^m \|x_i\|^p$ could be regarded as the $\ell^1$ norm. For example, if $m=n$ and $e \in \mathbb{C}^n$,  $e = (1, 1, \ldots, 1)$ and $e_i$ are the appropriate standard basis vectors, then $\|e\|_1 = \sum_{i=1}^n \|e_i\|_p^p$. Of course, this example can be extended beyond just $m=n$.

For example, if I suppose $m<n$ and I choose $x_1 = e \in \mathbb{C}^m$, then I get $\|x_1\|_p \leq m^{1/p - 1}\|x_1\|_1$, which looks suspicously close to the result I want. Furthermore, I could now consider $x_2 \in \mathbb{C}^n, x_2 = (x_1, 0, \ldots, 0)$. If I choose $y_i \in \mathbb{C}^n$ where $y_i = (e_i,0, \ldots,0)$ and $e_i$ is still the $i^{th}$ standard basis vector in $\mathbb{C}^m$, then

$$\|y_1 + \cdots + y_m\|_p \leq m^{1/p-1}\|y_1 + \cdots + y_m\|_1$$

which looks even closer to the result I want given $y_i \in \mathbb{C}^n$.

Somehow I just can't figure out how these facts piece together to give the final result. If anyone has any hints or solutions it would be great to see how it all comes together!

答:
This follows from the triangle inequality (1) and an application of Hölder's inequality with exponent $1/p.$ Indeed if $0 < p < 1$ by repeatedly applying (1) we have
$$ \lVert x_1 + \dots + x_m \rVert_p^p \leq \lVert x_1 \rVert_p^p + \dots + \lVert x_m \rVert_p^p, $$
and applying Hölder to the right-hand side with exponents $1/p$ and $1/(1-p)$ we get
$$ \sum_{i=1}^m\lVert x_i \rVert_p^p \leq \left(\sum_{i=1}^m\left(\lVert x_i \rVert_p^p\right)^{\frac1p} \right)^p \left(\sum_{i=1}^m 1^{\frac1{1-p}}\right)^{1-p} = m^{1-p} \left(\sum_{i=1}^m\lVert x_i \rVert_p \right)^p, $$
so raising both sides to the power $1/p$ gives
$$ \lVert x_1 + \dots + x_m \rVert_p \leq m^{\frac1p -1} \left( \lVert x_1 \rVert_p + \dots + \lVert x_1 \rVert_p \right),$$
as required.


问:
Inner product of matrices and direction
In vector $v\in \mathbb{R}^n$, we can use inner product to say something

if $u, v\in \mathbb{R}^n$ are in the same direction, $$\left\langle \frac{v}{\|v\|_2}, \frac{u}{\|u\|_2}\right\rangle = 1$$

if $\left\langle v,u \right\rangle>0$, then we know the angle between them are less then $90^\circ$.

Can we define such concept in matrix?  if we view a square matrix $X\in \mathbb{R}^{n\times n}$ as a vector, can we do the similar thing as above?

We know we have frobenius norm (for matrix) is the Euclidean norm of the vector by listing the coefficient of matricies.

答:
Yes, most certainly. Just like the degree of parallelity of two vectors is measured by the inner product of their norms,

$\langle \widehat{\pmb{u}}, \widehat{\pmb{v}} \rangle$,

The parallelity of two $n \times n$ matrices $A$ and $B$ is measured by their anticommutator,

$\{ A, B \} = AB+BA$

Similarly, the matrix operation which measures degree of orthogonality, corresponding to $\widehat{\pmb{u}} \times \widehat{\pmb{v}}$ in the case of vectors, is given by their commutator,

$\left[ A, B \right] = AB-BA$

This isn't surprising, considering the fact that the notion of cross products can be generalized using the wedge product, defined as,

$\pmb{u} \wedge \pmb{v} = \pmb{u} \otimes \pmb{v} - \pmb{v} \otimes \pmb{u}$

Where $\pmb{u} \otimes \pmb{v} = \pmb{u} \pmb{v}^T$

These are just general tricks which when applied to matrices, tell us about the 'angle' between them in the matrix space $\mathbb{R}^n \mapsto \mathbb{R}^n$

Note: Just as in the case of vectors, we consider their norms to find the pure angle between them; in the case of matrices $A$ and $B$, we really apply the above operations to $\frac{A}{\det \left( A \right)}$ and $\frac{B}{\det \left( B \right)}$ respectively.

Title: Inner product of matrices and direction


问:
Number of Walks in a graph in terms of the eigenvalues and eigenvectors of the adjacency matrix
I am reading "Algebraic Combinatorics" by Richard Stanley and had a question about one of the corollaries.

It states that given the adjacency matrix, $A(G)$ of the graph $G$, denote $\lambda_1, \lambda_2, \cdots \lambda_n$ as the eigenvalues and $v_1, v_2, \cdots v_n$ as the corresponding eigenvectors.

Then, if we set up the orthogonal matrix, $U$, with its columns being the eigenvectors, the value of the $(i, j)$ input in $A(G)^{\ell}$ can be expressed as: $$[A(G)^{\ell}]_{i,j} = \sum_{k=1}^n (u_{ik}u_{jk}\lambda_k^{\ell}).$$

However, when I tried to follow the proof, I got something a little different, namely: $$[A(G)^{\ell}]_{i,j} = \sum_{k=1}^n (u_{ki}u_{kj}\lambda_k^{\ell})$$ and cannot figure out where I went wrong.

Here is my work:

From basic linear algebra, we know that $U$ diagonalizes $A(G)$.

Hence, $$U^{-1} \cdot A(G) \cdot U = diag( \lambda_1, \lambda_2 \cdots \lambda_n).$$

Rewriting, we get: $$A(G) = U \cdot diag( \lambda_1, \lambda_2 \cdots \lambda_n) \cdot U^{-1}.$$

Now, we need $A(G)^{\ell}.$ It is a known fact that we can rewrite the above equation to: $$A(G)^{\ell} = U \cdot diag( \lambda_1^{\ell}, \lambda_2^{\ell} \cdots \lambda_n^{\ell}) \cdot U^{-1}.$$

Now, we need to do matrix multiplication to get the $(i, j)$ index.

When we do $U \cdot  diag( \lambda_1^{\ell}, \lambda_2^{\ell} \cdots \lambda_n^{\ell})$, the result is a matrix where the value at $(i, j)$ is: $$u_{i, j} \cdot \lambda_i^{\ell}.$$

Now, we must multiply this matrix by $U^{-1}.$ Since $U$ is orthogonal (this is another known fact that the matrix with columns being the eigenvectors are orthogonal), $U^t = U^{-1}.$

If we let matrix $W = U \cdot diag( \lambda_1^{\ell}, \lambda_2^{\ell} \cdots \lambda_n^{\ell})$, then we can write the $(i, j)$ index of $W \cdot U^{-1}$ as: $$\sum_{k=1}^n (W_{k, j} \cdot U^t_{i, k}) = \sum_{k=1}^n (W_{k, j} \cdot U_{k, i}) = \sum_{k=1}^n (u_{k, j} \cdot \lambda_k^{\ell} \cdot u_{k, i}).$$

The textbook didn't show any work for these computations but got: $$\sum_{k=1}^n (u_{j, k} \cdot \lambda_k^{\ell} \cdot u_{i, k}).$$

Where did I go wrong?

答:
In general, both of your mistakes seem to stem from an incorrect definition of matrix multiplication. The correct definition is that $$(AB)_{ij} = a_{i1}b_{1j} + a_{i2} b_{2j} + \dots + a_{in} b_{nj} = \sum_{k=1}^n a_{ik} b_{kj}.$$

Here's the computation in more detail. Let $\Lambda = \text{diag}(\lambda_1, \dots, \lambda_n)$.

If we multiply $U \Lambda^\ell$, the $(i,j)$ entry of the result is actually $$\sum_{k=1}^n u_{ik} (\Lambda^\ell)_{kj} = u_{ij} (\Lambda^\ell)_{jj} = u_{ij} \lambda_j^\ell$$ and not $u_{ij} \lambda_i^\ell$. The first expression is the definition of matrix multiplication. To go from the first expression to the second, note that $\Lambda^\ell$ is a diagonal matrix, and therefore $(\Lambda^\ell)_{kj} = 0$ unless $k=j$. The third is just simplification.

From there, the $(i,j)$ entry of $U \Lambda^\ell U^{-1} = U \Lambda^\ell U^{\mathsf T}$ can be found as
$$
    \sum_{k=1}^n (U \Lambda^\ell)_{ik} (U^{\mathsf T})_{kj} = \sum_{k=1}^n (U \Lambda^\ell)_{ik} u_{jk} = \sum_{k=1}^n (u_{ik} \lambda_k^\ell) u_{jk}
$$
which is what the book states. The first expression is the definition of matrix multiplication. To go from the first to the second, we replace the $(k,j)$ entry of $U^{\mathsf T}$ by the $(j,k)$ entry of $U$. To go from the second to the third, we substitute the expression obtained earlier.

There's another way to do the whole operation at once:
$$
    (U \Lambda^\ell U^{\mathsf T})_{ij} = \sum_{a=1}^n \sum_{b=1}^n u_{ia} (\Lambda^\ell)_{ab} (U^{\mathsf T})_{bj}.
$$
Here, $(\Lambda^\ell)_{ab}$ is $0$ unless $a=b$, so we can set $a=b=k$ and rewrite this as
$$
    \sum_{k=1}^n u_{ik} (\Lambda^\ell)_{kk} (U^{\mathsf T})_{kj} = \sum_{k=1}^n u_{ik} \lambda_k^\ell u_{jk}.
$$


问:
Evaluating $\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$

How can I find
$$\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$$

I know the domain of the function is any real number except 4, but how can I eliminate the zero denominator?

答: I personally believe you should use l'hôpital. Differentiate both numerator and denominator if you get 0/0 or inf/inf. Then, evaluate at 4. So, in this case we would get the square root of $\dfrac{x}{\sqrt{x^2-7}}$. Now, evaluate at 4 yielding 4/3. Doing the same with the denominator, we get $-\dfrac{3}{2\sqrt{3x+4}}$ which evaluates to -3/8 at 4. Dividing 4/3 by -3/8 yields -32/9.


问:
How can we prove that Gaussian integers are algebraic integers?
I heard in a lecture that every gaussian integer is in fact a root of a monic 2nd degree polynomial, it has been a day now and I can't figure this out.</p>

答:
There are a few things to say here:

It's a general fact that the sum of two algebraic numbers/integers is also one and clearly, $bi = b\sqrt{-1}$ and $a$ are algebraic integers so $a+bi$ is one too.

But in this example, we can be more explicit: $a+bi$ is the root of the polynomial $x^2 - 2ax + (a^2+b^2) = 0$.


问:
Asymptotics of the quantum dilogarithm
Fadeev and Kashaev define the quantum dilogarithm by
$$
\Psi(x) = \prod_{n=1}^\infty (1 - x q^n)
$$
for $|q| < 1$.
For $q = \exp(\epsilon)$, $\Re \epsilon < 0$, they say the asymptotic expansion
$$
  \Psi(x) = \frac{1}{\sqrt{1 - x}} \exp( \operatorname{Li}_2(x) / \epsilon)(1 + O(\epsilon))
$$ as $\epsilon \to 0$ is "easy to see," but I'm having trouble deriving it.

Here $\operatorname{Li}_2$ is the dilogarithm
$$
  \operatorname{Li}_2(x) = - \int_0^x \frac{\log( 1-t )}{t} \, dt = \sum_{n=1}^\infty \frac{x^n}{n^2}.
$$
I've tried expanding
$$
    \log \left( \Psi(x) \sqrt{1-x} \exp(- \epsilon^{-1} \operatorname{Li}_2(x)) \right) = \log \Psi(x) + \frac{1}{2} \log(1 - x) - \frac{1}{\epsilon} \operatorname{Li}_2(x).
$$
but this doesn't seem like it can work: if you expand $\log \Psi(x)$ in $\epsilon$ there's only positive powers, so I don't see how you can get a cancellation with the $\epsilon^{-1} \operatorname{Li}_2(x)$.

答:
Not exactly an answer, but this may help.

Define the coefficients $a_n(q)$ by
$$f_q(x)=\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}a_n(q)x^n.$$
Since $f_q(0)=1$, we have $a_0(q)=1$. Then note that
$$\begin{align}
\prod_{m\ge1}(1-xq^m)&=(1-xq)\prod_{m\ge2}(1-xq^m)\\
&=(1-xq)\prod_{m\ge1}(1-xq^{m+1})\\
&=(1-xq)\prod_{m\ge1}(1-(xq)q^{m})\\
&=(1-xq)f_q(xq).
\end{align}$$
Thus
$$\begin{align}
\sum_{n\ge0}a_n(q)x^n&=(1-xq)\sum_{n\ge0}a_n(q)q^nx^n\\
&=\sum_{n\ge0}a_n(q)q^nx^n-\sum_{n\ge0}a_n(q)q^{n+1}x^{n+1}\\
&=a_0(q)(qx)^0+\sum_{n\ge1}a_n(q)q^nx^n-\sum_{n\ge1}a_{n-1}(q)q^{n}x^{n}\\
&=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\
1+\sum_{n\ge1}a_n(q)x^n&=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\
\sum_{n\ge1}a_n(q)x^n&=\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}.
\end{align}$$
Equating coefficients, we have
$$a_n(q)=\left(a_n(q)-a_{n-1}(q)\right)q^{n},\qquad n\ge1.$$
This is
$$a_n(q)=\frac{q^n}{q^n-1}a_{n-1}(q).$$
And since $a_0(q)=1$, we have
$$a_n(q)=\prod_{k=1}^{n}\frac{q^k}{q^k-1}=\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n},$$
that is
$$\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n}x^n.$$
Hopefully this can help :)

Title: Asymptotics of the quantum dilogarithm


问:
Summation including Stirling numbers of the second kind
I would like to show the following: (I am pretty sure it is correct but was not able to prove it.)

$\begin{eqnarray*}
&&\sum_{j=1}^{n}j\binom{n}{j}S\left( n,j\right) j! =n(n^n-(n-1)^n) \\
&&
\end{eqnarray*}$

where $S(n,j)$ stands for the Stirling numbers of the second kind.

答:
Using the Stirling number EGF we find for $n\ge 1$

$$\sum_{j=1}^n j{n\choose j} {n\brace j} j!
= n \sum_{j=1}^n {n-1\choose j-1} {n\brace j} j!
\\ = n  \sum_{j=0}^{n-1} {n-1\choose j} {n\brace j+1} (j+1)!
\\ = n \times n! [z^n]
\sum_{j=0}^{n-1} {n-1\choose j}
\frac{(\exp(z)-1)^{j+1}}{(j+1)!} (j+1)!
\\ = n \times n! [z^n] (\exp(z)-1)
\sum_{j=0}^{n-1} {n-1\choose j}
(\exp(z)-1)^{j}
\\ = n \times n! [z^n] (\exp(z)-1) \exp((n-1)z)
\\ = n \times n! [z^n] (\exp(nz)-\exp((n-1)z))
= n (n^n - (n-1)^n).$$

as claimed. Here we have used the fact that set partitions have
combinatorial class

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z}))$$

which gives the EGF

$$\exp(u(\exp(z)-1))$$

so that

$$\sum_{n\ge j} {n\brace j} \frac{z^n}{n!} =
[u^j] \exp(u(\exp(z)-1)) = \frac{(\exp(z)-1)^j}{j!}.$$


问:
Abelian Elliptic Surfaces
By abelian surface we mean a 2-dimensional algebraic complex torus. Thus
$$ S=\Bbb{C}^2/\Gamma$$
where $\Gamma$ is a rank $4$ lattice in $\Bbb{C}^2$ and such that $S$ is algebraic. It has trivial canonical bundle $K=0$, so all plurigenera $P_n$ are equal to $1$. Also, topologically $S$ is a $4$-torus so one finds $q(S)=2$.

By elliptic surface we mean that there exists a fibration $S\rightarrow C$ on a smooth curve $C$, such that the generic fiber is an elliptic curve.

In examples (ix.4) on Beauville's book he claims: an abelian surface $S$ is elliptic if and only if there are two elliptic curves $E,F$ and an exact sequence
$$ 0\rightarrow E\rightarrow S\rightarrow F\rightarrow 0 $$

Any hint to understand why is this true? Thank you.

答:
The if part is obviously true, so we just need to prove the only if part.

Suppose an abelian surface $S$ is elliptic, so there is a fibration $S\rightarrow C$ s.t. the generic fiber is an elliptic curve. We pick a generic fiber $f$ which is an elliptic curve, then we have the closed embedding $f\rightarrow S$.

By Proposition V.12, any morphism between two complex tori is a composition of a translation and a group homomorphism, so there is a translation $E$ of $f$ such that $E$ is a subgroup of $S$. Since $S$ is an abelian surface, $S/E$ is a one-dimensional abelian variety, hence an elliptic curve $F$, which proves the only if part.

Remark: Actually you can get more. Since any curve is movable (in the sense of algebraically equivalence) by translation, the intersection number of any two fibers is zero implies that every fiber is a translation of $f$, so the elliptic fibration $S\rightarrow C$ must isomorphic to $S\rightarrow F$.

Title: Abelian Elliptic Surfaces


问:
What does group theory add to the understanding of error correcting codes?
I'm learning about group codes. Here's how I understand it (please correct me if I'm wrong):

$\mathbb{Z}_2^n$ under xor is an abelian group.

$\text{syn}: \mathbb{Z}_2^n \rightarrow \mathbb{Z}_2^n$, the syndrome function, is a homomorphism.

The kernel of this homomorphism is, by definition, the set of codewords. Let's call it $C$. Being a kernel, it's also a normal subgroup.

It can also be shown that $C$ is generated by the rows of the parity check matrix of the code.

By the Fundamental Homomorphism Theorem, $\mathbb{Z}_2^n / C \cong \text{Im}\ syn$. So, every coset corresponds to a different syndrome and the elements in a given coset all have the same syndrome.

From the item above, it follows that the received message and the error pattern that generated it (from the sent codeword) are in the same coset.

Finally, choose the element with smallest number of $1$s (the coset leader). Let's call it $\mathbf{e}$. And let's call the received word $\mathbf{r}$. The codeword to be reconstructed is $\mathbf{e} + \mathbf{r}$.

Now, we can explain the same thing without any reference to group theoretical concepts, and dealing only with matrices instead. Even in the explanation above, we have to use matrices anyway. So it doesn't seem to be adding anything to understanding of this type of code.

But, of course, I'm missing something. What is it? Is there any insight we can get by looking at this subject from a group theoretical perspective, that we wouldn't get otherwise? Does it make the implementation of algorithms easier?

答:
For error-correcting codes, the geometry is really the natural thing for which to ask. That is, we want code words to be sufficiently far apart, so that we can easily detect and correct errors. In order to measure distance, we have some sort of metric. This metric induces a geometry on the space.

Now if we want to compute things, it is often helpful to have additional structure on the space. Algebra is really about computation, and so asking for algebraic structure is a very natural thing to do. This motivates group codes. Namely, do our code words form a (sub)group? If the answer is yes, then we now have additional algebraic structure to help us make sense of our code!

In the case when the group is a field (such as $\mathbb{Z}_{2}$), we can (as you have pointed out) reduce to the case of linear algebra, which is very well understood both mathematically and algorithmically. In particular, reducing to linear algebra allows for very efficient computations. Syndrome decoding, for example, is very efficient in the linear algebraic setting. Because of all these reasons, it is very nice when we can reduce to linear algebra without any issues!

Now in your case, since $\mathbb{Z}_{2}$ is a field, the individual codewords generate one-dimensional subspaces of $\mathbb{Z}_{2}^{n}$, while the codespace is a vector subspace of $\mathbb{Z}_{2}^{n}$. Matrices are precisely linear transformations from one vector space to another. It is nice to think about the parity-check matrix $H$ (i.e., the syndrome map) as a linear transformation, for which the code space $C$ is the kernel (or the nullspace, if you prefer). So if $e$ is our error and $x$ is a codeword, then $H(x+e) = Hx + He$ by linearity. As $x \in \text{ker}(H)$, we have that $Hx = 0$. So $H(x+e) = He$.

In practice, we will compute $H(x+e)$. However, the linear algebra machinery provides that if $H(x+e) \neq 0$, then an error was made. Furthermore, we can look up the error corresponding to the output $H(x+e) = He$.

Now vector spaces are Abelian groups with some additional structure. So folks sometimes prefer to use the language of group theory. However, if you are comfortable with the linear algebra, it is perfectly fine to talk about these as vector spaces.

One natural question to ask is how much of this generalizes when we are working over an arbitrary cyclic group $\mathbb{Z}_{n}$ (i.e., when $n$ is not necessarily prime).  I am not entirely sure how much of this breaks. One thing to point out is that while $\mathbb{Z}_{n}$ and $\mathbb{Z}_{p}$ ($p$ is prime) have the same additive structure as groups, their ring structures (when incorporating addition and multiplication) are very different. $\mathbb{Z}_{p}$ is a field, while $\mathbb{Z}_{n}$ fails to even be an integral domain. Namely, if $n$ is not prime, there are non-zero elements $a, b \in \mathbb{Z}_{n}$ such that $ab = 0$. This does not happen when we have a field.

In the case of general rings $R$, we want our codes to be sub-modules of the $R$-module $R^{n}$ rather than just additive subgroups of $R^{n}$. The sub-module condition is closer in analogy to a vector subspace than an additive subgroup, as the sub-module requires closure under both addition and scalar multiplication. When dealing with $R$-modules, the goal is often times to see what linear algebraic tools we can apply.

I'm not an expert in general group codes. There is some work in generalizing linear codes (such as codes over $\mathbb{Z}_{2}^{n}$) to the setting of $R$-modules, when the ring $R$ is nicely behaved. One such class of ring is the class of Galois rings, which are used to construct Galois codes. This undergraduate honors thesis might serve as a gentle introduction to the area (https://www.whitman.edu/documents/Academics/Mathematics/2016/Holdman.pdf). It is helpful to have a group theoretic perspective, as well as a linear algebraic perspective, when looking at these more general settings.


答: In the second-degree polynomial $f(x) = \Big( x-(a+bi)\Big)\Big( x - (a-bi)\Big),$ the imaginary parts cancel out and the coefficients are then real integers.

Title: How can we prove that Gaussian integers are algebraic integers?

Body: I heard in a lecture that every gaussian integer is in fact a root of a monic 2nd degree polynomial, it has been a day now and I can't figure this out.</p>


问:
Matrix Derivative Product Rule $\frac{d}{dx} \langle x,y\rangle Ax $
Let $A\in\mathbb{R^{n\times n}}$,$x,y\in\mathbb{R^{n}}$. What is the derivative of the following expression?

$$
\frac{d}{dx} \langle x,y\rangle Ax
$$

What I tried:

Applying the product rule doesn't work because the first term (see next equation) is inconsistent dimensionally, or I am applying it in the wrong way. Let $f(x) = \langle x,y \rangle, g(x) = Ax$, then

$$ \frac{d}{dx} f(x)g(x) = \frac{df(x)}{dx} g(x) +  f(x)\frac{dg(x)}{dx} = xAx + \langle x,y \rangle A$$.

Here the first term in the last equation doesn't make sense dimensionally, any help or suggestion would be appreciated.

答:
Avoiding coordinates one may alternatively compute $d_pf(v)$, where $f(x)=\langle x,y\rangle Ax$.  For that purpose take a differential path $c$, defined on an interval containing $0$, satisfying $c(0)=p$ and $\dot c(0)=v$, where $\dot c(t)$ is a shorthand for $\frac{d}{dt}c(t)$.  Now compute
$$\begin{align}
d_pf(v)&=\frac{d}{dt}\big|_{t=0}f(c(t))\\
&=\frac{d}{dt}\big|_{t=0}\bigl(\langle c(t),y)\rangle Ac(t)\bigr)\\
&=\bigl(\langle \dot c(t),y\rangle A\cdot c(t)+\langle c(t),y\rangle A\cdot\dot c(t)\bigr)\big|_{t=0}\\
&=\langle v,y\rangle Ap+\langle p,y\rangle Av.
\end{align}$$

Edit: If you insist to write the differential without argument (but should one?), proceed as follows:
$$\begin{align}
d_pf(v)&=\langle v,y\rangle Ap+\langle p,y\rangle Av\\
&=A\bigl(\langle v,y\rangle p+\langle p,y\rangle v  \bigr).
\end{align}$$
Observe that
$$\langle v,y\rangle p=
p\langle y,v\rangle=py^Tv,
$$
hence
$$\begin{align}
d_pf(v)&=A\bigl(py^Tv+\langle p,y\rangle v  \bigr)\\
&=A\bigl(py^T+\langle p,y\rangle I\bigr)v
\end{align}
$$
and finally
$$d_pf=A\bigl(py^T+\langle p,y\rangle I \bigr)$$

But, to be honest, I like the expression $d_pf(v)$ more for symmetry reasons.  And it works for any inner product.


问:
A question regarding the power of a sinus function.
My calculus book says the following:

$\sin^{n}{x}$ is short for $(\sin{x})^n$, $n\neq -1$.

This is obviously trivial, however I don't get it why it says  $n\neq -1$. The above condition should for $n=-1$, right?

Thank you very much for your help!
Regards,
Charlie

答:
I would go as far as saying the notation
$$
\sin^n x = (\sin x)^n
$$
is only standard when $n$ is a positive integer. $\sin^{-1}$ means $\arcsin$, and I have never seen $\sin^{-2}$ or the like. It's for this very reason that people consider the $\sin^{-1}$ notation to be confusing, since it does not conform with how we interpret $\sin^n$ generally. There are other reasons this notation is illogical: strictly speaking, $\sin$ does not have an inverse, and so writing $\sin^{-1}$ is somewhat suspect to begin with. I go into more detail about this here.


答:
That's a good looking proof! The first thing I noticed though, is that you never explicity take care of the base case, i.e. when $n=1$. I know that in this case it is quite trivial, but you should at least mention that it is trivially true. Otherwise, you never knock the first domino over, to use an analogy.

For the inductive step, you could simply start with "assume this proposition is true for $n$". Then derive that the proposition is true for $n+1$, exactly the way you did, and that is sufficient to prove that $P(n)\rightarrow P(n+1)$.

Title: Would this proof by induction for the sequence of the product of odd numbers be correct? Is there something I can do better with it?

Body:
I'm new to proofs by induction, and I was challenged by my teacher to learn this when I came up with a form to represent this sequence. Would this be correct?

$$1\cdot3\cdot\ldots\cdot(2n-1)=\frac{(2n)!}{2^nn!}$$
$$\text{If $n=1$, then }1=\frac{(2\cdot1)!}{2^1\cdot1!}$$
$$1=1$$
$$\text{Suppose the proposition holds for $n=k$ so that }1\cdot3\cdot\ldots\cdot(2k-1)=\frac{(2k)!}{2^kk!}$$
$$\text{We show that the proposition holds for $n=k+1$ holds, i.e. }1\cdot3\cdot\ldots\cdot(2(k+1)-1)=\frac{(2(k+1))!}{2^{k+1}(k+1)!}$$
$$1\cdot3\cdot\ldots\cdot(2k+1)=1\cdot3\cdot\ldots\cdot(2k-1)\cdot(2k+1)=\frac{(2k)!}{2^kk!}\cdot(2k+1)$$
$$=\frac{(2k+1)!}{2^kk!}=\frac{(2k+1)!}{2^kk!}\cdot\frac{(2k+2)}{(2k+2)}=\frac{(2k+2)!}{2^kk!(2k+2)}=\frac{(2k+2)!}{2^kk!\cdot2\cdot(k+1)}$$
$$=\frac{(2k+2)!}{2^{k+1}(k+1)!}=\frac{(2(k+1))!}{2^{k+1}(k+1)!}$$
If so, is there anything I could improve on it format-wise?


答:
It's a tricky notation problem. Generally in mathematics, $f^{-1}(x)$ means the functional inverse--not literally $\frac{1}{f(x)}$. That's why there's the distinction for $n \neq -1$. Furthermore, there is a function for $\frac{1}{\sin(x)} = \csc(x)$ (the cosecant), same for cosine (secant), and same for tangent (cotangent).

Personally, I'm not a fan of the $\sin^{-1}(x)$ notation for the inverse sine function (for the exact reason you're asking about). When I taught calculus, I tried to convince my students that this was, indeed, a confusing notation and urged them to, instead, use $\arcsin(x)$, $\arctan(x)$, and $\arccos(x)$ for the inverses.

I will add that there is the pesky notation of $\frac{df}{dx} = f'(x)$, $\frac{d^2f}{dx^2} = f''(x)$, and so on...it's gets cumbersome at a point, i.e. $\frac{d^6f}{dx^6} = f''''''(x)$??? So the general notation is a parenthesis: $\frac{d^6f}{dx^6} = f^{(6)}(x)$.

Title: A question regarding the power of a sinus function.

Body:
My calculus book says the following:

$\sin^{n}{x}$ is short for $(\sin{x})^n$, $n\neq -1$.

This is obviously trivial, however I don't get it why it says  $n\neq -1$. The above condition should for $n=-1$, right?

Thank you very much for your help!
Regards,
Charlie


答:
$$\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}=\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}} \times \dfrac{4+\sqrt{3x+4}}{4+\sqrt{3x+4}}\times \dfrac{\sqrt{x^2-7}+3}{\sqrt{x^2-7}+3} \\= \dfrac{(x^2-16)(4+\sqrt{3x+4})}{3(4-x)(\sqrt{x^2-7}+3)}$$

Now cancel the factor $x-4$ from numerator and denominator and plug $x=4$ to get the answer.

Title: Evaluating $\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$

Body:
How can I find
$$\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$$

I know the domain of the function is any real number except 4, but how can I eliminate the zero denominator?


问:
If three corners of a parallelogram are known solve for the 3 possible 4th corners.
An example would be three corners being the points: (1,1), (4,2) and (1,3). I understand the specific solution for this example: (4,4), (4,0) or (-2,2). Which I reasoned when i drew it out. The example came from a linear algebra textbook and i'm curious if it can be done some other way.</p>

答: The three possibilities come from adding the coordinates of two points and subtracting the third.  So $(4,0)=(4,2)+(1,1)-(1,3)$.  There are three choices of which point to subtract.  You are translating the point you subtract to the origin, then adding the vectors to the other two points to find the opposite one, then translating back.  So you could look at it  as $(4,0)=[(4,2)-(1,3)]+[(1,1)-(1,3)]+(1,3)$</p>


问:
Law of large numbers for non-stationary but independent process?
This question is motivated by the following: Suppose we have a conditional distribution $p(Y|X)$, denoting a “channel” (in the information theory sense). Suppose we fix a particular sequence $x_1,...$, and think of it as some realization of a random sequence $X_1,...$. Then this induces a sequence of random variables $Y_1,...$, such that $Y_i$ is distributed independently by $p(Y|X=x_i)$. Note that this gives a different distribution for each $Y_i$, but nevertheless they are independent. Intuitively, $Y_1,...$ denotes a random output sequence, based on a fixed particular input sequence $x_1,...$ and where each output variable depends only on its particular input variable.

My sense is that even though the sequence is not iid, and therefore not stationary and not ergodic, there should still be a version of the law of large numbers that applies, so long as we make some not-very-restrictive assumptions on $p(Y|X)$ (maybe we need the domains to be finite?).

Is there a result analogous to the LLN for such a sequence?

答:
The question is open-ended, but one resolution is via the notion of a conditional typical set, which is a pretty standard concept.  See, for example, Section 2.3 of these notes (which turned up as one of the top things with a google search, but the concept is standard):

https://web.stanford.edu/class/ee376a/files/2017-18/lecture_15.pdf

This is made more precise with the method of types, etc.  Assuming the alphabets of X,Y are finite.


问:
Does the Collection of Graphs not Embeddable in a Fixed Orbifold have a Well-quasi-ordering?
Using the Robertson-Seymour Theorem, one can show that given a fixed surface $S$ the collection of graphs which can't be embedded in it are defined by a finite set of forbidden minors - just as the non-planar graphs have a finite set ($K_5$ and $K_{3,3}$) of forbidden minors for embeddability.

Is the same true for $2$-dimensional compact orbifolds?

答: The underlying topological space of a 2-dimensional orbifold is a topological manifold with boundary. The point is that the only allowed stabilizers are cyclic groups of rotations or dihedral groups, which you can calculate model quotients of explicitly.</p>


问:
Why do we need that $f$ is continuous at $L$ to have $\lim \limits_{n \to \infty} f(a_{n}) = f(L)$ when $\lim \limits_{n \to \infty} a_{n} = L$?

If $$\lim \limits_{n \to \infty} a_{n} = L $$
and the function $f$ is continuous, then

$$\lim \limits_{n \to \infty} f(a_{n}) = f(L)$$

I do not understand why do we have to indicate that $f$ is continuous at $L$. Can't we just say that $f$ is defined at $L$?

答:
Consider $f\colon \Bbb R \to \Bbb R$ defined as $f(x)=1$ if $x\neq 0$ and $f(0)=0$.

Then with $L=0$, your statement does not hold for every sequence $(a_n)$ such that $a_n\neq 0$ for all $n$. This is precisely because $f$ is not continuous at $0$ although it is well defined on the whole real line (including at $0$).


问:
Integral of an exponential with abosolute values in the exponent
I would like to know how can the following integral be computed

$\int_{-\infty}^{+\infty}dw'e^{-k\mid w-w'\mid} e^{-k\mid w'\mid}$

All information is really appreciated.

答:
Perhaps I am going nuts, but let's try.

Firs of all, the integrand depends on the sign of both $\omega - \omega'$ and $\omega'$. So we split initially into two cases:

When $\omega - \omega' > 0$ that is, $\omega' < \omega$ AND $\omega' > 0$

This means that $|\omega - \omega'| = \omega - \omega'$ and $|\omega'| = \omega'$ hence the integral becomes

$$\int_0^{\omega} e^{-k(\omega - \omega')}e^{-k\omega'}\ \text{d}\omega' = \int_0^{\omega} e^{-k\omega}\ \text{d}\omega = \boxed{\frac{1-e^{-k \omega }}{k}}$$

When $\omega - \omega' > 0$, that is again $\omega' < \omega$ BUT $\omega' < 0$

This means that $|\omega - \omega'| = \omega - \omega'$ and $|\omega'| = -\omega'|$ hence the integral becomes

$$\int_{-\infty}^{\omega} e^{-k(\omega - \omega'}e^{k\omega'}\ \text{d}\omega' = \int_{-\infty}^{+\omega} e^{-k\omega + 2k\omega'}\ \text{d}\omega' = \boxed{\frac{1}{2k} e^{k\omega}}$$

Now we have to face the two other cases, that is:

When $\omega - \omega' < 0$ that is, $\omega' > \omega$ AND $\omega' > 0$

When $\omega - \omega' < 0$ that is, $\omega' > \omega$ BUT $\omega' < 0$

In the first of the latter cases the integral becomes

$$\int_{\omega}^{+\infty} e^{k(\omega - \omega')}e^{-k\omega'}\ \text{d}\omega' = e^{k\omega}\int_{\omega}^{+\infty}e^{-2k\omega'}\ \text{d}\omega' = \boxed{\frac{1}{2k}e^{k\omega}}$$

In the last of the later cases the integral becomes

$$\int_{\omega}^{0} e^{k(\omega - \omega')}e^{k\omega'}\ \text{d}\omega' =  = \int_{\omega}^0 e^{k\omega}\ \text{d}\omega' = \boxed{\frac{1-e^{k \omega }}{k}}$$


问:
Variable change and calculation of derivative respect the change
Suppose I have a function $v(x,t)$ and set the coordinate change
\begin{align}
x&\to\frac{1}{2}(\xi_1+\xi_2)\\[5pt]
t&\to\frac{1}{2c}(\xi_1-\xi_2).
\end{align}

Let $V(\xi_1,\xi_2)$ be $v$ in the new variable.  How can I prove that $v_x=V_{\xi_1}+V_{\xi_2}$?

I have seen this trick in PDE's equation but I don't understand what it means and how it is connected to the chain rule. I mean $v_x$ and $V_{\xi_1}$ or $V_{\xi_2}$ are functions in different spaces.

If needed I can add details.

Any suggestion please?

答:
First, note that we can solve for $\xi_{1}$ and $\xi_{2}$ in terms of $x$ and $t$:
$$\begin{cases}
x = \frac{1}{2}(\xi_1+\xi_2)\\
t = \frac{1}{2c}(\xi_1-\xi_2)
\end{cases}\implies
\begin{cases}
\xi_{1} = x +ct\\
\xi_{2} = x - ct
\end{cases}$$
This tells us that $$v(x,t) = V(\xi_{1},\xi_{2}) = V(x+ct,x -ct).$$  So, to differentiate $v$ with respect to $x$ we have by the chain rule: $$v_{x} = V_{\xi_{1}}\frac{d}{dx}(x+ct) +V_{\xi_{2}}\frac{d}{dx}(x-ct) = V_{\xi_{1}} + V_{\xi_{2}}.$$


问:
Bijections between Manifolds of the same dimension
Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?

I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.

I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.

I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.

答:
It may happens that your manifolds are not homeomorphic but there is a continuous bijection between them.
However, if we assume that in your question that they are homeomorphic, then in Theorem 2 of the
paper

P.H. Doyle and J.G. Hocking, Continuous bijections on manifolds, J. Austral. Math. Soc. 22 (1976), 257-263.

it is proved that if $M$ is a manifold with compact boundary $\partial M$, then any continuous bijection $f:M\to M$ is a homeomorphism.

The proof follows from invariance domain theorem as mentioned in Gustavo's answer.

That paper also contains a lot of different results when a self bijection is a homeomorphism as well as counterexamples.


问:
Question on proof of fundamental theorem of calculus

Let be $F:[a,b]\to\mathbb{R}$, where $F(x):=\int\limits_a^xf(t)dt$. We assume $f$ to be continuous at point $x_0$ and Riemann-integrable over $[a,b]$. Then we know that $F'(x_0)=f(x_0)$.

My (edited) approach (we assume $x<x_0$):

Let's define the function $M:[a,x_0]\to\mathbb{R}$, where $M(x):=\sup\{\left|f(t)-f(x_0)\right|\mid t\in[x,x_0]\}$.

If we choose an arbitrary $\frac{\epsilon}{2}>0$ then by continuity of $f$ at $x_0$ there exists a $\delta>0$ such that for all $t\in [x,x_0]$  with $|t-x_0|<\delta$ we have $|f(t)-f(x_0)|<\frac{\epsilon}{2}$. By definition of the supremum there exists a $t'\in[x,x_0]$ such that $M(x)-\frac{\epsilon}{2}<|f(t')-f(x_0)|$ and therefore $M(x)<\frac{\epsilon}{2}+|f(t')-f(x_0)|<\epsilon$. So if we only admit $x\in [a,x_0]$ such that $|x-x_0|<\delta$ we see that $M(x)$ is continuous at $x_0$.

Keeping $|x-x_0|<\delta$ in mind, allows us to conclude:

$$
\left|\frac{\int\limits_x^{x_0}f(t)dt}{x-x_0}-f(x_0)\right|=\left|\frac{\int\limits_x^{x_0}f(t)-f(x_0)dt}{x-x_0}\right|\leq \left|\frac{M(x)(x_0-x)}{x-x_0}\right| = M(x) <\epsilon.
$$
The case "$x>x_0$" is basically is the same. Hence, $F'(x_0)=f(x_0)$.

Is this proof o.k.?

答:
You are fairly close, but as I note in the comments, there are a few problems with what you write. There is an additional issue at the end.  I had already written up a very similar approach, which is how I usually do it, but let me address how to fix your argument:

Let us assume $f(x)$ is integrable on $[a,b]$ (with $x_0\in (a,b)$) so that $F(x)$ makes sense.

To show continuity of $M(x)$ from the left at $x_0$ (equivalently, show its limit as $x\to x_0^-$ is $0$), proceeding as you do until you pick $t'$ (which was a problem prior to the edits as I noted in comments), let us do this instead: we know there exists $t'\in [x,x_0]$ such that
$$M(x) - \frac{\epsilon}{2} \lt |f(t')-f(x_0)| \leq M(x).$$
Hence $$0\leq M(x) \lt |f(t')-f(x_0)|+\frac{\epsilon}{2} \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$
This holds for all $\epsilon\gt 0$, so we conclude the limit is indeed $0$.

Finally, there is a small correction I think you need to make to the final chain of inequalities, namely $r\leq s$ does not imply that $|\int_x^{x_0} r\,dt| \leq |\int_x^{x_0}s\,dt|$. (For example, integrade $-3$ on $[0,1]$, compraed to the integral of $-1$ on $[0,1]$).

Instead, you should remember that in general
$$\left|\int_a^b f(t)\,dt\right| \leq \int_a^b |f(t)|\,dt,$$
assuming both integrals exist,
so that you can say
$$\begin{align*}
\left|\frac{\int_x^{x_0}f(t)\,dt}{x-x_0} - f(x_0)\right| &= \left|\frac{\int_x^{x_0}(f(t)-f(x_0))\,dt}{x-x_0}\right|\\
 &\leq \frac{\int_x^{x_0}|f(t)-f(x_0)|\,dt}{|x-x_0|}\\
 &\leq \frac{\int_x^{x_0}M(x)\,dt}{|x-x_0|}\\
 &= \frac{M(x)(x_0-x)}{x_0-x} = M(x),
\end{align*}$$
and then taking the limit as $x\to x_0^-$ you get the desired
$$\lim_{x\to x_0^-}\frac{\int_x^{x_0} f(t)\,dt}{x-x_0} = f(x_0).$$
And then a similar correction and fix is needed for the continuity from the right and the derivative from the right.

As I had already written up the stuff below, I keep it here. This is how I usually do this.

Rather than work with the supremum of the absolute value of the difference on the interval, I'm just going to work with the supremum and infimum of the values on that interval. It should be straightforward (though wordy) to convert this to your function.

I will restrict to $x\leq x_0$, as you do. Let me define functions that are close to your $M(x)$: for $x\lt x_0$,
$$\begin{align*}
m(x) &= \inf\{ f(t)\mid x\leq t\leq x_0\}\\
M(x) &= \sup\{ f(t)\mid x\leq t\leq x_0\}.
\end{align*}$$
Because $f(x)$ is integrable on $[x,x_0]$, we know it is bounded, so $m(x)$ and $M(x)$ both make sense.

I claim that
$$\lim_{x\to x_0^-}(M(x)-m(x)) = 0.$$
Indeed:  let $\epsilon\gt 0$. Since $f$ is continuous at $x_0$, we know that there exists $\delta\gt 0$ such that if $t\in [a,x_0]$ and $|t-x_0|\lt\delta$, then $|f(t)-f(x_0)|\lt \frac{\epsilon}{4}$. Shrinking $\delta$ if necessary, we may assume that $a\leq x_0-\delta$, so that if $0\leq x_0-t\lt\delta$, then $t$ is in $[a,x_0]$.

Let $x$ be such that $0\leq x_0-x\lt\delta$. Since $M(x)$ is a supremum, there exists $t_M\in [x,x_0]$ such that $M(x)-\frac{\epsilon}{4}\lt f(t)\leq M(x)$; and there exists $t_m\in [x,x_0]$ such that $m(x)\leq f(t_m)\lt m(x)+\frac{\epsilon}{4}$. Then
$$\begin{align*}
|M(x)-m(x)| &= |M(x) - f(t_M) + f(t_M) -f(x_0)+f(x_0)- f(t_m) + f(t_m)-m(x)|\\
&\leq |M(x)-f(t_M)| + |f(t_M)-f(x_0)| \\
&\qquad\mathop{+} |f(x_0)-f(t_m)| + |f(t_m)-m(x)|\\
&\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4}\\
&= \epsilon,
\end{align*}$$
where the first $\frac{\epsilon}{4}$ follows from $M(x)-\frac{\epsilon}{4}\lt f(t_M)$; the second from the continuity of $f$ at $x_0$ and the fact that $|t_M-x_0|\lt\delta$; the third from continuity of $f$ at $x_0$ and the fact that $|t_m-x_0|\lt\delta$; and the fourth from the fact that $f(t_m)\lt m(x)+\frac{\epsilon}{4}$.

Thus, the limit of $M(x)-m(x)$ is $0$ as $x\to x_0^-$.

(This argument can  be simplified if you assume $f$ is continuous on the entire interval and you take the Extreme Value Theorem for granted, since then we can actually achieve $M(x)$ and $m(x)$, so we don't have to approximate them.)

Likewise, since $m(x)\leq f(x_0)\leq M(x)$, we have
$$\begin{align*}
0&\leq \lim_{x\to x_0^-}(f(x_0)-m(x))\\
&\leq \lim_{x\to x_0^-}(M(x)-m(x))\\
&= 0,
\end{align*}$$
so $\lim_{x\to x_0^-}m(x) = f(x_0)$; similarly, $\lim_{x\to x_0^-}M(x)=f(x_0)$.

So now,
$$\lim_{x\to x_0^-}\frac{F(x)-F(x_0)}{x-x_0} = \lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt.$$
Since $m(x)\leq f(t)\leq M(x)$ for all $t\in [x,x_0]$,
$$m(x) \leq \frac{1}{x-x_0} \int_{x_0}^xf(t)\,dt \leq M(x).$$
Therefore,
$$0 \leq \left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) \leq M(x)-m(x).$$
Letting $x\to x_0^{-}$, the Squeeze Theorem tells us that
$$\lim_{x\to x_0^-}\left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) = 0.$$
Since $\lim_{x\to x_0^-}m(x) = f(x_0)$, then
$$\lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt$$
exists, and equals $\lim_{x\to x_0^-}m(x) = f(x_0)$, as required.

A similar argument works if $x_0\leq x\leq b$.


问:
Real commutative Banach algebra with identity
I am looking for example of real commutative Banach algebra with identity which does not admit a nonzero real multiplicative linear functional</p>

答: Let $\mathbb{C}$ with it's usual structure of real commutative unital Banach algebra. If $\varphi:\mathbb{C}\rightarrow\mathbb{R}$ is linear and multiplicative, then $\varphi(1)=\varphi(1)^2$, hence $\varphi(1)\in\left\{0,1\right\}$. Since $-\varphi(1)=\varphi(-1)=\varphi(i^2)=\varphi(i)^2$, then $\varphi(1)$ cannot be equal to $1$. Therefore, $\varphi(1)=0$ and $\varphi=0$.</p>


问:
Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

答:
I think this is elementary as it requires:

1-Definition of continuity and derivative

2-Binomial Theorem ( which I think is quite elementary)

3-Geometric series

4-Series convergence test by comparison

5-limit of Uniformly Convergent series is continuous

It suffice to prove $\frac{d}{dx}\log{x}=\frac{1}{x}$ using elementary results

$x > 0$ , $n \in N$

define $e=\lim_{n \to \infty}(1+\frac{1}{n})^n$ which exists according to binomial theorem and comparison with geometric series

define $g(x)=\lim_{n \to \infty}(1+\frac{x}{n})^n$  which exists as shown below

$h(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$ which exists by comparison with geometric series and is uniformly convergent

using binomial theorem $g(x)=h(x)$

because $x>0$, $g(x)$ is increasing on $[0,\infty)$

$k,m,j \in N$

$ y \in Q$

$y=\frac{k}{m}$

$g(y)=\lim_{n \to \infty}((1+\frac{y}{n})^{\frac{n}{y}})^y$

set $n=kj$

$g(y)=\lim_{j \to \infty}((1+\frac{1}{mj})^{mj})^y=e^y$

because $g(x)$ is increasing for every $x>0$ we can find two rational numbers $p,q$ and $q>p$ such that $g(p)\le g(x) \le g(q)$

since $h(x)$ is continuous the limit $\lim_{r_n \to r}e^{r_n}=\lim_{r_n \to r}h(r_n)=h(r)$ , where $\{r_n\}$ is a sequence of rational numbers converging to $r$

So it makes sense to define $e^r=\lim_{r_n \to r}e^{r_n}$

since we can let $ p \to q$ and $e^x$ is continuous ,this implies $e^x=h(x)$

$\frac{dh}{dx}=\lim_{t \to 0} e^x\frac{e^t-1}{t}=\lim_{t \to 0} e^x\frac{\sum_{n=1}^\infty \frac{t^n}{n!}}{t} =  e^x$

$x=e^{\log{x}}$

taking derivative of both sides and using chain rule

$1=e^{\log{x}}\frac{d}{dx}\log{x}$

$\frac{d}{dx}\log{x}=\frac{1}{e^{\log{x}}}=\frac{1}{x}$


问:
Picture of l'Hopital's rule at infinity
I know that when L'Hopital's rule is taken for the $\frac{0}{0}$ case, we get that the functions start looking at lines, like in this figure from Stewart's Calculus:

But what is the visual when the limit is the $\frac{\infty}{\infty}$ case?

The only thing that comes to mind that is easily visualized is a rational function's asymptotes (and that comes from dividing numerator and denominator by the largest exponent in the denominator)

答:
The counterpart of 0/0 to infinities can be seen graphed in the example below.   Although numerator and denominator individually must  necessarily go to $\infty,$ the  quotient goes to a constant in the limit, in this case to 1.

Applying L'Hospital's Rule for any $(a,b)$ values

$$ lim_{x\to \infty}\dfrac {\log(a+3x)}{\log(b+x)} = \dfrac31\dfrac{b+x}{a+3x}=\dfrac31\dfrac{b/x+1}{a/x+3}\to 1 $$

Title: Picture of l'Hopital's rule at infinity


问:
Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base.
While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about the Taylor expansion , and try to prove this in the expanded form only, how to will we prove it ?

It's already clearly  visible for $ x > 0$ as well as for $ -1<x<0 $ . But how to prove it for $x<-1$ ? I have tried much but not able to build up a strong convincing proof.

Any help would be highly appreciated. thanks!

P.S : prove it 'WITHOUT' using $e^{x}$ function.

答:
Here is a proof that uses only that $f'=f$, a fact that is easy to prove from the Taylor expansion of $f$.

Letting
$$
  f(x)=\sum_{n=0}^\infty\frac{x^n}{n!},
  $$
suppose by contradiction that there exists $x\in \mathbb R$, such that $f(x)<0$, and let
$$
  a= \sup\{x\in \mathbb R: f(x)<0\}.
  $$
Since $f$ is positive on $\mathbb R_+$, it is clear that $a$ is well defined.  Moreover, since $f$
is continuous, we have that $f(a)=0$.

Observing that $f$ is nonzero on the interval $(a,\infty )$, we see that
$$
  g(x) = \frac{f(x+1)}{f(x)},
  $$
is well defined on that interval.  Moreover,  using that $f'=f$,  one has for every $x$ in $(a,\infty )$ that
$$
  g'(x) = \frac{f'(x+1)f(x) - f(x+1)f'(x)}{f(x)^2} =
  \frac{f(x+1)f(x) - f(x+1)f(x)}{f(x)^2} = 0,
  $$
so $g\equiv c$, for some constant $c$.
For every $x$ in $(a,\infty )$, one then has that
$f(x+1) = c f(x)$, whence
$$
  f(a+1) = \lim_{x\to a^+}f(x+1) = \lim_{x\to a^+}cf(x) = cf(a) =0,
  $$
contradicting the choice of $a$.


问:
$M \longmapsto M^3 + M$ bijective for real symmetric matrixes

$\begin{array}{ccccc}
f & : & S_n{\mathbb{(R)}} & \to & S_n\mathbb{(R)} \\
 & & M & \mapsto & M^3+M \\
\end{array}\,\,$
prove that $f$ is a bijection

proof of surjectivity

let $A\in S_n{\mathbb{(R)}}.$ $\,\alpha_1,...,\alpha_n$ its eigenvalues. There exists $Q$ an real orthogonal matrix such as $A =Q^\top \begin{bmatrix}\alpha_{1} & & \\ & \ddots & \\ & & \alpha_{n}\end{bmatrix} Q$

Let $ i \in [\![1,n]\!]$ there exists $\lambda_i\in \mathbb{R}$ such as $\alpha_i = \lambda_i^3 + \lambda_i$ (surjectivity of $x \longmapsto x^3 + x$ for real numbers)

let's define $B:= Q^\top \begin{bmatrix}\lambda_{1} & & \\ & \ddots & \\ & & \lambda_{n}\end{bmatrix} Q\,\,$ We have clearly $B\in S_n{\mathbb(R)}$ and $f(B) =A$

and I stuck for the injectivity

答:
The function $g: \Bbb{R} \to \Bbb{R}$ given by $g(x) = x^3+x$ is a continuous bijection so there exists a continuous inverse $g^{-1} : \Bbb{R} \to \Bbb{R}$.

Now notice that your function $f$ can be understood as $A \mapsto g(A)$ in the sense of continuous functional calculus since the spectrum of $A$ is real. It makes sense to consider $A \mapsto g^{-1}(A) : S_n(\Bbb{R}) \to S_n(\Bbb{R})$ as well and we have
$$g^{-1}(g(A)) = (g^{-1}\circ g)(A) = \text{id}_{\sigma(A)}(A) = A, \quad g(g^{-1}(A)) = (g\circ g^{-1})(A) = \text{id}_{\sigma(A)}(A) = A$$
so this is in fact the inverse of $f$.


问:
How do you solve for Y?
I know there has got to be a way to solve for $Y$ but I just can't seem to figure it out. Does anyone know how to solve this? Please help :)

$$5(Y(8))=C$$

$$C(Y(4))=B$$

$$B(Y(2))=A$$

$$A(Y(1))=0.50$$

答:
Hint: Substitute for $C$ from the first to the second equation to get:

$160Y^2 = B$

Then substitute for $B$ into the third equation and so on.

Title: How do you solve for Y?


问:
Coloring 9 squares in a 9x9 grid
In a 9 x 9 grid, how many ways can you color 9 boxes red such that

No two red squares touch either adjacently or diagonally.

there is exactly one red square in each row, and there is exactly one red square in
each column

答:
This is a difficult combinatorial problem.

In https://oeis.org/A002464 you will find a list of the number of ways, $A(n)$, for $n\times n$ boards.

The answer given there for $A(9)$ is $47622$.

There is also a recurrence relation for these numbers.

$$A(n)=(n+1)A(n-1)-(n-2)A(n-2)-(n-5)A(n-3)+(n-3)A(n-4).$$

Title: Coloring 9 squares in a 9x9 grid


问:
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$
How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

答:
It is a basic result of field theory that if a binomial $\, x^n - c\,$ is reducible then it has a proper factor given by the Factor Theorem or difference of squares, of one of the following two forms

$$\begin{eqnarray} \text{Factor Theorem}\!:\ \ & x^n - a^n &=\,& (x\ -\ a)\,(\cdots) \\ \Rightarrow\ \ {\rm Form_1}\!:\ \ &(x^k)^p\!-a^p &=\,& (x^k- a)\, (\cdots)\quad \text{for prime}\ \  p\mid n \\
\Rightarrow\ \ {\rm Form_2}\!:\ \  & x^{4k}\! + 4a^4 &=\,& (x^{2k}\!+2a^2)^2 - (2ax^k)^2 = \ \cdots \end{eqnarray}$$

So $\, x^4\! - 1296\,$ reducible $\,\Rightarrow\, 1296 = a^2,\,a^4,\,{\rm or}\,\ {-}4a^4.\,$ $\,1296 = 4\cdot 324 = 4\cdot 4\cdot 81 = 2^4\cdot 3^4,\,$ hence we obtain factors of Form$_1,\,$ viz. $\, x^4-(\pm a)^4 =  (x\mp a)\,(\cdots)$

A proof of said theorem can be found in Karpilovsky, Topics in Field Theory, Theorem 8.1.6.


问:
Given that $x,y,z$ are positive reals such that $xyz=1$.What is the minimum value of $(x^2+4)(y^2+4)(z^2+4)$.
Help me to prove that minimum value of expression $ (x^2+4)(y^2+4)(z^2+4) $ is 125 when $ xyz = 1$. I think that minimum value achieved when $x=y=z=1$. But I can prove only $ (x^2+4)(y^2+4)(z^2+4) = xyz(x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) = (x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) \ge 2\sqrt{x \frac{4}{x}} \cdot 2\sqrt{y \frac{4}{y}} \cdot 2\sqrt{z \frac{4}{z}} = 2^6=64$

答:
Hint: Note that the (suspected) equality condition is when $ x = y = z = 1 $.

In your inequality, it is achieved only when $ x = \frac{4}{x}$, which gives us $x = 2$. So, your approach likely would not work.

How can we take the equality condition into account? We can weight the variables so that the AM-GM equality condition results in $x=1$.

 $ x^2 + 1 + 1 + 1 + 1 \geq 5 x^{\frac{2}{5} }$, with equality when $ x= 1$.


答:
A sketch:

You can expand the product
\begin{align}
(x^2+4)(y^2+4)(z^2+4)&=x^2y^2z^2+4(x^2y^2+y^2z^2+z^2x^2)+16(x^2+y^2+z^2)+64\\
&=4(x^2y^2+y^2z^2+z^2x^2)+16(x^2+y^2+z^2)+65.
\end{align}
Now use the AM-GM inequality: the geometric mean of $x^2,y^2, z^2$ is equal to $1$, so $\frac{x^2+y^2+z^2}3\ge 1$, and the minimum is attained when $w=y=z \:(=1)$.

Same observation for $x^2y^2, y^2z^2,z^2x^2$ and therefore
$$(x^2+4)(y^2+4)(z^2+4)\ge 4\cdot 3+16\cdot 3+65=125.$$

Title: Given that $x,y,z$ are positive reals such that $xyz=1$.What is the minimum value of $(x^2+4)(y^2+4)(z^2+4)$.

Body: Help me to prove that minimum value of expression $ (x^2+4)(y^2+4)(z^2+4) $ is 125 when $ xyz = 1$. I think that minimum value achieved when $x=y=z=1$. But I can prove only $ (x^2+4)(y^2+4)(z^2+4) = xyz(x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) = (x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) \ge 2\sqrt{x \frac{4}{x}} \cdot 2\sqrt{y \frac{4}{y}} \cdot 2\sqrt{z \frac{4}{z}} = 2^6=64$


问:
Pointwise convergence of Random walk to the Brownian motion
Let $\{X_i\}_{i \in \mathbb{N}}$ be a sequence of i.i.d random variables taking value in the set $\{-1,1\}$ with $P(X_i = 1) = \frac{1}{2}$.
Let $S_n = \sum\limits_{i=1}^n X_i$, and for $k \in \mathbb{N}$ define the random process $B^k$ as
$$
B^k(t) = \frac{S_{[kt]}}{\sqrt{k}} \qquad t \in \mathbb{R}
$$

where $[x]$ is the integer part of $x$.

I know that $(B^k(t))_{t \in \mathbb{R}}$ converges to the standard Brownian motion in distribution, does it also converges pointwise? i.e. is there a random process $\{B(t)\}_{t\in \mathbb{R}}$ so that
$$
\forall t \in \mathbb{R} \quad  \lim_{k \to \infty} B^k(t) = B(t) \quad \text{a.s}
$$
and further that $B(t)$ is the Brownian motion.

答: You cannot do this but if you allow constructing $B^k$ with different copies of i.i.d. random variables, then by Skorokhod's representation theorem you can find a space such that the point-wise convergence hold (for different $B^k$ the $X_i$ are different).


问:
For what $a$ and $b$ is  $9x^4-12x^3+28x^2+ax+b$ a perfect square?
If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)</p>

Body:
$9x^4-12x^3+28x^2+ax+b=(3x^2+cx+d)^2=9x^4+6cx^3+(c^2+6d)x^2+2cdx+d^2$

So $c=-2, d=4$ and $a=2cd=-16, b=d^2=16$


问:
Subobjects in the category of compact Hausdorff spaces
Why in the category of compact Hausdorff spaces the subspaces of a given object coincide with its subobjects?
This is not the case in the category of topological spaces.</p>

答: Let $X$ and $Y$ be compact Hausdorff spaces, and let $f:X\to Y$ be a continuous injection; then $f[X]$ is a compact subspace of $Y$ homeomorphic to $X$, and a monomorphism $g:Z\to Y$ is equivalent to $f:X\to Y$ iff $Z$ is homeomorphic to $X$ (and both are homeomorphic to $f[X]=g[Z]$). In short, the subobject is completely determined by a compact subspace of $Y$, and vice versa.


问:
Consider the sequence of functions and compute the limit of the integral
Consider the sequence of functions $f_n:[0,1] \to R$ given by

$$f_n(x) = \frac{e^{-(n + x^2)}}{n^2 + x}$$
for $x \in [0,1]$ and $n \geq 1$

Compute $$\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx$$

Maybe I'm overthinking this, but it seems like all I have to do is compute the integral of $f_n(x)$ and then take the limit. However, I am wondering if there is more to it since I can't seem to find an easy way to compute the integral. Can anyone lend me a hand?

Thanks!

答:
$$f_n(x)=\frac{e^{-(n+x^2)}}{n^2+x}=\frac{e^{-n}}{n^2}\frac{e^{-x^2}}{1+x/n^2}$$
now if we make the substitution $u=x/n^2\Rightarrow dx=n^2du$ and our integral becomes:
$$I_n=\int\limits_0^{1/n^2}e^{-n}\frac{e^{-n^4u^2}}{1+u}du=e^{-n}\int\limits_0^{1/n^2}\frac{(e^{-u^2})^{n^4}}{1+u}du$$
Now remember that the average of a function is defined as:
$$\hat{f}_{[a,b]}=\frac1{b-a}\int_a^bf(x)dx$$ now if $b\to a^+$ it is clear that $\hat{f}\to f(a)$ In our case the range of the interval is tending to zero, in other words:
$$\lim_{n\to\infty}n^2\int\limits_0^{1/n^2}\frac{e^{-n^4u^2}}{1+u}du=\lim_{n\to\infty}\lim_{u\to0^+}\frac{e^{-n^4u^2}}{1+u^2}$$
maybe you can use this?


问:
If three corners of a parallelogram are known solve for the 3 possible 4th corners.
An example would be three corners being the points: (1,1), (4,2) and (1,3). I understand the specific solution for this example: (4,4), (4,0) or (-2,2). Which I reasoned when i drew it out. The example came from a linear algebra textbook and i'm curious if it can be done some other way.</p>

答:
Given 3 vertices, these vertices define 2 vectors; the unknown endpoint corresponds to the sum of these 2 vectors.

Example: $(1,1)$ and $(1,3)$ form the vector $(0,2)$.  $(1,1)$ and $(4,2)$ form the vector $(3,1)$.  Sum these vectors to get $(3,3)$.  Then add the vertex $(1,1)$ back to get $(4,4)$ as the formerly unknown corner.


问:
Calculating area of cube that has same volume as sphere with known area
I am reading math book that has following statement:

Certain sphere and cube have same volume

Area of the cube can be calculated using formula

$$ A_{cube} = 6 * \Biggl(\sqrt[3]{\frac{4π}{3}*\biggl(\sqrt[2]\frac{A}{4π}\biggl)^3}\Biggl) $$

where $A$ is area of the sphere.

$ A = 3.847m^2 $

$ A_{cube} $ should be $ 4.773m^2$

When I try calculate result of the formula I get different result: $ A_{cube} = 5.351m^2 $. I get same result with my calculator and when using Wolfram Alpha.

I can get to expected result diffently:

Calculate volume of the sphere from area of the sphere

Since the cube must have same volume as the sphere I know volume of the cube and can use it to calculate its area

This leads to expected result of $ 4.773m^2$

I found that ratio between areas of cube and sphere should be close to $ 1.24 $. $ 1.24 * 3.847m^2 \approx 4.77m^2 $ which also is consistent with expected result.

So why do I get different result using the formula?

答:
That equation can be abbreviated as:
$6*V_{sphere}^{1/3}$

However, $V_{sphere}^{1/3}$ is the length of each edge on the cube-- you need the area of each face. The equation should be $6*V_{sphere}^{2/3}$. This equation will get you the expected answer. The unabbreviated version of the true equation is:

$$ A_{cube} = 6 * \Biggl(\sqrt[3]{\frac{4π}{3}*\biggl(\sqrt[2]\frac{A}{4π}\biggl)^3}\Biggl)^2 $$


问:
Linear map, prove that $T^2 = T$ is equivalent to existence of certain matrix representation
I'm trying to prove that if for a linear map $T$ from a finite vector $V$ space to itself we have $T^2 = T$ iff there exists a basis of $V$ such that the matrix representation $A$ of $T$ with regard to that basis looks like:

$$
\begin{pmatrix}
  I & 0 \\
  0 & 0
\end{pmatrix}
$$

Proving that the latter implies the former is easy, but I'm not sure about the other direction. I have tried the following:

Let $\lambda$ be an Eigenvalue of $A$ with corresponding eigenvector $v$, then we have $Av = \lambda v$ and $A^2v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda^2 v$. And since $A^2 = A$ this means that $\lambda^2 = \lambda \implies \lambda \in \{0, 1\}$. That means that all the diagonal elements of $A$ must be zero or one, but how do I show that they are consecutive?

EDIT: Further attempts, is this correct?

Assume $A$ has dimension $n \times n$ and k eigenvalues are one, the rest zero. Then for a Jordan normal form $J$ of $A$, it must be true that $\operatorname{rank} J = k$ and thus $\dim \ker J = \dim \ker{A} = \dim E(0) = n - k$ and $\dim \ker{J -I} = \dim \ker{A - I} = \dim E(1) = k$. Thus the algebraic and geometric multiplicities of the eigenvalues match and $A$ is diagonalisable.

答:
You're definitely on the right track. Once you know that the eigenvalues are $0$ or $1$, you know you can write the matrix with respect to some basis in Jordan normal form so the diagonal elements are $0$ or $1$ (if you try to diagonalize the matrix and the $1$s and $0$s are in the wrong order, you can just swap the orders of your basis elements). So, the only thing left to show is that the matrix can be made diagonal.

For this, say you have a transformation that has ones and zeroes along the diagonal in Jordan normal form but isn't diagonal. Can it satisfy $T^2=T$?


问:
Simple Linear Regression Question
Let $Y_{i} = \beta_{0} + \beta_{1}X_{i} + \epsilon_{i}$ be a simple linear regression model with independent errors and iid normal distribution.  If $X_{i}$ are fixed what is the distribution of $Y_{i}$ given $X_{i} = 10$?

I am preparing for a test with questions like these but I am realizing I am less up to date on these things than I thought.  Could anyone explain the thought process used to approach this kind of question?

答:
Let $\epsilon_i \sim N(0,\sigma^2)$. Then, we have:

$$Y_i \sim N(\beta_0 + \beta_1 X_i,\sigma^2)$$

Further clarification:

The above uses the following facts:

(a) Expectation is a linear operator,

(b) Variance of a constant is $0$,

(c) Covariance of a random variable with a constant is $0$ and finally,

(d) A linear combination of normals is also a normal.

Does that help?

Title: Simple Linear Regression Question


问:
A series of equations
ax+y-2=0
x-ay+3=0
2x+y-a=0

Find a if (a,a) lies inside the triangle formed by these three lines

Title: A series of equations

Body:
Assuming $a,b \in\mathbb{R}$, you have

$$ a^2+2ab+4b^2 = a^2+2ab+b^2+3b^2 =  (a+b)^2 +3b^2 \geq 0.$$


问:
minimum polynomial of $e^{2\pi i/p}$ for $p$ a prime number
I wanna find the minimum polynomial of $e^{2\pi i/p}$ for $p$ a prime number.

I tried to fix a prime number $p$ and so we can write it as $x^p-1=(x-1)(x^{p-1}+x^{p-2}+...+1$). But how to continue now?

答:
what you basically want to find is an extension of the rational numbers that contains the n-th root of unity $\alpha$ over $\mathbb{Q}$. Generally, for a positive integer $n$, $$T^n-1= \prod_{d\mid n} \Phi_d(T)$$ where $ \Phi_d(T)$ is the d-th cyclotomic polynomial of order $\phi(d)$, which is irreducible over $\mathbb{Q}$and gives you the field extension of $\mathbb{Q}$ containing precisely the d-th roots of unity.

$$\Phi_d(T) = \!\! \prod_{1 \leq k \leq d\atop \operatorname{ggT}(k, d) = 1}
  \left(T - e^{2 \pi \cdot \mathrm i k / d}\right)$$

Let $m_\alpha(T)$ be the minimal polynomial of $\alpha$. In your case $n=p$ and $\alpha = e^{2\pi i/p}$, thus here $m_\alpha(T)=\Phi_p(T)$.

The extension is then given by the quotient $\mathbb{Q}(\alpha) \cong\mathbb{Q}[T]/m_\alpha(T)$.

Now you have to convince yourself that $m_\alpha(T)=\frac{X^p-1}{X-1}=1+X+\dots X^{p-1}$ and that this polynomial is irreducible. Hint: Consider the automorphism of rings $\mathbb{Q}[T] \rightarrow \mathbb{Q}[T]$ sending $f(T) \mapsto f(T+1)$, then use Eisenstein criterium and argue that irreducibilty is persevered under isomorphisms.


答:
Specifically the eigenspace for $\lambda=0$ is the kernel of $T$, and the eigenspace for $\lambda=1$ is the range.

Observe also that $x=Tx+(Tx-x)$ with $(Tx-x)\in\ker T$ and that $Tx=x$ if $x$ is in the range of $T$.

Title: Linear map, prove that $T^2 = T$ is equivalent to existence of certain matrix representation

Body:
I'm trying to prove that if for a linear map $T$ from a finite vector $V$ space to itself we have $T^2 = T$ iff there exists a basis of $V$ such that the matrix representation $A$ of $T$ with regard to that basis looks like:

$$
\begin{pmatrix}
  I & 0 \\
  0 & 0
\end{pmatrix}
$$

Proving that the latter implies the former is easy, but I'm not sure about the other direction. I have tried the following:

Let $\lambda$ be an Eigenvalue of $A$ with corresponding eigenvector $v$, then we have $Av = \lambda v$ and $A^2v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda^2 v$. And since $A^2 = A$ this means that $\lambda^2 = \lambda \implies \lambda \in \{0, 1\}$. That means that all the diagonal elements of $A$ must be zero or one, but how do I show that they are consecutive?

EDIT: Further attempts, is this correct?

Assume $A$ has dimension $n \times n$ and k eigenvalues are one, the rest zero. Then for a Jordan normal form $J$ of $A$, it must be true that $\operatorname{rank} J = k$ and thus $\dim \ker J = \dim \ker{A} = \dim E(0) = n - k$ and $\dim \ker{J -I} = \dim \ker{A - I} = \dim E(1) = k$. Thus the algebraic and geometric multiplicities of the eigenvalues match and $A$ is diagonalisable.


问:
$f_n(x) = [x_n + g (x)]^2$ is not uniformly convergent, for every $x\in X$.
The statement is:

Let $(x_n)_{n\in\mathbb{N}}$ be a sequence of non-zero real numbers such that $x_n → 0$ when $n → + ∞$. Show
that if g: X → R is unlimited, then the sequence of functions $f_n$: X → R defined by

$$f_n(x) = [x_n + g (x)]^2$$
for every $x \in X$, it will not converge uniformly.

I am not sure that what I did is correct. Let's go:

As $f_n=x_n^2+2x_ng(x)+g(x)^2$, is that true that if one term of this sum is not uniformly convergent, then $f_n$ won't be? See:

$x_n$ is uniformly convergent due to the fact that $x_n  → 0$ when $n → + ∞$. And $g(x)$ is uniformly convergent due to the fact that for every $\varepsilon>0, \exists n_0\in\mathbb{N}$ such that
$$n>n_0\Rightarrow |g(x)^2-g(x)^2|=0<\varepsilon, \forall x\in X.$$

However, $x_ng(x)$ is not uniformly convergent (I will hide the procedure because it is a routine calculation, it just uses the fact that g is unlimited). So, $2x_ng(x)$ isn't uniformly convergent as well.

So, can I conclude that $f_n$ is not uniformly convergent based on the fact that one of the terms isn't uniformly convergent?

答:
If $f_n$  were unif. conv, then $x_ng(x)=f_n(x)-x_n^2-g(x)^2$ would be unif conv, and you have checked it is not possible.

So, can I conclude that fn is not uniformly convergent based on the fact that one of the terms isn't uniformly convergent?

NO, BUT, you can conclude that it is not unif. conv, since one term is NOT unif conv and the other terms are.


问:
Sign in Doob's Inequality
I am reading Durrett's Probability Theory ad Examples. In chapter 4 the Doob's inequality is define as follows:

I am wondering what is $X_{m}^{+}$ in this case? I am not sure why there is a $+$ sign. Does it indicates that we are only considering the positive values of $X_{m}$? But then why cant we just use the absolute value of $X_{m}$? Any idea is appreciated.

答: The usual definition is $X_m^+:=\max(X_m,0)$. This is also sometimes written as $X_m\lor 0:=\max(X_m,0)$.

Title: Sign in Doob's Inequality


问:
Prove that if $A \preceq B$, then $C^{A} \preceq C^{B}$
I am proving the following proposition from set theory:

If $A \preceq B$, then $C^{A} \preceq C^{B}$.

Here is my proof. Suppose $A \preceq B$. Then there exists a one-one mapping $f: A \to B$. Define a function $\theta: C^{B} \to C^{A}$ such that for an arbitrary $\alpha \in C^{B}$ and $x \in A$, the following relation holds:
\begin{equation*}
\left[\theta\left(\alpha\right)\right]\left(x\right) = \alpha\left(f\left(x\right)\right).
\end{equation*}
We need to prove that $\theta$ is onto. That is, for all $\beta \in C^{A}$, there exists an $\alpha \in C^{B}$, such that $\beta = \theta\left(\alpha\right)$. This is equivalent to saying that for an arbitrary $\beta \in C^{A}$, there is an $\alpha \in C^{B}$, for an arbitrary $x \in A$,
\begin{equation*}
\beta\left(x\right) = \left[\theta\left(\alpha\right)\right]\left(x\right) = \alpha\left(f\left(x\right)\right),
\end{equation*}
or,
\begin{equation*}
\langle f\left(x\right), \beta\left(x\right) \rangle \in \alpha.
\end{equation*}
It is equivalent to proving that for each $\beta \in C^{A}$, there exists an $\alpha \in C^{B}$, such that
\begin{equation*}
\left\{\langle f\left(x\right),\beta\left(x\right) \rangle: x \in A\right\} \subseteq \alpha.
\end{equation*}
As $f: A \to B$ is one-one, it is unlikely that $x_{1} \neq x_{2}$ and $f\left(x_{1}\right) = f\left(x_{2}\right)$. Thus, $\left\{\langle f\left(x\right),\beta\left(x\right) \rangle: x \in A\right\}$ is indeed a subset of a function in $C^{B}$. Thus, $\theta$ is onto, and there exists a one-one function $\gamma: C^{A} \to C^{B}$, and $C^{A} \preceq C^{B}$.

Is this argument solid?

答:
Unless you’re assuming the axiom of choice, showing that there is a surjection $\theta:C^B\to C^A$ does not prove that $C^A\preceq C^B$; you really do need to show that there is an injection from $\varphi:C^A\to C^B$. Fortunately, what you did in the last paragraph can easily be adapted to do this.

Fix $c_0\in C$. For each $\alpha\in C^A$ define $\varphi(\alpha)\in C^B$ as follows:

$$\big(\varphi(\alpha)\big)(b)=\begin{cases}
\alpha(a),&\text{if }b=f(a)\text{ for some }a\in A\\
c_0,&\text{otherwise.}
\end{cases}$$

In other words, $\alpha=\big(\varphi(\alpha)\upharpoonright f[A]\big)\circ f$ for each $\alpha\in C^A$. It’s straightforward to verify that $\varphi$ is injective.

(If $C=\varnothing$, we cannot choose $c_0\in C$, but in that case $C^A=\varnothing=C^B$.)


问:
Why do block matrices behave so similarly to regular matrices?
I've been using block matrices a bit in my numerical analysis course. They have many identities that mimic the identities for regular matrices. I understand the proofs, which involve boiling things down into summations. Is there some algebraic structure behind block matrices that is causing these nice theorems? Are block matrices linear maps on the vector space of matrices or something?</p>

答: One intuition as to why block matrices behave nicely (but not too nicely) is because the set of n by n matrices forms a ring, so block matrices act as linear operators on the module over the ring of matrices. This is explains a little bit why some of the algebraic properties are retained.</p>


答:
Your proof is correct, although it seems a bit hard to read. Here is an alternative shorter proof:

Given $f:A\to B$ injective, $f$ is a bijection onto its image, which we call $D$. Thus, $C^A$ and $C^D$ have the same cardinal (you can find a bijection from $C^D$ to $C^A$ by composing with $f$). Moreover, $C^{D} \preceq C^{B}$, because to each function $f:D\to C$ we can assign an extension $F:B\to C$, and obviously if two functions are different, their extensions are different too, so we have a one-one map from $C^D$ to $C^B$.

Edit: I assumed you could use the axiom of choice, but as
Brian M. Scott
says in his answer, if that is not the case it is not enough to find a surjection from $C^B$ to $C^A$.

Title: Prove that if $A \preceq B$, then $C^{A} \preceq C^{B}$

Body:
I am proving the following proposition from set theory:

If $A \preceq B$, then $C^{A} \preceq C^{B}$.

Here is my proof. Suppose $A \preceq B$. Then there exists a one-one mapping $f: A \to B$. Define a function $\theta: C^{B} \to C^{A}$ such that for an arbitrary $\alpha \in C^{B}$ and $x \in A$, the following relation holds:
\begin{equation*}
\left[\theta\left(\alpha\right)\right]\left(x\right) = \alpha\left(f\left(x\right)\right).
\end{equation*}
We need to prove that $\theta$ is onto. That is, for all $\beta \in C^{A}$, there exists an $\alpha \in C^{B}$, such that $\beta = \theta\left(\alpha\right)$. This is equivalent to saying that for an arbitrary $\beta \in C^{A}$, there is an $\alpha \in C^{B}$, for an arbitrary $x \in A$,
\begin{equation*}
\beta\left(x\right) = \left[\theta\left(\alpha\right)\right]\left(x\right) = \alpha\left(f\left(x\right)\right),
\end{equation*}
or,
\begin{equation*}
\langle f\left(x\right), \beta\left(x\right) \rangle \in \alpha.
\end{equation*}
It is equivalent to proving that for each $\beta \in C^{A}$, there exists an $\alpha \in C^{B}$, such that
\begin{equation*}
\left\{\langle f\left(x\right),\beta\left(x\right) \rangle: x \in A\right\} \subseteq \alpha.
\end{equation*}
As $f: A \to B$ is one-one, it is unlikely that $x_{1} \neq x_{2}$ and $f\left(x_{1}\right) = f\left(x_{2}\right)$. Thus, $\left\{\langle f\left(x\right),\beta\left(x\right) \rangle: x \in A\right\}$ is indeed a subset of a function in $C^{B}$. Thus, $\theta$ is onto, and there exists a one-one function $\gamma: C^{A} \to C^{B}$, and $C^{A} \preceq C^{B}$.

Is this argument solid?


问:
Help understanding notation of first fundamental form
I'm having a hard time understanding some stuff about the first fundamental form. Since I think I have decently understood what the first fundamental form means (inner product restricted to the tangent space of a surface) I think what confuses me is the notation.. Specifically I'm having a hard time understanding how to calculate the angle of two curves on a surface

Let's take this problem as an example

Let $g_{11} = 1, g_{12}=0, g_{22}=4+u^2$ be the coefficients of the first fundamental form of some surface $Σ$ calculate the angle between the curves that lie on the surface given by the equations:

$u+v=0$ and $u-v=0$

So typically a general type of solution for this would be to calculate the tangent vectors of each of their curves at their point of intersection and use their inner product to find their angle. Let's try to do this here:

Let $c_1(t)=(t,t)$ and $c_2(t)=(t,-t)$ be a parametrization of the two curves. These two curves have tangent vectors $x_1 = (1,1) $ and $x_2 = (1,-1)$.

Now let $θ$ be their angle, it can be found as:

$cosθ = \frac{\left \langle x_1,x_2 \right \rangle}{\left \| x_1 \right \| \left \| x_2 \right \|}$

Of course since we are talking about angles on a surface we have to use the first fundamental form. In my textbook this inner product is written as: $\left \langle a,b \right \rangle = g_{ij} a^i b^j$

Can someone please use the example to help me understand this notation?

Thanks in advance!

答:
This is Einstein summation notation. You are summing over indices which appear twice, once lowered and once raised. So

$$\langle a, b\rangle = g_{ij}a^ib^j = \sum_{i,j = 1}^ng_{ij}a^ib^j.$$

If $a$, $b$ are the column vectors with components $a^i$ and $b^j$ respectively, then we can also view this sum as computing $a^TGb$ where $G$ is the $n\times n$ matrix with $(i, j)^{\text{th}}$ entry $g_{ij}$. This should be reminiscent of the situation in $\mathbb{R}^n$ where every inner product is of this form for some positive-definite symmetric $n\times n$ matrix. The difference here is that the matrix $G$ is a matrix of functions (indicating that the inner product varies from tangent space to tangent space).

In this case, we see that

$$\langle x_1, x_2\rangle = \begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix} 1 & 0\\\ 0 & 4 + u^2\end{bmatrix}\begin{bmatrix} 1\\ -1\end{bmatrix} = 1 - (4 + u)^2 = -3 - u^2.$$

You can compute similarly to find $\|x_1\|$ and $\|x_2\|$ which you can then use to compute $\theta$. However, you only need to do this at the point where the two curves intersect, namely $u = v = 0$.