问:
Alternative form of Gradient Descent
We define the standard gradient descent as follows:

$$x^{(k+1)} = \arg\min_{x\in\mathbb{R}^n} f(x^{(k)}) + \langle \nabla f(x^{(k)}), x - x^{(k)}\rangle + \frac{1}{2\alpha_k} ||x - x^{(k)}||_2^2$$

Here, $x^{(k)}$ is the result of the $k$-th iteration of the gradient descent. Solving the above equation gives us $x^{(k+1)} = x^{(k)} - \alpha_k \nabla f(x^{(k)})$, which is our usual gradient descent.

How do we prove that the gradient descent problem is equivalent to the following version:

$$x^{(k+1)} = \arg\min_{x\in\mathbb{R}^n}\langle \nabla f(x^{(k)}), x - x^{(k)}\rangle $$ subject to $$||x - x^{(k)}||_2 \leq \alpha_k||\nabla f(x^{(k)})||_2$$

答:
Just find the solution of the latter problem.

Denote $x_k = x^{(k)}$ and $g_k = \nabla f(x^{(k)})$. We need to solve the following convex problem:
$$\min\ \langle g_k, x\rangle\quad \mbox{s.t. } \|x-x_k\|^2 \le \alpha_k^2 \|g_k\|^2.$$

If $\|g_k\| = 0$ then obviously $x=x_k$. Suppose now that $\|g_k\| > 0$. Slater's conditions hold because when $x=x_k$ we have $\|x-x_k\|^2 < \alpha_k^2 \|g_k\|^2$. Hence, the above problem is equivalent to its KKT system:
\begin{align}
g_k + 2\lambda(x-x_k) &= 0,\\
\lambda(\|x-x_k\|^2 - \alpha_k^2 \|g_k\|^2) &= 0,\\
\|x-x_k\|^2 &\le \alpha_k^2 \|g_k\|^2,\\
\lambda &\ge 0.
\end{align}
The first and the last equations yield $\lambda > 0$. The first equation now becomes $x = x_k - \frac{1}{2\lambda}g_k$. Plugging this into the second equation yields $\alpha_k = \frac{1}{2\lambda}$. We conclude that the optimal solution is $x = x_k - \alpha_k g_k$, QED.

Title: Alternative form of Gradient Descent


问:
Tangent line of a convex function
Let $V : \mathbb{R}^n \rightarrow  \mathbb{R}$ be (strictly) convex and continuously differentiable on $\mathbb{R}^n$. Show that for any $y \in \mathbb{R}^n$ and any $x \in \mathbb{R}^n$
$$
V(y) \geq V(x) + \nabla V(x)^\top(y-x)
$$
For $n=1$ this implies that for any $x \in \mathbb{R}$ and for any $y \in \mathbb{R}$, $V(y)$ lies above the tangent line to $V$ at $x$.</p>

答:
I am well aware that the question is old and already has very good answers. But, since as in the linked duplicate question here someone might also care about
the intuition on how to derive the proof and the other answers are more formal, let me just add some informal explanations together with some imaginery that visualize the relationship and build intuition in the one-dimensional case.

In my opinion, the underlying idea about the proof is the link between convexity of the function and the behavior of
its derivatives. When you know that $f$ is convex, you know the derivative $f'$ is monotonically increasing. This is a standard textbook result.
Now, the claim that the graph of a convex function always lies above the tangent line of any of its points, can be regarded as the statement that the graph curves away from the tangent line, positioning the graph of $f$ above it.

To see that relationship, consider the following graph I produced in TikZ.

We have to show the claim for arbitrary $x \in (a, b)$ and $p \in (a, b)$. Therefore, take some $p \in (a, b)$ like in the picture. First, we consider another point $x_2 \in (a, b)$ such that $x_2 > p$. We know that $f$ is convex. So, because it holds that $f'(p) < f'(x_2)$ since $p < x_2$ and $f'$ monotonically increasing due to its convexity, we have that
\begin{equation}
 f'(x_2) \geq \frac{f(x_2) - f(p)}{x_2 - p} \geq f'(p)
\end{equation}
The middle term just represents the green secant line in the picture, between $p$ and $x_2$.
The RHS inequality then immediately implies that $f(x_2) \geq f(p) + f'(p)(x_2 - p)$ and so the claim holds for all $x \in (p, b)$.

Now, what if we consider points $x$ for which $x < p$, i.e. $x \in (a, p)$? Take $x_1 \in (a, p)$ as in the picture. Then, we similarly have that
\begin{equation}
 f'(p) \geq \frac{f(p) - f(x_1)}{p - x_1} \geq f'(x_1)
\end{equation}
again due to the monotone increase of $f'$ which is due to its convexity.

But then
\begin{equation}
 f'(p) \geq \frac{f(p) - f(x_1)}{p - x_1} \iff f'(p) \geq \frac{(-1)}{(-1)}\frac{f(x_1) - f(p)}{x_1 - p} \iff f(p) + f'(p)(x_1 - p) \leq f(x_1)
\end{equation}
and voilà, we're done. Note that in the last step, the inequality sign changed because $x_1 - p < 0$.

ContentLicense: CC BY-SA 3.0

Title: Tangent line of a convex function

ContentLicense: CC BY-SA 4.0


问:
Geometry problem about area of triangle.
In a $\triangle{ABC}$, $D$ is the midpoint of $BC$ and $E$ is any point on $AC$. $F$ is a point on $BC$ such that $AF \parallel DE$. if $Ar(ABC)=256 cm^2$ then find $Ar(CEF)$.

I solved this by taking a special case where $E$ was the midpoint of $AC$. Then $F$ became coincident with $B$ because of midpoint theorem. So I could easily find out the area using the fact that $CF$ (or $CB$) was a median. So $Ar(CEF)=128 cm^2$.

I know that this is the answer for all cases but I don't know how to generalize it. Please help me with this.

答:
$$[\triangle ADC]=[\frac{\triangle ABC} {2}]=128$$

D is the midpoint of BC

$$[\triangle FDE]=[\triangle ADE]$$

(same base DE, equal heights - parallel lines)

$$[\triangle CEF]=[\triangle FDE]+[\triangle CDE]=[\triangle ADE]+[\triangle CDE]=[\triangle ADC]=128$$


问:
Alternative proof of $x^x \geq \sin x$ if $x>0$
My lecturer gave me this exercise: "Show that for $x>0$ it is $x^x \geq \sin x$."

This is my approach: since $x \geq \sin x$ for all $x \geq 0$, it is sufficient to show that $x^x \geq x$; since $x>0$ we can divide by $x$ and find out that $x^x \geq x \iff x^{x-1} \geq 1 \iff e^{(x-1) \log x} \geq 1 \iff (x-1) \log x \geq 0$ and this is true for all $x>0$.

However I was curious if another approach of the following kind is possible: define $D(x):=x^x-\sin x$, so the problem is equivalent to show that $D(x)>0$ for all $x>0$; since $D(x)\to \infty$ for $x \to \infty$ and $D(x)\to1$ for $x\to0^+$ we know that
$$\forall M>0 \ \exists K_M \ \text{s.t.} \ x> K_M \implies D(x)>M$$
$$\forall\varepsilon>0 \ \exists \delta_{\varepsilon}>0 \ \text{s.t.} \ 0<x<\delta_{\varepsilon} \implies1-\varepsilon<D(x)<1+\varepsilon$$
So for the arbitrarity of $\varepsilon>0$ and $M>0$, we can deduce that in the interval $(0,\delta_\varepsilon)\cup(K_M,\infty)$ it is $D(x)>0$.

So, if I'm not wrong, proving the inequality is equivalent to prove that $K_M<\delta_{\varepsilon}$; but I don't know if this is possible, since we know that those numbers exist but we don't know them and they depend on $M,\varepsilon$.

Thank you.

答:
The inequality is obvious for $x> 1$, so we restrict to $[0,1]$. It suffices to show $D(x)$ is convex, as convex functions have a unique global minimum, i.e. it suffices to show $D''(x)>0$. Calculating we have
$$
D''(x) = x^{x-1}+x^x(1+\log(x))^2+\sin(x)>0
$$Using Newton's Method or the bisection method or some such, the unique real root of $D'$ is $x_0\approx 0.820453$, with $D(x_0)\approx 0.118675$.


问:
Rate of change problem involving trigonometry

A and B are $100$ m apart. A runner R runs at a speed of 6 m/s from A to B. Given that $BC = 50$ m and $\angle CBA = \frac{\pi}{3}$. Let $BR$ = $x$ m and $\angle BCR = \theta$. Find the rate of change of $\theta$ when $R$ is $60$ m away from $A$.

Sine law shows:

$$x = \frac{50\sin\theta}{\sin(\frac{\pi}{3} + \theta)} $$

then
\begin{align}
\frac{d\theta}{dt}
&= \frac{dx}{dt}\frac{d\theta}{dx} \\
&= -6 \cdot \frac{\sin^2(\frac{\pi}{3} + \theta)}{25\sqrt{3}}
\end{align}
Also,
$$40 = \frac{50\sin\theta_0}{\sin(\frac{\pi}{3} + \theta_0)}$$
but I couldn't find the exact value of $\sin^2(\frac{\pi}{3} + \theta_0)$ from this. How should I proceed? Thanks in advance.

答:
You have $$\dfrac{\sin (60+\alpha)}{\sin \alpha}=\dfrac{5}{4}$$

$$\dfrac{\sin 60 \cos \alpha+\cos 60 \sin \alpha}{\sin \alpha}=\dfrac{5}{4}$$

$$\Rightarrow \cot \alpha = \dfrac{\sqrt{3}}{2}$$

$$\Rightarrow \sin \alpha = \dfrac{1}{\sqrt{1+\cot ^2 \alpha}}=\dfrac{2}{\sqrt 7}$$

$$\therefore \sin (60+\alpha) = \dfrac{5}{4} \cdot \dfrac{2}{\sqrt 7}$$


问:
Show that if a group $G$ of finite order has elements of order $d,$ then $G$ has subgroups of order of every divisor of $d.$
My attempt: Let $G$ be cyclic and let $a\in G$ has order $d.$
Since order of an element of a group divides order of that group, and since converse of Lagrange theorem is true for cyclic groups, so we can have cyclic subgroups of order of every divisors of $d.$
However, I am stuck to prove for an arbitrary group of finite order.

答: If $c$ divides $d$, the order of $a^{d\over c}$ is $c$.


答: If there's an element of order $d$, say $g\in G$, then $\langle g\rangle$ is cyclic of order $d$.

Title: Show that if a group $G$ of finite order has elements of order $d,$ then $G$ has subgroups of order of every divisor of $d.$

Body:
My attempt: Let $G$ be cyclic and let $a\in G$ has order $d.$
Since order of an element of a group divides order of that group, and since converse of Lagrange theorem is true for cyclic groups, so we can have cyclic subgroups of order of every divisors of $d.$
However, I am stuck to prove for an arbitrary group of finite order.


问:
$\lim\limits_{n \to\infty}\dfrac{n^{2}-n+2}{3n^{2}+2n-4}=\dfrac{1}{3}.$ Change $n$'s by $x$ and find $M$.
$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$.

With epsilon definition I get my answer as $N=\left[ \dfrac{5}{9\varepsilon }\right] +1$. But then I thought that how can I evaluate this sequence, in functions $\delta$-$\varepsilon$ definition by substituting $n$ by $x$. But I can't make it.

So here's my question:

What can I do is I have function like this one?
$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$.

I'm thinking about that function. But how to find.

$\forall \varepsilon>0$ be given, then $\exists M>0$ such that $x>M$ implies $| f\left( x\right) -l|<\varepsilon$

For function:
$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$

And this is where I stuck. Because $x\in \mathbb{R}$, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!

答: How is it any different from what you've already done? Renaming $n$ into $x$ and $N$ into $M$ doesn't really change anything. The same answer still works. Or you can simplify it a little by dropping the rounding, since $M$ doesn't have to be an integer anymore.


问:
Algorithm for checking if a given embedding is planar in less than $O(n^2)$ time.
Given an embedding of a graph (with that I mean a concrete spacial layout of the graph on the 2D plane with only $n$ straight edges. Is this the right terminology?), how can I determine if that embedding is planar?

Just for clarification, this would be a planar embedding:

while this would not be a planar embedding:

The graph itself is of course planar both times, but since AD and BC cross in the second embedding, that embedding is not planar.

A simple solution would of course be to test if any of the edges intersect, but that is an $O(n^2)$ operation, which may be too slow for my purposes. Since there are algorithms for determining if a graph is planar in linear time, I hoped that there would also be algorithms for determining if an embedding is planar (since that seems like an easier problem), but all algorithms I can find are either used to draw an embedding, generate an embedding, or check if a graph is planar.

I am not interested in creating a planar embedding, I need to determine if a given embedding is planar. It would also be useful if the algorithm could determine which edges violate the planarity, but that is not a requirement.

答: The sweep line algorithm sounds like what you need.</p>


问:
Does the decimal representation of $\pi$ contain almost all numbers?
Let $\Pi \in \Sigma^\infty$ be the decimal representation of $\pi$ where $\Sigma=\{0..9\}$. It is not known whether $\Pi$ contains each natural number. On the other hand, no number is known not to occur in $\Pi$. (Hope I've got that right)

First of all: is this property unique to $\pi$ or do all irrational numbers share it?
UPDATE: As pointed out by Brian and MJD, irrationality is completely unrelated: 1. there are irrational numbers which are known not to contain all numbers in their representation (Brian) 2. there are irrational numbers whose representation contains all numbers (MJD).

At any event, I think it should be much easier to show that almost all numbers are present, i.e. $ \exists N\in\mathbb{N}\forall k>N\exists p\in\Sigma^\star\exists s\in\Sigma^\infty: \Pi=pR(k)s$ where $R:\mathbb{N} \to \Sigma^\star$ maps each number to its decimal representation. So "allmost all" here means all except finitely many.

Has something like this been proven or disproven?

In general I don't think it's possible to construct a non-repetitive stream of symbols $\sigma \in \Sigma^\infty $ that does not contain almost all $w\in\Sigma^\star$ ("almost all" with respect to some enumeration of $\Sigma^\star$). (EDIT: This is possible as Brian points out)

答:
There are many irrational numbers known not to contain some particular number. There are uncountably many irrational numbers which do not contain the decimal digit 7, for example -- and therefore don't contain 17, 73, or any numbers with a 7. Since almost all numbers contain a given decimal digit, this shows that there are uncountably many irrational numbers which fail to contain most integers as substrings.

As far as I know it is compatible with current knowledge that all digits of $\pi$ after some point are either 3 or 7 (no significance to these choices, obviously). So we can't even prove that almost all numbers appear.


问:
Convexity of $ (a+b)^{1/n}$
How to prove that $(a+b)^{1/n} \le a^{1/n}+b^{1/n}$ by the convexity of $(a+b)^{1/n}$

答:
Welcome to MSE: as a hint take $f(x)=\sqrt[n]{x+1}$ and prove for concavinty of $f(x)$ then use $$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}$$  now it suffice to put $x=\frac ab$ and multiply both sides by $\sqrt[n]{b}$

extra hint: if you accept  $f(x)=\sqrt[n]{x+1}$ is concave function,so
$$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}\\
\text{ put } x=\frac ab \to
\\\sqrt[n]{\frac ab+1}\leq \sqrt[n]{\frac ab}+\sqrt[n]{1}\\\text{ multiply by} \sqrt[n]{b} \text{both sides} $$

Title: Convexity of $ (a+b)^{1/n}$


问:
What an implication means when stating the solutions of an equation?
I am having trouble understanding what the following statement means: $$x^2=4 \implies x=2 \hspace{1em} \text{or} \hspace{1em} x=-2.$$

How can we relate this statement to a material conditional?

I am trying to see it as a material conditional. So on the left hand side we have the equation $A(x)$ which takes different truth values for different $x$. The same goes for the right hand side which we can denote it as $B(x)$. Now does the original statement means that $$\forall x \in \mathbb{R}: A(x) \implies B(x).$$

If I substitute the values of $x$ in these two expressions then the conditionals are all true. So how can such a statement make sense when we are trying to solve an equation or when stating the solutions of the equation? Does it mean that in the case that the left hand is true then the value "we substituted" was $2$ or $-2$ in order to make the right hand side also true?

答:
I think the real problem here is psychological. In undergraduate mathematics you learn "logical reasoning" two ways. One is essentially simple common sense, writing statements in English and parsing their meaning. The other is formal reasoning with truth tables and implications.

They rarely mix well at an elementary level.

In this example I would think about the everyday (but precise) meaning of the statements.

If (for some number $x$) you know that $x^2 = 4$ then it follows that
$x$ must be either $2$ or $-2$.

In this particular case the reasoning works the other way too.

For each of $x=2$ and $x=-2$ you know $x^2$ = 4.

You can write these statements with $\implies$ and $\iff$ but I would recommend going no further with quantifiers and formal logic.

Note. The square root symbol often causes confusion in this context. $\sqrt{4} = 2$, not $\pm 2$. That symbol always refers to the nonnegative solution.


问:
Give some example of linear functional?
Give  some example  of linear functional ?

My attempt : I know  the definition of linear functional

linear  functional  on $V$ is a  function $ f : V \to F$ such that

$f(a_1 v_1 + a_2v_2)=a_1f(v_1) + a_2f(v_1)$ for all $ a_1,a_2 \in F$ and all $v_1,v_2 \in V$

Here   i don't  know how to find the example of linear functional

But i can find the example of linear transformation  take  $f : \mathbb{R}^3 \to \mathbb{R}^3$ defined

$f(x,y,z)= (x+y,y+z,z+x)$

答:
Common examples are integrals.  Another (apparently different) example is differentiation.  Take as your space the collection of (once) differentiable functions on some open interval, say $(0,1)$.  Then use linearity of the derivative:
$$  \frac{\mathrm{d}}{\mathrm{d}x} \left( a f(x) + b g(x) \right) = a \frac{\mathrm{d}f(x)}{\mathrm{d}x} + b \frac{\mathrm{d}g(x)}{\mathrm{d}x}  \text{.}  $$

If you demand finite dimensionality, use polynomials of bounded degree.

Title: Give some example of linear functional?


问:
Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?
It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of
\begin{align}
TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\
(v,p)&\mapsto p
\end{align}
(I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.

答:
Here's an explanation that will fit into the time allotted for your crash course:

Physicists will sometime say that two vectors are the same if they point in the same direction and have the same base point.

As the abstraction that mathematicians have adopted for vectors doesn't include the base point, "$\times \{p\}$" is how they label each vector with its base point. By the way, this is also probably how computer programmers would go about it too.


问:
Using Diophantine equation
I have the following problem:

An integer leaves a remainder of 15 when divided by 77 and a remainder of 16 when divided by 78.

What remainder leaves that integer when divided by 14?

I know that we can solve this problem by using the Chinese Remainder Theorem. However, we didn't cover this theorem in class, so the idea is to instead use the diophantine equation.

My attempt:

$a = 77x + 15$ and $a = 78y + 16$

$77x +15 = 78y + 16 \Rightarrow 77x - 78y = 1.$

Calculating GCD(77,-78) gives:

$-78 = (-2).77 + 76$

$77 = 1.76 + 1 $

$76 = 76.1 + 0$

Then applying the Extended Euclidean Algorithm:

$1=(1 . 77) + (-1 . 76)=(-1 . -78) + (-1 . 77)$

A particular solution is:

$x_{0} = -1 , y_{0}=-1$

The complete solution is:

$x = -1 - 78n$

$y = -1 - 77n$

For $n=0$ we get $x=-1$ and $y=-1$.

But now, I don't know how to proceed to find the remainder when dividing by 14.

答:
If $x\equiv16\bmod78$ and $x\equiv15\bmod77$, then $x\equiv0\bmod2$ and $x\equiv1\bmod 7$.

Can you take it from here?

But if you insist on doing it your way,

$a=77x+15=78y+16$ with $x=y=-1$

so $a\equiv 77(78n-1)+15\equiv 78(77m-1)+16\bmod (77\times78)\equiv5944\bmod6006$.

Now can you find the remainder mod $14$?

Title: Using Diophantine equation


问:
Cross section of cylinder $5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$
Prove that

$$5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$$

represents a cylinder whose cross-section is an ellipse of eccentricity $\frac{1}{\sqrt 2}$.

I know how to find the plane of a cross section if I know the direction ratios of at least one generator of the cylinder, but I don't know how to find that out from just the equation of the cylinder alone.

答:
Let
$$f(x,y,z)=5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6$$
The axis of the cylinder can be obtained via $f’_x=f’_y=f’_z=0$, which leads to its equation
$$(x,y,z)=(0,2,-1)+t(1,1,-1)$$
The elliptical axes of the cross section containing the center $(0,2,-1)$ can be obtained via
$$f’_x:f’_y:f’_z=(x-0):(y-2):(z+1)$$
which leads to the respective equations for the major and minor axes
$$(0,2,-1)+t(1,-1,0),\>\>\>\>\>\>\>(0,2,-1)+t(1,1,2)$$
Then, substitute  them into $f(x,y,z)= 0$ to get the major vertexes
$(\pm\frac1{\sqrt2},2\mp\frac1{\sqrt2},-1)$ and the minor vertexes
$(\pm\frac1{2\sqrt3},2\pm\frac1{2\sqrt3},-1\pm\frac1{\sqrt3})$, and in turn the lengths of the respective major and minor axes $a=2$ and $b={\sqrt2}$. Thus, the eccentricity is $\frac1{\sqrt2}$.


问:
Directional derivative in the direction of $u(t)=(t+,1+2t)$
I have to evaluate the directional derivative of the function
$$f(x,y)=\frac{xy}{x^{2}+y^{2}+1}$$
in the direction of the vector
$$u(t)=(t,1+2t)$$
at the point
$$P(1,-1)$$

I know
$$\nabla f(1,-1)=\left(-\frac{1}{9},\frac{1}{9}\right)$$
but my question is: how to normalize the vector $u$? I found out that for $t=1$
$$u(1)=(1,3)$$
and then
$$D_{u}f=\left(-\frac{1}{9},\frac{1}{9}\right)\cdot\frac{1}{\sqrt{10}}(1,3)=\frac{2}{9\sqrt{10}}$$
but I'm not sure if this is the right way to do it.

答:
What was the original statement of the problem?

If we interpret $u(t)$ as a vector, then it's not a single vector, but rather an infinite family of different vectors — one for each real value of $t$. And we can find the corresponding family of answers, but I doubt that's what this question means.

It looks more likely that the given $u(t)$ represents a line, and the question is to find the directional derivative in the direction of the directional vector of this line. From the given vector-parametric equation of the line, its directional vector is $\mathbf{u}=(1,2)$, and then you proceed as you did.


答:
For $x_0=-1$, $a=77(-1)+15=-62$ is indeed one of the infinitely many integers which are solutions to the question.

So just find $-62 \pmod {14}$.

Title: Using Diophantine equation

Body:
I have the following problem:

An integer leaves a remainder of 15 when divided by 77 and a remainder of 16 when divided by 78.

What remainder leaves that integer when divided by 14?

I know that we can solve this problem by using the Chinese Remainder Theorem. However, we didn't cover this theorem in class, so the idea is to instead use the diophantine equation.

My attempt:

$a = 77x + 15$ and $a = 78y + 16$

$77x +15 = 78y + 16 \Rightarrow 77x - 78y = 1.$

Calculating GCD(77,-78) gives:

$-78 = (-2).77 + 76$

$77 = 1.76 + 1 $

$76 = 76.1 + 0$

Then applying the Extended Euclidean Algorithm:

$1=(1 . 77) + (-1 . 76)=(-1 . -78) + (-1 . 77)$

A particular solution is:

$x_{0} = -1 , y_{0}=-1$

The complete solution is:

$x = -1 - 78n$

$y = -1 - 77n$

For $n=0$ we get $x=-1$ and $y=-1$.

But now, I don't know how to proceed to find the remainder when dividing by 14.


问:
Is every positive definite matrix a positive definite kernel
I'm having trouble with kernels:https://en.wikipedia.org/wiki/Positive-definite_kernel, and have two small questions

1.) Is it correct to say any positive definite matrix can make a kernel.

eg.this is the definition of a positive definite kernel,

$$
\sum_{i=1}^{n} \sum_{j=1}^{n} c_{i} c_{j} K\left(\mathbf{x_{i}}, \mathbf{x_{j}}\right) \geq 0
$$

can I set $$K\left(\mathbf{x_{i}}, \mathbf{x_{j}}\right) $$to $$\mathbf{x_{i}}^{T}\mathbf{Ax_{j}}$$, where A is positive definite. And the above hold?

If this is true can I write  $$
\sum_{i=1}^{n} \sum_{j=1}^{n} c_{i} c_{j} \mathbf{x_{i}}^{T}\mathbf{Ax_{j}} \geq 0
$$ as a new kernel.

$$
\sum_{i=1}^{n} \sum_{j=1}^{n}K_{n}\left(\mathbf{x_{i}}, \mathbf{x_{j}}\right) \geq 0
$$

where $$ K_{n} = c_{i} c_{j} \mathbf{x_{i}}^{T}\mathbf{Ax_{j}} \geq 0
?$$

答:
Yes, for any positive definite matrix $A$, the kernel as $K(x_i,x_j) = x_i^TAx_j$ is positive definite. The proof is quite simple. For any $c_i$, $i=1,2,\dots,n$, we have $$\sum_i\sum_j c_ic_jK(x_i,x_j) = \sum_i\sum_j c_ic_jx_i^TAx_j = s^TAs\ge 0,$$ where $s = \sum_i c_ix_i$.

Your question doesn't really make sense, because the $c_i$ variables in the definition are not fixed, predefined variables. The definition requires that a certain inequality holds for any vector $c = (c_1,\dots,c_n)$. Now if the question is: "For a given vector $c = (c_1,\dots,c_n)$, is the kernel $K(x_i,x_j) = c_ic_jx_i^TAx_j$ positive definite?" then we need to check whether the inequality $\sum_i\sum_j c'_ic'_jK(x_i,x_j) \ge 0$ holds for any vector $c'=(c'_1,\dots,c'_n)$ (which has nothing to do with vector $c$). The answer is also yes. Same proof as above, just use $s = \sum_i c_i'c_ix_i$.


问:
Orthogonal complement to a set in $L^2[0,1]$
Given the set $F=\{f\in C[0,1] | f(t)=0, \forall t\leq 1/2\}$ of the Hilbert space $L^2[0,1],$ how do I find the orthogonal complement?

答:
For a function $g$ to be orthogonal to $F$ we need $\left \langle f,g \right \rangle = 0\ \forall f \in F$ which reduces to
$$\int_{\frac{1}{2}}^1 f(t)g(t)\text{d}t = 0$$ This means we must have $g|_{[\frac{1}{2},1]}= 0$ (since $g$ is orthogonal to each $f(t)=\delta(t-\xi)$ for $\xi \geq 0.5$), and of course $g$ can be whatever it wants in $[0,\frac{1}{2}]$.

Edit: Dirachs are not in $L^2$ so that proof needs a different reasoning. This can be emulated by using indicator functions of intervals with size $2\epsilon$ around each $x$, and using the integral mean value theorem.


问:
How to construct Morse function
I am studying cellular homology theory and there is a theorem saying that critical points of morse function corresponds to cells. For example, there should be a morse function of $\mathbb{R}P^n$ with $n+1$ critical points, how to construct such funtion?

答:
Let $Q$ be the quadratic form on $\mathbb{R}^{n+1}$ defined by
$$ Q(x_0,\dots,x_n)=\sum_{i=1}^{n+1}\alpha_ix_i^2,$$
where $0<\alpha_1<\dots<\alpha_{n+1}$ is a sequence of $n$ positive real numbers. Now take the standard embedding $S^n\subset\mathbb{R}^{n+1}$ of the $n$-sphere and let $\widetilde f=Q|_{S^n}:S^n\rightarrow\mathbb{R}$ the restriction of $Q$. Then $\widetilde f$ is a Morse function on $S^n$ with $2(n+1)$ critical points $Cr(\widetilde f)=\{e_1,-e_1,\dots,e_{n+1},-e_{n+1}\}$.

Now, $\widetilde f$ is invariant under the antipodal $\mathbb{Z}_2$ action on $S^n$, so factors to over the quotient as a map $f=\widetilde f/\mathbb{Z}_2:\mathbb{R}^n\rightarrow\mathbb{R}$. The critical points of $f$ are the $\mathbb{Z}_2$ equivalence classes of the critical points of $\widetilde f$. In particular the critical set consists of the $n+1$ points
$$Cr(f)=\{[1,0,\dots,0],[0,1,0,\dots,0],\dots,[0,\dots,0,1]\}.$$

Title: How to construct Morse function


问:
How to find if a sequence is converging or diverging?
The question is as follows: Determine if the sequence:

$$\{1/1,\ 1/3,\ 1/2,\ 1/4,\ 1/3,\ 1/5,\ 1/4,\ 1/6,\ \ldots\}$$

diverges or converges. If it is convergent, find the limit.

I am not sure where to begin, I'd appreciate it if anybody could help me on this one.

Thankyou

答: The sequence $1, 1, \frac 12, \frac 12, \frac 13, \frac 13, \dots$ is a sequence that is greater than or equal to yours and converges to zero. The sequence $0, 0, 0, \dots$ is less than or equal to it and converges to zero, so by the squeeze theorem your sequence also converges to zero.


问:
Uniform convergence of $\sum\limits_{n=1}^{\infty}\frac{n^{2}x^{2}}{e^{n^{2}|x|}}$
I want to study the uniform convergence of $F(x):=\sum\limits_{n=1}^{\infty}\frac{n^{2}x^{2}}{e^{n^{2}|x|}}$ on $S=\mathbb{R}$.

For the root test, it's clear that $F(x)$ exists for all $x\in \mathbb{R}$. But I don't see if the convergence is or not uniform.

答:
$e^x>x^2$ for $x\in R_+$ and $n^2|x|\in R_+$, therefore we have that:
$$e^{n^2|x|}>n^4x^2$$
$$\frac{n^2x^2}{e^{n^2|x|}}<\frac{n^2x^2}{n^4x^2}=\frac1{n^2}$$
Since $\sum \frac1{n^2}$ converges, from Weierstrass's test for uniform convergence it follows that your series converges uniformly.


问:
Is it possible to get rid of a matrix in one side of an equation?
Let's say we have below equation:

$$ sA = B $$

Where $s$ is an scalar value and $A$ and $B$ are matrices. How can I find the value of s? If $A$ and $B$ were real numbers I could do

$$s = \frac{B}{A}$$

答: Assuming that such an $s$ exists, it may be found by dividing an entry of $B$ by the corresponding entry of $A$, as both matrices must be the same size for the dimensions to match. Any entry of $B$ may be chosen for this, provided that the corresponding $A$ entry is not zero.


问:
How many functions can be defined from $\textrm{A}$ to $\textrm{B}$?
Help me with the confusion here.

Suppose that set ${A}$ has $4$ elements and set ${B}$ has $2$ elements.

Let $f:{A}\to {B}$ be a function.

If each element from ${A}$ can make $2$ possible choices, shouldn't the number of functions be $2\times 4=8$?

Why does the formula suggest $2^4 = 16$? ($n^m$)

If possible, please give an elaborate answer.

答: When you go up the tree of your ancestors you have at each step $2$ choices: the male or the female parent. In this way you have $2$ parents, $4$ grandparents, $8$ great-grandparents, and $16$ great-great-grandparents. The $16=2^4$ persons on the uppermost level represent the $16$ different choice-sequences of length $4$ that are possible. Each such sequene is a function $f:\>\{1,2,3,4\}\to\{{\tt male, female}\}$.


问:
Does axiom schema of specification in ZFC states that any sub-set of any set exist?
According to axiom schema of specification in ZFC:

$\forall z \forall w_1 \forall w_2\ldots \forall w_n \exists y \forall x [x \in y \Leftrightarrow (( x \in z )\land \phi )]$,

where  $\phi$ can be any formula in the language of ZFC with all free variables among $x,z,w_{1},\ldots ,w_{n}$ ($y$ is not free in $\phi$ ).

Informally, the axiom states (as far as I understand) that for any set $z$ there is (exist) a sub-set $y$, if $y$ is constructed in the way described by the axiom schema of specification.

However, couldn't we just say that any subset of set $z$ exist? It looks to me that a set construction provided by the axiom schema of specification works like this: Loop over all elements of the set $z$ check if the current element satisfy condition $\phi$ and, if it is the case, include this element into $y$.

So, couldn't we just write the following?

$\forall z [\forall x ((x \in y) \Rightarrow (x \in z)) \Rightarrow \exists y]$

Moreover, for me it is not clear how this axiom is related to the Power Set Axiom, which states that for any set $A$ there is a set that contains all the subsets of $A$. Shouldn't this axiom (making a statement about a set of all sub-sets) first "prove" that those sub-sets exist?

答:
In standard first-order logic all elements of the universe "exist", so the axiom you've written doesn't really make sense: "$\exists y$" all on its own is not a formula: quantifiers have to govern a predicate, as in $\exists y(y \neq x)$.

However, it is possible to formalise a theory of classes, in which sets like those in ZF feature as a special kind of class. One of the best known systems of this kind is called NBG. In such a system, being a set is a definable property, sets are the things that are elements of some class. So you can define the notion of set-hood thus:
$$
M(x) \equiv \exists y(x \in y)
$$
("M" for "Menge" - German for set.) In such a framework, it is meaningful to write
$$
\forall z\forall y (M(z) \land (\forall x(x \in y \Rightarrow x \in z)) \Rightarrow M(y))
$$
which says that any class $y$ that is contained in a set $z$ is itself a set. This statement or some equivalent statement is one of the axioms of NBG.


答:
The purpose of the axiom schema of specification is to give us a way of writing down (specifying) subsets of a set $z$. That is, for every formula $\phi(x)$, we can construct a subset $\{x\in z\ |\ \phi(x)\}$. In particular every subset of $z$ can be specified by a specific formula.

Without this axiom schema we have no way of knowing how to construct subsets of $z$, let alone showing that they exist. The powerset axiom does not help us either in constructing subsets. The best we could do if we know how to construct a finite number of elements $x_1,\ldots,x_n\in z$ is to construct the finite subset $\{x_1,\ldots,x_n\}\subset z$ with the pairing axiom.

Note that the formula you write is not well formed because you refer to $y$ before it is quantified.

The following question is related: Is the Subset Axiom Schema in ZF necessary?

Title: Does axiom schema of specification in ZFC states that any sub-set of any set exist?

Body:
According to axiom schema of specification in ZFC:

$\forall z \forall w_1 \forall w_2\ldots \forall w_n \exists y \forall x [x \in y \Leftrightarrow (( x \in z )\land \phi )]$,

where  $\phi$ can be any formula in the language of ZFC with all free variables among $x,z,w_{1},\ldots ,w_{n}$ ($y$ is not free in $\phi$ ).

Informally, the axiom states (as far as I understand) that for any set $z$ there is (exist) a sub-set $y$, if $y$ is constructed in the way described by the axiom schema of specification.

However, couldn't we just say that any subset of set $z$ exist? It looks to me that a set construction provided by the axiom schema of specification works like this: Loop over all elements of the set $z$ check if the current element satisfy condition $\phi$ and, if it is the case, include this element into $y$.

So, couldn't we just write the following?

$\forall z [\forall x ((x \in y) \Rightarrow (x \in z)) \Rightarrow \exists y]$

Moreover, for me it is not clear how this axiom is related to the Power Set Axiom, which states that for any set $A$ there is a set that contains all the subsets of $A$. Shouldn't this axiom (making a statement about a set of all sub-sets) first "prove" that those sub-sets exist?


问:
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$?
Please help!

Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?

答:
There is no  $f$, analytic in a neighborhood $U$ of $0$, with $f\bigl({1\over n}\bigr)=2^{-n}$ for all $n\geq1$, let alone an entire function with this property.

Proof. Such an $f$ would be not identically zero. By general principles about analytic functions there would exist an $r\geq0$ and a function $g$, analytic in $U$, with
$$f(z)=z^r \>g(z)\quad(z\in U);\qquad g(0)=:a\ne0\ .$$
It follows that
$$f\bigl({\textstyle{1\over n}}\bigr)\ 2^n={2^n\over n^r}\> g\bigl({\textstyle{1\over n}}\bigr)\to\infty\qquad(n\to\infty)\ ,$$
contradicting our basic assumption about $f$.

(Ayman Hourieh beat me by 9 minutes.)


答:
The string you wrote is not a well-formed string, $$\forall z [\forall x ((x \in y) \Rightarrow (x \in z)) \Rightarrow \exists y]$$

For one, quantifiers can be added to existing formulae, they are not atomic. The goal is to bound a variable into a context. So $\exists y$ on its own is a syntax error. The range of the quantifier is also mishandled, since it leaves the $y$ in the first part free, so you're not even writing an axiom, but rather a formula which requires the assignment to say something meaningful about $y$ in advance.

But maybe it will be clearer if we clarify the terminology. Sets are exactly the things that exist, in the context of $\sf ZFC$, so talking about a sub-set is kind of funny, since it seems to presume the existence from the get go. Every subset of a set $x$ exists, since it's a set by definition. So what's the catch here, why do we even need this axiom?

The axiom of specification is saying that every subclass of a set is a set. In other words, given a set, and given a property, the collection of elements which correspond to the property inside the set will also form a set.


问:
Proving $2|c_1|^2|c_2|^2 \leq c_1c_2^* + c_1^*c_2$ for $|c_1|,|c_2|\leq 1$
For two complex numbers $c_1, c_2$ such that $|c_1|,|c_2|\leq 1$, is the following inequality true?

$$2|c_1|^2|c_2|^2 \leq c_1c_2^* + c_1^*c_2$$

Here $c^*$ is the complex conjugate of $c$. For real numbers, this is easily shown but I am not sure how to prove this for complex numbers. I have tried a few examples numerically and haven't found a counterexample so far.

答:
It's not true. Pick two complex numbers $c_1, c_2$ such that $c_1 \overline{c_2}$ is a non-zero real multiple of $i$, e.g. $c_1 = i$ and $c_2 = 1$. Then
$$c_1 \overline{c_2} + \overline{c_1} c_2 = 2\operatorname{Re}(c_1 \overline{c_2}) = 0.$$
If the inequality were true, then we would have $2|c_1|^2 |c_2|^2 \le 0$, which would imply that $c_1 = 0$ or $c_2 = 0$, and hence $c_1 \overline{c_2} = 0$, against assumption.


问:
Jacobian of F(X)=XBX involving kronecker product.
I'm trying to prove the following:

Let $X_{m\times m}$ be a symmetric matrix of variables. Find the Jacobian, defined as:
$$\frac{\partial}{\partial vec{X'}}vec{F}$$

where $F(X)=XBX$ , where $B_{m\times m}$ is a symmetric matrix of constants. Answer: ($XB\otimes I_m$)+($I_m\otimes XB$). Where $\otimes$ refers to the Kronecker product.

My attempt:

\begin{align*}
    vec{F} &= vec{XBX} \\
           &= (X'\otimes X)vec{B} \\
           &= (X\otimes X)vec{B}
\end{align*}

Where "vec" refers to the vec operator.
So, the jacobian is given by:

$$
\frac{\partial}{\partial(X)'}(X\otimes X)vec{B}
$$

But, I don't know what to do next. Any suggestion please? Thanks!

答:
Calculate the differential before applying the vec operator.
$$\eqalign{
dF &= dX\,BX + XB\,dX \\
df &= \big((BX)^T\otimes I\big)\,dx + \big(I\otimes XB\big)\,dx \\
   &= \big((XB\otimes I) + (I\otimes XB)\big)\,dx \\
\frac{\partial f}{\partial x}
   &= \big((XB\otimes I) + (I\otimes XB)\big) \\
}$$
where
$$f={\rm vec}(F),\quad x={\rm vec}(X)$$


问:
Sangaku: to prove one of the intangents is parallel to $BC$
Given an acute triangle $\triangle ABC$ whose incircle is $I(r)$. Let $O(R)$ be the circle through $B$ and $C$ and which touches $I(r)$ interiorly. Show that the circle $P(p)$ which is tangent to $AB$, $AC$ and $O(R)$ (externaly) is such that one intagent line from $P(p)$ and $I(r)$ is parallel to side $BC$.

I have seen a couple solutions for this online but I never really understood them and they all seem wrong to me. For example, this is the problem 2 in here. The solution they give doesn't ring a bell. When they say that $\angle ACB - \angle ADE = \angle AED - \angle ABC$ (which is correct) and then they claim that  $\angle ADE < \angle ACB$ and $\angle AED > \angle ABC$ imply that $\angle ADE=\angle ABC$ it just sounds wrong. They could just use the congruence of $ADE$ and $AFG$ but instead they use this confuse argument.

And things get worse after the co-axial system. They give a huge jump and conclude that $BCD'E'$ is cyclic. This is best solution I have found for this problem but I just can't agree with it.

答:
As @Blue has pointed out in the comments,

Ignore the incircle context, and consider disjoint circles $p$ & $q$,
with circle $r$ tangent to both (internally, externally, whatever).
Then, according to some ugly coordinate-bashing in Mathematica, I
believe we can say this: When $r$ meets the external tangents of $p$ &
$q$ in four points, two of the resulting chords in $r$ are parallel to
the internal tangents of $p$ & $q$; and vice-versa. (The result in
question corresponds to a special circumstance where a chord of $r$
happens to be tangent to one of the other circles.)

There's an interesting discussion of this claim at mathoverflow.  In particular, this answer cites Gueron's Two Applications of the Generalized Ptolemy Theorem, where it is referred to and proven as the Parallel Tangent Theorem.


问:
Prove arithmetic formula for the number of bits in $j=2^k-1$
If $k$ is an integer with $0\le k<155$, and $j=2^k-1$, then it holds:
$$k\,=\,5\left(\left\lfloor\frac j{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4$$

Why does this equation hold for such a large interval of $k$? How can we prove that?

This is a shortened version of more complex formula# posted by Hallvard B. Furuseth in a comp.c post on 2003-11-03, which reportedly works to $k$ of at least $3\cdot10^{10}$.

Update (not part of the bountied question): per this source, the same author gave that formula working for $0\le k<2040$:
$$k\,=\,8\left(\left\lfloor\frac j{255((j\bmod255)+1)}\right\rfloor\bmod 255\right)-\left\lfloor\frac{86}{(j\bmod255)+12}\right\rfloor+7$$

Try it online with Python or with Wolfram Mathematica! The code computes the upper limits for both formulas.

# Full formula, for reference (not in the above code or part of the bountied question):
$$\begin{align}k\,=\,&30\left(\left\lfloor\frac j{((j\bmod(2^{30}-1))+1)\,(2^{30}-1)}\right\rfloor\bmod (2^{30}-1)\right)+\\&5\left(\left\lfloor\frac{j\bmod(2^{30}-1)}{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4\end{align}$$

答:
There exist integers $m,r$ such that
$$k=5m+r,\qquad 0\le m\le 30,\qquad 0\le r\le 4$$

Since we have
$$j=2^k-1=2^{5m+r}-1=(2^5)^m\cdot 2^r-1\equiv 1^m\cdot 2^r-1\equiv 2^r-1\pmod{31}$$
there exists an integer $a$ such that $2^{5m+r}-1=31a+2^r-1$ which can be written as $$a=2^r(2^{5-r}a+1-2^{5m})$$ This implies that $a$ is divisible by $2^r$, so there exists an integer $b$ such that $j=2^{5m+r}-1=31\cdot 2^rb+2^r-1$.

We have $j\bmod 31=2^r-1$ and $$b\bmod 31=\frac{32^m-1}{31}\bmod 31=(32^{m-1}+32^{m-2}+\cdots +32+1)\bmod 31=m$$
Using these, we get
$$\begin{align}&5\left(\left\lfloor \frac{j}{31((j \bmod 31)+1)}\right\rfloor  \bmod 31\right)-\left\lfloor\frac{12}{(j \bmod 31)+3}\right\rfloor+4
\\\\&=5\left(\left\lfloor \frac{31\cdot 2^rb+2^r-1}{31\cdot 2^r}\right\rfloor  \bmod 31\right)\underbrace{-\left\lfloor\frac{12}{2^r+2}\right\rfloor+4}_{=\ r}
\\\\&=5\left(\left\lfloor b+\frac{2^r-1}{31\cdot 2^r}\right\rfloor  \bmod 31\right)+r
\\\\&=5\left(b  \bmod 31\right)+r
\\\\&=5m+r
\\\\&=k\end{align}$$


问:
Leibniz formula for determinants even/odd?
I am trying to get my head around the Leibniz formula for this $3 \times 3$ determinant. I know that

So for $n=3$ we get $$\det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} - a_{1,2}a_{2,1}a_{3,3} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2}$$

But I don't know how we work out the signs in each case above? Could someone show me how? I know the sign is $+$ when the permutation is even but how is $a_{1,1}a_{2,2}a_{3,3}$ even but $a_{1,2}a_{2,1}a_{3,3}$ is odd?

答:
An alternative odd/even property is the parity of the number of inversions. An inversion of a permutation $\sigma$ is defined to be the set $$Inv(\sigma)=\{(i,j):i<j\text{ and }\sigma_i>\sigma _j\}.$$
So, in your case $Inv(321)=\{(1,2),(1,3),(2,3)\}$ because $3>1$ and $3>2$ and $2>1$ is odd because it has $3$ inversions and $Inv(213)=\{(1,2)\}$ (the only one is $2>1$) also has $1$ so is odd. In general, the sign of a permutation $\sigma$ is $(-1)^{|Inv(\sigma)|}.$


问:
How should I start answering this chemical substance question?
Two  substances,  A  and  B,  are  being  converted  into  a  single  compound C. In  the  laboratory  it  has  been  shown  that,  for  these  substances,  the following  law  of  conversion  holds:    the  time  rate  of  change  of  the amount x of compound C is proportional to the product of the amounts of unconverted substances A & B.  If at time t = 0 there are CAO mols/li of substance A and CBO mols of substance B, and none of compound C present,  find  x(t).    Assume  that  1  mole  of  C  is  formed  from  the combination of 1 mole of A and one mole of B.

答:
Starting from @J. W. Tanner's answer, the easy way to solve
$$\dfrac{dC}{dt}=k(A_0-C)(B_0-C)$$ is to write it as
$$\dfrac{dt}{dC}=\frac 1{k(A_0-C)(B_0-C)}=\frac{1}{k (A_0-B_0)}\left( \frac 1{C-A_0}- \frac 1{C-B_0}\right)$$ which gives (since $C_0=0$
$$t =\frac{1}{k (A_0-B_0)}\,\log \left(\frac{C-A_0}{C-B_0}\right)$$ from which it is easy to express $C(t)$.


问:
Evaluate $T_k=\sum_{n\geq 1}\text{sech}^{2k}(n \pi)$
The following formula is well known due to theory of elliptic functions: $\sum _{n=1}^{\infty } \text{sech}^2(n \pi )=-\frac{1}{2}+\frac{1}{2 \pi }+\frac{\Gamma \left(\frac{1}{4}\right)^4}{16 \pi ^3}$.

Question $1$: How to evaluate the following $$T_k=\sum _{n=1}^{\infty }\text{sech}^{2k}(n \pi)$$
For instance Mathematica gives $T_2=\sum _{n=1}^{\infty } \text{sech}^4(\pi  n)=-\frac{1}{2}+\frac{1}{3 \pi }+\frac{\Gamma \left(\frac{1}{4}\right)^8}{192 \pi ^6}+\frac{\Gamma \left(\frac{1}{4}\right)^4}{24 \pi ^3}$ but I don't know how it's generated (it cannot evaluate $T_3$, etc).

Note that series of class $S_k=\sum _{n=1}^{\infty }\text{csch}^{2k}(n \pi)$ is evaluated in this post, by manipulating the normalized Eisenstein series $G_{2k}(i)=\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+ni)^{2k}}$. For $T_k$ the corresponding series should be $\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+\frac12+ni)^{2k}}$, but so far I can't see how to compute it for arbitrary $k$.

Question $2$: By Jacobi theta identity one have $\sum _{n=1}^{\infty } \text{sech}(\pi  n)=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \pi ^{3/2}}-\frac{1}{2}$ (Mathematica also 'remembers' this result), while no closed-form of $\sum _{n=1}^{\infty } \text{csch}(\pi  n)$ seems to exist. So, what causes the difference between two cases? More generally, is it possible to evaluate the odd-weight class $$\tilde T_k=\sum _{n=1}^{\infty }\text{sech}^{2k-1}(n \pi)$$

Any help will be appreciated.

答:
This is how I see the theory behind those things, which means there could very well be a simpler and more concrete way to solve it in this particular case.

The elliptic way

We can evaluate $$\sum_{m,n}\ \frac{1}{(m+\frac12+ni)^{2k}},\qquad \sum_{(m,n)\ne (0,0)}\ \frac{1}{(m+ni)^{2k}}$$
in term of $\wp_i^{(k-2)}(1/2)$ and $(\wp_i^{(k-2)}-z^{-1})(0)$ where
$\wp_i(z) = \sum_{m,n}\frac{1}{(z+m+ni)^2}-\frac{1_{(m,n)\ne (0,0)}}{(m+in)^2}$,

which reduces to show that $\wp'(z)^2-4\wp(z)^3-g_2(i) \wp(z)=0$ (bounded entire function) then find $g_2(i)$ from $1=\int_0^1 dz=\int_0^1 \frac{d\wp(z)}{\sqrt{4\wp(z)^3+g_2(i)\wp(z)}}=\int_C \frac{dx}{\sqrt{4x^3+g_2(i)x}}$ $=A g_2(i)\int_0^1 \frac{du}{\sqrt{u^3-1}}=2A g_2(i)B(1/2,3/4)=C g_2(i)\Gamma(1/4)^2$.

This way we'll obtain some recursive formulas for the derivatives of $\wp_i$ in term of $\wp_i,\wp_i'$ which will give for the values at $0,1/2$ a rational expression in term of $g_2(i)$.

The modular way

For $f$ a level $N$ modular form with integer coefficients then the coefficients of $F_z(X)=\prod_{\gamma\in SL_2(Z)/\Gamma(N)}(X-f|_k\gamma)$ are level $1$ weight $\le m$ modular forms thus in $\sum_{4a+6b=m} \Bbb{Q} E_4^aE_6^b$ so that $f(i)$ is algebraic over $E_4(i)$ (as $E_6(i)=0$), and the complex multiplication stuffs show that in $F_i(X)$ we can replace $SL_2(Z)/\Gamma(N)$ by an abelian group (the class group of $\Bbb{Z}[Ni]$) so that $f(i)$ is radical over $E_4(i)$ whose $\Gamma(1/4)$ expression is found from the elliptic way.


问:
Bilinear Forms: An Initial Condition Proof
Let $B$ be a bilinear form on a finite dimensional vector space $V$. Suppose that for any nonzero vector $v \in V$ there exists a $w \in V$ such that $B(v, w)\neq 0$. Prove that for any linear functional $f \in V^*$ there exists a $v \in V$ such that $f(w) = B(v, w)$ for all $w \in V$.

Some of these proofs seem really simple, but I'm making this overcomplicated.

答:
Here is an easy proof of this fact:

Look at the map $\varphi:V\rightarrow V^\star, v\mapsto (w\mapsto B(v,w))$.

$\varphi$ is a linear map and according to your assumption it is also injective :
$\ker(\varphi)=\{v\in V, \varphi(v)=0 \}=\{v\in V, \forall w\in V, B(v,w)=0\}=\{0\}$.

Since $V$ and $V^\star$ have the same dimension, $\varphi$ is an isomorphism and we are done.


问:
A continuous map that is not a proper map
I know that if a map is proper then it is continuous. But to disprove the converse I am trying to think of a counter example of a continuous map that is not proper.</p>

答: Let $X$ be any noncompact space. Then $X\rightarrow\ast$ is not proper.


问:
Solving recurrence relation with a sum of binomial coefficients
I am trying to solve this recurrence relation, I have researched on how to solve it by trying to get rid of the sum however I wasn't able to do so.

$$S(a,b)=\frac{\sum_{n=0}^b{\binom{b}{n}S(a,n)}}{2^a},a\in N, b\in N_0$$
$$S(a,0)=1$$

答:
Let $\sum_{j=0}^n\binom{n}{j}S(a,j)=2^a S(a,n)$ for $n>0$ and $F(x)=\sum_j S(a,j)\frac{x^j}{j!}.$

Then we have  $e^x F(x)=\sum_n{\sum_{j=0}^n\binom{n}{j}S(a,j)\frac{x^n}{n!}}=1+2^a(F(x)-1)=1-2^a+2^a F(x)$ and therefore $F(x)=\frac{2^a-1}{2^a-e^x}.$

Edit

Let $A(n,k)$ denote the Eulerian numbers and $a_n(t)=\sum_{j=0}^n A(n,j)t^j$ the Eulerian polynomials.
Then
$$\frac{t-1}{t-e^{z(t-1)}}= \sum_j a_n(t)\frac{z^n}{n!}.$$
Therefore $$S(a,n)=\frac{a_n(2^a)}{(2^a-1)^n}.$$


问:
Find general equation for the following sequence
Given the following sequence: $$a_1 = 1, a_{n+1}=2*a_{n} + 1$$

what is the general equation for $a_n$?
I couldn't find it because the sequence is not in the form of $a_{n+1} = a_n * c$ neither is it
in the form of $a_{n+1} = a_n + c$, but more like a combination of the two.

答:
Just another way to solve
$$a_{n+1}=2\,a_{n} + 1$$
Let $a_n=n_n+k$ and replace
$$b_{n+1}+k=2\,b_{n} +2k+ 1$$ If you make $k=2k+1$, that is to say $k=-1$ you have
$$b_{n+1}=2\,b_{n}$$

Solve it and, when done, go back to $a_n$.


问:
Prove the machine can make exactly $( 2x - y )$ cakes at a period of consecutive hours
Let $x, y$ be positive integers such that $x < y < 2x − 1$.

A machine has been making cakes for aperiod of $x$ hours.

The machine made at least one cake per hour and had made less than y cakes in total.

Prove that there is a period of consecutive hours during which the machine made exactly $(2x − y)$ cakes.

I've been stuck with this seemingly easy question for days. Can anyone share some thoughts?

答:
Suppose the machine made $a_i$ cakes during the $i^{th}$ hour, and denote $S_n = a_1 + a_2 + \cdots a_n$. Then $$1 \le S_1 < S_2 < \cdots < S_x \le y-1$$

It's easy to show that $S_{x-1} \le y-2, S_{x-2} \le y-3, \ldots, S_{y-x} \le 2y-2x-1$.

Now consider the following $x+(y-x)=y$ numbers:

$$\text{Group 1: }S_1 < S_2 < \cdots < S_x $$ and

$$\text{Group 2: } S_1 + (2x-y) < S_2 + (2x-y) < \cdots < S_{y-x} + (2x-y) $$

Note that $S_1+1 < S_1+(2x-y), S_{y-x} + (2x-y) \le 2y-2x-1+2x-y=y-1.$

All numbers are between 1 and $y-1$, so there must be two of these numbers having the same value. Since in each group the numbers are distinct so there exist $i, j$ such that $S_i = S_j + 2x-y$, therefore $a_{j+1} + \cdots + a_i = 2x-y$.


问:
question about complex analysis log function
I have this domain:

$$D = \{z \in\Bbb C: \text{Re}(z) \leq \text{Im}(z) < \text{Re}(z) + 2\pi\}.$$

We know that the exponential function restricted to D is one-to-one

The log is defined as: Log: C - {0} ---> C

And the question is:

Define log as the corresponding inverse function. Is so define log continuous at all points of its domain?

I was thinking about how to prove and I think I have to prove that is compact, i.e. that is closed and bounded, and I think it is, but I don't really know how to make this proof.

答:
Let $f$ be the restriction of the exponential function to $D$. As you observed, $f$  is injective on $D$. Therefore, it has the inverse function $f^{-1}:f(D)\to D$.

However, the inverse function is not continuous. The problem is that $f$ "glues" two sides of the infinite strip $D$ together. For example, $f(0)=1$ and $\lim_{y\to 2\pi-}f(iy)=1$. Therefore, in any neighborhood of $1\in f(D)$ the inverse function $f^{-1}$ will attain value $0$, as well as some values close to $2\pi i$. This implies that   $f^{-1}$ is not continuous.


问:
Heun's method: Why is it of second order?
I want to get an intuitive understanding of Heun's method and why it has order of consistency 2, i.e. the error lies is in $O(h^{3})$ if $h$ is the stepwidth., I have read it everywhere, my professor says it without proving it. All the resources I have seen so far on the web either omit the proof as trivial, and the only one that does the proof does it wrong. I am at a loss. Can someone explain it to me in as much detail as possible?

答:
Here is my attempt at a proof of the order of the Heun's method using Taylor expansions.

Let's suppose $y(t)$ is the true solution of the autonomous ODE $d_t y = f(y)$ and assume $y$ is scalar for the sake of simplicity.
Let's suppose we have $y_n = y(t_n)$. The value at time $t_{n+1}$ given by the Heun's method is:

$$y_{n+1} = y_n + \frac{\Delta t}{2} f(y_n) + \frac{\Delta t}{2} f\left( y_n + \Delta t f(y_n) \right)$$

Let's express the last term using a Taylor expansion:
$$ f\left( y_n + \Delta t f(y_n) \right) = f(y_n) + (d_y f)(y_n) (\Delta t f(y_n)) + (d_{yy} f)(y_n) \frac{(\Delta t f(y_n))^2}{2} + (d_{yyy} f)(y_n) \frac{(\Delta t f(y_n))^3}{6} + O(\Delta t^4)$$

Inserting this expansion back into the first equation yields:
$$y_{n+1} = y_n + \Delta t f(y_n) + (d_y f)(y_n) \dfrac{\Delta t^2 f(y_n)}{2} + (d_{yy} f)(y_n) \frac{\Delta t^3 f(y_n)^2}{4} + O(\Delta t^4)$$

Now let's perform a Taylor expansion of the true solution:
$$y(t_{n+1}) = y_n +\Delta t (d_t y)(y_n) + \dfrac{\Delta t^2}{2} (d_{tt} y)(y_n) + \dfrac{\Delta t^3}{6} (d_{ttt} y)(y_n) + O(\Delta t^4)$$

which we can develop into:

$$y(t_{n+1}) = y_n +\Delta t f(y_n) + \dfrac{\Delta t^2}{2} (d_{y}f)f(y_n) + \dfrac{\Delta t^3}{6} ( (d_y f)^2 f + f^2 (d_{yy} f))(y_n) + O(\Delta t^4)$$

Comparing the terms of equal power of $\Delta t$ in the expressions of $y_{n+1}$ and $y(t_{n+1})$ we see that the local truncation error is  is of order 3, therefore the method is of order 2.

The answer provided by @mason is more elegant, but here we see what the term in $O(\Delta t^3)$ is.


问:
How to change integration bounds
i solved the integration $\frac{1}{ab}\int_{-\infty}^{\infty}e^{\frac{-(x-z)^2}{2\sigma^2}}\cdot K_0(\frac{|z|}{b})\text{d}z$ and reach upto the following step: $\frac{e^{\frac{-x^2}{2\sigma^2}}}{ab}\int_{-\infty}^{\infty}K_0(\frac{|z|}{b})\cdot e^{\frac{-z^2}{2\sigma^2}}e^{\frac{xz}{\sigma^2}}\text{d}z$. But in research paper it is derived as: $\frac{e^{\frac{-x^2}{2\sigma^2}}}{ab}\int_{0}^{\infty}K_0(\frac{|z|}{b})\cdot e^{\frac{-z^2}{2\sigma^2}}(e^{\frac{xz}{\sigma^2}}+e^{\frac{-xz}{\sigma^2}})\text{d}z$. Where $K_0$ is modified bessel function of second kind. I am not getting how it is coming. Any help please. Thank you.

Title: How to change integration bounds

Body:
$$\int_{-\infty}^\infty f(x) \, dx
= \int_0^\infty f(x) \, dx + \int_{-\infty}^0 f(x) \, dx
= \int_0^\infty f(x) \, dx + \int_0^\infty f(-x) \, dx
= \int_0^\infty (f(x) + f(-x)) \, dx$$
where $f(x) = K_0(|z|/b) e^{-z^2/(2\sigma^2)} e^{xz/\sigma^2}$.


问:
A question about 3-D Rotations in a Coordinate Plane?
I was reading the following article about how to represent a $3$-D rotation in coordinates: link

I saw that in order to do a rotation around the Z-axis, the rotation matrix is: $$\begin{equation}R_Z(\theta) =
\begin{bmatrix}
\cos \theta  & -\sin \theta   & 0 \\
\sin\theta  & \cos \theta   & 0 \\
0 & 0 & 1
\end{bmatrix}\end{equation}$$

In order to do a rotation around the X-axis, the rotation matrix is: $$\begin{equation}R_X(\theta) =
\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos \theta  & -\sin \theta  \\
0 & \sin\theta  & \cos \theta
\end{bmatrix}.\end{equation}$$

In order to do a rotation matrix around the Y-axis, the rotation matrix is: $$\begin{equation}R_Y(\theta) =
\begin{bmatrix}
\cos \theta & 0 & \sin\theta \\
0 & 1 & 0\\
-\sin \theta & 0 & \cos \theta
\end{bmatrix}.\end{equation}$$

Let's say I wanted to rotate some point, $(x, y, z)$, $\alpha^{\circ}$ about the Z-axis, $\beta^{\circ}$ around the X-axis, and $\gamma^{\circ}$ around the $Y$-axis.

Then, could I represent the coordinates as: $$
\begin{equation}
\begin{bmatrix}
\cos \alpha & -\sin \alpha   & 0 \\
\sin\alpha  & \cos \alpha   & 0 \\
0 & 0 & 1
\end{bmatrix} \cdot \begin{bmatrix}
1 & 0 & 0 \\
0 & \cos \beta  & -\sin \beta  \\
0 & \sin\beta  & \cos \beta
\end{bmatrix} \cdot \begin{bmatrix}
\cos \gamma & 0 & \sin\gamma \\
0 & 1 & 0\\
-\sin \gamma & 0 & \cos \gamma
\end{bmatrix} \cdot \begin{bmatrix}x \\
y \\
z \end{bmatrix} \end{equation}?
$$

Is there some simpler way to do it so that I don't have to work with all of these matrices? Does anyone know of a general form that'll move everything into one matrix?

答: The Rodrigues rotation formula turns an axis and angle into a single rotation matrix, which in its most general form is an orthogonal matrix with determinant $1$. Alternatively, use quaternions.


问:
Demonstration of the impossibility to draw a parallel through a point using only a straightedge.
From the responses to this  question  , it appears to be well know that it is impossible to trace a parallel   to a straight line: $\ell$   through  a point: $P$,  using exclusively a straightedge.

Can you provide a demonstration of such fact?

答:
A construction that uses only a straightedge can be transformed via a projective transformation (aka homography).

Suppose that you had a straightedge construction for a line $m$ through point $P$ parallel to line $\ell$.  Suppose a projective transformation maps $P\rightarrow P'$ and $\ell\rightarrow \ell'$.  Then the same construction would produce a line $m'$ which in general is not parallel to $\ell'$.  So we have a contradiction, and there is no such straightedge construction.

The demonstration is a little more compelling if the projective transformation leaves $P$ and $\ell$ invariant.  In that case, the same construction would produce two different lines, when applied before and after to the same point and line.


问:
Determining whether a quadratic congruence is solvable using Legendre symbol
I'm trying to detect whether the quadratic congruence $2x^2 +  5x - 9$ is congruent to $0$ modulo $101$. I've think I'll be able to detect whether there is or there is no solution using Legendre symbol, but I can't figure out how.
I'll be grateful if someone could point me to the solution.

Thanks in advance.

答:
Edit: The original modulus was $89$. We keep the original calculation. A small appended modification deals with $101$.

The given congruence has a solution if and only if the congruence
$$16x^2+40x-72\equiv 0\pmod{89}$$
has a solution. (We multiplied through by $8$ as a preparation to completing the square.)

This congruence has a solution  if and only if
$$(4x+5)^2-25-72\equiv 0\pmod{89}$$
has a solution. Equivalently, we want to find out whether $(4x+5)^2\equiv 8\pmod{89}$ has a solution.

The congruence $w^2\equiv 2\pmod{89}$ has a solution. We can see this from the fact that the Legendre symbol $(2/89)$ is equal to $1$, since $89$ is of the shape $8k+1$.

Since for any $w$, the congruence $4x+5\equiv w\pmod{89}$ has a solution, our original congruence does.

Added: For the new modulus of $101$, we arrive in the same way at the congruence $w^2\equiv -4\pmod{101}$. We have $(-4/101)=(-1/101)$. But $101$ is of the form $4k+1$, so $(-1/101)=1$, and thus the congruence has a solution.


问:
Using Diophantine equation
I have the following problem:

An integer leaves a remainder of 15 when divided by 77 and a remainder of 16 when divided by 78.

What remainder leaves that integer when divided by 14?

I know that we can solve this problem by using the Chinese Remainder Theorem. However, we didn't cover this theorem in class, so the idea is to instead use the diophantine equation.

My attempt:

$a = 77x + 15$ and $a = 78y + 16$

$77x +15 = 78y + 16 \Rightarrow 77x - 78y = 1.$

Calculating GCD(77,-78) gives:

$-78 = (-2).77 + 76$

$77 = 1.76 + 1 $

$76 = 76.1 + 0$

Then applying the Extended Euclidean Algorithm:

$1=(1 . 77) + (-1 . 76)=(-1 . -78) + (-1 . 77)$

A particular solution is:

$x_{0} = -1 , y_{0}=-1$

The complete solution is:

$x = -1 - 78n$

$y = -1 - 77n$

For $n=0$ we get $x=-1$ and $y=-1$.

But now, I don't know how to proceed to find the remainder when dividing by 14.

Title: Using Diophantine equation

Body:
$$N = 77x + 15=78y + 16$$

$$N+62 = 77x + 77=78y + 78$$

$$LCM(78;77)=6006$$

$$N+62=6006k, k \in Z$$

$$6006k-62\equiv -62\equiv 8 \pmod {14}$$


问:
Differentiation with respect to complex matrix
I want to differentiate $f(X,Y) = \Vert XYA-b \Vert^2_2 $ with respect to $X$ and $Y$, $X$ and $Y$ are matrices with complex variable elements. In real cases, I know how to differentiate there is online tools such as http://www.matrixcalculus.org/#example-collapse7 ,But, I do not know how much differences exist between complex and real cases?

答:
$\def\bb{\mathbb} \def\p#1#2{\frac{\partial #1}{\partial #2}}$
Denote the transpose, complex, and hermitian conjugates of $X$ as $(X^T,X^*,X^H)$ respectively. And denote the trace/Frobenius product using a colon
$$A:B = \sum_{i=1}^m\sum_{j=1}^nA_{ij}B_{ij} \;\doteq\; {\rm Tr}(A^TB)$$
This product is also defined for vectors by considering them as rectangular matrices. The cyclic property of the trace permits terms in such a product to be rearranged in many equivalent ways. For example
$$\eqalign{
A:BC &= B^TA:C &= AC^T:B = I:A^TBC = I:BCA^T \\
B:C &= B^T:C^T &= C:B = C^T:B^T \\
}$$
A useful guide is that the quantity on either side of the colon must have exactly the same dimensions.

The Frobenius norm of $X$ can be written using the Frobenius product as
$$\big\|X\big\|^2_F = X^*:X$$
The last tool is the $\bb{CR}$-calculus
(aka Wirtinger derivatives), wherein the real and imaginary components of a variable are treated independently under differentiation.

Write the function using the above ideas, and calculate its real Wirtinger derivative.
$$\eqalign{
 f &= (XYA-B)^*:(XYA-B) \\
df &= (XYA-B)^*:dX\,YA \\
   &= A^TY^T(XYA-B)^*:dX \\
\p{f}{X}   &= A^TY^T(XYA-B)^* \\
}$$
Analogous calculations yield the gradient wrt the complex component of $X$
$$\eqalign{
\p{f}{X^*}   &= A^HY^H(XYA-B) \\
}$$
or wrt the components of $Y$
$$\eqalign{
\p{f}{Y}   &= X^T(XYA-B)^*A^T \\
\p{f}{Y^*} &= X^H(XYA-B)A^H \\
}$$


问:
Normal intersecting a sphere
Let $\textbf{x}$ and $\textbf{y}$ be two points on the sphere. Show that the normal to the plane determined by the great circle through $\textbf{x}$ and $\textbf{y}$ intersects the sphere at the points $\pm \textbf{z}$, where

$$\bf z = \frac{x \times y}{||x \times y||}$$

Now obviously, if we let the center of the sphere lie at the origin, $\bf x \times y$ will be the normal vector to the plane that crosses the sphere and make that great circle.

Since that plane divides the sphere into two equal parts, the points were the normal intersects the sphere will be opposites of one another, thus $\pm \textbf{z}$.

I don't really get why we need to adjust that point by  $\bf ||x \times y||$? Could somebody explain?

Also, the problem itself is easier, as I am not asked to find the points, just to show that they are valid. But I don't really know how could I do it?

I could reason, that if we place that sphere at the origin, the equation of the plane that cuts the sphere and makes the great circle is described by:

$${\bf w}\cdot({\bf x\times y})=0$$

But what next?

Thanks!

答:
I don't really get why we need to adjust that point by $\bf ||x\times y||$? Could somebody explain?

The magnitude of the cross product is $||{\bf x\times y}|| = ||{\bf x}|| \cdot ||{\bf y}|| \cdot \sin (\angle {\bf xOy})$. We're assuming that the sphere has unit radius, so $||{\bf x}|| = ||{\bf y}|| = 1$, and we want the points where the normal intersects the sphere. Therefore unless $\bf x \perp y$ we need to normalise.

I'm afraid I can't help with the second half, because I'm not sure what definitions you're working from. Obviously you don't already have a definition or theorem to the effect that the equation of a plane is ${\bf P . N} = 0$ where $\bf N$ is its normal, or you would have used it already.

Title: Normal intersecting a sphere


问:
Is $ \sup\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\}<1$?

I want to compute the following supremum of at least prove that
$$
\sup\left\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\right\}<1.$$

I used the following inequalities
$$|a\overline{b}|\leq \frac{1}{2}(|a|^2+|b|^2)\text{ and } |a|^2\leq |a|^2+|b|^2.$$
I get
$$|a|^4+|a\overline{b}|^2\leq \frac{5}{4},$$
for every $(a,b) \in \mathbb{C}^2$ such that $|a|^2+|b|^2=1$. So
$$
\sup\left\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\right\}\leq \frac{5}{4}.$$
But this not solve my problem.

答:
You have

$$|a|^4+|a\overline{b}|^2= \vert a \vert^4 + \vert a \vert^2 \vert b\vert^2= \vert a \vert^4 + \vert a \vert^2( 1- \vert a \vert^2)=\vert a \vert^2 \le 1$$

for all $a,b$ such that $\vert a \vert^2+ \vert b\vert^2=1$.


问:
Small angle approximation on cosine
The problem is

Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant

I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{x\to 0}{\approx}1-\frac{x^2}{2}$ to get $$3-2(1-\frac{x^2}{2})+4(1-x^2)=5-3x^2$$ hence $k=3$. But why does using $\cos^2(x)\overset{x\to 0}{\approx}(1-\frac{x^2}{2})^2$ doesn't work out? I originally tried plugging that into the $\cos^2(x)$ but got another complete thing. why?

答:
Directly subbing $\cos^2x=(1-x^2/2+\cdots)^2$ should work out, provided you expand properly:
$$\cos^2x=1-2(x^2/2)+\dots=1-x^2+\cdots$$
$$1-\sin^2x=1-(x-\cdots)^2=1-x^2+\cdots$$

Title: Small angle approximation on cosine


答: I think you simply forgot to plug $cos\theta \approx 1 - \frac{x^2}{2}$ in for both $cos\theta$ and $cos\theta^2$ in the expression.

Title: Small angle approximation on cosine

Body:
The problem is

Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant

I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{x\to 0}{\approx}1-\frac{x^2}{2}$ to get $$3-2(1-\frac{x^2}{2})+4(1-x^2)=5-3x^2$$ hence $k=3$. But why does using $\cos^2(x)\overset{x\to 0}{\approx}(1-\frac{x^2}{2})^2$ doesn't work out? I originally tried plugging that into the $\cos^2(x)$ but got another complete thing. why?


问:
Understanding the average slope of a curve
This question mainly asks: Is my understanding of the average slope correct? This question is somewhat related to my previous question. However, its different from the previous question to a certain extent.

If $P(x_0,y_0)$ and $Q(x_0+\Delta x, y_0+\Delta y)$ are two points on the curve of the graph of square function ($y=x^2$), then the average slope of the curve between these two points can be given by the equation: $\bar m = 2x_0+\Delta x$. If $\Delta x$ is infinitely small, it can be neglected and we're left with $m=2x_0$ which is the formula to find the slope of a point at the curve.

Let $P(2,4)$ and $Q(2.00005, 4.000200003)$ be two points on the curve. The average slope between these two points of the curve is $4.00005$. Now, let's take 3 points which are between the curve $PQ$, and find their slopes by the formula $2x_0$. Let those 3 points be $(2.00002, 4.00008),(2.00003, 4.0000120001)$ and $(2.00004, 4.000160002)$.

Calculating the slopes at the given points,

1) $m=4$ at $(2, 4)$

2) $m=4.00004$ at $(2.00002, 4.00008)$

3) $m=4.00006$ at $(2.00003, 4.0000120001)$

4) $m=4.00008$ at $(2.00004, 4.000160002)$

5) $m=4.0001$ at $(2.00005, 4.000200003)$

Now we shall calculate the average of above values:

$m_{av} = \frac{4+4.00004+4.00006+4.00008+4.0001}{5}$

$m_{av} = 4.000056$

Conclusion:
The average slope $\bar m = 2x_0 + \Delta x$ is called so, because it's value is approximately equal to the average slope found by dividing the sum of certain points at an interval by the total no. of points, or $\bar m ≐ m_{av}$.

I know that $\Delta x=0.00005$ is not infinitely small nor negligible, but for the sake of simplicity I've used it here.

$4.00005≈4.000056$. They indeed are approximately equal. My question is: Is the way I understand the average slope correct? The way I've given above by calculating those long values?

答:
I'm not clear as to what your question is, but you're conclusion is correct.  You will learn shortly why $y = x^2 \implies \;\text{slope}\; \approx 2x_0 + \Delta x$, and why as $\Delta x$ grows smaller, slope $\approx 2x$ at the point $(x, x^2),$ when you learn about derivatives.

As you calculated: $$ m_{av} = 4.000056,$$ and we can compute $$\overline m = 2 x_0 + \Delta x = 2\cdot 2 + (2.00005 - 2) = 4 + .00005 = 4.00005$$

So can you see how $\overline m \approx m_{av}$?

Added: Yes, your "experiment" using this function, and calculating slopes using $m = 2x$ at points various points for which the change in $x$ is very small and dividing by the number of slopes you calculated, indicates what average slope over a small interval seeks to estimate.


问:
The four straight lines given by the equation $12x^2+7xy-12y^2 =0$ and $12x^2+7xy-12y^2-x+7y-1=0$ lie along the side of the?
I know these equations are called general equation of second degree and also represent a pair of straight lines. I could extract lines from the equation $$12x^2+7xy-12y^2 =0 $$ (these are $$ 3x+4y=0$$ and $$4x-3y=0$$)  but I don't know how to separate lines from $$12x^2+7xy-12y^2-x+7y-1=0.$$ I think we have to find the solution by finding out the difference between all four straight lines but I have trouble separating them. Thank you!</p>

答:
For large $x$ and $y$, the equation approaches
$$12x^2+7xy-12y^2-x+7y-1\approx 12x^2+7xy-12y^2 = (3x+4y)(4x-3y)$$
Then, write

$$12x^2+7xy-12y^2-x+7y-1= (3x+4y+a)(4x-3y+b)$$

Solve to obtain $a=-b=-1$.


问:
What is chromatic number of the square of $C_n$?
What is chromatic number of the square of $C_n$, for $n \ge 3$ ?

By "the square of $C_n$" I mean graph obtained from $C_n$ (cycle on $n$ vertices) by adding edges between all pairs of vertices with distance 2 on the cycle.

What I figured out:

chromatic number is at least 3, because there are triangles

if $3|n$ then chromatic number is 3: we color consequitive vertices on the cycle: $0, 1, 2, 0, 1, 2, \ldots $ and so on

if $4|n$ then chromatic number is at most 4: we color consequitive vertices on the cycle: $0, 1, 2, 3, 0, 1, 2, 3, \ldots $ and so on

Now I'm stuck.
Can anyone help?

答:
Split the problem into a couple of cases. If $3 \mid n$, then as you rightly pointed out, the chromatic number is $3$ since the vertices can be colored according to the sequence $0,1,2,0,1,2,\ldots$ ($3$ colors).

Next, if $n=3k+1$ for positive integer $k$, if we want to color with $3$ colors, we are forced to use the three colors alternatively as in the first case. However, we get caught in the last vertex, proving that the chromatic number cannot be $3$. However, coloring the last vertex with a fourth color proves that the chromatic number is $4$.

Finally, if $n=3k+2$ for positive integer $k$, similar to above, the chromatic number cannot be $3$. It is clear that when $n=5$, the chromatic number must be $5$ since the graph is isomorphic to $K_5$. For $n>5$, we can color the first $8$ vertices according to the pattern $0,1,2,3,0,1,2,3$ and then color the rest of the $3(n-2)$ vertices in the pattern $0,1,2,0,1,2,\ldots$ which proves that the chromatic number is $4$. Thus, for your graph $G$, where $n \geqslant 3$, the chromatic number is as follows:

$$\chi(G)=
\begin{cases}
3 && \text{if $3 \mid n$} \\
5 && \text{if $n=5$} \\
4 && \text{otherwise}
\end{cases}$$


问:
What would be the immediate implications of a formula for prime numbers?
What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed Eureka?

答:
As others have mentioned there are many formulas for primes.

I can't pass up the opportunity to mention my favorite:

$$p_n=1+\sum^{2^n}_{m=1}\left\lfloor \sqrt[n]n \left( \sum^{m}_{x=1}\left\lfloor \cos^2\left( \pi \frac{(x-1)!+1}{x}\right) \right\rfloor \right)^{-1/n} \right\rfloor$$

Maybe it's just my favorite because it's so complicated and unwieldy!

I first found the formula in Hacker's Delight by Warren. The formula is cited as "Willan's Formula". The formula can be derived from the fact that $(x-1)!\equiv -1 (\mod x)$ iff x is 1 or prime. So if $((x-1)!+1)/x$ is an integer, $x$ is prime (or 1), so the cosine of this times $\pi$ will be -1 or 1 iff $x$ is prime or 1. $\cos^2$ gets rid of the negative, floor keeps only the values for which the $\cos^2$ is 1.

This formula appears to be described on the mathworld page for prime formulas.


问:
Topology: crosscaps and handles
I am struggling with the following exercise:

Let $A$ be a compact connected surface. $\quad$ We say that a crosscap $\otimes$ is inserted into $A$ if an open disc $D$ is removed from $A$ and opposite points of the resulting circular boundary $\partial D$ are identified.$ \quad $We say that a handle is attached to $A$ if two open discs are removed from $A$ and their boundaries are identified. $\quad$ Now, show that:

a) inserting a crosscap into $A$ is equivalent to gluing a Moebius strip into the hole resulting from removing a disc,

b) attaching a handle to $A$ is equivalent to forming the connected sum $A \# T^2$.

Unfortunately, I have no idea how to show this. Furthermore, I don't even know what is meant by "gluing a Moebius strip into the hole". Gluing it in in what way?

答:
This can all be done without recourse to the classification of surfaces or the Seifert-van Kampen theorem. Instead, I would say that a good understanding of this problem is an important step in (some treatments of) the proof of the classification of surfaces.

For (a), let's use $p : A - D \to F$ to be the quotient map. Also, let $D'$ be a larger open disc whose interior contains $D$ concentrically. Consider the boundary circles $C' = \partial D'$ and $C = \partial D$, and let $R$ be the compact annulus in $A$ with $\partial R = C' \cup C$.

Question: What is the image of $R$ under the quotient map $p$?

Answer: That image $p(R)$ is homeomorphic to the quotient space of $R$ where opposite points of $C$ are identified, and one can show by direct construction  that it is homeomorphic to a Möbius band which I'll denote $M = p(R)$. Thus, removing $D$ and gluing opposite boundary points of $C$ results in the same surface as removing $D'$ and gluing in the Möbius band $M = p(A)$ by identifying the circle $p(C') \subset p(A)$ with the boundary of $M$ by a homeomorphism.

For (b) the formulation is not quite correct: one needs to be careful with orientations. Let me assume that $A$ is orientable, in which case one needs to choose an orientation on $A$ which induces an orientation on $A-(D_1 \cup D_2)$ and thereby induces boundary orientations on each circle $C_i = \partial D_i$, and then one needs to require that the gluing homeomorphism $C_1 \mapsto C_2$ reverses boundary orientations. The danger is that if you instead preserve boundary orientations then the result is $A \# K$, where $K$ is the Klein bottle. (The case that $A$ is nonorientable is more complicated, leading to the conclusion that $A\#T$ and $A\#K$ are homeomorphic).

So, with that proviso about orientations, there is a similar answer but one needs a preliminary construction. Let $D_1,D_2$ be the two open discs that are removed, and denote their boundary circles $C_i = \partial D_i$. What you need next is that there exists an open disc $D \subset A$ which "engulfs" both of $D_1,D_2$, in the sense that in an appropriate coordinate system on $D$, the discs $D_1,D_2$ are disjoint round subdiscs of $D$. One way to see this, using path connectivity, is to construct an embedded path $\gamma$ in $A$ whose intersection with $D_i$ is a single point of $C_i$; one can then take $D$ to be a regular neighborhood of $D_1 \cup \gamma \cup D_2$, which one can show by direct construction is homeomorphic to a disc with subdiscs as said.

Once you have $D$, now let $P$ be the compact "pants" subsurface of $D$ with $\partial P = \partial D \cup \partial D_1 \cup \partial D_2$. One can show, again by direct construction, that when the quotient map $p : A - (D_1 \cup D_2) \to F$ is restricted to $P$, the image $p(P)$ is a torus.

Title: Topology: crosscaps and handles


问:
What an implication means when stating the solutions of an equation?
I am having trouble understanding what the following statement means: $$x^2=4 \implies x=2 \hspace{1em} \text{or} \hspace{1em} x=-2.$$

How can we relate this statement to a material conditional?

I am trying to see it as a material conditional. So on the left hand side we have the equation $A(x)$ which takes different truth values for different $x$. The same goes for the right hand side which we can denote it as $B(x)$. Now does the original statement means that $$\forall x \in \mathbb{R}: A(x) \implies B(x).$$

If I substitute the values of $x$ in these two expressions then the conditionals are all true. So how can such a statement make sense when we are trying to solve an equation or when stating the solutions of the equation? Does it mean that in the case that the left hand is true then the value "we substituted" was $2$ or $-2$ in order to make the right hand side also true?

答:
I am having trouble understanding what the following statement means:
$$x^2=4 \implies x=2 \hspace{1em} \text{or} \hspace{1em} x=-2$$
How can we relate this statement to a material conditional?

For any logical propositions $A$ and $B$, $A\implies B$ means precisely that it is not the case that both $A$ is true and $B$ is false.

Here is the truth table.

In your example, you have:

$A\equiv x^2=4$

$B\equiv x=2 \lor x=-2$

Applying the above definition, it is not the case that both $x^2=4$ is true and $x=2 \lor x=-2$ is false.


问:
Maximum value of $|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2$
I have a question the goes, if $|\overrightarrow{a}|=|\overrightarrow{b}| = 1$ and $|\overrightarrow{c}|=2$, then what is the maximum value of :

$|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2$ ?

Here's what I tried:

$|\overrightarrow{a}-2\overrightarrow{b}|^2+|\overrightarrow{b}-2\overrightarrow{c}|^2+|\overrightarrow{c}-2\overrightarrow{a}|^2 = 30-4\cos\theta_1-8\cos\theta_2-8\cos\theta_3$

where $\theta_1=$ angle between $\overrightarrow{a}$ and $\overrightarrow{b}$, $\theta_2=$ angle between $\overrightarrow{b}$ and $\overrightarrow{c}$, $\theta_3=$ angle between $\overrightarrow{c}$ and $\overrightarrow{a}$

and for maximum value, I put $\cos\theta_1=\cos\theta_2=\cos\theta_3=-1$, and got the maximum value of $50$ but the answer says $42$. Where did I go wrong and what would be the correct approach? Was I wrong in putting $\cos\theta_1=\cos\theta_2=\cos\theta_3=-1$? Why?

答:
By putting all $\cos \theta_i=-1$ you imply that all the vectors make an angle $\pi$ with each other !

start off with $${|\vec{a}+\vec{b}+\vec{c}|}^2\ge 0$$ to get $$-\sum \vec{a}\cdot \vec{b}\le 3$$ therefore $$\sum {|\vec{a}-\vec{2b}|}^2=30-4\sum \vec{a}\cdot \vec{b}\le 42$$


答:
By Taylor, as the function is even the development to second order is $f(0)+\dfrac{f''(0)}2x^2$ or

$$(3-2\cos(0)+4\cos^2(0))+(2\cos(0)+8\sin^2(0)-8\cos^2(0))\frac{x^2}2.$$

Title: Small angle approximation on cosine


问:
A proposition about closed graph of a bounded linear operator
Let $E$ be a normed space and $T: D\rightarrow E_1$ be a bounded linear operator, where $D\subset E$ is the domain of $T$ and $E_1$ is a normed space. If the graph of $T$ is closed, then $D$ is closed.

I can only prove that $D$ is closed when $E_1$ is a Banach space. But I am not sure whether the original proposition is true or not?

答: The identity mapping from $D$ to $E_1=D$ has a closed graph but $D$ might not be closed.


问:
Newton's method : Show inequality
I have written the formula of Newton's method to appoximate $a^{1/n}$, which is $$x_{k+1}=\left (1-\frac{1}{n}\right )x_k+\frac{a}{nx_k^{n-1}}$$

Now I want to show the inequality $x_{k+1}-a\leq \left (1-\frac{1}{n}\right )(x_k-a)$.

Since we subtract at the left side $a$ instead of $a^{1/n}$ we cannot consider it as the error, right?

I achieved to show the inequality but for $a\geq 1$:
$$x_{k+1}=\left (1-\frac{1}{n}\right )x_k+\frac{a}{nx_k^{n-1}} \leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n\left (a^{1/n}\right )^{n-1}} = \left (1-\frac{1}{n}\right )x_k+\frac{a}{na^{1-\frac{1}{n}}} = \left (1-\frac{1}{n}\right )x_k+\frac{a^{\frac{1}{n}}}{n} $$
Then subtracting at both sides "$a$" we get $$x_{k+1}-a\leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n}-a=\left (1-\frac{1}{n}\right )x_k-\left (1-\frac{1}{n}\right )a=\left (1-\frac{1}{n}\right )(x_k-a)$$
We have that $x_k\geq a^{1/n}$.

Could you give me a hint for the case $a<1$ ? Or is there an other way to show the inequality without taking cases for $a$ ?

答:
Modified to give a better estimate for the error and avoid the need for $n$ to be an integer.

Let $x_k$ be the $k$th iteration of the Newton Raphson procedure applied $x^n = a$.  Let us define $\xi = a^{1/n}$ and assume $x_k \geqslant \xi$.  We also introduce the symbol $\delta_k = x_k - \xi \geqslant 0$ and $\rho_k = \delta_k/\xi$, being the error and relative error respectively.

Then, calculating $\rho_{k+1}$, we have
\begin{align}
\rho_{k+1} &= \frac{1}{\xi}\Big( x_{k+1}-\xi \Big) \\
&= \frac{n-1}{n}x_k + \frac{a}{nx_k^{n-1}} - \xi \\
&= \rho_k - \frac{1}{n\xi}\Big ( x_k - \frac{\xi^n}{x_k^{n-1}} \Big) \\
&= \rho_k - \frac{1+\rho_k}{n} \Big(1 - (1+\rho_k)^{-n} \Big) \tag{1}\label{E1} \end{align}
Now concentrate on the second term.  In general, we write if $f(\theta) = \big(1+\theta\big)\big(1-(1+\theta)^{-n}\big)$ where $n \geqslant 1$ and $\theta \geqslant 0$ we can claim that
\begin{aligned}
\theta \leqslant f(\theta) \leqslant n\theta.
\end{aligned}
The left inequality is straightforward.  For the right observe $f'(\theta) = 1 +(n-1)(1+\theta)^{-n}$, $f'(0) = n$ and $f'(\theta)$ is decreasing.  Using the fact $f(\theta) = 0$ and integrating we can then see that $f(\theta) \leqslant n\theta$.

This now gives us what we need.  From the right hand side of the inequality we see that $\rho_{k+1} \geqslant 0$ so that the error remains positive.  The left hand side of the inequality shows,
$\rho_{k+1} \leqslant \Big(1-\frac{1}{n}\Big)\rho_k$, which is the required relation (expressed in relative terms).  So the error converges monotonically to $0$, and $x_k \searrow \xi$.  This is as far as the question went.

We can use \eqref{E1} to strengthen the error analysis.  If $\rho_k < 3/(n+1)$ then we can apply the general binomial theorem to say
\begin{aligned}
\rho_{k+1} &= \rho_k - \frac{1}{n}\Big( 1 +\rho_k - 1 + (n-1)\rho_k - \frac{(n-1)n}{2}\rho_k^2 + \frac{(n-1)(n)(n+1)}{3!}\rho_k^3 - \cdots \Big)\\
&=\frac{n-1}{2}\rho_k^2-\frac{(n-1)(n+1)}{3!}\rho_k^3+\cdots
\end{aligned}
The limit on the size of $\rho_k$ implies we have a convergent alternating series with reducing terms, so that the sum is less than the first term,
$$
\rho_{k+1} \leqslant \frac{n-1}{2} \rho_k^2
$$
It means we nearly double the number of significant figures on each iteration.  This also leaves a way forward to measure the error even more exactly if you should need.

Title: Newton's method : Show inequality


问:
Example of a space whose complex K-theory is not easily computable from singular cohomology
I am looking for a counterexample to the formula
$$ K^n(X) \cong \prod_{i\equiv n \mod 2} H^i(X) $$
where $K^*$ denotes complex topological $K$-theory, $H^*$ singular cohomology and $X$ a compact space.
I would hope for an example where $X$ is a finite $CW$-complex.

There should be such an example, as I was told that the $K$-theoretic Atiyah-Hirzebruch sprectral sequence does not collapse at $E^2$ in general.

答: Good example is $X=\mathbb RP^n$ (see e.g. $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$).


问:
Is there a known theory regarding decomposing trivariate functions into composition of two bivariate functions?
I would like to take a given trivariate function $f(x,y,z)$ and decompose it into two bivariate functions $g$ and $h$ like so:

$$f(x,y,z) = g(x, h(y,z))$$

Is there a known theory regarding when this is possible, given certain conditions, for example, all values are real, and $f$ is a polynomial (or rational, or analytic etc.) function, and g and h are (polynomial, rational, analytic etc.) I am especially interested in the case where $f$ is symmetric in $y$ and $z$. That is, $f(x,y_0,z_0) = f(x, z_0, y_0)$

I can randomly perform such decompositions by trial and error, but I am seeking a methodical understanding.

答:
Assuming that the domain of all the variables is the same (e.g. all of them are reals), in theory this is always possible if this domain is infinite. This is because for any infinite set $X$, there always exists a bijection from $X \to X \times X$. Let $h$ be such a bijection, and define
$$g(x, k) = f(x, h^{-1}(k))$$
We have in essence taken $y, z \in X$ and "condensed" them into one value $k \in X$, without losing any information.

With a similar counting argument, you can reason that it cannot always be possible if the domain is finite. If $f$ can be decomposed as $g(x, h(y, z))$, with $g: X^2 \to X$ and $h: X^2 \to X$, then we will know that $f$ can only take at maximum $|X|^2$ distinct values.

A couple remarks:

Unfortunately, this bijection won't always be nice (e.g. there does not exist a continuous bijection from $\mathbb{R}^2 \to \mathbb{R}$). Thus, your hopes of making $g$ and $h$ polynomials will require $f$ to have further qualifications. No sufficient criterion for $f$ come to mind at the moment for me, though.

You can use this same argument to reason about the case where the domains/ranges are not all the same set, e.g. $f: \mathbb{R} \times \mathbb{Q} \times \mathbb{Q} \to \mathbb{R}$.


问:
How to calculate modulo of two polynomials
I am having issues with calculating things like this:
$3x^2 + 2 \mod x^2 + 3$
How would I got about this? I somehow couldn't find information anywhere.

答:
Note that $a\equiv b\bmod c$ if $a-b$ is a multiple of $c$. Generally, the result of $a\bmod b$ is the unique value $x$ such that the degree of $x$ is less than the degree of $b$, and $a\equiv x\bmod b$.

To remove the quadratic term from $a$, we can subtract $3$ times the modulus. So, the answer is $$3x^2+2-3(x^2+3)=-7$$


问:
Given a positive definite bilinear form, show that there exists $\alpha > 0$ so that $a(x,x) \ge \alpha||x||_{2}^2$
Let $a : \mathbb{R}^n * \mathbb{R}^n \rightarrow \mathbb{R}$ be a bilinear form: $$a(x,y) = \sum_{i,j=1}^n a_{ij}x_{i}y_{j} \ \ \ \ \ \ \ a_{ij} \in \mathbb{R},\ \ x = \begin{pmatrix}x_1\\.\\.\\.\\x_n\\\end{pmatrix} ,\ \ y = \begin{pmatrix}y_1\\.\\.\\.\\y_n\\\end{pmatrix}$$

Show that if "a" is a positive definite bilinear form $( \ a(x,x) > 0 \ \ \ \forall x \in \mathbb{R}^{n } \backslash\{0\} \ ) $ then there exists $\alpha > 0$ so that $a(x,x) \ge \alpha||x||_{2}^2$

答:
Let $$\alpha = \min_{\|x\|_2=1}a(x,x).$$
Such $\alpha$ exists because $a$ is continuous and unit sphere is compact. From positive definitness of $a$ we have $\alpha\geq 0$. Case $\alpha=0$ is impossible. Otherwise, there would exist $x_0$, $\|x_0\|_2=1$, such that $a(x_0,x_0)=\alpha=0$, contradicting positive definitness of $a$. Therfore, $\alpha>0$.

Now, for every $x\neq0$ we have $a\left(\frac{x}{\|x\|_2},\frac{x}{\|x\|_2}\right)\geq\alpha$. Multiplying this inequality by $\|x\|_2^2$ and using bilinearity of $a$ we obtain
$$a(x,x)\geq \alpha\|x\|_2^2.$$
Inequality is trivially true for $x=0$.


问:
Finding the flow lines of a given vector field on $S^2$
Good morning!

I have a quick question which I couldn't find the answer to.

Say we have a vector field $\eta : S^2 \to TS^2$ which takes a point $(x,y,z) \in S^2$ to the tangent vector $(-y, x, 0)$.

I am trying to figure out the flow lines of $\eta$, ie, fixing $(x,y,z)$, I'm trying to find a curve $\gamma$ through $(x,y,z$) such that $\gamma'(0) = (-y, x, 0)$.

I was just thinking to set: $\gamma : t \mapsto (x - yt, y + xt, z)$, which does satisfy the property above... but I want it to be a curve in $S^2$, and I do not know how to enforce that, because unless $x \ne y$ or $t = 0$, $||\gamma(t)||^2 \ne 1$.

答:
Let's consider the point $p_0 = (x_0,y_0,z_0)$ at which the tangent vector is $\eta(p_0)=(-y_0,x_0,0)$.

Noticing that the $z$-coordinate of $\eta(p_0)$ is zero, one sees that $\eta(p_0)$ is parallel to the $xy$-plane, and in fact $\eta(p_0)$ is tangent to the horizontal plane $P_{z_0} = \{(x,y,z_0) \mid x,y \in \mathbb R\}$. The same is true for each $p \in S^2 \cap P_{z_0}$: the tangent vector $\eta(p)$ is tangent to the plane $P_{z_0}$; it follows that $\eta(p)$ is tangent to the "latitude circle" $S^2 \cap P_{z_0}$. But this proves that the latitude circle $S^2 \cap P_{z_0}$ is the flow line. And this argument is independent of $z_0$. Thus, the foliation of $S^2$ by flow lines is identical to the folation by latitude circles (with the north and south poles as stationary points of the flow).


问:
$\lim\limits_{n \to\infty}\dfrac{n^{2}-n+2}{3n^{2}+2n-4}=\dfrac{1}{3}.$ Change $n$'s by $x$ and find $M$.
$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$.

With epsilon definition I get my answer as $N=\left[ \dfrac{5}{9\varepsilon }\right] +1$. But then I thought that how can I evaluate this sequence, in functions $\delta$-$\varepsilon$ definition by substituting $n$ by $x$. But I can't make it.

So here's my question:

What can I do is I have function like this one?
$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$.

I'm thinking about that function. But how to find.

$\forall \varepsilon>0$ be given, then $\exists M>0$ such that $x>M$ implies $| f\left( x\right) -l|<\varepsilon$

For function:
$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$

And this is where I stuck. Because $x\in \mathbb{R}$, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!

答:
We have to prove that

$\forall\;\varepsilon>0\;\;\exists\;M>0\;$ such that $\;x>M\;$ implies $\;\left|f\left(x\right)-l\right|<\varepsilon\;.$

For any $\;\varepsilon>0\;,\;$ let $\;M=\max\left\{2,\dfrac{5}{9\varepsilon}\right\}>0\;.$

Moreover,

$x>M\ge2\implies|-5x+10|=5x-10<5x\;,$

$x>M\ge2\implies|9x^2+6x-12|=9x^2+6x-12>9x^2\;.$

Hence,

$x>M\implies\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5x}{9x^2}=\dfrac{5}{9x}\;.$

Consequently,

$x>M\implies|f(x)-l|=\left|\dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5}{9x}<\dfrac{5}{9M}\le\dfrac{5}{9\frac{5}{9\varepsilon}}=\varepsilon\;.$


问:
Limit in Polar Coordinates?
EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

Consider the following limit:
$$
\lim_{(x,y) \to (0,0)} \frac{x^2}{x + y^2}.
$$

It is easy to show, without using polar coordinates, that such a limit is $0$. Indeed,
$$
0 \leq\Bigg|\frac{x^2}{x + y^2}\Bigg| = \Bigg|\frac{x}{x + y^2}\Bigg||x| \leq |x|,
$$
and the argument follows by the squeeze theorem. Now assume that we do not know this fact and we start using polar coordinates instead. Write
$$
x = r\cos(\theta), y = r \sin(\theta).
$$
Then,
$$
\frac{x^2}{x + y^2} = \frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}.
$$
It is obvious that, for a fixed $\theta$, the above expression goes to zero as $r \to 0$, but of course, this is not enough to show what we want. What we have to see is that there exists a one-variable function $h = h(r)$, depending only on $r$, such that
$$
0 \leq \Bigg|\frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}\Bigg| \leq h(r)
$$
for every $\theta$ and $r$, and with $\lim_{r \to 0}h(r) = 0$.

The question is: how do we find the function $h = h(r)$ in our case?

EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

答:
The limit is NOT zero. It does not exist!

If $x=y^2$, as $y\to 0$, then $x\to 0$ and
$$\frac{x^2}{x + y^2}
=\frac{y^4}{y^2 + y^2}=y^2\to 0.$$
On the other hand, along the curve $x=-y^2+y^4$, as $y\to 0$, then $x\to 0$ and
$$\frac{x^2}{x + y^2}
=\frac{(-y^2+y^4)^2}{-y^2+y^4 + y^2}=\frac{y^4-2y^6+y^8}{y^4}=
1-2y^2+y^4\to 1.$$
Indeed, in your first approach, $\left|\frac{x}{x + y^2}\right|$ is NOT bounded in a neighbourhood of the origin.

Title: Limit in Polar Coordinates?


问:
Closed-form formula for $E_{x}[\max(u^\top x,0)\max(v^\top x ,0)]$ where $u,v$ are fixed vectors in $\mathbb R^d$ and $x$ is uniform on the sphere
Let $x$ be uniformly distributed on the unit-sphere in $\mathbb R^d$ and let $u,v$ be fixed nonparallel vectors in $\mathbb R^d$

Question. Is there a closed-form formula for $f(u,v) := \mathbb E_{x}[\max(u^\top x,0)\max(v^\top x ,0)]$ ?

Due to rotationational symmetry of the distribution of $x$, my guess is that $f(u,v)$ admits a simple expression in terms of the angle between $u$ and $v$.

答:
Disclaimer. This is just to simplify @chamd's answer and make it more "recognisable", and draw some links to standard machine-learning theory terminology (kernel methods).

So, @chamd's has proved that $f(u,v) = \|u\|\|v\|\phi(\frac{u^\top v}{\|u\|\|v\|})$,

where $\phi:[-1,1] \to \mathbb R$ is defined by $\phi(t) = \dfrac{\pi t + 2 \sqrt{1-t^2} + 2t\arctan(\frac{t}{\sqrt{1-t^2}})}{4(d+2)\pi}$.

We're in the $uv$-plane and $t \in [-1,1]$ is nothing but the height a point  $P=(\sqrt{1-t^2},t)$ on the unit circle in this plane. The angle pointint the positive $v$ axis ray $OP$ is given by $\theta = \arctan(t/\sqrt{1-t^2})$. By inspecting the geometry of the situation (basically drawing a diagram on a piece of paper...), it is evident that we can rewrite $\theta = \arccos(t)$. Thus, we can rewrite the function $\phi$ as

$$
\begin{split}
\phi(t) &= \dfrac{\pi t + 2 \sqrt{1-t^2} + 2t\arccos(t)}{4\pi} = \frac{t(\pi + 2\arccos(t)) + 2\sqrt{1-t^2}}{4\pi}\\
&= \frac{t(\pi - \arccos(t)) + \sqrt{1-t^2}}{2\pi}\\
&= \frac{t\arccos(-t) + \sqrt{1-t^2}}{2\pi}.
\end{split}
$$

Now is the time for the harvest.

The second line in the above display is nothing but the main ingredient in so-called arc-cosine kernels: these is the kernel for the Gaussian process which emerges when one looks at infinite-width limit of neural networks with ReLU activations, width the weights initialized at random. More precisely, $f \sim (1/(d+2))K_{ReLU}$.

Noting that, for large $d$, (1) The individual coordinates of a point sampled uniformy at random the unit sphere in $\mathbb R^d$ is roughly distributed as $\mathcal N(0,1/d)$, and (2) The pairwise correlations of the coordinates is approximately zero, we could have expected $f(u,v)$ to be approximately equal to $\mathbb E_{z \sim \mathcal N(0,1/d)}[\max(u^\top z, 0)\max(v^\top z, 0)] = (1/d)K_{ReLU}(u,v)$.


问:
$ 2^{\sqrt n} \leq |\operatorname{ tr}( A^n)| \leq 2020\cdot2^{\sqrt n} $
I am trying to prove the following claim

There does not exist an integer $r\geq1 $ and a symmetric matrix $ A \in M_r(\Bbb R) $ Such that for all $ n \in \Bbb N $ we have

$$ 2^{\sqrt n} \leq |\operatorname{tr}( A^n)| \leq 2020\cdot2^{\sqrt n} $$
Please give me some idea how to start this problem?

答:
I'm going to prove it by contradiction using the fact that real symmetric matrices have real eigenvalues and are diagonalizable. Let $\lambda_1,\lambda_2,\cdots,\lambda_r$ be the real eigenvalues of the matrix $A$. Then $|tr(A^n)|=|\lambda_1^n+\lambda_2^n+\cdots+\lambda_r^n|$.

Dividing both sides of your equation by $2^\sqrt{n}$. We get,
\begin{align*}
1\leq \biggl|{\sum_{i=1}^{r} \frac{\lambda_i^n}{2^\sqrt{n}}}\biggr| \leq2020
\end{align*}
Call the middle expression $S_n$

$\mathbf{Case 1}$: When $|\lambda_i| \leq 1 \: \forall i=1,2,\cdots, r$. Then taking limit $ n \rightarrow \infty $ in the above inequality we get $1\leq \lim_{n\rightarrow \infty}S_n = 0$. A contradiction

$\mathbf{Case 2}$: When at least one $|\lambda_i| > 1$. Wlog assume that to be $\lambda_1$. Then taking limit $n\rightarrow \infty$ in the above inequality we get $\lim_{n\rightarrow \infty}S_{2n} = +\infty \leq 2020$. Again a contradiction

$\mathbf{Edit}$: As pointed out in the comments the cases should be split accordingly whether $\lambda_i \geq 1$ and $\lambda_i < 1$.