问:
Evaluate $\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx$ with real methods
I started like this:
$$\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx\:
=\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+\overset {x=\frac{1}{x}}{\int _1^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx}$$
$$=2\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+5G$$
where $G$ is Catalan's constant. But the integral left is very hard to calculate. As suggested by Zacky one can use the sub $x=\frac{1-t}{1+t}$ and get these integrals.
$$\ln 2\int _0^1\frac{1}{1+t^2}\:dt+\ln 5\int _0^1\frac{1}{1+t^2}\:dt+\int _0^1\frac{\ln \left(t^2+1-\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt$$
$$+\int _0^1\frac{\ln \left(t^2+1+\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt-5\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$
To evaluate those one can use the identity
$$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x^2}\:dx=\frac{\pi }{2}\ln \left(\sqrt{a}+\sqrt{b}\right)+\text{Ti}_2\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)-G$$

In the end, the integral evaluates to
\begin{align}
\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx
&= -\frac{3\pi }{8}\ln 2+\frac{\pi }{4}\ln 5-2G\\
&+\frac{\pi }{2}\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)\\
&+\frac{\pi }{2}\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)
\end{align}

But, I have no idea how to simplify the $\text{Ti}_2\left(z\right)$ terms, which seems possible because I found that a similar version can be expressed without these
$$\int _0^{\infty }\frac{\ln \left(1+x^5\right)}{\left(1+x^2\right)^2}\:dx=-\frac{5\pi }{8}-\frac{7\pi }{40}\ln \left(2\right)+\frac{\pi }{5}\ln \left(4+\sqrt{10-2\sqrt{5}}\right)
+\frac{\pi }{10}\ln \left(43+7\sqrt{5}+4\sqrt{130+38\sqrt{5}}\right)+\frac{G}{10}$$

NB:

$\displaystyle x\in\mathbb{R},\text{Ti}_2(x)=\int_0^x \frac{\arctan t}{t}dt$

答:
Continue to reduce the result in the post as follows

\begin{align}
& \text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)\\
=&\ \text{Ti}_2(\tan\frac\pi{20}) - \text{Ti}_2(\tan\frac{3\pi}{20})
=\int_{\tan\frac{3\pi}{20} }^{\tan\frac{\pi}{20} }\frac{\tan^{-1}x}xdx\\ \overset{ibp}=&
\ \frac\pi{20}\ln \tan\frac\pi{20}- \frac{3\pi}{20}\ln \tan\frac{3\pi}{20}+\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt\\
=& \ \frac\pi{20}\ln \frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}- \frac{3\pi}{20}\ln \frac{1-\sqrt{1-\frac{2}{\sqrt{5}}}}{1+\sqrt{1-\frac{2}{\sqrt{5}}}}-\frac25G\\
=& -\frac25G -\frac\pi5 \ln\frac2{\sqrt5 }
-\frac\pi5\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)
+\frac{3\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)
\end{align}

where $\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt=-\frac25G$. Then, plug into the result to arrive at the elementary  expression for the integral

\begin{align}
\int_0^\infty\frac{\ln(1+x^5)}{1+x^2}
&= \frac15G -\frac{19\pi }{20}\ln 2+\frac{3\pi }{5}\ln 5\\
&\hspace{5mm}+\frac{4\pi}5 \ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\frac{8\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)\\
\end{align}


问:
How can I prove this question?
We know that it is not proved that $e^e$ is transcendental, so neither is the number that $e^{e\sqrt{2}}$. My question is, if one turns out to be, how can it be proved that the other is? Because there may be some connection between the two...

答:
Partial answer:

Let $a = \mathrm{e}^\mathrm{e}$.

If $a$ turns out to be algebraic (i.e., not transcendental), then $\mathrm{e}^{(\mathrm{e} \sqrt{2})} = a^\sqrt{2}$ is an algeraic number that is not zero or one, raised to an irrational algebraic power.  Then, by the Gelfond-Schneider theorem, $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is transcendental.

By a similar argument, if $a^\sqrt{2}$ turns out to be algebraic, then the Gelfond-Schneider theorem applies to $(a^\sqrt{2})^\sqrt{2} = a^2$, making $a^2$ transcendental.  Then $a$ is transcendental.

So at most one of $\mathrm{e}^{\mathrm{e}}$ and $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is algebraic.  Equivalently, at least one of $\mathrm{e}^{\mathrm{e}}$ and $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is transcendental.

The situation with (non-integer) powers of transcendental numbers is less well understood.  I would be surprised if there were any known consequences for $a^\sqrt{2}$ after assuming $a$ transcendental or any known consequences for $a$ after assuming $a^\sqrt{2}$ transcendental.

ContentLicense: CC BY-SA 4.0

Title: How can I prove this question?


问:
If the limit comparison test fails
Just want to make sure I'm applying the limit comparison test correctly. And if so and it does indeed fail, what should be my next step in determining if the series diverges?

$$\sum_{n=1}^\infty \frac{5n^3 + 7n^2 + 9}{n^6+2n^3+1}$$
Taking the dominant terms this reduces to $\frac{1}{n^3}$ which converges.
$$\frac{5n^3 + 7n^2 + 9}{n^6+2n^3+1} * \frac{n^3}{1}$$
$$= \frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}$$
$$= \frac{5 + 7n^5 + 9}{2+1}$$
$$= \frac{7n^5 + 14}{3}$$
$\frac{7n^5 + 14}{3}$ isn't finite for $n=\infty$, so the test fails. Where should I go from here?

答:
After this
$$ \frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}$$
you made errors.

The limit above is finite and it's values is $5$

Indeed
$$\lim_{n\to\infty}\frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}=\lim_{n\to\infty}\frac{n^6\left(5+\frac{7}{n}+\frac{9}{n^3}\right)}{n^6\left(1+\frac{2}{n^3}+\frac{1}{n^6}\right)}=\lim_{n\to\infty}\frac{5+\frac{7}{n}+\frac{9}{n^3}}{1+\frac{2}{n^3}+\frac{1}{n^6}}=5$$

Title: If the limit comparison test fails


问:
Why does the definition of topological space require that the empty set be open?
I'm taking my first course in general topology, and so obviously I came across the definition of topological space $(T, \tau)$:

$\tau \in \mathcal P \mathcal P (T)$

$\emptyset, T \in \tau$

$(A_i)_{i \in I} \in \tau \implies \bigcup_{i \in I} A_i \in \tau$

$A_1, \dots, A_n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau$

This is very abstract, so I tried to find a few examples, and these examples I found caught my eye:

$T = \mathbb R^n$, $\tau = \{ \text{every set whose complement is compact} \}$

$T = [0,1]$ $\tau = \{ \text{every set whose complement is a finite set}  \}$

Now these satisfy every part of the definition, except that one has to add the empty set to $\tau$ explicitly as well. This gives the natural question: why is this needed?

答:
The fact that $\emptyset \in \tau$ for any topology $\tau$ does not actually need to be demanded explicitly, it is a consequence of demanding that a topology is closed under arbitrary union. "Arbitrary union" includes taking the union of no open sets whatsoever, and that will be the empty set.

Similarly, if we are careful about the intersection property and demand that the open sets are closed under finite intersection, then we see that $X \in \tau$ follows, because $X$ is the result of taking the intersection of zero subsets of $X$.


问:
Finding unknown values in Matrix with eigenvalues
"Consider the companion matrix  $$A =  \begin{pmatrix}
    0 & 1 & 0\\
    0 & 0 & 1\\
    a & b & c
    \end{pmatrix}$$Knowing that  −1 ,  3  and  −2  are the eigenvalues of $A$ , what is $a$, $b$ and $c$?

I have tried to find $c$ with the trace of the Matrix: $0+0+c = tr(A) = -1 + 3 -2 \Rightarrow c = 0$ (sum of eigenvalues), and find $a$ using the determinant of A and by multiplying eigenvalues $det(A) = -1 \times a = (-1) \times (3) \times (-2) = 6 \Rightarrow a = -6$. But like it is wrong according to solutions.

How can I find the values for $a, b, c$?

答:
There exists a $\mathbb v$ such that $\begin{bmatrix} 0&1&0\\0&0&1\\a&b&c\end{bmatrix}\mathbb v = -\mathbb v$

$\begin{bmatrix} v_2\\v_3\\av_1+bv_2+cv_3\end{bmatrix} = \begin{bmatrix} -v_1\\-v_2\\-v_3\end{bmatrix}$

$\mathbb v = \begin{bmatrix} 1\\-1\\1\end{bmatrix}$ and $a-b+c = -1$

This together with what you know about $a = 6, c = 0$ would imply that if the matrix is diagonalizable $b$ must equal $7$


问:
Finding the Expected Value and Variance
The distribution function of a random variable $X$ is given by

$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}

Determine the mean and variance of $X$.

Workings:

I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.

\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}

答:
Although this can be solved by direct integration, it would be easier if we notice that $X$ follows a $\text{Beta}(k,1)$ distribution:

$$f_X(x)=kx^{k-1}\propto x^{k-1}$$

We can directly see that $$\mathbb E(X)=\frac{k}{k+1},\:\text{Var}(X)=\frac{k}{(k+1)^2(k+2)}$$

By the way it isn't uncommon for the expected value to be undefined (see Cauchy distribution), but in this case it is well defined.

Title: Finding the Expected Value and Variance


问:
Prove that an interval is NOT in the sigma algebra generated by a collection.
I have been thinking about this for a while and could not find an answer. This is a homework question so any hints are appreciated (HINTS ONLY).

Let $\mathcal{M}$ be the sigma-algebra in $\mathbb{R}$ generated by the collection
$$\mathcal{F} = \{ [n, n+1), n \in \mathbb{Z} \}.$$
Prove that the semi-open interval $[0, \frac{1}{2}) \notin \mathcal{M}$.

My approach was to find a sigma algebra $\mathcal{C}$ containing the collection $\mathcal{F}$ such that $[0, 1/2) \notin \mathcal{C}$.

答:
Your approach is correct. The obvious choice of $\mathcal{C}$ would of course be $\mathcal{M}$.

Hint: $\mathcal{M}$ must contain all unions of the form $\bigcup_{n=1}^\infty A_n$, where $A_1,A_2,\dots \in \mathcal{F}$. Is $\{\bigcup_{n=1}^\infty A_n \: | \: A_1,A_2,\dots \in \mathcal{F}\}$ a $\sigma$-algebra?


问:
Proving matrix norm to be submultiplicative?
Consider the matrix norm $\|\mathbf{A}\|_\chi := n\max|a_{ij}|$.

How would I prove it satsifies the property that $\|\mathbf{AB}\| \leq\|\mathbf{A}\|\cdot\|\mathbf{B}\|$.

答:
There are some $1\leq i,j\leq n$ such that:

$||AB||=n|\sum\limits_{k=1}^n a_{ik}b_{kj}|\leq n\sum\limits_{k=1}^n |a_{ik}|\cdot |b_{kj}|\leq n\sum\limits_{k=1}^n \frac{||A||}{n}\cdot\frac{||B||}{n}=n^2 \frac{||A||}{n}\cdot\frac{||B||}{n}=||A||\cdot ||B||$


问:
How to solve these trigonometric equations, using angle addition equations?
$$4\sin(x)+7\cos(x)=6$$

where $0 \le x \le 360^{\circ}$

I put the equation into the form $a\sin(x)+b\cos(x)=R\sin(x+a)$, but after determining that $R\cos(a)=4, R\sin(a)=7$ and $R\sin(x+a)=6$, I don't know how to proceed. Any help?

答:
Starting from $R=\sqrt{65}$, $a=\arcsin 7/\sqrt{65}$, we have

$$\sqrt{65}\sin(x+a)=6$$
$$\Rightarrow x=\arcsin \frac{6}{\sqrt{65}}-a=\arcsin \frac{6}{\sqrt{65}}-\arcsin \frac{7}{\sqrt{65}}$$

Using $$\arcsin u - \arcsin v=\arcsin (u\sqrt{1-v^2}-v\sqrt{1-u^2})$$
$$x=\arcsin\left(\frac{6}{\sqrt{65}}\cdot \frac{4}{\sqrt{65}}-\frac{7}{\sqrt{65}}\cdot \frac{\sqrt{65-6^2}}{\sqrt{65}}\right)$$
$$x=\arcsin \left(\frac{24-7\sqrt{29}}{65}\right)$$


问:
Prove that the area of a right triangle is no more than the square of the hypotenuse divided by 4
I'm struggling with this proof. Let $a,b,c$ denote the base, height, and hypotenuse of a right triangle. Let $A$ denote area of this triangle

We have the following relations:
$$
a^2 + b^2 = c^2  \\
A = a*b/2
$$

and also the triangle inequality, though I'm not sure if this inequality comes into play.

Here is some of my work and I'm unsure if it's in the right direction
$$
a^2 + b^2 = c^2 \\
a^2*b + b^3 = c^2*b \\
\implies A*a + b^3 = b * c^2 \\
\implies A = b*c^2/a - b^3/a
$$

I feel that this is not in the right direction as I don't know how to show that this last expression is $\leq c^2/4$. Can anyone advise?

答:
Well, the area the triangle is $\frac 12ab$ and the hypotenuse is $\sqrt{a^2 + b^2}$ and we need to show $\frac 12 ab \le \frac {a^2 + b^2}4$.

This is equivalent to proving $ab \le \frac {a^2 + b^2}4$.

Are you familiar with the AM-GM inequality?

If you have two positive (or non-negative... but if either is  $0$ it  is trivial) $m,n$ then the Arithmetic Mean,  $\frac {m+n}2$, is greater or equal to the Geometric Mean, $\sqrt{mn}$.

This is exactly that.  Let $m = a^2 > 0$ and $n = b^2 > 0$ then $\frac {m+n}2 \ge \sqrt{mn}$ so $\frac {a^2 + b^2}2 \ge ab$ and $\frac {a^2 + b^2}4 \ge \frac 12 ab$.

That's that.

....

Unless you aren't familiar with AM-GM theorem.

Pf:  Note that $(a-b)^2 \ge 0$ with equality holding only if $a = b$.

So $a^2 -2ab +b^2 \ge 0$ and so $a^2 + b^2 \ge 2ab$ and $\frac {a^2 + b^2}2 \ge ab$.

Replace $m =\sqrt a$ and $n= \sqrt b$ (assuming $a,b > 0$) to get the AM-GM inequality.

$\frac {m+n}2$ is the Arithemetic Mean (or average).  https://en.wikipedia.org/wiki/Arithmetic_mean

And for $m,n \ge 0$ we have $\sqrt{mn}$ is the Geometric Mean https://en.wikipedia.org/wiki/Geometric_mean

The way to think of the geometric mean to imagine you have a rectangle with sides $a,b$ and area $ab$ and you want to find the "average" of the sides so that the area stays the same but the sides are "typical" and about the same.  That is the "average" of $a,b$ is $s$ where $s\times s = ab$ or $s = \sqrt{ab}$.

That's actually the same philosophy as the arithemetic mean.  If we have $a + b = T$ some sum and we want to "average" the terms so that the sum $T$ stays the same but the terms are "typical" and about the same, we want the "average" of $a,b$ to be $m$ where $m + m = a+b$ or $m = \frac {a+b}2$.

The only difference is that for geometric mean we want a product to stay the same while the factors "average out".  ANd for the arithmetic mean we want a sum to stay the same while the addends "average out".

......

Another way to grok why the AM-GM theorem should feel right-- If I look and $\frac {a+b}2 \ge \sqrt ab$, I don't see anything that makes me think "well, duh, obvious". In fact I think, well why the heck should that be true.

But.... if we consider that for any $a,b$ we can consider that for if we assume $a < b$ then there is a midpoint $m$ between them and $a = m-d$ and $b= m+d$ where $d$ is have the distance between $a$ and $d$.

Then $ab = (m-d)(m+d) = m^2 - d^2$ and of course $m^2 - d^2 < m^2$.

So $\sqrt {ab} < m = \frac {a+b}2$.

Well, duh, obvious.  .... (no... not really).


问:
Tilings of a 5×5 square with smaller squares.
There is a $5 × 5$ array of lights, such that at each step, we may toggle all the lights in any
$2 × 2, 3 × 3, 4 × 4$ or $5 × 5$ sub-square. Initially all the lights are switched off. After a certain
number of toggles, exactly one light is switched on. Find all the possible positions of the
light

Here's what I have done. We can safely ignore the $4 × 4$ squares as they are just a bunch of $2 × 2$ squares put together. Also the sum of the number of $3$ sided squares and $5$ sided squares must be odd as they are the only squares that can change the parity of the total number of turned on lights. As each $3$ sided as well as each $5$ sided square must contain the middle square, it must be turned on after placing these squares. How do I proceed from here?

答:
You have made some good observations! As you have observed, we need to place an odd number of $3\times 3$ or $5\times 5$ squares, each of which toggles the central cell. We can also forget about the $4\times 4$ square, since it arises from $2\times 2$ square.

Now, observe that placing a $2\times 2$ does not change the parity of the cells in any row or column. So if we are to have a single cell at the end, we must have only one row and only one column with an odd number of cells after placing the $3\times 3$ and $5\times 5$ squares.

But all of these squares flip the parity of the central row and the central column, so since we've placed an odd number of them, those rows and columns must be odd, and hence have to contain our final square (if it exists).

So we've shown that the only possibility is the center cell. Can this actually be done? Yes it can! Flip a $5\times 5$, a $3\times 3$ nestled in two opposite corners, and a $2\times2$ at each of the other two corners. Here is a picture of everything but the $5\times 5$, showing how the central cell is covered twice:


问:
Let $f(x)=1$ if $x=0$ and $f(x)=0$ if $x>0$. Show that $f$ is integrable.
Let $f(x)=1$ if $x=0$ and $f(x)=0$ if $x>0$. Show that $f$ is Riemann integrable on $[0,1]$. I think for everyone this question is really basic, but I'm just training myself on proving the integrability of functions. Certainly, there are easiests ways to prove the statement, but I would like to prove it considering upper and lower Darboux sums. So, I would just like to know if my approach is correct, please x)

By definition, $f$ is integrable iff. $\exists$ a subdivision $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon>0$.

First, I remark that for all subdivisions, the lower Darboux sum is equal to $0$. So, we have to show that $\overline{S}_{\sigma}(f)<\epsilon \ \forall \epsilon>0$.

Then, consider $0<\epsilon<2$, the intervals $[0,\epsilon/2]$ and $(\epsilon/2,1]$ and two first terms of partition $x_0=0,x_1=\epsilon/2...$ We don't consider the rest of partition, as the upper Darboux sum is trivially $0$ whatever the partition on the interval $(\epsilon/2,1]$.

Thus, $\overline{S}_{\sigma}(f)=\sum_{i=0}^{0}1\cdot (\epsilon/2-0)+0<\epsilon$.

We conclude then that $f$ is Riemann integrable on $[0,1]$.

答: What you did is Ok. You could simplify by considering $\epsilon$ instead of $\epsilon/2$ which is not necessary here.


问:
Solution to matrix equation with exponentials
Assuming
$$
A, M \in \mathcal{R}^{n \times n }
$$
with A, M being invertible. Could someone tell me if it is possible to find a closed form solution for the matrix $M$ in the following equation and if so, provide me with some hints on how to obtain it
$$
A = e^M + e^{M^T}.
$$

I have tried a few things but can´t seem to make it work.

Thanks.

答:
The equation implies that $A/2$ is the symmetric part of $e^M$. In other words, $\frac12A+K$ has a real logarithm $M$ for some appropriate skew-symmetric matrix $K$.

Hence it is not always solvable. For example, when $n$ is odd, every skew-symmetric $n\times n$ matrix $K$ is singular. Therefore the equations $e^M=K$ and  $e^M+e^{M^T}=0$ are not solvable.

However, given that $A$ is real symmetric, the equation $e^M+e^{M^T}=A$ is solvable when $n$ is even or when $A$ has at least one positive eigenvalue. By an orthogonal similarity transform, we may assume that $A$ is a real diagonal matrix. Now:

When $n$ is even, let $K$ be the direct sum of $n/2$ copies of $\pmatrix{0&-1\\ 1&0}$. Then $\frac12A+cK$ has not any zero or negative eigenvalues when $c\in\mathbb R$ is sufficiently large. Hence $\frac12A+cK$ has a real logarithm $M$ and $e^M+e^{M^T}=A$.

When $n$ is odd but $A$ has a positive eigenvalue $a$, let $A=a\oplus A'$. Since the size of $A'$ is even, our previous argument shows that there exists a skew-symmetric matrix $K'$ such that $\frac12A'+K'$ has a real logarithm. Hence $\frac a2\oplus(\frac12A'+K')$ has a real logarithm $M$ and $e^M+e^{M^T}=A$.


答: There are a few problems with your approach: when you write that you have to show that$$\overline{S_\sigma}<\varepsilon\ \forall\varepsilon>0,\tag1$$you don't tell was which partition $\sigma$ is, but then you acto as if it was a concrete partition. For instance, how do you know that $\frac\varepsilon2$ belongs to the partition. Besides, what you need to prove is that, for each $\varepsilon>0$, there is some partition $\sigma$ such that $\overline{S_\sigma}<\varepsilon$, which is not the same thing as $(1)$. And it is easy to do: just take $\sigma=\left\{0,\frac\varepsilon2,1\right\}$, if $\frac\varepsilon2<1$; otherwise, $\sigma=\{0,1\}$ will do.

Title: Let $f(x)=1$ if $x=0$ and $f(x)=0$ if $x>0$. Show that $f$ is integrable.

Body:
Let $f(x)=1$ if $x=0$ and $f(x)=0$ if $x>0$. Show that $f$ is Riemann integrable on $[0,1]$. I think for everyone this question is really basic, but I'm just training myself on proving the integrability of functions. Certainly, there are easiests ways to prove the statement, but I would like to prove it considering upper and lower Darboux sums. So, I would just like to know if my approach is correct, please x)

By definition, $f$ is integrable iff. $\exists$ a subdivision $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon>0$.

First, I remark that for all subdivisions, the lower Darboux sum is equal to $0$. So, we have to show that $\overline{S}_{\sigma}(f)<\epsilon \ \forall \epsilon>0$.

Then, consider $0<\epsilon<2$, the intervals $[0,\epsilon/2]$ and $(\epsilon/2,1]$ and two first terms of partition $x_0=0,x_1=\epsilon/2...$ We don't consider the rest of partition, as the upper Darboux sum is trivially $0$ whatever the partition on the interval $(\epsilon/2,1]$.

Thus, $\overline{S}_{\sigma}(f)=\sum_{i=0}^{0}1\cdot (\epsilon/2-0)+0<\epsilon$.

We conclude then that $f$ is Riemann integrable on $[0,1]$.


问:
Pushing forward vectorfields to compute flows.
So, I talked to my professor today, asking him for a way to compute flows of vectorfields. An example was given by $ v = x\partial_x + y\partial_y$ living in $\mathbb{R}^2$.
Now I know that the way to go would be to solve the system of ODE's given by $x'(t) = x(t), y'(t)=y(t)$ given varying initial values.

However, drawing the vectorfield one sees how it acts on rays that turn radially, so it looks like an easy one-dimensional problem when only looking at the rays. And since it is all radial we went on to look at the problem in polar coordinates. Form here on I get a little lost:

Let $\Phi(r,\varphi) := (r\cos\varphi, r\sin\varphi)$. We next want to get a description of $v$ in these coordinates. As I see it we cannot compute $\Phi_*(v)$ simply because at first $v$ is given in the wrong coordinates. I thought what we want is somewhat $\Phi^{-1}_*(v)$ and then replacing $x=r\cos\varphi$ a.s.o.

What we did however was something along these lines:
$$\begin{align}x\partial_x + y\partial_y &= r\cos(\varphi)\partial(r\cos\varphi) + r\sin\varphi \partial(r\cos\varphi)\\
&= r\cos(\varphi)\left(\frac{\partial\Phi_1(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_1(r,\varphi)}{\partial \varphi}\partial_\varphi\right) + r\sin\varphi \left(\frac{\partial\Phi_2(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_2(r,\varphi)}{\partial \varphi}\partial_\varphi\right)\\
&= r\partial_r
\end{align}$$

And then we have found the "pushforward" of $v$ somewhat magically. Now I do not get why this works? Why is this exactly what we want? These seem like magical calculations that randomly give the right result to me. Maybe I remember his steps wrong, but in the end we never computed $\Phi^{-1}$ or the inverse of the jacobian and yet came to the right result.

(The next steps would be to get the flow in the coordinates for $\Phi$ and write it back in the coordinates of $(x,y)$ I guess).

I would be happy for clarifications on why this should work, where it comes from and also maybe a more formal proof on how to compute the flow of the given vector field when first changing coordinates and then getting back.

答:
It looks like your professor is exploiting the standard metric and the musical isomorphisms to avoid calculating the inverse matrix, because pulling back differential forms $(x,y)$ to $(r,\theta)$ works the same direction as we want.  We just need to be able to freely change vector fields to forms, and that is provided by the metric (= inner product, or if want indefinite signatures, = nondegenerate symmetric bilinear form) inducing a natural isomorphism between tangent and cotangent vectors $v\leftrightarrow v^\flat:=\langle v,-\rangle$: if the metric in $x^i$ coordinates is $\sum_i h^i(\mathrm{d}x^i)^2$ then $(\partial_{x_i})^\flat = h^i\,\mathrm{d}x^i$ and the other way is $(\mathrm{d}x^i)^\sharp=\frac1{h^i}\partial_{x_i}$.

Now what you have given looks like the usual calculation for $x\,\mathrm{d}x+y\,\mathrm{d}y=r\,\mathrm{d}r$.  Since the standard metric $\mathrm{d}x^2+\mathrm{d}y^2=\mathrm{d}r^2+r^2\,\mathrm{d}\theta^2$, both $(x,y)$ and $(r,\theta)$ are orthogonal coordinates making the calculation easy.  So "sharp"ing it gives $x\partial_x+y\partial_y=r\partial_r$ indeed.

If we try a similar approach on $x\partial_y-y\partial_x$, we have
$$x\,\mathrm{d}y-y\,\mathrm{d}x=r^2\,\mathrm{d}\theta$$
so when you sharp it, you need to remember the coefficient $r^2$ in front of $\mathrm{d}\theta^2$ in $\mathrm{d}r^2+r^2\mathrm{d}\theta^2$ means $(r^2\,\mathrm{d}\theta)^\sharp=\partial_\theta$ so the full calculation goes
$$
x\partial_y-y\partial_x=(x\,\mathrm{d}y-y\,\mathrm{d}x)^\sharp=(r^2\mathrm{d}\theta)^\sharp=\partial_\theta.
$$


问:
How to prove $\text{if } {B \subseteq A} \implies A^*B^* = A^*$
For context, A and B is a regular language, and related to Theory of Computation.

As the title says, it is so intuitive that
$$
\text{if } {B \subseteq A} \implies A^*B^* = A^*
$$

I don't know how to prove this formally. Is this an axiom or I'm missing some sort of key observation that can formally prove the statement?

The best I can do is:

We can see that $\{\varepsilon\} \in B^* \implies A^* \subseteq A^*B^*$.

We also know that $B \subseteq A \implies B^* \subseteq A^* \implies   A^*B^* \subseteq A^*A^* = A^* $

This proofs $A^* \subseteq A^*B^* \wedge A^*B^* \subseteq A^* \implies A^*B^* = A^*$

But, I don't think that's good enough. Because I feel like it's missing something or it has something wrong. Is there any better way to prove this instead of the way I'm doing it right now?

答:
The essence of the proof is that
$$A^*=A^*\{\epsilon \}\subseteq A^*B^* \subseteq A^*A^*=A^*.$$
Is this formal enough or do you want to break it down further into definitions?


问:
Solution of the Integral equation $y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt$
I want to find the solution for the integral equation: $$y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt $$
I tried finding the resolvent kernel of the equation and it is getting more complicated at each new iteration. Is there any other "clean" method to find the solution?
Any help is highly appreciated.

答:
$$y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt \tag 1$$
In the integral the function is multiplied by a polynomial of second degree of $x$. So three successive differentiations are necessary in order to get rid of the integral.
$$y'(x)=2-8x+3y(x) +\int_{0}^x \left[ 6 - 8(x-t) \right]y(t)dt \tag 2$$
$$y''(x)=-8+3y'(x)+6y(x) -8\int_{0}^x (x-t)y(t)dt \tag 3$$
$$y'''(x)=3y''(x)+6y'(x)-8y(x)\tag 4$$
The integral equation is transformed into a third order ODE :
$$y'''-3y''-6y'+8y=0\tag 5$$
which general solution is :
$$y(x)=c_1e^{-2x}+c_2e^x+c_3e^{4x} \tag 6$$
One have to determine the coefficients in putting (6) into (1) :
$$c_1e^{-2x}+c_2e^x+c_3e^{4x} = 1 - 2x -4x^2 +\int_0^x \left[ 3 + 6(x-t)-4(x-t)^2 \right] (c_1e^{-2x}+c_2e^x+c_3e^{4x}) dt  $$
After simplification
$$(-2c_1+4c_2+c_3-4)x^2+(5c_1+2c_2-c_3-2)x+(-c_1-c_2-c3+1)=0$$
$$\begin{cases}
-2c_1+4c_2+c_3-4=0\\
5c_1+2c_2-c_3-2=0\\
-c_1-c_2-c3+1=0
\end{cases} \quad\implies\quad
\begin{cases}
c_1=0 \\
c_2=1 \\
c_3=0
\end{cases}$$
The solution is :
$$y(x)=e^x$$


问:
Find all functions $f:\mathbb N\to\mathbb N$ such that $f\left(f(m)+f(n)\right)=m+n$, for all $m,n \in\mathbb N$.

Find all functions $f:\mathbb N\to\mathbb N$ such that $f\left(f(m)+f(n)\right)=m+n$, for all $m,n \in\mathbb N$.

I did not get any idea. I was just plugging out values for $m=n=1$ and so on
no idea.

答:
Here, we assume $\mathbb N$ doesn't contain $0$, although the analysis should be pretty similar in the case where $0\in\mathbb N$.

Firstly, if $f(x)=f(y)$, then
$$2x=f(2f(x))=f(2f(y))=2y,$$
so $x=y$; i.e., $f$ is injective. Now, consider any $a,b,c\in\mathbb N$ with $a+b>c$. Then
$$f(f(a)+f(b))=a+b=c+(a+b-c)=f(f(c)+f(a+b-c)),$$
which implies that
$$f(a)+f(b)=f(c)+f(a+b-c).$$
In particular, let $f(1)=t$ and $f(2)=t+u$. We claim that $f(n)=t+u(n-1)$ for all $n$, which can be proven by strong induction. It holds for $n\in\{1,2\}$, and for $n\geq 3$,
$$f(n)+f(1)=f(2)+f(n-1),$$
which implies
$$f(n)=f(n-1)+f(2)-f(1)=t+u(n-2)+t+u-t=t+u(n-1),$$
as desired. This means that
$$m+n=f(f(m)+f(n))+f(2t+u(m+n-2))=t+u(2t+u(m+n-2)).$$
Since $u\geq 0$, we must have $u=1$, since the left side is $m+n$ and the right side is $u^2(m+n)$ plus some constant independent of $m$ and $n$. Then, we need
$$m+n=t+2t+m+n-2,$$
which gives $t=1$. As a result, the only working function is the identity.


问:
What is the meaning of the words here?
I'm not sure what the meaning of the question is, and I need some help understanding what I'm even trying to do. The unitary operator in question is in $U(4)$

The question is: "Decompose the above unitary operator as a product of unitary operators, each which acts only on a two-dimensional subspace of the Hilbert space"

My confusion comes from "acts only on a two-dimensional subspace of the Hilbert space" - what does this mean, and what does it imply I should be doing in my question? Does it mean I need to use matrices with columns 3 and 4 filled with zeroes?

I'm confused and my lecturer is terrible. Please help.

答:
Fleshing out the comments in a bit of detail.

I assume $U(4)$ is the group of $4 \times 4$ unitary matrices acting on the Hilbert space $H = \mathbb{C}^4$.  I like using $U$ for unitary operators, though, so I won't write $U(4)$ again (When I say "unitaries" without other qualification, I just mean elements of that-named set.)

If $U$ is the unitary in your question, in the "likely scenario" identified in Semiclassical's comment above, the question is basically asking to to find a basis in which you can write $U$ as a block sum of $2 \times 2$ unitary matrices, i.e., a unitary matrix $V$ (representing a change of basis) for which
$$
U = V^* \begin{pmatrix} U_1 & 0 \\ 0 & U_2 \end{pmatrix} V,
$$
where $U_1$ and $U_2$ are $2 \times 2$ unitary matrices (and $0$ denotes the $2 \times 2$ zero matrix).

Put another way, the question is to find a coordinates in which $U$ acts separately as unitary transformations of the first two coordinates and the second two coordinates.

To connect this to the phrasing of the problem, if you have a factorization of this form, then $U$ is a product of two unitaries, $W_1 = V^* \begin{pmatrix} U_1 & 0 \\ 0 & I \end{pmatrix} V$, and $W_2 = V^* \begin{pmatrix} I & 0 \\ 0 & U_2 \end{pmatrix} V$.  In the coordinates of your new basis (which probably won't be the standard basis), $W_1$ leaves the second two coordinates alone, while $W_2$ leaves the first two coordinates alone.

Or in slightly more spatial terms: let $S$ be the span of the first two elements of your basis.  Then the span of the second two elements of the basis is $S^{\perp}$ (see comments above).  Since $W_1$ fixes $S^{\perp}$, it may make sense to say that $W_1$ "acts only on $S$," and similarly it may make sense to say that $W_2$ "acts only on" $S^{\perp}$.

It may mean that your instructor meant something less restrictive, where the two-dimensional subspaces for your unitaries may not need to be complementary to one another.  But I agree with Semiclassical that this is the most likely interpretation.

Title: What is the meaning of the words here?


问:
How do you generate the numbers from an empty set?
In a short 2007 Scientific American article, the former Harvard mathematician and author Dr Robert M Kaplan stated that:

in mathematics we can generate all numbers from the empty set

I've not been able to find the proof of that statement. I assume he means the natural numbers. I can see how you can get to the empty set by removing each number from a set of numbers, therefore, reverse that process. (Though I wonder if that applies to an infinite set?) But what is the start point? Is the start related to counting the span of the empty set? I.e. by counting the span of the empty set to get the first number then a proof can be shown that involves cardinality.

What is the straightforward proof that demonstrates that all numbers can be generated from the empty set?

答:
There is a standard set-theoretic method to use the empty set and its succession of iterated power sets to make a model of the natural numbers.

Once you have the naturals, you can model integers as pairs of natural numbers: $(a,b)$ represents the integer $a-b$.

Once you have the integers, you can model the rationals as pairs of integers: $(c,d)$ represents the rational number $c/d$.

Once you have the rationals, you can model the reals using Dedekind cuts.

Once you have the reals, you can model the complexes using a field extension by a root of $x^2 +1 = 0$ (so again, using pairs, $(e,f)$ representing $e + \mathrm{i}f$) or by constructing the algebraic closure of the reals.  (You get the same complex numbers either way, although the proof of equivalence is not a one-liner.)

Other sets of numbers, quaternions, octonions, et c., proceed similarly.  Cardinals and ordinals are addressed in the other answer(s) to this question.


问:
Mathematics for cryptography.
Besides number theory, what other areas should I study for crypto. I did my undergrad in Comp Sci so the crypto course didnt have much mathematical topics. But for a grad specilisation in crypto I need to get acquainted with the required mathematical skills.

Any sort of help is appreciated.

Regards
Rahul

答: See the book An Introduction to Mathematical Cryptography by Hoffstein, Pipher, and Silverman.</p>

ContentLicense: CC BY-SA 3.0

Title: Mathematics for cryptography.


答:
It is the same thing as saying that among all right triangles sharing a common hypotenuse the right isosceles triangle has maximum area:

If $a^2+b^2 $ is a constraint, maximum area $ab$ occurs when $a=b$, can be established by plugging either $a$ or $b$ into area $ab$ and differentiating...

The area of all other right triangles come out less than the gray triangle shown.

Title: Prove that the area of a right triangle is no more than the square of the hypotenuse divided by 4

Body:
I'm struggling with this proof. Let $a,b,c$ denote the base, height, and hypotenuse of a right triangle. Let $A$ denote area of this triangle

We have the following relations:
$$
a^2 + b^2 = c^2  \\
A = a*b/2
$$

and also the triangle inequality, though I'm not sure if this inequality comes into play.

Here is some of my work and I'm unsure if it's in the right direction
$$
a^2 + b^2 = c^2 \\
a^2*b + b^3 = c^2*b \\
\implies A*a + b^3 = b * c^2 \\
\implies A = b*c^2/a - b^3/a
$$

I feel that this is not in the right direction as I don't know how to show that this last expression is $\leq c^2/4$. Can anyone advise?


答:
HINT:

The R is related to

$\sqrt{4^2+7^2} = \sqrt{65},\text {so that} \sin a= \dfrac{7}{\sqrt{65}} ,\cos a= \dfrac{4}{\sqrt{65}}$

which is more convenient. Divide both sides by $\sqrt{65}$

$$ \sin ( x+ a) =\dfrac{6}{\sqrt{65}}$$

where

$$ \tan \alpha=\dfrac{7}{4}$$

Title: How to solve these trigonometric equations, using angle addition equations?

Body:
$$4\sin(x)+7\cos(x)=6$$

where $0 \le x \le 360^{\circ}$

I put the equation into the form $a\sin(x)+b\cos(x)=R\sin(x+a)$, but after determining that $R\cos(a)=4, R\sin(a)=7$ and $R\sin(x+a)=6$, I don't know how to proceed. Any help?


问:
Given algorithm, find and solve the recurrence relation
Consider the following recurrence relation:

$T(2) = 5$

$T(n) = 2T\left( {\left\lceil {{{2n} \over 3}} \right\rceil } \right) + T\left( {\left\lfloor {{n \over 3}} \right\rfloor } \right) + 3$

I got this recurrence relation by analysing the following algorithm:

I got $T(2) = 5$ as the base case because of 2 comparisons in the if statement plus 3 statements in swap() function.

The question requests to solve it using the characteristic equation. However, I don't know how to proceed with the constant $1$ at the end.

My attempt:

$a_n - a_{2n/3} - a_{n/3} = 3
\\
x^n - x^{2n/3} - x^{n/3} = 3
\\
x^{n/3}(x^3 - 2x^2 - 1) = 3$

Also, if I analysed incorrectly the algorithm, please tell me!
All help is appreciated!

edit: I incorrectly deduced the recurrence equation. I didn't count the cost of first and second if statement.

答:
Then you can solve your recurrence by putting
$$
n = 3m + k = 3\left\lfloor {{n \over 3}} \right\rfloor  + n\bmod 3
$$
and split the recurrence into a system three, one for each k.


问:
Factor-square polynomials with real coefficients
A polynomial $f(x)$ is a factor-square polynomial if $f(x)|f(x^2)$. For example, $x-1|x^2-1$, so $x-1$ is a factor-square polynomial.

Here is a list of polynomials degrees 1 and 2:
$$, $−1$; $x^2$, $x^2-1$, $x^2-x$, $x^2+x+1$, $x^2-2x+1$.

After generating some more of these types of polynomials, it appears that either all of the coefficients are real, or all complex and irreducible over reals.

Are there monic factor-square polynomials of some degree with real number coefficients, but some of those coefficients are not integers?

答:
First notice that if $\alpha$ is a root of $f$, then $\alpha^2$ is also a root of $f$ and, since $f$ has a finite number of roots, we should have $\alpha^{2^i}=\alpha^{2^j}$, thus $\alpha=0$ or it is a root of unity. This helps us narrow the search. For an example, let $\omega$ be a primitive root of unity with $\omega^{2^n}=1$.
Then consider
$$f(X)=(X-1)^2\prod_{i=1}^{n-1} (X-\omega^{2^i})(X-\overline{\omega}^{2^i}).$$

This obviously has real coefficients, but it will not have integer coefficients.
Assuming the contrary, note that it has $\omega^2$ as a root which is a primitive root of order $2^{n-1}$. Since the cyclotomic polynomial $\Phi_{2^{n-1}}(X)$ is the minimal polynomial of $\omega^2$ over $\mathbb{Z}[X]$, we should have $\Phi_{2^{n-1}}(X)(X-1)^2|f(x)$, thus looking at degrees $2^{n-2}=\varphi(2^{n-1})\leq 2n-2$, which is false once $n\geq 6$.


答:
$R\sin a = 7\\
R\sin a = 4\\
\tan a = \frac {\sin a}{\cos a} = \frac {7}{4}\\
a = \arctan \frac{7}{4}$

Now that we know $a$ what is $R$?

$\sin a = \frac {7}{R}\\
\cos a = \frac {4}{R}\\
\sin^2 a + \cos^2 a = 1\\
\left(\frac {7}{R}\right)^2 +  \left(\frac {4}{R}\right)^2 = 1\\
\frac {49 + 16}{R^2} = 1\\
R^2 = 65\\
R = \sqrt {65}$

$\sqrt {65} \sin (x + \arctan\frac{7}{4})  = 6\\
x = \arcsin{\frac {6}{\sqrt{65}}} - \arctan \frac {7}{4}$

or

$x = \arcsin{\frac {6}{\sqrt{65}}} - \arcsin \frac {7}{\sqrt {65}}$

And for a general solution we will include $x+2n\pi$ and $x+\frac {4n-1}{2}\pi$


问:
intersection of analytic set : codimension?
Hello. Let $A$ and $B$ be analytic sets of pure dimension in an analytic set X of $\mathbb{C }^{n}$. Suppose $A$ is a local complete intersection (i.e. for every point $a$ of $A$, there exists $\Omega$ an open neighbourhood of $a$ such that  $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorphs distincts and $p = codim_{X}(A)$).
Then, if $A \cap B \neq \emptyset$ why do we have $codim_{X}(A \cap B) \leq codim_{X}(A) + codim_{X}(B)$?

I tried to solve this by induction on $dim(A)$; here is the proof I first made (I realised afterwards the step $*$ is false). It's trivial if $dim(A) = 1$. Then, if $dim(A) > 1$, I choose $a \in A \cap B$ and I wright $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorph distincts and $p = codim_{X}(A)$ and $B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p'}(x) = 0\}$ where $g_{i}$ are holomorph and distincts. Since the problem is located in $a$, we will wright $dim_{a} = dim$ and the same for $codim$. Notice that I have $p' \leq codim_{X}(B) $. We can wright $A \cap B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\}$ where $p'' \leq p'$ is such that there is no repetition between $g_{j}$ and $f_{i}$. Now let's have a look at : $A' := \{x \in \Omega | g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\} \subset A$. We wright $B' := \{x \in \Omega | g_{1}(x) = ... = g_{p''-1}(x) = 0\}$. By looking$*$ the local irreductible components of $A$( which are all of dimensions $dim(A)$ since $A$ is of pure dimension) in $a$ we have $A'$ is of dimension $dim(A) - 1$ so of codimension $codim_{X}(A) + 1$. Thus, $A'$ is a local complete intersection and so by induction $codim_{X}(A \cap B) = codim_{X}(A' \cap B') \leq codim_{X}(A') + codim_{X}(B') = codim_{X}(A) + 1 + codim_{X}(B')$. But $1 + codim_{X}(B')$ doesn't necessary have to be $\leq codim_{X}(B)$ (imagine $B$ is an irreductible component of $B'$ in $a$).

$*$Here I use the (6.2) Theorem page 106 of this curse https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf . But it dosen't work since $g_{p''}$ can be constant on some irreductible components of $A$ in $a$.

Thank you for your help.

I wish you a good day.

答:
I simplified the problem : if $a$ is choose regular in $B$ (i.e. $B$ is a manifold in an neighbourhood of $a$), then the last inequality hold. However, I still don't know what to do in the case where $A \cap B_{reg} = \emptyset$. All your suggestions are welcomed.

Perhaps (4.33) Corollary page 102 can help? I don't know.


问:
How to prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$ diverges?

Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.

Please provide me a hint or a full solution.

答: I suggest that you read about the Cauchy condensation test.</p>


问:
Compute $\int_{\gamma} z\, dz$ for any smooth path $\gamma$ which begins at $z_0$ and ends at $w_0$

Compute the complex line integral$$\int\limits_{\gamma} z\, \mathrm{d}z$$ for any smooth path $\gamma$ which begins at $z_0$ and ends at $w_0$.

Confused as to how I am supposed to go about parametrizing any smooth path $\gamma$. I know how to solve this for a line from $z_0$ to $w_0$ but I am stumped by "any" smooth path.

答: $f(z)=z$ is an entire function in the complex plane, so $\int_\gamma z\,dz$ only depends on the endpoints of $\gamma$ and not the "interior" of it (like a conservative vector field). The result for any path from $z_0$ to $w_0$ is the same as the result for the straight line from $z_0$ to $w_0$: $\frac12(w_0^2-z_0^2)$.


问:
Kiselev's geometry ex 7,8
I need some help on two exercises from Kiselev's geometry, about straight lines.

Ex 7: Use a straightedge to draw a line passing through two points given on a sheet of paper. Figure out how to check that the line is really straight. Hint: Flip the straightedge upside down.

I would draw the first line, then flip the straightedge and draw the second line over the first. The two lines should coincide nicely iff the straightedge is straight. Because, this shows that there is no "unevenness" or "bumps" on the edge of the straightedge. There would be gaps between the two lines if there are "unevenness/bumps" on the edge of the straightedge.

Ex 8: Fold a sheet of paper and, using ex 7, check that the edge is straight. Can you explain why the edge of a folded paper is straight?

Ex 8 is marked as more difficult by the author. I'm completely clueless about this exercise.

Please provide insights and help me with these two exercises. I'd appreciate if they are more of an "experimental approach" than theoretical because exercises 7 and 8 are arranged in between the introduction and first chapter of the book.

Thank you. :)

答:
I’m late to the game but here is my argument for #8. (Incidentally, I don’t think Kiselev is suggesting to use #7 to prove #8. He says use the technique to check that the line is straight, and, separately, asks us to explain why it is true.)

When we fold the paper, we identify each point on the bottom 'half' with a unique point on the top 'half'.  (You may imagine the paper to expand as much as necessary to make this a reality!  Or, you can imagine we are folding a plane.)

Lay the paper on top of a plane and fold the paper.  The edge creates a curve in the plane.  Our goal is to convince ourselves that this curve is straight.

Unfold the paper.  Now the paper has a crease, which of course is the same curve as the edge.  The crease divides the paper into two halves — call them 'left' and 'right'.

Now, imagine any two points on this crease, and further imagine them connected by some curve which does not lie on the crease — say, a curve in either the left or right half of the paper.  What can we say about such a curve?

A curve like this cannot be straight.  The reason is, folding the paper would give us a congruent but different curve on the other side of the page, and we cannot have two different straight curves connecting a pair of points.

So we have shown that if two points on the crease are connected by a curve which travels off the crease, that curve is not straight.  But we know that any two points are connected by a straight line, and hence that straight line lies on the crease.  Applying this argument to the endpoints of the curve, we see that the whole crease/edge itself is straight.

Obviously I've left off some details and assumptions (for example, a line congruent to a straight line is straight, etc) but I think I've managed to find the thrust of a good argument.

It’s a wonderful problem that I’ve thought about for years.

Title: Kiselev's geometry ex 7,8


问:
What's the intuition behind the Co-Area formula?
I mainly work in statistics and I know only basic measure theory. I was trying to understand the Co-Area formula by Federer.

If $f:\mathbb{R}^M\to \mathbb{R}^N$ is a Lipschitz function with $M \geq N$ then
$$
\int_A J_N f(x) d\mathcal{L}^M x = \int_{R^N} \mathcal{H}^{M-N} (A\cap f^{-1}(y)) d\mathcal{L}^N y
$$

Could anyone please give me an intuition of what the formula implies and what it means practically? In particular, this is a simple example I am trying to wrap my head around: suppose we are in 2 dimensions and there is a curve defined by a function

What does the co-area formula tell us in this context?

Here are some practical questions:

My understanding is that we have some space of dimension $M$ and there a manifold in it of dimension $M-N$ and we would like to compute the area of this manifold using the Hausdorff measure. Is this what is happening?

I am completely lost as to why we need $A\cap f^{-1}(y))$ what's the intuition behind this?

On the right hand side we have $\mathcal{H}^{M-N}$. I am familiar with the Lebesgue measure $d\lambda$ however I am lost as to why we seem to have both the Lebesgue and the Hausdorff measure on the right hand side. I thought we should only have the Hausdorff measure on the RHS?

答:
First, it follows from approximation by characteristic functions that the formula in the OP generalizes to
$$
\int_{R^M} u(x) J_N(Df(x)) \, d\mathcal{L}^{M} = \int_{R^N} \Big(\int_{f^{-1}(z)}u\, d\mathcal{H}^{M-N}\Big)d\mathcal{L}^N(z)\, ,
$$
for every measurable $u\colon R^M \to [0,+\infty]$ . It is part of the claim that the different components are measurable and that the integrals are meaningful.

This means that to integrate a function $u$ we may first integrate along fibers coming from the function $f$. The most familiar example of this is the Fubini's theorem: If you take $f(x_1,\cdots,x_M) = (x_1,\cdots,x_N)$.

The next example is integration in spherical coordinates: take $f:R^M \to R^1$ be $f(x)=|x|$.

So, that must be a good start for intuition.

Important delicate issues:

Even for smooth $f$ we cannot guarantee that all fibers $f^{-1}(z)$ will be manifolds (let alone of dimension $M-N$). Example: $f(x,y)=x^2-y^2$, and level set of zero. In Lipschitz case, you can draw any shape and let $f(x)$ be distance of $x$ to that set. Obviously, $f$ is Lipschitz and level set of zero is the shape you started with. The reason coarea formula survives this is that if $f$ is Lipschitz, then for a.e. $z$ the sets $f^{-1}(z)$ will be countably $\mathcal{H}^{M-N}$-rectifiable and integration against Hausdorff measure will be meaningful.

Integrability of $z \to \mathcal{H}^{M-N}(f^{-1}(z)\cap A)$ is not easy! It requires the so-called coarea inequality.

The coarea formula is not obvious even for sets $A$ of measure zero!

There is a proof of the coarea formula that uses the implicit function theorem to "straighten" the fibers so that the proof reduces to Fubini's theorem.

Pictorially, if $U$ is a (measurable) set in $R^M$ (here in my picture $R^2$) then $F^{-1}(z)$ (here sample $z=a_1,a_2,\cdots$ drawn) "foliate" $U$. The coarea formula says that integration over $U$ can be done first by integration along these fibers/level sets.


问:
Lagrange Multiplier Problem - Distance from point With a Circle Constraint
Given a point $P:(x_0,y_0)$ in $\mathbb{R}^2$, and a constraint function
$$x^2+y^2=R^2$$where $R$ is the radius of the circle. The distance from $P$ to any point on the circle is to be minimized using the method of Lagrange Multiplier. The distance $d$ is given by
$$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}.$$
Then the Lagrange function becomes
$$\mathcal{L}(x,y,\lambda)=d(x,y)+\lambda\ (x^2+y^2-R^2)$$
with the optimality conditions $\partial_x\mathcal{L}=0$, $\partial_y\mathcal{L}=0$ and $\partial_\lambda\mathcal{L}=0$. These conditions yield:
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(x-x_0)+2\ \lambda\ x=0$$
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(y-y_0)+2\ \lambda\ y=0$$
$$x^2+y^2-R^2=0$$
How should I continue from this point?</p>

答:
HINT:

As $d(x,y)\ge 0, d(x,y)$ will be minimum if $d^2(x,y)$ is minimum

So, start with $$\mathcal{L}(x,y,\lambda)=d^2(x,y)+\lambda\ (x^2+y^2-R^2)=(x-x_0)^2+(y-y_0)^2+\lambda\ (x^2+y^2-R^2)$$


问:
How to solve these trigonometric equations, using angle addition equations?
$$4\sin(x)+7\cos(x)=6$$

where $0 \le x \le 360^{\circ}$

I put the equation into the form $a\sin(x)+b\cos(x)=R\sin(x+a)$, but after determining that $R\cos(a)=4, R\sin(a)=7$ and $R\sin(x+a)=6$, I don't know how to proceed. Any help?

答:
The  simplest way to solve this equation is to write the l.h.s. in the form
$$R\cos (x-\varphi)=6$$
Now  $R=\sqrt{4^2+7^2}=\sqrt{65}$, and we have to solve (in radians)
$$\begin{cases}
\sin\varphi=\frac 4{\sqrt{65}}\\
\cos\varphi=\frac 7{\sqrt{65}}
\end{cases} \iff \varphi=\arctan \tfrac 47$$
(because $\sin\varphi,\,\cos\varphi >0$).

Once $R$ and $\varphi$ have been determined, the equation becomes
$$\cos(x-\varphi)=\frac 6R\iff x\equiv\varphi\pm \arccos\frac 6R\pmod{2\pi}.$$
The remaind to find the solutions which live in the interval  $[0,2\pi]$, and translate it in degrees if you prefer a solution with this unit.


问:
Do (some) mathematicians think that beauty is evidence of truth in mathematics?
Some physicists have argued that (other things being equal) a beautiful theory is more likely to be true than an ugly one. Dirac famously took this view: "A theory with mathematical beauty is more likely to be correct than an ugly one that fits some experimental data."

It's not uncommon for mathematicians to talk about beauty in mathematics. (I always think of Hardy, "there is no permanent place in the world for ugly mathematics.") But are there mathematicians who think (as some physicists do) that beauty (or elegance, or whatever) is evidence of truth?

I've done some Googling, and I've found many discussions of beauty in maths, but none of the things I've seen so far deal with my question directly.

答:
It's empirically true that reality has so far cleaved rather well to neat models. See The Unreasonable Effectiveness of Mathematics in the Natural Sciences for much more. Since it's always been true so far that "every mystery we've ever solved has turned out to be not magic nicely modellable", we often tend to treat this as evidence that any other given mystery will be nicely modellable.

However, all our models so far have been very much approximations, and we know that they're not Truly Correct because they're not all compatible with each other (c.f. quantum mechanics and general relativity, both excellent but incompatible models). So it's quite possible that reality does not in fact admit any nice compact description.

Note also that there's a huge selection effect going on here: we know how to solve physical problems using mathematics, so of course the problems we have solved are overwhelmingly done using mathematics!

I will note, as a software engineer, that coming up with a solution to a software problem that in some sense has mathematical beauty (e.g. it has no edge cases) does seem to be a pretty good indicator that you've found The Right Answer. In my experience, such a solution will be extensible, low in bugs, require fewer rewrites, and so on, and it often turns out to model well entirely new requirements that I hadn't even thought of. But I would argue that this is more likely because an answer with a neat mathematical model is simpler and has fewer moving parts, all of which are more easily described and more easily proved correct, than because there is something inherently "correct" about mathematics as applied to the real world.


问:
I want to obtain partition of an integer with an initial value and
I want to obtain a partition of an Integer with an initial value and
the value following it is smaller and the value following it is smaller than the previous value and no value repeats itself.
within a limited number of terms
For example
Integer = 41
Intial Value = 11
number of terms = 8

11+9+6+5+4+3+2+1 = 41

11+8+7+5+4+3+2+1 = 41

let's say I am going to generate such partitions for number 42, how to do that?

答:
I am not sure whether a super direct approach would exist, these problems with partitions and restrictions are usually complicated to compute since closed formulas are not easy to find and a lot relies on recurrences and bijections between different kind of partitions.

I guess the easiest way to attack what you say is via generating functions: since you don't only want terms to be different, but also have a condition on the total number of terms used, you will need two variables. One, say $x$, will count the actual contribution of the term, while $z$ will count how many terms this contributes. For example, a term $z^2x^4$ would be saying a contribution of $4$ with two terms, i.e. a $2 + 2$ or a $3 + 1$, although we don't know which contributed this term in the final product.

If your biggest term is $N$, then that is a $zx^N$. After that you allow any term to appear, but maybe it won't, so each possible term has a generating function $1 + zx^j$, for $1\le j \le N - 1$. So, the generating function for the partitions whose largest term is $N$, the remaining ones do not repeat and in which you count how much contributions you have is:
$$zx^N(1 + zx^{N - 1})...(1 + zx).$$
To make an example, if $N = 5$ what you have to see is the product
$$\begin{align*}
&\;zx^5(1 + zx^4)(1 + zx^3)(1 + zx^2)(1 + zx)\\
&= zx^5(1 + zx + zx^2 + (z + z^2)x^3 + (z + z^2)x^4 + 2z^2x^5 + (z^2 + z^3)x^6 + (z^2 + z^3)x^7 + z^3x^8 + z^3x^9 + z^4x^{10})\\
&=zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + (z^3 + z^4)x^{11} + (z^3 + z^4)x^{12} + z^4x^{13} + z^4x^{14} + z^5x^{15}
\end{align*}
$$

For example, this would be telling us that for $10$ there are only $2$ ways to make this, each of them uses $2$ terms (that is the $z^2$), and sure enough these are $5 + 4 + 1$ and $5 + 3 + 2$, while for $9$ this would be saying that there is one way with $2$ terms and one with $3$ terms: $5 + 4$ and $5 + 3 + 1$.

These polynomials get complicated to manipulate as your parameters grow, nevertheless if you want to bound the number of terms, then you want to disregard any terms whose power of $z$ is bigger than some number. For example, if you did not want to count partitions with more than $3$ terms you take quotient modulo $z^4$ and the polynomial becomes:
$$
zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + z^3x^{11} + z^3x^{12}
$$
Finally, if what you want at this point is just to see how many there are of a given integer, without knowing how many terms they have since you already took care of your bound then you put $z = 1$ and you get:
$$
x^5 + x^6 + x^7 + 2x^8 + 2x^9 + 2x^{10} + x^{11} + x^{12}
$$
This, for example, would tell you: there are $2$ partitions of $10$ in which:

The biggest term is $5$,

All terms are different,

The number of terms is at most $3$.

These would be $5 + 4 + 1$ and $5 + 3 + 2$. For eleven, there would be only one: $5 + 4 + 2$.

For $N = 42$ you can do this, tho you would probably need a computer that makes the calculations for you, and you need to do everything in the correct modulus from the start to make sure computations remain doable.


问:
Image of path under fiber transport of a right G-principal cover is G-equivariant
I have seen many different definitions of $G$-principal covers (which I assume are largely equivalent) so in this context, a right $G$-principal cover is a properly discontinuous action of $G$ on a covering $p \colon E \to B$ with $p(xg) = p(x)$ such that $G$ acts transitively on the fibers.

The construction of the fiber transport I am working with is from Tom Dieck, which for a path $w \colon I \to B$ from $b$ to $c$ gives a map $T_p(w) \colon \pi_0(F_b) \to \pi_0(F_c)$, where $F_b$ and $F_c$ are the fibers of $b$ and $c$ respectively. If $x \in F_b$ then there is a lift $v \colon I \to E$ with initial condition $x$, and $T_p(w)$ sends the path component of $x$ to the path component of $v(1) \in F_c$.

In Tom Dieck's algebraic topology book, he claims that it is immediate from the construction that $T_p(w)$ is $G$-equivariant for a right $G$-principal cover, but I'm not really sure why this is true. If $v_x$ is the lift of $w$ at $x$ and $v_{gx}$ the lift at $gx$, it seems to me that this statement is equivalent to the existence of a path from $v_{gx}(1)$ to $v_x(1)g$. However I am not sure how proper discontinuity or transitivity can tell us anything about how compatible an action is with paths.

答: Using your notation, observe that $v_xg$ - that is, the map $t\mapsto v_x(t)g$ - is a lift of $w$ with $v_xg(0) = v_x(0)g=xg$. The unique path lifting property tells you that $v_xg=v_{xg}$. In particular, the endpoints are the same: $v_{xg}(1)=v_x(1)g$.


问:
Solvable Lie algebras. derived algebras and ideals
Let L be a vector subspace codimension $1$ in $g/ g'$, where g is a Lie algebra solvable and $g'=[g,g]$ derived algebra.

Have:

$g/g'$ is abelian

In fact, if $x, y \in g$
$$[x+g',y+g']=[x,y]+g'= g'=0+g'$$
because $[x,y]\in [g,g]=g'$.

Then L is a ideal in $g/g'$

I need help the second case.

答: Since $g/g'$ is abelian every vector subspace $I$ is an ideal. Since for every $x\in g/g', i\in I$ we have $[x,i]=0\in I$.


答:
It's poetic and all. and beauty $\cong$ simplicity and simple when true will probably have fundimental obvious reasons it is true.  But let's not get carried away.

IMO  $(a+b)^k = a^k + b^k$ is for more beautiful that then $(a+b)^k = \sum_{k=0}^n \frac {n!}{(n-k)!k!}a^kb^{n-k}$ and the statement "There is harmony in the universe and all values are ratios of whole numbers" is much more beautiful than "The values that can't be express by any finite combination of rational numbers is uncountable more than those that can be".

So here we have the truth is much uglier than the poetic symmetry.

Of course there is beauty in thinking and the reasoning that $(a+b)^k = \sum_{k=0}^n \frac {n!}{(n-k)!k!}a^kb^{n-k}$ contains some clever and "beautiful" thoughts.

BUt... ya know.... this is subjective and romantic but boils down to no real meaning.

I suppose it could be argued that I am being temporally chauvinistic in thinking romance and poetry are outdated and archaic and ...well, just a bit saccharine and twee..... But... I think is wiser to be objective and taking things on the faith that "the universe must have beauty in its method otherwise life is too grim" is just not objective.

......

Of course:

All mathematicians are romantics.  They find ideas more beautiful than things.

A cynic is just a romantic who lived to be forty years old.

Title: Do (some) mathematicians think that beauty is evidence of truth in mathematics?

Body:
Some physicists have argued that (other things being equal) a beautiful theory is more likely to be true than an ugly one. Dirac famously took this view: "A theory with mathematical beauty is more likely to be correct than an ugly one that fits some experimental data."

It's not uncommon for mathematicians to talk about beauty in mathematics. (I always think of Hardy, "there is no permanent place in the world for ugly mathematics.") But are there mathematicians who think (as some physicists do) that beauty (or elegance, or whatever) is evidence of truth?

I've done some Googling, and I've found many discussions of beauty in maths, but none of the things I've seen so far deal with my question directly.


问:
Minimum distance between parabolas
Is there a simple way to find the minimum distance between two parabolas?

For example, between y=-0.1(x-17)2 + 8.6 {7.726 < x < 19.134}
and y=-0.12(x-17.5)2 + 6.2 {10.313 < x < 18.829}

See graph of the two parabolas here (the exponential function simply determines the restricted domain for the parabolas)

Context: I have to ensure that there is a minimum of 1.8 units between these two parabolas, to ensure the "path" is at least 1.8 metres wide always.

Attempt: I thought about trying to find the closest distance visually then finding the normal and subbing in for x and y at that specific point, but that doesn't seem very scientific and I'm a bit stuck.

Thanks in advance

答:
This is not an answer. I just want to show you the graphical solution to your problem. This diagram was drawn to scale in GeoGebra. For easy viewing please download it to your computer.

The restricted domain is $EGKJ$. The absolute minimum of the non-restricted area is $MF$. The absolute minimum for the restricted are is $JR$. Since this is equal to 2.1569, a gap of 1.8 is guaranteed. The other distances are given for comparison only.

For the non-restricted case, there are two global minimums on either sides of the axis of the inner parabola (e.g. $MF$ and $QH$). One of them is always the absolute minimum. There is a construction to determine these two values. However, I could not prove it yet. Nevertheless, I hope to post an answer at a later date.

If neither of the above mentioned minimums occurs in the restricted area, then, the minimum inside this area is to be found at one of the boundaries as shown in the diagram. But, the minimum is not always the boundary line. In this diagram $JK$ is the boundary line, but the minimum is $JR$.

Can you please share your results with us, if you have already solved the problem?

Title: Minimum distance between parabolas


问:
What would be the diameter of this set?
Let, $t \in [0,1]$ and $x,y \in \mathbb{R}$. Let, $f(t) = tx + (1-t)y$, such that, $f:[0,1] \rightarrow \mathbb{R}$. What would $$\textrm{diam}(f([0,1])),$$
where $\textrm{diam}([a,b]) = \sup\{d(x,y):x,y \in [a,b] \}$ be? I think the answer is $f(1) - f(0)$, but I have no way of showing this.

答:
Suppose without loss of generality that $x<y$ (I'm assuming $x\neq y$, otherwise $f$ is constant and $f([0,1])$ has diameter $0$).

If $s,t\in [0,1]$, then
$$
|f(s)-f(t)|=|(s-t)x+(t-s)y|=|s-t||x-y|\le |x-y|
$$
Thus $|x-y|$ is an upper bound of your set. Since this value is attained, namely at $0$ and $1$, then the diameter is $|x-y|$.

Title: What would be the diameter of this set?


问:
Help to sum these two series $\sin(x)-\sin(2x)+\sin(3x)...$
I am reading Euler's paper entitled "Subsidium Calculi Sinuum" and he wrote down some "sums" for these trigonometric series:

\begin{align}S &= \sin(x)-\sin(2x)+\sin(3x)-\sin(4x)...\\\\
C &= \cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\
-1+C &= -1+\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\
-1+C+iS&=-1+[\cos(x)+\sin(x)]-[\cos(2x)+\sin(2x)+[\cos(3x)+\sin(3x)]-[\cos(4x)+\sin(4x)]...\end{align}

$$-1+e^{ix}-(e^{ix})^2+(e^{ix})^3-(e^{ix})^4+...$$

This is the geometric series, and I recognise that its "sum" is $-\dfrac{1}{1+e^{ix}}$

So, we have:

\begin{align}
& -\frac{1}{1+e^{ix}}=-\frac{1+e^{-ix}}{(1+e^{ix})(1+e^{-ix})} \\[8pt]
= {} & -\frac{1+\cos(x)-i\sin(x)}{(1+e^{ix})(1+e^{-ix})}=-\frac{1+\cos(x)-i\sin(x)}{2\cos(x)+2}
\end{align}

Equating the real and imaginary part with $C+iS$, we have:

$$-1+C=-\frac{1+\cos(x)}{2+\cos(2x)}\implies C=\frac{1}{2}$$

$$S=-\frac{\sin(x)}{2\cos(x)+2}$$

However, for the sum of $S$, there shouldn't be a minus sign, according to Euler.

So what is wrong with my derivation?

答: Your numerator should be $1+e^{-ix}$, not $1+e^{ix}$.


答:
My view (PhD Physics, former Visiting Professor, Mathematics), is that given any system of mathematical postulates, there are some resulting theorems (mathematical "truths") that many or even most mathematicians would describe with terms such as "elegant," "parsimonious," even "beautiful," and other theorems that might be called "convoluted," "perplexing," even "ugly."  It is the "beautiful" ones that are more easily remembered, more easily disseminated, more easily used in computations and thus more likely to become part of the canon taught and studied.  Thus there is an "evolutionary" force that favors these theorems and ideas.  The proverbial person-on-the-street knows $a^2 + b^2 = c^2$, in large part because it has properties associated with "beauty."

That being said, mathematicians' notions of "beauty" (and synonyms) certainly changes ("evolves") over time.  Some of mathematicians' early responses to Mandelbrot's fractals involved terms like "ugly," "horrendous," and such.  But through exploration and familiarity—and especially the representations through computer graphics—are now considered some of the most beautiful math.

Some (misguided) mathematicians even call certain fractal images "art."  I disagree... but that's a separate discussion.


问:
A Flatness Criterion of Dedekind Scheme
I'm reading Qingliu's book Proposition 4.3.9.

Definition: A Dedekind scheme is an irreducible normal locally Noetherian scheme of dimension 0 or 1.

Proposition 4.3.9. Let $Y$ be a Dedekind Scheme, and $f:X\rightarrow Y$ be a morphism with $X$ reduced. Then $f$ is flat if and only if every irreducible component of $X$ dominates $Y$.

In the proof of the 'if' part, assume $y\in Y$ closed point, $f(x) = y$ and $\pi\in\mathcal O_{Y,y}$ be a uniformizing parameter for $\mathcal O_{Y,y}$. The author says the assumption implies $\pi$ does not belong to any minimal prime ideal of $\mathcal O_{X,x}$.

My questions:

Why $\pi$ does not belong to any minimal prime ideal of $\mathcal O_{X,x}$.

Is there any intuitive meaning of uniformizing parameter?

Thank you in advance!

答:
If $y$ is a closed point of $Y$, then $\mathcal{O}_{Y,y}$ is a local ring that is not a field, and $\mathcal{O}_{Y,y}$ is a DVR (Proposition 3.1.12). This means $\mathcal{O}_{Y,y}$ is a local PID, i.e. the maximal ideal $\mathfrak{m}_y$ is principal. Its generator is called the uniformizing parameter for $\mathcal{O}_{Y,y}$. Of course it can't belong to any minimal prime of $\mathcal{O}_{Y,y}$ since $(\pi)$ is maximal (unless $\dim Y = 0$ in which case I'm concerned about the validity of Qing Liu's argument since all non-units of an Artinian ring are zero divisors, so it seems to be his argument is deficient. Perhaps he meant to assume $\dim Y >0$).

Regardless, if you are not comfortable with DVRs, chapter 4 of Qing Liu should probably wait until you have a stronger background in commutative algebra. This is a very challenging book (I have been struggling with it for over a year!) and having a really solid grasp of commutative algebra and basic scheme theory is essential for reading it. To learn about DVRs, I strongly recommend Atiyah-MacDonald Chapter 9 (make sure you understand integral extensions very well first), especially Proposition 9.2.

Title: A Flatness Criterion of Dedekind Scheme


问:
$\sum_{n=1}^{\infty} \frac{n^2 +1}{2n^2 \sqrt{n} +5}$
How can I show this series diverges $\sum_{n=1}^{\infty} \frac{n^2 +1}{2n^2 \sqrt{n} +5}$. I tried doing a root test, and end up with the limit $$\lim_{n\to\infty} \frac{(n^2+1)^{1/n}}{(2n^2\sqrt{n}+5)^{1/n}}$$

which ends up as $\frac{1}{\sqrt{2}}$ after my calculations, which indicates convergence.

答:
As $\lim_{n\to\infty}\frac{n^2+1}{n^2}=1$, there is some $N$ such that $n\ge N$ implies $\frac{n^2+1}{n^2}\ge \frac{1}{2}$. But then
$$
\sum_{n\ge N}\frac{n^2+1}{2n^{3/2}+5}\ge\sum_{n\ge N}\frac{1}{2}\frac{1}{2\sqrt{n}+\frac{5}{n^2}}\ge\sum_{n\ge N}\frac{1}{2}\frac{1}{2\sqrt{n}+2\sqrt{n}}=\frac{1}{8}\sum_{n\ge N}\frac{1}{\sqrt{n}}=\infty
$$


答:
Hint: think about the partial sums
$$ S_k = \sum_{n = 2^k+1}^{2^{k+1}} \frac{u_n}n$$

Notice that $$\sum_{n=1}^\infty \frac{u_n}n = u_1 + \frac{u_2}2 + \sum_{k=1}^\infty S_k$$
so it's enough to show that the sequence $\sum_{k=1}^\infty S_k$ diverges.

Notice that the terms in the partial sums $S_k$ are decreasing.  Therefore the smallest term in the sum is the last one, $\frac{u_{2^{k+1}}}{2^{k+1}}$.

As there are $2^k$ terms in each sum we must have
$$\sum_{n = 2^k+1}^{2^{k+1}} \frac{u_n}n \geq 2^{k} \frac{u_{2^{k+1}}}{2^{k+1}}=\frac{u_{2^{k+1}}}{2} $$

So $$\sum_{n=1}^\infty \frac{u_n}n \geq u_1 + \frac 12\left(u_2 + u_4 + u_8 + \dots\right)$$which we are given as divergent.

Title: How to prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$ diverges?

Body:
Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.

Please provide me a hint or a full solution.


问:
How to transform a line to circle curve.
I am facing a problem of the transformation of a line to a circle curve as shown in the attached figure.

The line segment $A1Z1$ which has many points$(A1(x_A,y_A), B1(x_B,y_B),...P1(x_P,y_P),...)$ is needed to transform to a circle curve $A2Z2 (\theta,R$).

So far I do not know how to get the coordinate of the corresponding points of the circle curve.

Hope for your help and thank you very much.
Dave.

答:
@ndhanson3, @Moti, Thank you very much for your comments. Parametrization is really a key word.
Let me answer myself.

The parametrization of a line segment is written as:
$$\begin{cases}
x = (x_Z-x_A)t + x_A, \quad where \; t\in[0,1]\\
y = (y_Z-y_A)t + y_A
\end{cases}$$
so
$$ t = \frac{x-x_A}{x_Z-x_A} , \quad where \; t\in[0,1]$$
while the parametrization of an arc is written as
$$\begin{cases}
\hat{x} = Rsin(\hat{t}), \quad where \; \hat{t} \in [0,\theta]\\
\hat{y} = Rcos(\hat{t})
\end{cases}$$
It is a transformation from $t\in[0,1]$ to $\hat{t}\in[0,\theta]$, so $\hat{t}$ is computed
$$\begin{aligned}
\hat{t} &= (\theta-0)\bigg(\frac{t-0}{1-0}\bigg) + 0 \\
        &= \theta t \\
        &= \theta \bigg(\frac{x-x_A}{x_Z-x_A}\bigg)
\end{aligned}$$
Subtitution into the arc equation, the transformation of the arc has the parametric equation
$$\begin{cases}
\hat{x} = Rsin\Big(\theta \big(\frac{x-x_A}{x_Z-x_A}\big) \Big), \\
\hat{y} = Rcos\Big(\theta \big(\frac{x-x_A}{x_Z-x_A}\big) \Big)
\end{cases}$$

Best,-Dave.


问:
Differentiable function and polynomials: Proof of $\phi (x)=ce^x + q(x)$ unclear to me.
I came across a proof which I can't quite understand:

If $p:\mathbb{R}\to\mathbb{R}$ is a polynomial of degree $n$ and $\phi:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $\phi'=\phi+p$, there exists a polynomial $q$ of degree $n$ with a number $c\in\mathbb{R}$ with $\phi(x)=ce^x+q(x)$ for all $x\in\mathbb{R}$.

The proof is given as follows:

$$\phi'=\phi+p$$
$$\Leftrightarrow\phi'-\phi = p$$
$$\Leftrightarrow\phi'*e^{-x}-\phi*e^{-x}=p*e^{-x}$$
$$\Leftrightarrow\left(\frac{d}{dx}\right)[\phi*e^{-x}]=p*e^{-x}$$
$$\int\left(\frac{d}{dx}\right)[\phi*e^{-x}]=\int p*e^{-x}$$
$$\Leftrightarrow\phi*e^{-x}=q*e^{-x}+c$$
$$\Rightarrow\phi(x)=ce^x+q(x)$$

Isn't the notation (especially the integrals) missing something?

Why exactly are we multiplying by $e^{-x}$ in line 3? Is this a special technique?

How does the integral with $p$ in it suddenly yield $q$? And why is the arbitrary constant $c$ on the left hand side of the equation missing?

Also, the conclusion is unclear to me. How exactly do the $e^{-x}$ and $c$ disappear and how does the $ce^x$ appear?

Thank you very much in advance!

答:
Do you remember the product rule from calculus ? Using that we have $${d \over dx} \left( \phi*e^{-x}\right) = {d \over dx}\phi*e^{-x}+{d \over dx}{e^{-x}}*\phi$$

Is the motivation of why we multiplied $e^{-x}$ to the equation.

The reason why $p$ sudenly seems like it turned into a $q$, is because you integrated

$$\int p(x)*e^{-x}dx = q(x)*e^{-x}+c$$

Here, when you integrate the polynomial becomes something different.
For example using integration by parts,$$\int x*e^{-x}dx = -(1+x)e^{-x}+c$$

where $p(x) = x$ and $q(x)=-(1+x)$

Finally, the reason why it seems like $e^{-x}$ suddenly dissapeared, is because it hasn't. When we look at the very last step, all it's doing is multiplying $e^x$ on both sides. And $e^{-x}*e^x = 1$.

Does this help ?


问:
Proof that in $\mathbb{R}^+$ the function $\frac{1}{1+x}$ is continuous
I want to proof that the function $\frac{1}{1+x}$ is continuous in the positive real numbers.
Therefore I use the epsilon delta definition of continuity. But when I make my calculations (to find epsilon in function of delta) I get stuck.

Can anybody help me to go further on $\left| \frac{x-a}{(1-x)(1-a)} \right| < \epsilon$.

答: It should read: $\left|\dfrac{x-a}{(1+x)(1+a)}\right| < |x-a| < \epsilon$ if you let $\delta = \epsilon > 0$.


问:
How do I calculate the shortest counter-clockwise angle to the closest incision made in an evenly-cut 2D plane?
I regularly use the equation $\operatorname{atan2}(\sin(k),\cos(k))$ to wrap an arbitrary angle $k$ to the range $[-\pi, +\pi)$. I think of this as expressing it counter-clockwise with respect to the $+x$ axis of the Cartesian plane. What if I instead wished to express the counter-clockwise angle with respect to the nearest of the four axes $+x, +y, -x, -y$ in the Cartesian plane? In this case the resulting angle should be in the range $[-\pi/4, +\pi/4)$. More generally, if I was to rotationally divide the Cartesian plane evenly into N angular slices, with the first cut being along $+x$ axis, is there a way to calculate the counter-clockwise signed distance to the nearest incision in the range $[-\pi/N, +\pi/N)$? I realize that this can probably be done programmatically with iterations and conditionals. I after the elegant trigonometric solution, assuming one exists  :)

答:
The points $e^{2ik\pi}$  for $k= 0, \ldots , n-1$ divide the circle into the $n$ equal segments you want. So find the value of $k$ that minimizes the distance from your angle to $2k\pi$.

Then use Euler's formula to get the Cartesian coordinates and the Euclidean distance.


问:
Can the length of the cross product actually be negative?
I'm trying to figure out why on earth the system of two vectors and their cross product must be positively oriented. Usually, this is just stated outright without further explanation.

I have however noticed that $||a x b||=||x||||y||sin \alpha $, which implies a somewhat strange solution to my problem. Since $ sin \alpha $ can be a negative value (for angles between 180 and 360 degrees) this would imply that $||a x b||$ could in fact be negative for those angles.

If by $||a x b||$ being positive we mean that it goes "up" from the plane and vice versa, this "would" in fact imply a positive orientation of the cross product and vectors a and b.

In that case the cross product vector would travel "upwards" from the plane if the angle was less than 180 degrees, making the entire triple of vectors positively oriented. It would travel "downwards" from the plane if the angle was between 180 and 360 degrees, but then we need only rotate the entire set of vectors 180 degrees and it would be plain that the system would still be positively oriented.

Am I onto something or am I grasping at straws?

答: The correct rule is $\Vert a\times b\Vert^2=\Vert a\Vert^2\Vert b\Vert^2\sin^2\alpha$, or equivalently $\Vert a\times b\Vert=\Vert a\Vert\Vert b\Vert|\sin\alpha|$ (note the $|\cdot|$ signs around $\sin\alpha$), because vectors cannot have negative length.


问:
Lie bracket and flows on manifold
Suppose that $X$ and $Y$ are smooth vector fields with flows $\phi^X$ and $\phi^Y$ starting at some $p \in M$ ($M$ is a smooth manifold). Suppose we flow with $X$ for some time $\sqrt{t}$ and then flow with $Y$ for this same time. Then we flow backwards along $X$ for the same time, and then flow backwards along $Y$. All in all, we can define a curve dependent on $t$ as follows $$\alpha(t):= \phi_{-\sqrt(t)}^Y \circ \phi_{-\sqrt(t)}^X \circ \phi_{\sqrt(t)}^Y \circ \phi_{\sqrt (t)}^X$$

It is an exercise to show that $\frac{d}{dt}|_{t=0} \alpha(t) = [X,Y](p)$. In theory, this should be workable with just the chain rule, (assuming I know how to do these derivatives properly, which is something I'd like some clarification on) but this process is going to be excruciatingly long and painful, and it's really just something I want to avoid if I can help it. Is there another way to do this computation that will be less painful and more illustrative of why exactly this works out?

答:
I've been working on this problem for a while. Even though this question is from quite a long time ago, I haven't seen an answer which is similar to this one, so I'm posting it mainly because this question appears rather commonly (I first saw it as an exercise in Warner's book).

I'm not going to prove that $\alpha'(0)$ exists, but rather that $\alpha$ defines a derivation at $t=0$ that is equal to the Lie bracket. More precisely, for any $f\in \mathfrak{F}(M)$:

$$
\lim_{t\to 0^{+}}\frac{f(\alpha(t))-f(p)}{t}=[X,Y]_{p}(f).
$$

If the derivative did exist, then this equation would imply that $\alpha'(0)=[X,Y]_{p}$, but paraphrasing Warner's book, the $\sqrt{t}$ makes it so that the derivative may not exist.

First, define $\beta(t)=\Phi_{-t}^{Y}\circ \Phi_{-t}^{X} \circ \Phi_{t}^{Y} \circ \Phi_{t}^{X}(p)$, so that $\alpha(t)=\beta(\sqrt{t})$ for every $t>0$ sufficiently small. Since the square root is a homeomorphism from $[0,\infty)$ to itself, it suffices to prove that

$$
\lim_{s\to 0^{+}}\frac{f(\alpha(s^{2}))-f(p)}{s^{2}}=\lim_{s\to 0^{+}}\frac{f(\beta(s))-f(p)}{s^{2}}=[X,Y]_{p}(f).
$$

By using L'Hôpital's rule twice, this limit can be rewritten as

$$
\lim_{s\to 0^{+}}\frac{(f\circ \beta)''(s)}{2}=\frac{1}{2}(f\circ \beta)''(0).
$$

So, in the end we have reduced the problem to computing the following second derivative:

$$
\frac{d^{2}}{dt^{2}}|_{t=0}f(\beta(t))=\frac{d^{2}}{dt^{2}}|_{t=0}f(\Phi_{-t}^{Y}\circ \Phi_{-t}^{X} \circ \Phi_{t}^{Y} \circ \Phi_{t}^{X}(p)).
$$

A common trick for evaluating this kind of derivatives is to "separate the $t$'s". If we define

$$
H(a,b,c,d)=f(\Phi_{a}^{Y}\circ \Phi_{b}^{X} \circ \Phi_{c}^{Y} \circ \Phi_{d}^{X}(p)), g(t)=H(-t,-t,t,t),
$$

then $(f\circ \beta)''(0)=g''(0)$. By using the chain rule twice, this equals

$$
g''(0)=H_{aa}(0)+H_{bb}(0)+H_{cc}(0)+H_{dd}(0)+2H_{ab}(0)-2H_{ac}(0)-2H_{ad}(0)-2H_{bc}(0)-2H_{bd}(0)+2H_{cd}(0).
$$

There are a lot of second derivatives to compute, but they are relatively easy (I can add an example if needed), and by the end you'll get

$$
g''(0)=YYf(p)+XXf(p)+YYf(p)+XXf(p)+2XYf(p)-2YYf(p)-2XYf(p)-2YXf(p)-2XXf(p)+2XYf(p) \\
=2XYf(p)-2YXf(p)=2[X,Y]_{p}(f).
$$

Now, plug this result into the original limit, and we get

$$
\lim_{t\to 0^{+}}\frac{f(\alpha(t))-f(p)}{t}=[X,Y]_{p}(f).
$$

Hope this helps! If anyone believes there's some error in the proof, I'd like to know, since I haven't seen this argument anywhere else.

EDIT: I'm aware that this solution uses explicitly the chain rule (which the author wanted to avoid), but since it is used in the simplest way I could think of (that is, using it with a function from $\mathbb{R}^{4}$ to $\mathbb{R}$), I hope it's forgivable.

Title: Lie bracket and flows on manifold


问:
Inductive proof parity
Let $a$ be an integer. I want to prove by induction that for all $n$ such that $n$ is a positive integer, that $a^n$ and $a$ have the same parity.

my work:

I know the base case is when $n = 1$, and $a^1 = 1 = 1$.

I know the induction step is when $n = k$ for some arbitrary positive integer $k$. And I want to prove that $a^{n+1}$ and $a$ have the same parity.

Where do I go from here?

答:
Assume that $a^n$ and $a$ have the same parity. Then, since\begin{align}a^{n+1}-a&=(a^{n+1}-a^n)+(a^n-a)\\&=a^n(a-1)+(a^n-a),\end{align}since $a^n(a-1)$ is even (one of the factors is even), and since $a^n-a$ is even too, $a^{n+1}-a$ is even, which means that $a^{n+1}$ and $a$ have the same parity.

But, of course, it is far more simple to say that if $a$ is odd, then $a^n$ is odd, and that if $a$ is even, then $a^n$ is even.

Title: Inductive proof parity


问:
Question about proof that $\delta(f(x)) = \sum_i\frac{\delta(x_i - x)}{|df/dx_i|}$
From an exercise and proof in Principles of Quantum Mechanics:

Question: The author instructs us to "expand $f(x)$ near such points in a Taylor series, keeping the first nonzero term.  This is what the proof answer provides when it assumes that $f(x) = f'(x_i)(x-x_i)$.  But why do we only keep the first non-zero term?  What about the other terms?  Do we have that $f^n(x_i) = 0$ for $n > 1$ If so, why?

答:
There is no need to expand in a series; indeed there is no need for $f$ to have more than one derivative at all in order to make sense of $\delta(f(x))$.

What is really happening is that you locally invert $f$. That is, $f$, being a differentiable function that is assumed to never have zero derivative at any of its zeros, can be locally inverted around each of its zeros. So you split the domain into $n+1$ pieces; on $n$ of them, $f$ has exactly one zero and is locally invertible, and on the last one $f$ has no zero. The last one should "obviously" not contribute to the integration.

For the other pieces, say $h_k$ is the local inverse of $f$ on $I_k$. Then you have $\int_{I_k} g(x) \delta(f(x)) dx = \int_{f(I_k)} g(h_k(u)) \delta(u) \left | \frac{dx}{du} \right | du$. (The inclusion of the absolute value can be justified by doing case work based on the sign of $\frac{dx}{du}$ on $I_k$, paying attention to the orientation of the domain in the two cases.)

From the inverse function theorem you have $\left | \frac{dx}{du}(u) \right | = \frac{1}{|f'(h_k(u))|}$ (the usual thing you do during $u$ substitution), so by the definition of the delta function you get $\frac{g(h_k(0))}{|f'(h_k(0))|}$ from the integration on $I_k$.


答: This formula only works for simple zeroes for which df/dx does not cancel.  A good proof can be found in Kanwal's book "Generalized Functions" pp. 49-50</p>

Title: Question about proof that $\delta(f(x)) = \sum_i\frac{\delta(x_i - x)}{|df/dx_i|}$

Body:
From an exercise and proof in Principles of Quantum Mechanics:

Question: The author instructs us to "expand $f(x)$ near such points in a Taylor series, keeping the first nonzero term.  This is what the proof answer provides when it assumes that $f(x) = f'(x_i)(x-x_i)$.  But why do we only keep the first non-zero term?  What about the other terms?  Do we have that $f^n(x_i) = 0$ for $n > 1$ If so, why?


问:
Vertex transitive graph with "intermediate dimensional" growth rate
Given a function $f : \mathbb{N} \to \mathbb{N}$, I'm interested in if we can find a graph $G$ such that if $g(n)$ is the number of vertices in $G$ within distance $n$ from a given vertex, we have $\lim_{n\to\infty} f(n)/g(n) = c$ for some constant c. I'm particularly interested in the case where $g(n) = n^r$ for $r>1$; when $r$ is a natural number, the natural Cayley graph from $\mathbb{Z}^r$ has this property, and I'm curious if there are natural graphs of "intermediate dimension" in this sense. (of course, without a condition like vertex transitivity, this problem is trivial. I'm also interested in loosening vertex transitivity to something like finitely many orbits under the action of the automorphism group, or even something looser, if necessary)

答:
If you look to Cayley graphs (which is a very particular class of vertex-transitive graph), the question you're asking is know as growth of groups.

A lot of things are known:

for all $r \in \mathbb{Z}$, the groups which have polynomial growth $\Theta(n^r)$ are the groups which are virtually nilpotent (nilpotency is, broadly speaking, a generalization of commutativity, and one can get the $r$, see https://en.wikipedia.org/wiki/Gromov%27s_theorem_on_groups_of_polynomial_growth). A detailed research article would be https://arxiv.org/pdf/1908.06044.pdf.
Notice that if a graph is of polynomial growth (meaning growth function $f_G(n)$ smaller that some polynomial), then there exists an integer $d$ and two constant $c_1,c_2 >0$ such that $c_1n^d \leq f_G(n) \leq c_2n^d$ (see  V.I. TROFIMOV, Graphs with polynomial growth theorem 21 and Can transitive graphs have non-integer growth dimension?)

there are groups with exponential growth $\Theta(a^n)$ for some $a>0$, you can think of the free group $\mathbb{F}_2$ (whose Cayley graph for the usual generating set is a $4$-regular tree, so growth function is $4.3^{n-1}$ for $n\geq 1$). Growth cannot be greater than exponential is the degree of your graph is bounded.

there are groups with intermediate growth, which is greater than polynomial yet smaller than an exponential. The most famous one is Grigorchuk group, which has growth $exp({n^{.767...}})$, but it has been proven that many behaviour are possible (e.g. https://arxiv.org/pdf/1110.3650.pdf, see also https://arxiv.org/pdf/math/0607384.pdf). An open question, known as the gap conjecture, ask whether it is possible for a group to have growth smaller than $\exp(\sqrt{n})$ and still not be of polynomial growth (actually, no group of intermediate growth smaller than the Grigorchuk group is known yet). I don't know if thing are very different for vertex-transitive graph.


问:
The Variance Rule of $V(cX) = c^2V(X)$
As a statistics major, I learned a while back that the $V(cX) = c^2 V(X)$, where $V(X)$ is the variance of $X$. However, I never quite understood why that was the case. Consider this: We have an event $X$ such that each event is independent of any previous events. Take for instance, a roll of copper wire. We know that the mean length is $150 \text{m}$ and the standard deviation is $4\text{m}$. If we want to find the variance of $5$ rolls of copper wire, we could write this as $V(5X) = 25V(X) = 400\text{m}^2$. However, an alternative way to write the problem is doing $V(X + X + X + X + X) = V(X) + \ ... \ + V(X)$ - as they are all independent. Thus, we can simplify to $5V(X)$ and get $80\text{m}^2$. This second way seems like it should be the same as the first way, but they produce different answers. Can anyone explain this? I was helping a friend out with a similar problem and the second way actually produces the correct answer, so can anyone explain why this is the case?

答:
The two examples aren't the same thing. $V[5X]$ represents the variance if you scale up your random variable $X$ by a factor of $5$, while $V[X_1+X_2+X_3+X_4+X_5]$ represents the variance of $5$ independent samples of $X$.

$V[X_1+X_2+X_3+X_4+X_5] = 5V[X]$ is the variance of five independent samples of wire which each have mean $150$ meters and standard deviation $4$ meters (your example). If you create a new random variable that represents the sum of the five rolls, its mean would be $750$ and its variance $5*4^2 = 80$.

$V[5X]$ is the expected variance if your roll of copper wire were scaled up 5 times, i.e., you now produce a single roll that has a mean length of $750$ meters and standard deviation $20$ meters. That roll's length then has a variance of $400$.

Title: The Variance Rule of $V(cX) = c^2V(X)$


答:
Your error is here:
$$
f(x)=\color{red}-\frac{1+\cos(x)\color{red}-i\sin(x)}{2\cos(x)+2}\implies
S=\Im f(x)=\color{red}+\frac{\sin(x)}{2\cos(x)+2}.
$$

Title: Help to sum these two series $\sin(x)-\sin(2x)+\sin(3x)...$

Body:
I am reading Euler's paper entitled "Subsidium Calculi Sinuum" and he wrote down some "sums" for these trigonometric series:

\begin{align}S &= \sin(x)-\sin(2x)+\sin(3x)-\sin(4x)...\\\\
C &= \cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\
-1+C &= -1+\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\
-1+C+iS&=-1+[\cos(x)+\sin(x)]-[\cos(2x)+\sin(2x)+[\cos(3x)+\sin(3x)]-[\cos(4x)+\sin(4x)]...\end{align}

$$-1+e^{ix}-(e^{ix})^2+(e^{ix})^3-(e^{ix})^4+...$$

This is the geometric series, and I recognise that its "sum" is $-\dfrac{1}{1+e^{ix}}$

So, we have:

\begin{align}
& -\frac{1}{1+e^{ix}}=-\frac{1+e^{-ix}}{(1+e^{ix})(1+e^{-ix})} \\[8pt]
= {} & -\frac{1+\cos(x)-i\sin(x)}{(1+e^{ix})(1+e^{-ix})}=-\frac{1+\cos(x)-i\sin(x)}{2\cos(x)+2}
\end{align}

Equating the real and imaginary part with $C+iS$, we have:

$$-1+C=-\frac{1+\cos(x)}{2+\cos(2x)}\implies C=\frac{1}{2}$$

$$S=-\frac{\sin(x)}{2\cos(x)+2}$$

However, for the sum of $S$, there shouldn't be a minus sign, according to Euler.

So what is wrong with my derivation?


问:
If $f(a)=a,f'(a)=1,f''(a)>0$ then $\exists \delta >0:(a,a+\delta)\subset W^c$

Let $f : \mathbb{R} \to \mathbb{R}$ be a $C^2$ increasing function.  If $a \in \mathbb{R}$ is a fixed point of $f$ (so that $f(a) = a$) and $f'(a)=1, f''(a)>0$, show that $\exists \delta > 0$ such that $(a,a+\delta)\subset W^c$, where $W = \left\{x : \lim_{n\to\infty}f^n(x) = a\right\}.$

Note $f^n(x) := \underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(x)$

I tried to show that if $a<x^*<a+\delta$ then $\lim_{n\to\infty} f^n(x^*)\neq a$ but I don't really know how to continue.

Any ideas?

答:
Since $f''$ is continuous, there exists $\delta>0$ such that $f''(x)>f''(a)/2>0$ whenever $|x-a|<\delta$. This implies $f'$ is increasing on the interval $(a-\delta,a+\delta)$; this is a standard exercise using the mean value theorem. Now fix $x\in(a,a+\delta)$. Applying the mean value theorem again, there exists $x^*\in(a,x)$ such that
$$\frac{f(x)-f(a)}{x-a}=f'(x^*)>f'(a) = 1.$$
Since $f(a)=a$, one can rearrange the above inequality to deduce $f(x)>x$ for all $x\in(a,a+\delta)$.

To finish off the proof, assume for a contradiction that there exists $x\in(a,a+\delta)$ for which $f^n(x)\to a$ as $n\to\infty$. Then there exists $N$ such that $f^n(x)\in(a,a+\delta)$ for all $n\ge N$ (here we also use the fact that $f$ is increasing and $f'(a)>0$). But from our above work, this means $f^{n+1}(x)=f(f^n(x))>f^n(x)$, so this sequence is increasing, contradicting our assumption that its limit is $a$. This completes the proof.


问:
Given a point on a circle, find it on a given square
Math is not my forte so I apologize. I have a seemingly simple problem and I'm unable to figure out the formula for this.

I'd describe the problem as given the center point of a circle and a point on the edge of the circle, find the same point on a square with the same length as the diameter of the circle.

The filled in points in this image are the ones I'm trying to find.

As an addition, I'd love some online resources on where I can learn how to do this kind of math and intuitively find the answer for myself in the future. Cheers.

答:
Suppose the circle and square is centered at $(0,0)$ the circle has radius $r$ and the square has sides $2r$.  The midpoint of the sides of the square are $(0,r), (r,0), (0, -r), (-r,0)$ and the corners are $(-r,r),(r,r), (r,-r)$ and $(-r,-r)$.

Okay a point on the circle is $(x_1,y_1)$

So the equation of the line from $(0,0)$ to $(x_1, y_1)$ will have slope $m=\frac {x_1}{y_1}$.  And the equation of the line is $y = \frac {x_1}{y_1}x$.

Now there are four cases to consider based on what side of the square the line will intersect.

Case 1:  The line goes through the right side of the square.

That means $x_1 > 0$ and $|y_1| \le x_1$ and so $ -1 \le m \le 1$.

So if $m = 1$ then the line will go through the right side of the square.

So let that point be $(r, y_2)$.

But theb we have the formula $y = mx$ so $y_2 = mr =\frac {y_1}{x_1}r$. and the point on the square is $(r, \frac {y_1}{x_1}r)$

Case 2:

The line passes through the top of the square.

Then $y_1 > 0$ and $|x_1| \ge y_1$ and so $|m| =|\frac {y_1}{x_1}| \ge 1$

So if $m > 1$ or $m < -1$ then the line goes through the top the square.

And the point is $(x_2, r)$.

I should point out that if the point of the circle was $(x_1, y_1) = (0, r)$ then we'd have a case where $m = \frac {y_1}{x_1} =\frac r0 =\infty$.  This is the case that the line is completely vertical.

In this case it's obvious the point on the square is the exact same point of the circle so the point is $(0,r)$.  Otherwise....

The equation of our line is $y = mx =\frac{y_1}{x_1}$ so $r = m x_2$ and $r = \frac {y_1}{x_1} x_2$ so $\frac {x_1}{y_1}r = x_2$.

And the point is $(\frac {x_1}{y_1}r, r)$>

....

The case where the line goes through the left and right side are similar.

If $-1 \le m \le 1$ but $x_1 < 0$ then the line goes through the left side of the square and

the point is $(x_2, y_2) = (-r, -r\frac {y_1}{x_1})$.

And if $m \le -1$ or $m \ge 1$ (or $x_1 = 0$ and $m = \infty$) but $y_1 < 0$ then the line goes through the bottom of the square and

the point is $(x_2, y_2) = (-r\frac {x_1}{y_1} ,-r)$


问:
How to prove $\lim_{n \to \infty}\frac{a_1+a_2+\dots+a_n}{n}=l.$
When Let ${\{a_n}\}$ be a convergent sequence, $\lim_{n \to \infty}a_n=l.$ Prove that $\lim_{n \to \infty}\dfrac{a_1+a_2+\dots+a_n}{n}=l.$
Hint : Prove that the sequence ${\{a_n}\}$ is bounded first.

Please provide me hint or full solution. Thanks.

答:
Since $a_n \to l$, given $\epsilon > 0$, there exists $N(\epsilon)$ such that for all $n>N(\epsilon)$, we have
$$\left\vert a_n - l \right\vert < \epsilon \implies l-\epsilon < a_n < l+\epsilon$$
We now have
\begin{align}
\dfrac{a_1+a_2+\cdots+a_{N-1}+a_N + a_{N+1} + \cdots + a_m}m &\in \left( \dfrac{a+(m-N)(l-\epsilon)}m, \dfrac{a+(m-N)(l+\epsilon)}m\right)\\
& = \left(l-\epsilon + \dfrac{a-N(l-\epsilon)}m, l+\epsilon + \dfrac{a-N(l+\epsilon)}m \right) \tag{$\star$}
\end{align}
where $a=a_1 + a_2 + \cdots + a_N$. Now given $\epsilon>0$, there exists $M(N,l,a,\epsilon)$ we have that for $m > M(N,l,a,\epsilon)$ such that
$$\left \vert \dfrac{a-N(l+\epsilon)}m \right \vert < \epsilon$$
This gives us that $(\star)$,
$ \left(l-\epsilon + \dfrac{a-N(l-\epsilon)}m, l+\epsilon + \dfrac{a-N(l+\epsilon)}m \right) \subseteq \left(l-2\epsilon,l+2\epsilon\right)$.

Hence, given $\epsilon > 0$, there exists $M$ such that for $m>M$, we have
$$\dfrac{\sum_{k=1}^m a_k}m \in \left(l-2\epsilon,l+2\epsilon\right)$$
Hence,
$$\dfrac{\sum_{k=1}^m a_k}m \to l$$


问:
How to properly write down $\frac{\partial}{\partial xy } f(xy, x^2+3y-2)$
I am wanting to take the derivative with respect to $x$ of $f(xy, x^2+3y-2)$, so, I use the chain rule to find
$$
\frac{\partial}{\partial x} f(xy, x^2-2) = y \frac{\partial}{\partial xy}f(xy, x^2-2) + 2x\frac{\partial}{\partial (x^2-2)}f(xy, x^2-2)
$$
With which I would then continue
$$
\frac{\partial}{\partial xy} f(xy, x^2-2) = \frac{1}{x} \frac{\partial}{\partial y} f(xy, x^2-2)
$$
However, when differentiating w.r.t $xy$, I am not communicating to the reader that $x^2-2$ is expected to remain a constant, and it would seem to them that I should rewrite $x^2-2$ as $\left(\frac{xy}{y}\right)^2-2$.

How do I use mathematical notation to communicate to another mathematician that $x^2-2$ remains constant when differentiating with respect to $xy$?

答:
Let $\alpha(x,y)=xy$ and $\beta(x,y)=x^2-2$. Then, the function you're actually looking at is the composition $F(x,y)=f(\alpha(x,y),\beta(x,y))$. The chain rule tells you
\begin{align}
(\partial_1F)_{(x,y)} &=(\partial_1f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\alpha)_{(x,y)} +
(\partial_2f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\beta)_{(x,y)}\\
&= (\partial_1f)_{(xy-x^2-2)}\cdot y +
(\partial_2f)_{(xy-x^2-2)}\cdot 2x.
\end{align}
This is the strictest way of writing down the chain rule, and also the least ambiguous: the subscripts $1,2$ tell you along which direction you're calculating the partial derivatives, and the subscripts $(xy,x^2-2)$ etc tell you the point of evaluation of the derivatives. So, if you want to be 100% precise, this is how you could write the chain rule calculation.

If you want a slight mix with the Leibniz notation, you could also write this:

\begin{align}
\frac{\partial}{\partial x} f(xy,x^2-2) &= (\partial_1f)_{(xy,x^2-2)}\cdot \frac{\partial (xy)}{\partial x} +
(\partial_2f)_{(xy,x^2-2)}\cdot \frac{\partial (x^2-2)}{\partial x}\\
&=(\partial_1f)_{(xy-x^2-2)}\cdot y +
(\partial_2f)_{(xy-x^2-2)}\cdot 2x.
\end{align}


问:
Justify if True or False $\forall x\in\mathbb{R}\left(-|x|\le x\le |x|\right)$.
I need some hints on how to solve this

$$\forall x\in\mathbb{R}\left(-|x|\le x\le |x|\right)$$

Body:
Hints:
$$|x|=\max(x,-x) \tag1$$
$$-\max(a,b)=\min(-a,-b) \tag2$$