答:
On the one hand, observe that there can be no arrangement of 4 or more consecutive coins that do not have a value of 3. Equivalently, let $a_j$ be the value of the $j$-th coin in the sequence; $j=1,2,\ldots,2001$. Then for each $i=0,1,\ldots,1997$ at least one of $a_{i+1},a_{i+2},a_{i+3},a_{i+4}$  must be 3. So the number $X$ of $j \in \{1,2,\ldots, 2001\}$ such that $a_j =$ 3 satisfies the inequality $X \ \ge \lfloor \frac{2001}{4} \rfloor$.

On the other hand, the condition that at least 3 other coins are between any coin of value 3 gives the inequality $X \le \lceil \frac{2001}{4} \rceil$.

So if there is a subsequence that satisfies the conditions of the problem, the number $X$ of coins of value 3, must satisfy $X \in \{ \lfloor \frac{2001}{4} \rfloor, \lceil \frac{2001}{4}\rceil \}$. Now in fact there is indeed such a sequence, and furthermore, there is such a sequence where $X$ $=$ $\lfloor \frac{2001}{4} \rfloor$, and there is also such a sequence where  $X=\lceil \frac{2001}{4} \rceil$; consider the infinite sequence 12131213121312131213.... Any contiguous subsequence of this of length 2001 satisfies the conditions of the problem and is a valid arrangement.

Title: Combinatorics problem IMO style, existence type problem

Body:
I've tried and failed solving this problem for a week now; it doesn't seem too difficult and I feel like I'm missing something obvious.  What is the solution?

EDIT:
I tried constructing some sequences that would work and would divide 2001, for example 23 divides 2001 and if you can make a sequence of coins with length 21 you could just use the sequence again and again, but I couldn't find such sequence that wouldn't interfere itself (by for example having a 3 on the outside); also I just noticed that my calculations were wrong and the length of the sequence doesn't perfectly divide 2001. I tried induction and strong induction but that didn't work; finally I tried using the hostile neighbors principle (which I read about in a book and I'm pretty sure the author made the name up)


问:
Laurent series of complex rational function.
I want to calculate the Laurent series of function :
$$
f(z) = \frac{z}{(z-1)^3(z+1)}$$
in $z =1$.

I know that point $z=-1$ is the pole of this function. So I have to calculate Laurent series for $|z-1|<2$ and $|z-1|>2$. I tried this way:

$f(z)=\frac{z}{(z-1)^3(z+1)}= \frac{1}{8(z+1)}-\frac{1}{8(z-1)}+\frac{1}{4(z-1)^2}+\frac{1}{2(z-1)^3}$.

$f_1(z)=-\frac{1}{8(z-1)}=\frac{1}{16}\frac{1}{1-\frac{z+1}{2}}=\frac{1}{16}\sum(\frac{z+1}{2})^n$.

By analogy,
$f_2(z)=\frac{1}{4}\frac{1}{(z-1)^2}=-\frac{1}{8}\sum(\frac{z+1}{2})^{2n}$ and $f_3(z)=\frac{1}{2}\frac{1}{(z-1)^3}=-\frac{1}{4}\sum(\frac{z+1}{2})^{3n}$. And finally, $$f(z)=\frac{1}{8(z+1)}+f_1(z)+f_2(z)+f_3(z)$$
Why I have to consider two cases $|z-1|<2$ and $|z-1|>2$ ? How is that different?

答:
$$\frac{z}{z+1}=\frac{z}{2+(z-1)}=\frac{1}{2} \frac{z}{1+(z-1)/2} = \frac{(z-1)/2}{1+(z-1)/2}+\frac{1}{2}\cdot\frac{1}{1+(z-1)/2}$$

If $|z-1|<2$, then using the geometric series (convergent since $|z-1|/2<1$), you get that

$$\frac{z}{z+1}=\frac{(z-1)/2}{1+(z-1)/2}+\frac{1}{2}\cdot\frac{1}{1+(z-1)/2}=\sum_{n=1}^\infty \frac{1}{2^n}(z-1)^n +\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}(z-1)^n$$
Now
$$\frac{z}{(z-1)^3)(z+1)}=\frac{1}{(z-1)^3}\cdot \frac{z}{z+1} = \sum_{n=1}^\infty \frac{1}{2^n}(z-1)^{n-3} +\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}(z-1)^{n-3}$$ and you can finish putting altogether in a single sum.

For $|z-1|>2$ you cannot use in this way the geometric series and need the next trick.
$$\frac{z}{z+1}=\frac{z}{2+(z-1)}=\frac{1}{z-1} \cdot \frac{z}{1+2/(z-1)} = \frac{z}{z-1} \cdot \frac{1}{1+2/(z-1)}= \left(1+\frac{1}{z-1}\right)\frac{1}{1+2/(z-1)}.$$
Now, since $|z-1|>2$, the geometric series ((convergent since $2/|z-1|<1$)) says that
$$\frac{1}{1+2/(z-1)}=\sum_{n=0}^\infty 2^n (z-1)^{-n}$$
and you probably can finish by yourself.

ContentLicense: CC BY-SA 4.0


问:
If $\sin(a)+2\sin(b)+3\sin(c)=\cos(a)+2\cos(b)+3\cos(c)=0$ then find $\sin(b+c)+2\sin(c+a)+3\sin(a+b)$
If $\sin(a)+2\sin(b)+3\sin(c)=\cos(a)+2\cos(b)+3\cos(c)=0$ then find $\sin(b+c)+2\sin(c+a)+3\sin(a+b)$

I think euler's theorem and De moivre's theorem are to be used

答:
The original problem has 2 parts I am sharing them both

This is our required answer

The images are the links in blue please do have a look


问:
Trivial Schur multiplier implies projective representations are linear?
Suppose $G$ is a simple non-abelian finite group with a trivial Schur multiplier and $V$ is a finite-dimensional $ \mathbf{F}_3$-vector space. Can one say that every projective representation $$\rho: G \rightarrow Aut_{\mathbf{F}_3}(V) $$ is linear? Since the Schur multiplier is trivial.

Thanks in advance!

答:
Over any field $k$ we can talk about $H^2(G, k^{\times})$ which we might call the "Schur multiplier over $k$"; it vanishes iff every projective representation of $G$ over $k$ lifts to a linear representation. The universal coefficient theorem gives a split exact sequence

$$0 \to \text{Ext}^1(H_1(G), k^{\times}) \to H^2(G, k^{\times}) \to \text{Hom}(H_2(G), k^{\times}) \to 0$$

which implies that $H^2(G, k^{\times})$ vanishes iff $\text{Hom}(H_2(G), k^{\times})$ and $\text{Ext}^1(H_1(G), k^{\times})$ vanish. If $G$ is perfect then $H_1(G)$ vanishes so for a perfect group (and in particular a non-abelian simple group) we only need to consider $\text{Hom}(H_2(G), k^{\times})$. If by "Schur multiplier" you mean $H_2(G)$ then the vanishing of $H_2(G)$ implies that $\text{Hom}(H_2(G), k^{\times})$ vanishes and hence $H^2(G, k^{\times})$ vanishes.

It's worth knowing that this argument fails if $H_1(G) \neq 0$.


问:
Spectrum of T in $B(\ell^2)$
Let $T:\ell^2 \to \ell^2$ be an operator on $\ell^2$ is defined as follows:
 $$T\{a_1,a_2,\dots\}=\{0,a_1,a_2,\dots\}$$
What is spectrum of T in $B(\ell^2)$?</p>

答:
First note that as $\|T\|= 1$, we have $\sigma(T) \subseteq \{\lambda \in \mathbb C: |\lambda| \le 1\}$.

Now for $|\lambda| < 1$, $(T-\lambda)\colon (a_1, \ldots) \mapsto (-\lambda a_1, a_1 - \lambda a_2, \ldots)$. Suppose there were an $a \in \ell^2$ with $(T -\lambda)(a) = (1, 0, \ldots)$, then $-\lambda a_1 = 1$, so $a_1 = -\lambda^{-1}$, $a_2 = \lambda^{-1} a_1 = -\lambda^{-2}$, hence we must have $a = (-\lambda^{-n})_n$. But that's not possible as $a \in \ell^2$. That gives $\{\lambda \in \mathbb C  : |\lambda| < 1\} \subseteq \sigma(T)$.

Now recall that $\sigma(T)$ is closed, and you are done.

ContentLicense: CC BY-SA 3.0

Title: Spectrum of T in $B(\ell^2)$


问:
Convergence of $\displaystyle\sum_{n=0}^{\infty} \left(\dfrac{n}{2n+1}\right)^{2n-1}x^n$
$\displaystyle\sum_{n=0}^{\infty} \left(\dfrac{n}{2n+1}\right)^{2n-1}x^n$ How can I solve this question? I tried Root and Ratio test, but I encounter with $\displaystyle\lim_{x\to\infty}$ $\left(\dfrac{n}{2n+1}\right)^{\dfrac{-1}{n}}$ so I can't solve this limit, are there anyway to evaluate this limit or sum?

答:
Using the root test, we have

$$L=\lim_{n\to\infty}\sqrt[n]{\left|x^n\left(\frac n{2n+1}\right)^{2n-1}\right|}\\
=\lim_{n\to\infty} |x|\left|\frac n{2n+1}\right|^{\frac{2n-1}n}\\
=|x|\lim_{n\to\infty}\left|\frac 12-\frac1{4n+2}\right|^{2-\frac 1n}\\
=|x|\left(\frac 12\right)^2\\
=\frac {|x|}4$$


问:
Taylor expansion of a function
I would like to approximate the function

$f(x)=\frac{2x}{1-e^{-2x}}$

analytically for both small and large $x$.
But when I use the formula for the Taylor expansion, I run into the problem that the function and its derivative are not defined for $x=0$. How can I get around this problem?

For large $x$, my idea was to substitute $y:=1/x$ and then expand the function $g(y)=\frac{2}{y(1-e^{-2x})}$ around $y=0$. However, here I run into the same problem as above.

How do you proceed in such a case?

Title: Taylor expansion of a function

Body:
As $e^{-2x} = 1-2x+2x^2+O(x^3)$, we get
\begin{align}
f(x) &= \frac{2x}{1-e^{-2x}} \\
&= \frac{2x}{1-1+2x-2x^2+O(x^3)} \\
&= \frac{2x}{2x-2x^2+O(x^3)} \\
&= \frac{1}{1-(x+O(x^2))} \\
&= 1+x+O(x^2)
\end{align}

Hence, about $x=0$, $f(x)\approx 1+x$.

For large $x$, as $e^{-2x}<<1$, we get $1-e^{-2x}\approx 1$. Hence, for large $x$, $f(x)\approx 2x$/


问:
What does the notation $\{0,1\}^{\mathbb{N}}$ mean?
This is coming up on one of my topology assignments and I have never seen this before. My best guess would be that it's the collection of binary strings of countable length. If anyone could explain this, I'd appreciate it a lot!</p>

答: More generally, $B^A$ is the set of functions from $A$ to $B$.  The motivation for this notation is that if $A$ and $B$ are finite, $|B^A|=|B|^{|A|}$.


问:
$\mathbb{Q}[\sqrt{2}]$ is not a complete space
I am trying to prove that $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$ is not a complete space (with the standard metric). For that purpose, I am looking for a Cauchy sequence in $\mathbb{Q}[\sqrt{2}]$ which is not convergent in $\mathbb{Q}[\sqrt{2}]$. Can somebody help me? Thanks!

答:
Not particularly elementary, but useful and also a continuation of my comment ...

Let's take an integer $k$ that is not a perfect square. Then $\sqrt{k}$ is algebraic and irrational (root of the polynomial $x^2 -k$). It turns out that $e^{\sqrt{k}}$ is transcendental (i.e. is not algebraic, i.e. not a root of any polynomial with rational coefficients), the proof is easy, given Lindemann–Weierstrass theorem

... by the second formulation of the theorem, if $\alpha$ is a non-zero algebraic number, then $\{0, \alpha\}$ is a set of distinct algebraic numbers, and so the set $\{e^0, e^{\alpha}\} = \{1, e^{\alpha}\}$ is linearly independent over the algebraic numbers and in particular $e^\alpha$ cannot be algebraic and so it is transcendental.

Any element of $\mathbb{Q}[\sqrt{k}]= \{a+b\sqrt{k}:a,b∈Q\}$ is algebraic. From $x=a+b\sqrt{k} \Rightarrow \frac{(x-a)^2}{b^2}-k$ is the polynomial, with rational coefficients. For $b=0$ the polynomial is $x-a$.

$\mathbb{Q}[\sqrt{k}]$is not complete, because

$1+\frac{\sqrt{k}}{n}\in\mathbb{Q}[\sqrt{k}]$ for $\forall n\in\mathbb{N}\setminus\{0\}$

$\left(1+\frac{\sqrt{k}}{n}\right)^n\in\mathbb{Q}[\sqrt{k}]$, easy to check using Binomial_theorem

But $\lim\limits_{n\to\infty}\left(1+\frac{\sqrt{k}}{n}\right)^n=e^{\sqrt{k}}$ transcendental, thus $e^{\sqrt{k}} \notin \mathbb{Q}[\sqrt{k}]$.


问:
Rank of eigenvector projections
Consider the following two $n\times n$ matrices:

$$A_1:=H_1\Sigma_1H_1' $$

$$A_2:=H_2\Sigma_2H_2'$$

where $\Sigma_1$,$\Sigma_2$ are positive definite $k\times k$ matrices ($k<n$), and $H_1$,$H_2$ are $n\times k$ matrices such that

$$H_1'H_1=H_2'H_2=I_k \hspace{1cm}\text{and} \hspace{1cm} H_1'H_2=\begin{bmatrix}
I_r & 0 \\
0 & 0
\end{bmatrix} .$$

Now I perform the eigendecomposition of $A_1$ and $A_2$ (with eigenvalues in decreasing order and orthonormal eigenvectors), but keep only the first $k$ eigenvectors  of $A_1$ and the first $s$ eigenvectors $(s>k)$ of $A_2$. Let $U_1$ and $U_2$ be the $n\times k$ and $n\times s$ matrices of eigenvectors for $A_1$ and $A_2$ respectively. I want to show that the matrix

$$U_1-U_2U_2'U_1=MU_1$$

has rank $n-r$, where $M:=I_n-U_2U'_2$. This is similar to here, but the columns of $H_1$ and $H_2$ are not fully independent.

Any help is greatly appreciated.

答:
The columns of $U_1$ form an orthonormal basis for the column space of $A_1$, which is in turn the column space of $H_1$. Let $S_1$ denote this column space.

The columns of $U_2$ form an orthonormal basis for the sum of the column space of $A_2$ and an arbitrary $(s-k)$-dimensional subspace of $\ker(A_2)$. Let $S_2$ denote the column space of $U_2$.

From the construction of $A_1$ and $A_2$, we know that the intersection of the column spaces of $A_1$ and $A_2$ is $r$-dimensional (in particular, it is equal to the column space of the matrix $J_0$ defined in the comments on your question). Let $S_0$ denote this intersection.

The rank of $(I - U_2U_2')U_1$ is equal to $\dim(S_1) - \dim(S_1 \cap S_2)$. Because $S_0 \subseteq S_2$, the dimension of this intersection is at most equal to
$$
\dim(S_1) - \dim(S_0) = k - r.
$$
However, if $S_0 \subsetneq S_2$, then this rank will be strictly smaller. In other words, if the arbtirary $(s-k)$-dimensional subspace of $\ker(A_2)$ has a non-trivial intersection with the column space of $A_1$, then the rank will be less than $k-r$.

If this $(s-k)$-dimensional subspace is "randomly" selected, then the probability of a non-trivial intersection occurring by chance is extremely small, which would explain your empirical results.

Illustrative counterexample:
$$
H_1 = \pmatrix{1&0\\0&1\\0&0\\0&0}, \quad H_2 = \pmatrix{1&0\\0&0\\0&1\\0&0}, \Sigma_1 = \Sigma_2 = \pmatrix{2&0\\0&1}.
$$
Note that the following is possible:
$$
U_1 = H_1, \quad U_2 = \pmatrix{1&0&0\\0&0&1\\0&1&0\\0&0&0}.
$$
We compute
$$
(I - U_2U_2')U_1 = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1} \pmatrix{1&0\\0&1\\0&0\\0&0} = 0,
$$
which has rank smaller than $k - r = 2 - 1 = 1$. On the other hand, if you take the final column of $U_2$ to be a "random" unit vector from the set $\{(0,x_2,0,x_4): x_2,x_4 \in \Bbb C\}$, you should find that the matrix $(I - U_2U_2')U_1$ has rank $1$ in practically all cases.

Title: Rank of eigenvector projections


问:
Is symmetric matrix has nonzero eigen values or is Positive (semi-)definite?
Let M be symmetric matrix. Its eigen values are 0 or real(but not complex)

$Mx = \lambda x$

taking transpose on both sides and multiplying by $Mx$ on both sides

$(Mx)^TMx = \lambda x^TMx, $

Let y = Mx then we have

$\lambda x^TMx = (Mx)^TMx = y^Ty$ --> this is nothing but squared sum. It means $y^Ty \geq 0$

It meaans $\lambda x^TMx \geq 0 \implies \lambda  \geq 0 , $ OR $x^TMx \geq 0$ which means symmetric matrix has either eigen values non-negative or positive semi-definite? May i know is this true? I was thinking of some other problem but got this which i do not know

In defination of positive (semi-)definiteness, there are 2 conditions

symmetric matrix

$x^TMx \geq 0$

答:
The reason that the eigen values of symmetric matrices are real is the following:

$\lambda=\langle v,\lambda v\rangle= \langle v,Av\rangle=\overline{\langle Av,v\rangle}=\langle \lambda v,v\rangle=\overline{\lambda}$


问:
Integral Involving Hermite Polynomial and rational function
We want to prove the formula

$$\frac{\int_{-\infty }^{\infty
   } \frac{e^{-x^2}
   H_n(x){}^2}{1+x^2} \,
   dx}{2^n n!}=\int_{-\infty
   }^{\infty }
   \frac{\left(\frac{1-x^2}{1+
   x^2}\right)^n
   e^{-x^2}}{1+x^2} \, dx$$
where $H_n(x)$ is the Hermite polynomial. The usual procedure of invoking the identity
$$\frac{\exp \left(\frac{2 x^2
   t}{1+t}\right)}{\sqrt{1-t^2
   }}=\sum _{n=0}^{\infty }
   \frac{t^n H_n(x){}^2}{2^n
   n!}$$
Leads to the relation
$$\frac{\int_{-\infty }^{\infty
   } \frac{e^{-x^2} \exp
   \left(\frac{2 t
   x^2}{1+t}\right)}{1+x^2} \,
   dx}{\sqrt{1-t^2}}=\sum
   _{n=0}^{\infty } \frac{t^n
   \int_{-\infty }^{\infty }
   \frac{e^{-x^2}
   H_n(x){}^2}{1+x^2} \,
   dx}{2^n n!}$$
It is not clear how to match powers of $t$ on both sides and obtain the desired formula.

答:
Hint: Upon invoking your summation identity, your new integrand contains the exponential factor $$\exp(-x^2)\exp\left(\frac{2tx^2}{1+t}\right)=\exp\left(-\frac{1-t}{1+t} x^2\right).$$

The coefficient $\dfrac{1-t}{1+t}$ may be eliminated by a judicious substitution. It remains to organize the new integrand so as to permit expanding in powers of $t$, and this is straightforward.


问:
Is this equality right?
I try solve that, but for me this equality is missing something:

For the denominator is wrong. I get this:

What is wrong?

答: $$\sum\limits_{i,j}\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum\limits_{i=j}\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} +\sum\limits_{i>j}\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} +\sum\limits_{i<j}\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}}$$ $$=0 +\sum\limits_{i>j}\dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} +\sum\limits_{i>j}\dfrac{a_j-a_i}{(1+a_i)\sqrt{a_j}} $$ $$= \sum\limits_{i>j}\dfrac{((1+a_i)\sqrt{a_j} - (1+a_j)\sqrt{a_i})(a_i-a_j)}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} $$ $$= \sum\limits_{i>j}\dfrac{(a_i\sqrt{a_j}-\sqrt{a_i}a_j - \sqrt{a_i}+\sqrt{a_j})(\sqrt{a_i}-\sqrt{a_j})(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} $$  $$= \sum\limits_{i>j}\dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}} $$

Title: Is this equality right?


答: Being more accurate it is the set of all infinite sequences with entries being $0$ or $1$. The general rule is given in the answer above. For Example $\mathbb{N}^{\mathbb{N}}$ is a set of all infinite sequances of natural numbers. The interesting think is that some set theorists would instantly call $\lbrace 0,1\rbrace^\mathbb{N}$ a Cantor set.

Title: What does the notation $\{0,1\}^{\mathbb{N}}$ mean?

Body: This is coming up on one of my topology assignments and I have never seen this before. My best guess would be that it's the collection of binary strings of countable length. If anyone could explain this, I'd appreciate it a lot!</p>


问:
Confusion regarding the action of $SU(2)$ on homogeneous polynomials
The group $SU(2)$ acts on the space of degree $d$ homogeneous polynomials on $\mathbb{C}^2$ as
$$(g \cdot f)(z) = f(g^{-1} z)$$
where $g \in SU(2)$ and $z = (z_1, z_2) \in \mathbb{C}^2$. To compute the action on the polynomial $z_1$, we can consider it as the map $\pi_1 : \mathbb{C}^2 \to \mathbb{C}$ defined by $\pi_1(z_1, z_2) = z_1$. Then, for
$$g = \begin{bmatrix}\alpha & -\overline{\beta}\\ \beta & \overline{\alpha}\end{bmatrix}$$
we have
$$(g \cdot \pi_1)(z) = \pi_1(g^{-1} z) = \pi_1 \left( \begin{bmatrix}\overline{\alpha} & \overline{\beta}\\ -\beta & \alpha\end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \right) = \overline{\alpha}z_1 + \overline{\beta}z_2$$
This is the same result obtained in Singer's Linearity, Symmetry and Prediction in the Hydrogen Atom, p. 138. So, to me, it seems like when computing the action of $g$ on some polynomial we replace $z_1 \mapsto \overline{\alpha}z_1 + \overline{\beta}z_2$ in the polynomial's expression.

However, in Woit's Quantum Theory, Groups and Representations, p. 113, it is said that one must replace $z_1 \mapsto \alpha z_1 + \beta z_2$ (here I adapted his notation to mine). So my question is: which one is right, and why?

答:
The first action you describe $(g\cdot f)(z):=f(g^{=1}z)$ is a group action.

The second formula seems to describe $(g\cdot f)(z):=f(g^Tz)$, also a group action.

(One could also have used $(f\cdot g)(z):=f(gz)$ as a right group action.)

Indeed, the first two actions are equivalent, because they are connected by an inner automorphism of $SU(2)$, namely $A\mapsto (A^T)^{-1}=JAJ^{-1}$ where $J=[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]$. Note that $(A^T)^{-1}$ is not an inner automorphism for $n\times n$ matrices for $n>2$.


问:
Why can a local taylor expansion tell how many derivatives are continuous?
Define $\chi=x(1+\cos(x))$ for $x\in[-\pi,\pi]$ and extend $\chi$ periodically to the entire real line. I want to check the continuity of $\chi'$ and $\chi''$ at $\pi$. A solution was shown as the following:

Since $\chi(\pi-\varepsilon)=(\pi-\varepsilon)(1-\cos(\varepsilon))$, we get
$$\chi(\pi-\varepsilon)=\frac{\pi}{2}\varepsilon^2-\frac{1}{2}\varepsilon^3+\cdots$$
showing that the first derivative is continuous but the second is not.

I am pretty confused by this statement. I tried to link the local expansion to the expansion of $\chi$ about $\pi$ to deduce that $\chi''$ is not continuous at $\pi$. But I couldn't get any further.
$$\chi(\pi-\varepsilon)=\chi(\pi)+\chi'(\pi)(-\varepsilon)+\chi''(\pi)(-\varepsilon)^2+\cdots$$
if we suppose $\chi''(\pi)$ exists. How does this tells us $\chi''$ is not continuous?

答:
Take

$$f(x)=x^3\sin(\frac 1x) \;if\; x\ne 0$$
and $ f(0)=0$.

$ f(x) $ can be seen as

$$f(x)=0+0.x+0.x^2+x^2(x\sin(\frac 1x))$$
$$=0+0.x+0.x^2+x^2\epsilon(x)$$

this says that $ f(x) $ has a second order Taylor expansion arround $ x=0 $

but, you can check that $ f''(0) $ doesn't exist.


问:
Determine the volume of the solid that's bounded by $(x^2+y^2+z^2)^2 = 2z(x^2+y^2)$.

Determine the volume of the solid that's bounded by $(x^2+y^2+z^2)^2 = 2z(x^2+y^2)$.

Transforming to spherical coordinates I have that $r^4 = 2r^3\cos(\varphi)\sin^2(\varphi)(\cos^2(\theta)+\sin^2(\theta)) \Longrightarrow r = 2\cos(\varphi)\sin^2(\varphi)$ so $r \in [0, 2\cos(\varphi)\sin^2(\varphi)].$ How can I find the bounds for $\varphi$ and $\theta$? I feel a bit lost here...

答:
First a help in visualization:

So surely $0 \leq \theta < 2 \pi$ and $0 \leq \phi \leq \pi/2$, as can be easily deduced from the equations once you realize $x^2 + y^2 = r^2$ and that $z \geq 0$.

Using the bounds on $r$ given above, the volume is then $\frac{2 \pi}{15}$.


问:
Galois action on a tensor product of fields
Let $K$ be a field, $F$ a finite field extension of $K$ and let $L$ be an algebraic closure of $K$. Let $G_K:=\operatorname{Gal}(K^{\text{sep}}|K)$ be the absolute Galois group of $K$. $G_K$ acts on $L$. We have
$$
F\otimes_K L=\prod_{\nu}{L}
$$
where the product is indexed over field embeddings $\nu:F\to L$. Any $\sigma\in G_K$ acts on the left-hand side of this equality by $1\otimes \sigma$. What is the corresponding action on the right-hand side in terms of Galois groups?

Many thanks!

答:
I think you assume that $F/K$ is separable, right ? For the tensor product to be reduced. Ok, so in order to obtain the induced action of $G_K$ on the right hand side, one needs first to fix a $K$-isomorphism that identifies the left hand side with the right hand side.

Writing $F$ as $K[X]/(f)$ for an irreducible polynomial $f$ (the minimal ideal of a primitve element), we can write the left hand side as $L[X]/(f)$ with the obvious $G_K$ action on it. Now take a complete factorization of $$f=(X-a_1)\cdots(X-a_n)$$ in $L[X]$ and apply the Chinese remainder theorem to obtain the canonical isomorphism

\begin{align*}
L[X]/(f) &\longrightarrow L[X]/(X-a_1)  \times \ldots \times  L[X]/(X-a_n)\\
g+(f) &\mapsto  (g + (X-a_1), ..., g + (X-a_n) )
\end{align*}

On the righthand side you have the cartesian product that you mentioned. If you want to know the induce $G_K$ action, you have to start with an arbitrary tuple on the right and side, pull it back to the left side, apply an element of $G_K$ and push the result back to the right hand side.

You will see that what you obtain is a superposition of a transitive permutation action of the components (corresponding to how the elements a_1,...,a_n are permuted) and a straight forward action of the elements of each component.

Sorry for not writing this up with more details provided. I just came across your question when I was looking for mentioning of a reference in the literature of this very observation.

Thus let me extend your question (hoping that maybe you have in the meantime the answer to my extended question): Does anybody know a reference in the literature where not only the equality of the tensor product and the cartesian product as k-algebras is stated, but also the induced $G_K$ action on the cartesian product is described ?


问:
Find scalar such that $B$ is a basis for $\mathbf R^3$
Let $B=\{v_1=(1,1,-1), v_2=(-1,3,1),v_3=(a,2-a,a+1)\} \subseteq \mathbf R^3$. Find the scalar $a$ such that $B$ is a basis for $\mathbf R^3$.
I tried taking a vector $(x,y,z) = α_1v_1 + α_2v_2 + α_3v_3$ but I got stuck in a point when I tried solving the system but I got $2a+1 = (x+z)/c$. Should I try a different approach?

答:
Hint:

The determinant
$$\det(v_1,v_2,v_3)= \begin{vmatrix}1&-1&a \\ 1&3&2-a \\ -1&1& a+1\end{vmatrix} \ne 0 $$.


问:
Can the limit of a sequence of functions in $L_p$ and the limit in almost everywhere convergence be different?
I'm sure this is an easy question, but I am somewhat confused. I'm considering here a measure space $(X, \mathcal{A}, \mu)$, and the functions are real valued.

Given a sequence of functions $(f_n) \in L_p$ that converges almost everywhere to a function $f$ and converges in $L_p$ to a function $g$, is it true that $f=g$ almost everywhere?

答: If $f_n \to g$ in $L^p$ there is some subsequence such that $f_{n_k}(x) \to g(x)$ for ae. $x$. Hence $g(x)=f(x)$ for ae. $x$.


问:
How do I write the rule for this sequence that is neither arithmetic nor geometric as an equation?
I have the sequence $3, 5, 9, 17, 33, 65,\dots$

I know the rule is multiply by $2$ then subtract one. $3$ times $2$ is $6$ minus $1$ is $5$ and so on. Easy enough.

How do I write the rule for this sequence as an equation?  I know it is neither arithmetic or geometric.

答:
You can write recursive function like this

$f(x)=\begin{cases}
\ 2(f(x-1))-1 ,  & \text{if x>1,} \\[2ex]
3, & \text{if x=1}
\end{cases}$


问:
If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $.
I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.</p>

Body:
$(1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+(abc)$

Applying AM≥GM on $(a+b+c)+(ab+bc+ca)+(abc)$

$$\frac{(a+b+c)+(ab+bc+ca)+(abc)}{7}≥(a^{4}b^{4}c^{4})^\frac{1}{7}$$

So $(a+b+c)+(ab+bc+ca)+(abc)≥7(a^{4}b^{4}c^{4})^\frac{1}{7}$

If we add 1 on LHS so now LHS>RHS

So $(1+a)^{7}(1+b)^{7}(1+c)^{7}>7^{7}(a^{4}b^{4}c^{4})$


答:
From the original equation, we have that $z>0$ which means $\varphi\leq\frac{\pi}{2}$. When the equation simplifies in spherical coordinates, there is no $\theta$ which means the shape is rotationally symmetric and thus $\theta\in[0,2\pi]$

Then looking at a graph of $r = 2\sin^2\varphi \cos\varphi$ on the restricted domain $\left[0,\frac{\pi}{2}\right]$, the graph is $0$ at $\varphi = 0, \frac{\pi}{2}$. Thus the integral is

$$\int_0^{2\pi}d\theta\int_0^{\frac{\pi}{2}}d\varphi\int_0^{2\sin^2\varphi\cos\varphi}dr\:r^2\sin\varphi$$

Body:
Determine the volume of the solid that's bounded by $(x^2+y^2+z^2)^2 = 2z(x^2+y^2)$.

Transforming to spherical coordinates I have that $r^4 = 2r^3\cos(\varphi)\sin^2(\varphi)(\cos^2(\theta)+\sin^2(\theta)) \Longrightarrow r = 2\cos(\varphi)\sin^2(\varphi)$ so $r \in [0, 2\cos(\varphi)\sin^2(\varphi)].$ How can I find the bounds for $\varphi$ and $\theta$? I feel a bit lost here...


问:
What does the golden spiral converge to?
I am looking at pictures of spirals associated with the Fibonacci sequence and the golden ratio and I am seeing several different spiral diagrams.

For example one image I saw shows the spiral with the following numbers $8, 5, 3, 2, 1, 1$.

Then another has $34, 21, 13, 8, 5, 3, 2, 1$.

So, since the Fibonacci sequence continues forever is it possible to figure out where the golden spiral converges to in an image with Fibonacci sequence of $F_n, F_{n-1}, F_{n-2}, \dots$ ?

答:
https://downloads.hindawi.com/journals/ddns/2019/3149602.pdf

I had this question too and this is a paper on it I came across. It gives (5-√5)/10 multiplied by the proportions of each side. This paper uses some weird math however using the relation that φ-1=1/φ and defines λ as 1/φ so a lot of the math is given by the inverse of the golden ratio in terms of λ.


问:
Echelon form of a Matrix, Rank, and Number of solution
My book defines a matrix A to be in row Echelon form as follows:

A matrix A is said to be in row Echelon form if:

The first non-zero element in each row is 1

Every non-zero row in A preceds every zero row

The number of zeroes before the first non zero element in 1st, 2nd, 3rd,... rows should be in >increasing order

Example:

$$
    \begin{bmatrix}
    1 & 2 & 3 \\
    0 & 1 & 4 \\
    0 & 0 & 1 \\
    \end{bmatrix}
$$

Note: The condition that a column containing the leading 1 must have all other elements above and below it as zero is NOT mentioned as a condition and I believe that it is only necessary for Reduced Row Echelon matrices.

Now, my book defines rank of a matrix to be the number of non-zero rows in a Row-Echelon Matrix
Further, it is related to number of solutions of equations as follows:

Let A be the coefficient matrix of a given system of equations, ie, AX=B, and C be the augmented matrix (Such that C = [A:B]), then

Consistent Equation $\rho(A)=\rho(C)$

Unique Solution: If $\rho(A)=\rho(C)=n $, where n is the number of unknowns

Infinite Solution: If $\rho(A)=\rho(C)= r$, such that $r < n$

Inconsistent Equation $\rho(A)\neq \rho(C)$, then it has no solution

Now, I have a question that says the follows:

Determine for what values of $\lambda, \mu$, the following system of equations

$$x+y+z=6$$
$$x+2y+3z=10$$
$$x+2y+ \lambda z = \mu$$

has (i) no solution, (ii) a unique solution, (iii) an infinite number of solutions

So, the solution is given, by first writing in the matrix form, then converting to the augmented form, which I am writing here....

$$ \left[
\begin{array}{ccc|c}
  1&1&1&6\\
  0&1&2&4\\
  0&0& \lambda-3 & \mu-10\\
\end{array}
\right] $$

Now, it is stated, that for (i) no solution, $\rho(A)\neq \rho(C)$

So,  $\lambda-3 = 0 $and $\mu-10 \neq 0$

which I think is so that rank of the matrix A is 2 and rank of the matrix C is 3, which means that the ranks of the two matrices are not equal.

But, shouldn't $\lambda-3 = 1???$ This just doesnt make sense to me, as the leading element must be 1 right???? Otherwise how do we get the Elchelon form to find the rank??? Also, if $\lambda-3=0$, then atleast $\mu-10 =1$ as $\mu-10 \neq 0$, right?? But thats not given as a condition.

Should there be a leading 1 necessary to get the rank or not??

(ii) A unique solution $\rho(A)=\rho(C)=3$

$\lambda -3 \neq 0$ and $\mu \in R$

(iii) Infinite solutions, $\rho(A)=\rho(C) < 3$

$\lambda -3 = 0$ and $\mu -10=0$

This I get, as now both the ranks of A and C is 2.

But Can someone please explain the 1st two cases to me, and also the doubts that I had which I explained mid-question.

答: The basic stategy is to find the leftmost column of the matrix that has a nonzero entry (usually that will be the first column). Choose any row with a nonzero entry in that column.  Use row interchange to make that row the first row of your new matrix. Multiply the first row by the reciprocal of its leading nonzero entry, producing a leading $1$ in the row. Add appropriate multiples of the first row to lower rows to get zeros under the leading $1.$ If a row has a $0$ under the leading $1$ you don't need to do anything to that row. Cover the first row of the matrix and repeat the process on the remaining part to produce a leading $1$ in the second row (it will occur in a column to the right of the first leading $1)$ and $0$'s below it. Keep going until you get a leading $1$ in as many rows as possible. The remaining rows should be all $0$'s at that point.


问:
Proving that $n^2 + n$ is even for any integer $n$
This type of question often shows up in introductory abstract math courses, computer science discrete math, or an introduction to logic course.

Show that $n^2 + n$ is even for any integer n.

答:
Here is a different take. Use Gauss's trick:

$S = 1 + 2 + \cdots + n$

$S = n +(n-1) + \cdots + 1$

$2S = (n+1) + (n+1) + \cdots + (n+1) = n(n+1)$

Thus, $n(n+1)=2S$ is even.


问:
Finding the Balance in a Math Question (Teaching)
As we try to work and teach in the midst of this pandemic, some problems arise when making online math exams. My questions is simple: What could be an interesting basic differentiation question such that students doing the online exam still have to work a bit to get it, rather than simply Googling it or using WolframAlpha/Mathematica to solve it for them?

Everything I can think of seems to be either too easy (meaning they can easily solve it online) or too demanding. I'm struggling to find a good balance, if there is any. This is a first-year math module I'm teaching (non-math students).

Any ideas?

答:
How about proving a more obscure result from first principles?

For example:

Prove using first principles that
$$\frac{d}{dx}(x\sin x)=\sin x +x\cos x$$
You may assume that $\lim\limits_{h\to0}\frac{\cos h-1}{h}=0,~~\lim\limits_{h\to0}\frac{\sin h}{h}=1.$

As a high school student myself I'd say that is potentially the most challenging part of differentiation (at least at high school level).

However, I think it also depends on what stage your students are at. Have they learnt the chain rule, product rule, quotient rule? Can they differentiate any logarithm? Can they differentiate any exponential and basic trigonometric functions? Can they differentiate inverse trigonometric and hyperbolic functions? Can they derive Maclaurin series for a given function? Can they use l'Hopital's rule?

Additionally, do you want them to use other skills in this question as well, such as knowledge of trigonometric identities?

Alternatively, perhaps you could ask them to prove certain results using differentiation. For example, if they are proficient in sigma notation and algebra:

Given the standard formula for the value of a geometric series, use differentiation to prove that
$$\sum_{r=1}^n ra^r=\frac{a(na^{n+1}-(n+1)a^n+1)}{(a-1)^2}$$
Hence evaluate
$$\sum_{r=1}^{13} r2^r$$

This may be too challenging, but it will hopefully be interesting and stimulating for your students.

Another example may be:

Use differentiation to prove that
$$\frac{1-\cos x}{\sin x}\equiv\tan\frac{x}{2}+A$$
for some constant $A$. Now find the value of $A$.


问:
For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable?
I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.

My idea was this. Since $G$ is solvable, there exists a normal abelian tower
$$
G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}.
$$
I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence
$$
G/H\supset K_1\supset\cdots\supset K_r=\{e\}.
$$
My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.

Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g'H_i\in K_i$, then I want to show $g'H_igH_{i+1}g'^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.

答:
You're approach looks good to me.
It might help to rewrite $K_i = G_i/(H\cap G_i)$ in the form $K_i = (G_i H)/H,$
so that you have a sequence of groups $G_i H$ all containing $H$, whose quotients are giving the $K_i$.  Now to verify that $K_{i+1}$ is normal in $K_i$, you just
have to verify that $G_{i+1} H$ is normal in $G_i H$, which isn't too hard to do.
(Essentially, the computations with cosets that were giving you trouble have all
been wrapped up once and for all into the isomorphism $G_i/(H\cap G_i) \cong
(G_i H)/H$, and the coset computations with $G_{i+1} H$ inside $G_i H$ will
be quite a bit easier.)

To see that $K_i/K_{i+1}$ is abelian, you can again use isomorphism theorems,
to rewrite it as $(G_i H)/(G_{i+1} H)
\cong G_i/(G_{i+1} H \cap G_i).$   You should be able to see that the latter group
is a quotient of $G_i/G_{i+1}$, and so, being a quotient of an abelian group,
is abelian.

As a general remark, when studying the image of a subgroup $G'$ under a quotient map $G \to G/H$,
passing back and forth between the descriptions $G'/G'\cap H$ and $G'H/H$
is a very standard method.  The former description helps you think about
the the image as a quotient of the given subgroup $G'$, while the latter
description is useful for bringing into play the fact that "the lattice of subgroups of
$G/H$ corresponds to the lattice of subgroups of $G$ containing $H$" --- it rewrites the
image as the quotient of a subgroup containing $H$, and so helps you understand the inclusion relations and so on between different images as $G'$ varies.


问:
Combinatorical permutation-cycles problem
How many permutations $\pi$ $\in$ $Sn$ ($n>2$) where $1$ and $2$ belong to the same cycle?

Here is what I have so far:

First I put $1$ and $2$ in the same cycle and make $j$ elements from the other $(n-2)$ join them in the same cycle: ${n-2 \choose j}$ with $0<j<n-2$

Then I must put the $(n-2-j)$ elements in other cycles. For that I use the Stirling numbers (which garantees that at least 1 element is put in a cycle) in  to distribuite those elements in the $k$ cycles, with $(1<k<j)$:

$S(n-2-j,k)$

My final answer is

$\sum_{j=0}^{\ n-2} {{n-2 \choose j} . \sum_ {k=0}^{\ j} {S(j,k)}}$

Is there something wrong with my resolution or a simpler way to solve it?

答:
We have from first principles the closed form

$$S_n = \sum_{j=0}^{n-2} {n-2\choose j} \frac{(j+2)!}{j+2} (n-2-j)!.$$

This is for n $\ge 2$

$$S_n = (n-2)! \sum_{j=0}^{n-2} (j+1) = \frac{1}{2} (n-2)! (n-1) n
= \frac{1}{2} n!.$$

We may think of the permutation in table notation to explain the factor $(n-2-j)!.$ Once we choose the $j$ extra elements for the cycle containing $1$ and $2$, the corresponding $j+2$ pre-image / image pairs are removed from the table. The table now has the remaining $n-2-j$ values in the pre-image and the image and we may permute the values in the image any way we like.


问:
Determining a function is absolutely integrable over the real line
I'm having a hard time understanding absolutely integrable functions and determining them. Can you please explain briefly what is this absolutley integrable over the real line and also how to determine whether a function is. ie. How to determine this?
$$\frac{sin(x^2)}{1+x^2}$$
Any help will be appreciated.

答:
An absolutely integrable function is a function whose absolute value is integrable.

To see that your function is absolutely integrable, we need to see if its absolute value is integrable over the real line:

$$\int_{-\infty}^\infty\left|\frac{\sin(x^2)}{1+x^2}\right|dx\overset * = 2\int_{0}^\infty\left|\frac{\sin(x^2)}{1+x^2}\right|dx \le 2\underbrace{\int_0^1 \frac{x^2}{1+x^2}dx}_{(1)} + 2\underbrace{\int_1^\infty\frac{1}{1+x^2}dx}_{(2)}$$
Since both $(1)$ and $(2)$ are convergent, the given integral is convergent as well.

Explanation:

$(*)$: Function is even

$(1)$: Used $\sin(x^2) \le x^2$.

$(2)$: Used $\sin(x^2)\le 1$.


答:
Your function is even, so you just need proving that the integral
$$\int_0^{+\infty}\frac{\sin(x^2)dx}{1+x^2}$$
is absolutely convergent.

For this, observe that

$$(\forall x\ge 1)\; |\frac{\sin(x^2)}{1+x^2}|\le \frac{1}{x^2}$$

and that

$$\int_1^{+\infty}\frac{dx}{x^2}$$
is convergent.
Now, use the Comparison criterion for positive functions.

Title: Determining a function is absolutely integrable over the real line

Body:
I'm having a hard time understanding absolutely integrable functions and determining them. Can you please explain briefly what is this absolutley integrable over the real line and also how to determine whether a function is. ie. How to determine this?
$$\frac{sin(x^2)}{1+x^2}$$
Any help will be appreciated.


问:
Is the identity map the only map from the positive integers to itself that simultaneously preserves multiplication, order, and primes?
Is the identity map the only map $f$ from the positive integers $\mathbb{Z}_{+}$ to itself that satisfies all three of the following properties?

$f(mn)=f(m)f(n)$ for all positive integers $m$ and $n$ (which implies that $f(1)=1$)

$f(m)<f(n)$ whenever $m<n$ (which implies that $f$ is injective)

$f(p)$ is prime for any prime number $p$

The above three properties (actually, for property 2, the weaker assumption that $f$ is injective suffices) imply that applying $f$ to any positive integer will not change the prime signature, because all that happens with $f(n)$ is that each prime $p$ dividing $n$ is replaced with $f(p)$ while maintaining the same exponent.

Here are some examples that suggest that the answer might be "Yes" (because they only satisfy properties 1 and 3 and injectivity, but are not order-preserving on all positive integers, although they are order-preserving on prime numbers):

Let $f(p)$ be the next prime following $p$, with $f$ extended to composite numbers using (complete) multiplicativity. Then, $8<9$, while $f(8)=f(2^3)=f(2)^3=3^3=27>25=5^2=f(3)^2=f(3^2)=f(9)$.

Let $f$(the $n$th prime) be the $(2n)$th prime. Then, $27<32$, while $f(27)=f(3^3)=f(3)^3=7^3=343>243=3^5=f(2)^5=f(2^5)=f(32)$.

Let $f(p)$ be the $p$th prime. Then, $f(8)=27>25=f(9)$, as in the first example.

Let $f$(the $n$th prime) be the $(n^2)$th prime. Then, $3<4$, while $f(3)=7>4=2^2=f(2)^2=f(2^2)=f(4)$.

But of course, specific examples (using inductive reasoning) are not enough to conclude that the answer is in fact "Yes".

答:
Building on the discussion in the comments of Teresa's answer, this follows from the six exponentials theorem. I will try to write a self-contained answer here.

First, as you say the given conditions imply that $f$ restricts to some strictly monotonic function on the primes, and is determined by its values on the primes. If $p_1, \dots p_k$ is any finite set of distinct primes, then strict monotonicity of $f$ means that if $\prod p_i^{e_i} < \prod p_i^{f_i}$ then $\prod f(p_i)^{e_i} < \prod f(p_i)^{f_i}$. Taking logarithms and subtracting, this gives

$$\sum q_i \log p_i < 0 \Rightarrow \sum q_i \log f(p_i) < 0, q_i \in \mathbb{Q}.$$

In other words, $f$ induces a strictly order-preserving linear map $F$ from the $\mathbb{Q}$-subspace of $\mathbb{R}$ spanned by $\{ \log p_i \}$ to the $\mathbb{Q}$-subspace of $\mathbb{R}$ spanned by $\{ \log f(p_i) \}$. This implies that $F$ is continuous (with respect to the subspace topology coming from $\mathbb{R}$), which gives that it must in fact be given by multiplication by some fixed real number $r$. Said another way:

Corollary: For real coefficients $r_i$, we have $\sum r_i \log p_i = 0 \Leftrightarrow \sum r_i \log f(p_i) = 0$. Equivalently, there is some $r$ such that $\log f(p_i) = r \log p_i$ for all primes $p_i$.

This is quite a strong condition. It's equivalent to the condition that $\frac{\log f(p)}{\log p} = \frac{\log f(q)}{\log q}$ for distinct primes $p, q$, which assuming Schanuel's conjecture (which implies that the logarithms of the primes are algebraically independent) gives $f(p) = p, f(q) = q$. (We can also assume the weaker four exponentials conjecture instead.)

But actually we can avoid Schanuel's conjecture by considering three distinct primes $p, q, r$ instead. The six exponentials theorem implies that if $f(p) \neq p$ then (using that unique prime factorization implies that the logarithms of the primes are linearly independent over $\mathbb{Q}$) the matrix

$$\left[ \begin{array}{cc} \log p & \log q & \log r \\ \log f(p) & \log f(r) & \log f(r) \end{array} \right]$$

has rank $2$. But we showed above that the second row must be a multiple of the first, so this matrix has rank $1$; taking contrapositives, it follows that $f(p) = p$, and since the argument works for any three distinct primes we get $f(p) = p$ for all primes $p$.


问:
Calculate area of a circle
\begin{align}
π &= 3.1415\dotsc\\
s&= πr^2
\end{align}
Because $π$ infinitely continues, does it means $s$ is not ever a right answer, does it mean we do not know exact $s$ of a circle? Is there any other way that we find exact value of $s$?

答:
In the math universe, $\pi$ times $r$ squared is an exact number whenever $r$ is known. Real numbers result from a logical construction and $\pi$ is a well-defined thing.

In the physical world, there are no perfect circles nor any length that can be exactly measured and the question is irrelevant.

If for you "to find the exact value" means to enumerate the decimals, you will be dead long before a substantial number has been listed. In fact, this will never arise. On the other hand, $3.1416$ is a better-than-enough approximation for all real-life needs.

Final remark:

The area of a circle of radius $\dfrac1{\sqrt\pi}$ is exactly $1$.

Title: Calculate area of a circle


答:
You can write the sequence recursively as

$$x_0 = 3$$

$$x_n = 2x_{n-1} - 1 $$

To write it non-recursively:

Consider a new sequence

$$3, 6, 12, 24, 48, 96, ...$$

and your original:

$$3, 5, 9, 17, 33, 65, ...$$

Taking the difference of the two sequences we have

$$0, 1, 3, 7, 15, 31, ...$$

which is $g(x) = 2^{x-1} - 1$.

The top sequence is $3 * 2^{x-1}$. Subtracting $3 * 2^{x-1} - (2^{x-1} - 1)$.

Therefore, your function is

$$ f(x) = 2^x + 1$$

Also, note that you could also add $1$ to $2^x$ and get the same sequence.

Title: How do I write the rule for this sequence that is neither arithmetic nor geometric as an equation?

Body:
I have the sequence $3, 5, 9, 17, 33, 65,\dots$

I know the rule is multiply by $2$ then subtract one. $3$ times $2$ is $6$ minus $1$ is $5$ and so on. Easy enough.

How do I write the rule for this sequence as an equation?  I know it is neither arithmetic or geometric.


问:
Change of Variable with Jacobian
Let $\Omega \subset  \mathbb{R}^m$ be some compactly contained domain in $\mathbb{R}^m.$ Let $Q_t(x) = x+t\zeta$ where $\zeta\in C^{\infty}_c(\Omega, \mathbb{R}^m)$ and $t$ is small enough so that $Q_t$ is a diffeomorphism on $\Omega.$ My goal is to compute,
$$\frac{d}{dt}\left(\int_{\Omega} |D(u(x+t\zeta))|^2dx\right)_{|t=0} = \int_{\Omega}\left(\frac{|Du|^2}{2}\operatorname{div}(\zeta) - \partial_{\alpha}u^i \partial_{\beta} u^i \partial_{\beta}\zeta^\alpha\right) dx$$
where $u\in H^{1}(\Omega, \mathbb{R}^n),$ and we implicitly sum over indices $1\leq \alpha, \beta \leq m$ and $1\leq i \leq n.$

I am reading a proof where the author considers the map, $u_t(x)=u(Q_t^{-1}(x))$ and argues that,
$$\int_{\Omega} |D(u(x+t\zeta))|^2dx =\int_{\Omega} |Du_t(x)|^2dx.$$
I am not sure how to prove this, I tried change of variable, by setting $y=Q_t(x)$ in the first integral, to get $x=Q_t^{-1}(y)\implies dx = |\det DQ_t^{-1}(y)| dy$ and so,
$$\int_{\Omega} |D(u(x+t\zeta))|^2dx = \int_{\Omega} |Du(x+t\zeta) DQ_t(x)|^2 dx \\
=\int_{\Omega} |Du(y)|^2 |DQ_t(Q_t^{-1}(y))|^2 |\det DQ_t^{-1}(y)| dy\\
=\int_{\Omega} |Du(y)|^2 |DQ_t(Q_t^{-1}(y))|^2 |\det DQ_t^{-1}(y)| dy.$$

On the other hand, the author claims that,
$$\int_{\Omega} |Du_t(x)|^2dx = \int_{\Omega} |Du(x)|^2 DQ_t^{-1}(Q_t(x))|^2 |\det DQ_t(x)| dx.$$

How do I show that the two expressions are the same?

答:
The equality
$$\int_{\Omega} |D(u(x+t\zeta))|^2\,\mathrm{d}x =\int_{\Omega} |Du_t(x)|^2\,\mathrm{d}x$$
follows immediately from the equality $D(u(x+t\zeta)) = Du_t(x)$ by virtue that $u_t(x) = u(x+t\zeta(x)).$ This is largely a matter of notation, and there is no change of variable involved (and this is not actually needed to show your desired identity).

Your change of variables $y = Q_t(x)$ is also not quite correct, as you should have
$$ D(u(x+t\zeta(x)) = DQ_t(x)Du(Q_t(x)) = Du(Q_t(x)) + tD\zeta^i(x)D_iu(Q_t(x)) $$
using the same summation convention. Here we see that $DQ_t(x)$ is a linear operator acting on $Du(Q_t(x)),$ so $$|DQ_t(x)Du(Q_t(x))|^2 \neq |DQ_t(x)|^2|Du(Q_t(x))|^2$$
in general. Note that it can be instructive at this point to write everything in coordinates, as it may clarify what's going on.

Hence a change of variables should give
$$ \int_{\Omega} |D(u(x+t\zeta))|^2\,\mathrm{d}x =\int_{\Omega} |DQ_t(Q_t^{-1}(y))Du(y)|^2 |\det DQ_t^{-1}(y)| \,\mathrm{d}y \\
=\int_{\Omega} |Du(y) + tD\zeta^i(Q_t^{-1}(y))D_iu(y)|^2 |\det DQ_t^{-1}(y)| \,\mathrm{d}y.$$
This is not quite what you write at the end, and I don't believe the equality you quote holds true.

However the above does allow you to derive the identity you seek to prove. Note that $y$ is just a integration variable now, so you are free to write it as $y$ or $x;$ in particular it is independent of $t$ (which is important in what follows). Now observe that we have
$$ \frac{\mathrm{d}}{\mathrm{d}t}\left.\left( tD\zeta^i(Q_t^{-1}(x))D_iu(x)\right)\right|_{t=0} = D\zeta^i(x)D_iu(x),$$
since if the derivative lands on the $\zeta$ term what you get vanishes as you set $t=0.$ Hence differentiating under the integral sign we have
$$ \frac{\mathrm{d}}{\mathrm{d}t}\left.\left(\int_{\Omega} |D(u(x+t\zeta))|^2\,\mathrm{d}x\right)\right|_{t=0}=\int_{\Omega} \frac{\mathrm{d}}{\mathrm{d}t}\left.\left(|Du(x) + tD\zeta^i(Q_t^{-1}(x))D_iu(x))|^2\right)\right|_{t=0} \,\mathrm{d}x + \int_{\Omega} |Du(x)|^2\frac{\mathrm{d}}{\mathrm{d}t}\left.\left(|\det DQ_t^{-1}(x)|\right)\right|_{t=0}  \,\mathrm{d}x \\
= \int_{\Omega} 2 D_ju^kD_j\zeta^iD_iu^k + |Du|^2 \operatorname{div} \zeta \,\mathrm{d}x,$$
as required (up to a factor of $\frac12$). I've used Jacobi's formula to differentiate the second term, which I presume your text also explains.

Also for reference to anyone else reading this, the above computes the inner variations for harmonic maps in Euclidean space.

Title: Change of Variable with Jacobian


问:
Prove that for $S$ a maximal mean less subset, $|S|=\aleph$.
We say that  $S \subset \mathbb{R}$  is a mean less subset if for all $a \in S,  b \in S$ we get that $(a+b)/2 \notin S$.

Prove that for $S$ a maximal mean less subset, $|S|=\aleph$.

Clearly we know that $|S|\leq\aleph$.

I need a direction how to prove that it can't be that $|S|< \aleph$.

Thank you for your help.

答:
For $A\subseteq\Bbb R$ let $M(A)$ be the set of means of $S$:

$$M(A)=\left\{\frac12(a+b):a,b\in A\text{ and }a\ne b\right\}\,.$$

Clearly $|A|\le|M(A)|\le|A\times A|$, so $|M(A)|=|A|$ whenever $A$ is infinite.

Now suppose that $S\subseteq\Bbb R$, $S\cap M(S)=\varnothing$, and $\le|S|<|\Bbb R|$. The finite case is easy, so I will assume that $S$ is infinite. Let

$$D=\{|x-y|:x,y\in S\text{ and }x\ne y\}\,;$$

then $|D|=|S|$. (The argument is the same as for $|M(S)|$.) Let

$$T=\{x\pm d:x\in S\text{ and }d\in D\}\,;$$

$|T|=|S|$, so $|S\cup M(S)\cup T|=|S|<|\Bbb R|$. Let $x_0\in\Bbb R\setminus\big(S\cup M(S)\cup T\big)$, and let $S'=S\cup\{x_0\}$; clearly $S\subsetneqq S'$, and it’s not hard to see that.

$$M(S')=M(S)\cup\left\{\frac12(x+x_0):x\in S\right\}\,.$$

Suppose that $x\in S'\cap M(S')$; then there are distinct $y,z\in S'$ such that $x=\frac12(y+z)$.

If $x\in S$, then $x\notin M(S)$, one of $y$ and $z$ is $x_0$, and the other is in $S$. Without loss of generality let $z=x_0$, and let $d=|x_0-x|$; $|x-y|=|x_0-x|$, so $d\in D$, and $x_0\in\{x-d,x+d\}\subseteq T$, which is impossible.

If $x=x_0$, then $x_0\in M(S)$, which is also impossible.

Thus, $S'\cap M(S')=\varnothing$, $S'$ is a mean-less subset of $\Bbb R$, and $M$ is not a maximal mean-less subset of $\Bbb R$.


问:
which of the followings are positive definite:
(CSIR-UGC NET Mathematical Sciences-2011)

Suppose $A,B$ are $n \times n$ positive definite. Then which of the followings are positive definite:

$A+B$

$ABA^{*}$

$A^2+I$

$AB$

答:
Partial answer:
Your complete answer can be given by combining these two links

link 1

link 2

and possible my partial answer

If $A, B \in \mathbb{R}^{n\times n}$ are positive definite then $x^t A x > 0$, $x^t B x > 0$,
$\forall x \neq 0  $. This implies that $x^t (A + B)x = x^t A x+ x^t B x > 0$. Hence sum of two positive definite matrices positive definite.


问:
$A_1\subseteq B(\mathscr H_1)$, $A_2\subseteq B(\mathscr H_2)$ are isomorphic $C^*$-algebras. $A_1$ is weakly closed. Is $A_2$ weakly closed?
I guess no?

Let $A$ be commutative and let $B$ be $C(X)\subseteq B(l^2(X))$, where $X$ is compact Hausdorff. Suppose $B$ is weakly closed, take an arbitary $x\in X$, there is a net of operators in $B$ which weakly converges to $1_{\{x\}}$:

$$\mathscr Y=\{Y\subseteq_{finite} X\backslash \{x\}\}\\f_Y(y)=\left\{\begin{array}{}1&,y=x\\0&,y\in Y\\ others&,y\notin Y\cup \{x\}\end{array}\right.$$

Therefore $1_{\{x\}}\in B$. However, $A$ may have no minimal projections.

Is this counter example correct?

But isn't it kind of wired that being isomorphic to a Von Neumann algebra doesn't make itself a Von Neumann algebra?

答:
I think your counter-example is not correct since your initial assumptions ended up implying that $X$ is discrete
$\big (1_{\{x\}}\in  B\big )$, hence $X$ is finite, so $C(X)$ is a von Neumann algebra, after all.

According to the original definition of von Neumann algebras it doesn't make sense to say that "$M$ is a von Neumann",
rather one would say that "$M$ is von Neumann algebra of operators on  $H$",  where $H$ is a Hilbert space.  In 1956 Sakai found an
intrinsic characterization of von Neumann algebras according to which a C*-algebra $A$ admits a faithful representation
on some Hilbert space $H$ whose range is a weakly closed algebra of operators on $H$ iff $A$ is a dual Banach space.

Sakai's result may then be interpreted as a definition of von Neumann algebras which is representation independent.

I  believe your question can be interpreted as asking whether or not an abstract von Neumann algebra admits a faithful
representation whose range is not a von Neumann algebra.

It is not hard to produce examples of such representations for any given infinite-dimensional von
Neumann
algebra $M$.  All you have to do is choose a non-normal state and consider the corresponding  GNS representation,   adding
to it a copy of
any old faithful representation to make the whole thing faithful.  If $\xi $ is the cyclic vector, then the map
$$
  T\in M\mapsto \langle T\xi , \xi \rangle \in \mathbb C
  $$
will fail to be normal, attesting to the fact that $M$ is not a von Neumann in that representation.


问:
Finite ultraproduct
I stucked when trying to prove:

If $A_\xi$ are domains of models of first order language and $|A_\xi|\le n$ for $n \in \omega$ for all $\xi$ in index set $X$ and $\mathcal U$ is ultrafilter of $X$ then $|\prod_{\xi \in X} A_\xi / \mathcal U| \le n$.

My tries:

If $X$ is finite set then $\mathcal U$ is principal. Then singleton $\{x\}\in \mathcal U$ and $|\prod_{\xi \in X} A_\xi / \mathcal U| = |A_x|$. If $\mathcal U$ is not principal then for $x \in X$ there is $S_x \in \mathcal U$ with $x \notin S_x$. Then for every $k \in \omega$ there exists equivalence class corresponding to $S_{x_1} \cap \dots \cap S_{x_k}$ with size greater $|A_1|\cdot \dots \cdot |A_k|$.

Can there be said anything about a structure of the ultrafilter if $X$ is infinite? And how to prove it?

答:
The statement you are trying to prove is a consequence of Łoś's theorem - if every factor satisfies "there are no more than $n$ elements", then the set of factors that satisfy it is $X$, which is in $\mathcal{U}$, so by Łoś's theorem the ultraproduct will satisfy that sentence as well. Note that "there are no more than $n$ elements" is the sentence
$$
(\exists x_1)\cdots(\exists x_n)(\forall y)[ y = x_1 \lor \cdots \lor y = x_n]
$$

Thus one way to come up with a concrete proof of the statement you want is to examine the proof of Łoś's theorem and specialize it to the situation at hand.

As a side note, if every factor is finite, but there is no bound on the sizes of the factors, then the ultraproduct will not be finite. The difference is that there is no longer a single sentence of interest that is true in all the factors, because finiteness is not definable in a first-order language.

I assume that the OP figured out the hint, so let me spell out the answer for reference. Because $|A^\xi| = k$ for all $\xi \in X$, we can write $A^\xi = \{a^\xi_1, \ldots, a^\xi_k\}$ for each $\xi$. For $1 \leq i \leq k$ define $\alpha_i$ by $\alpha_i(\xi) = a^\xi_i$. Then every $\beta$ in the ultraproduct is equal to $\alpha_i$ for some $1 \leq i \leq k$. Proof: For $1\leq i \leq k$ let $B_i = \{\xi : \beta(\xi) = a^\xi_i\}$. Then
$$X = B_1 \cup B_2 \cup\cdots\cup B_k.$$
Because $\mathcal{U}$ is an ultrafilter, one of the sets $B_i$ must be in $\mathcal{U}$, and if $B_i \in \mathcal{U}$ then $\beta = \alpha_i$ in the ultraproduct, QED. Thus we can explicitly name the $k$ elements of the ultraproduct: $\alpha_1, \alpha_2, \ldots, \alpha_k$.

Title: Finite ultraproduct


问:
Getting from $\frac{x-4}{x-5}$ to $1+\frac{1}{x-5}$
Pretty embarrassing question lol.

How do you get from $\;\dfrac{x-4}{x-5}\;$ to $\;1 + \dfrac{1}{x-5}\;$?

My calculus teacher did without explaining how, and its been 3 years since I took a non-stat focused class. Feel free to ridicule as necessary.

答:
Let's say I have a rational function $$f(x) = \frac{A(x)}{B(x)}$$ so that the denominator polynomial is of degree no greater than the numerator polynomial: $\deg A \geq \deg B$.

This is the rational function equivalent of an "improper fraction".

For many purposes, particularly to calculate indefinite integrals, it can be useful to write this rational function as a sort of "mixed function" analogous to a mixed number via polynomial division. "Improper" or not, any rational function $f(x)$ may always be uniquely expressed as $$f(x) = Q(x) + \frac{R(x)}{B(x)},$$ where $Q(x)$ is the quotient and $R(x)$ is the remainder when $A(x)$ is divided by $B(x)$. $R(x)$ is guaranteed to have degree strictly smaller than $B(x)$.

Once $f(x)$ is expressed in this form, it's much easier to antidifferentiate: $Q(x)$ can be very easily antidifferentiated using Power Rule, and there exist various techniques (such as partial fractions and/or inverse tangent substitution) to compute $R(x)/B(x)$ in general.

This also explains where the horizontal asymptote rule comes from: $f(x)$ has a horizontal asymptote if and only if $Q(x)$ is a constant, which implies $\deg(A) \leq \deg(B)$.

If $\deg(A) = \deg(B)$, $Q(x)$ is the ratio of the leading coefficients; if $\deg(A) < \deg(B)$, then $Q(x) = 0$.

In the specific case you mention, long division is overkill. We notice by inspection that the numerator is just the denominator plus a constant: $$(x - 4) = (x - 5) + 1,$$ and as someone pointed out in the comments, the result simplifies to $$\frac{(x-4)}{(x-5)} = \frac{(x-5)}{(x-5)} + \frac{1}{x - 5} = 1 + \frac{1}{x-5},$$ which is very easy to antidifferentiate using the natural log.


问:
Intersection of the images is trivial, then kernels span the full space?
Suppose $\mathbf{M}_i$ is a set of distinct (no matrix is a multiple or linear combinations of each other) $n$ by $n$ matrices for $i=1,...,n$ and each $\mathbf{M}_i$ is of rank $n-1$. Is it ture that $\bigcap_{i}^n \mathbf{M}_i$ is trivial if and only if the $\ker \mathbf{M}_1+...+\ker\mathbf{M}_n$ is the full vector space?

答: Define linearly independent rank one matrices $M_1,M_2\in \mathbb{R}^{2\times 2}$ by $M_1=\big(\begin{smallmatrix} 1 & 1\\ 2 & 2 \end{smallmatrix}\big)$ and $M_2=\big(\begin{smallmatrix} 1 & 1\\ 3 & 3 \end{smallmatrix}\big)$. Note that we have $$\ker (M_1) = \ker (M_2) =\text{span}\Big\{\big(\begin{smallmatrix} 1\\ -1 \end{smallmatrix}\big)\Big\}$$ $$\text{Col}(M_1) =\text{span}\Big\{\big(\begin{smallmatrix} 1\\ 2 \end{smallmatrix}\big)\Big\}$$ $$\text{Col}(M_2) =\text{span}\Big\{\big(\begin{smallmatrix} 1\\ 3 \end{smallmatrix}\big)\Big\}$$ Hence $\text{Col}(M_1)\cap \text{Col}(M_2)$ is trivial and $\ker(M_1) + \ker (M_2)\neq \mathbb{R}^2$. So your statement is false.


问:
How do I write the rule for this sequence that is neither arithmetic nor geometric as an equation?
I have the sequence $3, 5, 9, 17, 33, 65,\dots$

I know the rule is multiply by $2$ then subtract one. $3$ times $2$ is $6$ minus $1$ is $5$ and so on. Easy enough.

How do I write the rule for this sequence as an equation?  I know it is neither arithmetic or geometric.

答:
We are given $$a_{n+1}=2a_n-1\tag1$$

Equivalently, $$a_{n+1}-2a_n=-1\tag2$$

and $$a_n-2a_{n-1}=-1.\tag3$$

Subtracting $(2)-(3)$ yields
$$a_n-3a_n+2a_{n-1}=0\tag4$$

The characteristic equation is $r^2-3r+2=(r-1)(r-2)=0$.

Therefore, the solutions are $a_n=A+B\cdot2^n.$

Can you use the first two elements of the series to solve for $A$ and $B$?


问:
Differentiability of Sin inverse
My book says that trigonometric functions and their inverses are differentiable in their domain.

I am looking for $y=\sin^{-1}(x)$ to verify the same. $\sin^{-1}(x)$ has its  domain in interval [-1.1]. But if I look at the derivative of $\sin^{−1}(x)$ as follows

$$\frac{1}{\sqrt{(1-x^2)}}$$

If my $x$ is $-1$ or $+1$ then the expression will becomes infinity. So I can say that it is not differentiable at $+1$ and $-1$. But I am contradicting my book. Am I going wrong somewhere in understanding this?

答: Quoting from the book : Concepts, Problems and Solutions in School Calculus (by Lev Tarasov and Chandra Shekhar Kumar)

Title: Differentiability of Sin inverse


问:
Proof that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$ for open $H$
This appears in W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st ed. 1975), chapter 3, exercise 3.9: 27 (iii).

"If $H$ is open in $T$ and $K \subseteq T$ then:
$$H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$$"

(where $\operatorname{cl}$ denotes closure and $T$ is a topological space.)

It's obvious that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$ but then it does not follow that $\operatorname{cl}(H) \cap \operatorname{cl}(K) = \operatorname{cl}(H \cap K)$.

(In fact we have $\operatorname{cl}(H \cap K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$.)

So that line of approach does not look very fruitful.

So I drew it up as a Venn diagram to get inspiration, and I got:

$$H \cap \operatorname{cl}(K) = H \cap (K \cup \delta K)$$

where $\delta K$ is the boundary of $K$.

This leads us to:

$$H \cap \operatorname{cl}(K) = (H \cap K) \cup (H \cap \delta K) \subseteq (H \cap K) \cup (H \cap \delta K) \cup (\delta H \cap K)$$

and "all that remains to be done" is prove that $(H \cap \delta K) \cup (\delta H \cap K) \subseteq \delta (H \cap K)$ (because $(H \cap \delta K) \cup (\delta (H \cap K)) = \operatorname{cl}(H \cap K)$)

but it's starting to get messy and I'm unsure of my ground.

Besides, this does not make any use of the fact that $H$ is open.

I'm pretty sure there's an insight needed here which I have not been able to come up with.

答: Hint: Let $x\in H\cap\overline K$. Since $x\in\overline K$, every neighborhood of $x$ contains a point from $K$. Since $H$ is a neighborhood of $x$ [fill in the gaps] it follows that every neighborhood of $x$ contains a point from $H\cap K$.


问:
Finding a function with Fourier expansion
I have to find the function for a sawtooth wave form, and using that the line has these conditions:

$$
f(x)=
\left\{
  \begin{array}{rcr}
    x & 0\leq x \leq 1 \\
    x-0.5 & 1 \leq x \leq 2 \\
    x -1 & 2 \leq x \leq 3 \\
  \end{array}
\right.
$$
After $x=3$ the function just repeats itself.
I compute the integrals for the coefficients $a_n$ and $b_n$ and get the following for the full Fourier expansion.
$$F(x)=\frac{a_0}{2}+\sum_{n=1}^\infty \frac{(0.5n\sin(2n)+0.5n\sin(n)+2n\sin(3n)+\cos(3n)-1)\cos(nx)+(\sin(3n)-0.5n\cos(2n)-0.5n\cos(n)-2n\cos(3n))\sin(nx)}{n^2}$$

when I plot this I get the following function:

Why doesn't the function repeat after $x=3$? My function also seems quite complicated, is it possible theres a simpler method?

答:
So I solved my problem with the help of 'Semiclassical' in the comments and I get the correct periodic function now when I plot it. My mistake was using a $2\pi$ periodic function instead of a $3$-periodic function. To solve this I only had to add $2\pi$ to the sine and cosine in the Fourier expansion. The correct form of the series for my case was:

$$
f(x)=C+\sum_{n=1}^{\infty }\left( a_{n}\cos \frac{2{\pi}nx}{T}+b_{n}\sin \frac{2{\pi}nx}{T}\right), \tag{1}
$$

and the Integrals for the coefficients had the same change to the cosine and sine:
$$
a_{n} =\frac{1}{\pi }\int_{-\pi }^{\pi }f(x)\cos \left(\frac{2{\pi}nx}{T}\right)
$$

the same applies to $b_{n}$ and $a_{0}$.


问:
How To Learn Mathematics Basics
I wanna learn mathematics from zero to advanced which books should I read and which websites should I visit?</p>

答:
I took quick look at your profile and some of your other questions on Stack Overflow.

I admire your zeal, and your fearless attitude of just "jumping into" the material.

That being said, I want to stress that the world of Mathematics (just as with software development), is a vast vast ocean. Study any one of these for a lifetime, and you will barely have scratched the surface.

Read more and you will discover that for each answered question, ten new ones pop up, and you will realize soon enough that you only grasp a tiny drop in a vast sea of the unknown.

There is no way to learn "all of math -- zero to advanced". You learn a little about a particular topic, often you go back and learn the same topic over and over with ever-increasing depth. You dabble a little outside your area of expertise.

Start with math that might be interesting for web development (google discrete math), if that interests you, but be warned that a great deal of web dev work requires very little math, and your time might be better spent elsewhere.

All the best of luck!

Title: How To Learn Mathematics Basics


答:
I think your confusion is about how we represent numbers. The point is - as you know, ${\pi}$ is an irrational number. We cannot write it as a fraction, and thus we also cannot write down the full number as a decimal number. It is not going to be possible to write it out fully in this form.

Just because we cannot represent a number in a way we would like to, that doesn't mean the number "doesn't exist", and we have created these symbols to deal with the fact that numerical representation becomes a problem. Symbolic representation is much more powerful. If you want to represent the area (or $\pi$) as a decimal number, you are right in saying it will just be an approximation. And there is no way around that. We know $\pi$ is irrational.

Just as @Yves Daoust has pointed out though - in the real world, circles aren't perfect. So the area of any "real world" circle will never be actually known, using ${\pi r^2}$ means we are approximating the circle as a perfect Mathematical circle. This is also what they do in Physics - they do a lot of "assume this is perfect" approximations, but we still can quite accurately describe the world around us!

Title: Calculate area of a circle

Body:
\begin{align}
π &= 3.1415\dotsc\\
s&= πr^2
\end{align}
Because $π$ infinitely continues, does it means $s$ is not ever a right answer, does it mean we do not know exact $s$ of a circle? Is there any other way that we find exact value of $s$?


问:
Time complexity (easy)
I know that this platform is probably not the best place for this question, on stack overflow I didn't receive any answers. Consider the following code snippet:

```<code>for (int i = n; i >= 1; i = i / 2) {
    for (int j = 1; j <= i; j = 2 * j) {
        f();
    }
}
</code>```

We want to determine the number of times f() get's called. My solution:
$$
\sum_{i=1}^{\log _{2}(n)} \sum_{j=1}^{\log _{2}(i)} 1 = \sum_{i=1}^{\log _{2}(n)} \log _{2}(i)$$
Apparently the solution is $\Theta(\log^2(n))$, how can I reduce the above further?

答:
I agree with Ross's answer, the simplest is indeed to bound $$\sum\limits_{i=1}^{\log_2(n)}\log_2(i)\le\sum\limits_{i=1}^{\log_2(n)}\log_2(n)=\Theta(\log_2(n)^2)$$

The sum is not directly calculable but roughly speaking we can invoke the comparison sum and integral.

Take for instance $\sum\limits_{i=1}^n i=\dfrac{n(n+1)}{2}=\Theta(n^2)$ and the integral is $\displaystyle \int_1^n x\,dx=\bigg[\frac 12x^2\bigg]_1^n=\Theta(n^2)$ as well.

Similarly since an antiderivative of the natural logarithm is

$\displaystyle\int_1^{\ln(n)}\ln(x)\,dx=\bigg[x\ln(x)-x\bigg]_1^{\ln(n)}=\Theta\big(\ln(n)\ln(\ln(n))-\ln(n)\big)=\Theta\big(\ln(n)\ln(\ln(n)\big)$

We can expect a result $\Theta\big(\log_2(n)\log_2(\log_2(n))\big)$ for your nested loops, which is a bit lower than $\Theta(\log_2(n)^2)$, but the latter is good enough in first approximation.

Title: Time complexity (easy)


答:
I did not read your proof since I believe is too short by using the definition of $\text{cl}.$ Let $x\in H\cap \text{cl(K)}$ then $x\in H$ and $x\in\text{cl(K)}.$ We need to show that $x\in\text{cl}(H\cap K),$ that is, for any open set $G$ and $x\in G$ we $$G\cap(H\cap K)\neq\emptyset$$
Notice that $G\cap H$ open set and $x\in G\cap H$. But $x\in\text{cl}(K)$. So, we have $(G\cap H)\cap K\neq\emptyset.$ Hence, $H\cap\text{cl}(K)\subset \text{cl}(H\cap K)$

Title: Proof that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$ for open $H$

Body:
This appears in W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st ed. 1975), chapter 3, exercise 3.9: 27 (iii).

"If $H$ is open in $T$ and $K \subseteq T$ then:
$$H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$$"

(where $\operatorname{cl}$ denotes closure and $T$ is a topological space.)

It's obvious that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$ but then it does not follow that $\operatorname{cl}(H) \cap \operatorname{cl}(K) = \operatorname{cl}(H \cap K)$.

(In fact we have $\operatorname{cl}(H \cap K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$.)

So that line of approach does not look very fruitful.

So I drew it up as a Venn diagram to get inspiration, and I got:

$$H \cap \operatorname{cl}(K) = H \cap (K \cup \delta K)$$

where $\delta K$ is the boundary of $K$.

This leads us to:

$$H \cap \operatorname{cl}(K) = (H \cap K) \cup (H \cap \delta K) \subseteq (H \cap K) \cup (H \cap \delta K) \cup (\delta H \cap K)$$

and "all that remains to be done" is prove that $(H \cap \delta K) \cup (\delta H \cap K) \subseteq \delta (H \cap K)$ (because $(H \cap \delta K) \cup (\delta (H \cap K)) = \operatorname{cl}(H \cap K)$)

but it's starting to get messy and I'm unsure of my ground.

Besides, this does not make any use of the fact that $H$ is open.

I'm pretty sure there's an insight needed here which I have not been able to come up with.


问:
Calculating of derivative
I'm a little confused with using derivatives. I understand that derivatives respond to question

how result of function changes when input of function changes

However I do not understand why we calculate derivative of $f(x) = x^2$ instead of calculating $f(2)$ and $f(2.001)$ and subtracting these two values of $f(2.001)$ from $f(2.0)$

Could someone explain that?

答:
Indeed, these are almost the same thing. The calculation
$$\frac{f(2.001)-f(2)}{0.001}$$
is an approximation to the exact rate of change at $x=2$. Note that the rate of change isn't just $f(2.001)-f(2)$; that's just the change in the value of the function. You need to divide by the amount of change in $x$ to get the rate of change; this is often expressed $\Delta y / \Delta x$.

For a much better approximation, you could calculate
$$\frac{f(2.00000001)-f(2)}{0.00000001}$$

Ultimately, the derivative is defined to be the limit of such expressions:
$$f'(2) := \lim_{h\to0}\frac{f(2+h)-f(2)}{h}$$

Finally, note that in addition to being exact (and not just an approximation), it is much simpler to calculate $f'(x)$, since in this example it is just $2x$. That being said, in the "real world", we don't always have nice functions like $x^2$, but rather only numerical data. In that case, you can approximate the derivative as above.

Title: Calculating of derivative


问:
Is $f(x) = x^2$ an isometry?
I am wondering if $f(x) = x^2$ is an isometry? My definition of an isometry is the following:

Let $(X, d)$ and $(\dot{X}, \dot{d})$ be metric spaces. $f: X \rightarrow \dot{X}$ is an isometry if:

$\dot{d}(f(x), f(y)) = d(x,y),   \forall x,y \in X$

For my question I would choose:

$f(x) = x^2$

$X = \mathbb{R}$

$d(x, y) = |x-y|$

$\dot{X} = \mathbb{R}_{\geq0}$

$\dot{d}(x,y) = ||\sqrt{x}|-|\sqrt{y}||$

It is also true that every isometry is injective. However, $f(x)=x^2$ is not injective which is clear.

Could you tell me what I am missing here? Thanks in advance.

Title: Is $f(x) = x^2$ an isometry?

Body:
$d(1,-1) = 2$

$\dot{d}(f(1),f(-1)) = 0$


问:
Taylor expansion of a function
I would like to approximate the function

$f(x)=\frac{2x}{1-e^{-2x}}$

analytically for both small and large $x$.
But when I use the formula for the Taylor expansion, I run into the problem that the function and its derivative are not defined for $x=0$. How can I get around this problem?

For large $x$, my idea was to substitute $y:=1/x$ and then expand the function $g(y)=\frac{2}{y(1-e^{-2x})}$ around $y=0$. However, here I run into the same problem as above.

How do you proceed in such a case?

答:
$$ \frac{2x}{1-e^{-2x}} = 2x\frac{1}{1-e^{-2x}} $$

We can then use $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} $ to get

$$
2x\sum_{n=0}^\infty e^{-2nx}
$$

We then use the Taylor series of $e^x$ to get

$$
\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{m}2^{m+1}n^{m}x^{m+1}}{m!}=  2x+\sum_{m=1}^\infty \frac{(-1)^{m}2^{m+1}x^{m+1}}{m!}\sum_{n=0}^\infty n^m
$$

Where we take the m=0 case out for future avoidance of poles. We can then use $\sum_{n=0}^\infty n^m = \zeta (-m) $ then use the reflection formula for the Riemann-Zeta function.

$$
\zeta (-m) = -2^{-m}\pi^{-1-m}sin(\frac{\pi m}{2})\Gamma (m+1)\zeta (1+m)
$$

To get

$$
2x + 2\sum_{m=1}^\infty \frac{(-1)^{m+1} \zeta (2m+2) x^{2m+2}}{\pi^{2m+2}} = 2x+2\sum_{m=2}^\infty (-1)^m \zeta(2m) \frac{x^{2m}}{\pi^{2m}}
$$

This can also be expressed in terms of Bernoulli numbers

$$
2x-\sum_{m=2}^{\infty} \frac{B_{2m}4^m x^{2m}}{(2n)!}
$$

Title: Taylor expansion of a function


问:
Does differentiability in an open interval imply continuity at end points?
It is well known that "differentiability at a point implies continuity at that point". I am thinking about the following proposition:

A function $f: \left[a,b\right] \to \mathbb{R}$ is differentiable over the open interval $\left(a,b\right)$, then it is continuous at $a$ and $b$.

Intuitively, I cannot think of a counter example for this proposition. Can anyone provide a proof or hints for a proof?

答:
Your proposition is wrong. Differentiability in the interior does not give you anything at the boundary. Consider, for example $f \colon [a,b] \to \mathbf R$ with
$$ f(x) = \begin{cases} 0 & x \in \{a,b\}\\ 1 & x \in (a,b) \end{cases} $$
Then $f$ is even analytic on $(a,b)$ but not continuous at $a$ and $b$.


问:
$\dfrac{O(1)}{x} = O(\dfrac{1}{x})$?
I just saw a lecture where the professor did this simplification in a proof but did not comment it further. How would you go about provning the statement in the title?</p>

答:
For the $x\to\infty$ case, the following statements are equivalent:

$f\in\frac{O(1)}{x}$

$xf\in O(1)$

There exist $x_0,\,M$ with $M>0$ such that for all $x>x_0$, $|xf(x)|<M$

There exist $x_0,\,M$ with $M>0$ such that for all $x>x_0$, $|f(x)|<M|1/x|$

$f\in O(1/x)$

You can handle $x\to0$ or $x\to0^+$ similarly.

Title: $\dfrac{O(1)}{x} = O(\dfrac{1}{x})$?


问:
cutting $2$ rods to $2$ pieces probability that $3$ chosen pieces form a triangle
We have two rods of length $a$ and $b$ ($b> a$) we randomly cut each rod to $2$ pieces. Then we choose $3$ pieces of $4$ pieces. What is the probability that these pieces form a triangle?

I think if we consider first piece if a as $x$ then $x$ has uniform distribution $(0,a)$ and the other piece has length of $a-x$. For the other rod we consider one piece as y and the other as $b-y$. We have $4$ different combinations for choosing pieces . I think I have to consider each combination, then find the probability of making triangle using triangle inequality but I don't know how to apply it. Also I think I don't have to calculate for each of $4$ states (because of symmetry).

答:
WLOG, assume $a=1$.
There are two cases: the left out piece corresponds to the long rod (length $b$), or it corresponds to the short rod (length $1$). Both cases are equiprobable.

Denote by $x \in [0,1]$ the cutting position for the short rod, $y \in [0,b]$ the cutting position for the long rod.

Lets compute the probability of success (a triangle can be formed) assuming first case.

The lengths of the pieces are then $a=x$, $b=1-x$, $c=y$. The triangle inequalities ($a<b+c$, $b<a+c$, $c<a+b$) give the conditions

$$[y<1], [y>2x-1], [y>1-2x] $$

Graphing this on the $x,y$ plane we get a triangle of area $1/2$, over the total zone (of uniform probability) of size $b$. Hence the probability is $$p=\frac{1}{2b}$$

As similiar calculation for the other case gives the same result.

In conclusion, in general:

$$p=\frac{\min(a,b)}{2 \max(a,b)}$$


答:
Take
$$f:[0,1] \to \Bbb R$$
defined by

$$f(x)=\frac{1}{x(1-x)}\; if \; 0<x<1$$
and

$$f(0)=f(1)=0$$

$ f $ is clearly differentiable at $ (0,1)$ with
$$(\forall x\in(0,1))\;\;f'(x)=\frac{2x-1}{x^2(1-x)^2}$$
but not continuous neither at $ 0 $ nor at $ 1$.

Title: Does differentiability in an open interval imply continuity at end points?

Body:
It is well known that "differentiability at a point implies continuity at that point". I am thinking about the following proposition:

A function $f: \left[a,b\right] \to \mathbb{R}$ is differentiable over the open interval $\left(a,b\right)$, then it is continuous at $a$ and $b$.

Intuitively, I cannot think of a counter example for this proposition. Can anyone provide a proof or hints for a proof?


问:
Calculating limit $\lim \frac{1}{x}\sin \frac{1}{x} $
$$\lim \frac{1}{x}\sin \frac{1}{x} = \infty $$ when $x \to 0$

But by the graph I doubt we should get an infinite oscillating discontinuity and so the limit does not exist here. But I am unable to point out the flaw. Please help or suggest.

答:
As a hint , you can take $\frac 1x=a$ so $$\lim_{x \to 0} \frac{1}{x}\sin \frac{1}{x} =\\
\lim_{a \to \infty} \frac{1}{\frac1a}\sin \frac{1}{\frac1a} =\\
\lim_{a \to \infty} a\sin a =?$$


问:
Average degree of $k$-degenerate graph is $\leq 2k$
How to prove the following claim?

Average degree of $k$-degenerate graph is $\leq 2k$

Definition: Graph is $k$-degenerate if for every $\,G' = (V',E') \subset G$ there exists $v
\in V'$ such that $\deg(v) \leq k$.

Formalization of the statement: $\sum_{v \in V}{\deg(v)}/ \lvert V \rvert \leq 2k$.

It seems to me there is some knowledge about $k$-degenerate graphs I am missing.

答:
Take graph G, as it's k-degenerate, there is a vertex of at most degree k, call it v. Removing v reduces the number of the edges by k and therefore reducing the sum of the degrees by at most 2*k.

Now, take the vertex u with the minimum degree of G / {v} and remove it; degree(u)<= k in G / {V}  as G is k-degenerate, therefore you will remove at most k edges, i.e. the sum of the degrees of the remaining graphs will be reduced by at most 2k. If you repeat this procedure for n-1 times, the sum of the degrees will be reduced by 2k * (n-1) and essentially this will be an upper bound on the sum of the degrees of G. Hence, the average degree would be 2k * (n-1)/n <= 2k.


问:
what is the general step to solve $f(x)\pm f(g(x))=m(x)$
I try with a simple example

I put $$f(x)=2x$$

and $$f(2x)=4x$$

so

$$f(x)-f(2x)=-2x$$

and I try to solve the last equation

$$f(x)-f(2x)=-2x$$

by put $$f(x)=e^{mx}$$

then the solution $$e^{mx}=\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}\neq 2x$$

I know that $$e^{mx}-e^{2mx}=\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}-(\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}})^{2}=-2x$$

but $$\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}\neq 2x$$

where is the wrong

and any way how to solve equation like $$f(x)\pm f(g(x))=m(x)$$

答:
The easiest way to solve this equation is if we assume that $f$ is continuous at $x=0$.

Then we have: $$f(x)=x+f(\frac{x}2) =x+\frac{x}{2} +\frac{x}{4} + \dots + \frac{x}{2^n} + f(\frac{x}{2^n})$$

The limit of the right side as $n\to\infty$ is $f(0)+2x$, which gives the general solution, $f(x)=C+2x$.

The problem with a general $g(x)$ is almost certainly intractable.


问:
GCSE: Find the surface of a volleyball ball
Can anyone help me?!

How many cm$^2$ of skin is needed to make a volleyball ball with a diameter of $30$ cm, if for cutting and sewing $12\%$ of the material is lost?

My attempt was the following:

I used the sphere surface formula:
$S=4\pi r^2 = 4(3,14)(225\text{cm}^2) = 2826 \text{cm}^2$

So, $\frac{12}{100} \cdot 2826 = 339,12cm$
Then, 2826+339,12 = 3165,12 cm

I don’t know if this is the right answer! All help/solutions appreciated. Thanks

答:
Your solution is not quite right, unfortunately. What you have calculated is $12\%$ of the surface area of the volleyball and added that on. Unfortunately, you don't lose $12\%$ of the surface area of the volleyball by cutting and sewing, you lose $12\%$ of the material you started with. Adding $12\%$ of the latter does not account for a loss of $12\%$ of the former. What you would want to solve is:

$0.88x = 2826$

where $x$ is the original fabric. Can you solve it from here?

Aside from this final step, everything else in your solution looks good.


答:
Let
$$f(x)=\frac 1x\sin(\frac 1x)$$

$$u_n=\frac{1}{n\pi}$$
$$v_n=\frac{1}{\frac{\pi}{2}+2n\pi}$$

then

$$\lim_{n\to+\infty}u_n=\lim_{n\to+\infty}v_n=0$$

$$\lim_{n\to+\infty}f(u_n)=0$$
$$\lim_{n\to+\infty}f(v_n)=+\infty$$

can you conclude.

Title: Calculating limit $\lim \frac{1}{x}\sin \frac{1}{x} $

Body:
$$\lim \frac{1}{x}\sin \frac{1}{x} = \infty $$ when $x \to 0$

But by the graph I doubt we should get an infinite oscillating discontinuity and so the limit does not exist here. But I am unable to point out the flaw. Please help or suggest.


答:
$\newcommand{\cl}[1]{\overline{#1}}$
Let $x \in H \cap \cl{K}$. $x \in H$ and $x \in \cl{K}$. The goal is to show that for every neighborhood $N_1$ of $x$, there is an element $y$ of $N_1$ that is also in both $H$ and $K$.

There is a neighborhood $N_2$ of $x$ that is contained in the open set $H$. Let $N_3 = N_1 \cap N_2$. Then $N_3$ is also a neighborhood of $x$ contained in $N_1$. Since $x$ is in $\cl{K}$, by definition of closure, there is an element $y$ of $N_3$ that is also in $K$. From $N_3 \subseteq N_1 \subseteq H$, we know that $y$ is in $H$, by which we showed the existence of such $y \in N_1$ that is in $ H\cap K$. So, we reached the goal.


答:
It may seem a good idea to try $y=f(x)=e^{mx}$, since then $f(2x)=e^{2mx}=y^2$ and then you get your quadratic equation from $f(x)-f(2x)=-2x$ in terms of $y$ after rearranging: $$ y^2-y-2x=0.$$
Now if $y$ had nothing to do with $x$ this would have the solutions $y=(1/2) \pm \sqrt{2x+1/4}$ which you mention. However when you put $y=e^{mx}$ back into this, you see it cannot lead to a solution, since if it did then the exponential function $e^{mx}$ would be expressed in terms of radicals.

Title: what is the general step to solve $f(x)\pm f(g(x))=m(x)$

Body:
I try with a simple example

I put $$f(x)=2x$$

and $$f(2x)=4x$$

so

$$f(x)-f(2x)=-2x$$

and I try to solve the last equation

$$f(x)-f(2x)=-2x$$

by put $$f(x)=e^{mx}$$

then the solution $$e^{mx}=\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}\neq 2x$$

I know that $$e^{mx}-e^{2mx}=\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}-(\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}})^{2}=-2x$$

but $$\frac{1}{2}\pm \sqrt{2x+\frac{1}{4}}\neq 2x$$

where is the wrong

and any way how to solve equation like $$f(x)\pm f(g(x))=m(x)$$


问:
If B is a subset of inner product space V, show orthogonal complement of B is a subspace of V
Let V be a inner product space and B be a subset (not subspace) of V.

Is then B┴ a subspace of V ?

There is a guy here that allegedly proves so, but I can't quite follow. Any heads-up or a counter-example?

答:
This is indeed true and not very difficult to prove. Just take the definition of $B^\perp$:
$$v \in B^\perp \Longleftrightarrow \forall\, b \in B: \langle v,b \rangle = 0.$$
Now ask yourself, if $v,w \in B^\perp$ and $\lambda \in \mathbb{C}$, do we have $\lambda v + w \in B^\perp$?


问:
Definition of the topology on $\mathbb{R}$
So I've just picked up learning topology and metric spaces and i am very confused on the definition of open sets in a topology.
I always thought that every set in a topology was open by definition so I was quite confused to see that when considering the topology on $\mathbb{R}$ i saw on a youtube video that the singletons for example $[0]$ is closed so it cannot be in the topology.
So does this mean that singletons are closed?.
However we can have singleton sets in a topology when considering a different set not the reals so what is going on why are singletons not included in the topology of the reals and what topology would we use on $\mathbb{R}$?.
This is probably a stupid question for most of you but im completely new and all these concepts baffle me.
Thanks in advance.

答:
By definition of a topology , the sets in it are going to be the open sets. And by definition the closed sets are going to be the complements of the open sets but it all depends on the choice of topology. A space can have a lot of topologies. In your video most likely you were working with the metric topology and there yes we do have that singletons are closed sets. But I could very well be working in the discrete topology were every set is open and closed, it all depends on the choice of topology . You can have singletons closed or open or both or neither it all depends in the choice of topology.

But yes most of the times people are working on what is called the metric topology of $\mathbb{R}$, for example when we do standard analysis we use this topology, and most of the times don't even  mention it.

I would recommend for you to take a look at Munkres' Topology; in my opinion the concepts are well explained and the initial part is easy for self-learning.


问:
Correctly describe a function when its codomain is a set of functions
I saw the following way of describing a function $f: A \rightarrow B$, defined as $f(a) = a$. How to correctly define a function when its codomain is supposed to be a set of functions?

For example, I need to define a function $g: N \rightarrow N^N$ to prove that it's injective. I'm thinking of two ways of defining this function:

Let $g: N \rightarrow N^N$, defined as $g(n) = f(k) = n$, where $f: N \rightarrow N$.

Let $f_i: N \rightarrow N$, defined as $f_i(n) = i$ for any $i \in N$. Then $g(n) = f_n$

Could someone please explain if there's something wrong with my definitions? The first way seems to be incorrect or at least confusing, because $f(k)$ looks like a result of $f(k)$ for some $k$. So I'm especially worrying about the second way. Is it correct to define something with subscript and use that subscript as if it is something real (like an argument of the function)?

答:
Your first version, taken literally as written, makes no sense. Your second is much better, but it can be improved:

For each $i\in\Bbb N$ let $f_i:\Bbb N\to\Bbb N:n\mapsto i$, and let $g:\Bbb N\to\Bbb N^{\Bbb N}:n\mapsto f_n$.

Or if you prefer a little more verbiage:

For each $i\in\Bbb N$ let $f_i:\Bbb N\to\Bbb N$ be the constant function $f_i(n)=i$, and define $g:\Bbb N\to\Bbb N^{\Bbb N}$ by letting $g(n)=f_n$ for each $n\in\Bbb N$.


问:
Solve in $\mathbb{R}$ the following equation:
$$(\frac{1}{\sqrt{x}-1})^\frac{1}{\sqrt{x}-1}=(1+\sqrt{x})^{1+\sqrt{x}}$$

My attempt:

$$\Leftrightarrow(\frac{1+\sqrt{x}}{x-1})^{\frac{1+\sqrt{x}}{x-1}}=(1+\sqrt{x})^{1+\sqrt{x}}$$

I tried writing $1+\sqrt{x}=t$, then $x=t^{2}-2t+1$ and $x-1=t^{2}-2t$

$$(\frac{t}{t^{2}-2t})^{\frac{t}{t^{2}-2t}}=t^{t}$$
$$\Leftrightarrow (\frac{1}{t-2})^{\frac{1}{t-2}}=t^{t}$$
$$\Leftrightarrow (\frac{1}{t-2})^{\frac{1}{t-2}}-t^{t}=0$$
And i got stuck here. Any help would be appreciated. I know that sometimes this type of exercises can be solved using the monotony of functions, but i don't think that is the case here. I tried with $0$ and $1$ but they aren't solutions.

答:
Observe that $x^x$ is strictly increasing for $x \ge 1$, and $1+\sqrt x \ge 1$.

Note also that for $x < 1$, $x^x < 1$ and for $x > 1$, $x^x > 1$.

Therefore if the equation holds, we must have $1 + \sqrt x = \dfrac 1{\sqrt x- 1}$,  which gives the unique solution $x=2$.


问:
How to cut a cube out of a tree stump, such that a pair of opposing vertices are in the center?
I saw this picture of a cube cut out of a tree stump.

I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it.

One of the opposing vertices pair is on the center of the tree stump:

I've been struggling to find the numbers and angle needed to make the cuts.

Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.

答:
Since you said you were finding difficult to find the numbers, I have naturally indicated how the numbers are calculated from 3d geometry on this maths site..

Vertex V is not yet located. An imaginary cone is shown.

Cylinder axis is along OV =h. Line OB is along x-axis. A,B,C are marked on a base circle 120 degrees apart. OA=OB=OC=r and  V is vertex of the cone and VA,VB,VC are mutually perpendicular. OV=h and semi-vertical angle of cone is OVB. The coordinates of A,B,V are

$$ (r,0,0),(-r/2, - r\sqrt 3/2 ),(0,0,h)$$

Position vectors $AV,BV$ are respectively
$$(r\;i,0j,-h\;k),\;(-r\;i/2, - r\sqrt 3/2,-h\;k)$$

Since VA,VB are orthogonal, dot product vanishes
$$ \dfrac{-r^2}{2}+h^2=0\to \boxed{\dfrac{h}{r}= \dfrac{1}{\sqrt 2} }$$

which is the first essential part of answer.

Next the  in-circle radius of base triangle CAB is OM = OB/2 =r/2. Angles VOM, VOB and OMB are right angles.

$$\tan OVM =  \dfrac{r}{2 h}=\dfrac{1}{\sqrt 2 } \to \angle OVM =  35.2644^{\circ}$$

The band-saw is aligned in plane passing through AB making the above angle OVM  to wooden cylinder axis and a saw cut is made holding the wooden cylinder stationary; repeated for BC, CA by rotating cylinder by $120^{\circ} $for other two planar saw-cuts. If alignment is made on the saw machine with respect to base plane then we  use its complement  $54.7356^{\circ}.$ Alternately the saw can touch center V and chop off three pieces at this angle leaving behind the trihedral angle cube corner.

The new three cut planes have $120^{\circ} $ dihedrals along VA,VB and VC with  $V$ as the common apex.

So three arbitrary ( along axis) cuts at $\angle OVM$ for each of three intersecting planes produce a cube vertex/corner. Three other faces on other side are generated in same way at 60, 180, 300 degrees cylinder rotation.

The above is sufficient to machine produce the cube shown in the question. It can be generalized.. Perhaps changed suitably to even cut a Kohi-noor type of diamond (with suitable stone/lapidary grinding machines) to grind produce several polyhedral faces.


问:
Probability questions with balls in a vase
Can anyone help me to solve this exercise ?

A bag contains six white and four black balls

a) You take two balls out. What is the chance for two white balls ?

b) You take three balls, and you don't replace them. What is the
chance for two white balls ?

c) You take three times a ball and put it back. What is the chance for
two white balls now ?

All solutions appreciated. Thanks

My solution

a) 6/10 x 5/9 = 5/15

b) 6/10 * 5/9 * 4/8

c) 6/10 * 6/10 * 6/10

I would like to know is that correct ?

答:
In total we have $10$ balls in which $6$ are white.

The chance for picking first white ball is $6/10$ Now if we take it out we have a total of 9 balls in which 5 are just white so the chance for another white one is $5/9$ which means in total we have $6/10 * 5/9$

Can you continue from here?


问:
Action of 3x3 invertible matrices on $\mathbb{C}$?
One of the many beautiful properties of mobius transformations includes the fact they form a natural map of equivalence classes (related by scalar multiplication of $2 \times 2$ complex matrices

$$\begin{pmatrix} a & b\\ c&d \end{pmatrix}  \ \text{~} \  \frac{ax+b}{cx+d}$$

This had got me wondering.. the space of meromorphic functions is pretty massive. Does there exist then a natural action of equivalence classes $3 \times 3$ complex matrices (up to multiplication by scalar) into complex functions?

I tried to explore and find some, outside of the mobius transformations and came across the group $z^{\theta}$ where $\theta \in \mathbb{C}, |\theta| = 1$ as an action of the circle group into the complex plane which generalizes naturally to the group $e^{\frac{a\log z +b}{c \log z + d}}$ (where $c=b=0$, $|a/d| =1$) corresponds to the circle group. But the general form suggests this isn't anything "new" from the mobius transformations we have already seen.

I suspect it may be possible there is no "elementary" set of functions (as in involving algebraic, and exponents), but then the problem gets a lot more complex since the space of meromorphic functions has a lot of exotic and complicated infinite series.

答:
After spending more time thinking about this we have two of the following results:

There is a natural action of $3 \times 3$ real matrices on the euclidean plane $\mathbb{R}^2$ given by:

$$ \begin{bmatrix} x \\ y  \end{bmatrix} \rightarrow \frac{1}{gx+hy+i} \begin{bmatrix} ax+by+c \\ dx+ey+f \end{bmatrix} $$

(It's a lot of work algebraically but one can verify that composing such transformations respects matrix multiplication). This implies by considering a real part and imaginary part function, denoted as $\text{Re}(z)$  and $\text{Im}(z)$ then an action of $3\times 3$ real matrices can be realized as a mobius transformations is given as

$$ z \rightarrow \frac{ (a\text{Re}(z)+b\text{Im}(z)+c) + i (d\text{Re}(z)+e\text{Im}(z)+f)}{g\text{Re}(z)+h\text{Im}(z)+j}   $$

For your choice of 9 real numbers $a,b,c,d,e,f,g,h,j$  (note that $i$ is also used for the imaginary constant so we skip it in the previous list).

But the introduction of real parts and imaginary parts this way is a little less than ideal, and this doesn't let us consider complex valued matrices.

Developing a functional equation trick:

The mobius transformations can identified as a class of smooth solutions to the following functional equation

$$ f: \mathbb{C}^5 \rightarrow \mathbb{C} $$

$$ f(a,b,c,d,f(e,f,g,h,z)) = f(ae+bg, af+bh, ce+dg, cf+dh,z)$$

(I conjecture solution to this all are of the form $\Omega^{-1} \left(  \frac{a \Omega(z)+b}{c\Omega(z)+d} \right) $, for a choice of bijection $\Omega: \mathbb{C} \rightarrow \mathbb{C}$. The case of $\Omega$ being the identity are the usual mobius transforms we encounter).

This strategy should be able to easily scale to let us consider

$$ f: \mathbb{C}^{10} \rightarrow \mathbb{C} $$

$$ f(A, f(B,z)) = f(AB, z) $$

where $A,B$ are $3 \times 3$ complex matrices that represent our 9 other arguments to $f$.

Developing strategies to extract solutions from these functional equations currently remains a mystery to me.