问:
What is an example and a non-example of a left-invariant vector field?
I am going through a few different books reviewing Lie Groups and Lie Algebras and came across the notion of a left-invariant vector field. Given a diffeomorphism $L_ag : G \to G, L_a(g) = ag.$ We then have a map from $T_gG$ to $T_{ag}G$ such that $L_{a*} X |_g = X|_{ag}.$ Additionally, we have in coordinates that $$X^u(g) \frac{\partial x^v(ag)}{\partial x^u(g)}\frac{\partial}{\partial x^v} |_{ag} = X^v(ag)\frac{\partial}{\partial x^v} |_{ag}.$$ However I have two issues. The fist is this definition seems so intuitively obvious I cannot think of a non-example (Note: I've checked previous Stackexchange questions on this matter and the solutions amounted to short comments that didn't fully answer my question) and the second is I am a bit shaky as to how use the coordinate definition to check if a vector field is or isn't left invariant.

Therefore, I am hoping someone could provide two examples. The first being an example of a left invariant vector field and the second a non-example. If you could show why these examples satisfy and fail to satisfy the requirements respectively I would be greatly apppreciative!

答: If you take the case that $G=(\mathbb R^n,+)$ then vector fields on $G$ are smooth functions $G\to\mathbb R^n$. Since the derivative of a translation is the identity, the conditions of left invariance reads as $f(x+y)=f(x)$ for any $x,y\in\mathbb R^n$. Thus left invariant vector fields correspond to constant functions.


问:
Maximum distance from a point in hyperbola $ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes
The canonical hyperbola is
$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

I have already derive the equation for the distance from any point $(x_0,y_0)$ in hyperbola to the asymptote $ y=\frac{b}{a}x$ as
$$ d =\frac{|bx_0-ay_0|}{c}$$
where
$a^2+b^2=c^2$ and $x_0 \geq a$.

I know that the maximum distance is when $x_0=a$ but I don't know how to proof it. Can you help me? Thanks.

答:
Note that the distance to the pair of the asymptotes is defined as the shorter of the two

$$ d =\min \left(\frac{|bx_0-ay_0|}{c} , \frac{|bx_0+ay_0|}{c}\right)
$$

Let the point on the hyperbola be $x_0=a\cosh t$ and $y_0=b\sinh t$. Then

$$d = \frac{ab}c \min(|\cosh t - \sinh t|, |\cosh t + \sinh t|)
=\frac{ab}c\left\{ \begin{array}{c} \cosh t - \sinh t , \>\>\>t\ge0\\ \cosh t +\sinh t,\>\>\> t<0\ \end{array} \right.$$

which is an even function of $t$ and has its maximum at the peak $t=0$, i.e. $x_0=a$.

ContentLicense: CC BY-SA 4.0


问:
pullback of an Euclidian inner product on $\mathbb{R^{2n+1}}$
If $M$ is a differentiable manifold of dimension n and there is a diffeomorphism $g:M \rightarrow \mathbb{R^{2n+1}}$, and we have $\langle\cdot,\cdot\rangle$ is the Euclidian inner product on $\mathbb{R^{2n+1}}$. Then the pullback of $g$ defines a metric on $M$ by
$t:=g*(\langle\cdot,\cdot\rangle)$, my question is:how is $t$ defined by? I mean can you write it as a mapping and it maps what to what...

答:
Let $M,N$ be two smooth manifolds, $f:M\to N$ a smooth map and let $S$ be a tensor field on $N$ of type $(0,k)$ (so at every point it eats $k$ elements of the tangent space and spits out a real number). Then, the definition of $f^*S$ is as follows: for every $p\in M$ and every $\xi_1,\dots, \xi_k\in T_pM$, we define
\begin{align}
(f^*S)_p[\xi_1,\dots, \xi_p]:= S_{f(p)}[f_{*,p}(\xi_1), \dots, f_{*,p}(\xi_k)]
\end{align}
where $f_{*,p}:T_pM\to T_{f(p)}N$ is the induced linear mapping between tangent spaces.

Recall that $S_{f(p)}:(T_{f(p)}N)^k\to \Bbb{R}$ is a multilinear map, and we're using this to define the multilinear map $(f^*S)_p: (T_pM)^k\to\Bbb{R}$.
If this is confusing, then you should certainly review the linear algebra questions of how a linear map between vector spaces induces mappings between the various tensor spaces, because all we're doing in the differential geometry setting is doing the linear algebra operations pointwise at every tangent space.

Now, suppose $M$ is a smooth manifold, $(N,h)$  is a Riemannian manifold, and $f:M\to N$ is an immersion (i.e for every $p\in M$, $f_{*,p}:T_pM\to T_{f(p)}N$ is injective). Then, we can pull-back the Riemannian metric on $N$ (which by definition is a $(0,2)$ tensor field) to get $f^*h$, which is also a Riemannian metric (since $f$ is an immersion). By using the definition above, what this says is for every $p\in M$ and every $\xi,\eta\in T_pM$, we have
\begin{align}
(f^*h)_p[\xi,\eta]&= h_{f(p)}[f_{*,p}(\xi),f_{*,p}(\eta)]
\end{align}
Or, if you want to denote the Riemannian metric as $h=\langle \cdot,\cdot \rangle$, and the pull-back as $t=f^*(\langle\cdot ,\cdot \rangle)$then
\begin{align}
t_p[\xi,\eta]&=\langle f_{*,p}(\xi),f_{*,p}(\eta)\rangle_{f(p)}.
\end{align}

Now, if you want the expression in components, that's also simple. Let's write $t=f^*h$, fix a point $p\in M$, a chart $(U,x)$ around $p$, and a chart $(V,y)$ around $f(p)$. Then,
\begin{align}
t_{ij}(p)&:= t_p\left(\frac{\partial}{\partial x^i}\bigg|_{p},\frac{\partial}{\partial x^j}\bigg|_{p}\right)\\
&=h_{f(p)}\left(\frac{\partial (y^a\circ f)}{\partial x^i}\bigg|_{p}\cdot \frac{\partial }{\partial y^a}\bigg|_{f(p)},
\frac{\partial (y^b\circ f)}{\partial x^i}\bigg|_{p}\cdot \frac{\partial }{\partial y^b}\bigg|_{f(p)}
\right)\\
&= \frac{\partial (y^a\circ f)}{\partial x^i}\bigg|_{p}\frac{\partial (y^b\circ f)}{\partial x^j}\bigg|_{p} h_{ab}(f(p))
\end{align}
where I've used the summation convention. If you want to omit the points of evaluation, then this can be written as
\begin{align}
t_{ij}&= \frac{\partial (f^*y^a)}{\partial x^i}\frac{\partial (f^*y^a)}{\partial x^i} f^*(h_{ab})
\end{align}
Or, the more common way to write this is to suppress the pullback symbol by saying $y=f(x)$; this yields the memorable formula (summation enforced):
\begin{align}
t_{ij}&=\frac{\partial y^a}{\partial x^i}\frac{\partial y^b}{\partial x^j}h_{ab}.
\end{align}

Finally, if you take $h$ to be the Euclidean metric, then in terms of cartesian coordinates (i.e the identity chart), we have $h_{ab}=\delta_{ab}$, so the above formula reduces to $t_{ij}=\sum_{a}\frac{\partial y^a}{\partial x^i}\frac{\partial y^a}{\partial x^j}$.


问:
Determine the closest point(s) from a line to a parametric cubic curve where the distance is less than $L$
Given a parametric cubic curve in 2D space:
$$\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$$
and a line defined by the equation:
$$px + qy + k = 0$$

I can easily determine where these two equations intersect by splitting up the cubic curve into its components and substituting them into the line equation:
$$p(A_xt^3 + B_xt^2 + C_xt + D_x) + q(A_yt^3 + B_yt^2 + C_yt + D) + k = 0$$

and solving this cubic equation.

This will give me where they intersect exactly, however, I need to determine where they nearly intersect within a specified distance tolerance i.e. I need the ranges of $t$ values where the distance between the line and the parametric curve is less than $L$.

Edit: I thought I'd include this as it might make the solution easier. I don't actually need a range of $t$ values, but rather only need the $t$ value that is closest to the curve in this range of values i.e. the parameter where the curve has a tangent vector parallel to the line (I think?).

How would I accomplish this? First thing that comes to mind is the distance formula, but I'm not too sure how I'd substitute the line equation into that, and it seems like things might get "messy" quickly and that there may be a better alternative.

I also thought that (when solving the cubic equation) I could analyse allow turning points that do not cross the x-axis (indicating they are close to producing roots but aren't quite) and consider the distance away this turning point is from the x-axis. However, I'm not too sure how I would relate this distance of the equation to the distance in 2D space.

Thanks!

答: Hint: given a line $L$ in the plane and distance $d$, there are two lines parallel to $L$ and at distance $d$ from it.  The endpoints of your desired intervals are points where the curve crosses those lines.


问:
$\sqrt{a^2 +(1-a)^2}+ \sqrt{b^2 +(1-b)^2} \geq \sqrt{2} $
$a,b \in \mathbb{R}$

My attempt:

If i can prove that:
$$x^2 +(1-x)^2 \geq \frac{1}{2}$$
So therefore:
$$\sqrt{x^2 +(1-x)^2} \geq \frac{\sqrt{2}}{2} $$
But i can’t prove the first inequality .

答: Why not, square it and put everything on LHS, get rid of the denumerator.  You get $$(2x-1)^2\geq 0$$ which is true.


问:
Inverse image of prime ideal in noncommutative ring
We say an ideal $P\neq R$ in a ring $R$ is prime, if for any two ideals $I,J$ of $R$ the following implication holds: if $IJ\subseteq P$ then $I\subseteq P$ or $J\subseteq P$.

If $f\colon R \to S$ is a homomorphism of noncommutative rings and $P$ is a prime ideal in $S$, is $f^{-1}(P)$ a prime ideal in $R$?

Of course, the above statement is true for commutative $R$ and $S$. I managed to prove it for noncommutative $R$ and $S$ when $f$ is an epimorphism.

答: By looking at the induced maps on quotient rings, the question is equivalent to: Is every subring of a prime ring also prime? There are lots of counterexamples. For example, if $R$ is an integral domain, then $M_2(R)$ is a prime ring, but the subring of diagonal matrices $\cong R \times R$ is not.

ContentLicense: CC BY-SA 3.0


问:
How many anagrams with a specific subword?
I'm trying to find the number of anagrams of "ACKNOWLEDGEMENT" (15 characters) that contain the subword "EDGE".

I know the total number of anagrams is $\frac{15!}{3! \cdot 2!}$. I'm not sure of what to do after this. I know this is an overcount.

答:
Hopefully you understand that the number of anagrams you have correctly given has the division by $3!\cdot 2!$ because of the $3$ E's and $2$ N's in the available letter pool.

To get the number of full anagrams containing "EDGE", you can simply regard $\fbox{EDGE}$ as a single "letter" that you anagram with the other $11$ items in the letter pool. Then you have $12$ items to permute with only the $2$ copies of N to adjust for, using the same calculation as you already exhibited.


问:
Contour integral of absolute value of a function
How is the contour integral with absolute value performed?

Here
$$\mathfrak{I}=\int_{C_1} \frac{a|z|}{z-\gamma}\mathrm{d}z$$
where the contour $C_1$ is parallel to the real line but passes above the point $z=\gamma$ and $a\in \mathbb{C}$

Is taking $z=R e^{i\theta}$ and hence $\mathrm{d}z=R i e^{i\theta}$ and taking the limit from $0$ to $\pi$ a correct approach?

That gives
$$\mathfrak{I}=\int_0^\pi \frac{aR}{Re^{i\theta}-\gamma}R i e^{i\theta}\mathrm{d}\theta=a R (\log (-\gamma -R)-\log (R-\gamma ))$$

but it does not seem to converge in the limit $R\to \infty$.

答:
It's a little unclear.  Given the contour $C_1=x+i k$ for constant k, then looks like your integral is:
$$\int_{x_1}^{x_2} \frac{a\sqrt{x^2+k^2}}{(x+ik)-\gamma}dx$$
$$=\biggr\{a \left(-\sqrt{\gamma } \sqrt{-\gamma +2 i k} \tan ^{-1}\left(\frac{k^2-i k x+\gamma  x}{\sqrt{\gamma } \sqrt{-\gamma +2 i k} \sqrt{k^2+x^2}}\right)+(\gamma -i k) \tanh ^{-1}\left(\frac{x}{\sqrt{k^2+x^2}}\right)+\sqrt{k^2+x^2}\right)\biggr\}\biggr|_{x1}^{x_2}$$

If it were mine, I would evaluate it numerically say from 1 to 10,  then compare the results with the expression above. However the antiderivative has multi-valued functions which will cause problems with a definite integral


问:
Riemann and Ricci tensor
Let $(M,g)$ be a smooth manifold and $f:M \rightarrow \mathbb{R}$ a smooth function.

Can we get some expression of: $$ \partial_k f g^{jh} R^k_{hij}$$ in terms of the Ricci tensor?
(I use the notations $R^a_{cdb} \partial_a = R(\partial_c, \partial_d) \partial_b$ and $R_{bd}= R^c_{bcd}$)

In a paper, I have seen that $$\partial_k f g^{jh} R^k_{jih}= \partial_j fR^j_i$$

I have the following ansatz:
$$\partial_k f g^{jh} R^k_{hij} =  \partial_k f  g^{jh} g^{lk} R_{hijl}
=- \partial_k f  g^{jh} g^{lk} R_{ihjl}=-\partial_k f  g^{jh} g^{lk}  g_{hl}R^h_{ihj} \\
=-\partial_k f  g^{jh} \underbrace{g^{lk}  g_{hl}}_{= \delta_{kh}}R_{ij} = \partial_h f  g^{jh} R_{ij} = \partial_h f   R^j_i
$$
I pretty sure that the first $=$ is correct but the third $=$ just looks wrong to me, because there are three $h$s.

Can someone help me here? Thanks in advance!

答:
There's actually four $h$'s there. To correct that we use the Bianchi identity,
$$ R_{hijl} + R_{hjli} + R_{hlij} = 0.$$
Thus (using $\sigma^{jh} R_{hjli} = 0$)

\begin{align}
\sigma^{jh} R^k_{hij} &= \sigma^{jh} \sigma^{lk} R_{hijl}\\
&=-\sigma^{jh} \sigma^{lk} (R_{hjli} + R_{hlij}) \\
&=-\sigma^{jh} \sigma^{lk} R_{hlij} \\
&=\sigma^{jh} \sigma^{lk} R_{lhij}\\
&= \sigma^{lk} R^h_{lhi} = \sigma^{lk} R_{li}.
\end{align}

Title: Riemann and Ricci tensor


问:
Kronecker delta in multivariate Taylor expansion
I find the following equation in a quantum theory book where multivariate Taylor expansion was discussed. The equation uses Kronecker delta to change the subscripts which looks bizarre to me, and it goes

$$\phi_{\alpha \beta}\left(\boldsymbol{R}_{i}+\boldsymbol{d}_{\kappa}-\boldsymbol{R}_{j}-\boldsymbol{d}_{\nu}\right)\left.u_{i \kappa, \alpha} u_{i \kappa, \beta}\right.=\left[\sum_{j^{\prime} \nu^{\prime}} \phi_{\alpha \beta}\left(\boldsymbol{R}_{i}+\boldsymbol{d}_{\kappa}-\boldsymbol{R}_{j^{\prime}}-\boldsymbol{d}_{\nu^{\prime}}\right)\right] \delta_{i j} \delta_{\kappa \nu}u_{i \kappa, \alpha} u_{j \nu, \beta},\tag{1}$$

where $\phi$ is a function of the difference of two coordinates, $\boldsymbol{R}_i+\boldsymbol{d}_{\kappa}$ and $\boldsymbol{R}_j+\boldsymbol{d}_{\nu}$. And $\phi_{\alpha\beta}$ is
$$\phi_{\alpha \beta}\left(\boldsymbol{R_{i}+d_{\kappa}-R_{j}-d_{\nu}}\right)=\frac{\partial^{2} \phi\left(\boldsymbol{R_{i}+d_{\kappa}-R_{j}-d_{\nu}}\right)}{\partial R_{i \alpha} \partial R_{i \beta}}.\tag{2}$$

with $R_{i\alpha}$ being the $\alpha$-element of coordinate $\boldsymbol{R}_i$. If $u_{i\kappa,\alpha}$ is the displacement of the coordinate $\boldsymbol{R}_i+\boldsymbol{d}_{\kappa}$ at $\alpha-$direction, then where does the summation in (1) come from? I'm trying to derive (1) using the properties of Kronecker delta but so far not much progress has been made.

答:
The RHS of eq.(1) should be first written as:

$$\sum_{j^{\prime}\nu^{\prime}}\phi_{\alpha\beta}(\boldsymbol{R_i+d_{\kappa}-R_{j^{\prime}}-d_{\nu^{\prime}}})\delta_{jj^{\prime}}\delta_{\nu\nu^{\prime}}u_{i\kappa,\alpha}u_{i\kappa,\beta}$$

and then

$$\sum_{j^{\prime}\nu^{\prime}}\phi_{\alpha\beta}(\boldsymbol{R_i+d_{\kappa}-R_{j^{\prime}}-d_{\nu^{\prime}}})\sum_{j\nu}\delta_{jj^{\prime}}\delta_{\nu\nu^{\prime}}u_{i\kappa,\alpha}\delta_{ji}\delta_{\nu\kappa}u_{j\nu,\beta}$$

which gives

$$\sum_{j^{\prime}\nu^{\prime}}\phi_{\alpha\beta}(\boldsymbol{R_i+d_{\kappa}-R_{j^{\prime}}-d_{\nu^{\prime}}})\delta_{ij^{\prime}}\delta_{\kappa\nu^{\prime}}u_{i\kappa,\alpha}u_{j\nu,\beta}$$


问:
Show that $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$, where $\left( \frac{q}{p} \right)$ is the Legendre Symbol

Show that if $p$ is any odd prime then

$$\left( \frac{q}{p} \right) \equiv q^{\frac{p-1}{2}} \mod p.$$

stating any theory that you use. In particular, you may assume the existence of a primitive element in $G_p$.

Here $\left( \frac{q}{p} \right)$ is the Legendre Symbol and $G_p$ is the group of elements $g \mod p$ such that $\gcd(g,p) = 1$. I said that for some $a \in \mathbb{Z}$, we have

$$a \equiv q^{(p-1) / 2} \mod p \implies a^2 \equiv q^{(p-1)} \equiv 1 \mod p$$

by Fermat's little theorem. And so, by definition of the Legendre symbol, we have that if $q$ is a quadratic residue mod $p$ then $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$. I'm now stuck on how to show that it is $\equiv -1 $ if it isn't a quadratic residue. Obviously the hint with primitive elements comes into play somehow, but I can't see how it does.

Can someone help me please.

答:
Hint (assuming I have correctly guessed what the question really is):

Let $g$ be a primitive element. Find $a$ in such a way that $q\equiv g^a\pmod p$. Show that $x=q^{(p-1)/2}\equiv1$, iff $a$ is even. Show that $x^2\equiv1\pmod p$ irrespective of parity of $a$, so...


问:
Cross, Dot Product Properties - Volume of Brillouin Zone

(See image) I'm trying to understand how the final line of this answer is reached. I know that an identity is used to get from the 3rd last line to the 2nd last line, but what properties of dot & cross products are used in the final step?

Thank you

答:
First pull the scalar out:

$$ (a_2\times a_3)\cdot\color{Blue}{(a_3\cdot(a_1\times a_2))}a_1=\color{Blue}{(a_3\cdot(a_1\times a_2))}((a_2\times a_3)\cdot a_1) $$

Then use the fact the scalar triple product is cyclic to see both triple products above are identical.


问:
Residues of $\sin\left(\frac{z^2+1}{z}\right)$

Looking for assistance with the following function..
i need to calculate the Residues of $$f(z)= \sin\left(\frac{z^2+1}{z}\right)$$
regrading my way:
$$\sin\left(\frac{z^2+1}{z}\right)=\sin(z)\cos(\frac{1}{z})+\cos(z)\sin(\frac{1}{z}) $$
$$= \left(z-\frac{z^3}{6}+...\right)\left(1-\frac{1}{2z^2}+...\right)+\left(1-\frac{z^2}{2}+...\right)\left(\frac{1}{z}-\frac{1}{6z^3}+...\right) $$
look on the coefficient of $$\frac{1}{z}$$
we get
$$ -\frac{7}{12}$$

it seems right?

Thanks!

答:
I would do it this way.  $$\begin{align}
\sin\left(\frac{z^2+1}{z}\right)&=\sin\left(z+\frac1z\right)\\
&=\frac1{2i}\left(e^{i(z+1/z)}-e^{-i(z+1/z)}\right)
\end{align}$$
Now, $$\begin{align}
e^{i(z+1/z)}&=e^{iz}e^{1/z}\\
&=\sum_{n=0}^\infty\frac{(iz)^n}{n!}\sum_{m=0}^\infty\left(\frac iz\right)^m\frac1{m!}\\
\end{align}$$
so that $$\operatorname{Res}_{z=0}e^{iz}e^{i/z}=\sum_{n=0}^\infty\frac{i^n}{n!}\frac{i^{n+1}}{(n+1)!}=i\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}$$

A similar calculation gives
$$\operatorname{Res}_{z=0}e^{-iz}e^{-i/z}=-i\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}$$ so that $$\operatorname{Res}_{z=0}\sin\left(\frac{z^2+1}{z}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}=J_1(2)$$ where $J_1$ is the Bessel function of the first kind of order $1$.

Numerical computation gives approximately $0.5767248077568736$.


答: Alternatively use $(a\times b)\cdot(c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)$ in$$\begin{align}(a_2\times a_3)\cdot(a_3\times a_1)\times(a_1\times a_2)&=(a_2\cdot a_3\times a_1)(a_3\cdot a_1\times a_2)\\&-(a_2\cdot a_1\times a_2)(a_3\cdot a_3\times a_1)\\&=(a_1\cdot a_2\times a_3)^2-0^2\\&=(a_1\cdot a_2\times a_3)^2\end{align}$$to get from the first line to the third.

Title: Cross, Dot Product Properties - Volume of Brillouin Zone

Body:
(See image) I'm trying to understand how the final line of this answer is reached. I know that an identity is used to get from the 3rd last line to the 2nd last line, but what properties of dot & cross products are used in the final step?

Thank you


答:
The number of ways EDGE can be the first $4$ letters is $1$. The number of ways the remaining letters can be arranged is then $\frac{11!}{2!}.$

The number of ways EDGE can be the $2nd$ to $5th$ letters is $1$. The number of ways the remaining letters can be arranged is then $\frac{11!}{2!}.$

...

So the answer is $\frac{11!}{2!} \times 12.$

Title: How many anagrams with a specific subword?

Body:
I'm trying to find the number of anagrams of "ACKNOWLEDGEMENT" (15 characters) that contain the subword "EDGE".

I know the total number of anagrams is $\frac{15!}{3! \cdot 2!}$. I'm not sure of what to do after this. I know this is an overcount.


答:
Let $g$ be a primitive root of $p$, a generator of the group of $p-1$ units. Suppose that $a$ is a QR. Then $a\equiv b^2\pmod{p}$ for some $b$. But $b$ is congruent to a power $g^e$ of $g$. So $a$ is congruent to $g^{2e}$. And by Fermat's Theorem, $(g^{2e})^{(p-1)/2}\equiv 1\pmod{p}$.

Note that all $(p-1)/2$ numbers congruent to an even power of $g$ are automatically QR.

Since there are $(p-1)/2$ QR, the numbers congruent to an odd power of $g$ are all NR. An odd power of $g$ cannot be congruent to $1$ modulo $p$. Let $x=a^{(p-1)/2}$. Then $x^2\equiv 1\pmod{p}$, so $x\equiv \pm 1\pmod{p}$. If $a$ is an NR, then $a^{(p-1)/2}\not\equiv 1 \pmod{p}$, so it is $\equiv -1\pmod{p}$.

Title: Show that $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$, where $\left( \frac{q}{p} \right)$ is the Legendre Symbol

Body:
Show that if $p$ is any odd prime then

$$\left( \frac{q}{p} \right) \equiv q^{\frac{p-1}{2}} \mod p.$$

stating any theory that you use. In particular, you may assume the existence of a primitive element in $G_p$.

Here $\left( \frac{q}{p} \right)$ is the Legendre Symbol and $G_p$ is the group of elements $g \mod p$ such that $\gcd(g,p) = 1$. I said that for some $a \in \mathbb{Z}$, we have

$$a \equiv q^{(p-1) / 2} \mod p \implies a^2 \equiv q^{(p-1)} \equiv 1 \mod p$$

by Fermat's little theorem. And so, by definition of the Legendre symbol, we have that if $q$ is a quadratic residue mod $p$ then $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$. I'm now stuck on how to show that it is $\equiv -1 $ if it isn't a quadratic residue. Obviously the hint with primitive elements comes into play somehow, but I can't see how it does.

Can someone help me please.


问:
Examples of continuous functions that are not homotopic to each other.
Let $f,g:X \to Y$ be continuous functions. Then, consider $F:X \times I \to Y$ defined as follows:$$F(x,t) = f(1-t)+gt$$

How does one check whether $F$ is continuous(w.r.t the product topology)?
Is there any other way to define a homotopy between two functions?

As an example, take $f,g :  \mathbb{R} \to \mathbb{R}$ to be $f(x) = x, g(x) = sin(x)$. Then how does one PROVE that the homotopy map as defined above is continuous?

Moreover, what are some examples of maps that are not homotopic?

答:
Generally, this all comes down to a space being path connected. If the path the homotopy described is contained in the space then it will be continous. For example, if we want to see the circular sector of radii 1 and 2 is homotopic to the unit circumference, we might consider the following map:
$$
f(x)=\frac{x}{||x||}
$$
It normalizes every point of the sector. It is continous because $0$ is not contained in the sector, thus, it is a retraction into the unit sphere. We want to prove it is a deformation retract, so we can end our proof, concluding that the two sets have the same type of homotopy. We consider the following homotopy:
$$
H(x,t)=f(x)·t+x·(1-t)
$$
Because of what I said earlier, one can check it is radial, making it thus continious (the sector is path connected).


问:
Euclidean GCD and why does it work?
By the dupe, the table implies $\,(14441,3565) = (3565,189) = \ldots = (28,21) = (21,7) = (7,0) = 7\ \ $

I understand that Euclid's algorithm on GCD is based on doing division via subtraction $x = qy + r$. I also understand that the process is keep expressing the quotient in terms of the remainder.

Example to find GCD of $14441$, $3563$:

x    =
q
* y
+  r

14441
4
3565
189

3565
18
189
161

189
1
161
28

161
5
28
21

28
1
21
7

21
3
7
0

So the GCD is $7$.

So basically we try to divide the $2$ original numbers and then try to see if the remainder can express evenly $y$ and keep doing that recursively i.e. try to find the smallest number that divides the remainder.

But I am not sure I understand the intuition behind the idea. Why would that process lead to the smallest number that divides the original $x$?

I also read that a core idea is that $gcd(x,y) = gcd(y,r)$ but I didn't really get that part too.

Could someone please help?

答:
You might find it easier to consider just a single step:

For any two numbers $x$ and $y$ with $y<x$, consider $x$ and $x-y$.

Any divisor of $x$ and $y$ is also a divisor of $x-y$ and $y$. Similarly, any divisor of $x-y$ and $y$ is also  a divisor of $x$ and $y$. Thus gcd$(x,y)=$ gcd$(x-y,y)$.

We have thus reduced the pair of numbers and can repeat this over and over again.

Title: Euclidean GCD and why does it work?


问:
Is the limit to infinity of one over an expression whose limit is infinity zero?
Put more correctly, do the following rules regarding limits always hold?

$ \lim\limits_{x \to \infty} \dfrac{1}{f(x)} = 0 $ where $ \lim\limits_{x \to \infty} f(x) = \infty $

$ \lim\limits_{x \to - \infty} \dfrac{1}{g(x)} = 0 $ where $ \lim\limits_{x \to - \infty} g(x) = \infty $

答:
Hint.

Yes. They essentially follow from the fact that
if $f(x)>M>0$, then
$$
0<\frac{1}{f(x)}<\frac{1}{M}
$$


答:
Let's see the minimum value of $\sqrt{x^2+(1-x)^2}$
$$=\sqrt{x^2+x^2-2x+1}$$
$$=\sqrt{2x^2-2x+1}$$
This expression will be minimum if the thing inside the root is minimum.

Thus,
$$y=2x^2-2x+1$$
$$0=4x-2$$ $\text{"Differentiating both sides"}$
$$\implies x=\frac{1}{2}$$
Thus the function is minimum at $x=\frac{1}{2}$

Thus, minimum value of function is:$$2 \cdot(\frac{1}{2})^2-2 \cdot\frac{1}{2}+1$$
$$=\frac{1}{2}$$
Hence the minimum value of the expression $\sqrt{x^2+(1-x)^2}$ is $\sqrt{\frac{1}{2}}$ $=\frac{1}{\sqrt{2}}$

Thus the minimum value of $$\sqrt{a^2 +(1-a)^2}+ \sqrt{b^2 +(1-b)^2}$$ is : $$2 \cdot\frac{1}{\sqrt{2}}$$ $$=\sqrt{2}$$

Thus the required inequality has been proved.

Title: $\sqrt{a^2 +(1-a)^2}+ \sqrt{b^2 +(1-b)^2} \geq \sqrt{2} $

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

My attempt:

If i can prove that:
$$x^2 +(1-x)^2 \geq \frac{1}{2}$$
So therefore:
$$\sqrt{x^2 +(1-x)^2} \geq \frac{\sqrt{2}}{2} $$
But i can’t prove the first inequality .


答:
Using the Cauchy-Schwarz inequality:
$$
 \sqrt{1^2 + 1^2} \cdot \sqrt{x^2 + (1-x)^2} \ge 1\cdot x + 1 \cdot (1-x) = 1 \, .
$$

The geometric interpretation is that a point on the line $y=1-x$ in the 2D plane has at least the distance $1/\sqrt 2$ to the origin.

Yet another option is to substitute $x= \frac 12 + u$. Then
$$
 x^2 + (1-x)^2 = \left( \frac 12 + u\right)^2 + \left( \frac 12 - u\right)^2 = \frac 12 + 2 u^2 \ge \frac 12 \, .
$$


问:
I know $\sqrt{2}$ is a solution to the equation $x^{x^{x^{x^{\vdots}}}}=2$. But why it doesn't blow out?
I saw the trick to solve  $x^{x^{x^{x^{\vdots}}}}=2$, and solution is $\sqrt{2}$. I understand that $x^{x^{x}}$ is different from $(x^{x})^x$, but I still cannot see intuitively why the $x^{x^{x^{x^{\vdots}}}}$ doesn't blow out when $x$ is larger than 1.

I experimented a little, and found that all values that satisfy the equation are less than $\sqrt{2}$. What makes this range special?

$a=2:20$

$b=a^{\frac{1}{a}};b$

Thank you in advance for any help

答:
This question is analogous to asking the (equally good) question: why does the infinite series $\sum_{k=1}^\infty r^k$ blow up when $r>1$ but not when $0<r<1$? The answer, in both cases, is that the infinite expression is defined to be the limit of its finite truncations; and sometimes limits exist and sometimes they don't. In this case, the sequence
$$
x,\, x^x,\, x^{x^x},\, x^{x^{x^x}}, \dots
$$
happens to stabilize if $e^{-e}\le x\le e^{1/e}$ but not if $x>e^{1/e}$.


问:
A problem on matrices: Find the value of $k$

If $
        \begin{bmatrix}
        \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
        \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}  \\
        \end{bmatrix}^k
 =
        \begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}
$, then the least positive integral value of $k$ is?

Actually, I got no idea how to solve this.

 I did trial and error method and got 7 as the answer.

how do i solve this?
 Can you please offer your assistance? Thank you

答:
Powers of matrices should always be attacked with diagonalization, if feasible. Forget $2\pi/7$, for the moment, and look at
$$
A=\begin{bmatrix}
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{bmatrix}
$$
whose characteristic polynomial is, easily, $p_A(X)=1-2X\cos\alpha+X^2$. The discriminant is $4(\cos^2\alpha-1)=4(i\sin\alpha)^2$, so the eigenvalues of $A$ are
\begin{align}
\lambda&=\cos\alpha+i\sin\alpha\\
\bar{\lambda}&=\cos\alpha-i\sin\alpha
\end{align}
Finding the eigenvectors is easy:
$$
A-\lambda I=
\begin{bmatrix}
-i\sin\alpha & -\sin\alpha\\
i\sin\alpha & \sin\alpha
\end{bmatrix}
$$
and an eigenvector is $v=[-i\quad 1]^T$. Similarly, an eigenvector for $\bar{\lambda}$ is $w=[i\quad 1]^T$. If
$$
S=\begin{bmatrix}-i & i\\1 & 1\end{bmatrix}
$$
you get immediately that
$$
S^{-1}=\frac{i}{2}\begin{bmatrix}1 & -i\\-1 & -i\end{bmatrix}
$$
so, by well known rules,
$$
A=SDS^{-1}
$$
where
$$
D=
\begin{bmatrix}
\cos\alpha+i\sin\alpha & 0 \\
0 & \cos\alpha-i\sin\alpha
\end{bmatrix}.
$$
By De Moivre's formulas, you have
$$
D^k=
\begin{bmatrix}
\cos(k\alpha)+i\sin(k\alpha) & 0 \\
0 & \cos(k\alpha)-i\sin(k\alpha)
\end{bmatrix}.
$$
Since $A^k=S D^k S^{-1}$ your problem is now to find the minimum $k$ such that $\cos(k\alpha)+i\sin(k\alpha)=1$, that is, for $\alpha=2\pi/7$,
$$
\begin{cases}
\cos k(2\pi/7)=1\\
\sin k(2\pi/7)=0
\end{cases}
$$
and you get $k=7$.

This should not be a surprise, after all: the effect of $A$ on vectors is exactly rotating them by the angle $\alpha$. If you think to the vector $v=[x\quad y]^T$ as the complex number $z=x+iy$, when you do $Av$ you get
$$
Av=
\begin{bmatrix}
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix}
=\begin{bmatrix}
x\cos\alpha-y\sin\alpha\\x\sin\alpha+y\cos\alpha
\end{bmatrix}
$$
and
$$
(x\cos\alpha-y\sin\alpha)+i(x\sin\alpha+y\cos\alpha)=
(x+iy)(\cos\alpha+i\sin\alpha)=\lambda z
$$
(notation as above). Thus $z$ is mapped to $\lambda z$, which is just $z$ rotated by an angle $\alpha$.


答:
Maybe you're confusing $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ with $\left({\sqrt{2}}^{\sqrt{2}}\right)^{\sqrt{2}}. $ The latter is done by calculating ${\sqrt{2}}^{\sqrt{2}}$ and then doing this number to the power of ${\sqrt{2}}.\ $

In fact, $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ is ${\sqrt{2}}$ to the power of ${\sqrt{2}}^{\sqrt{2}},\ $ or to say it another way, you're doing ${\sqrt{2}}^{\sqrt{2}},\ $ and then doing ${\sqrt{2}}$ to the power of this number.

Expanding on this. Pretend ${\sqrt{2}} = 1.4.$

$1.4^{1.4} = 1.6.$

$1.4^{1.6} = 1.7.$

$1.4^{1.7} = 1.77.$

$1.4^{1.77} = 1.81.$

$1.4^{1.81} = 1.84.$

If the exponent is $< 2,\ $ then $1.4^{exponent} = new\_exponent$ will be less than $1.96$ which is less than $2$, so the next step in the iteration process is to do $1.4^{new\_ exponent}$, and this new exponent will again be be less than $1.96$ and so the result from the iteration process can never be greater than $1.96.$

Whereas,

$1.4^{1.4} = 1.6$

$1.6^{1.4}=1.93$

$1.93^{1.4}=2.5$

$2.5^{1.4}=3.6$

$1.93^{1.4}=6.0$

and this is growing without bound.

Title: I know $\sqrt{2}$ is a solution to the equation $x^{x^{x^{x^{\vdots}}}}=2$. But why it doesn't blow out?

Body:
I saw the trick to solve  $x^{x^{x^{x^{\vdots}}}}=2$, and solution is $\sqrt{2}$. I understand that $x^{x^{x}}$ is different from $(x^{x})^x$, but I still cannot see intuitively why the $x^{x^{x^{x^{\vdots}}}}$ doesn't blow out when $x$ is larger than 1.

I experimented a little, and found that all values that satisfy the equation are less than $\sqrt{2}$. What makes this range special?

$a=2:20$

$b=a^{\frac{1}{a}};b$

Thank you in advance for any help


问:
The matrix analogue of the Theorem about diagonalizable operators - Linear Algebra by Hoffman and Kunze, page 187
Here is the theorem and its "matrix analogue"  :

First of all, does the formulation of $X$'s follow from the fact that $T(X) = cX$ iff $AX = cX$, where $c$ is a characteristic value of $T$ and $A$ is the representative matrix of $T$? If so, I guess this is true only when $A$ represents $T$ relative to the standard basis of $V$ and $V=R^n$. Do I get this right?

Second, what is the relationship between the matrix $P$ and the similar matrices of $A$? I mean, what is the motivation for the construction of $P$? I think the writer's manner of telling is not explicit here.

答:
Yes, he's switching from the case of a linear transformation that might not be a matrix to a linear transformation that does operate on a set of vectors in the standard way by matrix multiplication. For some reason books use letters near the end of the alphabet (often $S$, $T$) for the former and letters near the beginning of the alphabet (often $A$, $B$) for the latter.  So the "column matrices" (most people would call these "column vectors" or just "vectors") $X$ for which $(T - c_i I)X = 0$ in the theorem become, after taking $T = A$, vectors $X$ for which $(A - c_i I) X = 0$ which as you note is the same condition as $AX = c_i X$.  (Separately, what he calls "characteristic values" are more commonly called "eigenvalues" these days, and the nonzero vectors $X$ for which $AX = c_iX$ are called "eigenvectors."  I will do this below, but it's exactly the same notion as what he's talking about in different terminology.)

Regarding the second part of the question, it really comes down to the relationship between matrices as formal objects and their properties as linear transformations.  When acting on the standard basis, any matrix $P$ can be thought of as a linear transformation sending the $i$th standard basis vector $e_i$ to the $i$th column of $P$ (call it $v_i$).  To repeat the essential point, the statements that "the $i$th column of $P$ is the vector $v_i$" and "$Pe_i = v_i$" are equivalent.  This is the heart of the matrix-theoretic aspect of the problem.  It might be the definition of matrix multiplication, or it might be a consequence of it, depending upon how your book has set this up.

So now if $v_i$, the $i$th column of $P$, was chosen to be an eigenvector of $A$ corresponding to the eigenvalue $c_i$ (as was done when the author constructed $P$ by making eigenvectors of $A$ for its columns) then you know what $A$ does to $v_i$: we have $A v_i = c_i v_i$.  And if $P$ is invertible, what would $P^{-1}$ do to that vector?  From $P e_i = v_i$ you can deduce $e_i = P^{-1} v_i$, so you know that $P^{-1}(A v_i) = P^{-1}(c_i v_i) = c_i P^{-1} v_i = c_i e_i$.

So the composite linear transformation $P^{-1} A P$ sends $e_i$ to $v_i$ (via $P$), then $v_i$ to $c_i v_i$ (via $A$), and then $c_i v_i$ to $c_i e_i$ (via $P^{-1}$).  And there's only one matrix that sends $e_i$ to $c_i e_i$ for all $i$, namely, the diagonal matrix with the list of numbers $c_i$ down the diagonal.

There's a separate question: if you weren't into matrix diagonalization in the first place, how would it occur to you to construct the matrix $P$ and even think to perform those calculations about $AP$ or $P^{-1} AP$.  I don't have a good answer, except that once you sit down long enough with the interaction between function composition and matrix multiplication, it seems like a very good or even natural idea.  And at one time it was not evident to anybody that it would be helpful to think in terms of matrix algebra at all.  We can benefit from the fact that a lot of this has been solved by others, and we can just come in later and understand what they've done. It really does fit together.


问:
Inner product of two matrix coefficient functions of an irreducible representation.
$\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\C}{\mathbf C}$

Let $G$ be a compact Hausdorff group and $\mu$ be the normalized Haar measure on $G$.
Let $\sigma:G\to \mathscr U(V)$ be a finite dimensional unitary representation of $G$.
We will write $\sigma_g$ to denote $\sigma(g)$.

Let $v, w\in V$ be fixed, and define a map $T:V\to V$ by writing $T(x) = \ab{x, v}w$.
Consider the linear operator
$$
\tilde T = \int_G \sigma_{g^{-1}} \circ T\circ \sigma_g\ d\mu(g)
$$
Then $\tilde T$ is a $G$-equivariant endomorphism of $V$, and therefore by Schur's lemma there is $\lambda\in \C$ such that $\tilde T = \lambda I$.
But then we have, for any $v', w'\in V$ that
$$
\ab{\tilde Tv', w'} = \lambda \ab{v', w'}
$$
giving
$$
\int_G \ab{\sigma_g v', v}\overline{\ab{\sigma_g w', w}}\ d\mu(g) = \lambda \ab{v', w'}
$$

Question.
Is $\lambda$ same as $\ab{v, w}$?

答:
The $G$-averaging process (Weyl's unitary trick) doesn't change the trace, since $\sigma_g$s are all unitary:

$$ \mathrm{tr}(\overline{T})=\int_G \mathrm{tr}(\sigma_g^{-1}T\sigma_g)=\int_G \mathrm{tr}(T)=\mathrm{tr}(T). $$

On the one hand, $\mathrm{tr}(\overline{T})=(\dim V)\lambda$. On the other hand, $\mathrm{tr}(T)=\mathrm{tr}(wv^\dagger)=\langle v,w\rangle$.

So, actually, $\lambda=\langle v,w\rangle/\dim V$.


问:
Find ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$
I'm currently learning how to use partial differentiation. Can someone give me a hint?

find  ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$Calculate
the differential of the entire equation
$$\eqalign{
0 &= d(xy + yz - xz) \\
  &= d(xy) + d(yz) - d(xz) \\
  &= (x\,dy + y\,dx) + (y\,dz + z\,dy) - (x\,dz + z\,dx) \\
(x-y)\,dz &= (y-z)\,dx + (x+z)\,dy \\
}$$
Holding $x$ constant (i.e. setting $\,dx=0$) yields
$$\eqalign{
(x-y)\,dz  &= (x+z)\,dy \quad\implies\quad  \p{z}{y} = \frac{x+z}{x-y} \\
}$$
While holding $y$ constant ($dy=0$) yields
$$\eqalign{
(x-y)\,dz  &= (y-z)\,dx \quad\implies\quad  \p{z}{x} = \frac{y-z}{x-y} \\
}$$


问:
Absolute convergance of function series
The question is for which values of $x\in \mathbb R$, the following series absolute/conditionally converge: $$\sum_{n=1}^{\infty}\frac{x^n}{(1+x)(1+x^2)...(1+x^n)}$$ I have no idea how to solev it except M-test of wirestrass but I don't know how to bound it . Forgive me if tagging the question as 'power-series' was wrong. </p>

答:
Let
$$u_n=\frac{x^n}{(1+x)(1+x^2)...(1+x^n)}$$
so we have
$$\left|\frac{u_{n+1}}{u_n}\right|=\left|\frac{x}{1+x^{n+1}}\right|\to\left\{\begin{array}\\
|x| \quad\text{if}\quad -1<x<1\\
\frac{1}{2} \quad\text{if}\quad x=1\\
0 \quad\text{if}\quad -1<x \,\text{or}\, x>1
\end{array}\right.$$
so we can see by the ratio test that the series converges absolutely for all $x\neq -1$</p>

Title: Absolute convergance of function series


问:
find $f$ such $f(f(x)+y)=2x+f(f(f(y))-x)$
Find all function $f:R\to R$,such for any real number $x,y$ have
$$f(f(x)+y)=2x+f(f(f(y))-x)$$

I have prove this one result:

$$f(f(0))=f(0)$$
proof:let $x=y=0$,then we have
$$f(f(0))=f(f(f(0)))$$
and take $x=f(f(0)),y=0$,we have
$$f(f(f(f(0))))=2f(f(0))+f(0)$$
so we have
$$f(f(0))=-f(0)$$

use this links Ivan Loh methods to prove two reslut:

$f(x)$ is also injective

proof: For convenience, let $P(x, y)$ represent the equation $f(f(x)+y)=2x+f(f(f(y))-x)$.

$P(x, -f(x))$ gives $f(f(-f(f(x)))-x)=f(0)-2x$. Thus $f(x)$ is surjective.

Suppose $f(a)=f(b)$ for some $a, b$. Then $P(x, a)$ and $P(x, b)$ give
$$f(f(x)+a)=2x+f(f(f(a))-x)=2x+f(f(f(b))-x)=f(f(x)+b)$$
Since $f(x)$ is surjective, $f(x+a)=f(x+b) \, \forall x \in \mathbb{R}$.

Now using the above equation and $P(a, y), P(b, y)$ give

\begin{align}
2a=f(f(a)+y)-f(f(f(y))-a) & =f(f(a)+y)-f(f(f(y))-(a+b)+b) \\
&=f(f(a)+y)-f(f(f(y))-(a+b)+a) \\
& =f(f(b)+y)-f(f(f(y))-b) \\
& =2b
\end{align}

Thus $f(x)$ is also injective.

Finally, $P(0, y)$ gives $f(y+f(0))=f(f(f(y)))$, so since $f(x)$ is injective, $$f(f(y))=y+f(0) \, \forall y \in \mathbb{R}$$.

then I can't  continue ,and I think this answer is $f(x)=x$.so How to prove it.and this problem is  from   my teacher to give me exercise .Thanks for you help!
Now when post this post,This site automatically displays a prompt for a similar question to me(But it's different problem) :links

答:
A partial answer for $f$ polynomial.

►Clearly $f(x)$ is not constant.

►Test for a linear polynomial $f(x)=ax+b$.

$$f(f(x)+y)=f(ax+b+y)=a(ax+b+y)+b=a^2x+ay+(ab+b)\\2x+f(fof(y)-x)=2x+f(a(ay+b)+b-x)=2x+a(a(ay+b)+b-x)+b\\2x+f(fof(y)-x)=(2-a)x+a^3y+(a^2+1)b$$ It follows
$$a^2=2-a\Rightarrow a=1\text {and } a=-2 $$ and $$\begin{cases}a=a^3\\(a+1)b=(a^2+a+1)b\end{cases}$$ therefore $a=1$ and $2b=3b\Rightarrow b=0$.

Thus the only possible polynomial of degree $1$ is $\boxed{f(x)=x}$

►Test for $f(x)=a_nx^n+\cdots+a_1x+a_0$ where $n\gt1$ ( and $a_n\ne0$ of course)

Since the equality is valid for all $x,y$ we take $y=0$ and look  for a contradiction. $$fof(x)=2x+f(fof(0)-x)\large?$$
We have $$ fof(x)=a_n(a_nx^n+\cdots+a_1x+a_0)^n+\cdots+a_1(a_nx^n+\cdots+a_1x+a_0)+a_0$$ Note that $fof(0)=f(a_0)=a_n(a_0)^n\cdots+a_1(a_0)+a_0$.
$$2x+f(f(a_0)-x)=2x+a_n(f(a_0)-x)^n+\cdots+a_1(f(a_0)-x)+a_0$$ $LHS$ is of  $n^2$ degree while $RHS$ is of $n$ degree which implies that all the coefficients from $x^{n^2}$ until $x^{n+1}$ should be zero. In particular we would have $a_na_n^{n}=0$. Contradiction.

$$**************************************$$

Is $f(x)=x$ the only solution $\large ?$

We have $$f(f(x)+y)=f(f(f(y))-x)+2x$$ for all $x,y$ so it is valid for $x=y=0$; therefore
$$fof(0)=f(fof(0))$$ which means that $fof(0)$ is a fixed point of $f$. Making now $x=0$ we have
$$f(f(0)+y)=f(fof(y))$$ so if $f(0)=0$ then $f(y)=fof(f(y))$ and $f(y)$ is a fixed point of $fof$ for all $y$ so $fof(x)=x$ for all $x$ in the codomain of $f$. It follows that if  $f$ is surjective and $f(0)=0$ then the only function required is the identity $\boxed{f(x)=x}$.

Note that for all fixed $y$ the function $F_y(x)$ defined by $$F_y(x)=f(f(x)+y)-f(f(f(y))-x)=2x$$
is a bijection because $2x$ so is. I stop here. Whoever wants, if desired, can finish the solution following this way.


问:
Lifting a map to a homeomorphism of coverings
This is a part of the proof of Lemma 4 of "Cobordism of classical knots" by Casson and Gordon.
Here, $\widetilde{X}$ denotes a prime-fold cyclic covering of $X$.

Let $h\colon X\to X$ generate the group of covering translations of $X$. Let $G$ be the image of $H_1(\widetilde{X},\mathbb{Z}_p)$ in $H_1(X,\mathbb{Z}_p)$.
Since $H_1(X,\mathbb{Z}_p)$ is finite, there is a non-zero integer $r$ such that $(h_\ast)^r(G)=G$.
It follows that $h^r\colon X\to X$ lifts to a homeomorphism $\tilde{h}\colon\widetilde{X}\to\widetilde{X}$, and $\widetilde{X}$ is an infinite cyclic covering of the finite complex $\widetilde{X}/\tilde{h}$.

Okay, the existence of $\tilde{h}$ comes from the lifting criterion for covering spaces. But how can we know that it is a homeomorphism?

答:
It is not true in general that for any covering map $f : \widetilde X \to X$ of any path connected, locally path connected space, and for any homeomorphism $h : X \to X$ and any lift $\tilde h : \widetilde X \to \widetilde X$ of $h$, the map $\tilde h$ is a homeomorphism.

However, if one adds the hypothesis that the given covering map $f$ has finite degree, which matches your situation, then the conclusion does hold, by an application of the general lifting lemma, as follows.

Pick $p \in X$, and $\tilde p \in \widetilde X$, and let $q=h(p)$ and $\tilde q = \tilde h(\tilde p)$. The fact that a lift $\tilde h : (\widetilde X,\tilde p) \to (\widetilde X,\tilde q)$ of $h$ exists implies that the group isomorphism $h_* : \pi_1(X,p) \to \pi_1(X,q)$ maps the subgroup $f_*(\pi_1(\widetilde X, \tilde p)) < \pi_1(X,p)$ isomorphically into the subgroup $f_*(\pi_1(\widetilde X,\tilde q)) < \pi_1(X,q)$. However, each of the subgroups $f_*(\pi_1(\widetilde X, \tilde p)) < \pi_1(X,p)$ and $f_*(\pi_1(\widetilde X,\tilde q)) < \pi_1(X,q)$ has finite index equal to $\text{degree}(f)$. Also, the group isomorphism $h_*$ preserves index of subgroups. The only way one subgroup of finite index can include into another subgroup of the same finite index is if those two subgroups are equal. Therefore, $h_*$ maps $f_*(\pi_1(\widetilde X, \tilde p))$ is isomorphically onto $f_*(\pi_1(\widetilde X,\tilde q))$.

The inverse homomorphism $(h^{-1})_* : \pi_1(X,q) \to \pi_1(X,p)$ therefore maps $f_*(\pi_1(\widetilde X,\tilde q))$ isomorphically onto $f_*(\pi_1(\widetilde X, \tilde p))$. An application of the general lifting lemma then produces a lift of $f^{-1}$ that takes $\tilde q$ to $\tilde p$, and the uniqueness clause of the general lifting lemma let's one conclude that this lift is a continuous inverse of $\tilde f$, hence $\tilde f$ is a homeomorphism.


问:
How to write the implicit equation of a surface when given the parametric equations?
How can I find the implicit equations of a surface if I have the parametric equations? For example, if the surface $(S)$ is given by:
$$x = u+\sin v$$
$$y=u+\cos v$$
$$z = u+a$$
what are the implicit equations of this surface?

答:
Hint.

Essentially you want to cancel out the parameter $u$ and $v$.

The parameter $u$ disappears immediately if you substitute $u=z-a$ into the first and second equations:
$$
x=z-a+\sin v,\quad y=z-a+\cos v\tag{1}
$$

This should remind you of a circle in poloar coordinates.

Now apply the identity $\sin^2 v+\cos^2 v=1$ to combine the two equations in (1).


问:
Does sequence converge in $C([0,1],\mathbb{R})$?
Does the sequence $(1-x)x^k$ converge uniformly on $[0,1]$? I am confident the answer is yes however I am having difficulty showing this, here is my attempt:
for any $\epsilon>0$ I need to find an $N$, $k \geq N \implies |(1-x)x^k|<\epsilon$ for all $x \in [0,1]$.

Attempt:

Here is my attempt to choose an $N$ independent of $x$.

Since the function is continuous on a closed interval there is a $c \in (0,1)$ such that $|(1-x)x^k|\leq |(1-c)c^k|$ for all $x \in [0,1]$. Then $|(1-c)c^k|<\epsilon \implies c^k<
\frac{\epsilon}{1-c}$. Now I am stuck, how am I supposed to choose $N$ here?

答:
To bound $f_k$, note that for $x\ge 1-\frac1{\sqrt k}$,
$$\tag1 f_k(x)\le 1-x=\frac1{\sqrt k}$$ and for $x\le 1-\frac1{\sqrt k}$,
$$\tag2f_k(x)\le x^k\le\left(1-\frac1{\sqrt k}\right)^k\approx e^{-\sqrt k}$$
so that $\sup f_k\to 0$ as desired.

To make $(2)$ more explicit, note that for $k\ge 2$ by Bernoulli's ineqaulity,
$$\frac1{\left(1-\frac1{\sqrt k}\right)^k}
=\left(1+\frac1{\sqrt k-1}\right)^k \ge 1+\frac k{\sqrt k-1}>\sqrt k$$
so that $$ f_k(x)<\frac1{\sqrt k}$$


问:
Correspondance between section of covering space and points in the preimage
Let $p:\tilde{X}\to X$ be a covering map (where $X$ is path conncetd and locally path connected). We call a continuous map $s:X\to\tilde{X}$ a section if $p\circ s=id_X$. Fix a point $x_0\in X$. Prove that there exists a bijective correspondance between the set of points $y\in p^{-1}(x_0)$ such that every loop $f$ with base point $x_0$ lifts to a loop $\tilde{f}$ with basepoint $y$ and the sections of $p$.

Attempt: I defined a map $s_y:X\to\tilde{X}$ with $s_y(x)=\tilde{f}(1)$ where $f$ is a path with $f(0)=x_0,f(1)=x$ and $\tilde{f}(0)=y$. Can we prove this correspondance is onto?

答:
Let $L(x_0)$ denote the set of points $y\in p^{-1}(x_0)$ such that every loop $f$ with basepoint $x_0$ lifts to a loop $\tilde{f}$ with basepoint $y$. Note that the latter means that the unique lift $\tilde{f}$ of $f$ with $\tilde{f}(0) = y$ has the property $\tilde{f}(1) = y$.

Let $S$ denote the set of sections $s : X \to \tilde X$ and define $\phi : S \to p^{-1}(x_0), \phi(s) = s(x_0)$.

$\phi$ is injective: Each section $s$ is a lift of $id_X$, thus $s$ is uniquely determined by the value $s(x_0)$. Here we use the fact that $X$ is connected.

$\phi(S) \subset L(x_0)$: For any loop $f$ with basepoint $x_0$ the loop $s \circ f$ is a lift of $f$ with basepoint $y$.

It remains to show that each $y \in L(x_0)$ is in the image of $\phi$. We need some preprations.

For each path component $C$ of $\tilde X$, the restriction $p\mid_C : C \to X$ is a covering map.

i) $p \mid_C$ is surjective: Let $x \in X$.  Pick $c_0 \in C$ and choose a path $u : I \to X$ such that $u(0) = p(c_0)$ and $u(1) = x$. There exists a lift $\tilde u$ of $u$ such that $\tilde u (0) = c_0$. Then $\tilde u(1) \in C$ and $p(\tilde u(1)) = u(1) = x$ which shows that $x \in p(C)$.

ii) Let $x \in X$. Choose a path connected open neigborhood $U$ of $x$ which is evenly covered by $p$. Then $p^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha$ with pairwise disjoint open $V_\alpha \subset \tilde X$ which are mapped by $p$ homeomorphically onto $U$. Thus each $V_\alpha$ is path connected and therefore contained in some path component of $\tilde X$. Hence $(p \mid_C)^{-1}(U) = \bigcup_{\alpha \in A_C} V_\alpha$ for a suitable nonempty subset $A_C \subset A$. This means that $U$ is evenly covered by $p \mid_C$.

The cardinality of $(p \mid_C)^{-1}(x_0)$ is the number of sheets of $p \mid_C$. Let $y \in p^{-1}(x_0)$ and $C$ its path component. We claim that

(a) If $p \mid_C$ has one sheet, then $y \in \phi(S) \subset L(x_0)$.

(b) If $p \mid_C$ has more than one sheet, then $y \notin L(x_0)$.

This clearly proves $\phi(S) = L(x_0)$.

(a) In this case $p \mid_C$ is an open bijection, i.e. a homeomorphism. Thus $s = (p \mid_C)^{-1} : X \to C \hookrightarrow \tilde X$ is a section such that $s(x_0) = y$ which means $y \in \phi(S)$.

(b) In this case there exists a point $y' \in C$ other than $y$ such that $p(y') = x_0$. There is a path $\tilde f$ in $C$ from $y$ to $y'$; it is no loop. On the other hand f = $p \circ \tilde f$ is a loop based at $x_0$. But since $\tilde f$ is the unique lift of $f$ with $\tilde f(0) = y$, we see that $y \notin L(x_0)$.

Update:

Instead of steps 3. and 4. we can invoke the following two well-known theorems:

(1) Let $p:\tilde{X}\to X$ be a covering map, $x_0 \in X$ and $y \in p^{-1}(x_0)$. The induced homomorphism $p_* : \pi_1(\tilde X,y) \to \pi_1(X,x_0)$ is injective. The image subgroup $p_*(\pi_1(\tilde X,y)) \subset \pi_1(X,x_0)$ consists of the homotopy classes of loops in $X$ based at $x_0$ whose lifts to $\tilde X$ starting at $y$ are loops.

(2) Lifting theorem. Let $p:\tilde{X}\to X$ be a covering map, $x_0 \in X$ and $y \in p^{-1}(x_0)$ and $g : (Z, z_0) \to (X,x_0)$ be a map  with $Z$ path-connected and locally path connected. Then a lift $\tilde g : (Z, z_0) \to (\tilde X, y)$ of $g$ exists if and only if $g_*(\pi_1((Y, y_0)) \subset p_*(\pi_1(\tilde X,y))$.

Now (1) says that $y \in L(x_0)$ if and only if $p_*(\pi_1(\tilde X,y)) = \pi_1(X,x_0)$ and (2) says that the identity $id_X : (X,x_0) \to (X,x_0)$ has a lift $s : (X,x_0) \to (\tilde X,y)$ (which is nothing else than a section of $p$ with $\phi(s) =y$) if and only if $\pi_1(X,x_0) = (id_X)_*(\pi_1((X, x_0)) \subset p_*(\pi_1(\tilde X,y))$. The latter condition means $\pi_1(X,x_0) = p_*(\pi_1(\tilde X,y))$. This shows $L(x_0) = \phi(S)$.


问:
Identical balls for non-identical people
I have 40 identical black balls, 23 identical red balls, and 31 identical yellow balls. I want to distribute this to 12 distinct people such that each person gets at least one of each color ball. How many ways can I make this distribution?

I started with the stars and bars method.

For the 40 black balls:
$\binom{39}{11}$

For the 23 red balls:
$\binom{22}{11}$

For the 31 yellow balls:
$\binom{30}{11}$

After this I'm unsure. Do I just apply multiplication rule and multiply all three of these combinations?

答: Since the distribution of one color of balls does not affect the other one, that means that the number of distributions for each color are independent, thus you can safely multiply all the numbers you got. I also checked your work, and I believe you got the right numbers to multiply together.</p>


答:
Call your sequence $f_k(x)$. Consider first finding the maximum as a function of $k$. The derivative gives:

$$-x^k+(1-x)kx^{k-1}=0$$

and when $x\neq 0$, $(1-x)k = x$, giving $x=k/(1+k)$. Using a second derivative test, show that this is indeed a maximum. So $f_k(x)\leq f_k(k/(1+k))$.

$$f_k((k/(1+k))=(1/(k+1))(k/(1+k))^k\rightarrow 0$$

which should give you uniform convergence in $x$.

Title: Does sequence converge in $C([0,1],\mathbb{R})$?

Body:
Does the sequence $(1-x)x^k$ converge uniformly on $[0,1]$? I am confident the answer is yes however I am having difficulty showing this, here is my attempt:
for any $\epsilon>0$ I need to find an $N$, $k \geq N \implies |(1-x)x^k|<\epsilon$ for all $x \in [0,1]$.

Attempt:

Here is my attempt to choose an $N$ independent of $x$.

Since the function is continuous on a closed interval there is a $c \in (0,1)$ such that $|(1-x)x^k|\leq |(1-c)c^k|$ for all $x \in [0,1]$. Then $|(1-c)c^k|<\epsilon \implies c^k<
\frac{\epsilon}{1-c}$. Now I am stuck, how am I supposed to choose $N$ here?


问:
A problem from "solving mathematical problems" by Terence Tao. Find the least no. of concerts
This is a problem from "Solving Mathematical Problems" by Terence Tao. I wasn't able to solve it, so I asked y'all. Please help me.

The complete problem-

Six musicians gathered at a music festival. At each concerts some musicians played in the concerts while others listened as part of the audience. What is the least number of concerts needed to be scheduled in order that each musician may listen, as part of the audience, to each other musician.

Hint, also from the book-

Hint: obviously not everyone can listen to every one else in one concert, so more than one concert is needed to exhaust all the listening possibilities think upon these lines and also of the point scoring ideas and you will get a reasonable lower bound for the number of concerts needed find an example satisfying the lower bound and you have solved it

Developments - I made a graph, represented the musicians with the nodes... Graph And it will be a directed graph where $A \rightarrow B$ would mean $A$ has sung and $B$ has listened as the audience... For example if $1$ sang in the first concert then the graph would be...Graph and so on. Now at last we will need a fully complete graph in which there are two edges (opp directions) b/w each pair of nodes so $n(n-1)$ edges in total

Observation- The order of concerts doesn't change the final graph, we can say this because anytime there is a concert and some edges have to be made, if they were already there then there is no change but if they were not there then they will be made so essentially we are adding the set of edges to be made to the set of edges already there to get the new graph's edges and because addition is commutative so therefore the order doesn't matter

i.e. $E_i + E_1 = E_f$ and $E_i +E_1 +E_2 =E_f$ is the same as $E_i+E_2+E_1$

As the hint said, I tried to find a lower bound by calculating the max no. of edges possible with one concert so in the graph if there are n nodes and in the concert $k$ musicians sing then then the max no. of edges increased is $k(n-k)$ And the max value of this is $\frac{n^2}{4}$ And now dividing this with the total no. of edges to be made $n(n-1)$ we get the lower bound to be $\frac{4(n-1)}{n}$ And for the specific case of $6$ musicians is $3.333$ i.e. $4$ but I am not able to make the concerts required for completion. Please help.

答:
Solution. Let the musicians be A, B, C, D,
E and F. Suppose there are only three
concerts. Since each of the six must
perform at least once, at least one
concert must feature two or more
musicians. Say both A and B perform in
the first concert. They must still perform
for each other. Say A performs in the
second concert for B and B in the third
for A. Now C, D, E and F must all
perform in the second concert, since it is
the only time B is in the audience.
Similarly, they must all perform in the
third. The first concert alone is not
enough to allow C, D, E and F to
perform for one another. Hence we need
at least four concerts. This is sufficient,
as we may have A, B and C in the first, A,
D and E in the second, B, D and F in the
third and C, E and F in the fourth.</p>


答:
Set $f_k(x)=(1-x)x^k$. Note that $f_k$ is positive and since $f_k$ is continuous on the compact interval, it admits its maximum value. We can solve $\frac{d}{dx}f_k(x)=0$ to find the maximum. We have that $\frac{df_k}{dx}(x)=kx^{k-1}-(k+1)x^k$, so $kx^{k-1}-(k+1)x^k=0$ if and only if $x=0$ or $x=\frac{k}{k+1}$. It is easily verified that the maximum value of $f_k$ occurs at $x_k=\frac{k}{k+1}$.

So
$$\sup_{x\in[0,1]}f_k(x)=f_k(x_k)=f_k(\frac{k}{k+1})=\frac{1}{k}\cdot\frac{k}{k+1}\cdot\big(\frac{k}{k+1}\big)^k=\frac{1}{k}\cdot\big(1-\frac{1}{k+1}\big)^{k+1}\to0$$

because $1/k\to0$ while $(1-\frac{1}{k+1})^{k+1}\to e^{-1}$. Since the supremum of $f_k$ tends to $0$ as $k$ becomes large, we have that $f_k$ converges to $0$ uniformly.


问:
Closure of a subset of $\mathbb{Q}$ under addition and multiplication

Let $A$ be the set of real numbers of the form $A=\left\{\left. a+b\sqrt{2}\right|a \in \mathbb{Q}, b\in \mathbb{Q}\right\}$. Prove that if $x \in A$ and $y \in A$, then $x+y\in A$ and $xy \in A$.

I am taking my first proof writing class and having a hard time with this proof. Any hints, guidance, or answer would be appreciated. Thank you

答:
Let $x,y$ be two elements of $\mathbb{A}$, they have to be written as follow:
$x=e+f\sqrt{2}$
$y=g+h\sqrt{2}$
with $e,f,g,h\in\mathbb{Q}$.
Now, $x+y=e+f\sqrt{2}+g+h\sqrt{2}=(e+g)+(f+h)\sqrt{2}$ $\in$ $\mathbb{A}$. Ideed, you have that $(e+g)\in \mathbb{Q}$ and $(f+h)\in \mathbb{Q}$ because they are sum of two numbers of $\in \mathbb{Q}$, that is close whit respect to the addition. So $x+y$ respects the conditions requested to stay in $\mathbb{A}$.
For $xy$ in the similar way: $xy=(e+f\sqrt{2})(g+h\sqrt{2})=(eg+2fh)+(eh+fg)\sqrt{2}$ $\in$ $\mathbb{A}$. Indeed $(eh+fg)\in \mathbb{Q}$ and $(eg+2fh)\in \mathbb{Q}$, because $\in \mathbb{Q}$ is closed with respect to the multiplication. So even $xy$ respects the conditions requested to stay in $\mathbb{A}$. So $\mathbb{A}$ is close under addition and multiplication as a consequence that $\mathbb{Q}$ is, too.


答:
For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\rightarrow(1-x)x^{n}$.
We go to show that $f_{n}\rightarrow0$ uniformly.

Proof: Let $\varepsilon>0$ be given. Choose $\delta\in(0,\min(\varepsilon,\frac{1}{2})).$
Since $(1-\delta)^{n}\rightarrow0$, there exists $N$ such that $(1-\delta)^{n}<\varepsilon$
whenever $n\geq N$. Let $n\geq N$ and $x\in[0,1]$. We assert that
$|f_{n}(x)-0|<\varepsilon$. If $x\in[0,1-\delta]$, then $0\leq f_{n}(x)\leq x^{n}\leq(1-\delta)^{n}<\varepsilon$.
Hence, $|f_{n}(x)-0|<\varepsilon$. If $x\in(1-\delta,1]$, then $0\leq f_{n}(x)\leq1-x<\delta\leq\varepsilon$.
We also have $|f_{n}(x)-0|<\varepsilon$. Therefore $f_{n}\rightarrow0$
uniformly.


问:
What is the name of a game in which if you both choose the same, you will lose, otherwise, one will win?
I am familiar with games like Prisoners Dilemma and Chicken game. However, I could not find out the name of this specific game that I believe should have interesting properties. This game has this payoff matrix:

Do you know whether this game has a famous name?

答: Turning my comment into an answer: This is an instance of an (anti-)coordination game. See this link.</p>


问:
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$

This is one of the questions from the Indian Institution of Technology Joint Entrance Examination question. I have been struck on this problem for hours.

I was solving some intriguing maths problems on combinatorics. Then I came across this wonderful questions and I was surprised looking at it. Firstly what I thought to give it a try was to Give it a BIJECTION of Distinct Balls distribution in Identical Boxes. For example $x_1+x_2+x_3=20$ can have $(20+3-1)C(3-1)$ solutions (including zero) but how does it work here?


答:
Case $n\ge2$

We can write (for $n\ge 2$)
\begin{align}
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})^n}dt=-\frac12\int_0^{\infty}\frac{tdt}{(1+t^2)^n}+\frac{1}{2}\int_0^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)^n},
\end{align}

The first integral can be calculated in terms of elementary functions:
$$\int_0^{\infty}\frac{tdt}{(1+t^2)^n}=\frac12\left[\frac{(1+t^2)^{1-n}}{1-n}\right]_0^{\infty}=\frac{1}{2n-2}.$$

In the second integral, we have an even function, so we can write it as $\displaystyle\frac12\int_{-\infty}^{\infty}$. Next it becomes easy to evaluate it by residues by pulling the integration contour to $i\infty$. In this way it will be given by the contribution of poles at $t_k=ik$, $k\in\mathbb{Z}_{>0}$, where in addition all the poles except $t_1=i$ are simple, and $t_1$ is of order $n+1$. Explicitly,
$$\int_{-\infty}^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)^n}=-\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^n}+2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^n}.$$
It is rather clear that the first sum can be expressed in terms of $\zeta$-function evaluated at integer values and the second term can be obtained by calculating a sufficient number of derivatives.

For example,
\begin{align}
&\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^2}=\frac{5}{8},\\
&\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^3}=\frac{7-8\zeta(3)}{16},\\
&\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^4}=\frac{45-32\zeta(3)}{128},\\
&\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^5}=\frac{77-40\zeta(3)-32\zeta(5)}{256},\\
&\\
&\ldots\;\ldots\;\ldots
\end{align}
and also
\begin{align}
2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^2}&=\frac{4\pi^2+3}{24},\\
2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^3}&=\frac{2\pi^2+3}{48},
\end{align}
and so on.

Combining all these results, one can calculate the initial integral for any integer $n\ge 2$:
$$\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})^n}dt=\frac14\left[2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^n}-\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^n}-\frac{1}{n-1}\right].$$

The case $n=1$ should be treated separately, I will do it  later (or maybe someone will come out with his own proof before).

Added: Case $n=1$

(I don't really like this derivation, there should be a simpler one; however, I am quite satisfied with my previous derivation for  $n\ge 2$)

Here the problem is that we cannot separate the integral into two parts as before. We will remedy this by considering a slightly more general integral
$$I(\alpha)=\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})(1+\alpha^2 t^2)}dt,\tag{1}$$
and recovering the result we are interested in in the limit $\alpha\rightarrow 0$. Repeating our  previous manipulations with (1), we find
\begin{align}
I(\alpha)=-\frac12\int_0^{\infty}\frac{tdt}{(1+t^2)(1+\alpha^2t^2)}+\frac{1}{4}\int_{-\infty}^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)(1+\alpha^2 t^2)},
\end{align}

Evaluating these integrals (the first directly, the second by residues), one obtains (for $\alpha>0$)
$$I(\alpha)=\frac{1}{2}\frac{\ln\alpha}{1-\alpha^2}-\frac14 \sum_{k=2}^{\infty}\frac{2k}{(1-k^2)(1-\alpha^2 k^2)}+\frac{1+3\alpha^2}{8(1-\alpha^2)^2}-\frac{\pi\coth\frac{\pi}{a}}{4(1-\alpha^2)},$$
where the last two terms correspond to the contributions of the poles at $t=i$ and $t=i/\alpha$. Hence the integral we are interested in is given by
$$I(0)=\frac{1}{2}\lim_{\alpha\rightarrow 0}\left[\ln\alpha-\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}\right]+\frac{1}{8}-\frac{\pi}{4}.\tag{2}$$
The sum inside the limit can be calculated in closed form for any $\alpha$ (recall that e.g. $\sum_{k=2}^N\frac{1}{k+a}=\psi(1+N+a)-\psi(2+a)$):
\begin{align}\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}&=-\frac{4\gamma-3+2\psi(2+\alpha^{-1})+2\psi(2-\alpha^{-1})}{4(1-\alpha^2)}=\\
&=-\frac{4\gamma-3+2\psi(2+\alpha^{-1})+2\psi(-1+\alpha^{-1})+2\pi\cot\frac{\pi}{\alpha}}{4(1-\alpha^2)}
\end{align}
Therefore, to compute the limit in (2), it suffices to know the asymptotics of digamma function for large argument. It is given by
$$\psi(\Lambda)=\ln\Lambda+O\left(\Lambda^{-1}\right),$$
and therefore we get that, as $\alpha\rightarrow 0$,
$$\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}=\ln\alpha-\frac{4\gamma-3+2\pi}{4}+O\left(\alpha\right).$$
Being combined with (2), this gives the result: $I(0)=\frac{\gamma}{2}-\frac14$. $\blacksquare$

Title: How to prove a generalized integral identity

Body:
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})}dt=-\frac{1}{4}+\frac{\gamma}{2}
$$
where $\gamma$ = Euler Gamma
$$
\int_{0}^{\infty }\frac{t}{( e^{2\pi t}-1)(1+t^{2}) ^{2}}dt=\frac{\pi^2}{24} -\frac{3}{8}
$$
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(
1+t^{2})^{3}}dt=\frac{\pi^2}{96} +\frac{\zeta(3)}{8} -\frac{7}{32}
$$</p>


问:
Is it silly to write $\phi : [a,b] \rightarrow \phi([a,b])$?
If it is rather silly, how would you express such a function? For example (to be specific), if $\phi : [a,b] \rightarrow \mathbb{R}$ is injective, how would you express the fact that the same function with the co-domain restricted to the range of $\phi$ is bijective?

EDIT: I am assuming I haven't previously defined what the function $\phi$ is in any way.

ANSWER FROM COMMENTS: It is not silly to write $\phi : [a,b] \rightarrow \phi([a,b])$.

答:
It doesn't quite sit so well with me as it seems to with people in the comments, but then I'm usually a bit of a stickler for notation. It strikes me as an abuse of notation.

A function $\phi$ ought be defined with a fixed, known domain and codomain (before you can even define where in the codomain it sends a point $x$ in the domain). In order to parse $\phi([a, b])$, I need to already know where $\phi$ sends the points in its domain, which means I already have some idea of the codomain (these points have to be mapping to somewhere, right?). This original codomain should technically be specified.

If I were doing it properly, I'd prefer to talk about $\phi : [a, b] \to \Bbb{R}$ (or whatever other codomain) first, then define some $\hat{\phi} : [a, b] \to \phi([a, b])$, or something like that.

Now, that said, being an abuse of notation doesn't mean that it's silly, or even that it's at all bad. We use abuses of notation all the time. So long as it is clear what you're saying (and I'd say it is pretty clear), then most people will be fine with it. In fact, it can often be preferable to abuse notation than to, say, define all your functions twice and leave hats over all of them!


问:
Three mutually perpendicular chords in a sphere with radius R
In a sphere, with radius R, there are three mutually perpendicular chords. They intersect at a point P (not the center). Point P divides the three chords into segments: a, b, c, d, e, f.

Is it true that: a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6R^2

答:
No, that's not true. More precisely,
$$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 2\,(3\,R^2-r^2),$$ where $r$ is the distance of $P$ from the center of the sphere.


问:
Side-limit of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$
The problem wants the value of left hand limit of the derivative of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ when $x\to3^-$.I graphed the function and the answer is $\frac{3}{2}$ but what's wrong with my method:

For any value $x\to3^-$ we have $[x]=2$ , $|x-1|=x-1$ so if $h\to0^-$, then $f(3+h)=\frac{2(3+h)^2}{2+h}=\frac{2h^2+12h+18}{2+h}$. So the value of the derivative is equal to:
$$\lim_{h\to0^-}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to0^-}\frac{\frac{2h^2+12h+18}{2+h}-\frac{27}{2}}{h}$$ which is infinity and clearly wrong with or without the graph. I'm not new into this but I can't figure out what's wrong here despite I thought about the method for one hour straight.

答: I would disagree with you that the answer is 3/2. The left hand limit is +inf, and the right hand limit is -inf, as you can see if you graph the function of the derivative.</p>


问:
Show that $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$, where $\left( \frac{q}{p} \right)$ is the Legendre Symbol

Show that if $p$ is any odd prime then

$$\left( \frac{q}{p} \right) \equiv q^{\frac{p-1}{2}} \mod p.$$

stating any theory that you use. In particular, you may assume the existence of a primitive element in $G_p$.

Here $\left( \frac{q}{p} \right)$ is the Legendre Symbol and $G_p$ is the group of elements $g \mod p$ such that $\gcd(g,p) = 1$. I said that for some $a \in \mathbb{Z}$, we have

$$a \equiv q^{(p-1) / 2} \mod p \implies a^2 \equiv q^{(p-1)} \equiv 1 \mod p$$

by Fermat's little theorem. And so, by definition of the Legendre symbol, we have that if $q$ is a quadratic residue mod $p$ then $\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p$. I'm now stuck on how to show that it is $\equiv -1 $ if it isn't a quadratic residue. Obviously the hint with primitive elements comes into play somehow, but I can't see how it does.

Can someone help me please.

答:
As you said, by Fermat's little theorem
$$
0 \equiv q^{p-1} - 1 = (q^{(p-1)/2} + 1)(q^{(p-1)/2} - 1) \pmod p.
$$
If $q$ is a quadratic residue, then $q^{(p-1)/2} \equiv 1 \pmod p$ again by Fermat, and if it is not, then $q^{(p-1)/2}+1$ must equals $0$ in $\mathbb{Z}_p$ since it is a field.</p>


问:
Question about closed knight's tours for n x m chessboard
Is there a simple mathematical algorithm where you can get a CLOSED knight' tour on an n x m chessboard? I need a way to prove that it is mathematically possible or impossible to have a closed knight's tour on such a board.

I have puzzled over many different websites such as this one, but I can't seem to understand the techniques used and how you can explain those. Maybe it's just me being daft, but I will really appreciate a simple, easy explanation of the techniques used in one of these puzzles!

I've understood that you cannot have a closed tour on a chessboard where n x m is odd because the knight always moves to a different-coloured square every move, and if there are an odd number of turns the knight won't be able to to return to the same coloured square that it started with.

After weeks, I still don't fully understand the problem!

How should I solve and attempt this difficult puzzle?

NOTE: This diagram is the one that confused me -- the dotted lines don't make any sense to me -- how can a knight get back to the dotted lines without getting in the way of the next steps and filling in all the gaps later:

答:
I'll explain the diagram you ask about.

It illustrates how to extend a closed or open tour on an $n\times m$ board to such a tour on an $n\times(m+4)$ board. To do this you need an open tour on an $n\times4$ board that starts and ends in the two adjacent squares just above the bottom left corner.

Here it is illustrated for the specific case of extending a $6\times5$ closed tour to a $6\times 9$ closed tour.

One move out of the corner of the board is removed, and the two squares of that move are connected to the two ends of the open tour on the $n\times 4$ board. This same method works for a tour on any $n\times m$ board with $n\ge 6$ because there are $n\times 4$ open tours with the required end points for all such sizes.


问:
Implicit Differentiation of 3 variables
Find $\frac{dy}{dz}$ when

$$(-5x+z)^4-2x^3y^6+3yz^6+6y^4z=10.$$

I got an answer of

$$\frac{-24y^3(z-3)z^6+2x^36y^5}{4(-5x+z)^3+8yz^5+6y^4}.$$

Is this correct?

答:
Calculate the differential of the entire equation
$$\eqalign{
0 &= 6\,d(y^4z) - 2\,d(x^3y^6) +3\,d(yz^6) + d(z-5x)^4 \\
\\
0 &= 6(y^4dz+4zy^3dy) - 2(3x^2y^6dx+6x^3y^5dy) \\
 &\quad+\; 3(z^6dy+6yz^5dz) + 4(z-5x)^3(dz-5\,dx) \\
}$$
Assuming $x$ is held constant (i.e. $\,dx=0$) allows us to drop some terms
$$\eqalign{
0 &= (6y^4dz+24zy^3dy) - (12x^3y^5dy)+ (3z^6dy+18yz^5dz) + 4(z-5x)^3dz \\
0 &= \Big(24zy^3+3z^6-12x^3y^5\Big)dy + \Big(6y^4+18yz^5+4(z-5x)^3\Big)dz \\
}$$
This can be rearranged into the desired partial derivative
$$\eqalign{
\frac{\partial y}{\partial z}
 &= \frac{6y^4+18yz^5+4(z-5x)^3}{12x^3y^5-24zy^3-3z^6} \\
}$$

Title: Implicit Differentiation of 3 variables


问:
Trying to prove this inequality involving exponential function
Given :

A continuous non-negative function $f(x)$ on $\mathbb R^{+} \cup \{ 0\}$

$\alpha > 0$

$h(x) = \displaystyle \int \limits_0^x f(u) du + \alpha$

$\forall x$, $f \leq h$

To prove : $$f(x) \leq \alpha e^x$$

Attempt :

$$\dfrac {d}{dx} [\ln h(x)] = \dfrac {h'(x)}{h(x)}$$

By the fundamental theorem of integral calculus, $h'(x) = f(x)$ and $\therefore $

$$\dfrac {d}{dx} [\ln h(x)] = \dfrac {f(x)}{h(x)}$$

From the 4th given point, $f \leq h$,

$$\dfrac {d}{dx} [\ln h(x)] \leq 1$$

i.e. the function $\ln [h(x)]$ never grows faster than the linear function, with slope 1.

From this step, I wrote $$\ln [h(x)] \leq x + C, \text{ for some } C$$

Is the above step correct?

$$\therefore h(x) \leq e^{x+C} = e^C e^x = B e^x$$

$$\therefore f(x) \leq h(x) \leq B e^x$$

Also, am unable to get in $\alpha$ in this inequality.

答: Your strategy is a good one. Since $\ln h(x)\le x+\ln h(0)$, $h(x)\le h(0)e^x$. But $h(0)=\alpha$.


问:
Is there a constant we can pick that will make this function holmomorphic on the disc of radius 2?
Given a function $f(z)=\frac{1}{z^5 -2z^4 +z^2+18z-40}-\frac{k}{z-2}$ is there a way we can find a constant $k$ that will make this function holmomorphic on an open set containing the disc $\{z:z \leq 2\}$ ?

My thoughts so far are as follows:

We can factor the first fraction and use partial fraction decomposition to get the expression $\frac{1}{38(z-2)}+\frac{-z^3 -9-2z^2-4z}{38(z^4 +z+20)}-\frac{k}{z-2}$.

So if we take $k=\frac{1}{38}$ then the first and last fractions cancel out leaving us with $f(z)= \frac{-z^3 -9-2z^2-4z}{38(z^4 +z+20)}$.

Using Roché Theorem we can show that the denominator of this function does not have any zeros in the disc ${z:z \leq 2}$. Is this enough to prove that it will be holmomorphic on an open set containing the disc $\{z:z \leq 2\}$ ? Showing this holds on an open set containing this disc is worrying me. I know that on the disc itself it will not have zeros but I'm not sure how to extend this to an open set containing the disc.

答: The denominator is a polynomial, so it has finitely many zeros, one of which must have the smallest absolute value $a$. You've showed that $a > 2$, so the open disk of radius $a$ is an open set which contains the closed disk of radius $2$ and does not contain any zeros of the denominator.


问:
$f(x)<+\infty$ ($x\to\infty$) $ \Rightarrow$ there exists {$x_n$}$ \subset \mathbb{R} $ s.t. $x_n\to\infty$ and $ x_n f'(x_n)\to0 $
How can I prove that

$f(x)$ is $C^1$ monotone increasing function on $[0,\infty)$ ,then

$f(x)<+\infty$ ($x\to\infty$) $ \Rightarrow$ there exists {$x_n$}$ \subset \mathbb{R} $ s.t. $x_n\to\infty$ and  $ x_n f'(x_n)\to0 $

答:
Let $l=\lim_{x\rightarrow\infty}f(x)<\infty$. We construct a sequence
$(x_{n})$ recursively as follow. Choose $x_{1}\in[0,\infty)$ arbitrarily.
Suppose that $x_{1},x_{2},\ldots,x_{n-1}$ have been chosen such that
$x_{k}-x_{k-1}\geq1$ for $k=2,3,\ldots,n-1$. For $\varepsilon=\frac{1}{4n}$,
there exists $X>x_{n-1}+1$ such that $|f(x)-l|<\varepsilon$ whenever
$x\in[X,\infty)$. In particluar, $|f(2X)-f(X)|\leq|f(2X)-l|+|l-f(X)|<2\varepsilon=\frac{1}{2n}$.
By mean-value theorem, there exists $x_{n}\in(X,2X)$ such that $f(2X)-f(X) = f'(x_n)(2X-X)$. We have that
\begin{eqnarray*}
 &  & \frac{1}{2n}\\
 & > & f(2X)-f(X)\\
 & = & f'(x_{n})(2X-X)\\
 & = & \frac{1}{2}(2X)f'(x_{n})\\
 & \geq & \frac{1}{2}x_{n}f'(x_{n})
\end{eqnarray*}
(observe that since $f$ is increasing, $f'\geq0$). Hence, $x_{n}f'(x_{n})<\frac{1}{n}.$

By recursion, we obtain a sequence $(x_{n})$ such that $x_{n}-x_{n-1}\geq1$
and $x_{n}f'(x_{n})<\frac{1}{n}$ for $n=2,3,\ldots$. In particular,
$x_{n}\rightarrow\infty$ and $x_{n}f'(x_{n})\rightarrow0$.


问:
Help with understanding "Assume for infinitely amount of $n$, the sequence $s_n=t_n$"
I was wondering whether anyone could help with the follow problem:

Let the sequence $s_n$ converge to $s$ and $t_n$ converge to $t$

Assume for infinitely amount of $n$, the sequence $s_n=t_n$.

Show the limit $s=t$

I am difficulty in understanding what is meant with the "assume for infinitely amount of $n$, the sequence $s_n=t_n$"

If they aren't the same sequence does it mean one of them is a subsequence? Or does it mean something different?

答: My guess is that who wrote that meant that, for infinitely many $n$'s, we have $s_n=t_n$. But it's not a good way of expressing that.


问:
Is the $n$th root of a positive real number a function?
Consider a simple case:

Let $f$ be a function from $\textbf{R}^+$ to $\textbf{R}$, such that $f(x) = \sqrt{x}$.

It is known that a positive real number can have two square roots. A basic case: $\sqrt{4} = \{+2, -2\}$. However, this goes against the very definition of a function:

"a function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set." (Source: wikipedia)

What am I missing?

答: $\sqrt{\cdot}$ is defined to be the positive square root. There are two numbers which square to $4$, but only one of them is $\sqrt{4}$. $\sqrt{\cdot}$ is a function.


问:
Counterexample to inverse Leibniz alternating series test
The alternating series test is a sufficient condition for the convergence of a numerical series. I am searching for a counterexample for its inverse: i.e. a series (alternating, of course) which converges, but for which the hypothesis of the theorem are false.

In particular, if one writes the series as $\sum (-1)^n a_n$, then $a_n$ should not be monotonically decreasing (since it must be infinitesimal, for the series to converge).

答:
If you want a conditionally convergent series in which the signs alternate, but we do not have monotonicity, look at
$$\frac{1}{2}-1+\frac{1}{4}-\frac{1}{3}+\frac{1}{6}-\frac{1}{5}+\frac{1}{8}-\frac{1}{7}+\cdots.$$
It is not hard to show that this converges to the same number as its more familiar sister. </p>


答: What you are missing is that $\sqrt x$ does not mean “square root of $x$”; instead, it means “non-negative square root of $x$”.

Title: Is the $n$th root of a positive real number a function?

Body:
Consider a simple case:

Let $f$ be a function from $\textbf{R}^+$ to $\textbf{R}$, such that $f(x) = \sqrt{x}$.

It is known that a positive real number can have two square roots. A basic case: $\sqrt{4} = \{+2, -2\}$. However, this goes against the very definition of a function:

"a function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set." (Source: wikipedia)

What am I missing?


问:
Every proper subgroup of $\mathbb{Z}(p^{\infty})$ is of the form $\left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$ for some $n>0$.
Let $\displaystyle \mathbb{Z}(p^{\infty})=\bigcup_{n>0} \left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$. Show the following:

a) Every proper subgroup of $\mathbb{Z}(p^{\infty})$ is of the form $\displaystyle \left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$ for some $n>0$.

b) The multiplication map $\mathbb{Z}(p^{\infty}) \to \mathbb{Z}(p^{\infty})$ by $p^t$ is a surjective group homomorphism. Also find the kernel of this map.

I have noted that if $z \in \mathbb{Z}(p^{\infty})$, then $z=a/p^m$ for some $m>0$. I am not sure how to proceed.

答:
Note that $\bar{x}$ denotes the residue class of $x \in \Bbb Q$ modulo $\Bbb Z \unlhd \Bbb Q$.

Assume that $H \lneq \Bbb Z(p^\infty).$ For $t \ge 0$, define $\Bbb Z(p^t) = \left\langle \overline{\dfrac{1}{p^t}} \right\rangle.$ Note that

$$\Bbb Z(p^0) \subset \Bbb Z(p^1) \subset \Bbb Z(p^2) \subset \cdots$$

and

$$\Bbb Z(p^\infty) = \bigcup_{t \ge 0}\Bbb Z(p^t).$$

Note that $\Bbb Z(p^0)$ is the trivial group and hence $\Bbb Z(p^0) \le H.$ Now, since $H \lneq \Bbb Z(p^\infty),$ there is some $t \ge 0$ such that $\Bbb Z(p^{t+1}) \not\subseteq H$. Fix the smallest such $t$.

Thus, we have $\Bbb Z(p^t) \subset H$. We now show the reverse inclusion.

For the sake of contradiction, assume that there exists $x \in H \setminus\Bbb Z(p^t)$. Then, $$x = \overline{\dfrac{r}{p^m}}$$
for some $r \in \Bbb N,$ $m \ge 0$ with $\gcd(r, p) = 1.$ However, note that since $x \notin \Bbb Z(p^t)$, we have $m \ge t + 1$.

Now, by Bézout's identity, there exist integers $a, b$ such that $$ar + bp^m = 1 \quad\text{or}\quad a\frac{r}{p^m} + b = \frac{1}{p^m}.$$
Going modulo $\Bbb Z$ shows that $$\overline{\frac{1}{p^m}} = a\overline{\frac{r}{p^m}} \in \left\langle \overline{\frac{r}{p^m}} \right\rangle = \langle x \rangle \subset H.$$

But the above contradicts that $\Bbb Z(p^{t + 1}) \not\subset H$.

Thus, we get that $H = \Bbb Z(p^t)$.

An aside. The above shows that all the proper subgroups of $\Bbb Z(p^\infty)$ are finite. In particular, all descending chains of subgroups stabilise. This shows that $\Bbb Z(p^\infty)$ is an Artinian $\Bbb Z$-module. However, the chain $$\Bbb Z(p^0) \subsetneq \Bbb Z(p^1) \subsetneq \Bbb Z(p^2) \subsetneq \cdots$$ shows that it is not Noetherian.

Showing that the map is subjective should not be too difficult. Forgetting the residue classes for a moment, think about dividing by $p^t$ in $\Bbb Q$. Conclude from there.

Now, note that the kernel is not the whole group, since the map is subjective. Thus, the kernel is of the form $\Bbb Z(p^n)$. Now you can just check for the largest $n$ such that $\overline{\dfrac{1}{p^n}} \mapsto 0.$


问:
what is the probability that the device will operate for at least another three hours?
The lifetime of a phone is $X$ with pdf $f_{X}(x)=\frac{18}{x^{3}}$ where $x>3$. If six hours have passed since the device was activated and it is still operating, what is the probability that it will operate for at least another three hours?

I'm trying to solve this question for some time. The first thing that came to my mind is that $X$ should has Exponential distribution but it's looks like it's not the case. It feels like I need to use $f_{X|X}(x|x)$ but what does it even mean? Never saw a formula for it. I know that $f_{X|Y}(x|y)=\frac{f_{X,Y}}{f_Y(x)(x,x)}$ so should it be $f_{X|X}(x|x)=\frac{f_{X,X}(x,x)}{f_X(x)}$? It also does not make sense because I can't calculate $f_{X,X}(x,x)$ as $X$ dependents to itself. How to solve it?

Also, as I remember, lifetime is usually memoryless. So shouldn't the answer just be:
$$
P(X>9|X>6)=P(X>9)=\int_9^{\infty}f_X(x)dx=\int_9^{\infty}\frac{18}{x^{3}}dx=\frac{1}{9}
$$
But how can I use it without actually proving it?

EDIT: Found out that the answer should be $\frac{36}{81}$. But why?

答: Don't overthink it. You know $$\text{prob}(X \geq 9\ |\ X \geq 6) = \text{prob}(X \geq 9\ \text{and}\ X \geq 6) / \text{prob}(X \geq 6).$$ And you're spot on that $\text{prob}(X \geq 9\ \text{and}\ X \geq 6) = \text{prob}(X \geq 9)$.


问:
The volume of a balloon has a radius with helium given by the function $f(r) = \frac{4}{3}\pi r^{3}$.
So my goal in this problem is to find the Domain as well as establish what the dependent and independent variables are.

I have established that the dependent variable is V for volume, and the independent variable is radius (in meters).

My problem now lies with the domain.

So a small constraint : The problem dictates that the balloon holds up to $1m^3$ of helium.

Now, back to the issue of domain:

Specifically, how to find the domain of something that is $f(r) = \frac{4}{3}\pi r^3$ such that the balloon can hold $1m^3$ of helium? What I would intuitively do is get the right side of the equation to equal 1, right? But I still have pi on that side to deal with.

答: You need to solve the equation $1=\frac 43 \pi r^3$ for $r$.  To do that, multiply both sides by $\frac 3{4\pi }$ and take the cube root.  The $\pi$ is just one more constant like the $3$ and $4$.


问:
Geometric Meaning of the Partial Derivative
Could anyone explain the geometric meaning behind the partial derivative?

I know that for a "normal" derivative, its geometric meaning is the slope of the tangent of a curve. From what I understand about the partial derivative, it is the slope of the tangent of a cross section of a function with two or more variables. If anyone could expand upon the definition and clarify, I would be grateful. Thank you.

答:
Think about a surface described by the equation $z = z(x,y)$. At a given point $(x_0,y_0)$, the partial derivative $\partial z/ \partial x$ is the derivative of the function $x \mapsto z(x,y_0)$ with respect to $x$. Of course, the graph of this function $x \mapsto z(x,y_0)$ is a "slice" through the original surface that is cut by the plane $y=y_0$.

You get similar things in other dimensions. Think about a parametric surface $\mathbf S = \mathbf S(u,v)$, where $\mathbf S$ is a mapping from $\mathbb R^2$ to $\mathbb R^3$. At given parameter values $(u_0,v_0)$, the partial derivative $\partial \mathbf S/ \partial u$ is the derivative of the the curve $u \mapsto \mathbf S(u,v_0)$ with respect to $u$. In other words, it's the derivative of a "constant parameter" curve on the surface.

Generally, a partial derivative is the derivative of a curve that's formed by changing one variable while holding all others constant.

Draw some pictures illustrating the two cases I described, and I think you'll get the idea. Then, once you understand how things work in $\mathbb R^2$ and $\mathbb R^3$, you can start pondering more general situations, if you need to.


问:
Find the Particular Integral of the following ODE using the variation of parameters?
The ODE is $$xy''-(1+x)y'+y=x^2$$ and the Complementary Function (homogenous solution) is $$y_{\small CF} =A y_1+By_2$$ where, $y_1=e^x$ and $y_2=x+1$.

However, while computing the particular integral by $$y_{\small p}=-y_1\int{\frac{y_2f(x)}{W(y_1,y_2)}}+y_2\int{\frac{y_1f(x)}{W(y_1,y_2)}}$$
(where $f(x)=x^2$)

the particular integral comes out to be $$y_{\small p} = -\frac{1}{2}(x^3+3x^2+6x+6)$$
and $$xy{\small p}''-(1+x)y{\small p}'+y{\small p}$$ comes out to be equal to $x^3$ instead of $x^2$.

Any help in this matter would be much appreciated.

答:
This answer is not correct:
$$y_{\small p} = -\frac{1}{2}(x^3+3x^2+6x+6)$$
Because before to apply the method of variation of parameters; you need to put the differential equation on the canonical form:
$$y''+P(x)y'+Q(x)y=f(x)$$
Now the DE:
$$xy''-(1+x)y'+y=x^2$$
Becomes:
$$y''-\left(1+\dfrac 1x\right)y'+\dfrac yx=\color{red}{x}$$
So that the function $f$ is $f(x)=x$ and not $f(x)=x^2$. Apply the method now and you will find the correct answer:
$$y_p=-x^2$$
And the solution is:
$$y(x)=c_1e^x+c_2(x+1)-x^2$$


问:
Implicit differentiation derivative not as expected
I am trying to solve the problem: $x^{2}+xy+y^{3}=0$ using implicit differentiation.

My workings:

$(1)\,\,\,{d\over dx}{[x^2]}\,+{d\over dx}{[xy]}\,+\,{d\over dx}{[y^3]}={d\over dx}{[0]}$

$(2)\,\,\,2x+y+{{dy^3}\over{dy}}{{dy}\over{dx}}=0$

$(3)\,\,\,2x+y+3y^2\,({dy\over dx})=0$

$(4)\,\,\,{dy\over dx}=\boxed{-{{2x+y}\over{3y^2}}}$

But the answer says it should be:

$(3)\,\,\,2x+y+{{dxy}\over{dy}}{{dy}\over{dx}} + 3y^2\,({dy\over dx})=0$

$(4)\,\,\,2x+y+{dy\over dx}(x+3y^2)=0$

$(5)\,\,\,{dy\over dx}=-{2x\,+\,y\over{x\,+\,3y^2}}$

Why?

答:
Differentials will make the process much more intuitive.
$$\eqalign{
&0 = d\Big(x^2 + xy + y^3\Big) \\
&0 = 2x\,dx + \big(x\,dy+y\,dx\big) + 3y^2dy \\
&\big(x+3y^2\big)\,dy = -\big(2x+y\big)\,dx \\
&\frac{dy}{dx} = -\left(\frac{2x+y}{x+3y^2}\right) \\
}$$


问:
Doubt - Optimal Control Theory Problem with transversality condition
I'm trying to solve this optimal control theory problem, but i'm having some problem finding the constants $k_1$ and $k_2$. I will show what I have developed so far.

The problem:

$$V(y)=\int_{0}^{1}u^2+2yu+2y^2\,dt$$ with $$y'=y+u;\quad y(0)=0;\quad y(1)=y_{1}$$ where $y(1)$ can vary and $T=1$ is given.

I will skip all the calculations, but after finding the Hamiltonian and applying the canonic equations and solving the system of differential equations I have:

\begin{align}
u&=-y-\frac{1}{2}\lambda\\
y&=-\frac{1}{2}k_1e^t+\frac{1}{2}k_2e^{-t}\\
\lambda&=k_1e^t+k_2e^{-t}
\end{align}

I should use the initial condition $y(0)=0$ and the traversality condition $\lambda(T=1)=0$ to find $k_1$ and $k_2$. But I don't know why, I can't solve these simple system:

\begin{align}
y(0)&=-\frac{1}{2}k_1+\frac{1}{2}k_2=0\\
\lambda(T=1)&=k_1e+k_2e^{-1}=0
\end{align}

I find $k_1=k_2=0$ which seems wrong to me.

Did I miss something in the process?

Thank you very much.

答: Your solution is fine.  Note that if $$H = u^2+2yu+2y^2 + \lambda(y + u)$$ then $$\frac{\partial H}{\partial u} = 2u + 2y + \lambda$$ and $$\frac{\partial^{2} H}{\partial u^{2}} = 2 > 0,$$ which means that this is a minimization problem.  It's not unreasonable to expect a minimization problem involving several squares and a product to have its minimum when the state and control are zero, particularly when the differential equation for $y$, along with the initial condition, tell us that $yu \ge 0$ at least initially.


问:
If $X_t=\int_0^t \sigma _s\,\mathrm d W_s$, how can I prove that $\mathrm d \left<X\right>_t=\sigma _t^2\,\mathrm d t$
Let $\sigma $ a very nice function. Let $X_t:=\int_0^t \sigma _s\,\mathrm d W_s$. We define $$\left<X\right>_t=\lim_{n\to \infty }\sum_{i=0}^{m_n-1}(X_{t_{i+1}^{(n)}}-X_{t_i^{(n)}})^2$$
where the limit is taken in the probability sense and $0=t_0^{(n)}<t_{1}^{(n)}<...<t_{m_n}^{(n)}=t$ is s.t. $\max_{i=0,...,m_n-1}|t_{i+1}^{(n)}-t_i^{(n)}|\to 0$ when $n\to \infty $.

Attempts : no need to read the steps 1 and step 2, just step 3 is relevant.

Step 1: Suppose $\sigma $ is of the form $\sigma _s=Y\boldsymbol 1_{[u,v)}$ where $Y$ is $\mathcal F_u-$measurable. Take $(t_i^{(n)})$ being a partition of $[u,v]$ s.t. $\max_{i}|t_{i+1}^{(n)}-t_i^{(n)}|\to 0$ when $n\to \infty $.

Then
\begin{align*}
&\mathbb E\left[\left(\sum_{i=0}^{m_n-1}(X_{t_{i+1}^{(n)}}-X_{t_i^{(n)}})^2-\int_0^t\sigma _s^2\,\mathrm d s\right)^2\right]\\
&=\sum_{i=0}^{m_n-1}\sum_{j=0}^{m_n-1}\mathbb E\left[Y^4\Big((W_{t_{i+1}^{(n)}}-W_{t_i^{(n)}})^2-(t_{i+1}^{(n)}-t_i^{(n)})\Big)\Big((W_{t_{j+1}^{(n)}}-W_{t_j^{(n)}})^2-(t_{j+1}^{(n)}-t_j^{(n)})\Big)\right]\\
&=\sum_{i=0}^{m_n-1}E\left[Y^4\Big((W_{t_{i+1}^{(n)}}-W_{t_i^{(n)}})^2-(t_{i+1}^{(n)}-t_i^{(n)})\Big)^2\right]\\
&=2\mathbb E[Y^4]\sum_{i=0}^{m_n-1}(t_{i+1}^{(n)}-t_i^{(n)})^2\\
&\leq 2\mathbb E[Y^4](v-u)\max_{i=0,...,m_n-1}|t_{i+1}^{(n)}-t_i^{(n)}|\underset{n\to \infty }{\longrightarrow }0.
\end{align*}

Step 2: Now, if $\sigma $ is a predictable process, the step 1 applies.

Step 3: If $\sigma $ is more general (i.e. s.t. $\mathbb E\int_0^t \sigma _s^2\,\mathrm d s<\infty $), there is a sequence $(\sigma ^n)$ of predictable process s.t. $\sigma _n\to \sigma $ in $L^2$. In particular $$\int_0^t \sigma ^n_s\,\mathrm d W_s\to \int_0^t \sigma _s\,\mathrm d W_s,\quad \text{in }L^2.$$

Let $(t_i^{(m)})$ a partition of $[0,t]$ s.t. $0=t_0^{(m)}<...<t_{k_m}^{(m)}=t$ s.t. $\max_{i=0,...,k_m-1}|t_{i+1}^{(m)}-t_i^{(m)}|\to 0$ when $m\to \infty $.

I know by step 2 that $$\lim_{m\to \infty }\mathbb E\left[\left(\sum_{i=0}^{k_m-1}\left(\int_{t_{i}^{(m)}}^{t_{i+1}^{(m)}}\sigma _s^{(n)}\,\mathrm d W_s\right)^2-\int_0^t(\sigma _s^n)^2\,\mathrm d s\right)^2\right]=0.$$

From that, how can I prove that

$$\lim_{m\to \infty }\mathbb E\left[\left(\sum_{i=0}^{k_m-1}\left(\int_{t_{i}^{(m)}}^{t_{i+1}^{(m)}}\sigma _s\,\mathrm d W_s\right)^2-\int_0^t(\sigma _s)^2\,\mathrm d s\right)^2\right]=0 \ \ ?$$

If this way doesn't work, maybe is there an other way ?

答:
Let's assume $E\int_0^T \sigma_s^2\,ds < \infty$. Then the usual way to show that $\langle X \rangle_t = \int_0^t \sigma_s^2\,ds =: A_t$ is by showing that $X_t := \int_0^t \sigma_s dW_s$ is a martingale, and that $X^2 -  A$ is a martingale. This property characterizes the quadratic variation and will prove your claim.

Take simple $\sigma^{(n)} \to \sigma$ converging in the sense that $E\int_0^T |\sigma_s-\sigma^{(n)}_s|^2\,ds \to 0$. For simple $\sigma^{(n)}$ the martingale and quadratic variation properties are easy to check. Let $I^{(n)}_t := \int_0^t \sigma^{(n)}_s\,dW_s$ and $A^{(n)}_t := \int_0^t|\sigma^{(n)}_s|^2 \,ds$.
First check the martingale property.
Fix $s \leq t \leq T$.

$$
E[I^{(n)}_t|\mathcal{F}_s] = I^{(n)}_s
$$
By the Ito isometry for simple functions, the RHS tends to $X_s$ in $L^2$, and since the integrand of the LHS converges to $X_t$ in $L^2$ and the conditional expectation is a contraction on $L^p$ for any $p \geq 1$ (use that $E[E[Y | \mathcal{F}]^p]] \leq E[Y^p]$ by conditional Jensen and tower property for any $Y, \mathcal{F}$), we find that $E[I^{(n)}_t|\mathcal{F}_s] \to E[X_t | \mathcal{F}_s]$ in $L^2$ as well, showing $E[X_t |\mathcal{F}_s] = X_s$. Thus $X$ is a martingale.

Similarly we can check the quadratic variation as follows.
By the properties for simple integrands, we have
$$
E[(I^{(n)}_t)^2 - A^{(n)}_t|\mathcal{F}_s] = (I_s^{(n)})^2 - A^{(n)}_s
$$
Then since $(I_s^{(n)})^2 - A^{(n)}_s \to X^2_s-A_s$ in $L^1$ (definition of $\sigma^{(n)} \to \sigma$ for the second term, isometry for simple processes for the first term) and
$(I_t^{(n)})^2 - A^{(n)}_t \to X^2_t-A_t$ in $L^1$ plus the fact that the conditional expectation is a contraction on $L^1$ shows that $X^2-A$ is a martingale.
By uniqueness of the quadratic variation, we find that $A = \langle X \rangle$.


问:
Proof by Induction: $f^{n}(x) = 2^n n!(5-2x)^{-(n+1)}$
I have trouble understanding the following problem:

"Consider the function $f(x)=(5-2x)^{-1}$:
Proof by induction, for all $n \in \mathbb{N}_0$:
$$f^{(n)}(x) = 2^n n!(5-2x)^{-(n+1)}$$
Additionally calclulate the taylorseries of $f$ for $x_0 = 1$".

What exactly should be proofed here? I have trouble understanding the notation.
For wich $f$ should the Taylor series be calculated?

答:
You have to show that the $n-$th derivative of $f(x)$ has that required form using induction. That means, you show that the first derivative is $f'(x)=2^1\cdot n!\cdot (5-2x)^{-(1+1)}$ and that if the $n-$th derivative of $f$ is given by the formula, then $f^{(n+1)}(x)$ has the desired formula. So, essentially, you assumed the formula for $f^{(n)}(x)$ and show it for $f^{(n+1)}(x).$ For that, just differentiate once.

Remember that taylor expansion around a point $x_0$ is $$f(x)=\sum _{n=0}^{\infty}f^{(n)}(x_0)\frac{(x-x_0)^n}{n!},$$
so you use the formula you showed at $x_0=1.$


问:
Find number of elements in the required set

Consider 3 sets A, B,C such that A contains all the 3 digit numbers that are multiples of 4, B contains 3 digit numbers that are even and multiples of 3, and C contains 3 digit numbers that are multiples of 5. Find number of elements in $(A\cup B\cup C )$

Individually, A, B, C contain 225,150,180 elements respectively

I am having trouble in removing the repeated elements. Had it been 2 sets, the process was fairly straight forward, ie. finding common multiples of both sets and subtracting from total elements. But having 3 sets complicates things, and while might eventually figure it out by manual calculations, I want to know if there is a more efficient way of solving.

答:
Hint:

Use the principle of inclusion and exclusion:  $$|A\cup B\cup C| = |A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$$

$A\cap B$ is a set of all 3 digit number that are divisible by $12$ and so on...


Title: Proof by Induction: $f^{n}(x) = 2^n n!(5-2x)^{-(n+1)}$

Body:
I have trouble understanding the following problem:

"Consider the function $f(x)=(5-2x)^{-1}$:
Proof by induction, for all $n \in \mathbb{N}_0$:
$$f^{(n)}(x) = 2^n n!(5-2x)^{-(n+1)}$$
Additionally calclulate the taylorseries of $f$ for $x_0 = 1$".

What exactly should be proofed here? I have trouble understanding the notation.
For wich $f$ should the Taylor series be calculated?

Body:
Welcome to MSE. If you try some first derivation you will find the clue.
$$f(x)=(5-2x)^{-1}\\f'=(-1)(5-2x)^{-2}(-2)=+2(5-2x)^{-2}(1)\\
f''=+2(-2)(5-2x)^{-3}(-2)=+2^2(5-2x)^{-3}(2.1)\\f'''=2^2(-3)(5-2x)^{-4}(-2)=2^3(5-2x)^{-4}(3.2.1)\\f^4=2^4(5-2x)^{-5}(4.3.2.1)\\\vdots$$
$$f^{(n)}(x) = 2^n n!(5-2x)^{-(n+1)}\\n\to n+1\\(f^{(n)}(x))' =( 2^n n!(5-2x)^{-(n+1)})'\\= 2^n n!(5-2x)^{-(n+2)}(-2)(-(n+1))\\=2^{n+1}(n!(n+1))(5-2x)^{-(n+2)}$$


Body:
For Taylor series arround $ x_0=1$.

Let $$g(x)=f(x+1)=\frac{1}{3-2x}=\frac13\frac{1}{1-\frac 23 x}$$
$$=\frac 13\frac{1}{1-X}$$
$$=\frac 13\sum_{n=0}^NX^i+o(X^N)$$

With $ X=\frac 23x $ .

So
$$f(x)=$$
$$\frac 13\sum_{n=0}^N(\frac 23)^n(x-1)^n+o((x-1)^N)$$


问:
Prove that exists a vector n-dimensional $x \neq 0$ such that $(A^2+B^2)x=0$ for two different square matrices $n\times n$.
Let $A$ and $B$ be two different square matrices $n\times n$. If $A^3=B^3$ and $A^2B=B^2A$. Prove that exists a vector n-dimensional $x \neq 0$ such that $(A^2+B^2)x=0$.

My first thought was that $(A^2+B^2)$ should be equal to $0$ as $x \neq 0$, but then I realize that with matrices this is not true. For example: if we have the matrix $$C=\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}$$
Then we should find a vector $x \neq 0$ such that $Cx=0$. $$\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=x_1\begin{bmatrix}1\\1\\1\end{bmatrix}+x_2\begin{bmatrix}2\\2\\2\end{bmatrix}+x_3\begin{bmatrix}3\\3\\3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ So a possible answer might be $$x=\begin{bmatrix}3\\3\\-3\end{bmatrix}$$ proving that the matrix is not necessarily equal to $0$.

I am not sure how I should use the fact that $A^3=B^3$ and $A^2B=B^2A$.

答:
HINT: Show that $$(A^2+B^2)A=(A^2+B^2)B$$
What could we then do if $A^2+B^2$ was invertible?


问:
Topology on Integers such that set of all Primes is open
In my topology homework we are asked to describe a topology on the Integers such that:

set of all Primes is open.

for each $x\in\mathbb Z$, the set $\{x\}$ is not open.

$\forall x,y \in\mathbb Z$ distinct, there is an open $U\ni x$ and an open $V\ni y$ such that $U\cap V=\emptyset$

i was looking at Furstenberg's topology as in this proof:
For $m, b \in\mathbb Z$ with $m > 0$ define $N(m,b):=\{mx + b : x ∈ Z\}$, an arithmetic progression streching towards infinity in both directions. A set $U$ is open if either:

$U = \emptyset$; or

For each $b\in U$ there is an $m>0$ such that $N(m,b)\subseteq U$.

but as I understand the set of Primes is not open in this topology. Now I'm not sure what should I do: is there a way to modify this topology to make set of Primes open or should I think of something completely different.

Any hints are appreciated!
thanks!

答:
A cheap way to go about it is the following:

Let $\mathbb{P}$ denote the set of prime numbers, and $\mathbb{P}^\prime = \mathbb{Z} \setminus \mathbb{P}$.

As both $\mathbb{P}$ and $\mathbb{P}^\prime$ are countably infinite, pick bijections $f : \mathbb{P} \to \mathbb{Q}$, $g : \mathbb{P}^\prime \to \mathbb{Q}$.

Define $U \subseteq \mathbb{Z}$ to be open iff $f [ U \cap \mathbb{P} ]$ and $g [ U \cap \mathbb{P}^\prime ]$ are both open subsets of $\mathbb{Q}$ (under the subspace topology).


问:
How to prove (analytically) that $\sin(\tan^{-1}(\sqrt{2}))=\sqrt{\frac{2}{3}}$?
I calculated $\sin(\tan^{-1}(\sqrt{2}))$ using Wolfram and got it as a result $\sqrt{\frac{2}{3}}$. However, I was unable to find an identity to obtain this result.

Is there any way to calculate this value without using Wolfram?

答:
Well, $\arctan x$ is an angle whose tangent function is $\frac{x}{1}$. Considering the sides of the right triangle. We have apposite side is $x$, adjacent side is $1$ and hypothenuse is $\sqrt{1+x^2}$. Therefore the sine of this angle is:

$$\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{x}{\sqrt{1+x^2}}\tag1$$

So, we have:

$$\sin\left(\arctan\left(x\right)\right)=\frac{x}{\sqrt{1+x^2}}\tag2$$

In your case:

$$\sin\left(\arctan\left(\sqrt{2}\right)\right)=\frac{\sqrt{2}}{\sqrt{1+\left(\sqrt{2}\right)^2}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}\tag3$$


问:
The inverse of a symmetric matrix
I am trying to simplify this expression where
$$
(X(X^TX)^{-1}X^T)^T = X(X^TX)^{-1}X^T
$$
should hold.
However, I am not sure what to do with $((X^TX)^{-1})^T$ once I make the transpose notation into the bracket. Any hint/help would be really appreiciated!

答:
For any invertible matrix $A$, we have:
$$(A^T)^{-1}=(A^{-1})^T\ .
$$

Now, let $A=X^TX$. Then since $A^T=A$, we have
$$
(A^{-1})^T=A^{-1}\ .
$$

Title: The inverse of a symmetric matrix