问:
$n$ lamps around a circle, find all $n$ such that the lamps can all be turned off after some moves.
Suppose there are $n \ge 3$ lamps around a circle, some switched on, the rest off. Let each move consist of toggling $3$ consecutive lamps. Find all $n$ such that it is possible to switch all lamps off after using a number of moves.

I've conjectured that all $n \equiv 0 \pmod{3}$ is impossible, while all other $n$ is possible through brute force. I'm trying to prove that for all $n$ not divisible by $3$ it is possible to execute a sequence of moves such that after the entire sequence, exactly one lamp has changed state, but I'm having some trouble. Any suggestions would be helpful.

答:
You are correct, you can always turn off all the lamps if and only if $n\not\equiv 0\pmod 3$. Since this is a fun puzzle, I would prefer to give hints.

If $n\equiv 0\pmod 3$, imagine dividing the lamps into three groups, red, green, and blue, by the repeating pattern $R, G, B, R, G, B,\dots$ Think about how the operation interacts with the lamps in each grouping, and find an invariant quantity.

If $n\not\equiv 0\pmod 3$, note that by toggling the block $(i,i+1,i+2)$, and then $(i+1,i+2,i+3)$, then net effect is that only $i$ and $i+3$ are toggled. Using this, you can also toggle $(i,i+6)$ alone, and $(i,i+9)$ alone, and in general $(i,i+3k)$ alone for any $k$, where the indices wrap around mod $n$. Since $3$ and $n$ are coprime, this allows you to toggle $(i,i+j)$ for any $j$ (how?). With this power, you should be able to figure out how how to turn off all the lights.


问:
Intersection of Closed and Compact Set is Closed
Fact

The intersection of a closed set $A$ and a compact set $B$ is closed.

Background

I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.

Relevant Facts

I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.

I also don't have that $B$ is closed otherwise it would be trivial.

Some Clarification

I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer Here.

So if $K$ is compact in an arbitrary topological space (which is just to say that
  it is a compact topological space when given its induced topology) and
   $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology

Is it true if $A$ and $B$ are not whole spaces?

I would appreciate any hints.

答:
I believe this constitutes a minimal counterexample to the modified question. Note that it is not a very "nice" space—it is neither $T_0$ nor preregular—but as requested, $A$ and $B$ are proper subsets of $X$:

Let $X = \{ 1, 2, 3 \}$.

Let $\tau = \{\varnothing, \{1\}, X \}$ be the topology for $X$.

Let $A = \{ 2, 3 \}$ and let $B = \{2\}$.

Then $A$ is closed because its complement is open, $B$ is compact because every finite set is compact, but their intersection, $\{2\}$, is not closed.

ContentLicense: CC BY-SA 3.0


问:
Is the product of metric spaces normal?
The product of a countable collection of metric spaces is always metric. Is it true in general that the product space of any collection of metric spaces is always normal?</p>

答: There is a theorem, due to Arthur Harold Stone, which states that the product of uncountably many non-compact metric spaces is never normal. So, take, for instance, an uncountable product of copies of $\Bbb R$, endowed with its usual topology. It will not be normal.

ContentLicense: CC BY-SA 4.0

Title: Is the product of metric spaces normal?


问:
Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$
I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.

Do you have any ideas? :)

EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to

$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$

By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$

And using the substitution: $t=\tan\left(x\right)$

$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$

$$t^2=\tan^2\left(x\right)$$

$$t^2=\frac{\sin^2x}{\cos^2x}$$

$$t^2=\frac{1-\cos^2x}{\cos^2x}$$

$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$

Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$

I don't think I got to the expected result but I can't seem to be able to find why…

答:
Double-check my arithmetic, but here's the strategy.

@Bernard's suggested substitution $t=\tan x$ gives
\begin{align}
& \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\
= {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align}
with $c:=\sqrt{2\sqrt{2}-1}$. You can do the rest with partial fractions.


问:
$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdWs$
I want to find such $\Theta_s$ for such the following equality holds

$$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdWs$$

My main question is:

Should I just choose $\Theta_s = e^\frac{t}{2}(e^{W_t-\frac t 2})$, because $e^\frac t 2$ does not depend on $W_s$ so I can bring it out of integral ?

答:
I assume you meant
$$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdW_s.$$
Set $$Y_t = \int_0^te^{W_s-\frac s2}dW_s.$$
Let's apply Ito formula to the equation, we obtain $$\Theta_tdW_t =\frac12e^{\frac t2}Y_tdt+e^{W_t}dW_t.$$
Since $Y\neq0$, it's well known that this equation doens't have a solution.

This is because $\int_0^t\Theta_s-e^{W_s}dW_s$ is a local martingale and $\frac12\int_0^te^{\frac s2}Y_sds$ is one iff $Y=0$.


问:
Probability distribution of a random variable X
$X:([0,1], \mathcal{B}([0,1]), \lambda) \rightarrow\left([0,2], \mathcal{B}([0,2])\right.$ a random variable  defined as $X(\sigma)=2 \sigma^{3}$, where $\lambda$ is the Lebesgue measure.

Give the probability distribution, i.e. $\mathbb{P}_X$, of X.
How do I have to find the probability distribution in this case? I can imagine that it is not difficult but I do not get it.

答:
For conciseness, denote $P_X(x)$ by $F(x)$ for $x \in \mathbb{R}$, and denote the Lebesgue measure by $\lambda$.

When $0 \leq x < 2$, direct application of distribution's definition gives that
\begin{align*}
F(x) = \lambda(X \leq x) = \lambda(2\sigma^3 \leq x) = \lambda(0 \leq \sigma \leq (x/2)^{1/3}) = \left(\frac{x}{2}\right)^{1/3}.
\end{align*}

For completeness, state the final result as
\begin{align*}
F(x) = \begin{cases}
0 & x < 0, \\
\left(\frac{x}{2}\right)^{1/3} & 0 \leq x < 2, \\
1 & x \geq 1.
\end{cases}
\end{align*}


问:
Show that $f(z)=0$.
Suppose that $f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic on the open unit disc and continuous on the closed unit disc. Assume that $f(z)=0$ on an arc of the circle $\{z\in\mathbb{C}:|z|=1\}$. Show that $f(z)=0$.</p>

答:
Here is a rough outline.

Conformally map the closed disk to the closed upper half plane so that the arc on which $f(z)$ is zero is sent to the real axis. (This is a standard exercise you should be able to do.) This transforms $f$ into an analytic function on the upper half plane with a continuous extension to the real line. Using the Schwarz reflection principle, we can extend $f(z)$ to be analytic on a neighborhood of some point on this arc. Then the identity theorem shows that $f$ is zero on the entire region, which means $f$ was identically zero originally.

Title: Show that $f(z)=0$.


问:
INMO 2021 polynomial question: Cubics with integer roots.
The question is the following:

Find all pairs of integers $(a, b)$ so that each of the two cubic polynomials $x^3 +ax+b$ and $x^3 +bx+a$ has all the roots to be integers.

Only $a=b=0$ is possible. It turns out that the sum of squares of roots is an even integer and this leads to infinite descent.

Can you tell me other methods to solve this problem?

答:
Let $x^3+ax+b=(x-u)(x-v)(x+u+v)$ with $u,v>0$, otherwise change some signs. Then
$$ a= -(u+v)^2+uv \Rightarrow |a|<(u+v)^2.$$
For the case $u,v\geq 2$ we get
$$ b=uv(u+v)\geq(u+v)^2$$
and in total $|a|<|b|$. It remains to show, that the case $u=1$ isn't possible.


问:
How to compute the time of upward movement of a Brownian motion
Suppose $B(t)$ is the standard Brownian motion. Does it make sense to compute the total time of upward movement
$$\int_0^t I(dB_\tau>0)d\tau$$
where $I$ is the characteristic function? What is a rigorous way to define it if it does make sense? Intuition dictates that the expectation of the above integral should be $\frac t2$. I am aware of Doob's upcrossing lemma. Perhaps we can use local time and Tanaka's formula. But I do not have a way to define my integral in similar terms.

答:
for almost any $\omega$, $ \tau \mapsto I_{B_\tau > 0}$ is a measurable function (since $B$ is continuous). You can therefore define and calculate $$\int_0^tI_{B_\tau>0}d\tau$$ for almost any $\omega$.

However, it is not constantly equal to $\frac t2$: why would it be?

What we can say is that its expected value is $\frac t2$, indeed $$ \mathbb E\Bigg[\int_0^tI_{B_\tau>0}d\tau\Bigg] = \int_0^t\mathbb E\Big[I_{B_\tau>0}\Big]d\tau = \int_0^t\frac12d\tau = \frac t2.$$


问:
Find the density function of X+Y, where X and Y are random variables
If $X$ and $Y$ have joint density function $f(x,y)=\frac{1}{2}(x+y)e^{-x-y}$ when $x,y>0$ (and $0$ otherwise), find the density function of $X+Y$.

Let $Z=X+Y$. Then from my textbook, it says that $f_Z(z)=\int_{-\infty}^{\infty} f_{x,y}(u,z-u)du$, where $u=x, u>0$. When I plug the joint density function into this equation, I get $f_Z(z)=\int_{0}^{\infty} \frac{1}{2}ze^{-z}du$. This integral doesn't evaluate because you would have a $u$ being evaluated at infinity. The answer in the book says $f_{X+Y}(u)=\frac{1}{2}u^2e^{-u}$ for $u>0$. I don't see how this is the answer.

答:
You made a small mistake:
$$ f_Z(z) = \int_0^{\color{blue}{z}} \frac{1}{2} ze^{-z} \ du = \frac{1}{2} z^2 e^{-z}$$

EDIT: One way to obtain this formula is to consider new variables $U = X$ and $Z = X+Y$. The Jacobian for this transformation is 1 and so
$$ f_Z(z) = \int_{-\infty}^\infty f_{U,Z}(u,z) \ du = \int_{-\infty}^\infty f_{X,Y}(u,z-u) \ du. $$


问:
A condition for a finite group $G$ be nilpotent
Is true that a finite group $G$ is nilpotent if and only if $[x,y]=1$ for all $x,y \in G$, such that $(\mid x\mid, \mid y \mid) = 1$, where $[x,y] = x^{-1}y^{-1}xy$, ie, is the  commutator of $x$ and $y$; and $(a,b)$ denotes the greatest common divisor of the integers $a$ and $b$?</p>

答:
Yes, this is true. The gist is that Sylows are normal in a nilpotent group.

If $P$ is a Sylow $p$-subgroup, then $N_G( N_G(P) ) = N_G(P)$ (abnormality). However, in a nilpotent group $N_G(Q) > Q$ as long as $G > Q$ (the normalizer condition). Hence in a nilpotent group, $N_G(P) = Q$ must be all of $G$, and all Sylow subgroups are normal. If $P,R$ are Sylow subgroups for different primes and are both normal, then $[P,R] \leq P \cap R = 1$ so $P$ and $R$ commute. Hence in a nilpotent group, $[x,y]=1$ if $\gcd(|x|,|y|)=1$.

Conversely, if $[x,y]=1$ whenever $\gcd(|x|,|y|)=1$, then each Sylow $p$-subgroup is normal (it is normalized by itself, and centralized by a supplement). Since $P$ and $R$ commute and are normal, $PR=RP = P \times R$, and doing this for all of the Sylows writes $G$ as a direct product of its (finitely many non-identity) Sylow subgroups. A direct product of nilpotent groups is nilpotent, and $p$-groups are nilpotent, so we get that $G$ is also nilpotent.


问:
Limit of ratio of $2$ functions
I need some help with my homework.
How do I prove that

$$\lim_{n\to\infty}\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}} =0\:?$$

I need to compare the growth rate of these functions and I know that $4^\sqrt{n}$ grows faster and thats why the limit should be $0$.

But how do I solve this? What steps should I show that lead this equation to being $0$?

Thank you for your help!

答:
Let $n\in\mathbb{N}$. Observe that $n^3\le4^{\sqrt n}$ and from this it follows that
$$\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}} =\frac{n^3}{\log_{2}(n)4^\sqrt{n}}\le \frac{1}{\log_{2}(n)}.$$

If we let $n\to\infty$ we can see that
$$\lim_{n\to\infty}\frac{1}{\log_{2}(n)}=0.$$

In conclusion:
$$\lim_{n\to\infty}\underbrace{\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}}}_{\text{non-negative}}\le0\Longrightarrow \lim_{n\to\infty}{\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}}}=0$$

Title: Limit of ratio of $2$ functions


问:
Is the set of sequences for which $\sum_{n=2}^\infty x_n=2x_1$ closed in $\ell_1$?
In $\ell_1$ (with the $\ell_1$-norm), consider the subspace $M=\{(x_n)_{n=1}^\infty:\displaystyle{\sum_{n=2}^\infty x_n=2x_1}\}$.

I am trying to determine whether this space is closed or not and if its codimension is 1.

I have tried to find a sequence contained in $M$ that does not converge on $M$. However, everything I have come up with is convergent on $M$. I started to suspect $M$ may actually be closed, but I have been unable to to prove it.

Any hints would be very helpful. Thanks in advance.

答:
Let $(x^{(n)})$ be a sequence in $M$ converging to some $x\in \ell^1$. We know that $2x_1=\lim_{n\to\infty} 2 x^{(n)}_1$, and that $\lim_{n\to\infty}\sum_{k=2}^\infty x_k^{(n)}=\sum_{k=2}^\infty x_k$. As $2 x^{(n)}_1=\sum_{k=2}^\infty x_k^{(n)}$ for every $n\in \mathbb N$, we must have $x\in M$.

EDIT:
$M$ does has codimension $1$. My previous argument was incorrect. Indeed as pointed out by KeeperofSecrets $M$ is just the kernel of a continuous linear functional, and thus must have codimension $1$.

This is actually intuitively clear too, as any element in $M$ can be constructed by freely varying all the coordinates except the first one, and then fixing the first coordinate to be $-2$ multiplied by the sum of all the other coordinates, so the "degrees of freedom" of $M$ is only one less than for all of $\ell^1$.

Rigorously, we can notice that $e_1=[(1,0,0,\dots)]$ is a basis vector for $\ell^1/M$. Indeed for any $(x_n)\in \ell^1$ we note that
$$(x_n)-\left(x_1-\frac{1}{2}\sum_{n=2}^\infty x_n,0,0,\dots\right)\in M,$$
so $[(x_n)]\in\operatorname{span}([e_1])$.


问:
Find the Galois group of a quotient field extension
The task is to determine
$Gal(L|K)$ for $L=\mathbb{F}_2(X)$ and $K=\mathbb{F}_2(X^2)$

I'm not sure what $K$ looks like. I assume $\mathbb{F}_2(X^2)\cong  \{ \frac{P(X)}{Q(X)}: P,Q \in F_2[X^2] \}$.

My first attempt was to guess a $K$-basis of $L$, namely $(1,X)$ but it didn't work because I can't get odd exponents in the denominator. Therefore I assume it is an extension of degree $>2$.

On the other hand I also realized that $irr(X,\mathbb{F}_2(X^2))=T^2-X^2$, hence $[\mathbb{F}_2(X^2)[X]:\mathbb{F}_2(X^2)]=2$ but here I get confused.

If $\mathbb{F}_2(X^2)[X]$ is a field, then it should be closed under inverse and thus it contains $\frac{1}{X}$. Does this mean $\mathbb{F}_2(X^2)[X]=\mathbb{F}_2(X)$?

I'd be very thankful if you could clarify the situation.

答:
You're right that $(1,X)$ is a $K$-basis of $L$.

First notice that $\mathbb{F}_2[X]\subset L$. For $P\in\mathbb{F}_2[X]$, one can write $$P=\sum_{}p_{2k}X^{2k} + \left(\sum_{}p_{2k+1}X^{2k}\right)X$$

Then remember that the Frobenius is $\mathbb{F}_2$-linear : this implies, among other things, that for $P\in\mathbb{F}_2[X]$, $P(X^2)=P(X)^2$.

The result follows : for $(P,Q)\in\mathbb{F}_2[X]^2$ and $Q\neq 0$, just write $$\frac{P(X)}{Q(X)}=\frac{P(X)Q(X)}{Q(X)^2}=\frac{P(X)Q(X)}{Q(X^2)}$$

which is therefore in the $K$ vector space spanned by $(1,X)$ inside $L$.


答:
Following your substitution $t= \tan x$, integrate the resulting as follows

\begin{align}
&\int \frac1{1+\sin^4x}dx\\
=&\int \:\frac{t^2+1}{2t^4+2t^2+1}dt
=\int \frac{1+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\
=&\frac{\sqrt2+1}{2\sqrt2} \int \frac{\sqrt2+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt
-\frac{\sqrt2-1}{2\sqrt2} \int \frac{\sqrt2-\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\
 =&\frac{\sqrt2+1}{2\sqrt2} \int \frac{d(\sqrt2t-\frac1{t})}{(\sqrt2t-\frac1t)^2 + 2(\sqrt2+1)}dt
-\frac{\sqrt2-1}{2\sqrt2} \int \frac{d(\sqrt2t+\frac1{t})}{(\sqrt2t+\frac1t)^2 -2(\sqrt2-1)} dt\\
 =&\frac{\sqrt{\sqrt2+1}}4\tan^{-1} \frac{t-\frac1{\sqrt2 t}}{\sqrt{\sqrt2+1}}
+\frac{\sqrt{\sqrt2-1}}4\coth^{-1} \frac{t+\frac1{\sqrt2 t}}{\sqrt{\sqrt2-1}}+C
\end{align}

Title: Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$

Body:
I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.

Do you have any ideas? :)

EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to

$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$

By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$

And using the substitution: $t=\tan\left(x\right)$

$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$

$$t^2=\tan^2\left(x\right)$$

$$t^2=\frac{\sin^2x}{\cos^2x}$$

$$t^2=\frac{1-\cos^2x}{\cos^2x}$$

$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$

Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$

I don't think I got to the expected result but I can't seem to be able to find why…


问:
What does it mean for an inner product to induce a norm?
I'm working on a problem set and am trying to understand the concept of "induced norms":

Let's say I have a space of $M \times N$ matrices (I'm interpreting this as the collection of all real-valued $M \times N$ matrices). Let's say I want to convert this space into an inner-product space using some inner product $\langle A, B\rangle$. I now have some inner-product vector space where each matrix pair has an associated value produced by the inner product. For those interested, the provided inner product is $\operatorname{trace}(A^{T}B)$.

The problem then goes on to state that the provided inner product induces a norm.

What does it mean to induce a norm? And is this an induced norm on the original $M \times N$ matrix space? Or the inner-product space?

答:
The norm induced by an inner product $\langle, \rangle$ is by definition $$||v|| = \sqrt{\langle v,v\rangle }.$$
You can check that the inner product properties imply that this really is a norm.


问:
Does the function $\sqrt{z}=|z|^{1/2} e^{i\frac{Arg(z)}{2}}$ extend continuously on $(-\infty, 0)$?
Generally, $z\mapsto Arg (z)\in (-\pi,\pi]$ is not a continuous function on $(-\infty ,0]$. However, we can extend it continuously on $(-\infty,0)$ by making a correction on the definition of the function on the left open complex plane as
$$f(z)=Arg(z),\ z=x+iy, x<0, y\geq 0 $$ and $$f(z)=Arg(z)+2\pi,\  z=x+iy, x<0, y< 0 .$$
So, I was wodering if the function $$z\mapsto \sqrt{z}=|z|^{1/2} e^{i\frac{Arg(z)}{2}}$$ can extend continuously on $(-\infty,0)$ exactly as  $z\mapsto Arg (z)$. I thought to make a correction only on the left open complex plane as
$$g(z)=\sqrt{z}, \ z=x+iy, x<0, y\geq 0 $$ anfd $$g(z)=-\sqrt{z},\ z=x+iy, x<0, y<0 .$$
Eventually, g is a continuous function on $(-\infty,0)$. What can I say now about the continuity of $g$ on $\mathbb{C}\setminus (-\infty,0)$? Is it possible the previous trick destroyed the continuity of $\sqrt{z}$ over $\mathbb{C}\setminus (-\infty,0)$? Thanks

答: Your correction leads to discontinuities on the negative $y$-axis. From the left we now approach $3\pi/2,$ from the right things are unchanged and we approach $-\pi/2.$ There's no way to avoid the $2\pi$-hiccup!


问:
How can we measure how "irrational" a number is?
I wonder, can we, for instance, say that $\pi$ is more irrational than $e$? Or that $e$ is more irrational than $\sqrt{2}$? What kind of irrationality measurement can we use to say that $A$ is (much) more irrational than $B$? (in particularly rigorous sense)</p>

答:
No rational number is irrational, but in the same way we can measure how "black" a shade of grey is, given that every irrational number has an infinite continued fraction, a fair measure of how irrational a rational number is, is by the length of its continued fraction.

As part of my studies of the Collatz conjecture I discovered (proof thanks to Torsten Schoeneberg) that every rational number has precisely one finitely-long continued fraction of the form $1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}$

provided the terms are measured by letting $\lvert x\rvert_2$ be the 2-adic value of $x$ and iterate $1-\dfrac1{x\lvert x\rvert_2}$ until you get to zero.  The number of steps taken is the length of the continued fraction and the measure.

What is more, if you wish more resolution, you can assign every rational number (up to bijection) an ordinal less than $\omega^\omega$.  To do this, simply identify $m_0, m_1,\ldots$ by counting the number of divisions by $2$ required at each step. Then use any bijection $f$ from $\Bbb Z\to \Bbb N$ to send $m_i\to n_i$ and:

$1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}\mapsto \displaystyle\sum_{i=0}^\infty\omega^i\cdot n_i=\displaystyle\sum_{i=0}^\infty\omega^i\cdot f(m_i)$

is a bijection from the rationals into the ordinal $\omega^\omega$.


问:
Every Lie algebra of dimension greater than $1$ has a non-zero Lie subalgebra
Looking at this question and its answer

Every Lie algebra contains a maximal proper Lie subalgebra

I asked myself what is an example of a finite-dimensional Lie algebra $L$ whose only proper Lie subalgebras are trivial?

Some thoughts: Clearly the dimension of $L$ has to be greater than $1$. Moroever, $L$ cannot be abelian. I tried $\frak{sl}_2$ but that clearly doesn't work . . . same goes for any simple Lie algebra. One could try solvable . . . but the derived series produces ideals, so no go there. Thus by Levi's theorem might one that every finite dim Lie algebra of dim greater than $1$ has a non-zero Lie subalgebra?

答: For any Lie algebra $\mathfrak g$ with $\dim\mathfrak g>1$ and for every $0\neq X\in\mathfrak g$, the set $\{\lambda X\mid\lambda\in F\}$, where $F$ is the field that you are working with, is a subalgebra of $\mathfrak g$, which is neither $\{0\}$ nor $\mathfrak g$.


问:
About the Proof of a lemma regarding representablility in Language of arithemetic
When I learn recursive theory, I see a lemma that says, in the language of arithmetic, $(0, S, +, \cdot)$, if a Theory $T$ which $T\vdash (0 \text{ and } 0 \neq S(0))$, and a relation $R$ belongs to $N^k$, then $R$ is representable in $T$ iff $\chi_R$ (aka its characteristic function) is representable (as a function) in $T$.

This lemma seems straightforward however I can't figure out a clear proof. Can anyone help to explain how to prove this lemma?

答:
Hint. If $\varphi(x)$ represents $R$ (and to reduce clutter, let's just consider the case where $R$ is one-place), then to represent the function $\chi_R$, put

$$\varphi'(x, y) =_{def} (\varphi(x)\wedge y =0) \vee (\neg\varphi (x)\wedge y = 1).$$

(if like Gödel you use '0' for true and '1' for false).

If that hint isn't enough, see §16.1 of my Intro to Gödel's Theorems which is freely downloadable from https://www.logicmatters.net/igt -- the result you want is Theorem  16.1 (I use "capturing" for "representing").


问:
Neyman-Pearson lemma on Normal distribution
We've got a random sample of iid $X_1,\dots,X_n$. We're testing the mean of $X \sim \mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. The size of the test $\alpha=0.05$.

$H_0: \mu=0$

$H_1: \mu=v$

By the Neyman-Pearson lemma the Most Powerful test is $\phi(X) = \mathbf{1}_A$ where the set $A =\{ x: \prod_{i=1}^n \frac{f(\mu_1,\sigma^2)}{f(\mu_0,\sigma^2)} > k \} $

simplyfying we can reduce the test to:
\begin{equation}
\frac{1}{n}\sum_{i=1}^n X_i > \frac{2\sigma^2(\log k+\mu_1^2 n)}{n \mu_1}
\end{equation}

where $\mu_1 = v$.
Calculating the critical value $k$ we evaluate:
$$
\begin{align}
\alpha=\mathbb{E}[\phi(X) |H_0] &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\phi(x) e^{-\frac{x^2}{2\sigma^2}} dx\\
&=\frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\mathbf{1}_A e^{-\frac{x^2}{2\sigma^2}} dx\\
&= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{k}^{\infty}e^{-\frac{(x-\mu_1)^2}{2\sigma^2}} dx\\
&= \frac{1}{\sqrt{2 \pi }} \int_{\frac{k-\mu_1}{\sigma}}^{\frac{\infty-\mu_1}{\sigma}}e^{-\frac{x^2}{2}} dx\\
&=1-\Phi\Bigg(\frac{k-\mu_1}{\sigma} \Bigg)
\end{align}
$$

therefore we can derive $k=\mu_1 + \sigma \Phi^{-1}(1-\alpha)$

Is my approach correct, or did I mess up the calculation of the critical value?

答:
You messed up the calculation of the critical value. First of all It is true that the test statistic you have is reduced to

$$T(X)=\frac{1}{n}\sum_{i=1}^n X_i>\gamma,\quad\quad \mu>0$$

Let $T(X)$ be a random variable $Y$. Then,

$Y\sim{\mathcal{N}}(0,\frac{\sigma^2}{n})$ if $\mathcal{H_0}$ is correct

$Y\sim{\mathcal{N}}(\mu,\frac{\sigma^2}{n})$ if $\mathcal{H_1}$ is correct

Now we simply have $$\alpha=P(Y>\gamma|\mathcal{H_0})=Q\left(\frac{\gamma}{\sqrt{\sigma^2/n}}\right)$$

As you can see $\alpha$ is independent of $\mu$. In a similar way you can calculate the detection probability as $$\beta=P(Y>\gamma|H_1)=Q\left(\frac{\gamma-\mu}{\sqrt{\sigma^2/n}}\right)$$

From here, since $Q$ is monotonically increasing as $1-Q$ is a CDF, it accepts an inverse form and eventually the critical value can be found as

$$\gamma=\sqrt{\sigma^2/n}Q^{-1}(\alpha)$$


问:
polynomial of degree $n$ with real coefficient takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued
Show that if $f(x)$ is a polynomial of $deg n$ with real coefficient ,which takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued.

How to prove this I thought of this like $P(x) =\sum_{i=0}^{\infty}a_{i}x^{i}$ where $a_{n}=\frac{P^{n}(0)}{n!}$
Your answers are really nice. But , I don't know or I want answer from strong principle of induction.

答:
Hint.

If $f(x)$ has degree $n$, show that it satisfies the recurrence equation
$$
\sum_{k=0}^{n+1} (-1)^k\binom{n+1}{k} f(x-k) = 0
$$
Use this, together with induction (forward and backward).


问:
pair homotopic maps induce the same homology
I'd like to prove that given $f,g : (X,A) \longrightarrow (Y,B)$ homotopic as map of pair, i.e $H(A\ \times I) \subset B$ then they induce the same homology.

I already know the theorem which states that homotopic map induce the same homology, what I'm interested in understanding how $f_*,g_* : H_n(X,A) \longrightarrow H_n(Y,B)$ are the same thanks to this in order to finish the proof. I should be able to conclude from the fact that called $H_{\sharp} : C_n(X) \longrightarrow C_{n+1}(Y)$ the natural homotopy between $f_{\sharp},g_{\sharp}$, this induces a natural homotopy $C_n(A) \longrightarrow C_{n+1}(B)$ which factors to an homotpy $C_n(X,A) \longrightarrow C_n(Y,B)$.

I don't understand how to prove the Lemma since I don't understand how the term natural is used here, even to obtain an homotopy from $C_n(A) \longrightarrow C_{n+1}(B)$.

My definition of natural trasformation is exactly the given on Wikipedia, any help would be appreciated.

答:
"Natural" here means that for the two chain homotopies $T:C_n(X) \rightarrow C_{n+1}(Y)$ and $T':C_n(A) \rightarrow C_{n+1}(B)$ we have the identity $Ti_{A,X} = i_{B,Y} T'$ where $i_{A,X}:C_*(A) \rightarrow C_*(X)$ and $i_{B,Y}:C_*(B) \rightarrow C_*(Y)$.

This would imply that we could find maps $S:C_n(X)/C_n(A) \rightarrow C_{n+1}(Y)/C_{n+1}(B)$ such that $S \pi_{X,A} = \pi_{Y,B} T$ for purely group theoretic reasons.

$\pi_{X,A}:C_*(X) \rightarrow C_*(X,A)$ and $\pi_{Y,B}:C_*(Y) \rightarrow C_*(Y,B)$ being the projection maps.

$S$ would then be the desired chain homotopy.

That is, $S$ is a chain homotopy between $f_{\#}$ and $g_{\#}:C_*(X,A) \rightarrow C_*(Y,B)$ since $T$ is a chain homotopy between $f_{\#},g_{\#}:C_*(X) \rightarrow C_*(Y)$


答:
One way to solve this is to transform the variables from $X,Y$ to $Z,W$ where $Z = X+Y$ and $W = X-Y$.  The Jacobian is $-2$.  So

$$f_{W,Z}(w,z) = \frac{1}{2} f_{X,Y}\big(\frac{1}{2}(z+w), \frac{1}{2}(z-w)\big)$$
Plugging in, we get
$$f_{W,Z}(w,z) = \frac{1}{2}\cdot \frac{1}{2}(z)e^{-z}$$
The limits of the domain are $z$ from $0$ to $\infty$, and for a given $z$, the limits of $w$ are ${-z}$ to ${z}$.  So if we integrate over $w$ to find the marginal distribution for $z$, we get
$$ \frac{1}{4}ze^{-z}\cdot 2z = \frac{1}{2}z^2e^{-z}$$

Title: Find the density function of X+Y, where X and Y are random variables

Body:
If $X$ and $Y$ have joint density function $f(x,y)=\frac{1}{2}(x+y)e^{-x-y}$ when $x,y>0$ (and $0$ otherwise), find the density function of $X+Y$.

Let $Z=X+Y$. Then from my textbook, it says that $f_Z(z)=\int_{-\infty}^{\infty} f_{x,y}(u,z-u)du$, where $u=x, u>0$. When I plug the joint density function into this equation, I get $f_Z(z)=\int_{0}^{\infty} \frac{1}{2}ze^{-z}du$. This integral doesn't evaluate because you would have a $u$ being evaluated at infinity. The answer in the book says $f_{X+Y}(u)=\frac{1}{2}u^2e^{-u}$ for $u>0$. I don't see how this is the answer.


答:
Another way to do it: First assume that the arc is $0 \le \theta \le \pi$, and define a new function $g(z) = f(z)f(-z)$. Then $g$ is analytic on $D$, continuous on $\bar D$ and equal to $0$ on the whole circle. What does that tell you about $g$, and consequently about $f$?

For the general case, put $g(z) = f(z)f(e^{i\alpha}z)f(e^{2i\alpha}z)\cdots$, where $\alpha$ is suitably chosen.

Title: Show that $f(z)=0$.

Body: Suppose that $f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic on the open unit disc and continuous on the closed unit disc. Assume that $f(z)=0$ on an arc of the circle $\{z\in\mathbb{C}:|z|=1\}$. Show that $f(z)=0$.</p>


问:
Recurrence formula
Could you please help me to understand the task.

The topic of the task is Recurrence formula.
There is a txt file with integers that represent the members of the row of a(k), starting with k = 0. It is required to calculate the value of S (100), starting with k = 3, if S (k) = S (k-1) + 2 * a (k + 1) + a (k) and S (2) = 5.

As far as I understand the conditions of the task, each namber is a(k), where k = 0, ..., n. If this is correct, then the first number is a(0) = 197, the second number is a(1) = 381, the third is a(2) = 171, etc. I don't understand what S(k) is.

By definition, a recurrence formula is a formula that expresses each term in a sequence through the previous terms. Does this mean that S (100) is the hundredth term in the sequence of numbers? If so, then why was it expressed through the new letter S(k), and not left a(k)? It turns out that the formula contains both S(k) and a(k), denoting the same thing, which seems to be nonsense.
And again, if S(k) is the k-th term, then why S(2) = 5? Where does 5 come from?

For some reason, the name of the text file with numbers contains the word sum - sum_recurr_rd_1.txt. Does this mean that S(k) is not the k-th term, but the sum of the k-th terms and I need to find the sum of the first 100 numbers but starting from a(3) = 385 instead of a(0) = 197 since it is said to start with k = 3. But even if we assume that this is the case (although I'm not sure), then the result of summ of firts 100 terms  is not the correct answer. I checked. Or maybe I need to sort the sequence of numbers before starting the calculation? Although nothing is said about this.

I see that my understanding of the assignment and perhaps the topic itself is not correct. Please help me figure out what is the S(k) that need to be find?

Thank you!

```<code>197
381
171
385
30
417
105
125
191
347
311
158
281
114
184
241
234
228
181
531
154
422
62
204
458
222
162
160
501
287
555
428
465
168
125
206
13
191
100
129
198
399
301
419
12
543
180
472
427
462
444
506
520
122
246
140
233
219
415
324
195
539
6
197
160
291
524
245
383
400
517
363
260
361
168
26
182
230
560
185
357
206
295
442
201
198
507
311
352
521
565
491
30
288
211
402
318
142
376
302
437
336
170
205
492
287
245
178
72
119
448
341
133
311
33
472
301
309
186
236
274
549
239
552
352
101
78
385
451
565
256
278
252
144
508
328
475
313
370
144
192
282
122
482
14
366
154
54
435
376
459
394
389
259
507
289
252
119
449
121
553
107
129
202
351
87
495
219
180
333
342
334
36
365
325
215
86
376
92
236
336
15
251
416
131
259
342
176
404
64
268
434
559
349
485
440
337
31
153
0
299
204
483
246
88
44
234
445
133
100
376
136
160
3
535
562
101
463
401
104
443
365
315
394
345
126
188
234
258
316
61
74
564
108
178
190
383
506
295
560
333
398
476
16
63
242
272
358
510
300
377
328
450
465
465
156
387
528
449
180
260
175
157
225
176
309
349
129
484
247
167
345
565
420
387
144
83
215
137
532
116
185
392
307
388
184
282
35
299
226
165
432
37
21
156
67
191
259
532
482
518
178
214
103
413
296
179
326
232
230
258
154
358
73
293
287
315
385
28
202
321
247
13
350
391
510
147
417
20
510
433
544
140
217
364
415
418
31
119
43
73
58
329
533
204
336
566
74
50
42
198
186
277
471
60
437
308
76
329
552
480
355
306
514
20
545
116
561
256
46
192
251
398
413
395
402
264
530
193
499
428
426
74
385
263
61
566
518
236
345
259
51
391
67
126
34
356
117
213
240
442
182
249
94
59
102
183
283
522
418
381
14
228
350
301
344
79
320
51
72
266
42
287
357
428
477
396
335
445
323
307
544
416
133
193
188
507
364
457
508
546
411
27
126
497
355
197
126
151
132
509
403
492
183
121
28
188
115
344
154
460
406
394
386
100
477
474
261
240
528
529
9
58
442
372
320
437
110
558
69
253
521
257
560
532
346
29
405
557
314
388
54
122
225
490
187
373
416
369
24
116
185
180
268
540
313
464
418
503
327
50
17
219
329
545
139
454
16
319
391
304
244
477
15
499
114
60
223
23
132
456
1
426
203
104
447
223
360
114
16
344
512
72
325
151
299
289
352
418
431
513
423
76
510
45
499
154
220
405
249
352
247
281
486
303
14
353
341
445
109
437
418
271
318
8
496
190
128
539
129
135
348
31
555
437
455
424
486
249
149
457
82
4
334
127
82
368
444
307
357
99
184
143
431
476
108
392
101
84
519
149
344
426
158
367
434
309
71
419
313
124
455
534
179
44
51
471
446
177
406
146
73
283
515
50
123
348
154
407
242
181
320
533
390
293
39
527
110
230
5
143
559
112
345
170
138
344
62
46
410
444
566
89
476
316
432
35
469
53
207
56
150
88
523
14
237
436
221
219
498
466
209
138
443
333
213
285
406
360
431
380
65
368
481
18
122
254
35
266
396
324
379
40
141
248
382
157
193
281
156
184
410
516
482
100
207
314
239
564
33
388
291
29
171
403
419
516
346
75
430
325
326
170
142
280
360
475
43
452
123
384
244
281
188
502
377
489
431
362
89
142
380
544
112
434
364
197
108
36
284
261
305
410
219
298
132
143
336
173
301
468
0
501
10
388
243
369
456
136
3
453
350
107
275
318
0
184
286
57
338
210
369
101
367
470
160
494
559
501
112
148
453
59
154
149
25
45
366
546
324
101
89
301
288
196
293
251
146
309
38
411
228
91
84
172
277
526
336
426
194
308
249
14
120
79
454
307
220
533
512
56
380
267
488
305
332
163
176
376
496
446
507
409
53
449
207
517
134
557
163
420
554
235
217
552
470
86
208
351
320
528
252
436
187
144
210
510
36
129
397
428
511
193
440
450
23
369
2
340
336
53
177
239
485
72
24
452
394
166
317
339
544
423
6
235
196
215
86
154
351
2
360
414
308
239
174
464
253
539
145
346
252
162
259
419
207
406
173
23
451
477
318
303
125
300
260
185
494
262
158
511
104
152
318
164
439
333
384
378
435
77
11
398
120
407
38
12
320
493
98
78
269
241
443
160
20
136
527
274
237
116
235
344
298
186
22
226
175
64
53
211
313
58
174
159
165
268
310
216
316
57
318
4
128
335
245
390
315
377
176
</code>```

Title: Recurrence formula

Body:
In the context, it is reasonable to assume S(k) is  the sum of first k terms and a(k) is each term.  We need to know S(2) (or a(1)+a(2)),a(3),a(4) and a(101) to get S(100)as proved below.  Not sure how to link this with the data, as you said it is not clear where S(2)=5  comes from.

S(k)=S(k-1)+2a(k+1)+a(k)

S(100)=S(99)+2a(101)+a(100)

S(99)=S(98)+2a(100)+a(99)

S(98)=S(97)+2a(99)+a(98)

...

S(4)=S(3)+2a(5)+a(4)

S(3)=S(2)+2a(4)+a(3)

So

S(100)=S(2)+2a(101)+2a(4)+a(3)+3(a(4)+a(5)+...+a(100))

S(100)=S(2)+2a(101)+2a(4)+a(3)+3S(100)-3(a(1)+a(2)+a(3))

S(100)=-$\frac{S(2)+2a(101)+2a(4)+a(3)-3(a(1)+a(2)+a(3))}{2}$

S(100)=S(2)-a(101)-a(4)+a(3)

S(100)=S(2)+a(3)-a(4)-a(101)


问:
Natural deduction - help solving
I have this issue in natural deduction:

B -> (A & ~A) ├ ~B

I am having a really hard time understanding the rules in natural deduction - but this is what I got so far - can someone help me to the end?

B / premis

(A&~A) /premis

B-> ^ /1, 2  ^E

B /1

^ /3->E

~B /4,5 ->I

Thank you!

答:
For a start, neither (1) nor (2) are given to you as premisses. Your premiss is the given conditional, and then your proof should proceed by assuming $B$ for reductio ...

To be honest, your proof attempt suggests you really haven't understood ND at all. For a gentle and I hope very accessible intro to ND (which is quite freely available) try the relevant chapters of my book Introduction to Formal Logic which you can download from https://www.logicmatters.net/ifl

Title: Natural deduction - help solving


答:
You have a hierarchy that's useful to remember when dealing with topology. From the most precise to the most general:

Inner product => norm => metric => topology

But none of these go the other way around (some topological spaces are not metric spaces, some metric spaces are not normed spaces, and some normed spaces are not euclidean spaces (IPSs)).

The standard topology (also called the euclidean topology) over $\mathbb{R}^n$ is that which is generated by the inner product, the 2-norm, the euclidean metric.

Inner product => norm:

$$\| v\| = \sqrt{\langle v, v \rangle}$$

Norm => metric:
$$ d(x, y) = \| x - y\|$$

Metric => topology (over a manifold $M$):

$$\tau \subset \mathcal{P}(M), \tau = \{ \bigcup_{i \in I} B_{r_i}(x_i) \; | \forall I\}$$

where $I$ is an arbitrary indexing set, and the $B_r(x)$ are the open balls centered in $x$ of radius $r$. The metric is what allows us to define this radius. So $\tau$ is the topology using as basis the open balls of $M$.

Title: What does it mean for an inner product to induce a norm?

Body:
I'm working on a problem set and am trying to understand the concept of "induced norms":

Let's say I have a space of $M \times N$ matrices (I'm interpreting this as the collection of all real-valued $M \times N$ matrices). Let's say I want to convert this space into an inner-product space using some inner product $\langle A, B\rangle$. I now have some inner-product vector space where each matrix pair has an associated value produced by the inner product. For those interested, the provided inner product is $\operatorname{trace}(A^{T}B)$.

The problem then goes on to state that the provided inner product induces a norm.

What does it mean to induce a norm? And is this an induced norm on the original $M \times N$ matrix space? Or the inner-product space?


问:
Why is $e^{-ax}u(x)$ not continuous for $a\in\mathbb{C}$?
Let $a\in\mathbb{C}$ with $\Re(a)>0$

Define the unit step function :
$$
u(x):=\begin{cases}
1&\text{if $x\geq0$}\\
0&\text{if otherwise}
\end{cases}
$$
We wish to study continuity of : $$f(x)=\frac{x^{k}}{k!}e^{-ax}u(x)$$
But I am only concerned regarding the case $k=0$ where we get $e^{-ax}u(x)$, with $a\in\mathbb{C}$ this would "virtually" mean that we have a decaying exponential multiplied with a complex sinusoidal but why would this imply that $f$ is not continuous for $k=0$? My textbook says $f$ must be continuous for all $k\geq1$

答:
When $k=0$ the term $x^k$ vanishes and thats the one that guarantees continuity at zero, so the discontinuity of $u$ at $0$ comes into play.
$$\lim_{x\rightarrow 0^-} e^{-ax}u(x)=0\neq 1 = f(0).$$


问:
Chance of winning when you are ahead a flip coin game
Two players, $A$ and $B$, alternately and independently flip a coin and the player wins only when he gets a head, and the other player gets the tail. Each round they will flip coin at the same time. If the coin is fair, and let’s assume there is infinitely round until one win. At the end, either player $A$ or $B$ will win, so the probability of player $A$ win plus probability of player $B$ win should equal to $100\%$.

The question I tried to solve is this:  In the above question, each win counts as one point, each loss counts as $-1$ point, whoever gets $3$ points first wins. (Game over when score is 3: -3).Now assume player $A$ wins one point at sometime. So the score are player A : player B is 1: -1; Then, what’s the probability player $A$ wins the game after that. Thanks in advance for any help.

答:
It is easy to see that the probabilities of A win, lose, and draw one game are

$\frac14,\; \frac14,\;$  and $\;\frac12$ respectively

Let there be $3\,$ states, D,W,L (in the last game you  drew, won and  lost respectively.)

Finally, let $d,w,l$ be the probabilities that you (ultimately) win from D, W and L Since the game starts with A up by $1$ game, we  seek the probability that we ultimately win starting from W

Note that from any stage you can draw and return to D with $Pr = 1/2$.

Thus the Pr equation to reflect various transitions that can take place to move to D, move to W or move to L will be given by the equation

$d = d/2 + w/4 + l/4\quad$   [I]

Similarly, $w = d/2 + (1/4)*1 +l/4\quad$ [II]

(Note that the second term , a win from W completes a win for A)

$l =  d/2 + w/4 + (1/4)*0 \quad$ [III]

(Note that a loss from l will end the game for A)

solving these $3$ linear equations gives us

$w = \frac35,\quad l = \frac25,\quad d = \frac12\;$(as expected)

The answer sought is $\frac35$

PS In a hurry, I read that A needs to be ahead of B by $2$ points to win, whereas you want her to win two more points. Anyway, having laid down the template, you should be able to solve on your own now. Just add  states, WW and LL and proceed as before.

PPS I'm more confused after reading some clarifications you have added. A had a difference of two $(1 - (-1))$ and by winning one more the scores are $2,-2$ so the difference is $4$, idk whether in your scheme of things, A has won or not, anyway you can proceed further on your own. I think you have devised this game yourself, giving plus and minus points together is not really a good idea, it doesn't add anything to the game.

Either games just have win or lose, or have a win-draw-lose system like chess, soccer, checkers.


问:
Mean Value Property of Harmonic Functions
I can't prove this theorem:

"Let $\Omega$ is a bounded domain, $u\in C^2(\Omega)$ satisfy $\Delta u=0(\geq0,\leq0)$, then for any ball $B=B_R(y)\subset \subset \Omega$, we have

$$u(y)=(\leq,\geq)\frac{1}{n\omega_nR^{n-1}}\int_{\partial B}u\ ds$$

$$u(y)=(\leq,\geq)\frac{1}{\omega_nR^n}\int_{B}u\ dx$$

Thanks

答:
Here's a hint to get you started:

Define $A(r) = \int_{\partial B(y,r)} u(x) \, dx$, and show that $A(r)$ is constant by differentiating with respect to $r$, and showing that the derivative is zero by using the divergence theorem to replace the integrand with a $\Delta u$.

For the subharmonic and superharmonic cases, the same technique leads to $A(r)$ increasing and decreasing, respectively.


问:
Show that there exist $n, m \in \mathbb{Z}\setminus\{0\}$ such that $\alpha^n = \beta^m$, without computing $n, m$.
Let $K = \mathbb{Q}(\sqrt{13})$, $ \alpha = (3+\sqrt{13})/2$ and $\beta = 23382 + 6485\sqrt{13}$.

Show that there exist $n, m \in\mathbb{Z}\setminus\{0\}$ such that $\alpha^n = \beta^m$, without computing $n, m$.

How would I show this? The norms I believe are both -1, so they must differ by a unit, but I do not know how to proceed, can I use Dirichlet's theorem or not? Any hints would be welcome

答: By Dirichlet's unit theorem, there is an element $x \in \mathcal O_K^{\ast}$ such that every unit in $\mathcal O_K^{\ast}$ is uniquely expressible as $\pm x^n$.  Write $\alpha$ and $\beta$ in this form.


问:
Define and calculate the transpose transformation $T^t(f)$ on dual space
Let $V=P_1(\mathbb{R})$ and $W=\mathbb{R}^2$ with the corresponding basis $b= \left \{ 1,x \right \}$ and $g= \left \{ e_1,e_2 \right \}$ We define the linear transformation $T:V\rightarrow W$ $$T(p)=(p(0)-2p(1),p(0)+{p}'(0))$$

if $f \in  W^{\ast } $ with: $$f(a,b)=a-2b$$

Calculate $T^t(f)$.

My first idea was to find the basis matrix of the transformation $T(f)$ and then find the transpose matrix, but I can't figure out how to define  $T(f)$ it seems like $f \in \mathbb{R}^2 $ but $T$ takes elements of $\mathbb{R}$ that really confuse me.
I calculated $[T]^{g}_{b}=\begin{pmatrix}
-1 &-2 \\
 1& 1
\end{pmatrix}$ and therefore $[T^t]^{b^{\ast }}_{g^{\ast }}=\begin{pmatrix}
-1 &1 \\
 -2& 1
\end{pmatrix}$ but i am unable to fully understand what  $T^t(f)$ means.

答:
I think I got it if $T^t(f)=f(T)$ then $f(T)=p(0)-2p(1)-2(p(0)+{p(0)}')$ plugging in the $b$ basis we get $-3,-4$ so  $T^t(f)=-3a-4b$

Is this correct?


问:
Tripos Type Integral
This is an old calculus question:

Integrate
$$\int x^{-2n-2}(1-x)^n(1-cx)^ndx$$ it comes from Edwards Treatise on Integral Calculus Vol I, pg. 65 and is designated as [Oxford I.P. 1917].

I guess my question is what do they have in mind as a solution to this ? Are they expecting a closed form ?

My solution is to expand the $1-cx$ factor to get $(-1)^{k}\binom{n}{k} \int\frac{x^k(1-x)^n}{x^{2n+2}}dx$ as the coefficient of $c^k$. This in turn can be integrated by the binomial theorem as
$\sum_{l=0}^n(-1)^{k+l}\binom{n}{k}\binom{n}{l}\frac{1}{2n-l-k+1}x^{-2n+k+l-1}$.
This almost seems too simple minded. But maybe it was what they intended. Any opinions ?

答:
This integral can be written as a simple summation. To simplify, we change $y=1/x$ to write
\begin{align}
 I_n=&\int x^{-2n-2}(1-x)^n(1-cx)^n\,dx\\
 &=-\int (y-1)^n(y-c)^n\,dy
\end{align}
and to have a symmetric expression, changing $y=z+(c+1)/2$ gives
\begin{align}
 I_n&=-\int \left( z^2-\left( \frac{c-1}{2} \right)^2 \right)^n\,dz
\end{align}
Now, expanding the polynomial,
\begin{align}
 I_n&=-\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{n-k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}z^{2k+1}+C\\
 &=(-1)^{n+1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}\left(1-\frac{(c+1)x}{2} \right)^{2k+1}x^{-2k-1}+C
\end{align}

Title: Tripos Type Integral


问:
Try to prove that this is a measure
Let (X, M) be a measurable space and f a non negative real function, prove that μ(A) = sum x∈A f(x) is a measure. I'm trying to prove that μ(U from n=1 to inf An)= sum from n=1 to inf (μ(An)). But I don't know how I can continue, I have μ(U from n=1 to inf An)= sum (xE (U from 1 to inf An)) f(x)

Thanks

答:
Presumably the $A_n$ are disjoint.
$$\mu \left(\bigcup_{n =1}^{\infty} A_n\right) = \sum_{x \in \bigcup_{n=1}^\infty A_n} f(x)
\overset{*}{=} \sum_{n=1}^\infty \sum_{x \in A_n} f(x) = \sum_{n=1}^\infty \mu(A_n)$$

The main step is to show the starred equality. In plain words, it is just stating that taking a sum over the elements of $\bigcup_{n=1}^\infty A_n$ is the same as taking the sum of elements in each set $A_n$, and then summing over all the sets; this relies on the fact that the $A_n$ are disjoint.

Title: Try to prove that this is a measure


问:
Manipulating Equations with Big-oh
Note: first part is all context to question labeled "The Question" below:

Working through the CLRS Introduction to Algorithms, 3rd ed, in Chapter 6.4 as they are talking about heaps they state:

Each call to MAX-HEAPIFY costs $O(lg\ n)$ and BUILD-MAX-HEAP makes $O(n)$ such calls.

Ok, I'm with them, the running time for for BUILD-MAX-HEAP is therefore $O(n\ lg\ n)$, they go onto say:

This upper bound, though correct, is not asyptotically tight.

I understand this as well because really MAX-HEAPIFY doesn't operate on $lg\ n$ levels each time it's called, but rather it operates on $h$ levels where h is the height of the node it was called on.  So they state:

Our tighter analysis relies on the properties that an $n$-element heap has height $\lfloor lg\ n \rfloor$ and at most $\lceil n/2^{h+1}\rceil$ nodes of any height $h$.

Ok, with them still as I know the entire tree has size $n$ nodes which is really using that formula when height=0: $n/2^0 = n$.

The time required by MAX-HEAPIFY when called on a node of height h is $O(h)$ and so we can express the total cost of BUILD-MAX-HEAP as being bounded from above by:

$\displaystyle\sum_{h=1}^{\lceil lg\ n\rceil} \lceil {\frac{n}{2^{h+1}}} \rceil O(h) = O(n\displaystyle\sum_{h=1}^{\lceil lg\ n\rceil} {\frac{h}{2^{h}}})$

The question

Here's where I'm lost, I don't get how the $^{+1}$ disappeared on $2^{h+1}$. Is that some type of unwritten assumption that can be made when you place the equation within the Big-oh? Or because they eliminated the ceiling function?

答: $O(2^{n+1})=O(2\cdot 2^n)=O(2^n)$ since for any function $f(n)$ and constant $C>0$ we have $O(C\cdot f(n))=O(f(n))$.</p>

Title: Manipulating Equations with Big-oh


问:
How to understand non defined limits graphically
I was recently introduced to the concepts of limits and I have a decent grasp on it however, I I think I still don’t have the right idea in my mind

For example :

How do I visual such a limit
I do understand that the two limits don’t meet at any point of the y axis basically there’s a hole I believe , howeverwhen I tried plotting a graph but I just don’t quite get what I just did :/

Here’s the plot i obtained

答:
It's called a jump discontinuity.  As $x$ approaches $1$ from the left, $f(x)$ approaches $0$, but as $x$ approaches $1$ from the right, $f(x)$ approaches $-2$.

A limit would require approaching the same number from both sides.


问:
Is there a function $h$ and a constant $c$ to satisfy $h(x-c) \neq O(h(x))$
Is there a function $h$ and a constant $c$ to satisfy $h(x-c) \neq O(h(x))$?

My attempt:

This claim isn't true. we shall show an equivalent claim to the negation of our claim. We will show that  for each function $h(x)$ and a constant $c$, from a certain $x$, we get $h(x-c)\le c_1\cdot h(x)$. Now I don't know how to proceed, however, my intuition says that $h(x-c)$ is the same function $h(x)$ but when you move it horizontally $c$, so by that fact we can pick $c_1$ such that will be big enough to overcome $h(x-c)$, because we have a bigger slope of $h(x)$. Therefore, we get that: $$h(x-c)\le c_1\cdot h(x)$$

I am not sure that this is formal, and probably I am missing details, so I will be glad to get some help. Thanks!

答:
Do we allow $c<0$? In that case maybe pick $h(x) = e^{x^2}$, then

$$ h(x-c) = e^{x^2 -2cx + c^2} > e^{2|c|x} e^{x^2} \neq O(h(x)).  $$

In fact, for $c>0$ simply pick $h(x) = e^{-x^2}$, then

$$ h(x-c) = e^{-x^2 +2cx - c^2} = e^{-c^2} e^{2cx} e^{-x^2} \neq O(h(x)).$$
To see why this is true, recall that $f(x) = O(g(x))$ iff there are constants $x_0$ and $M$ such that for all $x\geq x_0$ we have $f(x) \leq M g(x)$. Suppose for contradiction that $h(x-c) = O(h(x))$, and let $x_0$ and $M$ be the corresponding constants. Then for all $x\geq x_0$ we have
$$  e^{-c^2 }e^{2cx} e^{ -x^2}\leq M e^{-x^2}. $$
Cancelling $e^{-x^2}$ from both sides, we now have constants $x_0$ and $M$ such that $x\geq x_0$ implies $e^{-c^2 }e^{2xc} \leq M$, but the left hand side of this goes to infinity! So this is a contradiction, and indeed we have $h(x-c) \neq O(h(x))$.

I think some of the confusion may come from the fact that $h(x)$ and $h(x-c)$ both go to zero as $x\to \infty$. But in this example $h(x-c)$ goes to infinity much slower, as there is an additional $e^{2cx}$ term which will be quite large.

Another counter example: This is kinda fun so I thought of another example of $h(x)$ and $c$ with $h(x-c) \neq O(h(x))$. Let $h(x)$ defined as
$$h(x) = \begin{cases}0 &\text{ if }x\in \mathbb{Q},\\
 x& \text{otherwise}.\end{cases} .$$
If $c$ is any irrational number then for all rational $x$ we have $h(x-c)=x-c$ while $h(x) = 0$. So, fixing such an irrational $c$, there are rational $x > c$ for which $h(x-c)-h(x)$ exceeds any bound.

Similarly, we can use e.g. $h(x)= 1/(\sin (\pi x) + 1.1)$ to achieve a similar effect.


问:
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?

It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?

Its answer is easy if it were $x.y=12^3$ , because there are $28$ possible $(a,b)$ pairs by $x.y=2^6 \times 3^3$ and $C(8,2) \times C(4,1) =28$.

Moreover, half of $28$ will give us the number of $x+y$ thanks to symmetry property.

Let's come to original question:

Firtly , i found the number of how many possible $x,y,x$ there exist by combination with repetition such that $C(6+3-1,6) \times C(3+3-1,3)=280$.

However , when i want to find the number of different values of $x+y+z$ , i stuck in.Because , i thought that if i divide $280$ by $3$ , i can obtain the result.However , it did not work and i dont know what i should do.

Therefore, i hope to find tricks or solutions for my problem. Moreover , what can i do for the expanding versions of this question such as $x+y+z+t$

答:
There are indeed $280$ possible ordered triples $[x,y,z]$ with $xyz=12^3$, and these have only $49$ different sums.  I don't see any easy way of counting these other than
explicit enumeration (which is easy on a computer).


问:
Numerical differentiation proof - Taylor polynomial?
Prove the following claim: Let $[a, b] ⊂ R $ and $x ∈ (a, b)$
be given. And E is the error of numerical differentation. If $f ∈ C^3 [a, b]$, then it holds that:

$| f'(x) - \frac{f(x+h)-f(x-h)}{2h}| \leq max_{E \in [a,b]} \frac{|f'''(E)|}{3!} h^2$

As I see it: I have to show the left side is less or equal to the right side. LHS is differentiated one time, RHS differentiated 3 times. My guess is to use Taylor polynomial, is that right? And if, then how to apply as a proof?

答:
hint

$f $ is $ C^3([a,b],\Bbb R) $ .
By Taylor-Lagrange formula, and for very small $ h $,

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{6}f'''(c_1)$$
and
$$f(x\color{red}{-h})=f(x)\color{red}{-h}f'(x)+....$$


问:
How to find change of basis matrices
Suppose $U$ has a basis $e_1$, $e_2$, $e_3$ and $V$ has a basis $f_1$, $f_2$ and that $T:U \rightarrow V$ is a linear transformation whose matrix $A$ with respect to the given bases is $A=\begin{pmatrix}
0 & 1 & 2\\
3 & 4 & 5
\end{pmatrix}$.

The first part of the questions asks to show that $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$ forms a basis of $U$ and that $f_1$, $f_1$ + $f_2$ forms a basis of $V$. I've shown that. Now there are two things I need help with.

Write a matrix $A'$ of $T$ with respect to $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$ and $f_1$, $f_1$ + $f_2$.

Write a change of basis matrix $P$ from $e_1$, $e_2$, $e_3$ to $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$.

$$P=\begin{pmatrix}
1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 & 1
\end{pmatrix}$$

The answer is given, but I do not know how to work it out. Could you help me? Also, are $e_1$, $e_2$, $e_3$ and $f_1$, $f_2$ supposed to be standard bases? Thank you for your time.

答:
I'm not sure it's necessary to say the bases are standard at all.

One of the big problems I perceive with writing out matrices as arrays is that people lose the ability to think about them abstractly--they no longer know what the matrices mean.  I think this is part of the way matrices are taught, so it's not your fault.

Let's look at the matrix $A$.  What does it mean?  Well, any matrix can be thought of as a linear function.  You can equivalently describe $A$ as so:

$$A(e_1) = 3 f_2, \quad A(e_2) = f_1 + 4 f_2, \quad A(e_3) = 2 f_1 + 5 f_2$$

You'll notice each of these is a column, but with the bases written out.

How does this approach help?  It makes problem (1) really trivial.  Let $f_2' = f_1 + f_2$.  Then $f_2 = f_2' - f_1$.  Similarly, let $e_2' = e_1 + e_2$ and $e_3' = e_1 + e_2 + e_3$.  Now, you should be able to evaluate the components.  Consider $A(e_2')$:

$$\begin{align*}A(e_2') &= A(e_1) + A(e_2) \\ &= 3f_2 + f_1 + 4 f_2 \\ &= f_1 + 7 f_2 \\ &= f_1 + 7 (f_2'- f_1) \\&= -6 f_1 + 7 f_2'\end{align*}$$

All it takes to evaluate that is simple substitution.

Now, for (2), what is it we want to do when we use the change of basis matrix?  We want to feed in a vector in the new (primed) basis, convert it back to the old basis, use the map $A$, and the convert back to the primed basis. The change of basis map should then take a basis vector in the primed basis and convert it to a basis vector in the original, unprimed basis.  So like this:

$$P(e_1') = e_1, \quad P(e_2') = e_2, \quad P(e_3') = e_3$$

I'll handle only one of these cases, the last one.  You should observe that $e_3' = e_2' + e_3$, or $e_3 = -e_2' + e_3'$.  That basically does it--this is exactly the third column of $P$ that you're given. All you have to do for this part is express $e_1, e_2, e_3$ as linear combinations of $e_1', e_2', e_3'$.  Since $e_1 = e_1'$, you only really have to do 2 equations, at that.

Title: How to find change of basis matrices


问:
How to prove that $\Gamma(x) \leq x^x$ for $x \geq 1$
Where can I find a (simple?) proof that $\Gamma(x) \leq x^x \,\,\,\, \forall x \geq 1$?

$\Gamma(x)$ is the Gamma function defined as

$$ \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, \mathrm{d}t
$$

答:
For $1 \le x < 2$ we have
$$
 \Gamma(x) \le 1 \le x^x \, .
$$

For $x \ge 2$ and $n  = \lfloor x \rfloor$ is
$$
\Gamma(x) \le \Gamma(n+1) = n!  \le n^n \le x^x
$$
because the Gamma function is increasing on $[2, \infty)$, see for example How one can show that the gamma function is a strictly increasing function on the interval $(1.4616,+∞)$.


问:
Using the First Isomorphism Theorem in Universal Property statements.
I’m looking at some universal properties in the categories of abelian groups, specifically, those of the free products, the direct sums, and abelianization of groups. Most of the statements I’ve seen are along the line of:

Given maps $\phi_i: G_i \to H$,$\dots$, there exists a unique homomorphism $\phi$: [some object built from the $G_i$] $ \to H$ such that [some property is satisfied].

An example is the universal property of the abelianization of groups:

Given a group $G$, an abelian group $H$, and a homomorphism $f: G \to H$, there exists a unique homomorphism $F: G/[G,G] \to H$ such that $f = \pi \circ F$, where $G/[G,G]$ is the quotient of $G$ by its commutator subgroup $[G,G]$, and $\pi$ is the canonical projection.

I’m trying to prove some claims that involves isomorphisms of the groups involved in the above statement. One thing I could think of is to use the first isomorphism theorem. But to do so, I’d need a map $f: G \to H$ that has $[G,G]$ as its kernel (so that $G/[G,G] \cong H$). How do I know that such map exists?

答:
It's not clear to me what you're asking. $G/[G,G]\cong H$ does not always apply, the statement only says there is a homomorphism $G\rightarrow H$  rather than an isomorphism (for example $\mathbb Z\rightarrow0$). All you can say is that $[G,G]\subset\ker f$.

If you want the commutator subgroup to be a kernel: By definition $[G,G]$ is normal, and any normal subgroup forms a quotient, so just take the quotient map $\pi:G\rightarrow G/[G,G]$. But this seems quite unnecessary in your question.

I would not say this is something that uses first isomorphism theorem. It must use the property of abelian group and commutator subgroup. Maybe mimic the proof of first iso theorem, using the value of $f$ vanishes on $[G,G]$, etc. If you really want to use universal properties, I would say universal property of quotients is much better than first iso theorem.


问:
Knowing $\cos\theta$, can $\cos(n\theta)=\cos(\pi k)$?
$\cos(\pi k)=1$ or $-1$. After expressing $\cos(n\theta)$ in terms of $\cos\theta$, I have found that $\cos(n\theta)=\sum^{\lfloor{\frac{n}{2}}{\rfloor}}_{l=0}{n\choose2l}(-\frac{8}{9})^{l}(-\frac{1}{3})^{n-2l}$ but I cannot proceed. Is this the right approach? Thanks very much.

答:
I think I found another way to prove the result for the particular value of $\cos x=-\frac13$.

Generally we have the following recurrence relation:
$$
\cos(n+2)x=2\cos (n+1)x\cos x-\cos nx\tag1
$$

If $\cos x=\frac pq$ with $p,q\in\mathbb Z, q>0, (p,q)=1$, then  $$\cos nx=\frac{a_n}{q^n},\tag2$$
where $a_n$ is an integer. Substituting $(2)$ into $(1)$ one obtains:
$$
\frac{a_{n+2}}{q^{n+2}}=2\frac{a_{n+1}}{q^{n+1}}\frac pq-\frac{a_{n}}{q^{n}}
\implies a_{n+2}=2p\cdot a_{n+1}-q^2\cdot a_{n}\tag3
$$
in the case $\cos x=-\frac13$ the equation $(3)$ reads:
$$
a_{n+2}=-2a_{n+1}-9\cdot a_{n},\quad\text{with}\quad a_0=1,\ a_1=-1.\tag4
$$

It follows from $(4)$ immediately by induction:
$$
a_n\mod 3=\begin{cases}
1,& n\text{ even}\\
2,& n\text{ odd}
\end{cases}\tag5$$

This means that $\cos nx=\frac{a_n}{3^n}$ cannot be integer, particularly it cannot be equal to $1$.


问:
Proving $\oint{u \nabla v d\vec{r}}$ = $-\oint{v \nabla u d\vec{r}}$
My approach was to assign $dv =\frac{\partial{v}}{\partial{x}}dx + \frac{\partial{v}}{\partial{y}}dy + \frac{\partial{v}}{\partial{z}}dz$ , $du =\frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy + \frac{\partial{u}}{\partial{z}}dz$ , $u=u$, and $v=v$

then using the partial fraction formula:
$$\oint{v \nabla u d\vec{r}} = u.v - \int{v du} = u.v - \int{v . \nabla u d\vec{r}} = -v\nabla ud\vec{r}$$

this assumes that $ du = \nabla u .d \vec{r} $ and I am not sure if that assumption is corrrect as I am unsure if me assuming $\iiint \frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy + \frac{\partial{u}}{\partial{z}}dz= u$ is correct.

Can I get any insight if what I am doing is incorrect and maybe a hint on how to apprroach this correctly?
Thank you.

答: It follows from the fundamental theorem of line integrals that $\oint \nabla(uv) d\vec{r} = 0$.  Since $\nabla(uv) = u \nabla(v) + v \nabla(u)$, you get the result by integrating both sides.


问:
Find the Image of $\tanh{x}$
Problem

Find the Image of $\tanh{x}$

Attempt

I know that $\tanh{x} \rightarrow 1$ for $x \rightarrow \infty$ and $\tanh{x} \rightarrow -1$ for $x \rightarrow -\infty$. I know that $\tanh{x}$ is continuous over the real numbers and I know that I have to argue from the intermediate value thereom. However, I do not know how to apply the IVT to show that the image is indeed $(-1,1)$. I now that I have to pick an element $y\in (-1,1)$ and....

This is where I am stuck. I do not see how the argue for the double inclusion for

$$\tanh{((-\infty,\infty))}=((-1,1))$$

Title: Find the Image of $\tanh{x}$

Body:
hint

$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}
$$

$$=1-\frac{2}{e^{2x}+1}=y$$

$$\implies x=\frac 12\ln(\frac{1+y}{1-y})$$


问:
A slight generalization of the Sieve of Sundaram that might shed light on the $6n \pm 1$ phenomenon of sequence A002822.
There's the $n$ such that $6n \pm 1$ is a twin prime pair sequence: https://oeis.org/A002822
It contains all twin prime averages (divided by $6$) other than $4$.

Notice this sequence: Positive numbers $n = 6xy + x + y$ such that $x,y \in \Bbb{Z}$.  That sequence is the complement (all entries found in $\Bbb{N}$ that are not in A002822, by its very definition).

But notice the $6xy + x + y$ part.  That reminded me of the Sieve of Sundaram's $2ij + i + j$ form. Thus:

Start with a list of integers from $81$ to $N$, and cross out all elements of the form $6ij + ki + lj$ such that $k, l \in \{1,3,5\}$ and $i,j \geq 1$.  Then taking the remaining elements, multiplying them by $6$ and adding respectively $lk = 1,3,5,9,15, $ or $25$; and of course we union the result of each case / cross-out form.

We have the following possibilities:

$$
6(6ij + i + j) + 1 = (6i + 1)(6j + 1) \\
6(6i + i + 3j) + 3 = (6i + 3)(6j + 1) \\
6(6ij + i + 5j) + 5 = (6i + 5)(6j + 1) \\
6(6ij + 3i +3j) + 9 = (6i + 3)(6j + 3) \\
6(6ij + 3i + 5j) + 15 = (6i+5)(6j+3) \\
6(6ij + 5i +5j) + 25 = (6i + 5)(6j + 5) \\
$$

The above will cover all odd numbers $\geq  81$ which we find by setting $i=j = 1$ and computing the smallest formula above which is $(6i + 1)(6j + 1) = 81$.

Thus we've filtered out all and only composites $\geq 81$, clearly, and via inspection of the above factorizations.

See: Sieve of Sundaram.

However, all primes $\gt 3$ are of the form $6n + 1$ or $6n + 5$.  Thus we can cross out lines $2,3,5$ in the above 6 formulas, or namely the lines with added $3,9,15$.  But $6n + 25 = 1 \pmod 6$ so it is left and the list of cross-out forms in the generalized Sundaram sieve becomes:

$$
6ij + i + j \\
6ij + i + 5j \\
6ij + 5i + 5j \\
i,j \geq 1
$$

Thus, that is the Sundaram sieve method genralized to a multiple of $6$ instead of just $2$.  You could probably do similarly, with other multiples.

Question. Can we remove even more unecessary cross-out forms so that there would be less than $3$?

答:
No, you can't remove more, $3$ is the minimum. In fact, there are $4$, but one of them is superfluous because of symmetry. Before getting to the point, a first observation: the $6xy+x+y$ form you mention at the top of your question is equivalent to, and better known as $6xy \pm x \pm y$, and if you take your $3$ final forms modulo 6, you'll realize they're in fact identical to that form.

And there's a good reason for that: the proof that positive integers $n$ that can't be written in the $6xy \pm x \pm y$ form are bijectively related to twin primes is mostly just a simple change of coordinates for an elliptic curve equation. The derivation should prove a fun exercise. And essentially, you'll arrive at
$$
6xy \pm x \pm y = n \quad \text{has no solutions for $(x,y) \in \mathbb{Z}^2$.}\\
\big\Updownarrow \\
X^2-Y^2=6n \pm 1 \quad \text{both have only one solution.}
$$
So yeah, I'm afraid you do need all $3$ of them.


答:
As

$$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1},$$

$$\frac{1+\tanh x}{1-\tanh x}=e^{2x}>0$$ and this occurs for $-1<\tanh x<1$ by resolution of the inequation. And for all $\tanh x$ in that range, $x$ is defined.

Title: Find the Image of $\tanh{x}$

Body:
Problem

Find the Image of $\tanh{x}$

Attempt

I know that $\tanh{x} \rightarrow 1$ for $x \rightarrow \infty$ and $\tanh{x} \rightarrow -1$ for $x \rightarrow -\infty$. I know that $\tanh{x}$ is continuous over the real numbers and I know that I have to argue from the intermediate value thereom. However, I do not know how to apply the IVT to show that the image is indeed $(-1,1)$. I now that I have to pick an element $y\in (-1,1)$ and....

This is where I am stuck. I do not see how the argue for the double inclusion for

$$\tanh{((-\infty,\infty))}=((-1,1))$$


问:
why $\sin 2πx$ is strictly monotonic?
I have some confusion in Munkres book

My confusion given below  marked in red colour

My confusion :why $\sin 2πx$ is strictly monotonic on such an interval?.

My thinking : Take $x=0 $  and $x=1$ , then $f(x)= \sin 2 \pi x$  will not be strictly monotonic  since $f(0)=f(1)=0$

答: Note that$$\overline{V_n}=\overline{\left(n-\frac14,n+\frac14\right)}=\left[n-\frac14,n+\frac14\right]$$and that $x\in\left[n-\frac14,n+\frac14\right]\iff2\pi x\in\left[2\pi n-\frac\pi2,2\pi n+\frac\pi2\right]$. Can you take it from here?

Title: why $\sin 2πx$ is strictly monotonic?


答:
hint

Let $ f(x)=\sin(2x\pi)$.
$$x\in V_n \implies$$
$$n-\frac 14< x <n+\frac 14\implies$$

$$-\frac{\pi}{2}< 2\pi x-2n\pi < \frac{\pi}{2}$$

$$\implies \cos(2\pi x)=\cos(2x\pi-2n\pi)>0$$

$$\implies f'(x)>0$$

$ f$ is continuous at $ \overline{V_n} \implies$

$ f $ is strictly increasing at $\overline{V_n}$.

Title: why $\sin 2πx$ is strictly monotonic?

Body:
I have some confusion in Munkres book

My confusion given below  marked in red colour

My confusion :why $\sin 2πx$ is strictly monotonic on such an interval?.

My thinking : Take $x=0 $  and $x=1$ , then $f(x)= \sin 2 \pi x$  will not be strictly monotonic  since $f(0)=f(1)=0$


问:
How is the answer to $5x + 10 = \frac{x}{3} + 24$ obvious without solving the equation or testing any of the choices (multi-choice answer)
\begin{align*}
5x + 10 = \frac{x}{3} + 24 \\
(A) 1 \\
(B) 2 \\
(C) 3 \\
(D) 4 \\
(E) 5 \\
\end{align*}
The problem implies that I can deduce the answer $C, 3$ without solving the equation or testing the five choices.
I can't see how.

答: Given that all the possibilities are integers, the solution must be divisible by $3$


答:
Hint: You can use the fact that $\tanh x$ is an increasing function:
$$
(\tanh x)'=\frac1{\cosh^2 x}>0.
$$

Title: Find the Image of $\tanh{x}$


问:
Variance of a sample with regards to $E[x]$
I still get a bit confused when calculating the variance of a sample:

Suppose I am given a value for $\sum_{i=1}^n x_i$, $\sum_{i=1}^n x_i^2$ and  $n$. Which formula is the best to use as I seem to get confused whether to divide through by $n$ or $n-1$ ?

To be more clear,

Is either of the two below preferred, or do you know of a more accurate equation for estimate of sample variance?

$[(\sum_{i=1}^n x_i^2)/n]$ $-$  $[(\sum_{i=1}^n x_i)/n)^2]$

$[(\sum_{i=1}^n x_i^2)/(n-1)]$ $-$  $[(\sum_{i=1}^n x_i)/(n-1)^2]$

答:
To estimate the mean you simply use

$$\hat{\mu}=\frac{1}{n}\sum_{i=1}^nx_i$$

Using this estimated mean, you get an unbiased estimator of the variance by computing

$$\hat{\sigma}^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\hat{\mu})^2$$

It can be shown that if you divide by $n$ instead of by $n-1$, the estimator for the variance is biased.


问:
Problem in notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ versus $L_v(\vec{v})$
I've seen more than one time the use of the following notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ to say $L_v(\vec{v})$ where the matrix $L_v$ is defined as follows: $(L_v)_{ij} := \partial_j v_i$.

For example this happens in the usual law of conservation of momentum of an eulerian fluid:
$$\rho\Big(\partial_t\vec{v} + (\vec{v}\cdot\vec{\nabla})\vec{v}\Big) = \rho\vec{b} + \vec{\nabla}p$$

My question is: how that notation could "help" us to understand "how things work"?

I think this is strange, but physicists are known to have strange notations that in some ways help to understand the nature of things. Take for example the divergence of a vector field $\vec{v}\ $: this is written in some texts as $\vec{\nabla}\cdot\vec{v}$ because if we think at $\vec{\nabla}$ as the vector $(\partial_1, \partial_2, \partial_3)$ then the scalar product $$\vec{\nabla}\cdot\vec{v} = (\partial_1, \partial_2, \partial_3) \cdot (v_1,v_2,v_3) = \sum_{i=1,2,3} \partial_i \ v_i = \text{div}(\vec{v})$$

Now, if I take for example a $2$ dimensional field $\vec{v} = (v_1,v_2)$ we have (correct me if I'm wrong)
$$L_v(\vec{v}) =
\begin{pmatrix}
\partial_1 v_1 & \partial_2 v_1 \\
\partial_1 v_2 & \partial_2 v_2
\end{pmatrix}
\begin{pmatrix}
v_1 \\
v_2
\end{pmatrix}
=
\begin{pmatrix}
(\partial_1 v_1)v_1 + (\partial_2 v_1)v_2 \\
(\partial_1 v_2)v_1 + (\partial_2 v_2)v_2
\end{pmatrix}
$$

How could I derive "intuitively" the last vector using the notation $(\vec{v}\cdot\vec{\nabla})\vec{v}$?

答:
Okay, I think I got it.

I have to consider the scalar product $\vec{v}\cdot \vec{\nabla}$ as the "scalar" $v_1 \partial_1 + v_2\partial_2$ that multiplies the vector $(v_1,v_2)$:

$$(\vec{v}\cdot \vec{\nabla})\vec{v} = \big(v_1 \partial_1 + v_2\partial_2 \big)
\begin{pmatrix}
v_1 \\v_2
\end{pmatrix} =
\begin{pmatrix}
v_1 \ \partial_1v_1 + v_2\ \partial_2 v_1 \\
v_1 \ \partial_1v_2 + v_2\ \partial_2 v_2
\end{pmatrix} = L_v(\vec v)
$$

This is really odd.


答:
You already understand that in 2D we are to treat $\overrightarrow{\nabla}$ as $\left\langle \partial_{1},\partial_{2}\right\rangle$. The key for this and many other related notations is basically to treat something like $\partial_1$ as a scalar that's distributive and commutes with scalars (it's a linear operator, after all), and associative where it makes sense, but not commutative.

We have $\overrightarrow{v}\cdot\overrightarrow{\nabla}=\left\langle v_{1},v_{2}\right\rangle \cdot\left\langle \partial_{1},\partial_{2}\right\rangle =v_{1}\partial_{1}+v_{2}\partial_{2}$.

And then:
\begin{align}\left(\overrightarrow{v}\cdot\overrightarrow{\nabla}\right)\overrightarrow{v}&=\left(v_{1}\partial_{1}+v_{2}\partial_{2}\right)\left\langle v_{1},v_{2}\right\rangle \\&=\left(v_{1}\partial_{1}\right)\left\langle v_{1},v_{2}\right\rangle +v_{2}\partial_{2}\left\langle v_{1},v_{2}\right\rangle \\&=\left\langle \left(v_{1}\partial_{1}\right)v_{1},\left(v_{1}\partial_{1}\right)v_{2}\right\rangle +\left\langle \left(v_{2}\partial_{2}\right)v_{1},\left(v_{2}\partial_{2}\right)v_{2}\right\rangle \\&=\left\langle v_{1}\left(\partial_{1}v_{1}\right),v_{1}\left(\partial_{1}v_{2}\right)\right\rangle +\left\langle v_{2}\left(\partial_{2}v_{1}\right),v_{2}\left(\partial_{2}v_{2}\right)\right\rangle \\&=\left\langle v_{1}\left(\partial_{1}v_{1}\right)+v_{2}\left(\partial_{2}v_{1}\right),v_{1}\left(\partial_{1}v_{2}\right)+v_{2}\left(\partial_{2}v_{2}\right)\right\rangle \\&=\left\langle \left(\partial_{1}v_{1}\right)v_{1}+\left(\partial_{2}v_{1}\right)v_{2},\left(\partial_{1}v_{2}\right)v_{1}+\left(\partial_{2}v_{2}\right)v_{2}\right\rangle \\&=\begin{bmatrix}\left(\partial_{1}v_{1}\right)v_{1}+\left(\partial_{2}v_{1}\right)v_{2}\\
\left(\partial_{1}v_{2}\right)v_{1}+\left(\partial_{2}v_{2}\right)v_{2}
\end{bmatrix}\\&=\begin{bmatrix}\partial_{1}v_{1} & \partial_{2}v_{1}\\
\partial_{1}v_{2} & \partial_{2}v_{2}
\end{bmatrix}\begin{bmatrix}v_{1}\\
v_{2}
\end{bmatrix}\end{align}

Title: Problem in notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ versus $L_v(\vec{v})$

Body:
I've seen more than one time the use of the following notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ to say $L_v(\vec{v})$ where the matrix $L_v$ is defined as follows: $(L_v)_{ij} := \partial_j v_i$.

For example this happens in the usual law of conservation of momentum of an eulerian fluid:
$$\rho\Big(\partial_t\vec{v} + (\vec{v}\cdot\vec{\nabla})\vec{v}\Big) = \rho\vec{b} + \vec{\nabla}p$$

My question is: how that notation could "help" us to understand "how things work"?

I think this is strange, but physicists are known to have strange notations that in some ways help to understand the nature of things. Take for example the divergence of a vector field $\vec{v}\ $: this is written in some texts as $\vec{\nabla}\cdot\vec{v}$ because if we think at $\vec{\nabla}$ as the vector $(\partial_1, \partial_2, \partial_3)$ then the scalar product $$\vec{\nabla}\cdot\vec{v} = (\partial_1, \partial_2, \partial_3) \cdot (v_1,v_2,v_3) = \sum_{i=1,2,3} \partial_i \ v_i = \text{div}(\vec{v})$$

Now, if I take for example a $2$ dimensional field $\vec{v} = (v_1,v_2)$ we have (correct me if I'm wrong)
$$L_v(\vec{v}) =
\begin{pmatrix}
\partial_1 v_1 & \partial_2 v_1 \\
\partial_1 v_2 & \partial_2 v_2
\end{pmatrix}
\begin{pmatrix}
v_1 \\
v_2
\end{pmatrix}
=
\begin{pmatrix}
(\partial_1 v_1)v_1 + (\partial_2 v_1)v_2 \\
(\partial_1 v_2)v_1 + (\partial_2 v_2)v_2
\end{pmatrix}
$$

How could I derive "intuitively" the last vector using the notation $(\vec{v}\cdot\vec{\nabla})\vec{v}$?


问:
Use Jensen's inequality to show that $e^{\frac{\sum^3_{i=1} X_i}{3}}$ is not an unbiased estimator of $e^{\theta}$
My problem:

Let $X_{1}, X_{2},X_{3}$ be I.I.D. Poisson variables with common mean $\theta > 0$.

Use Jensen's inequality to show that $e^{\frac{\sum^3_{i=1}X_i}{3}}$ is not an unbiased estimator of $e^{\theta}$ and then use moment generating functions to compare this result.

My work so far:

With Jensen's inequality, I know I am trying to show that

$$E(h(x)) \geq h(E(x)).$$

I have also noted that for $e^{\frac{\sum^3_{i=1}X_i}3}$, the sum of three independent Poison variables will be an independent Poisson variable.

I also know that $E\left(e^{\frac{\sum^3_{i=1}X_i}3}\right)=e^{ \lambda(e-1)}$

From here I start to get lost. I am having quite some trouble with this one. I appreciate the help.

答:
To show that your estimator is biased it is very simple. In fact, using Jensen's inequality you get that

$$\mathbb{E}\left[e^{\overline{X}_3}\right]\ne e^{\mathbb{E}[\overline{X}_3]}=e^{\theta}$$

...it is not necessary to identify if your expectation is greater or lower than $e^{\theta}$


答: The left side is always an integer because the input answers are also always integers. Thus, the right side must be an integer, and for $\frac x3$ to be an integer, $x$ must be divisible by $3$ as marla stated.

Title: How is the answer to $5x + 10 = \frac{x}{3} + 24$ obvious without solving the equation or testing any of the choices (multi-choice answer)

Body:
\begin{align*}
5x + 10 = \frac{x}{3} + 24 \\
(A) 1 \\
(B) 2 \\
(C) 3 \\
(D) 4 \\
(E) 5 \\
\end{align*}
The problem implies that I can deduce the answer $C, 3$ without solving the equation or testing the five choices.
I can't see how.


问:
Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$
Please help to evaluate this limit

$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$

where $0 \leq a$ and $a \not= 1$.

I tried to logarithm from both sides, and apply taylor series but so far without success.

答: The limit of $\left(\frac1x\right)^{\frac1x}$ equals one (by taking logs). If $a>1,$ then $(a-1)^{\frac1x} \to 1,$ while $(a^x-1)^{\frac1x} \to a,$ so the limit equals $a.$ If $a<1,$ then the limit equals $1$ (exercise).


问:
How to solve $x\log(x) = 10^6$
I am trying to solve

$$x\log(x) = 10^6$$

but can't find an elegant solution. Any ideas ?

答: There is no simple solution (as you have found out), your only way out is using a numerical method to solve the equation, e.g. Newton's method.</p>

Title: How to solve $x\log(x) = 10^6$


问:
Integration by parts question: why does the constant of integration get swallowed up?
To integrate $e^x\cos x$ we can use integration by parts twice:
\begin{align}
\int e^x \cos x \, dx &= e^x\sin x - \int e^x\sin x \, dx \\
&= e^x\sin x - \left(-e^x\cos x + \int e^x\cos x \, dx\right) \\
&= e^x\sin x + e^x\cos x - \int e^x \cos x \, dx \\
2 \int e^x\cos x \, dx &= e^x(\sin x + \cos x) \\
\int e^x \cos x \, dx &= \frac{e^x(\sin x + \cos x)}{2}
\end{align}
But this solution is incomplete: there should be a constant of integration at the end. Why does this happen? I think it might have something to do that you are adding a family of functions to itself, but I can’t quite pin down what the problem is.

答:
This solution to your problem makes use of a very basic fact. Note that $\int f(x)~dx=\int f(x)+0~dx$
$$\begin{align}\int e^x \cos x \, dx&= e^x\sin x - \left(-e^x\cos x + \int e^x\cos x \, dx\right)\\
&=e^x\sin x+e^x\cos x-\int e^x\cos x \, dx\\
&=e^x\sin x+e^x\cos x-\int e^x\cos x \, dx+\int0~dx\end{align}$$
Hence,
$$2\int e^x\cos x \, dx=e^x\sin x+e^x\cos x+C$$


问:
Convergence in $L^p$ space
Let $(X, M, \mu)$ be a $\sigma$-finite measure space.

Let $1<p < \infty$. Suppose $(f_n)_{n \in \mathbb{N}}$ is a bounded sequence in $L^{p}(\mu)$ such that $\lim_{n \rightarrow \infty} f_n g d\mu$ exists for every $g \in L^{q}(\mu)$.

Find $f \in L^p(\mu)$ so that $\lim_{n\rightarrow \infty}\int f_n g d\mu = \int fg d\mu$.

I thought that the last part requires the Lebesgue Dominated Convergence Theorem. But the assumptions don't meet the assumptions to apply the theorem here.

Any help will be appreciated!

答:
If we allow the limit of $\int f_n gd\mu$ to depend on $g$, then the function $f$ could be any element of $L^p\pr{\mu}$ (take $f_n=f$).

If $\lim_{n\to\infty}\int f_n gd\mu$ is not allowed to depend on $g$, then (replacing $g$ by $-g$), one gets that $\lim_{n\to\infty}\int f_n gd\mu=0$ and only $f=0$ can work, since $\int fgd\mu=0$ for each $g\in L^q(\mu)$ implies that $g=0$.

Title: Convergence in $L^p$ space


问:
Ring $R'$ with $R\subset R'\subset T$ where $R$, $T$ integrally closed and obtained by a purely inseparable extension
Let $R$ be an integrally closed domain with field of fractions $K$. Let $M/K$ be a purely inseparable extension and let $T$ be $R$'s integral closure in $M$. Consider a subring $R\subset R'\subset T$ and put $L$ its field of fractions and $S$ the integral closure of $R$ in $L$. We have $R\subset R'\subset S\subset T$ and purely inseparable extensions $K\subset L\subset M$.

Question. is $R'$ automatically integrally closed, i.e. is $R'=S$ ?

$
\newcommand{\P}{\mathfrak{P}}
\newcommand{\p}{\mathfrak{p}}
\newcommand{\Spec}{\mathrm{Spec}}
\newcommand{\N}{\mathbb{N}}
$I don't remember exactly how the question came about but I know it came up when reading the proof of the theorem below from Gerald J. Janusz's Algebraic Number Fields. For more details see the final sentence of the post.

Theorem 6.1 Let $R$ be a Dedekind ring with field of fractions $K$ and let $L$ be a finite field extension of $K$. Then the integral closure of $R$ in $L$ is a Dedekind domain.

The proof is in two steps. Considering the intermediate field $K\subset E\subset L$ of all separable elements of $L$ over $K$ one reduces the question to two special cases: separable extension and purely inseparable extension. In the purely inseparable case the author proves something that can be generalized as follows:

Lemma. Let $R$ be integrally closed with field of fractions $K$ of characteristic $p$ and consider a purely inseparable extension $L/K$. Let $S$ be the integral closure of $R$ in $L$ and consider a subring $R\subset R'\subset S$. Then the inclusion $i:R\to R'$ induces a bijection on spectra, i.e. $i^*:\Spec(R')\to\Spec(R),\P\mapsto\P\cap R$ is a bijection.

Proof. One first notices that
$$S=\{x\in L\mid \exists n\in\N,~x^{p^n}\in R\}$$
Indeed, if $x\in L$ satisfies $x^{p^n}\in R$ then $x$ is integral over $R$ so that $x\in S$ by definition of $S$. Conversely, if $x\in L$ then, $L/K$ being purely inseparable, $x^{p^n}\in K$ for some $n$. If furthermore $x\in S$, then $x$ is integral over $R$ and so is $x^{p^n}\in K$. Since $R$ is integrally closed we have $x^{p^n}\in R$, and so both inclusions are proven.

$\mathbf{i^*}$ is one to one. First of all, if $\P$ is nonzero prime ideal of $R'$ then so is $\p=i^*(\P)$. Indeed, if $x\in\P$ is nonzero then for some $n$ we have
$$x^{p^n}\in R\setminus 0$$
and thus $x^{p^n}$ is a nonzero element of $\p$. Furthermore, we can reconstruct $\P$ from $\p$. Indeed, one shows
$$\P=\{x\in R'\mid \exists n\in \N,~x^{p^n}\in\p\}$$
Indeed, if $x\in R'$ satisfies $x^{p^n}\in\p\subset\P$, then, $\P$ being prime, $x\in\P$. And conversely, if $x\in\P$ then for some $n$, $x^{p^n}\in R$ so that $x^{p^n}\in\P\cap R=\p$.

$\mathbf{i^*}$ is onto. Let $\p$ be a prime of $R$ and put
$$\p'=\{x\in R'\mid\exists n\in\N,~x^{p^n}\in\p\}$$
$\p'$ is a prime ideal of $R'$. Indeed, if $x,y\in\p'$ and $r\in R'$ then there exist exponents $a,b,c\geq 0$ with $x^{p^a},y^{p^a}\in\p$ and $r^{p^c}\in R$. Thus if $n=\max\{a,b,c\}$ we have $(x+ry)^{p^n}\in \p$ so that $\p'$ is an ideal of $R'$. And if $xy\in\p'$ with $x,y\in R'$ and $y\notin\p'$ then for some $n\in\N$, $x^{p^n}\cdot y^{p^n}=(xy)^{p^n}\in\p$. With $\p$ prime and $y\notin\p'$ we have $x^{p^n}\in\p$ i.e. $x\in\p'$, and so $\p'$ is prime. Now one easily sees
$$\p=R\cap\p'=i^*(\p')$$
Indeed, $\p\subset\p'$ thus $\p\subset R\cap\p'$, and if $x\in R\cap\p'$ then $x^{p^n}\in\p$ with $x\in R$ and with $\p$ being a prime ideal of $R$, $x\in\p$. Therefore $i^*$ is onto.$\quad\blacksquare$

I was surprised that there would be a bijection on spectra without further hypothesis. The author of the book doesn't state the result above, rather he proves and uses it only for the subring $R'=S$ and I wondered whether this generalization is indeed a generalization or whether all subrings $R'$ are automatically integrally closed (whence my question, I think.)

答:
If $R'/R$ is integral and $Frac(R')/Frac(R)$ is purely inseparable then it is immediate that $\mathfrak{p}\to \mathfrak{p}\cap R$ is a bijection $Spec(R')\to Spec(R)$.

$\Bbb{F}_p[x^p,x^{p+1}]$ is not integrally closed in its fraction field and it is between the PID $\Bbb{F}_p[x]$ and $\Bbb{F}_p[x^p]$.

The problem of non-separable extensions $L/K$ is that $O_L$ doesn't have to be a finitely generated $O_K$-module (can't use the trace map which is identically $0$) and the $n=\sum_i f_i e_i$ stuffs may fail.


问:
Matrix inequality of inverses of sum of two matrices
$A$ and $B$ are two real symmetric non-singular matrices but not necessarily sign definite. Suppose $A+B$ is non-singular. I want to find a $\lambda \in \Bbb R$ such that $$(A+B)^{-1}\leq(A+\lambda I)^{-1}$$ where $A\leq B$ means $A-B$ is negative semi-definite. Does such $\lambda$ exist? How can I find it?

I have tried the identity $(A+B)^{-1}=A^{-1}-A^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$. But it led me to

$$(A^{-1}+\lambda^{-1} I)^{-1}\leq(A^{-1}+B^{-1})^{-1}$$

Thanks in advance!

答:
For positive Definite symmetric matrices (2/3 down on that page) one has $M\geq N > 0$ iff $N^{-1}\geq M^{-1} >0$ which under your hypotheses translates to
$$ 0< (A+B)^{-1} \leq (A +\lambda I)^{-1}$$
iff
$$ A+B\geq A +\lambda I > 0$$
or $B\geq \lambda I$. So if $B$ has a non-trivial kernel then you need $\lambda \leq 0$ (and $A$ should be strictly positive for the above to make sense).


问:
Projections in Graph Products
I was going throug the topic Projections in Graph Products. There a term weak homomorphism was used. Although I am familiar with the term Isomorphism of graphs, but have no idea what this term "weak homomorphism" means here.

PROJECTION: Let * represent either cartesian, the direct or strong product of graphs, and consider a product $G_1 * G_2* ....*G_k$. For any index i 1$\leq$*i*$\leq$*k*, a Projection map is defined as :

$p_i$ : $G_1 * G_2* ....*G_k$ $\rightarrow$ $G_i$ where $p_i$($x_1,x_2,...,x_k$)=$x_i$.

Can anybody help me out with the term weak homomorphism.

Thanks a lot.

Following links will help for definiton of product graphs...
http://en.wikipedia.org/wiki/Tensor_product_of_graphs
http://en.wikipedia.org/wiki/Cartesian_product_of_graphs
http://en.wikipedia.org/wiki/Strong_product_of_graphs

答:
(We don’t actually need to know about the graph products in order to answer the question.)

If $G$ and $H$ are graphs with vertex sets $V(G)$ and $V(H)$, respectively, a homomorphism from $G$ to $H$ is a map $\varphi:V(G)\to V(H)$ that preserves adjacency: if $uv$ is an edge in $G$, then $\varphi(u)\varphi(v)$ is an edge in $H$. A weak homomorphism from $G$ to $H$ is a map $\varphi:V(G)\to V(H)$ such that if $uv$ is an edge in $G$, then either $\varphi(u)\varphi(v)$ is an edge in $H$, or $\varphi(u)=\varphi(v)$. In other words, if $\varphi$ is a weak homomorphism from $G$ to $H$, and $e$ is an edge in $G$, $\varphi$ either preserves $e$ as an edge by collapses it to a single vertex. Clearly every homomorphism is automatically a weak homomorphism.

In the diagram below let $G$ be the path graph on the left and $H$ the triangle graph on the right.

```<code>           1---2---3                a---b
                                     \ /
                                      c
</code>```

The map

$$\left\{\begin{align*}
1&\mapsto a\\
2&\mapsto b\\
3&\mapsto c
\end{align*}\right.$$

is a homomorphism from $G$ to $H$ is a homomorphism. The map

$$\left\{\begin{align*}
1&\mapsto a\\
2&\mapsto b\\
3&\mapsto b
\end{align*}\right.$$

is not a homomorphism, because it does not take the edge $23$ of $G$ to an edge of $H$, but it is a weak homomorphism: it collapses that edge to the single vertex $b$ of $H$.

Title: Projections in Graph Products

Body:
(We don’t actually need to know about the graph products in order to answer the question.)

If $G$ and $H$ are graphs with vertex sets $V(G)$ and $V(H)$, respectively, a homomorphism from $G$ to $H$ is a map $\varphi:V(G)\to V(H)$ that preserves adjacency: if $uv$ is an edge in $G$, then $\varphi(u)\varphi(v)$ is an edge in $H$. A weak homomorphism from $G$ to $H$ is a map $\varphi:V(G)\to V(H)$ such that if $uv$ is an edge in $G$, then either $\varphi(u)\varphi(v)$ is an edge in $H$, or $\varphi(u)=\varphi(v)$. In other words, if $\varphi$ is a weak homomorphism from $G$ to $H$, and $e$ is an edge in $G$, $\varphi$ either preserves $e$ as an edge by collapses it to a single vertex. Clearly every homomorphism is automatically a weak homomorphism.

In the diagram below let $G$ be the path graph on the left and $H$ the triangle graph on the right.

           1---2---3                a---b
                                     \ /
                                      c

The map

$$\left\{\begin{align*}
1&\mapsto a\\
2&\mapsto b\\
3&\mapsto c
\end{align*}\right.$$

is a homomorphism from $G$ to $H$ is a homomorphism. The map

$$\left\{\begin{align*}
1&\mapsto a\\
2&\mapsto b\\
3&\mapsto b
\end{align*}\right.$$

is not a homomorphism, because it does not take the edge $23$ of $G$ to an edge of $H$, but it is a weak homomorphism: it collapses that edge to the single vertex $b$ of $H$.


问:
Check if two rotated cuboids overlap?
Is there a general approach towards finding the intersection over union between two rotated cuboids? Or even a method that checks whether two rotated cuboids overlap?

At the moment, I sample points along each of the 12 edges of a cuboid and then check whether any of these points are inside the other cuboid. However, this is not an elegant solution and I was looking for something more mathematically straightforward. Is there any such method in math that can tell whether two rotated cuboids overlap?

答:
Since cuboids are convex polytopes you can use the Separating Axis Theorem. From wikipedia:

"The separating axis theorem (SAT) says that:

Two convex objects do not overlap if there exists a line (called axis) onto which the two objects' projections do not overlap.

SAT suggests an algorithm for testing whether two convex solids intersect or not.

Regardless of dimensionality, the separating axis is always a line. For example, in 3D, the space is separated by planes, but the separating axis is perpendicular to the separating plane.

The separating axis theorem can be applied for fast collision detection between polygon meshes. Each face's normal or other feature direction is used as a separating axis, as well as the cross products."

https://en.m.wikipedia.org/wiki/Hyperplane_separation_theorem

Title: Check if two rotated cuboids overlap?


问:
$n$ by $n$ Primally Magic Squares
(Again copied verbatim from a September 2009 thread I made.)

A Primally Magic Square (PMS) is exactly like a traditional magic square with a change of criteria. Where a traditional magic square is one where the sum of each row, column, and both diagonals are some number, usually 15 (uses 1-9). By contrast, a PMS is one where all rows, columns, and both diagonals form a prime number. Note: if a number is prime either forward or back, it counts.

Example:

$\displaystyle \begin{matrix} 1 & 3 & 7 \\ 3 & 9 & 7 \\ 7 & 7 & 9 \end{matrix}$

137, 397, 977, 797, and 199 are all prime, so this square is a PMS.

How many 3x3 PMS's are there?

How many 4x4 PMS's are there?

How many 5x5 PMS's are there?

How many 9x9 PMS's are there?

Side note: Feel free to calculate the number of PMS's for some other $n$.

答:
Following a comment, let's look at the $2×2$ case.  I shall consider the case where all digits are distinct and to avoid issues with convention, nonzero.

Within these limits we select a set of four digits such as $\{1,2,3,4\}$.  If each of the six two-element permutations generates a prime in either order, we have three primary magic squares apart from rotations and reflections (depending on which element is opposite a given one).

The set $\{1,2,3,4\}$ fails because $42$ and $24$ are both composite, and similarly there cannot be any two even digits, nor the digit $5$ (only two numbers in the 50s are prime and we would need three).  We also cannot have two digits with  residues $1,2\bmod 3$ or two multiples of $3$ (which lead to multiples of $3$), nor specifically $4$ together with $9$ (the odd number $49$ is $7×7$).

Here are the sets that work:

$\{1,3,4,7\}$

And that seems to be it!  Thus three squares (plus trivial operations):

$\begin{array}&1&3\\4&7\end{array}$

$\begin{array}&1&4\\7&3\end{array}$

$\begin{array}&1&7\\3&4\end{array}$

Title: $n$ by $n$ Primally Magic Squares


问:
Prove that $x^4 \equiv a^2 \pmod p$ is solvable for all $a$
Let $p$ be a prime such that $p \equiv 3 \pmod{4}$

Prove that $x^4 \equiv a^2 \pmod p$ is solvable for all $a$.

I'm asked to work out this problem and I'm wondering if my approach is correct?

My solution:

If $p|a$ then letting $x=0$ is a solution.

If $p \nmid a$, first factorize to get $(x^2 -a)(x^2-(-a)) \equiv 0 \pmod p$

and then $\left(\frac{a}{p}\right) = 1$ or $\left(\frac{a}{p}\right) = -1$.

If $\left(\frac{a}{p}\right) = 1$ then there is a solution.

If $\left(\frac{a}{p}\right) = -1$ then $\left(\frac{-a}{p}\right)= \left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)= (-1)(-1) = 1$ because $\left(\frac{-1}{p}\right)=-1$ if $p\equiv 3 \pmod4$ and there is a solution.

答:
Your solution is fine.  Here's an another way to see it that doesn't rely on knowing how to manipulate quadratic residues.

Since $\Bbb Z_p^*$ is cyclic of order $p-1$, let $b$ be a generator.  Then $a^2=b^{2k}$ for some $k \in \Bbb N$.  If $k$ is even, we're done.  If $k$ is odd, then $b^{2k}=b^{2k+p-1}$ and since $p \equiv 3 \pmod 4, p-1=2m$ for some $m$ odd, so that $k+m$ is even, $4 \mid 2(k+m)$ and again we're done.


答:
Let $y=\log(x)$. Then the equation is
$$ye^y=10^6.$$
The Lambert W-function is defined such that this means $y=W(10^6)$, and therefore $x=e^{W(10^6)}$.
(This is effectively just a notational trick; it doesn't make anything more explicit).</p>

Title: How to solve $x\log(x) = 10^6$

Body:
I am trying to solve

$$x\log(x) = 10^6$$

but can't find an elegant solution. Any ideas ?


问:
Is $y\lim\limits _{n\to +\infty }x^n$ where $(0\leq x\leq 1)$ continuous?
Let
$$y=\lim _{n\to +\infty }x^n,$$ where $x\in[0,1]$. Is this function continuous or in what points is it discontinuous.

I am thinking that because multiplication of continuous functions is continuous than this function should be continuous.But the answer is that point $x=1$ is  first kind discontinuous point.

答:
Let $$f(x)=\lim_{n\to\infty}x^n$$

$$n>0\implies 0^n=0\implies f(0)=0$$

$$n\ge 0\implies 1^n=1\implies f(1)=1$$

$$0<x<1\implies x^n=e^{n\ln(x)}\to 0$$

So $ f $ is defined at $ [0,1] $ by

$$f(x)=0 \text{ if } x\ne 1\text{ and } f(1)=1$$

So
$$\lim_{x\to 1^-}f(x)=0\ne f(1)$$


问:
Confused about Fourier's Inversion Theorem with another Proposition
Consider the space $L^{1}(\mathbb{R})$.

The Inversion theorem states :

If $f$ and $\widehat{f}$ are in $L^{1}(\mathbb{R})$, then $\overline{\mathscr{F}}\widehat{f}(t)=f(t)$ at all points where $f$ is continuous.

There is another proposition that is similar which states :

If $f$ is continuous and integrable and if $\widehat{f}\in L^{1}(\mathbb{R})$, then for all $x\in\mathbb{R}$ :
$$
\mathscr{F}\widehat{f}(x)=f(-x)
$$

I am lost in trying to differentiate between these two results. Is the proposition (second statement) saying that the inverse of Fourier is now just $f(-x)$ given that $f$ is also integrable? because I have seen examples where this proposition is being used to compute inverse Fourier of some functions.

答:
The notation of the first theorem is slightly nonconventional. The $\overline{\mathscr{F}}$ is meant to be the inverse Fourier transform, given by the formula
$$
\overline{\mathscr{F}}g(x) = \int_{\mathbb{R}} e^{2\pi i x \cdot \xi} g(\xi)\,d\xi.
$$
This is what you get if you "complex conjugate" the formula for the Fourier transform
$$
\mathscr{F} f(\xi) = \int_{\mathbb{R}} e^{-2\pi i x \cdot \xi} f(x)\,dx.
$$
This is not true complex conjugation because $g$ could be complex valued, but the bar didn't get applied to $g$.

The first theorem thus tells you that $\overline{\mathscr{F}}$ actually inverts the Fourier transform and gives you the function back at points of continuity of the function.

The second theorem is not about applying $\mathscr{F}$ then $\overline{\mathscr{F}}$, but rather about applying $\mathscr{F}$ twice. Essentially, applying the Fourier transform twice just reflects the function. And hence applying it four times gets you back to the original function. In other words, under suitable continuity and integrability assumptions: $\mathscr{F}^{-1} = \mathscr{F}^3 = \overline{\mathscr{F}}$.

Thus, to answer your question, NO $\mathscr{F} \hat f(x) = f(-x)$ is NOT saying that if $f$ is integrable then the inverse of the Fourier transform is just $f(-x)$. It is saying that the square of the Fourier transform is the reflection map, the map that sends $f(x) $to $f(-x)$.


问:
Show $ \int_0^{\frac\pi2} \frac{(1+\sec^2t)\sqrt{\sec t}}{ (1+\sec t)^2-2}dt= \frac\pi{\sqrt2}$
I try to figure out how to deal with the trigonometric integral

$$ \int_0^{\frac\pi2} \frac{(1+\sec^2t)\sqrt{\sec t}}{ (1+\sec t)^2-2}dt
$$
which is supposed to be equal to $\frac\pi{\sqrt2}$. Despite the simplicity of the result, it does not seem easy. What is troublesome is the square-root part in the integrand, which renders known techniques, such as half-angle substitution, ineffective.

I also threw it into WA, which churns out a complicated antiderivative expression in elliptical functions not all that helpful.

Body:
$$ I = \int_0^{\frac\pi2} \frac{(1+\sec^2t)\sqrt{\sec t}}{ (1+\sec t)^2-2}dt $$

$1.$ Put $\sec t = x^2$

$$I = 2 \int_1^\infty \frac{(x^4 + 1)}{\sqrt{x^4-1}(x^4 + 2x^2 - 1)}dx$$

$2.$ Put $x^2 - \frac{1}{x^2} = y^2$

$$I = 2 \int_0^\infty \frac{dy}{y^2+2} = \frac{\pi}{\sqrt{2}}$$


问:
A consequence of Rank theorem
I am reading a free book on Hodge theory. I am having a slight issue understanding this line: how is it true that $z_1 = \ldots = z_k = 0$. Aren't those variables the variables that are not zero by rank theorem?

答: But $X = F^{-1}(q)$ and $\psi(q) = 0$. This means that $\psi\circ F\circ\phi^{-1}(z_1,\dots,z_n) = (z_1,\dots,z_k,0,\dots,0) = (0,\dots,0,0,\dots,0)$ precisely when $z_1=\dots=z_k=0$, and $U\cap X$ is parametrized by $(z_{k+1},\dots,z_n)$.

Title: A consequence of Rank theorem