问:
Find the volume inside the paraboloid $x^2+4z^2+8y=16$ and on the positive side of xz-plane
No answer was given for this question, so wanted to confirm my answer. Steps followed for solving:

V=$4\int_0^2\int_0^{\sqrt{16-4z^2}}(2-\frac{x^2}{8}-\frac{z^2}{2})dxdz$

After solving this I got V = $8\pi$. Is this answer correct?

答:
Integral - $$v = 4\int_0^2 \int_0^\sqrt{16-4z^2} 1/8(16-x^2 +4z^2) \,dx\,dz$$

Answer is $$ {8\pi} $$

below is the Exact solution of the given question -


问:
Nice limit: $\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$
Find

$$L=\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

Not much experienced with limits of this kind, I have tried to reduce this limit to a form without exponent:

$$\ln L=\lim_{x \to 0}\ln\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

$$\ln L=\lim_{x \to 0}\frac{\ln\arcsin x - \ln x}{x^2}$$

...and then I tried to apply L'Hospital to numerator and denominator. However, the solution becomes a complete mess and you can repeat derivation as many times as you want without ever reaching a conclusion.

I also tried to introduce substitution $x=\sin y$ which leads to:

$$L=\lim_{y \to 0}\left(\frac{y}{\sin y}\right)^{\frac1{\sin^2 y}}$$

This makes manipulation slightly easier but the same approach (L'Hospital after log) is still unable to generate a solution.

答:
As you did take $x=\sin(y)$, $$\ln L=\lim_{y\to 0}\frac{\ln y-\ln(\sin(y))}{\sin^2(y)}\\=\lim_{y\to 0} \frac{\frac 1 y-\cot(y)}{2\sin(y)\cos(y)}\\=\frac 1 2 \lim_{y\to 0} \frac{\frac {-1} {y^2}+ \csc^2(y)}{\cos(y)}\\=\frac 1 2 \lim_{y\to 0} \frac 1 {\sin^2(y)} - \frac {1} {y^2}\\=\frac 1 2 \lim_{y\to 0} \frac {1} {y^2} \left(\frac {y^2} {\sin^2(y)} - 1 \right)\\=\frac 1 2 \lim_{x\to 0} \frac {1} {\sin^2(x)} \left(\frac {\sin^2(x)} {x^2} - 1 \right)=\frac 1 2 \lim_{x\to 0} \frac {1} {\sin^2(x)}\left(1-\frac {2x^2}{3!}+\cdots-1\right)=\frac 1 6.$$

Note:- I used L'Hospital's rule without mentioning and also note that I used it even when the ratio is not in $\frac 0 0 , \frac {\pm\infty}{\pm\infty}$ so it may not satisfy you. I will post a more correct solution later, so you can downvote if you like. Sorry.

ContentLicense: CC BY-SA 4.0


问:
Binomial Identity and Counting
A string in $\{0, 1\}*$ has even parity if the symbol $1$ occurs in the word an even number of times; otherwise, it has odd parity.

(a) How many words of length $n$ have even parity?

(b) How many words of length $n$ have odd parity?

It seems to me that correct approach will be to use summations of combinations, and correct answer will yield $2^{n-1}$ in both cases. Is there any way to prove separately that each of them is equal to $2^{n-1}$? Would be happy to know your ideas!

答:
The number of words of length $n$ with exactly $k$ ones is
$$\binom{n}{k}$$
There are $2^n$ binary strings of length $n$ since there are two possible choices for each of the $n$ entries.  Moreover, by the Binomial Theorem,
$$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^{k} = \sum_{k = 0}^{n} \binom{n}{k}$$
and
$$0 = 0^n = (1 - 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}(-1)^k = \sum_{k = 0}^{n} (-1)^k\binom{n}{k}$$
Notice that adding the positive terms in the second summation counts all the binary strings of length $n$ with an even number of ones, while adding the negative terms gives the additive inverse of the number of strings of length $n$ with an odd number of ones.  Since this sum is equal to zero, the number of binary strings of length $n$ with an even number of ones is equal to the number of binary strings of length $n$ with an odd number of ones.  Since every binary string of length $n$ must have an even number of ones or an odd number of ones, exactly half the strings of length $n$ have an even number of ones.  Since there are $2^n$ such strings, the number of binary strings of length $n$ with an even number of ones is
$$\frac{1}{2} \cdot 2^n = 2^{n - 1}$$
as is the number of binary strings of length $n$ with an odd number of ones.

This can also be seen algebraically.  If we add the two equations above, we find that
$$2^n = 2\sum_{k = 0}^{n} \binom{n}{2k}$$
since the terms with an odd number of ones cancel.  Hence, the number of binary strings of length $n$ with an even number of ones is
$$2^{n - 1} = \sum_{k = 0}^{n} \binom{n}{2k}$$
where
$$\binom{n}{2k} = 0$$
if $2k > n$.  The number of binary strings of length $n$ with an odd number of ones is found by subtracting $2^{n - 1}$ from $2^n$, which yields
$$2^n - 2^{n - 1} = 2^{n - 1}(2 - 1) = 2^{n - 1}$$

Title: Binomial Identity and Counting


答:
Notice that $\frac{\arcsin x}{x} \to 1$ as $x \to 0$, then

$$\ln L = \lim_{x \to 0} \frac{\ln\left(1+\frac{\arcsin x}{x}-1\right)}{x^2} = \lim_{x \to 0} \frac{\frac{\arcsin x}{x}-1}{x^2} = \lim_{x \to 0} \frac{\arcsin x-x}{x^3}.$$

Taking $x = \sin t$, the limit becomes

$$\ln L = \lim_{t \to 0} \frac{t - \sin t}{\sin^3 t} = \lim_{t \to 0} \frac{t - \sin t}{t^3} = \lim_{t \to 0} \frac{1 - \cos t}{3t^2} = \frac{1}{6},$$

and hence $L = e^{1/6}$.

Note: Such method is useful to most of $1^{\infty}$-form limits.

Title: Nice limit: $\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$

Body:
Find

$$L=\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

Not much experienced with limits of this kind, I have tried to reduce this limit to a form without exponent:

$$\ln L=\lim_{x \to 0}\ln\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

$$\ln L=\lim_{x \to 0}\frac{\ln\arcsin x - \ln x}{x^2}$$

...and then I tried to apply L'Hospital to numerator and denominator. However, the solution becomes a complete mess and you can repeat derivation as many times as you want without ever reaching a conclusion.

I also tried to introduce substitution $x=\sin y$ which leads to:

$$L=\lim_{y \to 0}\left(\frac{y}{\sin y}\right)^{\frac1{\sin^2 y}}$$

This makes manipulation slightly easier but the same approach (L'Hospital after log) is still unable to generate a solution.


问:
why there's not a fourier transform to the function $x(t) = -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$
I know that the Laplace transform of a signal is

$X(s)= \frac{1}{s^2+6s+8}$

$ ROC =  $ {s|-4 < Re{s} < -2}

therefore $x(t) =  -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

$u(t)$ is the heaviside function.

I need to find its Fourier transform (or to prove why there is not such transform).

by wolfram alpha I see that the Fourier transform doesn't exist for this signal.
Someone here can explain why ?

Thank you :)

答:
First of all, notice that:

$$\mathcal{L}^{-1}\left[\frac{\alpha}{(s+\beta)^2 + \alpha^2}\right] = e^{-\beta t} \sin(\alpha t)u(t),$$

where $u(t)$ is the Heaviside step function.

Given $X(s)= \frac{1}{s^2+2s+5}$, observe that:

$$s^2 + 2s +5 =s^2 + 2s + 1 + 4 = (s+1)^2 + 2^2.$$

Hence: $$X(s) = \frac{1}{2}\frac{2}{(s+1)^2 + 2^2} \Rightarrow \mathcal{L}^{-1}\left[X(s)\right] = \frac{1}{2}e^{-t}\sin(2t)u(t).$$

Secondly, letting $g(t) = \frac{1}{2}e^{-t}u(t)$, we observe that:

$$\begin{align*}
\mathcal{F}\left[g(t)\right]= &\int_{\mathbb{R}}\frac{1}{2}e^{-t}u(t) e^{-i2\pi ft} dt \\
=& \int_{0}^{+\infty}\frac{1}{2}e^{-t} e^{-i2\pi ft} dt \\
=& \frac{1}{2}\int_{0}^{+\infty} e^{-(i2\pi f+1)t} dt \\
=& \frac{1}{2}\left[\frac{e^{-(i2\pi f+1)t}}{-(i2\pi f+1)}\right]_{t=0}^{t=+\infty} \\
=& \frac{1}{-2(i2\pi f+1)}[0-1]\\
=& \frac{1}{i4\pi f+2} = G(f).
\end{align*}$$

Finally:

$$\begin{align*}
\mathcal{F}\left[g(t) \sin(2t)\right]=& \frac{G\left(f - \frac{2}{2\pi}\right) - G\left(f + \frac{2}{2\pi}\right)}{2 i} = \ldots\end{align*}$$


问:
Evaluating real integral using complex analysis.
I'm trying to compute the following integral:
$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx$$

I'll not write down everything I've done, but choosing the branch cut on the positive real axes we have that:

$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{\pm \sqrt{i},\pm\sqrt{-i}\}$$

So we have to compute four residues.
My thought was changing the branch cut by putting it on the negative imaginary axes. We can do it by choosing $arg(z) \in (-\frac{\pi}{2},\frac{3\pi}{2}]$. So we have that:

$$(1+i)\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=2\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{e^{i\frac{\pi}{4}},e^{i\frac{3\pi}{4}}\}$$

By doing this, we now need to compute only two residues. But I'm really finding difficulties in computing those residues: in fact I can't obtain the result I'm expecting. Can you please show me the computation and tell me if my argument was clear and correct?

Thanks in advance.

答:
The calcus of the residues are relatevely simple when you have simple poles.

Infact, if $z_0$ is a simple pole then $f(z) = a_{-1}(z-z_{0})^{-1}+ \sum\limits_{n \geq 0}a_n(z-z_0)^n$

So $(z-z_{0})f(z) = (z-z_{0})^{-1}+ \sum\limits_{n \geq 0}a_n(z-z_0)^n$ which implies

$$\text{Res}(f,z_{0}) = a_{-1} = \lim\limits_{z \to z_0}(z-z_{0})f(z)$$

This result to be useful when we condiser $f$ of the form $\frac{f}{q}$ with $p,q$ holomorphic function, $p(z_0) \ne 0$ and $z_0$ a simple pole of $q$ since

$$\text{Res}(f,z_{0})= a_{-1} = \lim\limits_{z \to z_0}(z-z_{0})\frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$$

In general :

For higher order poles a strategy could be : If $f$ has a pole of order $k$ in $z_0$, $g(z) = (z-z_0)^k f(k)$ extends to an holomorphic function in $z_0$ (I'm gonna call it improperly by $g$ as well)

With this setting $$f(z) = a_{-k}(z-z_0)^k + \cdots + a_{-1}(z-z_0)^{-1} + \sum\limits_{n \geq 0}a_n(z-z_0)^n$$

$$g(z) = a_{-k} + \cdots + a_{-1}(z-z_0)^{k-1} +  \sum\limits_{n \geq 0}a_n(z-z_0)^{n+k}$$

So $a_{-1}$ is the coefficient of $(z-z_0)^{k-1}$ in the expansion of $g$ which is holomorphic. Knowing that $a_{n} = \frac{f^{(n)}(z_0)}{n!}$ we have $$\text{Res}(f,z_{0}) = a_{-1} = \frac{g^{(k-1)}(z_0)}{(k-1)!}$$

Hope this helps with your calculations.


问:
Why higher mathematics books don't contain solutions and what to do about them?
Due to some unavoidable reason I cannot go to university and I don't have access to good teachers where I live. I have completed my high school with good grades and determined to self-study undergraduate mathematics including calculus, combinatorics, number theory, graph theory, probability, etc.

The problem I face while studying from books is whenever I cannot solve a particular question, I have no access to step by step solutions.

Why do most mathematics books don't contain solutions? How can I deal with this situation?

When you were in your undergrad years, How did you cope with lack of solutions in textbooks? Did your professors help you?

And as most books have huge number of questions asking them all here on StackExchange does not seem feasible. How to deal with this situation??

答: I think I can speak to this. First, for the traditional calculus sequence in the U.S. most step-by-step solutions can be found either on chegg study (monthly subsciription cost) or slater. Additionally, for a first course in differential equations and linear algebra, the same can be found. Most mathematics textbooks contain solutions to odd-numbered problems but not step-by-step solutions, that's when these other resources can be useful. Considering you don't have a professor to show you the way, I would definitely recommend these resources, as they greatly helped me. Sometimes, they even offer a different way of doing a problem that is easier to follow. However, I'd definitely recommend trying your hardest to figure out how to solve the problem before looking at step-by-step solutions, as it will pay off in the long run. Good luck!</p>


Title: why there's not a fourier transform to the function $x(t) = -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

Body:
I know that the Laplace transform of a signal is

$X(s)= \frac{1}{s^2+6s+8}$

$ ROC =  $ {s|-4 < Re{s} < -2}

therefore $x(t) =  -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

$u(t)$ is the heaviside function.

I need to find its Fourier transform (or to prove why there is not such transform).

by wolfram alpha I see that the Fourier transform doesn't exist for this signal.
Someone here can explain why ?

Thank you :)

Body:
Given $X(s)= \frac{1}{s^2+6s+8} = \frac{1}{2}\left[\frac{1}{s+2} - \frac{1}{s+4}\right]$, observe that:
$$\mathcal{L}^{-1}[X(s)] = \frac{1}{2}\left(e^{-2t} - e^{-4t}\right)u(t) = x(t).$$

Since:

$$\mathcal{F}[e^{-\alpha t}u(t)] = \frac{1}{\alpha + 2\pi i f},$$

then:

$$\mathcal{F}[x(t)] = \frac{1}{2}\left[\frac{1}{2 +2 \pi i f} - \frac{1}{4 +2 \pi i f}\right].$$


问:
Trace vanishes for powers of a given matrix implies that the matrix is nilpotent — Algebraically closed condition needed?
I’m currently considering the question in this popular post: Traces of all positive powers of a matrix are zero implies it is nilpotent, namely:

Let $F$ be a field with characteristic zero and $A$ an $n \times n$ matrix with entries in $F$. If $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.

Almost all comments and answers related to this question considered eigenvalues and applied conclusions that proved by theories related to Jordan canonical forms (as far as the proofs that I know).

My question is: Why such methods can be used when we haven’t required that the base field $F$ to be algebraically closed?

Since on non-algebraically closed fields, the eigenvalues may not exists, or the number of them may be less then the size of the matrix $A$, even if we count multiplicity. Moreover, it seems that the theories on JNF cannot be applied in non-algebraically closed fields.

Thank you for your answers and comments! :)

答:
There is a finite extension $K$ of $F$ in which the characteristic polynomial of $A$ splits. Therefore, $A$ is (in $K$ similar to a triangular matrix $T$). Thus for each $n \geq 1$, $T^n$ has trace zero.

Now, let $\lambda_1,\ldots,\lambda_r$ be the distinct eigenvalues of $T$ with multiplicities (on the diagonal) $a_1,\ldots,a_r$.

We know that for any polynomial $P$ with $P(0)=0$, $\sum_{i=1}^r{a_iP(\lambda_i)}=0$ (it’s the trace of $P(T)$).

Assume that $\lambda_i \neq 0$. Then there is a polynomial $P$ such that $P(\lambda_j)=0$ if $j \neq i$, $P(\lambda_i) \neq 0$, $P(0)=0$. So $0=\sum_{j=1}^r{a_jP(\lambda_j)}=a_iP(\lambda_i)$ so $a_i=0$, a contradiction since we are in characteristic zero. So every $\lambda_i$ is zero.

Therefore, $T^n=0$. So $A^n=0$.


问:
Is there direct notation for this sum
for this boy here I don't know how to deal with the dependence to the outer sum or the floor $$f(x)=2+\sum_{i=0}^{x-1}\left(4+1+\sum_{j=1}^{\lfloor\frac{x-1}{2}\rfloor}(2+1)+1\right)+3$$

答:
Note that $\sum_{k=a}^b c=(b-a+1)c$ (perhaps only under the premise that  $b\ge a-1$). Thus
$$ \sum_{j=1}^{\lfloor \frac{x-1}2\rfloor}(2+1)=3\cdot\left\lfloor \frac{x-1}2\right\rfloor$$
and
$$\begin{align}f(x)&=5+x\cdot\left(6+3\cdot\left\lfloor \frac{x-1}2\right\rfloor\right) \\&=5+6x+3x\left\lfloor \frac{x-1}2\right\rfloor\\
&=\begin{cases}\frac32x^2+\frac92x+5&\text{if $x$ is odd}\\
\frac32x^2+3x+5&\text{if $x$ is even.}\end{cases}\end{align}$$

Title: Is there direct notation for this sum


问:
Finding two numbers when having their sum and product
I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad  ab=238,\quad  238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance

答:
We use the identity
$$(a-b)^2=(a+b)^2-4ab.$$
Thus in our case $(a-b)^2=(41)^2-4(238)=729$.

It follows that $a-b=\pm 27$. Using $a+b=41$, we solve the system of two linear equations for $a$ and $b$. We get $a=34$, $b=7$ or $a=7$, $b=34$.

ContentLicense: CC BY-SA 3.0


问:
Conditional probability (cards question).
Given a piece of cards with $52$ total cards, $13$  of each one from the $4$ options ($\heartsuit, \spadesuit, \diamondsuit, \clubsuit$) . You and your friend getting randomly $7$ cards each,

what is the probability that your friend have one heart card given that you have three heart cards?

*note that I mean to the amount of cards from type hearts  not the number 1$\heartsuit$.

My attempt

Let $A$ and $B$ be two events :

$A$- my friend have $1$ heart cards

$B$- I have $3$ heart cards

we need to find $$P(A|B)$$
I want to use the formula $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

now I am stuck how to calculate $P(B)$ and $P(A\cap B)$

In the book there is 4 options $$1. \ \frac{12!}{3!\cdot9!}\cdot\frac{32!}{5!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$
$$2.\ \frac{12!}{3!\cdot9!}\cdot\frac{33!}{4!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$ $$3.\ \frac{10!}{1!\cdot9!}\cdot\frac{34!}{7!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$ $$4.\ \frac{10!}{1!\cdot9!}\cdot\frac{35!}{6!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$

I did not understand this pattern with the factorial, how did they build it?

答:
Actually, you don't want to treat this as a conditional probability question.  Just count combinations.  You have $45$ cards available for your friend to choose of which $10$ are hearts.  So of the possible combinations, what fraction have exactly $1$ heart?

There are $\binom{45}{7}$ possible hands your friend could hold.  For a "good" hand, there are $10$ different hearts he could hold and $\binom{35}{6}$ ways to choose the remaining cards.  So the answer is:

$$\frac{10 \binom{35}{6}}{\binom{45}{7}}.$$

That's choice $4$.


问:
Does $g$ always normalize $H \cap H^g$?
Let $G$ be any group, $H$ a subgroup, $g$ an element of $G$ and $H^g:=g^{-1}Hg$ the conjugate of $H$ by $g$. Is it true that $$g\in N_G(H\cap H^g)?$$

答:
No.

As a counterexample take $G$ to be the split extension of an elementary abelian group of order $16$ by a cyclic group of order $4$ with this action
$$
G=\left\langle x_1,x_2,x_3,x_4, g\mid x_i^2=1, x_i x_j=x_j x_i,   g^4=1, x_i^g=x_{i+1 \mod 4}\right\rangle
$$
and let $H=\langle x_1,x_2\rangle$.

Then $H\cap H^g=\langle x_2\rangle$ is not normalised by $g$.

Title: Does $g$ always normalize $H \cap H^g$?


问:
Continuous open metrizable image of a Polish space is Polish
Let $X$ be a Polish space and $Y$ a metrizable space. Assume that there exists a continuous mapping $f \colon X \to Y$ such that $f(X)=Y$ and $f(G)$ is open in $Y$ for every open set $G \subseteq X$.

Under these conditions I'm trying to prove that $Y$ is actually a Polish space. Since $X$ is Polish (and thus separable), it is clear that $Y$ is separable (being a continuous image of a separable space). Unfortunately, I am not able to prove that $Y$ is completely metrizable. I tried the following:

Since $Y$ is assumed to be metrizable, there is a metric $d$ on $Y$ which is compatible with the topology of $Y$. Let $(\hat{Y}, \hat{d})$ be the completion of $(Y,d)$. Then $(\hat{Y}, \hat{d})$ is a separable complete metric space. Therefore, it suffices to show that $Y$ is $G_{\delta}$ in $(\hat{Y}, \hat{d})$. But I don't know how to show this and how to use the openness of $f$.

I will appreciate any help.

答:
I actually needed this fact earlier today to show an unrelated result (namely that the quotient of a Polish group by a closed subgroup is Polish) and I remembered this question since I spent some time thinking about it when it was posted. The answer by Henno Brandsma is quite nice, but I had a feeling there had to be an elementary approach, so I did some digging and I found a proof in Hjorth's book Classification and Orbit Equivalence Relations which I reproduce below with minimal changes. This is theorem 7.5 in the book, where it is attributed to Hausdorff.

Theorem: Let $X$ be Polish and $Y$ metrizable. If $\pi\colon X\to Y$ is open, continuous and onto, then $Y$ is Polish.

Proof: Let $\widehat{Y}$ be the completion of $Y$, which is Polish since $Y$ is separable, so we only need to show that $Y$ is $G_\delta$ in $\widehat{Y}$.

We build a Lusin scheme on $X$, that is a set of open subset of $X$ indexed over $\Bbb N^{<\Bbb N}$ with the following properties:

$N_\varnothing=X$.

$\mathrm{diam}(N_s)<2^{-|s|-1}$.

$N_t=\bigcup_{|s|=|t|+1}\{N_s\mid t\subset s\}$.

$\overline{N_s}\subset N_t$ for $t\subset s$.

Now for every $s\in\Bbb N^{<\Bbb N}$ pick $W_s$ open in $\widehat{Y}$ such that $W_s\cap Y=\pi(N_s)$ (here we are using that $\pi$ is an open map). Using that open sets are $F_\sigma$, since we are in a metric space, we can now build open $V_s\subseteq\widehat{Y}$ indexed over $\Bbb N^{<\Bbb N}$ with the following properties:

$\bigcup\{V_s\mid s\in\Bbb N^1\}=\bigcup\{W_s\mid s\in\Bbb N^1\}$.

$V_s\subseteq W_s$ for all $s\in\Bbb N^{<\Bbb N}$.

For all $k\in\Bbb N$, $s\in\Bbb N^k$, $$\bigcup\{V_t\mid t\in\Bbb N^{k+1},t\supset s\}=V_s\cap\bigcup\{W_t\mid t\in\Bbb N^{k+1},t\supset s\}.$$

For all $y\in\widehat{Y}$, $k\in\Bbb N$ and $s\in\Bbb N^k$, there are only finitelymany $t\supset s$ with $t\in\Bbb N^{k+1}$ with $y\in V_t$, and there are only finitely many $t\in\Bbb N^1$ with $y\in V_t$.

$4$ is the only difficult condition, and it is where we use that open sets are $F_\sigma$. Namely write each $W_t$ as $\bigcup_n C_{n,t}$ for closed sets $C_{n,t}$ with $C_{n,t}\subseteq C_{n+1,t}$ for all $n\in\Bbb N$. Then we can demand that for $t_0,t_1\in\Bbb N^{k+1}$,$s\in\Bbb N^k$ with $s\subset t_0,t_1$, if $$t_0(k)<t_1(k),$$ then $$W_{t_1}\cap C_{n,t_0}=\varnothing$$ for $n<t_1(k)$. As a consequence for every $y\in\widehat{Y}$ and $k\in\Bbb N$ there are only finitely many $s\in\Bbb N^k$ with $y\in V_s$.
After this is all done it's clear that $$Y\subseteq\bigcap_{k\in\Bbb N}\bigcup_{s\in\Bbb N^k}V_s,$$ so showing the revere inclusion is enough to conclude that $Y$ is $G_\delta$ in $\widehat{Y}$. To show the reverse inclusion fix $y$ in the RHS and consider $$T_y=\{s\in\Bbb N^{<\Bbb N}\mid \text{ there exist infinitely many $t\supset s$ with $y\in V_t$}\},$$ which is a tree in the sense that $s\subset t\in T_y\implies s\in T_y$. By 3 if $s\in T_y$, then $y\in V_s$ and by 4 for each $s\in T_y$ there is a proper extension $s'\supset s$ also in $T_y$, thus we have infinite branches in the tree and we can find $f\in\Bbb N^{\Bbb N}$ such that

$f\upharpoonright k\in T_y$ for all $k\in\Bbb N$.

$y\in V_{f\upharpoonright k}$, for all $k\in\Bbb N$.

And since $X$ is complete we can also find an $x\in\bigcap_{k\in\Bbb N}N_{f\upharpoonright k}$. Since $\{N_{f\upharpoonright k}\mid k\in\Bbb N\}$ is a neighbourhood basis for $x$ and $\pi$ is open and continuous, it follows that $y$ is contained in every open set containing $\pi(x)$, thus $y=\pi(x)$ and we obtain the reverse inclusion, which concludes the proof.


问:
If $A^3$ diagonalizable, then $A$ diagonalizable
I can prove this statement wrong by considering $A\neq\mathbf O$ and $A$ not diagonalizable while $A^3=\mathbf O$.

But what's wrong with my intuitive proof:

$A^3=PDP^{-1}$, and we just consider $A=PD^{1/3}P^{-1}$, it seems that everything is satisfied. And furthermore, we can also show uniqueness by considering in certain eigenspace $V_\lambda$, $A^3=\lambda I$ so the eigenvalue of $A$ must be $\sqrt[3]\lambda$.

Could anyone help me to figure out what's wrong here? And why does the converse hold?

答:
Let us consider the problem in the field $\mathbb{C}$ of complex numbers.

If $0$ is an eigenvalue of $A^3$, the result is false in general as the following example shows
\begin{equation}
A = \begin{pmatrix}0&1&0\cr 0&0&1\cr 0&0&0\cr\end{pmatrix}
\qquad A^3=0
\end{equation}

If $0$ is not an eigenvalue of $A^3$, the result is true by the following argument: assuming $A^3$ is diagonalizable, let
$P(X) = \prod_i (X-\mu_i)$ be the minimal polynomial of $A^3$, with $\mu_i\not = \mu_j$ for $i\not = j$, and let $Q(x)=\prod (X-\lambda_i)^{m_i}$ be the minimal polynomial of $A$. Obviously,
\begin{equation}
R(X) = P(X^3) = \prod_i (X^3-\mu_i)
\end{equation}
is a multiple of $Q(X)$ in $\mathbb{C}[X]$ because $R(A)=0$. It follows that if $\lambda$ is a root of $Q$, then $R(\lambda) = 0$ and one has $\lambda^3 = \mu_k$ for some index $k$. In particular one has $\lambda\not = 0$. If $\lambda$ is a multiple root of $Q$, then it must also be a multiple root of $R$. But one has
\begin{equation}
R'(\lambda) = \sum_i 3 \lambda^2 \prod_{j\not = i} (\lambda^3-\mu_j) =
3 \lambda^2 \prod_{j\not = k} (\mu_k-\mu_j) \not = 0
\end{equation}
Thus $\lambda$ cannot be a multiple root of $R$, hence $Q$ has only simple roots and $A$ is diagonalizable.


问:
Slope of $xe^{-(\frac{1}{x}+ \frac{1}{|x|})}$ at x =0
Taking the right-hand differential to calculate the slope of $xe^{-(\frac{1}{x}+ \frac{1}{|x|})}$ to the right of x=0, I got $e^{-\frac{2}{x}}(1 + \frac{2}{x})$, after simplifying it to $xe^{-\frac{2}{x}}$. Every other solution I found was by using the first principle; $\frac{he^{-\frac{2}{h}}-0}{h}$, h~0. This simplifies to $e^{-\frac{2}{h}}$, which is 0 at x=0.

Why did my normal differential fail here? And either way, how do I solve the differential I got, because it's $0\cdot\infty$?

答:
After letting $f(x)=xe^{-(\frac{1}{x}+ \frac{1}{|x|})}$, and $t=\frac{2}{x}$, it follows that
$$\lim_{x\to 0^+}f'(x)=\lim_{x\to 0^+}(e^{-\frac{2}{x}}(1 + \frac{2}{x}))=\lim_{t\to +\infty}e^{-t}(1 + t)=\lim_{t\to +\infty}\frac{1 + t}{e^t}=0$$
which is equal to your evaluation of $f_+'(0)$ (after extending $f$ at $x=0$ with the value $0$).

It is interesting to note that
$$f'_-(0)=\lim_{h\to 0^-}\frac{h}{h}=1\not=0$$
and therefore such function is not differentiable at $x=0$, that is $f$ has not a "proper" slope at $x=0$.


问:
If $z\in\mathbb C$ with $|z|\leqslant\frac{4}{5}$, then $\sum_{n\in S}z^n\neq-\frac{20}{9}$
Let $z$ be a complex number with $|z|\le\tfrac{4}{5}$. If $S\subset\mathbb N^+$ is a finite set, then I'd like to show  that
$$\sum_{n\in S}z^n\neq-\frac{20}{9}\,.$$
This problem is from an exam in Bejing as of today, and I find this is a very nice problem.

I have no idea how to access it, maybe use Newton's identity to  solve it.

Could someone help me to solve it?

答:
Not a final answer, but some lead I hope:

I can see that for $r=\frac 45$ then $-\frac{20}{9}=\sum\limits_{k=0}^{\infty} (-r)^{2k+1}$ so as to accumulate the maximum negative values possible.

And this is more than that if either $r<\frac 45$ or sum is incomplete.

I guess we have to prove that negative contributions cannot exceed this for a $z$ with random argument.


问:
Necessity of probability in $P \ll \mu \iff \forall \varepsilon > 0 \exists \delta > 0 : \forall A, \mu(A) < \delta , P(A) \leq \epsilon$
Given the following proposition :

Let $P$ be a probability and $\mu$ a $\sigma$ finite measure. Then $P \ll \mu \iff \forall \hspace{0.1cm} \varepsilon > 0 \hspace{0.1cm} \exists\hspace{0.1cm}  \delta > 0 : \forall A$ with $\mu(A) < \delta$ we have $P(A) \leq \epsilon$

I do understand the proof but I'd like to see the necessity of $P$ being a probability, so I'd like to prove that the result doesn't hold if we take a $\sigma-$finite measure instead of $P$, i.e $P \not\ll \mu$.

Here I'm taking $P = \lambda$ the lebesgue measure on $\mathbb{R}$ and $\mu = f.\lambda$ where $f(x) = \frac{1}{1+x^2}$.

In this way we have $\lambda(A) = \int_{A} 1+x^2 \frac{1}{1+x^2}d\lambda(x)$, i.e $\lambda \ll \mu$.

The idea is that $\mu(]x_1,x_2[) = \int_{x_1}^{x_2} \frac{1}{1+x^2}dx = \text{arctg}(x_2)-\text{arctg}(x_1)$ and $\lim\limits_{x \to +\infty}\text{arctg}(x) = \frac{\pi}{2}$.

This is where I got stuck, trying to prove that $\exists \hspace{0.1cm} \varepsilon > 0 \hspace{0.1cm} \forall\hspace{0.1cm}  \delta > 0$ we have $\mu(]x_{1},x_{2}[) < \delta$ but $\lambda(]x_1,x_2[)
 \geq \varepsilon$.

Any help or hint would be appreciated.

答:
What about taking the sets $A_n=[n,\infty)$ instead. Since $\frac{1}{1+x^2}$ is integrable then, $\mu(A_n)\to 0$ as $n\to \infty$. But $\lambda(A_n)=\infty$. Hence, for example if you pick $\epsilon_0=1$. Then, for every $\delta>0$ you can find some $n_\delta$ with $\mu(A_{n_{\delta}})<\delta$ but $\lambda(A_{n_\delta})\geq 1$.

You can also take the sets $[n,n+1]$. Then again $\mu([n,n+1])=\arctan(n+1)-\arctan(n)\to 0$. But $\lambda([n,n+1])=1$, so again for $\epsilon_0=1$ you can get what you want.


问:
Formula of the sum
$$\sum_{i=1}^{n} \frac{\sin(\frac{\pi i}{2n})}{n}$$
Any ideas on how to evaluate this? Basically, I need to find a limit from this sum, when $n \to \infty$ and the only approach I see is to find a formula from this sum.

答:
By the definition of the definite integral$$\lim_{n\to \infty}\frac 1n\sum_{i=1}^{n} \sin(\frac{\pi i}{2n}) = \int_0^1\sin\frac{\pi x}{2}dx = -\frac 2\pi\cos\frac{\pi x}{2}\bigg|_0^1=\frac{2}{\pi}$$

Also, see this page.

If you really want to find a formula for your sum, see this page. This should help.

Title: Formula of the sum


问:
The sum of digits in a 2-digit number

The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number.

My attempt:

Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at unit's place. According to question:

$$x+y=8$$

I could not figure out the other condition. Please help. Thanks in advance.

答:
Let the $2$-digit number be $\overline{ab} = 10a + b$.

If $9$ is added to the number then both the digits become equal.

This means that $a = b$, so the number plus $9$ is $10a + a = 11a$, which is always $0 \text{ mod } {11}$ (divisible by $11$).

Therefore $\overline{ab} + 9 \equiv 0 \text{ mod } {11} \Rightarrow \overline{ab} \equiv 2 \text{ mod } {11}$.

But $a + b \equiv 8 \text{ mod } {11}$. Thus $(10a+b)-(a+b) \equiv 9a \equiv -2a \equiv 2-8 \text{ mod } {11}$, so $a \equiv 3 \text{ mod } {11}$.

As $a+b = 8$, hence $\overline{ab} = \boxed{35}$.

Title: The sum of digits in a 2-digit number


问:
From submartingale to martingale
Let's consider submartingale, so variable that satisfies :

$$E[X_{n+1}|\mathbb{F}_n] \ge X_n$$

Also let's assume that $\forall_{i, j}:E[X_i]=E[X_j]$

I want to prove that $X_n$ is martingale.

My work so far

To prove martingale, we only need to prove that X is supermartingal i.e.
$$E[X_{n+1}|\mathbb{F}_n] \le X_n$$
From very first assumption we have that :

$$E[E[X_{n+1}|\mathbb{F}_n]] \ge E[X_n] \Leftrightarrow X_{n+1} \ge E[X_n]$$

Now from second assumption (equality of expectations) we have that $$X_n \ge E[X_n]$$

And now I stacked. Could you please give me a hand ?

答:
Denote $Z_n:=\mathbb E[X_{n+1}\mid \mathcal F_n]$. You have that $$Z_n-X_n\geq 0.$$ Since $(X_n)$ has constante expectation, $$\mathbb E[Z_n-X_n]=\mathbb E[X_{n+1}]-\mathbb E[X_n]=0,$$
and thus $Z_n=X_n$ a.s. as wished.

Title: From submartingale to martingale


问:
Prove that matrix is invertible by knowing that other matrix is invertible
Let A, B be matrices of the order nxn,
Prove that if AB^2-A is invertible, so BA-A is also invertible.

How can I prove that without using determinants?

Thanks!

答:
If $A(B^2 - I)=A(B-I)(B+I)$ is invertible, so all three factors are invertible.

You need to know that $AB$ is invertible iff $A$ and $B$ are both invertible. One implication follows from $(AB)^{-1} = B^{-1} A^{-1}$ and the other from $A\big(B(AB)^{-1}\big)=I$ etc.


问:
Help requested, conditional probability question, What is my mistake?
Can anyone help me with this question

If the real answer is "no", then give 50% yes and 50% no
If the real answer is "yes“, then give 60% yes and 40% no

Suppose that the noisy answers consist of 55% yes and 45% no
Estimate the percentage of "yes" in the original answer

This was my attempt

$p(\text{yes | the real answer is "no"}) = 0.5$

$p(\text{no | the real answer is "no"}) = 0.5$

$p(\text{yes| the real answer is "yes"}) = 0.6$

$p(\text{no | the real answer is "yes"}) = 0.4$

$p(\text{yes}) = 0.55$

$p(\text{no}) = 0.45$

p(the real answer is "yes | yes) = p(yes| the real answer is "yes") / [p(yes| the real answer is "no") +  p(yes| the real answer is "yes")]
$= 0.6 / [0.5 + 0.6] = 0.54$

p(the real answer is "yes ^ yes) = p(the real answer is "yes | yes) x p(yes)
$= 0.54 \times 0.55
                             = 0.3$

I feel what I am doing is wrong.

is $p(A|B) = p(B|A)/[p(B|A) + p(B|\text{not }A)]$ ?

答:
No, the formula at the end is not correct, and should be
$$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}.$$
However, this formula isn't immediately useful here, as if $A$ is "real answer yes" and $B$ is "yes" then both $P(A\mid B)$ and $P(A)$ are unknown. Instead, try using $$P(B)=P(A)P(B\mid A)+(1-P(A))P(B\mid \neg A).$$


问:
How can I count the number of colored combinations in a set of regions?
Let me first start out by saying as you might have guessed, this is a homework problem. Therefore I am not looking for an answer to the question. I am looking for help in how to analyze it (My textbook is horrid).

I have a grid of 9 squares surrounded by a circle.
I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`

I am asked to find the number of
  possible different colorings.

I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.

I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.

No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.

答:
I'm not sure about this, maybe I don't understand the problem, but here it goes. If you colour the region outside the grid A, the 8 external cells of the grid must be coloured alternately with B and C, and the one in the center, with A. Thus you have two possible colourings. Same thing if you colour the region outside B, or C. Thus the total number of colourings would be 6.

Update Indeed, there are more. If the external cells are coloured B and C, then the one in the centre may be A, and either B or C, so there are 12 different colourings now.

ContentLicense: CC BY-SA 2.5


问:
smallest number of vertices of degree $1$ in tree with $3$ vertices of degree $4$ and $2$ of degree $5$

Find, with proof, the smallest number of vertices of degree $1$ in a tree with $3$ vertices of degree $4$ and $2$ of degree $5$. Provide an example of such a tree.

I'm not sure how to find this. I know that every tree with at least $2$ vertices has at least $2$ leaves, trees are bipartite, trees have no cycles. Also, every tree is connected. But I'm not sure how to make use of these properties to get the desired proof. So far, the smallest value I've been able to come up with is $14,$ though I think that number can be made smaller. My basic idea is to make it so that as many vertices are as "tightly joined or connected" as possible, so as to maximize the number of vertices with degree greater than $1$.

答:
Hint: The main property of a tree you want to use here is its number of edges. A tree with $n$ vertices has $m=n-1$ edges. You can use the fact that the sum of the degree is $2m$ (twice the number of edges) to prove how many degree $1$ vertices you need.

Click the spoiler for a full answer:

 Let $T=(V,E)$ be a tree. Consider the sum $\sum_{u\in V} d(v) = 2|E| = 2n-2$. From this we get that $\sum_{u\in V}(d(u) - 2) = -2$. Note that vertices of degree $1$ contribute $-1$ to the left and vertices of degree $2$ contribute $0$. All other vertices add a positive amount to the left. If we have three vertices of degree $4$ and two of degree $5$, this gives us a sum $3\cdot(4-2) + 2\cdot(5-2) = 12$. Thus your figure of $14$ was correct ($-14 + 12 = -2$), you need at least $14$ degree $1$ vertices to make the total left hand side sum to $-2$.

Since your minimum number was correct, I assume you've found a tree that works as an example (any tree with exactly three degree $4$, two degree $5$ and 14 degree $1$ vertices will work).


答:
Let $X$ be the event the real answer is "yes" and $Y$ be the event the noisy answer is "yes".  The question tells you:

$\mathbb P(Y \mid X^c)=0.5$ and  $\mathbb P(Y^c \mid X^c)=0.5$

$\mathbb P(Y \mid X)=0.6$ and  $\mathbb P(Y^c \mid X)=0.4$

$\mathbb P(Y)=0.55$ and  $\mathbb P(Y^c )=0.45$

Since $$\mathbb P(Y) = \mathbb P(Y \mid X)\mathbb P(X)+ \mathbb P(Y \mid X^c)\mathbb P(X^c)$$ i.e. $$\mathbb P(Y) = \mathbb P(Y \mid X)\mathbb P(X)+ \mathbb P(Y \mid X^c)(1-\mathbb P(X))$$ you can say

$$\mathbb P(X)=\frac{\mathbb P(Y)- \mathbb P(Y \mid X^c)}{\mathbb P(Y \mid X)-\mathbb P(Y \mid X^c)}$$

which in this case is $\dfrac{0.55-0.5}{0.6-0.5}=0.5$

Title: Help requested, conditional probability question, What is my mistake?

Body:
Can anyone help me with this question

If the real answer is "no", then give 50% yes and 50% no
If the real answer is "yes“, then give 60% yes and 40% no

Suppose that the noisy answers consist of 55% yes and 45% no
Estimate the percentage of "yes" in the original answer

This was my attempt

$p(\text{yes | the real answer is "no"}) = 0.5$

$p(\text{no | the real answer is "no"}) = 0.5$

$p(\text{yes| the real answer is "yes"}) = 0.6$

$p(\text{no | the real answer is "yes"}) = 0.4$

$p(\text{yes}) = 0.55$

$p(\text{no}) = 0.45$

p(the real answer is "yes | yes) = p(yes| the real answer is "yes") / [p(yes| the real answer is "no") +  p(yes| the real answer is "yes")]
$= 0.6 / [0.5 + 0.6] = 0.54$

p(the real answer is "yes ^ yes) = p(the real answer is "yes | yes) x p(yes)
$= 0.54 \times 0.55
                             = 0.3$

I feel what I am doing is wrong.

is $p(A|B) = p(B|A)/[p(B|A) + p(B|\text{not }A)]$ ?


问:
A differentiable function whose derivative is $0$ almost everywhere, but not everywhere
In another thread, I remarked that if
$$\int_0^1 \vert f'(x)\vert\mathrm dx=0,$$
then $f'(x)=0$ for all $x\in(0,1)$ can only be concluded if $f'$ is continuous. This would certainly be correct if we were talking about a general integrable function $(0,1)\to\mathbb R$. Take the characteristic function of a singleton set as a counterexample where the integral vanishes, but the function doesn't. But such a function is not a valid derivative of any function, since it doesn't have the mean value property. And I couldn't find any functions where the integral of the absolute value vanishes and which have the mean value property. That's why I'm wondering:

Is there a differentiable function $f:(0,1)\to\mathbb R$ such that $f'$ is not identically $0$, but
$$\int_0^1\vert f'(x)\vert\mathrm dx=0,$$
or equivalently, $f'$ is $0$ almost everywhere, but not everywhere.

答: If $f$ is differentiable at every point of $(0,1)$ and $\int_0^{1} |f'| <\infty$ then $f$ is absolutely continuous. This theorem is proved in Rudin's RCA. So if we also assume that $f'=0$ a.e. then $f$ is  a constant and $f'(x)=0$ for every $x$.


问:
"Explicitization" of Implicit Euler
Consider the ODE $\dot y=f(t,y)$.

Explicit Euler says: $$y_{n+1}=y_n+hf(t_n,y_n)\tag{1}$$

Implicit Euler says: $$y_{n+1}=y_n+hf(t_{n+1},y_{n+1})\tag{2}$$

The above schemes are well-known. In a way similar to what is done in Heun's method, we could make Explicit the Implicit Euler scheme as follows:

Plug (1) in the right handside of (2) to get: $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_n,y_n))=0\tag{3}$$

We could also plug (2) (possibly multiple times) into the right handside of (2) to get $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1}))=0\tag{4}$$
or $$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_{n+1})))=0\tag{5}$$

We could finally make (5) explicit by plugging (1) into its right handside, as follows:
$$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_{n+1},y_n+hf(t_n,y_n))))=0\tag{6}$$
I have never heard of the schemes above. Maybe they bear no specific good numerical features. Are there known results on this?

答:
With 3 evaluations of the ODE function you can implement an explicit 3-stage order 3 method. Your combinations of Euler steps however only give order 1. This should be the main reason that you do not find them widely discussed.

This pattern may seem negative for implicit methods in general, but for higher order methods the explicit methods need more stages than the order number, and implicit methods can be constructed to need only close to half of the order as stages. With step size control there is also a tendency that implicit methods allow larger steps, so that over the same time interval the number of overall function evaluations can become competitive.

Title: "Explicitization" of Implicit Euler


问:
Linked Legendre diophantine equations are they solvable?
Given two equations

$$n_1\cdot{d_1^2} + n_2\cdot{d_2^2} = n_3\cdot{d_3^2}$$

and

$$n_1\cdot{d_1^2} + 2\cdot{n_2}\cdot{d_2^2} = n_4\cdot{d_4^2}$$

Where $n_1,n_2,n_3,n_4$ are known relatively prime square free integers and $d_1,d_2,d_3,d_4$ are unknown integers.

Is there a general method to solve them together so $n_1=n_1$ and $n_2=n_2$ and $d_1=d_1$ and $d_2=d_2$?

答:
To make things a bit easier, I'm replacing $n_1$ with $a$, $n_2$ with $b$, $n_3$ with $c$, $n_4$ with $d$, $d_1$ with $e$, $d_2$ with $f$, $d_3$ with $g$, and $d_4$ with $h$.

We have the equations:

$$ae^2+bf^2=cg^2$$

$$ae^2+2bf^2=dh^2$$

We can subtract the first equation from the second. The $ae^2$ cancels out, and $2(bf^2)-bf^2=bf^2$.

Our newly formed equation:

$$bf^2=cg^2-dh^2$$

Now we can subtract the above equation from the first equation.

We get:

$$ae^2=dh^2$$

Now in the second equation, we can replace $ae^2$ with $dh^2$.

$$2bf^2=dh^2-dh^2=0$$

Now we know $2bf^2=0$. Since $0/2=0, bf^2=0$. Since $b$ is square-free, and $0$ is a square number, $f$ must equal $0$. This means $b$ can be any real, square-free number.

Back to the original equations. We can get rid of $bf^2$, because it is $0$, and get this:

$$ae^2=cg^2$$

We know $ae^2=dh^2$, so $cg^2$ must equal $dh^2$.

It is impossible to go any farther than that, because we can get rid of $bf^2$ in all of the equations because it equals zero. Now we are dealing with the same variable. Say we replace $ae^2, cg^2$, and $dh^2$ with $x$. (This works because they are all equal.) Eventually, all our equations would be $x=x$ because our only integer that we can use gets multiplied by zero (the two in the second equation).

So the most we can deduce is:

$$\boxed{ae^2=cg^2=dh^2}$$

$$\boxed{b=\text{any real, square-free number}}$$

$$\boxed{f=0}$$


问:
Lie subalgebra, Lie subgroup and membership
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $H$ be a connected Lie subgroup with Lie algebra $\mathfrak{h}$.

We have that $X \in \mathfrak{h} $ iff $exp(tX) \in H \ \ \  \forall  t \in \mathbb{R} $ .

I have seen a proof of this but I don't understand why we need that $\forall t \in \mathbb{R}$.
An answer to this question  shows that it is equivalent to ask $\forall t \in \mathbb{R}$ or just in $\forall t \in \, ]-t_0,t_0[$ for some $t_0$. Therefore I'm not worried about "for all $t$" versus "for $t$ small enough", but about "for just one $t$" versus "for more than one $t$" (of course $t\neq 0$).

I'd like to see an example in which you have some vector $X \in \mathfrak{g}$ such that $exp(X) \in H$ but $X \not \in \mathfrak{h}$. Is this possible?

答: Take $G=\Bbb C^\times$, $H=\{1\}$ so $\mathfrak h=\{0\}$ and $X=2\pi\mathbf i\notin\mathfrak h$ but $\exp(X)\in H$.</p>


问:
Pointwise and uniform convergence - solution verification
Consider the sequence of functions defined by the following
$$ f_n(x) = x^n \sin \left(\frac{1} {x^n} \right),
  0\lt x \lt \frac {1} {\sqrt[n] \pi},  f_n(x)=0,$$
elsewhere  in  $\mathbb R$.
What can we say about its convergence?
I think the sequence converges pointwise but not uniformly.

I have made an attempt I now write as an answer not to make the question too long. I'm not completely sure about it. Please check it and feel free to suggest hints or other ways to solve this problem. Thank you.

答:
The "interesting" part of the function is the open interval where it is not defined as the 0 function. For every positive integer we have a different interval, although they all are subsets of $(0,1)$.
However, for every x and n we can say
$$ - x^n \le x^n sin \left(\frac{1} {x^n} \right) \le x^n $$ so the sequence converges pointwise to the 0 function everywhere in $\mathbb R$ by the squeeze theorem.
Now I want to prove it doesn't converge uniformly, though. For any fixed $n$, I consider the point $\frac {1} {\sqrt[n] \frac {5\pi}{2} } $ which is in the interval $(0, \frac {1} {\sqrt[n] \pi}) $ because $\pi \lt \frac {5\pi}{2}$ and so their n-th root.
$$ \sup _ {x \in  (0, \frac {1} {\sqrt[n] \pi})} | f_n (x)| \ge  f_n\left(\frac {1} {\sqrt[n] \frac {5\pi}{2} } \right ) =  \frac {2} {5\pi} $$ thus $$ lim _ {n \to \infty} \sup _ {x \in  (0, \frac {1} {\sqrt[n] \pi})} | f_n (x)| \neq 0$$


问:
Calculus: Isn't the velocity & tangent at a critical point = 0, and therefore a function is not increasing at that point?
I encountered a question from Khan Acad that asked, at what intervals of x does the function increase. My intuition is that all ranges except for 0 and 1, because tangents at those points are flat per green graph below.
Backed up by the red graph, showing h'(1)=0 & h'(0)=0, zero velocity.

I need help figuring out my knowledge gap:

Am I misinterpreting what "increasing" here?

Am I missing the point of what a critical point is..?

答:
A function $f$ is increasing if $f(b) ≥ f(a)$ with $b ≥ a$, for all $a,b$ in the domain, which is $x \in \mathbb R$ in your case. Note the $≥$ sign instead of $>$.

What you are thinking of is a strictly increasing function, otherwise called a monotonically increasing function. These functions can have (* see edit) $f'(x) = 0$, whereas an increasing function can $f'(x) = 0$ for a finite number of $x$ values.

Note that the derivative does not need to be defined at all points for a function to be an increasing function. For example, the function $g(x) = x^{1/3}$ increases for all $x \in \mathbb R$, despite the derivative not being defined at $x = 0$:

It is also possible for a function to be increasing in a restricted domain such as $(a,b)$, with other variations depending on whether the interval is open or closed.

Edit: Strictly increasing functions can have $f'(x) = 0$, as long as $f(x)$ is increasing in the neighbourhood of that point, say $(x - \epsilon, x + \epsilon)$ as mentioned by Adam Rubinson.


问:
Conditional probablity , how should I continue from here? (attempt added)
Given that a family have $n$  childern for $n\in \mathbb{N}$ is in the probability $$C\cdot\left({\frac{9}{17}}\right)^n$$ for $C>0$. the probability for getting male or female is equal.

What is the probablity  that there is $7$ childrens in  the family given that there is no females

My attempt

First of all I found $C$ in the following way:
$$P(\mathbb{N})=1=\large\Sigma_{n=0}^{\infty}C\cdot\left({\frac{9}{17}}\right)^n$$
$$C=\frac{8}{17}$$

Let $A$ and $B$ be an events :

$A$ - there is 7 childrens in the familiy

$B$ - there is no females in the familiy

I want to calculate $$P(A|B)$$
for this I want use the formula $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
now calculate $P(B)$ is easy it is $\frac{1}{2}$

Now how should I calculate $P(A\cap B)$? is it true to argue that $A$ and $B$ are independent and than use the formula $$P(A\cap B)= P(A) \cdot P(B)?$$
not sure how should I do this step

答:
$P(B)$ is not $\frac{1}{2}$, that would be the probability of one of the children being male or female, but from what you said I think $B$ is the event where none of the children are female (so all of them are male). You need to consider how many children are (and you cannot say 7 because $B$ has nothing to do with $A$ in $P(B)$).

If you call $C_i$ for $i=0,1,2,\ldots$ the event that there are $i$ children, then $P(B)=\sum\limits_{i=0}^\infty P(B\mid C_i)P(C_i)=\sum\limits_{i=0}^\infty\frac{1}{2}^i\frac{8}{17}\frac{9}{17}^i=\frac{8}{17}\sum\limits_{i=0}^\infty(\frac{1}{2}\frac{9}{17})^i=\ldots$

Note that I substituted $P(B\mid C_i)=\frac{1}{2}^i$, since that's the probability of having no females given that there are $i$ children. From that summation you get $P(B)$.

Now, for the rest, you just need one more step:

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B\mid A)P(A)}{P(B)}$

Note $A=C_7$.


问:
Are the two definitions of the complementary Young function equivalent?
The description of the problem:

For a Young function I would refer the reader to the book "Function spaces" by Pick Luboš, Kufner Alois, John Oldrich and Fucík Svatopluk, and published by de Gruyter. The following Definition 1 is about the complementary Young function which would be found in this book (see p.115).

Definition 1. Let $\Phi $ be a Young function generated by a function $\varphi$, that is,
$$
\Phi(t)=\int_{0}^{t}\varphi(s)ds, \ t \in \left [ 0, \infty\right ).
$$
We set
$$\psi(t)=\underset{\varphi(s)\leq t}{ \sup}s$$

and
$$
\Psi(t)=\int_{0}^{t}\psi(s)ds, \ t\in \left [0,\infty  \right ).
$$
The function $\Psi$ is called the complementary function to $\Phi$.

However, I have found another kind of definition of $\Psi$ in many occasions. See below:

Definition 2. Geven a Young function $\Phi$, the complementary Young function $\Psi$ is defined by
$$
\Psi(t)=\underset{s>0}{\sup}\left \{ st-\Phi(s) \right \}, \ t>0.
$$

My question is:

Are these two definitions equivalent? And if they are, how can we proof this result?

Thank you for your help.

答:
The is YES if $\varphi$ is invertible. In this case, we clearly have
$$\psi= \varphi^{-1}.$$

The second definition
$$\Psi(t)= \sup_{s>0} [st-\Phi(s)]= at-\varphi(a) \quad\text{for some }a>0$$

Note that $a$ exists by convexity and we have $\Psi(0)=0$. Since $\Phi'=\varphi$ even have

$$\frac{d}{ds}[st-\Phi(s)](a)=0\Longleftrightarrow t=\varphi(a)\Longleftrightarrow a=\varphi^{-1}(t)$$

which means that $$\Psi(t) = at-\varphi(a)= t\varphi^{-1}(t)-\Phi(\varphi^{-1}(t))$$

That is we have $$\Psi'(t) =  t(\varphi^{-1})'(t)+\varphi^{-1}(t)-(\varphi^{-1})'(t)\Phi'(\varphi^{-1}(t))= t(\varphi^{-1})'(t)+\varphi^{-1}(t)-(\varphi^{-1})'(t)\varphi(\varphi^{-1}(t))= \varphi^{-1}(t). $$

This shows that
$$\Psi(t)=\sup_{s>0} [st-\Phi(s)]= \Psi(0)+\int_0^t \Psi'(s) ds=\int_0^t \varphi^{-1}(s) ds=\int_0^t \psi(s) ds.$$


问:
$P(S^2 > 3579)$?
Suppose

$$\bar{X} = \frac{\sum_{i=1}^n}{n}$$

and

$$S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2$$

Let $X_1, ... X_{20}$ a sample of size 20 with $N(300, 2500)$

The probability that the sample mean is less than $281.61$ is equal to $5$%.

The probability that the sample variance is greater than $3579$ is equal to $90$%.

For me, I think only 2 might be true. Intuitively, am I wrong?

I have tried $P(\bar{X} < 281.61) = 1 - \phi(\frac{300-281.61}{\sqrt{2500}}) = 1 - 0.64058 = 0.35942$. So the first choice can't be true. I have no clue where to start with $S^2$.

答:
You can use the following statement (it is not difficult to prove it)

$$\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(n-1)}^2$$

But the correct statement is the first.

$$\mathbb{P}[\overline{X}_{20}<281.61]=\mathbb{P}\Big[Z<\frac{281.61-300}{50}\sqrt{20}\Big]=\Phi(-1.64485)=0.05=5\%$$

The second statement is

$$\mathbb{P}[S^2>3579]=\mathbb{P}\Bigg[\chi_{(19)}^2>\frac{3579\times 19}{2500}\Bigg]=0.1=10\%$$

The first result has been derived using Z-table, the second using $\chi^2$ table

Title: $P(S^2 > 3579)$?


问:
bifurcation value
I tried to understand how to locate the bifurcation value for the one-parameter family. From my understanding the bifurcation value is the maximum or minimum point of a parabola, so I set the differential equals to zero to find the equilibrium points. From two equilibrium point, I'll find the midpoint and try to find the output value that corresponding to that midpoint. For example: I got this differential equation

$dy/dt=y^2-ay+1$

I let $dy/dt=0$ then solve for $y$. I got $y= (a+\sqrt{a^2-4})/2$ and  $y= (a-\sqrt{a^2-4})/2$ . I find the midpoint of these two value and got $y=a/2$. Plug back to the differential equation, solve for $a$ and got $a=2$ and $a=-2$. What do I do from here if I want to use the values of $a$ to draw the phase line ?

答:
1.You're using a somewhat circuitous way to find bifurcation points. Most of the work is computing the equilibria of this differential equation,
$$
y = \frac{a \pm \sqrt{a^2- 4}}{2} ~~~~~ (*)
$$
Inspecting these forms, the parameter values $a = \pm 2$ correspond to the points at which equilibria are 'created' or 'destroyed', and so correspond to parameter values for which the qualitative aspects of the system change drastically. More specific to this one-dimensional problem, the phase line changes as $a$ crosses $\pm 2$ and remains the same elsewhere.

What you're doing, effectively, is solving for the parameter value(s) when these equilibria coincide; maybe in a more complicated system, it might be useful to pursue this method, but here you ought to be able to read off this behavior directly from $(*)$.

2.The phase line will depend on the value of $a$, and so a complete answer would include a phase line for each of the 5 cases: $a < -2, a = -2, -2 < a < 2, a = 2, a > 2$.

For each of $a < -2, a > 2$, there are two equilibria. I can tell you that in each case, one of them is a sink and other a source; you have to figure out which is which by writing
$$
\dot{y} = (y - \lambda_1) ( y - \lambda_2)
$$
where $\lambda_i$ are the equilibria given by $(*)$, and figuring out the sign of $\dot{y}$ for each of $y \in (-\infty, \lambda_1), (\lambda_1, \lambda_2), (\lambda_2, \infty)$.

For $-2< a < 2$ there are no equilibria, and the sign of $\dot{y}$ is positive for all $y$. What phase line corresponds to this situation? (Hint: it's pretty simple).

For $a = \pm 2$, there is a single equilibrium at $y = \pm 1$ that is neither a sink nor a source, because the sign of $\dot{y} = (y \mp 1)^2$ is always positive.

Title: bifurcation value


问:
Positive measurable function multiplied with indicator function is measurable
Suppose a measured space (X, $A$, $\mu$) and a A-measurable function: $f: X\longrightarrow [0,\infty]$ on X. I have to show, that $\int_A$ f =:$\nu(A)$ defines a measure. $\nu(\emptyset)$ = 0 is easily shown.
However, I am stuck with: $ \nu(\bigcup_{n=1}^{\infty} (A_n))$ =  $\int_{\bigcup_{n=1}^{\infty} (A_n)}fd\nu$ = $\int_{X} 1_{\bigcup_{n=1}^{\infty}} fd\nu$ = $\int_{X} \sum_{n=1}^{\infty} f1_{A_n}d\nu$. If I could show, that $f1_{A_n}$ is A-measurable for every n, then I have a lemma, that allows to me to bring the sum out of the integral, so that I'm done. Therefore the question:
How can I show, that a function multiplied with an indicator function is (A)-measurable?

$1_{A_n}$ is the indicator function and of course all $A_n$ are disjoint to each other.

答:
$\{x: f(x)1_{A_n} (x) <a\}=\emptyset$ if $a \leq 0$ and $\{x: f(x)1_{A_n} (x) <a\}=\{x\in A_n: f(x) <a\} \cup A_n^{c}$ of $a >0$.  In either case $\{x: f(x)1_{A_n} (x) <a\} \in \mathcal A$. This is true for every real number $a$. Also $\{x: f1_{A_n}=\infty\} =A_n \cap \{x: f(x)=\infty\} \in \mathcal A$. So $f1_{A_n}$ is measurable.

According to the definition you gave in the comments below a subset $B$ of $[0,\infty]$ is measurable if and only if it is either a Borel subset of $[0,\infty)$ or it has the form $\{\infty\} \cup C$ where $C$ is a Borel subset of $[0,\infty)$. Since $f_n^{-1} (\{\infty\} \cup C)=f_n^{-1}(C) \cup f_n^{-1} (\{\infty\})$ it is enough to show that $f_n^{-1}(C) $ and $f_n^{-1} (\{\infty\})$  are measurable for all Borel sets $C$ in $[0,\infty)$. But the latter condition follow from the fact that $f^{-1} [0,a)$ is measurable for each real number $a \in [0,\infty)$.


问:
Indian National Olympiad $2002$, Problem $6$

Suppose the $n^{2}$ numbers $1,2,3, \ldots, n^{2}$ are arranged to form an $n$ by $n$ array consisting of $n$ rows and $n$ columns such that the numbers in each row (from left to right) and each column(from top to bottom) are in increasing order. Denote by $a_{j k}$ the number in $j$-th row and $k$-th column. Suppose $b_{j}$ is the maximum possible number of entries that can occur as $a_{j j}, 1 \leq j \leq n$ Prove that
$$
b_{1}+b_{2}+b_{3}+\cdots b_{n} \leq \frac{n}{3}\left(n^{2}-3 n+5\right)
$$
(Example: In the case $n=3,$ the only numbers which can occur as $a_{22}$ are $4$,$5$ or $6$ so that $ b_{2}=3.)$

My solution-

I applied Induction on this problem and it reduces to proving that

$b_{k+1}\leq k^2-k+1$

But it follows directly from the fact that $b_{k+1}=1$ because $a_{k+1,k+1}$ has only one option i.e it has to be $(k+1)^2$

But I think this is too easy solution for problem $6$, so someone can pls point out my mistake...

Thankyou

答:
$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
\color{green}{1} & \color{green}{s} & \color{green}{s} & \color{green}{s} &   &   &    \\ \hline
\color{green}{s} & \color{green}{s} & \color{green}{s} & \color{green}{s} &   &   &    \\ \hline
\color{green}{s} & \color{green}{s} & \color{green}{s} & \color{green}{s} &   &   &   \\ \hline
\color{green}{s} & \color{green}{s} & \color{green}{s} & \color{red}{a_{jj}} & \color{blue}{l} & \color{blue}{l} & \color{blue}{l} \\ \hline
  &   &   &  \color{blue}{l} & \color{blue}{l} & \color{blue}{l} & \color{blue}{l}  \\ \hline
  &   &   &  \color{blue}{l} & \color{blue}{l} & \color{blue}{l} & \color{blue}{l}  \\ \hline
 &   &   & \color{blue}{l} & \color{blue}{l} & \color{blue}{l} & \color{blue}{n^2}\\ \hline
\end{array}
$$

I indicate the necessarily smaller entries than $a_{jj}$ by $s$ and necessarily larger entries by $l$. The ends of main diagonal are fixed.

It should be clear that on arranging first $j^2$ numbers $(1,2,3,\ldots, j^2)$ until $a_{jj}$ gives its minimum possible value i.e., $j^2$. E.g.,

$$
\begin{array}{|c|c|c|}
\hline
\color{green}{1} & \color{green}{2} &     \\ \hline
\color{green}{3} & \color{red}{4} &     \\ \hline
  &   & \color{blue}{9}  \\ \hline
\end{array}
$$

While arranging numbers from last $(n^2, n^2-1, n^2-2, \ldots)$  should give maximum possible value

$$
\begin{array}{|c|c|c|}
\hline
 \color{green}{1} &   &     \\ \hline
  & \color{red}{6} & \color{blue}{7}    \\ \hline
  &  \color{blue}{8} & \color{blue}{9}  \\ \hline
\end{array}
$$

which would be $n^2 - ((n-j+1)^2-1)$.

Thus $a_{jj}$ can take possible values in the set $${j^2, j^2+1, j^2+2, \ldots, n^2 - (n-j+1)^2+1} $$

whose cardinality is defined as $b_{j,{\text{max}}}$.

Can you proceed from here?


答:
Note that the function $g(u)=u\sin {1\over u}$ has a maximum in $u_0$ such that $u_0$ is the largest positive solution to $$\tan{1\over u}={1\over u}$$which means that
$$
\pi<{1\over u_0}<{3\pi\over 2}\implies {2\over 3\pi}<{u_0}<{1\over \pi}
$$
hence the function $f_n(x)=x^n\sin{1\over x^n}$ always has a fixed-value extremum at $x=\sqrt[n]{u_0}$ which lies within $(0,{1\over \sqrt[n]{\pi}})$, therefore meaning that the convergence is not uniform.

Title: Pointwise and uniform convergence - solution verification

Body:
Consider the sequence of functions defined by the following
$$ f_n(x) = x^n \sin \left(\frac{1} {x^n} \right),
  0\lt x \lt \frac {1} {\sqrt[n] \pi},  f_n(x)=0,$$
elsewhere  in  $\mathbb R$.
What can we say about its convergence?
I think the sequence converges pointwise but not uniformly.

I have made an attempt I now write as an answer not to make the question too long. I'm not completely sure about it. Please check it and feel free to suggest hints or other ways to solve this problem. Thank you.


问:
LQR control for Quadrotor
Recently, I have found a paper which applies different linear and nonlinear controllers on quadrotor. So, I have started to read and apply it.

I created Matlab simulation for my quadrotor

I have read and applied LQR controller to the quadrotor

I have been successful in applying both of the aforementioned steps. I can say that based on the plots of mine in Matlab and the plots of the paper. I understood the derivation of the dynamics of the quadrotor, its linearization and construction of the LQR controller. However, one thing I didn't understand is while I can control my quadrotor perfectly along x, y, z translational directions and yaw (z) rotational direction, I cannot control it around x and y (roll and pitch) directions. The quadrotor becomes unstable and does stupid things. I referred the plots of the paper, but interestingly the author of the paper mentioned the plot of only yaw, but not roll and pitch. I checked the controllability of the system by using A and B matrices I got from linearization of the dynamics and the controllability matrix is full matrix. So, the system is controllable and I should be able to control it in every direction. In conclusion, I would like to learn what I am missing and I need your help for that. Thank you very much :)

(I added some related equations in any case)

Approximated linear model
A matrix
B matrix

答:
When a system is (locally) controllable does not mean that you can hold that system at a constant state. Instead, controllability only requires that one can bring a system to any state for only one instance. For example a double integrator, $\ddot{x} = u$, can be held a constant position $x$ with $\dot{x}=0$, but when trying to hold it at a constant velocity then the position will change linearly with time. However, it is possible to bring the double integrator to any position and velocity simultaneously, but only at one moment in time.

Similarly, when you try to hold a quadrotor at a non-zero pitch or roll then the thrust vector is pointing partially horizontal and would indefinitely accelerate the quadrotor sideways (or eventually at a constant velocity if you also include drag from air). But it is possible to drive a quadrotor to a certain orientation and position simultaneously, as is demonstrated in this video.

Title: LQR control for Quadrotor


问:
$\int_\Omega fd\mu_n\to\int_\Omega fd\mu,\ \forall\ f\in C_0(\Omega)$ implies $\mu_n(\Omega)\to \mu(\Omega)$?
Let $\Omega\subset\mathbb{R}^N$ be a bounded open smooth domain and $C_0(\Omega)$ the set of bounded continuous functions with compact support. It is know that $C_0(\Omega)^\star =M(\Omega)$, where $M(\Omega)$ denotes the set of Radon measures (regular, bounded and infinitely additive). Suppose that $\mu_n\to \mu$ ($\mu_n,\mu\geq 0$) in the weak star sense, i.e. $$\int_\Omega fd\mu_n\to\int_\Omega fd\mu,\ \forall\ f\in C_0(\Omega)$$

Is it true that $\mu_n(\Omega)\to \mu(\Omega)$?

Thank you

答:
If $\Omega:=(0,1)$ and $\mu_n:=\delta_{n^{-1}}$, then we have the assumption with $\mu=0$, but not $μ_n(\Omega)=1\to \mu(\Omega)=0$.

One can generalize this taking $x\in \overline{\Omega}\setminus\Omega$ and $(x_n,n\in\Bbb N)\subset \Omega$ converging to $x$.


问:
Paired t-test null hypothesis correct wording
I am studying paired t-tests from the Handbook of Biological Sciences and I am not sure about the following claim: "The null hypothesis is that the mean difference between paired observations is zero. When the mean difference is zero, the means of the two groups must also be equal." Does that mean the following null hypotheses are equivalent or just that the first implies the second and how do you easily see that?

H0: the mean difference between the paired observations is 0.

H0: the difference between the paired population means is equal to 0.

Intuitively, those seem like quite different claims to make. When I looked into this, it seems that the first is used as the null hypothesis of paired t-tests frequently, but I also have found websites write the hypotheses in the format of the second null (such ashttps://libguides.library.kent.edu/SPSS/PairedSamplestTest). I was hoping somebody could shed some light on this.

Thank you.

答:
If the null hypothesis $H_0$ is $\mathbb E[X_{1,j}-X_{2,j}]=0$ for the $j$th pair then this is equivalent to $\mathbb E[X_{1,j}]-\mathbb E[X_{2,j}]=0$, i.e. $\mathbb E[X_{1,j}]=\mathbb E[X_{2,j}]$.

That is a statement for the paired observations, but if  $\mathbb E[X_{1,j}]=\mathbb E[X_{2,j}]$ is true for each of the paired observations then $\mathbb E[X_1]=\mathbb E[X_2]$ across all the observations


问:
What is $(-1)^{\frac{1}3}$?
I was surfing Facebook and I ran into this question posted by a high school student:

$$\text{Which value equals } (-1)^{\frac{1}3}?\quad\text{Is it } 1\text{ or }-1?$$

He said that he asked this because he did it in two ways, both of which seem valid but they generate different answers.

Way #1:
$$(-1)^{\frac{1}3}=\sqrt[3]{-1}=-1$$ since $(-1)^3=-1$. However, he also did something else.

Way #2:
$$(-1)^{\frac{1}3}=(-1)^{\frac{2}6}=\left((-1)^2\right)^{\frac{1}6}=\sqrt[6]1=1$$
and this solution seems valid as well. He's confused, and after reading his question, I became somewhat confused as well.  I know that $1^3\ne-1$, but I can’t see why way#2 is wrong.

Which solution above is invalid? Or is $(-1)^{\frac{1}3}$ undefined? Thanks in advance. I apologize if this is a stupid question.

答:
The first solution is valid. Note that it is natural to define $(-1)^{1/3}$ as $\sqrt[3]{-1}$ and that indeed $-1$ is the only (real) cube root of $-1$.

However, the second solution is based upon the formula $a^{bc}=(a^b)^c$, which only holds for $a>0$.

Title: What is $(-1)^{\frac{1}3}$?


答:
Try cubing $1$ and you will see that it is not a cube root of $-1$. It is a sixth root of $1$ though. Others will comment more on the issues involved in defining roots.

Since cubing is bijective on the real numbers, a single real cube root can be defined for each real number.

What you have done is involve squaring too, and squaring real numbers is neither injective nor surjective (neither one-to-one nor onto). And that is why the second method does not work.

When you move to the complex numbers the question of Nth roots is different - each non-zero complex number has N different Nth roots, and you have to be really careful if you need to pick just one.

Title: What is $(-1)^{\frac{1}3}$?

Body:
I was surfing Facebook and I ran into this question posted by a high school student:

$$\text{Which value equals } (-1)^{\frac{1}3}?\quad\text{Is it } 1\text{ or }-1?$$

He said that he asked this because he did it in two ways, both of which seem valid but they generate different answers.

Way #1:
$$(-1)^{\frac{1}3}=\sqrt[3]{-1}=-1$$ since $(-1)^3=-1$. However, he also did something else.

Way #2:
$$(-1)^{\frac{1}3}=(-1)^{\frac{2}6}=\left((-1)^2\right)^{\frac{1}6}=\sqrt[6]1=1$$
and this solution seems valid as well. He's confused, and after reading his question, I became somewhat confused as well.  I know that $1^3\ne-1$, but I can’t see why way#2 is wrong.

Which solution above is invalid? Or is $(-1)^{\frac{1}3}$ undefined? Thanks in advance. I apologize if this is a stupid question.


答:
The fact is that $$a^{mn}={(a^m)}^n$$ need not be always valid.
The correct way is $$x={(-1)}^{1/3}$$
$$x^3+1=0$$
$$(x+1)(x^2-x+1)=0$$
if x is real $$x=-1$$

Title: What is $(-1)^{\frac{1}3}$?


问:
Rank of a block-matrix which implies a zero block
Why if I have a matrix like this one:
$$
\left(\begin{array}{cc}
 I_k & 0 \\
 B & C
\end{array} \right),
$$
which have rank k I can automatically say that $C=0$?

答: Because the span of the first $k$ rows (row rank = column rank, remember?) lies entirely in the span of $e_1, \ldots, e_k$ (where these are standard unit ROW vectors). If there were a nonzero entry in $C$, then the row containing that nonzero entry, say $v$, could not be written as a linear combination of the first $k$ rows, hence $\{e_1, \ldots, e_k, v \}$ would be linearly independent, and the rank would be at least $k+1$, hence not $k$.


问:
$\ker A = \{ 0\}$ where $A \in \mathbb{K}^{n\times m}$
I struggle with the understanding of how an n x m matrix A can have $ker A = \{ 0\}$. Can someone provide me with an example or an explanation as to why this might be the case for a non quadratic matrix?

答: $KerM=\{0\}$ is equivalent to: linear mapping is injective. On the other hand, $KerM=\{0\}$ of a matrix $M\in\mathbb K^{m\times n}$ is equivalent to: $rank(M)=n$. But $rank(M_{m, n})\leq \min\{m,n\}$, so $\mathbb K^{m\times n}$ has some matrices with $KerM=\{0\}$ if and only if $m\geq n$. In the language of vector spaces, $Kerf=\{0\}$ for linear mapping $f:V\to W$ means, that the vector space has the same dimension as its image: $\dim V=\dim f(V)$, that is $n$. See wiki article about rank.


问:
Help solving $2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right)$
I'm at the begginig of a differential equation course, and I'm having trouble finishing this problem. I'm asked to solve the equation
$$2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right). \ \ \ \text{(I)}$$
I'm used the variable change $p=y'$, $\ \dot p=\frac{dp}{dy}$, $ \ y''=\dot pp$, I get the equation:
$$p=2y\dot p+\frac{1}{2}\log \dot p. \ \ \ \text{(II)}$$
That last one is a Lagrange differential equation. I solved it finding a parametric solution using the variable change $\dot p = u$, and using that I got the first order linear equation
$$\frac{dy}{du}+\frac{2y}{u}=\frac{-1}{2u^2}.$$
Solving it by integrating factor, I finally found the parametric values for $y$ and $p$, and they are
$$
\left\{
\begin{array}{rcl}
     y(u,C)&=&\frac{C}{u^2}-\frac{1}{2u}
  \\ p(u,C)&=&\frac{2C}{u^2}-1+\frac{1}{2}\log u
\end{array}
\right.
$$
This two equations are the solution to the equation (II), but I'm asked to solve the equation (I). Since $p=y'(x)$ in the original equation (I), I integrate the expression of $p$ with respect to $x$, and I get
$$y=\int\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right)\ dx + K$$
$$\Downarrow$$
$$\boxed{y(x)=x\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right) + K}$$
This may be my solution, but something feels wrong about it, since my original $y$ in equation (I) is some $y(x)$, but what I get is a single equation $y(x,u)$ ($C,K$ are constants), and I'm not sure if I'm forgetting another expression to make it have sense. I feel that the solution to (I) must be a parametric one, but I only get one equation for $y$ in function of $x$ and $u$. Is the boxed equation the right solution? If not, where am I wrong? Any help will be appreciated.

答:
I'll try to formally trace your solution. First you apply the trick $y'(x)=p(y(x))$ with the dot indicating the $y$ derivative. This gives correctly equation (II). Next you set $\dot p(y)=u$ and compute $y$ as a function of $u$, $y=v(u)$, giving the combined equation $y(x)=v(\dot p(y(x)))$, that is, $v$ is the inverse function to $u$. This transforms (II) into
$$
p(v(u))=2v(u)u+\frac12\ln(u)
$$
and taking the $u$ derivative
\begin{align}
\dot p(v(u))v'(u)=2v'(u)u+2v(u)+\frac1{2u}
&\implies 0=v'(u)+\frac{2v(u)}{u}+\frac1{2u^2}\tag{IIIa}
\\
&\implies 0=[u^2v'(u)+2uv(u)]+\frac12
\\
&\implies C=u^2v(u)+\frac{u}2\tag{IIIb}
\end{align}
and thus
$$
p(v(u))=2\left(\frac{C}u-\frac12\right)+\frac12\ln(u)
$$
At this point you may note that there is a different power in the denominator as $uv(u)=\frac Cu-\frac12$.

Though the connection may be weak, there is still dependence between $x$ and $u$. You can not set $u$ constant in the $x$-integration of $y'(x)=p(y(x))$. Rather you would attempt to get $x$ also as a function $x=w(u)$, so that from $y(w(u))=v(u)$ one gets the equation
$y'(w)w'(u)=p(v(u))w'(u)=v'(u)$ where $v'(u)$ is known from (III), giving
$$
x=\int w'(u)du=\int\frac{v'(u)}{p(v(u))}du=\int\frac{-\frac{2C}{u^3}+\frac1{2u^2}}{2\frac{C}u-1+\frac12\ln(u)}du
$$
which does not give the impression of an easy solution.


问:
How to define a finite set S which is a non-abelian group under binary operation without commutativities except the trivial ones (see Cayley table)?
By "without trivial commutativities" I mean "$bc\neq cb$ for any $b,c \in S \setminus \{a\}$, where $a$ is the identity and $b\neq c$.

I found out how to calculate the element $x_{ij}$ of the $n\times n$ Cayley table for the general* symmetric case: $x_{ij} = mod(i+j,n)$.
However, I am clueless about how to calculate it for the asymmetric case (manually permuted here to achieve minimal commutativity).

Probably, defining the general* asymmetric case itself is part of the solution.
The result of the formula above is given in the Figure for each pair of elements,
and the correct elements are marked as ok in the last table.

general* -> general for any finite number of elements that is suitable to define such a group

答:
The only groups that satisfy the condition $bc \ne bc$ for all $b,c \ne 1$ and $b \ne c$ are the groups of orders $1$ and $2$, which are abelian and satisfy the condition trivially.

Otherwise, if there exists and element with $g \ne g^{-1}$ then the condition fails with $b=g$ and $c=g^{-1}$. If there is no such element, then by a well-known result $G$ is abelian.

A more interesting condition would be $bc = cb$ if and only if $b$ is a power of $c$ or $c$ is a power of $b$. That holds in some examples, such as dihedral groups of twice odd order, and also in some infinite groups, such as free groups.


问:
Equivalent characterisation of hypothesis & conclusion in Grothendieck's inequality
The basic version of Grothendieck's inequality states that for a $m\times n$ matrix $[a_{i, j}]$, if

$$\bigg|\sum_{i, j}a_{i, j}x_i y_j\bigg|\le 1,$$

for all $x_i, y_i\in \{-1, 1\}$, then

$$\bigg|\sum_{i, j}a_{i, j}\langle u_i, v_j\rangle\bigg|\le K,$$

for any unit vectors $u_i, v_j$ in a Hilbert space $H$ (i.e. $\|u_i\| = \|v_j\| =1$), with $K$ an absolute constant.

The following conditions are equivalent to the hypothesis and conclusion of the prior statement respectively:

hypothesis:
if $$\bigg|\sum_{i, j}a_{i, j}x_i y_j\bigg|\le \max_i |x_i| \max_j|y_j|,$$

for all $x_i, y_i\in \mathbb{R}$.

conclusion:
then
$$\bigg|\sum_{i, j}a_{i, j}\langle u_i, v_j\rangle\bigg|\le K\max_i\|u_i\|\max_j\|v_j\|,$$

for any vectors $u_i, v_j\in H$, a Hilbert space, with $K$ an absolute constant.

This equivalence seems like it ought to be very simple to prove, however I seem to be having a mental block which is preventing me from proceeding & presume that I'm just not seeing something pretty obvious here. I'd be grateful for any hints or suggestions on how to approach the proof.

答:
To see the equivalence, consider the following lemmas, the first of which is well-known.

Lemma 1. $|\sum_i\pm a_i|\le 1\iff |\sum_it_ia_i|\le1,\quad \forall t_i\in[-1,1]$.

Proof: $\Leftarrow$ is obvious. $\Rightarrow$: $|\sum_it_ia_i|\le\sum_i|a_i|=|\sum_i\pm a_i|\le1$ where a choice of signs is made so that $|a_i|=\pm a_i$.

Lemma 2. $|\sum_i\langle u_i,w_i\rangle|\le K, \forall u_i$ unit $\iff |\sum_i\langle u_i,w_i\rangle|\le K$, $\forall u_i$, $\|u_i\|\le1$.

Proof: $\Leftarrow$ is obvious. $\Rightarrow$: The inequality is equivalent to $|\sum_i\pm\langle u_i,w_i\rangle|\le K$ by considering $\pm u_i$ instead of $u_i$. So by lemma 1, one can take $t_i=\|u_i\|$ and write $u_i=t_i\hat{u}_i$.

Both lemmas can be extended to multiple indices:

Proposition 1. $|\sum_{ij}a_{ij}x_iy_j|\le 1,\ \forall x_i,y_j\in\{-1,+1\}\iff |\sum_{ij}a_{ij}x_iy_j|\le1\ \forall x_i,y_j\in[-1,1]$.

Proof: $\Rightarrow$: $|\sum_ix_i\sum_ja_{ij}y_j|\le1$ by lemma 1. when $y_j\in\{-1,1\}$.

So $|\sum_j\pm\sum_ia_{ij}x_j|\le1$. Again by lemma 1., $|\sum_jy_j\sum_ia_{ij}x_j|\le1$ when $y_j\in[-1,1]$.

Proposition 2. $|\sum_{ij}\langle u_i,a_{ij}v_j\rangle|\le K,\ \|u_i\|=1=\|v_j\|\iff |\sum_{ij}\langle u_i,a_{ij}v_j\rangle|\le K,\  \|u_i\|,\|v_j\|\le1$.

Proof: $\Rightarrow$: When $v_j$ are unit, but $\|u_i\|\le1$, lemma 2. implies $|\sum_i\langle u_i,\sum_ja_{ij}v_j\rangle|\le K$.

So $|\sum_j\langle\sum_ia_{ij}u_i,v_j\rangle|\le K$. Hence again by lemma 2., this last inequality is valid for $\|v_j\|\le1$.

The equivalence between the two sets of statements should now be clear.


问:
How does the spherical Laplacian commute with the euclidean convolution?
INTRODUCTION.

It is a standard and well-known property that for smooth, compactly supported $f, g\colon \mathbb R^d\to \mathbb R$, the euclidean Laplacian
$$\tag{1}\Delta=\sum_{j=1}^d \frac{\partial^2}{\partial x_j^2}$$ commutes with the convolution
$$
f\ast g(x):=\int_{\mathbb R^d} f(x-y)g(y)\, dy. $$
We can rephrase this in terms of the linear operator
$$\tag{2}
T_g f:=f\ast g; $$
indeed, the aforementioned property is equivalent to the vanishing of the commutator
$$
[T_g, \Delta]=T_g\Delta - \Delta T_g=0.
$$

THE QUESTION.

Let us define the spherical Laplacian $\Delta_{\mathbb S^{d-1}}$ via the formula
$$
\frac{1}{r^2}\Delta_{\mathbb S^{d-1}}f := \Delta f - \frac{1}{r^{d-1}}\frac{\partial}{\partial r}\left(r^{d-1}\frac{\partial f}{\partial r}\right),$$
where $r=\lvert x \rvert$ denotes the radial coordinate of $\mathbb R^d$.   $^{[1]}$

Big Question. Can you compute the commutator $[T_g, \Delta_{\mathbb S^{d-1}}]\ ?$

This question is probably quite ambitious, and I would already be happy with some partial answer. For example,

Baby Question. Can you compute $[T_g, \Delta_{\mathbb S^{d-1}}]$ in the case $$g(x)=\phi(r)H_n(x),$$ where $\phi$ is smooth and compactly supported and $H_n$ is a homogeneous harmonic polynomial of degree $n$? If the general case is too hard, can you compute the commutator for small values of $n$, say $n\in \{0, 1, 2\}$?

SOME THOUGHTS.

If $g$ is radial, then $[T_g, \Delta_{\mathbb S^{d-1}}]=0$. This answers the Baby Question for $n=0$.

To prove this, we note that the radial symmetry of $g$ implies that
$$\tag{3}
T_g(f\circ R)=(T_g f)\circ R,\qquad \forall R\in SO(d).$$
In turn, this implies the following commutation;
$$\tag{4}
[T_g, Z_{ij}]=0,\qquad \text{where } Z_{ij}:=x_i\partial_{x_j}-x_j\partial_{x_i}, $$
and $1\le i<j\le n$. To prove (4), we observe that the operators $Z_{ij}$ are infinitesimal rotations in the following sense. We let
$$
R_{12}(\theta):=\begin{bmatrix} \cos \theta & -\sin\theta & 0 & 0 & \ldots & 0 \\
\sin \theta & \cos \theta & 0 & 0 & \ldots & 0 \\
0 & 0 & 1 & 0 &\ldots & 0 \\
0 & 0& 0 & 1 & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots& \ldots \\
0 & 0 & 0 & \ldots & 0 & 1\end{bmatrix}, $$
observing that $R_{12}(\theta)\in SO(d)$. We have the differential relation
$$
\left.\frac{\partial}{\partial \theta}f\left( R_{12}(\theta)x\right) \right|_{\theta=0}= Z_{12} f(x),$$
and similarly for the other operators $Z_{ij}$. So (3) implies (4) by differentiation.

Now, the property (4) implies $[T_g, \Delta_{\mathbb S^{d-1}}]=0$, because
$$
\Delta_{\mathbb S^{d-1}}=\sum_{1\le i<j\le d} Z_{ij}^2. $$
(We remark that this last formula is the spherical analogue of (1)).

PERSPECTIVES.

The previous section shows that the knowledge of the commutators (4) gives  $[T_g, \Delta_{\mathbb S^{d-1}}]$. Indeed, assuming that
$$
[T_g, Z_{ij}]=C_{ij}, $$
it holds that
$$
[T_g, \Delta_{\mathbb S^{d-1}}]=\left[T_g, \sum Z_{ij}^2\right]=\sum_{1\le i<j\le d}\left( C_{ij}Z_{ij}^2 + Z_{ij}C_{ij}Z_{ij}\right).$$
Can these $C_{ij}$ be explicitly computed, at least for $g$ of the form $g_0(r)H_n(x)$, as in the Baby Question?

[1] This operator is often known as the Laplace-Beltrami operator.

答:
The problem turned out to have a simpler solution than I expected. It is  a consequence of the following proposition.

Main Proposition. For all smooth and compactly supported $f, g$,

$$\tag{1}\Delta_{\mathbb S^{d-1}} (f \ast g) = (\Delta_{\mathbb S^{d-1}} f) \ast g +2 \sum_{1\le i<j\le d} Z_{ij}f\ast Z_{ij}g + f \ast \Delta_{\mathbb S^{d-1}}g.$$

Remark. Essentially, the algebra behind this is the same as the one of the usual formula
$$
\frac{d^2}{dx^2}( t(x)s(x))= \frac{d^2 t}{dx^2} s  +  2\frac{dt}{dx}\frac{ds}{dx}+t\frac{d^2 s}{dx^2} .$$

Proof of the Main Proposition. As observed in the main question, the spherical derivative $Z_{ij}$ satisfies
$$
Z_{ij} (f\ast g)(x)=\lim_{\epsilon\to 0} \frac1{\epsilon} \int_{\mathbb R^d} \left[ f(R_{ij}(\epsilon)x -y)g(y) -f(x-y)g(y)\right]\, dy.$$
We claim that this implies the Leibniz formula

$$\tag{2}
Z_{ij}(f\ast g)=(Z_{ij}f)\ast g + f\ast Z_{ij}g. $$
Indeed,
$$
\begin{split}
&\frac1\epsilon \left(f\ast g(R_{ij}(\epsilon)x)-f\ast g(x)\right) = \frac1\epsilon \int_{\mathbb R^d} f(R_{ij}(\epsilon)x-y)g(y)-f(x-y)g(y)\, dy \\
&=\frac1\epsilon\int_{\mathbb R^d} f(R_{ij}(\epsilon)(x-y))g(R_{ij}(\epsilon)y) - f(x-y)g(y)\,dy \\
&= \frac1\epsilon\int_{\mathbb R^d} f(R_{ij}(\epsilon)(x-y))g(R_{ij}(\epsilon)y) - f(x-y)g(R_{ij}(\epsilon)y) +f(x-y)g(R_{ij}(\epsilon)y) -f(x-y)g(y)\, dy\\
&=\frac1\epsilon\int_{\mathbb R^d} \left( f(R_{ij}(\epsilon)(x-y))-f(x-y)\right)g(R_{ij}(\epsilon)y) + f(x-y)\left( g(R_{ij}(\epsilon)y)-g(y)\right)\, dy \\
&=\frac1\epsilon \left(f(R_{ij}(\epsilon)\cdot) \ast g - f\ast g\right) +\frac1\epsilon \left( f\ast g(R_{ij}(\epsilon)\cdot) -f\ast g\right)\to Z_{ij}f\ast g + f\ast Z_{ij}g.
\end{split}
$$
Sorry for the long set of equations; it looks terrible, but it actually is only a straightforward adaptation of the proof of the usual Leibniz formula for the one-dimensional derivative $\frac{d}{dx}$.

Once (2) is proved, (1) immediately follows from
$$
\Delta_{\mathbb S^{d-1}}=\sum_{1\le i<j\le d} Z_{ij}^2.\quad \Box$$

Now that the Main Proposition is proved, the commutator can be computed as
$$\begin{split}
[T_g, \Delta_{\mathbb S^{d-1}}]f
&=\Delta_{\mathbb S^{d-1}}f\ast g -\Delta_{\mathbb S^{d-1}}(f\ast g) \\
&=-2\sum_{1\le i<j\le d} Z_{ij}f\ast Z_{ij}g -f\ast \Delta_{\mathbb S^{d-1}}g.
\end{split}
$$
This answers the "Big Question".


问:
Find an exponential equation based on a grapth - my solution not matching textbooks solution
Screen shot of the question from my textbook:

My solution is $f(x)=5(\frac{1}{5})^1$ whereas my textbooks solution is $f(x)=-2(3)^x+7$

Here's my working:

General form: $y=ab^x$.

Y intercept is $(0,5)$ so $a = 5$.

Another point on the graph with integer values is $(1,1)$.

Substituting my values into the general form:
$$f(x)=ab^x$$
$$f(x)=5b^x$$
$$1=5b^1$$
$$\frac{1}{5}=b^1$$
$$b=\frac{1}{5}$$

Thus the function is: $f(x)=5(\frac{1}{5}^x)$

Where did I go wrong and how can I arrive at $f(x)=-2(3)^x+7$?

答:
Notice how the graph has a horizontal asymptote at $y = 7$. This is from the graph being shifted up by 7. This means we should use the form:

$y = ab^x + c$, where c is the horizontal asymptote. This form has technically been used by you before, it's just that c normally equals $0$ so there's no need to write it. E.g., if you graph $y = 2^x$, you'll notice the horizontal asymptote is at $y = 0$ (so the graph hasn't been shifted).

So we have $y = ab^x + 7$. Now for this, because the graph has been shifted you can't get $a$ immediately just by looking at the graph's y-intercept. Instead you need to plug in the y-intercept's x and y coordinates in order to solve for $a$:

$5 = ab^0 + 7$

$5 = a(1) + 7$

$a = -2$

Notice how the $b$ was eliminated above since it was raised to $0$. This is thanks to the x-coordinate of the y-intercept being $0$. If we instead tried to use the other point on the graph first, you'd be left with both $a$ and $b$ in the equation.

Anyway, now that we've solved for $a$, plug -2 into the equation:

$y = -2b^x + 7$

There's only one variable left to solve: $b$. So use the (1,1) coordinate on the graph:

$1 = -2b^1 + 7$

$-6 = -2b$

$b = 3$

So we get: $y = -2(3)^x + 7$

Note - the following video explains a similar problem to yours. The graph is flipped over the y-axis but the method is the same.

https://www.youtube.com/watch?v=12LDAqnsH_w


问:
Finding fundamental set of solutions of a given differential equation

Suppose that $y_1,y_2$ is a fundamental set of solutions of this equation $t^2y''-3ty'+t^3y=0$ such that $W[y_1,y_2](1)=4$  , Find $W[y_1,y_2](7).$

I can't evaluate faundamental solutions.Seems difficult . Don't understand where use this result $W[y_1,y_2](1)=4$

答:
$$t^2y''-3ty'+t^3y=0$$
$$y''-\dfrac 3ty'+ty=0$$
$$y''+py'+qy=0$$
You can certainly use Abel's identity:
$$W'+pW=0$$
$$tW'- 3 W=0$$
Where $W$ is the wronskian of the differential equation. You need $W[y_1,y_2](1)=4$ for the constant of integration.


问:
Binomial Identity and Counting
A string in $\{0, 1\}*$ has even parity if the symbol $1$ occurs in the word an even number of times; otherwise, it has odd parity.

(a) How many words of length $n$ have even parity?

(b) How many words of length $n$ have odd parity?

It seems to me that correct approach will be to use summations of combinations, and correct answer will yield $2^{n-1}$ in both cases. Is there any way to prove separately that each of them is equal to $2^{n-1}$? Would be happy to know your ideas!

答:
Say you want an odd parity. For the first $n-1$ digits, you can choose however you like, but for the last digit, you cannot choose because you need to make the number of $1$s odd. Hence $2^{n-1}$.

The same for even parity.

Title: Binomial Identity and Counting


问:
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$?
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ?
i know that its "1", but why?
How can i calculate this?

Thank you very very much =)

答:
In order to compute this limit you simply make the substitution $t=\log x$, which is possible because $\log$ is a continuous invertible function. Then the limit is just
$$
\lim_{t\to0^+}t^t
$$
that's quite easy to compute. Write it
$$
\lim_{t\to0^+}t^t=\lim_{t\to0^+}e^{t\log t}
$$
and compute
$$
\lim_{t\to0^+}t\log t=\lim_{t\to0^+}\frac{\log t}{1/t}
$$
As you see, this has the form "$0/0$", so it's not computable just by substitution.</p>


问:
Issue finding the models of a formula using semantic tableux method.
Using semantic tableux method, decide what kind of formula U is. If it is consistent, find its models.

U = (p V q → r) → (p V r → q).

This is the semantic tableux I got, is it correct?

DNF(U) = ( ¬ r ^ p) V (¬ r ^ q) V ( ¬ p ^ q) V ( ¬ r ^ q) by absorption law

≡ ( ¬ r ^ p) V (¬ r ^ q) V ( ¬ p ^ q)

Is it correct so far? Then how do I find its models?

答:
Yes, your tableaux is correct.

To find a model, simply take any open branch and see what literals (atomic statements or negations of atomic statements) you find on that branch.

For example, the right most branch contains $q$ as a literal, meaning that you have model by setting $q$ to true ... apparently it does not matter whether $p$ and $r$ are true of false: as soon as $q$ is true, the original sentence is true as well (it's always good to verify that fact ... if you find that this is not the case, then you know something went wrong with your tableaux)


问:
Can I find length of bisector by knowing the position vectors?

$\vec A$ = $(\vec i - \vec j-3\vec k)$

$\vec B$ = $(2\vec i +\vec j-2\vec k)$

$\vec C$ = $(-5\vec i +2\vec j-6\vec k)$

To find the length of angle bisector of bac I marked the points $A(1,-1,-3)$ $B(2,1,-2)$ $C(-5,2,-6)$. How can I use the fact that the angle between the bisector and two adjacent sides is equal?

答:
Method 1Just find the lenth of the sides of the triangle amd use the formula given here

Method 2: By the angle bisector theorem we know the ratio in which the bisector divides the side opposite to the bisected angle. Use the section formula too find point of intersection and then use distance formula


问:
MAGMA: semidirect product of finite matrix group in matrix form
Apologies for the very novice question. I am trying to find conjugacy classes and representatives (in matrix form) of a finite group of matrices $G$, isomorphic to $$ (\mathbb{Z}/n\mathbb{Z})^3 \rtimes S_3. $$
The isomorphism is as follows: the $i$-th copy of $\mathbb{Z}/n\mathbb{Z}$ is generated by the diagonal matrix with entries $\omega = e^\frac{2\pi \sqrt{-1}}{n}$ in the $(i,i)$ entry, and 1 elsewhere, and $S_3$ acts by conjugation, permuting the copies of $\mathbb{Z}/n\mathbb{Z}$.

I set this problem up in Magma using SemidirectProduct (an example follows with $n=3$).
The command Classes(G); seems to be what I want, however, it returns representatives expressed in cycle notation since SemidirectProduct presents $G$ as a permutation group. Is there a way to preserve the matrix form of the elements of the group when invoking SemidirectProduct (getting elements of the semidirect product as pairs of matrices)? Or an alternative way to construct the group $G$?

```<code>//Three copies of Cyclic group
F<w> := CyclotomicField(3);
H := MatrixGroup< 3, F | [w,0,0, 0,1,0, 0,0,1], [1,0,0, 0,w,0, 0,0,1], [1,0,0, 0,1,0, 0,0,w] >;

//Permutation group
K := MatrixGroup<3, F | [0,1,0, 1,0,0, 0,0,1], [0,0,1, 1,0,0, 0,1,0]>;
k1 := elt <K | 0,1,0, 1,0,0, 0,0,1>;
k2 := elt  <K | 0,0,1, 1,0,0, 0,1,0>;

//Conjugacy action
A := AutomorphismGroup(H);
p1 := elt<A|hom<H->H | g :-> g^k1>>;
p2 := elt<A|hom<H->H | g :-> g^k2>>;
f := hom<K->A| k1 -> p1, k2-> p2>;

G := SemidirectProduct(H,K,f);
</code>```

答:
The SemidirectProduct intrinsic actually produces several maps as output that you can use for this.  For example we can define

```<code>G, m1, m2, m3 := SemidirectProduct(H,K,f);
</code>```

Then we have that

$m_{1} : H \to G$ by $h \mapsto (h,1)$,

$m_{2} : K \to G$ by $k \mapsto (1,k)$ (here of course the outputs of $m_{1}$ and $m_{2}$ are actually permutations in $G$ corresponding to these matrix pairs),

$m_{3} : G \to K$ by $g \mapsto k$ where $g = m_1(h)*m_2(k)$.

This give a nice way to take a matrix pair $(h,k) : h \in H, k \in K$ as an element of $G$ by

```<code>g := m1(h)*m2(k)
</code>```

(I know you are asking about the other direction).

If you would like to turn a permutation in $G$ to a pair of matrices $(h,k)$, we can write a function

```<code>function PermutationToMatrixPair(g)
    k := m3(g);
    h := (m1^-1)(g*m2(k^-1));
    return <h,k>;
end function;
</code>```

This will take $g$ as a permutation, and you want to write it as a corresponding ordered pair of matrices $h \in H$, $k \in K$;

$k$ we can get directly from the projection map $m_{3}$, k := m3(g).

Then if we multiply $g$ by (the permutation corresponding to)
$k^{-1}$, we have a new permutation $g^{\prime}$ that corresponds to
$(h,1)$; ie gprime := g*m2(k^-1);

to recover the matrix $h$ from the permutation, we can take the
inverse image under our embedding map $m_{1}$ (since this map has as
its image the set of permutations in $G$ corresponding to matrix
pairs of the form $(h,1)$ with $h \in H$); h := (m1^-1)(gprime).

(Note: you will either need to define this function at an appropriate place in your code where m1, m2, m3 have been previously defined, or else define the function as taking m1, m2, m3 as arguments.)

As an alternate way to do this, you can define a map with specified inverse to convert back and forth between permutations and matrix pairs.

```<code>phi := map<G-> CartesianProduct(H,K) | g  :-> <(m1^-1)(g*m2(m3(g)^-1)), m3(g)>,
                                       hk :-> m1(hk)*m2(hk) >;
</code>```

Body:
The SemidirectProduct intrinsic actually produces several maps as output that you can use for this.  For example we can define

G, m1, m2, m3 := SemidirectProduct(H,K,f);

Then we have that

$m_{1} : H \to G$ by $h \mapsto (h,1)$,

$m_{2} : K \to G$ by $k \mapsto (1,k)$ (here of course the outputs of $m_{1}$ and $m_{2}$ are actually permutations in $G$ corresponding to these matrix pairs),

$m_{3} : G \to K$ by $g \mapsto k$ where $g = m_1(h)*m_2(k)$.

This give a nice way to take a matrix pair $(h,k) : h \in H, k \in K$ as an element of $G$ by

g := m1(h)*m2(k)

(I know you are asking about the other direction).

If you would like to turn a permutation in $G$ to a pair of matrices $(h,k)$, we can write a function

function PermutationToMatrixPair(g)
    k := m3(g);
    h := (m1^-1)(g*m2(k^-1));
    return <h,k>;
end function;

This will take $g$ as a permutation, and you want to write it as a corresponding ordered pair of matrices $h \in H$, $k \in K$;

$k$ we can get directly from the projection map $m_{3}$, k := m3(g).

Then if we multiply $g$ by (the permutation corresponding to)
$k^{-1}$, we have a new permutation $g^{\prime}$ that corresponds to
$(h,1)$; ie gprime := g*m2(k^-1);

to recover the matrix $h$ from the permutation, we can take the
inverse image under our embedding map $m_{1}$ (since this map has as
its image the set of permutations in $G$ corresponding to matrix
pairs of the form $(h,1)$ with $h \in H$); h := (m1^-1)(gprime).

(Note: you will either need to define this function at an appropriate place in your code where m1, m2, m3 have been previously defined, or else define the function as taking m1, m2, m3 as arguments.)

As an alternate way to do this, you can define a map with specified inverse to convert back and forth between permutations and matrix pairs.

phi := map<G-> CartesianProduct(H,K) | g  :-> <(m1^-1)(g*m2(m3(g)^-1)), m3(g)>,
                                       hk :-> m1(hk)*m2(hk) >;


问:
Finding parabola tangents equations given parabola equation and a point
I have parabola equation (notice it's "sideways", given x =)

$\displaystyle x = \frac{y^2}{2}-4y+3$ and a point $(2, -1)$.

Find parabola tangents (equations) that go through point $(2,-1)$.

Ive watched several videos and looked similar solved problems, however I'm getting it wrong somewhere.
Maybe it's because I cant take into account that parabola is given through $x$ and not $y$. Im not sure.
But here is what I tried to do.

I take derivative of parabola equation : $x'= y-4$
so $x'$ is the slope of tangents

I can take some point on parabola $x,y$ and replace $x$ with equation so point is $(\frac{y^2}{2}-4y+3, y)$
now I put that point into slope forumla with given point

$\displaystyle \frac{y-(-1)} {\frac{y^2}{2}-4y+3} - 2 = y - 4$

if I slove this I should get y coordinates of points where tangents touch parabola ?
However the results do not make sense and the resulting formulas I make for tangents are not correct.

There is correct answer given to this problem, which is tangents equations are either

```<code>y = −0,1202...x − .0, 7595..
or
y = −0,5940...x + 0,188
</code>```

I cant reach them, what am I missing here?

答:
Your derivative is incorrect and hence you are getting wrong answer.

Parabola equation is $x = \frac{y^2}{2}-4y+3$

you should differentiate wrt. $x$

$1 = y \frac{dy}{dx} - 4 \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{y-4}$

So equation of tangent line is

$y + 1 = \frac{1}{y_0-4} (x -2)$ ...(i)

(if the tangent goes through point $(x_0, y_0)$ on the parabola).

As the point $(x_0, y_0)$ is both on the line and the parabola,

$y_0 + 1 = \frac{1}{y_0-4} (x_0 -2) \implies y_0^2 - 3y_0 -2 = x_0$

and also $x_0 = \frac{y_0^2}{2}-4y_0+3$

Solving both,

$y_0 = \pm \sqrt 11 - 1$ gives you slopes $(\frac{1}{y_0-4}) \approx -0.59404, -0.12024$

Substituting both values in equation $(i)$, you get two tangents to parabola passing through point $(2, -1)$.


问:
find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$
find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$

My approach was
$$x^{3(x^x-1)} \lt x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$$
and if x is 5 -> $5^{3(5^5-1)}=5^{9372} \gt 1990^{1990}$

so $x\in \{1,2,3,4\}$,and I tried to check in cases when x=1, x=2,x=3,x=4 but I couldn't proceed.

答:
First note the prime factorization of $1990 = 2 \times 5 \times 199$.

Hence at least one of $x,y,z$ must be divisible by $199$, from the equation $x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$.

Ignore the order for now and suppose $199 \mid x$. Then $x^{x^y-1} \ge 199^{199^y-1}$ and $z^{z^x-1} \ge z^{z^{199}-1}$.

For $y \ge 2$, $199^{199^y-1} \ge 199^{199^2-1} \gg 1990^{1990}$ (by observing logarithms.) Hence $y=1$.

Similarly we see that for $z \ge 2$, $z^{z^{199}-1} \ge 2^{2^{199}-1} \gg 1990^{1990}$. Hence $z = 1$ as well.

This reduces the equation to

$$x^{x-1} = 1990^{1990}$$

which has no solutions, since $2,5,199 \mid x$.


答:
Substituting $t = \log x$ you obtain
$$
\lim_{x \to 1^+} \log(x)^{\log(x)} = \lim_{t \to 0^+} t^t = \lim_{t \to 0^+} e^{t \log(t)} = e^{\lim_{t \to 0^+} t \log(t)}.
$$
To compute $\lim_{t \to 0^+} t \log(t)$ use, for example, de l'Hospital rule.</p>

Title: How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$?

Body:
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ?
i know that its "1", but why?
How can i calculate this?

Thank you very very much =)


问:
Confidence interval for exponential distribution with MLE
Assume that $Y_1, Y_2,\dots, Y_n$ are iid samples from the exponential distribution with the density function $f_{\theta}(y) = \frac{1}{\theta}e^{\frac{-y}{\theta}}$. Assume that $\hat{\theta}$ is the MLE estimate for $\theta$. How can we given a approximately confidence interval for $\theta$ based on the asymptotic distribution of $\hat{\theta}$?

We know that $\hat{\theta}$ indicated above is $\frac{Y_1+Y_2+\dots+Y_n}{n}$ and I know that $Y1+Y2+\dots+Y_n$ has a $Gamma(n,\frac{1}{\theta})$ distribution. But I do not know how to relate this to the confidence interval?

答:
How can we given a approximately confidence interval...

This is done with the asymptotic distribution of $\hat{\theta}_{ML}$ that is Gaussian

Your (correct) attempt leads to the exact confidence interval

What do you wanna do?

Asymptotic CI

$\hat{\theta}=\overline{Y}_n$

It is known that $\overline{Y}_n\dot{\sim}N\Big(\theta;\frac{\theta^2}{n}\Big)$

First way to proceed

Estimate the standard deviation of  $\overline{Y}_n$ with $\frac{\overline{Y}_n}{\sqrt{n}}$ and solve the following double inequality in $\theta$

$$-z < \frac{\overline{Y}_n-\theta}{\overline{Y}_n}\sqrt{n}<z$$

2. Second method

Solve in $\theta$ the following double inequality

$$-z < \frac{\overline{Y}_n-\theta}{\theta}\sqrt{n}<z$$

I prefer the third method, the exact one your posted in your attempt (is easier, related with a $\chi^2$

In fact, as you showed $\Sigma_i Y_i\sim Gamma(n;\frac{1}{\theta})$ thus

$\frac{2n}{\theta}\overline{Y}_n\sim Gamma(n;\frac{1}{2})=\chi_{(2n)}^2$

Thus you can solve the following doubel inequality in $\theta$

$$a<\frac{2n}{\theta}\overline{Y}_n<b$$

where $a,b$ are the quantiles of the $\chi_{(2n)}^2$


问:
Convergence of $\{na_n\}$ when $a_{n+1}=f(a_n)$.
I solved if $a_0>0, a_{n+1}=\ln(1+a_n)$ then $\lim_{n \to \infty} {na_n}=2$

And, if $0<a_0<1, a_{n+1}=a_n -a^2_{n}$ then $\lim_{n \to \infty} {na_n}=1$

So, I think if real function $f$ has continuous second derivative in $[0,a]$ for positive real number $a$

and $f(0)=0,f'(0)=1,f''(x)<0$ when $x \in [0,a] $ and $0<a_0<a,a_{n+1}=f(a_n)$, then $\lim_{n \to \infty} {na_n}= \frac 2 {- f''(0)}$

I want to prove it. So I use Taylor Theorem to prove this.

First, I want to prove that there is a positive real number $m, M$ such that $m \leq na_n \leq M$ for all $n$

 But I cannot prove this. How to prove it? There is another different idea to prove this?

Body:
If $f\in C^3$, $f(0)=0$, $f'(0)=1$, $f''(0)<0$, and $a_0>0$, $a_{n+1}=f(a_n)$, then
$$
na_n\to -\frac{2}{f''(0)} \tag{1}
$$
To show that, first observe that $a_n\to 0$, and then
$$
\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{a_n-f(a_n)}{a_na_{n+1}}=\frac{a_n-\big(a_n+\frac{1}{2}f''(0)a_n^2+{\mathcal O}(a_n^3)\big)}{a_n\big(a_n+f''(0)a_n^2+{\mathcal O}(a_n^3)\big)} \\=
\frac{-\frac{1}{2}f''(0)+{\mathcal O}(a_n)}{1+\frac{1}{2}f''(0)a_n+{\mathcal O}(a_n^2)}\to -\frac{1}{2}f''(0).
$$
Stolz-Cezàro implies that
$$
\lim_{n\to\infty}\frac{\frac{1}{a_n}}{n}=\lim_{n\to\infty}\frac{\frac{1}{a_{n+1}}-\frac{1}{a_n}}{(n+1)-n}=-\frac{1}{2}f''(0),
$$
which implies $(1)$.

Note. If $\,f(0)=0$, $f'(0)=1$, $f''(0)=\cdots=f^{(k-1)}(0)=0\,$ and $f^{(k)}(0)<0$, then
$$
n^{1/(k-1)}a_n\to \left(-\frac{k!}{(k-1)f^{(k)}(0)}\right)^{1/(k-1)}.
$$
For example, if $a_0>0$ and $a_{n+1}=\sin a_n$, then $\sqrt{n}a_n\to \sqrt{3}$.