答:
(Rewriting this answer)

That's fine. The requirement for continuity
$$\lim\limits_{x\to x_0}f(x) = f(x_0)$$
implicitly means that as $x\to x_0$, it does so through values of $x\neq x_0$ that are in the domain of $f$.

So if $x_0 = 1$, all of the domain values of $f$ are the to left of $x_0$. You are therefore implicitly computing a one-sided limit, and $x\nearrow 1$.

The limit depends only on values of $f(x)$ for $x<1$. If this happens to equal the value of $f(x)$ for $x=1$, then $f$ is continuous at $x=1$.

Title: Limit Find the value of C given function

Body:
Limit Find the value of $c$ given function

Find $c$, such that the function

$$f(x) = \begin{cases}
 & {1 - \sqrt{x} \over x - 1},&0\le x < 1\\
 & c,& x= 1
\end{cases}$$

is continuous for all $x \in [0, 1 ]$.

I try to solve that question on interval $x = 0$ i get answer $-1$, with $f(0)$, and get undefined on limit so this is discontinous.

but when the $x$ is $1$ how can i solve? and How i get value of $c$?

i try the following approach

on left hand limit we get $- 1 / 2 $ but on right hand limit we have nothing!. to me this function seems to  not complete. this was the assignment i told our teacher to upload solution of this, but he didnt do that till yet.

Sorry, if it is nonsense question; i am software engineer student and on 2nd sem we're studying calculus subject.


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答:
Consider the cube roots of unity.  You can write them in the plane with $x$ being the real part and $y$ being the imaginary part.  They give three vectors all with unit length, equally spaced around the unit circle.  Now let's add the two vectors that are not 1 (ie, the two that face to the left) and you get $-1$ as shown in the second picture below (the light vectors are gone and their sum is the left-pointing vector which represents $-1$).  Of course you can add any two vectors you like, but it will always give you a vector that is the negative of the remaining one.  Adding the final two will give 0.  This happens for any number of vectors when they are a complete set of all roots of unity.

Note: This is far from a rigorous proof, but I'm trying to appeal to your intuition.  The picture isn't nearly so nice for 5-ths roots of unity.

To address your edited question, yes, $2\cos(72) + 2\cos(144) = -1$.  It's not obvious, so I don't think you're being silly.  It comes from the general formula

$$ \sum_{k=1}^n \cos \frac{2 \pi k}{n} = 0 $$

which, with a tiny amount of manipulation, gives you the formula above.  This identity gives another proof that the $n$-th roots of unity sum to 0: the $\sin$ values will cancel each other out leaving a cosine sum equivalent to the above.  The normal proof for this formula uses complex numbers, but you can also obtain a more elementary proof using trig identities.  However, I would argue that your request for an intuitive understanding is better served by the respondents using "center of mass" type arguments than this trigonometry stuff.

ContentLicense: CC BY-SA 3.0


问:
Value of coefficients of the power series when radius of convergence is "less than 1" and "greater than or equal to 1"
Let $\sum_{n=1}^\infty c_n (x-a)^n $ be a power series. As "n" approaches infinity,the value of the coefficients "$ c_n $" may or may not be 0 when Radius of convergence R is such that 0< R <1.But when R >= 1, the value of the coefficients of the power series "$ c_n $" must be "0" when "n" approaches infinity.I know this fact,but i am not able to visualize why this is so.I know its an easy one,but still i cant figure it out.
I applied the root test on the power series which gave :

$ |c_n * (x-a)^n|^{1/n} $ < 1

or

$ |c_n|^{1/n} * |x| $ < 1

or

|x| < 1/$ |c_n|^{1/n} $

Thus R = 1/$ |c_n|^{1/n} $

when 0< R <1 :

0 < 1/$ |c_n|^{1/n} $ <1

and when R >= 1 :

1/$ |c_n|^{1/n} $ >=1

I know these facts.But i am not able to derive what i asked in the question through these facts.i tried "ratio test" too,but to no avail there too.Can someone help me out to understand how to come to the conclusion.Some hint or link will also do.I was not able to find any link based on this particular fact.Thanks.

答:
Page 69 - Theorem 3.39 - Rudin - Principles of Mathematical Analysis

For any power series $\sum c_n z^n$ the radius of convergence is the following formula:

$\displaystyle R=\frac{1}{\underset{n\to\infty}{\text{limsup}}\ \sqrt[n]{|c_n|}}$

Note that limsups always exist. Also another interesting fact is that:

$\underset{n\to\infty}{\text{liminf}}\ \frac{|c_{n+1}|}{|c_n|}\le \underset{n\to\infty}{\text{liminf}}\ \sqrt[n]{|c_n|}\le \underset{n\to\infty}{\text{limsup}}\ \sqrt[n]{|c_n|}\le \underset{n\to\infty}{\text{limsup}}\ \frac{|c_{n+1}|}{|c_n|}$

This means that if the "ratio test" works then the "root test" will work as well and will have the same limit, and that the reciprocal of that limit will be the radius of convergence. In general the "root test" is better than the "ratio test".


问:
$\lVert x \rVert_1^k \leq C \left(|x_1|^k + \cdots + |x_n|^k \right)$
I'm trying to prove that given $k \in \mathbb{N}$ there exists a constant $C >0$ such that for all $x \in \mathbb{R}^n$

$$\lVert x \rVert_1^k \leq C \left(|x_1|^k + \cdots +|x_n|^k \right)$$

where

$$\lVert x \rVert_1=|x_1|+ \cdots + |x_n|$$

Any ideas?

答:
If your problem is to find a constant, it is direct, my friend.

$\Vert x\Vert_1^k \le n^k \Vert x \Vert_{\infty}^k \le n^k( |x_1|^k+...+|x_n|^k)$

ContentLicense: CC BY-SA 4.0


问:
Are the singular values of matrices in $SL_2(Z)$ dense in $(0,1)$?
Given a $2 \times 2$ real matrix $A$, we denote by $\sigma_1(A) \le \sigma_2(A)$ its singular values. If $\det A=1$, then $0<\sigma_1(A)\le 1$.

Define $X=\{\sigma_1(A) \, | \, A \in SL_2(Z)\}$. Is $X$ dense in $(0,1]$?

I know that $1 \in X$ (since $A=\text{Id} \in SL_2(Z)$), and that $0 \in \overline X$. Indeed, consider
$
A_n=\begin{pmatrix} 1 & 0 \\\ n & 1 \end{pmatrix}.
$
We have $\lim_{n \to \infty}\sigma_2(A_n)=\infty$, so $\lim_{n \to \infty}\sigma_1(A_n)=0$.

Here $SL_2(Z)$ is the group of $2 \times 2$ matrices with integer values and determinant $1$.

答:
No, the squared singular values are the roots of $X^2-Tr(AA^\top)X+1$ where $Tr(AA^\top)$ is an integer, and given $r\ne 0$ there are only finitely many values of $n$ such that $|r^2-nr+1|\le 1$, thus only finitely many values of $n$ such that $\exists s, s^2-ns+1=0$ and $|s-r|< \frac{r}{2}$.

Whence for $\sqrt{r}\ne 0$ to be in the closure of the set of singular values of all $SL_2(\Bbb{Z})$ matrices then $r$ must be a root of $X^2-nX+1$ for some $n$.


问:
Projection onto projective space is injective iff $\Bbb K=\Bbb F_2$
Consider $\Bbb K^{n+1}$ a vector space over a field $\Bbb K$ and $\Bbb P^n(\Bbb K)\colon\!=\dfrac{\Bbb K^{n+1}\setminus\{0\}}{\sim}$, where $\Bbb \sim $ is defined by: $v\sim w\Leftrightarrow \exists\,\lambda\in\Bbb K\setminus\{0\}:v=\lambda w\quad \forall v,w\in\Bbb K^{n+1}\setminus\{0\}$. I am trying to proof that the natural projection $\pi:\Bbb K^{n+1}\setminus\{0\}\to\Bbb P^n(\Bbb K)$ is injective if and only if $\Bbb K=\Bbb F_2$.

I think I have got the right to left implication, since $\pi(v)=\pi(w)\Leftrightarrow \exists\,\lambda\neq0:v=\lambda w$, and in the case $\Bbb K=\Bbb F_2$ we must have $\lambda=1$.

My problem lays with the left to right implication, any hints will be appreciated.

Thanks in advance!

答:
If you know $\pi$ is injective then for all $v,w\in\Bbb K^{n+1}$ you have $\pi(v)=\pi(w)\Rightarrow v=w$.

You know $\pi(v)=\pi(w)\Leftrightarrow \exists\, \lambda\in\Bbb K\setminus\{0\}:v=\lambda w$, so $v=w=\lambda w$ and then (try to figure it out first!)

$w-\lambda w=0\Rightarrow (1-\lambda)w=0$, and since $w\neq0$ this means $\lambda=1$.

Therefore every nonzero element $\lambda$ of $\Bbb K$ is actually $1$, so $\Bbb K=\Bbb F_2$.


问:
Length of Triangle BCD

Hey, well I'm doing some higher level revision and I'm stuck...

In the diagram triangle BCD is mathematically similar to triangle ACE.

So what is the length of BD? How do you work it out?

答: You can easily find the length of DB from $\triangle ACE\sim \triangle BCD$, the coefficient of similarity is evident from the side $CE$. On the other hand, you can't recover the length $BC$ unless you have any additional information. You may draw several triangles satisfying those lengths but with different $\angle CEA$ (and therefore, different lengths of $AC$).</p>

Title: Length of Triangle BCD


问:
Finding distribution of $Y=-1/\log X$
Having some difficulty with this particular question:
Let the pdf of $X$ be $f(x)=\lambda*x^{\lambda-1}$, $x=[0,1]$

Let $Y=-1/\log(X)$

Solving for inverse of $Y$ we get

$ \log(X)=-1/Y$ and $
X=e^{-1/Y}$

Now, $dY/dX=1/(x\log^2(X))$ $>0$ for $x$

Now for the inverse function $dX/dY=-e^{-1/Y}$

After substituting into the general formula for transformation of variables we receive

$f(y)=\lambda*e^{-1/y*(\lambda-1)}*e^{-1/y}=\lambda*e^{-\lambda/y}$, $y=[0,1]$

However the answer should be $f(y)=\lambda/y^2*e^{-\lambda/y}$ for  $y=[0,\infty]$

I have no idea at all where the $1/y^2$ comes from. Would be very grateful to receive some help.

Kind regards,

答:
$\frac{dX}{dY}=X\log^2(X)=e^{-1/Y}\left(\frac{-1}Y\right)^2=\frac{e^{-1/Y}}{Y^2}$

Or you can do $X=e^{-1/Y}\implies \frac{dX}{dY}=\frac1{Y^2}e^{-1/Y}$.

The exponent is $-1/Y,$ not $-Y$.

$Y=-1/\log X$ ranges from $0^+\to\infty$ as $X$ ranges from $0^+\to1^-$.

Title: Finding distribution of $Y=-1/\log X$


问:
Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|.$
Let $A=(a_{ij})$ be normal. Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|,$ where $\rho(.)$ is the spectral radius of $A$.

I know that $\rho(A) \leq \sqrt{\lambda_{\text{max}}(A^{^*}A)}:=\sigma_{\text{max}}(A)=||A||.$ Is there any relation between diagonal element and the eigenvalue? I know the relation with the trace but can not connect with each diagonal element.

答:
Correct me if I'm wrong. Don't we have the fact that :

$ \rho(A) = \max_{ \Vert v \Vert \le 1} \Vert \langle  v, Av \rangle\Vert $  for any normal matrix $A$ ?

If this is true, isn't your problem solved?

Disclaimer: I don't know, just asking because to me it is feasible due to the fact that $A$ is unitarily diagonalizable.


问:
finding $x^6+y^6$ given $x+y$ and $xy$
Here is the question

If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.

This is from a previous timed competition, so fastest answers are the best answers.

I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

Title: finding $x^6+y^6$ given $x+y$ and $xy$

Body:
$(x+y) ^2 =16$$\Rightarrow$$ x^2 +y^2 +2xy=16$$\Rightarrow$$ x^2 +y^2 =12$

Because $xy=2$

$x^2 +y^2 =12$ $\Rightarrow $ $ (x^2 +y^2)^3 =12^3$$\Rightarrow $$x^6+y^6+3(xy)^2(x^2 +y^2) =1728$$\Rightarrow $$x^6+y^6=1728 - 3(4)(12)=1584$

Finally :

$x^6+y^6 =1584$


答:
I have no idea at all where the $1/y^2$ comes from.

Using fundamental transformation theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

... you need also to calculate

$$\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|=e^{-\frac{1}{y}}\frac{1}{y^2}$$

Title: Finding distribution of $Y=-1/\log X$

Body:
Having some difficulty with this particular question:
Let the pdf of $X$ be $f(x)=\lambda*x^{\lambda-1}$, $x=[0,1]$

Let $Y=-1/\log(X)$

Solving for inverse of $Y$ we get

$ \log(X)=-1/Y$ and $
X=e^{-1/Y}$

Now, $dY/dX=1/(x\log^2(X))$ $>0$ for $x$

Now for the inverse function $dX/dY=-e^{-1/Y}$

After substituting into the general formula for transformation of variables we receive

$f(y)=\lambda*e^{-1/y*(\lambda-1)}*e^{-1/y}=\lambda*e^{-\lambda/y}$, $y=[0,1]$

However the answer should be $f(y)=\lambda/y^2*e^{-\lambda/y}$ for  $y=[0,\infty]$

I have no idea at all where the $1/y^2$ comes from. Would be very grateful to receive some help.

Kind regards,


问:
If $f:[0,a] \rightarrow \mathbb{R}$ is continous for all $a>0$ then $f:\mathbb{R}_+ \rightarrow \mathbb{R}$ is continuous?
Let $f:[0,a] \subset \mathbb{R}_+ \longrightarrow \mathbb{R}$ be a continuous function.

Question. If $f$ is well defined and continuous for all $a>0$, then can I assert that $f:\mathbb{R}_+ \longrightarrow  \mathbb{R}$ is also well defined and continous?

I think so, since $a>0$ is arbitrary. But I'm not sure.

答:
Assume $f \notin C(\mathbb R^+)$. Then, there exists some $x_0>0$ such that $$\lim_{x\to x_0}f(x) \neq f(x_0)$$
But since $a$ is arbitrary, we can choose $a = x_0+1$. Then, as the statement provides $f:[0, x_0+1] \to \mathbb R$ should be continuous at $x_0$. Contradiction!

Hence, our assumption was wrong and $f\in C(\mathbb{R^+})$.


问:
Probability of having the sequence (H,H,T) as a part of a 6 time coin flip.
How can I calculate the probability of having the sequence (H,H,T) as a part of a 6 time coin flip?

H - stands for heads.
T - stands for tails.

I know that $\left|\varOmega\right|=2^{6}$, and I know that possibly, if I look at this 3-tuple, I can arrange it in 4 different indexes on the general 6-tuple, then I have 3 more "open spots" which can have 2 results.

So I computed $P(A)=\frac{\left|A\right|}{\left|\varOmega\right|}=\frac{2^{3}\cdot2^{2}}{2^{6}}=\frac{2^{5}}{2^{6}}=\frac{1}{2}$ where A is the wanted probability.

However, this isn't the right answer, because I count some events twice. How can I calculate this properly?

答:
The only outcome you are counting twice is HHTHHT. So $P(A) = \frac{2^5-1}{2^6}$.

To see this, note that the possibilities are as shown in the following table:

$$
\begin{array}{rcccccc}
1)&H&H&T&*&*&*\\\
2)&*&H&H&T&*&*\\
3)&*&*&H&H&T&*\\
4)&*&*&*&H&H&T
\end{array}
$$

Clearly you can't fit another HHT in rows 2) and 3) and there is exactly one way of inserting another HHT into row 1) and row 4). So if you count the possibilities as $4\cdot 2^3$ you have counted HHTHHT twice.


问:
$E[X]$ finite iff $\sum\limits_{n} P(X>an)$ converges
Show that:

$$\sum\limits_{n \in N } P(X>an) < \infty\ \text{for some}\ a > 0 \Rightarrow E[X] < \infty \Rightarrow \sum\limits_{n \in N } P(X>an) < \infty\ \text{for every}\ a > 0.$$

We have an unknown probability space ($\Omega,\mathbb{F},P)$, where $\mathbb F$ is a sigma-algebra, and $P:\mathbb F \rightarrow [0,1]$ is a probability measure, $X$ is a $\mathbb{F}/\mathbb{B}(\mathbb{R})$-measurable random variable and $E[X]$ is the expectation of the variable $X$.

I've been working through past exam papers, and am stuck on this. Would appreciate a solution!

答:
Here's a hint: for $X>0$,
$$E[X] = \int_0^\infty x f(x) \mathrm d x = \sum_{n=0}^\infty \int_{na}^{(n+1)a} x f(x) \mathrm d x$$

Can you bound these integrals in terms of integrals of $f$? What can you do with the factors?


问:
How to formally write "Let $X$ take on the value of TRUTH"?
I have to argue that a given logical expression is satisfiable and not universally applicable. This is quite easy because all I have to do is find for which values is the expression true and false. However, I want to know how to write the solution formally. For example, let's say that we have the following expression:

$$\alpha_i := X \lor \lnot X$$

Due to the the laws of contradiction, we know that the expression is always true. But if I want to say that: "Let $X=T$ , $T$ is shorthand for TRUTH". How can I let $X$ take on the value of TRUTH?

This might be badly explained, but essentially I want to know how to formally write "let $X$ take on the value of TRUTH".

答:
If you want to assume that the formula takes the value $True$ under some valuation, but may not be always true, write "Let $V(X) := True$", or "$[\![X]\!]_V := True$", where $V$ is the valuation function assigning truth values to formulas. (Depending on your textbook, it may call and symbolize that function differently, but it surely has been introduced somewhere.)

Then, after calculating the truth values of the compound formulas, you can say something like "With $V_1$ such that $V_1(X) = 1$, there exists a valuation under which the formula $X \land Y \to Z$ is true, so the formula is satisfiable, but under $V_2$ with $V_2(X) = \ldots$ the formula is false, so it is not universally valid".

If you want to say that $X$ is a formula which is tautological, i.e. logically equivalent to a statement of the form $A \lor \neg A$, then you can use the symbol $\top$ and write "Let $X := \top$". $\top$ is an atomic formula which always evaluates to $True$.


问:
Find the smallest $x$ for $9x \equiv 3 \pmod {23}$
$9x \equiv 3 \pmod {23}$

How to derive the smallest $x$. I understand I can use the extended euclidean algorithm for eg $19x = 1 \pmod {35}$.

However, I not too sure how to work on it when it is $3 \pmod {23}$.

I am able to reach the step of $1 = -5(9) + 2(23$) after going thru the euclidean algorithm.

答:
We employ an algorithm that was discussed here.

$\alpha-$Solve:

$\;9x \equiv 3 \pmod{23}, \text{ and } \; 23 = 9 \cdot 2 + 5,\quad -3 + 1 \cdot 9 = 6$

$\alpha-$Solve:

$\;5x \equiv 6 \pmod{9}, \text{ and } \; 9 = 5 \cdot 1 + 4,\quad -6 + 2\cdot 5 = 4$

$\alpha-$Solve:

$\;4x \equiv 4 \pmod{5},  \text{ ANS: } \; x = 1 \text{ is the least residue solution}.$

Propagating backward,

Solve $5x = 6 + 1 \cdot 9, \text{ ANS: } \; x = 3$

Solve $9x = 3 + 3 \cdot 23, \text{ ANS: } \; x = 8$

$\text{ANSWER: }  x = 8 \text{ is the least natural number satisfying }$

$\quad 9x \equiv 3 \pmod{23}$


问:
Is this map an homeomorphism with its image?
Suppose $\varphi:\mathbb{R}^2 \to \mathbb{R}^3$ is defined as:

$$\varphi(u,v)=(u-\frac{u^3}{3}+uv^2,v-\frac{v^3}{3}+vu^2,u^2-v^2)$$

Can I conclude that $\varphi$ is an homeomorphism with its image?

I tried defining $x=u-\frac{u^3}{3}+uv^2$, $y=v-\frac{v^3}{3}+vu^2$ and $z=u^2-v^2$ finding $u=\frac{3x}{3-z}$ and $v=\frac{3y}{1+z}$ but I do not know if this is enough to conclude.

答: Considering the plot it doesn't seem so:


问:
Ways of making 10 letter combinations out of 3 given letters
Find the number of 10 letter combinations from ${a, b, c}$ that contain:

At least one of each letter

At least two of each letter

At least one A, two Bs, three Cs

Any number of each letter

(Also repetition is allowed)

For number 1 I think that the answer is $\binom{9}{2}$, since there's n-1 places that I can place a bar from stars&bars.
For number 2 though I'm not sure how I can apply this in order to place the bars so that I get at least 2 of each element $a,b,c$.

Edit 2: I did some thinking about question 2 and I think I might have an answer which is correct. Since I need at least 2 of each letter I decided to group 1 of $AA, BB, CC$ so that $AA$ was one object meaning that now in total I had 7 objects. Which then means that I would have $\binom{6}{2}$ places to divide this group up. Please let me know if this approach is correct. Or if I've made a mistake. Thank you.

答:
For (1), start with "abc" so that you have fulfilled  the "at least one of each letter" requirement.  You have 7 letters that can be any of the three "a", "b", or "c" so there are 3^7 possible choices.  For each such 10 letter combination there are 10! possible arrangements of the letters.  There are 10!(3^7) such combinations.

For (2), start with "aabbcc".  There are then 4 letters left so 3^4 possible choices.  Since there are again 10! different arrangements, there are 10!(3^4) such combinations.

For (3), start with "abbccc" and finish in the same way.

For (4), with just 10 letters chosen from "a", "b", and "c", there are 3^10 such combinations.


答:
Try to integrate the pointwise inequality $X\leqslant a\sum\limits_{n=0}^{+\infty}\mathbf 1_{X\gt an}$ to get one implication. For the other implication, try to find a similar pointwise inequality, only reversed.

Edit: Consider $a\sum\limits_{n=1}^{+\infty}\mathbf 1_{X\gt an}$ (note that the sum is starting at $n=1$, not $n=0$), can you compare it to $X$?

Title: $E[X]$ finite iff $\sum\limits_{n} P(X>an)$ converges

Body:
Show that:

$$\sum\limits_{n \in N } P(X>an) < \infty\ \text{for some}\ a > 0 \Rightarrow E[X] < \infty \Rightarrow \sum\limits_{n \in N } P(X>an) < \infty\ \text{for every}\ a > 0.$$

We have an unknown probability space ($\Omega,\mathbb{F},P)$, where $\mathbb F$ is a sigma-algebra, and $P:\mathbb F \rightarrow [0,1]$ is a probability measure, $X$ is a $\mathbb{F}/\mathbb{B}(\mathbb{R})$-measurable random variable and $E[X]$ is the expectation of the variable $X$.

I've been working through past exam papers, and am stuck on this. Would appreciate a solution!


问:
Why is it not sufficient to only check the third condition when verifying equality of functions?
I have been told that two functions $f$ and $g$ are equal if and only if the domain and the subset of the cartesian product of the two functions is the same. My question is, given that a function is a special case of a relation, both $f$ and $g$ are sets, why is it not sufficient to just verify if the cartesian product is the same between the two functions?. In other words, doesn't the third condition implies that both domains are equal?

答:
In foundations of mathematics, especially set-theoretic foundations, a function is nothing more than a certain kind of set of ordered pairs, from which you can easily extract its domain and its range, and there is no such thing as a codomain of a function. Two functions are equal exactly when they are the same set, which is equivalent to when they have the same domain and the same output on every input from the domain.

This behaviour is very important when you actually want to build foundations of mathematics, especially in a proper proof of a basic recursion theorem, and for a proper proof of the full recursion theorem. In such proofs, you glue together approximations of a desired function by simply taking the union, and there is no need nor benefit in requiring functions to have "codomains" in such foundations.

So if you really want to understand how everything is built based on ZFC, indeed for two functions to be equal it suffices for them to be the same set of ordered pairs, as it implies that their domains are equal.

Concerning the arguments people are having in other comments, I will just say that it is actually standard terminology to say "$f$ surjects onto $T$" when one wants to be consistent with set-theoretic foundations, since "$f$ is a surjection" makes no sense in this setting. Foundationally, "$f : S→T$" simply means that $f$ is a function with domain $S$ whose range is a subset of $T$.

In practice people who do not have a background in foundations of mathematics tend to mean not just that but also that $f$ is 'tagged' with a codomain $T$. Foundationally, we would need to represent such a 'tagged function' by something like a pair $(f,T)$. Then again, I would say that even these people are often inconsistent with their notation, because they frequently define functions without specifying a codomain, and also they consider any function from $ℕ$ to $ℝ^+$ as also a function from $ℕ$ to $ℝ$.


问:
Does inequality sign remain same?
Let say we have inequality $y< x^{-1}$. We take log on both side And we get
[log( y ).... -1 \log(x)].
What will be the inequality sign in between. Will it remain same or flip ?

答: Assuming $0<y<x^{-1}$, applying $\log$ to both sides does not change the inequality because the function $\log$ on $(0,+\infty)$ is a monotonically (strictly) increasing function. So it stays: $\log y<\log x^{-1}=-\log x$.

Title: Does inequality sign remain same?


问:
Spivak Calculus Chapter 4, Appendix 3. Polar Coordinates (Comprehension Question)
This appendix is for the most part very straightforward. Nevertheless, there exists one sentence that really confuses me (marked bold in the quotation).

Spivak Calculus, page 84:

...In many situations it is more convenient to introduce polar coordinates, which are illustrated in Figure 1.

To the point $P$ we assign the polar coordinates $(r,\theta)$, where $r$ is the distance from the origin $O$ to $P$, and $\theta$ is the angle between the horizontal axis and the line from $O$ to $P$. This angle can be measured either in degrees or in radians (Chapter 15), but in either case $\theta$ is not determined unambiguously. For example, with degree measurement, points on the right side of the horizontal axis could have either $\theta=0$ or $\theta=360$; moreover $\theta$ is completely ambiguous at the origin $O$. So it is necessary to exclude some ray through the origin if we want to assign a unique pair $(r,\theta)$ to each point under consideration.

What is the meaning of the marked sentence? What ray is he talking about?

I think this is supposed to be an example of one ray (that contains all points on the right side of the horizontal axis and the origin) which cannot have a unique pair $(r,\theta)$ assigned to any of its points. And by rotating this ray it follows that no point in the plane can have a unique pair $(r,\theta)$ assigned to it, but isn't this already obvious at this point? I'm not so sure...

答:
I think this is supposed to be an example of one ray (that contains all points on the right side of the horizontal axis and the origin) which cannot have a unique pair (,) assigned to any of its points.

Yes, I think this is what he means. I was confused by this paragraph as well.

He doesn't really say it, but of course our choice of $\theta$ is somewhat arbitrary, so long as it covers the full 360°-$2\pi$ radians.

We can use $\theta$'s defined by
$$0 \leq \theta \leq 360$$
or
$$73 \leq \theta \leq 433$$
or
$$-\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}$$

(Analogously, we could choose any ray through the origin against which we measure $\theta$. The choice to use the positive $x$-axis is somewhat arbitrary.)

In general we can use

$$\theta_0 \leq \theta \leq\theta_0 + 360$$

where $\theta_0$ is any initial angle.

I think what he's getting at with his "exclude some ray through the origin"
is that we exclude all points along the ray defined by $(r,\theta_0)$ where $r \geq 0$.

For any choice of $\theta_0$, we exclude the points $(r,\theta_0)$, along with the origin, leaving us only with points that can be unambiguously assigned to unique $(r, \theta)$, where $\theta_0 < \theta < \theta_0 +360$ and $r > 0$.

Analogously, We can chose to measure our $\theta$ against the positive $y$-axis, or the ray that's 76° counterclockwise from the negative $x$-axis, or any other ray through the origin, but regardless of which one we chose we must exclude points along this ray including the origin if we want to be able to assign unique $(r, \theta)$ coordinates.

This of course is not really satisfactory and will leave us with points we can't describe regardless of our choice of $\theta_0$

We could instead, restrict our $\theta$ to either

$$\theta_0 < \theta \leq \theta_0 + 360$$

or

$$\theta_0 \leq \theta < \theta_0 + 360$$

and define the origin point as having some fixed $\theta$, for example $\theta_0$ or $\theta_0 + 360$, depending on our choice of allowed $\theta$. We can easily avoid the ambiguity Spivak's worried about, without excluding any rays through the origin.

Fortunately, it's not really important. He goes on to say that, despite any difficulties in picking points on the plane and then assigning to them unique $(r, \theta)$ coordinates, going the other way is no problem at all.

Given any $(r, \theta)$, this maps to a unique $(x,y)$ in the plane, and the rest of the chapter involves this process.


问:
how to prove $\nabla_AtrABA^T=AB+AB^T$
how do we prove $\nabla_Atr(ABA^T)=AB+AB^T$? where B is square, and $\nabla_A$ is the derivative according to every element in A.
I'm getting lost with all the matrix indices... is there a neat way to show this? even an intuitive explanation of why this is true will do for me

答:
$\def\p{\partial}$
First note the behavior of the trace function under transposition
$$\eqalign{
f = {\rm Tr}(XBY^T) &= {\rm Tr}\Big((XBY^T)^T\Big) = {\rm Tr}(YB^TX^T) \\
}$$
Next calculate the gradient of the function with respect to $X$ and $Y$
$$\eqalign{
\frac{\p f}{\p X} &= YB^T,\qquad
\frac{\p f}{\p Y} &= XB \\
}$$
Finally, set $\,X=Y=A\,$ and use the chain rule
$$\eqalign{
\frac{df}{dA}
 &= \left(\frac{\p f}{\p X}\right)\left(\frac{\p X}{\p A}\right)
  + \left(\frac{\p f}{\p Y}\right)\left(\frac{\p Y}{\p A}\right) \\
 &= AB^T + AB \\
}$$

The above derivation assumes the following result is already known
$$\eqalign{
\frac{\p \,{\rm Tr}(BY^T)}{\p Y} &= B \\
}$$

Title: how to prove $\nabla_AtrABA^T=AB+AB^T$


问:
How to find $x$ of this equation?
I'm doing some equations and I'm not sure how to find $x$. Any idea? Step by step answers please, it's like my first time.

$$2(3x^2+x-5)(2x^2-7x+3)=0.$$

Title: How to find $x$ of this equation?

Body:
$$2(3x^2+x-5)(2x^2-7x+3)=0$$
so either $(3x^2+x-5)=0$ or $(2x^2-7x+3)=0$
so take it one by one.

part1

formula is for: $(ax^2+bx-c)=0$ ,$x=\dfrac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ ,
$\dfrac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$
$$(3x^2+x-5)=0$$

$$x=\frac{-1+\sqrt{1^2-4\cdot3\cdot(-5)}}{2\cdot3}$$
$$x=\frac{-1+\sqrt {1+60}}{6}$$
$$x=\frac{-1+\sqrt {61}}{6}$$
and also
$$x=\frac{-1-\sqrt {61}}{6}$$

part2

$$(2x^2-7x+3)=0$$
from different method
$$(2x^2-6x-x+3)=0$$
$$2x(x-3)-1(x-3)=0$$
$$(x-3)(2x-1)=0$$
$$(x-3)=0 ,(2x-1)=0$$
so $x=3,x=\frac12$
so this euqtion have 4 roots $$x=\frac{-1+\sqrt {61}}{6},\frac{-1-\sqrt {61}}{6},3,\frac12$$


答: For the transformation to be univocal, you must decide if the points on axis $x$ (non-negative side) have $\theta=0$ or $2\pi$ radians.

Title: Spivak Calculus Chapter 4, Appendix 3. Polar Coordinates (Comprehension Question)

Body:
This appendix is for the most part very straightforward. Nevertheless, there exists one sentence that really confuses me (marked bold in the quotation).

Spivak Calculus, page 84:

...In many situations it is more convenient to introduce polar coordinates, which are illustrated in Figure 1.

To the point $P$ we assign the polar coordinates $(r,\theta)$, where $r$ is the distance from the origin $O$ to $P$, and $\theta$ is the angle between the horizontal axis and the line from $O$ to $P$. This angle can be measured either in degrees or in radians (Chapter 15), but in either case $\theta$ is not determined unambiguously. For example, with degree measurement, points on the right side of the horizontal axis could have either $\theta=0$ or $\theta=360$; moreover $\theta$ is completely ambiguous at the origin $O$. So it is necessary to exclude some ray through the origin if we want to assign a unique pair $(r,\theta)$ to each point under consideration.

What is the meaning of the marked sentence? What ray is he talking about?

I think this is supposed to be an example of one ray (that contains all points on the right side of the horizontal axis and the origin) which cannot have a unique pair $(r,\theta)$ assigned to any of its points. And by rotating this ray it follows that no point in the plane can have a unique pair $(r,\theta)$ assigned to it, but isn't this already obvious at this point? I'm not so sure...


问:
First Isomorphism Theorem - hole in the proof from my book?
I have a book that contains a proof for their so-called "First Isomorphism Theorem", which esentially states that if $f$ is a surjective group morphism $f:G\to G'$ and $H$ is a normal subgroup of $G$, then $G/H$ is isomorphic to $G'/f(H)$.

Their proof goes like this:

They consider $\pi: G' \to G'/f(H)$ to be the canonical map, $\pi(x) = xf(H)$, which is surjective. Then they take $f':G\to G'/f(H), \ f'=\pi \circ f$. Since $f$ and $\pi$ are surjective, $f'$ is also surjective. Thus, by a previously proven theorem in the book (called the 'Fundamental Isomorphism Theorem'), we have that $G/\ker f'$ is isomorphic to $G'/f(H)$.

Now they go on to prove that $\ker f' = H$ in order to finalize the proof.

The following sequence of equivalences holds:

$x \in \ker f' \Leftrightarrow \pi(f(x)) = 1 \Leftrightarrow f(x) \in \ker \pi = f(H)$. (the $1$ here is the unity for $G'/f(H)$, so $f(H)$).

So now we have $f(\ker f') =f(H)$ and now comes the part where I get stuck. They now call on a previously proven theorem (called the Correspondence Theorem for Normal Subgroups), which states that if we have a surjective group morphism $f:G\to G'$, then the map $H\to f(H)$ from the set of normal subgroups of $G$ that contain $\ker f$ to the set of normal subgroups of $G'$ is bijective), but this does not look correct as we have no knowledge of whether $H$ actually contains $\ker f$ ($\ker f'$ does).

I also thought of $\pi$'ing the equality and instead using $f'(\ker f') = f'(H)$, but we don't know that $\ker f'$ belongs to $H$ either).

**EDIT: ** Source of the faulty theorem: The Romanian book "Bazele Algebrei, vol. I" (English would probably be "Fundamentals of Algebra) by C. Năstăsecu, C. Niță and C. Vraciu, Bucharest, 1986. The page is 54 and the theorem I mentioned is Proposition 3.19.

答:
It looks like the authors forgot to add the assumption that $H$ contains $\ker (f)$ to the statement of the result.

One can find many cases where the assertion fails without an assumption like this. For example let $G$ be $\mathbb{Z}$ and let $H$ be the trivial subgroup. For any $n\geq 1$, we have a surjective homomorphism $f$ from $\mathbb{Z}$ to $G'=\mathbb{Z}/n\mathbb{Z}$, and $f(H)$ is still the trivial subgroup. In particular, $G/H$ and $G'/f(H)$ are not isomorphic.

Taking $n=1$ in the previous situation shows how drastic the error is. Indeed, given any nontrivial group $G$, consider the surjective homomorphism $f:G\to G'$, where $G'$ is trivial group. If $H$ is the trivial subgroup of $G$, then $G/H=G$ and $G'/f(H)$ is trivial.

More generally if this formulation of the First Isomorphism Theorem were true then it would imply that anytime there is a surjective homomorphism from $G$ to $G'$, then $G$ and $G'$ are isomorphic, which is absurd.

I might also suggest you edit your question to add the name of the book and page number. This could help future users who run into the same problem.


答:
Let frequency of $a,b,c$ be $x,y,z$ respectively. You're looking for integral solutions of following cases :

$x \ge 1, y \ge 1, z \ge 1$ and $x+y+z=10$

$x \ge 2, y \ge 2, z \ge 2$ and $x+y+z=10$

This is same as $u \ge 1, v \ge 1, w \ge 1$ and $u+v+w=7$

$x \ge 1, y \ge 2, z \ge 3$ and $x+y+z=10$

This is same as $x \ge 1, p \ge 1, q \ge 1$ and $x+p+q=7$

$x \ge 0, y \ge 0, z \ge 0$ and $x+y+z=10$

Hence answers would be $\binom{9}{2}, \binom{6}{2},\binom{6}{2}, \binom{12}{2}$ respectively.

Title: Ways of making 10 letter combinations out of 3 given letters

Body:
Find the number of 10 letter combinations from ${a, b, c}$ that contain:

At least one of each letter

At least two of each letter

At least one A, two Bs, three Cs

Any number of each letter

(Also repetition is allowed)

For number 1 I think that the answer is $\binom{9}{2}$, since there's n-1 places that I can place a bar from stars&bars.
For number 2 though I'm not sure how I can apply this in order to place the bars so that I get at least 2 of each element $a,b,c$.

Edit 2: I did some thinking about question 2 and I think I might have an answer which is correct. Since I need at least 2 of each letter I decided to group 1 of $AA, BB, CC$ so that $AA$ was one object meaning that now in total I had 7 objects. Which then means that I would have $\binom{6}{2}$ places to divide this group up. Please let me know if this approach is correct. Or if I've made a mistake. Thank you.


问:
Coloring Two Faces of an Icosahedron
Suppose that you have a solid regular icosahedron (a polytope with 20 sides all of which are
equilateral triangles), and all the sides are white.

(a) In how many ways can exactly two sides be painted red?

(b) In how many ways can one side be painted red and another one side painted blue?

Two colorings considered the same if one can be converted into the other by
rotating the icosahedra in space.

I tried doing it by mapping which sides connected to others, but it didn't seem to help. Any help would be greatly appreciated!

答:
As Christian Blatter said in the comments, coloring faces of the regular icosahedron is the same as coloring vertices of its dual, the regular dodecahedron.

Now, the regular dodecahedron $D$ is a so-called distance-transitive polytope (as all the other Platonic solids too). This means, that for any two pairs $(v,w),(v',w')$ of vertice with pair-wise identical graph-theoretical distance along the edge-graph (that is, $\mathrm{dist}(v,w)=\mathrm{dist}(v',w')$), there is a symmetry of $D$ that maps $v$ to $v'$ and $w$ to $w'$.

In other words, for any $\delta\in\Bbb N$, there is, up to symmetry, at most one pair $(v,w)$ of vertices (same or different color, does not matter) with distance $\delta$.
And since the edge-graph of the dodecahedron has diameter five, $\delta$ can take on the values $1,2,3,4$ and $5$.

So the answer to (a) and (b) is five.

As a bonus, you got an answer for all the Platonic solids: the answer to (a) and (b) is always the diameter of the edge-graph of its dual. And the same holds for all regular polytopes in all dimensions, excluding the 4-dimensional exceptions (which are not distance-transitive).

Title: Coloring Two Faces of an Icosahedron


问:
The sum of dihedral angles of an $n-$hedral angle is bigger than $(n-2)\pi$
Given a polyhedral angle $O.A_1A_2\dots A_n$ (where $O$ is the vertex), I need to show that the sum of dihedral angles with sides $OA_1,\dots,OA_n$ is bigger than $(n-2)\pi$.

My try is to draw a plane meeting $OA_i$ at $B_i$ and show that the sum of dihedral angles is bigger than the sum of the angles of polygon $B_1B_2\dots B_n$. But it seems that each dihedral angle is not bigger than the corresponding angle of the polygon and I do not know how to continue.

Can someone help me? Thanks a lot!

答: Thanks to the hint of @timon92, we only need to show that $\sum_{i=1}^n \angle P_iPP_{i+1}<2\pi$. Without loss of generality, one can assume that $PP_1=PP_2=\cdots=PP_n$. Let $Q$ be the perpendicular projection of $P$ on the plane $P_1P_2\dots P_n$ then $QP_1=QP_2=\cdots QP_n$. Consider two isosceles triangles $PP_1P_2$ and $QP_1P_2$: they have common side $P_1P_2$ and $PP_1>QP_1$, thus $\angle P_1PP_2<\angle P_1QP_2$. Similarly for the remaining angles, we get $$\sum_{i=1}^n \angle P_iPP_{i+1}<\sum_{i=1}^n \angle P_iQP_{i+1}=2\pi.$$


问:
Which values of $a$ and $b$ will give $x^{2}+2$ as a factor of $x^{17}+ax+b$ over $\mathbb{Z}_{3}$?
The question:

Let $x^{17}+ax+b$ be a polynomial over $\mathbb{Z}_{3}$. For which values of $a$ and $b$ will $x^{2}+2$ be a factor?

I know that I can find a solution by a brute force technique, but is there a more efficient way to solve this problem? Thanks.

答:
Here's the efficient method :-)

In $\Bbb{Z}_3$ we have $2=-1$, so $x^2+2=x^2-1$. This is clearly a factor of $x^{2n}-1$ for all positive integers $n$. In particular $x^2-1\mid x^{16}-1$. Consequently
$$x^2+2\mid x^{17}-x.$$
We can deduce that the remainder of $x^{17}+ax+b$ in long division by $x^2+2$ is thus
$$
r(x)=(a+1)x+b.
$$
Obviously this vanishes if and only if $a=-1=2$ and $b=0$.


问:
What are the pros and cons of natural deduction relative to Hilbert-style systems?
What are the pros and cons of natural deduction relative to Hilbert-style systems?

From Wikipedia, I get the impression that natural deduction proofs tend to be shorter and closer to how humans do it. If so, why do some proof-checkers, like Metamath, use Hilbert-style systems?

答:
Indeed natural deduction (and in particular Fitch-style) is simpler for humans to use. That does not imply that proof assistants would use what is simpler for humans... In fact, Coq is widely known as a "write-only system", meaning that you can write Coq proofs but (typically) cannot read them. In any case, you simply have to actually sit down and look at a proof of a theorem of some complexity (say of Zorn's lemma) from the axioms alone (say ZFC), in various formal system styles, to see that pure Hilbert-style proofs are much more difficult to read and understand than Fitch-style proofs or tree-style proofs (as in Gentzen's natural deduction system) or sequent-style proofs (as in Gentzen's LK or similar). The advantage of Fitch-style over tree-style or sequent-style is that it is very similar to programming style, and can easily be represented in indented text format, whereas representing tree-style or sequent-style in text format tends to make it become like Fitch-style in the end.

As you should already know, the advantage of Hilbert-style is that there is only one inference rule, so of course it is (slightly) easier to write a proof checker for Hilbert-style than for Fitch-style. And of course theorems about FOL are somewhat easier if proofs are defined as Hilbert-style proofs. It is nevertheless easy to translate any Fitch-style system into a Hilbert-style system, so that is not a big deal actually.


问:
asymptotic behavior of the solution to an ODE
Given

$$y(t) = \frac{d_2 y_0 e^{d_2 t/\epsilon}}{d_2-\epsilon \, d_1 y_0 (e^{d_2 t/\epsilon}-1)}$$

I think that $y = O(1/\epsilon) $ as $\epsilon \to 0$. But as this is important for what I am doing and I am not used to asymptotic I wanted to double-check. Thx!

答:
(I'm assuming that $\epsilon > 0$.)

If $y_0 > 0$ then $y(t)$ has a singularity at

$$
T(\epsilon) =\frac{\epsilon}{d_2} \log\left(\frac{d_2+\epsilon d_1 y_0}{\epsilon d_1 y_0}\right).
$$

For any fixed $t > 0$ it will eventually be true that $T(\epsilon) < t$ for $\epsilon$ small enough.  That is, each fixed $t > 0$ will eventually leave the maximal interval of existence as $\epsilon \to 0^+$.  If your application requires that you only consider $t$ in the maximal interval of existence then the asymptotics for fixed $t$ will be useless to you.

Note that there is no such singularity if $y_0 < 0$.

Asymptotics for fixed $t > 0$.

We have

$$
\begin{align}
y(t) &= \frac{d_2 y_0 e^{d_2 t/\epsilon}}{d_2-\epsilon \, d_1 y_0 (e^{d_2 t/\epsilon}-1)} \\
&= \frac{d_2}{\epsilon d_1} \cdot \frac{1}{\epsilon^{-1} d_1^{-1} d_2 y_0^{-1} e^{-d_2 t/\epsilon} - 1 + e^{-d_2 t/\epsilon}} \\
&\sim -\frac{d_2}{\epsilon d_1}
\end{align}
$$

as $\epsilon \to 0^+$ since

$$
\lim_{\epsilon \to 0^+} \frac{1}{\epsilon^{-1} d_1^{-1} d_2 y_0^{-1} e^{-d_2 t/\epsilon} - 1 + e^{-d_2 t/\epsilon}} = -1.
$$

Asymptotics in a subinterval of $[0,T(\epsilon))$ when $y_0 > 0$.

The singularity at $t=T(\epsilon)$ precludes the existence of an $O(f(\epsilon))$ bound which holds uniformly with respect to $t$ in the interval $[0,T(\epsilon))$, so if you're interested in such a bound then you must restrict $t$ to some proper subinterval of $[0,T(\epsilon))$.

One way to do this is to fix some $0 < \alpha < 1$ and consider $t \in [0,\alpha T(\epsilon)]$.

Since $y$ is increasing on $[0,T(\epsilon))$ we have

$$
y_0 \leq y(t) \leq y(\alpha T(\epsilon)) = \frac{1}{(\epsilon d_1 y_0/d_2)^{\alpha}(d_1 \epsilon+d_2/y_0)(1+\epsilon d_1 y_0/d_2)^{-\alpha} - d_1 \epsilon}.
$$

Since $\alpha < 1$ we have

$$
\frac{1}{(\epsilon d_1 y_0/d_2)^{\alpha}(d_1 \epsilon+d_2/y_0)(1+\epsilon d_1 y_0/d_2)^{-\alpha} - d_1 \epsilon} \sim \frac{y_0}{d_2}\left(\frac{d_2}{\epsilon d_1 y_0}\right)^{\alpha} = O(\epsilon^{-\alpha}),
$$

so that $y(t) = O(\epsilon^{-\alpha})$ as $\epsilon \to 0^+$ uniformly for $t \in [0,\alpha T(\epsilon)]$.

Uniform asymptotics when $y_0 < 0$.

The function $y(t)$ will be decreasing when $\epsilon < -\frac{d_2}{d_1 y_0}$, and since we're taking $\epsilon \to 0^+$ we'll assume that this is indeed the case.  As such we have

$$
y_0 \geq y(t) \geq y(+\infty) = - \frac{d_2}{\epsilon d_1}
$$

so that $y(t) = O(\epsilon^{-1})$ uniformly for $t>0$.

In summary,

For each fixed $t \geq 0$ we have $y(t) = O(\epsilon^{-1})$ as $\epsilon \to 0^+$.

If $y_0 > 0$ and $0 <\alpha < 1$ then $y(t) = O(\epsilon^{-\alpha})$ as $\epsilon \to 0^+$ uniformly with respect to $t$ on $[0,\alpha T(\epsilon)]$.

If $y_0 < 0$ then $y(t) = O(\epsilon^{-1})$ as $\epsilon \to 0^+$ uniformly with respect to $t$ on $\mathbb R^+$.


问:
find the probability that more than 50 of the observations of the random sample are less than 3
Let X1,X2,...,Xn be a random sample of size n=72 from a distribution with pdf

f(x)={1/x2    1<x<infinity
0      otherwise

Find the probability that more than 50 of the observations of the random sample are <3 (less than 3).

答:
First calculate the probability that a single rv is less than 3, i.e.

$$\mathbb{P}[X_1<3]=p$$

Then use CLT to find the requested probability to get at least 51 successes on 72 independent trials


问:
Is a diagonalizable matrix also injective?
This question seems to be very basic, but I couldn't find the answer on SE.

I want to know if a diagonalizable matrix $A$, i.e. $A = SDS^{-1}$, is also an injective matrix, i.e. $Av = 0$ if and only if $v = 0$ or also $\ker A = \{ 0 \}$.

I believe it can be simply proven by

"a composition of injective functions is also injective" and

"$S, S^{-1}$ are injective because they are invertible and $D$ is injective because it is diagonal",

but I want to know if I am correct.

I'm not sure about the injectiveness of $D$. If this is the dealbreaker, what are criteria for the injectiveness of $D$ proceeding from the properties of $A$?

I am asking because this would imply, that the only fixed point of a dynamical system
$ \dot{\vec{x}} = A \vec{x}$ with a diagonalizable matrix $A$ would be only $\vec x = 0$.

答:
No, (the linear transformations given by) diagonal matrices are not injective in general, since a diagonal matrix can have zero entries on the diagonal (and a diagonalizable matrix can have $0$ as an eigenvalue). In fact, being diagonal or diagonalizable is irrelevant when it comes to being injective.

A square matrix describes an injective map exactly when $0$ is not an eigenvalue, or, equivalently, exactly when the determinant is nonzero. I don't think there is any simpler criterion.


问:
Product homomorphism
Let $(G,+)$ and $(H,\star)$ be groups (with respectively noted relations $+$ and $\star$). Let $f$ and $g$ be group homomorphisms $f,g:G \to H$.

Claim: If $H$ is abelian, then the product map $f \cdot g,\,x \mapsto f(x) \star g(x)$ is also a homomorphism.

I was able to prove this, but now I am wondering if there exists a concrete but simple counterexample for when $H$ is not abelian... (e.g. with whole numbers under addition or similar...)

答:
Well, you could consider the left multiplications $\ell_\pi:S_n\rightarrow S_n:\sigma\mapsto\pi\sigma$ with given $\pi\in S_n$, where $S_n$ is the symmetric group on $n$ places.

The set of left multiplications $L(S_n)$ forms a group under function composition, since
$\ell_\pi\ell_\sigma=\ell_{\pi\sigma}$, $(\ell_\pi)^{-1} = \ell_{\pi^{-1}}$ and $\ell_e= id$ is the identity mapping, where $e$ is the identity permutation.

Consider the mapping $F:S_n \rightarrow L(S_n):\pi\mapsto \ell_\pi$, which is a homomorphism, since $F(\pi\sigma) =\ell_{\pi\sigma} = \ell_\pi\ell_\sigma = F(\pi)F(\sigma)$.

Title: Product homomorphism


问:
Find the least positive integer $x$ that satisfies the congruence $90x\equiv 41\pmod {73}$

Find the least positive integer $x$ that satisfies the congruence $90x\equiv 41\pmod{73}$.

It is clear that an attempt to write this out as $90x-41=73n,\exists n\in \mathbb{Z}$ won't be very efficient. Is there another way of solving this problem? Could I receive a bit of guidance?

EDIT: I followed anorton's advice as follows:

To find $90^{-1}\pmod{73}$, I use the Euclidean Algorithm to find $s$ and $t$ such that $90s+73t=1$. If my calculations are correct, the values $s=-30$ and $t=37$ work.

This means that $(90)( -30)\equiv 1 \pmod{73}$.

Now if I just multiply $-30$ by $41$, I will get

$$(90)(-30)(41)\equiv 41 \pmod{73}$$

$$(90)(-1230)\equiv 41 \pmod{73} \implies x=-1230$$

But $x$ isn't positive nor the smallest possible in magnitude, so we compute $-1230\pmod{73}$, which is $11$ I believe.

Does this look alright?

答:
We employ an algorithm that was discussed here.

To start the algorithm observe that $90 \equiv 17 \pmod{73}$

$\alpha-$Solve:

$\;17x \equiv 41 \pmod{73}, \text{ and } \; 73 = 17 \cdot 4 + 5,\quad -41 + 3 \cdot 17 = 10$

$\alpha-$Solve:

$\;5x \equiv 10 \pmod{17},  \text{ ANS: } \; x = 2 \text{ is the least residue solution}.$

Propagating backward (just one step),

Solve $17x = 41 + 2 \cdot 73, \text{ ANS: } \; x = 11$

$\text{ANSWER: }  x = 11 \text{ is the least natural number satisfying }$

$\quad 17x \equiv 41 \pmod{73}$


问:
A $\Delta ABC$ has vertices lying on a circle $w$ of radius $1$, with $\angle BAC = 60^\circ$. A circle with center $I$ is inscribed in $\Delta ABC$.

A $\Delta ABC$ has vertices lying on a circle $w$ of radius $1$, with $\angle BAC = 60^\circ$. A circle with center $I$ is inscribed in $\Delta ABC$. The line $AI$ meets circle $w$ at $D$. Find $ID$.

What I Tried: I am assuming the circle with center $I$ is the incircle of $\Delta ABC$. So here is a picture :-

Let $O$ be the center of the circumcircle. Then all the red-marked angles are $30^\circ$ each, and the green angles are $120^\circ$ each. So, simply from properties of cyclic quadrilaterals, one can find that $\square BOCD$ is a rhombus  with :- $BO = OC = OD = CD = DB$ .

But that does not seem to help on how to find $ID$ , I thinking I am missing something here which I could use. Also after checking this through Geogebra I find $ID$ to be miraculously $1$ cm, the same length of the radius of the circumcircle.

Can anyone help me how are we finding this?

答:
Hint. Prove that $ID=DB=DC$ using angle-chasing. (This lemma is actually known as Fact 5 in the Olympiad context). You should then have $ID=DB=R=1$, since $\triangle OBD$ is equilateral.

 Proof of the Lemma "Fact 5".
Consider the triangle $\triangle IDB$ and notice that $$\angle BID=\angle BAI+\angle IBA=\frac{\alpha+\beta}2$$ At the same time, observe that $$\angle DBI=\angle DBC+\angle CBI=\angle DAC+\angle CBI=\frac{\alpha+\beta}2$$ Thus, $\triangle IBD$ is isosceles with $ID=DB$.


答:
It is easy to verify that a diagonal matrix is injective if and only if all diagonal entries are nonzero.

Your argument does show that $D$ is injective if and only if $A$ is injective. In general, a matrix $A$ is injective if and only if $\det A\neq0$, or equivalently if and only if $Av=0\ \Rightarrow\ v=0$.

Title: Is a diagonalizable matrix also injective?

Body:
This question seems to be very basic, but I couldn't find the answer on SE.

I want to know if a diagonalizable matrix $A$, i.e. $A = SDS^{-1}$, is also an injective matrix, i.e. $Av = 0$ if and only if $v = 0$ or also $\ker A = \{ 0 \}$.

I believe it can be simply proven by

"a composition of injective functions is also injective" and

"$S, S^{-1}$ are injective because they are invertible and $D$ is injective because it is diagonal",

but I want to know if I am correct.

I'm not sure about the injectiveness of $D$. If this is the dealbreaker, what are criteria for the injectiveness of $D$ proceeding from the properties of $A$?

I am asking because this would imply, that the only fixed point of a dynamical system
$ \dot{\vec{x}} = A \vec{x}$ with a diagonalizable matrix $A$ would be only $\vec x = 0$.


问:
All smooth periodic functions with bounded derivatives
I am currently working on Fourier series and the following questions arose in this context:

Which smooth $2\pi$-periodic functions (smooth on $\mathbb{R}$) exist, which themselves ($n = 0$) and for every $n$-th derivative and all $x\in \mathbb{R}$ satisfy the inequality $-1\leq f^{(n)}(x) \leq1$?

These functions can all be represented in a Fourier series. Obviously this is fulfilled by sin and cos. For trigonometric polynomials $f(x)=a_0+\sum_{k=0}^d (a_k \cos(kx)+ b_k \sin(kx))$  with degree $d>2$ it seems like the inequality cannot be fullfilled since the coefficients are exponentially growing, when we take derivatives. So what about other fourier series? I know that for example $\sin$ can be represented as fourier series of $\cos$, so this series would fullfill the inequality. Are there any other series, which are not converging to $\sin$ or $\cos$ or a shifted version of them? And how could I proof that?

答:
The notation will be simpler with the complex version of Fourier series:

Say $$f(x)=\sum_kc_ke^{ikx}.$$Then for each $k_0$ we have$$|k_0|^{2n}|c_{k_0}|^2
\le\sum_k|k|^{2n}|c_k|^2=||f^{(n)}||_2^2\le||f^{(n)}||_\infty^2\le1.$$

If $|k_0|>1$ then letting $n\to\infty$ shows that $c_{k_0}=0$. So $$f(x)=\sum_{|k|\le1}c_ke^{ikx}=a\cos(x)+b\sin(x)+c.$$Now you just have to figure out what $a,b,c$ actually give $||f^{(n)}||_\infty\le 1$. (Start with some trig, rewriting $a\cos(x)+b\sin(x)=A\cos(x+\phi)$; the condition I get is $(a^2+b^2)^{1/2}+|c|\le1$.)


答:
The last determinant can be $0$. This is the case for:

$$\det\left(\begin{array}{rrrr}1&0&1&-\sqrt{2}\\-1&0 &1 &-\sqrt{2}\\0 &-1 &1 &-\sqrt{2}\\0 &1& 1 &-\sqrt{2}\end{array}\right)=0$$

Title: How to prove that the tetrahedron exist unique incentre?

Body:
My approach:

Let the equations of 4 faces be

$\alpha_1:A_{11}x_1+A_{12}x_2+A_{13}x_3=B_1,$
$\alpha_2:A_{21}x_1+A_{22}x_2+A_{23}x_3=B_2$

$\alpha_3:A_{31}x_1+A_{32}x_2+A_{33}x_3=B_3,$
$\alpha_4:A_{41}x_1+A_{42}x_2+A_{43}x_3=B_4$

We want to find $X_1,X_2,X_3,d$ such that

$\frac{|A_{11}X_1+A_{12}X_2+A_{13}X_3-B_1|}{\sqrt(A_{11}^2+A_{12}^2+A_{13}^2)}=\frac{|A_{21}X_1+A_{22}X_2+A_{23}X_3-B_2|}{\sqrt(A_{21}^2+A_{22}^2+A_{23}^2)}   =\frac{|A_{31}X_1+A_{32}X_2+A_{33}X_3-B_3|}{\sqrt(A_{31}^2+A_{32}^2+A_{33}^2)}=\frac{|A_{41}X_1+A_{42}X_2+A_{43}X_3-B_4|}{\sqrt(A_{41}^2+A_{42}^2+A_{43}^2)}=d$

which is equivalent to solve the homogenous equation
\begin{align*}
  A_{11}X_1+A_{12}X_2+A_{13}X_3 -d*\sqrt(A_{11}^2+A_{12}^2+A_{13}^2) &= \,B_1 \\
  A_{21}X_1+A_{22}X_2+A_{23}X_3 -d*\sqrt(A_{21}^2+A_{22}^2+A_{23}^2) &= \,B_2 \\
  A_{31}X_1+A_{32}X_2+A_{33}X_3 -d*\sqrt(A_{31}^2+A_{32}^2+A_{33}^2) &= \,B_3 \\
  A_{41}X_1+A_{42}X_2+A_{43}X_3 -d*\sqrt(A_{41}^2+A_{42}^2+A_{43}^2) &= \,B_4 \\
\end{align*}

so,if I could prove that the determinant

\begin{vmatrix} A_{11} & A_{12} & A_{13} & -\sqrt(A_{11}^2+A_{12}^2+A_{13}^2)
 \\ A_{21} & A_{22} & A_{23} & -\sqrt(A_{21}^2+A_{22}^2+A_{23}^2)
\\A_{31} & A_{32} & A_{33} & -\sqrt(A_{31}^2+A_{32}^2+A_{33}^2)
\\A_{41} & A_{42} & A_{43} & -\sqrt(A_{41}^2+A_{42}^2+A_{43}^2)
\\ \end{vmatrix} not equal to zero than by cramer's rule the equations has only 1 roots,which means that the tetrahedron has only 1 incentre.Any hints for me to prove the determinant is not zero?


问:
What is $dW^a_t dW^b_t$ for two various Wiener processes?
I need to simplify $dW^a_t dW^b_t$ where $W^a_t$ and $W^b_t$ are two different Wiener processes.
If I had only one process $W_t$, I could use the following properties:

$$ (dt)^2 = 0 \\ dt \text{ } dW_t = 0 \\ (dW_t)^2 = dt$$

Is it possible to do something similar with $dW^a_t dW^b_t$

答: Well, you can't have anything. But like, if you have the independence of two processes, the product is necessarily zero.</p>


问:
Show that $(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$
I want to show that $$(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$$ for arbitrary sets $A,B,A',B'$.

I have done the following:

Let $x\in (A\cap A')\setminus (B\cap B')$.

Then we have that
\begin{align*}x\in &(A\cap A')\setminus (B\cap B') \iff x\in (A\cap A') \land x\notin  (B\cap B') \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B\cap B')^c \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B^c\cup B'^c) \\ &\iff ((x\in A)\land (x\in A')) \land   ((x\in B^c)\lor (x\in B'^c)) \\ &\iff ((x\in A)\land (x\in A')\land (x\in B^c))    \lor ((x\in A)\land (x\in A') \land (x\in B'^c)) \end{align*}

Is everything correct so far? How could we continue?

答:
I believe the equality that you mentioned

$$(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B') \tag{1}$$

is not true, in general.

Consider the following.

$$\begin{align}
x \in (A \cap A’) \setminus (B \cap B’) & \iff x \in A \cap A’ \wedge x \notin B \cap B’\\
& \iff \big(x \in A \wedge x \in A’\big) \wedge \big( x \notin B \vee x \notin B’ \big)\\
& \iff \big( x \in A \wedge x \in A’ \wedge x \notin B \big)\\
& \quad \quad \quad \vee \big(x \in A \wedge x \in A’ \wedge x \notin B’\big)\\
& \implies \big( x \in A \wedge x \notin B \big) \vee \big( x \in A’ \wedge x \notin B’ \big)\\
& \iff (x \in A \setminus B) \vee \big( x \in A’ \setminus B’ \big)\\
& \iff x \in (A \setminus B) \cup (A’ \setminus B’)
\end{align}$$

From this I can just get a counter example that shows that $(1)$ doesn’t hold.

$$A = \{1,2\} \quad A’ = \{1\} \quad B = \{1\} \quad B’ = \{2\}$$

$$\begin{align}
1 \in A,A’,B \wedge 1 \notin B’ & \implies 1 \in A \cap A’ \wedge 1 \notin B \cap B’\\
& \implies 1 \in (A \cap A’) \setminus (B \cap B’)
\end{align}$$

$$1 \in A,B \implies 1 \notin A \setminus B \implies 1 \notin (A\setminus B) \cap (A’ \setminus B’)$$


问:
How to prove that there exists a $z_0 \in U_{1} [0]$ such $ \prod_{k=1}^{n} |z_0 - a_k | \geq 1 $ for $a_1, \dots , a_n \in U_{1} [0] $?
Let $a_1 , \dots , a_n $ be points in the unit circle/ball in $\mathbb{C}$ around $(0,0)$ (also known as $U_{1} [0]$), which do not necessarily differ from one another. How to prove that there exists some $z_0 \in U_{1} [0] $ such that $$ \prod_{k=1}^{n} |z_0 - a_k| \geq 1 ? $$
Here's what I tried thus far: First, pick a $z_0$ such that $|z_0 - a_j| \geq 1 $ for some $1 \leq j \leq n$. Then pick another point $a_k$ on the unit cirle. By the triangle inequality, we know that $|a_j - a_k| + |z_0 - a_k | \geq |z_0 - a_j| \geq 1 $, so $|z_0 - a_k | \geq 1 - |a_j - a_k| $. Therefore, we find that $$ \prod_{k=1}^{n} |z_0 - a_k | \geq |z_0 - a_j| \cdot \prod_{k=1 , k \neq j }^{n} (1 - |a_j - a_k| ) \geq  \prod_{k=1 , k \neq j }^{n} (1 - |a_j - a_k| ) .$$

I don't know how to proceed from here though. Any suggestions?

答:
The conclusion holds for any $a_1,\dots,a_n\in\mathbb{C}$, not necessarily to be in the closed unit disk(ball).

To see this, let $f(z)=\prod_{k=1}^n(z-a_k)$, and it suffices to show that
$$M:=\max_{|z|=1}|f(z)|\ge 1.\tag{1}$$
By Cauchy's integral formula,
$$n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)}{z^{n+1}}dz.\tag{2}$$
From $(2)$ we know that
$$1=\big|\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)}{z^{n+1}}dz\big|\le\frac{1}{2\pi }\int_{|z|=1}|f(z)||dz|\le M,\tag{3}$$
which completes the proof.

Remark: $(3)$ also implies that $M=1$ if and only if $f(z)=z^n$, i.e. $a_k=0$ for $k=1,\dots,n$.


问:
$x^2 - y^2=9$, $2y-x=3$, Find maximum value of $\dfrac{y}{x}$
I came across this question and have no idea what to do. Given,
$$
x^2-y^2=9 \\
2y-x=3
$$

find maximum value of $\dfrac{y}{x}$

答: Solve the system of equations, for instance, with $2y-x=3\iff x=2y-3$; insert this value in the first equation to obtain $(2y-3)^2-y^2=9\iff ...\iff y\cdot(y-4)=0$. Thus the solutions $(x,y)$ are $(3, 0)$ and $(5, 4)$. Which one of them yields the maximum ratio $\frac yx$?


Title: $x^2 - y^2=9$, $2y-x=3$, Find maximum value of $\dfrac{y}{x}$

Body:
I came across this question and have no idea what to do. Given,
$$
x^2-y^2=9 \\
2y-x=3
$$

find maximum value of $\dfrac{y}{x}$

Body:
Because $x = 2y - 3$, we have:

$$(2y-3)^{2} - y^{2} = 9$$

$$4y^{2} - 12y + 9 - y^{2} = 9$$

$$3y^{2} - 12y = 0$$

$$3y(y - 4) = 0$$

$$y = 0, 4$$

$$(x, y) = (-3, 0), (5, 4)$$

$$\frac{y}{x} = 0, \frac{4}{5}$$

$$\boxed{\max(\frac{y}{x}) = \frac{4}{5}}$$


问:
A martingale with uncorrelated components
Let $\zeta_{i}$ be a sequence of random variables such that the partial sums
\begin{align*}
X_{n}=\zeta_{0}+\zeta_{1}+\cdots+\zeta_{n},\quad n\geq 1,
\end{align*}
determine a martingale. Show that the summands are mutually uncorrelated, i.e. that $\mathbb {E}(\zeta_{i}\zeta_{j})=\mathbb {E}(\zeta_{i})\mathbb {E}(\zeta_{j})$ for $i\neq j$. I'm seeking for some hints to start with since I don't quite see what's the connection between the given condition and the conclusion. Thanks.

答:
I guess that $X_0$ should also be defined as well, otherwise there would be no reason for $\zeta_0$ and $\zeta_1$ to be uncorrelated. (Let for example n=1 and $\zeta_1=\zeta_0$ be integrable correlated random variables. The $1$-step process $X$ is then clearly a martingale as it must only satisfy integrability.)

Let $(\mathcal{F}_i)_{0\leqslant i \leqslant n}$ be the corresponding filtration. First notice that $\mathbb{E}(\zeta_i)=0$ for $i\geqslant 1$, as, using tower property of conditional expectation and the Martingale property of $X_i$,
$$\mathbb{E}(\zeta_i)=\mathbb{E}(X_i-X_{i-1})=\mathbb{E}(\mathbb{E}(X_i-X_{i-1}|\mathcal{F}_{i-1}))=0.$$

W.l.o.g. let $i<j$. Now the result directly follows as
$$ \mathbb{E}(\zeta_i \zeta_j)=\mathbb{E}(\zeta_i (X_j-X_{j-1}))=\mathbb{E}(\mathbb{E}(\zeta_i (X_j-X_{j-1})|\mathbb{F}_i))=\mathbb{E}(\zeta_i \mathbb{E}(X_j-X_{j-1}|\mathbb{F}_i))=\mathbb{E}(\zeta_i (\mathbb{E}(X_j|\mathcal{F}_i)-\mathbb{E}(X_{j-1}|\mathcal{F}_i)))=\mathbb{E}(\zeta_i(X_i-X_i))=0.$$


问:
How much to charge for being a math tutor?
I have been requested by some people I know fairly well to tutor their son in math. He is going into 9th grade, so it's not super-fancy stuff.

Some background: I graduated fairly recently from college with a bachelor's in engineering, and I have a programming job that pays well enough. They approached me, not the other way around, and I've never tutored anyone in anything before.

How much should I charge?

Edit: I ought to mention that they will be paying for transportation, so that's not an issue here.

I hope this isn't too far offtopic.

答:
This is a highly location dependent question.

What I suggest is to ask around more locally. Is there a near by university?  What does the average tutor there charge (there are usually signs or sites online).  Also what do you feel your time is worth?  Maybe you require it to pay as much per hour as your programming job, maybe not, maybe more.

Also, if you google this question you get quite a few results and already existing forum posts.

(For what its worth, if I remember, the posters at my university usually charged somewhere between 25-35 dollars, and those students had bachelor degrees.  Again different places are certainly very different.)


问:
Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$
Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$

The way I attempted to the question is the following:

I state that $a=\sqrt[3]{x+1}$ hence we have that $3a^3-3+a+a^2=-1$. Hence $3a^3-2+a+a^2=0$, so $a^2(3a+1)+a-3=0$. And this is where I got stuck.

Could you please explain to me how to finish off the question with this train of thought and out of 10, what would be logical to assume that I would get for these workings out?

Body:
$$
3 a^3-2+a+a^2=(3 a - 2) (a^2 + a + 1)
$$


答:
$3a^3+a^2+a-2=0$

$(3a-2)(a^2+a+1)=0$

We can find this by using Rational Root theorem, or another way is by noting that $a^2+a+1$ is also a root because $a^2+a+1=0 \implies a^3=1$ (roots of unity) so
$$3a^3+a^2+a-2=3+a^2+a-2=a^2+a+1=0$$

Body:
Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$

The way I attempted to the question is the following:

I state that $a=\sqrt[3]{x+1}$ hence we have that $3a^3-3+a+a^2=-1$. Hence $3a^3-2+a+a^2=0$, so $a^2(3a+1)+a-3=0$. And this is where I got stuck.

Could you please explain to me how to finish off the question with this train of thought and out of 10, what would be logical to assume that I would get for these workings out?


问:
Determine how many group homomorphisms from $\Bbb{Z}_3$ (with addition as the group operation) to $S_4$ are there
First of all, is  $\Bbb{Z}_3$ with addition a set that contains ${0,1,2}$ with $\bmod(3)$?
Then, $h(1)=$ identity element is a possible homomorphism.
If $h(1)$ is not the identity element, are we trying to find how many element in S4 has a order of $3$?
If so, given $S_4$ is isomorphic to the symmetry group of a cube, we have the elements included rotation $270^\circ$ are of order $3$. If that's the case the answer equal to $13$ since there are 13 rotational symmetry axis in a cube?

答:
As the order of $\mathbb Z_3$ is equal to $3$, if $h$ is a non trivial requested homomorphism, $h(1)$ must be an element of order $3$, i.e. a $3$-cycle.

A $3$-cycle of $S_4$ fixes one element. And that element being chosen, the are $2$ $3$-cycles. Therefore in total $8$ $3$-cycles. Adding the trivial homomorphism gives in total $9$ homomorphisms.


问:
A be a matrix satisfying ${(AA^t)}^r=I$ then when it is invertible
This question was asked in my linear algebra mid term and I was unable to solve it.

Let m ,n ,r be natural numbers . Let A be an m*n matrix with real entries such that ${(AA^t)}^r =I $ where I is m * m identity matrix. Then which  of following is / are true:

A m =n

B $ AA^{t}$ is invertible

C $A^{t} A $ is invertible

D if m=n , then A is invertible

I am really confused in this question due to $A A^{t}$a nd was unable to think which option are true or any contradictions in options . It's my humble request to guide you through this question.

I followed Hoffman and Kunze Linear algebra.

Thanks!!

答:
Correction: as it is written, $I$ must be an $m \times m$ matrix.

$M = A\,A^T$ is an $m\times m$ symmetric matrix.

Since it is symmetric, it is diagonalizable, with orthonormal eigenvectors: $M = V\,D\,V^T$

$$M^r = V\,D^r\,V^T = I \Rightarrow D^r = V^T\,V = I$$

Therefore, the eigenvalues of $M$ will all be $\pm 1$, which means $M$ is invertible.

A very simple example that shows that $m$ may be different than $n$; and that $A^T\,A$ may be singular is:

$$A = \begin{bmatrix}1&0\end{bmatrix}$$

If $m=n$ and $B = (A\,A^T)^{-1}$

$$(A\,A^T)\,B = I \Rightarrow A\,(A^T\,B) = I$$

If a square matrix has a right inverse then it has a left inverse (and  they are the same). So $A$ must be invertible and $A^{-1} = A^T\,B$.

$\boxed{\text{Answer: A=False; B=True; C=False; D=True}}$


问:
Convergence in Sobolev spaces implies classical convergence?
Let $p>1$ and $\Omega$ be an open bounded domain in $\mathbb{R}^N$. Take $(u_n)\subset W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$ and $u\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$.

If $$u_n\longrightarrow u \quad\mbox{ in } W_0^{1, p}(\Omega),$$
it it true that $u_n\longrightarrow u$ in the classical sense (up to subsequences)?

Could anyone please help me to understand?

Thank you in advance!

答: If $u_n\rightarrow u$ in $W^{1,p}_0,$ then (in particular) it converges in $L^p$. If $u_n\rightarrow u$ in $L^p$, then there exists a subsequence $(u_{n_j})$ of $(u_n)$ with the property that $u_{n_j}\rightarrow u$ pointwise almost everywhere. The proof of the latter fact comes in when demonstrating that $L^p$ is complete.


问:
Find matrix $R$ such that $(R^TPR-P)x=y$ or that $PRx = y$
Consider an invertible matrix $P\in\mathbb{R}^{n\times n}$ and two vectors $x,y\in\mathbb{R}^n$ ($P,x,y$ are given). Now consider the following two problem.

Problem 1: Find a matrix $R\in\mathbb{R}^{n\times n}$ (if it exists) such that $(R^TPR-P)x=y$.

Problem 2: Find a matrix $R\in\mathbb{R}^{n\times n}$ (if it exists) such that $PRx = y$

Lately I've been encountering problems of this type. I've been trying to use some vectorization tricks without any success. I wanted to see how would you solve these equations so I can make a general idea of what would be the general strategy if it exists.

答:
I don't think there are any general strategies. As shown by the answer below, the  methods for solving your two problems are quite different.

Problem 1. I suppose $n\ge2$. Rewrite the equation as $R^TPRx=z$ where $z=Px+y$. Clearly it is not solvable when $x=0\ne z$. On the other hand, when $z=0$, an obvious solution is given by $R=0$.

Now suppose $x$ and $z$ are both nonzero. By absorbing some constant into $P$, we may assume that $x$ and $z$ are unit vectors. From $R^TPRx=z$, we obtain $x^TR^T(P+P^T)Rx=2x^Tz$. Hence the equation is solvable only if one of the following conditions is satisfied:

$x^Tz=0$ and $P+P^T$ is indefinite or singular,

$x^Tz>0$ and $P+P^T$ has a positive eigenvalue,

$x^Tz<0$ and $P+P^T$ has a negative eigenvalue.

We will show that these conditions are also sufficient. Since the third case reduces to the second one when we negate both $P$ and $x$, we shall omit it.

If $P+P^T$ is indefinite or singular, $u^T(P+P^T)u=0$ for some unit vector $u$. Therefore there exists an orthogonal matrix $U$ such that $\left(U^TPU\right)_{11}=\frac12\left(U^T(P+P^T)U\right)_{11}=0$. Since $P$ is invertible, the first column of $U^TPU$ must contain a nonzero entry. Therefore, by composing $U$ with a permutation matrix, we may further assume that
$$
U^TPUe_1=(0,c,\ast,\cdots,\ast)^T \text{ for some } c\ne0.
$$
Since $x^Tz=0$, $\{x,z\}$ is an orthonormal set. Thus $Qx=e_1$ and $Qz=e_2$ for some orthogonal matrix $Q$. Now let $D=\operatorname{diag}\left(c^{-1},1,0,\ldots,0\right)$. Then
$$
DU^TPUDQx=DU^TPUDe_1=c^{-1}DU^TPUe_1
=c^{-1}D\pmatrix{0\\ c\\ \ast\\ \vdots\\ \ast}
=e_2=Qz.
$$
Hence $Q^TDU^TPUDQx=z$ and $R=UDQ$ is a solution.

When $P+P^T$ has a positive eigenvalue, there exists an orthogonal matrix $U$ such that $\left(U^TPU\right)_{11}=\frac12\left(U^T(P+P^T)U\right)_{11}>0$. That is,
$$
U^TPUe_1=(c,\ast,\cdots,\ast)^T \text{ for some } c>0.
$$
Let $Q$ be an orthogonal matrix such that $Qz=e_1$. The first entry of $Qx$ is $x^Tz$ because $\langle Qx,e_1\rangle=\langle Qx,Qz\rangle=\langle z,x\rangle$. Let $D=\operatorname{diag}\left(\frac{1}{\sqrt{cx^Tz}},0,\ldots,0\right)$. Then
$$
DU^TPUDQx=DU^TPUD\pmatrix{x^Tz\\ \ast\\ \vdots\\ \ast}
=\sqrt{\frac{x^Tz}{c}}DU^TPUe_1
=\sqrt{\frac{x^Tz}{c}}D\pmatrix{c\\ \ast\\ \vdots\\ \ast}
=e_1
=Qz.
$$
Hence $Q^TDU^TPUDQx=z$ and $R=UDQ$ is a solution.

Problem 2. If $x=0$, every matrix $R$ is a solution when $y=0$ and the equation is insolvable otherwise. If $x\ne0$, just pick any matrix $R$ such that $Rx=P^{-1}y$. As noted by the other answer, one particularly simple solution is given by the rank-one matrix $R=P^{-1}y\frac{x^T}{x^Tx}$.


问:
Old Math Books Request
I'm looking for books that are written in an old style but are also concise and rigorous. Some examples of what I mean:

Calculus on Manifolds by Spivak

Complex Analysis by Ahlfors

Mechanics by Landau and Lifshitz (not a math book, but consistent with the style I'm looking for)

All of these books are written very concisely (and as a plus, they're written in a very nice old-style font!). Any other recommendations people would have of math books that resemble these? Thanks for the help!

答: Euclid's Elements are written in an old style (if not very old style) but also concise and rigorous.</p>

Title: Old Math Books Request


问:
What is the definition of faces in Graph Theory? Why does 3f≤2|E| hold for any connected planar graph?
From Euler's characteristic, we know any connected planar graph $G = (V, E)$ satisfies $|V|-|E|+f=2$ where $f$ is the numbers of faces in the drawing of $G$. Why do trees have 1 face? Because you can't form more than 1 face without a Jordan Curve and trees do not contain cycles. So I want know the strict definition of faces mathematically, and why $3f≤2|E|$ holds for any connected planar graph?

答:
It's a double counting argument.

Each face has at least 3 edges, and each edges is in (at most) 2 faces.

Do you see how to set it up?


问:
Why does this function split up like this? (Rudin, Real and Complex Analysis)
This is regarding Theorem 1.8 on page 11 of Rudin's Real and Complex Analysis, we have a measurable space $X$ and two real valued functions on it $u$ and $v$. We then have a continuous mapping $\Phi$ from the real plane into a topological space $Y$. We define $f(x)=(u(x),y(x))$ and $h=\Phi\circ f$, and from here we need to show that $h$ is measurable and we know that it suffices to show that $f$ is measurable.

We then take a rectangle $R$ in the plane with sides parallel to the axes, which we write as the Cartesian product of $I_1$ and $I_2$ which are two segments, presumably two adjacent sides of the rectangle. We then say that: $$f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2) \tag{1},$$ and this is what's confusing me. $I_1$ and $I_2$ are essentially just lines in the plane, why would the intersection of their preimage be equal to the preimage of the entire rectangle and not just two "lines" in $X$?

Disclaimer: I am a physicist not a mathematician please have mercy on my soul if this is a trivial error on my part.

答:
As far as I understand, we have $R=I_1\times I_2$.

Now we want to show $f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2)$.

This is a straight forward proof:

$f^{-1}(R)=\{x\in X: f(x)\in R\}=\{x\in X: (u(x),v(x))\in I_1\times I_2\}=\{x\in X: u(x)\in I_1~\text{and}~ v(x)\in I_2\}$

$=\{x\in X: u(x)\in I_1\}\cap \{x\in X: v(x)\in I_2\}=u^{-1}(I_1)\cap v^{-1}(I_2)$


答:
A correct formula would be the following:
$$ (A\smallsetminus B)\cap(A'\smallsetminus B')=(A\cap A')\smallsetminus(B\cup B'), $$
which is easy to prove:
\begin{align}
(A\smallsetminus B)\cap(A'\smallsetminus B')&=(A\cap B^c)\cap(A'\cap B'^c) \\
&=(A\cap A')\cap (B^c\cap B'^c)\\
&=(A\cap A')\cap (B\cup B')^c=(A\cap A')\smallsetminus(B\cup B').
\end{align}

Title: Show that $(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$

Body:
I want to show that $$(A\cap A')\setminus (B\cap B')=(A\setminus B)\cap (A'\setminus B')$$ for arbitrary sets $A,B,A',B'$.

I have done the following:

Let $x\in (A\cap A')\setminus (B\cap B')$.

Then we have that
\begin{align*}x\in &(A\cap A')\setminus (B\cap B') \iff x\in (A\cap A') \land x\notin  (B\cap B') \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B\cap B')^c \\ &\iff ((x\in A)\land (x\in A')) \land x\in  (B^c\cup B'^c) \\ &\iff ((x\in A)\land (x\in A')) \land   ((x\in B^c)\lor (x\in B'^c)) \\ &\iff ((x\in A)\land (x\in A')\land (x\in B^c))    \lor ((x\in A)\land (x\in A') \land (x\in B'^c)) \end{align*}

Is everything correct so far? How could we continue?


问:
Find UMVUE in a uniform distribution setting
Let $X_1, ..., X_n$ be independent and uniformly distributed on $(\theta_1-\theta_2,\theta_1+\theta_2)$ for $\theta_1 \in \mathbb{R}$, $\theta_2>0$.

Find UMVUEs for

a) $\theta_1$ and $\theta_2$,

b) $\theta_1/\theta_2$.

Naturally, I would like to use the Lehmann-Scheffé theorem that says:

If $V$ is a complete, sufficient statistic for $\theta$ and $\mathbb{E}_\theta[g(V)]=h(\theta)$ holds. Then $g(V)$ is an UMVUE for $h(\theta)$.

So first, I have to find such a statistic for any of my $\theta$s.

In the lecture we learned that $(X_{(1)}, X_{(n)})$ sufficient for $(a,b)$ if the $X_i$ are uniform on $(a,b)$ (where $X_{(1)}<...<X_{(n)}$ is the order statistic for $X_1, ..., X_n$).

So it follows that in my setting $(X_{(1)}, X_{(n)})$ is sufficient for $(\theta_1-\theta_2,\theta_1+\theta_2)$? But looking at the proof, $(X_{(1)}, X_{(n)})$ is already a sufficient statistic for $(\theta_1, \theta_2)$, too, right?

So I have to check for completeness and thus show that for all functions $g$ mapping from the range of $V$ to $\mathbb{R}$ from $\mathbb{E}_\theta[g(X_{(1)}, X_{(n)})]=0$ for all $\theta=(\theta_1, \theta_2)$ follows $\mathbb{P}_\theta(g(X_{(1)}, X_{(n)})=1)=1$ $\mathbb{P}_\theta$-almost-surely.

Here I'm stuck: How can I show that?

答:
I don't think you can. At least if X is distributed on $(\theta,\theta + 1)$, then $(X_{(1)}, X_{(n)})$ is sufficient, but not complete.

Lets say $g(X_{(1)}, X_{(n)})$ is a function of your statistic. Then

$\mathbb{E}_{\theta}[X_{(n)} - X_{(1)}] = \mathbb{E}_{\theta}[X_{(n)}] - \mathbb{E}_{\theta}[ X_{(1)}] = (\theta + n/(n+1)) - (\theta + 1/(n+1)) = (n-1)/(n+1)$

Then $g$ defined as

$g(X_{(1)}, X_{(n)}) := X_{(n)} - X_{(1)} - (n-1)/(n+1)$

would be 0 $\forall \theta$ and thus not complete.

As I said, the setting is slightly different though. Let me know if you find a definite solution - I'm looking at the same problem.


问:
When is the image of $\mathbb{R}^n$ a smooth submanifold?
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be an injective $C^{k}$-function, for some positive integer $k$.  Under what conditions is the complement of the open set (I know its open by Brower's domain invariance theorem) a finite union of smooth submanifold of $\mathbb{R}^n$?

答:
Here is a partial answer. Recall that every simply connected open subset of $\mathbb{C}$ is biholomorphically equivalent to the unit disk $\Delta = \{z\in \mathbb{C}~|~ |z| <1\}$.

Let $\gamma : \mathbb{S}^1 \to \mathbb{C}$ be a continuous simple curve that is nowhere differentiable (for example, the Von Koch curve), and let $\Omega$ be its interior (by Jordan's theorem). Then $\Omega$ is a simply connected open subset of $\mathbb{C}$ different from $\mathbb{C}$ : there exists a biholomorphism $g : \Delta \to \Omega$. Moreover, $\Delta$ is diffeomorphic to $\mathbb{R}^2$, say $h : \mathbb{R}^2 \to \Delta$ is such a diffeomorphism. Then
\begin{align}
f = g\circ h : \mathbb{R}^2 \to \mathbb{R}^2
\end{align}
is an injective smooth function with image $\Omega$. But $\mathbb{R^2}\setminus \Omega$ is not a smooth manifold as otherwise, its boundary would be smooth, that is not the case.

Here has been shown that even if $f$ is $\mathcal{C}^{\infty}$, there is no reason for smoothness to extend to the boundary.

Notice that we could have choosen $\Omega$ to be the interior of a triangle, and the boundary would not be smooth either because of corners. My example shows that it can happen everywhere.

Edit It is asked in comment if, whenever $f$ has a dense image, $\mathbb{R}^{n} \setminus \mathrm{Im}(f)$ would be a $n-1$ dimensional smooth manifold. The answer is no. Here is a counter example.

Let $\gamma : \mathbb{R}_+ \to \mathbb{C}$ be an injective curve such that $\gamma$ is nowhere differentiable, and goes to infinity (imagine the trajectory of a brownian motion). Then $ \Omega =\mathbb{C}\setminus\gamma(\mathbb{R}_+)$ is a simply connected open subset of $\mathbb{C}$ different from $\mathbb{C}$. The above construction show there is a smooth function $f : \mathbb{R}^2 \to \mathbb{R}^2$ with image $\Omega$, which is dense in $\mathbb{R}^2$ but with complement that is nowhere smooth.


问:
counterexample of $\rho(AB) \leq \rho(A)\rho(B)$
Suppose $A$ and $B$ are normal complex $n \times n$ matrices. Then we know that $\rho(AB) \leq \rho(A)\rho(B)$ where $\rho(.)$ is the spectral radius of a matrix.
Is there any counterexample if $A$ or $B$ is not normal?

答: This almost always occurs in the Cholesky factorisation $LL^T$ of a positive semidefinite matrix. When all entries in the strictly lower triangular part of $L$ are nonzero, we have $\rho(LL^T)>\rho(L)^2$.


问:
Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|.$
Let $A=(a_{ij})$ be normal. Show that $\rho(A) \geq \max_{1\leq i \leq n}|a_{ii}|,$ where $\rho(.)$ is the spectral radius of $A$.

I know that $\rho(A) \leq \sqrt{\lambda_{\text{max}}(A^{^*}A)}:=\sigma_{\text{max}}(A)=||A||.$ Is there any relation between diagonal element and the eigenvalue? I know the relation with the trace but can not connect with each diagonal element.

答:
It $A$ is a bounded operator on a Hilbert space (including the case of $n\times n$ matrices), one defines:

the norm: $\|A\|=\sup_{\|x\|\leq1}\|Ax\|$,

the spectral radius: $\rho(A)=\sup_{\lambda\in\sigma(A)}|\lambda|$, and

the numerical radius: $w(A)=\sup_{\|x\|\leq1}|\langle Ax,x\rangle|$.

For every $A$
$$
  \rho(A)\leq w(A)\leq \|A\| \leq 2w(A).
  $$
but when $A$ is normal one has that
$$
  \rho(A) = w(A) =\|A\|.
  $$

Since a diagonal entry of an $n\times n$ matrix is given by $A_{ii}=\langle Ae_i,e_i\rangle$, we then have
$$
  |a_{ii}| \leq w(A) = \rho(A),
  $$
provided $A$ is normal.

EDIT

Lemma.  If $A$ is a normal matrix, then the numerical radius of $A$ coincides with its spectral radius.

Proof. Let $\{e_i\}_{i=1}^n$ be an orthonormal basis consisting of eigenvectors, with corresponding eigenvalues
$\{\lambda _i\}_{i=1}^n$.  For $\|x\|\leq 1$, we then have
$$
  \langle Ax, x\rangle  =
  \Big\langle A\Big(\sum_{i=1}^n x_ie_i\Big),\sum_{i=1}^n x_ie_i\big\rangle  =
  \Big\langle \sum_{i=1}^n x_i\lambda _ie_i,\sum_{i=1}^n x_ie_i\big\rangle  =
  \sum_{i=1}^n |x_i|^2\lambda _i.
  $$
Therefore
$$
  |\langle Ax, x\rangle | \leq  \sum_{i=1}^n |x_i|^2|\lambda _i| \leq  \big (\sup_{1\leq i\leq n} |\lambda _i|\big ) \sum_{i=1}^n |x_i|^2 = \rho (A)\|x\|^2 \leq \rho (A).
  $$
Taking the supremum over $x$ we get
$$
  w(A) = \sup_{\|x\|\leq 1}|\langle Ax, x\rangle | \leq  \rho (A).
  $$
On the other hand, for every $i≤n$ we have that
$$
  w(A) \geq  |\langle Ae_i, e_i\rangle | = |\langle \lambda _ie_i, e_i\rangle | = |\lambda _i|.
  $$
Taking the supremum over $i$ we get
$$
  w(A) \geq  \sup_{1≤i≤n} |\lambda _i| = \rho(A).
  $$
QED


问:
I need help to evaluate this integral with a constant $c$
Find the value of the constant $c$ for which the integral converges, and evaluate the integral:

$$\int_0^\infty \left(\frac{9x}{x^2+1}-\frac{9c}{2x+1}\right)dx$$

答:
We have
\begin{align}
\int_0^a \left(\dfrac{9x}{x^2+1} - \dfrac{9c}{2x+1}\right) dx & = \left. \left(\dfrac92 \log\left(x^2+1 \right) - \dfrac{9c}2 \log(2x+1) \right) \right \vert_0^a\\
& = \dfrac{9}2 \log \left(\dfrac{a^2+1}{(2a+1)^c}\right)
\end{align}
Given this, as $a \to \infty$, what happens if $c<2$ or $c>2$? Once you get the $c$ from this, use this $c$ and let $a \to \infty$, to get the answer.</p>


问:
High-Level math subjects applicable to contest math
For context, I'm a sophomore high school student. I'll be done with undergrad Abstract Algebra, Topology, and Analysis by the end of the school year, and I'll be burning through the necessary graduate Analysis, (Rudin) and Algebra (Lang) over the summer for Hartshorne, and doing some Differential Geometry from Tu along with Hartshorne.

I'm hoping that by the time I make it to Diophantine Geometry, I'll be able to use it to solve Olympiad number theory problems (Diophantine Equations in particular). However, that's something I can't really find out for myself until I actually learn basic Diophantine Geometry, which will take at least two years.

So, in order to get around that, I thought "Why not ask people who know Diophantine Geometry?"

Although this question is mainly about Arithmetic Geometry, answers about other high-level math subjects applicable to Olympiad problems are welcome.

And here we are. By the way, I also enjoy the Algebra by itself, it's not like I want to slog through more for the express purpose of solving competition math problems. I just wanted to know if it was an alternative or supplement to studying classical methods.

答:
Algebra and analysis: you should definitely know about the derivative and limits of sequences, though I doubt that you will need integration; continuity can help with functional equations. It doesn't hurt to have an understanding of fields and rings if you want to really understand what is going on with polynomials, like Gauss's lemma and Bezout's lemma for polynomials and irreducibility criteria.

Geometry: projective and inversive geometry; Barycrentric coordinates can be made formal using higher mathematics. Euclidean geometry is not a part of higher math. Be familiar with determinants.

Combinatorics: generating functions, including the derivative. There is occasionally a bit of graph theory though nothing very advanced. On that note, I guess knowing the general version of Ramsey's theorem isn't a bad idea. The probabilistic method is good to know. Solving homogeneous linear recurrences in general requires some linear algebra. Combinatorial Nullstellensatz and Polya enumeration theorem are probably as sophisticated as olympiad combinatorics gets.

Number theory: There are some non-trivial number theory theorems that pop up in olympiads, such as Thue's lemma, Lifting the Exponent lemma, Hensel's lemma, Zsigmondy's theorem, the solution to Pell's equation. Gaussian integers is also a possibility, as are aspects of algebraic number theory. You might consider checking out some of Titu Andreescu's books on number theory as he likes to occasionally bring in higher level stuff into his curriculum.


问:
How can I prove that any function $f: \mathbb{N} \rightarrow \mathbb{R}$ is continuous?
I have to find all functions of the type:

$$f: \mathbb{N} \rightarrow \mathbb{R}$$

that are continuous. My claim is that all such functions are continuous. If we think about it, $f$ would only have isolated points and we know that a function is always considered continuous at an isolated point. So, by this reasoning, any function $f:\mathbb{N} \rightarrow \mathbb{R}$ is continuous. I hope my reasoning is correct. What I am confused about is how could I prove this formally. We know that continuity is defined like this:

A function $f: A \rightarrow \mathbb{R}$ is continuous at a point $c \in A$ if

$$\forall V \in \mathcal{V}(f(c)), \exists U \in \mathcal{V}(c) \text{ such that } \forall x \in U \cap A \text{ we have } f(x) \in V$$

How could I possibly prove my point using the definition of continuity? It's really not that difficult to find the answer intuitively, but I don't see how I could make my argument more formal.

答: For each $c\in\Bbb N$, take $\mathcal V(c)=\{c\}$. You can do it, since it is a neighborhood of $c$.


问:
Inverse transform of a modified Abel transform
I have been struggling for 6 months on finding the analytical inverse transform of a transformation below:

$$F(y,k) = 2 \int_y^{\infty}\cos\left(ka\sqrt{r^2-y^2}\right) f(r,k) \frac{r}{\sqrt{r^2-y^2}}\ \mathrm{d}r$$

where $a$ is a known constant and $\lim_{r\to\infty}f(r,k) = f'(r,k) = 0$ and all functions are continuous.

It has a similar form of Abel transform, except that it has the cosine factor in the integrand. Any help will be appreciated!

答:
You have a sequence of operators $A_k$ indexed by $k$ (integers, I suppose) given by
$$
(A_kg)(y)
=
2 \int_y^{\infty}\cos\left(ka\sqrt{r^2-y^2}\right) g(r) \frac{r}{\sqrt{r^2-y^2}}\ \mathrm{d}r.
$$
If you can invert every $A_k$, then you can find $f(r,k)$ from $F(t,k)=[A_kf(\cdot,k)](y)$ by applying $A_k^{-1}$ at each $k$.

In Abel transforms with low regularity with applications to X-ray tomography on spherically symmetric manifolds I proved with de Hoop that each $A_k^{-1}$ exists assuming you are compactly supported in $r$ — your case appears to easily satisfy our conditions.
The argument works for pretty general integral equations of this kind but gives no explicit inversion formula.
It does give an inversion method, though: if you divide the relevant interval in a large but finite amount of layers and use a Neumann series in each one, you get an explicit procedure.
See section 2 of the linked paper for details.


问:
Arc length of logarithm function
I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$.

So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$
I am having trouble calculating the integral. I tried to do substitution, but I still fail to think of a way to integrate it. This is what I have done so far:
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}$$
$$x=\frac{2u}{u-1}$$
$$dx=\frac{2}{(u-1)^2}du$$
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}=u-\frac{1}{2}+\frac{1}{2u}$$
$$\int\sqrt{1+\frac{1}{x^2}}dx=2\int\frac{u-\frac{1}{2}+\frac{1}{2u}}{(u-1)^2}du$$
And I am stuck.

答: You're doing great, now split this into three integrals; one by partial fractions.</p>

Title: Arc length of logarithm function


问:
Relative Dimension of Vector Space Over Two Fields
Let $K$ and $L$ be fields, with $L$ a vector space over $K$, and $V$ a vector space over both $K$ and $L$. Suppose also that their actions are compatible, so
$$k\cdot(l\cdot v)=(k\cdot l)\cdot v$$

I am trying to show the following:

If $\dim_KL=\dim_KV=n$, then $\dim_LV=1$.

We must have isomorphisms $\varphi:L\to K^n$ and $\psi:K^n\to V$ of $K$-modules. Then $\psi\circ\varphi:L\to V$ is also an isomorphism of $K$-modules.

However I can't seem to show that it is $L$-linear also, and I feel like I'm missing something obvious.

Any help would be much appreciated.

答:
I'm paraphrasing Ahr's comment as an answer so that I can accept it, if they would like to post one themselves then I am happy to remove it.

Let $v\in V$ be any non-zero vector. Then $Lv$ is a $K$-vector subspace of $V$ since the actions are compatible.

We have $L\cong Lv$ as $K$-vector spaces via $l\mapsto lv$, this map is $K$-linear again since the actions are compatible.

Then
$$\dim_K(Lv)=\dim_K(L)=\dim_K(V)$$
and so we must have $Lv=V$ since $Lv\subseteq V$.

This shows that $v$ generates $V$ as an $L$-vector space, and so $\dim_L(V)=1$ as desired.


问:
How to calculate this scalar triple product?
I have been trying for over an hour to calculate this scalar triple product but I just can't succeed, I always get some crazy long expression that I can't do anything with.

We have $$\overrightarrow{b_1}=\frac{\overrightarrow{a_2}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_2}=-\frac{\overrightarrow{a_1}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_3}=\frac{\overrightarrow{a_1}\times \overrightarrow{a_2}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$

I have to calculate $\overrightarrow{b_1}\cdot(\overrightarrow{b_2}\times\overrightarrow{b_3})$.

The solution given in the book is $$\frac{1}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$

答:
$$\vec u=\frac{\vec a \times \vec b}{[\vec a, \vec b, \vec c]}, ~\vec v=\frac{\vec b \times \vec c}{[\vec a, \vec b, \vec c]},~\vec w=\frac{\vec c \times \vec a}{[\vec a, \vec b, \vec c]}~~~~(1)$$
$$\vec u. (\vec v \times \vec w)=\frac{1}{[\vec a, \vec b, \vec c]^3}[(\vec a \times \vec b).(\vec b \times \vec c)\times (\vec c \times \vec a)]~~~~(2)$$
Let $(\vec b \times \vec c)=\vec p $,
Then$$(\vec b \times \vec c)\times (\vec c \times \vec a)=\vec p \times (\vec c \times \vec a)=(\vec p .\vec a)\vec c-(\vec p. \vec c)\vec a=[\vec b, \vec c, \vec a]\vec c-[\vec b, \vec c, \vec c]\vec a=[\vec a, \vec b, \vec c]\vec c~~~(3)$$
Using this (2) becomes:
$$\vec u. (\vec v \times \vec w)=\frac{[\vec a, \vec b, \vec c]^2}{[\vec a, \vec b, \vec c]^3}=\frac{1}{[\vec a, \vec b, \vec c]}.$$


问:
Length of Triangle BCD

Hey, well I'm doing some higher level revision and I'm stuck...

In the diagram triangle BCD is mathematically similar to triangle ACE.

So what is the length of BD? How do you work it out?

Title: Length of Triangle BCD

Body: $$4:DB=10:7\frac{1}{2}\Rightarrow DB=\frac{7\frac{1}{2}\cdot 4}{10}=3$$


答:
First, note that $$b_2\times b_3=\frac 1{[a_1,a_2,a_3]^2}((a_1\times a_2)\times (a_1\times a_3))$$

since $(-y)\times x=x\times y$ (cross product is anticommutative) and $(\alpha u\times\beta v)=\alpha\beta(u\times v)$ where $\alpha,\beta$ are scalars and $u,v$ are vectors.

We have, by a property of cross-product, $$(a_1\times a_2)\times (a_1\times a_3)=(a_1\cdot(a_2\times a_3))a_1=[a_1,a_2,a_3]a_1$$

So, we have $b_2\times b_3=\frac 1{[a_1,a_2,a_3]}a_1$ and finally,

$$[b_1,b_2,b_3]=b_1\cdot \frac 1{[a_1,a_2,a_3]}a_1=\frac 1{[a_1,a_2,a_3]}(a_1\cdot b_1)=\frac 1{[a_1,a_2,a_3]}$$

since $a_1\cdot b_1=\dfrac {[a_1,a_2,a_3]}{[a_1,a_2,a_3]}=1$

Title: How to calculate this scalar triple product?

Body:
I have been trying for over an hour to calculate this scalar triple product but I just can't succeed, I always get some crazy long expression that I can't do anything with.

We have $$\overrightarrow{b_1}=\frac{\overrightarrow{a_2}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_2}=-\frac{\overrightarrow{a_1}\times \overrightarrow{a_3}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})} \qquad \overrightarrow{b_3}=\frac{\overrightarrow{a_1}\times \overrightarrow{a_2}}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$

I have to calculate $\overrightarrow{b_1}\cdot(\overrightarrow{b_2}\times\overrightarrow{b_3})$.

The solution given in the book is $$\frac{1}{\overrightarrow{a_1}\cdot(\overrightarrow{a_2}\times\overrightarrow{a_3})}$$


答:
Notation, $\vec a \cdot \vec b \times \vec c = [\vec a \, \vec b \, \vec c]$

We use vector quadruple product formula

$$(\vec a \times \vec b)\times(\vec c \times \vec d)=[\vec a \, \vec b \, \vec d]\vec c - [\vec a \, \vec b \, \vec c]\vec d$$

to obtain $$\vec b_2 \times \vec b_3 = \dfrac{-(\vec a_1 \times \vec a_3)\times(\vec a_1 \times \vec a_2)}{[\vec a_1\, \vec a_2 \, \vec a_3 ]^2}$$
$$ = \dfrac{-[\vec a_1\, \vec a_3 \, \vec a_2 ] \, \vec a_1}{[\vec a_1\, \vec a_2 \, \vec a_3 ]^2}$$
$$ = \dfrac{\vec a_1}{[\vec a_1\, \vec a_2 \, \vec a_3 ]}$$

Hence $$[\vec b_1 \, \vec b_2 \, \vec b_3] = \dfrac{(\vec a_2 \times \vec a_3 \cdot \vec a_1)}{[\vec a_1 \, \vec a_2 \, \vec a_3]^2}$$
$$ = \dfrac{1}{[\vec a_1 \, \vec a_2 \, \vec a_3]}$$


问:
A be a matrix satisfying ${(AA^t)}^r=I$ then when it is invertible
This question was asked in my linear algebra mid term and I was unable to solve it.

Let m ,n ,r be natural numbers . Let A be an m*n matrix with real entries such that ${(AA^t)}^r =I $ where I is m * m identity matrix. Then which  of following is / are true:

A m =n

B $ AA^{t}$ is invertible

C $A^{t} A $ is invertible

D if m=n , then A is invertible

I am really confused in this question due to $A A^{t}$a nd was unable to think which option are true or any contradictions in options . It's my humble request to guide you through this question.

I followed Hoffman and Kunze Linear algebra.

Thanks!!

答:
The question has already been answered but you can also prove (B) and (D) in a different way as follows:

Let $(AA^T)^r=I$. If $r=1$ then $(AA^T)=I$ , since $I$ is invertible so $(AA^T)$ is invertible.

If $r>1$ then $(AA^T)(AA^T)^{r-1}=I$. So $(AA^T)^{r-1}$  is the inverse of $(AA^T)$ and hence $(AA^T)$ is invertible. This proves (B).

Now suppose that $n=m$ then both $A$ and $A^T$ are $n \times n$ matrices.

Again if $r=1$ then $AA^T=I$ , so $A^T$ is the inverse of $A$ and hence $A$ is invertible.

Now suppose $r>1$. Since $(AA^T)(AA^T)^{r-1}=I$ $\Rightarrow$  $AA^T(AA^T)^{r-1}=I$ .

Take $B=A^T(AA^T)^{r-1}$ then $B$ is also an $n \times n$ matrix and it satisfies $AB=I$ so $B$ is the inverse of $A$ and hence $A$ is invertible. This proves (D).


答:
Use $x=\tan\theta$. Or, if you are familiar with hyperbolic functions, use $x=\sinh t$.

A somewhat less common substitution that works nicely is $x=\frac{1}{2}\left(t-\frac{1}{t}\right)$.  Then $\sqrt{x^2+1}=\frac{1}{2}\left(t+\frac{1}{t}\right)$.

Title: Arc length of logarithm function

Body:
I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$.

So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$
I am having trouble calculating the integral. I tried to do substitution, but I still fail to think of a way to integrate it. This is what I have done so far:
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}$$
$$x=\frac{2u}{u-1}$$
$$dx=\frac{2}{(u-1)^2}du$$
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}=u-\frac{1}{2}+\frac{1}{2u}$$
$$\int\sqrt{1+\frac{1}{x^2}}dx=2\int\frac{u-\frac{1}{2}+\frac{1}{2u}}{(u-1)^2}du$$
And I am stuck.


问:
How do I approximate the function?
Let $f$ be a continuous function on the unit circle $T=\{|z|=1\}$. Show that $f$ can be approximated uniformly on $T$ by a sequence of the polynomial in z if and only if $f$ has an extension $F$ that is continuous on the closed unit disk.

Can anyone suggest to me how I do the approximation of complex function?

答:
Let $0<r<1$ and define $F_r(z)=F(rz)$, for $|z| \le 1$.

Since $F$ is analytic in the interior of the unit disk, it can be uniformly approximated on compact subsets by polynomials in $z$, in particular, on $\overline{D_r}(0)$. And since $F$ is continuous on the closed unit disk, it is also uniformly continuous there, so $F_r \to F$ uniformly on the closed unit disk as $r \to 1^-$. Glueing these two pieces should give the desired approximation.

Title: How do I approximate the function?


问:
Evaluate $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}$
$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}$$

I got asked this question in a group, and solved it the following way:

$$\lim\limits_{n\to\infty}\left(
\int\limits_1^{n+1}\frac{x}{x^2+n^2}\space dx\leq \sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq \int\limits_1^n \frac{x}{x^2+n^2}\space dx + f(1)\right)
$$

$$\frac{\ln2}{2}\leq \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq\frac{\ln2}{2}$$

Therefore:

$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2} = \frac{\ln2}{2}$$

The guy who asked me the question said that the answer key said $\frac{\ln\left(\frac{5}{2}\right)}{2}$ but also said that it might be incorrect, so I don't have the answer.

What I want to ask is that how come this sum is not equal to $0$? We were able to use the squeeze theorem because the limit for $f(1)$ goes to zero. Every other value after $f(1)$ also goes to $0$ (even faster?) and that had me thinking about how this summation can be equal to such a value.

Can anyone please explain how this happens? And it would be great if you could also verify or correct my solution.

Body: $$S=\lim_{n \to \infty} \frac{k}{n^2+k^2}=\lim_{n \to \infty} \sum_{k=0}^{n}\frac{1}{n} \frac{k/n}{1+(k/n)^2}=\int_{0}^{1}\frac{x}{1+x^2} dx=\frac{1}{2} \ln(1+x^2)|_{0}^{1}$$ $$=\frac{1}{2} \ln 2$$


问:
Show $(ab)^{1/n}=a^{1/n} b^{1/n}$ for any $a,b >0$ and $n \in \mathbb N$
So, what I have in hand is the definition of the nth root of a real number, that is $y=x^{1/n}$ is the number that satisfies $y^n=x$. And I also have $(ab)^n=a^nb^n$ for $n\in \mathbb N$, $a,b >0$.

How can I go from here to $(ab)^{1/n}=a^{1/n} b^{1/n}$?

I tried to write things like $ab=((ab)^n)^{1/n}$ using the definition. But I can only get $((ab)^n)^{1/n}=(a^n)^{1/n}(b^n)^{1/n}$ and I am stuck.

Body: $(a^{1/n}b^{1/n})^n=\underbrace{(a^{1/n}b^{1/n})\cdot(a^{1/n}b^{1/n})\cdot\cdot\cdot(a^{1/n}b^{1/n})}_{n\text{ times}}=\underbrace{(a^{1/n}\cdot\cdot\cdot a^{1/n})}_{n\text{ times}}\times\underbrace{(b^{1/n}\cdot\cdot\cdot b^{1/n})}_{n\text{ times}}=ab$