问:
$a^b+b^c+c^a$ is a multiple of 8 if and only if $a+b+c$ is multiple of 8
Let $a, b, c$ are odd natural numbers. Is "$a^b+b^c+c^a$ is a multiple of 8 if and only if $a+b+c$ is multiple of 8" right?

I don't know how to begin, please help me.

答:
Examine the ‘if’ requirement of the proof (this is usually the easiest part of a proof). For this we require in the first instance that $8\mid a+b+c$. Now $a$, $b$, and $c$ are odd numbers, so have the form $a=2l+1$, $b=2m+1$, $c=2n+1$. Look at their sum modulo $8$:
$$a+b+c=2(l+m+n)+3\equiv\pm1,\pm3\pmod{8}$$
and so $8\nmid a+b+c$ as this congruence is never $0$ modulo $8$. (To see why allow $l+m+n$ take the values $0$ through $7$ in the congruence and check whether the answer is divisible by $8$-note you only need to actually check the multiples $2\times0$ to $2\times3$, after which the congruences repeat.) So a false start ‘disproving’ the claimed hypothesis!

ContentLicense: CC BY-SA 4.0


问:
Given f is analytic in D and satisfy |f(z)|->1 as |z|->1 then prove that f is rational
Let $f$ be analytic in $D$ and satisfy $|f(z)|\rightarrow 1$ as $|z|\rightarrow 1$. Prove $f$ is rational. I don't know how to go about this. I don't need Blachke products to verify this. Is there a simpler way? I do know that if we have $\phi$ is one-to-one analytic map of $D$ onto $D$, where $\phi (z)=\frac{z-a}{1-wz}$, then would that serve as a rational function that satisfies the conditions?

I am not sure how to proceed. Anyone know how to solve it?

答:
The standard proof of this involves noticing that the hypothesis implies that $f$ has only finitely many zeroes in the unit disc (as $|f| > 1/2$ for $|z|$ close enough to $1$), so one cn take the corresponding Blaschke product $B$ with the same zeroes and notice that $f/B, B/f$ do not vanish and $|f/B| \to 1, |B/f| \to 1, |z| \to 1$ hence by maximum modulus $|f/B| \le 1, |B/f| \le 1$ in the unit disc, hence $|f/B|=1$ there, $f=\alpha B, |\alpha|=1$

Edit later - as per comments, I thought of a different way of doing it as I realized that the hypothesis implies that $u=\log |f|$ is harmonic in a neighborhood $r <|z| <1$ (as before $f$ has only finitely many zeroes in the closed disc) and $u(z) \to 0, |z| \to 1$ which means by reflection that $u$ is extendable to a harmonic function on some annulus containing the circle $r <|z| <R, R>1$, but now $u=\Re F + c\log z$ for some analytic $F$ and some $c$ on the larger annulus and by analytic continuation $F=f$ on the original part inside the unit disc and $c=0$ hence $f$ extends to an analytic function on the larger annulus so is continuous on the closed unit disc; but now clearly we can extend $f$ to the whole plane as a meromorphic function with finitely many poles, while $\infty$ is a pole or removable depending whether $f(0)=0, \ne 0$ by the usual $f(w)=\frac{1}{\bar f(1/\bar w)}, |w| >1$ and then since $\infty$ is not essential singualarity it immediately follows that $f$ is rational so done!


问:
Convergence of definite integrals given unknowns.
Given the following equation,
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx
$$
I am supposed to find the condition on the constant $a$ such that the above integral is convergent.

Substituting $u = \ln(x)$, we get

$$
\frac{du}{dx} = \frac{1}{x}
$$

Where the integral

$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx = \int_{0}^{\ln 2} \frac{du}{dx}\frac{1}{u^{a}}dx = \int_{0}^{ln2} \frac{1}{u^{a}}du
$$

simplifies to

$$
\frac{(\ln 2)^{1-a}}{1-a}
$$

When $a = 1$, the above diverges. However, I am unsure of what values the above equation converges to, i.e when $a>1$ or $a<1$

Body:
Take $x=e^t$, then
$$I=\int_{1}^{2} \frac{dx}{x \ln^a x}=\int_{0}^{\ln 2} \frac{dt}{t^a}=\frac{1}{1-a}|_{0}^{\ln 2}= \frac{1}{1-a}[(\ln 2)^{1-a}-0^{1-a}]= \text{finite if}~ a<1.$$


问:
Testing for Orthogonality
When testing for orthogonality, why is it that when $\vec{v} \cdot \vec{w}=0$ do we say that these vectors are perpendicular? What does the dot product do for us that we end up with that conclusion?</p>

答:
It may be the inspiration for such a name comes from what happens in the plane $\,\Bbb R^2\,$, where two vectors (let's anchor them say at the origin) are perpendicular iff the product of the slopes of the lines on which they lay is minus one , so

$$(a,b)\perp(x,y)\iff \frac ba\frac yx=-1\iff ax+by=0$$

and since the usual, euclidean dot product in the real plane is $\,\;(a,b)\cdot (x,y):=ax+by\,\;$ there you have a relation between the two things.

ContentLicense: CC BY-SA 3.0

Title: Testing for Orthogonality


答:
Let's be a bit more formal here. First of all, it indeed suffices to consider the transformed integral
$$
\int_0^{\ln(2)} \frac 1{u^a}\,du = \int_0^{\ln(2)} u^{-a}\,du.
$$
Note that because of the possible division by zero, this is an improper integral. In other words, the quantity that we're after is the limit
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du.
$$
In the case that $a \neq 1$, this can be written as
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du = \lim_{b \to 0^+} \left(\frac{\ln(2)^{1-a}}{1-a} - \frac{b^{1-a}}{1-a}\right).
$$
What can we say about this limit when $a > 1$? When $a < 1$? What does the integral become in the case that $a = 1$?

Title: Convergence of definite integrals given unknowns.

Body:
Given the following equation,
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx
$$
I am supposed to find the condition on the constant $a$ such that the above integral is convergent.

Substituting $u = \ln(x)$, we get

$$
\frac{du}{dx} = \frac{1}{x}
$$

Where the integral

$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx = \int_{0}^{\ln 2} \frac{du}{dx}\frac{1}{u^{a}}dx = \int_{0}^{ln2} \frac{1}{u^{a}}du
$$

simplifies to

$$
\frac{(\ln 2)^{1-a}}{1-a}
$$

When $a = 1$, the above diverges. However, I am unsure of what values the above equation converges to, i.e when $a>1$ or $a<1$


问:
Calculate integral of a closed curve without Green's Theorem
I used Green's Theorem to calculate the integral of this curve, I got $4i\sqrt{2}$ . However, I also want to calculate it through the definition of the integral $\int_{\gamma}{\bar{z}}dz$, by splitting the line segments and assigning bounds. Can someone please help me out?

For Green's Theorem, I did $\frac{1}{2} \int_{2e^{-i\pi/4}}^{2e^{i\pi/4}}{(2cos\theta*2cos\theta - {2sin\theta*(-2sin\theta))d\theta}}$ = $4i\sqrt{2}$.

Am I headed in the right direction, also am I applying Green Theorem properly? Of course, both answers should be equivalent.

答:
$\newcommand{\i}{\mathrm{i}}$
$\newcommand{\C}{\mathbb{C}}$
$\newcommand{\e}{\mathrm{e}}$
$\newcommand{\d}{\mathrm{d}}$

Let $\zeta^s=\e^{\i \pi s/4}$, $\zeta = \zeta^1$.

Write $I$ for $[0,1]$.
$$\begin{align}
f_0 &: I \to \C & f_0(t) &= 2\zeta^{-1}t \\
f_1 &: I \to \C & f_1(t) &= 2\zeta^{-1-2t} \\
f_2 &: I \to \C & f_2(t) &= 2\zeta t\text{.}
\end{align}$$

Write $C_i$ for the image of $f_i$. Then

$$\gamma = C_0 + C_1 + C_2\text{.}$$

Note that $\bar{z}=-\i z$ on $C_0$, $\bar{z}=4/z$ on $C_1$, and $\bar{z}=\i z$ on $C_2$. Then

$$\begin{split}
\int_{\gamma}\bar{z}\d z
  &= \int_{C_0}\bar{z}\d z
   + \int_{C_1}\bar{z}\d z
   + \int_{C_2}\bar{z}\d z \\
  &= -\i \int_{C_0}z\d z
   + 4 \int_{C_1}\frac{\d z}{z}
   + \i \int_{C_2}z \d z \\
  &= -\i \int_0^{2\zeta^{-1}} z \d z
   + 4\i \int_{-\pi/4}^{-7\pi/4}\d\theta
   + \i \int_{2\zeta}^0 z \d z \\
  &= \left.-\i\frac{z^2}{2}\right\rvert_{z=0}^{2\zeta^{-1}}
   + 4\i \left.\theta\right\rvert_{\theta=-\pi/4}^{-7\pi/4}
   + \left.\i\frac{z^2}{2}\right\rvert_{z=2\zeta}^{0} \\
  &= -2\i\zeta^{-2}
   - 6\pi \i
   - 2\i\zeta^2 \\
  &= -6\pi\i
\end{split}$$

Using Green's theorem to convert to an area integral: write A for the region whose boundary is $-\gamma$. Then

$$\begin{split}
\int_{\gamma}\bar{z}\d z
  &= -\int_A \d\bar{z}\wedge \d z\\
  &= -2\i\int_A \d x\wedge \d y \\
  &= -2\i\lvert A\rvert \\
  &= -2\i(3\pi) \\
  &= -6\pi \i\text{.}
\end{split}$$


问:
Conditional combination
I have 3 objects A, B and C

and 3 baskets 1, 2 and 3

A can go to any of the basket

But B can only go to the same basket with A or in the baskets that has higher number than what A chose.

C can only go to the same basket with A or B or in the baskets with higher number than what A and B chose

For example, if A is in basket 1, B can go to any of the 1,2,3. If B chose basket 2 then C can either go to 2 or 3. If A is in basket 3, B and C then have no choice and will have to go to basket 3.

I was struggling how to formulate it in a general way to find the total number of combination, so that I can do it for any number of objects and baskets.

答:
The general problem is:  "how many non-decreasing sequences of length $n$ can be formed from $\{1,\cdots, k\}$".

Call the answer $F(n,k)$

This is a Stars and Bars problem in disguise.  Indeed, every such sequence can be written (uniquely) as $$1^{a_1}2^{a_2}\cdots k^{a_k}$$

Where the $a_i$ are non-negative integers and $a_1+a_2+\cdots +a_k=n$.

(Note:  as to notation, the $a_i's$ just tell us how often each character $i$ appears.  Thus $1^32^2$ denotes the sequence $11122$, for example).

But Stars and Bars enables us to count the number of the good $k-$tuples $(a_1, \cdots, a_k)$, and we get the final answer $$F(n,k)=\binom {n+k-1}n$$

Check:  the stated problem has $n=3,k=3$ so we get $\binom 53=10$ which is easily confirmed (by just writing out all the cases, say).

Title: Conditional combination


答:
We have that

$$\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx=\int_{0}^{1} \frac{1}{(1+x)(\ln (1+x))^{a}} dx$$

with $(\ln (1+x))^{a} \sim x^a $ as $x \to 0^+$ and therefore

$$\frac{1}{(1+x)(\ln (1+x))^{a}} \sim \frac1{x^a}$$

and we can conclude by limit comparison test.


问:
non-recursive function

Give a direct proof that the set $\{x|\Phi_x(1) \downarrow\}$ (which is a set of program numbers that halt on input $1$) is not recursive.

I've got an idea that indirect proof must work. Assuming that the set is recursive and ... But I have no idea how to go on from here.

The course source is Computability, Complexity, and Languages, Second Edition: Fundamentals of Theoretical Computer Science by Davis. So I'm familiar with notation used there and understand it better, if possible.

答:
Try reduce the problem to Halting problem ,

Halting problem: given M, Turing machine and input x,decide if M(x) halts .

so you want to find a computable functions that take instance of Halting problem and convert it to instance of this problem .

Title: non-recursive function


问:
Why is an element that belongs to a set that belongs to another set not an element of the second set?
If $A \in B$ and $a \in A$, does $a \in B$?
I've been told it does not.
Why does it not?

答:
Put a sandwich in a bag, and put the bag in a box. The box contains a bag.

Put a sandwich in a box. The box contains a sandwich.


问:
Bijection of sets with cartesian product?
Show that given two bijections:
$$f: A \rightarrow \{ 1,...,m \}$$
$$g: B \rightarrow \{ 1,...,n \}$$
there exists a bijection:
$$A \times B\rightarrow \{ 1,...,mn \}$$

What have I tried so far?

So, I know that for there to be  a bijection the function must be both injective and surjective. I figured to prove this, I must show both. So I thought I may need to construct the bijection using a table and showing these but I'm not sure if this would be best?

答:
Hint : Consider $h : A \times B \rightarrow \lbrace 1, ..., mn \rbrace$ defined for all $(a,b) \in A \times B$ by
$$h(a,b) =(f(a)-1)n + g(b) $$

Show that $h$ is well-defined and bijective.


问:
If $I=\int_{0}^{\frac 12} \frac{1}{\sqrt{1-x^{2n}}} dx$, then prove that $I<1$ for $n\ge 1$
The maximum value of $\frac{1}{\sqrt{1-x^{2n}}}$ is $\frac{2^n}{\sqrt{2^{2n}-1}}$

So the maximum value for is the integral is
$$\frac 12 \frac{2^n}{\sqrt{2^{2n}-1}}$$

Now I don’t know how to find the max value from here. I tried to use $\lim_{n\to \infty} $ but that just gave the answer $\frac 12$

Body: For $n\ge1$ and $0\le x\le1$, $x^{2n}\le x^2$ and $\tfrac{1}{\sqrt{1-x^{2n}}}\le\tfrac{1}{\sqrt{1-x^2}}$. So$$I\le\int_0^{1/2}\tfrac{dx}{\sqrt{1-x^2}}=\arcsin\tfrac12=\tfrac{\pi}{6}<1.$$


问:
$X_r$ is the number of cycles of length $r$ in a random permutation of $n$ elements.
$X_r$ is the number of cycles of length $r$ in a random permutation of $n$ elements. Find $EX_r, DX_r$.

I tried to find $EX_r$ and get it = $\frac{1}{r}$.

$DX_r = E(X_r^2) - (EX)^2$. How can i find $E(X_r^2)$?

答:
Using combinatorial classes we have the following class $\mathcal{P}$
of permutations with cycles of length $r$ marked:

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \textsc{SET}(
\textsc{CYC}_{=1}(\mathcal{Z})  + \cdots
+ \textsc{CYC}_{=r-1}(\mathcal{Z})
+ \mathcal{U}\times  \textsc{CYC}_{=r}(\mathcal{Z})
+ \textsc{CYC}_{=r+1}(\mathcal{Z})
+ \cdots).$$

This gives the EGF

$$G(z, u) = \exp\left(z+\frac{z^2}{2}
+ \cdots
+ \frac{z^{r-1}}{r-1}
+ u \frac{z^r}{r}
+ \frac{z^{r+1}}{r+1}
+ \cdots \right)
\\ = \exp\left(\log\frac{1}{1-z} + (u-1)\frac{z^r}{r}\right)
= \frac{1}{1-z} \exp\left((u-1)\frac{z^r}{r}\right)
\\ = \frac{1}{1-z} \exp\left(-\frac{z^r}{r}\right)
\exp\left(u\frac{z^r}{r}\right).$$

We have

$$\mathrm{E}[X_r] =
[z^n] \left. \frac{\partial}{\partial u} G(z,u) \right|_{u=1}
\\ = [z^n] \frac{1}{1-z} \exp\left(-\frac{z^r}{r}\right)
\exp\left(\frac{z^r}{r}\right)
\frac{z^r}{r}
= [z^{n-r}] \frac{1}{r} \frac{1}{1-z}
= \frac{1}{r}$$

where $r\le n.$ We also get

$$\mathrm{E}[X_r (X_r-1)] =
[z^n] \left.
\left(\frac{\partial}{\partial u}\right)^2 G(z,u) \right|_{u=1}
\\ = [z^n] \frac{1}{1-z} \exp\left(-\frac{z^r}{r}\right)
\exp\left(\frac{z^r}{r}\right)
\frac{z^r}{r} \frac{z^r}{r}
= [z^{n-2r}] \frac{1}{r^2} \frac{1}{1-z}
= \frac{1}{r^2}$$

where $2r\le n$ and zero otherwise. This is because when $2r\gt n$ we
must have for a given permutation that $X_r = 0$ or $X_r = 1$ so that $X_
r (X_r-1) = 0.$

 We obtain at last

$$\mathrm{Var}[X_r] =
\mathrm{E}[X_r (X_r-1)] + \mathrm{E}[X_r] - \mathrm{E}[X_r]^2
\\ = \begin{cases}
\frac{1}{r^2} + \frac{1}{r} - \frac{1}{r^2} = \frac{1}{r}
\quad\text{when}\quad 2r\le n \\
\frac{1}{r} - \frac{1}{r^2} = \frac{r-1}{r^2}
\quad\text{otherwise.}
\end{cases}.$$


答:
For example, $ 1 \in \mathbb{N} \in $ S={subsets of $\mathbb{R}$}, but $1 \notin S$.

Also, in maths we have precise meanings for subset of and member of. You are conflating the two.

Title: Why is an element that belongs to a set that belongs to another set not an element of the second set?

Body:
If $A \in B$ and $a \in A$, does $a \in B$?
I've been told it does not.
Why does it not?


问:
Biharmonic version of Poisson Integral

Background

The Poisson Integral

$$ \tilde h(r e^{i \theta}) = \frac{1}{2 \pi}\int_{-\pi}^\pi h(e^{i \phi}) P_r(\theta - \phi) d\phi$$

takes an arbitrary continuous complex-valued function $h(e^{i \theta})$ defined on the unit circle and extends it to a function $\tilde h(z)$ defined inside the unit disk such that $\tilde h(z)$ is harmonic ($\Delta \tilde h(z) = 0$) and has boundary values $h(z)$ on the unit circle.

$P_r(\theta)$ is the so-called Poisson Kernel and, among other representations,

$$P_r(\theta) = \Re \left( \frac{1 + r e^{i \theta}} {1 - r e^{i \theta}} \right) , \ \ \ \ r e^{i \theta} \in \mathbb{D}.$$

Question

Is there a different kernel function I could use which would make $\tilde h(z)$ biharmonic instead of just harmonic inside the unit disk?  That is, I'd like to find something to replace $P_r(\theta)$ with so that $\Delta^2 \tilde h(z) = 0$.

答:
After a tip from Conrad I was able to find a different paper (Tangential limit values of a biharmonic poisson integral in a disk) which gives the "Biharmonic Poisson Kernel" $\tilde P$ in terms of the regular Poisson Kernel:

$$ \tilde P_r(\theta) = \frac{1 - r^2}{2} (P_r(\theta) + P_r^2(\theta)) $$

Title: Biharmonic version of Poisson Integral


问:
A and B play a series of games. Find the probability that a total of 4 games are played.
A and B play a series of games. Each game is independently
won by A with probability $p$ and by B with probability $1 - p$. They stop when the
total number of wins of one of the players is two greater than that of the other player.

The player with the greater number of total wins is declared the winner of the series.

(a) Find the probability that a total of 4 games are played.

(b) Find the probability that A is the winner of the series.

I though I could do this using conditional probability and independent trials/bernoulli trials but I am really confused.

let's say I want to find the probability that A wins three games and B wins 1 game and add that to the probability of B wins 3 games and A wins 1 game, but I am not sure of how to do that because B has to win at least 1 game before A wins 2.

If I set $n$ = number of games played would the sample space equal $p^n * (1-p)^n$?

How do I solve this? is there an easier way?

答:
I can see that the upvoted answer is incorrect to the answer b, so I give a way to solve this type of problem recursively.

Let us call $p_A$ probability that team A wins  and $q=1-p$. First, $p_A \neq \frac{p^2}{p^1 + (1-p)^2}$.

In fact, A can win at first two stages, or B one one of the two games. In the latter case, the game restart. Translated mathematically :
$$
p_A = p^2 + 2pq \cdot p_A
$$

Solve it for $p_A$ we can find as in other correct answers : $p_A = \frac{p^2}{1 - 2pq}$


问:
Converting between coordinate systems with integrals/infinitesimals
I recently discovered for myself an interesting trick that I didn't know you were allowed to do.

Let's say I have the expression:

$ydx$

And I want to convert it to cylindrical coordinates. When I was solving this specific problem assigned to me, I did this:

$(rsin(\theta)) d(rcos\theta)$

$(rsin(\theta)) (-rsin(\theta)) d\theta$

$ydx = -r^2sin^2(\theta)d\theta $

When I went about solving the problem, I got the correct answer.

Can someone explain why this is allowed and how it works? Or if it isn't allowed, why it worked in this specific scenario? I have never seen this done before nor have I ever been taught that this is an acceptable thing to do when converting between coordinate systems.

Here is the problem that I had to work on for those who are curious:

A magnetic field intensity vector, $H$ , is described by the vector function. $H (x,y,z)=9y\hat{x}$ [A/m]. Calculate the circulation of $H$  around a circle of radius $r=7$ [meters] in the $x-y$ plane, traversed in the $+\hat{\phi}$-direction. Give the answer in units of [A]. Give three significant figures.

答:
No you are not

$d(rcos\theta)=((-rsin(\theta))+r^`cos\theta))d\theta$

it worked in that case because r is constant


问:
Consider the equation $y'(x)+ay(x)=f(x)$ where $a<0$ and $|f(x)|<M$ for all $x\in(-\infty,\infty),M>0)$
Consider the equation
$$y'(x)+ay(x)=f(x)$$
Where $a<0$ and $|f(x)|<M$ for all $x\in(-\infty,\infty),M>0$

(a) Prove that there exists a unique solution which is bounded on the whole real line.

(b) Prove that the solution of (a) is periodic when $f(x)$ is assumed to be periodic.

I've managed to show that
$$y(x)=e^{-ax}\left(\int_{0}^{x}e^{as}f(s)ds+c\right)$$
But I needed to assume that $f$ is integrable in the domain and that show existence but the part of bounded I got stuck.

答:
The conclusion is wrong. For example, let $a=-1,f(x)=0$ and then the equation becomes
$$ y'-y=0. $$
This one has solution $y=ce^x$ which is unbounded in the whole real line for $c\neq0$.

Update: If $f$ is nonconstant, let $a=-1, f(x)=\sin(x)$. The the equation becomes
$$ y'-y=\sin(x). $$
The solution is
$$ y=ce^x-\frac12[\sin(x)+\cos(x)]. $$
Thus the solution is unbounded for the whole real line for $c\neq0$.


问:
Do not understand step in solving differential equation

Hey, I am doing revision before exam and I do not understand the
following differential equation:

$y^{''}+\frac{2(y^{'})^{2}}{1-y}=0$

Solution:

We multiply it by $\frac{1}{y^{'}}$ so we got:

$\frac{y^{''}}{y^{'}}+\frac{2y^{'}}{1-y}=0$

But then we did some step and we got:

$(\ln \left ( y^{'} \right ))^{'}-2(\ln (1-y))^{'}=0$

Can someone please explain me how?

Body: Hint: $\int\frac{f'(x)}{f(x)}dx=\ln f(x)$


问:
Little oh notation in the exponent
So I saw this notation in my analytic number theory course: $|f(n)|\leq n^{o(1)}$. So I am wondering whether this is the same as the notation $|f(n)|\leq o(n)$?

答:
$o(1)$ represents an unknown function that is tending to $0$ as $n\to\infty$; so, for every $\varepsilon>0$, we have $|o(1)|<\varepsilon$ when $n$ is large enough in terms of $\varepsilon$. Therefore $|f(n)| \le n^\varepsilon$ when $n$ is large enough in terms of $\varepsilon$, which is actually much stronger than $f(n) = o(n)$.

One example of a function satisfying $|f(n)| \le n^{o(1)}$ is $f(n) = \log n$.

Title: Little oh notation in the exponent


答:
HINT:
Let us denote $[k] = \{1, 2, \cdots, k\}$. Note that $\alpha: A \times B \to [m] \times [n]$ defined by $\alpha(a, b) = (f(a), g(b))$ is a bijection (why?). Thus, we only need to find a bijection from $[m] \times [n]$ to $[mn]$.

To do this, let us order elements of $[m] \times [n]$ with the rule that
$$(a, b) < (a', b) \iff a < a' \text{ or } (a = a' \text{ and } b < b')$$
in other words, we first compare the first elements, and if they are equal, we compare the second elements. Let us write the elements of $[m] \times [n]$ in order according this rule:
$$(1, 1) \\
(1, 2) \\
\vdots \\
(1, n) \\
(2, 1) \\
\vdots \\
(m, n)$$
...and now we simply can simply map the first element in our list to $1$, the second element to $2$, and so on:
$$(1, 1) \iff 1 \\
(1, 2) \iff 2 \\
\vdots \\
(1, n) \iff n \\
(2, 1) \iff n + 1 \\
\vdots \\
(m, n) \iff mn $$
And in fact you can find a closed form for this bijection (call it $\beta$). To do this, you can note that $\beta(a, b)$ is always $b$ greater than a multiple of $n$, and that $\beta(a + 1, b)$ and $\beta(a, b)$ always differ by $n$, for any $a$ and $b$. Can you proceed from here?

In summary: We first found a bijection from $\alpha: A \times B$ to $[m] \times [n]$. Then, we found a bijection from $[m] \times [n]$ to $[mn]$. It follows that $\beta \circ \alpha$ is a bijection from $A \times B$ to $[mn]$.

Addendum: Why must $\alpha$ be a bijection?

Answer: Suppose it weren't injective. Then we could find $(a, b), (a', b') \in A \times B$ such that $\alpha(a,b) = \alpha(a', b')$. This means that
$f(a) = f(a')$ and $g(b) = g(b')$, but this contradicts the fact that $f$ and $g$ are bijections (and are hence injections).

Suppose it weren't surjective. Then we could find $(x, y) \in [m] \times [n]$ such that no element of $A \times B$ mapped to $(x, y)$. For this to be the case, it would have to be that either there existed no element $a \in A$ such that $f(a) = x$, or that there existed no element $b \in B$ such that $g(b) = y$. But this contradicts either $f$ or $g$'s surjectivity (and thus bijectivity).

Title: Bijection of sets with cartesian product?

Body:
Show that given two bijections:
$$f: A \rightarrow \{ 1,...,m \}$$
$$g: B \rightarrow \{ 1,...,n \}$$
there exists a bijection:
$$A \times B\rightarrow \{ 1,...,mn \}$$

What have I tried so far?

So, I know that for there to be  a bijection the function must be both injective and surjective. I figured to prove this, I must show both. So I thought I may need to construct the bijection using a table and showing these but I'm not sure if this would be best?


问:
Infinite intersection of finitely generated ideals in a coherent ring.
It has been claimed without proof in several answers that an intersection of two finitely generated ideals in a coherent ring is finitely generated. Thus, the finitely generated ideals in a coherent ring form a lattice. However, can an infinite intersection of finitely generated ideals fail to be finitely generated? The typical examples of coherent rings which are not Noetherian (polynomial ring in infinitely many variables over $\mathbb{Z}$, entire functions on $\mathbb{C}$) seem to have the property that infinite intersections happen to be finitely generated, but for non-general reasons. Does anyone have a nice counterexample?

My motivation is that, given the existence of such a pathological intersection, the abelian category of finitely presented modules can fail to admit certain colimits. Without a reference for the proof of the statement about finite intersections, I don't know where to begin.

[commutative and non-commutative examples accepted]

答:
I think this is a counterexample.

Let $F_2$ be the field of two elements, and consider the subring of $\prod_{i\in \mathbb N} F_2$ of "eventually constant sequences."

It is commutative and von Neumann regular, so the finitely generated ideals are all principle.  It's coherent because it is semihereditary.

Now, you can take "unit vectors" $e_i$ in this ring, and $1-e_i$ generate a cyclic maximal ideal.

But $\bigcap_{i\in\mathbb N}(1-e_{2i})R$ is an ideal whose elements have to be zero on the "even indices". Therefore every element in it is eventually zero. But clearly it's not generated by a single element, because it contains elements that are nonzero on indices in a cofinal subset of $\mathbb N$.


问:
What is the best way to express the set of sets that has no element in common with another set?
I am trying to express the following: I have a set $A$ and the powerset (set of all subsets of $A$) $P(A)$. I have another set $S \in P(A)$, and I want to get the sets in $P(A)$ with no element in common with $S$.

The way I express this is $$X \in P(A): X \cap S = \{\}$$

Is there a more concise way or a special notation from set theory?

答: So $P(A\setminus S)$ could do the job ...


答:
We have by the chain rule $$\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}$$ then $(\ln(y'(x))'=\frac{y''(x)}{y'(x)}$ and $-2(\ln(1-y(x)))'=-2\frac{-y'(x)}{1-y(x)}=\frac{2y'(x)}{1-y(x)}$ as required.

See derivatives of logarithmic functions.

Title: Do not understand step in solving differential equation

Body:
Hey, I am doing revision before exam and I do not understand the
following differential equation:

$y^{''}+\frac{2(y^{'})^{2}}{1-y}=0$

Solution:

We multiply it by $\frac{1}{y^{'}}$ so we got:

$\frac{y^{''}}{y^{'}}+\frac{2y^{'}}{1-y}=0$

But then we did some step and we got:

$(\ln \left ( y^{'} \right ))^{'}-2(\ln (1-y))^{'}=0$

Can someone please explain me how?


问:
Find positive definite $X$ such that $AXA^T=\alpha X$.
Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$

Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $(1-\alpha)>0$ this looks like the Lyapunov equation here. However, the analytic solution to the Lypunov equation has the form
$$
X = \sum_{k=0}^\infty A^kQ(A^T)^k = (1-\alpha)\sum_{k=0}^\infty A^kX(A^T)^k
$$
So, even if such equation have a solution, I don't have a clue on how to obtain it from here.

Do you have other thoughts/suggestions?

答:
If the equation is solvable, by taking determinants on both sides, we must have $\alpha=\det(A)^2$. If this is the case, we may put $B=\det(A)^{-1}A$ and equation reduces to $BXB^T=X$.

It is solvable if and only if $B$ is similar to a real orthogonal matrix.

Suppose $BXB^T=X$. Then $(X^{-1/2}BX^{1/2})(X^{-1/2}BX^{1/2})^T=I$. Hence $X^{-1/2}BX^{1/2}$ is real orthogonal.

Conversely, suppose $B=PQP^{-1}$ for some real orthogonal matrix $Q$. Let $X=PYP^T$ (i.e. $Y=P^{-1}X(P^{-1})^T$). The equation then further reduces to $QY=YQ$. Hence any positive definite matrix $Y$ that commutes with $Q$ will give rise to a solution $X$. In particular, we can always take $Y=I$ to obtain a solution $X=PP^T$.


问:
What is the covariance of the outer product of a vector with itself?
Given some $n \times 1$ vector $\mathbf{x}$, with the corresponding $n \times n$ covariance matrix $\Sigma_x$, is there a straightforward way to find the $n^2 \times n^2$ covariance matrix of:

$\mathbf{x} \mathbf{x}^\mathrm{T}$

in terms of the original vector $\mathbf{x}$ and covariance matrix $\Sigma_x$? The elements of $\mathbf{x}$ cannot be assumed to be uncorrelated.

答:
Note that the vectorization of the matrix $V = \mathbf x \mathbf x^T$ is the $n^2 \times 1$ vector $\mathbf v = \mathbf x \otimes \mathbf x$, where $\otimes$ denotes a Kronecker product. With that, we have
$$
\mu_\mathbf v = \operatorname{vec}(\mu_{\mathbf x \mathbf x^T}) = \operatorname{vec}(\Sigma_{ \mathbf x}),
$$
and
$$
K_{\mathbf v \mathbf v} = E(\mathbf v \mathbf v ^T) - \mu_\mathbf v \mu_\mathbf v^T\\
= E((\mathbf x \otimes \mathbf x)(\mathbf x \otimes \mathbf x)^T) - \operatorname{vec}(\Sigma_{ \mathbf x})\operatorname{vec}(\Sigma_{ \mathbf x})^T \\
= E((\mathbf x \mathbf x^T) \otimes (\mathbf x \mathbf x^T)) - \operatorname{vec}(\Sigma_{ \mathbf x})\operatorname{vec}(\Sigma_{ \mathbf x})^T. \\
$$

Note that the components of the first expectation will consist of $4$th moments which cannot be expressed in terms of the entries of $\Sigma_{\mathbf x}$. However, these can perhaps be expressed in terms of the kurtosis and cokurtosis.


答: This happens because the derivative of the log of a function is its reciprocal and then to complete we must still apply the Chain Rule of differentiation, multiplying the previous result with its derivative.</p>


答: You could use union of disjoint subsets to achieve the same thing: $X, Y \in P(A), X \sqcup Y \equiv A$

Title: What is the best way to express the set of sets that has no element in common with another set?

Body:
I am trying to express the following: I have a set $A$ and the powerset (set of all subsets of $A$) $P(A)$. I have another set $S \in P(A)$, and I want to get the sets in $P(A)$ with no element in common with $S$.

The way I express this is $$X \in P(A): X \cap S = \{\}$$

Is there a more concise way or a special notation from set theory?


问:
How to compute ratio of decreasing value
I have a value that at 40 -> 80%, I want to reduce 80% to 50% and compute x such that x -> 50% would simple ration do this? Note percentage here is CPU usage and I want to increase resource from 40 to some value so that CPU usage can come down from 80% to 50%. Is my reasoning correct here?

```<code>    40  -> 80%
X + 40  -> 50%
</code>```

I want to increase 40 to some value so that percentage is reduced to 50%.I did simple ratio which gives me 25, would adding 25 to 40 the right approach here?

答:
Assuming that the flow of data to be processed is constant, we have that

$\#40$ corresponds to an utilization rate of $\to 80 \%$

which means that

$\#32$ corresponds to an utilization rate of $\to 100 \%$

and more in general

$\#N$ corresponds to an utilization rate of $\to P \%$ such that $N\cdot P=3200$

then we are looking for an utilization rate of $\to 50 \%$ that is

$$N\cdot 50=3200 \implies N=64$$

therefore we need to increase of $x=24$.


问:
How to find the number of a students using sets when different clauses are stated using a Venn diagram or a table?
The problem is as follows:

From a group of $90$ students it is known that: $12$ prefer
mathematics, but not literature; $27$ prefer literature, but are not
$18$ years old; $18$ who do not prefer literature do not prefer
mathematics and are $18$ years old; $7$ prefer literature and are $18$
years old, but do not prefer mathematics, $4$ prefer mathematics and
literature and are $18$ years old. How many students who are not $18$
years old, do not prefer mathematics and not literature?

The alternatives given in my book were:

$\begin{array}{ll}
1.&22\\
2.&24\\
3.&25\\
4.&21\\
5.&20\\
\end{array}$

I'm confused exactly how to arrange this information in a table or in a Venn diagram to find the requested information. Can someone help me?.

What I've attempted so far was to make up a table as this:

$$\begin{array}{|c|c|c|c|}
\hline
\, &\textrm{Maths}  &\textrm{Literature}  &\textrm{Neither maths nor literature} \\ \hline
\textrm{More than 18 years} &x  &y  &z\\ \hline
\textrm{Less than 18 years} &a  &b  &c\\ \hline
\end{array}\\$$

But that's how far I went. Does it exist a way to solve this using this approach or a Venn diagram?.

I really would appreciate someone could include a drawing or table so I can better understand this and avoid getting confused with the equations.

Can someone help me here please?.

答:
Here is one way to lay out the information in the problem statement.

Explanation: For example, "FT" (think False-True) under Math-Literature means that in that column students do not prefer math and they do prefer literature.  The "T" to the right of 18 means that in that row students are 18.  So the $7$ in the column with "FT" and the row with "T" means that $7$ students fall in the category of being 18 years old, do not like math, and do like literature. The column to the right of the 2 by 4 grid is for row totals (but none are given in this problem), and the row below the grid is for column totals; so we see $12$ is the sum of the two numbers in the Math-Literature "TF" column. The cell to the lower right of the grid is for a grand total; so the sum of all the cells in the 2 by 4 grid is $90$.  You could also say that $90$ is the sum of the row totals or the sum of the column totals.

The only number given in the problem statement that is awkward in the sense of not having a well-designated cell is the $27$ in the 18 "F" row which is placed between the Math-Literature "FT" and "TT" columns. This is meant to show that the sum of the two cells in the "FT" and "TT" columns is $27$.

The problem asks us to find the $x$ in the upper left corner of the 2 by 4 grid, in the intersection of the 18 "F" row and the Math-Literature "FF" column, i.e. the number of students who are not 18, do not like math, and do not like literature.

Hint: You might think about what you can do about filling out the missing data in the column subtotals--the row below the 2 by 4 grid.


问:
$ X = A \cup B $ where $ A $ and $ B $ are closed and $ A \cap B $ is locally connected. Show that $A$ and $B$ are locally connected.

Let $(Y, \tau)$ a locally connected topological space. suppose $ Y = A \cup B $ where $ A $ and $ B $ are closed and $ A \cap B $ is locally connected. Show that $A$ and $B$ are locally connected.

let's see that A is locally connected. Let $x \in A$ and $U \subset A$ open, then $U = A \cap W$, where $ W $ is an open of $ X $.

We have two cases if $x \in A \setminus B$ and if $x \in A\cap B$.

If $x \in A \setminus B$. Since $ A \setminus B \subset A $ is open in $X$, let's take the component $C$ of $x$ in $ (A \setminus B) \cap W$, $C$ is connected and is also open for $ X $ is locally connected and $C \subset A \cap W=U$.

If $x \in A \cap B$, I don't know how to prove this case, I would appreciate any help.

答:
Let us prove the following

Theorem. Let $Y$ be locally connected and $C,D$ be closed subspaces such that $C \subset D$, $D \setminus C$ is open in $Y$ and $C$ is locally connected. Then $D$ is locally connected.

Before we give a proof, let us come to corollaries.

Corollary 1. Let $Y$ be locally connected and $A,B$ be closed subspaces such that $A \cup B  = Y$ and $A \cap B$ is locally connected. Then $A$ is locally connected.

This is the contents of your question.

Proof: Set $D = A$ and $C = A \cap B$. Then $D \setminus C =  A \setminus (A \cap B) = A \setminus B = (A \cup B) \setminus B = Y \setminus B$. Hence $D \setminus C$ is open in $Y$.

Corollary 2. Let $Y$ be locally connected and $A \subset Y$ be a subset such that the boundary $\partial A = \overline A \setminus \operatorname{int}(A)$ is locally connected. Then $\overline A$ is locally connected.

Note that this is the contents of Local connectedness in the boundary implies local connectedness in the closure.

Proof. Set $D = \overline A$ and $C = \partial A$.

Proof of Theorem.

Clearly $D$ is locally connected at all points of $D \setminus C$ since this set is an open subset of the locally connected space $Y$.

It remains to show that $D$ is locally connected at all points of $C$.

So let $x \in C$ and let $U_D$ be an open neighborhood of $x$ in $D$. Let $U$ be an open subset of $Y$ such that $U \cap D = U_D$. Then $U_C= U_D \cap C = U \cap C$ is an open neighborhood of $x$ in $C$.

Since $C$ is locally connected, there exists a connected open subset $V_C$ of $C$ such that $x \in V_C \subset  U_C$. Let $V$ be an open subset of $Y$ such that $V \cap C = V_C$. W.l.o.g. we may assume $V \subset U$ (otherwise $V' = V \cap U$ satisfies $V' \subset U$ and $V' \cap C = V \cap U \cap C = V_C \cap U = V_C$ since $V_C\subset U_C \subset U$).

For each $y \in V_C$ let $W_y$ be a connected open subset of $Y$ such that $y \in W_y \subset V$. Define $W = \bigcup_{y\in V_C}W_y$. Since $V_C$ is connected and each $W_y$ has a non-empty intersection with $V_C$, the set $W$ is a connected open subset of $Y$ with $x \in V_C \subset W \subset V$. Note that $W \cap C = V_C$ because $V \cap C = V_C$.

Define $W_D = W \cap D$. This is an open subset of $D$ such that $x \in V_C \subset W_D \subset U_D$. We claim that $W_D$ is connected which will prove that $D$ is locally connected at $x$.

Assume there exist non-empty disjoint open subsets $R, S$ of $W_D$ such that $R \cup S = W_D$. Note that $R,S$ are also closed in $W_D$.  W.l.o.g. $x \in R$. Thus $V_C \cap R \ne \emptyset$. It is impossible that also $V_C \cap S \ne \emptyset$ because $V_C$ is connected. Hence $V_C \subset R$. This implies $S \subset W_D \setminus V_C$.

$S$ is open in $W$ : The set $S$ is open in $W_D$, thus also open in $W_D \setminus V_C = W \cap D \setminus W \cap C = W \cap (D \setminus C)$. Claim 1. follows because $W \cap (D \setminus C)$ is open in $W$.

$S$ is closed in $W$: The set $S$ is closed in $W_D$ and $W_D = W \cap D$ is closed in $W$.

Thus $S$ is a non-empty clopen subset of $W$ such that $S \ne W$. This is a contradiction because $W$ is connected.


问:
Rademacher functions independent random Variables
You call $r_n$=sign($sin^{2n}(\pi t))$ Rademacher functions.

I read " the $r_n's$ are independant random variables"

Why is this true ? I don't get it...

答: because the probability of its values are independent such that $p ( r_1=e_1 , r_2=e_2 , ... , r_n=e_n ) = p ( r_1=e_1 )  p ( r_2=e_2 ) .... p ( r_n=e_n )$ where $p$ is the probability measure and $e$ are borel subsets of the space , if this satisfied then $r_n$'s are independent random variables , you can take some specific values of n it would be much easier to prove it , GOOD LUCK.


问:
Why is an element that belongs to a set that belongs to another set not an element of the second set?
If $A \in B$ and $a \in A$, does $a \in B$?
I've been told it does not.
Why does it not?

答:
How about:

There is a word document on "My Desktop"

The word document contains several pictures, one of which is a picture of a rabbit.

However, there is not a picture of a rabbit on My Desktop.

$$$$

This is hopefully a clearer illustration because you can see the word document is there when looking at "My Desktop". However, you cannot see the picture of the rabbit when looking at "My Desktop". In other words, the word document acts like a black box, which is what we have when it comes to sets.

$$$$

Addendum:

However, it must be noted that sets do not always have a "meaning" and a "purpose" behind them. Sets are mathematical objects and the most important thing is knowing the precise mathematical definitions of things related to sets, e.g. you should know the definition of member of, subset of, complement of, intersection, union etc. The "meaning" and "purpose" of sets, for most people, I imagine, is slowly illuminated to them, the more they work with sets.


问:
Find the formula of a sequence using generating functions
The given sequence is $\sum_{k=1}^{n}kx^{k}$.

How can I do this? I think I should start with the already known generating function:

$A(x) = \sum a^{n}x^{n} = \frac{1}{1 - ax}$

And then I should do something with this function so that I finally come up with the given one. But well, how can I do this?

$\sum_{k=1}^{n}kx^{k} = \sum_{k>=1} [A(x) - 1]k$

I could only get something like this, but first of all it's not the formula without sums, and second I'm not sure it's correct.

Could you help me?

答:
An idea without (properly) generating functions:

$$\forall\,|x|<1\;,\;\;\frac1{1-x}=\sum_{k=0}^\infty x^k\stackrel{\text{derivative}}\implies\frac1{(1-x)^2}=\sum_{k=1}^\infty kx^{k-1}\;\;\ldots$$


问:
relationship quadratic program and linear program
Consider the following quadratic programming problem

$$
\min\limits_{x \in S} f(x) = c^{\text{T}}x + \frac{1}{2}x^{\text{T}}Qx,
$$
where $S \subseteq \mathbb{R}^n$ is a convex compact set, $Q$ is an $n \times n$ symmetic matrix and $c \in \mathbb{R}^n$. Suppose $x^*$ is the global solution of the above problem. I have found without proof that $x^*$ is also optimal for the linear program $\min\limits_{x \in S} \nabla f(x^*)^{\text{T}}x$? Can someone please provide a proof? What is the intuition behind this result?

答:
I will use following theorem:

$P$: $\min f(x)$ subject to $x \in S$

Theorem 1.1 If $\bar{x}$ is a local solution to the problem $P$, then $f'(\bar{x};d)\leq 0$ for all feasible directions $d$ for $S$ at $\bar{x}$ for which $f'(\bar{x};d)$ exists.

If $x^*$ is optimal for the quadratic problem, then $(c + Q x^*)^Td \leq 0$ for all feasible directions $d$ at $x^*$. For the linear problem, a solution $\bar{x}$ is optimal if $(c + Q x^*)^Td \leq 0$ for all feasible directions $d$ at $\bar{x}$, and it is clear that $\bar{x}=x^*$ satisfies this condition.


问:
Characterization of $L^\infty(\Omega)$ norm using $L^p(\Omega) $ norm.
Let $\Omega \subset \mathbb{R}^d$ and $f \in L^\infty(\Omega)$. We know that if $\lambda(\Omega) < +\infty$ with $\lambda$ the Lebesgue measure on $\mathbb{R}^d$, we have the inclusion

$$L^\infty(\Omega) \subset L^p(\Omega), \quad \forall p \in [1,+\infty[.$$
and that this injection is continuous.

I am looking for references or a proof for the following property : $||f||_{L^\infty(\Omega)} = \underset{p \rightarrow + \infty}{\liminf} ||f||_{L^p(\Omega)}.$

Any help is welcomed.

答:
Assume $f\in L^{\infty}$. Then, $|f|\leq \|f\|_{\infty}$ almost everywhere and we get that

$$
\|f\|^p_p=\int |f|^p\textrm{d}\lambda\leq  \int \|f\|_{\infty}^p\textrm{d}\lambda=\lambda(\Omega)\|f\|^{p}_{\infty},
$$
implying that $\limsup_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}$.

Similarly, for any $\|f\|_{\infty}>\varepsilon>0$, we have that $|f|\geq 1_{\{|f|\geq \|f\|_{\infty}-\varepsilon\}} (\|f\|_{\infty}-\varepsilon)$ and we get that
$$
\|f\|_p\geq \lambda(\{|f|\geq \|f\|_{\infty}-\varepsilon\})^{1/p}(\|f\|_{\infty}-\varepsilon)
$$
As $\lambda(\{|f|\geq \|f\|_{\infty}-\varepsilon\})>0,$ we get that
$$
\liminf_{p\to\infty}\|f\|_p\geq \|f\|_{\infty}-\varepsilon,
$$
Since $\varepsilon>0$ was arbitrary, we get the result.


问:
Taking a complicated partial derivative
Mostly I believe in math. However I have trouble in my economic textbook (which really should be right).

I have the following equation:

$$ u(c,d)=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}$$

where $\theta=(1-\gamma)/(1-1/\psi)$, $a\in (0,1),b\in (0,1), c>0, d>0, \gamma>0$, and $\psi\in(0,\infty)$.

Is the derivative of $u$ wrt. $c$ really
  $$ u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{1/\psi}$$

If so, can someone tell me why? I would write it as

$$ u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{-1}$$

Hope to hear from someone,
thanks in advance.

Title: Taking a complicated partial derivative

Body:
$$\frac{\partial}{\partial c}u(c,d)=\frac{\partial}{\partial c}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}$$
$$\Rightarrow \left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}\frac{\partial}{\partial c}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)$$
$$\Rightarrow \left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}ac^{\frac{1-\gamma}{\theta}-1}$$
Since
$$u=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}\Rightarrow u^{\frac{1-\gamma}{\theta}}=ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}$$
And
$$u_c=\left(u^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}ac^{\frac{1-\gamma}{\theta}-1}$$
$$\Rightarrow u_c=u^{\frac{\theta+\gamma-1}{\theta}}ac^{\frac{1-\gamma-\theta}{\theta}}=u^{\frac{\theta+\gamma-1}{\theta}}ac^{-\frac{\gamma+\theta-1}{\theta}}$$
Because
$$\theta=(1-\gamma)/(1-1/\psi)\Rightarrow \frac1{\psi}=\frac{\theta+\gamma-1}\theta $$
$$\Rightarrow u_c=u^{\frac 1{\psi}}ac^{-\frac1{\psi}}$$


问:
Second Order Finite Difference Scheme to Solve Initial Value Problem
I would appreciate help in a problem that I cannot figure out where am I doing the mistake. The mistake could be something I am unaware of so please excuse my lack of knowledge.

I am trying to solve the equation of motion,
$$M \ddot{x} + K x = F,$$
where $M$, $K$, and $F$ are variables independent of $x$, and $\ddot{x}=\frac{d^2x}{dt^2}$. Using a central difference finite scheme, I rewrote the equation as,

$$M \frac{ x_{i+1} - 2x_i + x_{i-1} }{ {\Delta t} ^2 } + Kx = F.$$

Then, I am solving this equation for $x_{i+1}$ at every time step $i$ as,

$$ x_{i+1} = \frac{{\Delta t}^2}{M} \cdot \left( F - K x + \frac{M}{{\Delta t}^2} \cdot 2 x_i - \frac{M}{{\Delta t}^2} \cdot x_{i-1} \right).$$

Now to solve this problem, one needs two initial conditions, to my knowledge, namely, $x(0)$ and $\dot{x}(0)$. At the first time step, I am setting $x_i$ to the $x(0)$ initial condition. Now I am stuck with two problems:

What value should I set $x_{i-1}$ at first time step?

How can I apply the initial condition $\dot{x}(0)$?

Body: $x_{i-1}=x_{i}-\dot x(0)\Delta t$


问:
Annihilator of various generators of a module
Is is possible for $m,n$ to be two generators of a module $M$ over a ring $R$ such that
$$\operatorname{Ann}_R(m) \not \cong \operatorname{Ann}_R(n)\,\,?$$

We know that $R/\operatorname{Ann}_R(m) \cong R/\operatorname{Ann}_R(n)$.

It's not possible in the commutative case, since then they are both equal to $\operatorname{Ann}_RM$.

Truth be told, I don't even know an example of two left ideals $L, L' \triangleleft R$ where $L \not \cong L'$ but
$$R/L \cong R/L'$$
as left $R$ modules.

答: Hint: Let $k$ denote any field and $R:=\mathbb M_n(k)$ denote the ring of $n\times n$ matrices over $k$. What are the maximal left ideals $I$'s of $R$? What does $R/I$ look like?


问:
Show $\vdash (\phi \to \psi) \land (\lnot \phi \to \psi) \to \psi$.
Note: $\lnot \phi$ is an abbreviation of $\phi \to \bot$. Using Dirk van Dalen. "Logic and Structure (Universitext)" as reference book.

Derivation:

$
\def\be{\mathsf{\tiny{\leftrightarrow} Elim}}
\def\bi{\mathsf{\tiny{\leftrightarrow} Intro}}
\def\ce{\mathsf{\tiny{\land} Elim}}
\def\ne#1{\mathsf{\tiny\neg Elim^{#1 }}}
\def\ni#1{\mathsf{\tiny\neg Intro^{#1}}}
\def\ii#1{\mathsf{\tiny{\to}Intro^{#1}}}
\def\ie{\mathsf{\tiny{\to}Elim}}
\def\RAA#1{\mathsf{\tiny RAA^{#1}}}
$

$
\dfrac{
  \dfrac{
      \dfrac{
        \dfrac{
          \dfrac{
            \dfrac{}{[\lnot \psi]_3}\dfrac{
              \dfrac{}{[\phi]_2}\dfrac{
                [(\phi \to \psi) \land (\lnot \phi \to \psi)]_1
              }{\phi \to \psi}\ce
            }{\psi}\ie
          }{\bot}\ie
        }{\lnot \phi}\ni 2
      \dfrac{
        [(\phi \to \psi) \land (\lnot \phi \to \psi)]_1
      }{\lnot \phi \to \psi}\ce
    }{\psi}\RAA 3 \dfrac{}{[\lnot \psi]_4}
  }{\psi}\ne 4
}{(\phi \to \psi) \land (\lnot \phi \to \psi) \to \psi}\ii 1
$

My questions are:

Is the hypotesis of $\lnot \psi$ labeled with a sub-index "4" correctly discharged ?

Is this proof correct ?

EDIT:

Revised derivation:

$
\dfrac{
  \dfrac{
      \dfrac{
        \dfrac{
        \dfrac{
          \dfrac{
            \dfrac{}{[\lnot \psi]_3}\dfrac{
              \dfrac{}{[\phi]_2}\dfrac{
                [(\phi \to \psi) \land (\lnot \phi \to \psi)]_1
              }{\phi \to \psi}\ce
            }{\psi}\ie
          }{\bot}\ie
        }{\lnot \phi}\ii 2
      \dfrac{
        [(\phi \to \psi) \land (\lnot \phi \to \psi)]_1
      }{\lnot \phi \to \psi}\ce
    }{\psi}\ie \dfrac{}{[\lnot \psi]_3}
      }{\bot}\ie
  }{\psi}\RAA 3
}{(\phi \to \psi) \land (\lnot \phi \to \psi) \to \psi}\ii 1
$

答: Using quite another proof system -- apologies -- but this may help...


问:
Why is the unit circle not homeomorphic to the closed unit disk?
I know that the unit circle = $\{(x,y): x^2+y^2 =1\}$ is not homeomorphic to the closed unit disk = $\{(x,y): x^2+y^2 \leq 1\}$, but I'm not sure how to prove it. I've tried with arguments with cut-points and with (path)connectedness, but still not getting a good argument. Any help?

答:
If you remove two points from the circle, it becomes disconnected.

The disk on the other hand remains connected after such removal. To see the latter observe that there are infintely many pairwise disjoint (apart from the ends) paths between any two points of the disk, and removing finitely many poitns can only destroy finitely many of them.


问:
Is $f(x) \in \mathbb{R}$ if $x \in \mathbb{R}$?
Let $f:\Omega\subset\mathbb{C}\rightarrow\mathbb{C}$ be holomorphic with $\Omega$ open. Assume there is an infinite sequence $(x_n)_n$ in $\Omega \cap \mathbb{R}$ that converges to $x_0 \in \Omega \cap \mathbb{R}$ and assume that for every $n$, $f(x_n) \in \mathbb{R}$. Let $D_r$ be the open disk of radius $r>0$ and centre $x_0$ such that $D_r \subseteq \Omega$. Is it true that $f(x) \in \mathbb{R}$ for every $x \in D_r \cap \mathbb{R}$?

I tried proving this by constructing a holomorphic function $g$ such that $g(z)$ is real whenever $z$ is real and $g(x_n)=f(x_n)$. Then by the identity theorem it follows that $g=f$. Unfortunately, I was unable to find such $g$.

Any help would be welcome. Thank you in advance.

答:
First of all, $D_r$ seems not to be part of the question or problem?

Secondly: your assumptions is that there exists a particular sequence $(x_n)$ converging to a particular point $x$ and you want to show a general statement that would be "$f(x)$ is real whenever $x$ is real". There is a clear lack of assumptions to show that.

This is the only thing that can be shown with your assumptions:

Let $f : \Omega \to \mathbb{C}$ be continuous. If $x \in \Omega$ is the limit of the sequence $(x_n) \in (\Omega)^{\mathbb{N}}$ and if for every $n$, $f(x_n) \in \mathbb{R}$, then $f(x)\in \mathbb{R}$. This is because $f$ is continuous at $x$ and because $\mathbb{R}$ is a closed subset of $\mathbb{C}$.


问:
price of two bonds given the period, par value and coupon being (i + 0.04) and (i - 0.04) and that one bond is x amount greater than the other
Given the following determine the price of each bond and the yield rate i.

i- each bond is 10-year bond with semiannual coupons redeemable at its
par value 10,000 and is bought to yield an annual nominal rate i, convertible
semi-annually.

ii- bond A has annual coupon rate of (i+.04) paid semi-annually
iii- bond B has an annual coupon rate of (i-0.04) paid semiannualy.

iv - Price of bond A is 5,341.12 greater than the price of Bond B.

I solved most of the questions in my textbook for this section but I went on to take a practice exam and was dropped a bomb with this question.. In fact every question on the practice exam appears to be layered with increasing difficulty and are the type of questions I was never exposed to in any of my textbook problems.. I dont even know where to begin and this is extremely discouraging as I've already spent a great deal of time and effort studying up to this point and feel as if I havent taken any significant step.

答:
Let $A$ and $B$ denote the prices of bonds $A$ and $B$, respectively. Part (i) is just giving us information. Parts (ii) and (iii) are intended for you to set up two equations and two unknowns.

Starting with bond A, we know $r=i+.04$. Since the face amount is 10,000, the coupons will be of size $10,000\frac{r}{2}=10,000(\frac{i+.04}{2}),$ where we divide by $2$ because of compounding semi-annually. These coupons are paid at times $t=1,2,\dots 20$, where time is in half-years. The face amount is then paid at time $t=20$. Hence, denoting $v=\frac{1}{1+i/2}$ the present value of $A$ and $B$ are:
\begin{align}
A&=10,000v^{20}+10,000\left(\frac{i+.04}{2}\right)(v+v^2+\dots+v^{20})\\
B&=10,000v^{20}+10,000\left(\frac{i-.04}{2}\right)(v+v^2+\dots+v^{20})
\end{align}
Subtracting, we find
\begin{align}
A-B&=10,000(.04)(v+v^2+\dots+v^{20}) \\
&=400(v+v^2+\dots+v^{20})
\end{align}
From (iii), we know $A-5341.12=B$, or equivalently, $A-B=5341.12$. It should be clear how to solve for $i/2$. Once you know $i/2$, just plug in to find the prices of $A$ and $B$.


答: Hint: $g(z) = \frac12\big( f(z) + \overline{f(\overline z)} \big)$. (This is the same construction as in a proof of the Schwarz reflection principle.)

Title: Is $f(x) \in \mathbb{R}$ if $x \in \mathbb{R}$?

Body:
Let $f:\Omega\subset\mathbb{C}\rightarrow\mathbb{C}$ be holomorphic with $\Omega$ open. Assume there is an infinite sequence $(x_n)_n$ in $\Omega \cap \mathbb{R}$ that converges to $x_0 \in \Omega \cap \mathbb{R}$ and assume that for every $n$, $f(x_n) \in \mathbb{R}$. Let $D_r$ be the open disk of radius $r>0$ and centre $x_0$ such that $D_r \subseteq \Omega$. Is it true that $f(x) \in \mathbb{R}$ for every $x \in D_r \cap \mathbb{R}$?

I tried proving this by constructing a holomorphic function $g$ such that $g(z)$ is real whenever $z$ is real and $g(x_n)=f(x_n)$. Then by the identity theorem it follows that $g=f$. Unfortunately, I was unable to find such $g$.

Any help would be welcome. Thank you in advance.


答:
Let $A = \{\{1,2\}, \{3,4\}\}$.  By definition, the only elements of $A$ are those in the comma-separated list: $\{1,2\}$ and $\{3,4\}$.  So $\{1,2\} \in A$ and $\{3,4\} \in A$ but $1 \notin A$.  This is just true by definition; $\in$ is not a transitive relation in standard set theory.  You can't justify this by appeals to intuition; it's just a rule of the game.

We (that is, the mathematical community) could define $\in$ so that it is transitive, but we don't, because it would not be very convenient.  With the definition as it is, it's very easy to tell whether anything is an element of a comma-separated list: it has to be one of the things in the list.  Otherwise, to say for sure that $x \notin A$ you would have to check whether there happens to be any $B \in A$ with $x \in B$, or any $B \in A$ with a $C\in B$ so that $x \in C$, etc., which would create a headache.  With the standard definition, you can say that a set $\{A,B\}$ with $A \neq B$ has exactly two elements: $A$ and $B$, and no more.


Title: Find the formula of a sequence using generating functions

Body:
The given sequence is $\sum_{k=1}^{n}kx^{k}$.

How can I do this? I think I should start with the already known generating function:

$A(x) = \sum a^{n}x^{n} = \frac{1}{1 - ax}$

And then I should do something with this function so that I finally come up with the given one. But well, how can I do this?

$\sum_{k=1}^{n}kx^{k} = \sum_{k>=1} [A(x) - 1]k$

I could only get something like this, but first of all it's not the formula without sums, and second I'm not sure it's correct.

Could you help me?

Body:
$$\frac{\mathrm{d}x^{k}}{\mathrm{d}x}=kx^{k-1}$$

and$$\dfrac{d(\dfrac{1}{1-x})}{dx}=\dfrac{1}{(1-x)^{2}}$$

So when $n\rightarrow \infty$,$$1+x+x^{2}+\cdots=\dfrac{1}{1-x}$$
On differentiating,
$$1+2x+3x^{2}+\cdots=\dfrac{1}{(1-x)^{2}}$$
Multiply with $x$,
$$x+2x^{2}+3x^{3}+\cdots=\dfrac{x}{(1-x)^{2}}$$


问:
The equivalent definition of singular poaitive linear functional on a von Neuamann algebra

The screenshot is from Takesaki's book.

In Theorem 3.8. Suppose the condition in (ii) is satisfied, how to get the contadiction by uaing Lemma 3.6?

By Lemma 3.6, we only know that $\omega_n=s(\omega_n)\omega_n$. If $e=s(\omega_n)\geq e_0$, we have $ee_0=e_0$.

答:
We also have that $\omega_n$ is faithful on $eAe$ by Lemma 3.6, and that $e_0=ee_0e\in eAe$.   Since $e_0$ is nonzero, it then follows that $\omega_n(e_0)>0$.  However,
$$0=\omega(e_0)\geq\omega_n(e_0)>0,$$
a contradiction.


问:
Show that $M_n = \sum_{i=1}^n \alpha_k(X_k -X_{k-1})$ is a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$.
$\textbf{question}$

Let $\{F_n, \ n\in \mathbb{N}\}$ be a filtration and $\{X_n, \ n\in \mathbb{N}\}$ a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$, with $X_0 = 0$ , and such additionally there exists $C \in (0,\infty)$ such that for all $n \in\mathbb{N}$:

$E((X_n - X_{n-1})^2) \leq C$

Additionally, $\{\alpha_k, \ k \in \mathbb{N}\}$ denotes a sequence of real numbers.

Show that $M_n = \sum_{i=1}^n \alpha_k(X_k -X_{k-1})$ is a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$.

$\textbf{problem}$

I don't understand how to show $M_n$ is a square integrable martingale $\textbf{with respect to}$ $\{F_n, \ n\in \mathbb{N}\}$.

Do I have to show that $E(M_n^2|F_{n-1})=M_{n-1}^2$ in which case I'll be fine.

Or do I have to show:

$sup_{n \geq 0}E(M_n^2)<\infty$

if so can I write:

$sup_{n \geq 0}E[(\sum_{i=1}^n \alpha_k(X_k -X_{k-1}))^2] \leq sup_{n \geq 0}E[(\sum_{i=1}^n \alpha_k^2(X_k -X_{k-1})^2]\leq M sup_{n \geq 0}E[(\sum_{i=1}^n(X_k -X_{k-1})^2]\leq ME[m(X_m -X_{m-1})^2]\leq MmC\leq \infty$

答:
For a process $\{Y_n\}_{n\in\mathcal{N}}$ to be a martingale wrt to a filtration $\mathcal{F}=\{\mathcal{F}_n\}_{n\in\mathcal{N}}$ you have to show that :

$Y_n$ is adapted to $\mathcal{F}$

$\mathbb{E}[|Y_n|]$ exists for all $n$

$\mathbb{E}[Y_n|\mathcal{F}_{n-1}] = Y_{n-1}$

So you have to show both basically to answer you question. The inequality shows that your process is $L^1$ for all $n$. The equality that it is a martingale.

Please do not ignore the first bullet point which justify the fact that you can write the left term of my third bullet point.

EDIT:

Sorry I misread the question and realize that only after struggling to prove Bullet 3.

What you have to show is that $\{M_n\}$ is a martingale and justify that it belongs to $L^2$.

Appendix to show integrability of $Mn^2$:

The two following propositions are not equivalent :

$\sup_n \mathbb{E}[M_n^2] < \infty$

$\mathbb{E}[M_n^2] < \infty,$  $\forall n \in \mathbb{N}$

The first is more restrictive and tells you that as $n$ goes larger, the quantity remain finite which is equivalent to saying that $\lim_{n\rightarrow\infty}\mathbb{E}[M_n^2] < \infty$

The second only tells you that the process is $L^1$ for all integers. So $1 \implies 2$ but nothing else.

To prove (2) $\forall n \in \mathbb{N}$:

$$
\mathbb{E}[(\sum_{k=1}^{n} \alpha_k(X_k - X_{k-1}))^2] \leq \sum_{k=1}^n \alpha_k^2 \cdot \mathbb{E}[\sum_{k=1}^n (X_k - X_{k-1})^2]
$$

From this I let you continue.


问:
prove the trigonometry identity:
prove the trigonometry identity:

$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$

of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.

答:
$1.  \ \cot^3A = \dfrac{1}{\tan^3A}\\$

$2.  \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\$

$3. \  \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A$

So, the RHS becomes,

$\begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac{1}{\tan^3A}\left(\tan^2A-\tan A+1\right)} \\&= \tan^3A\dfrac{\tan A(\tan^2A+1-\tan A)}{\tan^2A+1-\tan A} \\&= \tan^4A\end{align}$

Title: prove the trigonometry identity:


答:
The result is true for $U_0$ the connected component of $\Omega \cap \mathbb R$ that contains $x_0$

(to show that it may not be true in general even in $\Omega$ connected - assuming, of course, the hypothesis just for the single point $x_0$ as above), one can take the branch of square root that is say positive on $(1,2)$ in a tube that connects $(1,2)$ with $(-2,-1)$ through the upper plane (eg the half annulus $1<|z|<2, \Im z \ge 0$) extended a bit below the real axis at both ends to make it open) and consider $\sqrt z$ which is clearly real on $(1,2)$ but not on $(-1,-2)$)

To show the result we let $A$ the set of points $x \in \Omega \cap \mathbb R$ which have an open neighborhood $V_x \subset \Omega$ for which $f(z) \in \mathbb R, z \in V_x \cap \mathbb R$ and we claim that $A$ is both open and relatively closed (in $\Omega \cap \mathbb R$ of course) while containing $x_0$; trivially $A$ is open by definition, while the proof that $x_0 \in A$ will apply to any limit points of $A$ that are in $\Omega$.

Let $f(z)=\sum a_k(z-x_0)^k$ the Taylor expansion of $f$ on a small neighborhood of $x_0$ and we show that we can take $V_{x_0}$ precisely that by showing that the coefficients $a_n$ are real; since $f(x_0)$ is real as the limit of real numbers, $a_0$ is so; but now $f_1(z)=\frac{f(z)-a_0}{z-x_0}=\sum_{k \ge 0} a_{k+1}(z-x_0)^k$ satisfies the same properties as $f$ since $f_1(x_n)$ is clearly real as the ratio of real numbers for $x_n$ close enough to be in $V_{x_0}$ of course and $f_1(x_n) \to f_1(x_0)$ since $f_1$ is analytic hence continuous, hence $a_1$ is real and clearly we can iterate the procedure so by induction $a_n$ real and we are done (noticing that if $y \in \Omega$ is a limit of point of points in $A$, $y$ definitely satisfies the same hypothesis as $x_0$ hence same proof works to show that it is in $A$)


答:
Let $t=\tan(A)$.

Multiply both sides by rhs denominator and let $\cot(A)=1/t$. Then lhs $=t^4(1-1/t)(1/t)+t=t^3-t^2+t$, while rhs $=t^3+(1-t)t=t^3+t-t^2$.

Therefore equality.

Title: prove the trigonometry identity:

Body:
prove the trigonometry identity:

$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$

of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.


答: Any open cover of the circle has a refinement such that the intersection of any three distinct sets is empty. But there is an open cover of the disc that admits no such refinement:

Title: Why is the unit circle not homeomorphic to the closed unit disk?

Body: I know that the unit circle = $\{(x,y): x^2+y^2 =1\}$ is not homeomorphic to the closed unit disk = $\{(x,y): x^2+y^2 \leq 1\}$, but I'm not sure how to prove it. I've tried with arguments with cut-points and with (path)connectedness, but still not getting a good argument. Any help?


问:
Why can't set of discontinous points with volume zero intersect with the boundary of a rectangle?
In Multivariable Mathematics by Dr. Shifrin, Prop. 1.8 in chapter 7 says

Suppose $f:R\to \mathbb{R}$ is a bounded function and the set $X=\{\mathbf{x}\in R:f \text{ is not continuous at }\mathbf{x}\}$ has volume zero. Then $f$ is integrable on $R$.

Here $R\subset\mathbb{R}^n$ is a closed rectangle. The proof then covers $X$ by finitely many closed rectangles $R_1', \dotsc, R_s'$ due to volume zero and the critical line where I'm stuck is "We can also ensure that no point of $X$ is a frontier point of the union of these rectangles".

My question is why can't a point of discontinuity of $f$ occur on the frontier of the rectangle $R$ and thus necessarily on the frontier of the union of $R_1', \dotsc, R_s' \subset R$?

答: Good catch! I think you are correct, in fact, but it won't affect the proof that follows. If a frontier point of $R$ is actually in $X$, covered, say, by the rectangle $R_\ell'$, then when we remove $\cup R_j'$ and take the closure, the frontier of $R_\ell'$ does not show up, and $f$ is still continuous on the complement.


问:
Exploring underdetermined linear system with non-negative solution
I haven't had much luck searching for this specific problems. Any pointers would be greatly appreciated.

I have an underdetermined system where $ A $ and $ b $ are known. $ x $ is a real vector with $x_i > 0 $.

$$ Ax = b $$

I'm not looking to optimize a function but to explore the solution space.

I'd like to be able to incrementally move around the solution space, using the "current" solution to evaluate a completely different problem (that would be absolutely horrible and infeasible to integrate into this). How would I be able to define this space?

Could anyone give an idea of where to start? Thanks!

答:
Say you have $x_1,\ldots,x_{n+1}$ with $x_k > 0$ and $Ax_k = b$ for all $1 \leq k \leq n+1$. Then every convex combination of these vectors, i.e. every sum $$
  y = \sum_{k=1}^{n+1} \lambda_kx_k
  \quad\text{where}\quad
  \sum_{i=1}^{n+1} \lambda_k = 1
  \quad\text{and for all $k$}\quad
  \lambda_k \geq 0
$$
also obeys $Ay=b$ and $y > 0$.

Now, the question becomes whether you can find "enough" such $x_i$ to guarantee that every solution $y$ can be obtained this way. The strictness in $x > 0$ causes trouble here, I think. Take for example the system $$
    \left(\begin{matrix}1&1\\0&0\end{matrix}\right)x
  = \left(\begin{matrix}1\\0\end{matrix}\right) \text{,}
$$ i.e. the system whose solution is a line through $(1,0)$ and $(0,1)$. Due to the strictness in your requirement, you have to $x_1=(1-\epsilon_1,\epsilon_1)$, $(\epsilon_2,1-\epsilon_2)$ for $\epsilon_1,\epsilon_2 > 0$ as your convex basis. But that means you'll miss e.g. the solution $(1-\frac{\epsilon_1}{2},\frac{\epsilon_1}{2})$.

So fix this, you'll have to replace $x > 0$ with $x \geq 0$ in your requirement, and then pick only convex combinations which satisfy the stricter condition. Usually simply requirement $\lambda_i > 0$ instead of $\lambda_i \geq 0$ should be enough, but there might cases where it's not. I have the feeling that in these cases, the stricter system has no solution at all, but I haven't proved this so I might be wrong.

You'll also have to figure out what "enough" means. $n$ can quite obviously be at most $\dim \ker A$, but there are cases where you won't find that many affinely independent vectors with only positive coordinates. Take for example the system $$
    \left(\begin{matrix}1&1&0&0\\0&0&1&1\\0&0&0&0\\0&0&0&0\end{matrix}\right)x
  = \left(\begin{matrix}1\\0\\0\\0\end{matrix}\right) \text{.}
$$
If the third or forth component of a solution is strictly greather than zero, the other one needs to be strictly less than zero, and you thus get only the two vectors $x_1=(1,0,0,0)$ and $x_2=(0,1,0,0)$.

I'd take a look at the simples algorithm, it should help to find suitable $x_k$.


问:
Why is the $\lim_{x \to \infty} \frac{(-1)^x}{x}= 0$?
I know that the answer is $0$ by using Wolfram. But I'd appreciate it if someone shows the solution to this problem. If I changed it to $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$ where $n$ takes on positive integers including $0$, would the answer change?

答:
The function given by $f(x) := \frac{(-1)^x}{x}$ is defined only for $x \in \mathbb{Z}$ and $x\neq 0$, hence $\operatorname{Dom}(f) = \mathbb{Z} - \{ 0\}$.

The "points" at positive or negative infinity, i.e. $\pm \infty$, are both cluster points for $\operatorname{Dom}(f)$, therefore you can try to evaluate both $\displaystyle \lim_{x\to +\infty} f(x)$ and $\displaystyle \lim_{x\to -\infty} f(x)$.

Since $-\frac{1}{x} \leq f(x) \leq \frac{1}{x}$, both the aforementioned limits evaluate to $0$ by the Squeeze Theorem.


Title: prove the trigonometry identity:

Body:
$$t^4\left(\dfrac{1 -t^{-1}}{t} +t^{-3}\right)=t^3 + \dfrac{1 - t}{t^{-1}}$$
is

$$t^3-t^2+t=t^3+t-t^2.$$


问:
Intuition for dimensions
I got confused on the web and on my previous answers.

Sometimes a line got 1 dimension because you just need one point to define every part of it, sometimes it's in 2 dimensions if you consider it in a plan and sometimes it's in 3 dimensions if you plot it considering x, y and z. So how many dimensions has a line for example ?

Also, I found out that an inequality changes the number of dimensions, for example a line x+y=1 is in 2 dimensions if I plot it considering x and y but as soon as I write x+y>1 it's not a line anymore but a plan, so the number of dimensions increased by 1 so it is now a plan with 3 dimensions ? I think that is not correct but I do not know how..

Can you give me some intuitions or method to understand when I got how many dimensions considering an equation.

NB:

答:
You need to think about how many independent things you need to specify in order to fully specify a point on your line (or other mathematical object).

A line itself (at least, at the high school level) is a 1D object with equation of the form $y = mx + b$.  Once we specify $x$, we automatically specify a $y$ value.

If we now choose to plot that line in the plane, then it is a 1D object that lives in 2D space (or 3D, or whatever).  It is still a 1D object, though, and we can see that by parameterizing it.  A line in the plane can be thought of as two equations:

$$x(t) = t$$
$$y(t) = mt + b$$

Now, although there are technically two functions, they both rely on a single variable, and as such the object is 1D.  The two functions allow us to essentially draw it in our 2D space. As such, the "set of points that lie on the line" is a set of 2D mathematical objects known as points, but the line itself is a 1D object.   You are free to think of it in either 1D (a single equation that produces y values) or 2D (a set of equations that produces points) whenever it is convenient.

When we include an inequality, let's think about what happens:  $x + y > 3$.  Here, we may consider this to be a region in the plane or a higher dimensional object, depending on what we want.

As a planar segment: $x+y > 3 \implies y > 3 - x$, therefore this specifies a region in the plane that contains, for all $x$ values, vertical line segments who start at $y = x-3$ and go to infinity.

As a segment of a higher dimensional space: if it is convenient, we might claim that $z = x+y$ is a 3D object, in this case a plane living in 3D*. Then $x+y>3$ specifies all points on that plane whose $z$ values are greater than 3.

*and if it is instead convenient, we may consider it to be a 2D object!

The actual answer is: it depends on what you need it to be.  If you need the line to be in 1D, then so be it.  If you need to plot that line in 2D or something else, then you need to think of it as an object living in that space, appropriately.

Title: Intuition for dimensions


问:
Pattern of divisibility within $2^n+1\equiv0\pmod{n}$
In recent days, I have been studying:the properties of $m^n+h\equiv0\pmod n$ where $m,n\in\mathbb{N}$ and $h\in\mathbb{Z}$, and I am trying to calculate all solutions for $2^n+1\equiv0\pmod n$ for all $n\leq10^{15}$.

Obviously, if one were to calculate the solutions by checking the numbers one by one, it would have taken an extremely long amount of time to do so. Therefore, I have tried to reduce the numbers needed to check by finding the divisibility patterns within the solutions.

From Proposition 3 of the paper "Primitive solutions of $n|2^n+1$", if $n$ is not a power of $3$ and $n=3^kq$ where $k\in\mathbb{Z}$, $k\geq0$, $k|2^k+1$ and $q\neq1$(In other words, $3^k|n$ and $n\neq3^k$), then $q$ is also divisible by one of the primes $p$ such that $p|2^{3^k}+1$. For example, if $81|n$ and $n\neq81$, then one of {$243,1539,13203,7064091,10970073,22032887841$} divides $n$.

However, I have also noticed that the same argument seems to be true for some other divisors. An example is that if $171|n$ and $n\neq171$, then it seems that one of {$513,3249,97641,29884473,27439598619,3279345384337785847895810090688081$} divides $n$.

Is it true for all divisors, and how can I prove it?

References:

Toby Bailey and Christopher J. Smyth (2008). Primitive solutions of $n|2^n+1$,  https://www.maths.ed.ac.uk/~chris/papers/n_divides_2to_nplus1.pdf

答:
Using the idea of the proof of Proposition 3, we have the following :

If $n$ is of the form $mN^k$ where $k$ is a positive integer, and $N$ is a prime, and $m\ (\gt 1)$ is an integer satisfying $\gcd(m,N)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{N^{k}\gcd(m,\frac{N-1}{2})}+1$.

The proof is written at the end of the answer.

Letting $N=19$, we have the following :

If $n$ is of the form $m\cdot 19^k$ where $k$ is a positive integer, and $m\ (\gt 1)$ is a positive integer satisfying $\gcd(m,19)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{9\cdot 19^{k}}+1$.

(Since the paper shows that if $n\gt 3$, then $9|n$, we have $\gcd(m,\frac{19-1}{2})=9$.)

The prime factorization of $2^{9\cdot 19^1}+1$ is given by
$$2^{9\cdot 19^1}+1$$
$$=3^3×19^2×571×174763×160465489×19177458387940268116349766612211$$

So, we can say that if $171|n$ and $n\not=171$, then $n$ is divisible by at least one of
$$513\ (=3^3\times 19)$$
$$3249\ (=3^2\times 19^2)$$
$$97641\ (=3^2\times 19\times 571)$$
$$29884473\ (=3^2\times 19\times 174763)$$
$$27439598619\ (=3^2\times 19\times 160465489)$$
$$3279345384337785847895810090688081\ (=3^2\times 19\times 19177458387940268116349766612211)$$

Finally, let us prove the following :

If $n$ is of the form $mN^k$ where $k$ is a positive integer, and $N$ is a prime, and $m\ (\gt 1)$ is an integer satisfying $\gcd(m,N)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{N^{k}\gcd(m,\frac{N-1}{2})}+1$.

Proof : Let $p$ be a prime such that $p|m$. Then $2^{2n}\equiv 1\pmod{pN^k},2^{(p-1)(N-1)N^{k-1}}\equiv 1 \pmod{pN^k}$ (Euler’s Theorem), so $2^{2\gcd(n,\ (p-1)\frac{N-1}{2}N^{k-1}\ )} \equiv 1 \pmod{pN^k}$,$$2^{2N^{k−1}\ \gcd(mN,\ (p-1)\frac{N-1}{2})}\equiv 1 \pmod p.$$ Now, we have $\gcd(mN, (p-1)\frac{N-1}{2})=\gcd(m,\frac{N-1}{2})$ or $N\gcd(m,\frac{N-1}{2})$. If $\gcd(mN, (p-1)\frac{N-1}{2})=\gcd(m,\frac{N-1}{2})$, then we get $2^{2N^{k−1}\gcd(m,\frac{N-1}{2})}\equiv 1 \pmod p.$ If $\gcd(mN, (p-1)\frac{N-1}{2})= N\gcd(m,\frac{N-1}{2})$, then we get $2^{2N^{k}\gcd(m,\frac{N-1}{2})}\equiv 1 \pmod p$, so the latter congruence certainly holds. So, we have $(2^{N^{k}\gcd(m,\frac{N-1}{2})}+1)(2^{N^{k}\gcd(m,\frac{N-1}{2})}-1)\equiv 0\pmod p$. Since $2^{N^{k}\gcd(m,\frac{N-1}{2})}-1\not\equiv 0 \pmod p$, we have $2^{N^{k}\gcd(m,\frac{N-1}{2})}+1\equiv 0\pmod p$.$\quad\blacksquare$


问:
Combinatorics - inclusion exclusion principle
i have some question and i'm not sure i solved it well.

How much numbers between 1 to 800 (Naturals) are devided at least by one of the following numbers: 3,4,6.

This is my solution:

Total devided by 3: 800/3 = 266

Total devided by 4: 800/4 = 200

Total devided by 6: 800/6 = 133

Total devided by 3*4 = 800/12 = 66

Total devided by 3*6 = 800/18 = 44

Total devided by 4*6 = 800/24 = 33

Total devided by 3* 4*6 = 800/72 = 11

Total Answer = 266+200+133-66-44-33+11=467

Am I right?
Thanks!

答:
I agree with the other responses.  Further, the OP specifically asked
about using Inclusion-Exclusion against this problem.  Therefore, if
the other responses had not already answered the OP's specific question,
I would not have posted this response.

The problem can easily (and validly) be attacked without Inclusion-Exclusion.

To see this, first see this link, including its addendum.

Number of positive integers $\le n$ which are a multiple of $p$

Using the ideas from the link, you can attack the problem as follows:

The problem is to identify all numbers that are multiples of 3, 4, or 6
from the set $\{1,2, \cdots, 800\}.$

The number $6$ can be discarded from the investigation because any number that is a multiple
of $6$ must also be a multiple of $3$.  Therefore, the computation is the
same if you restrict the computation to all numbers that are multiples of
$3$ or $4$.

$3$ and $4$ are relatively prime.  Further, $792$ is a multiple of
$3 \times 4$.

This means that with attention temporarily restricted to the subset
$\{1,2,\cdots,792\}$, divisibility by $3$ and divisibility by $4$ can
be regarded as independent events.

This means that in this subset, the chance of a number not being
divisible by either $3$ or $4$ is

$$\frac{4-1}{4} \times \frac{3-1}{3} = \frac{2}{4} = \frac{1}{2}.$$

This means that in the set $\{1, 2, \cdots, 792\},$ exactly
$\frac{1}{2} \times 792 = 396$ numbers will be multiples of either
$3$ or $4$.

The next step is to check the numbers $\{793, 794, \cdots, 800\}$
manually.

In this small subset, the numbers $796$ and $800$ are divisible by
$4$ and the numbers $795$ and $798$ are divisible by $3$.

This means that the final computation is $396 + 2 + 2 = 400.$

Addendum

It occurs to me, after the fact, that the approach that I am suggesting should be formally proven.

Consider the numbers $k, m, n \in \mathbb{Z^+}$, where $m,n$ are relatively prime.

Let the set $S \equiv \{1, 2, \cdots, (k\times m \times n)\}.$

(Ironically), using Exclusion-Inclusion, the number of elements in $S$ that are either multiples of $m$ or multiples of $n$ are

$$(kn) + (km) - k.$$

This means that the number of elements in $S$ that are not multiples of
either $m$ or $n$ is

$$kmn - kn - km + k = k(mn - n - m + 1) = kmn \times
\frac{mn - n - m + 1}{mn} = kmn \times \frac{n-1}{n} \times \frac{m-1}{m}.$$


问:
How to write 3 matrix multiplications as a sum?
Imagine I have three $N\times N$ matrices: A, B and C. How could I write the matrix multiplication $X = ABC$ as a sum s.t. $X_{ij} = \sum something ?$.

I tried to write out $AB = D$ first: $D_{ij} = \sum_k A_{ik}B_{kj}$ and then use the associative property s.t. $X_{ij} = [DC]_{ij} = \sum_{l}D_{il}C_{lj} = \sum_{l}(\sum_{k}A_{ik}B_{kl})C_{lj}$

Is this correct? Is there an easier way to write it down?

I ask because I have a question about finding the derivative of the trace of $ABC$ w.r.t $B$ and I need this as my first step.

答:
This is correct.  There's not really any easier way of writing it down.  There is some intuition, though:  $A_{ik}B_{kl}{C_{lj}}$ can be thought of as a "walk" from $i$ to $j$ with three steps $i \rightarrow k \rightarrow l \rightarrow j$. Then $(ABC)_{ij}$ is given by taking the sum over all "walks" from $i$ to $j$, where each walk is "weighted" by the entries $A_{ik}B_{kl}{C_{lj}}$.  Similarly, $$(ABCD)_{ij} = \sum_{i_1} \sum_ {i_2}\sum_{i_3} A_{i, i_1} B_{i_1, i_2} C_{i_2, i_3} D_{i_3, j}$$ is given by taking the sum over all walks from $i$ to $j$ with four steps.

If you write it as a single sum, and generalize it to arbitrary products, the notation gets a little trickier.  Let $A^1, A^2, \ldots, A^l$ be $n \times n$ matrices; note that the superscripts here are not powers, just indices on the tuple - we need them because we'll reserve subscripts for the row and column indices. Define a walk $\gamma$ from $i$ to $j$, of length $l$, as a tuple $(i, i_1, i_2, i_3, \ldots, i_{l-1}, j)$, with each entry between $1$ and $n$, and let $P_l$ be the set of all walks $\gamma$ of length $l$.  Define the weight of $\gamma \in P_l$ to be $$w(\gamma) = A^1_{i,i_1} A^2_{i_1, i_2} A^3_{i_3, i_4} \cdots A^{l}_{i_{l-1}, j}.$$  Then $$(A^1 A^2 \cdots A^l)_{ij} = \sum_{\gamma \in P_l} w(\gamma).$$

This idea is used frequently in graph theory (the incidence matrix) and probability (Markov chains).


答: The unit disk $B=\{x^2+y^2\leq1\}$ has fundamental group $\pi_1(B)=\{0\}$, while the circle $S^1=\{x^2+y^2=1\}$ has fundamental group $\pi_1(S^1)=\Bbb Z$. And a necessary condition for two topological spaces to be homeomorphic is that they have the same fundamental group (or better isomorphic groups).


答:
may you accept like below
$$\forall x\to \infty :\lfloor x\rfloor \leq x \leq\lfloor x\rfloor+1$$now rewrite the limitation
$$-1 \leq (-1)^{\lfloor x\rfloor} \leq 1\\
-1 \leq (-1)^{\lfloor x\rfloor+1} \leq 1\\\to
\frac{-1}{x} \leq \frac{(-1)^x}{x}\leq \frac1x$$apply the limit to both sides

Title: Why is the $\lim_{x \to \infty} \frac{(-1)^x}{x}= 0$?

Body: I know that the answer is $0$ by using Wolfram. But I'd appreciate it if someone shows the solution to this problem. If I changed it to $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$ where $n$ takes on positive integers including $0$, would the answer change?


问:
Electric Field and Direction of Field
I want to measure the magnitude, and the direction of the electric field at point P induced by a rod that has a charge of $-22.0\mathrm{\mu C}$. The problem has been accurately dimensioned. The distance of the rod to point P has to be calculated, because the dimensions given where to the centroid of the rod. My ultimate problem is how do I know the direction of the field? Here is my work for the problem.
\begin{equation}\vec{E}=\int\frac{k\ dq}{r^2}\hat{r} \end{equation}
\begin{equation}\frac{dq}{dx}=\frac{Q}{L}=\lambda \end{equation}
\begin{align}E&=\int_{.29}^{.43} \frac{k\lambda}{x^2}dx\\ &=\frac{kQ}{L}[-\frac{1}{x}]\Bigg|_{0.29}^{.43}\\&=-1.59\times 10^6 \frac{N}{C} \end{align}
Can anyone please help explain why the answer is that the electric field is point to the right according to my physics professors answer sheet?

答:
<s>First of all, if I read the figure right, the limits of integration are $36\pm 0.14/2$, not $0.36\pm 0.14/2$. But this will only change the numerical value. </s>

In your first equation, where you define the electric field,you have two quantities that will inform you about the direction of the electric field. $\hat r$ is a vector from the charge creating the electric field toward the point where you measure the electric field. So in your case $\hat r$ points to the left. The other quantity of interest is the charge. Yours is negative, so the electric field will point in a direction opposite to $\hat r$. Therefore, in this problem the electric field points to the right.

If you want a physical intuition, the direction of the electric field is the direction of the force acting on a positively charged particle. A positive charge at $P$ is attracted by a negatively charged rod, so the direction of the electric field is to point from $P$ to the negatively charged rod

Title: Electric Field and Direction of Field


答:
First, let's consider some terms of this sequence. Let $x\in\mathbb{N}\{1,2,3,...\}$, then $\frac{(-1)^x}{x}=(-1, \frac{1}{2},-\frac{1}{3},\frac{1}{4},...)$. Let's call the sequence $a_n=\frac{(-1)^x}{x}$.

Formally we define convergence of a sequence in this manner: "for all $\epsilon>0$, there exists some $n,M\in\mathbb{N}$ so that if $M>0$, then for all choices $n>M$ we have $|a_n-0|<\epsilon$. (NOTE: I have $0$ as the point inside the absolute value bars since this is the value we're trying to show is the limit of the sequence.)

I'm sorry if you're unfamiliar with $\epsilon$-proofs of a sequence. What this means is that if I choose any teeny-weeny number called $\epsilon$, I can find some other number $M$ that acts as a boundary for this sequence (think Gandalf from the "You shall not pass!" scene.) That way, if I have some other number $n$ so that $n>M$, then I'm guaranteed this sequence will be smaller than that teeny-weeny number $\epsilon$. (NOTE: this is why it's important to have a limit sign here. Obviously the sequence will never become $0$ because it is undefined at $0$, but it comes close enough so that we may say its limit is $0$.)

Now, for the fun part. Let's choose $\epsilon>0$ and $x\in\mathbb{N}$, then $|a_n-0|<\epsilon\Rightarrow|\frac{(-1)^x}{x}-0|<\epsilon\Rightarrow|\frac{1}{x}|<\epsilon\Rightarrow x>\frac{1}{\epsilon}$. We're going to choose $M=\frac{1}{\epsilon}$.

Now that we know our boundary point, we can say with confidence that if $x>M=\frac{1}{\epsilon}$, then it is always true that $|a_n-0|<\epsilon$, which is the same as saying $\lim_{x\rightarrow\infty}\frac{(-1)^x}{x}=0$.

Please don't hesitate to ask questions if this doesn't make sense.


答:
To begin with when we say $\lim_{x\to \infty} a_x = K$, we are assuming that $x$ is restricted to a set only where $a_x$ is defined.  As $(-1)^x$ is only defined on integers[*] to claim $\lim_{x\to \infty} \frac {(-1)^x}x=whatever$ we must assume $x \in \mathbb Z$.  And as $x \to \infty$ we can limit ourselves to only considering $x \in \mathbb N$.

By convention "$x$" is a variable for a real number.  If we are doing a limit restricted to $\mathbb N$ it is more conventional to use something like "$n$".

[**]I should also point out that even if we took a limit good on the reals, say $\lim_{x\to \infty} a_x  = L$ then as $\mathbb N\subset \mathbb R$ then $\lim_{n\to \infty}a_n = L$ as well because if $x\to \infty; x\in \mathbb R \implies a_x \to L$ then $n\to \infty; n \in \mathbb N\subset \mathbb R$ so $n \in \mathbb R$ as well, the same result must follow.

Anyhow.  If any $\epsilon > 0$ if $x > \frac 1{\epsilon}$ and $x\in \mathbb N$ then $(-1)^x  = \pm 1$ and $|\frac {(-1)^x}x - 0| = |\pm \frac 1x| =|\frac 1x|=\frac 1x < \epsilon$.   So that is the definition of $\lim_{x\to \infty} \frac {(-1)^x}x=0$.

=======

[*] Although $x^{3} = -1$ is solvable by $x=-1$ we do not define $(-1)^{\frac 13} = x$ because .... we just don't. there is no point in defining $b^x$ for $b < 0$ and all $x\in \mathbb R$ as such a function would not be consistent of $x\not \in \mathbb Z$, and wouldn't but continuous, and wouldn't work for any rational $\frac ab$ where $b$ is odd.

...unless you are using a text that does allow rationals with odd denominators.  Some texts do.  Oh well.

But in that case hee my paragraph marked [**].


问:
What square matrices cannot be expressed as the sum of symmetric and skew-symmetric parts
I thought I knew that any square matrix can be written as the sum of symmetric and antisymmetric matrices since we have the property that any $n\times n$ matrix $A$ can be expressed as $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$. However, I have come across the following statement and was told this is not always true and asked if I could find a counterexample where this isn't true but I have been struggling to find a counterexample (perhaps when $\mathbb{F}=\mathbb{Z}_2$?). Here is the statement:

Let $\mathbb{F}$ be a field. Then any $A\in \mathbb{F}(n,n)$ may be written as a sum of a symmetric and an anti-symmetric matrices in $\mathbb{F}(n,n)$.

I would greatly appreciate if someone could show me a counterexample to this.

答:
You have already made the key observation. The formula $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$ is not valid for a field of characteristic $2$. Look at the Galois field $F_4$ which has 4 elements $(0, 1, \alpha, 1+\alpha)$. The matrix $$\begin{pmatrix}0, 1 \\ \alpha, 0 \end{pmatrix}$$ has the desired form. Actually now that I think of it
$$\begin{pmatrix}0, 1 \\ 0, 0 \end{pmatrix}$$
is an example in $\mathbb{Z}_2$ as well.


答:
In case you are in $\mathbb{N}$ ;  $n \to \infty$ means:

either there is a $k$ such as $n = 2k$ with $k$ in $\mathbb{N}$ ($n$ is even.)

or there is a $k$ such as $n = 2k+1$ with $k$ in $\mathbb{N}$ ($n$ is odd.)

Moreover: when $n \to \infty$ you have certainly $n>1$

This means:

either $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{(-1)^{2k}}{k}$

or $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{(-1)^{2k+1}}{k+1}$

Which means

either $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{1}{k} = 0$

or $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{-1}{k+1} = 0$

Therefore $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$


问:
Finding upper and lower bounds of the harmonic series
I want to prove that, for any $n \in \mathbb {Z}^+$, $$\ln(n + 1) \leq 1 + \frac 1 2 + ... + \frac 1 n \leq 1 + \ln n\ ,$$ using the fact that, for any $i \in \mathbb {Z}^+$, $$\frac 1 {i + 1} \leq \ln(i + 1) - \ln i \leq \frac 1 i\ .$$

Some background

I am taking an introductory module to calculus in my university and this is actually one of the questions of my homework for this week. There are two parts to the question and this is the second part, which I have been stuck at for hours. The first part required me to prove that $$\frac 1 {i + 1} \leq \ln(i + 1) - \ln i \leq \frac 1 i\ ,$$ which I have finally been able to do. However, I am not getting any inspiration as to how I can use this to prove that $$\ln(n + 1) \leq 1 + \frac 1 2 + ... + \frac 1 n \leq 1 + \ln n\ .$$

I did a little research online about the middle portion of the inequality and found that it is a special series called the harmonic series (no series are taught in my module). The Wikipedia page on the harmonic series also showed how to prove the lower and upper bounds that I am looking for, but they are far beyond my current ability.

Any intuitions/suggestions on how to carry on this problem will be greatly appreciated :)

答:
For the second part, you in fact have two inequalities:
\begin{align}
\forall i \geqslant 1,~ \dfrac{1}{i+1} &\leqslant \ln(i+1) - \ln i &\text{and}~ \forall i \geqslant 1,~ \ln(i+1) - \ln i &\leqslant \dfrac{1}{i}
\end{align}
Sum the first for $i \in \{1,\ldots,n-1\}$, you find $\frac{1}{2} + \ldots + \frac{1}{n} \leqslant \ln n$. Add $1$ to both sides and you have
\begin{align}
\sum_{i=1}^n \frac{1}{n} \leqslant \ln n +1
\end{align}

Sum the second inequality for $i \in \{1,\ldots,n\}$ and you have
\begin{align}
\ln( n+1 ) \leqslant \sum_{i=1}^n \frac{1}{n}
\end{align}


答:
Note that

$$\tag 1\ln (n+1) = \int_1^{n+1}\frac{dt}{t} = \int_1^{2}\frac{dt}{t} + \int_2^{3}\frac{dt}{t} + \cdots +\int_n^{n+1}\frac{dt}{t}.$$

Take the first integral on the right of $(1).$ On the interval $[1,2],1/t\le 1.$ Thus

$$\int_1^{2}\frac{dt}{t} \le 1\cdot 1 = 1.$$

The same thing works for the second integral in the sum:

$$\int_2^{3}\frac{dt}{t} \le \frac{1}{2}\cdot 1 = \frac{1}{2}.$$

Continue to see that the sum on the right of $(1)$ is bounded above by

$$1+\frac{1}{2}+\cdots + \frac{1}{n}.$$

Title: Finding upper and lower bounds of the harmonic series

Body:
I want to prove that, for any $n \in \mathbb {Z}^+$, $$\ln(n + 1) \leq 1 + \frac 1 2 + ... + \frac 1 n \leq 1 + \ln n\ ,$$ using the fact that, for any $i \in \mathbb {Z}^+$, $$\frac 1 {i + 1} \leq \ln(i + 1) - \ln i \leq \frac 1 i\ .$$

Some background

I am taking an introductory module to calculus in my university and this is actually one of the questions of my homework for this week. There are two parts to the question and this is the second part, which I have been stuck at for hours. The first part required me to prove that $$\frac 1 {i + 1} \leq \ln(i + 1) - \ln i \leq \frac 1 i\ ,$$ which I have finally been able to do. However, I am not getting any inspiration as to how I can use this to prove that $$\ln(n + 1) \leq 1 + \frac 1 2 + ... + \frac 1 n \leq 1 + \ln n\ .$$

I did a little research online about the middle portion of the inequality and found that it is a special series called the harmonic series (no series are taught in my module). The Wikipedia page on the harmonic series also showed how to prove the lower and upper bounds that I am looking for, but they are far beyond my current ability.

Any intuitions/suggestions on how to carry on this problem will be greatly appreciated :)


问:
Riemann surface arising as a quotient of the upper half-plane.
Let $H$ be the upper half-plane $\{z \in \mathbb C \mid \Im(z) > 0\}$. For a fixed real $\lambda > 0$, let be the automorphism $$d_\lambda : H \to H, z \mapsto \lambda z .$$

Denote $\Gamma$ the subgroup of $\mathrm{Aut(H)}$ generated by $d_\lambda$. The goal is to determine the Riemann surface $H/\Gamma$.

Let's discard the trivial case $\lambda=1$. [Edit, false : $H/\Gamma$ is open as the open set $\{ z \in H \mid |z|<1 \}$ is a fundamental domain of the holomorphic projection.] One have $\pi_1(H/\Gamma) \simeq \Gamma \simeq \mathbb Z$. Hence, as no compact Riemann surface has $\mathbb Z$ as fundamental group, $H/\Gamma$ must be an open Riemann surface.

As I don't know a lot of open Riemann surfaces, I'm tempted to test $\mathbb C^\ast$ and the punctured unit disk $\mathbb D^\ast$ that matches those properties. But $\mathbb C^\ast$ has universal cover $\mathbb C \not\simeq H$, and so $H/\Gamma \not\simeq \mathbb C^\ast$. This leaves me with $\mathbb D^\ast$. How can I show that $H / \Gamma$ is (or isn't) biholomorphic to $\mathbb D^\ast$ ? If it isn't, does this quotient look like a well-know surface ?

P.S. Does that kind of quotient have a name ? I'm aware of the name of modular curve when $\Gamma$ is a subgroup of $PSL(2,\mathbb Z)$ which is not the case here.

答:
Without loss of generality, assume that $\lambda>1$. Then $\mathbb{H}/\Gamma$ is biholomorphic to the annulus
$$A(\lambda):=\{z\in\mathbb{C}\mid e^{-\frac{2\pi^2}{\log\lambda}}<|z|<1\}.$$
To see this, let us fix a branch of the logarithm function on $\mathbb{H}$ as follows:
$$\log(re^{i\theta})=\log r+i\theta,\quad~\forall r>0,~ \theta\in(0,\pi)~.$$

Define
$$f:\mathbb{H}\to \mathbb{C},\quad z\mapsto e^{\frac{2\pi i\log z}{\log \lambda}}.$$
By definition, $f$ is holomorphic. It is easy to verify that $f(\mathbb{H})=A(\lambda)$ and  $f:\mathbb{H}\to A(\lambda)$ is a covering map with deck transformation group $\Gamma$. Since $\mathbb{H}$ is simply connected, the conclusion follows.

Remark: On the one hand, the generator $d_\lambda\in Aut(\mathbb{H})$ of $\Gamma$   is hyperbolic. When considered as an element in $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$, the conjugate class of $d_\lambda$ is uniquely determined by $(\mathrm{tr}~d_\lambda)^2$. On the other hand, up to biholomorphic equivalence, a hyperbolic Riemann surface with fundamental group $\mathbb{Z}$ is uniquely determined by its modulus. For $\mathbb{H}/\langle d_\lambda\rangle\cong A(\lambda)$, there is a $1$-$1$ correspondence between $(\mathrm{tr}~d_\lambda)^2$ and the modulus of  $A(\lambda)$.


问:
How to find $\frac{dP}{dt}$ when $P=r\cos t$ and $r\sin t =2te^r$ (Boas, 3ed, Chapter 4, Problem 7.2)
We want to find $\frac{dP}{dt}$ (not $\frac{\partial P}{\partial t}$), with $$P=r\cos t$$ $$r\sin t =2te^r$$

The book usually takes the differentials like this:

$$ dP = \cos t\,dr-r\sin t\,dt$$
$$ \sin t\,dr+r\cos t\,dt = 2e^r\,dt+2te^r\,dr $$

Then eliminating $dr$ (with help from Mathematica) gives:

$$ \frac{dP}{dt}=
\frac{2e^r\cos t - r + 2r\,t\,e^r\sin t}
{\sin t-2te^r} ~~~\textrm{(wrong)}$$

Boas gives a very different answer:
$$ \frac{dP}{dt}=
\frac{2e^r\cos t-r+r^2\sin^2t}
{(1-r)\sin t} ~~~\textrm{(correct)} $$

EDIT after getting 2 answers (thanks!): I plugged in different $(r,t)$ values in the two formulas and got different answers... but then $r$ and $t$ are not independent.

答: Using $2te^r=r\sin t$ in, you will have the same answer.


问:
Proof of cyclicity using power of a point

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $AB$ intersects
altitude $CC'$ and its extension at points $M$ and $N$, and the circle with diameter $AC$ intersects altitude $BB'$ and its extensions at $P$ and $Q$. Prove that the points $M, N, P, Q$ lie on a common circle. (USAMO, 1990)

$MPNQ$ is cyclic if and only if $CBC'B'$ is cyclic. This is because $MPNQ$ is cyclic if and only if $MX\cdot XN=PX\cdot XQ$ by the power of $X$ on the circle whose existence is to be proven. Notice that $B'$ is on the circle with diameter $AB$, because $\angle AB'B=90^{\circ}$, thus $B'X\cdot XB=MX\cdot XN.$

Also, $PX\cdot XQ=CX\cdot XC',$ because $C'$ must be on the circle with diameter $AC$. Therefore, $MX\cdot NX=PX\cdot QX\Longleftrightarrow B'X\cdot XB=CX\cdot XC'$, which happens if and only if $CBC'B'$ is cyclic.

$CBC'B'$ is cyclic. This is because $\angle B'CC'=\angle C'BB'$, since $\triangle ACC'\sim\triangle ABB'.$ Therefore, $MPNQ$ is also cyclic.

I think my solution is likely correct, but my handout gives a different solution, and I'm not very familiar with geometry. Is there anything that I'm missing?

答:
Your solution is correct though it could be made neater.

Let circle with diameter AC be $\omega_C$ and circle with diameter AB be $\omega_B$. As the diameter subtends a right angle on the semicircle, $C'$ lies on $\omega_C$ and $B'$ lies on $\omega_B$.

Now

Power of orthocenter $X$ wrt $\omega_C$ is $$CX\cdot C'X = PX\cdot QX$$

Power of orthocenter $X$ wrt $\omega_B$ is $$BX\cdot B'X = MX\cdot NX$$

Since $$\angle BB'C = 90^{\circ} = \angle BC'C$$ $BC'B'C$ is cyclic.

Power of X wrt circumcircle of $BC'B'C$ is  $$CX\cdot C'X = BX\cdot B'X$$

from which follows $$PX\cdot QX = MX\cdot NX$$

Hence proved, $MPNQ$ is cyclic.


答: Both formulas (correct) and (wrong) are the same. Just use $2te^r = r\sin(t)$ so $2rte^r \sin(t) = r^2 \sin^2(t)$

Title: How to find $\frac{dP}{dt}$ when $P=r\cos t$ and $r\sin t =2te^r$ (Boas, 3ed, Chapter 4, Problem 7.2)

Body:
We want to find $\frac{dP}{dt}$ (not $\frac{\partial P}{\partial t}$), with $$P=r\cos t$$ $$r\sin t =2te^r$$

The book usually takes the differentials like this:

$$ dP = \cos t\,dr-r\sin t\,dt$$
$$ \sin t\,dr+r\cos t\,dt = 2e^r\,dt+2te^r\,dr $$

Then eliminating $dr$ (with help from Mathematica) gives:

$$ \frac{dP}{dt}=
\frac{2e^r\cos t - r + 2r\,t\,e^r\sin t}
{\sin t-2te^r} ~~~\textrm{(wrong)}$$

Boas gives a very different answer:
$$ \frac{dP}{dt}=
\frac{2e^r\cos t-r+r^2\sin^2t}
{(1-r)\sin t} ~~~\textrm{(correct)} $$

EDIT after getting 2 answers (thanks!): I plugged in different $(r,t)$ values in the two formulas and got different answers... but then $r$ and $t$ are not independent.


问:
Matrix Column scaling
I wonder is there any way to scale the column of the matrix by multiplying the matrix on the left? In the other words, let $A = \begin{bmatrix} 1& 2 \\ 3&4 \end{bmatrix}$ and $D = \begin{bmatrix} 2& 0 \\ 0 & 3 \end{bmatrix}$. $AD = \begin{bmatrix} 2 & 6 \\ 6 & 12 \end{bmatrix}$. Is there exist a $D'$ where $D'A = \begin{bmatrix} 2 & 6 \\ 6 & 12 \end{bmatrix}$?

答:
Yes, as long as $A$ is square and invertible it's quite simple to find:

$$D'A = A_s$$

therefore

$$D' = A_s A^{-1}$$

Title: Matrix Column scaling


问:
Show that the Force vector points from the object towards the point (1,0-1)
An object of mass 10 moves with position function $r(t) = < 1+3\cos(t), 4\cos(t), −1+5\sin(t) >$
Find the force vector acting on this object at time t. Show that it points from the object towards the point $(1, 0, −1)$.

I believe that I have found the correct force vector:
$F(t) = 10<-3\cos t,-4\cos t,-5\sin t>$

However, I am not sure how to figure out how to show that it points from the object towards the point $(1,0,-1)$.

Any help would be greatly appreciated.

答:
What you need to do is describe the line along $\vec F$ at time $t$. You know that it is going through $\vec r(t)$. So the equation of this line is $$\vec l_\alpha=\vec r(t)+\alpha\vec F(t),\ \alpha\in\mathbb R$$
Plugging in what you calculated so far,
$$\vec l_\alpha(t)=<1+3\cos t-30\alpha \cos t, 4\cos t-40\alpha\cos t, -1+5\sin t-50\alpha\sin t>$$
Now let's calculate the intersection by saying that $\vec l_\alpha(t_1)=\vec l_\beta(t_2)$. Then $$1+3(1-10\alpha)\cos t_1=1+3(1-10\beta)\cos t_2\\4(1-10\alpha)\cos t_1=4(1-10\beta)\cos t_2\\-1+5(1-10\alpha)\sin t_1=-1+5(1-10\beta)\sin t_2$$
If we want this intersection to be time independent, we see that $\alpha=\beta=0.1$
Plugging in this value of $\alpha$, you get a single point $<1,0,-1>$.