答:
Your induction proof looks fine.

Alternatively, you can use uniqueness of inverses and just check that
$$
(a_1*\cdots*a_n)(a_n^{-1}*\cdots*a_1^{-1}) = e
$$
using associativity.

Title: If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Body:
If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Attempt:
For the simplify and the convenience,
write $a*b=ab$.

I'll prove it by induction on $n$.

Let $a_1,a_2,\ldots,a_n \in G$.
Let $P(n)$ be the statement:
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Base Case:
Clearly $P(1)$ is true since $(a_1)^{-1} = a_1^{-1}$. (Is it enough? What if I take $P(1)$ and $P(2)$ for the base case?)

Inductive Step:
Assume that $P(k)$ is true. That is,
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.
To show that $P(k+1)$ is also true, here it is:
\begin{align*}
(a_1a_2\ldots a_{k+1})^{-1} &=
((a_1a_2\ldots a_k)a_{k+1})^{-1} \\
&= (a_{k+1})^{-1}(a_1a_2\ldots a_k)^{-1} \\
&= a_{k+1}^{-1}(a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1}) \\
&=a_{k+1}^{-1}a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1}
\end{align*}
as desired.

Hence, $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$ for all
$a_1,a_2,\ldots,a_n \in G$. $\Box$

Is the approach above true?


问:
The course of the function $f(x) = \sin^3 (x) + \cos^3 (x)$.
First, sorry for my english, hope you understand.

so I'm examining the course of this function $f(x) = \sin^3 (x) + \cos^3 (x)$ where I need to find everything I can about this function (e.g. definition field, range of values, inflection points, asymptots, global and local max/min, etc.). Now I'm struggling with 3 tasks.

Prove there are no limits in $+\infty$ by showing there are different limits of $f(\pi/2+2k\pi)$ and $f(2k\pi)$ where $k$ goes to $+\infty$. Is this right?
 Are the results of the limits different?

How to get inflection points? I know I get them by solving the 2nd derivative of $f(x)$ which I think I have but I don't know how to get the exact values.
graph and inflex points

Find intersection of $f(x)$ with axes. I got to the point where I have two equations and I need to prove why is it like that.

Thank you in advance

edit: Thank you all for helping me with this problem. I've used @Robert Lee solution in the end.

答:
1)

Notice that if $k \in \mathbb{Z}$ we have that
$$
\sin\left( 2k\pi - \frac{\pi}{2} \right) = -\cos\left(2k\pi \right)= -1
$$
and
$$
\cos\left( 2k\pi - \frac{\pi}{2} \right) = \sin\left(2k\pi \right)= 0
$$
for the first limit. For the second, notice that
$$
\sin\left( 2k\pi  \right) = 0
$$
and
$$
\cos\left( 2k\pi  \right) =1
$$
Finally, notice that
$$
\lim_{k \to \infty} (-1)^3 + 0^3 = -1 \color{red}{\neq} 1 = \lim_{k \to \infty} (0)^3 + 1^3
$$

2)

This one's a bit tricky. By repeated application of the chain rule and then using trigonometric identities we get the following:
\begin{align*}
\frac{d^2}{dx^2}\sin^3(x) + \cos^3(x) &= 3 \left[2 \sin(x) \cos^2(x) - \underbrace{\sin^3(x)}_{\color{blue}{\left(1-\cos^2(x)\right) \sin(x)}} + 2 \cos(x) \sin^2(x) - \underbrace{\cos^3(x)}_{\color{blue}{\left(1-\sin^2(x)\right) \cos(x)}}\right] \\
& =  3 \left[2 \sin(x) \cos^2(x) \color{blue}{-\sin(x) + \cos^2(x) \sin(x)} + 2 \cos(x) \sin^2(x) \color{blue}{-\cos(x) + \sin^2(x) \cos(x)}\right]\\
& = 3 \left[\color{blue}{3} \sin(x) \cos^2(x)  + \color{blue}{3}  \cos(x) \sin^2(x) - \sin(x) - \cos(x)\right]\\
& =\frac{3}{2} \left[6 \sin(x) \cos^2(x)  + 6  \cos(x) \sin^2(x) -2 \sin(x) - 2\cos(x)\right]\\
& =\frac{3}{\sqrt{2}\sqrt{2}}\left( \sin(x) + \cos(x)\right)\left(6 \  \underbrace{\sin(x) \cos(x)}_{\color{purple}{\frac{\sin(2x)}{2}}} -2 \right)\\
& =\frac{3}{\sqrt{2}}\left( \sin(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\cos\left(\frac{\pi}{4}\right)}} + \cos(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\sin\left(\frac{\pi}{4}\right)}}\right)\left(\color{purple}{3}\sin(2x) -2 \right)\\
& = \frac{3}{\sqrt{2}}\sin\left(x + \frac{\pi}{4}\right)\left(3\sin(2x) -2 \right)
\end{align*}
We can set this last equation equal to $0$, which would them imply that each of the factors is equal to $0$ separately. Recalling that
$$
\sin(\theta) = \alpha \color{blue}{\iff} \theta = (-1)^n\arcsin\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z}  \tag{1}
$$
We then see that the inflection points are at
\begin{align}
\sin\left(x + \frac{\pi}{4}\right) = 0 &\iff x + \frac{\pi}{4} = (-1)^n\arcsin\left(0\right) + \pi n\\
&\iff x + \frac{\pi}{4} = \pi n\\
&\iff \boxed{x  = \pi n -  \frac{\pi}{4}}, \quad n \in \mathbb{Z}
\end{align}
And also at
\begin{align}
3\sin(2x) -2 = 0 & \iff \sin(2x) = \frac{2}{3}\\
&\iff 2x = (-1)^n\arcsin\left(\frac{2}{3}\right) + \pi n\\
&\iff x = \boxed{\frac{(-1)^n}{2}\arcsin\left(\frac{2}{3}\right) + \frac{\pi}{2} n},\quad n \in \mathbb{Z}
\end{align}

3)

Your solution is correct. The zeros of the function are at the points where
$$
\sin(x) = - \cos(x) \color{blue}{\implies}\tan(x) = -1
$$
And recalling that
$$
\tan(\theta) = \alpha \color{blue}{\iff} \theta = \arctan\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z}  \tag{2}
$$
We see that the zeros are at the points
\begin{align}
\tan(x) = -1 & \iff  x = \arctan(-1) + \pi n\\
& \iff  x = \boxed{-\frac{\pi}{4} + \pi n}, \quad n \in \mathbb{Z}
\end{align}
Which is exactly the same set that you propose in your solution since, if we start counting the integers from one over by setting
$$
n = k+1 \in \mathbb{Z}
$$
we get
\begin{align}
x &=-\frac{\pi}{4} + \pi n\\
&=-\frac{\pi}{4} + \pi (k+1)\\
&=-\frac{\pi}{4} + \pi k+ \pi\\
&=\pi\left(1-\frac{1}{4}\right) + \pi k\\
&=\frac{3\pi}{4} + \pi k, \qquad k \in \mathbb{Z}
\end{align}

Also, if you want to argue why $1 = \sin(x)\cos(x)$ has no solutions, recall the previously used identity that $\sin(x)\cos(x) = \frac{\sin(2x)}{2}$. So you want to show that
$$
\frac{\sin(2x)}{2} = 1 \color{blue}{\implies} \sin(2x) = 2
$$
but recalling that the sine function can only output values in the interval $[-1,1]$, since $2 \notin [-1,1] $ the above equation has no solutions.

If you have any questions about the solutions please let me know!

ContentLicense: CC BY-SA 4.0


问:
Find the derivative of the function $f(x)=(P(x))^a$ where $a\in\Bbb{R}$ and $P(x)$ is a real polynomial with no real roots
I need help to clarify this statement. I found it in a friend's old homework notebook and I think that the answer is

$$f'(x)=aP'(x)\cdot(P(x))^{a-1}$$
However, I just don't get why the fact of $P(x)$ having no real roots is relevant. Is there a way to state the answer using only $P(x)$ instead of $P'(x)$? Am I missing something?

I think his teacher was trying to make this problem look more complex than it is, but I just wanted to check if maybe I am the one who is mistaken.

答:
If you want to have defined $P(x)^a$ for all $a$, you need $P(x) > 0$ for all $x$.

Let us omit the case $a = 0$ which leads to $P(x)^0 \equiv 1$. Okay, it may be questionable what $0^0$ is (occurring if $P(x)$ has real zeros), but either we take $0^0 = 1$ or let it undefined which gives us a function not defined on all of $\mathbb R$.

So let $a \ne 0$. It is of course no problem to define $y^a$ for

all $y \in \mathbb R$ and all $a \in \mathbb N$,

all $y \in \mathbb R \setminus \{0\}$ and all $a \in -\mathbb N = \{-1,-2,-3,\ldots\}$.

For $y > 0$ one usually defines $y^a = e^{a \ln y}$. This definition does not make sense for $y \le 0$. Okay, if $a > 0$ and $y = 0$ we may take $0^a = 0$, but for $a < 0$ there is no $0^a$. For $y < 0$ there is no general method to give a reasonable interpretation of $y^a$ in case $a \notin \mathbb Z$. You might argue that it works for $a = p/q$ with $q$ odd because we could take $y^a = \sqrt[q]{y^p}$, but this is only a very special case and the function $y^a$ is not differentiable at $y  = 0$ (which transfers to $P(x)^a$ if $P(x)$ has real zeros).

Therefore, if $p(x)$ has a (real) zero $x_0$, then $P(x_0)^a$ is undefined for $a < 0$. This is the reason for excluding polynomials $P(x)$ with real zeros. Polynomials without zeros are either positive or negative, and the negative case gives problems as shown above.


问:
How to compute the integral $\int_0^\infty \arctan\left(\frac 4x\right)\sin(x)\,\mathrm dx$
I want to compute the definite integral $$\int_0^\infty \arctan\left(\frac 4x\right)\sin(x)\,\mathrm dx.$$

My answer is below.

答:
Let $$I\overset{\text{Def.}}=\int_0^\infty \arctan\left(\frac 4x\right)\sin(x)\,\mathrm dx.$$ Note that the integral is indeed well-defined, as for instance the substitution $y=\arctan\left(\frac4x\right)$ shows.

Integration by parts gives $$I=\left[-\cos(x)\arctan\left(\frac4x\right)\right]^{\infty}_0-4\int_0^\infty \frac{\cos(x)}{x^2+16}\,\mathrm dx = \frac\pi2-4\int_0^\infty \frac{\cos(x)}{x^2+16}\,\mathrm dx.$$

Now, since $\int_{-\infty}^\infty\frac{\sin(x)}{x^2+16}\,\mathrm dx=0$,
\begin{equation}\begin{split}
\int_0^\infty \frac{\cos(x)}{x^2+16}\,\mathrm dx
&= \frac12\int_{-\infty}^\infty\frac{\cos(x)}{x^2+16}\,\mathrm dx
\\&=\frac12\left(\int_{-\infty}^\infty\frac{\cos(x)}{x^2+16}\,\mathrm dx+i \int_{-\infty}^\infty\frac{\sin(x)}{x^2+16}\,\mathrm dx\right)
\\&=\frac12\int_{-\infty}^\infty\frac{\exp(ix)}{x^2+16}\,\mathrm dx.
\end{split}\end{equation}

After Jordan's Lemma, if $\gamma_R$ denotes a semi-circular contour with radius $R$, i.e. $$\gamma=\gamma_1 \text{ and then }\gamma_2,$$ where $\gamma_1$ is the path for $[-R,R]$ and $\gamma_2$ is the path for $\{z\in\mathbb C: \lvert z\rvert  = R, \operatorname{Im}(z)\geq0\}$, then

$$\frac12\int_{-\infty}^\infty\frac{\exp(ix)}{x^2+16}\,\mathrm dx=\frac12\lim_{R\to\infty}\int_{\gamma_R}\frac{\exp(iz)}{z^2+16}\,\mathrm dz\overset{\text{Residue Theorem}}=\frac12\operatorname{Res}\left(\frac{\exp(iz)}{z^2+16},z=4i\right)=\frac{\pi}{8e^4}.$$

In conclusion,

$$\bbox[15px,border:1px groove navy]{I=\frac\pi2-\frac\pi{2e^4}.}$$


问:
I solved the problem. But I have some doubts on the solution.
Given the sequence $a_n = \sin(n)$, where $n$ is an integer. Prove that limit superior is $1$.

My solution: by previous theorems we know that if $x,y \in \Bbb R$ and the ratio of $x$ and y is irrational, then there exists the integers $a$ and $b$ such that $$\forall \epsilon, \quad  0<ax+by<\epsilon$$ So we can put integrs $m$ ands $n$ such that $$0<\frac{n}{m}-\frac{\pi}{2}<\epsilon$$ And we are done.

But here m should be in form of $m=4k+1,\, k\in \Bbb Z$.

My question is, how we can be sure that we can find such $m$?

答: if you want From the theorem above You can deduce that you can find infinitely pairs of $n,k \in Z$ such that $0< \frac{\pi}{2}+2* \pi*k-n< \varepsilon$. And dont worry about form of the denominator like $ \frac{\pi}{2} \approx \frac{n}{4k+1}$ becasuse actually what you need is the ratio and with infinitely many n, k you are able form any kind of ratio. Because in $Q$ class of ratio infinitely many. for example $\frac{1}{2}$=${\frac{1}{2}, \frac{2}{4}, \frac{4}{8} .....}$


答:
COMMENT.-(1)Any periodic function defined over $\mathbb R$ has a limit when $x\to\infty$$\space (f$ non-constant).

(2) $f(x)=(\sin(x)+\cos(x))(1-\sin(x)\cos(x))$ so the first factor gives the zeros of the function $x=-\dfrac{\pi}{4}+k\pi$ where $k\in\mathbb Z$. (the second factor clearly has

non-real roots).

(3) Inflection points: you have $f''(x)=-3(\cos(x)-\sin(x))^2(\cos(x)+\sin(x))=0$ find yourself $f'''(x)$ and determine this thirds question

Title: The course of the function $f(x) = \sin^3 (x) + \cos^3 (x)$.

Body:
First, sorry for my english, hope you understand.

so I'm examining the course of this function $f(x) = \sin^3 (x) + \cos^3 (x)$ where I need to find everything I can about this function (e.g. definition field, range of values, inflection points, asymptots, global and local max/min, etc.). Now I'm struggling with 3 tasks.

Prove there are no limits in $+\infty$ by showing there are different limits of $f(\pi/2+2k\pi)$ and $f(2k\pi)$ where $k$ goes to $+\infty$. Is this right?
 Are the results of the limits different?

How to get inflection points? I know I get them by solving the 2nd derivative of $f(x)$ which I think I have but I don't know how to get the exact values.
graph and inflex points

Find intersection of $f(x)$ with axes. I got to the point where I have two equations and I need to prove why is it like that.

Thank you in advance

edit: Thank you all for helping me with this problem. I've used @Robert Lee solution in the end.


问:
cake cutting puzzle: why do finitely many cuts suffice?
Puzzle from http://www2.maths.bris.ac.uk/~majwm/compendium/cakeslice.php

A piece of angle $x$ is cut from a cake, which is purple on top and yellow underneath, and turned upside down. Then another piece of angle $x$ is cut from where the last cut is made, and the cuts carry on going clockwise around the cake. Why will the cake become all purple on top after finitely many moves, whether or not $x$ is rational?

Also, will the cake still return to purple after finitely many moves if the cake has 3 or more layers and instead of the pieces being turned upside down the layers are cyclically permuted? Or if the angles of consecutive cuts cycle through some finite sequence $x_1...x_k$?

答:
It should have been obvious that merely permuting the pieces wouldn't work.

If a line between a yellow and a purple section was reached on the cut that ended at angle $nx$, then if the cut ending at angle $mx$ is the next piece cut that contains that line, the line will be flipped to the position that will be reached on the $m+(m-1)-n \ th$ cut. The way all the lines get flipped over and turned into lines that will be reached later forces the cake to return to purple eventually, and the same kind of thing happens for when there are two angles $x_1,x_2$.

When there are three angles this argument does not work because when the cut that reaches $(n+1)x_1+nx_2+nx_3$ goes over a line at $(m+1)x_1+(m+1)x_2+mx_3$ it will flip it to $(2n-m)x_1+(2n-m-1)x_2+(2n-m)x_3$ which is not a cut that is guaranteed to be reached later.

ContentLicense: CC BY-SA 3.0


问:
Is it true that $\Gamma(x+1) \leq x^x$ for every $x>0$?
I've seen the inequality $\Gamma(x+1) \leq x^x$ for $x > 0$ used without proof or reference and I'm not sure it's true.

Thanks for your help.

答:
Here is a solution that only uses simple properties of the Gamma function:

For $x\geq 1$, write $x=n+t$ with $0\leq t<1$. Then by property $\Gamma(x+1)=x\Gamma(x)$, we see that
$$\Gamma(x+1)=x \Gamma((n-1)+t)=x(x-1)\Gamma((n-2)+t)=\\
x(x-1)(x-2)\cdots (x-(n-2))\Gamma(1+t)$$

It is easy to see that $\Gamma(1+t)\leq 1$ for $t\in [0,1)$, for instance by noting that for $t=0$ and $t=1$, $\Gamma(1+t)=1$ and $\Gamma(1+t)$ is convex. This shows that
$$\Gamma(x+1)\leq x(x-1)(x-2)\cdots(x-(n-2))\cdot 1\leq x^n\leq x^x$$
since there are $n$ terms in the product, each less than $x$.


问:
Interesting Theorems on Finitely Generated Abelian Groups?
First time teaching algebraic topology, probably gonna be related to most of my questions on here for a while.

I was wondering if anyone knows of particularly interesting theorems or examples from the theory of finitely generated abelian groups, beyond the standard auxiliary stuff and the classification/primary decomposition.  By auxiliary I mean stuff like the basic results on subgroups, quotients, torsion/rank uniqueness and change of base.  Doesn't have to be something that can quickly be PROVEN from first principles, but as long as it's something digestible and believable (or even better, unbelievable).  Perhaps certain actions on manifolds or trees or the like?  Something for topology-minded folks rather than algebra-minded would be ideal, but whatever you think is cool I'd love to hear about!

It could also be surprising internal/structural features, or results about their automorphism groups.

Thanks, hope someone finds this topic interesting!

答:
Somewhat related: All subgroups of finitely generated abelian groups are finitely generated, however this is not necessarily so for subgroups of finitely generated groups.

Perhaps the classic example of the free group of 2 generators has subgroups that are not finitely generated, and its relation to algebraic topology: Commutator subgroup of rank-2 free group is not finitely generated.


问:
Is This Proof of Pythagorean Theorem?

Given image:

$\triangle\mathit{BDE}$ is an altitude right triangle as given by the image.

$\triangle\mathit{BDE}$ and $\triangle\mathit{FDE}$ both contain 90° given by the image; therefore, $\triangle\mathit{BDE}$ and $\triangle{\mathit{FDE}}$ are both right triangles.

$\triangle\mathit{BDE}$ is an altitude right triangle with $\triangle\mathit{FDE}$ as given by the image; therefore, $\triangle\mathit{BDE}\sim\triangle\mathit{FDE}$

If $\triangle\mathit{BDE}$ is similar to $\triangle\mathit{FDE}$ then:

If these triangles are similar, then the ratio of each of the sides of $\triangle\mathit{BDE}$ to the corresponding sides of $\triangle\mathit{FDE}$ is common; therefore, the equations below can be derived:

$\overline{\mathit{DE}}^{2}=\overline{\mathit{BE}}\overline{\mathit{EF}}$

$\overline{\mathit{DF}}^{2}=\overline{\mathit{EF}}\overline{\mathit{BF}}$

$\overline{\mathit{BE}}+\overline{\mathit{EF}}=\overline{\mathit{BF}}$ as given by the image

Given these equations algebra can then be performed to solve for $\overline{\mathit{DF}}$. Algebra:

Rearranging:$\overline{\mathit{DE}}^{2}=\overline{\mathit{BE}}\overline{\mathit{EF}}$ to get $\frac{\overline{\mathit{DE}}^{2}}{\overline{\mathit{EF}}}=\overline{\mathit{BF}}.$

$\overline{\mathit{DF}}^{2}=\overline{\mathit{EF}}\overline{\mathit{BF}}$

Substituting:$$\overline{\mathit{DF}}^{2}=\overline{\mathit{EF}}(\overline{\mathit{BE}}+\overline{\mathit{EF}})$$

Substituting Again:$$\overline{\mathit{DF}}^{2}=\overline{\mathit{EF}}(\frac{\overline{\mathit{DE}}^{2}}{\overline{\mathit{EF}}} + \overline{\mathit{EF}})$$

Simplifying:$$\overline{\mathit{DF}}^{2}=\overline{\mathit{DE}}^{2}+\overline{\mathit{EF}}^{2}$$

Which is the familiar form of Pythagorean Theorem.

Is this a mathematically correct derivation?

Is this proof general enough to be used as a general proof?

答:
OK here's a bit of criticism, meant in the spirit of improvement. Reworking the steps of your attempted proof with the same numbering (plus some commentary):

(Your diagram is hard to read; the letters are very small relative to the diagram. I suspect you were sometimes misreading it yourself. The square angle mark is a useful device for avoiding too many numbers:)

$\triangle BDF$ is a right triangle with right angle at $D \quad$ - (introduce the main triangle)

$DE\;$ is the altitude from $D$ to $E$ on $BF$, perpendicular to $BF \quad$ - (key construction line)

$ $ - (I didn't really understand what you were trying to set up in point 2)

$\triangle BDE$ and $\triangle FDE $ both contain $90^\circ$ by construction; therefore, $\triangle BDE$ and $\triangle FDE$ are both right triangles.

$ $- (basically OK, could go into next point)

$\triangle BDE$ shares the angle at $B$ with $\triangle BDF$ and both are right triangles, therefore $\triangle BDE \sim \triangle BDF$. In the same way $\triangle DEF$ shares the angle at $F$ with $\triangle BDF$ so $\triangle DEF\sim \triangle BDF$. Thus also $\triangle BDE \sim \triangle DEF$.

$ $ - (You did not establish this similarity - it works via the main triangle, and the fact that a right triangle can be cut into two similar triangles is key)

Thus the ratio between sides of $\triangle DEF$ and the corresponding sides of both $\triangle BDE$ and $\triangle BDF$ are equal, giving:

$\frac{|DE\,|}{|BE\,|} = \frac{|EF\,|}{|DE\,|} \implies |DE|^2 = |EF|\cdot |BE|$

$\frac{|DF\,|}{|BF\,|} = \frac{|EF\,|}{|DF\,|} \implies |DF|^2 = |EF|\cdot |BF|$

$ $ - (No need to say "if $P$" for something you've established; "since $P$" works but it's implied anyway. You neglected to note that one of your ratios here was with the main triangle. Having said you're equating ratios, you should probably do that explicitly. You were using vector-style notation for scalar length.)

Note that $|BE|+|EF| = |BF| \quad$

$ $- (I might have said this back in point 2 when we split $BF$ )

Then: \begin{align}|DF|^2 &= |EF|\cdot \left(|BE|+|EF|\right) \\[1ex]
|DF|^2 &= |EF|\cdot \left(\frac{|DE|^2}{|EF|}+|EF|\right) \\[1ex]
|DF|^2 &= |DE|^2+|EF|^2
\end{align}
as required since $\triangle DEF \sim \triangle BDF$.

$ $- (You can usually just let the equations speak for themselves but linking words here or there are OK. For more involved arguments, equation numbering is very helpful. Probably the link back to the main triangle through similarity should be stated here.)

Clean version with a variation from your point 5 to solve direct for the main triangle:

$\triangle BDF$ is a right triangle with right angle at $D$ - construct altitude $DE\;$ from $D$ to $E$ on $BF$; this is perpendicular to $BF$ and note that $|BE|+|EF| = |BF|$.

$\triangle BDE$ and  $\triangle BDF$ share the angle at $B$ and both are right triangles, therefore $\triangle BDE \sim \triangle BDF$. In the same way $\triangle DEF$ shares the angle at $F$ with $\triangle BDF$ so $\triangle DEF\sim \triangle BDF$.

Thus the ratio between sides of $\triangle BDF$ and the corresponding sides of both $\triangle BDE$ and $\triangle DEF$ are equal, giving:

$\frac{|DB\,|}{|BF\,|} = \frac{|BE\,|}{|DB\,|} \implies |DB|^2 = |BF|\cdot |BE|$

$\frac{|DF\,|}{|BF\,|} = \frac{|EF\,|}{|DF\,|} \implies |DF|^2 = |BF|\cdot |EF|$

Then: \begin{align}
|DB|^2 + |DF|^2 &= |BF|\cdot |BE| + |BF|\cdot |EF| \\
 &= |BF|\cdot \left(|BE|+|EF|\right) \\
 &= |BF|^2
\end{align}
as required.

Title: Is This Proof of Pythagorean Theorem?


问:
Exponential Generating Function For Derangements
I have been introduced to the concept of exponential generating functions a few days ago. However, my understanding of them are still quite limited, and I would like to see some examples. Earlier this term, I derived a formula for the number of derangements of size $n$ using the inclusion/exclusion principle, namely that $D_n = n!\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}$. How would I go about deriving this result using exponential generating functions, without using the inclusion/exclusion principle to derive $D_n$. The formula we are using for the these generating functions are $\Phi_D(x) = \sum_{n=0}^{\infty}|D_n|\frac{x^n}{n!}$.

If anyone could walk me through this example, I would greatly appreciate it :)
Thanks!

答:
Another way: There are $n!$ permutations in all. A permutation with $k$ fixed points shuffles the other $n - k$ elements around without fixed points, the $k$ fixed points can be selected in $\binom{n}{k}$ ways. So:
$$
n! = \sum_{0 \le k \le n} \binom{n}{k} d_{n -k}
$$
This is a binomial convolution, using Mike Spivey's notation that gives:
$$
\begin{align*}
\frac{1}{1 - x} &= G_D(x) e^x \\
G_D(x) &= \frac{e^{-x}}{1 - x}
\end{align*}
$$</p>


问:
Finding change of a function of another function of another variable with derivatives/integrals
I've recently concocted an innocent looking problem that I'm struggling to solve myself.

The question is as follows: some wheel generates energy, in Joules, by spinning, at $W(\Theta)$ Joules per radian. The same wheel moves at an angular velocity $\omega (t)$, in radians per second. How many Joules are generated from $0 \le t \le 10$?

I've thought of multiplying $W(\Theta)$ and $\omega(t)$ because the units would cancel accordingly and allow for integration, but they are functions of different variables. I don't even have an answer because the problem is made up, but I am still curious. How would it be done? Is it even possible?

答:
Since $W(\Theta)$ is in units of joules per radian and we want joules, we need only know through how many radians the wheel turns, then we can pass that angle through $W$ to get the number of joules.

We are told that the wheel turns at $\omega(t)$ radians per second.  So over the interval $0 \leq t \leq 10$, the wheel turns through
$$  \int_0^{10} \omega(t) \,\mathrm{d}t  $$
radians.

Passing this angle to $W$, we compute
$$  W \left( \int_0^{10} \omega(t) \,\mathrm{d}t \right)  \text{,}  $$
the number of joules generated during the interval $[0,10]$.


问:
Transforming of Chi Square Distribution and F Distribution
Let X and Y be two independent Chi-Square-distributed radom variables each with n d.f. Then
$ Z = \frac{\sqrt n}{2} \frac{X - Y}{\sqrt {X.Y}}$ has the t-Distribution. (I proved this already).

But I can't prove how Z can be written as $ Z = \frac{\sqrt n}{2} (\sqrt{F_{n,n}}-\frac{1}{\sqrt{F_{n,n}}})$ with $F_{n,n}$  is F distribution with n and n d.f. Someone can help me please. Thanks in advance!

答:
If $X$ has chi-square($m$) distribution and $Y$ has chi-square($n$) distribution, and the two are independent, then it is either a theorem or a definition that
$(X/m)\big/(Y/n)$ has $F(m,n)$ distribution. So defining $W:=X/Y$, we have $$Z=\frac{\sqrt n}2\left(\sqrt W-\frac1{\sqrt W}\right).$$


问:
Functions $f:\mathbb Z^2\to\mathbb R$ with $f(x,y)=\frac{f(x-1,y)+f(x,y-1)}{2}$
Find all functions $f:\mathbb Z^2\to[0,1]$ such that for any $x,y\in\mathbb Z$, we have
$$f(x,y)=\frac{f(x-1,y)+f(x,y-1)}{2}.$$
Any constant function works, and I'm tempted to believe that they are the only solutions.

By inductively applying the relation, I have shown that
$$f(x+n,y+n)=2^{-n}\cdot\sum_{k=0}^n\binom nkf(x+k,y+n-k).$$
I have not really made more progress from here.

答:
Starting at $(x,y)$, take a random walk $S_n$ down and left. The condition implies that $M_n=f(S_n)$ is martingale. As $f$ is bounded, by the martingale convergence theorem, $M_n$ converges (say to $M_\infty$) almost surely.

So let $X_1,X_2\dots$ be the individual left/down steps of the random walk. By the Hewitt-Savage zero-one law, for any interval $[a,b]\subset\mathbb R$ we have
$$\mathbb P\left(\lim_{n\to\infty} f((x,y)+X_1+\dots+X_n)\in[a,b]\right)\in\{0,1\}.$$
Thus there exists $\lambda\in\mathbb R$ such that
$$\mathbb P\left(\lim_{n\to\infty} f((x,y)+X_1+\dots+X_n)=\lambda\right)=1.$$
As $f$ (and hence $M_n$) is bounded, it follows by the dominated convergence theorem that
$$f(x,y)=\mathbb E[M_0]=\lim_{n\to\infty}\mathbb E[M_n]=\mathbb E[M_\infty]=\lambda$$

Finally, note that for any $(x',y')\neq(x,y)$, the two random walks starting at $(x,y)$ and $(x',y')$ meet with positive probability, so we must also have $f(x',y')=\lambda$ too, i.e. $f$ is constant.

This is similar to the problem that a lower-bounded discrete harmonic function on $\mathbb Z^2$ is constant (see e.g. here). That problem could be solved by high-tech probabilistic arguments, which motivated me trying the same for this one.

The argument actually ended up being a little more fiddly for this problem.


问:
How do you read an expression such as e^{-rt}
I heard people say that
the exponential to the power of R times T

e^{rt}

Is it normal?

What about other similar things?

e^{-rt}

Is it read

the exponential to the negative power of R times T

or

the exponential to the power of negative R times T

or

the exponential to the negative exponent of R times T

What do you think about?

答:
The expression e^rt is sloppy. Either write it as $e^{rt}$ or e^{rt} so that you won't confuse it with $e^r\cdot t$. (The expression $\exp(rt)$ is also used very often.)

As long as the context is clear, "$e$ to the power of $rt$" is a usual way to read it.

Similarly, I would write e^{-rt} or $e^{-rt}$ to avoid any confusion. "$e$ to the power of negative $rt$" is reasonable.


问:
Preconditions of the definition of a multivariate real-valued function being real-analytic
Recently I'm reading a text and got confused about preconditions of the definition of a multivariate real-valued function being real-analytic. Writing the coordinates on $\Bbb R^n$ as $x=(x^1,\dots,x^n)$ and let $p=(p^1,\dots,p^n)$ be a point in an open set $U$ in $\Bbb R^n$. Following is the definition of real-analytic of a function $f:U\to\Bbb R$ in this text:

The function $f$ is real-analytic at $p$ if in some neighborhood of $p$ it is equal to its Taylor series
at $p$: $$f(x)=f(p)+\sum\limits_i\frac{\partial f}{\partial x^i}(p)(x^i-p^i)+\frac{1}{2!}\sum\limits_{i,j}\frac{\partial^2f}{\partial x^i\partial x^j}(p)(x^i-p^i)(x^j-p^j)\\+\cdots+\frac{1}{k!}\sum\limits_{i_1,\dots,i_k}\frac{\partial^k f}{\partial x^{i_1}\cdots\partial x^{i_k}}(p)(x^{i_1}-p^{i_1})\cdots(x^{i_k}-p^{i_k})+\cdots,$$ where the general term is summed over all $1\le i_1,\dots,i_k\le n$.

My question is, do we need to assume as preconditions in the definition of real-analytic: 1) all partial derivatives of all orders in any order exist and 2) the mixed partial derivatives are equal no matter what order with regard to which the variables are taken. As I see it, If 1) does not hold, the Taylor series would not be well-defined, not to mention it being equal to $f(x)$. If 2) is missing, I would be totally confused about what the summands in general term look like. The continuity of mixed partial derivatives can guarantee 2), but I choose not to assume these continuities to keep the list of preconditions to minimum. However, the above preconditions 1) and 2) are not mentioned explicitly in the text. I also checked the definition in some other textbooks and wikipedia, but no source states these preconditions explicitly. So I'd like to ask here for a confirmation: should we assume the above two conditions 1) and 2) in the definition of real-analytic? Two derivative questions follow: do the text and other sources just assume the above preconditions implicitly and is it a kind of convention in math community that when we write down something, all the preconditions that make sense of what is written are assumed to hold in advance and it is the reader's responsibility to figure them out? I'm self-studying so I don't have math professor or TA to ask; something that is obvious to math students may be a big obstacle for me, so I would be greatly grateful for any of your help.

答:
At this point I'm essentially just spouting out stuff which is readily available in Henri Cartan's text on complex analysis. Let us first consider a formal power series in $n$-variables with complex coefficients,
\begin{align}
S(X_1,\dots, X_n)=\sum_{I\in \Bbb{Z}_{\geq 0}^n} a_I X^I \equiv \sum_{i_1,\dots, i_n\geq 0}
a_{i_1,\dots, i_n}X_1^{i_1}\cdots X_n^{i_n}
\end{align}
So far this is just a formal series, in the sense that this is just a bunch of symbols, with no meaning attached to the sum notation. Let us now denote by $\Gamma_S$ the set of $(r_1,\dots, r_n)$ such that each $r_i\geq 0$ and such that the following sum is finite:
\begin{align}
\sum_{i_1,\dots, i_n\geq 0}|a_{i_1,\dots, i_n}|(r_1)^{i_1}\cdots (r_n)^{i_n}<\infty.
\end{align}
The set $\Gamma_S$ is of course not empty, since it contains the origin $(0,\dots, 0)$. We define the domain of convergence of $S$ to be $\Delta_S := \text{int}(\Gamma_S)$. Note that $\Delta_S$ is always an open set, but may or may not be empty. If the domain of convergence is non-empty, this means there exist $r_1,\dots, r_n>0$ such that the above sum is finite, and furthermore it is easy to show (I think it's called Weirstrass's M-test?) that if $r_1,\dots, r_n>0$ are such that the above sum is finite, then the series
\begin{align}
S(z_1,\dots, z_n)&=\sum_{i_1,\dots, i_n\geq 0}a_{i_1,\dots, i_n}(z_1)^{i_1}\cdots (z_n)^{i_n}
\end{align}

converges absolutely and uniformly whenever $|z_i|\leq r_i$ for $i\in\{1,\dots, n\}$ (also, you can show that if $(|z_1|,\dots, |z_n|)$ doesn't lie in the closure of $\Gamma$ then the series $S(z_1,\dots, z_n)$ diverges).

That was a brief intro to the basic definitions we need to talk about analyticity. Given an open set $U\subset \Bbb{R}^n$, a function $f:U\to \Bbb{R}$ is real analytic at $p$ if there exists a formal series $S$ with real coefficients, having non-empty domain of convergence $\Delta_S$, such that for $x\in U$ close to $p$ (i.e $(|x_1-p_1|,\dots, |x_n-p_n|)\in \Delta_S$), we have $f(x)= S(x_1-p_1,\dots, x_n-p_n)$.

With this, you can try to extend the usual $1$-D argument to show that if $f$ is analytic at $p$, then $f$ in some small neighbourhood of $p$, $f$ has continuous partial derivatives of all orders (so the order of differentiation doesn't matter), and that the coefficients are determined uniquely as
\begin{align}
a_{i_1,\dots, i_n}&=\frac{1}{i_1!\cdots i_n!}\frac{\partial^{i_1+\dots i_n}f}{\partial x_1^{i_1}\dots \partial x_n^{i_n}}(p)
\end{align}
Of course, since we have absolute convergence, we can rearrange the sums however we wish, so this means in a small neighbourhood of $p$, $f$ will equal its Taylor series about $p$.

Anyway, I would highly suggest you review the relevant definitions of analyticity in $1$-dimension and see what that implies about differentiability, relation to Taylor series etc, because the $n$-dimensional case simply amounts to rewording the definitions slightly and reusing the $1$-variable theorems carefully (so I hope I haven't made any typos lol).


问:
Absolute value of an element in a C*-algebra
Is absolute value of a partial isometry a partial isometry itself?</p>

Body:
We have, as square roots are self-adjoint $\def\abs#1{\left|#1\right|}$
\begin{align*}
  \abs v^*\abs v &= [(v^*v)^{1/2}]^*(v^*v)^{1/2}\\
              &= (v^*v)^{1/2}(v^*v)^{1/2}\\
              &= v^*v
\end{align*}


问:
Solution to second order homogeneous ODE
I have the 2nd order homogeneous ODE:
$$y(x)''+\frac{2}{x}y'(x)-By=0 $$
Where $B$ is a constant. Since I am a bit rusty with analytical solutions, I plugged it in MatLab's symbolic solver which found the following solution.
$$y(x)=C_1\frac{e^{\sqrt{B}x}}{x}-C_2\frac{e^{-\sqrt{B}x}}{2\sqrt{B}x}$$
Unofrtunately, the constant B takes both positive and negative values, but since the ODE is derived from a physical problem I'd like the solution to always be real.
Any suggestion/comment is highly appreciated.

答:
$$y''+\frac{2}{x}y'-\beta y=0$$
$$xy''+{2}y'-\beta xy=0$$
Substitute $z=\sqrt \beta x  ( \beta >0)$:
$$x{\beta} y''(z)+{2}{\sqrt {\beta}}y'(z)-\beta xy(z)=0$$
$$z y''+2y'-zy=0$$
$$(zy)''-zy=0$$
Charactristic polynomial is:
$$r^2-1=0 \implies r= \pm 1$$
$$\implies zy=c_1e^{-z}+c_2e^{z}$$
$$\boxed {y(x)=\dfrac {1}{x} \left (k_1e^{-\sqrt {\beta}x}+k_2e^{\sqrt {\beta}x} \right)}$$

For  $\beta <0 $ :
$$xy''+{2}y'+\alpha xy=0$$
Where $\alpha=-\beta $.
Substitute $z=\sqrt \alpha x $:
$$x{\alpha} y''(z)+{2}{\sqrt {\alpha}}y'(z)+\alpha xy(z)=0$$
$$z y''+2y'+zy=0$$
$$(zy)''+zy=0$$
$$ \implies r^2+1=0 \implies r=\pm i$$
$$\implies zy=c_1\cos z+c_2 \sin z$$
Unsubstitute $z= \sqrt {\alpha } x$.
$$\boxed {y(x)=\dfrac {1}{x} \left (k_1\cos {(\sqrt {\alpha}x)}+k_2 \sin ({\sqrt {\alpha}x}) \right)}$$


问:
If $(a) = (b)$, does there exists a unit $u$ such that $a = ub$ for commutative ring?

$R$ is a commutative ring. Let $(x)$ denote the ideal generated by $x \in R$. If $(a) = (b)$, is it true that there exists a unit $u$ such that $a =ub$?

If $R$ is integral domain, then it's simple. Simply choose $a = ub$ and $va =b$ so that $b(uv - 1) = 0$. But what if $R$ is not integral domain? I wrote a short code to run all $\mathbb Z/n\mathbb Z$ for $n < 10000$ and can't find any contradiction. I also tried construction like $\mathbb Z/n\mathbb Z \times \mathbb Z/n\mathbb Z$ and let $b = (0, x)$ and $uv = (y, 1)$ so that $b(uv-1) = 0$. None of them works. Any suggestions?

答: Let $R$ be the ring of all polynomials in $x$ vanishing at zero (and having integer coefficients, or rational coefficients, or real coefficients, doesn't really matter). Then $x$ and $-x$ generate the same ideal, but there's no unit $u$ such that $-x=ux$. Indeed, there's no identity element, so there are no units at all.


问:
Metric space is Hausdorff
I was reading Is this proof that all metric spaces are Hausdorff spaces correct? to see how the statement is proved, but I have a question. Is it possible to generalize that any ball in metric space is open? How do we know the "open set" in metric space? Is it a convention to say that topology of metric space is topology generated by every ball around every point? I can't think it right now but would there be any topology in metric space that some ball around some point is not open?</p>

答:
Given a metric space $(X,d)$, the topology induced by the metric $d$, i.e., the topology generated by the open balls:
$$
B(x,r)=\{y\in X\mid d(y,x)<r\}
$$
is called "metric topology" of $X$.

By definition, any open ball is an open set under this topology.

but would there be any topology in metric space that some ball around some point is not open?

As a set, you can of course define other topology $\tau$ on $X$. An "open ball" of $(X,d)$ is then not necessarily an "open" set in $\tau$.

Title: Metric space is Hausdorff


问:
Graph of Sample Proportion Distribution vs Binomial Distribution
I was reading my textbook and had a question,

$P\left(\hat{p}=\frac{x}{n}\right)=P(X=x)={ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}$, so the various graphs of a sample proportion distribution are just the graphs of the corresponding binomial distribution stretched horizontally by a factor of $\frac{1}{n}$.

I understand why the graph of sample proportion are stretched horizontally by a factor of $\frac{1}{n}$ because you are dividing by $n$.

But to stretch the graph horizontally by a factor of $\frac{1}{n}$ wouldn't you have to replace $x$ by $\frac{x}{\frac{1}{n}}=nx$? This is what I learned as the rule for dilating a function, so I am a bit confused. Thanks..

答:
If $y=\frac xn$ then $P\left(\hat{p}=y\right)={ }^{n} \mathrm{C}_{ny} \,\,p^{ny} q^{n-ny}$ replacing $x$ with $ny$.

Is this what you meant?


答:
That solution is also valid for negative $B$, although it will use complex numbers. To find the solution without complex numbers using that solution, I first made the substitution $\frac{C_2}{2\sqrt{B}} \to C_2$ just for convenience: $$y(x)=\frac{C_1e^{\sqrt{B}x} - C_2e^{-\sqrt{B}x}}{x}$$

Then for negative $B$, this is $$\frac{C_1e^{i\sqrt{-B}x} - C_2e^{-i\sqrt{-B}x}}{x}$$

Using that $e^{ix} = \cos(x)+i\sin(x)$, this becomes $$\frac{C_1\left(\cos(x\sqrt{-B}) + i\sin(x\sqrt{-B})\right) - C_2\left(\cos(x\sqrt{-B}) - i\sin(x\sqrt{-B})\right)}{x}$$

Making the substitution $C_1 - C_2 = c_1$ and $iC_1 + iC_2 = c_2$ makes it $$\frac{c_1\cos(x\sqrt{-B}) + c_2\sin(x\sqrt{-B})}{x}$$

Title: Solution to second order homogeneous ODE

Body:
I have the 2nd order homogeneous ODE:
$$y(x)''+\frac{2}{x}y'(x)-By=0 $$
Where $B$ is a constant. Since I am a bit rusty with analytical solutions, I plugged it in MatLab's symbolic solver which found the following solution.
$$y(x)=C_1\frac{e^{\sqrt{B}x}}{x}-C_2\frac{e^{-\sqrt{B}x}}{2\sqrt{B}x}$$
Unofrtunately, the constant B takes both positive and negative values, but since the ODE is derived from a physical problem I'd like the solution to always be real.
Any suggestion/comment is highly appreciated.


问:
How to calculate a combinatorial sum
I have a combinatorial sum in hand which I suspect equals zero. But I do not know how to prove it. Can you guys help me? (I am even not sure if this is a hard question or not)

Is
$$
\sum_{k=0}^n (-1)^k k^{n-1} \left(\begin{array}{l}
n \\
k
\end{array}\right) = 0
$$
?

答: HINT: Start with $$f(x)=(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k.$$ Differentiate to show that $f'(-1)=0$. Differentiate again to show that $f''(-1)=0$. Keep going (inductively) until the $(n-1)$-th derivative.

Title: How to calculate a combinatorial sum


问:
Find $\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$ without resorting to series

Find $$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$$

Firstly, note that the upper limit makes this an improper integral.

The method I employed against this problem was integration by parts, as shown below:
$$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx=2\int_0^1\text{artanh} x ~dx=2\left[x~\text{artanh}~x+\frac{1}{2}\ln\lvert 1-x^2\rvert\right]_0^1$$
However, I am unsure how to find the value that this converges to.

I then tried to find a series representation of the value of the integral using the Maclaurin expansion of $\text{artanh}~x$, which was successful; I found that the value of the integral is
$$2\sum_{r=0}^\infty\frac{(2r)!}{(2r+2)!}=2\sum_{r=0}^\infty\left(\frac{1}{2r+1}-\frac{1}{2r+2}\right)=2\sum_{r=0}^\infty\frac{(-1)^r}{r+1}$$
which I know is $2\ln2$, which is the answer.
However, I would like to know how to use a regular method of integration such as my first method and still calculate the limit it converges to.

Thank you for your help.

答:
You already found the indefinite integral as $$2x\operatorname{arctanh}\left(x\right)+\ln\left(1-x^{2}\right)$$ Evaluated at $x = 0$, it is clearly $0$. This function can be rewritten as $$\left(1+x\right)\ln\left(1+x\right)+\left(1-x\right)\ln\left(1-x\right),$$ so the answer is $$\lim_{x \to 1}\left( \left(1+x\right)\ln\left(1+x\right)+\left(1-x\right)\ln\left(1-x\right) \right) = 2\ln(2)+\lim_{x \to 1}\left(1-x\right)\ln\left(1-x\right)$$

The last limit is equivalent to $\lim_{x \to 1}\frac{\ln\left(1-x\right)}{\frac{1}{1-x}}$, which by L'Hopital is $\lim_{x \to 1}(x-1) = 0$. Therefore, the final integral is $$2\ln(2) + 0 = 2\ln(2)$$


问:
Argand Diagram - Quadrants help
I forgot the rules of adding angles when it comes to argand diagrams.

In the first quadrant, you add 90 degrees to whatever angle you get, what about Q2 Q3 Q4 ?

This picture will explain what i mean :

notice how 18.43 was added with 180 .. can someone list the rules for the other quadrants as well ?

what if i have -3i -3j for example ?

Thanks

答:
NLed,  Sorry to be so late with the following answer.  (I found your posting because I am looking for software to construct an Argand diagram just like the one in the gray area in the original posting and need help.  I hope you will contact me at onekevATgmailDOTcom.)

The decision to add 180 degrees to the inverse tangent is based on the sign of the denominator "inside" the inverse tangent. Add 180 degrees only if denominator < 0. Also, for this procedure to work, use angles (in degrees) in (-180, 180] instead of the more typical [0, 360).

The angle interval (-180, 180] is more commonly used in the study of functions of a complex variable and electrical engineering than the (perhaps more intuitive) interval [0, 360) used in trigonometry and pre-calculus.

But if you prefer an angle in [0, 360), use the scheme described above then add 360 degrees only if the angle value found is negative.

Title: Argand Diagram - Quadrants help


问:
show that $-0 = 0$ in any vector space
Show that $-\vec{0} = \vec{0}$ in any vector space.

I know this is a seemingly obvious statement but is the following justification correct:

Assume $-\vec{0} \neq \vec{0}$.

$$(4): \vec{u} + \vec{0} = \vec{u}$$
$$ \vec{u} + \vec{0} + (-\vec{0}) = \vec{u} + (-\vec{0})$$
$$(5): \vec{u} -\vec{u} + \vec{0} + (-\vec{0}) = \vec{u} - \vec{u} + (-\vec{0})$$
$$(5): \vec{0} + (-\vec{0}) =  (-\vec{0}) \rightarrow -\vec{0} + \vec{0} =  \vec{0}$$

Hence we clearly see that $-\vec{0} = \vec{0}$ QED Is this reasoning correct?

答:
$$\begin{align}
-\vec{0} &= -\vec{0}  + \vec 0 &(\vec 0 \text{  is additive identity)} \\
&= \vec 0 &\text{(property of additive inverse)}\\
\end{align}$$

Title: show that $-0 = 0$ in any vector space


问:
Proof of the Multinomial Theorem Using Mathematical Induction
I am trying to prove the following proposition:

Suppose that $n \in \mathbb{Z}^{+}$ and that $r_{1}, r_{2}, \dots, r_{n}$ are nonnegative integers such that
\begin{equation}
r_{1} + r_{2} + \dots + r_{n} = r \geq 0.
\end{equation}
Then the following relation holds:
\begin{equation}
\left(z_{1} + z_{2} + \dots + z_{n}\right)^{r} = \sum_{r_{1}+\dots+r_{n} = r}\frac{r!}{r_{1}!r_{2}!\dots r_{n}!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}.
\end{equation}

The case for $n = 1$ is obviously correct. The case for $n = 2$ reduces to nothing but
\begin{equation}
\left(z_{1} + z_{2}\right)^{r} = \sum_{r_{1} = 0}^{r}\binom{r}{r_{1}}{z_{1}}^{r_{1}}{z_{2}}^{r-r_{1}},
\end{equation}
which is the famous binomial formula, and can be accepted without proof. However, I am not sure how to use the induction hypothesis at $n$ to lead to the conclusion at $n+1$. Can anyone provide hints or a complete proof?

P.S. I am looking at this formula for understanding the expansion of the differential of a multivariate function.

I have come up with the proof. The proof goes as follows.

The case for $n = 1$ and $n = 2$ can be easily verified. To use mathematical induction, we assume that the formula holds at an arbitrary $n \geq 2$. That is to say, we are of the next induction hypothesis:
\begin{equation}
\forall s \geq 0, \left(z_{1} + z_{2} + \dots + z_{n}\right)^{s} = \sum_{r_{1}+\dots +r_{n} = s}\frac{s!}{r_{1}!r_{2}!\dots r_{n}!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}.
\end{equation}
Then at $n+1$, we have
\begin{equation}
\begin{aligned}
\left(z_{1} + z_{2} + \dots + z_{n} + z_{n+1}\right)^{r} &= \sum_{s=0}^{r}\frac{r!}{s!\left(r-s\right)!}\left(z_{1} + \dots + z_{n}\right)^{s}{z_{n+1}}^{r-s}\\
&= \sum_{s=0}^{r}\frac{r!}{s!\left(r-s\right)!}\left(\sum_{r_{1}+\dots +r_{n} = s}\frac{s!}{r_{1}!r_{2}!\dots r_{n}!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}\right){z_{n+1}}^{r-s}\\
&= \sum_{s=0}^{r}\sum_{r_{1}+\dots + r_{n} = s}\frac{r!}{s!\left(r-s\right)!}\frac{s!}{r_{1}!r_{2}!\dots r_{n}!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}{z_{n+1}}^{r-s}\\
&= \sum_{s=0}^{r}\sum_{r_{1}+\dots +r_{n}=s}\frac{r!}{r_{1}!r_{2}!\dots r_{n}!\left(r-s\right)!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}{z_{n+1}}^{r-s}\\
&= \sum_{r_{1}+r_{2}+\dots +r_{n}+r_{n+1}=r}\frac{r!}{r_{1}!r_{2}!\dots r_{n}!r_{n+1}!}{z_{1}}^{r_{1}}{z_{2}}^{r_{2}}\dots{z_{n}}^{r_{n}}{z_{n+1}}^{r_{n+1}},
\end{aligned}
\end{equation}
where the induction hypothesis is used for the second equality, and a change of variable with $r_{n+1} = r - s$ is used for the last equality. As a result, we may conclude that the formula holds for all $n \in \mathbb{N}$.

答:
This is called the "multinomial formula".  Look it up ... for example HERE

As a proof by induction, use something like
$$
(x_1+x_2+\dots+x_n+x_{n+1})^r = (y+x_{n+1})^r
$$
say, with $y=x_1+x_2+\dots+x_n$.
Use the binomial theorem for that, then use your induction  hypothesis on all the powers of $y$.


答:
By definition (axiom of the inverse element), $-\vec 0$ is a vector such that
$$
\vec 0+(-\vec 0)=\vec 0\tag{1}
$$
But $\vec 0$ is the additive unit of the vector space (axiom of the identity element of addition) and thus
$$
\vec 0+(-\vec 0)=-\vec 0\tag{2}
$$
Combining (1) and (2), we are done.

Title: show that $-0 = 0$ in any vector space

Body:
Show that $-\vec{0} = \vec{0}$ in any vector space.

I know this is a seemingly obvious statement but is the following justification correct:

Assume $-\vec{0} \neq \vec{0}$.

$$(4): \vec{u} + \vec{0} = \vec{u}$$
$$ \vec{u} + \vec{0} + (-\vec{0}) = \vec{u} + (-\vec{0})$$
$$(5): \vec{u} -\vec{u} + \vec{0} + (-\vec{0}) = \vec{u} - \vec{u} + (-\vec{0})$$
$$(5): \vec{0} + (-\vec{0}) =  (-\vec{0}) \rightarrow -\vec{0} + \vec{0} =  \vec{0}$$

Hence we clearly see that $-\vec{0} = \vec{0}$ QED Is this reasoning correct?


问:
Solve Surface Area Formula For Radius
Ok.  I have a Surface Area Formula for a cylinder, but I am wanting to solve for the radius.  The only information I have is the height of the cylinder, which is 8 inches.  I know I can use the Quadratic Formula to convert the formula to solve for the radius, but I get stuck when doing the math.

$$ SA=2\pi r^{2}+2\pi rh$$

Any help?

答:
If you have

$$A=2\pi r^{2}+2\pi rh$$

then rearrange it:

$$2\pi r^{2}+2\pi hr - A = 0$$

This is quadratic in $r$ with solutions:

$$r = \frac{-2\pi h \pm \sqrt{4\pi^2h^2 + 8\pi A}}{4\pi}.$$

Since the expression under the radical is greater than $2\pi h$ you choose the "plus" root so that the radius is positive:

$$r = \frac{-2\pi h + \sqrt{4\pi^2h^2 + 8\pi A}}{4\pi}.$$

But to solve for the numerical value of $r$ given $h=8$ inches, you'll need more information. You'll need the surface area also to get the value of $r$.

Title: Solve Surface Area Formula For Radius


答:
I'd just complete the square:

$$ A = 2\pi r^2 + 2\pi rh $$

$$ \frac{A}{2\pi} = r^2 + rh $$

To complete the square on the right-hand side, we need to add $\frac{h^2}{4} $ to both sides:

$$ \frac{A}{2\pi} + \frac{h^2}{4} = r^2 + rh + \frac{h^2}{4} $$

$$ \frac{A}{2\pi} + \frac{h^2}{4} = \left(r + \frac{h}{2} \right)^2 $$

Since $r > 0$ and $h > 0$, we take the positive root:

$$ r + \frac{h}{2} = \sqrt{\frac{A}{2\pi} + \frac{h^2}{4}} $$

$$ r = -\frac{h}{2} +\sqrt{\frac{A}{2\pi} + \frac{h^2}{4}} $$

Title: Solve Surface Area Formula For Radius

Body:
Ok.  I have a Surface Area Formula for a cylinder, but I am wanting to solve for the radius.  The only information I have is the height of the cylinder, which is 8 inches.  I know I can use the Quadratic Formula to convert the formula to solve for the radius, but I get stuck when doing the math.

$$ SA=2\pi r^{2}+2\pi rh$$

Any help?


问:
Pursuing (pure) math while going to medical school and being a doctor?
I LOVE pure math, it sparks joy and excitement for me. At the same time, I want to be a trauma surgeon for a variety of reasons (it interests me, has more job prospects, academia doesn't seem like a great environment for me, I don't love teaching, it pays better). Would it be possible for me to research pure math without getting: 1) A job in academia. 2) A PhD? I would at the very least major in math in college (probably a double major with biology). I am really at a crossroads, and I am hoping it would be possible to at least in some capacity do research-level math so that I don't have to abandon one for the other.</p>

答:
(In my personal Opinion)

( This don´t pretend presure you to take a desition I only pretend tell to you my little experience)

I don´t like broke dreams and in my special case ago a time I had your same situation, but in my case the papers be switch I´m studying Pure Maths in the university and want be a surgeon, but I discover that simply I can´t do both carriers.

My advice here is think and decide which of the two thinks your love more and when you select one think if is possibly apply the other, in my case for example I take Math and since I like medicine I decide make Apply maths and do all my researches in Pharmacology .

If you can apply one over the other nice, but in otherwase you should take one of them and
think of what form your apply your love to your carrier.


答:
There are great opportunities for MDs with deep interests and abilities in math.  Integrating trauma surgeon with math seems awkward and unrealistic, but if you're a bit more flexible, I'm pretty sure you can find an intellectual home where you'd be happy.

A few examples:

Sophisticated mathematical analysis in epidemiology

Mathematical and computational modeling of neural functions... everything from perception to memory

Mathematical genetics and epi-genetics

There are opportunities where you don't have to teach.  Think of NIH, Howard Hughes, and several labs such as Salk Institute.

The best work is always by researchers who understand deeply the problems.  As an MD you'll have a great advantage over the "mere" modelers, despite the fact that they might be a bit more mathematically savvy.

An important problem solved well is far more valuable than an unimportant (or misguided problem) solved to perfection.

Title: Pursuing (pure) math while going to medical school and being a doctor?

Body: I LOVE pure math, it sparks joy and excitement for me. At the same time, I want to be a trauma surgeon for a variety of reasons (it interests me, has more job prospects, academia doesn't seem like a great environment for me, I don't love teaching, it pays better). Would it be possible for me to research pure math without getting: 1) A job in academia. 2) A PhD? I would at the very least major in math in college (probably a double major with biology). I am really at a crossroads, and I am hoping it would be possible to at least in some capacity do research-level math so that I don't have to abandon one for the other.</p>


问:
Proof of $(ab)^{-1} = a^{-1}b^{-1}$
I wonder if you would consider this proof of my hypothesis of

$(ab)^{-1} = a^{-1}b^{-1}$ with $a,b\ne0$ correct.

By definition of the inverse of $ab$: $$(ab)^{-1}(ab)=1$$

Multiplying the equation with $a^{-1}$ and $b^{-1}$ gives $$(ab)^{-1}aba^{-1}b^{-1} = 1a^{-1}b^{-1}=a^{-1}b^{-1}$$ using property $1x=x$.
On the left hand side I rearrange the terms using the property $xy=yx$ to $$(ab)^{-1}aa^{-1}bb^{-1}= a^{-1}b^{-1}$$ which gives by using again the definition of the inverse $$(ab)^{-1}1\cdot1= a^{-1}b^{-1}$$
Using the property $1x=x$ on the left twice I have the answer.

Title: Proof of $(ab)^{-1} = a^{-1}b^{-1}$

Body: $$(ab)^{-1}(ab)=1$$ $$ (ab)^{-1}abb^{-1}=b^{-1}$$ $$(ab)^{-1}a=b^{-1}$$  $$(ab)^{-1}aa^{-1}=b^{-1}a^{-1}$$ $$(ab)^{-1}=b^{-1}a^{-1}.$$


问:
Median from Probability distribution formula
Can any one tell, how i can get formula for median of geometric distribution with this probability mass function formula
$P_{r}=\frac{ (1-\alpha) \alpha^{CW}} { 1- \alpha^{CW} }. \alpha^{-r}$

答:
Utilize the following:

Find $m$ such that

(1) $\displaystyle P(X \le m) \ge \frac{1}{2}$

(2) $\displaystyle P(X \ge m) \ge \frac{1}{2}$

The median will be any such $m$. This can be calculated using the PMF.

Edit

According to Truncated Distributions

The median of a truncated distribution will be

$\displaystyle F^{-1}\bigg( \frac{F(a) + F(b)}{2} \bigg )$


问:
calculate $\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$
How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get

$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?

答: In the original expression, use the half angle equation in the denominator, and the double angle equation in the numerator. Then $ cos(\pi t/2) $ cancels, and the original expression equals minus $\sqrt 2 sin(\pi t/2) $</p>


问:
If $S$ is dense in $X$ and $S\subset T$, then $T$ is dense in $X$
Problem is from Topology without Tears by Morris

If $S$ is dense in $X$ and $S\subset T$, then  $T$ is dense in $X$.

Try

Assume $S$ is dense in $X$. Suppose $S\cap X=\varnothing$. Then every element of $X\setminus S$  is a limit point of $S$. By hypothesis  $S\subset T$  so every element of $X\setminus T$  is a limit point of $T$.

Thus $T$ is dense in $X$.

I am not too good with this topic. I basically tweaked a theorem and hope it works

Help

答: This should be pretty easy, since closure respects containment, so to speak.  That is, $S\subset T\implies\bar S\subset\bar T$.  That is by the definition of closure, say as the smallest closed set containing the given set.


问:
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$ , $\forall x\in G$ . Prove that $H$ is a normal subgroup of $G$
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.

I have tried to using the definition but failed. Can someone help me please.

答:
As I began to correct my former post: Hints

$$\begin{align*}\bullet&\;\;\;G^2:=\langle x^2\;;\;x\in G\rangle\lhd G\\
\bullet&\;\;\;G^2\le H\\
\bullet&\;\;\;\text{The group}\;\;G/G^2\;\;\text{is abelian and thus}\;\;G'\le G^2\end{align*}$$


问:
Integral Along Y-Axis
I wish to integrate the following along the Y-axis:$ x = \frac{1}{0.088(log (y)}dy$ from $y= 0$ to $y = 0.42$. I get $y \log y - y$ evaluated between these two limits.  Since $\log y$ can't be $04$, I was wondering if this equation is not capable of being integrated in "closed form" because at $y = 0 $I will have $0 \cdot \log 0$. And $\lim \log y$ as $y$ approaches $0$ is, I think, + infinity. If integral can't be done in closed form, is there another way to do it? Not an expert in math, but wish to learn more. Any help is greatly appreciated. Fred

答: $ylogy\to 0$ as $y \to 0$  Use L'Hopital's rule on$\frac{logy}{1/y}$ get $\frac{1/y}{-1/y^2}=-y\to 0$.

Title: Integral Along Y-Axis


问:
Analytic Proof of Wilson's Theorem
I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:

We start by considering the Maclaurin series of the function $f(x)=\ln \left( \frac{1}{1-x} \right)$, that is,
$\displaystyle \ln \left( \frac{1}{1-x} \right)= \sum_{n=1}^{\infty} \frac{x^n}{n} \tag*{}$
for $x \in [-1, 1).$
By simply using the definition of $\ln$ on the equation above, we get
$\displaystyle e^{x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots}=\frac{1}{1-x}=1+x+x^{2}+ \cdots + x^{p} + \cdots \tag{1}$
We can then rewrite the leftmost side of $(1)$ as
$\displaystyle \begin{align} e^{x+\frac{x^2}{2}+\frac{x^3}{3}+…} & = e^{x}e^{\frac{x^2}{2}}e^{\frac{x^3}{3}} \cdots \\ & = \prod_{n=1}^{\infty} e^{\frac{x^n}{n}} \\ & = \prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\left( \frac{x^n}{n} \right)^k}{k!} \\ & = \left( 1+\frac{x}{1!}+\frac{x^2}{2!}+ \cdots \right) \cdot \left( 1+\frac{x^{2}/2}{1!}+\frac{(x^{2}/2)^2}{2!} + \cdots \right) \cdots \left(1+\frac{x^{p}/p}{1!}+\frac{(x^{p}/p)^2}{2!}+ \cdots \right) \cdots \\ & = 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots \end{align} \tag{2}$
Combining the rightmost sides of $(1)$ and $(2)$ together gives us
$\displaystyle 1+x+x^{2}+ x^{3}+ \cdots + x^{p} + \cdots= 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots , \tag{3}$
from which follows that the coefficient of $x^{p}$ on the RHS of $(3)$ is $1$. That is,
$\displaystyle \frac{1}{p!}+\frac{1}{(p-2)!} \frac{1/2}{1!}+ \cdots + \frac{1/p}{1!} = 1. \tag{4}$
Notice that such a coefficient is of the form $\frac{1}{p!}+\frac{r}{s}+\frac{1}{p}$, where $\frac{r}{s}$ is the sum of a finite number of rationals, none of which has the factor $p$ in its denominator. Therefore, if $\frac{r}{s}$ is irreducible, then $p$ doesn’t divide $s$. We then proceed by rewriting $(4)$ as
$\displaystyle \frac{1}{p!}+\frac{r}{s}+\frac{1}{p}=1. \tag*{}$
Equivalently,
$\displaystyle 1-\frac{r}{s}=\frac{1}{p!}+\frac{1}{p}=\frac{1+(p-1)!}{p!}, \tag*{}$
and
$\displaystyle (s-r) \cdot (p-1)! = \frac{s(1+(p-1)!)}{p}. \tag*{}$
Hence, $\frac{s(1+(p-1)!)}{p}$ is an integer. Moreover, since $p$ doesn’t divide $s$, $\frac{1+(p-1)!}{p}$ is also an integer. Equivalently, $(p-1)! \equiv -1 \text{ (mod } p)$, as desired.

答:
I don't have a reference but I would not call this an "analytic" proof; although you are using power series every step of the argument works fine in formal power series, so really you are doing combinatorics with generating functions. The combinatorial content of this argument can be extracted as follows.

The identity $\exp \log \frac{1}{1 - x} = \frac{1}{1 - x}$ expresses the uniqueness of cycle decompositions of permutations; it generalizes to an identity called the permutation form of the exponential formula which expresses the same uniqueness but in more detail. Comparing the coefficient of $x^n$ (by the way, I don't think your version of this expansion has enough terms - there should be one for every partition of $n$) expresses the uniqueness of cycle decompositions of permutations in the symmetric group $S_n$, and can be interpreted as its class equation

$$n! = \sum_{[\pi]} \frac{n!}{|C_{S_n}(\pi)|}$$

which expresses $n!$ as the sum of sizes of all conjugacy classes in $S_n$. In more detail, the correct form of this expansion is

$$n! = \sum_{\sum ic_i = n} \frac{n!}{\prod_{i=1}^p i^{c_i} c_i!}.$$

where the term indexed by $\{ c_i \}$ in the sum corresponds to the conjugacy class of permutations with $c_i$ cycles of length $i$. The denominators here are the sizes of the centralizers of each conjugacy class. In terms of the generating function, the factor of $i^{c_i} c_i!$ in the denominator is contributed by the expansion of $\exp \left( \frac{x^i}{i} \right)$. Note that once you've proven this identity in any way, for example by calculating the sizes of the centralizers of every conjugacy class of permutations, you can throw away the generating function (which is equivalent to it).

Now we specialize $n$ to a prime $p$ and take the class equation $\bmod p$. From the above equation it's not hard to see that the denominator can only be divisible by $p$ if either $c_1 = p$ or $c_p = 1$, corresponding to the conjugacy classes of the identity and the $p$-cycles. This can be proven conceptually as follows. $S_p$ has the property that its only elements of order $p$ are the $p$-cycles, which are all conjugate and which generate self-centralizing subgroups: that is, the centralizer of a $p$-cycle is just the cyclic subgroup it generates. It follows that the only conjugacy classes in $S_p$ whose centralizer has order divisible by $p$ (equivalently, whose centralizer contains a $p$-cycle) are the identity and the $p$-cycles. Taking the class equation $\bmod p$, every term except the terms corresponding to the identity and the $p$-cycles vanishes, and we conclude that the number of $p$-cycles plus $1$ is congruent to $0 \bmod p$. But since we calculated the centralizer already, it follows that there are $(p-1)!$ $p$-cycles, so we conclude that

$$(p-1)! + 1 \equiv 0 \bmod p$$

as desired.

So, stripped of the power series trappings, this is a straightforward counting argument in $S_p$. It is related to counting arguments which can be used to prove the Sylow theorems, especially Sylow III (although one of the proofs of Sylow I also uses the class equation). Sylow III applied to $S_p$ actually gives $(p-2)! \equiv 1 \bmod p$ (this is the number of conjugacy classes of cyclic subgroups of order $p$) which is straightforwardly equivalent to Wilson's theorem.

There is also a simpler version of the counting argument which directly uses the $p$-group fixed point theorem, which is just the easy statement that if $P$ is a finite $p$-group acting on a finite set $X$ then

$$|X^P| \equiv |X| \bmod p.$$

We apply the PGFPT to the action of a $p$-cycle on the set of $p$-cycles in $S_p$. As above, there are $(p-1)!$ of these, and the centralizer of any $p$-cycle is the cyclic subgroup it generates. It follows that this action has exactly $p-1$ fixed points, which immediately gives

$$(p-1)! \equiv (p-1) \bmod p.$$

A similarly quick and easy counting proof can be given for Fermat's little theorem and a few other fun facts; see the linked blog post for more.

Just for fun, here is a formal power series proof that $\exp \log \frac{1}{1 - x} = \frac{1}{1 - x}$, which requires no "actual" analysis. As above, this identity is equivalent to a combinatorial statement about permutations and can be proven that way, but it can also be proven as follows. Namely, it's straightforward to show by induction that $\exp(x)$ is the unique formal power series $f(x)$ satisfying $f(0) = 1, f'(x) = f(x)$ (just differentiate to get that every coefficient of the Taylor series is $1$). On the other hand, $\log \frac{1}{1 - x}$ is the unique formal power series $g(x)$ satisfying $g(0) = 0, g'(x) = \frac{1}{1 - x}$ (just compare terms on both sides). (Both of these facts can be given combinatorial interpretations.)

It follows that $f(g(x)) = h(x)$ satisfies $h(0) = 1$ and

$$\frac{d}{dx} \exp \log \frac{1}{1 - x} = \left( \exp \log \frac{1}{1 - x} \right) \frac{1}{1 - x}$$

and there is a unique formal power series satisfying $h(0) = 1$ and $h'(x) = \frac{h(x)}{1 - x}$ (by induction; this follows from a formal version of the Picard-Lindelof theorem, which is much easier to prove), namely $h(x) = \frac{1}{1 - x}$.

Title: Analytic Proof of Wilson's Theorem


问:
What is the probability that 4 employees complete their task in a total of 10 hours or more.
A company expects its employees to finish a task at an average of 12 hours. What is the probability that 4 employees complete their tasks in a total of 10 hours or more?

Would this be a Poisson, binomial, or hypergeometric? It was posted by the instructor after learning about the aforementioned probability distributions

I am leaning toward poisson, solved in this manner:

$$ \lambda = 4 \text{ employees} * 12 \text{ average hours} = 48 \text{ hours} , $$
and the solution be $P(X > 10)$?

Thanks

答:
It is important you understand that, in most practical situations, a Poisson random variable counts the number of times a particular event occurs in a specific interval of time or a specific region of space. These events must be independent, two events cannot occur simultaneously, and the average number of events that occur in a specific interval of time or a specific region of space is approximately constant. In your scenario, an "event" corresponds to an employee completing his or her work task.

That being said, if $X$ denotes the number of tasks a single employee finishes in a one hour period, then $X \sim \text{Poisson}(1/12)$ assuming aforementioned conditions are met.

If $T$ represents the time it takes for an employee to complete his or her first task, you can show that $T\sim \text{Exponential}(1/12)$. The argument goes like this. For $t> 0$ we have $$P(T\leq t)=1-P(T>t)=1-P\Big(\text{No tasks completed in } [0,t]\Big)$$

The number of tasks an employee finishes in $[0,t]$ is $\text{Poisson}(t/12)$ so $$P\Big(\text{No tasks completed in } [0,t]\Big)=e^{-t/12}$$ Hence $P(T \leq t)=1-e^{-t/12}$ which shows $T\sim \text{Exponential}(1/12)$.

Now assume that the employees complete their tasks independently of each other. Let $T_1,T_2,T_3,T_4$ denote the times it takes for the employees to finish their first tasks. You're looking to compute the probability that $\max\{T_1,T_2,T_3,T_4\}\geq 10$. We get $$P\Big(\max\{T_1,T_2,T_3,T_4\}\geq 10\Big)=1-P\Big(\max\{T_1,T_2,T_3,T_4\}< 10\Big)=1-\Big(1-e^{-5/6}\Big)^4$$


问:
epsilon delta question limit questions
I have two quick questions. In calculus books when proving limits, I always see something like, let $\delta = \min(\delta_{1}, \delta_{2})$. Does this mean that we take a $\delta$ such that it contains both $\delta_{1}$ and $\delta_{2}$?</p>

答: You take the smaller of $\delta$s. Usually one of them is a "safety", protecting against too large an $\epsilon$ and the other depends explicitly on $\epsilon$</p>

Title: epsilon delta question limit questions


答: For a combinatorial proof, note that both sides count the number of surjections of an $(n-1)$-set onto an $n$-set. The RHS is obvious, and the LHS is the inclusion-exclusion formula, where the $n$ properties to be avoided are that $i$ is not in the image of the function. More generally, this proof shows that the identity holds if you replace the power $n-1$ with any number $<n$.

Title: How to calculate a combinatorial sum

Body:
I have a combinatorial sum in hand which I suspect equals zero. But I do not know how to prove it. Can you guys help me? (I am even not sure if this is a hard question or not)

Is
$$
\sum_{k=0}^n (-1)^k k^{n-1} \left(\begin{array}{l}
n \\
k
\end{array}\right) = 0
$$
?


问:
Selecting points of intersection given a point and equations of lines

I have an $x$-axis and a $y$-axis both beginning from the origin. I also have $n$ straight lines with known equation ($n$=4 in the diagram). If all intercepts of the lines to the axes are positive, then at most I have ${n \choose 2} + 2n$ points of intersection.

The question is, of these points of intersection, how do I determine the ones that form the boundary of the region that contains a point $M$ with known coordinates? In the diagram they are coloured red. I am interested in knowing how these points can be selected from all points.

答:
Two hints to help solve this problem

The lines divide the plane into mutually non-intersecting regions, and the point $M$ must lie in exactly one of these regions

Each of these regions has a unique representation through the vertices of that region

Hence, if we can find the region, and figure out how to map the region to it's vertices, we can essentially find the vertices you need for any point $M$

Further, if you substitute he point $M$ into the equation of each line, you will get a set of values - some positive and some negative. Each region is uniquely determined by the signs of these numbers


问:
In a Locally Compact Hausdorff and second countable space every compact subset is a $G_\delta$ set
Let $X$ be a second countable LCH (Locally Compact Hausdorff) then every compact subset of $X$ is a $G_{\delta}$ set.

A set $E$ is $G_{\delta}$ if is a countable intersection of open sets.

I have two questions,

How to prove this result,

I am also wondering if that would be true for the closed sets in this space, or what would be necessary to be true for closed sets.

Edit:

I posted a resolution, which I found very interesting, of question 1.

答:
This is not an answer but recommendation for you to see the following theorem
Theorem In a locally compact Hausdorff space, the intersection of an open set with a closed set is locally compact. Conversely, a locally compact subset  of a Hausdorff space is the intersection of an open set and a closed set.

You can find the proof Willard's book page 130.


问:
Is there a general formula for the eccentricity $e$ of ellipsoid?
For an ellipse of eccentricity $e$ the formulas are:

${x^2 \over a^2} + {y^2 \over b^2} = 1 \\ e = \sqrt {1-\left({b \over a} \right)^2}$

what about the "3D case"?

答:
In the aspect of measure of the boundary

Perimeter of an ellipse

\begin{align}
  1 &= \frac{x^2}{a^2}+\frac{y^2}{b^2} \tag{$a \ge b$} \\
  P &= 4aE(e_{ab}) \\
  e_{ab} &= \frac{\sqrt{a^2-b^2}}{a} \\
\end{align}

Surface area of an ellipsoid

\begin{align}
  1 &= \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \tag{$a \ge b \ge c$} \\
  S &= 2\pi
  \left[
    c^2+\frac{bc^2}{\sqrt{a^2-c^2}}F(\theta,k)+b\sqrt{a^2-c^2} E(\theta,k)
  \right] \\
  \theta &= \cos^{-1} \frac{c}{a} \\
  k &= \frac{a}{b} \sqrt{\frac{b^2-c^2}{a^2-c^2}} \\
  &= \frac{e_{bc}}{e_{ac}} \\
  k' &= \sqrt{1-k^2} \\
  &= \frac{c}{b} \sqrt{\frac{a^2-b^2}{a^2-c^2}} \\
  &= \frac{e_{ab}}{e_{ac}}
\end{align}

For prolate spheroid

$$b=c \implies (k,k')=(0,1)$$

For oblate spheroid

$$a=b \implies (k,k')=(1,0)$$

$k$ or $k'$ is not quite well-defined for the case of sphere.

In the aspect of confocal system

Confocal conics

\begin{align}
  1 &= \frac{x^2}{a^2+s}+\frac{y^2}{b^2+s} \\
  e^2 &= \frac{a^2-b^2}{a^2+s}
\end{align}

Confocal quadrics

\begin{align}
  1 &= \frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}+\frac{z^2}{c^2+s} \\
  \kappa^2 &= \frac{a^2-c^2}{a^2+s}
\end{align}

For $s\to \infty$, $\kappa \to 0$ which close to a sphere.

For $s\to -c$, $\kappa \to 1$ which shrinks to focal ellipse namely

$$0=z=\frac{x^2}{a^2-c^2}+\frac{y^2}{b^2-c^2}-1$$

This can not analogous to paraboloids.

See also another posts of mine here and here.


问:
Weak tangent but not a strong tangent
Question: Show that $\alpha(t)=(t^3,t^2)$, $t\in \Bbb R$, has a weak tangent but not a strong tangent at $t=0$.

Definitions from this answer:

(Weak tangent) $\alpha: I \to \Bbb R^3$ has a weak tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0)$ has a limit position when $h \to 0$.

(Strong tangent) $\alpha: I \to \Bbb R^3$ has a strong tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0 + k)$ has a limit position when $h \to 0$ and $k \to 0$.

My query:

I'm not really clear what argument to use to demonstrate this. The weak tangent is the line joining $\alpha(t_0)$ and $\alpha(t_0+h)$, which is
$$
(\lambda(x(t_0+h)-x(t_0))+x(t_0), \lambda(y(t_0+h)-y(t_0))+y(t_0))
$$
If $t_0=0$ then $x(t_0)=0$ so this becomes
$$
(\lambda h^3, \lambda h^2)
$$
The strong tangent is
$$
(\lambda(x(t_0+h)-x(t_0+k))+x(t_0+k), \lambda(y(t_0+h)-y(t_0+k))+y(t_0+k))
$$
$$
=(\lambda (h^3-k^3)+k^3,\lambda (h^2-k^2)+k^2)
$$
As $h,k\rightarrow0$ this seems badly defined. But how can I make this argument precise?

Also, what is the intuitive meaning of the strong and weak tangents?

[This is exercise 1-3-7 of Differential Geometry of Curves and Surfaces by Do Carmo.]

答: Maybe one comment is worth here. Observe that the curve is $y = x^{2/3}$ and $dy/dx = (2/3)x^{-1/3}$ goes to $\infty$  when $x \rightarrow 0$, so the derivative does not exist at $x=0$ and $\alpha(t)$ is not regular at $t=0$. The same result should be obtained by the definition of weak and strong tangent if they exist. As mentioned before, the slope of the line determined by $\alpha(t_0+h)$ and $\alpha(t_0)$ goes to $\infty$ when $h\rightarrow 0$. So the weak tangent exists. But this does not happen for any $h$ and $k$ in the case of the strong tangent (the slope is $0$ if you choose $h=-k$).

Title: Weak tangent but not a strong tangent


问:
Question About The Intersection Of 3 Equations
I have a question about the following simultaneous equation problem:

I was able to reduce this system to: $y + 2z = 1$ and $(k^2 - 4)z = k - 2$. However, the case that is confusing me is when $k = 2$. If $k = 2$, there will be an infinte amount of solutions, as an infinte amount of pairs satisfy $y + 2z = 1$ and $0 = 0$.

So this means that for every x-value, there is an infinite amount of pairs that satisfy all 3 equations (the pairs being all that satisfy: $y + 2z = 1$).

My question is, would this mean that the intersection between these 3 equations would be a plane, because for each x-value, the equations are satisfied by $y + 2z = 1$? And if so, how is that possible? I can't visualize how the intersection between 3 non-identical planes can be a plane?

And if not, why wouldn't the above mean that the intersection between these 3 equations would be a plane? Because for every x-value, we have an infinite amount of x and y values that satisfy the syste.

Thanks in advance.

答:
What happens to the third equation when you put $k=2$? It basically becomes the first equation, hence you only have two unique equations on three variables $(x,y,z)$

The intersection of these two planes is the line $2y+z = 1$, think of it like the hinge between the two planes which are described by the plane equations

If you had a third equation which was not linearly dependent, the intersection of the plane and the above line could either be a single point or no solutions(if parallel).


问:
If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism?
$X$ and $Y$ denote Hilbert spaces. If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism? Homeomorphism means continuous map with continuous inverse.

I think the answer is yes, the only thing I am unable to show is what is the inverse of $T^*$ and if it is continuous?

答:
Let $S$ be inverse to $T$. Then $ST=1_X$ and $TS=1_Y$. Apply to this equalities $^*$ functor to get $T^*S^*=1_{X^*}$ and $S^*T^*=1_{Y^*}$. This means that $T^*$ is invertible and what is more $(T^*)^{-1}=S^*$

In this proof I implicitly assumed that $T^*\in \mathcal{B}(Y^*,X^*)$ whenever $T\in\mathcal{B}(X,Y)$. This is indeed true. For the begining recall one of the corollaries of Hahn-Banach theorem
$$
\Vert g\Vert=\sup\{|g(y)|:y\in\operatorname{Ball}_{Y}\}
$$
where $g\in Y^*$. Then
$$
\begin{align}
\Vert T^*\Vert
&=\sup\{\Vert T^*(g)\Vert: g\in\operatorname{Ball}_{X^*}\}\\
&=\sup\{\Vert T^*(g)(x)\Vert: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\
&=\sup\{| g(T(x))|: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\
&=\sup\{\Vert T(x)\Vert: x\in\operatorname{Ball}_{X}\}=\\
&=\Vert T\Vert
\end{align}
$$


问:
Find the bijection between Kummer's field and Galois subgroup

Let $\mathbb{k}$ is field and $\exists \xi \in \mathbb{k}^*,\ O(\xi) = n$   I need find the bijection between k-isomorphic finite separable
  normal extension $\mathbb{k}$ (i.e. Kummer's field with an exponent n) with abelian
  Galois group with exponent n and finite subgroup
  $\mathbb{k}^*/\mathbb{k}^{*n}$

So, I've seen this theorem (Robert B. Ash, "Abstract Algebra: The Basic Graduate Year", theorem 6.7.5, chapter 6, p.21):

Let $E/F$ be a finite extension, and assume that F contains a
  primitive n-th root of unity $\xi$. Then E/F is a Kummer extension
  whose Galois group G has an exponent dividing n iff there are
  nonzero elements $a_1, ..., a_r \in F$ such that E is a splitting
  field of $(X^n - a_1)\cdot \cdot \cdot (X^n - a_r)$ over F.

I think it's very similar to my problem. If it is, may I use this theorem for my problem? I just have no other ideas how I may prove my problem.

答:
Notations:

Let k be a field containing m-th roots of unity, where m is coprime to the characteristic of k (if not $0$). Denote by $a^{1/m}$ any m-th root of unity, and the group of m-th roots of unity by $\mu_m$. Let $B$ be a subgroup of $k^*$ containing $(k^*)^m$. Denote by $k(B^{1/m})$ the compositum of all $k(a^{1/m})$ where $a\in B$.

Theorem:

Let $K_B=k(B^{1/m})$, then $K_B$ is abelian of exponent $m$ over $k$. The map $\text{Gal}(K_B/k)\times B\rightarrow \mu_m$, defined as $(\sigma, a)\mapsto \sigma\alpha/\alpha$ where $\alpha^m=a$, is a bi-group homomorphism with trivial left kernel and right kernel $(k^*)^m.$ $K_B/k$ is finite iff $B/(k^*)^m$ is finite and $B/(k^*)^m\cong \text{Hom}(\text{Gal}(K_B/k),\mu_m), [K_B:k]=(B:(k^*)^m)$

Proof.  By definition, $K_B$ contains all roots of $x^m-a$ where $a\in B$, which has no multiple root. Hence $K_B$ is the splitting field of a family of separable polynomials and is thus Galois.

For any $\sigma \in\text{Gal}(K_B/k)$ and $\alpha$ an m-th root of some $a\in B, \sigma\alpha=\alpha\mu_{\sigma}$ where $\mu_{\sigma}$ is an m-th root of unity determined by $\sigma$. Therefore, $\text{Gal}(K_B/k)$ is an abelian group and thus $K_B/k$ is abelian. In addition we have $\sigma^m\alpha=\mu_{\sigma}^m\alpha=\alpha$ for each $\alpha\in K_B$, so $K_B/k$ is of exponent $m$.

The map defined above is well-defined because m-th roots of unity are contained in $k$ and thus are fixed by any element of $\text{Gal}(K_B/k)$. It is a homomorphism in $\sigma$ since for $\sigma,\tau\in \text{Gal}(K_B/k),(\sigma\circ\tau,a)\mapsto\sigma(\tau\alpha)/\alpha=\sigma(\alpha\mu_{\tau})/\alpha=\mu_{\sigma}\mu_{\tau}$ and $(\sigma,a)\mapsto\mu_{\sigma}$, similarly for $\tau$. It is a homomorphism in $a$ since $\sigma$ is a homomorphism.

If $\sigma\alpha/\alpha=1$ for any $\alpha^m=a\in B, \sigma$ fixes all elements of $K_B$ and is thus the identity homomorphism. On the other hand, if $\sigma\alpha/\alpha=1$ for any $\sigma\in \text{Gal}(K_B/k)$, then $\alpha$ must lie in $k^*$, so its m-th power $a$ lies in $(k^*)^m$.

The above bi-group homomorphism induces an injective homomorphism from $\text{Gal}(K_B/k)$ to $\text{Hom}(B/(k^*)^m,\mu_m)$, and similarly an injective homomorphism from $B/(k^*)^m$ to $\text{Hom}(\text{Gal}(K_B/k),\mu_m)$. Therefore $K_B/k$ is finite iff $B/(k^*)^m$ is. Since $\text{Gal}(K_B/k)$ is finite abelian, it is isomorphic to its dual $\text{Hom}(\text{Gal}(K_B/k),\mu_m)$. Suppose $k^*$ has exponent $r$ strictly less than its order as a group, then any element in $k^*$ is a root of the polynomial $x^r-1$, so the number of roots in $k^*$ alone exceeds the degree of the polynomial, contradiction. Hence $r=|k^*|$ and thus $k^*$ is cyclic and a fortiori abelian, and so is $B$ since it's a subgroup of $k^*$. Then $|\text{Hom}(\text{Gal}(K_B/k),\mu_m)|=[K_B:k]\ge (B:(k^*)^m)$, and the other injection gives the reverse inequality. Hence the injection is in fact an isomorphism, that is, $\text{Hom}(\text{Gal}(K_B/k)\cong B/(k^*)^m$, and in particular, $[K_B:k]= (B:(k^*)^m)$.

Theorem:

The map $B\mapsto K_B=k(B^{1/m})$ defines a bijection between the set of subgroups of $k^*$ containing $(k^*)^m$ and the set of abelian extensions of $k$ of exponent $m$.

Proof.  Let $B_1, B_2$ be two subgroups of $k^*$ containing $(k^*)^m.$ If $B_1\subset B_2$, clearly $K_{B_1}\subset K_{B_2}$. Conversely, suppose $K_{B_1}\subset K_{B_2}$.  Given any element $b\in B_1, k(b^{1/m})$ is contained in a finitely generated subextension of $K_{B_2}$, so we may assume without loss of generality that let $K_{B_2}$ is finitely generated, hence finite. Now, let $B_b$ be the smallest subgroup of $k^*$ generated by $B_2$ and $b$. Then $K_{B_b}=K_{B_2}$ and hence by the previous theorem, $(B_2:(k^*)^m)=[K_{B_2}:k]=[K_{B_b}:k]=(B_b:(k^*)^m)$. Hence $B_2=B_b$ and in particular $b\in B_2$.  Since we may pick $b$ to be any element in $B_1$, this implies that $B_1\subset B_2$.

Given any abelian extension $K/k$ of exponent $m$ and any $b\in K, k(b^{1/m})$ is contained in a finitely generated subextension of $K$, which we denote by $K_b$. Then $\text{Gal}(K_b/k)$ is finite abelian and thus by the fundamental theorem of finite abelian groups, $\text{Gal}(K_b/k)\cong \Bbb Z/d_1\Bbb Z\times\cdots\times \Bbb Z/d_r\Bbb Z$ where $d_i\mid d_{i+1}$. Let $K_i$ be the fixed field of the group obtained from replacing the i-th position in the direct product by $\{1\}$, then by the fundamental theorem of Galois theory, $K_i/k$ is Galois because its corresponding automorphism group is a normal subgroup of $\text{Gal}(K_b/k)$. Then the restriction of $\text{Gal}(K_b/k)$ on $\text{Gal}(K_i/k)$, which is a surjective homomorphism with kernel $\text{Gal}(K_b/K_i)$, induces an isomorphism $\text{Gal}(K_i/k)\cong(\text{Gal}(K_b/k))/(\text{Gal}(K_b/K_i))$ by the first isomorphism theorem. But $K_i$ is the fixed field of $\Bbb Z/d_1\Bbb Z\times\cdots\times\Bbb \{1\}\times\Bbb Z/d_{i+1}\Bbb Z\times \cdots\times \Bbb Z/d_r\Bbb Z$ and so $\text{Gal}(K_b/K_i)\cong \Bbb Z/d_i\Bbb Z$, which implies that $K_i/k$ is a cyclic extension.

Since the primitive m-th root of unity $\mu$ is in $k$, it is fixed by any automorphism in $\text{Gal}(K_i/k)$ and hence has norm $1$. Therefore by Hilbert's theorem $90,$ the generator of $\text{Gal}(K_i/k)$ sends any $\beta\in K_i\setminus k$ to $\mu\beta$. Then for $j=1,\cdots,m/d_i, \mu^j\beta$ must be the roots of the minimal polynomial of $\beta$ in $k$, and they are all distinct, so $K_i=k(\beta)$ for some $\beta\in K_i\setminus k$ such that $\beta^{m/d_i}\in k$. The compositum of all $K_i$ is the fixed field of the intersection of all of their corresponding Galois group, which is clearly the identity. Hence $K_b=K_1\cdots K_r$ where each $K_i$ is generated by an element $\beta_i$ such that $\beta_i^{m/d_i}\in k$, and thus $K_b=k(\gamma)$ where $\gamma$ has minimal polynomial $x^m-c$ over $k$. In other words, $K_b=k(c^{1/m})$ and then $K$ is the compositum of all $K_b$, so $K=K_B$ where $B=\langle c^{1/m}: c\in k^*\rangle(k^*)^m$, which concludes our proof.


答:
Here is a related argument with power series.

For $|x| < 1$ there is an identity
$$
e^x = \prod_{n \geq 1} (1 - x^n)^{-\mu(n)/n}.
$$
This can also be viewed as an identity of formal power series, and checked by verifying that the logarithmic derivatives ($f \leadsto f'/f$) of both sides are the same and that they start with the same constant term.

If we remove each factor on the right side where $p \mid n$, we get a different identity:
$$
e^{x + x^p/p + \cdots + x^{p^k}/{p^k} + \cdots} = \prod_{\substack{n \geq 1\\ p\nmid n}} (1 - x^n)^{-\mu(n)/n}.
$$
Look at the coefficient of $x^p$ on both sides.  On the left side, it is the coefficient of $x^p$ in $e^xe^{x^p/p}$, which is $1/p! + 1/p = (1 + (p-1)!)/p!$. On the right side, the coefficients of $(1 - x^n)^{\pm 1/n}$ are rational with no $p$ in the denominator when $p \nmid n$, so all coefficients on the right are rational with no $p$ in the denominator. Thus $(1 + (p-1)!)/p!$ is rational with no $p$ in the denominator, which means $p \mid (1 + (p-1)!)$, so $(p-1)! \equiv -1 \bmod p$.

This argument can be found in books on $p$-adic analysis where $p$-adic integrality of all the coefficients of the Artin-Hasse exponential is discussed.

This is related to the identity
$$
\log\left(\frac{1}{1-x}\right)  = \sum_{n \geq 1} \frac{x^n}{n}
$$
by extracting from each $n$ the highest power of $p$ dividing it:
$$
\log\left(\frac{1}{1-x}\right)  = \sum_{\substack{n \geq 1\\ p \nmid n}} \frac{g(x^n)}{n}
$$
where $g(x) = \sum_{k \geq 0} x^{p^k}/p^k$, so
$$
g(x) = \sum_{\substack{n \geq 1}{p \nmid n}} \frac{\mu(n)}{n}\log\left(\frac{1}{1-x^n}\right).
$$

Title: Analytic Proof of Wilson's Theorem

Body:
I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:

We start by considering the Maclaurin series of the function $f(x)=\ln \left( \frac{1}{1-x} \right)$, that is,
$\displaystyle \ln \left( \frac{1}{1-x} \right)= \sum_{n=1}^{\infty} \frac{x^n}{n} \tag*{}$
for $x \in [-1, 1).$
By simply using the definition of $\ln$ on the equation above, we get
$\displaystyle e^{x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots}=\frac{1}{1-x}=1+x+x^{2}+ \cdots + x^{p} + \cdots \tag{1}$
We can then rewrite the leftmost side of $(1)$ as
$\displaystyle \begin{align} e^{x+\frac{x^2}{2}+\frac{x^3}{3}+…} & = e^{x}e^{\frac{x^2}{2}}e^{\frac{x^3}{3}} \cdots \\ & = \prod_{n=1}^{\infty} e^{\frac{x^n}{n}} \\ & = \prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\left( \frac{x^n}{n} \right)^k}{k!} \\ & = \left( 1+\frac{x}{1!}+\frac{x^2}{2!}+ \cdots \right) \cdot \left( 1+\frac{x^{2}/2}{1!}+\frac{(x^{2}/2)^2}{2!} + \cdots \right) \cdots \left(1+\frac{x^{p}/p}{1!}+\frac{(x^{p}/p)^2}{2!}+ \cdots \right) \cdots \\ & = 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots \end{align} \tag{2}$
Combining the rightmost sides of $(1)$ and $(2)$ together gives us
$\displaystyle 1+x+x^{2}+ x^{3}+ \cdots + x^{p} + \cdots= 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots , \tag{3}$
from which follows that the coefficient of $x^{p}$ on the RHS of $(3)$ is $1$. That is,
$\displaystyle \frac{1}{p!}+\frac{1}{(p-2)!} \frac{1/2}{1!}+ \cdots + \frac{1/p}{1!} = 1. \tag{4}$
Notice that such a coefficient is of the form $\frac{1}{p!}+\frac{r}{s}+\frac{1}{p}$, where $\frac{r}{s}$ is the sum of a finite number of rationals, none of which has the factor $p$ in its denominator. Therefore, if $\frac{r}{s}$ is irreducible, then $p$ doesn’t divide $s$. We then proceed by rewriting $(4)$ as
$\displaystyle \frac{1}{p!}+\frac{r}{s}+\frac{1}{p}=1. \tag*{}$
Equivalently,
$\displaystyle 1-\frac{r}{s}=\frac{1}{p!}+\frac{1}{p}=\frac{1+(p-1)!}{p!}, \tag*{}$
and
$\displaystyle (s-r) \cdot (p-1)! = \frac{s(1+(p-1)!)}{p}. \tag*{}$
Hence, $\frac{s(1+(p-1)!)}{p}$ is an integer. Moreover, since $p$ doesn’t divide $s$, $\frac{1+(p-1)!}{p}$ is also an integer. Equivalently, $(p-1)! \equiv -1 \text{ (mod } p)$, as desired.


问:
How do I calculate marginal cost?
Link to excel Marginal calculation

I want to know if I did the calculations for the terms mentioned in the question correctly.

I am trying to see how much the organization saves in costs if the organization does mass production.

I am really awful at Maths. I am supposed to analyze a business, suggest a strategy, but as part of the analysis, I am supposed to look at discounted cash flow which I have done and now I decided to look at marginal cost with mass production.

To explain what I have done for the marginal cost. I did a change in unit produced/change in total cost.

So for sell I3 I did =($E3-$E4)/($A3-$A4). Then I did the same for each sell going down? Is that right? This looks like (£28,872-26.760)/(200-100) = £21 but on excel it equals £17,60 I don't  know what I am doing wrong

I need this for an assignment due in 5 days

Thanks

答:
Your definition is correct. Think of marginal cost as the cost of producing the next good. Let $C$ represent total cost and $Q$ quantity. Then, we define marginal cost ($MC$) as:

$$ MC = \frac{\Delta C}{\Delta Q} . $$

In your spreadsheet, at $Q = 100$,

$$ \Delta C = \text{E}4 - \text{E}3 , $$

$$ \Delta Q = \text{A}4 - \text{A}3 . $$

So, at $Q = 100$,

$$ MC = \frac{\text{E}4-\text{E}3}{\text{A}4-\text{A}3} . $$

Note, because of the identity $C = FC + VC$, where $FC$ is fixed cost and $VC$ is variable cost, you also have the following:

$$ MC = \frac{\Delta C}{\Delta Q} $$

$$ = \frac{\Delta (FC + VC)}{\Delta Q} $$

$$ = \frac{\Delta FC + \Delta VC}{\Delta Q} . $$

Because fixed costs are fixed for all quantities, $\Delta FC = 0$ is always true, so we can further simplify to

$$ MC = \frac{\Delta VC}{\Delta Q} $$

So, instead of using column E in your spreadsheet, you could also have used column C. You'd end up with the same answer for marginal cost.

One more note: there shouldn't be a value in cell I12, since you don't have any data on the cost of producing an additional good at a quantity of 1000. I'm also not sure where you see a value of 28,872 - in the spreadsheet, it looks like the value in cell E4 is 28,520, so the calculation in Excel is correct.

Title: How do I calculate marginal cost?


问:
The number of ways to order 26 alphabet letters, no two vowels occurring consecutively
What is the shortest solution to the following problem?

What is the number of ways to order the 26 letters of alphabet so that
  no two of the vowels a,e,i,o,u occur consecutively?

What I thought is to subtract permutations consisting of 2 vowels, 3 vowels, 4 vowels and 5 vowels occurring consecutively from all permutations 26!. But I could not even find an reasonable answer with that method. It seems to be too long and I always make mistakes.

答:
the question should be seen in different way:

number of alphabets = 26

no. of vowels = 5

total no of ways in which 26 letters can be arrange in 26 positions is = 26p26 = 26!

so we have 26! ways in which we can arrange this letter. In this 26! way we got both the ways in which vowels occur together & vowel don't occur together.

so no of ways in which vowels occur together is :
no. of positions is = 22 (21 are alphabets & 1 is group of 5 alphabets)
so no of ways 22 positions is = 22p22 = 22!
so no of ways in 5 vowels is = 5p5 = 5!

so total no of way in which vowel occur = 22! * 5!

so total no of ways in which vowel don't occur together = 26! - 22! * 5!


问:
Ratio test is inclusive; Discuss the convergence of $\sum_{n = 1}^\infty n^re^{-k\sum_{m = 1}^n \frac{1}{m}}$.

Investigate the convergence of $\sum_{n = 1}^\infty n^re^{-k\sum_{m = 1}^n \frac{1}{m}}$ when $r > k$ and when $r < k$.

Unfortunately, the ratio test failed. I was trying to use the comparison test instead. My proof is not complete, but I am going to show my work anyway.

It is not hard to see that $\log(n+1)\le \sum_{m = 1}^n \frac{1}{m} < 1+\log(n)$. I consider two cases. Assume $k\ge 0$, then
$k\log(n+1)\le k\sum_{m = 1}^n \frac{1}{m} \le (k+1)(1+\log(n))\Rightarrow \frac{n^r}{e^{(k+1)(1+\log(n))}} \le n^re^{-k\sum_{m = 1}^n \frac{1}{m}} \le \frac{n^r}{e^{k\log(n+1)}} \Rightarrow \frac{1}{e^{k+1}}\frac{1}{n^{k-r+1}} \le n^re^{-k\sum_{m = 1}^n \frac{1}{m}} \le \frac{1}{(n+1)^kn^{-r}}$. From here we can conclude that the original series diverges if $k < r$ and converges if $k -1 > r$. Well, I guess we can say the series converges when $k > k-1 > r$. But I am interested in knowing what happens to the series when $k > r \ge k-1$. I ran into the similar problem when I consider the case $k < 0$.

Looks like I need to use a different approach to solve the problem. Any help will be appreciated.

答:
We know that $H_n = \sum_{m=1}^n \frac{1}{m}$ is asymptotic to $\log n +\gamma$, i.e. $H_n \sim\log n+ \gamma$. Thus, your summand is asymptotically
$$n^r e^{-k\log n}e^{-k\gamma}=n^{r-k}e^{-\gamma k}.$$
Summing over $n$, we see that the series will only converge if $k-r>1$, and will diverge otherwise.


答: Your solution used the bounds $$\log(n+1) \le \displaystyle\sum_{m = 1}^{n}\dfrac{1}{m} \le \log(n)+1.$$ It is possible to prove tighter bounds $$\log(n)+\gamma \le \displaystyle\sum_{m = 1}^{n}\dfrac{1}{m} \le \log(n)+\gamma+\dfrac{1}{2n}$$ where $\gamma \approx 0.5772156649$ is the Euler-Mascheroni constant. If you use these tighter bounds, you'll can bound the series between $\displaystyle\sum_{n = 1}^{\infty}e^{-k\gamma}n^{r-k}$ and $\displaystyle\sum_{n = 1}^{\infty}e^{-k\gamma-\tfrac{k}{2n}}n^{r-k}$. This will let you determine convergence for all values of $k$.

Title: Ratio test is inclusive; Discuss the convergence of $\sum_{n = 1}^\infty n^re^{-k\sum_{m = 1}^n \frac{1}{m}}$.

Body:
Investigate the convergence of $\sum_{n = 1}^\infty n^re^{-k\sum_{m = 1}^n \frac{1}{m}}$ when $r > k$ and when $r < k$.

Unfortunately, the ratio test failed. I was trying to use the comparison test instead. My proof is not complete, but I am going to show my work anyway.

It is not hard to see that $\log(n+1)\le \sum_{m = 1}^n \frac{1}{m} < 1+\log(n)$. I consider two cases. Assume $k\ge 0$, then
$k\log(n+1)\le k\sum_{m = 1}^n \frac{1}{m} \le (k+1)(1+\log(n))\Rightarrow \frac{n^r}{e^{(k+1)(1+\log(n))}} \le n^re^{-k\sum_{m = 1}^n \frac{1}{m}} \le \frac{n^r}{e^{k\log(n+1)}} \Rightarrow \frac{1}{e^{k+1}}\frac{1}{n^{k-r+1}} \le n^re^{-k\sum_{m = 1}^n \frac{1}{m}} \le \frac{1}{(n+1)^kn^{-r}}$. From here we can conclude that the original series diverges if $k < r$ and converges if $k -1 > r$. Well, I guess we can say the series converges when $k > k-1 > r$. But I am interested in knowing what happens to the series when $k > r \ge k-1$. I ran into the similar problem when I consider the case $k < 0$.

Looks like I need to use a different approach to solve the problem. Any help will be appreciated.


问:
Computation of $\omega$- limit set of the given system.
\begin{equation*}
\begin{bmatrix}
x'   \\
y'
\end{bmatrix}
=F\begin{bmatrix}
x   \\
y
\end{bmatrix}
=
\begin{bmatrix}
y(y^2+(x^2-1)^2)+x(1-x^2-y^2)   \\
-x(y^2+(x^2-1)^2)+y(1-x^2-y^2)
\end{bmatrix}
\end{equation*}

I need to calculate $\omega(p)$ $\forall p \in \mathbb{R}^2$. ($\omega(x) = \{y \in X: \text{there is}\ t_n \to +\infty\ \text{such that}\ \phi(t_n,x) \to y   \}$ )

My hint is: $
$. I do not know how to do this, I found $$
=||p||^2(1-||p||^2).$$

Any help would be great.

答: The vector $p$ points in the radial direction, so the sign of the inner product $\langle p,F(p) \rangle$ tells you whether the vector $F(p)$ is pointing “outwards” or “inwards”. And from the expression you computed, it's clear that this depends on whether you're inside or outside the unit circle. Can you take it from there?


问:
Is the small resolution of a double point an lci morphism?
Let $X = \{xy = zw\} \subset {\mathbb C}^4$ be the cone over a quadric surface and $f:Y \to X$ its usual small resolution. Is $f$ a local complete intersection morphism?

More generally (and this is mainly out of curiosity), are there any reasonable criteria to tell when a birational morphism is lci?

答: An lci morphism has finite $\mathrm{Tor}$-dimension, while a resolution of a (non-trivial) singularity has infinite $\mathrm{Tor}$-dimension. So, the answer is no.


问:
How to solve $\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx$

\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx
\end{equation}

This is a MCQ question and there are 5 options to choose which are "A.$2\sqrt{3}\pi+9\ln 3$, B.$2\sqrt{7}\pi+8\ln 3$ , C.$2\sqrt{3}\pi+8\ln 3$ , D.$2\sqrt{2}\pi+2\ln 3$, E.Other solution."

How do you solve this integral?

This appears in MCQ test, so I think there should be a trick to solve this without using too much force. The test have 30 questions and 2 hours, so each question should be finished under 4mn ( I doubted it though). Here is my solution that I spend around 3h to solve it. ( Sorry for not using latex and sorry for the inconveniences )Photo

I appreciate any solution tricks. Thanks!

答:
Here are a few shortcuts. With $t =\sin x$

\begin{equation}
I=\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36\int_0^1 \frac {1}{(t^4+t^2+1)^2}dt\tag1
\end{equation}
Note
$$\left(\frac{t-t^3}{t^4+t^2+1}\right)’
= \frac{t^6+5}{(t^4+t^2+1)^2}-\frac{4}{t^4+t^2+1}
$$
Integrate both sides over $(0,\infty)$

\begin{align}
\int_0^\infty \frac{4}{t^4+t^2+1}dt &= \int_0^\infty \overset{t\to 1/t} {\frac{t^6}{(t^4+t^2+1)^2} }dt
 + \int_0^\infty \frac{5}{(t^4+t^2+1)^2}dt\\
&= \int_0^1\frac{6}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{6}{(t^4+t^2+1)^2}dt\tag2
\end{align}

Similarly, integrate over $(0,1)$

\begin{align}
\int_0^1\frac{4}{t^4+t^2+1}dt=\int_0^1\frac{5}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{1}{(t^4+t^2+1)^2}dt \tag3
\end{align}
Combine (2) and (3) to obtain
\begin{align}
\int_0^1\frac{dt}{(t^4+t^2+1)^2}= \int_0^1\frac{dt}{t^4+t^2+1}
-\frac14 \int_0^\infty \frac{dt}{t^4+t^2+1}= \frac14\ln3+\frac{\pi}{6\sqrt3}
\end{align}
Plug into (1)
$$I= 9\ln3+ 2\sqrt3{\pi}$$

P.S. The last integral is carried out as follows

$$\int \frac{1}{t^4+t^2+1}dt= \frac12 \int\frac{1-t^2}{t^4+t^2+1}dt
 + \frac12 \int\frac{1+t^2}{t^4+t^2+1}dt \\
= \frac12 \int\frac{d(t+\frac1t)}{(t+\frac1t)^2-1}
 + \frac12 \int\frac{d(t-\frac1t)}{(t-\frac1t)^2+3}
$$