问:
subscheme where two morphisms agree is points where they agree on residue fields
Let $X, Y, Z$ be schemes, where $X, Y$ are $Z$ schemes.  I know the definition of "the locally closed subscheme of $X$ where two $Z$-  morphisms $\pi, \pi': X\rightarrow Y$ agree" from its universal property.  Also I can define it as the fiber product of the diagonal $$\delta :  Y\rightarrow Y\times_Z Y$$ with $$(\pi, \pi'): X\rightarrow Y\times_Z Y.$$

My question:  how to prove that the underlying set of "the locally closed subscheme where the two morphisms agree" is the same as the set of points where the two morphism agree on the residue field.

It is probably clear thatthe former is contained in the latter, but why is it all of them?  That is, why is a point where $\pi, \pi'$ agree on the residue field necessarily contained in "the subscheme where $\pi, \pi'$ agree"?

答: An alternative  to the excellent accepted answer: If $\pi, \pi'$ agree at $ p$ let $ \varphi: Spec(\kappa(p))\rightarrow X.$ Then $ \pi \circ \varphi=\pi'\circ \varphi$ thus $ \varphi$ factors through the scheme where $\pi, \pi'$ agree. Thus $ p$ is in the latter.


问:
Finding the radius of convergence of $\frac{z+1}{z^2+z+1}$ about $z=3$
So the problem asks to find the radius of convergence without finding the power series itself. I thought it would be best to begin by factoring the denominator. That can by done by the quadratic formula and we obtain that:

$z = \frac{-1 \pm \sqrt{3}i}{2}$. If we want the radius of convergence about $z=3$ can we just say since $z = \frac{-1 \pm \sqrt{3}i}{2}$ then $z-3 = \frac{-1 \pm \sqrt{3}i}{2} -3 = \frac{-1 \pm \sqrt{3}i-6}{2}$. But from here where do I need to take this? if this is the right track to take.

答: The singularities are at:  $z=(-1\pm\sqrt3i)/2$.  You could note that the nearest singularity to $z=3$ is at a distance $d=\sqrt{(49+3)/4}=\sqrt{52/4}=\sqrt{13}$.

ContentLicense: CC BY-SA 4.0


问:
Derivative of Trig Functions (Intuition Help?)
Looking for some intuition help here.

I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$

$$
u=\frac{1}{x}
$$

$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$

$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}
$$

$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}
$$
This is incorrect but intuitively I want to multiply it this way.
$$
cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}
$$

But the correct answer is:
$$
-\frac{cos-\frac{1}{x}}{x^2}
$$

Help me absorb the why so I can intuitively solve problems like these.

答:
Here is where parentheses come in handy:

You found, correctly, $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$.

But the scope of $\cos$ is restricted to its argument: $\left(\dfrac 1x\right)$ ONLY:

The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$. That is not what you did. You multiplied argument of the $\cos$ function by the function $\dfrac{du}{dx} = -\dfrac{1}{x^2}$.

$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\left[cos\left(\frac{1}{x}\right)\right]\times\left(-\frac{1}{x^2}\right) $$ $$= -\frac 1{x^2} \cos\left(\frac 1x\right)
= -\dfrac{\cos\left(\frac 1x\right)}{x^2}$$

ContentLicense: CC BY-SA 3.0


问:
Finite sum which sums to $x e^x$
In my analysis class, we have to prove this and another two related sums that I think I could prove if I knew this. However, I don't know how to begin solving this; any hints would be appreciated.
$$\sum_{n=0}^{\infty} \left( e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right)=xe^x$$

One method of proving this mentions summation by parts, and the only reference I found in relation to that is Abel's lemma which I tried to use but doesn't help me. What other solutions are there?

答:
Let's write the sum in question using summation notation:

\begin{align*}
S&=\sum_{n=0}^\infty \left(e^x-\sum_{k=0}^n\frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{x^k}{k!}-\sum_{k=0}^n \frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!}
\end{align*}
Let's switch the order of summation:
$$S=\sum_{k=1}^\infty\sum_{n=0}^{k-1}\frac{x^k}{k!}$$
But note: the value of the summand doesn't depend on $n$. Therefore, we can treat that inner sum like we have a constant summand, in which case we're just multiplying by $k=(k-1)-0+1$. So,
$$S=\sum_{k=1}^\infty k\cdot\frac{x^k}{k!}=\sum_{k=1}^\infty\frac{x^k}{(k-1)!}$$
Let $k=m+1$:
$$S=\sum_{m=0}^\infty \frac{x^{m+1}}{m!}=x\sum_{m=0}^\infty\frac{x^m}{m!}=xe^x$$
as required.

Title: Finite sum which sums to $x e^x$


问:
Finding last digit of $\,625703^{43898^{614961^{448629}}}\!\!$ using Euler's Theorem
I'm a little bit stuck with this problem I hope you can help. I want to find the last digit of a power tower using Euler's theorem:
\begin{align}
q &= 10, \\ \varphi(q) &= 4, \\ \varphi(\varphi(q)) &= 2, \\\varphi(\varphi(\varphi(q))) &= 1.
\end{align}
\begin{align}
625703 ^{\displaystyle 43898 ^{\displaystyle 614961 ^{\displaystyle 448629}}} &\equiv (625703 \bmod 10)^{\displaystyle (43898 \bmod \varphi(10))^{\displaystyle (614961 \bmod \varphi(\varphi(10)))^{\displaystyle (448629 \bmod \varphi(\varphi(\varphi(10))))}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1^{\displaystyle 0}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1}} \mod 10 \\ &\equiv 3^{\displaystyle 2} \mod 10 \\ &\equiv 9 \mod 10
\end{align}
According to this approach the last digit of the power tower must be 9. However, the right solution is 1 (see here) - what am I doing wrong?

This approach is based on the following two answers

computing ${{27^{27}}^{27}}^{27}\pmod {10}$

What's a general algorithm/technique to find the last digit of a nested exponential?

Body:
\begin{align}
625703^{43898^{\scriptstyle614961^{\scriptstyle448629}}}\mkern-18mu\bmod 10&= (625703\bmod 10)^{43898^{\scriptstyle614961^{\scriptstyle448629}}\bmod\varphi(10)}\\
&= 3^{43898^{\scriptstyle614961^{\scriptstyle448629}}\mkern-12mu\bmod4}=3^{2^{\scriptstyle614961^{\scriptstyle448629}}\mkern-12mu\bmod4}=3^0.
\end{align}


问:
Is the following boolean expression a tautology?
I have the following boolean Expression:

```<code>!(x && z) || (x && !y && z) || (z && !x && y) || !(!x && y) || (!y && !x && z)
</code>```

or

```<code>(¬(x ∧ z) ∨ ((x ∧ (¬y ∧ z)) ∨ ((z ∧ (¬x ∧ y)) ∨ (¬(¬x ∧ y) ∨ (¬y ∧ (¬x ∧ z))))))
</code>```

So im asked to simplify this Expression and later build the conjunctive normalform.

For me it seems like !(!x && y) can be converted to x || !y and !(x && z) can be converted to !x || !z
After that you can sum up both expressions and !x || x will give you 1. So now there is:

```<code>1 || !z || (x && !y && z) || (z && !x && y) || !y || (!y && !x && z)
</code>```

And this equals to 1. But is 1 the minimal Expression of this boolean expression?

So i'm almost 100% sure this is a tautology and i also checked it with a truth table. What i'm missing here?

答:
Yes. You are right. Here's the derivation.

$$(xz)' \lor (xy'z) \lor (x'yz) \lor (x'y)' \lor (x'y'z)$$

Applying De Morgan's Law
$$\Leftrightarrow \color{red}{(x' \lor z')} \lor (xy'z) \lor (x'yz) \lor \color{red}{(x\lor y')} \lor (x'y'z)$$

Applying associativity law to remove brackets
$$\Leftrightarrow x' \lor z' \lor xy'z \lor x'yz \lor x \lor y' \lor x'y'z$$

$$\Leftrightarrow  \color{red}{x'} \lor z' \lor xy'z \lor x'yz \lor \color{red}{x} \lor y' \lor x'y'z$$

Applying commutativity of $\lor$ to rearrange
$$\Leftrightarrow \color{red}{(x \lor x')} \lor {\text {rest of expression}}$$

$$\Leftrightarrow \color{red}{TRUE} \lor {\text {rest of expression}}$$

$$\Leftrightarrow TRUE$$

So, TRUE is the equivalent minimal expression.


问:
Why discontinuous derivatives exist?
I stumbled upon a question involving discontinuous derivative but I couldn't understand how could it happen that the derivative of a function (let's say $f(x)$) is defined at point (let's suppose $c$) but then it is discontinuous (or undefined) at the same point $c$. Is there a problem in calculus?

(The function $f(x)$ is also continuous at $c$)

答:
The function
$$
f(x) = \begin{cases}
x^2\sin(1/x), &\text{if } x\ne0, \\
0, &\text{if } x=0
\end{cases}
$$
is differentiable everywhere (in particular, $f'(0)=0$) but its derivative is not continuous at $x=0$.

Whether this is a "problem" in calculus is in the eye of the beholder! (By way of analogy, the school of Pythagoras discovered that a real number could be defined but could not be written as a quotient of two integers. To them that might have seemed like a problem, but today we're pretty comfortable with the idea of irrational numbers.)

Certainly the drive to formalize these concepts in analysis in the 19th century was motivated in large part by "pathological" examples like this one. In the end, the definitions of continuity and differentiability we settled one do permit functions to have derivatives that are defined everywhere but not continuous everywhere, just as they permit functions that are themselves defined everywhere but not continuous everywhere.

Mathematical definitions develop in order to reflect our intuitions as much as possible within mathematics. But once we have settled on mathematical definitions, if ever their consequences run counter to our intuitions, we must be prepared to update our intuitions.

Title: Why discontinuous derivatives exist?


问:
What is the probability of having 5 green and white capplets and all 5 of them have the same color in the same side?
What is the probability of having 5 green and white caplets and all 5 of them have the same color on the same side?Image to Ilustrate</p>

答: The reason it's $\frac{1}{16}$ is because you originally have $\frac{1}{2^5}=\frac{1}{32}$, but there are two cases: either the greens are on the left or the greens are on the left. Each has equal probability $\frac{1}{32}$, so the total probability is $\frac{1}{32}+\frac{1}{32}=\frac{1}{16}.$


问:
Can any smooth manifold be realized as the zero set of some polynomials?
Is any real smooth manifold diffeomorphic to a real affine algebraic variety? (I.e. is there an "algebraic" Whitney embedding theorem?)

And are all possible ways of realizing a manifold $M$ as an algebraic variety equivalent? I.e. suppose $M$ is diffeomorphic to varieties $V_1$ and $V_2$, are these isomorphic in the algebraic category?

Admittely I'm just asking out of curiosity after reading this question: Can manifolds be uniformly approximated by varieties?

答:
A quick Google search found this paper, where it is stated that the answer to the first question is yes in the compact case (due to Tognoli): this result is called the Nash-Tognoli theorem. In general, the answer is no: a real affine variety has finite-rank homology groups, and it's easy to construct non-compact manifolds for which this is false (e.g. a surface of infinite genus). In fact, apparently there is a bound due to Milnor for the sum of the Betti numbers of a real variety. </p>

<p>The answer to the second question is certainly not: just take two elliptic curves with slightly different $j$-invariants. </p>

Body:
A quick Google search found this paper, where it is stated that the answer to the first question is yes in the compact case (due to Tognoli): this result is called the Nash-Tognoli theorem. In general, the answer is no: a real affine variety has finite-rank homology groups, and it's easy to construct non-compact manifolds for which this is false (e.g. a surface of infinite genus). In fact, apparently there is a bound due to Milnor for the sum of the Betti numbers of a real variety.

The answer to the second question is certainly not: just take two elliptic curves with slightly different $j$-invariants.


答: Remember that it’s entirely possible for a function to be continuous at a point where the derivative is not. Moreover, take care not to confuse continuity with existence (or being well defined for that matter). There is no prerequisite for differentiability of a function based on the differentiability of its derivative. A thorough discussion can be found here: Discontinuous derivative.</p>

Title: Why discontinuous derivatives exist?

Body:
I stumbled upon a question involving discontinuous derivative but I couldn't understand how could it happen that the derivative of a function (let's say $f(x)$) is defined at point (let's suppose $c$) but then it is discontinuous (or undefined) at the same point $c$. Is there a problem in calculus?

(The function $f(x)$ is also continuous at $c$)


问:
Flat principal bundles over a simply connected manifold are trivial
I am looking for a proof of the following result.

Let $p:P\to M$ be a $G$-principal bundle over a connected and simply connected manifold.
  Suppose there is a flat connection $A\in\Omega^1(P,\mathfrak{g})$, that is $F^A\equiv 0$.
  Then $P$ is trivial and there exists a global section $s:M\to P$ such that $s^*A \equiv 0$.

That a global section on $P$ implies that it is trivial is well known, so the first task is showing that $P$ is trivial.
It seems that this is a "well known fact", but I could find no concise proofs.
Wikipedia suggests that this is closely related to the holonomy of $A$.

Even then, I don't know how do we hit the $s^*A$ condition.

答:
A connection in a principal $G$-bundle $P$ is said to be flat at $u\in P$ if there is a local trivialization of the bundle in a neighborhood of u, $(U_{\alpha},\psi_{\alpha})$,  under which the given horizontal subspaces map to the horizontal subspaces of the trivial connection on the corresponding trivial bundle.

Consider the local section $s(m)=\psi_{\alpha}(m,e)$, being $e$ the identity element in $G$. Since the horizontal space at $s(m)$ consists of those $V\in T_{s(m)}P$ such that $A_{s(m)}(V)=0$  it is clear that
$$
s^*(A)_{m}(X)=A_{s(m)}(ds_m(X))=0
$$
and so $s^*(A)=0$.

On the other hand, I don't know how to solve the part that $P$ being trivial, that is, that $s$ is global. Sorry.


问:
How to begin this problem in abstract algebra? What information do I have from fixed things?

Let $a = (x_1, x_2,... , x_m)(x_{m+1})\dots(x_n)\in S_n$ (so $x_{m+1},\dots, x_n$ are fixed). Find all elements of $S_n$ that commute with $a$.

Do fixed thing commute with anything or do I need to focus only on $(x_1,x_2,...,x_m)$?

I kind of understand how to do this for $S_5$ but this problem has me questioning everything.

I am a beginner and have just gotten into this.

答:
I will use the usual cycle notation $a = (1\ 2\ \dots\ m)$ for the permutation $a\in S_n$ with $1\mapsto 2\mapsto 3 \mapsto \dots \mapsto m\mapsto 1$ and $k\mapsto k$ for $k>m$. We are looking for $b\in S_n$ with $ab=ba$.
Evaluating this at $k>m$ we get
$$
a(b(k)) = b(a(k)) = b(k),
$$
so $b(k)$ is a fixed point of $a$ and hence $b(k)>m$ as well. So all $k>m$ have images $b(k)>m$ and $b$ restricts to separate bijections on $\{1,\dots,m\}$ and $\{m+1,\dots,n\}$. What happens on $\{m+1,\dots,n\}$ is irrelevant to $ab=ba$ being satisfied.

For $k\le m$ let us equip $\{1,\dots,m\}$ with addition modulo $m$ so that $a(k)=k+1$ and $ab=ba$ yields
$$
b(k)+1 = b(k+1).
$$
Setting $q=b(1)$ this yields:
$$
b(2)=q+1,\ b(3)=q+2,\ \dots\,b(m) = q+m-1.
$$

We conclude that $b\in S_n$ commutes with $a$ if and only if it cyclically shifts the elements of $\{1,\dots,m\}$ and arbitrarily permutes the elements of $\{m+1,\dots,n\}$.


答:
What am I doing wrong?

Euler's theorem assumes that the base and the modulus are relatively prime.

In your problem, that is not the case:  $43898$ is not relatively prime to $4$.

In fact, $43898^n\equiv0\pmod4$ for $n\ge2$.

Can you solve it now?

Title: Finding last digit of $\,625703^{43898^{614961^{448629}}}\!\!$ using Euler's Theorem

Body:
I'm a little bit stuck with this problem I hope you can help. I want to find the last digit of a power tower using Euler's theorem:
\begin{align}
q &= 10, \\ \varphi(q) &= 4, \\ \varphi(\varphi(q)) &= 2, \\\varphi(\varphi(\varphi(q))) &= 1.
\end{align}
\begin{align}
625703 ^{\displaystyle 43898 ^{\displaystyle 614961 ^{\displaystyle 448629}}} &\equiv (625703 \bmod 10)^{\displaystyle (43898 \bmod \varphi(10))^{\displaystyle (614961 \bmod \varphi(\varphi(10)))^{\displaystyle (448629 \bmod \varphi(\varphi(\varphi(10))))}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1^{\displaystyle 0}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1}} \mod 10 \\ &\equiv 3^{\displaystyle 2} \mod 10 \\ &\equiv 9 \mod 10
\end{align}
According to this approach the last digit of the power tower must be 9. However, the right solution is 1 (see here) - what am I doing wrong?

This approach is based on the following two answers

computing ${{27^{27}}^{27}}^{27}\pmod {10}$

What's a general algorithm/technique to find the last digit of a nested exponential?


问:
People in theater probability
There are $200$ seats in a theater, $10$ of which are reserved for the press. $150$ people come to
the show one night, and are seated randomly. What is the probability of all the seats reserved
for the press to be occupied? Can somebody give me some ideas, please? I can't figure out what probabilistic model to use.

答:
This question utilises the hypergeometric probability distribution. We have $N=200$ seats, $n=150$ seats that are occupied on the given night, and $r=10$ seats reserved for the press.

Let $\mathbb{X}$ be a discrete random variable representing the number of press seats occupied, where $\mathbb{X}=0,1,...,10$. Thus $\mathbb{X}\sim$ Hyp$(200,150,10).$

We want to know the probability when all of the press seats of occupied, $\mathbb{P}(\mathbb{X}=10)=p(10)$.

Using the formula for the hypergeometric distribution, we obtain $p(10)=\frac{{10\choose10 }\cdot{190\choose140}}{200\choose150}=0.0521$.

Title: People in theater probability


问:
Example of skew (non-intersecting) lines in $\mathbb{P}^3$
In Hartshorne's "Algebraic Geometry", exercise I-3.7(a) asks us to show that any two curves in $\mathbb{P}^2$ have non-empty intersection. This lead me to think about whether this was generalizable to higher dimensional projective space, $\mathbb{P}^n$ for $n \geq 3$.

In particular, I read in an article that one can already find skew lines in $\mathbb{P}^3$ so this non-empty intersection principle does not hold for higher dimensions. I was wondering whether there were any examples of this?

Hartshorne later suggests in I-7 Theorem 7.2 a criterion for this not to be the case.

However, assuming we took two lines (1-dimensional varieties) in $\mathbb{P}^3$, of the general form:

$L_1=V(a_1X+b_1Y+c_1Z+d_1W)$

$L_2=V(a_2X+b_2Y+c_2Z+d_2W)$

wouldn't it be true that the point $[0:0:0:0] \in V(L_1,L_2)$, so the intersection would be non-empty? Is there an example where this would not be the case?

Thanks in advance and sorry!

答: Remember, $\Bbb P^n$ is $\Bbb A^{n+1}\setminus\{(0,\dots,0)\}$ up to scaling, so $[0:\dots:0]$ is not an element of $\Bbb P^n$. One easy example of two skew lines in $\Bbb P^3$ are $\{[a:b:0:0]\mid [a:b]\in\Bbb P^1\}$ and $\{[0:0:c:d]\mid [c:d]\in\Bbb P^1\}$.


问:
Evaluating $\lim_{n \to +\infty} ((n+1)^a-n^a)$
I am trying to determine the limit of the following expression which arose when looking at correlations for a stochastic process.

$\lim_{n \to +\infty} ((n+1)^a-n^a)$ for values of $a\in (0,2)$.

I'm not sure how to approach this. I am aware that for $a>1$ the limit equals $\infty$ and for $a<1$ the limit is $-1$ (and $a=0 \implies$ limit $=0$).

Any ideas of what to try here would be greatly appreciated.

Body:
For $a>1$ $$\lim_{x\to 0^+}\frac{(1+x)^a-1}{x^a}=\lim_{x\to0^+}\frac{1}{x^{a-1}}\cdot\frac{(1+x)^a-1}{x}=\left[\infty\cdot a\cdot(1)^{a-1}\right]=\infty$$
Therefore $\lim_{n\to\infty}(n+1)^a-n^a=\lim_{n\to\infty}\frac{\left(1+n^{-1}\right)^a-1}{n^{-a}}=\lim_{x\to0^+}\frac{(1+x)^a-1}{x^a}=\infty$


问:
Does ${\rm card}({\rm Aut}(\mathbb{Z}_p\times\mathbb{Z}_p ))$ equal to $(p^2 -1)(p^2-p)$? $(p \in \mathbb{P})$
Suppose that we have $G=\mathbb{Z}_p\times \mathbb{Z}_p$, where $p\in \mathbb{P}$. I know that every arbitrary permutation of the elements of $G$ such that $ \forall x \in G: {\rm ord}(f(x))=o(x) \Rightarrow f\in {\rm Aut}(G)$. Therefore, we can write $G$ group as follows:$$G=\{x^i,y^j,x^ky^l\mid i,j,k,l\in \{0,...,p-1\}\}=\langle x,y\rangle .$$ where $x=(1,0)$ and $y=(0,1)$. I know that $f(e)=e$ and $(\forall x\in G) \land (\forall f\in{\rm Aut}(G)):{\rm ord}(f(x))={\rm ord}(x)$ But, $\forall x\in G:{\rm ord}(x)\mid|\mathbb{Z}_p|\Rightarrow ({\rm ord}(x)=1) \lor ({\rm ord}(x)=p)$. Therefore,$\forall x \in G: ({\rm ord}(x)=p) \lor (x=e)$.

If I can define the correspondent elements of $f(x)=\bar x$ and $f(y)=\bar y$ ($\bar x,\bar y \in G$). I have $p^2-1$ candidate elements for $\bar x$ and $p^2-p$ candidate elements for $\bar y$ (because $ord(\bar x)=p,\forall \bar x \in G $ and $f(\bar y)\notin \langle\bar x\rangle$. Thus,$|{\rm Aut}(G)|=(p^2-1)(p^2-p)$.

Is it correct? If it's not, can you please indicate my mistakes? Thank you very much in advance!!

答: To expand on Greg's comment, your answer is indeed correct. An automorphism of $\mathbb{Z}_p\times\mathbb{Z}_p$ can be thought of as an invertible 2×2 matrix with coefficients in $\mathbb{Z}_p$. Thus $|\text{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)|=|\text{GL}_2(\mathbb{Z}_p)|$. So how many of these matrices are there? Giving a 2×2 invertible matrix is the same as finding two linearly independent vectors over $\mathbb{Z}_p$. Well, let's take any nonzero vector $v$. There are $p^2-1$ such choices. Now we need to see how many vectors are linearly independent from $v$. But it's actually much easier to see how many linearly dependent vectors there are to $v$. Such vectors  must of course be of the form $av$ for some $a\in \mathbb{Z}_p$. So there are $p$ vectors linearly dependent to $v$, and hence $p^2-p$ vectors linearly independent to $v$. Putting it all together, there are $(p^2-1)(p^2-p)$ pairs of linearly independent vectors.


问:
An statement about non-negative trace and positive semi-definite operators.
Let $\rho \in \mathfrak{L}(A)$ and $\sigma \in \mathfrak{L}(A)$. Now I want to know can we say that

for all $\sigma \geqslant 0:$$\quad \quad Tr(\rho \sigma) \geqslant 0 \quad \quad$  if and only if $\quad \quad\rho \geqslant 0 $

$\rho \geqslant 0 $ and $\sigma \geqslant 0 $ means that both of them are positive semi-definite operator and $Tr$ means trace. $A$ is a Hilbert space and $\mathfrak{L}(A)$ means all linear maps on $A$.

Could you please help me how can I show it?

答:
The "if" direction does hold:

Let $\rho,\sigma$ be positive-semidefinite (matrices, linear operators). Denote by $\rho^{1/2}$
the positive-semidefinite square root of $\rho$ ("the", as it is uniquely determined). Then
$$\operatorname{Tr}(\rho\sigma) \;=\; \operatorname{Tr}(\rho^{1/2}\sigma\rho^{1/2})\;\geqslant\; 0$$
by cyclicity of the trace, and because $\,\rho^{1/2}\sigma\rho^{1/2}\,$ is positive-semidefinite.

Regarding the other direction, the "only if":
$$\text{With}\quad\sigma = \begin{pmatrix}1&0\\0&1\end{pmatrix},\quad
\rho =\begin{pmatrix}2&0\\0&-1\end{pmatrix}
\text{ you get } \operatorname{Tr}(\rho\sigma)=1\,.$$


问:
Primary submodule implies zerodivisors act nilpotently
Let $M$ be a finitely generated module over a commutative Noetherian ring $R$. Let $N\subseteq M$ be a primary submodule. That is, $\text{Ass}_R(M/N)=\{\mathfrak{p}\}$ for some prime ideal $\mathfrak{p}\subseteq R$. I want to prove:

Let $r$ be a zerodivisor of $M/N$. Then $r^kM\subseteq N$ for some $k\in\mathbb{N}$.

My thoughts: I know that we have a finite filtration
$$
0=F_0\subseteq\cdots\subseteq F_n=M/N
$$
where each $F_{i+1}/F_i\cong R/\mathfrak{p_i}$ for some prime $\mathfrak{p_i}\subseteq R$. Moreover, since $\mathfrak{p}$ is an associated prime of $M/N$, I know that $R/\mathfrak{p}$ appears as some of the $R/\mathfrak{p_i}$ in the filtration. Now if $r$ is a zerodivisor of $M/N$, then $r\in\bigcup_{\mathfrak{p}\in\text{Ass}_R(M/N)}\mathfrak{p}=\mathfrak{p}$. I'm not sure where to go next. How can I use the filtration to show that $r^kM\subseteq N$ for some $k\in\mathbb{N}$, i.e. $r$ acts nilpotently on $M/N$?

答:
[Gopalakrishnana's commutative algebra, Proposition 4 of Page 72 and  Proposition 6 of Page 74] Let $R$ be a Noetherian ring and $M$ a finitely generated $R$-module. Then $\sqrt{Ann(M)}=\cap_{p\in Ass(M)}p$, and then the set of zero divisors of $M$ is equal
to  $\cup_{p\in Ass(M)}p$.

By your assumptions and above propositions for the $R$-module $M/N$ we have $\sqrt{Ann(M/N)}=p=$then the set of zero divisors of $M/N$. Thus, if  $r$ is a zerodivisor of $M/N$, then $r\in \sqrt{Ann(M/N)}$, that is, $r^kM\subseteq N$ for some $k\in N$.


问:
I am puzzled with which one is right.

I am puzzled with which one is right.If my work is wrong.please give me a right explanation in detail.

答:
The problem with recovering the polar angle $\theta$ from the equation $\tan \theta=y/x$ is that the solution is not unique modulo $2\pi$: both $\theta$ and $\theta+\pi$ have the same tangent (as Glen O wrote). Programmers got so fed up with this difficulty that they invented a new function atan2, which takes both $y$ and $x$ as arguments and gives $\theta=\mathrm{atan2}\,(y,x)$, taking in to account the signs of both $y$ and $x$.

But people can take the signs of $x,y$ into account as well. Here the real part of $1-z$ is positive (unless $z=1$, when the argument of $1-z$ is undefined). Therefore, the argument of $1-z$ lies in   $(-\pi/2,\pi/2)$, up to integer multiples of $2\pi$. If $t$ is in the range $(0,2\pi)$, then $(t-\pi)/2$ is in $(-\pi/2,\pi/2)$, so this is the value we should take.

But if $t$ was in the range $(2\pi,4\pi)$, then $(\pi+t)/2$ would be the correct answer.

Title: I am puzzled with which one is right.


问:
Is there meaning in the minimal instances of a variable you need to write a rational expression?
This is something I've never thought about before. Given a rational function $f \in \mathbf{k}(x)$, the minimum number of $x$ you need to write down a formula for $f$ on its domain is well-defined. My cute example is that the rational function
$$f(x) = \frac{x}{x+1} \;\;=\;\;1-\frac{1}{x+1}$$
only needs one $x$ to be expressed, which means here that $f$ is invertible on its domain: there exists an $f^{-1}$ with range equal to the entire domain of $f$. But is this interesting? Like, can we say something general relating the minimal number of $x$s needed to express $f$ and the number of sections of $f$?

答:
Doing this is sometimes convenient for either of two purposes:

(As you noted) finding the inverse function quickly;

In particular, it sometimes makes it obvious whether the function increases or decreases as $x$ increases;

It makes differentiation with respect to $x$ a bit simpler.

And maybe in some contexts the fact that $\dfrac 1 {x+1}$ is immediately recognizable as the sum of a geometric series is useful.


问:
Showing that $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ is continuous in the open unit disk $|z| <1$
So I figured since its a power series that maybe finding the radius of convergence here would be useful. So as $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ we can find the radius of convergence as$$R= \lim_{n \to \infty} \left|\frac{(n+1)^2}{(n+2)^2}\right| = 1.$$Since the radius of convergence is greater than $0$ we have that the function is analytic. Since the function is analytic it is continuously differentiable, so since the function is differentiable, can we just conclude that the function itself is continuous as well in the unit open disk? Or am I going about it wrong here.

答: It's fine, but why all those intermediate steps? When the radius of convergence of a power series centered at $a$ is $R$, its sum defines a continuous function on $D(a,R)$. That's all.


答:
You can use the Generalized Binomial Theorem to learn more about the behavior of the function. You may know the normal Binomial Theorem: for positive integers $n$, we have
$$(x+y)^n=\sum_{k=0}^n \binom nk x^{n-k}y^k$$
It turns out that this formula works for non-integer $n$, as long as we can generalize the binomial coefficients properly. And we do this by defining $\binom rk = \frac{(r)_k}{k!}$, where $(r)_k$ is the falling Pochhammer symbol, $(r)_k=r(r-1)\cdots(r-(k-1))$. With this, we can say that
$$(x+y)^r=\sum_{k=0}^\infty\binom rkx^{r-k}y^k$$
We can use this here to say that $(n+1)^a=\sum_{k=0}^\infty\binom ak n^{a-k}$. Keep in mind that $\binom a0=1$ for any $a$. Then we can substitute this into the limit:
$$L=\lim_{n\to\infty}\left(\sum_{k=0}^\infty\left(\binom ak n^{a-k}\right)-n^a\right)=\lim_{n\to\infty}\sum_{k=1}^\infty\binom ak n^{a-k}$$
The behavior of this limit will be dominated by the $k=1$ term, because that is the term with the largest exponent on $n$. So $L$ behaves like $\binom a1 n^{a-1}=an^{a-1}$. When $a>1$ this approaches $\infty$; for $a<1$ this will actually approach $0$, and for $a=1$ the limit also approaches $0$.

Title: Evaluating $\lim_{n \to +\infty} ((n+1)^a-n^a)$

Body:
I am trying to determine the limit of the following expression which arose when looking at correlations for a stochastic process.

$\lim_{n \to +\infty} ((n+1)^a-n^a)$ for values of $a\in (0,2)$.

I'm not sure how to approach this. I am aware that for $a>1$ the limit equals $\infty$ and for $a<1$ the limit is $-1$ (and $a=0 \implies$ limit $=0$).

Any ideas of what to try here would be greatly appreciated.


问:
Let $M$ be the only maximal ideal of the local ring $\Bbb{Z}_{(p)}$, show that $\Bbb{Z}_{(p)}/M$ is isomorphic to $\Bbb{Z}/(p)$.
Question: Let $M$ be the only maximal ideal of the local ring $\Bbb{Z}_{(p)}$, show that $\Bbb{Z}_{(p)}/M$ is isomorphic to $\Bbb{Z}/(p)$.

I have proved that $M=\{\dfrac a b\in\Bbb{Z}_{(p)}:p|a\}$ is only maximal ideal of $\Bbb{Z}_{(p)}$. So $\Bbb{Z}_{(p)}/M$ is a field. I saw somewhere that, this field is isomorphic to $\Bbb{Z}/(p)$, but the proof is not trivial. Can any one give an outline of the proof of this? Thank you.

答: Let $S:=\mathbb{Z}\setminus (p)$. Thus, $\mathbb{Z}_{(p)}=S^{-1}\mathbb{Z}$ and $M=S^{-1}(p)$. Hence, $\mathbb{Z}_{(p)}/M= S^{-1}\mathbb{Z}/S^{-1}(p)\cong S^{-1}(\mathbb{Z}/(p))=S^{-1}\mathbb{Z}_{(p)}$. Now since the elements of $S$ are unite in the ring $\mathbb{Z}_{(p)}$ we have $S^{-1}\mathbb{Z}_{(p)}=\mathbb{Z}_{(p)}$, and so $\mathbb{Z}_{(p)}/M\cong \mathbb{Z}_{(p)}$.


问:
maximum and minimum of the function $x^n\ln^k x$ in the interval $\bigl[\frac{1}{e},e\bigr]$
I have to find maximum and minimum of the function $x^n\ln^k x$ in the interval $\bigl[\frac{1}{e},e\bigr]$ with $k$ and $n \in N$

and I'm lost in the calculus.

$y'= x^{n-1} \ln^{k-1}x (n\ln x+k)$

$y'=0 \Leftrightarrow x=1 \lor x= e^{-\tfrac{k}{n}}$

While  $1 \in\bigl[\frac{1}{e},e\bigr]$, for $e^{-\tfrac{k}{n}}$ it depends.

If $k<-n$ then $x=e^{-\tfrac{k}{n}}<e$

If $k>n$ then $x=e^{-\tfrac{k}{n}}<\dfrac{1}{e}$

If $-n<k<n$ then $\dfrac{1}{e}<x=e^{-\tfrac{k}{n}}<e$

From this moment I'm lost and I can't decide where $y'>0$
Can someone help me?

答:
I think you solved the problem already because $k \ge 0$ and $n > 0$

To be the second optimum $e^{-\tfrac{k}{n}}$ must satisfy that $e^{-1} \le e^{-\tfrac{k}{n}}\le e$ which means $-n \le k \le n$ or in other words $0 < n \le k$

I would just ignore the other cases where $e^{-\tfrac{k}{n}}$ is not in the interval $[e^{-1},e]$

What could be still missing from your point of view?

Whether maximum or minimum in the case that you have 2 optima depends on the comparison between $e^{-\tfrac{k}{n}}$ and $1$.

Why did you want to investigate $y'>0$ if you already have the zeros of it?


问:
Finding an upper limit on a continuous function
I'm tasked with finding a smallest constant $c$ so that for a function $f\in C^4([0,1])$ and $x,y\in[0,1]$:
$$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert\le c\vert x-y\vert ^2\max_\limits{\xi\in[0,1]}\vert f'''(\xi)\vert$$
Since this is an excercise I was given on the Topic of Taylor series, my first thought was making use of Taylor's theorem and somehow set the upper bound using the Lagrange remainder but I'm not quite sure that's the way to go. Alternatively I thought about using the fact, that $f(x)$ is lipschitz continuous. Any nudge in the right direction would be greatly appreciated. Thanks in advance.

Body:
For $x \ne y$, there exists $\xi_1, \xi_2$ between $x,y$ such that:

\begin{align}
f(x)&=f(y)+f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3\\
f'(x)&=f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2\end{align}

Substituting:

\begin{align}&\quad\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert
\\&=\left\vert\frac{f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3}{x-y}-\frac{2f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2}{2}\right\vert\\
\\&=\left\vert f'(y)+\frac {f''(y)}2(x-y)+\frac{f'''(\xi_1)}6(x-y)^2-f'(y)-\frac {f''(y)}2(x-y)-\frac {f'''(\xi_2)}4(x-y)^2\right\vert
\\&=\left\vert \frac{f'''(\xi_1)}6(x-y)^2-\frac {f'''(\xi_2)}4(x-y)^2\right\vert
\\&\le\left(\left\vert \frac{f'''(\xi_1)}6\right\vert+\left\vert\frac {f'''(\xi_2)}4\right\vert\right)(x-y)^2
\end{align}


问:
$X \subseteq \mathcal{P}(B)$ contains filters on $\mathcal{B}$, are $\bigcap_{F\in X}F$ and $\bigcup_{F\in X}F$ filters too?

Consider a Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ and $\phi \neq X \subseteq \mathcal{P}(B)$ whose elements are filters on $\mathcal{B}$. Show that:

$\bigcap_{F\in X}F$ is also a filter on $\mathcal{B}$. However, $\bigcup_{F\in X}F$ may not be a filter.

If $X$ is totally ordered by the inclusion relation $\subseteq$, then $\bigcup_{F\in X}F$ is a filter on $\mathcal{B}$.

I'm working with the definition: $F \subseteq B$ is a filter if -

$F \neq \phi$

If $x,y \in F$ then $x\land y\in F$

If $x\in F$ and $x\leq y$ then $y\in F$

To start with, I need to show that $\bigcap_{F\in X}F \neq \phi$ - which I'm unable to do. Can we find an element common to all filters on $\mathcal{B}$, which may help us conclude that the intersection is not empty? I'm thinking in this direction because $X$ might as well be the set of all filters on $\mathcal{B}$.

Next, I want to show: if $x,y \in \bigcap_{F\in X}F$ then $x\land y\in \bigcap_{F\in X}F$. This seems easy, since $x,y \in \bigcap_{F\in X}F$ means that $x$ and $y$ are contained in every filter in $X\subseteq\mathcal{P}(B)$, and so is $x\land y$ (property of filters). Similarly for the last property, i.e. if $x \in \bigcap_{F\in X}F$ then $x$ is in every filter in $X$, and we know that filters are upwards closed - so if $x\leq y$ then $y$ is in every filter in $X$ (and hence in $\bigcap_{F\in X}F$)

Next, I want to show that $\bigcup_{F\in X}F$ (non-empty, of course) may not always be a filter - which calls for a counterexample? I'm unable to think of one. So, when will $\bigcup_{F\in X}F$ not be a filter? From the 2nd part, it seems that this may have something to do with ordering?

For the last part, since X is totally ordered, we could probably start off with $X= \{X_1,X_2,...\}$ (X may not be finite, who knows?), and w.l.o.g assume that $X_1 \subseteq X_2 \subseteq ...\subseteq X_i\subseteq X_{i+1}...$ (that's the total ordering defined by inclusion, yes?). How do I take it from here?

TL;DR I have shared my thoughts and work for every part of the question, and it would be a great help if I could get hints or insights that could help me complete my solution (happy to see other solutions also, though)! Thanks!

答:
For the second question you cannot assume that $\langle X,\subseteq\rangle$ is even countable, let alone that it can be ordered like the positive integers: it might be ordered like $\Bbb R$, for instance. All that you can assume is that if $F_1,F_2\in X$, then either $F_1\subseteq F_2$, or $F_2\subseteq F_1$.

Let $G=\bigcup_{F\in X}F$. It’s clear that $G\ne\varnothing$. Suppose that $x,y\in G$; then there are $F_x,F_y\in X$ such that $x\in F_x$ and $y\in F_y$. Without loss of generality we may assume that $F_x\subseteq F_y$. Can you finish it from there and go on to show that $G$ is upward closed?

This proof should suggest how to find a counterexample when $X$ is not linearly ordered by inclusion: when you’ve finished it, you’ll see that we used the linear order only to show that $G$ was closed under $\land$. For a counterexample, then, we probably want an $X$ that contains filters $F_x$ and $F_y$ containing elements $x$ and $y$, respectively, but no filter containing both $x$ and $y$. The simplest way to do that is to let $X=\{F_x,F_y\}$, where $x,y\in B$, $x\in F_x\setminus F_y$, and $y\in F_y\setminus F_x$, and if we can ensure that $x\land y=0$, we’ll make certain that $F_x\cup F_y$ is not a filter.

Clearly we need $B$ to have at least two elements, and they have to be incomparable. (Otherwise, the larger one will be in the filter containing the smaller one.) If we set $x\lor y=1$, $x\land y=0$, $x^c=y$, and $y^c=x$, we have simple Boolean algebra whose partial order has this Hasse diagram:

```<code>                     1
                    / \
                   x   y
                    \ /
                     0
</code>```

(It’s really just the power set algebra on a $2$-point set, as in amrsa’s comment.) And we can take $F_x=\{x,1\}$ and $F_y=\{y,1\}$ to get the desired counterexample: $F_x\cup F_y=\{x,y,1\}$, which is clearly not a filter, precisely because it doesn’t contain $x\land y$.

Body:
For the second question you cannot assume that $\langle X,\subseteq\rangle$ is even countable, let alone that it can be ordered like the positive integers: it might be ordered like $\Bbb R$, for instance. All that you can assume is that if $F_1,F_2\in X$, then either $F_1\subseteq F_2$, or $F_2\subseteq F_1$.

Let $G=\bigcup_{F\in X}F$. It’s clear that $G\ne\varnothing$. Suppose that $x,y\in G$; then there are $F_x,F_y\in X$ such that $x\in F_x$ and $y\in F_y$. Without loss of generality we may assume that $F_x\subseteq F_y$. Can you finish it from there and go on to show that $G$ is upward closed?

This proof should suggest how to find a counterexample when $X$ is not linearly ordered by inclusion: when you’ve finished it, you’ll see that we used the linear order only to show that $G$ was closed under $\land$. For a counterexample, then, we probably want an $X$ that contains filters $F_x$ and $F_y$ containing elements $x$ and $y$, respectively, but no filter containing both $x$ and $y$. The simplest way to do that is to let $X=\{F_x,F_y\}$, where $x,y\in B$, $x\in F_x\setminus F_y$, and $y\in F_y\setminus F_x$, and if we can ensure that $x\land y=0$, we’ll make certain that $F_x\cup F_y$ is not a filter.

Clearly we need $B$ to have at least two elements, and they have to be incomparable. (Otherwise, the larger one will be in the filter containing the smaller one.) If we set $x\lor y=1$, $x\land y=0$, $x^c=y$, and $y^c=x$, we have simple Boolean algebra whose partial order has this Hasse diagram:

                     1
                    / \
                   x   y
                    \ /
                     0

(It’s really just the power set algebra on a $2$-point set, as in amrsa’s comment.) And we can take $F_x=\{x,1\}$ and $F_y=\{y,1\}$ to get the desired counterexample: $F_x\cup F_y=\{x,y,1\}$, which is clearly not a filter, precisely because it doesn’t contain $x\land y$.


问:
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$
I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$

Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.

答:
Here is a different approach: Consider the change of variable $b_n=a_{n+4},$ so that the only initial condition is given by $b_0=0.$ Notice that the equation becomes
$$(n+5)a_{n+5}-(n+4)a_{n+4}+3a_{n+4}-12(n+5)H_{n+5}=0,$$
$$(n+5)b_{n+1}-(n+4)b_{n}+3b_{n}-12(n+5)H_{n+5}=0,$$
take $f(x)=(x+4)b_x$ so that $$\Delta (f)=f(x+1)-f(x)=(x+5)b_{x+1}-(x+4)b_x,$$
so we get that $$\Delta (f(x))=\frac{-3}{x+4}f(x)+12(x+5)H_{x+5}.$$
This looks like variation of parameters, check Remark 10.
We get then that the solution becomes
$$f(x)=\sum _{u=1}^{x-1}\left (\prod _{t=u+1}^{x-1} \left (1-\frac{3}{t+4}\right )\right )\cdot \left (12(u+5)H_{u+5}\right ),$$
notice that the product becomes
$$\left (\prod _{t=u+1}^{x-1} \left (1-\frac{3}{t+4}\right )\right )=\left (\prod _{t=u+1}^{x-1} \left (\frac{t+1}{t+4}\right )\right )=\frac{(u+2)(u+3)(u+4)}{(x+1)(x+2)(x+3)},$$
replacing we get
$$f(x)=12\sum _{u=0}^{x-1}\frac{\binom{u+5}{4}4!H_{u+5}}{(x+1)(x+2)(x+3)}=\frac{12\cdot 4!}{(x+1)(x+2)(x+3)}\left (\sum _{u=0}^{x+4}\binom{u}{4}H_{u}-\sum _{u=0}^4\binom{u}{4}H_{u}\right )$$
using your professors hint(which is a good exercise of integration by parts), we get that
$$f(x)=\frac{12\cdot 4!}{(x+1)(x+2)(x+3)}\left (\binom{x+5}{5}H_{x+5}-\frac{1}{5}\binom{x+5}{5}-\frac{25}{12}\right )$$
Taking back the change of variable, meaning plugging at $x-4,$ we get
$$na_n=f(n-4)=\frac{12\cdot 4!}{(n-3)(n-2)(n-1)}\left (\binom{n+1}{5}H_{n+1}-\frac{1}{5}\binom{n+1}{5}-\frac{25}{12}\right )$$


问:
$\|Tx\|\ge c\|x\|\iff\ker T=\{0\}\,\,\text{and}\,\,\text{ran}\,T$ is closed.

Let $B,C$ be Banach spaces and $T:$ $B\to C$ be bounded and linear. Show that there is a $c>0$ such that $\|Tx\|_C\ge c\|x\|_B$ for all $x\in B$, if and only if $\ker T=\{0\}$ and $\text{ran}\, T$ is closed.

$\implies$ In this direction, showing that $\ker T=\{0\}$ is easy enough. Let $x\in\ker T$, then
\begin{align*}
&0=\|0\|_C=\|Tx\|_C\ge c\|x\|_B\\
\implies &0\ge\|x\|_B\\
\implies&0=\|x\|_B\\
\implies&0=x\\
\implies&\ker T=\{0\}.
\end{align*}
But I am having more difficulty in showing that the range is closed.

$\Longleftarrow$ In this direction I am trying to go by way of contradiction and assuming that for any $c>0$ there exists an $x\in B$ such that
\begin{align*}
\|Tx\|_C<c\|x\|_B,
\end{align*}
then contradict the fact that $\ker T=\{0\}$ or that $\text{ran}\,T$ is closed but I am having some trouble there as well.

Any help here would be greatly appreciated.

答:
Let $T(x_n)$ be a Cauchy sequence, $\|T(x_n)-T(x_m)\|=\|T(x_n-x_m)\|\geq c\|x_n-x_m\|$ implies that $x_n$ is a Cauchy sequence and converges towards $x$ since the space is Banach. $T(x)=lim_nT(x_n)$.

For the other side, $T(B)$ is Banach. This implies that $T':B\rightarrow T(B)$ defined by $T'(x)=T(x)$ is bijective. The open mapping theorem implies that it is invertible. ${T'}^{-1}$ is continuous. This implies that there exists $c$ such that $\|{T'}^{-1}(y)\|\leq {1\over c}\|y\|$. Write $y=T(x)$, we deduce that $\|T(x)\|\geq c\|x\|$.


问:
Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$
Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\on{I}\pars{a,b} \equiv \left.\int_{0}^{\infty}{b\ln\pars{1 + ax} - a\ln\pars{1 + bx} \over x^{2}}\,\dd x
\,\right\vert_{a,\, b\ \in\ \mathbb{R}_{\, >\, 0}}}}$

Lets consider
$\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{\ln\pars{1 + px}\,x^{\nu - 2}}\,\,\dd x
\,\right\vert_{%
\substack{p\ >\ 0 \\[1mm] 0\ <\ \nu\ <\ 1}}}}$ which I'll evaluate by means of the
Ramanujan's Master Theorem. Note that
\begin{align}
\ln\pars{1 + px} & =
-\sum_{k = 1}^{\infty}{\pars{-px}^{k} \over k}
\\[2mm] & =
\sum_{k = 0}^{\infty}\braces{\color{red}
{-\bracks{k \not= 0}\Gamma\pars{k}p^{k}}}
{\pars{-x}^{k} \over k!}
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{\ln\pars{1 + px}\,x^{\pars{\color{red}{\nu - 1}} - 1}}\,\,\dd x
\,\right\vert_{\substack{p\ >\ 0 \\[1mm] 0\ <\ \nu\ <\ 1}}}
\\[5mm] = &\
\Gamma\pars{\nu - 1}
\braces{-\bracks{1 - \nu \not= 0}\Gamma\pars{1 - \nu}p^{1 - \nu}}
\\[5mm] = &\
-{\Gamma\pars{\nu} \over \nu - 1}\,\,
{\Gamma\pars{1 - \nu}p^{1 - \nu}} =
{p^{1 - \nu} \over 1 - \nu}\,{\pi \over \sin\pars{\pi\nu}}
\end{align}

\begin{align}
\on{I}\pars{a,b} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{b\ln\pars{1 + ax} - a\ln\pars{1 + bx} \over x^{2}}\,\dd x
\,\right\vert_{a,\, b\ \in\ \mathbb{R}_{\, >\, 0}}}
\\[5mm] & =
\lim_{\nu\ \to\ 0^{+}}\,\,\bracks{%
b\,{a^{1 - \nu} \over 1 - \nu}\,{\pi \over \sin\pars{\pi\nu}}
-
a\,{b^{1 - \nu} \over 1 - \nu}
\,{\pi \over \sin\pars{\pi\nu}}}
\\[5mm] & =
\pi\lim_{\nu\ \to\ 0^{+}}\,\,
{b\,a^{1 - \nu} - a\,b^{1 - \nu} \over \sin\pars{\pi\nu}} \\[5mm] = &\
\pi\lim_{\nu\ \to\ 0^{+}}\,\,
{-b\,a^{1 - \nu}\,\ln\pars{a} + a\,b^{1 - \nu}\,\ln\pars{b} \over \cos\pars{\pi\nu}\pi}
\\[5mm] = &\
\bbx{ab\ln\pars{b \over a}} \\ &
\end{align}


问:
Is a normal line to a constant function defined?
This is probably more so an attempt to clarify definitions rather than prove anything, but my question is roughly as follows: in $2D$ space, the naive way to define a normal line to some function $f(x)$ at a given point is through perpendicularity to the tangent line at that point. Algebraically, this is expressed as the negative reciprocal of the slope of the tangent line. But since the slope of a constant function is always $0$, such a reciprocal is clearly undefined. Now, mathematically that makes sense because this normal line wouldn't be a function, but graphically it doesn't, as it's really easy to draw a line perpendicular to a constant function at any given point. So, my first question is whether there exists a way to deal with such an object in a formal manner? And secondly, does the normal even have to be a function in the rigorous sense? I'm not well-versed in formal geometry, so I'm not sure what alternative definitions exist that could bridge this gap between mathematical definitions and graphical intuition.

答:
A normal is a straight line and a straight line needn't be described by a function.

$$x=x_0$$ is the equation of a vertical line.

By the way, a curve needn't be described by a function either.

$$(x-x_0)^2+(y-y_0)^2=r^2$$ is the equation of a circle.


问:
Prove $\text {Dom}(f) \subseteq \text {Im} (g)$ for $g \circ f(x)=x$
Given functions $f$ and $g$ such that $g \circ f(x)=x$ then show that $f$ is an injection and $\text{Im}(f) \subseteq \text{Dom} (g)$ and $\text{Dom}(f) \subseteq \text{Im} (g)$.

Clearly $g \circ f(x)=\text{I}_X$ where $\text{I}_X:X \to X$ such that $x \mapsto x$,since identity mapping is a bijection, it follows that it's also an injection and from here we see that $f$ is an injection.

On the other hand, $\text {Im}(f) \subseteq \text {Dom} (g)$ is necessarily and sufficient for $g \circ f$ to be defined.

Finally it's easy to see that the number of injections from $A$ with $\text{Card}(A)=n$ to the set $B$ with $\text{Card}(B)=k$ is given by  $$\binom{k}{n}n!\left[n \le k\right]$$

So it follows that if there is an injection from $A$ to $B$ then $n \le k$.
Applying this to the main problem we see that $\text{Card}(\text{Dom}(f)) \le \text{Card}(\text{Cod}(g))$ ,however this does imply that  $\text {Dom}(f) \subseteq \text {Im} (g)$.

So how can I prove the last one?

答: If $x$ is in $Dom(f)$ then $g\circ f(x)=x$ so in particular x is the image of $f(x)$ under $g$.


问:
Inverse of cosine in Head First Physics
I am reading Head First Physics, and on page 362 it discusses using the inverse of sin, cos, or tan to work out an angle.

I understand that without using Pythagoras I must circle the table where I can use the inverse of either sin, cos, or tan to get the angle. This is determined by what the available ratios are.

I can see that the second row (triangle B) does not have cos circled for number 0.913 as the ratio is missing and we were told not to use Pythagoras. My confusion however, is why the inverse of cos 0.913 gives 24.076 when the answer is 24.2? The rest of the table, whether circled or not, seems to give the correct answer for the angle column...

Here is a picture of the table:

答: It looks like either a typo or rounding error.  For triangle b, if you actually use the Pythagorean Theorem, you find that the length of the hypotenuse is $\sqrt{481}$, so $\cos(\theta) = \frac{20}{\sqrt{481}} \approx 0.912$ instead of $0.913$.  If you them compute $\cos^{-1}(0.912)$ you get approximately $24.2^{\circ}$.

Title: Inverse of cosine in Head First Physics


问:
Proving that the improper integral $\int_1^\infty\frac{f(x)}{\int_0^x f(t)dt}dx=\infty$
Let f be a positive continious function such that $\displaystyle \int_0^\infty f(x)dx=\infty$. Prove that $$\int_1^\infty\frac{f(x)}{\int_0^x f(t)dt}dx=\infty$$</p>

答:
For any $x$, since $\displaystyle \int_0^{\infty} f(t) dt$ diverges, there exists $y>x$ such that $$\int_0^y f(t) dt \geq 2 \int_0^x f(t) dt$$ Hence, we have
$$\overbrace{\int_x^y \dfrac{f(s)}{\displaystyle \int_0^s f(t) dt} ds \geq \int_x^y \dfrac{f(s) ds}{\displaystyle \int_0^y f(t) dt}}^{\because\text{Integrating a positive fn in denominator over a larger set}} = \dfrac{\displaystyle \int_x^y f(s) ds}{\displaystyle \int_0^y f(t) dt} = 1 - \dfrac{\displaystyle \int_0^x f(s) ds}{\displaystyle \int_0^y f(t) dt} \geq \dfrac12$$
Now conclude from this.

PS: This is adapted from Did's answer to this question.


问:
It can be shown that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?
It can be show that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?

What I have tried:

I have calculated the order of $(\mathbb{Z}_{10})^\times$ to be 4. Hence I think it is isomorphic to $\mathbb{Z}_{4}$.

Would someone be able to verify if my answer is correct?

Thank you for taking the time to respond to my question.

答:
Note that $(\mathbb{Z}/10\mathbb{Z}), \times)$ has 4 elements in it. So it is isomorphic to $(\mathbb{Z}/4\mathbb{Z}, +)$, indeed any cyclic group with 4 elements. Note that for each $m$ there is exactly one cyclic group of order $m$, up to group isomorphism.

Meanwhile to show that $(\mathbb{Z}/10\mathbb{Z}), \times)$ is indeed cyclic, note that 3 generates every element in this group.


答:
You are correct that the order of $(\mathbb Z_{10})^\times$ is $4$,

and there is only one cyclic group of order $4$ up to isomorphism.

The set of elements of $(\mathbb Z_{10})^\times$ is $\{[1],[3],[7],[9]\}$, and

you could make the isomorphism from  $(\mathbb Z_{10})^\times$ to $\mathbb Z_4$ explicit

by mapping $[1]\mapsto0$, $[3]\mapsto1$, $[9]\mapsto2$, and $[7]\mapsto3$, for example.

Title: It can be shown that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?

Body:
It can be show that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?

What I have tried:

I have calculated the order of $(\mathbb{Z}_{10})^\times$ to be 4. Hence I think it is isomorphic to $\mathbb{Z}_{4}$.

Would someone be able to verify if my answer is correct?

Thank you for taking the time to respond to my question.


问:
Closed, unbounded subset of a cardinal.
I missed two lectures in my set theory course, and now I don't understand the homework problems.

One is this: let $\kappa$ be a regular uncountable cardinal. Show that the following sets are closed and unbounded in $\kappa.$

$\{\alpha<\kappa\,:\,\alpha \text{ is a limit ordinal}\},$

$\{\alpha<\kappa\,:\,\mathrm{cf}(\alpha)=\omega\}$ for $\kappa=\omega_1,$

$\{\lambda<\kappa\,:\,\lambda\text{ is a cardinal}\}$ for $\kappa$ inaccessible.

I don't understand what closed or unbounded subsets of $\kappa$ is. Could you explain this to me and give some pointers on solving the problem?

答:
Let $\kappa$ be a regular uncountable cardinal. $X \subseteq \kappa$ is unbounded if $\sup X = \kappa$.

A $\alpha$ is a limit point of $X$ if and only if  $\alpha$ is a limit ordinal and $\sup(X \cap \alpha) = \alpha$.

$X \subseteq \kappa$ is closed if it contains all its limit points less than $\kappa$.

1 Let $X = \{\alpha < \kappa : \alpha \text{ is a limit ordinal}\}$. Limit points of $X$ must be limit ordinals. $X$ is closed. Now to show that $X$ is unbounded. Let $\alpha < \kappa$. Since $\kappa$ is a cardinal, $|\alpha| < \kappa$. Then $\alpha + \omega$ is a limit ordinal with cardinality less than $\kappa$. So $\alpha + \omega \in X$. $\alpha + \omega > \alpha$. $X$ is unbounded.

2 Let $X' = \{\alpha < \omega_1 : cf(\alpha) = \omega\}$. At least in Jech, the cofinality is only defined for limit ordinals. Using this convention, $X' \subseteq X$, where $X$ is the set of ordinals form part 1. The cofinality of any limit ordinal is always a cardinal (in fact regular cardinal). The only cardinality less than $\omega_1$ is $\omega$. It has been shown that if $\alpha < \omega_1$ is a limit ordinal, then $cf(\alpha) = \omega$. So $X' = X$. It is closed and unbounded by part 1.

3 Suppose $\kappa$ is (weakly) inaccessible. Let $X'' = \{\alpha < \kappa : \alpha \text{ is a cardinal}\}$. It is unbounded since by definition of weakly inaccessible, $\kappa$ is a limit cardinal. Suppose $\alpha$ is a limit point of $X''$. $\alpha$ is a cardinal since $\sup$ of a set of cardinals is always a cardinal.

Title: Closed, unbounded subset of a cardinal.


问:
Solving Poisson's equation on $B_1(0)\subset \mathbb{R}^2$
I am trying to solve a specific Poisson equation on the following set $B_1 =\left \{ (x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1 \right \}$

\begin{cases}
\Delta u = y & \text{in}\quad B_1\\
u = 1 & \text{on}\quad \partial B_1
\end{cases}

I have studies Green's functions but I don't understand them very well - I don't know how to apply them to a specific case. I have tried changing coordinates to polar, but I don't know how to handle the $y$ term. Any help would really help me understand these problems better.
Hints welcome too!

答:
General method

The standard result to use here asserts that the solution $w$ to
\begin{cases}
\Delta w = 0 & \quad \text{in}\quad B_1(0)\\
w=P_m(x,y) & \quad \text{on} \quad \partial B_1(0)
\end{cases}
where $P_m(x)$ is a polynomial of $\mathbb{R}^2$ restricted to $\partial B_1(0)$, is another polynomial of degree $P_{m-2}$ and it has the form $$w(x,y)=(1-(x^2+y^2))q(x,y)+P_m(x,y),$$ where $q$ has degree $m-2$. For example, you can find the proof in Theorem 5.1 in Chapter 5 in the book "Harmonic Function Theory" of Axcler, Bourdon, Ramey.

Application

We note that $\Delta \frac{y^3}{6}=y$ and we reduce to the previous case defining $$w(x,y):=u(x,y)-\frac{y^3}{6}$$
In our case $P_3(x,y)=1-\frac{y^3}{6}$ and hence we search for $$q(x,y)=a+bx+cy.$$
Imposing $\Delta w = 0$ we compute $a=b=0$ and $c=-1/8$. Hence  we obtain
$$
u(x,y)=\frac{x^2+y^2}{8}-\frac{y}{8}+1.
$$


问:
Let $f: (\mathbb{Z}_{24}, +) \rightarrow (\mathbb{Z}_{36}, +)$ be a group homomorphism. How many elements are there in the kernel of $f$?
Let  $f: (\mathbb{Z}_{24}, +) \rightarrow (\mathbb{Z}_{36}, +)$
be a group homomorphism. If the image of $f$ contains exactly $2$ elements, how many elements are there in the kernel of $f$?

What I know:

Let $f : G \to H$ be a group homomorphism between two finite groups. Then we
have $$|G| = |\ker(f )| \times |{\rm im}(f )|$$
$$|G|= |\ker(f )|  \times  2$$

However I don't know how to work out $|G|$ for my question.

答: The order of $G$ is 24 [why is this], so plugging this into your equation $|G| =$ $|$ker$(f)| \times |$Im$(f)|$ yields $24 =$ $|$ker$(f)| \times 2$ or $|$ker$(f)| = 12.$


问:
Prove the formula for Jacobian determinant for this coordinate transform for general $n$
Let $n$ be some integer $n \geq 2$, $(x_1, \cdots, x_n)$ are the old coordinates and $(r, \phi_1, \phi_2, \cdots, \phi_{n-1})$ are some new coordinates. I want to transform $(x_1, \cdots, x_n)$ to $(r, \phi_1, \cdots, \phi_{n-1})$ in the following way
$$
 \begin{aligned}
  x_1 &= r \cos^2 \phi_1 \\
  x_2 &= r \sin^2 \phi_1 \cos^2 \phi_2 \\
  x_3 &= r \sin^2 \phi_1 \sin^2 \phi_2 \cos^2 \phi_3 \\
  x_{n-1} &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \cos^2 \phi_{n-1} \\
  x_n &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \sin^2 \phi_{n-1}
 \end{aligned}
$$
so that $x_1 + \cdots + x_n = r$ (yes, I just took the generalized spherical coordinates and squared the angular part). These coordinates are only defined for $x_1, \cdots, x_n \geq 0$, so the angles only move in range $\left\langle 0, \pi/2 \right)$ each and $r > 0$. It's okay, as I'm only interested in these coordinates when they're positive. I want to prove that the Jacobian determinant of this transformation is (for a general $n$)
$$
J (r, \phi_1, \cdots, \phi_n) = 2^{n-1} r^{n-1} \cos \phi_1 \cdots \cos \phi_{n-1} \sin \phi_1^{2 n - 3} \sin \phi_2^{2n-5} \cdots \sin \phi_{n-1}
$$
so for some small $n$
$$
\begin{aligned}
J (r, \phi_1) &= 2 r \cos \phi_1 \sin \phi_1 \\
J (r, \phi_1, \phi_2) &= 4 r \cos \phi_1 \cos \phi_2 \sin^3 \phi_1 \sin \phi_2 \\
J (r, \phi_1, \phi_2, \phi_3) &= 8 r \cos \phi_1 \cos \phi_2 \cos \phi_3 \sin^5 \phi_1 \sin^3 \phi_2 \sin \phi_3 \\
\end{aligned}
$$

What's my best shot at proving this formula for general $n$? I only spotted this pattern while playing with Mathematica, so I'm not even sure it's correct, but I hope it is.

I wanted to try induction, but adding one more coordinate $x_{n+1}$ actually changes the form for $x_n$ ($x_n$ will end with $\cdots \sin^2 \phi_{n-1} \cos^2 \phi_n$ instead of just $\sin^2 \phi_{n-1}$), so it does not help that much. I wonder if it's useful to look at the metric tensor first, rather than the Jacobian determinant itself, as Jacobian determinant = det $g$ anyway. I didn't spot any useful pattern in the metric tensor itself for small $n$, as it has all off-diagonal elements in the new coordinates (because the new coordinates are not orthogonal).

答:
The proof can be done inductively using the volume forms. Let $\omega_n$ be
$$
\omega_n = \mathrm{d} x_1 \wedge \cdots \wedge \mathrm{d} x_n = J_n (r, \phi_1, \cdots, \phi_{n-1}) \, \mathrm{d} r \wedge \mathrm{d} \phi_1 \wedge \cdots \wedge \mathrm{d} \phi_{n-1}
$$

Now let's add another coordinate so we have $n+1$ coordinates, in the following form
$$
 \begin{aligned}
  x_1 &= r \cos^2 \phi_1 \\
  x_2 &= r \sin^2 \phi_1 \cos^2 \phi_2 \\
  x_3 &= r \sin^2 \phi_1 \sin^2 \phi_2 \cos^2 \phi_3 \\
  &\;\;\vdots \\
  x_{n-1} &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \cos^2 \phi_{n-1} \\
  x_n &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \sin^2 \phi_{n-1}
 \end{aligned} \quad \to \quad
 \begin{aligned}
  x_1^\prime &= x_1 \\
  x_2^\prime &= x_2 \\
  x_3^\prime &= x_3 \\
  &\;\;\vdots \\
  x_{n-1}^\prime &= x_{n-1} \\
  x_n^\prime &= x_n \cos^2 \phi_n \\
  x_{n+1}^\prime &= x_n \sin^2 \phi_n \\
 \end{aligned}
$$

Now let's calculate the form $\omega_{n+1}$
$$
\omega_{n+1} = \mathrm{d} x^\prime_1 \wedge \cdots \wedge\mathrm{d} x^\prime_{n+1} = \left( \mathrm{d} x_1^\prime \wedge \cdots \wedge \mathrm{d} x_{n-1}^\prime \wedge \mathrm{d} x_n^\prime \right) \wedge \mathrm{d} x_{n+1}^\prime = \\=\left( \mathrm{d} x_1 \wedge \cdots \wedge \mathrm{d} x_{n-1} \wedge \mathrm{d} x_n \right) \wedge \left( \frac{\partial}{\partial x_n} \left( \mathrm{d} x^\prime_n \wedge \mathrm{d} x^\prime_{n+1} \right) \right) = \omega_n \wedge \left( \frac{\partial x_n^\prime}{\partial x_n} \mathrm{d} x^\prime_{n+1} - \frac{\partial x_{n+1}^\prime}{\partial x_n} \mathrm{d} x^\prime_n \right)
$$

The one-form at the end of this expression is a matter of simple calculation
$$
 \mathrm{d} x_n^\prime = \mathrm{d} x_n \cos^2 \phi_n - 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n = \cos \phi_n \left( \mathrm{d} x_n \cos \phi_n - 2 x_n \sin \phi_n \, \mathrm{d} \phi_n \right)
$$
$$
 \mathrm{d} x_{n+1}^\prime = \mathrm{d} x_n \sin^2 \phi_n + 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n = \sin \phi_n \left( \mathrm{d} x_n \sin \phi_n + 2 x_n \cos \phi_n \, \mathrm{d} \phi_n \right)
$$
$$
\frac{\partial x_n^\prime}{\partial x_n} \mathrm{d} x^\prime_{n+1} - \frac{\partial x_{n+1}^\prime}{\partial x_n} \mathrm{d} x^\prime_n = \cos^2 \phi_n \left( \mathrm{d} x_n \sin^2 \phi_n + 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n \right) - \\-\sin^2 \phi_n \left( \mathrm{d} x_n \cos^2 \phi_n - 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n \right) = 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n
$$
and the factor $2 x_n \sin \phi_n \cos \phi_n$ is exactly what we find when we divide $J_{n+1}$ by $J_n$.

$$
\frac{J_{n+1}}{J_n} = \frac{2^n r^n \cos \phi_1 \cdots \cos \phi_{n-1} \cos \phi_n \sin \phi_1^{2 n - 1} \sin \phi_2^{2n-3} \cdots \sin^3 \phi_{n-1} \sin \phi_n}{2^{n-1} r^{n-1} \cos \phi_1 \cdots \cos \phi_{n-1} \sin \phi_1^{2 n - 3} \sin \phi_2^{2n-5} \cdots \sin \phi_{n-1}} = \\=2 r \cos \phi_n \sin^2 \phi_1 \sin^2 \phi_2 \cdots \sin^2 \phi_{n-1} \sin \phi_n = 2 x_n \sin \phi_n \cos \phi_n
$$


答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{u} & =
{1 \over 6}\,y^{3} +
\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 6}\,r^{3}\sin^{3}\pars{\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 8}\,r^{3}\sin\pars{\theta} -
{1 \over 24}\,r^{3}\sin\pars{3\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D
\\[1mm] \mbox{Polar Coordinates}}}
\\[5mm] & =
{1 \over 8}\,r^{3}\sin\pars{\theta} -
{1 \over 24}\,r^{3}\sin\pars{3\theta}
\\[2mm] & +
\bracks{1 + a_{1}r\sin\pars{\theta} +
a_{3}r^{3}\sin\pars{3\theta}}
\\[5mm] & =
1 + \pars{{1 \over 8}r^{3} - a_{1}r}\,\sin\pars{\theta} +
\pars{-{1 \over 24}\,r^{3} + a_{3}r^{3}}\sin\pars{3\theta}
\end{align}
Since it must be $\ds{\theta}$-independent in
$\ds{\partial B_{1}}$:
\begin{align}
\on{u} & =
1 +
{1 \over 8}\pars{r^{3} - r}\,\sin\pars{\theta}
\\[5mm] & =
1 +
{1 \over 8}\pars{x^{2} + y^{2} -1}\,y
\end{align}

Title: Solving Poisson's equation on $B_1(0)\subset \mathbb{R}^2$

Body:
I am trying to solve a specific Poisson equation on the following set $B_1 =\left \{ (x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1 \right \}$

\begin{cases}
\Delta u = y & \text{in}\quad B_1\\
u = 1 & \text{on}\quad \partial B_1
\end{cases}

I have studies Green's functions but I don't understand them very well - I don't know how to apply them to a specific case. I have tried changing coordinates to polar, but I don't know how to handle the $y$ term. Any help would really help me understand these problems better.
Hints welcome too!


问:
Isomorphism id+f where f is a linear map

Let $V$ be a Vector Space on a field $\mathbb{K}$, and let $f:V\rightarrow V$ be a linear map such that $f\circ f\circ f\circ f=\underline{0}$. Prove that $id+f$ is an isomorphism.

First of all, I proved that $id+f$ is injective. In fact, $ker(id+f)=\lbrace v\in V\mid f(v)=-v\rbrace$

$\Rightarrow  ker(id+f)=\lbrace\underline{0}\rbrace$. I have some problmes to prove that $f$ is surjective because I don't know how to prove that $w\in V$ can be wrote as $v+f(v)$. Can someone give me an advice? Thanks before!

答:
Hint :  For every $x \in V$, you have
$$x = v + f(v)$$
where $v = x -f(x) + f^2(x) - f^3(x)$.


问:
Finding last digit of $\,625703^{43898^{614961^{448629}}}\!\!$ using Euler's Theorem
I'm a little bit stuck with this problem I hope you can help. I want to find the last digit of a power tower using Euler's theorem:
\begin{align}
q &= 10, \\ \varphi(q) &= 4, \\ \varphi(\varphi(q)) &= 2, \\\varphi(\varphi(\varphi(q))) &= 1.
\end{align}
\begin{align}
625703 ^{\displaystyle 43898 ^{\displaystyle 614961 ^{\displaystyle 448629}}} &\equiv (625703 \bmod 10)^{\displaystyle (43898 \bmod \varphi(10))^{\displaystyle (614961 \bmod \varphi(\varphi(10)))^{\displaystyle (448629 \bmod \varphi(\varphi(\varphi(10))))}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1^{\displaystyle 0}}} \mod 10 \\ &\equiv 3^{\displaystyle 2^{\displaystyle 1}} \mod 10 \\ &\equiv 3^{\displaystyle 2} \mod 10 \\ &\equiv 9 \mod 10
\end{align}
According to this approach the last digit of the power tower must be 9. However, the right solution is 1 (see here) - what am I doing wrong?

This approach is based on the following two answers

computing ${{27^{27}}^{27}}^{27}\pmod {10}$

What's a general algorithm/technique to find the last digit of a nested exponential?

答:
Hint: $ $ notice that: $\ n\ge 2\,\Rightarrow\, \color{#c00}{(2k)^n\bmod 4 \,\equiv\, 0}\ $ so by $ $ modular order reduction

$\!\!\bmod 10\!:\ 3^{\large \color{#c00}4}\equiv 1 \Rightarrow\ 3^{\large \color{}{(2k)^{\large n}}}\!\!\!\equiv 3^{\large \color{#c00}{(2k)^{\large n}\bmod 4}}\!\equiv 3^{\:\!\large\color{#c00} 0}\equiv 1,\ $ and

$\!\!\bmod 10\!:\  625703\equiv 3\Rightarrow 625703^N\!\equiv 3^N\,$ by the Congruence Power Rule.

Remark $ $ The oversight is: Euler's theorem $\,a^{\phi(m)}\equiv 1\pmod{\!m}\,$ has hypothesis $\,\gcd(a,m)= 1\,$ so it does not apply to $\,(2k)^n\pmod{\! 4}.\,$ In such cases we can pull out the gcd of $\,a^N$ and $\,m\,$ using the mod Distributive Law, reducing to the coprime case where Euler applies, e.g. see here. Though it is overkill in this case, we could apply it above to factor out the common factor $2^2$ as follows

$$\quad\ \color{#0a0}{n\ge 2}\,\Rightarrow\,\color{#c00}{(2k)^{\large n}\!\bmod 4} \,=\, 2^2 (k^2 (2k)^{\large \color{#0a0}{n-2}}\!\bmod 1) \:\!=\:\! 2^2(0) \:\!=\:\! 0$$

Using this idea in your recursive algorithm then allows you to handle arbitrary power towers.


问:
Finding the basis for a subspace of P3
How would I find a basis for the set $V \subset P_{3} \subset C[-1,1]$ defined as all $f \in P_{3}$ with $f(-1)=0$ and $\int_{-1}^{1}f(x)dx = 0$? Should I be looking for a specific polynomial that satisfies both of the conditions of $V$?

答:
Yes, you should. We know that every $f\in P_3$ is of the form $a+bx+cx^2+dx^3$; then $f(-1)=0$ implies that $a-b+c-d=0$.

Then $\int_{-1}^1f(x)dx=ax+\frac b2x^2+\frac c3 x^3+\frac d4x^4\Bigg|_{-1}^1=2a+\frac{2c}3=0$

We can simplify these down by finding the null space of the corresponding matrix $\begin{bmatrix}1&-1&1&-1\\2&0&\frac23&0\end{bmatrix}\sim\begin{bmatrix}1&0&\frac13&0\\0&1&-\frac23&1\end{bmatrix}$, and so the null space is spanned by $\begin{bmatrix}1\\0\\-3\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}$.

Therefore, a basis for $V$ is the polynomials $1-3x^2-2x^3$ and $x-x^3$.

Title: Finding the basis for a subspace of P3


问:
If $pq=1$, then $p=q=1$ for $p,q \in \mathbb {Z}, p,q >0$
If $pq=1$, then $p=q=1$ for $p,q \in \mathbb {Z}$,  $p,q >0$

I tried to do this by contradiction
and I get

$(pq=1) \land (p\neq 1 \lor q \neq 1)$

then I have no ideas how to continue with a formal proof. What I know is if i choose q not equal to 1 and it is greater then 0, I will get

$p = \frac{1}{q}$

and p will be in the interval (0-1) and I get a contradiction.

Do I have to do (1) if p is not 1 and (2) both p and q is not 1?

答:
Suppose $p$ and $q$ are two positive integers such that $pq=1$.

If $p>1$, then $q=1/p<1$ and this is impossible because $q$ is a positive integer. As $p$ cannot be less than $1$, again because it is a positive integer, it can only be equal to $1$. Of course, $q$ is also $1$ in that case.


问:
Does any irreducible polynomial in $\mathbb{R} [x]$ with degree $n$ necessarily have $n$ equally spaced roots around a circle on the Argand diagram?
More precisely: Does any polynomial $f \in \mathbb{R} [x]$ which is irreducible in $\mathbb{R}$ with $deg(f) = n$ necessarily have roots in $\mathbb{C}$ of the form $\{Ke^{2(k/n)\pi i}, Ke^{2(k+1/n)\pi i}, \cdots, Ke^{2(k+(n-1)/n)\pi i}\}$ where $k \in [0,1)$ and $K \in \mathbb{R}$?

Thanks!

答: No.  Consider $f(x)=x^2+x+1$, for a counterexample.


问:
Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$
One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$

I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$

Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$

I dont know what to do from here, can you tell me what is the next step?

答:
Here is a variation using the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.

We start with $f(x)$ and obtain for $n\geq 3$:
\begin{align*}
\color{blue}{[x^n]f(x)}&=[x^n]x^3\sum_{k=0}^\infty\binom{k+4}{4}(-1)^kx^k\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{1}\\
&=[x^{n-3}]\sum_{k=0}^\infty\binom{k+4}{4}(-1)^kx^k\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{2}\\
&=\sum_{k=0}^{n-3}\binom{k+4}{4}(-1)^k[x^{n-3-k}]\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{3}\\
&=\sum_{k=0}^{n-3}\binom{k+4}{4}(-1)^k\sum_{l=0}^\infty\binom{l+5}{5}[x^{n-3-k}]x^l\\
&\,\,\color{blue}{=\sum_{k=0}^{n-3}\binom{k+4}{4}\binom{n-k+2}{5}(-1)^k}\tag{4}
\end{align*}

Comment:

In (1) we write the sums using different indices $k$ and $l$ which helps to not mix them up.

In (2) we absorb $x^3$ by applying the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ again. Note the upper index is set to $n-3$, since other indices do not contribute to the coefficient of $x^{n-3}$.

In (4) we select the coefficient of $x^{n-3-k}$ which implies choosing $l=n-3-k$. Here we use $[x^p]x^q=\begin{cases}1&p=q\\0&p\ne q\end{cases}$.


答: I would rather mark (easier to find a mistake) $$z+{1\over z} = c$$ where $c$ is real and now solve $$z^2-zc+1=0$$ It discriminant is $c^2-4$ so $$z_{1,2}= {c\pm \sqrt{c^2-4}\over 2}$$

Title: Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good?

Body:
I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:

$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$

so, the solutions are:

$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$

I am not sure if that should be done like I did it.


答:
For an indeterminate $x$ we have the polynomial identity

$(1 - x)(1 + x + x^2 + x^3) = 1 + x + x^2 + x^3- x -x^2 - x^3 - x^4 = 1 - x^4; \tag 1$

taking

$x = -y, \tag 2$

we obtain

$(1 + y)(1 - y + y^2 - y^3) = 1 - y^4, \tag 3$

which may also be directly derived as is (1);  thus since

$f^4 = 0, \tag 4$

we have

$(I + f)(I - f + f^2 - f^3) = I - f^4 = I, \tag 5$

which shows that $I + f$ is invertible with inverse

$(I + f)^{-1} = I - f + f^2 - f^3, \tag 6$

hence an isomorphism.

Note that in accord with (5), for any $w \in V$ we have

$w = Iw = (I + f)(I - f + f^2 - f^3)w = (I + f)v, \tag 7$

where

$v = (I - f + f^2 - f^3)w. \tag 8$

Title: Isomorphism id+f where f is a linear map

Body:
Let $V$ be a Vector Space on a field $\mathbb{K}$, and let $f:V\rightarrow V$ be a linear map such that $f\circ f\circ f\circ f=\underline{0}$. Prove that $id+f$ is an isomorphism.

First of all, I proved that $id+f$ is injective. In fact, $ker(id+f)=\lbrace v\in V\mid f(v)=-v\rbrace$

$\Rightarrow  ker(id+f)=\lbrace\underline{0}\rbrace$. I have some problmes to prove that $f$ is surjective because I don't know how to prove that $w\in V$ can be wrote as $v+f(v)$. Can someone give me an advice? Thanks before!


问:
Function with $|f(x)-\int^{\delta}_{-\delta}f(x+u)du|<\epsilon$
I am looking for a function $f:\mathbb{R}\to \mathbb{R}$ and $\epsilon>0$ such that there is no $\delta>0$, for him any $x\in\mathbb{R}$: $|f(x)-\int^{\delta}_{-\delta}f(x+u)du|<\epsilon$

Clearly, $f$ must be unbounded.

答:
Let $f(x)=x^2$ and $\varepsilon=1/12$. For any $\delta>0$ and any $x\in\mathbb{R}$,
$$|f(x)-\int_{-\delta}^\delta f(x+u)du|=|x^2-\int_{-\delta}^\delta (x^2+2xu +u^2)du|=|x^2(1-2\delta)-(2\delta^3/3)|$$
If $\delta=1/2$, let $x=x(0)=0$. Then $|f(x)-\int_{-\delta}^\delta f(x+u)du|=1/12=\varepsilon$. If $\delta\neq 1/2$, then let $x=x(\delta)\in\mathbb{R}$ such that $x^2|1-2\delta|-2|\delta|^3/3\geq 1/12=\varepsilon$. (you can choose $x(\delta)=\dfrac{\sqrt{1/12+2|\delta|^3/3}}{\sqrt{|1-2\delta|}})$

Therefore, $f$ and $\varepsilon>0$ are such that for every $\delta>0$ there exists $x=x(\delta)\in\mathbb{R}$ satisfying $|f(x)-\int_{-\delta}^\delta f(x+u)du|\geq \varepsilon$


问:
Prove the convergence of $\sum_{i=3}^\infty \ln(1- \frac{1}{i} - \frac{1}{i \ln i})$
As part of the question that I'm solving, I need to prove the convergence of the sum $$\sum_{i=3}^\infty \ln(1- \frac{1}{i} - \frac{1}{i \ln i})$$
I feel that 1 possible way of approaching the problem is using the Taylor expansion of $\ln (1-x)$, $$\ln (1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$$
which will give us $$\ln(1- \frac{1}{i} - \frac{1}{i \log i})=-\sum_{n=1}^{\infty} \frac{1}{n} (\frac{1}{i} + \frac{1}{i \ln i})^n$$
But I don't know how to continue from here. Any help is appreciated,
thanks in advance!

Edit: changed from $\sum_{i=2}^\infty \ln(1- \frac{1}{i} - \frac{1}{i \ln i})$ to $\sum_{i=3}^\infty \ln(1- \frac{1}{i} - \frac{1}{i \ln i})$, since otherwise the argument to \ln can be negative.

答:
hint

You just need the limit comparison test:

$$\ln(1-X)\sim -X \;(X\to 0)$$

So,

$$\ln\Bigl(1-\frac 1i-\frac{1}{i\ln(i)}\Bigr)$$
$$\sim \frac{- 1}{i}\Bigl(1+\frac{1}{\ln(i)}\Bigr)$$

$$\sim \frac{- 1}{i }\;\;(i\to \infty)$$

The series will diverge because  $-\frac 1i $ has a constant sign and $ \sum \frac 1i $ is divergent.


答:
Your solution is correct if you restrict $a \ne 0$ in the case
$$
 z = a + 0i
$$
and $|a| \le 1$ in the case
$$
 z = a \pm i \sqrt{1-a^2} \, .
$$

So for nonzero $z$,  $z + 1/z$ is real if $z$ is real or $|z| = 1$.

As alternative solution you can use that a complex number is real if and only if it equal to its complex conjugate:
$$
 z + \frac 1 z \in \Bbb R \\
\iff z + \frac 1 z = \overline z + \frac{1}{\overline z} \\
\iff z^2 \overline z + \overline z = z \overline z^2 + z \\
\iff (z - \overline z)(|z|^2 - 1) = 0 \, .
$$


问:
Is the series converging?
$$\sum_{n=1}^{\infty} n^{1/3}*e^{-n^{1/2}} $$ How to prove the convergence of this series?  Without using an integral comparison criterion

答:
In fact $\sum_{n=1}^{\infty} n^a e^{-n^{b}}$; $a,b$ nonzero real numbers,
converges for all choice of nonzero real $a$ and real positive $b$. Including $a=1/3$ and $b=1/2$.

Taking logarithms:

$$e^{n^{-b}} = n^{-\frac{n^{b}}{\ln n}}.$$ One can check that $\frac{n^{b}}{\ln n} > 2-a$ for $n \ge n'(a,b)=4^{5+5/b^2+|a|}$, for example. Thus,

$$0 \le \sum_{n=1}^{\infty} n^a e^{-n^{b}} \le \sum_{n=1}^{n'(a,b)}n^a e^{-n^{b}} +\sum_{n \ge n'(a,b)} n^{-2}.$$

Can you finish from here.

Title: Is the series converging?


答:
If $z$ is real, you know what happens.

Now if $z$ is not real, the vectors $z$, $1/z$, and $z+1/z$ form a triangle. You would like the complex angle of $z+1/z$ to be $0$.

The vectors $z$ and $1/z$ have complex angles that are negative of each other. So now this triangle has two equal angles with $z+1/z$ representing the edge between them. It's an isosceles triangle, and it follows that the magnitude of $z$ equals that of $1/z$.

So the only $z$ for which this can hold are the $z$ on the unit circle. And you can then verify that for all such complex numbers, the sum is real: $e^{it}+e^{-it}=2\cos(t)$.


问:
I need help or some advice to solve this exercise about functions random variables.
I've thought about it for a while using the cumulative distribution, but I've not concluded.

Three people $A$, $B$ and $C$ arrive at the same time at a telephone
booth that has two telephone sets. Both devices are used immediately
by A and B. Person C replaces the first person to end their call and
each person leaves the telephone booth after their call is finished.
Let $X_1$, $X_2$ and $X_3$ be the connection times of $A$, $B$ and
$C$, respectively. Suppose that $X_1$, $X_2$ and $X_3$ are independent
and identically distributed random variables with exponential
distribution with parameter $\lambda$.  Determine the probability
density function of $Z = max \{X_1, X_2\} - min \{X_1, X_2\}$ and
calculate $P(Z<X_3)$.

答:
I will try to give you a start: $X_1, X_2 \stackrel{iid}{\sim}\mathsf{Exp}(\lambda).$
Two 'parameterizations' of exponential distributions are in common use: rate and mean.
I will assume $\lambda$ is the rate (so that the mean is $1/\lambda.$

Then let $W = \max(X_1,X_2).$ Then
$$F_W(t) = P(W \le t) = P(X_1\le t,\,X_2\le t) = P(X_1\le t)P(\,X_2\le t)
=(1 - e^{-\lambda t})^2,$$
for $t > 0,$ from which you can find $F_W(t)$ the density function $f_w(t)$. and $E(W).$

Somewhat similarly, for  $V = \min(X_1,X_2),$ you have
$$1 - F_V(t) = P(V > t) = P(X_1 > t, X_2 > t),$$ from which you can deduce that
$V \sim \mathsf{Exp}(2\lambda).$

Here is a simulation for $\lambda = 2,$ which gives good approximations of the
distributions of $W,V,$ and $Z.$ You can check your exact analytic (and perhaps intuitive) answers against these
approximate ones. [With a million iterations, it is reasonable to expect
about three-place accuracy.]

```<code>set.seed(111)
x1 = rexp(10^6, 2);  x2 = rexp(10^6, 2)
v = pmin(x1, x2);  w = pmax(x1, x2)
z = w - v

summary(v); sd(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000001 0.071874 0.173258 0.250102 0.346770 3.383253
[1] 0.2505        # both mean and sd exactly 1/4
summary(w); sd(w)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000295 0.346604 0.613705 0.750089 1.004757 7.450897
[1] 0.5595383     # not exponential, mean not = sd
summary(z); sd(z)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.0000  0.1435  0.3463  0.5000  0.6927  7.0444
[1] 0.5001754

par(mfrow=c(1,3))
 hist(v, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 4), add=T, col="red", lwd=2)
 hist(w, prob=T, br=40, col="skyblue2")
 hist(z, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 2), add=T, col="red", lwd=2)
par(mfrow=c(1,1))
</code>```

Finally, what is the (non-exponential) distribution of the sum of two exponential distributions with different rates? Consider these additional simulation results:

```<code>x3 = v + rexp(10^6, 2)
summary(x3); sd(x3)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000316 0.346711 0.614066 0.750727 1.006107 7.288232
[1] 0.5598703
mean(z < x3)
[1] 0.66665   # aprx P(Z < X3)
mean(w > x3)
[1] 0.500151  # aprx P(W > X3)
</code>```

Body:
I will try to give you a start: $X_1, X_2 \stackrel{iid}{\sim}\mathsf{Exp}(\lambda).$
Two 'parameterizations' of exponential distributions are in common use: rate and mean.
I will assume $\lambda$ is the rate (so that the mean is $1/\lambda.$

Then let $W = \max(X_1,X_2).$ Then
$$F_W(t) = P(W \le t) = P(X_1\le t,\,X_2\le t) = P(X_1\le t)P(\,X_2\le t)
=(1 - e^{-\lambda t})^2,$$
for $t > 0,$ from which you can find $F_W(t)$ the density function $f_w(t)$. and $E(W).$

Somewhat similarly, for  $V = \min(X_1,X_2),$ you have
$$1 - F_V(t) = P(V > t) = P(X_1 > t, X_2 > t),$$ from which you can deduce that
$V \sim \mathsf{Exp}(2\lambda).$

Here is a simulation for $\lambda = 2,$ which gives good approximations of the
distributions of $W,V,$ and $Z.$ You can check your exact analytic (and perhaps intuitive) answers against these
approximate ones. [With a million iterations, it is reasonable to expect
about three-place accuracy.]

set.seed(111)
x1 = rexp(10^6, 2);  x2 = rexp(10^6, 2)
v = pmin(x1, x2);  w = pmax(x1, x2)
z = w - v

summary(v); sd(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000001 0.071874 0.173258 0.250102 0.346770 3.383253
[1] 0.2505        # both mean and sd exactly 1/4
summary(w); sd(w)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000295 0.346604 0.613705 0.750089 1.004757 7.450897
[1] 0.5595383     # not exponential, mean not = sd
summary(z); sd(z)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.0000  0.1435  0.3463  0.5000  0.6927  7.0444
[1] 0.5001754

par(mfrow=c(1,3))
 hist(v, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 4), add=T, col="red", lwd=2)
 hist(w, prob=T, br=40, col="skyblue2")
 hist(z, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 2), add=T, col="red", lwd=2)
par(mfrow=c(1,1))

Finally, what is the (non-exponential) distribution of the sum of two exponential distributions with different rates? Consider these additional simulation results:

x3 = v + rexp(10^6, 2)
summary(x3); sd(x3)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.000316 0.346711 0.614066 0.750727 1.006107 7.288232
[1] 0.5598703
mean(z < x3)
[1] 0.66665   # aprx P(Z < X3)
mean(w > x3)
[1] 0.500151  # aprx P(W > X3)


问:
Are soluble/nilpotent lie algebras always isomorphic to a subalgebra of upper triangular matrices?
For soluble lie algebras, every representation has some basis such that the image is a subalgebra of upper triangular matrices (if you assume the field is algebraically closed). Then by Ado's theorem, over (an algebraically closed field of) characteristic zero, any soluble lie algebra is a subalgebra of some upper triangular matrices.

For nilpotent lie algebras, (it seems to me) you can't quite apply the same idea directly, because for a representation of a nilpotent lie algebra, you can only conclude there is a basis where the image is represented by strictly upper triangular if you additionally assume that all elements act by nilpotent endomorphisms too (but here you don't need to assume the field is algebraically closed). However, there is a form of Ado's theorem for nilpotent lie algebras, which says that over characteristic $0$, you can always find a faithful representation such that this additional assumption is true, and so nilpotent lie algebras over characteristic $0$ are a subalgebra of some strictly upper triangular matrices.

Firstly, I would like to know if this reasoning is correct and if so that this is the easiest way of getting to these conclusions.

Secondly, it seems that Ado's theorem holds for characteristic $p$ by a proof by Iwasawa. Is there a good reference for this? Can you still take the faithful representation of a nilpotent lie algebras to act by nilpotent endomorphisms?

If not then if $k$ has characteristic $p$, is there a nilpotent lie algebra not isomorphic to any subalgebra of strictly upper triangular matrices?

Additionally, if $k$ is not algebraically closed of arbitrary characteristic, then is there a soluble lie algebra not isomorphic to any subalgebra of upper triangular matrices? I'm sure that Lie's Theorem doesn't hold over non-algebraically closed fields, but that doesn't mean that this is automatically the case.

Thanks, please let me know if I'm missing anything obvious!

答:
I don't follow your particular reasoning, but here's a proof (Ado being granted) that every nilpotent Lie algebra in char zero can be represented by strictly upper triangular matrices, over a field $K$ of characteristic zero.

First suppose $K$ is algebraically closed. Using Ado, take a faithful $\mathfrak{g}$-module $V$. Then (see e.g. Bourbaki), $V=\bigoplus V_\chi$, where $\chi$ ranges over $\mathrm{Hom}(\mathfrak{g},K)$ (Lie algebra homomorphisms) and $V_\chi=\{v\in V:\forall g\in\mathfrak{g}:(g-\chi(g))^{\dim(V)}v=0\}$. That is, on $V_\chi$, the action is scalar + nilpotent, the scalar being given by $\chi$. Then we can modify the module structure on $V_\chi$, to remove the scalar (i.e., act with the same matrices, but with zero diagonal). The resulting representation is unchanged on $[\mathfrak{g},\mathfrak{g}]$, hence its kernel has trivial intersection with $[\mathfrak{g},\mathfrak{g}]$. Hence, adding an abelian nilpotent action, we can produce a faithful nilpotent module.

Next, if $K$ is arbitrary (of char. zero), start with a faithful representation; on a finite extension $L$ of $K$ it can be made upper triangular, and the previous argument applies. Eventually we have a faithful nilpotent $L\otimes\mathfrak{g}$-module, and by (Weil) restriction of scalars this yields a faithful $\mathfrak{g}$-module.

I don't think this adapts to char $p$. However, let me emphasize that Ado in characteristic $p$ is known as much easier than its char 0 counterpart (due to the enveloping algebra being PI). I think it's covered in Jacobson's book. I'd actually guess that the proof actually provides directly a nilpotent module when the Lie algebra is nilpotent.


问:
Tell me which of $691$ and $697$ is prime (no calculator)
For two numbers, $691$ and $697$, tell me which (or both) is prime.

Without using calculator, what is the strategy for tackling this? I tried all the numbers up to 8 which are easy to test. How do you deal with better numbers?

答:
$$29^2 - 697 = 144 = 12^2$$
so $$ 697 = 29^2 - 12^2 = 17 \cdot 41 $$

This is Fermat's method, it works quickly if there are two factors that are pretty close together. Beginning with $27^2 = 729,$  subtract each number from the square. If the result is a square you get two factors. Otherwise take the next square...


问:
Centralizer of one element on a compact connected Lie group
Exercise 16.2 from Daniel Bump - Lie Groups.

Let $G$ be a compact connected Lie group and let $g\in G$. Show that
the centralizer $C_{G(g)}$ of $g$ is connected.

I have some problems verifying this, i have tried to use that in that case the exponential map is surjective, and things of maximal torus, but i have not achieved it, i would appreciate some answer.

答:
This is not true in general. Consider $SO(3)$ and the matrix $$ \begin{bmatrix}
  1 & 0 & 0\\ 0 & -1 & 0 \\ 0&0&-1
\end{bmatrix} $$
Then $C_G(g)=\left\{ \begin{bmatrix}
  a & 0 & 0\\ 0 & \pm \cos \theta & \mp \sin \theta \\ 0&\sin\theta&\cos\theta
\end{bmatrix}: a\in \{\pm 1\}, \theta\in \mathbb{R}  \right\}   $.

We can fix the problem statement by adding that $G$ is simply connected. Then we can argue using a long exact sequence in homotopy groups.


问:
How to explicitly show $2$ angles are distinct in this circle geometric construction?
I'm solving a homework problem and I need to explicitly show that 2 angles are distinct. The problem is as follows:

Given a circle of radius $4r$ which is tangent to a circle of radius $2r$, take a line parallel to the axis which connects both circle centers that is $r$ distance apart from said axis. We call the $2$ points from the intersection of this secant line with both circles that are furthest apart to be $A$ and $B$, and we call the other pair of points $C$ and $D$. Show that the angle between $A,B$ and the tangency point is greater than the angle between $C,D$ and the tangency point.

I made the following diagram of the situation.

It seems fairly obvious that the triangle$\triangle PAB$ contains the smaller triangle $\triangle PDC$, but I don't know how to justify this. I was thinking of explicitly calculating the equations for both circles and the secant line to calculate the coordinates of the points, but it seems like overkill to me. Does anyone know a simple argument I could use to show that $\angle APB \neq \angle CPD$?

答:
It does seem like overkill, but unless you are required to use the numbers given, which my figure reflects but I do not make use of, an explicit proof might be to draw the tangent $PK$.

Then since $K$ lies between $D$ and $C$,$$\angle APK>\angle DPK$$Likewise,$$\angle BPK>\angle CPK$$Therefore by addition$$\angle APB>\angle DPC$$


问:
Several questions about d.e $y'(t)+\sin^2(t) y(t)= \left\{\begin{matrix} 6-2t & , & t\in [0,3] \\ 0 & , & t>3 \\ \end{matrix}\right .$
True or false? Let the linear differential equation of first class:
$$y'(t)+\sin^2(t) y(t)= \left\{\begin{matrix} 6-2t & , & t\in [0,3] \\  0 & , & t>3 \\  \end{matrix}\right .$$
$a)$ There does not exist a solution such that $y(2020)=1010.$

$b)$ For every solution $y$ of the equation holds $\displaystyle\lim_{x\to \infty} y(x)=0$.

$c)$ There's exist a $y$ oscillating solution (i.e. $y$ has infinite roots).

$d)$ Every solution $y$ is finally constant (i.e there's exist a $t_0$ such that, for all $t\geq t_0$ $y(t)=const.)$

I understand that it doesn't need to solve the equation to answer these quastions (and I can't solve it from the other side because we can not compute the integral) but its difficult for me to give the correct answer in these questions only from the equation. For example to $(d)$ if every solution $y$ is finally constant to $[0,+\infty)$ then  there's exist a $t_0$ such that, for all $t\geq t_0$ $y(t)=c,$ thus $y'(t)=0,t\geq t_0$. Then, by replacing to equation (and w.l.o.g supposing that $t_0> 3$) we have:
$0+c\sin^2 (t)=0, \ \forall t\geq t_0$ then what? Am I missing something? The same problem appearing to all other quastions, I end up to a point that I can't continue. Any help? Thanks

答:
It is easy to see if $t\in[0,3]$, the solution the equation is
$$ y(t)=e^{\frac14(-2t+\sin(2t))}\bigg(y_0+\int_0^te^{\frac14(2s-\sin(2s))}(6-2s)ds\bigg) $$
and if $t>3$, the solution the equation is
$$y(t)=e^{\frac14(-2t+\sin(2t))}(y_0+a) \tag{1}$$
where
$$ a=\int_0^3e^{\frac14(2s-\sin(2s))}(6-2s)ds. $$
Since $2020>3$, then
$$ y(2020)=e^{\frac14(-4040+\sin(4040))}(y_0+a)=e^{-1010+\frac14\sin(4040)}(y_0+a). $$
Note that $e^{-1010+\frac14\sin(4040)}$ is very small positive number. If $y(2020)=1010$, the initial condition $y_0$ satisfies
$$ y_0=1010e^{1010-\frac14\sin(4040)}-a $$
which is a very big number. This answers (a). From (1), it is easy to see $\lim_{t\to\infty}y(t)=0$ which is (b). Clearly $y(t)$ is not oscillating and not finally constant, which answers (c) and (d).

Update: If $y_0+a>0$, then (1) implies $y(t)>0$ and if $y_0+a>0$, then (1) implies $y(t)<0$. Thus $y(t)$ is not changing signs and hence is not oscillating. This answers (c).


问:
Let $f: (\mathbb{Z}_{24}, +) \rightarrow (\mathbb{Z}_{36}, +)$ be a group homomorphism. How many elements are there in the kernel of $f$?
Let  $f: (\mathbb{Z}_{24}, +) \rightarrow (\mathbb{Z}_{36}, +)$
be a group homomorphism. If the image of $f$ contains exactly $2$ elements, how many elements are there in the kernel of $f$?

What I know:

Let $f : G \to H$ be a group homomorphism between two finite groups. Then we
have $$|G| = |\ker(f )| \times |{\rm im}(f )|$$
$$|G|= |\ker(f )|  \times  2$$

However I don't know how to work out $|G|$ for my question.

答: By the first isomorphism theorem, $G/\rm{ker}f\cong\rm{im}f$.  So, $24/|\rm{ker}f|=2$.  Thus $|\rm{ker}f|=12$.


问:
Proving that the improper integral $\int_1^\infty\frac{f(x)}{\int_0^x f(t)dt}dx=\infty$
Let f be a positive continious function such that $\displaystyle \int_0^\infty f(x)dx=\infty$. Prove that $$\int_1^\infty\frac{f(x)}{\int_0^x f(t)dt}dx=\infty$$</p>

答:
Let $F(x):=\int_0^xf(t)dt$. Then $F$ is positive, $C^1$, and $\displaystyle\lim_{A\rightarrow +\infty}F(A)=\int_0^{+\infty}f(t)dt=+\infty$. Hence
$$
\int_1^A\frac{f(x)}{\int_0^xf(t)dt}dx=\int_1^A\frac{F'(x)}{F(x)}dx=\log F(x)\Big|_1^A=\log F(A)-\log F(1)\longrightarrow +\infty
$$
when $A$ tends to $+\infty$.</p>


问:
Rate of increase of depth of water pumped into conical vessel
I cannot get the answer shown in my text book for this problem:

Water runs into a conical vessel fixed with its vertex downwards at the rate of $3 \pi$ cubic cm per second, filling the vessel to a depth of 15 cm in a time of one minute. Find the rate at which the depth of water in the vessel is increasing when the water has been running for 7.5 seconds.

I have said let V = volume, h = depth of water and r = radius of surface of water:

$V = \frac{\pi r^2h}{3}$

When $V = 180 \pi, h = 15$ so $180 \pi = \frac{\pi r^2h}{3}, 540 = r^2h = 15r^2$ so $r = 6$

Therefore the ratio of r to h is 6:15 = 2:5, so $r = \frac{2}{5}h$

So $V = (\frac{2}{5}h)^2\frac{\pi h}{3} = \frac{4h^3\pi}{75}$ and $\frac{dV}{dh} = \frac{12h^2\pi}{75}$

Now to find $\frac{dh}{dt}$

After 7.5 seconds $V = 7.5$ x $3\pi = 22.5$

I need to find h, so:

$22.5 = \frac{4 \pi h^3}{75}$ so $h = 5.12$

$\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt}$

$\frac{dV}{dt} = 3\pi, \frac{dV}{dh} = \frac{12h^2\pi}{75} $

but when I substitute these values, plus that for h, I do not get the correct answer, which is $\frac{1}{3}$ cm per sec.

Where am I going wrong?

答:
You method of solution is correct, but you have miscalculated the value of $h$ when $t=7.5$.  The correct calculation is

$h^3=\frac{22.5*75}{4}$ so $h=7.5$ and it follows that $\frac{dV}{dt}=1/3$.


问:
If we define $x := y$, is it true that $P(x) \iff P(y)$ for any property $P$?
The question should be trivial, but I still can't get my head around it:

Question. Suppose we have an object $y$ belonging to the type $T$ (e.g $y$ may be an integer, matrix, set etc). Now suppose we define $x := y$. Is it true for arbitrary property $P$, $P(x)$ is true if and only if $P(y)$ is true?

In an attempt to further clarify my question, let me give more concrete example (definitions below are taken from the Terence Tao's Real Analysis book):

Definition 1 (Integer).  We define integer to be an expression of the form $a $$−−$$b$ , where $a,b$ are natural numbers. Furthermore, integers $a$$−−$$b$ and $c$$−−$$d$ are equal, if and only if $a+d = c + b$.

Now suppose that without verifying whether the substitution axiom holds for the aforementioned definition of equality, we proceed by defining a sum of integers:

Definition 2 (Sum of integers). For arbitrary integers $a$$−−$$b$ and $c$$−−$$d$, we define the sum of an integers as $(a$$−−$$b)+(c$$−−$$d) := (a+c) $$−−$$ (b+d)$

Based on the definitions above, can I safely assume that:

$(a$$−−$$b)+(c$$−−$$d)$ is an integer

$P\bigr((a$$−−$$b)+(c$$−−$$d)\bigl) = P\bigr((a+c) $$−−$$(b+d)\bigl) $ for all functions and operations $P$

?

In other words, for arbitrary mathematical statement or expression, can  I substitute $(a+c) $$−−$$ (b+d)$ for $(a$$−−$$b)+(c$$−−$$d)$ (and vice versa) without  either changing the truth value or the meaning?

答: Yes, this is the case. However, the rules governing equality are "one level lower" than the more "mathematical" axioms like commutativity, associativity, etc. - they're logical rules, at the same level as modus ponens, "From '$A$ and $B$' we can deduce $A$ and we can deduce $B$," and so forth. This level of logic is often not made explicit in texts, which can lead to some confusion; any good book on logic however will present them explicitly.


问:
Higher Order Implicit Differentiation
How do you find the $y'''$ of $y^2 + x^3 = 4$?

Title: Higher Order Implicit Differentiation

Body:
$$
2yy'+3x^2=0\\
2y'^2+2yy''+6x=0\\
4y'y''+2y'y''+2yy'''+6=0\\
y'''=-\frac{3+3y'y''}{ y}
$$


问:
$4$ points in order $A,B,C,D$ lie on a circle with the extension of $AB$ meeting the extension of $DC$ at $E$ and that of $AD$ and $BC$ at $F$.

$4$ points in order $A,B,C,D$ lie on a circle with the extension of $AB$ meeting the extension of $DC$ at $E$ and that of $AD$ and $BC$ at $F$. Let $EP$ and $FQ$ be tangents to this circle with points of tangency $P$ and $Q$ respectively. Suppose $EP = 60$,$FQ = 63$. Find $EF$.

What I Tried: Here's a picture :-

Used Power of a Point to get :-
$$EA * EB = ED * EC = 3600$$
$$FA * FD = FB * FC = 3969$$
Yet this does not help to proceed and neither finds the value of $EF$ .

I thought of Pythagoras Theorem once. Join OP and OQ and let them be $x$ each . Then we can find $OE$ and $OF$ in terms of $x$ . Not sure if that helps in the problem though.

Can anyone help me? Thank You.

答:
Hatton's Projective Geometry, pg 156 gives two versions of a theorem:

[T]he square of the distance between a pair of conjugate points with
respect to a circle is equal to the sum of the powers of the points.

and

If a quadrangle be inscribed in a circle, the square of the distance
between two of its diagonal points external to the circle equals the
sum of the square of the tangents from these points.

The second version uses your setup.

Here's a screen capture of the theorems and their proofs.

The proofs should be easy to follow, except for perhaps the last sentence.  A more explicit version would be

$$
\begin{align}
EF^2 &= (EK+KF)^2 \\
     &= EK^2+2EK\cdot FK+FK^2 \\
     &= EK(EK+KF)+(EK+FK)FK \\
     &= EK\cdot EF + EF\cdot FK \\
     &= EC\cdot ED + FB\cdot FC \text{ (using powers of $E,F$ wrt circles $CDF,BCE$)}\\
     &= \text{sum of powers of $E$ and  $F$ wrt original circle $ABCD$}
\end{align}
$$


问:
Area of triangle bounded by a tangent to a curve

Find a curve such that the surface of the triangle bounded by the line going through the tangent point and perpendicular to the x-axis and the tangent line to the graph is equal to $a^2$.

I didn't understand the question at first glance, and I found related answers like this one - but this question also assumes the triangle is bounded by a perpendicular line to the $x$ axis.

This is what I got so far, but I don't think I got the DE entirely correct. I think I need some help with interpreting the question.

The tangent line to the curve at any point $x$ is given by:

$y-xy^{\prime}=0$

This line intersects the $x$ axis at some point $x_{0}$ s.t. $y(x_0)=0$.
Then, for any $x>x_{0}$, a $\perp$ line of height $y\left(x\right)$ intersects the tangent line.

So the triangle is defined by:

$A\left(x_{0},0\right)$

$B\left(x,0\right)$

$C\left(x,y-xy^{\prime}\right)$

Thus, the area of $\triangle ABC$ is given by:

$\frac{1}{2}\left(x-x_{0}\right)\left(y-xy^{\prime}\right) =a^{2}$

答:
Let the point of tangency on the curve $y=g(x)$ be $(X,g(X))$.

Equation of tangent is $\frac{y-g(X)}{x-X}=g'(X)$.

$x-$intercept of tangent is $X-g(X)/g'(X)$.

The area of the triangle in question is $\frac12\times|X-(X-g(X)/g'(X))|\times|g(X)|=a^2$.

This gives the ODE$$|g'|=g^2/2a^2$$

Since we want to find any one solution, we let $g'\ge0$. Thus$$\int\frac{dg}{g^2}=-\frac1g=\frac x{2a^2}+c$$You can assume $g'\le 0$ and you would get $1/g=x/2a^2+c_1$ which is also admissible.