问:
On the Hurwitz Zeta Function
In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like
$\zeta(2k+1,\frac{1}{4})=C(k)\zeta(2k+1)$. With $C(k)$ some elementary function of $k$.

After some basic calculations I found the following

$\zeta(2k+1,\frac{1}{4})=2^{2k+1}(2^{2k+1}-1)\zeta(2k+1)-\zeta(2k+1,\frac{3}{4})$

but I don't know what to do about $\zeta(2k+1,\frac{3}{4})$. Trying the same method on this function, I stumble upon $\zeta(2k+1,\frac{1}{8})$. And I'm going in a loop getting nothing like my prof. said. Also I've looked through books and articles on the Hurwitz function and found nothing.

答:
It can't be written as a scaled up version of the zeta function by elementary functions, however
$$\zeta(2k+1,\frac{1}{4})=2^{2k}(2^{2k+1}-1)\zeta(2k+1)+\frac{(-4)^kE_{2k}\pi^{2k+1}}{2(2k)!}$$

Where $E_k$ are euler numbers, with the first few

$E_0=1$

$E_2=-1$

$E_4=5$

$E_6=-61$

Title: On the Hurwitz Zeta Function


问:
Is there a non-negative finite measure $\mu$ on a $\sigma$-compact metric space $X$ s.t. $\int_X f d \mu >0$?
Is there a non-negative finite measure $\mu$ on a $\sigma$-compact metric space $X$ s.t. for any $f \in C_c(X), f \geq 0, f \not\equiv 0$ we have $\int_X f d \mu >0$?

I think it is right, but I have no idea to construct such a measure.

I have tried to consider the condition "$\sigma$-compact", which means $X= \cup_{n=1}^{\infty} K_n$ for $K_n$ compact. Note that $X$ is a metric space. I wanted to construct such a measure by $\rho(x,K_n)$.

I am stuck here.

答:
Choose a dense set $\{x_n\}_n$ in $X$ and let
$$
  \mu = \sum_n {\delta_{x_n}\over 2^n},
  $$
where the Dirac measure $\delta_{x}$ is defined by $\delta_{x}(E)=0$ or $1$, acccording to whether $x\in E$ or not.

ContentLicense: CC BY-SA 4.0


问:
Combinatorial Proof for $\binom{3n}{n}$
I am trying to prove the following equality:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \binom{3n}{n}$

This is what I have so far:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \sum_{k=0}^{n}\binom{n}{k} \sum_{l=0}^{n}\binom{n}{l} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{n} \cdot 2^{n} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{2n} \cdot \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

I got stuck here, and I'm having trouble manipulating the second term. I would appreciate any help on this proof!

答:
Here's a combinatorial proof:

Let $X$ be a set of $3n$ elements. Let's count the number of subsets of $X$ that have exactly $n$ elements. This is of course $\binom{3n}{n}$. However, here's another way to count them that yields your LHS:

Decompose $X$ into three disjoint subsets $X=X_1\cup X_2\cup X_3$ such that $|X_i|=n$ for $i=1,2,3$. A subset $U\subseteq X$ now corresponds to a triple of subsets $U_i\subseteq X_i$ for $i=1,2,3$. When $U$ has $n$ elements, we have $|U_3|=n-k-l$ for $k=|U_1|$ and $l=|U_2|$. Hence, counting all the triples $U_1,U_2,U_3$ that yield $n$ element subsets $U\subseteq X$ we get
$$
\sum_{k=0}^n \sum_{l=0}^n \binom{n}{k} \binom{n}{l} \binom{n}{n-k-l}.
$$
Since $\binom{n}{n-k-l}=\binom{n}{k+l}$ (by considering complements in $X_3$), this proves your identity.

Title: Combinatorial Proof for $\binom{3n}{n}$


问:
Existence of Fredholm operators of a given index in a non-separable Hilbert space
Let $H$ be an infinite-dimensional Hilbert space. The (Fredholm) index of a bounded operator $T : H \to H$ is defined as $$ \mathrm{index}(T) = \dim \ker(T) - \dim \mathrm{coker}(T), $$ whenever the kernel and cokernel of $T$ are finite-dimensional. Operators satisfying these conditions are called Fredholm operators. When $H$ is separable (equivalently, has a countably infinite orthonormal basis), $F_n(H) \neq \emptyset$ for all $n \in \mathbb Z$, which can be seen by picking a countable orthonormal basis and calculating the index of left and right shift operators with respect to this basis.

If $H$ is non-separable, is it possible that for some $n \in \mathbb Z$, the set $F_n(H)$ of Fredholm operators of index $n$ is empty?

Edit: I was able to simplify the problem with a little algebra. Any bounded invertible operator is a Fredholm operator, so $F_0(H) \neq \emptyset$. If there is a Fredholm operator $T$ of index $1$, then, given an integer $n \neq 0$, we can (much like the separable case with the shift) apply powers of $T$ and its adjoint to find an operator of index $n$. Specifically,

If $n > 0$, then $$ \mathrm{index}(T^n) =  \mathrm{index}(T \circ \cdots \circ T) = n \cdot \mathrm{index}(T) = n.$$

If $n < 0$, then $$ \mathrm{index}(T^{*(-n)}) = -n \cdot \mathrm{index}(T^*) = -n \cdot (-1) = n. $$

So my question will be rephrased as follows: If $H$ is non-separable, is it possible that $F_1(H) \neq \emptyset$?

答:
If $H$ is infinite dimensional then it may be written as $H_0\oplus H_1$, with $H_0$ separable infinite dimensional,
and for any Fredholm operator $T$ on $H_0$ one has that
$T\oplus I$ is Fredholm on $H$ and
$\text{ind}(T\oplus I)=  \text{ind}(T)$.

Picking a Fredholm operator $T$ on $H_0$ with a given index $n$, one then gets a Fredholm operator $T\oplus I$ on $H$ with the same index.


问:
How to find the dot product using the law of cosines
I'm working with the following problem:

We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$

Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.

$$AC^2=AB^2 +BC^2 -2AB\cdot BC\cos x$$
$$\implies AC^2=3^2 +2^2 -2\cdot2\cdot3\cos 60^\circ$$
$$\implies AC^2=9 +4 -12\cdot0.5$$
$$\implies AC^2=13-6=7$$
$$\implies AC=\sqrt 7.$$

My next step is to calculate the angle $BAC$; we should be able to use the law of cosines here as well:

$$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cos x$$
$$\implies4=9+(\sqrt 7)^2 -2\cdot3\sqrt 7\cos x$$
$$\implies4=9+7 -6\sqrt 7\cos x$$
$$\implies-2=-6\sqrt 7\cos x$$
$$\implies\frac{1}{3}=\sqrt 7\cos x$$
$$\implies\frac{1}{3\sqrt 7}=\cos x.$$

If we want to calculate our dot product using only the vector lengths, we would use the fact that $A\cdotp B=|A||B|\cos x$, which in this case would mean that:

$$AB \cdot AC=3\sqrt7\frac{1}{3\sqrt 7}=1,$$

which is quite wrong since the answer is supposed to be $12$.

Can someone please tell me where I've made a mistake?

In the proposed solution another method seem to be used. The textbook claims that $AB \cdotp AC = AB \cdotp (AB + BC)=AB \cdotp AB + AB \cdotp BC = |AB|^2 + |AB||BC|cos(60)=9+6*\frac{1}{2}=12$

答:
$$\vec{AB}=\vec{AC}+\vec{CB}$$

and

$$\vec{AC}\cdot\vec{AB}=\vec{AB}^2-\vec{CB}\cdot\vec{AB}=9-3\cdot2\cdot\cos60°=6.$$


问:
Finding representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Z}_p)$.
The representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Q})$ can be found as mentioned in https://www.math3ma.com/blog/rational-canonical-form-example-1

However, if I think of the representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Z}_p)$, where $p(>2)$ is a prime, then how can I get the conjugacy classes?
What differences should be present when obtaining the conjugacy classes for matrices of finite order in $GL_2(\mathbb{Z}_p)$?

Can someone please help to obtain the results similar to that in the above website, where the matrix satisfies the polynomial $x^n-1$ (same as in the website) for $n \geq 2$ (and $n$ is an odd prime less than $p$, $n|p^2-1$)?

Thanks a lot in advance.

答:
I am assuming that $\Bbb{Z}_p$ are the $p$-adic integers.

For $p\ge 5$ and $A\in GL_2(\Bbb{Z}_p)$ of finite order then the $GL_2(\Bbb{Z}_p)$-conjugacy class is determined by its characteristic polynomial $f\in \Bbb{Z}_p[x]$.

$A$ is diagonalizable over the splitting field,

Either $f$ has a double root and $A$ is diagonal-scalar

Or $f=(x-\mu)(x-\xi)$ where $\mu,\xi$ are two distinct roots of unity in $\Bbb{Q}_p(\zeta_{p^2-1})$. Thus $f\bmod p \in \Bbb{F}_p[x]$ is separable.

In that case $A\bmod p$ has a cyclic vector $v$, ie. such that $v,Av$ is a basis of $(\Bbb{F}_p)^2$

(here any $v$ which is not an eigenvector)

taking any $u\in (\Bbb{Z}_p)^2,u\equiv v\bmod p$ we get that $u,Au$ is a $\Bbb{Z}_p$-basis of $(\Bbb{Z}_p)^2$.

For another matrix $B\in GL_2(\Bbb{Z}_p)$ with the same charasteristic polynomial, $w,Bw$ is a $\Bbb{Z}_p$-basis of $(\Bbb{Z}_p)^2$, then the matrix $P$ sending $u,Au$ to $w,Bw$ is in $GL_2(\Bbb{Z}_p)$ and satisfies $B=PAP^{-1}$.

If you meant the finite field with $p$-elements then I almost answered too, the only remaining case is when $A\bmod p$ is not diagonalizable, $f=(x-a)^2$, $A-aI$ is nilpotent, from its kernel and an element not in the kernel we get $A = Q \pmatrix{a&b\\0&a}Q^{-1}$ with $b\ne 0$ and $Q\in GL_2(\Bbb{F}_p)$, and $\pmatrix{b&0\\0& 1}^{-1}\pmatrix{a&b\\0&a}\pmatrix{b&0\\0& 1}=\pmatrix{a&1\\0&a}$ whose conjugacy class are every element with characteristic polynomial $(x-a)^2$ and different from $\pmatrix{a&0\\0&a}$.

For the size of the $GL_2(\Bbb{F}_p)$ conjugacy classes, count the number of matrices with a given characteristic polynomial. Given $charpoly(\pmatrix{a&b\\c&d})=x^2-rx+s$ we must have $d=r-a,a(r-a)-s=bc$

if $r^2-4s$ is not a square then $p$ choices for $a$ give $bc\ne 0$ so we have $p-1$ choices for $b$.

If $r^2-4s$ is a non-zero square. $p-2$ values of $a$ give $bc\ne 0$ thus $p-1$ choices for $b$.

$2$ values of $a$ give $bc=0$ thus $2p-1$ choices for $b,c$.

if $r^2-4s=0$ then $p-1$ values of $a$ give $bc\ne 0$ thus $p-1$ choices for $b$.

$1$ value of $a$ gives $bc=0$ thus $2p-1$ choices for $b,c$.

Substract one ( the conjucacy class of the scalar matrix $\pmatrix{r/2&0\\0&r/2}$).


Title: Combinatorial Proof for $\binom{3n}{n}$

Body:
I am trying to prove the following equality:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \binom{3n}{n}$

This is what I have so far:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \sum_{k=0}^{n}\binom{n}{k} \sum_{l=0}^{n}\binom{n}{l} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{n} \cdot 2^{n} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{2n} \cdot \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

I got stuck here, and I'm having trouble manipulating the second term. I would appreciate any help on this proof!

Body:
\begin{eqnarray*}
\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} &=& [x^0]: \sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} x^{-k-l} (1+x)^n \\
&=& [x^0]: \left(1+ \frac{1}{x} \right)^{2n} (1+x)^n \\
&=& [x^{2n}]:  (1+x)^{3n} = \binom{3n}{n}.\\
\end{eqnarray*}


问:
Prove that $\operatorname{Null}(A) = \operatorname{Null}(A^T)$ when $A = A^2$
So I first let x belong to $\operatorname{Null}(A) $ i.e $Ax = 0$. Now I tried using the fact that if the length of $A^Tx = 0$ then $A^Tx = 0$.

But I don't know how to prove that $x^TA^TAx = 0$.

Any help will be appreciated.

答:
Consider
$A=\pmatrix{1 & 1\\0 & 0}$ and $x=\pmatrix{1\\-1}$. Are you sure you have the question correctly posted? Am I not understanding something correctly?


答:
As per given question, here is the parallelogram -

$\angle ABC = 60^0$. That means the angle between $\vec{AB}$ and $\vec{BC}$ is $120^0$ (basically the angle between $\vec{AB}$ and $\vec{AD}$.

Now the given answer would have been right if $\angle ABC$ was $120^0$. See the below diagram. You can see the angle between $\vec{AB}$ and $\vec{BC}$ is now $60^0$.

You can also visually compare size of $AC$ in both cases.

Title: How to find the dot product using the law of cosines

Body:
I'm working with the following problem:

We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$

Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.

$$AC^2=AB^2 +BC^2 -2AB\cdot BC\cos x$$
$$\implies AC^2=3^2 +2^2 -2\cdot2\cdot3\cos 60^\circ$$
$$\implies AC^2=9 +4 -12\cdot0.5$$
$$\implies AC^2=13-6=7$$
$$\implies AC=\sqrt 7.$$

My next step is to calculate the angle $BAC$; we should be able to use the law of cosines here as well:

$$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cos x$$
$$\implies4=9+(\sqrt 7)^2 -2\cdot3\sqrt 7\cos x$$
$$\implies4=9+7 -6\sqrt 7\cos x$$
$$\implies-2=-6\sqrt 7\cos x$$
$$\implies\frac{1}{3}=\sqrt 7\cos x$$
$$\implies\frac{1}{3\sqrt 7}=\cos x.$$

If we want to calculate our dot product using only the vector lengths, we would use the fact that $A\cdotp B=|A||B|\cos x$, which in this case would mean that:

$$AB \cdot AC=3\sqrt7\frac{1}{3\sqrt 7}=1,$$

which is quite wrong since the answer is supposed to be $12$.

Can someone please tell me where I've made a mistake?

In the proposed solution another method seem to be used. The textbook claims that $AB \cdotp AC = AB \cdotp (AB + BC)=AB \cdotp AB + AB \cdotp BC = |AB|^2 + |AB||BC|cos(60)=9+6*\frac{1}{2}=12$


问:
Kaprekar's constant is 6174: Proof without calculation
How to prove that by performing Kaprekar's routine on any 4-digit number
repeatedly, and eventually we will get the 4-digit constant $6174$ rather than get stuck in a loop, without really calculating anything?

I've seen a 11-page paper that did this job, and with a little modification it's capable to prove that we can get $6174$ within 7 steps, and that such constant doesn't exist for numbers with 2, 5, 6, 7, 8, 9, 10, and 15 digits.

(In that paper the author proposed to divide the 4-digit numbers into some 70 groups, and Kaprekar's routine always goes from a group to another so we can redraw the state-transition diagram with only 70 vertices. And the new diagram has a longest path of 7. The paper is in Chinese so I won't post it here.)

Is there a simpler proof?

答:
Kaprekar's routine is a mapping from the set $F_0=\{0000,0001,0002,\dots,6174,\dots,9999\}$ of four digit natural numbers to itself. It only depends on the digits of a number and not on their positions within the number. Kaprekar's routine consists in

forming an integer by arranging the digits in descending order,

forming a smaller integer by arranging the digits in ascending order,

eventually taking the difference of both.

Hence if $\,d_3,d_2,d_1,d_0\,$ are the given digits satisfying $\,9\geqslant d_3\geqslant d_2\geqslant d_1\geqslant d_0\geqslant 0$, then Kaprekar's routine yields
$$1000d_3+100d_2+10d_1+d_0\;-\;(1000d_0+100d_1+10d_2+d_3)\\[2ex]
=\;999(d_3-d_0) + 90(d_2-d_1)\,.
\tag{0}$$
So the result is completely described by the pair $(D,d)$ with $D=d_3-d_0, d=d_2-d_1$, and $\,9\geqslant D\geqslant d\geqslant 0$.

Target of this post is to prove without explicitly calculating all cases that Kaprekar's routine possesses a unique fixed point which is the famous $6174$. Writing 'unique' assumes that we ignore the boring (in the sense of uninteresting) case $0000$.

It is consequent and helpful to not distinguish between numbers in $F_0$ having the same $(D,d)$ value, since Kaprekar's routine generates the same image number from them. Hence partition the set $F_0$ into corresponding subsets labelled by $(D,d)$. The smallest representative in the respective subset is the four digit number $10d+D$, it has at least two leading zeros.

The mapping $K:(D,d)\mapsto (D\,',d\,')$ induced by Kaprekar's routine is referred to as a Kaprekar step. Look at the arrows in the the nice directed graph in Peter Taylor's answer.

$(D=0,d=0)\,$ labels the subset $\{0000, 1111,\ldots, 9999\}\subset F_0$ of numbers with all digits equal. Kaprekar's routine produces $0000$ out of them, thus $K(0,0)=(0,0)$.

If $D\geqslant 1$, then Kaprekar's routine generates numbers ranging from $0999$ to $9801$ which are divisible by $9$, whence $K(D,d)\neq(0,0)$ in that case.

In the sequel attention is restricted to pairs $(D,d)$ with $D\geqslant 1$.

Lemma$\;$ There is a collection of formulae for the Kaprekar step $K(D,d)=(D\,',d\,')$, but case distinction is needed:

If $\,D>d=0\,$ then
$$\left.\begin{matrix}D\,' \\ d\,'\end{matrix}\right\}\;=\; 4.5 \pm |D-5.5|\tag{1}$$
Note the invariance $K(11-D,d) = K(D,d)$ for $D\geqslant 2$, and that $D\,'>d\,'$.

If $\,D=d\geqslant 1\,$ then
$$\begin{align} D\,' & \;=\;\; 2\,|D-5|+1\\[1ex] d\,' & \;=\: \big|\:2\,|D-5|-1\,\big| \end{align}\tag{2}$$
Note that $D\,'=d\,'+2\,$ if $\,D\neq 5$.

If $\,D>d\geqslant 1\,$ then
$$\begin{align} D\,' & \;=\;\: |D+d-10| + \delta_{10,D+d} + (D-d)\\[1ex]
d\,' & \;=\; \big| |D+d-10| + \delta_{10,D+d} - (D-d)\,\big|\end{align}\tag{3}$$
where $\delta_{p,q}$ denotes the Kronecker delta.

Proof: The relevant values $\,9\geqslant D\geqslant d\geqslant 0\,$ form a triangle in the first octant in the $D,d$-plane.

corresponds to the lower edge of the triangle. Kaprekar's routine yields
$$999D\;=\; 1000\,(D-1) + 100\cdot 9 + 10\cdot 9 + (10-D)\,.$$
Reading off the digits and using
$$\left.\begin{matrix}\min(p,q) \\ \max(p,q)\end{matrix}\right\}
\;=\; \frac12\big(p+q\mp |p-q|\big)\tag{4}$$
with $\,p=D-1, q=10-D$, the formula is obtained.

corresponds to the upper "diagonal" edge.

Kaprekar's routine, not yet specialising to $D=d$ in view of case 3, yields
$$999D+90d\;=\; 1000\,D + 100\,(d-1) + 10\,(9-d) + (10-D)\,.\tag{5}$$
Setting $d=D\,$ now, and rewriting the digits with the help of $(4)$ gives $5\pm |D-5|, 4\pm |D-5|$. This leads to
$$\begin{align} D\,' & \;=\; 5+|D-5| - \big(4-|D-5|\big)\\[1ex]
d\,' & \;=\; \big|\, 5-|D-5| - \big(4+|D-5|\big)\big|\,. \end{align}$$

corresponds to the interior of the triangle including the RHS vertical edge.

From $(5)$ we read off the digits which satisfy $9-d\geqslant 10-D$ in the present case, and $D>d-1$ anyway. Thus,
$$\begin{align} D\,' & \;=\; \max(D,9-d) - \min(d-1,10-D)\\[1ex]
d\,' & \;=\; \big|\min(D,9-d) - \max(d-1,10-D)\,\big|\,, \end{align}$$
and using
$$\begin{align}
\begin{matrix}\max\\ \min\end{matrix}\big(D,9-d\big) &\;=\; \frac{9+D-d\pm |D+d-9|}2\\[1.67ex]
\begin{matrix}\max\\ \min\end{matrix}\big(d-1,10-D\big) & \;=\; \frac{9-(D-d)\pm |D+d-11|}2\\[2ex]
\frac{|D+d-9|+|D+d-11|}2 & \;=\; |D+d-10| +\delta_{10,D+d}
\end{align}$$
one deduces the desired.$\quad\blacksquare$

Just an observation regarding the set of pairs $(D,d)$ with $d\geqslant 1$, as covered by cases 2. and 3. in the lemma: The mapping $(D,d)\mapsto (10-d,10-D)$ defines an involution, and the formulae $(2)$ and $(3)$ are invariant under it.

Fixed point theorem$\;$ The Kaprekar step has precisely one fixed point, namely $K(6,2)=(6,2)$.

Proof: Looking for fixed points we proceed along the cases in the lemma.

If $\,d=0\,$ then $\,0\stackrel{!}{=}d\,'=4.5-|D-5.5|\implies D=1$, but $(1)$ would require $D\,'=9\,$.

If $\,D=d\,$ then
$$K(D,D) \;=\; \begin{cases}(1,1) & D=5\\ (d\,'+2,d\,') & \text{else.}\end{cases}$$
Hence again, no $(D,D)$ is a fixed point.

Now we assume $\,D>d\,$ and the fixed point conditions read
$$\begin{align} D & \;\stackrel{!}{=}\;\: |D+d-10| + \delta_{10,D+d} + (D-d)\iff d\;=\;\: |D+d-10| + \delta_{10,D+d}\tag{6}\\[1ex]
d & \;\stackrel{!}{=}\; \big| |D+d-10| + \delta_{10,D+d} -D +d\,\big|\tag{7}\end{align}$$
Inserting $(6)$ into $(7)$ implies $d =|2d-D|$, thus $D=3d$ as $\,D>d\,$ needs to be fulfilled.

Insert this into $(6)$, the Kronecker delta will vanish, and the resulting equation $\,d=|4d-10|\,$ is uniquely solved by $d=2\implies D=3d=6\,$.$\quad\blacksquare$

The fixed point $(6,2)$ of $K$ determines the unique fixed point of Kaprekar's routine as $\,999\cdot 6+90\cdot 2=6174\,$.


问:
effective version of Mertens Theorem for the Euler product
I'm referring to the theorem given here, which is

$$\displaystyle\lim_{n\to \infty} \:\: \left(\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right)\right) \;\;\; = \;\;\; \exp(\hspace{.01 in}\gamma)$$

where $p$ ranges over the primes and $\gamma$ is the Euler-Mascheroni constant.

What I'm really interested in is a good (effective) lower bound for $\;\; \displaystyle\prod_{p\leq n} \: \left(1-\frac1p \right) \;\;$,

so if you have another way to get such a bound, that would work too.

Otherwise, is it explicitly known how large and $n$ is sufficiently large to get $$\frac1{\ln(n)} \cdot \left(\displaystyle\prod_{p\leq n} \frac1{1-\frac1p}\right) \; < \; c$$

where, for example, $\: c=2 \:$ or $\: c = \frac95 \;$?

答:
See Theorem 6.12 in Pierre Dusart's Estimates of some functions over primes without RH.  In particular for all $x \ge 2973$ one has the effective unconditional lower bound:
$$\prod_{p\le x}\left(1 - \frac1p\right) > \frac{e^{-\gamma}}{\ln x}\left(1 - \frac{0.2}{\ln^2 x}\right).$$

So, already for $x$ in this range it is true that one can take $c = 1.78666$, which is much smaller than $c=\tfrac95$.  A small amount of computation would be all it takes to get the smallest $x$ that works.

Update: it looks like $c=2$ is true for $n\ge 14$, while $c=\tfrac95$ holds for $n \ge 469$.

ContentLicense: CC BY-SA 3.0


问:
Parallel lines in 3D and their distance?
Show that the two lines are parallel. Determine the distance between them and the plane containing both lines.

$$L_{1}\left\{\begin{matrix}
x=3-2t\\y=4+t
\\z=1-t
\end{matrix}\right.\quad
L_{2}\left\{\begin{matrix}
x=5+2t\\y=1-t
\\z=7+t
\end{matrix}\right.$$

I'm feeling lost.

EDIT:
They are parallel because the direction vectors are the same. $(2,-1,1)$

To calculate the normal of the plane i take $\begin{pmatrix}-2&1&-1\end{pmatrix}\times \begin{pmatrix}2&-1&1\end{pmatrix}=\begin{pmatrix}0&0&0\end{pmatrix}.$ That is wrong. The normal shall be $(3,10,4)$. How do i calculate this?

Arbitrary point on $L_{1}(3,4,1)$.
Equation of the plane: $(x,y,z)=(3,4,1)+t(-2,1,-1)+s(2,-1,1)$

答:
The equation $L:\mathbf r(t)=\mathbf v_0+t\mathbf v_1$ of line $L$ requires a point $\mathbf v_0$ on $L$ and a vector $\mathbf v_1$ parallel to $L$.
$$L_1:\mathbf f(t)=\langle 3,4,1\rangle+t\langle 2,-1,1\rangle\;\;\;\;\;L_2:\mathbf g(t)=\langle 5,1,7\rangle+t\langle 2,-1,1\rangle$$
Therefore, $L_1,L_2$ are parallel because their direction vectors are the same and thus parallel. Let $d$ be the distance between $L_1,L_2$. Construct a parallelogram of area $A$ using points $\mathbf f(0),\mathbf f(1)$ on $L_1$ and points $\mathbf g(0),\mathbf g(1)$ on $L_2$.
$$A=d\cdot\vert\mathbf f(1)-\mathbf f(0)\vert=\vert(\mathbf f(1)-\mathbf f(0))\times(\mathbf g(0)-\mathbf f(0))\vert$$
$$\therefore\;\;d=\frac{\vert(\mathbf f(1)-\mathbf f(0))\times(\mathbf g(0)-\mathbf f(0))\vert}{\vert\mathbf f(1)-\mathbf f(0)\vert}=\frac{5}{6}\sqrt{30}$$


问:
How to compute derive a complex number from a square root of another complex number?
In plane wave in a lossy medium, the complex propagative constant is given to be

$$\gamma = \alpha\ + j\beta $$
$$= j \omega \sqrt {\mu \epsilon_c}$$
$$= j \omega \sqrt { \mu \left(\epsilon - j\frac{\sigma}{\omega}\right) }$$

Which will eventually result in a square root of a complex number like

$$\gamma = j\omega \sqrt {X+jY}$$

how do I proceed on to find $\gamma$ in terms of $\alpha$ and $\beta$ ?

答: Note that if $(X+jY)^2=a+jb$ with $a:=-\tfrac{\alpha}{\omega^2},\,b:=-\tfrac{\beta}{\omega^2}$ then $X^2-Y^2=a,\,2XY=b$. Squaring, suming and square-rooting, $X^2+Y^2=\sqrt{a^2+b^2}$. Averaging $X^2\pm Y^2$, $X^2=\frac{\sqrt{a^2+b^2}+a}{2}$ and $Y^2=\frac{\sqrt{a^2+b^2}-a}{2}$. While $X^2$ ($Y^2$) is consistent with two values of $X$ ($Y$), only two solutions exist, not four; the other two have a sign error in the value of $XY$.


问:
Showing the inequality holds (binomial coefficient)
I'm struggling to understand how to show that the inequality holds for $ m,n \in \mathbb{N} \; \text{with} \; m<n $
$$\frac{1}{m^k} {m \choose k} < \frac{1}{n^k} {n \choose k} \;\; \text{for all} \; k=2,...,m$$

Any pointers of how to approach this would be very helpful.

答:
There's a potentially interesting combinatorial interpretation to your question.

The number of functions from $[k] = \{1,\dots,k\}$ to $[n]$ is $n^k$. The number of injective functions from $[k]$ to $[n]$ is $\binom{n}{k} k!$ (first choose the range, then the bijection to the range). Thus your inequality reduces to the following: the probability a random function from $[k]$ to $[m]$ is injective is less than the probability a random function from $[k]$ to $[n]$ is injective.

This makes some intuitive sense, since the larger the codomain, the 'easier' it is for a random function to be injective.


问:
Differential equations why doesnt this work?
I have to solve the equation $tx'=x+\tan\frac xt$. It is homogeneous. I wrote $x=u\cdot t$ and I have $x'=\frac{x+\tan\frac xt}t$ and then $x'=\frac{ut+\tan\frac{ut}t}t$ in the end I get $x'=u+\frac{\tan u}t$. So $u't +u=u +\frac{\tan u}t$. So $u'=\frac{\tan u}{t^2}$,
$\frac{\mathrm du}{\mathrm dt}=\frac{\tan u}{t^2}$. $\int\frac{\mathrm d u}{\tan u}=\int\frac{\mathrm dt}{t^2}$, so $\ln|\sin u|=-\frac1t$, but the answer in my book is
$\sin\frac xt=\frac t2$. How come?</p>

答:
Your solution is correct, except you should have  $-\frac{1}{t}+C$ after integration. Either your textbook has a wrong answer or you looked up the answer to a wrong problem.

I double-checked with Maple: it said  $\sin(x/t)=Ce^{-1/t}$.


答:
Let $\sqrt{a+ib}=x+iy$, or $a+ib=(x+iy)^2=x^2-y^2+i2xy$.

So after identification,

$$ax^2=x^4-x^2y^2=x^4-\frac{b^2}4$$

and solving that biquadratic equation,

$$x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}2}$$ and $$y=\frac b{2x}.$$

Title: How to compute derive a complex number from a square root of another complex number?

Body:
In plane wave in a lossy medium, the complex propagative constant is given to be

$$\gamma = \alpha\ + j\beta $$
$$= j \omega \sqrt {\mu \epsilon_c}$$
$$= j \omega \sqrt { \mu \left(\epsilon - j\frac{\sigma}{\omega}\right) }$$

Which will eventually result in a square root of a complex number like

$$\gamma = j\omega \sqrt {X+jY}$$

how do I proceed on to find $\gamma$ in terms of $\alpha$ and $\beta$ ?


问:
Confusion on the proofs of convolution theorem of Fourier Transform.
The convolution theorem of Fourier transform is stated as follows:

Define $h(x):=f(x)*g(x)$, then we have $$\hat{h}(k)=\hat{f}(k)\hat{g}(k).$$

I have a confusion of the proofs of this theorem. Most proofs go in the following way:

Define $h(x):=f(x)*g(x)$.  By definition, we have $$\hat{h}(k)=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx.$$
Then, plugging in the integral expression of $f*g$ into the above integral yields us
\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx\\
&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)g(s)ds\Big)e^{-ikx}dx\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx.\ \ \ \ \ \ (*)
\end{align*}

Then, most proofs directly change integral order in the question $(*)$, so that

\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dxds\\
&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds.
\end{align*}

What remains is standard: note that by the shifting property, the inner integral is $$\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx=\hat{f}(k)e^{-iks},$$ and thus
\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds\\
&=\hat{f}(k)\int_{-\infty}^{\infty}g(s)e^{-iks}ds\\
&=\hat{f}(k)\hat{g}(k),
\end{align*}
where the last equality was obtained by definition. The proof is concluded.

However, why could we change the integral order of (*) in the first place? Don't we need Fubini? I tried to use Fubini as follows: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)e^{-ikx}|dsdx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dsdx$$ but I do not know how to proceed further. I had expected that I can split the integral so that I would have two $L^{1}-$norm, but since the first integral is $ds$, I could not split the integral..

Do we directly assume the function is nice enough so that we can interchange the integral order? If so, what type of functions do we assume?

Thank you!

答:
Thanks to demonakos' help, it turned out that as long as $f,g\in L^{1}$, the integral order can be interchanged.

The proof is as follows: note that by Tonelli's theorem (in the third equality),
\begin{align*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)e^{-ikx}|dsdx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dsdx&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dxds\\
&=\int_{-\infty}^{\infty}\Big(|g(s)|\int_{-\infty}^{\infty}|f(x-s)|dx\Big)ds\\
&=\int_{-\infty}^{\infty}\Big(|f(x-s)\|g\|_{L^{1}(\mathbb{R})})dx\\
&=\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}<\infty,
\end{align*}
the $L^{1}-$norm is finite since $f$ and $g$ were hypothesized to be in $L^{1}$.

Hence, it follows from Fubini's Theorem that we can interchange integration order in $(*)$. That is
\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dxds\\
&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds.
\end{align*}

In fact, one could see the Fubini-Tonelli Theorem here: https://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli's_theorem_for_non-negative_measurable_function.

Instead of using Tonelli and Fubini seperately, one could directly check $\int\int|f(x-s)g(s)|dxds<\infty$ as above, and using Fubini-Tonelli to directly conclude that the integral order can be interchanged.

Although equivalent, there is still a little bit difference.


问:
Integral Identity for Permanent of a Matrix
Say we have an $N\times N$ matrix $A$ with elements $A_{ij} = \alpha_i^{(1- \delta_{ij})}$. For example, for $N=3$ we have

$$ A = \left[ \begin{array}{ccc}
1 & \alpha_1 & \alpha_1 \\
\alpha_2 & 1 &  \alpha_2 \\
\alpha_3 & \alpha_3 &  1  \\
 \end{array}\right]$$

How can we show that $A$ satisfies the identity

\begin{equation}
 \text{perm}(A) =  \int^{\infty}_{0} ds\, e^{-s} \prod_{\ell=1}^{n} \Big[1+(s-1) \alpha_i\Big]
\end{equation}

More generally are there any other known integral identities for the permanent of a (perhaps more general) matrix?

答:
For the particular question, do something like this
$$Perm(A)=\sum _{\sigma \in S_n}\prod _{i=1}^nA_{i,\sigma (i)}=\sum _{\sigma \in S_n}\prod _{i=1}^n\alpha _i^{1-\delta _{i,\sigma (i)}},$$
so we have to know exactly where there are fixed points, for a permutation $\sigma, $call $Fix(\sigma)=\{i:\sigma (i)=i\}.$. Notice then than fixing a set $X\subseteq [n]=\{1,\cdots ,n\}$ we can iterate over this.
$$perm(A)=\sum _{X\subseteq [n]}\sum _{\substack{\sigma \in S_n\\ Fix(\sigma)=X}}\prod _{i=1}^{n}\alpha _i^{1-\delta _{i,\sigma (i)}}=\sum _{X\subseteq [n]}\sum _{\substack{\sigma \in S_n\\ Fix(\sigma)=X}}\prod _{x\in [n]\setminus X}\alpha _x=\sum _{X\subseteq [n]}D_{n-|X|}\prod _{x\in [n]\setminus X}\alpha _x,$$
where the last step is noticing that the argument of the sum does not depend really on the sum index. The number of permutations with exactly $|X|$ fixed points are counted by Derangements, denoted by $D_n$.
Now, in the alternative universe of the right hand side,
notice that the product can be seen as
$$\prod _{i=1}^n(1+(s-1)\alpha _i)=\sum _{X\subseteq [n]}(s-1)^{|X|}\prod _{x\in X}\alpha _x,$$
so the integral becomes
$$\sum _{X\subseteq [n]}\prod _{x\in X}\alpha _x\int _{0}^{\infty}(s-1)^{|X|}e^{-s}ds,$$
so the only thing left to check is that $$D_n=\int _{0}^{\infty}(s-1)^{n}e^{-s}ds.$$
I will suggest either expanding $(s-1)^n$ using the binomial expansion and deducing that these are indeed the Derangements or trying integration by parts to deduce a recursion.

For your follow up question, this will help.


问:
Showing an equivalence relation on $\mathbb Z$
Using the fact that \begin{align*}(a|b \ \land \ a|c) &\implies a|(b+c) \\ a|b &\implies a|
bc \\ (a|b \ \land \ b|c) &\implies a|c \end{align*}

then I want to prove $\equiv_d$ is an equivalence relation on $\mathbb Z$ for every positive integer $d$

I'm pretty confused on this, but so far what I have done is attempt to show reflexivity and symmetry, but not yet transitivity and I don't know how to apply the predicates above.
For reflexivity I said $dRd \implies d-d \in \mathbb Z, d\in \mathbb Z^+$, and since $0\in \mathbb Z$ then there is a reflexive relation. For symmetry I said if we take any $k\in \mathbb Z$, then $dRk \implies kRd$. So then $d-k \in \mathbb Z, k-d \in \mathbb Z$. Since $d-k=c, k-d=-c$ then symmetry is satisfied. And it is this point I don't know how to continue further.

答:
Reflexive:  For every $a \in \mathbb Z$, $a\equiv a\pmod d$.

Pf: $d= 1\cdot d$ so $d|d$.  So $d|d\times 0 = 0=a -a$ for any $a\in \mathbb Z$.  So $a\equiv a \pmod d$.

Definition of $a\equiv a \pmod d$ is that $d|(a-a)$ or in  other words $d|0$.  Does $d|0$?

Well $d|d$ so $d|d\cdot 0=0=a-a$.  So yes it does.  So $d|a-a$ so $a \equiv a \pmod d$ for all $a\in \mathbb Z$.  So $\equiv_d$ is reflexive.

Symmetric:  For every $a,b \in \mathbb Z$ where $a\equiv b\pmod d$ then $b \equiv a \pmod d$.

Proof: $a\equiv b\pmod p$ means $d|a-b$.  So $d|-1(a-b) = b-a$.  So that means $b\equiv \pmod d$.  SO if $a \equiv b \pmod d$ then $b\equiv a \pmod d$.  So $\equiv_d$ is symmetric.

Transitive.  If $a \equiv b\pmod d$ and $b\equiv c \pmod d$ then $a\equiv c\pmod d$.

Proof.  If $a \equiv b \pmod d$ then $d|(a-b)$.  And if $b\equiv c\pmod d$ then $d|(b-c)$.

So $d|(a-b) + (b-c) = a-c$.

So $a\equiv c \pmod d$.  So $\equiv_d$ is transitive.


问:
Conditon for $f(x)$ such that $f(d(x, y))$ induces the same topology with $d(x, y)$
On Munkres' Topology, Sec 20, Q 11, for $f(x)=\frac{x}{1+x}$, $f(d(x, y))$ induces the same topology with $d(x, y)$.

And I have some question here : what conditions for $f(x)$ could make same result?

My attempt : If $f(x)$ satisfies

$f(a+b) \leq f(a)+f(b)$,

$f(0)=0$, and $0< f(x) \leq 1$ for $0 < x$

edit) Maybe, latter of condition 2) can be replaced by "$f$ is bounded"!

But, I can't prove that it induces the same topology. Does it need some specified condition?

答:
This can be seen as part of a more general problem. We say that two metrics $d_1$ and $d_2$ are equivalent if they induce the same underlying topology. The basic theorem characterizing equivalent metrics is the following.

Theorem: Let $d_1$ and $d_2$ be metrics on some space $X$. Then $d_1$ and $d_2$ are equivalent if and only if for all $\epsilon > 0$ and $x\in X$, there exists $\delta_1 > 0$ and $\delta_2 > 0$ such that
$$B_1(x,\delta_1) \subseteq B_2(x,\epsilon),\qquad\text{and}\qquad B_2(x,\delta_2) \subseteq B_1(x,\epsilon),$$
where $B_i(x,r)$ is the open ball of radius $r$ centered on $x$, as determined by the metric $d_i$.

I won't prove this theorem here. It's not difficult, and you should be able to find a proof rather easily when you search for "topologically equivalent metrics". Roughly speaking, the theorem says that two metrics are equivalent if and only if you can find an open ball of one inside any open ball of the other.

What does this mean for the functions $f$? First, let us require that $f(d(x,y))$ remain a metric. This means that we should have $f(r) \ge 0$ for all $r\in \mathbb{R}^+$, with $f(r) = 0$ if and only if $r=0$. We should also require that $f$ be a non-decreasing function such that $f(r+s) \le f(r) + f(s)$ for all $r,s\in \mathbb{R}^+$. This will ensure that the triangle inequality is satisfied.

Fix some $\epsilon > 0$ and $x \in X$. Let $d_1 = d$ and $d_2 = f\circ d$. Then the condition $B_1(x,\delta_1) \subseteq B_2(x,\epsilon)$ is equivalent to requiring the existence of some $\delta_1$ such that $d(x,y) < \delta_1$ implies $f(d(x,y)) < \epsilon$. This is simply continuity of $f$ at $0$.

The condition $B_2(x,\delta_2) \subseteq B_1(x,\epsilon)$ is equivalent to requiring the existence of some $\delta_2$ such that $f(d(x,y)) < \delta_2$ implies $d(x,y) < \epsilon$. This latter condition is automatically satisfied for any non-decreasing function. Indeed, pick $\delta_2 = f(\epsilon)$. Then $f(d(x,y)) < \delta_2 = f(\epsilon)$ implies $d(x,y) < \epsilon$.

Therefore the following requirements are sufficient for $f:[0,\infty)\rightarrow [0,\infty)$ to determine an equivalent metric:

$f$ is non-decreasing.

$f(r) = 0$ if and only if $r=0$.

$f$ is sub-additive, i.e., $f(r+s) \le f(r) + f(s)$.

$f$ is continuous at $0$.

Notice that the first 3 conditions are just generic conditions for $f\circ d$ to be a metric again. Any metric determined by such a function will define a topology which is at least as fine as the original topology. The new topology will be equivalent if and only if $f$ is continuous at $0$. In particular, note that all four of the conditions above are satisfied by the function $f(r) = r/(1+r)$, so that it determines an equivalent topology.


问:
Need a counterexample of limit of average integral
Assume that $f(x)\geq 0$ is integrable in $[0,1]$ and

$$\lim\limits_{h\to 0^+}\frac{1}{h}\int_{0}^{h}f(x)dx=0$$

I feel that $\lim_{x\to 0}f(x)=0$ is incorrect, but I don’t know the example. I could only think about $1+\sin\frac{1}{x}$, but it seems that it’s not work.

In general, what do we know of $f$ as $x\to 0$ ?

答:
For $n>3,$ define the intervals $I_n =[1/n,1/n +1/n!].$ Then let $f(x) = 1$ on each $I_n,$ $f=0$ elsewhere. Note that $f$ is integrable, and does not have limit $0$ at $0.$

The idea here is that the $I_n$ are very small in length, with "lots" of space between them. So much space in fact that we expect

$$\tag 1 \frac{1}{h}\int_0^hf(x)\,dx \to 0\,\text{ as } h\to 0^+.$$

Let's try the sequence $h_n = 1/n +1/n!$ We get

$$\frac{1}{h_n}\int_0^{h_n}f(x)\,dx=\frac{1}{1/n +1/n!}\int_0^{1/n +1/n!} f(x)\,dx$$ $$< \frac{1}{1/n}\sum_{k=n}^{\infty}\frac{1}{k!} = n\left(\frac{1}{n!}+ \frac{1}{(n+1)!}+ \cdots \right).$$

Now the last expression $\to 0.$ That suggests that $(1)$ holds as desired. There are details remaining to verify, which I'll leave to you for now.


问:
Find P(C) - Mutually Exclusive Events and Conditional Probability
Events $A, B, C$ are such that $B$ and $C$ are mutually exclusive and $P(A) = 2/3, P(A \cup B) = 5/6$ and
$P(B \cup C) = 4/5$. If $P(B|A) = 1/2$ and $P(C|A) = 3/10$, calculate $P(C)$.

答:
You correctly guessed that we need $P(B)$ first. Since $P(A)$ is known, we should look for relations between $A,B$.

They are $P(A\cup B)=5/6,P(B|A)=1/2$.

By definition$P(B|A)=P(A\cap B)/P(A)$ which gives $P(A\cap B)=\frac12\times P(A)=1/3$.

We know $P(A\cup B)=P(A)+P(B)-P(A\cap B)=5/6$ using which find $P(B)$.

Finally $P(B\cup C)=P(B)+P(C)$ since $P(B\cap C)=0,$ using which find $P(C)$.


问:
Derivative of function w.r.t to vector where matrix elements are functions itself
Consider the function $f: \mathbb{R}^N \to \mathbb{R}$ with

$$
f(r) = ||Ap - b||_2^{2} = p^{\top}A^{\top}Ap - 2p^{\top}A^{\top}b + b^{\top}b
$$

where

$A \in \mathbb{R}^{N \times N}$ and  $p,b \in \mathbb{R}^N$

The elements of the matrix $A$ are linear functions of the variables $r = (r_1, \ldots, r_N)$, i.e.
$a_{i,j} : \mathbb{R}^{N} \to \mathbb{R}$ and $A = (a_{i,j}(r_1, \ldots, r_N))_{i = 1, \ldots, N, j = 1, \ldots, N}$.

I know the partial derivatives of each matrix element, i.e. $ \frac{\partial a_{i,j}}{\partial r_k}$ is known.

Is there any chance for a closed form of $\nabla f$? Any hints or help is really appreciated!

PS: This question is similar to this one. But there the matrix depends only on one variable.

答:
Let's use a colon to denote the trace/Frobenius product, i.e.
$$\eqalign{
X,Y &\in {\mathbb R}^{m\times n} \\
X:Y &= {\rm Tr}(X^TY) = {\rm Tr}(Y^TX) = Y:X \\
}$$
or in terms of components
$$\eqalign{
X:Y &= \sum_{i=1}^n\sum_{j=1}^m X_{ij}Y_{ij}
 \qquad&\big({\rm explicit\,summation}\big) \\
    &= X_{ij}Y_{ij}
 \qquad&\big({\rm Einstein\,convention}\big) \\
}$$
The colon product can also be applied to vectors by setting $m=1$ and treating them as rectangular matrices.

The following matrix will turn out to be useful
$$G = 2\left(Ap-b\right)p^T$$

Use this product to write the function and calculate its differential
$$\eqalign{
 f &= (Ap-b):(Ap-b) \\
df &= 2(Ap-b):(dA\;p) \\
   &= 2\left(Ap-b\right)p^T:dA \\
   &= G:dA \\
}$$
Switch to index notation and substitute the known derivative
$\;H_{ijk} = \left(\frac{\partial A_{ij}}{\partial r_k}\right)$
$$\eqalign{
dA_{ij} &= H_{ijk}\,dr_k \\
df &= G_{ij}\,H_{ijk}\,dr_k \\
\frac{\partial f}{\partial r_k} &= G_{ij}\,H_{ijk}
  &\big({\rm gradient\,wrt\,}r\big) \\
}$$
or, using explicit summations, the components of the gradient are
$$\eqalign{
\frac{\partial f}{\partial r_k}
 &= \sum_{i=1}^N\sum_{j=1}^N G_{ij}
    \left(\frac{\partial A_{ij}}{\partial r_k}\right) \\
}$$


问:
Integration by parts help
Integration by parts

$$x\tan^{-1}x~dx=\frac12\left(x^2+1\right)\tan^{-1}x-\frac12x+c$$

and

$$x^2e^{-3x}dx =-\frac13e^{-3x}\left(x^2+\frac23x+\frac29\right)+c$$

I have

$$d(uv)=u~dv+v~du$$

Can anyone please help me?

答:
Hold onto your hat:
For the integral $$\int x^2e^{-3x}dx$$ we will let $u= x^2$ and $dv=e^{-3x}dx$. Thus, $du = 2x dx$ and $v = \dfrac{e^{-3x}}{-3}$.

Thus we have  $$\int x^2e^{-3x}dx = x^2 \frac{e^{-3x}}{-3} - \int 2x \dfrac{e^{-3x}}{-3} dx.$$

For the latter integral above, we simplify and factor out the constants so that we have $$\dfrac{2}{-3} \int x e^{-3x}dx.$$

Now we let $u= x$, $dv = e^{-3x}dx$. Then $du=dx, v=\dfrac{e^{-3x}}{-3}$. Hence we have $$\dfrac{2}{-3} \int x e^{-3x}dx = \dfrac{2}{3} \left( x \dfrac{e^{-3x}}{-3} - \int  \dfrac{e^{-3x}}{-3} dx\right) = \dfrac{2}{-3} \left(x \dfrac{e^{-3x}}{-3}  + \dfrac{e^{-3x}}{-9} \right).$$

Putting this all together,

$$\int x^2e^{-3x}dx = e^{-3x} \left(\frac{1}{-3} x^2 - \frac{2}{9}x -\frac{2}{27} \right) + C$$

Title: Integration by parts help


问:
Simultaneous linear two variable congruences
I am stuck with this system of linear congruences and could use some guidance as to what I'm doing wrong. This is what I've tried:

I have a system of linear congruences:

12x+4y≡16(mod 174)

9x+9y≡27(mod 174)

Now, avoiding non-regular transformations, I do as following:

-I add second congruence multiplied by -1 to the first one

-then first one multiplied by -3 to the second one

And I get:

3x-5y≡-11(mod 174)

24y≡60(mod 174)

I solve the second one as a Diophantine equation and get y≡17(mod 29) as a solution.

I write y as y=17+29t and plunge it back into the first equation and get:

3x-5(17+29t)≡-11(mod 174)

3x≡74+145 t(mod 174)

For this congruence to be solvable, 74+145t has to be divisible by 3. Since that is not obvious (at least to me), I write: 74+145t=3s

Then I try solving that equation for s and t and I get:

t=-74-3p

s=-3552-145p

Now I tried plunging this t I got back into 74+145t and after limiting it between 0 and 174 I get t=679,t=676,t=673,t=670

I plunge t=679 back into 3x+174k=74+145t and get x≡0(mod 58)

These solutions however make no sense when I check them.

I am sorry if this question is too simple or a mistake I'm making is obvious, but I tried solving this in so many ways, and neither produced correct solutions. (I did get few correct solutions for x and y, along with too many excess solutions(because I used non-regular transformations I guess).

答:
Your approach in solving for $y$ was right, but got it wrong with $x$; you could reduce them all to modulo $29$ to obtain
$$2x+20y\equiv 22\mod 29\,,$$
$$16x+16y\equiv 19\mod 29\,.$$
Solving these gives us
$$x\equiv 15\mod 29\,,$$
$$ y\equiv 17\mod 29\,.$$
You then extend to the possible solutions modulo $174$:
$$x\equiv 15,44,73,102,131,160\mod 174\,,$$
$$y\equiv 17,46, 75,104,133,162\mod 174\,.$$
Recognizing that the first equation demands that $y\equiv 1\mod 3$, we either have $y\equiv 46~or~133\mod 174$; now, subtracting the second equation from the first gives
$$3x\equiv 5y-11 \mod174\,.$$
Solving for $x$ using the two values of $y$ gives us the respective solution $x\equiv 15~or~44\mod 58$; thus, the system has solutions
$$(x,y)\in\left\lbrace(15,46)\,,(73,46)\,,(131,46)\,,(44,133)\,,(102,133)\,,(160,133)\right\rbrace\,.$$


问:
What is the smallest multiple of 77 that is greater than 70,000? 700,000?

What is the smallest multiple of $77$ that is greater than $70000$? $700000$?

How would you approach this question? My first thought was to factorize the $77$ into $7$ and $11$ but then how would you choose the exponents?

答: In general you can do this by first dividing by $77$, obtaining some remainder $r$. If we then add $77-r$ to the number, we obtain the smallest such multiple.


问:
The quotient ring of polynomials over $\mathbb{C}$ is meromorphic
I would like to ask about some help for proving that the quotient ring of polynomials with coefficients in $\mathbb{C}$ is meromorphic ( $Q(\mathbb{C}[z]) \subset  M(\mathbb{C})$, where $M(\mathbb{C})$ is the set of all meromorphic functions on $\mathbb{C}$ ).

Good, i know that if $f$ and $g$ are holomorphic functions, then the function $h:=\frac{f}{g}$ is meromorphic with the condition that $g\neq 0$. The polynomials with coefficient in $\mathbb{C}$ are obviously holomorphic.

Do you have any ideas how to move on if the idea is correct at all?

Thank you in advance!

答:
Terminological remark: I think you meant quotient field (also known as the field of fractions). Quotient ring is what you get by taking the quotient by some ideal in the ring.

And: yes, there is not much to say here: the quotient $f/g$ of two polynomials (with $g$ not identically zero) is  a meromorphic function, because polynomials are holomorphic.


问:
Degrees of Freedom for Student's t-Distribution
The student's T statistic is given by the formula - t = [X(bar) - Mu]/ [S/sqrt(n)]

X(bar) is sample mean,
Mu is population mean,
S is the mean square

I have interpreted the degrees of freedom in the following way, please let me know if I am making a mistake.

In all the exact sampling distributions, there's one assumption that's always constant i.e. parent population is always normal. Going by this assumption, the statistic [X(bar) - Mu]/[Sigma/sqrt(n)] would follow a normal distribution without the aid of Central limit theorem that necessitates the size of sample to be large. The degrees of freedom of this statistic would be n since n values can vary independently. But for small samples like this one, a possible estimator of Population variance Sigma^2 can be the mean square, S^2. If we estimate sigma by S , we'll have to reduce our degrees of freedom by 1 and hence the degrees of freedom for the statistic -

t = [X(bar) - Mu]/[S/sqrt(n)] becomes (n-1)

Is this interpretation of Student's t statistic correct?

答:
Theoretically, the degrees of freedom parameter $\nu = n-1$ is due to the definition of Student's t distribution. You are correct that this distribution converges standard normal for increasing $n.$

Intuitively, people sometimes think of DF in terms of
dimensionality: A sample of size $n$ exists in $n$-dimensional space. One dimension is "used" to find the sample mean $\bar X = \frac 1n\sum_{i=1}^n X_i.$ Then the remaining $n-1$ dimensions are "used" to find the sample variance $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)^2.$
The use of $n-1$ in the definition of $S^2$ implies that
$E(S^2) = \sigma^2,$ the population variance.

If I take many samples $m$ of size $n=5$ from $\mathsf{Norm}(\mu=100,\sigma=15)$ and then compute
t-statistics $T$ for each sample under the null hypothesis
that $\mu=100,$ then a histogram of the $m$ values of $T$ will closely match the density function of Student's t distribution with $\nu = n-1 = 4.$

In R, I can find the t-statistic for one sample of size $n=5$ as follows:

```<code>set.seed(111)
x = rnorm(5, 100, 15);  x
[1] 103.52831  95.03896  95.32564  65.46482  97.43686
t = t.test(x, mu=100)$stat; t
        t
-1.299141
</code>```

The R code below computes $m$ such t-statistics and
puts them into a vactor t.stat of length $m,$ makes
a histogram of the $m$ values, and shows the density
function of $\mathsf{T}(\nu=4)$ for comparison.
[The t-statistic shown above is the first of the $m$ statistics sampled in this way.]

```<code>set.seed(111)
m = 10^5;  n = 5;  mu.0 = 100;  sg = 15
t.stat = replicate(m, t.test(rnorm(n,mu.0,sg),mu=mu.0)$stat)
summary(t.stat)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max.
-24.220863  -0.741238  -0.001904  -0.000634   0.735395  25.473347
var(t.stat)
[1] 1.992431   # aprx Var(T) = 2

hist(t.stat, prob=T, br=100, ylim=c(0,.4), col="skyblue2")
 curve(dt(x,n-1), add=T, lwd=2, col="red", n=10001)
</code>```

Notice that the distribution of $\mathsf{T}(\nu=4)$ has a higher variance $(2)$ than does
standard normal $(1).$

Body:
Theoretically, the degrees of freedom parameter $\nu = n-1$ is due to the definition of Student's t distribution. You are correct that this distribution converges standard normal for increasing $n.$

Intuitively, people sometimes think of DF in terms of
dimensionality: A sample of size $n$ exists in $n$-dimensional space. One dimension is "used" to find the sample mean $\bar X = \frac 1n\sum_{i=1}^n X_i.$ Then the remaining $n-1$ dimensions are "used" to find the sample variance $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)^2.$
The use of $n-1$ in the definition of $S^2$ implies that
$E(S^2) = \sigma^2,$ the population variance.

If I take many samples $m$ of size $n=5$ from $\mathsf{Norm}(\mu=100,\sigma=15)$ and then compute
t-statistics $T$ for each sample under the null hypothesis
that $\mu=100,$ then a histogram of the $m$ values of $T$ will closely match the density function of Student's t distribution with $\nu = n-1 = 4.$

In R, I can find the t-statistic for one sample of size $n=5$ as follows:

set.seed(111)
x = rnorm(5, 100, 15);  x
[1] 103.52831  95.03896  95.32564  65.46482  97.43686
t = t.test(x, mu=100)$stat; t
        t
-1.299141

The R code below computes $m$ such t-statistics and
puts them into a vactor t.stat of length $m,$ makes
a histogram of the $m$ values, and shows the density
function of $\mathsf{T}(\nu=4)$ for comparison.
[The t-statistic shown above is the first of the $m$ statistics sampled in this way.]

set.seed(111)
m = 10^5;  n = 5;  mu.0 = 100;  sg = 15
t.stat = replicate(m, t.test(rnorm(n,mu.0,sg),mu=mu.0)$stat)
summary(t.stat)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max.
-24.220863  -0.741238  -0.001904  -0.000634   0.735395  25.473347
var(t.stat)
[1] 1.992431   # aprx Var(T) = 2

hist(t.stat, prob=T, br=100, ylim=c(0,.4), col="skyblue2")
 curve(dt(x,n-1), add=T, lwd=2, col="red", n=10001)

Notice that the distribution of $\mathsf{T}(\nu=4)$ has a higher variance $(2)$ than does
standard normal $(1).$


问:
Maximal domain question related to the function $f(x) = \ln((x-1)^{2} - m)$
Let $f(x) = \ln((x-1)^{2}-m)$, where $m$ is a constant.

Question: state the maximal domain over which $f$ is defined for the case where $m > 0$?

Issue: how to go about it?

Thought process so far:

-Inside function must be greater than zero. We hence end up with $((x-1)^{2}-m)> 0$. Define $f(x)$ on CAS, but I get lost at what domain to use.

答: You would have, $$x>1\pm\sqrt{m}$$ if you solved the above equation. Putting this into set builder notation, we have $$\{x|x\in\mathcal R \quad and\quad x>1\pm\sqrt m\}$$ This is the only area of the real numbers where the logarithm is defined. Unless you extend it to the complex plane (i.e. use the principle logarithm so that it isn't multivalued) and set every negative integer $-\ell$ equal to $|\ell |e^{i\pi}$ then the principal logarithm is $$\operatorname{Log}(-\ell)=\log(|\ell|)+i\pi$$ this allows us to redefine it in the complex plane as $$\{z|z\in \mathcal R \}$$ or $$\{z| \arg z \in\{\pi,2\pi\}\}$$ by the definition of the principal logarithm. If you want to learn more about complex logarithms, check out this question: Log of a negative number.


问:
Why some cases can be ignored when dealing with probability
Simultaneously roll a blue die and a red die repeatedly until the values on the dice agree. Let X denote the number of rolls for which the dice total seven. Find the probability mass function of X.

The solution says we can ignore rolls in which the dice do not total seven and are not equal. I am wondering why we can do that.

Thanks!

答:
You can do that as a result of the fact that the probabilities associated with rolling a given amount are independent of the results of any previous rolls (i.e. the system is Markovian).

Consequently, you can ignore rolls that neither trigger the “end condition” (values are equal) nor alters the value of the random variable in question (no. of times the total equals seven), because the number of “null” rolls doesn’t affect the probabilities of either ending the rolls or rolling a net seven.


问:
Is the change of basis matrix a linear transformation?
Consider a change of basis matrix T on the vector space $R^3$. Can one say that there is a Isomorphism from $R^3 \rightarrow R^3 $ given by this matrix T?

答:
An isomorphism from $\Bbb R^3$ into $\Bbb R^3$ is a bijective linear map. And matrices are not linear maps. Therefore, the answer is negative.

However, if $T$ is a change of basis matrix, the map$$\begin{array}{ccc}\Bbb R^3&\longrightarrow&\Bbb R^3\\v&\mapsto&M.v\end{array}$$is a linear map, which happens to be an isomorphism.


问:
Calculating acceleration based on a starting and ending distance
I have been trying to figure this problem out and just can't.

The answer is supposed to be (A): 1.0 m/s/s.

Can anyone tell me how they got that answer?  I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer.  What am I doing wrong? Or is the correct answer not listed?  Thanks for any help.

Problem:
  A Pinewood Derby car rolls down a track.   The total distance travel is measured each second.   Use the chart to calculate the acceleration of the car.

Time(s) |  Dist(m)

0       |     0

1       |     1

2       |     3

3       |     6

4       |    10

A) 1.0  meters/sec/sec

B) 1.5  meters/sec/sec

C) 2.0 meters/sec/sec

D) 2.5 meters/sec/sec

答:
Remark: Since the data is discrete, we are going to assume mean velocities in each time intervals, otherwise the movement's law would have to be given explicitly.

Another approach is to compute the velocity first and then the acceleration, without using equations explicitly.
Since you are given five pairs $(t,d)$, where $t$ is the time and $d$ is the
distance, we can evaluate the average velocity $v=\frac{\Delta d}{\Delta t}$ over each period $\left[t_{i},t_{i+1}\right] $, starting at $t=0$ s, and $\Delta t_{i}=d_{i+1}-d_{i}$. The measurements are equally spaced, with $\Delta t=1$ s. We get the following average velocities:

$i=0\quad t\in \left[ 0,1\right] $ s,$\quad v=\frac{1-0}{1}=1$ ms$^{-1}=1 \text{m/s}$,

$i=1\quad t\in \left[ 1,2\right] $ s,$\quad v=\frac{3-1}{1}=2$ ms$^{-1}$,

$i=2\quad t\in \left[ 2,3\right] $ s,$\quad v=\frac{6-3}{1}=3$ ms$^{-1}$,

$i=3\quad t\in \left[ 3,4\right] $ s,$\quad v=\frac{10-6}{1}=4$ ms$^{-1}$.

Now we compute the average acceleration from one period to the next $a=\dfrac{\Delta v}{%
\Delta t}$:

from $\left[ 0,1\right] $ to $\left[ 1,2\right] $ s,$\quad a=%
\frac{2-1}{1}=1$ ms$^{-2},$

from $\left[ 1,2\right] $ to $\left[ 2,3\right] $ s,$\quad a=%
\frac{3-2}{1}=1$ ms$^{-2},$

from $\left[ 2,3\right] $ to $\left[ 3,4\right] $ s,$\quad a=%
\frac{4-3}{1}=1$ ms$^{-2}.$

Thus the average acceleration $a$ is constant: $a=1$ $\text{ms}^{-2}=1\text{m/s}^{2}$.

In summary we get the following table with data and computations:

$$\begin{array}{ccccccccc}
i & t_{i}\text{ (s)} &  & d_{i}\text{ (m)} &  & v_{i}\text{ (ms}^{-1}%
\text{)} &  & a_{i}\text{ (ms}^{-2}\text{)} &  \\
&  &  &  &  &  &  &  &  \\
& 0 &  & 0 &  &  &  &  &  \\
0 & 1 &  & 1 &  & 1 &  &  &  \\
1 & 2 &  & 3 &  & 2 &  & 1 &  \\
2 & 3 &  & 6 &  & 3 &  & 1 &  \\
3 & 4 &  & 10 &  & 4 &  & 1 &
\end{array}.$$

Notes:

Since the measurements are discrete, the velocity and accelerate
values are mean values and not instant one.

The acceleration unit ms$^{-2}$ (or $\text{m}/\text{s}^2$) is the same as meters/sec/sec in the question, tought this is an abuse of notation, since the meaning is (meters/second)/second.


答:
Of course, $T$ is linear.

Now suppose there are some vectors $v,w$ such that $T(v)=T(w)\Rightarrow T(v-w)=0$. Express $v,w$ in your initial basis $u_1,u_2,u_3$: $v=\alpha_1u_1+\alpha_2u_2+\alpha_3u_3, w=\beta_1u_1+\beta_2u_2+\beta_3u_3$. Then $T(v-w)=T((\alpha_1-\beta_1)u_1+(\alpha_2-\beta_2)u_2+(\alpha_3-\beta_3)u_3)=(\alpha_1-\beta_1)T(u_1)+(\alpha_2-\beta_2)T(u_2)+(\alpha_3-\beta_3)T(u_3)=0$.

We have that $\{T(u_1),T(u_2),T(u_3)\}$ is a basis by hypotheis, so they are linear independent, implying $(\alpha_i-\beta_i)=0$ for $i=1,2,3$, hence $v=w$ and $T$ is one-to-one.

Now take some vector $w$. We can express it with the final base $w=\gamma_1T(u_1)+\gamma_2T(u_2)+\gamma_3T(u_3)$. Consider the vector $v=\gamma_1u_1+\gamma_2u_2+\gamma_3u_3$. Then $T(v)=\gamma_1T(u_1)+\gamma_2T(u_2)+\gamma_3T(u_3)=w$, so $T$ is surjective.

We conclude $T$ is an isomorphism.

Note that this can be generalized for a change of basis in $R^n$ given by some matrix $T$.

Title: Is the change of basis matrix a linear transformation?

Body: Consider a change of basis matrix T on the vector space $R^3$. Can one say that there is a Isomorphism from $R^3 \rightarrow R^3 $ given by this matrix T?


答:
We know $70000 \equiv 0 {\pmod 7}$. Next we compute its residue under mod 11.

$$
70000=7\cdot 10^4 \equiv 7 \cdot (-1)^4\pmod{11}
$$

I will leave the rest for you.

Title: What is the smallest multiple of 77 that is greater than 70,000? 700,000?

Body:
What is the smallest multiple of $77$ that is greater than $70000$? $700000$?

How would you approach this question? My first thought was to factorize the $77$ into $7$ and $11$ but then how would you choose the exponents?


问:
Solution of |x| + 3x = 2 + 6i. Where is an error in my thinking?
So I solved the equation: $|x| + 3x = 2 + 6i$ but I don't know where my error is. And I know there is an error because Wolfram Alpha shows that the only solution is $ x = 2i $ . In my calculations I have 2 solutions.

$$|x| + 3x = 2 + 6i$$
$$\sqrt{a^2 + b^2} +3(a + bi) = 2 +6i$$
$$\sqrt{a^2 + b^2} = 2 +6i - 3a -3bi$$
$$\sqrt{a^2 + b^2} = 2-3a +(6 -3b)i$$
$$6 -3b=0$$
$$b=2$$
$$\sqrt{a^2 + 4} = 2-3a $$
$$a^2 + 4 = 4-12a+9a^2 $$
$$8a^2-12a=0 $$
$$4a(2a-3)=0 $$
$$a = 0 || a=3/2 $$

Why $x = 3/2 + 2i$ is not an answer?

答:
When you square both sides of $\sqrt{a^2+4} = 2-3a$, an extra solution is generated.

In particular, the solution $a = 3/2$ is the solution to $\sqrt {a^2+4} = 3a-2$, which is a perfectly valid solution for the squared equation $a^2+4 = (2-3a)^2$.

However for this $a$, $\sqrt{a^2+4} > 0$ but $2-3a < 0$, rendering it invalid for the original equation. For this reason, we must check whether all our solutions satisfy the original equation. There are no invalid steps in your calculations otherwise.


问:
Show that if $P$ is an invertible $m$ x $m$ matrix, then$rank(PA)=rank(A)$, then show if $Q$ is invertible, then rank $AQ$= rank $A$.
Did I show this properly? Any mistakes, any steps left out? Thank you.
$$$$

Show that if $P$ is an invertible $m$ x $m$ matrix, then$rank(PA)=rank(A)$.

solution:

$col(PA)=PA$x

$col(PA)=P(A$x$)$

$col(A)⊆col(PA)$

$rank(A)≤rank(PA)$

$nul(PA)=$x such that $PA$x$=0$

$nul(PA)=$x such that $P(A$x$)=0$

$nul(A)⊆nul(PA)$

$n-rank(A)≤n-rank(PA)$

$rank(PA)≤rank(A)$

$rank(PA)=rank(A)$

Part two: Now, use the above exercise to show that: if $Q$ is invertible, then rank $AQ$= rank $A$

From the above, we know that $rank(Q^TA^T)=rank(A^T)$.

$rank(Q^TA^T)=rank((AQ)^T)=rank(AQ)$

$rank(A^T)=rank(A)$

so, rank $AQ$= rank $A$.

答: By the rank theorem it suffices to show $\dim(\ker(\mathbf A))=\dim(\ker(\mathbf{PA}))=\dim(\ker(\mathbf {AP}))$. $$\ker(\mathbf{PA})=\{\mathbf x:\mathbf 0=\mathbf{PA}(\mathbf x)=\mathbf P(\mathbf {Ax})\}=\{\mathbf x:\mathbf{A}(\mathbf x)=\mathbf 0\}=\ker(\mathbf A)$$ $$\therefore \dim(\ker(\mathbf {PA}))=\dim(\ker(\mathbf A))$$  Finally, because $\mathbf P^T$ is invertible and the row and column space of $\mathbf{AP}$ have the same dimension then $$\dim(\ker(\mathbf{AP}))=\dim(\ker(\mathbf{P}^T\mathbf{A}^T))=\dim(\ker(\mathbf A^T))=\dim(\ker(\mathbf A))$$ where $^T$ denotes matrix transposition.


问:
Regarding supremum of Mobius distance with function values from the Disc algebra
Let $\mathbb{D}$ be the open unit disc in
$\mathbb{C}.$

Let $w_1, w_2 \in \overline{\mathbb{D}}$ with $w_1\neq w_2.$

Let $\tilde{c}(w_1, w_2)= \sup_f\left\{\left|\frac{f(w_1)-f(w_2)}{1-\overline{f(w_2)}f(w_1)}\right| : f\in \tilde{A}(\mathbb{D})\right\}$, where

$\tilde{A}(\mathbb{D})= \{ f\in$ Disc algebra$ : f(\overline{\mathbb{D}})\subset \mathbb{D}\}$.

Where the disc Algebra is the space of all complex valued functions that are analytic on the unit disc $\mathbb{D}$ and are continuous on $\overline{\mathbb{D}}.$ Is it true that  $\tilde{c}(w_1, w_2)$ is always strictly less than 1?

This question is similar to a previous question. The difference is that in the previous question the supremum was taken over bounded continuous functions, but here the supremum is taken over bounded analytic functions.

答:
The Schwarz-Pick theorem states that
$$
\left |\frac{f(w_1) - f(w_2)}{1 - \overline{f(w_2)} f(w_1)} \right| \le \left |\frac{w_1 - w_2}{1 - \overline{w_2} w_1} \right|
$$
for all holomorphic functions $f: \Bbb D \to \Bbb D$. It follows that
$$
\tilde{c}(w_1, w_2) \le \left |\frac{w_1 - w_2}{1 - \overline{w_2} w_1} \right| < 1 \, .
$$


问:
Sequence of strongly convergent operators $\{T_n\}$ satisfies $\sup_{x\in K }\| T_n(x) - T(x)\| = 0$ for compact $K$
Exercise: Let $E,F$ be Banach Spaces and $T,T_1,T_2$,... operators in $L(E,F)$ such that $T_n(x)\rightarrow T(x)$, $\forall x\in E$ . prove that for all compact $K\subset E$
\begin{equation}
\sup_{x\in K}{||T_n(x)-T(x)||}\rightarrow 0
\end{equation}

the book gives the following suggestion:
proceed by contradiction and use the  Banach-Steirhauss theorem to guarantee that $\sup\{||T||,||T_1||, ||T_2||,...\}< \infty $.

Proof:
Suppose that $\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}\nrightarrow 0$ then we have that exist an $\varepsilon>0$  such that $\forall n\in N$ exist $T_n$ such that
\begin{equation*}
||T_n-T||=\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}>\varepsilon
\end{equation*}
By hypothesis $T_n(x) \rightarrow T(x)$  for all $x \in E$, i.e,  $\forall \epsilon=1$ exists $N \in \mathbb{N}$ s.t $n \geq N$
\begin{equation*}
 ||T_n(x)-T(x)|| < 1
\end{equation*}
then $||T_n(x)||< 1 +||T(x)|| \leq 1+c$, i.e, $\sup\{||T(x)||,||T_1(x)||, ||T_2(x)||,...\}< c_x $.  By  Banach-Steirhauss theorem

$$sup\{||T||,||T_1||, ||T_2||,...\}< \infty.$$

In this moment I have two ideas to continue.

1.Since $K$ is compact, we know that  all sequences $\{T_n\}\in K$ admits a subsequence convergent.I don't know how to relate this fact with the negation and how to get to the contradiction.

2.On the other hand, I was thinking in prove that the space generated by the sequence of operators is Cauchy, but since $L(E,F)$ is Banach then all Cauchy sequences converge. However, I don't know how I can apply in this prove that $K\subset E$ is compact.

I would be very grateful if someone could help me.

答:
Suppose to the contrary that $$\lim_n\sup_{x \in K}\|T_nx - Tx \| \neq 0.$$

By going to a subsequence, we may assume that $$c_n:=\sup_{x \in K}\|T_n x- Tx \|\geq \epsilon$$
for some $\epsilon > 0$ and all $n \geq 1$.

Choose a sequence $(x_n)_n$ in $K$ with $$\|T_n x_n- Tx_n \| = c_n. \quad \quad(*)$$ This exists by continuity and compactness of $K$. Again, by choosing a subsequence, we may assume that $(x_n)_n$ converges to some $x\in K$.

Note that $\lim_n T_n x_n = Tx$. Indeed
$$\|T_n x_n - Tx\| \leq \|T_n x_n - T_n x\| + \|T_n x -Tx\| \leq \|T_n\| \|x_n- x \| + \|T_n x  - Tx\|$$
and by an application of the uniform boundedness principle, $\sup_n \|T_n\| < \infty$.

Letting $n \to \infty$ in $(*)$, we thus obtain
$$\lim_n c_n=0$$
which is impossible since $c_n \geq \epsilon$ for all $n \geq 0$. This is the desired contradiction.


问:
Finding the space of solutions of a single equation?
If the equation is $$3x+3y+2z=0$$

How would you be able to determine a vector/pair of vectors which describes the space of solutions?

I'm quite new to linear algebra so apologies if this is a very dumb question.

答:
You are looking for vectors perpendicular to $(3,3,2)$.

Two are $(3,3,2)\times(0,0,1)$ and $(3,3,2)\times(1,0,0)$.

They are linearly independent and span the space of solutions.


问:
Finding triangular matrix
If I have the matrix:

$$M=\begin{pmatrix}-1 & 1 & 1 \\ -10 & 5 & 3 \\ -2 & 1 & 3  \end{pmatrix} $$

I know the matrix $M$ is triangularisable, but I am not sure how to find $A$ such that $M=P^{-1}AP$ to find the corresponding triangular matrix $A$.

So far I have found the eigenvalues are $2,3$, then once I do $M-3I$,  I get that the basis for that matrix is $\begin{pmatrix}1 & 2 & 1 \end{pmatrix}^T$ and $\begin{pmatrix}1 & 3 & 1 \end{pmatrix}^T$. However I am not sure how to go any further? Can someone show me how? Thanks!

答: The vector $(1,2,2)$ is an eigenvector of $M$ corresponding to the eigenvalue $3$ and the vector $(0,-1,1)$ is an eigenvector of $M$ corresponding to the eigenvalue $2$. Now, consider the equation$$M.(x,y,z)=2(x,y,z)+(0,-1,1).$$One solution of this system is $(1,3,0)$. So, if $v_1=(1,2,2)$, $v_2=(0,-1,1)$, and $v_3=(1,3,0)$, you have $M.v_1=3v_1$, $M.v_2=2v_2$, and $M.v_3=2v_3+v_2$. So, take$$P=\begin{bmatrix}1&0&1\\2&-1&3\\2&1&0\end{bmatrix}$$(the columns of $P$ are the vectors $v_1$, $v_2$, and $v_3$). Then$$P^{-1}MP=\begin{bmatrix}3&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$

Title: Finding triangular matrix


问:
Prove $f(x)=x^2+3x$ is Continuous at $x=a$
I am attempting to give a $\delta,\epsilon$ proof that the following function, $f(x)=x^2+3x$ is continuous at $x=a$, where $a$ is any real number. I have the following:

Given any $\epsilon > 0$, find a $\delta > 0$ such that if $|x-a|<\delta$ then $|(x^2+3x)-((a^2)+3a)|<\epsilon\\ \\$
$\text{We have that:}\\ \\
\quad\quad \begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3| <\epsilon \end{align}\\$
$\text{Attempt to Bound $|x+a+3|$: Let $\delta \leq 1$}\\ \\
\quad\quad \begin{align} &\Rightarrow |x-a|<1\\
&\Rightarrow -1 <x-a<1\\
&\Rightarrow a-1<x<a+1 ;\text{add $a+3$ in order to obtain $x+a+3$}\\
&\Rightarrow 2a+2<x+a+3<2a+4\\\end{align}$

The issue seems to be my attempt to bound $|x+a+3|$ because I am supposed to reach the following:

$\begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3|< \epsilon\\
&=|x-a|*(2|a|+3)< \epsilon\\
&=|x-a|< \frac{\epsilon}{2|a|+3}\\ \\
&\text{Choose $\delta=min(1,\frac{\epsilon}{2|a|+3})$} \end{align}\\$

I am not exactly sure how we arrive at $2|a|+3$. An idea that I have is if $2a+2<x+a+3<2a+4$, and $a$ is a real number, then $2a+3$ is in between $2a+2$ and $2a+4$.

Any help would be greatly appreciated!

答:
Use the sequential criterion for continuity at $a$. Let $(a_n)_{n=1}^\infty$ be a real sequence with limit $a$. Now, it suffices to show $\lim_{n\to\infty}f(a_n)=f(a)$.
$$\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a_n(a_n+3)=\lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}(a_n+3)=a(a+3)=f(a)$$


问:
Basic algebra: linear equations
$$0 = \frac{1}{6}(-X) + \frac{5}{6}(7.5) \implies X = 5 \times 7.5 = 37.5.$$

I cannot see how you get X = 5 x 7.5 in the equation above; could someone make the steps to this conclusion explicit for me?

答:
$$0=\dfrac16(-X)+\dfrac56(7.5).$$

Multiply both sides by $6$:

$$0=-X+5\times7.5.$$

Add $X$ to both sides:

$$X=5\times7.5.$$

Title: Basic algebra: linear equations


问:
Derivative of $L_2$ Norm function
I have an equation of the form $f(x) = ||x - x_0 ||^2$ where $x$ and $x_0$ is in $R^n$.

What I want to do is Taylor expand this about a point - lets call this $x_a$. I know that:
$f(x)$ is approximated $f(x_a) + (\operatorname{grad}(f(x_a))*(x-x_a)$ which is equivalent to $||x_a - x_0 ||^2 + (\operatorname{grad}(f(x_a))*(x-x_a)$.

The problem I have, is that I'm very unsure of how to calculate what $\operatorname{grad}(f(x))$ is.

I'd appreciate any help. (If someone can fix all the notation up, I'd appreciate it too!)

答:
For typing convenience, drop the subscripted $x$ variables in favor of
$$a=x_a,\qquad z=x_0$$
Expand the norm using the dot product and write the function as
$$\eqalign{
f &= (x-z)\cdot(x-z) \\
}$$
and calculate its gradient
$$\eqalign{
df &= 2(x-z)\cdot dx \\
g(x) \;=\; \frac{\partial f}{\partial x} &= 2(x-z) \\
}$$
Therefore at $x=a,\,$ the gradient is $\,g(a)=2(a-z)\,$ and the first-order Taylor expansion becomes
$$\eqalign{
f(a+h)
 &\approx f(a) + g(a)\cdot h \\
 &= f(a) + 2(a-z)\cdot h \\
}$$
where $0<\|h\|\ll 1\;$ (i.e. $\,h$ is very small).

Title: Derivative of $L_2$ Norm function


问:
Confusion on the proofs of convolution theorem of Fourier Transform.
The convolution theorem of Fourier transform is stated as follows:

Define $h(x):=f(x)*g(x)$, then we have $$\hat{h}(k)=\hat{f}(k)\hat{g}(k).$$

I have a confusion of the proofs of this theorem. Most proofs go in the following way:

Define $h(x):=f(x)*g(x)$.  By definition, we have $$\hat{h}(k)=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx.$$
Then, plugging in the integral expression of $f*g$ into the above integral yields us
\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}h(x)e^{-ikx}dx\\
&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)g(s)ds\Big)e^{-ikx}dx\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx.\ \ \ \ \ \ (*)
\end{align*}

Then, most proofs directly change integral order in the question $(*)$, so that

\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dsdx\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-s)g(s)e^{-ikx}dxds\\
&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds.
\end{align*}

What remains is standard: note that by the shifting property, the inner integral is $$\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx=\hat{f}(k)e^{-iks},$$ and thus
\begin{align*}
\hat{h}(k)&=\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{\infty}f(x-s)e^{-ikx}dx\Big)g(s)ds\\
&=\hat{f}(k)\int_{-\infty}^{\infty}g(s)e^{-iks}ds\\
&=\hat{f}(k)\hat{g}(k),
\end{align*}
where the last equality was obtained by definition. The proof is concluded.

However, why could we change the integral order of (*) in the first place? Don't we need Fubini? I tried to use Fubini as follows: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)e^{-ikx}|dsdx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|f(x-s)g(s)|dsdx$$ but I do not know how to proceed further. I had expected that I can split the integral so that I would have two $L^{1}-$norm, but since the first integral is $ds$, I could not split the integral..

Do we directly assume the function is nice enough so that we can interchange the integral order? If so, what type of functions do we assume?

Thank you!

答:
Without changing the order of integration:

$ikx=ik(x-s)+iks$  and $u=x-s$ so $h(k)=\int_{-\infty} ^\infty \int_{-\infty} ^\infty f(u)e^{-iku}g(s)e^{-iks}dsdu$$=\hat{f}(k)\hat{g}(k)$.


问:
Deriving formula for function F(g,h,n) such that (g^a)%N = h%N
I have a maths / coding assignment in which I am required to write a code to find the value of a for given values of g,h and N such that:

(g^a)%N = h%N

where % operator returns the remainder after dividing the 2 numbers (example: 5%2 = 1)

Since there is no other information given,I can't convert the equation to g^a = N.m + h%N where m is another integer used to reverse the modulus process.

How can I derive a formula for a?

答:
Your equation (g^a)%N = h%N is equivalent to

$$g^a \equiv h \pmod N$$

The problem of finding $a$ is known as the discrete logarithm, and no efficient algorithm is known. The simplest way to find the smallest solution is to write a loop checking each (g^a)%N until it is equal to h%N.

However $g^a$ gets large quite quickly. You can use the fact that (g*((g^a)%N))%N = g^(a+1)%N to make the numbers involved smaller.


问:
Expected number of maximum difference between heads and tails
Let's say a fair coin is tossed $n$ times. Let $H(i)$ denote the total number of heads by the $i$-th toss and $T(i)$ denote the total number of tails by the $i$-th toss.

Define $D(i) = |H(i) - T(i)|$, the difference at $i$-th toss.

The maximum is defined as $M(n) = max^n_{i=1}D(i)$.

Now, what'd be an asymptotic bound for $E(M(n))$?

答:
By Doob's L^2 inequality (see e.g. https://arxiv.org/pdf/1202.0447.pdf) we have
$E(M(n)^2) \le 4E(D(n))^2)=4n $, so $E(M(n)) \le 2 \sqrt{n}$.

On the other hand, denote  by $Z$ a standard normal variable. By the Central limit Theorem,
$$E(M(n)) \ge E|D(n)|=\sqrt{n}E|Z|(1+o(1))=\sqrt{\frac{2n}{\pi}}(1+o(1)) \,.$$


问:
What is the smallest multiple of 77 that is greater than 70,000? 700,000?

What is the smallest multiple of $77$ that is greater than $70000$? $700000$?

How would you approach this question? My first thought was to factorize the $77$ into $7$ and $11$ but then how would you choose the exponents?

答:
A multiple of $77$ is multiple of $7$ and $11$ simultaneously.

$70070$ is clearly multiple of $11$. (The two $7$'s are placed in odd and even places; the alternating sum is zero. )

It is obviously multiple of $7$. Hence answer is $\boxed{70070}$.

By exactly same logic, the other number is $\boxed{700007}$.


问:
Prove the complement of $\operatorname{Col}(A) = \operatorname{Null}(A)$ if $A = A^2$.
I tried by proving that when $A=A^2$, the complement of $\operatorname{Col}(A)$ is a subset of $\operatorname{Null}(A)$ and vice versa. But I don't know how to use the face that $A$ is idempotent.
How should I approach this question?

答:
Hint :

$$x =  x - A(x) + A(x)$$

with $x-A(x) \in \mathrm{Ker}(A)$ and $A(x) \in \mathrm{Im}(A)$.


问:
How can I use the Box-Mueller method to generate generic normal distributions?
Using the box-mueller method, we can generate standar normal distributions from two uniformly distributed random variables.

However, how do we use this method to generate normal distributions with means and variances other than 0 and 1 respectively?

答: If X is a standard normal random variable (mean $0$ and variance $1$), then $\sigma X+\mu$ is a normal random variable with mean $\mu$ and variance $\sigma^2$. So, just apply this transformation to the samples obtained from the Box-Mueller method. </p>


问:
Why are $e^{-4\ln(-x)+4}$ and $e^{-4\ln(-x)}$ positive?

Algebraically, why is $e^{-4\ln(-x)+4}$ a positive result but $e^{-4\ln(-x)}$ a positive result?

It seems like with logarithmic properties, they should both be negative, but Wolfram Alpha shows the graph and states that it must be positive. Is there a logarithmic identity to explain this discrepancy? Thanks much!

答:
You can write the former as $$e^{-4\ln(-x) + 4} = e^{-4\ln(-x)} \cdot e^4 = e^{\ln((-x)^{-4})} \cdot e^4 = \frac{e^4}{x^4}.$$ Notice that this will always be nonnegative.

Similarly, you can write $$e^{-4\ln(-x)} = e^{\ln((-x)^{-4})} = \frac{1}{x^4},$$ which will still be nonnegative.


问:
In how many ways can 15 identical math books be distributed to six students?
question: In how many ways can 15 identical math books be distributed to six students?
i try do this: P(15,6) could be this the answer or i should check the question, i find out other? help please</p>

答:
One particular distribution of the $15$ identical books to $6$ students could be

$$B B B | B B | B B B | B B | B B B | B B$$

Note that this has needed $5$ dividers, and these could be placed anywhere in the configuration of $15 + 5 = 20$ symbols.
Thus the number of ways of distributing them = ${20}\choose 5$


问:
Gram-Schmidt Process: How does it fail?
I am confused about the Gram-Schmidt process for a linearly dependent set. I can't seem to grasp why it would cause the linearly dependent vectors to equal zero. Can someone please explain?</p>

答:
In the middle of Gram-Schmidt, you have your current list of (orthogonal) vectors $\{e_i\}$, and a new vector $v$. Then you subtract off the part of $v$ which can be constructed from the $\{e_i\}$, and if $v$ is not in the span of $\{e_i\}$, you are left with the part which is orthogonal to all of the $\{e_i\}$ vectors.

But if $v$ is in the span of the $e_i$, then the "part of $v$ which can be constructed from the $\{e_i\}$" is in fact just $v$ itself, so you get a 0 vector when you subtract that off.

Title: Gram-Schmidt Process: How does it fail?


问:
Do three value complex numbers exist?
My understanding of complex numbers is that it is the use of an extra axis of dimension. By plotting a signal on polar axis, the real number, x, is the sine of the signal, and the imaginary number, y, is the cosine of the signal. This understanding is derived from its use in the fourier transform.

(By this understanding, one could take the cosine of a value to get its imaginary component, so my understanding it limited as it doesn't explain the logic behind √(−1)=i.)

So, I have the question, is it possible to have three value complex numbers?

答:
You can have 1, 2, 4, and 8-dimensional real normed division algebras.

What does that mean? It means that if you want to find a way to multiply lists of $n$ numbers (or $n$-dimensional vectors, alternatively), you need $n=1,2,4,$ or $8$ to get a meaningful answer. It is difficult to say why only these numbers work. As one elementary indication that perhaps 3-dimensions is spooky, note that multiplication of complex numbers multiplies length; that is,

$$|a+bi| \cdot |c+di| = |(a+bi) \cdot (c+di)|.$$

The length of the product is the product of the lengths. This leads to an identity

$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2,$$

where a product of two sums of two squares is also a sum of two squares.

However, a product of a sum of three squares is not always a sum of three squares, so no such identity exists for triples.

Title: Do three value complex numbers exist?


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:
As I showed in this post, given a circle $\Gamma$ and two points $X,Y$ such that $Y$ lies on the polar of $X$ w.r.t. $\Gamma$, we have

$$XY^2=\text{Pow}_{\Gamma}(X)+\text{Pow}_{\Gamma}(Y)$$

Back to your case, it is well-known that $E$ lies on the polar of $F$ w.r.t. the circumcircle of $ABCD$ (recall, for example, how to prove Brokard's Theorem) and, hence, $$EF^2=60^2+63^2=7569\iff EF=87$$


答: $e^y > 0$   for all $y$ which means that it is also the case for $y=-4ln(-x)+4$ and $y=-4ln(-x)$ means $e^{-4ln(-x)+4} > 0$ and  $e^{-4ln(-x)>0} >0$ there where they are defined.

Title: Why are $e^{-4\ln(-x)+4}$ and $e^{-4\ln(-x)}$ positive?

Body:
Algebraically, why is $e^{-4\ln(-x)+4}$ a positive result but $e^{-4\ln(-x)}$ a positive result?

It seems like with logarithmic properties, they should both be negative, but Wolfram Alpha shows the graph and states that it must be positive. Is there a logarithmic identity to explain this discrepancy? Thanks much!


问:
Isomorphism type of order 16 subgroup of $S_6$
[This question is prompted by Robinson ex. 1.6.17 which I am self-studying.

The question is mostly about improving my style and rigour.]

I needed to find the isomorphism type of a Sylow 2 subgroup of $S_6$.

Obviously such a subgroup is of order 16 (the highest power of 2 dividing 6!)
Since all the Sylow 2 subgroups are isomorphic, I just need to pin one down, call it $S$, and investigate it.

First I looked at the representation of $S_6$ as permutations, and spotted that (some) permutations on elements 1..6 elements can be split into a direct sum of permutations on 1..4 and 5..6.  So $S_4 \times S_2$ is a subgroup of $S_6$.

Question: how to make the previous remark more rigorous/elegant?

By Sylow again, $S_4$ has a subgroup of order $8$. So there is a subgroup of order 16 within $S_4 \times S_2$.

Again, looking at $S_4$ as symmetries of 1,2,3,4 and playing around I eventually found elements $a = (1 2)$ and $b = (1 4 2 3)$ with $abab=1$, I realised that $a$ and $b$ generate $D_8$, and so $S \cong D_8\times Z_2$.

Question: how to remove some of the guesswork from this step.  For example, I know $D_8$ is a candidate of the right order, how can I show that it is or isn't the right group more easily than guessing permutations?  [I can usually understand the general theories, but when confronted by specific groups I go to pieces and end up tabulating.]

Thanks.

答:
For the remark that $S_6$ contains a subgroup ismorphic to $S_4\times S_2$, I believe the neatest way of saying it is that the standard embedding of $S_4$ in $S_6$ (i.e. permutations fixing 5 and 6) commutes elementwise with $H=\langle (56)\rangle\simeq S_2$ and their intersection is trivial, so that their product $S_4\cdot H=\{\sigma\tau\ |\ \sigma\in S_4, \tau\in H\}$ is indeed isomorphic to their direct product.

The rest of the argument is fine in itself: moreover, you will have to do some handiwork when dealing with concrete groups. For example, knowing that $D_8$ has the right order is not enough.  Even if you exclude abelian groups of order 8 (which seem unlikely to be 2-Sylows of $S_4$), you are left with two options, namely $D_8$ or $Q_8$ (quaternions), and it is concrete exploration of $S_4$ to give you the answer on which of the two is the correct Sylow group.

Surely, though, there are ways to pick the right permutations that are more clever than others: in this case, moreover, there is a way you can get (somewhat) naturally to the answer, i.e. costruct a $2$-Sylow of $S_n$ inductively on $n$. For a positive integer $n$, call $\mu_2(n)$ the exponent of 2 in the prime factorization of $n!$, $P_n$ the (isomorphism class) of a 2-Sylow of $S_n$.

Clearly, $P_2\simeq C_2$ (where $C_n$ is the cyclic group of order $n$), and this holds for $P_3$ as well since $S_2$ embeds in $S_3$ and $\mu_2(3)=\mu_2(2)$.

As $\mu_2(4)=\mu_2(2)+2$, you must enlarge $\langle(12)\rangle<S_4$ by a factor 4. One factor 2 comes naturally by embedding $S_2$ in $S_4$ as $\langle(34)\rangle$: for the same argument used in the first paragraph, you may clearly see that $$H=\langle(12)\rangle\cdot\langle(34)\rangle\simeq C_2\times C_2.$$ Now, you have no more disjoint $C_2$ to multiply to your $H$, but you can still exchange $(12)$ and $(34)$, namely by conjugating by the double transposition $(13)(24)$. If $K=\langle(13)(24)\rangle$, this gives the product $HK$ (which is a subgroup, since $H$ and $K$ commute) a natural structure of semidirect product $H\rtimes K$ or, if you prefer, the isomorphism structure of $D_8$, so that $P_4\simeq (C_2\times C_2)\rtimes C_2$.

At this point, since $\mu_2(6)=\mu_2(4)+1$, just multiplying $P_4<S_6$ by the disjoint $C_2$ given by $\langle (56)\rangle$ does the trick, and leaves you with $P_6\simeq P_4\times P_2$.

Also, this construction has the advantage of showing a pattern. In fact, if you compute $\mu_2(n)$, you will easily see that:

$P_{2^{n+1}}\simeq (P_{2^{n}}\times P_{2^{n}})\rtimes C_2$ as in the case of $P_4$, which is normally written $P_{2^{n}}\wr C_2$ and called a wreath product;

if $n=a_0+a_1\cdot 2+\cdots+a_k\cdot 2^k$ in base 2, $P_n\simeq \prod_{j=1}^k (P_{2^i})^{a_i}$, as with $P_6$;

this generalizes to other primes in the obvious way.


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?

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

Body:
Hint :

$$\sum_{n=0}^{+\infty}  \sum_{k={n+1}}^{+\infty} \frac{x^k}{k!}=\sum_{k=1}^{+\infty}  \sum_{n=0}^{k-1} \frac{x^k}{k!} = \sum_{k=1}^{+\infty}  k\times\frac{x^k}{k!} = \sum_{k=0}^{+\infty}  \frac{x^{k+1}}{k!} = x e^x$$


问:
Graph theory proving question on connected graphs
Let $G_1$ and $G_2$ be two vertex disjoint (that is $V(G_1)\cap V(G_2) = \emptyset$) connected graphs of order at least $2$. Let $H = G_1 + G_2$. Prove that $\text{Cen}(H)$ is always a connected graph.

答:
Case 1: For $i=1,2$ no vertex of $G_i$ is adjacent to every other vertex of $G_i$.

Case 2: For $i=1,2$, some vertex of $G_i$ is adjacent to every other vertex of $G_i$.

Case 3: For  $i=1$ or $2$ no vertex of $G_i$ is adjacent to every other vertex of $G_i$, but for $j=1$ or $2$ some vertex of $G_j$ is adjacent to every other vertex of $G_j$.

In all cases note that the distance between two vertices is at most $2$, as you can get from one to the other by going to a vertex in the other graph, and (if necessary) back.


问:
Sequence of strongly convergent operators $\{T_n\}$ satisfies $\sup_{x\in K }\| T_n(x) - T(x)\| = 0$ for compact $K$
Exercise: Let $E,F$ be Banach Spaces and $T,T_1,T_2$,... operators in $L(E,F)$ such that $T_n(x)\rightarrow T(x)$, $\forall x\in E$ . prove that for all compact $K\subset E$
\begin{equation}
\sup_{x\in K}{||T_n(x)-T(x)||}\rightarrow 0
\end{equation}

the book gives the following suggestion:
proceed by contradiction and use the  Banach-Steirhauss theorem to guarantee that $\sup\{||T||,||T_1||, ||T_2||,...\}< \infty $.

Proof:
Suppose that $\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}\nrightarrow 0$ then we have that exist an $\varepsilon>0$  such that $\forall n\in N$ exist $T_n$ such that
\begin{equation*}
||T_n-T||=\sup_\limits{x\in K}^{}{||T_n(x)-T(x)||}>\varepsilon
\end{equation*}
By hypothesis $T_n(x) \rightarrow T(x)$  for all $x \in E$, i.e,  $\forall \epsilon=1$ exists $N \in \mathbb{N}$ s.t $n \geq N$
\begin{equation*}
 ||T_n(x)-T(x)|| < 1
\end{equation*}
then $||T_n(x)||< 1 +||T(x)|| \leq 1+c$, i.e, $\sup\{||T(x)||,||T_1(x)||, ||T_2(x)||,...\}< c_x $.  By  Banach-Steirhauss theorem

$$sup\{||T||,||T_1||, ||T_2||,...\}< \infty.$$

In this moment I have two ideas to continue.

1.Since $K$ is compact, we know that  all sequences $\{T_n\}\in K$ admits a subsequence convergent.I don't know how to relate this fact with the negation and how to get to the contradiction.

2.On the other hand, I was thinking in prove that the space generated by the sequence of operators is Cauchy, but since $L(E,F)$ is Banach then all Cauchy sequences converge. However, I don't know how I can apply in this prove that $K\subset E$ is compact.

I would be very grateful if someone could help me.

答:
$\newcommand{\F}{{\mathscr{F}}}$

This post already has an excellent answer by @MathQED. Nevertheless, since this question pops up in many situations in Analysis, I
think it is interesting to highlight some of the ingredients involved as this might  help to see things from a better
perspective and, most importantly, to identify the phenomenon when it shows up in other situations.

Let me phrase this in terms of one definition and three exercises for the interested reader:

1. Definition.  Given a metric (or topological space) $X$, a metric space $M$, and a set $\F$ of functions from $X$ to $M$,  one says
that $\F$ is equicontinuous at a given point $x_0\in  X$  if, for all $\varepsilon >0$,  there exists an open set $U\subseteq X$ containing
$x_0$,  such that
$$
  d(f(x), f(x_0))<\varepsilon ,  \quad \forall x\in  U,  \quad \forall f\in  \F.
  $$
(The crucial point here is that the same $U$ can be chosen for all functions $f$ in $\F$).
If $\F$ is equicontinuous at every point of $X$, we simply say that $\F$ is equicontinuous.

2. Exercise.
A collection $\F$ of linear transformations from a normed space $E$ to a normed space $F$ is equicontinuous if and
only if $\F$ is equicontinuous at $0$, if and only if
$$
  \sup_{T\in \F}\|T\|<\infty .
  $$

3. Exercise.  Given $X$ and $M$ as in (1), given an equicontinuous sequence  $\{f_n\}_n$ of functions from $X$ to $M$, and given another
function $f$ from $X$ to $M$,  TFAE:

$f_n\to f$ pointwise,

$f_n(x)\to f(x)$, for every $x$ in some dense subset $D\subseteq X$,

$f_n|_K\to f|_K$, uniformly, for every compact subset $K\subseteq X$.

4. Exercise.  Answer the question posed by the OP based on the above exercises!


答:
We can start with the well known Maclaurin series for

$e^x$

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

Body:
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?


问:
Why does P(A,B) notation sometimes cause an addition and sometimes cause a multiplication?
I am trying to understand these notes on a whiteboard

$P(A,B) = P(A)+P(B) - P(A\cap B)$

if $A,B$ are independent $P(A\cap B)= P(A) P(B)$

I can understand the first equation by thinking of Venn Diagram with circles A and B overlapping.

For the second equation I think of a Venn Diagram with the circles not overlapping. But why would I be multiplying instead of adding?

To add to my confusion, I sometimes see $A\cap B$ written as $AB$

答:
If this is on one whiteboard, the instructor made a typographical error, or someone transcribed or interpreted the whiteboard incorrectly.  There is variation in the meaning of $P(A,B)$, but those are clearly two different meanings.

The first "$A,B$" means "$A$ or $B$", or $A \cup B$, i.e. $A$ union $B$, for events that are not necessarily disjoint.

The second "$A,B$" means "$A$ and $B$" or $A\cap B$, i.e. $A$ intersect $B$, in the case where $A$ and $B$ are independent.

Sometimes people write "$A$ and $B$" or $A\cap B$ as $AB$.  Sometimes people use the comma for this.  Sometimes comma is used for conditional probability.  I've never seen the comma for disjunction or union, as in the first example, but as I said, there is variation.

(To visualize the second equation, don't use circles.  Use a rectangle divided vertically and horizontally.  What's on one side of the vertical line is $A$, and what's on one side of the horizontal line is $B$.  That's independence.  Non-independence would be represented using a diagonal line or a line with steps in it.)


问:
If $\sum_{x=1}^{|A|} |\psi_x\rangle\langle \psi_x|=I$, then $|\psi_x\rangle\langle \psi_x|$ is a basis projectoer.
Let $\left\{E_{x}\right\}_{x=1}^{|A|}$ is a rank 1 matrix in $\operatorname{Herm}(A)$ which $\sum_{x=1}^{|A|} E_{x}=I^A$. $A$ is our Hilbert space and $I^A$ is the identity matrix.Then I want to show that $E_{x}=p_x|y\rangle\langle y|$ which $\left\{|y\rangle\right\}_{x=1}^{|A|}$ are the basis of Hilbert space $A$ and $p_x$ is a number.

$\textbf {Note}:$

All of them are in the Dirac notation. It means that $|y\rangle$ is a column vector and $\langle y|$ is hermitian conjugate of $|y\rangle$.

$\textbf {Attempt}:$

I know $E_x$ is rank-$1$, so it has a form like $|\psi_x\rangle\langle \psi_x|$. We assumed that $\sum_{x=1}^{|A|} E_{x}=I^A$, So $\sum_{x=1}^{|A|} |\psi_x\rangle\langle \psi_x|=I^A$. Then I wrote $|\psi_x\rangle=\sum_{y=1}^{|A|} a_{x,y} |y\rangle$, So $\sum_{x=1}^{|A|}|\psi_x\rangle\langle \psi_x|= \sum_{x=1}^{|A|} \sum_{y=1}^{|A|} \sum_{z=1}^{|A|} a_{x,y}a_{x,z}^* |y\rangle\langle z|=I^A$. On the other hand we know that $\sum_{y=1}^{|A|} |y\rangle\langle y|=I^A$. But now I have no idea to conect them with each other to conclude that $E_{x}=p_x|y\rangle\langle y|$. Could you please help me ?

答:
First off, you mean $E_x=p_x|x\rangle\langle x|$, not $p_x|y\rangle\langle y|$. Second off, you're misinterpreting the problem.

If the basis $\{|x\rangle\}$ is already given, you can't say $E_x=p_x|x\rangle\langle x|$, because that's certainly not true! Rather we need to say there exists a basis $\{|x\rangle\}$  for which the formulas for $E_x$ are true. Plus we can go further and show it is an orthogonal basis, or even an orthonormal basis (so the $p_x$s are $1$).

Here is a version of the proposition that ought to have been given:

If $\{E_x\}$ is a set of $d$ rank-$1$ Hermitian matrices (on a Hilbert space of dimension $d<\infty$) for which the sum $\sum_x E_x$ is the identity matrix $I$, then $E_x=|x\rangle\langle x|$ for some orthonormal basis $\{|x\rangle\}$.

First off, we already know each $E_x$ may be written as $|\psi_x\rangle\langle\psi_x|$ for some vectors $|\psi_x\rangle$, or by factoring out the square magnitude $p_x$ of $|\psi\rangle$ we could write these as $E_x=p_x|x\rangle\langle x|$ for some unit vectors $|x\rangle$. Now we want to show $\{|x\rangle\}$ is an orthonormal basis and the $p_x$s are $1$.

To see that it's a basis, note that the image of $I=\sum_x p_x|x\rangle\langle x|$ is the whole space and it is also contained in the span of $\{|x\rangle\}$, so this is a spanning set, and since it's the same number of elements as the Hilbert space's dimension (which is finite) this must be a basis.

Second, if we apply $I$ to a specific $|y\rangle$ from the basis we get

$$ |y\rangle=\sum_x p_x|x\rangle \langle x|y\rangle $$

Equating coefficients gives $1=p_y$ and $\langle x,y\rangle=0$ for $x\ne y$. Since $y$ is arbitrary, that means all $p_y$s are $1$, and the unit vectors $\{|x\rangle\}$ are orthogonal so it's an orthonormal basis.


答:
\begin{align}
& \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) = \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
& \begin{array}{cccccccccc}
= & x & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & & & + & \dfrac{x^5}{120} & + & \cdots
\end{array} \\[10pt]
= {} & x + x^2 + \frac{x^3} 2 + \frac{x^4} 6 + \frac{x^5}{24} + \cdots \\[8pt]
= {} & x\left( 1 + x + \frac{x^2} 2 + \frac{x^3} 6 + \frac{x^4}{24} +\cdots \right) = xe^x.
\end{align}

Postscript: Maybe it is useful to express it in a way in which the general form is explicit rather than being indicated by three dots that mean "continue with the same pattern."
\begin{align}
& \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
= {} & \sum_{n,k\,:\,k\,\ge\,n+1\,\ge\,1} \frac{x^k}{k!} \\[8pt]
= & \sum_{k=1}^\infty \left( \sum_{n=0}^{k-1} \frac{x^k}{k!} \right)
\end{align}
But no $\text{“}n\text{”}$ appears in this sum in which $n$ goes from $0$ to $k-1$; therefore the sum is just $x^k/k!$ multiplied by the number of terms, which is $k$:
$$
\sum_{k=1}^\infty k\cdot\frac{x^k}{k!} = x\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!} = xe^x.
$$

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


问:
Calculate limit of $f_n(x)$ for $x \in (0, 1]$
I want to calculate the limit as $n \longrightarrow \infty$ of
$$f_n(x)=-\frac{1}{x\ln n+1} + \frac{nx}{(n\ln n)x^2+1}$$
when $x \in (0, 1]$ (in fact I want to prove that it is $0$).

I am having trouble distinguishing the various possible values of $x$. For example, if $x$ is very close to $0$, then
$$\lim_{n\to \infty} f_n(x)=-\frac{1}{x\ln n+1} = -1$$
But if it is very close to $1$ then
$$\lim_{n\to \infty} f_n(x)=-\frac{1}{x\ln n+1} = 0$$
The other limit is even more problematic, because it diverges if $x$ is very close to $0$. I don't know how to manage this and, even more, I am very far from proving that the whole limit is actually $0$

答:
Keeping in mind that $x$ is held constant as $n\to\infty$, we see that

For the first term, $\lim_{n\to\infty}\frac{-1}{x\ln{n}+1}=0$ as $\ln{n}\to\infty$

For the second term, since both $nx$ and $(n\ln{n})x^2+1$ both tend to $\infty$, we may apply L'Hospital's Rule.

$\lim_{n\to\infty}\frac{nx}{(n\ln{n})x^2+1}=\lim_{n\to\infty}\frac{x}{(\ln{n}+1)x^2}=0$

We now conclude that $\lim_{n\to\infty}\frac{-1}{x\ln{n}+1}+\frac{nx}{(n\ln{n})x^2+1}=0$


问:
CLT for independent, but non-identically distributed Poisson random variables

Show that

$$\frac{\sum_{i=1}^n(Y_i-\lambda_i)}{\sqrt{\sum_{i=1}^n\lambda_i}}\overset{d}{\rightarrow}\mathcal
 N(0,1)$$

where $\{Y_i\}\sim\mathsf{Pois}(\lambda_i)$ are independent and
$\sum_{i=1}^n\lambda_i\rightarrow\infty$

My Try:

I tried using Lyapunov CLT but could not get a sharp enough bound. I next tried moment generating functions where we'd need $M_{\frac{\sum(Y_i-\lambda_i)}{\sqrt{\sum\lambda_i}}}(t)\rightarrow e^{\frac{t^2}{2}}$ but we have

$$\begin{align*}
M_{\frac{\sum(Y_i-\lambda_i)}{\sqrt{\sum\lambda_i}}}(t)
&=M_{\frac{Y_1}{\sum\lambda_i}}(t)M_{\frac{-\lambda_1}{\sum\lambda_i}}(t)\cdots M_{\frac{Y_n}{\sum\lambda_i}}(t)M_{\frac{-\lambda_n}{\sum\lambda_i}}(t)\\\\
&=M_{Y_1}\left(\frac{t}{\sum\lambda_i}\right)\exp\left(\frac{-\lambda_1}{\sum\lambda_i}t\right)\cdots M_{Y_n}\left(\frac{t}{\sum\lambda_i}\right)\exp\left(\frac{-\lambda_n}{\sum\lambda_i}t\right)\\\\
&=\exp\left(\lambda_1\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)\cdots\exp\left(\lambda_n\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)e^{-t}\\\\
&=\underbrace{\exp\left(\sum\lambda_i\left(e^{\frac{t}{\sum\lambda_i}}-1\right)\right)}_{\rightarrow e^t}e^{-t}\\\\
&\rightarrow 1
\end{align*}$$

so either I made a mistake or this approach doesn't work. I think the property that $\sum_{i=1}^n Y_i\sim\mathsf{Pois}\left(\sum_{i=1}^n \lambda_i\right)$ is of importance but I'm not sure how I'd use that here. Perhaps the Lindeberg condition would be the way to go. It's used in a similar problem here for the case of independent, but non-identically distributed exponential random variables. Any help would be appreciated!

答:
For simplicity, let $Y=\sum Y_i$ and $\lambda=\sum\lambda_i$. Then the MGF of $(Y-\lambda)/\sqrt{\lambda}$ is
$$
\exp(\lambda(e^{t/\sqrt{\lambda}}-1))\cdot e^{-t\sqrt{\lambda}}
$$
You can prove this approaches $e^{t^2/2}$ using the MacLaurin series of $e^{t/\sqrt{\lambda}}$.