问:
Pattern for all the binary chains divisible by 5
For instance, $x = 101$ is divisible by $5$ because it is the integer 5. Same thing for $x=1111$ is also divisible by 5 as it is the integer 15. However, $x=1100$ is not divisible by $5$ as it is the integer 12.

Is there a pattern the recognise the binary chains divisible by 5?

答:
I use a similar approach to rogerl, but using hexadecimal (base 16), rather than base 4. The divisibility test for 5 in base 16 works like the divisibility test for 3 in base 10, because 3 divides (10-1) and 5 divides (16-1).

The base 16 representation of a positive integer $n$ is of the form
$$n = a_0 + a_1\cdot16 + a_2\cdot16^2 + a_3\cdot16^3 + a_4\cdot16^4 + \dots$$
with $0 \le a_i<16$.

But $16\equiv 1 \pmod{15}$, so
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{15}$$
And because 3 and 5 both divide 15, we also have
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{3}$$
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{5}$$

Thus we can test divisibility by 5 of a hex number by simply adding its hex digits, mod 5. (And of course we can get the hex digits of a binary number by grouping the bits in blocks of 4).

For example, $123_{10}$ is  $$0111 1011_2$$ in binary and $$\mathrm{7b}_{16}$$ in hex. The values of those two blocks are $7$ and $11$. $7 + 11 = 18 \equiv 3\pmod{5}$, so $$0111 1011_2$$ leaves a remainder of 3 when divided by 5.

Here are some larger examples.

```<code>
dec 1000
bin 0011 1110 1000
val    3   14    8
sums   3    2    0 mod 5
remainder = 0

19136214
0001 0010 0011 1111 1110 1101 0110
   1    2    3   15   14   13    6
   1    3    1    1    0    3    4
remainder 4

</code>```

Here's a link to a small live Python script which does these calculations.

As I said earlier, you can also use this method for divisibility by 3, just add mod 3 instead of 5. Or add mod 15, and test divisibility by 15, 5, and 3 at the same time.

Note that you can do this test in any order, but if you start from the left, don't forget to pad the bit string with zeroes (if necessary) to make the total number of bits divisible by 4.

Body:
I use a similar approach to rogerl, but using hexadecimal (base 16), rather than base 4. The divisibility test for 5 in base 16 works like the divisibility test for 3 in base 10, because 3 divides (10-1) and 5 divides (16-1).

The base 16 representation of a positive integer $n$ is of the form
$$n = a_0 + a_1\cdot16 + a_2\cdot16^2 + a_3\cdot16^3 + a_4\cdot16^4 + \dots$$
with $0 \le a_i<16$.

But $16\equiv 1 \pmod{15}$, so
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{15}$$
And because 3 and 5 both divide 15, we also have
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{3}$$
$$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{5}$$

Thus we can test divisibility by 5 of a hex number by simply adding its hex digits, mod 5. (And of course we can get the hex digits of a binary number by grouping the bits in blocks of 4).

For example, $123_{10}$ is  $$0111 1011_2$$ in binary and $$\mathrm{7b}_{16}$$ in hex. The values of those two blocks are $7$ and $11$. $7 + 11 = 18 \equiv 3\pmod{5}$, so $$0111 1011_2$$ leaves a remainder of 3 when divided by 5.

Here are some larger examples.

dec 1000
bin 0011 1110 1000
val    3   14    8
sums   3    2    0 mod 5
remainder = 0

19136214
0001 0010 0011 1111 1110 1101 0110
   1    2    3   15   14   13    6
   1    3    1    1    0    3    4
remainder 4

Here's a link to a small live Python script which does these calculations.

As I said earlier, you can also use this method for divisibility by 3, just add mod 3 instead of 5. Or add mod 15, and test divisibility by 15, 5, and 3 at the same time.

Note that you can do this test in any order, but if you start from the left, don't forget to pad the bit string with zeroes (if necessary) to make the total number of bits divisible by 4.


问:
Assume a real-valued net has a cluster point $a\in \mathbb R$. Is there a cofinal subnet converging to $a$?
Let $(D, \geq)$ be a directed set and let $(n_d)_{d\in D}$ be a real-valued net. Assume $a\in \mathbb R$ is a cluster point of $(n_d)$, i.e., for every neighborhood $U$ of $a$ and every $d\in D$ there exists $d'\geq d$ such that $n_{d'}\in U$.

It is a standard fact that there exists a subnet of $(n_d)$ that converges to $a$.

But is it possible to find a cofinal subnet (see for example wikipedia) with the same property? To rephrase: is there a cofinal subset $D'\subset D$ such that $\lim_{d\in D'}n_d = a$?

If $(n_d)$ takes value in an arbitrary topological space $X$, then this is not true. See for example this answer and this one. My question is specific to the case $X=\mathbb R$, and for this space I could not find any counter-example nor proof.

答:
Not necessarily.

Let $D=\omega_1\times\omega$ ordered lexicographically: $\langle\alpha,m\rangle\preceq\langle\beta,n\rangle$ iff $\alpha<\beta$, or $\alpha=\beta$ and $m\le n$; this is clearly a directed set. Define a net

$$\nu:D\to\Bbb R:\langle\alpha,n\rangle\mapsto 2^{-n}\,;$$

clearly $0$ is a cluster point of $\nu$.

Let $C$ be a cofinal subset of $D$. For each $n\in\omega$ let $C_n=C\cap(\omega_1\times\{n\})$; $|C|=\omega_1$, so there is some $m\in\omega$ such that $|C_m|=\omega_1$. Then $C_m$ is cofinal in $C$, but $\nu[C_m]=\{2^{-m}\}$ is disjoint from the nbhd $(-2^{-m},2^{-m})$ of $0$. Thus, no cofinal subnet of $\nu$ converges to $0$.

ContentLicense: CC BY-SA 4.0


问:
Radius of convergence of a Taylor series
I'm currently working my way through Lanczos' Linear Differential Operators, and had a question from the same.

Find the radius of convergence of the Taylor expansion of $e^{-1/x^2}$, if the center of expansion is at the point $x=4$. The answer is stated to be $R=4$.

My understanding is that this function has no singularities, on the real line or in the complex plane. Hence the radius of convergence must be infinite. Is this incorrect? What subtlety am I missing?

答:
It's not true that $lim_{x \longrightarrow 0} \ e^{-1/x^{2}} = 0$ in the complex plane.

\begin{equation}
lim_{n \rightarrow \infty} \frac{i}{n} = 0
\end{equation}

\begin{equation}
lim_{n \rightarrow \infty} \ e^{-1/(\frac{i}{n})^{2}} = lim_{n \rightarrow \infty} \ e^{n^{2}} = \infty
\end{equation}

Thus, function has a singularity on $x = 0$ so $R=4$.


问:
one of the following statements is true.Identify:-
one of the following statements is true.Identify:-

(A) Every subset of $\Bbb{Q}$ which is bounded above has a unique supremum rational number.

(B)Every subset of $\Bbb{Q}$ which is bounded above has a unique supremum real number.

(C)The set of rational numbers has least least upper bound property.

(D)The set of real numbers does not have the greatest lower bound property.

I am confused because I think both B and D are correct but they say only one option is correct

答:
They are all wrong.

(A) $\{x\in \mathbb Q\mid x^2<2\}$ is bounded from above (e.g. $42$ is an upper bound), but has no rational number as supremum

(B) $\emptyset$  is bounded from above (e.g. $42$ is an upper bound), but has no real number as supremum (of course this example also works for (A))

(C) Least upper bound property means that every nonempty set that is bounded from above has a least upper bound (in the set). As the exmaple to (A) shows, $\mathbb Q$ does not have this property

(D) $\mathbb R$ is specifically constructed to have the least upper bound property!

ContentLicense: CC BY-SA 3.0


问:
Convergence of $\sum_{n=0}^{\infty}a_nb_n$ if $a_n>0 \ \forall n\in \mathbf{N}$, $\sum_{n=0}^{\infty}a_n$ converges and $b_n$ bounded
Let $(a_n)$ a sequence such that $a_n>0 \ \forall n\in \mathbf{N}$. Suppose that $\sum_{n=0}^{\infty}a_n$ converges and the sequence $(b_n)$ is bounded. Show that $\sum_{n=0}^{\infty}a_nb_n$ converges.

I would like to have a feedback on my proof and to know if everything holds. Thanks in advance.

Proof.

First of all, as  $\sum_{n=0}^{\infty}a_n$ converges, we can note that $\lim_{n\to \infty}a_n=0$. So, $\exists N \ \forall n\ge N$: $0<a_n<1$ .

Then, as $(b_n)$ is a bounded sequence, we have by definitiot that:

$\exists M>0: \ |b_n|\le M \ \forall n\in \mathbf{N}$.

With all these preliminary results we can write that:

$|b_n|\le M \iff |b_n|\cdot a_n\le M\cdot a_n \underbrace{\iff}_{a_n>0 \ \forall n} |b_na_n|\le M|a_n|$.

As $\sum_{n=0}^{\infty}a_n$ converges and $a_n>0 \ \forall n \in \mathbf{N}$ (so there is no difference between $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}|a_n|$), $\sum_{n=0}^{\infty}|a_n|$ converges and so $\sum_{n=0}^{\infty}M|a_n|$ too.

By comparaison we conclude that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely so it converges.

答:
Your answer is correct. You can use Cauchy criterion.

Assume $ M>0$.
Given $ \epsilon>0$,

For $ m $ and $ n $ large enough
$$|\sum_{k=n}^ma_k|=\sum_{k=n}^ma_k< \frac{\epsilon}{M}$$
and
$$|\sum_{k=n}^ma_kb_k|\le M\sum_{k=n}^ma_k<\epsilon$$

This proves the convergence of the series $\sum a_nb_n$.


问:
How to prove $1+\frac{1}{2} \ln(x) \leq \sqrt{x}$ when $x \ge 1$
Please help proving that $1+\frac{1}{2}\ln(x) \leq \sqrt{x}$ when $x \ge 1$.

I tried with $\frac{1}{2}\ln(x)=\ln{\sqrt{x}}$ then use $X=\sqrt{x}$. So we need to prove: $1+\ln{X} \leq X$ but then I'm stuck

Thank you so much.

答:
Welcome to MSE,you can take $$f(x)= 1+\frac12 \ln x -\sqrt x$$over $[1,+\infty)$now check $f'(x)$
$$f'=\frac{1}{2x}-\frac{1}{2\sqrt x}=\\\frac12 (\frac{1}{\sqrt x})(\frac1{\sqrt x}-1)\\=\frac1{2\sqrt{x}}(\frac{1-\sqrt x}{\sqrt x})\leq 0 $$ so $f(x) $ is decreasing function over $[1,+\infty)$   can you take over now ?


问:
Finding Jordan basis of a matrix $(4\times 4)$
I'm facing a problem finding a Jordan basis for this ($4 \times 4$) matrix:
$$\left(\begin{matrix}3&-1&1&7\\9&-3&-7&-1\\0&0&4&-8\\0&0&2&-4\end{matrix}\right)$$

I know that the characteristic polynomial is $\lambda^4=0$ and it's minimal polynomial is: $x^2$. This means that $(A-0I)^2=0$. Also the Jordan form for this matrix is
$$\left(\begin{matrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{matrix}\right)$$ But I'm facing problems in the general procedure to compute the basis for such a matrix?

答:
First, the minimal polynomial is not $\lambda^2$, and it seems to me that Ben's row reduction is wrong.

We have that the characteristic polynomial is $p(\lambda)=\lambda^4$, so, we already know that the jordan matrix will only have the value $0$ on the main diagonal.

Now we want to find the kernel of $(A-0I)=A$, so, reducing matrix $A$, we get

$$\pmatrix{
3&-1&1&7\\
9&-3&-7&-1\\
0&0&4&-8\\
0&0&2&-4} \leadsto
\pmatrix{
3&-1&0&0\\
0&0&1&0\\
0&0&0&1\\
0&0&0&0}
$$

Solving the system is not difficult,
$$
\pmatrix{
3&-1&0&0\\
0&0&1&0\\
0&0&0&1\\
0&0&0&0}
\cdot
\pmatrix{
x_1\\
x_2\\
x_3\\
x_4}
=
\pmatrix{
0\\
0\\
0\\
0}
$$
we get, $x_3=x_4=0$ and $3x_1=x_2$, if $x_1=1$, then $x_2=3$, then the base of the kernel is given by the vector, $(1,3,0,0)$. From this, we conclude that the form of jordan has only one block, so it is of the form,

$$\left(\begin{matrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{matrix}\right)$$

as the exponent of the minimal polynomial can be given by the size of the largest block, we have that the minimal polynomial is $p_m(\lambda)=\lambda^4$

You can go on and find the Jordan base, as follows:

You want to find a vector $v$, so that, $A^3v,A^2v,Av,v$, they are all non-null. Calculating the exponents, we obtain,

$$A^2=\pmatrix{
0&0&28&-14\\
0&0&0&126\\
0&0&0&0\\
0&0&0&0}
\qquad
A^3=\pmatrix{
0&0&84&-168\\
0&0&252&-504\\
0&0&0&0\\
0&0&0&0}
$$

note that $e_3$ and $e_4$ are suitable vectors for what we want (because the third and fourth columns do not cancel and $A^4$ is the null matrix), let's choose $e_4$. So,

$$A^3e_4=(-168,-504,0,0)\qquad A^2e_4=(-14,126,0,0)\qquad Ae_4=(7,-1,-8,-4)$$

this way a Jordan base is

$$\{(0,0,0,1),(7,-1,-8,-4),(-14,126,0,0),(-168,-504,0,0)\}$$

it is not a pretty base, but this is a way to resolve the exercise quickly. We have so,

$$P=
\pmatrix{
0&7&-14&-168\\
0&-1&126&-504\\
0&-8&0&0\\
1&-4&0&0}
$$

And you can see that this solves the problem, $J=P^{-1}AP$.


问:
Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$
Find the range of:
$$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$

What I tried:

Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$
then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$
$$y_{max}at X=\pi/4$$

The rest is too complicated. I am stuck up.

Can someone please give a analytical solution.

Even Wolfram alpha doesn't help.

答:
A related problem. First, we study the expression

$$ \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1} \longrightarrow_{|x|\to\infty} 1,  $$

which implies $ y(x)\longrightarrow_{|x|\to \infty} 0 $. To find the maximum of the function $y(x)$, Let's study the function

$$ h(t)=\sqrt{\sin(t)}+\sqrt{\cos(t)}.  $$

The maximum of the above function is attained at $t=\frac{\pi}{4}$ which can proved using derivative test. So, this implies our function attains its max when

$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} \implies x=0.6632987771, -0.6632987771. $$

Plugging back in the function y(x) gives the max which is $y=1.681792830$. So the range is

$$ 1 < y \leq 1.681792830.$$

Note: You can solve $ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} $ easily as,

$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4}\implies \frac{x^2+e}{x^2+1} =e^{\frac{\pi}{4}}\implies x^2+e= e^{\frac{\pi}{4}}x^2+ e^{\frac{\pi}{4}}=\dots.$$

I think you can finish it.


问:
Degrees or radian in Euler's identity?
In Euler's identity, should values in the RHS (the sin and cos part) always be in radians? like $e$ to the power of imaginary $\pi$ is negative one, so if I had $e$ to just the power of $i$, then I would take the radian values of $\cos(1)$ and $\sin(1)$ right?

答:
This is a good question, because the answer gets to the heart of what an angle actually is. There are many ways to formally define this concept, but the simplest is that an angle denotes the ratio of an arc length to a radius. This idea is particularly important in the context of unit circle. Since the unit circle has a radius of $1$, $\cos\theta$ and $\sin\theta$ correspond to the $x-$ and $y$-coordinates that you arrive at after tracing an arc of $\theta$ units counterclockwise around, where your starting position is $(1,0)$. When angles are understood in this way, the question of how they are measured becomes rather odd. An angle is a dimensionless quantity, a pure number that represents a ratio. The degrees symbol is a bizarre scale factor of $\pi/180$. For instance,
$$
\sin(90^{\circ})=\sin(90 \cdot \pi/180)=\sin(\pi/2)=1 \, .
$$
Euler's formula, $e^{it}=\cos t+i\sin t$ has a very simple geometric interpretation. If a particle is moving around the unit circle centred at the origin of the complex plane, then at time $t=0$, the particle has position $(1,0)$. The particle then moves at unit speed anticlockwise around the unit circle, so that at time $t$ the particle has moved $t$ units along the circle's arc. Hence, by the geometric definitions of sine and cosine, the position of the particle is $(\cos t,\sin t)$.

Title: Degrees or radian in Euler's identity?


答:
First, we define the function $f(x)= \sqrt{x}-1-\frac{1}{2}\ln x$. Notice that
$$f'(x) = \frac{1}{2\sqrt{x}}-\frac{1}{2x} = \frac{\sqrt{x}-1}{2x}\geq 0,\phantom{a} \forall x\in[1,+\infty)$$

Now, notice that
$$f(1)=\sqrt{1}-1+\frac12\ln(1)=0,$$
so $f(1)\geq 0$ and $f$ is an increasing function in the interval $[1,+\infty)$, so from this it is concluded that
$$f(x)\geq 0, \forall x\in[1,+\infty),$$
hence that $\sqrt{x}\geq 1+\frac12\ln x$ for all $x\geq1$.

Title: How to prove $1+\frac{1}{2} \ln(x) \leq \sqrt{x}$ when $x \ge 1$

Body:
Please help proving that $1+\frac{1}{2}\ln(x) \leq \sqrt{x}$ when $x \ge 1$.

I tried with $\frac{1}{2}\ln(x)=\ln{\sqrt{x}}$ then use $X=\sqrt{x}$. So we need to prove: $1+\ln{X} \leq X$ but then I'm stuck

Thank you so much.


问:
Prove subset of function space is open
Prove $S=\{f \in C[0,1]|f(1)<3\}$ is open.With the sup metric.$C[0,1]$ is space of continuous functions.

Attempt let $\epsilon=\text{min} \{3-f(x)|x \in [0,1]\}$

Let $g \in B_{\epsilon}(f)$. Then $|g(x)|<|f(x)-g(x)|+|f(x)|<3-|f(x)|+|f(x)|<3$ so $|g(1)|<3$ and $g \in S$.

Comment: Thinking about this, it seems to me the choice of $\epsilon$ may be wrong. Would it be better to choose $\epsilon=3-f(1)$? Is this proof okay, or what?

答: Consider a function $\xi :C[0,1]\to\mathbb{R},$ $\xi (f) =f(1)$ since $$|\xi (f) -\xi (g) |=|f(1) -g(1)|\leqslant ||f-g||$$ the function $\xi $ is continuos hence $S=\xi^{-1} ((-\infty ,3))$ is open as a inverse image of open set by continuous function.

Title: Prove subset of function space is open


问:
Prove $(\frac{-\pi}{2},\frac{\pi}{2})$ with $d(x,y)=|\tan x-\tan y|$ is complete
I was trying to prove that this metric space is complete Prove $(\frac{-\pi}{2},\frac{\pi}{2})$ with $d(x,y)=|\tan x-\tan y|$ is complete.

Attempt let $x_n$ be cauchy in the metric space with respect to the metric $d$. Then for $\epsilon>0$  there is an $N$ such that $n,m \geq N$ $\implies$ $|\tan (x_n)-\tan(x_m)|<\epsilon$. Then $|\tan(x_n)|$ is a cauchy sequence of real numbers, and converges to $x \in \mathbb{R}$. So $x_n$ converges to a $y \in (\frac{-\pi}{2},\frac{\pi}{2})$. So the metric space is complete.

Comments: I am having quite some difficulty determining the validity of the last few sentences. Can anyone say if this is correct? If not how should I approach this?

答: The map $\arctan\colon\Bbb R\longrightarrow\left(-\frac\pi2,\frac\pi2\right)$ is an isometry if in $\Bbb R$ you consider the usual metric and in $\left(-\frac\pi2,\frac\pi2\right)$ you consider the metric $d$. So, since $\Bbb R$ is complete with respect to the usual metric, then so is $\left(\left(-\frac\pi2,\frac\pi2\right),d\right)$.


问:
Rigid transformations
Given that $T(x,y)= (y-2,x+3)$ is a rigid transformation how can i express $T = S \circ U$ where S is a translation and $U$ is an orthogonal transformation.

It may be a little simple, but it´s been a long time since i dive into geometry and i´m trying to be on the track again.
Thanks :)

答:
It's indeed simple:

Let $U$ be the reflection through the diagonal line $x=y$, that is,
$$U(x,y)=(y,x)$$
and let $S$ be the translation  by vector $(-2,3)$, i.e.
$$S(a,b)= (a-2,\,b+3)\,.$$
Then
$$S\circ U(x,y)=S(y,x)=(y-2,\,x+3)=T(x,y)$$
for any vector $(x,y)\in\Bbb R^2$.

Bonus: can you find the other translation $S'$ that makes $T=U\circ S'$?

Title: Rigid transformations


问:
Intersection of two arcs on sphere
I have two arcs on a sphere that are defined as pair of points: $(\theta_0, \varphi_0)$, $(\theta_1, \varphi_1)$. I need to find a point where they intersect, or some indication if they don't. What is important is that they are arcs, not circles, so it is important to find intersection that is on the arcs specifically (not on projected circles). Arcs are meant to be shortest possible arcs. I am failing to find any specific equations on the internet. Thanks!</p>

答: As far as I can tell to verify that two arcs intersect on the same sphere they must have an angular range that has at least shares one particular value of phi and theta, for instance if one arc covers (30,30 ) to (60, 60) and the other covers (0, 0) to (0, 60) they wouldn't intersect even if theta agrees. Basically the condition is that the intervals for both sets of phi and theta have nonzero set theoretic intersection at the same time, that is the ranges for phi must have at least one value in common at the same time as both ranges for theta have at least one value in common.</p>

Title: Intersection of two arcs on sphere


答:
Picking up where the OP left off, having reduced the inequality to showing that $1+\ln X\le X$ for $X\ge1$, note that

$$\ln X=\int_1^X{dt\over t}\le\int_1^Xdt=X-1$$

since $1/t\le1$ for $t\ge1$.


问:
Proof by induction: L'Hospital rule doesn't work, what next?
Show through induction:
$\lim_{x\to\infty} \frac{\ln^k x}x=0$

Well the base case of 1 is rather trivial, but when substituting k for k+1 for the inductive hypothesis step we get,

$\lim_{x\to\infty} \frac{\ln^{k+1} x}x=0$

Applying L'Hospital's rule gets is not helpful as we will get

$\lim_{x\to\infty} \frac{(k+1)\ln^{k} x}x$

Applying l 'hospital's again gets:

$\lim_{x\to\infty} \frac{-(k+1)(\ln^{k-1} x) ({\ln (x)} -k)}{x^2} $

Applying l 'hospital's rule runs into an issue of always having an infinity/infinity issue.

Sometimes when l'hospital's rule gets stuck, there is an algebraic manipulation that can be done to overcome the problem. I am not seeing anything. I am wonder what would be another course of action to prove the k+1 case of induction.

答:
My diagnosis of the OP's difficulty is not yet being comfortable with mathematical induction. Induction tells us that if we have any statement $P(k)$ about positive integers $k$, and we can prove both

$P(1)$ is true, and

for all positive integers $m$, if $P(m)$ is true then $P(m+1)$ is true;

then that comprises a proof that $P(k)$ is true for all positive integers $k$.

In our case, let $P(k)$ be the statement: $\displaystyle\lim_{x\to\infty} \frac{\ln^k x}x = 0$.

The OP claims that proving $P(1)$ is done, and indeed it follows directly from an application of l'Hopital's rule.

We now need to prove that for all positive integers $m$, if $P(m)$ is true then $P(m+1)$ is true. Notice that this task is not proving that $P(m)$ is true, nor is it proving that $P(m+1)$ is true. It is literally proving that the implication $P(m) \implies P(m+1)$ is true. And indeed, L'Hopital's rule can be used to show that if $\displaystyle\lim_{x\to\infty} \frac{\ln^m x}x = 0$, then $\displaystyle\lim_{x\to\infty} \frac{\ln^{m+1} x}x = 0$; indeed, this is what was done in manofwar's answer.

Together, these two steps prove (by mathematical induction) that $\displaystyle\lim_{x\to\infty} \frac{\ln^k x}x = 0$ is true for all positive integers $k$.

It's perfectly reasonable to be confused by induction at first, to feel that it's circular or incomplete, or to wonder why proving both the base case and the induction implication suffices to prove the whole universal statement. That's all ok! But I do want to emphasize that this really is a complete proof (and, in particular, that manofwar's answer does successfully overcome the difficulty stated by the OP).


问:
Integration in physical forms
Suppose we have a straight uniform rod and we divide it into infinitesimal sections.The total mass of the rod is $M$.Now each section has mass $dM$.Then the total mass in integration form is given by $\int_{0}^{M}dM$.My question is what do we mean by the limits $0$ and $M$ on the integration sign.If it was length $dx$ instead of mass,then $\int_{0}^{x}dx$ would have made sense to me since we are slicing from the top to bottom in length but i don't get it how $0$ and $M$ plays the role of top to bottom in this case.

答: It should be $M=\int_0^MdM^\prime$ vs. $x=\int_0^xdx^\prime$. If you seek a physical interpretation, imagine the gradual addition of small amounts of mass.

Title: Integration in physical forms


问:
Regular n-simplex with unit n-volume
Given are the coordinates of a regular $n$-simplex in $\mathbb{R}^n$ where all edge lengths are 1. For symmetry the vertices are centered on the origin.

The volume of that unit $n$-simplex is $V_n=\frac{\sqrt{n+1}}{n! \sqrt{2^n}}$ (see Wikipedia for definitions and formulas).

As the $n$-simplex is centered we can multiply the coordinates by a factor $\alpha_n$ without changing symmetry. The factor only affects the volume. By which factor $\alpha_n$ the coordinates of an $n$-simplex have to be multiplied to get $V_n=1$?

The first (approximate) values are
$\alpha_1=1, \alpha_2\approx1.5194, \alpha_3\approx 2.0397, \alpha_4=2.5598$.

答: The volume scales with the $n$th power of the scale factor, so $\alpha_n$ is the reciprocal of the $n$th root of the volume you know.

Title: Regular n-simplex with unit n-volume


问:
Cross product of two sums of three vectors?
This seems like a simple question but I couldn't find anywhere to verify

Is it true that:

$$(\mathbf{a}+\mathbf{b}+\mathbf{c})\times (\mathbf{d}+\mathbf{e}+\mathbf{f})
=
(\mathbf{a}\times \mathbf{d})+(\mathbf{a}\times\mathbf{e})+(\mathbf{a}\times\mathbf{f})
+
(\mathbf{b}\times \mathbf{d})+(\mathbf{b}\times\mathbf{e})+(\mathbf{b}\times\mathbf{f})
+
(\mathbf{c}\times \mathbf{d})+(\mathbf{c}\times\mathbf{e})+(\mathbf{c}\times\mathbf{f})
$$

Just wondering whether the following identity is true.

答:
This was already ansered in comments, so I'll just write it in answers from Colescu's comment.
We know that the cross product is distributive, meaning $(a+b)\times c=a\times c + b \times c$ and $c×(a+b)=c×a+c×b$

We can do this a bunch of times from the left to arrive to the right side of the equation $(a+b+c)×(d+e+f)=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$

So we start with $(a+b+c)×(d+e+f)$, We'll keep using the distributive propiety until we arrive at the desired equation.

$$(a+b+c)×(d+e+f)$$
$$=(a+[b+c])×(d+e+f)$$
$$=(a×(d+e+f))+([b+c]×(d+e+f))$$
$$=(a×(d+e+f))+(b×(d+e+f))+(c×(d+e+f))$$
$$=(a×(d+[e+f]))+(b×(d+[e+f]))+(c×(d+[e+f]))$$
$$=(a×d+a×[e+f])+(b×d+b×[e+f])+(c×d+c×[e+f])$$
$$=(a×d+a×e+a×f)+(b×d+b×e+b×f)+(c×d+c×e+c×f)$$
$$=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$$


答:
Induction start/inition:

$k=0$: $\lim_{→∞}\frac{\ln^{0}}{}=\lim_{→∞}\frac{1}{}=0$

or

$k=1$: $\lim_{x\to\infty}\frac{\ln^{1}x}{x}=\lim_{x\to\infty}\frac{\ln x}{x}$

This is the case l'Hospital $\frac{\infty}{\infty}$!

So
$\lim_{x\to\infty}\frac{\ln^{1}x}{x}\approx_{x \to \infty}\frac{\frac{1}{}}{1}=\lim_{x\to\infty}\frac{1}{x}=0$

This induction start already shows some part of the induction step. The start for $k=1$ makes use of the one for $k=0$.

In words the derivation of the nominator is $\frac{1}{x}$, the one of the denominator is $1$. The limit exists and is $0$.

Induction step:

$k$ is some arbitrary integer. We assume that for this chosen $k$ the identity for the limit to $\infty$ is true.

We now look at $k+1$ following exactly in the Integers the $k$ for which we know the identity holds.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}$

This is indeed a case for the rules of l'Hospital.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}\approx_{x \to \infty}\frac{\infty}{\infty}$

It if allowed to differentiate nominator and denominator separate.
The derivative of the nominator is $(k+1)\ln^{k}x\frac{1}{x}$. This is chain rules applied twice. First the outer polynomial of the degree k+1 is derivated, then the in Logarithm is derivated. The derivative of the denominator is $1$.

This quotient is the $k+1$ of the quotient of the induction step premise.

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=(k+1)\lim_{x \to \infty}\frac{\ln^{k}x}{x}$

$k$ in integer and fixed so:

$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=0$.

For $k=1$ this is simply the extented induction start.

There is nothing more necessary then the derivation of polynomials without coefficients of integer degree and $0$ and the derivation of the $\ln$ (natural logarithm) and the rules of l'Hospital for the case $\frac{\infty}{\infty}$ (quotient of infinity by infinity both positive).

The answer presents the indcution schema following Mathematical_induction. That is the biggest ambiguation overall. I did write explicitly that the induction step premise did assump that the equation, limit holds for all other natural $k$ up to the selected $k$ for the induction step already and that everyone was able to convince him- or herself that this is true.

I have not specially not written that the last equation of manofwar holds. I already factored and wrote the limit identity. This holds only in the limit of infinity:

$$\lim_{x \to \infty}\frac{\ln^{k+1}x}{x}=(k+1)\lim_{x \to \infty}\frac{\ln^{k}x}{x}$$

Since $k+1$ is a constant for the limit this is true for all integer $k$ after applied the rule of l'Hospital and calculating rules for limits (Limit of a function section Properties).

I did not restrict the presentation to reals.

Title: Proof by induction: L'Hospital rule doesn't work, what next?

Body:
Show through induction:
$\lim_{x\to\infty} \frac{\ln^k x}x=0$

Well the base case of 1 is rather trivial, but when substituting k for k+1 for the inductive hypothesis step we get,

$\lim_{x\to\infty} \frac{\ln^{k+1} x}x=0$

Applying L'Hospital's rule gets is not helpful as we will get

$\lim_{x\to\infty} \frac{(k+1)\ln^{k} x}x$

Applying l 'hospital's again gets:

$\lim_{x\to\infty} \frac{-(k+1)(\ln^{k-1} x) ({\ln (x)} -k)}{x^2} $

Applying l 'hospital's rule runs into an issue of always having an infinity/infinity issue.

Sometimes when l'hospital's rule gets stuck, there is an algebraic manipulation that can be done to overcome the problem. I am not seeing anything. I am wonder what would be another course of action to prove the k+1 case of induction.


问:
Recovering the CDF of a Laplace Distribution with a specified PDF
I'm working on finding the cumulative distribution function of a Laplace distribution with the following PDF:
$$f(x) = \frac{\lambda}{2}e^{-\lambda\mid x-\theta\mid}, \ -\infty \lt x \lt \infty$$
I have attempted the following solution, however I am not sure whether it is correct:
$$
F_{X}(x) = \int_{-\infty}^{x}\frac{\lambda}{2}e^{-\lambda\mid y-\theta\mid}dy
$$
Firstly find the indefinite integral via the substitution method:
$$
\frac{\lambda}{2}\int e^{u}du,\ u=-\lambda\mid y-\theta\mid \\
$$
The differentiation of $u$, $du$, works out to be $\frac{-\lambda y(y-\theta)}{\mid y- \theta \mid}$. Rewrite the integral in terms of $u$ and include $du$:
$$\frac{\mid y-\theta\mid}{-2y(y-\theta)}\int e^{u}du \\
=\frac{\mid y-\theta\mid}{-2y(y-\theta)}(e^{-\lambda\mid y-\theta\mid} + C)
$$
With that done, the definite integral I think is as follows:
$$F_X(x) = \frac{\mid x-\theta\mid}{-2x(x-\theta)}(e^{-\lambda\mid x-\theta\mid} + C)$$
Could someone help with verifying this? I am unsure if I made a mistake in finding the indefinite integral.

答:
Assume first that $x < \theta$, then

\begin{align*}
F_{X}(x) &= \int_{-\infty}^{x}\frac{\lambda}{2}e^{-\lambda(y-\theta)}dy
=\frac{1}{2}e^{\lambda ( x- \theta)}.
\end{align*}

Then, for $x \ge \theta$, we have

\begin{align*}
F_{X}(x) &= \int_{-\infty}^{x}\frac{\lambda}{2}e^{-\lambda|y-\theta|}dy\\
&= \int_{-\infty}^{\theta}\frac{\lambda}{2}e^{\lambda(y-\theta)}dy
+ \int_{\theta}^{x}\frac{\lambda}{2}e^{-\lambda(y-\theta)}dy\\
&= \frac{1}{2} + \frac{1}{2}(1-e^{-\lambda(x-\theta)})\\
&= 1-\frac{1}{2} e^{-\lambda (x - \theta)}
\end{align*}


问:
Integral of matrix exponential: $\int_0^T e^{At}\: e^{A^{\top}t} \: \mathrm{d}t $
Consider a matrix $A \in \mathbb{R}^{n \times n}$, is there any way to derive a closed-form solution of:

$$\int_0^T e^{At}\:  e^{A^{\top}t} \: \mathrm{d}t $$

In general, $A$ and $ A^{\top}$ do not commute so $e^{At}\:  e^{A^{\top}t} \neq e^{(A+A^{\top})t}$.

答:
A late answer, but here is one observation. If we want the limit as $T \to \infty$ (assuming this limit exists), i.e. the matrix
$$
M = \int_0^\infty e^{At} e^{A^\top t}dt,
$$
then one could note that $M$ is the solution to the continuous time Lyapunov equation
$$
AM + MA^T = -I.
$$
With vectorization, this solution can be expressed as
$$
M = -\operatorname{vec}^{-1}[(I \otimes A + A \otimes I)^{-1}\operatorname{vec}(I)].
$$


问:
Trace of $BAB^{\top}$
Given two square matrices $A$ and $B$, is there any nice way to simplify $\operatorname{Tr} (BAB^{\top})$?

答:
If you know something about $B^TB$ then yes, using the cyclic property of the trace:
$$
\text{Tr}(BAB^T) = \text{Tr}(AB^T B) = \text{Tr}(B^TB A).
$$

Title: Trace of $BAB^{\top}$


问:
Properties of independence and conditional independence
Recently, I see some properties from conditional independence wiki page https://en.wikipedia.org/wiki/Conditional_independence

I don't quite understand the properties of "Rules of conditional independence" in the wiki page.

Definition: $X\perp A$ means random variables $A$ and $X$ are independent from each other. $X\perp A\ |\ B$ means random variables $A$ and $X$ are conditionally independent given random variable $B$.

Question 1: how to prove "Contraction-weak-union-decomposition" property as follows:
\begin{equation*}
\left.
\begin{array}{rl}
X\perp A\ |\ B\\
X\perp B
\end{array}
\right\}
and\ \ \ \ \ \ \Leftrightarrow
\ \ \ \ \ \ X\perp A,B
\ \ \ \ \ \ \Rightarrow
\ \ \ \ \ \ and\left\{
\begin{array}{ll}
X\perp A\ |\ B\\
X\perp B\\
X\perp B\ |\ A\\
X\perp A
\end{array}
\right.
\end{equation*}

Question 2: how to prove the "Intersection" rule as follows:
\begin{equation*}
\left.
\begin{array}{rl}
X\perp A\ |\ B\\
X\perp B\ |\ A
\end{array}
\right\}
and\ \ \ \ \ \ \Rightarrow
\ \ \ \ \ \ X\perp A,B
\end{equation*}

Question 3: Is the inverse proposition of "Decomposition" rule still true? how to comprehend it, please give me an example.

Original proposition:
\begin{equation*}
X\perp A,B\ \ \ \ \ \ \Rightarrow
\ \ \ \ \ \ and
\left\{
\begin{array}{rl}
X\perp A\\
X\perp B
\end{array}
\right.
\end{equation*}
Inverse proposition:
\begin{equation*}
\left.
\begin{array}{rl}
X\perp A\\
X\perp B
\end{array}
\right\}
and\ \ \ \ \ \ \Rightarrow
\ \ \ \ \ \ X\perp A,B
\end{equation*}
Thanks very much!

答:
The above answer and comments give proofs for Q1 and Q2. For Q3, the answer is no:

Let A, B be two independent Bernoulli random variables $\mathcal{B}(\frac{1}{2})$ (e.g. representing unbiased coin flips), and let $X = A \oplus B$ (i.e. $X = 1$ whenever $A \neq B$, $X=0$ otherwise).

Then for $x, a \in \{0,1\}$:

$P(X=x, A=a) = P(B=|x-a|, A=a) = 1/2 \cdot 1/2 = P(X=x) \cdot P(A=a)$.

Therefore, $X⊥A$. Similarly, $X⊥B$, but since $X$ is entirely known if we know $A$ and $B$, we do not have $X⊥A,B$.


问:
Difference between upper and lower tails of multiplicative Chernoff bounds
I'm struggling with the intuition behind why Chernoff bounds differ in the upper and lower tails. That is, for the lower tail we have:
$$
    Pr(X \le (1 - \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{2}}
$$
Whereas for the upper tail:
$$
    Pr(X \ge (1 + \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{3}}
$$

Both bounds are found in roughly the same manner, so at a high level, why would they differ in value when we don't know that the distribution is necessarily asymmetric?

答:
The following intuition builds on the previous answer by user79365.

Suppose we fix $\delta$, and consider the case where $\mu$ is small (close to 0). Consider what this probability distribution would look like. The probability mass is more concentrated below the mean, since $\mu$ is close to 0. Therefore the same fixed $\delta mu$ deviation below the mean would be a rarer event than above. In this case the bounds do make sense.

Now consider the complementary case when $\mu$ is close to $\max (X)$. We can even take $p' = 1-p.$ It seems like we should get the same argument now for a deviation above $\mu$ with $\delta$ still fixed as before. However, $\mu$ is now much larger. So $\delta \mu$ is much larger. That is we consider much larger deviations from $\mu$ in this complementary case, so, it is believable that the previous bounds from the case where $\mu$ is small still hold, and they do.

So the lack of symmetry of the bounds boils down to the multiplicative effect of $\delta$ on $\mu$.


问:
$(a \lor b) a a^*(a \lor a a^*b) \lor (a \lor b)aa^*aa^*$ is wrong?

Here is the non-deterministic finite automaton. I want to find the reguluar expression related to this diagram. I found $(a \lor b) a a^*(a \lor a a^*b) \lor (a \lor b)aa^*aa^*$, but it seems my answer is wrong. Where am I making an error?

Body: $(a \cup b) (a \ (a \cup b)) ^*   aa^*\{b \ (a \cup b) (a \ (a \cup b)) ^*   aa^*\}^*$


问:
Why are de Rham and Hodge cohomology groups coherent if the underlying morphism is proper?
I am reading a paper on Hodge cohomology in order to make myself more acquainted with the topic, and the authors take a few facts for granted that are unknown to me.

In particular: If $X\longrightarrow S$ is a proper morphism, why are the sheaves $H^{q}_{*}(X/S)$, where $*\in\left\{\operatorname{de Rham, Hodge}\right\}$, coherent? Why is $\Omega^{q}_{X/S}$ coherent? (In my case $S=\operatorname{Spec}(k)$for a field $k$, but this seems to hold more generally.) Please explain it to me in as much détail as possible, since I am new to the topic. A reference would also be nice.

答:
If $S$ is locally Noetherian and $f:X\to S$ is a morphism of finite type, then the sheaf $\Omega_{X/S}$ of differentials is coherent (and hence so are all of its exterior powers).  You should be able to find a proof of this in any introductory text on schemes; for instance, it is Remark II.8.9.1 in Hartshorne's Algebraic Geometry.  If you define $\Omega_{X/S}$ locally in terms of modules of differentials in the affine case, the proof is very simple: if $A$ is a Noetherian ring and $B$ is a finitely generated $A$-algebra, then $\Omega_{B/A}$ is a finitely generated $B$-module since it is generated by the elements $db_1\dots,db_n$ if $b_1,\dots,b_n$ generate $B$ as an algebra.  (For any $b\in B$, we can write it as a polynomial in $b_1,\dots,b_n$, and thus write $db$ in terms of $db_1,\dots,db_n$ since $d$ is a derivation.)

The other fact you need is that if $S$ is locally Noetherian, $f:X\to S$ is a proper morphism, and $F$ is a coherent sheaf on $X$, then the derived pushforwards $R^if_*F$ are coherent on $S$ for all $i$.  This is a harder theorem and is one of the major foundational theorems in the theory of cohomology of coherent sheaves; see for instance the Stacks project, or Hartshorne Theorem III.8.8 for the special case of a projective morphism.


问:
Is each power of a prime ideal a primary ideal?
I want to show that each power of a prime ideal is a primary ideal or I have to think about a counterexample?</p>

答:
Alex Becker answered this question completely, but I would like to give another example, which shows the failure can be even more acute: there in fact exist principal prime ideals with powers that are not primary. Indeed, consider the ideal $P=(\overline{x})$ in the ring $R:=F[x,y]\big/(x^2y)$, where $F$ is your favorite field. $P$ is prime, since $(x)$ is a prime ideal of $F[x,y]$ containing $(x^2y)$. However, we claim the ideal $I=P^3=(\overline{x^3})$ is not primary. To see this, let $a=\overline{x^2}$ and $b=\overline{y}$. Then $ab=\overline{0}$ and thus lies in $I$. Note that $a\notin I$, since $x^2$ does not lie in the ideal $(x^3,x^2y)<F[x,y]$, which is the preimage of $I$ under the projection map $F[x,y]\to R$. (Every non-zero element of that ideal has degree at least $3$.) However, we also have $b\notin\sqrt{I}=P$, since $y\notin (x)$ in $F[x,y]$. Thus $I$ is a power of a principal prime ideal, but not primary, and so gives the desired counterexample. I find this behavior somewhat surprising (for instance, it cannot occur in integral domains), which I've why I've included this example.

Exercise: Show that, if $R$ is an integral domain, then every power of a principal prime ideal of $R$ is primary.


问:
Recursive function with multiple recursive terms
I have a questions about recursive functions. What I usually find online are discrete functions where a new term is computed from the previous one starting from an initial condition.

What I'm generally interested in is, in general, something like this:

$$f(x) = g(x) + kf(x + p)$$

with $p > 0$ and no need of any initial conditions. This translates into an infinite expression that may or may not converge:

$$f(x) = g(x) + kg(x + p) + k^2g(x + 2p) + x^3g(x+3p)+...$$

More specifically, if $g(x)=ax+b$ is a linear function and if $k<1$, it's easy to prove (though I won't prove it here) that the recursive function converges to this explicit expression:

$$f(x) = \frac{1}{1 - k}(ax + b + \frac{apk}{1 - k})$$

So here's my question: what if the recursive (linear) function has more than one recursive term?

$$f(x) = ax + b + k_1f(x + p_1) + k_2f(x + p_2) + ... + k_nf(x + p_n)$$

I suppose that if $k_1, k_2,... k_N$ are all smaller than 1, $f(x)$ will converge for all $x$ (but I haven't proved it). I searched a bit online but I couldn't find much in the existing literature. I guess I can try computing it myself but it doesn't look trivial (since the number of terms explode quickly) and likely beyond my abilities... Do you have any hints on how to compute this (or know if there are existing results somewhere in the literature)?

Thanks

答:
If the $p_k$ are in rational ratios, you can rescale $x$ so that they are all integer. Now we have the discrete recurrence relation

$$f(m)-\sum_{i=1}^n k_if(m+p_i)=am+b.$$

The solution of the homogeneous part of the equation can be written in terms of the roots of the polynomial

$$1-\sum_{i=1}^n k_iz^{p_i}=0,$$ let $z_1,z_2,\cdots z_{p_n}$ (where $p_n$ is the largest coefficient). The solution is

$$f(m)=\sum_{i=1}^{p_n}c_i z_i^m.$$

Then a particular solution of the non-homogeneous equation is a linear ansatz,

$$f(m)=um+v$$ and

$$um+v-\sum_{i=1}^n k_i(u(m+p_i)+v)=am+b$$

and by identification,

$$u\left(1-\sum_{i=1}^n k_i\right)=a$$
$$v\left(1-\sum_{i=1}^n k_i\right)=b+u\sum_{i=1}^n k_ip_i.$$

Finally, the general solution

$$f(m)=um+v+\sum_{i=1}^{p_n}c_i z_i^m.$$

Note that this is only valid for integer values of $m$, and a similar relation holds for all sets of values of $x$ that are one unit apart. Finally we can write

$$f(x)=u(\{x\})\lfloor x\rfloor+v(\{x\})+\sum_{i=1}^{p_n}c_i(\{x\}) z_i^{\lfloor x\rfloor}.$$ The functions $u,v$ and $c_i$ are completely arbitrary on $[0,1)$.

As you see, the solution with a linear RHS is complex even for a moderate number of terms. For more general RHS, finding particular solutions is tremendous. And if the $p_i$ are incommensurable, I guess that the number of terms can be infinite.


问:
$1 + \log_2 (n + 1) = n + \cos \frac {n\pi}{6}$
I have encountered this equation in a preparatory test for a contest in Romania.

Solve for every integer $n$:

$$1 + \log_2 (n + 1) = n + \cos \frac {n\pi}{6}$$

I am not sure how to deal with it. I tried trying to find some values and try to establish some orders regarding those values but did not find success. Of course, one of the solution is 0, but that is all I can find.

答:
$n+1$ must be positive since we use it in a logarithm. So
we have that $n \geq 0$. Now we have
$$ 1 + \log_2(n+1) = n + \cos(n\pi/6) $$
$$ \geq n - 1$$

So we can rearrange to get $2^n \leq 4n+4$, so it must be that $n \leq 4$, and we can manually check from here that the only solutions are $n=0$ and $n=3$.


问:
What is this diagram?
This diagram plots every irreducible fraction between 0 and 1 as points, where a point's x coordinate is the fraction's value (from 0 to 1) and its y coordinate is the fraction's denominator (from 2 to whatever the maximum is).

It makes really nice patterns the further the denominators go and I'd really like to know more about it. I think I remember finding it on Wikipedia or something a long time ago with a name along the lines of "the stars of x" or "the towers of y".

Maybe it's just nothing, though.

答:
This is the graph of the reciprocal of Thomae's
function.

https://en.wikipedia.org/wiki/Thomae's_function

Title: What is this diagram?


问:
How can prove this inequality(8)
Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2<AB$, prove that
$$
\frac{A}{C+\sqrt{2(AB-C^2)}}<\frac{a+b+c}{x+y+z}<\frac{C+\sqrt{2(AB-C^2)}}{B}.
$$
I created this inequality. Are there any nice proofs?</p>

答:
It seems the following.

The inequality is not strict. For instance, put $a=b=c=x=1$, $y=z=0$. Then
$A=3$, $B=C=1$, and $C^2<AB$. But

$$\frac{a+b+c}{x+y+z}=3=\frac{C+\sqrt{2(AB-C^2)}}{B}.$$

The non-strict inequality can be easily proved by means of spherical trigonometry.
Consider the following vectors of unit length in $\mathbb R^3$. Let $u=(a,b,c)/\sqrt{A}$,
$v=(x,y,z)/\sqrt{B}$, and $w=(1,1,1)/\sqrt{3}$. Let $\alpha=\angle (v,u)$, $\beta=\angle(u,w)$,
and $\gamma=\angle(w,v)$. Since $a,b,$ and $c$ are positive, we can easily check that $\cos\beta=(a+b+c)/\sqrt{3A}=(a+b+c)/\sqrt{3(a^2+b^2+c^2)}\ge
1/\sqrt{3}$. Then $\tan\beta\le\sqrt{2}$. By spherical law of cosines, we have that
$\cos\gamma=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos\angle u$. Then
$\cos\gamma/\cos\beta\le \cos\alpha+\sin\alpha\tan\beta\le\cos\alpha+\sqrt{2}\sin\alpha$.
Substituting the cosines by the inner products, we obtain

$$\frac {(x+y+z)/\sqrt{3B}}{(a+b+c)/\sqrt{3A}}\le\frac{C}{\sqrt{AB}}+\sqrt{2-2\left(\frac{C}{\sqrt{AB}}\right)^2}.$$

The equality should hold only if $\cos\angle u=1$  (that is, the vectors $u,v,w$ and $0$ are
coplanar) and $\cos\beta=1/\sqrt{3}$, that is one of $a,b,$ and $c$ is equal to $1$, and the others
are equal to $0$.

From the another law of cosines
$\cos\beta=\cos\alpha\cos\gamma+\sin\alpha\sin\gamma\cos\angle v$,
we can similarly obtain the inequality
$$\frac{a+b+c}{x+y+z}\le \frac{C+\sqrt{2(AB-C^2)}}{B}.$$

PS. Maybe the $n$-dimensional version of the inequality holds, if we change in it "$2$" to "$n-1$".

Title: How can prove this inequality(8)


问:
Fourier series of $|x|$ on $[-\pi,\pi]$ and sum of series
So I have been solving various problems on Fourier series and this particular one got me struggling a bit.

Given a function $f(x) = |x|$ find a Fourier series on $[-\pi,\pi]$ and find the sum of following series: $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$.

So I successfully found a Fourier series of given function which is: $f(x) = \frac{\pi}{2} - \sum_{0}^{\infty}{\frac{1}{(2n+1)^2}}\pi\cos{nx}$. I have no problem of finding sum of $\frac{1}{(2n+1)^2}$ but how could I find the sum of $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$ ? I googled before posting this and all I could find is Parseval identity but that's not something I am familiar with. Can someone help me finding the sum or at least explain how could I do it with/without Parseval identity?

Body:
$$f(x)=|x| \sim \frac{\pi}{2} - \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}\cos (2n-1) x.$$

Evaluating this at $x=0$:

$$\frac{\pi}{2} =  \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}$$

$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$

$$\sum_{n=0}^\infty \frac{1}{(2n-1)^2}=1+\frac{\pi^2}{8}$$


问:
Prove that $\operatorname{res}(A(ax),B(ax)) = a^{de}\operatorname{res}(A(x),B(x))$, using the fact that $\sum(i-\sigma(i))=0$
I have been asked to prove the following result:

$$\operatorname{res}(A(ax),B(ax)) = a^{de}\operatorname{res}(A(x),B(x))$$

where $a,e$ are the degrees of the polynomials $A(x)$ and $B(x)$, using the fact that $\sum(i-\sigma(i))=0$. Furthermore, we are given the hint
that the product on the main diagonal $a_{1,1}\cdot . . . \cdot a_{n+m,n+m}$ has degree nm.

I understand it has something to do with the permutation with the entries. However, there is clearly a point I am missing.
Very grateful if someone could thoroughly explain their solution.

答:
Before we begin, a note on my convention for Sylvester matrices: I take the version with $e$ columns of $a_i$s followed by $d$ columns of $b_j$s.

Recall that $\det M = \sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n m_{i\sigma(i)}$ for a $n\times n$ matrix $M$. Letting $S$ be the Sylvester matrix of the polynomials $A(x)$ and $B(x)$ and $T$ be the Sylvester matrix of the polynomials $A(ax)$ and $B(ax)$, I claim that $$\prod_{i=1}^n t_{i\sigma(i)} = a^{de}\prod_{i=1}^n s_{i\sigma(i)}.$$

The big observation to make here is that when $s_{ij}\neq 0$, then $t_{ij}=a^{i-j}s_{ij}$ or $t_{ij}=a^{i-j+e}s_{ij}$, depending on whether $j\leq e$ or not. Here's an example to help you see this, where $A$ is of degree $3$ and $B$ is of degree $2$: $$S=
\begin{pmatrix} a_0 & & b_0 & & \\
 a_1 & a_0 & b_1 & b_0 & \\
 a_2 & a_1 & b_2 & b_1 & b_0 \\
 a_3 & a_2 & & b_2 & b_1 \\
 & a_3 & & & b_2 \end{pmatrix},
T=
\begin{pmatrix} a_0 & & b_0 & & \\
 aa_1 & a_0 & ab_1 & b_0 & \\
 a^2a_2 & aa_1 & a^2b_2 & ab_1 & b_0 \\
 a^3a_3 & a^2a_2 & & a^2b_2 & ab_1 \\
 & a^3a_3 & & & a^2b_2 \end{pmatrix}.$$

We therefore obtain $$\prod_{i=1}^n t_{i\sigma(i)} = \left(\prod_{i=1}^{e} a^{i-\sigma(i)}s_{i\sigma(i)}\right)\left(\prod_{i=e+1}^{d+e} a^{i-\sigma(i)+e}s_{i\sigma(i)}\right)=a^{de} \prod_{i=1}^n a^{i-\sigma(i)}s_{i\sigma(i)},$$ which simplifies to $$\prod_{i=1}^n t_{i\sigma(i)} = a^{de} \prod_{i=1}^n s_{i\sigma(i)}$$ from the fact $\sum i-\sigma(i)=0$.


问:
Do lexicographic orders require the choosing of an a priori order?
I bet that the answer to my question is simply yes, but I'm unsure. Consider Example 20.3 in Maschler, Solan & Zamir (pages 803 & 804). In that example, four vectors are presented: namely,
\begin{align}
\theta(\mathbf{x})&=(1/3,1/3,0,0,-1/3,-1/3,-1/3,-2/3)\\
\theta(\mathbf{y})&=(0,0,0,0,0,0,0,-1)\\
\theta(\mathbf{z})&=(1/3,1/6,0,0,-1/6,-1/3,-1/2,-1/2)\\
\theta(\mathbf{w})&=(1/3,1/3,0,0,0,-1/3,-1/3,-1)\\
\end{align}

Clearly, the elements of the vectors above are written in decreasing order. When comparing these four vectors lexicographically in this particular order, we conclude that $\theta(\mathbf{y})<_L\theta(\mathbf{z})<_L\theta(\mathbf{x})<_L\theta(\mathbf{w})$.

My question is simple: is it possible to order the elements within each vector in a different manner and still compare them lexicographically? Or is this decreasing ordering a necessary requisite to compare them lexicographically?

答:
I think the answer to your question is "no".

You can compare lists of numbers like these lexicographically whether or not they are decreasing. The meaning of "lexicographic" is that if the coordinates match in the first place you compare the second place elements, and so on. If you reorder the coordinates in the items that will change the order in which they appear lexicographically.


问:
Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
 + \frac{b}{c+a} + \frac{c} {a+b}$$

Using Nesbitt's inequality: $I \ge \frac 32$

I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c
}^{3}+{c}^{2}a-2\, \left( b+c \right)  \left( c+a \right)  \left( a+b
 \right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2
}a+{c}^{2}b+abc
$$
and stuck.

EDIT: Look like prove $I \le 2$ not a good thinking :P

答:
This not a complete solution, just a 'case' of a problem.

When $\gcd(a+b,a+c,b+c)=1 \implies \text { } a,b,c  \text { are not all odd or not all even}$

Your expression is

$\dfrac{a(c+a)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)}{(a+b)(b+c)(c+a)}$

Denoting $a+b=k, b+c=l$ and $c+a=m$

$$klm |akm+blk+cml \implies a=p_1l, b=p_2m, c=p_3k$$

$a=p_1l, b=p_2m,$ and $ c=p_3k$ is not possible. So, we have no solution if we consider $\gcd(a+b,a+c,b+c)=1$.


问:
A simple quadratic reformulation of the twin prime conjecture?
Due to https://oeis.org/A024702 we have that $p^2 - 1 ≡ 0$ (mod 24).

For twin primes, we then must also have that in this case $(p+2)^2 - 1 ≡ 0$ (mod 24), which is the same as saying that $p^2 + 4p + 3 ≡ 0$ (mod 24).

Adding these and dividing both sides by 2, one would then have a simple quadratic reformulation of the twin primes conjecture like so:

There are infinitely many primes p such that $p^2 + 2p + 1 ≡ 0$ (mod 12).

Is that correct? Would it possibly be of any practical benefit, like extending it to prime gaps of arbitrary size that have been proven already, and then "proving backwards" towards a gap of size 2?

答: No, it's not equivalent to the twin prime conjecture.  Any $p$ such that $p \equiv 5$ or $11 \mod 12$ satisfies $p^2+2p+1 \equiv 0 \mod 12$.  But most of them are not twin primes.  Counterexamples are easy to find.


问:
$z$- score probability

The proportion of people who like playing basketball is 2%, so a
student randomly sample 1000, but find the proportion is 4%, if the
true proportion is 2%. What the probability that this student detect
at least 4% in random sample?

I used the formula, but found the $z$ score is $4.51$ which is impossible I think.

答:
This is related to testing [and there is nothing wrong with showing that, as in @stacker's Answer (+1)], but the fundamental problem is to find $$P(X \ge 40) = 1 - P(X \le 39)= 0.000043,$$ when
$X \sim \mathsf{Binom}(n=1000, p=0.02),$
The exact value (nearly $0)$ can be found using R, where 'pbinom'
is a binomial CDF:

```<code>1 - pbinom(39, 1000, .02)
[1] 4.339876e-05
</code>```

This is indeed a very unlikely outcome. With proportion $p = 0.02$ who like to play basketball, it is very likely
that there will be fewer than 31 in a random sample of 1000 is over 99%. [The code 'qbinom' is a binomial quantile function (inverse CDF).]

```<code>qbinom(.99, 1000, .02)
[1] 31
pbinom(31, 1000, .02)
[1] 0.992492
</code>```

The plot below shows the two binomial distributions.

R code for figure above:

```<code>x = 0:60;  pdf.02 = dbinom(x, 1000, .02)
pdf.04 = dbinom(x, 1000, .04)
hdr = "PDF of BINOM(1000,p); p=.02 [bars], p-.04 [dots]"
plot(x, pdf.02, type="h", lwd=2, ylab="PDF", main=hdr)
 points(x, pdf.04, pch=19, col="red")
 abline(h=0, col="green2")
</code>```

Note: If you try to use a normal approximation, you will get an answer to the main problem that is very nearly $0,$ but your z-score will be far beyond
the values usually included in printed normal CDF tables.

```<code>z = 20/sqrt(1000*.02*.98);  z
[1] 4.51754
1 - pnorm(z)
[1] 3.128118e-06  # 0.00000313
</code>```

Title: $z$- score probability

Body:
This is related to testing [and there is nothing wrong with showing that, as in @stacker's Answer (+1)], but the fundamental problem is to find $$P(X \ge 40) = 1 - P(X \le 39)= 0.000043,$$ when
$X \sim \mathsf{Binom}(n=1000, p=0.02),$
The exact value (nearly $0)$ can be found using R, where 'pbinom'
is a binomial CDF:

1 - pbinom(39, 1000, .02)
[1] 4.339876e-05

This is indeed a very unlikely outcome. With proportion $p = 0.02$ who like to play basketball, it is very likely
that there will be fewer than 31 in a random sample of 1000 is over 99%. [The code 'qbinom' is a binomial quantile function (inverse CDF).]

qbinom(.99, 1000, .02)
[1] 31
pbinom(31, 1000, .02)
[1] 0.992492

The plot below shows the two binomial distributions.

R code for figure above:

x = 0:60;  pdf.02 = dbinom(x, 1000, .02)
pdf.04 = dbinom(x, 1000, .04)
hdr = "PDF of BINOM(1000,p); p=.02 [bars], p-.04 [dots]"
plot(x, pdf.02, type="h", lwd=2, ylab="PDF", main=hdr)
 points(x, pdf.04, pch=19, col="red")
 abline(h=0, col="green2")

Note: If you try to use a normal approximation, you will get an answer to the main problem that is very nearly $0,$ but your z-score will be far beyond
the values usually included in printed normal CDF tables.

z = 20/sqrt(1000*.02*.98);  z
[1] 4.51754
1 - pnorm(z)
[1] 3.128118e-06  # 0.00000313


问:
Product of test function and function in first order Sobolev space is also in first order Sobolev space

Suppose $1 \leq p \leq \infty$ , $\phi \in C^{\infty}_c (\Omega) $
and $u \in  W^{1,p}(\Omega)$ ,where $Ω$  is an open subset of
$\mathbb{R}^n$. Prove that $\phi u \in W^{1,p} (Ω)$  and $D_{x_i}(\phi u) = \phi D_{x_i} u + u D_{x_i} \phi $ , where $D_{x_i}$ denotes
the weak i-th partial derivative.

I am trying to understand why I need to show the product rule is true for $D_{x_i}(\phi u)$ when the question already states that $\phi$ is infinitely differentiable and $u$ has weak derivatives. Am I asked to show the product rule is valid for the function $\phi u$ or am I missing something? Thanks for any help.

Edit : Follow up question on how does the weak product rule imply $\phi u \in W^{1,p} (\Omega)$ ?

答: You don't necessarily know that the "weak" product rule holds. So, you need to show that if $\psi\in C_c^\infty(\Omega)$ and $D_{x_i}u=v,$ then $$\int\limits_{\Omega}\phi  uD_{x_i}\psi\, dx=-\int\limits_{\Omega}(\phi v+uD_{x_i}\varphi)\psi\, dx.$$ I used here that the weak derivative of something smooth is the same as its strong derivative.


问:
What's the cardinality of this set?
let $A_K= \{ a\in \mathbb{N}^{\mathbb{N}} \mid \sum_{100k}^{100k +99}a_i=200 \}$. and let $A=\cap^\infty_{k=1}A_k$.

What is the cardinality of $A$?

I can't think of how to figure this out, the intersection bugs me alot, I am not able to understand what will $A$ be like?

Any help is really appreciated.

答:
HINT: For each $S\subseteq\Bbb N$ define a sequence $a^S\in\Bbb N^{\Bbb N}$ as follows: for each $k\in\Bbb N$,

$$\begin{align*}
a_{100k}^S&=\begin{cases}
200,&\text{if }k\notin S\,,\\
0,&\text{if }k\in S
\end{cases}\\\\
a_{100k+1}^S&=\begin{cases}
0,&\text{if }k\in S\\
200,&\text{if }k\notin S\,,\text{ and}
\end{cases}\\\\
a_{100k+\ell}^S&=0\text{ for }\ell=2,\ldots,99\,.
\end{align*}$$

Show that $a^S\in A$ for each $S\subseteq\Bbb N$.

Show that the map $\wp(\Bbb N)\to\Bbb N^{\Bbb N}:S\mapsto a^S$ is a bijection.

From this you should be able to determine the cardinality of $A$.

Title: What's the cardinality of this set?


问:
Seeking clue for inequality proof
I'm attempting to show that no variable $\lambda$ can satisfy both of the following inequalities:

$\dfrac{x_{1_1}-x_{2_1}+x_{d_2}+x_{2_2}}{x_{d_2}-x_{1_2}} <\lambda$

$\lambda < \dfrac{x_{1_1}-x_{2_1}}{x_{2_2}-x_{1_2}}$

where $x_{1_1}>x_{2_1}>x_{d_1}>0$ and $x_{2_2}>x_{d_2}>x_{1_2}>0$. To do so, I plan on proving that the LHS of the first inequality is greater than or equal to the RHS of the second inequality. Does anyone have a clue as to the best way to proceed? Things that I've noticed are that there is a constant expression in both numerators, and that the denominator of the LHS of the first inequality is smaller than the denominator of the RHS of the second inequality.

-------edit: I am aware that $x_{d_1}$ doesn't appear in either inequality. It is not a typo.

Title: Seeking clue for inequality proof

Body:
You could use that $x_{d_2}-x_{1_2}<x_{2_2}-x_{1_2}$ and therefore:
$$\dfrac{x_{1_1}-x_{2_1}+x_{d_2}+x_{2_2}}{x_{d_2}-x_{1_2}} >
\dfrac{x_{1_1}-x_{2_1}+x_{d_2}+x_{2_2}}{x_{2_2}-x_{1_2}}=\dfrac{x_{1_1}-x_{2_1}}{x_{2_2}-x_{1_2}}+\frac{x_{d_2}+x_{2_2}}{x_{2_2}-x_{1_2}}.$$

Furthermore $\frac{x_{d_2}+x_{2_2}}{x_{2_2}-x_{1_2}}>0$ and hence

$$\dfrac{x_{1_1}-x_{2_1}+x_{d_2}+x_{2_2}}{x_{d_2}-x_{1_2}} >
\dfrac{x_{1_1}-x_{2_1}}{x_{2_2}-x_{1_2}}+\frac{x_{d_2}+x_{2_2}}{x_{2_2}-x_{1_2}}>\dfrac{x_{1_1}-x_{2_1}}{x_{2_2}-x_{1_2}} .$$


问:
Solution of a functional equation $f(x+y)=f_1(x)+f_2(y)$
I am looking for a proof that the solution of the functional equation:

$$f(x+y)=f_1(x)+f_2(y)$$

is:

$$f(z)=az+(b_1+b_2)$$
$$f_1(z)=az+b_1$$
$$f_2(z)=az+b_2$$

Assume that $x,y,z$ are real numbers, that the domain of $f$ is the real numbers, and that $f$ is continuous. Feel free to substitute some other domain/regularity assumption if they are necessary.

I am interested in this functional equation because it comes up in social choice theory. An accessible proof would be an advantage. Thanks!

答:
Suppose the functions $f,f_1,f_2$ are all $C^0$. Then by setting $y=0, x=x$ and $x=0, y=y$

$$f(x)=f_1(x)+f_2(0)~~,~~f(x)=f_1(0)+f_2(x)$$

We substitute back and after setting $g(x)=f(x)-f_1(0)-f_2(0)$ we see that it satisfies the well known functional relation

$$g(x+y)=g(x)+g(y)$$

Since $g$ is continuous, it can be shown using some standard manipulations (see some related tricks and results for reference in the question here) that $g(x)=ax$. Thus we conclude that with $f_{1,2}(0)=b_{1,2}$

$$f(x)=ax+b_1+b_2~~,~~f_1(x)=ax+b_1~~,~~ f_2(x)=ax+b_2$$

NOTE: If the functions are supposed to be $C^1$ showing this becomes trivial since we can rearrange the last equation, divide by $y$ and take the limit $y\to 0$

$$\lim_{y\to 0}\frac{g(x+y)-g(x)}{y}=\lim_{ y\to 0} \frac{g(y)}{y}\Rightarrow g'(x)=g'(0)=a$$

which in turn implies that $g$ is linear, as above.


答:
I will consider $\mathbb{N} = \{0, 1, 2, \dots \}.$If I understand correctly, we consider $\mathbb{N}^\mathbb{N}$ as a direct product and define the sets $A_k,$ which are the sets of all sequences of natural numbers such that the sum of the elements on positions $100 k$ through $100 k + 99$ is exactly $200$ (thus the sum of certain $100$ elements of the sequence is $200$).

We would want to determine how many elements of $\mathbb{N}^{100}$ satisfy the condition that the sum of their elements is exactly $200.$ Let us denote $b \in \mathbb{N}^{100}$ as $b = (b_1, b_2, \dots, b_{100}).$ Let $n_k := \# B_k,$ where $B_k := \{b \in \mathbb{N}^{100}: \sum_{1 \leq j \leq 100} b_j = k\}.$ Clearly, $n_1 = 100.$ Moreover, any sequence $b \in B_{k + 1}$ can be obtained from a sequence in $c \in B_k$ by adding $1$ to one of the elements $c_j, \, 1 \leq j \leq 100.$ There are usually many ways to do this (depending on the number of non-zero elements of $b \in B_{k + 1}$) and counting the exact number would be pretty tedious. However, it is not hard to see that $n_k \leq n_{k + 1} \leq 100 n_k$ (the lower bound is obtained by the previous remark, the one about adding a $1$ on a certain position and the upper one follows from the fact that the number of ways we can add a $1$ to an element $c \in B_k$ to obtain a fixed element $b \in B_{k + 1}$ is at most $100$ since these entities themselves have exactly $100$ positions). Thus, $100 = n_1 \leq n_{200} \leq 100^{200} n_0 = 100^{200}.$ It follows that there are injections $\mathbb{100} \hookrightarrow B_{200} \hookrightarrow \mathbb{100^{200}}$ (where by $\mathbb{n}$ i denote a set with $n$ elements).

Then $A := \bigcap_{k \geq 0} A_k$ is the set of all sequences of natural numbers such that the sum of the elements with indices $100 k$ through $100 k + 99$ is exactly 200. Note that $A \simeq \prod_{k \in \mathbb{N}} B_{200} = B_{200}^\mathbb{N}.$ It follows that there are injections $\mathbb{100^N} \hookrightarrow A \hookrightarrow (\mathbb{100^{200}})^\mathbb{N}.$ However, we know that $\mathbb{n}^\mathbb{N}$ has cardinality $2^{\aleph_0}$ for any $n \in \mathbb{N},$ so it follows from an immediate application of Cantor-Bernstein-Schroder that the cardinality of $A$ is $2^{\aleph_0}.$ Hope this helps. :)

Title: What's the cardinality of this set?

Body:
let $A_K= \{ a\in \mathbb{N}^{\mathbb{N}} \mid \sum_{100k}^{100k +99}a_i=200 \}$. and let $A=\cap^\infty_{k=1}A_k$.

What is the cardinality of $A$?

I can't think of how to figure this out, the intersection bugs me alot, I am not able to understand what will $A$ be like?

Any help is really appreciated.


问:
Is a dice more random than a coin?
My roommate and I have gotten into an argument.  I presented him with the question is a dice, with half of the sides numbered 1 and the other half numbered 2, more random than just flipping a coin?  Statistically we agree that the chance is 1/2 either way.  But I am arguing that a dice is better as a random generating object then the coin.  My idea is that if both the coin and the dice were thrown, the coin would land and be finished "rolling" before the dice was done generating its number.  With more computation time, the dice would be more random.  My question then becomes, is a dice or any other object just better at randomly generating numbers? Would a person have a much easier time creating a random event with one object over another?</p>

答:
Randomness of these objects are due to the chaotic nature of their motion, especially when bouncing on a surface. Chaotic motion amplifies the initial conditions exponentially, so that even thermal noise could make a change to the outcomes.

Anyway, I would not be surprised that a well trained player, in repeatable conditions and able to choose the initial position could obtain a bias in the drawings. (Especially if catching the coin in hand, or throwing the dice on a short distance.)

But I have no insight on which could be most easily tricked.

It is believed that quantum effects can generate perfectly unpredictable events. Note that modern Intel Processors include a RNG based on such effects.

Title: Is a dice more random than a coin?


问:
trigonometric representation of a complex number.
Let $z=e^{it}+1$ where $0\leq t\leq \pi$, Find the trigonometric representation of $z^2+z+1$.

(The trigonometric representation should be in the form of : $r(\cos \theta +i \sin \theta)$, where $r,\theta \in \mathbb{R}$ and $r>0$.

What I have done : $$z^2+z+1=e^{2it}+3z=(\cos (t)+i \sin (t)) (-2 i \sin (t)+4 \cos (t)+3).$$
The problem is that the expression in the right bracket is not real always.

答:
Another approach, perhaps a little more elementary:

$$a^3-b^3=(a-b)(a^2+ab+b^2)\implies a^2+ab+b^2=\frac{a^3-b^3}{a-b}$$

Well, now just put $\,a=z\;,\;b=1\,$ above , and get

$$z^2+z+1=\frac{z^3-1}{z-1}$$

and since $\,z=e^{it}+1\,$ , we get

$$z^2+z+1=\frac{(e^{it}+1)^3-1}{e^{it}}=e^{2it}+3e^{it}+3=$$

$$=(\cos 2t+3\cos t+3)+(\sin 2t+3\sin t)i$$


问:
$f(n) = k\times 124^n +1$
Let $f(n) = 1 + k \times124^n$ where $k$ is a positive integer. Assume there exists distinct integers $m,n \ge 2$ such that either $m,n \mid f(i)$ for all $i$. Prove there exists integer $x \ge 2$ such that $x \mid f(n)$ for all $n$.

I tried some small cases like $k=1$, but I didn't see any patterns. For $k=1$, I noticed that $f(1) = 125 = 5^3$ and that $f(2)$ was prime. Could that mean that one of the numbers must be $f(2)?$ and what could $r$ be. I also noticed that $f(1) \mid f(3)$ if $k=1$. However, I couldn't see this in the $k=2$ case, at least for the terms I could compute. However, for $k=2$ I noticed divisibility by $3$. I tried finding a contriadiction and find possible $x$, but I can't seem to find any.

答:
Without loss of generality, we may assume that $m, n$ are prime numbers, which are both prime to $k$ and $124$.

Now consider $f(1)$, $f(2)$ and $f(3)$. We may assume that two of them are divisible by $m$, say $f(a), f(b)$ with $1 \leq a < b \leq 3$.

We have then $m \mid f(b) - f(a) = k(124^b - 124^a)$.

If $b - a = 1$, then we have $m \mid 123$. It follows that for any integer $c$, we have $f(c) \equiv 1 + k \cdot 124^c \equiv 1 + k\mod m$. Therefore $f(c) \equiv f(a) \equiv 0\mod m$ for any $c$.

If $b - a = 2$, then $a, b = 1, 3$ and we have $m \mid 124^2 - 1 = 123 \times 125$. The case $m \mid 123$ is already covered above, hence it only remains to look at the case $m \mid 125$, which forces $m = 5$.

In that case, we have $5 \mid 1 + k \cdot 124^c$ for any odd integer $c$, which implies that $5$ does not divide $1 + k \cdot 124^c$ for any even integer $c$.

Thus we must have $n \mid f(2)$ and $n \mid f(4)$. However, this leads to the same situation as above, now with $a, b = 2, 4$. The same argument then tells us that either $n \mid 123$ (which leads to $n \mid f(c)$ for all $c$) or $n \mid 125$ (which leads to $n = 5$, a contradiction with the assumption $m \neq n$).

As can be seen from the proof, one may replace the number $124$ with any number of the form $p^e - 1$ with $p$ prime.

Title: $f(n) = k\times 124^n +1$


问:
$\sigma$ algebra and complete lattice
If Ω is not finite, let’s say it is countable, then is any σ-algebra defined on it a complete lattice?
Here is what I have been thinking suppose $\Omega$ is finite then the set has Greatest Element and the least element if I am able to prove that any subset has a lower bound then it would be a complete lattice. But I am not able to figure out how to do that and also when the $\Omega$ is not finite.</p>

答:
$\hspace{0.44cm}$Below is my attempt. Although it is too late to answer your question but it would be helpful if any mistakes are spotted.

$\hspace{0.44cm}$Let $S$ be the given infinite set and $\Sigma$ be the given $\sigma$-algebra. Assume there is a finite positive real-valued set function $\mu$ defined on $\Sigma$ such that $(S, \Sigma, \mu)$ is a finite measure space. Let $\{A_{\lambda}\} \subset \Sigma$. Since $\{\mu(A_{\lambda})\} = \{\|\chi_{A_{\lambda}}\|_1\}$, we then want to show the function $\chi_A$ where $A = \bigcup A_{\lambda}$ is $\mu$-measurable and hence $A \in \Sigma$. Then we can find a sequence $\{A_n\}\subseteq\{A_{\lambda}\}$ such that $\mu(A_n) \uparrow \mu(A) = \sup_{A_{\lambda}}\mu(A_{\lambda})\,\implies\,\|\chi_{A_n} - \chi_A\|_1 \rightarrow 0$ (since $\mu$ positive) $\implies\,\chi_{A_n}$ converge to $\chi_A$ in $\mu$-measure (definition can be found here).

$\hspace{0.44cm}$One can refer to this paper for definitions of totally measurable functions and measurable functions. Briefly speaking in a positive finite measure space, a sequence of simple functions can only converge in measure to a totally-measurable function. If that limit function is also a characteristic function, then the set where it obtain $1$ will be measurable. Hence here we can conclude $A \in \Sigma$.

$\hspace{0.44cm}$Pick $B \in \Sigma$ and fix $b \in B$. Then the Dirac delta measure $\delta_b$ is a finite positive real-valued measure (the idea here comes from this post) and the existence of $\sup\{A_{\lambda}\}$ in $\Sigma$ follows the remark above and so does the existence of $\inf\{A_{\lambda}\} = S \backslash\sup\{A_{\lambda}^c\}$.

$\hspace{0.44cm}\Large\spadesuit$Now I am also curious if $\Sigma$ admits any other complex measures $\lambda$ (i.e. $\Sigma$ is the set of all $\lambda$-measurable sets). If so, then $\arctan \vert\lambda\vert$ is also a positive real-valued measure.

Title: $\sigma$ algebra and complete lattice


问:
How to show that $\frac{1}{(1-\frac{1}{4}z^{-1})(1-\frac{1}{4}z)} = \frac{-4z^{-1}}{(1-\frac{1}{4}z^-1)(1-4z^{-1})}$
Can anyone help me clarify what rule is used in this rewriting of this fraction?

$$\frac{1}{\left(1-\dfrac{1}{4}z^{-1}\right)\left(1-\dfrac{1}{4}z\right)} = \frac{-4z^{-1}}{\left(1-\dfrac{1}{4}z^{-1}\right)\left(1-4z^{-1}\right)}$$

答:
Multiply both numerator and denominator with $-4z^{-1}$. For the term $(1-z/4)$ in the denominator you get

$$-4z^{-1}(1-z/4)=-4z^{-1}+1=(1-4z^{-1})$$


问:
Diffeomorphic manifolds with the same constant curvature but not isometric
I am looking for a two Riemannian manifolds which are diffeomorphic, have the same constant curvature but they are not isometric.

I thought about the product manifold $(-\frac{\pi}{2}, \frac{\pi}{2})^n$ and $\mathbb{R}^n$, with the diffeomorphism:
$f:(-\frac{\pi}{2}, \frac{\pi}{2})^n \rightarrow \mathbb{R}^n$ given by $f(x_1,\dots,x_n)=(\tan(x_1), \dots, \tan(x_n))$.
Both manifolds have seccional curvature zero, since they are flat manifolds, but they are not isometric ($(-\frac{\pi}{2}, \frac{\pi}{2})^n$ is not complete).

Is that example right? Anyway, is there any "more intuitive" example?

答:
To answer your question, this example is right, and your proof that it is right is right.

But I would say that there is a simpler and more intuitive proof that this example is right: the diameter of $(-\pi/2,\pi/2)^n$ is finite, whereas the diameter of $\mathbb R^n$ is infinite, and "finite diameter" is an isometry invariant.

Of course the diameter of a Riemannian manifold $M$ must be formalized, and this is done by defining
$$\text{diam}(M) = \sup_{x,y \in M} d(x,y)
$$
where $d(x,y)$ is the infimum of the lengths of all paths having endpoints $x$ and $y$. This formula yields a value $\text{diam}(M) \in [0,\infty) \cup \{\infty\}$ which is an invariant of Riemannian isometries.


问:
Difference between upper and lower tails of multiplicative Chernoff bounds
I'm struggling with the intuition behind why Chernoff bounds differ in the upper and lower tails. That is, for the lower tail we have:
$$
    Pr(X \le (1 - \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{2}}
$$
Whereas for the upper tail:
$$
    Pr(X \ge (1 + \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{3}}
$$

Both bounds are found in roughly the same manner, so at a high level, why would they differ in value when we don't know that the distribution is necessarily asymmetric?

答:
The question is targeted towards the multiplicative Chernoff bounds for th tails of distribution. I do not claim to have solved the problem but I can provide a better reference on how to do it:

Five Proofs of Chernoff’s Bound with Applications by Wolfgang Mulzer.

The advance to the already mentioned answers and proofs is that this source has the restriction of the question to the multiplicative cases of the upper tail and lower tail. This is really proof and honors the scientific history of the problems in the question.

I am able to give another link wherein there is a set formulas for working this through for more interesting distributions with the attributve multiplicative:
CumulantGeneratingFunction.

This provides some impression on the relation of the surfaces under consideration:

.

Yellow is $\frac{1}{2}$. Blue is $\frac{1}{3}$.


问:
$\phi(t) = 1 + o(t^2)$ as $t \downarrow 0$, then $\phi(t) = 1$?
In the title $\phi(t)$ is the characteristic function of some random variable. My question here is not really about how to prove this. As this is a question posted by my professor, I am wondering if there can be a typo in the question.

Namely, the part I am suspicious about being incorrect is the $t \downarrow 0$ part, my suspicion is that it should instead be $t \to 0$. Since if I have a $\phi$ such that $\phi(t) = 1$ for $t \geq 0$ and some other functional form so that $\phi(t) = 1 + o(t^2)$ is not satisfied when $t < 0$, then apparently this claim does not need to hold. For example, I can let $\phi(t)$ be some function that has an $o(t)$ term in it when $t < 0$. However, I am not sure if I am correct on this.

Hence, I am wondering if this $t \downarrow 0$ is enough to prove the claim or it should instead be $t \to 0$? Thank you!

答: You might note that $\phi(-t) = \overline{\phi(t)}$, so if $\phi(t) = 1 + o(t^2)$ as $t \downarrow 0$, it is also true as $t \to 0$.


问:
Finding the value of distribution function of a converging random variable
There is this example in a note that I think this is supposed to be a simple problem, but I still find it not as straightforward.

Consider a sequence of random variables $X_n\equiv1/n,X\equiv0$. Then $F_{X_n}(t)\to F_X(t)$ at all $t\not=0$ (as $n\to\infty$), but how can we show that $F_X(0)=1$ and $F_{X_n}(0)=0$? (in other words, how can we compute $F_X(t)$ and $F_{X_n}(t)$ at $t=0$?)

(these values are given, I just don't know how they come up with these).

I tried to reason as follows:

1. Since $X$ and $X_n$ are constant random variables, then both $F_X(0)$ and $F_{X_n}(0)$ should equal to $0$, am I right? or at least they should equal in value?

2. It was not said explicitly, but I suppose $t\in \mathbb{R}$. My question is does it affect $F_{X_n}(t)$ and $F_X(t)$ afterall?

The purpose of this example is I think to illustrate convergence in distribution only applies at $t\in \mathbb{R}$ where $F_X(t)$ is continuous.

I am so confused about this problem for so long. I hope my question is clear enough. Thanks very much in advance for any help.

答:
I assume that by $X_n \equiv \frac{1}{n}$ you mean that $X_n$ always takes the value $\frac{1}{n}$. $X_n$ then corresponds to a probability measure $\mathbb{P}_n$ on $\mathbb{R}$ with $\mathbb{P}(A) = 1$ iff $\frac{1}{n} \in A$. $X$ similarly corresponds to a probability measure with $\mathbb{P}(A) = 1$ iff $0 \in A$.

The distribution function $F_n$ of $X_n$ is thus $$
  F_n(t) = \begin{cases}
    0 &\text{if $t < \frac{1}{n}$} \\
    1 &\text{if $t \geq \frac{1}{n}$.}
  \end{cases}
$$
and the distribution function $F$ of $X$ is $$
  F(t) = \begin{cases}
    0 &\text{if $t < 0$} \\
    1 &\text{if $t \geq 0$.}
  \end{cases}
$$

Now, if you pick a $t > 0$, then for all $n > \frac{1}{t}$ you get $F_n(t) = 1$. Thus, for all $t > 0$, you have $F_n(t) \to 1 = F(t)$. For $t < 0$, you have for all $n$ that $F_n(t) = 0 = F(t)$. In other words, $F_n \to F$ pointwise on $\mathbb{R}\setminus\{0\}$.

For $t=0$, $F_n(t) = F_n(0) = 0 \neq 1 = F(0)$ for all $n \in \mathbb{N}$.

Does that mean that the $X_n$ do not converge in distribution? No, because $F_n$ only needs to converge pointwise to $F$ for those points $x$ where $F$ is continuous. And $F$ isn't continuous at $0$.

So, why is convergence in distribution defined this way? One way to look at this is to ask whether requiring the $F_n$ to converge only at continuity points of $F$ somehow makes the limit ambiguous. It turns out it does not, because one you know the value of $F$ at those points, you can recover the full $F$ by using that since $F$ is a distribution function, it's always  right-continuous, i.e., $\lim_{x\to a^+} F(x) = F(a)$. (I've cheated a bit here, because two different $F$ may have two different sets of continuity points, but this still explains the general idea..)

Take your example. You know that $F_n$ goes to $1$ for $t > 0$, and to $0$ for $t < 0$. There then is only one right-continuous $F$ which fits, since the right-continuity at $0$ requires that $F(0)=1$ too. So the $F_n$ uniquely defined $F$, even though there's a point where the pointwise convergence fails.


答:
Unless you specify in mathematical terms what you mean by "random" and "better" and "easier time generating," and unless you are talking about an idealized die and idealized coin, then your question is not really about mathematics.

Here are some ways in which we might be able to formulate your question mathematically:

If we specify the geometry and mass of the coin and the die, and for a fixed velocity choose a random angle and orientation to throw them, on average, what is the time for them to "settle?"  This is a physics question and depends on a number of additional model parameters such as the proportion of energy lost per collision.

Assuming the coin and die each have a known and idealized probability distribution, and ignoring the amount of time to generate realizations from each,  would a third observer who is furnished with samples of arbitrarily large but equal size from each object but not the information about which object generated which sample, be able to distinguish the two?  This is a mathematical question.

Title: Is a dice more random than a coin?

Body: My roommate and I have gotten into an argument.  I presented him with the question is a dice, with half of the sides numbered 1 and the other half numbered 2, more random than just flipping a coin?  Statistically we agree that the chance is 1/2 either way.  But I am arguing that a dice is better as a random generating object then the coin.  My idea is that if both the coin and the dice were thrown, the coin would land and be finished "rolling" before the dice was done generating its number.  With more computation time, the dice would be more random.  My question then becomes, is a dice or any other object just better at randomly generating numbers? Would a person have a much easier time creating a random event with one object over another?</p>


问:
Finding the limit by using Maclaurin series
$$ \lim_{x \to 0}\frac {x\cdot \sin x}{1-\cos(2x)} $$

I'm supposed to find the limit by using the terms, and figuring out $ x\times \sin x$ is easy enough, just multiply $\sin x$ with $x$, but when it's divided by $1-\cos(2x)$, I'm really lost. I can and have calculated what the first 5 terms of $1-\cos(2x)$ is, but I have no idea what I do with those two to get one series, so I can get the limit. If WolframAlpha isn't lying to me, the limit should be $\frac{1}{2}$.

From all my lecturer has written, I can't see anywhere that it is mentioned how I divide power series of any kind in the notes he wrote. Now, I'm studying over the internet, so I just follow the lectures online, and don't have any proper help around me, so hence, I'm asking here, for the second time in 24hrs.

So where do I go from having the first 5 terms of the Maclaurin series of $ x\times \sin x$ and $1-\cos(2x)$?

答: We have that $\cos 2x = 1 - \frac12 (2x)^2 + \frac1{24} (2x)^4 + \dots,$ so $1 - \cos 2x = 2 x^2 - \frac{2}{3} x^4 + \dots$ and $x \sin x = x^2 - \frac16 x^4 + \dots.$ Thus $\lim_{x \to 0} \frac{x \sin x}{1 - \cos 2x} = \frac{x^2 - \frac16 x^4 + \dots}{2 x^2 - \frac23 x^4 + \dots} = \lim_{x \to 0} \frac{1 - \frac16 x^2 + \dots}{2 - \frac23 x^2 + \dots} = \frac12$ simply by evaluating the limits of the numerator and denominator. Hope this helps. :)


问:
Is there a curve for which the area underneath between n and n+1 is exactly double the area underneath the curve between n+1 and n+2?
For example:
$$\int_x^{x+1} f(x)\ dx = \frac{1}{2^x}$$
So for x = 0, the integral would be from 0 to 1 and would return 1.
For x = 1, the integral would be from 1 to 2 and would return 0.5 . And so on...

答:
This could be a solution:

First off apply the fundamental theorem of calculus

$$
\frac{{\rm d}}{{\rm d}x}\int_{x}^{x+1}{\rm d}z~f(z) = f(x+1) - f(x)
$$

therefore, the function $f$ should satisfy

$$
f(x+1) - f(x) = -\frac{\ln 2}{2^x} \tag{1}
$$

A solution for this is

$$
f(x) = \ln 4 - 2^{1-x}(2^x - 1)\ln 2 \tag{2}
$$

The term $\ln 4$ is actually arbitrary, I just selected that number to keep $f(x) > 0$


答:
Let $X$ be any (real) random variable. Then,
$$
\phi_{X}(0)=\mathbb{E}\exp\left\{ iX\cdot0\right\} =1.
$$
Moreover,
$$
\left|\phi_{X}(t)-\phi_{X}(s)\right|=\left|\mathbb{E}\left[\exp\left\{ iXs\right\} \left(\exp\left\{ iX\left(t-s\right)\right\} -1\right)\right]\right|\leq\mathbb{E}\left|\exp\left\{ iX\left(t-s\right)\right\} -1\right|.
$$
By the DCT, the quantity on the right hand side of the inequality
above goes to zero as $t-s$ vanishes. Therefore, $\phi$ is continuous
(actually, uniformly continuous) and hence
$$
\lim_{t\rightarrow0}\phi(t)=1.
$$

Title: $\phi(t) = 1 + o(t^2)$ as $t \downarrow 0$, then $\phi(t) = 1$?

Body:
In the title $\phi(t)$ is the characteristic function of some random variable. My question here is not really about how to prove this. As this is a question posted by my professor, I am wondering if there can be a typo in the question.

Namely, the part I am suspicious about being incorrect is the $t \downarrow 0$ part, my suspicion is that it should instead be $t \to 0$. Since if I have a $\phi$ such that $\phi(t) = 1$ for $t \geq 0$ and some other functional form so that $\phi(t) = 1 + o(t^2)$ is not satisfied when $t < 0$, then apparently this claim does not need to hold. For example, I can let $\phi(t)$ be some function that has an $o(t)$ term in it when $t < 0$. However, I am not sure if I am correct on this.

Hence, I am wondering if this $t \downarrow 0$ is enough to prove the claim or it should instead be $t \to 0$? Thank you!


问:
Not sure what kind of equation I need (7 question with 4 answers, how many possible combinations of answers can I get)?
Very limited math skills here.  I have been trying to search for what kind of equation I would need to solve this and have been stuck.

I need to calculate how many possible combinations there would be for answers if I have 7 questions each with the same 4 answers.

So for example, the 4 answers are 100, 75, 50, 25 to each question, and there are 7 total questions.

How can I calculate the total number of possible combinations for a result where all 7 have been answered?

So a possible answer combination could look like

100,75,25,25,50,100,100

Only one answer per question.

答:
Assuming only one answer is correct to each of the questions, there are exactly four ways to answer each of the questions, right? Just pick one of them. Since we have seven questions to answer, it follows that the number of ways to answer all of them is exactly $4^7 = 2^{14} = 16384.$ Hope this helps. :)

P.S.: By the way, if you were allowed to choose any number of answers to the questions, you would have $2^4 - 1 = 15$ ways to answer each of the question (you either choose an answer or you don't, so you have a binary choice for each of the four answers and you cannot not choose any answer at all- hence the $-1$), so there would be $15^7 = 170859375$ ways of answering the questionnaire.


Title: Finding the limit by using Maclaurin series

Body:
$$ \lim_{x \to 0}\frac {x\cdot \sin x}{1-\cos(2x)} $$

I'm supposed to find the limit by using the terms, and figuring out $ x\times \sin x$ is easy enough, just multiply $\sin x$ with $x$, but when it's divided by $1-\cos(2x)$, I'm really lost. I can and have calculated what the first 5 terms of $1-\cos(2x)$ is, but I have no idea what I do with those two to get one series, so I can get the limit. If WolframAlpha isn't lying to me, the limit should be $\frac{1}{2}$.

From all my lecturer has written, I can't see anywhere that it is mentioned how I divide power series of any kind in the notes he wrote. Now, I'm studying over the internet, so I just follow the lectures online, and don't have any proper help around me, so hence, I'm asking here, for the second time in 24hrs.

So where do I go from having the first 5 terms of the Maclaurin series of $ x\times \sin x$ and $1-\cos(2x)$?

Body:
hint

$$\sin(x)=x+x\epsilon(x)$$

$$1-\cos(2x)=2\sin^2(x)$$

$$\frac{x\sin(x)}{1-\cos(2x)}=\frac{x}{2\sin(x)}$$
$$=\frac 12\frac{1}{\frac{\sin(x)}{x}}$$

$$=\frac 12\frac{1}{1+\epsilon(x)}$$
the limit is $ 0.5$.


答:
Hint:

Use Taylor-Young's formula at order $2$:
$$\cos u=1-\frac{u^2}2+o(u^2),$$
and substitute $2x$ to $u$.


问:
L Hopital rule and side limits
Today I was in the process of solving a side limit, to be precise this limit:
$$ \lim_{x\rightarrow 0^{+}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} $$
I was supposed to use L'hopital and found that the answer was $1$. However, this function is not differentiable at some interval $(0,a)$ so L'hopital hypothesis is not true here. Based on the fact that I was supposed to use the rule here I assume that it can be used here, but if that was the case then it must be true that:$$\lim_{x\rightarrow 0^{+}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \lim_{x\rightarrow 0^{-}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \lim_{x\rightarrow 0}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \dfrac{\dfrac{d}{dx}\ln(\sin(2x)))}{\dfrac{d}{dx}\ln(\sin(x))}$$
Because that is what I believe L'hopital states.

What is happening here? L'hopital should not be used here? Is it real that the rule may be true for JUST one side limit in case it is a limit point of the domain? Or even.. Is $$\lim_{x\rightarrow 0}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = 1$$ ?

I am pretty confused with this fact about side limits and limit points of the domain, and now with L'hopital rule into play it is even more confusing. I hope that I expressed myself the best way possible.

Thanks for your time and sorry if there are grammar mistakes, I do not speak english.

答: Why use L'Hopital's rule?  We have $$\sin 2x = 2\sin x \cos x,$$ and since in a sufficiently small neighborhood of $0$, we have $\cos x > 0$, it follows that $$\log \sin 2x = \log \sin x + \log (2 \cos x).$$  Then $$\lim_{x \to 0^+} \frac{\log \sin 2x}{\log \sin x} = 1 + \lim_{x \to 0^+} \frac{\log (2 \cos x)}{\log \sin x} = 1.$$

Title: L Hopital rule and side limits


问:
doing proofs of implications
Hi this question is regarding a proof from the book "INTRODUCTION TO MODERN CRYPTOGRAPHY" (https://eclass.uniwa.gr/modules/document/file.php/CSCYB105/Reading%20Material/[Jonathan_Katz%2C_Yehuda_Lindell]_Introduction_to_Mo(2nd).pdf?fbclid=IwAR1hf1OTKAhf4ZHvswERpcZ3ZVDQMxHuP2FWRg2tvlo3-tUMSdFIPLWZR_8) by katz & lindell page 30. The question does contain some specific cryptography notation that will be implicit, it is explained very well in the book.
First we have this definition of perfect secrecy:

```<code>Pr[M=m | C = c] = Pr[M = m]
</code>```

And then we have this definition of perfect secrecy:

```<code>Pr[EncK(m)=c] = Pr[EncK(m') =c].
</code>```

and now I have the following lemma:

```<code>An encryption scheme(Gen,Enc,Dec) with message spaceMis perfectly secret if and only if Equation
(2.1) holds for everym, m′∈Mand every c∈C.
</code>```

so we want to improve indistinguishability implies perfect secrecy. indistinguishability -> perfect secrecy.

There then is a proof that starts with perfect secrecy, and ends up at

```<code> Pr[M=m | C = c] =
                 =  ....
                 =  ....
                 = Pr[M = m]
</code>```

The proof is not so much what I want to ask about, it is moreso the general structure.

If I want prove indistinguishability -> perfect secrecy or A -> B, wouldnt it make the most sense to start with A, and then do a bunch of algebra, until I reach B, such that I can prove that they are the same. Such that my proof would look like
A =
= ...
= ...
= B
in the proof for perfect secrecy, in my mind all we seem to be doing is showing at perfect secrecy is.. perfect secrecy?

So could someone put into natural language what the general idea in a proof like this is

答:
I'll try an educated guess.

$A==\ldots =\ldots =B$ would be a proof of $A=B$, not of $A\to B$. Here we are rather in a situation where the proposition $B$ itself is of the form $X=Y$. So the proof that $A\to B$ looks like $X=\ldots =\ldots =Y$ and the argument for some of the intermediate "$=$"  uses the hypothesis $A$. Thus overall we prove that $X=Y$ (i.e., $B$) holds under the hypothesis $A$. In other words $A\to B$.

Title: doing proofs of implications


问:
Proof of greedy algorithm.
Given n numbers find the way to assign them to blocks of 3 (and possible one block of 1 or 2 if n is not divisible by 3) so that sum of smallest elements from each full block is maximal.

ie.
numbers 400, 350, 300, 250, 200, 150, 100, can be divided into [400, 350, 300], [250, 200, 150], [100] with above mention sum of 450.

I know that correct approach is just to assign numbers to block in sorted order(so first block contains first, second and third largest element and so on), but i can't find a way to prove it correctness.

答:
We sort all the blocks according to the descending order of their minimal elements. We name the blocks $B_1, B_2, \dots$ in that order.

For every $i\geq 1$, let $m_i$ be the smallest element of $B_i$. Thus $m_j \geq m_i$ whenever $j \leq i$.

For any $1 \leq j \leq i$ and any element $x \in B_j$, we know that $x \geq m_j \geq m_i$.

This means that, for any $i$, there are at least $3i$ elements (i.e. all elements from $B_1$ to $B_i$) larger than (or equal to) $m_i$.

In other words, if we sort the original $n$ numbers in descending order as $x_1, x_2, \dots$, then we must have $m_i \leq x_{3i}$.

Adding everything together, we see that the sum $m_1 + m_2 + \dots$ is at most $x_3 + x_6 + \dots$.

On the other hand, this upper bound can be achieved simply with the greedy algorithm: group them as $\{x_1, x_2, x_3\}$, $\{x_4, x_5, x_6\}$, etc.

Title: Proof of greedy algorithm.


答: Hint: Write $\left(\ln(\sin(kx))\right)' = \dfrac{k\cos(kx)}{\sin(kx)}$,and plug in $k =1,2$ into the equation to get the derivatives and take it from there.

Title: L Hopital rule and side limits

Body:
Today I was in the process of solving a side limit, to be precise this limit:
$$ \lim_{x\rightarrow 0^{+}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} $$
I was supposed to use L'hopital and found that the answer was $1$. However, this function is not differentiable at some interval $(0,a)$ so L'hopital hypothesis is not true here. Based on the fact that I was supposed to use the rule here I assume that it can be used here, but if that was the case then it must be true that:$$\lim_{x\rightarrow 0^{+}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \lim_{x\rightarrow 0^{-}}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \lim_{x\rightarrow 0}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = \dfrac{\dfrac{d}{dx}\ln(\sin(2x)))}{\dfrac{d}{dx}\ln(\sin(x))}$$
Because that is what I believe L'hopital states.

What is happening here? L'hopital should not be used here? Is it real that the rule may be true for JUST one side limit in case it is a limit point of the domain? Or even.. Is $$\lim_{x\rightarrow 0}\dfrac{\ln(\sin(2x)))}{\ln(\sin(x))} = 1$$ ?

I am pretty confused with this fact about side limits and limit points of the domain, and now with L'hopital rule into play it is even more confusing. I hope that I expressed myself the best way possible.

Thanks for your time and sorry if there are grammar mistakes, I do not speak english.


问:
Permutation of Matrices on a Vector

We have the following two matrices:
$A=\left( \begin{matrix}1 & 1 \\ 0 & 1\end{matrix} \right)$ and $B=\left( \begin{matrix}1 & 0 \\ 1 & 1\end{matrix} \right).$ Take any finite product of them and multiply this by the vector $x=\left<1,1\right>$. What possible vectors $(x',y')$ can we get?

E.g. $ABx=\left<3,2 \right>$.

答:
You could start by showing that
$$A^n = \begin{bmatrix}
1 & n\\
0 & 1
\end{bmatrix},\quad B^m = \begin{bmatrix}
1 & 0\\
m & 1
\end{bmatrix}.$$
Then,
$$A^nB^m = \begin{bmatrix}
1+nm & n\\
m & 1
\end{bmatrix}$$
and
$$A^nB^m x = \begin{bmatrix}
1+(1+m)n\\
1+m \end{bmatrix}.$$

Title: Permutation of Matrices on a Vector


问:
one of the following statements is true.Identify:-
one of the following statements is true.Identify:-

(A) Every subset of $\Bbb{Q}$ which is bounded above has a unique supremum rational number.

(B)Every subset of $\Bbb{Q}$ which is bounded above has a unique supremum real number.

(C)The set of rational numbers has least least upper bound property.

(D)The set of real numbers does not have the greatest lower bound property.

I am confused because I think both B and D are correct but they say only one option is correct

答: Greatest lower bound for reals is $+\infty$, similarly, the supremum of reals is $-\infty$. I assume that whoever gave you that question meant that (B) should be correct(is there a non-empty assumption?), however, as $\emptyset$ is always a subset, and has $-\infty$ as upper bound (which is NOT real by the way), (B) is also false.</p>


问:
Continuity of matrix operator norm
Suppose $A=(a_{i,j})$ is an $n\times n$ real matrix, and consider the operator norm

$$ \|A\|_{op} = \sup_{\|x\| = 1} A \cdot x$$

where $\|\cdot\|$ is some norm on $\mathbb{R}^{n}$. Suppose that by $A \rightarrow B$ we mean that each $|a_{i,j} - b_{i,j}| \rightarrow 0$ according to the standard norm on $\mathbb{R}$.

Hypothesis: $A \to B$ implies that $\|A\|_{op} \to \|B\|_{op}$.

Proof Attempt

Consider the Frobenius norm

$$ \| M \|_{F} = \sqrt{  \sum_{i} \sum_{j} |m_{i,j} |^{2}},$$

where again $|m_{i,j} |$ is the absolute value of $m_{i,j}$ in $\mathbb{R}$. Then by finite dimensions and the equvalence of norms, we have that for some constant $K>0$,

$$ \|\cdot \|_{op} \leq K \cdot \| \cdot \|_{F}.$$

Hence, by the reverse triangle inequality and the above observation,

$$| \| A \|_{op} - \| B \|_{op} | \leq \| A -B \|_{op} \leq K\cdot \| A-B \|_{F} $$

That the right hand side tends towards $0$ is then immediate from the definition of the Frobenius norm and the fact that each $|a_{i,j} - b_{i,j}| \rightarrow 0$. QED.

Does this look okay?

答:
Your proof is correct, I just wanted to add a more general description of the situation.

In general, if $A$ is any $C^*$-algebra and $n\geq1$ we have the inequalities

$$\sup_{i,j}\|a_{i,j}\|_A\leq\|[a_{i,j}]_{i,j}\|_{M_n(A)}\leq\sum_{i,j}\|a_{i,j}\|_A$$
where the norm of $M_n(A)$ is the operator norm. So, if a sequence of matrices converges to $0$ in the way you are describing, the right inequality yields trivially that the operator norm tends to $0$

Title: Continuity of matrix operator norm


问:
Does there exist such holomorphic bijection?
Let $A=\{z\in\mathbb{C} \mid \mathrm{Re}(z)>0\}$ and $B=\{z\in\mathbb{C} \mid 0<\mathrm{Re}(z)<1\}$. Is there a holomorphic bijection $f:A\rightarrow B$?

答: Hint: Try a constant multiple of an appropriate branch of $\log$.


问:
Integral of product of Standard Normals
Hopefully someone can help me understand the solution to the following integral which is the final step in a problem I'm looking at:

$\int^{\infty}_{0}\phi(x)\Phi(x)dx=\left[\frac{1}{2}\{\Phi(x)\}^2\right]^{\infty}_{0}=\frac{1}{2}\left[1^2-\left(\frac{1}{2}\right)^2\right]=\frac{3}{8}$

$\phi(x)$ is the probability density function of a Standard Normal and $\Phi(x)$ the distribution function. Obviously $\Phi(x)$ is the integral of $\int\phi(x)dx$ but I don't follow the first step in the evaluation of the above integral. The solution it's trivial. If anyone could explain the above it would be appreciated.

Thanks

答: This is just standard integral substitution (sometimes called $u$-substitution).  The fact that $\phi$ is the density and $\Phi$ is the cumulative distribution for a standard normal implies $$\Phi'(x) = \phi(x).$$  Therefore the integral can also be written as $$\int \Phi(x) \Phi'(x) \, dx.$$  And with the choice of substitution $$u = \Phi(x), \quad du = \Phi'(x) \, dx,$$ we see it becomes $$\int u \, du = \frac{u^2}{2} + C = \frac{(\Phi(x))^2}{2} + C.$$

Title: Integral of product of Standard Normals


问:
Group Testing Problems Where the Number of Defectives is Given By a Probability Distribution
I was looking at Wolves and Sheep and had an interesting thought.

The original problem says that out of $100$ sheep, it is known that there are $5$ wolves and they want to minimize the number of tests used. Let's say that instead of giving a discrete number such as $5$ wolves, they said that each sheep has a $10 \%$ chance of secretly being a wolf.

In that case, how would we do the testing? All of the solutions on Wolves and Sheep rely on the fact that there are exactly $5$ wolves.

I wish I had some sort of insight or steps that I had taken thus far to approach this problem, but I simply haven't been able to do anything. I don't get how to solve this whatsoever.

答:
You would need a multistage testing and scoring mechanism and only have a probabilistic answer at the end so you only know (say with probability 0.99) you have found all the defectives.

Assuming noiseless group tests, if $T(A)$ for a subset $A$ returns “defect present” you use the conditional probability distribution that the number of defectives in $A$ is at least 1, and score each element equally in terms of likelihood of being defective.

Then apply $T(A’)$ to another subset.

If both $T(A)$ and $T(A’)$ say “defective present” any element belonging to $A \cap A’$ gets a higher score.

You can determine based on $p=0.1$ that the most informative test is to test an untested subset of size $k$ where the probability of “defect absent” is roughly $1/2$. This means $(1-p)^k=1/2$ and you take the nearest integer to determine $k.$ This should in general minimize the total number of tests.

In this vein, Chan et al. (2011) introduced COMP, a probabilistic algorithm that requires no more than
$t=ed(1+\delta )\ln(n)$
tests to find up to $d$ defectives in
$n$ items with a probability of error no more than
$n^{-\delta }$.
This is within a constant factor of the
$t=O(d\log _{2}n)$ lower bound.

See wikipedia for more, the above quote is from there in the “non adaptive and probabilistic testing” section.


问:
Integrating $ a\,f(x) +b\,y(x)=\frac{dy}{dx}$
Can somebody put me on the right track for integrating the following equation? How do I separate the variables?

$$ a\,f(x) +b\,y(x)=\frac{dy}{dx}$$

答: Hint: Write down a differential equation solved by $z(x)=\mathrm e^{-bx}y(x)$.