问:
Are unit quaternions and quaternion group different?
I am studying real representation from the book Representation theory by Fulton and Harris. Here I find an example which demonstrates the fact that if the character of a representation is real then the underlying representation need not be real. For instance the author take $G$ to be a finite non-abelian subgroup of $SU(2).$ To find this he identified $SU(2)$ with the unit quaternions. Now my question $:$ What are unit quaternions? What is difference of it with quaternion group? Any help will be warmly appreciated.

Thanks for your time.

答:
Unit quaternions are the multiplicative group of the quaternions of norm $1$, where by norm we mean $$\lVert q\rVert=\sqrt{q_1^2+q_i^2+q_j^2+q_k^2}=\sqrt{q^*q}=\sqrt{qq^*}$$

and $(a+bi+cj+dk)^*=a-bi-cj-dk$ is the conjugate quaternion. The conjugate quaternion works in similar fashion to the complex conjugate, with the main thing to care about being that $(pq)^*=q^*p^*$. So, for instance, in the complex representation of the quaternion we have $(z_1+z_jj)^*=z_1^*-jz_j^*=z_1^*-z_jj$.

The quaternion group $Q_8$ usually indicates its finite subgroup $\{1,-1,i,-i,j,-j,k,-k\}$.


问:
Indices confusion
Consider the equation $3^{y} = 9^{x}$

It follows that $3^{y} = 3^{2x}$

But $3^{2x} \equiv (3^{x})^{2} \equiv (3^{2})^{x}$ (I think? Since e.g. $(x^{2})^{3} \equiv x^{2 \cdot 3} \equiv (x^{3})^{2}$ right?)

So which of the following is correct? $y = 2x$ or $y = x^2$ or $y = 2^x$?

Thanks!

答:
$y=2x$ is correct. $(3^2)^x=3^y$ is also correct but does not imply $y=2^x$ or something similar.

The confusion might arise from the fact that powering is not associative: i.e. in general it is not true that $a^{(b^c)}=(a^b)^c$, e.g. $3^{(2^3)}\ne (3^2)^3$.

ContentLicense: CC BY-SA 3.0

Title: Indices confusion


问:
Normal distribution, finding the lowest grade point

The grades on the final examination in a large statistics section are
normally distributed with a mean of $78$ and a standard deviation of $8$.

The teacher wants to give the grade $A$ to $10\%$ of the students.

What is the lowest grade point that can be designated an $A$ on the
final?

I tried to find the $z$-score and probability, but I couldn't understand what the question was asking. I need some help on what to do. Thanks!

答:
You want the 90th percentile of the distribution $\mathsf{Norm}(\mu=78,\sigma=8).$ You can find the
answer by standardizing and using printed normal CDF
tables. You will find that the 90th percentile of
the standard normal distribution is about 1.28.

In R, you can find the answer directly, using the
normal quantile function qnorm as follows:

```<code>qnorm(.9, 78, 8)
[1] 88.25241
</code>```

If the lowest A is 89, then very slightly fewer
than 10% may get A's. If the lowest A is 88, then
very slightly more than 105 may get A's.

Simulation of a hypothetical class of size $n = 100.$
In any one particular such class there may be gaps
or ties at or near 88 and 89, making it impossible
to be fair and also to give exactly 10% A's.

```<code>set.seed(2021)
x = round(rnorm(100, 78, 8)
quantile(x, .9)
90%
 90

sort(x)
  [1] 60 61 61 62 63 63 65 65 65 66 66 66 66 66 67 67 67 68 68 69
 [21] 69 70 70 70 70 70 70 70 71 72 72 72 73 73 73 74 74 75 75 75
 [41] 75 75 76 76 76 76 77 77 77 77 77 77 77 77 78 78 78 78 78 78
 [61] 78 79 79 79 79 79 80 80 80 81 81 81 81 81 82 82 82 82 83 83
 [81] 84 84 84 85 85 86 86 87 88 90 90 90 90 90 91 91 91 92 92 95
</code>```

ContentLicense: CC BY-SA 4.0

Body:
You want the 90th percentile of the distribution $\mathsf{Norm}(\mu=78,\sigma=8).$ You can find the
answer by standardizing and using printed normal CDF
tables. You will find that the 90th percentile of
the standard normal distribution is about 1.28.

In R, you can find the answer directly, using the
normal quantile function qnorm as follows:

qnorm(.9, 78, 8)
[1] 88.25241

If the lowest A is 89, then very slightly fewer
than 10% may get A's. If the lowest A is 88, then
very slightly more than 105 may get A's.

Simulation of a hypothetical class of size $n = 100.$
In any one particular such class there may be gaps
or ties at or near 88 and 89, making it impossible
to be fair and also to give exactly 10% A's.

set.seed(2021)
x = round(rnorm(100, 78, 8)
quantile(x, .9)
90%
 90

sort(x)
  [1] 60 61 61 62 63 63 65 65 65 66 66 66 66 66 67 67 67 68 68 69
 [21] 69 70 70 70 70 70 70 70 71 72 72 72 73 73 73 74 74 75 75 75
 [41] 75 75 76 76 76 76 77 77 77 77 77 77 77 77 78 78 78 78 78 78
 [61] 78 79 79 79 79 79 80 80 80 81 81 81 81 81 82 82 82 82 83 83
 [81] 84 84 84 85 85 86 86 87 88 90 90 90 90 90 91 91 91 92 92 95


问:
Prove: Let A be a finite set with at least 8 elements. Then A can be partitioned into subsets where each subset has size 3 or 5.
I am trying proof this by induction.

Base Step I choose n=8,9,10
which When n=8 (3,5)  it's a combination of 3 and 5
When n=9 (3,3,3)It's just 3
When n=10 (5,5) it's only 5

In the induction step, I assume that Set A has K elements and Card(k)=3m+5n where m,n are integer.

I need to prove k+1 can be partition by 3 and 5.

Now I stuck at Card(k+1)=3m+5n+1.

答:
Prove it manually for $n \in \{8,9,10,11,12,13,14,15\}.$

Then, if there is a partition for $n$, there must be one for $(n+8)$, because you can simply add one subset of size $3$ and one subset of size $5$.


问:
I am having trouble finding $f'(x)$ given $f(x) = \frac{4}{x}$?
I am having trouble finding $f'(x)$ given $f(x) = \frac{4}{x}$. I used the definition of a derivative but I got stuck. I think the right answer is $\frac{-4}{x^2}$ but I am stuck at $\frac{-4h}{x(x+h)}$. My work is posted down below. Thank you.

答:
Just posting an answer for completeness's sake!

$\displaystyle f(x) = \frac{4}x$, then
$$\begin{align}
f'(x) &= \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\\
&= \lim_{h\to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}\\
&= \lim_{h\to 0} \frac{-4h}{hx(x+h)}\\
&= \lim_{h\to 0} \frac{-4}{x(x+h)}\\&= \frac{-4}{x^2}
\end{align}$$


答:
Assume $P(k), P(k+1), P(k+2)$ is true, now we want to verify $P(k+3)$.

To partition $k+3$, we consider the partition that we can obtain from induction hypothesis $P(k)$ union it with the remaining set of size $3$.

Title: Prove: Let A be a finite set with at least 8 elements. Then A can be partitioned into subsets where each subset has size 3 or 5.

Body:
I am trying proof this by induction.

Base Step I choose n=8,9,10
which When n=8 (3,5)  it's a combination of 3 and 5
When n=9 (3,3,3)It's just 3
When n=10 (5,5) it's only 5

In the induction step, I assume that Set A has K elements and Card(k)=3m+5n where m,n are integer.

I need to prove k+1 can be partition by 3 and 5.

Now I stuck at Card(k+1)=3m+5n+1.


问:
Polynomial linear transformations
This is probably a simple question:

How does f(x) = 3x become f(x+y) = 3(x+y) in this example:

f(x) = 3x => f(x+y) = 3(x+y) = 3x+3y = f(x) + f(y)

Where does the y come from?

Thanks.

答:
Good question! I understand how this could be confusing for a beginner. In this context, $f(x) = 3x$ really means
$$f(x) = 3x \text{ for all }x\in D$$
where $D$ is your domain of interest. In most cases, you'd be happy with $D = \mathbb R$, so let's use that. The reason why $\forall x\in D$ is implicit in such cases, is because $f(x) = x$ is the function definition with you. To define a function, you must know what values it takes on the entire domain!

If $x,y\in \Bbb R$, then $x+y\in\Bbb R$ too, and you can use your definition of $f$ to say that $$f: x+y\mapsto 3(x+y)\\ \text{i.e. } f(x+y) = 3(x+y) = 3x+3y$$
In fact, feel free to conclude things like $$f(x^2) = 3x^2, f(e^x) = 3e^x, f(\text{anything in the domain}) = 3(\text{anything in the domain})$$
given $f(x) = 3x$.

Title: Polynomial linear transformations


问:
Generalized eigenspaces of a compact operator are finite dimensional
Let $T : H\rightarrow H$ be a compact operator on a Hilbert space $H$. Say that $\lambda \in \mathbb C$ is a generalized eigenvalue of $T$ if there is some $n \geq 1$ such that $(\lambda - T)^n$ is not injective. Define the generalized eigenspace corresponding to $\lambda$ to be the space $V$ of vectors $x\in H$ such that $(\lambda - T)^n x = 0$ for some $n$. I am trying to show that $V$ is necessarily finite dimensional if $\lambda\not=0$. I can show that the kernel of $(\lambda  - T)^n$ is finite dimensional for each $n$. But, I am having trouble extending this to the union of all of these kernels. Does anyone have any suggestions?</p>

答:
This is answered in Banach Algebra Techniques in Operator Theory by Ronald G. Douglas, as part of Theorem 5.22 in the second edition.  The result you stated does not hold for $\lambda = 0$ (for instance consider the zero operator).

You already know that $\operatorname{ker}(T - \lambda)^n$ is finite dimensional for each $n$, so we need only show that the generalized eigenspace for $\lambda$ is equal to $\operatorname{ker}(T - \lambda)^n$ for some $n$. If we suppose this is not the case, then we may form an orthonormal sequence of vectors $e_k \in H$ and a strictly increasing sequence of positive integers $n_k$ such that $e_k \in \operatorname{ker}(T - \lambda)^{n_k}$ and $e_k$ is orthogonal to $\operatorname{ker}(T - \lambda)^{n_k - 1}$. In particular, $e_k$ is orthogonal to $(T - \lambda)e_k$, so

$||Te_k||^2 = ||(T - \lambda)e_k + \lambda e_k||^2 = ||(T - \lambda)e_k||^2 + |\lambda|^2 \geq |\lambda|^2.$

Because the sequence $e_k$ is orthnormal, $e_k$ tends weakly to zero, and by compactness of $T$ we have that $Te_k$ tends to zero in norm. This is a contradiction since $\lambda \not= 0$.


问:
How to solve the equation $k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$
I've been struggling a little on this problem for a while. I'm not sure what I'm missing. It goes like this:

$$k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$$

Here $k, a$ are constants and $v, t$ represent velocity and time respectively. I want to know $v$ as a function of $t$. Is it possible to solve this? Any help would be appreciated!

答:
Set $V(t)=\int_0^tv(s)ds$. Then your equation reads as
$$
k=V''(a+2V)^2.
$$
this can be separated and integrated once
$$
V''=\frac{k}{(a+2V)^2}\implies V'^2=c-\frac{k}{a+2V}.
$$
In principle you can now apply separation again, the resulting integral with square root might be difficult or impossible to solve.


问:
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$? (Singapore 1970)

I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $2^{2001}\mod{10}\equiv 4$ and $2^{2001}\pmod{10}\equiv 5$. Could you please explain to me how to solve this question? This question is multiple choice with options $1999, 2003, 4002, 6003, 2002$

答:
Generalization of the problem:

What is the sum of number of digits of the numbers $2^N$ and $5^N$?

$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$

$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$

$$10^{m+n}<10^{N}<10^{m+n+2}$$

$$N= m+n+1$$

$$ m+n=N-1$$

The sum of digits of the numbers $2^{N}$ and $5^{N}$ will be equal :

$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=N-1+2 \\\qquad \qquad \qquad \thinspace=N+1.}}}\end{align}$$

Short answer: $N+1$ digits.


问:
A morphism that is monic and epic but does not have a retraction
I’m reading the Basic algebra II written by Jacobson

and I’m wondering In Top. a morphism that is monic and epic but does not have a retraction

But I can’t come up with some good examples, can anyone give some hints?

答:
Let $f(x)=x$ from a non-trivial (not empty or singleton) set $X$ in the discrete topology to the same set in the indiscrete topology.

It's monic (as it's 1-1), epic (as it's onto), but has no retraction.


问:
If m things are distributed among a men and b women, find the probability that the number of things received by men is odd ?( given, $a$ $\leq$ $b$)
If m things are distributed among '$a$' men and '$b$' women, find the probability that the number of things received by men is odd ? ( given, $a$ $\leq$ $b$)

If the expression is known, then it can be shown easily by induction. But for this question how can I derive the expression.

答:
The formula is here:

https://www.chegg.com/homework-help/probability-even-number-successes-n-independent-bernoulli-tr-chapter-5.1-problem-32e-solution-9780131453401-exc

Probability of even number of successes in a series of independent trials

In your notation, it's shown that the formula is:

$$\frac{1}{2}\Big[1+(1-\frac{2a}{a+b})^m\Big]$$

In their notation, your $m$ is their $n$.  To convert your problem to a Bernoulli Experiment, let $a+b = c$ and let $P= \frac{a}{c}$.  Then assuming that the things are distributed randomly with uniform probability across your $c$ people, that's the probability of a 'success,' i.e. the probability of distributing an item to one of the men.


问:
question about stereographic projection
Suppose that $S^1 = \{(x,y)\in \mathbb{R}^2|x^2+y^2=1\}$, and let $x \in S^1$. How can I go about finding a homeomorphism from $S^1 \to S^1$ that sends $x$ to $(0,1)$. I thought about defining the line from $x$ to $p$ as $l(t) = tx+1$. But I'm unsure what to do. I just want a hint , and not the answer to this question.

Thank you.

答:
If $(x,y) \in S^1$ we can find such a homeomorphism by a rotation of the plane (restricted to $S^1$), in essence a multiplication by $e^{i\phi}$ for a suitable $\phi$, looking at the plane as $\Bbb C$, or the linear map with matrix $\begin{bmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi\end{bmatrix}$ etc.

This has in itself nothing to do with a stereographic projection. You can rotate any point of the circle onto any other one.

Title: question about stereographic projection


问:
Given $\lim \limits_{x \to c}f(g(x)) = f(\lim \limits_{x \to c}g(x))$ provided what 2 conditions are met?
The problem that I am having trouble with is $\lim \limits_{x \to c}f(g(x)) = f(\lim \limits_{x \to c}g(x))$. It seems like the limit of function $g$ is wrapped inside of function $f$. I need to provide two conditions that make this statement true. I am guessing that I have to use two of the three rules from "How to determine whether or not a function is continuous" but I am not sure. However, I know that $f(c)$ is not defined.

答: The conditions are: $f\circ g$ and $g$ are both continuous at $x = c$. To see this is the case, we need to have: $\displaystyle \lim_{x\to c} f(g(x)) = \displaystyle \lim_{x \to c} (f\circ g)(x) = (f\circ g)(c) = f(g(c)) = f\left(\displaystyle \lim_{x \to c} g(x) \right)$. So to make it happen, you need what is stated above.


问:
solve the equation $\tanh(\mu \pi)=\frac{2\mu}{a}$
I am trying to find what condition it must meet for the equation to have a real solution for $\mu$.

I understand that the function $\tanh(x)$ is bounded in the interval $ (- 1,1) $ if I make use of this I find the following:

$-1<\tanh(\mu \pi)<1$ then $-1< \frac{2\mu}{a}< 1$

where do i get that

$a>2\mu$

But I was reading the procedure for this and it tells me the following:

A simple sketch shows that this equation can only have a real solution for $\mu$ if
the slope of $f(\mu) = \tanh(\mu \pi)$ at $\mu = 0$ is greater than the slope of $g(\mu)=\frac{2µ}{a}$ at
the same place. The former slope is π and the latter $\frac{2}{a}$. Thus the condition for
negative eigenvalues of the original equation is $a > \frac{2}{\pi}$.

But the latter I cannot fully understand. Could you explain me?

答:
Consider that you look for the zero of funtion
$$f(\mu)=\tanh(\mu \pi)-\frac{2\mu}{a}$$ Let $x=\mu \pi$ and $k=\frac{2a}\pi$ and, instead, consider the function
$$g(x)=\frac{\tanh (x)}{x}-k$$ The maximum value of $\frac{\tanh (x)}{x}$ is $1$; so in order to have a solution, you must have $0 < k \leq 1$ that is to say $0 < a \leq \frac \pi 2$. Outside this range, there will not exist any real solution.

Now, if you want to solve for $x$ $g(x)=0$, there is no analytical solution and you will need some numerical method (such as Newton which is the simplest) and, as usual, a reasonable starting guess.

It is possible to do it building around $x=0$ the $[4,4]$ Padé approximant

$$\frac{\tanh (x)}{x}\sim \frac{x^4+105 x^2+945}{15 \left(x^4+28 x^2+63\right)}$$ which means that the estimate will be otained at the price of a quadratic equation in $t=x^2$
$$(15 k-1) t^2+105(4k-1) t-945 (1-k)=0$$ giving as estimate
$$x_0=\sqrt{\frac{3 \left( \sqrt{35(380 k^2-88 k+23)}-140 k+35\right)}{2 (15 k-1)}}$$ which would work only for $k >\frac 1 {15}$ corresponding to $x>15$.

A few values for comparison

$$\left(
\begin{array}{ccc}
k & x_0 & \text{solution} \\
 0.1 & 11.7601 &  10.0000 \\
 0.2 & 5.03871 &  4.99955 \\
 0.3 & 3.32735 &  3.32471 \\
 0.4 & 2.46434 &  2.46406 \\
 0.5 & 1.91504 &  1.91501 \\
 0.6 & 1.51223 &  1.51222 \\
 0.7 & 1.18376 &  1.18376 \\
 0.8 & 0.88802 &  0.88802 \\
 0.9 & 0.58381 &  0.58381
\end{array}
\right)$$


问:
Could $(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) $ be an integer?
Let $ a,b  \in \mathbb{R}_{>0} \backslash \left \{ \tfrac{2N-1}{2M-1} : N,M \in \mathbb{N} \right \} $ and $n \in \mathbb{N}$.

I want to prove that the quantity
\begin{equation}\label{key}
 (2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right)
\end{equation}
cannot be an integer.

I got the following:

Assume that it is an integer, $N$, say. Then
\begin{equation}
 \dfrac{Nab}{a-b}
\end{equation}
is odd.
However, I get stuck here.

答:
Let $a = b$.  Then the unnamed quantity is the integer $0$.

Possibly the strongest signal that your argument cannot work is your line "odd plus odd is odd", which is false.  The sum of two odd numbers is an even number.  Easy example: $1+1$.


问:
$t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t.
$t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t and what happens to the value of x when t becomes larger. this is the full question I was able to solve the fist part which was $\frac{1}{(2-x)(x+1)}$ and I will attach it below but I can figure out how to continue the question any help is much appreciated. I know the answers are $x=2(e^{3t}-1)/e^{3t}+2$ and x arrow to 2 but can get to it.
Any help is much appreciated .Thank you

答: Hint: $\dfrac{dx}{(2-x)(x+1)} = dt$. This is known as "separable method". Can you take it from here?


问:
Estimating $\frac{\sum_{r = 1}^{N} r^n}{N^n}$, getting a good numerical approximaion for large $N$
I am trying to find an accurate estimate for

$$f(n,N) = \frac{\sum^{N}_{r = 1} r^n}{N^n}$$

for large values of $N$.

I know the result that as ${N\to \infty}$, the $\frac{f(n,N)}{N}$ assumes value of $\frac{1}{n+1}$.

So I try to approximate $f(n,N)$ as $\frac{N}{n+1}$. Now what I have found running some python program that a better estimation for the sum is $\frac{N}{n+1} + \frac{1}{2}$ which fits much more exactly with correct value.

```<code>from math import *

n = 40
N = 1000000

s = 0
for i in range(N+1):
    s += i**n

s /= N**n
print("s:  ", s)

s1 = N/(n+1)
print("s1: ", s1)

s2 = (N)/(n+1) +1
print("s2: ", s2)

av = (s1+s2)/2
print("av: ", av)

perr_s_av = fabs(av-s)/s * 100
print("error (av): ", perr_s_av)

perr_s_s1 = fabs(s1-s)/s * 100
print("error (s1): ", perr_s_s1)
</code>```

The output is this

```<code>s:   24390.743905772357
s1:  24390.243902439026
s2:  24391.243902439026
av:  24390.743902439026
error (av):  1.3666376215028742e-08
error (s1):  0.0020499716419575546
</code>```

It is clear that the value $\frac{N}{n+1} + \frac{1}{2}$ is numerically much better approximation, why is this so, am I missing something

EDIT:

My assumption is that the next term of the polynomial $f(n, N)$ must be $\frac{1}{2}$ which is leading to this result, ie the polynomial form of $f(n, N) = \frac{N}{n+1} + \frac{1}{2} + \text{lower powers of N}$

答:
Using generalized harmonic numbers
$$f(n,N) = \frac{\sum^{N}_{r = 1} r^n}{N^n}= \frac{H_N^{(-n)} }{N^n}$$ Using asymptotics for large $N$
$$f(n,N) = \frac{N}{n+1}+\frac{1}{2}+\frac{n}{12 N}-\frac{n(n-1)(n-2)}{720   N^3}+O\left(\frac{1}{N^5}\right)$$

The above approximation gives
$$f(40,10^6) \sim \frac{1000020500136666666663291}{41000000000000000000} $$ showing a relative error equal to $1.07\times 10^{-29}$%.


问:
betting and probability
I got problem answering this puzzle, would really appreciate your help. It goes like this:

We are racing, and can at any time signal to the other that we would
  like to double our bet. We've put down 100 to start and during the
  race I signal to you I want to double the bet, What is the minimum
  probability of winning for you to accept to continue?

答:
If, by declining, the player immediately loses. Then the value of declining is $-100$

The expected value of accepting is (where $p$ is the probability of winning) $200p-(1-p)200$.

$200p-(1-p)200\geq-100$

$p\geq\frac{1}{4}$

If, on the other hand, by declining, the race continues at the current stakes, then we have:

$p\geq\frac{1}{2}$

Title: betting and probability


问:
Is the set of periodic sequences a vector subspace of the space of sequence?
A sequence $(a_n)_{n\geq0}$ is periodic if there exists an integer $t\geq1 $ such that $a_{n+t}=a_n$ for all $n \geq0$. Is the set of periodic sequences a vector subspace of the space of sequences ?

Please kindly tell a hint for me or something recommend for me. Thank in advance !

答:
You ask for a hint...

Check each condition for a vector space:  Is the zero sequence a periodic sequence and does it act as the identity for the addition operation?  Is the scalar multiple of a periodic sequence a periodic sequence, is the sum of two periodic sequences a periodic sequence?  and so on ...

For example:

$(0)_{i=1}^\infty$ is the zero sequence.  It is periodic with period $1$ (meaning, "with $t = 1$" in the definition you recite).  This is the zero in the vector space of sequences and is the zero in the space of periodic sequences.

Let $a$ be a scalar and $b = (b_i)_{i=1}^\infty$ be a periodic sequence with period $t$, so that $b_{i+t} = b_i$ for all $i \geq 1$.  Then $ab = (ab_i)_{i=1}^\infty$ is a periodic sequence with period $t$ because $ab_{i+t} = ab_i$ for all $i \geq 1$.  [ Here, we are using the equality $b_{i+t} = b_i$ to replace $b_{i+t}$ with $b_i$, obtaining that $(ab_i)_{i}$ is also periodic with period $t$. ]


问:
How can i solve this LCM puzzle?
I have this puzzle "A woman went to the market and a horse broke the eggs she had in her basket. The owner of the horse offered her to pay for the damage caused. He asked her how many eggs were broken by his horse. She said she didnt knew , but remebered that when she arrenged them in pairs, there was one left. Same thing happened when she arrenged them in groups of 3, 4, 5 and 6. But when she arrenged them in groups of 7 there was none left. what is the minimum quantity of eggs that were in the basket?" I tried showing the equations

2a+1=x

3b+1=x

4c+1=x

5d+1=x

6e+1=x

7f=x

But I really dont know what to do with them

答:
You can rewrite your equations as

$x - 1 = 2a$

$x - 1 = 3b$

$x - 1 = 4c$

$x - 1 = 5d$

$x - 1 = 6e$

$x = 7f$

The first five equations imply that $x-1$ is a multiple of $2,3,4,5$, and $6$.

The least common multiple of $2,3,4,5$, and $6$ is $60$

So $x = 60n + 1$ for some integer, $n$.

The last equation implies that $x$ is a multiple of $7$.

Since $56$ is a multiple of $7$ we can say $x = 56n + (4n + 1)$.

So $4n + 1$ needs to be a multiple of $7$.

So just keep adding $4$ to $1$ until you get a multiple of $7$

$$1, 5, 9, 13, 17, \color{red}{21}$$

where $21 = 1 + 4(5)$ implies that $n = 5$

So $x = 60(5)+1 = 301$

CONFIRMATION

$301 - 1 = 2 \cdot 150$

$301 - 1 = 3 \cdot 100$

$301 - 1 = 4\cdot 75$

$301 - 1 = 5 \cdot 60$

$301 - 1 = 6 \cdot 50$

$301 = 7 \cdot 43$

Title: How can i solve this LCM puzzle?


问:
Prove that the integral of a monotone function is between a given area interval
So I have to prove the following:

If $f$ is monotone in an interval $I:\{a\leq x\leq b\}$, $\int_a^bf(x)dx$ is between $f(a)(b-a)$ and $f(b)(b-a)$.

I clearly see the geometric interpretation of this. The previous statement is saying that the area represented by the integral is "between" the area of a rectangle with height $f(a)$ and base $(b-a)$, and a rectangle with height $f(b)$ and base $(b-a)$. Obviously the function could be strictly increasing, decreasing or just being a constant (as far as I know that's the definition of a monotone function), as such I can visualize that the statement holds, however I don't seem to be able to put this into math notation and actually prove it.

I have tried to use a Riemann sum to solve it, I've also tried to use upper and lower integrals but I didn't really know what to do with those. Any help or insight on how to tackle this would be truly appreciated.

EDIT 1:

So, from the comment I came up with the following (for the case in which we have a strictly increasing function):

If $f(b)>f(a)$, then $\int_a^b[f(b)-f(a)dx]\geq 0$, since the integrand is positive, while $\int_a^b[f(a)-f(x)]dx\leq 0$, because the integrand is negative. From this we get the following inequality:

$\int_a^b[f(a)-f(x)]dx\leq 0\leq\int_a^b[f(b)-f(a)]dx$

By using properties of the integrals, we simplify as:
$\int_a^bf(a)-\int_a^bf(x)dx\leq 0\leq\int_a^bf(b)dx-\int_a^bf(x)$

It is then obvious,

$\int_a^bf(a)\leq\int_a^bf(x)\leq\int_a^bf(b)$

For this last part I am not really sure how to convert the integrals of the function evaluated at the lower and upper bound of the integral into $f(a)(b-a)$ and $f(b)(b-a)$. I can see that the height will always be the same and the base is just going to be $(b-a)$. Maybe a Riemann sum of the previous integral would help? I have tried doing this:

$\int_a^bf(a)=\lim_{x\to \infty}\sum_{k=1}^nf(\xi)\Delta_kx$. Since we have that for every $\xi$, $f(\xi)$ will be the same as $f(a)$, then by properties of limits and sums (because f(a) is a constant), $\lim_{n\to \infty}\sum_{k=1}^nf(a)\Delta_kx=f(a)\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$, and since the sum of all the little sub-intervals that compose $\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$ is $(b-a)$, we finally have $f(a)(b-a)$. We then do the same for $f(b)$. This leads us to,

$f(a)(b-a)\leq\int_a^bf(x)\leq f(b)(b-a)$

答:
If $f$ is increasing, then by the mean value theorem for integrals, there is $ \mu \in [f(a),f(b)]$ such that

$$\int_a^b f(x)dx= \mu(b-a).$$

Since $f(a) \le \mu \le f(b)$, we have $f(a) (b-a) \le \mu(b-a) \le f(b)(b-a)$ and the result follows.

If $f$ is decreasing, then by the mean value theorem for integrals, there is $ \mu \in [f(b),f(a)]$ such that

$$\int_a^b f(x)dx= \mu(b-a).$$

Since $f(b) \le \mu \le f(a)$, we have $f(b) (b-a) \le \mu(b-a) \le f(a)(b-a)$ and the result follows.


问:
What's the probability of having n equal numbers in different slots between 00000 and 600000
Say you're trying to figure out which lottery ticket to buy for the weekend. Although, the chances of winning are slim, you think that maybe knowing how likely it is that there are $n\in\{0,1,2,3,4,5\}$ instances of the same number in different slots on a ticket with numbers from 00000 to 60000 will help.

What are the chances of having $n\in\{0,1,2,3,4,5\}$ instances of the same number in a random ticket with values between 00000 and 60000?

Is there a way to generalize this for chains of slots that are $k$ digits in length going from $000\dots00n$  to  $m00\dots000$ with $0\leq n \leq 9$ and $1\leq m \leq  9$ ?

答:
Taking it to mean you want to count the number of times a particular digit repeats in the string:

For k repetitions of a particular digit:

(a): The particular digit occurs in the first digit and (k-1) others

$\binom1 1\binom 5 {k-1}9^{(6-k)}$

(b): The particular digit doesn't occur in the first digit

$\binom51\binom5{k}9^{(5-k)}$

Work out and add up


问:
If $M_{Y,Z} (t_1,t_2) = \frac{e^{\frac{t_1^2}{1-2t_2}}}{1-2t_2}, t_2 < 1/2$, what is $\text{Corr}(Y,Z)$ and $M_{X = Y\pm Z}(t)$?

Let the joint m.g.f (moment generating function) of $(Y,Z)$ be $$\color{blue}{M_{Y,Z} (t_1,t_2) = \frac{e^{\frac{t_1^2}{1-2t_2}}}{1-2t_2}, t_2 < 1/2}$$
(a) Find $\text{Corr}(Y,Z)$

(b) Are $Y,Z$ independent?

(c) Find the m.g.fs of $Y-Z$ and $Y+Z$.

(a) $$\text{Corr}(Y,Z) = \frac{\text{cov}(Y,Z)}{\sqrt{\text{var}(Y)}\sqrt{\text{var}(Z)}}$$
From the joint m.g.f, I can find the individual m.g.fs as
$$M_Y(t_1) = M_{Y,Z}(t_1,0) \text{ and } M_Z(t_2) = M_{Y,Z}(0,t_2)$$
which gave me $E[Y] = 0, E[Z] = 2, \text{var}(Y) = 2, \text{var}(Z) = 4$. Now, $\text{cov}(Y,Z) = E[YZ] - E[Y]E[Z]$. How do I find $E[YZ]$?

(b) Is the following condition necessary and sufficient to check that $Y,Z$ are independent?
$$M_{Y,Z}(t_1,t_2) = M_Y(t_1) M_Z(t_2)$$
In our case, clearly, $M_{Y,Z}(t_1,t_2) \ne M_Y(t_1) M_Z(t_2)$, so $Y,Z$ shouldn't be independent. Please let me know if there are any flaws in this reasoning.

(c) The m.g.f of $X = Y-Z$ would be $$E[e^{tX}] = E[e^{tY-tZ}]$$
Had $Y,Z$ been independent (they aren't, right?), I could've figured this out instantly. The expectation would've split and our required m.g.f would be the ratio of the m.g.fs of $Y,Z$. This doesn't seem to be the case! How do I proceed now?

Thank you very much.

答:
a) Differentiate partially w.r.t $t_1$ and $t_2$ (in any order) and put $t_1=t_2=0$. This gives $E[YZ]$.

b) Yes. $M_{Y,Z}(t_1,t_2)=M_Y(t_1)M_Z(t_2)$ for all $t_1,t_2$ is necessary and sufficient for indepedence

c) $E[e^{tY-tZ}]$ is nothing but $M_{Y,Z}(t_1,t_2)$ when $t_1=t$ and $t_2=-t$.


问:
Divergence theorem in volume integral
We have a partial differential equation
 \begin{equation}
\nabla \cdot (p_1^2\nabla\alpha)=0\,.
\end{equation}

Question: from this equation how can I write the following condition?

\begin{equation}
\int_\Omega\alpha\nabla \cdot(p_1^2\nabla\alpha)=
\int_{\partial\Omega}\alpha p_1^2 n\cdot\nabla\alpha
-\int_\Omega p_1^2(\nabla\alpha)^2=0 \,.
\end{equation}
$p_1$ and $\alpha$ are position dependent variable.

答:
It appears this is an application of the Divergence theorem, assuming everything is smooth:
$$
\int_{\Omega} \nabla \cdot F  = \int_{\partial \Omega} F\cdot n\,dS,
$$

Let $F = \alpha p \nabla \alpha$, where $p = p_1^2$. The product rule reads for a vector $v$ and a scalar $\phi$:
$$
\nabla \cdot (\phi v) = \nabla \phi \cdot v + \phi\nabla \cdot v,
$$
therefore we have:

$$
\int_{\Omega} \nabla \cdot (\alpha p \nabla \alpha) = \int_{\Omega} \Big(\alpha \nabla \cdot (p \nabla \alpha) + p \nabla \alpha \cdot \nabla \alpha \Big) = \int_{\partial \Omega} \alpha p \nabla \alpha\cdot n\,dS,
$$
and this is
$$
\int_{\Omega}\alpha \nabla \cdot (p \nabla \alpha) = \int_{\partial \Omega} \alpha p (\nabla \alpha\cdot n)\,dS - \int_{\Omega}p \nabla \alpha \cdot \nabla \alpha ,
$$
rewrite $\nabla \alpha \cdot \nabla \alpha  = |\nabla \alpha|^2$, and by the original equation $\nabla \cdot (p \nabla \alpha) = 0$, the left hand side of above vanishes, hence we have:
$$
\int_{\partial \Omega} \alpha p (\nabla \alpha\cdot n)\,dS - \int_{\Omega}p |\nabla \alpha|^2 = 0 .
$$

Or you can just memorize the integration by parts formula derived from the divergence theorem:
$$
\int_{\Omega} u\nabla \cdot v = -\int_{\Omega} \nabla u\cdot v + \int_{\partial \Omega} u(v\cdot n)\,dS.
$$

Title: Divergence theorem in volume integral


问:
Trying to prove that the set of global minimisers of a convex function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a convex set

Prove that the set of global minimisers of a convex function
$f:\mathbb{R}^n\rightarrow\mathbb{R}$ over the convex set $\Omega$ is
a convex set

I have been told to consider the set $$S=\{\underline{x}\in\Omega : f(\underline{x})\leq \gamma\},$$ where $\gamma$ is value of $f$ at the minimising points. Let both $\underline{x},\underline{y}\in S$ and $\theta\in [0,1]$ be arbitrary, such that $f(\underline{x})\leq\gamma$ and $f(\underline{y})\leq\gamma \hspace{8pt} (1)$.  Then
\begin{align*}
f(\theta\underline{x}+(1-\theta)\underline{y})&\leq\theta f(\underline{x})+(1-\theta)f(\underline{y}) \hspace{20pt}\text{($f$ is convex)} \\
&=\theta\gamma+(1-\theta)\gamma \hspace{47pt}\text{(by $(1)$)}\\
&=\gamma
\end{align*}
Hence $\theta\underline{x}+(1-\theta)\underline{y}\in S \ \ \forall \underline{x},\underline{y}\in\Omega$ and $\theta\in [0,1]$.

Though I don't fully understand how the set $S$ represents the set of global minimizers of $f$.

答: If $\gamma$ is the infimum of the function then $\gamma \leq f(x)$ is satisfied for any $x$. So $x \in S$ iff $f(x)=\gamma$ which menns $f$ is minimized at $x$.


问:
sqrt of dB number
If the variance is $\sigma^2= 8 {\bf \, dB}$, and I want to calculate standard deviation $\sigma=?$ in dB. Which of the following is correct

a. $\sigma = 8^{0.5} = 2.82 $ dB

b.  $\sigma =10\log_{10}( \sqrt{10^{0.8}})= 4$ dB

答:
I'll go first on a tangent to (properly this time) address the very good point made by lulu in the comments. The decibel is not a proper unit. It's only a signal that the associated numerical value is based on a (specific) logarithmic scale.

Imagine you were a budding scientist 200 years ago. You make measurements of something that feels like "power". Let's call the unit PU. Your measurements have a very high dynamic range. You take notes on pen and paper, and you realize that even your previous neat trick of using powers of $10$ isn't good or convenient enough, because you can't remember if you meant $10^4$ or $104$. That's the point where you start using the decibel logarithmic transformation.

$$
f_{dB}(x) = 10\log_{10}(x) \quad\text{and}\quad f_{dB}^{-1}(x)=10^{x/10}
$$

And so, your $1000$ PU become $30$ dB$_{(PU)}$. Later, you measure standard deviations and variances, and they are respectively in pseudo-units of dB$_{(PU)}$ and dB$_{(PU^2)}$, but rapidly in the literature, everything is just dB, and everyone is supposed to know which underlying hidden unit is being used, and to which dimension.

In conclusion, as Ninad correctly pointed out in the comments, you must always do the $f \circ g \circ f^{-1}$ "unwrapping/rewrapping" when dealing with decibels. So, definitely option b. Yes, there are circumstances where you can take a shortcut. To give a stupid example, you could technically calculate a momentum measured in dB$_{(\text{kg.m.s$^{-1}$})}$ by just adding a mass in dB$_{(\text{kg})}$ and a velocity measured in dB$_{(\text{m.s$^{-1}$})}$. But the OP's case is clearly not one of those.

Title: sqrt of dB number


问:
Mixed strong-weak convergence problem
Let $\Omega$ be an open bounded domain in $\mathbb{R}^N$. Let $J \;$ be a $\mathcal{C}^1$ class functional and $(u_n)_n\subset W_0^{1, 2}(\Omega)$. Suppose that
$$ u_n\rightharpoonup u \quad\mbox{ in } W_0^{1, 2}(\Omega)\quad\mbox{ and }\quad J^{\prime}(u_n)\to 0 \quad \mbox{ in } \; W^{-1, 2}(\Omega).$$

How to show that
$$J^{\prime}(u_n)(u_n - u)\to 0?$$

On my notes, it seems to be an immediate thing, but I don't know how to justify it. Could anyone please explain me why the above convergence holds?

Thank you in advance!

答:
This follows from the general fact that in a Banach space $X,$ if $x_n \rightharpoonup x$ weakly in $X$ and $f_n \to f$ strongly in $X',$ then $f_n(x_n) \to f(x)$ as $n \to \infty.$ Indeed we can write
$$ |f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n-x)|, $$
and by the principle of weak boundedness, $\lVert x_n\rVert_X \leq C$ for all $n$ so we have $|f_n(x_n)-f(x_n)| \leq C \lVert f_n - f\rVert_{X'} \to 0$ as $n \to \infty.$ Also by weak convergence $|f(x_n-x)| \to 0$ as $n \to \infty,$ so it follows that $f_n(x_n) \to f(x)$ as $n \to \infty.$

In your setting take $X = W^{1,2}_0(\Omega)$ $x_n = u_n$ and $f_n = J'(u_n).$

Title: Mixed strong-weak convergence problem


问:
Derivation of the summation
I am trying to use gradient descent to find two straight lines that fit a discrete set of points (2D).
One obvious way is to use a brute force search algorithm (that is already implemented) but I am trying to improve the performance of the code. In reality, at this point, I am trying to better understand the underlying math.

My cost function is defined as follows:

$$
J(k) = \frac{1}{k}\sum_{i=1}^{k}(pred_i - y_i)^2+\frac{1}{n-k}\sum_{i=k}^{n}(pred_i - y_i)^2
$$

I want to find the point k where the two lines are connected, the next step would take the derivative of the cost function. But the $k$ parameters is in the summation limits.
In this formulation, the two linear models are computed from the first and $k$ point for the first line and $k$ to the last point for the second line.

Is this possible, any ideas? Some preliminary tests showed that this cost function is not convex (at least it depends on the discrete set of points).

答:
You want to minimize
$$J(k)=\frac 1 k \sum_{i=1}^k(\hat y_i-y_i )^2+\frac 1{n-k}\sum_{i=k}^n(\hat y_i-y_i )^2$$ If you use ordinary least squares for straight lines, you have the symbolic values for each of the sums.

Now, use any integer programming solver to find $k$.

I have been working this kind of problems for years and never faced any serious difficulty.

Title: Derivation of the summation


问:
Can a non-linear transformation, specifically a translation, have eigenvectors?
In trying to grasp the basics of linear algebra, I've encountered the following question:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the transformation
defined by $(x,y) \mapsto (x+1,y+1)$. Does $f$ have any eigenvectors?
Explain. What type of transformation is $f$?

I know that $f$ is a translation, which is a non-linear transformation. Does this mean that it has no eigenvectors? Do eigenvectors only apply to linear transformations?

Or can you actually have eigenvectors for non-linear transformations? If so, would all vectors be eigenvectors for this transformation? The transformation causes points to be translated 1 unit right and 1 unit up, but don't vectors stay the same? For example, the vector $\begin{bmatrix}3\\2\end{bmatrix}$ remains $\begin{bmatrix}3\\2\end{bmatrix}$. Before the transformation, it has initial point $(0,0)$ and terminal point $(3,2)$; after the transformation, it has initial point $(1,1)$ and terminal point $(4,3)$.

Hope you'll answer the quoted question, and also let me know if I've misunderstood anything. Thanks!

答: Yes, eigenvectors and eigenvalues are defined only for linear maps. And that map $f$ maps elements of $\Bbb R^2$ into elements of $\Bbb R^2$. In particular, it maps $(3,2)$ into $(4,3)$. Thinking about it as acting on vectors, by acting on their initial and final points, is misleading.


问:
prove $\left\{\begin{array}{l}X^{\prime \prime}+\lambda X=0,0<x<1 \\ X(0)=0, X(1)+X^{\prime}(1)=0\end{array}\right.$ has $\lambda >0$ without solving
Can you help me some idea to prove it or some similar question about it. I know prove $\lambda>0$ is easy when solving it, it's just like normal differential equation.

答:
This is a problem of eigenvalues.
We attach the characteristic polynomial: $r^2+\lambda=0\Rightarrow r^2=-\lambda.$
We talk about the parameter $\lambda.$

If $\lambda>0\Rightarrow r_{1}=i\sqrt{\lambda}, r_{2}=-i\sqrt{\lambda}\in \mathbb{C}.$
In this case the solution will be: $X(x)=c_1cos(\sqrt{\lambda} x)+c_2sin(\sqrt{\lambda} x)$.
We set the conditions to the limit and we have: $X(0)=c_1=0$ and $X(1)=c_1cos(\sqrt{\lambda})+c_2sin(\sqrt{\lambda})$ and $X'(x)=-c_1sin(\sqrt{\lambda}x)\sqrt{\lambda}+c_2cos(\sqrt{\lambda}x)\sqrt{\lambda}\Rightarrow X'(1)=-c_1sin(\sqrt{\lambda})\sqrt{\lambda}+c_2cos(\sqrt{\lambda})\sqrt{\lambda}.$

$X(1)+X'(1)=c_1cos(\sqrt{\lambda})+c_2sin(\sqrt{\lambda})+c_1sin(\sqrt{\lambda})\sqrt{\lambda}+c_2cos(\sqrt{\lambda})\sqrt{\lambda}=0$, $c_1=0 $ so
$X(1)+X'(1)=c_2sin(\sqrt{\lambda}) +c_2cos(\sqrt{\lambda})\sqrt{\lambda}=0$ and solve this equation and find the eigenvalues.

If $\lambda=0\Rightarrow r_1=r_2=0$. In this case the solution will be:$X(x)=c_1+c_2x.$

$X(0)=c_1=0$ and $X(1)=c_2$, $X'(x)=c_2\Rightarrow X'(1)=c_2.\Rightarrow X(1)+X'(1)=c_2+c_2=0\Rightarrow c_1=c_2=0$$\Rightarrow X(x)=0$
so $\lambda=0$ it's not a eigenvalue.

If $\lambda<0\Rightarrow r_1=\sqrt{-\lambda}, r_2=-\sqrt{-\lambda}.$
In this case the solution will be: $X(x)=c_1e^{r_1}+c_2e^{r_2}.$
And we put the initial conditions in this case as well and we see if $\lambda<0$ is a eigenvalue.


问:
Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
On the interval [0,1].  Define $f(x)=x\sin(1/x)$ for $x\in(0,1]$ and $f(0)=0$.  I didn't work out the exact details but I'm pretty sure that then $$\Big |\int_0^xf'(t)dt\Big |=\infty,$$ due to a process similar to something of the form $1-2+3-4+5-...$ , as one approaches zero from above.

However according to the measure-theoretic definition of absolute continuity, there should in fact be some set of measure zero $E\in[0,1]$ such that $$\Big |\int_Ef'd\mu\Big | > 0.$$

I wasn't under the impression that this was even possible.

Edit:  Maybe I wasn't clear about what my question is.  What I want is a proof (constructive or not) that there exists a set of measure zero $E$ such that $\Big |\int_Ef'd\mu\Big | > 0.$  Or if that's not possible then for someone to explain to me what my misconception is concerning the measure theoretic definition of absolute continuity:

For $v(E)=\int_Efd\mu$.

If $\mu(E)=0$ then $v(E)=0$.

link to definition

definition can also be found in Royden's Real Analysis

答:
Update: Corrected the definition type mistake, but it seems the proof is not measure based, as OP said what he/she needed. So further work is needed.

Given positive number $\epsilon$, for every $\delta>0$, if you pick up points $$a_{k}=\frac{1}{2km\pi},a_{k+1}=\frac{1}{(2k+1)m\pi}$$for example, then you have $$f(a_{k})=\frac{1}{2km\pi},f(a_{k+1})=\frac{-1}{(2k+1)m\pi},|f(a_{k})-f(a_{k+1})|\ge \frac{2}{(2k+1)m\pi}$$Here $m\in \mathbb{N}$ is an odd number large enough such that $$\sum_{k=1}^{\infty}|a_{k}-a_{k+1}|<\delta,\forall k\in \mathbb{N}$$
This is possible because we are essentially taking the partial sums of the alternating series. So if we choose $m$ to be large enough, we can "squeeze" the sum to be less than $\delta$.

Now if you pick up points $\{a_{k}\}_{k\rightarrow \infty}$, then $$\sum_{k=1}^{\infty}|f(a_{k})-f(a_{k+1})|>\epsilon$$since the left hand side essentially diverges.

For your question in the comment, the derivative is only undefined when $x=0$. Otherwise it is a perfectly well-defined function. So it is defined almost-everywhere.


问:
Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$
Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?

答:
Let $ab\neq 0$
$$a^3+b^3=(a+b)(a^2-ab+b^2)=1$$ Therefore $a,b$ cannot have the same signs, then $$ab < 0$$ Hence $$a^2-ab+b^2>1$$ which is a contradiction.


问:
$t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t.
$t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t and what happens to the value of x when t becomes larger. this is the full question I was able to solve the fist part which was $\frac{1}{(2-x)(x+1)}$ and I will attach it below but I can figure out how to continue the question any help is much appreciated. I know the answers are $x=2(e^{3t}-1)/e^{3t}+2$ and x arrow to 2 but can get to it.
Any help is much appreciated .Thank you

答:
From the first part, by integration,

$$\frac13(\log(x+1)-\log(2-x))=t+c$$

or

$$\frac{x+1}{2-x}=Ce^{3t}$$ and using the initial condition,

$$\frac12=C.$$

Now solve for $x$.


问:
Let $P$ be a polynomial with integer coefficients satisfying $P(0)=1, P(1)=3, P'(0)=-1, P'(1)=10$. What is the minimum possible degree of $P$?
I am preparing for GRE MATH SUBJECT TEST, I have reviewed many problems, some of these problems are not in GRE Math Practice books, but they are in some other books. I found the following problem, but not sure if it can be a GRE problem or no. And I do not even know how to start;

Let $P$ be a polynomial with integer coefficients satisfying $P(0)=1, P(1)=3, P'(0)=-1, P'(1)=10$. What is the minimum possible degree of $P$?

(A) $3$

(B) $4$

(C) $5$

(D) $6$

(E) No such $P$ exists.

答: Since you are given two points of $P'(x)$ , start with a linear function for $P'(x)$, let it be $ax+b$, plug the values to get $P'(x)=11x-1$, now integrate it to get $P(x)=\dfrac{11x^2}{2}-x+C$, you have $P(0)=1$, so $C=1$, but at $x=1$, we are getting $P(1)=\dfrac{11}{2}\ne 3$, so this suggests start with a second degree equation for $P'(x)$, and since this will contain one extra variable, this will give you the answer.


问:
Evaluating $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}$ (and another)

Compute
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$ and
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}\sin{\frac{2a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$
where $a$ is a positive real number.

I think the second limit may be solved by using the comparison criteria(limit comparison test), putting $-1 \le \sin{\frac{2a}{3^k}} \le 1$ and using the first limit. I suspect the first limit is $0$, so the second one should be $0$ too, but I might be wrong. I'm not sure how would you do the first one - I tried using the comparison test on the first one, but I didn't succeed. Can you help?

答:
The second sum is telescoping, due to $$\frac{\sin x\sin 2x}{\sin 3x}=\frac14(\cot x-3\cot 3x).$$ This easily gives $$\sum_{k=1}^n\frac{\sin\frac{a}{3^k}\sin\frac{2a}{3^k}}{3^k\sin\frac{a}{3^{k-1}}}=\frac14\sum_{k=1}^n\left(\frac{1}{3^k}\cot\frac{a}{3^k}-\frac{1}{3^{k-1}}\cot\frac{a}{3^{k-1}}\right)=\frac14\left(\frac{1}{3^n}\cot\frac{a}{3^n}-\cot a\right).$$
The first sum doesn't seem to. Trying to replace $\cot$ by $-\csc$ yields$$\frac14(3\csc 3x-\csc x)=\frac{\sin^2 x}{\sin 3x},$$ so it would telescope with the sine in the numerator squared.


问:
Is there a continuous surjective map from a 2-cell to a 2-sphere?
I think there is, tho there probably isn't a differentiable one. My idea is to consider squares centered at origin in $[-1,1]^2$ and send them to circles cut by a plane in $S_2:= \{p \in \mathbb{R}^3 : |p| = 1\}$. (The origin point would go to the point when the plane intersects the sphere at only 1 point, and the outer square would go the the other one). Hopefully if the way the squares are sent is "consistent" (so that near squares don't for example send points in same face of square to opposite sides of the circle), then gluing this all together we'd have the desired map. But i have trouble formalizing and proving whether this works...

答:
Ok, here's a very step-by-step proof.

I assume your definition of $2$-cell is the convex hull $H$ of the points $(0,0), (1,0)$ and $(0,1)$ in the plane.  Inscribe a circle $C$ in $H$ and let $x$ be the center of the disc inside $C$.  Let $t_h$ be the line segment from $x$ to $h$ for each $h$ in the boundary of $H$, denoted $\partial(H)$.  Then since $H$ is convex, $t_h$ intersects $\partial(H)$ in a single point and is contained in $H$.

If $y$ is in $H$ then there is some $t_h$ containing $y$ by convexity, and since these segments only intersect in the point $x$, each point in $H$ but outside of $C$ is contained in exactly one such $t_h$.  For each $y$, let $b(y)$ be the unique point on $C$ sharing such a segment containing $y$.

Define a function $f$ on $H$ that's equal to the identity inside of $C$ and its boundary, and otherwise is equal to $b(y)$.  It's surjective by definition, and continuous on $C$ and the disc $D$ inside it, since it's just the identity there.  We just need to show it's continuous outside of $\overline{D}$.  Let $y$ be such a point and let $y_n \rightarrow y$.

Then by convexity there are line segments $s_n$ between $y_n$ and $y$ respectively, whose lengths converge to zero since $y_n \rightarrow y$.  Thus the angle $\alpha_n$ of the triangle between $x, y$ and $y_n$ at the vertex $x$ converges to zero.  If $R_n$ is one of the other sides of this triangle, then $\alpha_n \rightarrow 0$ implies that $R_n \cap C \rightarrow t_y \cap C = b(y)$.  But that's the definition of $f$, and this shows that it's continuous since these are compact metric spaces.

This shows that the $2$-cell has a continuous surjection onto the closed disc.  Then the one-point compactification of the open disc is the sphere, and this extends continuously to a function on the closed disc by mapping $\partial(D)$ to $\infty$.  Composition gives your function.

Smooth surjections do exist, as well, but the details involved to construct it explicitly are more complicated.  A sketch: Take an open disc in the triangle away from the boundary and do the one-point compactification there as well, extending it beyond the boundary to the rest of the $2$-cell by having the function 'continue' along the $t_h$'s which will be mapped to (subarcs of) great circles through infinity.  Basically 'overwrap' the sphere with the plane so that your $2$-cell spills over $\infty$.


答:
Picking up from where you left off:

$$\int \frac{dx}{3(2-x)}+\int \frac{dx}{3(x+1)}=\int dt \Rightarrow \frac 13 \ln \frac {x+1}{2-x} + C =t \\
\ln \frac{x+1}{2-x}+C=3t \Rightarrow \ln \frac 12+C=0 \Rightarrow C= \ln 2 \\
\ln \frac {2(x+1)}{2-x}=3t \Rightarrow e^{3t}=\frac{2(x+1)}{2-x}=-2+\frac{6}{2-x} \\
e^{3t}+2=\frac{6}{2-x} \Rightarrow \frac{2-x}{6}=\frac{1}{e^{3t}+2} \Rightarrow 2-x=\frac{6}{e^{3t}+2} \Rightarrow x=2-\frac{6}{e^{3t}+2}.$$

Title: $t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t.

Body:
$t=0$ and $x=0$ for the equation: $\frac{dx}{dt}=(2-x)(x+1)$? what is the differential equation and obtain an expression for x in terms of t and what happens to the value of x when t becomes larger. this is the full question I was able to solve the fist part which was $\frac{1}{(2-x)(x+1)}$ and I will attach it below but I can figure out how to continue the question any help is much appreciated. I know the answers are $x=2(e^{3t}-1)/e^{3t}+2$ and x arrow to 2 but can get to it.
Any help is much appreciated .Thank you


问:
Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true?
Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.

$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$

Where he defines:

$$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$

Question: I can't see how that statement is true.

I would appreciate replies that don't assume university level training in mathematics.

Body:
This is just an idea, why don't you view it like this
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$

Where $$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
But this in terms of a function also say $p(n)$ is the function of the $n$th prime
$$ \exp[ \sum_{k=1}^{\infty } ( \frac{1}{k} \cdot    \sum_{n=1}^{\infty} ( \frac{1}{p(n)^k}  ) )] = \sum_{n = 1}^{\infty}  ( \frac{1}{n} ) $$


问:
Is every sigma-compact space is lindelof space?
Is every sigma-compact topological  space is Lindelof space ?
I know about $\sigma$-compact and Lindelof space.Say about converse is true or not.

答:
Suppose that $X= \bigcup_{n \in \Bbb N} K_n$ where each $K_n$ is compact.

Let $\mathcal{U}$ be any open cover of $X$.

We have for each $n$ that $K_n \subseteq \bigcup \mathcal{U}$ so there is a finite subcover $\mathcal{U}_n \subseteq \mathcal{U}$ such that $K_n \subseteq \bigcup \mathcal{U}_n$ as $K_n$ is compact.

Then $\bigcup_{n \in \Bbb N} \mathcal{U}_n$ is a countable (a countable union of finite sets is countable) subcover of $\mathcal{U}$.

As $\mathcal{U}$ was arbitrary, $X$ is Lindelöf.


问:
Angle between Harmonic functions
I have a question about a practice prelim problem:
What is the angle between the curves $Re(z^3) = 1$ and $Re(z^3) = Im(z^3)$?

Also, What is the angle between the curves $Re(z^3) = 0$ and $Re(z^3) = Im(z^3)$?

So for the first part I believe that we can use conformal maps preserve angles so we are just finding the angle between $x=1$ and $x=y$ in the $x-y$ plane where $z^3 = w = x+iy$.  However, this only works for $f(z) =z^3 $ and $f'(z) \neq 0$.  Thus, we cannot use this in the second part.  Any ideas for the second part would be greatly appreciated.

Thanks!

答:
What is the angle between the curves $\operatorname{Re}(z^3)=1$ and $\operatorname{Re}(z^3)=\operatorname{Im}(z^3)$?

The problem is poorly phrased, since these sets are not curves. The first one is the disjoint union of three simple unbounded curves. The second one is the union of three lines passing through the origin.

Your idea to use conformality is correct, though.  The lines $\operatorname{Re}(w)=1$ and $\operatorname{Re}(w)=\operatorname{Im}(w)$ cross at an angle $\pi/4$ and since $w=z^3$ is a conformal transformation outside of the origin, at every intersection of  $\operatorname{Re}(z^3)=1$ and $\operatorname{Re}(z^3)=\operatorname{Im}(z^3)$ the angle is $\pi/4$.

What is the angle between the curves $\operatorname{Re}(z^3)=0$ and $\operatorname{Re}(z^3)=\operatorname{Im}(z^3)$?

Again, I object to the statement. Each set is the union of three lines through $0$. The    common point of these sets is also $0$. There is an assortment of angles to choose from. The smallest of them is indeed $\pi/12$. There are also $\pi/3\pm \pi/12$, $2\pi/3\pm \pi/12$, and $\pi-\pi/12$ angles there (not counting those over $\pi$ radian).

Title: Angle between Harmonic functions


问:
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone help?

答:
The closed form using polylogarithms is given elsewhere.  An alterantive for (upper and lower) bounds...

I note that for $n > 4$, $\displaystyle \frac{2}{2^{n/2}} < \frac{\sqrt{n}}{2^{n/2}} < \frac{1}{2^{n/4}}$.

The left inequality is easy: For $4 < n$, $2 < \sqrt{n}$, then divide through by $2^{n/2}$.  Using the first three terms of the original series and then the lower bounding term subsequently, we get the lower bound $(1/4)(4 + 6\sqrt{2} + \sqrt{6} = 3.733{\dots}$.

For the right inequality, start with $n = 4$, observing $\displaystyle \sqrt{4} = 2 = 2^{4/4}$.  Then $\frac{\mathrm{d}}{\mathrm{d}n}\sqrt{n} < 1/4$ for $n > 4$ and $\frac{\mathrm{d}}{\mathrm{d}n} 2^{n/4} > 1/4$ when $2^{n/4} \ln 2 > 1$, which is certainly the case for $n > 4$.  Therfore,
$$  \frac{\sqrt{n}}{2^{n/4}} < 1 \qquad (n > 4)  \text{.}  $$
Now divide through by $2^{n/4}$.  Splicing the three terms from the original series into the series of these upper bounding terms gives the upper bound $\sqrt{3/8} + \sqrt{2} + \frac{1}{2-2^{3/4}} = 5.169{\dots}$.


问:
Basis of neighbourhoods on cofinite topology
Let $(E,\tau)$ be a topological space where $\tau=\{ \emptyset \}\cup \{G\subset E, |E\setminus G|<\infty\}$

How I can write $W_x$ the basis of neighbourhoods  of $x \in E$?

Thank you

答: If $\mathcal{T}$ is a topology on a set $X$, the set $\mathcal{O}_x = \{O \in \mathcal{T}\mid x \in O\}$ is always a base for the topology at $x$, by definition. Of course, smaller bases at $x$ also exist, in general, but this is one you can always use.


问:
Problem on motion in a straight line
Question: A particle starts from rest at a distance b(>a) from a fixed point O under the action of a force through a fixed point, the law of which at a distance x from O is
$F= \mu (1-a/x)$ towards O when $x>a$ but
$F= \mu(a^2/x^2 - a/x)$ away from O when $x<a$.
Show that the velocity of the particle will again vanish when $x= a^2/b$

I've taken the differential equations
$v dv/dx = -\mu(1-a/x)$, limit $v:0 \rightarrow V$, $x:b \rightarrow x$ and
$v dv/dx=\mu(a^2/x^2 - a/x)$, limit $v:V \rightarrow 0$, $x:a \rightarrow x$
After that I'm stuck.

答:
It would be preferable to compare the work done on the particle along each path to $ \ x = a \ . $  For $ \ x > a \ , $ the force acts "to the left", so the change in kinetic energy of the particle is $$ \ -\int_b^a \mu·(1  - \frac{a}{x}) \ dx \ \ = \ \  \mu \ · \ (a \ln x \ - \ x)|_b^a \ \ . $$  When $ \ x < a \ , $ the force acts "to the right", so the change in KE along this path is $$ \ \int_X^a \mu·(\frac{a^2}{x^2}  - \frac{a}{x}) \ dx \ \ = \ \  \mu · \left( -\frac{a^2}{x}  - a \ln x \right)|_X^a \ \ , $$ where $ \ X < a \ $ is the starting point.  Since the particle is (momentarily) at rest at each end, these work integrals will be equal.

While you could solve the equation for $ \ X \ , $ you are just asked to "verify" that setting $ \ X \ = \ \frac{a^2}{b} \ $ makes the two amounts of work equal.

Title: Problem on motion in a straight line


问:
Number of +ve Integer Solutions for an Equation with Inequality condition
How many different pairs $(a,b)$ of $+ve$ integers are there such that $a \leqq b$ and $\frac{1}{a}+\frac{1}{b} =\frac{1}{9} ?$

I approached this problem by simplifying the equation in term of $b$ which is $b=\frac{9a}{a-9}$ and then substituting the value of $b$ in $a \leqq b$ and I got $a \leqq 18$ and similarly doing for $b$, I got $b \geqq 18$.

So from here I can say that one such pair is $(18,18)$ and the rest I found out by manually putting the value of $a$ less than $18$ till $1$ as $a \in [1,18]$ in the equation $\frac{1}{a}+\frac{1}{b} =\frac{1}{9} $ and finding $b$ and by doing so I found two another pairs i.e. $(12,36)$ and $(10,90)$.

But this approach is very tedious and takes a lot of time by putting the value in the equation and then finding the value of the another variable which satisfies the condition. Can someone please explain to me a better way to solve this question so that I do not have to find the pairs by manual putting and finding the value method?

Thanks in Advance!

答:
Your equation is equivalent to $$(a-9)(b-9)=81=3^4,$$ and $a\le b$ implies $a-9=\sqrt{(a-9)(a-9)}\le\sqrt{(a-9)(b-9)}=\sqrt{81}=9$. So $a-9$ must be in $\{1,3^1,3^2\}$ (all divisors of $3^4$ not greater than $9$).

Obviously, we don't have to consider products of two negative divisors of $81$, since one of them would have to be $\le-9$, and thus $a$ or $b$ not positive.


答:
It can be shown that
\begin{align*}
\sum\limits_{n = 1}^\infty  {\frac{1}{{n^s }}} & = \prod\limits_p {\frac{1}{{1 - \frac{1}{{p^s }}}}}  = \exp \left( { - \sum\limits_p {\log \left( {1 - \frac{1}{{p^s }}} \right)} } \right) \\ &= \exp \left( {\sum\limits_p {\sum\limits_{n = 1}^\infty  {\frac{1}{{np^{ns} }}} } } \right)=
\exp \left( {\sum\limits_{n = 1}^\infty \frac{1}{n} {\sum\limits_p {\frac{1}{{p^{ns} }}} } } \right),
\end{align*}
whenever $\Re s>1$. Taking the limit $s\to 1+$ yields Euler's formal identity. I say formal, because the expressions on both sides will diverge. Note that I am not making an attempt to describe this in terms of non-university level mathematics.

Title: Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true?

Body:
Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.

$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$

Where he defines:

$$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$

Question: I can't see how that statement is true.

I would appreciate replies that don't assume university level training in mathematics.


答:
The question has already been answered in the straightforward way, but I just wanted to point out another perspective. There is a way to view this as a linear map in some sense. It is an affine map, which can be represented as a matrix by embedding it in $\mathbb{R}^3$ as follows.
$$ \begin{pmatrix} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ x+1 \\ y+1 \end{pmatrix} $$
Thinking about the eigenvectors and eigenvalues of this matrix may be another instructive perspective on the original problem.

Title: Can a non-linear transformation, specifically a translation, have eigenvectors?

Body:
In trying to grasp the basics of linear algebra, I've encountered the following question:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the transformation
defined by $(x,y) \mapsto (x+1,y+1)$. Does $f$ have any eigenvectors?
Explain. What type of transformation is $f$?

I know that $f$ is a translation, which is a non-linear transformation. Does this mean that it has no eigenvectors? Do eigenvectors only apply to linear transformations?

Or can you actually have eigenvectors for non-linear transformations? If so, would all vectors be eigenvectors for this transformation? The transformation causes points to be translated 1 unit right and 1 unit up, but don't vectors stay the same? For example, the vector $\begin{bmatrix}3\\2\end{bmatrix}$ remains $\begin{bmatrix}3\\2\end{bmatrix}$. Before the transformation, it has initial point $(0,0)$ and terminal point $(3,2)$; after the transformation, it has initial point $(1,1)$ and terminal point $(4,3)$.

Hope you'll answer the quoted question, and also let me know if I've misunderstood anything. Thanks!


答:
As a polynomial of degree $3$ $p(x) = a x^3+b x^2 + c x +d$ has 4 coefficients, this may be sufficient providing that the equations induced by the conditions are compatible. Namely, that would be the case if the determinant

$$\begin{vmatrix}
0  & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
1 & 1  & 1 & 1\\
3 & 2 & 1 & 0
\end{vmatrix}$$ doesn't vanishes. As this determinant is equal to one, a polynomial of degree $3$ is solution and answer (a) is the right one.

Body:
I am preparing for GRE MATH SUBJECT TEST, I have reviewed many problems, some of these problems are not in GRE Math Practice books, but they are in some other books. I found the following problem, but not sure if it can be a GRE problem or no. And I do not even know how to start;

Let $P$ be a polynomial with integer coefficients satisfying $P(0)=1, P(1)=3, P'(0)=-1, P'(1)=10$. What is the minimum possible degree of $P$?

(A) $3$

(B) $4$

(C) $5$

(D) $6$

(E) No such $P$ exists.


问:
prerequisites for understanding game theory
I am from programming background but with very limited knowledge of maths.

I am very much eager to learn and apply game theory to understand dynamics of International Politics and economics. But I am facing difficulties to understand the maths involved in it.

So,

What are prerequisites for understanding the math in the game theory?

What are the authentic sources to understand game theory mathematically and
and its application in the international politics and economics?

答:
In economics, the classic source for game theory is:

Fudenberg, D. and Tirole, J. (1991). Game Theory

At least with Game Theory applied to economics, you can begin with a minimal knowledge of mathematics (applied calculus and some set theory will usually suffice).


问:
Can I define such an average norm of an infinite-dimensional vector?
Suppoes that there is an infinite-dimensional vector $x = \{x_1,x_2,\dots,x_{\infty}\}$. Can I define such a norm
$$
\|x\| =  \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^{N} x_k,
$$
which implies that the limit must exist.

I believe that it satisfies the definition of a norm, but I have never seen it in any contexts before. Is this norm properly defined?

答:
If $x_k=k$ for all $k$, then

$$\frac{1}{N} \sum_{k=1}^{N} x_k= \frac{N+1}{2},$$

so $\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^{N} x_k$ does not exist.


问:
Is it true that $\pi = \lim_{x \to \infty} 10^{x+2}\sin (1.8\times 10^{-x})$? (in degrees)
When I was doing experiments, I found this shocking results(!). Am I hallucinating? Why I found that the decimal digit expansions of constant $\pi$ is NOTHING MORE than the decimal digit expansions of a sine function.

I found that
$$\pi = \lim_{x \to \infty} (10^{x+2}\sin (1.8\times 10^{-x})) \quad\text{(in degrees, not radians)}$$

Is this true?

答:
In radians your formula reads:
$$
\lim_{x\to\infty}\frac{\sin{(\pi\cdot 10^{-x-2})}}{10^{-x-2}}=\pi
$$
by the well-known limit:
$$
\lim_{x\to0}\frac{\sin{(ax)}}{x}=a.
$$

OwnerDisplayName: Antonius Sugianto


答:
As @JordanMahar mentions, Fudenberg and Tirole is the standard graduate-level text. But I would start with Game Theory for Applied Economists by Gibbons. It is very readable.

Prerequisites for Gibbons are minimal. A little algebra and probability will do just fine.

Title: prerequisites for understanding game theory

Body:
I am from programming background but with very limited knowledge of maths.

I am very much eager to learn and apply game theory to understand dynamics of International Politics and economics. But I am facing difficulties to understand the maths involved in it.

So,

What are prerequisites for understanding the math in the game theory?

What are the authentic sources to understand game theory mathematically and
and its application in the international politics and economics?


问:
Which estimator is better?
$X=\left(X_{1},...,X_{n}\right), Y=\left(Y_{1},...,Y_{n}\right)$ - independent sample from distributions $\mathcal{N}\left(m_{x},\sigma^{2}\right)$ and $\mathcal{N}\left(m_{y},\sigma^{2}\right)$. Which of these two estimators of $m_xm_y$:
$$T_{1}\left(X,Y\right) = \overline{X}\overline{Y}\\
T_{2}\left(X,Y\right) = \frac{1}{n}\sum_{i=1}^{n}X_{i}Y_{i}$$
is better when we apply quadratic loss function? Can I ask for a hint?

答:
Both estimators are unbiased thus, according to the Classical Statistical Theory,  you can evaluate them comparing their variance.

After some calculations I get

$$\mathbb{V}[T_1]=\frac{\sigma^2}{n}\left[\frac{\sigma^2}{n}+m_X^2+m_Y^2\right]$$

$$\mathbb{V}[T_2]=\frac{\sigma^2}{n}[\sigma^2+m_X^2+m_Y^2]$$

thus I prefer $T_1$ as its variance is always lower that the one of $T_2$ for any $n>1$

To compare two estimators in Classical Statistical Theory, the one having the lower MSE is preferred.

By definition,

$$MSE_{T_1}=\mathbb{E}[T_1-\mathbb{E}[T_1]]^2$$

In your statement you assumed immediately that $\mathbb{E}[T_1]=\mathbb{E}[T_2]=m_X\cdot m_Y$

... which is correct but it has to be verified first...

$$\mathbb{V}[T_1]=\mathbb{V}[\overline{X}\overline{Y}]=\mathbb{E}\left[(\overline{X})^2\right]\cdot \mathbb{E}\left[(\overline{Y})^2\right]-m_X^2\cdot m_Y^2$$

now observe that $\overline{X}\sim N(m_X ;\sigma^2/n)$

thus

$$\mathbb{V}[T_1]=\left(\frac{\sigma^2}{n}+m_X^2\right)\left(\frac{\sigma^2}{n}+m_Y^2\right)-m_X^2\cdot m_Y^2$$

Title: Which estimator is better?


问:
Solving for the volume of a tetrahedron
Can anyone explain why I was incorrect in this problem:

I used a tetrahedron solver online, http://rechneronline.de/pi/tetrahedron.php, and it says I have the right answer.

答: Assuming that $S$ (or is it $S'$?) indicates side length and not surface area, you're correct.</p>

Title: Solving for the volume of a tetrahedron


问:
Given $\lim \limits_{x \to c}f(g(x)) = f(\lim \limits_{x \to c}g(x))$ provided what 2 conditions are met?
The problem that I am having trouble with is $\lim \limits_{x \to c}f(g(x)) = f(\lim \limits_{x \to c}g(x))$. It seems like the limit of function $g$ is wrapped inside of function $f$. I need to provide two conditions that make this statement true. I am guessing that I have to use two of the three rules from "How to determine whether or not a function is continuous" but I am not sure. However, I know that $f(c)$ is not defined.

答:
Let $A,B, C$ be non empty subsets of $\mathbb R$ and $g:A\to B, f:g(A)\to C$, be two functions, where $g(A)=\{g(x):x\in A\}$. Clearly, $g(A)\subseteq B$.

With this condition $fog:A\to C$ is defined.

Now let $c\in \mathbb R$ be a limit point of $A$. If

$\lim_{x\to c}g(x)$ exists.

and

$f$ is continuous at $\lim_{x\to c}g(x)$.

Then, $\lim_{x\to c}f(g(x))=f(\lim_{x\to c}g(x))$


答: Assuming that $S$ denotes the length of a side, your answer is correct.  See here.</p>

Title: Solving for the volume of a tetrahedron

Body:
Can anyone explain why I was incorrect in this problem:

I used a tetrahedron solver online, http://rechneronline.de/pi/tetrahedron.php, and it says I have the right answer.


问:
Proof outline of a construction of injective function
Let $A$ be nonempty s.t. there exists $f:A\to A$ injective but not subjective. Show there is a $g:\mathbb{N} \to A$ that is injective.

Below is my proof, is this a correct approach? Anything I'm missing?

答:
No, it is not correct. Why do you say that if $a_3=a_1$ then $f$ is surjective? If, say $f\colon\Bbb Z\longrightarrow\Bbb Z$ is defined by$$f(n)=\begin{cases}-n&\text{ if }n\in\{-1,0,1\}\\n+1&\text{ if }n>1\\-n-1&\text{ if }n<-1,\end{cases}$$then $f$ is injective and not surjective ($\pm2\notin f(\Bbb Z)$). But, if you define $(a_n)_{n\in\Bbb N}$ as you did with $a_1=1$, then $a_3=a_1$.

You can do it as follows. Let $A_0=A$ and let $A_1=f(A_0)=f(A)$. Since $f$ is not surjective, $A_1\ne A_0$. So, $A_1\varsubsetneq A_0$. Now, let $A_2=f(A_1)$. Since $A_1\subset A_0$,$$A_2=f(A_1)\subset f(A_0)=A_1.$$In fact, $A_2\varsubsetneq A_1$. In order to see why, take $y\in A_0\setminus A_1$; clearly, $f(y)\in A_1$, and I will prove that $f(y)\notin A_2$. If $f(y)\in A_2$, then $f(y)=f\bigl(f(x)\bigr)$, for some $x\in A$. But, since $f$ is injective,$$f(y)=f\bigl(f(x)\bigr)\implies y=f(x),$$which is impossible, since $y\notin A_1=f(A)$. By the same argument, if $A_3=f(A_2)$, then $A_3\varsubsetneq A_2$ and so on. Now, for any $n\in\Bbb N$, let $g(n)$ be an element of $A_{n-1}\setminus A_n$ and then $g$ is an injective map from $\Bbb N$ into $A$.


问:
Vector trapezoid

In the trapezoid $ABCD$ the sides $AB$ and $CD$ are parallel. In what ratio do the
  diagonals cut each other if $|AB|=3|CD|$ ?

Not too sure how to approach this question. So far I have attempted to find the intersection of the diagonals and called it point $M$.

Hence, I have gotten vector $\vec{AM} =\vec{AC} + t\cdot\vec{CB}$ where $t$ is a real value for example.
vector $\vec{AM}$ however can be rewritten as $s\cdot\vec{AD}$ where $s$ is a real value and this is equivalent to $s\cdot (\vec{AC} + \vec{CD})$

Not too sure what to do now, any ideas??

答:
Let the diagonals' intersection point be $X.$

Triangle $XAB$ is similar to triangle $XCD.$ So

 $$\frac{XA}{XC}=\frac{AB}{CD}=3\\ XA:XC:3:1$$ and $$\frac{XB}{XD}=\frac{AB}{CD}=3\\ XB:XD:3:1.$$

However, if a vector method is required:

Let the diagonals' intersection point be $X,$ and $$\vec{DC}=\mathbf{p},\\
\vec{DA}=\mathbf{q},\\
\vec{DX}=\lambda\,\vec{DB},\\
\vec{AX}=\mu\,\vec{AC}.$$

Now we form an equation relating $\lambda, \mu, \mathbf{p},$ and $\mathbf{q}:$

 $$\vec{DX}=\vec{DA}+\vec{AX}\\ \lambda\,\vec{DB}=\mathbf{q}+\mu\,\vec{AC}\\ \lambda(\mathbf{q}+3\mathbf{p})=\mathbf{q}+\mu(\mathbf{p}-\mathbf{q})\\ (3\lambda-\mu)\mathbf{p}+(\lambda+\mu-1)\mathbf{q}=\mathbf{0}.$$

Since $\mathbf{p}$ and $\mathbf{q}$ are not collinear, the coefficients of both vectors are $0.$

 Therefore, $$\lambda=\frac14 \\ DX=\frac14 DB \\ DX:XB=1:3$$ and $$\mu=\frac34 \\ AX=\frac34 AC \\ AX:XC=3:1.$$

Title: Vector trapezoid


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
Peek-a-boo mentioned in his answer "$\delta(0)$  has to equal $\infty$ in just the right way", and it can be illuminating to see what can go wrong. Lets look at a different function which also "converges pointwise" to $\delta$:

$$ G_\epsilon(x) = \begin{cases} \epsilon & x = 0 \\ 0 & x \neq 0\end{cases}$$

Now $\lim_{\epsilon \to \infty} G_\epsilon(0) = \infty$ so we would expect that this is the Dirac delta function. But if we integrate over $G_\epsilon$, it behaves very differently:

$$\int f(x) G_\epsilon(x) dx = 0$$

Which can be seen by using the Riemann definition and never using $0$ as an evaluation point:

$$\int f(x) G_\epsilon(x) dx = \sum f(x_i) (x_i-x_{i+1}) = \sum f(x_i) (0-0) = 0$$

(alternatively it is a well known theorem from Lebesgue integration that changing a single point cannot change the value of the integral)

So defining the Delta distribution simply via stating that it is infinite at 0 does not work, as having two different functions ($F_\epsilon$ and $G_\epsilon$) "converging pointwise" to this definition leads to different values when integrating over them. It matters "how you converge to $\infty$".

Why do we then even care analysing $F_\epsilon$, if defining distributions that way is tricky (compared to the "direct way" of defining a distribution as a function from functions to a number)? Functions are dense in their distributions, which means that every distribution can be approximated by such a series of functions. Thus it is very natural to look at approximations of distributions.


问:
Isaac's character theory of finite groups, Theorem 6.9.
I am trying to understand Theorem 6.9 in Isaac's Character Theory of finite groups:

Suppose $\chi(1)$ is a power of the prime $p$ for all irreducible characters $\chi$ of $G$. Then $G$ has a normal abelian $p$-complement. According to the proof, proceeding by induction on $|G| = p^km$, it suffices to show that $G$ has a normal subgroup $N$ of index $p$. The stated reasoning is that $N$ will have a normal abelian $p$-complement by the inductive hypothesis, and this same subgroup will be a normal abelian $p$-complement for $G$. However, I can't seem to prove that it is indeed normal in $G$.

Any ideas?

答: If $A$ is a normal abelian $p$-complement of $N$, then gcd$(|A|,|N:A|)=1$, implying that $A$ is characteristic in $N$. So $A$ char $N \lhd G$, hence $A \lhd G$. And $|G:A|$ is a $p$-power, $p \nmid |A|$, so at this point you could apply the theorem of Schur-Zassenhaus, providing a complement in $G$ to $A$. But you do not have to invoke that heavy theorem: $G/A$ is a $p$-group, so if $S \in Syl_p(G)$, then $SA/A=G/A$. Hence $G=SA$ and of course $S \cap A=1$.


问:
Two coins with heads probabilities 1/3 and 1/4 are alternately tossed, starting with the 1/3 coin, until one of them turns up heads.
Two coins with heads probabilities  1/3  and  1/4  are alternately tossed, starting with the  1/3  coin, until one of them turns up heads. Let  X  denote the total number of tosses, including the last. Find:

E(X)

I found that using 1/p the E(X) of the first coin will be approx 3 and the E(X) of the second coin will be approx 4, but since these coins are tossed alternately I don't understand what the E(X) could be.

答:
You need to solve the following equation.

$\mathbb{E}(X) = \displaystyle 1 + \frac{2}{3} (1 + \frac{3}{4} \mathbb{E}(X))$

Explanation: Say the initial state is $I$. You get $H$ with probability $\frac{1}{3}$ in the first toss or continue with probability $\frac{2}{3}$ to the second coin. We get $H$ on the second coin with probability $\frac{1}{4}$ or we get back to initial state $I$ with probability $\frac{3}{4}$.


问:
a trace class operator problem
Could someone help me with this

Prove that If $A$ and $B$ are positive trace class operators on a Hilbert space, then so is $A^zB^{(1-z)}$ for a complex number $z$ such that $0 <Re(z)< 1$.

An operator $A$ is called trace class if
$\sum_{n=1}^{\infty}\langle |A|e_n,e_n\rangle < \infty  $

|A| is positive square root of $ A^*A$

答:
Write $z=a+ib$. Then $A^zB^{1-z}=A^aA^{ib}B^{1-a}B^{-ib}=A^{ib}A^aB^{1-a}B^{-ib}$. Since the trace-class operators form an ideal, it is enough to show that $A^aB^{1-a}$ is trace-class. Even more, it is enough--by the polar decomposition--to show that $|A^aB^{1-a}|$ is trace class.

(julien has suggested a much shorter and efficient way to the answer; so I'm following his suggestion here, and leaving the rest of my original answer towards the end)

Now we can use Hölder's inequality $\|ST\|_1\leq\|S\|_p\,\|T\|_q$, which holds independently of whether $ST$ is trace-class or not. So, with $p=1/a$, $q=1/(1-a)$,
$$
\|A^aB^{1-a}\|_1\leq\|A^a\|_{1/a}\,\|B^{1-a}\|_{1/(1-a)}=\|A\|_1^a\,\|B\|_1^{1-a}<\infty.
$$
Thus, $|A^aB^{1-a}|$ is trace-class.

(here is the rest of my original answer)

To this goal, we can use Young's inequality for compact operators (proven in a paper by Erlijman-Farenick-Zeng, from around 2001): there exists a partial isometry $U$ with initial space $(\ker|A^aB^{1-a}|)^\perp$ such that
$$\tag{1}
U|A^aB^{1-a}|U^*\leq \frac1p\,|A^a|^p+\frac1q\,|B^{1-a}|^q=aA+(1-a)B
$$
(where we are using $p=1/a$, $q=1/(1-a)$; this is where we use the requirement $0<a<1$). So $U|A^aB^{1-a}|U^*$ is trace class. As $|A^aB^{1-a}|U^*U=|A^aB^{1-a}|$, we get
$$
\mbox{Tr}(|A^aB^{1-a}|)=\mbox{Tr}(|A^aB^{1-a}|U^*U)=\mbox{Tr}(U|A^aB^{1-a}|U^*)<\infty.
$$
This shows that $|A^aB^{1-a}|$ is trace class, since it is positive.

Title: a trace class operator problem


问:
Estimate of the L2 projection error
https://pdfs.semanticscholar.org/097d/1f110ae8d39e67545bffca0e7835e4a2679d.pdf

This paper (3.5) says, for any $\nu\in[0,1]$, the $L^2$ orthogonal projection operator $\pi_h$ has the stability: $\forall \psi \in H^{\nu}(\Gamma_C)$,

$$
\left\|\pi_{h} \psi\right\|_{H^{\nu}\left(\Gamma_{C}\right)} \leq C\|\psi\|_{H^{\nu}\left(\Gamma_{C}\right)}
$$

It also says the following approximation result holds: for any $\nu\in[0,1]$ and $\mu\in[0,\nu+\frac12]$ we have, $\forall \psi\in H^{\frac12+\nu}(\Gamma_C)$,

$$
h^{-\frac{1}{2}}\left\|\psi-\pi_{h} \psi\right\|_{H^{-\frac{1}{2}}\left(\Gamma_{C}\right)}+h^{\mu}\left\|\psi-\pi_{h} \psi\right\|_{H^{\mu}\left(\Gamma_{C}\right)} \leq C h^{\nu+\frac{1}{2}}\|\psi\|_{H^{\frac{1}{2}+\nu}\left(\Gamma_{C}\right)}$$

It mentions that the estimate in $H^\nu(\Gamma_C)$ is a direct consequence of the stability result and the boundedness of the error w.r.t the $H^{-\frac12}(\Gamma_C)$-norm is obtained by duality.

Could somebody explain how we get the error estimate result from the stability?

答:
I'm assume that $\pi_h$ is the $L^2$-projection onto $\mathcal{P}_p$, the set of piecewise polynomials of degree $p$ over $\Gamma_C$, and that $\Gamma_C$ is some shape-regular union of simplices.

The result follows from scaling and the Bramble--Hilbert Lemma, but the main trick is to realise that since $\pi_h(q)=q$ for any $q\in \mathcal{P}_p$, and so using the stability estimate,
$$
\|\psi-\pi_h\psi\|_{H^\mu(\Gamma_C)}=\|(\psi-q)-(\pi_h\psi-q)\|_{H^\mu(\Gamma_C)}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\|(\psi-q)-\pi_h(\psi-q)\|_{H^\mu(\Gamma_C)}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\le\|\psi-q\|_{H^\mu(\Gamma_C)}+\|\pi_h(\psi-q)\|_{H^\mu(\Gamma_C)}
$$
$$~~~~~~~~~~~~~~~~\le(1+C)\|\psi-q\|_{H^\mu(\Gamma_C)}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~\le(1+C)\inf_{q\in\mathcal{P}_p}\|\psi-q\|_{H^\mu(\Gamma_C)},$$
where the inf holds due to the fact that we can pick any $q\in \mathcal{P}_p$. Now we can apply the Bramble--Hilbert Lemma and scaling on each simplex to get the error estimate. Essentially the stability of the projection in the $H^\mu$-norm implies that the $L^2$ projection is (up to a constant) as accurate as the best approximation in the $H^\mu$-norm.

Title: Estimate of the L2 projection error


答:
I'm not sure why it hasn't yet been pointed out that your definition of $δ$ is wrong. You certainly cannot defined it by pointwise limits as you attempted to do. Otherwise you will indeed get $δ(0) = ∞$ but it will also be completely useless and not the desired dirac-delta (e.g. you would get $2·δ(0) = ∞$ as well).

Here is another way to think of $δ$. Instead of taking limits 'prematurely', you simply refrain from taking the limit for $δ$ at all. That is, you treat $δ$ as a suitable sequence of functions, such as $(F_{1/n})_{n∈ℕ^+}$, meaning $δ$ is literally nothing more than the sequence $(F_1,F_{1/2},F_{1/3},\cdots)$. In doing so, you can then rigorously manipulate expressions involving $δ$. We first define arithmetic and integrals to apply point-wise on a sequence. For example, for sequences $p,q$ we define $p·q$ to be the point-wise product of $p$ and $q$, namely the sequence $(p_0·q_0,p_1·q_1,p_2·q_2,\cdots)$. And given sequence $r(x)$ for every $x∈ℝ$, we define $\int_ℝ r(x)\ dx$ to be $\big( \int_ℝ r(x)_0\ dx, \int_ℝ r(x)_1\ dx, \int_ℝ r(x)_2\ dx, \cdots \big)$.

Now look. For any continuous $g : ℝ→ℝ$, we have that $\int_ℝ g(x)·δ(x)\ dx$ is also a sequence, namely $\big(\int_ℝ g(x)·F_1(x)\ dx, \int_ℝ g(x)·F_{1/2}(x)\ dx, \int_ℝ g(x)·F_{1/3}(x)\ dx, \cdots \big)$ and we can easily show that this sequence tends to $g(0)$. Your $F$ works for this, but if you want further nice properties you may need smoother ones than the ones you chose (e.g. infinitely differentiable).

~ ~ ~ ~ ~ ~ ~

[Edit: Here is my response to the later edit that quotes from a text.]

The text's definition of $δ$ is indeed completely wrong. As many people here (not just myself) have already pointed out, if you define $δ(x) = \lim_{a→0} δ_a(x)$ then definitely $δ(x) = 0$ everywhere except when $x=0$, and $δ(0) = ∞$ (using the affinely-extended reals), and definitely $\color{blue}{\int_{-∞}^∞ δ(x)\ dx = 0}$ (using the Lebesgue integral), NOT $1$, contrary to that text's erroneous claim. Note that if you do not use the affinely-extended reals, then the definition of $δ(0)$ is simply meaningless, and the whole wrong business stops there. And if you do not use the Lebesgue integral or something capable of handling isolated infinities, then you cannot even integrate $δ$. But there is no meaningful way that you can ever get $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$ from your text's definition, and here is why:

If $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$, then $\int_{-∞}^∞ δ(2x)\ dx = \int_{-∞}^∞ δ(x)\ dx = 1$ because $δ(2x) = δ(x)$ for every $x∈ℝ$ (by your text's definition of $δ$). But also $\int_{-∞}^∞ δ(2x)\ dx = \frac12 · \int_{-∞}^∞ δ(x)\ dx = \frac12$ by a simple stretch, so contradiction. This shows that if you use your text's definition of $δ$, then you do not have the desired properties for $δ$ at all.

~ ~ ~ ~ ~ ~ ~

If, in contrast, you use the sequence-based definition I gave you, you get no such nonsense. In this section I will again use my definition. See, $\int_{-∞}^∞ δ(2x)\ dx$ is truly equal with $\frac12 · \int_{-∞}^∞ δ(x)\ dx$, because each of them is a sequence and they have identical terms! The latter is $(\frac12,\frac12,\frac12,\cdots)$, which makes complete sense.

Furthermore, if you choose your $F$ to be a sequence built from a function that is infinitely differentiable and nonzero only on a finite interval, such as a bump function (as briefly mentioned here), then you get further nice properties. For instance, $δ'$ (i.e. the term-wise derivative of $δ$) would be well-defined, and for any $g : ℝ→ℝ$ with continuous derivative and any $y∈ℝ$ we have:

  $\int_ℝ δ'(x)·g(y-x)\ dx$

  $ = \big[ δ(x)·g(y-x) \big]_{x=-∞}^∞ + \int_ℝ δ(x)·g'(y-x)\ dx$

  which converges to $g'(y)$.

To explain the last line, consider a suitable example for $δ$: Let $δ_0$ be an infinitely-differentiable function on $ℝ$ that is nonzero on $[-1,1]$ such that $\int_ℝ δ_0(x)\ dx = 1$. For each $k∈ℕ^+$ let $δ_k : ℝ→ℝ$ defined by $δ_k(x) = δ_0(x·2^k)·2^k$ for every $x∈ℝ$, so we have $δ_k$ is nonzero on $[-1/2^k,1/2^k]$ and $\int_ℝ δ_k(x)\ dx = 1$. Intuitively, $δ_k$ is nonzero on a shrinking interval as $k→∞$, but its integral over $ℝ$ remains exactly one.

Then $\big[ δ(x)·g(y-x) \big]_{x=-∞}^∞$ is a constant $0$ sequence because $\big[ δ_k(x)·g(y-x) \big]_{x=-∞}^∞ = 0$ for each $k∈ℕ$. And $\int_ℝ δ(x)·g'(y-x)\ dx$ is a sequence that converges to $g'(y)$ because:

  $\int_ℝ δ_k(x)·g'(y-x)\ dx$

  $= \int_{-1/2^k}^{1/2^k} δ_k(x)·g'(y-x)\ dx$

  $= \int_{-1}^1 δ_k(x/2^k)/2^k·g'(y-x/2^k)\ dx$

  $= \int_{-1}^1 δ_0(x)·g'(y-x/2^k)\ dx$

  $∈ \int_{-1}^1 δ_0(x)·[m_k,M_k]\ dx = [m_k,M_k]$

  where $m_k,M_k$ are the min,max of $g'(y-x/2^k)$ over all $x∈[1,1]$.

The result follows since continuity of $g'$ implies that $m_k,M_k → g'(y)$ as $k→∞$.

Title: Is it meaningful to define the Dirac delta function as infinity at zero?

Body:
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$


问:
Show that $a^2 - 2021b^2 = 13$ has no integer solutions

Show that $a^2 - 2021b^2 = 13$ has no integer solutions.

I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A square is $0$ or $1 \bmod 4$, so we get $a^2\equiv 1\bmod 4$ and $b^2 \equiv 0 \bmod 4$. Thus, $2|b$ and $a^2 = 4k+1$. I do not see a contradiction yet. What else should I try?

Also tried to use the fact that squares have units digit only $0,1,4,5,6,9$, so if $a,b$ exist that satisfy the equation, we must have $a^2$ ending in $\{0,1,4,5,6,9\} \cap \{3,4,7,8,9,2\} = \{4,9\}$. Moreover, we must have $b^2$ ending in $\{0,1,4,5,6,9\} \cap \{7, 8, 1,2,3,6\} = \{1,6\}$.

If $b^2$ ends in $1$, then $a^2$ ends in $4$ - but we just saw that $a^2 = 4k+1$, so this is not a possibility.

If $b^2$ ends in $6$, then $a^2$ ends in $9$. How do I contradict this? Thank you!

答: $a^2=13+2021b^2 \Rightarrow a^2 \equiv 6b^2 \mod 13 \iff \Big(\frac{a}{b}\Big)^2 \equiv 6 \mod 13 $ for $b\neq0$. But as $\genfrac(){}{}{6}{13}=-1$ then $6$ is not a quadratic residue so the equation does not have solutions unless $a \equiv b \mod 13$ and $a \equiv 0 \mod 13$. But even in this case, $13^2$ divides the LHS but does not divide the RHS.


问:
Convergence of eigenvalues and spaces of sequence of compact, szmmetric and positive-semi-definite operators in Hilbert spaces.
I have a compact operator, which is uniformly approximated by (finite-dimensional) compact operators and I am concerned with the question, how the eigenvalues and eigenspaces of the former are approximated by the ones of the latter. Specifically, let $K_m,K:X\to X$ be compact symmetric and positive semi-definite with $X$ a separable Hilbert-space and $||K_m-K||\to 0$ as $m\to\infty$.
Let $\lambda_1\ge\lambda_2\ge...$ and $\lambda_1^m\ge \lambda_2^m\ge ...\ge \lambda_m^m$ be the eigenvalues of $K$ and $K_m$ and $v_1,v_2,...$ and $v_1^m,...,v_m^m$ corresponding Eigenvectors, such that $(v_j)$ form a basis of $\mathcal{N}(K)^\perp$ (the orthogonal komplement of the kernel of $K$) and $(v_j^m)$ forms a (finite) basis of $\mathcal{N}(K_m)^\perp$.
Define iteratively $I_1:=[k~:\lambda_k=\lambda_1]$ and $I_j:= [ k~:~\lambda_k=\lambda_{\max(I_{j-1})+1}]$ for $j\ge 2$. Define the subspaces $E_i:=span(v_k~:~k\in I_i)$ and $E_i^m:=span(v_k^m~:~k\in I_i)$ (thus $E_i$ is the space spanned by the eigenvectors of the i-th largest eigenvector of K) and $E_i^m$ is only defined for $m$ sufficiently large. Denote by $P_{E_i}$ and $P_{E_i^m}$ the orthogonal projections onto the subspaces $E_i$ and $E_i^m$. I cannot find reference for the following statements, which I belive should be true:

1: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}\lambda_i^m = \lambda_i$.

2: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}||P_{E_i}-P_{E_i}^m||=0$.

May someone provide a proof or a counter example for the statements?

Does someone have a proof at least for the finite-dimensional case?

答:
Thank you for the answer! I am still not sure, if I am able to use the statement to show the claim for compact operators. I'll try first the finite-dimensional case and assume for simplicity that the multiplicity of each eigenvalue is one ($\lambda_1>\lambda_2>\lambda_3>...$). I prove $\lambda_i^m\to \lambda$ as $m\to\infty$ via induction over $i\in\mathbb{N}$. For $i=1$ this holds, since $\lambda_1^m=||K_m||\to ||K|| = \lambda_1$ as $m\to\infty$. Now assume that $\lambda_j^m\to\lambda_j$ as $m\to\infty$ for all $j\le i-1$. Let $\varepsilon<\lambda_i/2$ be arbitrary. For $m$ large enough, there holds $|\lambda_j - \lambda_j^m|<\varepsilon$ for all $j\le i-1$. Moreover, by your statement, for $m$ sufficiently large there exists an eigenvalue $\mu^m$ of $K_m$ such that $|\mu^m-\lambda_i^m|<\varepsilon$ and $\mu^m$ is the only eigenvalue of $K_m$ which has this property. It remains to show, that for $m$ large enough there holds $\mu^m=\lambda_i^m$. We argue by contradiction and assume, that for infinitely many $m\in\mathbb{N}$ there are $j_m\neq i$ with $\mu^m=\lambda_{j_m}^m$. Clearly, we then must have $j_m<i$ because of the induction hypothesis and the choice of $\varepsilon$. Then it follows, that for such $m$ we have $\lambda_i^m<\lambda_{i-1}-\varepsilon$ and $\lambda_i^m>\lambda_i+\varepsilon$. W.l.o.g. (restrict to a subsequence of the $\lambda_i^m$ if necessary) there is a $\lambda$ with $\lambda_i+\varepsilon <\lambda<\lambda_{i-1}-\varepsilon$ and $\lambda_i^m\to\lambda$. Let $v_i^m$ be such that $K_m v_i^m = \lambda_i^m v_i^m$. W.l.o.g., there is a $v$ with $v_i^m\to v$ as $m\to\infty$ (because of finite dimension). Finally,

$$||Kv - K_m v_i^m|| \le || (K-K_m)v|| + ||K_m(v-v_i^m)|| \to 0$$

as $m\to\infty$, thus $Kv=\lambda v$. This is a contradiction, since $\lambda_i+\varepsilon<\lambda<\lambda_{i-1}+\varepsilon$ implies that $\lambda$ is not an eigenvector of $K$. Thus it holds that $\mu^m=\lambda_i^m$ for $m$ sufficiently large.

Is the above reasoning correct? If yes, how to handle the infinite-dimensional case? The proof feels awkward to me...


问:
Number of roots of the equation $\sum_{i=1}^{2020}\frac{i^2}{x-i}=0$
What is the number of roots of the equation $\sum_{i=1}^{2020}\frac{i^2}{x-i}=0$?

I suspect that the equation has a single root. This is my guess based on the observation that we could choose some integer $x$ from $1$ to $2020$ so that the positive part of sum cancels with the negative part of the sum. But, the problem, is we are allowed to choose any real values for $x$. The function is not a polynomial. So, how should I proceed? Is intermediate value theorem helpful here? Any hints? Thanks beforehand.

答:
Keep in mind that $f$ has vertical asymptotes at $x = 1,2,3, \cdots, 2020$ and that at each interval $[i, i+1]$ the function is strictly monotone and takes any real value.

This means that the equation has exactly one solution in each interval $[i, i+1]$, $i = 1, \cdots, 2019$.

Finally, since the function is negative for $x<1$ and positive for $x>2020$, we can conclude that the equation has exactly 2019 solutions in its domain of definition.