答: Based on your comments, I would suggest revising the concepts of expectation values before attempting this question. Nevertheless, I have provided a solution below with some references and definitions to guide you towards the correct path. I haven't provided the explanation for everything so that you spend some time understanding the concepts and finally figure it out on your own(if you need some more explanation or are stuck somewhere, please reach out via the comment section). Hope it helps.

Title: Conditional expectation- biased coin, fair coin and die

Body:
Suppose you have a fair six-sided die, one fair coin, and one biased coin. The biased coin has probability 0.25 of showing Heads. First you roll the die. If the number on the die is less than three, you then toss the fair coin until you see a Head; otherwise, you toss the biased coin until you see a Head. Compute the expectation of the number of tosses, giving your answer correct to two decimal places.

I have no clue how to approach this.any hints would be helpful


问:
Matrix for a linear map between 2 different basis
Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$

and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the basis $\begin{pmatrix}
2\\
3
\end{pmatrix}, \begin{pmatrix}
1\\
2
\end{pmatrix}$ for $K^2$

Question

Find the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain.

My effort so far

I am thinking that I first have to evaluate the vectors in the basis for $K^3$ and thus get some scalar. I then multiply the scalar on the basis for $K^2$. I.e If I evaluate $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}$ I get $(5,-1)$. I would then have to multiply that onto both my basis vectors i.e $5(2,3)+-1(1,2)=(10,15)+(-1,-2)=(9,13)$. That would give my first column in the matrix for the representation and thus I would end up with a matrix with 2 rows and 3 columns. This sounds wrong to me and I just want to have a quick check.

答:
Let's consider our bases.

For $K^3$, we have the base $\mathcal A = \{a_1,a_2,a_3\}$, where $a_1 = (-1,0,2)$, $a_2 = (3,1,-1)$ and $a_3 = (1,-1,1)$.

For $K^2$, we have the base $\mathcal B = \{b_1,b_2\}$, where $b_1 = (2,3)$ and $b_2 = (1,2)$.

The matrix $M$ that represents $f$ must satisfy the following property:

$$M[v]_{\mathcal A} = [f(v)]_{\mathcal B},$$

where $[u]_{\mathcal C}$ denotes $u$ written in base $\mathcal C$.

For this kind of exercise, I think it's good to consider the following:

A linear transformation $K^n\mapsto K^m$ is represented by an $m\times n$ matrix;

If $M$ is an $m\times n$ matrix and $e_i$ is the  $n\times 1$ vector with all entries equal to $0$ except that the $i$-th entry is $1$, then $Me_i$ is the $i$-th column of $M$;

If $\mathcal C = \{c_1,\dots, c_n\}$ is an ordered basis of some vector space, then $[c_i]_{\mathcal C} = e_i$.

From $(1)$, we know that $M$ will be $2\times 3$ (so you are correct in this regard, that's 2 rows and 3 columns).

Of course, to find out $M$, all we need to do is find $M$'s columns.
$(3)$ implies that

$$M[a_i]_{\mathcal A} = Me_i = [f(a_i)]_{\mathcal B},$$

or in other words, the $i$-th column of $M$ is given by $f(a_i)$ written in base $\mathcal B$.

Do you think you can finish?
You answer is incorrect in this last regard.

ContentLicense: CC BY-SA 4.0


问:
Integration of function help
I'm having problems integrating this function $\displaystyle E(X)=\int^ \infty_0 x\lambda e^{-\lambda x} dx$. I did the integration by parts and had $-xe^{-\lambda x}- \lambda e^{-\lambda x}$. However the solution gives $-xe^{-\lambda x} - \dfrac{1}{\lambda}e^{-\lambda x}$. I can't find any mistakes. What am I doing wrong?</p>

答: Consider $$\begin{align*}\int\lambda xe^{-\lambda x}\,\mathrm{d}x&=\lambda\int\underbrace{x}_u\,\underbrace{e^{-\lambda x}\,\mathrm{d}x}_{\mathrm{d}v}\\&=\lambda\left(-\frac1\lambda xe^{-\lambda x}+\frac1\lambda\int e^{-\lambda x}\,\mathrm{d}x\right)\\&=\lambda\left(-\frac1\lambda xe^{-\lambda x}-\frac1{\lambda^2}e^{-\lambda x}\right)\\&=-e^{-\lambda x}\left(x+\frac1\lambda\right)\end{align*}$$... and so$$\begin{align*}\int_0^\infty\lambda xe^{-\lambda x}\,\mathrm{d}x&=\lim_{c\to\infty}\int_0^c\lambda xe^{-\lambda x}\,\mathrm{d}x\\&=\lim_{c\to\infty}\left[-e^{-\lambda x}\left(x+\frac1\lambda\right)\right]_0^c\\&=\lim_{c\to\infty}\left(-e^{-\lambda c}\left(c+\frac1\lambda\right)+\frac1\lambda\right)\\&=\frac1\lambda\end{align*}$$</p>

ContentLicense: CC BY-SA 3.0

Title: Integration of function help


问:
5 digit numbers from 1 3 3 0 0
How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?

First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.

Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.

$30-3-9=18$

Is $18$ the correct answer, or where did i thought wrong?

答:
I would just compute the total permutations , and multiply it by the probabilty $\frac{3}{5}$ of starting with a non-zero

$\dfrac{5!}{2!2!}\cdot\dfrac35 = 18$

Title: 5 digit numbers from 1 3 3 0 0


问:
How do I solve the following recursive equation?
$$\begin{cases}
    p(1) = \frac{1}{100} \\
    p(t) = p(t-1)\left(1+\frac{1}{101-t}\right)
\end{cases}$$

I feel like I could use the Telescoping Technique, but I do not quite see how.

答:
Write as $$(101-t)p(t)=(101-(t-1))p(t-1).$$

As this works for all $t$, the two members are a constant, equal to

$$(101-1)\frac1{100}.$$


问:
Motivation for Open Mapping Theorem
In complex analysis, the open mapping theorem says:

A holomorphic function carries open sets to open sets.

I keep hearing that this is a deep, profound theorem. But I struggle to see why it is significant. How do I interpret it intuitively? For example, does it suggest something about how a holomorphic function locally stretches space? Does it place some sort of bound on the rate of growth of a holomorphic function near a point?

The closest I have to an intuition comes from the real analysis example,  $f(x) = x^2$, which is continuous but not open (since $f$ maps the open interval $(-1,1)$ to the interval $[0,1)$, which is not open). This happens because $f$ is not injective at $x=0$, but I'm not sure if this intuition carries over to the complex case.

I think I am misinterpreting the theorem here, so any insight would be much appreciated. Thanks!

答:
I came across a satisfactory intuition a few days back.

A holomorphic function can never map something $2$-dimensional (like an open set) onto something $1$-dimensional (like a line or a curve). The function either maps the $2$-dimensional object to a single point (in which case, the function is constant), or it maps it to something else that's $2$ dimensional.

Title: Motivation for Open Mapping Theorem


问:
Lee - Introduction to Smooth Manifolds 2nd ed: exercise 21-5
I am trying to show the hint to this problem. The set-up is as follow

Suppose a Lie group $G$ acts smoothly and freely on a smooth manifold $M$; and the orbit space $M/G$ has a smooth manifold structure such that the quotient map  $\pi:M\rightarrow M/G$ is a smooth submersion.

and the hint is

show that for any smooth local section $\sigma: U \rightarrow M$ of $\pi$, the map $\left(g, x\right)\mapsto g\cdot\sigma(x)$ is a diffeomorphism from $G\times U$ to $\pi^{-1}(U)$.

I am able to show that $\left(g, x\right)\mapsto g\cdot\sigma(x)$ is a smooth (left-) equivariant bijection so I know what follows is that its inverse $g\cdot\sigma(x) \mapsto \left(g, x\right)$ is well-defined and equivariant. However, I am not sure how to show the smoothness of this inverse.

In my solution leading up to the smoothness and equivariance of $\left(g, x\right)\mapsto g\cdot\sigma(x)$, I have used smoothness and freeness of the action $G\times M\rightarrow M$. As such my guess would be to use the smooth structure on $M/G$ that makes $\pi$ a submersion. But besides admitting the existence of such section $\sigma$, I am not sure how else to manipulate this smooth structure of $M/G$'s.

答:
Given $p\in M$, you have $G\to M\to M/G$, where the first map is $g\mapsto g\cdot p$ (call this map $f$). Taking the differential
$$\mathfrak{g} \to T_pM\to T_{\pi(p)}(M/G).$$
Since $\pi$ is a submersion the second map is surjective, and the composition is obviously zero, so the first map is injective, and its image is in $\ker d\pi$, and so is $\ker d\pi$ by dimensional considerations.

Differentiating a local section of $\pi$ at $\pi(p)$ you get a section of $d\pi$, whose image must be transversal to $\ker d\pi= df(\mathfrak{g})$. So the map $(g,x)\mapsto g\cdot\sigma(x)$ has a differential at $(e,\pi(p))$ that takes $\mathfrak{g}$ to $f(\mathfrak{g})$ and $T_{\pi(p)}(M/G)$ to a subspace transversal to it. So it's an isomorphism, and therefore the function is a local diffeomorfism


答:
Integrating by parts using $u=x \Rightarrow du=dx$ and $dv= e^{-\lambda x} dx \Rightarrow v = \dfrac{e^{-\lambda x}}{-\lambda}$, we have

$$\begin{align}
\int x\lambda e^{-\lambda x} dx & = \lambda \int x e^{-\lambda x} dx \\
& = \lambda \left [\frac{xe^{-\lambda x}}{-\lambda} - \int \frac{e^{-\lambda x}}{(-\lambda)} dx\right] \\
& = \lambda \left [\frac{-xe^{-\lambda x}}{\lambda} + \int \frac{e^{-\lambda x}}{\lambda} dx\right] \\
& = \lambda \left [\frac{-xe^{-\lambda x}}{\lambda} + \frac{e^{-\lambda x}}{\lambda(-\lambda)} dx\right] \\
& = \lambda \left [\frac{-xe^{-\lambda x}}{\lambda} - \frac{e^{-\lambda x}}{\lambda^2} \right] \\
& = -xe^{-\lambda x} - \frac{e^{-\lambda x}}{\lambda}. \\
\end{align}$$

Therefore we have
$$\begin{align}
\int_0^\infty x\lambda e^{-\lambda x} dx & = \left[-xe^{-\lambda x} - \frac{e^{-\lambda x}}{\lambda}\right]_0^\infty \\
& = \left[\left(0 - 0\right) - \left(0 - \frac{1}{\lambda}\right) \right] \\
& = \frac{1}{\lambda}.
\end{align}$$

Title: Integration of function help

Body: I'm having problems integrating this function $\displaystyle E(X)=\int^ \infty_0 x\lambda e^{-\lambda x} dx$. I did the integration by parts and had $-xe^{-\lambda x}- \lambda e^{-\lambda x}$. However the solution gives $-xe^{-\lambda x} - \dfrac{1}{\lambda}e^{-\lambda x}$. I can't find any mistakes. What am I doing wrong?</p>


问:
Solve equation $\cos (a)\cos(b)\cos(a+b) = -\frac18$ over $0 < a,b < \frac\pi2$
I have tried this question as follows:

$$4[2(\cos(a)\cos(b))\cos(a+b)] = -1$$

Then converted the $2\cos(a)cos(b)$ in $\cos(a+b) + \cos(a-b)$.

And so on but this got messy and big.

Another approach that I tried was that as $a$ and $b$ lie in the first quadrant then their cosine is positive this implies $\cos(a+b)$ is negative.

And without losing generality we can say each term be equal to 1/2 which implies

$a = b = π/3$.

I googled it but didn't find any solution.

Wolfram alpha after converting into the same thing as I have converted used complex numbers which was neat, but my question is:  can we solve this with a short method (using trigonometry)?

答:
As you did, rewrite the equation
$\cos a\cos b\cos(a+b) = -\frac18$ as

$$f(a,b)=\cos(2a)+\cos(2b)+\cos(2a+2b)= -\frac32$$

By examining $f_a’=f_b’=0$, it can be shown that $f(a,b)\ge -\frac32$; hence the solutions coincide with the minima at
$$a=b+ n_1\pi =\pm \frac\pi3+n_2\pi$$

Given the domain restriction $0 < a,b < \frac\pi2$, the only valid solution is
$$a=b=\frac\pi3$$


问:
Visual intuition for the definition of "asymptotically equivalent"
I'm trying to intuitively grasp the following definition:

The real-valued functions $f$ and $g$ are asymptotically equivalent as $x \to \infty$ if $$\lim_{x \to \infty} \dfrac{f(x)}{g(x)}=1.$$ We write this as $f \sim g$.

My question is: how do we visually interpret this in terms of the graphs of $f$ and $g$? Does this mean that the graphs of $f$ and $g$ get closer to each other as $x$ gets larger and larger?

My only intuition for this comes from the following example: we know that $\sin x \sim x$ as $x \to 0$ (since $\lim_{x \to 0} (\sin x)/x = 1$). And as we can see below, the graphs of $\sin x$ (the green line) and $x$ (the black line) get closer and closer as $x$ goes to $0$.

But this intuition does not seem to hold for functions asymptotically equivalent at $\infty$. I graphed $x^2 + x$ (black line) and $x^2$ (green line) and their graphs do not appear to be getting closer at all! In fact, it looks like there's a "gap" between the two graphs.

This leads me to believe that I'm not interpreting "asymptotically equivalent" in the right way. I've come across the idea that $f \sim g$ means that $f$ and $g$ have the "same rate of growth", but that feels very unintuitive for me. Is there are a way to see that in the graphs?

Any guidance would be greatly appreciated! Thanks.

答:
$x^2+x$ and $x^2$ are asymptotically equivalent, since
$$
\lim_{x \to \infty}\frac{x^2+x}{x^2}=\lim_{x \to \infty}1+\frac{1}{x}=1 \, .
$$
So it is probable that you simply didn't choose $x$ values that were large enough to make your intuitions work. For example, $x=100$ gives
$$
\frac{100^2+100}{100^2}=\frac{10100}{10000}=1.01 \, ,
$$
and you can see that they are very close to each other in relative terms. Notice the use of the word relative here. The example you gave helps illustrate what I mean. Let $f(x)=x^2+x$ and $g(x)=x^2$. $f(x)-g(x) \to \infty$ as $x \to \infty$. This means that in absolute terms, the two functions are growing apart. However, the ratio between them—what you need to multiply $x^2$ by to get $x^2+x$—is approaching $1$. This is what the notion of 'asymptotically equivalent' is trying to capture.

To visualise this, it would be better to use a logarithmic scale:

Look at $x=10$, for instance. $x^2=100$, whereas $x^2+x=110$. This discrepancy looks small when we use a logarithmic y-axis, since the gap of $10$ is small relative to how large the functions are. Or, as Paramanand Singh has suggested, we could plot $y=(x^2+x)/x^2$:


问:
Basic facts about subsequences of a sequence $(x_n)$
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I am doing exercise 2.5.2 of Understanding Analysis by Stephen Abbott. Is my justification for the below propositions sound?

Decide whether the following propositions are true or false, providing a short justification for each conclusion.

(a) If every proper subsequence of $(x_n)$ converges, then $(x_n)$ converges as well.

(b) If $(x_n)$ contains a divergent subsequence, then $(x_n)$ diverges.

(c) If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits.

(d) If $(x_n)$ is monotone and contains a convergent subsequence, then $(x_n)$ converges.

Proof.

(a) This proposition is false. As a counterexample, consider the sequence
\begin{align*}
 \left(1,1,\frac{1}{2},1,\frac{1}{2},\frac{1}{3},1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right)
\end{align*}

The subsequences $(1,1,1,1,\ldots)$, $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots\right)$, $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\ldots\right)$ are all convergent subsequences with different limits. Hence, $(x_n)$ is not a sequence.

The Monotone Convergence Theorem only guarantees: a bounded sequence of real numbers has atleast one convergent subsequence. These subsequences may or may not converge to the same limiting value.

If every proper subsequence of $(x_n)$ converges to the same limit $a$, then $(x_n)$ converges as well.

(b) This proposition is true.

Justification. If a subsequence of $(x_n)$ diverges, it implies that, given any $\epsilon >0$, there are an infinite number of real numbers in the tail that do not lie in an $\epsilon$-neighbourhood of a limit point. So, the original sequence is divergent.

(c) This proposition is true.

Justification. Assume that the sequence $(x_n)$ is bounded, but divergent. By definition, there exists large number $M$, such that $\absval{x_n} \le M$ for all $n \in \mathbf{N}$. Consider the subsequences
\begin{align*}
(x_1,x_2,x_3,x_4,\ldots) \\(x_2,x_3,x_4,\ldots) \\(x_3,x_4,\ldots) \\\vdots
\end{align*}
Each of these subsequences are bounded subsets of $\mathbf{R}$ and therefore by AoC, it is reasonable to talk about their supremum and infimum. Define the subsequences,

\begin{align*}
 u_n := \sup \{x_k:k \ge n\}\\
 d_n := \inf \{x_k:k \ge n\}
\end{align*}

$u_n$ is bounded and monotonically decreasing. $d_n$ is bounded and monotonically increasing. By the Montone Convergence theorem, $u_n$ and $d_n$ are convergent.

(d) This proposition is true.

Justification. Assume that the subsequence $(x_{n_k})$ converges to $x$. Then, given an $\epsilon > 0$, there exists $N$ such that $x_{n_k}$ lies in the interval $(x-\epsilon,x+\epsilon)$ for all $n \ge N$. Because $(x_n)$ is monotone, the values of the sequence $(x_n)$ between $x_{n_k}$ and $x_{n_{k+1}}$ also lie in $(x-\epsilon,x+\epsilon)$. Consequently, the same $N$ can be used for the original sequence.

答:
For a) I think it's true.
First you counter-example does not work : it has many non-convergent subsequences, for example $(1, 1/2, 1, 1/2, ...)$.
Then it's actually true even if you only suppose that $(x_{2n})$, $(x_{2n+1})$ and $(x_{3n})$ converge. In fact, then if we note $a, b, c$ the limit of each, we have $(x_{6n})$ that can be extracted from both $(x_{3n})$ and $(x_{2n})$, so $a=c$, and considering $(x_{6n+3})$, we also have $b=c$. So $a=b$.
Thus $(x_n)$ converges.

For b) you can also proceed by contraposition : if $(x_n)$ converges, then all subsequences converge.

For c), as $(x_n)$ is bounded, we can extract $(x_{\varphi(n)})$ that converges. Let's call $\ell$ its limit. As $(x_n)$ does not converge to $\ell$, there exists an $\varepsilon$ such that for all $N\in \mathbb{N}$, there exists a $n \geqslant N$ such that $|x_n - \ell| > \varepsilon$. So we can extract another subsequence that does not converge to $\ell$. This subsequence being bounded too, we can extract from it a another subsequence that converges to $\ell ' \neq \ell$. So we found two subsequences that converge to different limits.

For d), you proof is fine.


问:
Find constant term in the expansion of $\left (1+\frac x2 -\frac 2x \right)^4$
Finding the constant term for

$$\left (1+\frac x2 -\frac 2x \right)^4$$

is easy, but that would require converting the expression into a binomial.

However, I have no idea about how to do that.   Completing squares doesn’t work because we have a negative sign on $\frac 2x$.

Body:
Hint: Let $$ p(x)= \left (1+\frac x2 -\frac 2x \right)^4$$ then $$16x^4 p(x) = (2x+x^2-4)^4$$ so you have to find the term in $(2x+x^2-4)^4$ which has $x^4$.

 $$(2x+x^2-4)^2  = x^4+4x^3-4x^2-16x+16$$

   \begin{align} (\color{red}{x^4}+\color{blue}{4x^3}-4x^2\color{blue}{-16x}+\color{red}{16})^2 &= ... 2\cdot \color{red}{16x^4} -  2\cdot \color{blue}{64x^4} + (-4x^2)^2 +...\\  &= ...-80x^4+... \end{align}


问:
Changing variables for delta function in two dimensions
I have an integral over a two-dimensional plane like

$$\int d^2 \mathbf{x} \, \delta(g(\mathbf{x})) f(\mathbf{x}).$$

The delta function restricts the integral to a curve implicitly given by $g(\mathbf{x}) = 0$. I can parametrize this curve in polar coordinates $(r,\theta)$; that is, the delta function is supported on some $r=R(\theta)$.

I need to get rid of the delta function, i.e. change variables to get an integral over $\theta$. This should give me something like
$$ \int \sigma (\theta) d\theta \frac{f(R \cos \theta, R \sin \theta)}{|\nabla g|}.$$

But what measure do I need — what is $\sigma(\theta)$? Is it correct to take the arc length measure
$$ \sigma(\theta)^2 = R'^2 + R^2?$$

答:
Suppose that the curve $C$ given by $g(\mathbf{x})=0$ is smooth and can be parameterized by arc length as $\mathbf{x} = \mathbf{p}(s) = (p_1(s), p_2(s)).$ Let $\mathbf{n}(s) = (n_1(s), n_2(s))$ be the unit normal field to $C,$ pointing to the left when going in the positive direction of $C$. Then a neighborhood of the curve can be parameterized by $\mathbf{x}=\mathbf{p}(s)+t\mathbf{n}(s)$ and we have
$$
|\mathbf{p}'| = 1, \quad
|\mathbf{n}| = 1, \quad
\mathbf{n}' = -\kappa \mathbf{p}
$$
where $\kappa$ is the curvature at the actual position on the curve.

The local metrics is given by
$$
d\mathbf{x}\cdot d\mathbf{x} = (1-\kappa t)^2 \, ds^2 + dt^2
$$
and the area measure by
$$
d^2\mathbf{x}
= \left| \frac{\partial\mathbf{x}}{\partial s} \times \frac{\partial\mathbf{x}}{\partial t} \right| \,ds\,dt
= (1-\kappa t) \, ds \, dt
.
$$

It should also be noted that $\nabla g = \pm|\nabla g| \mathbf{n},$ where the sign is $+$ if $g$ grows to the left, $-$ if $g$ grows to the right. Then, to first order,
$$
g = t\,\nabla g\cdot \mathbf{n} = \pm t |\nabla g|
$$
so
$$
\int d^2 \mathbf{x} \, \delta(g(\mathbf{x})) f(\mathbf{x})
= \iint ds\,dt\,(1-\kappa t) \, \delta(\pm t |\nabla g|) f(\mathbf{x})
= \iint ds\,dt\,\frac{1-\kappa t}{|\nabla g|} \, \delta(t) f(\mathbf{p}(s)+t\mathbf{n}(s))
\\= \int ds\,\frac{1}{|\nabla g|} \, f(\mathbf{p}(s))
$$
This result is basically of the same form as you have, so $\sigma(\theta)\,d\theta$ should be the arc length measure.


答:
$\displaystyle  \int x\lambda e^{-\lambda x} dx$

$\displaystyle \lambda\int x e^{-\lambda x} dx$

Integrating by parts we have

$\displaystyle \lambda \left(x\int  e^{-\lambda x} dx-\int\dfrac{dx}{dx}\left(\int  e^{-\lambda x} dx\right) dx\right)$

$\displaystyle \lambda \left(\frac{xe^{-\lambda x}}{-\lambda}-\int \frac{e^{-\lambda x}}{-\lambda} dx\right)$

$\displaystyle -xe^{-\lambda x}+\int e^{-\lambda x} dx$

$\displaystyle -xe^{-\lambda x}+\frac{e^{-\lambda x}}{-\lambda}$

$\displaystyle -xe^{-\lambda x}-\frac{e^{-\lambda x}}{\lambda}$

Title: Integration of function help


Title: Find constant term in the expansion of $\left (1+\frac x2 -\frac 2x \right)^4$

Body:
Finding the constant term for

$$\left (1+\frac x2 -\frac 2x \right)^4$$

is easy, but that would require converting the expression into a binomial.

However, I have no idea about how to do that.   Completing squares doesn’t work because we have a negative sign on $\frac 2x$.

Body: Let $[x^k]f(x)$ denote the $x^k$ coefficient in $f$, so $[1]f$ is $f$'s constant term. Then$$\begin{align}[1](1+x/2-2/x)^4&=\tfrac{1}{16}[x^4](x^2+2x-4)^4\\&=\tfrac{1}{384}\left.[(x^2+2x-4)^4]^{(4)}\right|_{x=0}\\&=\tfrac{1}{384}\left.[(y^2-5)^4]^{(4)}\right|_{y=1}\\&=\tfrac{1}{384}\left.[y^8-20y^6+150y^4-o(y^4)]^{(4)}\right|_{y=1}\\&=\tfrac{1}{384}\left.[1680y^4-7200y^2+3600]\right|_{y=1}\\&=\tfrac{1}{384}[1680-7200+3600]\\&=-5.\end{align}$$


答:
Say you were to use the binomial theorem. Then only terms containing even powers of the factor $x/2-2/x$ would contain any constants because getting a constant would require the exponent of $x/2$ and $2/x$ to be equal. The constant in $(x/2-2/x)^k$ is thus$$\binom k{k/2}(x/2)^{k/2}(-2/x)^{k/2}=(-1)^{k/2}\binom k{k/2}$$ This gets multiplied by $\binom4k$ which is the coefficient from the outer binomial expansion.

Can you sum this for even $k\in\{0,1,2,3,4\}$?

$\binom40+\binom42(-1)^1\binom21+\binom44(-1)^2\binom42=-5$


Body:
Using the hint of @Aqua :-) we can see that
\begin{eqnarray*}
\left( 1+\frac{x}{2}-\frac{2}{x}\right)^{4}&=&\left(\frac{x^{2}}{2x}+\frac{2x}{2x}-\frac{4}{2x} \right)^{4}\\
&=&\left(\frac{x^{2}+2x-4}{2x} \right)^{4}\\
&=&\frac{(x^{2}+2x-4)^{4}}{16x^{4}}
\end{eqnarray*}
Now, since that
$$16x^{4}p(x)=(x^{2}+2x-4)^{4}$$
but since that $$(x^{2}+2x-4)^{4}=x^{8}+8x^{7}+8x^{6}-64x^{5}-80x^{4}+256x^{3}+128x^{2}-512x+256$$
so, we have
$$\mathbf{constant}\left( 1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\boxed{-5}$$


问:
Generate a frequency distribution function with given Mean, Median, Std Dev, 1st and 3rd Quartile
I want to generate plot of a function whose Mean, Median, Std Dev, 1st and 3rd Quartile are given.
How do I generate one?

I tried Gaussian Distribution but there Mean = Median and we don't have control over quartiles.

Edit 1:-
I'm helping my 10th grade cousin with statistics. In one of the question I need to compare the skewness of functions with two different sets of variables (Mean, Median, Std Dev, 1st and 3rd Quartile). I want to show him how the frequency distribution will look like.

I did some research and found some skewed function here https://www.desmos.com/calculator/ptnf1tbdgf

答:
Both the question of whether it is possible and how to do it will depend on the actual values.  For example if you know the mean $\mu$ and standard deviation $\sigma$ and median $q_{2}$ and quartiles $q_{1}$ and $q_{3}$, then having $\sigma < \frac12(q_{3}-q_{1})$ will be impossible to satisfy.

Here is an a piecewise density function which works if the standard deviation is large enough:

Let $q_0 = \frac{16\mu -3q_3-
4q_2-5q_1 -
\sqrt{96\sigma^2-160\mu^2+32(3\mu-q_2)(q_1+q_2+q_3)+32(\mu q_2-q_3^2-q_1^2)+9(q_3-q_1)^2}}{4}$ and $q_4 = \frac{16\mu -5q_3-
4q_2-3q_1 +
\sqrt{96\sigma^2-160\mu^2+32(3\mu-q_2)(q_1+q_2+q_3)+32(\mu q_2-q_3^2-q_1^2)+9(q_3-q_1)^2}}{4}$.

and then let $$f(x)=\begin{cases}
      0 &  \,\,\,\,\,\,\,\,\,\,\,\,\,x\lt q_0 \\
      \frac{1}{4(q_1-q_0)} & q_0\lt x\lt q_1 \\
      \frac{1}{4(q_2-q_1)} & q_1\lt x\lt q_2 \\
      \frac{1}{4(q_3-q_2)} & q_2\lt x\lt q_3 \\
      \frac{1}{4(q_4-q_3)} & q_3\lt x\lt q_4 \\
      0 & q_4 \lt x
   \end{cases}$$

This does not work if the square root is of a negative number or if it leads to $q_0 \gt q_1$ or $q_4 \lt q_3$, but otherwise it should.

As an illustration, if say $\mu =10$, $\sigma=3$, $q_1=8$, $q_2=9$, $q_3=12$ then this gives $q_0 = 12-\sqrt{45} \approx 5.2918$ and $q_4 = 10+\sqrt{45} \approx 16.7082$ and that seems to work


问:
Concrete examples of valuation rings of rank two.
Let $A$ be a valuation ring of rank two. Then $A$ gives an example of a commutative ring such that $\mathrm{Spec}(A)$ is a noetherian topological space, but $A$ is non-noetherian. (Indeed, otherwise $A$ would be a discrete valuation ring.)

Is there a concrete example of such a ring $A$?

答:
I'm late to this game, and Robin Chapman's elaboration of Qiaochu's example is good -- so good I just used it as a (counter)-example in my answer here. Whilst thinking about that, I realised there is another example which feels slightly more natural to me as someone who works with $p$-adics a lot:

Let $K$ be the field $\mathbb Q_p((X))$ of Laurent series over the $p$-adics. We have the obvious rank-one discrete valuation

$$w_1(\sum a_i X^i) = \min\{i: a_i \neq 0\}$$

like over any other field of Laurent series, just treating the coefficient field as being of zero valuation and identifying it with the residue field of the valuation ring $R_1 = \{x \in K: w_1(x) \ge 0\} = \mathbb Q_p[[X]]$ with respect to its maximal ideal $\mathfrak{m}_1 = \{x \in K: w_1(x) \ge 1\}= X \cdot \mathbb Q_p[[X]]$.

So far we have not used that $\mathbb Q_p$ has a valuation $v_p$ as well. Let's do it. Define a rank-two valuation $w_2 : K \rightarrow \mathbb Z \times \mathbb Z \cup \{\infty\}$ as follows: $w_2(0) := \infty$ and for $0 \neq x = \sum a_i X^i$ set

$$w_2(x) := \left(w_1(x), v_p(a_{w_1(x)})\right)$$

i.e. we "refine" the rank-one valuation $w_1$ by also keeping track of the $p$-adic valuation of the leading coefficient. Of course we give the value group the lexicographic order here. Check that the valuation ring to this is
$$R_2 = \{x \in K: w_2(x) \ge (0,0)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in \mathbb Z_p\};$$

and the maximal ideal of $R_2$ is

$$\mathfrak m_2= (p) = \{x \in K: w_2(x) \ge (0,1)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in p\mathbb Z_p\}.$$

Note that we have proper inclusions $R_2 \subsetneq R_1$ and $\mathfrak m_1 \subsetneq \mathfrak m_2$ (indeed, "the other way around"!), exactly as in the answer linked above. I found it worthwhile to think about how the valuation rings and their maximal ideals relate for this kind of "refinement" of a given valuation. One could obviously iterate the procedure.

It is also worthwhile to think how this example and Robin Chapman's / Qiaochu's are "almost the same". Actually, these are standard first examples of higher(-dimensional) local fields which have been studied in recent decades for their class field theory (Kato, Fesenko), and in connection to $p$-adic Langlands.


问:
Convergence of serie
Does the series $\displaystyle\sum_{k=1}^{\infty}\dfrac{k\sin(2kx)}{4k^2-1}$ converge? I tried to use the tests, but I couldn't get anything. Can anybody help me?

答:
It converges, by Dirichlet's test, since:

the sequence $\left(\frac k{4k^2-1}\right)_{k\in\Bbb N}$ is monotonic and converges to $0$;

the series $\sum_{k=1}^\infty\sin(2kx)$ has bounded partial sums.

Title: Convergence of serie


问:
Prove convergence of a sequence and find it's limit if $x_{0} = 2$
I'm studying for my end of semester exam. And I'm having trouble understanding how to prove convergence of a sequence and finding it's limit.
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) $$
I need to prove that this sequence converges and find it's limit if $x_{0} = 2$. $$$$ I had an idea to prove it by Cauchy's criteria, but I'm not sure how to do it here.

答:
Update: Just realized an almost identical question  was asked a while ago (no wonder it looks familiar to me) and there is a way to get the closed form solution of the sequence.

I just copied and pasted my solution here and made a few changes (but I'll keep the $a_n$s from that problem):

$$\frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n+\frac{1}{a_n}+2}{a_n+\frac{1}{a_n}-2}=\left(\frac{a_n+1}{a_n-1}\right)^2 \\\implies \frac{a_n+1}{a_n-1}=\left(\frac{a_0+1}{a_0-1}\right)^{2^n}=3^{2^n} \\ \implies
a_n =\frac{3^{2^n}+1}{3^{2^n}-1} = 1+\frac{2}{3^{2^n}-1} \to 1$$

Original answer:

Via AM-GM $x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) \ge 1$

Then $$x_n-1 = \frac 12 \left(x_{n-1}+\frac{1}{x_{n-1}} - 2\right) = \frac{(x_{n-1}-1)^2}{2x_{n-1}} \le \frac{(x_{n-1}-1)^2}{2}\\
\implies |x_n-1| = x_n-1 \le \frac{1}{2^n} (x_0-1)^{2^n} \to 0$$

Therefore $x_n \to 1$, as $n \to +\infty$.


问:
Estimating Poisson rate and using the resulting distribution.
The following table shows the frequency of arrival of patients at a health clinic ina 15-minute time interval. What is the probability that more than 6 patients will arrive in next half an hour?

```<code>Number of customers  : 0 1 2 3 4 5 6
Observed frequency   : 1 4 5 6 4 3 2
</code>```

The problem I face is here random variable is defined for number of intervals but we have to find the probability for number of customers How is that possible?

I am new to Statistics Please help me with ideas or examples

答:
EDIT: This solution does not use a Poisson distribution, it uses the observed/empirical distribution to estimate the answer. For the solution using a Poisson assumption, see BruceET's solution.

You want to think about the next half hour as two distinct events, each of 15 minute intervals.

Let $X$ be the number of arrivals in the first 15 minutes, and $Y$ be the number of arrivals in the second 15 minutes. Then, the number of arrivals in the next 30 minutes can be represented by $ Z = X + Y $.

First, we need the distributions of $X$ and $Y$. We can get to that using the number of observations for number of observed patients. For example, out of a total of 25 observations, 0 patients were observed 1 time, so the probability of 0 is $1/25  = 0.04$. One patient was observed 4 times, for a probability of $4/25 = 0.16$. Continuing on through all observations, we arrive at the following probability distribution:

Number of Patients
Frequency
Probability

0
1
4%

1
4
16%

2
5
20%

3
6
24%

4
4
16%

5
3
12%

6
2
8%

Since $Y$ is just another 15 minute interval, we can assume it has the same distribution as $X$.

Then, we get to the distribution of $Z$ by considering possible combinations of $X$ and $Y$. For example, $Z = 0$ can only happen if $X = 0$ and $Y = 0$:

$$ P[Z = 0] = P[X = 0 \cap Y = 0]. $$

But, $Z = 2$ can happen if $X = 2$ and $Y = 0$ OR $X = 1$ and $Y = 1$ OR $X = 0$ and $Y = 2$:

$$ P[Z = 2] = P[(X = 2 \cap Y = 0) \cup (X = 1 \cap Y = 1) \cup (X = 0 \cap Y = 2)]. $$

Assuming $X$ and $Y$ are independent events, you can break this up like so:

$$ P[Z = 2] = P[(X = 2 \cap Y = 0) \cup (X = 1 \cap Y = 1) \cup (X = 0 \cap Y = 2)] $$
$$ = P[X = 2 \cap Y = 0] + P[X = 1 \cap Y = 1] + P[X = 0 \cap Y = 2] $$
$$ = P[X = 2]*P[Y = 0] + P[X = 1]*P[Y = 1] + P[X = 0]*P[Y = 2] $$
$$ = 0.20*0.04 + 0.16*0.16 + 0.04*0.20 $$
$$ = 0.0416 $$

Once you go through the different possible combinations, you arrive at the following distribution for $Z$:

Number of Patients
Probability

0
0.16%

1
1.28%

2
4.16%

3
8.32%

4
12.96%

5
15.68%

6
16.64%

7
15.04%

8
11.52%

9
7.68%

10
4.00%

11
1.92%

12
0.64%

Then, what you're looking for is

$$ P[Z > 6] = 15.04\% + 11.52\% + ... + 0.64\% = 40.80\%. $$


问:
How can it be proved that the limit definition and series definition of $e$ are equivalent, how do they model continuous growth?
The limit definition of $e$ gives the best intuition toward how $e$ can model continuous growth.

$$\lim\limits_{n \to \infty} \left( 1 + \frac1n\right)^n$$

Let n represent n moments of growth that scale the initial value by $ 1 + \frac1n $.

The key to me seems to be that the $\Delta x $ approaches 1 but does not reach it. This allows for the best approximation $\Delta x = 1 $ at each moment. In other words, at each moment of growth the value x experiences $\Delta x = 1 + \frac1n $, or as $n \to \infty$... $$\lim\limits_{n \to \infty} 1 + \frac1n = 1 $$

Growth with n = 2 moments:

$1 \times 1\frac12 \times 1\frac12 $ = 2.25

insight: If 2 moments of growth occur to x where $\Delta x = 1\frac12$, this amounts to discrete growth by $\Delta x = 1\frac12$ twice.

Growth with three moments:

$1\times 1\frac13 \times 1\frac13 \times 1\frac13 $ = 2.35

insight: If 3 moments of growth occur to x in where $\Delta x = 1\frac13$, this amounts to discrete growth by $\Delta x = 1\frac13$ three times.

Growth at seven moments:

$1\times \frac17 \times 1\frac17  \times 1\frac17 \times 1\frac17 \times 1\frac17 \times 1\frac17 \times 1\frac17 $ = 2.55

insight: If 7 moments of growth occur to x in this period where $\Delta x = 1\frac17$, this amounts to discrete growth by $\Delta x = 1\frac17$ seven times.

$\lim\limits_{n \to \infty} \left( 1 + \frac1n\right)^n $ is a definition of $e$ as it models n moments of growth occurring where at each instant the initial value is scaled by a factor $\lim\limits_{n \to \infty} 1 + \frac1n = 1 $.

To me this seems intuitive to a small extent. If we want to grow continuously we scale by an infinitely small increase an infinite number of times and this converges onto the number $e$. Why it must converge is mysterious to me, and besides seeing a proof I can't intuit to something that is satisfying.

Now there is also the Taylor series representation of $e$

$$ \sum_{n=0}^\infty \frac1{n!} = 1 + 1 + \frac12 + \frac1{3\times2\times1} + \frac1{4\times3\times2\times1} ... $$

Speaking loosely it seems to get $e$ you can sum a whole and itself, with $ \frac12$ that whole, and $ \frac13$ of $ \frac12$ of that whole and so on...and I have no intuition for this.

Again, the Taylor Series converges to $e$ but I don't see the same intuition for how this can model continuous growth. Furthermore, with my limited math skills I wouldn't know how to prove:

$$\lim\limits_{n \to \infty} \left( 1 + \frac1n\right)^n = \sum_{n=0}^\infty \frac1{n!}$$

My 3 questions I am seeking to answer are

1. By what application or intuition can it be seen that $ \sum_{n=0}^\infty \frac1{n!} $ models continuous growth?

2. What are some proofs that:

$$\lim\limits_{n \to \infty} \left( 1 + \frac1n\right)^n = \sum_{n=0}^\infty \frac1{n!}$$

3. Is euler's number inextricably linked to time as $\pi$ is to the circle?

When discussing $e$ it seems necessary to talk of periods of continuous growth $ p = kt $ being represented by $ e = e^p $ where $ p = 1 $ in the fundamental case. I do not know how to think of $e$ without seeing it as one period of continuous growth proven to be $e$ by the convergence of the intuitive definition I gave. I cannot see this same inextricable link to a period in the Taylor Series nor in other definitions of $e$.

Thanks for any comments, I have been trying to wrap my head around $e$ at the most fundamental level for a long time so that I can more deeply understand differential equations.

How was it arrived upon that e best modeled continuous growth? Was it an experimental observation or was it a proof that convinced mathematicians and scientists that $e$ embodies the idea of continuous growth i.e. growth at all instances?

I have so many questions about Taylor Series and $e^{i\theta}$ I'm trying to answer as well..

Best Wishes,

答:
You're asking multiple questions here, and you've asked them in a certain order. But I'm gonna take a liberty and answer them in a more intuitive order.

In your third question, you asked if Euler's number is inextricably linked to time. It is, in fact, not. Euler's number is inherently a mathematical constant, not a physical one. While it may be useful regarding time dependent functions, that doesn't necessarily mean that it is fundamentally tied to time dependent functions as $\pi$ is to circles.

You see, fundamentally the number $e$ is about $2$ things. Exponential growth, and rate of change. The first thing to consider is the derivative of any exponential. Let's begin by differentiating a random exponential $a^x$.

$$\frac{d(a^x)}{dx} = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}$$

As $a^{x+h} = a^x a^h$,

$$\frac{d(a^x)}{dx} = \lim_{h \to 0} \frac{a^x a^h - a^x}{h}$$

Factoring out $a^x$ from the numerator, and outside the limit gives us

$$\frac{d(a^x)}{dx} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}$$

From this, we can intuitively see that the derivative of any exponential is going to be directly proportionate to the exponential itself, as the derivative of $a^x$ is equal to $a^x$ times some constant. The question then arises: does there exist a base for which this proportionality constant is equal to $1$? Is there a function such that the function is its own derivative?

The short answer to this is yes. And we have with us the means to find it. The first thing to do is to take $\lim_{h \to 0} \frac{a^h - 1}{h}$ to be our proportionality constant. But we also know that our proportionality constant must be $1$. Hence, we get

$$\lim_{h \to 0} \frac{a^h - 1}{h} = 1$$

multiplying $\lim\limits_{h \to 0} h$ on both sides, we get:

$$\lim_{h \to 0} \frac{a^h - 1}{h}  \lim_{h \to 0} h = \lim_{h \to 0} h$$

Collecting the limits on the left hand side gives us

$$\lim_{h \to 0} \frac{a^h - 1}{h} h = \lim_{h \to 0} a^h - 1 = \lim_{h \to 0} h $$

bringing the $1$ on the left hand side to the right hand side gives us

$$\lim_{h \to 0} a^h = \lim_{h \to 0} h + 1$$

Some amount of finagling of the expression will hence give us

$$a = \lim_{h \to 0} (h + 1)^{\frac{1}{h}}$$

let us now state that $\frac{1}{h} = n$. This is equivalent to stating $\frac{1}{n}=h$. When $n\to\infty$, $h\to0$. Hence, we can now rewrite out original limit as being

$$a = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$$

This is where the limit definition of Euler's number comes from. From this, we can know for certain a fundamental (if not the defining property of) the number $e$. And more importantly, it helps us understand why the derivative of $e^x$ is $e^x$. It isn't a coincidence, it's the very definition of $e$! If you're determined to find a connection between $e$ and some property as $\pi$ is to circles, then here it is: the holy grail. Euler's number is fundamentally and inextricably linked to rate of change problems. Although this is an idea you seem to have grasped intuitively, even before considering the underlying calculus.

Questions $1$ and $2$ can be answered simultaneously. By proving that the limit and series definitions of $e^x$ are equivalent, we can show definitively that there is a link between the Taylor series and exponentiation. The intuition will come naturally once you can see how the two can be equated directly without using the exponential function as an explicit intermediate.

This is slightly more difficult to prove (or rather, more difficult on my stubby little fingers as I do my best to work out the LaTex)

Let us begin by only considering the expression

$$\left(1+\frac{1}{n}\right)^n$$.

We can expand the brackets by using the binomial theorem as follows:

$$\left(1+\frac{1}{n}\right)^n = \sum_{k=1}^n \frac{n!}{k!(n-k)!n^k}$$

By doing some simplifications we get

$$\left(1+\frac{1}{n}\right)^n = 1 + \sum_{k=1}^n \frac{\prod\limits_{j=1}^{k-1} (n-j)}{k!n^{k-1}}$$

Simplification is a relative term. It comes together later though, I swear. Given $x\in\mathbb N$, we can state that

$$n^x = \prod_{i=1}^x n$$

We can hence use this information to state that

$$\left(1+\frac{1}{n}\right)^n = 1 + \sum_{k=1}^n \frac{1}{k!} \prod\limits_{j=1}^{k-1} \frac{(n-j)}{n}$$

This is equal to

$$1 + \sum_{k=1}^n \frac{1}{k!} \prod_{j=1}^{k-1} \left(1 - \frac{j}{n}\right)$$

Therefore, we can state that

$$\left(1+\frac{1}{n}\right)^n = 1 + \sum_{k=1}^n \frac{1}{k!} \prod_{j=1}^{k-1} \left(1 - \frac{j}{n}\right)$$

Let us now take the limit as $n\to\infty$ on both sides

$$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = 1 + \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k!} \prod_{j=1}^{k-1} \left(1 - \frac{j}{n}\right)$$

Since $\frac{j}{n} \to 0$ when $n\to\infty$, the product term simply turns into the product of a whole bunch of $1$s. As a result, regardless of what $n$ or $k$ is, the whole product becomes $1$. Hence

$$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = 1 + \sum_{k=1}^\infty \frac{1}{k!}$$

Because $0!=1$, $\frac{1}{0!} = 1$. As a result, we can bring the $1$ that's in front of the summation back into the summation. Hence

$$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = \sum_{k=0}^\infty \frac{1}{k!}$$

QED.

I would recommend trying this for yourself, by attempting to prove that this holds true for all $e^x$. Here's a hint to get you started: $e^x = \lim\limits_{n \to 0} \left(1 + \frac{x}{n}\right)^n$

Finally, you ask the big boy. The mega question. The real thinker. What does $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ mean?

There are two common approaches to this. Firstly, there is the method that uses the Taylor series of $\sin(x)$, $\cos(x)$, and $e^{ix}$. But if I'm being perfectly honest... I kinda really don't want to go through the trouble of doing all that LaTex again. Which is why we move on to the second way: the Picard-Lindelöf theorem.

This theorem states that given that there are $2$ functions which satisfy

The same differential equation and

The same initial conditions

The two functions will be equal

Let us look at the function $f(x) = e^{ix}$. The derivative of this function becomes $ie^{ix}$. Hence, this is a solution to the differential equation $f(x) = if'(x)$. In addition, let us take note of the fact that $f(0) = 1$. This will become important later.

Next, let us take a look at the function $g(x) = \cos(x) + i\sin(x)$. In this case, $g'(x) = -\sin(x) + i\cos(x) = i(\cos(x) + i\sin(x))$. Therefore, $g'(x) = ig(x)$. in addition, $g(0) = \cos(0) + i\sin(0) = 1 + 0i = 1$.

Therefore, we know that the two functions have the same initial values. We know that they both are solutions to the same differential equation. As a result, we can state (by Picard-Lindelöf) that $e^{ix} = \cos(x) + i\sin(x)$. (Please note that this is not my proof. This is something I retrieved from this video; see Epic Math Time's segment on the proof of Euler's Formula). If you have the time, feel free to attempt the other method as well. It is actually simple enough if you do it by hand, given that you expand everything out rather than leaving it in terms of sigma notation.

Now onto the more pressing question: what does this mean?

Let us begin by opening the complex plane (take the $y$ axis to be the imaginary axis and the $x$ axis to be the real axis)

Given that you have a complex number $z = a + ib$ (where $a$ and $b$ are both members of the real numbers) $a$ will be the real part of $z$, and $b$ will be the imaginary part of $z$. We can plot $z$ using rectangular coordinates.

Given $z = \cos(t) + i\sin(t)$, $\cos(t)$ is the real part of $z$, and $\sin(t)$ is the imaginary part of $z$. We can hence parameterize the real and imaginary parts of $z$ in terms of an arbitrary parameter $t$ (which, as is customary with calculus, is measured in radians). Drawing this on the graph gives us the following;

As we can see, this image shows a unit circle. So the graph of outputs of $\cos(x) + i\sin(x)$ forms a unit circle on the complex plane. But we proved earlier that $\cos(x) + i\sin(x) = e^{ix}$. Therefore, given that we have a real number input $x$, the output of the function $e^{ix}$ will be a unit circle on the imaginary plane. I just spent over an hour on just the LaTex. Please send help. Peace out.


问:
2D random walk - first hits
A 2D symmetric random walk $(X_k,Y_k)$ $k\ge 0$ (Markov chain) where $(X_{k+1},Y_{k+1})$ takes one value of the next ones: $(X_k,Y_k+1),(X_k,Y_k-1),(X_k+1,Y_k),(X_k-1,Y_k)$, all have the same probability and the initial values of X and Y are zero.

a) Find the value of $P(X_T=3,Y_T=0)$ and $E[T]$ if $T=inf(k\ge 0:\left| X_k \right|+\left| Y_k \right|=3)$

b) Find the value of $P(X_T=-2,Y_T=0)$ and $E[T]$ if $T=inf(k\ge 0:max(-X_k,\left| Y_k \right|=2))$

Any help is greatly appreciated!

答:
I am not sure my answer for part a) is right, just for your reference.

a) First, it is important to observe that the point $(X_{T-2}, Y_{T-2})$ must be one of the four:
\begin{align}
(1,0),\,(-1,0),\,(0,1),\,(0,-1).\tag{1}
\end{align}
Now, if $(X_T, Y_T)=(3,0)$, then $(X_{T-2}, Y_{T-2})=(1,0)$. Consider (1), and by symmetry,
\begin{align}
P((X_{t-2}, Y_{T-2})=(1,0))=\frac{1}{4}.
\end{align}
On the other hand, if we start the random walk at $(X_{T-2}, Y_{T-2})=(1,0)$, then the point $(X_T,Y_T)$ must be one of the following five:
\begin{align}
\begin{array}{ll}
(1,2)&\mbox{with $1$ path},\\
(2,1)&\mbox{with $2$ paths},\\
(3,0)&\mbox{with $\color{red}{1}$ path},\\
(2,-1)&\mbox{with $2$ paths},\\
(1,-2)&\mbox{with $1$ path}.
\end{array}\tag{2}
\end{align}
Since each path has equal probability, we get that
\begin{align}
P(X_T=3, Y_T=0)=P((X_{t-2}, Y_{T-2})=(1,0))\cdot\frac{\color{red}{1}}{1+2+1+2+1}=\frac{1}{28}.
\end{align}

To calculate $E[T]$, we omit the detail, just note that $T\ge 3$, $T$ must be odd, and moreover, $(X_3,Y_3), (X_5,Y_5),\ldots,(X_{T-2},Y_{T-2})$ are one of the following four:
\begin{align}
(1,0),\,(-1,0),\,(0,1),\,(0,-1).\tag{3}
\end{align}
From the above observations, we may obtain
\begin{align}
P(T=3)&=\frac{28}{4^3}=\frac{7}{16},\\
P(T=5)&=\frac{9}{16}\cdot\frac{7}{16},\\
P(T=7)&=\left(\frac{9}{16}\right)^2\cdot\frac{7}{16},\\
&\vdots
\end{align}
and therefore,
\begin{align}
E[T]&=\sum_{n=3,5,7,\ldots} nP(T=n)\\
&=\frac{7}{16}\left(3+5\cdot\frac{9}{16}+7\cdot\left(\frac{9}{16}\right)^2+\cdots\right)\\
&=\frac{39}{7}.
\end{align}

Title: 2D random walk - first hits


问:
Radical of an ideal equals the intersection of all prime ideals containing it.

I am trying to prove that for a ring $R$ and an ideal $I\leq R$ we have
  $$
\sqrt{I}=\bigcap_{I\leq\mathfrak p}\mathfrak p,
$$
  the intersection of all prime ideals containing $I$.

The definition of the radical of an ideal I am working with is
$$
\sqrt{I}=\{x\in R\mid\exists m\in\mathbb{N} \text{ with }x^m\in I\}.
$$
I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?

答:
The one sentence answer is, for any $x \notin \sqrt{I}$,  consider the ring homomorphism

$$R \to (R/\sqrt{I}) \to (R/\sqrt{I})_x$$
pick any maximal ideal in $(R/\sqrt{I})_x$ it pulls back to a prime ideal of $R$ containing $\sqrt{I}$ and not containing $x$, which is what we wanted to find.

This answer is in some sense the same as the other given answer by Kobe but hides the Zorn's lemma argument in the fact that any nonzero ring has a maximal ideal:

Here is the argument in excessive detail:

To avoid having to write $\sqrt{I}$ all the time, let $J=\sqrt{I}$ so that $J$ is radical ideal (if $x^n \in J$ then $x \in J$). If $\mathfrak{p}$ is a prime ideal containing $I$ then it contains $J$ (if $x \in J$ then $x^n \in I$ so $x^n \in \mathfrak{p}$ so $x \in \mathfrak{p}$). So we want to show $$J \supseteq \bigcap_{J \subseteq \mathfrak{p}} \mathfrak{p}$$
where the intersection is over prime ideals.

As in Kobe's proof we show this by showing if $x \notin J$ then $x \notin \bigcap \mathfrak{p}$, which means we have to find a prime ideal $\mathfrak{p}$ containing $J$ but not containing $x$.

So assume $x \notin J$. The key insight is that $J$ being radical means $(R/J)_x$ is nonzero ring: if we had $0=1$ in this localization, then by definition of equality in localization, $\overline{x}^n\cdot 1=0$ in $R/J$ so $x^n \in J$, and since $J$ is radical, $x \in J$ contrary to assumption.

Now consider the composite of the ring canonical homomorphism
$$ R \to R/J \to (R/J)_x$$
Since $(R/J)_x$ is nonzero ring, it has a maximal ideal $\mathfrak{m}'$ (using Zorn's lemma), and consider its preimage $\mathfrak{p}$ in $R$...it need not be a maximal ideal (although it will be if $R$ is finite type over a field...that's some form of a bunch of results that go by Zariski's lemma) but it will a prime ideal since pre-images of prime ideals are primes and maximal ideals are prime. And by the corresponding between prime ideals in a ring and it quotients/localizations, $\mathfrak{p}$ is prime ideal of $R$ containing $J$ and not containing $x$ which is what we wanted to find.


问:
Product of independent random variables property
I need to prove that the joint distribution function of a vector random variable is equal to the product of distribution functions if and only if the random variables are independent

$$\xi_1, \xi_2  \text{- discrete random variables. } \vec{\xi} = (\xi_1, \xi_2). \text{ Prove that } F_\vec{\xi}(a_1,a_2) = F_{\xi_1}(a_1)F_{\xi_2}(a_2),\ a_1, a_2 \in \mathbb{R}\iff\ \xi_1, \xi_2 \text{ - independent.}$$

答:
It follows immediately from the definition of independence.  $X_1$ and $X_2$ are independent if $P(X1\in S_1 \cap X_2\in S_2)=P(X1\in S_1)\times P( X_2\in S_2)$ where $S_1$ and $S_2$ are arbitrary.

For your case $F_{X_k}(a_k)=P(X_k\le a_k)$ is given where $S_k$ consists of all possibilities $\le a_k$.

For the other direction, use let $S_k$ be a particular vale of $X_k$ and $T_k$ the next lower value of $X_k$. Then $P(S_1\cap S_2)=F(S_1,S_2)-F(S_1,T_2)-F(T_1,S_2)+F(T_1,T_2)$

where $F$ is the joint distribution.  Using $F(a_1,a_2)=F_{X_1}(a_1)\times F_{X_2}(a_2)$, we get $P(S_1\cap S_2)=P(S_1)\times P(S_2)$.


问:
Verifying the hyperbolic law of cosines and law of sines..
I am trying to understand hyperbolic triangles. These two laws are supposedly universal in the hyperbolic plane.

However, whenever I've tried to verify them using applets which let me construct hyperbolic triangles either in the Poincaré disk model, or the Poincaré half-plane model, they never seem to be valid. The construction below serves as an example:

$\sin(72.60867°) / \sinh(1.07)$ $=$ $0.74195\dots $ whereas $\sin(32.02868°) / \sinh(0.84)$ $=$ $0.56280\dots $

Similarly, you will find an inequality when you try to verify the cosine law.

Have I fundamentally misunderstood something? Why do these laws never seem to hold up on any triangle?

答:
Expanding a comment ...

This Hyperbolic Canvas applet uses a number of under-the-hood constructions to get at the hyperbolic distance between two points in the Poincaré model. Here's a screenshot of the constructions based on "input" points $A$ and $B$. (I relabeled various elements to make sense of them as I was going through the process.)

In a nutshell: We take a few steps to construct the orthogonal circle through $A$ and $B$ to determine ideal points $A'$ and $B'$. (There are conditional constructions to accommodate the circle being a diameter, etc.) Then the hyperbolic distance is calculated as
$$d(A,B) = \left|\log\frac{|AB'||A'B|}{|AA'||BB'|}\right| \tag{1}$$

where $|PQ|$ indicates the Euclidean distance (marked eucPQ in the screenshot) between $P$ and $Q$ in the model. This is all geometrically correct. Equation $(1)$ is the first formula given in the Distance section of Wikipedia's "Poincaré disk model" entry.

A more-direct route to the distance, also given in Wikipedia, is

$$d(A,B) = \operatorname{arccosh}\left(1 + \frac{2|AB|^2}{(1-|OA|^2)(1-|OB|^2)} \right) \tag{2}$$

which requires nothing more than coordinates of points $A$ and $B$. Showing that $(1)$ and $(2)$ are equivalent is a nice exercise.

Now, even though $(1)$ and $(2)$ are equivalent, something seems to go wrong in the applet's implementation of $(1)$. Accumulated computational error from the numerous inherently-approximate constructions are at least partly to blame, especially since the input (model) points are already "close" by virtue of living inside the unit circle. The error may be compounded by the construction's conditional logic, but I haven't looked that hard. (Why the applet doesn't use the direct calculation of $(2)$ —especially since this is invisible to the user— is beyond me.)

To see how bad things get, consider a triangle with points $A=(0.1,0.2)$, $B=(0.3,0.0)$, $C=(0.4, 0.5)$. Here it is in the applet:

The applet's "hyperbolic distance" tool gives
$$a := d(B,C) = 0.56397 \qquad b := d(C,A) = 0.46909 \qquad c := d(A,B) = 0.26031 \tag3$$
but calculating the Euclidean distances directly from the coordinates and entering them into $(2)$ gives
$$a = 1.29859 \qquad b = 1.08011 \qquad c = 0.59939 \tag4$$
Taking the calculated angles as correct, the Law of Sines applied to this second set of lengths gives
$$\frac{\sin 91.5^\circ}{\sinh 1.29859} = \frac{\sin 50.1^\circ}{\sinh 1.08011} = \frac{\sin 22.02^\circ}{\sinh 0.59939} = 0.5895 \tag5 $$
as expected from the Hyperbolic Law of Sines.

I can't actually say for sure that the applet's "hyperbolic distance" tool is using the construction shown at the top of this answer. (This was my first experience peering into a GeoGebra applet, and its author's under-the-hood labeling conventions weren't terribly clear. There are a handful of functions called "HyperbolicDistance". The construction shown was the first one.) Whatever the applet is doing to generate the distances in $(3)$, though, is obviously problematic because those distances are way off. I'll note that the applet also has trouble calculating angles when a vertex is at the origin.

So, all-in-all, the numerics of this particular applet are just plain untrustworthy, and the applet should probably only be used for a qualitative understanding of the Poincaré disk; eg, appreciating the broader nature of parallels, the non-constant angle sum, the curvy-looking (but not-actually-curvy) lines, and so forth.

P.S. When I want to draw a hyperbolic figure, I don't bother with special purpose macros in GeoGebra. I just construct what I need "from scratch" using standard GeoGebra tools. Perhaps the limitations of the applets will motivate OP to do the same thing; you're going to want to get good at this stuff, anyway. :)

For instance, while an applet might give you a one-click "hyperbolic line through two points" command, it's not at all complicated to construct the h-line from scratch: invert/reflect one of the points in the boundary circle, then invoke "circle through three points". Done! ... Well, there's an extra couple of steps to draw just the interior arc of the circle, but the point here is that a one-click command doesn't necessarily save you all that much work. (Indeed, the applet's path to the h-line is pretty complicated —just look at all those auxiliary elements!— so it may actually be doing a lot of unnecessary work.)


答:
Substitute $a=\frac x2-\frac2x $ into $(a+1)^4 = a^4+4a^3 + 6a^2 +4a +1 $ and note that only the three even terms contain constants, which are their middle coefficients

$$\left( \frac x2-\frac2x\right)^4 \implies 6,\>\>\>\>\>\>\>
6\left( \frac x2-\frac2x\right)^2 \implies -12,\
$$
Thus, the constant is $6-12+1=-5$.

Edit:

Same logic, the constant term for the general expression $\left (1+\frac x2 -\frac 2x \right)^{2n}$ is
$$c_{2n} = \sum_{k= 0}^{n}(-1)^k \binom {2n}{2k}\binom {2k}{k}
$$
which yields $c_4=-5$, $c_6= 41$, $c_{12}=3301$, $c_{20}=-1813055$, ... The summation has a close-form, but only in hypergeometric function $c_{2n} = {}_2F_1(1/2-n,-n;1,-4)$.


问:
Be $X$ a topological space and $D_1$, $D_2 \subseteq X$ dense in $X$.
Be $X$ a topological space and $D_1$, $D_2 \subseteq X$ dense in $X$. Can it be said that $D_1 \cap D_2$ is dense in $X$? And if $D_1$, besides dense, is also open in $X$.

If $D_1$ is dense in $X$ this means that $\overline{D_1}=X$. In the same way that $D_2$ is dense in $X$ this means that $\overline{D_2}=X$, so the intersection is also dense in X.

Is that correct? Can someone help me please?

答:
What about $X= \mathbb R$, $D_1 = \mathbb Q$ and $D_2= \mathbb R \setminus Q$?

... Where $D_1 \cap D_2 = \emptyset$.


问:
Is this problem NP-hard?
Is the following problem NP-hard?

Let $Ax=b$, where $a_{i,j} \in \{0,1\}, b_i \in \{0,1\}, x_j \in \{0,1\}$.

Decide, wether there is a solution or not.

答: Assuming you are given $A$ and $b$ and want to find $x$, this is the exact cover problem, which is one of the original 21 NP-complete problems.  Row $i$ corresponds to an element, and column $j$ corresponds to a subset.  The (constant) value of $a_{i,j}$ indicates whether element $i$ appears in set $j$.  The (variable) value of $x_j$ indicates whether set $j$ is selected.  If $b_i = 0$, then $x_j=0$ for all $j$ with $a_{i,j}=1$, and row $i$ can be removed.

Title: Is this problem NP-hard?


问:
Chain rule proof

Let $a \in E \subset R^n, E \mbox{ open}, f: E \to R^m, f(E) \subset U
 \subset R^m, U \mbox{ open}, g: U \to R^l, F:= g \circ f.$ If $f$ is
   differentiable in $a$ and $g$ differentiable in $f(a)$, then $F$ is
   differentiable in $a$ and $F'(a)=g'(f(a)) f'(a)$.

I have been given a short proof, but I do not understand every step of it.

We have $f(a+h) = f(a) + f'(a)h + o(|h|)$ and $g(f(a)+k) = g(f(a)) + g'(f(a))k + o(|k|)$ as $h,k \to 0$. Thus

$\begin{align} F(a+h)=(g \circ f)(a+h) &= g(f(a+h)) \\
&= g(f(a)+f'(a)h+o(|h|)) \\
&= g(f(a)) + g'(f(a))(f'(a)h+o(|h|)) + o(|f'(a)h+o(|h|)|) \\
&= (g\circ f)(a) + g'(f(a))f'(a)h+o(|h|) + o(O(|h|)) \\
&= F(a) + g'(f(a))f'(a)h+o(|h|) + o(|h|).
\end{align}$

That's all. I feel like there is some trickery involved with the little o en big o. My problems are:

Why does $f'(a)h+o(|h|) \to 0$ when $h\to 0$. I see that the first term tends to $0$, but how come $o(|h|) \to 0$?

$g'(f(a))o(|h|))=o(|h|)$.

$o(|f'(a)h + o(|h|)|)=o(O(|h|))$

$o(|h|) + o(|h|) = o(|h|)$.

This last one isn't included in the proof, but I expect that's would have been something trivial. I've been trying to work this out with the definitions of big o en little o, but to no prevail. It's confusion to me how I can do computations with big o and little o, since you can't really perform algebraic operations with them and it's more like a function property.

If anyone could show me how this would be done I'd be so thankful.

答:
The symbol $o(|h|)$ is like a variable for a function $g(h)$ that satisfies $$\lim_{h \rightarrow 0} \frac{g(h)}{|h|} = 0.$$ Multiplying any such function by a constant gives you another function in that class, so we write $co(|h|) = o(|h|).$ Similarly, adding any two functions in that class gives another function in that class, so $o(|h|) + o(|h|) = o(|h|).$

$O(f(x))$ stands for any function $g(x)$ so that there is some $M$ so that $|g(x)| \leq M|f(x)$ for $x$ near 0. We want to show $f'(a)h + o(|h|)$ is $O(|h|)$. What this means is that if $g(h)$ is a function in $o(h)$, then we need a constant $M$ so that $|f'(a)h + g(h)| \leq |h|$. But $$|f'(a)h + g(h)| \leq |f'(a)||h| + |g(h)|,$$ and since the ratio $\frac{g(h)}{h}$ goes to zero, there must be a neighborhood of $0$ for which $|g(h)| \leq |h|$. So $o(|f'(a)h + o(|h|)|)$ is $o(O(|h|)$.

Now to show $o(O|h|)$ is $o(|h|)$. Suppose that $g$ is $o(O|h|)$. Then there is a function $p$ that is $O(|h|)$ with $\lim \frac{g(h)}{p(h)} = 0$. Choose $M$ so that $|p(h)| \leq M|h|$ near $0$. Then $\frac{g(h)}{|h|} \leq \frac{Mg(h)}{p(h)} \rightarrow 0$.

If you wanted to, you could easily go through the proof and replace each instance of $o$ or $O$ by an actual function with the required properties.

Title: Chain rule proof


问:
How did this shorter method of finding $x^r$ in equation come?

Q)What is coefficient of $x^{30}$ in $$\left(\frac{1-x^{11}}{1-x}\right)^3\left(\frac{1-x^{21}}{1-x}\right)$$

I did this question by expanding $\frac{1}{1-x}$ and multiplying and got a large equation which in the end came to the correct solution of 1111.However, in the book the following steps are provided:

=Coefficient of $x^{30}$ in $(1-x^{11})^3(1-x^{21})(1-x)^{-4}$

=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22})(1-x^{21})(1-x)^{-4}$

=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22}-x^{21})(1-x)^{-4}$
Here,(according to my understanding) the terms greater than $x^{30} $are excluded as only the terms lower than $x^{30}$ can give the desired result(Expansion of $\frac{1}{1-x}=1+x+x^2+x^3+....$). However, in the next step, it equates the above to:

$$\frac{33!}{30!3!} +(-3)\frac{22!}{19!3!}+(3)\frac{11!}{8!3!}+(-1)\frac{12!}{9!3!}$$which also provides the correct solution. How did this equation come?

答:
$(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+...$

Basically, the coefficient of $x^r$ in $(1-x)^{-n} = \binom{n+r-1}{r}$ where n is a natural number

In your case $n=4$

What we want is:

(Coefficient of $x^{30}$ in $(1-x)^{-4}) -(3 \cdot $ Coefficient of $x^{19}$ in $(1-x)^{-4})+(3 \cdot$ Coefficient of $x^8$ in $(1-x)^{-4}) -($ Coefficient of $x^9$ in $(1-x)^{-4}$)

You can take it from here.


问:
Confusion about the definition of the Fisher information for discrete random variables
Suppose that $X$ has a absolutely continuous and positive probability density function $f(x)$, I know its Fisher information is defined through $$I(X) := \int_{\mathbb R} \frac{|f'|^2}{f} \mathrm{d}x.$$ However, I am very confused about the definition of $I(X)$ when $X$ is a non-negative integer-valued random variable with probability mass function $(p_0,p_1,\ldots,p_n,\ldots)$. In most references I have found so far, it seems that the definition of the Fisher information contains some unknown parameter $\theta$ (for example, if $X$ is Bernoulli distributed, then the parameter is the success probability in each trial). Since I am not familiar with statistics, I am very confused as to how should we define Fisher information $I(X)$ when $X$ is a non-negative integer-valued random variable with (unknown) probability mass function $(p_0,p_1,\ldots,p_n,\ldots)$. Basically, my motivation is that I have a Fokker-Planck equation for the evolution of the probability mass function of a non-negative integer-valued stochastic process, and I want to know how its Fisher information evolves in time. Thanks for any help or reference!

答:
For a discrete known probability mass function, there is no parameter $\theta$—you know the full distribution.  If however you know just the type or form distribution (such as a Gaussian, Bernoulli, etc.), you need to know the parameters (such as the sufficient statistics) in order calculate the Fisher Information (and other measures).  After all, the Fisher Information (and the mean, and the variance, and...) of a Gaussian distribution depends upon the mean and the standard deviation, which in your terminology is $\theta$.

In the discrete case, every textbook on information theory will give the discrete version of the definition, in which an integral is replaced by a sum, for instance,

$$I[{\bf P}] = \sum\limits_{i=1}^n \frac{(p_{i} - p_{i-1})^2}{p_i}$$

as shown here.


答:
When you expand out $(1-x)^{-4}$ you get a sum of the form
$$
(1+x+x^2+x^3+\cdots)^4=\sum_a\sum_b\sum_c\sum_d x^{a+b+c+d}\tag1
$$
where the integer indices $a,b,c,d$ run from $0$ to $\infty$. To multiply this on to the four terms in $(1-3x^{11}+3x^{22}-x^{21})$ and collect only powers of $x^{30}$, you'll want to fetch the coefficient of $x^{30}$ in (1) as well as the coefficient of $x^{19}$, and that of $x^8$ and $x^9$.

Think about how many ways you can achieve $x^{30}$ in (1). You need to find four nonnegative integers that sum to $30$. Using a stars-and-bars argument, the answer is ${33\choose 3}$. In general the coefficient on $x^r$ in (1) is ${r+3\choose 3}={r+3\choose r}$.

Title: How did this shorter method of finding $x^r$ in equation come?

Body:
Q)What is coefficient of $x^{30}$ in $$\left(\frac{1-x^{11}}{1-x}\right)^3\left(\frac{1-x^{21}}{1-x}\right)$$

I did this question by expanding $\frac{1}{1-x}$ and multiplying and got a large equation which in the end came to the correct solution of 1111.However, in the book the following steps are provided:

=Coefficient of $x^{30}$ in $(1-x^{11})^3(1-x^{21})(1-x)^{-4}$

=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22})(1-x^{21})(1-x)^{-4}$

=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22}-x^{21})(1-x)^{-4}$
Here,(according to my understanding) the terms greater than $x^{30} $are excluded as only the terms lower than $x^{30}$ can give the desired result(Expansion of $\frac{1}{1-x}=1+x+x^2+x^3+....$). However, in the next step, it equates the above to:

$$\frac{33!}{30!3!} +(-3)\frac{22!}{19!3!}+(3)\frac{11!}{8!3!}+(-1)\frac{12!}{9!3!}$$which also provides the correct solution. How did this equation come?


问:
How do I solve the following recursive equation?
$$\begin{cases}
    p(1) = \frac{1}{100} \\
    p(t) = p(t-1)\left(1+\frac{1}{101-t}\right)
\end{cases}$$

I feel like I could use the Telescoping Technique, but I do not quite see how.

Body:
$$p(t)=\frac{102-t}{101-t}\, p(t-1)$$
for $t=1$ initial condition is $p(1)=\frac{1}{100}=\frac{1}{101-1}$

for $t=2$ we have $p(2)=\frac{100}{99}\,p(1)=\frac{100}{99}\,\frac{1}{100}=\frac{1}{99}=\frac{1}{101-2}$

for $t=3$, $p(3)=\frac{99}{98}\,p(2)=\frac{100}{99}\,\frac{99}{98}\frac{1}{100}=\frac{1}{98}=\frac{1}{101-3}$

for general $t$ we have $$p(t)=\frac{1}{101-t}$$


问:
Derive Rodrigues’ formula for Laguerre polynomials

Derive Rodrigues’ formula for Laguerre polynomials
$$
L_n(x)=\frac{e^x}{n!}.\frac{d^n}{dx^n}(x^ne^{-x})
$$

The Rodrigues’ formula for Hermite polynomials  can be obtained by taking $n^{th}$ order partial derivatives of its generatig function
$$
g(x,t)=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}=1+tH_1(x)+\frac{t^2}{2!}H_2(x)+\cdots\cdots\cdots+\frac{t^n}{n!}H_n(x)+\cdots\cdots\cdots\\
\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)=H_n(x)+\frac{(n+1)n(n-1)\cdots2}{(n+1)!}tH_{n+1}(x)+\cdots\\
H_n(x)=\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)\Bigg]_{t=0}=e^{x^2}\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}\\
=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-x^2}\Big)\Bigg]
$$

Generating function for laguerre polynomials is $g(x,t)=\dfrac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^\infty L_n(x) t^n$

I do not think the same technique applies in the case of Laguerre polynomials. So how do I derive that for Laguerre polynomials ?

答:
If you are allowed to use the explicit representation for Laguerre polynomials,
$$ L_n(x) =\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{k!}  \quad ,$$
then the result follows easily from the Leibniz differentiation rule for products
$$ (f(x) \ g(x) )^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x) .$$

Refer to the Rodrigues formula you want to prove. Let $f(x)=x^n$ and $g(x)=e^{-x}.$ Then $f^{(k)}(x) = x^{n-k}\ n!/(n-k)! $ and
$g^{(n-k)}(x) = (-1)^{n-k}e^{-x}.$  Now clean it up and redo the sum in reverse order, i.e., $k \to n-k,$ and you'll get the top formula.


问:
A connected manifold can be covered by finite open subsets
In Vick's $\textit{An  Introduction to Algebraic Topology}$,

6.5 theorem states that “if M is a connected, noncompact n-manifold without boundary, then $H_n(M) = 0$".

Consequently, I have a question:
If M is a connected n-manifold without boundary, can M be covered by finite open subsets $\{U_1, \dots , U_k\}$ such that every $U_i$ is homeomorphic to $ \mathbb{R}^n$?

Even though I have tried many examples, but I didn't construct a counterexample. I don't know whether it is right.
Here are some related results:
(1)https://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover

(2)Surface where number of coordinate charts in atlas has to be infinite

答:
Good question. The answer is positive.

First of all, as in my answer here, there exist $n+1$ families ${\mathcal F}_1,..., {\mathcal F}_{n+1}$ of subsets in $M$ such that:

Each ${\mathcal F}_i$ is a union of closed (and tame) pairwise disjoint $n$-balls $B_{ij}, j\in J_i$, where each $J_i$ is (at most) countable.

$$\bigcup_{i=1}^{n+1} \bigcup_{j\in J_i} int(B_{ij})=M.$$

Each compact $K$ in $M$ intersects only finitely many closed balls in the above collections.

I will show that each
$$
V_i= \bigcup_{j\in J_i} int(B_{ij})
$$
is contained in an (open) subset $U_i$ of $M$ homeomorphic to $R^n$. This will do the job. To construct $U_i$, I will treat each $B_{ij}$ as a $0$-handle in $M$. I will identify each $J_i$ with an interval in ${\mathbb N}$ (finite or infinite). Next, for each pair of consecutive indices $j, j+1\in J_i$, connect $B_{ij}, B_{ij+1}$ by a 1-handle $H_{ij}$ in $M$ so that distinct 1-handles are pairwise disjoint and intersect only the balls $B_{ij}, B_{ij+1}$ in  ${\mathcal F}_i$ and only along disks in their boundaries. A 1-handle is a thickened (tame, simple) arc $a_{ij}$ in $M$ connecting boundary spheres of $B_{ij}, B_{ij+1}$. Making these arcs $a_{ij}$ pairwise disjoint is easier if $M$ has dimension $\ge 3$ (first take any locally finite collection of tame arcs and then away any accidental intersection). Constructing these arcs in the case of surfaces is not hard but tedious, I can explain how to do so if you like. (Connectivity of $M$ is used to ensure the existence of an arc connecting $B_{ij}, B_{ij+1}$.)

Informally, the union of the 0-handles $B_{ij}$ and $1$-handles $H_{ij}$ is a "chain of closed balls" (finite or infinite). Arguing inductively, one sees that each finite chain
$$
W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik}
$$
is homeomorphic to the closed $n$-ball. Similarly, each infinite chain
$$
W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik} \cup ...
$$
is homeomorphic to the closed $n$-dimensional half-space. The interior $U_i$ of each chain is homeomorphic to ${\mathbb R}^{n+1}$.

Thus,
$$
M= U_1\cup ...\cup U_{n+1}.
$$

A similar argument works in PL and smooth categories.


Title: What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$?

Body:
I did

\begin{align}
& \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t^3k+4tk}{(t^2+2)^2} \\[8pt]
= {} & \frac{-4tj+4tk}{(t^2+2)^2}
\end{align}

Is this true?

Body:
$(\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$

$\frac {1}{(t^2 + 2)^2} ((-2(t^2+2)+4t^2) \mathbf i + (4t^2+t^2(t^2+2))\mathbf j + (-2t^3-4t) \mathbf k)$

$\frac {1}{(t^2 + 2)^2} ((2t^2 - 4) \mathbf i + (t^4+6t^2) \mathbf j -(2t^3+4t) \mathbf k)$

$\frac {2t^2 - 4}{(t^2 +2)^2} \mathbf i + \frac {t^4+6t^2}{(t^2 + 2)^2} \mathbf j - \frac {2t^3+4t}{(t^2 + 2)^2} \mathbf k$


问:
Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$
How do I prove that:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$

Do I use induction?

答:
Prove the following claim using induction on $n$:
$$\sum_{k=1}^n \dfrac1{\sqrt{k}} < 2 \sqrt{n}$$

In the induction, you will essentially need to show that
$$2\sqrt{n} +\dfrac1{\sqrt{n+1}} < 2 \sqrt{n+1} \tag{$\star$}$$

To prove $(\star)$, note that
$$\sqrt{n} < \sqrt{n+1} \implies \sqrt{n} + \sqrt{n+1} <2 \sqrt{n+1} \implies \dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}$$
Multiplying and divding the right hand side by $(\sqrt{n+1} - \sqrt{n})$, we get
$$\dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}\cdot \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = 2({\sqrt{n+1} - \sqrt{n}})$$
which gives us  $(\star)$.


问:
Question regarding conditional probabilities
Suppose 10% of the people are good at art, and that 40% of the people who are good at art are also good at math. If a person is not good at art, they have only a 30% chance of being good at math. What is the probability that a person who is good at math will also be good at art?

Is my solution correct?:

Let M be person being good at math and A be good at art. Then A’ is being not good at art

P(A)=0.1

P(M|A)= 0.4

P(M|A’)=0.3

So P(M)=0.4+0.3=0.7

By bayes theorem: P(A|M)=0.4*0.1/0.7=0.06

答:
Sometimes a figure is worth 1000 words:

So consider just the "good at math" area.  What proportion of that is also "good at art"?


问:
Can someone help me with this sets question? (How many students knew no coding language?)
Q:
In a survey of a group of 100 computing students, it was found that
32 knew Java, 33 knew C, 32 knew Python,
5 knew both Java and C, 7 knew both C and Python while none of the
students knew both Java and Python. How many students knew none of
the three languages?

Attempt:

First attempt was I guessed 3 because it only adds up to 97 so I
assumed 3 students knew none of them.

Then I tried it out and got 27 for java, 21 for C and 25 for python,
added those up and got 72. Minus 72 from 100 and got 28 as my final
answer for students who know none.

My teacher has no marking scheme for this question, this is a practice
q. Anyone know the solution?

答:
Since no student knows both Java and Python, no student knows all three. Inclusion/exclusion then gives the number of students knowing at least one of the languages as
$$32+33+32-5-7-0+0=85$$
Subtracting from $100$ gives $15$ students not knowing any language.


答:
\begin{align}
i\times j & = k = -j\times i \\
j\times k & = i = -k\times j \\
k\times i & = j = - i\times k \\
i \times i & = j \times j = k\times k = 0
\end{align}


问:
Calculating Correlation given Two discrete random variables
Let $U$ and $V$ be discrete random variables with $U = −(V+1)/2$. What
is $Corr(U,V)$?

Do I substitute in $u$ as $(-v+1)$ and go from there?

答:
Yes.

We have

$$corr(U,V) = \frac{cov(U,V)}{\sqrt{DU \cdot DV}} = \frac{cov(-\frac{V+1}2,V)}{\sqrt{D(\frac{-V-1}{2}) \cdot DV}} = \frac{cov(-\frac{V}2,V)}{\sqrt{D(\frac{-V}{2}) \cdot DV}} =  \frac{-\frac12cov(V,V)}{\sqrt{\frac14 DV \cdot DV}} =$$
$$= \frac{-\frac12 DV}{\frac12 DV} = -1.$$


问:
How to prove that this series of functions is uniformly convergent?
Let $\{f_n\}$ be a sequence of functions on $[a, b]$ such that:
$f_n(x) \le 0$ if $n$ is even, $f_n(x) \ge 0$ if $n$ is odd;
$|f_n(x)| \geq |f_{n+1}(x)|$ for all $x$;
$f_n$ converges to $0$ uniformly.

I have to prove that the summation of $f_n$ is uniformly convergent.

How do I approach this problem? Where should I start?

答:
Ishraaq Parvez gave you the answer but I'm going to detail.
First the theorem about Alternating Series :

Hypothetis :

$(u_n)$ is alternate, which means either $((-1)^n u_n)_{n\in \mathbb{N}}$ or $((-1)^{n+1} u_n)_{n\in \mathbb{N}}$ is positive

$(|u_n|)$ is decreasing

$|u_n|\rightarrow 0$ when $n\rightarrow +\infty$

Result :

$\sum u_n$ converges

if we note $R_n = \sum\limits_{i=n+1}{+\infty} u_i$, then for all $n\in \mathbb{N}$, $|R_n|\leqslant |u_n|$

Then for your problem : it is clear that for all $x$, $(f_n(x))$ is alternate.
We also know that $\sum f_n$ uniformly convergent is equivalent to $R_n = \sum\limits^{+\infty}_{i=n+1} f_i $ uniformly converges to $0$.
By the theorem on alternate series, we have first that $\sum f_n$ exists, and also that for all $x$, $|R_n|\leqslant |f_n(x)|$. So we have that $\sup |R_n| \leqslant \sup |f_n|$. (you will have remarked that the sup are defined since we have convergence)

Since $(f_n)$ uniformly converges to $0$, $\sup |f_n| \rightarrow 0$.
So $\sup |R_n| \rightarrow 0$.
Thus $\sum f_n$ uniformly converges.


问:
$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$
Suppose that $X$ is an infinite set of cardinality $\alpha$.  Also, suppose that, for some $A \subseteq X$, we have that $|A| = |X\backslash A|$.  I want to show that $|A| = |X|$.

When, for example, $X = \mathbb{N} = \{1, 2, 3, \dots \}$, and $A = \{2n: n \in \mathbb{N}\}$, the requisite bijection $X\rightarrow A$ is about as obvious as it can be (since the definition of $A$ already provides one such bijection; in fact, in this case, the fact $|A| = |X\backslash A|$ is not even used.)  But I have a hard time pinning down the proof of the more general case stated above.

答:
It does require some of the axiom of choice, which may explain why you have a bit of a hard time proving that in the general case.

But if you assume that $X$ is a well-ordered set, then it's easy. Note that for infinite well-ordered sets $A,B$ it's true that $|A\cup B|=\max\{|A|,|B|\}$.

Note that the above follows from having $|A|+|A|=|A|$ for every well-ordered $A$ (and the fact that well-ordered cardinalities are all comparable). To prove that $|A|+|A|=|A|$ simply write $A=\{a_\alpha\mid\alpha<\kappa\}$ for some limit ordinal $\kappa$, then note that $$A_1=\{a_\alpha\mid\alpha=\delta+2n\text{ for a non-successor }\delta,\text{ and }n\in\omega\},\text{ and}\\
A_2=\{a_\alpha\mid\alpha=\delta+2n+1\text{ for a non-successor }\delta,\text{ and }n\in\omega\},$$

are both equipotent with $A$.


问:
What software/website can help me graph and get a diagram of an equation $f(x, y)$
Below is the type of equation I want a digital diagram/graph for:

$f(x,y) = a_0x^3 + a_1x^2y + a_2xy^2 + a_3y^3 + a_4x^2 + a_5xy + a_6y^2 + a_7x + a_8y$

To clarify, $a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7 ,a_8$ are known coefficients.

Please pardon my ignorance as I have just started.

Any help regarding how I can graph the solution (any website or software that can be useful) would be highly appreciated.

This is the actual polynomials with the coefficients substituted:
$$f(x, y) = -2.8553 x^3-10.431 x^2 y+0.28077 x^2-4.82487 x y^2+1.89407 x+\\+65.8761 x y-2.68436 y^3+14.7628 y^2-40.3066 y$$

These are the two new polynomials I need to graph:
$$ p(x, y) = -1.87019(x^3) -3.95625(x^2)(y) - 4.23593(x)(y^2) – 0.783805(y^3) + 4.39398(x^2) + 29.5375xy + 2.64509(y^2) – 4.64281x – 6.76092y $$

$$ q(x, y) = 8.53686(x^3) + 3.95625(x^2)(y) + 4.23593(x)(y^2) + 0.783805(y^3) – 34.394(x^2) – 29.5375xy – 2.64509(y^2) + 41.3095x + 6.76092y $$

Please teach me how to operate CalcPlot or GeoGebra because when I try to graph them myself, the scale is blown out of proportion and I do not know the right way to do it. Or if the graph could be shared by someone, I would really appreciate it as well.

答: Geogebra 3d is the easiest one, it is free and online. Otherwise you can use Matlab (or Octave) which can do much more than geogebra, but you have to learn their programming language</p>


问:
interior, closure and border on the Sorgenfrey line
[General Topology] Display, for each following case, int$(A)$, $\overline{A}$ and $\operatorname{Fr}(A)$ in the topology of the Sorgenfrey line.

Could you see if that's correct? And can you help me?
$\DeclareMathOperator{\int}{int}$

答:
The set $A=[0,1)$ is open in the Sorgenfrey topology, so $\operatorname{int}A=A$. $A$ is also closed, because its complement is open:

$$\Bbb R\setminus A=\bigcup_{n\in\Bbb Z^+}[-n,-n+1)\cup\bigcup_{n\in\Bbb Z^+}[n,n+1)\,.$$

Because $A$ is open, its complement is closed, so

$$\operatorname{bdry}A=(\operatorname{cl}A)\cap\operatorname{cl}(\Bbb R\setminus A)=A\cap(\Bbb R\setminus A)=\varnothing\,.$$

Now consider $\Bbb Z$. Clearly $\Bbb Z$ does not contain any set of the form $[x,y)$ with $x<y$, so as you say, $\operatorname{int}\Bbb Z=\varnothing$. But $\operatorname{cl}\Bbb Z$ cannot be empty, because it contains $\Bbb Z$. In fact $\Bbb Z$ is closed in the Sorgenfrey topology, so $\operatorname{cl}\Bbb Z=\Bbb Z$. One way to see this is to prove that if $\tau_S$ is the Sorgenfrey topology, and $\tau_E$ is the usual topology, then $\tau_E\subseteq\tau_S$. This isn’t hard: if $a<b$, then $(a,b)=\bigcup_{a<x<b}[x,b)$, which is open in $\tau_S$. And since every set that is open in the usual topology is also open in the Sorgenfrey topology, it follows that every set that is closed in the usual topology is closed in the Sorgenfrey topology. $\Bbb Z$ is closed in the usual topology, so it is closed in the Sorgenfrey topology. Finally, it’s clear that for any $n\in\Bbb Z$, every basic open nbhd $[n,x)$ of $n$ intersects $\Bbb R\setminus\Bbb Z$, so $\operatorname{cl}(\Bbb R\setminus\Bbb Z)=\Bbb R$, and $\operatorname{bdry}\Bbb Z=\Bbb Z$.

We can consider $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$ together. Clearly $\Bbb Q$ also does not contain any set of the form $[x,y)$ with $x<y$, so $\operatorname{int}\Bbb Q=\varnothing$. The same is true of $\Bbb R\setminus\Bbb Q$. Every basic open set $[x,y)$ contains a non-empty open interval $(x,y)$, which contains both rational and irrational numbers, so $[x,y)\cap\Bbb Q\ne\varnothing\ne[x,y)\cap(\Bbb R\setminus\Bbb Q)$. That is, both $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$ are dense in $\Bbb R$ in the Sorgenfrey topology, so $\operatorname{cl}\Bbb Q=\Bbb R=\operatorname{cl}(\Bbb R\setminus\Bbb Q)$. And that implies that $\operatorname{bdry}\Bbb Q=\operatorname{bdry}(\Bbb R\setminus\Bbb Q)=\Bbb R$.

Now let $A=(0,1)$. $A$ is open in the usual topology, so it is open in the Sorgenfrey topology as well, and $\operatorname{int}A=A$. $\Bbb R\setminus[0,1]$ is open in the usual topology, so it is also open in the Sorgenfrey topology; this means that no point of $\Bbb R\setminus[0,1]$ can be in the Sorgenfrey closure of $(0,1)$. We know that $A\subseteq\operatorname{cl}A$, so the only points that remain to be considered are $0$ and $1$. $[1,2)$ is a Sorgenfrey open nbhd of $1$ disjoint from $A$, so $1\notin\operatorname{cl}A$. However, you can easily verify that every Sorgenfrey open nbhd of $0$ intersects $A$, so $0\in\operatorname{cl}A$. Thus, $\operatorname{cl}A=[0,1)$. The only point of $[0,1)$ that is in the Sorgenfrey closure of $\Bbb R\setminus(0,1)$ is $0$, which is actually in $\Bbb R\setminus(0,1)$, so $\operatorname{bdry}A=\{0\}$.

Now that I’ve gone through the first five in some detail, see if you can do (f) yourself.


问:
Beginner's question: points on exponential curves
I'm new here, pretty much a maths beginner but with a huge interest in this art, English not being my first language, and I have a lot of questions that I hope people will not laugh at me for (especially for the terminology).

My very first one is:

I wanted to learn how to calculate an exponential function based on certain points on the curve, and how to determine a value on an exponential curve based on that underlying function. In this case, I wanted a curve where y increases five-fold everytime x increases by $20$. Example: $[x = 20; y = 5], [x = 40; y = 25], [x = 60; y = 125], [x = 80; y = 625], [x = 100; y = 3125]$.

But I neither understand how to put that into a formula, nor how to, i.e., calculate $y$ for, say, an $x$ of $25$ or $30$.

答: You want $y$ to get multiplied by $5$ when $x$ increases by $20$, or $\frac{x}{20}$ increases by $1$. This is can be achieved by the function $$y=5^{\frac{x}{20}}$$ Just input an $x$-value to get the corresponding $y$-value in this.


问:
Evaluate the integral $\int_0^{\pi/2}\frac{\tan x}{\tan(\frac{x}{2})} \, dx$
Evaluate the integral.
$$\int_0^{\frac{\pi}{2}}\frac{\tan{x}}{\tan(\frac{x}{2})}\,dx$$

I tried to solve it with $u = \tan{x/2}$, but i got divergent part of the solution. How can I integrate it, such that, when boundaries are plugged in, the result will be able to be calculated.

Body:
hint

with $ t=\tan(\frac x2) $,

$$dx=2\frac{dt}{1+t^2}$$

$$\tan(x)=\frac{2t}{1-t^2}$$

$ I $ becomes
$$I=4\int_0^1\frac{dt}{(1-t^2)(1+t^2)}$$

$$=2\int_0^1(\frac{1}{1-t^2}+\frac{1}{1+t^2})dt$$

$$=\int_0^1(\frac{1}{1-t}+\frac{1}{1+t}+\frac{2}{1+t^2})dt$$
$$=+\infty$$
$ I $ is then divergent.


答:
Hint (almost solution):

Put $u = \tan \frac{x}2$. Hence $x = 2 \arctan u$ and

\begin{align}
\text{answer} & = \int_0^{\tan \frac12} \frac{\left( \frac{2u}{1-u^2} \right)}{u} \cdot \frac{2\,du}{1+u^2} \\[6pt]
& = 4\int_0^{\tan \frac12} \frac{du}{(1+u^2)(1-u^2)} \\[6pt]
& =\int_0^{\tan \frac12} \frac{A\,du}{1+u^2} + \int_0^{\tan \frac12} \frac{B\,du}{1+u^2}
\end{align}
for some $A$ and $B$.

Title: Evaluate the integral $\int_0^{\pi/2}\frac{\tan x}{\tan(\frac{x}{2})} \, dx$

Body:
Evaluate the integral.
$$\int_0^{\frac{\pi}{2}}\frac{\tan{x}}{\tan(\frac{x}{2})}\,dx$$

I tried to solve it with $u = \tan{x/2}$, but i got divergent part of the solution. How can I integrate it, such that, when boundaries are plugged in, the result will be able to be calculated.


答:
Take y = $ae^{\alpha x} + c$ as it is an exponential function

y increases by the power of 5 whenever x increases by 20. So,

$\alpha$ is $\frac{1}{20}log_e(5)$. ($log_e5$ was multiplied to change the base from e to 5. If it increased by a power of 10, multiply by $log_e10$, which is roughly 2.303)

or, y = a.$5^{x/20}$ + c

Plug the numbers you gave in this equation, and you will find that a and c are 1 and 0 respectively. Now you can choose any value of x to get y.

Also don't ever feel shy of asking questions. We are here to help :)

Title: Beginner's question: points on exponential curves

Body:
I'm new here, pretty much a maths beginner but with a huge interest in this art, English not being my first language, and I have a lot of questions that I hope people will not laugh at me for (especially for the terminology).

My very first one is:

I wanted to learn how to calculate an exponential function based on certain points on the curve, and how to determine a value on an exponential curve based on that underlying function. In this case, I wanted a curve where y increases five-fold everytime x increases by $20$. Example: $[x = 20; y = 5], [x = 40; y = 25], [x = 60; y = 125], [x = 80; y = 625], [x = 100; y = 3125]$.

But I neither understand how to put that into a formula, nor how to, i.e., calculate $y$ for, say, an $x$ of $25$ or $30$.


问:
Equation for number of permutations
Not quite sure about the terminology here, but I just used permutation. Imagine you have a amount of dots and b amount of loops, where $a \gt b$. Now imagine you put those loops around the dots, and see how many permutations/combinations you get. Is there an equation to give you the answer if both a and b are  known? Here are some pictures to illustrate what I mean.

Left image: $a=4, b=3$.

Right image: $a=5, b=3$

So, is there an equation where these answers could be attained without drawing them out? (please excuse my bad drawing skills)

答:
The answer is:  $${a \choose b}$$

Read "$a$ choose $b$."  (Look up "combinations.")

Title: Equation for number of permutations


问:
1-forms vs k-forms vs Vector Fields
I'm trying to get a comprehensive understanding of similarities and differences of differential forms and vector fields. As I understand, 1-forms are analogous to vector fields in that for a vector field $X \in \Gamma (TM)$ and a 1-form $\xi \in \Gamma (T^*M)$ defined on an n-dimensional manifold $M$ we have

$X: M \rightarrow TM$

$\xi: M \rightarrow T^{*}M$

where $X$, $\xi$ are sections of $TM$ and $T^{*}M$ (resp.), and $X(p)$, $\xi (p)$ are tangent vectors and cotangent vectors (resp) for any $p \in M$.

With this correspondence of 1-forms and vector fields, I'm wondering if there is an equivalent object from the world of vector fields which corresponds to $k$-forms for $k>1$. For a $k$-form $\omega$ defined over $M$ we have

$\omega: M \rightarrow \Lambda^{k}T^{*}M$.

This $k$-form is created by some operation on 1-forms (wedge product), is there an analogous operation defined on vector fields to create some "$k$-vector field"?

I find that in differential geometry there are so many different objects living in different spaces, I am trying to simplify and draw similarities where possible. This is what motivates the above question.

答: A section of $\bigwedge^kTM$ is called a polyvector field (or multivector field) of degree $k$. Just as you can take the wedge product of differential forms, you can take the wedge product of polyvector fields. Furthermore, there is also the Schouten-Nijenhuis bracket which extends the Lie bracket of vector fields.

Title: 1-forms vs k-forms vs Vector Fields


答: Write $$\frac{\tan x}{\tan \frac{x}{2}} = \frac{\sin x}{\cos x} \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\sin\frac{x}{2} \cos\frac{x}{2}}{\cos x}\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\cos^2\frac{x}{2}}{\cos x} = \frac{1 + \cos x}{\cos x}$$Then your integral reduces to $$\int_0^{\frac{\pi}{2}} (\sec x + 1)\, dx$$


问:
A basic doubt on Lebesgue integration
Can anyone tell me at a high level (I am not aware of measure theory much) about Lebesgue integration and why measure is needed in case of Lebesgue integration? How the measure is used to calculate the horizontal strip mapped for a particular range? </p>

答: The idea of (exterior) measure is required to define the integral of the characteristic function of a measurable set. Without this basic building block you cannot even define the Lebesgue integral.</p>

Title: A basic doubt on Lebesgue integration


答:
You're simply selecting $b$ dots out of $a$. So number of ways is $$\binom{a}{b}=\dfrac{a!}{b!(a-b)!}$$

where $k!=k\times (k-1)\times \cdots \times2\times1$

Thus for your illustrations $$\binom{4}{3}=\dfrac{4!}{3!1!}=4$$ and $$\binom{5}{3}=\dfrac{5!}{3!2!}=10$$

Title: Equation for number of permutations

Body:
Not quite sure about the terminology here, but I just used permutation. Imagine you have a amount of dots and b amount of loops, where $a \gt b$. Now imagine you put those loops around the dots, and see how many permutations/combinations you get. Is there an equation to give you the answer if both a and b are  known? Here are some pictures to illustrate what I mean.

Left image: $a=4, b=3$.

Right image: $a=5, b=3$

So, is there an equation where these answers could be attained without drawing them out? (please excuse my bad drawing skills)


问:
Limiting Poisson behaviour in an urn-ball problem
This is problem from K.R. Parthasarathy's book  "Introduction to probability and measure."

Consider the placement of r indistinguishable balls in n urns. The sample space here consists of the $\displaystyle {n+r-1 \choose r}$ n-tuples of non-negative integers that add up to r, with the ith entry of the n-tuple denoting the number of balls in urn i. Each point in the sample space is assigned an equal probability of $\displaystyle \dfrac{1}{{n+r-1 \choose r}}.$

Let $p_{n,r}(k)$ denote the probability that exactly $k$ urns are empty.

It can be shown that for each $n$ and $r$ we have
$$
p_{n,r}(k) = \sum_{j = k}^n (-1)^{j-k} { j \choose k} {n \choose j} \frac{ {n-j+r-1 \choose r}  }{{n+r-1 \choose r}}.
$$

For each positive integer $n$, we are given a positive integer $r_n$ such that $\displaystyle \frac{n^2}{r_n} \to \lambda > 0$ as $n \to \infty$.

We are asked to show $\displaystyle p_{n,r_n}(k) \to \exp(-\lambda) \frac{\lambda^k}{k!}$, and I need some help in proving this limiting behavior.

Some Observations

One can see that
$$
p_{n,r}(k) = \frac{1}{k!} \sum_{j = k}^\infty \frac{(-1)^{j-k}}{(j-k)!} q_{n,r}(j),
$$
where $$q_{n,r}(j) =
\begin{cases} \dfrac{n!}{(n-j)!}\frac{{\genfrac {(} {)} {0pt}{0} {n-j+r-1} {r} }} {{\genfrac {(} {)} {0pt} {0} {n+r-1}{r} }}, \mbox{} \text{ for } 0 \leq j \leq n-1 \\
0, \mbox{} \text{ otherwise}
\end{cases}
$$
So it is sufficient to show that $q_{n,r_n}(j) \leq M^j$ for some $M > 0$ independent of $n$ and $j$ and $q_{n,r_n}(j) \to  \lambda^{j}$ as $n \to \infty$, for then we can take limit inside the summation sign and see that $\lim_{n\to\infty} p_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lim_{n \to \infty} q_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lambda^j = \exp(-\lambda) \frac{\lambda^k}{k!}.$

Body:
\begin{align}q_{n,r}(j)
&= \frac{n!}{(n-j)!}\frac{(n-j+r-1)!}{(n+r-1)!} \frac{(n-1)!}{(n-1-j)!}
\\
&= \frac{[n(n-1) \cdots (n-j+1)][(n-1)(n-2) \cdots (n-j)]}{(n+r-1)(n+r-2) \cdots (n+r-j)}
\\
&= \prod_{m=1}^j \frac{(n+1-m)(n-m)}{n+r-m}
\\
&= \prod_{m=1}^j \frac{(1 + \frac{1-m}{n})(1-\frac{m}{n})}{\frac{r}{n^2} + \frac{n-m}{n^2}}
\\
&\overset{n \to \infty}{\longrightarrow} \prod_{m=1}^j \frac{1}{1/\lambda}
\\
&= \lambda^j.
\end{align}


问:
Proving that a function is surjective and the equality holds
This is exercise 2.76 (ii) from Rotman's "Introduction to the theory of Groups".

I was able to do (i) but I have no idea how to proceed to prove (ii).

答:
You know that:
$$[G:H_1\cap\dots \cap H_n]=[G:H_i][H_i:H_1\cap\dots \cap H_n]$$

Hence $[G:H_i] \;\; | \;\; [G:H] \; \forall i$, but they are coprime so:

$$\prod_{i=1}^{n} [G:H_i] \;\; | \;\; |G/H_1\cap\dots \cap H_n| $$

But $G/(H_1\cap\dots \cap H_n)\sim Im(\varphi) \leq G/H_1\times\dots \times G/H_n \;$ by first isomorphism theorem. So $|G/H_1\cap\dots \cap H_n|=|Im(\varphi)|\;\; | \;G/H_1\times\dots \times G/H_n;|= \prod_{i=1}^{n} [G:H_i] \implies \prod_{i=1}^{n} [G:H_i] = |G/H_1\cap\dots \cap H_n|$.


问:
Markov's Chain stationary matrix determination
I have the following transition matrix $P$

\begin{equation}
P=
\begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix}
\end{equation}

I have tried to calculate the stationary matrix as
\begin{equation}
P^n= C D^n C^{-1}
\end{equation}

With $D=C^{-1} P C$ the diagonal matrix. Calculating $C$ with the eigenvalues and eigenvectors

\begin{equation}
C=
\begin{pmatrix}
v_1 & w_1 & z_1\\
v_2 & w_2 & z_2\\
v_3 & w_3 & z_3
\end{pmatrix}
\end{equation}

where the eigenvectors are

\begin{align}
\vec{v} =\langle (v_1,v_2,v_3)\rangle &= \langle(1,1,1)\rangle \\
\vec{w} =\langle(w_1,w_2,w_3)\rangle &= \langle\left(\frac{2i\sqrt{39}+3}{6},\frac{-3i\sqrt{39}-7}{16},1\right)\rangle \\
\vec{v} =\langle(v_1,v_2,v_3)\rangle &= \langle\left(\frac{-2i\sqrt{39}+3}{6},\frac{3i\sqrt{39}-7}{16},1\right)\rangle \\
\end{align}

so the diagonal matrix is
\begin{equation}
D=
\begin{pmatrix}
1 & 0 & 0\\
0 & \frac{-i\sqrt{39}+39}{120} & 0\\
0 & 0 & \frac{i\sqrt{39}+39}{120}
\end{pmatrix}
\end{equation}

how can i calculate $D^{n}$? is the diagonalization correct?

I tried by other way using $BP=B$ where B is the final vector state, then

\begin{align}
[p_1 \quad p_2 \quad p_3]  \begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix} = [p_1 \quad p_2 \quad p_3]
\end{align}

using Gauss-Jordan elimination for solving this system it gives me infinite solutions.

I dont know if it is a problem with the transition matrix.

答:
If you want to get a stationary distribution then it's needed to solve the system, consisting of these relations:
$$(p_1, p_2, p_3) P = (p_1, p_2, p_3)  $$
$$p_1 + p_2 + p_3 =1$$
$$p_i \ge 0, 1 \le i \le 3.$$
If you will get more than one solution it's normal: e.g. if $P$ is identity matrix then every distribution is stationary.

Everything that you did with $D^n$ may be useful for limit distribution but not for stationary one.


问:
Matrix for a linear map between 2 different basis
Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$

and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the basis $\begin{pmatrix}
2\\
3
\end{pmatrix}, \begin{pmatrix}
1\\
2
\end{pmatrix}$ for $K^2$

Question

Find the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain.

My effort so far

I am thinking that I first have to evaluate the vectors in the basis for $K^3$ and thus get some scalar. I then multiply the scalar on the basis for $K^2$. I.e If I evaluate $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}$ I get $(5,-1)$. I would then have to multiply that onto both my basis vectors i.e $5(2,3)+-1(1,2)=(10,15)+(-1,-2)=(9,13)$. That would give my first column in the matrix for the representation and thus I would end up with a matrix with 2 rows and 3 columns. This sounds wrong to me and I just want to have a quick check.

答:
HINT: You should not multiply your basis vectors with $(5,−1)$, but rather express $(5,-1)$ in terms of your basis vectors.

SOLUTION:

 $f(-1,0,2)=(5,-1)=\textbf{11} \cdot (2,3) + \textbf{(-17)} \cdot (1,2)$
 $f(3,1,-1)=(2,10)=\textbf{(-6)} \cdot  (2,3) + \textbf{14} \cdot (1,2)$
 $f(1,-1,1)=(2,2)=\textbf{2} \cdot (2,3) + \textbf{(-2)} \cdot (1,2)$
 Let $B_1=((-1,0,2),(3,1,-1),(1,-1,1)), B_2=((2,3),(1,2))$.
 Then, $[f]^{B_1}_{B_2}=\begin{bmatrix} 11 & -6 & 2\\ -17 & 14 & -2 \end{bmatrix}$

EDIT:
How does one express $(5,-1)$ in terms of $(2,3)$ and $(1,2)$?

Since $\text{span}(\{(2,3),(1,2)\})=K^2$ and $(5,-1)\in K^2$, we know there exist $a,b \in K$ such that $$(5,-1) = a \cdot (2,3) + b \cdot (1,2)$$
Applying the scalar multiplication, we have the following: $$(5,-1)=(a \cdot 2, a \cdot 3) + (b \cdot 1, b \cdot 2)$$
Adding the two 2-tuples together, we have $$(5,-1)=(a \cdot 2 + b \cdot 1, a \cdot 3 + b \cdot 2)$$
This implies:
$$2\cdot a+b=5$$
$$3\cdot a + 2 \cdot b = -1$$

I'll leave the rest to you.


问:
Find generating function for linear recurrence
Let $a_0=0, a_1=1,$ and $a_{n+2}=2a_{n+1}+a_n.$ Find the closed form for this sequence using generating functions.

I'm not sure what to do, I tried getting some cancellation like when solving for Lucas/Fibonacci numbers but I'm not sure how to.

答:
characteristic equation:

$x^2=2x+1$

$x = 1 \pm \sqrt 2$;

the homogeneous form: will be $$a_n = r(1 + \sqrt 2)^n + s(1 - \sqrt 2)^n$$

the particular solution: since $a_0$ = 0 and $a_1$ = 1 then $r + s = 0$ and $(1 + \sqrt 2)r + (1 - \sqrt 2)s = 1$

$$r=\frac{\sqrt 2}{4}, s=-\frac{\sqrt 2}{4}$$

the general solution:

$$a_n = \frac{\sqrt 2}{4} \left(1 + \sqrt 2\right)^n -\frac{\sqrt 2}{4} \left(1 - \sqrt 2\right)^n$$