问:
Understanding how EM algorithm actually works for missing data
I am currently studying EM algorithm for handling missing data in a data set. I understand that the final goal of EM algorithm is not to impute data, but to calculate the parameters of interest. However, when I am thinking of a practical example I am really struggling to make sense of the E-step: calculate $Q(\theta, \theta^{(k)}) =  \operatorname{E_{Z|X,\theta^{(k)}}} [\ln{p_{XZ}(x, z| \ \theta)}] \nonumber = \int_{\mathcal{Z}} {\ln{p_{XZ}(x, z| \ \theta)} \  p_{Z|X}(z| \ x, \theta^{(k)}) \ dz}$, where $x$ are the realization of $X$ (i.e. the observed variables) and $z$ are the realization of $Z$ (i.e. the variables with missing values).

Now suppose that we have a data set, where all the instances come from a random variable $Z$ which has a bivariate normal distribution, and also suppose that several data are missing in this data set. How will E-step will work in this case (where $\ln{p_{XZ}(x, z| \ \theta)} = \ln{p_{Z}(z| \ \theta)}$ and $p_{Z|X}(z| \ x, \theta^{(k)})=\ln{p_{Z}(z| \ \theta)}$)? Does an imputation for missing values takes place in E-step and if yes how they are imputed? Will they be imputed by their current value of mean parameter?

答:
Having studied this, I would remark that there is quite a lot going on with the EM algorithm if it's unfamiliar. Yours is an interesting question - some of the presentations I have seen on the EM algorithm sometimes don't explicitly specify how imputation of missing data is handled.

Having seen the presentations in some of the more well-known machine learning/statistics textbooks and tutorial papers, I would say that the clearest way in which you can see how imputation of missing values occurs in the EM algorithm is the viewing how it works through expected sufficient statistics.

Missing data imputation using the EM algorithm.

You are entirely correct that the EM algorithm is for maximum-likelihood estimation in the presence of latent variables (which can defined to be missing data), and that imputation/inference of these latent variables is a subroutine for parameter estimation.

And yes, you are correct that you use current estimates of the parameters $\theta^{k}$ (or an initial guess when $k=0$) to impute what the latent variables might be.

There are two senses in which the imputation of the missing data occurs:

In the E-step, when you compute the conditional distribution of the missing data $Z$, given the observed variables $X = x$, and current parameter estimates $\theta = \theta^{k}$, that is $p_{Z | X}(Z | X = x, \theta = \theta^{k})$. This is imputation in a probabilistic sense, and using Bayesian semantics, you are computing a posterior distribution on the values that missing data $Z$ can take given observed variables and current parameter estimates.

However, if you mean imputation in the sense of "filling in the missing data" using an appropriate point prediction, that occurs in both the 2nd part of the E-step and also the M-step.

In the E-step, when you come to compute the expected complete log-likelihood under the conditional distribution $p_{Z | X}(Z | X = x, \theta = \theta^{k})$, you will need to compute expected sufficient statistics, which will require not only filling in the missing values $Z$ with the mean of the conditional distribution/posterior $p_{Z | X}(Z | X = x, \theta = \theta^{k})$, but also requires the variance of the posterior.

In the M-step, these expected sufficient statistics are used to update your parameters.

If you are unfamiliar with the expected sufficient statistic viewpoint, I refer you to pages "Section 11.6 Fitting models with missing data", p374-p376, of Machine Learning: A Probabilistic Perspective by Kevin Murphy, which shows you how to deal with the specific context of missing data EM in the multivariate Normal rather than bivariate case (the multivariate case seems to be the norm in most of the main ML textbooks)

Lastly, I would remark that your confusion may be arising from a faulty partitioning of your data into missing and observed variables; and the formulae you have specified in brackets are incorrect due to this faulty partitioning. Again, I refer you to Murphy, but here is the gist of what happens for multivariate Normal distributions. The exposition I've added is from my handwritten notes as Murphy is not entirely clear, in my view, on how the partitioning works.

Our data matrix, is say $\mathbf{Y} \in \mathbb{R}^{D \times N}$, where the $n$th observation of $D$ input variables is a D-dimensional vector $y_n \in \mathbb{R}^D$. The case where $y_n$ is a random vector that has a multivariate Normal distribution, $y_n \sim N(\mu, \Sigma)$ where $\mu \in \mathbb{R}^{D}$ and $\Sigma \in \mathbb{R}^{D \times D}$ is often used to show how this works analytically.

Each of the $y_n$ will have missing components, which we denote as $z_n$; and also observed components which we denote as $x_n$ (you can think of $x_n$ and $z_n$ as sub-vectors of $y_n$). We can now consider partitioning the $y_n$ into an observed sub-vector $x_n$, and missing data sub-vector $z_n$. We assume that each $x_n$ and $z_n$ have a joint Normal distribution, meaning that we can specify the following:

$$y_n =
\begin{bmatrix}
x_n \\
z_n \\
\end{bmatrix}
\sim N \left(
\begin{bmatrix}
\mu_{n, x} \\
\mu_{n, z} \\
\end{bmatrix},
\begin{bmatrix}
\Sigma_{n, xx}, \Sigma_{n, xz} \\
\Sigma_{n, zx}, \Sigma_{n, zz} \\
\end{bmatrix} \right)$$

Where $\mu_{n, x}$ and $\mu_{n, z}$ are sub-vectors of $\mu$ and $\Sigma_{n, xx}, \Sigma_{n, xz}, \Sigma_{n, zx}, \Sigma_{n, zz}$ are sub-matrices of $\Sigma$ using the relevant dimensions defined by $x_n$.

We can now use standard formulae for multivariate Normal conditional distributions to specify the posterior used in the E-step:

$$p(z_n | x_n, \mu, \Sigma) \sim N(m_n, V_n)$$

where the posterior mean and variance are:

$$m_n = \mu_{n, z} + \Sigma_{n, zx}\Sigma_{n, xx}^{-1}(x_n - \mu_{n, x})$$

$$V_n = \Sigma_{n, zz} - \Sigma_{n, zx}\Sigma_{n, xx}^{-1}\Sigma_{n, xz}$$

We now use the posterior mean $m_n$ and variance $V_n$ to compute the expected sufficient statistics $\mathbb{E}[x_n]$ and $\mathbb{E}[x_n x_n^T]$. These are used to compute the $Q(\theta, \theta^{k})$ function in the E-step, and to compute maximum likelihood parameter updates in the M-step, that is, $\mu^{k+1}$ and $\Sigma^{k+1}$.

Further references.

Whilst the expected sufficient statistics viewpoint, in my opinion, is excellent for seeing how data imputation is working in the EM algorithm, it is not as appropriate for understanding why it is we use the expected complete log-likelihood in the first place, nor why we can iteratively maximise this as a surrogate for the observed data/incomplete log-likelihood. The clearest exposition I have seen on this aspect of the EM algorithm is in Chapter 11 of a unpublished book callsed "Probabilistic Graphical Models" by the legendary Michael Jordan, which you can find here.


问:
Split the matrix 8-ways PUZZLE
I have a problem that has been bugging me for the last month, there is a matrix with 8x8 squares, so 64 squares, and with 8 balls placed randomly each in a square. I need to find the solution of how the matrix should be split into 8 parts such that each ball is in a different part and each part has exactly 8 squares.
By the way, each "puzzle" can have multiple, one, or no solutions.

Before:

After:

答:
You can model this as a set partitioning problem as follows.  Let $S$ be the set of all connected sets of cardinality 8 that contain exactly 1 ball.  For the sample puzzle above, it turns out that $|S|=23617$.  Let $N=\{1,\dots,8\}$.  For $(i,j) \in N \times N$, let $S_{i,j} \subset S$ be the subset of sets that contain cell $(i,j)$.  For $s\in S$, let binary decision variable $x_s$ indicate whether set $s$ is selected.  The constraints are:
$$\sum_{s\in S_{i,j}} x_s = 1 \quad \text{for $(i,j)\in N \times N$}$$

There are multiple solutions, including this one:

ContentLicense: CC BY-SA 4.0

Title: Split the matrix 8-ways PUZZLE


问:
Alternative approach to prove that if a sequence is a Cauchy Sequence, then it converges.
I am familiar with the traditional proof that if a sequence of real numbers is a Cauchy sequence, then it is convergent. However, I am stuck on a problem that asks me to prove the implication using the fact that if all the subsequences of a sequence converge to a real number then the sequence itself converges.

Question: (i) Let $(a_n)$ be a bounded sequence of real numbers and let $x \in \mathbb{R}$ have the property that every subsequence of $(a_n) \neq (a_n)$ converges to $x$. Show that $(a_n) \to x$. (ii) Furthermore, use this result to prove that if a sequence is a Cauchy sequence, then it converges.

Proving part (i) was fairly easy for me. Outlining a proof for part (ii) is tripping me up. I am having a hard time applying the properties of a Cauchy sequence to the proof other than the fact that this implies the sequence is bounded.

Any help is greatly appreciated.

答:
For (i), simply let $b_n:=a_{n+1} ,\forall n\ge1.$ Then by the assumption, $(b_n)$ is a convergent subsequence of $(a_n)$ , and it implies obviously that $(a_n)$ is a convergent sequence by the definition of limits.

For (ii), try to argue by contradiction with the definitions of convergent sequences, limits, and Cauchy sequences.

 The basic idea is that, if a sequence $(a_n)$ is divergent, then no matter how rear in the "tail"(i.e. $\{a_n\,\vert n\ge k, k$ large$\}$, there always exist two elements of it which are not close enough. However, every two terms with sufficiently large indexes of a Cauchy sequence can get sufficiently close. This is the contradiction.


问:
Trigonometry Addition Thereom
Using the expansion of

a. $\sin(+)$, prove that $\sin75°=\sqrt 6+\sqrt{24}$

b. $\sin(+)$, prove that $\tan75°=2+\sqrt 3$

Where to start? draw up triangle of sin 75? find other values? help please.

答:
Hint

$\sin(75^\circ) = \sin(45^\circ+30^\circ) = \sin(45^\circ) \cdot \cos(30^\circ) + \cos(30^\circ) \cdot \sin(45^\circ)$.

$\displaystyle \tan(75^\circ) = \frac{\sin(75^\circ)}{\cos(75^\circ)} = \frac{\sin(45^\circ+30^\circ)}{\cos(45^\circ+30^\circ)}$

ContentLicense: CC BY-SA 3.0

Title: Trigonometry Addition Thereom


问:
Strong Topological Conjugacy and Reparameterization of Flows
In discrete dynamics, we say that two systems $f:X \to X$ and $g:Y \to Y$ are topologically conjugate if there exists a homeomorphism $p:X \to Y$ such that $p \circ f = g \circ p$.

My professor was discussing what the corresponding equivalence relation should be for flows. She said that a "naive" definition would be to say that the flows $f: X \times \mathbb{R} \to X$ and $g: X \times \mathbb{R} \to X$ are "equivalent" if there exists a homeomorphism $p:X \to Y$ such that for all $t$ we have $p \circ f_t = g_t \circ p$. This is called being "strongly topologically conjugate", but my professor said that this is too strong; if a flow is a parameterization of another one, then the two flows are not strongly topologically conjugate.

To see why this is true, I investigated an example. Consider the flows $f_t(x) = x + tv$ and $g_t(x) = f_{2t}(x) = x + 2tv$ where $x, v \in \mathbb{R}^2$. It seems to me that if we choose the homeomorphism $p(x) = 2x$, then the equation $(p \circ f_t) (x)= (g_t \circ p)(x)$ becomes

$$p(x+tv) = p(x) + 2tv \implies 2x + 2tv = 2x + 2tv,$$

so these systems are indeed strongly conjugate. Can someone please give me an example, or an intuitive explanation, of why reparameterizing does not necesarily give strongly conjugate systems?

Thank you!

EDIT: Throughout this I assumed that a reparameterization of $f_t(x)$ is a function of the form $f_{kt}(x)$ where $k$ is a real number. Is this the actual definition of a reparametrization?

答: $\mathbb{R}^2$ is very special because it has $\mathbb{R}^{\times}$ acting on it by dilation. Do the same calculation but for a vector field on $S^1$, which doesn't. Alternatively, pick a vector field which has a zero at some smallest finite time $t_0$; the value of $t_0$ doesn't change under strong conjugacy but it does change under reparameterization.


问:
Nine identical gemstones. Four of these are diamonds and five are not. What is the probability that at least one diamond is selected?
I was stuck on this question in a school assignment. The question is as follows:
In a box are nine identically shaped gemstones. Four of these are diamonds and five are not.What is the probability that at least one diamond is selected, if a person took out two gems blindfolded?

I tried doing $\frac{(4)(8)}{{9}\choose{2}}\frac{1}{2!} = \frac{16}{36}$, where the $4$ is the first selected diamond, and the $8$ represents the remaining gems after the first is selected. The fraction $\frac{1}{2!}$ is for correcting the overcounting. However when I tried drawing an extended tree diagram, I found that the answer is $\frac{26}{36}$. Where did I go wrong?

答:
It's not necessarily the first gem, or just one gem, that's the diamond. $4 \cdot 8$ counts all ordered pairs of gems in which the first gem is the diamond, and the second gem is anything. A set of two diamonds is counted twice, but set of two gems with only one diamond is only counted once, so you don't correct the overcount when you divide by $2$.

Instead, we can count $\binom 42$ cases with two diamonds and $\binom 41 \binom 51$ cases with one diamond, for a total of $26$. Divide this by the $\binom 92$ total cases, and we get the answer your tree diagram gave you.


答: I suggest, you start by listing all angles of which you know exact values for $\sin$ and $\cos$ (should include $0°$, $30°$, $45°$, $60°$, $90°$ at least), then see how you can express $75°$.</p>

Title: Trigonometry Addition Thereom

Body:
Using the expansion of

a. $\sin(+)$, prove that $\sin75°=\sqrt 6+\sqrt{24}$

b. $\sin(+)$, prove that $\tan75°=2+\sqrt 3$

Where to start? draw up triangle of sin 75? find other values? help please.


问:
Can one define a Fourier transform of $\log(a+|x|)$ with positive a?
I would like to know if one can define the Fourier transform of $\log(a+|x|)$ with positive $a$ in some distributional sense. I tried to compute it using Mathematica, but the answer I get seems to be wrong since the inverse Fourier transform of it is not consistent. Does anyone have a trick to massage this function that allows the computation of the Fourier transform by hand? Thanks

答:
Since tonight I'm on a roll: Per the discussion in the answer here we introduce a convergence factor in the integral and we consider

$$F_{\epsilon}(\omega)=\int_{-\infty}^{\infty}dx \log(|x|+a)e^{-i\omega x}e^{-\epsilon|x|}=e^{za}\int_{a}^{\infty}dx \log(x)e^{-zx}+ e^{z^* a}\int_{a}^{\infty}dx \log(x)e^{-z^*x}$$

where $z=\epsilon+i\omega$. Both integrals can be conveniently rewritten after an integration by parts as

$$e^{za}\int_{a}^{\infty}\log(x)e^{-zx}dx=\log a\frac{e^{za}}{z}+\frac{1}{z}\int_{a}^{\infty}e^{-zx}\frac{dx}{x}=\log a\frac{e^{-za}}{z}+\frac{e^{za}}{z}E_1(za)$$

where $E_1$ denotes the conventional exponential integral. It is well known that this function has only a log-type branch point at $z=0$ so if we are going to encounter interesting distributions in the $\epsilon\to 0$ limit, it is going to be at $\omega=0$. To isolate the distributional behavior at $\omega=0$ we exploit the series expansion of the special function around $x=0$ and we define $G(z)=\frac{E_1(z)+\gamma+\log(z)}{z}$, which is entire. The final result is

$$F_\epsilon(\omega)=\left((-\gamma-\log z)\frac{e^{za}}{z}+(-\gamma-\log z^*)\frac{e^{z^*a}}{z^*}\right)+a(e^{za}G(za)+e^{z^*a}G(z^*a))$$

which upon taking the limit gives (exercise)

$$F(\omega)=-2\pi\gamma\delta(\omega)-2\gamma\frac{\sin(\omega a)}{\omega}+2\pi\mathcal{P} \frac{\cos(a|\omega|)}{|\omega|}-2\mathcal{P}\frac{\sin(a|\omega|)\log|\omega|}{|\omega|}+a(e^{i\omega a}G(i\omega a)+e^{-i\omega a}G(-i\omega a))$$


问:
$\lim_{n \to \infty} \frac{1}{n} (\sum_{k=1}^n x_ky_k)=\lim_{n \to \infty}x_n \lim_{n \to \infty}y_n$

if $\{x_n\}$ and $\{y_n\}$ converge, then $$\lim_{n \to \infty}
\frac{1}{n} \bigg(\sum_{k=1}^n x_ky_k\bigg)=\lim_{n \to \infty}x_n
\lim_{n \to \infty}y_n$$

Let's say that $\{x_n\} \to x$ and $\{y_n\} \to y$.

Intuitively, I can see that for a very large $n$, the summation will look very similar to $\frac{nxy}{n}=xy$. However, I can't seem to find a proper proof for this (how could I translate this intuitive thought into a formal proof?). I tried using the fact that the sequences are bounded but that leaves me with

$$
\inf (x_n)\inf(y_n)\le\lim_{n \to \infty}
\frac{1}{n} \bigg(\sum_{k=1}^n x_ky_k\bigg)\le\
\sup (x_n)\sup(y_n)$$

I'd love to see some hints. By the way, I'm only using basic aspects about sequences: Cauchy, Weierstrass theorem, Definition of limits, Squeeze theorem and so on. That is, no derivatives or integrals.

答:
Given $\epsilon>0$ choose $N\in\Bbb N$ such that $k>N\implies|x_ky_k-xy|<\epsilon$.  $$\big|\lim_{n\to\infty}\big(\frac{1}{n}\sum_{k=1}^nx_ky_k\big)-\lim_{n\to\infty}x_ny_n\big|=\big|\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n(x_ky_k-xy)\big|\leq\lim_{n\to\infty}\frac{1}{n+N}\sum_{k=1}^{n+N}|x_ky_k-xy|$$ $$\leq\lim_{n\to\infty}\frac{1}{n+N}(n+N)\epsilon=\epsilon$$
$$\therefore\;\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}|x_ky_k-xy|=xy=\lim_{n\to\infty }x_ny_n$$


问:
What does $k\in\mathbb{Z}$ in the general solutions of trigonometric equations mean?
I wish to understand the meaning of the term $k\in\mathbb{Z}$, in solving trigonometric equations, for example, it is written

$\theta=2k\pi+\frac{\pi}{2}$, for all $k\in\mathbb{Z}$.

答:
It means that $k$ is an integer.

$\mathbb{Z}$ represents the set of all integers. “$\in$” means “is an element of ”. So, $k\in\mathbb{Z}$ means $k$ is an element of the set of all integers.


答: $\mathbb{Z}$ denotes the set of all integers. The symbol $\in$ means "belongs to". So the statement $k\in\mathbb{Z}$ simply means that $k$ belongs to the set of integers, i.e. $k$ is some unspecified integer. For an example of how this is used: $\cos 2\pi k=1$ for any $k\in\mathbb{Z}$.

Title: What does $k\in\mathbb{Z}$ in the general solutions of trigonometric equations mean?

Body:
I wish to understand the meaning of the term $k\in\mathbb{Z}$, in solving trigonometric equations, for example, it is written

$\theta=2k\pi+\frac{\pi}{2}$, for all $k\in\mathbb{Z}$.


答:
Although you can proceed by adding Pr($1$ diamond) + Pr($2$ diamond), Pr questions of the "at least one" type are best tackled by finding the complement, viz. $Pr = 1 - \frac{\dbinom5 2}{\dbinom8 2}$

Or, using probabilities directly, $1 -\left(\frac{5}{9}\cdot\frac{4}{8}\right)$

Title: Nine identical gemstones. Four of these are diamonds and five are not. What is the probability that at least one diamond is selected?

Body:
I was stuck on this question in a school assignment. The question is as follows:
In a box are nine identically shaped gemstones. Four of these are diamonds and five are not.What is the probability that at least one diamond is selected, if a person took out two gems blindfolded?

I tried doing $\frac{(4)(8)}{{9}\choose{2}}\frac{1}{2!} = \frac{16}{36}$, where the $4$ is the first selected diamond, and the $8$ represents the remaining gems after the first is selected. The fraction $\frac{1}{2!}$ is for correcting the overcounting. However when I tried drawing an extended tree diagram, I found that the answer is $\frac{26}{36}$. Where did I go wrong?


答:
The set $\mathbb{Z}$ essentially means the set of all integers, while the symbol $\in$ means belongs to. Hence, $k\in\mathbb{Z}$ means $k$ is an integer.

In the quoted statement, it means that $\theta$ takes all possible values of $2k\pi+\frac{\pi}{2}$, where you can safely put any integer as $k$.

Hope this helps :)


问:
Hatcher n-Torus Cohomology
On pg 210 and 211(section 3.2, example 3.11), Hatcher attempts to prove that $H^{k}(T^{n},R)$ has basis the cup products $\alpha_{i_{1}} \smile \cdots \smile \alpha_{i_{k}}$ where $i_{1}< \cdots < i_{k}$ and $T^{n} $ is the n-Torus. He first shows that if $\alpha \in H^{1}(I,\partial I;R)$ is a generator, the map given by the external cross product:

$H^{n}(Y;R) \to H^{n+1}(I \times Y,\partial I \times Y;R)$ given by $\beta \mapsto \alpha \times \beta$ is an isomorphism. To do this, he establishes the simple fact of naturality with respect to coboundary maps in the long exact sequence of relative cohomology. He mentions the following split short exact sequence obtained for the pair $(I \times Y,\partial I \times Y)$

$0 \rightarrow H^{n}(I \times Y;R) \rightarrow H^{n}(\partial I \times Y;R) \xrightarrow{\delta} H^{n+1}(I \times Y,\partial I \times Y;R) \rightarrow 0$. I did some calculations to verify that it is indeed split. However, I can't understand any of the explanation following that

How is $\delta$ an isomorphism on the restriction and how exactly is $\beta \mapsto \alpha \times \beta$ an isomorphism?
 Of course, the entire problem becomes much easier using the Kunneth formulas but I'd like to understand this proof too.

答:
I will assume that all cohomology groups have coefficients in $R$ and omit it from the notation. Let $1_0$ the $0$-cocycle defined as in the example, and let $1_1$ be defined analogously. Then it is not hard to show that every cocycle in $H^n(\partial I \times Y)$ can be written uniquely as a sum $1_0 \times \beta_0 + 1_1 \times \beta_1$, where $\beta_0, \beta_1 \in H^n(Y)$.

Let's start by proving that $\delta$ is injective when restricted to elements of the form $1_0 \times \beta$ for $\beta \in H^n(Y)$. Suppose that $\delta(1_0 \times \beta) = \delta(1_0) \times \beta = 0$. Let $\sigma : [v_1, \dots, v_{n+1}] \to Y$ be any singular $n$-simplex in $Y$. There is a singular $(n+1)$-simplex $\tilde{\sigma} : [v_0, \dots, v_{n+1}] \to I \times Y$ such that $p \circ \tilde{\sigma}|_{[v_1, \dots, v_{n+1}]} = \sigma$ where $p : I \times Y \to Y$ is the projection, and such that $\tilde{\sigma}(v_0) \in \{1\} \times Y$ and $\tilde{\sigma}([v_1, \dots, v_{n+1}]) \subseteq \{0\} \times Y$. Let $q : I \times Y \to I$ be the other projection. Then
$$
0 = (\delta(1_0) \times \beta)(\tilde\sigma) = (q^*\delta(1_0) \smile p^*\beta)(\tilde\sigma) = q^*\delta(1_0)(\tilde\sigma|_{[v_0,v_1]}) \cdot p^*\beta(\tilde\sigma|_{[v_1, \dots, v_{n+1}]}) = \beta(\sigma).
$$
Therefore $\beta = 0$, and $\delta$ is injective.

To prove surjectivity, first note that $\delta(1_1) = -\delta(1_0)$. To see this, we extend $1_0 + 1_1$ to all of $I$ by setting equal to $1$ everywhere. Then $\delta(1_0 + 1_1) = 0$, proving the claim. Since the sequence at the bottom of page 210 is exact, we have that $\delta$ is surjective when its domain is taken to be all of $H^n(\partial I \times Y)$. Hence, for every $\psi \in H^{n+1}(I \times Y, \partial I \times Y)$, we have some $\beta_0, \beta_1$ such that
$$
\psi = \delta(1_0 \times \beta_0 + 1_1 \times \beta_1) = \delta(1_0) \times \beta_0 + \delta(1_1) \times \beta_1 = \delta(1_0) \times (\beta_0 - \beta_1) = \delta(1 \times (\beta_0 - \beta_1).
$$
this concludes the proof of surjectivity, and therefore that $\delta$ is an isomorphism when restricted to a copy of $H^n(Y)$.

It follows from the above that $\beta \mapsto \delta(1_0) \times \beta$ is an isomorphism. Since for any unit $r \in R$, $\delta(1_0) \times \beta \mapsto (r\delta(1_0)) \times \beta$ is an isomorphism, we conclude that $\beta \mapsto \alpha \times \beta$ is an isomorphism for any $\alpha$ generating $H^1(I, \partial I)$.

Title: Hatcher n-Torus Cohomology


问:
Understanding big O notation
I'm not a mathematician by any stretch and I'm trying to translate some maths terms into simple maths terms. Please don't laugh, I do consider this complicated!

The equations in question are

```<code>O(n) and O(n ^ 2)
</code>```

Now, I have read up on Wiki about this but it has been written (IMO) for people who already understand it!

I believe n ^ 2 translates to the power of, in this case it is also the equivalent of squaring it (i.e. n * n).

However, I can't get my head around O in terms of what it is describing. Wiki says it's the limiting behaviour. So, does this mean O is more of a description than a function or command? In my understanding, the following 2 equations are the same

```<code>O(n^2)

n^2
</code>```

答: Quickly, $O(n^2)$ is any function $f=f(n)$ such that $$\left| \frac{f(n)}{n^2} \right|$$ remains bounded as $n \to +\infty$. It may be $n^2$ itself, but it may also be $n$, or $\sin \cos n$, etc.

OwnerDisplayName: Dave Rook

Title: Understanding big O notation


问:
How does $(abc) = (ac)(ab)$?
I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots  \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$

However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.

Thank you in advance!

答:
There are different conventions, but here we are performing the permutation on the right first.

So $a\mapsto b$ by $(ab)$, and the result $b$ is unaffected by $(ac)$;

$b\mapsto a\mapsto c$ by $(ab)$ followed by $(ac)$;

and $c$ is unaffected by $(ab)$ but $c\mapsto a$ by $(ac)$.

Title: How does $(abc) = (ac)(ab)$?


答:
These cycles are effectively functions (think $f\circ g$). Since you are asking what the composition of functions $(ac)(ab)$ does to the elements $a$, $b$, and $c$, you need to read right to left, just as you would for $f(g(x))$.

$\bullet$ $(ab)$ sends $a$ to $b$. Then $(ac)$ fixes $b$. So we get $a\mapsto b$.

$\bullet$ $(ab)$ sends $b$ to $a$. Then $(ac)$ sends $a$ to $c$. So we get $b\mapsto c$.

$\bullet$ $(ab)$ fixed $c$. Then $(ac)$ sends $c$ to $a$. So we get $c\mapsto a$.

So we have $(ac)(ab)=(abc)$.

Title: How does $(abc) = (ac)(ab)$?

Body:
I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots  \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$

However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.

Thank you in advance!


问:
$2n\times 4$ $(0,1)$-matrix such that each row has two $0$'s and $1$'s and each column has $n$ $0$'s and $1$'s.
How many ways to fill a $2n\times 4$ $(0,1)$-matrix such that each row has two $0$'s and $1$'s and each column has $n$ $0$'s and $1$'s.

Equivalently, how many ways to fill a $2n\times 4$ $(-1,1)$-matrix such that each row and each column have zero sum.

By any chance, is there an asymptotic result for large $n$ ?

答:
There are six possibilities for each row.  The numbers of
$(0,0,1,1), (0,1,0,1), (0,1,1,0), (1,1,0,0), (1,0,1,0),$ and $(1,0,0,1)$, call these $n_1, \ldots, n_6$, must satisfy
$$ \eqalign{n_4 + n_5 + n_6 &= n\cr
 n_2+n_3+n_4 &=n\cr n_1+n_3+n_5 &=n\cr n_1+n_2+n_6&=n\cr}$$
Using some linear algebra one finds that $n_1 = n_4$, $n_2 = n_5$, $n_3 = n_6$, so the
number of solutions is just the number of ways to write
$n = n_1 + n_2 + n_3$ with nonnegative integers $n_1, n_2, n_3$.  And using generating functions, that is the coefficient of $x^n$ in the Maclaurin series of
$$\left(\sum_{k=0}^\infty x^k\right)^3 = \frac{1}{(1-x)^3}$$
which is $${n+2 \choose 2}$$


问:
Showing that Equality Holds for $\mathbb{P} ( X + Y + Z \leq 15) = \mathbb{P}(X+Y+Z \geq 26)$ for R.V.s $X, Y, Z$
I am a bit struggling to prove the following:

Let $X$ be the outcome of a standard die roll, $Y$ the outcome of an icosahedral die roll - 20 faces - and $Z$ the outcome of a dodecahedral die roll - 12 faces. If the three dice are rolled simultaneously, how can I show that$\mathbb{P} ( X + Y + Z \leq 15) = \mathbb{P}(X+Y+Z \geq 26)$?

So far, I have been trying to algebraically manipulate the probabilities, but I got stuck. This is what I have:
\begin{align*}
\mathbb{P} (X+Y+Z \leq 15) &= \sum_{k=3}^{15} \mathbb{P}(X+Y+Z = k) \\
&= \sum_{k=3}^{15} \sum_{i=1}^{k} \sum_{j=1}^{k} \mathbb{P}(X=i, Y=j, Z=k-(i+j))
\end{align*}
With this expression, I am not entirely sure how to proceed to show that the inequality holds. I would appreciate any help on how to prove this!

答:
Let $W:=X+Y+Z$, then as these three r.v. are independent and symmetric w.r.t. it own mean then it is enough to show that addition of independent and symmetric r.v. defines a symmetric random variable respect to it mean, that is, we want to show that
$$
\Pr[X+Y\leqslant  \mu-c]=\Pr[X+Y\geqslant\mu+c],\quad \text{ for all }c\text{ with }\mu:=\mathrm E[X+Y]
$$
whenever $\Pr[X\leqslant\mathrm E[X]-c]=\Pr[X\geqslant \mathrm E[X]+c]$ and $\Pr[Y\leqslant  \mathrm E[Y]-c]=\Pr[Y\geqslant  \mathrm E[Y]+c]$ for all $c$. WLOG we can assume that $\operatorname{E}[X]=\operatorname{E}[Y]=0$ as the r.v. $X-\operatorname{E}[X]$ and $Y-\operatorname{E}[Y]$ will be symmetric respect to zero, then
$$
\begin{align*}
\Pr [X+Y\leqslant  c]&=\int_{\{(x,y)\in \mathbb{R}^2:x+y\leqslant c\}}\mathop{}\!d F_{X,Y}(x,y)=\int_{\mathbb{R}}\left(\int_{(-\infty ,c-y]}\mathop{}\!d F_X(x)\right)\mathop{}\!d F_Y(y)\\
&=\int_{\mathbb{R}}\Pr [X\leqslant c-y]\mathop{}\!d F_Y(y)\overset{*}{=}\int_{\mathbb{R}}\Pr [X\geqslant y-c]\mathop{}\!d F_Y(y)\\
&=\int_{\mathbb{R}}\left(\int_{[y-c,\infty )}\mathop{}\!d F_X(x)\right)\mathop{}\!d F_Y(y)=\int_{\{(x,y)\in \mathbb{R}^2:y-c\leqslant x \}}\mathop{}\!d F_{X,Y}(x,y)\\
&=\int_{\{(x,y)\in \mathbb{R}^2:y\leqslant x+c \}}\mathop{}\!d F_{X,Y}(x,y)=\int_{\mathbb{R}}\left(\int_{(-\infty ,x+c]}\mathop{}\!d F_Y(y)\right)\mathop{}\!d F_X(x)\\
&=\int_{\mathbb{R}}\Pr [Y\leqslant x+c]\mathop{}\!d F_X(x)\overset{*}{=}\int_{\mathbb{R}}\Pr [Y\geqslant  -x-c]\mathop{}\!d F_X(x)\\
&=\int_{\mathbb{R}}\left(\int_{[-x-c,\infty )}\mathop{}\!d F_Y(y)\right)\mathop{}\!d F_X(x)=\int_{\{(x,y)\in \mathbb{R}^2:-x-c\leqslant y\}}\mathop{}\!d F_{X,Y}(x,y)\\
&=\int_{\{(x,y)\in \mathbb{R}^2:-c\leqslant x+y\}}\mathop{}\!d F_{X,Y}(x,y)=\Pr [X+Y\geqslant -c]
\end{align*}
$$
where in the steps marked with $*$ I used the symmetry of $X$ and $Y$ respect to zero, and where I used it independence in the identity $\mathop{}\!d F_{X,Y}=\mathop{}\!d F_X \mathop{}\!d F_Y$. Therefore we conclude that $W$ is symmetric respect to it mean, and $\operatorname{E}[W]=\operatorname{E}[X]+\operatorname{E}[Y]+\operatorname{E}[Z]=\frac{7}{2}+\frac{21}{2}+\frac{13}{2}=\frac{41}{2}$. By last note that $26=\frac{41}{2}+\frac{11}{2}$ and $15=\frac{41}{2}-\frac{11}{2}$.∎


问:
Taylor series expansion about point $x_0 + \epsilon$
The taylor series expansion about point $a$ for function $f(x)$ is given by $$f(x) \approx f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 +...$$

As more and more terms are added to the taylor polynomial, the interval of approximation over $f(x)$ increases as well.

For a taylor series expansion about point $x_0 + \epsilon$, shouldn't the expansion be $$f(x) \approx f(x_0 +\epsilon) + f'(x_0+\epsilon)(x-(x_0+\epsilon)) + ... $$

Why do they write it as
$$f(x_0 + \epsilon) = f(x_0) + f'(x_0)\epsilon + \frac{1}{2}f''(x_0)\epsilon^2$$
This is taken from wolfram alpha

答:
It's kind of a weird mixed series obtained from the Taylor series. Start with the Taylor series about $x=x_0$, namely

$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+ \dots$$

Now substitute $x-x_0=\varepsilon$ on the RHS, but leave the evaluations of $f$ and its derivatives on the RHS at $x_0$. Then on the LHS, write $f(x)=f(x_0+\varepsilon)$. This gives

$$f(x_0+\varepsilon)=f(x_0)+f'(x_0)\varepsilon+\frac{1}{2}f''(x_0)\varepsilon^2+ \dots$$


问:
Find $ \bigcap\limits_{n \in \mathbb{Z}} \bigcup\limits_{x \in \mathbb{R}} A_{n,x}$ for $A_{n,x} = \{z \in \mathbb{R}: |n-z|>|x-z|\}$
The family of sets $\langle A_{n,x}: n \in \mathbb{Z}, x \in \mathbb{R} \rangle $ is given by:  $A_{n,x} = \{z \in \mathbb{R}: |n-z|>|x-z|\}$ find:
$$ \bigcap\limits_{n \in \mathbb{Z}} \bigcup\limits_{x \in \mathbb{R}} A_{n,x}$$

My idea was to show that the answer is $\emptyset$. I thought that I can just take n = z to make the first module equal to $0$. Then I need my second module to be not equal to $0$, but now I see that there are values z equal to x that solve the inequaluty.

答:
HINT: First show that

$$A_{n,x}=\begin{cases}
\left(\leftarrow,\frac{n+x}2\right),&\text{if }x<n\\
\varnothing,&\text{if }x=n\\
\left(\frac{n+x}2,\to\right),&\text{if }x>n\,.
\end{cases}$$

Then show that

$$\begin{align*}
\bigcup_{x\in\Bbb R}A_{n,x}&=\bigcup_{x<n}\left(\leftarrow,\frac{n+x}2\right)\cup\bigcup_{x>n}\left(\frac{n+x}2,\to\right)\\
&=(\leftarrow,n)\cup(n,\to)\\
&=\Bbb R\setminus\{n\}\,.
\end{align*}$$

Once you’ve done that, it’s not too hard to see what $\bigcap\limits_{n\in\Bbb Z}\bigcup\limits_{x\in\Bbb R}A_{n,x}$ is.


答: For fixed $n \in \mathbb N, x \in \mathbb R$, the set $A_{n,x}$ is all the points that are closer to $x$ than they are to $n$. Take any point $z \in \mathbb R$ with $z \neq n$. Then there is some $x^* \in \mathbb R$ such that $z$ is closer to $x$ than $n$, and so $z \in A_{n,x^*}$. By contrast, the point $n$ itself is always at least as close to $n$ as it is to any other $x$, and so $n \not \in A_{x,n}$ for any $x$. Therefore $$ B_n := \bigcup_{x \in \mathbb R} A_{x,n} = \mathbb R \setminus \{n\}.$$ That is, the inner union is all points except $n$. Then when you take the intersection of all $B_n$, you remove every integer $n$, and so $$\bigcap_{n \in \mathbb Z} B_n = \bigcap_{n \in \mathbb Z} \bigcup_{x \in \mathbb R} A_{n,x} = \mathbb R \setminus \mathbb Z.$$

Title: Find $ \bigcap\limits_{n \in \mathbb{Z}} \bigcup\limits_{x \in \mathbb{R}} A_{n,x}$ for $A_{n,x} = \{z \in \mathbb{R}: |n-z|>|x-z|\}$

Body:
The family of sets $\langle A_{n,x}: n \in \mathbb{Z}, x \in \mathbb{R} \rangle $ is given by:  $A_{n,x} = \{z \in \mathbb{R}: |n-z|>|x-z|\}$ find:
$$ \bigcap\limits_{n \in \mathbb{Z}} \bigcup\limits_{x \in \mathbb{R}} A_{n,x}$$

My idea was to show that the answer is $\emptyset$. I thought that I can just take n = z to make the first module equal to $0$. Then I need my second module to be not equal to $0$, but now I see that there are values z equal to x that solve the inequaluty.


问:
Why are these numbers coprime?
I have encountered the following argument in a number theory class:

Let $p$ be an odd prime and let $n,x$ and $y$ be integers with $\gcd(x,y) = 1$, $n > 0$ and $p \not \mid n$. Suppose that $p \mid x^2+ny^2$, it then follows that $\gcd(y,p) = 1$.

I do not understand why $\gcd(y,p) = 1$ should hold here. Could you give me a hint?

答:
We have:

$$x^2+ny^2=kp$$

Or:

$$x^2=kp-ny^2$$

Due to definition of question:

$(x, y)=1$

If $(p, y)=a$, that is p and y have a common divisor like a, then $a|x$ which in turn means $a|y$ or x and y have a common divisor like a, this contradict the definition of question.

Title: Why are these numbers coprime?


问:
Infimum of inner product over a closed subspace
Let $M\subset H$ a closed linear subspace of a Hilbert space H. Suppose $f \in H$ but $f \not\in M^{\perp}$. Show that
$$ \inf_{x\in M,\|x\|=1} (f,x)$$
is attained and the min is unique.

So what I've tried so far was to use extreme value theorem. I know that the closed unit circle is compact with respect to $\sigma(H,H^*)$ due to reflexivity and $M$ is weakly closed as well so the intersection would be weakly compact, EVT would come in here until I realized the infimum is only over the shell intersect M, so my idea stops there.

I also tried constructing a sequence $(f,x_n)$ converging to the infimum, since M is also reflexive and each $x_n$ has norm 1, there exists a weakly convergent subsequence. If $x$ is the weak limit and $\|x\|=1$, I would be done. But I don't know how to show $\|x\|=1$, I do know however $\|x\| \leq 1$ simply by weak convergence.

Just looking for a hint!!
Thank you!

答:
You are almost there. As you said we know that the unit ball is weak* compact, so by the EVT there exists $u$ in the closed unit ball so that $\inf_{\|x\|\leq 1}(f,x)=(f,u)$. We also note that $f\neq 0$ so $u\neq 0$. Set $\alpha=1/\|u\|$. We claim that $\inf_{\|x\|=1}(f,x)=(f,\alpha u)$. Assume not: Then there exists a $v$ on the unit sphere such that $(f,v)<(f,\alpha u$). Then $\frac{1}{\alpha}(f,v)<(f,u)$, meaning that $(f,\|u\|v)<(f,u)$, which contradicts the choice of $u$.

This approach, however, is not the best approach in order to see uniqueness. We need to exploit the geometry of the Hilbert space. Note that we can write $f=P_Mf+P_{M^\perp}f$. Thus for any $x\in M$ we have $(f,x)=(P_Mf,x)$. Combining this with Cauchy-Schwarz, we see that
$$\inf_{\|x\|=1}(f,x)\geq-\|P_Mf\|.$$
Clearly this infimum is achieved at $x=\frac{-P_Mf}{\|P_Mf\|}.$ Finally we remember that equality is achieved in Cauchy-Schwarz if and only if the two elements are linearly dependent. I.e. for any $x,y\in H$ $|(y,x)|=\|y\|\|x\|$ if and only if $x=\alpha y$. As any ray only intersects the unit sphere at a unique point, we are done.


问:
$\Delta ABC$ is right angled triangle. $AP$ and $AQ$ meet $BC$ and $BC$ produced in $P$ and $Q$ and are equally inclined to $AB$.
$\Delta ABC$ is right angled triangle at $A$. $AP$ and $AQ$ meet $BC$ and $BC$ produced respectively in $P$ and $Q$ and are equally inclined to $AB$ ($\angle BAP=\angle BAQ$). Show that $\frac{BP}{BQ}=\frac{CP}{CQ}$.

I used angle bisector theorem and found that $\frac{BP}{BQ}=\frac{AP}{AQ}$. But I don't know how to proceed further. Can this be solved without trigonometry? Thanks in advance.

答:
HINT: Construct $QQ'$ with $Q'$ on $AC$ extended such that $QQ'=QA$

From angle bisector theorem,
$$\frac{BP}{BQ}=\frac{AP}{AQ}$$
In $\Delta CPA$ and $\Delta CQQ'$

$$\angle BAP=\angle BAQ \Rightarrow 90-\angle BAP=90-\angle BAQ \Rightarrow \angle PAC=\angle QAQ'=\angle QQ'C$$
$$\angle C = \angle C$$
Therefore,
$$\Delta CPA \sim \Delta CQQ'$$
Hence, by similarity,
$$\frac{BP}{BQ}=\frac{AP}{AQ}=\frac{AP}{QQ'}=\frac{CP}{CQ}$$
Therefore, the proof is done without trigonomery.


问:
Integral closure of k-algebra
Let $k$ be a field and $A$ a finitely generated algebra over $k$ that doesn't have zero divisors. Why is the integral closure of $A$ a finitely generated module over $A$ ? </p>

<p>(edited)</p>

答:
As Martin said in the comments, the result is not true in general but holds if $A$ is finitely generated over $k.$ This follows from Proposition 16 (page 46) of Serre's Local Algebra:

Proposition 16. Let $A$ be a domain which is a finitely generated algebra over a field $k,$ let $K$ be its field of fractions, and let $L$ be a finite extension of $K.$ Then the integral closure $B$ of $A$ in $L$ is a finitely generated $A$-module (in particular it is a finitely generated $k$-algebra).

The Google Books preview of Serre's book lets one see page 46 (and surprisingly, most of the book).

Title: Integral closure of k-algebra


问:
Triple Integrals Find the mass of the solid S made out of material with density f.
Find the mass of the solid S made out of material with density f in maple.
S is a ball bounded by the sphere (x−2)^2+(y+3)^2+(z−4)^2 =5; f(x, y, z)=x^2y^2.

How can I solve without changing it into a cylindrical coordinate? I tried the following but I am not sure if I did it right.
z limit from 4-sqrt(5-(x-2)^2 -(y+3)^2) to 4+sqrt(5-(x-2)^2 -(y+3)^2)
y limit from 3-sqrt(5-(x-2)^2) to 3+sqrt(5-(x-2)^2)
and x limit from -3 to 7.

Body:
Check this for mistakes on my part...

restart;

VectorCalculus:-int(x^2*y^2,
                    [x,y,z] = Sphere(<2,-3,4>,sqrt(5)));

         2320     (1/2)
         ---- Pi 5
          7

Int(x^2*y^2,
    [ z = 4-sqrt(5-(x-2)^2-(y+3)^2) .. 4+sqrt(5-(x-2)^2-(y+3)^2),
      y = -3-sqrt(5-(x-2)^2) .. -3+sqrt(5-(x-2)^2),
      x = 2-sqrt(5) .. 2+sqrt(5) ]):

 simplify(value(%));

         2320     (1/2)
         ---- Pi 5
          7


问:
Prove statement is equivalent to little-o notation
This question has already been asked here: Little o notation - equivalence, however, non of the answers I found were correct/thorough enough.

The question is:

I need to help with mathematical problem. Let's have function $g(n) : \mathbb{N} \to \mathbb{R}^+$ and $2$ sets defined:
$$o(g(n)) = \{f(n) : \mathbb{N} \to \mathbb{R}^+\;|\; \forall c \in \mathbb{R}^
+\; \exists n_0 \in \mathbb{N}\; ∀n \geq n_0 : f(n) < c · g(n)\}
\\ and \\
\bar{o}(g(n)) = \{f(n) : \mathbb{N} \to \mathbb{R}^
+\;|\;\forall c \in \mathbb{R}^
+\; \exists n_0 \in \mathbb{N}\; \forall n \geq n_0 : f(n) \leq c · g(n)\}$$
How can I prove that $o(g(n))$ is equal to $\bar{o}(g(n))$? Which consequence does it have to definition of little o ($o$)?

I know that $o(g(n)) \implies \bar{o}(g(n))$ side of equivalence should be trivial.

The only answer is this:
Let $f\in\bar o(g)$. To prove $f\in o(g)$, let $c$ be any positive number. Since $f\in \bar o(g)$, there exists a natural number $n_0$ such that, for all $n\ge n_0$,  we have
$$f(n)\le \frac c2g(n)<cg(n).$$

My question is, is the answer correct? And if it is, why can you divide one $c$ by 2? If it is not, what is the correct answer?

答:
To prove $f\in o(g)$, all we need to show is that for any positive number $\color{blue}c$, $\exists \color{blue}{n_0}$ such that $\forall n\ge \color{blue}{n_0}$, we have $f(n) \color{red} < \color{blue}c · g(n)$. For this, we first choose a positive number $\color{blue}c$.

Now if $f\in \bar o(g)$, then for any positive number $\color{green}{c'}$, $\exists \color{green}{n_0'}$ such that $\forall n\ge \color{green}{n_0'}$, we have $f(n) \color{red}\le \color{green}{c'} · g(n)$. Since the statement says we can pick any positive number, we cleverly choose $c'=\dfrac{c}{2}$. With this we have,
$$
f(n) \le \frac{c}{2} · g(n)\quad\forall n\ge n_0'
$$
Now, we know that $\dfrac{c}{2}<c$ which means $\dfrac{c}{2}\cdot g(n)<c\cdot g(n)$ so we write the above expression as
$$
f(n) \le \frac{c}{2} · g(n)<c\cdot g(n)
$$
which implies,
$$
f(n)<c\cdot g(n)\quad \forall n\ge n_0'
$$
We started with any $\color{blue}{c}$ and found a corresponding $\color{blue}{n_0}(=\color{green}{n_0'})$ such that $f(n) \color{red} < \color{blue}c · g(n),\forall n\ge \color{blue}{n_0}$. Hence $f\in o(g)$.

So it doesn't matter that we have used $\dfrac{c}{2}$ in the condition of $f\in \bar o(g)$ (we could have chosen $\dfrac{c}{3},\dfrac{c}{4},\dfrac{c}{5}$ etc.). We just wanted to arrive at a strict inequality $<$ from $\le$.


问:
Quick question about the definition of a random variable in a baseball pitches example

I'm just wondering, in the last sentence of the third paragraph, why can X not be larger than 6? Is that because the number of innings is 9? I'm really confused, I thought X goes from 3,4,5,6,...to infinity?

答:
Each pitch results in either a ball or a strike. If you get to three strikes you are out. If you get to four balls then you walk. If you have thrown five pitches and have neither struck out nor walked then you must have three balls and two strikes. On the next pitch you either get a ball in which case you have four balls and you walk or you get a strike in which case you have three strikes and you are out.

The batter is either out or has walked by the sixth pitch.


问:
Given linear map, find nonzero polynomials such that...

Consider the linear map
\begin{align*}
      \theta : P_1 &\to P_1 \\
      a+bx &\mapsto (3a + 4b) - (2a + 3b)x
\end{align*}

Find a non zero polynomial $f \in P_1$ such that $\theta(f) = f$.

Find a non zero polynomial $g \in P_1$ such that $\theta(g) = -g$.

Show that $f,g$ is a basis of $P_1$, and find the matrix representing $\theta$ in the basis of $f, g$.

I am unsure how to go about solving 1 and 2 without just plugging in numbers for $a$ and $b$ until the two sides match. Is there a better way to go about solving 1 and 2 that is possibly faster than just plugging in numbers? These problems are on a worksheet dedicated to a certain section of my textbook that explains matrix similarity and diagonalizability so I feel there is something that I am missing that would allow me to solve 1 and 2 using similarity or diagonalizability.

答:
Just use the definition of the map $\theta$. Let $f(x)=a+bx$. Then, for $1$, you want

$$\theta(a+bx)=(3a+4b)-(2a+3b)x=a+bx.$$

We can equate the coefficients on each side. This gives the system of equations

$$3a+4b=a$$
$$-(2a+3b)=b,$$

which you can solve for $a,b$ to ultimately get $f$. After you solve the system, use the definition of $\theta$ to verify that it works. Same process for number 2.


问:
How do I find the proportion of 2 different sub groups within one group?
I am trying to find the proportion of nhl 2k (video game) players that play more than 3 games or play on playstation. 40% of players use xbox, 50% use playstation, and 10% use pc. Of those who play on xbox, 60% play more than 3 games. Of those who play on playstation, 15% play more than 3 games. Of those who play on pc, 0% play more than 3 games.

What proportion of players play more than 3 games OR play on playstation?

I'm not sure if I am supposed to add these 2 proportions together and if my math is correct:

Probability of players play on XBOX = 0.4

Probability of players play on Playstation = 0.5

Probability of players play on pc = 0.1

Probability of players play more than 3 games on XBOX = 0.6

Probability of players play more than 3 games on Playstation = 0.15

Probability of players play more than 3 games on pc = 0.0

Probability of players play more than 3 games = 0.4 * 0.6 + 0.5 * 0.15 = 0.315

So 0.315 proportion of players play more than 3 games.

proportion of players who play on playstation = 0.5 * 0.315 = 0.1575

Should I be adding the probability of players that play more than 3 games to the proportion of players that play on playstation?

答:
Firstly, some semantical notes. You state 'Probability of players play more than 3 games on XBOX = 0.6' which I am taking to mean 'Probability that a player plays more than three games on XBOX = 0.6' or 'Proportion of players that play more than three games on XBOX = 0.6'. Both of these are not correct. It would be correct to say 'Proportion of XBOX players that play more than three games = 0.6' or 'Probability that a player plays more than three games GIVEN that they play on XBOX = 0.6'. The key difference is the conditionality. The problem is saying that OF THE 40% that play XBOX, 60% play more than three games. This means that 'Probability that a player plays more than three games on XBOX' or 'Proportion of players that play more than three games on XBOX' is 0.4$\cdot$0.6 = 0.24. The same goes for your statements about Playstation and Other. You seem to understand this even if your statements are incorrect because you correctly calculate 'Probability that a player plays more than three games' to be 0.4$\cdot$0.6 + 0.5$\cdot$0.15 + 0.1$\cdot$0 = 0.315. I am confused by your last calculation. I don't know  of any reason you would multiply those two figures together. You state that the problem asks for the probability that a player plays Playstation OR more than three games. To calculate this you should use the following formula:
$$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$$
Here we are looking to calculate $P(\text{plays Playstation or plays more than three games})$ so our formula tells us this is
$$P(\text{plays Playstation})+P(\text{plays more than three games})-P(\text{plays Playstation and plays more than three games})$$
We are given that,
$$P(\text{plays Playstation})=0.5$$
And we computed,
$$P(\text{plays Playstation and plays more than three games})=0.5\cdot0.15=0.075$$
$$P(\text{plays more than three games})=0.315$$
Therefore
$$P(\text{plays Playstation or plays more than three games})=0.5+0.315-0.075=0.74$$
which is our answer.


问:
sequence characterization of first countable space
Let $X$ equipped with two different first countable topology $\tau_1$ and $\tau_2$ ,assume $x_n\to x$ in $\tau_1$ iff $x_n\to x$ in $\tau_2$,prove $\tau_1 = \tau_2$.

My attempt:$x\in \bar{A} \iff \text{exist}\ x_n\to x$ then we know any closure $\text{cl}_1 A = \text{cl}_2 A$ for any set $A$.

But I don't know to show then topology are the same?

答:
Let $A$ be a closed set in $X$ with respect to $\tau_1$ and $x \in \bar{A}=A$ , then there exists ${x_n\to x}$ in $A$ in $\tau_1$ ,but as  ${x_n\to x}$ in $\tau_2$ which is in $A$, it implies $x \in \bar{A}$ with respect to $\tau_2$. Therefore Closure of $A$ in $\tau_1$
equals to Closure of $A$ in $\tau_2$.

$\therefore$ cl$_2A$=cl$_1A$=$A$. Therefore $A$ is closed in $\tau_2$. Likewise we can prove that, if $B$ be a closed set in $\tau_2$ then it is closed in $\tau_1$. Which implies $\tau_1$=$\tau_2$


问:
Using $p\supset q$ instead of $p\implies q$
I saw that a use for the notation $p\supset q$ instead of $p\implies q$
that got me a bit confused.

One occurrences is in this Wikipedia link.

It seems to me opposite than what it should be, let me explain what
I mean:

If $A,B$ are sets s.t $A\subset B$, $p$ is for $x\in A$, and $q$
is for $x\in B$ then we can identify (in some way) $$A\text{ with }p$$
$$B\text{ with }q$$

We have it that $p\implies q$, since $A\subset B$, but in the above
notation we have $q\subset p$ which to me looks like $B\subset A$

which is the opposite of what we wanted to express.

Can someone please explain to me the logic behind this notation ?

答: Notice that in the line in the link numbered 2, they say the symbol $p\supset q$ is often confused for set inclusion. But it is not set inclusion. It's just a symbol used to denote material implication. So, it seems you just exhibited how often it actually happens that this confusion arrises. </p>


问:
Is this a correct way to show that $\sum_{k=2}^{\infty} \frac {1}{\sqrt k+k(-1)^k}$ converges?
Show that $\sum_{k=2}^{\infty} \frac {1}{\sqrt k+k(-1)^k}$ converges or diverges. My attempt is ot first rewrite and see if the sequence is convergent according to the Leibniz test:

$\frac {1}{\sqrt k+k(-1)^k}= \frac{(-1)^k}{\sqrt k} \frac{1}{\sqrt k + (-1)^k} \le\frac{(-1)^k}{\sqrt k} \frac{1}{\sqrt k -1}$

Now see if it holds that:

$a_{k+1} \le a_k$

$lim_{k \rightarrow \infty} a_k = 0$

The first criterion holds because the denominator is an increasing functions and thus an increasing sequence:
$f(x) = x-\sqrt x \implies f'(x)=1-\frac{1}{2\sqrt x} \ge 0  \forall x>2$

The second criterion also holds because: $lim_{k \rightarrow \infty} \frac{1}{\sqrt k} \frac{1}{\sqrt k-1} \rightarrow 0 * 0=0 $. So according to the Leibniz test the sum converges

答:
It is true that $a_k=\frac1{\sqrt{k}+(-1)^kk}$ is less than $b_k=\frac{(-1)^k}{\sqrt{k}(\sqrt{k}-1)}$.  But each of these sequences is alternating in sign.  And even if $\sum_{k=2}^\infty b_k$ converges, the dominance of $b_k$ over $a_k$ does not guarantee the convergence of $\sum_{k=2}^\infty a_k$.

However, we can proceed as follows.  Note that we can write

$$\begin{align}
\frac1{\sqrt{k}+(-1)^kk}&=\frac{\sqrt{k}-(-1)^kk}{k(1-k)}\\\\
&=\frac{1}{\sqrt{k}(1-k)}-\frac{(-1)^k}{1-k}
\end{align}$$

Hence, we can write

$$\begin{align}
\sum_{k=2}^K \frac1{\sqrt{k}+(-1)^kk}&=\sum_{k=2}^K \frac{1}{\sqrt{k}(1-k)}-\sum_{k=2}^K\frac{(-1)^k}{1-k}\tag1
\end{align}$$

The first sum on the right-hand side of $(1)$ converges as $K\to \infty$ by comparison to $\sum_{k=1}^\infty \frac1{k^{3/2}}$ and the second sum converges as guaranteed by Leibniz's Test.


问:
How do you solve the equation $3x^2+2\sin x=2$?
This seems like a simple equation to solve, but I can't find a way to do it and wolfram alpha can't give me an exact answer.</p>

答:
This is a transcendental equation; so, you will need numerical methods and then good starting guesses.

Consider that you look for the zero's of function $f(x)=3x^2+2\sin x-2$.

For example, using Taylor series built around $x=0$
$$f(x)=-2+2 x+3 x^2-\frac{x^3}{3}+O\left(x^5\right)$$ Ignoring the term in $x^3$, solving the quadratic would give as estimates
$$x_1=-\frac{\sqrt{7}+1}{3} \qquad \qquad x_2=\frac{\sqrt{7}-1}{3}$$ We could solve the cubic but this is not very pleasant.

But what we can do is to take quite many terms in the series expansion and use series reversion. Looking at the values of $x_1$ and $x_2$, reasonable base points could be $-\frac \pi 3$ and $\frac  \pi 6$. This would give as approximation
$$x_1=-\frac{\pi }{3}+t+\frac{\left(6+\sqrt{3}\right) }{2 (2 \pi
   -1)}t^2+\frac{\left(59+18 \sqrt{3}-\pi \right) }{3 (2 \pi
   -1)^2}t^3+O\left(t^4\right)$$ where $t=\frac{\pi ^2-3 \left(2+\sqrt{3}\right)}{3(2 \pi -1)}$. Numerically, this gives $x_1\approx -1.12638$ while Newton method would give as solution $x_1= -1.12630$.

Doing the same for the second root
$$x_2=\frac{\pi }{6}+t-\frac{5 }{2 \left(\sqrt{3}+\pi \right)}t^2+\frac{\left(78+\sqrt{3}
   \pi \right) }{6 \left(\sqrt{3}+\pi \right)^2}t^3+O\left(t^4\right)$$ where $t=\frac{1-\frac{\pi ^2}{12}}{\sqrt{3}+\pi }$.  Numerically, this gives $x_2\approx 0.559374$ while Newton method would give as solution $x_2= 0.559372$.

For sure, at the price of more messy terms, we could be even much closer to the solution.


问:
Convergence of chi-squared distribution to standard normal
I want to prove that $\frac{(X_n-n)}{\sqrt{2n}}\xrightarrow{{L}}Z\sim N(0,1)$ as $n\rightarrow\infty$ where $X_n\sim\chi^2_n$.

I don't want to use the CLL directly. So far this is what I got, but I am stuck at the end:

Considering sequence of rvs of $Z_1,Z_2,...\sim iid N(0,1)$, $\chi^2_n\equiv\Sigma_{i=1}^nZ_i^2\sim Chisq(n)$. We know that $E(\chi^2_n)=n$ and $Var(\chi^2_n)=2n$.

Let $Y_i=(X_i-n)/\sqrt{2n}$ with mean $n$ and variance of $2n$. (Should I consider n and 2n or just 1 and 2 (assume n=1) for mean and var?)

So $(X_i-n)/\sqrt{2n}=\Sigma_{i=1}^nY_i=Z_n$.

By Taylor series $Y_1$ has mgf:

$$M_{Y_1}(s)=E[e^{sY_1}]=\int_Re^{sY}dP(Y_1\leq y)=\int_R(1+sy+\frac{1}{2!}s^2y^2+o(s^2y^2))dF_{Y_1}(y)$$
$$=1+n+s^2n+o(2s^2n)$$

Since $X_i$s are iid:
$$E[e^{sZ_n}]=E[e^{s\Sigma_{i=1}^nY_i}]=(E[e^{sY_1}])^n=(1+n+s^2n+o(2s^2n))^n$$

To prove the statement mentioned, this should converge to $e^{s^2/2}$ as $n\rightarrow\infty$ which is the mgf of $Z$, but it doesn't! Can someone please help me find out which part I am doing wrong and how to do this correctly?

Thank you in advance

答:
The way you approximated the mgf of $Y_1$ is not quite right. Moreover, you wrote $Z_n = \sum_{i=1}^n (X_i - n)/\sqrt{2n},$ which should actually be $Z_n = \sum_{i=1}^n (X_i - 1)/\sqrt{2n}.$

An easier way is to use the usual mgf of $X_1$, which is given by $$\mathbb{E}[e^{tX_1}] = (1-2t)^{-1/2}, \text{ for }t < 1/2.$$ This gives $$\mathbb{E}[e^{sY_1}] = \mathbb{E}[e^{s(X_1 - 1)/\sqrt{2n}}] = e^{-s/\sqrt{2n}} (1-2s/\sqrt{2n})^{-1/2}.$$ Note that this is valid only for $s/\sqrt{2n}< 1/2.$ This is why it is better to use characteristic functions instead of moment generating functions. The characteristic function of $Z_n$ is given by $$\mathbb{E}[e^{isZ_n}] = e^{-is\sqrt{n/2}} (1-2is/\sqrt{2n})^{-n/2}.$$ The advantage here is that it is defined for all $s\in \mathbb{R},$ and now you should show that it converges to the required characteristic function, which is $e^{-s^2/2}.$
This can be shown by taking log of the characteristic function of $Z_n,$ as shown below.
$$-is\sqrt{\frac{n}{2}} -\frac{n}{2}\log \left(1 - \frac{2is}{\sqrt{2n}}\right)
= -is\sqrt{\frac{n}{2}} -\frac{n}{2} \left(- \frac{2is}{\sqrt{2n}} + \frac{2s^2}{n} + o\left(\frac{1}{n}\right)\right) = - \frac{s^2}{2} + o(1).$$


问:
$(F\ast G)(x)\leq F(x)G(x)$
Consider the CDFs $F$ and $G$ on $\mathbb R_+$.

How can I show that $$(F\ast G)(x)\leq F(x)G(x),\ \ \ x \in \mathbb R_+$$

$|(F\ast G)(x)|\leq\int_{\mathbb R_+}|F(x-y)G(y) |dy$ but how do I continue? Do we have that $F$ is bounded? (because it is a cdf, so it is bounded by $1$?) Then I could write $...\leq \|F\|_\infty\|G\|_1=\|G\|_1$

?

How would a 'probabilistic' (I don't know if this is the correct word) look like? (i.e. using appropriate random variables and probabilities?

Edit:

Apparently, for distribution functions, the convolution is defined as $$\int_{-\infty}^{\infty}F(x-y)dG(y)$$ which is $$\int_{-\infty}^{\infty}F(x-y)G'(y)dy$$ for absolute continuous $G$.

Instead of $$\int_{-\infty}^{\infty}F(x-y)G(y)dy$$

Okay, so this should be the solution:

$\int_{-\infty}^{\infty}F(x-y)dG(y)=\int_{0}^{x}F(x-y)dG(y)\leq\int_{0}^{x}F(x)dG(y)=F(x)G(x),\quad x \geq 0$

答:
Before dealing with the problem itself, we may need one more setting.

Because we have talked about convolution, i.e. $\int_{\mathbb R^+}F(x-y)G(y)dy$, we need that $F$ is also define on $\mathbb R^-$. The reason is that, for any fixed $x\in\mathbb R^+$, when $y$ ranges over $\mathbb R^+$, $x-y$ will range over $(-\infty, x)$. Since your functions $F$ and $G$ are CDFs, we may assume that they are zero on $\mathbb R^-$(i.e. we identify the zero extensions of them with themselves).

With this setting in mind, we can get started now.

The statement in the question is not true, at least for large x.

Counter-Example:

First, because $F$ and $G$ are CDFs, we know that $\lim_{x\to\infty} F(x)=1=\lim_{x\to\infty} G(x)$. And since CDFs are non-decreasing, there exists a real number $a$ s.t. $F(a)>0.9$ and $G(a)>0.9$ .

Pick an $x\in\mathbb R^+$ so large that $x-(a+100)>a.$ So,  $\forall y\in(a,a+100),$ $x-y>a.$

Then, $$\int_a^{a+100} F(x-y)G(y)dy>\int_a^{a+100} F(a)G(a)dy=F(a)G(a)\cdot 100 >81.$$

Because $F$ and $G$ are non-negative, we have that, $$\int_{\mathbb R^+}F(x-y)G(y)dy >\int_a^{a+100} F(x-y)G(y)dy>\int_a^{a+100} F(a)G(a)dy=F(a)G(a)\cdot 100>81.$$

But $F(x)\le 1$ and $G(x)\le 1,$ which implies that for this $x$, we have that $(F\ast G)(x) >81>1 \ge F(x)G(x),$ i.e. $(F\ast G)(x) > F(x)G(x).$

A weaker result (maybe also the best result) we can prove will be: $(F\ast G)(x)\le xG(x)F(x) ,\,\forall x\in \mathbb R^+.$

Fix $x \in \mathbb R^+.$

$|(F\ast G)(x)|\leq\int_{\mathbb R^+}|F(x-y)G(y) |dy=\int_{A\cup B}|F(x-y)G(y) |dy,$ where $A:= \mathbb{R^+} \cap \{x\geq y\}$ and $B:= \mathbb{R^+} \cap \{x\lt y\}$.

On $B$, $F=0,$ and so the integral over $B$ is zero.

On A, because $\{ x\ge y\}$, we see $G(y)\le G(x), \forall y \in A.$ This is the non-decreasing property of CDF.

So, $\int_A|F(x-y)G(y) |dy\le G(x)\int_A|F(x-y) |dy\le G(x)\int_A|F(x)|dy$, where the last inequality holds because CDF is non-negative and non-decreasing and $x-y\in [0,x)$ on $A$.

Now, $G(x)\int_A|F(x)|dy\le G(x)|F(x)|\int_A dy= G(x)|F(x)|x=G(x)F(x)x$.

So we have shown that, $(F\ast G)(x)\le xG(x)F(x).\,\forall x\in \mathbb R^+.$

$$\tag*{$\blacksquare$}$$

Remark 1: By the estimate we proved in the answer, we can see that the statement in the problem is true for $x\in(0,1].$ But as the counter-example shows, it is not true for large $x$.

Remark 2: Actually we don't need any absolute sign in the proof, because every CDF and the convolution of CDFs are non-negative. I didn't realize this when I started writing the proof. I am lazy to change it but you should be aware of this.

Title: $(F\ast G)(x)\leq F(x)G(x)$


问:
Condition on norms to be equivalent using inverse operator theorem
I am trying to prove the following:

Suppose $\left(X,\|\cdot\|_{1}\right)$ and $\left(X,\|\cdot\|_{2}\right)$ are Banach spaces over $\mathbb{F}$. Prove that if $\left\|x_{j}\right\|_{1} \rightarrow 0$ implies $\left\|x_{j}\right\|_{2} \rightarrow 0$ then there are two constants $c_{1}, c_{2}>0$ such that $c_{1}\|x\|_{1} \leq\|x\|_{2} \leq c_{2}\|x\|_{1}$ for all $x \in X$.

If I have one of the inequalities, the inverse operator theorem gives me the other one, but I am not sure how to get one of those, I saw a post that shows that if a sequence is Cauchy in one norm implies that is Cauchy in the other norm, that implies that the norms are equivalent, as I am in Banach Spaces, this are complete and therefore Cauchy sequences converge, for example $x_n\to x \in X$, I can consider maybe $\hat{x}_n=(x_n-x)\to 0$ and somehow use the result from that post. Am I in a good way? I am not sure how to conclude here. Or if I am wrong about this idea can you give me a hint please? Thanks.

答:
There is no completeness involved in  this result so there is no point in looking at Cauchy sequences.

A linear map $T$ from one normed linear space to another is continuous if and only if it  is continuous at $0$ iff $\|Tx\| \leq C\|x||$ for some $C \in (0,\infty)$ (and all $x$).

By hypothesis, the identity map from $(X,\|.\|_1)$ to $(X,\|.\|_2)$  is continuous. By Open Mapping Theorem its inverse is also continuous. This proves the existence of $c_1$ and $c_2$.


问:
Why Edmonds-Karp Shortest Augmenting Path Has Cubic Time
It was given that Edmonds-Karp algorithm bound has cubic running time if it uses the shortest flow-augmenting path:

$|E(G)| \times (|V(G)|+1)/2$

$|E(G)|$ is quadratic in terms in $|V(G)|$, why? For each edge, we have 2 vertices, so for $n$ edges, we have $2n$ not quadratic, is not it?

Second, could you please tell me what is the shortest flow-augmenting path here that we should use in Edmonds-Karp algorithm and how it's different from the f-augmenting path?

Thanks.

答:
You are correct that for each edge we have two endpoints, but that doesn't mean that the number of edges $|E(G)|$ is bounded by $2|V(G)|$. For example, consider a graph where there are $n$ nodes and every possible edge is present (the complete graph $K_n$). Then the number of edges is $\binom{n}{2} = \Theta(n^2)$, since each pair of nodes contributes one edge.

The general idea behind Edmonds-Karp is similar to that of Ford-Fulkerson: find an augmenting path and increase the flow along the path. However, Edmonds-Karp specifically does this by finding, of all possible augmenting paths, the shortest such augmenting path in the graph. This can be done using something like a breadth-first search. (In that sense, you can think of Edmonds-Karp as a very simple optimization on Ford-Fulkerson.)


答:
Bisection method

$$ f= 3x^2 + 2 \sin x -2$$

We can use the bisection method for finding the root, the first is to guess an interval. I will start with an interval $[0,1]$.

Note that:

$$ f(1) = \text{positive quantity}$$

$$f(0) = \text{negative quantity}$$

Hence there must be a root or two in an interval of $[0,1]$ since there is a sign flip at the end of the interval. Now, consider the middle point of the interval which is $.5$, now the value of function at this point is given as:

$$ f(.5) = 3 (.5)^2 + 2 \sin(.5)-2 = \text{negative quantity}$$

Now, we have contracted the original interval to $[.5,1]$, now consider a new point which is the midpoint of this interval, and keep repeating that algorithm till you reach the root!

Title: How do you solve the equation $3x^2+2\sin x=2$?

Body: This seems like a simple equation to solve, but I can't find a way to do it and wolfram alpha can't give me an exact answer.</p>


答:
There is indeed a connection between this and diagonalisability, but nothing that will help you solve this problem faster than the method laid out by @innerproduct.

Question 1 is asking you for an eigenvector $f$ of $\theta$, corresponding to eigenvalue $1$. That is, a non-zero vector $f$ such that $\theta(f) = 1f$. Question 2 is asking you, similarly, for an eigenvector $g$ of $\theta$, corresponding to eigenvalue $-1$.

If you can find such vectors, then $1$ and $-1$ are indeed eigenvalues of $\theta$. Because eigenvectors of distinct eigenvalues are automatically linearly independent, $f$ and $g$ would have to be linearly independent. Since $P_1$ has dimension $2$, this would make $B = (f, g)$ a basis for $\theta$, and in fact, a basis of eigenvectors. As such, the corresponding $2 \times 2$ matrix of $\theta$ corresponding to $B$ must be diagonal, with the corresponding eigenvalues down the diagonal. In particular,
$$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

I reckon this exercise is to illustrate two things:

To see diagonalisation in action, and verify that it works, and

To see that diagonalisation applies not just to linear operators on $\Bbb{R}^n$, but to linear operators on other finite-dimensional vector spaces.

I would suggest finding polynomials $f$ and $g$ exactly as @innerproduct suggests, by turning the problem into two systems of two linear equations of $a$ and $b$, and solving simultaneously. You'll find that the two equations in each system are actually just multiples of each other, and you can choose any non-zero value of $b$ (e.g. just use $b = 1$, or whatever you want) and find a suitable (non-zero) value of $a$. These values are the coefficients of $1$ and $x$ in your polynomial.

Alternatively (and to be clear, I wouldn't recommend this approach when solving this problem for the first time) you could also turn the whole operator into a matrix. If we pick a basis $S = (1, x)$ for $P_1$, we can compute the corresponding matrix like so:
\begin{align*}
\theta(1) = 3 - 2x &\implies [\theta(1)]_S = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \\
\theta(x) = 4 - 3x &\implies [\theta(x)]_S = \begin{pmatrix} 4 \\ -3 \end{pmatrix}.
\end{align*}
Therefore,
$$[\theta]_S = \begin{pmatrix} 3 & 4 \\ -2 & -3 \end{pmatrix}.$$
The eigenvalues of this matrix are precisely the eigenvalues of $\theta$ (which, if you compute them, should be $\pm 1$). The eigenvectors of this matrix are the coordinate vectors of the eigenvectors of $\theta$. So, if you're more comfortable with matrices, you can do all the same calculations, leaving polynomials behind for the most part.

Anyway, I hope that helps.

Title: Given linear map, find nonzero polynomials such that...

Body:
Consider the linear map
\begin{align*}
      \theta : P_1 &\to P_1 \\
      a+bx &\mapsto (3a + 4b) - (2a + 3b)x
\end{align*}

Find a non zero polynomial $f \in P_1$ such that $\theta(f) = f$.

Find a non zero polynomial $g \in P_1$ such that $\theta(g) = -g$.

Show that $f,g$ is a basis of $P_1$, and find the matrix representing $\theta$ in the basis of $f, g$.

I am unsure how to go about solving 1 and 2 without just plugging in numbers for $a$ and $b$ until the two sides match. Is there a better way to go about solving 1 and 2 that is possibly faster than just plugging in numbers? These problems are on a worksheet dedicated to a certain section of my textbook that explains matrix similarity and diagonalizability so I feel there is something that I am missing that would allow me to solve 1 and 2 using similarity or diagonalizability.


问:
$F_{1} \times F_{2}$ is a closed subset then $F_{1}$ is a closed subset of $\mathbf{R}^{m}$ and $F_{2}$ is a closed subset of $\mathbf{R}^{n}$
I am reading Axler's MIRA and found the following question:

Suppose $F_{1}$ is a nonempty subset of $\mathbf{R}^{m}$ and $F_{2}$ is a nonempty subset of $\mathbf{R}^{n}$ Prove that $F_{1} \times F_{2}$ is a closed subset of $\mathbf{R}^{m} \times \mathbf{R}^{n}$ if and only if $F_{1}$ is a closed subset of $\mathbf{R}^{m}$ and $F_{2}$ is a closed subset of $\mathbf{R}^{n}$

答:
HINT:

Let $X$, $Y$ be topological spaces. Then the map
$$x\mapsto (x, y)$$
from $X$ to $X\times \{y\}\subset X\times Y$ (with subspace topology) is a homeomorphism, with inverse $(x,y) \mapsto x$.

Now consider a section $F_1\times \{y\}$ of $F_1\times F_2$, which equals $(F_1\times F_2) \cap (X\times \{y\})$, so a closed subset of $X\times \{y\}$.


答:
Suppose $F_1 \times F_2$ is closed. Fix any $y \in F_2$. Then $(F_1 \times F_2) \cap (\mathbb R^{m} \times \{y\})$ is closed. This means $F_1 \times \{y\}$ is closed which implies that $F_1$ is closed.

[The projection to the first coordinate becomes a homeomorphism when restricted to $\mathbb R^{m} \times \{y\}$]

Title: $F_{1} \times F_{2}$ is a closed subset then $F_{1}$ is a closed subset of $\mathbf{R}^{m}$ and $F_{2}$ is a closed subset of $\mathbf{R}^{n}$

Body:
I am reading Axler's MIRA and found the following question:

Suppose $F_{1}$ is a nonempty subset of $\mathbf{R}^{m}$ and $F_{2}$ is a nonempty subset of $\mathbf{R}^{n}$ Prove that $F_{1} \times F_{2}$ is a closed subset of $\mathbf{R}^{m} \times \mathbf{R}^{n}$ if and only if $F_{1}$ is a closed subset of $\mathbf{R}^{m}$ and $F_{2}$ is a closed subset of $\mathbf{R}^{n}$


问:
Find a definite integral and calculate the average temperature
A geographer measures the temperature in a place with a mild climate. He sees that the temperature is very well approximated by the function $y=0.25T_0+T_0sin^2\left(x\pi \right)$. He has taken measurements for two full years,
$0 ≤ x ≤ 2.$

Make a qualitative sketch of the temperature function and use a suitable definite integral to calculate the average temperature over the whole measurement period.

Like I need someone to guide me through this, I think I need to integrate y, but idk how? Do I take T as x? but there is already x in the equation. Do I replace T with 2 since hat is the duration of the measurement? What do I do with T??

答:
It is given that $y = 0.25T_0 + T_0\sin^2(x\pi)$. Note that $T_0$ is just a fixed constant. This is a function of $x$, so we find it natural to write $y$ using function notation as $y = T(x)$. You should ask yourself what the variable $x$ represents in this question based on the problem data. To calculate the average temperature, there is a useful formula that says if $f$ is a function on the interval $[a,b]$ to $\mathbb{R}$, then the average value of $f$ over $[a,b]$ (call it Avg) is  $$\text{Avg} = \frac{1}{b-a}\int_a^b f(x)dx$$
See if you can apply this result to your problem.


问:
How to compute partial derivatives by definition?
For $$f(x,y)=\begin{cases}
\frac{x^{2}y}{x^{4}+y^{2}}, & x\ne0,y\neq0\\
0 & ,x=0,y=0
\end{cases}$$

I'm trying to compute $f_x(0, 0)$ and $f_y(0, 0)$. I'm trying to apply the definition but in both cases I get indeterminate forms:

$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{h^{2}*0}{h^{4}+0^{2}}}{h}=\lim_{h\to0}\frac{h^{2}*0}{h^{5}+h*0^{2}}=\lim_{h\to0}\frac{0}{h^{5}}=\frac{0}{0}=?$$

$$\frac{\partial f}{\partial y}(0,0)=\lim_{h\to0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{0h}{0+h^{2}}}{h}=\lim_{h\to0}\frac{0}{h^{3}}=\frac{0}{0}=?$$

What am I doing wrong here?

答:
Note that $0$ divided by anything except $0$ is $0$. Also, note that infinite small value is not considered to be $0$. So, you do not get $\cfrac{0}{0}$ but $\cfrac{0}{\text{something very small}}$.

Therefore, your limit is:
$$\lim_{h\to 0} \frac 0 h =\lim_{h\to 0} 0 = 0 $$


问:
If $X$ is homeomorphic to dense subspace $Y\subseteq X$, then $X=Y$
It is a basic fact of topology that if $X$ is a topological space and $Y\subseteq X$ is homeomorphic to $X$, it does not need to occur that $X=Y$ (for example, $X=\mathbb{R}$, $Y=(0,1)$). My question is, if I add the requirement that $Y$ is dense in $X$. Is this still the case? Or is the following true?

If $X$ is homeomorphic to a dense subspace $Y\subseteq X$, then $X=Y$.

答:
It's not true.  Consider $Y=\mathbb N$ and $X=\mathbb Z$, both with the trivial (indiscrete) topology.

Then $Y$ is dense in $X$ (every non-empty subset is dense in this topology) and homeomorphic to $X$

(there is a bijection between $Y$ and $X$, and every map to an indiscrete space is continuous),

but clearly $X\ne Y$.


问:
Show the following integral is finite
Given $f$ a probability density function that is symmetric around $0$, and assume that
$$
\int_R x^2f(x)dx < \infty,
$$
and
$$
\int_R f^2(x)dx < \infty.
$$
I want to show
$$
\int_R x^2f^2(x)dx < \infty.
$$
My way of tackling this is by separating the real line into
$$
A = \{x: f(x) > 1\}
$$
and $A^c$. Then I have
$$
\int_R x^2f^2(x)dx = \int_Ax^2f^2(x)dx + \int_{A^c}x^2f^2(x)dx < \int_Ax^2f^2(x)dx + \int_{A^c}x^2f(x)dx.
$$
So
$$
\int_R x^2f^2(x)dx < \int_Ax^2f^2(x)dx + \int_R x^2f(x)dx.
$$
I am thinking that since $f$ is a PDF, then $A$ must have finite measure. Through this I can somehow show
$$
\int_Ax^2f^2(x)dx < \infty,
$$
but I am failing in showing this inequality. Can anyone provide some hint or maybe a different way of tackling this? Thank you!

答:
This is false. Let $g(x)=n$ for $n-\frac 1 {n^{5}}<x<n+\frac 1 {n^{5}}$ for $n =2,3,4...$ and $g(x)=0$ for all other positive numbers $x$. Let $g(-x)=g(x)$ for $x >0$ and $g(0)=0$.

Then $\int x^{2}g(x) dx <\infty$ and  $\int [g(x)]^{2}dx <\infty$ but $\int x^{2} [g(x)]^{2}dx =\infty$. Take $f$ to be a constant multiple of $g$ so that it becomes a probability density function.

Title: Show the following integral is finite


问:
Let Q(n) be the sum of digits of n. Prove that Q(n) = Q(2n) implies 9|n
I don't know how to approach this question any hint would be appreciated .</p>

答:
Hint
$$Q(n) \equiv n \pmod{9}$$

Therefore $Q(n)=Q(2n)$ implies
$$n \equiv 2n \pmod{9}$$


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答: I propose Sage. In my opinion it is the best free open-source mathematics software system.</p>

Title: Free online mathematical software


问:
Find limit of $\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}$
Given that for each $a\in (0, 1)$,  $\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}$ exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$. Find the value of $g(\frac{1}{2})$

My approach

$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^a}{{\left( {1 - t} \right)}^{ - 1}}dt} $

$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{{\left( {\frac{{1 - t}}{t}} \right)}^a}{{\left( {1 - t} \right)}^{ - 1}}dt}$

$g(a)=\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\frac{1}{{1 - t}}{{\left( {\frac{{1 - t}}{t}} \right)}^a}dt} $

From here onward not able to approach

答:
$$g\left(\frac{1}{2}\right) = \int_0^1 \frac{dx}{\sqrt{x(1-x)}}$$

Now, if $u = \sqrt{x} \implies du = \frac{dx}{2\sqrt{x}}$

$$\implies g\left(\frac{1}{2}\right) = 2\int_0^1\frac{du}{\sqrt{1-u^2}} = 2[\arcsin(u)]_0^1 = \pi$$


问:
Normal distribution comparing 3 or more variables
Let's say we had 3 students about to run race and their previous run times each follow a normal distribution,

$X \sim N(10, 2)$

$Y \sim N(14, 3)$

$Z \sim N(5, 1)$

Where $X$, $Y$ and $Z$ are independent.

I want to know if it is possible to work out the probability that $X$ will run the fastest time?

I feel like the question boils down to $P((X>Y) \cap (X>Z))$. I can work out $P(X>Y)$ and $P(X>Z)$  by creating a new distribution for each and just calculating mean and variance. But I am struggling to go this extra step to determine the probability that both scenarios are true.

答: If $\phi_1,\phi_2,\phi_3$ are the density function sof $X,Y$ and $Z$ then $$P((X>Y)\cap (X>Z))$$ $$=\int_{-\infty}^{\infty} \int_z^{\infty}\int_y^{\infty} \phi_1(x)\phi_2(y)\phi_3(z) dx dydz$$ $$+ \int_{-\infty}^{\infty}  \int_{-\infty}^{z}\int_z^{\infty} \phi_1(x)\phi_2(y)\phi_3(z) dx dydz$$


答:
Counterexample:

$$\mathbb{Q}\backslash \{0\} \subset \mathbb{Q}$$

$\bf{Added:}$  It turns out that if $A$, $B$ are dense, countable subsets of $\mathbb{R}$, then there exists a homeomorphism $f\colon \mathbb{R}\to \mathbb{R}$, such that $f(A)=B$.

Perhaps an explicit example is best. Consider $A=\mathbb{Q}$, and $B = \mathbb{Q}^3\subset \mathbb{Q}$. The map $x\mapsto x^3$ is a homeomorphism of $\mathbb{R}$ taking  $A$ to $B$.

Title: If $X$ is homeomorphic to dense subspace $Y\subseteq X$, then $X=Y$

Body:
It is a basic fact of topology that if $X$ is a topological space and $Y\subseteq X$ is homeomorphic to $X$, it does not need to occur that $X=Y$ (for example, $X=\mathbb{R}$, $Y=(0,1)$). My question is, if I add the requirement that $Y$ is dense in $X$. Is this still the case? Or is the following true?

If $X$ is homeomorphic to a dense subspace $Y\subseteq X$, then $X=Y$.


答:
For a more interesting example, let $X=\Bbb Q$ with the usual topology, and let

$$Y=\left\{\frac{m}{2^n}\in\Bbb Q:m,n\in\Bbb Z\text{ and }n\ge 0\right\}$$

be the set of dyadic rationals. Then $Y$ is a proper subset of $X$ that is both dense in $X$ and homeomorphic to $X$.


问:
global section of a locally free sheaf
This is from Hartshorne:

Let $s\in \Gamma(X,\mathcal{E})$ be a global section of a locally free sheaf $\mathcal{E}$ of rank $r$ on $X$. Then $s$ defines a homomorphism from $\mathcal{O}_X$ to $\mathcal{E}$ by sending $1$ to $s$.

Can someone explain in more detail how this homomorphism should be defined?

答:
Whenever $(X, \mathcal{O}_X)$ is a ringed space and $\mathcal{F}$ is an $\mathcal{O}_X$-module, you can pick a global section $s \in \Gamma(X,\mathcal{F})$ to define the multiplication by $s$ morphism $s \colon \mathcal{O}_X \rightarrow \mathcal{F}$, $f \mapsto sf$. Concretely it is given by multiplying with the restriction of $s$, i.e. by $\left.s\right|_U \colon \Gamma(U,\mathcal{O}_X) \rightarrow \Gamma(U,\mathcal{F})$, $f \mapsto \left.s\right|_U \cdot f$ on the level of local sections.

On the other hand, a morphism $\mathcal{O}_X \rightarrow \mathcal{F}$ also determines such a global section, by taking the image of $1 \in \Gamma(X,\mathcal{O}_X)$.

Title: global section of a locally free sheaf


问:
Difference of sets with size of them
$X$ is an uncountable set and $\mathcal B$ is the sigma-algebra of all the subsets $A\subseteq X$ such that either $A$ is countable or $X\setminus A$ is countable. How to prove that $f: X \to [0,\infty]$ is a $\mathcal B$-measurable function if and only if there is an uncountable set $A_f\in\mathcal B$ such that $\left.f\right\rvert_{A_f}$ is constant?

答:
First assume that there is an uncountable set $A_f\in\mathcal{B}$ such that $f|_{A_f}=k$ a constant. Then by definition $X\setminus A_f$ is countable. This means that $f$ can only take countably many values ($k$ and $f(x)$ for each of the countably many $x\in X\setminus A_f$). Thus any Borel set $B\subset[0,\infty]$ interects at most countably many of these values. If $k\notin B$ then $f^{-1}(B)\subset X\setminus A_f$ and thus must be countable and so $f^{-1}(B)\in\mathcal{B}$. If $k\in B$ then consider $B\setminus\{k\}$, which is itself a Borel set. We have shown that $f^{-1}(B\setminus\{k\})\in\mathcal{B}$. Additionally $f^{-1}(\{k\})=A_f\cup\{\text{some (countable) subset of }X\setminus A_f\}\in\mathcal{B}$. Thus $f^{-1}(B)=f^{-1}(B\setminus\{k\})\cup f^{-1}(\{k\})$ which is clearly in $\mathcal{B}$. Therefore $f$ is $\mathcal{B}$-measurable.

Now to show the other implication assume that $f$ is $\mathcal{B}$-measurable. Then for any Borel set $B\subset[0,\infty]$, $f^{-1}(B)$ is countable or $X\setminus f^{-1}(B)=f^{-1}([0,\infty]\setminus B)$ is countable. Notice $[0,\infty]=\cup_{k=1}^\infty[\frac{k-1}{m},\frac{k}{m}]$ for all $m\in\mathbb{Z}^+$. Fix $m$. If $f^{-1}([\frac{k-1}{m},\frac{k}{m}])$ is countable for all $k$ then $f^{-1}([0,\infty])=X$ would be countable since the countable union of countable sets is countable. Thus there exists $k$ such that $f^{-1}([\frac{k-1}{m},\frac{k}{m}])$ is not countable, and thus $X\setminus f^{-1}([\frac{k-1}{m},\frac{k}{m}])=f^{-1}([0,\infty]\setminus[\frac{k-1}{m},\frac{k}{m}])$ is countable. Notice that this means that there can be only one such $k$ for each $m$ since $[0,\infty]\setminus[\frac{k-1}{m},\frac{k}{m}]=\cup_{l=1,l\neq k}^\infty[\frac{l-1}{m},\frac{l}{m}]$. Thus we get a sequence $\{k_m\}_{m=1}^\infty$ such that $f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}])$ is uncountable. Notice also that these intervals must be nested. A decreasing nested sequence of closed compact sets has non-empty intersection and because the length of these intervals goes to zero $\cap_{m=1}^\infty[\frac{k_m-1}{m},\frac{k_m}{m}]=\{h\}\subset\mathbb{R}$. Thus
$$f^{-1}(\{h\})=f^{-1}(\cap_{m=1}^\infty [\frac{k_m-1}{m},\frac{k_m}{m}])=\cap_{m=1}^\infty f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}])$$
$$=\cap_{m=1}^\infty X\setminus(X\setminus f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}]))=X\setminus(\cup_{m=1}^\infty (X\setminus f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}])))$$
$X\setminus f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}])$ is countable and the countable union of countable sets is countable so $\cup_{m=1}^\infty (X\setminus f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}]))$ is countable. Thus since $X$ is uncountable, $X\setminus(\cup_{m=1}^\infty (X\setminus f^{-1}([\frac{k_m-1}{m},\frac{k_m}{m}])))$ is uncountable and in $\mathcal{B}$ by definition of $\mathcal{B}$. Therefore $f^{-1}(\{h\})$ is uncountable and in $\mathcal{B}$.

Title: Difference of sets with size of them


问:
Find the number of ways to get a total of 10.
My math teacher was explaining a question in the chapter Combinatorics and I met with a hiccup in the solution. The question is given below and the solution (as given by my teacher) is also given in the form of a screenshot of the zoom meeting.

$3$ dices are rolled simultaneously once and the resulting numbers are added. Find the number of ways of getting a total of $10$.

The solution

The hiccup is that I cant understand how the last line is equal to the one before it.

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

\begin{align}
&\bbox[5px,#ffd]{\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}
\bracks{z^{10}}z^{d_{1}\ +\ d_{2}\ +\ d_{3}}} =
\bracks{z^{10}}\pars{\sum_{d_{1} = 1}^{6}z^{d}}^{3} =
\bracks{z^{10}}\pars{z\,{z^{6} - 1 \over z - 1}}^{3}
\\[5mm] = &\
\bracks{z^{7}}\pars{1 - z^{6}}^{3}\pars{1 - z}^{-3} =
\bracks{z^{7}}\pars{1 - 3z^{6}}\pars{1 - z}^{-3}
\\[5mm] = &\
\bracks{z^{7}}\pars{1 - z}^{-3} -
3\bracks{z^{1}}\pars{1 - z}^{-3} =
{-3 \choose 7}\pars{-1}^{7} - 3{-3 \choose 1}\pars{-1}^{1}
\\[5mm] = &\
{9 \choose 7} - 3{3 \choose 1} = 36 - 3 \times 3 = \bbx{27} \\ &
\end{align}


问:
Proving the Fourier equation: $\frac{1}{2} L - x = \frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin\frac{2n\pi x}{L}$
The following question is about Fourier series, specifically about extension of the Fourier Series to arbitrary intervals:

"Show that:

$$\frac{1}{2} L - x = \frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin\frac{2n\pi x}{L},\quad 0<x<L$$"

The question also came with the following hint:

"Note that the series on the right-hand side has the period L. Therefore, it should be viewed as a Fourier series on an interval of length L; in this case, it is $[0, L]$. Think about how
to reduce this scenario into those we have discussed before (e.g. the classic case on $[-\pi, \pi])$ by a suitable change of variable"

I thought to use the change of variable: $t = \frac{2 \pi x}{L} - \pi$, so that as x runs from $0$ to $L$, $t$ runs from $-\pi$ to $\pi$.

After substituting in the variable, the right-hand side reduced to:

$$\frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin(nt + n\pi),\quad -\pi<t<\pi$$

Furthermore, since:

\begin{align*}
\sin(nt + n\pi) &= \sin(nt)\cos(n\pi) + \cos(nt)\sin(n\pi)\\
&= \sin(nt)(-1)^n + 0 \quad \text{since n is an integer}
\end{align*}

Hence, the above expression simplifies to:

$$\frac{L}{\pi}\sum^\infty_{n=1}\frac{(-1)^n}{n} \sin(nt),\quad -\pi<t<\pi$$

I got stuck here. I'm not sure if there is some identity that is just evading me so that this reduces to the desired expression $\frac{1}{2}L - x$, and if not, I'm not sure how to get from here to the required expression.

Any help would be appreciated, thank you!

答:
$$f(x)=\Im \frac{L}{\pi} \sum_{n=1}^{\infty} \frac{e^{2in\pi x/L}}{n}=-\Im\frac{L}{\pi}\ln[1-e^{2i \pi x/L}]=-\Im \left (\frac{L}{\pi} \ln [e^{-i\pi x/L}-e^{i  \pi x/L}]+\frac{L}{\pi} \ln e^{i\pi x/L}\right)$$
$$\implies f(x)=-\Im\frac{L}{\pi} \ln(-2i\sin \pi x L)+ix]=\frac{L}{2}-x.$$
Here, we have used $\sum_{n=1}^{\infty} \frac{z^n}{n}=-\ln(1-z), |z| \le 1$ and $e^{iz}-e^{-iz}=2i\sin z$


问:
Using $p\supset q$ instead of $p\implies q$
I saw that a use for the notation $p\supset q$ instead of $p\implies q$
that got me a bit confused.

One occurrences is in this Wikipedia link.

It seems to me opposite than what it should be, let me explain what
I mean:

If $A,B$ are sets s.t $A\subset B$, $p$ is for $x\in A$, and $q$
is for $x\in B$ then we can identify (in some way) $$A\text{ with }p$$
$$B\text{ with }q$$

We have it that $p\implies q$, since $A\subset B$, but in the above
notation we have $q\subset p$ which to me looks like $B\subset A$

which is the opposite of what we wanted to express.

Can someone please explain to me the logic behind this notation ?

答:
One way to rationalise this notation is to think of a proposition as having a certain information content.

Then $p \supset q$ can be thought of as "the information content of $q$ is contained in that of $p$". A particular piece of information that can be obtained from $q$ is "$q$ is true".

Thus we see that $p \supset q$ naturally gives rise to the statement $p \implies q$, and conversely.

However, I agree with Hagen von Eitzen's comment that $p \to q$ should be the notation of choice, at least in symbolic logic. One could still use $\implies$ in a meta-context, i.e. as more or less a part of the natural language we discuss mathematics in.


问:
Probability of university students studying mathematics
At MIT, forty percent of the students are in Mathematics.

Thirty percent of Mathematics students study Probabilistic Systems, and fifty percent of the students studying Probabilistic Systems are in Mathematics.  What percentage of students at MIT are either in Mathematics or studying Probabilistic Systems?

work:

assuming no. of students=100

m denotes student in mathematics, p denotes student in probabilistic systems

n(m) + n(p) - n(m and p)

=40+24-12

=52 = 52%

答:
$40$ students out of $100$ are in Maths. $0.3 \cdot 40 = 12$ of them are in PS too. And this $12$ seems to be a $50\%$ of PS students. Hence there are $24$ students in PS course. Hence, $$n(\{ \text{Math students}\} \cup \{ \text{PS students}\}) = 40 + 24 -12 = 52$$

So, the answer is $52\%$.


问:
Finding the norm of g
Consider polynomials $f,g \in P^5(-1,1)$ satisfying $||f|| = \sqrt{6}$, $||f+g||= 4$ and $||f-g||=2$. Where the norm is the $L^2$ norm.

What is $||g||$?

Originally I though to use the triangle inequality on $||f-g||=2$ to get that $||f||= \sqrt{6} \leq ||g||+2$ but I'm unsure that this is the right way to go. I don't think we can use the triangle equality on $||f+g||$, we can but I'm unsure that it will help here. What can I do to find $||g||$

答: $\int f^{2}=6, \int (f+g)^{2}=16$ and $\int (f-g)^{2}=4$. Adding the second and the third we get $2(\int f^{2}+\int g^{2})=20$. Combined with the first we get $\int g^{2}=4$ so $\|g\|=2$.

Title: Finding the norm of g


问:
Finite Sum of Infinite Sums is Infinite Sum of Finite Sums?
If I have a finite sequence of $N$ functions $f_n\colon\mathbb{N}\to\mathbb{C}$ and a sequence of complex numbers $z_k$, must it be true that

$$\sum_{n=1}^{N} \sum_{k=1}^\infty f_n(z_k) = \sum_{k=1}^\infty \sum_{n=1}^N f_n(z_k)?$$

It seems that a similar question is addressed at Summation Symbol: Changing the Order, but this question addresses only the case where both sums are finite or both are infinite and doesn't seem to address what happens when we're considering the finite sequence of functions.

Motivation for this question

It seems like an equality of this form is used to prove Lemma 5.4 in the proof of Dirichlet's Theorem on Arithmetic Progressions in http://people.csail.mit.edu/kuat/courses/dirichlet.pdf, but the use of the identity isn't explicit so I'm not sure if I'm understading this right. I think I could make sense of the lemma's proof if the above formula always holds, but I don't know if that is a valid assumption or not.

Any assistance is greatly appreciated!

答: Without any assumption the statement doesn't hold in general. Let $a_n=1$ and $b_n=-1$ for all $n\in \mathbb{N}$, then we have that $\sum_{n\geqslant 1}(a_n+b_n)=\sum_{n\geqslant 1}0 =0$ but $\sum_{n\geqslant 1}1=\infty $ and $\sum_{n\geqslant 1}(-1)=-\infty $, therefore the sum $\sum_{n\geqslant 1}a_n+\sum_{n\geqslant 1}b_n$ doesn't exists.


答: Here is a drawing of the solution. Your solution is correct but to check it, and to face more difficult examples, a Venn diagram like the following is very useful (IMHO)

Title: Probability of university students studying mathematics

Body:
At MIT, forty percent of the students are in Mathematics.

Thirty percent of Mathematics students study Probabilistic Systems, and fifty percent of the students studying Probabilistic Systems are in Mathematics.  What percentage of students at MIT are either in Mathematics or studying Probabilistic Systems?

work:

assuming no. of students=100

m denotes student in mathematics, p denotes student in probabilistic systems

n(m) + n(p) - n(m and p)

=40+24-12

=52 = 52%


问:
Showing that $\mathbb{P}[X\geq a]\leq \exp[-ta]\mathbb{E}[\exp[tX]]$
The problem is to show that $\mathbb{P}[X\geq a]\leq \exp[-ta]\mathbb{E}(\exp[tX])$ given $\exp(tX)<\infty$ for $t\in \mathbb{R}$ where $X$ is a random variable.

Then to show that $\mathbb{P}[X\geq a]\leq \inf\limits_{t\in\mathbb{R}} \exp[-ta]\mathbb{E}(\exp[tX])$. But I imagine this shouldn't be too difficult after the first part.

I tried to use Jensen's inequality.

$\psi(X)=\exp(tX)$ is a convex function so

$\psi(\mathbb E (X)) \leq\mathbb  E (\psi (X)) \equiv\mathbb E( \exp[tX])$

So I am left with $\mathbb P [X\geq a]\exp[ta]$ on the LHS which seems to have no relation to $\exp(\mathbb E[tX])$ i.e. $\psi(\mathbb E(tX))$

答:
Hint: The inequality is equivalent to showing that:
$$
{\rm E}\left[1_{\{X\geq a\}}e^{ta}\right]\leq {\rm E}\left[e^{tX}\right].
$$


问:
Probability Distribution vague problem
A clerk at a convenience store leaves her counter unattended at 3:02 PM. When she returns at 3:08 PM, there is an impatient customer waiting at the counter. Given that the impatient customer did not arrive in the first 2 minutes, what is the probability that he arrived in the last 2 minutes of this interval?

Is there even a method to do this. Because there should be information about mean or the type of distribution. Or am i missing something?

答:
the interval of unattended counter is 6 minutes long, that is the union of 3 equal intervals of 2 minutes long.

Given the fact that in the first 2 minutes nobobody arrived, the probability that the customer arrived in the last 2 minutes is $\frac{1}{2}$

EDIT: assuming uniform distribution of arrival times

Title: Probability Distribution vague problem


问:
Can $PSL(2, p)$ be a subgroup of $PSL(2,q)$ if $p$ doesn't divide $q$?
Let $p$ be a prime and let $q$ be a power of another prime. Suppose $p\nmid q$, then is it possible that $PSL(2, p)$ can be isomorphic to a subgroup of $PSL(2,q)$?

Any suggestion or reference is welcome.

答:
This can happen if $p\in \{2,3,5\}$.

Look here for a list of the subgroups of $\mathrm{PSL}(2,q)$: https://www.staff.ncl.ac.uk/o.h.king/KingBCC05.pdf

You will see for example that $\mathrm{PSL}(2,5)(\cong A_5)$ occurs when $q\equiv \pm 1 \pmod {10}$.

You can do the same for $p\in\{2,3\}$ by remembering that $\mathrm{PSL}(2,3)\cong A_4$ and $\mathrm{PSL}(2,2)$ is a dihedral group of order $6$.


问:
Need help on with this problem on linear mapping of co-ordinates of square.
This is a homework question. If I could get a hint or solution for the first part, it would be really helpful. I couldn't attend the lecture due to my illness.

答:
I'll give you the concept of finding $A$ and $\vec{v}$, and see whether you can apply it

$A$ represents deformations. These type of transformations include stretching and rotations. $\vec{v}$ represents translations, which will be the same for all points of the figure.

Since $A$ cannot remap the origin, to find $\vec{v}$ is trivial, just see the difference of the origin (or in this case, the first point of the square)

For $(1)$, $\vec{v} = (1,1)$ and for (2), $\vec{v} = (2,0)$

Now, you can subtract $\vec{v}$ from each point to see what just the $A$ transformation will give you. For tthe first part, you get the original square, so $A$ is identity

For the second part, you get a parallelogram which shares the base of the square, but shifts outward in linear proportion to $y$ as you go up. This can be written as $(x,y) \to (x+y,y)$

Hence $A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$


问:
Heat Equation on a Cylinder
Please note this is a homework question, and I just want some discussion on the choice of the separation parameter.

Suppose we have a cylinder which satisfies the following steady-state heat equation:

$$u_{rr}+\frac{1}{r}u_r+u_{zz}=0$$,

with the following boundary conditions:

\begin{align}
u(r,0)&=0\\
u(r,20)&=20\\
[u_r+u]|_{r=4}&=0
\end{align}

Separation gives two very standard ODEs - the radial part gives a parametric Bessel equation of order $0$, and the other gives either a hyperbolic equation or a trigonometric equation. In other examples available to me it is stated that making $Z(z)$ be hyperbolic "makes sense in the context," with no explanation of what exactly is meant by that. I'm pretty sure the choice here depends entirely on how we expect $Z(z)$ to behave. If we expect the $Z$ part to be hyperbolic I have absolutely no idea why we expect it to be hyperbolic.

答:
If one insists that the $Z(z)$ function is oscillatory, then $R(r)$ obeys the modified Bessel equation of order 0 instead which does have solutions $I_0(\lambda x), K_0(\lambda x)$. The second solution diverges at the origin so we can't use it to build a set of solutions. The problem with the first solution is that, unfortunately, it doesn't exhibit oscillatory behavior and in fact it is always positive. As a result the 3rd boundary condition cannot be satisfied or has  a finite number of functions that can satisfy it, which in turn stiffens the boundary condition in the z-direction, in the sense that most boundary problems you come up with have no solution.

All in all, there is a give and a take in elliptical equations in which direction is oscillatory and which is not. We always choose critically which one will be given the oscillatory behavior given how constraining the boundary conditions are. In this case, the radial direction needs the oscillatory behavior of a Bessel function so that it can satisfy the radial boundary condition. In other cases one has to think about both possibilities to make a decision, since there is no general rule of thumb.

Title: Heat Equation on a Cylinder


问:
Problem on multiplicative subsets 1
Let $f:A\to{B}$ be a homomorphism of rings. If $f$ is surjective, is it true that every multiplicative subset $\Gamma\subseteq{B}$ has the form $f(\Sigma)$ with $\Sigma\subseteq{A}$ a subset multiplicative of $A$?

Please, I need help with this question. Thank you.

答: Hint: $f^{-1}(Γ)$ is multiplicative.

Title: Problem on multiplicative subsets 1

Body: Hint: $f^{-1}(Γ)$ is multiplicative.