问:
Proof that a sequence $(x_n)_{n \ge 1}$ such that $|x_{n+1} - x_n| < \frac{1}{n}$ converges.
I'm studying for a test and I came across this exercise:

Let $(x_n)_{n \ge 1}$ be a sequence such that

$x_{2n-1} \le x_{2n+1} \le x_{2n+2} \le x_{2n}, \hspace4ex |x_{n+1} - x_n| < \frac{1}{n}$

Prove that the sequence converges.

I've been stuck on this question for a while now, by doing some arithmetics I've gotten a few inequalities, namely:

(1) $x_{2n} - x_{2n+1} < \frac{1}{2n}$

(2) $x_{2n+2} - x_{2n+1} < \frac{1}{2n+1}$

(3) $x_{2n} - x_{2n-1} < \frac{1}{2n-1}$

But I have no idea where to go from here, or even if those inequalities are useful at all.

答: Argument may need some minimal corrections We have the sequence $x_{2n+1}$ increasing and bounded above by $x_2$, you could try by induction as $x_1\leq x_3\leq x_5\leq x_7\leq x_9 \leq x_{10}\leq x_8\leq x_6\leq x_4\leq x_2$, so for any $n$, $x_{2n+1}\leq x_2$. Then by the Bolzano theorem, it has a subsequence (the odd subsequence has a subsequence $y_{{n}_k}$ which converges to $X$). Then if $x_{2n+1}$ is convengent it should converges to $X$. Notice that $y_{n_k}=x_{2n+1+2p}$ for some $p(n)=p>0$, and $$|x_{2n+1}- X|\leq |x_{2n+1}-y_{n_k}|+|y_{n_k}-X|$$ here you know how to make $|y_{n_k}-X|$ small, it remains to deal with $$|x_{2n+1}-y_{n_k}|=|x_{2n+1}-x_{2n+2}+x_{2n+2}-x_{2n+3}+x_{2n+3}-\cdots+x_{2n+1+2(p-1)}-x_{2n+1+2p}|\leq \frac{2p}{2n+1}\to 0$$ then the next step is to notice $|x_{2n+1}-x_{2n}|<\frac{1}{2n}\to 0$ then $x_{2n}\to X$, so $x_{n}$ also converges to $X$. You could try to do this more rigurous.


问:
Why is $y=b^x$ not continuous for $b>0$ and $b\ne 1$?

The lines graphed by such do not have an interrupted domain.

So why is my textbook telling me this?

E: Typo in title. See photo.

答:
$b$ is a fixed number in the function, not the variable.

What your textbook says is that, for every (fixed) real number $b$ such that $b>0$ and $b\neq 1$, the function $y=b^x$ is a continuous function on the real line. The domain of this function is the set of all real numbers. No interruptions.

For exponential function (and the logarithm) the fixed number $b$ is called the basis. We do not consider the case $b=1$ simply because it is trivial (a constant function) and this trivial case behaves rather differently from the nontrivial cases.

ContentLicense: CC BY-SA 4.0


答:
$y=b^x$ is continuous, the definition of continuous function is
Let $f: [a,b] \to \mathbb{R}$ is continuous in a point $c \in I$ if
$\forall \varepsilon >0$ there exists $\delta>0$ such that
When $|x-c|<0$ then $|f(x)-f(c)|< \varepsilon$
Or actually true
$f $ is continuous function in $c$ if
$\lim_{x \to c} f(x)=f(c)$
And ass other comments say the function
$y=b^x$ is continuous for $b\neq 0$
And for $b=1$ the function is constant

Title: Why is $y=b^x$ not continuous for $b>0$ and $b\ne 1$?

Body:
The lines graphed by such do not have an interrupted domain.

So why is my textbook telling me this?

E: Typo in title. See photo.


问:
Does $\sum_{k = 1}^{n-1} \frac{\binom{2k}{k}\binom{2n-2k}{n-k}}{k(n-k)}$ have a simpler expression?

The following constants have appeared in my recent research and I was wondering if they have a simpler expression $?$.

I have calculated several values but I don't see any obvious combinatorial answer.

$$
\sum_{k = 1}^{n-1}
\frac{\displaystyle\binom{2k}{k}\binom{2n - 2k}
{n - k}}{k\left(n-k\right)} =\ ?
$$
and more general
$$
\sum_{\substack{k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{r}\ =\ n
\\[1mm] k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}}
\frac{\displaystyle\binom{2k_{1}}{k_{1}}
\binom{2k_{2}}{k_{2}}\cdots\binom{2k_{r}}{k_{r}}}{k_{1}\,k_{2}\ldots k_{r}} =\ ?
$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{%
\sum_{\substack{k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{r}\ =\ n
\\[1mm] k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}}
{\ds{{2k_{1} \choose k_{1}}
{2k_{2} \choose k_{2}}\cdots{2k_{r} \choose k_{r}}} \over k_{1}\,k_{2}\ldots k_{r}}}
\\[5mm] = &
\sum_{k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}
{\ds{{2k_{1} \choose k_{1}}
{2k_{2} \choose k_{2}}\cdots{2k_{r} \choose k_{r}}} \over k_{1}\,k_{2}\ldots k_{r}}\,\bracks{z^{n}}
z^{\sum\limits_{j = 1}^{r}\,\, k_{j}}
\\[5mm] = &
\bracks{z^{n}}\bracks{\sum_{k = 1}^{\infty}
{\ds{2k \choose k} \over k}\,z^{k}}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\sum_{k = 1}^{\infty}
\ds{-1/2 \choose k}\pars{-4}^{k}\, z^{k}\int_{0}^{1}t^{k - 1}
\,\,\dd t}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\int_{0}^{1}\sum_{k = 1}^{\infty}
\ds{-1/2 \choose k}\pars{-4zt}^{k}
\,\,{\dd t \over t}}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\int_{0}^{1}
{\pars{1 - 4zt}^{-1/2} - 1 \over t}
\,\dd t}^{\, r}
\\[5mm] = &\
\bbx{2^{r}\bracks{z^{n}}\bracks{\vphantom{\LARGE A}%
\ln\pars{2} - \ln\pars{1 + \root{1 - 4z}}}^{\, r}} \\ &
\end{align}
I guess it would be a quite difficult task to extract the coefficient of $\ds{z^{n}}$ from the last expression. I leave it here while we find some sort to handle the above mentioned task.


问:
Verification that a function satisfies a PDE
Consider the problem
\begin{equation}
\begin{cases}
\frac{\partial v^{\epsilon}(t,x)}{\partial t}= L^{\epsilon}v^{\epsilon}(t,x)+c(x)v^{\epsilon}(t,x)+g(x); \quad t>0, \: x \in \mathbb{R}^r \\
v^{\epsilon}(0,x)=f(x) \\
\end{cases}
\end{equation}
for $\epsilon > 0$,together with the problem for the first-order operator which is obtained for $\epsilon = 0$:
\begin{equation}
\begin{cases}
\frac{\partial v^{0}(t,x)}{\partial t}= L^{0}v^{0}(t,x)+c(x)v^{0}(t,x)+g(x); \quad t>0, \: x \in \mathbb{R}^r \\
v^{0}(0,x)=f(x) \\
\end{cases}
\end{equation}
Consider then the SDE: $dX_{t}^{\epsilon,x}=b(X_{t}^{\epsilon, x})dt+\epsilon \sigma(X_{t}^{\epsilon,x})dW_t$, where $W$ is a Wiener process and

$
L^{\epsilon}= \frac{\epsilon^2}{2} \sum_{i,j=1}^{r}a^{i,j}(x)\frac{\partial^2}{\partial x^i \partial x^j} + \sum_{i=1}^{r}b^i(x)\frac{\partial}{\partial x^i}
$

with $\sigma(x)\sigma^t(x)=a^{i,j}(x)$.

I proved that
\begin{equation*}
\begin{split}
\lim_{\epsilon \to 0} v^{\epsilon}(t,x) &= f(X_{t}^{0,x})\exp\biggl\{ \int_{0}^{t}c(X_{s}^{0,x})ds \biggr\} \\
&+ \int_{0}^{t} g(X_{s}^{0,x})\exp \biggl\{ \int_{0}^{s}c(X_{v}^{0,x})dv \biggr\} ds
\end{split}
\end{equation*}
and now I have to show that the right-hand side is the solution of the second Cauchy problem (the one with $epsilon = 0$). I know that I have to compute the derivatives and substitute but I'm having some trouble. Here's what I came up with:

$
\frac{\partial v^0}{\partial t}= f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int_{0}^{t}c(X_{s}^{0,x})ds}+g(X_{t}^{0,x})e^{\int_{0}^{t}c(X_{s}^{0,x})ds}
$

$
L^0v^0 = \frac{\partial f}{\partial X}(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds}+f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds}+g(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds}
$

If I substitute in the second problem and sum all the terms up, I'm left with
\begin{equation}
\frac{\partial f}{\partial X}(X_{t}^{0,x})e^{\int_{0}^{t} c(X_{s}^{0,x})ds}+f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int_{0}^{t} c(X_{s}^{0,x})ds}+c(X_{t}^{0,x})\int_{0}^{t}g(X_{s}^{0,x})e^{\int_{0}^{s} c(X_{v}^{0,x})dv}ds+g(X_{t}^{0,x})=0
\end{equation}

Is what I did until now correct? If so how do I conclude? If not how do I fix it to get the desired solution? Thank you in advance.

答: When you compute $\frac{\partial v^0(t,x)}{\partial t}$, you clearly forgot the term $\nabla f(X^{0,x}_t) \cdot b(X^{0,x}_t) \exp\left[\int_0^t c(X^{0,x}_s) \mathrm{d}s\right]$, where the gradient is taken with respect to the space variable only...


问:
Algebraic manipulation of a quadratic equation to an alternative form
I have a quadratic equation that I have attempted to manipulate algebraically but still don't have a reasonable solution.

Given a quadratic equation as follows

$a(x-k)^2 + b(k-c)^2$

how could this be expressed in the form

$d(e-k)^2 + f$ such that the terms $d$, $e$ and $f$ do not involve $k$, preferably using completing the square approach.

答:
Given $ a(x-k)^2 + b(k-c)^2 = d(e-k)^2+f$, you're right, the easiest way is to complete the square.

First, we need to expand the expression:

$$ a(x - k)^2 + b(k - c)^2 = ax^2 - 2axk + ak^2 + bk^2 - 2bck + bc^2 $$

Then, we group within different powers of $k$:

$$ = ak^2 + bk^2 - 2axk - 2bck + ax^2 + bc^2 $$
$$ = (a+b)k^2 - 2(ax + bc)k + ax^2 + bc^2 $$

Next, we factor out $(a+b)$ from the first two terms and complete the square using just those two terms:

$$ = (a+b)\left[k^2 - 2\left(\frac{ax + bc}{a+b}\right)k\right] + ax^2 + bc^2 $$

$$ = (a+b)\left[k^2 - 2\left(\frac{ax + bc}{a+b}\right)k + \left(\frac{ax + bc}{a+b}\right)^2 \right] + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b} $$

$$ = (a+b)\left(k - \frac{ax + bc}{a+b}\right)^2 + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b} $$

At this point, we can reverse the order of the expression within the square (to look more like the expression we're trying to replicate):

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b} $$

With a bit more manipulation, we can further simplify the expression after the square:

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{(a+b)(ax^2 + bc^2) - (ax + bc)^2}{a+b} $$

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{a^2x^2 + abx^2 + abc^2 + b^2c^2 - a^2x^2 - 2abcx + b^2c^2}{a+b} $$

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{abx^2 + abc^2 - 2abcx}{a+b} $$

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{ab}{a+b}(x^2 - 2abcx + c^2) $$

$$ = (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{ab}{a+b}(x - c)^2 $$

Comparing against the form we want: $d(e-k)^2+f$, we have:

$$ d = a + b , $$
$$ e = \frac{ax + bc}{a+b} , $$
$$ f = \frac{ab}{a+b}(x - c)^2 . $$


问:
What's the relationship between t test and t distribution?
This question always confuses me. I know the Student's t distribution is defined as follows: let $Z \sim N(0,1)$ and $V \sim \chi^2(v)$. If Z and V are independent, then the distribution of

$$T=\frac{Z}{\sqrt{V/v}}$$ has the Student's t distribution with v degrees of freedom.

I also know parametric method t-test. Assume $F=N(\mu, \sigma^2)$, $X_1, X_2,...,X_n \sim N(\mu, \sigma^2)$ , iid. Test statistic:

$$T= \frac{\bar{X}-\mu}{S/\sqrt{n}} \sim t_{n-1}$$

What's the relationship between the two above?

答:
The issue is making the connection between the test statistic

$$T_{\text{test}}=\frac{\bar{X}\,-\,\mu}{S/\sqrt{n}}$$

to assess the deviation of the sample mean from the population mean when the population variance $\sigma$ is not known (in real life situations), and the t-distribution:

$$T_{\text{test}}=\frac{\bar{X}\,-\,\mu}{S/\sqrt{n}}\sim t_{n-1}$$

Perhaps the following simple tricks could help see it a bit more intuitively. Starting off from the intended statistic:

\begin{align} \frac{\bar{X}\,-\,\mu}{\frac{S}{\sqrt{n}}} &= \frac{\bar{X}\,-\,\mu}{\frac{\sigma}{\sqrt{n}}} \frac{1}{\frac{S}{\sigma}}\\[2ex]
&\sim Z\,\frac{1}{\frac{S}{\sigma}}\\[2ex]
&\sim\frac{Z}{\sqrt{\frac{\color{blue}{\sum_{i=1}^n(X_i-\bar X)^2}}{\,\color{blue}{\sigma^2}}\frac 1{(n-1)}}}\\[2ex]
&\sim\frac{Z}{\sqrt{\frac{\color{blue}{\chi_{n-1}^2}}{n-1}}}
\end{align}

where $\small X_1,\dots,X_n$ are iid $X_i \sim N(\mu,\sigma^2),$ and $S=\small \frac{\displaystyle\sum_{i=1}^n(X_i-\bar X)^2}{n-1}.$

The question then boils down to getting an idea of the step in blue above:

$$\small \frac{\displaystyle\sum_{i=1}^n(X_i-\bar X)^2}{\sigma^2}\sim \chi_{n-1}^2$$

From Wikipedia, the chi-square distribution with $k$ degrees of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables.

However, and also from Wikipedia in here, the iid normally distributed random variables from which the sample is obtained geometrically form a random vector $\small \begin{bmatrix}X_1,\dots,X_n \end{bmatrix}^\top,$ and likewise the error from the mean of these iid normal  rv's, $\small X_i-\bar X,$ i.e. $$\small\begin{bmatrix}
X_1 - \bar X\\ \vdots\\ X_n-\bar X
\end{bmatrix}$$

can be seen as its projection on an $(n-1)$-dimensional subspace of $\mathbb R^n$ orthogonal to the span of the vector $\small \begin{bmatrix}1,\dots,1 \end{bmatrix}^\top$

which will result in a predictable loss of one degree of freedom. The $n-1$ degrees of freedom also can be seen from the fact that $\sum_{i=1}^n (X_i-\bar X)=0.$

And the squared norm of this vector will logically follow a $\chi^2_{n-1}$ scaled by a $\sigma ^2$ factor.


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答:
I always keep Viete's formula in mind, it's particularly easy to apply to quadratics. It states that given a polynomial of form $ax^{2}+bx+c$ the roots $x_1$ and $x_2$ will satisfy the following equations:

$$x_1+x_2=-\frac{b}{a}$$ and

$$x_1 \times x_2=\frac{c}{a}$$

This allows for a quick and easy 'mental' calculation.

ContentLicense: CC BY-SA 3.0

Title: How to factor quadratic $ax^2+bx+c$?


问:
Generating functions to solve recurrence relation
Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you

答: $A_n = 2*3^{n} - 1 $.  This should do it</p>

Body: $A_n = 2*3^{n} - 1 $.  This should do it


答:
Note that, for $b=1$ the function $y = \log_1 x$ is actually not defined for any $x \not =1$, as $1^y =1$ for all real $y$. For real $x \not =1$ there is no $y$ satisfying $1^y = x$. However for every other positive $b >0$ and every other positive $x$ the function $y = \log_b x$ is well-defined; equivalently there is a real number $y$ such that $b^y = x$.

However, $b^x$ is indeed defined for all $b >0$ including $b=1$. It is the horizontal line $y=1$.

The likely reason why $b =1$ was excluded even for the function $y = b^x$ is that for precisely $b>0$; $b \not =1$ the functions $y=b^x$ and $y=\log_b x$ are invertible; in fact they pair up as each other's inverse; if $y=b^x$ then $x = \log_b y$.


问:
How to prove that $\mathbb{N}$, the set of natural numbers, is well ordered?
I am reading classic set theory by D.C. Goldrei. I am stuck on understanding an argument for proving $\mathbb{N}$ is well ordered (every non-empty subset is of a least element). The author assumed that $B$ is a nonempty subset with no least element. Then the following subset of $\mathbb{N}$ is constructed:
\begin{equation}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{m} \not\in B\ \textrm{for all}\ \boldsymbol{m} \leq \boldsymbol{n}\right\},
\end{equation}
and the author proved that $A = \mathbb{N}$, so that $B$ is empty, which leads to a contradiction. I am not quite sure about how this argument leads to the fact that $B$ is empty, and what is the relation between the assumption that $B$ is nonempty with no least element and the construction of set $A$. I have no idea why $A$ has to be constructed as such. Could anyone provide hints?

答: Assume $k \in B$.  Then $k+1 \notin A$ because there is a natural number ($k$) that is less than $k+1$ that is also in $B$.  If $A= \Bbb N$, then $k+1 \notin A$ can't happen, so your initial assumption that $k \in B$ must be false.  But $k$ was an arbitrary natural number, so that means no natural numbers can be elements of $B$; i.e., $B$ is empty.


问:
How to prove $ E(|X-Y|) \le E(|X+Y|)$ when $X,Y$ are i.i.d variables?
Let $X$ and $Y$ be independent random variables having the same distribution and the finite mathematical expectation. How to prove the inequality $$ E(|X-Y|) \le E(|X+Y|)?$$</p>

答:
Edit: Question has changed. Will give answer when time permits.

By the linearity of expectation, the inequality $E(X-Y)\le E(X+Y)$  is equivalent to $-E(Y)\le E(Y)$, which in general is false. It is true precisely if $E(Y)\ge 0$.

Independence is not needed for the argument. Neither is the hypothesis that the random variables have the same distribution.


问:
$\{U_j\}_{j\ge1}$ a sequence of independent and identically distributed. Prove convergence almost surely
Let $\{U_j\}_{j\ge1}$ a sequence of independent and identically distributed random variables such that $U_1$ is a uniform random variable on $[0,1]$. Show that
$$X_n:=\Big(\prod_{j=1}^nU_j\Big)^{1/n}\rightarrow1/e$$
almost surely.

Since $U_1$ is uniform in [0,1] then the expected value is finite. My idea is to define a new sequence of random variables and apply the strong law of large numbers

答: $\log(X_n)=\frac{\sum_{j=1}^n \log(U_j)}{n}$. However $E\left(\log(U_j)\right)=\int_0^1\log(x)dx=-1$, so sum converges to $-1$ and $X_n\to e^{-1}$.


问:
How can I prove that $\nabla \times ( f^2 \nabla f)=0$?
I can see that $\nabla \times (f^2 \nabla f)=\nabla f^2 \times \nabla f+f^2\nabla \times \nabla f$ where the second term is $0$ but I don't have any idea about the first term...What should I do with the $f^2$ here?

答:
You can do the same thing you would if you encountered the ordinary derivative of $f(x)^2$, or more generally, $f(x) g(x)$, that is, apply product rule.

In this case, it would simply be:
$$
\nabla (fg) = f\nabla g + g\nabla f
$$

The first term will then be evaluated to zero, because $A\times A = 0$, for any $A$.

=].


问:
Construct a frame $\mathfrak{F} = (W,R)$ that refutes $\square \bot$

(Blackburn, Modal Logic, Section 1.3) Construct a frame $\mathfrak{F} = (W,R)$ that refutes $\square \bot$, to show that $\square \bot$ is not a valid formula.

I think I've solved this, but I'm not sure if my understanding of $\bot$ is correct. My solution is as follows -

Consider $W = \{a\}$, and $R$ such that $Raa$ (self-loop). If $\square \bot$ holds at $a$, i.e. $\mathfrak{F},a \Vdash \square \bot$ then $\mathfrak{F},a \Vdash \bot$ also holds (by definition of $\square$). However, $\bot$ cannot hold anywhere, so $\square\bot$ is not a valid formula (We have constructed a frame in which it is not valid).

Is my understanding of $\bot$ right? Is $\bot$ something that can never be true, i.e. is always false?

答:
Yes.   $\bot$ is valued as false by its definition in propositional logic, so cannot be valued as true by any world in any frame of modal logic.

When $\mathsf W=\{a\}$ and $\operatorname R=\{\langle a,a\rangle\}$, then $a\Vdash\square\varphi$ if and only if $a\Vdash\varphi$.   Since $a\nVdash\bot$ by definition, therefore $a\nVdash\square\bot$ in such a frame and thus $\square\bot$ is not a valid formula there in.

More.

The meaning of $~a\Vdash\varphi~$ is that "formula $\varphi$ is valued true in world $a$".

The meaning of $~a\Vdash\square\varphi~$ is that "formula $\varphi$ is valued true in all worlds accessible from $a$."

A valid formula is one that is valued true in all worlds of the frame.

$\bot$ is valued false.  No worlds can value $\bot$ as true.

Thus $\square\bot$ is only valued true in worlds which have no accessible worlds.

Therefore $\square\bot$ is only a valid formula in frames which have no accessible worlds.


问:
Big Greeks and commutation
Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering?

Clearly if $\mathbf{x}_i$ is a matrix then:

$$\prod_{i=0}^{n} \mathbf{x}_i$$

depends on the order of the multiplication. But, even if one accepts that it has a sequence, it is not clear if it should mean $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ or $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$.

A similar question, is there a "big" wedge product convention?

$$\overset{n}{\underset{i=0}{\Huge\wedge}} \;{}^{\Large{\mathbf{x}_i} \;=\; \mathbf{x}_0 \wedge \mathbf{x}_1 \;\cdots \mathbf{x}_{n-1}\; \wedge \mathbf{x}_{n}} $$

答: If I wanted $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$ I would write it as $$\prod_{i=0}^{n} \mathbf{x}_{n-i}$$</p>

Title: Big Greeks and commutation


问:
Pairs of Functions + How to find derivative of the inverse function.
1.Find a pair of function $f(x)$ and$g(x)$ such that$\frac{d(f(x))}{dx}= ln (2)*2^x*cos (g(x))$.

I can't figure out how to begin to solve this question. I know that there should be multiple pairs of solutions just don't know how to rearrange this equation to solve for them.
I'm not sure if this is right?
$f(x)=2^x\cos(g(x))\ln(2)dx$

2.Given$f(t) =t+ ln(t)$, find $\frac{d(f^{-1})}{dt}$ at $t= 1 +e$. Exact values. Note that the symbol$\frac{d(f^{-1})}{dt}$means the derivative of the inverse function with respect to t.

I've solved for the inverse of $f$ but have gotten stuck at $e^yy=e^x $
so I don't know how  I can apply this to find$\frac{d(f^{-1})}{dt}$ at $t= 1 +e$.

Body:
For the second one you can use $$f(f^{-1})(t) = t$$

$$\implies \frac{d}{dt}[f(f^{-1}(t))] = 1$$

$$\implies f'(f^{-1}(t)) \cdot \frac{d}{dt}[f^{-1}(t)] = 1$$

Use the fact that $f^{-1}(1+e) = e$, we have

$$\frac{d}{dt}[f^{-1}(t)] = \frac{1}{f'(e)}$$


问:
Polish group actions: if an orbit is non-meager in itself, it is a Baire space?
Assume that $G$ is a Polish group continuously acting on a Polish space $X$. Let $x \in X$ be a point such that $G \cdot x$, the orbit of $x$, is non-meager in its relative topology. I would like to know why this implies that $G \cdot x$ is a Baire space (as a topological subspace of $X$). I'm reading a proof of the Effros theorem and the author seems to be using this observation.

Note: We say that a topological space $X$ is Baire iff every non-empty open subset of $X$ is non-meager in $X$.

答: Suppose a nonempty open subset $U\subseteq Gx$ is meager in $Gx$.  We may assume $x\in U$ (otherwise, rename $x$ to be an element of $U$).  Let $V=\{g\in G:gx\in U\}$; then $V$ is an open neighborhood of $1$ in $G$.  Now if $\{g_n:n\in\mathbb{N}\}$ is a countable dense subset of $G$, the translates $g_nV$ cover all of $G$ (proof: for any $g\in G$, there is some $g_n$ in the open set $gV^{-1}$ which means $g\in g_nV$).  It follows that $Gx$ is covered by the sets $(g_nV)x=g_nU$.  But since $U$ is meager, each $g_nU$ is also meager, so $Gx$ is covered by countably many meager sets and is meager.


答: I always thought $\prod_{i=1}^n x_i$ as a notational shortcut for $x_1\cdot x_2\cdot\dots\cdot x_n$, but now that you make me think about it, I don't recall having seen the formal definition anywhere.</p>

Title: Big Greeks and commutation

Body:
Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering?

Clearly if $\mathbf{x}_i$ is a matrix then:

$$\prod_{i=0}^{n} \mathbf{x}_i$$

depends on the order of the multiplication. But, even if one accepts that it has a sequence, it is not clear if it should mean $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ or $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$.

A similar question, is there a "big" wedge product convention?

$$\overset{n}{\underset{i=0}{\Huge\wedge}} \;{}^{\Large{\mathbf{x}_i} \;=\; \mathbf{x}_0 \wedge \mathbf{x}_1 \;\cdots \mathbf{x}_{n-1}\; \wedge \mathbf{x}_{n}} $$


问:
How to use substitution to evaluate $ \int \frac{3}{\sin^2{x}} dx~ $?
I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

$$ \int \frac{3}{\sin^2{x}} dx $$

I tried using a method where I set something to $u$ and $dv$ and go from there, but I don't end up anywhere with this problem.
 I know you can use substitution method and then integrate by parts, but I'm not sure which part of the integral I should begin substituting.

答:
Another solution (even if the previous answers are simpler).

Use the tangent half-angle substitution
$$I=\int \frac{3}{\sin^2{x}} dx=\frac{3}{2}\int  \left(\frac{1}{t^2}+1\right)\,dt=\frac{3 t}{2}-\frac{3}{2 t}+C$$ Back to $x$
$$I=-3\cot(x)+C$$


答:
Okay.

Let's do a sort of but not quite proof by contradiction.

The  well-ordered principal says:  Every non-empty subset of $\mathbb N$ has a least element.

Let's assume $B\subset \mathbb N$ without a least element.

If we can prove that $B$ can't be non-empty we are done.  Because if we can prove for $B$ to not have a least element we must have $B=\emptyset$ than that means for any subset that isn't empty, it can't be possible that it doesn't have a least element (because if it doesn't have a least element we have proven it must be empty.).

Okay, let's prove $B$ must be empty.

If we can prove that for any $k \in \mathbb N$ that $k \not \in B$ we will be done.  $B\subset \mathbb N$ but none of the $k \in \mathbb N$ can actually be in $B$ than nothing can be in $B$.  So $B$ would have to be empty.

So let's do that.  Let's prove that if $k \in \mathbb N$, then $k\not \in \mathbb N$.

So your text considers the set $A = \{n\in \mathbb N| $ if $m \le n$ then $m\not \in B\}$.

We can prove $1\in A$.  If $m\le 1$ then $m = 1$.  ANd if $m \in B$ then $m$ is the least element in $B$. But $B$ doesn't have a least element. So $1\not \in B$. so all $m\le 1$ are not in $B$.  So $1 \in A$.

Suppose we have a $k$ so that if $n \le k$ then $n\in A$.  We can prove $k+1\in A$.

If $m \le k+1$ then either $m \le k$ or $m = k+1$.  If $m\le k$ then $k \in A$ by our assumption.  If $m=k+1$ the either $k+1 \in B$ or $k+1\not \in B$.  Because $k\in A$ then all $n< k+1$ then $n \le k$ so $n\not \in B$.  So if $k+1 \in B$ then $k+1$ is the least element of $B$.  But $B$ has no least element. So $k+1 \not \in B$.  So for all $n \le k+1$ we have $n \not \in B$.  So $k+1 \in A$.

Now the natural numbers have the induction principal.  So if $1 \in A$ and if $k\in A \implies k+1 \in A$ we can conclude all $n \in \mathbb N$.

So for all $n \in \mathbb N$ we have $n \in A$.

But that means $B$ is empty.  Why?  Because if $m \in \mathbb N$ then $m \in A$.  And if $m \in A$ then all $n\le m$ are not in $B$.  And $m \le m$ so $m \not \in B$.  So no $m\in \mathbb N$ are in $B$.

So $B$ is empty.

And we have proven everything we needed to prove.

We have proven that if $B$ is any subset of $\mathbb N$ without a least element then $B$ must be empty.  That means for any non-empty set, the set can't not have a least element.  So every non-empty set of natural numbers has a least element.

Note, this will hold for any set that has the induction principal.  Note $\mathbb Z, \mathbb Q, \mathbb R$ etc. don't have the induction principal so $\mathbb Z, \mathbb Q, \mathbb R$ need not be well-ordered.

=====

Actually if it were me I wouldn't bother with the set $A$.

I'd just show

$1 \not \in B$.  (Because otherwise $1$ is the minimal element in $B$)

If $n \le k$ not in $b$ then $k+1$ is not it $B$ (else $k+1$ would be the least element.)

So by induction there are no $n \in \mathbb N$ that are in $B$.  So $B$ is empty.

Title: How to prove that $\mathbb{N}$, the set of natural numbers, is well ordered?

Body:
I am reading classic set theory by D.C. Goldrei. I am stuck on understanding an argument for proving $\mathbb{N}$ is well ordered (every non-empty subset is of a least element). The author assumed that $B$ is a nonempty subset with no least element. Then the following subset of $\mathbb{N}$ is constructed:
\begin{equation}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{m} \not\in B\ \textrm{for all}\ \boldsymbol{m} \leq \boldsymbol{n}\right\},
\end{equation}
and the author proved that $A = \mathbb{N}$, so that $B$ is empty, which leads to a contradiction. I am not quite sure about how this argument leads to the fact that $B$ is empty, and what is the relation between the assumption that $B$ is nonempty with no least element and the construction of set $A$. I have no idea why $A$ has to be constructed as such. Could anyone provide hints?


问:
positive semidefinite why is theinequality
What are the conditions on $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$ to be true?

I started the service today, so I don't have any reputation points and can't make any comments. So, let me ask a question here.

I don't understand the part of this question that says tr(^1/2^1/2)≤tr(^1/2()^1/2).

if A<=aI is true, why is the inequality holds?

答:
The trace of a PSD matrix is non-negative. This means that the trace function is increasing with respect to the PSD partial order $\preceq$ (the Löwner order). If $A \preceq B$ then $B - A$ is PSD so

$$ \operatorname{tr}(B - A) = \operatorname{tr} B - \operatorname{tr} A \ge 0. $$

(The trace is also linear.) Therefore $\operatorname{tr} A \le \operatorname{tr} B$.


问:
Convergence of sequence of functions $ f_n(x)=\begin{cases} n^{p+1}x & \text{if $0≤x≤1/n$}\\ \frac{1}{x^p} & \text{if $1/n<x≤1$} \end{cases} $
Question: Let $p$ be positive real number and consider the sequence $\{f_n\}_{n=1}^\infty$ defined by

$$
f_n(x)=\begin{cases}
         n^{p+1}x & \text{if $0≤x≤\displaystyle\frac1n$}\\
        \displaystyle\frac{1}{x^p} & \text{if $\displaystyle\frac1n<x≤1$}
     \end{cases}
$$

Let $f(x)=\lim_{n\to\infty} f_n(x)$ where $x\in[0,1].$
Then

(i) find $f(x)$

(ii) Is $f(x)$ is Riemann integrable?

(iii) Is the convergence uniform?

My attempt: As $n\to\infty$ we have $1/n\to0$ and hence $x=0$. So I thought $f(x)=\lim_{n\to\infty} f_n(x)=0$ and hence $f$ is Riemann integrable. But, key says I am wrong. How to find $f(x)$? Please help.

答: Notice that for $n$ large enough, $1/n\to 0$ so your function $f(x)=\frac{1}{x^p}$ for $0< x\leq 1$. In $0$ the function should be define as zero. So we have a function which is not bounded in its interval of definition so couldn't be Riemann integrable, for $p\geq 1$. For the case $0<p<1$ we have that the improper $\int_0^1 \frac{1}{x^p}dx$ converges (not really Riemann integrable, but I think for your purposes could be say yes it is.) Notice that for uniform convergence we should normally have $$ \lim_{n\to\infty} \sup_{x\in[0,1]}|f(x)-f_n(x)|=0$$ but in this case, each $f_n$ is bounded, but $f(x)$ is not bounded, it means that $$\lim_{n\to\infty} \sup_{x\in[0,1]}|f(x)-f_n(x)|=\infty$$ and the convergence is not uniform. Indeed, faster: as each $f_n$ is Riemann integrable, but its limit $f$ is not, then the convergence is not uniform.


问:
Prove that there exists a monochromatic shape in $K_{17}$ with each edge being one of 4 different colours
Prove that there exists a monochromatic odd cycle (thanks to Misha Lavrov for correcting me) in a complete graph of a 17-sided shape ($K_{17}$) with each edge being one of 4 different colours

I've done some work on Ramsey theory on this question, but I haven't gotten very far.
I've found that for any point A, there are 16 edges. At least 1 colour (say red) has 4 or more edges from Point  to another that are that colour. There are then 6 remaining edges that connects those 4 points. If any of those edges are red, then there is a triangle, so we are finished.

However, if none of those edges are red, ten there is 3 possible colours remaining. If any of those colours connects 3 flights or 5 flights, we are done. However, worst case it that each colour only connects 2, which doesn't really prove anything...

P.S. For the record, this is from a textbook that I've been using to study for an exam, but they don't have answers. :(

答:
The key to this problem is the following result: a graph that does not contain any odd cycle is bipartite. That is, its vertices can be partitioned into two sets $A$ and $B$ such that all edges go between $A$ and $B$.

We can prove a more general claim: if the complete graph on $2^k+1$ vertices is edge-colored with $k$ colors, then there is a monochromatic odd cycle.

This is easy for $k=1$: the complete graph on $3$ vertices contains an odd cycle, and there is only one color. From there, induct on $k$.

Pick your favorite color, and consider the subgraph of all edges of that color. If there is no odd cycle of that color, then it is bipartite: we can split the $2^k+1$ vertices into two sets $A,B$ such that all edges of your favorite color go between $A$ and $B$.

One of $A$ or $B$ (without loss of generality, $A$) has size at least $\lceil \frac{2^k+1}{2}\rceil = 2^{k-1}+1$. Within $A$, your favorite color never gets used. So the coloring restricted to $2^{k-1}+1$ vertices of $A$ only uses $k-1$ colors, and by the inductive hypothesis, there is a monochromatic odd cycle there.


问:
Is the following definition of an elliptic curve correct?
Im new to algebraic geometry so I want to make sure im getting my definitions right. I know there are a few ways to state what an elliptic curve is (ex a smooth projective curve of genus one with distinguished $K$-rational point). But I am wondering if the following is equivalent. For simplicity lets just work over $\mathbb{C}$.:

$\textbf{Definition:}$ An elliptic curve $E$ is a non-singular projective curve in $\mathbb{P}^2$ of the form
$$E: Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3 $$

I am wondering if this is sufficient to define an elliptic curve?

答:
This is subtle. It is true that every elliptic curve can be written in this form, and that every smooth projective curve of this form has genus $1$.

But an elliptic curve is not a smooth projective curve of genus $1$. An elliptic curve (over a field $K$) is a smooth projective curve of genus $1$ together with a distinguished $K$-rational point, and this is key to the theory because it's the point that forms the identity for the group law. You need to say something about what this distinguished point is: with an equation of this form it's conventional to take it to be the point at infinity, with coordinates $(X : Y : Z) = (0 : 1 : 0)$. But this needs to be said explicitly in the definition; it is in fact possible to pick another point, which changes the group law. This is important for the following reasons among others:

There are smooth projective curves of genus $1$ over non-algebraically closed fields $K$ which have no $K$-rational points, and hence there is no choice of point which turns them into elliptic curves.

Elliptic curves have endomorphisms and automorphisms, and these are required to preserve the distinguished point (which turns out to imply that they preserve the group law). If you don't keep this in mind you will be confused when you read statements about endomorphisms and automorphisms of elliptic curves in the literature (which in fact happened recently on MO). For example, endomorphisms of an elliptic curve form a ring, but only if you require that endomorphisms preserve the distinguished point.


答:
If your elements commute with eachother, then there is no need for an ordering in the case of finite sums/products. In the non-commutative case things are more complicated.

Anyhow, IMO there is no need for an ordering if the sum/product doesn't depend on the order. And this covers many non-commutative cases too. Otherwise, it is clear that one should explain the order.

If you write $_{i=1}^n$ by convention the order is understood to be $1,2,.., n$.

Title: Big Greeks and commutation


问:
differentiability of f:$\mathbb {R} \rightarrow \mathbb {R}$ at $x=x_0$ imply continuous of f' at $x=x_0$
For the fucntion $f:\mathbb{R}\rightarrow \mathbb{R}$ which are differentiable at $x=x_0$
imply $f'$ is continuous at $x=x_0$?

$f$ is differentiable at $x=x_0$ when $\forall \epsilon >0,\exists \delta>0, c\in \mathbb{R} \ s.t$
$$0<|x-x_0|<\delta \ \ imply \ \ |\frac{f(x)-f(x_0)}{x-x_0}-c|<\epsilon $$

in this case we let $c=f'(x_0)$

if $f'(x)$ is defined $\forall x \in\mathbb{R}$ from the above definition of derivative.
can we prove $\forall \epsilon >0,\exists \delta>0,\ s.t$
$$|x-x_0|<\delta \ \ imply \ \ |f'(x)-f'(x_0)|<\epsilon $$?

I don't have any proof of above problem or counterexample, please help me!

答:
Differentiability does not necessarily imply the derivative is continuous. For a counterexample, see Differentiable but not continuously differentiable..

The class of differentiable functions that do have a continuous derivative are denoted $C^1$.


问:
Show that for every countable subset A of R , Int(A)=0
Show that for every countable subset A of R ,Show Int(A)=$\emptyset$

Try

Let A=[0,1]
The closed interval is uncountable

But since  A$\subset $ R is countable and A is not ,each set
can’t be uncountable  so Int(A)=0

Help

答:
I assume we're using the standard topology on $\Bbb R$.

The interior of any set $A$ is simply the union of all open sets that are contained within $A$.  So if $A$ has non-empty interior, it must contain some open interval.  But every non-empty open interval in $\Bbb R$ is uncountable, so if $A$ is countable, it cannot possibly contain any uncountable sets, so it cannot contain any non-empty open intervals.

That means there are no open sets contained within $A$, so $A$ has empty interior.


问:
probability - straw drawing
There are five straws and two of them are winners. Three people, draw one straw each - straw A first, then straw B, then straw C.  Find the probability that straw C is a winning straw.

Besides using a tree diagram, what other easier method can I use to solve this?

I have found the answers using tree diagram as $2/5$.

答:
Hint: have you noticed your answer is just the ratio of number of winning straws to the total number of straws? Can you find a simple argument to prove this?

 You basically draw find straws one by one, and want to know the probability of the third straw being a winner. By symmetry, the probability should be the same for the first, second, third, fourth or fifth straw, therefore it's just $\frac 25$.

Title: probability - straw drawing


答: $f(x)=x^{2}\sin (\frac1 x)$ for $x \neq 0$, $ f(0)=0$ defines a function which is differentiable at $0$ but its deriavtive is not continuous at $0$. In fact, $f'(x)$ does not even have a limit as $x \to 0$.

Title: differentiability of f:$\mathbb {R} \rightarrow \mathbb {R}$ at $x=x_0$ imply continuous of f' at $x=x_0$

Body:
For the fucntion $f:\mathbb{R}\rightarrow \mathbb{R}$ which are differentiable at $x=x_0$
imply $f'$ is continuous at $x=x_0$?

$f$ is differentiable at $x=x_0$ when $\forall \epsilon >0,\exists \delta>0, c\in \mathbb{R} \ s.t$
$$0<|x-x_0|<\delta \ \ imply \ \ |\frac{f(x)-f(x_0)}{x-x_0}-c|<\epsilon $$

in this case we let $c=f'(x_0)$

if $f'(x)$ is defined $\forall x \in\mathbb{R}$ from the above definition of derivative.
can we prove $\forall \epsilon >0,\exists \delta>0,\ s.t$
$$|x-x_0|<\delta \ \ imply \ \ |f'(x)-f'(x_0)|<\epsilon $$?

I don't have any proof of above problem or counterexample, please help me!


问:
On the definition of a tensor in the study of manifolds
In the book An introduction to manifolds by Tu, Loring W, it defines the $k$-tensor as follows

Denote by  $V^{k}=V \times \cdots \times V$  the Cartesian product of  $k$  copies of a real vector space  $V $.  A function $f: V^{k} \rightarrow \mathbb{R}$  is $ k  $-linear if it is linear in each of its $ k $ arguments:
$$
f(\ldots, a v+b w, \ldots)=a f(\ldots, v, \ldots)+b f(\ldots, w, \ldots)
$$
for all $ a, b \in \mathbb{R} $ and $ v, w \in V $.  Instead of $2$-linear and $3 $-linear, it is customary to say "bilinear" and "trilinear." A  $k $-linear function on  $V$  is also called a $ k  $-tensor on $ V $. We will denote the vector space of all $ k  $-tensors on $ V $ by $ L_{k}(V) $.  If  $f $ is a $ k $ -tensor on $ V $, we also call $ k $ the degree of $ f $.

However in Wikipedia, it defines the $(p,q)$-tensor as follows

In this approach, a type $(p,q)$ tensor $T$ is defined as a multilinear map,
$$
T: \underbrace{ V^* \times\dots\times V^*}_{p \text{ copies}} \times \underbrace{ V \times\dots\times V}_{q \text{ copies}} \rightarrow \mathbf{R},
$$
where $V^*$ is the corresponding dual space of covectors, which is linear in each of its arguments.

It is obvious that the definition of a $k$-tensor in An introduction to manifolds is just the $(0,q)$-tensors in the definition of Wikipedia.

My question is, why do not we define the tensors to be $(p,q)$-tensor in Tu, Loring W?

EDIT: I have noticed that there are no $(p,q)$-tensors in the whole book of Loring Tu.

Is it because that the elements from dual space will never be used in tensors in manifolds in the following study (so $p$ always equals to $0$)? If not, where do we use $(p,q)$-tensors in the study of manifolds? (maybe it will appear in tensor fields?)

I would really appreciate it if anyone could tell me the connections of these things.

答:
Note that even Wikipedia's definition is not the most general. The most general definition (that I know of) for a (real) tensor is a multilinear map
$$ T: V_1 \times V_2 \times \cdots \times V_k \rightarrow \mathbb{R},$$
where $V_1, \dots, V_k$ are (not necessarily the same) vector spaces. And such generalized tensors definitely do appear in differential geometry and topology.

I also want to note that these simple definitions are also very misleading in the sense that the tensors that arise in geometry and topology all have symmetries or anti-symmetries in the indices. This leads to most of the headaches when learning how to deal with tensors. In fact, if you get serious with tensors, the notation can become quite a challenge. Penrose developed a particularly creative approach to dealing with this.

However, when you're first learning differential geometry and topology, almost all of the tensors encountered are indeed $(0,q)$ tensors. That's because either you're working with only exterior differential forms or you're working with a Riemannian manifold, where the metric can be used to convert any $(p,q)$-tensor into $(0,p+q)$-tensor.

It's good to know the definitions of general tensors (as described above) and $(p,q)$ tensors, because they are sometimes necessary or useful. But it's probably best to avoid devoting much attention to them in introductory courses and textbooks.

ADDED: An extremely useful canonical isomorphism is
$$
\operatorname{Hom}(V,W) = V^*\otimes W
$$
where $V$ and $W$ are vector spaces and $\operatorname{Hom}(V,W)$ is the space of linear maps from $V$ to $W$. Used with care, this means that you can view a $(p,q)$ tensor $T \in V_1^*\otimes\cdots\otimes V_q^*\otimes V_{q+1} \otimes \cdots \otimes V_{p+q}$ as a multilinear map
$$
T: V_{1}\times\cdots\times V_{q} \rightarrow V_{q+1}\otimes\cdots\otimes V_{p+q}.
$$
In short, to avoid working with $(p,q)$-tensors directly, it can be useful to view them as maps from $(q,0)$-tensors to $(p,0)$-tensors instead. Or, equivalently, as $(V_{q+1}\otimes\cdots\otimes V_{p+q})$-valued $(0,q)$-tensors.


问:
Martingale with expectation
Let a probability space defined on $\Omega = [0 \ 1]$ and assume the probability of any interval be defined as the length of that interval. Let $\\$
$Y_i(w) = \begin{cases}
1 & \quad 0 \leq w \leq 1/i, \\
0 & \quad otherwise
\end{cases}
$ and $X(w) = w$. I am trying to find $Z = E[X | Y_1, ..., Y_k]$ to show that the process Z is a martingale. To eliminate $w$, I write the probabilities as:
$Y_i = \begin{cases}
1 & \quad w.p. \quad 1/i, \\
0 & \quad w.p. \quad 1-1/i
\end{cases}
$ and $X \sim U[0,1]$. To show Z is a martingale we need to show $E[Z_k | Z_{k-1}, ..., Z_1] = Z_{k-1}$ but I'm struggling to find Z. Any ideas to approach this problem?

答: Hints: You have to show that $\int_A Z_kdP=\int_A Z_{k-1}dP$ if $A \in \sigma (Y_1,Y_2,...,Y_{k-1})$. [This will automatically imply that the equation also holds for $A \in \sigma (Z_1,Z_2,...,Z_{k-1})$]. Verify that  $\sigma (Y_1,Y_2,...,Y_{k-1})$ is simply the collection of all possible unions of the intervals $[0,\frac 1 {k-1}), [\frac 1 {k-1},\frac 1 {k-2}),....,[\frac 1  2,1)$. So it is enough to verify that $\int_A Z_kdP=\int_A Z_{k-1}dP$  for each of these $k-1$ intervals. This verification is straightforward.

Title: Martingale with expectation


问:
Closed formula for sum of exponentials
I'd like to have a closed formula for $\sum\limits_{i=0}^n e^{\sqrt{i}x}=1+e^{\sqrt{1}x}+...+e^{\sqrt{n}x}$, something similar to the formula $\sum\limits_{i=0}^ne^{ix}=\frac{e^{(n+1)x}-1}{e^x-1}$. The main problem is that the last sum is geometric, while the first one isn't. I tried some sum packages in Maple, but it seems most of the theory is geared towards hypergeometric sums, which the first sum above isn't.

Any help would be tremendously appreciated. Thanks!

Title: Closed formula for sum of exponentials

Body:
There is no formula (even using special functions) but you could use tbe bounds
$$I_1=\int_0^{n} e^{\sqrt{i}x}di<\sum\limits_{i=0}^n e^{\sqrt{i}x} <\int_0^{n+1} e^{\sqrt{i}x}di=I_2$$

$$I_1=\frac{2 e^{x\sqrt{n} } \left(x\sqrt{n} -1\right)+2}{x^2}$$
$$I_2=\frac{2 e^{x\sqrt{n+1} } \left(x\sqrt{n+1} -1\right)+2}{x^2}$$

Trying for $x=\frac 12$ and a few values of $n$
$$\left(
\begin{array}{ccccc}
n & I_1 & I_2 & \frac 12(I_1+I_2) & \text{summation} \\
 5  & 10.8884 & 14.1190 & 12.5037 & 12.8314 \\
 10 & 30.5969 & 35.6515 & 33.1242 & 33.4443 \\
 15 & 59.9519 & 67.1124 & 63.5322 & 63.8415 \\
 20 & 100.522 & 110.143 & 105.332 & 105.629 \\
 25 & 154.190 & 166.680 & 160.435 & 160.717 \\
 30 & 223.108 & 238.930 & 231.019 & 231.285 \\
 35 & 309.698 & 329.369 & 319.533 & 319.781 \\
 40 & 416.659 & 440.755 & 428.707 & 428.934 \\
 45 & 546.993 & 576.151 & 561.572 & 561.777 \\
 50 & 704.021 & 738.946 & 721.483 & 721.664 \\
 55 & 891.409 & 932.878 & 912.143 & 912.296 \\
 60 & 1113.19 & 1162.06 & 1137.63 & 1137.75 \\
 65 & 1373.81 & 1431.02 & 1402.42 & 1402.51 \\
 70 & 1678.14 & 1744.70 & 1711.42 & 1711.47 \\
 75 & 2031.49 & 2108.55 & 2070.02 & 2070.03 \\
 80 & 2439.70 & 2528.48 & 2484.09 & 2484.06 \\
 85 & 2909.14 & 3010.97 & 2960.05 & 2959.98 \\
 90 & 3446.73 & 3563.08 & 3504.90 & 3504.78 \\
 95 & 4060.02 & 4192.47 & 4126.25 & 4126.07 \\
 100 & 4757.22 & 4907.50 & 4832.36 & 4832.12
\end{array}
\right)$$

So, it seems that

$$\frac 12(I_1+I_2)=\frac{e^{x\sqrt{n} } \left(x\sqrt{n} -1\right)+e^{x\sqrt{n+1} } \left(x\sqrt{n+1}-1\right)+2}{x^2}$$ could be a decent approximation.


问:
The functional equation $\delta(n) = \delta(2n) + \kappa(n)$
I am interested in finding a function $\delta(n)$ that solves a functional equation of the form—$$\delta(n) = \delta(2n) + \kappa(n),$$for all integers $n \ge 1$ that are powers of 2, where $\kappa(n)$ is a known function that depends on $n$.  This equation came up when seeking answers to a question of mine about converging polynomials.  For example (Nacu and Peres 2005, proposition 10):

If $\kappa(n) = \dfrac C {\sqrt{2n}}$, then $\delta(n) = \dfrac{(1+\sqrt{2})C}{\sqrt{n}}$, where $C > 0$.

If $\kappa(n) = \dfrac M {4n}$, then $\delta(n) = \dfrac M {2n}$, where $M > 0$.

Unfortunately, I don't know how to solve this functional equation for $\delta(n)$ when $\kappa(n)$ is arbitrary. An example is $\kappa(2^m) = \dfrac M 2 \cdot \dfrac 1 {8\cdot2^m-4}$, which is not a hypergeometric function as rsolve requires. How can I do that?

REFERENCES:

Nacu, Şerban, and Yuval Peres. "Fast simulation of new coins from old", The Annals of Applied Probability 15, no. 1A (2005): 93-115.

答:
If $\kappa(2^m)$ is a sum of hypergeometric functions, the functional equation in my question has to be converted to a linear recurrence of the form—$$\eta(m) = \eta(m+1) + \kappa(2^m), $$ and the solutions found correspond to those in the original functional equation.

Here is an example in SymPy:

```<code>In [250]: rsolve(Eq(f(n), f(n+1) + z/sqrt(2*2**n)), f(n)).simplify()
Out[250]:
 -n  ⎛ n              ⎞
 ─── ⎜ ─              ⎟
  2  ⎜ 2              ⎟
2   ⋅⎝2 ⋅C₀ + z + √2⋅z⎠

In [251]: rsolve(Eq(f(n), f(n+1) + z/(4*2**n)), f(n)).simplify()
Out[251]:
      -n
     2  ⋅z
C₀ + ─────
       2
</code>```

EDIT:

As I found out, this works only if $\kappa(2^m)$ is a sum of hypergeometric functions, not necessarily for more general functions. Thus, my question remains open.

EDIT (Apr. 2):

In the meantime, I have managed to settle this question.

For the equation $\delta(n) = \delta(2n) + \kappa(n)$, it's enough to sum $\delta(n) = \kappa(n) + \kappa(2n) + \kappa(4n) + ...$ to get the solution. For this solution, it doesn't matter what the initial $c_1$ is; all that matters is whether the sum converges. I can assume that the sum does converge, and in that case, that sum is the solution I need.

Body:
If $\kappa(2^m)$ is a sum of hypergeometric functions, the functional equation in my question has to be converted to a linear recurrence of the form—$$\eta(m) = \eta(m+1) + \kappa(2^m), $$ and the solutions found correspond to those in the original functional equation.

Here is an example in SymPy:

In [250]: rsolve(Eq(f(n), f(n+1) + z/sqrt(2*2**n)), f(n)).simplify()
Out[250]:
 -n  ⎛ n              ⎞
 ─── ⎜ ─              ⎟
  2  ⎜ 2              ⎟
2   ⋅⎝2 ⋅C₀ + z + √2⋅z⎠

In [251]: rsolve(Eq(f(n), f(n+1) + z/(4*2**n)), f(n)).simplify()
Out[251]:
      -n
     2  ⋅z
C₀ + ─────
       2

EDIT:

As I found out, this works only if $\kappa(2^m)$ is a sum of hypergeometric functions, not necessarily for more general functions. Thus, my question remains open.

EDIT (Apr. 2):

In the meantime, I have managed to settle this question.

For the equation $\delta(n) = \delta(2n) + \kappa(n)$, it's enough to sum $\delta(n) = \kappa(n) + \kappa(2n) + \kappa(4n) + ...$ to get the solution. For this solution, it doesn't matter what the initial $c_1$ is; all that matters is whether the sum converges. I can assume that the sum does converge, and in that case, that sum is the solution I need.


问:
Formula or Algorithm to Draw curved lines between points
I'm developing a script to connect point with curved lines.

The points are in ascendent order in x-axis like this:

I'm studying Bezier Curves, but I don't think it's the best solution (see https://www.geogebra.org/m/qcnExXbn):

But, instead of straight lines, I just would like to connect these points in a smooth way.

Could anyone help me with a formula or algorithm?

答:
I created my own code to generate what you're looking for using the math in this video: https://www.youtube.com/watch?v=WCGKqJrf4N4

I required this code for a flutter project using Dart.

```<code>class LinearInterpolationPainter extends CustomPainter {
  final List<double> data;
  final double spacing;

  LinearInterpolationPainter({
    this.data = const [],
    this.spacing,
  });

  @override
  void paint(Canvas canvas, Size size) {

    //Used to determine the pixels between each data point
    double _spacing = spacing ?? size.width / (data.length - 1);

    var paint = Paint();
    // TODO: Set properties to paint

    paint.color = Colors.green.shade800;
    paint.style = PaintingStyle.stroke;
    paint.strokeWidth = 2.0;

    var path = Path();

    /// Set the path to initial data point
    path.moveTo(0, size.height - data[0]);

    // Draw path
    for (int x = 0; x < data.length - 1; x++) {
      for (int space = 0; space < _spacing; space++) {
        path.lineTo(
          (x * _spacing) + space,
          size.height - interpolate(x + (space / _spacing)),
        );
      }
    }

    /// Draw path to final data point
    path.lineTo(
      (data.length - 1) * _spacing,
      size.height - data.last,
    );

    canvas.drawPath(path, paint);
  }

  @override
  bool shouldRepaint(CustomPainter oldDelegate) {
    return true;
  }

  /// Calculates the y-value for an x-value given a set of points
  double interpolate(double a) {
    double result = 0;

    for (int j = 0; j < data.length; j++) {
      double product = data[j];

      for (int i = 0; i < data.length; i++) {
        if (i != j) product *= (a - i) / (j - i);
      }

      result += product;
    }

    return result;
  }
}
</code>```

Body:
I created my own code to generate what you're looking for using the math in this video: https://www.youtube.com/watch?v=WCGKqJrf4N4

I required this code for a flutter project using Dart.

class LinearInterpolationPainter extends CustomPainter {
  final List<double> data;
  final double spacing;

  LinearInterpolationPainter({
    this.data = const [],
    this.spacing,
  });

  @override
  void paint(Canvas canvas, Size size) {

    //Used to determine the pixels between each data point
    double _spacing = spacing ?? size.width / (data.length - 1);

    var paint = Paint();
    // TODO: Set properties to paint

    paint.color = Colors.green.shade800;
    paint.style = PaintingStyle.stroke;
    paint.strokeWidth = 2.0;

    var path = Path();

    /// Set the path to initial data point
    path.moveTo(0, size.height - data[0]);

    // Draw path
    for (int x = 0; x < data.length - 1; x++) {
      for (int space = 0; space < _spacing; space++) {
        path.lineTo(
          (x * _spacing) + space,
          size.height - interpolate(x + (space / _spacing)),
        );
      }
    }

    /// Draw path to final data point
    path.lineTo(
      (data.length - 1) * _spacing,
      size.height - data.last,
    );

    canvas.drawPath(path, paint);
  }

  @override
  bool shouldRepaint(CustomPainter oldDelegate) {
    return true;
  }

  /// Calculates the y-value for an x-value given a set of points
  double interpolate(double a) {
    double result = 0;

    for (int j = 0; j < data.length; j++) {
      double product = data[j];

      for (int i = 0; i < data.length; i++) {
        if (i != j) product *= (a - i) / (j - i);
      }

      result += product;
    }

    return result;
  }
}


问:
Reality check for my understanding of map space of infinity category
So the map space is $\operatorname{Map}_C(a,b)$, and it is the pullback of $\operatorname{Fun}(\Delta^1,C)\to C\times C$ along $\Delta^0\to C\times C$.

In my understanding $\pi_0(\operatorname{Map}_C(a,b))$ is the homotopy classes of maps so $\operatorname{Map}_C(a,b)$ should carry higher order data, right?

Then $\operatorname{Fun}(\Delta^1,C)$ should also carry higher order data, right?

The way $\operatorname{Fun}(\Delta^1,C)$ carries higher order data should be that for a degenerate simplex  the functor maps it to , right?

Please help me to do these reality checks, big thanks.

答:
Okay so I think I get it now and here's a note for anyone who has the same inquiry: the $\operatorname{Fun}$ does not mean $\operatorname{Hom}_{\mathrm{wKan}}$, but rather $\operatorname{Hom}_{\infty\text{-Cat}}$.

The latter does not have face and degeneracy maps, so the simplicial identities does not need to be satisfied.


问:
An abelian group $G$ of order $35$ with $g^{35}=e$ for all $g\in G$ is cyclic.
I'm reading "Contemporary Abstract Algebra," by Gallian.

This is Exercise 4.20.

Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?

There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $h\in H$, we have $h^{\lvert H\rvert}=e$; indeed: the cyclic subgroup $\langle h\rangle$ of $H$ has the same order as a group as the order $\lvert h\rvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $\lvert h\rvert$ divides $\lvert H\rvert$, so that then $h^{\lvert H\rvert}=e$.

So what gives?

I get the feeling that it's something obvious.

答:
A more general version:

If $G$ is an abelian group of order $pq$, where $p<q$ are primes and $p-1$ doesn’t divide $q-1$, then G must be cyclic.

Proof:
The elements have possible orders $1,p,q,pq$.(by Lagrange theorem)

Now if there is an element of order $pq$, we’re done because we have found an element whose order is the order of the group hence the group is cyclic.

If there exists elements $a,b$: $|a|=p$ and $|b|=q$, then $|ab|$ divides $pq$. [Why? Because in an abelian group $$(ab)^{lcm(|a|,|b|)}=a^{lcm(|a|,|b|)}b^{lcm(|a|,|b|)}=e$$ so $|ab|$ divides $pq$]. But $|ab|≠p$ and $|ab|≠q $ so $|ab|=pq$. Again we're done.

If every non-trivial element has the same order say, $p$, then  we have $pq-1$ elements of order $p$, but $\phi(p)=p-1$ should divide $pq-1=p(q-1)+(p-1)$, so $p-1$ should divide $q-1$ which isn’t possible.

I have used the fact that: In a finite group, the number of elements of order $d$ is divisible by $\phi(d)$.

This doesn't work with 33 as $33=3\times 11$ but $(3-1)$ divides $(11-1)$


问:
Generating functions to solve recurrence relation
Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you

答:
Let $h_n=a_n3^{-n}$. What is $h_n-h_{n-1}$? Now telescope. This gives that $$h_{n}-h_0=2\sum_{k=1}^n\frac{1}{3^k}$$

One can look at generating functions, but it proves much more tortuous. In general, a recurrence of the form $$x_{n+1}=ax_n+b$$ can be reduced by $y_n=x_n a^{-n}$ by $$y_{n+1}=y_n+\frac{b}{a^{n+1}}$$ and upon telescoping to $$y_{n+1}-y_0=b\sum_{k=0}^n \frac{1}{a^{k+1}}$$ that is $$x_{n+1}=b\sum_{k=0}^n a^{n-k}+a^{n+1}x_0$$

$$x_{n+1}=b\sum_{k=0}^n a^{k}+a^{n+1}x_0$$


问:
Pairs of Functions + How to find derivative of the inverse function.
1.Find a pair of function $f(x)$ and$g(x)$ such that$\frac{d(f(x))}{dx}= ln (2)*2^x*cos (g(x))$.

I can't figure out how to begin to solve this question. I know that there should be multiple pairs of solutions just don't know how to rearrange this equation to solve for them.
I'm not sure if this is right?
$f(x)=2^x\cos(g(x))\ln(2)dx$

2.Given$f(t) =t+ ln(t)$, find $\frac{d(f^{-1})}{dt}$ at $t= 1 +e$. Exact values. Note that the symbol$\frac{d(f^{-1})}{dt}$means the derivative of the inverse function with respect to t.

I've solved for the inverse of $f$ but have gotten stuck at $e^yy=e^x $
so I don't know how  I can apply this to find$\frac{d(f^{-1})}{dt}$ at $t= 1 +e$.

答:
For the first question, notice that $\ln(2) * 2^x$ is the derivative of $2^x$.  Consider:

$$
\frac{d(f(x))}{dx}= \left(ln (2)*2^x\right)*cos (g(x)) \\
\frac{d(f(x))}{dx}= g'(x) * cos (g(x))
$$


问:
Differential equation of a central orbit
I am solving dynamics(central orbits) and have a doubt in differential equation of a particle moving in central orbit.

Suppose a particle moves in central orbit under influence of a central force.
Let a particle move in a plane with an acceleration P , always directed towards a fixed point O which is also the centre of force.Let (r,θ) be the polar coordinates of the position P of the moving particle at any instant t.

Radial acceleration of particle is given by :-

d2r/dt2 - r (dθ/dt)2 = -P

Transverse acceleration of particle is given by :-

$\frac{1}{r}$$\frac{d}{dt}$(r2$\frac{dθ}{dt}$)

I know about radial and transverse accelerations in circular motion. Central orbit motion seems similar . But still not able to relate how the author presented these equations.

Any help is appreciated !!

答:
the core of a Central Forces problem is that F(r) = F(r)$ \hat{r}$. Due to this, the choice of spherical coordinates is pretty obvious.

So now, $$\mathbf{\dot{r}} = r\,\mathbf{\hat{r}} + r\dot{\varphi} \,\boldsymbol{\hat{\varphi}} \tag{*}$$

How do we get this? I would suggest checking out a classical mechanics book, like Taylor or Goldstein. Basically, this equation is a vector representation of radial and transverse velocities.

Differentiating (*) you would easily yield both the radial and transverse accelerations. I would highly suggest you try doing it yourself. Once you are done differentiating you would be getting two components, one of them would be the radial acceleration and the other the transverse acceleration.

hint on how to get equation (*).

$\mathbf{r} = (x, y) = (r\sin{\varphi}, r\cos{\varphi}) $

$\mathbf{\dot{r}}=\frac{d\mathbf{r}}{dt}= {\mathbf  {v}}={\dot  {r}}(\cos \varphi ,\ \sin \varphi )+r{\dot  {\varphi }}(-\sin \varphi ,\cos \varphi )   $

From here on hopefully you can clearly see what the unit vectors are.


问:
If $|G|$ does not divide $k!$, show that there is $N$ a normal subgroup in $G$, different from $\{e\}$ and $G$.
Let $G$ be a finite group and $H ≤ G$ with index $k > 1$. If $|G|$ does not divide $k!$, show that there is a normal subgroup $N$ of $G$, different from
$\{e\}$ and $G$.

P.D. I have been trying to solve this problem, and I have found similar problems on the site. I think that maybe I could take some ideas from those solutions but I still do not understand what role does the fact that $|G|$ does not divide $k!$ plays on this problem as a comparison to the question that I linked above where it is not specified.

If anyone can give me a hint or advice I would be very grateful.

答: Hint:  Let $G$ act on the set of cosets $G/H$.  This gives a morphism into $S_k$.  (The kernel of that morphism will be normal.)


问:
Find the first non-prime
Find the first non-prime $x(d)$ in Pell’s equation using the procedure below.

For every prime number $p$, construct a (possibly empty) series of natural numbers using the following procedure (start with $p=2$): Solve Pell’s equation
$$xx-dyy=1 \qquad \text{for } d=p-1$$
We note that $x$ (and $y$) depend on $d$ and rewrite the equation
$$
(x(d))^2 – d (y(d))^2=1
$$ for clarity.

If $p-1$ is a square (as it is for $p=2$) and consequently the equations will not have a solution, continue to the next higher prime $p=3$.
Solve Pell-s equation for $d= p-1=2$ and find the solution $x(2)=3$. Now continue to solve the equation for a new constant
$d_2=x(d-1)-1=2$.

In this case the equation
$$(x(2))^2-2(y(2))^2=1$$
has the solution $x(2)=3$. This presents us with the second reason to stop the process, circularity in the occurrence of numbers, since $2=d3=d2$.

Continue to the next prime $p=5$ (and dismiss it on the ground that $p-1$ is a square).
Note that our solution for $x$, $x(2)=3$ has been a prime. The problem is to find the first non-prime $x$ following this procedure.

答:
If I understood the procedure correctly, it seems that the process looks as follows:

$2$ (square plus 1)

$3\Rightarrow 3$ (repetition)

$5$ (square plus 1)

$7\Rightarrow 5$ (repetition)

$11\Rightarrow 19\Rightarrow 17$ (square plus 1)

(here I'm not sure if you want to go to next prime after $17$ or to the smallest not-yet-tried prime; it doesn't matter much in this case anyway)

$13\Rightarrow 7$ (repetition)

$17$ and $19$ (repetition)

$23\Rightarrow 197$ (square plus 1)

(the same dilemma here... but in this case, it matters)

$29\Rightarrow 127\Rightarrow 449\Rightarrow 127$ (repetition)

$31\Rightarrow 11$ (repetition)

$37$ (square plus 1)

$41\Rightarrow 19$ (repetition)

$43\Rightarrow 13$ (repetition)

$47\Rightarrow 24335$ (composite at last!)

If, instead of proceeding to $29$, you intended to go to the prime following $197$, it'd look like this:

$199\Rightarrow 197$ (repetition)

$211\Rightarrow 29\Rightarrow 127\Rightarrow 449\Rightarrow 127$ (repetition)

$131\Rightarrow 6499$ (composite!)

Title: Find the first non-prime


问:
question about parametric derivative.
In thomas calculus, if y=g(t), x=f(t), f,g is differentiable at t, and dy/dx exist and f'(t)$\neq 0$
imply $$dy/dx=g'(t)/f'(t)$$

I wonder then there exist a example s.t f,g is differentiable at t, and f'(t)is not zero but dy/dx is not exist?

答:
There is no such example. If $f$ is differentiable and $f'(t)\neq 0$ in some neighborhood of $t=t_0$, there is an inverse function defined in a neighborhood of $f(t_0)$, that is $t=h(x)$ for some differentiable $h$. To show this, just use the contrapositive of Rolle's Theorem. Then the composition $y=g(t)=g(h(x))$ is a differentiable function of $x$ in a neighborhood of $x_0=f(t_0)$ and $y'(x)$ is given by the chain rule.

Observe that you need to assume $f'(t)\neq 0$ in some interval, not just at the point $t_0$, since you are not assuming continuous differentiability.

Title: question about parametric derivative.


问:
Numerical integration: Solve upper bound given integral
Given the answer L to the integral of a function f(x), how can I solve for the upper bound b using numerical integration?

$$\displaystyle \int_0^{b} f(x) dx = L$$

The anti-derivative of f(x) is of course not trivially solvable (unless I've missed something - in my case it is the square root of a quartic function)

答:
You want to solve $F(b) = L$ for $b$ with $$F(b) = \int_0^bf(x)\,dx\,.$$
There are many methods to numerically solve an equation. As you know the derivative of $F$, I would suggest Newton's method.

Given a starting guess $b_0$, you simply iterate
$$b_{n+1} = b_n - \frac{F(b_n)}{F'(b_n)}=b_n - \frac{1}{f(b_n)} \left(\int_0^{b_n}f(x)\,dx -L\right)\,.$$


答:
Well there are actually many ways of estimate your integral, provided that it is typical Riemann integral, you can use the Trapezoidal rule, Simpson's rule, Gauss quadrature. Wikipedia has info, https://en.wikipedia.org/wiki/Trapezoidal_rule but it is better if you consults a book on numerical analysis.

For example refer to section 3.2 of Numerical Analysis Walter Gautschi.
https://books.google.co.cr/books?id=-fgjJF9yAIwC&printsec=frontcover#v=onepage&q&f=false

Title: Numerical integration: Solve upper bound given integral

Body:
Given the answer L to the integral of a function f(x), how can I solve for the upper bound b using numerical integration?

$$\displaystyle \int_0^{b} f(x) dx = L$$

The anti-derivative of f(x) is of course not trivially solvable (unless I've missed something - in my case it is the square root of a quartic function)


问:
Is it true that for any subset of real numbers there is a sequence in the real numbers that converges to it?
So I am defining the set convergence of a sequence $\{a_n\} \subset \mathbb{R}$ as follows:

$$
L = \{p \in \mathbb{R} \quad | \quad \exists \{b_n\} \subset \{a_n\}  \quad b_n \longrightarrow p\}
$$
and I note this as $a_n \longrightarrow L$

My question is:

Is it true that for any subset $A$ of $\mathbb{R}$ there is some $\{a_n\} \subset \mathbb{R}$ such that $a_n \longrightarrow A$?

答:
It is not true. If $L$ is the limit of a sequence, then $L$ is closed. This is easiest to see from a topological point of view. Your condition for set convergence can be reformulated this way:

$a_n$ converges to the set $L$ if for every point $p\in L$, every open neighborhood of $p$ contains elements of $a_n$.

Now let $x$ be a boundary point of $L$, that is, every open neighborhood $U_x$ of $x$ intersects $L$. Let $y\in U_x\cap L$. Now since $U_x$ is open, there is an open neighborhood $U_y$ of $y$ contained in $U_x$. And since $y\in L$, this neighborhood contains elements of $a_n$. But then so does $U_x$, since it's a superset of $U_y$. Since $U_x$ was an arbitrary neighborhood, this holds for all neighborhoods, and thus $x\in L$. So $L$ must contain all of its boundary points and is thus closed. Unless a set is closed, it can't be the limit of a sequence.

Uncountable set limits are possible, though. Enumerations of the rationals have the reals as their limit, since the rationals are dense in the reals.

Also, $L$ being closed is a sufficient condition for being the limit set of a sequence. Take a dense, countable subset of $L$ and enumerate it with the sequence $a_n$. Using this, define the sequence $b_n$ which goes like this:

$$a_1,~a_1,a_2,~a_1,a_2,a_3,~a_1,a_2,a_3,a_4,~a_1,a_2,a_3,a_4,a_5,~\dots$$

You get the pattern. Now any point in $L$ is a limit point of $b_n$, since $a_n$ enumerates a dense subset, meaning that all points in $L$ are limits of sequences whose elements are elements of $a_n$, and any such sequence can be constructed as a subsequence of $b_n$. No points outside of $L$ are limit points, since $L$ is closed and thus already contains all limit points of any sequence contained in $L$ (which $b_n$ is by construction).


答:
Here is a start

$$ \sum_{n=0}^{\infty}a_{n+1}x^n = 3 \sum_{n=0}^{\infty} a_{n}x^n + 2\sum_{n=0}^{\infty} x^n. $$

Now, recall this

$$  \sum_{n=0}^{\infty} x^n =\frac{1}{1-x}.$$

I think you can finish it now. Just follow the technique in this problem.

Title: Generating functions to solve recurrence relation

Body:
Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you


问:
calculating the stream function
Given that the flow is found incompressible (previous question here), is the approach above the right way towards finding the stream function? given that:
$\mathbf{u}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}$

$$\tag{B}
\sin{x}\sin{y}=\frac{\partial\phi}{\partial y}\qquad \cos{x}\cos{y}=-\frac{\partial\phi}{\partial x}$$

I have thought of approaching it this way:

$\frac{\partial\phi}{\partial y}=\sin{x}\sin{y} = \int\sin{x}\sin{y}$ and integrating with respect to y:

$-\cos{y}\sin{x} + f(x)$ where f(x) is a constant of integration,  and substituting the second part of (B).

$\frac{\partial\phi}{\partial x}(-\cos{y}\sin{x} + f(x)) = -f'(x) ; f(x) =-\sin{x}\cos{y} + C $

Therefore when adding both integrated factors we get :   $\phi=-\cos{y}\sin{x}-\sin{x}\cos{y}$ where $\phi$ is a constant.

Which is equal to:
$-2\cos{y}\sin{x}?$ when we simplify the equation on the right-side on 3.

答:
Step one gives you

$$ \phi(x,y) = -\cos y\sin x+f(x)$$
for some $f(x)$,
and step two gives
$$ \phi(x,y) = -\cos y\sin x+g(y),$$
for some $g(y)$. These are both correct.

It does not follow that $\phi$ is constant, i.e. your step 3 is wrong. Instead, combine the first two steps, and you should see $f\equiv0\equiv g$. (get some paper and actually write it out carefully, if it is not clear.)

You can check by direct computation that your guess for $\phi$ is wrong: Well, firstly, $\phi$ is not a constant, and neither is $-2\cos y \sin x$, but if you meant to say that

$$\phi(x,y)=-2\cos y \sin x$$

then just differentiate in either $x$ or $y$ to see that you have an extra factor of $2$.

Title: calculating the stream function


问:
Distance from the center of a regular unit simplex to a vertex
I am working through Lectures in Discrete Geometry, and one of the questions goes as follows:

Any point set of diameter $1$ in $\mathbb{R}^2$ can be enclosed in a disc of radius $1/\sqrt{3}$. Prove this statement, then find a $r = r(d)$ for which a point set of diameter $1$ in $\mathbb{R}^d$ can be enclosed by a $d$-disc of radius $r$. Prove it.

I know the structure of the general proof, using Helly's Theorem. However, a key step in the proof is noting that any equilateral triangle with side length $1$ will always have a distance of $1/\sqrt{3}$ from its center to its vertices.

The $d$-dimensional generalization should have the exact same structure as the $2$-dimensional proof, but I cannot find the $r$. There is a hint in the back of the book which says:

$\sqrt{2d/(d+1)}$

But I cannot parse the hint. Is this supposed to be $r$? In that case, why is $r(2) = \sqrt{4/3}$, rather than the correct $1/\sqrt{3}$?

I believe that $r$ should be the distance I ask for in the title, but I am having trouble coming up with an argument for a distance dependent only on $d$, and I still don't understand the hint.

答:
There is a nice embedding of the regular $d$-dimensional simplex into $\mathbb{R}^{d+1}$, by taking the vertices to be the unit vectors along every axis, i.e., $(1,0,0,\ldots,0), (0,1,0,\ldots,0), \ldots, (0,0,\ldots,0,1)$.

This tetrahedron has a center of $(a,a,a,\ldots,a)$, where $a=\frac1{d+1}$. It has a side length of $\sqrt{2}$.

Taking the squared differences in each coordinate, we see that the distance from this center to any vertex is $\sqrt{d\cdot a^2+(1-a)^2} = \sqrt{\frac{d+d^2}{(d+1)^2}}=\sqrt{\frac{d}{d+1}}$.

Scaling down by a factor of $\sqrt{2}$ so that we normalize to a unit tetrahedron, our final answer is $\sqrt{\frac{d}{2(d+1)}}$. You can verify that at $d=1$ this is $1/2$, and at $d=2$ is it $1/\sqrt{3}$, as expected.

Perhaps the book misplaced this factor of $2$? I am not sure how they arrived at that value.


问:
Young's generalizated inequality
If $a_{1},a_{2},\dots,a_{n}$ are non-negative numbers, $p_{1},p_{2},\dots,p_{n}, \ p_{i}>1,$ for all $i=1,\dots,n$ and  $\sum_{i=1}^{n}\dfrac{1}{p_{i}}=1,$ then $$\prod_{i=1}^{n} a_{i} \leq \sum_{i=1}^{n} \frac{a_{i}^{p_{i}}}{p_{i}}.$$

Could anyone help me? For case $n=2$ is classic and I know. But I couldn't generalize.

答:
A nice and not so difficult way to prove Young's generalized inequality is by induction on $n \geq 2$.

The base case for $n = 2$ is classic and you said you already know how to prove it so let's assume the desired result is valid for $n=2$.

Now, suppose that the inequality is true for some $n \geq 2$. That is, given any nonnegative real numbers $b_{1}, \ldots , b_{n}$ and any real numbers $q_{1}, \ldots, q_{n}$ such that $q_{i} > 1$ for $i = 1, \ldots , n$ and $\sum_{i=1}^{n} \frac{1}{q_{i}} = 1$, then $\prod_{i=1}^{n} b_{i} \leq \sum_{i=1}^{n} \frac{b_{i}^{q_{i}}}{q_{i}}$.

For the inductive step, assume that $a_{1}, \ldots , a_{n}, a_{n+1}$ are nonnegative real numbers and $p_{1}, \ldots , p_{n}, p_{n+1}$ are real numbers satisfying the conditions $p_{i} > 1$ for $i = 1, \ldots , n+1$ and $\sum_{i=1}^{n+1} \frac{1}{p_{i}} = 1$. We want to show that $\prod_{i=1}^{n+1} a_{i} \leq \sum_{i=1}^{n+1} \frac{a_{i}^{p_{i}}}{p_{i}}$. Let $a= \prod_{i=1}^{n} a_{i} \geq 0$ and $x = \sum_{i=1}^{n} \frac{1}{p_{i}} = 1 - \frac{1}{p_{i+1}}$. Notice that $x \in (0,1)$. Then $y = \frac{1}{x} > 1$, $p_{n+1} > 1$, and $\frac{1}{y} + \frac{1}{p_{n+1}} = x + \frac{1}{p_{n+1}} = 1$. Since $a$ and $a_{n+1}$ are nonnegative real numbers, by Young's inequality (which is the base case for $n=2$), we know that:

$\prod_{i=1}^{n+1} a_{i} = a \cdot a_{n+1} \leq \frac{a^{y}}{y} + \frac{a_{n+1}^{p_{n+1}}}{p_{n+1}}$.

Therefore, it is enough to prove that $\frac{a^{y}}{y} \leq \sum_{i=1}^{n} \frac{a_{i}^{p_{i}}}{p_{i}}$. For this purpose observe that $a_{1}^{y}, \ldots , a_{n}^{y}$ are nonnegative real numbers, $\frac{p_{1}}{y}, \ldots , \frac{p_{n}}{y}$ are real numbers greater than $1$ (because clearly $0 < \frac{1}{p_{i}} < x$ for $i = 1, \ldots , n$) and $\sum_{i=1}^{n} \left( \frac{p_{i}}{y} \right)^{-1} = y \sum_{i=1}^{n} \frac{1}{p_{i}} = yx = 1$. Our induction hypothesis lets us conclude that:

$a^{y} = \prod_{i=1}^{n} a_{i}^{y} \leq \sum_{i=1}^{n} \frac{(a_{i}^{y})^{\frac{p_{i}}{y}}}{\frac{p_{i}}{y}} = y \sum_{i=1}^{n}\frac{a_{i}^{p_{i}}}{p_{i}}$,

which leads us immediately to the desired inequality.

Title: Young's generalizated inequality


问:
$X$ is an independent random variable from any other random variable then $X$ has to be a constant.
I had to show that the constant random variable is independent of any other random variable.

Now I'm looking if $X$ is an independent random variable from any other random variable then $X$ has to be a constant.

Is the following proof correct?

Edit: I have developed a little my proof but am still unsure if it is enough to show what is asked

$F_{X,X}(x,x)=\mathbb{P}(X \leq x, X \leq x) = \mathbb{P}(X \in A, X \in A)= \mathbb{P}(X \in A) \cdot \mathbb{P}(X \in A)$

$=[\mathbb{P}(X \in A)]^2 = F_{X}(x) \cdot F_{X}(x)$

Therefore, $\mathbb{P}(X \leq x)=0$ or $\mathbb{P}(X \leq x)=1$

Hence, the event:

\begin{equation}
     (X \in A) = \left\{
        \begin{array}{ll}
   1      & \mathrm{if\ } x \in A \\
   0     & \mathrm{if\ } x \notin A
        \end{array}
      \right.
   \end{equation}

then

\begin{equation}
     (X \leq x) = \left\{
        \begin{array}{ll}
   \emptyset      & \mathrm{if\ } X < x \\
   \Omega     & \mathrm{if\ } x \ge X
        \end{array}
      \right.
   \end{equation}

and therefore, $X$ is a constant

答:
If you're interested in an other solution,
try consider the random variable $Y:=$ arctan$(X)$ and notice that var$(Y) = 0$, what can you conclude  ?


问:
Transitivity of Equivalence Relation of Tangent Vectors on Manifold
I'm just starting to learn differential topology, and came across a notion when defining tangent vectors on a smooth manifold that confused me a bit. Let $M$ be a smooth ($C^\infty$) manifold, with $p \in M$, and define a smooth curve at $p$ as a smooth map $\gamma \colon (-1, 1) \to M$ such that $\gamma(0) = p$. Let $(U, \phi)$ be any coordinate chart on $M$ containing $p$. Two smooth maps at $p:\,$ $\gamma_1, \gamma_2$ are called equivalent if $(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0)$; this is denoted $\gamma_1 \sim_p \gamma_2$. A tangent vector is defined as an equivalence class of all smooth curves at $p$ under this relation.

The problem I'm having with this is proving that it is an equivalence relation. Obviously, reflexivity and symmetry follow immediately from the definition, but I'm trying to prove the transitivity of this relation. I wrote out a proof but wasn't sure if it was valid because of the use of the chain rule outside Euclidean space. Here it is:

Let $\gamma_1, \gamma_2, \gamma_3 \colon (- 1, 1) \to M$ be smooth curves at $p$. Assume that $\gamma_1 \sim_p \gamma_2$ and $\gamma_2 \sim_p \gamma_3$. Then there exists coordinate charts $(U, \phi), (V, \psi)$ containing $p$ such that:

$$(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0) \,\,\text{ and } \,\, (\psi \circ \gamma_2)'(0) = (\psi \circ \gamma_3)'(0)$$

Using the chain rule, the first equality implies that

$$ \phi'(\gamma_1(0)) \cdot \gamma_1'(0) = \phi'(\gamma_2(0)) \cdot \gamma_2'(0)$$
As $\gamma_1(0) = \gamma_2(0) = p$, this means that $\gamma_1'(0) = \gamma_2'(0)$. Likewise, we have that

$$(\psi'(\gamma_2(0)) \cdot \gamma_2'(0) = (\psi'(\gamma_3(0)) \cdot \gamma_3'(0) \Rightarrow \gamma_2'(0) = \gamma_3'(0)$$

This means that $\gamma_1'(0) = \gamma_3'(0)$, meaning that

$$\phi'(\gamma_1(0))\cdot \gamma_1'(0) = \phi'(\gamma_3(0)) \cdot \gamma_3'(0),$$

so that $$(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_3)'(0) \Rightarrow \gamma_1 \sim_p \gamma_3$$

I feel like it isn't legal to use the chain rule there because the codomain of our smooth maps isn't Euclidean space, but I can't think of any other way to prove the transitivity of this relation. Is my proof correct? If not, could someone give me a hint as to how to prove transitivity? Thanks!

答:
Indeed, at this point the use of the chain rule is still illegal because the domains of charts are open subsets of the manifold and you still "don't know" what a derivative means there. Here's how to fix it.

If $\gamma_1\sim_p\gamma_2$ and $\gamma_2\sim_p \gamma_3$, there are $(U,\varphi)$ and $(V,\psi)$ around $p$ with $(\varphi\circ \gamma_1)'(0) = (\varphi\circ \gamma_2)'(0)$ and $(\psi\circ \gamma_2)'(0)=(\psi\circ \gamma_3)'(0)$, alright. Compute $$\begin{align} (\varphi\circ \gamma_3)'(0) &= (\varphi\circ\psi^{-1}\circ \psi\circ\gamma_3)'(0) \\ &= {\rm d}(\varphi\circ \psi^{-1})_0((\psi\circ\gamma_3)'(0)) \\ &= {\rm d}(\varphi\circ \psi^{-1})_0((\psi\circ\gamma_2)'(0)) \\ &= (\varphi\circ \psi\circ \psi^{-1}\circ\gamma_2)'(0) \\ &= (\varphi \circ \gamma_2)'(0) \\ &= (\varphi\circ \gamma_1)'(0).\end{align}$$My use of the chain rule here was legal because I made sure to apply it only for the transition between charts, which is a map between open subsets of Euclidean spaces, and composing a curve in the manifold with a chart gives a curve in Euclidean space.


问:
Derivative of quadratic form for matrices and vectors
I'm not very familiar with multivariable calculus as it relates to matrices. Could someone explain, in detail, why
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = (A + A^T)x$$
In the case of a symmetric matrix and
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = 2Ax$$
if the matrix is not symmetric. I'm mainly confused about how we even arrive at the first derivative. However, I understand how the first derivative simplifies to the second in the case that A is symmetric.

答:
We have
$$
\begin{split}
\frac{\partial}{\partial x} \left[ x^T A x \right]v
&=
\lim_{h\to0}\frac{(x+hv)^T A (x+hv)-x^T A x}{h} \\
&=
\lim_{h\to0}\frac{x^T A hv+(hv)^T A x+(hv)^T A hv}{h} \\
&=
x^T A v+v^T A x \\
&=
x^T A v+x^T A^T v \\
&= x^T(A + A^T)v
\end{split}
$$
and so
$$
\frac{\partial}{\partial x} \left[ x^T A x \right]= x^T(A + A^T).
$$
You can see the row vector $x^T(A + A^T)$ as the column vector $(x^T(A + A^T))^T=(A + A^T)x$, but strictly speaking it is not the same.


问:
How can a discrete stochastic integral be used to show that $(X_n)$ submartingale, $S\le T\implies E(X_S)\le E(X_T)$
I am just learning about martingales and discrete stochastic integrals. For completeness, we define a discrete stochastic integral as
$$Y_n=\sum_{k=1}^nC_k(X_k-X_{k-1})=(C\bullet X)_n$$
for $(X_n)$ adapted and $C_n$ previsible in this context.

An exercise from this section says we can use a discrete stochastic integral to show that if $(X_n)$ is submartingale and $S, T$ are a.s. bounded stopping times with $S\le T$ then $E(X_S)\le E(X_T)$.

Because of the boundedness of the stopping times, I thought this looked more like an optional stopping problem, but the exercise clearly states that this can be shown using a suitable discrete stochastic integral.

Does someone know how the stochastic integral relates to this problem?

EDIT: The only result I already have for the discrete stochastic integral is
if $C_n\ge 0$ then $(C\bullet X)_n$ is a submartingale as well. Perhaps I need to define a discrete stochastic integral and then apply optional stopping to that?

答:
The result for the discrete stochastic integral that you mention, is exactly what is required. Furthermore the required stochastic integral is also easy to construct.

Let's take two a.s. bounded stopping times $S \leq T$. There is an implicit filtration w.r.t which $X_n$ is adapted and a submartingale, that I call as $\mathcal F_n$.

First, note that $X_S$ and $X_T$ are well-defined random variables, because $S$ and $T$ are stopping times. Since $T$ is a.s. bounded there is a natural number $M>0$ such that $T \leq M$ a.s. and because of this, we know that $|X_T|= \sum_{i=0}^M 1_{\{T = i\}} |X_i|$ a.s. , but the RHS is integrable by linearity and has integral at most equal to $ME[|X_0|] < \infty$ since $X_0$ is part of a submartingale hence is by definition integrable. Therefore, $E[|X_T|]< \infty$ and so $E[X_T] < \infty$.

In a similar fashion, $E[X_S] < \infty$.

Now, like the optional stopping theorem, we introduce a $t_0 > 0$, and wish to prove $E[X_{T \wedge t_0}] \geq E[X_{S \wedge t_0}]$ before taking $t_0 \to \infty$ and ensuring things work out.

To do this, we prove that $X_{T \wedge t_0} - X_{S \wedge t_0}$ equals a discrete stochastic integral of $X_n$ by a non-negative random variable. Hence it will be a sub-martingale, and we can use the optional stopping theorem.

But such a sub-martingale is easy to create : let $C_n = 1_{T \geq n} - 1_{S \geq n}$. Then note that $\sum_{i=1}^{t_0} C_i (X_{i}-X_{i-1}) = X_{T \wedge t_0} - X_{S \wedge T_0}$ by an easy $\omega$-by-$\omega$ check. $C_n$ is non-negative a.s. since $S \leq T$, and $C_n$ is previsible since each of $T \geq n$ and $S \geq n$ are $\mathcal F_{n-1}$ measurable, and $C_n$ is just a linear combination of their indicators.

Therefore, the conclusion is that $Y_{t_0} = X_{T \wedge t_0} - X_{S \wedge t_0}$ is a well defined submartingale, with $Y_0 = 0$. But then, for a submartingale it is well known that the sequence $E[Y_n]$ is an increasing sequence. Therefore $E[Y_0] \leq E[Y_n]$, which gives us that $E[X_{T \wedge t_0}]  \geq E[X_{S \wedge t_0}]$ for all $t_0$.

Finally, note that as $t_0 \to \infty$, the a.s. boundedness of $T$ and $S$ imply that for $t_0 >M$ we have $T \wedge t_0 = T$ and $S \wedge t_0 = S$ a.s. , so the pointwise limits $X_{T \wedge t} \to X_T$ and $X_{S \wedge t} \to X_S$ hold a.s.

It is now about the dominated convergence theorem, but then that is clear since $|X_{T \wedge t}| \leq \sum_{i=0}^M 1_{\{T=i\}} |X_i|$ a.s. , and the latter has finite expectation. Similarly for $S$. Now we conclude that $E[X_T] \geq E[X_S]$.

In continuous time, more regularity must be imposed on both the filtration and the process itself if we are to obtain a result of the kind above. But it does hold.


问:
Showing a topology is finer/coarser
The following theorem is given:
"Let $(X, \tau_1)$ and $(X, \tau_2)$ be topological spaces. Then $\tau_1$ is coarser than $\tau_2$ if and only if for each $x \in X$ and every $U \in \tau_1$ with $x \in U$ there exists a $V \in \tau_2$ such that $x \in V \subseteq U$."

I was wondering whether the theorem is still true if I only apply them for basis elements. Since every open set is unions of basis, if I can show that the theorem above is true for basis elements, then I can show one topology is finer/coarser than another correct?

答:
Yes we can restrict to bases:

If $(X,\tau_1)$ has a base $\mathcal{B}_1$ and $(X.\tau_2)$ has a base $\mathcal{B}_2$ then $\tau_1$ is coarser than $\tau_2$ iff

$$\forall x \in X: \forall U \in \mathcal{B}_1: (x \in U) \to (\exists V \in \mathcal{B}_2: x \in V \subseteq U)\tag{1}$$

That (1) holds if $\tau_1 \subseteq \tau_2$ is clear: the basic $U$ is open in $\tau_1$ and so also in $\tau_2$ so the last fact follows from $\mathcal{B}_2$ being a base for it.

And if $(1)$ holds, write  $U \in \tau_1$ as a union of basic sets $B' \in \mathcal{B}_1$ and then we write each such $B'$ as a union of elements from $\mathcal{B}_2$ using $(1)$ (for each $x$ in $B'$ take some $B_x \in \mathcal{B}_2$ with $x \in B_x \subseteq B'$ and union up those $B_x$) which
is thus $\tau_2$-open and so $\mathcal{B}_1 \subseteq \tau_2$ from which $\tau_1 \subseteq \tau_2$ follows directly.

Title: Showing a topology is finer/coarser


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答:
Depending on the number of levels that you have in your game, it may make more sense to hard code the values. Really depends if these experience points are never going to change, or if in a couple weeks your leader has a new set of values (and thus a new function you have to produce).

Ask your leader how he/she came up with the values and they have a heuristic that they are using. Likely it will be something like "$x$ times as many points as the previous level required".

Otherwise, you can go to wikipedia and read up on Regression analysis.

Title: How to "invent" a function?


问:
Irreducibility of the standard representation of $S_n$.
The permutation representation of $S_n$ is $\mathbb C^n$ with elements of $S_n$ permuting the basis vectors $\{e_1, e_2, \ldots, e_n\}$. It has a trivial subrepresentation spanned by the vector $v = \sum_i e_i$. By Maschke's theorem there is a complement subrepresentation (given by the condition $\sum_i x_i = 0$, where $x_i$ is the i'th coordinate). This last representation (the standard representation of $S_n$) is irreducible. Why is this?

If we try characters, we can get the following: denote the standard representation by $V$ (dimension n-1), and the full permutation representation by $Perm$. Then
$$
 Perm = V\oplus Triv,
$$
where the latter is the trivial representation. This means that $\chi_V(g) = |X^g| - 1$, and
$$
 \langle\chi_V, \chi_V\rangle = \frac{1}{|G|}\sum(|X^g|-1)^2|C(g)| = 1+\frac{1}{|G|}\sum (|X^g|^2 - 2|X^g|)|C(g)|,
$$
where $X = \{1,2,\ldots, n\}$ and $X^g$ is the set of fixed points of permutation $g\in S_n$ and $C(g)$ is the conjugacy class of $g$. The sum is taken over all different conjugacy classes of $S_n$.

Now, since I know that $V$ is irreducible, the last sum must be zero. I don't see, however, why that should be.

Perhaps there is a direct proof or irreducibility?

Thank you.

答:
Another nice proof comes the following fact:

if a finite group $G$ acts on a set $X$ two-transitively, then the obvious  representation of $G$ on the vector space with $X$ has a basis is irreducible,

whose proof, in turn, I will leave as an exercise.


问:
“Generalized” Fourier transform
When solving boundary value problems, I know that a lot of the time a solution can be found using an eigenfunction expansion, where the eigenfunctions are discrete. I am wondering if there is a similar theory for operators which have continuous eigenfunctions. To be precise, I’m trying to justify that given a continuous spectrum of eigenfunctions $\psi_f$ of an operator
f, a function $\Psi$ may be written as
$$\Psi(x)=\int a_f \psi_f(x) df$$
where the $\psi_f$ are “normalized” such that
$$\int |a_f|^2df = 1$$
and
$$a_f=\int \Psi(x)\psi_f^*(x)dx$$
That statement is taken out of a quantum mechanics textbook and I am trying to find the mathematical theory which makes it precise. Any explanation or reference recommendation would be greatly appreciated.

答:
The Fourier transform is one such example. However, the eigenfunctions are not normalizable, even though integrals of them are normalization. If $L=\frac{1}{i}\frac{d}{dx}$, then $L\psi_s = s\psi_s$ where $\psi_{s}(x)=\frac{1}{\sqrt{2\pi}}e^{isx}$. However $\psi_s(x)$ is not normalizable on $(-\infty,\infty)$ even though integrals with respect to the parameter $s$ are integrable:
$$
                \psi(x)=\int_a^b a(s)\psi_{s}(x)ds
$$
This function is square integrable if the coefficient function $a$ is square integrable, and
$$
            \int_{-\infty}^{\infty}|\psi(x)|^2dx=\int_a^b|a(s)|^2ds.
$$
This is justified using properties of the Fourier transform.

Title: “Generalized” Fourier transform


问:
Pullback of Fubini-Study form on $\mathbb{C}P^1$ under stereo graphic map
There is a well-known result that pullback of Fubini-Study form on $\mathbb{C}P^1$ under stereo graphic map gives one quarter of the standard metric on $S^2$, for example see exercise 3.2 (Ⅴ) on page 3 of this note.

However, when I was trying to prove this I got exactly the same result as this question and I can't see where the process went wrong. This is not pointed out in the answer of that question. Can anybody tell me where I went wrong on this?

答: Hint: Instead of using the stereographic projection on $\Bbb{C}P^1$ use it on $S^2$. So one has the following map $$\phi : \Bbb{R}^2 \to \Bbb{S}^2  \\ \phi(x_1,x_2) = (\frac{2x_1}{1+x_1^2 + x_2^2}, \frac{2x_2}{1+x_1^2 + x_2^2}, \frac{-1+x_1^2 + x_2^2}{1+x_1^2+x_2^2})$$ Now pullback the standard 2 form on the sphere to $\Bbb{R}^2$


问:
Two forms of Generalization Rule in logic
I have a problem with two forms of the generalization rule in logic.

Shoenfield proves in his book that if $T\vdash\phi$ where $T$ is a set of formulas and $\phi$ is a formula, then $T\vdash\forall x\phi$.

His proof is, "from $T\vdash\phi$ and the tautology $\phi\to(\lnot\forall x\phi\to\phi)$, $T\vdash\lnot\forall x\phi\to\phi$. Then by the $\forall$-introduction rule, $T\vdash\lnot\forall x\phi\to\forall x\phi$ ; so $T\vdash\forall x\phi$ by the tautology $(\lnot\forall x\phi\to\forall x\phi) \to\forall x\phi$" ( The $\forall$-introduction rule can be deduced by the $\exists$-introduction rule which is considered as one of rules of inference in the book).

On the other hand, Enderton shows that if $T\vdash\phi$ and $x$ do not occur free in any formula in $T$, then $T\vdash\forall x\phi$, and says the restriction that $x$ not occur free in $T$ is essential, whereas Shoenfield does not restrict anything.

So $Px\vdash\forall x (Px)$ by Shoenfield and may not by Enderton. I wonder why it happens.

My opinion is that it happens because of the difference between their definition of '$\vDash$ for sets of formulas'. Enderton defines that $Px\vDash\forall x Px$ iff for every structure $\mathfrak A$ and every $a\in A$, the universe of $\mathfrak A$, such that $\mathfrak A\vDash Px[a]$, $\mathfrak A\vDash\forall x Px$. But Shoenfield writes $Px\vDash\forall x Px$ when for every structure $\mathfrak A$ such that $\forall a\in A(\mathfrak A\vDash Px[a])$, $\mathfrak A\vDash\forall x Px$. Obviously, $Px\not\vDash\forall x Px$ with Enderton's definition and $Px\vDash\forall x Px$ with Shoenfield's one.

Are there the others which expain why it happens? and Which definition is used in the modern mathematics?

答:
You are right.

Here is a proof of the "full" Generalization Theorem.

First of all, we can "relax" the $\forall$-Introduction rule (page 31): $A \to B \vdash A \to \forall x B$, provided that $x$ is not free in $A$.

Thus:

$\Gamma \vdash A$

$\vdash A \to (y=y \to A)$ --- it is an instance of a tautology: $y$ new

$\Gamma \vdash y=y \to A$

$\Gamma \vdash y=y \to \forall x A$ --- by "relaxed" $\forall$-Introduction

$\vdash y=y$ --- identity axiom (page 21)

$\Gamma \vdash \forall x A$.

Regarding $\vDash$, there is a difference.

According to Shoenfield (page 20) formula $A$ is a logical consequence of $\Gamma$ if $A$ is valid in every structure in which all formulas of $\Gamma$ are valid.

Consider the case of $(x=0)$ and the structure $\mathbb N$. Formula $(x=0)$ is not valid in it because it is not true that every instance of the formula is true in $\mathbb N$.

If we choose the new constant $c_1$ to name the number $1$, we have that $(x=0)[c_1]$ is false, and thus: $\mathbb N \nvDash (x=0)$.

Thus, "if $(x=0)$ is valid, then $\forall x (x=0)$ is valid" is vacuously true, and thus in Shoenfield's system we have:

$Px \vDash \forall Px$,

and this is consistent with the "unrestricted" Generalization rule:

For any $\Gamma$ and any $A$, if $\Gamma \vdash A$, then $\Gamma \vdash (∀x)A$.

See Tourlakis, page 43 and page 52.

In a nutshell, for Shoenfield $Px$ and $\forall x Px$ are semantically equivalent. This is not so in Enderton's system.

This fact induces some other differences: in Enderton, thanks to the restriction on Generalization, we have an "unrestricted" Deduction Theorem, while in Shoenfield and Tourlakis we have that "for any closed formula $A$ and arbitrary formula $B$, if $\Gamma, A \vdash B$, then $\Gamma \vdash A → B$."

In both cases, what is avoided is to "validate" the invalid: $Px \to \forall x Px$.

In Enderton's system the "move" $Px \vdash \forall x Px$ is not allowed.

In Shoenfiled's system we have it but we cannot use DT to derive $Px \to \forall x Px$.

See also this post as well as this one.


问:
Prove the metric distance function is continuous using only topology?
Let $(X,d)$ be a metric space and $τ$ the corresponding topology on $X$.  Fix $a∈X$.  Prove that the map $f:  (X,τ)→R$ defined by $f(x)=d(a,x)$ is continuous.

Is there a way that this can be proved using only topology without using the epsilon-delta definition?

答:
We can use that $\Bbb R$ is an ordered topological space as well:

Let $O=(\beta,\rightarrow)=\{r \in \Bbb R\mid r > \beta\}$ be a subbasic set in $\Bbb R$. Let $x \in f^{-1}[O]$ be arbitrary (so we know that $d(x,a) > \beta$).

Set $r=d(x,a)-\beta > 0$ and show that $B(x,r) \subseteq f^{-1}[O]$. So $f^{-1}[O]$ is open.

If $O'=(\leftarrow, \gamma)=\{r \in \Bbb R\mid r < \gamma\}$ is another type of subbasic open set, let $x \in f^{-1}[O']$ and set $r' = \gamma-d(x,a)>0$. Then show $B(x, r') \subseteq f^{-1}[O']$ so the latter set is also open.

From these two facts it follows that $f$ is continuous.


问:
How to derive the kth coefficient standard error?
Given a multiple regression with the usual assumptions satisfied, with $X \in R^{n \times p}$

$$ y = X \beta + e $$

I know that the estimated variance is given by $\sigma^2 (X^TX)^{-1}$. But what I want to know is the estimated variance for the $k^{th}$ coefficient out of this whole covariance matrix.

I've seen from the resource A (cited below) that this value is the $k^{th}$ diagonal term of the $\sigma^2 (X^TX)^{-1}$. But more interestingly (AND THIS IS WHAT I WANT TO PROVE), this value is:

$$ \dfrac{\sigma^2}{(1-R^2_k) \sum_{i=1}^n (x_{ik} - \bar{x_k})^2 } $$

Here, $x_k$ is the $k^{th}$ column of $X$. $R^2_k$ is the $R^2$ from regressing $x_k$ on $X_{(k)}$(=$X$ after taking $k^{th}$ column out).

Here is my (failed) attempt at this derivation:

Using resource B (cited below), the $k^{th}$ diagonal value of $\sigma^2 (X^TX)^{-1}$ is:

$$\sigma^2 [x_k^Tx_k - x_k^T X_{(k)} (X_{(k)}^TX_{(k)})^{-1}X_{(k)}^T x_k ] ^ {-1}$$

Since $X_{(k)} (X_{(k)}^TX_{(k)})^{-1}X_{(k)}^T x_k$ can be seen as projection of $x_k$ onto column space of $X_{(k)}$, we can say:

$$ X_{(k)} (X_{(k)}^TX_{(k)})^{-1}X_{(k)}^T x_k = \hat{x_k} $$, which is the regression prediction after regressing $x_k$ on $X_{(k)}$.

So our diagonal value simplifies to:

$$ \sigma^2 [x_k^Tx_k - x_k^T X_{(k)} (X_{(k)}^TX_{(k)})^{-1}X_{(k)}^T x_k ] ^ {-1}= \sigma^2 [ x_k^Tx_k - x_k^T \hat{x_k} ] ^ {-1}$$

Now, since (I will omit subscript $i$ so $x_k = x_{ik}$)

$$ R^2_k = 1 - \dfrac{ \sum (x_k - \hat{x_k} )^2 }{ \sum (x_k - \bar{x_k} )^2 } $$

$$ \dfrac{\sigma^2}{(1-R^2_k) \sum (x_{k} - \bar{x_k})^2 } $$

will simplify to:

$$ \dfrac{\sigma^2}{ \sum (x_{k} - \hat{x_k})^2 } $$

So in summary, we want to show that:

$$ \sigma^2 [ x_k^Tx_k - x_k^T \hat{x_k} ] ^ {-1} = \dfrac{\sigma^2}{ \sum (x_{k} - \hat{x_k})^2 }$$

But then,

$$ \sigma^2 [ x_k^Tx_k - x_k^T \hat{x_k} ] ^ {-1} = \dfrac{\sigma^2}{\sum (x_k)^2 - \sum x_k \hat{x_k} } $$

where as,

$$ \dfrac{\sigma^2}{ \sum (x_{k} - \hat{x_k})^2 } = \dfrac{\sigma^2}{\sum (x_k)^2 - 2\sum x_k \hat{x_k} + \sum x_k^2 } $$

Could someone please help me find where I am making a mistake, and if this is not the right approach, help me derive it correctly?

Resource A: http://people.stern.nyu.edu/wgreene/MathStat/GreeneChapter4.pdf (page 40)

Resource B: About the diagonal entries of an inverse matrix

答:
You are on the right track. Putting the subscript $i$ back into your equation, you are trying to show that
$$x_k^T x_k-x_k^T \hat x_k =\sum_i (x_{ik}-\hat x_{ik})^2. \tag1$$
The RHS of (1) can be written in matrix form as $(x_k-\hat x_k)^T(x_k-\hat x_k)$. So
comparing this to the LHS, it is enough to show that
$$\hat x_k^T(x_k-\hat x_k)=0.\tag2$$ But (2) is a consequence of:

Claim: In least squares regression with the usual assumptions, the predicted response vector $\hat y$ is orthogonal to the residual vector $y-\hat y$.

Proof: Write $\hat y=Hy$ where $H:= X(X^TX)^{-1}X^T$ is symmetric (i.e., $H^T=H$) and idempotent (i.e., $H^2=H$). Therefore
$$\hat y^T(y-\hat y)=(Hy)^T(y-Hy)=y^TH^T(I-H)y=y^TH(I-H)y=0.$$