问:
How do Taylor Series represent analytic functions when the function has a singularity?
So I'm reading Taylor's Theorem:

Let $f(z)$ be analytic in a domain $D$, and let $z=z_0$ be any point in $D$. Then there exists precisely one Taylor series (1) with center $z_0$ that represents $f(z)$. This representation is valid in the largest open disk with center $z_0$ in which $f(z)$ is analytic.

And my book's text re: Taylor and Power series:

We see that Taylor series are power series. From the last section we know that power series represent analytic functions. And we now show that every analytic function can be represented by power series, namely, by Taylor series (with various centers). This makes Taylor series very important in complex analysis. Indeed, they are more fundamental in complex analysis than their real counterparts are in calculus... They are the most important series in complex analysis because their sums are analytic functions (Theorem 5, Sec. 15.3), and every analytic function can be represented by power series

But the geometric series can be represented by this function:

$$f(z) = \frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + \dots$$ which converges if $\left | z \right | < 1$

$f(z)$ is singular at $z = 1$. Is it still analytic?

My definitions of analytic:

A function $f(z)$ is said to be analytic in a region $R$ of the complex plane if $f(z)$ has a derivative at each point of $R$ and if $f(z)$ is single valued.

If the geometric function is undefined at $z = 1$, is it still analytic?

答:
One of the "magic" things about complex analysis is that
$$
f:U\to\mathbb{C} \text{ is (complex) differentiable at } z_0\in U
$$
is equivalent to
$$
f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n \text{ in some neighborhood of } z_0
$$
(where $a_n=f^{(n)}(z_0)/n!$ as they have to be).

Now, $f(z)=\frac{1}{1-z}$ is differentiable on its domain $U=\mathbb{C}\backslash\{1\}$, so at every $z_0\in U$, $f$ can be written as a power series (which will have radius of convergence $R=|z_0-1|$, the distance to the nearest/only singularity).  Namely
$$
\frac{1}{1-z}=\frac{1}{1-z_0-(z-z_0)}=\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{1-z_0}}=\frac{1}{1-z_0}\sum_{n=0}^{\infty}\frac{1}{(1-z_0)^n}(z-z_0)^n.
$$


问:
Complex number $|z|$=1
Are there only finitely many complex numbers $z$ with $|z|=1$?

I think there are infinitely many such numbers.

答: Ummm... sure.  Here's a graph of them:

ContentLicense: CC BY-SA 4.0

Title: Complex number $|z|$=1

Body: Ummm... sure.  Here's a graph of them:


问:
Name for cuts which result in shapes that are rotatable images of one another
I'm looking for the technical term to describe any line that cuts through a shape such that the two pieces are a rotation of one another.

Obviously any line of reflective symmetry fulfils this property, but there are often other lines that have this property as well, e.g. 4th example from the left in the image below. Indeed for a square there are an infinite number of such lines.

答: Any such cut must pass through a twofold rotation axis of the object, so they may be simply called cuts through a twofold axis.


问:
Positive defnite matrices
Hei guys,

I would like to check with you a prove I have

Let the matrix $X_1 = \begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1\\
  \end{bmatrix}$ be positive definite. Based on this I would like to show that the matrix $X_2 = \begin{bmatrix}
    X_1 & B_2\\
    B_2 ^ T & A_1\\
  \end{bmatrix}$, where $B_2 = \begin{bmatrix}
    0\\
    B_1\\
  \end{bmatrix}$ is also positive definite.

The prove I have looks like this:

Based on the assumptions on $X_1$ and the Schur Lemma we get that $A_1, A_1 - B_1 ^ T \times A_1 ^ {-1} \times B_1$ and $A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T$ are positive definite.

In order for $X_2$ to be positive definite one has to show (based on the same lemma) that $X_1 - B_2 \times A_1 ^ {-1} \times B_2 ^ T$ is positive definite.
This is equivalent with showing that the matrix:

$\begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T\\
  \end{bmatrix}$ is positive definite.

For me this is true on if the matrix $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ is positive definite.

I would like to ask you if is possible $X_2$ to be positive definite with the $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ being positive definite?

Thanks,
Bogdan.

答:
I have to say I got lost somewhere on the way, but was thinking about the following alternative proof:

Assuming that $X_1$ is positive definite we would get that $\begin{bmatrix}
    x & y
  \end{bmatrix} \times \begin{bmatrix}
    A & B\\
    B^T & A\\
  \end{bmatrix} \times \begin{bmatrix}
    x ^ T\\ y ^ T
  \end{bmatrix} \geq 0$. This is equivalent with: $x A x^T + 2x B y^T + y A y^T \geq 0, \forall x, y$ (vectors with proper sizes).

For $X_2$ to be positive definite (based on the same definition) would imply that: $u X_1 u^T + 2u B_2 v^T + v A v^T \geq 0, \forall u, v$. Choosing $u = (x, y), v = z$ we get that $x A x^T + 2x B y^T + y A y^T + 2y B z^T + z A z^T \geq 0$.

I fail to see why the second inequality is true $\forall x, y, z$ assuming that the first one is true $\forall x, y$.

This makes me to think that $X_2$ in general is not positive definite, but rather for some specific B with some specific properties.

ContentLicense: CC BY-SA 3.0

Title: Positive defnite matrices


问:
finding the population in Statistics
I have the following question:

Netflix would like to estimate the proportion of their current subscribers who would pay extra for a premium membership including access to more movies and TV shows. To do this they plan to calculate a 95% confidence interval to estimate the proportion. They would like a margin of error of 0.05. How many subscribers must they sample to obtain this interval?

I know the answer is 385 but I keep getting an answer of 36.8 when I use this formula: n = ((z*/m)^2)p(1-p*) and it doesn't work when I use this formula either: n = ((z* * sigma)/m)^2.
Can anyone tell me where I'm going wrong?

答:
For a Wald-type confidence interval for a population proportion $p$, the confidence limits are $$\hat p - z_{\alpha/2}^* \sqrt{\frac{\hat p (1- \hat p)}{n}}, \quad  \hat p +  z_{\alpha/2}^* \sqrt{\frac{\hat p (1- \hat p)}{n}}$$ where $\hat p$ is the sample proportion observed, $n$ is the sample size, and $z_{\alpha/2}^*$ is the upper $\alpha/2$ quantile of the standard normal distribution for a $100(1-\alpha)\%$ confidence interval.  If a $95\%$ CI is desired, this corresponds to $\alpha = 0.05$, hence $$z_{\alpha/2}^* \approx 1.96.$$  The quantity $$ME = z_{\alpha/2}^* \sqrt{\frac{\hat p (1 - \hat p)}{n}}$$ is known as the margin of error.  Since we also desire $ME \le 0.05$, we solve for $n$ to obtain $$n \ge \hat p (1 - \hat p) \left(\frac{z_{\alpha/2}^*}{ME}\right)^2.$$  The only unknown quantity is $\hat p$; but since $\hat p(1 - \hat p) \le 1/4$ for all $\hat p \in [0,1]$ (see note below), we know that a sample size of at least $$n \ge \left(\frac{z_{\alpha/2}^*}{2ME}\right)^2$$ guarantees the Wald interval will have a margin of error less than the desired margin no matter the sample proportion observed.  Substituting the given values results in $$n \ge \left(\frac{1.96}{2(0.05)}\right)^2 = 384.16,$$ for which we must round up to guarantee the sample size is sufficient, hence $n = 385$.  But notice that this is a crude bound; if the sample proportion is extreme, then $n$ may be substantially smaller.  Also note that this is based on the Wald interval, which in extreme cases does not have good coverage properties.  But a discussion of this is beyond the scope of the question.

Note.

Why is $\hat p (1 - \hat p) \le 1/4$ on $\hat p \in [0,1]$?  This may be shown in many ways; an elementary one is to write $$\hat p (1 - \hat p) = \frac{1}{4} - \left( \frac{1}{4} - \hat p + \hat p^2 \right) = \frac{1}{4} - \left( \hat p - \frac{1}{2} \right)^2.$$  Since the square of a real number is never negative, the RHS is never greater than $1/4$.

Title: finding the population in Statistics


问:
Given the function g(x). Determine $g^n(x)$ in terms of $n$. Next, find the value of $g^n(x)$.
Help me.

Given that the function $g(x)=\frac{x+1}{x-1}$

Determine the function $g^n(x)$ such that $n$ is a positive integer. Next, find the value of $g^{37}(x)$.

答:
HINT

\begin{align*}
g(x) = \frac{x+1}{x-1} = \frac{(x-1) + 2}{x-1} = 1 + \frac{2}{x-1} & \Rightarrow g'(x) = -\frac{2}{(x-1)^{2}}\\\\
& \Rightarrow g''(x) = \frac{2\times 2}{(x-1)^{3}}\\\\
& \Rightarrow g'''(x) = -\frac{2\times 2\times 3}{(x-1)^{4}}\\\\
& \Rightarrow g^{(4)}(x) = \frac{2\times 2\times 3\times 4}{(x-1)^{5}}
\end{align*}

Thus it can be proven by induction that
\begin{align*}
g^{(n)}(x) = (-1)^{n}\times\frac{2\times n!}{(x-1)^{n+1}}
\end{align*}

I am sure you can take it from here.

EDIT

Based on the same expression as above, we have that
\begin{align*}
g^{2}(x) = 1 + \frac{2}{1 + \frac{2}{x-1} - 1} = x \Rightarrow g^{3}(x) = g(x) = 1 + \frac{2}{x-1} \Rightarrow g^{4}(x) = g^{2}(x) = x \Rightarrow \ldots
\end{align*}

Based on such pattern, we conclude that
\begin{align*}
g^{n}(x) =
\begin{cases}
\displaystyle 1 + \frac{2}{x - 1} &\text{when}\,\,n = 2k-1\\\\
x &\text{when}\,\,n = 2k
\end{cases}
\end{align*}
for $\mathbb{N}\ni k\geq 1$.


问:
Bourwer degree zero degree lemma
I was reading Milnor's differential topology book,in the Brouwer degree chapter.there is a famous lemma state as follows:

If $f:M\to N$ extend to smooth map $F:X\to N$ then $\text{deg}(f,y) = 0$ for every regular value(where $\partial X =M$,all manifold are oriented,M is compact,N is connected)

In the proof,we take the $F^{-1}(y)$ as regular value
submanifold,it's one-dimension compact manifold,hence exist two
cases arc and circle,in the book,case of arc is considered just
prove:$\text{sgn}(df_a)+\text{sgn}(df_b) = 0$ ,I don't know why
don't consider the case of circle.(since no intersection withh
$\partial X$) is my understanding correct?

the second question is the induced orientation for the arc,I don't understand why sgn of $df_{a/b}$ is induced by the map $F:X\to N$? I guess something to do with extension,but I can't make it clear

答:
For $1$): Indeed the boundary points of $X$, $\partial X= M$ only intersect with the simple curves

For $2$): the orientation of an arc $A$ is induced by both $X$ and $N$: let $x\in A$
and let $(v_1,...,v_{n+1})$ be a positively oriented basis for the tangent space of $X$, with $v_1$ tangent to the simple curve $A$. Then the vector $v_1$ determines the appropriate orientation for the tangent space of the simple curve $A$ if the differential map $dF_x$ send $(v_2,...,v_{n+1})$ into a positively oriented basis for $N$. Hence the sign of $df_a$ induced by the map $F$.

Title: Bourwer degree zero degree lemma


问:
Evaluating contour integral of complex conjugate
This is part of a homework assignment. Any hints will be useful, I haven't made any progress. I need to evaluate:
$$
\int_{\left\vert z - 1\right\vert = 1}
\overline{z}^{\,n}\,\mathrm{d}z\,,\quad n \in \mathbb{Z}
$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\,
\overline{z}^{\,n}\,\dd z
\,\right\vert_{\,n\ \in\ \mathbb{Z}}}:\ {\Large ?}}$.

Note that $\ds{1 =\pars{z - 1}\pars{\overline{z} - 1} =
\pars{z - 1}\overline{z} - z + 1 \implies \overline{z} = {z \over z - 1}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\,
\overline{z}^{\,n}\,\dd z
\,\right\vert_{\,n\ \in\ \mathbb{Z}}} =
\int_{\verts{z - 1}\ =\ 1}\,\,\,{z^{n} \over
\pars{z - 1}^{n}}\,\dd z
\\[5mm] = &\
\int_{\verts{z}\ =\ 1}\,\,
{\pars{1 + z}^{n} \over z^{n}}\,\dd z
\end{align}

$\ds{\large n < 0:}$ The integrand is singular at the "contour point" $\ds{z = -1}$.

$\ds{\large n = 0:}$ The integral vanishes out: There's not any pole inside the contour.

$\ds{\large n > 0:}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\,
\overline{z}^{\,n}\,\dd z
\,\right\vert_{\,n\ \in\ \mathbb{Z}_{\,\geq\ 1}}} =
\int_{\verts{z}\ =\ 1}\,\,\,{\pars{1 + z}^{n} \over z^{n}}
\,\dd z
\\[5mm] = &\
2\pi\ic{n \choose n - 1} = \bbx{2n\pi\ic} \\ &
\end{align}


问:
A two-dimensional set of measure zero
I have a 2D domain $[0,1]\times[0,1]$. This domain contains some set of measure zero $A$, the last understood as the Lebesgue measure in $\mathbb{R}^{2}$.

Is the following true: for almost all $t\in[0,1]$ the set $A_{t}:=\{x\in[0,1]|(t,x)\in A\}$ has measure zero, in standard $\mathbb{R}^{1}$ sense? Actually, I'm hoping to get a negative answer. This question can be more general, for $A$ being $\Omega\times[0,1]$ in $\mathbb{R}^{n}$ with $\Omega$ being a bounded domain in $\mathbb{R}^{n-1}$.
Thanks for your response!

答:
Your statement is true. This can, for instance, be seen from Fubini's theorem: Let $\chi_A$ denote the characteristic function of $A$. Then
$$
0 = \int_{[0,1]^2} \chi_A(t,x)dtdx = \int_0^1 \bigg( \int_0^1 \chi_A(t,x)dx \bigg) dt
$$
and hence for almost every $t\in[0,1]$, the function $\int_0^1 \chi_A(t,x)dx$ vanishes, which implies that $A_t$ as defined by you has measure zero.</p>

Title: A two-dimensional set of measure zero


问:
Precise meaning of $\ll_{n, \varepsilon}$ in number theory paper
Upon reading this paper of Dietmann, I came across the following line

$N_n(H;G) \ll_{n, \varepsilon} H^{n - 1 + \delta_G + \varepsilon}$

that appears in the statement of Theorem $1$. What precisely does the symbol $\ll_{n, \varepsilon}$ mean in this context?

Dietmann does not explain what this notation means, and I have never seen this notation before.
The left hand side of this "inequality" does not depend on $\varepsilon$, contrary to this question, but from reading the answer there my guess is

For all $\varepsilon > 0,$ there exist constants $M, K > 0$ such that for all $n > M$, we have that $N_n(H;G) \leq K H^{n - 1 + \delta_G + \varepsilon}$.

After reading this blog post of Terence Tao and looking at his statement of the ABC conjecture (which uses the notation $\ll_\varepsilon$), and looking at the corresponding Wikipedia page, which expresses the ABC conjecture in terms of quantifiers, I think that $N_n(H;G) \ll_{n, \varepsilon} H^{n - 1 + \delta_G + \varepsilon}$ could also mean

For all integers $n \geq 1$, $\varepsilon > 0$, there exists a constant $K$ such that $N_n(H;G) \leq K H^{n - 1 + \delta_G + \varepsilon}.$

答: $X \ll_{n,\epsilon} Y$ typically means that there's a "constant" $C$ which depends on the parameters $n$ and $\epsilon$ such that $X \leq C \cdot Y$. This is meaningful when you consider $X$ and $Y$ as functions of some other variable besides $n$ and $\epsilon$, and treat $n$ and $\epsilon$ as parameters.


问:
Show that f a holomorphic map, is either infinite or a rational function
Let $f :C \cup {\infty} \to C \cup {\infty}$ be a holomorphic map from the Riemann sphere to itself. Show that f
is either identically $\infty$, or is a rational function in the sense that there exist polynomials $P(z), Q(z)$ of one complex variable, with Q not identically zero, such that $f(z) = P(z) / Q(z) \forall {z \in {\bf C}} $ with $Q(z) \neq 0$.

My first question is how do I show f has finitely many poles.

答: We denote the Riemann sphere as $\widehat{\mathbb{C}}$. If $f:\widehat{\mathbb{C}}\rightarrow\mathbb{C}$ is holomorphic then $f$ is constant, indeed $f$ being continuous at $\infty$, there exists a compact $K\subset\mathbb{C}$ such that $f$ is bounded on $\widehat{\mathbb{C}}\setminus K$. Since $f$ is also bounded on $K$, $f$ is bounded on $\widehat{\mathbb{C}}$ and thus is constant. Now let $f:\widehat{\mathbb{C}}\rightarrow\widehat{\mathbb{C}}$ holomorphic which is not a constant function, then $f$ has a finite number of zeroes and singularities, $\widehat{\mathbb{C}}$ being compact. Let $F\in\mathbb{C}(X)$ such that $F$ has the same zeroes and singularities (counted with multiplicity) than $f$, then $f/F:\widehat{\mathbb{C}}\rightarrow\mathbb{C}$ is holomorphic and thus is constant so that $f$ is a rational function.


问:
What is the transitive closure of the following digraph?
What is the transitive closure of the following digraph ? To find reach-ability matrix and adjacency matrix.

My approach:

The adjacency matrix is \begin{bmatrix} 0 &1 &0 &0 \\ 0 & 0&0 &1 \\ 0 & 0 &0 &0 \\1 & 0 & 1 & 0 \end{bmatrix}.

Using Warshall's  algorithm, the reach-ability  matrix  for  transitive  closure  is  \begin{bmatrix} 1 & 1 & 1 &1 \\ 1 & 1 & 1& 1\\ 0 & 0 & 0 & 0\\1 & 1 & 1 & 1 \end{bmatrix}.

So the transitive closure is the graph:

Am I right  ?

Please help

答:
Not quite.  Here's the transitive closure graph:

Recall that the transitive closure of graph $g$ has the same vertices as $g$ but with (directed) edges between vertices $u$ and $v$ if there is a path—any path—between $u$ and $v$ in $g$.  Just examine your graph and you can construct the transitive closure "by hand."  Notice, for instance, there is no path from $c$ to $b$ in $g$... and hence no edge between those vertices in the transitive closure graph.  But there is a (long) path from $a$ to $c$ in $g$, and hence there is an edge from $a$ to $c$ in the transitive closure.

You can check all the others...

A note to the specialist:  Transitive closures are most properly defined on directed acyclic graphs (DAGs).  If one admits cycles ("loops") then there will be some vertices with paths to themselves, and thus a transitive closure should include a "self loop."


问:
Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$
Evaluate the following integral :

$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$

I was tried use change variable ,

If I use $x=y^2$ integral becomes :

$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$

From here I have one idea  the derivative under sing integral but I got I difficult integration :

$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$

I already to see you hints or solution!

Body:
Note
$$1+x^8=(1+e^{i \frac\pi4}x^2)(1+e^{-i \frac\pi4}x^2)(1+e^{i \frac{3\pi}4}x^2)(1+e^{-i \frac{3\pi}4}x^2)$$

$$J(a)=\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12})
$$
Then
\begin{align}
I&=\int\limits_0^{\infty}\frac{\ln (1+x^{4})}{\sqrt{x}(1+x)}dx\overset{x\to x^2}
=2\int\limits_0^{\infty}\frac{\ln (1+x^{8})}{1+x^{2}}dx\\
&= 2\left[J(e^{i \frac\pi4})+ J(e^{-i \frac\pi4})+J(e^{i \frac{3\pi}4}) +J(e^{-i \frac{3\pi}4})
\right]\\
&=2\pi \ln\left[(1+e^{i \frac\pi8})(1+e^{-i \frac\pi8})(1+e^{i \frac{3\pi}8})(1+e^{-i \frac{3\pi}8})\right]\\
&=2\pi \ln\left[4(1+\cos\frac\pi8)(1+\cos\frac{3\pi}8) \right]\\
&=4\pi\ln (\sqrt2+\sqrt{2+\sqrt2})
\end{align}


问:
Solve the roots of a cubic polynomial?
I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:

"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

答:
It means the roots are of the form $a, ar, ar^2$.

Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients
$$
14 = 2 + 4 + 8,
\qquad
56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.
$$

Title: Solve the roots of a cubic polynomial?


问:
Derivation of 2D Korteweg-de-Vries equation
Coming from engineering rather than mathematics, I am recently dealing with non-linear partial differential equations e.g. like the well known Korteweg-de-Vries equation:

$$u_{t} + uu_x + u_{xxx} = 0$$

In the literature I found the two- and three-dimensional versions of the KdV-equation:

$$(u_{t} + uu_x + u_{xxx})_x + u_{yy} = 0 $$
$$(u_{t} + uu_x + u_{xxx})_x + u_{yy} + u_{zz} = 0$$

What bugs me now is how one derives the two- respectively three-dimensional versions from the original version? In all of the literature I have found so far, the multi-dimensional version seems to fall from heaven, nowhere I could find a complete derivation or any intuition behind it. This is unfortunately not the first time I can not grasp how one author jumps from the 1-D to a n-D version.

So generally asked: How does one derive a multi-dimensional version of a (non-linear)PDE given a 1 + time-dimensional form like the first equation? It would be more clear to me if the original version would be written in vector notation (with $\nabla$, $\cdot$, $\times$, $\Delta$, etc. operators). Would be great if someone with a more profound mathematical background could educate me.

Thanks a lot!

Edit:

The 2D version is also called Kadomtsev-Petviashvili equation, if that rings a bell for anyone.

答:
The KP equation was derived from physical principles, so you should be able to find this derivation somewhere. Worst case scenario, you may have to refer to the original paper of Kadomtsev and Petviashvili, who derived it for plasma waves; however, I believe that paper is in Russian. And it's probably not helpful if you aren't familiar with plasma physics. Later, Ablowitz and Segur derived it for surface and internal water waves, and Pelinovsky, Stepanyants and Kivshar derived it in nonlinear optics, so you could look for references in those directions. It is better to think of the KP equation as physical equation in its own right, rather than just as a purely mathematical generalization of the KdV equation. After all, KdV can be generalized in any number of ways depending on which of its properties you intend to preserve.

Typically one does not just generalize a PDE to higher dimensions without first thinking about what feature of the PDE you are trying to generalize. For example, there is another generalization of KdV to two spatial dimensions called the Novikov-Veselov equation, which is related to but not the same as the KP equation. I understand the Novikov-Veselov equation to have originally arisen more out of some mathematical investigations on inverse scattering theory rather than from physical principles. Thus I would also expect the 3D case to arise either from physical applications or from attempting to generalize some aspect of the inverse scattering theory for the KdV or KP equations, as this is likely the area of research on these equations that has received the most attention. But I cannot say for sure, not being an expert on integrable systems.

See here for a brief history of the KP equation, where I found the references for the physical derivations.


问:
How to solve a numerical reasoning grid pattern with numbers and letters.
I'm having trouble solving a question in which there is a 6x6 grid pattern. One of the rows have been erased and I need to find what number or letter is in the grid square with the question mark.

2 A 4 A 6 A

C B 4 B 6 B

4 4 4 C 6 C

```<code>  ?
</code>```

6 6 6 6 6 E

K J I H G F

The answer is e, yet I cannot find why that would be the answer. Please help.

The question

答: Not sure you still need the answer, but I've found the pattern. It's a reverse $L$, starting from the top right corner, going down, then to the left. If the column $n$ starts with a number, you repeat that number in the entire pattern: go down $n-1$ rows, then go left $n-1$ columns. Similar pattern fro the letters, but you fill with $A,B,C,D, ...$. To check the result, the erased row would be $GFED6D$. Then the question mark is letter $E$, option $D$ in the question.


问:
Convert a polygone in space, to a plane
I have a surface defined by a list of 3D points (X,Y,Z). (so all vectors are coplanars)
I would like to convert that on a plane (X,Y).
I am looking for a formula to convert that polygon using vectors, but until now only found a solution offering this :

Calculate length of each segment.

Calculate angles between each of them

Then I can find the X,Y coordinates with that.

But I think there may exist a vectorial solution for that?

Thanks in advance

答: Find two vectors parallel to the plane by taking differences of points in the plane, such as $\vec v_1=(x_1,y_1,z_1)-(x_0,y_0,z_0),\;\vec v_2=(x_2,y_2,z_2)-(x_0,y_0,z_0)$, then use the Gram-Schmidt process on $\vec v_1,\vec v_2$ to get an orthonormal basis $\hat u_1,\hat u_2$ for the plane, and finally take dot products to get points $(x',y')$ in 2D: $x'=(x,y,z)\cdot\hat u_1,\;y'=(x,y,z)\cdot\hat u_2$.

Title: Convert a polygone in space, to a plane


问:
If $z\in\mathbb C$ with $|z|\leqslant\frac{4}{5}$, then $\sum_{n\in S}z^n\neq-\frac{20}{9}$
Let $z$ be a complex number with $|z|\le\tfrac{4}{5}$. If $S\subset\mathbb N^+$ is a finite set, then I'd like to show  that
$$\sum_{n\in S}z^n\neq-\frac{20}{9}\,.$$
This problem is from an exam in Bejing as of today, and I find this is a very nice problem.

I have no idea how to access it, maybe use Newton's identity to  solve it.

Could someone help me to solve it?

答:
Some generic observations.

$$\min S \leq 3.$$
If not, then
$$\left|\sum_nz^n\right|\leq\sum_n|z|^n\leq r^4+r^5+\dots r^{n+3} = \dfrac{r^4}{1-r}(1-r^{n}) < 5\cdot\dfrac{4^4}{5^4} = \dfrac{256}{125} < \dfrac{20}{9}$$
, a contradiction.

On the other hand, let $|S| = n.$ Then,

$$\dfrac{20}{9} \leq r+r^2+\dots +r^n\implies n\geq 4.$$
So, the set is $S$ has at least $4$ elements, the least of which is no larger than $3.$


问:
Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$
Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?

Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$

Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$

Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$

We have removed one square root.

Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?

Simplify: $2x-5 = 2\sqrt{(x-1)} + x$

Simplify more: $x-5 = 2\sqrt{(x-1)}$

Now do the "square root" thing again:

Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$

Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$

Square root removed

Thank you in advance for your help

答:
$$2x-5 = (1 + \sqrt{x-1})^2$$

to expand RHS use this formula or simple mulipty it with itself(to do square).
formula is:

$(a+b)^2=a^2+b^2+2\times a\times b$

so your expansion will be
$$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$
$$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$
$$2x-5 = x+2\sqrt{x-1}$$
$$x-5 = 2\sqrt{x-1}$$
now you have your way.

OwnerDisplayName: Kalina Kama

Title: Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$


问:
Let $\Sigma(k)$ be the sum of all numbers whose largest proper divisor is $k$. Does an odd $k$ for which $\Sigma(k)$ is divisible by 6020 exist?
I have noticed that since the numbers whose largest proper divisor is $k$ are bounded by $k$ and $k^{2}$,the number of summands is smaller than or equal to $k^{2}-k$. $\Sigma(k)$ has to have factors of $2,2,5,301$ but I don't know how to proceed with that.

Any help would be appreciated.

答:
Update: The answer is yes. See the results at the end of the answer.

Original answer:
I don't know the answer but here's some thoughts.

Suppose the smallest prime divisor of $k$ is $p_m$, the $m^{th}$ prime number. Note that $k$ is the largest proper divisor of a number $N$ if and only if $N=pk$ where $p$ is the smallest prime divisor of $N$. Therefore
$$\Sigma (k) = (2+p_2+\cdots + p_m)k$$
where $p_j$ is the $j^{th}$ prime number.

Denote $S(n):=\sum_{j=1}^n p_j$. Note that $6020=2^2 \times 5 \times 7 \times 43$.
If $k$ is odd and $6020 | \Sigma(k)$, then $4|S(m), m$ is odd.

$5|k \implies m=3, \Sigma(k)=S(3)k=10k$, but $4\nmid 10k, \Rightarrow \Leftarrow.$

Therefore $5\nmid k, 20 | S(m)$. The first such $S(m)$ is $S(23)=100$, so $7\nmid k$, $140|S(m)$.

Then we can check there is no such $S(m)$ that is a multiple of $140$ if $m\le 50$, therefore $43\nmid k$, we must have $6020|S(m)$.

I have checked the first 1000 $S(m)$'s and found no multiples of $6020$. I found this place but it doesn't help.

Maybe you can write a program to check many more primes, and ask in MathOverflow if you still can't find a $S(m)$ divisible by $6020$.

Update

Using some prime number function in R I checked all prime numbers under one million, and did find some solution. The smallest  $k$ that has the property you wanted is 197893, the $17811^{st}$ prime number.

The prime number function I used is:

```<code>is.prime <- function(n) n == 2L || all(n %% 2L:max(2,floor(sqrt(n))) != 0)
</code>```

However it returns TRUE for $n=1$ so I have to exclude it from the list. Here's my program and the results:

```<code>N=10^6;
M=6020;
temp=rep(NA,N)

for(i in 1:N) temp[i]=as.integer(is.prime(i));

temp[1]=0; # exclude 1
pn=which(temp==1); # prime number list
pnr=pn %% M; # avoid overflow
S = cumsum(pnr); # cumulative sum

which(S %% M ==0);
</code>```

[1] 17811 20121 25159 41437 55023 60667 67033 68509 69863 76257

```<code>pn[1:25] # sanity check
</code>```

[1]  2  3  5  7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

```<code>pn[17811] #  The first <span class="math-container">$k$</span> is 197893, the <span class="math-container">$17811^{st}$</span> prime number.
</code>```

[1] 197893

```<code>sum(pn[1:17811]) # sanity check.
</code>```

[1] 1675185400

```<code>1675185400/6020 # sanity check
</code>```

[1] 278270

Body:
Update: The answer is yes. See the results at the end of the answer.

Original answer:
I don't know the answer but here's some thoughts.

Suppose the smallest prime divisor of $k$ is $p_m$, the $m^{th}$ prime number. Note that $k$ is the largest proper divisor of a number $N$ if and only if $N=pk$ where $p$ is the smallest prime divisor of $N$. Therefore
$$\Sigma (k) = (2+p_2+\cdots + p_m)k$$
where $p_j$ is the $j^{th}$ prime number.

Denote $S(n):=\sum_{j=1}^n p_j$. Note that $6020=2^2 \times 5 \times 7 \times 43$.
If $k$ is odd and $6020 | \Sigma(k)$, then $4|S(m), m$ is odd.

$5|k \implies m=3, \Sigma(k)=S(3)k=10k$, but $4\nmid 10k, \Rightarrow \Leftarrow.$

Therefore $5\nmid k, 20 | S(m)$. The first such $S(m)$ is $S(23)=100$, so $7\nmid k$, $140|S(m)$.

Then we can check there is no such $S(m)$ that is a multiple of $140$ if $m\le 50$, therefore $43\nmid k$, we must have $6020|S(m)$.

I have checked the first 1000 $S(m)$'s and found no multiples of $6020$. I found this place but it doesn't help.

Maybe you can write a program to check many more primes, and ask in MathOverflow if you still can't find a $S(m)$ divisible by $6020$.

Update

Using some prime number function in R I checked all prime numbers under one million, and did find some solution. The smallest  $k$ that has the property you wanted is 197893, the $17811^{st}$ prime number.

The prime number function I used is:

is.prime <- function(n) n == 2L || all(n %% 2L:max(2,floor(sqrt(n))) != 0)

However it returns TRUE for $n=1$ so I have to exclude it from the list. Here's my program and the results:

N=10^6;
M=6020;
temp=rep(NA,N)

for(i in 1:N) temp[i]=as.integer(is.prime(i));

temp[1]=0; # exclude 1
pn=which(temp==1); # prime number list
pnr=pn %% M; # avoid overflow
S = cumsum(pnr); # cumulative sum

which(S %% M ==0);

[1] 17811 20121 25159 41437 55023 60667 67033 68509 69863 76257

pn[1:25] # sanity check

[1]  2  3  5  7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

pn[17811] #  The first $k$ is 197893, the $17811^{st}$ prime number.

[1] 197893

sum(pn[1:17811]) # sanity check.

[1] 1675185400

1675185400/6020 # sanity check

[1] 278270


问:
Looking for a function that approximates a parabola
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?

答:
Obligatory "not an answer but I wanted to share the image".

How about a transformation of the $\text{sinc}$ function?


问:
Question abut poisson distribution in university stats
Given that new fabric getting of a loom has an average of 1.3 tear every 10 feet. What is the probability that:

a) There is more than 10 feet between the next two tears? I have absolutely no idea how to solve this. Do i reformulate? Do i use exponential function?

b)There is no tear in the 5 feet getting out of the loom? for this one i got 52.20% because i did (=POISSON.DIST(0,0.65,FALSE)) on excel

I’d appreciate any help understanding how I can properly answer this question!

答:
Poisson distribution:  $p_k=e^{-\lambda}\frac{\lambda^k}{k!}$, where $p_k$ is the probability of exactly $k$ events during the interval in question.

For a) $\lambda=1.3$ and you want $p_0=e^{-1.3}=0.2725$.

For b) $\lambda=0.65$ and you want $p_0=e^{-0.65}$, which is your answer.


答:
Let the roots be $a,a\cdot r,a\cdot r^2$

Using Vieta's formula  $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$

and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$

On division, $ar=4$ as $a\cdot r\ne0$

$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{  or }\frac12$$

Alternatively,
 using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$

So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$

Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation

$\implies a\cdot r=4\iff a=\frac4r$

Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$

Title: Solve the roots of a cubic polynomial?

Body:
I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:

"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.

Any help would be appreciated.


问:
Show that $w_3=\frac{-1+i\sqrt3}2$ without $\cos$ and $\sin$
I struggle a lot to show that $w_3 = e^{i2\pi/3}$ is equal to $\frac{-1+i\sqrt{3}}{2}$  without using cos and sin, knowing only that $f :(\mathbb R, +) \to (\mathbb C(1), \times)$ that from an angle $\theta$ gives the complex number $e^{i\theta}$ is a morphism so $\operatorname{ker} f = 2\pi \mathbb Z$, also knowing $\cos(\theta) = \Re(e^{i\theta})$, $\sin(\theta) = \Im(e^{i\theta})$, knowing $e^{i\pi}=-1$ all summation and duplication rules of $\cos$ and $\sin$. The indication tells to verify that $w_3$ is a solution of $X^{2} + X + 1 = 0$ then compute the complex solutions with $\Delta$ and find that since $w_3$ is a solution so it must be one of them, without knowing what $\cos(\pi/3)$ is. Tried all possible methods but still need to find that value of $\cos(\pi/3)$ but the whole goal of the exercice is to do it without knowing it. I also wrote $f(\pi) = f(2\pi/3 + \pi/3) = f(2\pi/3)\cdot f(\pi/3)$ but leads me nowhere... I would like your help to show that $w_3^2 + w_3 + 1 = 0$.

答: When viewed as vectors in the complex plane, $1,w_3,w_3^2$ are equally spaced around the origin and are of the same length. Thus their sum must be zero and we have $w_3$ as a solution to $x^2+x+1=0$.


问:
principal value of improper integrals
How can I find the principal value of the following? </p>

<p>$$PV\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+x+1)} dx$$ </p>

<p>I'm able to evaluate integrals which have trig identities in them or just polynomials but I can't quite get my head around how to do them with exponentials. Any help would be wonderful please! </p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{p \equiv {-1 + \root{3}\ic \over 2}}$:
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{-\infty}^{\infty}
{\expo{\ic x} \over x\pars{x^{2} + x + 1}}\,\dd x}
\\[5mm] = &\
\int_{-\infty}^{\infty}
{\expo{\ic x} \over x^{2} + x + 1}
\bracks{{1 \over x + \ic 0^{+}} + \ic\pi\delta\pars{x}}\,\dd x
\\[5mm] = &\
\int_{-\infty}^{\infty}
{\expo{\ic x} \over
\pars{x + \ic 0^{+}}\pars{x - p}\pars{x - \overline{p}}}\,\dd x +
\ic\pi
\\[5mm] = &\
\ic\pi +
2\pi\ic\,{\expo{\ic p} \over p\pars{p - \overline{p}}} =
\ic\pi +
2\pi\ic\,{\expo{-\ic/2 - \root{3}/2} \over
\bracks{\pars{-1 + \root{3}\ic}/2}\pars{\ic\root{3}}}
\\[5mm] = &\
\ic\pi +
2\pi\,{\root{3} \over 3}
\expo{-\ic/2 - \root{3}/2}\,\,\,{-1 - \root{3}\ic \over 2}
\\[5mm] = &\
\bbx{\pars{\ic - {\root{3} + 3\ic \over 3}\,\expo{-\ic/2 - \root{3}/2}}\pi} \approx -1.3030 + 2.3477\,\ic \\ &
\end{align}

Title: principal value of improper integrals


问:
Discontinuous function by epsilon delta
How do I prove that the function $ \frac{x ^ 2 y} {| x ^ 3 | + y ^ 2} $ is not continuous on $ \{0,0\} $ using epsilon-delta?

I have tried it using sequences that have $ \{0,0 \} $, but I would like an idea to test it by epsilon-delta, since it seems that the solution is complicated by that means

答:
$ \frac{x ^ 2 y} {| x ^ 3 | + y ^ 2} $  is what you typed in.

Note $$  0 \leq \left( |y| -|x|^{\frac{3}{2}}  \right)^2 = y^2 - 2 |y| |x|^{\frac{3}{2}}  + |x|^3  $$

so
$$    2 |y| |x|^{\frac{3}{2}} \leq y^2 + |x|^3 $$
and
$$ \frac{ 2 |y| |x|^{\frac{3}{2}}  }{ |x|^3 + y^2 }   \leq 1   $$
as long as $x,y$ are not both zero.
Thus
$$ \frac{ 2 |y| x^2  }{ |x|^3 + y^2 }   \leq \sqrt{|x|}   $$
This means that assigning the value $0$ as the value of the function at $(0,0)$ creates a continuous function.

Title: Discontinuous function by epsilon delta


问:
$\mathbb{E}$[$X_{\mathbb{E}(T)}]$
Consider an experiment consisting of a sequence of trials, each of which yields a success or a failure. Let $X_{n}$ be the number of successes after n trials, and let $T$ be the expected number of trials for the first success to occur.

Then I propose the following conjecture:

$\mathbb{E}$[$X_{\mathbb{E}(T)}$] $\leq$ $1$

We have equality when the probability of success is fixed (does not change from trial to trial). If we let this probability be $p$, then $\mathbb{E}(T) = \frac{1}{p}$, and so $\mathbb{E}$[$X_{\mathbb{E}(T)}] = \mathbb{E}$[$X_{\frac{1}{p}}] = p.\frac{1}{p} = 1$

We can also achieve $\mathbb{E}$[$X_{\mathbb{E}(T)}$] < $1$, as in the following experiment: Let a trial consist of throwing a fair six-sided die, and let a success be obtaining an outcome that we have obtained before (a duplicate). Then:

$\mathbb{E}(T) = 1.0 + 2.\frac{1}{6} + 3.\frac{5}{6}.\frac{2}{6} + 4.\frac{5}{6}.\frac{4}{6}.\frac{3}{6} + 5.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{4}{6} + 6.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{5}{6} + 7.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{1}{6}.1 \approx 3.77 $

However $\mathbb{E}[X_{3.77}] \leq \mathbb{E}[X_{4}] = 1.{4 \choose 2}.\frac{6.5.4}{6^4} + 2[{4 \choose 3}.\frac{6.5}{6^4} + \frac{1}{2}.{4 \choose 2}.\frac{6.5}{6^4}] + 3.\frac{6}{6^4} \approx 0.89 < 1$

Can anyone provide some insight into what is going on here? Is the conjecture true?

Thanks

答:
Without more assumptions on the trials, this isn't true.  Consider the following example:

Let $E_k$ be the event that trial #$k$ is a success.  Let the joint distribution be such that $E_{1}, E_{4}, E_{9}, E_{16}, \dots$ are all independent with probability $1/2$, and $E_2, E_3$ are successes if and only if $E_1$ is a success, $E_5, E_6, E_7, E_8$ are successes if and only if $E_4$ is a success, etc.  Then we note that the time $T$ of first success takes the value $n^2$ with probability $2^{-n}$, so
$$
E[T] = \sum_{n \geq 1} n^2 2^{-n} = 6.
$$
But we can also compute $E[X_6] = 0\cdot \frac14 + 3 \cdot \frac14 + 3 \cdot \frac14 + 6 \cdot \frac14 > 1$.

Title: $\mathbb{E}$[$X_{\mathbb{E}(T)}]$


问:
I need help modeling the velocity profile of a river's current
I am trying to solve Zermelo's Navigation Problem.

One of the cases I'm looking at is when the river's current is a function of the $x$-position only.

From what I learned in Fluid Mechanics courses, I know that at the two ends when (i.e. the river banks) the velocity should be zero. Then in the center the velocity is at its maximum value.

In other words: $v(x=0) = v(x=L) = 0$, and $v(x=0.5L) = V_{\text{max}}$

Everything I learned in the past was these velocities as function of radius, which makes sense for pipes and tubes, but since this can be thought of a $2D$ rectangular flow, I can't figure this out.

I know it should be a quadratic expression.

Any help is appreciated. Thanks

Here is a sketch of the function I am trying to model:
Plot of River's Current vs x-Position

答:
Any polynomial has factors corresponding to its roots. You assumed that the function is quadratic, so we should have

$$v=k(x−0)(x−L)=-kx(L-x)$$

for some constant $k$. By the nature of the problem, $0\leq x\leq L$, which implies $L-x\geq0$ and thus $x(L-x)\geq0$. This quadratic is maximized at $x=L/2$ (you can use calculus or just complete the square to prove this), with value $(L/2)^2$. So $v$ is maximized or minimized (depending on the sign of $k$) with value $v_m=-k(L/2)^2$. Now we have $k=-4v_m/L^2$, and

$$v=\frac{4v_m}{L^2}x(L-x).$$


答:
Here is a family of real analytic functions that does the job; making the parameter $N=1,2,3,4,\dots$ larger makes the approximation better. Lets just say the quadratic is $q(x) = 1-x^2$, so the function is $f(x)=\max(0,q(x))$. Its easy to do a more general case. Then set
$$f_N(x) = \frac1{1+x^{2N}}q(x).$$
This works because for $N$ very big, when $|x|<1$, $x^{2N}\approx 0$, so $\frac1{1+x^{2N}}\approx 1$. But the moment $|x|>1$, $x^{2N}$ becomes huge, so then $\frac1{1+x^{2N}} \approx 0$. It's not exactly 0 outside $|x|<1$, but instead decays like $O(|x|^{-2N+2})$ (again, crank up $N$ to get a better approximation)

This is what it looks like for $N=10$:

The general case: the quadratic with height $h$ above $y=c$, base $d$ centered around $x_0$ is
$$ q_{h,c,d,x_0}(x)=h q\left(\frac{x-x_0}{d/2}\right)+c.$$
The  quadratic truncated at $y=c$ is then
$$ f_{h,c,d,x_0}(x) = h f\left(\frac{x-x_0}{d/2}\right)+c,$$
and the corresponding approximation I am proposing is
$$ f_{N,h,c,d,x_0}(x) = h f_N\left(\frac{x-x_0}{d/2}\right)+c.$$

You can adjust these parameters in this interactive Desmos graph.

A further variation is using $1/(1-x^{2N})$ instead to define $f_N$; this makes $f_N$ everywhere positive (because it changes sign exactly where $q$ does, and $f_N$ is differentiable at the truncation points by l'Hopital), which could be useful. But the approximation is worse near the boundary.

Here's a graph of this last variant-

The first variant is exactly 1 at $x=0$, and exactly 0 at $x=\pm 1$. It has the same sign as the original polynomial $1-x^2$ everywhere.

The second is exact only at $x=0$, (but even if $N$ is small), and is everywhere positive. (In particular, its a slight mischaracterisation to say that "unlike Calvin Khor's suggestions, this avoids ever going negative" :) but I didn't include a graph of this variant earlier, and their solution is great too!)

Title: Looking for a function that approximates a parabola

Body:
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?


问:
Interchanging Iterated Limits for a compactly convergent sequence of functions
Let $\mathcal{A}\left(\mathbb{D}\right)$ denote the space of all holomorphic functions $f:\mathbb{D}\rightarrow\mathbb{C}$, where $\mathbb{D}$ is the open unit disk in $\mathbb{C}$, let $\omega$ be a positive real number, let $\left\{ f_{n}\right\} _{n\geq0}$ be a sequence in $\mathcal{A}\left(\mathbb{D}\right)$ converging uniformly on every compact subset of $\mathbb{D}$ to a limit $f\in\mathcal{A}\left(\mathbb{D}\right)$, and suppose that:$$\lim_{x\uparrow1}\left(1-x\right)^{\omega}f_{n}\left(x\right)=1,\textrm{ }\forall n\geq0$$

Is it then true that:$$\lim_{x\uparrow1}\left(1-x\right)^{\omega}f\left(x\right)=1$$ or, do I need more assumptions on the convergence of the $f_{n}$s (if so, what)? Also, note here that there is no guarantee that $f$ is holomorphic, or even finite-valued at $1$.

答: There is a sequence of polynomials $g_n$ that converge uniformly on compact subsets of $\mathbb D$ to the principal branch $g$ of $1/(1-x)^\omega$.  Take $f_n = g + g_n$ and $f = 2 g$; you have $\lim_{x \uparrow 1} (1-x)^\omega f_n(x) = 1$ but $\lim_{x \uparrow 1} (1-x)^\omega f(x) = 2$.


问:
Logic of infinite statements

Background: I have learned a little about mathematical logic and set theory from only Tao's Analysis I.

I am wondering the following question. Suppose we have a property $P$ pertaining to a set. And we know that $P(X) \wedge P(Y) \to P(X \cap Y)$. If we have infinitely many sets for which $P$ is always true, then can we say that for the intersection of all these sets $P$ is still true?

I think it is, but I don't know from what axioms can we conclude it. I think that if we can build a injection from a subset of $\mathbb{N}$ to the set containing those infinitely many sets, then we can use the mathematical induction axiom to prove this. However, we might not be able to build the injection. I don't know if my question is somehow related to mathematical logic. Do we conventionally admit that such a conclusion is true?

答:
In general it is not true. For instance, the intersection of two open sets in a topological space is always open, but the intersection of infinitely many open sets need not be. An example in $\Bbb R$ is the family of open intervals $\left(-\frac1n,\frac1n\right)$ for $n\in\Bbb Z^+$: any intersection of finitely many is in fact one of these intervals, but the intersection of all of them is $\{0\}$, which is not open.

An example that needs no topology is obtained by letting $A_n=\Bbb N\setminus\{n\}$ for each $n\in\Bbb N$ and taking $P(A)$ to be the property that $\Bbb N\setminus A$ is finite. Clearly the intersection of any finite number of the sets $A_n$ has $P$, but the intersection of all of them is $\varnothing$, which does not.

What one can conclude is that for any finite family $\mathscr{F}$ of sets with property $P$, $\bigcap\mathscr{F}$ will have property $P$: this is a simple proof by induction on $|\mathscr{F}|$.

Title: Logic of infinite statements


问:
Suppose that $g: A \rightarrow B$, $h: B \rightarrow C$. Prove that if $g$ is onto and $h$ is not one-to-one, then $h \circ g$ is not one to one.
How can I start to approach this question? I know that if $g$ and $h$ are one-to-one, then $h \circ g$ is one-to-one, and also if $h$ and $g$ are both onto, and $h \circ g$ is onto. But this question didn't fit into both cases...

答:
Since $h$ is not one to one, there is $b_1, b_2 \in B$ distinct such that $h(b_1) = h(b_2)$. Since $g$ is onto, there is $a_1, a_2 \in A$ such that $g(a_1) = b_1$ and $g(a_2) = b_2$.

Then notice that $h(g(a_1)) = h(b_1) = h(b_2) = h(g(a_2))$. So then $h\circ g$ is not one to one.


答: I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.</p>

Title: Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$

Body:
Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?

Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$

Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$

Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$

We have removed one square root.

Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?

Simplify: $2x-5 = 2\sqrt{(x-1)} + x$

Simplify more: $x-5 = 2\sqrt{(x-1)}$

Now do the "square root" thing again:

Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$

Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$

Square root removed

Thank you in advance for your help


问:
Applying differential forms to non-vectors $\omega(X,Y)$
$\omega=2xdx\wedge dy + y^2dx\wedge dz$

$X=x^2y\frac{\partial}{\partial y} + x\frac{\partial}{\partial z}$

$Y=x\frac{\partial}{\partial y}$

Calculate $\omega(X,Y)$

I understand how to apply the differential form onto parameters given vectors $(v1,v2)$, where you take  the determinant. Would I perform the same for this operation?

I am trying this, but I'm not sure where to go from here. Should I be taking the partials as each parameter?

$2xdet\begin{bmatrix}
0&0\\ x^2y&x\\ \end{bmatrix} + y^2det\begin{bmatrix}
0&0\\ x&0\\ \end{bmatrix} = 0$

I have looked through Do Carmo Differential Forms, and Faris' Vector fields and differential forms, but have not been able to find an explanation or example.

答:
We wish to compute: $$\omega(X,Y) = (2xdx\wedge dy+y^2dx\wedge dz)(x^2y\partial_y+x\partial_z,x\partial_y).$$
For the first term,
$$2x[dx(x^2y\partial_y + x\partial_z)dy(x\partial_y)-dy(x^2y\partial_y + x\partial_z)dx(x\partial_y)] = 2x(0-0)=0.$$

So the final answer is just the second term, which is
$$y^2[dx(x^2y\partial_y + x\partial_z)dz(x\partial_y)- dz(x^2y\partial_y + x\partial_z)dx(x\partial_y)] = y^2(0-x\cdot0) = 0.$$
So the whole thing is zero (unless I've made a mistake, in which case please comment). I hope this makes the method clear - with practice, you should be able to do these computations fast (skipping the LHS and just writing down the middle and final terms).


问:
Writing Fourier transform integral in terms of real-valued functions
I am interested in the following integral.

$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$

One way (for example, by using Mathematica) to solve it is in the following way.

$$ f(x,t) = \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}} $$

I believe that $f(x,t)$ should be real valued, but it is not clear to me how to simplify given expression so that it consists only of real valued functions.

答:
The exponential term can be recast as $2cos(kx)cos(pt+qk^2t)$

a)$e^{ikx}=cos(kx)+isin(kx)$.  The rest of the integrand is a function of $k^2$, so integrating over the real line eliminates the $sin(kx)$ term.

b) $e^{-i(pt+qk^2t)}+e^{i(pt+qk^2t)}=2cos(pt+qk^2t)$.  Note the $sin$ terms cancel.


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?

I know that $289 = 17 \times 17$

so would it be $2^8\equiv 256\bmod17 =1$

and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

答:
No.

The order of $\bar 2$ in $\mathbb Z_{17}^\times$ is $8$ because $2^8\equiv1\pmod{17}$.

However, $2^8\not\equiv1\pmod{289}$, so $8$ is not the order of $\bar2$ in $\mathbb Z_{289}^\times$.

The order of $\bar 2$ in $\mathbb Z_{289}^\times$, i. e. the smallest positive integer $k$ such that $2^k\equiv1\pmod{289}$, is $136$. (I used my computer to get this.)

Fact:

Let $\operatorname {ord}_n(a)$ be the order of $\bar a$ in $\mathbb Z_{n}^\times$. Then, for prime $p$ and positive integers $k<l$,
$$
\operatorname {ord}_{p^k}(a)\mid\operatorname {ord}_{p^l}(a).
$$
For example, $8\mid136$.


答:
I'd start with considering that is $h$ is not one to one considering the $x\ne y$ so that $h(x) = h(y)$ and see what happens.

$g$ is onto so there are $a,\alpha \in A$ so that $g(a) = x$ and $g(\alpha) = y$.  And.... well, seems the obvious thing to try is to  $h(g(a)) = h(x)$ while $h(g(\alpha) = h(y) = h(x)$.

So if $a \ne \alpha$ then $h\circ g$ is not one-to-one.  And clearly $a \ne \alpha$ because if $a = \alpha$ then $g(a) = g(\alpha)$ but $g(a) = x\ne y = g(\alpha)$.

So that's that.

Title: Suppose that $g: A \rightarrow B$, $h: B \rightarrow C$. Prove that if $g$ is onto and $h$ is not one-to-one, then $h \circ g$ is not one to one.

Body: How can I start to approach this question? I know that if $g$ and $h$ are one-to-one, then $h \circ g$ is one-to-one, and also if $h$ and $g$ are both onto, and $h \circ g$ is onto. But this question didn't fit into both cases...


答:
First observe that we have: $$\displaystyle e^{ix^2/(-4i+4qt)}$$ is the same as $$e^{ix^2(4qt+4i)/(16(q^2t^2+1))}$$ which then becomes $$e^{-4x^2/(16(q^2t^2+1))}e^{i4x^2qt/(16(q^2t^2+1))}$$

Similarly $$\displaystyle e^{-ix^2/(4i+4qt)}$$ really is just $$e^{4x^2/(16(q^2t^2+1))}e^{i4x^2qt/(16(q^2t^2+1))}$$

Next observe that $$\sqrt{1-iqt}=\left(\sqrt{1+q^2t^2}\right)e^{-i\theta}e^{-i\pi n}$$ where $\theta=2\arctan(qt)$. Similarly observe that $$\sqrt{1+iqt}=\left(\sqrt{1+q^2t^2}\right)e^{i\theta}e^{i\pi n}$$ with again $\theta$ as above.

At any rate we then have $f(x,t)= \displaystyle \frac{\sqrt{\pi}}{2\sqrt{1+q^2t^2}}\left[e^{-ipt}e^{i4x^2qt/(16(q^2t^2+1))}e^{-4x^2/(16(q^2t^2+1))}\sqrt{1+q^2t^2}e^{-i\theta}e^{-i\pi n}+e^{ipt}e^{-i4x^2qt/(16(q^2t^2+1))}e^{4x^2/(16(q^2t^2+1))}\sqrt{1+q^2t^2}e^{i\theta}e^{i\pi n}\right]$ which then reduces to:

$f(x,t)= \displaystyle \frac{\sqrt{\pi}}{2}\left[e^{-ipt}e^{i4x^2qt/(16(q^2t^2+1))}e^{-4x^2/(16(q^2t^2+1))}e^{-i\theta}e^{-i\pi n}+e^{ipt}e^{-i4x^2qt/(16(q^2t^2+1))}e^{4x^2/(16(q^2t^2+1))}e^{i\theta}e^{i\pi n}\right]$

This expression is absurd, thus let $$A=\frac{4x^2}{16(q^2t^2+1)}$$

We thus have: $$f(x,t)= \displaystyle \frac{\sqrt{\pi}}{2}\left[e^{-i(pt+\theta+\pi n)}e^{i4x^2qt/(16(q^2t^2+1))}e^{-A}+e^{i(pt+\theta+\pi
 n)}e^{-i4x^2qt/(16(q^2t^2+1))}e^A\right]$$

I'll let you maybe take it from here using Euler's formula....lol. I'm sure there's a faster way of doing this...smh

Title: Writing Fourier transform integral in terms of real-valued functions

Body:
I am interested in the following integral.

$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$

One way (for example, by using Mathematica) to solve it is in the following way.

$$ f(x,t) = \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}} $$

I believe that $f(x,t)$ should be real valued, but it is not clear to me how to simplify given expression so that it consists only of real valued functions.


问:
Find centroid of shape given by a circle and a cardoid
I'm stuck on finding the centroid of this shape (the shaded area):

With the sphere having a radius of $r=2$, and the cardioide being given by $r=1+\cos\phi$

I found the Area, which is
$A=\frac12(2\pi r^2-\frac32 \pi)=\frac{13}{4} \pi$

Now I don't know how to proceed further. Could anyone point me in the right direction?
Thank you :)

答:
Your calculated area is slightly wrong; it should be $\frac12\left(\pi r^2-\frac32\pi\right)=\frac{5\pi}4$.

We use Green's theorem to calculate the $x$- and $y$-moments, which are the sum of line integrals over the three segments of the shape.
$$M_x=-\frac12\int_a^b y^2(t)x'(t)\,dt$$
$$M_y=\frac12\int_a^b x^2(t)y'(t)\,dt$$

The three segments are

the line at the lower left: $(t,0),a=-2,b=0$. This gives $M_x=M_y=0$ of course.

the cardioid arc: $((1+\cos t)\cos t,(1+\cos t)\sin t),a=\pi,b=0$. This gives $M_x=-\frac43,M_y=-\frac{5\pi}8$.

the circular arc: $(2\cos t,2\sin t),a=0,b=\pi$. This gives
$M_x=\frac{16}3,M_y=0$.

So the moments are $M_x=4,M_y=-\frac{5\pi}8$. The centroid is then given by $(M_y/A,M_x/A)$ (note the swapping of arguments):
$$K=\left(-\frac12,\frac{16}{5\pi}\right)$$


问:
Prove or disprove that the set  is closed in metric space ( [ −1 , 1 ] , ∞ ) where ∞(  ,  ) = sup −1≤  ≤1 | () − ()
Let  = {  ∈ [−1 , 1] ∶ 2 ≤ (0) ≤ 4 }

My idea is to use the fact that for each closed set  in (  , ) , the preimage −1[  ] is closed in metric space (,).  We have the set A is all the continuous function in the interval [-1,1], how can we use this in the proof?

By looking at similar questions, is the boundary set is A itself? Otherwise , how do we find the boundary set?

答: Define $\phi : C[-1,1] \to \mathbb R$ by $\phi(f) = f(0)$. Note that $|\phi(f)-\phi(g)| \leq d_\infty(f,g)$ for every $f$ and $g$ in $C[-1,1]$. Thus, $\phi$ is continuous and then $A = \phi^{-1}([2,4])$ is closed in $(C[-1,1],d_\infty)$.


答: Let $A = \{f \in C[-1, 1] : 2 \leq f(0) \leq 4 \}$. We'll show this is closed by showing that the complement is open. Suppose $f \in A^{c}$. Then $f(0) > 4$ or $f(0) < 2$. In either case, there is some $\epsilon > 0$ such that $(f(0) - \epsilon , f(0) + \epsilon) \cap [2 , 4] = \emptyset$. This should be intuitive, but rigorously, this follows because $[2, 4]$ itself is a closed subset of the reals. Anyway, then consider the ball $B(f, \epsilon)$. If $g \in B(f, \epsilon)$, then $sup_{|t| \leq 1}|f(t) - g(t)| < \epsilon.$ In particular, then, $|f(0) - g(0)| < \epsilon$. So then $g(0) \in (f(0) - \epsilon,f(0) + \epsilon)$, so that $g(0) \not \in [2, 4]$, as $(f(0) - \epsilon, f(0) + \epsilon) \cap [2, 4] = \emptyset$. This means $B(f, \epsilon) \subseteq A^c$. As $f$ was arbitrary, this means that $A^c$ is open, and thus $A$ is closed.

Title: Prove or disprove that the set  is closed in metric space ( [ −1 , 1 ] , ∞ ) where ∞(  ,  ) = sup −1≤  ≤1 | () − ()

Body:
Let  = {  ∈ [−1 , 1] ∶ 2 ≤ (0) ≤ 4 }

My idea is to use the fact that for each closed set  in (  , ) , the preimage −1[  ] is closed in metric space (,).  We have the set A is all the continuous function in the interval [-1,1], how can we use this in the proof?

By looking at similar questions, is the boundary set is A itself? Otherwise , how do we find the boundary set?


问:
Does there exist English letter frequency charts based on the assumed age of the reader?
Has anyone done work to construct letter frequency charts based on the assumed age of the reader?

One would expect that letter usage would be different in books targeted at (for example) a 3rd grader vs a college student.

My first grade daughter loves the show "Pokemon".  In that show the Pokemon characters only speak sounds that are made form pieces of their name.  For example a Pikachu pokemon only speaks words made from combinations of the sounds "Pi", "Ka", and "Chu".  She thought it would be cool to make a real Pikachu language.  And I think its a good opportunity to teach her about encoding schemes.

The obvious choice is encoding letters of the alphabet using these three sounds.  Ideally one would want the length of the words to be minimized.  We have three sounds, therefore a ternary Huffman code based on an English letter frequency chart would provide an optimal code.

I have seen many charts (like this one) ...

http://pi.math.cornell.edu/~mec/2003-2004/cryptography/subs/frequencies.html

Based on that chart, here is the code I came up with so far.

... but this chart is based on generic data and therefore wouldn't be optimal in terms of the words a first grader would choose speak.

This would be a spoken language only, so there is no need to encode special symbols or different letter cases.  I am only interested in the frequency of English letters English letters A-Z.

Ideally I want to find a table based on 1st grade books.  But tables based on text from any elementary school grade would be acceptable.

答:
Of course the letter frequencies would vary a lot from book to book, especially in those with a very limited vocabulary.  A 1st grade reader about Dick and Jane would have an unusually high frequence of "J".  I don't know if the texts of many of those readers
are readily available online, but if you can find them it's easy to compute the letter frequencies.

EDIT: Well, Project Gutenberg has lots of old readers and other children's books available for plain-text download.


问:
with which values parameter 'a' series converges and with which diverges
Which values for $a$ will make series convergent or divergent?

$$
\sum_{k=1}^{\infty}\frac{1}{(2k+1)\ln^a(2k+1)}
$$

I have been following this tutorial
http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx
  But now I'm stuck. I haven't seen examples as this one above. There is additional parameter $a$ which values I have to find myself. Could anyone suggest me how should I work this time then?
  Maybe I just choose randomly some values for $a$? Like $a=-2$, $0$ and $2$?

答:
If you are comfortably familiar with the tutorial series that you have been studying from, look at their discussion under integral test.

By following what they say word for word, you can see that the given series converges precisely if
$$\int_{x=1}^\infty \frac{1}{(2x+1)\ln^{a}(2x+1)}dx$$
converges.

The integral is easy to deal with.  To find an antiderivative, just make the substitution $u=\ln(2x+1)$.  Then $du=2dx/(2x+1)$, so for the antiderivative you want to find
$$\int \frac{du}{2u^a}$$
and the definite integral is
$$\int_{u=\ln 3}^\infty \frac{du}{2u^a}$$

I imagine that you are familiar with the fact that this "improper" integral diverges if $a \le 1$ and converges if $a>1$.  If you are not, just calculate the integral from $\ln 3$ to $M$, and examine what happens as $M\to\infty$. This is quite easy, since you know very well how to integrate $u^{-a}$.

We conclude that the original series converges if $a>1$ and diverges otherwise.


问:
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.

p.s. The answers are $\pm 6\sqrt 3$.

Body:
$\displaystyle\sqrt[3]{x+10}-\sqrt[3]{x-10}=2---(1)$

Let $y=x+10$

$(1)$ becomes

$\displaystyle\sqrt[3]{y}-\sqrt[3]{y-20}=2$

$\displaystyle\sqrt[3]{y}=2+\sqrt[3]{y-20}$

Cube both sides to get

$\displaystyle{y}=8+12  (y-20)^{\frac{1}{3}}+6(y-20)^{\frac{2}{3}}+y-20$

$\displaystyle{0}=-12+ 6  (y-20)^{\frac{1}{3}}(2+(y-20)^{\frac{1}{3}})$

$\displaystyle{12}= 6  (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$

$\displaystyle{2}= (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$

Cube both sides

$\displaystyle{8}= (y-20)(y)$

$\displaystyle{y^2-20y-8}= 0$

Solving the above equation yields $y=10\pm 6\sqrt{3}$ which implies $x=\pm 6\sqrt{3}$.


问:
IMO 2019 Problem N5 Solution 2
IMO 2019 Problem N5 Solution 2.

https://www.imo-official.org/problems/IMO2019SL.pdf

Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.

Solution 2. $\quad$ We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p\mid a$ for all primes $p\leqslant b$. We do this with Lucas' theorem.
$\quad$ Suppose that we have $p\leqslant b$ with $p\not\mid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p\leqslant b$. Suppose it is the $p^t$ digit for $t\geqslant 1$.
$\quad$ Now, as $n$ varies over the integers, $an+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $an\gt b$) such that the $p^0$ digit of $an$ is $p-1$ (so $p\mid an+1$) and the $p^t$ digit of $an$ is $0$. Consequently, $p \mid an+1$ but $p\mid \binom{an}{b}$ (by Lucas’ theorem) so $p\not\mid\binom{an}{b}-1$. Thus $b$ is not $a$-good.
$\quad$ Now we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $an+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\binom{an}{b+2}$ and $\binom{an}{b}$ is $(b+1)(b+2)/(an-b)(an-b-1)$. We know that there must be a choice of $an+1$ such that the former binomial coefficient is $1$ modulo $an+1$ but the latter is not, which means that the given ratio must not be $1\ \text{mod}\ an+1$. If $b+1$ and $b+2$ are both coprime to $an+1$ then the ratio is $1$, so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.
$\quad$ However, we can observe that $b$ must be even by insisting that $an+1$ is prime (which is possible by Dirichlet’s theorem) and hence $\binom{an}{b}\equiv (-1)^b=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.

I struggle with "$an+1$ is prime (which is possible by Dirichlet’s theorem)". For Dirichlet's theorem to be applicable here, there has to be an arithmetic progression of numbers $an+1$ for which $\binom{an}{b}\equiv1$ and $\binom{an}{b+2}\not\equiv1$ modulo $an+1$. Because $b$ is $a$-good and $b+2$ is not, we are guaranteed that at least one such $n$ exists. But I don't see how a single $n$ leads to a whole arithmetic progression of such numbers.

Any ideas? Thanks.

答:
I think that if $n$ satisfies $an\ge b+2,\binom{an}{b}\equiv 1\pmod{an+1}$ and $\binom{an}{b+2}\not\equiv 1\pmod{an+1}$, then $an+1$ is a composite number.

To prove this, it is sufficient to prove that if $n$ is such that $an+1$ is prime satisfying $an\ge b+2$ and $\binom{an}{b}\equiv 1\pmod{an+1}$, then $\binom{an}{b+2}\equiv 1\pmod{an+1}$.

We can have
$$(an+1)\mid (b+1)(b+2)\bigg(\binom{an}{b+2}-1\bigg)\tag1$$
since we have
$$\begin{align}&(b+1)(b+2)\bigg(\binom{an}{b+2}-1\bigg)
\\\\&=(b+1)(b+2)\cdot\frac{an(an-1)\cdots (an-b+1)(an-b)(an-b-1)}{(b+2)(b+1)b!}-(b+1)(b+2)
\\\\&=(an-b)(an-b-1)\cdot\frac{an(an-1)\cdots (an-b+1)}{b!}-(b+1)(b+2)
\\\\&=(an-b)(an-b-1)\binom{an}{b}-(b+1)(b+2)
\\\\&\equiv (-1-b)(-b-2)\times 1-(b+1)(b+2)\pmod{an+1}
\\\\&\equiv (b+1)(b+2)-(b+1)(b+2)\pmod{an+1}
\\\\&\equiv 0\pmod{an+1}\end{align}$$

It follows from $an+1\ge b+3$ that $(an+1)\not\mid (b+1)(b+2)$.

So, from $(1)$, we get $(an+1)\mid (\binom{an}{b+2}-1)$, i.e.
$$\binom{an}{b+2}\equiv 1\pmod{an+1}.\quad\blacksquare$$

Title: IMO 2019 Problem N5 Solution 2


问:
Evaluating Contour Integral of square
Evaluating $\frac{1}{2i\pi} \oint_C \frac{z^2}{z^2+4} dz$ where $C$ is the square with vertices at $\pm 2$ and $\pm 2 + 4i$

So, I rewrote the contour integral:$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$

then used the Cauchy Integral Theorem:
$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$ = $\frac{1}{2i\pi} \oint_C \frac{\frac{z^2}{z+2i}}{z-2i} dz$ + $\frac{1}{2i\pi} \oint_C \frac{\frac{z^2}{z-2i}}{z+2i} dz$ and my answer for the two contour integrals is $0$.

Is it done correctly or I did incorrectly.

答:
You have done the integration incorrectly; the answer is $i$ (backed up by numerical evaluation using mpmath).

The contour encircles a simple pole of the integrand at $2i$. By Cauchy's integral formula applied to $f(z)=\frac {z^2}{z+2i}$
$$f(2i)=i=\frac1{2\pi i}\oint_C\frac{f(z)}{z-2i}\,dz=\frac1{2\pi i}\oint_C\frac{z^2}{z^2+4}\,dz$$

Title: Evaluating Contour Integral of square


问:
2D heat equation with Neumann boundary
Homogenous Heat equation on 2d rectangle, $[0,a]\times[0,b]$, with time independent initial conditions and homogenous Neumann boundaries

$$\frac{\partial{u}}{\partial{t}} = \alpha^2 (\frac{\partial^2{u}}{\partial^2{x}} + \frac{\partial^2{u}}{\partial^2{y}}), \quad0<x<a, \quad 0<y<b, \quad 0<t$$
$$IV: u(x,y,0) = f(x,y) , \quad0<x<a, \quad 0<y<b$$
$$BC: \begin{cases} \frac{\partial{u}}{\partial{x}}(0,y,t) = \frac{\partial{u}}{\partial{x}}(a,y,t) = 0 , \quad 0<y<b, \quad 0<t \\ \frac{\partial{u}}{\partial{y}}(x,0,t) = \frac{\partial{u}}{\partial{y}}(x,b,t) = 0 , \quad 0<x<a, \quad 0<t \end{cases}$$

I tried to find a solution using separation of variables to get Fourier series. I could not find it anywhere so am asking, is this correct?

$$u(x,y,t) = \sum^\infty_{m=1}{\sum^\infty_{n=1}{A_{nm}\cos{(\frac{ n\pi x}{a})}\cos{(\frac{ m\pi y}{b})}e^{-\alpha^2 \pi^2 (\frac{n^2}{a^2}+\frac{m^2}{b^2})t}}}$$
where
$$A_{nm} = \frac{4}{ab} \int^a_0 {\int^b_0 {f(x,y) \cos{(\frac{ n\pi x}{a})}\cos{(\frac{ m\pi y}{b})} dy}dx}$$

答:
The separation of variables leads to
$$
\left\{ \begin{array}{lcc}
     X'' + \lambda  X & = & 0
  \\ Y'' + \mu Y & = & 0
             \end{array}
   \right.
$$
The boundary conditions
$$ \left.
\begin{array}{rcl}
         u_x(0,y,t) = 0 \ , &  u_x(a,y,t)=0
\\ u_y(x,0,t) = 0 \ , & u_y(x,b,t)=0
\end{array}
\right\} $$

Imply

$$ \left.
\begin{array}{rcl}
         X'(0)=0 \ , &  X'(a)=0
\\ Y'(0) = 0 \ , & Y'(b)=0
\end{array}
\right\} $$

These are problem of Sturm Liouville

The eigenvalues are $ \lambda_0 = 0 \ , \lambda_n = \frac{n^2 \pi^2}{a^2} $ and $ \mu_0 = 0 \ , \mu_n = \frac{n^2 \pi^2}{b^2} $

The eigenfunctions are $ X_0 = 1 , \ X_n = cos(\frac{n\pi}{a}x)  $ and $ Y_0 = 1 , \ Y_n = cos(\frac{n\pi}{b}y) $

On the other hand

$$ T(t) = e^{ -\alpha^2 (\lambda + \mu)t } $$

By the superposition principle

$$ u(x,y,t) = A_{00} + \sum_{n=1}^{\infty} A_{n0} e^{ -\alpha^2\frac{ n^2 \pi^2 }{a^2}t }cos\left( \frac{ n \pi }{a} x \right ) + \sum_{m=1}^{\infty} A_{0m} e^{ -\alpha^2\frac{ m^2 \pi^2 }{b^2}t }cos\left( \frac{ m \pi }{b} y \right ) $$
$$ + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} A_{nm} e^{ -\alpha^2\pi^2(\frac{n^2}{a^2} + \frac{m^2}{b^2} )t } cos\left(  \frac{n\pi}{a} x \right)cos\left(  \frac{m\pi}{b} y \right) $$

I suppose that you can compute the coefficients of the double cosine series.

Title: 2D heat equation with Neumann boundary


问:
Discrete Approximation of Continuous Distribution, $P(X<Y)$
I dont really get continuous probability, but we were introduced to it, and I'm struggling with this question:

Suppose $X \sim \text{Geometric}(p)$ and $Y \sim \text{Poisson}(\lambda)$ are independent and random variables. Find $P(X>Y)$

I know that for $X \sim \text{Geo}(p)$, $$P[X\ge i] = (1-p)^{i-1}$$ which means $$P[X<i] = 1- (1-p)^{i-1}$$ I also know that for $Y \sim \text{Poisson}(\lambda)$, $$P[Y = k] = P[X = i] =(e^{-\lambda}) \lambda^i/i! $$

My guess is to create some kind of summation equivalent, but Im not sure how to set up and simplify it.

Maybe something like:
$$\sum\limits_{x}^{\infty} \sum\limits_{y}^{\infty} (1- (1-p)^{y-1})(e^{-x}) \lambda^x/x!$$

Am I on the right path? How do I set the limits for the summation?

答: $P(X>Y)=\sum_{i>j} P(X=i)P(Y=j)=\sum_{i>j} p(1-p)^{i-1} e^{-\lambda}\frac {\lambda^{j}} {j!}$. You can write this as $ \sum\limits_{j=0}^{\infty} \sum\limits_{i=j+1}^{\infty} p(1-p)^{i-1} e^{-\lambda}\frac {\lambda^{j}} {j!}$. The inside sum is a geometric sum and its value is $(1-p)^{j}$. Finally you end with then series for $e^{\lambda (1-p)}$ multiplied by $e^{-\lambda}$ so the answer is $P(X>Y)=e^{-\lambda p}$.


问:
K(G,1) space is unique up to homotopy equivalence
I am reading Hatcher's Proposition 1B.9 page 90. He is trying to prove that if $X$ is a connected CW complex, $Y$ is $K(G,1)$ then every homomorphism $\pi_1(X,x_0) \rightarrow \pi_1(Y,y_0)$ is induced by a map $(X,x_0) \rightarrow (Y,y_0)$. Specifically, I fail to see why in order to extend $f$ over a cell $e^2_\beta$ with attaching map $\omega_\beta: S^1 \rightarrow X^1$, we need $f\omega_\beta$ to be nullhomotopic.

答:
Let's say that we're given a map $g: S^1 \rightarrow X$ and we want to know when it extends to a map $D^2 \rightarrow X$. This happens precisely when $g$ is nullhomotopic.

If $g$ is nullhomotopic then there is a homotopy $H:S^1 \times I \rightarrow X$ so that $H(-,0) = g$ and $H(-,1)$ is a constant map. This means that $H$ descends to a map $H':(S^1 \times I)/(S^1 \times \{1 \}) \rightarrow X$ and since $(S^1 \times I)/(S^1 \times \{1 \}) \approx D^2$ we're done. The proof is easily reversible as well so we see that $g$ extends to a map $D^2 \rightarrow X$ iff $g$ is nullhomotopic.

Does this answer your question?


问:
If $p^k m^2$ is an odd perfect number, then what is the optimal constant $C$ such that $\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$?
The topic of odd perfect numbers likely needs no introduction.

The question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then what is the optimal constant $C$ such that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)

MY OWN PROOF FOR $C = 1$

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This implies that
$$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$
since $\sigma(m^2)/p^k$ is always odd.

Now, suppose to the contrary that
$$\sigma(m^2) > p^k m^2 - p^{2k}.$$
$$p^{2k} + \sigma(m^2) > p^k m^2$$
$$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$
where we have used the lower bound $\sigma(p^k) \geq p^k + 1$.
But from the paper Dris (2012), we have the lower bound
$$3p^k \leq \sigma(m^2)$$
while we also have the upper bound
$$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$

This implies that
$$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$
$$p^k - 2 < \frac{2}{p^k - 1}$$
$$(p^k - 1)(p^k - 2) < 2.$$
This last inequality is a contradiction, as $p^k \geq 5$.

We therefore conclude that the inequality
$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$
holds when $C=1$.

Does the inequality still hold when the constant $C > 1$?  If so, then what is the optimal value for $C$?

答:
This is a partial answer.

Does the inequality still hold when the constant $C > 1$?

Yes, one can get
$$\frac{\sigma(m^2)}{p^k} \le \frac{m^2 - p^k}{\color{red}{4/3}}\tag1$$

I don't know if this is the optimal value for $C$.

Proof for $(1)$ :

Note that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}\iff \frac{p^k(m^2 - p^k)}{\sigma(m^2)}\gt C$$

Let $F:=\dfrac{p^k(m^2 - p^k)}{\sigma(m^2)}$. We can write $F$ as

$$F=\frac{p^k(m^2 - p^k)\sigma(p^k)}{\sigma(m^2)\sigma(p^k)}=\frac{p^k(m^2 - p^k)\sigma(p^k)}{2p^km^2}=\frac{m^2\sigma(p^k)-p^{k}\sigma(p^k)}{2m^2}$$

$$=\frac{\sigma(p^k)}{2}-\frac{p^{k}\sigma(p^k)}{2m^2}=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}$$

Using $3p^k\le \sigma(m^2)$, we get

$$F=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}\ge \frac{\sigma(p^k)}{2}-\frac{p^{k}}{3}=\frac{p^{k+1}-1}{2(p-1)}-\frac{p^{k}}{3}=\frac{p^{k+1}+2p^k-3}{6(p-1)}$$

Let $g(k):=\dfrac{p^{k+1}+2p^k-3}{6(p-1)}$. Since $g(k)$ is increasing, we get
$$F\ge g(k)\ge g(1)=\frac{p+3}{6}\ge \frac{5+3}{6}=\frac 43$$
to have
$$F\ge \frac 43$$
which is equivalent to $(1)$.$\quad\blacksquare$


问:
Simple example of uniformly continuous and continuous
I'm struggling somewhat with the difference between 'uniformly continuous' and 'continuous'. I understand the intuition, but I need to see it in an example. </p>

<p>Say $f=2x$, or some other easy function on the compact set $[1,2]$. Then is it correct to say that the function is uniformly continuous on that set? My question is what the difference is between the proofs for uniformly continuous, and just plain continuous, using the delta epsilon definition. I can't wrap my head around the proof at all. </p>

答:
The difference is in the fact that the $\delta$ you find does not depend on $x$, but only on $\varepsilon$.

As you've said, $x \mapsto ax$ (for some $a \ne 0$) is uniformly continuous on $[1,2]$ but also on the whole real line. Indeed, pick a point $x\in\mathbb R$ and fix $\varepsilon > 0$: then
$$
\vert f(x)-f(y) \vert = a \vert x -y \vert < \varepsilon
$$
whenever $\vert x-y \vert < \frac{\varepsilon}{a}=: \delta$. Such a $\delta$ is independent on the point $x$, as you easily can see.

On the contrary, take the function $x \mapsto \frac{1}{x}$ with $x\in (0, 1]$.
I show that this function is not uniformly continuous, i.e. I want to show that there exists a $\varepsilon>0$ s.t for every $\delta>0$ there are $x_1, x_2$ such that $\vert x_1-x_2\vert <\delta$ but $\vert f(x_1)-f(x_2)\vert \ge \varepsilon$. Pick a $\delta \in (0,1)$: if we choose
$$
x_1 = \frac{\delta}{5}, \qquad x_2=\frac{\delta}{10}
$$
we have
$$
\vert x_1-x_2 \vert = \frac{\delta}{10} < \delta
$$
but
$$
\vert f(x_1)-f(x_2) \vert =\frac{5}{\delta}>5.
$$
So, this is enough to take $\varepsilon \in (0, 5]$ to get our claim.

Some final notes

1. Every continuous function on a compact set is uniformly continuous (Heine-Cantor theorem).

2. If a continuous function $f \colon [0,+\infty) \to \mathbb R$ is s.t. $\lim_{x\to +\infty} f(x)$ exists and it is finite then $f$ is uniformly continuous.


问:
Find the coefficients and convergence of radius of $\frac{e^{-x}}{1-x} = \sum_{n=0}^{\infty} a_n x^n$.
I am trying to find the coefficients and convergence of radius of $\frac{e^{-x}}{1-x} = \sum_{n=0}^{\infty} a_n x^n$.

From the Taylor series, I know around $x=x_0$
\begin{align}
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n
\end{align}
and I know for the power series
\begin{align}
\sum_{n=0}^{\infty} a_n x^n  \quad \textrm{converges} \quad |x|<R,
\end{align}
where $R$ is the radius of convergence given by $\alpha = \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|$. i.e.,
\begin{align}
R = \begin{cases}
&\frac{1}{\alpha}, \quad 0<\alpha <\infty \\
& \infty, \quad \alpha = 0 \\
& 0 ,\quad \alpha = \infty
\end{cases}
\end{align}

First my starting point is find the explicit form of $a_n$ of $\frac{e^{-x}}{1-x}$. Simply denote $g(x) = \frac{e^{-x}}{1-x}$ then

After some computations I have
\begin{align}
g'(x) = \frac{e^{-x} x}{(1-x)^2}, \quad g''(x) =   \frac{e^{-x}(1+x^2)}{(1-x)^3},
\quad g'''(x) = \frac{e^{-x} (2+3x+x^2)}{(1-x)^4},
\end{align}
I can put $x=x_0$ but couldn't figure out the general form of $a_n$.

Is there any smart method to compute $a_n$?

For the radius of convergence, I guess only after figuring out explicit or asymptotic behaviour of $a_n$ it can be determined. ...

答:
Multiplication of a power series $(a_0,a_1,\dots)$ by $\frac1{1-x}=1+x+x^2+\dots$ produces the series $(a_0,a_0+a_1,a_0+a_1+a_2,\dots)$ Since
$$e^{-x}=\frac1{0!}x^0-\frac1{1!}x^1+\frac1{2!}x^2-\frac1{3!}x^3+\cdots$$
it follows that
$$\frac{e^{-x}}{1-x}=\frac1{0!}x^0+\left(\frac1{0!}-\frac1{1!}\right)x^1+\left(\frac1{0!}-\frac1{1!}+\frac1{2!}\right)x^2+\cdots$$
The radius of convergence around $0$ is $1$, the same as that of $\frac1{1-x}$.


答:
Your phrase "apply the differential form onto parameters" suggests you're not aware of the isomorphism $T_p(\Bbb R^n) \cong \Bbb R^n$. This explicit correspondence means, for instance, that $$T_{(x,y,z)}(\Bbb R^3) \ni x^ 2y \partial_y + x\partial_z \stackrel{\cong}{\mapsto} (0, x^2y, x) \in \Bbb R^3,$$and similarly for $Y$. In other words, if you know how to evaluate $\omega$ at $(0,x^2y,x)$ and $(0,x,0)$, then you know how to evaluate it at $x^2y\partial_y+x\partial_z$ and $x\partial_y$, there's no excuse. Namely, $\{{\rm d}x, {\rm d}y, {\rm d}z\}$ is dual to $\{\partial_x,\partial_y,\partial_z\}$, so that you have ${\rm d}x(\partial_x) = 1$, ${\rm d}x(\partial_y) = 0$, etc. So $$\begin{align} \omega(X,Y) &= 2x({\rm d}x\wedge {\rm d}y)(X,Y) + y^2({\rm d}x\wedge {\rm d}z) \\ &=2x \begin{vmatrix}{\rm d}x(X) & {\rm d}x(Y) \\ {\rm d}y(X) & {\rm d}y(Y) \end{vmatrix}+ y^2\begin{vmatrix} {\rm d}x(X) & {\rm d}x(Y) \\  {\rm d}z(X) & {\rm d}z(Y) \end{vmatrix} \\ &= 2x \begin{vmatrix} 0 & 0
 \\ x^2y & x\end{vmatrix} + y^2 \begin{vmatrix} 0 & 0 \\ x & 0\end{vmatrix}  \\ &= 0.\end{align}$$

Title: Applying differential forms to non-vectors $\omega(X,Y)$

Body:
$\omega=2xdx\wedge dy + y^2dx\wedge dz$

$X=x^2y\frac{\partial}{\partial y} + x\frac{\partial}{\partial z}$

$Y=x\frac{\partial}{\partial y}$

Calculate $\omega(X,Y)$

I understand how to apply the differential form onto parameters given vectors $(v1,v2)$, where you take  the determinant. Would I perform the same for this operation?

I am trying this, but I'm not sure where to go from here. Should I be taking the partials as each parameter?

$2xdet\begin{bmatrix}
0&0\\ x^2y&x\\ \end{bmatrix} + y^2det\begin{bmatrix}
0&0\\ x&0\\ \end{bmatrix} = 0$

I have looked through Do Carmo Differential Forms, and Faris' Vector fields and differential forms, but have not been able to find an explanation or example.


问:
Minkowski's Integral Inequality when $p=\infty$
My version of the Minkowski's Integral Inequality is that, suppose $(X_1,\mathcal{M}_1,\mu_1)$ and $(X_2,\mathcal{M}_2,\mu_2)$ are complete $\sigma$-finite measure spaces and $f$ is non-negative measurable function on $X_1\times X_2$ with respect to the product measure $\mu_1\times \mu_2$, then for any $p\in [1,\infty]$ there is
$$\left\lVert \int f(x_1,x_2)dx_2\right\rVert_{L^p(X_1)}\leq \int \lVert f(x_1,x_2)\rVert_{L^p(X_1)}dx_2. $$

When $p=\infty $, this yields to prove that $\int f(x_1,x_2)dx_2\leq \int \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}dx_2$ almost everywhere $x_1$, so (I think) we need to show that for almost everywhere $x_1$,
$$ f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^{\infty}(X_1)},\ \ a.e. x_2.$$
What we have by hand is that for any fixed $x_2\in X_2$, $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ almost everywhere $x_1$, hence for any $x_2\in X_2$ there exists a nullset $Z(x_2)\subset X_1$ such that $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $x_1\in X_1\setminus Z(x_2)$. It follows that for $Z= \bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}\subset X_1\times X_2$ we have $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $(x_1,x_2)\in X_1\times X_2\setminus Z$. If we can show that $Z$ is measurable in $\overline{X_1\times X_2}$ the completion of $X_1\times X_2$, then by the Tonelli's Theorem $\overline{\mu_1\times \mu_2}(Z)=\int \mu_1(Z(x_2))dx_2=0$, whence for almost everywhere $x_1$ we have the inequality for almost everywhere $x_2$ as desired.

So the question reduces to show that $\bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}$ is measurable in the completion of $X_1\times X_2$ if each $Z(x_2)$ is of measure $0$. Is this true? If so, how to prove that? If not, then how to prove the Minkowski's Integral Inequality when $p=\infty$?

Thanks in advance.

答:
I figured out the solution after several discussions. For my argument, it remains only to show that the function $(x_1,x_2)\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $\overline{X_1\times X_2}$ and it suffices to show that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$, so that the set $\{(x_1,x_2)\in X_1\times X_2\mid f(x_1,x_2)>\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)} \}$ would be measurable and by "what we have by hand" we know that each of its slice has measure $0$, whence by Tonelli's Theorem this set is of measure $0$ and the conclusion follows. The condition that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$ is in fact essential for this question, otherwise the integral $\int \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}dx_2$ is never defined.

To show that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable, observe that for any simple function $\varphi$ on $X_1\times X_2$, $\varphi\in L^P(X_1)$ for all $x_2\in X_2$ and any $1\leq p<+\infty$, it follows (trivially by direct calculation) that $\lVert \varphi(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{p\to+\infty}\lVert \varphi(x_1,x_2)\rVert_{L^p(X_1)}$ for all $x_2$. Since by Tonelli's Theorem each $\lVert \varphi(x_1,x_2)\rVert_{L^p(X_1)}$ is measurable as a function on $X_2$, we know that $\lVert \varphi(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$ for any simple $\varphi$. Now for any measurable $f$ on $X_1\times X_2$ with each of its slice $f(-,x_2)$ having finite measure support in $X_1$, choose a sequence of simple functions $\{\varphi_n\}$ that converges to $f$ pointwisely and Egorov's Theorem provides a simple proof for the fact that $\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{n\to\infty}\lVert \varphi_n(x_1,x_2)\rVert_{L^\infty(X_1)}$ for all $x_2\in X_2$ as one can argue by contradiction. Thus $\lVert f(x_1,x_2)\rVert_L^\infty(X_1)$ is measurable in this case. To pass to general measurable functions $f$ on $X_1\times X_2$, write by $\sigma$-finite condition $X_1=\bigcup_{n=1}^\infty B_n$ with $B_n\subset B_{n+1}$ and $\mu_1(B_n)<\infty$ for all $n$. Observe again by contradiction that $\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{n\to\infty}\lVert f\cdot\chi_{B_n\times X_2}(x_1,x_2)\rVert_{L^\infty(X_1)}$ and the fact that each $\lVert f\cdot\chi_{B_n\times X_2}(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable finishes the proof.


问:
General approach of pricing an European option with specific payoff & instance
I have a question regarding the general approach to questions like the following and the solution in this special case.
I'm referring to the question 2.6 from the book "Quant Job Interview Questions And Answers" from Mark Joshi, i.e.

"In the Black Scholes world, price a European option with payoff of $ \max(S_T^2 - K, 0)$ at time $T$."

His solution and the application of the Black formula (using the forward price) aren't lucid to me.
The methodology applied by Joshi is basically to translate the problem first into the standard assumptions in the Black-Scholes theory, i.e. to assume the stock moves lognormally, $dS_t = \mu S_t dt +\sigma S_t dW_t$ with $W_t$ denoting Brownian motion. He goes on to translate the equation into the risk-neutral measure, i.e. replacing $\mu$ with $r$ in the equation above.

By stating the solution to said equation with $S_t = S_0 \cdot \exp \{(r - 0.5\sigma^2)t + \sigma W_t\}$ he simply squares this for our problem to achieve $S_t^2 = S_0^2 \cdot \exp \{(2r - \sigma^2)t + 2\sigma W_t\}$ (first question: why can we just do that?).
He then goes on to say that since we are dealing with a geometric Brownian motion here, we're able to apply the Black formula. He finishes his explanation by stating that "we simply have to write the $S_T^2$ as $ F_T (0)e^{-\nu^2T/2 + \nu W_T}$ and compute the values for $F_T(0)$ and $\nu$ (second question: why and how can we do that?).

There has to be a general approach and recipe regarding pricing and translating such simple problems into the Black-Scholes framework, maybe outside the proposed solution by Joshi.
Answers are highly appreciated.

答:
First I believe you mean you want a European option with payoff of $\max(S^2_T- K, 0)$, not $\max(S^2_T, K)$. As to why the solution seems so specific and not very general, that book is for job interview questions, so it assumes the reader is very comfortable with its contents already and wants solutions that are fast, easy to remember, and do not need a lot of complex calculations.

For the first question, $S_t = S_0 \cdot \exp((r - \frac{\sigma^2}{2})t + \sigma W_t)$ is the solution to Geometric Brownian motion for any time $t$. Nothing says if we want $S_t^2$ we can't just square both sides giving $S_t^2 = S_0^2 \exp((2r -\sigma^2)t + 2\sigma W_t))$. We could also find this using Ito's lemma with $f(t, x) = x^2$. To see this let $S_t$ follow
$$dS_t = rS_tdt + \sigma S_tdW_t.$$

Using $f(t, x) = x^2$ by Ito's lemma we have

$$df = (2rf + \sigma^2 f)dt + 2 \sigma fdW_t.$$

Then if we let $x = S_t$ we have

$$dS_t^2 = (2rS_t^2 + \sigma^2 S_t^2)dt + 2 \sigma S_t^2dW_t.$$

This then has the solution for $S_t^2$ given above.

For the second question $F_T(0)$ is the forward price of $S_t$ at time $t=0$ and $\nu$ would be equal to $2\sigma$. The use of the Black model is not very intuitive and most derivations for this do not use that approach. I think one of the more intuitive approaches would be to see that $S_t^2$ is a Geometric Brownian motion with drift $2r + \sigma^2$ and diffusion $2\sigma$ and use the Black-Scholes formula with the proper adjustments. The adjustments would be use the Black-Scholes formula with dividends and in it let $\hat{\sigma} = 2\sigma$, $\hat{r} = r$, and $\hat{q} = -(\sigma^2 + r)$.

As far as a more general framework for option pricing your best bet is to consult a book on it. Two that come to mind are Martingale Methods in Financial Modelling by Musiela and Rutkowski or Arbitrage Theory in Continuous Time by Bjork, but almost any will suffice they tend to just vary in how deep mathematically they go.


问:
Prove that there is a quadratic irreducible in $\mathbb Z_p[x]$ for $p$ prime.
I am kind of stuck at this problem. I know how to find irreducible polynomials in fields like $\mathbb Z_1,\mathbb Z_2$ or where the specific field is given.

I can't figure out a way to prove this. I am thinking to use theorems like:
If an irreducible polynomial $p(x)$ belongs to a field $F$, then $F[x]/\langle p(x)\rangle$ is a field.

Or should I use arguments like if $\mathbb Z_p$ is $\mathbb Z_7$, then polynomials like $x^2 + 3$ are irreducible. So, there exists a quadratic irreducible in $\mathbb Z_p[x]$?

Or is there any other method that I should follow? Any suggestions will be helpful.

答:
There are $p^2$ monic quadratic polynomials in $\Bbb Z/p \Bbb Z[x]$ but only $p+ \binom p2$ distinct products of monic linear polynomials, so since $p+ \binom p2 \lt p^2$, there is some monic quadratic polynomial that is not a product of linear factors.  That quadratic polynomial must be irreducible.

To see that the inequality holds, $p+ \binom p2 = \binom p1 + \binom p2 =\binom{p+1}{2}$ and a little arithmetic shows that $p^2-\binom{p+1}{2}=\binom p2 \gt 0$ for all primes $p$.


问:
Showing almost-sure convergence, given condition.
Let $(\Omega, F, P)$ be a probability space with $X_1,X_2,...:\Omega\to R$ independent random variables. Take $E[X_i]=0$ for all $i\in N$ and
$\sum_i E[X_i^2\cdot \chi_{\{|X_i|\le 1\}} + |X_i|\cdot \chi_{\{|X_i|>1]}]< \infty$

show $\sum_i X_i$ converges $P$- almost everywhere.

I tried splitting the sample space into two - On $\{|X_i|> 1\}$ we can use Kolmogorov's three series theorem - the expectations converge because the sum of the expectations of the absolute values are finite, for example, and the other two series can be shown to converge.
I don't know how to deal with $\{|X_i|> 1\}$. I can't use the theorem which guarantees a.s. convergence given expectations equal to zero and given $\sum_i E[X_i^2]<\infty$, because after restricting the sample space I can't be sure that the expectations are zero.

答:
$\sum X_i \chi_{|X_i| \leq 1}$ converges almost surely by Kolmogorov's Three Series Theorem. Note that $\sum EX_i \chi_{|X_i| \leq 1}=-\sum EX_i \chi_{|X_i| > 1}$ since $EX_i=0$ for all $n$. And it is given that $\sum EX_i \chi_{|X_i| > 1}$ is convergent.

Since $\sum E|X_i|\chi_{|X_i| > 1} <\infty$ it follows that $\sum |X_i|\chi_{|X_i| > 1} <\infty$  almost surely. Adding these two series we get almost sure convergence of $\sum X_i$.


问:
Finding the matrix of the adjoint transformation
I have the following exercise:

Let $B=\{b_1,b_2\}$ a orthonormal basis of a Euclidean space $V$. Let $C=\{b_1,b_1+b_2\}$, and the matrix of $f:V\to V$ with respect to $C$ is:
$$A_{f,C}=\begin{pmatrix}
1 & 2 \\ 1 & -1
\end{pmatrix}.$$
If we let $f^*$ be the adjoint transformation, find $A_{f^*,C}$. My attempt:

We know that the relation with of the basis is given by $C=BT$ with $T$ the matrix $\begin{pmatrix}1&1\\0&1 \end{pmatrix}$. Also we know that $A_{f,C}=T^{-1}A_{f,B}T$, and therefore $A_{f,B}=TA_{f,C}T^{-1}$, this is:
$$\begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & 2 \\ 1 & -1
\end{pmatrix}\begin{pmatrix}
1 & -1 \\ 0 & 1
\end{pmatrix}= \begin{pmatrix}
2 & -1 \\ 1 & -2
\end{pmatrix}, $$
Now, as $B$ is orthonormal, we have $A_{f^*,B}=A_{f,B}^*=\begin{pmatrix}
-2 & 1 \\ -1 & 2
\end{pmatrix}$. Now, we conjugate with $T$ to change the basis:
$$
\begin{pmatrix}
1 & -1 \\ 0 & 1
\end{pmatrix}\begin{pmatrix}
-2 & 1 \\ -1 & 2
\end{pmatrix}\begin{pmatrix}
1 & 1 \\0 & 1
\end{pmatrix}=\begin{pmatrix}
-1 & -2 \\ -1 & 1
\end{pmatrix}
$$

This is the procedure that is more logical to me, but I have are two questions:

This is ok? I think the calculations are ok, but my professor told us that the answer was $\begin{pmatrix}
3 & 6 \\ -1 & -3
\end{pmatrix}$, but i don't know why.

2)I can notice that the matrix was exactly the adjoint of the initial matrix $A_{f,C}$, so, this implies that in this problem I just needed to calculate de adjoint matrix of this? Or is this just a coincidence?

Thanks.

答:
After you changed basis back to $B$, you wrote:

Now, as B is orthonormal, we have $A_{f∗,B} = A^∗_{f,B} = \begin{pmatrix} −2 & 1 \\ -1 & 2 \end{pmatrix}$.

This was not correct. You switched the signs on the diagonal entries of the matrix. This is not the transpose of the matrix you had on the line before. It should instead be $\begin {pmatrix} 2 & 1 \\ -1 & -2 \end{pmatrix}$. Now when you conjugate by $T$ you will get the matrix that your teacher said should be the answer.


问:
How many scalar multiplications are possible on a $\mathbb{Q}$-Vector Spaces?
Let $V$ be a $\mathbb{Q}$ Vector-Space. Then from the definition of Vector-Spaces, we have $1.x=x$ for all $x \in V$. From this one can show $n.x=x+x+...+x(\text{n times})$ for all $n \in \mathbb{Z}$.

So any scalar multiplication defined on a $\mathbb{Q}$-Vector Space, when restricted to $\mathbb{Z}$ becomes identical. Now the question is, is there only one possible way to define scalar multiplication on a $\mathbb{Q}$-Vector Space?

This can be generalized to ask

How many scalar multiplications are possible on $\mathbb{R}$-Vector Space and $\mathbb{C}$-vector space?

答:
$\newcommand{\Z}{\mathbb Z}$
$\newcommand{\Q}{\mathbb Q}$
$\newcommand{\R}{\mathbb R}$
$\newcommand{\C}{\mathbb C}$

Yes, the scalar multiplication on a $\Q$-vector space is unique.

We only need to show that for every fixed non-zero vector $v\in V$, the scalar multiplication on the subspace $\Q v$ is unique. And, $\Q v$ is isomorphic to $\Q$ as additional groups, so we only need to prove the statement for $V=\Q$ and $v$ is a rational number.

Let $f:\Q\to \Q$ be a function such that $f(q)=q*v$ for every $q\in\Q$. ($xv$ is the common multiplication on $\Q$ and $x*v$ is the scalar multiplication on the vector space.) Then, $f(1)=v$ and $f(kq)=k*f(q)=kf(q)$ for $k\in\Z$. So $f(k)=kv$ for integer $k$. Therefore, if $q=\dfrac mn\in\Q$, then $nf(q)=n*f(q)=f(m)=mv$, so $f(q)=\dfrac{mv}{n}$ is uniquely determined.


答:
Set $x+10=t^3$ and $x-10 = u^3$. So we want to solve $t - u = 2$.

However $t^3 - u^3 = (t-u)^3 + 3tu(t-u) = 20$.

That is $2^3 + 6tu = 8 + 6tu =20$.

Now we have $tu = 2$. This means $x^2 -100 = 8$ and thus $x = \pm6\sqrt{3}.$

Body:
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.

I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.

p.s. The answers are $\pm 6\sqrt 3$.