问:
Analogy between prime numbers and singleton sets?
While trying -- in vain -- to write an alternative answer for another question (If $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.), I discovered the following property for sets: $$A \textrm{ is a singleton set} \;\equiv\; \langle \forall B : B \subseteq A : B = \emptyset \;\not\equiv\; B = A \rangle$$ and I noticed the similarity to the following definition for positive whole numbers: $$n \textrm{ is prime} \;\equiv\; \langle \forall d : d \textrm{ divides } n : d = 1 \;\not\equiv\; d = n\rangle$$  So it looks like singleton sets act a bit like prime numbers.  Which is not strange, come to think of it, given that both are 'indivisible atoms' of some sort.

So what common theory underlies both concepts of indivisibility?

答:
Let $P$ be the set of all prime numbers, and let $N$ be the collection of all multisets whose elements are in $P$.  Notice the fundamental theorem of arithmetic gives a bijection $N\leftrightarrow \mathbb{N}$, where $1$ corresponds to the empty multiset.

Under this bijection, the notion of divisibility of natural numbers corresponds to the notion of containment of sets, and the prime numbers correspond to singletons in $N$.  Also, notice that $\varnothing$ is contained in all elements of $N$, just as $1$ divides all natural numbers.


问:
$\lim_{n\to\infty}n^{2020}\int^{\pi/2}_0 \left[(\sin x)^n \cdot (\pi/2-x)^n\right]\,dx$

Calculate:
$$\lim_{n\to\infty}n^{2020}\int^{\pi/2}_0 \left[(\sin x)^n \cdot (\pi/2-x)^n\right]\,dx$$

The idea that I had was to let
$ y=x-\pi/2$ and to substitute it in the integral. But I get to
-$\lim_{n\to\infty}n^{2020}\int^{\pi/2}_0 \left[(\cos)^n \cdot (y)^n\right]\,dy $ which I do not know how to calculate.

答:
Hint: For $x\in [0,\frac{\pi}{2}]$ one has $$  \left|\sin(x)\cdot (\frac{\pi}{2}-x)\right|\le \left|x\cdot (\frac{\pi}{2}-x)\right| \le \left|\frac{\pi^2}{4^2}\right|< 1. $$
What is $ \lim_{n\to \infty}n^{2020} a^{n}$ for $|a| <1$?

ContentLicense: CC BY-SA 4.0


问:
Finding the nth derivative of functions, in particular y=tan(x)

For $y=\cos2x$

Since $\frac{d y}{d x} = -2 \sin2x$, and $\frac{d^2 y}{d x^2} = -4 \cos2x$, then

$\frac{d^2 y}{d x^2} = -4y$, so

$\frac{d^3 y}{d x^3} = -4\frac{d y}{d x}$, $\frac{d^4 y}{d x^4} = -4\frac{d^2 y}{d x^2}$, etc.

Therefore, the nth differential of $y=\cos2x$ is $\frac{d^n y}{d x^n} = -4 \frac{d^{n-2} y}{d x^{n-2}}, n\geq 2$

For $y=\ln(x)$

Same as above, but this time the differentials are expressed in terms of $x$, not each other.

$ y=\ln(x) \Rightarrow \frac{d y}{d x} = x^{-1} \Rightarrow \frac{d^2 y}{d x^2} = -x^{-2}$, etc.

So for $y=\ln(x)$,

$\frac{d^n y}{d x^n} = (n-1)!(-1)^{n-1}x^{-n}, n\geq 2$

For $y=\tan(x)$

So the question is, does an equation such as that exist for $y=\tan(x)$?

I have tried two approaches:

Differentials of $\tan(x)$ in terms of $y$ and $\frac{dy}{dx}$:

\begin{array}{c}
 \frac{d y}{d x} & = & 1+y^2 \\
 \frac{d^2 y}{d x^2} & = & 2y\frac{d y}{d x} \\
 \frac{d^3 y}{d x^3} & = & 4(\frac{d y}{d x})^2 & - & 2\frac{d y}{d x} \\
 \frac{d^4 y}{d x^4} & = & 20y(\frac{d y}{d x})^2 & - & 4y\frac{d y}{d x} \\
 \frac{d^5 y}{d x^5} & = & 20(\frac{d y}{d x})^3 & + & 68(\frac{d y}{d x})^2 & - & 72 \frac{d y}{d x} \\
\end{array}

There are some patterns you can spot but it didn't seem very promising so I moved on.

Differentials of $\tan(x)$ in terms of other differentials:

\begin{array}{c}
 \frac{d y}{d x} & = & 1+y^2\\
 \frac{d^2 y}{d x^2}& = & 2y\frac{d y}{d x}\\
 \frac{d^3 y}{d x^3}& = & 2y\frac{d^2 y}{d x^2} & + & 2(\frac{d y}{d x})^2\\
 \frac{d^4 y}{d x^4}& = & 2y\frac{d^3 y}{d x^3} & + & 6\frac{d^2 y}{d x^2}\frac{d^2 y}{d x^2}\\
 \frac{d^5 y}{d x^5}& = & 2y\frac{d^4 y}{d x^4} & + & 8\frac{d^3 y}{d x^3}\frac{d y}{d x} & + & 6(\frac{d^2 y}{d x^2})^2\\
 \frac{d^6 y}{d x^6}& = & 2y\frac{d^5 y}{d x^5} & + & 10\frac{d^4 y}{d x^4}\frac{d y}{d x} & + & 20\frac{d^3 y}{d x^3}\frac{d^2 y}{d x^2}\\
 \frac{d^7 y}{d x^7}& = & 2y\frac{d^6 y}{d x^6} & + & 12\frac{d^5 y}{d x^5}\frac{d y}{d x} & + & 30\frac{d^4 y}{d x^4}\frac{d^2 y}{d x^2} & + & 20(\frac{d^3 y}{d x^3})^2\\
 \frac{d^8 y}{d x^8}& = & 2y\frac{d^7 y}{d x^7} & + & 14\frac{d^6 y}{d x^6}\frac{d y}{d x} & + & 42\frac{d^5 y}{d x^5}\frac{d^2 y}{d x^2} & + & 70\frac{d^4 y}{d x^4}\frac{d^3 y}{d x^3}\\
 \frac{d^9 y}{d x^9}& = & 2y\frac{d^8 y}{d x^8} & + & 16\frac{d^7 y}{d x^7}\frac{d^1 y}{d x^1} & + & 56\frac{d^6 y}{d x^6}\frac{d^2 y}{d x^2} & + & 112\frac{d^5 y}{d x^5}\frac{d^3 y}{d x^3} & + & 70(\frac{d^4 y}{d x^4})^2\\
\end{array}

Table of coefficients expressed in terms of prime factors:

\begin{array}{c|c c c c c }
&1&2&3&4&5\\\hline
2&2\\
3&2&(2)\\
4&2&2\times3\\
5&2&2\times2\times2&(2\times3)\\
6&2&2\times5&2\times2\times5\\
7&2&2\times2\times3&2\times3\times5&(2\times2\times5)\\
8&2&2\times7&2\times3\times7&2\times5\times7\\
9&2&2\times2\times2\times2&2\times2\times2\times7&2\times2\times2\times2\times7&(2\times5\times7)\\
\end{array}

This is much more exciting. I have worked out formulae up to the third term:

\begin{array}{ c c|c c c}
&term& 1 & 2 & 3 & m & last\\
&& 2y\frac{d^{n-1} y}{d x^{n-1}} & 2(n-1)\frac{d^{n-2} y}{d x^{n-2}}\frac{d y}{d x} & (n-1)(n-2)\frac{d^{n-3} y}{d x^{n-3}}\frac{d^2 y}{d x^2} & (?)\frac{d^{n-m} y}{d x^{n-m}}\frac{d^{m-1} y}{d x^{m-1}}*^1 & 2(?)\frac{d^{\frac{n-1}{2}} y}{d x^{\frac{n-1}{2}}}*^2\\
\end{array}

$*^1$Note: $(n-m)$ and $(m-1)$ add up to $n-1$, i.e. one less than the differential that is being calculated.

$*^2$ Only appears when $\frac{n-1}{2}$ is a whole number, i.e. for $n$ odd.

The problem I am encountering is that there doesn't seem to be an expression that works for every terms of every differential. Then again I have only calculated up to the 3rd term. Any ideas?

(Also I'd be very interested if you found any other nth differentials of this sort.)

答:
I actually have two answers for this question.  The first is a series I came up with a while back when I was trying to implement the polygamma function and needed a generalized reflection formula.  The second I obtained by entering my coefficients into google and looking at the resulting OEIS page.

First formula:

For positive integer n, the n-th derivative of tan(x) can be represented as

$$\sec^2(x) \sum_{k=0}^{n-1} C_{n,k} \tan^k (x)$$

With $C_{n,k}$ being the integer at the n-th row and k-th column of the following table:

(The empty cells represent 0, and are omitted for visual clarity)

The numbers in this triangle can be generated by the recursive definition:

$$C_{n+1,k} = (k+1)(C_{n,k-1}+C_{n,k+1})$$

For the sake of the recurrence relation, all cells outside the triangle are assumed to be 0.  In fact, $C_{n,k}$ is 0 for all

•k<0 (left of the triangle)

•k>n (right of the triangle)

•k+n is even (checkerboard pattern)

This was going to be the end of the post, but I then typed in these numbers into google, and an OEIS (online encyclopedia of integer sequences) page came up, which I've linked below:

https://oeis.org/A185896

At some point on the page, it mentions that these polynomials can actually be generated via what are known as Eulerian polynomials.  After rearranging some of the variables, I came up with the following formula:

$$ \frac{d^n}{dx^n} \tan(x) = 2\sec^{n+1}(x)  \sum _{m=0} ^{n/2-1} (-1)^{m+1-n/2} A(n,m)\sin((n-1-2m)x)$$
for all even n>0

$$ \frac{d^n}{dx^n} \tan(x) = \sec^{n+1}(x)  \left( A(n,\frac{n-1}2) + 2 \sum _{m=0} ^{\frac{n-3}2} (-1)^{m+(1-n)/2} A(n,m) \cos((n-1-2m)x) \right)$$
for all odd n>0

Where A(n,m) is another triangle of numbers which, unlike the last one, has an official name: the Eulerian numbers.  Also unlike the last triangle of numbers, you can actually generate each row without knowing the numbers on the previous row, via the following formula:

$$A(n,m) = \sum _{k=0} ^{m+1} (-1)^k {n+1\choose k} (m+1-k)^n $$

Both of these formulas have some pros and cons.  They both have about the same number of terms.  The second formula has the benefit that you don't need to generate all the coefficients for the previous derivatives to find a particular derivative.  However, it also has the downside that its coefficients alternate in sign, which could potentially lead to roundoffs errors.

I hope this answers your question. :)

ContentLicense: CC BY-SA 3.0


问:
Algebra II online courses
Can anyone give me recommendations for Algebra II online courses. I do know about Khan academy but looking for other courses that are also really good for learning Algebra II.

Thank you

答:
How about edX?

They have thousands of courses on virtually every subject you could ever want.

Title: Algebra II online courses


问:
Hey, how do you calculate this the easiest (not necessarily quickest) way? $\lim_{ x\to1-} (1-x)^{\sin (\pi x)}$
Please help me, I have no idea and exams are coming soon :( what i mean</p>

答: It depends on what you already know. For example, setting $y=1-x$ and taking logarithms is quick if you know $\lim_{y\to0^+}y\ln y$ and $\lim _{y\to0}\frac{\sin(\pi y)}{\pi y}$ (both links found in caase you don't with approach0).


问:
How do you integrate $x ^ 2 + y ^ 2$ over the area where $x \geq 0, y \geq 0,$ and $3x + 4y <10$?
How do you integrate $x ^ 2 + y ^ 2$ over the area where $x \geq 0, y \geq 0,$ and $3x + 4y <10$？

I thought I should use polar coordinates because of the form of the function $x^2+y^2$, but I can't set up the integral properly. How can I calculate this integral?

Thanks in advance for your help!

答:
Hint.

Your integral can be written in the form of
$$
\int_{a}^b\int_{0}^{d(y)}f(x,y)dxdy
$$

You usually don't use polar coordinates when the region of the integral cannot be represented nicely by $r=\sqrt{x^2+y^2}$ even though your integrand is a radial function.

When you calculate double integrals on a normal domain, you simply use iterated integrals.


问:
Enumerate partitions of identical objects
I have a problem concerning enumerating partitions of a set of identical objects. I know, now someone is going to talk about Stirling number of second kind, but I'm quite sure this is not the answer.

I will describe the problem with an example. Suppose to have a set of $n$ identical objects. The problem is to enumerate and to generate all the partitions obtained with $k$ subsets, like in the following example.

Let $n=5$, $k=3$. Define $B_i$, with $i=1,\dots,k$ the set of the partition. Hence, I can partition the set of $n$ objects by constructing the following partitions:

$|B_1|=3$, $|B_2|=1$, $|B_3|=1$.

$|B_1|=2$, $|B_2|=2$, $|B_3|=1$.

Notice that: $(i)$ must result $B_i\neq \emptyset$; $(ii)$ partitions like, for instance, $|B_1|=1$, $|B_2|=3$, $|B_3|=1$ or $|B_1|=1$, $|B_2|=1$, $|B_3|=3$ must not be considered, since are in practice already considered in the first partition (i.e. $|B_1|=3$, $|B_2|=1$, $|B_3|=1$).

Since objects are all identical, I only express the sets cardinality and I doesn't care about the specific object allocated to a set of the partition.

The problem is particular useful in the formulation of algorithms for assignment problems.

答:
You want the function $P(n,k)$ described after formula $(58)$ of this MathWorld article, giving the number of partitions of $n$ into exactly $k$ parts. This is always the same as the number of partitions of $n$ with largest part $k$. For this function, unlike the intermediate function in the Wikipedia article, $P(5,3)=2$; you’ve exhibited the two partitions of $5$ into $3$ parts, and the two partitions of $5$ with maximum part $3$ are $3+2$ and $3+1+1$. The link has a nice recurrence for $P(n,k)$, formula $(59)$:

$$P(n,k)=P(n-1,k-1)+P(n-k,k)\;,$$

with initial conditions

$$\begin{align*}
&P(n,k)=0\quad\text{if}\quad k>n\\
&P(n,n)=1\quad\text{for}\quad n\ge 0\\
&P(n,0)=0\quad\text{for}\quad n\ge 0\;.
\end{align*}$$

The recurrence is easily verified. Every partition of $n$ into $k$ parts, at least one of which is $1$, is obtained by adding a $1$ part to a partition of $n-1$ into $k-1$ parts. Every partition of $n$ into $k$ parts, all of which are at least $2$, is obtained from a partition of $n-k$ into $k$ parts by adding $1$ to each part.


问:
Chain Rule in Calculus of Variations
When you take the derivative of the function $F(y+a\eta, y'+a\eta, x)$ with respect to $a$, you first take the derivative of $F$ with respect to $y+a\eta$ then multiply it with the derivative of $y+a\eta$ with respect to $a$, right? But how do you take the derivative of $F$ with respect to $y+a\eta$? We do not know the function exactly. In the book I am studying (Taylor Classical Mechanics), the result is given as η.f'(y)+η.f'(y') </p>

答:
On the previous page in Chapter 6.2, Taylor mentions "we must just check that the derivative dS/d$\alpha$ is zero when $\alpha=0$," where $S=\int_{x_1}^{x_2}f\left[y\left(x\right),\ y'\left(x\right),\ x\right]dx$. So I believe that in addition to applying the chain rule, Taylor evaluates $\frac{\partial f\left(y+\alpha \eta,\ y'+\alpha \eta',\ x\right)}{\partial \alpha}$ at $\alpha=0$. Then $Y(x)=y(x)+\alpha \eta(x)$ gives
$$\frac{\partial f\left(y+\alpha \eta,\ y'+\alpha \eta',\ x\right)}{\partial \alpha}\Big|_{\alpha=0}=\frac{\partial Y}{\partial \alpha} \frac{\partial f}{\partial Y}\Big|_{\alpha=0}+\frac{\partial Y'}{\partial \alpha} \frac{\partial f}{\partial Y'}\Big|_{\alpha=0}=\eta \frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}.$$

Title: Chain Rule in Calculus of Variations


答:
You haven't really tried anything, you've just read your dataset. Nevermind, you can compute $g'(x)$ using the very definition of what a derivative is:

$g'(x) = \lim\limits_{h \to 0} \frac{g(x+h) - g(x)}{h}$

Since, the step size in your dataset is quite small, you'll get a nice approximation by doing: $\frac{g_{i+1} - g_i}{x_{i+1} - x_i}$, where $(x_i, g_i)$ are the pair of numbers on the $i^{th}$ line.

Regarding the integral, you can approximate it by summing the areas of all the rectangles whose height is $g(x_i)$ and width is $x_{i+1} - x_i$. More formally, you need to compute:

$
\sum_{i} g(x_i)(x_{i+1} - x_i)
$

Hope this helps.

Title: How to calculate Integral estimation?

Body:
How can I do the following in python3 on the provided data set listed below?

Problem

Knowing that the data.txt has 2 columns:

xValues, where   ≤  ≤  , with  and  being some constants

gOfXValues

1Compute second order estimates of  ′()

2Compute second order estimates of  $g'(x)$ and $\int_a^b g(x)dx.$

Generally, we don't know that    values given from a random data sample is evenly separated.

3Plot  ()  and  ′() , print the integral.

4Based on the graph, what function do you think  ()  is?

5Verify this by qualitatively comparing the exact derivative and integral of your supposed  ()  with the numerical results obtained previously.

What I have done so far

```<code>import pandas as pd

dataFrame = pd.read_csv('/Users/Files/data.txt', sep="\s+", names=['xValue','gOfXValue'])
dataFrame.info
dataFrame.head()
  for index, row in dataFrame.iterrows():
    print(row['xValue'], row['gOfXValue'])

    xValue = row['xValue']
    gOfXValue = row['gOfXValue']
</code>```

DataSet

```<code>0.000000000000000000e+00 1.000000000000000000e+00
3.157379551346525814e-02 1.031568549764810605e+00
6.314759102693051629e-02 1.063105631312673660e+00
9.472138654039577443e-02 1.094579807794844983e+00
1.262951820538610326e-01 1.125959705067717476e+00
1.578689775673262907e-01 1.157214042967250833e+00
1.894427730807915489e-01 1.188311666489717755e+00
2.210165685942568070e-01 1.219221576847691280e+00
2.525903641077220652e-01 1.249912962370308467e+00
2.841641596211873511e-01 1.280355229217014390e+00
3.157379551346525814e-01 1.310518031874168710e+00
3.473117506481178118e-01 1.340371303404112702e+00
3.788855461615830977e-01 1.369885285416546861e+00
4.104593416750483836e-01 1.399030557732340974e+00
4.420331371885136140e-01 1.427778067710209653e+00
4.736069327019788444e-01 1.456099159207016047e+00
5.051807282154441303e-01 1.483965601142838819e+00
5.367545237289094162e-01 1.511349615642326727e+00
5.683283192423747021e-01 1.538223905724288576e+00
5.999021147558398770e-01 1.564561682511917962e+00
6.314759102693051629e-01 1.590336691936528268e+00
6.630497057827704488e-01 1.615523240908179226e+00
6.946235012962356237e-01 1.640096222927107217e+00
7.261972968097009096e-01 1.664031143110431099e+00
7.577710923231661955e-01 1.687304142609184154e+00
7.893448878366314814e-01 1.709892022391332755e+00
8.209186833500967673e-01 1.731772266367076707e+00
8.524924788635619421e-01 1.752923063833377038e+00
8.840662743770272280e-01 1.773323331215339138e+00
9.156400698904925139e-01 1.792952733082778582e+00
9.472138654039576888e-01 1.811791702421020833e+00
9.787876609174229747e-01 1.829821460135725886e+00
1.010361456430888261e+00 1.847024033772298957e+00
1.041935251944353436e+00 1.863382275431223256e+00
1.073509047457818832e+00 1.878879878861460462e+00
1.105082842971284007e+00 1.893501395714874080e+00
1.136656638484749404e+00 1.907232250945481322e+00
1.168230433998214579e+00 1.920058757338174438e+00
1.199804229511679754e+00 1.931968129152434877e+00
1.231378025025145151e+00 1.942948494867437148e+00
1.262951820538610326e+00 1.952988909015839214e+00
1.294525616052075501e+00 1.962079363094462847e+00
1.326099411565540898e+00 1.970210795540986215e+00
1.357673207079006072e+00 1.977375100766707305e+00
1.389247002592471247e+00 1.983565137236369846e+00
1.420820798105936644e+00 1.988774734587002602e+00
1.452394593619401819e+00 1.992998699778669724e+00
1.483968389132867216e+00 1.996232822271006846e+00
1.515542184646332391e+00 1.998473878220378808e+00
1.547115980159797566e+00 1.999719633693477938e+00
1.578689775673262963e+00 1.999968846894156327e+00
1.610263571186728138e+00 1.999221269401275647e+00
1.641837366700193535e+00 1.997477646416338626e+00
1.673411162213658709e+00 1.994739716020657028e+00
1.704984957727123884e+00 1.991010207442792002e+00
1.736558753240589281e+00 1.986292838338002742e+00
1.768132548754054456e+00 1.980592311082403967e+00
1.799706344267519631e+00 1.973914308085537694e+00
1.831280139780985028e+00 1.966265486126021811e+00
1.862853935294450203e+00 1.957653469715929573e+00
1.894427730807915378e+00 1.948086843500509424e+00
1.926001526321380775e+00 1.937575143700825064e+00
1.957575321834845949e+00 1.926128848607841171e+00
1.989149117348311346e+00 1.913759368137436745e+00
2.020722912861776521e+00 1.900479032456751760e+00
2.052296708375241696e+00 1.886301079693208704e+00
2.083870503888706871e+00 1.871239642738459663e+00
2.115444299402172490e+00 1.855309735160411755e+00
2.147018094915637665e+00 1.838527236237377238e+00
2.178591890429102840e+00 1.820908875129262583e+00
2.210165685942568015e+00 1.802472214201578105e+00
2.241739481456033189e+00 1.783235631518890418e+00
2.273313276969498808e+00 1.763218302525168424e+00
2.304887072482963983e+00 1.742440180929283100e+00
2.336460867996429158e+00 1.720921978814716535e+00
2.368034663509894333e+00 1.698685145993306556e+00
2.399608459023359508e+00 1.675751848623608709e+00
2.431182254536824683e+00 1.652144947115186779e+00
2.462756050050290302e+00 1.627887973340858441e+00
2.494329845563755477e+00 1.603005107179614530e+00
2.525903641077220652e+00 1.577521152413588590e+00
2.557477436590685826e+00 1.551461512003107668e+00
2.589051232104151001e+00 1.524852162764468444e+00
2.620625027617616620e+00 1.497719629475682934e+00
2.652198823131081795e+00 1.470090958436002904e+00
2.683772618644546970e+00 1.441993690505579018e+00
2.715346414158012145e+00 1.413455833652134119e+00
2.746920209671477320e+00 1.384505835032010967e+00
2.778494005184942495e+00 1.355172552633428618e+00
2.810067800698408114e+00 1.325485226510211501e+00
2.841641596211873289e+00 1.295473449634670038e+00
2.873215391725338463e+00 1.265167138398678670e+00
2.904789187238803638e+00 1.234596502792368877e+00
2.936362982752268813e+00 1.203792016290152311e+00
2.967936778265734432e+00 1.172784385474099356e+00
2.999510573779199607e+00 1.141604519424951558e+00
3.031084369292664782e+00 1.110283498911275091e+00
3.062658164806129957e+00 1.078852545407476660e+00
3.094231960319595132e+00 1.047342989971558280e+00
3.125805755833060751e+00 1.015786242013636764e+00
3.157379551346525925e+00 9.842137579863630137e-01
3.188953346859991100e+00 9.526570100284420528e-01
3.220527142373456275e+00 9.211474545925236734e-01
3.252100937886921450e+00 8.897165010887251313e-01
3.283674733400387069e+00 8.583954805750482198e-01
3.315248528913852244e+00 8.272156145259004223e-01
3.346822324427317419e+00 7.962079837098479107e-01
3.378396119940782594e+00 7.654034972076313448e-01
3.409969915454247769e+00 7.348328616013215520e-01
3.441543710967712943e+00 7.045265503653301842e-01
3.473117506481178562e+00 6.745147734897881664e-01
3.504691301994643737e+00 6.448274473665716044e-01
3.536265097508108912e+00 6.154941649679892546e-01
3.567838893021574087e+00 5.865441663478661027e-01
3.599412688535039262e+00 5.580063094944210933e-01
3.630986484048504881e+00 5.299090415639968743e-01
3.662560279561970056e+00 5.022803705243168437e-01
3.694134075075435231e+00 4.751478372355316671e-01
3.725707870588900406e+00 4.485384879968926652e-01
3.757281666102365580e+00 4.224788475864115211e-01
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3.852003052642761549e+00 3.478550528848133316e-01
3.883576848156226724e+00 3.242481513763912915e-01
3.915150643669691899e+00 3.013148540066936665e-01
3.946724439183157074e+00 2.790780211852837978e-01
3.978298234696622693e+00 2.575598190707169000e-01
4.009872030210087424e+00 2.367816974748317982e-01
4.041445825723553043e+00 2.167643684811096927e-01
4.073019621237018661e+00 1.975277857984218954e-01
4.104593416750483392e+00 1.790911248707376391e-01
4.136167212263949011e+00 1.614727637626225398e-01
4.167741007777413742e+00 1.446902648395883562e-01
4.199314803290879361e+00 1.287603572615404479e-01
4.230888598804344980e+00 1.136989203067911847e-01
4.262462394317809711e+00 9.952096754324846195e-02
4.294036189831275330e+00 8.624063186256325508e-02
4.325609985344740060e+00 7.387115139215894022e-02
4.357183780858205679e+00 6.242485629917493561e-02
4.388757576371671298e+00 5.191315649949035382e-02
4.420331371885136029e+00 4.234653028407053821e-02
4.451905167398601648e+00 3.373451387397807810e-02
4.483478962912066379e+00 2.608569191446230562e-02
4.515052758425531998e+00 1.940768891759592218e-02
4.546626553938997617e+00 1.370716166199725805e-02
4.578200349452462348e+00 8.989792557207887391e-03
4.609774144965927967e+00 5.260283979342972316e-03
4.641347940479392697e+00 2.522353583661263166e-03
4.672921735992858316e+00 7.787305987243531291e-04
4.704495531506323047e+00 3.115310584367314561e-05
4.736069327019788666e+00 2.803663065220618478e-04
4.767643122533254285e+00 1.526121779621192331e-03
4.799216918046719016e+00 3.767177728993265085e-03
4.830790713560184635e+00 7.001300221330386542e-03
4.862364509073649366e+00 1.122526541299739833e-02
4.893938304587114985e+00 1.643486276363004261e-02
4.925512100100580604e+00 2.262489923329280561e-02
4.957085895614045334e+00 2.978920445901367398e-02
4.988659691127510953e+00 3.792063690553715283e-02
5.020233486640975684e+00 4.701109098416056398e-02
5.051807282154441303e+00 5.705150513256296296e-02
5.083381077667906922e+00 6.803187084756523451e-02
5.114954873181371653e+00 7.994124266182545124e-02
5.146528668694837272e+00 9.276774905451867781e-02
5.178102464208302003e+00 1.064986042851256975e-01
5.209676259721767622e+00 1.211201211385396492e-01
5.241250055235233241e+00 1.366177245687767439e-01
5.272823850748697971e+00 1.529759662277010435e-01
5.304397646262163590e+00 1.701785398642742253e-01
5.335971441775628321e+00 1.882082975789789447e-01
5.367545237289093940e+00 2.070472669172214175e-01
5.399119032802559559e+00 2.266766687846610839e-01
5.430692828316024290e+00 2.470769361666227404e-01
5.462266623829489909e+00 2.682277336329232931e-01
5.493840419342954640e+00 2.901079776086668005e-01
5.525414214856420259e+00 3.126958573908157346e-01
5.556988010369884989e+00 3.359688568895683458e-01
5.588561805883350608e+00 3.599037770728926722e-01
5.620135601396816227e+00 3.844767590918209965e-01
5.651709396910280958e+00 4.096633080634712876e-01
5.683283192423746577e+00 4.354383174880819274e-01
5.714856987937211308e+00 4.617760942757109799e-01
5.746430783450676927e+00 4.886503843576732731e-01
5.778004578964142546e+00 5.160343988571614027e-01
5.809578374477607277e+00 5.439008407929837308e-01
5.841152169991072896e+00 5.722219322897904581e-01
5.872725965504537626e+00 6.009694422676585823e-01
5.904299761018003245e+00 6.301147145834531393e-01
5.935873556531468864e+00 6.596286965958874093e-01
5.967447352044933595e+00 6.894819681258308464e-01
5.999021147558399214e+00 7.196447707829856100e-01
6.030594943071863945e+00 7.500870376296912001e-01
6.062168738585329564e+00 7.807784231523084983e-01
6.093742534098795183e+00 8.116883335102823560e-01
6.125316329612259914e+00 8.427859570327489447e-01
6.156890125125725532e+00 8.740402949322825243e-01
6.188463920639190263e+00 9.054201922051545726e-01
6.220037716152655882e+00 9.368943686873262289e-01
6.251611511666121501e+00 9.684314502351897280e-01
6.283185307179586232e+00 9.999999999999997780e-01
</code>```


问:
Positive integer solution to $a^3=3b^2-2$
The equation $a^3=3b^2-2$ seem to only have one positive integer solution $(a,b)=(1,1)$, but I am unable to prove that.

What I did:

Google told me that the elliptic curve theory might help me, but I don't really know anything about it.

This thread in MO looks similar to what we have, but I was unable to apply it in my case.

答:
Let $a$ and $b$ be integers such that $a^3=3b^2-2$. First note that $a$ is not divisible by $3$, and that $a$ and $b$ are both odd. In the UFD $\Bbb{Z}[\sqrt{6}]$ we have the factorization
$$3a^3=9b^2-6=(3b+\sqrt{6})(3b-\sqrt{6}),$$
where the gcd of the two factors divides their difference $2\sqrt{6}$ as well as their product $3a^3$. As noted $a$ is odd, and so the two factors are coprime. The prime factorization $3=(3+\sqrt{6})(3-\sqrt{6})$ in $\Bbb{Z}[\sqrt{6}]$ shows that, after changing the sign of $b$ if necessary, we have
$$3b+\sqrt{6}=u(3+\sqrt{6})(c+d\sqrt{6})^3,\tag{1}$$
for some unit $u\in\Bbb{Z}[\sqrt{6}]$ and some integers $c$ and $d$. The unit group of $\Bbb{Z}[\sqrt{6}]$ is generated by $-1$ and $5+2\sqrt{6}$, meaning that $u=\pm(5+2\sqrt{6})^k$ for some integer $k$. Adjusting $c$ and $d$ if necessary, we may assume without loss of generality that $k\in\{-1,0,1\}$ and that we have the $+$-sign. Then we are left with solving
$$3b+\sqrt{6}=(5+2\sqrt{6})^k(3+\sqrt{6})(c+d\sqrt{6})^3,$$
for some integers $c$ and $d$, and some $k\in\{-1,0,1\}$. Expanding the product for each of the three values of $k$ and comparing coefficients of $\sqrt{6}$, we end up with the following three Diophantine equations:
\begin{eqnarray*}
1&=&-c^3+9c^2d-18cd^2+18d^3\\
1&=&c^3+9c^2d+18cd^2+18d^3\\
1&=&11c^3+81c^2d+198cd^2+162d^3\\
\end{eqnarray*}
These are three cubic Thue equations, for which there exist effective methods for finding all integral solutions. The first has the unique solution $(c,d)=(-1,0)$, the second has the unique solution $(c,d)=(-1,0)$, and the third has no integral solutions. These two solutions correspond to $b=-1$ and $b=1$, respectively, and this shows that these are the only integral solutions.


问:
Change from differentiation wrt to matrix to wrt to inverse of matrix for symmetric matrices
For the rule below:

$$
\frac{\partial J}{\partial \mathbf{A}}= -\mathbf{A}^{-T} \frac{\partial J}{\partial \mathbf{W}} \mathbf{A}^{-T}
$$

where $\mathbf{A}$ is an invertible square matrix, $\mathbf{W}$ is the inverse of $\mathbf{A}$, and J is a function (see end of section 2.2 in matrix cookbook https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf)

Does this rule hold if $\mathbf{A}$ is a symmetric matrix?

答:
$
\def\b{\bullet}
\def\e{\varepsilon}
\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}
\def\p#1#2{\frac{\partial #1}{\partial #2}}
$I
really like The Matrix Cookbook but the section on structured matrices is not very good, so here's a different approach to the subject.

Given a vector of parameters $\{p\}$ and matrix basis $\{B_i\}$
$$\eqalign{
p &= \m{\alpha \\ \beta},\qquad
B_1 = \m{1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0},\qquad
B_2 = \m{0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0}
\\
}$$
create a structured matrix $\{A\}$ and cost function $\{\phi\}$
$$\eqalign{
A &= \sum_{i=1}^2\;p_iB_i
 \;=\; \m{\alpha & \alpha & 0 & 0 \\ 0 & \beta & 0 & 0 \\ 0 & \beta & \beta & 0},\qquad
&\phi = \tfrac 12\Big\|AX-Y\Big\|_F^2 \\
}$$
Note that $(\alpha,\beta)$ are the only independent variables in the entire problem.

When $A$ is unconstrained it's easy to calculate the gradient/differential of the cost
$$\eqalign{
G = \p{\phi}{A} = (AX-Y)X^T \quad\implies\quad d\phi = G\b dA \\
}$$
where the bullet denotes the matrix inner product, i.e.
$$\eqalign{
G\b dA
 &= \sum_{i=1}^3\sum_{j=1}^4 G_{ij}\;dA_{ij} \;=\; {\rm Tr}(G^TdA) \\
}$$
Because of the structure which was imposed on $A$,
its differential is also structured
$$dA = \sum_{i=1}^2 B_i\,dp_i$$
Substituting this expression leads to the parametric gradient
$$\eqalign{
d\phi &= \sum_{i=1}^2\;G\b(B_i\,dp_i)
 = \sum_{i=1}^2\left(\p{\phi}{p_i}\right)dp_i \quad\implies\quad
\p{\phi}{p_i} = G\b B_i \\
}$$
At this point, one would do all further calculations
in terms of the $p$-vector.

Now comes the weird part...

Every basis $\{B_i\}$ has a dual basis $\{B_i^\delta\}$ which spans the same subspace $\cal S$, but is orthonormal with respect to the inner product
$$B_i\b B_j^\delta \;=\; \delta_{ij}$$
Some bases are self-dual, such as the canonical vector basis
$\{\e_i\}$, but in general determining the dual basis requires a pseudoinverse calculation
$$\eqalign{
&\;b_k = {\rm vec}(B_k) \qquad &\;b_k^\delta = {\rm vec}(B_k^\delta) \\
&\m{b_1 & b_2 &\ldots & b_p}^+ = &\m{b_1^\delta & b_2^\delta &\ldots & b_p^\delta}^T \\
}$$
In the vector case, the gradient with respect to the $p$-vector can be written as the sum of each component multiplied by the corresponding vector from the dual basis, i.e.
$$\eqalign{
\p{\phi}{p}
  &= \sum_{i=1}^2 \left(\p{\phi}{p_i}\right)\e_i \\
}$$
Many authors extend this idea and define the structured gradient
as the matrix
$$\eqalign{
\left(\p{\phi}{A}\right)_S
 &= \sum_{i=1}^2\left( \p{\phi}{p_i} \right) B_i^\delta \\
 &= \sum_{i=1}^2\left(G\b B_i\right) B_i^\delta \\
 &= G\b\left(\sum_{i=1}^2 B_i B_i^\delta \right) \\
 &= G\b{\cal B} \\
}$$
where $\cal B$ is a fourth-order tensor with components
$${\cal B}_{jk\ell m}
 = \sum_{i=1}^2\;\left(B_i\right)_{jk}\,\left(B_i^\delta\right)_{\ell m} \\$$
The $\cal B$ tensor is a projector into the subspace
$\big(\,{\cal B}\b X\in{\cal S}\;\;{\rm for}\;X\in{\mathbb R}^{3\times 4}\big)$
where it also acts as an identity tensor for the subspace
$\big({\cal B}\b M=M\b{\cal B} = M\;\;{\rm for}\;M\in{\cal S}\big)$ .

If the basis spans the whole space $\,{\cal S}\equiv{\mathbb R}^{3\times 4}\,$ then $\cal B$ becomes the true identity tensor $\cal I$, and the structured gradient is identical to the full unstructured gradient $G$
(as expected).
$$\eqalign{
{\cal B}_{jk\ell m} \;&\to\; {\cal I}_{jk\ell m}
 = \delta_{j\ell}\delta_{km} \\
(G\b{\cal B}) \;&\to\; (G\b{\cal I}) = G \\
}$$

As a concrete example, let's examine a symmetrically constrained $2\times 2$ matrix.
$$\eqalign{
p &= \m{\alpha \\ \beta \\ \lambda},\qquad
B_1 = \m{1 & 0 \\ 0 & 0},\qquad
B_2 = \m{0 & 0 \\ 0 & 1},\qquad
B_3 = \m{0 & 1 \\ 1 & 0} \\
A &= \m{\alpha & \lambda \\ \lambda & \beta}
 \quad=\quad \alpha B_1 + \beta B_2 + \lambda B_3,\qquad
B_k^\delta = \frac{B_k}{B_k\b B_k} \\
}$$
The structured gradient calculation then goes as follows
$$\eqalign{
\left(\p{\phi}{A}\right)_S
 &= \frac{(G\b B_1)B_1}{B_1\b B_1}
  + \frac{(G\b B_2)B_2}{B_2\b B_2}
  + \frac{(G\b B_3)B_3}{B_3\b B_3} \\
 &= G_{11}\,B_1 +G_{22}\,B_2 +\tfrac 12(G_{12}+G_{21})\,B_3 \\
 &= \m{G_{11} & \tfrac 12(G_{12}+G_{21}) \\ \tfrac 12(G_{12}+G_{21}) & G_{22}} \\
 &= \left(\frac{G+G^T}{2}\right) \;\doteq\; {\rm Sym}(G) \\
}$$
But The Matrix Cookbook uses the regular basis instead of the dual basis
which results in the following miscalculation
$$\eqalign{
\left(\p{\phi}{A}\right)_{S^*}
 &= \left(G\b B_1\right)B_1
  + \left(G\b B_2\right)B_2
  + \left(G\b B_3\right)B_3 \\
 &= G_{11}\,B_1 +G_{22}\,B_2 +(G_{12}+G_{21})\,B_3 \\
 &= \m{G_{11} & (G_{12}+G_{21}) \\ (G_{12}+G_{21}) & G_{22}} \\
 &= G+G^T-{\rm Diag}(G) \\
}$$

The skew-symmetric case is similar but is seldom mentioned.

There is only one parameter and one matrix in the basis
$$\eqalign{
p &= \m{\alpha},\qquad B = \m{0 & 1 \\ -1 & 0},\qquad
  B^\delta = \frac{B}{B\b B} \\
A &= \m{0 & \alpha \\ -\alpha & 0} \;\;=\;\; \alpha B \\
}$$
and the structured gradient is
$$\eqalign{
\left(\p{\phi}{A}\right)_S
 &= \frac{(G\b B)B}{B\b B} \\
 &= \tfrac 12(G_{12}-G_{21})\,B \\
 &= \m{0 & \tfrac 12(G_{12}-G_{21}) \\ \tfrac 12(G_{21}-G_{12}) & 0}  \\
 &= \left(\frac{G-G^T}{2}\right) \;\doteq\; {\rm Skew}(G) \\
}$$
If you use $B$ instead of $B^\delta$ in this case, the gradient has the right direction but the wrong length, i.e.
$$\eqalign{
\left(\p{\phi}{A}\right)_{S^*}
 &= \left(G-G^T\right) \;=\; 2\;{\rm Skew}(G) \\
}$$


问:
Does ``Converge to'' and ``Strict Equality'' Always Mean the Same Thing? If Not, is This a Paradox?
Consider the geometric series with a = 1 and r = 1/10.

Then, we have

$$
\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = \sum_{n=0}^{\infty}9\left(\frac{1}{10}\right)^{n} - 9 = 9\left[ \lim_{k\to\infty}\left(\frac{ar^{k}}{r - 1}\right) + \frac{a}{1-r}  - 1 \right] = 9\left[ \lim_{k\to\infty}\left(\frac{1\cdot(1/10)^{k}}{1/10 - 1}\right) + \frac{1}{1-1/10}  - 1 \right] = 9\left(0 + \frac{10}{9} - 1 \right) = 9 \left( \frac{1}{9} \right) = 1.
$$

Hence, we say that the geometric series with a = 1 and r = 1/10 converges to 1.

Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to 2. I conclude such because the limit of the quotient after the second equals sign never achieves its limit of 0.

Furthermore, there are many generally accepted ways to show that $0.\overline{9} = 1$; that is identically equal to 1.

One conventional way, of course, is to let $x=0.\overline{9}$ and subtract it from $10x$ giving $9x = 9$, and thus $x = 1$. And I've heard nobody complain about that---even though we must subtract an infinite number of terms in the exact same manner as we do a finite number of terms.

But, that means

$$
\underbrace{1 = 0.\overline{9} = \sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n}}_{\text{exactly equal to}} = \overbrace{\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = 1}^{\text{arbitrarily close to}}.
$$

Doesn't it?

It seems to me that the only way the above equalities can hold is if converges to always means exactly equal to.

Otherwise, do we have a paradox here?

Please explain.

Thank you.

答:
Your doubts about: Does Converge to and Strict Equality Always Mean the Same Thing? might look quite reasonable at first sight.

When we say something like: Let $(a_n)_{n\geq 0}$ be a sequence of real numbers and  $a\in\mathbb{R}$ so that
\begin{align*}
\lim_{n\to \infty}a_n=a\tag{1}
\end{align*}
Is it really so that the limiting value represented by the left-hand side of (1) is equal to $a$, so that the equality sign $=$ is justified?

We know from definition of the limit of a convergent sequence: For each $\varepsilon>0$ there is an index $N\in\mathbb{N}$, so that for all $n>N$ we have
\begin{align*}
|a_n-a|<\varepsilon \tag{2}
\end{align*}
and from that we conclude equality. But this is not obvious. In fact we need to recall an axiom in order to do so.

Here we use behind the curtain the axiom of trichotomy which states that each real number is either greater than, or equal to or less than zero.

So, in fact we axiomatically cope with the situation. However small we choose $\varepsilon>0$ we can find an index $N\in\mathbb{N}$ so that $|a_n-a|<\varepsilon$ for all $n>N$. But this means that the limiting value is neither greater than nor smaller than $a$. Thanks to the axiom of trichotomy we can conclude that equality holds.


问:
Non-uniqueness of MLE of multivariate Laplace distribution?
Suppose $\{X\}_{i=1}^n\overset{i.i.d}{\sim}X$, and $X\in\mathbb{R}^d$ has density,$$f_{\theta}(x)=c\exp\left\{-||x-\theta||\right\},\theta\in\mathbb{R}^d,$$
where $||\cdot||$ denotes the Euclidean norm.

Show that the MLE $\hat{\theta}$ exists but is not unique when $n$ is even.

I know how to prove that the MLE $\hat{\theta}$ exists, by noticing $\underset{\theta\rightarrow\partial\mathbb{R}^d}{\lim} \log f(\theta)=-\infty$, however I don't know how to prove it is not unique. I understand that the log-likelihood function, $l(\theta)=-\sum_{i=1}^n||x_i-\theta||$ is a convex, but not strictly convex, function, but this does not guarantee the nonuniqueness. I also tried taking the first and second gradient of $l(\theta)$. But even I have a non-negative definite $l(\theta)$, I still don't have the non-uniqueness...

---Update---

The original question is here. This is from Mathematical Statistics (Bickel).

答:
The minimizer of such set of points is given by the Geometric Median.

For example, for $ d = 1 $ you get the Median which is not unique for an odd set of different numbers.

For higher dimensions you need to take care of the case the points are collinear, which basically means that the problem, is again, equivalent to 1D.


问:
Calculating divergence for parabolic coordinates.
I'm trying to calculate divergence for parabolic coordinates with a goal to check if i understand all notions of vector analysis correctly.

Definition of parabolic coordinates:

$$x = uv \\ y = \frac{1}{2}(u^2 - v^2)$$

Firstly, I calculate $du$ and $dv$:
$$
dx = v du + udv \\
dy = udu-vdv
$$

$$
dv = \frac{u}{u^2+v^2}dx - \frac{v}{u^2+v^2}dy\\
dv = \frac{v}{u^2+v^2}dx + \frac{u}{u^2+v^2}dy
$$
Then I calculate $e_u$ and $e_v$ - basis vectors for tangent plane in new variables:
$$
x_u = v \\
y_u = u \\
e_u = \frac{(x_u,y_u)}{||(x_u,y_u)||} = (\frac{v}{\sqrt{u^2+v^2}}, \frac{u}{\sqrt{u^2+v^2}})
$$

$$
x_v = u \\
y_v = -v \\
e_v = \frac{(x_v,y_v)}{||(x_v,y_v)||} = (\frac{u}{\sqrt{u^2+v^2}}, -\frac{v}{\sqrt{u^2+v^2}})
$$

Calculating $du(e_u)$ and $dv(e_v)$:
$$
du(e_u) = \frac{u^2+v^2}{\sqrt{u^2+v^2}(u^2+v^2)} = \frac{1}{\sqrt{u^2+v^2}} (=dv(e_v))
$$
Noting that $du(e_v) = dv(e_u) = 0$, let $(F, H)$ be a vector field.

Now I need to calculate area form:
$$dx\wedge dy = (u^2 + v^2)du\wedge dv$$
And take interiour product with my field:
$$(u^2 + v^2)((du)(F,H) dv - dv(F,H) du) \\
= (u^2 + v^2)(\frac{F}{\sqrt{u^2+v^2}}dv - \frac{H}{\sqrt{u^2+v^2}}du) \\
= \sqrt{u^2+v^2}(Fdv-Hdu)$$
Then I apply differential on top of it to get my divergence as coefficient:
$$
d(\sqrt{u^2+v^2}(Fdv-Hdu)) \\
= (\frac{u}{\sqrt{u^2+v^2}}du + \frac{v}{\sqrt{u^2+v^2}}dv)\wedge(Fdv-Hdu) + \sqrt{u^2+v^2}(F_u du\wedge dv - H_v dv \wedge du) \\
= \frac{1}{\sqrt{u^2+v^2}}(uFdu \wedge dv - vH dv \wedge du) + \sqrt{u^2+v^2}(F_u + H_v) du \wedge dv \\
= (\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)) du \wedge dv
$$
So divergence of $(F, H)$ should be of form
$$
\frac{uF + vH}{\sqrt{u^2+v^2}} + \sqrt{u^2+v^2}(F_u + H_v)
$$
But there I can see that another $\frac{1}{u^2 + v^2}$ was lost somewhere. But I can't find where it was lost, so I'm asking for help to find it.

答:
Thanks for clarifying. OK, so the vector field is dual to the $1$-form $\omega = F\omega_1 + H\omega_2$, where $\omega_1 = \sqrt{u^2+v^2}\,du$ and $\omega_2=\sqrt{u^2+v^2}\,dv$ give an orthonormal coframe. [Note that the metric $dx^2 + dy^2 = \omega_1^2 + \omega_2^2$.]

Of course we have $\star\omega_1 = \omega_2$ and $\star\omega_2=-\omega_1$. So
\begin{align*}
d(\star\omega) &= d(F\omega_2) - d(H\omega_1) = d\big(F\sqrt{u^2+v^2}dv\big) - d\big(H\sqrt{u^2+v^2}du\big)\\
&= \left(\sqrt{u^2+v^2}(F_u+H_v) + F\frac u{\sqrt{u^2+v^2}}+H\frac v{\sqrt{u^2+v^2}}\right)du\wedge dv \\
&= \left(\frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}\right)\omega_1\wedge\omega_2.
\end{align*}
Finally, using $\star(\omega_1\wedge\omega_2) = 1$, we have
$$\text{div} (F,H) = \frac{F_u+H_v}{\sqrt{u^2+v^2}} + \frac{uF+vH}{(u^2+v^2)^{3/2}}.$$


问:
Finding seven harmonic cubics
I am given an exercise of my thesis which is about "Computational Algeberaic Geometry", but I dont have enough knowldege to do it or even think about it. The exercixe is:

A ternary cubic form is a homogeneous degree 3 polynomials in three variables and the space of ternary cubic has dimension 10. I need to find a basis of seven harmonic cubics which ones killed by the Laplace operator. In this way, we can also compute the dimension of the space of harmonic homogeneous polynomials of degree $d$ in $n+1$ variables, for each of the seven cubics $f$, we should find the seven eigenvectors, that is the vectors $v \in  \mathbb{C}^3$, up to scalar, such that the gradient $(f_x, f_y, f_z)$ is proportional to $(x,y,z)$.

In some ways it is related to $3$-Veronese Embedding of $\mathbb{P}^2$. Any help, hint and guidance are greatly appreciated.

答:
Let $\mathbb{P}_k^{9}$ parametrise the space of ternary cubics over $k$, where the $i^{th}$ coordinate gives the coefficient of the $i^{th}$ monomial in, say, lexigraphical order. The Laplacian is a linear operator which takes the space of ternary cubics to the space of homogeneous linear forms on $k$. That is, we may view $\nabla^2$ a linear map
$$\nabla^2 : \mathbb{P}_k^{9} \to \mathbb{P}_k^{2}$$
on the parameter spaces. Say given by multiplying by a matrix $A \in \operatorname{Mat}_{3 \times 10}(k)$ (defined up to scalar multiplication by a unit).

The kernel of this matrix gives corresponds to the linear subspace of $\mathbb{P}_k^{9}$ parametrising ternary cubics vanishing under $\nabla^2$. Given the coefficient space it is obvious how to write down the original cubics.

If you are allowing rings it should be clear how to generalise this, as with higher degree polynomials. The above is also more or less an algorithm for computing this space too.

Title: Finding seven harmonic cubics


问:
Why is the characteristic function of the sign function $2i/t$?
A book I'm reading says:

Let $\theta(x)$ be the Heaviside function... Since $\operatorname{sgn}(x)=2\theta(x)-1$, its characteristic function will be:
$$\chi_{\operatorname{sgn}(x)}(t)=\frac{2i}{t}$$

I understand why $\operatorname{sgn}(x)=2\theta(x)-1$, but I don't follow the claim about the characteristic function. If the CDF of $X$ is the Heaviside function, then $X$ is a.s. zero, and the sign of $X$ is a.s. zero, and the characteristic of a random variable that is a.s. zero is $1$. How does $2i/t$ come into it?

答:
The word characteristic function is misleading.

Here $\theta(x)$ is not the CDF of a random variable, we are looking at the Fourier transform of $\theta(x)$ not of $d\theta(x)=\delta(x)dx$.

This Fourier transform exists only in the sense of distributions, it will be $PV(i/t)+\pi \delta(t)$ ($PV$ for principal value).


问:
Question on colimit over the category of simplices
I get slightly confused on the colimit definition:

$$X \cong \lim_{\Delta^n \rightarrow X} \Delta^n $$ where the symbol $\lim$ is actually for denoting colimit.

I know that colimit is defined for a diagram $F: J \rightarrow C$ as the initial object of the category $\text{CoCone}(F)$. What is the index category $J$ for the above-mentioned case?

In the category of simplices, $\Delta \downarrow X$, the objects are given by natural transformations $\Delta^n \rightarrow X$ or $\Delta^m \rightarrow X$ and the morphisms  by natural transformations such as $\Delta^n \rightarrow \Delta^m$ induced by $[n] \rightarrow [m]$. This should serve as a CoCone(?) of something. Is this way of thinking correct?

答:
You are entirely that the index category is the category of simplicies (Grothendiecks category of elements is an alternative name) of $X$.

I think you have a misconception in what the data is of a colimit depends on. Here we are missusing notation. A colimit depends on both the index category and functor.

Now the category of simplicies $Elt(X)$ has forgetful functor $U:Elt(X)\to \mathrm{Set}_{\Delta}$ which is given on objects by $$
(\Delta^n \to X) \mapsto \Delta^n.
$$

Now lets explore what a cocone for this functor is. By definition a cocone on $U$ is an object $Y$ and natural transformation $U\to \delta(Y)$, where $\delta$ is the functor $\mathrm{Set}_\Delta\to \mathrm{Fun}(Elt(X),\mathrm{Set}_\Delta)$. Using the Yoneda this can be seen to be the same as an $n$-simplex of $Y$ for all $n$-simplicies of $X$. Satisfying some naturality condition, which heuristically says that this collection defines a simplicial subset of $Y$.

Now it is clear that $X$ defines a cocone for the forgetful functor. So suppose $(Y,\alpha)$ is another cocone for $U$. We want to build map $X\to Y$ making everything commute. We then basicly build this by $x \in X_n$ maps to $\alpha_{x}$ seen as a $n$-simplex in $Y$ by the Yoneda lemma. I leave it to you to show that this is welldefined.


问:
In how many different ways can these fruits be eaten in succession considering only the type of fruit?
A basket contains 3 mangoes, 2 papayas and 2 kiwis. In how many different ways can these fruits be eaten in succession considering only the type of fruit?

For me the answer is $3! C^7_3C^4_2C^2_2 = 1260$, but the answer seems $7\times6 \times 5=210$.

Can you explain to me where have I done the error?

答:
Note that we are "considering only the type of fruit".

The answer here is basically the number of arrangements of MMMPPKK. This is a multinomial coefficient,
$$\binom{7}{3,2,2} = \frac{7!}{3!\;2!\;2!} = \frac{7\cdot 6\cdot 5\cdot 4}{2\cdot 2} = 7\cdot 6\cdot 5$$


问:
Convergence of: $\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$
Convergence of:

$$\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$$

Tried ratio test but somehow it doesn't work for me.

Body:
For this series,

$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}(n+1)^{n+1}}{(n+1)!^2}\cdot\frac{n!^2}{3^{n}n^{n}} = \frac{3(n+1)(1+1/n)^n}{(n+1)^2}<\frac{3e}{(n+1)}\to 0.$$

By the ratio test, this series converges (with gusto).


答: Integrating by parts viz. Riemann–Stieltjes,$$\int_{\Bbb R}e^{itx}\operatorname{sgn}x\mathrm dx=-\frac{1}{it}\int_{\Bbb R}e^{itx}2\delta(x)\mathrm dx=\frac{2i}{t},$$but this still requires the question to intend convergence in a distributional sense.

Title: Why is the characteristic function of the sign function $2i/t$?

Body:
A book I'm reading says:

Let $\theta(x)$ be the Heaviside function... Since $\operatorname{sgn}(x)=2\theta(x)-1$, its characteristic function will be:
$$\chi_{\operatorname{sgn}(x)}(t)=\frac{2i}{t}$$

I understand why $\operatorname{sgn}(x)=2\theta(x)-1$, but I don't follow the claim about the characteristic function. If the CDF of $X$ is the Heaviside function, then $X$ is a.s. zero, and the sign of $X$ is a.s. zero, and the characteristic of a random variable that is a.s. zero is $1$. How does $2i/t$ come into it?


问:
Non Countable set - All points are boundary points
I am looking for some examples on R for non countable set that all points are boundary points.

Definition for boundary point (Bartle book): x is a boundary point of A if every neighborhood of x contains points in A and points in Ac

One example that I can intuitively think of are the irrational numbers, but I am not really sure how to formaly prove this.

Thank you

答:
HINT

For the Irrationals, take into account that between any to different real numbers always lies a rational and an irrational number.


答:
You could prove a general result:

If $A\subseteq\Bbb R$, every point of $A$ is a boundary point of $A$ if and only if $\Bbb R\setminus A$ is dense in $\Bbb R$.

Once you have that, you just need to find uncountable subsets of $\Bbb R$ whose complements are dense in $\Bbb R$. The irrationals are one such set. As noted in the comments, the middle-thirds Cantor set is another. If $a,b\in\Bbb R$ and $a<b$, $(a,b)\setminus\Bbb Q$ is yet another.

Title: Non Countable set - All points are boundary points

Body:
I am looking for some examples on R for non countable set that all points are boundary points.

Definition for boundary point (Bartle book): x is a boundary point of A if every neighborhood of x contains points in A and points in Ac

One example that I can intuitively think of are the irrational numbers, but I am not really sure how to formaly prove this.

Thank you


答:
Here is a way using Taylor series. First, let $f(x) = sin(\pi x)$.

Around a number $a$, a function $f$ can be approximated by:

$
\begin{aligned}
f(x) = \sum_n \frac{f^{(n)}(a)}{n!}(x-a)^n
\end{aligned}
$

In our case, around $a=1$ (since $x \to 1^-$), and stopping the expansion at order 1 we get:

$
\begin{aligned}
f(x) &= \sum_n \frac{f^{(n)}(1)}{n!}(x-1)^n \\
&\approx f(1) + f'(1)(x-1)
\end{aligned}
$

Moreover,

$
\begin{aligned}
&f(1) = sin(\pi) = 0 \\
&f'(1) = \pi cos(\pi) = - \pi
\end{aligned}
$

Therefore,

$
\begin{aligned}
f(x) = sin(\pi x) \approx \pi (1 - x)
\end{aligned}
$

You can verify this is true for a value of $x$ close to 1.

e.g. $sin(\pi * 1.001) \approx \pi (1 - 1.001)$

Finally,

$
\begin{aligned}
\lim\limits_{x \to 1^-} (1-x)^{sin(\pi x)} &\approx \lim\limits_{x \to 1^-} (1-x)^{\pi (1 - x)} \\
&= \lim\limits_{x \to 1^-} \big((1-x)^{(1 - x)}\big)^\pi \\
&= \lim\limits_{x \to 1^-} \big(e^{(1-x)ln(1 - x)}\big)^\pi \\
&= 1^\pi \\
&= 1
\end{aligned}
$

Title: Hey, how do you calculate this the easiest (not necessarily quickest) way? $\lim_{ x\to1-} (1-x)^{\sin (\pi x)}$

Body: Please help me, I have no idea and exams are coming soon :( what i mean</p>


问:
How do we find finite difference schemes to numerically solve this B.V.P?

\begin{equation}
\frac{\partial P}{\partial t}=\frac{\partial}{\partial x}\left[D(x)\frac{\partial P}{\partial x}\right]-v\frac{\partial P}{\partial x}-\tau P+S
\end{equation}
and
\begin{equation}
\frac{\partial Q}{\partial t}=\frac{\partial}{\partial x}\left[D(x)\frac{\partial Q}{\partial x}\right]-v\frac{\partial Q}{\partial x}-\tau P,
\end{equation}
where $t$ is time, $x$ is the spatial variable along the column, $D(x)$ is the diffusion rate, $v$ is the advection rate, $\tau$ is the decay rate, and $S(x)$ is the source rate. We will be looking for the late-time, steady state solution to these equations with $\partial P/\partial t=\partial Q/\partial t = 0$. The diffusion rate and source rate may be assumed to be smooth and non-negative. We will also assume $v$ is non-negative.

The $x$ domain is $x\in[0,L]$. Boundary conditions for $P$ and $Q$ are mixed at $x=0$, representing zero influx
\begin{equation}
\left.\left(D\frac{\partial P}{\partial x}-vP\right)\right|_{x=0}=\left.\left(D\frac{\partial Q}{\partial x}-vQ\right)\right|_{x=0}=0.
\end{equation}
At $x=L$ the boundary condition will be $P=Q=c$, where $c$ is a constant.

The question is we wish to find a finite difference scheme (break up the domain into $N$ gridpoints ranging from $0$ to $N-1$ and form index equations for each point) in order to create matrices such as $A$ where $AP = S$ then numerically solve this equation. I'm wondering which schemes would be most useful. I have thought of using centred/forward/backward finite difference scheme but I'm very confused about how to input the boundary conditions and which to use?

We require a tridiagonal matrix A. e.g as produced in this example. values of D(x) and S(x) are provided at all points on the index grid.

https://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node9.html

答:
1) Premise

Consider the function $P(x,t)$ to correspond to the matrix
$$
P(x,t) \Rightarrow {\bf P}  = \left( {\matrix{
   {P(0,0)} & {P(0,\Delta t)} & {P(0,2\Delta t)} &  \cdots   \cr
   {P(\Delta x,0)} & {P(\Delta x,\Delta t)} & {P(\Delta x,2\Delta t)} &  \cdots   \cr
   {P(2\Delta x,0)} & {P(2\Delta x,\Delta t)} & {P(2\Delta x,2\Delta t)} &  \cdots   \cr
    \vdots  &  \vdots  &  \vdots  &  \ddots   \cr
 } } \right)
$$
where the starting point has been taken to be $(0,0)$, but could of course be shifted.

Then consider ${\bf E}, \overline {\bf E}$ to be the shift matrices
$$
{\bf E} = \left( {\matrix{   0 & 0 & 0 &  \cdots   \cr   1 & 0 & 0 &  \cdots   \cr
   0 & 1 & 0 &  \cdots   \cr    \vdots  &  \vdots  &  \ddots  &  \ddots   \cr } } \right)\quad
 \overline {\bf E}  = \left( {\matrix{ 0 & 1 & 0 &  \cdots   \cr  0 & 0 & 1 &  \cdots   \cr
   0 & 0 & 0 &  \ddots   \cr  \vdots  &  \vdots  &  \vdots  &  \ddots   \cr
 } } \right)
$$
(we denote the transpose with an overline)

So we will have
$$
\left\{ \matrix{
  {\partial  \over {\partial x}}P(x,t) \Rightarrow {{P_{\,n,m}  - P_{\,n - 1,m} \,} \over {\Delta x}}
 = {1 \over {\Delta x}}\nabla _x P(x,t) = {1 \over {\Delta x}}\left( {{\bf I} - {\bf E}} \right){\bf P} \hfill \cr
  {\partial  \over {\partial x}}P(x,t) \Rightarrow {{P_{\,n + 1,m}  - P_{\,n,m} \,} \over {\Delta x}}
 = {1 \over {\Delta x}}\Delta _x P(x,t) = {1 \over {\Delta x}}\left( {\overline {\bf E}  - {\bf I}} \right){\bf P} \hfill \cr
  {\partial  \over {\partial t}}P(x,t) \Rightarrow {1 \over {\Delta t}}\nabla _t P(x,t)
 = {1 \over {\Delta t}}{\bf P}\left( {{\bf I} - \overline {\bf E} } \right) \hfill \cr
  {\partial  \over {\partial t}}P(x,t) \Rightarrow {1 \over {\Delta t}}\Delta _t P(x,t)
 = {1 \over {\Delta t}}{\bf P}\left( {{\bf E} - {\bf I}} \right) \hfill \cr}  \right.
$$

It is to be underlined that, with matrices of finite dimension ($[0,h]^2$) we get
$$
\left( {{\bf I} - {\bf E}} \right){\bf P}
 = \left( {\matrix{   {P_{\,0,\,0}  - 0} & {P_{\,0,\,1}  - 0} &  \cdots  & {P_{\,0,\,h}  - 0}  \cr
   {P_{\,1,\,0}  - P_{\,0,\,0} } & {P_{\,1,\,1}  - P_{\,0,\,1} } &  \cdots  & {P_{\,1,\,h}  - P_{\,0,\,h} }  \cr
    \vdots  &  \vdots  &  \ddots  &  \vdots   \cr
   {P_{\,h,\,0}  - P_{\,h - 1,\,0} } & {P_{\,h,\,1}  - P_{\,h - 1,\,1} } &  \cdots  & {P_{\,h,\,h}  - P_{\,h - 1,\,h} }  \cr
 } } \right) \Rightarrow {\partial  \over {\partial x}} H(x)P(x,t)
$$
meaning that it actually reproduces the Backward difference of $P(x,t)$ with $P(x,t)=0 \; |x<0$, i.e. for $P(x,t)$ multiplied by the
Heaviside step function.

While the Forward difference is suited to  $P(x,t)$ cropped for $h < x$
$$
\left( {\overline {\bf E}  - {\bf I}} \right){\bf P}
 = \left( {\matrix{   {P_{\,1,\,0}  - P_{\,0,\,0} } & {P_{\,1,\,1}  - P_{\,0,\,1} } &  \cdots
  & {P_{\,1,\,h}  - P_{\,0,\,h} }  \cr   {P_{\,2,\,0}  - P_{\,1,\,0} } & {P_{\,2,\,1}  - P_{\,1,\,1} }
 &  \cdots  & {P_{\,2,h}  - P_{\,1,\,h} }  \cr     \vdots  &  \vdots  &  \ddots
  &  \vdots   \cr   {0 - P_{\,h,\,0} } & {0 - P_{\,h,\,1} } &  \cdots  & {0 - P_{\,h,\,h} }  \cr
 } } \right) \Rightarrow{\partial  \over {\partial x}} \left( {1 - H(x - h - 1)} \right)P(x,t)
$$
and analogously for the differences in $t$.

The Central difference expressed as
$$
\eqalign{
  & \mathop \diamondsuit \nolimits_x P(x,t) = {{P_{\,n + 1,m}  - P_{\,n - 1,m} \,} \over {2\Delta x}}
 = {1 \over 2}\left( {\nabla _x  + \Delta _x } \right)P(x,t)\; \Rightarrow   \cr
  &  \Rightarrow \;{1 \over {2\Delta x}}\left( {\overline {\bf E}  - {\bf E}} \right){\bf P}
 = {1 \over {2\Delta x}}\left( {\matrix{ {P_{\,1,\,0}  - 0} & {P_{\,1,\,1}  - 0} &  \cdots  & {P_{\,1,\,h}  - 0}  \cr
   {P_{\,2,\,0}  - P_{\,0,\,0} } & {P_{\,2,\,1}  - P_{\,0,\,1} } &  \cdots  & {P_{\,2,h}  - P_{\,0,\,h} }  \cr
    \vdots  &  \vdots  &  \ddots  &  \vdots   \cr
   {0 - P_{\,h - 1,\,0} } & {0 - P_{\,h - 1,\,1} } &  \cdots  & {0 - P_{\,h - 1,\,h} }  \cr  } } \right)\; \cr}
$$
uses the matrix $\overline {\bf E}  - {\bf E}$, which however is not invertible, and is suitable for a "boxed" $P(x,t)$ .

I am not going to list and discuss here the other formulations ("stencils") that may be adopted, and the important aspects of stability and convergence which are well
summarized in these three Wikipedia articles:Finite diff., FD method, FD coefficients.

I will focus on the matrix formulation.

2) Example

Before coming to your specific equation let's analyze the simpler case of the [Heat equation]
$$
\left\{ \matrix{
  {\partial  \over {\partial t}}T(x,t) = {{\partial ^{\,2} } \over {\partial x^{\,2} }}T(x,t) \hfill \cr
  T(0,t) = T(h\Delta x,t) = 0 \hfill \cr
  T(x,0) = T_{\,0} (x) \hfill \cr}  \right.
$$

Solution by the Explicit Method

Using the so called FTCS scheme, we put
$$
\left\{ \matrix{
  {\partial  \over {\partial t}}T(x,t) \Rightarrow {1 \over {\Delta t}}\Delta _t T(x,t)
 = {1 \over {\Delta t}}{\bf T}\left( {{\bf E} - {\bf I}} \right) \hfill \cr
  {{\partial ^{\,2} } \over {\partial x^{\,2} }}T(x,t) \Rightarrow
 {{T_{\,n + 1,m}  - 2T_{\,n,m}  + T_{\,n - 1,m} \,} \over {\Delta x^{\,2} }} =  \hfill \cr
   = {{\Delta _x T(x,t) - \nabla _x T(x,t)} \over {\Delta x^{\,2} }}
 = {1 \over {\Delta x^{\,2} }}\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right){\bf T} \hfill \cr}  \right.
$$
thereby the PDE translates to
$$
{\bf T}\left( {{\bf E} - {\bf I}} \right)
 = {{\Delta t} \over {\Delta x^{\,2} }}\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right){\bf T}
 = r\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right){\bf T}
$$
which, due to the presence of both left and right multiplications on the variable $\bf T$, is solved by progressively
determining the column vectors of $\bf T$ as
$$
{\bf TE} = r\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right){\bf T} + {\bf T}
 = \left( {1 - 2r} \right)\left( {{\bf I} + {r \over {\left( {1 - 2r} \right)}}
\left( {\overline {\bf E}  + {\bf E}} \right)\,} \right){\bf T}
$$
that is
$$
{\bf T}_{\,m + 1} \left( x \right) = \left( {1 - 2r} \right)
\left( {{\bf I} + {r \over {\left( {1 - 2r} \right)}}\left( {\overline {\bf E}  + {\bf E}} \right)\,} \right)
{\bf T}_{\,m} \left( x \right)\quad \left| {\,0 \le m \le h - 1} \right.
$$
where the two ends of the vector
$$
{\bf T}_{\,m + 1} \left( 0 \right),\;{\bf T}_{\,m + 1} \left( {h\Delta x} \right)
$$
are not actually computed, yet set to the imposed boundary values (here taken to be null).

Solution by the Implicit Method

This method , that takes for ${\partial  / {\partial t}}$ the backward difference, would lead to
$$
{\bf T}_{\,m + 1} \left( x \right)
 = {1 \over {1 + 2r}}\left( {{\bf I} - {r \over {\left( {1 + 2r} \right)}}
\left( {\overline {\bf E}  + {\bf E}} \right)} \right)^{\, - \,{\bf 1}} {\bf T}_{\,m} \left( x \right)
$$
3) Conclusion

Coming to the specific equations you are interested in, we put $D(x)$ as a diagonal matrix
$$
D(x) \Rightarrow
 \left( {\matrix{ {D(0)} & 0 & 0 &  \cdots   \cr  0 & {D(1\,\Delta x)} & 0 &  \cdots   \cr
   0 & 0 & {D(2\,\Delta x)} &  \cdots   \cr     \vdots  &  \vdots  &  \vdots  &  \ddots   \cr
 } } \right) = \left( {D(n\,\Delta x) \circ {\bf I}} \right)
$$
then
$$
D(x){\partial  \over {\partial x}}P(x,t) \Rightarrow
 {1 \over {\Delta x}}\left( {D(n\,\Delta x) \circ {\bf I}} \right)
\left( {\overline {\bf E}  - {\bf I}} \right){\bf P}
$$
if the forward difference is taken.

For the term
$$
{\partial  \over {\partial x}}\left( {D(x){\partial  \over {\partial x}}P(x,t)} \right)
$$
we have a even wider selection of "stencils" that can be applied, noting that there are slight differences
between them, because for instance we have
$$
\eqalign{
  & {{\partial ^{\,2} } \over {\partial x^{\,2} }}
 \Rightarrow {1 \over {\Delta x^{\,2} }}\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right)
 = {1 \over {\Delta x}}\left( {{1 \over {\Delta x}}\left( {\overline {\bf E}  - {\bf I}} \right) - {1 \over {\Delta x}}
\left( {{\bf I} - {\bf E}} \right)} \right) \approx   \cr
  &  \approx {1 \over {\Delta x}}\left( {{\bf I} - {\bf E}} \right){1 \over {\Delta x}}
\left( {\overline {\bf E}  - {\bf I}} \right) = {1 \over {\Delta x^{\,2} }}
\left( {\overline {\bf E}  - {\bf I} - {\bf E}\overline {\bf E}  + {\bf E}} \right) \approx   \cr
  &  \approx {1 \over {\Delta x}}\left( {\overline {\bf E}  - {\bf I}} \right){1 \over {\Delta x}}\left( {{\bf I} - {\bf E}} \right)
 = {1 \over {\Delta x^{\,2} }}\left( {\overline {\bf E}  - {\bf I} - \overline {\bf E} {\bf E} + {\bf E}} \right) \cr}
$$
since ${\bf E}\overline {\bf E}$ and $\overline {\bf E} {\bf E}$ differ from $\bf I$ by having one diagonal term null (the first at bottom, the second on top).

In practice one adopts the stencil whose (a-)symmetry matches that of $D(x)$ and of  (the expected) $P(x)$.

Also, if $D(x)$ is known as a continuous function, the term can be expanded into
$$
{\partial  \over {\partial x}}\left( {D(x){\partial  \over {\partial x}}P(x,t)} \right)
 = \left( {{\partial  \over {\partial x}}D(x)} \right){\partial  \over {\partial x}}P(x,t)
 + D(x){{\partial ^{\,2} } \over {\partial x^{\,2} }}P(x,t)
$$

Assuming we can do that, and taking the forward difference for the first derivative in $t$, while  for that in $x$ we take the central one
supposing that we expect an almost central-symmetric $P$ vs. $x$, the equation
$$
{\partial  \over {\partial t}}P(x,t)
 = \left( {D'(x) - v} \right){\partial  \over {\partial x}}P(x,t)
 + D(x){{\partial ^{\,2} } \over {\partial x^{\,2} }}P(x,t) - \tau P(x,t) + S\left( x \right)
$$
translates into
$$
\eqalign{
  & {\bf P}_{m + 1} (x) = \left( {{{\Delta t} \over {2\Delta x}}\left( {\left( {D'(n\Delta x) - v} \right) \circ {\bf I}} \right)
\left( {\overline {\bf E}  - {\bf E}} \right) + {{\Delta t} \over {\Delta x^{\,2} }}
\left( {D(n\Delta x) \circ {\bf I}} \right)\left( {\overline {\bf E}  - 2{\bf I} + {\bf E}} \right)
 + \left( {1 - \tau \,\Delta t} \right)\,{\bf I}} \right){\bf P}_m (x) + {\bf S}\left( x \right) =   \cr
  &  = \left( {\left( {a(n) + b(n) \circ {\bf I}} \right)\overline {\bf E}  + \left( {c(n) \circ {\bf I}} \right)
 + \left( { - a(n) + b(n) \circ {\bf I}} \right){\bf E}} \right){\bf P}_m (x) + {\bf S}\left( x \right) \cr}
$$
where, as in the example , the edges ${\bf P}_{m + 1} (0),\;{\bf P}_{m + 1} (h\Delta x)$ shall be set to the prescribed boundary values.


问:
Homotopy composition Hatcher exercise
Show that composition of paths satisfies the following cancellation property: if $f_0 \cdot g_0 \simeq f_1 \cdot g_1 $ and $g_0 \simeq g_1$, then $f_0 \simeq f_1$.

So I have two homotopies.

So say $g_0,g_1: X \rightarrow Y$ and $f_0,f_1:Y \rightarrow Z$.

Then we know that $G:X \times I \rightarrow Z$ s.t. $G(x,0)=f_0 \cdot g_0(x)$ and $G(x,1)=f_1 \cdot g_1 (x)$. Also, $H:X \times I \rightarrow Y$ s.t. $H(x,0)=g_0(x)$ and $H(x,1)=g_1(x)$.

I was wondering how do you construct the homotopy for f? or is there a simplier way.

I would think you construct a homotopy, but can't see how to.

答:
Composition of two paths $f\cdot g$ implies that these maps are defined on same domain (which is unit interval $I=[0, 1]$) and range, in other words: $f, g:I\rightarrow X$. As I understood, in your question you defined $f$ and $g$ on different domain and range which is incorrect.

There are few ways to solve the original problem, I will show two:

In the Hatcher's book author defined inverse path $\overline{f}$ and proved that $f\cdot\overline{f}$ is homotopic to a constant path. Using that we can solve the problem:
$$f_0 \simeq f_0\cdot\underset{\text{constant path}}{\left(g_0\cdot\overline{g_0}\right)} \simeq \left(f_0\cdot g_0\right)\cdot\overline{g_0} \simeq \left(f_1\cdot g_1\right)\cdot\overline{g_0} \simeq f_1\cdot (g_1\cdot\overline{g_0}) \simeq f_1\cdot (g_0\cdot\overline{g_0}) \simeq f_1.$$

Another approach is to define homotopy between $f_0$ and $f_1$ directly. The fact that $g_0\simeq g_1$ implies $g_0(0) = g_1(0)\Rightarrow f_0(1) = f_1(1)$. It is clear that $f_0(0) = f_1(0)$. If $F:I\times I\rightarrow X$ is assosiated map of homotopy between $f_0\cdot g_0$ and $f_1\cdot g_1$ then the function
$$G(s, t) = F\left(\frac s 2, t\right)$$
is assosiated map of homotopy between $f_0$ and $f_1$. Basically we ignore the second part of homotopy between $f_0\cdot g_0$ and $f_1\cdot g_1$ which is responsible for the homotopy between $g_0$ and $g_1$.

Title: Homotopy composition Hatcher exercise


问:
Analogy between prime numbers and singleton sets?
While trying -- in vain -- to write an alternative answer for another question (If $\cup \mathcal{F}=A$ then $A \in \mathcal{F}$. Prove that $A$ has exactly one element.), I discovered the following property for sets: $$A \textrm{ is a singleton set} \;\equiv\; \langle \forall B : B \subseteq A : B = \emptyset \;\not\equiv\; B = A \rangle$$ and I noticed the similarity to the following definition for positive whole numbers: $$n \textrm{ is prime} \;\equiv\; \langle \forall d : d \textrm{ divides } n : d = 1 \;\not\equiv\; d = n\rangle$$  So it looks like singleton sets act a bit like prime numbers.  Which is not strange, come to think of it, given that both are 'indivisible atoms' of some sort.

So what common theory underlies both concepts of indivisibility?

答:
We can unify the two definitions by passing to their common generalisation in order theory. To begin, note that the set $\mathscr{P} (X)$ of all subsets of a fixed set $X$ is partially ordered by inclusion, and the set $P$ of all positive integers can be partially ordered by divisibility, and that both $\mathscr{P} (X)$ and $P$ has a bottom element ($\emptyset$ and $1$, respectively).

Moreover, both $\mathscr{P} (X)$ and $P$ have the property that, for any two elements $a$ and $b$, there exists a unique element $a \vee b$ such that, for all $c$ with $a \le c$ and $b \le c$, $a \vee b \le c$. In the case of $\mathscr{P} (X)$, this is the operation of taking the union of two subsets, and in the case of $P$, this is the operation of taking the l.c.m. of two numbers. A partially ordered set with a bottom element $\bot$ and this binary operation $\vee$ is said to be a join semilattice.

Definition. An indecomposable element of a join semilattice $L$ is an element $c$ such that, if $c = a \vee b$, then either $a = \bot$ or $a = c$ (but not both!).

As you have observed, the indecomposable elements of $\mathscr{P} (X)$ are precisely the singleton subsets, and the indecomposable elements of $P$ are precisely the prime numbers.


问:
Finding $\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$ where $S$ is part of a sphere

Let $S$ be the subset of the sphere $x^{2} + y^{2} + z^{2} = 1,  z > 0$. Calculate the integral $$\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$$

So I know that this is a surface integral. I used these parameters:

$$\boldsymbol{\mathbf{}\Phi} (\varphi ,\theta )=(\sin \varphi \cos \theta, \sin \varphi \sin \theta, \cos \varphi) , 0<\varphi < \frac{\pi}{2}, 0<\theta<2\pi$$ I also found $$\left \| \Phi_{\phi} \times \Phi_{\theta} \right \| = \sin \varphi$$
So I got the double integral $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} \sin^{4}\varphi\cos^{4}\theta \sin(\sin^{3}\varphi \cos^{3}\theta \cos^{5}\varphi)\sin\varphi \,d\varphi \,d\theta $$ but I don't think that it's a good idea.

Body:
Denote $f(x,y,z) = x^{4} \sin (x^{3}z^{5})$. We have for all $(x,y,z) \in S$

$$f(x,y,z) = -f(-x,y,z)$$ and therefore

$$\int_{S}^{} x^{4} \sin (x^{3}z^{5})dxdydz=0.$$


Title: Finding $\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$ where $S$ is part of a sphere

Body:
Let $S$ be the subset of the sphere $x^{2} + y^{2} + z^{2} = 1,  z > 0$. Calculate the integral $$\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$$

So I know that this is a surface integral. I used these parameters:

$$\boldsymbol{\mathbf{}\Phi} (\varphi ,\theta )=(\sin \varphi \cos \theta, \sin \varphi \sin \theta, \cos \varphi) , 0<\varphi < \frac{\pi}{2}, 0<\theta<2\pi$$ I also found $$\left \| \Phi_{\phi} \times \Phi_{\theta} \right \| = \sin \varphi$$
So I got the double integral $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} \sin^{4}\varphi\cos^{4}\theta \sin(\sin^{3}\varphi \cos^{3}\theta \cos^{5}\varphi)\sin\varphi \,d\varphi \,d\theta $$ but I don't think that it's a good idea.

Body:
Note

$$\int_{S}^{} x^{4} \sin (x^{3}z^{5})dxdydz
\\= \int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz+
\int_{S, x<0}^{} x^{4} \sin (x^{3}z^{5})dxdydz\\
 = \int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz-
\int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz=0
$$


问:
Is area under logarithmic curve identical regardless of "dosing" as pharmacokinetics claims?
In pharmacokinetics, it is assumed that the area under a curve is the same even if a substance is dosed gradually. This applies when "first order kinetics" is used. Is this mathematically true?

I graphed two examples, and used a simple app to estimate the area under the curve, and it is close to identical. But would be good to get a mathematical point of view.

答:
I'll take the governing equation for first-order kinetics to be

$$\frac{dN}{dt}=-KN+\frac{dN_{in}}{dt} \tag{1}$$

where $N(t)$ is the amount of drug in the system at time $t$, $K$ is the elimination constant, and $N_{in}(t)$ is the amount of drug infused up to time $t$. I'll assume $N(0)=N_{in}(0)=0$ (no drug is present or has been infused by time $t=0$) and that that $N_{in}(t)=0$ for $t\geq T$ (the drug infusion has ceased by time $T$). As such, for $t\geq T$ we have $$\frac{dN}{dt}=-KN$$ and therefore $N(t)=Ae^{-kt}$ which vanishes as $t\to\infty$.

With these assumptions, suppose we rearrange eq. $(1)$ as $$N(t)=\frac{1}{K}\frac{dN_{in}}{dt}-\frac{1}{K}\frac{dN}{dt}=\frac{1}{K}\frac{d}{dt}(N_{in}-N).$$ Then the area under the $N(t)$ curve is obtained by integrating from $0$ to $\infty$, and the fundamental theorem of calculus yields

\begin{align}
\int_0^\infty N(t)\,dt
&=\frac{1}{K}\int_0^\infty \frac{d}{dt}(N-N_{in})\,dt\\
&=\frac{1}{K}[N_{in}-N]_0^\infty \\
&= \frac{1}{K}N_{in}(\infty).
\end{align}
Note that this only depends on the final value of $N_{in}(t)$ and not the infusion schedule.


问:
How can I evaluate $\int_0^\infty e^{\pm i \xi} d \xi $
I have encountered this problem while trying to evaluate
$$
\int_0^\infty e^{ix^2}dx
$$
by squaring it and using Fubini's theorem.

I just treat $i$ as a constant:
$$
\int_0^\infty e^{i \xi} d \xi = \frac{1}{i} \left[ e^{\pm i\infty} -1 \right]
$$
But what is '$i\infty$'?

答:
This integral does not converge. We can associate a "meaning" to it by stepping into the world of generalised functions (distributions) via the Dirac Delta generalised function as

$$\delta (k) = \int_0^{+\infty} e^{\pm ikx}\ \text{d}x$$

but this is a completely different story.

Edit, for the sake of completeness

The Dirac delta is the distribution (generalised function) $\delta \in \mathcal{D}(\mathbb{R}) := C_c^\infty(\mathbb{R})^\prime$ defined by
$$
 \forall f \in C_c^\infty(\mathbb{R}), \quad \delta(f) := f(0).
$$
For any $a \in \mathbb{R} \setminus \{0\}$, define a linear transformation $S_a : C_c^\infty(\mathbb{R}) \to C_c^\infty(\mathbb{R})$ by
$$
 \forall f \in C_c^\infty(\mathbb{R}), \; \forall x \in \mathbb{R}, \quad (S_a f)(x) := f(ax).
$$
For any $a \neq 0$, one can use the change of coordinates $x^\prime = ax$ to show that for all $f,g \in C_c^\infty(\mathbb{R})$,
$$
\begin{align}
 \int_{-\infty}^\infty (S_a f)(x)\, g(x)\,dx
&= \int_{-\infty} f(ax)\,g(x)\,dx\\
&= \int_{-\infty}^\infty f(x^\prime) g(a^{-1}x^\prime)\frac{1}{\lvert a \rvert}dx^\prime\\
&= \int_{-\infty} f(x) \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}g\right)(x) \,dx,
\end{align}
$$
i.e., that $S_a^\ast = \lvert a \rvert^{-1} S_{a^{-1}}$ with respect to the $L^2$ inner product on $\mathbb{R}$, so, by duality, we're obliged to define $S_a\delta$ (i.e., $\delta(ax)$) to be the distribution
$$
 S_a\delta := \delta \circ \frac{1}{\lvert a \rvert}S_{a^{-1}} \in \mathcal{D}(\mathbb{R}).
$$
But now, for any $f \in C_c^\infty(\mathbb{R})$,
$$
 (S_a\delta)(f)
= \delta\left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right)
= \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right)(0)
= \frac{1}{\lvert a \rvert}f(a^{-1}0)
= \frac{1}{\lvert a \rvert}f(0)
= \left(\frac{1}{\lvert a \rvert}\delta\right)(f),
$$
so that, indeed,
$$
 S_a\delta = \frac{1}{\lvert a \rvert}\delta,
$$
which is the rigorous statement of the identity $\delta(ax) = \frac{1}{\lvert a \rvert}\delta(x)$.

For what concerns how to represent Dirac Delta, then we have the following:

Assume $h_n \in L^1_{\text {loc}} $ are such that

$h_n \geq 0$

$\int h_n (x) \, dx = 1$

$\int x h_n (x) \, dx \to 0$ as $n \to \infty $

Then $h_n \to \delta $.

Proof

First notice that
$$|\phi (x) - \phi (0)| = |\int_0^x \phi'(t) \, dt|
\leq \int_0^x |\phi'(t)| \, dt
\leq \int_0^x \|\phi'\| \, dt
= \|\phi'\| x $$
where $\|\cdot\|$ is uniform/infinity norm.

Then notice that
$$\langle \delta, \phi \rangle = \phi (0)
= \phi (0) \int h_n (x) \, dx
= \int h_n (x) \, \phi (0) \, dx $$

Therefore,
$$|\langle h_n, \phi \rangle - \langle \delta, \phi \rangle|
= | \int h_n (x) \, \phi (x) \, dx - \int h_n (x) \, \phi (0) \, dx|
\leq \int h_n (x) |\phi (x) - \phi (0)| \, dx \\
\leq \int h_n (x) \, \|\phi'\| x \, dx
= \|\phi'\| \int x h_n (x) \, dx
\to 0$$
since $\|\phi'\| < \infty $.

Thus, $h_n \to \delta $ as $n \to \infty $.

Dirac Delta Representations*

For what concerns the integral the OP presented, we have when you compute it,

$$∫_{-L}^L e^{i\alpha x} \ \text{d}\alpha = \frac{e^{iLx} - e^{iLx}}{ix} = \frac{\sin(Lx)}{2x} = \frac{L}{2} \frac{\sin(Lx)}{Lx}$$
you see that there isn't a limit as $L\to +\infty$. The expression is more subtle than that.

In your case it's $\int_0^{+\infty}$ which has no problems in zero but when you compute $L\to +\infty$... eh!

You may want to see the following related question, Dirac Delta function and normal distribution
Roughly, the Gaussian distribution with variance tending to zero, has the 'pointwise' behaviour one asks of the 'dirac delta function'.

Also, the following PDF may be helpful.
http://www.cse.yorku.ca/~kosta/CompVis_Notes/fourier_transform_Gaussian.pdf

Advances

I find it to be most natural to define the Dirac delta $\delta_{x'}$ on a manifold $M$ so that for any smooth function $f:M\to\mathbb C$ the formal integral $\int_Mf(x)\delta_{x'}(x)dV(x)$ (or more precisely, the duality pairing $\langle\delta_{x'},f\rangle$) equals $f(x')$.
Here $V$ denotes the volume measure of the manifold (which may be Riemannian or Lorentzian, for example).
In fact, I find this the most natural way to go even over a Euclidean space.
I would call this the definition of the Dirac delta.

I wrote $\delta_{x'}(x)$ instead of $\delta(x-x')$ on purpose.
The difference $x-x'$ doesn't mean anything on a general manifold.
However, $x-x'$ does mean something in local coordinates, but then both $x$ and $x'$ have to be within the same coordinate chart.
In local coordinates $dV(x)=\sqrt{|g(x)|}dx$.
Once you are in a local coordinate chart, you get
$$
\int\sqrt{|g(x)|} \delta_{x'}(x)f(x)dx=f(x'),
\tag{2}
$$

(Again, this is just a formal integral.)
However, I would not treat this as the definition, but as a local coordinate representation of the Dirac delta.

Once you are on a local coordinate chart, you can indeed formally write
$$
\delta_{x'}(x)
=
|g(x')|^{-1/2}\int_{\mathbb R^d}\frac{1}{(2\pi)^d}e^{ik\cdot (x-x')}dk.
$$
Here $x$ and $x'$ are vectors in $\mathbb R^d$ and are identified with two points on $M$ via a coordinate chart.
(If you plug this representation to (2) with a smooth function $f$ supported in the coordinate chart, you obtain the correct result.)
The integral is taken over the entire Euclidean space.
I doubt this representation is of much use, since the Fourier transform tends to be less useful on manifolds.
(It can be defined on coordinate charts, but it does not have the nice and simple global properties the usual Fourier transform does.)

Euclidean spaces have a lot of special structure.
This structure allows us to speak of the difference $x-x'$ and also allows for (simple) Fourier analysis.

On general manifolds you lose both, so a Fourier representation and difference of two points are kind of lost at the same time.
Locally a manifold can be identified with (a subset of) a Euclidean space, but only locally.


问:
Notation on isomorphism showing that a vector space is isomorphic to its double dual.
Let $V$ be a finite dimensional vector space, and let $V^{\prime\prime}$ be the double dual space.

From S. Axler's linear algebra text, he defines a function $\Lambda : V \rightarrow V^{\prime\prime}$ by
$$(\Lambda v)(\varphi) = \varphi(v).$$

But I cannot make heads nor tails of this definition. I am confused on both sides of the equation.

For the LHS, $\Lambda v$ makes sense because that is what I would expect to appear when defining a function; for example, let $f(x) = \text{whatever}$. But then we tack on $\varphi$, which I know is in $V^{\prime}$; but if the LHS is supposed to deal with the domain $V$, then I don't understand how it is valid to juxtapose this $\varphi$.

The RHS is also concerning because $\varphi(v)$ is a scalar. Assuming that $\varphi(v)$ is supposed to be the output of $\Lambda,$ it should follow that $\varphi(v) \in V^{\prime}$.

I appreciate all clarification and insight concerning this notation and, perhaps, an explanation as to why such notation is valid (I think this last bit will follow once clarification is provided).

Thank you =)

答:
Lets write out the map in question in more detail. It is
$$\begin{array}{rcl}
\Lambda: V & \longrightarrow & V‘‘=\operatorname{Hom}(\operatorname{Hom}(V,K),K)\\
v & \longmapsto & \left(\begin{array}{rcl}
\operatorname{Hom}(V,K) & \xrightarrow{\Lambda v} & K\\
\varphi & \mapsto & \varphi(v)
\end{array}\right)
\end{array}$$

So we map a vector $v$ to a map $\Lambda v$, which takes in maps $\varphi$ and evaluates them at $v$. As a function is determined on its values the shortest form to write this is the equation
$$(\Lambda v)(\varphi) = \varphi(v)$$
but with this it is hard to keep the overview over which map does what...


问:
What happens if $f \circ f = f$ and $f$ is a real function.
So suppose that $f(f(x)) = f(x).$ $f$ is not necessarily invertible. Is it true that f must be the identity function?

I think that it is true, since if we let $u = f(x),$ then $f(u) = u,$ and so $f$ is the identity function, but I am not sure for some reason if this is valid.

答: Constant functions, $f(x) = |x|$, $f(x) = -|x|$, etc.... there are many functions like this that are not the identity one.


问:
Link between thetrahedral numbers and combinatorics
The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is  $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1}{2}$ pairs. But you could also count the pairs differently (see here): You can match the first person with $n$ different persons, the second person can be paired with $n-1$ different persons, the third with $n-2$, $\ldots$. Hence $\binom{n+1}{2}=n+(n-1)+(n-2)+\ldots$.

Now the tetrahedral numbers $1, 4=1+3, 10=1+3+6,\ldots$ can be written as $\binom{n+2}{3}$. Is there also an combinatorical explaination as for triangular numbers? I just found this explanation but I can't see clearly why $\binom{n+2}{3}$ should be the same as $1+3+6+\ldots+\frac{n(n+1)}{2}$.

答:
Your last link says

The numbers you're seeing are binomial coefficients. To go from $(0,0)$ to $(3,n)$ you have to take $3+n$ steps: $3$ right and $n$ up. You have to choose which $3$ of the $3+n$ steps that are going to go to the right, and the number of ways to choose is $\binom{3+n}{3}$

Your question is why $\binom{n+3}{3} =\sum\limits_{m=0}^n  \binom{m+2}{2}$

In the context of the link, this says the number of ways of going $3$ right and $n$ up is equal to the number of ways of going $2$ right and $m$ up in any pattern, followed immediately by $1$ right and then by $n-m$ up, summed over all the possible $m$. In other words all the possible ways, measured by how many steps up before the last step right.


问:
If each slice of $f : X \times Y\rightarrow Z$ is continuous then $f$ is continuous?
I'm having trouble to conclude my proof of the next statement.

If $f:X \times Y \rightarrow Z$ is a function such that, $f_x : Y \rightarrow Z$ given by $f_x(y) = f(x,y)$ is continuous for each $x \in X$ and $f_y : X \rightarrow Z \ $ given by $\ f_y(x) = f(x,y)$ is continuous for each $y \in Y\ $ then $f$ is continuous

This is what I've tried, given $(x,y) \in X \times Y$ and $\ U \in \tau_Z$ such that $f(x,y) \in U$ we take $\ f_x, \ f_y$ continuous functions, so $f_x^{-1}(U) \in \tau_Y \text{ and } f_y^{-1}(U) \in \tau_X$ so we take $V = f_y^{-1}(U) \times f_x^{-1}(U)$ wich is open in $X \times Y$ and $(x,y) \in V$.

My problem is when I have to show that $f(V) \subseteq U$. I need some help

Edit: I've saw something similar here and uses strongly compactness, so this is likely false, But I don't know how can be a function wich each are continuous but not the whole function.

答:
This is not correct in general. Take for example the function $f = \begin{cases} \frac{xy^2}{x^2+y^4} \ , \quad &(x,y)\neq (0,0) \\ 0 \ , \quad &\text{otherwise} \end{cases}$.

Then, $f_x$ is continuous for every $x \in \mathbb{R}$ and $f_y$ is continuous for every $y \in \mathbb{R}$. But $f$ fails to be continuous at $(0,0)$, since $f(x,0)=0 \to 0$ but $f(y^2,y)=\frac{1}{2} \to \frac{1}{2}$ (as $(x,y)\to (0,0)$).


问:
Comparing the size of the sets
Let $A, B, C, D$ be sets. Prove that if $|A|=|B|$ and $|C|=|D|$, then $|A^C|=|B^D|$

I am asking for help in solving the task.

答:
HINT: Since $|A|=|B|$, there is a bijection $f:A\to B$. Similarly, there is a bijection $g:C\to D$. Define a function $h:A^C\to B^D$ as follows: for each $\varphi\in A^C$ let

$$h(\varphi)=\{\langle g(c),f(a)\rangle:\varphi(c)=a\}\,.$$

Verify that $h(\varphi)$ is indeed a function from $D$ to $B$, and show that $h$ is a bijection. It may help to realize (or even prove) that $h(\varphi)=f\circ\varphi\circ g^{-1}$.

Added: Suppose that $\varphi,\psi\in A^C$, and $\varphi\ne\psi$; then there is a $c\in C$ such that $\varphi(c)\ne\psi(c)$. Let $a=\varphi(c)$; then $\langle g(c),f(a)\rangle\in h(\varphi)$, but $\langle g(c),f(a)\rangle\in h(\psi)$, since $\psi(c)\ne a$, so $h(\varphi)\ne h(\psi)$, and $h$ is injective.

Now let $\psi\in B^D$; we want to find a $\varphi\in A^C$ such that $h(\varphi)=\psi$. In other words, we want to find a $\varphi\in A^C$ such that

$$\psi=\{\langle g(c),f(a)\rangle:\varphi(c)=a\}\,.\tag{1}$$

Let $c\in C$; then $g(c)\in D$, and $\psi\big(g(c)\big)\in B$. The map $f$ is a bijection, so there is a unique $a\in A$ such that $f(a)=\psi\big(g(c)\big)$; let $\varphi(c)=a$. Then $\varphi(c)=a$ if and only if $f(a)=\psi\big(g(c)\big)$, which is true if and only if $\langle g(c),f(a)\rangle\in\psi$, so $(1)$ holds, i.e., $h(\varphi)=\psi. Thus, $h$ is also surjective and hence a bijection.

Title: Comparing the size of the sets


问:
Calculate $\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$
Calculate
$\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$.

The only idea that I have is to substitute $t=x+1$, but I do not think it is a good one.

答: By the order-reversing involution $x\mapsto\frac{2-x}{1+2x}$,$$I=\int_0^2\frac{\arctan\frac{2-x}{1+2x}dx}{x^2-4x-1}.$$(This can be found by substituting $y=\arctan x$ in the original definition of $I$, then verifying a suspicion that the new integrand is of the form $yf(y)$ with $f(\arctan2-y)=f(y)$.) Averaging,$$I=\frac{\arctan2}{2}\int_0^2\frac{dx}{x^2-4x-1}=-\frac{\arctan2\operatorname{artanh}\frac{2}{\sqrt{5}}}{2\sqrt{5}}.$$According to Wolfram Alpha this is numerically correct, $-0.357395$.


问:
local property of tangent vector
Given a smooth manifold $M$ and a point $p \in U$, $U \subset M$ open, my book says there is a linear isomorphism between vector spaces $T_pU$ and $T_pM$ by:

$v \rightarrow \tilde{v}$ where for any $f \in C^\infty(M)$, $\tilde{v}(f) = v(f|_U)$. Now I can see the linear part, but I cannot see the bijection.

Edit: Background lemma that might help:
Let $v \in T_pM$. If $f,g \in C^\infty(M)$ are equal on a neighborhood of $p$, then $v(f) =  v(g)$. If $h \in C^\infty(M)$ is constant on a neighborhood of $p$, then $v(h) = 0$

答:
You can use the following lemma to prove the surjectivity of $ \bar {v} $:

Let $M$ be a differentiable manifold, $U \subset M$ open and $g: U \longrightarrow \mathbb{R}$, with $g \in C^{\infty}(U)$. If $p \in U$ then there exists a function $h \in C^{\infty} (M)$ such that $h = g$ in a neighborhood of $p$ contained in $U$.

And use that if $ f, g \in C^{\infty} (M) $ are equal on a neighborhood of $ p $, then $ v (f) = v (g) $ to show the injectivity of $ \bar {v} $.

Title: local property of tangent vector


问:
Completness of norm in $L_\infty$
How can I prove that $L_\infty$ is a complete space with the following standard norm?

$||f||_\infty=\inf\{M\in\mathbb{R}^+:|f(x)|\leq M\text{ a.e.}\}$

This proof is left to the reader in the text I'm following, but it does not give any hint on how the proof should be carried out.

答:
Some ideas: (not a complete solution but too long for a comment)

Do you know the fact that the infimum in your definition is actually a minimum? You can find a set $A$ of measure 0 so that $\|f\|_{\mathcal{L}^\infty} = \|f|_{X\setminus A}\|_\infty$ with the classical uniform norm.
Next you can look at an arbitrary absolute convergent series $\sum_{n=1}^\infty\|[f_n]\|_{L^\infty} < \infty$ of functions $[f_n] \in L^\infty$ and try to show that this series converges in $L^\infty$ (a space is Banach iff absolute convergent series are convergent). Now you can use the previous fact, that the infimum is a minimum, together with the fact that the space of all bounded functions is a Banach space (giving you convergence of the series a.e. since countable unions of sets of measure 0 have measure 0).

Title: Completness of norm in $L_\infty$


问:
Does Euler totient function gives exactly one value(answer) or LEAST calculated value(answer is NOT below this value)?
I was studying RSA when came across Euler totient function. The definition states that- it gives the number of positive values less than $n$ which are relatively prime to $n$.

I thought I had it, until I came across this property:-

Euler Totient function is multiplicative function, that is: $\varphi(mn) = \varphi(m)\varphi(n)$

Now, if $p$ is a prime number, $\varphi(p)=p-1$.

Putting values of $p$ as 11 and 13 one by one,

$$\varphi(11)=10$$

$$\varphi(13)=12$$

Applying above stated property,

$$\varphi(11\cdot 13)=\varphi(11)\varphi(13)$$

$$\varphi(143)=12 \cdot 10$$

$$\varphi(143)=120$$

Is it correct? Does that mean we have $23$ values between $1$ and $143$ which are not relatively prime to $143$? Sorry if its something basic I'm missing. I'm not some genius at maths and came across this during study of RSA Algo. Thanks.

答: Yes there are $23$ values that are not relatively prime. The classic way to count these is to look at multiples of $11$, $\{1\cdot 11,...,12\cdot11, 13 \cdot 11 \}$, and multiples of $13$, $\{1 \cdot 13,...,10 \cdot 13, 11 \cdot 13\}$. You then look at the size of the union which is $12+10+1=23$.</p>


问:
Uniform convergence of $(f(x+h)-f(x))/h$ to $f^{\prime}(x)$
Let $f:\mathbb{R}\to\mathbb{R}$ be continuously differentiable and let $f^{\prime}$ be Lipschitz continuous on $[0,1]$. Why is it that
\begin{equation}
  \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f^{\prime}(x)\ \
  \text{uniformly on $[0,1]$}\ ?
\end{equation}

I don't see how the Lipschitz continuity of $f^{\prime}$ helps to conclude the uniform convergence. Help is very much appreciated!

Body:
$f(x+h)-f(x)=h f'(z_h)$ where $z_h\in[x,x+h]$ (or $[x+h,x]$ if $h<0$). So $$\frac{f(x+h)-f(x)}{h}=f'(z_h).$$

Now $|f'(z_h)-f'(x)|\le L |z_h-x|\le L|h|$.


问:
Dirichlet L functions
I would like to know more special Dirichlet L functions (like Zeta function for instace). Despite Zeta, Beta, Eta, Lambda and Hurwitz zeta are there more special Dirichlet L functions? I went to the wiki and DLMF pages on L functions and could not find any others.

Thanks

答:
$$L(s,\chi_5)=\sum_{n=1}^\infty \chi_5(n)n^{-s}=\sum_{n=0}^\infty (5n+1)^{-s}+i(5n+2)^{-s}-i(5n+3)^{-s}-(5n+4)^{-s}$$

$\chi_5$ is a character $\Bbb{Z/5Z^\times\to C^\times}$. Every Dirichlet character comes from a character $\Bbb{Z}/N\Bbb{Z^\times\to C^\times}$ for some $N$.

"their integral forms": it works mostly the same way as for $\zeta(s)$.

https://beta.lmfdb.org/Character/Dirichlet/5/2

Title: Dirichlet L functions


答:
$$\left|\frac{f(x+h ) -f(x)}{h} -f'(x) \right|=|f'(\theta ) -f'(x)|\leq M|\theta -x| \leq M|h|$$
Since $\theta \in [x, x+h]$ by Lagrange theorem.

So $$\lim_{h\to 0} \sup_{x} \left|\frac{f(x+h ) -f(x)}{h} -f'(x) \right| \leq \lim_{h\to 0} M|h| =0$$
which means that the convergence is uniform.

Title: Uniform convergence of $(f(x+h)-f(x))/h$ to $f^{\prime}(x)$

Body:
Let $f:\mathbb{R}\to\mathbb{R}$ be continuously differentiable and let $f^{\prime}$ be Lipschitz continuous on $[0,1]$. Why is it that
\begin{equation}
  \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f^{\prime}(x)\ \
  \text{uniformly on $[0,1]$}\ ?
\end{equation}

I don't see how the Lipschitz continuity of $f^{\prime}$ helps to conclude the uniform convergence. Help is very much appreciated!


问:
Scale values between -1 and 1 with a sum of 0
I need a function that scales array of values (from -inf to inf) between -1 and 1 with a sum of 0. The closest thing that comes to mind is the softmax function. But I don't know how to convert it to the form I need.</p>

答:
Matti P.'s comment shows that there is no function  which always meets this condition.

If you have the data $x_1,x_2,\ldots,x_n$ with minimum $a$, mean $b$ and maximum $c$ with $a <b< c$ then $$f(x_i) = \dfrac{x_i - b}{\max(c-b,b-a)}$$ will transform the data linearly so it has mean of $0$, a maximum of $1$ or less and a minimum of $-1$ or more, and at least one of $1$ and $-1$ being achieved


问:
Why does model theory treat well-ordered but uncountable languages?
In model theory, results such as the Compactness theorem and the Löwenheim–Skolem theorem hold true, not only in countable languages, but in well-ordered languages.

I'm not questioning this idea in principle, but I am wondering why it is significant. That is, besides the fact that mathematical theorems ought to be as general as possible, what is the practical effect of extending such theorems to well-ordered languages? Are there even any examples of uncountable but well-ordered languages? If so, what do we gain by a study of such languages?

答:
Now in the usual axioms for set theory every set is well-orderable - so "all well-orderable languages" just amounts to "all languages," and you're asking for examples of uncountable languages which are somehow useful. Nonstandard analysis provides one such example, where we look at a language naming every function on the real numbers (so this is a language of size $2^{2^{\aleph_0}}$, much larger than necessary). Another type of example is gotten by taking a structure $\mathcal{M}$, some subset $A$ of the domain of $\mathcal{M}$ (possibly the whole thing), and adding a constant symbol for each element of $A$; this type of expansion occurs frequently within model theory, even to study structures in "small" languages.

However, if we weaken our ambient set theory to $\mathsf{ZF}$ (= set theory without the axiom of choice), there is a genuine difference between "all well-orderable languages" and "all languages." It's worth noting first of all that model theory, even over $\mathsf{ZF}$ alone, does not confine itself to well-orderable languages; the situation is simply that there are theorems which only apply to well-orderable languages, just as (even in $\mathsf{ZFC}$) there are theorems only applicable to finite languages, or to countable languages, or to languages satisfying some other condition.

Interestingly, over $\mathsf{ZF}$ most of the natural uncountable languages are not provably well-orderable - e.g. the language of nonstandard analysis above. There are still examples which are important, but they tend to be more technical.

And as usual, in my opinion the answer to

if so, what do we gain by a study of such languages?

is "The resulting theory is interesting." There are certainly (e.g. via nonstandard analysis) uncountable languages with applications outside pure model theory, but I think that's secondary: part of the takeaway of model theory is exactly the rich and surprising behavior of structures in "large" languages.


答:
For each $v \in V$ define a map $\Lambda v : V' \to \mathbf F$ whose value in a functional $\varphi \in V'$ is $(\Lambda v)(\varphi) = \varphi(v)$. Then it is easy to see that $\Lambda v$ is linear$^1$, and then $\Lambda v$ is an element of $V'' := (V')'$. Thus, $\Lambda$ is a map from $V$ to $V''$ that sends each $v \in V$ to the map $\Lambda v \in V''$.

Hope this helps :)

$^1$By definition of the operations in $V'$ we have that for any $\varphi_1,\varphi_2 \in V'$ and $\lambda \in \mathbf F$, $$(\lambda \varphi_1+\varphi_2)(v) = \lambda \varphi_1(v)+\varphi_2(v)$$ that is, $(\Lambda v)(\lambda \varphi_1+\varphi_2) = \lambda (\Lambda v)(\varphi_1)+(\Lambda v)(\varphi_2)$.

Title: Notation on isomorphism showing that a vector space is isomorphic to its double dual.

Body:
Let $V$ be a finite dimensional vector space, and let $V^{\prime\prime}$ be the double dual space.

From S. Axler's linear algebra text, he defines a function $\Lambda : V \rightarrow V^{\prime\prime}$ by
$$(\Lambda v)(\varphi) = \varphi(v).$$

But I cannot make heads nor tails of this definition. I am confused on both sides of the equation.

For the LHS, $\Lambda v$ makes sense because that is what I would expect to appear when defining a function; for example, let $f(x) = \text{whatever}$. But then we tack on $\varphi$, which I know is in $V^{\prime}$; but if the LHS is supposed to deal with the domain $V$, then I don't understand how it is valid to juxtapose this $\varphi$.

The RHS is also concerning because $\varphi(v)$ is a scalar. Assuming that $\varphi(v)$ is supposed to be the output of $\Lambda,$ it should follow that $\varphi(v) \in V^{\prime}$.

I appreciate all clarification and insight concerning this notation and, perhaps, an explanation as to why such notation is valid (I think this last bit will follow once clarification is provided).

Thank you =)


问:
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$

I found the following proof:
$\frac{1}{(-q,q)_\infty}=\frac{(1-q)(1-q^2)\cdots}{(1-q^2)(1-q^4)\cdots}=(q,q^2)_\infty$
However, this seems to lack formality since I am treating the infinite product as something that can  "cancel out" infinitely. I feel like I would have to make this argument for the partial product, but this method no longer works. Is there a way to formalize what I did here?

Body: Note that $$(-q,q)_\infty=\prod_{k=0}^\infty(1+q^{k+1})=\prod_{k=0}^\infty\frac{1-q^{2k+2}}{1-q^{k+1}}$$ and $$(q,q^2)_\infty=\prod_{k=0}^\infty(1-q^{2k+1})$$ so $$(-q,q)_\infty(q,q^2)_\infty=\prod_{k=0}^\infty\frac{(1-q^{2k+2})(1-q^{2k+1})}{1-q^{k+1}}=\prod_{j=0}^\infty\frac{1-q^{j+1}}{1-q^{j+1}}=1.$$


问:
Prove the convergence of the series.
Let r > 1 be a real number. Prove that the following series is convergent.
$$\sum_{n = 1}^{\infty}\frac{1}{n^r}$$</p>

答: You can use the integral test.</p>

Title: Prove the convergence of the series.

Body: You can use the integral test.


答:
Well you need to work with the definition of a complete space. i.e., take a Cauchy sequence $\{ f_n \}_{n=1}^\infty$ w.r.t $\Vert \cdot \Vert_ {L^\infty}$, find a limit function $f$ and show that $\Vert f_n-f\Vert_{L^\infty}\to 0$.

As a hint, consider the set

$$ E:=\Big \{ x: \underset{n\to \infty}{\lim} f_n(x) \; \text{exists} \Big\},   $$
and define

$$ f(x)=\begin{cases} \underset{n\to \infty}{\lim} f_n(x) & , x\in E\\
0 & , x\notin E \end{cases} $$

Now show that $\mu(E^c)=0$, and that for all $\epsilon>0$ there exists $N_\epsilon$ such that

$$ \vert f_n(x)-f(x)\vert <\epsilon \quad \text{for all} \quad n>N_\epsilon \quad \text{and} \quad x\in E. $$

Recall also that

$$E = \{ x: \big( f_n(x) \big)_{n=1}^\infty \; \text{is a Cauchy seq.} \Big\}. $$

Title: Completness of norm in $L_\infty$

Body:
How can I prove that $L_\infty$ is a complete space with the following standard norm?

$||f||_\infty=\inf\{M\in\mathbb{R}^+:|f(x)|\leq M\text{ a.e.}\}$

This proof is left to the reader in the text I'm following, but it does not give any hint on how the proof should be carried out.