问:
How can I solve this ODE? $2(G')^2 - GG'' < 0$
I know it maybe not unique, but how can I tell the class of functions $G(x)$ for which

$2(G'(x))^2 - G(x)G''(x) < 0$

Suppose I also know

$G(0) = 1$

and

$G'(x) < 0$

答:
The function $G(x)$ is twise differentiantable in the interval $\Delta=(a,b)$ with continuous secod derivative and $0\in \Delta$. Also $G'(x)<0$, $G''(x)G(x)-2G'(x)^2>0$ and $G(0)=1$. Since $G(0)=1\neq 0$, then we may assume that $G(x)\neq 0$ in $\Delta$ (or better $G(x)>0$ in $\Delta$ from the continuity of $G$). However let us only assume that $G(x)\neq 0$. Set then
$$
f(x)=\frac{1}{G(x)}
$$
Then we have
$$
f'(x)=-\frac{G'(x)}{G(x)^2}\Rightarrow
$$
$$
f''(x)=-\frac{G''(x)G(x)^2-2G(x)G'(x)^2}{G(x)^3}
$$
Hence
$$
f''(x)=-\frac{G''(x)G(x)-2G'(x)^2}{G(x)^3}.
$$
and
$$
f''(x)G(x)<0\tag 1
$$
Also
$$
f'(x)>0\tag 2
$$
Hence $f$ is increasing in $\Delta$. Also $G(x)>0$ or $G(x)<0$ in $\Delta$ since if exists $x_0$ such that $G(x_0)=0$, then from (1) we have $f''(x_0)G(x_0)=0<0$, which is imposible. Hence $f''(x)>0,\forall x\in\Delta$ or $f''(x)<0,\forall x\in\Delta$.

If $f''(x)>0,\forall x\in\Delta$, (then it is from (1) $G(x)<0$). Hence $f(x)<0$ in $\Delta$. But $0>f(0)=1$ which is imposible. Hence we have $f''(x)<0$ and $f'(x)>0$ and $f(x)>0$.

Hence every $f$ that passes through the pont $\{0,1\}$ and is positive and increasing and have negative curvature ($f''(x)<0$) in any interval $(a,b)$ that contains $0$, define a $G(x)=1/f(x)$ with the proerties you want.


问:
positive Integer value of $n$ for which $2005$ divides $n^2+n+1$

How Can I calculate positive Integer value of $n$ for which $2005$
  divides $n^2+n+1$

My try:: $2005 = 5 \times 401$

means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$

because $2005 = 5 \times 401$

now $n^2+n+1 = n(n+1)+1$

now $n(n+1)+1$ contain last digit $1$ or $3$ or $7$

$\bullet $ if last digit of $n(n+1)+1$ not contain $5$. So it is not divisible by $5$

Now how can I calculate it? please explain it to me.

答:
A number of the form $n^2+n+1$ has divisors of the form 3, or any number of $6n+1$, and has a three-place period in base n.

On the other hand, there are values where 2005 divides some $n^2+n-1$, for which the divisors are of the form n, 10n+1, 10n+9. This happens when n is 512 or 1492 mod 2005.

ContentLicense: CC BY-SA 3.0


问:
If $|a_{ij}| \leq L$ for all $i,j$, then $\| A \| \leq L\,$?
This is my attempt:
Let $A \in M(n, \mathbb R)$, $v \in \mathbb R^n$.
Note that $\| A \| = \sup\limits_{v, \|v\| \leq 1} \|Av\|$.

I may assume WLOG that $\|v\| = 1$.

Take any $v \in \mathbb R^n$. Then, $Av = y$, where $y_i = \sum\limits_{j=1}^n a_{ij}v_j$.
Then,
\begin{align*}
\|y\|^2 &= \Big(\sum\limits_{j=1}^n a_{1j}v_j\Big)^2 + \cdots + \Big(\sum\limits_{j=1}^n a_{nj}v_j\Big)^2 \\
&\leq\Big(\sum_{j=1}^n a_{1j}^2\Big) \cdot \sum_{j=1}^n v_j^2 + \cdots + \Big(\sum_{j=1}^n a_{nj}^2\Big) \cdot \sum_{j=1}^n v_j^2 \quad \text{(by Cauchy-Schwarz)}\\
&\leq\sum_{j=1}^n a_{1j}^2 + \cdots + \sum_{j=1}^n a_{nj}^2 \\
&= nL^2 + \cdots +nL^2 \\
&= (nL)^2
\end{align*}
Hence, $\|A\| \leq nL$. However, my lecture notes state that $\|A\| \leq L$, and I am wondering what should be done to achieve that inequality.

答:
Your question is answered with a No.

Please calculate the norm of $\begin{pmatrix}1 &1\\0&1\end{pmatrix}$. Or consider its action on the vector $\begin{pmatrix}1 \\1\end{pmatrix}$.

A symmetric or self-adjoint example would be the orthogonal projection
$\begin{pmatrix}0.5 &0.5\\ 0.5 & 0.5\end{pmatrix}$ which turns into
$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$ upon diagonalisation.

ContentLicense: CC BY-SA 4.0


答:
$$ae^a=\sum_{n=1}^{\infty}\frac{a^n}{(n-1)!}$$ differentiate $$ae^a+e^a=\sum_{n=1}^{\infty}\frac{na^{n-1}}{(n-1)!}$$
multiply $a$ on bs  $$a^2e^a+ae^a=\sum_{n=1}^{\infty}\frac{na^n}{(n-1)!}$$
Now diffrentiate again what do you get??

Body:
How to prove $\sum _ { n = 1 } ^ { \infty } \frac { n a ^ { n - 1 } } { ( n - 1 ) ! } = ( a + 1 ) e ^ { a }$ and $\sum _ { n = 1 } ^ { \infty } \frac { n ^ { 2 } a ^ { n - 1 } } { ( n - 1 ) ! } = ( a ^ { 2 } + 3 a + 1 ) e ^ { a }$?

I know $\sum _ { n = 1 } ^ { \infty } \frac { a ^ { n - 1 } } { ( n - 1 ) ! } = e ^ { a }$ using Taylor's Formula. How can I apply it to the above two?


答:
Take the derivaty:

$\sum _ { n = 1 } ^ { \infty } \frac { (n-1)a ^ { n -2} } { ( n - 1 ) ! } =  e ^ { a }$

multiply it by $a$:

$\sum _ { n = 1 } ^ { \infty } \frac { (n-1)a ^ { n -1} } { ( n - 1 ) ! } = a \cdot e ^ { a }$

distribute in $(n-1)$ and pass the term multiplied in the $1$ to the other side and you will get your answer.

Same principle works for second equality, take the derivative and isolate the $n^2$

Title: $\sum _ { n = 1 } ^ { \infty } \frac { n a ^ { n - 1 } } { ( n - 1 ) ! } = ( a + 1 ) e ^ { a }$


问:
Relationship between number of trials and complementary CDF of a binomial distribution
I have a question related to Binomial distribution. Can someone help me?

Consider the following complementary CDF of a binomial distribution with parameters n and p :

$$
\bar{F}(x) = \sum_{k=x+1}^{n}(^n _k)p^k(1-p)^{n-k}
$$

Is the complementary CDF non-decreasing in n? If yes, how does one prove it?

My intuition is that the complementary CDF is non-decreasing in n. But I am not sure how to prove it.

答:
Is the complementary CDF non-decreasing in n

$n$ is a given number in a Binomial law. $k$ is the variable.

It is non- increasing.

A property of any CDF is that it is non-decreasing, and it is evident as, by definition, the CDF is the cumulative probability function.
Its complement, 1-CDF, is obviously non-increasing in $k$.


问:
Manipulation of summation index - power series
Problem

Consider the differential equation: $y''+ty'+2y=t^2e^t $

By setting $y(t)=\sum_{n=0}^{\infty}c_nt^n$ show that the differential equation can be rewritten as
$$\sum_{n=0}^{\infty} \big((n+2)(n+1)c_{n+2}+(n+2)c_n \big)t^n=\sum_{n=2}^{\infty}\frac{t^n}{(n-2)!} $$

An attempt

$$y''=\sum_{n=2}^{\infty}c_n(n-1)nt^{n-2}=\sum_{n=0}^{\infty}c_{n+2}(n+2)(n+1)t^n$$

$$ty'=t\sum_{n=1}^{\infty}c_nnt^{n-1}=t\sum_{n=0}^{\infty}c_{n+1}(n+1)t^n $$

$$2y=2\sum_{n=0}^{\infty}c_nt^n=\sum_{n=0}^{\infty}2c_nt^n $$

Combining all these terms on the left hand side gives

$$\sum_{n=0}^{\infty} \Bigg[c_{n+2}(n+2)(n+1)+tc_{n+1}(n+1)+2c_n \Bigg]t^n=t^2e^t$$

As you can see, I'm almost getting to the answer, but there are some things that don't match up. I'm also unsure how

$$t^2e^t=\sum_{n=2}^{\infty}\frac{t^n}{(n-2)!} $$

Can anyone help me out and perhaps spot if I have made a mistake?

答:
I'm also unsure how
$$t^2e^t=\sum_{n=2}^{\infty}\frac{t^n}{(n-2)!} $$

Basically they just shifted the index:

$$ \sum_{n=2}^{\infty} \frac{t^n}{(n-2)!} =\frac{t^2}{0!} + \frac{t^3}{1!} ... +O(t^4) = \sum_{i=0}^{\infty} \frac{ t^{i+2} }{i!}$$

$$ty'=t\sum_{n=1}^{\infty}c_nnt^{n-1}=t\sum_{n=0}^{\infty}c_{n+1}(n+1)t^n $$

Indexing is 'non-ideal' here. If you had rather multiplied it out, then you'd get:

$$ t y' = \sum_{n=1}^{\infty} n c_n t^n$$

What you did is correct but to get the simplest form it'd be ideal to first multiply and then re-index. Now, put this new quantity into your differential equation and it should give you the answer.


问:
Can anyone help me get this Laplace transform of
$f(t) = a + b  \exp(-c \cdot t ^ d) $, where $a,b,c,d$ are constants, and $d$ is power of $t$.</p>

答:
It boils down to computing $J = \int_0^\infty \exp(-st - t^d)\, dt$.
I'll assume $d > 1$.  Write as a series in $-s$: $J = \sum_{n=0}^\infty \int_0^\infty \frac{(-s)^n t^n}{n!} \exp(-t^d)\, dt = \sum_{n=0}^\infty \frac{\Gamma((n+1)/d)}{n!\, d} (-s)^n$.  I don't know if there is a "closed form" for this, but for each positive integer $d$ it can be written in terms of hypergeometric
functions.</p>


问:
How can I make my figure resemple more the example?
Currently I'm learning about SDE's. In my course notes the following example was given for the following SDE:
This results in the following image:

\begin{array}{l}
d B_{t}=\left(-K_{1} B_{t}+s_{t}+\frac{1}{2} B_{t} \sigma^{2}\right) d t-B_{t} \sigma d W_{t} \\
B_{t_{0}}=B_{0}
\end{array}

With Python I tried to recreate the figure:

```<code>import numpy as np
import matplotlib.pyplot as plt

T = 100
N = 10000
dt = T/N
K1 = 1
st = 3
Bt = 20
sigma = 1
y = np.zeros(N*2)
x = np.linspace(0,T,N*2)

for i in range(N*2):
    y[i] = Bt
    Bt += (-K1*Bt + st + 0.5*Bt*sigma**2)*dt - Bt*sigma*np.random.normal(0, dt)

plt.plot(x, y)
</code>```

The values for $K_1$, $s_t$ and $\sigma$ were not given, but I think I did a OK job.

Now I want to create this figure that shows the variance and mean. How do I modify my code in order to achieve this goal?

答: Under the light of Monte Carlo, you can just simulate more random paths and the calculate the respective empirical mean and std.</p>


答:
Like Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$

for $n\ge2,$

$$\dfrac n{(n-1)!}=\dfrac{1+n-1}{(n-1)!}=\dfrac1{(n-1)!}+\dfrac1{(n-2)!}$$

$$\implies\sum_{n=1}^\infty\dfrac{na^{n-1}}{(n-1)!}=\sum_{n=1}^\infty\dfrac{a^{n-1}}{(n-1)!}+\sum_{n=2}^\infty\dfrac{a^{n-1}}{(n-2)!}$$

Set $n-1=m$ to find the first infinite sum $ =e^a$

The second sum $$=a\sum_{n=2}^\infty\dfrac{a^{n-2}}{(n-2)!}=ae^a$$

Similarly for $n^2,$  write $$n^2=(n-1)(n-2)+a(n-1)+b$$ so that for $n-3\ge0,$

$$\dfrac{n^2}{(n-1)!}=\dfrac1{(n-3)!}+\dfrac a{(n-2)!}+\dfrac b{(n-1)!}$$

Now comparing the coefficients of $n,$  $$0=-a+3\iff a=3$$

Compare the constants, $0=b-a+2\iff b=a-2=?$

Similarly for $n^3,$ write $$n^3=(n-1)(n-2)(n-3)+c(n-1)(n-2)+d(n-1)+e$$


答:
Generally to do these kind of problems see how to map from the first summation to second,

$$ \sum_{n=1}^{\infty} \frac{na^{n-1} }{(n-1)!} \xrightarrow[]{?} \sum_{n=1}^{\infty} \frac{n^2 a^{n-1}}{(n-1)!}$$

We can find by inspection that the operation we have to do is:

$$ Q() = \frac{d}{da} \left[a \left( \right)\right]$$

i.e: multiply by 'a' then differentiate. Consider our original series,

$$ \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!}= (a+1)e^a$$

Apply $Q$ on both sides:

$$ Q \left( \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!} \right)= \frac{d}{da} \left[ a( a+1) e^a \right]$$

By the premise, we can directly right LHS and simplifying the RHS by triple product rule:

$$ \sum_{n=1}^{\infty} \frac{n^2 a^{n-1}}{(n-1)!} =  (a+1) e^a + a e^a + a(a+1) e^a$$

You'll reach the answer after simplifying RHS using algebra :)

Oh, and for deriving the first summation, consider LHS:
$$ \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!}$$

And do a change of variables $n-1 = i$,

$$ \sum_{i=0}^{\infty} \frac{ (i+1)a^i }{(i)!} =\sum_{i=1}^{\infty} \frac{a^i}{(i-1)!} + \sum_{i=0}^{\infty} \frac{a^i}{i!} = a \sum_{i=1}^{\infty} \frac{a^{i-1} }{(i-1)!} + e^a  = a e^a + e^a$$

Note:

$$? = \sum_{i=1}^{\infty} \frac{a^{i-1}}{(i-1)!} = \sum_{j=0}^{\infty} \frac{ a^j}{j!}$$

By reindexing


问:
Dimension of the linear system of conics passing through four points.
Let $P_1$,$P_2$,$P_3$,$P_4$ $\in \mathbb{P}^2$. Let $V$ be the linear system of conics passing through these points. Then show that if  $P_1$,$P_2$,$P_3$,$P_4$ lie on a line, then $\dim(V)=2$ and if not, then $\dim(V)=1$.

[my attempt]

The definition of $V$ ,the linear system of conics passing through given points, is  $$V:= \{ [a:b:c:d:e:f] \in \mathbb{P}^5 \mid ax^2+by^2+cz^2+dxy+eyz+fxz=0 \text{  passing through  } P_1 , P_2 ,P_3, P_4\}$$

Now, let's denote $P_i=:[l_i:m_i:n_i]$ and consider the followings :

$$ \begin{pmatrix}
l_1^2 & m_1^2 & n_1^2 & l_1m_1 & m_1n_1 & l_1n_1 \\
l_2^2 & m_2^2 & n_2^2 & l_2m_2 & m_2n_2 & l_2n_2 \\
l_3^2 & m_3^2 & n_3^2 & l_3m_3 & m_3n_3 & l_3n_3 \\
l_4^2 & m_4^2 & n_4^2 & l_4m_4 & m_4n_4 & l_4n_4 \\
\end{pmatrix}
\begin{pmatrix}
a \\ b \\ c \\ d \\ e \\ f
\end{pmatrix}=
\begin{pmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0
\end{pmatrix}$$

Suppose $P_1$,$P_2$,$P_3$,$P_4$ lie on a line (say, $\alpha x+ \beta y +\gamma z=0$).

Then $$ \begin{pmatrix}
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2 \\
l_3 & m_3 & n_3 \\
l_4 & m_4 & n_4 \\
\end{pmatrix}
\begin{pmatrix}
\alpha \\ \beta  \\\gamma
\end{pmatrix}=
\begin{pmatrix}
0 \\ 0 \\ 0
\end{pmatrix}$$

Therefore the rank of this matrix is less than equal $2$. (This is beacause ($\alpha , \beta , \gamma $) is a non-trivial solution of this linear system) so $P_1$,$P_2$,$P_3$,$P_4$ are linearly dependent.

However, I stuck in this point. How to solve this?

Thank you.

答:
This is best understood without equations IMO. The first two points are completely free choices that cut us down to a $\mathbb P^3$ of conics (i.e. the only way $P_2$ can fail to be general is if it is equal to $P_1$). If we add a third point lying on the line between the first two (call this Step 3a), then the line must be part of the conic (otherwise Bezout is violated); in this case, the remaining two points determine a second line, so we get a (generally) unique reducible conic.

If, instead, we take $P_3$ to not lie on the line $L := \overline{P_1 P_2}$ (call this Step 3b), there is no issue with Bezout so we still have a $\mathbb P^2$ of conics whose general member is irreducible. Continuing with one more general point, we get a $\mathbb P^1$ of conics whose general member is irreducible.

On the other hand, if we continue from Step 3a and impose a fourth point on the line, then we don't "lose" any conics this time; since all conics from Step 3a contain the line $L$, they all pass through $P_4$, so we still have a $\mathbb P^2$ of conics by choosing any line to be the second component.

Finally (on the third hand, I guess?), if we continue from Step 3a but let $P_4$ be a general point, then we are cut down to a $\mathbb P^1$ of reducible conics: choose a line passing through $P_4$ (there is a $\mathbb P^1$ of these) and take its union with $L$.


问:
Show a function from a submanifold is a local diffeomorphism
Let $M = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$. I want to show that the function $f: M \to \mathbb{R}^2$ defined by $f(x, y, z) \to (x, y+ z)$ is a local diffeomorphism at the point $p = (0, 1, 0)$.

I can use the implicit function theorem with $g(x, y, z) = x^2 + y^2 +z^2$ to show that $M$ is a submanifold of $\mathbb{R}^3$ (it is an ellipsoid).

I have also found $dg = \begin{bmatrix} 2x & 2y & 2z\end{bmatrix}$, $dg_p = \begin{bmatrix}0 & 2 & 0 \end{bmatrix}$, $\ker(dg_p) = \{(x, 0, z) : x, z,  \in \mathbb{R}\} = T_p\mathbb{R}^3$.

I want to use the invertible function theorem to show $f$ is a local diffeomorphism at $p$, but I'm confused about what exactly I need to show.

Obviously I need to show $df_p$ is bijective, and I found the matrix for $df$ as $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}$.

This is where I am stuck, how do I show this is bijective at $p$, and where does the submanifold come into it?

答:
Let us use stereographic projection to show that $f$ is a local diffeomorphism.

Let $h_S : (x,y,z)\in M\setminus\{N\} \mapsto \left(\frac{x}{1-z},\frac{y}{1-z} \right)\in \mathbb{R}^2 $ be the stereographic projection with pole $S=(0,0,-1)$ the south pole. It is a diffeomorphism from $M \setminus \{N\}$ to $\mathbb{R}^2$, where $N =(0,0,1)$ is the north pole.

Let $g  : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $g = f\circ  {h_S}^{-1}$. If you can show that $g$ is a local diffeomorphism, the chain rule will show that $f$ is a local diffeomorphism at any point $p \in M \setminus \{N\}$. This is because
$$
f = g\circ h_S
$$
and thus, if $p=(x,y,z) \in M\setminus\{N\}$, one has
$$
\mathrm{d}f_{p} = \mathrm{d}g_{h_S(p)}\circ \mathrm{d}{h_S}_{p}
$$

The same study with $h_N$ the stereographic projection with pole $N$ will show that $f$ is a local diffeomorphism on $M\setminus \{S\}$ and the result will follow.


问:
How can I practice Jean-Robert Argand idea of the rotation of a square root of -1
I am studying complex numbers and I really need an intuition on how they work.
I found the following video of Mathematician named Adrien Douady

https://www.youtube.com/watch?v=2kbM96Jr4nk

He explains complex numbers in a algebraic geometry way, as he mentions using his protractor and turning it by 180 degrees to represent multiplication by (-1). But then at 4:26 he mentiond a mathematician Robert Argand http://en.wikipedia.org/wiki/Jean-Robert_Argand, and that he had a great Idea. "He said to himself, since multiplying by -1 is a 180 degree rotation, the square root is 1/2 of 180 degree rotation. How is this? how can I practice this? from what philophy of mathematics does this come from?

Thank and forgive me if I sound to dumb.

答:
For vectors on the plane, the mapping $v\mapsto c\cdot v$ is zooming the vector by scalar $c$, if $c\in\Bbb R$.

When we compose two such maps, $f:=v\mapsto c\cdot v$ and $g:=v\mapsto d\cdot v$, we have that $g\circ f=v\mapsto d\cdot c\cdot v$.

In particular, if $c=d$, we get $v\mapsto c^2\cdot v$.

The mapping  $s:v\mapsto -v$  can be indeed viewed as the rotation by $180^\circ$. Let us find its square root(s)!

Both rotations $r_{1,2}$ by $\pm90^\circ$ are such that $r_1\circ r_1=s=r_2\circ r_2$.

This can be set in a correspondence with the complex imaginary units $\pm i$.

This way, complex numbers can be identified with 'zooming-and-rotating' transformations of the plane:

The complex number $1$ corresponds to the identity, i.e. zoom by $1$ and rotate by $0^\circ$.

$0$ corresponds to the degenerated zooming by $0$, i.e. the mapping $v\mapsto 0$

A positive real number $c$ corresponds to the zooming by $c$ (i.e. $v\mapsto c\cdot v$) and no rotation (i.e. rotating by $0^\circ$)

A negative real number $-c$ (with $c>0$) corresponds to the zooming by $c$ and rotating by $180^\circ$.

The imaginary unit $i$ corresponds to rotation by $+90^\circ$ (and no zooming, i.e. zoom by $1$).

Addition of complex numbers corresponds to 'pointwise addition' of these transformation:
$$(f+g):=v\mapsto f(v)+g(v)$$
(There's an important point hidden here: try to prove that $f+g$ is again a 'zooming-and-rotating' transformation.)

Multiplication of complex numbers corresponds to composition of these transformation.


问:
$\lim_{E\ni(x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y}$, where $E=\{(x,y):y>x^2\}$
By intuition, since the numerator has a higher degree, the value of the limit should be zero. Is there a rigid proof for this statement? Particularly, how does one solve this particular limit? I can't use polar coordinates and I have tried using the squeeze theorem but they don't seem to work(at least, I haven't found a way that does)

any help would be appreciated!

Body:
For $y>x^2$ you have for $0<x^2<y<1$

\begin{eqnarray*} \frac{|x^3+y^3|}{x^2+y}
& \leq & \frac{|x^3|}{x^2+y} +  \frac{|y^3|}{x^2+y} \\
& \stackrel{0<x^2<y<1}{\leq} &  \frac{|x^3|}{x^2} +  \frac{|y^3|}{y} \\
& \leq &|x|+ y^2 \stackrel{(x,y)\to (0,0)}{\longrightarrow} 0\\
\end{eqnarray*}


问:
I am not able to simplify this vector problem
$\vec { a } ,\vec { b } ,\vec { c } $ are three unit vectors such that $|\vec { a } +\vec { b } +\vec { c } |=\sqrt { 3 } $  then the maximum value of $\left( \vec { a } \times \vec { b }  \right) .\left( \vec { b } \times \vec { c }  \right) +\left( \vec { b } \times \vec { c }  \right) .\left( \vec { c } \times \vec { a }  \right) +\left( \vec { c } \times \vec { a }  \right) .\left( \vec { a } \times \vec { b }  \right) $

I tried it by opening the given product using $\left( \vec { a } \times \vec { b }  \right) .\left( \vec { c } \times \vec { d }  \right) =\begin{vmatrix} \vec { a } .\vec { c }  & \vec { a } .\vec { d }  \\ \vec { b } .\vec { c }  & \vec { b } .\vec { d }  \end{vmatrix}$ but i am not able to find the maximum value .
Please help

答:
Hint:

$|\vec {a} +\vec { b } +\vec { c } |=\sqrt { 3 }$

$(\vec { a } +\vec { b } +\vec { c }) \cdot (\vec { a } +\vec { b } +\vec { c }) = 3$

So $ \sum \vec { a } \cdot\vec { b } = 0 \,$ (as $\vec {a}, \vec {b}, \vec {c} \,$ are unit vectors)

$(\vec { a } \times \vec { b }) \cdot (\vec {b} \times \vec {c}) = (\vec {a} \cdot \vec {b}) (\vec{b} \cdot \vec {c}) - \vec {a} \cdot \vec {c} \,$ (since $\vec{b} \cdot \vec{b} = 1,$ being unit vector)

Expand others similarly and add. If you get stuck, let me know.


问:
Create unique identifier from close coordinates
Suppose that I have the following list of coordinates:

```<code>[(35.0, 811.0), (37, 820), (66, 80), (88.0, 674.0), (97, 343), (98, 343), (126, 984), (132.0, 784.0), (133.0, 457.0), (165.0, 481.0), (193, 92), (206, 645), (206, 646), (256, 409), (304, 255), (305, 255), (360, 889), (456.0, 689.0), (500, 147), (501, 147), (528.0, 822.0), (536, 483), (536, 509), (542, 872), (560, 512), (605, 755), (605, 756), (663, 144), (664, 144), (727.0, 426.0)]
</code>```

These coordinates are the centeroid of some shapes, and some of them are very close (different in x or y) like (304, 255), (305, 255). All I want to do is categorizing such coordinates in one set. And for that aim I want to create a unique identifier from this points.

First of all you might think about subtraction or multiplication or pow or etc. but since there is a lot of coordinate in my array and some of them might have same result for aforementioned operations I can't use this trick.

So the question is that, is there any mathematical solution for this task? so that I can create a unique identifier for (304, 255) and  (305, 255) ?

PS. I'm using python programming language and Numpy.

答:
I suppose the coordinates have their own practical limits here in terms of range.
If they do, a simple function can generate a unique number for each coordinate, essentially mapping the 2 dimensional coordiantes into one dimension.

```<code>#!/usr/bin/python3

min_x = 0
max_x = 10
coords = [(5,5),(8,9),(4,5),(3,2)]

def coord_to_hash(x,y):
    return (y * (max_x - min_x)) + x

for c in coords:
    print(coord_to_hash(c[0],c[1]));
</code>```

The mapping looks like this:

Body:
I suppose the coordinates have their own practical limits here in terms of range.
If they do, a simple function can generate a unique number for each coordinate, essentially mapping the 2 dimensional coordiantes into one dimension.

#!/usr/bin/python3

min_x = 0
max_x = 10
coords = [(5,5),(8,9),(4,5),(3,2)]

def coord_to_hash(x,y):
    return (y * (max_x - min_x)) + x

for c in coords:
    print(coord_to_hash(c[0],c[1]));

The mapping looks like this:


问:
Identity Symbol in Craig's Interpolation Theorem
"
In each of the following, $\varphi$ and $\psi$ are sentences such that
the identity symbol occurs in at most one of them, and $\varphi \vDash \psi$; however,
$\varphi$, $\psi$ have no Craig interpolant in which the identity symbol does not occur:

(i) $\varphi$ is $(\exists x)(P(x) \land \neg P(x))$, $\psi$ is $(\exists x)Q(x)$;

(ii) $\varphi$ is $(\exists x)Q(x)$, $\psi$ is $(\exists x)(P(x) \lor \neg P(x))$;

(iii) $\varphi$ is $(\forall xy)(x \equiv y)$, $\psi$ is $(\forall xy)(P(x) \leftrightarrow P(y))$. "

答:
It may be useful to state the Theorem in slightly different form [see Raymond Smullyan, First Order Logic, page 127]: if $\varphi, \psi$ have at least one predicate in common, then there is an interpolant sentence. If they have no predicate in common, then either $\psi$ is valid or $\varphi$ is unsatisfiable.

The two cases can be reduced to one, using $\top$ and $\bot$. If $\varphi, \psi$ have no predicate in common, we have that:

if $\psi$ is valid, the interpolant is $\top$, while if $\varphi$ is unsatisfiable, then the interpolant is $\bot$.

Thus, we can adapt this "trick" to the above examples using $\exists x(x \ne x)$ for (i):

$(∃x)(P(x) ∧ ¬P(x)) \vDash \exists x(x \ne x)$ and $\exists x(x \ne x) \vDash (∃x)Q(x)$,

because $(∃x)(P(x)∧¬P(x))$ is a contradicition (it is unsatisfiable).

The same for (ii), where $(∃x)(P(x)∨¬P(x))$ is valid, using $\forall x (x=x)$.


问:
Integral of $\int^1_0\frac{dx}{\sqrt{x+3}-1}$
I want to solve this integral and need some directions.
$$\int^1_0\frac{dx}{\sqrt{x+3}-1}$$
I decided to call $x+3 = t^2  \rightarrow 2tdt = dx$ then : $$\int^1_0 \frac {2tdt}{t^2-1}$$ Now what should I do? call $t^2-1 = u$ ? and do the same thing on u?

Thanks!

答:
Hints:

Plug in $\sqrt{x+3}=t^2$, $dx=4t^3 dt$

$\int \dfrac{4t^3}{t^2-1}=\int\dfrac{2t^2}{t-1}+ \int\dfrac{2t^2}{t+1}$

Now use integration by parts for the each of the expression.


问:
Diagonalization of a block hermitian matrix
Suppose $A\in\mathbb{C}^{m\times m}$ has a singular value decomposition: $A=U \Sigma V^H$. Find diagonalization of a block hermitian matrix $B\in \mathbb{C}^{2m\times 2m}$, $B=[0 \; \; A^H; A \; \; 0]$.

I am kind of lost on this problem. I was thinking about writing $B=[0 \; \;  (U\Sigma V^H)^H; U\Sigma V^H\; \; 0]$, but not sure where to go from there, I hope someone can help me get on the right track.

答:
I've been searching on a similar question and found this approach helpful:

(Note that in the link, the $A^H$ and $A$ block matrices are swapped.)

I'm still a student myself, but since there are no other answers here yet, perhaps this will help you as well.

Good luck!


问:
Can someone help me with this proof
Prove that, for any integer $n$, if $n-2$ is divisible by $4$,then $n^2-4$ is divisible by $16$.

*I don't know what kind of method to use.

Title: Can someone help me with this proof

Body:
Hint:

$$n^2-4=(n-2)(n-2+4)=(n-2)^2+4(n-2)$$


问:
Proof for $\det(A)\det(B)=\det(AB)$ from ProofWiki
I am looking through some proof on ProofWiki and found this for the identity $\det(A)\det(B)=\det(AB)$. I follow the proof but having trouble understanding the "hence the result". Why does it follow from the previous line that $\det(A)\det(B)=\det(AB)$?

答: The Cauchy-Binet formula gives the determinant of any square $AB$. The proof's final paragraph addresses the case where $A,,B$ are individually square. Since $m=n$, there is only one way to choose $m$ strictly increasing indices from $1$ to $n$, viz. $j_l=l$. So the formula's right-hand side is a single term, which is $\det A\det B$.


问:
Suppose A is an m-by-n matrix with $A\ne0$. Prove that the rank of A is 1 if and only if ....
Suppose $A$ is an $m \times n$ matrix with $A\ne 0$. Prove that the rank of A is $1$ if and only if there exist $(c_1,...,c_m) \in {F^m}$ and $(d_1,...,d_n) \in {F^n}$ Such that ${A_{j,k}}$ = ${c_j}{d_k}$ for every $j = 1,...,m$ and every $k = 1,...,n$.
It is problem from linear algebra done right(3.F.11)

I think for firs part if we take all entries $1$ and all i,j ${c_{j}=1}$ , ${d_{k}=1}$  it will work.
but for other direction??

答:
I will give you a proof of one of the implications and leave the other one for you. Assume that $A \neq 0$ and $\text{rank} A = 1$. That means that the column space of $A$ is one-dimensional. Let $\mathbf{c}$ be a vector spanning $\text{Col} (A)$. Then there exists constant $d_1, d_2, \ldots, d_n$ so that

$$A = [d_1 \mathbf{c} \quad d_2 \mathbf{c} \quad \cdots \quad d_n \mathbf{c} ] = \mathbf{c}[d_1 \quad d_2 \quad \cdots \quad d_n]=\mathbf{c} \mathbf{d}^T.$$

$A = \mathbf{c} \mathbf{d}^T$ is equivalent to $A_{j, k} = c_j d_k$ (see why?), which proves the first implication.


答:
You are told that $n-2$ is divisible by $4$, which means that there exists an integer $k$ such that $n-2=4k$.

And now it's easy:

$n-2=4k\implies n=4k+2\implies n^2-4=(4k+2)^2-4=\ldots$ what?

Title: Can someone help me with this proof

Body:
Prove that, for any integer $n$, if $n-2$ is divisible by $4$,then $n^2-4$ is divisible by $16$.

*I don't know what kind of method to use.


问:
Can I convert to polar coordinates when calculating multivariate limits with three variables
When working on limits of functions with two variables, $f(x,y)$, I like to convert the problem to polar coordinates a lot of the time, by changing the question from $$\lim_{(x,y)\to (0,0)}f(x,y)$$ to $$\displaystyle\lim_{r\to 0}f(r\cos\theta,r\sin\theta).$$ I was just doing some problems in my book when I encountered a limit of a function with three variables, $f(x,y,z)$. I was just wondering if there was a way to calculate such a limit with polar coordinates.

An example being: $$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$$

Converting it into polar coordinates gives me:

$\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta+r\sin\theta \cdot z^2+r\cos\theta\cdot z^2}{r^2(\sin^2\theta+\cos^2\theta)+z^4}=\displaystyle\lim_{r\to 0}\dfrac{r(r\sin\theta\cos\theta+\sin\theta\cdot z^2+\cos\theta\cdot z^2)}{r^2+z^4}$

Can I proceed or is polar coordinates strictly for use with two variables only?

答:
The easy way to show that a limit does not exist is to show that we get two different answers approaching (0, 0, 0) in different directions.
In each case, let u -> 0
Case 1 $x = u$; $y = u$; $z=\sqrt u$  Substituting, we get $3u^2/3u^2 = 1$
Case 2 $x = u$; $y = -u$; $z = \sqrt u$  Substituting, we get $u^2/3u^2 = 1/3$
Two different limits, so no limit.


问:
How to esitimate the multi-index sum below
Let $\alpha,\beta$ be multi index in $\mathbb{N}^d$. $m$ is some positive integer.

Define $\alpha\le \beta$ when each component $\alpha_i\le \beta_i$  and $|\alpha| = \sum_i\alpha_i$

Given $c_\beta \ge 0$ .How can we esitmate the following sum has same order?

$$\sum_{|\alpha|\le m} \sum_{\beta\le \alpha}c_\beta \le C\sum_{|\beta|\le m}c_\beta$$

I don't know how to compute it,since so may terms in it.

答:
$\sum_{|\alpha|\leqslant m}\sum_{\beta\leqslant\alpha}c_\beta=\sum_{|\beta|\leqslant m}n_\beta c_\beta$ where $n_\beta=\#\{\alpha\in\mathbb{N}^d : |\alpha|\leqslant m,\,\alpha\geqslant\beta\}$.

The greatest $n_\beta$ is at $\beta=0$ (assuming $0\in\mathbb{N}$), so you're fine with $C=n_0=\binom{m+d-1}{m}$.


问:
how to compute the matrix polynomial
Let $A$ be the matrix $$A = \begin{pmatrix} 1 & \sqrt{2}\\ -\sqrt{2} & -1\\ \end{pmatrix}$$

Compute the matrix $B = 3A -2A^2 - A^3 -5A^4 + A^6$.

Could any one give me any hint for this one? I have calculated the eigenvalues they are $(1+\sqrt{2}i),(1-\sqrt{2}i)$

答:
We have the characteristic polynomial $ch_A(x)=\begin{vmatrix}1-x&\sqrt{2}\\-\sqrt{2}&1-x\end{vmatrix}=(x-1)^2+2=x^2-2x+3$.

Then by Cayley-Hamilton theorem we know that $A^2-2A+3=0$.

So if you divide your polynomial $p(x)$ by $(x^2-2x+3)$ and you get $p(x)=q(x)(x^2-2x+3)+r(x)$, you only need to calculate $r(A)$, i.e., to plug the matrix $A$ into the remainder.

Title: how to compute the matrix polynomial


问:
Rewriting a less than or equal to equation with multiple absolute values
So I have the following equation where a is a constant

$$|(x_1 - a)| + |(x_2 - a)| \leq 2$$

Can I rewrite this as the following:  $$|(x_1 - 2a + x_2)| \leq 2$$

答:
Certainly the first inequality implies the second, by an application of the triangle inequality: if the first holds then
$$|(x_1-2a+x_2)| = |(x_1-a)+(x_2-a)| \le |(x_1-a)| + |(x_2-a)| \le 2
$$
And sometimes just one direction of implication is what you actually need in proofs.

But when you reverse the direction of the triangle inequality what you get is not generally true. And therefore, when you reverse the direction of the implication, and ask whether the second inequality implies the first, the answer is "no", as shown in the comments.


问:
Prove rigorously that $4n^2 - 2 \geq 2n$ for $n \geq 1$
Apologies that this is a real low-level question, I'm having a bit of a brain-fart as they say.

I would like to show rigorously that $4n^2-2 \geq 2n$ for $n \geq 1$. This is obvious by inspection but for some reason I'm struggling to show it.

Body: $4n^2 - 2n - 2 = (2n + 1)(2n - 2) = 2(n-1)(2n+1) >= 0$ for $n >= 1$


问:
$A$ def. pos.$\implies a_{ii} >0, \;\max_{1\leq i, j\leq n}|a_{ij}| = \max_{1\leq i, j\leq n}|a_{ii}| >0.$ When you can reverse the inequality?
Let $A\in\mathbb{R}^N$ be a square matrix. It is well known that if $A$ is positive definite, thus
$$a_{ii} >0 \quad\mbox{ and }\quad \max_{1\leq i, j\leq n}|a_{ij}| = \max_{1\leq i, j\leq n}|a_{ii}| >0.$$

My question is: does a sort of "viceversa" hold? I mean, if $a_{ii} >0$ and $\max_{1\leq i, j\leq n}|a_{ij}| = \max_{1\leq i, j\leq n}|a_{ii}| >0$ it is true that $A$ is positive definite?

About me, the answer is not, unless you deal with some special case (for example, if $A$ is a diagonal matrix).

Could anyone please help me saying if a viceversa holds?

Thank you in advance!

答:
I guess you mean strictly diagonally dominant matrix,
$$|a_{ii}|>\sum_{j\neq i}|a_{ij}|\qquad i=1,\ldots, n.$$

If a real symmetric matrix $A$ is strictly diagonally dominant with $a_{ii}>0$,
that is
$$a_{ii}>\sum_{j\neq i}|a_{ij}|\qquad i=1,\ldots, n.$$

then it is positive definite.

For the proof, we only need to show all its eigenvalues are positive.

Assume $Ax=\lambda x$, with $x_i=\max_j |x_j|>0$ (if necessary, consider $-x$).

Then $\lambda x_i=a_{ii} x_i+\sum_{j\neq i}a_{ij}x_{j}\geq (a_{ii}-\sum_{j\neq i}|a_{ij}|)x_i>0$, that is $\lambda>0$.

Here we use the fact $x+\sum_j y_j\geq x-\sum_j |y_j|$ for $x,y_j\in \mathbb R$, that is
$$\begin{aligned}
a_{ii} x_i+\sum_{j\neq i}a_{ij}x_{j}&\geq a_{ii} x_i -\sum_{j\neq i}|a_{ij}|\cdot |x_j|,\qquad &&a_{ii},x_i>0\\
&\geq a_{ii} x_i -\sum_{j\neq i}|a_{ij}| x_i,\qquad &&x_i\geq |x_j|,\quad j\neq i\\
&=(a_{ii}-\sum_{j\neq i} |a_{ij}|)x_i,\qquad &&a_{ii}>\sum_{j\neq i}|a_{ij}|\\
&>0
\end{aligned}$$


答:
Let $f(x) = 4x^2-2x-2$.

We first observe that $f(1)=0$ and $f(2)>0$. Then the derivative of $f$ is $f'(x) = 8x - 2 > 0$ for all $x \geq 1$. So $f$ is increasing function. Since $f(2)>0$ , $$f(x) \geq 0$$ for all $x \geq 1$

Edit:

Here is a method which does not make use of derivatives.
We know that for any positive $a$ and $b$ by AM - GM inequality, $$a+b >2\sqrt(ab)$$

Take $a=2n^2$ and $b= 2n^2-2$, which would imply that $4n^2-2 \geq 4n\sqrt{n^2-1} \geq 2n$ for all $n\geq 1$.

Title: Prove rigorously that $4n^2 - 2 \geq 2n$ for $n \geq 1$

Body:
Apologies that this is a real low-level question, I'm having a bit of a brain-fart as they say.

I would like to show rigorously that $4n^2-2 \geq 2n$ for $n \geq 1$. This is obvious by inspection but for some reason I'm struggling to show it.


问:
Find the pedal equation of the curve $y^2=4a(x+a)$
Find the pedal equation of the curve $y^2 = 4a(x+a)$

My Attempt:
Given curve is
$$y^2 = 4a(x+a)$$
$$y = \sqrt {4a(x+a)}$$
Differentiating both sides w.r.t $x$
$$y^{'} = \frac {1}{2\sqrt {4a(x+a)}} \cdot 4a$$

The equation of tangent to the curve at point $(x,y)$ is
$$Y - y = \frac {2a}{\sqrt {4a(x+a)}} \cdot (X - x)$$
$$\sqrt {4a(x+a)} Y - y\sqrt {4a(x+a)} = 2aX - 2ax$$
$$2aX - \sqrt {4a(x+a)}.Y + (y\sqrt {4a(x+a)} -2ax) = 0$$
Let $p$ be the length of perpendicular drawn from $(0,0)$ to this tangent so
$$p = \frac {y\sqrt {4a(x+a)} - 2ax}{\sqrt {4a^2 + 4a(x+a)}}$$

答:
Pedal equation of $\gamma:y^2=4a(x+a)$ wrt origin $O(0,0)$ is $p^2=|a|r$, where $r=\sqrt{x^2+y^2}$ is the distance of $P\in\gamma$ from the origin $0$ and $r$ is the distance from $O$ to the tangent to $\gamma$ passing through $P$.

Derivative $y'$ can be found by implicit differentiation
$$2yy'=4a\to y'=\frac{2a}{y}$$
The equation of the tangent is
$$Y-y=\frac{2a}{y}(X-x)$$
$$yY-y^2=2aX-2ax\to 2aX-yY+y^2-2ax=0$$
the distance from the origin $(0,0)$ is
$$p=\frac{y^2-2ax}{\sqrt{4a^2+y^2}}$$
while the distance of the generic point of $\gamma$ from the origin is
$$r=\sqrt{x^2+y^2}$$
From the equation of $\gamma:y^2=4ax+4a^2$, substitute in the equation of $p$
$$p=\frac{4ax+4a^2-2ax}{\sqrt{4a^2+4ax+4a^2}}\to p=\frac{2ax+4a^2}{\sqrt{4ax+8a^2}}\to p=\sqrt{ax+2a^2}$$
and finally
$$p^2=ax+2a^2\to x=\frac{p^2-2a^2}{a}$$
Now substitute in $r$
$$r=\sqrt{x^2+4ax+4a^2}\to r=\sqrt{(x+2a)^2}\to r^2=(x+2a)^2$$
Now plug $x$ in the equation of $r^2$
$$r^2=\left(\frac{p^2-2a^2}{a}+2a\right)^2\to r^2=\frac{p^4}{a^2}\to p^4=a^2r^2$$
The requested equation is $p^2=|a|r$.


问:
To talk about positivity of Lebesgue measure of projections onto axes of a set in $\mathbb{R}^2$.
Let $A\subset\{(x,y)\in\mathbb{R}^2:x=y\}$ such that one dimensional Lebesgue measure of $A$ is positive.

Is it true that projection of $A$ onto $x$-axis and $y$-axis is also a set of positive one dimensional Lebesgue measure?

答:
Denote by $D$ the diagonal $\{(x,y) \in \mathbf R^2 \mid x=y\}$ and denote by $\pi(x,y) = x$ be the projection from $D$ to $\mathbf R$.  You need to be a bit careful defining the one-dimensional Lebesgue (outer) measure of a subset of $\mathbf R^2$, but for any set $A \subset D$ a reasonable definition is
$$\lambda_D^*(A) = \inf \sum_k \ell (J_k)$$
where the infimum is taken over all countable families $\{J_k\}$ of open intervals in $D$ whose union covers $A$.

Let $\{I_k\}$ be a covering of $\pi(A)$ by open intervals. Then $\{\pi^{-1}(I_k)\}$ is a covering of $A$ in $D$, and moreover each set $\pi^{-1}(I_k)$ is an interval in $D$ with $\ell(\pi^{-1}(I_k)) = \sqrt{2} \ell(I_k)$. Thus
$$\lambda_D^*(A) \le \sum_k \ell(\pi^{-1}(I_k)) = \sqrt{2}  \sum_k \ell(I_k),$$and by taking the infimum over all covers $\{I_k\}$ of $\pi(A)$ it follows that
$$\lambda_D^*(A) \le \sqrt{2} \lambda^*(\pi(A)).$$

A simple consequence is that $\lambda_D^*(A) > 0$ implies $\lambda^*(\pi(A))) > 0$.


问:
A question about Sobolev space and the relationship between the weak convergence on it and the strong convergence on L^p space.
Suppose that $\Omega$ is a bounded open set in $\mathbb{R}^n$, $H_0^1(\Omega)$ is the Sobolev space $W_0^{1,2}(\Omega)$. If $u_n$ convergent to $u$ weakly in $H_0^1(\Omega)$, can I get the conclusion that $u_n$ convergent to $u$ strongly in $L^2(\Omega)$ and how to prove it? Or we can only get the conclusion with sense of a subsequence?

Thanks a lot for anyone and any help!

答:
Suppose that $u_n$ does not converge to $u$ in $L^2(\Omega)$.
Then, there exists $\varepsilon > 0$ and a subsequence such that
$$
\|u_{n_k} - u\| \ge \varepsilon \qquad\forall k.$$
By compactness, there exists a further subsequence $u_{n_{k_\ell}}$
with $u_{n_{k_\ell}} \to v$ in $L^2(\Omega)$.
By weak convergence in $H_0^1(\Omega)$, $u = v$.
However, this contradicts the first inequality.


问:
In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$, $BC = 100$ and $EC + ED = 150$, find $CF$.

What I Tried: Here is the diagram :-

I have assigned variables for the different lengths. From there we have $3$ equations :-
$$x + y + a + z = 250$$
$$b + c = 100$$
$$y + z = 150$$
Which also implies :- $$x + a = 100$$
Now, $\Delta ABC \sim \Delta EFC$ . From here :-
$$\frac{AC}{EC} = \frac{BC}{FC} = \frac{AB}{EF}$$
$$\rightarrow \frac{(x+y)}{y} = \frac{(b+c)}{c}$$
We also have $\Delta DBC \sim \Delta EBF$. From there we get :-
$$\frac{DB}{EB} = \frac{CB}{FB} = \frac{CD}{EF}$$
$$\rightarrow \frac{(a+z)}{a} = \frac{(b+c)}{b}$$
$$\rightarrow \frac{250 - (x + y)}{a} = \frac{100}{100 - c}$$

After doing this I am completely lost and I am not finding any way to proceed. Can anyone help?

答:
From $\triangle DEC \sim \triangle BEA$ $$\dfrac{DE}{CE}=\dfrac{BE}{AE}$$

$$\dfrac{150-y}{y}=\dfrac{100-x}{x}$$

$$\Rightarrow \dfrac{y}{x}=\dfrac{3}{2}$$

So $$CF=\dfrac{3}{3+2}BC=60$$


问:
Forcing Bezier Interpolation
I found this very informative site that discusses forcing bezier interpolation and the site gives formulae for calculating the control points so that the curve goes through a set of four points, y0, y1, y2 and y3, for t=0, 1/3, 2/3, 1 respectively.

The formulae given on this site are as follows:

```<code>P0 = y0
P1 = ( -5 * y0 + 18 * y1 -  9 * y2 + 2 y3 ) / 6
P2 = (  2 * y0 -  9 * y1 + 18 * y2 - 5 y3 ) / 6
P3 = y3
</code>```

What I am looking for is a more general form of the formulae for calculating P1 and P2. So instead of having fixed values of t of 1/3 and 2/3, rather to express them as a function of t1 and t2, where t1 is the value of t for point y1 and t2 is the value of t for y3. Can someone please help in deriving these more general formulae?

答:
Without going into the full theoretical background, Bézier curves use a set of polynomials called Bernstein polynomials to weight the points. For a cubic Bézier, which seems to be what your question is about, this results in the parametric definition $$P(t) = (1-t)^3 P_0 + 3t(1-t)^2 P_1 + 3t^2(1-t) P_2 + t^3 P_3$$

When $t = 0$ this gives simply $P(0) = P_0$, and similarly $P(1) = P_3$. So we only have two degrees of freedom in forcing the curve to pass through your intermediate control points. As the site you linked mentioned, we can set up simultaneous equations:

$$\begin{eqnarray*} & P_0 & & & & & = & y_0 \\
(1-t_1)^3 & P_0 & + 3t_1(1-t_1)^2 & P_1 + 3t_1^2(1-t_1) & P_2 + t_1^3 & P_3 & = & y_1 \\
(1-t_2)^3 & P_0 & + 3t_2(1-t_2)^2 & P_1 + 3t_2^2(1-t_2) & P_2 + t_2^3 & P_3 & = & y_2 \\
& & & & & P_3 & = & y_3 \end{eqnarray*}$$

which easily reduce to

$$\begin{eqnarray*}3t_1(1-t_1)^2 & P_1 + 3t_1^2(1-t_1) & P_2 & = & y_1 - (1-t_1)^3 y_0 - t_1^3 y_3 \\
3t_2(1-t_2)^2 & P_1 + 3t_2^2(1-t_2) & P_2 & = & y_2 - (1-t_2)^3 y_0 - t_2^3 y_3 \end{eqnarray*}$$

I assume you can take it from here.

Title: Forcing Bezier Interpolation


Title: In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

Body:
In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$, $BC = 100$ and $EC + ED = 150$, find $CF$.

What I Tried: Here is the diagram :-

I have assigned variables for the different lengths. From there we have $3$ equations :-
$$x + y + a + z = 250$$
$$b + c = 100$$
$$y + z = 150$$
Which also implies :- $$x + a = 100$$
Now, $\Delta ABC \sim \Delta EFC$ . From here :-
$$\frac{AC}{EC} = \frac{BC}{FC} = \frac{AB}{EF}$$
$$\rightarrow \frac{(x+y)}{y} = \frac{(b+c)}{c}$$
We also have $\Delta DBC \sim \Delta EBF$. From there we get :-
$$\frac{DB}{EB} = \frac{CB}{FB} = \frac{CD}{EF}$$
$$\rightarrow \frac{(a+z)}{a} = \frac{(b+c)}{b}$$
$$\rightarrow \frac{250 - (x + y)}{a} = \frac{100}{100 - c}$$

After doing this I am completely lost and I am not finding any way to proceed. Can anyone help?

Body:
We also have,

$\frac{x}{y} = \frac{b}{c} = \frac{a}{z} = t$ (say)

So, $x + y + a + z = (y+z)(1+t) = 250$

We also have $b +c = c(1+t) = 100, y + z = 150$

Using $\, (y+z)(1+t) = 250, y + z = 150, c(1+t) = 100,$

$c = 60$.


问:
limit of $\cos{x}^{\frac{1}{\sin{x}}}$
I need to solve this limit:

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}}$$

I feel that it should be a rather easy limit, but I find myself struggling with the answer. I have tried substituting $\cos{x}$ and $\sin{x}$ with their series expansion, $\cos{x} \approx 1-\frac{x^2}{2}+\dots$ and $\sin{x} \approx x-\frac{x^3}{6}+\dots$, but the indeterminate form still remains.

Title: limit of $\cos{x}^{\frac{1}{\sin{x}}}$

Body:
Hint:

$$\lim_{x\to0}(\cos x)^{1/\sin x}$$

$$=\lim_{x\to0}(1-\sin^2x)^{1/2\sin x}$$

$$=\left(\lim_{x\to0}(1-\sin^2x)^{-\frac1{\sin^2x}}\right)^{-\lim_{x\to0}\frac{\sin x}2}$$


问:
generating function for k-combinations
Let $S$ be a set with $n$ elements. I want to compute the following generating function:

$$G(x)=\sum_{A\subseteq S}x^{|A|}.$$ I want to understand the following equality: $G(x)=(1+x)^n$.

In a book, it says "each of the $n$ elements in the set contributes the term $(1+x)$ to the generating function."

Is there a better explanation?

答:
Notice the following.
When you place
$$(1+x)^n=(1+x)(1+x)\cdots (1+x),$$ and you unfold the product, every multiplicand will contribute with either a $1$ or an $x.$ Call the multiplicands $(1+x)^n=p_1\cdots p_n,$ where $p_i=1+x.$ If you decide to pick the $x$ in the $i-$th multiplicand, is like you are considering one more (the exponent of the $x$) in the size of the sets that you want. So this corresponds to adding $i$ to the set. In that way, the coefficient of $x^k$ in the expansion corresponds to the number of ways to create a set of size $k$ out of $n$ elements, or $\binom{n}{k}.$

For example, if $n=3$ you have
$$(1+x)(1+\color{red}{x})(1+\color{blue}{x})=\underbrace{1\cdot 1\cdot 1}_{\emptyset}+\underbrace{x\cdot 1\cdot 1}_{\{1\}}+\underbrace{x\cdot \color{red}{x}\cdot 1}_{\{1,2\}}+\cdots +\underbrace{1\cdot 1\cdot \color{blue}{x}}_{\{3\}}+\cdots +\underbrace{x\cdot \color{red}{x}\cdot \color{blue}{x}}_{\{1,2,3\}}.$$
Notice that in general, unfolding the product gives rise to $2^n$ summands (that is the beauty of the binomial theorem, it puts together the $n+1$ size possibilities).

Title: generating function for k-combinations


问:
Does a Riemannian submersion map geodesics to geodesics?
Let's suppose I have a Riemannian submersion (projection) $\pi: G \rightarrow G/K$, with $G$ a matrix Lie group and where $G/K$ is a reductive homogeneous space. I have a decomposition of Lie algebras $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{m}$.

I know how geodesics look like in a reductive homogeneous space. This is in the book of O'Neill. If we write $o := eK$, they are of the form $\gamma_{d\pi X} (t) = \alpha(t) o = \pi (\alpha(t))$ where $\alpha(t)$ is the one-parameter subgroup of $X \in \mathfrak{m}$.

I also know that horizontal geodesics map under $\pi$ to geodesics.

But let's say I have a geodesic in $G$, this is of the form $e^{tX}$, where $X \in \mathfrak{g}$. Is the image of this geodesic under $\pi$ also a geodesic? Or do I have to figure out what $\pi$ does with the vertical component?

答: There is no reason for this to be true. If you are looking for a counter example, consider $G=SL(2,\mathbb{R})$ and  $X=\begin{pmatrix}1&1\\-1&-1\end{pmatrix}\in \mathfrak{sl}(2,\mathbb{R})$. Then the Cartan decomposition of $X$ is $$X=\begin{pmatrix}0&1\\-1&0\end{pmatrix}+\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ Note that $\pi_\ast(Id)$ has kernel $\mathfrak{l}$. So the geodesic passing through $eK$ in the direction $\pi_\ast(Id)(X)$ is $t\to \pi\begin{pmatrix}e^t&0\\0&e^{-t}\end{pmatrix}$. Also note that $t\to \pi(e^{tX})$, is not this geodesic, as at time $t=1$ it can be checked that their difference is not in $K$. For this, note that $e^X=Id+X$, as $X$ is nilpotent.


问:
How to solve the homogeneous system of a differential equation
I want to solve the system $\dot{x}=Ax$ where $ A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right).$ I found the eigenvalues of $A$ as $\lambda_1=4,\lambda_2=-1$ with correspongind eigenvectors $v_1=(2,3)^\intercal,v_2 =(1,-1)^\intercal $ respectively. Then, the solution of the system will be
$$x(t)=e^{At}x_0$$
so that I obtained $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
where $x_0$ is the initial condition. Now, if I say that $x_0 =(c_1,c_2) $ for some constants $c_1,c_2$ I obtain a sum which is very mixed as you can guess from the above matrix. (Note that I verified the matrix via Wolfram).

However, Paul's Online Notes and an online solver says that the solution can be found as

$$x(t)=c_1 e^{-t} (-1,1)^\intercal+ c_2 e^{4t}(2,3)^\intercal$$

that is not equal to the solution that I found. So, what is the problem with my solution?

Final note: The second solution takes the first eigenvector as $-v_1$.

答:
$$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
$$x(t)=\left( \begin{matrix} \frac{1}{5}e^{-t}(3c_1-2c_2)  + \frac{2}{5}e^{4t} (c_1+c_2) \\ - \frac{1}{5}e^{-t}(3c_1-2c_2)  +  \frac{3}{5}e^{4t} (c_1+c_2) \end{matrix} \right) $$
$$x(t)=\left( \begin{matrix} e^{-t}k_1 +  {2}e^{4t} k_2 \\ - e^{-t}k_1  +  {3}e^{4t} k_2 \end{matrix} \right) $$
This agree with the answer you posted.
$$x(t)=k_1e^{-t}\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2e^{4t}\left( \begin{matrix}2 \\ 3\end{matrix} \right) $$
Where :$$k_1= \frac {3c_1-2c_2 }5, k_2= \frac {c_1+c_2}5 $$
There is nothing wrong with your solution.
And for initial conditions $(x_0)$ you have:
$$x(0)=k_1\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2\left( \begin{matrix}2 \\ 3\end{matrix} \right)$$
You have the same formula as the one in Paul's notes.

found the eigenvalues of $A$ as $λ_1=4,λ_2=−1$ with correspongind eigenvectors $v_1=(2,3)^t,v_2=(1,−1)^t$ respectively. Then, the solution of the system will be:
$$x(t)=c_1e^{4t}v_1+c_2e^{-t}v_2$$


问:
Let $~k, ~l ∈ ℝ~$ be s.t. every solution of the given ODE satisfies $~\lim_{→∞}() = 0~.$ Then which option/s is/are true?

Question: Let $~k, ~l ∈ ℝ~$ be such that every solution of $$\dfrac{^2}{^2} + 2\dfrac{}{}+  = 0$$ satisfies $~\lim_{→∞}() = 0~.$ Then

$1.~~  \, 3^2 +  < 0  ~~\text { and } ~~ > 0 \\ 2.~~ \, ^2 +  > 0 ~~~~ \text { and } ~~ < 0 \\3.~~  \, ^2 −  ≤ 0  ~~~~ \text { and } ~~ > 0 \\ 4. ~~\, ^2 −  > 0, ~~~ > 0 ~~\text { and }~~  > 0$

My thought: Clearly every solution of the given differential equation is of the form $~y(x)=Ae^{\lambda_1 x}+Be^{\lambda_2 x}~,$ where $A, B$ are constants and $~\lambda_1~$ and $~\lambda_2~$ are solution of the auxiliary equation $~m^2+2km+l=0~.$

Also since $~\lim_{→∞}() = 0~,$ so there are three cases,

Case I: $~\lambda_1~$ and $~\lambda_2~$ are both negative and the constant $A, B$ take any value.

Case II: One of $~\lambda_1~$ and $~\lambda_2~$ is negative and the other is positive then the constant with positive $\lambda_i$ must be zero.

Case III: Both $~\lambda_i~$ are positive and then the constant $A, B$ both zero.

Now from here how to proceed the further to get the desired conclusion? Please help.

答:
Starting from your equation, the characteristic equation is $$m^2+2km+l=0$$
with solutions $$\lambda_{1,2}=-k\pm\sqrt{k^2-l}$$
You get two real solutions if $k^2-l\ge 0$. If you want both of them to be negative, the largest of them should be negative:
$$-k+\sqrt{k^2-l}<0$$
This means $k>0$ and $k^2-l<k^2$, meaning $l>0$. So for two negative real solutions for $\lambda$ you get answer 4.

But there is another option. You can have complex solutions. If $k^2-l<0$, then $$\lambda_{1,2}=-k\pm i\sqrt{|k^2-l|}$$
The solutions for the differential equations are then $$y=e^{-k}(A\sin\sqrt{|k^2-l|}x+B\cos \sqrt{|k^2-l|}x)$$
This go to zero is the exponential part go to zero. So $k^2-l<0$ and $k>0$. This is answer option 3.

Therefore the solution to your problem is: either 3 or 4 verify the conditions for the differential equation, but not at the same time.


问:
Relationship between ends of a hyperbolic 3-manifold and ideal boundary components
I've been reading Sullivan's paper "Quasiconformal homeomorphisms and dynamics II: Structural stability implies hyperbolicity for Kleinian groups", and I was hoping I could get some clarification regardig concepts that come up in section 7, where he sets out to show Kleinian groups satisfying certain properties are convex cocompact (i.e. geometrically finite and without parabolics). I would be happy with a reference to somewhere where this material is presented in more detail, since I suspect I'm missing background knowledge.

Let $\Gamma \subset \mathrm{PSL}(2,\mathbb{C})$ be a finitely generated non-elementary Kleinian group, and $\Lambda(\Gamma) \subset \partial \mathbb{H}^3, \Omega(\Gamma) = \partial \mathbb{H}^3 - \Lambda(\Gamma) \subset \partial \mathbb{H}^3$ be the limit and ordinary sets for $\Gamma$, respectively. Then, $\Gamma$ acts properly discontinuously by Moebius transformations on $\Omega$, so the quotient $\Omega/\Gamma$ is a hyperbolic Riemann surface, which is of finite type by Ahlfors' finiteness theorem. That is, $\Omega/\Gamma = S= \bigcup_{i=1}^r S_{g_i,n_i}$, where $g_i$ and $n_i$ are the genus and number of punctures of $S_{g_i,n_i}$. I'll refer to S as the ideal boundary at infinity of $M = \mathbb{H}^3/\Gamma$.

On the other hand, by the Scott core theorem, there exists a compact submanifold $C \subset M$ (the compact core of $M$) such that the inclusion $C \to M$ is a homotopy equivalence. Let $E = \bigcup_{j=1}^n E_j$ be the closure of the connected components of $M - C$ (where the $E_j$ are the connected components of $E$). Then, we have that $\partial E_j$ is a compact surface, for each $j = 1, \dots, n$. Using excision, you can show that  the homology groups of $E_j$ coincide with those of $\partial E_j$, so all the $\partial E_j$ are connected surfaces. I'll refer to the $E_j$ as the ends of $M$.

So, the overarching question is (I'll number them for convenience):

How do the surfaces $\partial E_j$ relate to the $S_{g_i,n_i}$?

For example, I know that if $M$ has rank two cusps, then you cannot "see" these in the ideal boundary (the intersection of a lift of the cusp with a fundamental domain only intersects $\partial \mathbb{H}^3$ at a limit point), but you should be able to "see" it in the ends of $M$, i.e. some of the $\partial E_j$ should be tori corresponding to these rank two cusps.

More specifically:

How do the ideal boundary components $S_{g_i,n_i}$ where $n_i \neq 0$ look in terms of the ends of $M$? Since all the $\partial E_j$ are compact surfaces, none of them can be homeomorphic to $S_{g_i,n_i}$, but it seems to me that they should show up in some way.

Do all ideal boundary components of the form $S_{g_i,0}$ have a unique component $\partial E_{j(i)}$ corresponding to them?

Thanks in advance!

答:
Here's a rough idea of what is going on. And I'm going to restrict my comments to the torsion free case which simplifies the language (and, I think, is implicit in your question anyway).

Let $\mathcal H$ be the convex hull of $\Lambda(\Gamma)$, and consider the quotient $\mathcal H / \Gamma$ which is embedded in $M$. Outside of some exceptional situations (e.g. Fuchsian groups), $\mathcal H / \Gamma$ is homeomorphic to $M$, and the difference between them, by which I mean the closure $\overline{M - \mathcal H / \Gamma}$, is homeomorphic to $\partial M \times [0,1]$. So, in particular, we have a rather natural homeomorphism $S \approx \partial \mathcal H / \Gamma$. (One can even get around the exceptions by replacing $\mathcal H$ with its $\epsilon$-neighborhood, as is explained in work of Thurston that one can find developed in work of Epstein and Marden; but I will stick with this simpler notation.)

Of course, under the assumption that $S$ is not compact, i.e. the number of punctures $n_i$ is nonzero for some $i$, it follows that $\mathcal H / \Gamma$ is not compact. So $\mathcal H / \Gamma$ is the "convex core" of $M$, but not yet its compact core.

To get the actual compact core of $M$, you now look at the "cusps" of $\mathcal H / \Gamma$. The compact core itself is obtained by cutting off each cusp and adding a new boundary portion associated to that cusp. Cusps are of two types: $\mathbb Z \oplus \mathbb Z$ cusps; and $\mathbb Z$ cusps. When you cut off a $\mathbb Z \oplus \mathbb Z$ cusp of $\mathcal H / \Gamma$, you just add a new boundary component homeomorphic to the torus. But when you cut of a $\mathbb Z$ cusp of $\mathcal H / \Gamma$, what you add is a cylinder that connects up two circle cusps of the Riemann surface $S = \partial \mathcal H / \Gamma$.

So to summarize, via the inclusions $C \subset \mathcal H / \Gamma \approx M$, one can think of $S$ as embedded in $\partial C$, in such a way that the complement $\partial C - S$ has components in one-to-one correspondence with  the cusps of $\mathcal H / \Gamma$: a torus component for each $\mathbb Z \oplus \mathbb Z$ cusp; and an annulus component for each $\mathcal Z$ cusp.


问:
value of fraction
Can we find out the value of

$$\frac{1+i}{1-i}$$

I have tried to solve it by multiplying $(1+i)$ to both sides and in the end see that the result is still $i$. Am I correct,or is there a different solution?

答:
As OP has answer I want to write a basic thing to solve this type of problems of complex number.Whenever you saw terms containing $i$ in denominator,try to get rid of them either using the rationalizing process(used in this question) or by partial fraction (using this method is based on question type).some useful values of $i$:

$i^2=-1; \,i^3=-i;\,i^4=1;\,\dfrac{1}{i}=-i$

Any power greater than 4 can reduced to remainder of that power divides by 4.

Title: value of fraction


答:
Call this function $f(x)$

Evaluate the limit $L = \lim_{x \to 0} \log f(x) = \lim_{x \to 0}\log \left( \left(\cos x\right)^{1/ \sin x} \right) = \lim_{x \to 0} \left(\log \cos x\right) / \sin x$ - this is a $0/0$ indeterminate form but you can apply L'Hôpital to get $\lim_{x \to 0} \left(\left(1 / \cos x\right) \sin x\right) / \cos x = \lim_{x \to 0} \sin x / \cos^2 x = 0$

The answer is $e^L = e^0 = 1$

Title: limit of $\cos{x}^{\frac{1}{\sin{x}}}$

Body:
I need to solve this limit:

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}}$$

I feel that it should be a rather easy limit, but I find myself struggling with the answer. I have tried substituting $\cos{x}$ and $\sin{x}$ with their series expansion, $\cos{x} \approx 1-\frac{x^2}{2}+\dots$ and $\sin{x} \approx x-\frac{x^3}{6}+\dots$, but the indeterminate form still remains.


Title: limit of $\cos{x}^{\frac{1}{\sin{x}}}$

Body:
When $$\ (cos{x})^{\frac{1}{\sin{x}}}$$ is meant

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}} = e^{\lim_{x\to 0}\,\frac{ln{cox{x}}}{\sin{x}}} = e^{\lim_{x\to 0}\,\frac{x*ln{cos{x}}}{\ x*sin{x}}} = e^{\lim_{x\to 0}\,\frac{ln{cos{x}}}{\ x}} = e^{\lim_{x\to 0}\,\frac{(cos{x} - 1) *ln{(cos{x} + 1 - 1)}}{\ (cos{x} - 1) * x}} = e^{\lim_{x\to 0}\,\frac{(cos{x} - 1)}{\ x}} = e^0 = 1$$


问:
Is $\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k=\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k, a \in \mathbb{R}$ true? If so, how?
I come here from this answer actually: https://math.stackexchange.com/a/3743805/765852

Due to re-indexing,
$$\sum_{p=1}^{k+1}(-1)^p\binom{k}{p-1}p^{k+1}$$
should equal
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+1)^k,$$
yet it's written as
$$-\sum_{p=0}^{k}(-1)^p\binom{k}{p}p^k.$$ I tested some results with my calculator and it seems that this is not only true for $(p+1)^k$, but also for $(p+a)^k$, where $a \in \mathbb{R}$. Is this the case? If so, how would one go about proving this?

I actually tried my hand at proving this (it's just using simple algebra). However it requires a property which I also tried proving but I am not sure about its proof. I'm going to shamelessly plug that in here (it's open): How would I prove $\sum_{p=1}^{n}\binom{n}{p}p^m(-1)^p=0, m \in \mathbb{N}, 1\leq m < n $ via induction?

Ok here's the proof:

Statement:
$$\sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k=\sum_{p=0}^{k}(-1)^p\binom{k}{ p}p^k, a \in \mathbb{R}$$

Proof:
\begin{align*}
 \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p+a)^k &= \sum_{p=0}^{k}(-1)^p\binom{k}{p}(p^k+kp^{k-1}a+\binom{k}{2}p^{k-2}a^2+...kpa^{k-1}+a^k) \\
      &= \sum_{p=0}^{k}(-1)^p \binom{k}{p}p^k+ (-1)^p\binom{k}{p}kp^{k-1}a + (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}kp^{k-1}a+ \sum_{p=0}^{k} (-1)^p\binom{k}{p} \binom{k}{2}p^{k-2}a^2... \\
      &= \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+ka \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-1}+a^2 \binom{k}{2} \sum_{p=0}^{k} (-1)^p\binom{k}{p}p^{k-2}...
        \end{align*}

Then, by the property linked above, it should be
$$\sum_{p=0}^{k} (-1)^p\binom{k}{p}p^k+0+0+....+0+ \sum_{p=0}^{k} (-1)^p\binom{k}{p}$$

Simply by the binomial theorem, the last term should also be zero. This completes the proof.

答:
Using
\begin{eqnarray*}
[x^k]: k! e^{(a+p)x} = (a+p)^k.
\end{eqnarray*}
We have
\begin{eqnarray*}
\sum_{p=0}^{k} (-1)^p \binom{k}{p}(a+p)^k &=& [x^k]: k! \sum_{p=0}^{k} (-1)^p \binom{k}{p} e^{ax} (e^x)^p \\
&=& [x^k]:k!  e^{ax} (1-e^x)^k= (-1)^k k! \\
\end{eqnarray*}
and so the sum is independent of $a$.


问:
What's the best current proof of the Poincaré conjecture?
Is there a current "gold standard" proof of the Poincaré Conjecture? I am interested in reading through the full proof, and some day perhaps reading Perelman's proof/notes. For now, though, I was wondering what the most "polished" version is, which would represent any advances during the 17 years since his publications and the 14 years since the formalization by various groups. I can see that Terry Tao has published some notes recently, but I don't believe he has published a full proof.</p>

答:
Still, the only (essentially) self-contained treatment of Perelman's proof of the Poincare Conjecture is

Morgan, John; Tian, Gang, Ricci flow and the Poincaré conjecture, Clay Mathematics Monographs 3. Providence, RI: American Mathematical Society (AMS); Cambridge, MA: Clay Mathematics Institute (ISBN 978-0-8218-4328-4/hbk). xlii, 521 p. (2007). ZBL1179.57045.

The only major part of the proof which they only sketch but do not provide details of is the short-time existence theorem for RF. One thing they do not address is the equivalence  between the topological and smooth category for 3-dimensional manifolds, but, as long as one is only interested in classifying smooth manifolds up to diffeomorphism, this does not matter.

Since then, there were numerous advances in understanding Ricci Flow in dimension 3 as well as in higher dimensions, but these did not lead to a substantial simplification of the proof. Needless to say, I do not recommend reading either the above book or other accounts of the proof of the 3-dimensional Poincare Conjecture (and, more generally, the Geometrization Conjecture) to an undergraduate student. My suggestion instead would be to read a textbook on Differential (Riemannian) Geometry.


问:
Question similar to Collatz conjecture
Let $a_0$ be an positive integer, define

$$  a_{n+1} =
\begin{cases}
a_n/2,  & \text{if $a_n$ is even} \\
a_n+2a_0-1, & \text{if $a_n$ is odd}
\end{cases}$$

Now form a sequence $S(a_0)_{n\ge 0}$ by performing this operation repeatedly.

Example: let $a_0=9$ then sequence $S(9)=\{9,26,13,30,15,32,16,8,4,2,1\}$

1 Question: For every $a_0$, sequence $S(a_0)$ will eventually reach the number $1$?

More example

S(1):  1

S(2):  2 1

S(3):  3 8 4 2 1

S(4):  4 2 1

S(5):  5 14 7 16 8 4 2 1

S(6):  6 3 14 7 18 9 20 10 5 16 8 4 2 1

S(7):  7 20 10 5 18 9 22 11 24 12 6 3 16 8 4 2 1

S(8):  8 4 2 1

S(9):  9 26 13 30 15 32 16 8 4 2 1

S(10):  10 5 24 12 6 3 22 11 30 15 34 17 36 18 9 28 14 7 26 13 32 16 8 4 2 1

Extending above claim

Define Sequence $S_t(a_0)$ where $$  a_{n+1} =
\begin{cases}
a_n/2,  & \text{if $a_n$ is even} \\
a_n+2^ta_0-1, & \text{if $a_n$ is odd}
\end{cases}$$

With $a_0$ and $t$ are  given positive integers.

Note $S_1(a_0)=S(a_0)$

Example $S_2(9)=\{9, 44, 22, 11, 46, 23, 58, 29, 64, 32, 16, 8, 4, 2, 1\}$

2 Question : for every $a_0,t\in\mathbb{N}$, sequence $S_t(a_0)$ will eventually reach the number $1$?

Source python

```<code>n=1
while n<11:

    def collatz(n):
        k=n
        while (n > 1):
            print(n, end=' ')
            if (n % 2):
                # n is odd
                n = n + 2*k-1
            else:
                # n is even
                n = n//2
        print(1, end='')

    #n = int(input('Enter n: '))
    print('\n',n,'Sequence: ', end=' ')
    collatz(n)
    n=n+1
'''

</code>```

答:
Indeed, the sequence will reach $1$ eventually. Here is how you can see it. Let's look into a slightly modified sequence:

$$b_{n+1}=\begin{cases}b_n/2&b_n\text{ even}\\\frac{b_n-1}{2}+a_0&b_n\text{ odd}\end{cases}$$

$$b_0=1$$

This is the same sequence as the previous one, except that it is "shifted" by one place ($b_1=a_0$) and from the original sequence we drop the first (necessarily even) number following any odd number.

Example: $a_0=6$, $(a_n)_{n\in\mathbb N}=(6, 3, 14, 7, 18, 9, 20, 10, 5, 16, 8, 4, 2, 1,\ldots), (b_n)_{n\in\mathbb N}=(1, 6, 3, 7, 9, 10, 5, 8, 4, 2, 1,\ldots)$.

The sequence $(b_n)$ is easier to look at, for example it immediately follows that its members are all smaller than $2a_0$, in fact they are bound by $2a_0-2$ (induction on $n$):

If $b_n$ is even, then $b_{n+1}=b_n/2\lt b_n\le 2a_0-2$

If $b_n$ is odd, then $b_{n+1}=\frac{b_n-1}{2}+a_0\le \frac{2a_0-3}{2}+a_0=2a_0-3/2$, but as the LHS is an integer, it must further be $\le 2a_0-2$.

(By the way, this also means that $(a_n)$ will be bound by $4a_0-4$.)

Now, this means that the sequence $b_n$ will eventually start to repeat.

On the other hand, the sequence $b_n$ can be also "replayed backwards":

$$b_{n-1}=\begin{cases}2b_n,&b_n\lt a_0\\2(b_n-a_0)+1,&b_n\ge a_0\end{cases}$$

This is because, if the following item $b_n$ is $b_n\ge a_0$, it could've only been produced by the second rule (or otherwise, if produced by halving, we would have $b_{n-1}\ge 2a_0$). Also, if the following item $b_n$ is $b_n\lt a_0$, then it could've only been produced by halving, because the second rule produces only values $\ge a_0$.

From the two facts shown in italics above, we can conclude that one of $b_n$'s will eventually be $1$. Namely, the sequence will repeat eventually, and let $b_m=b_n$ be the first repetition (lexicographically), with $m<n$. If $m=0$, then $b_n=b_0=1$ - problem solved! If $m\ne 0$, then, because of the "replay backwards" rule we have $b_{m-1}=b_{n-1}$ - a contradiction with $b_m=b_n$ being the first repetition.

Title: Question similar to Collatz conjecture


问:
How can I construct such set?
Is it possible to construct a uncountable set $S\subset (0,1]$ which satisfies:

$I_0$ - Every sequence $(x_{n})\subset S$ has the property $\sum_{n=1}^\infty x_{n}<\infty$

$I_1$ - Take any sequence of disjoint sequences in $S$, i.e. if $x_{k,n}$ are sequences in $S$ (for fixed $k$), then $x_{k,n}\neq x_{s,m}$ for all $n,m,s,k$. If we enumarate the sum of the sequences, i.e. $x_{1,k}=\sum_{n=1}^\infty x_{k,n}$ then $\sum _{k=1}^{\infty} x_{1,k}<\infty$

Is it possible to construct a uncountable set $R\subset (0,1]$ which satisfies: $I_0$, $I_1$ and :

$I_2$ - We repeat the process in $I_1$, but now with the sequences of sums formed in $I_1$.

...

$I_{n+1}$ - We repeat the process $I_1$, but now with the sequences of sums formed in $I_n$.

...

I tried to make things all clear, but it is really hard to write what I am thinking. Also I could not find a better title, so feel free to change it to a better one.

Thank you

答:
There is no uncountable set satisfying $I_0$.

Let $S$ be some set satisfying $I_0$, then for every $\varepsilon>0$ there are only finitely many $x\in S\cap \left(\varepsilon,\infty\right)$. If there were infinitely many I could choose a countable sequence $x_i>\varepsilon$ and $\sum_{i=1}^\infty x_i = \infty$.

Therefore I may arrange the elements of $S\cap \left(\varepsilon,\infty\right)$ in decreasing order, and letting $\varepsilon\to 0$ this gives me a bijection between $S$ and $\mathbb N$.

Title: How can I construct such set?


问:
How to prove by contrapositive
By proving the contrapositive, prove that, if n^2 is not divisible by 5, then n is not divisible by 5.</p>

答:
You want to prove $5\nmid n^2\Rightarrow 5\nmid n$. The contrapositive of this statement is $5\mid n\Rightarrow 5\mid n^2$. Clearly this holds. Take for example, $5\mid 10$ as $5\times 2=10$, therefore $5\times 2\times 10=100=10^2$.

You have to remember that the contrapositive of $A\Rightarrow B$ is $\overline{B}\Rightarrow \overline A$. Sometimes, you can make a prove easy by taking the contrapositive, your question is an example of this.

Title: How to prove by contrapositive


问:
$p=\infty$ case for Minkowski's inequality.
Prove that $||f+g||_{\infty} = ||f||_{\infty} + ||g||_{\infty}$ iff either one of $f$ and $g$ is $0$ a.e., or $A_f \cap A_g \neq \varnothing$ and for some $\{x_n\}_{n \in \mathbb{N}} \subset A_f \cap A_g$, there exists $\lambda>0$ such that $b=\lambda \, a$, where $f(x_n) \to a$ and $g(x_n) \to b$.

Here, we set $A_f=\{\{x_n\}_{n \in \mathbb{N}}\}$ where $\{x_n\}_{n \in \mathbb{N}} \subset X$ such that $|f(x_n)| \to ||{f}||_{\infty}$.

I have proved the if part, but still doubtful about the only if part.

答:
There is specific and a wider definition of the Minkowski's inequalities. For example Minkowskis Inequalities states a rather different definition. This is another definition Minkowski inequality. This reference offers a proof and references to other types of proofs. This is another informative and reliable reference minkowski inequality. The ask for properties of $f$ and $g$.

This page focues on proving on the Inequality for Sums. This is a answer set to comparable or closely related question on this community: prove minkowskis inequality directly in finite dimensions. But this is restricted for $f$ and $g$ to be convex and somewhat polynomial. And this minkowskis-inequality used the Hölder's inequaltiy twice. This one names it homogenization: proving minkowskis inequality with homogenization. And another one using Holder and Minkowski minkowskis and holders inequality confusion making use to the $=$ concept. This can be continued.

Your concept with the sequences and  function sequences is for most of the proofs a lit to much effort, but addmitantly this is sensible if used correct. The first equation is on or in and that has to be taken into account in proper proof theory. But for that $A_f$, $A_g$ and X have to be defined as in the courses or referenced correct. For Minkowski $|(_)|→||||_∞$ is inappropriate. $|(_)|_p→_n→_p||||_∞$ and so on.


问:
If $M \cong T \oplus F$, then $T \cong \text{Tors}(M)$ where $T$ is torsion and $F$ is free.
Define $\text{Tors}(M)$ to be the torsion submodule of a module $M$ over an integral domain $R$. The claim is now the following.

Claim: Let $M$ be a module over an integral domain $R$. If $M \cong T \oplus F$ where $T$ is torsion and $F$ is free, then $T \cong \text{Tors}(M)$.

I can't even see a clear direction for the proof but apparently notes I found online put it off as obvious which is definitely not to me.

My attempt stuck just after writing down the definitions. Since $T$ is torsion, then $T = \text{Tors}(T)$ i.e. for every element $t \in T$, there is a nonzero $r \in R$ such that $rt = 0_T$. Now an element $\text{Tors}(M)$ are elements $m \in M$ such that there exists nonzero $s \in R$ such that $sm = 0_M = (0_T, 0_F)$. From here, I don't see how to construct the isomorphism. Anyone care to point me in the right direction?

答:
$\textbf{Hint}$

First, notice that if $x \in Tors(M)$, if $\phi$ denotes the isomorphism $M \simeq T \oplus F$ then $\phi(x)$ is a torsion element.

For a generic element $(a, b) \in T \oplus F$ in order to be a torsion element there should exist a $k \in R$ such that $k \cdot (a,b)= (ka, kb)=(0,0)$. Since $F$ is a free module over an integral domain this means that $b$ must be equal to $0$. It is not difficult to conclude from this point.


问:
Basis for complement of a subspace
Consider we have a vector space $V$ with a given basis $\{v_i\}_{i \in I}$, and a subspace $U$. Show that we can always find a subset $J \subset I$ such that $\{v_i\}_{i \in J}$ is a basis for a complement of $U$ in $V$. For me intuitively it is clear but I cannot make it rigorous. By a complement $W$ of $U$ in $V$ I mean a subspace $W \subset V$ such that $V=U + W$ and $U \cap W =\{0\}$.

答: Let $\mathcal F$ be the set of those subsets $J$ of $I$ such that its $\operatorname{span}\bigl(\{v_i\mid i\in I\}\bigr)\cap W=\{0\}$. You can deduce from Zorn's lemma that $\mathcal F$ has a maximal element $M$. Then $\{v_i\mid i\in M\}$ is a basis of a complement of $W$ in $V$.

Title: Basis for complement of a subspace


答:
First of all, let $x\in S\subseteq(0,1]$, then the sum $\sum_{n=1}^\infty x$ cannot converge. Hence we need to restrict $I_0$ to the requirement that any injective sequence $x_n$ has a convergent sum.

Now note the following:
$$S = \bigcup_{n=1}^\infty S\cap\left(\frac{1}{n},1\right]$$
since this is a countable union and since $S$ is uncountable, there needs to be an $n\in\mathbb{N}$ such that $S\cap\left(\frac{1}{n},1\right]$ is uncountable.
But then there is an injective sequence $(x_m)\subseteq S\cap\left(\frac{1}{n},1\right]\subseteq S$ and clearly $\sum_{m=1}^\infty x_m < \sum_{m=1}^\infty \frac{1}{n} = \infty$.
Thus even for injective sequences $I_0$ can never be satisfied.

Title: How can I construct such set?

Body:
Is it possible to construct a uncountable set $S\subset (0,1]$ which satisfies:

$I_0$ - Every sequence $(x_{n})\subset S$ has the property $\sum_{n=1}^\infty x_{n}<\infty$

$I_1$ - Take any sequence of disjoint sequences in $S$, i.e. if $x_{k,n}$ are sequences in $S$ (for fixed $k$), then $x_{k,n}\neq x_{s,m}$ for all $n,m,s,k$. If we enumarate the sum of the sequences, i.e. $x_{1,k}=\sum_{n=1}^\infty x_{k,n}$ then $\sum _{k=1}^{\infty} x_{1,k}<\infty$

Is it possible to construct a uncountable set $R\subset (0,1]$ which satisfies: $I_0$, $I_1$ and :

$I_2$ - We repeat the process in $I_1$, but now with the sequences of sums formed in $I_1$.

...

$I_{n+1}$ - We repeat the process $I_1$, but now with the sequences of sums formed in $I_n$.

...

I tried to make things all clear, but it is really hard to write what I am thinking. Also I could not find a better title, so feel free to change it to a better one.

Thank you


问:
Sampling distribution of sample mean
We are sampling $Y_1,\dots,Y_n$ without replacement from the population of size N with $Y\sim$ Bernuolli($\mu$). So what will the asymptotic distribution of $\bar(Y_n)$ would be?

Since CLT only applies to indepedent $Y_i$'s but clearly our $Y_i$'s are not independent. So I don't know how to get the whole distribution

答:
If the population has $\mu N$ values of $1$ and $(1-\mu)N$ values of $0$ then the distribution of $\sum\limits_1^n Y_i$ is a hypergeometric distribution with mean $n\mu$ and variance $n\mu(1-\mu)\frac{N-n}{N-1}$, and the distribution of $\bar Y_n = \frac1n \sum\limits_1^n Y_i$ is a scaled version of this so with mean $\mu$ and variance $\frac1n \mu(1-\mu)\frac{N-n}{N-1}$

If the population has a binomial distribution for the number of $1$s with parameters $N$ and $\mu$ with the rest being $0$s then the distribution of $\sum\limits_1^n Y_i$ is a binomial distribution with parameters $n$ and $\mu$ with mean $n\mu$ and variance $n\mu(1-\mu)$, and the distribution of $\bar Y_n = \frac1n \sum\limits_1^n Y_i$ is a scaled version of this so with mean $n\mu$ and variance $\frac1n \mu(1-\mu)$

You ask for an asymptotic distribution, but that depends on what limits you are taking.  If it is as $n \to \infty$ and $N \to \frac{n}k$ for some $k$ with $0 < k < 1$ then in both cases I would expect a CLT result, so with suitable adjustment to location and scale towards a normal distribution:

so in the hypergeometric case $\frac{1}{\sqrt{n}}\left(\bar Y_n - \mu\right)$ will converge in distribution towards $\mathcal N\left(0, \mu(1-\mu)(1-k)\right)$

while in the binomial case $\frac{1}{\sqrt{n}}\left(\bar Y_n - \mu\right)$ will converge in distribution towards $\mathcal N\left(0, \mu(1-\mu)\right)$

Title: Sampling distribution of sample mean


答:
$$ ( \cos x)^{\frac{1}{ \sin x} }= ( 1 + ( \cos x -1) )^{\frac{1}{ \sin x} } = 1 + \frac{\cos x -1}{\sin x} + \frac{1}{\sin x} \left( \frac{1}{ \sin x} -1 \right) \frac{( \cos x -1)^2}{2} + \text{stuff}$$

Taking the limit as $ x \to 0$ and using the fact that:

$$ \lim_{ x \to 0} \frac{ (\cos x -1)^j}{ (\sin x)^k} =0$$

for $ j,k,n \in \mathbb{N}$  and $j \geq k$

$$ \lim_{x \to 0} ( \cos x)^{\frac{1}{ \sin x } } = 1 $$

Explanation:  The exponent is exploding off to infinity, but even for such a large exponent, we can use binomial series if the $ ( \cos x -1) \leq 1$  which it is for some interval around $ x=0$. Use that and the second limit result I've written, you get the answer.

*:You can prove this by some maclaurain expansions.

Title: limit of $\cos{x}^{\frac{1}{\sin{x}}}$


问:
Why does $a^{\log_a(x)}=x$?
I'm currently learning about logs and yesterday made a post on here.

In that post I was told that $a^{\log_a(x)}=x$

From a commenter:

"It should be clear that..."

Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $a^{log_a(x)}=x$?

答:
By definition, the $\log-$function is the inverse of the exponential function. It means that, if $f:\mathbb R\to \mathbb R^+$ such that,
$$f(x)=a^x$$
then its inverse is a function $f^{-1}:\mathbb R^+\to \mathbb R$ such that,

$$f(f^{-1}(x))=x.$$

We then define $f^{-1}$ as $f^{-1}(x)=\log_a x$. So,

$$f(f^{-1}(x))=x\Leftrightarrow a^{\log_a x}=x.$$

Title: Why does $a^{\log_a(x)}=x$?


问:
Splitting of Singular Homologies
In Singular homology, let $C_n(X)$ be the free abelian group generated by all the $n$-siimplices of the topological space $X$. Let $U$ be a subspace of $X$, then we have a spliting sequence
$0\rightarrow C_n(U) \rightarrow C_n(X) \rightarrow C_n(X,U) \rightarrow 0$.

But why we do not have the splitting $H_n(X) \cong H_n(U) \oplus H_n(X,U)$ of homologies?

答:
Adding an answer here that I found helpful when thinking about why the homology groups split in the case of retractions: a retraction $r:X \to A$ is left inverse for the inclusion $i$. Functors always preserve mutual left/right inverse-ness (in other words they preserve split epi/mono pairs) so there is a left inverse for the induced map $H_\bullet(i):H_\bullet(A) \to H_\bullet(X)$. On the other hand, the pointwise splitting of a chain complex is not necessarily a chain-map. More precisely, the short exact sequence of chain groups
$$ 0 \to C_n(A) \to C_n(X) \to C_n(X, A) \to 0$$ splits in $\mathrm{Ab}$ for any given $n$, but the short exact sequence of chain-complexes
$$ 0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$ may not split in the category of chain complexes. If the latter sequence splits (such as when there is a retraction $r:X \to A$, or when $X$ is the direct sum of $A$ and some other space), then by applying the homology functor you do in fact get a splitting of the homology groups.

Title: Splitting of Singular Homologies


问:
Show that the only solution to $\phi(n) =n-2$ is $n=4$
Came across this question in Number Theory.

Let $\phi$ denote Euler's totient function;  Show that the only
  solution to $\phi(n) =n-2$ is $n=4$

My workings so far have included, firstly proving that $4$ is indeed a solution, I have also noted that the solution cannot be prime because $\phi(p)=p-1$.

From here, could I note that my solution must be composite, and the only way in which it could have only 2 numbers which are not coprime would be if n is a square number?

Not sure how to prove this completely though.

Also, could I incorporate the fact that  $\phi(p^k)=p^{k-1}(p-1)$ in any way?

All help greatly appreciated. Thanks.

答: Hint: We know that $(n,n-1)=1$ and $(1,n)=1$.  For $n >4$, can you use primes to come up with a further coprime integer less than $n$?


问:
Prove the inequality $4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$
$$4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$$,  $a,b,c\ge0$

I tried to work out this inequality the way I did
So we have to show  $$ 4(a + b + c) ^ 3 \ge 27 (a ^ 2b + b ^ 2c + c ^ 2a + abc) $$

One way is to use cyclic symmetry, and WLOG assumes $ a $ is $ a, b, c $ minute. Then we can write $ b = a + x, c = a + y $, where $ x, y \ge 0 $. Now the inequality reduces to 9 $$ (x ^ 2-xy + y ^ 2) + (x-2y) ^ 2 (4x + y) \ge 0 $$ which is obvious. Also from the above, we get then equality is possible iff $ x = y = 0 $ or when $ a = 0, x = 2y $, i.e. when $ (a, b, c) = (1, 1, 1) $ or permutation $ (0, 2, 1) $.

I think it's a mistake please help

答:
The last inequality should be
$$9(x^2-xy+y^2)a+4x^3-15x^2y+12xy^2+4y^3 \geqslant 0,$$
or
$$9(x^2-xy+y^2)a+(4x+y)(x-2y)^2 \geqslant 0.$$
It's called BW method.


答: Assume that $a^b = x$. If we take the logarithm of both sides in base $a$, we will get $\log_a(a^b) = \log_a(x)$. By the definition of logarithm, $log_a(a^b) = b$, since logarithm function returns the exponent of an input number in the given base. Thus, returning to $\log_a(a^b) = \log_a(x)$, we can see that $b = \log_a(a^b) = \log_a(x)$, and $b = \log_a(x)$. Since our original equation is  $a^b = x$, we can just substitute $b$ in the original equation to find the expression $a^{\log_a(x)} = x$. Hope this is clear!

Title: Why does $a^{\log_a(x)}=x$?

Body:
I'm currently learning about logs and yesterday made a post on here.

In that post I was told that $a^{\log_a(x)}=x$

From a commenter:

"It should be clear that..."

Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $a^{log_a(x)}=x$?


问:
Calculating whether infimum optimization problem is convex
Is the following optimization problem convex?

$$\text{inf}\quad 5x_1^2 - 4x_1x_2 + 2x_2^2 + x_1x_3 \quad \text{ s.t.} \\
\quad x_1 - x_3 = 1 \\
\quad 4x_1^2 + 9x_2^2 - x_3^2 \leq 25
$$

I know that the first constraint is convex as it is linear. The second constraint is also convex as it is a cone. The intersection of two sets is convex, hence the feasible region is convex.

I calculated the Hessian of the problem and got this:

$$H(x_1, x_2, x_3 ) = \begin{bmatrix}
10 & -4 & 1 \\
-4 & 4 & 0 \\
1 & 0 & 0
\end{bmatrix}$$

The determinant of the matrix is $-4$, hence it is negative definite and the problem is therefore not convex. Is this correct?

答:
You almost got it right. A function is convex if and only if its Hessian is PSD. And a matrix is PSD iff all it's eigenvalues are non-negative.

The objective function is not convex since it has one negative eigenvalue ($\approx$-0.158). For $2\times 2$ matrices only, a matrix is PSD iff its determinant is non-negative.

The first constraint is indeed linear, hence it is convex.

The last constraint is again non-convex, as one of its eigenvalues is -2. This one is easy since the Hessian of the third constraint is a diagonal matrix, therefore it's eigenvalues are simply the diagonal entries.


问:
Finding the condition number of a matrix $A$, and the values of b such that $A$ is ill-conditioned
Let $A=\begin{pmatrix} 1 & b \\ 0 & 1\\\end{pmatrix}$, what is the condition number of A, for which  values of b  the matrix  A is ill-conditioned.

My trial:
Since it does not specify in which norm to calculate the cond number, therefore I computed it for the $l^1, l^{\infty}, l^2$ norms, using the formula :
$cond(A)= ||A|| ||A^{-1}||$.

I got

For $l^1$ $cond_1(A)= (1+|a|)^2$

For $l^{\infty}$ $cond_{\infty}(A)= (1+|a|)^2$

For $l^2$ $cond_2(A)\leq (2+a^2)$

And the

I don't know if this is what I shoud compute, or there are more options that I shoud consider.

For the second part, i am not sure about it but I think that the matrix would be ill-conditioned when it has a  big or large condition number so it happens for big values of b.

Can you correct me, or give any tips.

答:
You computed correctly $K_1(A)$ and $K_{\infty}(A)$.

Try to solve $A x=c$ when $c=[1,1]$ and then compute it for a perturbed r.h.s. $\tilde{c} = c + \delta c$. The key thing point in this case is that $$\frac{||\delta x||}{||x||} \leq K(A) \frac{||\delta c||}{||c||}$$
i.e. the condition number is a magnification factor for the relative error. Note also that it's not so important in this case which norm you use to compute $K(A)$.

For $b=2 \cdot 10^{6}$ you obtain $x=[-1999999,1]$ to be the exact solution.

Let's perturb of the r.h.s. $c$ with a $\delta c$ s.t. $||\delta c||=10^{-3}$: the solution is

$$\tilde{x} =x + \delta x=[-2.001998999000000\cdot 10^{6},1+10^{-3}]$$ where the first component is dramatically different from the first one of $x$. Thus your linear system has an high sensitivity to a small variation of the data, and this behaviour is justified by the large condition number you got in this case: $K_\infty(A)=(1+2 \cdot 10^{6})^2$.

You can try by yourself playing with the following Python snippet

```<code>    import numpy as np

    b = 1e6;
    A = np.array([[1,b],[0,1]])
    print("Condition number compute with linalg",np.linalg.cond(A))
    c = np.array([1,1])

    deltac = np.array([1e-3,1e-3]) #perturbation
    cc = c + deltac #perturbed r.h.s.

    x = np.linalg.solve(A,c)
    xx = np.linalg.solve(A,cc)

    print("Solution of original system is",x)
    print("Solution with perturbed r.h.s.",xx)
</code>```

Body:
You computed correctly $K_1(A)$ and $K_{\infty}(A)$.

Try to solve $A x=c$ when $c=[1,1]$ and then compute it for a perturbed r.h.s. $\tilde{c} = c + \delta c$. The key thing point in this case is that $$\frac{||\delta x||}{||x||} \leq K(A) \frac{||\delta c||}{||c||}$$
i.e. the condition number is a magnification factor for the relative error. Note also that it's not so important in this case which norm you use to compute $K(A)$.

For $b=2 \cdot 10^{6}$ you obtain $x=[-1999999,1]$ to be the exact solution.

Let's perturb of the r.h.s. $c$ with a $\delta c$ s.t. $||\delta c||=10^{-3}$: the solution is

$$\tilde{x} =x + \delta x=[-2.001998999000000\cdot 10^{6},1+10^{-3}]$$ where the first component is dramatically different from the first one of $x$. Thus your linear system has an high sensitivity to a small variation of the data, and this behaviour is justified by the large condition number you got in this case: $K_\infty(A)=(1+2 \cdot 10^{6})^2$.

You can try by yourself playing with the following Python snippet

    import numpy as np

    b = 1e6;
    A = np.array([[1,b],[0,1]])
    print("Condition number compute with linalg",np.linalg.cond(A))
    c = np.array([1,1])

    deltac = np.array([1e-3,1e-3]) #perturbation
    cc = c + deltac #perturbed r.h.s.

    x = np.linalg.solve(A,c)
    xx = np.linalg.solve(A,cc)

    print("Solution of original system is",x)
    print("Solution with perturbed r.h.s.",xx)


问:
What is the meaning of k-th largest absolute value?
Let $A$ be a matrix. What is the meaning of $λ_k$ is the eigenvalue with $k$-th largest absolute value?

答: After the eigen decomposition, you can order the eigenvalues. The k-th largest eigenvalue is $\lambda_k$, which is usually the k-th most important eigenvalue of the decomposed matrix.


问:
Differntiability of multivariable function
Let $ S\subset\mathbb{R}^{2} $ be the set defined by :

$ S=\left\{ \begin{pmatrix}x\\
|x|
\end{pmatrix}:x\in[-1,1]\right\}  $

let $ \gamma:[a,b]\to\mathbb{R}^{2} $ be a continuous 1-1 function such that $ \gamma\left(a\right)=\begin{pmatrix}-1\\
1
\end{pmatrix}\thinspace\thinspace\thinspace,\thinspace\thinspace\thinspace\gamma\left(b\right)=\begin{pmatrix}1\\
1
\end{pmatrix} $.

Let $ t_{0}\in[a,b] $ such that

$ \gamma\left(t_{0}\right)=\begin{pmatrix}0\\
0
\end{pmatrix} $

I have to prove or disprove:

If $ \gamma $ is differntiable at $ t_0 $ then $ \gamma'\left(t_{0}\right)=\begin{pmatrix}0\\
0
\end{pmatrix} $.

I tried to assume for contradiction that $ \gamma $ is diiferentiable at $ t_0 $ and that $ \gamma'\left(t_{0}\right)\neq\begin{pmatrix}0\\
0
\end{pmatrix} $

But couldnt reach a contradiction. Any ideas would help, thanks.

答:
I assume, you want that $\{\gamma(t),t\in[a,b]\}=S$

Solution assuming $\gamma$ is differentiable everywhere, which is not true

If $\gamma(t)=(\gamma_1(t), \gamma_2(t))^\intercal$ is differntiable, then both $\gamma_1(t), \gamma_2(t)$ have derivatives at every point $t\in[a,b]$.

Consider, $\gamma_2(t)$. It is a differentiable function that reaches minimum at $t_0$, since $\gamma_2(t_0)=0$ and $\gamma_2(t)\ge0$ by the construction of set $S$. Thus, $\gamma'_2(t_0)=0$.

Let's assume $\gamma'_1(t_0)\neq0$, then there exists a neighbourhood $U=(t_0-\epsilon, t_0+\epsilon)$ where $\gamma_1(t)$ is monotonous. Together with continuity it gives you the invertibility of $\gamma_1(t)$ in $U$, so there is a continuous differentiable function $t=g(\gamma_1)$. However, function $\gamma_2(t)=\gamma_2(g(\gamma_1))=|\gamma_1|$ doesn't have a derivative at $\gamma_1=0, g(\gamma_1)=t_0$ despite being a composition of differentiable functions.

Solution if $\gamma$ is differentiable only at $t_0$

There exist $U=(t_0-\epsilon,t_0+\epsilon)$ and $g(t)$, that
$$\gamma(t_0+h) = \gamma(t_0) + \gamma'(t_0)h + g(h),$$
where $\lim_{h\to 0} |g(h)|/h \to 0$. Assume $|\gamma'(t_0)|=C\neq0$. There exist some $h_0>0$ that both $g(h_0)$ and $g(-h_0)$ are smaller than $h_0C/8$.

Let's denote $a=\gamma(t_0+h)$, $b=\gamma(t_0-h)$, $c=h_0\gamma'(t_0)$. Triangle inequality:

$$
|a| > |c| - |a-c| = |h_0\gamma'(t_0)| - |g(h_0)| = Ch_0 - \frac18Ch_0 = \frac78 Ch_0
$$

Thus, for each point $|\gamma_1|=\gamma_2$: $\gamma_2(t_0+h_0)>Ch_0 7/8\sqrt2 > 0.6 Ch_0$

Another triangle inequality gives us:

$$
|b-(-a)| < |b-(-c)| + |(-c)-(-a)| = |g(-h_0)| + |g(h_0)| < \frac14 Ch_0
$$

In other words:
$$|\gamma_2(t_0-h_0)+\gamma_2(t_0+h_0)| < \frac14 Ch_0$$

Since $\gamma_2(t_0+h_0) > 0$:
$$
\gamma_2(t_0-h_0) < \gamma_2(t_0+h_0) + Ch_0/4 < −0.35 Ch_0.
$$

Which is impossible since $\gamma_2\ge0$


问:
limit of $\cos{x}^{\frac{1}{\sin{x}}}$
I need to solve this limit:

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}}$$

I feel that it should be a rather easy limit, but I find myself struggling with the answer. I have tried substituting $\cos{x}$ and $\sin{x}$ with their series expansion, $\cos{x} \approx 1-\frac{x^2}{2}+\dots$ and $\sin{x} \approx x-\frac{x^3}{6}+\dots$, but the indeterminate form still remains.

Title: limit of $\cos{x}^{\frac{1}{\sin{x}}}$

Body:
$$y=\big[\cos({x})\big]^{\frac{1}{\sin({x})}}\implies \log(y)={\frac{1}{\sin({x})}}\log(\cos(x))$$
Now, composing Taylor series
$$1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\log(y)=\frac{-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right) } {x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) }=-\frac{x}{2}-\frac{x^3}{6}+O\left(x^5\right)$$
$$y=e^{\log(y)}=1-\frac{x}{2}+\frac{x^2}{8}+O\left(x^3\right)$$


问:
In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

What I Tried: Here is the figure :-

I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.

Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.

Can anyone help me? Thank You.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain where you're stuck.)

Hint: There is a degree of freedom in the problem.

Hence conclude that the area is $5\times 5 = 25$ by considering a special case (and assuming the area is an invariant).

This motivates how I thought of the next step.

Hint: Rotate $\Delta AED$ $90^\circ$ about $A$. What do we get?

(Note: There is slight work involved in proving this.)

Hence, conclude that the area is indeed 25.


问:
$[h_i,{(ad\ f_i)^t\ f_j}]=-(t\alpha_i+\alpha_j)(h_i){(ad\ f_i)^t\ f_j}$ in a Kac-Moody algebra
In Carter's Lie algebras of finite and affine type, when constructing the Weyl group in the setting of a Kac-Moody algebras, Carter uses an identity $[h_i,{(ad\ f_i)^t(f_j)}]=-(t\alpha_i+\alpha_j)(h_i){(ad\ f_i)^t(f_j)}$. I'm having trouble seeing why this is the case? I've tried showing it by induction, but it doesn't seem to get me anywhere (the basecase is just from the defintion).

答:
It seems that induction on $t$ works. As you noted, the case $t=0$ follows from the definition. Suppose $[h_i, (ad f_i)^t (f_j)]=-(t\alpha_i + \alpha_j)(h_j)(ad f_i)^t(f_j) $ and write $x=(ad f_i)^t (f_j)$ for simplicity.

Then $(ad f_i)^{t+1}(f_j) = [f_i, x] $ and
\begin{align} [h_i, (ad f_i)^{t+1}(f_j)] &= [f_i, [h_i, x]]-[x,[h_i, f_i]] &&\text{Jacobi identity} \\ &=[f_i, -(t\alpha_i+\alpha_j)(h_i)x]+[x,\alpha_i(h_i)f_i] &&\text{induction hypothesis}\\ &=-(t\alpha_i+\alpha_j)(h_i)[f_i,x]-\alpha_i(h_i)[f_i, x] \\ &=-\left(  (t+1)\alpha_i + \alpha_j \right)(h_i)[f_i, x] \\ &= -\left(  (t+1)\alpha_i + \alpha_j \right)(h_i)(ad f_i)^{t+1}(f_j)
\end{align}

as desired.


问:
Cholesky decomposition -
I am trying to understand the reason why the author of my book resort to Cholesky decomposition in this particular case.

He starts assuming that we know the historical prices of each asset, so (for the relationship between price and return $R_t:=\frac{S_t}{S_0}$) the conditions

(2.16) $\rightarrow \mu-\Gamma_t\sigma\leq \widetilde{r}_t^S \leq \mu+\Gamma_t\sigma$

(2.17) $\rightarrow e^{t\mu_{\operatorname{log}}-\Gamma \sqrt{t}\sigma_{\operatorname{log}}}\leq \widetilde{R}_t^S\leq e^{t\mu_{\operatorname{log}}+\Gamma \sqrt{t}\sigma_{\operatorname{log}}}$

that represent respectively the uncertainty set for single-period returns and cumulative returns (with $\mu,\mu_{\operatorname{log}},\sigma, \sigma_{\operatorname{log}},\Gamma \in \mathbb{R}^+$) allow us, like he says, to estimate the upper bound and lower bound for $\widetilde{S}_t^m$ (with $\sim$ that denotes the randomness of variable and $m$ is the number of assets included in the underlying basket of derivative). In fact, knowing that

$U:=\begin{Bmatrix}
\widetilde{R}_t^S: \underline{R}_t^S \leq \widetilde{R}_t^S \leq \overline{R}_t^S
\end{Bmatrix}=\begin{Bmatrix}
\widetilde{R}_t^S: (1+\underline{r}_{t-1}^S)\widetilde{R}_{t-1}^S \leq \widetilde{R}_t^S \leq (1+\overline{r}_{t-1}^S)\widetilde{R}_t^S
\end{Bmatrix}=\begin{Bmatrix}
\frac{\widetilde{S}_t^S}{S_0}: \frac{\underline{S}_t^S}{S_0} \leq \frac{\widetilde{S}_t^S}{S_0} \leq \frac{\overline{S}_t^S}{S_0}\end{Bmatrix}=\begin{Bmatrix}
\frac{\widetilde{S}_t^S}{S_0}: (1+\underline{r}_{t-1}^S)\frac{\widetilde{S}_{t-1}^S}{S_0} \leq \frac{\widetilde{S}_t^S}{S_0} \leq (1+\overline{r}_{t-1}^S)\frac{\widetilde{S}_{t-1}^S}{S_0}
\end{Bmatrix}$

for $S$ the single asset, we can extend this consideration to $m$ assets and obtain

$\Rightarrow\left\{\begin{matrix}
\frac{\widetilde{S}_{t}^m}{S_0}\leq \frac{\overline{S}_{t}^m}{S_0}\Rightarrow \widetilde{S}_{t}^m\leq \overline{S}_{t}^m,\forall t=1,...,T,\forall m=1...,M\\
\frac{\widetilde{S}_{t}^m}{S_0}\geq \frac{\underline{S}_{t}^m}{S_0}\Rightarrow \widetilde{S}_{t}^m\geq \underline{S}_{t}^m,\forall t=1,...,T,\forall m=1...,M\\
\frac{\widetilde{S}_{t}^m}{S_0}\leq \frac{\widetilde{S}_{t-1}^m}{S_0}(1+\overline{r}_{t-1}^S)\Rightarrow \widetilde{S}_{t}^m\leq \widetilde{S}_{t-1}^m(1+\overline{r}_{t-1}^S),\forall t=1,...,T,\forall m=1...,M\\
\frac{\widetilde{S}_{t}^m}{S_0}\geq \frac{\widetilde{S}_{t-1}^m}{S_0}(1+\underline{r}_{t-1}^S)\Rightarrow \widetilde{S}_{t}^m\geq \widetilde{S}_{t-1}^m(1+\underline{r}_{t-1}^S),\forall t=1,...,T,\forall m=1...,M
\end{matrix}\right.$

for $m=1,...,M$ the basket of assets. Thus we can estimate $\underline{S}_t^m$ and $\overline{S}_t^m$ precisely because of those conditions. In particular we extract samples from time series of prices of $m$ assets, i.e. to say from time series of log-returns, and under a condition of invariance of statistic properties of samples (and hence of parameters of mean and variance) due to second order stationarity of processes related to the single time series we proceed to estimate their respective sample means and standard deviations. This condition, obviously, is based on the assumptions of normality and independence of log-returns that apply per every singly series. (We are in a Black and Scholes environment). Then he says that we have to consider the correlation between the assets and to do it we have to consider the correlation between the asset returns. So we'll use the  covariance matrix of the single period returns. Therefore we consider, one by one, all possible couple of assets, and for any $j$-th realization of returns of assets (for $j=1,...,n$ and $n$ the sample size) we'll calculate the sum of cross products of deviations in order to complete the covariance matrix. We also know that $\sum$ is symmetric and positive definite. Well.

At this point, author introduces Cholesky decomposition. You can also see here (pag. 31-33).

My question is: why the Cholesky decomposition? To what end?

Thanks in advance just for reading.

答:
It says in the paper you linked. If a matrix is symmetric positive definite then it has a Cholesky decomposition. It is the LU decomposition but for symmetric positive definite matrices.

It says on page 848

...Let $\Sigma$ be the covariance matrix of the single period returns.
Given the $\Sigma$ is asymmetric and positive definite, it has a
Cholesky decomposition and we can compute matrix $C = (\Sigma^{-1})$

It then sets up a relationship for the correlation between the asset returns

$$ \bigg\|  \mathbf{C} \bigg( \tilde{\mathbf{R}}_{1} - \hat{\mathbf{R}}_{2} \bigg)\bigg\|
 \leqslant  \mathbf{\Gamma}  $$

It is quicker than the standard LU decomposition and there is a connection between it and other decompositions.

Title: Cholesky decomposition -


问:
Compute $\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$
Here is an integral.

$\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$

Obviously, the function is not analytic in $D(0,4)$, thus, we should apply residue theorem

$\oint_{c} f(z) d z=2 \pi i \sum_{j=1}^{n} b_{1 j}$

where $b_{1j}$ is the first coefficient in Laurent series at the point of singularity.

So,

at the point $z_1 = 2$ we have a pole of the first order.

$Res_{z_1} = e^1 $

at the point $z_2 = 1$ we have an essential singularity. Compute Laurent series of the function in the point $z=1$. WolframAlpha:

$\frac{e^{1 /(z-1)}}{z-2} = \sqrt[x-1]{e}\left(-1-(z-1)-(z-1)^{2}-(z-1)^{3}+O\left((z-1)^{4}\right)\right)$

Thus, $Res_{z_2} =  -1$

And we get:

$\oint_{c} f(z) d z=2 \pi i (e - 1)$

It does not make sense, for a correct answer is $2i$

What is wrong? Tell me, please.

答:
The computation of the residue is wrong as it has to be done term by term multiplying the Laurent series of the exponential with the geometric series expansion of the denominator in powers of $z-1$; straightforward but tedious.

A much easier way is to use Cauchy and note that for $R>4$ we have (noting that if $z=Re^{it}, dz=iRe^{it}dt=izdt$):

$\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} dz=\oint_{|z|=R} \frac{e^{1 /(z-1)}}{z-2} dz=\int_0^{2\pi}\frac{ize^{1 /(z-1)}}{z-2} dt \to 2\pi i, |z|=R \to \infty$ as $\frac{iz}{z-2} \to i$ and $e^{1 /(z-1)} \to 1$ for $|z|=R \to \infty$.

so the integral is actually $2\pi i$


问:
An Interesting Number Theory Problem: Solve $2520=(x+y+xy)^2+2xy+2y-3x, x, y \in \mathbb Z$
Solve:

$2520=(x+y+xy)^2+2xy+2y-3x$

Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.

Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers

答:
Denote $z=x+1, w=y+1$ then your result $$t^2+2t-5x=2520 \implies (t+1)^2-5(z-1)=2519 \\\implies w^2z^2-5z=2516=z(w^2 z - 5) \implies z|2516=2^2\cdot 17\cdot 37$$

If $z$ is even then $w^2z-5$ is odd, so $4|z, z=4u, 4u^2w^2=5u+629 \implies u\equiv -1 \pmod 4$, and $u|17\cdot 37$, therefore $u=-1, -17, -37, -629$, none of which yields a perfect square for $5u+629$.

If $z$ is odd then $wz$ is odd, $(wz)^2 \equiv 1 \pmod 4$, so $z\equiv 1 \pmod 4, z=1, 17, 37, 629$. We have the following four cases:

$z=1, w^2-5=2516$, no solution;

$z=17, 17w^2-5=148 \implies w=\pm 3 \implies x=16, y=2 \text{ or } -4;$

$z=37, 37w^2-5=68$, no solution;

$z=629, 629w^2-5=4$, no solution.

So the only solutions are $(16, 2)$ and $(16, -4)$.