问:
Probabilistic modelling
This question is about how to make sense of a probabilistic model I'm reading about:

You have three random variable $A$, $B$ and $C$ that are all real-valued. You want to model an input-output relationship with these random variables, by assuming the following relationships between them:

It is assumed that $A$ is the constant random variable that only takes the value $a\in \mathbb
{R}$, where $a$ is thought of as input.

Furthermore, $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ and  $P(C=c | B=b)=\mathcal{N}(c;2\cdot b^2,1)$, where $\mathcal{N}(x;\mu,1)$ denotes the probability density of a normal distribution with mean $\mu$ and variance $1$.

My questions are:

Does an equation like $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ even make sense?

If we takes samples, on the left we have a number between $0$ and $1$ but on the right we have a number that can be larger, as the pdf does not need to stay within $[0,1]$, right?

What is the output?

In the (not online available) text I'm reading, the above is all that is mentioned. I would assume that to arrive at the output, one needs to proceed in the following way:

Given $a$, draw a number $b$ from a normally distribution with mean $3\cdot e^a$ and unit variance and then similarly draw $c$, the output, from a normally distribution with mean with mean $2\cdot b^2$ and unit variance. But I'm not sure if that's correct; also, it is not clear to me, in case this interpretation is indeed correct, why conditional probabilities in the model formulation are needed.

答:
Firstly, the following

It is assumed that $A$ is the constant random variable that only takes the value $a\in \mathbb{R}$, where $a$ is thought of as input.

does not really make sense. If something is assumed to be constant, then there is no need to model it as a random variable. Moreover if $A$ is going to model the input, it must vary from time to time. So let us assume that this is rather poorly written - $A$ is a random variable with some (maybe unknown) non-constant distribution. Maybe it should be discrete, or taking only two values? It might make sense.

Nextly, we have

Furthermore, $\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1)$ and $\mathbb{P}(C=c|B=b)=\mathcal{N}(c;2⋅b^2,1)$, where $\mathcal{N}(x;\mu,1)$ denotes the probability density of a normal distribution with mean $\mu$ and variance $1$.

This is also weirdly written, and makes little sense. You have written

...on the left we have a number between 0 and 1 but on the right we have a number that can be larger...

You are write, we should not use $\mathbb{P}$ sign when the probability density function is really what we are referring to.

What the author probably wanted to say, was that conditionally on $A$, the distribution of $B$ is normal with given ($A$ dependent) parameters. Later on, after $(A, B$) have been sampled, the distribution of $C$ is also normal with the assigned ($B, A$ dependent) parameters.

The above is only my educated guess, as in the cited text it is written

$$\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1),$$
\begin{equation} \mathbb{P}(C=c|B=b)=\mathcal{N}(c;2⋅b^2,1), \tag{1} \end{equation}
and I assume that it should have been
$$\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1),$$
\begin{equation}\mathbb{P}(C=c|B=b, A=a)=\mathcal{N}(c;2⋅b^2,1). \tag{2} \end{equation}
My guess is based on the fact, that as You say, we are actually going to model some process. As long as $(1)$ might be valuable piece of information, it does not define a unique model.

Moreover, this should be written in a more clear way though. For example:

$$B|A \sim \mathcal{N}(3\cdot e^A,1),$$
and
$$C|B,A \sim \mathcal{N} (2\cdot B^2,1).$$

Note that this is exactly what You have written in Your interpretation

Given $a$, draw a number $b$ from a normally distribution with mean $3\cdot e^a$ and unit variance and then similarly draw $c$, the output, from a normally distribution with mean with mean $2\cdot b^2$ and unit variance.

I think it is OK. Saying it once again in different words:

we start by sampling $A$,

given $A$, we sample $B$ from $\mathcal{N}(3\cdot e^A,1)$,

given $B, A$, we sample $C$ from $\mathcal{N} (2\cdot B^2,1).$

Such models - known as hierarchical models - are very important  in modern statistics.
Basically, Bayesian Statistics is devoted almost entirely to the study of such models.

Title: Probabilistic modelling


问:
Does convergence of a trace imply that of an eigenvalue?
I came up with the following conjecture:

Let $A_n$ be an $n \times n$ positive definite matrix, and suppose that $0 < \lambda_1^{(n)} \leq \cdots \leq \lambda_n^{(n)}$ are the eigenvalues of $A_n.$
Then $\frac{1}{n}\mathrm{tr}(A_n) \to 1$ and $\frac{1}{n}\mathrm{tr}(A_n^{-1}) \to 1$ hold if and only if $\lambda_1^{(n)} \to 1$ and $\lambda_n^{(n)} \to 1.$

The "if" part is obvious because all eigenvalues of $A_n$ get close to one, but I cannot find a good way to show the "only if" part. I would appreciate it if you could give me a hint. You can assume some additional conditions on the order of the convergence if necessary.

答:
Let
$ A_n$ be the diagonal matrix with diagonal entries $\frac12, 1,\ldots, 1$. Then $$\frac1n\operatorname{tr} A_n=1-\frac1{2n}\to 1$$ and $$\frac1n\operatorname{tr} A_n^{-1}=1+\frac1n\to 1$$
but of course $\lambda_1^{(n)}\not\to 1$.

ContentLicense: CC BY-SA 4.0


问:
Help regarding odd/even functions and limits.

Hey I was wondering if I'm missing something in my proof or if there are any gaps of logic.. Specifically I'm not sure about my justification of the chain rule and overall accuracy..

b.

Claim: $g(x) = -g(-x)$ ($g$ is odd)

Proof:

$-g\left(-x\right)\ =\ f\left(-x\right)-f\left(x\right)\ =>g\left(-x\right)=f\left(x\right)-f\left(-x\right)$

Then we have

$g\left(x\right)=f\left(x\right)-f\left(-x\right) = -g(-x)$

Therefore $g$ is odd.

Now we show there exists a point $x_{0} ∈ R $ such that $g'(x_{0}) = 0$.

Note that by the algebra of differentiable functions $g$ is differentiable.

By (3a) we know that since $g$ is odd, we have

$\lim _{x\to \infty }\left(g\left(x\right)\right)=0 =>\:\lim _{x\to -\infty }\left(g\left(x\right)\right)=-0 = 0$

Therefore by HW10 question (5b) we know that $g$ has an extreme point in $R$, we shall denote it $x_{0}$.

Since $g$ is differentiable at $x_{0}$, and $x_{0}$ is an extreme point of $g$, by Fermat's theorem we have
$g'(x_{0}) = 0$.

c.

Note that since $f$ is differentiable at $x_{0}$, and $g'$ is differetiable at $f(x_{0})$ by the chain rule we have

$g'\left(x_{0}\right)=f'\left(x_{0}\right)-\left(-f'\left(-x_{0}\right)\right)\ =f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$

$=>f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$

Which mean $f'\left(x_{0}\right)$ and $f'\left(-x_{0}\right)$ are additive inverses of each other.

Then either $f'\left(x_{0}\right)=f'\left(-x_{0}\right)=0$, in which case there exist $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.

or we have $f'\left(x_{0}\right)\cdot f'\left(-x_{0}\right)<0$, in which case by Darboux's theorem there exist a point $c∈R$ such that

$x_{0}<c<-x_{0}$ or $-x_{0}<c<x_{0}$ and $f'(c)=0$.

Therefore there must exist an $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.

答: If there is no such $x_1$ then $f'(x) >0$ for all $x$ or $f'(x) <0$ for all $x$. [This is because derivatives always have intermedite value property. See Darboux Theorem: https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)]. Suppose $f'(x) >0$ for all $x$. Then $f$ is strictly incerasing. This implies that $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ both exist (possibly $\pm \infty$). However, $g(n)=f(n)-f(-n)\to 0$. So $f$ has the same limits at $\pm \infty$ and $f$ is strictly increasing. Do you see a contradiction? [The case when $f$ is strictly decreasing is handled by changing $f$ to $-f$].


答:
$f(x)$ doesn't have to be bounded, so no, $\lim_{x \to \infty}f(x)$ doesn't necessarily exist.

$$$$

I would approach proving $(c)$ as follows:

If there does not exist $x_1$ such that $f'(x_1) = 0,$ then this means that $f(x)$ is strictly monotone - i.e. either strictly increasing or strictly decreasing. Suppose that $f(x)$ is strictly increasing (the proof for strictly decreasing is similar). Let $c \in \mathbb{R}.$

By the definition of strictly increasing, $f(x) > f(c+1) > f(c)$ for all $x > c+1$ and $f(x) < f(c)$ for all $x < c$.

Then $\lim_{x \to -\infty}f(x) < f(c) < f(c+1) < \lim_{x \to \infty}f(x) \\ \implies \lim_{x \to \infty} \left(f(x)-f(-x)\right) \geq f(c+1)-f(c) > 0.$

This contradicts the property: $\lim_{x \to \infty} \left( f(x) - f(-x) \right)  = 0$.

Title: Help regarding odd/even functions and limits.

Body:
Hey I was wondering if I'm missing something in my proof or if there are any gaps of logic.. Specifically I'm not sure about my justification of the chain rule and overall accuracy..

b.

Claim: $g(x) = -g(-x)$ ($g$ is odd)

Proof:

$-g\left(-x\right)\ =\ f\left(-x\right)-f\left(x\right)\ =>g\left(-x\right)=f\left(x\right)-f\left(-x\right)$

Then we have

$g\left(x\right)=f\left(x\right)-f\left(-x\right) = -g(-x)$

Therefore $g$ is odd.

Now we show there exists a point $x_{0} ∈ R $ such that $g'(x_{0}) = 0$.

Note that by the algebra of differentiable functions $g$ is differentiable.

By (3a) we know that since $g$ is odd, we have

$\lim _{x\to \infty }\left(g\left(x\right)\right)=0 =>\:\lim _{x\to -\infty }\left(g\left(x\right)\right)=-0 = 0$

Therefore by HW10 question (5b) we know that $g$ has an extreme point in $R$, we shall denote it $x_{0}$.

Since $g$ is differentiable at $x_{0}$, and $x_{0}$ is an extreme point of $g$, by Fermat's theorem we have
$g'(x_{0}) = 0$.

c.

Note that since $f$ is differentiable at $x_{0}$, and $g'$ is differetiable at $f(x_{0})$ by the chain rule we have

$g'\left(x_{0}\right)=f'\left(x_{0}\right)-\left(-f'\left(-x_{0}\right)\right)\ =f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$

$=>f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$

Which mean $f'\left(x_{0}\right)$ and $f'\left(-x_{0}\right)$ are additive inverses of each other.

Then either $f'\left(x_{0}\right)=f'\left(-x_{0}\right)=0$, in which case there exist $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.

or we have $f'\left(x_{0}\right)\cdot f'\left(-x_{0}\right)<0$, in which case by Darboux's theorem there exist a point $c∈R$ such that

$x_{0}<c<-x_{0}$ or $-x_{0}<c<x_{0}$ and $f'(c)=0$.

Therefore there must exist an $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.


问:
maximum eigenvalue of rank 1 update matrix
I am stuck with the following problem. The ingredients are:

$A=diag(a_1,\dots,a_n)$, $a_i \in (0,1)$ and $a_1+\dots+a_n=1$,

$B=diag(b_1,\dots,b_n)$, $b_i \in (0,1)$ and $b_1<b_2<\dots<b_n$,

the entries of the matrices $A$ and $B$ are related by $\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$.

Denote ${\bf b}^T=(\sqrt{a_1} b_1,\dots,\sqrt{a_n}b_n)$, $c={1 \over \sum_{i=1}^n a_i (1-b_i)}$. The problem is to prove that the maximum eigenvalue of the matrix $D=B+ c {\bf b} {\bf b}^{T}$ is less than 1.

I've been studying the steady-state distribution of the two-dimensional Markov chain. Its steady-state vector turned out to be ${\vec p}_k={\vec p}_1 D^{k-1}$, $k \ge 1$. As usual, ${\vec p}_1$ is found from the normalization condition, which requires evaluation of $\sum_{k=0}^\infty D^k$. To make it $(I-D)^{-1}$, the aforementioned proof is required. My numerous numerical expreiments show that the maximum eigenvalue is $<1$. But I cannot prove it.

What I've tried and failed:

interative methods (like power iteration) don't help since the power $D^k$ does not have a nice expression.

Gershgorin theorem does not help, since some entries of $D {\vec 1}$ may be $>1$.

the matrix $c {\bf b} {\bf b}^{T} $ is of rank 1 and has one eigenvalue $\rho=c \sum_{i=1}^n {a_ib_i^2}$. This matrix is symmetric.

the eigenvalues of $D$ are all different and do satisfy the well-known secular equation $f(\lambda)=0$, where:
$$
f(\lambda)=1+ c \sum_{i=1}^n {a_i b_i^2 \over b_i - \lambda}.
$$

denote the roots of $f(\lambda)=0$ by $\lambda_1,\dots,\lambda_n$. According to the theory, the roots satisfy the inequalities:
$$
b_1<\lambda_1<b_2<\lambda_2<\dots<b_n<\lambda_n.
$$
This does not help since the upper bound for the maximum eigenvalue $\lambda_n$ remains unknown. The theory says that $\lambda_i=b_i+m_i \rho$, where $m_i \in (0,1)$ and $\sum_{i=1}^n m_i=1$. But I can't prove that $\lambda_n=b_n+m_n\rho<1$.

$\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$ is equivalent to $1<\sum\limits_{i=1}^n {a_i \over 1-b_i}<2$. Since $\sum_i (u_i/v_i) \sum_i (u_iv_i) \ge 1$ we have $\sum\limits_{i=1}^n a_i (1-b_i)>{1 \over2}$ and thus $1<c<2$.

one of the classical inequalities gives
$$
\sum_{i=1}^n a_i (1-b_i)+ (1-b_1)(1-b_r) \sum_{i=1}^n {a_i \over 1-b_i} \le 2- b_1-b_r,
$$
and thus
$$
\sum_{i=1}^n a_i (1-b_i) \le 2- b_1-b_n - (1-b_1)(1-b_n) = 1-b_1 b_n <1,
$$
and thus, instead of $1<c<2$ we have ${1 \over 1-b_1 b_n}<c<2$. Moreover, since $\sum_{i=1}^n a_i (1-b_i)>{1\over 2}$, from the last inequality we get
${1\over 2} \le 1-b_1 b_n$ or $b_1 b_n < {1\over 2}$. Since $b_1<b_2<\dots<b_n$ we have
$$
b_1 b_i < {1\over 2}.
$$

Unfortunately, all this did not help me to make a step towards the proof and I would be grateful for any useful suggestions.

答:
Seems that I have found the solution. It is given by the Example 7.10.3 on page 620 in here link.
Specifically, if a matrix $H$ is non-negative then its maximum eigenvalue, say $\rho(H)$, satisfies $\rho(H)<r$ if and only if $(rI-H)^{-1}$ exists and $(rI-H)^{-1}\ge 0$.

Let us consider the matrix $(I-D)^{-1}$. Due to Sherman–Morrison formula
this inverse is equal to
$$
(I-D)^{-1}=(I-B)^{-1} + c {(I-B)^{-1} {\bf b} {\bf b}^{T} (I-B)^{-1} \over 1 - c {\bf b}^T (I-B)^{-1} {\bf b} }
= \underbrace{(I-B)^{-1}}_{\ge 0 \, component-wise\!\!\!\!\!\!\!\!} + {c \over \alpha} (I-B)^{-1} \underbrace{{\bf b} {\bf b}^{T}}_{\ge 0} (I-B)^{-1},
$$
and exists if and only if $\alpha \neq 0$. By direct inspection the value of $\alpha$ is equal to
$$
\alpha=1-c\sum_{i=1}^n {a_i b_i^2 \over 1-b_i}
$$
and clearly $\alpha\neq 0$, and thus $(I-D)^{-1}$ exists.

Now notice that $(I-D)^{-1}\ge 0$ if ${c\over \alpha}>0$. The value of $\alpha$ can be rewritten:
$$
\alpha=1-c\sum_{i=1}^n {a_i (b_i-1+1)^2 \over 1-b_i}=
c \left (2-\sum_{i=1}^n {a_i \over 1-b_i} \right )
=
c \left (1- \sum_{i=1}^n {a_i b_i \over 1-b_i} \right ).
$$
Thus ${c\over \alpha}>0$ if ${1 \over \left (1- \sum\limits_{i=1}^n {a_i b_i \over 1-b_i} \right )}>0$, which is equivalent to $\sum\limits_{i=1}^n {a_i b_i \over 1-b_i}<1$.

Thus we can use the statement of the Example 7.10.3 and thus the maximum eigenvalue of $D$ is less than 1.


问:
Linearization of an implicitly defined function.
Problem:

Given the equation: $xz^{2}+y^{2}z^{5}=19$

Also given: (3,4,1) is a solution to the equation.  This point is not the only solution.

1) Find dz/dx and dz/dy (through implicit differentiation) evaluated at (3,4).

2) Find the linearization L(x,y) of z(x,y) at (3,4)

3)  Use L to approximate z(3.01, 4.02).  Plug this (approximate) result into the given equation to see if it satisfies it.

My attempted solution:

1) dz/dx = $(-z^{2})/(2xz+5y^{2}z^{4})$

dz/dx evaluated at (3,4) = $(-z^{2})/(6z+80z^{4})$

dz/dy = $(-2yz^{5})/((2xz+5y^{2}z^{4})$

dz/dy evaluated at (3,4) = $(-8z^{5})/(6z+80z^{4})$

2) Formula for linearization: L(x) = f(a) + Df(a) (x-a)

f(a) = $3z^{2} + 16z^{5}=19$

$(z-1)(16z^{4} + 16z^{3} + 16z^{2} +19{z} +19)=0$

z=1 is the only real solution

I'm not entirely sure I am solving f(a) correctly.  Up until now, I have only dealt with explicitly defined functions.

Df(a) = Dz(a)[dz/dx evaluated at a  dz/dy evaluated at a]

Dz= \begin{bmatrix}
(-z^{2})/(6z)+80z^{4}),(-8z^{5})/((6z)+80z^{4})&\end{bmatrix}

(x-a) =
\begin{bmatrix}
x-3,y-4\\ \end{bmatrix} (Transposed so that it is a column vector).

L(x) = $1 + (x-3)*((-z^{2})/(6z+80z^{4})+(y-4)*((-8z^{5})/(6z+80z^{4}))$

3) $1+((-0.01z^{2})/(6z+80z^{4})) + ((-0.16z^{5}/(6z+80z^{4}))$

Is this the approximation for z(3,4)?  I'm not sure how to proceed from here.  What do I plug into the original equation to see that it approximately satisfies the equation?

Thanks!  Sorry about the formatting. I tried to tidy up the formatting a bit.  I'm not too familiar with latex though.

答:
1) Your differentiation is correct. Plug $z=1$ to have numerical values.

2) You have a function in the form of
$$L(x,y,z)=xz^2+y^2z^5-19=0$$
The linearization around $(x_0,y_0,z_0)$ is as follows
$$\hat L(x,y,z)=L(x_0,y_0,z_0)+\frac{\partial L}{\partial x}|_{x_0,y_0,z_0}(x-x_0)+\frac{\partial L}{\partial y}|_{x_0,y_0,z_0}(y-y_0)+\frac{\partial L}{\partial z}|_{x_0,y_0,z_0}(z-z_0)=0$$
By some manipulation and since $L(x,y,z)=0$ by definition
$$z=z_0-\frac{\frac{\partial L}{\partial x}|_{x_0,y_0,z_0}(x-x_0)+\frac{\partial L}{\partial y}|_{x_0,y_0,z_0}(y-y_0)}{\frac{\partial L}{\partial z}|_{x_0,y_0,z_0}}$$
Here you can find partial derivatives and plug $(x_0,y_0,z_0)=(3,4,1)$ to have a linear eqaution.
$$\frac{\partial L(3,4,1)}{\partial x}=z^2=1$$
$$\frac{\partial L(3,4,1)}{\partial y}=2yz^5=8$$
$$\frac{\partial L(3,4,1)}{\partial z}=2xz+5y^2z^4=86$$
$$\Rightarrow z=1-\frac 1{86}(x-3)-\frac 8{86}(y-4)$$

3) For $(x,y)=(3.01,4.02)$ the linear equation gives for $z$
$$z=1-\frac 1{86}(0.01)-\frac 8{86}(0.02)=0.998$$
If you plug it into original equation
$$L(3.01,4.02,0.998)=-0.00259$$
which is the linearization error.

ContentLicense: CC BY-SA 3.0


问:
Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?
When I get around at park , I made up a question to hang out by myself.However , I stuck in it.

Question: Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?

Where I stuck in :

$1-)$ I know how to write $75$ in the base of $3$ , but I do not know writing $75!$ in the base of $3$.

$2-)$ To find the last two digits, we utilize from $(\text{mod }100)$, then which modulus should I use to find its last two digits?

Thanks for your helps..

答:
Hint: $75!$ is a multiple of $3^2$.

More generally, the number of trailing zeros in the representation of $n!$ in base $3$ is
$$
f(n)=\sum _{{i=1}}^{\infty}\left\lfloor {\frac  {n}{3^{i}}}\right\rfloor
$$
but note that there are only finitely many nonzero terms.

So, $75!$ has $35$ trailing zeros in base $3$.


问:
How to calculate the Area of a "Polygon" including Arcs
I have a set of points forming a polygon. However, any 3 points in this polygon can also be represented as an arc (starting at point 1, through point 2, to point 3).

I need to find the area of this polygon (which technically is not a polygon but an area formed by straight or curved lines from point to point).

My idea was, to exclude the arcs (say the middle point of each arc) and calculate the area of the resulting polygon separately and then add the areas of the arcs:

```<code>sum = 0
loop through points
  if next three points contain arc middle point
    add arc area to sum
    get rid of middle point of arc and use next point instead

  add area of points to sum
</code>```

This will work for convex shapes but I don't know how to solve it for concave shapes.

How would I go about this?

//EDIT: The arcs are always circular arcs (~circle segments). The radius and the angle of the segment is known (it can be calculated from the 3 points).

答:
Can't you dissect it into convex pieces as indicated below?

EDIT

In response to the OP's comment, if you had an inward bulge, you could connect the two extreme points, and compute the area of a convex shape minus the area of another convex shape.

If you are trying to write a computer program to do this, you must first specify exactly what kinds of shapes you will be dealing with, and how they will be presented.  For example, if the shapes are presented as a list of points in counterclockwise order, with line segments or circular arcs joining them, then lhf's suggestion of Green's theorem seems hard to beat.  It's not at all clear to me how a "curved line" will be presented to the program, nor what kinds of curves you want to be prepared to deal with.

I definitely would want to break the area up into regions bounded by simple closed curves, that is, curves that do not intersect themselves.  In your second diagram, I'd want to break the area into two pieces, which would require first finding the point where the boundary curve crosses itself.

I don't really know much about this.  I answered because I thought you were asking a simpler question.  I've added the computational-geometry tag in hopes of attracting an expert.


答: Since $75!= 1\times 2\times \dots \times 9 \times \dots \times 75$, it is divisible by $9$, which means that it can be written as $$75!= 3^2(a_0 + 3a_1 +3^2a_2 +\dots +3^n a_n) =3^2 a_0 +3^3a_1 +3^4 a_2 +\dots+3^{n+2}a_n\\ =(a_n \dots a_2 \ a_1 a_0 \ 0\ 0)_3 $$

Body:
When I get around at park , I made up a question to hang out by myself.However , I stuck in it.

Question: Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?

Where I stuck in :

$1-)$ I know how to write $75$ in the base of $3$ , but I do not know writing $75!$ in the base of $3$.

$2-)$ To find the last two digits, we utilize from $(\text{mod }100)$, then which modulus should I use to find its last two digits?

Thanks for your helps..


问:
Interconversion between Complex Number Forms
Suppose we give an oscillatory force $\Re{(F)}$, where $F=F_0 \text{e}^{\text{i}\omega t}$, onto a system, which responds with a velocity $\Re{(v)}$, where $v=v_0 \text{e}^{\text{i}\omega t}$. ($F_0$ and $v_0$ are complex numbers as well.)

The power input $P$ is therefore $P=\Re{(F)}\Re{(v)}$. How can I write $P$ in the form $\Re{(A\text{e}^{\text{i}\alpha(t)}+B)}$ instead, where $A$ and $B$ are constants?

$\Re{(x)}$ denotes "$\text{Re}(x)$."

I tried the identity $\Re{(z)}=\frac{1}{2}(z+z^*)$, but that doesn't work.

答:
Hint: a first attempt is to represent each complex number in the form $a + bi$, and if it's a function, then to represent as $x(t) + iv(t)$, where $x(t)$ and $v(t)$ are real functions, as well as $a, b$.

Another Hint: to simplify, it's good also to remember that $(a + bi)(c+di) = (ac-bd)+ i(ad+bd)$.


答: Hint: Find a parametrization of the boundary and use Green's theorem.

Title: How to calculate the Area of a "Polygon" including Arcs

Body:
I have a set of points forming a polygon. However, any 3 points in this polygon can also be represented as an arc (starting at point 1, through point 2, to point 3).

I need to find the area of this polygon (which technically is not a polygon but an area formed by straight or curved lines from point to point).

My idea was, to exclude the arcs (say the middle point of each arc) and calculate the area of the resulting polygon separately and then add the areas of the arcs:

```<code>sum = 0
loop through points
  if next three points contain arc middle point
    add arc area to sum
    get rid of middle point of arc and use next point instead

  add area of points to sum
</code>```

This will work for convex shapes but I don't know how to solve it for concave shapes.

How would I go about this?

//EDIT: The arcs are always circular arcs (~circle segments). The radius and the angle of the segment is known (it can be calculated from the 3 points).


问:
Proving a Covariant Derivative is Torsion Free
Let $(M,g)$ be a metric manifold and $\phi:M\to N$ a diffeomorphism, where $N$ is another manifold. Let $\nabla$ be the Levi Civita connection with respect to the metric $g$, and we define a connection in $(N,\phi_*(g))$ by:

$$\tilde{\nabla}_XY=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)$$

I am trying to prove that $\tilde{\nabla}$ is the  Levi Civita connection of $(N,\phi_{*}(g))$. So there is a step in which I evidently make a mistake and I am not being able do find it. Im trying to prove that it is torsion free (I don't want an alternative way to do this, I simply want to understand what am I doing wrong) so lets go. We want to see that

$$\tilde{\nabla}_XY-\tilde{\nabla}_YX=[X,Y]$$

So, since $\phi$ is a diffeo, we have that $\phi^*=(\phi^{-1})_*$ and thus,

\begin{equation}
\begin{split}
\tilde{\nabla}_XY-\tilde{\nabla}_YX &=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)-\phi_*\left(\nabla_{\phi^*(Y)}\phi^*(X)\right)\\
&=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)-\nabla_{\phi^*(Y)}\phi^*(X)\right)\\
&=\phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right)
\end{split}
\end{equation}
where in the second line I used linearity of $\phi_*$ and in the third line that $\nabla$ is the LC connection and thus Torsion Free. So now comes the problem, lets act on a function $f:N\to \mathbb{R}$ to get:

\begin{equation}
\begin{split}
\phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right)(f)&=\left[\phi^*(X),\phi^*(Y)\right]\left(\phi^*(f)\right)\\
&=\phi^*(X)\left\{\phi^*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{\phi^*(X)(\left(\phi^*(f)\right))\right\}\\
&=\phi^*(X)\left\{(\phi^{-1})_*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{(\phi^{-1})_*(X)(\left(\phi^*(f)\right))\right\}\\
&=\phi^*(X)\left\{Y(f)\right\}-\phi^*(Y)\left\{X(f)\right\}
\end{split}
\end{equation}

where in the second line I have used that $[X,Y](f)=X(Y(f))-Y(X(f))$, in the third that $\phi^*=(\phi^{-1})_*$ and in the last line I used the definition of the pushforward of $X$ and $Y$ and the pullback of $f$ via $\phi^{-1}$ and $\phi$ respectively. However, in this last step there should be a mistake I am not being able to find since $\phi^*(X)$ should act on functions $h:M\to \mathbb{R}$ while $Y(g):N \to \mathbb{R}$.

Any help will be appreciated.

答:
The mistake is not only in the last line but also in the first line of your last mathematical paragraph. Let $Z$ be a vector field on $M$ and $f \colon N \rightarrow \mathbb{R}$ be a smooth function. In your derivation, you are using the "identity"

$$ \left( \phi_{*} \left( Z \right) \right)(f) = Z(\phi^{*}(f)) $$
(where $Z = [\phi^{*}(Z),\phi^{*}(Y)]$) but this identity clearly cannot hold as the left hand side is a function on $N$ while the right hand side is a function on $M$.

More explicitly, for $p \in M$ and $q \in N$ we have
$$ (\phi_{*}(Z)(f))(q) = df|_{q} \left( d\phi|_{\phi^{-1}(q)} \left( Z|_{\phi^{-1}(q)} \right) \right), \\
(Z(\phi^{*}(f)))(p) = d(\phi^{*}(f))|_{p}(Z_p) = d(f \circ \phi)|_{p}(Z_p) = df|_{\phi(p)} \left( d\phi|_{p} \left( Z_p \right) \right) $$
so a correct identity is
$$ \left( \phi_{*}(Z) \right)(f) = \phi_{*} \left( Z(\phi^{*}(f)) \right) $$
(this is an identity between functions on $N$ and you can also write another one in terms of functions on $M$).

Similarly, the identity

$$ \phi^{*}(X) \left( \phi^{*}(f) \right) = X(f) $$

is wrong but

$$ \phi^{*}(X) \left( \phi^{*}(f) \right) = \phi^{*} \left( X(f) \right) $$
is correct.

Thus,

$$ \phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \right)(f) =
\phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \left( \phi^{*}(f) \right) \right)=
\\
\phi_{*} \left( \phi^{*}(X) \left( \phi^{*}(Y) \left( \phi^{*}(f) \right) \right) - \phi^{*}(Y) \left( \phi^{*}(X) \left( \phi^{*}(f) \right) \right) \right) =
\\
\phi_{*} \left( \phi^{*}(X(Yf)) - \phi^{*}(Y(Xf)) \right) = X(Yf) - Y(Xf) = [X,Y](f). $$


问:
Fourier series of Bernoulli polynomials
We know that Bernoulli polynomials $\phi_n(x)$ can be defined as below:$$\phi_0(x)=1,\dot\phi_n(x)=n\phi_{n-1}(x),\int^1_0{\phi_n(x)=0}$$
and I want to use this to prove:$$\phi_k(x)=(-1)^{1+\frac{k}{2}}\frac{2(k!)}{(2\pi)^k}\sum_{n=1}^{\infty}{\frac{cos2\pi nx}{n^k}}$$where k is even and$$\phi_k(x)=(-1)^{\frac{k+1}{2}}\frac{2(k!)}{(2\pi)^k}\sum_{n=1}^{\infty}{\frac{sin2\pi nx}{n^k}}$$
where k is odd , but I have no idea about it , can anyone help me?

答:
I'll use the more familiar notation (for me) $B_n(x)$ for the Bernoulli polynomials.  The defining characteristics should make clear for which $n$ they apply, namely,
\begin{align}
B_0(x) &= 1 \\
\frac{d}{dx} B_n(x) = n B_{n-1}(x) \quad&\text{and} \quad
\int_0^1 B_n(x) dx = 0 \quad\text{when } n \geqslant 1
\end{align}
It follows $B_1(x) = x - \frac{1}{2}$ and $B_n(1) = B_n(0)$ when $n \geqslant 2$ because
\begin{align}
0 &= \int_0^1 B_{n-1}(x) dx \\
&= \Big[ \frac{1}{n} B_{n}(x) \Big]_0^1 \\
&=B_n(1)-B_n(0)
\end{align}
Therefore, each $B_n$ is continuously differentiable in $[0,1]$, being a  polynomial, and its end point values are the same (except for $B_1$) so its Fourier series will converge pointwise.

Then, if $B_n(x)$ has Fourier series $\displaystyle \sum_{k=-\infty}^{\infty}a_{n,k}e^{2\pi i k x}$, we have
\begin{align}
a_{n,k} = \int_0^1 B_n(x) e^{-2\pi i k x} dx
\end{align}
By direct calculation, when $n=0$, $a_{0,k} = \left\{ \array{0&\text{when }k \neq 0 \\ 1 &\text{when }k = 0}\right.$ and when $n\neq 0, k=0$ $a_{n,0} = 0$.

Otherwise, for $n \geqslant 2$,
\begin{align}
a_{n,k} &= \int_0^1 B_n(x) e^{-2\pi i k x} dx \\
&= -\frac{1}{2\pi i k}\Big[ B_n(x) e^{-2\pi i k x}\Big]_0^1 + \frac{1}{2\pi i k }\int_0^1 nB_{n-1}(x) e^{-2\pi i k x} dx.
\end{align}
If $n=1$ and $k \neq 0$ this becomes,
\begin{align}
a_{1,k} = -\frac{1}{2\pi i k}
\end{align}
and otherwise, we derive, $n \geqslant 2, k\neq 0$
\begin{align}
a_{n,k}=\frac{n}{2\pi i k}a_{n-1,k}.
\end{align}
Applied repeatedly and using the special cases derived above, we can see that
\begin{align}
a_{n,k} = -(-i)^n \frac{n!}{(2\pi k)^n}
=\left\{
\array{
(-1)^{n/2+1} \frac{n!}{(2\pi k)^n}&\text{when } n \geqslant 1 \text{ and even} \\
-(-1)^{(n+1)/2} \frac{n!}{(2\pi k)^n} \cdot i &\text{when } n \geqslant 1 \text{ and odd} \\
1 & \text{when } n = 0 \text{ and } k = 0 \\
0 &\text{otherwise.}
} \right.
\end{align}
These are expressed in terms of Fourier terms in $e^{2\pi i k x}$.  To convert to cosine and sine series simply pair the $k$ and $-k$ coefficients.  Even numbered pairs give the cosine terms with a factor of $2$ and odd numbered pairs the sine terms with a factor of $-2$.

Title: Fourier series of Bernoulli polynomials


问:
which of the followings are true for bijective functions
which of the followings are true:-

1. There is a continuous bijection from $\mathbb{R}^2\to \mathbb{R}$.

2. There is a bijection between $\Bbb{Q}$ and $\mathbb{Q}\times \mathbb{Q}$.

Can somebody help me please.I am totally stuck on it.

答:
1) Suppose such a continuous bijection $f$ exists. Let $p$ be any point in $\mathbb {R}^2$. Now consider the spaces $A=\mathbb R^2-\{p\}$ and $B=\mathbb R-\{f(p)\}$. Argue why $g:A\to B$ given by the restriction of $f$ is a continuous bijection. Now, any two points in $A$ can be joined by a curve, but not every two points in $B$ can be joined by a curve. Use that to arrive at a contradiction. </p>

<p>2) Both sets are countable. This can be shown in various ways, like the general theorem that the cartesian product of any two countable sets is countable (this will help also to prove that $\mathbb Q$ is countable, solving the problem). </p>

Body:
1) Suppose such a continuous bijection $f$ exists. Let $p$ be any point in $\mathbb {R}^2$. Now consider the spaces $A=\mathbb R^2-\{p\}$ and $B=\mathbb R-\{f(p)\}$. Argue why $g:A\to B$ given by the restriction of $f$ is a continuous bijection. Now, any two points in $A$ can be joined by a curve, but not every two points in $B$ can be joined by a curve. Use that to arrive at a contradiction.

2) Both sets are countable. This can be shown in various ways, like the general theorem that the cartesian product of any two countable sets is countable (this will help also to prove that $\mathbb Q$ is countable, solving the problem).


答:
$f(x)$ is defined on all $\mathbb{R}$ and so is $g(x)$.

$g(-x)=f(-x)-f(x)=-g(x)$ thus $g(x)$ is an odd function so $g(x)\to 0$ as $x\to -\infty$

This means that $g(x)$ is not increasing or decreasing on all $\mathbb{R}$. There is at least one point $x_0$ where $g'(x_0)=0$

$g'(x_0)=f'(x_0)+f'(-x_0)=0\to f'(x_0)=-f'(-x_0)$

As $f(x)$ is differentiable, $f'(x)$ is continuous and $f'(x)$ has opposite values at $x_0$ and $-x_0$, then for the IVT there exists at least one $x_1\in(-x_0,x_0)$ such that $f'(x_1)=0$


问:
Proving that $f(x) = (1-\cos x)\sin\frac1x$ is equally continuous on $(0,\infty)$
When dealing with a proof that a given function is a equally continuous function, and the function involves trigonometry, I find it hard if not impossible to prove it via the definition of an equally continuous function. I am looking for a better method.

Here is the function I have problems with:

Prove that
$$f(x) = (1-\cos x)\sin\frac1x$$
is equally continuous in the section $(0, \infty)$

How can I solve this problem?

答: Hint: Using the facts that $\frac {1-\cos x} {x^{2}} \to \frac1  2$ and $|\sin (\frac 1 x)| \leq \frac 1 {|x|}$ show that $f(x) \to 0$ as $ x\to 0+$. Also, using boubndedness of $1-\cos x$ show that $f(x) \to 0$ as $ x\to \infty$. These two facts and continuity of $f$ yield uniform continuity.


答:
$$-(1-\cos x)\le  (1-\cos x)\sin(1/x) \le (1-\cos x)$$
Then by sandwich theorem
$$\lim_{x \to 0} (1-\cos x) \sin(1/x)=0$$
As $\lim_{x \to 0} \pm (1-\cos x)=0$
Further,  $f(x)$ being finite in $(0,\infty)$, it is continuos everywhere

Title: Proving that $f(x) = (1-\cos x)\sin\frac1x$ is equally continuous on $(0,\infty)$

Body:
When dealing with a proof that a given function is a equally continuous function, and the function involves trigonometry, I find it hard if not impossible to prove it via the definition of an equally continuous function. I am looking for a better method.

Here is the function I have problems with:

Prove that
$$f(x) = (1-\cos x)\sin\frac1x$$
is equally continuous in the section $(0, \infty)$

How can I solve this problem?


问:
Prove $ax \equiv1 \textrm{ mod } 10^k \implies aN(x) \equiv1 \textrm{ mod } 10^{2k}$

For every $a \in Z$, consider the function $N:\mathbb{Z} \rightarrow \mathbb{Z}$:
$$N(x)=x(2-ax)$$
Prove for every $k \in \mathbb{N}$ that
$$ax \equiv1 \textrm{ mod } 10^k \implies aN(x) \equiv1 \textrm{ mod } 10^{2k}$$

Thoughts so far:
We can make the following computation and find that
$$aN(x)-1 = ax(2-ax)-1=-(ax-1)^2$$
which shows that $aN(x) \equiv1 \textrm{ mod } 10^k$, right? But where does $10^{2k}$ enter the picture?

答:
You're almost there:

$$ax-1=10^k q \implies (ax-1)^2=10^{2k} q^2 \equiv 0 \bmod 10^{2k}$$


问:
Hawaiian Earrings are covering space for shrinking wedge of circles

for any $P=(x,y) \in \mathbb{R}^2$,  Let $C_{P,n}$ denote a circle centred at $\big(x+\frac1n, y \big)$ of radius $\frac1n$. Let $W_{P}$ denote the shrinking wedge of circle $\bigcup_{n \in \mathbb{Z}} W_{(0,0)} $..

Define the space
$$X=\{0\} \times \mathbb{R} \cup \big(\bigcup_{i \in \mathbb{Z}}  W_{(0,4i)} \big)$$.
Show that X is a covering space for $W_{(0,0)}$.

Hatcher claims that $X$ is a covering space for $W_{(0,0)}$ but he did not provide any hint. I tried some maps but I wonder if they will be covering space. I was thinking that the number of circles in $X$ are countable and this is the case with $W_{(0,0)}$. Hence there is a bijection between them. Hence a circle in $X$ would map to a circle in $W_{(0,0)}$, but then I doubt this would be continous. Please help me with this problems.

答:
Let us prove a slightly more general result:

Let $(X,x_0) \vee (Y,y_0)$ be the wedge (= one point union) of two pointed spaces, let $p : \tilde X \to X$ be a covering map and $F = p^{-1}(x_0)$. On the disjoint union $D =  \tilde X + (Y \times F)$ define an equivalence relation $\sim$ by identfying $\xi \in F \subset \tilde X$ with $(y_0,\xi) \in Y \times F$ for all $\xi \in F$. Let $\pi : D \to Z = D/\sim$ denote the quotient map. Intuitively $Z$ is obtained from $\tilde X$ by gluing at copy of $Y$ to each $\xi \in F$ via identifying $y_0$ and $\xi$.

Define $r : D \to (X,x_0) \vee (Y,y_0), r(\xi) = p(\xi)$ for $\xi \in \tilde X$ and $r(y,\xi) = y$ for $(y,\xi) \in Y \times F$. This induces a unique map $\rho : Z \to (X,x_0) \vee (Y,y_0)$ which is clearly surjective. We claim that it is a covering map.

Let $U$ be an open subset of $X$ which is evenly covered by $p$. Write
$p^{-1}(U) = \bigcup V_\alpha$ with pairwise disjoint open $V_\alpha \subset \tilde X$ which are mapped by $p$ homeomorphically onto $U$.

Now let $x \in X$.

If $x$ has an open neigborhood $U$ evenly covered by $p$ such that $x_0 \notin U$, then $U$ is open in $(X,x_0) \vee (Y,y_0)$ and $\rho^{-1}(U) = \bigcup V_\alpha$ which shows that $U$ is evenly covered by $\rho$.

If $x$ has an open neigborhood $U$ evenly covered by $p$ such that $x_0 \in U$, then $(U,x_0) \vee (Y,y_0)$ is open in $(X,x_0) \vee (Y,y_0)$ and $\rho^{-1}((U,x_0) \vee (Y,y_0)) = \bigcup (V_\alpha,\xi_\alpha) \vee (Y,y_0)$ which is the union of pairwise disjoint open subsets of $Z$ which are mapped by $\rho$ homeomorphically onto $(U,x_0) \vee (Y,y_0)$. Thus $(U,x_0) \vee (Y,y_0)$ is evenly covered by $\rho$.

Similarly, let $y \in Y$. Since $x_0$ has an open neigborhood $U$ evenly covered by $p$, the argument of 2. above shows that $(U,x_0) \vee (Y,y_0)$ is an open neigborhood of $y$ evenly covered by $\rho$.

Let us come to your question. Let $H$ denote the Hawaiian earring with basepoint $h_0$ = point where all circles meet. Then $(S^1,*) \vee (H,h_0)$ is homeomorphic to $H$ and our above construction with the standard covering map $p : \mathbb R \to S^1$ gives us the desired covering of the Hawaiian earring,


问:
totally unimodular matrix - negative identity matrix
Theorem: The inverse of a (non-singular) totally unimodular matrix is totally unimodular.

The proof will use the following lemmas.

Lemma 1: Permuting rows and columns preserves total unimodularity.

Lemma 2: Matrix A is totally unimodular if and only if the matrix [AI]
is totally unimodular (where I is the identity matrix of the appropriate size).

The slides provided as a support material for my course say that:

Given A $\in M_{m x n}$

such affermations are equivalent

$A$ is TU

$[A, I]$ is TU

$A^T$ is TU

$[A, -A]$ is TU

$-A$ is TU

$\begin{bmatrix}
       A\\
       I
     \end{bmatrix}$ is TU

$[A, A]$ is TU

$\begin{bmatrix}
       A\\
       -I
     \end{bmatrix}$ is TU

Based on such assumption, is it correct to assume that matrix A is totally unimodular if and only if the matrix [A -I] is totally unimodular (where I is the identity matrix of the appropriate size).

答:
I am not sure how to use those two lemmata, but it seems easier to consider the minors of $B=A^{-1}$ directly. In general, if $\mathcal I,\mathcal J\subseteq\{1,2,\ldots,n\}$ are such that $|\mathcal I|=|\mathcal J|=p$ for some $1\le p\le n$, then
$$
\det B(\mathcal I,\mathcal J)
=\pm\frac{\det A(\mathcal I^c,\mathcal J^c)}{\det A}.\tag{1}
$$
(See e.g. Prasolov, Problems and Theorems in Linear Algebra, theorem 2.5.1.) Therefore, if $A$ is a nonsingular totally unimodular matrix, the RHS of $(1)$ must lie inside $\{0,1,-1\}$. Hence $B=A^{-1}$ is totally unimodular.


问:
$\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$

Body:
\begin{align*}
\int_0^1 \frac{x{\rm d}x}{x^2+2x+1}&=\int_0^1 \frac{x{\rm d}x}{(x+1)^2}\\
&=\int_1^2 \frac{t-1}{(t-1+1)^2}{\rm d}(t-1)\\
&=\int_1^2 \frac{1}{t}-\frac{1}{t^2} {\rm d}t\\
&=\left[\ln t+\frac{1}{t}\right]_1^2\\
&=\ln 2-\frac{1}{2}.
\end{align*}


问:
Expand recursive equation to convert it into a normal formula
The Problem:

I am faced with the following recursive equation:

$$V(n) =\begin{cases}
2V(n/2)+n& \text{for n > 1}\\
0 &\text{for n = 1}
\end{cases}$$

I am trying to expand the function entirely and find a formula that is not recursive and is only dependent on n.

While I have proven by induction, that n log n works for that purpose, the expansion of the formula gives me a bit of trouble, since it doesn't seem even remotely related to n log n.

Question: How do I expand the recursive function above in a way, that clearly results in the formula n log n?

Here is what I have so far:

\begin{eqnarray*}
V(n) &=& 2*V(n-1)+n\\
&=& 2*(2*V(n-2)+(n-1))+n\\
&=& 2*(2*(2*V(n-3)+(n-2))+(n-1))+n \\
&\vdots \\
&=& n\hspace{1mm} log\hspace{1mm} n
\end{eqnarray*}

I appreciate any help given!

答:
A: V(n) = 2 * V(n/2) + n

Replacing n by n/2 in equation A gives:

B: V(n/2) = 2 * V(n/(2^2)) + n/2

Replacing n by n/(2^2) in equation A gives:

C: V(n/(2^2)) = 2 * V(n/(2^3)) + n/(2^2)

Now, plugin the value of B in A

D: V(n) = 2 * [2 * V(n/(2^2)) + n/2] + n

Now, plugin the value of C in D

D: V(n) = 2 * [2 * [2 * V(n/(2^3)) + n/(2^2)] + n/2] + n

Simplifying

D: V(n) = 2^3 * V(n/(2^3)) + (2^2) * n / (2^2) + 2 * n / 2 + n

D: V(n) = 2^3 * V(n/(2^3)) + n + n + n

D: V(n) = 2^3 * V(n/(2^3)) + 3n

So the general form of the equation is:

V(n) = 2^k * V(n/(2^k)) + kn

We need to figure out what value of k will get is to the termination condition V(1)

n/(2^k) = 1

n = 2^k

k = log2 n

Substitute k in the general form of the equation gives:

V(n) = 2^(log2 n) * V(1) + (log2 n)n

Simplifying:

V(n) = n * V(1) + n (log2 n)

V(n) = n * 0 + n (log2 n)

V(n) = n (log2 n)


问:
Irreducible representation of tensor product
Let $\varphi_n$ denote the standart irreducible representation of $SU(2)$ group with highest weight $n$.

What are the irreducible representaions of $\varphi_2 \otimes \varphi_3$?

答:
A result commonly attributed to Clebsch and Gordan gives you the answer:
$$
\varphi_2\otimes \varphi_3\cong \varphi_5\oplus\varphi_3\oplus\varphi_1.
$$
You can study the dimensions of weight spaces to see this. The weight five space is 1-dimensional (spanned by the tensor product of the highest weight weight spaces of both $\varphi_2$ and $\varphi_3$. The weight three space is two dimensional (= direct sum of two tensor products. That of weight zero space of $\varphi_2$ and weight three space of $\varphi_3$, and that of weight two space of $\varphi_2$ and weight one space of $\varphi_3$). The summand generated by the weight five submodule consumes only one of the two, leaving us with a high weight vector of weight three unaccounted for - resulting in a summand $\varphi_3$ on the r.h.s.. Continuing in the same vein leaves us with an extra $\varphi_1$ component.

Let's check the dimensions. $\dim\varphi_n=n+1$, so the tensor product $\varphi_2\otimes\varphi_3$ is of dimension $3\cdot4=12$. On the r.h.s we have summands of dimensions $6+4+2=12$. Check.

In quantum mechanics this result is known as the quantum mechanical addition of angular momenta.

The general formula for the tensor product of two irreducible reps of $SU_2$ (or $sl_2$ reads: (assume $m\ge n$)
$$
\varphi_m\otimes\varphi_n=\varphi_{m+n}\oplus\varphi_{m+n-2}\oplus\varphi_{m+n-4}\oplus\cdots\oplus\varphi_{m-n+2}\oplus\varphi_{m-n}
$$


答:
If $[A, -I]$ is totally unimodular then also $A$.

Conversely, let $A$ be totally unimodular. Let $B$ be a square sub-matrix of $[A, -I]$. We do a proof by induction of the size of the sub-matrix. If $B$ is $1\times 1$ then $B=\{\pm1,0\}$. Now assume that the determinant of all sub-matrices of size strictly less than $k$ are in $\{\pm1,0\}$.
Let $B$ be of size $k\times k$.
Let $(b_1\dots b_k)$ denote the columns of of $B$. If one such column is zero, then $\det B=0$. If one of these $b_i$ is non-zero, and this $b_i$ is a column of $I$ then $b_i$ is a unit vector. We can develop the determinant of $B$ with respect column $i$ and have to compute the determinant of a submatrix of size $k-1$. By induction assumption this is in $\{\pm 1,0\}$, so is the determinant of $B$.

Title: totally unimodular matrix - negative identity matrix

Body:
Theorem: The inverse of a (non-singular) totally unimodular matrix is totally unimodular.

The proof will use the following lemmas.

Lemma 1: Permuting rows and columns preserves total unimodularity.

Lemma 2: Matrix A is totally unimodular if and only if the matrix [AI]
is totally unimodular (where I is the identity matrix of the appropriate size).

The slides provided as a support material for my course say that:

Given A $\in M_{m x n}$

such affermations are equivalent

$A$ is TU

$[A, I]$ is TU

$A^T$ is TU

$[A, -A]$ is TU

$-A$ is TU

$\begin{bmatrix}
       A\\
       I
     \end{bmatrix}$ is TU

$[A, A]$ is TU

$\begin{bmatrix}
       A\\
       -I
     \end{bmatrix}$ is TU

Based on such assumption, is it correct to assume that matrix A is totally unimodular if and only if the matrix [A -I] is totally unimodular (where I is the identity matrix of the appropriate size).


问:
Is $f$: $(\mathbb{R}^m$, $\vert \vert \circ \vert \vert)$ $\rightarrow$ $(\mathbb{R}$, $\vert \circ \vert)$ a linear map?
Given that $f$ is continuous between two normed spaces, is it a linear map?

Because I think it is a map between two normed linear spaces, I think that it is but I'm not sure if that is true.

Also, if this is the case, would it be correct to say that $f$ is bounded because  it is continuous?

答:
It is correct to say that a linear map between normed spaces is bounded when continuous. In fact, in this situation being bounded and continuous are equivalent for linear maps.

This gives a way to disprove your first claim: find a continuous but unbounded function between normed spaces. Then, by what we have just observed, it will necessarily fail to be linear.


问:
How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$
I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.

So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.

Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.

How would I start with this integration?

Body:
\begin{align*}
\int_0^1 \sqrt{1+\frac{1}{x}}\,{\rm d}x&=\int_0^1 \sqrt{1+\frac{1}{t^2}}\,{\rm d}t^2\\
&=2\int_0^1 \sqrt{t^2+1}\,{\rm d}t\\
&=2\int_0^{{\rm arsinh}\ 1} \sqrt{\sinh^2 u+1}\,{\rm d}(\sinh u)\\
&=2\int_0^{{\rm arsinh}\ 1} \cosh^2u\,{\rm d}u\\
&=\int_0^{{\rm arsinh}\ 1} \cosh 2u+1\,{\rm d}u\\
&=\left[\frac{1}{2}\sinh 2u+u\right]_0^{{\rm arsinh}\ 1}\\
&=\sqrt{2}+\ln(1+\sqrt2).
\end{align*}


问:
How to solve linear combination of matrices?
If A, B and C are rectangular matrices of same shape. x and y are two scalars.
And if, xA + yB = C.
How to find x and y?
I know how to do this when A, B and C are vectors. But not sure about matrices.</p>

答:
If $A$, $B$ and $C$ are $n\times m$ matrices, the matrix equation $xA+yB=C$ is equivalent to a system of $n\cdot m$ linear equations
$$
a_{ij} x + b_{ij} y = c_{ij}
$$
for $i=1,\dots,n$ and $j=1,\dots, m$.

That's the same situation as if $A$, $B$ and $C$ would have been column vectors with $n\cdot m$ entries each.


问:
I need to find the formula for the truth table by using ∨ ∧ ¬
This is the process I did, but I’m not sure if the procedure is correct:

P | Q | ?

F | F | F

F | T | T

T | F | T

T | T | F

(¬P ∧ Q) ∨ (P ∧ ¬Q)

=> ((¬P ∧ Q) ∨ P) ∧ ((¬P ∧ Q) ∨ ¬Q)

=> (P ∨ Q) ∧ (¬P ∨ ¬Q)

答: My apologies if this should be a comment, I don't have the rep. It's been a while since I learned this, but isn't it going to be $P \oplus Q$ or $P \veebar Q$, the XOR/exclusive or operator? Unless there is more information that's not available, that fits all of your datapoints.


问:
Can little-o be composite with itself? $o(o(1))=o(1)$
(I know this example here uses the $o_p$ which is a bit different than $o$, but I'm guessing the same rule applies to both).

In one proof my professor used the following:

$$\hat \theta_n =\theta + o_p (\frac{1}{\sqrt n})
$$
$$o_p(\sqrt n(\hat \theta_n-\theta))=o_p(1)
$$
Technically, if I derive it, then
$$o_p(\sqrt n(\hat \theta_n-\theta))= o_p(\sqrt n(\theta +o_p(\frac{1}{\sqrt n}) - \theta))=o_p(\sqrt n o_p(\frac{1}{\sqrt n}))=o_p(o_p(1))$$

Intuitively I understand we are talking magnitudes here, but I'm missing the formal/rigor/definition understanding why this is equal to $o_p(1)$

答:
By definition of $o(g)$, we have that
$$ f(n) = o(g(n)) \iff \lim_{n \to \infty} \frac{f(n)}{g(n)} = 0.$$
Read it as '$f$ is eventually much smaller than $g$'.
What your professor is essentially claiming, is that if $f = o(g)$ and $g = o(h)$, then $f = o(h)$. This is saying that you can compose little-oh with itself. Indeed, $f = o(h)$ because
$$ \lim_{n \to \infty} \frac{f(n)}{h(n)} = \lim_{n \to \infty} \frac{f(n)}{g(n)}\frac{g(n)}{h(n)} = 0.$$
In fact, more is true: if we weaken $g = o(h)$ to $g = O(h)$, it still works.


问:
Mean curvature of graph over its tangent plane
Let $S$ be a regular surface in $\mathbb{R}^3$ and $p\in S$ a point on the surface.
By the implicit function theorem $S$ can be locally written as a graph of a function, e.g. $V\cap S = \{ (x,y,z) \in \mathbb{R}^3 : (x,y)\in U, z=f(x,y)\}$ for some open neighbourhood $V$ of $p$, open set $U\subset \mathbb{R}^2$ and some smooth function $f: U \rightarrow \mathbb{R}.$
By choosing local coordinates we can identify $U$ as part of the tangent plane of $S$ at $p$, furthermore we can set $f^{-1}(p)=(0,0)$. In this case, the mean curvature at $p$ is given by $H=\frac{f_{xx}\;\;+f_{yy}}{2}$ (average of second derivatives at $p$) and principal curvatures are $f_{xx},f_{yy}$.

Is this correct? If it is, how can one describe this more precisely than "choosing local coordinates..."? If it is not, how could I achieve a similar result (surface as graph over its tangent plane and easy formula of $H$)?

Thank you.

答:
Your formula is correct. The mean curvature can be defined as half of the trace of the shape operator (minus the differential of the unit normal vector). If the surface is the graph of a function, the normal vector field at the point $(x,y,z)$ is given by
\begin{align*}
\frac{\nabla(z-f(x,y))}{|\nabla (z-f(x,y))|} &= \frac{\nabla z- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}} \\ &= \frac{e_3- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}}\, ,
\end{align*}
with $\nabla=(\partial_x,\partial_y,\partial_z)$. Note that, $\nabla f=(\partial_x f,\partial_y f,0)$.

The trace of the differential of this unit normal vector field is nothing but the divergence of it. Therefore
\begin{align*}
2H&=-\mathrm{div}\left(\frac{e_3- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}}\right)\\ &=\frac{\Delta f}{\sqrt{1+|\nabla f|^2}}+(\nabla f(x,y)-e_3)\cdot \nabla\left(\frac{1}{\sqrt{1+|\nabla f|^2}}\right) \, .
\end{align*}
Now, if the graph parameterization is with respect to the tangent space at $(x_0,y_0,z_0)\in S$ then $\nabla f(x_0,y_0)=0$ and, therefore, $H$ reduces to
$$2H(x_0,y_0,z_0)=\Delta f(x_0,y_0) \, .
$$
This is the formula you were writing and, in general, it is true at every point of $(x,y,f(x,y))\in S$ where $\nabla f(x,y)=0$.


问:
Figuring out how many revolutions a bicycle tire rotates for every pedal revolution..
I'm trying to figure out this problem on my assignment, goes as follows:

A bicycle works using two gears and a chain. Turning the pedals causes the front gear to rotate and the chain then causes the rear gear to rotate which cause the bike tire to turn. The bike has a front gear with radius of 4 inches and the rear gear has radius 2 inches.

Determine the number of revolutions that the bicycle tire rotates when the pedals are rotated one revolution.

I truly have no clue.. Could someone help me out?

Any help or even clues are appreciated.

Thank you.

答:
When front gear makes 1 rotation the rear gear will rotate 2 times of that. Ratio of the circumference(2×pi×R/ 2xpi×r) = 4÷2=2.
Rear gear attached  to back wheel so same rotation as rear gear</p>


问:
Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?
When I get around at park , I made up a question to hang out by myself.However , I stuck in it.

Question: Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?

Where I stuck in :

$1-)$ I know how to write $75$ in the base of $3$ , but I do not know writing $75!$ in the base of $3$.

$2-)$ To find the last two digits, we utilize from $(\text{mod }100)$, then which modulus should I use to find its last two digits?

Thanks for your helps..

答:
You can answer with Legendre's formula and calculate the $3$-valuation of $75!$:
$$ v_3(75!)=\biggl\lfloor\frac{75}3\biggr\rfloor+\biggl\lfloor\frac{75}9\biggr\rfloor + \biggl\lfloor\frac{75}{27}\biggr\rfloor=25+8+2=35,$$
so you know that, in bases $3$, the last $\mathbf{35}$ digits will be $0$.


问:
Inverse Laplace transform after derivative of transform.
I've been using the following theorem in my intro ODE course:

If $F(s) = \mathscr{L}\left\{ f(t) \right\}$ and $n\in\mathbb{N}$, then $\mathscr{L}\left\{t^n f(t) \right\} = (-1)^n\frac{d^n}{ds^n}F(s).$

What is the inverse of this transform? For instance, say I wanted to take $$ \mathscr{L}^{-1}\left\{ \frac{s}{(s^2+16)^2} \right\}.$$ I can see that the result would be $\frac{1}{8}t\sin(4t)$. But, how would one go about manually working this out (and particularly in general)?

答:
ILTs like this may be evaluated using the residue theorem: the ILT is, in the this case, the sum of the residues at the poles to the left of the vertical line of integration.  Here, the poles at $s_{\pm}=\pm 4 i$ are double poles.

The residue of a double pole $s=s_0$ of a function of the form

$$f(s) = \frac{p(s)}{q(s)^2}$$

is given by the expression

$$\text{Res}_{s=s_0} \frac{p(s)}{q(s)^2} = \frac{p'(s_0)}{q'(s_0)^2} - \frac{q''(s_0) p(s_0)}{q'(s_0)^3}$$

In this case, $p(s)=s e^{s t}$ and $q(s)=s^2+16$.  Using the above ingredients, it is not hard to show that the ILT is

$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s^2+16)^2} e^{s t}\\ &= \left [\frac{1+i 4 t}{-64} - \frac{8 i}{-512 i} \right ] e^{i 4 t} +  \left [\frac{1-i 4 t}{-64} - \frac{-8 i}{512 i} \right ] e^{i 4 t} \\ &= \frac18 t \, \sin{4 t}\end{align}$$


答:
For a normed space $(V, \Vert \cdot \Vert)$, there is an immediate non linear map... The map $N: x \mapsto \Vert x \Vert$ itself as for $x \neq 0$

$$\Vert - x \Vert = \vert -1 \vert \Vert x \Vert = \Vert x \Vert \neq - \Vert x \Vert.$$

Title: Is $f$: $(\mathbb{R}^m$, $\vert \vert \circ \vert \vert)$ $\rightarrow$ $(\mathbb{R}$, $\vert \circ \vert)$ a linear map?

Body:
Given that $f$ is continuous between two normed spaces, is it a linear map?

Because I think it is a map between two normed linear spaces, I think that it is but I'm not sure if that is true.

Also, if this is the case, would it be correct to say that $f$ is bounded because  it is continuous?


问:
Prove that an abelian group is direct sum of two subgroups.
I am trying to do this exercise:

Let $G$ be an abelian group of order $mn$, where $m$ and $n$ are relatively prime. If
$H_1 =\{x \in G: mx = 0\}$ and $H_2 =\{x \in G: nx = 0\}$, prove that $G=H_1  \oplus H_2$.

By proving that the intersection of $H_1$ and $H_2$ is the trivial subgroup; I have managed to prove that the sum (it is an additive group, althoght that is not important) is direct. However I can not prove the equality of $G$ and the direct sum. How should I approach that? (What I struggle with is the case in which the order of an element of $G$ does not divide either $m$ nor $n$).

Thanks in advance.

答: The number of element of the direct sum of $H_1$ and $H_2$ is the number of elements of $G$


问:
Find the closure of $\{\frac{(-1)^n}{n+a} |n \in \mathbb{N}\}$
Let $A=\{\frac{(-1)^n}{n+a} |n \in \mathbb{N}\}$ ($a$ is some positive real number). I want to find the closure of $A$.

I could show that $B:=A\cup \{0\}\subset \overline{A}$ and now I only need to show that $B$ is closed.

I would like to do this by finding the complement of $B$, but I can't write it down explicitely. I know that it will be a countable union of open intervals, yet I cannot formally write this.

EDIT: I think I found a way to show that $B$ is closed the way I wanted to go about it. Consider the sequence $(x_n)_{n\ge 1}$, $x_n=\frac{(-1)^n}{n+a}$. The subsequence $(x_{2n})_{n\ge 1}$ is decreasing and positive, whilst the subsequence $(x_{2n+1})_{n\ge 1}$ is increasing and negative. Then we may write that $\displaystyle \mathcal{C}_B=(-\infty, x_1) \cup \bigcup_{n\ge 1} (x_{2n-1},x_{2n+1})\cup \bigcup_{n\ge 1}(x_{2n+2}, x_{2n})\cup (x_2, \infty)$, so $\mathcal{C}_B$ is open and we are done. Am I right?

答: To find closure it is enough to find limit points of $A$. Observe that for any $n$ and for all $\epsilon$ the interval $(- \epsilon , \epsilon)$ contains a point of $A$ since $ \lim_{n\to \infty} \frac{1}{n+a} = 0$. So $0$ is a limit point of $A$. It is easy to see that there are no other limit points by choosing suitable $ \epsilon$. So closure of $A$ is $\{0\} \cup A$.


问:
Is it possible to fit more than 23 dots in a 7 by 8 rectangle?
For practical reasons I would like to know how many dots you can fit in an $7 \times 8$ box with no two dots closer than 2 metres from each other.

The simplest arrangement has 4 dots along each row and 5 rows.  Can you do any better?

One method is to expand the box to $9 \times 10$ and fit circles of radius $1$.  We can fit $23$ with:

But is it possible to fit more?

答:
Yes, in this example fitting 23 circles of radius=1 into an 9x10 rectangle is optimal.

This hexagonal packing arrangement plus the close cropped edges yields the highest packing efficiency for circles into a rectangle.

Packing efficiency for 23 circles = 23xPi/(9x10) = 80.3%

Packing efficiency for 24 circles = 24xPi/(9x10) = 83.7%

Referring to this graph:
https://i.stack.imgur.com/ziRBe.jpg
it would appear that 83.7% is not possible.
(Note: this graph allowed for a slight spacing between circles, I could recalculate the graph for zero spacing as per the original question but I think you get the idea)

For further details on similar problems:
How many circles of a given radius can be packed into a given rectangular box?


问:
How to get the distance between two anchoring points in a tv mast?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{4 m}\\
2.&\textrm{5 m}\\
3.&\textrm{6 m}\\
4.&\textrm{7 m}\\
\end{array}$

How exactly should this be solved relying only in euclidean geometry?.

I'm confused exactly if this should use congruence or what?. The part which I'm stuck is what to do with that the double of the angle on $\angle ACD$?.

I'm stuck can someone help me here?. Please include a drawing in your answer because I'm not very savvy with understanding where exactly congruence could be

答:
Reflect $\triangle ABD$ over segment $AB$ so that point $D$ ends up in a new position,  point $E$.

Observe that ,in $\triangle AED$ and $\triangle CAE$, $\angle EAD=2\alpha=\angle ACE$ and both the triangle share the same angle $\angle E$. Hence, they are similar and thereafter $\triangle CAE$ is isosceles as well.

Thus, $CE=10$ and $CD=CE-DE=10-2\cdot 2=10-4=\boxed {6}$


问:
If $\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m]$ for all $n,m$, then $X$ and $Y$ are independent.
I had an exam this morning, and I had to prove that for $X$ and $Y$ bounded, if for all $n$ and all $m$, $$\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m],$$then $X$ and $Y$ are independents. Using Characteristic function I think that I could justify it using the fact that $$\cos(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k}X^{2k}}{(2k!)}\quad \text{and}\quad \sin(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k+1}X^{2k+1}}{(2k+1)!},$$
but how can I do without characteristic function, and out measure theory ? (it's a lecture of introduction of probability, so we don't have tools of measure theory as DCT or MCT, neither characteristic function).

My definition of the expectation is : for $X$ s.t. $\mathbb E|X|<\infty $, $$\mathbb E[X]=\lim_{n\to \infty }\mathbb E[X_n]=\sum_{k\in\mathbb Z}\frac{k}{2^n}\mathbb P\left(X_n=\frac{k}{2^n}\right),$$
where $X_n=2^{-n}\lfloor 2^nX\rfloor$.

答:
It would be interesting to known if there is a short proof of this.

Here is a (not so short) one...
Let $X,Y$ be bounded random variables.  We start with
$$
\mathbb E[X^nY^m] = \mathbb E[X^n]\mathbb E[Y^m]\qquad \text{for all }n,m \in \mathbb N
\tag1$$
Using linear combinations, we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all polynomials }f,g
\tag2$$
Using Weierstrass approximation (uniform convergence) we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all continuous functions }f,g
\tag3$$
Using dominated convergence (pointwise a.e. convergence) we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all bounded measurable functions }f,g
\tag4$$
In particular, $(4)$ holds for all indicator functions.  That is the definition of "$X$ and $Y$ are independent".


问:
Finding all complex solutions of equations
I don't know if I am too stupid or if the question was incorrectly typed. The prof and the tutors seem to have decided they ain't gonna answer their emails. :/

So,
$$
Im\left(\frac{\bar1}{z}\right) = \left|\frac{1}{z}\right|^2+\frac{1}{Im(z)}
$$

Yes. If the bar term on the LHS is related to 1 or the the whole thing was my first question too. I know. I mean, if it covered the entire parenthenses, then it'd be clear, but just over the 1 got me wondering.

In any case I tried both and I still can't get to the following answer:

$y=-x^2$

(this would be analytically)

And then, regarding the graphical part, it says:
"An open down parabola, through the origin, but undefined at the origin."

How to get to these 2 conclusions? Well, I didn't get to this point haha

The equations/set up I came up with were:

$\frac{1}{-y}=x^2-2xyi-y^2+\frac{1}{y}$

$y=x^2-2xyi-y^2+\frac{1}{y}$

$\frac{1}{-y}=\frac{1}{x^2+y^2} +\frac{1}{y}$

Another time I even got $0$ for the LHS.

None of them led to the described parabola. Some of them get cubic, some of them contain an "i", some of them have a shift, and so on.
Yes, I am feeling really stupid. I don't even know which point I am missing here: argghhh

I appreciate any help.
Thanks.

答:
Since $\ \overline{1}=1\ $, it would appear to be more  sensible to take the bar over the left side of the equation to apply to the whole fraction rather than the numerator alone.

In his first comment, Ivan Neretin has pinpointed your error, but since you may find his hint a little cryptic, I'll take the liberty of providing some more detail.  Assuming you're taking $\ z=x+iy\ $, you appear to have set $\ Im\left(\frac{1}{\overline{z}}\right)=\frac{1}{-y}\ $ in your third equation.  This is incorrect. Multiplying the numerator and denominator of $\ \frac{1}{\overline{z}}\ $ by $\ z\ $ gives you
$$
\frac{1}{\overline{z}}=\frac{z}{z\overline{z}}\ .
$$
What is this in terms of $\ x\ $ and $\ y\ $? What is its imaginary part?  If you can answer these questions correctly, the solution of your equation should fall out with no further trouble.


问:
For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.
Reading a linear algebra textbook, I encounter

For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.

I noticed in Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd., the proof without determinant are posts.

It seems for me with the trick of determinant, this problem might be easy.  The first thing I came up with is $\det(A_n-\lambda I_n)$ but this does not guarantee that the solution has at least one real eigenvalues. i.e., setting characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we do not know one of $\lambda$ should be real.

What can be the simple proof with determinants?

答:
Setting the characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we know that $f_A$ has at least one real root when $n$ is odd by the intermediate value theorem.

As $\lambda \to -\infty$, $f_A(\lambda) \to - \infty$; as $\lambda \to +\infty$, $f_A(\lambda) \to +\infty$. We know both of these, because that's the behavior of $\lambda^n$ for large $|\lambda|$, and $\lambda_n$ will eventually dominate over the other terms.

Therefore there must be some very negative value of $\lambda$ where $f_A(\lambda)<0$, and some very positive value of $\lambda$ where $f_A(\lambda)>0$. Between these, there must be a value of $\lambda$ where $f_A(\lambda)=0$, and this gives us a real eigenvalue.


答:
If $n$ is odd, then

$$f_A( \lambda) \to \infty$$

as $ \lambda \to \infty$

and

$$f_A( \lambda) \to -\infty$$

as $ \lambda \to -\infty.$

The IVT shows now that there is $ \lambda_0 \in \mathbb R$ such that $f_A( \lambda_0=0.$

Body:
Reading a linear algebra textbook, I encounter

For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.

I noticed in Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd., the proof without determinant are posts.

It seems for me with the trick of determinant, this problem might be easy.  The first thing I came up with is $\det(A_n-\lambda I_n)$ but this does not guarantee that the solution has at least one real eigenvalues. i.e., setting characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we do not know one of $\lambda$ should be real.

What can be the simple proof with determinants?


答: For odd $n$, the characteristic polynomial of an $n\times n$ real matrix has real coefficients and odd degree, but an odd degree polynomial with real coefficients has a real root by the intermediate value theorem.

Title: For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.


问:
An upper bound of binary entropy
Binary entropy is given by $$H_{\mathrm b}(p) = -p \log_2 p - (1 - p) \log_2 (1 - p), \hspace{6 mm} p \le \frac{1}{2}$$
How can I prove that $$H_{\mathrm b}(p) \le 2 \sqrt{p(1-p)}$$</p>

答:
You may use the inequality $\ln(x)\le \sqrt{x-1}$ for any $x\ge 1$. Precisely speaking, note that
\begin{equation}
\begin{aligned}
H(p)&=p\ln\frac{1}{p} +(1-p)\ln\frac{1}{1-p} \\
&\le p\sqrt{\frac{1}{p}-1} +(1-p)\sqrt{\frac{1}{1-p}-1}\\
&= 2\sqrt{p(1-p)}
\end{aligned}
\end{equation}
Here the logarithm is to base natural exp..

Title: An upper bound of binary entropy


答:
I think you can use your characteristic polynomial. In fact, if you factorize $f_A$, you get a result like this :

$$f_A(\lambda)=\prod \underbrace{(\lambda-\lambda_i)^{a_i}}_{\text{real eigenvalues}}\times\prod \underbrace{(b_i\lambda^2+c_i\lambda+d_i)^{2e_i}}_{\text{corresponds to complex eigenvalues}} $$

If you had only complex eigenvalues, it would mean that $\forall i, a_i=0 $, so that $\sum 2e_i=n $. But n is odd, and $2e_i$ is always even, so $\sum 2e_i $ is also even, which is a contradiction.
$$ $$
So you get that $ \exists i, a_i\neq 0$, so that there is at least one real eigenvalue


问:
calculate the distribution and density of the variable $Y=X^2$
$$ \mbox{Let X have the normal distribution: } N(0,\sigma^{2}) \to E(X)=0, Var(X)=\sigma^{2}. \mbox{ Let's define a new variable: }Y=X^2. \mbox{  Calculate the distribution and density of Y} $$
Is that solution correct?
$$f_{X}(x)= {{1}\over{(2\pi)^{1/2} \sigma}}e^{{-x^2}\over{2\sigma^2}}$$ $$F_{X}(x)= {\phi({{x}\over{\sigma}})} $$ $$ F_{Y}(y)=P(Y<y)=P(X^2<y)=P(-{y}^{1/2} < X < {y}^{1/2})=F_{X}({y}^{1/2})-F_{X}(-{y}^{1/2})=\phi({{y}^{1/2}\over\sigma})-\phi({{-y}^{1/2}\over\sigma})= 2\phi({{y}^{1/2}\over{\sigma}}) $$ But I still don't know how to calculate the density of $Y$.

答:
Could you please elaborate last step? How did $\sigma^2$ fit in distribution?

Let's have

$$X\sim Gamma(n;\theta)$$

(I mean $\theta$ is the rate parameter)

$$f_X(x)=\frac{\theta^n}{\Gamma(n)}x^{n-1}e^{-\theta x}$$

Let's set

$$y=\sigma^2 x$$

$$x=\frac{y}{\sigma^2}$$

$$x'=\frac{1}{\sigma^2}$$

thus

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|=\frac{\theta^n}{\Gamma(n)}\Bigg(\frac{y}{\sigma^2}\Bigg)^{n-1}e^{-\frac{\theta}{\sigma^2} y}\cdot \frac{1}{\sigma^2}$$

$$f_Y(y)=\frac{\Big(\frac{\theta}{\sigma^2}\Big)^n}{\Gamma(n)}y^{n-1}e^{-\frac{\theta}{\sigma^2}y}$$

that is

$$Y\sim Gamma(n;\theta/\sigma^2)$$


答: The characteristic polynomial is of odd degree.  Since complex roots come in conjugate pairs, and by the fundamental theorem of algebra there are an odd number of roots over $\Bbb C$ (counting multiplicity), there must be a real root.


答: Hint: Write $1=am+bn$ for some integers $a,b$. Then, for each $x \in G$, $x=m(ax)+n(bx)$ and $mG \subset H_2$, $nG \subset H_1$.

Title: Prove that an abelian group is direct sum of two subgroups.

Body:
I am trying to do this exercise:

Let $G$ be an abelian group of order $mn$, where $m$ and $n$ are relatively prime. If
$H_1 =\{x \in G: mx = 0\}$ and $H_2 =\{x \in G: nx = 0\}$, prove that $G=H_1  \oplus H_2$.

By proving that the intersection of $H_1$ and $H_2$ is the trivial subgroup; I have managed to prove that the sum (it is an additive group, althoght that is not important) is direct. However I can not prove the equality of $G$ and the direct sum. How should I approach that? (What I struggle with is the case in which the order of an element of $G$ does not divide either $m$ nor $n$).

Thanks in advance.


问:
P(D^c|A^c) when given P(D|A^c)
So to my understanding, my teacher just told me that the following is incorrect:

If P(D|A^c)=0.4, then P(D^c|A^c) = 1-0.4 which is 0.6?

I was very sure this is true but she just said it's not as I need to work out the intersection and use a formula. Is what I put not correct??

答:
What's written in the post is correct, though of course the instructor might have expected some argument for it.

One such argument would be to remark that $$(D\cap A^c)\cup (D^c\cap A^c)=A^c\implies P(D\cap A^c)+P(D^c\cap A^c)=P(A^c)$$

where we have used the fact that $D\cap A^c$ and $D^c\cap A^c$ are disjoint.

But then we have $$P(D\,|\,A^c)+P(D^c\,|\,A^c)=\frac {P(D\cap A^c)+P(D^c\cap A^c)}{P(A^c)}=1$$

which is the claim that you relied on.

Title: P(D^c|A^c) when given P(D|A^c)


问:
Multiplication in $\mathcal D'(R)$.
I was reading in my text book that multiplication of elements of $\mathcal D'(R)$ is not a continuous function although it is defined and can be seen from the convergence of $\sin(nx)$ to $0$ as $n$ goes to infinity in $\mathcal D'(R)$. And not in the case of $\sin^2(nx)$ as $n$ goes to infinity. Please explain me what does he mean. Thank you in advance.</p>

答:
In the sense of distributions, $\sin nx \to 0$ as $n\to\infty$, since

$$ \lim_{n\to\infty}\int_{-\infty}^\infty \phi(x)\,\sin nx\,dx = 0$$

for every test function $\phi$. (This follows for example from Riemann-Lebesgue's lemma.)

On the other hand $\sin^2 nx = \frac12 - \frac12\cos 2nx$ so
$$ \lim_{n\to\infty}\int_{-\infty}^\infty \phi(x)\,\sin^2 nx\,dx = \frac12 \int_{-\infty}^\infty \phi(x)\,dx,$$
which means that $\sin^2 nx$ tends to $1/2$ in the sense of distributions as $n \to \infty$.

Hence, the operation $f \mapsto f^2$ can't be continuous on $\mathcal{D}'$. If it were, it would map a sequence of distributions tending to $0$ to another sequence of distributions tending to $0$.

(Multiplication of two arbitrary distributions can't even be defined in general: what would $\delta^2$ be?)

Title: Multiplication in $\mathcal D'(R)$.


问:
Proving that a number is divisible by $2$
How can I prove that:

$$\frac{1}{2} \left(3 n^2-2\right) \left(3 n^2-1\right) \left(3 n^3-3 n+1\right)\in\mathbb{N}$$

When $n\in\mathbb{N}$?

答: Every product of two consecutive integers is even. So $(3n^2 - 2)(3n^2 - 1)$ is even, and the whole product is even (before you divide by 2).


问:
$M||\textbf{x}||$ $\leq$ $f(\textbf{x})$?
Given a function $f$: $(V, || \circ ||_{\infty})$ $\rightarrow$ $(\mathbb{R}, |\circ |)$

how can you then show that there exists  $M$ $>$ $0$ such that $M||\textbf{x}||$ $\leq$ $f(\textbf{x})$ for every $\textbf{x} \in \mathbb{R}$?

I thought maybe that it could be done by showing that $f$ is a bounded linear operator but i'm not sure that it is.

答: You can just take $\alpha=\min\{f(y)\,:\,\lVert y\rVert=1\}$ because $S^{m-1}$ is compact.


问:
To sum $1+2+3+\cdots$ to $-\frac1{12}$
$$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

答:
there are many ways to see that your result is the right one. What does the right one mean?

It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.

It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.

Exponential regulator method

Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
$$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$
Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.

The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
$$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$
We will only need some Laurent expansion around $\epsilon = 0$.
$$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$
We have
$$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$
You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command

1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]

and the result is
$$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$
In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.

In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$.

There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.

Euler's method

Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.

We will work with two sums,
$$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$
Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
$$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$
Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
$$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$
so $S=-T/3=-1/12$.

A zeta-function calculation

A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found in the Pictures of Yellows Roses, a Czech student journal. The website no longer works, although a working snapshot of the original website is still available through the WebArchive (see this link). A 2014 English text with the same evaluation at the end can be found at The Reference Frame.

The comments were in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.

OwnerDisplayName: perplexed

Title: To sum $1+2+3+\cdots$ to $-\frac1{12}$


问:
Show that for all $x\in(0,1)$, $(1.5-x)(2-x)^{\frac{1}{\varepsilon}}-(x^2-3x+2)^{\frac{1}{\varepsilon}}>0$
Let $\varepsilon>0$. Show that for all $x\in(0,1)$

\begin{equation}
(1.5-x)(2-x)^{\frac{1}{\varepsilon}}-(x^2-3x+2)^{\frac{1}{\varepsilon}}>0.
\end{equation}

Body:
I suspect that claim is incorrect.
Let $\frac{1}{\epsilon} = m \gt 0$.
$$(1.5-x)(2-x)^m \gt (x^2-3x+2)^m \iff(1.5-x)(2-x)^m \gt (2-x)^m (1-x)^m \\ \iff 1.5-x \gt (1-x)^m \\ \iff x+(1-x)^m \lt 1.5$$

But for $m\approx 0$ and near $x=1$, $$x+(1-x)^m \approx 1+1=2$$


问:
What's the right property for sections in the structure sheaf of the spectrum of a ring?
In Hartshorne's Algebraic Geometry one reads that

For an open set $U \subseteq Spec A$,
we define $\mathscr{O}(U)$ to be the set of functions $s: U \rightarrow \displaystyle\coprod_{\mathfrak{p}\in U} A_{\mathfrak{p}}$ such that $s(\mathfrak{p}) \in A_{\mathfrak{p}}$
for each $\mathfrak{p}$, and such that $s$ is locally a quotient of elements of $A$: to be precise, we require that for each $\mathfrak{p} \in U$ there is a neighborhood $V$ of $\mathfrak{p}$, contained in $U$, and elements $a, f \in A$, such that for each $\mathfrak{q} \in V$, $f \notin \mathfrak{q}$, and $s(q) = \frac{a}{f}$ in $A_{\mathfrak{q}}$.

Does Hartshorne mean a neighborhood $V$ or an open neighborhood $V$? I'm not familiar with the author's conventions.

答: It does not matter. Here neighborhood and open neighborhood are interchangeable, they lead to equivalent definitions.


答:
Hint

By definition, if $x\in \mathbb R$, then
$$\sqrt{x^2}=|x|.$$

Title: Find $\sqrt{(-1)^2}$

Body:
What is the value of $\sqrt{(-1)^2} ;$ 1 or -1?

$\sqrt{(-1)^2} = \sqrt 1= 1 $

$\sqrt{(-1)^2} = {((-1)^2)}^{1/2} = (-1)^1 = -1$

Or is it both?


问:
Find the error of the interpolation (either Lagrange or Newton) of the following set of points

Find the polynomial that interpolates a function given by $f(x) =
 x(\ln(x)-1)$ on the points 2 and 3. Use values for f with 4 correct
decimal places (?) and give an estimation for the error of the
interpolation.

I'm not sure what "4 correct decimal places" means here, I'll assume it means simply "with 4 decimal places". Using my calc I got:
$$f(2) = -0.6137 \\
f(3) = 0.2958$$

For the interpolations, I got $0.9095x-2.4327$ for both.

I'm having trouble applying the formulas for the error.

Lagrange:

What is $\xi (x)$?

What is $f^{(n+1)}$?

Newton:

What is $c$?

What is $f^{(n+1)}$?

How do I apply these formulas in this case?

答:
I agree with the calculations you did.

$c$ and $\xi(x)$ are some unknown point in the range $[x_0, x_n]$.

$f^{(n+1)}$ is the $(n+1)^{th}$ derivative of the function.

Since $n = 1$ (two points - linear interpolation), for the error, we have $f^{(2)}(x) = f''(x)$

$$E_1 = |f(x) - P_1(x) | \le \max_{[2,3]} \left|\dfrac{f''(\xi)}{2}\right|\max_{[2,3]} \left|(x-x_0)(x-x_1)\right|$$

The second derivative $f''(x) = \dfrac{1}{x}$.

The error is given by

$$E_1 \le \max_{[2,3]} \left|\dfrac{1}{2x}\right|\max_{[2,3]} \left|(x-2)(x-3)\right|$$

Finding the max over those ranges

$$E_1 \le \left(\dfrac{1}{4}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{16}$$


问:
A and B have negative correlation, so -A and -B have positive?
If A and B have negative correlation, would I be correct to assume -A and -B have positive correlation?

eg taking an example of -ve correlation: A student who has many absences has a decrease in grades.

so taking -A and -B: A student who is always present has an increase in grades - which is positive correlation.

So is my assumption in the 1st line true?

答:
If $A$ and $B$ are negatively correlated, increasing $A$ decreases $B$.

But since increasing $A$ is equivalent to decreasing $-A$ and decreasing $B$ is equivalent to increasing $-B$, the previous statement is equivalent to “Decreasing $-A$ increases $-B$”, which means $-A$ and $-B$ are also negatively correlated.


问:
Quick Question- is This valid?,
quick question I was looking to find the intersection of $3$ events $A,B$ and $C$( I couldn't find anything online to affirm this)

just wanted to know is $P(A\cup B\cup C)$ the same as $P(A\cup B)\times P(B\cup C)$

thanks!

答: No, this is not true. Consider the following counter-example. Let $A=B=C$ and let's say $P(A)=P(B)=P(C)=\frac{1}{3}$. Then $A \cup B \cup C = A$ as well as $B \cup C = A$ and $B \cup C = A$ but $$ P(A \cup B \cup C) = \frac{1}{3} \neq \frac{1}{3}\cdot \frac{1}{3} = P(A\cup B) \cdot P(B\cup C).$$

Title: Quick Question- is This valid?,


答:
If $z$ is a non-zero complex number and $m,n$ are positive integers prime to each other, then $(z^{1/n})^m=(z^m)^{1/n}$

If gcd$(m,n)\neq 1$ then $(z^{1/n})^m\neq (z^m)^{1/n}$, in general.

Example:  Let $z=i,m=4,n=6.$ Then $(z^{1/n})^m=(i^{1/6})^4 \qquad \qquad \;\;=\cos {4r\pi+\pi \over 3}+i\sin {4r\pi+\pi \over 3}$ where $r=0,1,2,...,5$. It has three distinct values. But $(z^m)^{1/n}=1^{1/6}=\cos {2k\pi \over 6}+i\sin {2k\pi \over 6}$, where $k=0,1,...5$. It has six distinct values.

In your question $\sqrt {(-1)^2}=\sqrt 1=1$

[$\sqrt$ is the positive square root function].

Title: Find $\sqrt{(-1)^2}$


问:
Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.

答:
If $I(r) $ is the integral then we have $I(0)=0$ and we can show that $I'(r) =0$ for all $r\in(-1,1)$ and then we get $I(r) =0$ for all $r\in(-1,1)$.

Clearly we have $$I'(r) =\int_{0}^{\pi}\frac{2r-2\cos t} {1-2r\cos t +r^2}\,dt$$ Clearly this is $0$ for $r=0$. So let $r\neq 0$ and then we can write $$I'(r) =\frac{1}{r}\int_0^{\pi}\left(1+\frac{r^2-1}{1-2r\cos t+r^2}\right)\,dt$$ and this equals $$\frac{1}{r}\left(\pi+\frac{\pi(r^2-1)}{\sqrt {(1+r^2)^2-4r^2}}\right)=0$$ as $r^2<1$.

We have used the standard formula $$\int_0^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt {a^2-b^2}},a>|b|$$


问:
What is the Kronecker Product of two vectors?
In my numerical methods course we got a homework problem that has a definition of a function

$\phi(x) = vec(M) - x \otimes x $

where $x\otimes x$ is the kronecker product of an n-vector and $ M $ is an $n\times n$ Matrix that is vectorized (flattened) in column-major by the $vec()$ operator.

I got confused as I thought the kronecker product would produce an $ n\times n$ matrix.
But according to the instructor it's a vector?

How would I compute the kronecker product of two vectors?
I thought it would be the entries of the first vector times the second vector appended in a matrix.

Thanks for anyone who can share some enlightenment. I tried to find some on Wikipedia but the examples there confirm my confusion.

I looked at this already: Kronecker product and outer product confusion

Thank you!

答: The vector $x$ should be regarded as a matrix consisting of a single column. It then makes sense to consider the Kronecker product $x \otimes x$.


问:
Question about the chain rule and derivates.
I've seen someone claim that $g(x) = f(x) - f(-x) => g'(-x) = f'(x) -(-1 * f'(x))$ by the chain rule.
I just can't grasp my head around why is this true?

答:
No, that (was originally $f'(-x)=-1·f'(x)$) is not true, just take $f(x)=x$ where you have a constant derivative $f'(x)=1$.

What you can however do is define $f_-(x)=f(-x)$ where then $$f_-'(x)=-f'(-x)$$ and changing the point of the derivative evaluation $$f_-'(-x)=-f'(x).$$

In the modified question, the odd function $g(x)=f(x)-f(-x)$ has the even-symmetric derivative
$$
g'(x)=f'(x)+f'(-x)
$$
which remains unchanged under changing $x$ to $-x$ in all places.


问:
which of the following is $T_1$ space?
From Topology without tears

A Topological space $(X,\tau)$ is said to be $T_1$ space if every singleton set is closed in $(X,\tau)$.
Show that precisely two of the following 9 topological spaces are $T_1$.

1.a discrete space

(I believe discrete is $T_1$ because every subset in discrete space is open and closed both.So singleton are also closed.)

2.An indiscrete space with atleast two points

(Since its topology contains  only $X$ and $\phi$ I believe  singleton will not be neither closed nor open.)

3.An infinite set with cofinite topology

(I have a example $\tau$ consisting of $\mathbb{N}$,$\phi$ and every set $\{n,n+1,...\}$ for any positive integer.Here $\mathbb{N}$ is infinite set and this is cofinite topology since every closed set will be finite here.But here singletons need not be closed.)

4.$X=\{a,b,c,d,e,f\}$ and $\tau=\{ X,\phi,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}\}$

(Here I can see every singleton are not closed.)

5.$\tau_1$ consist of $\mathbb{R},\phi$, and every interval $(-n,n)$ for any $n$ positive integer.

(for $\{n\}$ to be closed its complement $(-\infty,-n] \cup [n,\infty)$ must belong to $\tau_1$.Which is not the case.So  $\tau_1$ is not $T_1$)

$\tau_2$ consist of $\mathbb{R},\phi$, and every interval $[-n,n]$ for any $n$ positive integer.

(Similarly I can show $\tau_2$ not $T_1$)

7.$\tau_3$ consist of $\mathbb{R},\phi$, and every interval $[n,\infty)$ for any $n$ positive integer.

(Similarly $\tau_3$ is also not $T_1$.

8.$\tau_1$ consists of $\mathbb{N}$,$\phi$ and every set $\{1,2,...,n\},$for $n$ any positive integer.

(Here also all singleton will not be closed.because its complement need not be in $\tau_1$)

9.$\tau_2$ consists of $\mathbb{N}$,$\phi$ and every set $\{n,n+1,...\},$for $n$ any positive integer.

(Here also all singleton will not be closed.because its complement need not be in $\tau_1$)

So,I got only one $T_1$ space.where I am wrong?

Thanks in advance!

答:
$3$ is also a $T_1$ space. The complement of singleton $\{a\}$ is $X-\{a\}$ which is open since its complement $\{a\}$ is finite (a non-empty set is open in $\color{red}{\text{co}}$finite topology iff its complement is finite).

The example you have taken is not the cofinite topology on $\Bbb N$. It is a subset (strictly coarser) of the cofinite topology. Note that $\Bbb N-\{2\}\in$ the cofinite topology on $\Bbb N$ but not your example topology.

Title: which of the following is $T_1$ space?


问:
Diffeomorphism from Inverse function theorem
I often heard that it is possible to show by using the inverse function theorem that if a function is smooth (i.e. arbitrarily often differentiable), a bijection between open sets, and has a non-singular jacobian, then it is a smooth diffeomorphism. but somehow the inverse function theorem that I know and that wikipedia seems to know, only states that if it is a continuously differentiable bijection with nonzero jacobian, then its inverse function is also continuously differentiable. But how do you get from this, to the statement that I proposed above? I don't see the implication.</p>

答:
Let $f(p)=q$, $J_f(p)\ne 0$. Then $f$ maps a suitable neighborhood $U$ of $p$ diffeomorphically onto a neighborhhod $V$ of $q$. Let
$g:\ V\to U$ be the inverse of $f\restriction U$. The statement of the inverse function then contains the formula
$$dg(y)=\left(df\bigl(g(y)\bigr)\right)^{-1}\qquad(y\in V)\ .$$
This can be interpreted as
$$dg=\iota\circ df\circ g\ ,$$
where $\iota:\ L\mapsto L^{-1}$ denotes inversion in $GL({\mathbb R}^n)$.

Assume that $g$ is $r$ times continuously differentiable for some $r\geq1$. As both $f$ and $\iota$ are infinitely differentiable it follows by the chain rule that $dg$ is $r$ times continuously differentiable as well, and this implies that $g$ is in fact $r+1$ times continuouslay differentiable.


问:
Align a vector as close as possible with given vector with a given rotation plane
I want to create an object that always points towards gravity. Imagine a compass, but instead of being magnetic, it has a weight on one end of the indicator. Tilting this compass slightly would make the weighted end point "as downwards as possible" with the given rotation plane (which would be the compass's face).

So essentially I have a rotation plane defined by a normal vector around which I can rotate my indicator vector. The indicator vector is perpendicular to the normal vector. Then I have a target vector to which I want to get as close as possible, in this case, this is my gravity vector.

I want to get the result in either a rotation matrix or in a quaternion.

I have found this question where the accepted answer seems to be what I seek but after implementing it I tested it with the simplest case: rotation axis parallel to the ground and rotating the object around the rotation axis. I got okay results for angles between 80° and -80°, but with 90° or more my indicator starts moving all over the place for some reason.

So at this point, I'm unsure if the solution is what I seek so I hope someone could confirm to me that the solution given in the other question can be applied to my case or, if not, point me in the direction of the solution I'm looking for.

答: A 90 degree angle corresponds to the weight of the “compass” being held vertically at the top. The weight can fall in either direction (CW or CCW) with no preference. Unless you tell your simulation which way you want it to fall in this special case, you’re going to get weird looking answers.</p>


问:
Simple physical intuition for differential 1-form of a scalar function acting on a tangent vector?
I'm trying to gain some very basic physical intuition as to how the differential 1-form $df$ of a smooth function $f$ acts on a tangent vector $\mathbf{v}$. Say the temperature of a hot plate is given by $f\left(x,y\right)=100-\left(x^{2}+y^{2}\right)$. I'm walking along a curve on the hot plate defined by some parametric equation $\Phi\left(t\right)$. When I get to the point $p\left(3,7\right)$ the tangent vector $\mathbf{v}$ is given by $$\mathbf{v}=\frac{d\Phi}{dt}=\left(\frac{dx}{dt},\frac{dy}{dt}\right)=\left(1,3\right).$$

At that point am I entitled to say that the rate the temperature is changing in the direction of $\mathbf{v}$ is given by$$df\left(\mathbf{v}\right)=\left(-2x\,dx-2y\,dy\right)\left(1,3\right)=-\left(2\times3dx+2\times7dy\right)\left(1,3\right)=-48.$$So if $t$ was time in minutes, and $f\left(x,y\right)$ was in Celsius, that would mean the temperature was changing at the rate of -48 degrees Celsius per minute in the direction of $\mathbf{v}$. Have I got that right?

EDIT 10 Jan 2021

Apologies, I messed up the arithmetic by computing for the wrong point (hopefully now fixed). Also, as pointed out by @Semiclassical the units of temperature are unknown. Having said that, I would like to ask again if my interpretation is correct that at point $p\left(3,7\right)$ the instantaneous rate of change of temperature is equal to -48 units of temperature (whatever they are) per minute in the direction of $\mathbf{v}$?

答:
Setup

that would mean the temperature was changing at the rate of -48 degrees Celsius per minute in the direction of $\mathbf v$. Have I got that right?

I would not use that exact phrasing to refer to any relevant quantity, but it's quite possible that what you had in mind was correct. There are three different calculations you might be interested in, that I might phrase as follows:

The rate of temperature change in the direction of $\mathbf v$.

The rate of temperature change along my path/the curve (taking into account my/its speed).

The thing you get when you calculate $\mathrm df(\mathbf v)$.

Calculation 1

The rate of temperature change in the direction of $\mathbf v$.

Without units

If $\mathbf v=\langle1,3\rangle$, then I would say "the direction of $\mathbf v$" is essentially $\mathbf w=\dfrac{\mathbf v}{\Vert\mathbf v\Vert}=\left\langle\dfrac1{\sqrt{10}},\dfrac{3}{\sqrt{10}}\right\rangle$. And then we can take a directional derivative of $f$ at $(3,7)$ in the direction $\mathbf w$. Since $f$ is differentiable, this is just a dot product: $\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf w$ $=\left\langle-2(3),-2(7)\right\rangle\bullet\left\langle\dfrac1{\sqrt{10}},\dfrac{3}{\sqrt{10}}\right\rangle$ $=\dfrac{-48}{\sqrt{10}}$.

With units

The meaning of this is difficult to understand properly without units.

Let's say $f(x,y)=100^{\circ}\mathrm{C}-\dfrac{1^{\circ}\mathrm{C}}{\mathrm{m}^2}(x^2+y^2)$. Then we could write $\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3\,\mathrm m,7\,\mathrm m)}$ as $-14^{\circ}\mathrm{C}/\mathrm{m}$. A geometrically relevant direction vector would be unitless, since dividing a vector* by its norm would cancel any units. Therefore, the final answer here is like $\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}$; the "rate" is with respect to meters on the hot plate, not time.

*For clarity, $\mathbf v$ might be $\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$ and then $\Vert\mathbf v\Vert$ would be $\sqrt{\left(1\,\mathrm m/\mathrm{min}\right)^2+\left(3\,\mathrm m/\mathrm{min}\right)^2}=\sqrt{10}\,\mathrm m/\mathrm{min}$.

Calculation 2

The rate of temperature change along my path/the curve (taking into account my/its speed).

Without units

In the calculation above, we found $\dfrac{-48}{\sqrt{10}}$ as the rate of temperature change in a direction. And the speed of the path at $(3,7)$ is defined to be $\Vert\mathbf v\Vert$, so the rate of change of temperature for you as you're traveling along the curve at a speed of $\sqrt{10}$ would be $\sqrt{10}*\dfrac{-48}{\sqrt{10}}=-48$.

With units

In the calculation above, we found a geometric rate of $\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}$. And if your velocity at $(3,7)$ is $\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$, then your speed is $\Vert\mathbf v\Vert=\sqrt{10}\,\mathrm m/\mathrm{min}$, so the rate of change of temperature at your location with respect to time (not position) is $\sqrt{10}\dfrac{\mathrm m}{\mathrm{min}}*\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}=-48^{\circ}\mathrm{C}/\mathrm{min}$.

Calculation 3

The thing you get when you calculate $\mathrm df(\mathbf v)$.

Without units

As you found in the question, this comes out to be $\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf v=-48$.

With units

$\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf v$ $=\left\langle-6^{\circ}\mathrm C/\mathrm m,-14^{\circ}\mathrm C/\mathrm m\right\rangle\bullet\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$ $=-48^{\circ}\mathrm{C}/\mathrm{min}$

Conclusion

The calculation of $\mathrm df(\mathbf v)$ gives you the rate of temperature change along the curve (taking into account its parametrization/speed) at the given point. This is not the same thing as what some would call "how the temperature was changing in the direction of $\mathbf v$" (that's calculation 1), even though the direction you're traveling at that point is indeed "the direction of $\mathbf v$". The reason is that your speed matters for $\mathrm df(\mathbf v)$.


问:
Why is a subrepresentation a *closed* invariant subspace?
If $(\rho,V)$ is a topological representation of a topological group $G$, we usually say that a subrepresentation consists of a closed subspace $U\subset V$ which is $G$-invariant. I don't understand why it has to be closed. If
\begin{align*}
G\times V &\to V\\
(g,v)&\mapsto \rho(g)v
\end{align*}
is continuous, then so is the restriction $G\times U\to U$ even if $U$ is not closed. In other words, if we omit the need for $U$ to be closed, $(\rho,U)$ is still a topological representation. So why we impose this condition?

答: The motivation is that closed subspaces allow nice decompositions in infinite dimensions (obviously in finite dimensions all subspaces are closed). If you have a Hilbert space $H$ and a closed subspace $V \subset H$ then $H = V \oplus V^\perp$. If $V$ were not closed this wouldn't be true. For example, suppose $V$ were a dense subspace of $H$ not equal to $H$. Then $V^\perp=0$ and so clearly $V \oplus V^\perp \neq H$.


问:
If $\sum_{n=1}^{\infty} x_n$ $\lt$ $\infty$, $x_n$ must be $\lt$ $1$ for all $n$?
If $\sum_{n=1}^{\infty}$$x_n$ $\lt$ $\infty$,

does that mean that $x_n$ must be $\lt$ $1$ for all $n$?

Body: $$x_n = 2, 0,0 ,0, 0, 0\dots$$


答:
For example:$$\sum_{n=1}^\infty \dfrac{2}{n^2}=\dfrac{\pi^2}{3}\lt\infty$$
and the first term is $2$.
In general:
$$\sum_{n=1}^\infty \dfrac{k}{n^2}=\dfrac{k\pi^2}{6}\lt\infty$$
and many terms are greater than one

Title: If $\sum_{n=1}^{\infty} x_n$ $\lt$ $\infty$, $x_n$ must be $\lt$ $1$ for all $n$?

Body:
If $\sum_{n=1}^{\infty}$$x_n$ $\lt$ $\infty$,

does that mean that $x_n$ must be $\lt$ $1$ for all $n$?