问:
Using the converse if Cayley Hamilton theorem
My textbook says

: Let $M$ be a $3 \times 3$ Hermitian matrix which satisfies the matrix equation
$$
M^{2}-5 M+6 I=0
$$
Where $I$ refers to the identity matrix. Which of the following are possible eigenvalues of the matrix $M$
(a) (1,2,3)
(b) (2,2,3)
(c) (2,3,5)
(d) (5,5,6)

Then it proceeds as:

According to Cayley-Hamilton theorem, we can write $\lambda^{2}-5 \lambda+6=0 \Rightarrow \lambda=2,3$ Correct option is (b)

It's clear that the author has used the Cayley Hamilton theorem but in reverse but how can we use the converse Cayley Hamilton theorem? I've read that the converse of Cayley Hamilton theorem doesn't hold in general so what's the author doing here?

I'd be glad if someone pointed out my mistake. Much thanks.

答:
The Cayley-Hamilton theorem has nothing to do with this business.

If $\lambda$ is an eigenvalue of $M$, then $Mv=\lambda v$, for some $v\ne0$. It follows that
$$
(M^2-5M+6I)v=(\lambda^2-5\lambda+6)v
$$
and since this is $0$ by assumption, we conclude that $\lambda^2-5\lambda+6=0$. Therefore $\lambda=2$ or $\lambda=3$.

After this you cannot conclude anything else from the data, because it could well be that $M$ has just the eigenvalue $2$ or just $3$: indeed $M=2I$ and $M=3I$ satisfy the given condition and are Hermitian. You can say nothing about the multiplicities of the eigenvalues, because also
$$
\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}
$$
satisfy the condition (and are Hermitian).

However, you can certainly exclude (a), (c) and (d)


问:
Finding the (not) median from a histogram
This question is related to How to find median from a histogram?

A common problem is to estimate a median from a histogram, where 50% of the numbers are below the median, and 50% are above.

I would like to do this, but for some fraction other than 50%. I don't know if this is still called the median, or if it has some other name. (let's call it the Fractian for the purposes of this question).

I would like to estimate the value where f% of the numbers are below the value, and (100-f)% are above the value:

For example, here I am showing an estimate of the 33% fractian in these three histograms:

答: No, it is certainly NOT called the median!  Determining some other value is just a matter of finding where P(x) is equal to whatever fraction you want.  And how you do that depends on the probability function itself!</p>

ContentLicense: CC BY-SA 4.0


问:
Primitive function of $x^3 \sin x^2$
I'm trying to find the primitive function of $x^3 \sin x^2$, and I've come to a variable exchange ($t = x^2$) which led me to $\frac{1}{2} \int t \sin t dt$.

According to my text book, the primitive function is $\frac{1}{2}(-t \cos t + \sin t) + C$, but I can't see why. Isn't the derivative of that $\frac{1}{2}(t \cos t \sin t + \sin t \cos t)$?

答: $\int t \sin(t) dt= \int t (-\cos(t))'dt\overset{\ast}{=}-t\cos (t) + \int -\cos(t) dt=-t\cos(t) +\sin(t) + C$, where equality $\ast$ is integration by parts.</p>

ContentLicense: CC BY-SA 3.0

Title: Primitive function of $x^3 \sin x^2$


问:
Given $f : U \to V \otimes W$, what is the minimal subspace $W' \subseteq W$ such that $\mathrm{im}(f) \subseteq V \otimes W'$?
Let $(u_i)_{i \in I}$, $(v_j)_{j \in J}$ and $(w_k)_{k \in K}$ be bases of the three vector spaces (which are not necessarily finite dimensional). We can determine $f$ uniquely as $u_i \mapsto \sum_{j,k} f_{ijk} \cdot (v_j \otimes w_k)$ for some $f_{ijk}$ in the ground field. I claim that the span of $\left\{\sum_k f_{ijk}\cdot w_k  \mid (i,j) \in I \times J\right\}$ is the desired subvector space $W'$ of $W$. It clearly does satisfy $\mathrm{im}(f) \subseteq V \otimes W'$, but how to show that it is the unique minimal subspace with this property?

答:
Your question reduces to the following:

Lemma:

For any vector $u \in U \otimes V$ , which can be expressed as the following expression:
$$u = \sum_{i}^n v_i \otimes x_i$$
where $(v_i)$ are linearly independent and $x_i$ are non null vector.

If for a subspace $W' \subset W$ we have :
$$u \in U \otimes W' $$
then
$$x_i \in W' \quad \forall i  \quad$$
$\square$

The proof for the above lemma is based on the essential unicity of that kind of expression.

To be more precise, if there are some vectors $v'_1,v'_2,...,v'_m \in V$ and $x'_1,..., x'_m \in W'$ such that:
$$ u = \sum_{i=1}^m v'_i \otimes x'_i$$
and $ (v'_i)$ are linearly independent, $(x'_i)$ non null

Then firstly,  by changing base, we can assume $v'_i$ and $v_i$ are identical  for $i \le min(m,n)$.

Secondly, $m$ and $n$ must be equal  ( = rank of u)

And by unicity of expansion of $u$ in $U\otimes W$, $x'_i$ must be equal to $x_i$

Thus $x_i \in W'$.

Hence, conclusion.

P/s Hope there is not any fatal error in my argument.


问:
Can someone explain to me the last step in this example problem? Using operators in a DE?
How does $$\begin{pmatrix}-2 & D-1 & t^2\\\ 2(D-1) & -4 & 2t\end{pmatrix}$$

become

$$\begin{pmatrix}-2 & D-1 & t^2\\\ 0 & D^2-2D-3 & -t^2+4t\end{pmatrix}$$

If someone could explain this to me I would really appreciate it.

答:
The same way it would if $D$ were a real number: Multiply the first row by $(D-1)$, and add the result to the second row. This in order to make the first element in the second row equal to $0$.

Multiplying the first row by $(D-1)$ yields
$$
\begin{pmatrix}
-2(D-1)&(D-1)^2&(D-1)t^2
\end{pmatrix}\\
= \begin{pmatrix}
-2(D-1)&D^2-2D+1&2t-t^2
\end{pmatrix}
$$
Adding this to the second row gives you the new matrix.


答: You multiply the first line by (D-1) and then you sum Line 1 and Line 2</p>

Title: Can someone explain to me the last step in this example problem? Using operators in a DE?

Body:
How does $$\begin{pmatrix}-2 & D-1 & t^2\\\ 2(D-1) & -4 & 2t\end{pmatrix}$$

become

$$\begin{pmatrix}-2 & D-1 & t^2\\\ 0 & D^2-2D-3 & -t^2+4t\end{pmatrix}$$

If someone could explain this to me I would really appreciate it.


问:
How do I find the joint density function of a vector (uniformly distributed) defined by the area between 4 straight lines?
I have a random vector (X,Y) uniformly distributed in the region defined by the following functions:

y = 1 + x

y = -1 + x

y = 1 - x

y = -1 - x

I need to answer the following questions, but I really don't know how to proceed:

What is the joint density function of the vector?

The marginal density function of X is fX(x) = 1 - |x| when -1 < x < 1. True or false?

P(X > 0, Y > 0) = 1/8. True or false?

Are the variables X and Y independent?

I'm stuck and quite panicking. Any and all help is appreciated!

答:
To get you started, first, just identify the shape.  It is the square with vertices $\def\<{\langle}\def\>{\rangle}\<-1,-1\>,\<0,-1\>,\<1,1\>,\<0,1\>$.

$$\{\<x,y\>: -1\leqslant x\leqslant 1 \,,\, (-1+\lvert x\rvert)\leqslant  y\leqslant (1-\lvert x\rvert)\}$$

So, since this has an area of 2, and the distribution is uniform over this support: $$f_{\small X,Y}(x,y)= \tfrac 12\,\mathbf 1_{-1\leqslant x\leqslant 1 , -1+\lvert x\rvert\leqslant  y\leqslant 1-\lvert x\rvert}$$


问:
Showing the group of integers modulo m with multiplication is a group
I've just started group theory and I don't know how to show this. Are you supposed to take examples of two elements from the group and an example modulo m (say 2 and 5, and modulo 3) and show the group axioms hold? That doesn't seem sufficient to me, it seems like I should be taking $a, b, c$ as elements of the group and proving it in general. I can see how to show that G has an identity element in that case but how do I show it has an inverse and that associativity holds?

Let $a, b, c \in G, G=(\mathbb{Z}/m\mathbb{Z})$

Identity:

$a * 1 = a$ so $1$ is the identity

Inverse:

I don't know how to show this...

Associativity:

I don't know how to show this either...

答: For every a s.t. gcd(a,n) = 1 there exists a multiplicative inverse; use Extended Euclidean Algorithm to find it.</p>


问:
Legendre operator's eigenvalues and eigenfunctions
Consider the Lengendre operator which is a Sturm-Liouville operator defined in [-1,1] give as:
$$
\begin{equation}
\mathcal{L} u=-\frac{d}{d x}\left[\left(1-x^{2}\right) \frac{d u}{d x}\right]=-\left(1-x^{2}\right) \frac{d^{2} u}{d x^{2}}+2 x \frac{d u}{d x}
\end{equation}
$$
Where $u\in \mathcal{C}^\infty [-1,1] $. I want to find the operator's eigenvalues and eigenfunctions. So I'm trying to solve next equation using Frobenius method:
$$
\begin{equation}
-\left(1-x^{2}\right) \frac{d^{2} u_{n}(x)}{d x^{2}}+2 x \frac{d u_{n}(x)}{d x}=\lambda_{n} u_{n}(x)
\end{equation}
$$
I got that $u=u_1 + u_2$, where $u_1=\sum_{k=0}^{\infty} C_k\ t^{k+1}$ and $ u_2=\sum_{k=0}^{\infty} b_k\ t^{k}+d\sum_{k=0}^{\infty} C_k\ t^{k+1}\cdot log(t)$.
I've tried to sustitute in the equation but I'm not able to reach to the solution, which is $\lambda_n =n(n+1)$ and $u_n=p_n$, which are the Legendre polynomials.

答:
There are $2$ independent solutions of $\mathcal{L}u=\lambda u$ for all $\lambda\in\mathbb{C}$. Take $\lambda=0$ for example. The solutions are found by solving directly:
$$
           \frac{d}{dx}\left((1-x^2)\frac{du}{dx}\right) = 0
$$
$$
            (1-x^2)\frac{du}{dx} = C
$$
$$
             \frac{du}{dx} = \frac{C}{1-x^2}= \frac{C}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)
$$
$$
                 u = \frac{C}{2}\ln\frac{1+x}{1-x}+D
$$
The classical Legendre polynomials are uniquely determined by the requirement that the solutions remain bounded near $x=\pm 1$. Then you get solutions only when $\lambda=n(n+1)$ for $n=0,1,2,3,\cdots$, and these are the classical Legendre polynomials. $\lambda=0$ is one such value of $\lambda$, and, as you see, one solution is bounded--namely, the constant function $u=1$, which is the solution where $C=0$ and $D=1$. The solution where $C=1$ and $D=0$ is not bounded near $\pm 1$.


问:
Price for a soda has a standard deviation of $\$1.00$ with an expected price of $\$3.00$
At a shop, the price doesn't vary but it varies independently between shops. If I buy two sodas from the same shop what the expected price and standard deviation? If my friend buys two sodas from two different shops what's the expected price and standard deviation?

What I have attempted

I am thinking my expected price is $E(3)+E(3)=\$6.00$ for my expected price of soda,

my standard deviation is: $1^2+1^2=\sqrt 1$ so $1.414$

My friend's is:
expected price $E(3)+E(3)=\$6.00$ for her expected price of coffee,

her standard deviation is: $1^2+1^2=\sqrt1$ so 1.414

答:
The prices $\ F_1, F_2\ $ your friend pays for her cups of coffee are independent random variables each with mean $\ \$3\ $ and standard deviation $\ \$1\ $, so the total cost $\ F_1+F_2\ $ she pays will have mean $\ \$6\ $ and standard deviation $\ \$\sqrt{2}\ $, as you have surmised.

The prices $\ Y_1,Y_2\ $ you pay for your cups of coffee, however, are not independent, because $\ Y_2=Y_1\ $.  While the total cost $\ 2Y_1\ $ you pay for your cups of coffee also has a mean of $\ \$6\ $, its standard deviation is $\ \$2\ $, not $\ \$\sqrt{2}\ $.


问:
Does $n$ divide $2^{n-1} - 1$ for an infinite number of composite $n$?
Yes, $p$ always divides $2^{p-1}-1$ for any prime $p \ge 3$. Are there any odd composite $n$ for which $n$ divides $2^{n-1}-1$?

And on a related note, $p$ never divides $2^{p-1}+1$ for any prime $p \ge 3$. Are there any odd composite $n$ for which $n$ divides $2^{n-1}+1$?

I ask these questions because they are natural spin-offs from some well-known questions. It is well-known that no integer $n>2$ divides $2^n-1$. And it is basic to establish $p$ always divides $2^{p-1}-1$ for any prime $p \ge 3$. But for general $n$, the relationship between the integers $\varphi(n)$ and $n-1$ is not clear-cut.

Thanks for considering!

答:
The answer to your first question is yes.

If a composite integer $n$ divides $ 2^{n−1} − 1$, then $n$ is called a Fermat pseudoprime to base $2$.

There are infinitely many.

You could see some in the The On-Line Encyclopedia of Integer Sequences.


问:
Requirements for Korn's inequality on piecewise $H_1$ vector field
I am looking at the Korn's inequality on $H^1$ vector fields, as described in this paper by Brenner. In particular, I am looking at how the seminorms defined in examples 2.3 - 2.5 satisfy the conditions of Lemma 2.2.

To give a big picture, let $\Omega$ be a bounded connected open polyhedral domain in $\mathbb{R}^d$ and $\mathcal{T}$ is a triangulation of $\Omega$ by simplexes (triangles or tetrahedrals). Let $\mathbf{RM}(\Omega)$ be the space of infinitesimal rigid motions on $\Omega$ defined by:
$$
\mathbf{RM}(\Omega) = \{ \mathbf{a} + \mathbf{\eta} \mathbf{x}: \mathbf{a} \in \mathbf{R}^d \text{ and } \mathbf{\eta} \in \mathfrak{so}(d) \}
$$
where $\mathbf{x}$ is the position vector function on $\Omega$ and $\mathfrak{so}(d)$ is the Lie algebra of anti-symmetric $d \times d$ matrices.

Define the space $[H^1(\Omega, \mathcal{T})]^d$ as:
$$
[H^1(\Omega, \mathcal{T})]^d = \{ \mathbf{v} \in [L_2(\Omega)]^d: \mathbf{v}_D = \mathbf{v}|_D \in [H^1(D)]^d \quad \forall D \in \mathcal{T} \}.
$$

We need to look at the seminorm $\Phi : [H^1(\Omega, \mathcal{T})]^d \rightarrow \mathbb{R}$ which has the following properties:

$ | \Phi (\mathbf{w}) | \leq C || \mathbf{w} ||_{H^1(\Omega)}, \quad \forall \mathbf{w} \in [H^1(\Omega)]^d$,

$\Phi(\mathbf{m}) = 0$ and $\mathbf{m} \in \mathbf{RM}(\Omega) \Leftrightarrow \mathbf{m} = \text{ a constant vector}$.

I want to try to understand how the following choices of $\Phi(\cdot)$ satisfy the conditions above, i.e.:

$ \Phi_1(\mathbf{v}) = || Q \mathbf{v} ||_{L_2(\Omega)}, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T}]^d$, where $Q$ is the orthogonal projection from $[L_2(\Omega)]^d$ onto the orthogonal complement of the constant vector fields.

$ \Phi_2(\mathbf{v}) = \sup_{\mathbf{m} \in \mathbf{RM}(\Omega); || \mathbf{m} ||_{L^2(\Gamma)} = 1; \int_\Gamma \mathbf{m} ds = 0} \int_\Gamma \mathbf{v} \cdot \mathbf{m} ds, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T})]^d$.

$ \Phi_3(\mathbf{v}) = |\sum_{T \in \mathcal{T}} \int_T \nabla \times \mathbf{v} dx|, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T})]^d$.

Coming from an engineering background, I have a hard time visualising what the seminorm of $\Phi_i$ is representing. Any suggestions or pointers?

答:
I can give you a partial answer, because I couldn't check all the details, especially for $\Phi_2$.

Let's start with $\Phi_1$, First note $Q$ is an orthogonal projector on $L^2(\Omega)$ then $Q^2v=Qv$ then $ $
\begin{align}
|\Phi_1(v)|  = \|Qv\|_{L^2(\Omega)}\leq\|v\|_{L^2(\Omega)} \leq \|v\|_{H^1(\Omega,\mathcal{T} )} \quad \forall v \in H^1(\Omega)
\end{align}
if $\Phi_1(\mathbf{m})=Q\mathbf{m}=0$ and $\mathbf{m} \in \mathbf{RM}(\Omega)$ (This asumption is unnecessary here) , then must be $\mathbf{m}$ is orthogonal to orthogonal complement of constant functions. Hence, $\mathbf{m}$ must be a constant. The converse if $\mathbf{m}$ is constant so $\Phi_1(\mathbf{m})=Q\mathbf{m}=0$, because is not in the orthogonal complement of constant functions. In particular constant vectors are rigid movements.

Now we proceed for $\Phi_2$, by Cauchy Schwarz inequality for any $\mathbf{m} \in \mathbf{RM}(\Omega)$ with $\|\mathbf{m}\|_{L^2(\Gamma)}=1$ and $\int_{\Gamma}\mathbf{m}~ds = 0$ (We don't use this yet)
\begin{align}
\int_\Gamma \mathbf{v} \cdot \mathbf{m}~ds &\leq \|\mathbf{v}\|_{L^2(\Gamma)}\|\mathbf{m}\|_{L^2(\Gamma)}\\
& \leq \|\mathbf{v}\|_{L^2(\partial \Omega)} = \|\gamma_0(\mathbf{v})\|_{L^2(\partial \Omega)} \leq C\|\mathbf{v}\|_{H^1(\Omega)} \quad \forall \mathbf{v} \in H^1(\Omega)
\end{align}
Where $\gamma_0:H^1(\Omega) \to L^2(\partial \Omega)$ is the trace operator i.e $\gamma_0(\mathbf{v})=\mathbf{v}|_{\partial \Omega}$. This operator is continuous, but is quite complicated to show that See Trace theorem Ch 1.6 on [1]. So taking the supreme we get $|\Phi_2(v)|_{L^2(\Omega)}\leq C\|\mathbf{v}\|_{H^1(\Omega)}$.

Now we proceed for $\Phi_3$, Observe in two dimensions (a similar trick can be use in three dimesions) we have
\begin{align}
|\nabla \times \mathbf{u} |^2
&= \left| \frac{\partial u_2}{\partial x_1 }-\frac{\partial u_1}{\partial x_2 } \right |^2 \\
&\leq 2\left |\frac{\partial u_2}{\partial x_1 } \right |^2
+ 2\left |\frac{\partial u_1}{\partial x_2 } \right |^2
\\
&\leq 2\left |\frac{\partial u_2}{\partial x_1 } \right |^2
+ 2\left |\frac{\partial u_1}{\partial x_2 } \right |^2
+ 2\left |\frac{\partial u_1}{\partial x_1 } \right |^2
+ 2\left |\frac{\partial u_2}{\partial x_2 } \right |^2
= 2|Du|_2^2,
\end{align}
Where $|Du|_2$ is the frobenius norm and $D$ the jacobian matrix. Now because $\mathbf{v} \in H^1(\Omega)$ we bound
\begin{align}
\Phi_3(\mathbf{v})
&=\left | \sum_{T \in \mathcal{T} } \int_{T} \nabla \times \mathbf{v} \right |
=\left | \int_{\Omega} \nabla \times \mathbf{v} \right | \\
&\leq \|\Omega\|^{1/2} \left ( \int_{\Omega}|\nabla \times \mathbf{v} |^2 \right)^{1/2} \\
&\leq   \|\Omega\|^{1/2} \sqrt{2} \left ( \int_{\Omega}|D\mathbf{v} |^2 \right)^{1/2} \\
& \leq   \|\Omega\|^{1/2} \sqrt{2} \|\mathbf{v}\|_{H^1(\Omega)}.
\end{align}

I left to check what happens when $\mathbf{m}$ is a rigid motion. I don't know how to deal with that. But I hope this will be useful.

References: [1] BRENNER, Susanne; SCOTT, Ridgway. The mathematical theory of finite element methods. Springer Science & Business Media, 2007.


问:
$\int_0^1 f(t)e^{nt} dt = 0 \implies f \equiv 0 $ an application of Stone-Weierstrass
In one of the questions on my problem sheet, I've been asked to prove that if there is an $f \in C[0,1]$ such that $ \int_0^1 f(t) e^{nt} dt = 0 $ for all $n \in \mathbb{N} $ then $ f \equiv 0 $.

Disclaimer: I have looked at a lot of solutions that discuss a similar problem for example here and all its related links but I still can't figure this one out, even though I've spent quite some time on it. If I've missed something and this question is a duplicate I apologise.

This is what I have worked out so far (which I don't think is very much):

If $f \in C[0,1]$ is such that $ \int_0^1 f(t) e^{nt} dt = 0 $ for all $n \in \mathbb{N} $ then let us consider the partial sum $ S_m(nt) := \sum_{k=0}^m \frac{(nt)^k}{k!}$.

Now we know that this partial sum converges to $e$ uniformly so, for a fixed $ \epsilon > 0 $ there is an $ N \in \mathbb{N} $ such that $\forall m \geq N$
$$\|S_m - e \|_{sup} < \epsilon $$ (I think I've proved it before but if this is incorrect please let me know)

Now that we have this consider:

$$ | \int_0^1 f(t) (S_m(nt) - e^{nt}) dt | \leq \int_0^1 |f(t)| | (S_m(nt) - e^{nt} | \leq \| f \| \| S_m - e \|  < \| f \| \epsilon $$

Since epsilon was arbitrary, therefore we have that (I've missed out a few justifications here)
$$ \lim_{m \to \infty} \int_0^1 f(t) (S_m(nt) - e^{nt}) dt = 0 $$

and from our assumption, we have that $ \int_0^1 f(t) e^{nt} dt = 0 $ so the above inequality gives us that:

$$ \lim_{m \to \infty} \int_0^1 f(t) S_m(nt) dt = 0 $$

which means that $$ \lim_{m \to \infty} \int_0^1 \sum_{k=0}^m f(t) S_m(nt) dt = \lim_{m \to \infty} \sum_{k=0}^m \int_0^1 f(t)S_m(nt)dt = 0 $$

Now this is where I'm stuck, the last part looks a lot like the elementary application of Weierstrass's Approximation theorem but it's not quite that because in that we assume that $\int_0^1 t^n f(t) \, dt=0, n=1,2,\ldots$ and then prove that $f \equiv 0$ but this is kind of going in the other direction. Am I doing something wrong?

Also I note that if I can somehow prove that $\int_0^1 f^2 (t) dt = 0 $ then we would have that $f \equiv 0$ but I can't seem to figure out any way to get there.

Finally, since this is a problem sheet question I very much would appreciate if you could only give hints rather than the whole answer. Any hints and correction would be much appreciated.

答: Let $g(t)=f(e)e^{t}$. Then $\int g(t)e^{nt} dt=0$ for all $n \geq 0$. By SWT linear combinations of the functions $e^{nt}: n \geq 0$ are dense in $C[0,1]$. Hence $\int g(t)h(t)dt=0$ for all $h \in C[0,1]$. Hence $g \equiv 0$ and $f \equiv 0$.


问:
How to show that the continued fraction of $\sqrt{a^2+1}$ is $[a,2a,2a,2a,\dots]$?
I thought about the following, as $a^2\leq a^2+1 \leq (a+1)^2$, then we have $a\leq \sqrt{a^2+1}\leq a+1$ so $\lfloor \sqrt{a^2+1}\rfloor=a$ but I am having trouble doing the same for the $2a,2a,\dots$ and even If I am able to do that, I don't know what I'd use to prove that all other elements are equal to $2a$.

答:
Note that $(\sqrt{a^2+1} +a) (\sqrt{a^2+1} -a)=1$. This can be rearranged to give
\begin{eqnarray*}
\sqrt{a^2+1}=a+ \frac{1}{a+\sqrt{a^2+1}}.
\end{eqnarray*}


答:
You get then that the fractional part of $\sqrt{a^2+1}$ is $\sqrt{a^2+1}-a,$ multiply and divide by $\sqrt{a^2+1}+a$ you get
$$\sqrt{a^2+1}=a+\frac{1}{\sqrt{a^2+1}+a}=a+\frac{1}{2a+(\sqrt{a^2+1}-a)},$$
so you will always get the same expression for the integer part.

Title: How to show that the continued fraction of $\sqrt{a^2+1}$ is $[a,2a,2a,2a,\dots]$?

Body: I thought about the following, as $a^2\leq a^2+1 \leq (a+1)^2$, then we have $a\leq \sqrt{a^2+1}\leq a+1$ so $\lfloor \sqrt{a^2+1}\rfloor=a$ but I am having trouble doing the same for the $2a,2a,\dots$ and even If I am able to do that, I don't know what I'd use to prove that all other elements are equal to $2a$.


问:
Evaluating Gibbs state in the second quantized formalism
First, let us fix our notation. If $A:\mathcal H \to \mathcal H$ is a linear operator on a single-particle Hilbert space $\mathcal H$, we can lift $A$ on the Fock space $\mathfrak F_{s/a}(\mathcal H)$ in the following two ways. Here the subscript indices $s/a$ stand for symmetrized/antisymmetrized versions corresponding to bosons/fermions respectively. Define $\Gamma, \text{d}\Gamma: \mathcal H^N_{s/a} \to \mathcal H^N_{s/a}$ on $N$-particle Hilbert spaces $\mathcal H^N_{s/a}$ by

$\Gamma(A) := \bigotimes_{i=1}^N A$

$\text{d}\Gamma := \sum_{j=1}^N 1\!\!1 \otimes...\otimes 1\!\!1 \otimes A \otimes 1\!\!1 \otimes ...\otimes 1\!\!1 \, ,$ where $A$ is in the $j$-th position, and $1\!\!1$ is the identity.

Note that $\text{d}\Gamma$ is a linear operator, whereas $\Gamma$ is not.

It follows from these definitions that if $U:\mathcal H \to \mathcal H$ is unitary, $$\Gamma(U)b^*(\varphi)=b^*(U\varphi)\Gamma(U) \tag 1 $$ where $b^*(\varphi): \mathcal H^{N-1}_{s/a} \to \mathcal H^N_{s/a}$ is the creation operator of state $\varphi \in\mathcal H\,.$

Moreover, $$\Gamma(e^{\alpha A})=e^{\alpha \ \text{d}\Gamma(A)} \,. \tag 2$$

Now, let $H$ be a self-adjoint Hamiltonian over the one-particle Hilbert space $\mathcal H$, governing the dynamics of a gas of non-interacting fermions at inverse temperature $\beta \in \mathbb R$ and chemical potential $\mu \in \mathbb R \,.$

If $K_\mu$ denotes the modified Hamiltonian

$$ K_\mu := \text{d}\Gamma(H - \mu 1\!\!1) = \text{d}\Gamma(H) - \mu \mathcal N \, , \tag3$$

then the Gibbs state is defined by

$$ \omega(A) := Z^{-1} \mathrm {Tr}(e^{-\beta K_\mu}A) \tag4$$

where we assume that $e^{-\beta K_\mu}$ is trace-class and in particular, $\mathrm {Tr} (e^{-\beta K_\mu}) =: Z$. Furthermore, $\mathcal N$ represents the number operator.

I want to know how the following two equalities follow from the above considerations.

First, $$\boxed{ e^{-\beta K_\mu} b^*(\varphi) = \lambda \ b^*(e^{-\beta H}\varphi) e^{-\beta K_\mu}} \,, \tag5$$ where $\lambda := e^{\beta \mu}$ is the activity.

Secondly, $$ \boxed{\omega \big( a^*(\varphi)a(\psi) \big) = \omega \big(a(\psi)a^*(\lambda e^{-\beta H}\varphi) \big)} \,. \tag6$$

I would appreciate if you do not skip any steps. For your reference, I am studying Operator Algebras and Quantum Statistical Mechanics II Proposition 5.2.22 (pp.46-48). The steps necessary to my understanding have been skipped from the book and I am struggling to connect the dots.

Kindly help.

答:
I know this a pretty old question, but I would like to give an incomplete answer, in the sense, that I will assume that $\mathcal{H}$ is finite-dim and that we are dealing with the anti-symmetrized Fock space so that everything is finite-dimensional. The generalization should be relatively straightforward, except that we would have to drown in details such as domains and convergence. I would love to see a complete answer though.

Also, let me assume that $\mu=0$ so that $K_\mu = d\Gamma(H)$, since the proof is pretty much the same with nonzero $\mu$.

Let $\varphi_n$ be an orthonormal basis of $\mathcal H$ and $a^*_n =a^*(\varphi_n)$. It's easy to check that $d\Gamma(\mathcal H)=\sum_n a^*(H\varphi_n) a(\varphi_n)$. Hence, by the Adjoint/adjoint representation, we see that
$$
e^{-\beta d\Gamma(H)} a^*(\varphi) e^{\beta d\Gamma(H)} = \exp{(-\beta [d\Gamma(H),\cdot])} a^*(\varphi) = a^*(e^{-\beta H}\varphi)
$$
where we used the fact that $[d\Gamma(H),a^*(\varphi)]=a^* (H\varphi)$. The second equality follows from the first equality and the fact that the trace is commutative.


问:
What is a Gauge symmetry, intuitively (string theory)?
I'm writing an essay for a popular (but mathematically mature) audience on the history of mathematical physics, wherein I have a section devoted to string theory. Unfortunately, neither I (nor my audience) have the background to understand the material, however (remaining in theme with the rest of the paper) I'd like to present something substantial and quantitative (i.e. be able to present some sort of hands-on mathematical concept(s)).

While I lack the necessary foundations (I've followed the traditional undergraduate Calculus sequence, and have some number theory and group theory) I feel that I have the mathematical maturity to understand the concepts. I keep coming across these vague references to "symmetries" and "Gauge symmetries", and the notation $SU(n)$. In my extensive search, I've come across only very advanced/terse material, or very watered down and vague material (which is to be expected, I suppose).

What are "symmetries" and "Gauge symmetries"? What is this notation (which I've seen extensively even in popular books for the general population, without much explanation): $SU(n)$? Why are these symmetries important (particularly as they relate to string theory?).

答:
I recommed you first look a the simplest gauge theory, that of Maxwell theory, which is understandable in terms of vector calculus. This describes the electric field E and the magnetic field B in terms of a scalar potential $\phi$ and a vector potential $A$.

In particular the magnetic field is given by $B=\nabla \times A$. We can see that if we change $A$ by a adding a term $\nabla f$ the magnetic field remains the same. We call this a gauge symmetry and refer to $A$ as a gauge potential/field. This is a gauge thery based on the group $U(1) = SU(1)$ where we are free to change $A$ by adding a gradient of a real valued function $f$.

Theories of this form can then be generalized by instead of having a scalar field $f$ we can have matrices. In particular $SU(n)$ are a group consisting of nxn dimensional matrices of a particular form (unitary with determinant 1). In particular the theory of the electroweak force is a gauge theory described by $SU(2)$ while the strong force is described by $SU(3)$. The entire standard model of particle physics is a gauge theory based on the product group $U(1) \times SU(2) \times SU(3)$.

This can be further generalized to other continous groups (Lie groups) but this is less interesting to physicists. Note that gauge theories are important in physics in general and are by no means particular to string theory.


问:
Probability Game Optimal Strategy
You're playing a game where you have a .45 probability of winning and .55 probability of losing. You start out with 2000 chips and the winning condition is that if you bet a cumulative total of 10000 chips, then you win. Otherwise if you run out of chips, you lose.

For example, if you bet 20 chips, no matter the outcome, that will contribute 20 to the total. And you'll be left with either 1980 or 2020 current chips.

The min bet size is 1$ and max is as much as you currently have. What is the best strategy to maximize your probability of winning?

What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.

Thank you!

答:
First, see the comment/assertion of lonza leggiera.  I strongly suspect that he is right.  I have added an Addendum to (superficially) explore his assertion.

Assuming so, the answer shown below is wrong.  The answer was based on my lack of intuition re Statistics, and is an example of my going off the rails.

I don't know enough about statistics to rigorously prove my answer.
However, I am fairly certain this is the optimal approach.

Try to bet 2000 three consecutive times.

If you lose the first bet, you lose.

If you win the first bet but lose the next two, you lose.

If however, you win the first bet, and then win either the 2nd or 3rd bet, then you automatically win.

This is because you will have at least 4000 and you will have already bet 6000.

Therefore, make the 4th bet for 4000.

Even if you lose, and it busts you, you still win overall because you succeeded in wagering 10000.

Your chance of losing is

$$ (0.55) + \left[(0.45) \times (0.55)^2\right].$$

Addendum

Responding to the comment/assertion of lonza leggiera.

First of all, his assertion hit me in my blind spot.  With my general ignorance of statistics, my intuition was that small bets will almost certainly deplete one's bankroll, given the [55%-45%] underdog constraint.  That intuition is not conclusive however, because the goal here is not to show a profit in the bets, but rather to succeed in wagering \$10,000.

In my initial answer, the probability of successfully wagering a cumulative total of \$10,000 is clearly less than 0.45.  Consider the following strategy that is very similar to the one proposed by lonza leggiera.

Strategy A:

Bet \$1 until you either go broke or have made the bet 10,000 times.

You start with \$2000.  As a [55%-45%] underdog, your expected return using Strategy A is to convert your \$2000 bankroll into only \$1000.  In order for this strategy to be (somehow) worse than the strategy proposed in my initial answer, your chances of going broke (instead of hovering around \$1000) would have to be greater than 55%.  It is hard to believe that your chances of losing \$2000 (instead of only losing \$1000) would be that high.

Now consider the strategy proposed by lonza leggiera.

Strategy B:

Bet \$1 until you either go broke or

(the cumulative amount that you have wagered + your remaining bankroll)

equals or exceeds \$10,000.

If you reach the point where the sum equals or exceeds \$10,000

then bet your remaining bankroll.

Actually, Strategy B is (in effect) equivalent to Strategy A.

Once you reach the point where

(the cumulative amount that you have wagered + your remaining bankroll)

equals or exceeds \$10,000,

then wagering your remaining bankroll \$1 at a time will (also) guarantee victory.

Title: Probability Game Optimal Strategy


问:
Product rule of stochastic exponents
we know that for standard exponents, $(e^x)(e^y)=e^{(x+y)}$. What is the product rule for stochastic exponents?

$E_n(U)E_n(V)=E_n(U+V+[U,V])$ where $U$ and $V$ are stocchastic sequences, $E_n$ is the stochastic exponent and $[U,V]_n$=$ \sum_{k=1}^n(\Delta_Uk*\Delta_Vk)$.

That is the rule, but whats the proof!? thx :)

答:
I assume that by "stochastic exponents", you mean stochastic exponentials, as in Doleans-Dade exponential semimartingales (at least, this definition is consistent with the product rule you mention). The proof of the property

$$
\mathcal{E}(X)\mathcal{E}(Y) = \mathcal{E}(X+Y+[X,Y])
$$

for semimartingales $X$ and $Y$ can be found as the proof of Theorem II.38 of P. Protter's book "Stochastic integration and Differential equations", 2nd edition.

Title: Product rule of stochastic exponents


问:
Recursively defined sequences exercises.
The sequence $ \{a_n \} $ defined recursively as,

$ a_1 = 1, a_2 = 2 $ and $ a_ {n + 2} = \dfrac{1}{2} (a_n + a_ {n + 1}) $ for $ n> 3 $.

I tried and found that the limit of that series is given by $ \dfrac {2} {3} \cdot 2+ \dfrac {1} {3} $ but I don't know if that result is good.

答: You can use characteristic equation to solve it. $$2x^2-x-1=0 \implies x = 1 \text{ or } x = -\frac{1}{2}$$ Therefore $a_n = (1)^n \cdot c_0+(-\frac{1}{2})^n\cdot c_1 \implies 1 = a_1 = c_0-\frac{1}{2}c_1$ and $2 = a_2=c_0+\frac{1}{4}c_1 \implies c_0=\frac{5}{3}$ and $c_1=\frac{4}{3}$, which means that $$a_n=\frac{5}{3}+(-\frac{1}{2})^n\cdot\frac{4}{3} \implies \lim_{n \to\infty}(a_n)=\frac{5}{3}$$


答:
Using a standard method for

solving homogeneous linear recurrence relations with constant coefficients,

it can be seen that $a_n=\dfrac53+\dfrac43\left(-\dfrac12\right)^n$, so you are correct that $\lim\limits_{n\to\infty}a_n=\dfrac53$.

Title: Recursively defined sequences exercises.

Body:
The sequence $ \{a_n \} $ defined recursively as,

$ a_1 = 1, a_2 = 2 $ and $ a_ {n + 2} = \dfrac{1}{2} (a_n + a_ {n + 1}) $ for $ n> 3 $.

I tried and found that the limit of that series is given by $ \dfrac {2} {3} \cdot 2+ \dfrac {1} {3} $ but I don't know if that result is good.


答:
The first thing to always do in any problem is to graph it and take a look at whats going on! So lets look at the region that the vector is defined in:

We can see that the region is this square in the middle of side length $ \sqrt{2}$ and hence area $2.$ If you look closely you can think of this region instead of the 4 linear functions but you can combine the two bottom lines/sides of the square as $y = |x| -1 $ and the top two sides as $y = 1 - |x| $ Hence the region can bee seen as $R = \{ (x,y) \in \mathbb{R} : |x| -1 < y <  1 - |x| \} $

The joint probability density function of a uniformly distributed vector is $\frac{1}{Area} = \frac{1}{2}. $ (This is identically analogous to the density of uniform random variable on an interval $(a,b)$ which is simply $\frac{1}{Length} = \frac{1}{b-a} $ )

For the marginal density of X we simply "integrate out" Y on its range from the joint density. As shown in our alternative deffiniton of the region R we have: $|x| - 1  < y < 1 - |x|$ Hence marginal of $X$ =  $f_X = \int_{|x| - 1}^{1 - |x|} f_{X,Y}(x,y) dy$
(Where $ f_{X,Y}(x,y)$ is the joint density = $ \frac{1}{2}$ ) Hence: $f_X = \int_{|x| - 1}^{1 - |x|} \frac{1}{2} dy  =[ \frac{y}{2} ]_{|x| - 1}^{1-|x|} = 1 - |x| $

So the answer is true.

Lets look at our picture for this one, no real need for numbers. The region where both x and y are greater than zero is that little top right hand right  triangle. If we split the square (our total sample space) into the four triangles it makes with the x and y axis we see that there are four of equal area and only in one of them do we have x and y both positive. Hence $P(X > 0 , Y > 0) = \frac{1}{4}$

I will provide two ways of doing this. First without numbers then with. The answer is quite clearly no, they are not independent as the value of $X$ quite clearly impacts $Y$. For example if we were told $X < -0.8 $ then we know for sure $Y$ cannot be greater than $0.9 $ As in the region of our sample space, the square, that $X < - 0.8  $ ,this little tiny wedge on the left, the maximum value of $Y$ is $0.2$ However if we were told that $ <-0.3<X<0.3$, this thin arrow like shape in the middle, then $Y$ can definitely be greater than $0.9$.

A more formal and rigorous proof with numbers would be to show that $f_X(x) \cdot f_Y(y) \not = f_{X,Y}(x,y) $ As the product of the marginal density is the joint density if and only if they are independent. We worked out before that $f_X(x) = 1 - |x| $ and via symmetry of this problem we have $f_Y(y) = 1 - |y| $ Checking $f_X(x) \cdot f_Y(y)$ yields $( 1 - |x| - |y| + |x||y| ) $ which is quite clearly not $\frac{1}{2} $ (the joint density of $X,Y$) hence they are not independent

I really hope this helped, if you would like any further explanation please ask away! Also for what it is worth I have been studying these topics for a while and this question is definitely very challenging so don't beat yourself up at all!

Title: How do I find the joint density function of a vector (uniformly distributed) defined by the area between 4 straight lines?

Body:
I have a random vector (X,Y) uniformly distributed in the region defined by the following functions:

y = 1 + x

y = -1 + x

y = 1 - x

y = -1 - x

I need to answer the following questions, but I really don't know how to proceed:

What is the joint density function of the vector?

The marginal density function of X is fX(x) = 1 - |x| when -1 < x < 1. True or false?

P(X > 0, Y > 0) = 1/8. True or false?

Are the variables X and Y independent?

I'm stuck and quite panicking. Any and all help is appreciated!


问:
Prove that the number of solutions to $x^a = 1 \pmod p$ in $\mathbb Z_p^\times$ is $\gcd(a,p-1)$
Let $p$ be a prime number and $a\geq 1$ an integer.  Show that the number of solutions to $x^a = 1 \pmod p$ in $\mathbb  Z_p^\times$ is $\gcd(a,p-1)$.</p>

答:
Hint Let $b$ be a primitive root mod $p$. Then $x=b^i$ for some $0 \leq i \leq p-2$.

Now, $x^a \equiv 1 \pmod p \Leftrightarrow (p-1)|ai \,.$


问:
For $0 \xrightarrow{f_0} \mathbb{C}^n \xrightarrow{f_1} \mathbb{C}^\ell \xrightarrow{f_2} \mathbb{C}^r \xrightarrow{f_3} 0$, what is $\ell$?
So the question is as follows:

Let $0, \mathbb{{C}}^{n}, \mathbb{{C}}^{\ell}$ and $\mathbb{{C}}^r$ be $\mathbb{{C}}$-vector spaces (where $0$ is the trivial vector space), and let $f_i$, $(i=0,…,3)$, be $\mathbb{{C}}$-linear maps:
$$0 \xrightarrow{f_0} \mathbb{{C}}^{n} \xrightarrow{f_1} \mathbb{{C}}^{\ell}
         \xrightarrow{f_2} \mathbb{{C}}^{r} \xrightarrow{f_3} 0$$
satisfying $\ker f_{i+1} =\operatorname{im} f_i$ for $i=0,1,2$.

Any tips on how to find the value of ${\ell}$?

答:
Since the sequence is exact, $f_1$ is injective. So $\dim\operatorname{im}f_1= n$. By exactness at $\mathbb{C}^l$, we know that $\dim\ker f_2=n$. By rank-nullity for $f_2$, we know that $\dim\ker f_2+\dim\operatorname{im}f_2=l$. But since $f_2$ is surjective, $\dim\operatorname{im}f_2= r$. So $l=n+r$.

In other words, every exact sequence in the category of finite dimensional $\mathbb{C}$-vector spaces is split exact.


问:
Stochastic difference equation
I am a newbie in studying time series. Could anyone help solve the following problem:

Consider the second-order stochastic difference equation: $y_t=1.5y_{t-1}-0.5y_{t-2}+\varepsilon_t$. Given initial conditions for $y_0$ and $y_1$, find $y_t$ in terms of $\{\varepsilon_t\}$ sequence.

Find the forecast function for $y_{T+s}$ given the value of $y_T$ and $y_{T-1}$.

答: I don't have time to give a full answer, but for part (1), it's just an AR(2) process, see e.g. Autoregressive model on wikipedia.  So you can solve by writing in terms of the lag operator, etc.</p>

Title: Stochastic difference equation


问:
Integer partitions of 12 given odd and even conditions to the components.
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4=12$ with $x_1$ and $x_2$ odd and $x_3$ and $x_4$ even?

So I know that for the integer $12$ there are $p(12)=77$ partitions and I have also deduced there are $15$ partitions of the integer $12$ of size $4$. Where I'm stuck is determining how many of these have $x_1$ and $x_2$ odd and $x_3$ and $x_4$ even.

答: Subtract $1$ from the last two, giving four even numbers that sum to $10$.  Divide them all by $2$, giving four numbers that sum to $5$.  Do you know how to solve that?


问:
Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$
I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.

答:
You can do the famous "proof by definition" as follows:

$a,b,c\in (0, \pi)$ such that

$$x = \tan\frac a2, y = \tan\frac b2, z = \tan\frac c2$$
and then you will get:
$$\dfrac{(x^2+1)(y^2+1)(z^2+1)}{8xyz} = \dfrac{1}{\sin a\sin b\sin c}\geq 1$$


问:
$a-b$ divides $a^n-b^n$
I wish to prove that for integers $a,b$ it is always true that $a-b$ divides $a^n-b^n$. I want to do this via induction. For our base case $n=1$ it holds that $a-b|a-b$. Our hypthesis would be that for some $k$ we know $a-b|a^k-b^k$ or equivalently for some integer l:
$$a^k-b^k =l (a-b) $$ How would I use this to prove that:
$$a^{k+1}-b^{k+1} =l' (a-b) $$

I was thinking I could use some clever factoring or something like the binomial theorem, but not quite sure. Could someone provide a tiny hint?

答:
Strong induction may be more natural here.

The base case is $n=1,2$. This is true since $a-b|a-b$ and $a-b|a^2-b^2=(a-b)(a+b)$.

Assume that it's true for $n=1,2,..k$. Then,
$$a^{k+1}-b^{k+1}=(a+b)(a^k-b^k)-ab(a^{k-1}-b^{k-1})$$

Title: $a-b$ divides $a^n-b^n$


问:
$X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find $\sigma$
$X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find the standard deviation $\sigma$.

If the question was: find $P(9<X<15) = 0.3$, then this would be easy using the inverse normal function. But I don't see how to use the inverse normal function for $P(3<X<15)$. I'm fairly sure the value of $\sigma$ is unique. If I were to guess, I would say $\sigma \approx 15$ or $20$.

It's not a book question: it's a question I came up with. And I don't see how to do it without maths I'm not familiar with, but I may just be lacking imagination for how to solve this.

答:
It can be shown that for random variable $X \sim N(\mu,\sigma^2)$

$$P(a < X < b)= \frac{1}{2} \left[\text{Erf} \left( \frac{b - \mu}{\sigma \sqrt 2} \right) - \text{Erf} \left( \frac{a - \mu}{\sigma \sqrt 2} \right) \right]. $$
Then it is just a classic root finding problem.


问:
Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions.
Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions, both solutions are negative. I want to prove this by contradiction.

Proof: If we let $a,b,c = 1$, we have $x^{2} + x + 1 = 0$

or $x^{2} + x = -1$. If $x$ is positive the sum of two positive terms can not be $< 0$

hence $x$ must be negative. But no matter the $x$ we pick there will be no solution to $x^{2} + x = -1$.

So $a,b,c$ must also be $ < 0$. If $a,b,c < 0$ we have $-ax^{2} -bx -c = 0$

or  $-ax^{2} -bx = c$. If $x$ is positive then resulting equation is $negative - negative = positive$, which is not possible. So $x$ must be negative.

Have I shown by contradiction, both solutions are negative? How can I improve my answer?

答:
Alternative approach.

Let $D = b^2 - 4ac ~\implies~$ the roots are

$$\frac{-b \pm \sqrt{D}}{2a}.$$

Confine attention to the numerator.

Since it is assumed that both roots real, and that $a,b,c > 0:$

$$b > |\sqrt{D}| \geq 0.$$

Therefore, both $(-b + \sqrt{D})$ and $(-b - \sqrt{D})$ are negative.


答:
$$a_{n+2}-a_{n+1} = -\frac 12 (a_{n+1}-a_n) \implies a_{n+1}-a_n=\left(-\frac 12\right)^{n-1} (a_2-a_1) = \left(-\frac 12\right)^{n-1}\tag 1
$$

$$a_{n+2}+\frac 12 a_{n+1} = a_{n+1} + \frac 12 a_n \implies a_{n+1} + \frac 1 2 a_n= a_2+\frac 12 a_1 = \frac 52 \tag 2
$$

(2) - (1)
$$
a_n = \frac 23 \left(\frac 52 - \left(-\frac 12 \right)^{n-1} \right)
$$

Check here for an explanation: Finding explicit formula for recursive relation


问:
Why can disjoint subsets be picked out in these sets?
Still another exercise on Reinhard Diestel Graph Theory, GTM 173, edition 3 (on page 51)

Let $A$ be a finite set with subsets $A_1, \cdots, A_n$, and let $d_1, \cdots, d_n \in \mathbb N$. Show that there are disjoint subsets $D_k \subseteq
A_k$, with $|D_k| =d_k$ for all $k \leq n$, if and only if $|\cup_{i\in I}A_i| \geq \sum_{i \in I} d_i$ for all $I \subseteq \{ 1,\cdots,n\}$.

This "if and only if" statement is clear in one direction since for the properly chosen $D_k$, $|\cup_{i\in I}A_i| \geq |\cup_{i\in I}D_i| = \sum_{i \in I} d_i$. But the proof of the other direction seems more complicated to me. I tried to prove it by induction, but failed.

Since I am a beginner of graph theory, I just take it as an exercise of set theory, and cannot think it in the way of graphs. Maybe graph theoretic methods are favorable. Longing for your advice. Thanks very much.

答:
If such $D_k$'s exist, then for all I,
$|\bigcup_{i\in I}A_i|\geq \sum_{i\in I}d_i$. Conversely, suppose
$|\bigcup_{i\in I}A_i|\geq \sum_{i\in I}d_i$ for all I. Construct
a bipartite graph G with bipartition (B1, B2) such that $B_1$
consists of $d_1$ copies of $A_1$, $d_2$ copies of $A_2$, ...,
$d_k$ copies of $A_k$, and $B_2 = A$. Let $U\in B_1$ be adjacent
to $v\in B_2$ if $v\in U$. Now consider $X\subseteq B_1$. Let J be
the set of indices of $A_j$'s that appear (in any number of
copies) in X. Then $|X| \leq \sum_{j\in J}d_j \leq |\bigcup_{j\in
J}A_j| = |N(X)|$. Hence by Hall's Theorem there is a complete
matching M of $B_1$ in G. Define $D_i = \{u\in B_2 : u$ is matched
with $A_i$ in M}.</p>


问:
Any good techniques for solving 2nd degree (sort of), 3 variable simultaneous equations?
So I came across this question the other day asking me to solve the following system of equations:

$$
\left\{
\begin{array}{c}
x^2-yz=1 \\
y^2-zx=2 \\
z^2-xy=3
\end{array}
\right.
$$

Normally when solving these 3 variable systems of equations I would use the method where you put the coefficients into a matrix. However, to me it seems like that wouldn't work here, unless there is some trick that needs to happen. Are there any easy-ish ways of solving this? Or do I just need to use good old substitution until I can isolate some variables? Substitution is what I tried originally, it just seemed to get really long and I gave up.

答:
HINT

The proposed system of equations can be rewritten as
\begin{align*}
\begin{cases}
x^{2} - yz = 1\\\\
y^{2} - zx = 2\\\\
z^{2} - xy = 3
\end{cases} & \Longleftrightarrow
\begin{cases}
x^{2} - yz = 1\\\\
y^{2} - x^{2} + yz - zx = 1\\\\
z^{2} - y^{2} + zx - xy = 1
\end{cases} \Longleftrightarrow
\begin{cases}
x^{2} - yz = 1\\\\
(y-x)(x + y +z) = 1\\\\
(z-y)(x + y + z) = 1
\end{cases}\\\\ & \Longleftrightarrow
\begin{cases}
x^{2} - yz = 1\\\\
(y-x)(x+y+z) = 1\\\\
(z-y)(x+y+z) = 1\\\\\
(z - 2y + x)(x+y+z) = 0
\end{cases}
\end{align*}

The last equation tells us that $x = 2y -z$ or $x = -y - z$. But $x+y+z$ cannot be equal to zero (why?). Thus it remains just the first option.

Can you take it from here?


问:
Simple Triangle-like Inequality for a general norm
I came up with this inequality as I was thinking about norms. I don't know if it's true though.

Given three points $x,y,z$ in a normed space $(X,\|\cdot\|)$, is it true that if

$$
\|x-y\|\le\|y\|
$$

and

$$
\|x-z\|\le\|z\|
$$

then

$$
\|x-(y+z)\|\le\|y+z\| ?
$$

In other words, for a given vector $x$ in a normed space $X$, is the set $K:=\{y:\|x-y\|\le\|y\|\}$ additive?

First, if it were true, it almost looks like we could change the $y$ and $z$ around so that, e.g. the 1st and 3rd inequalities imply the 2nd. As it is, given any arrangement $x+y=z$ (none of which is $0$), we can tell which is which by simply adding.

I thought proving $X=\mathbb{R}$ would easily turn into $X=l_p$, but that seems naive.

Inner Product Space

The first two inequalities reduce to

$$
\langle x,x\rangle\le 2\langle y,x\rangle
$$

and

$$
\langle x,x\rangle\le 2\langle z,x\rangle
$$

Adding these together gets

$$
\|x\|^2\le \langle y+z,x\rangle\le 2\langle y+z,x\rangle
$$

Working the same algebra backwards then gets the third inequality.

Other norms?

Note that if $y\in K$, then $cy\in K$ for all $c\ge 1$. So there is a curve/surface in $X$, symmetric about $0$ (AKA odd), that divides $X$ into two parts, one part being $K$. It looks like if the unit ball $B$ of $X$ is oblong and $x$ is not perpendicular to an axis of symmetry of B, the curve/surface will not be linear and hence not convex (symmetric about $0$ and convex implies linear).

So, the factor of $2$ cannot be dismissed for general $X$ as it was for inner product spaces (?). I feel like that argument says the 2 cannot be dismissed for some inner product spaces as well. Mistake?

Thanks!

答:
Update: Total rewrite! … And then partial retraction. Hopefully, some part of this is still helpful and I will continue to think about how to get a possibly complete answer. This is a great question. I've spent quite a while thinking about it now, as well as perusing Dan Amir's Characterizations of Inner Product Spaces for possibly related conditions. I thought I had a solution, but I made a sign error that you (Aaron Goldsmith) pointed out.

First, some definitions. Let $(V,||\cdot||)$ be a normed vector space. For any $x\in V$, define
$$K_x=\big\{y\in V:||x-y||\leq ||y||\big\}\qquad\text{and}\qquad J_x=\big\{y\in V:||x-y||\leq ||x+y||\big\}.$$
For any $x\in V$, notice that $K_{2x}=x+J_x$ and thus $\partial K_{2x}=x+\partial J_x$. Note also that $$\partial J_x=\big\{y\in V:||x-y||=||x+y||\big\}=J_x\cap J_{-x},$$ for any $x\neq 0$.
Now, I can state my partial result.

Proposition. Consider the following conditions:

$||\cdot||$ arises from an inner product on $V$;

$J_x$ is convex for all $x\in V$;

$\partial J_x$ is convex for all $x\in V$;

$J_x$ is additive for all $x\in V$;

$\partial J_x$ is additive for all $x\in V$;

$K_x$ is convex for all $x\in V$;

$\partial K_x$ is convex for all $x\in V$;

$K_x$ is additive for all $x\in V$.

I conjecture that $(8) \Longrightarrow (1)$, but have yet to prove it. We do know that $$(1)\Longleftrightarrow(2) \Longleftrightarrow (3) \Longleftrightarrow (4) \Longleftrightarrow (5) \Longleftrightarrow (6) \Longleftrightarrow (7) \Longrightarrow (8).$$

Proof. The equivalences $(1)\Longleftrightarrow (3)$ and $(1)\Longleftrightarrow(5)$ are conditions $(3.2)$ and $(4.13)$ in Amir's book, respectively. Since convexity is preserved by translation, we also have $(2)\Longleftrightarrow(6)$ and $(3)\Longleftrightarrow(7).$ To complete the proof, we will show that $(2)\Longrightarrow (3)$ and $(4)\Longrightarrow (5)$, and that $(1)\Longrightarrow (2),$ $(4)$, $(6)$ and $(8)$.

$(2)\Longrightarrow (3)$ and $(4)\Longrightarrow (5)$: Suppose that $J_x$ is convex (resp. additive) for all $x\in V$. First note that $\partial J_0=\partial V=\emptyset$ is vacuously convex and additive. If $x\neq 0$, then $\partial J_x=J_x\cap J_{-x}$ is convex (resp. additive) because convex (resp. additive) sets are closed under intersection.

$(1)\Longrightarrow (2)$, $(4)$, $(6)$ and $(8)$: Suppose that $||x||^2=\langle x,x\rangle$ for all $x\in V$, where $\langle \,\cdot\,,\cdot\,\rangle$ is an inner product on $V$. Then we have $$J_x=\big\{y\in V:\langle x,y\rangle \geq 0\big\}\qquad\text{and}\qquad K_x=\big\{y\in V:2\langle x,y\rangle \geq ||x||^2\big\}.$$ These sets are both convex and additive, since they are inverse images of sets that are convex and additive under the linear map $y\mapsto \langle x,y\rangle$. $\square$

The proof of $(8)\Longrightarrow (1)$ still eludes me at this time, though I thought for a minute that I had figured it out. It’s more difficult because additivity is not preserved by translation, unlike convexity. For another instance of this, we could consider the condition “$\partial K_x$ is additive for all $x\in V$.” This is never true when $\dim V>0$, because if $x\neq 0$, then $x\in\partial K_{2x}$, but $2x$ is in the interior of $K_{2x}$.


问:
Let a, b, c $\in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.
I'm lost on how to prove this statement.

Let $a, b, c \in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.

So far, I have

There exists $x_1, x_2, x_3 \in \mathbb{Z}$ and $y_1, y_2, y_3 \in \mathbb{Z}-{0}$ such that $a = x_{1} // y_{1}$, $b = x_{2} // y_{2}$, and $c = x_{3} // y_{3}$.
Suppose $a + c < b + c$. Then, we have

$a + c < b + c \iff (x_1//y_1) + (x_3//y_3) < (x_2//y_2) + (x_3//y_3)$....

How do I proceed next? I don't even know if my idea is correct or not. Thank you.

答:
$a+c < b + c \implies (a+c) + (-c) < (b+c) + (-c) $

$\implies a+(c+(-c) < b+(c+(-c)) \implies a + 0 < b + 0 \implies a < b.$

Since $-c$ is the element $x$ such that $c+x = 0$.


答:
It depends on your axioms.

You probably have an axiom that if $\alpha < \beta$ then $\alpha + \gamma < \beta + \gamma$ for all $\gamma$.

So you can either prove this by contradiction:

If $a=b$ then $a+c = b+c$ and we know that isn't true; and if $b < a$ then $b+c < a+c$ and we know that isn't true.  So the only option left is $a< b$.

Or we can simply use the $-c$.

$a + c < b+c$ so $(a+c) + (-c) < (b+c)+(-c)$ so $a+0 < b+0$ and $a < b$.

(More thorough details:

$a +c < b+c$ was a given

$-c$ is a number that exists (by definition and existence of additive inverses)

$(a+c) + (-c) < (b+c) + (-c)$ by axiom

$a+(c+(-c)) < b +(c + (-c))$ by associativity

$a+ 0 < b + 0$ by definition of additive inverse

$a < b$ by definition of additive identity.)

.... but ... that's assuming that $a<b \implies a+c < b+c$ is your axiom.  If you introduced you math system by different methods, you'd prove this in a different manner.

Title: Let a, b, c $\in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.

Body:
I'm lost on how to prove this statement.

Let $a, b, c \in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.

So far, I have

There exists $x_1, x_2, x_3 \in \mathbb{Z}$ and $y_1, y_2, y_3 \in \mathbb{Z}-{0}$ such that $a = x_{1} // y_{1}$, $b = x_{2} // y_{2}$, and $c = x_{3} // y_{3}$.
Suppose $a + c < b + c$. Then, we have

$a + c < b + c \iff (x_1//y_1) + (x_3//y_3) < (x_2//y_2) + (x_3//y_3)$....

How do I proceed next? I don't even know if my idea is correct or not. Thank you.


问:
Is the eventown problem solution unique up to isomorphism
Suppose we have a town with a set of residents $X$, where $|X| = n$ (suppose $n$ even for simplicity here). The residents like forming clubs, and we have clubs $C_1,C_2,\ldots,C_m \subseteq X$. The eventown problem is interested in the maximum number of clubs $m$ that can be formed with these specific rules:

All clubs are distinct.

A club must have an even number of members. $\forall i, |C_i|$ is even.

Any pair of two clubs shares an even number of members: $\forall i \neq j, |C_i \cap C_j|$ is even.

It is easy to show that the largest number of clubs possible is $m=2^{n/2}$. An easy construction to attain this bound goes as follow : Take any partition of $X$ into sets of size 2, $B_1,\ldots,B_{n/2}$, and consider the family $\mathcal{F}=\{\bigcup_{i\in S}B_i \ :\ S\subseteq [n/2] \}$

For any partition, this matches the required bound, and all these possibilities are pairwise isomorphic.

Is the solution unique up to isomorphism, or are there some other constructions?

答:
Just writing down some details from the counter-example found by Calvin Lin.

Note that in a maximum solution derived from a partition of $X$, for any 3 clubs $C_1,C_2,C_3$, we have $\vert C_1\cap C_2 \cap C_3 \vert$ is also even.

Let $C_1,C_2,C_3$ be the following clubs in $\mathbb{F_2}^n$ (where the $\dots$ are all $0$ )

$$\begin{cases}
C_1 &= &( & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & \dots &)\\
C_2 &= &( & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & \dots &)\\
C_3 &= &( & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & \dots &)\\
\end{cases}$$

Note that they form an Eventown, but their global intersection is odd, therefore the will never be from a partition of the ground set $X$.

Now, we use the fact that in Eventown, any maximal solution (not extendable) is maximum (of size $2^{n/2}$).

Therefore we will have a maximal solution not isomorphic to the partition solution.


问:
Limit involving $\sum^n_{k=1} k^2$ in both the nominator and denominator.
How would you solve the following limit without using the $n^2$ sum formula:

$\lim\limits_{n \to \infty} \frac{(\sum_{k=1}^n k^2)+(n+1)^2}{\sum_{k=1}^n k^2}$

or written in another way:

$\lim\limits_{n \to \infty} 1 + \frac{(n+1)^2}{\sum_{k=1}^n k^2}$

答:
Hint: write it as an approximation to an integral.

 As $n \to \infty$ $$\frac{(n+1)^2}{\sum_{k=1}^n k^2}=\frac{\frac{(n+1)^2}{n^3}}{\frac 1n \sum \frac{k^2}{n^2}} \to \frac{0}{\int_0^1 x^2 dx}$$


问:
Markov chains with continuous time - is $P(2)$ having a given form even possible?

Is it possible that
$$
\begin{vmatrix}
0 & 1 \\
1 & 0
\end{vmatrix} = P(2)
$$
for any Markov semigroup $\{P(t), t \geq 0\}$?

Recalling what properties must hold for transition matrix $P$:

$\forall t>0$ every row of matrix must sum to 1 and all entries must be non negative

$P(0) = I$ and $\lim_{t \to 0^+} P(t) = I$

$P(s+t) = P(s)P(t)$

Any hints appreciated. I suppose we're looking for some kind of contradiction but I couldn't find any clues.

答:
Let us abbreviate

$$
  A :=
  \begin{pmatrix}
    0 & 1 \\
    1 & 0
  \end{pmatrix}.
$$

Claim: A Markov semigroup $(P(t))_{t \in [0,\infty)}$ with $P(2) = A$ does not exist.

Here are three different proofs, so you can choose your favourite one. :-)

Proof 1 ($A$ has no real square root).

There is no matrix $B \in \mathbb{R}^{2 \times 2}$ such that $B^2 = A$; indeed, the equality $B^2 = A$ would imply $-1 = \det(A) = (\det(B))^2 \ge 0$.

Proof 2 (Perron-Frobenius theory).

For Markov semigroup $(P(t))_{t \in [0,\infty)}$ on $\mathbb{R}^n$ it follows from Perron-Frobenius theory that each matrix $P(t)$ has only one eigenvalue on the complex unit circle - namely the number $1$.

However, the matrix $A$ has the eigenvalues $1$ and $-1$.

Proof 3 (The isometry group with respect to the infinity norm is discrete).

Endow $\mathbb{R}^n$ with the infinity norm and assume that $(P(t))_{t \in [0,\infty)}$ is a Markov semigroup on $\mathbb{R}^n$ such that $P(2)$ is a permutation matrix. We show that this implies $P(t) = I$ (:= identity matrix) for all $t \ge 0$:

Each $P(t)$ acts contractively with respect to the infinity norm on $\mathbb{R}^n$ (via $x \mapsto P(t)x$). But $P(2)$ is an isometry with respect to this norm, which readily implies that each $P(t)$ is an isometry. Hence, each $P(t)$ is a signed permutation matrix (see this MathOverflow post). Since every entry of each $P(t)$ is $\ge 0$, this implies that each $P(t)$ is a permutation matrix.

From $P(t) \to I$ as $t \downarrow 0$ we now conclude that $P(t) = I$ for all sufficiently small times $t$. The semigroup law then implies that $P(t) = I$ for all times $t$.

In particular, the permutation matrix $P(2)$ must be equal to $I$, too.

Remark. Proofs 2 and 3 are, admittedly, quite an overkill if you are merely interested in the concrete example from the question. But on the other hand, proofs 2 and 3 give us more structural insight and work in more general situations than proof 1.


答:
By the distributive property of limits,
$$1 + \lim_\limits{n\to\infty}{\frac{(n + 1)^2}{\sum_{k=1}^nk^2}}$$
Now, as we are in a limit that tends to infinity, we don't care cases of small numbers.

See that
$$f(n) = (n-1)^2 + (n-2)^2 +  n^2 - (n + 1)^2 = (n-1)^2 + (n-2)^2 - 2n-1$$
So,
$$f'(n)>0$$
So $f$ is increasing function for greater values of $n$. Now as $n$ tends to infinity, this difference is also tending to infinity.

Hence, we see that numerator is infinitely smaller than denominator and when $n$ is infinity, $\lim_\limits{n\to\infty}{\frac{(n + 1)^2}{\sum_{k=1}^nk^2}}$ is $0$.
So, the original limit is $1$.

Note: $(n-1)^2$, $(n-2)^2$ and  $n^2$ are the last few terms of $\sum_{k=1}^n{k^2}$.

Title: Limit involving $\sum^n_{k=1} k^2$ in both the nominator and denominator.

Body:
How would you solve the following limit without using the $n^2$ sum formula:

$\lim\limits_{n \to \infty} \frac{(\sum_{k=1}^n k^2)+(n+1)^2}{\sum_{k=1}^n k^2}$

or written in another way:

$\lim\limits_{n \to \infty} 1 + \frac{(n+1)^2}{\sum_{k=1}^n k^2}$


答: Use L'Hopitals rule on the original equation, to get $lim_{n\rightarrow \infty}\frac{(n+1)^2}{n^2}=1$. The derivative of $(\sum_{k=1}^n k^2)$ is $n^2$.


问:
Show that $R^* = R\setminus \mathfrak{m}.$
Here is the question I want to answer:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$

Hint: The quotient $R/ \mathfrak{m}$ is a field.

Here is a trial which I feel that it is not correct specifically because I did not use the hint:

Let $R$ be a commutative ring and let $(R, \mathfrak{m})$ be a local ring, we want to show that $R^* = R\setminus \mathfrak{m}.$

Let $x \in R\setminus \mathfrak{m}$ s.t. $X \notin R^*.$ Then $\langle x \rangle$ is a non-zero proper ideal which is contained in a maximal ideal different from $M,$ which is a contradiction. Therefore $R^* = R\setminus \mathfrak{m}.$

So could anyone criticize this solution to me and tell me the correct and inaccurate/incorrect statements in it please?

Another trial which I feel like it is not written logically correct(specifically the last paragraph and how all the proof before it proves the required):

Suppose that $R$ is a local ring with maximal ideal $\mathfrak{m}$. Then $R/\mathfrak{m} = \{r + \mathfrak{m} \mid r \in R \}$ and the zero element of $R/\mathfrak{m}$ is $0 + \mathfrak{m} = \mathfrak{m}.$\

We will show that every non-zero element in $R / \mathfrak{m}$ has a multiplicative inverse in $R / \mathfrak{m}$ i.e. $R / \mathfrak{m}$ is a field.

Let $ r + \mathfrak{m} \neq \mathfrak{m}$ be a non-zero element in $R / \mathfrak{m}.$Then $r \notin \mathfrak{m}.$\

Clearly, $rR$ is an ideal of $R$ and $ \mathfrak{m} + rR$ is an ideal of $R$ containing $\mathfrak{m}$ properly.

Then $\mathfrak{m} \subset \mathfrak{m} + rR \subset R$ and since $\mathfrak{m}$ is maximal and $r \notin \mathfrak{m}$ then $\mathfrak{m} \neq \mathfrak{m} + rR $ and  $\mathfrak{m} + rR = R.$\

Now, since $1 \in R,$ then $1 = m + rs$ for some $m \in \mathfrak{m}$ and $s \in R.$ So, $1 - rs = m \in \mathfrak{m} $ which means $1 - rs + \mathfrak{m} = \mathfrak{m}.$ Therefore, $$1   + \mathfrak{m} = rs + \mathfrak{m} = (r + \mathfrak{m})(s + \mathfrak{m})$$

Hence, for every $\mathfrak{m} \neq  r + \mathfrak{m}$ in $R / \mathfrak{m},$ there exists  $ s + \mathfrak{m}$ in $R / \mathfrak{m},$ such that $(r + \mathfrak{m})(s + \mathfrak{m}) = 1 + \mathfrak{m}$ where  $1 + \mathfrak{m}$ is the multiplicative identity element of $R / \mathfrak{m}.$ And since $R$ is commutative then $$(r + \mathfrak{m})(s + \mathfrak{m}) = (s + \mathfrak{m})(r + \mathfrak{m}) = 1 + \mathfrak{m}$$

Hence, every non-zero element of $R / \mathfrak{m}$ is a unit and since the zero element of $R / \mathfrak{m}$ is $\mathfrak{m}$ then  $R^* = R\setminus \mathfrak{m}.$

In this second proof I do not see if I really proved the required or not, so could anyone help me in proving the required if I did not do that or correct my mistakes if I was doing that?

A trivial question on the second proof, why $rR$ is clearly an ideal? could anyone explains that to me please?

In general, I would appreciate if I saw a smart, elegant, simple and logically correct proof of the required.

答:
Given any commutative ring $A$, the relation $A \setminus \mathrm{U}(A)=\displaystyle\bigcup \mathscr{Max}(A)$ holds, where $\mathrm{U}(A)$ indicates the units of the ring and $\mathscr{Max}(A)$ the maximal spectrum, i.e. the collection of all maximal ideals.

On the one hand, it is elementary that any proper ideal must be disjoint from the subset of units, which applies in particular to maximal ideals. This justifies the inclusion $\displaystyle\bigcup\mathscr{Max}(A) \subseteq A \setminus \mathrm{U}(A)$.

Conversely, consider a nonunit $x \in A \setminus \mathrm{U}(A)$. The principal ideal $Ax$ will then be proper and thus included in a certain maximal ideal $M \in \mathscr{Max}(A)$ by virtue of Krull's theorem. This proves the converse inclusion $A \setminus \mathrm{U}(A) \subseteq \displaystyle\bigcup\mathscr{Max}(A)$.

When $A$ is local $\mathscr{Max}(A)$ is a singleton, and the above general relation reduces to precisely the claim you seek to prove.


答:
You're solving the equation
$$
\begin{aligned}
0.3 &= P\left( 3<X<15\right)\\
&=P\left(\frac{3-12}\sigma < \frac {X-12}\sigma<\frac{15-12}\sigma\right)\\
&=P\left(\frac{-9}\sigma<Z<\frac3\sigma\right)\\
&=\Phi\left(\frac3\sigma\right)-\Phi\left(\frac{-9}\sigma\right)
\end{aligned}
$$
for $\sigma$, where $\Phi$ denotes the distribution function for the standard normal. This has to be solved numerically. For example, you can use Newton's method
$$
s_{\rm new} = s-\frac{f(s)}{f'(s)}
$$
to solve $f(s)=0$ where
$$
f(s):=\Phi\left(\frac3s\right)-\Phi\left(\frac{-9}s\right)-0.3
$$
with derivative
$$
f'(s)=-\frac3{s^2}\phi\left(\frac3s\right)-\frac9{s^2}\phi\left(\frac{-9}s\right);
$$
as usual $\phi$ denotes the standard normal density.

Here is an R script to carry out the iteration. The solution is approximately $s=15$, as you've guessed.

```<code>f <- function(s) pnorm(3/s) - pnorm(-9/s) - 0.3
df <- function(s) (-3 * dnorm(3/s) -9 * dnorm(-9/s)) / (s*s)

s <- 10
for (i in 1:10)
{
    print(s)
    s <- s - f(s) / df(s)
}
</code>```

with output

```<code>[1] 10
[1] 13.78225
[1] 15.14899
[1] 15.27338
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428
</code>```

Title: $X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find $\sigma$

Body:
$X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find the standard deviation $\sigma$.

If the question was: find $P(9<X<15) = 0.3$, then this would be easy using the inverse normal function. But I don't see how to use the inverse normal function for $P(3<X<15)$. I'm fairly sure the value of $\sigma$ is unique. If I were to guess, I would say $\sigma \approx 15$ or $20$.

It's not a book question: it's a question I came up with. And I don't see how to do it without maths I'm not familiar with, but I may just be lacking imagination for how to solve this.

Body:
You're solving the equation
$$
\begin{aligned}
0.3 &= P\left( 3<X<15\right)\\
&=P\left(\frac{3-12}\sigma < \frac {X-12}\sigma<\frac{15-12}\sigma\right)\\
&=P\left(\frac{-9}\sigma<Z<\frac3\sigma\right)\\
&=\Phi\left(\frac3\sigma\right)-\Phi\left(\frac{-9}\sigma\right)
\end{aligned}
$$
for $\sigma$, where $\Phi$ denotes the distribution function for the standard normal. This has to be solved numerically. For example, you can use Newton's method
$$
s_{\rm new} = s-\frac{f(s)}{f'(s)}
$$
to solve $f(s)=0$ where
$$
f(s):=\Phi\left(\frac3s\right)-\Phi\left(\frac{-9}s\right)-0.3
$$
with derivative
$$
f'(s)=-\frac3{s^2}\phi\left(\frac3s\right)-\frac9{s^2}\phi\left(\frac{-9}s\right);
$$
as usual $\phi$ denotes the standard normal density.

Here is an R script to carry out the iteration. The solution is approximately $s=15$, as you've guessed.

f <- function(s) pnorm(3/s) - pnorm(-9/s) - 0.3
df <- function(s) (-3 * dnorm(3/s) -9 * dnorm(-9/s)) / (s*s)

s <- 10
for (i in 1:10)
{
    print(s)
    s <- s - f(s) / df(s)
}

with output

[1] 10
[1] 13.78225
[1] 15.14899
[1] 15.27338
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428
[1] 15.27428


问:
How to show $T$ is diagonalizable?
Let $T\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$.  Show that $T$ is diagonalizable.

It seems as if this is a very simple question but I'm a bit confused about how to start!

答:
You need to prove that $\mathbb{R}^3$ has a basis of eigenvectors.  The key is that if you have vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ that are eigenvectors for distinct eigenvalues, then the vectors must be linearly independent.

To prove this latter fact, assume that
$$c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 = \vec{0}.$$
Apply $(T - \lambda_1 I)(T -\lambda_2 I) = (T -\lambda_2 I)(T - \lambda_1 I)$ to both sides to find that $c_3 = 0$.  (Here $I$ is the identity operator.)  Note that you must use $\lambda_3 \neq \lambda_1, \lambda_2$ here.  Similar operators can be applied to show that $c_1$ and $c_2$ must equal $0$ as well.

In this case, the argument gives you three linearly independent eigenvectors in $\mathbb{R}^3$, which are therefore a basis.

Title: How to show $T$ is diagonalizable?


问:
Is not first order axiomatizable?
I have the following question: ¿Being a: connected graph, Archimedean field, characteristic field 0,  well order, is not definable first order?, if the answer were true we would have to show that there is no first order $\varphi$  statement (in proper language) such that $\mathfrak{A}⊨\varphi$ if and only if $\mathfrak{A}$ is connected graph, characteristic field 0, etc. I need help to prove or disprove this question.

答:
The case of well-orders is a good example of this this kind of argument usually runs. Let $LO$ be the theory of linear orders, $(c_i)_{i\in\omega}$ be a new set of constants and for each $i\in\omega$ let $\varphi_i$ be the sentence in $c_i<\ldots <c_{1}<c_0$ in the expanded language.

If there were a sentence $\psi$ that a structure satisfied exactly when it's a well-order, then $(\omega,<)$, for example, would satisfy it; but we can also interpret any finite number of the $\varphi_i$ in $(\omega,<)$, which means that $LO\cup\{\psi\}\cup\{\varphi_i:i\in\omega\}$ is satisfiable, so there's a model $M$ of this theory.

But notice that $M$ cannot be well-founded on account of there being an infinite strictly decreasing sequence through the ordering. Yet it must satisfy $\psi$, a contradiction; so there is no such $\psi$.

The other examples work similarly; you use compactness to show that there is a structure in which your special property (well-foundedness, connectedness, being archimedean) fails, but which still satisfies all the same sentences as some structure which does satisfy that property.

Title: Is not first order axiomatizable?


问:
Linear system of equations with an equality constraint on some of the unknowns

Background

I have a system of equations which can be written in matrix form as
\begin{align}
  \mathbf{A}\mathbf{x} = \mathbf{0},\label{eq:system}\tag{1}
\end{align}
where $\mathbf{A}$ is invertible and has size $N \times N$, and the values of some of the entries of $\mathbf{x}$ (let's call them $x_i$) have to satisfy
\begin{align}
  x_i = x_j - c_iy_i - d,\label{eq:constraint}\tag{2}
\end{align}
where $x_j$ is some other entry of $\mathbf{x}$. Also, $y_i\in\mathbf{y}$ which is another vector of length $M < N$, and is known. (in reality it is unknown, and I have to add in another set of equations, but I'm simplifying the situation here).
Equation (\ref{eq:constraint}) can be written in matrix form as
\begin{align}
  \mathbf{P}_i\mathbf{x} = \mathbf{P}_j\mathbf{x} - \mathbf{C}\mathbf{y} - \mathbf{d},\label{eq:matconstraint}\tag{3}
\end{align}
where $\mathbf{P}_i$ "selects" the $x_i$ entries of $\mathbf{x}$, $\mathbf{P}_j$ "selects" the $x_j$ entries of $\mathbf{x}$, and $\mathbf{C}$ "selects" the $y_i$ entries of $\mathbf{y}$. Assuming there are $M < N$ such constraints, then the size of $\mathbf{P}_i$ and $\mathbf{P}_j$ is $M \times N$, the size of $\mathbf{C}$ is $M \times M$, and $\mathbf{d}$ has length $M$, as does $\mathbf{y}$.

The goal is to solve (\ref{eq:system}) subject to (\ref{eq:matconstraint}).
In particular, I want to end up with one linear system of equations, solvable with standard linear techniques, e.g. LU factorization, GMRES.

My Approach

To satisfy the constraints, I decided to delete the columns of (\ref{eq:system}) which correspond to $x_i$, and correspondingly delete those $x_i$ from $\mathbf{x}$. I plan to do this using an "incidence" matrix $\mathbf{Q}$ with ones and zeros, to get
\begin{align}
  \mathbf{A}\mathbf{Q}^T\mathbf{Q}\mathbf{x},
\end{align}
where $^T$ means transpose.
Now I need to to add in the contributions of my "known" $x_i$, which I can do as
\begin{align}
  \mathbf{A}\mathbf{Q}^T\mathbf{Q}\mathbf{x} + \mathbf{A}\mathbf{P}_i\mathbf{x} = \mathbf{0},
\end{align}
where $\mathbf{P}_i$ "selects" the $x_i$ entries of $\mathbf{x}$.
Finally, subbing in (\ref{eq:matconstraint}), I can write
\begin{align}
  \mathbf{A}\mathbf{Q}^T\mathbf{Q}\mathbf{x} + \mathbf{A}\mathbf{P}_i\left(\mathbf{P}_j\mathbf{x} - \mathbf{C}\mathbf{y} - \mathbf{d}\right) = \mathbf{0}.\label{eq:final}\tag{4}
\end{align}

Questions

Is (\ref{eq:final}) the correct requisite linear system that I must solve? I may have made some silly indexing mistakes in the above, but are the concepts correct?

Will the resulting system (\ref{eq:final}) be invertible, given $\mathbf{A}$ is invertible? Alternatively, are there any conditions which guarantee that (\ref{eq:final}) is uniquely solvable?

Is my approach a "standard" approach for how one would solve such a problem? I have found that some similar problems require least squares solvers or some kind of optimization (e.g. this and this), but am I correct in assuming that those aren't needed here, and a linear solver is enough? I don't fully understand when and why one would pick an optimization routine over the above approach. In particular, this question seems similar to mine, but apparently there the answer is to simply invert the original system, and the boundary conditions are ignored, according to the only answer. The right-hand side in that question is not zero though - is that the key difference?

Thank you!

答:
Add a row to $x$, $x_{n+1}$; add a row to $A$ that looks like
$\pmatrix{0&0&\dots&0}$. Add a "1" at the bottom of the $0$ vector on the right-hand side. You now have the same system, except you have $x_{n+1} = 1$.

Rewrite this equation:
\begin{align}
  x_i = x_j - c_iy_i - d,
\end{align}
as
\begin{align}
  1 \cdot x_i - 1 \cdot x_k + d \cdot x_{n+1} =  c_iy_i
\end{align}
and define $z_i =c_i y_i$ to get rid of an extraneous variable.

Now let $h$ be a vector with $+1$ in the $i$th slot, a $-1$ in the $j$th slot, and $d$ in the $n+1$th slot, and zeroes everywhere else. You can append the vector $h$ to the bottom of $A$, and append $z_i$ to the bottom of the right-hand side. Do this once for each relation of the form you've written above.

Now $A$ has more rows than columns, so you can't just invert it. But that's OK, because in general there won't be any solutions to your enlarged system.

Indeed, there's generally only one solution to the initial system. Because
$$
Ax = 0
$$
and $A$ is invertible, we have
$$
A^{-1}Ax = A^{-1} 0 \\
Ix = A^{-1}0 \\
x = 0
$$
So...if you start constraining the $x_i$ entries to be nonzero, you're going to get no solutions.


问:
Using co-factor expansion, how can i prove that $ Det(E_i) = Det(E_{i-1}) - Det(E_{i-2}) $?

Okay so I am not exactly sure how to answer the question the way it wants me to.... for instance.. the determinant for E1=1 and E2=0 and for E3= -1 which satisfies the formula it wants me to prove for part a... but just showing an instance would not suffice in proving the formula so how can i go about it in a more general manner?

答:
I am going to denote by $\overline{(E_n)_{1,1}}$ the $i,j$ minor of $E_n$. In general, you can define $E_n$ as an $n\times n$ matrix such that $(E_n)_{i,j}=1$ if and only if $|i-j|<2.$ So just the $3$ inner diagonals survive. So, consider then the determinant (laplace expansion) of $E_n$ from the first row
$$det(E_n)=\sum _{k=1}^n(-1)^{k+1}(E_n)_{1,1}det(\overline{(E_n)_{1,n}})= det(\overline{(E_n)_{1,1}})-det(\overline{(E_n)_{1,2}})$$
notice that $\overline{(E_n)_{1,1}}=E_{n-1}$ and $\overline{(E_n)_{1,2}}$ has a unique $1$ in the first column, so you can also eliminate the first column(again by the Laplace expansion of the determinant) show that this matrix that results is $E_{n-2}$ Hint: Check that you are basically eliminating the first two rows and columns of the original matrix.


问:
If $f_n \xrightarrow{\mu} f$ and $f_n$ is increasing $f_n \xrightarrow{a.e}f$
I have this problem

Let $f$, $f_n:X \to \mathbb{R}$ be measurable and $f_n  \xrightarrow{\mu} f$. Show that if $f_n$ is monotone increasing then $f_n \xrightarrow{a.e}f$

My idea is by contradiction, to suppose that $f_n \not\xrightarrow{a.e}f$ so, $\{x:\lim f_n(x)\neq f\}$ it is not of measure zero, and how $f_n  \xrightarrow{\mu} f$ there is a subsequence such that $f_{n_k} \xrightarrow{a.e}f$. But I don't know how finish. Could someone help me? Thank you in advance.

答: As you already pointed out, the hypothesis of convergence in measure implies there is a subsequence $(n_k)$ such that $f_{n_k} \to f$ almost everywhere.  Now fix any $x$ such that $ f_{n_k}(x) \to f(x)$.  The sequence $(f_n(x))$ is a monotonically increasing sequence, so it must converge to something (a priori, possibly $\infty$).  But we already know that it has a subsequence converging to $f(x)$ so therefore the whole sequence $(f_n(x))$ must converge to $f(x)$.


问:
Vanishing direct limit of vector spaces
Let $k$ be a field. Let $\{V_n\}_{n\in \mathbb{N}}$ be a direct system of (possibly infinite-dimensional) $k$-vector spaces. Is it true that if $\varinjlim_{n\in \mathbb{N}}V_n=0$, then $V_m=0$ for some $m\in \mathbb{N}$? If not, can you give an example?

答:
Here's a counterexample along similar lines to the one Mars mentioned in the comments: For all $n \in \mathbb{N}$, let $V_n$ be the vector space of sequences of elements of $k$ that are eventually zero (i.e., sequences with finite support). For all $m \leq n$, let the morphism $f_{mn}\colon V_m \to V_n$ be the "left shift" map that sends a sequence $(v_0, v_1, v_2, \dots)$ to the sequence $(v_{n-m}, v_{n-m+1}, v_{n-m+2}, \dots)$. This forms a direct system of vector spaces over $k$.

All of the morphisms are nonzero, but any element is "eventually zero" in this direct system, i.e., for any $m \in \mathbb{N}$ and $v \in V_m$, there exists $n \geq m$ such that $\varphi_{mn}(v) = 0$. So $\varinjlim_{n \in \mathbb{N}} V_n = 0$. This shows how it's crucial to think about the morphisms, not just the objects, when working with (co)limits.

Title: Vanishing direct limit of vector spaces


答:
In case you want to keep things discrete, and are allowed to use the $n$ sum formula, note that for $m=n/2$

$$\frac{\sum_{h=0}^{m}(n-h)^2}{n^2}=\frac{\sum_{h=0}^{m}(n^2-2nh+h^2)}{n^2}\geq m-2\frac{m^2-m}{2n}=\frac{n+2}{4}$$

Thus

$$\frac{\sum_{k=1}^nk^2}{n^2}=\frac{\sum_{k=1}^{n-m-1}k^2+\sum_{h=0}^{m}(n-h)^2}{n^2}>\frac{n+2}{4}$$

and

$$\lim_{n\rightarrow \infty}\frac{\sum_{k=1}^nk^2}{n^2}=\infty$$

Finally, note

$$1+ \lim_{n\rightarrow \infty} \frac{(n+1)^2}{\sum_{k=1}^nk^2}=1+\lim_{n\rightarrow \infty}\Big(\frac{n^2}{\sum_{k=1}^nk^2}-\frac{2n}{\sum_{k=1}^nk^2}+\frac{1}{\sum_{k=1}^nk^2}\Big)=1$$


问:
Rewriting SDEs - "Multiplication on both sides"
I have a question concerning a calculus "trick" sometimes used in stochastic calculus (e.g. in the Book on Arbitrage Theory in Cont. Time of Bjoerk). There they do the following in the proof of Prop. 29.6:

$\frac{1}{B_t}dF_t=h_tdW_t$, multiply on both sides with $B_t$ to obtain
$dF_t=B_th_tdW_t$. The step seems rather obvious. $B_t$ is a prob. a finite variation process. The question is: Can one apply this also in general?

I.e. $F$ is a solution to

$dM_t = F_td\frac{1}{S_t}$ iff it solves $S_tdM_t = S_tF_t \frac{1}{S_t}$

$M$ depends on F, but is a local martingale. Say $F_td\frac{1}{S_t}$ is a (strict) local martingale. Multiplying $S$ gives $S_t F_td\frac{1}{S_t}$ which might well be a true martingale. So this transformation can have significant impact. Is it allowed?

Thank you for you attention!

答:
Written out in detail, the statement $\frac{1}{B_t}dF_t = h_tdW_t$ means

$$
  \int_0^t \frac{1}{B_s}dF_s = \int_0^t h_s dW_s \qquad\textrm{for all }t\ge0,
$$
so that $B^{-1}\cdot F = h\cdot W$, in process notation, where $(h\cdot W)_t = \int_0^t h_sdW_s$, et cetera. Using the composition rule for stochastic integration, namely $K\cdot (H\cdot X) = KH\cdot X$ (see for example Theorem II.19 of Protters book), yields that the above equality implies $F = B \cdot B^{-1} \cdot F = B\cdot (h\cdot W) = Bh\cdot W$, meaning $dF_t = B_th_tdW_t$. Whether $B$ has finite variation is irrelevant, it suffices that $B$ for example is predictable and locally bounded (this is in particular the case if $B$ is continuous). If $B$ is everywhere positive, the other implication holds as well, as we may then simply use the composition rule with $B^{-1}$.

By similar arguments in the general case, the following result holds: Let $X$ and $Y$ be some semimartingales, and let $H$ be some positive, predictable and locally bounded process. It then holds that

$$X = Y \qquad \textrm{if and only if}\qquad H \cdot X = H \cdot Y$$

or, in other words,

$$dX_t = dY_t \qquad \textrm{if and only if}\qquad H_tdX_t = H_tdY_t.$$

Hope this answers your question.


问:
What is the answer to $17^{16} \pmod {10}$? Is it equal to $ 9 $ or $1$
I encounter a modular arithmetic problem, which says:

"Find the last Digit of $17^{16}$, by intuition the last digit of a number is the remainder of the number divided by $10$."

So the statement is:  $17^{16} \pmod {10}.$

According to my knowledge, I figured out that the solution to this problem can be "solved" this way:

** $17^{16}\pmod {10}$ = $(17^8\pmod{10} * 17^8\pmod{10})\pmod{10}$**

then we get $(7 * 7)\pmod{10}$, which is equal to $49\pmod{10}$, and we get a result of  $ 9$.

The problem is that when I go to modular calculator around the internet I get a result of 1. and I certainly don't know why..
check for yourself: https://www.mtholyoke.edu/courses/quenell/s2003/ma139/js/powermod.html

Body:
$17^2 =289$

$\Rightarrow$
$17^2 =-1[5]$ and $ 17^2 =1[2]$

$(17^{2}) ^8=1[5] $$\Rightarrow$$ 17^{16} =1[5]$

So $17^{16} =1[5]$ and $17^{16}=1[2] $

$gcd(5,2)=1$

$\Rightarrow$ $17^{16} =1[10]$

Note:$a=b[n] $ and $a=b[m] $ and $gcd(m, n) =1$$\Rightarrow$ $a=b[mn]$


问:
Unit Conversions with standard deviation
When converting units of standard deviation, should I do the calculations in variance then convert the variance back?

For reference, I am trying to convert the standard deviation from fahrenheit to celsius. I have a standard deviation of 0.9 so I am doing $\frac{5}{9}*0.9^2$=0.45. Then I did $\sqrt45$ to get my new standard deviation; since the equation for converting is: $celsius=\frac{5}{9}(farenheit-32)$. But I think I am over complicating it and should just directly convert the standard deviation to celsius without converting to the variance.

答: The units of variance are the square of the units used for the distribution.  You can verify that from the defining equation for variance.  If you want to convert the variance of a number measured in $^\circ F$ to a number measured in $^\circ C$ you need to multiply by $\left(\frac 59\right)^2$.  The units of standard deviation are the same as the units of the distribution, so if you are converting a standard deviation you multiply by $\frac 59$


答:
Letting $a_n=(n+1)^2$ and $b_n=\sum_{k=1}^{n}k^2$, you could also run Stolz-Cesaro on

$$\frac{a_{n}-a_{n-1}}{b_n-b_{n-1}}=\frac{2n-1}{n^2}$$


问:
Let $G$ a connected graph. If $x$ is a cut vertex of $G$, then $\lambda(G)\le\dfrac{\delta(x)}{2}$.
Let $G$ a connected graph. If $x$ is a cut vertex of $G$, then $\lambda(G)\le\dfrac{\delta(x)}{2}$.

Where $\lambda(G)=\min\{|A| : A \text{ is a cut set of edges}\}$

It is not clear to me how to demonstrate that result. I know that the cardinal of the set of edges whose one of its end points is x, $|E_G(x)|=\delta(x)$ and this set is a cut set of edges, then $$\lambda(G)\le\delta(x)$$

答:
Let's assume that we generate 2 connected components by removing the vertex $x$. denote these components as $G_1$ and $G_2$, define $\delta_i(x)$ as the number of edges between $x$ and $G_i$, $i = 1,2$. Clearly $\delta_1(x) + \delta_2(x) = \delta(x)$.

If $\delta_1(x) < \delta_2(x)$ then we can remove $\delta_1(x) < \delta(x) / 2$ edges from $G$ to disconnect it from $G_1$. Which means $\lambda(G) \leq \delta_1(x) < \delta(x) / 2$, where $\lambda(G) \leq \delta_1(x)$ because it can be another cut vertex with less edges. Note that we don't use $\delta_2(x)$ because we remove less edges by considering $\delta_1(x)$ and we are searching for a minimum.

In the worst case, $\delta_1(x) = \delta_2(x)$ which would mean $\lambda(G) \leq \delta_1(x) = \delta(x) / 2$.

By the above reasoning, we have proved the proposition for the case where $x$ generates 2 connected components. If $x$ generates $n$ components $G_1, ..., G_n$, then again if one component has more edges to $x$ than any of the other, we can choose the edges from the component with the least number of connections to $x$, which should be less than or equal to $\delta(x) / n \leq \delta(x) / 2$ (observe that $n =1$ is not a valid case).

Thus, as $x$ is an arbitrary cut vertex we can only ensure $\lambda(G) \leq \delta(x) /2$.


问:
If $ \int_{0}^{b}f(x) \; dx =0$ then $ \int_{0}^{b}f(x+y_0) \; dx =0 $?
Let $f: \mathbb{R}_+ \longrightarrow \mathbb{R}$ be a continuous and periodic function, with period $b>0$. Let $ y_0 \in [0,b]$ be a fixed number.

Question. If
$$\frac{1}{b}\int_{0}^{b}f(x) \; dx =0 \tag{1}$$
then
$$\int_{0}^{b}f(x+y_0) \; dx =0 \tag{2}?$$

In order to prove $(2)$ I thought: making a variable change $z=x+y_0$ in $(2)$ we have
\begin{eqnarray}
\int_{0}^{b}f(x+y_0) \; dx&=&\int_{y_0}^{b+y_0}f(z) \; dx =\int_{y_0}^{b}f(z) \; dx+\int_{b}^{b+y_0}f(z) \; dx \\
&=& \int_{0}^{b}f(z) \; dx+ \int_{y_0}^{b}f(z) \; dx+ \int_{b}^{y_0}f(z) \; dx+ \int_{y_0}^{y_0}f(z) \; dx \\
&=&\int_{y_0}^{b}f(z) \; dx+ \int_{b}^{y_0}f(z) \; dx= \int_{y_0}^{y_0}f(z) \; dx=0.
\end{eqnarray}

Is this right? Moreover, I think that this holds for any $y_0>0$.

答:
Yes, you are correct.

Alternatively, you could write

$$\begin{align}
\int_0^bf(x+y_0)dx&=\int_{y_0}^{y_0+b}f(z)dz\\
&=\int_{y_0}^{y_0+b}f(z)dz\\
&=\int_{y_0}^bf(z)dz+\int_{b}^{y_0+b}f(z)dz\\
&=\int_{y_0}^bf(z)dz+\int_{0}^{y_0}f(z)dz\\
&=\int_{0}^bf(z)dz\\
&=0.
\end{align}$$


答:
Yes, keep it simple.

If your standard deviation is $0.9^\circ\mathrm F$, then it is $(\tfrac 450.9)^\circ\mathrm C$.  ... that is: $0.72^\circ\mathrm C$.

We do not include the term of $-32$ because we are measuring in displacement from the mean temperature (well, the square root of the expected square of the displacement from the mean).   So we are only concerned about the scale.

Title: Unit Conversions with standard deviation

Body:
When converting units of standard deviation, should I do the calculations in variance then convert the variance back?

For reference, I am trying to convert the standard deviation from fahrenheit to celsius. I have a standard deviation of 0.9 so I am doing $\frac{5}{9}*0.9^2$=0.45. Then I did $\sqrt45$ to get my new standard deviation; since the equation for converting is: $celsius=\frac{5}{9}(farenheit-32)$. But I think I am over complicating it and should just directly convert the standard deviation to celsius without converting to the variance.


问:
is this a mistake or am I missing something?
In my material, I saw this:
let there be $a_n$ a sequence such that $lim_{n\to \infty} a_n = L$ and $b_n=\frac 1n\sum_{j=1}^n a_j $, prove that $\lim_{n\to\infty}b_n=L$
and it was proven like this: Let ther be $\epsilon \gt 0$ we choose $N_0$ such that for every $n\gt N_0$ such that $\vert a_n - L\vert \lt \frac\epsilon 2$ then :
$$\vert b_n-L\vert\leq\frac1n \sum_{j=1}^n \vert a_j-L\vert\leq\frac1n \sum_{j=1}^{N_0} \vert a_j-L\vert+\frac1n \sum_{j=N_0+1}^n \frac\epsilon 2$$
now if we define $M_0=\frac1n \sum_{j=1}^{N_0} \vert a_j-L\vert$ then we get $\vert b_n-L\vert\leq\frac {M_0} n + \frac\epsilon2$ for every $n\gt N_0$ and we coose $N_1$ such that for every $n\gt N_1$ such that $\frac {M_0} n\lt \frac \epsilon 2$ then choose $N_\epsilon = (N_0,N_1)$ and we're done. my question is about this part $$\vert b_n-L\vert\leq\frac1n \sum_{j=1}^n \vert a_j-L\vert$$ shouldn't it be $$\vert b_n-L\vert\leq\vert(\frac1n \sum_{j=1}^n  a_j)-L\vert$$

答:
Simple algebra shows:

\begin{align}
\vert b_n-L\vert&=\vert(\frac1n \sum_{j=1}^n  a_j)-L\vert\\
&=\vert(\frac1n \sum_{j=1}^n  a_j)-\frac 1n \sum_{j=1}^nL\vert\\
&=\frac 1n\vert\sum_{j=1}^n  (a_j-L)\vert \\
&\leq \frac 1n\sum_{j=1}^n  |a_j-L|.
\end{align}


问:
Tangent Line Intercept
The question is like this: there is a curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$, where $c$ is a real number and $c > 0$. If $L$ is a tangent line, any tangent line, with only 1 x-intercept and 1 y-intercept, prove that the sum of the x, y-intercepts of $L$ is $c$.

First of all, I used implicit differenciation on the equation. $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$

Then I also put a graph here (The curve when c=4)

But I still do not have idea. Thank you for looking at this question.

答:
Suppose tangent line intersect the curve at $(x_{0},y_{0})$. Then the line equation will be

$$
\frac{x}{\sqrt{x_{0}}}+\frac{y}{\sqrt{y_{0}}}=\sqrt{c}
$$

Were you able to get this line equation? From here you substitute $x=0$ to find $y$ intercept and vise versa.

Title: Tangent Line Intercept


问:
Distribution of $ \tan \theta$ when $\theta$ is uniform on $[0,2\pi)$
Suppose you have $X$ and $Y$, which are two random variables whose joint distribution is the standard Gaussian distribution. Let random variables $R \geq 0 $ and $\theta \in [0, 2 \pi)$ satisfy: $X=R\cos\theta$ and $Y=R\sin\theta$.

Let $Z=\frac{Y}{X}$.

Thus, to find the density of $Z$, I did:
$Z= \tan \theta$.

And so, $$F_Z(z)=P(Z\leq z) = P(\tan \theta \leq z) = P( \theta \leq \arctan z)$$

Since $ \theta\sim \text{Uniform} [0, 2\pi)$ due to the fact that $X$ and $Y$ are standard Gaussian and deducing the distribution of $\theta$ from the joint distribution of $ R$ and $ \theta $, we get that this is equal to:

$ \int_{0}^{\arctan z} \frac{1}{2} dt$, which then works out to be equal to $\frac{\arctan(z)}{2} $.

After differentiating this to obtain the density function of $Z$, I get:

$$ f_Z(z)= \frac{1}{2\pi(1+{z^2})}$$

I am not sure about what the limits of $Z$ or $\theta$ should now be.

Any feedback will be appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Besides notation details, the answer is reduced to the evaluation of the following integral:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}{1 \over 2\pi}\,\delta\pars{z - \tan\pars{\theta}}\dd\theta} =
{1 \over 2\pi}\int_{-\pi}^{\pi}\delta\pars{z - \tan\pars{\theta}}\dd\theta
\\[5mm] = &\
{1 \over 2\pi}\int_{0}^{\pi}\bracks{\delta\pars{z - \tan\pars{\theta}} +
\delta\pars{z - \tan\pars{-\theta}}}\dd\theta
\\[5mm] = &\
{1 \over 2\pi}\int_{-\pi/2}^{\pi/2}\bracks{\delta\pars{z + \cot\pars{\theta}} +
\delta\pars{z - \cot\pars{\theta}}}\dd\theta
\\[5mm] = &\
{1 \over \pi}\int_{0}^{\pi/2}\bracks{\delta\pars{z + \cot\pars{\theta}} +
\delta\pars{z - \cot\pars{\theta}}}\dd\theta
\\[5mm] = &\
{1 \over \pi}\int_{0}^{\pi/2}\delta\pars{\verts{z} - \cot\pars{\theta}}\dd\theta =
{1 \over \pi}\int_{0}^{\pi/2}
{\delta\pars{\theta - \on{arccot}\pars{\verts{z}}} \over
\verts{-\bracks{-\csc^{2}\pars{\theta}}}}\dd\theta
\\[5mm] = &\
{1 \over \pi}\int_{0}^{\pi/2}
{\delta\pars{\theta - \on{arccot}\pars{\verts{z}}} \over
\cot^{2}\pars{\on{arccot}\pars{\verts{z}}} + 1}\dd\theta
\\[5mm] = &\
{1 \over \pi}\,{1 \over \verts{z}^{2} + 1} =
\bbx{{1 \over \pi}\,{1 \over z^{2} + 1}}\\ &
\end{align}


问:
Finding the marginal distribution
I have that $$ X_2|X_1 =  Bin(X_1,p)$$

and

$$ X_1 = Geo(\frac 1{B+1}) $$

I need to find the marginal distribution of $$X_2$$

I know that the joint distribution is given by $$P(X_1,X_2) = P(X_2|X_1)P(X_1)$$

I'm thinking to take the marginal I need to take the sum over X1 however it is getting messy very quickly.

$$ P(X_1,X_2) = {x_1 \choose x_2} p^{x_2}(1-p)^{x_1-x_2} \frac{1}{B+1}\left(\frac{B}{B+1}\right)^{x_1}$$

I know that to find the marginal for X2 I need to take the sum over X1. But I am stuck with the following.

$$ P(X_2) = \sum_{x_1=x_2}^\infty{x_1 \choose x_2} \left(\frac{p}{1-p}\right)^{x_2} \frac{1}{B+1}\left(\frac{B(1-p)}{B+1}\right)^{x_1}$$

Any hints or tips would be appreciated.

答:
You are on the right track. To evaluate your sum, use the formula
$$
\sum_{n=k}^{\infty} \binom{n}{k} x^n = \frac{x^k}{(1-x)^{k+1}}
$$
which you can prove by induction.

Title: Finding the marginal distribution