问:
Find angle in triangle with incenter

In a triangle $ABC$ if $AD$, $BE$, $CF$ are the angle bisectors of $\angle A$, $\angle B$ $\angle C$ respectively with incenter $I$ and $\angle A=120^\circ$ Find $\angle DFI$

It's $30^\circ$. If $BE$ meet $DF$ at $G$ and $CF$ meet $DE$ at $H$ it seem that $\triangle GEI$ and $\triangle HFI$ is an isosceles triangle. And also $GA$ and $HA$ is the angle bisector of $\angle BAD$ and $\angle CAD$ respectively. I dun know how to prove all of these, I just put it in GeoGebra and see the accurate picture, all I could find is $\angle BIC=150^\circ$, $\angle BIF=30^\circ$ and $\angle DFI+\angle DEI=60^\circ$. Please help

答:
In $\triangle BCD, \, $ $CA$ is the external angle bisector of $\angle DCH \, $ and $BE \,$ is the internal angle bisector of $\angle DBC$ and that makes point $E$ as the ex-center. Hence $DE$ is the external angle bisector of $\angle ADC$.

(As the bisectors of the external angles on one side of a triangle and the bisector of the opposite internal angle are concurrent and in this case the concurrency point is $E$).

say $\angle BEC = \alpha, \angle DEB = \beta$

Then $\angle CDE = 180^0 - (60^0 + \alpha + \beta) = 120^0 - \alpha -\beta \,$ (in $\triangle CED$)

Also $\angle ADE = \angle DEB + \angle DBE = \beta + \angle CBE = 60^0 - \alpha + \beta$

As $\angle CDE = \angle ADE, \angle DEB = \beta = 30^0$

By the way, in any triangle with $\angle C$ as $120^0$, the angle bisector of $\angle C$ meeting at point $D$ on $AB$ will make a right angle with the points where other two angle bisectors meet the other two sides of the triangle.

Title: Find angle in triangle with incenter


问:
Can you simplify an equation to use explicit differentiation instead of implicit differentiation?
Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get: $$\frac{12+18y}{(1-3x)^{2}}$$
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)?
By doing that I got: $$\frac{48}{(1-3x)^{3}}$$
I'm not sure if this is a correct answer as I am new to differentiation.
I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?

Thank you.

答:
We differentiate once,

$$y'-2-3y-3xy'=0$$
and twice,

$$y''-3y'-3y'-3xy''=0.$$

Now we can eliminate $y'$, using

$$(1-3x)y'=3y+2$$

and

$$(1-3x)y''=6y'=6\frac{3y+2}{1-3x}.$$

You can also eliminate $y$ using the initial equation.

ContentLicense: CC BY-SA 4.0


问:
In unitary space $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$.

Prove that in unitary space if $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$.

Now, so far I have just written

$\|x_n\| \to \|x\| \implies \lim_\limits{n\to \infty} |\|x_n\|-\|x\|| = 0$

and

$\langle x_n, x\rangle \to \|x\|^2 \implies \langle x_n, x\rangle \to \langle x, x\rangle \implies \lim_\limits{n\to \infty}  | \langle x_n, x\rangle - \langle x, x\rangle | = \lim_\limits{n\to \infty} | \langle x_n - x, x\rangle |$

I am however, to show that $\|x_n - x \| \to 0$.

How can I do that?

答:
It can be done easier. Observe that
\begin{equation*}
\begin{split}
\|x_n-x\|^2 & =  \langle x_n-x, x_n-x\rangle = \langle x_n, x_n\rangle - \langle x, x_n\rangle - \langle x_n, x\rangle + \langle x, x\rangle \\
& = \|x_n\|^2 - \langle x, x_n\rangle - \langle x_n, x\rangle + \|x\|^2
\end{split}
\end{equation*}
From the hypothesis,
$$\|x_n\| \to \|x\|, \quad \langle x_n, x\rangle \to \|x\|^2$$
so we conclude that $$\|x_n-x\|^2  \to 0 \Rightarrow x_n \to x. $$


问:
What's wrong with my proof ??
This question was on my exam but apparently my answer was wrong as I only got half of the point:
Suppose $p:X \to Y$ is an open map, prove if $A$ is open in $X$ then $q:A \to p(A)$ obtained by restricting $p$ is also an open map.

My proof: Let $U$ be open in $A$. We have that $q(U) = p(U)$ is open in $Y$, hence it's open in $p(A)$. So q is an open map.

This sounded like a quick excercise but apparently my proof needs improvement. What's wrong about my proof?

答:
Let $U$ be an open subset in $A$. Then for some open in $X$ like $O$, we have $U=O \cap A$.
Since $A$ is open in $X$ then $O  \cap A$ is also open in $X$. Hence, $q(U)=p(U)=p(O \cap A)$ is open in $Y$.
We conclude that $q(U)=p(U)$ is open in $p(A)$ since open sets in $p(A)$ are open sets in $Y$ contained in $p(A)$.

Title: What's wrong with my proof ??


问:
Finding numbers from LCM and HCF
Let two numbers be $x$, $y$ and their LCM and HCF be $L$ and $H$ respectively. Then,

$7H=L$ and
  $x+y=392$
How can I find $x$ and $y$? I have tried to use $7 (\gcd)^2=xy$ but nothing came out of it. Any hints?

答: Let $x = a H, y = b H$, where $\gcd(a,b) = 1 $. Then $L = abH$, so $ab=7$. Thus, we must have $\{a, b \} = \{1, 7\}$. WLOG, $x=H, y = 7H$.</p>

Title: Finding numbers from LCM and HCF

ContentLicense: CC BY-SA 3.0


问:
Solving a constrained optimization problem
I have the following problem to optimize:

\begin{align}
\begin{split}
\hat{\mathbf{x}}, \hat{\delta} = \underset{\mathbf{x}, \,\delta}{argmin} \, \sum_{i=1}^l \, \sum_{j=1}^m \, a_{i,j}\left(x_i + \delta \right)^2
\\
s.t. \, \, \, \, 1 - x_i \leq 0 \, \, \forall \, i
\\
\text{and} \,\, \delta \leq 0
\end{split}
\end{align}
where $a_{i,j}$ is just a variable that doesn't depend of $x_i$.

I am thinking to solve this problem by minimizing the following Lagrangian function:
\begin{align}
\underset{\mathbf{x}, \, \delta}{argmin} \, L\left(\mathbf{x}, \, \delta, \, \lambda_1, \, \lambda_2\right) = \sum_{j=1}^m \, \sum_{i=1}^l \, a_{i,j}\left(x_i + \delta \right)^2 + \lambda_1 \left(1 - x_i\right) +\lambda_2 \, \delta
\end{align}
and then find the derive =0.

I am sure I am wrong. Can someone provide me with brief mathematical details about how this problem can be solved?

Any help will be very appreciated.

答:
I assume that $a_{ij}$ are nonnegative. Define $q_i = \sum_j a_{ij}$, $p_i = q_i / \sum_i q_i$ and $c = -\delta$, then the problem is:

$$\min_{x,c} \left\{ \sum_i p_i(x_i - c)^2 : x_i \geq 1, c \geq 0 \right\}$$
You recognize the formula for the variance in the objective function. Let me write the problem as:
$$\min_x \min_c \left\{ \sum_i p_i(x_i - c)^2 : x_i \geq 1, c \geq 0 \right\},$$
For fixed $x$, the inner problem is $\min_c \left\{ \sum_i p_i(x_i - c)^2 : c \geq 0 \right\}$. This is the minimum of a convex quadratic function with a nonnegativity constraint. The unconstrained minimum of the quadratic function is at $c=\sum_i p_i x_i$. This solution happens to satisfy the constraint $c\geq 0$ (because $x_i \geq 0$), so this solution is also optimal to the constrained problem.

Now you can write the original problem as:
$$\min_x \left\{ \sum_i p_i(x_i - \sum_j p_j x_j)^2 : x_i \geq 1\right\},$$
or as
$$\min_x \left\{ x^T(\textrm{Diag}(p) - pp^T)x : x_i \geq 1\right\},$$
where $\textrm{Diag}(p)$ is a matrix with $p$ on the diagonal and zeros elsewhere.
If all $x_i$ are equal, the objective value is $0$. It is clear that you cannot get a value less than $0$, so $0$ is optimal.


问:
The 'ratio' of a 2x2 matrix
Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$

to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.

My instinct is to go with a proof by induction, but I really can't see a way to prove this.

答:
In the sense of the above definition the matrix

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

has ratio $=0$. Since $A^2=0$, the ratio for $A^2$ is not defined.

It seems, that "ratio" is not a good definition.

Title: The 'ratio' of a 2x2 matrix


答:
Hint:

$\gcd(x,y)=h \implies x=kh, y=mh$ and LCM=$kmh$, $7\gcd=LCM \implies km=7$

Title: Finding numbers from LCM and HCF

Body:
Let two numbers be $x$, $y$ and their LCM and HCF be $L$ and $H$ respectively. Then,

$7H=L$ and
  $x+y=392$
How can I find $x$ and $y$? I have tried to use $7 (\gcd)^2=xy$ but nothing came out of it. Any hints?


问:
What's wrong with just adding a top and bottom element to a poset to convert it into a lattice?
I'm a novice in order theory. This question (Smallest lattice containing a poset) and its answers indicate that it is not so straightforward. And I want to know where my reasoning is going wrong.

Following are my steps to convert a poset to a lattice. I have not been able to prove them correct but unfortunately I haven't found a counter-example either.

To a set with a partial order $(S, \leq)$ add two special elements $\top$ and $\bot$ to make $S'$ and extend $\leq$ to make $\leq'$ such that $\forall s \in S,~ \bot \leq' s \leq' \top$

Define the meet operation $a \land b$ as the least common ancestor of a and b in a graph where $\leq'$ forms directed edges, and the join operation $a \lor b$ as the first common descendant. Both of these exist (due to $\top$ and $\bot$) and so these operations are well defined.

Now $(S', \lor, \land)$ form a lattice.

Is there a counter-example to help me see that this definition would lead to violation of one of the laws for join and meet operations as I've defined them? Or is it the case that this is actually a valid lattice?

答:
Because just adding a greatest and a least element doesn't fix pairs that have upper (or lower) bounds, but no least upper (greatest lower) bound.

For instance, take the set $\{a,b,c,d\}$ with partial order given by
$$
a<c\\
b<c\\
a<d\\
b<d
$$
Then the subset $\{a,b\}$ is bounded above; both $c$ and $d$ are upper bounds. However, $c$ and $d$ are incomparable, so there is no least upper bound.

Adding a new element that is larger than both $c$ and $d$, and a new element that is smaller than both $a$ and $b$, doesn't remedy this. You would also need to insert an element "between" $\{a,b\}$ and $\{c,d\}$. Inserting such elements for a general poset (rather than perhaps the simplest example there is as I've done here), and at the same time proving that there is a minimal way of doing so, is not an easy task.


问:
Find a sequence that converges uniformly with g is differentiable
The question asks to find sequence of differentiable continuous functions $g_n:\mathbb{R}\rightarrow\mathbb{R}$ such that:

$g_n\rightarrow g$ uniform on $\mathbb{R}$, but $g$ is not differentiable on $\mathbb{R}$

I have the that $g_k(x)=\sqrt{x^2+1/k} \rightarrow |x|$, but I need to be able to prove this is true.

I have $| \sqrt{x^2+1/k} - (|x|)|<\varepsilon$,

答:
For $0<a<b$ we have $0<\sqrt{b}-\sqrt{a}<\sqrt{b-a}$ (which can be easily shown by squaring the inequality and using $0<x<y \iff 0<x^2<y^2$). Therefore, using this and $|x|=\sqrt{x^2}$ for all $x\in \mathbb{R}$, we have

$| g_n (x)-|x|| =\sqrt{x^2+\frac{1}{n}}-\sqrt{x^2}\leq \sqrt{x^2+\frac{1}{n}-x^2}=\frac{1}{\sqrt{n}}$ for all $x\in \mathbb{R}$ and $n\in \mathbb{N}$.

Can you take it from here?


问:
All theorems that should be considered before proving Fermat's Little theorem?
Hello fellow mathematicians. I was allowed to choose any topic that we learned about in our proof-writing/number theory class. And I decided to go with Fermat's little theorem because I find it really interesting and its applications in cryptography are also cool.

The goal of this assignment is to basically write a 3 page paper proving the theorem but it must be extremely explicit. Meaning I need to provide and prove theorems, lemmas, etc, before actually proving Fermat's Little theorem.

For now all I have considered is proving unique multiplicative inverses (modulo). But that is really all that I can think of currently. Is there anything else I should consider showing to a reader who may not be the most mathematically adept?

My goal with this question is to build a flowchart of how my paper should be laid out. Your help and guidance would be super appreciated!

答:
My favorite is the group theory proof, where by considering the multiplicative group $\Bbb F_p^*$ of the field, we know by Lagrange's theorem that any element has order dividing the order of that group, $p-1$.

Or you can do induction on the base, and invoke the freshman's binomial, more from the point of view of number theory, I guess.

(Both of these nice proofs were mentioned in the comments.)

Perhaps it is worth mentioning that Euler's theorem is a generalization of it.


问:
Do I need to analyse sequence given by $ x_{1+n} = \frac{1}{2 + x_{n}}$ without an equation with $0$?
I have a problem with exercises with sequences given by recursion when I need to "prove convergence and find limit if it exists" and I am given a recursion of that kind:

$$ x_{1+n} = \frac{1}{2 + x_{n}}, x_1 \in (0 ; \infty)$$

It is fairly easy to find the limit - I just assume that the limit exists in $ \mathbb{R}$ and then use arithmetic properties of limits:
$$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} x_{n}$$
$$\lim_{n \to \infty} x_{n} = l, l \in \mathbb{R}>0$$

Taking my recursion:
$$l = \frac{1}{2 + l}$$
$$l^2 +2l - 1 = 0$$
$$l_1 = \sqrt{2} - 1 \in D$$
$$l_2 = -1 - \sqrt{2} \notin D$$

So my only possible limit in $ \mathbb{R}$ is $l = \sqrt{2} - 1$. That is if I can actualy prove that the limit exists - that is: the sequence is monotonous and bounded. And here is my problem - it is just impossible to analyse without computer the difference of:

$$x_{1+n} - x_{n} = \frac{1}{2 + x_{n}} - x_{n}$$

In search of limits I just multiply both sides of equation by $ \lim_{n \to \infty} x_{n} = l$ and it is impossible to do so here, so I get:
$$x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$$

Then I can't tell when it is bigger than $0$ to analyse monotonicity and I can't the see for which values o $n$ which values of $n+1$ i get (to get the boundary) because min value gets crazy.

So I just waneted to ask - am I missing something? Is it possible to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function (red one on the picture)?

答:
This is a Möbius transformation. Once you get the roots $l_1, l_2$ of the characteristic function $l^2+2l-1=0$, it follows that $1-2l_1=l_1^2$ and $1-2l_2=l_2^2$. Then

$$
x_{n+1}-l_1 = \frac{1}{2+x_n}-l_1 = \frac{1-2l_1-l_1 x_n}{2+x_n} = \frac{l_1^2-l_1 x_n}{2+x_n} = -l_1 \frac{x_n-l_1}{2+x_n} \tag 1
$$

Similarly
$$
x_{n+1}-l_2 = -l_2 \frac{x_n-l_2}{2+x_n} \tag 2
$$

$(1) \div (2)$ (you can do this because $x_n>0>l_2$),
$$
\frac{x_{n+1}-l_1}{x_{n+1}-l_2} = \frac{l_1}{l_2}\cdot \frac{x_n-l_1}{x_n-l_2}
$$

Therefore $\frac{x_n-l_1}{x_n-l_2}$ is a geometric sequence,

$$
\frac{x_n-l_1}{x_n-l_2} = \left(\frac{l_1}{l_2} \right)^{n-1} \cdot \frac{x_1-l_1}{x_1-l_2} \tag3
$$

Then $$x_n=\frac{l_1-\frac{x_1-l_1}{x_1-l_2}\left( \frac{l_1}{l_2}\right)^{n-1} \cdot l_2}{1- \frac{x_1-l_1}{x_1-l_2}\left(\frac{l_1}{l_2}\right)^{n-1}}$$

As $n\to \infty, \left(\frac{l_1}{l_2}\right)^{n-1} \to 0, x_n \to l_1 = \sqrt 2 - 1$.

To solve using matrices, see here for an example.


答:
Let $g(x)=|x|$. Then

$$|g_n(x)-g(x)|= \frac{|(g_n(x)-g(x))(g_n(x)+g(x))|}{g_n(x)+g(x)}=\frac{1/n}{g_n(x)+g(x)} \le \frac{1}{\sqrt{n}}.$$

Hence, $(g_n)$ converges uniformly to $g$ on $ \mathbb R.$

Title: Find a sequence that converges uniformly with g is differentiable

Body:
The question asks to find sequence of differentiable continuous functions $g_n:\mathbb{R}\rightarrow\mathbb{R}$ such that:

$g_n\rightarrow g$ uniform on $\mathbb{R}$, but $g$ is not differentiable on $\mathbb{R}$

I have the that $g_k(x)=\sqrt{x^2+1/k} \rightarrow |x|$, but I need to be able to prove this is true.

I have $| \sqrt{x^2+1/k} - (|x|)|<\varepsilon$,


答:
I haven't checked the following, but this is more in line of a comment than an answer but as my phone isn't allowing me to add comments, I'm just adding it here.

I think you've shown how to complete any poset into a lattice. In terms of category theory there's an inclusion functor of Lat into Pos. This has a left adjoint, called the reflector, and this adds top and bottom to a given poset in the way that you've indicated. We also say, that Lat is a reflective subcategory of Pos.

Title: What's wrong with just adding a top and bottom element to a poset to convert it into a lattice?

Body:
I'm a novice in order theory. This question (Smallest lattice containing a poset) and its answers indicate that it is not so straightforward. And I want to know where my reasoning is going wrong.

Following are my steps to convert a poset to a lattice. I have not been able to prove them correct but unfortunately I haven't found a counter-example either.

To a set with a partial order $(S, \leq)$ add two special elements $\top$ and $\bot$ to make $S'$ and extend $\leq$ to make $\leq'$ such that $\forall s \in S,~ \bot \leq' s \leq' \top$

Define the meet operation $a \land b$ as the least common ancestor of a and b in a graph where $\leq'$ forms directed edges, and the join operation $a \lor b$ as the first common descendant. Both of these exist (due to $\top$ and $\bot$) and so these operations are well defined.

Now $(S', \lor, \land)$ form a lattice.

Is there a counter-example to help me see that this definition would lead to violation of one of the laws for join and meet operations as I've defined them? Or is it the case that this is actually a valid lattice?


问:
Convert SVM into convex QP
I have the following minimization:

$$\min\limits_{\mathbf{x}}\frac{1}{2}\|\mathbf{x}\|^{2}_{2}+\frac{1}{n}\sum_{i=1}^{n}\max(0,1-\mathbf{x}^{T}\mathbf{a}_{i})$$

The problem asks me to show that it is convex, which is fine and I did without any problems, and then it asks me to rewrite this problem as a convex QP, in order to find its Lagrangian and KKT conditions, The last step of the exercise is to show that the dual problem can also be written as a QP, but with box constraints.

Here are the steps I took in order to convert the primal as a QP:

Let us define $p_i=\max(0,1-\mathbf{x}^{T}\mathbf{a}_{i})$, which can be translated into the constraints:

$$\begin{align}
-p_i&\leq0\\
1-p_i&\leq\mathbf{x}^{T}\mathbf{a}_{i}
\end{align}$$

or

$$\begin{align}
-\mathbf{p}&\leq\mathbf{0}\\
-\mathbf{p}-\mathbf{A}\mathbf{x}&\leq\mathbf{1}
\end{align}$$

And define the vector $\mathbf{z}=\left[\mathbf{x} \quad \mathbf{p}\right]^{T}$

Then we can rewrite the problem as:

\begin{align}
&\min\limits_{\mathbf{z}}\frac{1}{2}\mathbf{z}^{T}\mathbf{P}\mathbf{z}+\mathbf{q}^{T}\mathbf{z}\\
&\text{ s.t. }\quad\mathbf{C}\mathbf{z}\leq \mathbf{b}
\end{align}

Where

$$\mathbf{P}=\begin{bmatrix}\mathbf{I} & \mathbf{0}\\
\mathbf{0} & \mathbf{0}\end{bmatrix},$$

$$\mathbf{q}=\begin{bmatrix}\mathbf{0}\\
\frac{1}{n}\mathbf{1}\end{bmatrix},$$

$$\mathbf{C}=\begin{bmatrix}\mathbf{0} & -\mathbf{I}\\
-\mathbf{A} & -\mathbf{I}\end{bmatrix},$$

and

$$\mathbf{b}=\begin{bmatrix}\mathbf{0}\\
-\mathbf{1}\end{bmatrix}.$$

The constraint is convex, but $\nabla_{z}^{2}f(\mathbf{z})=\mathbf{P}\succeq0$, which is only positive semi-definite. However, in order for my QP to be convex, I would need $\nabla_{z}^{2}f(\mathbf{z})=\mathbf{P}\succ0$, right?

What did I do wrong? Is there another way to formulate this problem in order to achieve the positive definiteness of my Hessian?

答:
A function is convex if its Hessian is positive semidefinite, and strictly convex if its Hessian is positive definite. Your problem is strictly convex in $x$, but linear (and hence, convex) in $p$.

The second block element of $\mathbf{b}$ should be $-\mathbf{1}$.

Title: Convert SVM into convex QP


问:
Exponential and Uniform Distribution Problems

Questions

I have the following problems on my probability module:

Qu1

Suppose that X is uniformly distributed over $[−3, 3]$.

(a) Find the probability that the first digit of $X$ after the decimal point is $3$.

(b) Find the distribution function of the random variable $|X|$ and use it to identify the
distribution of $|X|$.

(c) Find the density function of the random variable $1/|X|$.

Qu2

Let $X$ be a standard exponential random variable, meaning that $X ∼ Exp(1)$. Prove
that the random variable $Y = \lceil X \rceil$ is $Geom(\lambda)$ and find $\lambda$.

My answers and ideas

Qu1

(a)
$\mathbb P($First digit after deciaml point of X is 3$)$$=\mathbb P(-2.4\lt X\lt -2.2)+\mathbb P(-1.4\lt X\lt -1.2)+\mathbb P(-0.4\lt X\lt -0.2)+\mathbb P(0.2\lt X\lt 0.4)+\mathbb P(1.2\lt X\lt 1.4)+\mathbb P(2.2 \lt X\lt 2.4)=6\mathbb P(0.2\lt X\lt 0.4)=6\frac{0.2}{3-(-3)}=\frac{1}{5}$

Im not 100% sure this is correct be great if someone could check my workings :)

(b)
$|X| \Rightarrow$ $f(x)=\begin{cases}
\frac{1}{3-0},  & \text{if $0\le x \le 3$} \\
0, & \text{o/w}
\end{cases}$

$\Rightarrow F(x)=\int_{-\infty}^{x}f(t)dt=\int_{-\infty}^{x}\frac{1}{3}dt=\frac{1}{3}(x-0)=\frac{x}{3}$

Even less sure with this one :/

(c)
Not confident tackling as I'm unhappy with (b)

Qu2

I am clueless on how to approach this, tried drawing various distribution graphs. But they havent seemed to help me :(

Be great if anyone could help me by checking solutions and/or helping me with the questions I cannot do :)

答:
For the decimal point to be 3, the number should be between 0.3 and 0.4. Not 0.2 and 0.4.
But this one you can solve with a very intuitive argument: between 0 and 1, each decimal "range" (0-0.1, 0.1-0.2, ...) has the same probability, so the answer should not depend on the fact that the question is about "3". Out of symmetry, the answer should be 0.1.

Your solution for the second part is correct. Again, the probability of $\vert X \vert$ to be in the range $[a,a+\Delta]$ (within $[0,3]$, of course) depends only on $\Delta$ and not the $a$ (it should be $\tfrac{2\Delta}{6}$ since $X$ is uniform), so $\vert X \vert$ is uniform on $(0,1)$.

For the second question, simply calculate the distribution. For natural $t$:

$$\Pr(Y=t)=\Pr(t-1<X<t)=\int\limits_{t-1}^t f_X(u)du =\cdot $$

Can you continue from here and show that the resulting probability looks like the one of Geom?


问:
Matrix Algebra Over the Opposite Ring
I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what there is to verify...

Clearly $\varphi^{-1} = \varphi$, so it is bijective, $\varphi (A+B) = (A+B)^{T} = A^{T} + B^{T}$, and $\varphi (A \star B) = \varphi (BA) = (BA)^{T} = A^{T} B^{T} = \varphi (A) \varphi (B)$, where $\star$ is the opposite product of $M_n(R)^{opp}$.

Is that it? Seems like a cheesy proof. I tried consulting some of the users in the chatroom. The only one to proffer a critique was Thorgott. He said that "I don't think $\varphi^{-1} = \varphi$, given that codomain and domain are different." After thinking about this for some time, I don't think this is a problem, because although $M_n(R^{opp})$ and $M_n(R)^{opp}$ denote different rings, the underlying sets are the same.

答:
It's usually best to avoid unnecessary detail, but in this case, since there are so many multiplication operations to confuse (in $R$, $R^{op}$, $M_n(R^{op})$, $M_n(R)$ and $M_n(R)^{op}$) it might be worth just writing the details.  (There's no ambiguity with the addition operation, of course.)

I'd like to use, respectively, juxtaposition, $\ast$, $\star$, $\circ$, $\bullet$ for these operations.

We'll abbreviate matrices by writing the $i,j$ entry between brackets.
Given two matrices $[a_{ij}]$, $[b_{ij}]$ we would have, by definition

$\phi([a_{ij}])=[a_{ji}]$

$[a_{ij}]\circ[b_{ij}]=[\sum_k a_{ik}b_{kj}]$

$[a_{ij}]\star[b_{ij}]=[\sum_k a_{ik}\ast b_{kj}]$

With that in mind

$$
\phi([a_{ij}]\bullet[b_{ij}])=\phi([b_{ij}]\circ[a_{ij}])=\phi([\sum_k b_{ik}a_{k,j}])\\ =[\sum_k b_{jk}a_{ki}]=[\sum_k a_{ki}\ast b_{jk}]=\phi([a_{ij}])\star\phi([b_{ij}])
$$

One computation, five operations!  Possibly not aesthetically pleasing, but at least straightforward bookkeeping.

Title: Matrix Algebra Over the Opposite Ring


问:
How to solve this ODE (answer is given as implicit equation)
This is from Boas 9.2, Problem 1

$$ \frac{d}{dx}\left(\frac{\sqrt x y'}{\sqrt{1+y'^2}}\right) = 0 $$

And therefore

$$ \frac{\sqrt x y'}{\sqrt{1+y'^2}} = k, ~~ k\textrm{ constant} $$

But then what? The answer is given implicitly as

$$(y-b)^2=4a^2(x-a^2), ~~ a,b\textrm{ constants} $$

She has several similar problems, (as part of solving Euler-Lagrange equations), but doesn't give details on how to solve this part.

答:
First note that $y'$ has the same sign as $k$, as the square root as function always takes non-negative values.

Then square and isolate $y'$,
$$
xy'^2=k^2(1+y'^2)
\\
(x-k^2)y'^2=k^2
\\
y'=\frac{k}{\sqrt{x-k^2}}
$$
This also contains the implicit condition $x>k^2$.
Now integrate
$$
y=m+2k\sqrt{x-k^2}.
$$
This can now be transformed into something similar but not equal to the claimed form.


问:
Let f is differentiable and has the property f(xy) = f(x)+f(y). Show that f'(x) =a/x ,where a is a constant.
Let f : R+ -> R such that f is differentiable and has the property f(xy) = f(x)+f(y).</p>

答:
Denote $g(x) = f(\exp(x))$ for $x \in \mathbb{R}$. Plugging $\exp(x)$ and $\exp(y)$ in the functional equation gives you $f(e^{x+y})=f(e^x)+f(e^y)$, so:

$$\forall x,y\in\mathbb{R},\ g(x+y)=g(x)+g(y)$$

You also know that $g$ is differentiable, so looking at standard results on the Cauchy functional equation, you find that $g(x) = a x$ for some constant $a$.

That gives you $f(x) = g(\ln(x)) = a \ln(x)$, so $f'(x) = \frac{a}{x}$.

Alternative answer using that $f$ is differentiable (and not only continuous).

First note that $f(1\cdot 1)=f(1)+f(1)$, so $f(1)=0$. Then take $x>0$, and $h$ close to $0$. Then using the equation with $y = 1+\frac{h}{x}$ yields $f(x+h) = f(x) + f\big(1 + \frac{h}{x}\big)$, so subtracting $f(1)=0$ and dividing by $h$, we get: $$\frac{f(x+h)-f(x)}{h} = \frac{1}{x} \cdot \frac{f\big(1 + \frac{h}{x}\big) - f(1)}{\frac{h}{x}}$$

When $h$ goes to zero, the RHS goes to $\frac{1}{x}\cdot f'(1)$, and the LHS goes to $f'(x)$, so $f'(x)=\frac{f'(1)}{x}$. Your constant $a$ is $f'(1)$.


问:
Total and partial derivatives
I have a function $Y(x,t)$ and another function $\eta(x,t)$.

I have the total derivative $\dfrac{dY}{d\eta}$.

So, what is the correct?

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{2\partial Y}{\partial x}$$

or

$$\dfrac{dY}{d\eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}\dfrac{\partial x}{\partial \eta}\dfrac{\partial \eta}{\partial x}+\dfrac{\partial Y}{\partial t}\dfrac{\partial t}{\partial \eta}\dfrac{\partial \eta}{\partial x}=\dfrac{\partial Y}{\partial x}+\dfrac{\partial Y}{\partial t}\underbrace{\dfrac{\partial t}{\partial x}}_{0}=\dfrac{\partial Y}{\partial x}$$

?

Thank you so much.

答:
In the described situation there is no such thing as ${dY\over d\eta}$.

When $$Y: \quad(x,t)\mapsto Y(x,t),\qquad \eta:\quad (x,t)\mapsto\eta(x,t)$$
are two real valued functions defined in the $(x,t)$-plane, and a point $(x_0,t_0)$ is given, then the gradients of $Y$ and $\eta$ at $(x_0,t_0)$ determine the increments of $Y$ and $\eta$ when moving away from $(x_0,t_0)$:
$$\eqalign{\Delta Y&=\nabla Y(x_0,t_0)\cdot(\Delta x,\Delta t)+o\bigl(\sqrt{\Delta x^2+\Delta t^2}\bigr),\cr
\Delta \eta&=\nabla \eta(x_0,t_0)\cdot(\Delta x,\Delta t)+o\bigl(\sqrt{\Delta x^2+\Delta t^2}\bigr)\ .\cr}$$
It follows that $\Delta Y$ and $\Delta\eta$ depend on the direction of the increment vector $(\Delta x,\Delta t)$. The same is true for the quotient ${\Delta Y\over\Delta\eta}$ and its limit (called ${dY\over d\eta}$ by you), when $\sqrt{\Delta x^2+\Delta t^2}\to0$.

It is another thing when $x$ and $t$ are not independent variables, and $x$ is in fact a function of $t$, given as $x=\psi(t)$. In this case $Y$ and $\eta$ also become composed functions of $t$:
$$\hat Y(t):=Y\bigl(\psi(t),t\bigr),\qquad \hat \eta(t):=\eta\bigl(\psi(t),t\bigr)\ .$$
In this case it makes sense to talk about $${d\hat Y\over d\hat\eta}={\hat Y'(t)\over\hat\eta'(t)}=\ldots\quad .$$

Title: Total and partial derivatives


问:
Solve polynomial interpolation $f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$
I think the polynomial

$$f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$$

may be easy to find, but I am a bit unexperienced. So, I would be glad to get some help!

$$f(a) = bc,\>\>\>f(b) = ac,\>\>\>f(c) = ab$$

which is what the polynomial should satisfy.

答:
The opening is a standard way but is not easy to simplify.  May try in symmetrical manner,

$$af(a)=bf(b)=cf(c)=abc$$

Now,

$$xf(x)-abc=k(x-a)(x-b)(x-c)$$

For $f(x)$ being a polynomial, $k=1$ which is what @Raffaele commented.

Furthermore, you may observe that

$\dfrac{bc}{(a-b)(a-c)}
  +\dfrac{ca}{(b-c)(b-a)}
  +\dfrac{ab}{(c-a)(c-b)} \equiv 1$

$\dfrac{bc(b+c)}{(a-b)(a-c)}
  +\dfrac{ca(c+a)}{(b-c)(b-a)}
  +\dfrac{ab(a+b)}{(c-a)(c-b)} \equiv a+b+c$

$\dfrac{(bc)^2}{(a-b)(a-c)}
  +\dfrac{(ca)^2}{(b-c)(b-a)}
  +\dfrac{(ab)^2}{(c-a)(c-b)} \equiv bc+ca+ab$


问:
Prove that $x^{p^{n}}-x-a$ is reducible over GF(p).
Let $p$ is a prime,$n≥1$ and $n \nmid p^{n}$.Suppose $a\in GF(p)$.Prove that $x^{p^{n}}-x-a$ is reducible over $GF(p)$.

I just have no idea. Please help me,thanks!

答:
$\DeclareMathOperator{\GF}{GF}$It seems that a slightly more general result holds true. This states that

for a prime $p$, if $n > 2$, or $n = 2$ and $p > 2$, then the polynomial $x^{p^{n}}-x-a$ is reducible over $\GF(p)$.

Suppose $f = x^{p^{n}}-x-a$ is irreducible over $F = \GF(p)$. (In particular, of course, $a \ne 0$.)

Then, if $\beta$ is a root of $f$, we have that $E = F[\beta]$ is a field with $p^{p^{n}}$ elements.

Since $f(\beta) = 0$, we have
$$
\beta^{p^{n}} = \beta + a,
\quad
(\beta^{p^{n}})^{p^{n}} = \beta^{p^{2 n}} = \beta^{p^{n}} + a = \beta + 2 a,
\quad
\dots
\quad
\beta^{p^{p n}} = \beta + p a = \beta.
$$
It follows that $\beta$ lies in the field $L$ with $p^{p n}$ elements, as $\beta$ is a root of the polynomial $x^{p^{p n}} - x \in F[x]$. But then $E \subseteq L$, so that $p^{n} \mid p n$, and $p^{n-1} \mid n$. This is only possible (if I'm not mistaken) for $n = 1$, and any $p$, or for $p = n = 2$.

And in fact, for $a \ne 0$, the polynomial $x^{p} - x - a$ is irreducible over $F$. And the polynomial $x^{4} - x - 1$ is irreducible over $\GF(2)$.


答:
Define $g\colon \Bbb R\to\Bbb R$ as $g(x)=f(e^ x)$. Then we arrive at the Cauchy functional equation$$g(x+y)=f(e^{x+y})=f(e^xe^y)=f(e^x)+f(e^y)=g(x)+g(y).$$
Recall that the only continuous $g$ with ths property are of the form $g(x)=ax$.
Then $$ a=g'(x)=e^xf'(e^x)$$
for all $x\in\Bbb R$.

Title: Let f is differentiable and has the property f(xy) = f(x)+f(y). Show that f'(x) =a/x ,where a is a constant.

Body: Let f : R+ -> R such that f is differentiable and has the property f(xy) = f(x)+f(y).</p>


问:
Doubt about why I can't treat this as a Bernoulli process
I know the title is not descriptive enough, but I don't know how else to say it. I don't know why I can't use the Binomial distribution to get the result I'm looking for. The teacher solved it long ago with a more straightforward method, but I tried today to solve the same problem using Binomial distribution and the result is not the same.

So, the problem is:

A total of $17$ balls are distributed in $20$ baskets, where every basket can hold as many balls as I want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishsable from another. What is the probability of every ball falling into the first basket?

The teacher solved it through combinatorics, using:

$$\dfrac1{\dbinom{20+17-1}{17}}$$

I wanted to solve it using Binomial distribution:

$$\dbinom{n}k p^k (1-p)^{n-k} = \dbinom{17}{17} \left(\dfrac1{20}\right)^{17} \left(\dfrac{19}{20} \right)^0$$

Knowing that there are $17$ throws, and I want $17$ successes. And as there are $20$ baskets, and only one is the successful one, the probability of each ball of going into that basket is $1/20$.

The problem is, it doesn't give the same result. And I don't know why, because I think I'm not using the binomial distribution wrong.

Thank you for your time!

答:
The key is that the balls are indistinguishable.

Let us consider an example, where we can work out explicitly. Consider $2$ balls and $3$ baskets.

If the balls are distinguishable, say colored red and blue, then the first ball can go into any of the three baskets, while the second ball can also go into any of the three baskets. This thereby gives us a total of $3 \times 3$ options.
$$\{\color{blue}1,\color{red}1\},\{\color{blue}1,\color{red}2\},\{\color{blue}1,\color{red}3\},\{\color{blue}2,\color{red}1\},\{\color{blue}2,\color{red}2\},\{\color{blue}2,\color{red}3\},\{\color{blue}3,\color{red}1\},\{\color{blue}3,\color{red}2\},\{\color{blue}3,\color{red}3\} \tag{$\star$}$$
where $\{\color{blue}i,\color{red}j\}$ means blue ball chose basket $i$ and the red ball chose basket $j$. Hence, the probability that both balls fall in the first basket is $\color{green}{\dfrac19}$.

If the balls are non distinguishable, then let us list out the options.
$$\{2,0,0\},\{0,2,0\},\{0,0,2\},\{1,1,0\},\{0,1,1\},\{1,0,1\}$$
where $\{m_1,m_2,m_3\}$ means the baskets $i$ has $m_i$ balls. Essentially we are counting $(\star)$, but we now consider $\{\color{blue}1,\color{red}2\}$ and $\{\color{blue}2,\color{red}1\}$ to be the same since there is no color to distinguish between them. Hence, the probability that both balls fall in the first basket is the same as the probability that the first basket contains both the balls, which in this case is $\color{green}{\dfrac16}$.

Hence, in general, if we want to distribute $m$ balls in $n$ (distinguishable) baskets, the total number of ways is

$n^m$, if the balls are distinguishable.

$\dbinom{n+m-1}{n-1}$, if the balls are indistinguishable.

Clearly, both intuitively and algebraically, we have $\dbinom{n+m-1}{n-1} < n^m$.

Hence, in general, the probability that all the $m$ balls fall in the first basket is

$\dfrac1{n^m}$, if the balls are distinguishable.

$\dfrac1{\dbinom{n+m-1}{n-1}}$, if the balls are indistinguishable.


问:
Borel $\sigma$-algebra is equivalent to the smallest $\sigma$-algebra containing (a,$\infty$)
I'm trying to prove that the Borel $\sigma$-algebra is equivalent to the smallest $\sigma$-algebra containing (a,$\infty$). The approach I'm trying to use is to show that if A$\subset$B and B$\subset$A, then A=B. So far I think I have the first part:

(a, $\infty$) = $\cup$(a, n) and since each (a, n) is a finite open interval, then they are included in the Borel $\sigma$-algebra and since a $\sigma$-algebra is closed with respect to countable union, then the smallest $\sigma$-algebra containing (a, $\infty$) $\subset$ Borel $\sigma$-algebra.

Looking at (a,b) it seems obvious that it would be contained within (a,$\infty$), but I'm not sure if it is sufficient to just say that for the second part of this proof or if there is a formal way to show that (a,b) is contained in (a,$\infty$).

答:
Let $\mathcal A$ be the smallest $\sigma$-algebra containing sets of the form $(a, \infty)$. It remains to show that the Borel $\sigma$-algebra $\subset \mathcal A$.

Every open subset of $\mathbb R$ can be written as countable union of open intervals. Thus we are done if we show that all open intervals belongs to $\mathcal A$.

Since $(a, \infty) \in \mathcal A$, we have $(-\infty, a] \in \mathcal A$.
Thus $(a, b]= (-\infty, b] \cap(a,\infty) \in \mathcal A$ and consequently $(a, b)= \cup (a, b-\frac{1}{n}]\in \mathcal A$.

Since $a,b \in \mathbb R$ are arbitary, it follows that all open intervals belongs to $\mathcal A$. As discussed above this proves that the Borel $\sigma$-algebra $\subset \mathcal A$.


问:
Compute a marginal density function from a joint density function
I am in the middle of an exercice dealing with this joint density function :

$$ f_{x,y}(x,y) = \frac{x^2+y}{a}$$ for $ 0 < x < y < 2$

I have been asked to compute the marginal densities and I know how to do it but I am struggling with the bounds.
To verify that this joint density function is legit, I found the value of a using this boundaries :

$$ \int_{0}^{2}\int_{x}^{2} \frac{x^2+y}{a} \,dydx $$

This bounds makes perfect sense and I found $a = 4$

My problem is when it comes to the marginal

I started like this using the same bounds for x :
$$ \int_{x}^{2} \frac{x^2+y}{4} \,dy$$

Then, I was about to do this :

$$ \int_{0}^{2} \frac{x^2+y}{4} \,dx$$

But the solutions say that we have to do it on the interval from 0 to y.

Could you explain me why ?

答:
For a given value of $Y$, by definition, the joint distribution is non-zero in the region $0<X<Y$. So, when you fix $Y$ and sum over all values of $X$, it must go until $Y$ and not until $X$.

Another idea: symmetry. If you understand why in the marginal of $X$ the limits of the integral are $x$ to $2$, then the same argument should apply for the marginal of $Y$. Remember that the double integral you did in the beginning only goes over the entire region the marginal distributions go over only over lines in which one of the variables is fixed.


问:
Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)
The problem statement is as follows:

Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem.  The gist of my proof is presented below.

If possible, I would like to find a simpler proof.  I was hoping that there might be some sort of manipulation along these lines, noting that
$$
\langle Av_1,A v_2\rangle
= \langle v_1,A^*Av_2\rangle
= \langle v_1,AA^*v_2\rangle
= \langle A^* v_1,A^* v_2 \rangle
$$

Any ideas here would be appreciated.

My proof:

Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that
$$
A v_{\lambda,i} = \lambda v_{\lambda,i}
$$
Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$.  We may write
$
v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda}
.$
We then have
$$
0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}
$$
So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$.  Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$.  From there, we have
$$
\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}
$$
the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

答:
linear_algebra_done_right
This is from linear algebra done right. btw, it's the greatest book about linear algebra I've ever seen!

Proof: Suppose $\alpha,\beta$ are distinct eigenvalues of $T$, with corresponding eigenvectors $u,v$. Thus, $Tu = \alpha u$ and $Tv = \beta v$. From 7.21 we have $T^*v = \bar \beta v.$ Thus,
\begin{align}
(\alpha - \beta)\langle u,v \rangle &= \langle \alpha u,v \rangle  - \langle u, \bar \beta v \rangle
\\ &=
\langle Tu,v \rangle - \langle u, T^*v \rangle
\\ & = 0.
\end{align}
Because $\alpha \neq \beta$, the equation above implies $\langle u,v \rangle = 0$. Thus, $u$ and $v$ are orthogonal, as desired.


问:
How to find roots of $X^5 - 1$?
How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)</p>

答:
It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$.
Here is an algebraic solution.  First factor the polynomial as  $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$
The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation
$$ x^4+x^3+x^2+x+1=0.$$
Divides both sides by $x^2$, and use the fact that $(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$, we get
$$ (x+\frac{1}{x})^2 +(x+\frac{1}{x})-1=0$$
Set $y=x+\frac{1}{x}$, and we get a quadratic equation
$$  y^2+y-1=0,$$
which one may easily solve using quadratic formulas. Once we have solved for $y=y_0, y_1$, the equation reduces into two other quadratic equations
$$ x+ \frac{1}{x}= y_i, \qquad i=0, 1 $$
or equivalently,
$$ x^2-y_i x+1=0.$$
We just apply the quadratic formula again. </p>

Title: How to find roots of $X^5 - 1$?


问:
Polynomials not dense in holder spaces
How to prove that the polynomials are not dense in Holder space with exponent, say, $\frac{1}{2}$?</p>

答:
I should have reported this long ago.
We prove, in fact, that even the larger class $C^1[0,\ 1]$ of continuously differentiable functions on $[0,\ 1]$ is not dense in $H_{\alpha}^0$.

We record that $C^1 \subset H_{\alpha}^0$. For, if $g \in C^1$ and if $\Arrowvert g' \Arrowvert_C = \Lambda$, then $\delta_{c,\ \alpha}(g) \le \Lambda c^{1 - \alpha} \rightarrow 0$.

For $a \in (0,\ 1)$, let  $x(t) = |t - a |^{\frac{1}{2}} $. Let $0 < \alpha < \frac{1}{2}$. We have, if $s < t$,
 $\frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - s|^{\alpha}}$
 $= \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - a - (s - a)|^{\frac{1}{2}}}\frac{|t - s|^{\frac{1}{2}}}{|t - s|^{\alpha}}$
 $\le \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - a - (s - a)|^{\frac{1}{2}}}$
$\le  \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{\big||t - a| - |s - a|\big|^{\frac{1}{2}}}\le 1$.  Hence $x \in H_{\alpha}$.

$\delta_{c,\ \alpha}(x) = \sup\limits_{|s - t| < c}\frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{\big||t - a| - |s - a|\big|^{\frac{1}{2}}}|t - s|^{\frac{1}{2} - \alpha} \le c^{\frac{1}{2} - \alpha} \rightarrow 0$\ as $c \rightarrow 0$. Hence $x \in H_{\alpha}^0$. And then $g - x \in H_{\alpha}^0$

Now, $\Arrowvert g - x \Arrowvert_{\alpha}$
$ = \sup\limits_{0\le t,s\le 1,\ s \neq t} \frac{| g(t) - x(t) - \{g(s) - x(s)\} |}{|t - s|^{\alpha}}$
$ \ge \sup\limits_{0\le t\le 1,\ t \neq a} \frac{| g(t) - x(t) - \{g(a) - x(a)\}|}{|t - a|^{\alpha}}$
$= \sup\limits_{0\le t\le1,\ t \neq a} \big|
\frac{| g(t) - g(a)|}{|t - a|^{\alpha}} - \frac{|x(t) - x(a)|}{|t - a|^{\alpha}} \big|$ $= \sup\limits_{0\le t\le1,\ t \neq a} \big|
\frac{| g(t) - g(a)|}{|t - a|^{\alpha}} - |t - a|^{\frac{1}{2} - \alpha}\big|$
$\ge \sup\limits_{\substack{0\le t\le1,\ t \neq a\\|t - a| < (\frac{\varepsilon}{\Lambda})^{\frac{1}{1 - \alpha}}}} \big||g'(t^*)|{|t - a|^{1 - \alpha}} - \max\big( (1 - a)^{\frac{1}{2} - \alpha},\  a^{\frac{1}{2} - \alpha} \big)\big|$
$\ge \max\big( (1 - a)^{\frac{1}{2} - \alpha},\  a^{\frac{1}{2} - \alpha} \big) - \varepsilon$.

Hence the sub set of all functions which are continuously differentiable is not dense in $H_{\alpha}^0$.

Title: Polynomials not dense in holder spaces


问:
Parametric Equations for a $2$-torus
I know that for a torus (with one hole) the parametric equations describing it are $x= (c + a\cos v)\cos u, y= (c + a\cos v)\sin u, z= a\sin v$, where $c$ is the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. My question is what would be the corresponding parametric equations describing a $2$-torus, i.e. a torus with $2$ holes. How about an $n$-torus?</p>

答:
$0$ holes, $3$ donuts

$$\Big(\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)^2+\\
\big(-\sin(\theta)-h+2\cos(\theta)\sin(\theta)\big)\big(\sin(\theta)+h-2\cos(\theta)\sin(\theta)\big)\text,\\
2\sin(\theta)\cos(\phi)\big(1+h-2\cos(\theta)\big)\\
\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)\text,\\
2\sin(\theta)\sin(\phi)\big(1+h-2\cos(\theta)\big)\\
\big(-\cos(\theta)^2+\sin(\theta)^2\cos(\phi)^2+\sin(\theta)^2\sin(\phi)^2+h+\sin(\theta)\big)\Big)$$

$h$ is the height.

$3$ donut:

Title: Parametric Equations for a $2$-torus


问:
Sum of exponential growth and decay
Suppose we have the equation:

$$y(t_i) = C_0 + C_1 e^{-\lambda_1 t_i} + C_2 e^{\lambda_2 t_i}$$

where $C_0$, $C_1$, $\lambda_1$, $C_2$, and $\lambda_2 \ge 0$

This is equivalent to the summation of an exponential decay curve with that of exponential growth.

Thus, one can see that

$C_0$ is equivalent to the sum of the asymptotic value for the decay part as $t \rightarrow \infty$ and the asymptotic values for the gowth part as $t \rightarrow -\infty$

$y(0) = C_0 + C_1 + C_2$

If I have a sample of n observations, what are some good starting approximations for the estimates of $C_1$, $\lambda_1$, $C_2$, and $\lambda_2$ which I can then pass into a non-linear optimization algorithm.  I've explored ODE solutions to these types of equations, but cannot find a good "least squares" type function of the observed data.

I am considering partitioning the observations close to the position of the inflection point in the data, and estimating the $\lambda_1, \lambda_2$ from the least squares solution to the log transform of the data points to the left and the right, and then solving for $C_0, C_1,$ and $C_2$ after substituting $\hat{\lambda_1}$ and $\hat{\lambda_2}$ into the above equation.

Any alternative solutions would be greatly appreciated.

UPDATE:

@JJacquelin gave an excellent answer.  I worked out the solution myself to confirm that the equations hold.

Here is an illustration:

However, there is at least one instances where the approach breaks down.

The expression for $p \ or \ q = \frac{B \pm \sqrt{B^2 + 4A}}{2}$ can't be calculated when $B^2 + 4A < 0$.

<img src="https://i.stack.imgur.com/EzzmU.png" alt="Example of a fit with B^2 + 4A < 0">

答: In addition :  With your data publish latter, the numerical calculus and result is :

Title: Sum of exponential growth and decay


问:
Can this differential equation exclusively have periodic solutions?

Can the differential equation
$$y^{(3)}+2y''+ay'+y=0, \: a\in \mathbb{R} $$
have exclusively periodic (like $\cos(\omega t)$ and $\sin(\omega t)$) as solutions?

I think the answer is no. I wasn't able to show this rigorously, but I used Maple's "dsolve" command to find the solutions to this differential equation.

It turns out that there is always a term $e^{-t}$ multiplied in the solution making it not a exclusively periodic solution. Is there a way to show that this always has to be the case?

Here is an example where I set $a=2$, and you can see the term $e^{-t}$ in the solution.

答: As the characteristic polynomial is real and of degree 3, there will always be a real root. As this can not be $0$ (see constant term), you always get an exponential basis solution.


问:
Irreducible components of Spec$(k[t] \otimes_{k[t^2]} k[t])$
This has already been asked in Question about irreducible components of the fiber product $\operatorname{Spec}(k[t]) \times_{\text{Spec}(k[t^2])} \operatorname{Spec}(k[t])$, but the answer there differs from mine and I don't know what I did wrong. My reasoning was that
$$ k[t] = k[t^2] \cdot 1 \oplus k[t^2] \cdot t $$
and therefore
$$ k[t] \otimes_{k[t^2]} k[t] \cong (k[t^2] \cdot 1 \otimes_{k[t^2]} k[t]) \oplus (k[t^2] \cdot t \otimes_{k[t^2]} k[t]) \cong k[t] \oplus k[t]. $$
This way the fibered product is the disjoint union of two lines. I also don't know where the hypothesis on the characteristic comes into play.

答:
(for $char(k)\ne 2$)

$R=k[t] \otimes_{k[t^2]} k[t]\cong k[t,u]/(u^2-t^2)$. The two minimal prime ideals are $(t-u)$ and $(t+u)$.

The two affine lines $Spec(R/(t-u)),Spec(R/(t+u))$ are not disjoint, they are glued along the maximal ideal $(t,u)$.

$R\cong k[x,y]/(xy)$.


问:
Prove that there exist a tensor whose components relative to the product basis of $\pmb{any}$ basis are the values of the generalized kronecker delta
ACHTUNG!!!

To follow I adopt the Einstein notation!!!

Definition

The dual space $V^*$ of a vector space $V$ over the field $\Bbb F$ is the vector space of the linear functions from $V$ to $\Bbb F$.

Definition

Give a basis $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ of a vector space $V$ the dual set of $\mathcal E$ is the set $\mathcal E^*:=\{\vec e^{\,1},...,\vec e^{\,n}\}$ of $V^*$ whose elements are defined through the equality
$$
\vec e^{\,i}(\vec e_j)=\delta^i_j
$$
for each $i,j=1,..,n$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis for a vector space $V$ then the dual set $\mathcal E^*$ of $\mathcal E$ as above defined is a basis of the dual space $V^*$ of $V$.

Theorem

Let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ and $\mathcal B:=\{b_1,..,b_n\}$ two basis of a vector space $V$ so that
$$
\vec b_k=B^s_k\vec e_s\,\,\,\text{and}\,\,\,\vec e_q=E^k_q\vec b_k
$$
for each $k,q=1,...,n$. So if this holds then
$$
\vec b^{\,\,q}=E^q_k\vec e^{\,\,k}\,\,\,\text{and}\,\,\,\vec e^{\,\,q}=B^q_k\vec e^{\,\,k}
$$
for each $k,q=1,...,n$ and moreover $B^q_kE^k_s=\delta^q_s$ for each $q,s=1,...,n$

Definition

If $V_1,...,V_s$ is a collection of vector spaces over the same field $\Bbb F$ then a $s$-linear function is a function $f$ form $V_1\times...\times V_s$ in $\Bbb F$ that is linear in each of its variables while the other variables are held costant. In particular if $V_i=V^*$ for some $r<s$ and $V_i=V$ otherwise we say that the function $f$ is a tensor of contravariant order $p:=r$ and covariant order $q:=(s-r)$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis of $V$ then a tensor $\tau$ of contravariant order $p$ and covariant order $q$ is completely defined by the equality
$$
T^{i_1,...,i_p}_{j_1,..,j_q}:=\tau\big(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},...\vec e_{j_q}\big)
$$
for $i_1,...,i_p,j_1,...,j_q=1,...,n$. Furthermore if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis of $V$ then
$$
\tau(\vec b^{\,i_1},..,\vec b^{\,i_p},\vec b_{j_1},...,\vec b_{j_q})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_q}_{j_q}T^{h_1,...,h_p}_{k_1,....,k_q}
$$
for each $i_1,...,i_p,j_1,...,j_q=1,...,n$.

Now let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ a basis of a vector space $V$ so that we can define a tensor $\tau$ of contravariant and covariant order p letting that
$$
\tau(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},..,\vec e_{j_p})=\delta^{i_1,...,i_p}_{j_1,..,j_p}
$$
for each values of the indices $i_1,...,i_p,j_1,...,j_p$ from $1$ to $n$.
So I ask to prove that for any vector space and for any natural number $p$ there exist a tensor $\tau$ of order $p$ whose components relative to the product basis of any basis are the values of the generalized kronecker delta as here defined, that is more explicitly I ask to prove that if $\tau$ is the tensor above defined and if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis for the space $V$ above defined then
$$
\tau(\vec b^{\,i_1},...,\vec b^{\,i_p},\vec b_{j_1},..,\vec b_{j_p})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}\delta^{h_1,...,h_p}_{k_1,...,k_p}
$$
for each values of the indices $h_1,...,h_p,k_1,...,k_p$.

So could someone help me, please?

答:
Warning: The following uses the invariant definition of a tensor product which can be found in any good book of algebra but is often written in such a way that it hides the essential clarity of the definition.

Let $V \in FinVect$.

Define $\delta:=e^i \otimes \e_i$. Here a sum is implied by Einsteims convention - said to be, by some (ironically), his greatest contribution to maths.

Let $v \in V$

Then $\delta.v = (e^i.v)e_i = v^i.e_i = v$
Hence we see that $\delta = Id: V \rightarrow V$

This is the invariant definition of the Kronecker delta in its simplest form and is simply the identity, which is obviously invariant. To check that it is, let us take components, then

$\delta^i_j$
$ = e^i(\delta)e_j$
$=  e^i(e^k\otimes e_k)e_j$
$= e^i((e^k.e_j).e_k)$
$= e^i(\delta^k_j.e_k)$
$= \delta^k_j(e^i.e_k)$
$=\delta^k_j.\delta^i_k$
$=\delta^i_j$

And which matches.


问:
Random variable transformation in $ \mathbb{R}^{2}$
Let $X, Y$ be random variables with joint probability density function
$$
f_{X,Y}\,\left(x,y\right) = \frac{1}{2}\,\mathrm{e}^{-x}\,\,
\mathbb{\large I}_{\,\left[\,{x\ >\ \left\vert\,{y}\,\right\vert}\,\right]}\,\,\,,
\quad (x,y) \in \mathbb{R}^{2}
$$

Now we define $U =X$ and $V=X- |Y|$ and we would like to find joint probability density of $U,V$ and marginal distribution of $V$.

We want to apply the theorem about transformation of random variables and since there is $|Y|$, we need to separate it into two subsets.

Let $ S = \{ x>0, \ x> |y| \}$ and $G_{1} = S \cap [y>0]$, $G_{2} = S \cap [ y<0]$. Then for $$ \varphi_{1} \begin{cases} u = x \\ v=x-y \end{cases} \text{ and } \varphi_{2} \begin{cases} u = x \\ v=x+y \end{cases} $$ we that the absolute value of their Jacobian matrix determinant is one and $$ \varphi_{1} (G_{1}) = \varphi_{2} (G_{2}) = (0, \infty )^{2}$$

Also $$ \mathbb{I}_{[ x > |y| ]} = \mathbb{I}_{[u>|u-v|]} = \mathbb{I}_{[u> |v-u|]}, $$ hence $$f_{U,V} (u,v) = \frac{1}{2} e^{-u} \mathbb{I}_{[u> |v-u|]} \mathbb{I}_{[u>0]} \mathbb{I}_{[v>0]} \cdot 2 =  e^{-u} \mathbb{I}_{[2u>v>0]} $$

However, by intergrating $f_{U,V}$ over $ \mathbb{R}^{2}$ we get $2$, so there must be $ \frac{1}{2}$ lost somewhere. I would appreciate some help at this point.

答:
Sorry @Peter, but without doing a lot of calculations, it is enough to observe that in the following system

$$\begin{cases}
u=x \\
v=x-|y|
\end{cases}$$

all the functions are linear with coefficients $\pm1$, thus without calculations the jacobian is 1 and thus the joint density is

$$f_{UV}(u,v)=\frac{e^{-u}}{2}$$

The support you calculated is correct and integrating you get

$V\sim Exp(\frac{1}{2})$

$U\sim Gamma(2;1)$

EDIT

I understood my error, $F_V(v)$ can be derived also in the following way

$$F_V(v)=1-\int_v^{\infty}\Bigg[\frac{e^{-x}}{2}\int_{v-x}^{x-v}dy\Bigg]dx=1-e^{-v}$$

That is

$$V\sim Exp(1)$$


问:
Probability of drawing an ace or 2 after an ace - intuition
This question is a follow-up question on this one: Probability of drawing an ace or 2 after an ace

Background Out of a standard deck you draw cards (without replacement) one by one until you reach an ace. Then you continue to draw cards until you reach another ace or a "2" card. What is the probability that you reached a "2" card and not an ace?

The solution To shorten things up, let's assume the deck has only two aces and two "2" cards. Clearly, all that matters are the relative position of these two cards, not all the other cards. So we can think of the permutations of 4 items in a row.

The event "the card following the first ace is ace" occurs when the two aces are together. The sample space is ${4 \choose 2}$ (choosing 2 places for the aces) and there are 3 locations in which they are together (1-2,2-3,3-4) so the solution is $\tfrac{3}{{4\choose 2}}=0.5$.

Surprisingly, the answer $0.5$ remains correct even if there are 4 aces and 4 "2"s in the original deck and in fact, as @InterstellarProbe shows in the comments to the original question, if there are $n+1$ of each type, the answer is:
$$\sum_{k=0}^{n+1} \frac{n\left(\begin{array}{c}
n+k \\
k
\end{array}\right)}{(n+k)\left(\begin{array}{c}
2 n+2 \\
n+1
\end{array}\right)}=\frac{1}{2}$$

The question The above result surprises me a bit. There is no obvious symmetry in the question and it is not obviously clear why there is the same number of arrangements in which after the first ace comes another one as arrangements in which after the first ace comes a "2". I spent a couple of months thinking, asked also fellow probability teachers, but no-one had a good idea.

So, returning the ball to this court. What is the intuitive explanation to this result? Why the permutations are symmetric and how can we get this $\tfrac{1}{2}$ without the algebra?

答:
I'm not sure that the following solution is any more intuitive than others (in my experience, some probability problems are simply non-intuitive, at least with my intuition), but at least it is a different approach than those given so far.

To start with, we can ignore all the cards in the deck except for aces and twos. So suppose we have a deck of four aces and four twos, which can be arranged in $8!$ ways, all of which we assume are equally likely.  We want to find the probability that the card following the first ace in the deck is also an ace.

As a further simplification, let's see if we can find the probability that the card immediately following the first ace is the ace of spades. We want to count the number of such arrangements.  To do so, suppose we remove the ace of spades from the deck and deal out the remaining cards; this can be done in $7!$ ways. There is only one place where the ace of spades can now go so that it is immediately following the first ace, so the number of arrangements in which the ace of spades follows the first ace is $7!$, and the probability of this event is $$\frac{7!}{8!} = \frac{1}{8}$$

If we don't care which ace follows the first ace in the deck, there are $4$ possible choices, all of which have the same probability of $1/8$. So the probability that the card following the first ace is also an ace is
$$4 \cdot \frac{1}{8} = \frac{1}{2}$$


问:
Minimum area of whole quadrilateral given areas of parts

I've attempted to mark segment $AO$ as $a$ and $CO$ as $b$.

Now, if draw a line from $D$ that is perpendicular to $AC$, and draw another line from B that is perpendicular to $AC$, then we can use some variables to express the area of $\triangle BOC$ and $\triangle AOD$.

However, when I try to make a equation, I found that I have too many variables.

How can I move further?

答:
Altitude from $B$ to $AC$, $h_1$ is $\dfrac{2[AOB]}{AO}=\dfrac{2\triangle_{1}}{a}$

Altitude from $D$ to $AC$, $h_2$ is $\dfrac{2[COD]}{CO}=\dfrac{2\triangle_{2}}{b}$

So
$$[ABCD] = [AOB]+[COD]+[BOC]+[DOA]$$

$$ = \triangle_{1}+\triangle_{2}+\dfrac{1}{2}\cdot b \cdot \dfrac{2\triangle_{1}}{a}+\dfrac{1}{2}\cdot a \cdot\dfrac{2\triangle_{2}}{b}$$

$$ = \triangle_{1}+\triangle_{2}+ \dfrac{b}{a}\triangle_{1}+ \dfrac{a}{b}\triangle_{2}$$

which is minimum when (their product is constant so their sum is minimum when both equal)

$$\dfrac{b}{a}\triangle_{1} = \dfrac{a}{b}\triangle_{2}$$
$$\Rightarrow \dfrac{\triangle_{1}}{\triangle_{2}} = \dfrac{a^2}{b^2}$$

Obtain same expression by taking altitudes on other diagonal.

Since $\angle AOB = \angle COD$, we conclude $[ABCD]$ is minimum when $$\triangle AOB \sim \triangle COD$$

As a result $\angle BAO = \angle DCO \Rightarrow AB || CD$

Thus $ABCD$ is a trapezium.

You can substitute back to find the minimum value is

$$\bbox[#FFFF33,5px]{[ABCD]_{\text{min}} = (\sqrt{\triangle_{1}}+\sqrt{\triangle_{2}})^2}$$


问:
How to place a limit that it's inside the integral, outside.
I did this:
$$\int_{1}^t x^{-1}dx=\int_{1}^t\lim_{n\rightarrow -1}{x^n}dx =\lim_{n\rightarrow -1}\int_{1}^t{x^n}dx $$ just to have a way to approximate $\ln t$. $$\ln{t}=\lim_{h\rightarrow 0}\frac{x^{h}-1}{h}$$

The second expression may be correct, but I was told I cannot say $\int_a^b \lim_{n\rightarrow -1}x^{n}dx=\lim_{n\rightarrow -1}\int_a^b x^n dx$ without previously prooved some statements. If so, what are those statement and where can I find information about this?

答:
In general, you cannot interchange two limits. In this case, recall that the integration is a actually a limit, for instance, the limit of a Riemann sum. Hence, you need to be careful, in general:
$$\int_a^b \lim_{n \to \infty} f_n(x) dx \neq \lim_{n \to \infty} \int_a^b f_n(x) dx$$

You may want to read through the following posts to figure out under what conditions, you can swap two limits.

When can the order of limit and integral be exchanged?

Can a limit of an integral be moved inside the integral?

When can we exchange order of two limits?


答:
We need to use the fact that the ratio of triangles with the same height is equal to the ratio of their bases.

Suppose $AO : OC = 1: k$. Then $[\triangle AOD] : [\triangle COD] = AO:OC = [\triangle AOB] : [\triangle COB] = 1:k$.

This gives $[\triangle AOD] = \frac4k$ and $[\triangle COB] = 9k$.

Now the area of the quadrilateral is equal to

$$4 + 9+\frac4k+9k=4+9+12+\frac4k-12+9k=25+\left(\frac{2}{\sqrt k}-3\sqrt{k}\right)^2 \ge 25$$

Equality holds when $k = \frac23$, so the minimum area is $25$.

We can use $AM \ge GM: a+b \ge 2\sqrt {ab}$ to obtain $\dfrac 4k + 9k \ge 12$ as well.

Title: Minimum area of whole quadrilateral given areas of parts

Body:
I've attempted to mark segment $AO$ as $a$ and $CO$ as $b$.

Now, if draw a line from $D$ that is perpendicular to $AC$, and draw another line from B that is perpendicular to $AC$, then we can use some variables to express the area of $\triangle BOC$ and $\triangle AOD$.

However, when I try to make a equation, I found that I have too many variables.

How can I move further?


问:
How to prove if a mapping is an affine map.
Suppose A,( A is not equal to B)B ∈ A2. Is it true that
f: X → centre (A, B, X) is an affine map? Is it bijective?

As per my understanding, we have to show that F(x) = Mx + c to prove the affine map condition(like some translation is happening). But I do not understand how to prove it here. Can anyone please suggest ways to solve it?

答:
Supposed by 'center' you mean the center of mass (intersection of medians), then
$$f(x)=\frac{A+B+x}3=\frac13x+\frac13(A+B)$$
and it has an inverse
$$x\mapsto 3\left(x-\frac{A+B}3\right)\,.$$


问:
Does the equality coefficients of linear regression of X onto Y and Y onto X imply coincidence of the lines?
Let's assume that we consider the model without an intercept $\hat{y_i} = x_i\hat{\beta}$. So, the formula for $\hat{\beta}$ is  $\frac{ \sum\limits_{i=1}^n x_i y_i}{\sum\limits_{j=1}^n x_j^2}$. I know that $\hat{\beta}_{XY}$ = $\hat{\beta}_{XY}$ if and only if $ \sum\limits_{j=1}^n x_j^2 = \sum\limits_{j=1}^n y_j^2$ (Here, $\beta_{XY}$ means regression coefficient of $X$ onto $Y$). If some of the squares indeed equal, we have that $\hat{\beta}_{XY}$ = $\hat{\beta}_{XY}$, however I don't understand whether it implies  the coincidence of the lines.

答: No. Assume that slope is $\sqrt{3}$. As you can see, the lines $y = \sqrt{3}x$ and $x = \sqrt{3}y$ do not coincide.


问:
$\chi _M + \chi _N$ iff $M$ and $N$ are measurable
Let $(\mathbb {R}^d, \mathcal {M}, \mu )$ be a measure space with $\mathcal {E} \subset \mathcal {M}$. $M$ and $N$ are disjoint subset in $\mathbb {R}^d$.

$\chi _M + \chi _N$ and $\chi _M + 2 \chi _N$ should be measurable iff $M$ and $N$ are measurable right?

($\mathcal {E}$ is the set of all unions of finite intervals.)

答: Yes.  If $\chi_M + \chi_N$ and $\chi_M + 2\chi_N$ are both measurable, then $(\chi_M + 2\chi_N) - (\chi_M + \chi_N) = \chi_N$ is also measurable, which then implies $\chi_M$ is measurable as well.  So $M$ and $N$ are measurable.


问:
Confusion on uniform convergence and subset uniform convergence
If we say $$(f_n)_{n = 1}^\infty \ a \ sequence \ of \ functions\  f_n: D \to \mathbb{C}\ \ converge\  uniformly\  \ on  \ f: D \to \mathbb{C}$$

For any subset $$ S \subseteq D$$ , is it tru that for the  functions $$f_n: S \to \mathbb{C}  \ and f: S \to \mathbb{C}\  have \ f_n \to f$$ uniformly on S ?

In other words, if we restrict the domain of the function that is uniformly convergent , does it remain uniformly convergent ?

答:
Yes. Recall the definition. Uniform convergence in $D$ means the following:

$\forall(\epsilon>0) \exists (n_0\in\mathbb{N})\ \forall (n\geq n_0)\ \forall (z\in D)\ [|f_n(z)-f(z)|<\epsilon]$

So if the condition in the end holds for each $z\in D$ then in particular it holds for all $z\in S$ when $S\subseteq D$. For a given $\epsilon>0$, the same index $n_0$ will work. Thus $f_n\to f$ uniformly in $S$ as well.


问:
What is the probability that at least one head is obtained from the unbiased coin?
A loaded die is tossed once; if N is the result of the toss, then $P$($N = i$) = $p_i$
,
$i = 1, ..., 6$. If $N = i$, an unbiased coin is tossed independently i times.

What is the probability that at least one head is obtained from the unbiased
coin? Given that at least one head is obtained, what is the probability that N
will be odd?

I know that a loaded die has probabilities 1/21, 2/21,3/21, 4/21, 5/21, 6/21 of showing 1,2,3,4,5,6. But, I am confused as to what $p_i$ is.

答:
If you flip an unbiased coin $k$ times, the chance of getting only tails is $\frac{1}{2^k}$, so the chance of getting heads at least once is $1-\frac{1}{2^k}$.

According to the way your die is biased, you have a $\frac{1}{21}$ chance of flipping one coin, $\frac{2}{21}$ of flipping two coins, etc.

Therefore, your probability of getting at least one heads is:
$$\sum_{k=1}^6\frac{k}{21}(1-\frac{1}{2^k})=\frac{1}{21}\frac{1}{2}+\frac{2}{21}\frac{3}{4}+\frac{3}{21}\frac{7}{8}+\frac{4}{21}\frac{15}{16}+\frac{5}{21}\frac{31}{32}+\frac{6}{21}\frac{63}{64}=\frac{51}{56}$$


问:
Can log 1 to the base 1 be any number?
If log 1 to the base 1 can be any number. So we can invent a new mathematical idea called the "any". If log 1(1) = "that idea", so is that wrong?</p>

答:
$\log_1 1$ is undefined. By the change of base formula, we can see this is equal to $\frac {\ln 1} {\ln 1} = \frac 0 0$, which is itself undefined (an indeterminate form).

Undefined or indeterminate does not mean "any number", it simply means that it is not possible to assign a proper meaning to the expression.

Title: Can log 1 to the base 1 be any number?


问:
The 'ratio' of a 2x2 matrix
Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$

to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.

My instinct is to go with a proof by induction, but I really can't see a way to prove this.

答:
Diagonalize the matrix.
If it can be diagonalized, you get $A =PDP^{-1}$.  The ratio depends on $P$, not on the diagonal matrix $D$.  All powers of $A$ use the same $P$.

If it can't be diagonalized, then $D$ becomes an upper triangular matrix with equal numbers on the diagonal, and again the ratio is independent of $D$

Title: The 'ratio' of a 2x2 matrix


问:
Doubt about why I can't treat this as a Bernoulli process
I know the title is not descriptive enough, but I don't know how else to say it. I don't know why I can't use the Binomial distribution to get the result I'm looking for. The teacher solved it long ago with a more straightforward method, but I tried today to solve the same problem using Binomial distribution and the result is not the same.

So, the problem is:

A total of $17$ balls are distributed in $20$ baskets, where every basket can hold as many balls as I want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishsable from another. What is the probability of every ball falling into the first basket?

The teacher solved it through combinatorics, using:

$$\dfrac1{\dbinom{20+17-1}{17}}$$

I wanted to solve it using Binomial distribution:

$$\dbinom{n}k p^k (1-p)^{n-k} = \dbinom{17}{17} \left(\dfrac1{20}\right)^{17} \left(\dfrac{19}{20} \right)^0$$

Knowing that there are $17$ throws, and I want $17$ successes. And as there are $20$ baskets, and only one is the successful one, the probability of each ball of going into that basket is $1/20$.

The problem is, it doesn't give the same result. And I don't know why, because I think I'm not using the binomial distribution wrong.

Thank you for your time!

答:
I think you may have some confusion (as do we) about what you mean by the first basket.

If you mean basket number 1 (or 7 or 12 or any other particular number) then your formula will give the correct result $\left(\frac{1}{20}\right)^{17}$.

If you mean the basket that the first ball drops into then you are interested only that the remaining 16 balls fall into that basket $\left(\frac{1}{20}\right)^{16}$.


问:
Solution of the differential equation
I think it is a linear form of differential equation but i had problem integrating the rhs.

Solve: $y'+x^2y= xe^x$

答:
Very often you won't be able to find an analytic solution for an integral.

\begin{align*}
y'+x^2y &= x e^x \newline
(e^{x^3/3} y)' &= x e^{x+x^3/3} \newline
e^{x^3/3} y &= \int x e^{x+x^3/3} dx + C \newline
\end{align*}

Apparently, this $\int x e^{x+x^3/3} dx$  doesn't have an analytic solution, so I don't think you can get any better than that.

Title: Solution of the differential equation


答:
We can solve the DE using the series method
$$
y' = xe^x  - x^2 y;\;y\left( 0 \right) = 0
$$
Let $y = \sum\limits_{n = 0}^\infty  {a_n x^n } $ therefore we have $x^2 y = \sum\limits_{n = 0}^\infty  {a_n x^{n + 2} } $ and $
xe^x  = \sum\limits_{n = 0}^\infty  {\frac{{x^{n + 1} }}{{n!}}} $
The DE equation can be written as
$$
\sum\limits_{n = 1}^\infty  {na_n x^{n - 1} }  = \sum\limits_{n = 0}^\infty  {\frac{{x^{n + 1} }}
{{n!}}}  - \sum\limits_{n = 0}^\infty  {a_n x^{n + 2} }
$$
As $y(0)=0$ we have $a_0$ and
$$a_1  + 2a_2 x + 3a_3 x^2  + 4a_4 x^3  + 5a_5 x^4  +  \ldots  = \\=x + x^2  + \frac{1}{2}x^3  + \frac{1}{6}x^4  + \frac{1}{{24}}x^5  \ldots  - a_0 x^2  - a_1 x^3  - a_2 x^4  - a_3 x^5  -  \ldots $$
Now setting equal the coefficients of the same powers of $x$ we get
$$
  a_1  = 0 \\
  2a_2  = 1 \to a_2  = \frac{1}{2}  \\
  3a_3  = 1 \to a_3  = \frac{1}{3} \\
  4a_4  = \frac{1}{2} \to a_4  = \frac{1}{8}  \\
  5a_5  = \frac{1}{6} - a_2  \to 5a_5  = \frac{1}{6} - \frac{1}{2} \to a_5  =  - \frac{1}
{{15}} \\
\ldots\\
\ldots
$$
Finally the solution can be written as
$$y=\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{8}-\frac{x^5}{15}+\ldots$$

Title: Solution of the differential equation

Body:
I think it is a linear form of differential equation but i had problem integrating the rhs.

Solve: $y'+x^2y= xe^x$


答:
Theorem

There are $x_n$ such that
$$ A_n :=
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
= A^n
$$

Proof by induction

For $n=1$ it is trivial: $x_1 := 1$.

Let the theorem be true for a $n$.

Then on the one hand
$$ A^{n+1} = A^n
\cdot
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
\cdot
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
aa_n + cbx_n & b(a_n + dx_n) \\
c (d_n + ax_n) & dd_n + bcx_n
\end{pmatrix}.
$$

On the other hand
$$ A^{n+1} =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\cdot
A^n
=
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\cdot
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
=
\begin{pmatrix}
aa_n + cbx_n & b(d_n + ax_n) \\
c (a_n + dx_n) & dd_n + bcx_n
\end{pmatrix}.
$$

Thus $a_n + dx_n = d_n + ax_n =: x_{n+1}$. $\square$

Title: The 'ratio' of a 2x2 matrix

Body:
Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$

to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.

My instinct is to go with a proof by induction, but I really can't see a way to prove this.


问:
What exactly is $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$ supposed to be?
I need help with understanding the space $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$. I am unable to really comprehend what this is actually supposed to mean.

To begin, I found the following definition. $R/I$ is a set of equivalence classes of elements of $R$, where $a,b\in R$ are in the same equivalence class, say $[a]$, if the following holds: $$a,b\in [a]\in R/I~~~ \Leftrightarrow ~~~ (a-b)\in I.$$

So in this specific case, $$a,b\in [a]\in \mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2~~~ \Leftrightarrow ~~~ (a-b)\in (\mathbb{Q}_p^*)^2.$$

But here I fail to see which property some $a\in\mathbb{Q}_p^*$ has to satisfy, such that $a\in(\mathbb{Q}_p^*)^2$ , specifically, how this is dependent on $p$. I would think that a square in $\mathbb{Q}_2$ is also a square in $\mathbb{Q}_3$ for instance, as I do not understand how the property of being a square can be impacted by some valuation on the set.

I hope my question has become somewhat clear and I would be glad, if someone could offer me a very explicit definition for when $$a=p^\alpha\frac{u}{v}\in(\mathbb{Q}_p^*)^2$$ and by extention when $$a,b\in [a]\in \mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$$

Thank you very much!

答:
I would think that a square in $\mathbb{Q}_2$ is also a square in $\mathbb{Q}_3$ for instance, as I do not understand how the property of being a square can be impacted by some valuation on the set.

Remember that $\mathbb{Q}_p$ is not just $\mathbb{Q}$ endowed with the $p$-adic norm; rather, it is the completion of $\mathbb{Q}$ with respect to that norm. As we vary $p$ we're not looking at different norms on the same ring, but rather genuinely different rings. So there's no mystery to the definition of $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$, it's exactly what it seems to be - but it is after all dependent on $p$.

An important point here is that even the set of rationals which are squares in $\mathbb{Q}_p$ (that is, $(\mathbb{Q}_p^*)^2\cap\mathbb{Q}$) depends on $p$: if $\alpha\in\mathbb{Q}$ is a square in $\mathbb{Q}_{p_1}$, its square roots in the sense of $\mathbb{Q}_{p_1}$ might not be in $\mathbb{Q}$ and so $\alpha$ might not be a square in the sense of some other $\mathbb{Q}_{p_2}$.

See also Hesel's lemma and various MSE questions (e.g. 1, 2).


问:
Limit $\lim_{n\to\infty} \frac{n!}{n+2^n}$
In an exercise I'm asked to evaluate the following limit:

$$\lim_{n\to\infty} \frac{n!}{n + 2^n}$$

I arrived at the conclusion that:

$$\frac{n!}{n + 2^n} \geq \frac{n!}{2^n}$$

So my plan is to prove that $\frac{n!}{2^n}$ diverges and as a consequence show that $\frac{n!}{n + 2^n}$ also diverges, But I'm not being able to do so.

How can I prove that $\lim n! / 2^n$ is infinity?

答:
Notice that for $n \ge 2$, we have $\frac{n!}{2^n} = \frac{n}{2} \cdot \frac{n-1}{2} \cdot \dots \frac {2}{2} \cdot \frac{1}{2} \ge \frac{n}{2} \cdot \frac{1}{2} = \frac{n}{4}$, which diverges to $\infty$.

I'll point out that your inequality is wrong, since $n +2^n \ge 2^n$, though. You could fix this by establishing something like $n + 2^n \le 2^{n+1}$.


问:
Complex integral help involving $\sin^{2n}(x)$
Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$

So far I have came up with:

$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:

$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$

but I'm not sure how to get the conclusion. Any help will be greatly appreciated.

答:
This integral is easily done using residue theory.  Rewrite the integral as

$$\begin{align} \frac12 \int_0^{2 \pi} d\theta \, \sin^{2 n}{\theta} &= \frac12 \frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} \frac{dz}{z} (z-z^{-1})^{2 n}\\ &= \frac{(-1)^n}{2^{2 n+1}} (-i) \sum_{k=0}^{2 n} (-1)^k \binom{2 n}{k} \oint_{|z|=1} \frac{dz}{z} z^{2 n-2 k} \end{align}$$

Note that the contour integral in the RHS of the last equation is zero unless $k=n$, in which case it is equal to $i 2 \pi$.  Therefore,

$$\int_0^{\pi} d\theta \, \sin^{2 n}{\theta} = \frac{(-1)^n}{2^{2 n+1}} (-i) (i 2 \pi) (-1)^n \binom{2 n}{n} = \frac{\pi}{2^{2 n}} \frac{(2 n)!}{(n!)^2}$$

as was to be shown.


问:
Find all triplets (x, y, z) of positive integers such that 1/x + 1/y + 1/z = 4/5
Here's what i did :-

i wrote Find all triplet $$1/x + 1/y + 1/z = 4/5$$ as $$1/x + 1/y = 4/5 - 1/z$$

$(4z-5)/5z = 1/x + 1/y$ which can be written as $1/x + 1/y = 1/(5z)/(4z-5)$

Let $(5z)/(4z-5) = n$ therefore $1/x + 1/y = 1/n$ which on factorization gives $(n-x)(n-y) = n^2$ and $n$ should be an integer , i am not sure about the proof but i got $(5z)/(4z-5)$ is never an integer except $z = <o,1>$ but z cannot be 0 .

Here's my problem :-
if we take $n = 1$ we get 0 solutions of $<x,y,-1>$
is what i did correct ?

答:
Hint: Say  $x\leq y\leq z$ then ${1\over z}\leq {1\over y}\leq {1\over x}$ so $${3\over x}\geq {4\over 5} \implies x\leq {15\over 4} \implies x\leq 3$$

So do the cases $x=1$ (clearly we can rule out this case), $x=2$ and $x=3$ (for each repeat the procedure I just did on two variables.)


答:
I am puzzled that  you set $\frac{5z}{4z-5}=n$.

Assuming $x \le y \le z$, we have

$$\frac45 \le \frac3x$$

and hence we can conclude that $x \le \frac{15}{4}<4.$ Furthermore, clearly, we need $x > 1$.

Hence $x$ is either $2$ or $3$.

Case $1$:

If $x=3$,$\frac1y + \frac1z = \frac45-\frac13=\frac7{15}$.

$$15(y+z)=7yz$$

$$7yz-15y-15z=0$$
$$7(7yz-15y-15z)=0$$
$$(7y-15)(7z-15)=225$$

Consider the factors of

\begin{align}
225 &= 1 \times 225 \\
&= 3 \times 75 \\
&= 5 \times 45 \\
&= 9 \times 25 \\
&= 15 \times 15
\end{align}

Since $x=3$, $7y-15 \ge 7x-15 \ge 6$. Also, it is easy to check that $7y-15$ can't take values $9$ or $15$. Hence we conclude that $x \ne 3$.

Hence $x=2$.

Case $2$:

$x=2$.

$$\frac1y+\frac1z=\frac45-\frac12=\frac3{10}$$

$$3yz-10y-10z=0$$

$$3(3yz-10y-10z)=0$$

$$(3y-10)(3z-10)=100$$

\begin{align}
100 &= 1 \times 100 \\
&= 2 \times 50 \\
&= 5 \times 20\\
&= 10 \times 10
\end{align}

Note that $3y-10\ge -4$ and $3z-10 \ge -4$.

It is easy to see that $3y-10 \in \{2,5\}$.

Hence $y \in \{4,5\}$.

If $y=4$, $z=20$. If $y=5, z=10.$

Now examine $x=2, y=4$, we have $\frac1x+\frac1y=\frac34$. The assumption of it is of the form of $\frac1n$ is not valid.

Title: Find all triplets (x, y, z) of positive integers such that 1/x + 1/y + 1/z = 4/5

Body:
Here's what i did :-

i wrote Find all triplet $$1/x + 1/y + 1/z = 4/5$$ as $$1/x + 1/y = 4/5 - 1/z$$

$(4z-5)/5z = 1/x + 1/y$ which can be written as $1/x + 1/y = 1/(5z)/(4z-5)$

Let $(5z)/(4z-5) = n$ therefore $1/x + 1/y = 1/n$ which on factorization gives $(n-x)(n-y) = n^2$ and $n$ should be an integer , i am not sure about the proof but i got $(5z)/(4z-5)$ is never an integer except $z = <o,1>$ but z cannot be 0 .

Here's my problem :-
if we take $n = 1$ we get 0 solutions of $<x,y,-1>$
is what i did correct ?


问:
Showing that the set of finite $\mu$-measurable sets is a ring
$\mathcal{M}_{\mathcal{e}}(\mu)$ is the set of finite $\mu$-measurable sets.

$A$ is $\mu$-measurable if a $B_n \in \mathcal{M}_{\mathcal{e}}(\mu)$ exists such that $A=\bigcup_{n=1}^\infty B_n$. We write $\mathcal{M}(\mu):=$ $\{A \subset \mathbb{R}^d | A$ is $\mu$-measurable$\}$.

How do I show that $\mathcal{M}_{\mathcal{e}}(\mu)$ is a ring?

答:
You need to prove 2 things:

$A,B\in \mathcal{M}_e(\mu) \implies A\cup B \in \mathcal{M}_e(\mu)$

$A,B\in \mathcal{M}_e(\mu) \implies A\setminus B \in \mathcal{M}_e(\mu)$

In fact, $\mu(A\cup B) = \mu(A)+\mu(B)-\mu(A\cap B)<\infty$ and $\mu(A\setminus B) = \mu(A)-\mu(A\cap B)<\infty$.


问:
Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction
I need to prove

$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$

I've seen other demonstrations of this, but they use some identities that I don't understand. So I asked my calculus teacher and he told me I can prove that by induction. I've already done the case $n=1$, but I can do the rest, could you help me please?

答:
Here's another way to do it (though it doesn't use induction). Write $\sin{x} = (e^{ix} - e^{-ix})/2i$. Then
$$
(\sin{x})^{2n+1} = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} e^{i(2n+1-2k)x}.
$$
Now, since your integrand is even, we can write
$$
\begin{align}
2\int_0^\infty {(\sin{x})^{2n+1}\over x}\,dx & = \int_{-\infty}^\infty {(\sin{x})^{2n+1}\over x}\,dx \\
& = \lim_{\delta\to 0}\int_{|x|>\delta} {(\sin{x})^{2n+1}\over x}\,dx \\
& = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} \lim_{\delta\to 0}\int_{|x|>\delta} {e^{i(2n+1-2k)x}\over x}\,dx.
\end{align}
$$
The real part of $e^{i(2n+1-2k)x}/x$ is $\cos\{(2n+1-2k)x\}/x$, which is an odd function of $x$; hence the integral of this real part over $|x| > \delta$ vanishes for each fixed $R$. So in fact
$$
\begin{align}
\lim_{\delta\to 0}\int_{|x|>\delta} {e^{i(2n+1-2k)x}\over x}\,dx &= i\lim_{\delta\to 0}\int_{|x|>\delta} {\sin\{(2n+1-2k)x\}\over x}\,dx \\
& = i\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx.
\end{align}
$$
A simple substitution shows that (using the value of your integral when $n = 0$)
$$
\begin{align}
\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx & = \pi\operatorname{sign}(2n+1-2k) = \left\{\begin{array}{ll}
\pi & 0\leq k\leq n\\
-\pi & k\geq n+1.
\end{array}\right.
\end{align}
$$
Finally, using the fact that $$(-1)^k{2n+1\choose k} = -(-1)^{2n+1-k}{2n+1\choose 2n+1-k},$$ we get
$$
\begin{align}
2\int_0^\infty {(\sin{x})^{2n+1}\over x}\,dx & = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} i\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx \\
& = {i\pi\over(2i)^{2n+1}}\left(\sum_{k = 0}^{n} - \sum_{k = n+1}^{2n+1}(-1)^k{2n+1\choose k}\right) \\
& = {2\pi(-1)^n\over2^{2n+1}}\sum_{k = 0}^{n}(-1)^k{2n+1\choose k}.
\end{align}
$$
Ron Gordon has already proved that the sum in this expression is equal to $(-1)^n{2n\choose n}$, so I'll leave it at that.</p>


问:
Fourth Order Nonlinear ODE
I was looking at an ode $w^{(4)} + w^3 = 0$ with initial conditions $[w'''(0),w''(0),w'(0),w(0)]=[1,0,0,0]$. I can see via maple that there is a blowup around 3.7. I was wondering if there was a way to show there is a blowup without using a computer, or if someone could direct me to some material on how to prove blowups. I'm having trouble finding literature on it. Thanks so much for your consideration.  </p>

答:
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0506.pdf

By integrating, we obtain

$2w'w'''-(w'')^2=-\dfrac{w^4}{2}+c$

Substitute $w'''(0)=1$ , $w''(0)=0$ , $w'(0)=0$ , $w(0)=0$ :

$c=0$

$\therefore2w'w'''-(w'')^2=-\dfrac{w^4}{2}$

Let $z=(w')^\frac{3}{2}$ ,

Then $\dfrac{d^2z}{dw^2}=-\dfrac{3w^4}{8z^\frac{5}{3}}$

Which reduces to an Emden–Fowler equation.

Title: Fourth Order Nonlinear ODE


问:
A $4 \times 4$ matrix over $\mathbb{C}$ such that rank$(A)=2$ and $A^3 =A^2 \neq 0$ and $A$ is not diagonalizable
This question was asked in my linear algebra quiz and I am having Difficulty in it . SO, I am asking for help here.

Edit : I am looking for an answer which solves it without using Jordan Forms. I have seen some Duplicate which use Jordan forms and which don't answer my 1 part. So, I asked it.

Let $A$  be  a $4 \times 4$ matrix over $\mathbb{C}$ such that rank (A)=2 and $A^3 =A^2 \neq 0$ and $A$ is not diagonalizable. (1) Then what is characteristic polynomial of $A$? (2) does there exists a vector v such that $Av\neq 0$ but $A^2 v=0$?

I have proved that minimal polynomial = $x^2 (x-1)$ but unable to deduce characteristic polynomial and (2) also.

Edit :For minimal polynomial I used that $A$ is not diagonalizable and $A^2 \neq 0$. So , (x) (x-1)( Diagonalizability) , x , $x^2$ can't be minimal polynomial. Also , rank =2 implies $x-1$ can't be .

Hope it's fine now.

Can you please help?

答:
Note that the minimal polynomial $m(x) = x^2(x-1)$ must divide the characteristic polynomial, and any zeros of the characteristic polynomial must also be zeros of the minimal polynomial. Since $A$ has size $4$, the characteristic polynomial must have degree $4$.

Thus, there are two possible characteristic polynomials: either we have $p(x) = x^2(x-1)^2$ or $p(x) = x^3(x-1)$. In order to eliminate the first possibility, we need to use the fact that $\operatorname{rank}(A) = 2$.

Suppose for contradiction that $x^2(x-1)^2$ is the characteristic polynomial. Because the exponent of $x$ is $2$, we note that $\dim \ker A^2 \leq 2 = \dim \ker A$. Since it always holds that $\ker A \subseteq \ker A^2$, we have $\ker A = \ker A^2$. However, the fact that $A^2(A-I) = 0$ implies that the image of $A - I$ is a subset of the kernel of $A^2$. Thus, the image of $A - I$ is a subset of the kernel of $A$, which means that $A(A-I) = 0$. This implies that the minimal polynomial is $m(x) = x(x-1)$, which means that $A$ is diagonalizable, contradicting the given information.

As we see in the above proof, it must hold that $\dim \ker A \subsetneq \dim \ker A^2$. This means that there exists a vector $v \in \ker A^2 \setminus \ker A$, which is to say that $Av \neq 0$ but $A^2 v = 0$.


问:
Bayes Probability/ Conditional Probability
A casino uses a regular die on Monday to Friday which is uniform on all faces and a weighted die on Saturdays and Sundays which is 50% likely to come out “6” and then 10% equally likely for the other rolls from 1 to 5. A player remember that they rolled a “6” but not the day played.

i. What is the chance that the player's "roll of a 6" came on Saturday?

ii. What is the chance that the “6” came from the regular die?

To use Bayes in this problem, how do you calculate the probability of a 6 regardless of the day it was rolled?

答:
To use Bayes in this problem, how do you calculate the probability of a 6 regardless of the day it was rolled?

$$\mathbb{P}[X=6]=\frac{5}{7}\times \frac{1}{6}+\frac{2}{7}\times \frac{1}{2}$$


问:
Working out Vector Variances using index notation
I was hoping someone could provide some intuition to some proofs of vector variances. The expectation case seems rather simple but I get confused when trying to work it out for the variance.

In the case of $$Var_x(a^T x) = \int \sum_i a_i (x_i - \bar x_i) \sum_j a_j(x_j - \bar x_j)p(x_i, x_j) d(\bar x)$$

$$= \sum_i \sum_j a_i a_j\int (x_i - \bar x_i)(x_j - \bar x_j)p(x_i, x_j) d(\bar x)$$

The result is $$a^TV_x(x)a$$ butI do not quite see how this works?

Similarly for the matrix for $$Var_x(Ax)$$ I don't know how to get to the result $$A \Sigma A^T$$

I would really appreciate if someone could explain how to see the transposes, specifically for index case as that is the method we are following in our class.

Thanks

答:
How well-versed are you in linear algebra ? What follows isn't a direct answer to your question (since balaji's answer seems satisfactory on the technical front), but understanding analogies between normal vectors and random variables should help you greatly when trying to figure out the role of the various pieces in what you're trying to prove, IMO.

Here is a series of analogies:

A random variable is a function. Function spaces are vector spaces. Random variables are thus a form of vector.

The covariance of two random variables is a dot product between vectors.

The variance of a random variable is the covariance of a random variable with itself. This makes it "the dot product of a vector with itself". This is the norm squared (also called a quadratic norm).

The standard deviation is the square root of the variance. The square root of a quadratic norm is a "simple" norm. The standard deviation is thus the vector norm of a random variable.

The Pearson correlation coefficient is defined as $\frac{cov(X,Y)}{\sigma_X \sigma_Y}$, which corresponds analogically to the cosine vector formula $\frac{}{|u||v|}$. The Pearson correlation coefficient is a cosine (ie, perfect colinearity/correlation for cosine 1, no correlation for cosine 0, anticorrelation for cosine -1).

A random vector is a list of random variables. A matrix is a list of column vectors, representing the "target basis" of a linear transformation. When multiplying two matrices together, each coordinate of the resulting product is defined as the dot product of a line of the left operand with a column of the right operand. When multiplying two random vectors, this dot product is covariance.

This makes the covariance matrix of a random vector the matrix product of a random vector with its (conjugate) transpose (which is an involution); what is usually called the gramian matrix in linear algebra. The gramian is a sort of "squared" matrix, that generalizes to non-square matrices. I say "sort of" squared, because the product of a matrix with its transpose $A^T A$ is not the same as $A^2$. To understand this difference, a much simpler example is $z^2$ (normal square) vs $z \bar{z} = |z^2|$ ("sort of" square) in the complex numbers. [As a side note, the relation between a matrix, its Frobenius norm and its gramian seem relevant to understanding this analogy further, but I haven't completely figured it out myself.]

The gramian is a symmetric matrix, and symmetric matrices $S$ represent dot products in the form $v^T S v$. The covariance matrix can thus be used as a form of dot product at the "random vector" level, rather than the "random variable" level.

SVD is just an improved Jordan decomposition of the gramian, and PCA is its equivalent on a covariance matrix.

PS: I think expectancy can be linked to covectors of probability spaces in some way, given that dot products of function spaces are often of the form $<f, g> = \int f(t)g(t)dt$; but I haven't figured that one out either completely. If someone more savvy than myself can help further this series of analogies, they are more than welcome to do so. I will add their contribution to this post if it is worthwhile.

Body:
How well-versed are you in linear algebra ? What follows isn't a direct answer to your question (since balaji's answer seems satisfactory on the technical front), but understanding analogies between normal vectors and random variables should help you greatly when trying to figure out the role of the various pieces in what you're trying to prove, IMO.

Here is a series of analogies:

A random variable is a function. Function spaces are vector spaces. Random variables are thus a form of vector.

The covariance of two random variables is a dot product between vectors.

The variance of a random variable is the covariance of a random variable with itself. This makes it "the dot product of a vector with itself". This is the norm squared (also called a quadratic norm).

The standard deviation is the square root of the variance. The square root of a quadratic norm is a "simple" norm. The standard deviation is thus the vector norm of a random variable.

The Pearson correlation coefficient is defined as $\frac{cov(X,Y)}{\sigma_X \sigma_Y}$, which corresponds analogically to the cosine vector formula $\frac{<u,v>}{|u||v|}$. The Pearson correlation coefficient is a cosine (ie, perfect colinearity/correlation for cosine 1, no correlation for cosine 0, anticorrelation for cosine -1).

A random vector is a list of random variables. A matrix is a list of column vectors, representing the "target basis" of a linear transformation. When multiplying two matrices together, each coordinate of the resulting product is defined as the dot product of a line of the left operand with a column of the right operand. When multiplying two random vectors, this dot product is covariance.

This makes the covariance matrix of a random vector the matrix product of a random vector with its (conjugate) transpose (which is an involution); what is usually called the gramian matrix in linear algebra. The gramian is a sort of "squared" matrix, that generalizes to non-square matrices. I say "sort of" squared, because the product of a matrix with its transpose $A^T A$ is not the same as $A^2$. To understand this difference, a much simpler example is $z^2$ (normal square) vs $z \bar{z} = |z^2|$ ("sort of" square) in the complex numbers. [As a side note, the relation between a matrix, its Frobenius norm and its gramian seem relevant to understanding this analogy further, but I haven't completely figured it out myself.]

The gramian is a symmetric matrix, and symmetric matrices $S$ represent dot products in the form $v^T S v$. The covariance matrix can thus be used as a form of dot product at the "random vector" level, rather than the "random variable" level.

SVD is just an improved Jordan decomposition of the gramian, and PCA is its equivalent on a covariance matrix.

PS: I think expectancy can be linked to covectors of probability spaces in some way, given that dot products of function spaces are often of the form $<f, g> = \int f(t)g(t)dt$; but I haven't figured that one out either completely. If someone more savvy than myself can help further this series of analogies, they are more than welcome to do so. I will add their contribution to this post if it is worthwhile.


答:
To use Bayes in this problem, how do you calculate the probability of a 6 regardless of the day it was rolled?

Represent the events with $W$ for a weekday, $U$ for a Sunday, $V$ for a Saturday, and $S$ for "success" (or "six"). Then just use the Law of Total Probability:$$\def\P{\mathop{\sf P}}\P(S)=\P(S\mid W)\P(W)+\P(S\mid U)\P(U)+\P(S\mid V)\P(V)$$

Title: Bayes Probability/ Conditional Probability

Body:
A casino uses a regular die on Monday to Friday which is uniform on all faces and a weighted die on Saturdays and Sundays which is 50% likely to come out “6” and then 10% equally likely for the other rolls from 1 to 5. A player remember that they rolled a “6” but not the day played.

i. What is the chance that the player's "roll of a 6" came on Saturday?

ii. What is the chance that the “6” came from the regular die?

To use Bayes in this problem, how do you calculate the probability of a 6 regardless of the day it was rolled?


问:
Why is there a condition on L'Hopital's Rule?
Why is the presence of an indeterminate form necessary for L'Hopital's rule to work? Why can't it work on functions not containing an indeterminate form? Is there an intuitive way to understand why this happens?</p>

答:
L'hospital comes from linear approximation.

$$f(x+h) \approx f(x) + f^{\prime}(x)h. $$

So for a limit of the ratio of two functions you should have

$$\lim_{h \to 0} \frac{f(x+h)}{g(x+h)} = \lim_{h \to 0} \frac{f(x) + f^{\prime}(x)h}{g(x) + g^{\prime}(x)h}.  $$

See that if $f(x) = g(x) = 0$  (the condition you're asking about) then the above is

$$= \lim_{h \to 0} \frac{f^{\prime}(x)h}{ g^{\prime}(x)h}= \lim_{h \to 0} \frac{f^{\prime}(x)}{ g^{\prime}(x)}.$$

If you don't have the condition, the $f(x)$ and $g(x)$  don't disappear and you can't conclude the last expression.


问:
what does $\exists \varepsilon>0, \ \ F_X(t+\varepsilon) \leq F_X(t-\varepsilon)$ imply
Given $X$ random variable with the given distribution function $F_X(t)$. given that $$\exists \varepsilon>0, \ \ F_X(t+\varepsilon) \leq F_X(t-\varepsilon)$$

what is true for sure from the following:

A.$\ \mathbb{P}[X=t]=0$

B. $\ \forall \varepsilon>0, \ \ F_X(t)\geq F_X(t-\varepsilon)+\frac12$

C.  $0<t<1$

D. $\ \mathbb{P}[X>t]=0$

I dont understand how the given can help me to infer something, I would like for a small hint how do I even start to approach this problem

答:
Note that
$$
F_X(t+\epsilon) - F_X(t-\epsilon) = \mathbb{P}[X\in(t-\epsilon,t+\epsilon]]\geq\mathbb{P}[X=t].
$$
Therefore, if $\mathbb{P}[X=t]>0$, then $F_X(t+\epsilon)-F_X(t-\epsilon)>0$, contradicting the given condition: $F_X(t+\varepsilon) \leq F_X(t-\varepsilon)$. So it must be that $\mathbb{P}[X=t]=0$.


问:
Extend inequality from dense subset to $H^{1}$
I know that the following inequality is true for all $g\in C^{\infty}_{0}(\mathbb{R}^{3})$:

$$\biggl\Vert\frac{g}{\vert\cdot\vert}\biggr\Vert_{L^{2}(\mathbb{R}^{3})}\leq2\Vert \nabla g\Vert_{L^{2}(\mathbb{R}^{3})}$$

My goal is to show that this inequality is still true for some $f\in H^{1}(\mathbb{R}^{3})$.

My Ansatz so far:
I use the fact that ${C^{\infty}_{0}(\mathbb{R}^{3})}$ is dense in $H^{1}(\mathbb{R}^{3})$, which means that we can find for each $\varepsilon\in\mathbb{R}_{>0}$ a function $g\in C_{0}^{\infty}(\mathbb{R}^{3})$, such that
$$\Vert f-g\Vert_{H^{1}(\mathbb{R}^{3})}=\sqrt{\Vert f-g\Vert_{L^{2}(\mathbb{R}^{3})}^{2}+\sum_{i=1}^{3}\Vert \partial_{i}(f-g)\Vert_{L^{2}(\mathbb{R}^{3})}^{2}}\leq\varepsilon.$$
This means that we must have that

$$\Vert f-g\Vert_{L^{2}(\mathbb{R}^{3})}\leq \varepsilon\hspace{1cm}\mathrm{and}\hspace{1cm}\Vert \nabla (f-g)\Vert_{L^{2}(\mathbb{R}^{3})}\leq \sum_{i=1}^{3}\Vert \partial_{i}(f-g)\Vert_{L^{2}(\mathbb{R}^{3})}\leq\varepsilon.$$

Now we have (using the notation $\Vert\cdot\Vert_{L^{2}(\mathbb{R}^{3})}:=\Vert\cdot\Vert_{2}$ from now on):

$$\bigg\Vert\frac{f}{\vert\cdot\vert}\bigg\Vert_{2}=\bigg\Vert\frac{f-g+g}{\vert\cdot\vert}\bigg\Vert_{2}\leq \bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}+\underbrace{\bigg\Vert\frac{g}{\vert\cdot\vert}\bigg\Vert_{2}}_{\leq 2 \Vert \nabla g\Vert_{2}}\leq \bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}+2(\Vert\nabla (f-g)\Vert_{2} + \Vert\nabla f\Vert_{2})$$

and hence

$$\bigg\Vert\frac{f}{\vert\cdot\vert}\bigg\Vert_{2}\leq 2\Vert\nabla f\Vert_{2} + 2\varepsilon+\bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}$$

First of all, is this right so far? Secondly, how can I proceed? If I somehow manage to relate $\bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}$ to $\Vert f-g\Vert_{2}\leq \varepsilon$ then I would recover the claimed inequality plus some $\varepsilon$, which would than prove the claim since this holds for all $\varepsilon>0$.

Thanks a lot!

答:
The problem with your argument is that you need the claim to work for $f-g$ in order to proof that it works for $f$.

Instead, approximate $f$ by a sequence of smooth functions $(g_n)$. Then by the inequality for these smooth functions, one finds that
$
\left( \frac{g_n}{|\cdot|} \right)
$
is a Cauchy sequence in $L^2$. Then it remains to prove that the  limit of the Cauchy sequence ($L^2$ is complete) is equal to $\frac{f}{|\cdot|}$.


问:
Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.
Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.

After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.

$z = \frac{7\pm3\sqrt{6}}{10}$.

I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Things worked out.

Am I still not done? Must I check for extraneous solutions in this situation?
I wonder if it might not require it because if ever we know one of the roots of a quadratic with rational coefficients is irrational, we know the other.

I suppose I'm also trying to avoid going through another round of substitution and test with these irrational conjugates. I'm not very trusting of my algebra yet. So I end up rewriting the same equations time and time again.

Body:
$$\sqrt{5 z+5}=2 \sqrt{z}+\sqrt{3-3 z}\tag{1}$$
Square both sides
$$5z+5=4z+3-3z+4\sqrt{3z-3z^2}$$
$$2+4z=4\sqrt{3z-3z^2}\to 1+2z=2\sqrt{3z-3z^2}$$
Square again both sides
$$1+4z+4z^2=12z-12z^2$$
$$16 z^2-8 z+1=0\to (4z-1)^2=0\to z=\frac14$$
Which is actually a solution of $(1)$ (verify!)