问:
Irrationality of ${\pi}$ and $e^{x/y}$
I attempt to prove that ${\pi}$ is an irrational number. For this, I use the beautiful continued fraction given by Brouncker who rewrote Wallis' formula as a continued fraction, which Wallis and later Euler (1775) proved to be equivalent.

The continued fraction given by Brouncker is
$$\frac{4}{\pi} =
1 + \cfrac{1}{3 + \cfrac{4}{5 + \cfrac{9}{7 + \dots}}}$$

Now we know that every infinite continued fraction is irrational. Does it mean that since RHS has an infinite continued fraction the LHS $\frac{4}{\pi}$ must also be irrational? And hence ${\pi}$ must be an irrational number?

EDIT 1: From this post General Continued Fractions and Irrationality If, in the continued fraction
$$\cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \dots}}}$$
the values $a_{i}, b_{i}$ are all positive integers, and if we have $a_i \geq b_i$ for all $i$ greater than some $n$, then the value of the continued fraction is irrational(Is this a necessary and sufficient condition? I am in doubt since the golder relation $\phi$ has $a_i \geq b_i=1$ for all $n\geq1$). As the condition is not satisfied the proof is actually wrong.

EDIT 2:
How to prove the following continued fraction of  $e^{x/y}$

$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $e^{x/y}$ will be an irrational number!

答:
To see that not every generalized infinite continued fraction is irrational, note that if
$$
x=1 + \frac{2}{1 + \frac{2}{1 + \frac{2}{1 + ...}}}
$$
then $x=1+{\large{\frac{2}{x}}}$, so $x=2$.

Of course that assumes that the continued fraction converges.

To prove $x=2$ more rigorously, we can argue as follows . . .

By definition, $x$ is the limit, if it exists, of the infinite sequence $x_0,x_1,x_2,...$ defined recursively by $x_0=1$ and
$$
x_n=1+\frac{2}{x_{n-1}}
$$
for $n\ge 1$.

Clearly we have $x_n \ge 1$ for all $n$.

For $n\ge 1$ we have
$$
x_n=1+\frac{2}{x_{n-1}}\le 1+\frac{2}{1}=3
$$
hence, since $x_0=1$, we have $x_n\le 3$ for all $n$.

Then for all $n\ge 1$ we get
$$
x_n=1+\frac{2}{x_{n-1}}\ge 1+\frac{2}{3}=\frac{5}{3}
$$
From the recursion we get
$$
x_n-2=\frac{2-x_{n-1}}{x_{n-1}}
$$
for all $n\ge 1$, hence for all $n\ge 2$ we get
$$
\left|x_{n}-2\right|\le\frac{\left|x_{n-1}-2\right|}{\left({\large{\frac{5}{3}}}\right)}
$$
and then an easy induction yields
$$
\left|x_n-2\right|\le\Bigl(\frac{3}{5}\Bigr)^{\large{n-1}}
$$
for all $n\ge 2$.

It follows that
$$
\lim_{n\to\infty}x_n = 2
$$
so $x=2$.

As to the question you raised in your edit, consider the generalized continued fraction
$$
1 + \frac{3}{1 + \frac{3}{1 + \frac{3}{1 + ...}}}
$$
which can be shown to be equal to
$$
\frac{1+\sqrt{13}}{2}
$$
Thus the condition for irrationality that you referenced is a sufficient condition, but not a necessary condition.

Title: Irrationality of ${\pi}$ and $e^{x/y}$


问:
Notation for vector with values $0$ and $1$?
For a real vector I know we write $a=(a_1,a_2,a_3)\in\mathbb R^3$, which means $a_1\in\mathbb R$, $a_2\in\mathbb R$ and $a_3\in\mathbb R$.

Suppose instead I have a vector $b=(b_1,b_2,b_3)$ where $b_1$, $b_2$ and $b_3$ only takes the values $0$ or $1$. What is the correct notation for this?

答:
Short Answer

$$(b_1,b_2,b_3)\in\{0,1\}^3$$

Description

For a smaller case of only one element, we write $b\in\{0,1\}$. For a two-element case, we can then write
$$(b_1,b_2)\in\{0,1\}\times\{0,1\}\equiv\{0,1\}^2$$
And likewise we get for the three-element case
$$(b_1,b_2,b_3)\in\{0,1\}\times\{0,1\}\times\{0,1\}\equiv\{0,1\}^3$$

Hope this helps. Ask anything if not clear :)

ContentLicense: CC BY-SA 4.0


问:
Problem based on proving equal triangles in area
In the figure, $ABCD$ is a parallelogram. If $O$ be any point on $BD$ then prove that $$\triangle OAB=\triangle OAD+\triangle OAC$$

My Attempt
$$\triangle ADB=\triangle BDC$$
$$\triangle OAM=\triangle OMC$$

Now, please help me to move further..

答:
Note: I will use the following notation $[\text{Figure}]$ for the area of the figure. Such as $[XYZ]$ denotes the area of $\triangle XYZ$.

Note that $\triangle ABD$ and $\triangle CDB$ are congruent, thus their areas are also equal, but since they share a common base $BD,$ the length of their altitudes $AQ$ and $CP$ must be equal for the areas of the triangles to be equal.

Thus we know that $\triangle AOD$ and $\triangle COD,$ which have the same base $OD$ are equal heights $AQ=CP,$ must have the same area. Thus, we have;
$$[OAD]=[OCD]$$
$$\implies [OAD]+[OAC]=[OCD]+[OAC]=[OACD].$$

Hence, the right hand side of the desired equation is the area of the quadrilateral $OACD.$

But quadrilateral $OACD$ can be bisected into $\triangle MCD$ and $\triangle OAM,$ so we have;
$$[OAD]+[OAC]=[OACD]=[MCD]+[OAM].$$

Since $\triangle MCD$ and $\triangle MAB$ are congruent, we know that $[MCD]=[MAB],$ and we have;
$$[OAD]+[OAC]=[MAB]+[OAM]=[OAB].$$

This is what was required.

ContentLicense: CC BY-SA 3.0


问:
soft question: solution manual for Demidovich Calculus
Context: I'm taking multivariable calculus this semester and my professor takes exercises from Problems in Mathematical Analysis by Demidovich - a book from a few decades ago, written by russian mathematicians. The problems in this book are of much higher difficulty than, say, Thomas Calculus (probably the most used textbook for calculus courses). I'm finding the course pretty difficult and it would be useful to have a solutions manual. However, there isn't any official one, so I thought I'd ask here if anyone has worked on the book and has solutions to the problems.

Thanks

答:
The latest editions with couple different publishing houses and actually the textbook itself in Chinese is right here. https://archive.org/details/jimiduoweiqi

It is much higher quality (not vector pdf already, alas) and OCR'ed.


问:
Are these functions orthonormal?
Are the following set of functions orthonormal over the interval $0$ to $1$?
$$Y_r(x) = \sin{\beta_r x}-\sinh{\beta_r x}-\frac{\sin\beta_r-\sinh\beta_r}{\cos\beta_r-\cosh\beta}\left(\cos\beta_r x-\cosh\beta_r x\right)$$
where $\beta_r$ are the positive solutions to:

$$1-\cos\beta_r\cosh\beta_r=0$$
I know that the functions $Y_r(x)$ are orthogonal. I went to check if they were orthonormal by evaluating $\int_0^1 Y_r Y_r \, dx$ numerical and checking if the integral equaled to $1$. For $\beta_r = 4.7300$ I get that the integral is equal to $1.03593$. For Larger values of $\beta_r$ the integral approaches $1$ so that makes me think that the set of functions is orthonormal and I am just encountering numerical error. Any thoughts?

答:
I agree with your computations; these functions are not normalized.  In fact, Mathematica is able to compute the exact value of of the the squared norm of the first function to be

$$\int_0^1 Y_1^2(1,x) = \frac{(\sin (\beta_1 )-\sinh (\beta_1 ))^2}{(\cos (\beta_1 )-\cosh (\beta_1))^2}.$$

Since $\beta_1 \approx 4.73$, we get about $1.0359$ for the integral.

I assume this computation is coming out of a Sturm-Liouville problem and I don't find this type of behavior to be at all unusual in that context.

Title: Are these functions orthonormal?


问:
To solve a geometric proof, do we need an image of the shape described in a problem to solve it?
To solve a geometric proof, for example, a two-column proof, do you need an image of the shape being described in the problem? If you are only told the "givens" of the problem and what you are trying to prove, would you be able to solve it? Given the proper algorithms, could something like a computer program solve a geometric proof without an image of the shape described in the problem?

For an extremely basic example, say you have a quadrilateral named ABCD, and you have to prove it is a parallelogram. You are given the fact that the sides AB||CD, and the side AB is congruent to the side CD. Would a computer program (or would we) need an image to solve this? And if yes, for more complex problems would we also not need an image?

答:
You can solve a geometric problem with only a description and not an image.

E.G.:

Nontrivial Problem:

If ABC is a right triangle with right angle at C, and if BD is any line cutting AC at point D, prove that $BD^{2}+AC^{2}=AB^{2}+DC^{2}$.

Solution:

DBC is also a right triangle with a right angle at C so from the Pythagorean Theorem:

$$DB^{2}=CD^{2}+CB^{2}$$

$$AB^{2}=CA^{2}+CB^{2}$$

So,

$$DB^{2}+CA^{2}=(CD^{2}+CB^{2})+(AB^{2}−CB^{2})=CD^{2}+AB^{2}$$

P.S.: Images can help you understand though. Nothing wrong with them.

Side Note: Also how would we do geometry that can't be immersed in 3d if we relied on pictures all the time.


问:
I would like some suggestions for notating a matrix minor.
I wish to show the steps in caculating the $ij$th minor of a matrix. I was thinking of writing $$\begin{vmatrix}
      de^{-x_1}+(1-x_1)e^{-2x_1} & (1-x_2)e^{-(x_1+x_2)} & \cdots &
      (1-x_\ell)e^{-(x_1+x_\ell)}\\
      (1-x_1)e^{-(x_2+x_1)} & de^{-x_2} +(1-x_2)e^{-2x_2}& \cdots &
      (1-x_\ell)e^{-(x_2+x_\ell)}\\
      \vdots & \vdots & \ddots & \vdots\\
      (1-x_1)e^{-(x_\ell+x_1)} &  (1-x_2)e^{-(x_\ell+x_2)} & \cdots &
      de^{-x_\ell} +(1-x_\ell)e^{-2x_\ell}
    \end{vmatrix}_{ij}$$
to indicate that the $i$th row and $j$th column had been removed. I was wondering if there was an already existing convention.

Body:
Two conventions are $M_i,_j$ and $M_{ij}$.

So,

$M_i,_j=\begin{vmatrix}
     de^{-x_1}+(1-x_1)e^{-2x_1} & (1-x_2)e^{-(x_1+x_2)} & \cdots &
      (1-x_\ell)e^{-(x_1+x_\ell)}\\
      (1-x_1)e^{-(x_2+x_1)} & de^{-x_2} +(1-x_2)e^{-2x_2}& \cdots &
      (1-x_\ell)e^{-(x_2+x_\ell)}\\
      \vdots & \vdots & \ddots & \vdots\\
      (1-x_1)e^{-(x_\ell+x_1)} &  (1-x_2)e^{-(x_\ell+x_2)} & \cdots &
      de^{-x_\ell} +(1-x_\ell)e^{-2x_\ell}
    \end{vmatrix}$

If you wish to express it on the determinant, one suggestion is

$\begin{vmatrix}
      de^{-x_1}+(1-x_1)e^{-2x_1} & (1-x_2)e^{-(x_1+x_2)} & \cdots &
      (1-x_\ell)e^{-(x_1+x_\ell)}\\
      (1-x_1)e^{-(x_2+x_1)} & de^{-x_2} +(1-x_2)e^{-2x_2}& \cdots &
      (1-x_\ell)e^{-(x_2+x_\ell)}\\
      \vdots & \vdots & \ddots & \vdots\\
      (1-x_1)e^{-(x_\ell+x_1)} &  (1-x_2)e^{-(x_\ell+x_2)} & \cdots &
      de^{-x_\ell} +(1-x_\ell)e^{-2x_\ell}
    \end{vmatrix}_{M_{i,j}}$


问:
Show: $\ker(b) \cong \ker(a)/U$.
I am working on the following task:
Let $V,W$ be $K$-vector space, $a:V\to W$ is a homomorphism and $U \subset V$ a subspace, $U \subset \ker (a)$.
Let $b: V/U\to W$, $v+U \mapsto a(v)$ be a homomorphism.
Show: $\ker(b) \cong \ker(a)/U$.
My idea:
$pi: V \to V/U$, $v \mapsto v+U$.

Thanks for your help!

答:
Hint : Show that the homomorphism
$$\pi_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a) \rightarrow \mathrm{Ker}(b)$$

is well-defined, surjective, and that its kernel is $U$. It will then induce an isomorphism
$$\tilde{\pi}_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a)/U \rightarrow \mathrm{Ker}(b)$$

Title: Show: $\ker(b) \cong \ker(a)/U$.


问:
Isolated singularities of this function?
I have a function $\sin (\frac 1 z) \sin(wz)$ where $w=\exp (\frac {i\pi} {4} )$. I need to find out the singularities of this function.

My progress :-

This function will have singularities on points where it isn't defined i.e. $z=0$.

This singularity may be removable, pole, or essential.

I'm unable to see how to go further. Computations on WolframAlpha returned limit as 0 along the real axis. It returned an infinity when $z$ was assumed purely imaginary. Hence I guessed it is a essential singularity.
Could someone tell me a non WolframAlpha solution/insight? Thank you :)

答:
You can characterize the different kinds of singularities by limits of the function. If it has a pole in $z_0 \in \mathbb{C}$, then we have that $\lim_{z \to z_0}|f(z)| = \infty$. If it has a removable singularity, then it has a finite limit (the value which you can give the function to expand it to a holomorphic function at that point if your initial function was already holomorphic in its domain). An essential singularity is "any other kind of singularity", which happens to have specific characteristics. Well, if it is just a singularity which is not removable and not a pole, then it is enough to find sequences $\{z_n\}_{n \in \mathbb{N}},\{z'_n \}_{n \in \mathbb{N}} \subset D$ (where $D$ is the domain of the given function) such that $z_n \to z_0$, $z'_n \to z_0$ with $|f(z_n)|\to \infty$ and $f(z'_n) \to a \in \mathbb{C}$.

(Actually, having an essential singularity at $z_0$ is actually equivalent to having a sequence $z_n \to z_0$ such that $lim_n f(z_n)$ does not exist.)

In this case, you can check that $z_n := \frac{i}{n}$ is such that $|f(z_n)|\to \infty$ and that $z'_n:= \frac{1}{\pi n}$ is such that $f(z'_n) \to 0$.

Hope this helps :)


问:
Understanding the proof if the union and intersection of two sets are connected, then the individual sets are connected
Proposition: "Let X be a topological space. Let A and B be two subsets of X such that: A and B are either both closed or both open, if A∪B and A∩B are connected, then A is connected and B is connected."

Here is the proof:

Suppose for a contradiction that A is disconnected. Assume both A and B are closed. Then there are closed, disjoint, non-empty subsets S,T of A such that A=SuT.
Since A∩B is connected, one of B∩S and B∩T must be empty, otherwise A∩B=(B∩S)∪(B∩T). WLOG assume B∩S=∅.
Now, since S and T are closed in A and A is closed in X, S and T are closed in X. Therefore, B∪T and S are closed non-empty subsets of A∪B. Obviously, (B∪T)∩S=∅ and A∪B=(B∪T)∪S, so A∪B is disconnected, contradiction.

Ok my question is, where was it used in the proof that A and B are closed and why does it matter that A and B are both closed or open? It seems like the proof works without that.

答:
You use that $A$ is closed in $X$ to conclude that $S,T$ closed in $A$ implies they're closed in $X$. The same applies to open (and then you'd have used that $S,T$ were open in $A$ and then open in $X$ as $A$ is open etc. (using the standard fact that "closed in closed is closed" and "open in open is open" for subsets and subspaces)

An alternative proof uses functions and the pasting lemma, which holds for two open or two closed sets:

Suppose $f: A \to \{0,1\}$ is continuous, where $\{0,1\}$ is discrete. As $A \cap B \subseteq A$ is continuous, $f\restriction_{A \cap B}$ is constant with value $i_0 \in \{0,1\}$. Now extending $f$ to $A \cup B$ by $f(x)=i_0$ for $x \in B$ as well we get a continuous function on $A \cup B$ by the pasting lemma (where we use that $A$,$B$ are either both closed or both open in $X$) and the connectedness of $A \cup B$ tells us the extended $f$, and so a fortiori the original $f$ too, is constant. QED


问:
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$
let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$

maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks

答:
Since at the point $x=y$, equality holds, and  multiple derivatives are $0$, it appears likely that a solution with derivatives is needed. Here is one:

Following @River Li's calculations, it suffices to prove, for $x \ge 1$,  that (let $n$ be a parameter)
$$
f(x) = \ln (x^{n+1} + 1) - \ln (x^n + 1)- \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2} \ge 0
$$
This holds with equality for $x=1$, so we prove that $f(x)$ is increasing with $x$. The derivative is
$$
f'(x) = \frac1x \Big[ \frac{(n+1)x^{n+1}}{x^{n+1} + 1} - \frac{nx^{n}}{x^{n} + 1}- \frac{x^{2n +1}}{x^{2n +1} + 1} \Big]
$$
Multiplying out gives

$$
f'(x) = \frac{x^{n-1}}{ (x^{n+1} + 1) (x^{n} + 1) (x^{2n +1} + 1)} \Big[x - n + nx - x^{2n +1} + n x^{2n +2} - n x^{2n +1}\Big]
$$
Hence it suffices to prove that the last term in brackets is $\ge 0$, slightly rewritten:
$$
g(x) = x(n+1) - n  + x^{2n +1}(-1 -n + n x)\ge 0
$$
Again we have equality for $x=1$. Further, note that $x(n+1) - n \ge 0$ holds always for $x \ge 1$, and $-1 -n + n x \ge 0 $ holds for $x\ge \frac{n+1}{n}$, so we only need to consider the range $x \in [1 \; \frac{n+1}{n}]$. In that range, we need to prove
$$
 \frac{x(n+1) - n  }{1 + n - n x} \ge x^{2n +1}
$$
Taking logarithms gives
$$
h(x) =  \ln (x(n+1) - n  ) - \ln (1 + n - n x) - (2n + 1 ) \ln x \ge 0
$$
Since we have equality for $x=1$, we use the above method again and prove that $h(x)$ is increasing. Taking derivatives gives that we need to prove
$$
h'(x) =   \frac{n+1}{x(n+1) - n  }+  \frac{n}{1 + n - n x} - (2n + 1 ) \frac1x \ge 0
$$
or, multiplying out,
$$
h'(x) =   \frac{1}{x(x(n+1) - n)  (1 + n - n x)} \Bigg[ n(2n^2 + 3n + 1)(x - 1)^2 \Bigg] \ge 0
$$
But this is clearly true for $x \in [1 \; \frac{n+1}{n}]$, which proves the claim. $\qquad \Box$


问:
When does a surjective module map admit a section such that the diagram commutes?
Let $A$ be a polynomial ring over $\mathbb{R}$. Let $M, N$ be free $A$-modules.
Let $L$ be a finitely generated $A$-module.
Suppose we have $A$-module maps $f\colon M\to N$, $g\colon M\to L$ and $h\colon L\twoheadrightarrow N$ such that the following diagram commutes.

Do we have an $A$-linear section $s\colon N\to L$ such that the diagram commutes (i.e., $h\circ s = id$ and $s\circ f = g$)?

答:
Assumption: $h \circ g = f$. If $L$ is a projective $A$-module, there is always an $A$-linear section $s:N \rightarrow L$ with $h \circ s = id_N$. Hence

$$ h \circ ( s \circ f- g)=h \circ s \circ f - h \circ g = f-f =0. $$

Hence

$$P1. \text{   } s(f(x))-g(x) \in ker(h)$$

for all $x \in M$. Hence $s(f(x))=g(x) + m $ with $m \in ker(h)$ and

$$ s(f(x))\equiv g(x) mod(ker(h)).$$

Hence if $L$ is projective a section $s$ with property P1 exist. If $L$ is finitely generated this is not the case.

Question: "Do we have an A-linear section $s:N\rightarrow L$ such that the diagram commutes?"

Answer: Not in general.


问:
Should I use an equality sign or "iff" for a definition?
In many mathematical textbooks the definitions are given in the following form (considering even numbers as an example):

"A natural number n is called even iff it is a multiple of 2".

Now as we know a definition should serve only as an abbrevation. That is I can use the following form for defining even numbers.

"Even number = a natural number that is a multiple of 2".

Is there any difference between the above definitions? I was thinking that the first makes the definition more strict. For example, suppose that I state:

"$1.5$ is an even number". According to the first definition this statement is meaningless because the term "even" is defined for natural numbers. According to the second definition the statement should be false because $1.5$ isn't a natural number. Which of the above definitions is more appropriate and why?

Edit
I have read the question with the proposed answer before I asked this question. But there is a difference because I am asking what notation should I use and if there is a difference between equality sign and "iff" whereas the other question is about the difference between "if" and "iff".

答:
The first one is more appropriate, and "if" is more conventional than "iff" in definitions (i.e. sentences introducing terminology with "is called" / "we say that" / etc.).

Your concern is that if our definition defines "even" only for natural numbers, then writing "1.5 is even" is a sort of "type error". But English text isn't typechecked computer code. We understand "x is even" in context to mean either "x is a natural number and x is even" or "(since here we've established that x is a natural number) x is even". The risk of miscommunication is small.

It's misleading to apply formal symbols like "=" to natural-language phrases, because the language isn't mathematically formalized and doesn't reliably follow simple compositional rules. (Is a sentence just a string of symbols? If two strings of symbols are "equal", does that mean you can substitute one for the other anywhere? And how can you state the answers to these questions if they haven't been answered yet?)


问:
$\mathbb{Z}[i]$ is a Dedekind domain
I know that $\mathbb{Z}[i]$ is a PID, and that every PID is a Dedekind. But I want to show that $\mathbb{Z}[i]$ is a Dedekind, without using PID.

One strategy coul be to show that $\frac{\mathbb{Z}[i]}{\mathfrak{P}}$ is a field for every nonzero prime ideal $\mathfrak{P}$. Could someone suggest me how to prove this?
Furthermore, what else am I supposed to prove to ensure that $\mathbb{Z}[i]$ is a Dedekind?

答:
Besides proving that every nonzero prime ideal is maximal, you need to prove that $\mathbb{Z}[i]$ is Noetherian and integrally closed.

To prove that $\frac{\mathbb{Z}[i]}{\mathfrak{P}}$ is a field for every nonzero prime ideal $\mathfrak{P}$, prove that it is finite, because every finite domain is a field. See the spoiler below.

 To prove that $\frac{\mathbb{Z}[i]}{\mathfrak{P}}$ is finite, consider $I=\mathfrak{P} \cap \mathbb{Z}$. Then $I$ is a prime ideal of $\mathbb{Z}$ and so $I=p\mathbb{Z}$ for some prime $p$. This implies that $\mathfrak{P}$ contains $p\mathbb{Z}[i]$ and so $\frac{\mathbb{Z}[i]}{\mathfrak{P}} \subseteq \frac{\mathbb{Z}[i]}{p\mathbb{Z}[i]}$ has at most $p^2$ elements.

Title: $\mathbb{Z}[i]$ is a Dedekind domain


问:
Algebraic division of very large number and finding the remainder
Basically, I got this from a olympiad practice test and I still don't understand the logic of it, so the question asks us to find the remainder of the following:

$$\frac{x^{74} + 23}{x + 1}$$

At first I thought I could just use the big brain shortcut of replacing x with 1, but the remainder turned out to be $0$, and the multiple choice options only has the answers $74$, $25$, $24$, and $23$. If you're going to use something like the modular operation please explain it to me because I'm not familiar to it.

Any help would be extrememly appreciated :)

答: When $f(x)$ is divided by $x-a$, the remainder is $f(a)$. Here, $f(x)=x^{74}+23$ so the remainder when it is divided by $(x+1)$ is $f(-1)=24.$ This is known as remainder theorem.


答:
$\frac{p(x)}{x+1}=c(x)+\frac{r(x)}{x+1}\iff p(x)=c(x)(x+1)+r(x)$ where $\text{deg}(r(x))<\text{deg}(x+1)=1$, so $\text{deg}(r)=0$, $r(x)=r\in\Bbb R$. So $r=p(x)-c(x)(x+1)$ for all $x$ and, in particular, $r=p(-1)-c(-1)(-1+1)=p(-1)$.

This is essentially the proof of the Remainder Theorem.

Title: Algebraic division of very large number and finding the remainder

Body:
Basically, I got this from a olympiad practice test and I still don't understand the logic of it, so the question asks us to find the remainder of the following:

$$\frac{x^{74} + 23}{x + 1}$$

At first I thought I could just use the big brain shortcut of replacing x with 1, but the remainder turned out to be $0$, and the multiple choice options only has the answers $74$, $25$, $24$, and $23$. If you're going to use something like the modular operation please explain it to me because I'm not familiar to it.

Any help would be extrememly appreciated :)


问:
If $Q^TQ$ is an orthogonal projection then $QQ^T=I$
Consider an $n\times m$ matrix $Q$. If $Q^TQ$ is an orthogonal projection in $\mathbb{R}^m$ onto a subspace of dimension $n$ then prove that $QQ^T=I$.

Proof: Using, the idempotent property of projection we can write:
$$Q^TQQ^TQ=Q^TQ$$ or  $Q^T(QQ^T-I)Q=0$. How should this imply, $QQ^T-I=0$?

答:
If
$$\dim {\rm im\,} Q^TQ  = n$$
it means that
$$\dim {\rm im\,} Q\ge n$$
(because multiplying by another operator cannot increase the dimension of the image). But since $Q: \mathbb{R}^m \rightarrow \mathbb{R}^n$, it just means that
$$ {\rm im\,} Q = \mathbb{R}^n $$
The equation
$$ Q^T(QQ^T -I)Q = 0 $$
can be read as
$$ Q^T(QQ^T -I)|_{{\rm im\,}Q} = 0 $$
but since $ {\rm im\,} Q = \mathbb{R}^n $, it just means that
$$ Q^T(QQ^T -I) = 0 $$
By transposing this equation we have
$$ (QQ^T -I)Q = 0 $$
Using the same argument again, we get
$$ QQ^T - I = 0$$


问:
Give a non-differentiable fixed-point function
Give an example of a function $\phi:[-1,1]\to[-1,1]$ with unique fixed point $\phi(0)=0$, that is not differentiable at $x=0$ and where the fixed point iteration $\alpha_{j+1}=\phi(\alpha_j)$ for every $\alpha_1\in[-1,1]$ results in $\alpha_3=\phi^2(\alpha_1)=0$.
Could somebody help me out? Thanks!

答:
Define
$$
\phi(x)= \left\{
    \begin{array}{ll}
       0 & \mbox{if } x=0 \text{ or } x=1 \\
       1  & \mbox{otherwise.}
    \end{array}
\right.
$$


问:
Is there a way to save my method for solving this differential equation?
$y' + y = y^{-2}$

I divided everything by $y^{-2}$ and let $v = y^3$, then solved as normal. (I relaced the $y^2$ with $y^{2}=\frac{1}{3}\frac{dv}{dy}$ after differentiating the $v$. Then I solved the "ordinary linear first-order differential equation". (Except I know technically its not). I know this was not correct because of the $y^2$ in front of $y'$ after division by $y^{-2}$ AND I know you can solve it the simple way by moving $y$ over...
What I am looking for is a fix for my method, if any exists.

After $v$ substitution, $\frac{1}{3}\frac{dv}{dx} + v = 1$

After integration, $ln|1-v| = 3x + c$

After raising both sides to $e$ and resubstitution, $|1-y^3|=e^c e^{3x}$

And that's the problem. $e^{3x}$, when it should be $e^{-3x}$

Body:
$$y'+y=y^{-2} \implies y^2y'+y^3=1 \implies \frac{1}{3}(y^3)'+y^3=1$$
$$\frac{dz}{dx}+3z=3 \implies \frac{de^{3x}z}{dx}=3 \implies e^{3x} z=3x+C.$$
Finally, $$z=3xe^{-3x}+Ce^{-3x}\implies y^3=3xe^{-3x}+Ce^{-3x}$$


问:
Is every prime number less than twice the previous prime number?
And if so, how do you prove it? (for example 7 is less than 2 times 5, 11 less than 2 times 7, and so on).</p>

答:
This is a Postulate(Betrand's) which says for every $n>3$, you can find a prime $p$ such that $n<p<2n$.

So, yes. If you any $n=p$, you can find prime $q$ such that it is less than $2p$.

At a very young age of $17$(As my guide says), Erdos' gave the most elementary proof to this. About which you can read here.


问:
Find the width of a triangle at a certain point
Is there a way to find the width of a triangle at a certain point? I'm not very good at maths so I don't know many maths words, so I'm going to explain it in a diagram.
Is there any way to find the length of p?

答: The only way to know $p$ is to know the overall height from the base that is not $x$ or $y$ up to the tip. Using that, $x$, $z$, $y$ you could find $p$, otherwise it's impossible. Also next time provide angles and measurements because there is a lot of assumptions banking on this. :)


问:
Prove that $\sup(S^2)=\sup(S)^2$ where $S$ is a nonempty bounded set of nonnegative real numbers.
It seems like this statement is common sense so I don't know how to prove it using mathematical logic. S={x : x \in S }, and S^2={x^2 : x \in S }. Would it suffice to say that because the numbers in the set are nonnegative, that taking the supremum before or after squaring the set is irrelevant? How do I say this as a proof?</p>

答:
The fact that the numbers are nonnegative is essential: indeed
$$
(\sup\{-2,1\})^2=1,\qquad \sup\{(-2)^2,1^2\}=4
$$

Suppose $S$ consists of nonnegative numbers (and is not empty). We can assume that $S$ contains positive numbers, for the case $S=\{0\}$ is trivial. Let $c=\sup S$, so $c>0$.

Let $b$ be an upper bound of $S$. Then, for $a\in S$, we have $a\le b$, so also $a^2\le b^2$. Hence $b^2$ is an upper bound of $S^2$. In particular, $c^2$ is an upper bound of $S^2$.

Is it the least upper bound? Let $\varepsilon>0$: we need to find $a\in S$ such that $a^2>c^2-\varepsilon$, which is the same as $c^2-a^2<\varepsilon$.

How can we do it? We need $(c+a)(c-a)<\varepsilon$. Since, for every $a\in S$ it holds $a\le c$, we can first find $a\in S$ such that
$$
a>c-\frac{\varepsilon}{2c}
$$
so
$$
c^2-a^2=(c+a)(c-a)\le 2c(c-a)<2c\frac{\varepsilon}{2c}=\varepsilon
$$
and we're done.


问:
Taylor series expansion of $\sin^2(x)$
I'm looking at a solution and I'm having a hard time understanding this equality (I'm not great at Sigma notation),

$
\sin^2(x) = \dfrac{1-\cos(2x)}{2}=\dfrac{1}{2}\left[1-\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right] = \dfrac{1}{2}\left[1-1-\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right]
$

It's unclear to me how to go from this

$
\dfrac{1}{2}\left[1-\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right]
$

to this

$
\dfrac{1}{2}\left[1-1-\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right]
$

答: The sum initially goes from $n=0$ to $n=\infty$. In the second line, the first term ($n=0$), which is $1$, is expanded apart, and now the sum starts at $n=1$. And don't be confused by the minus "-" in front :)

Title: Taylor series expansion of $\sin^2(x)$


问:
Gateaux and Fréchet differentials in $\ell^1$
I am in trouble trying to solve the following:

Let $X = \ell^1$ with the canonical norm and let $f \colon \ell^1 \ni x\mapsto \Vert x \Vert$. Then $f$ is Gateaux differentiable at a point $x = (x_1, x_2, \ldots )$ if and only if $x_i \ne 0$ for every $i \in \mathbb N$ with Gateaux differential given by

  $$
\Phi_G f(x) = \left( \frac{x_1}{\vert x_1 \vert}, \ldots , \frac{x_n}{\vert x_n \vert}, \ldots \right )
$$

Well, I show you what I've done.

1 If for some $i$ we have $x_i = 0$ then I am quite sure the limit
$$
\lim_{t \to 0} \frac{f(x + te_i) - f(x)}{t}
$$
does not exist (in an analogous way to the one variable function $x \mapsto \vert x \vert$) but I don't manage to prove it rigorously (how can I evaluate the left and the right limits?).

2 Secondly, I do not know how to show that $f$ is Gateaux differentiable if $x_i \ne 0$ for every $i$. How can I do?

3 Eventually, here there is a self-posed question: is the function $f$ Fréchet differentiable at some point $x$? My guess is "no", since Gateaux differential is non-linear but I have some doubts...

I thank you for any help.

答:
Fix some $v\in \ell^1$ and note that $$\tag{1}\frac{\|x+tv\|-\|x\|}{t}=\frac{\sum_{i=1}^\infty(|x_i+tv_i|-|x_i|)}{t}$$

It follows from $(1)$ that $$\tag{2}\lim_{t\to 0}\frac{\|x+tv\|-\|x\|}{t}=\sum_{i=1}^\infty \lim_{t\to 0}\frac{(|x_i+tv_i|-|x_i|)}{t}$$

Now the question is: In what points the function $|\cdot|:\mathbb{R}\to\mathbb{R}$ is differentiable? and what is its derivative in those points?

For your third question: the fact that Gateaux differential is non-linear has nothing to do with $f$ being or not Frechet differentialbe. Try to verify if in the points where $f$ is Gateaux differentiable the derivative is continuous.


问:
Throwing fair dice, what can I say about the following two Random variables?
Droping 2 times fair dice, Let $X$ be the number of drops that got even number, Let $Y$ be the number of drops got a number that is divided by $3$ what is true from the following?

A. $\mathbb{E}[X^2Y]= 1$ and $X$ , $Y$ are independant

B.$\mathbb{E}[X^2Y]= \frac{2}{3}$ and $X$ , $Y$ are independant

C.$\mathbb{E}[X^2Y]= \frac{2}{3}$ and $X$ , $Y$ are Not independant

D.$\mathbb{E}[X^2Y]= 1$ and $X$ , $Y$ are Not independant

How should I approach this question? Do I really need to calculate the probability function or there is an efficient way to approach it?

答:
To calculate the expectation is is easy to find out the distribution but it is very immediate...

First observe that

$$\mathbb{P}[\text{Even}]=\frac{1}{2}$$

$$\mathbb{P}[\text{Div 3}]=\frac{12}{36}=\frac{1}{3}$$

and

$$\mathbb{P}[\text{Div 3}|\text{Even}]=\frac{6}{18}=\frac{1}{3}$$

Thus being

$$\mathbb{P}[\text{Div 3}|\text{Even}]=\mathbb{P}[\text{Div 3}]$$

These events are independent.

Given this property also $X\perp \!\!\!\!\!\!\perp Y$ and $X^2\perp \!\!\!\!\!\!\perp Y$

Thus

$$\mathbb{E}[X^2Y]= \mathbb{E}[X^2]\cdot\mathbb{E}[Y]=\Bigg(\frac{2}{4}+1 \Bigg)\Bigg(\frac{4}{9}+\frac{2}{9}\Bigg)=1$$


问:
inequality:$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$

If $a,b,c,d>0$ such that $abcd=1$ prove that $$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$$

My first idea was to find counter examples and also check the bahaviour of inequality in the edge cases.Easy to see the equality occurs when $a=b=c=d=1$ However there is a big problem when $a\to 0^+ b\to 0^+,c\to 0^+,d\to \infty$ In this case the RHS $\to \infty$ and the LHS although appears  $\to 0$ we see that as $b+d\to \infty$ so we cannot say much.

Mixed variables doesnt help much as we cannot take WLOG $a\ge b\ge c\ge d$

I also tried Holder $$(a+c)(b+d)(bd+ac)\ge {(\sqrt[3]{ab^2d}+\sqrt[3]{ac^2d})}^3=ad{(b^{2/3}+c^{2/3})}^3$$ which  has only complicated it further

Looking forward to solutions without lagranges multiplier.

Body:
WLOG assume that $bd\ge 1$ and $ac\le 1$. The inequality rewrites as
$$(a+c)((b+d)(ac+bd)-3)+4\ge 3(b+d).$$
AM-GM and $bd\ge 1$ yield
$$(b+d)(ac+bd)-3\ge 2\sqrt{bd}\cdot 2\sqrt{acbd}-3 \ge 2\cdot 2-3=1>0,$$
hence by AM-GM
$$(a+c)((b+d)(ac+bd)-3) \ge 2\sqrt{ac}((b+d)(ac+bd)-3) = 2(b+d)((ac)^{3/2}+(bd)^{1/2})-6(bd)^{-1/2}.$$
Hence it is enough to prove that
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) + 4 \ge 6(ac)^{1/2} .$$
Using AM-GM again we obtain $$(ac)^{3/2}+(bd)^{1/2} \ge 4\sqrt[4]{(ac)^{3/2}\cdot \left(\frac{(bd)^{1/2}}{3}\right)^3}=4\cdot 3^{-3/4}>\frac 32,$$
hence by AM-GM
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) \ge 4(bd)^{1/2}\left((ac)^{3/2}+(bd)^{1/2}-\frac 32\right) = 4ac+4bd-6(bd)^{1/2}$$
and we are left to prove that
$$4ac+4bd+4\ge 6(ac)^{1/2} + 6(bd)^{1/2}.$$
This follows from AM-GM:
$$4ac+4bd+4\ge 3ac+3bd+6 = 3ac+3 + 3bd+3 \ge 6\sqrt{ac}+6\sqrt{bd}$$


问:
Prove that at least one of the matrices is invertible
The question is to show that if there is a matriz $C \in \mathbb{M}_3 (\mathbb{C})$, then at least one of $C + I_3$, $C + 2I_3$, $C + 3I_3$, $C + 4I_3$ is invertible.

My first approach was by eigenvalues. $C + I_3 = C - (-I_3)$ and $-1$ is the only eigenvalue of $-I_3$, so
I tried to prove that $det(C + I_3 - \lambda I_3) = det(C - (-I_3) - \lambda I_3) = det(C- (-1 + \lambda)I_3)$ was equal to $0$ or not.

So, if $\lambda = 1$, then $det(C - (-1 + \lambda)I_3) = det(C)$, but we don't know anything about $C$ to know if $det(C) = 0$ or not. If $\lambda \neq 1$, then $det(C - (-1 + \lambda)I_3) = det(C - \gamma I_3)$, $\gamma = -1 + \lambda$, which inplies that $C - \gamma I_3 \neq 0$, so that $C + I_3$ would be invertible. Anyay, I didn't get to any conclusion. The idea as to do this for each of the four matrices.

My second approach was by contradiction. So, I supposed that any of these matrices was invertible. Then, $det(C + I_3) = det(C + 2I_3) = det(C + 3I_3) = det(C + 4I_3) = 0$. But, this implies that $C + I_3 = \cdots = C + 4I_3$, which is the same as saying that $C = -I_3 = \cdots = -4I_3$, which is not true. So, we would have that at least one of the four matrices is invertible. Anyway, I don't think my proof is correct.

Any suggestions? Thanks.

答:
I am assuming that the correct statement is the next one:

Given any matrix $C\in\Bbb M_3(\Bbb C)$, at least one of the matrices  $C+kI_3$ ($k=1,2,3,4$) should be invertible.

By way of contradiction, assume that there is some matrix $C\in\Bbb M_3(\Bbb C)$ such that $C+kI_3$ is not invertible for $k=1,2,3,4$.

Then $\det(C+k I_3)=0$ $k=1,2,3,4$ and this would imply that $M$ has 4 differents eigenvalues (say, $\lambda=-1,-2,-3,-4$). BUt this is impossible since $M$ is a 3x3 matrix.

So at least one of these matrices would have nonzero determinant.


答:
Hint :
$$\det(C+XI_n)$$ is a nonzero polynomial of degree $3$, hence it cannot have four distinct roots.

Title: Prove that at least one of the matrices is invertible

Body:
The question is to show that if there is a matriz $C \in \mathbb{M}_3 (\mathbb{C})$, then at least one of $C + I_3$, $C + 2I_3$, $C + 3I_3$, $C + 4I_3$ is invertible.

My first approach was by eigenvalues. $C + I_3 = C - (-I_3)$ and $-1$ is the only eigenvalue of $-I_3$, so
I tried to prove that $det(C + I_3 - \lambda I_3) = det(C - (-I_3) - \lambda I_3) = det(C- (-1 + \lambda)I_3)$ was equal to $0$ or not.

So, if $\lambda = 1$, then $det(C - (-1 + \lambda)I_3) = det(C)$, but we don't know anything about $C$ to know if $det(C) = 0$ or not. If $\lambda \neq 1$, then $det(C - (-1 + \lambda)I_3) = det(C - \gamma I_3)$, $\gamma = -1 + \lambda$, which inplies that $C - \gamma I_3 \neq 0$, so that $C + I_3$ would be invertible. Anyay, I didn't get to any conclusion. The idea as to do this for each of the four matrices.

My second approach was by contradiction. So, I supposed that any of these matrices was invertible. Then, $det(C + I_3) = det(C + 2I_3) = det(C + 3I_3) = det(C + 4I_3) = 0$. But, this implies that $C + I_3 = \cdots = C + 4I_3$, which is the same as saying that $C = -I_3 = \cdots = -4I_3$, which is not true. So, we would have that at least one of the four matrices is invertible. Anyway, I don't think my proof is correct.

Any suggestions? Thanks.


问:
Statement about Weil representations being unramified over a certain extension
Let $K = \mathbb{Q}_p$ and $\rho: W_K \to \operatorname{GL}(V)$ be a Weil representation over a finite-dimensional complex vector space $V$ (where $W_K$ denotes the Weil group over $K$).

Let $L/K$ be a finite cyclic extension of degree $n$ and ramification index $e$ such that the restriction $\rho|_L: G_L \to \operatorname{GL}(V)$ is unramified (equivalently, $\rho(I_L)=1$ where $I_L$ is the inertia group over $L$).

Let $L'/K$ be a totally ramified of degree $e$ such that $LL'/L'$ is unramified of degree $n$.

Question: Is $\rho|_{L'}$ unramified, i.e. $\rho(I_{L'})=1$?

I am certain that if this is true, the only chance to show it is by using the fact that $LL'/L'$ is unramified. But I am not sure how this should work. Could you please help me with this problem?

答:
This is true, because $I_{L'} \subset I_{L}$. Hence, if $\rho(I_L) = 1$, then $\rho(I_{L'}) = 1$.

Indeed, let $L^{\mathrm{ur}}$ be the maximal unramified extension of $L$ and let $(L')^{\mathrm{ur}}$ be the maximal unramified extension of $L'$. Because $LL'/L'$ is unramified, $(L')^{\mathrm{ur}}$ contains $L$ and hence $L^{\mathrm{ur}}\subset (L')^{\mathrm{ur}}$.  But
$$\begin{align*}I_{L'} = \mathrm{Gal}(\overline K/(L')^{\mathrm{ur}})\subset \mathrm{Gal}(\overline K/L^{\mathrm{ur}}) = I_L.\end{align*}$$


问:
How to calculate the viewing angle for curved monitors?
Some higher-end computer monitors and TVs nowadays offer curved screen, which curvature is provided.

However, the viewer often sits closer to the screen than the centre of curvature, resulting in a viewing angle which is not a right angle anymore for every point offset from the center.

What is the formula to obtain the viewing angle as function of radius of curvature of the monitor and distance from the center of the monitor (or distance from the center of curvature) of the viewer? The coordinates on the monitor can be provided as linear distance along the curve.

答:
$$ w= r \sin \gamma = h \tan \beta $$

where $h$ is distance before mirror nearest end points shown and $w$ the half-width.

$$\tan \beta = \dfrac{r \sin \gamma}{h}$$

which is higher


问:
How is a change of variables performed in Lotka-Volterra normalizations?
How is the change of variables really performed in Lotka-Volterra?

I'm trying to understand what happens in the following document:

https://www.maths.dur.ac.uk/~ktch24/term1Notes(10).pdf, p. 17

I tried plugging the variables in, but I don't see how it produces the result.

答:
Let's try it from a slightly different angle: Decompose every variable into a constant scale factor and a "normalized" variable, $x=x_*\hat x$ etc. Then insert into the equations$\newcommand{\dd}{{\rm d}}$
\begin{align}
\frac{\dd(x_*\hat x)}{\dd(t_*\hat t)} &= a\,x_*\hat x-b\,x_*\hat x\,y_*\hat y\\
\frac{\dd(y_*\hat y)}{\dd(t_*\hat t)} &= c\,x_*\hat x\,y_*\hat y-d\,y_*\hat y
\end{align}
Now remove the scale factors from the left side and collect the scale factors
\begin{align}
\frac{\dd\hat x}{\dd\hat t} &= (at_*)\,\hat x-(bt_*y_*)\,\hat x\hat y\\
\frac{\dd\hat y}{\dd\hat t} &= (ct_*x_*)\,\hat x\hat y-(dt_*)\,\hat y
\end{align}
Now set the combined coefficients to $1$ as long as that is possible:
$$
at_*=1\implies t_*=\frac1a,\\
bt_*y_*=1\implies y_*=\frac{a}{b},\\
ct_*x_*=1\implies x_*=\frac{a}{c},\\
dt_*=\frac{d}{a}=\hat d\text{ has no free parameter},\\
$$
This gives the normalized system
\begin{align}
\frac{\dd\hat x}{\dd\hat t} &= \hat x-\hat x\hat y\\
\frac{\dd\hat y}{\dd\hat t} &= \hat x\hat y-\hat d\,\hat y
\end{align}

The other variant to simplify the second equation is to have both coefficients equal, so that the resulting factorization and thus the stationary points is/are normalized.
$$
ct_*x_*=dt_*\implies x_*=\frac{d}{c}.
$$
The normalized system is then
\begin{align}
\frac{\dd\hat x}{\dd\hat t} &= \hat x-\hat x\hat y\\
\frac{\dd\hat y}{\dd\hat t} &= \hat d\,(\hat x\hat y-\hat y)
\end{align}


问:
find the space left in a triangle with an almost inscribed circle
is there a way to find the length of the gap that is left if a triangle is almost inscribed by a circle then the top bit of the circle is cut off? (sorry I'm bad at explaining stuff) Here's what I mean: is there a way to find p?</p>

答:
HINT

$e$ is gap angle balance between side and tangent to circle of length T from A. The gap

$$ g =  T \tan e= \sqrt{p^2-\frac{d^2}{4}}\tan (\pi-x-y- 2 \sin^{-1} \frac{d}{2p})$$


问:
On primality of an element in the ring $\mathbb Z[\omega]$
The problem is: Show that if $p(\neq 3)$ is a prime in $\mathbb Z$ and $p \neq a^2+b^2-ab$ for any $a,b \in \mathbb Z$, then $p$ is a prime element in  the ring $R = \mathbb Z[\omega]$.

My approach: I was trying by contradiction that if $p$ isn't prime in $R$, then there exists  $(a+b\omega)$ and $(c+d\omega)$ such that $p\mid(ac-bd)$ and $p\mid(ad+bc-bd)$ but $p\nmid(a+b\omega)$ and $p\nmid(c+d\omega)$; but then how to use $p$ is prime in $\mathbb Z$?

Am I approaching correctly?

A small hint is warmly appreciated.

答: If $x|p$, where $x \in \mathbb{Z}[\omega]$, then $x \bar{x} | p \bar{p} = p^2$. Since $x \bar{x} \in \mathbb{Z}_+$, you have only three options for $x \bar{x}$. Try to rule out these three posibilities (for example, when $x \bar{x}=1$, $x$ is a unit, so our alleged factorisation wasn't quite proper).


问:
For what $z\in\mathbb{C}$ is $\sum_{n=0}^\infty \frac{z^n}{(1+z)^n}$ convergent

For what $z\in\mathbb{C}$ is $\sum_{n=0}^\infty \frac{z^n}{(1+z)^n}$ convergent?

I tried the ratio test, $\frac{u_{n+1}}{u_n}=\frac{z^{n+1}}{(1+z)^{n+1}}\frac{(1+z)^n}{z^n}=\vert\frac{ z}{1+z}\vert$ I'm not sure what I can say about this now, since $z$ is complex.

答:
If you write $w=\frac{z}{1+z}$ you get the power series $\sum w^n$ which is (trivially) convergent for $|w|<1$.

So your series converges whenever $$\left|\frac{z}{z+1}\right|<1$$

Can you finish?

HINT $|z|<|1+z|\iff d(z,0)<d(z,-1)$.


答:
Chebyshev said it, and I say it again - There is always a prime between $n$ and $2n$.

N. J. Fine's description of Erdos's new proof of Betrand's postulate.

Title: Is every prime number less than twice the previous prime number?

Body: And if so, how do you prove it? (for example 7 is less than 2 times 5, 11 less than 2 times 7, and so on).</p>


问:
Showing that an open unit disk is homeomorphic to the surrounding space
I was wondering that can I show that an open unit disk in $\mathbb{R}^n$ is homeomorphic to the surrounding space $\mathbb{R}^n$ by arguing that all of the half-axes of the space are homeomorphic to the half-axes of the unit disk. By half-axes of the space I refer to the two parts, $(-\infty, 0], [0, \infty)$ of each coordinate axis, and by half-axes of the open unit disk I refer to the intervals $(-1, 0], [0, 1)$ along all of the coordinate axis of the surrounding space.

So essentially I would argue that by some univariate functions $f$ and $-f$ all $[0, 1) \approx [0, \infty)$ and $(-1, 0] \approx (-\infty, 0]$. Then the mapping $g(x_1, x_2,\dots,x_n)$ from the $n$-dimensional open disk to the $\mathbb{R}^{n}$ could be constructed by considering the sign of the arguments, so that if $x_i \in [0, 1)$ then the $i$th coordinate of the image of $g$ is determined by $f$, and if $x_i \in (-1, 0]$ then it is determined by $-f$.

答:
You say "if $x_i\in[0,1)$". But not all such coordinates combined give a point in an open disk (assuming the standard Euclidean norm). For example for $n=2$ the point $(0.9,0.9)$ does not belong to the unit disk. Your approach would be ok for $(-1,1)^n$, but not for unit disk.

The simpliest method for $B\to\mathbb{R}^n$ homeomorphism is to map $v\mapsto f(\lVert v\rVert)\cdot v$ where $f:[0,1)\to[0,\infty)$ is a homeomorphism, say $f(x)=\frac{\pi}{2}\cdot\tan(x)$. That's because it has a simple inverse: $v\mapsto f^{-1}(\lVert v\rVert)\cdot v$. As a nice bonus: this approach works with any norm.


问:
Three questions that I failed to approach properly (statements, true or false).
I have these three questions: one about power series (true or false), and one about a statement about derivative and continuity (that I'm still not sure why it's true), and one about limits, that ruined my confidence in solving these questions and I'll explain my thought process for each one, I will appreciate any tips about how to approach these type of questions so I can improve myself.

True or false: If  $\sum^\infty_{n=0}a_nx^n$  converges exactly at $[-2,2)$, then $\sum^\infty_{n=0}a_n(x^2+1)^n$ converges exactly at $(-1,1)$.

$\textbf {My Work}$: I said let $t=x^2+1$, then $\sum^\infty_{n=0}a_nt^n$ converges at $[-2,2)$, now $-2\le t < 2$, then $-3\le x^2 < 1$, Now I know that $x^2$ can never reach $-3$, for some reason I went with that and inferred that $x$ can be whatever in the negative side $(-\infty,0$], ( I guess I understand my mistake right now but I will put this here maybe I can get some tips about how I approached the question).

Second question: If the right and left derivatives of $f(x)$ exist at $0$, then $f(x)$ is continuous at $0$.

$\textbf {My Work}$:  I know that if the derivative exists at point, then that means the function is continuous at that point, but what went through my mind is this: If the right and left derivatives exist but they're not equal, that means $f(x)$ isn't differentiable at $x=0$, which means it's not surely continuous, what surprised me that this is true, still can't get why.

Third and last question about limits: let $f(x)$ be differentiable for all $x$, then which of these is true?

a) if $\lim_{x\to\infty}\frac {f(x)}{x}=\infty $ ,then $\lim_{x\to\infty}f'(x)=\infty$.

b) if $\lim_{x\to\infty}f(x)=1$ then $\lim_{x\to\infty}f'(x)=0$.

c) if $\lim_{x\to\infty}f'(x)=8$ then $\lim_{x\to\infty}f(x)=\infty$.

$\textbf {My Work}$:

a) I didn't know how to approach, it looked very true but it was false in the end.

b) I couldn't find a counter example, but from intuition, from drawing a graph, a function can converge to a limit in a weird way (like it goes up and down fast), and the derivative limit might not exist, couldn't find a counter example (thought of functions with $sin$ or $cos$ but didn't succeed).

c) It's the answer, but I couldn't figure out why it's true rather than intuition which fails me most of the time sadly.

It would mean alot to me to get tips about my thought process and how I should approach these questions, (especially the third one), any tip is really appreciated, thanks in advance to everyone.

答:
(1) We want to know which values of $x$ will make the second series converge. If $x\le1$ or $x\ge1$, then $x^2+1\ge2$, so we're plugging a value into the first series that's not in $[-2,2)$, so it diverges. On the other hand, if $x\in(-1,1)$, then $x^2+1\in[1,2)$, which is a subset of $[-2,2)$, so we're plugging in a value that makes the series converge. In other words, $(-1,1)$ is exactly the set of $x$ values for which $x^2+1\in[-2,2)$. This question isn't really about power series at all, just sets and functions.

(2) You're right that $f$ might not be differentiable at $0$, but it is necessarily continuous. The left and right derivatives both take into account the value $f(0)$ (recall the limit definition of the derivative), and for both derivatives to exist, $f(x)$ must approach $f(0)$ as $x$ approaches $0$ from either side. That's exactly what it means for $f$ to be continuous at $0$.

(3) Your idea for (b) is exactly right! A counterexample is $f(x)=1+\sin(e^x)/x$. For a similar reason, (a) is false: picture something like $f(x)=x^2+\sin(e^x)$. Your mental image for (c) is probably right too. Since $f'(x)$ converges to 8, the slope of $f$ at some point becomes larger than 7 and stays that way forever, so $f$ increases without bound.


问:
Projective Special Linear and Projective Special Orthogonal Groups
Ergodic Theory with a view towards Number Theory, Chapter 9, Page 279, 280.

Lemma 9.1. The action of $\mathrm{PSL}_2(\mathbb{R}$) on $\mathbb{H}$ defined by (9.1) has the following properties.

(1) The action is isometric, meaning that
$$\mathsf{d}(g(z_0),g(z_1))=\mathsf{d}(z_0,z_1)$$
for any $z_0,z_1\in\mathbb{H}$ and $g\in\mathrm{PSL}_2(\mathbb{R}$). Moreover, the action of $\mathrm{PSL}_2(\mathbb{R})$ on $\mathrm{T}\mathbb{H}$ defined by the derivative $\mathrm{D}g$ of the action of $g\in\mathrm{PSL}_2(\mathbb{R})$ on $\mathbb{H}$ preserves the Riemannian metric.

(2) The action is transitive on $\mathbb{H}$: given any two points $z_0,z_1\in\mathbb{H}$ there is a matrix $g\in\mathrm{PSL}_2(\mathbb{R})$ with $g(z_0)=z_1$.

(3) The stabilizer
$$\mathrm{Stab}_{\mathrm{PSL}_2(\mathbb{R})}(\mathrm{i})=\{g\in\mathrm{PSL}_2(\mathbb{R})\mid g(\mathrm{i})=\mathrm {i}\}$$
of $\mathrm{i}\in\mathbb{H}$ is the projective special orthogonal group
$$\mathrm{PSO}(2)=\mathrm{SO}(2)/\{\pm I_2\}$$
where
$$\mathrm{SO}(2)=\left\{\left(\begin{array}{rr} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{array}\right)\mid\theta\in\mathbb{R}\right\}.$$
Notice that property (3) gives an identification
$$\mathbb{H}\cong\mathrm{PSL}_2(\mathbb{R})/\mathrm{PSO}(2),$$
and under the identification the coset $g$ $\mathrm{PSO(2)}$ is sent to $g(\mathrm{i})$.

Where $$\mathbb{H}=\{x+iy \in \mathbb{C}, y>0\}.$$
How do we have that $$\mathbb{H} \cong\mathrm{PSL}_2(\mathbb{R}) / \mathrm{PSO}(2).$$

Is $\mathrm{PSO}(2)$ a normal subgroup of $\mathrm{PSL}_2(\mathbb{R})$? (Of course, it is a subgroup).

答:
If $H\le G$ then the coset space $G/H$ carries a (left) $G$-action, $g(xH)=(gx)H$.

If $G$ is a topological group and $H$ is closed, there is a quotient topology on $G/H$, even a smooth structure if $G$ is a Lie group.

In any action $G\curvearrowright\Omega$, the orbit $\mathrm{Orb}(\omega)$ of a point $\omega\in\Omega$ is also a $G$-set contained in $\Omega$.

The orbit-stabilizer theorem says $\mathrm{Orb}(\omega)\cong G/\mathrm{Stab}(\omega)$ are equivalent $G$-sets for any $\omega\in\Omega$. Indeed, there is a $G$-equivariant one-to-one correspondence $g\omega\leftrightarrow g\mathrm{Stab}(\omega)$. Under nice conditions, this will also be a homeomorphism (indeed, diffeomorphism if applicable).

The OST applies here for $G=\mathrm{PSL}_2(\mathbb{R})$, $\Omega=\mathbb{H}$, $H=\mathrm{PSO}(2)$.

And no, $\mathrm{PSO}(2)$ is not normal in $\mathrm{PSL}_2(\mathbb{R})$ (can you show this?).


问:
Additive group of rational numbers

Let $\mathbb{Z}[\frac{1}{p}]$ be the additive group of rational numbers of the form $mp^n$ where $m$, $n$ are elements of $\mathbb{Z}$ and $p$ is a fixed prime. Describe $\text{End}(\mathbb{Z}[\frac{1}{p})]$ and $\text{Aut}(\mathbb{Z}[\frac{1}{p}])$.

Unfortunately I have never done this type of exercise and do not know where to start: I have great difficulties in general because I do not know how to describe these types of groups. Can you help?

Sorry if my English is not exactly correct.

答:
This answer is basically combining what Jack Schmidt has remarked and condensing Lubos Motl's answer.

First we notice that $\mathbb{Z}[\frac{1}{p}]$ has a ring structure. This allows us to view $\mathbb{Z}[\frac{1}{p}]$ as a free $\mathbb{Z}[\frac{1}{p}]$-module of rank 1. We show that any endormorphism of $\mathbb{Z}[\frac{1}{p}]$ as an abelian group is in fact an endomorphism of  $\mathbb{Z}[\frac{1}{p}]$-modules as well.  For a commutative ring with unit, $\mathrm{End}_R(R)=R$ since it only depends on the image  of $1$.

Last, $\mathrm{Aut}(\mathbb{Z}[\frac{1}{p}])$ is all invertible elements in $\mathrm{End}(\mathbb{Z}[\frac{1}{p}])$. In other words, we look for the units in the ring $\mathbb{Z}[\frac{1}{p}])$.  One can easily show that the units are $p^{\mathbb{Z}}\times \langle \pm 1\rangle$.

Title: Additive group of rational numbers


问:
Chance that every player got five
Four players play a game with rolling dice. They play 100 rounds. What's the probability that there was a round where every of them got five?

My solution:
$$\sum_{k=1}^{100} {100 \choose k} (\frac{1}{6})^{400 \cdot k}(\frac{5}{6})^{400-4k}$$ where $k$ is the number of rounds where every of them got 5. Is it correct? I tried to do this with opposite event but it turned out to not be so trivial.

答:
I want at least one round.

Thus  the answer is simply

$$1-\Bigg(1-\frac{1}{6^4}\Bigg)^{100}\approx 7.4\%$$

Title: Chance that every player got five


问:
Finding the explicit formula of a given recurrence relation
This is the recurrence relation: $a_n=2a_{n-1}+2n, a_2=2$.The way we have found the explicit formulas in my course in the past was to notice some pattern either by finding several values or using the next terms to define $a_n$.However, I'm completely lost on the equation above.

答:
This is a linear difference equation with constant coefficients. You can get the solution as follows

Find the general solution of the homogeneous equation $a_n = 2a_{n-1}$, which yields $a_n^h = k 2^n$.

Find a particular solution to the equation: Try something similar to $2n$, say $a_n^* = \alpha n +\beta$. replacing into the equation you get $a_n^* = -2n-4$

The general solution is $a_n =a_n^h + a_n^* = k 2^n -2n -4$.

Using the initial condition you compute $k$, to get the particular solution
$$
a_n = 5\cdot 2^{n-1}-2n-4
$$


问:
Convexity of open balls in a metrizable subset of a locally convex t.v.s.
The following result is well known (Rudin, Functional Analysis, Theorem 1.24): If $X$ is a locally convex topological vector space with a countable local base then there is a compatible metric $d$ such that all open balls are convex.
My question is: Assume that $X$ is a locally convex t.v.s. which is not metrizable and $C$ is a compact, convex and metrizable subset of $X$. Is there a metric compatible with the topology of $C$ such that all open balls are convex?

答: $Y=\operatorname{span}(C)$ is a locally convex tvs which is still metrisable I think (using that $C$ has a countable base, being compact metrisable). So the first result would apply there.


问:
Given Orthocenter of a triangle and length of three segments, need to find area.
Let, $O$ be an orthocenter of the triangle $\triangle ABC$, where $AO = 7, AB = 9, CO = 4$. Find the combined area of the triangles $\triangle AOC$ and $\triangle AOB$. Options given are $7\sqrt{3}, 14\sqrt{3}, 21\sqrt{3}, 28\sqrt{3}$.

I started off by assigning $AE =x$. Observing $\Delta AOE$, I can say $OE = \sqrt{49-x^2}$.
Then, $BE = 9-x$ and $OB = \sqrt{130-18x}$. Furthermore, $BC$ would be $\sqrt{146+8\sqrt{49-x^2}}$. But not sure if I'm going in right direction.

答:
$$\triangle=[AOB]+[AOC]=\frac{1}{2}\cdot BC\cdot AO$$

So we need to find $BC=a$

We use the fact that perpendicular bisector from circumcenter to any side is half the length of segment joining the opposite vertex to the orthocenter.

In the diagram below, not to scale, we see that

$$\left( \frac{c}{2} \right)^2 + \left( \frac{OC}{2} \right)^2 = R^2 = \left( \frac{a}{2} \right)^2 +\left( \frac{AO}{2} \right)^2 $$
$$\Rightarrow a^2 = 4^2+9^2-7^2=48$$
$$\therefore \triangle=14\sqrt{3}$$


问:
Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?

Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.

答: You have two numbers $\sqrt{x+4}$ and $\sqrt{x+1}$.  Consider their squares, which are (x+4) and (x+1).  So, their squares differ by 3, with the larger square number equaling (x+4).  Do we know of two square numbers which differ by 3?  As you might recall, the square numbers belong to the sequence (1, 4, ...).  So, the larger square number (x+4) equals 4, and the smaller square number (x+1) equals 1.  Thus, x=0.</p>

Title: Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$


问:
If $A$ is compact, $\lambda\ne0$ and $y$ is fixed, is there a unique $x$ with $(\lambda-A)x=y$?
If $B$ is a compact linear operator on a Banach space $X$, we know that $\dim\mathcal N(1-B)\in\mathbb N_0$ and $\mathcal R(1-B)$ is closed.

Thus, if $\lambda\ne0$ and $$B:=\frac A\lambda$$ such that $\dim\mathcal N(1-B)=0$, then for all $y\in X$, there is at most one $x\in X$ with $(\lambda-A)x=y$. But can we guarantee that there always is such a $x$?

答:
A bounded operator $T$ is said to be Fredholm if $\text{dim}(\mathcal N(T))<\infty $,  and
$\text{codim}(\mathcal R(T))<\infty $.  In this case the index of $T$ is defined by
$$
  \text{ind}(T)=
  \text{dim}(\mathcal N(T))-\text{codim}(\mathcal R(T)).
  $$
It is a well known result,  sometimes called the Fredholm alternative,  that $I-B$ is a Fredholm
operator with
index zero  for every compact operator $B$.

Thus,  if $\mathcal N(I-B) = \{0\}$,  it follows that $\mathcal R(I-B) = X$ so, yes, the desired $x$ always exists.


Title: Finding the explicit formula of a given recurrence relation

Body: This is the recurrence relation: $a_n=2a_{n-1}+2n, a_2=2$.The way we have found the explicit formulas in my course in the past was to notice some pattern either by finding several values or using the next terms to define $a_n$.However, I'm completely lost on the equation above.

Body:
$$A_n=2A_{n-1}+2n$$
The solution of $A_n=2A_{n-1}$ is $A_n= C~~ 2^n$
Next, for $A_n-2A_{n-1}=2n$ take $A_n=pn+q$, we get
$(pn+q)-2[p(n-1)+q]=2n \implies p=-2, q=-4.$
The total solution is $A_n=2^nC-2n-4$. $A_2=2$ gives $C=5/2.$
WE finally get $A_n=5~. 2^{n-1}-2n-4.$


问:
Find the width of a triangle at a certain point
Is there a way to find the width of a triangle at a certain point? I'm not very good at maths so I don't know many maths words, so I'm going to explain it in a diagram.
Is there any way to find the length of p?

答: Suppose that $x,y,$ and $z$ are fixed values.  As the (presumably acute) angle between sides $x$ and $y$ increases, $p$ will increase.  Therefore, there is insufficient information to determine $p$.


问:
Prove that for any natural number $ \geq 2$ $ \langle x_1, ..., x_n \rangle = \langle y_1, ..., y_n\rangle $ iff $ x_1 = y_1, ..., x_n = y_n $
My first step was to figure out which set I am trying to show is $(\{2\}, F)$-inductive.
So I came up with this:

$ A = \{ n \geq 2 : \langle x_1, ..., x_n \rangle = \langle y_1, ..., y_n\rangle $ iff $ x_1 = y_1, ..., x_n = y_n\} $

From here, I figure I need to show first the base case, where n = 2. Or, prove that:

$ \langle x_1,x_2 \rangle = \langle y_1, y_2\rangle $ iff $ x_1 = y_1, x_2 = y_2$

My question isn't about this proof. But for reference I prove this by reducing it into set theory using Kuratowski's definition:

$ \langle x, y \rangle $ = {{x}, {x, y}}.

And then prove the relevant set identity using the principle of extensionality for sets.

I figure my next step is to prove the inductive step. Which I formulate as follows:

If $ n \in A $, then $n + 1 \in A $

As a conditional, I begin this proof by assuming the truth of the antecedent for some arbitrary n:

$  k \in A$.

Given the definition of A above, this gives me:

$ \langle x_1,..., x_k \rangle = \langle y_1,..., y_k\rangle $ iff $ x_1 = y_1 ,..., x_k = y_k$

And now show how from this we can get to this:

$ \langle x_1,..., x_k, x_{k+1} \rangle = \langle y_1,..., y_k, y_{k+1}\rangle $ iff $ x_1 = y_2 ,..., x_k = y_k, x_{k+1} = y_{k+1}$

This is the part I am a bit confused on. For in many of the examples of mathematical induction I've been able to track down, this step is executed using algebraic operations. But I'm not entirely sure what the analog of those operations are here.

I tried my best to mirror these examples and started by just supposing I can take both of the sequences and add an item to them. So, for example,

$ \langle x_1,..., x_k \rangle $

becomes

$ \langle x_1,..., x_k, x_{k+1} \rangle $

I then exploited the recursive definition of n-tuples to rephrase this as:

$ \langle \langle x_1,..., x_k \rangle x_{k+1} \rangle $

And finally, I exploited the results of the Kuratowski definition and rephrased this as:

$\{\{\langle x_1, ... , x_k\rangle\},\{\langle x_1, ...x_k \rangle , x_{k+1} \}\}$

And this is where I am a bit stuck. I feel like you can just keep applying the Kuratowski definition until you get rid of all the angle brackets. At which point you work through the proof as in the base case of establishing the identity of the two sets. But I am not sure again what operations are permissible or necessary to complete this step.

Going back to the very beginning, I guess I don't know what functions A is supposed to be closed under. Like which function takes me from 2 to 3 in A, from 3 to 4, etc.  and so on?

Thanks for any and all help. I really appreciate it.

答:
Well, for pairs define $(x,y) = \{\{x\},\{x,y\}\}$.
Then its clear that

$(x_1,x_2) = (y_1,y_2)$ iff $x_1=y_1$ and $x_2=y_2$.

Define $n$-tuples inductively as follows:

$(x_1,\ldots,x_n,x_{n+1}) := ((x_1,\ldots,x_n),x_{n+1})$, $n\geq 3$.

Suppose the result was proved for $n\geq 2$. Then

$(x_1,\ldots,x_n,x_{n+1}) = (y_1,\ldots,y_n,y_{n+1})$ iff by definition

$((x_1,\ldots,x_n),x_{n+1}) = ((y_1,\ldots,y_n),y_{n+1})$ iff by using equality of pairs,

$(x_1,\ldots,x_n) = (y_1,\ldots,y_n)$ and $x_{n+1} = y_{n+1}$ iff by induction hypothesis,

$x_1=y_1,\ldots,x_n=y_n,x_{n+1}=y_{n+1}$.


问:
Exp. Stability of perturbed system with temporally vanishing perturbation
I have a perturbation problem for which I can't find a fitting theorem in Khalil's Nonlinear Systems. Maybe someone can point me in the right direction:

Given a nominal system

$\dot x(t) = A(t)x(t)$

which is exponentially stable and a perturbed system

$\dot x(t) = A(t)x(t)+g(t)$

where $g(t)$ is bounded and as $t\rightarrow \infty$ converges exponentially to zero.

I imagine that the perturbed system is exponentially stable as well.

Any pointers appreciated.

答:
Ok, it is not true.

Consider $\dot x = -x +e^{-3t}$.

The solution is $x=c_1e^{-t}-0.5e^{-3t}$ with $c_1=x_0+0.5$.

So $\|x\|$ has a part that is independent from $x_0$, which means that for arbitrarily small $\|x_0\|$, the exponential stability condition breaks.


问:
Is the sum of the first N Laguerre polynomials (with alternating signs) always positive?
I have noticed that the following simple sum of Laguerre polynomials (weighted with alternating signs) seems to be positive for any $N$ when $x>0$:

$$\sum_{k=0}^{N}\;(-1)^{k}\;L_{k}(x)$$

More precisely, when $N$ is even, the sum is positive $\forall\;x$ and when $N$  is uneven, the sum has only one zero in $x=0$.

I have checked this numerically up to $N=10{}^{2}$.

I cannot find any reference to this sum in the literature. Are my observations correct?   If this is a known result, could someone point me in the right direction to find a reference or some documentation for this?

答:
Here is an additional remark.

Let us denote by
$$
g_N(x) := \sum_{n=0}^N (-1)^n L_n(x)
$$

1) When $x≤0$. First, since $L_n(x) = \sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{k!}$, we immediately have for $x≤0$
$$
g_N(x) = \sum_{n=0}^N \sum_{k=0}^n \binom{n}{k} \frac{|x|^k}{k!} ≥ 0
$$

2) When $x≥0$.

When $m = 2N+1$ is odd, by the answer of @mathreadler, one has
$$
g_{2N+1}(x) = \sum_{n=0}^N L_{2n}(x)-L_{2n+1}(x) = \int_0^x \sum_{n=0}^{N} L_{2n}(x)\,\mathrm d x
$$
Now, recall the formula $\sum_{n=0}^{m} L_{n} = L_m^{(1)} = -L'_{n+1}$ (see e.g. wikipedia). Thus since $\frac{1+(-1)^n}{2} = \mathbf{1}_\text{n even}$,
$$
2\,g_m(x) = \int_0^x \sum_{n=0}^{m} (1+(-1)^n)\, L_{n}(x)\,\mathrm d x =  \int_0^x  g_{m}(x) + L^{(1)}_{m}(x)\,\mathrm d x
$$
and so since $L_{2N+2}(0)=1$, for $m=2N+1$ any odd positive number, one has
$$
2\,g_{m}(x) = \int_0^x  g_{m}(t) \,\mathrm d t - L_{m+1}(x) + 1
$$
Taking the derivative, it tells that
$$
g_m' = \frac{1}{2} \,g_m + \frac{1}{2} L^{(1)}_{m}.
$$
Since $g_m(0)=0$, we can then solve this differential equation to get
$$
2\,e^{-x/2}\,g_m(x) = \int_0^x L^{(1)}_{m}(t) \,e^{-t/2}\,\mathrm d t
$$

$\bullet$ Similarly, when $n=2N$ is even, removing the first term $L_0(x) =1$ of the sum,
$$
g_{2N}(x) = 1- \sum_{k=1}^N L_{2k-1}(x)-L_{2k}(x) = 1-\int_0^x \sum_{k=1}^{N} L_{2k-1}(x)\,\mathrm d x
\\
= 1-\int_0^x \sum_{k=0}^{2N} \frac{1-(-1)^k}{2}\,L_{k}(x)\,\mathrm d x
$$
so
$$
g_n(x) = 1 + \frac{1}{2}\int_0^x (g_n - L_{n}^{(1)})
$$
and so
$$
g_n' = \frac{1}{2} (g_n - L_{n}^{(1)})
$$
and so since $g_n(0)=1$,
$$
2\,e^{-x/2}\,g_n(x) = 2 - \int_0^x L^{(1)}_{n}(t) \,e^{-t/2}\,\mathrm d t
$$
which is positive if $\int_0^x L^{(1)}_{n}(t) \,e^{-t/2}\,\mathrm d t < 2$.

3) Conjecture.

When $N\to\infty$, $2\,e^{-N|x|^2/2}\,g_N(N|x|^2)$ approaches $\mathbb 1_{[-2,2]}$.


问:
Conditional regression points
Suppose I have a model $$y=\beta_0+\beta_1x+\beta_3\sqrt{x}e_1,$$ where the "beta" coefficients are known and $e_1$ is Normal(0,1). Now I simulate $n=100$ points from that model and have the scatter-plot. The question is: how could I get (and plot) the points conditioned to a specific value of $x$ in R? That is, given the simulated points $(x_i,y_i)$, how could I extract those $y_i|x=0.5$, for example? Thank you very much in advance!

答: The probability to observe a specific $y$ given $x$ is $P(y|x)=N({(y-\beta_0 -x\beta_1) \over \beta_2 {\sqrt x}}),1)$. So the $y's$ you are looking for are ones for which this expression is as large as possible.

Title: Conditional regression points


问:
Stationary point on graph of $y=(\ln x)/x$
This is the question 8

I tried solving it for an hour but I don't know how to get to the answer,1/e and nature is maximum this is my working I don't know how to get further from here. How do I get to the coordinates (e,1/e)

Any help is very much appreciated. Thank you!

答:
Use the formula for the derivative of a quotient:
$$y'=\frac{x \ln'x-x'\ln x}{x^2}=\frac{x \cdot\cfrac1x-1\cdot\ln x}{x^2}=\frac{1-\ln x}{x^2}.$$

Some more details:

As the denominator is positive, the sign of the derivative is the sign of its numerator, namely

$f'(x)>0\iff 1=\ln\mathrm e>\ln x\iff \mathrm e> x$,

$f'(x)<0\iff 1 >\ln x\iff \mathrm e< x$.

Therefore, $f\,$ has a maximum at $x=\mathrm e$.


问:
Commutative Artinian ring with infinitely many ideals.
Intro: An immediate example of Artinian ring is any ring with finitely many ideals.
Recently, when I was thinking about Artinian rings, I realized that this is actually the only example I really know. I always implicitly assumed there are lots of Artin rings with infinitely many ideals, but I can't really find such a ring.

Question: Please, can you point out an example of unital, commutative Artinian ring with infinitely many ideals?

Non-commutative example is welcomed too, but commutative one is preferred.

My thoughts (here I consider only commutative case) Usually when I think about some concrete ring, it is an integral domain. But only Artin domain is a field so this is not a good way to go.

I've read that Artinian ring is a finite product of Artin local ring. So it seems to be a good idea to look at local Artin rings with the maximal ideal that is not principal. However, I am still not really used to localizations (good enough to understand some proof but not very experienced with actually using this technique) so I can't use this to actually produce a concrete example.

Your thoughts and examples are really appreciated. Thanks!

答: Take your favourite infinite field $k$ and consider the Artinian ring $k[x,y] / (x^2,xy,y^2)$. Then for any $c$ in $k$, there's an ideal $(x - cy)$.


Body:
The following manipulations yield a homogeneous linear recurrence relation with constant coefficients:

$$a_n=2a_{n-1}+2n$$

$$a_{n-1}=2a_{n-2}+2(n-1)$$

$$a_n-a_{n-1}=2a_{n-1}-2a_{n-2}+2$$

$$a_n=3a_{n-1}-2a_{n-2}+2$$

$$a_{n-1}=3a_{n-2}-2a_{n-3}+2$$

$$a_n-a_{n-1}=3a_{n-1}-5a_{n-2}+2a_{n-3}$$

$$a_n=4a_{n-1}+5a_{n-2}-2a_{n-3}$$

Can you take it from here?


问:
How to evaluate the definite integral $ \int_{-\pi/3}^{\pi/3} \sqrt{1-\cos^2(x)}dx $
I'm trying to evaluate this integral. Here are some steps: $$ \int_{-\pi/3}^{\pi/3} \sqrt{1-\cos^2(x)}dx = \\  =\int_{-\pi/3}^{\pi/3} \sqrt{\sin^2(x)}dx = \\ =\int_{-\pi/3}^{\pi/3} |\sin(x)|dx  $$ Now, how do I continue? I'm thinking about integrate two times the $\sin (x)$ between $0$ and $\pi/3$, but it isn't an even function. How can I do?

Body: This is $2\int_0^{\pi/3}\sin xdx=2(1-\cos(\pi/3))=1$.


问:
Test dependency for two arbitrary random variables
I want to know if there is a way to calculate test the dependency of two random variables $X, Y$ no matter what their distributions are.

I know that if the probability distribution is known, we can check the value of $W=P(X, Y)-P(X)P(Y)$ and if it is equal to zero then we conclude they're independent.

What if we don't know the distribution? Is there any general way to learn the probability distribution?

答:
There are several "non parametric tests" to verify dependence between rv's

Just as an example

$\chi^2$ test

Spearman's $\rho$

Kendall's $\tau$


问:
If M and N are Neighbourhoods of point p, prove that M union N is also Nbd. of p
I was going through the topic Neighbourhoods (Nbds) as part of Real Analysis and there i happen to ponder over some properties of Nbds, properties pertaining to intersections was discussed but no discussion of properties related to unions. i.e. if M and N are Nbds of p then M union N is Nbd of p.

I just need help to find out if such a property even exists and if it does then what is it's proof?
I couldn't find this property in any of the books i have.

答:
Neighborhood of p means open set containing p. An arbitrary union of open sets in $\mathbb{R}$ is open in $\mathbb{R}$ (the same holds for $\mathbb{R}^n$ and any topological space). To see this in the case of $\mathbb{R}^n$, $\mathbb{R}^n$ is a metric space also, so the typical definition for open set in $\mathbb{R}^n$ is that $U\subseteq \mathbb{R}$ is open if for any $x\in U$ there is a ball of some positive radius $\epsilon$ of x contained in U, that is:

$B_{\epsilon}(x) := \{y\in \mathbb{R}^n \mid\text{ } \mid \mid y - x\mid \mid < \epsilon\}\subseteq U$.

If you have a union of two open sets $U$ and $V$ and $a\in U\cup V$, then $a\in U$, so $\exists \epsilon > 0$ s.t. $B_{\epsilon}(a)\subseteq U\subseteq U\cup V$. Hence $U\cup V$ is open and contains $a$, so $U\cup V$ is a neighborhood of $a$.


问:
Inverse | Modulo | Power
Describe the inverse of $5$ modulo $18$ as a positive power of $5\pmod{18}$.

I've got that the inverse of $5$ is $11$, but is this question asking to find a $t$ such that
$$ 11=5^t\pmod{18}?$$

Title: Inverse | Modulo | Power

Body: mod $\,18\!:\ 5^3\! \equiv 5(25)\equiv 5(7) = -1 \,\Rightarrow\, 5^6\! \equiv (-1)^2\!\equiv 1,\ $ i.e. $\ 5\cdot \color{#c00}{5^5} \equiv 1,\,$ so $\, 5^{-1}\! \equiv \color{#c00}{5^5}$


问:
Inversibility of a block matrix in schur decomposition
Let $A'$ be a given, n × n, real, positive definite matrix partitioned as
follows:
\begin{pmatrix} A & B \\ B^T & C \end{pmatrix}

show that $C − B^TA^{−1}B$ is positive definite.
I know that I have to do something like this
. But when I started writing the solution, I come up with a question. Can I show that $A$ is invertible or is it a necessary condition?

答: If $A'$ is symmetric positive definite, its principal submatrix $A$ must also be symmetric positive definite, because $x^TAx=\pmatrix{x^T&0}A'\pmatrix{x\\ 0}>0$ for every nonzero vector $x$. Hence $A$ is necessarily nonsingular.


问:
Verifying some properties of Hamiltonian Graphs
I want to verify these two properties of Hamiltonian graphs:

Graph having multiple Hamiltonian cycles can have different cycle lengths or all are of same length?
i.e. All Hamiltonian cycles have same length or they can have different lengths.

Does Hamiltonian cycle is longest length cycle in graph?

For first property Complete graph comes to my mind but they have all same length Hamiltonian cycles.
For second property I think Hamiltonian is longest as it passes through every vertex.

Please can someone provide me more clarification about my ideas?

答:
Let's look at the definition of a Hamiltonian cycle:

It is a cycle  (i.e. closed path. You see, no repetition of vertices except for the starting point, no repetition of edges) that goes through all the vertices.

So if we fix $n$ to be the number of vertices, each Hamiltonian cycle has the length $n$.

Now suppose there is a cycle of length more than $n$. Then it has to visit some vertex multiple times, but this is in contradiction with the definition of a cycle.


问:
If $f$ is holomorphic at $z_0$, then $\partial f/\partial \bar{z}=0$ and $f'(z_0)=\partial f/ \partial z(z_0)=2\partial u/\partial z(z_0)$

If $f$ is holomorphic at $z_0$, then $\frac{\partial f}{\partial \bar{z}}=0$ and $f'(z_0)=\frac{\partial f}{\partial z}(z_0)=2\frac{\partial u}{\partial z}(z_0)$

Also if we write $F(x,y)=f(z)$, then $F$ is differentiable in the sense of real variables and $\det J_F(x_0,y_0)=\vert f'(z_0)\vert ^2$.

Where $\frac{\partial}{\partial z}= \frac{1}{2}(\frac{\partial}{\partial} z +\frac{1}{i}\frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \bar{z}}=\frac{1}{2}(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial }{\partial y})$ are differential operators.

This is from Stein and Sharkarchi, Complex analysis. Previous to this the book showed the Cauchy Riemann equations $\frac{\partial f}{\partial x}(z_0)=\frac{1}{i}\frac{\partial f}{\partial y}(z_0)$

I see that $\frac{\partial f}{\partial \bar{z}}(z_0)=\frac{1}{2}(\frac{\partial f}{\partial x}-\frac{1}{i}\frac{\partial f}{\partial y})=0$ directly from the cauchy riemann equations.

And that $f'(z_0)==\frac{\partial f}{\partial z}(z_0)$ also from cauchy riemann equations.

Then also $\frac{\partial f}{\partial z}=2\frac{\partial u}{\partial z}$

Which I think comes from this:

$\frac{1}{2}(\frac{\partial f}{\partial x}(z_0)+\frac{1}{i}\frac{\partial f}{\partial y}(z_0))=\frac{1}{2}\biggr((u_x (x,y),v_x(x,y))+i(u_y(x,y), v_y(x,y))\biggr)$

But you would write it as $u_x+iv_x+-iu_y+v_y=(u_x+v_y)+i(v_x-u_y)$ which by Cauchy Riemann equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

So I end up with $(u_x+v_y)+i(v_x-u_y)=2u_x-2iu_y$ which I thought I should have $2\frac{\partial u}{\partial x}=2u_x+2iu_y$

Im not sure if when $\frac{\partial}{\partial z}$ is written is this supposed to mean the partials written as a complex number with $z=x+iy$? Or does this mean the partial derivative as a vector of the component functions partial derivatives?

答:
We have (omitting the arguments for brevity)
$$
\frac{\partial}{\partial z} f = \frac{\partial}{\partial z}(u +iv) = \frac12 \left((u_x +iv_x) + \frac 1i (u_y + iv_y)\right) =
\frac12 \left(u_x +v_y + i(v_x-u_y)\right) \, .
$$
Using the Cauchy-Riemann equations this is equal to
$$
\frac12 \left(u_x +u_x + i(-u_y-u_y)\right) = u_x + \frac 1i u_y = 2\frac{\partial}{\partial z} u \, .
$$


问:
Convexity of open balls in a metrizable subset of a locally convex t.v.s.
The following result is well known (Rudin, Functional Analysis, Theorem 1.24): If $X$ is a locally convex topological vector space with a countable local base then there is a compatible metric $d$ such that all open balls are convex.
My question is: Assume that $X$ is a locally convex t.v.s. which is not metrizable and $C$ is a compact, convex and metrizable subset of $X$. Is there a metric compatible with the topology of $C$ such that all open balls are convex?

答:
Let
$
  s: X\times X\to X
  $
be the subtraction map, namely
$$
  s(x,y) = x-y, \quad \forall x,y\in  X,
  $$
and put
$$
  D:= s(C\times C) =  \{x-y: x, y\in  C\}.
  $$

We next claim that $D$ is homeomorphic to a quotient of $C\times C$.  Indeed, consider the equivalence relation $R$ on $C\times C$
according to which two points are equivalent iff they have the same image under $s$.   Since $s$ obviously respects $R$,  there is a unique
continuous map
$$
  \tilde s:(C\times C)/{R}\ \to \ D,
  $$
such that $s=\tilde s\circ  q$, where $q$ is que quotient map.

It is not hard to see that $\tilde s$ is a bijective map,
and since $(C\times C)/{R}$ is compact and $D$ is Hausdorff, we conclude that $\tilde s$ is a homeomorphism, thus proving the claim.

Every Hausdorff quotient of a compact metrizable space is itself metrizable, so $D$ is metrizable and hence also   first countable.

Noticing that  $0$ lies in $D$,  we may then find a countable open neighborhood base
for $0$ in the relative topology of $D$,  say  $\{V_n\}_n$.   Write each $V_n$ as $W_n\cap D$,  where $W_n$ is an open subset of
$X$, and choose an open,  convex,  balanced set $U_n$,  with
$$
  0\in U_n\subseteq W_n.
  $$

It is then evident that
$$
  V'_n := U_n\cap D \subseteq  W_n\cap D = V_n,
  $$
so the $V'_n$
form an open neighborhood base for $0$ in the relative
topology of $D$.

For each $n$, let $p_n$ be the Minkowsky functional associated to $U_n$ and consider the (not necessarily
Hausdorff), locally convex topology $\tau'$ on $X$ generated by the $p_n$.

We next claim that $(C, \tau ')$ is Hausdorff.  To see this, let $x$ and $y$ be  distinct points  in $C$,
and notice that
$$
  z:= y-x\in  C-C = D.
  $$
Choose an open set $A\subseteq X$, such that $0\in A$ and
$z\notin A$,  and  let $n$ be such that $V'_n\subseteq  A\cap D$.

Notice that $z\notin U_n$ since otherwise
$$
  z\in  U_n\cap D = V'_n\subseteq  A\cap D\subseteq A.
  $$
It follows that
$$
  1\leq p_n(z)=p_n(y-x),
  $$
so $x$ and $y$ may be separated in the topology generated by $p_n$, and hence also in $\tau '$, proving our
claim.

Clearly $\tau '$ is coarser than the default
topology $\tau $ of $X$,   so the identity  map
$$
  \text{id} : (C, \tau ) \to  (C, \tau ')
  $$
is continuous.

Given that $(C, \tau)$ is compact, and that we now know that $(C, \tau ')$ is Hausdorff, we have that
the above map is a homeomorphism.

In case $(X,\tau ')$ happens to be Hausdorff, an application of the result by Rudin mentioned by the OP would provide the
desired metric, but a slight adaptation of it also works even if $\tau '$ is not Hausdorff.  By that I mean that Rudin's
proof carries through without the assumption that $(X,\tau ')$ is Hausdorff, as long as we drop the
requirement that $d$ satisfies
$$
  d(x,y)=0\Rightarrow x=y,
  $$
that is, as long as we satisfy ourselves with a pseudo-metric, rather than a metric.

The resulting pseudometric $d$ would of course be compatible with the topological space  $(C, \tau ')$, and hence also $(C, \tau )$,
but since this is a Hausdorff space,  as seen above,  $d$ must be a metric when restricted to $C$.


问:
Definition of continuous time Markov Process/Chain
I have some doubts about the definition of continuous time Markov chains. The book I'm reading uses the following definition.

A random process $\{X(t) : t \ge 0\}$ is a Markov process if for all $0 \le s < t$, $\{x(u) : 0 ≤ u ≤ s\}$ and $x(t)$, $$P(X(t) ≤ x(t) | X(u) = x(u), 0 ≤ u ≤ s) = P(X(t) ≤ x(t) | X(s) = x(s)).$$

A Markov chain is a discrete value Markov process.

But for (continuous time) Markov chains I also have this definition:
A discrete value random process $\{X(t) : t \ge 0\}$ is a Markov chain if
$$t_0 < … < t_n ⇒ P(X(t_n) = x_n | X(t_{n-1}) = x_{n-1} ,…, X(t_0) = x_0) = P(X(t_n) = x_n | X(t_{n-1}) = x_{n-1})$$

Are these two formulations equivalent or is the first more restrictive?

答:
There are weird things that can happen for continuous time (so I prefer discrete time).

Here is one weird thing: Let $Z(t)$ be a 2-state homogeneous CTMC with state space $\{0,1\}$ and with transition rates $q_{01}=q_{10}=1$. Let $U$ be a random variable that is uniformly distributed over $[0,1]$ and that is independent of the $Z(t)$ process.   Define:
$$ X(t) = \left\{\begin{array}{cc}
Z(t) & \mbox{if $t \neq U$} \\
Z(100) & \mbox{if $t = U$} \end{array}\right.$$

Then $X(t)$ and $Z(t)$ have the same joint distribution when sampled over any deterministic times $0\leq t_0<t_1<...t_n$ (for any positive integer $n$). This is because, with probability 1, these times miss the crazy time $U$. So $X(t)$ satisfies your second definition.

However, $X(t)$ does not satisfy your first definition because if we know the sample path of $X(t)$ over an interval of time $t \in [0, 5]$, that includes knowledge of $X(U)$, which reveals the value of the future state $X(100)$.  But if we only know $X(5)$ then we have no knowledge of the future state $X(100)$.


问:
Minimizing trigonometric expression with the hint of substitution
I have asked this question earlier for minimization of this expression.

$$
\left[\cos ^{2}\left(\dfrac{\theta }{2}\right) +
\,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{4}\,\sin^{2}\left(\theta\right)\quad
\mbox{where}\quad 0 \le \gamma \le 1.
$$

But then I am not sure if we can reach any relevant minimization (closed form) from the strategy mentioned here. What is the way to minize this with the following hint?

Hint is given in the first question itself.

答:
You are minimizing$$\left(\frac{1+\cos\theta}2+\sqrt{1-\gamma}\,\frac{1-\cos\theta}2\right)^2+\frac\gamma4(1-\cos^2\theta).$$

After expansion, you get a quadratic trinomial in $\cos\theta$.

find the location of the minimum by cancelling the derivative; if $\cos\theta$ falls in $[-1,1]$ it corresponds to the global minimum.

otherwise, compute the trinomial for $\cos\theta=\pm1$ and keep the smallest value.


问:
Proving that a function is continuous in an interval
I need to prove that $f(x)=x^2 \cdot \sin(\frac{1}{x})$ is continuous in $(0,\infty  )$.

My attempt:
Let $\varepsilon>0$. We define $\delta=\varepsilon$ and then for each $x,y\in(0,\infty)$, such that: $\left|x-y\right|<\delta$, therefore:
$$\left|f(x)-(y)\right|=\left|x^2\cdot \sin\bigg(\frac{1}{x}\bigg)-y^2\cdot \sin\bigg(\frac{1}{y}\bigg)\right|$$Since $\sin(x)<x$ then: $$\left|x^2\cdot \sin\bigg(\frac{1}{x}\bigg)-y^2\cdot \sin\bigg(\frac{1}{y}\bigg)\right|<\left|x^2\cdot \frac{1}{x}-y^2\cdot \frac{1}{y}\right|=\left|x-y\right|<\varepsilon$$

Did I got it right?

答:
I would prove continuity as follows:

The function $f$ is continuous on $\mathbf R$, except possibly at $0$, as a product and composition of continuous functions.

We also see that $f$, which is defined on $\mathbf R^*$, has a continuous continuation at $0$, which means $\lim_{x\to 0}f(x)$ exists.

Indeed,
$$|f(x)|=x^2\biggl|\sin\frac1x\biggr|\le x^2\cdot 1=x^2\to 0.$$


问:
Scaling normal distribution by Y = 2X
I know that when you scale a normal random variable by $a$, you get $a\mu$ and $a^2 \sigma^2$. In this example I have $y=2x$.

$f{_x}(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^-{\frac{(x-\mu)^2}{2\sigma^2}}$

and my solution is

$f{_y}(y)=\frac{1}{\sqrt{8\pi\sigma^2}}e^-{\frac{(\frac{y}{2}-2\mu)^2}{8\sigma^2}}$

is it correct?

Thanks

答:
You actually did the job twice. Let's start from the beginning.
$$f{_x}(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$
Now we substitute $x=\frac{y}2$. We also have to divide the solution by 2, because the integral of the distribution over whole real axis has to stay 1.
$$f{_y}(y)=\frac12\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{\left(\frac{y}2-\mu\right)^2}{2\sigma^2}\right)=
\frac{1}{2\sqrt{2\pi\sigma^2}}\exp\left(-\frac{\frac14\left(y-2\mu\right)^2}{2\sigma^2}\right)=
\frac{1}{\sqrt{8\pi\sigma^2}}\exp\left(-\frac{\left(y-2\mu\right)^2}{8\sigma^2}\right)
$$
This is almost the same as you wrote but the $y$ isn't divided by 2.

Title: Scaling normal distribution by Y = 2X


问:
Complexity Of A Recurrence Summation.
I was trying to find all possible full trees as a recurrence formula and I found it but now I want to find the complexity of it as $\theta$ relation. I found that it is something like this.

$$f(n)=\sum_{i=0}^{n-1}f(i)*f(n-i-1) \text{ and } f(0)=1,f(1)=1$$

I know it is not exactly this function (I simplified it) but if I can learn how to solve this I can apply it to all possible full trees formula.

TLDR; I want to know how I can find time complexity of $f(n)$

答:
Turns out Catalan numbers as @kelalaka said was the answer. If modeled correctly many problems like number off all possible binary tree's could be looked in this Catalan numbers perspective.
here is more detail.</p>

Title: Complexity Of A Recurrence Summation.


问:
If $A$ is compact, $\lambda\ne0$ and $y$ is fixed, is there a unique $x$ with $(\lambda-A)x=y$?
If $B$ is a compact linear operator on a Banach space $X$, we know that $\dim\mathcal N(1-B)\in\mathbb N_0$ and $\mathcal R(1-B)$ is closed.

Thus, if $\lambda\ne0$ and $$B:=\frac A\lambda$$ such that $\dim\mathcal N(1-B)=0$, then for all $y\in X$, there is at most one $x\in X$ with $(\lambda-A)x=y$. But can we guarantee that there always is such a $x$?

答:
Let $ \lambda \ne 0$. If $ \lambda $ is an eigenvalue of $A$, then

$$0< \dim \mathcal{N}(\lambda -A) < \infty,$$

since $A$ is compact. If $y \in \mathcal{R}(\lambda -A)$, then there is $x_0 \in X$ with $(\lambda -A)x_0=y$.

Then we have $(\lambda -A)(x+x_0)  =y$ for all  $x \in  \mathcal{N}(\lambda -A).$


答:
So you started off correctly:
$$\int_{-\pi/3}^{\pi/3}\sqrt{1-\cos^2x}\,dx=\int_{-\pi/3}^{\pi/3}|\sin x|\,dx$$
but notice that $|\sin x|$ is an even function since it is the absolute value of an odd function, which gives you:
$$2\int_0^{\pi/3}|\sin x|\,dx$$
Now since $\sin x$ is positive in $0\le x\le\pi/3$ this is just:
$$2\int_0^{\pi/3}\sin x\,dx=2\left[-\cos(x)\right]_0^{\pi/3}=2(1-\cos(\pi/3))$$
and since $\cos(\pi/3)=0.5$ your integral evaluates to $1$

Title: How to evaluate the definite integral $ \int_{-\pi/3}^{\pi/3} \sqrt{1-\cos^2(x)}dx $

Body: I'm trying to evaluate this integral. Here are some steps: $$ \int_{-\pi/3}^{\pi/3} \sqrt{1-\cos^2(x)}dx = \\  =\int_{-\pi/3}^{\pi/3} \sqrt{\sin^2(x)}dx = \\ =\int_{-\pi/3}^{\pi/3} |\sin(x)|dx  $$ Now, how do I continue? I'm thinking about integrate two times the $\sin (x)$ between $0$ and $\pi/3$, but it isn't an even function. How can I do?


问:
Runge-Kutta method and step doubling
I am studying Runge-Kutta and step size control and came up with a few doubts. Because they are related with this integration method, I will divide it in two parts. First, allow me to introduce the problem.

$1^{st}$ part - questions about Runge-Kutta Method

Consider a $2^{nd}$ order Runge-Kutta with general form:

$k_{1}=hf(x_{n},y_{n})$

$k_{2}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{1})$

$y_{n+1}=y_{n}+k_{2}+O(h^{3})$

where $f(x_{n},y_{n}) = y'(x_{n})$

Now, if we are to consider $4^{th}$ order Runge-Kutta we would get

$k_{1}=hf(x_{n},y_{n})$

$k_{2}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{1})$

$k_{3}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{2})$

$k_{4}=hf(x_{n}+h,y_{n}+k_{3})$

$y_{n+1}=y_{n}+\frac{1}{6}k_{1}+\frac{1}{3}k_{2}+\frac{1}{3}k_{3}+\frac{1}{6}k_{4}+O(h^{5})$

This leads me to the following questions:

I understand why the 2nd order method has the $y_{n+1}$ indicated above. Yet, shouldn't the $4^{th}$ order version of the method be expressed only as $y_{n+1}=y_{n}+k_{4}+O(h^{5})$ given that $k_{4}$ implicitly has the values of $k_{1}$ to $k_{3}$?

Why does the $4^{th}$ order version have those fractional coefficients?

$2^{nd}$ part - questions about step-size control

Consider the exact solution for an advance from $x$ to $x+2h$ by $y(x+2h)$ and the two approximate solutions by $y_{1}$ (one step 2h) and $y_{2}$ (two steps each of size $h$). Considering the $4^{th}$ order method we have:

$y(x+2h)=y_{1}+(2h)^{5}\phi+O(h^{6})+...$

$y(x+2h)=y_{1}+2(h)^{5}\phi+O(h^{6})+...$

The difference between these estimates permits estimating the truncation error:

$\Delta=y_{2}-y_{1}$

Which we then use to improve the numerical estimate of the true solution:

$y(x+2h)=y_{2}+\frac{\Delta}{15}+O(h^{6})$

This brings me to the following questions:

In the two expressions of y(x+2h) at the top, does $\phi$ include the terms $k_{1}...k_{4}$ in the first part of this post?

Why is the final expression $5^{th}$ order, if the original problem was $4^{th}$ order?

Where does the coefficient $\frac{1}{15}$ come from?

Thank you for all the insight!

答:
Answer to Part 1

The fractional coefficients in RK4 sum to 1. This is in part to satisfy what is known as the order conditions of the integrator. Basically, we want RK4 to be a 4th order method. For this to be true, it has to satisfy a special relationship using the coefficients of its Butcher Tableau (see at the top here), namely

$$\mathbf{b}^T A^kC^{l-1}\mathbf{1} = \frac{(l-1)!}{(l+k)!}$$

where $A$ is the coefficients in the main part of the tableau, $\mathbf{b}$ is a vector of the $b$ coefficients (in RK4, this is 1/6, 1/3, 1/3, 1/6), and $C$ is a diagonal matrix formed by the $C$ coefficients.

We need to craft these coefficients to get $0$-stability. It is provable that if the method is $0$-stable, it will converge to the order of accuracy -- in this case, order 4, or $h^4$.

The reason that we carry $k_1$, $k_2$, etc. through to the final solution is because $k_4$ is not simply a sum of the previous $k$'s. Each $k$ can be thought of as a projection of a finite difference. The RK method essentially averages these projections in a non-standard way. If you do out the math formally, you find that with these weights, you get optimal accuracy for your number of function evaluations.

Title: Runge-Kutta method and step doubling


问:
expected value of failures before success in n experiments
let $Y \sim B(15, p)$ be the number of successful experiments out of 15 and let $X$ be the number of failed experiments before the first success. I need to calculate $E[X|Y=1]$ and get a final answer without $p$.

I defined $W$ as the number of experiments until the first success so $W \sim G(p)$ hence $X=W-1$ so $E[X|Y=1] = E[W|Y=1 - 1] = E[W|Y=1] - 1$ but when I try to calculate $E[W|Y=1]$ I get
$$E[W|Y=1] = \sum_{i=1}^{15}ip(1-p)^{i-1}=\sum_{i=1}^{15}\binom{i}{1}p(1-p)^{i-1}$$ and I don't know how I can get a solution without p from this

Body:
$$\mathbb{E} (X|Y=1) = \sum\limits_{i = 0}^{14} i \cdot \mathbb{P} (X=i|Y=1) =  \sum\limits_{i = 0}^{14} i \cdot \frac{ \mathbb{P} ( (X=i) \cap (Y=1) )}{  \mathbb{P} (Y=1)  }\  $$
$$\mathbb{P} ( (X=i) \cap (Y=1) ) = p \cdot (1-p)^{14}, \ \forall i \in \{0,1, \dots, 14\}, \ \ \ \mathbb{P} (Y=1) = 15p\cdot(1-p)^{14}.$$
Therefore,
$$\mathbb{E} (X|Y=1) = \frac{ \sum\limits_{i=0}^{14} i}{15} = 7 $$


问:
Solution Verification: If $\{I_k\}_{k = 1}^n$ (open, bounded intervals) cover $\mathbb{Q} \cap [0, 1]$, then $\sum_{k = 1}^n m^\ast(I_k) \geq 1$.
I have a proof of Problem #8 from Royden & Fitzpatrick's Real Analysis but I am suspicious whether it works, because I can't find any similar approaches to the problem. So the set up is:

Let $\{I_k\}_{k = 1}^n$ be a finite collection of open, bounded, non-empty intervals (for $n \in \mathbb Z_+$) that covers $\mathbb Q \cap [0, 1]$. Then show that $\sum_{k = 1}^n m^\ast (I_k) \geq 1$.

I would appreciate if someone verified (i) my general approach to the problem and (ii) the proof itself. If the proof is too cumbersome, verifying my general approach is sufficient. Also, any improvements/simpler approaches are welcome.

Tentative Approach: My general idea is that $\{I_k\}_{k = 1}^n$ "essentially" covers the whole of $[0, 1]$. In fact, I believe: the set of uncovered points $[0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be a subset of the right (or left) endpoints of the $I_k$s. Is this correct or am I missing a simple counterexample? Assuming it is, we move on:

Tentative Proof Sketch: As the $I_k$s are open, bounded, non-empty intervals, write each as $(a_k, b_k)$ for reals $a_k, b_k$. So let $x \in [0, 1] \setminus \bigcup_{k = 1}^n I_k$ and note that $0 < x < 1$. Consider the two sets:
\begin{gather*}
 A_\ast(x) = \{a_k \in \mathbb R : x \leq a_k,\ 1 \leq k \in \mathbb Z \leq n\} \\
 B^\ast(x) = \{b_k \in \mathbb R : b_k \leq x,\ 1 \leq k \in \mathbb Z \leq n\}
\end{gather*} $A_\ast(x)$ is non-empty. Indeed, one of the intervals $I_j = (a_j, b_j)$ covers $1$ (for $1 \leq j \in \mathbb Z \leq n$). Hence, $x < 1 < b_j$; but then, $x \leq a_j$, otherwise $x \in I_j$ would be true. Similarly, $B^\ast(x)$ is also non-empty. So, as these are finite sets, $a_\ast := \min(A_\ast(x))$ and $b^\ast := \max(B^\ast(x))$ exist, and, furthermore $b^\ast \leq x \leq a_\ast$.

Now, in the case of $x = b^\ast$, we are done, as $x$ is then a right endpoint of one of the $I_k$s. The remaining case: $b^\ast < x$ is impossible. Indeed, for a contradiction, assume $b^\ast < x$. Then $\max\{b^\ast, 0\} < x \leq \min\{a^\ast, 1\}$, and, by the density of the rationals, there is some rational $q$ such that $\max\{b^\ast, 0\} < q < x \leq \min\{a^\ast, 1\}$. So, since $q \in \mathbb Q \cap [0, 1]$, some $I_k$ contains the $q$ i.e. $a_k < q < b_k$. In particular, we have $b^\ast < q < b_k$ so that $b_k \not\in B^\ast(x)$. Therefore $x < b_k$ by definition of $B^\ast(x)$, and $a_k < q < x < b_k$ as a result. But this means $x \in I_k$, which is not possible. Hence, every $x \in [0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be the right endpoint of some $I_k$ i.e. $x \in B := \{b_k \in \mathbb R : 1 \leq k \in \mathbb Z \leq n\}$. So $[0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be a finite set $\{x_i\}_{i = 1}^{n'}$ (for $n' \in \mathbb N$), as $B$ is finite.

The rest of the proof is trivial and therefore omitted: I essentially throw in the singleton sets $[x_i, x_i]$ of uncovered points into our existing interval collection $\{I_k\}_{k = 1}^n$. The new collection obviously covers $[0, 1]$. So by properties of $m^\ast$ (finite subadditivity, monotonicity, interval-length preservation) the conclusion follows.

答:
Your argument is correct. Here is an alternative approach using induction.

Let us prove by induction the slightly more general result:

Let $\{I_k\}_{k = 1}^n$ be a finite collection of open, bounded, non-empty intervals (for $n \in \mathbb Z_+$) that covers $\mathbb Q \cap [a, b]$. Then show that $\sum_{k = 1}^n m^* (I_k) \geq b-a$.

Proof: Case $n=1$. Then $I_1 \supset \mathbb Q \cap [a, b]$ and it is clear that $\sum_{k = 1}^1 m^* (I_k)=m^* (I_1) \geq b-a$.

Now suppose we have proved the result for any collection of $n$ open bounded non-empty intervals.

Let $\{I_k\}_{k = 1}^{n+1}$ be a finite collection of open, bounded, non-empty intervals that covers $\mathbb Q \cap [a, b]$. Then there is at least one  $k\in \{1, \dots n+1\}$ such that $b \in I_k$. Since the collection is finite, we can re-index the collection so that this interval is numbered as $I_{n+1}$ (and such re-indexing does not change the value of $\sum_{k = 1}^{n+1} m^* (I_k)$).

Now, there are $c, d \in \mathbb R$, such that  $I_{n+1} =(c,d)$. Since $b \in I_{n+1}$, we have $c < b < d$.  We have now two possibilities:

if $c< a$ then, $[a,b] \subset (c,d)= I_{n+1}$ and it is immediate that
$$\sum_{k = 1}^{n+1} m^* (I_k) \geq  m^* (I_{n+1}) >  b-a$$

if $a \leq c$ then we have that $\{I_k\}_{k = 1}^{n}$ covers  $\mathbb Q \cap [a, c]$. Then, by the induction hypothesis, we have that $\sum_{k = 1}^n m^* (I_k) \geq c-a$. So we have
$$\sum_{k = 1}^{n+1} m^* (I_k) = \sum_{k = 1}^n m^* (I_k) + m^* (I_{n+1}) \geq (c-a)+(d-c) = d-a  > b-a$$


答:
No.

First of all, it does not make sense to denote the random variable and the variable for the density function both as $x$.

If you have a random variable $X\sim N(\mu,\sigma^2)$, the density function for $X$ is
$$
f_X(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}
$$
If you let $Y=2X$, then $Y\sim N(\mu',\sigma'^2)=N(2\mu,4\sigma^2)$ and thus the density function for $Y$ is
$$
f_Y(y)=\frac{1}{\sqrt{2\pi\sigma'^2}}e^{-\frac{(y-\mu')}{2\sigma'^2}}
=\frac{1}{\sqrt{2\pi\cdot 4\sigma^2}}e^{-\frac{(y-2\mu)}{2\cdot 4\sigma^2}}
$$

Title: Scaling normal distribution by Y = 2X

Body:
I know that when you scale a normal random variable by $a$, you get $a\mu$ and $a^2 \sigma^2$. In this example I have $y=2x$.

$f{_x}(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^-{\frac{(x-\mu)^2}{2\sigma^2}}$

and my solution is

$f{_y}(y)=\frac{1}{\sqrt{8\pi\sigma^2}}e^-{\frac{(\frac{y}{2}-2\mu)^2}{8\sigma^2}}$

is it correct?

Thanks


问:
Why isn't sgn'(0) = 0? Related to the theorem lim f'(c) = C => f'(c) = C
I don't know the name of this theorem but I learned it in class and it says: If function f is defined at (a, b) and there exists c such that a < c < b, and function f has derivatives on (a, c) U (c, b), then if
lim f '(x) = C when x -> c then f '(c) exists and f '(c) = C. Now I know that sgn(x) doesn't have a derivative at x = 0, but sgn'(x) = 0 when x is not equal to 0. And it is defined at these points so it has left and right limits when x -> 0+/- and they equal 0. So it has limit when x -> 0, i.e. lim sgn'(x) = 0 when x -> 0 so why isn't sgn'(0) = 0 rather then being undefined?</p>

答: That theorem requires you to assume $f$ is continuous at $c$. But $\text{sgn}$ is not continuous at $0$ (so of course it cannot be differentiable at $0$). In fact this example shows you why you need to assume continuity at the point.


问:
Finding convergence of the next function: $f(x)=\sum_{n=1}^\infty\frac{(\ln n)^3}{n}x^n$
How can i find whether the next function converges: $f(x)=\sum_{n=1}^\infty\frac{(\ln n)^3}{n}x^n$?

I thought about this question for quite a while, What's the trick?

答:
The ratio test here will work nicely:

Recall that we apply the ratio test to the the entire summand:

$$a_n=\left({(\ln n)^3\over n}|x|^n\right)$$

and doing so here will give you the result $\;1\cdot |x|.\;$ So the series will converge $\forall x: |x| \lt 1$, and will diverge $\forall x:|x| > 1$.

What happens at $x = +1, x = -1\,$ remains to be determined, so be sure to test for convergence/divergence at $\;x = 1,\;$ and also at $\; x = -1.\;$


问:
x mod 2 = x mod p, where p odd prime. What can be said about x and p.
What sort of solutions do we get. What can be said about x and the odd prime p?
Any insights?</p>

答:
I am afraid there is not much to be said if I understand your question correctly.

Counting modulo two is just determining whether the number is even or odd.

So if we have $x \equiv 1 \pmod{2} $ then it tells it is odd. Now $x \equiv 1 \pmod{p}$, so $x$ has to be of form $x=kp+1$ where $k$ is some integer. But we also know that $x$ is odd, so we conclude that $k$ is even.

Similarly you discuss for $x$ even (i.e. $x \equiv 0 \pmod{2}$).

In conclusion: you have some information about $x$ but now information about $p$.