问:
Has the congruence $x^3\equiv 84 \mod (139)$ more than one solution?
Has the congruence $x^3\equiv 84 \mod (139)$ more than one solution?

I could find $x=27$ by brute force. I could not find more solutions:

$x^{138}=(x^3)^{46}\equiv 1 \mod (139)$

Is this correct:

$x^3\equiv 1 \mod (139)$

If yes how should we continue to find more solutions?

答:
This is my experiment on this question:

Consider equation $X=k.a+r$. This equation can have infinite solutions for given R  and a. But the number of solutions are limited when $X=x^n$ , in this case magnitude of R is also limited to certain values.

A-The goal is to find out  whether for a number like a  there exist primitive roots for (mod a)  for a certain power.To find a primitive root of congruence $x^n\equiv r \mod (a)$:

1- take the power n of two integer $b^n$ and $c^n$  such that:$b^n< a < c^n $.

2-Find differences $r_1=b^n-a$ and $r_2=c^n-a$.

3- Check $R=r_1$ and $R=r_2$ for primitive roots.

Example: for $a=139$, $n=3$, the closest numbers are $b=5$ and $c=6$ such that $5^3<139<6^3$ and we have:

$r_1=5^3-136=-14$

We check these:

$x^3\equiv -14 \ mod 139$ gives primitive roots (68, 76, 134). $x^3\equiv 14 \mod 139$ gives (63, 71, 144).

Now notice that $84=6\times 14$ and $R=84$ gives primitive roots (22, 27, 90). That is in congruence $x^n\equiv R \ mod (a)$ if you find a number like b such that $b^n<a$ and $r_1=b^n-a$,  if there is primitive roots for $r_1$ then there will also be primitive roots for $R =\ k\cdot|r_1| $.

For $r_2= 6^3-139=77$ we have primitive roots (20, 113, 145).

B- The sum of primitive roots is equal to a; for example $20+27+90=139\equiv 0 \mod (139)$.

C- The product of primitive roots is congruent $R \ mod (a)$ ; for example $22\times 27\times 90=53460\equiv 84 \ mod (139)$

D- If there exist a primite root then there exist $n-1$ other roots, or for power n there exist n roots.

E- If there exist primitive roots then there exist infinite cyclic roots which are generated by this formula:

$x=x_p+ m\cdot a$; $m∈\mathbb N$

For example for 139 ;

$ (22, 27, 90)$, gives $(22+139m), (27+139m), (90+139m) $

D- For n>3 mod (a)  minus or plus a primitive root $x_1$  gives another primitive root. Examples:

$x^4\equiv 80 \mod (139)$, gives $(62,77, 201, 340)$

$77= 139-62$, $(62+139=201)$and $201+139=340$

$x^5\equiv 7 \mod (139)$, gives $(51, 190, 329, 468, 607), 746, 885, . . .$

$51+139=190$, $329+139=468$, $468+139=607$

$x^6\equiv 6 \mod (139)$, gives $(11,45,56,83,94,122), 150, 184, . . .$

$11+139=122$

we may conclude that if n is even primitive roots make a cyclic group in addition to cyclic group of all roots.


问:
Expected Value Problem with a 10 Sided Die
I have a $10$ sided die numbered $1-10$. I keep rolling it until it lands on a prime number. Let $X$ be the number of times I roll the die.

What is the expected value $E(X)$?

What I have so far is the probability of rolling a prime is $P(\text{prime}) = \frac{4}{10}$.

How do I calculate the expected value? I tried the formula: $$\sum_{s\in S} P(s) \cdot X(s) = P(1)X(1) + P(2)X(2) +P(3)X(3)+P(4)X(4)$$

Where $P(s)$ is the probability of rolling a prime and $X(s)$ is the number of rolls. I plug in $\frac{4}{10}$ for each $P$ and $X$ is incremented from $1$. I am getting $4$ and the answer is $\frac{10}4$. Can someone guide me to what is going wrong here?

ContentLicense: CC BY-SA 4.0

Body:
Hint

$$E(X) = \frac{4}{10} + 2 \cdot \frac{6}{10} \cdot \frac{4}{10} + 3 \cdot \left(\frac{6}{10}\right)^2 \cdot \frac{4}{10} +.... + (n+1) \cdot \left(\frac{6}{10}\right)^n \cdot \frac{4}{10} + ... \infty$$


问:
Inverting a Large Symmetric Matrix
I just did a past paper and came across this matrix that I need to get an inverse for. Obviously it has a lot of deliberate zeros to help but I do not know how to solve it?

答:
You should use matrix block multiplication.

$A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$

$B=\begin{pmatrix}2&3\\3&5\end{pmatrix}$

$C=\begin{pmatrix}1&0\\0&\cfrac{1}{2}\end{pmatrix}$

$M=\begin{pmatrix}A&0&0\\0&B&0\\0&0&C\end{pmatrix}$

So $M^{-1}=\begin{pmatrix}A^{-1}&0&0\\0&B^{-1}&0\\0&0&C^{-1}\end{pmatrix}$

And to inverse those matrix, you can use $X^{-1}=\cfrac{Adj(X)}{det(X)}$ where $Adj(X)=\,^t C$ with $C$ the matrix of cofactors. (http://en.wikipedia.org/wiki/Adjugate_matrix)

ContentLicense: CC BY-SA 3.0

Title: Inverting a Large Symmetric Matrix


问:
prove that f is a surjection
Let C be a partition of a given set A.

Prove that there exists a set B and $ f : A \rightarrow B$ such that

$ \quad $ (i) f is surjective

$\quad $ (ii) $ C = \{\{x \in A : f(x)=y\} : y \in B \}.  $

Let $ C = A_1 \cup A_2 ... \cup A_n$, n = infinity if A is infinite set.

Let f be a function s.t. $ \forall x,y \in A, \exists i \in \{1,2,....n\} $ where $ x,y \in A_i \iff f(x) = f(y) = i. $

$ B = \{1,2,...n\}. $

Take any $ y \in B,  f(x) = y, \forall x  \in A_y $. Therefore, f is surjective

I am not sure if this proof is correct?

答:
There are several problems with your attempt. First, $C$ is a set of subsets of $A$, not the union of those sets. Next, there is no reason to think that $C$ can be enumerated by the natural numbers: we might, for instance, have the partition

$$\big\{\{x,-x\}:x\in\Bbb R\big\}$$

of $\Bbb R$, which has uncountably many members. Finally, there may not be any natural way to index the members of $C$ by numbers of any kind. Fortunately, there is no need to do so: we can in a sense index them by themselves.

Just let $B=C$, and for each $a\in A$ let $f(a)$ be the unique member of $C$ containing $a$. Clearly $f$ is surjective, since every member of $C$ is non-empty. Let $c\in C$; then for each $a\in A$ we have $f(a)=c$ iff $a\in c$, so $c=\{a\in A:f(a)=c\}$. Thus,

$$\begin{align*}
C&=\{c:c\in C\}\\
&=\big\{\{a\in A:f(a)=c\}:c\in C\big\}\,.
\end{align*}$$

Title: prove that f is a surjection


答:
The key thing you are doing wrong is that you need $P(s)$ to be the probability of rolling a prime on turn $s$ and not before. So $P(1)=\frac{4}{10}$, $P(2)=\frac{6}{10}\times\frac{4}{10}$, etc. It appears you just calculated $P(s)$ as the probability of getting a prime on turn $s$.

You also need to add up all the terms, not just the first four (you are not guaranteed to get a prime within four goes).

Title: Expected Value Problem with a 10 Sided Die

Body:
I have a $10$ sided die numbered $1-10$. I keep rolling it until it lands on a prime number. Let $X$ be the number of times I roll the die.

What is the expected value $E(X)$?

What I have so far is the probability of rolling a prime is $P(\text{prime}) = \frac{4}{10}$.

How do I calculate the expected value? I tried the formula: $$\sum_{s\in S} P(s) \cdot X(s) = P(1)X(1) + P(2)X(2) +P(3)X(3)+P(4)X(4)$$

Where $P(s)$ is the probability of rolling a prime and $X(s)$ is the number of rolls. I plug in $\frac{4}{10}$ for each $P$ and $X$ is incremented from $1$. I am getting $4$ and the answer is $\frac{10}4$. Can someone guide me to what is going wrong here?


问:
Derivative of a function of the form $f=f(x,g(x))$
Consider a function of the following form $f=f(x,g(x))$. For the derivative of this function, we do the following.
$$
\frac{df}{dx} = \frac{df}{dx} \bigg\rvert_{g}+\frac{df}{dg} \frac{dg}{dx}
$$
I have the following question about the above equation.

Since $g$ itself is a function of $x$, does it even make sense to say "$f$ is a function of $x$ and $g$"?

Does the first term of the equation read "derivative of $f$ with respect to $x$ at a fixed $g$"? If so, how does this make sense since if $x$ is changed, $g$ also changes and the derivative is no longer calculated at a fixed $g$.

Please let me know if you need me to make myself more clearer. Thank you for answering.

答:
I would handle all this differently.

Let a function $f(x,y)$ of two arguments. We have the partial derivatives

$$f_x=\frac{\partial f}{\partial x},f_y=\frac{\partial f}{\partial y},$$ which are themselves functions of two variables.

Now for some $g(x)$, define $h(x):=f(x,g(x))$, and by using differentials,

$$\frac{dh}{dx}=f_x\frac{dx}{dx}+f_y\frac{dg}{dx}=f_x+f_yg'.$$

More precisely,

$$\frac{dh(x)}{dx}=f_x(x,g(x))+f_y(x,g(x))g'(x).$$

$h$ is a function of $x$.

the partial derivatives are taken formally, with the other argument held fixed.

The variation of $h$ has two terms: a direct one due to the variation of $x$ as the first argument of $f$, and an indirect one due to the variation of the second argument via $g$.


问:
Determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open
I want to determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open. Here is my attempt at the problem, but I am having some trouble with this idea so please point out any errors. Thank you!

Claim: ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open.

Let $x \in {\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$. By the definition of an intersection we have that $x$ is an element of every set. That is,
\begin{equation*}
0 < x < \frac{1}{n},  \forall n\in\mathbb{N}^+.
\end{equation*}
By the density of the rationals we have that there exists $r_1,q_{n}\in\mathbb{Q}$ such that
\begin{equation*}
0 < r_1 < x < q_{n} < \frac{1}{n}, \forall n\in\mathbb{N}^+.
\end{equation*}
That is, for every $x\in A$ there exists $h > 0$ such that
\begin{equation*}
x - h = r_1
\end{equation*}
and
\begin{equation*}
x + h = q_{n}
\end{equation*}
so we have that
\begin{equation*}
(x - h, x + h) = (r_1, q_{n}) \subset (0, \frac{1}{n}).
\end{equation*}
The set is open.

答: The set is open. Also closed. Indeed, it is the empty set.</p>


答:
For proving that a set $X$ is open, you need to show $\forall x\in X,\exists p,q$ such that $x\in(p,q)\subseteq X$, where $p,q$ are constants.

In your construction, $r_1,r_2$ depend $n$. A choice of $r_1,r_2$ that works for $n_1$ may not work for $n_2$.

It is true that $\forall x\in(0,1/n),x\in(r_1,r_2)\subseteq(0,1/n)$, which indicates that $(0,1/n)$ is open. But for given $r_1,r_2$, the set $(r_1,r_2)\not\subseteq\cap_{n\in\Bbb N}(0,1/n)$ as $\exists m\in\Bbb N$ such that $1/m<r_1$ so $(r_1,r_2)\not\subseteq(0,1/m)$.

In fact you can prove that your set is empty. For every $x>0,\exists n\in\Bbb N$ such that $1/n<x$ i.e. $x\notin(0,1/n)$ so $x\notin\cap_{p\in\Bbb N}(0,1/p)$.

Title: Determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open

Body:
I want to determine whether ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open. Here is my attempt at the problem, but I am having some trouble with this idea so please point out any errors. Thank you!

Claim: ${\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$ is open.

Let $x \in {\bigcap_{n=1}^\infty \left(0,\frac{1}{n}\right)}$. By the definition of an intersection we have that $x$ is an element of every set. That is,
\begin{equation*}
0 < x < \frac{1}{n},  \forall n\in\mathbb{N}^+.
\end{equation*}
By the density of the rationals we have that there exists $r_1,q_{n}\in\mathbb{Q}$ such that
\begin{equation*}
0 < r_1 < x < q_{n} < \frac{1}{n}, \forall n\in\mathbb{N}^+.
\end{equation*}
That is, for every $x\in A$ there exists $h > 0$ such that
\begin{equation*}
x - h = r_1
\end{equation*}
and
\begin{equation*}
x + h = q_{n}
\end{equation*}
so we have that
\begin{equation*}
(x - h, x + h) = (r_1, q_{n}) \subset (0, \frac{1}{n}).
\end{equation*}
The set is open.


问:
When the Jacobian is time-dependent, how to analyse stability?
Suppose the jacobian matrix of an equilibrium point is time-dependent, as in the second case of Carlos' answer on this question: Jacobian matrix of time dependant system.

Did you know about some reference to deal with stationary points on this case? How to measure stability? I mean, I have a stationary point $X$ and the Jacobian is $t$-dependent. When this does not occur, I calculate the Jacobian on $X$. But now it's not sufficient. Which $t$ should I choose to analyse stability?

Thank you in advance for any comment.

答:
For definition of stability for non-autonomous systems, see page 44 to page 45 of Mathematical Definition of Stability of Non-autonomous Systems.

For criterion for such stability relating to Jacobian, see Hartman–Grobman theorem, A generalization of Hartman's linearization theorem, and On the linearization theorem for nonautonomous differential equations.


答:
You can easily see that the given intersection is empty. Suppose that it isn’t. Then, there is an $x$ such that:

$$\forall n \in \mathbb{N}: x \in (0,\frac{1}{n})$$

However, we know, by the Archimedean Property, that:

$$\exists N \in \mathbb{N}: N > \frac{1}{x}$$

That is, $x > \frac{1}{N}$. So, $x \notin (1,\frac{1}{N})$ and that is impossible. Hence, the given intersection is empty. So, it is, indeed, open.

With regards to your specific argument, I can definitely say that it’s vacuously correct. On the other hand, you’ve introduced an “$A$” in there without explaining what that is. Moreover, we’re led to believe that there is a $h > 0$ such that $x+h$ and $x-h$ are both rational. This doesn’t have to be true because it would imply that $x$ is rational and you haven’t explained why that has to be true.

The gist of your argument (showing that every point in the given set has an open ball centered on it that is a subset of the given set) makes sense and is a good way to approach it. You should reformulate your argument so it’s more precise.


答:
There is an abuse of notation here, insofar as the letter $x$ is used for various purposes. In reality you have a function
$$(x,y)\mapsto f(x,y)$$
of two variables, giving, say, the temperature at the point $(x,y)$ in the plane. Furthermore you have a path
$$\gamma:\qquad t\mapsto \left\{\eqalign{x(t)&:=t \cr y(t)&:=g(t)\cr}\right.$$
in the plane. Now you want to know to deal with the temperature felt at the moving point, as a function of $t$. This is the function
$$\phi(t):=f\bigl(x(t),y(t)\bigr)\qquad\bigl(\> =f(t,g(t))\ \bigr)\ .$$
The chain rule gives
$$\eqalign{\phi'(t)&=f_x\bigl(x(t),y(t)\bigr)x'(t)+f_y\bigl(x(t),y(t)\bigr)y'(t)\cr  &=f_x\bigl(t,g(t)\bigr)\cdot1+f_y(t,g(t)\bigr)g'(t)\ .\cr}$$

Title: Derivative of a function of the form $f=f(x,g(x))$

Body:
Consider a function of the following form $f=f(x,g(x))$. For the derivative of this function, we do the following.
$$
\frac{df}{dx} = \frac{df}{dx} \bigg\rvert_{g}+\frac{df}{dg} \frac{dg}{dx}
$$
I have the following question about the above equation.

Since $g$ itself is a function of $x$, does it even make sense to say "$f$ is a function of $x$ and $g$"?

Does the first term of the equation read "derivative of $f$ with respect to $x$ at a fixed $g$"? If so, how does this make sense since if $x$ is changed, $g$ also changes and the derivative is no longer calculated at a fixed $g$.

Please let me know if you need me to make myself more clearer. Thank you for answering.


问:
Resnick's A Probability Path Exercise 10.7
Suppose that X1, X2 are i.i.d. exponential random variables with mean 1, i.e., their densities are given by $e^{-x}$ for x > 0. Compute

$E[X_1|X_1 + X_2]$.

$P(X-1 < 3|X-1 + X-2)$.

$E[X-1| \min(X-1, t)]$.

$E[X-1| \max(X1, t)]$.

My work:For the first one, I think since $X_1$ is measurable on $\sigma(X_1+X_2)$ then $E[X_1|X_1+X_2]=E[X_1]$. And this imply that $X_1$ is independent to $X_1 + X_2$, which is not make sense. I am wondering where I am wrong here.

答:
My work:For the first one

My solution for the first one

$$\mathbb{E}[X_1|X_1+X_2]+\mathbb{E}[X_2|X_1+X_2]=\mathbb{E}[X_1+X_2|X_1+X_2]=x_1+x_2$$

As per independence the rv are also exchangeable, thus

$$\mathbb{E}[X_1|X_1+X_2]=\frac{x_1+x_2}{2}$$


问:
What is law of random variable $ S_n = X_1 + \cdots + X_n $
We consider a sequence $ (X_n)_{n \leq 1} $ of mutually independent random variables defined on a probability space $ (\Omega, \mathcal{A}, \mathbb{P}) $  with value in $\{-1,1\}$ and such that, for all $ k \geq 1$ $$ \mathbb{P} (X_k = -1) = \mathbb{P} (X_k = 1) = \frac{1}{2} $$ For all $ n \geq 1 $, we set $ S_n = X_1 + \cdots + X_n $

Problem

What is law of random  variable $S_n$?

An idea please.

答:
Ignoring order, all drawings with $p$ positives are equiprobable with probability $\dfrac1{2^p}\dfrac1{2^{n-p}}=\dfrac1{2^n}$, and the values sum to $p-(n-p)=2p-n$. Observe that all sums are distinct, hence a function of $p$.

Now if we restore order, the $p$ positives can appear in $\displaystyle\binom np$ different ways, and this gives us the distribution:

$$\mathbb P(S_n=2p-n)=\binom np\frac1{2^n}$$ for $p=0,1,\cdots n$.


答:
Your formula should be

$E(X) = \sum_{x=1}^{\infty} xP(x)$

The probability that you only have one dice roll (x=1) is equal to the probability that you roll a prime on the first roll. That's $\frac{4}{10}$

The probability that you have two dice rolls (x=2) is equal to the probability that you first roll a non-prime and then roll a prime. That's $\frac{6}{10} \times \frac{4}{10}$

The probability that you have $n$ dice rolls is equal to the probability that you first roll a non-prime for the first $n-1$ times and then roll a prime. That's $(\frac{6}{10})^{n-1} \times \frac{4}{10}$.

So using that in the formula for $E(X)$ gives:

$E(X) = \sum_{x=1}^{\infty} x(\frac{6}{10})^{x-1} \times \frac{4}{10}$


问:
Locally bounded function definition
I don't get the idea of locally bounded functions. I know that locally bounded means that the specific function is bounded around every point. Also i know that all bounded functions are locally bounded by definition. However i can not prove that $f(x)=\frac{1}x$(unbounded) is not locally bounded or $f(x) = 2x+3 $(unbounded) is locally bounded.

答:
Let $D$ be a non-empty subset of $ \mathbb R$ and let $f:D \to \mathbb R$ be a function.

$f$ is locally bounded id for each $x_0 \in D$ there is a neighborhood $N_{x_0}$ and a constant $c_{x_0}$ such that

$$|f(x)| \le c_{x_0}$$

for all $x \in D \cap  N_{x_0}$.

For the function $f$ in your first example we have $D= \mathbb R \setminus \{0\}$ and $f$ is locally bounded.

For the function $f$ in your second example we have $D= \mathbb R $ and $f$ is locally bounded.

Title: Locally bounded function definition


问:
If f and g are maps such that $g \circ f$ and f are continuous, also $f$ is surjective. Then show that $g$ is continuous.
This question came into my mind while I was seeing the proof of Jordan separation theorem and came across through the following,

Let $f:[0,1] \rightarrow U$ be a loop in $U$ with $f(0)=f(1)=x_{0} ,$ where U is a topological space. Let $p:[0,1] \rightarrow S^{1}$ be the map given by
$$
p(x)=(\cos (2 \pi x), \sin (2 \pi x))
$$
Note that $p$ is continuous and is a bijection except at $0$ and $1$.

$p$ induces a map $h: S^{1} \rightarrow U$ such that
$$
h(x)=\left\{\begin{array}{ll}
f(0) & \text { if } x=(1,0) \\
f\left(p^{-1}(x)\right) & \text { if } x \neq(1,0)
\end{array}\right.
$$

Here $f=h \circ p$. How h is continuous?

答: For $C \subset U$ we have $h^{-1}(C) = p(f^{-1}(C))$. If $C$ is closed in $U$, then $f^{-1}(C)$is closed in $[0,1]$, thus compact. Hence its image under $p$ is compact, hence closed.


答:
Tl;dr:

$f(x,y)$ is a surface.

Substituting $y(x)$ i.e: y as a function of x means that we are moving on some curve in the input plane, corresponding to this curve we trace out a curved path on the surface.

So actually this function is now dependent on only one variable $x$, the rate of change as we 'change' $x$ is equal to the rate of change of the function if x and y were independent plus the rate of of change of the function with y if x and y were independent multiplied by the rate of change amplification which comes from the $y$ being parameterized by $x$

i.e:

$$ \frac{df}{dx} = \frac{ \partial f}{\partial x}  + \frac{ \partial f}{\partial y} \frac{dy}{dx}$$


问:
Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$?
As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:

$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$

The lower estimate

$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$

which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate

$$R_k<\ln\left(1+\frac{1}{k}\right)$$

is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?

Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.

答:
In the same spirit as @Robert Israel in his answer, we could use
$$\sum_{n=N+1}^\infty \frac{1}{n^2}=\psi ^{(1)}(N+1)$$ and use the series expansion of the rhs
$$\psi ^{(1)}(N+1)=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{30
   N^5}+\frac{1}{42
   N^7}-\frac{1}{30 N^9}+O\left(\frac{1}{N^{11}}\right)$$ which, since alternating, allows to propose as sharp bounds as required.

What could be interesting is to look at the expansion of @Robert Israel's results
$$\frac{1}{N+\frac 12} - \log \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{8
   N^4}+O\left(\frac{1}{N^5}\right)$$
$$\frac{1}{N+\frac 12} - \log \left(\frac{2N+3}{2N+1}\right) + \frac{1}{N+1}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}+\frac{1}{8
   N^4}+O\left(\frac{1}{N^5}\right)$$

Much less accurate : in my former group, we used for numerical purposes the simple double inequality
$$\sinh \left(\frac{1}{N+1}\right)<\psi ^{(1)}(N+1)<\frac{1}{2} \sinh \left(\frac{2}{N}\right)$$

Later, playing with Padé-like approximants, I found  better bounds (much better for the lower than for the upper)
$$\color{blue}{\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}<\psi ^{(1)}(N+1)}<\frac{2 N^2+7 N+7}{2 (N+1)^2 (N+2)}$$
$$\Delta=\psi ^{(1)}(N+1)-\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}=\frac{1}{45 N^5}+O\left(\frac{1}{N^6}\right)$$

Continuing for answering this question, a better one
$$\color{blue}{\frac{5N(1302 N^2+573 N+697) } {6(1085 N^4+1020 N^3+910 N^2+285 N+27 ) }<\psi ^{(1)}(N+1)}$$ and for this one $$\Delta=\frac{207}{15190 N^8}+O\left(\frac{1}{N^9}\right)$$


问:
Is $x^2$ differentiable at $0$ if we change the domain to $\{0\}\cup\{\frac1n\mid n\in\mathbb{N}\}$?

We define the function $x^2$ to be from $\{0\}\cup\{\frac1n\mid n\in\mathbb{N}\}$ to the real numbers. Is the function differentiable at $0$?

I think the answer is yes because the function is continuous at $0$, since $0$ is in the domain, and there is a sequence $(1/n)$ that converges to $0$, so $0$ is also a limit point.

Also, I think this function is not differentiable at any other point in the domain since we do not have any other limit point besides $0$.

The part that I'm not so sure about is that we only have the right-hand limit. There is no left-hand limit so I'm not sure if we can apply the definition of the derivative. But I remember in Darboux's theorem that functions can be defined to be differentiable at their endpoints.

答:
Let $f$ be your function. Yes, $f$ is differentiable at $0$ because:

$0$ is a non-isolated point of the domain of $f$;

the limit $\displaystyle\lim_{x\to0}\frac{f(x)-f(0)}x$ exists (it is equal to $0$).

And it is not differentiable anywhere else, since every other element of the domain of $f$ is an isolated point.


问:
an inequality over integers
Let $x$, $y$ and $z$ be integers different from 1.

Prove that if $x+y\leq xy$ and $y+z \leq yz$ then $x+z\leq xz$

My attempt:

$x+2y+z\leq (x+z)y$   $~~$so $2y \leq (x+z)(y-1)$

and I'm stack here.

I need just some hint.

Thanks.

答:
We have
$$\left\{\begin{aligned} & x+y \leqslant xy \\& y+z \leqslant yz \end{aligned}\right. \Rightarrow \left\{\begin{aligned} & (x-1)(y-1) \geqslant 1\\& (y-1)(z-1) \geqslant 1\end{aligned}\right. \Rightarrow (x-1)(z-1)(y-1)^2 \geqslant 1 > 0.$$
Therefore
$$ (x-1)(z-1) > 0 $$
But $x,\,y,\,z$ are integers, so
$$(x-1)(z-1) \geqslant 1 \Rightarrow x+z \leqslant xz.$$
Done.

Title: an inequality over integers


问:
Let  be a root of $f ∈ [, ]$ and $f$ is differentiable and $_$ be a sequence of iterates obtained in bisection method. Then show that ...
Then show that $ − _n = −f(_) /f'(_)$ where $ < _ < _n$

I know this statement represents the mean value theorem, I think It can be proved using proof of the mean value theorem but Are there any other methods to prove it using the bisection method?

答:
The result comes directly from Lagrange's theorem in the interval $[\alpha, x_n]$:

$$
\underbrace{f(\alpha)}_{=0}-f(x_n)= f'(\xi_n)(\alpha -x_n) \Leftrightarrow \alpha -x_n = -\frac{f(x_n)}{f'(\xi_n)}, \quad \xi_n \in (\alpha, x_n)
$$

This is not specific to the bisection method. If $x_n$ is a sequence that is supposed to converge to $\alpha$, this just tells you that  $|f(x_n)|< \varepsilon$ is not a good stopping criteria when $|f'|$ is small.


答:
It is known that for any $N,M\geq 1$
$$
\sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  = \frac{1}{N} - \frac{1}{{2N^2 }} + \sum\limits_{m = 1}^{M - 1} {\frac{{B_{2m} }}{{N^{2m + 1} }}}  + \theta _M (N)\frac{{B_{2M} }}{{N^{2M + 1} }},
$$
where $B_m$ denotes the Bernoulli numbers and $0<\theta _M (N)<1$ is a suitable number depending on $N$ and $M$. From this you can obtain, for example,
$$
\sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  > \frac{1}{N} - \frac{1}{{2N^2 }},\quad \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  > \frac{1}{N} - \frac{1}{{2N^2 }} + \frac{1}{{6N^3 }} - \frac{1}{{30N^5 }}.
$$

Title: Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$?

Body:
As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:

$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$

The lower estimate

$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$

which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate

$$R_k<\ln\left(1+\frac{1}{k}\right)$$

is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?

Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.


问:
Probability to have defective items
An engineer tests the quality of produced computers. Suppose that 5% of computers have
defects and defects occur independently of each other. Find the probability:

a) of exactly $3$ defective computers in a shipment of $20$ (event A);

b) that the engineer has to test at least $5$ computers in order to find $2$ defective ones (event B).

Cand somebody give me some ideas, please? I don't know what probabilistic model to use?

答:
You can model this problem as a Bernoulli trial. Since defects occur independent of each other, the probability that any chosen computer is defective is $\frac{5}{100}$.

Part 1
First, out of the $20$ computers, we choose any $3$ which can be defective with a probability $\frac{5}{100}$ each, and the rest of the $17$ computers are non-defective with a probability $\frac{95}{100}$ each.

Thus, the required probability is given by:

$$P = C(20,3)\left(\frac{5}{100}\right)^3\left(\frac{95}{100}\right)^{17}$$

Here I have used $C(20,3)$ to represent $\binom{20}{3}$

Part 2 For this part, you need to find the probability that at least $5$ tries are required to detect $2$ faulty computers. It will be a tad complicated to calculate this probability directly, so calculate the cases in which the $2$ defects are detected in $4$ tries or less. This too is a Binomial distribution. I encourage you to calculate this part on your own.

Title: Probability to have defective items


问:
Proving $(\mathbb{R}, \star)$ is a group using ordered field axioms of $\mathbb{R}$
For example, suppose we are trying to prove that $(\mathbb{R}, \star)$ is a group equipped with a binary operation $\star$. In the proof, is it permitted to use the fact that $\mathbb{R}$ is an ordered field? Does this extend to other common sets ($\mathbb{N}, \mathbb{Z_5}$, etc,) and their properties?

答:
As I mentioned in the comment, the operation $\star$ should be defined explicitly in order to determine whether $(\Bbb{R},\star)$.

For example, take $\star$ to be the usual multiplication.

Then $(\Bbb{R},\star)$ is clearly not a group since the set has $1$ as the identity but the element $0$ has no inverse in this set.

From the information that $\Bbb{R}$ is a field, you can only conclude that $(\Bbb{R},+)$ and $(\Bbb{R}-\{0\},\cdot)$ are groups.


问:
Among all shapes with the same area, a circle has the shortest perimeter
Is it true that "Among all shapes with the same area, a circle has the shortest perimeter"  ? and how to prove it ?</p>

答:
This is true and classical, and not quite as trivial as one might think. This also works in higher dimensions.

For a real-life argument - look at the shape water drops take! Or even more convincingly: look at soap bubbles (thanks to fgp for this example).

For an intuition why this should be true mathematically, try the following reasoning:

Let $F$ be the figure with the shortest perimeter from among figures with a given area. Try cutting the figure $F$ by a line $l$ in such a way that both parts have the same area. Call these parts $F_1 $ and $F_2$.

The illuminating observation is that you could now construct another figure $F'$ with same area as $F$. This is achieved by replacing $F_2$ by the reflection of $F_1$ with respect to $l$; i.e. take $F' := F_1 \cup R_l(F_1)$, where $R_l$ is the reflection with respect to $l$. Of course, we can also take $F'' := F_2 \cup R_l(F_2)$. Because the areas $\operatorname{Area}(F_1)$ and $\operatorname{Area}(F_2)$ are equal (and have the value $\frac{1}{2} \operatorname{Area}(F)$), the new figures have areas $\operatorname{Area}(F') = \operatorname{Area}(F'') = \operatorname{Area}(F) $. Likewise, we have $\operatorname{Perimeter}(F') = 2 \operatorname{Perimeter}(F_1)$ (here, we are not counting the part of the perimeter that lies on $l$), $\operatorname{Perimeter}(F'') = 2 \operatorname{Perimeter}(F_2)$ and of course:
$$\operatorname{Perimeter}(F) = \operatorname{Perimeter}(F_1) + \operatorname{Perimeter}(F_2) = \frac{\operatorname{Perimeter}(F') + \operatorname{Perimeter}(F'')}{2} $$
If $F_1$ and $F_2$ had different perimeters, then one of $F'$, $F''$ would have a smaller perimeter than $F$, and the same area. But we assumed that $F$ was optimal, so $\operatorname{Perimeter}(F_1) = 2 \operatorname{Perimeter}(F_2)$.

Let us push this idea a little further. Note that $F'$ and $F''$ are now again optimal, in the sense that they have the same area as $F$. This means, they have to be convex (if a figure is not convex, you can make a "shortcut" at its concavity, making the perimeter shorter and area larger; but it is easy to get rid of the extra area without increasing the perimeter, for instance by scaling). This means, among other things, that the lines perpendicular to $l$ through the intersection points with $F$ are tangent to $F$ (so if $F$ has a well defined tangent at this points, the tangent is perpendicular to $l$). This doesn't yet show that $F$ has to be a circle, but we see that $F$ has a lot of symmetry that only the circle seems to have.

With a little more work, we can almost get from here to the solution. Note that instead of working with $F$, can work with $F'$, i.e. we can assume that the figure is symmetric with respect to $l$. But this is equivalent to just considering $F_1$. Let us denote the two points where $l$ meets $F$ by $A,B$, and let $C$ be a point at the boundary of $F_1$. The lines $AB,BC,CA$ cut $F_1$ into three regions: triangle $ABC$ and two irregular shapes touching lines $AC$ and $BC$. We can do two things now. The idea from Cut the knot is to transform $ABC$ by changing the length of $AB$ Because $\operatorname{Area}(ABC) = \frac{1}{2} |AC| \cdot |CB| \cdot \sin \angle C$, we can assume that $\angle C = \pi/2$. Another possibility is to reflect the area adjacent to $AC$ an conclude that the tangent to $F$ at $C$ meets $AC$ at the same angle as the tangent at $A$ meets $AC$. Doing the same for $BC$ an counting angles, we again conclude that $\angle C = \pi/2$. But $\angle C = \pi/2$ means that $C$ lies on the half-circle based on $F_1$. So $F_1$ is a half-circle, an we are done!

Of course, there is some cheating involve in assuming the best figure exists. Also, at a few steps I assumed existence of tangents. This can be explained away, but Im not sure if it's of interest here.

For a detailed discussion, this seems relevant: Cut the knot (which I used it here strongly, but most of the ideas are mathematical folklore).


问:
Strange proposition in probability book for conditional probability
I found the following proposition (15.1) in the probability book of Heinz Bauer:

Let us given that $X$ is a numeric random variable on
  $(\Omega,\mathcal{A},P)$ which is non-negative / integrable. Then to
  any sub-algebra $\mathcal{C} \subset \mathcal{A}$ it exists a almost
  surly unique non-negative / integrable random variable $X_0$, which is
  $\mathcal{C}$ measurble and it holds $$ \int_C X_0 \,dP = \int_C X \,dP
 \quad \text{for all }C\in \mathcal{C}.$$ When $X$ is integrable and
  non-negative, then $X_0$ is almost surly non-negative.

I do not get the addition at the end

When $X$ is integrable and
  non-negative, then $X_0$ is almost surly non-negative.

Isn't that already clear? Because part of the above statement is if $X \geq 0 \, \Rightarrow \, X_0 \geq 0 $? I think he is actually proving the same statement twice in one proof with different techniques - or am I missing something?

答:
Being familiar with Bauer's notation from his book on Measure Theory, I believe Bauer wants to make the following distinction:

For a numerical (meaning extended real-valued for Bauer) non-negative (everywhere!) random variable $X$ (for such $X$ the integral always exist in $[0,\infty]$), Bauer defines the conditional expectation $X_0$ as a non-negative (everywhere!) numerical random variable which is $\mathcal{C}$ measurable and has the property

$$\int_C X_0 \,dP = \int_C X \,dP
 \quad \text{for all }C\in \mathcal{C}.$$

Such an $X_0$ exists (by the first part of the proof) and is $P$-a.s. unique, in the sense that if $X'_0$ is another non-negative (everywhere!) $\mathcal{C}$ measurable numerical random variable with the above property then $X=X_0$ $P$-a.s.

The requirement that $X_0$ be non-negative everywhere comes from the first part of the proof, which involves the existence of a density from the Radon-Nikodym theorem, and Bauer defines a density to be non-negative everywhere. Essentially this is the standard machinery in Lebesgue integration: you start with non-negative functions, and then for signed functions you consider the positive and negative parts. Hence the first part of the proof is a preliminary step for the following general case:

For a numerical (meaning extended real-valued) integrable  random variable $X$, Bauer defines the conditional expectation $X_0$ as a numerical integrable random variable which is $\mathcal{C}$ measurable and has the property

$$\int_C X_0 \,dP = \int_C X \,dP
 \quad \text{for all }C\in \mathcal{C}.$$

Such an $X_0$ exists (by the second part of the proof) and is $P$-a.s. unique, in the sense that if $X'_0$ is another $\mathcal{C}$ measurable integrable numerical random variable with the above property then $X=X_0$ $P$-a.s.

Now what about the case where $X$ is integrable and almost surely non-negative? Strictly speaking we cannot apply Bauer's first result since it concerns only everywhere non-negative random variables. This is why Bauer gives the extra argument at the end of proof.

But we can obtain this result from the first step as follows:

If $X$ is integrable and almost surely non-negative, then we can alter it, and call it say $X'$, on a null set so that $X'$ is measurable, non-negative everywhere and integrable with $\int X' dP=\int X dP$  (setting $X':=0$ on $\{X<0\}\in \mathcal{A}$ will do the job). Hence from the firt part we know there exist a non-negative (everywhere!) numerical random variable $X'_0$ which is $\mathcal{C}$ measurable and has the property

$$\int_C X'_0 \,dP = \int_C X' \,dP =\int_C X \,dP
 \quad \text{for all }C\in \mathcal{C}.$$

The special case $C=\Omega$ says that such an $X'_0$ is in addition integrable. But then from the uniqueness assertion of the second part we know that $X_0=X'_0$ $P$-a.s., that is $X_0\geq 0$  $P$-a.s.


问:
Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible
The question is this-

Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible.

I can show that there cannot exist a sub-matrix with rank more than the actual rank of the matrix. But I cannot show the other way around, i.e when the rank of the actual matrix is $r$, then there exist a principal sub-matrix of the same rank (though I can prove that there exist a sub-matrix with rank $r$).

I was thinking like this: if the rank of the matrix is $r$, we can find r linearly independent rows of the matrix, say $a_1, a_2, ..., a_r$ -th rows are linearly independent. Then the corresposding columns $a_1^t, a_2^t, ..., a_r^t$ are also linearly independent. But how to show that the submatrix that they produce is of rank r?

I hope my question is clear. Any hints or help would be highly appreciated.

答: Assume that the leading $r\times r$ minor is $0$. Then the $r$ columns of this minor are linearly dependent. Since the first $r$ rows  of the big matrix generate all the other rows, the same linear relation will hold for the first $r$ columns of the big matrix, contradiction.


问:
Confusion over calculus notation (differentials/derivatives)
I have read from multiple sources that dy/dx is not to be interpreted as a ratio as the idea of 'dy' and 'dx' themselves will lead to logical difficulties.

However, I have seen in many areas (e.g. thermodynamics) where notations involving dx etc. are used constantly. For example, in http://en.wikipedia.org/wiki/Total_differential, it seems to tread df/dt as a ratio by multiplying dt on both sides to cancel out the denominator.

So in what cases are we allowed to manipulate the notation as if they are quotients?

答:
Counterexample:

Consider the polar coordinates:

$$x=r\cos\theta$$
$$y=r\sin\theta$$
Then $$\dfrac{\partial x}{\partial r}=\cos\theta ,\dfrac{\partial x}{\partial \theta}=-r\sin\theta $$
However, what is $\dfrac{\partial r}{\partial x}$? is it $\dfrac{1}{\frac{\partial x}{\partial r }}$?

No:

$$x^2+y^2=r^2\Rightarrow 2x+0=2r\dfrac {\partial r}{\partial x}\Rightarrow $$
$$\dfrac {\partial r}{\partial x}=\frac{x}{r}=cos\theta $$
Can you find the other partial derivatives?

Generally, you are not allowed to treat differentials as fractions. Even in the example in Wikipedia, the rule that holds is the Chain Rule, and not a rule of fractions(http://en.wikipedia.org/wiki/Chain_rule).

In case of 1 variable, the differentials are used like fractions, but in more varialbes it is completely wrong to do that.


问:
Confidence interval for $n$ of $X$~Bin$(n,0.3)$
I want to calculate a 90% confidence interval for $n$ (from $X$~Bin$(n,0.3)$) where the number of successes is known ($x=12$). I have found the smallest $n$ so that $P(X \le 12)\le 0.05$ is 61 and the largest $n$ so that $P(X \ge 12)\le 0.05$ is 25. However I can't find any examples online or in my textbook that construct CIs for $n$, only ones that do so for $p$ so I am a bit lost on where to go from here. Any hints would be greatly appreciated.

Update: I think I must use some pivot variable. For example, I know that $0.9 = P(-1.64 \le N(0,1) \le 1.96)$. Also, for large $n$, $X$~Bin$(n,p)$ has an approximate $N(np, np(1-p))$ distribution. If I let $\hat n=\frac{X}{p}$ denote the estimator of $n$, then $\hat n$ has an approximate $N(n, \frac{n(1-n)}{p})$ distribution (although I'm not 100% sure about this latter part). Then, I should be able to say $\frac{\hat n -n}{\sqrt{n(1-n)/p}}$~$N(0,1)$ and so I can substitute this into my first equation and get: $$0.9 = P(-1.64 \le \frac{\hat n -n}{\sqrt{n(1-n)/p}} \le 1.96)$$ and then rearrange for $\hat n$ by itself in the middle of the inequality. My only concern is that $\hat n$ does not an approximate $N(n, \frac{n(1-n)}{p})$ distribution rendering this incorrect. Does anyone have any ideas?

Update 2: Just realised the $n(1-n)$ will be negative (since $n \ge 12$) and so we will get an imaginary number in the square root, thus my proposed solution is wrong. Back to square one :(

答:
With n large enough (here $n$ looks like around $\frac{12}{0.3}=40$, so far so good) you can approximate

$$Bin(n;0.3)\approx N(0.3n;0.21n)$$

Thus your interval becomes

$$\Bigg| \frac{12-0.3n}{\sqrt{0.21n}}\Bigg|\leq 1.64$$

Solving this double inequality (it's an easy quadratic inequality) it results to me

$$[27;59]$$

I think that this post can be useful for your reasoning


问:
The function $\Psi:G \to G$ with $g \mapsto g^m$ is an automorphism.
Let $G$ be an Abelian group of order $n$ and for a positive integer $m$ we have that $\text{gcd}(m,n)=1$.

Then the function $\Psi:G \to G$ with $g  \mapsto g^m$ is an automorphism.

My try:

First I show that the function is homomorphism:
Let $g_1,g_2 \in G$ and consider $\Psi(g_1g_2)=(g_1g_2)^m$ and this is true since $G$ is a group,besides it's Abelian and so $\Psi(g_1g_2)=g_1^{m}g_2^{m}=\Psi(g_1)\Psi(g_1)$.

The function is clearly surjective.

To show that it's injective we need to show $\text{ker}(\Psi)=\left\{e_G\right\}$, if we assume there exists $e_G\ne g  \in\text{ker}(\Psi)$ then $\Psi(g)=e_G=g^m$,but I don't know how to continue.

Any help is appreciated.

答:
Any $f\colon X\to X$ is bijective if it is surjective/injective when $X$ is finite. In this case, you can also use the following. Since $\gcd(m, n) = 1$ there are integers $r, s$ such that $rm+sn = 1$, so if $g^m = 1$, then $g^{rm+sn} = g^{sn} = 1$ because $|G| = n$, but at the same time $g^{rm+sn} = g^1 = g$, hence $\Psi$ is injective.

To obtain surjectivity, observe that by $rm+sn=1$, we have for any $g\in G, (g^r)^m=g$.


问:
Intuition behind the construction
This is one of my old homework questions, and my instructor gave a solution. I think I understand why this is true, but its intuition is not obvious. I mean this is kinda magic :) I couldn't think in that way. If you share your ideas or give different answers, it would be great.

Here is the question:

Let $(\mathfrak{M}_n)_{n\in \mathbb{N}}$ be a family of infinite well-orderings, considered in $\mathcal{L}_{ord}=\{<\}$. Let $U$ be a non-principal ultrafilter on $\mathbb{N}$, and let $\mathfrak{M}_U$ be the ultraproduct of the $\mathfrak{M}_n$ with respect to $U$. Prove that there is strictly decreasing sequence in $\mathfrak{M}_U$ of length $\aleph_1$. In particular, $\mathfrak{M}_U$ is not a well-ordering.

This is the sketch proof:

Wlog, we may assume each $\mathfrak{M}_n$ is $(\mathbb{N},\leq)$. We claim that if the sequence $f_i\in \prod \mathfrak{M}_n$ be monotone and unbounded, then there is $f^*\in \prod \mathfrak{M}_n$ monotone and unbounded such that $ [f^*] <_U [f_i] $ for all $i$. From this, we can get $(f_{\alpha})_{\alpha < \omega_1}$, decreasing in $<_U$.

To prove this, we will make sure for each $i$, $\{n| f^*(n)<f_i(n)\}$ is cofinite so that it will be in the ultrafilter $U$, and we are done.

Set $a_0=0$, let $a_k$ be the least such that $a_k>a_{k-1}$ and

$(\forall n \geq a_k) f_0(n), \cdots, f_{k-1}(n)>k$. (1)

Set $f^*(n):=$ least $k$ s.t. $n\geq a_k$. Then we have for $n\in [a_k,a_{k+1})$; $f^*(n)=k$, $f_i(n)>k$ for $i<k$ by (1). So $f^*(n)< f_i(n)$.

Thanks in advance.

答:
It’s hard for me to know what to say, because to me that does seem the natural thing to do: if you have only countably many functions, you can take care of them (i.e., get ‘under’ them) one at a time — not completely, but from some point on, which is good enough. Natural or not, the basic idea is a pretty standard one that you will likely see again.

It might seem a little more natural if you saw a simpler application of the same idea.

Proposition. ${^\omega}\omega$ is the family of functions from $\omega$ to $\omega$. Define a relation $<^*$ on ${^\omega}\omega$ by $f<^*g$ iff $\{n\in\omega:f(n)\ge g(n)\}$ is finite. There is a family $F=\{f_\alpha:\alpha<\omega_1\}\subseteq{^\omega}\omega$ such that $f_\alpha<^*\beta$ whenever $\alpha<\beta<\omega_1$.

Note that $f<^*g$ says that $f(n)<g(n)$ for almost every $n\in\omega$, where almost all means all but finitely many; we might say that $f$ is almost strictly less than $g$. The proposition then says that there is an almost strictly increasing $\omega_1$-sequence in ${^\omega}\omega$. This may at first seem surprising, since there clearly is no strictly increasing $\omega_1$-sequence in ${^\omega}\omega$. But it turns out that almost gives us a great deal of leeway.

The idea of the proof is to construct the functions $f_\alpha$ recursively — one at a time, so to speak — in such a way that when we construct $f_\alpha$, we ensure that $f_\xi<^*f_\alpha$ for each $\xi<\alpha$. We’re able to do this because there are only countably many functions $f_\xi$ with $\xi<\alpha$.

Say there are countably infinitely many of them, and we temporarily enumerate them as $\{g_n:n\in\omega\}$ instead of $\{f_\xi:\xi<\alpha\}$. The idea is to define $f_\alpha$ so that

$f_\alpha(k)>g_0(k)$ for all $k\in\omega$,

$f_\alpha(k)>g_1(k)$ for all $k\ge 1$,

$f_\alpha(k)>g_2(k)$ for all $k\ge 2$,

and so on. This is actually pretty easy: just let

$f_\alpha(0)=g_0(0)+1$,

$f_\alpha(1)=\max\{g_0(1),g_1(1)\}+1$,

$f_\alpha(2)=\max\{g_0(2),g_1(2),g_2(2)\}+1$,

and so on. At each $k\in\omega$ we can ensure that $f_\alpha$ ‘rises above’ one more of the functions $g_n$, and since there are only countably many of those functions, we can force $f_\alpha$ to be above each of them eventually. It’a bit like the diagonal argument for proving the uncountability of the reals: we have countably infinitely many ‘things to take care of’, and we have just enough things to define — here the values $f_\alpha(k)$ — to ‘take care of’ each of them.

Of course and so on won’t do for a proper proof, but now that we have the basic idea, writing it up properly is mostly a matter of experience and practice. Here’s one possible version.

Proof. For $n\in\omega$ let $f_n(k)=n$ for each $k\in\omega$; clearly $f_m<^*f_n$ whenever $m<n<\omega$.1 We construct $f_\alpha$ for $\omega\le\alpha<\omega_1$ by recursion. Suppose that $\omega\le\alpha<\omega_1$, and $f_\xi$ has been defined for each $\xi<\alpha$. We temporarily re-index $\{f_\xi:\xi<\alpha\}$ as $\{g_n:n\in\omega\}$ and define $f_\alpha$ by setting $$f_\alpha(k)=1+\max\{g_i(k):i\le k\}$$ for each $k\in\omega$. If $\xi<\alpha$, there is some $i\in\omega$ such that $f_\xi=g_i$, and $f_\alpha(k)>g_i(k)=f_\xi(k)$ for all $k\ge i$, so $f_\xi<^*f_\alpha$. Clearly we can carry out this construction as long as $\alpha$ is countable, so in this way we can construct the desired family $F$. $\dashv$

1 It isn’t actually necessary to start by defining the functions $f_n$ for $n\in\omega$, but it makes matters just a little simpler by letting me start the recursion at $\alpha=\omega$: that way I don’t have to worry about whether $\{f_\xi:\xi<\alpha\}$ is finite or countably infinite. This doesn’t really make the argument any simpler, but it does make the explanation a bit simpler.

Title: Intuition behind the construction


问:
Number of essentially different Rubik's cube permutations, or rather arrangements
I can find everywhere (e.g. wikipedia, ruwix.com and MIT) the information that the standard $3\times3$ Rubik's cube can be scrambled in $4.3 \times 10^{19}$ different configurations. These are calculated keeping the center cubies in a defined position (say, white on top and green in front), so that rotations of the cube are already ruled out. But I cannot find anywhere the number of "essentially different" arrangements, in the sense I'm going to (try to) explain.

Suppose that from the solved state I perform just one $90°$ clockwise rotation of the face on the right (the move called "R"). Let me call the new configuration A. Then, starting again from the solved state, I perform a single $90°$ clockwise rotation on the face on the left (move "L"), obtaining configuration B. Now, among the $43$ quintillions, A and B are counted separately as different permutations; but I don't consider them to be essentially different: after a $180°$ rotation about the vertical axis and a reassignment of the colors (swapping green $\leftrightarrow$ blue and red $\leftrightarrow$ orange, if the cube respects the standard color scheme) the two configurations become the same. I'd call this arrangment "cube with one face turned $90°$ clockwise". In fact another way to see the equivalence is that, after a proper rotation of the cube in my hands, I can reach the solved state from either A or B with the same sequence of moves (in this simple case, one single counterclockwise rotation).

There are four other configurations that I consider to be equivalent to A and B, being essentially the same arrangement, for a total of six (one for each of the faces that can undergo the initial rotation). Yet I cannot just divide by $6$ the $43$ quintillion figure hoping to get the number of arrangements, since not all of them has the same multiplicity. For example if, from the starting position, I turn clockwise both the right and the left face I find an arrangement (if needed, let's call this C) that has only two other equivalents; therefore three configurations among the $43$ quintillions (one for each of the main axes of the cube) that should count as one arrangement (= "two opposite faces turned $90°$ clockwise").

It reminds me the caution needed while counting the possible polyominoes, depending on whether one considers "free", "one-sided" or "fixed" ones (see here). I don't know shortcuts to avoid checking each element's geometrical symmetries to understand its multiplicity. Also related is the assessment of the possible latin squares (or their widespread counterpart, sudokus): it's not the actual numbers/symbols (that can always be relabelled, see this article and the cyted paper) that matter, but their "pattern". It seems to me that my Rubik's cube question mixes both concerns, and the number of possible configurations prevents any hope to brute-force an answer. Are their smart arguments to tackle the problem? Is there someone who has already asked (and answered?) my question?

答: This message on the Cube Lover's Archive calculates this number as 901,083,404,981,813,616 but note that they factor out not only rotations (which would be almost a factor 24) but also reflections (for a factor of almost 48).</p>


问:
Showing that $2C(x)^2 = C(2x)+1$ using Cauchy product Formula
I'm stuck trying to calculate $2C(x)^2 = C(2x)+1$

with $C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$ (Cosine Power series)

So I've seen proofs for this identity before, but never using the Cauchy product formula.

So this is what I've tried:

Plugging in the power series we get:

$2\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}\right)^2$

$=$ $2$$\left(\sum_{n=0}^\infty\sum_{k=0}^n (-1)^{n-k}\frac{x^{2(n-k)}}{(2(n-k))!}(-1)^k\frac{x^{2k}}{(2k)!}\right)$

$=$ $2\left(\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}\right)\\$ I hope this is correct at least.

For the other side of the equation I'm actually not quite sure how to write down.

It seems to me that this is going to be quite a long equation in general and I'm not able to calculate this further. Is it possible that I'm missing something or how can I proceed further to prove this? (Using the Cauchy product formula)

答:
By definition, $$C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$$
The series is absolutely convergent, so you can consider the Cauchy product with itself, which gives you :

$$C(x)^2 = \left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)=\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}(-1)^{n-k} \frac{x^{2(n-k)}}{(2(n-k))!}$$

which simplifies to give :
$$C(x)^2 =\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^n \frac{x^{2n}}{(2n)!}{2n \choose 2k}= \sum_{n=0}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!}$$

By developping $(1+i)^{2n}$, you can see that for all $n \geq 1$, one has $$\sum_{k=0}^n{2n \choose 2k}=2^{2n-1}$$

Plugging this in the expression, you get $$C(x)^2 = 1 +\sum_{n=1}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!} = 1 + \sum_{n=1}^{\infty} (-1)^n 2^{2n-1}\frac{x^{2n}}{(2n)!}$$

So $$2C(x)^2 = 2 + \sum_{n=1}^{\infty} (-1)^n 2^{2n}\frac{x^{2n}}{(2n)!} = 2 +\sum_{n=1}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!}$$

Finally, reincorporing the term for $n=0$, you get
$$2C(x)^2 = 1 + \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} = 1+ C(2x)$$


问:
Prove using the $\epsilon-\delta$ definition that $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$ for any $a > 0$, $a \in \mathbb{R}$.
Scratch Work

Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$. Then for every $\epsilon > 0$, we aim to find a $\delta > 0$ such that

$$
0 < |x-a| < \delta \Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| < \epsilon
$$

Note that

\begin{align*}
\sin x &\le x \\
\cos x &\le 1
\end{align*}

for every $x \ge 0$.

Also

\begin{align*}
|x-a| &= |\sqrt x - \sqrt a| \cdot |\sqrt x + \sqrt a|
\end{align*}

Proof

Choose $\delta = \min\{a, \epsilon \sqrt {a}\}$. We have

\begin{align*}
0 < |x-a| < \delta &\Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| \\
&= \left|2\cos\left(\frac {\sqrt{x} + \sqrt{a}} 2\right) \cdot \sin\left(\frac {\sqrt{x} - \sqrt{a}} 2 \right) \right| \\
&\le 2 \cdot 1 \cdot \frac {\left|\sqrt {x} - \sqrt {a} \right|} 2 \\
&= \frac {|x-a|} {|\sqrt x + \sqrt a|} \\
&\le \frac 1 {\sqrt a} \cdot \epsilon \sqrt a \\
&= \epsilon
\end{align*}

I am more concerned with my workings which led me to the proof. Is there any mistake in my manipulations? Any advice would be greatly appreciated!

答:
I agree with user837206’s comment: your post is correct.

I suggest some tweaks:

You start with “Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$.” That’s the statement you are trying to prove. It’s just some notes, not the actual proof, so it’s not technically wrong. But make sure you are clear about what you have already proved and what you still need to prove.

You follow with “for every $\epsilon > 0$, we aim to find a $\delta > 0$ …”. I would say we can find such a $\delta > 0$.

Turning to the actual proof, you can strengthen most of your inequalities:

For every $x > 0$, $\sin x < x$, so

$$\left| 2 \cos\left(\frac{\sqrt{x} + \sqrt{a}}{2}\right) \sin\left(\frac{\sqrt{x} - \sqrt{a}}{2}\right) \right| < 2 \cdot 1 \cdot \frac{\left|\sqrt{x} - \sqrt{a}\right|}{2}.$$

By using either of these inequalities:

\begin{align*}
|x - a| < \delta \leq a &\implies x > 0 \\
&\implies \frac{1}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \tag{1} \\
|x - a| < \delta \leq \epsilon\sqrt{a} &\implies |x - a| < \epsilon\sqrt{a} \tag{2}
\end{align*}

We can make this inequality strict:

$$\frac{|x - a|}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \cdot \epsilon\sqrt{a}.$$

Actually, do we really need the absolute value in $\left| \sqrt{x} + \sqrt{a} \right|$?


问:
Prove that the zeros of an analytic function are finite and isolated
Let us assume that the zeros of $f = \{Z_1,\ldots,Z_n,a\}$ are infinite and converge towards $a$.

The book which I am reading says that any neighborhood of $a$ will contain infinite zeros. Since $f$ is continuous, $f$ must be identically zero in this neighborhood. .....(A)

Then as the neighborhood grows, it will be identically $0$ in that neighborhood too until it engulfs the whole of this region. Hence, the function becomes identically zero in the whole region.

I don't quite understand (A). I am confused over (A) with this analogy. Consider $f(z) = z-1$. This has a zero at $z=1$. Since $f$ is continuous in the $z$ plane, this means that as per (A) points in the neighborhood of $z =1$ should also have mapped value $= 0$.

I am confused; where could I be making a mistake?
Thanks

答:
I think that problem is that continuity is not enough. (Although I can't think of a counter-example off-hand). What you really need is for the function to be holomorphic in some neighbourhood of it's zeroes.

I know of the related result (known as the "principle of isolated zeroes"):

Suppose $f:B_a(r)\rightarrow \mathbb{C}$ is a non-zero holomorphic function on $B_a(r)$with $f(a) = 0$. Then $\exists 0 <\rho<r$ such that $\forall z \in B_a(\rho)\text{ \ }\{a\}, f(z) \not = 0$.

To prove this, we need Taylor's theorem. This says that on $B_a(r)$, $f$ can be written as

$$f(z) =\sum_{n=0}^\infty a_n(z-a)^n$$
 If $N$ is the largest integer such that $a_n = 0$ for all $n \leq N$ then we can write $f(z) = (z-a)^Ng(z)$ where $g$ is a holomorphic function and $g(a)$ is not zero. By the continuity of $g$ there is some open ball around $a$ (call it $B_a(\rho)$) where $g$ is not zero, so since $(z-a)^N$ is non-zero in $B_a(\rho)\text{ \ }\{a\}$, $f$ is also non-zero there.


答:
A (purely) group-theoretical argument here:

Suppose $g^m=e_G$, then $\text{ord}(g)|m$. By Lagrange's theorem we have $\text{ord}(g)|n$. Therefore $\text{ord}(g)$ is a common divisor of two relatively prime numbers, so $\text{ord(g)}=1$ and $g=e_G$.

Title: The function $\Psi:G \to G$ with $g \mapsto g^m$ is an automorphism.

Body:
Let $G$ be an Abelian group of order $n$ and for a positive integer $m$ we have that $\text{gcd}(m,n)=1$.

Then the function $\Psi:G \to G$ with $g  \mapsto g^m$ is an automorphism.

My try:

First I show that the function is homomorphism:
Let $g_1,g_2 \in G$ and consider $\Psi(g_1g_2)=(g_1g_2)^m$ and this is true since $G$ is a group,besides it's Abelian and so $\Psi(g_1g_2)=g_1^{m}g_2^{m}=\Psi(g_1)\Psi(g_1)$.

The function is clearly surjective.

To show that it's injective we need to show $\text{ker}(\Psi)=\left\{e_G\right\}$, if we assume there exists $e_G\ne g  \in\text{ker}(\Psi)$ then $\Psi(g)=e_G=g^m$,but I don't know how to continue.

Any help is appreciated.


Body:
To show injectivity, suppose that $g^m = g_{1}^{m}$.

Then $g^mg_{1}^{-m} = 1 \implies g^mg_{-1}^m = 1 \implies (gg_{1}^{-1})^{m}=1$. Since $(m,n)=1, mx+ny=1$ and so $1=(gg_{1}^{-1})^{m} = (gg_{1}^{-1})^{1-ny} = gg_{1}^{-1} \implies g = g_{1}$ because $|G| = n$.


问:
Show that $h_n(x)=x^{1+\frac{1}{2n-1}}$ converges uniformly on $[-1, 1]$.
This is a question about the uniform convergence of a sequence of functions.

First, note that $\lim_{n\to\infty}h_n=|x|$ is known already. We simply need to get an appropriate bound on $|h_n(x)-h(x)|$.

Here's my attempt (note crucially that the domain is $[-1, 1]$ instead of $\mathbb{R}$):

$$
|h_n(x)-h(x)|=|x^{1+\frac{1}{2n-1}}-|x||=|x\cdot x^{\frac{1}{2n-1}}-|x||\leq|x||x^{\frac{1}{2n-1}}|+|x|=(|x^{\frac{1}{2n-1}}|+1)|x|\leq |x^{\frac{1}{2n-1}}|+1\leq2.
$$

which obviously isn't helpful because the final expression does not involve $n$. In other words, I can't choose $n$ in response to some $\epsilon$-bound on the error $|h_n(x)-h(x)|$.

What are the required modifications? Your help is much appreciated.

答: It is enough to show that $h_n(x) \to x$ uniformly on $[0,1]$ since $h_n$ is even. Let $\epsilon >0$ and note that $|h_n(x)-x| \leq |x||x^{\frac 1  {2n-1} }-1|\leq 2x<\epsilon $ if $x <\frac {\epsilon} 2$. Now let $x \geq \frac {\epsilon} 2$. Then $|h_n(x)-x| \leq |x^{\frac 1  {2n-1} }-1|=|e^{\frac 1  {2n-1} \ln x }-1|$. So $|h_n(x)-1| \leq e^{t_n} \frac 1 {2n-1} |\ln x|$ for some $t_n$ between $0$ and $\frac  1{2n-1} \ln x$ by MVT applied to the exponential function and observe that $\ln x$ is bounded for $\geq \frac {\epsilon} 2$. Can you finish the proof?


问:
Prove that Gauss map on M is surjective
Let $M$ be a closed, orientable, and bounded surface in $\mathbb{R}^3$.

(a) Prove that the Gauss map on $M$ is surjective.

(b) Let $K_+(p) = \max \{0, K(p)\}$.  Show that
$$
\int K_+dA \ge 4\pi.
$$
in the area of $M$.  Do not use the Gauss-Bonnet Theorem to prove this.

答: For (b), you need to interpret $\displaystyle\int_M K\,dA$ as the area (counting sign and multiplicity) of the image of the Gauss map. This is the change of variables theorem for integrals, together with the (more or less) definition of $K$ as the determinant of the derivative of the Gauss map.</p>

Title: Prove that Gauss map on M is surjective


问:
how to prove that $f_n(x)=e^{-nx^2+x}$ is Lebesgue integrable for all $n\geq 1$
I have tried using this definition: $$ f  \text{ integrable (measureable)} \Leftrightarrow \int_{\mathbb{R}}|f(x)| d\mu < \infty$$

We know that $f$ is measurable because it is a continuous function.

I think that I need to bound $f_n(x)$ with an integrable function... but I don't know which one I should choose.

I have done the following: $$|f_n(x)|= e^{-nx^2+x} =e^{-nx^2}e^{x} \leq e^{-x^2}e^x\leq h(x), \forall n \in \mathbb{N}$$
$$h(x)=\begin{cases}
   e^{-x^2},& x \leq 0\\
    ???????& x>0
    \end{cases} $$
Thank you

答:
$\int e^{-x^{2}+x}dx=e^{1/4}\int e^{-(x-\frac1  2)^{2}}dx=e^{1/4}\int e^{-y^{2}}dy=e^{1/4}\sqrt {\pi}$

[$\int_{\mathbb R} e^{-y^{2}} dy=\sqrt {\pi}$. This is a standard result in Calculus and it is proved by going to polar coordinates].


问:
Retraction $D^n\times I\times I\to (S^{n-1}\times I\times I)\cup (D^n\times\{0,1\}\times I)\cup (D^n\times I\times \{0\})$
The title says it all. I want to find a retraction
$D^n\times I\times I\to (S^{n-1}\times I\times I)\cup (D^n\times\{0,1\}\times I)\cup (D^n\times I\times \{0\})$

You can also view this as $(S^{n-1}\times I\times I)\cup (D^n\times U)$, where $U=\{0,1\}\times I\cup I\times \{0\}$ is in the form of a letter U in the unit square.

I have been given that there is a retraction $r:D^n\times I\to S^{n-1}\times I\cup D^n\times\{0\}$ (so I don't need a specific retraction, but preferably one expressed in terms of $r$)

This last condition is the same as $S^{n-1}$ being a cofibration of $D^n$.

答:
This is a follow-up question to Homotopy on $S^{n-1}$ extends to homotopy on $D^n$ .

I doubt that the retraction $r_n:D^n\times I\to S^{n-1}\times I\cup D^n\times\{0\}$ will help you in the sense you expect - but of course we can use it. There exists a homeomorphism
$h : D^n \times I \to D^{n+1}$ such that $h(D^n \times \{0,1\} \cup S^{n-1} \times I) = S^n$. Note that $D^n \times \{0,1\} \cup S^{n-1} \times I$ is the boundary of $D^n \times I$.

The desired retraction is
$$R_n : D^n \times I \times I \stackrel{h \times id_I}{\to} D^{n+1} \times I \stackrel{r_{n+1}}{\to} S^n \times I \cup D^{n+1} \times \{0\} \stackrel{h^{-1} \times id_I}{\to} \\ h^{-1}(S^n) \times I \cup h^{-1}(D^{n+1}) \times\{0\} = \\  (D^n \times \{0,1\} \cup S^{n-1} \times I) \times I \cup D^n \times I \times \{0\} = \\  (S^{n-1}\times I\times I)\cup (D^n\times\{0,1\}\times I)\cup (D^n\times I\times \{0\}) .$$

Let us come to the construction of $h$. Of course this is equivalent to the construction of a homeomorphism $H : D^n \times [-1,1] \to D^{n+1}$ such that $H(D^n \times \{-1,1\} \cup S^{n-1} \times [-1,1]) = S^n$.

Let $\lVert - \rVert^{(k)}$ denote the standard Euclidean norm on $\mathbb R^k$. On $\mathbb R^{n+1}$ define $\lVert (x_1,\ldots,x_{n+1}) \rVert = \max(\lVert (x_1,\ldots,x_n) \rVert^{(n)},\lvert x_{n+1} \rvert)$. This a norm on $\mathbb R^{n+1}$ which is equivalent to $\lVert - \rVert^{(n+1)}$. The closed unit ball with respect to $\lVert - \rVert$ is $D^n \times [-1,1]$. Define
$$\phi : \mathbb R^{n+1} \to \mathbb R^{n+1}, \phi(x) = \begin{cases} \dfrac{\lVert x \rVert}{\lVert x \rVert^{(n+1)}} x & x \ne 0 \\ 0 & x = 0 \end{cases}$$
This a homeomorphism such that $\phi(D^n \times I) = D^{n+1}$. See my answer to Homeomorphism between $k$-simplex and a product of $k$ unit intervals .

Restriction to $D^n \times I$ gives the desired $H$.


问:
A set in the Borel $\sigma$-algebra over $[0,1]$ that isn't in the algebra generated by open sets
The Borel algebra on $[0,1]$ is by defintion a $\sigma$-algebra, the smallest one containing every open subset of $[0,1]$. I'm wondering how the Borel algebra differs from the algebra generated by the open subsets of $[0,1]$.

What's an example of a set in the Borel algebra that you can't obtain by closing open subsets of $[0,1]$ under complements and finite unions?

Added. Does any countable dense set work? Obviously we have to choose a Borel set that is neither open nor closed. The set of rationals in $[0,1]$ fits the bill here, and my intuition is that it cannot be written using finitely many set operations with open sets, but I'm not quite sure how to prove it.

答:
Claim: For any set $X$ obtainable the boundary is nowhere dense.

Proof: Boundary of open set is nowhere dense. Now for the operations: 1) complement - since boundary is the same for a set and its complement, this is automatic 2) union - boundary of a union is subset of union of boundaries, and so it is a subset of a union of two nowhere dense sets, so is a subset of a nowhere dense set, and so is nowhere dense.

Thus we can not obtain the rationals, since their boundary is all of $[0,1]$.  The same is true for all countable dense sets.


问:
A "convergence theorem" in measure theory?
Let $A_n$ ($n \in \mathbb{N}$) be a sequence of measurable subsets of some measure space with $A_1 \subseteq A_2 \subseteq...$ and let $A:= \bigcup_{n < \infty} A_n$. Let $f:A \to \mathbb{R}$ a measurable map which is integrable on every $A_n$. Moreover, assume that the limit
$$\lim_{n \to \infty} \int_{A_n} f d\mu$$
exists (and is a real number). Can we conclude that $f$ is integrable on $A$ and that
$$\lim_{n \to \infty} \int_{A_n} f d\mu=\int_A f d\mu\phantom{aa}\text{?}$$

The motivation for the question is a bit tedious to explain in detail but I just found some sources (i.e. handwritten lecture notes) where this type of inference seems to be implicitly used without any explanation. The problem in these sources is that the classical convergence theorems from measure theory cannot be applied in an obvious way since the integrals $\int_{A_n} |f| d\mu$ have no (obvious) upper bound. So, I was just wondering whether the above might be true.

答: On the real line with Lebesgue measure take $A_n=(-n,n), A=\mathbb R$ and $f(x)=x$ for  a counter-example.


答:
No, this is essentially the same as saying that if the improper Riemann integral $\int_0^\infty f(t)\,dt:=\lim_{x\to\infty}\int_0^x f(t)\,dt$ converges, then $\int_{[0,\infty)}\lvert f(t)\rvert\,dt<\infty$. This is famously not the case. For instance, see $\int_0^\infty \frac{\sin x}x\,dx=\frac\pi2$ and $\int_0^\infty\left\lvert \frac{\sin x}x\right\rvert\,dx=\infty$.

The claim holds for $f\ge0$ by Beppo Levi.

Title: A "convergence theorem" in measure theory?

Body:
Let $A_n$ ($n \in \mathbb{N}$) be a sequence of measurable subsets of some measure space with $A_1 \subseteq A_2 \subseteq...$ and let $A:= \bigcup_{n < \infty} A_n$. Let $f:A \to \mathbb{R}$ a measurable map which is integrable on every $A_n$. Moreover, assume that the limit
$$\lim_{n \to \infty} \int_{A_n} f d\mu$$
exists (and is a real number). Can we conclude that $f$ is integrable on $A$ and that
$$\lim_{n \to \infty} \int_{A_n} f d\mu=\int_A f d\mu\phantom{aa}\text{?}$$

The motivation for the question is a bit tedious to explain in detail but I just found some sources (i.e. handwritten lecture notes) where this type of inference seems to be implicitly used without any explanation. The problem in these sources is that the classical convergence theorems from measure theory cannot be applied in an obvious way since the integrals $\int_{A_n} |f| d\mu$ have no (obvious) upper bound. So, I was just wondering whether the above might be true.


问:
Product of two algebraic varieties is affine... are the two varieties affine?
Let $X_1$ and $X_2$ two algebraic varieties such that their product $X_1\times X_2$ is affine. Are $X_1$ and $X_2$ affine then?

If this is not true, could you give a counterexample?

答:
Let $X,Y$ be two $k$-schemes. If $X \times_k Y$ is affine and $X \neq \emptyset$, then $Y$ is affine.

Proof: Assume $x \in X$ is a $k$-rational point. The two morphisms $\mathrm{id}_{X \times_k Y}, (x,\mathrm{id}_Y) : X \times_k Y \to X \times_k Y$ have equalizer $Y$. But affine schemes are closed under finite limits of schemes, so that we are done. If $X$ has no $k$-rational point, then $X_K$ has a $K$-rational point for some field extension $K/k$ and we conclude that $Y_K$ is affine. But then $Y$ is also affine by fpqc descent. $\square$


问:
Is it possible to fill a table (64 rows and 6 columns) with zeros and ones so that the conditions are met?
Is it possible to fill a table (64 rows and 6 columns) with zeros and ones so that the following conditions are met:

All binary words of length 6 are written in the rows of the table;

Not a single row in the table can be obtained from any higher row by replacing ones with zeros;

And does the 57th line contain the binary word 011100?

Every help will be appreciated!

答:
If you list the binary numbers in order from $000000$ to $111111$, you get a list whose 57th entry is $111000$, which satisfies your conditions.

But now you can take this list and simply swap the first and fourth columns, to transform $111000$ into $011100$.


问:
Is there an example of a convex differentiable function which is not continuously differentiable?
It is easy to see that any convex differentiable function of one variable is also continuous differentiable. But the proof is based on monotony of the derivative and doesn't work in multiple dimensions. So the question is, has any convex differentiable function of many variables to be also continuously differentiable? Or is there some counterexample?

Thanks in advance.

答:
I think the derivative will be continuous.

Here is an intuitive argument. Consider a closed convex body ( the epigraph of the convex function). Assume that the function is differentiable at every point. This means that the boundary of this convex body has a unique supporting hyperplane at each boundary point. We would like to show that the dependence of this hyperplane ( the directions of its normal) is continuous on the point $P$ on the boundary.

It is easier to think in the case of a compact convex body $C$ Surround it completely with a sphere $S$. The map from $S$ to the boundary
$\partial C$ of $C$ given by $x \mapsto \phi(x)$, the closest point in $C$ to $x$ is a contraction, so continuous, and surjective. The vector $\phi(x)-x$ is perpendicular to a supporting hyperplane at $\phi(x)$.

Now assume that every point on the boundary of $C$ has a unique supporting hyperplane. Then the map $x\mapsto \phi(x)$ is bijective from $S$ to $\partial C$, and so a homeomorphism, since $S$ is compact.  We conclude that the inverse map $\phi(x) \mapsto x$ is also continuous. Now note that $\frac{\phi(x) - x}{\|\phi(x) - x\|}$ is the unit normal vector to the supporting hyperplane at $\phi(x)$.


问:
Group with topology which is not topological group
What will example of a group $G$ with topology $\tau$ such that $f: G \to G$ such that $f(x) =x^{-1}$ and $g: G \times G \to G$ such that $g((x,y)) = x * y$ (where $*$ is binary operation on $G$) both are not continuous.</p>

答:
$(\mathbb R,+)$ with the Zariski topology.

If $g:\mathbb R\times \mathbb R\to \mathbb R$ is given by $g(x,y)=x+y$, then the preimage of, say, $(0,1)$ is the set $-x<y<-x+1$ which is not open for the product of the zariski topologies.


问:
$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$
Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$

I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it gave a factorised form.
$$(ab+bc+ca).((ab)^2+(bc)^2+(ca)^2-abc.(a+b+c))=abc.(a+b+c).(a^2+b^2+c^2-ab-bc-ca)$$
How can I proceed after that?

答:
This one is a long-way of solving (With a very obvious result).
\begin{gather*}
\implies(ab+bc+ca)^3-abc(a+b+c)^3=0\\
\end{gather*}
\begin{multline*}
\implies (a^3b^3+b^3c^3+c^3a^3+3a^2b^3c+3a^2bc^3+3a^3b^2c+3ab^2c^3+3a^3bc^2+3ab^3c^2)-\\abc(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2)=0
\end{multline*}

After simplifying
\begin{equation}
\implies a^3b^3+b^3c^3+c^3a^3-a^4bc-ab^4c-abc^4=0
\end{equation}
Factorising this will give
\begin{equation}
\implies(ab-c^2)(bc-a^2)(ca-b^2)=0
\end{equation}
Therefore, a,b,c must be in G.P.


问:
Two indistinguishable stochastic processes are equivalent
I have the following definitions.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probabilised space.
Let's take two stochastic processes $(X(t))_{i\in I}$ and $(Y(t))_{i\in I}$ with values in a separable Banach space $E$ endowed with its borelian sigma-algebra $\mathcal{E}=\mathcal{B}(E)$, where $I$ is an arbitrary set of indices.

$Y$ is a modification of $X$ if for all $t\in I$, $\mathbb{P}(X(t)=Y(t))=1$.

$X$ and $Y$ are equivalent if for all $n\in\mathbb{N}^*$, for all $(t_1,\dots,t_n)\in I^n$, the random variables $(X(t_1),\dots,X(t_n))$ and $(Y(t_1),\dots,Y(t_n))$ with values in $E^n$ follow the same distribution.

$Y$ and $Y$ are indistinguishable if there exists $\Omega_0\subset\Omega$ with $\mathbb{P}(\Omega_0)=1$ such that for all $\omega\in\Omega_0$, for all $t\in I$, $X(t)=Y(t)$, or equivalently $\mathbb{P}(\sup\limits_{t\in I}\vert X(t)-Y(t)\vert>0)=0$.

I have shown that if two stochastic processes are indistinguishable, then they are modification of each other. But I am not able to prove that if they are modification of each other, then they must be equivalent.

From the definition, I can only consider the different probabilies for each $t_1,\dots,t_n$,but how do I assemble this to make sure that the laws are the same ?

Body: Let $A=\bigcap_i \{X(t_i)=Y(t_i)\}$ Then $P(A)=1$ and $(X_{t_1},X_{t_2},\dots,X_{t_n})=(Y_{t_1},Y_{t_2},\dots,Y_{t_n})$ on $A$. Hence $P((X_{t_1},X_{t_2},\dots,X_{t_n}) \in C)=P((Y_{t_1},Y_{t_2},\dots,Y_{t_n}) \in C)$ for any Borel set $C$ in $ E^{n}$.


问:
Prove $P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$?
In class we proved that:

If $A_1$ is in $A_2$ and $A_2$ is in $A_3$ and ..., then

$P(\cup_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$

By declaring the following groups: $B_n=A_n\smallsetminus A_{n-1}$

Now I want to prove that if $A_1$ contains $A_2$ and $A_2$ contains $A_3$ and so on, then:

$P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$

What groups should I declare? Using the same $B_n$ doesn't help.

答:
$...A_3\subseteq A_2\subseteq A_1\implies  P(\cap_{n=1}^\infty A_n)=1-P(\cup_{n=1}^\infty A_n^C)$ where $A_1^C\subseteq A_2^C\subseteq A_3^C...$

$=1-\lim_{n\to\infty} P(A_n^C)=\lim_{n\to\infty}(1-P(A_n^C))=\lim_{n\to\infty} P(A_n)$


答:
$
\newcommand{\P}{\mathrm{P}}
\newcommand{\bb}[1]{\left( #1 \right)}
$
$$
\P\bb{\bigcap_{i=1}^\infty A_n} = 1 - \P\bb{\bigcup_{i=1}^\infty A_n^c} = 1 - \lim_{n \to \infty} \P(A_n^c) = \lim_{n \to \infty} \bb{1 - P(A_n^c)} = \lim_{n \to \infty} P(A_n)
$$

Title: Prove $P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$?

Body:
In class we proved that:

If $A_1$ is in $A_2$ and $A_2$ is in $A_3$ and ..., then

$P(\cup_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$

By declaring the following groups: $B_n=A_n\smallsetminus A_{n-1}$

Now I want to prove that if $A_1$ contains $A_2$ and $A_2$ contains $A_3$ and so on, then:

$P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$

What groups should I declare? Using the same $B_n$ doesn't help.


问:
Minimal value of a diophantine expression
Given the expression $$x^a - y^b>0$$ what is the minimum positive value it can have given $x>y > 1$ and $a,b>1$. For example, if I have $4^a - 3^b$ I would conjecture that the smallest value would be for $4^2 - 3^2 = 7$. Another close solution would be $4^4 - 3^5 = 13$. But how would I formally prove it, and if there is a smaller value how would I find the smallest one?

答:
For $4^a-3^b$, by Mihăilescu's theorem (in reference to solutions of the equation $x^a - y^b = 1$)
we know that $8,9$ are the only consecutive powers and so $4^a-3^b > 1$. Working $\mod 2,3$ it is clear that $4^a-3^b \neq 2,3,4,6$. Rechecking $\mod 3$ we have for the equation $4^a-3^b = 5$ that $1 \equiv 2 \mod 3$ meaning that $5$ is not a solution either and $7 = 4^a - 3^b$ is the minimal solution.


问:
A question about a.s. uniqueness of a conditional expectation
Consider a non-negative random variable defined om $(\Omega,\mathcal A, P)$. A conditional expectation $E[X|Y]$ of $X$ under $\sigma (Y)$ ($Y$ is another r.v.) is a $\sigma (Y)$ measurable r.v. such that $$ \int_{C} E[X|Y] dP = \int_{C} X dP, \forall C \in \sigma (Y)$$.

In my book (Probability Theory by Heinz Bauer) the author firstly shows existence of $E[X|Y]$. He shows how to construct a function $X_0$ which has all the needed properties. Then the author proves uniqueness by saying "Let's consider $X_1$ which is $\sigma (Y)$ measurable and satisfies the condition for integrals above, then $X_1 = X_0$ a.s. follows from the fact that $\{X_1 = X_0 \} \in \sigma (Y)$".

I clearly see why $\{X_1 = X_0 \} \in \sigma (Y)$ but I do not understand why from that one can conclude that $X_1 = X_0$ a.s. Any hints?

Edit

One can solve it if assumes that $X_1 \ge 0$, then $X_1$ can be viewed as a density for a measure defined on $\sigma (Y)$ and given by $\int_C X_1 d P_{|\sigma (Y)}, \forall C \in \sigma(Y)$, where $P_{|\sigma (Y)}$ is the restriction of $P$ for $\sigma (Y)$.

Now we get $$\int_C X_1 d P = \int_C X_1 d P_{|\sigma (Y)} = \int_C X_0 d P_{|\sigma (Y)} = \int_C X_0 d P, \forall C \in \sigma(Y)$$

So $X_0, X_1$ as densities define the same measure on $\sigma(Y)$ and thus are identical $P_{|\sigma (Y)}$-a.s. and consequently $P$-a.s.

So we can show that $X_0$ is a.s. unique among non-negative $\sigma(Y)$ measurable random variables (satisfying the equation for conditional expectations) and consequently among $\sigma(Y)$ measurable random variables which are a.s. non-negative and satisfy the equation.

答:
As you said in your edit, the fact that  the conditional expectation $X_0\geq 0$ of $X\geq0$ is $P_0$-a.s. unique is because the density from the Radon-Nikodym theorem is itself $P_0$-a.s. unique, where $P_0$ denotes the restriction of $P$ to the sub-$\sigma$-algebra under consideration.

This is what Bauer means when he writes the following on page $111$:

" Being a density, $X_0$ is moreover $P_0$-a.s. determined ( cf. MI, 17.11)"

where MI is his book on Measure Theory and theorem 17.11 is the theorem showing uniqueness.

Now, the fact that $A$ holds $P_0$-a.s. does not imply that that $P_0(A)=1$ in general, since the event $A$ may not be measurable. This the point I think Bauer is trying to make here:

Since $\{X_0=X'_0\}$ is measurable we can say that $P_0\{X_0=X'_0\}=P\{X_0=X'_0\}=1$.


答:
Define the sets $B_n = A_n^c$, where $A^c$ denotes the complementary of $A$. Since $A_1 \supset A_2 \supset \dots$, we have that $A_1^c \subset A_2^c \subset \dots$ and so $B_1 \subset B_2 \subset \dots$.

Notice as well that $\bigcup_{i=1}^\infty B_i = (\bigcap_{i=1}^\infty A_i)^c$. Now apply your first result to these new sets to get:

$$\lim\limits_{n\to \infty} P(B_n) = P(\bigcup_{n=1}^\infty B_n) = P((\bigcap_{i=1}^\infty A_i)^c) = 1 - P(\bigcap_{i=1}^\infty A_i)$$

But we also have that:

$$\lim\limits_{n\to \infty} P(B_n) = \lim\limits_{n\to \infty} (1 - P(A_n)) = 1 -  \lim\limits_{n\to \infty} P(A_n)$$

And so $1 -  \lim\limits_{n\to \infty} P(A_n) = 1 - P(\bigcap_{i=1}^\infty A_i)$, and hence $\lim\limits_{n\to \infty} P(A_n) = P(\bigcap_{i=1}^\infty A_i)$.


问:
Finding the rate of water rising a vase with a square base and sloping sides
How do you solve this rates of change question?

A vase with has a square base side length $6$cm and flat sides sloping outwards at and angle of $120^\circ$ with the base. Water flows in at $12\text{cm}^3/\text{s}$. Find the rate of the height of water rising for $3$ seconds.

So I have my volume formula
$$v = 36h + \frac{12h^2}{\sqrt3} \tag1$$
Also,
$$\frac{dh}{dt} = \frac{dh}{dv}\frac{dv}{dt} = \frac{12}{36+\frac{24h}{\sqrt3}}\tag2$$
since given $12\text{cm}^3/\text{s}$.

However, I am having trouble finding $h$. I tried using the condition that at $3 $ seconds, the volume is $=36$, but I get an incorrect value and hence incorrect answer when I sub it into the final $dh/dt$ equation.

What do I do?

答:
The way you are calculating volume may be leaving certain parts of it in the $4$ corners.

Please note that at height $h$, you get a square with side length of $6 + \frac{2h}{\sqrt3}$ ($\frac{h}{\sqrt 3}$ adds on both ends of each side).

So what you have is a truncated square pyramid and the volume is given by the below formula (or you have to calculate and add extra volume in $4$ corners) -

$V = \frac{1}{3}(6^2 + (6 + \frac{2h}{\sqrt 3})^2 + 6(6 + \frac{2h}{\sqrt 3}))h$

This leads to a cubic -

$V = \frac{4h^3}{9} + 4 \sqrt3 h^2 + 36h$

At $t = 3$,

$36 = \frac{4h^3}{9} + 4 \sqrt3 h^2 + 36h$

and leads to $h \approx 0.8525$.

Now $\frac{dV}{dt} = (\frac{4h^2}{3} + 8 \sqrt3 h + 36) \frac{dh}{dt}$

Can you take it from here?


问:
Show that every element of $G$ can be written as $a^{-1}\Psi(a)$ for some $a \in G$.
Assume $G$ is a finite group and $\Psi \in \text{Aut(G)}$ such that for every $g \in G:\Psi(g)=g \implies g=e_G$.

Show that every element of $G$ can be written as $a^{-1}\Psi(a)$ for some $a \in G$.

I tried to use some basic results, however, these were not helpful, besides I don't know how the condition
$g \in G:\Psi(g)=g \implies g=e_G$ will help us.

Any hint is appreciated.

答:
The additional condition is required, as without it the map $\phi: G\rightarrow G$ defined by $g\mapsto\Psi(g)g^{-1}$ is not injective.

We want to prove that the map $\phi: G\rightarrow G$ is surjective. As $G$ is a finite set, this map is surjective if and only if it is injective. We shall prove that the map is injective.

To see that it is injective, suppose $\phi(g)=\phi(h)$, so $\Psi(g)g^{-1}=\Psi(h)h^{-1}$. This rearranges to give $\Psi(h^{-1}g)(h^{-1}g)^{-1}=1$, and so (by the condition) $h^{-1}g=1$, so $h=g$. Hence, the map is injective. It is therefore surjective, by the above, as required.


问:
Is this considered a valid proof that, if $A$ and $I+AB$ are invertible, then so is $I+BA$, and $(I+AB)^{-1}A=A(I+BA)^{-1}\,?$
(Newbie Question Warning :D)

Prove that if $A$ and $I + AB$ are invertible, so is $I+BA$, and that $$(I+AB)^{-1}A=A(I+BA)^{-1}$$

I managed to work this out with the following calculation:

$$\begin{align}
(I+AB)^{-1}A &= A(I+BA)^{-1} \\
\rightarrow \quad A &= (I+AB)A(I+BA)^{-1} \\
\rightarrow \quad A &= (A+ABA)(I+BA)^{-1} \\
\rightarrow \quad A &= A(I+BA)(I+BA)^{-1} \\
\rightarrow \quad A &= A
\end{align}$$

But I have no idea how to specify my proof with words... the only thing I did is deconstruct the formula I was given, is that considered a legit proof by itself?

答:
No, your proof is not valid.

You are asked to prove that (1) $I + BA$ is invertible and (2) $(I + AB)^{-1} A = A (I + BA)^{-1}$. But your proof assumes that both of these statements are true to begin with, a common beginner error. A valid proof starts with statements that are already known to be true, and works towards the statements that need to be proved.

It so happens that (assuming (1) has already been proved) your proof of (2) works if you reverse all the steps:

\begin{align*}
A = A &\implies A = A (I + BA) (I + BA)^{-1} \\
&\implies A = (A + ABA) (I + BA)^{-1} \\
&\implies A = (I + AB) A (I + BA)^{-1} \\
&\implies (I + AB)^{-1} A = A (I + BA)^{-1}
\end{align*}

(Warning: This is not always the case. Sometimes, statement A implies statement B but statement B does not imply statement A.)

You still need to prove (1). Use the hint provided by Kavi Rama Murthy:

$A^{-1} (I + AB)^{-1} A$ is a candidate for $(I + BA)^{-1}$.

More specifically: Multiply $(I + BA)$ by $A^{-1} (I + AB)^{-1} A$ and see if you can reduce it to $I$. (Since you are aiming for $I$, it does not matter which way around the multiplication is done.) The required manipulations are similar to the ones in your proof of (2).

Your question says:

But I have no idea how to specify my proof with words

In general, the presence or absence of words does not determine whether a proof is valid. Words are very useful in mathematics, but they are not required here.


问:
Confidence interval for $n$ of $X$~Bin$(n,0.3)$
I want to calculate a 90% confidence interval for $n$ (from $X$~Bin$(n,0.3)$) where the number of successes is known ($x=12$). I have found the smallest $n$ so that $P(X \le 12)\le 0.05$ is 61 and the largest $n$ so that $P(X \ge 12)\le 0.05$ is 25. However I can't find any examples online or in my textbook that construct CIs for $n$, only ones that do so for $p$ so I am a bit lost on where to go from here. Any hints would be greatly appreciated.

Update: I think I must use some pivot variable. For example, I know that $0.9 = P(-1.64 \le N(0,1) \le 1.96)$. Also, for large $n$, $X$~Bin$(n,p)$ has an approximate $N(np, np(1-p))$ distribution. If I let $\hat n=\frac{X}{p}$ denote the estimator of $n$, then $\hat n$ has an approximate $N(n, \frac{n(1-n)}{p})$ distribution (although I'm not 100% sure about this latter part). Then, I should be able to say $\frac{\hat n -n}{\sqrt{n(1-n)/p}}$~$N(0,1)$ and so I can substitute this into my first equation and get: $$0.9 = P(-1.64 \le \frac{\hat n -n}{\sqrt{n(1-n)/p}} \le 1.96)$$ and then rearrange for $\hat n$ by itself in the middle of the inequality. My only concern is that $\hat n$ does not an approximate $N(n, \frac{n(1-n)}{p})$ distribution rendering this incorrect. Does anyone have any ideas?

Update 2: Just realised the $n(1-n)$ will be negative (since $n \ge 12$) and so we will get an imaginary number in the square root, thus my proposed solution is wrong. Back to square one :(

答:
When $n$ is unknown for IID Bernoulli trials, then an appropriate inferential model would be negative binomial:  Let $N \sim \operatorname{NegativeBinomial}(X = 12, p = 0.3)$ with PMF $$\Pr[N = n \mid X, p ] = \binom{n-1}{r-1} p^X (1-p)^{n-X}, \quad n \in \{X, X+1, X+2, \ldots \}.$$  Then we can construct a two-sided, equal-tailed, exact confidence interval for $n$ as follows.  For an overall confidence level of $100(1-\alpha)\%$, the lower confidence limit $L$ is the largest $L$ such that $$\Pr[N < L \mid X = 12, p = 0.3] \le \alpha/2.$$  For $\alpha = 0.1$ and a $90\%$ confidence interval,  $L = 26$, since $\Pr[N < 26 \mid X = 12, p = 0.3] \approx 0.0442465$ but $L = 27$ exceeds $0.05$.  For the upper confidence limit $U$, we similarly solve $$\Pr[N > U \mid X = 12, p = 0.3] = \alpha/2$$ to obtain $U = 57$, since $\Pr[N > 57 \mid X = 12, p = 0.3] \approx 0.0486239$ but $U = 56$ exceeds $0.05$.  Therefore your equal-tailed (to the extent possible for a discrete parameter) interval is $N \in [26, 57]$.

Again, this is a calculation for an exact interval--"exact" in this case meaning it is not based on any asymptotic approximations.  The coverage probability strictly exceeds the nominal confidence level.


答:
$x$ is an eigenvalue of $AB$ iff $x$ is an eigenvalue of $BA$. Since $AB+I=AB-(-I)$ is invertible, $-1$ is not an eigenvalue of $AB$, hence $-1$ is not an eigenvalue of $BA$ and $|BA-(-I)|\ne0$. Hence $I+BA$ is invertible.

Note that you have started working by assuming $(I+AB)^{-1}A=A(I+BA)^{-1}$, which is needed to be proven and is not known. To prove the matrix equality, just use your chain of equations backwards, i.e. start from $A=A$, multiply RHS by $I=(I+BA)(I+BA)^{-1}$ to get $A=A(I+BA)(I+BA)^{-1}...$ and so on.

Title: Is this considered a valid proof that, if $A$ and $I+AB$ are invertible, then so is $I+BA$, and $(I+AB)^{-1}A=A(I+BA)^{-1}\,?$

Body:
(Newbie Question Warning :D)

Prove that if $A$ and $I + AB$ are invertible, so is $I+BA$, and that $$(I+AB)^{-1}A=A(I+BA)^{-1}$$

I managed to work this out with the following calculation:

$$\begin{align}
(I+AB)^{-1}A &= A(I+BA)^{-1} \\
\rightarrow \quad A &= (I+AB)A(I+BA)^{-1} \\
\rightarrow \quad A &= (A+ABA)(I+BA)^{-1} \\
\rightarrow \quad A &= A(I+BA)(I+BA)^{-1} \\
\rightarrow \quad A &= A
\end{align}$$

But I have no idea how to specify my proof with words... the only thing I did is deconstruct the formula I was given, is that considered a legit proof by itself?


答:
There was a good answer posted by @Albusdumbledore that was deleted. I am just  reproducing that answer and will delete mine if the other is undeleted.

The suggestion was to write $a = q, b = q r$ and $c = q r k$. Naturally, since $a,b,c \ne 0$, we also have $q,r,k \ne 0$. Plugging this into the relation, one concludes that $r, k$ must satisfy the relation
$$
(k-r) \left(k^2 r-1\right) \left(k r^2-1\right)=0
$$

All solutions lead to GP's. More specifically,

Case 1. If $k=r$, $a=q$, $b = r q$ and $c= r^2 q$ are consecutive terms in a GP with ratio $r$.

Case 2. If $k=\frac{1}{r^2}$ then $c=\frac{1}{r^2} q$, $a = q$ and $b=rq$ are (nonconsecutive) terms in a GP with ratio $r$

Case 3. If $k=\frac{1}{\sqrt{r}}$ then $c= \frac 1r q$, $a=q$ and $b= rq$ are (nonconsecutive) terms in a GP with ratio $\sqrt{r}$.

Case 4. If $k=-\frac{1}{\sqrt{r}}$ then $a=q$, $c=-\sqrt{r} q$, $b = rq$ are consecutive terms in a GP with ratio $-\sqrt{r}$

Title: $(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$

Body:
Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$

I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it gave a factorised form.
$$(ab+bc+ca).((ab)^2+(bc)^2+(ca)^2-abc.(a+b+c))=abc.(a+b+c).(a^2+b^2+c^2-ab-bc-ca)$$
How can I proceed after that?


问:
Finding basis for a space
I apologize for my easy question but  this is  new to me
My question is why another basis for $T_{a}^{(2)}$ is the phrase that I denote it in the picture?

答:
It's just to mention that in addition to $\mathrm d_ax\wedge\mathrm d_ay$ expressed in $x$ and $y$ being a basis you also have a basis expressed in $z$ and $\overline z$, namely $\mathrm d_az\wedge\mathrm d_a\overline z$ and they just differ by a factor of $-2i$.

This is the result of an easy calculation:
\begin{align*}
\mathrm d_az\wedge \mathrm d_a\overline z &=
\mathrm d_a(x+iy) \wedge \mathrm d_a(x-iy) \\&=
(\mathrm d_ax+i\,\mathrm d_ay) \wedge (\mathrm d_ax-i\,\mathrm d_ay) \\&=
\underbrace{\mathrm d_ax\wedge\mathrm d_ax}_{0} -i\,\mathrm d_ax\wedge\mathrm d_ay+i\,\underbrace{\mathrm d_ay\wedge\mathrm d_ax}_{-(\mathrm d_ax\wedge\mathrm d_ay)}-i^2\,\underbrace{\mathrm d_ay\wedge\mathrm d_ay}_{0} \\&=
-2i\,\mathrm d_ax\wedge\mathrm d_ay.
\end{align*}

Title: Finding basis for a space


问:
Topological and algebraic groups
I learnt that an algebraic group $G$ is not a topological group because Zariski topology on $G$ x $G$ is not the product topology. Are there any other topologies, other than Zariski, that do not generate a topological group? If so, is it for the same reason (topology on the product is not the product topology)?

答:
It's unclear to me what the actual question means.  That said:

A group object in a category-with-finite-products $\mathsf{C}$ is an object $G \in {\rm Ob}({\mathsf{C}})$ together with $\mathsf{C}$-morphisms $\mu : G \times G \to G$, $e : 1 \to G$, and $\sigma : G \to G$ satisfying the group axioms.

If we take $\mathsf{C}$ to be:

the category $\mathsf{Set}$ of sets and functions, we get the usual definition of groups;

the category $\mathsf{Top}$ of topological spaces and continuous maps, we get the notion of a topological group;

the category $\mathsf{Var}_k$ of algebraic varieties over $k$ and regular maps, we get the notion of an algebraic group.

What do we mean when we say that "an algebraic group is not a topological group"?  Well, we mean that there's a forgetful functor $F : \mathsf{Var}_k \to \mathsf{Top}$ that takes a variety to its underlying Zariski topological space, and that if you apply this functor to an algebraic group the resulting thing is not (necessarily) a topological group, because $F$ doesn't preserve products.

So I'll interpret the question to mean "are there other categories-with-finite-products $\mathsf{C}$, functors $F :\mathsf{C} \to \mathsf{Top}$, and group objects $\mathbf{G} = (G, 1, \mu, \sigma)$ in $\mathsf{C}$ such that $F(\mathbf{G})$ is not a topological group?"  (Maybe we want to put some condition on $F$ to make it resemble a forgetful functor, like requiring that it have a left adjoint.)

If $F$ preserves products, then functoriality guarantees that everything else goes through, so if we interpret the question in this sense then the answer is "this is the only thing that can go wrong."

But of course we can cook up all sorts of bizarre, useless categories $\mathsf{C}$ and functors $F : \mathsf{C} \to \mathsf{Top}$ which don't preserve products, and then come up with group objects in these categories that don't get sent by $F$ to topological groups.

Title: Topological and algebraic groups


问:
Poisson process, mean is random variable
What is the probability that group of $5$ items will last in warehouse for more than $1$ day, if sale of this item can be described by the Poisson process:

a) with mean of $5$ items per day

b) with mean being geometrically distributed random variable with parameter $\frac{1}{2}$

a) $P(N(1)<5)=\sum\limits_{k=0}^4 P(N(1)=k)=\sum\limits_{k=0}^4 e^{-4} \frac{4^k}{k!}$

Is a) correct? How to do b)?

答:
a)

if you sell 5 items per day, that means your group of 5 items has an average duration of one day. Their random survival is an $Exp(1)$ density thus the answer is

$$P(X>1)=e^{-1}$$

b)

Y is a geometric distribution with support $y \in\{0,1,2,3..\}$ thus

$p(Y=y)=\frac{1}{2^{y+1}}$

Considering the group of 5 items, $X|Y$ is an exponential distribution with mean $\frac{5}{y}$

(if you sell i.e. 10 items per day, the expected duration of your 5 items' group is half day...)

Thus

$$f_{X|Y}(x|y)=\frac{y}{5}e^{-\frac{y}{5}x}$$

The requested probability is thus

$$\mathbb{P}[X>1]=\sum_{y=0}^{\infty}\frac{1}{2^{y+1}}\int_1^{\infty}\frac{y}{5}e^{-\frac{y}{5}x} dx=\sum_{y=0}^{\infty}\frac{e^{-\frac{y}{5}}}{2^{y+1}}$$

Without doing a lot of math reasoning, the result is $\approx 0.847$; it is enough to calculate the first 9-10 addends....the rest are all $\approx 0$


问:
Probability of breast cancer
I'm having trouble with a probability problem I've been trying to solve for a while. It's about the accuracy of breast cancer testing. Relevant probabilities are listed below, where:

"$\text{cancer}$" is the event "has breast cancer".

"$+$" is the event "tests positive for breast cancer".

$P(\text{cancer}) = \frac{12}{1000}$

$P(+|\text{cancer}) = \frac{11}{12}$

$P(+) = \frac{31}{1000}$

$P(\text{cancer}|+) = 0.355$

This last line is a result from a previous problem. The next part involves updating the probability of having cancer, but I'm having trouble figuring out what the answer is.

In the next part of the question, there's a woman who has tested positive and her doctor says she's part of a population for which there's a 40% chance of breast cancer.

I need to find the probability that the woman has cancer.

I am confused by this update to the cancer probability, but i will assume that this means $P(\text{cancer})$ has changed.

I also assume this means I need to find a new value for $P(\text{cancer}|+)$, but I'm not getting this right.

$P(+ | \text{cancer}) = \frac{11}{12} = \frac{P(\text{cancer} | +)\cdot P(+)}{P(\text{cancer})} = \frac{P(\text{cancer} | +) \cdot \frac{31}{1000}}{0.40}$

$P(\text{cancer} | +) = \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} = 11.828$

The result can't be correct because it's way over 1.

How can i fix this? Thank you in advance for any insight.

答:
It is natural to assume that in this case the previous value of $P(+)$ is not applicable (it is a very bed test, which has $3.1\%$ chances to be positive in a population where a real chance is $40\%$). Moreover, this contradicts the condition $P(+ | cancer) = \frac{11}{12}$, because then $P(+)$ is at least $\frac{11}{12}\cdot 0.4>0.031$.

It is natural to assume that values of $P(+|cancer)$ and $P(cancer|+)$ reflect efficiency and reliability of the test. I expect that a  testing procedure deals with an isolated sample, so it is independend on cancer spread. But if we keep these values then the probability $P’(cancer|+)$ that the woman has a breast cancer is $P(cancer|+)$, and the information $P’(cancer)=0.4$ is redundant.

So we assume that the testing procedure provides only  $P(+|cancer)$ and $P(+|\neg cancer)$. Then from the given probabilities we have

$$\frac{31}{1000}=P(+)=P(+|cancer)P(cancer)+ P(+|\neg cancer)P(\neg cancer)=$$ $$ \frac{11}{12}\cdot \frac{12}{1000}+ P(+|\neg cancer) \cdot \frac{988}{1000},$$

so $P(+|\neg cancer)=\frac 5{247}$.

Then

$$P’(+)=P(+|cancer)P’(cancer)+ P(+|\neg cancer)P’(\neg cancer)= \frac{11}{12}\cdot 0.4+\frac 5{247}\cdot 0.6=\frac {2807}{7410}.$$

Since $P’(cancer|+) P’(+)=P’(cancer\, \&\, +)= P’(+|cancer) P’(cancer),$
we have

$$P’(cancer|+)=\frac{ P’(+|cancer) P’(cancer)}{P’(+)}=\frac{\frac{11}{12}\cdot 0.4}{\frac {2807}{7410}}=\frac {2717}{2807}\approx 0.968.$$

Title: Probability of breast cancer


问:
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.

(2) Assuming that the condition hold for $n=k.$

(3) Proving that it holds for $n=k+1$

$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$

Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that

$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$

Thus $5 \mid 3^{4(k+1) + 2} + 1$.

Is this a valid proof? I'm not entirely sure I'm correct with this...

Body: Hint: $3^4 \cdot 3^{4k+2} + 1= 3^4(3^{4k+2}+1)+1-3^4=3^4(3^{4k+2}+1)-80.$


问:
Gauss-Markov Theorem assumption of normality.
Under the 6th assumption of Gauss-Markov Theorem, it states that if the conditional distribution of random errors is normal, then the conditional distribution of the least squares estimator will be normal aswell.

Why is this true?

答:
If the errors are conditionally normal, then the vector $y$ is conditionally joint normal. Now write
$$\hat{\beta} = (X'X)^{-1} X'y.$$
Conditional on $X$, this is a linear function of $y,$ so it is also conditionally joint normal.