问:
Augmenting a matrix with a column and a row — what happens to the determinant?
Let $\mathbb A = (a_{i,j})_{1\leq i \leq n, 1\leq j \leq n}$ be a real matrix such that $\det(\mathbb A)\neq0$. Suppose we add a column $(a_{i,0})_{1\leq i \leq n+1} $ and a row  $(a_{n+1,j})_{0\leq j \leq n+1} $ to $\mathbb A$.

Under what sufficient conditions on the newly added numbers do we still have $\det(\mathbb A)\neq0$ ?

Under what sufficient and necessary conditions on the newly added numbers do we still have $\det(\mathbb A)\neq0$ ?

答:
Essentially, you're considering $A'=\begin{pmatrix} a_0 & b^\top\\ c & A \end{pmatrix}$ for scalar $a_0$, column vectors $b,c$ and a nonsingular matrix $A$. A tool which is applicable for matrices of this form is the Schur complement. This is based on performing block-Gaussian elimination. In particular, we note the following block-diagonalization of $A'$:

$$\begin{pmatrix}1 & -b^\top A^{-1} \\ 0 & I_n\end{pmatrix}
\begin{pmatrix} a_0 & b^\top\\ c & A \end{pmatrix}
\begin{pmatrix}1 & 0\\ -A^{-1}c& I_n\end{pmatrix}
=\begin{pmatrix}a_0-b^\top A^{-1}c & 0 \\ 0 & A\end{pmatrix}$$
which applies so long as $A$ is invertible. One may then check that the block-triangular matrices have determinant $1$, so $$\det(A')=(a_0-b^\top A^{-1}c)\det(A).$$ Hence $A'$ is invertible if and only if $a_0 \neq c^\top A^{-1}b$.


问:
Examples of non-cyclic group with a cyclic automorphism group
In introduction to algebra we got the exercise:

Let $G$ be a group. Show that when $\operatorname{Aut}(G)$ is cyclic $G$ is abelian.

This doesn't make that much trouble. Denote the center (all commuting elements) with $Z$.
Then $G/Z$ is isomorphic to $\operatorname{Int}(G)$ where $\operatorname{Int}(G)$ denotes the subgroup of inner automorphisms. As every subgroup of a cyclic group is cyclic we have $G/Z$ is cyclic and hence $G$ is abelian (As a group which is cyclic of the center is abelian).

So the question is:

Is there a non cyclic group with a cyclic automorphism group?

We discussed the question already in chat and I found (thanks to google) a solution there, but I would enjoy other Examples.

答: $\mathbb{Q}^*$ is non cyclic, but the only automorphism is the identity.</p>

ContentLicense: CC BY-SA 3.0


问:
How to solve $\int{ \sqrt{ 1+ \frac{ m }{ { a }^{ 2 }- { x }^{ 2 } } } }~\mathrm{d} x$

$$\int{  \sqrt{ 1+ \frac{ m  }{ { a  }^{ 2  }- { x  }^{ 2  }    }    }    }~\mathrm{d} x.$$
$m$ and $a$ are both constants and I have already simplified $m$ and $a$ all the way.  And I can't simplify it anymore.
Is this integration even possible? My calculator just shows a math error.

I saw that an ellipse didn't have any formula for its circumference and so I wanted to try making a formula for the circumference of an ellipse using some calculus.
Here, $m = ((ba)^2)/(a^2-b^2)$

Where $a$ and $b$ are the two radius of the ellipse
There are some other constants but I was able to put it outside the integration and so this is the only part that I don't know how to proceed further with.

I personally don't think that this integration is possible cause if it was then people way smarter than me would've already calculated a formula for the circumference already. But I still wanted to try anyway.

The starting integral is
$$2 \int{  \sqrt{  \frac{ b  ^ { 2  }  x  ^ { 2  }    }{ a  ^ { 2  }  (a  ^ { 2  }  -x  ^ { 2  }  )  }  +1   }    }\,d x   $$

I got this starting integral using the formula of an ellipse and the formula for the length of a graph

答:
This smells like an Elliptic integral... Let's see.

Call $$x = a\sin(p) ~~~~~ \text{d}x = a\cos(p)\ \text{d}p$$

The integral becomes

$$\int \sqrt{1 - \frac{m}{a^2\cos^2(p)}}\ a\cos(p)\ \text{d}p$$

Which can be easily arranged like

$$a\int \sqrt{\cos^2(p) - \beta}\ \text{d}t ~~~~~~~ \beta \equiv \frac{m}{a^2}$$

The result of this is indeed an Elliptic integral:

$$a\frac{\sqrt{-2 \beta+\cos (2 p)+1} \color{red}{E\left(p\left|\frac{1}{1-\beta}\right.\right)}}{\sqrt{-\frac{-2 \beta+\cos (2 p)+1}{\beta-1}}}$$

Consider that being $\beta$ unknown, we can simplify the previous result iff $\beta < 1$ which would yield

$$\sqrt{1 - \beta}\color{red}{E\left(p\left|\frac{1}{1-\beta}\right.\right)}$$

You can pullback to $p\to x$ easily now.

As I smelled, $\color{red}{\text{Elliptic integral}}$.

More on Elliptic integrals here: https://en.wikipedia.org/wiki/Elliptic_integral

ContentLicense: CC BY-SA 4.0


问:
Prove that ${\mathbb{Q}}/{\mathbb{Z}}$ is isomorphic to $\bigoplus_p \mathbb{Z}[p^{\infty}]$
Prove that ${\mathbb{Q}}/{\mathbb{Z}}$ is isomorphic to $\bigoplus_p \mathbb{Z}[p^{\infty}]$

I know we will use $A_p$'s for solution. But i dont know how to do the isomorphism between ${\mathbb{Q}}/{\mathbb{Z}}$ and $\bigoplus_p \mathbb{Z}[p^{\infty}]$.

答:
Just wanted to provide some interesting context.

You may be familiar with partial fraction decomposition for rational functions. Any rational function $f/g$ is expressible as $f(T)/g(T)=p(T)+\sum a_k(T)/\pi_k(T)^{e_k}$ where $p(T)$ is a polynomial, $g(T)=\prod_k \pi_k(T)^{e_k}$ is a factorization into irreducible polynomials, and $a_k(T)$s are polynomials with $\deg a_k< e_k\deg \pi_k$ (so none of the summands can be partial-fractioned more in this fashion). One could choose to reduce $a(T)/\pi(T)^e$ to the form $\sum_{1<j\le e} a_j/\pi(T)^j$ for some scalar coefficients $a_j$.

Ultimately, if $\mathbb{F}$ is a field, this can be phrased as $\mathbb{F}(T)/\mathbb{F}[T]=\bigoplus \mathbb{F}[\pi(T)^{-1}]/\mathbb{F}[T]$.

Similarly, $\mathbb{Q}/\mathbb{Z}=\bigoplus \mathbb{Z}[1/p]/\mathbb{Z}$ can be phrased as follows: every rational number $f/g$ is uniquely expressible as $f/g=n+\sum a_k/p_k^{e_k}$ for some integer $n$, factorization $g=\prod p_k^{e_k}$, and coefficients $0<a_k<p_k^{e_k}$, and we can rewrite $b/p^e$ as $\sum_{1\le j\le e} b_j/p^j$ for coefficients $0<b_j<p$ if desired.

You can solve for the $a_k$s by multiplying $f/g$ by $p_k^{e_k}$ and reducing mod $p_k^{e_k}$.


问:
The relation $a^n|b^n$ implies that $a|b$, where $a$ and $b$ are positive integers and $n\ge1$.
Let $gcd(a, b) = d$

Let $a = rd$ and $b = sd$, where $gcd(r, s) = 1$.

$\therefore gcd(r^n, s^n) = 1$

Now, how do I show that $r = 1$?

答:
Okay.  It shouldn't be hard to prove $\gcd(m^k, n^k) = [\gcd(m,n)]^k$.

You can do it by considering prime factors, or by considering $$\gcd(m^k,n^k)=\\\gcd(m^{k-1}\frac m{\gcd(m,n)}\gcd(m,n), n^{k-1}\frac n{\gcd(m,n)}\gcd(m,n))=\\\gcd(m,n)\gcd(\frac m{\gcd(m,n)}m^{k-1},\frac m{\gcd(m,n)}n^{k-1})=\\\gcd(m,n)\gcd(m^{k-1},n^{k-1})$$ and using induction.

Or we could do it your way.

If $\gcd(m,n) = d$ and $m=rd;n=sd$ with $\gcd(r,s) =1$. then $\gcd(m^k,n^k)=\gcd(d^kr^k,d^ks^k)=d^k\gcd(r^k,s^k) = d^k*1 = d^k$.

.......

So now bear in mind $g|h \iff \gcd(g,h) = g$.  (That's too obvious to require any discussion isn't it?  If $g$ is a common divisor $g$ and $g$ then $g|h$.  ANd if $g|h$ than $g$ is a common divisor and there can't be any greater divisor of $g$ than $g$ itself... right?)

So we are done.

......

$a^n|b^n\implies a^n = \gcd(a^n,b^n) = d^n\gcd(r^n,s^n) =d^n*1 =d^n\implies a=d=\gcd(a,b) \implies a|b$.

That's it.


问:
Square root of 9 from a technical perspective.
When we say $$  \sqrt{9}= x $$
then $\;x = 3,\;$ right?

So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$

答:
$a=b\implies a^2=b^2$ (merely by replacement of $b$ by $a$ in the right side of the implication, assuming the left side holds), but the converse is wrong in the sets $\mathbb R$, $\mathbb Z$, $\mathbb Q$, $\mathbb C$, and $\mathbb Z/p\mathbb Z$ for prime $p>2$. Proof by example: $a=1$, $b=-1$.

However, $a^2=b^\implies a=b$ in some other sets, including $\mathbb R_+$, $\mathbb N$, and the Booleans $\mathbb Z/2\mathbb Z=\{0,1\}$.

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x$ ?

No in $\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$, where a square root is non-negative by definition. Yes in sets where $-3=3$; or for some unusual definition of square root.


问:
Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$.
Here is the problem statement:

Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$
Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.

Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:

Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that
$$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$
Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$.
$$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$
Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have
$$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$
Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$

However, I'm stuck in finding the limit. Any hints would be appreciated.

答:
$$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
a_{n-1}\\
a_{n-2}
\end{bmatrix}
$$

$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$

$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}^{n}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$

$$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix} ^ {n}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
$$


问:
3D Circle/ground intersection
This one stumps me: A circle in 3D space given by its center = $(0.15, 0.5, 1.0)$, its radius $=64$ and an orientation vector that points away from the circle's plane $(0.251, -0.796, 0.551)$

How would I go about finding the $2$ point coordinates where it touches the ground plane?

As a I am math beginner (and always will be), I have given some example numbers because I hoped I'll understand better when I see how they're transformed into the solution.

I'll reply to myself here in case I find the solution.

Title: 3D Circle/ground intersection

Body:
Is the ground plane the plane $z=0$?

The plane of the circle is given by
$r. \left( \begin{array} \\
0.251\\
-0.796\\
0.551
\end{array} \right) =
 \left( \begin{array} \\
0.15\\
0.5\\
1.0
\end{array} \right)
. \left( \begin{array} \\
0.251\\
-0.796\\
0.551
\end{array} \right) =0.19065
$

The problem is asking for a point $(x,y,0)$ that is on the plane i.e.

$\left( \begin{array} \\
x\\
y\\
0
\end{array} \right).\left( \begin{array} \\
0.251\\
-0.796\\
0.551
\end{array} \right)=0.19065$

and of distance $64$ from the centre, i.e.

$\left| \begin{array} \\
x-0.15\\
y-0.5\\
-1.0
\end{array} \right|=64$

$x^2-0.3x+0.0225+y^2-y+0.25+1=4096$

$x=\frac{0.19065+0.796y}{0.251}=0.759562+3.171315y$

$10.057237y^2+4.817621y+0.576934-0.227868-0.951395+0.00225+y^2-y+1.25=4096$

$11.057237y^2+3.817621y-4095.3501=0$

$y=\frac{-3.817621 \pm \sqrt{3.817621^2+4\times 11.057237 \times 4095.3501}}{2 \times 11.057237}=-19.4186,19.0733$

The points are $(-60.82,-19.42,0)$ and $(61.25,19.07,0)$


问:
Notation in the definition of random variable
I am trying to understand definition of random variable.

Given a probability space $(\Omega, \mathcal F, P),$ a random variable $X$ is a function from the sample space $\Omega$ to the real numbers $\mathbb R.$ Once the outcome $\omega \in \Omega$ of the experiment is revealed, the corresponding $X(ω)$ is known as realization of the random variable.

Consider two sample spaces $\Omega_1$ and $\Omega_2$ and a sigma-algebra $\mathcal F_2$ of sets in $\Omega_2.$ Then, for $X$ to be a random variable, there must exist a sigma-algebra $\mathcal F_1$ in $Ω_1$ such that for any set $S$ in $\mathcal F_2$ the inverse image of $S,$ defined by:

$$X^{-1}(S) := \{ω \mid X(ω) ∈ S\}$$

I do not know how to read this and I am hoping some one can explain each part to me.

答:
It may be pertinent to frame the answer defining a random variable as a measurable function $X$ from the sample space of a probability space $(\Omega, \mathcal F, \mathbb P)$ to the space $(\mathbb R, \mathcal B)$ of the real numbers endowed with the Borel sigma algebra (i.e. the smallest sigma algebra which contains the open sets of $\mathbb R$), such that

$$X^{-1}(A)=\{\omega \in \Omega \mid X(\omega)\in A\}\in \mathcal F \;\; \forall A \in \mathcal B.$$

Constructing the sigma-field (or sigma-algebra) generated by two random variables will bring home the idea as follows:

The random experiment is tossing a fair die once, which has as possible outcomes or sample space $\Omega=\{1,2,3,4,5,6\}.$

It is up to the definition of the probability space to determine which events will have assigned probabilities. We could consider the events even or odd outcome and contrapose it to the events prime or not-prime. Thus we define two random variables: $X: \Omega \to \{0,1\}$ maps the outcome to $1$ if the die lands on an odd number, and otherwise $0.$ On the other hand, the second random variable $Y: \Omega\to \{0,1\}$ maps to $1$ when the die lands on a prime number.

Now these two random variables will generate two different sigma algebras. In the first case, the pre-image of the random variable $X$ will include the set $V=\{1,3,5\}$ corresponding to the value of the random variable $1$ in the Borel set $A_1=\{1\}:$

$$V=X^{-1}(A_1)=\{\omega\in\Omega \mid X(\omega) \in A_1\}=\{1,3,5\}$$

This reads: The pre-image (not the inverse function!) of the set $A_1$ for the random variable $X$ is the set $\{\text{...}\}$ containing those outcomes ($\omega$) in the big set $\Omega$ such that (that is the little bar $\mid$) the random variable maps to elements in $A_1.$ And the fact that $\{1,3,5\}\in \mathcal F$ (see below) makes the function $X$ measurable.

Logically, there would be the set $W =\{2,4,6\}$ corresponding to the pre-image of the value $0$ in the Borel set $ A_2=\{0\},$ or

$$W=X^{-1}(A_2)=\{\omega\in\Omega \mid X(\omega) \in A_2\}=\{2,4,6\}$$

The sigma algebra will also contain complementary, union and finite intersections, ending up with the sigma algebra $\mathcal F=\{\{1,3,5\} , \{2,4,6\}, \emptyset,\Omega \},$ of which $\{1,3,5\}$ and $\{2,4,6\}$ are called atoms (an atom of $\mathcal F$ is a set $F \in \mathcal F$ such that the only subsets of $F$ which are also in $\mathcal F$ are the empty set $\emptyset$ and $F$ itself). And the probability space will determine (in a fair die) that $\frac12$ is assigned to each one of them by the PMF: $\mathbb P\left(\{1,3,5\}\in \mathcal F\right )= \sum_{\mathbf 1_{1,3,5}} p(\omega_i )=3\frac 1 6.$

Here is the construct:

All Borel sets of interest have pre-images corresponding to sets in the sigma algebra $\mathcal F.$

If instead the events of interest were powers of 2 and powers of 3, the sigma algebra would contain $\small \left\{\emptyset, \Omega,\{1,2,4\},\{3,5,6\},\{1,3\},\{2,4,5,6\},\{1,2,3,4\},\{1,3,5,6\},\{1,2,4,5,6\},\{2,3,4,5,6\}\right\}$, the function simply mapping the face of the die to the numbers $1,\dots,6$ would not be a random variable, because the set formed by the pre-image of the singleton Borel set $\small \{5\}$ is not in the sigma algebra (not measurable): $\small \text{pre-image}(\{5\})=\{5\}\notin \mathcal F$.

Going back to the random variable $X,$ i.e. (odd, even), it will not contain information allowing separation of $1$ from $3$ for example, because they are both outcomes belonging to the same event ("odd").

The same exercise for $Y$ will result in two atomic sets $\{2,3,5\}$ and $\{1,4,6\},$ and a sigma algebra $\mathcal G=\{\{2,3,5\} , \{1,4,6\}, \emptyset,\Omega \}.$

This random variable $Y,$ in contradistinction to $X,$ contains information allowing us to separate the outcomes $1$ and $3.$


答:
The characteristic equation is
$$\lambda^2=\frac{1}{2}(\lambda+1)$$
Its solutions are
$$\lambda=1;\;\lambda=-1/2$$
The general solution of the recurrence is
$$a_n=m+n\left(-\frac{1}{2}\right)^n$$
for $n=0$ we get $a_0=m+n=a$

for $n=1$ we have $a_1=m-\frac12n=b$

so we get
$$m= \frac{1}{3} (a+2 b),n= \frac{2 (a-b)}{3}$$
and the recurrence can be written as
$$a_n=\frac{1}{3} (a+2 b)+ \frac{2 (a-b)}{3}\left(-\frac{1}{2}\right)^n$$
as $n\to\infty$ we have $a_n\to \frac{1}{3} (a+2 b)$.

Hope this is useful

Title: Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$.

Body:
Here is the problem statement:

Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$
Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.

Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:

Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that
$$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$
Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$.
$$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$
Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have
$$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$
Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$

However, I'm stuck in finding the limit. Any hints would be appreciated.


问:
about noetherian topological space
Definition. A topological space $X$ is called noetherian if it satisfies the descending chain condition for closed subsets: for any sequence $Y_{1} \supseteq Y_{2} \supseteq \ldots$ of closed subsets, there is an integer $r$ such that $Y_{r}=Y_{r+1}=\ldots$.

Show that the following conditions are equivalent for a topological space $X$ :

(i) $X$ is noetherian;

(ii) every nonempty family of closed subsets has a minimal element;

(iii) $X$ satisfies the ascending chain condition for open subsets;

(iv) every nonempty family of open subsets has a maximal element.

proof : i can show $(i) \iff (iii)$ (because complement of open set is close and if $A \subset B \implies B^c \subset A^c $ ) and $(ii) \implies (i)$ :

Suppose $(ii)$ holds and let $\{F_n\}_{n \in \mathbb N}$ be a sequence of decreasing closed subsets. By hypothesis, there is $k \in \mathbb N$ such that $F_k \subset F_n$ for all $n$. But since the sequence is decreasing, we have $F_n \subset F_k$ for all $n \geq k$. From here it follows directly $X$ is noetherian.

I think we can use  Zorn's Lemma for $(iii) \implies (iv)$.

答:
Your proof that (ii) implies (i) isn’t quite right: the hypothesis guarantees only that $\{F_n:n\in\Bbb N\}$ has a minimal element, not that it has a minimum element. Thus, there is a $k\in\Bbb N$ such that no $F_n$ is a proper subset of $F_k$. But we know that $F_n\subseteq F_k$ for $n\ge k$, so we must have $F_n=F_k$ for $n\ge k$.

To show that (i) implies (ii), suppose that $\mathscr{F}$ is a family of closed subsets of $X$ with no minimal element. Then for each $F\in\mathscr{F}$ there is an $F'\in\mathscr{F}$ such that $F'\subsetneqq F$, and you can recursively construct a strictly decreasing infinite sequence of closed sets.

You can then prove the equivalence of (ii) and (iv) the same way that you proved the equivalence of (i) and (iii).

Title: about noetherian topological space


答:
Hint:

$$a_n=\frac12(a_{n-1}+a_{n-2})\implies a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$
Can you use this to find a closed form for $a_n-a_{n-1}$, and then for $a_n$?


问:
Multiple Choice: What is the design for this experiment?
The prompt is the following: A biology student wants to determine if using a fertilizer would help promote the growth of new babies in spider plants. The student has access to 90 baby spider plants of three varieties: green, variegated, and curly. There are 30 plants of each variety. They all are potted in the same amount and type of soil, given the same amount of water, and exposed to the same amount of light. The numbers 1–30 are written on slips of paper, placed in a hat, and mixed thoroughly. A plant is selected and a slip of paper is drawn. If the slip has the numbers 1–15, then the plant will receive fertilizer. If the slip has the numbers 16–30, the plant will not receive fertilizer. A green spider plant is selected and a slip of paper is drawn. This plant is placed in the treatment group indicated by the number, and the slip is not put back in the bag. The slips are mixed again, the next green spider plant is selected, and a slip is drawn. The plant is placed in the treatment group indicated by the number. This procedure is repeated until all 30 green spider plants are assigned to treatments. The numbered slips are placed back in the bag and this procedure is repeated for the remaining types of spider plants. After one year, the shoots will be counted for each plant.

The answer choices are as follows:

A. observational study

B. matched pairs design

C. randomized block design

D. completely randomized design

My solution: My guess is C. In this case, the matched pair is when each experimental unit receives both treatments in random order, and the participants were separated into the yoga or meditation group by a flip of the coin. In A, I do not think that it is a matched pair because each person was asked their stress level. I rule out D as well, because I do think this is a matched pairs design.

What do you think is the correct answer? Explain your reasoning, please.

答:
Your solution appears to be for a different question. In this case, it's a randomized block design. The "randomized" comes from the fact that the assignment to either treatment or control is random for each plant. The "block" refers to the fact that there are three varieties of the plant, and randomization occurs within those blocks (15 treatment + 15 control for the 30 plants in each block), rather than across the entire sample of the study (45 treatment + 45 control across all 90 plants).

An observational study would be one where the plants had already been planted in fertilizer and you weren't able to randomly assign treatment/control.

A matched pairs design typically involves taking pairs of plants that are similar in several characteristics, then assigning one to treatment and one to control.

Completely randomized would be if you just assigned 45 to treatment and 45 to control without accounting for the three varieties.


问:
why $F(a^2) \subset F(a)$?
I have some  doubt in this post

Let $E$ be an extension field of $F$. If $a \in E$ has a minimal polynomial of odd degree over $F$, show that $F(a)=F(a^2)$.

let $n$ be the degree of the minimal polynomial $p(x)$ of $a$ over $F$ and $k$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$

Since $a^2 \in F(a)$, We have $F(a^2) \subset F(a)$, then $k\le n$

I don't understand  why  $F(a^2) \subset F(a)$?

My thinking: Take $ a  \in \mathbb{R} $, $a \subset a^2 \implies F(a) \subset F(a^2)$

答:
Credit : Atticus Stonestrom( comment section)

For  $a = \sqrt 2$, $ F(a)=\mathbb{Q}({\sqrt 2})$ contain basis  $\{1, \sqrt 2\}$ but  $ F(a^2)=\mathbb{Q}(2)$ contain basis $1 \implies F(a^2) \subset F(a)$

Title: why $F(a^2) \subset F(a)$?


问:
Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$
Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:

Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
      &\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}

Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$

Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$

So $x = 13$ or $x = 11$.

Is it correct that way?

答:
There is a direct way to solve modulo $35$ with the traditional quadratics method.

$\Delta^2=b^2-4ac=2^2-4\times3\times 9=-104\equiv 1\pmod {35}$

However as you can see it requires you to know how to solve $\Delta^2\equiv 1\pmod{35}$, but let assume you have a precomputed table of the squares then:

$\Delta\in\{1,6,29,34\}=\{-1,1,-6,6\}$

Then use $x=(-b\pm\Delta)\times (2a)^{-1}\pmod {35}$

$(2a)^{-1}\equiv 6^{-1}\equiv 6\pmod{35}$ and we do not have to worry about the $\pm$ in the formula since $\Delta$ already covers additive inverses.

Therefore for each $4$ values of $\Delta$ we get same number values of $x\in\{6,11,13,18\}$


问:
Finding $\int_{-\infty}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx$ by symmetry
I can easily show that (substituting $\frac{x^2}{2} = t $ using the identity for Gamma function of $n+\frac{1}{2}$, then further expanding $\Gamma(n+\frac{1}{2})=\dfrac{(2n-1)!! \sqrt{\pi}}{2^n}$ and so on that

$$
I_1 =\int_{0}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx = \frac{(2n-1)!!\sqrt{\pi}}{\sqrt{2}}
$$

My question is, can (and how) can I use symmetry to show that

$$
I_2 = \int_{-\infty}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx = 2I_1 = (2n-1)!!\sqrt{2 \pi}
$$

答:
The below is probably as formal as it gets (substituting $-x$ for the first equality):

$$\begin{align}
\int_0^\infty e^{-x^2/2}x^{2n}\,\mathrm dx &= \int_0^{-\infty} e^{-(-x)^2/2}(-x)^{2n} \cdot (-1) \,\mathrm dx\\
&= \int_{-\infty}^0 e^{-x^2/2}x^{2n}\,\mathrm dx
\end{align}$$

which, together with $\displaystyle \int_{-\infty}^\infty e^{-x^2/2}x^{2n}\,\mathrm dx = \int_{-\infty}^0 e^{-x^2/2}x^{2n}\,\mathrm dx + \int_0^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx$, implies the result immediately.


问:
Malthus Model - Solution Differential Equation
I have this equation of a time dependent Malthus model with a term representing a time dependent immigration:
$$N'(t)=r(t)N(t) + m(t)$$
with $r(t)$ and $m(t)$ both continuous and periodic with Period $T$.
I have to prove that the function
$$N_\infty (t)=\int_{-\infty}^t exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(s)ds$$
is a solution of the ODE above.

I tried to derive $N_\infty (t)$ to obtain our ODE, but unsuccessful. I think that I do somethings wrong on the calculation.
I know that the derivative of an integral is so computed:
Let $G(t):=\int_a^x f(t)dt$. Then $G'(t)=f(x)$

So in our case I did:
$$N'_\infty (t)= exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(t)$$
But I think I miss something. Someone can help me please?

答:
With $a$ in the domain of definition let $R(t) = \int_{a}^{t}r(s)ds$ so that $R'(t) = r(t)$. Then if we use the theory of an integrating factor

$$\frac{d}{dt}\left(N(t)\exp(-R(t))\right) = \exp(-R(t))m(t).$$

Integrating this with respect to $t$ gives us
$$N(t)\exp(-R(t))-N(a) = \int_{a}^{t}\exp(-R(s))m(s)ds.$$
Therefore the solution is given by

$$N(t) = N(a)\exp(R(t))+\exp(R(t))\int_{a}^{t}\exp(-R(s))m(s) = N(a)\exp\left(\int_{a}^{t}r(s)ds\right)+\exp\left(\int_{a}^{t}r(s)ds\right) \int_{a}^{t}\exp(-R(s))m(s)ds.$$


答:
When solving $ye^y=x$ for real $y$ (given real $x$) there are two cases:

If $x\ge0$ then there is only one solution $W_0(x)$.

If $-1/e\le x<0$ then thre are two solutions $W_0(x)$ and $W_{-1}(x)$.

In this problem $-\frac{1}{2}e^{-1/2}$ is in the second case, so we find the solutions to be

$$ -2W_0(-\frac{1}{2}e^{-\frac{1}{2}})=1, \qquad -2W_{-1}(-\frac{1}{2}e^{-\frac{1}{2}})\approx 3.51286. $$

For these I typed the following two things into Mathematica:

N[-2 ProductLog[0, -0.5 Exp[-0.5]]]

N[-2 ProductLog[-1, -0.5 Exp[-0.5]]]

Title: Find the two solutions of $\log x=\frac{1}{2}(x-1)$

Body:
Question: solve $\log x=\frac{1}{2}(x-1)$

With the following I only get one solution (apparently with $W_0$), but I can't find the solution with $W_{-1}$ this way. What is my mistake?

$$\log x=\frac{1}{2}(x-1)$$
$$\log x-\frac{1}{2}x=-\frac{1}{2}$$
$$e^{\log(x)-\frac{1}{2}x}=e^{-\frac{1}{2}}$$
$$xe^{-\frac{1}{2}x}=e^{-\frac{1}{2}}$$
$$-\frac{1}{2}xe^{-\frac{1}{2}x}=-\frac{1}{2}e^{-\frac{1}{2}}$$
$$-\frac{1}{2}x=W(-\frac{1}{2}e^{-\frac{1}{2}})$$
$$x=-2W(-\frac{1}{2}e^{-\frac{1}{2}})$$


问:
Positive integers $(x,y)$ satisfying $x^2-y^2= 12345678$?
I saw this question in one of the previous year Olympiad questions.

Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ satisfying the condition : $x^2-y^2= 12345678$

Then what can be said about the number of elements in set $S$ ?

The solution of this question doesn't seem to be correct. Look at it :

Can someone provide me a reasonable approach ?

答:
$x$ and $y$ are both even or both odd, since $x^2-y^2$ is even.

Suppose they are both even.

$x=2m;\;y=2n$

$4m^2-4n^2=4(m^2-n^2)$ is a multiple of $4$, but $12345678$ is not divisible by $4$.

$x,y$ can't be both even.

Suppose now they are both odd

$x=2m+1;\;y=2n+1$

$(2m+1)^2-(2n+1)^2=4m^2+4m+4n^2+4n=4(m^2+m+n^2+n)$ is a multiple of $4$ and this can't be true so $x,y$ can't be both odd.

This means that there are no numbers satisfying the given property.

Hope this can be useful

Edit

Actually the solution given in the image OP posted is not very clear. Why both odd? And why $x^2-y^2$ must be a multiple of $8$?

This last one is actually true because
$$4(m^2+m+n^2+n)=4 (m-n) (m+n+1)$$
as one of the two factors $m-n$ or $m+n+1$ is even, then the result is divisible by $8$


问:
If $G$ is simple and finite, and H subgroup then $|G|$ divides $[G:H]!/2$
I'm having trouble to prove that if $G$ is a finite simple group and $H$ is a subgroup of finite index, then $|G|$ divides $\frac{k!}{2}$, where $k$ is the index of H.

I've prove that in this conditions $G$ is isomorphic to a subgroup $F$ of the symmetric group of $k$, $S_k$. As we are trying to prove that $|G| = |F|$ divides $\frac{k!}{2}$ I think that I need to show that $F \leq A_k$ where $A_k$ is the alternating group, but I don't get to see clearly why this is true.

Also, I saw a similar question here but don't completely understand the answer. I would appreciate any help.

答:
You need to assume $H\neq G$, of course... Also, the statement is false for $G=C_2$ and $H=\{e\}$. Then $k=2$, but $|G|$ does not divide $\frac{2!}{2}=1$. So assume $G\neq C_2$.

Let $G$ act on the cosets of $H$, which gives us a homomorphism $\varphi\colon G\to S_k$; the kernel is normal in $G$, but cannot equal $G$ because $H\neq G$ and the action is transitive. Hence, $G$ is isomorphic to a subgroup of $S_k$.

Now, $\varphi(G)\cap A_k$ is normal in $\varphi(G)$, since $A_k$ is normal in $S_k$. Thus, either $\varphi(G)\cap A_k=\varphi(G)$, or else $\varphi(G)\cap A_k=\{e\}$.

In the latter case, let $g\in G$, $g\neq e$. Then $\varphi(g^2)=e$, since $\varphi(g)^2\in A_k\cap \varphi(G)$. Thus, $\varphi(G)$ has exponent $2$ and is abelian, hence being simple you must have $G$ cyclic of order $2$, contradicting our working assumption that $G\neq C_2$. Thus, $\varphi(G)\cap A_k$ cannot be trivial, and so $\varphi(G)\cap A_k=\varphi(G)$, proving that $\varphi(G)\subseteq A_k$. Thus, $|G|=|\varphi(G)|$ divides $|A_k| = \frac{k!}{2}$, as claimed.


问:
max min is less than min max proof
I saw the following proof that max min of a function is $\leq$ than min max of a function on Max Min of function less than Min max of function, pasted below for your reference

 Let $ f(x_{0}, y_{0}) = \max_x\min_y f(x, y)$  and  $f(x_{1},
> y_{1}) = \min_y\max_x f(x, y)$.

By this definition the problem is to prove that $f(x_{0}, y_{0}) \leq
> f(x_{1}, y_{1})$ provided that they exist.

By definition of min and max function we have:

$\min_yf(x, y) = f(x, y_{0}) \leq f(x, y) \forall y$. Here $\min_yf(x,
> y)$ would be a function only of x.

$\max_xf(x, y_{0})=f(x_{0}, y_{0}) \geq f(x, y_{0})\forall x$. Here
$\max_xf(x, y_{0})$ is a scalar.

$\max_xf(x, y)=f(x_{1}, y) \geq f(x,y) \forall x$. Here
$\max_xf(x_{1}, y)$ is a function of y.

$\min_yf(x_{1}, y)=f(x_{1},y_{1}) \leq f(x_{1}, y) \forall y$. Here
$\min_yf(x_{1}, y)$ is a scalar.

From all this equation you get the following inequalities:

$f(x, y_{0}) \leq f(x_{0},y_{0}) \leq f(x,y) \leq f(x_{1}, y_{1}) \leq
> f(x_{1}, y)$

My confusion is on the last line, where did the $f(x_{0},y_{0}) \leq f(x,y) \leq f(x_{1}, y_{1})$ come from? I don't see how this conclusion was made from the prior steps.

答: $f(x_0, y_0) \leq f(x, y)$ because $f(x, y_0) \leq f(x, y)$ for all $x$ and $y$, and $f(x_0, y_0) \geq f(x, y_0)$ since $x_0$ is a maximizer. Similar logic applies to $f(x_1, y_1)$.

Title: max min is less than min max proof


问:
Finding range value given normal distribution probability, mean, and variance
I am attempting to find an unknown value given the probability of the range of X.  I am unsure of how to calculate the value given the added constants defined in the problem and which values I need to use to look up the z-scores to plug into the x = z + $\mu (\sigma)$. I have found z given probability of 0.95 to be 1.644853.  Not really sure if that is the correct value to use or how to go from here?  Here is the problem:

A random variable X has the distribution $N(1500, (200)^2)$.  Find B, where $P(1500 - B < X < 1500 + B) = 0.95$

答:
You can normalize the distribution and convert N(1500, 200^2) to N(0, 1):

P(1500-B < X < 1500+B) = 0.95

P(-B/200 < (X - 1500)/200 < B/200) = 0.95

Then you should be able to look up the value for B/200 in the table to find where

P(-B/200 < Z < B/200) = 0.95 where Z = $\frac{X-\mu}{\sigma}$= N(0, 1).

Then you can solve for B by setting that value you found in the table equal to B/200.


答:
Edited my answer.

As J. W. Tanner said in the comments, $\sqrt{9} = 3$. So $x = 3,\ $ not $-3$.

Therefore $x^2 = 9,\ $ because $3^2 = 9$.

Now the notation $t = \pm 3$ means that $\left(t=3\ \ or\ \ t = -3\right).$

So it is true that $x = \pm 3$ here because $\left(x=3\ \ or\ \ x = -3\right) $ is true.

Just because $x^2 = 9,\ $ doesn't mean that $x$ equals both $3$ and $-3$. It just means that $x = 3$ or $x = -3$ (or both). In this case, $x=3$ only, because $x$ was defined to be equal to $3$ only.

Title: Square root of 9 from a technical perspective.

Body:
When we say $$  \sqrt{9}= x $$
then $\;x = 3,\;$ right?

So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$


问:
Show that a function is continuous
Let K be bounded and continuous and bounded on $\mathbb{R}^{n}$ and let $f$ be Lebesgue integrable on $\mathbb{R}^{n}$.

Show that the function $g$ defined on $\mathbb{R}$ by

$g(t) = \int_{\mathbb{R}^{n}} K(tx)f(x)dx$

is well defined and continuous on $\mathbb{R}$.

Well defined:

$g(t) \leq |g(t)| \leq |\int K(tx)f(x)dx| \leq \int |K(tx)||f(x)|dx \leq N\int|f(x)|dx < \infty $,

where $N = \sup \{ |K(x)| : x \in \mathbb{R}^{n} \}$.

Now prove it is continuous, I think I have come up with a raher complicated approach.

Since $f$ is integrable, we have for a positive number $M \in \mathbb{R}$ that

$\int_{B(0,M) \cap\mathbb{R}^{n}} |f(x)|dx = \int |f(x)| dx - \epsilon'$,

where $B(0,M)$ is a ball in $\mathbb{R}^{n}$ with center $0$ and radius $M$.

Now, since $K$ is continuous on $\mathbb{R}^{n}$, it is uniformly continuous on the compact set $B(0,M) \cap\mathbb{R}^{n}$.

Letting $|\frac{t_{1} - t_{0}}{M} | < \frac{\delta}{M}, \delta > 0$,

we get that on this compact subset

$|g(\frac{t_{1}}{M}) - g(\frac{t_{0}}{M}) | = |\int_{B(0,M) \cap\mathbb{R}^{n}} (K(\frac{t_{1}x}{M}) - K(\frac{t_{0}x}{M}))f(x)dx| $.

Since $||\frac{t_{1}x}{M} - \frac{t_{0}x}{M}|| = |\frac{t_{1}}{M} - \frac{t_{0}}{M}|$ $||x|| < \delta $ ,

$|K(\frac{t_{1}x}{M}) - K(\frac{t_{0}x}{M})| < \epsilon$  $\forall x \in B(0,M)\cap \mathbb{R}^{n}$ and we get

$|\int_{B(0,M) \cap\mathbb{R}^{n}} (K(\frac{t_{1}x}{M}) - K(\frac{t_{0}x}{M}))f(x)dx| \leq \int_{B(0,M) \cap\mathbb{R}^{n}} \epsilon|f(x)| dx =\epsilon (\int_{\mathbb{R}^{n}} |f(x)|dx  - \epsilon')$.

Since we can make $\epsilon$ and $\epsilon'$ arbitrary small, $g$ is continuous.

Now this seems to me like a complicated solution (if it is valid), but at the moment I can't come up with something simpler. I appreciate feedback on my approach, whether my proof is correct or if something is wrong.

A simpler solution would also be nice!

Thanks in advance!

/Erik

答:
For continuity you can apply the Dominated convergence theorem:

Consider a $t\in\mathbb R$ and a sequence $t_n\to t$. Then $$g(t_n) = \int_{\mathbb{R}^{n}} K(t_nx)f(x)dx$$
Let $h_n(x)=K(t_nx)f(x)$. It is apparent that $h_n\to h=K(tx)f(x)$ since $K$ is continuous.

Also $|h_n(x)|=|K(t_nx)||f(x)|\leq \sup|K(x)|\cdot |f(x)|=N|f(x)|$. Now note that $N|f(x)| $ is integrable, therefore the dominated convergence theorem applies:

$$g(t) = \int_{\mathbb{R}^{n}} K(tx)f(x)dx=\int_{\mathbb R^n}
\lim\limits_{n\to\infty}K(t_nx)f(x)dx=$$
$$\lim\limits_{n\to\infty }\int_{\mathbb R^n}K(t_nx)f(x)dx=\lim\limits_{n\to\infty}g(t_n)$$thus $g$ is continuous.

Title: Show that a function is continuous


问:
No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$
How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with

$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1,  & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$

I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.

Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq  \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.

Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?

答: Hint Use Darboux Theorem. If you are not familiar with it, mimic and simplify (due the fact that you have a very simple function) one of its proofs.


问:
the (seeming) loop caused by defining Natural numbers through inductive sets.
I'm a first year student in mathematics, and we recently defined Natural numbers as following:

"For all numbers, a number is an element of the set Natural numbers, if and only if it is an element of every inductive set".
An inductive set is defined as "A subset of the Real numbers that contains 1 and for each element, that element + 1 is also an element of the set'.

One's natural instinct is to ask why not define $\mathbb{N}$ = {1, 2, 3 ...} which is easy to dismiss because we cannot define an infinite set in Roster notation when there is no set builder notation. We have not yet defined what natural numbers are and therefore these numbers are not related by any means (The only property that they share is being Natural which yet has no definition in our stack. even if one wishes to define this set inductively, through something call a successor function if I'm not mistaken, the domain of the said function must be the natural numbers which...defeats the purpose? Someone might argue for a definition through sequences, which...can one have sequences before defining Natural numbers?).

I suppose the reason why we are being so rigorous in defining natural numbers in such way is that we wish to prove things by induction. When we prove through induction we basically define a set like M to be the subset of $\mathbb{N}$ of all numbers who satisfy P(n) which P(n) is usually our the algebraic conclusion in our theorem. We prove 1 $\in$ M and $\forall n (n+1 \in M)$. From this we conclude M = $\mathbb{N}$.

This triviality is in need of proving, so we naturally (no pun intended) resort to making a more rigorous definition of $\mathbb{N}$ in order to be able to use induction.
We define inductive sets, then define a Natural number as said above, see that according to definition $\forall \hspace{1mm} inductive A, \hspace{1mm}(\mathbb{N} \subseteq A)$.

So it is sufficient to know a subset of Natural numbers is inductive to prove they are equal (which we apply in M for the above). It is easy now to prove that 1, 2, 3 and so on are natural numbers individually, and I don't know if I'm being nitpicky about this, but I have the following concern: when we prove things through induction we essentially have in mind that we want to prove that a theorem holds for {1, 2, ...} and that happens to be called $\mathbb{N}$. We can't have {1, 2, 3, ...} because of the reasons above so we take $\mathbb{N}$ as nothing but a name and define it as we did above. we can prove that 1 $\in \mathbb{N}$ and so on, but that is a stretch of it being equal to the infinite set of {1, 2, 3, ...}. Instantly, it seems ridiculous to call it a stretch for it not being equal to something that we cannot define or is defined through something we earlier defined. does it suffice to say {n| n $\in \mathbb{N}$} = {1, 2, 3, ...}, or we should find a separate way to define the algebraic form of Natural numbers and then prove it to be equal to the $\mathbb{N}$ we defined above through inductive sets.
It also seems that we can prove 1 $\in \mathbb{N}$, 2 $\in \mathbb{N}$ and ... . But it seems that proving {1, 2, ...} to be equal to $\mathbb{N}$ is (in addition to needing to define {1, 2, ..} in a way it makes sense) in itself in need of induction for proving.

I would be most thankful if someone would help clarify my perspective.

答:
"Induction" is actually a much more general concept, and here, it's being applied specifically to $\mathbb{N} \subset \mathbb{R}$ in a way that rather obfuscates what's going on.

A notion of the "natural numbers" arises in this way from any sufficiently-large structure with a nice enough relation on it. For example, we can conduct exactly the same construction as you originally did, but in the universe of sets. There is a nice relation on the universe of sets, $\in$, and we can define an "inductive set" $I$ to be a set which contains the empty set $\emptyset$ and which also has the property that for every set $x \in I$, $I$ also contains $\{x\}$. Then the analogous "natural numbers" is the intersection of all the inductive sets, i.e. $\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \dots\}$.

All that is needed for this construction to work is that there is some nice operation that produces a bigger thing from a smaller thing, and that there is somewhere to start. In your example, the starting place was $1$; in my sets example, it was $\emptyset$. In your example, the operation was $x \mapsto x+1$; in my example, it was $x \mapsto \{x\}$.

In this way, we can basically "build a thing that looks like the natural numbers" anywhere we want. (There are some technical restrictions.) The natural numbers aren't really this privileged subset of the real numbers, though they can certainly be defined that way. Rather, they are the abstract concept of the smallest place where you can keep making something bigger forever. Your course has identified one rather natural way that this abstract concept shows up in the reals, and it has shown you a method that will let you identify instances of that concept in other places. The method is extremely general, because it didn't rely on anything: it just said "this is what it means to be inductive; now take the intersection of all the inductive things".

The crucial reason we're interested in $\mathbb{N}$ is because it's the fundamental thing you can do induction on. Induction on your conception of "$\mathbb{N}$ as a subset of the reals" is all very well, but is a bit weird. It is much better to think of induction as happening on "anything you can climb up", i.e. "anything with an operation that makes things bigger". If a property $P$ is such that "$P$ is true on everything less than $x$" always implies "$P$ is true for $x$" whenever $x$ is in your inductive structure, then $P$ is true for everything in the inductive structure.

The natural numbers happen to be especially nice because if you're working in the natural numbers, then there's nothing outside the inductive structure (by the definition of the naturals as the smallest inductive structure), so this actually tells you that $P$ is true everywhere. By contrast, showing $P$ is true for the real number 1, and that if $P$ is true for all real naturals less than $x$ then $P$ is true for $x$, you deduce a fact that is true for all real naturals but probably not for all reals.


答:
The important feature that the formal definition gives is that it specifies what is in not in $\mathbb N$ as much as it specifies what is. For instance, you have an intuitive notion that $\pi\notin\mathbb N$.  The definition lets us confirm that logically.  Even though there are some inductive sets that contain $\pi$ (heck, all of $\mathbb R$ is an inductive set), there are also inductive sets that do not (like $\mathbb Q$).  You can also point out inductive sets that don't contain negative numbers or rational numbers with a denominator greater than 1, and by the time you're done you can say things like $\{1,2,3,...\}$ in a mathematical statement without having to sheepishly add "You know what I mean".

The same issue comes up when the natural numbers are defined by the Peano axioms, where we don't have addition defined ahead of time.  It's important there because it is essential to know that there isn't some special "inaccessible" natural number that we might call $\infty$ that can't be reached by counting up from $1$.  The Peano axioms settle this with a slightly different notion of induction, but it is the same idea that we can have a set with an infinite number of elements but still be able to prove truths about all of them with a finite amount of work.

Title: the (seeming) loop caused by defining Natural numbers through inductive sets.

Body:
I'm a first year student in mathematics, and we recently defined Natural numbers as following:

"For all numbers, a number is an element of the set Natural numbers, if and only if it is an element of every inductive set".
An inductive set is defined as "A subset of the Real numbers that contains 1 and for each element, that element + 1 is also an element of the set'.

One's natural instinct is to ask why not define $\mathbb{N}$ = {1, 2, 3 ...} which is easy to dismiss because we cannot define an infinite set in Roster notation when there is no set builder notation. We have not yet defined what natural numbers are and therefore these numbers are not related by any means (The only property that they share is being Natural which yet has no definition in our stack. even if one wishes to define this set inductively, through something call a successor function if I'm not mistaken, the domain of the said function must be the natural numbers which...defeats the purpose? Someone might argue for a definition through sequences, which...can one have sequences before defining Natural numbers?).

I suppose the reason why we are being so rigorous in defining natural numbers in such way is that we wish to prove things by induction. When we prove through induction we basically define a set like M to be the subset of $\mathbb{N}$ of all numbers who satisfy P(n) which P(n) is usually our the algebraic conclusion in our theorem. We prove 1 $\in$ M and $\forall n (n+1 \in M)$. From this we conclude M = $\mathbb{N}$.

This triviality is in need of proving, so we naturally (no pun intended) resort to making a more rigorous definition of $\mathbb{N}$ in order to be able to use induction.
We define inductive sets, then define a Natural number as said above, see that according to definition $\forall \hspace{1mm} inductive A, \hspace{1mm}(\mathbb{N} \subseteq A)$.

So it is sufficient to know a subset of Natural numbers is inductive to prove they are equal (which we apply in M for the above). It is easy now to prove that 1, 2, 3 and so on are natural numbers individually, and I don't know if I'm being nitpicky about this, but I have the following concern: when we prove things through induction we essentially have in mind that we want to prove that a theorem holds for {1, 2, ...} and that happens to be called $\mathbb{N}$. We can't have {1, 2, 3, ...} because of the reasons above so we take $\mathbb{N}$ as nothing but a name and define it as we did above. we can prove that 1 $\in \mathbb{N}$ and so on, but that is a stretch of it being equal to the infinite set of {1, 2, 3, ...}. Instantly, it seems ridiculous to call it a stretch for it not being equal to something that we cannot define or is defined through something we earlier defined. does it suffice to say {n| n $\in \mathbb{N}$} = {1, 2, 3, ...}, or we should find a separate way to define the algebraic form of Natural numbers and then prove it to be equal to the $\mathbb{N}$ we defined above through inductive sets.
It also seems that we can prove 1 $\in \mathbb{N}$, 2 $\in \mathbb{N}$ and ... . But it seems that proving {1, 2, ...} to be equal to $\mathbb{N}$ is (in addition to needing to define {1, 2, ..} in a way it makes sense) in itself in need of induction for proving.

I would be most thankful if someone would help clarify my perspective.


问:
Square root of 9 from a technical perspective.
When we say $$  \sqrt{9}= x $$
then $\;x = 3,\;$ right?

So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$

答:
If $P$ implies $Q$, then this does not necessarily mean that $Q$ implies $P$. For example, all zebras are white-striped animals, but not all white-striped animals are zebras:
$$
\require{cancel}
\text{Zebra} \implies \text{Stripeyness}
$$
but
$$
\text{Stripeyness} \cancel{\implies} \text{Zebra} \, .
$$
The same principle applies here. We have
$$
x=3 \implies x^2=9
$$
but
$$
x^2 = 9 \cancel{\implies} x=3 \, . \\
$$
This is because every positive number has two square roots, and so just from the fact that $x^2=9$, we can't deduce that $x=3$. However, we certainly can deduce from $x=3$ that $x^2=9$.


答: If such a $g$ existed, we could take without loss of generality $g(0)=0$. Then $g(x)=|x|$, but this achieves a contradiction: $g^\prime(0)$, rather than being $0$, does not exist because the left- (right-)derivative is $-1$ ($1$).

Title: No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$

Body:
How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with

$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1,  & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$

I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.

Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq  \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.

Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?


问:
what does the inverse membership symbol means?
I know that the symbol $$ \in $$ stands for membership, but what does the symbol $$ \ni $$ stand for?

Because I know that in the set membership, one symbol stands for subset and the other ones for superset, but I cannot find the meaning of the other symbol

Thanks

答:
Because $a < b$ means the same thing as
$b > a$ and $A \subset B$ means the same thing as
$B \supset A$, one should not use the mirror image of $\in$ to mean "such that". It is a bad, confusing notation.</p>


答:
Just because $A$ is something that does $P$, does not mean that $A$ is the ONLY thing that does $P$.

It is true that $3^2 = 9$ but $3$ is not the only thing that when squared will equal $9$.

so $3^2 =9$ and $w^2 = 9$ does not in any way mean that $w = 3$.

This is no more valid or reasonable than saying $(1)^2- 3(1) +7 = 5$ and $(2^2 - 3(2) + 7 = 5$ and therefor $1 = 2$.

Or saying if Fred is a member of the chess club, and Betty is a member of the chess club, then Fred and Betty are the same person.

....

Your confusion is based on what is the definition of $\sqrt{9}$.

The definition of $\sqrt{k}$ has TWO parts.  If $\sqrt{k} = m$ then TWO things must be true:  1)  $m^2 = k$....  It's a common mistake to think this is the only thing that must be true but we must also have  2)  $m \ge 0$.

So although $3^2 = 9$ and $(-3)^2 = 9$ so both have 1) true.   $3>0$ but $-3< 0$ so only $3$ has 2) true and $-3$ has 2) false.

So $\sqrt 9 = 3$ and $-3\ne \sqrt 9$.

Now just because $3 = \sqrt 9$.  ANd $(\sqrt 9)^2 = 9$, it does not follow that $\sqrt 9$ is the ONLY thing that when squared equals $9$.  $\sqrt 9$ is only one thing that when squared equals $9$.

Just like Fred isn't the only member of the chess club and Betty can also be a member; $\sqrt 9$ is not the only thing that when squared is equal to $9$.  $-3$ when squared is also equal to $9$ when squared.


答:
Suppose that such a function $g$ existed.  Then based on your definition of $g'$ we can conclude that
$$g(x) = \begin{cases}
-x+c_1, &x<0\\
a, &x = 0\\
x+c_2, &x > 0
\end{cases}$$ where we don't know the value of $a$.  The only question is whether or not there are some choice of constants to make $g'(x) = \operatorname{sgn}(x)$.  If that were to work, then following would work out:
$$0 = g'(0) = \lim_{h\to 0}\frac{g(h) - g(0)}{h}.
$$
Because the definition of $g$ differs between the left and right of $0$, we need to consider the left and right limits, though as we will see it suffices to consider the right-hand limit:
$$\lim_{h\to 0^{+}}\frac{g(h) - g(0)}{h} = \lim_{h\to 0}\frac{h+c_2 - c_2}{h} = \lim_{h\to 0}\frac{h}{h} = 1 \ne 0.$$  Thus, we can conclude that there is no such function $g$ with $g'(x) = \operatorname{sgn}(x)$.


答:
The derivative of $N_\infty(t)$ is incorrect. To calculate its derivative the Leibniz integral rule should be used because the variable $t$ is in the integral sign as well as in the integrand.

$$ N^\prime_\infty (t) = exp\left(\int_t^t r(\sigma)\,d\sigma\right) m(t) + \int_{-\infty}^t r(t)\, exp\left(\int_s^t r(\sigma)d\sigma\right) m(s)\,ds $$

Now the second term is equal to r(t)N(t)
$$ r(t)N(t) = \int_{-\infty}^t r(t)\, exp\left(\int_s^t r(\sigma)d\sigma\right) m(s)\,ds $$

Moreover
$$ exp\left(\int_t^t r(\sigma)\,d\sigma\right) = exp(0) = 1 $$

Then, replacing in $N^\prime_\infty(t)$
$$ N^\prime_\infty (t) = m(t) + r(t)N(t) $$

which is the ODE. So that solution satisfies it.

Title: Malthus Model - Solution Differential Equation

Body:
I have this equation of a time dependent Malthus model with a term representing a time dependent immigration:
$$N'(t)=r(t)N(t) + m(t)$$
with $r(t)$ and $m(t)$ both continuous and periodic with Period $T$.
I have to prove that the function
$$N_\infty (t)=\int_{-\infty}^t exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(s)ds$$
is a solution of the ODE above.

I tried to derive $N_\infty (t)$ to obtain our ODE, but unsuccessful. I think that I do somethings wrong on the calculation.
I know that the derivative of an integral is so computed:
Let $G(t):=\int_a^x f(t)dt$. Then $G'(t)=f(x)$

So in our case I did:
$$N'_\infty (t)= exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(t)$$
But I think I miss something. Someone can help me please?


问:
Minimal value of the product $(x+y)(y+z)$

Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.

I want to solve this problem using geometry of a triangle:

Let $a,b,c$ be the sides of the triangle and we can substitute
$a = x + y\,,\,\, b = y + z\,,\,\, c = z + x$.

$A^{2} = xyz\left(x + y + z\right)$ is just a squared area of a triangle and $\left(x + y\right)\left(y + z\right)$ is a part of another formula of triangle  $$
\mbox{i.e.}\quad A = {1 \over 2}\,ab\sin\left(\theta\right)\quad \mbox{or}\quad
A = {1 \over 2}\left(x + y\right)\left(y + z\right)\sin\left(\theta\right)
$$
I' ve tried the same with other formulas but I don't know how to "extract" inequality.

Before the solution I would appreciate that you give a hint, and then a solution. Any help appreciated.

答:
$$(x+y) (y+z) = 2\csc \theta$$

So what is the minimum value of $2\csc \theta$? What is the corresponding value of $\theta$? What kind of relationship does this indicate among $x, y, z$?


答:
"Can I use the contraposition from Darboux's Theorem for this specific case?"

Yes. Darboux's theorem says that if $f:[a,b]\to\mathbb{R}$ is a differentiable function and suppose $f'(a)<\lambda<f'(b)$, then there is a point $x\in(a,b)$ such that $f'(x)=\lambda$.

The contraposition says that if $f'$ does not satisfy the "intermediate property above", then $f$ cannot be differentiable on $[a,b]$.

The existence of $g$ in your question would contradict this theorem: you may consider for instance $\lambda=\frac12$.


问:
What is the correct mapping notation for this set-valued function?
Let $P \subset \mathbb{N}$ be a finite index set. Given $i \in P$, suppose

$c(i) \subset P$ returns a particular subset of $P$

and $e(i) \in c(i)$ returns are particular element of that subset.

How are $c$ and $e$ expressed in arrow notation? Since $c$ is a set-valued function, presumably
$$
c : P \to \mathcal{P}(P),
$$
where $\mathcal{P}(P)$ is the set of all subsets of $P$.

For $e$, naturally
$$
e: P \to P
$$
is correct, but, it doesn't convey that $e(i) \in c(i) : i \in P$. So, would
$$
e: P \to c[P],
$$
where $c[P]$ denotes the image of $P$ under $c$, be correct?

答:
Yes, $c\colon P \to \mathcal{P}\left(P\right)$, but we cant write $e\colon P\to c\left(P\right)$. Set $c\left(P\right)$ is subset of $\mathcal P\left(P\right)$ and we don't need that set.

Function $e$ takes index $i$ and returns element of $c\left(i\right)$. As $c\left(i\right)\subseteq P$, it follows that $e\left(i\right)\in P$, so type of function $e$ is $$e\colon P\rightarrow P.$$


问:
What is the relation of $\int f dx^1\wedge dx^2\wedge ...\wedge dx^n=\int f dx^1...dx^n$
In a book "calculus on manifolds" it is defined that $\int f dx^1\wedge dx^2\wedge ...\wedge dx^n=\int f dx^1...dx^n$ but how it is possible the relate the integrand of a multilinear function (n-differential form) with the remann integral. when i am learning remann integration, i considered $dx$ as an infinitestimal distance although the lecturer didn't explain isolately what $dx$ is. I don't quite see the relationship between remann integral and tensors </p>

答:
This is a definition. See bottom of pg. 100 of Spivak's book.

If $\omega$ is a $k$-form on $[0,1]^k$, there is a unique $f$ such that $\omega = f dx^1 \wedge \cdots \wedge dx^k$. Then define
$$
\int_{[0,1]^k} \omega := \int_{[0,1]^k} f
$$
or
$$
\int_{[0,1]^k} f dx^1 \wedge \cdots \wedge dx^k= \int_{[0,1]^k} f(x^1,\cdots,x^k)dx^1\cdots dx^k
$$


问:
When is a Markov of order 1 a Martingale?
I have two questions and I am very confused about the concepts

Can a Markov process of order one also be a a Martingale?
Is any Markov process of order one also a Martingale?
For 1. I would say yes, since a Markov of order one is also an AR(1) it would be a martingale if c=0 and phi=1. However, I am very confused about the second. My answer for 2 would be no, but I cannot explain it properly.

Can anyone maybe help me?

答:
The answer by @RobertIsrael shows that the Markov property does not imply the martingale property, even when the Markov chain has order one.

As for the relevance to autoregressive processes:

An AR(1) process $(X_t)$ (where $t$ is a discrete time index) satisfies
$$
X_{t+1} = c + \phi X_t + \varepsilon_{t+1}\tag1
$$
It follows that every AR(1) process is Markov of order one, since the distribution of $X_{t+1}$ given $X_1,X_2,\ldots ,X_t$ is the same as the distribution of $X_{t+1}$ given $X_t$. Taking conditional expectations of (1) given $X_t$, we get:
$$
E(X_{t+1} \mid X_t) = c + \phi X_t\tag2
$$
This tells us that an AR(1) process is not a martingale except in the special case when $c=0$ and $\phi=1$.


问:
If $(a,b)=1$ then prove $(a+b, ab)=1$.

Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.

$(a,b)=1$ means $a$ and $b$ have no prime factors in common

$ab$ is simply the product of factors of $a$ and factors of $b$.

Let's say $k\mid a+b$ where $k$ is some factor of $a$.

Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.

So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].

But $(a,b)=1$ so $a$ cannot divide $b$.

We have a similar argument for $b$.

So $a+b$ is not divisible by any factors of $ab$.

Therefore, $(a+b, ab)=1$.

Would this be correct? Am I missing anything?

答:
Suppose that gcd(a,b)=1.

Let d=gcd(a+b,ab).

Now d|(a+b) and d|ab.

Suppose that d|a.

Then a=dqa.

Now examining a+b=dq1 we have dqa+b=dq1 b=d(q1−qa).

So then d|b and d=1.

A similar argument shows d|b implies d|a.

So we know that d|a or d|b implies d=1.

Suppose that d∤b and d∤a.

We have a+b=dq1 a=dq1−b.

Here we have that gcd(a,−b)=1=gcd(a,d) (this is from a Lemma used in the proof of the Euclidean Algorithm).

We can also derive that gcd(b,−a)=1=gcd(b,d).

So we know that a and d and b and d are both relatively prime.

Considering d|ab we know by Euclid's Lemma that since a and d are relatively prime implies that d|b.

This contradicts our assumption, so this case is not possible.

Therefore, we may conclude that d=1. □

Title: If $(a,b)=1$ then prove $(a+b, ab)=1$.


问:
Intersection of two $L^p$ spaces is complete.
Given $1 \leq p < q < \infty$, I wish to show that $L^p \cap L^q$ is a Banach space given the norm $\|f\| = \|f\|_p + \|f\|_q$. I have shown that this space is a vector space and that the norm is indeed a norm, all I have left is showing that this space is complete and am have some trouble. I was able to deduce that if a sequence is Cauchy in $L^p \cap L^q$ then it must be Cauchy in $L^p$ and $L^q$, but I am not sure where to go from here. Any advice would be appreciated.

答:
It suffices to show that given a sequence $(f_n)_n$ in $L^p \cap L^q$ such that $\sum_n \lVert f_n \rVert < \infty$, the sequence $(\sum_{k=1}^n f_k)_n$ is convergent in $L^p\cap L^q$. (see Theorem 5.1 of Folland's Real Analysis)

Let $(f_n)_n$ be a sequence in $L^p\cap L^q$ such that $\sum_n \lVert f_n \rVert < \infty$. Due to the Monotone Convergence Theorem, there exists a function $g$ defined a.e. with values in $[0,\infty]$ such that $g=\sum_n \lvert f_n \rvert$. $\lVert \cdot \rVert$ is a sum of norms, hence a norm. A limit process with the triangle inequality assures that $g \in L^p \cap L^q$.

In this situation, we claim that there exists an a.e. defined function $f$ such that $f=\sum_n f_n$. Indeed: for a.e. $x$, $\left(\sum_{k=1}^n \lvert f_k(x) \rvert \right)_n$ is a Cauchy sequence, then so is $\left(\sum_{k=1}^n f_k(x) \right)_n$. $g \in L^p\cap L^q$, so $f \in L^p \cap L^q$.


问:
Clarification: Expected number of flips required for T,H,H in any order.
If we have a coin such that $P(Heads)=p,P(Tails)=1-p$. And we continually toss until we get a at least two heads and a tails.

This is more of a clarification of a specific portion of the problem, we can condition on the outcomes of previous flips:

Let $X:=$ number of flips required to get at least two heads and one tails.

Now,

$\displaystyle E[X] = E[X\mid H]p + E[X\mid T](1-p) = \bigg( E[X \mid H,H]p + E[X\mid T,H](1-p)\bigg)p+ E[X\mid T,T](1-p)(1-p) $.

Now if we break this up:

$\displaystyle E[X\mid H,H] = 2+ \frac{1}{1-p}$ and $\displaystyle E[X\mid T,H] = 2+ \frac{1}{p}$

Thus we have

$\displaystyle \bigg( E[X \mid H,H]p + E[X\mid T,H](1-p)\bigg)p = \bigg(2p+ \frac{p}{1-p}\bigg)p+\bigg(2(1-p)+ \frac{(1-p)}{1-p}\bigg)p$

This part can be simplified further but it is not the issue I am needing clarified.

Returning to the top we have the term

$\tag{1} \displaystyle E[X\mid T,T](1-p)(1-p) $

And herein lies my confusion. The book has this term listed as being equal to

$\tag{2} \displaystyle \bigg(1+\frac{2}{p}\bigg )(1-p)$

It seems a bit subtle, as we don't require the two heads to be consecutive; hence, it seems the $\displaystyle \frac{2}{p}$ comes from linearity of expectation. That is expected 'number of flips requires to get a heads + 'number of flips to get a heads'. And the $+1$ in their expression (2) makes sense as we have been given that a tails has occurred. However, what occurs to the extra factor of $(1-p)$ I have in my expression (1) above?

答:
$E[X\mid T] = E[X\mid H T] p \ + E[X\mid TT] (1-p)$

$E[X\mid TT] = 1 + E[X\mid T]$ because we waste one flip and get back to the place from where we still need to get $2$ heads.

So, $E[X\mid T] = E[X\mid H T] p \ + (1+E[X\mid T]) (1-p)$

$\implies E[X\mid T] = \frac{1 - p}{p} + E[X\mid H T] = 1 + \frac{2}{p}$

$ \therefore (1-p)E[X\mid T] = (1 + \frac{2}{p}) (1-p)$


问:
Uniform convergence of $\sum n^\alpha \sin{\frac{x}{n}}$
I want to find the range of $a \in \mathbb{R}$ such that $\displaystyle \sum_{n = 1}^\infty n^a \sin{\frac{x}{n}}$ converges uniformly on $[-\pi, \pi]$.

If $a < -1$, then I can prove it's uniform convergence with $\displaystyle \sum_{n = 1}^\infty n^a \Big|\sin{\frac{x}{n}}\Big| \geq \displaystyle \sum_{n = 1}^\infty n^a$.

Also, If $a \geq 1$, then I can say it does not uniformly converge. It is because

$n^{a}\Big|\sin{\frac{x}{n}}\Big| \geq n\Big|\sin{\frac{x}{n}}\Big| = |x|\Big| \frac{\sin{\frac{x}{n}}}{\frac{x}{n}}\Big| \rightarrow |x| \quad (n \to \infty)$

and If $x \neq 0$, then $\displaystyle \sum_{n = 1}^\infty n \sin{\frac{x}{n}}$ does not converge.

So, I should to consider weather it uniformly convergences or not at $a \in [-1,1)$. Please some idea.

答: Hint: by Taylor's theorem with Cauchy's form of the remainder $$ \left|\sin\left(t \right) - t \right| \le t^2/2 $$


问:
Proof that $| e^x-e^y|\leq | x-y|$ for $\Re(x), \Re(y) < 0$
I am studying a proof in complex analysis which uses this estimation but does not prove it and I have been unsuccessful in trying to find a proof on the internet. My professor gave me a proof before christmas but I forgot some details. First he looked at the integral $\int_a^b e^z dz = e^b - e^a$. In my next line I have simply written a+t(b-a), a line, and then $\int_0^1 e^{a+t(b-a)}(b-a) dt = e^b - e^a$

What happens between the evaluation of the first integral and where I wrote the line? Are we applying some kind of transformation?

答:
Since $$f(b)-f(a) = \int_{\gamma}f^{'}(z)dz$$ where
$\gamma$ is a line segment $ta-(1-t)b$  and $t \in [0,1]$
one gets, sort of a mean value theorem for complex functions
$$|f(b) - f(a)| <= max_{t\in [0,1]}|f^{'}(ta + (1-t)b)|*|b-a|$$
The rest follows from the formula on $f(z) = e^{z}$ that is $f^{'}(z) = e^{z}$ and the fact that Re(x) < 0 and Re(y) < 0.

Note that the last assumption implies that Re(z) on the line segment above is negative as well.


问:
Find all $f(x)$ so that $[y(x)]^2\cos x + y(x)f(x)y'(x)=0$ is exact
Finde all functions $f(x)$, so that the ODE is exact...
$$[y(x)]^2\cos x + y(x)f(x)y'(x)=0$$
First I rewrote the ODE slightly...
\begin{align*}
    &[y(x)]^2\cos x + y(x)f(x)y'(x)=0 \\
    \Longrightarrow \quad & [y(x)]^2\cos x + y(x)f(x)\frac{dy}{dx}=0 \\
    \Longrightarrow \quad & ([y(x)]^2\cos x)dx + y(x)f(x)dy=0 \\
\end{align*}
Then I identified $P$ and $Q$...
\begin{align*}
    &P_x(x,y) = [y(x)]^2\cos x\\
    &Q_y(x,y) = y(x)f(x) \\
    \\
    \Longrightarrow \quad &P_{x,y}(x,y) = 2y(x)\cos x \\
    \Longrightarrow \quad &Q_{x,y}(x,y) = y'(x)f(x) + y(x)f'(x) \\
\end{align*}
Since $P_{x,y}$ must be equal to $Q_{x,y}$ for the ODE to be exact, I put them equal to find out $f(x)$ but following equations doesn't look solveable...
\begin{align*}
    y'(x)f(x) + y(x)f'(x) \overset{!}{=}2\cos(x) y(x)
\end{align*}
Any suggestions where I went wrong?

答:
Basically, partial w.r.t $x$ treats $y$ as a constant. In long, assume ODE is an exact differential:
\begin{align}
y^2(x)\cos x+y(x)f(x)\frac{\mathrm dy}{\mathrm dx}&=\frac{\mathrm d}{\mathrm dx}\lambda(x,y)=\frac{\partial \lambda}{\partial x}+\frac{\partial \lambda}{\partial y}\frac{\mathrm dy}{\mathrm dx},
\end{align}
Equate like terms
\begin{align}
y^2\cos x&=\frac{\partial \lambda}{\partial x}\\
yf(x)&=\frac{\partial \lambda}{\partial y},
\end{align}
Integrate the first (arbitrary which of the two you choose, and you can also differentiate the two for the same result, as you did),
\begin{align}
-y^2\sin x+g(y)=\lambda,
\end{align}
take a derivative and set equal to the latter,
\begin{align}
-2y\sin x+\frac{\mathrm dg}{\mathrm dy}=yf(x),
\end{align}
we care about $f(x)$ so let $g(y)$ be zero, and
\begin{align}
f(x)=-2\sin x.
\end{align}
Notice that if I took second order partial derivatives and set them equal to each other I would end up with
\begin{align}
2y\cos x=y\frac{\mathrm df}{\mathrm dx},
\end{align}


问:
Intuition behind the Poisson distribution parameter lambda
Why do we define $\lambda = np$ when deriving the Poisson distribution from the Binomial distribution? How do you come up with this definition? I couldn't find a good intuitive explanation of why this is done. My textbook and professors just stated this definition and then made a derivation.

Body:
You are not 'deriving' the Poisson distribution from the binomial
distribution. In some circumstances (especially when binomial $p$ is small and binomial $n$ is large) it is possible to find a Poisson distribution
that approximates the a binomial distribution. Such an approximation works best if the Poisson mean $\lambda$ matches the binomial mean $np.$

Here is a table, made using R, comparing selected probabilities for
$\mathsf{Binom}(n = 200, p = 0.1)$ and $\mathsf{Pois}(\lambda = 20).$

(You can ignore line numbers in [ ]s.)

k = 5:35
b.pdf = round(dbinom(k, 200, .1), 4)
p.pdf = round(dpois(k, 20), 4)
abs.dif = abs(b.pdf - p.pdf)
cbind(k, b.pdf, p.pdf, abs.dif)

Table:

       k  b.pdf  p.pdf abs.dif
 [1,]  5 0.0000 0.0001  0.0001
 [2,]  6 0.0001 0.0002  0.0001
 [3,]  7 0.0003 0.0005  0.0002
 [4,]  8 0.0009 0.0013  0.0004
 [5,]  9 0.0021 0.0029  0.0008
 [6,] 10 0.0045 0.0058  0.0013
 [7,] 11 0.0087 0.0106  0.0019
 [8,] 12 0.0153 0.0176  0.0023
 [9,] 13 0.0245 0.0271  0.0026
[10,] 14 0.0364 0.0387  0.0023
[11,] 15 0.0501 0.0516  0.0015
[12,] 16 0.0644 0.0646  0.0002
[13,] 17 0.0775 0.0760  0.0015
[14,] 18 0.0875 0.0844  0.0031
[15,] 19 0.0931 0.0888  0.0043
[16,] 20 0.0936 0.0888  0.0048
[17,] 21 0.0892 0.0846  0.0046
[18,] 22 0.0806 0.0769  0.0037
[19,] 23 0.0693 0.0669  0.0024
[20,] 24 0.0568 0.0557  0.0011
[21,] 25 0.0444 0.0446  0.0002
[22,] 26 0.0332 0.0343  0.0011
[23,] 27 0.0238 0.0254  0.0016
[24,] 28 0.0163 0.0181  0.0018
[25,] 29 0.0108 0.0125  0.0017
[26,] 30 0.0068 0.0083  0.0015
[27,] 31 0.0042 0.0054  0.0012
[28,] 32 0.0024 0.0034  0.0010
[29,] 33 0.0014 0.0020  0.0006
[30,] 34 0.0008 0.0012  0.0004
[31,] 35 0.0004 0.0007  0.0003

In the following graphical comparison, differences in probabilities less than 0.003
are difficult to distinguish.

hdr="PDFs of BINOM(200, .1) [bars] and POIS(20)"
k  = 0:40
plot(k, dbinom(k,200,.1), type="h", lwd=2,
      col="blue", main=hdr)
 points(k, dpois(k, 20), col="red")
 abline(h = 0, col="green2")
 abline(v = 0, col="green2")

Note: Binomial probabilities $X\sim\mathsf{Binom}(n,p)$ have $P(X > n) = 0.$ Technically, Poisson probabilities never reach $0,$ no
matter how far into the right tail you go. But for practical
purposes the probabilities become too small to be of importance
beyond some finite point. In particular, for $Y \sim\mathsf{Pois}(\lambda= 20),$ we have $P(Y > 40) \approx 0.$ Markov's inequality does not give
'tight' bound, but does guarantee $P(Y \ge k\lambda) \le \lambda/k\lambda = 1/k.$

1 - ppois(40,20)
[1] 2.542632e-05


答:
We need to verify that the mixed partial derivatives are equal for the ODE to be exact, or in other words, it needs to hold that
$$
\frac{\partial}{\partial y} \left[y^2 \cos(x)\right] = 2y \cos(x) = y f'(x) = \frac{\partial}{\partial x} \left[yf(x)\right]
$$
Which in turn implies
$$
f'(x) = \frac{df}{dx} = 2 \cos(x)
$$
assuming $y(x)$ is not the trivial $0$ solution. From here you just integrate
$$
f(x) = \int 2 \cos(x) \ dx = \boxed{2 \sin(x) + C}
$$

In your solution, you incorrectly used the chain rule on $\frac{\partial}{\partial x} \left[yf(x)\right]$. The reason it's wrong is that in the exact ODE method you assume that you have a parent function $F(x,y)$ of which $x$ and $y$ are independent variables, so when you take the total derivative of $F$ you get
$$
dF = F_x  dx + F_y dy
$$
where here $F_x = P_x$ and $F_y = Q_y$. In turn, if this $F$ exists (and if it's smooth enough) its mixed partials should be equal, in other words
$$
F_{x,y} = F_{y,x} \color{blue}{\implies}  P_{x,y} = Q_{y,x}
$$
which is the reason we equal the partial derivatives.

Title: Find all $f(x)$ so that $[y(x)]^2\cos x + y(x)f(x)y'(x)=0$ is exact

Body:
Finde all functions $f(x)$, so that the ODE is exact...
$$[y(x)]^2\cos x + y(x)f(x)y'(x)=0$$
First I rewrote the ODE slightly...
\begin{align*}
    &[y(x)]^2\cos x + y(x)f(x)y'(x)=0 \\
    \Longrightarrow \quad & [y(x)]^2\cos x + y(x)f(x)\frac{dy}{dx}=0 \\
    \Longrightarrow \quad & ([y(x)]^2\cos x)dx + y(x)f(x)dy=0 \\
\end{align*}
Then I identified $P$ and $Q$...
\begin{align*}
    &P_x(x,y) = [y(x)]^2\cos x\\
    &Q_y(x,y) = y(x)f(x) \\
    \\
    \Longrightarrow \quad &P_{x,y}(x,y) = 2y(x)\cos x \\
    \Longrightarrow \quad &Q_{x,y}(x,y) = y'(x)f(x) + y(x)f'(x) \\
\end{align*}
Since $P_{x,y}$ must be equal to $Q_{x,y}$ for the ODE to be exact, I put them equal to find out $f(x)$ but following equations doesn't look solveable...
\begin{align*}
    y'(x)f(x) + y(x)f'(x) \overset{!}{=}2\cos(x) y(x)
\end{align*}
Any suggestions where I went wrong?


问:
How to prove that any finite extension field for this field would be cyclic
This question was asked in mid term exam of Field Theory quite earlier this year and I was unable to completely solve and so I am looking for help here.

Let $K$ be a field, $\bar K$ an algebraic closure of $K$, and $\sigma \in \operatorname{Aut}_K(\bar{K})$. Let
$F = \{u \in \bar{K} : \sigma(u)=u\}$.
Then $F$ is a field and every finite dimensional extension of $F$ is cyclic.

I have proved $F$ to be a field but unable to prove that any finite dimensional extension would be cyclic. I took the extension to be $K = F(u_1,\dots,u_k)$ but I am unable to get any ideas why it should be cyclic and I would require some help.

答: Take any finite Galois extension $E/F$. Verify that its Galois group must be generated by $\sigma|_E$. Then given any finite separable extension $E/F$, apply the previous sentence to its Galois closure $\hat{E}/F$, and use that cyclic groups are abelian to conclude that $E=\hat{E}$ was already Galois over $F$. Finally, note that $F$ must be perfect, since if $\sigma(x^p)=x^p$, then $\sigma(x)=x$.


问:
Similar statements for expressions
Is there an easy way to find out which 3 are similar from the left and right side, it will be nice with some tricks to find it out, or if you have some rules that can be followed.

$$
{lg\,n +\frac12} \qquad \qquad \frac {(n^2 + lg\,n)}n
$$
$$
\frac {n^2} 2 + n\, lg \,n \qquad \qquad n^{lg \,10}
$$
$$
\sum_{i=1}^{lg \, n} 2^i \qquad \qquad \sum_{i=1}^{100} \frac12
$$
$$
100(4 + \frac1n) \qquad \qquad 5n(4 + 2^{lg\,n})
$$

ps.
i'm a noob at math

答: If you’re talking about rates of growth, $\sum_{i=1}^{100}\frac12=50$ is $\Theta(1)$, as is $100\left(4+\frac1n\right)$. Assuming that $\lg$ is the binary log, $n\le\sum_{i=1}^{\lg n}2^i\le 2n$ for $n\ge 1$, so $\sum_{i=1}^{\lg n}2^i$ is $\Theta(n)$, as is $\frac{n^2+\lg n}n=n+\frac{\lg n}n$. And $\frac{n^2}2+n\lg n$ is $\Theta(n^2)$, as is $5n(4+2^{\lg n})=5n(4+n)=5n^2+20n$. The unmatched functions are $\lg n+\frac12$, which is $\Theta(\lg n)$, and $n^{\lg 10}$, which is already the simplest function in its growth class, falling strictly between $\Theta(n^3)$ and $\Theta(n^4)$.</p>

Title: Similar statements for expressions


问:
Is the Schwartz Kernel of the Identity an $L^2$ function?
The answer must be no, otherwise the identity would be a H.S. operator. However, whats wrong with the following:

We have by Schwartz kernel theorem

$k_I( \phi\otimes\psi)=(I\phi,\psi)_{L^2}$

Therefore, for $\psi,\phi \in D(x)$ we have

$|k_I( \phi\otimes\psi)|=|(I\phi,\psi)_{L^2}|\le ||\phi||_{L^2}||\psi||_{L^2}=||\phi\otimes\psi||_{L^2\otimes L^2}$

Since $D(X)\otimes D(X)$ is dense in $L^2(X)\otimes L^2(X)$ and $L^2(X)\otimes L^2(X)\cong L^2({X^2})$ the above lines seems to show that $k_I$ is in $(L^2({X^2}))^*=L^2({X^2})$.

答:
I believe the problem is that the Schwartz Kernel Theorem simply does not hold in the $L^2$-setting: You will not be able to find a $k_I \in L^2(X^2)^* = L^2(X^2)$ with $k_I(\phi \otimes \psi) = \langle \phi, \psi \rangle_{L^2}$ for all test functions $\phi, \psi$. In fact, I think your reasoning constitutes a proof of this fact.

There are many intuitive reasons for this: An expression like $\phi \otimes \psi \mapsto \langle \phi, \psi \rangle_{L^2}$ is supported in the diagonal of $X^2$, which is a set of measure zero, so this can't be a well-defined functional on $L^2(X^2)$.

What you can find is a distribution $k_I \in \mathcal D'(X^2)$ fulfilling the property, simply the delta distribution $\delta(x-y)$ along the diagonal. But if you stay in $L^2$, this will not be possible. If you want to get into the functional-analytic details: The problem is that $L^2(X)$ is not a nuclear space.


问:
Truncated alternating binomial sum
It is easily checked that
$\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem.

I'm trying to figure out what happens with the truncated sum
$\displaystyle\sum_{i\ =\ 0}^{D}\left(\, -1\,\right)^{i}\binom{n}{i}$.

How far away from $0$ can this get, as a function of $D$ ?.

I'm mostly interested in the case of when $D \ll n$, such as
$D \sim \,\sqrt{\,n\,}\,$.

Thanks !

答:
Just to show another way
$$
\eqalign{
  & \sum\limits_{i = 0}^D {\left( { - 1} \right)^{\,i} \left( \matrix{  n \cr   i \cr}  \right)}
  = \sum\limits_{i = 0}^D {\left( \matrix{  i - n - 1 \cr   i \cr}  \right)}  =   \cr
  &  = \sum\limits_i {\left( \matrix{  D - i \cr   D - i \cr}  \right)
\left( \matrix{  i - n - 1 \cr   i \cr}  \right)}  =   \cr
  &  = \left( { - 1} \right)^{\,D} \sum\limits_i {\left( \matrix{   - 1 \cr   D - i \cr}  \right)
\left( \matrix{  n \cr   i \cr}  \right)}  =   \cr
  &  = \left( { - 1} \right)^{\,D} \left( \matrix{  n - 1 \cr   D \cr}  \right)
\quad \left| \matrix{
  \;n \in \mathbb C \hfill \cr
  \;0 \le D \in \mathbb Z \hfill \cr}  \right.
 \cr}
$$
where the steps are:

upper negation;

replacing the bounds on the sum with a binomial;

upper negation on both binomials;

Vandermonde convolution.

Title: Truncated alternating binomial sum


问:
proof verification on stone weierstrass problem
Let $f$ be a continuous function in $[a,b]$ stone-weierstrass theorem tell us that it exist a polynomial sequence {$p_n$} such that $p_n \rightarrow f$ uniformly in $[a,b]$. Suppose $m_n$ is the degree of the polynomial $p_n$ if $f$ a polynomial show that $m_n \rightarrow \infty$.

Proof:

let  $m_n \rightarrow k$ , $k \in \mathbb{N}$ consider also $p_n= a_0x^{b_0}+a_1x^{b_1}+...+a_nx^{b_n}$ where $a_i \in \mathbb{R}$  , $b_i \in \mathbb{N}$, $0\leq i\leq n$ and

$b_i<b_{i+1} $ now since $m_n \rightarrow k \Rightarrow $  $lim p_n\rightarrow a_0x^{b_0}+a_1x^{b_1}+...+a_nx^{k}$ and $a_n \neq0 $ now since $p_n \rightarrow f$

$\left | a_0x^{b_0}+a_1x^{b_1}+...+a_nx^{k}-f \right |<\varepsilon ,\: \forall \varepsilon >0$ and this is a contradiction because $a_0x^{b_0}+a_1x^{b_1}+...+a_nx^{k} $ is a finite sum wich means i can find one $\varepsilon$  such that $\left| a_0x^{b_0}+a_1x^{b_1}+...+a_nx^{k}-f \right |> \varepsilon $.

I am sure this isn't right,  but is the idea correct ? Can someone help me prove it formally? Thanks.

PS Where can I find more problems on the  stone-weierstrass theorem?

答:
Assume that $f$ is not a polynomial. Here is a proof using that every norm on a finite dimensional normed space is equivalent.

Assume that we can find a subsequence of the $m_n$ such that $m_n\leq N$ where $\deg p_n = m_n$ and $\sup_{x\in [a,b]}|p_n(x)-f(x)|\rightarrow 0$. We let $p_n$ and $m_n$ denote the subsequence so we assume that $m_n\leq N$ for all $n$.

Clearly the sequence $\sup_{x\in [a,b]}|p_n(x)|$ is bounded. An alternative norm on the polynomials of degree at most $N$ is given by
$$\|\sum_{k=1}^{N}a_kx^k\|_{\mathrm{new}} = \sum_{k=1}^{N}|a_k|.$$

This implies that the sequence of coefficients are bounded. By repeated use of the Bolzano-Weierstrass theorem (applied $N$ times) we can pick out a subsequence $p_{n_k}$ such that the coefficients of the polynomial converge so that $p_{n_k}$ converges to a polynomial $p$ of degree at most $N$. This implies that $\sup_{x\in [a,b]}|p(x)-f(x)| = 0$ which is a contradiction since $f$ was not a polynomial.


问:
Does the 'arrow' internal category forgetful functor have a right adjoint?
Let $\mathcal{E}$ be a category with finite limits, and let $Arr : Cat(\mathcal{E}) \to \mathcal{E}$ be the forgetful functor that sends an internal category to its object of arrows/morphisms. Does this functor always have a right adjoint? In particular, does $Arr : Cat \to Set$ have a right adjoint?

One conjecture I have is that if $\mathcal{E}$ also has binary coproducts and $X$ is an object of $\mathcal{E}$, then the right adjoint to $Arr$ would send $X$ to the indiscrete internal category on the $\mathcal{E}$-object $X + X$ of objects.

I am particularly interested in knowing whether $Arr : Cat(\mathcal{E}) \to \mathcal{E}$ preserves (binary) coproducts in general, which is true in the case $\mathcal{E} = Set$.

答:
No, $Arr$ does not even have a right adjoint in the case $\mathcal{E}=\mathsf{Set}$. The reason is that it fails to preserve coequalizers.

Any monoid $M$ can be viewed as a category with one object (so the monoid is the set of arrows under composition). The coequalizer of a pair of monoid homomorphisms in the category $\mathsf{Mon}$ of monoids agrees with the coequalizer of the corresponding pair of functors in $\mathsf{Cat}$. But the forgetful functor from $\mathsf{Mon}$ to $\mathsf{Set}$ does not preserve coequalizers, so neither does $Arr$.


问:
What to do with an expression of the form: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$?
We are looking at the following expression:

$$\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$$

The solution is straightforward for this: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(t) dt$. Do we evaluate the given expression in like manner? Do we treat the $f(x) dx$ as if it were $f(t) dt$?

答:
You should write $\int_{u(x)}^{v(x)}f(t)dt$. Then use the fundamental theorem of calculus by first rewriting

$$\int_{u(x)}^{v(x)}f(t)dt = \int_{c}^{v(x)}f(t)dt+\int_{u(x)}^{c}f(t)dt=\int_{c}^{v(x)}f(t)dt-\int_{c}^{u(x)}f(t)dt$$
for any constant $c$. The notation $\int_{u(x)}^{v(x)}f(x)dx$ does not make sense.


问:
Confused with a proof involving minimal polynomials
I am currently reading a proof of Galois's Great Theorem and there is a certain part that I do not understand. I will not present the whole proof given in my text. I will instead try to present the necessary parts to understand my question. Here's how it goes:

Let $K$ be a splitting field of $f(X)$ over $F$ and assume that $G \equiv \text{Gal}(K/F)$ is a soluble group. Assume that $G$ is non-trivial and let
$$\{e\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_k = G$$
be a chain of subgroups such that $G_{i+1}/G_i$ is abelian. In fact, every quotient
is cyclic of prime order. Let the order of $G_{k-1}/G$ be $p$ for some prime $p$.
Let
$$L = G_{k-1}^* \equiv \text{Fix}_K(G_{k-1}).$$
The proof uses induction to conclude that there is a radical extension $E$ of $L$ containing $K$. By adjoining an element $\varepsilon \neq 1$ such that $\varepsilon^p = 1$ to $E$, if necessary, we may also assume that $X^p -1$ splits in $E$. The proof goes on and states other things that are unnecessary for this question. Here's the part that I do not understand: Now, since $\varepsilon^p = 1$ and $\varepsilon \neq 1$, the minimum polynomial of $\varepsilon$ over $F$ divides
$$X^{p-1} + X^{p-2} + \cdots + X +1.$$
How do we know that the minimum polynomial of $\varepsilon$ divides the above polynomial? I know that this must be true if and only if $\varepsilon$ is a root to the above polynomial, but I can't seem to understand how the above polynomial evaluates to zero when we replace the indeterminant with $\varepsilon$.

答: You are almost there. Note that the polynomial in question is simply $(X^p-1)/(X-1)$, and $\epsilon$ is a root of the numerator but not of the denominator.


问:
Non-regular Continuous-Time Markov Chain
Consider a Continuous-Time Markov Chain (CTMC) with $P_{i,i+1}=1$ and $v_i=i^2$. Here, $P_{i,i+1}$ is the probability of jumping from state $i$ to state $i+1$, and $v_i$ is the rate of holding times. We know a regular CTMC is defined to be a CTMC such that with probability $1$, the number of jumps in any finite time interval is finite. Prove that the above CTMC is not regular.
I appreciate any comment/hint.

my attempt: I want to use the fact that the holding time in state $i$, denoted by $\tau_i$ has an exponential distribution with rate $v_i$. Now, consider the interval $[0,t]$. The probability that the number of jumps is finite (say less than a finite number $n$) is equivalent to the probability that that $\sum_{i=1}^n \tau_i > t$. Now, $\sum_{i=1}^n \tau_i$ is the sum of exponential random variables with rates $v_i$. I need to find the pdf of $\sum_{i=1}^n \tau_i$. I do not know how to proceed.

答:
As suggested by John Dawkins, we can show that $\sum_{i=1}^\infty \tau_i < \infty$. Note that
\begin{align}
\mathbb{E}[\sum_{i=1}^\infty \tau_i] = \sum_{i=1}^\infty \mathbb{E}[\tau_i] = \sum_{i=1}^\infty \frac{1}{i^2} = \frac{\pi^2}{6} < \infty.
\end{align}
Since the expectation $\mathbb{E}[\sum_{i=1}^\infty \tau_i] $ is finite, we have $\mathbb{P}(\sum_{i=1}^\infty \tau_i < \infty) = 1$, and this completes the proof.


答:
Hint: In addition to @RobertIsrael's hint, one may proceed as follows:
Using the fact that $t\mapsto\frac{\sin t}{t}$ decreases on $[0,\pi/2]$, one gets that for all $0<x\leq M$ and $n\geq M$
$$
0<\sin1 \frac{x}{n^{1-a}}\leq n^a\sin\tfrac{x}{n}\leq  \frac{x}{n^{1-a}}
$$
Divergence of the series follows for $a\geq0$. Uniform convergence follows for $a<0$.

Title: Uniform convergence of $\sum n^\alpha \sin{\frac{x}{n}}$

Body:
I want to find the range of $a \in \mathbb{R}$ such that $\displaystyle \sum_{n = 1}^\infty n^a \sin{\frac{x}{n}}$ converges uniformly on $[-\pi, \pi]$.

If $a < -1$, then I can prove it's uniform convergence with $\displaystyle \sum_{n = 1}^\infty n^a \Big|\sin{\frac{x}{n}}\Big| \geq \displaystyle \sum_{n = 1}^\infty n^a$.

Also, If $a \geq 1$, then I can say it does not uniformly converge. It is because

$n^{a}\Big|\sin{\frac{x}{n}}\Big| \geq n\Big|\sin{\frac{x}{n}}\Big| = |x|\Big| \frac{\sin{\frac{x}{n}}}{\frac{x}{n}}\Big| \rightarrow |x| \quad (n \to \infty)$

and If $x \neq 0$, then $\displaystyle \sum_{n = 1}^\infty n \sin{\frac{x}{n}}$ does not converge.

So, I should to consider weather it uniformly convergences or not at $a \in [-1,1)$. Please some idea.


答:
This expression is badly written. It only makes sense to understand it as
$$
\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt
$$
where the dummy variable for the definite integral should not be the same as any variable in the bounds.

The general method for dealing with such derivative is called the Leibniz integral rule, which tells you how to find
$$
\frac{d}{dx}\int_{a(x)}^{b(x)}g(x,t)\,dt
$$

Your question is a special case where $g(x,t)=f(t)$.

Title: What to do with an expression of the form: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$?

Body:
We are looking at the following expression:

$$\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$$

The solution is straightforward for this: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(t) dt$. Do we evaluate the given expression in like manner? Do we treat the $f(x) dx$ as if it were $f(t) dt$?


问:
Differentiable Operator Continuous?
Consider the space $C^{\infty}[a,b]$ with norm $||f|| = max_{[0,1]} |f(x)|$, with $f\in C^{\infty}[a,b]$. Is the differentiation operator $\frac{d}{dx}$ continuous on $C^{\infty}[a,b]$?

I'm very confused because it seems almost trivial-- there are plenty of examples of the derivative not being a continuous operator-- that's why it's so hard to study.. but in this space, isn't it defined to be only those functions whose derivatives are infinitely differentiable? How do I show this formally?

答: Consider $f_n(x) = \sin(n \pi x)$. $||f_n|| = 1$, but $df_n/dx$ is unbounded.</p>

Title: Differentiable Operator Continuous?


问:
Finding the intervals of increase and decrease and concavity of $y=\frac{\sqrt{1+|x-2|}}{1+|x|}$
I have the function $f(x)=\frac{\sqrt{1+|x-2|}}{1+|x|}$, which I need to analyze. I have already proven that $\lim\limits_{x \to +-\infty}{f(x)}=0$, therefore the function has only a horizontal asymptote at $y=0$. Then I looked at three cases for the derivative

$x>2 \Longrightarrow f'(x)=\frac{1-3x}{2\sqrt{x-1}(1+x)^2}$

$x \in[0,2]\Longrightarrow f'(x)=\frac{x-7}{x(1+x)^2\sqrt{3-x}}$

$x<0 \Longrightarrow f'(x)=\frac{5-x}{2(1-x)^2\sqrt{3-x}}$

Now I need to look at the intervals of concavity of the function. I suppose I have to differentiate all derivatives for a second time and look at the results in each of the corresponding intervals. But this would require too much calculations. Isn't there a simpler way to find the intervals of increase/decrease and of concavity?

答:
hint

The derivative you gave at $ [0,2] $, is false. You should find
$$\forall x\in [0,2] \;\;f'(x)=\frac{x-7}{\color{red}{2}(1+x)^2\sqrt{3-x}}$$

Your derivative at $[2,+\infty) $ is also false


问:
Compute $E(2B_T e^{-B_T}+ 4e^{-B_T})$ for $B = (B_t)_{t \geq 0}$ a standard Brownian motion and $T=\inf\{t \geq 0 : |B_t|=1\}$.
Let $B = (B_t)_{t \geq 0}$ a standard Brownian motion, and $T=\inf\{t \geq 0 : |B_t|=1\}$. I then have to calculate the following expectation

$$E(B_T + \int_0^T B_s e^{-B_s} I(B_s \geq -1)ds)$$

where $I$ is the indicator function. By symmetry I can see that $E(B_T)=0$, thus I just have to calculate the expectation of the integral. Also between $0$ and $T$ the brownian motion clearly lies above $-1$ thus I can drop the indicator function. Through an application of Ito's lemma I found that

$$\int_0^t B_s e^{-B_s}ds = 2B_t e^{-B_t}+ 4e^{-B_t} -4 + 2 \int_0^t (B_se^{-B_s} + e^{-B_s}) dB_s$$

How would I continue from here? Should I try to show the integral is a martingale and potentially use the optional stopping theorem? If so I'd still need to calculate $E(2B_T e^{-B_T}+ 4e^{-B_T})$ which I'm not sure how to go about either? Or is there a completely different way to go about this problem?

Edit:

To show the integral is a martingale I believe it is sufficient to show that

$$\int_0^t E(B_s^2e^{-2B_s})ds \ \text{and} \int_0^tE(e^{-2B_s})ds$$

are both finite. How could I be sure of this? And then to use the optional stopping theorem I'd need to show that

$$\int_0^{t \wedge T} (B_se^{-B_s} + e^{-B_s}) dB_s$$

is bounded. How could I show this also?

答:
From the definition of $T$ we know that $B_T = \pm 1$, and by the optional stopping theorem $\mathbb{E}[B_T] = 0$ so if we let $p := P(B_T = 1)$ then we know $p = \frac 12$.  Hence
\begin{align*}
\mathbb{E}[2B_T e^{-B_T} + 4 e^{-B_T}] &= \frac 12 (2e^{-1}+4e^{-1}) + \frac 12 (-2e + 4e) \\
&= 3e^{-1} + e.
\end{align*}

Now we will show $\int_0^{t \wedge T} (B_s e^{-B_s} + e^{-B_s})dB_s$ is a uniformly integrable martingale, so we can apply the optional stopping theorem to it.  We have

\begin{align*}
\mathbb{E}[\int_0^{t \wedge T} (B_s e^{-B_s} + e^{-B_s})^2ds] &= \mathbb{E}[\int_0^{t \wedge T} (B_s +1)^2e^{-2B_s}ds] \\
&\le \mathbb{E}[\int_0^{t \wedge T} (1 +1)^2e^{2}ds] \\
&= 4e^2 \mathbb{E}[t \wedge T] \\
&\le 4e^2 \mathbb{E}[T] < \infty
\end{align*}

so the quadratic variation is bounded in $L^2$ and hence $\mathbb{E}[\int_0^{T} (B_s e^{-B_s} + e^{-B_s})dB_s]=0$.


问:
Generating functions to solve recurrence relation
Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you

Body:
Suppose
$$
\begin{align}
f(x)
&=\sum_{n=0}^\infty a_nx^n\\
&=1+\sum_{n=1}^\infty(3a_{n-1}+2)x^n\\
&=1+x\sum_{n=0}^\infty(3a_n+2)x^n\\
&=1+3xf(x)+\frac{2x}{1-x}\\
(1-3x)f(x)&=\frac{1+x}{1-x}\\
f(x)&=\frac{1+x}{(1-x)(1-3x)}\\
&=\frac2{1-3x}-\frac1{1-x}\\
&=\sum_{n=0}^\infty(2\cdot3^n-1)x^n
\end{align}
$$
Equating coefficients yields
$$
a_n=2\cdot3^n-1
$$


问:
Is this subset of $\mathbb{R^2}$ compact?
Let the set to be $\{(x,y):2x^2+y^2=1\}$. Is this subset compact? I was thinking about totally bounded and complete. So that is why I could not prove.

答:
Consider the function $f(x,y) = 2x^{2} + y^{2} - 1$ which is continuous.

Consequently, $\{(x,y)\in\mathbb{R}^{2} \mid 2x^{2} + y^{2} = 1\} = f^{-1}(\{0\})$.

Since $\{0\}$ is closed (because its complement is open), we conclude the proposed set is closed.

Moreover, it is also bounded since $\{(x,y)\in\mathbb{R}^{2} \mid 2x^{2} + y^{2} = 1\}\subseteq B(0,1)$.

Indeed, if $(x,y)$ belong to such set, we have that
\begin{align*}
\|(x,y) - (0,0)\| & = \sqrt{x^{2} + y^{2}}\\\\
& = \sqrt{x^{2} + 1 - 2x^{2}}\\\\
& = \sqrt{1 - x^{2}} < 1
\end{align*}

since $x\in[-1/\sqrt{2},1/\sqrt{2}]$. Hence it is compact.

Hopefully this helps!


问:
Expected number of maximum difference between heads and tails
Let's say a fair coin is tossed $n$ times. Let $H(i)$ denote the total number of heads by the $i$-th toss and $T(i)$ denote the total number of tails by the $i$-th toss.

Define $D(i) = |H(i) - T(i)|$, the difference at $i$-th toss.

The maximum is defined as $M(n) = max^n_{i=1}D(i)$.

Now, what'd be an asymptotic bound for $E(M(n))$?

答:
$$\frac{E M(n)}{\sqrt{n}} \rightarrow \sqrt{\frac{\pi}2}\approx 1.25$$
see here


问:
what the limit of $\lim_{x→0}f(\sqrt{4 +x^2})$
I want to use the following graph to explain one-sided limits and their relationship to two sided limits.

But I am not sure how should I make it organised and clear.

Let's say determining the limit:  $$\lim_{x→0} f(\sqrt{4 +x^2})$$

The graph of function $f$ on the intervals [-5,5] is shown below.

答:
hint

After the graph, we have

$$\lim_{t\to 2^+}f(t)=-3$$

on the other hand

$$\lim_{x\to 0}\sqrt{4+x^2}=2^+$$

Proof

Given $ \epsilon>0 $.

$$\lim_{t\to 2^+}f(t)=-3\implies$$
$$\exists \eta>0\;:\;\forall t\in\color{red}{(2,2+\eta)}\; $$
$$|f(t)+3|<\epsilon$$

and

$$\lim_{x\to0}\sqrt{4+x^2}=2\implies$$
$$\exists \delta>0\;:\;\forall x\in(-\delta,\delta) $$
$$\sqrt{4+x^2}\in\color{red}{(2,2+\eta)}$$

The combination gives

$$\forall x\in(-\delta,\delta)\;\; |f(\sqrt{4+x^2})+3|<\epsilon$$

we conclude that
$$\lim_{x\to0}f(\sqrt{4+x^2})=-3$$


问:
How to transform a matrix $A$ so that $\operatorname{norm}(A \times B) = 0$
I have two matrices, $A$ and $B$. I would like to find the transform $f$ makes the minimum change to $A$ to satisfy the constraint $\operatorname{norm}(f(A) \times B) = 0$, where norm is the $L_1$ norm.

In other words, I want to find $f$ so that $f(A) \times B$ is a matrix with all zeros.

The trivial solution would be to set all of the elements of $A$ to zero, but that is not an acceptable solution in this case.

I apologize in advance if the notation is not clear. Thanks for your help!

答:
You don't specify what you mean by a "minimum" change, so for my own convenience I'll assume that you want to minimize $\|A - f(A)\|_F$, where $\|\cdot\|_F$ denotes the Frobenius norm.

Note that a matrix $M$ will satisfy $MB = 0$ if and only if each row $r$ of $M$ satisfies $rB = 0$. In fact, for any row-vector $r$, the nearest vector for which $rB = 0$ is given by $r(I - BB^+)$, where $B^+$ denotes the Moore-Penrose pseudoinverse.

So, we should take $f(A)$ to be the matrix obtained by replacing each row $r$ with $r(I - BB^+)$. That is,
$$
f(A) = A(I - BB^+).
$$


问:
Prove or disprove: If $v_1,v_2,...,v_n$ are vectors in $\mathbb{R}^n$...
and for every $v_i$ , $i=1,..,n$ the sum of coordinates of $v_i$ is $0$, then $v_1,...,v_n$ and linearly dependent.

What I tried to do: This questions was given after I learnt about matrix similarity and eigen vectors and eigen values, so I tried to think in that direction, I couldn't really get the question saying "the sum of coordinates of $v_i$ is $0$), because I always saw coordinates come with basis so I can use them? (would be appreciated if someone explains in what way should I understand that sentence).

But I proceed to think in that direction and reached something: I chose matrix $A$ to be all ones, ($a_{ij}=1\space \forall i,j) $, and let $T(v)=Av$ a linear transformation, and here I got stuck but my idea was to try to say by contradiction that they're linearly  independent and reach contradiction if I find that $A$, is similiar to the zero matrix (if I use the sum of coordinates), and then explain that they can't be similar with different ranks, and prove the statement. But  I have no idea how to continue and my weak understanding of the question scares me if I'm in the right direction so I came here to ask.

Appreciate all the help, Thanks in advance to everyone.

答:
There's a far simpler proof. Note that if the vectors were linearly independent, they would span $\mathbb R^n$.

I claim that $\bar 1$ is not in the span of these vectors.

To see why, note that $\bar1\cdot\bar1=n$, but $$\bar1\cdot (a_1\cdot v_1+\ldots +a_n\cdot v_n)=a_1\cdot(\bar1\cdot v_1)+\ldots+a_n\cdot(\bar1\cdot v_n)=0+\ldots+0=0$$ for any choice of coefficients $a_1,\ldots,a_n$.