问:
Why does a finitely oscillating sequence always has at least $2$ limit points?
Why is it not possible to find a finite oscillating sequence which has only $1$ limit point or no limit point at all?

Why must it always have at least $2$?

I'm just not able to picture it.

Thanks

EDIT : I think I should've started out by saying how do you even picture a finite oscillating sequence? What is an intuitive way to imagine a finite oscillating sequence?

答:
We can always say that there always exist at least one limit point of a bounded sequence (why??)[ If It's infinite then we can apply Bolzano weierstrass theorem, otherwise it must be finite and must have a limit point].

Let $(a_n)$ be a finite oscillating sequence. Suppose there exists a limit point say $\alpha$ and since $(a_n)$ is not convergent implies there exists a subsequence which doesn't converge to $\alpha$ and the subsequence thus obtained is also a bounded subsequence, again by Bolzano Weierstrass theorem we get a new limit point.

Hence $(a_n)$ has at least 2 limit points


问:
quadratic formula and simplification
Consider the following equation:

$$x=\frac{(p+x(1-p))\cdot c}{1-(p+x(1-p))(1-c)}$$

Solution should be $$x=\frac{cp}{(1-c)(1-p)}$$

I tried to rewrite the first equation in such that:$$x^2[c(1-p )-(1-p)] + x[2cp - c + 1-p]-pc=0$$

Applying quadratic formula leads to something, since the term under the root cannot be simplified.

Is there another way to compute that?

ContentLicense: CC BY-SA 4.0

Title: quadratic formula and simplification

Body:
$$x^2 (c+p-cp-1)-x (c+p-2 c p-1)-c p=0$$
Discriminant is $$D=(c+p-2 c p-1)^2-4cp (c+p-cp-1)=c^2+p^2+1+2 c p-2 c-2 p=(c+p-1)^2$$
so $$x=\frac{(c+p-2 c p-1)\pm(c+p-1)}{c+p-cp-1};\;c\ne 1\land p\ne 1$$
denominator $p-cp+c-1=p(1-c)-1(1-c)=(1-c)(p-1)$
$$x_1=\frac{c+p-2 c p-1-c-p+1}{2(1-c)(p-1)}=\frac{-cp}{(1-c)(p-1)}=\frac{cp}{(1-c)(1-p)}$$
and
$$x_2=\frac{c+p-2 c p-1+c+p-1)}{2(1-c)(p-1)}=\frac{-2 (c-1) (p-1)}{2(1-c)(p-1)}=1$$
if $p=1\lor c=1$ then $x=1$


问:
Riemann integrability of step functions
Suppose the step function is defined as follows. A function $f$ is a step function on $[a,b]$, if there exists a finite partition $P$ of $[a,b]$ such that $f$ is constant on the interior of each subinterval $I_k$ of $P$, where $I_k = [x_{k-1}, x_k]$ for $k=1,\dotsc,n$. Also suppose $a = x_0$ and $b = x_n$.

Questions:

(a)
Since the function is constant on the interior of each subinterval, discontinuities occur at the endpoints. Even though there are finitely many endpoints, is it right to say that the value of $f$ at these endpoints may not be finite?

(b)
If the value of $f$ at these endpoints are finite, then it is possible to show that it is Riemann integrable. But if they are not finite, would the step function also be Riemann integrable?

(c)
A related question: Based on the above definition of the step function, is $f$ bounded on $[a,b]$?

答:
You don't say this, so to avoid foolishness: suppose throughout $a < b$.

(a) Your definition makes no requirement of $f$ on $\{x_0, x_1, x_2, \dots, x_N\}$, so $f$ is not required to exist or be defined at those points.  Since the value of a function (from any set to the reals) is always a number, if it is defined, the value is finite.  However, $f$ is free to be undefined at the various $x_i$.

(b) No.  The partition of the Riemann sum is not required to be a refinement of the partition $P$ given in the definition.  If $f$ is undefined somewhere, say at $x_i$ for some $i \in [0,N]$, then every Riemann sum partition that does not have $x_i$ as a boundary has it interior to some $I_k$ (using the notation in your definition).  Consequently, a Riemann sum may sample the undefined point, making the sum undefined.  This continues through the forest of partitions all the way through the limit as the diameter of the partition goes to zero.  Consequently, no matter how far along the limit of diameter we go, there are still Riemann sums that are undefined -- this obstructs the existence of the limit.  Consequently, the limit does not exist and such a step function is not integrable.

(c) No.  $f$ isn't even required to be defined on $[a,b]$.  We can ask: Is the restriction of $f$ to the intersection of its domain and $[a,b]$ bounded?  Yes.  You have $N$ constant segments, say of height $c_1$, ..., $c_N$, and up to $N$ finite values $f(x_1)$, ..., $f(x_N)$.  Of course a collection of between $N$ and $2N$ numbers is bounded.

Title: Riemann integrability of step functions


问:
If $T \in \mathcal{B}_{00}(\mathcal{H}, \mathcal{H})$, show $\dim(\operatorname{ran}(T)) = \dim(\operatorname{ran}(T^*))$
Taken from Conway's A course in functional analysis chapter 2 section 4 problem 3

Problem Statement: If $T \in \mathcal{B}_{00}(\mathcal{H}, \mathcal{H})$, show that $T^*\in \mathcal{B}_{00}(\mathcal{H}, \mathcal{H})$ and $\dim(\operatorname{ran}(T)) = \dim(\operatorname{ran}(T^*))$.

Below is my following attempt:

We just have to show $\dim(\operatorname{ran}(T)) = \dim(\operatorname{ran}(T^*))$ and $T \in \mathcal{B}_{00}(\mathcal{H}, \mathcal{H})\Rightarrow T^*\in \mathcal{B}_{00}(\mathcal{H}, \mathcal{H})$ will follow immediately. By theorem 2.19, we know that $Ker(T) = (\operatorname{ran}(T^*))^\perp \Rightarrow \operatorname{ran}(T^*) = Ker(T)^\perp$. Similarly, $\operatorname{ran}(A) = Ker(T^*)^\perp$. By definition of the adjoint operator, we know for any $x, y \in \mathcal{H}$ we have: $\langle Tx, y\rangle = \langle x, T^*y\rangle$.

Not sure how I would finish this proof.

答:
Let $U:=(\ker T)^\perp$. We have $H=\ker T\oplus U$ and $T|_U:U\to\mathop{\rm ran}T$ is a linear isomorphism, so based on what you already have, we get
$$\dim(\mathop{\rm ran}T^*)\,=\, \dim(U)\,=\, \dim(\mathop{\rm ran}T)\,.$$


问:
Does there exist a $GL(V)$-invariant element which is dependent on the choice of basis?
The motivating example is the element $C = \sum b_i \otimes b_i^*$, where $b_i$ a basis of some finite-dimensional vector space $V$ and $b_i^*$ its dual basis.
This element is independent of the choice of basis, I think because if $c_i$ were some other basis then there would be an invertible transformation $ g \in GL(V)$ between the two bases, however $C$ is assumed to be $GL(V)$ invariant, so this gives the same element.

But if this reasoning is correct, then it seems that all $GL(V)$-invariants are independent of basis choice, which seems both plausible but maybe wrong. What kind of counterexample is there?

答:
It depends on what you mean by "dependent on the choice of basis." The simplest definition is "not $GL(V)$-invariant," in which case the answer is no by definition.

On the other hand, your example has the funny property that it's unclear whether there's a way to define $\sum b_i \otimes b_i^{\ast} \in V \otimes V^{\ast}$ in a way that doesn't require picking a basis first. Here is one such definition: there's a natural map

$$V \otimes V^{\ast} \to \text{End}(V)$$

and $V$ is finite-dimensional iff this map is an isomorphism (in general its image is the space of finite rank operators). So when $V$ is finite-dimensional we can talk about the preimage of $\text{id}_V \in \text{End}(V)$ under this map, which turns out to be $\sum b_i \otimes b_i^{\ast}$. But this is a definition that doesn't require a choice of basis (except insofar as bases may be required to define what "finite-dimensional" means and to prove facts like the above about finite-dimensional vector spaces).


问:
A question about atlas in smooth differentiable
I know this is definitely an easy question. However, I just cannot figure out.

I am using Do Carmo's Riemannian Geometry to teach myself Differential Geometry.

In his book, he gave a definition of differentiable manifold:

A differentiable manifold of dimension $n$ is a set $M$ and a family of injective mappings $x_\alpha:U_\alpha\subset \mathbb{R}^n\to M$ of open sets $U_\alpha$ of $\mathbb{R}^n$ into $M$ such that:

(1)$\bigcup_\alpha x_\alpha(U_\alpha)=M$

(2)for any pair $\alpha,\beta$, with $x_\alpha(U_\alpha)\cap x_\beta(U_\beta)=W\neq\emptyset$, the sets $x_\alpha^{-1}(W)$ and $x_\beta^{-1}(W)$ are open sets in $\mathbb{R}^n$ and the mappings $x_\beta^{-1}\circ x_\alpha$ are differentiable.

(3) The family $\{(U_\alpha,x_\alpha)\}$ is maximal relative to the conditions (1) and (2).

Furtuhermore, he define $A\subset M$ to be an open set in $M$ iff $x_\alpha^{-1}\left(A\cap x_\alpha(U_\alpha)\right)$ is an open set in $\mathbb{R}^n$ for every $\alpha$.

I can show that this is a topology, and that $x_{\alpha}(U_\alpha)$ is open and $x_\alpha$ is continuous. However, I have no idea how to prove that $x_\alpha$ is actually a homeomorphism. It suffices to prove that $x_\alpha$ is a open mapping, but I have no idea.

Any help will be appreciated. Thanks.

答:
We check that $x_{\alpha}:U_{\alpha}\rightarrow M$ is an open mapping. So given $U\subset U_{\alpha}$ open, we only need to see that $x_{\alpha}(U)\subset M$ is open according to the definition you gave.

For any $\beta$, we have
\begin{align}x_{\beta}^{-1}\left(x_{\alpha}(U)\cap x_{\beta}(U_{\beta})\right)&=(x_{\beta}^{-1}\circ x_{\alpha})(U)\cap (x_{\beta}^{-1}\circ x_{\beta})(U_{\beta})=(x_{\alpha}^{-1}\circ x_{\beta})^{-1}(U)\;\cap\; U_{\beta}\end{align}

which is  open because $x_{\alpha}^{-1}\circ x_{\beta}$ is differentiable.

ContentLicense: CC BY-SA 3.0


问:
In how many ways can $kn$ people be arranged around $k$ tables such that each table has $n$ people?
Find a general formula to arrange $kn$ people around $k$ tables such that each table has $n$ people?

The way I dealt with it was to form $k$ groups of $n$ and got ans as (kn)!/(n!)^k which is wrong please tell the mistake

答:
Here’s a slightly different approach to counting the arrangements.

Imagine that we line up the $kn$ people at the door and let them into the room one at a time. The first $n$ of them are seated in order clockwise around the first table, the second $n$ in order clockwise around the second table, and so on. There are $(kn)!$ ways to line them up, so there are $(kn)!$ different possible seatings if both the tables and the seats at each table are considered distinct (e.g., if the tables and the seats at each table are numbered).

However, the default assumption is that two arrangements of people around a table are considered the same if one is a rotation of the other. Thus, when the first $n$ people take their seats around the first table, it doesn’t matter where the first person sits, as long as the others fill the remaining seats in order clockwise around the table. Thus, what we originally counted as $n$ distinct arrangements, one for each place at which the first person could sit, is really only one. The same is true for each of the $k$ tables, so we need to divide the original answer of $(kn)!$ by $n^k$, a factor of $n$ for each table. If this is the intended interpretation, there are $\frac{(kn)!}{n^k}$ arrangements.

Finally, if the individual tables are not considered distinct, we must divide by $k!$, and there are only $\frac{(kn)!}{k!n^k}$.


问:
Finite extension of $\mathbb Q_p$
Let $\mathbb K/\mathbb Q_p$ be a finite extension of $p-$adic field $\mathbb Q_p$. Let ${\mathcal O}=\{x\in K\;:\;|x|\leq1\}$ and ${\mathcal P}=\{x\in K:\;|x|<1\}$, here $|\cdot|$ is the absolute value. Show that the quotient ring $\mathcal O/\mathcal P$ is a finite field. What is the cardinal of its and show a complete system of representatives of the residue classes of this quotient ring.</p>

答:
${\cal{O}}$ is a very distinguished ring inside $\mathbf{K}$...what is it?

Try studying the situation first when $\mathbf{K}=\mathbf{Q}_p$.

Title: Finite extension of $\mathbb Q_p$


问:
Definite Integral of square root of polynomial
I need to learn how to find the definite integral of  the square root of a polynomial such as:
    $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$

EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.

This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

Body:
$$\int {\sqrt{36x+1}}\,dx$$

use this formula

$$\int {\sqrt{x}}\,dx=\frac {x^{\frac{3}{2}}}{\frac32}\implies \frac {2x^\frac32}{3}+C$$

so
$$\int {\sqrt{36x+1}}\,dx=\frac {2(36x+1)^\frac32}{(3)(36)}$$
$$\frac {(36x+1)^\frac32}{54}+C$$

Q2

$$\int\sqrt{2x^2 + 3x + 7}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72+\frac{3^2}{4^2}-\frac{3^2}{4^2})}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} +\frac{3^2}{4^2}+ \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + (\frac{\sqrt47}{4})^2}\,dx $$

now use this formula
$$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$
I hope you can take it now on your own


问:
Diameter of graph where $V = \mathbb{N}$ and two vertices are connected if they sum to a square
After watching the Numberphile video The Square Sum Problem, I explored the idea further and came up with this conjecture.

Let $G$ be a graph where $V(G)=\mathbb{N}$ and $E(G)=\{(a,b)\in\mathbb{N}:a+b \space$ is a perfect square$\}$. The diameter of $G$ is $3$.

Exploring finite subgraphs of this with Mathematica seems to suggest the conjecture is true. For example, the shortest path between $1$ and $2$ is $(1,15,34,2)$, and I can confirm that the maximum distance between any two integers $\le 100$ is $3$.

I can also rephrase the conjecture in number-theoretic terms:

$\forall n,m\in\mathbb{N}, \exists x,y\in\mathbb{N}:n+x,x+y,$ and $y+m$ are all perfect squares. This would imply the above conjecture, as it would yield the path $(n, x, y, m)$ for any two vertices $n,m$.

How might one prove this conjecture?

答:
Note that $$(1+h)^2-h^2-1=2h,\qquad(1+h)^2-h^2-4=2h-3,$$
so any integer (even or odd) may be written in the form $a^2+b^2-c^2$ for integers $a,b,c$.

In particular, given integers $m,n$ we have integers $a,b,c$ such that $$a^2+b^2-c^2=m+n.$$

Then $m+(a^2-m)$ is a perfect square, $n+(b^2-n)$ is a perfect square, and $$(a^2-m)+(b^2-n)=a^2+b^2-(m+n)=c^2$$ is a perfect square.

Thus we can get from $m$ to $n$ via the sequence:$$m,\quad a^2-m,\quad b^2-n,\quad n.$$

Note this is working over the integers so not quite the same as your question if you only want to consider naturals.


问:
If $\lim x_{n} = 0$, then $\frac{1}{|x_{n}|}$?
If $x_{n}$ is a sequence such that $\lim x_{n} = 0$, then $\frac{1}{|x_{n}|} = \infty$. Is this true?

What about if we have another sequence $y_{n}$ such that $\lim y_{n} = 0$, then either $\lim \frac{1}{y_{n}} = +\infty$ or $\lim \frac{1}{y_{n}} = -\infty$

Is there a way to prove/disprove some of these?

答:
Let $ (x_n)$ be a real sequence such that for $ n $ great enough $ x_n\ne 0 $ and $\lim_{n\to+\infty}x_n=0$

then

$$(\forall A>0) (\exists N\in\Bbb N)\;\;:\;$$
$$n\ge N \;\implies \; 0<|x_n|<\frac 1A$$

thus

$$n\ge N \;\implies \;\; \frac{1}{|x_n|}>A$$

which proves that
$$\lim_{n\to+\infty}\frac{1}{|x_n|}=+\infty$$

For the second, take $$y_n=\frac{-1}{n+1}$$
then
$$\lim_{n\to+\infty}\frac{1}{y_n}=-\infty$$


问:
A question on subgroups, more specifically, a finite cyclic group
I define a binary operation $\oplus$ on $\mathbb{Z} _n$ as follows: $$[a] \oplus [b] = [a + b],$$ for all $[a], [b] \in \mathbb{Z} _n$. Then $(\mathbb{Z} _n, \oplus )$ is a finite cyclic group with order n.

This proposition in my notes says that the proof involves showing that the value of $[a] \oplus [b]$ is independent of the choice of equivalence class representatives.

Could someone please clarify to me why this is the case?

Kindest regards.

答:
<<Could someone please clarify to me why this is the case?>>

Because you are trying to define an operation between classes ("$[a] \oplus [b]$") by using (arbitrary) representatives of them and the group operation ($a$, $b$ and "$+$" in the writing "$ = [a + b]$"). So, you want the result to be the same (i.e. the class $[a + b]$) if you pick any other representatives of the two classes-term (say $a'\in [a]$ and $b'\in [b]$).

This is the fee to pay to the impossibility to handle with the classes if not by means of their representatives.


问:
Does uniform convergence of distribution functions imply convergence of moments?
Suppose $X_n \sim F_n$, $X \sim F$ with $\Vert F_n - F\Vert_\infty \rightarrow 0$.

Using $EX = EX^+-EX^-$, and the integral tail formula for expectation, I have that
\begin{align*}
|EX_n - EX| =&\ \left|\int_0^\infty P(X^+_n > x)\ dx-\int_0^\infty P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)\ dx- \int_0^\infty P(X^-_n > x)\ dx \right|\\
=&\ \left|\int_0^\infty P(X^+_n > x)-P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)- P(X^-_n > x)\ dx \right|\\
=&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx +\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\
\leq&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx\right| +\left|\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\
\rightarrow 0
\end{align*}
since cdfs of $X_n^\pm$ also converge uniformly to cdf of $X^\pm$, and since, the cdfs are themselves bounded between 0 and 1, so we can take their difference to be very small outside some interval and bound them inside the interval using uniform continuity.

But how can I generalize this to higher order moments?

答:
This is not true as the following example suggests. Let $\mathsf{P}(X_n=e^n)=n^{-1}=1-\mathsf{P}(X_n=0)$ and $X=0$. In this case,
$$
\|F_{X_n}-F_X\|_{\infty}=n^{-1}\to 0.
$$
However, for each $r>0$,
$$
\mathsf{E}X_n^r=\frac{e^{rn}}{n}\to\infty.
$$


问:
Scalar times Zero Vector = Zero Vector
Is the following an incorrect argument to prove $a\vec{0} = \vec{0}$?

$$\vec{u} + \vec{0} = \vec{u} \implies\vec{0} = \vec{0} + \vec{0}$$

Multiply B.S by $a$.

$$a\vec{0} = a(\vec{0} + \vec{0})$$

Using $a(\vec{u} + \vec{v}) = a\vec{u} + a\vec{v}$

$$\implies a\vec{0} = a\vec{0} + a\vec{0}$$

Applying "1st axiom" used $\vec{u} + \vec{0} = \vec{u} \implies a\vec{0} + \vec{0} = a\vec{0}$

$$\implies a\vec{0} = \vec{0}$$

答:
Once you know
$$
a0 = a0 + a0
$$
you can use the fact that $a0$ has an additive inverse  (presumably that's one of your axioms). Add that to both sides.

Title: Scalar times Zero Vector = Zero Vector


答:
I think you are trying to find a link to your limit objects, given immense info you have on the distribution of an RVs sequence. However, your attempt this time is also wrong.

Taking $X_n$ and $X$ such that:
$$\mathcal{L}(X_n)= \left(1-\frac{1}{n} \right) \delta_0+\frac{1}{n}\delta_{n^2}$$
$$ \text{and } \mathcal{L}(X)= \delta_0$$
and you see, this time, we don't even have the convergence of the first moment, while we have the uniform convergence for CDFs.

So what is the problem here? Also, what is the problem with your above "solution"? That is you did not give a proper treatment to the tails of these random variables. More precise, even if their difference is uniformly small. There integral can always explode because you are integrating on a domain of measure infinity.

To sum up : In order for you  to have to desired results, you need to have a control on the tails of $F_n$

And in fact, this is also enough, as given a suitable control, you can imply the uniform integrability of  $|X_n|^p$

For example:: $\exists \epsilon> 0$ such that :
$$ \lim_{x \rightarrow +\infty} e^{\epsilon x}\sup_n\Vert F_n(x) -1\Vert =0 $$
(together with another limit for the minus infinity tail)

Remark :This condition is closely related to the notion "Exponential tension", which has demonstrated its importance in theory of Large deviation.

P/s: I forgot to put minus one into the bracket, last time.

Title: Does uniform convergence of distribution functions imply convergence of moments?

Body:
Suppose $X_n \sim F_n$, $X \sim F$ with $\Vert F_n - F\Vert_\infty \rightarrow 0$.

Using $EX = EX^+-EX^-$, and the integral tail formula for expectation, I have that
\begin{align*}
|EX_n - EX| =&\ \left|\int_0^\infty P(X^+_n > x)\ dx-\int_0^\infty P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)\ dx- \int_0^\infty P(X^-_n > x)\ dx \right|\\
=&\ \left|\int_0^\infty P(X^+_n > x)-P(X^+ > x)\ dx +\int_0^\infty P(X^- > x)- P(X^-_n > x)\ dx \right|\\
=&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx +\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\
\leq&\ \left|\int_0^\infty P(X^+ \leq x)-P(X^+_n \leq x)\ dx\right| +\left|\int_0^\infty P(X^-_n \leq x)- P(X^- \leq x)\ dx \right|\\
\rightarrow 0
\end{align*}
since cdfs of $X_n^\pm$ also converge uniformly to cdf of $X^\pm$, and since, the cdfs are themselves bounded between 0 and 1, so we can take their difference to be very small outside some interval and bound them inside the interval using uniform continuity.

But how can I generalize this to higher order moments?


问:
Does every restriction on word formation constitute a formal grammar?
Consider an alphabet $\Sigma$ from which I can generate words, which are just concatenations of symbols from the set $\Sigma$. I then define some "rule" that indicates whether or not a word belongs to another set called a formal language.

Does that inclusion/exclusion rule always constitute a formal grammar? If not, is there a formal name for such a rule?

Motivation: Most definitions of formal grammars are explicitly dependent on production rules, which are statements about how to "rewrite" words in a language into other words in the language. This seems far more restrictive than a simple inclusion/exclusion rule—perhaps both of these are equivalent?

答:
Definition by a production grammar and definition by a total Turing machine are equivalent definitions of recursive languages.

There exist nonrecursive languages.  The language accepted by a partial Turing machine is such a language.  ("Total" and "partial" are used in their usual function theoretic senses.  A total Turing machine halts on all finite inputs from the alphabet of the language, so implements a total function.  A partial Turing machine need not halt on all such inputs, so implements a partial function.)  Any undecidable problem gives an example.  The existence of non-recursive languages (that are still recursively enumerable) has been addressed at CS.SE and the answers there are based on undecidable problems.

So, a way to give a rule that is not equivalent to a production grammar is to say that your language is positive (xor negative) problem instances of an undecidable problem.  For instance, your language is the collection of all "sets of Diophantine equations having a solution".


问:
Is it possible to sharpen and turn this heuristic argument into a proof of Stirling's formula?
We assume that $n$ is a very large number. Then

\begin{align*} (1+n)^n &\simeq n^n + \binom{n}{1} n^{n-1} + \binom{n}{2}n^{n-2} + \cdots \\
              &\simeq n^n(1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots) \\
              &\simeq n^n e.
\end{align*}

Therefore

$$ n^n = n n^{n-1} = n (1 + n-1)^{n-1} \simeq n e (n-1)^{n-1}. $$

Also, of course $e^n = e e^{n-1}$. We thus deduce that

$$ \left( \frac{n}{e} \right)^n \simeq n \left( \frac{n-1}{e} \right)^{n-1}.$$

So if we let $f(n) = (n/e)^n$, we see that if $n$ is very large, then

$$f(n) \simeq n f(n-1),$$

which is the functional equation of the factorial (actually of the Gamma function, up to some translation by $1$).

I know this is far from being a proof, but I wonder if there is any truth in this heuristic kind of argument. Is it perhaps possible to build on it and turn it into a proof perhaps?

答:
Here's a version of the argument being a bit more careful about error terms. Write $e_n = \left( 1 + \frac{1}{n} \right)^n = \frac{(n+1)^n}{n^n}$. We have $\frac{(n+1)^{n+1}}{n^n} = (n+1) e_n$ and iterating this gives

$$n^n = n! \prod_{i=1}^{n-1} e_i$$

and hence $n! = \frac{n^n}{\prod_{i=1}^{n-1} e_i}$. Since $e_i$ converges rapidly to $e$ we expect that this gives, approximately, $n! \approx \left( \frac{n}{e} \right)^n$, and we can be more precise about this by bounding the error more precisely. As a rough first pass we have

$$\log e_n = n \log \left( 1 + \frac{1}{n} \right) = 1 - \frac{1}{2n} + O \left( \frac{1}{n^2} \right)$$

(using the first three terms of the Taylor series expansion of $\log (1 + x)$), which gives

$$\log \prod_{i=1}^{n-1} e_i = \sum_{i=1}^{n-1} \log e_i = (n-1) - \frac{H_{n-1}}{2} + O(1) = n - \frac{\log n}{2} + O(1)$$

and exponentiating this gives

$$n! = e^{O(1)} \sqrt{n} \left( \frac{n}{e} \right)^n$$

which gets us within a constant factor of Stirling's approximation. A more careful version of this argument would replace the $O(1)$ term with a constant given by the sum of the error terms in the Taylor series approximations above, which converges, which would get us Stirling's approximation with a fixed but unknown multiplicative constant (several other techniques can also prove this), up to a multiplicative error of order $e^{ O \left( \frac{1}{n} \right)} = 1 + O \left( \frac{1}{n} \right)$ (which can itself be estimated and bounded more precisely).

By the way, this is a nice observation! I don't think I've seen someone explain Stirling's approximation this way before. It is a sort of disguised discrete version of the argument based on comparing $\sum_{i=1}^n \log i$ to the integral $\int_1^n \log x \, dx$, where we instead compare the sum to $n \log n - n$ by taking finite differences.

You may also be interested in reading Terence Tao's notes on Stirling's approximation which I found very helpful. In particular the $\sqrt{2 \pi n}$ factor can be interpreted conceptually as coming from the central limit theorem.


问:
Canonical Form and DE
If we have the differential equation $x'' = x - \cos(x')$, then

In part a) Compute the corresponding non-linear 2D system and and its (unique) equilibrium

part b) compute the linearized system at the equilibrium and classify its equilibrium.

so in part a, I let y = x' and x'' = y' so I get then:

F(x,y) = (x',y') = (y, x-cos(y))

and the equilibrium is y = 0 and back-substituting, I get x = 1, so the unique equilibrium is (x,y) = (1,0)

in part b, I get the Jacobian matrix of F(x,y) and evaluate it at (1,0) and get the eigenvalues and realize that the system is NOT hyperbolic.

Now, how can we do part c, d, and e?

c) What can we tell about the non-linear system from the linearized one at the equi-
librium; draw an approximative phase-portrait of the non-linear system around the equilibrium

d) computing the Jordan form of the linearized system matrix and and the linear map
that puts it in Jordan canonical form

e) Is this system a gradient system, a hamiltonian system, or none of these? If so,
what is the hamiltonian/potential function then?

So, for part a), I introduced $y' = x$ and $y'' = x''$,

I need help especially with part c), d), and e). I did parts a) and b).

b) I get hyperbolic equilibria after computing the Jacobinan matrix and evaluating it at (1,0). The jacobian matrix I get is:

(0, 1  as the first row

(1, (x-cos(y))*sin(y)) as the second row

and evaluating at (1,0) I get hyperbolic equilibriia

答:
We are given the second-order nonlinear system:

$$\tag 1 x'' = x - \cos x'$$

Part a

We are asked to compute corresponding non-linear 2D system and and its (unique) equilibrium.

So, lets write this as two equations (linearize it sort of speak). Let $x' = y$, so $x'' = y'$ and substituting this back into $(1)$ yields the system:

$$x' = f(x, y) =  y$$

$$y' = g(x, y) = x -\cos y$$

The equilibrium points are given by the points that simultaneously satisfy $x' = 0$ and $y' = 0$, thus:

$$x'  = y = 0, \text{and}$$

$$y' =  x - \cos y = 0$$

We know $y = 0$, so $x - \cos 0 = 0 \rightarrow x = 1$. Thus, we have one critical point at $(x, y) = (1, 0)$.

Part b

Compute the linearized system at the equilibrium and classify its equilibrium.

Here we would calculate the Jacobian of the system and then classify that matrix for each critical point. The Jacobian is given by:

$$\displaystyle J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix}0 & 1\\1 & \sin y\end{bmatrix}$$

Evaluating $J$ at the critical point $(1, 0)$, yields the matrix:

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

The eigenvalues for this matrix are given by:

$$|A - \lambda I| = 0 \rightarrow (\lambda-1) (\lambda+1) = 0 \rightarrow \lambda_1 = -1, \lambda_2 = 1.$$

Since these eigenvalues are real and opposite sign, we classify this critical point as a saddle node. Generally speaking, we know that this is a robust case. Also note that we could have several critical points and would need to classify EACH one using this process!

Part c

What can we tell about the non-linear system from the linearized one at the equilibrium; draw an approximative phase-portrait of the non-linear system around the equilibrium?

Lets draw a direction field and plot solutions in a phase portrait for this system. Of course, we expect to see an unstable saddle - node at the critical point $(1,0)$, so look for that in the diagram below.

Part d

Compute the Jordan form of the linearized system matrix and and the linear map that puts it in Jordan canonical form.

To find the Jordan Normal Form, we need to find the eigenvalues and eigenvectors for the system.

We already found the two eigenvalues and the corresponding eigenvectors are given by:

$$\lambda_1 = -1, v_1 = (-1, 1)$$

$$\lambda_2 = 1, v_2 = (1, 1)$$

So, to write the Jordan Normal Form, using the eigenvalues and eigenvectors, we form:

$$A = PSP^{-1} = \begin{bmatrix}-1 & 1\\1 & 1 \end{bmatrix} \cdot \begin{bmatrix}-1 & 0\\0 & 1 \end{bmatrix} \cdot \begin{bmatrix}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}$$

Title: Canonical Form and DE


答:
First statement is only true if there exists some N such that $x_n \ne 0$ for $n \ge N$. Otherwise $\frac{1}{|x_n|}$ is undefined for some n and the limit does not exist.

Second statement is not true.
Here's a counterexample:

$y_n = \frac{(-1)^n}{n}$

whose limit is 0 as $n \to \infty$

$z_n = \frac{1}{y_n} = (-1)^nn$

whose limit as $n \to \infty$ is not $\infty$ or $-\infty$

Title: If $\lim x_{n} = 0$, then $\frac{1}{|x_{n}|}$?

Body:
If $x_{n}$ is a sequence such that $\lim x_{n} = 0$, then $\frac{1}{|x_{n}|} = \infty$. Is this true?

What about if we have another sequence $y_{n}$ such that $\lim y_{n} = 0$, then either $\lim \frac{1}{y_{n}} = +\infty$ or $\lim \frac{1}{y_{n}} = -\infty$

Is there a way to prove/disprove some of these?


问:
Complement of a cyclic subgroup in a finite, abelian p-Group (Proof-Verification)
I'm trying to prove following statement, can anyone check if my proof works? In Isaacs, I.M. Algebra - A Graduate Course the proof is really long and complicated(but needs less theory), so I'm afraid I made a mistake, as the statement seems to be quite trivial with the use of the fundamental theorem of finitely generated abelian groups.

Lemma 1

In a finite, abelian p-group $G$, a cyclic subgroup $H$ of maximal possible order has a complement.

The following proof is incomplete, see my answer for a correct one.

Proof. Let $G\cong \Bbb{Z}_{p^{r_1}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_s}}$ with $r_1 = \cdots =r_l \geq  ... \geq r_s$.
As $H$ is cyclic and of maximal order, $H=\Bbb{Z}_{p^{r_i}}$ for $i\in\{1,\cdots,l\}$.
Let $X$ be the subgroup consisting of all summands, but the $i$-th. Meaning $X=\bigoplus \Bbb{Z}_{p^{r_k}}$, $k \in \{1,\cdots,s\} \setminus \{i\}$.

Then $G=H \oplus X$ and $H$ has a complement in $G$.

q.e.d.

答:
As I see it, your proof is incomplete.

The claim to be proved is this:

In a finite abelian $p$-group, every cyclic subgroup of maximum order has a complement.

You chose a cyclic subgroup of maximum order from among those of maximum order in a particular direct sum decomposition. But there can be others.

In fact, if a given direct sum decomposition has more than one summand, there will be other cyclic subgroups of maximum order besides those which appear in the direct sum decomposition.


Title: Derive that $\sum_{k=0}^{n-1} k \ 2^{n-(k+1)} = 2^n-n-1$.

Body:
$\sum_{k=0}^{n-1} k \ 2^{n-(k+1)} = 2^n-n-1$.

How does one show a sum like this?

Body:
Hint

Observe that your sum can be written as

$$S=2^{n-1}\sum_{k=1}^{n-1}k(\frac 12)^k$$

Consider
$$F(x)=1+x+x^2+...+x^{n-1}=$$
$$\frac{1-x^n}{1-x}$$
$$F'(x)=\frac{x^{n-1}((n-1)x-1)}{(1-x)^2}$$

$$xF'(x)=x+2x^2+3x^3+...+kx^k+...$$

So,
$$S=2^{n-2}F'(\frac 12)$$


答:
There are many ways. Induction is one, and another uses a bit of calculus, but a direct calculation reducing it to sums of geometric series is also possible, starting like this:

$$\begin{align*}
\sum_{k=0}^{n-1}k2^{n-(k+1)}&=\sum_{k=1}^{n-1}\sum_{i=1}^k2^{n-(k+1)}\\
&\overset{(1)}=\sum_{i=1}^{n-1}\sum_{k=i}^{n-1}2^{n-(k+1)}\\
&\overset{(2)}=\sum_{i=1}^{n-1}\sum_{k=i+1}^n2^{n-k}\\
&\overset{(3)}=\sum_{i=1}^{n-1}\sum_{\ell=0}^{n-(i+1)}2^\ell
\end{align*}$$

$(1)$ is just reversing the order of summation, analogous to reversing the order of integration; $(2)$ is shifting the index of the inner summation up by $1$; and $(3)$ is substituting $\ell=n-k$ in the inner summation.

The inner summation is now a finite geometric series, which you should be able to sum, and once you’ve done that, you’ll find that completing the calculation is mostly just summing another finite geometric series.


问:
Irreducible polynomial with degree 2 over $\mathbb{R}$

Let $f(x) \in \mathbb{R}[x]$ with $\deg f=2$. Show that $f(x)$ is irreducible over $\mathbb{R}$ if and only if $f(x)=(x-a)^2 +b^2$, where $a,b \in \mathbb{R}$ and $b \neq 0$.

I don't know how to do the first part of this exercise and I need some help.

For the way back I did:

If $f(x)=(x-a)^2 +b^2$, to get $f(x)=0$, We need $a=0$ and $b=0$. But, by hypothesis, $b \neq 0$. So $f(x)$ doesn't have roots in $\mathbb{R}$, therefore $f(x)$ is irreducible.

Is this part right?

答:
Hint:

If $f(x)$ is irreducible, it has two conjugate roots $\zeta,\bar\zeta$, so it factors over $\bf C$ as
\begin{align}
f(x)&=(x-\zeta)(x-\bar\zeta)=x^2-(\zeta+\bar\zeta)x+\zeta\bar\zeta \\
&=x^2-2\operatorname{Re}(\zeta)x+|\zeta|^2=\bigl(x-\operatorname{Re}(\zeta)\bigr)^2-\bigl(\operatorname{Re}(\zeta)\bigr)^2+|\zeta|^2 \\
&=\cdots
\end{align}


问:
Construct center of circle without using its interior
I need to find the center of a circle using compass and ruler (or preferably just a straightedge) without using the interior of the circle. In other words, no line or point can be used which is inside the circle. All lines and points used in the construction must be outside the circle.

The center can be defined by points outside the circle. For example, you could identify 4 points outside the circle. The intersection of the lines going through each pair of points indicates the center.

答:
Choose any two points on the edge of your circle. Connect these two points with a line segment, this forms a chord in your circle.

You can show that the perpendicular bisector of any chord of a circle goes through the center of the circle.

Therefore, to find the circle center, construct two different chords of the circle, and find the perpendicular bisector of each chord. The circle center is where these bisectors meet.


问:
Conditional Random Variable and Iterated Expectation
First, I can't seem to find a definition of a "conditional random variable"; I understand the concepts of conditional expectation and conditional probability, but is there a notion for the random variable defined by $X|A$ for $A$ an event? Or more generally, is there a notion for the r.v. $X|Y$ for r.v.s $X$ and $Y$?

I ask because I argued in a problem that a desired expectation is $\mathbb{E}( \mathbb{E}( X | A) | B ) $ for $A$ and $B$ two events, but I didn't actually compute $\mathbb{E}( \mathbb{E}( X | A) | B ) $  mathematically per say, but argued that it was equal to $\mathbb{E}( X | A \cap B)$ based on the logical construct of the problem and the idea of total expectation.

My second question is, is it generally true that $\mathbb{E}( \mathbb{E}( X | A) | B ) = \mathbb{E}( X | A \cap B)$ for $A,B$ events?

答:
One usually doesn't define "conditional random variables", but you could say: given random variable $X$ and event $A$ with $P(A) > 0$, consider the new sample space $A$ with probability
measure $P'$ given by $P'(B) = P(B|A)$ for $B \subset A$, and the random variable $Y$ on this sample space
such that $P(Y \ge y) = P(X \ge y|A)$.

Since ${\mathbb E}(X|A)$ is a number (if it is defined at all) rather than a random variable,
your ${\mathbb E}({\mathbb E}(X|A)|B)$ can only be ${\mathbb E}(X|A)$.


答:
Take $2^{n-2}$ outside the sum.

Use $\sum_0^{n-1}kx^{k-1}=\frac{d}{dx}\sum_0^{n-1}x^k=\frac{d}{dx}\frac{1-x^n}{1-x}$. Then let $x=\frac{1}{2}$ to finish off.


答:
Choose any two points on the circle.  Set the radius of your compasses larger than the diameter of the circle, and draw a circle centered on each point.  These circles intersect in two points that form a line through the required center.

If the compasses are too small, or collapse when you lift them off the paper, construct a hexagonal grid of points until you have two points diametrically opposite each other.

Title: Construct center of circle without using its interior

Body:
I need to find the center of a circle using compass and ruler (or preferably just a straightedge) without using the interior of the circle. In other words, no line or point can be used which is inside the circle. All lines and points used in the construction must be outside the circle.

The center can be defined by points outside the circle. For example, you could identify 4 points outside the circle. The intersection of the lines going through each pair of points indicates the center.


问:
Proving $a^2+b^2+c^2+\cdots+z^2<(a+b+c+\cdots+z)^2$
I know and can prove that, $a^2 + b^2 < (a+b)^2$ where $a > 0$ and $b > 0$

Now, how can I prove the generalization, i.e.

$$
a^2 + b^2+ c^2 + \cdots + z^2 < (a +b + c+ \cdots + z)^2
$$

What is the way to reach that conclusion from the above statement?

Body:
Let $a_1,...,a_n\geq 0$. Then\begin{align}
(a_1+\cdots+a_n)^2&=\left(\sum_{i=1}^na_i\right)^2\\
&=\sum_{i=1}^n\sum_{j=1}^na_ia_j\\
&=a_1^2+\cdots+a_n^2+\sum_{i\neq j}a_ia_j\\
&\geq a_1^2+\cdots+a_n^2.
\end{align}


问:
Do copulas preserve uniform convergence?
Suppose I have a distribution $Q$ of $\left(Y_i, X_i\right)$, such that $Q(y,x)=C(F_Y(y), F_X(x))$, for some copula $C$ and marginal CDFs $F_X$ and $F_Y$, by Sklar's theorem.

Now, suppose I have sequences of CDFs $F^{(y)}_n$ and $F^{(x)}_n$ such that $\Vert F^{(y)}_n - F_Y \Vert_\infty \rightarrow 0$, $\Vert F^{(x)}_n - F_X \Vert_\infty \rightarrow 0$.

Now, we can define $Q_n = C(F^{(y)}_n, F^{(x)}_n)$. My question is, do we have $\Vert Q_n - Q\Vert_\infty \rightarrow 0$?

答:
Concerning the limit of copulas, I think you may have forgotten the fact that every copula is Lipschitz wrt to $l^1$-norm, with Lipschitz coefficient is $1$. That is, for any $u,v \in [0,1]^N$, we have:
$$ |C(u)-C(v)| \le \sum_{k=1}^N |u_k-v_k|$$

Your desired limit is just a direct consequence.


答:
Induct.  For instance,
\begin{align*}
a^2 + b^2 + c^2 &= (a^2 + b^2) + c^2  \\
    &< (a+b)^2 + c^2  &  &  \text{base case on $a$, $b$}\\
    &< ((a+b)+c)^2  &  &  \text{base case on $a+b$, $c$}  \\
    &= (a+b+c)^2  \text{.}
\end{align*}

You should be able to see how to adapt this to a generic induction step...

Title: Proving $a^2+b^2+c^2+\cdots+z^2<(a+b+c+\cdots+z)^2$

Body:
I know and can prove that, $a^2 + b^2 < (a+b)^2$ where $a > 0$ and $b > 0$

Now, how can I prove the generalization, i.e.

$$
a^2 + b^2+ c^2 + \cdots + z^2 < (a +b + c+ \cdots + z)^2
$$

What is the way to reach that conclusion from the above statement?


问:
Let $p$ be any prime greater than or equal to $5$. Prove that $a^p ≡ a \pmod{3p}$
I am unsure how to prove the following statement, although I am fairly sure it has something to do with Fermat's Little Theorem, which states that $a^p \equiv a \pmod{p}$ for a prime, $p$. Any help is greatly appreciated.

答: By Fermat's little theorem we have $a^{p} \equiv a \pmod {p}.$ We need to show that $a^p \equiv a \pmod {3}$. If $3$ does not divide $a$, we have $a^2 \equiv 1 \pmod {3}$ which implies that $a^{2j} \equiv 1 \pmod {3}$. Since the exponent is an odd prime, $p = 2j+1$ meaning that $a^p=a^{2j+1} \equiv a \pmod {3}$. Otherwise, we have $a= 3k$ for some $k \in \mathbb{Z}$ so that $a^{p}-a = (3k)^p - 3k = 3(3^{p-1}k^{p} - k)$. Since $(3,p) = 1$, by the chinese remainder theorem we have that $a^{p} \equiv a \pmod {3p}$.


答:
It all boils down to

$$0<|x_n|<\epsilon\iff \dfrac1{|x_n|}>M=\frac1\epsilon.$$

But neither

$$0<|y_n|<\epsilon\iff \dfrac1{y_n}>M=\frac1\epsilon$$

nor

$$0<|y_n|<\epsilon\iff \dfrac1{y_n}<-M=-\frac1\epsilon$$

do hold.


问:
Let f be an embedding of disk $D$ in Real Projective plane $P$. Show that $P- int(f(D))$ is homeomorphic to Mobius strip.
This question is partially answered here: How to see the real projective plane is a Möbius band glued to a disk?

In that answer it is shown that for a very specific embadding $f: D \rightarrow P$, space $P - int(D)$ is homeomorphic to Mobious strip and it can be
(rigorously) proved constructively.

I want to show that that is the case for every embedding.

I would like to somehow use Jordan curve theorem and Schoenflies theorem but I must just stuck on how to prove that for a non specific embedding of a disk $D$.

Anyone any ideas?

答:
I am hesitant to use the Jordan curve theorem here but perhaps it is possible. Instead what I would use is the important fact that the space $Emb(D^n , M)$ is equivalent to $Fr(M)$, the total space of the fiber bundle over $M$ where the fiber over $x$ is the space of bases of the tangent bundle. Here $M$ is n-dimensional. This equivalence just comes from pushing forward the standard basis along the derivative.

For a nonorientable connected manifold, this space is path connected because the base is path connected and there exists a path in the total space that connects any two points in a specific fiber (this is equivalent to being nonorientable).

From this we conclude that any two embeddings of the disk are isotopic. A very nice result called isotopy extension then implies that there is a diffeomorphism of the manifold taking one of these disks to the other. This implies that as pairs (the manifold and the image of the embedded disk), they are exactly the same topologically (as well as smoothly). This implies that removing the interior gives homeomorphic (as well as diffeomorphic) manifolds.


问:
When two numbers of a certain form are coprime.
Suppose I am dealing with some $r \in \mathbb{N}$ and want to know when the numbers $(r + 2)$ and $(r - 1)$ are coprime. When these two numbers are coprime then $\exists x, y \in \mathbb{Z}$ where $(r + 2)x + (r - 1)y = 1$. Is there a general algebraic way to find all solutions for $(x, y)$ without using something like Fermat's Little Theorem? Maybe something related to Bezout's Theorem?

答:
$(r+2)$ and $(r-1)$ are coprime if $r$ is not congruent to $1$ $\mod 3$

Indeed $GCD(r+2,r-1)=GCD(r-1,3)$ thus if $r-1$ is a multiple of $3$ then $GCD\ne 1$

Now suppose that $r\equiv 0\mod 3$ that is $r=3k$

The equation becomes

$$(3k+2)x+(3k-1)y=1$$
Whose solution are
$$x=h (3 k-1)+k;\;y=h -k-1-(3 k+2));\;\forall h\in\mathbb{Z}$$
Suppose $r\equiv 2\mod 3$, that is $r=3k-1$

The equation becomes
$$(3k+1)x+(3k-2)y=1$$
Whose solutions are
$$x=h (3 k-2)+2 k-1,\;y=h (-(3 k+1))-2 k-1;\;\forall k\in\mathbb{Z}$$


问:
Imaginary integration constants
I have following two PDEs:

$\frac{\partial u_1'}{\partial t}+f(u_2')=\nu(\frac{\partial^2 u_1'}{\partial x_2^2}+\frac{\partial^2 u_1'}{\partial x_3^2})$

$u_2'\frac{\partial u_1'}{\partial x_2}+u_3'\frac{\partial u_1'}{\partial x_3}=0$

I already obtained $u_2'$ and $u_3'$ from other PDES wherer both $u_2'$ and $u_3'$ contain of in total 4 constants $C_1, C_2, C_3, C_4$, occuring from some integration. Now I have these two PDES to solve for $u_1'$. I used the first PDE to solve for $u_1'$ now I want to check if the second PDE is fullfilled, which it should be. However this is only the case if I choose some of the constant $C_i$ to be imaginary. I dont know if this makes much sense however, especially since $u_i$ represent velocity components.

Is it ok for constants of PDES to be imaginary even for a "real" physical problem?

答:
$$y''+y=0$$ has the general solution

$$y=c_+e^{ix}+c_-e^{-ix}.$$

With the initial conditions $y(0)=0,y'(0)=1$, $c_-+c_+=0$ and $i(c_+-c_-)=1$ and finally

$$y=-\frac i2e^{ix}+\frac i2e^{-ix}=\sin x.$$

This could answer your question.

Title: Imaginary integration constants


问:
Definite Integral of square root of polynomial
I need to learn how to find the definite integral of  the square root of a polynomial such as:
    $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$

EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.

This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

答:
For a linear polynomial under the radical a $u$-substitution will do. If you wish to solve this problem for higher order polynomials in the general case then you will be appealing to a lot of algebra, hyperbolic trig functions, and elliptic integrals (which are similar in that they are arclength problems). In short there is no easy short cut to solve this problem for arbitrary degree polynomials.

For degree one polynomials, do as the other answers have advised. For degree two polynomials, like $\sqrt{2x^{2}+3x+7}$, you should employ trigonometric substitutions.

Step 1: Given $\int\sqrt{ax^{2}+bx+c}$ first complete the square into something of the form $k\int\sqrt{\pm u^{2}\pm l}$.

Step 2: Using one of the three substitutions from this article on the matter to get the integral into the form of $k\int\sqrt{f(x)^{2}}$ for some trig function $f(x)$.

Step 3: Remove the radical and solve using known integrals.

If you wish to venture into solving cubics (or higher degree polys) in the general case (i.e. not reducible by a substitution) then do read up on elliptic integrals.


问:
To what extent is compactness inherited by subspaces?
In Valter Moretti's "Spectral Theory and Quantum Mechanics," Remark 4.2 (1), he claims:

Compactness is hereditary, in the sense that it is passed on to induced topologies.

I am not sure exactly what he means by this. In particular, one might interpret this in the most strong/general sense possible:

Let $(X,\mathcal T_X)$ be a topological space, and $(A,\mathcal T_A$) be a subspace, i.e., $A \subseteq X$ and $\mathcal T_A$ is the induced topology from $\mathcal T_X$. Let $K⊆ X$ be compact. Then $K\cap A$ is compact in $(A,\mathcal T_A)$.

Is this true? Please explain why or why not.

答:
False. $X=K=[0,1]$ is compact but $K \cap A=A=(0,1)$ is non-compact.

It is true that compactness is inherited by closed subsets. Indeed, let $A\subseteq X$ be a closed set. If $\{U_i\}$ is an open cover of $A$, then $\{U_i\}\cup\{A^c\}$ is an open cover of $X$ and by compactness we may extract a finite subcover for $X$, which also gives a finite subcover for $A$.


问:
Proof of Von Neumann Ergodic Theorem
I am trying to understand the line that says: and hence $U_nf \rightarrow 0$ as $n \rightarrow \infty$. I understand everything up until that claim.

答:
Since $U$ is an isometry, $\|U\|=1$ and thus
$$\|U_nf\|=\frac 1n\left\|g-U^ng\right\|\leq\frac2n\|g\|\to0$$
as $n\to\infty$.

Title: Proof of Von Neumann Ergodic Theorem


问:
Rare cases of bivariate probability distributions
What would be a special case of a discrete bivariate probability distribution that two events are dependent but also have a covariance of zero?</p>

答:
Suppose each of the following pairs have probability $\frac13$

```<code>X Y
1 1
2 2
3 1
</code>```

Body:
Suppose each of the following pairs have probability $\frac13$

X Y
1 1
2 2
3 1


答:
Since $U$ is an isometry, \begin{align*}
||g - U^n g|| &\leq ||g|| + ||U^n g||  \\
    &= ||g|| + ||g||  \\
    &= 2||g||  \text{.}
\end{align*}

So $|| \sum_{i=0}^{n-1} U^n f || \leq 2||g||$ and when you divide through by $n$, making the left-hand side $U_n f$, the right-hand side goes to zero as $n \rightarrow \infty$.

Title: Proof of Von Neumann Ergodic Theorem

Body: I am trying to understand the line that says: and hence $U_nf \rightarrow 0$ as $n \rightarrow \infty$. I understand everything up until that claim.


问:
Do constraints influence the domain of the objective function in optimization problems?
I'm trying to devise a general method for finding the domain of objective functions, but it's unclear to me whether or not the constraints are taken into consideration.

It seems the constraints will affect the domain of an objective function, but there are instances in James Stewart's Calculus (8th edition) where it's unclear if the author is deriving the domain of the objective function from the function itself, without considering the constraints, or if the constraints are being considered.

For example,

Example 3) Find the point on the parabola $y^2 =2x$ that is closest to point $(4, 1)$

Constraint: $y^2 = 2x \Rightarrow x = \frac{1}{2}y^2$

Objective function: $d= \sqrt{(x-4)^2 + (y-1)^2} \Rightarrow d(y) = \sqrt{(\frac{1}{2}y^2 -4)^2 + (y-1)^2}$

The author implies that the domain of $d(y)$ is $(-\infty, \infty)$, but he doesn't explain how he arrived at that. It seems to me that the author used the objective function, without considering the constraint, to derive its domain since looking at it we can see that $d(y)$ would be defined for every value of $y$.

Looking at the constraint, it was rendered a polynomial function of $y$ which will be defined over $(-\infty, \infty)$. So maybe the constraint wasn't considered because it had no bearing on the domain of the objective function due to being defined everywhere and thus unbounded when written in terms of $y$. If the constraint were written in terms of $x$, $y = \pm \sqrt{2x}$, then the domain of the constraint becomes $[0, \infty)$. When substituting the constraint $y = \pm \sqrt{2x}$ into the objective function, $d(x) =\sqrt{(x-4)^2 + (\pm \sqrt{2x}-1)^2}$, the domain of $d(x)$ becomes $[0, \infty)$. Thus it would seem to be the case that the domain of the constraint does affect the domain of the objective function, and furthermore depending on which variable the constraint is written in terms of also affects the domain.

If the constraint is viewed as a function, then substituting it into the objective function renders the objective function a composite function. Viewing it this way makes sense that the constraint's restricted domain would also restrict the objective function's domain.

Therefore, it seems that to find the overall domain of the objective function we would use the same process or method employed when finding the domain of any composite function. In other words, when finding the domain of the objective function we DO consider the constraint (more precisely its domain).

But I'm uncertain if my thoughts on this are correct and would appreciate some guidance and/or input.

Does the constraint affect the domain of the objective function?

Tips or general thought process on how to find the domain of the objective function

答: From a pedagogical standpoint, my advice to students is to consider the domain of the quantity to be optimized and then restrict it further based upon the larger context of the problem. I do not believe there is a universal algorithm for this process as this class of problems is highly varied.</p>


问:
When are two $L^p$ spaces equal
Q) Consider a probability space i.e. $\mu(X)=1$. Under what conditions is $L^p(\mu)=L^q(\mu)$ for $0<p<q\leq\infty$?

I can see when $||f||_p=||f||q$ by looking at the fact the Holder's inequality is an equality if and only if $\alpha |f|^q=\beta$ a.e. for some $\alpha,\beta$ both not zero which leads to the fact that $f$ is constant a.e.

But I'm not sure under what conditions the whole of $L^p$ space is equal to $L^q$?

答:
If I am not wrong, they are not equal, except trivial case in which $\mu$ is atomic and  has only a finite number of atoms.

If it is not the case, you can choose a sequence of disjoint measurable sets $E_1,E_2,..$ such that:
$$ \mu(E_n)=a_n>0 \forall n$$
So it's only the work of creating a measurable function , which belongs to $L^p$, but not $L^q$, of form:
$$ f= \sum_{n \ge 1} b_n \mathbb{1}_{E_n}$$
For simplicity, let $p=1$; $q=2$,  I can choose :
$$b_n= \frac{1}{ \sqrt{R_n} } $$
where $R_n$ is the residu of the series $\sum_{n \ge 1} a_n$, that is:
$$R_n= \sum_{k \ge n} a_n$$
Then clearly ,
$$ \int f d\mu \le \sum_{n \ge 1} 2\left( \sqrt{R_{n}}-\sqrt{R_{n+1}}\right) < \infty$$
and
$$ \int f^2 d\mu = \sum \dfrac{a_n}{R_n} = \infty$$

So the inclusion $ L^q( \mu ) \subset L^p( \mu)$ is strict.

The same idea can so apply for other $p,q$, we just need to adjust the power of $R_n$

Disclaimer: I'm just not so sure about my argument about finite atom measures in the beginning, but I guess having well presented the essence of solution.

Title: When are two $L^p$ spaces equal


答:
Following my comment, here is a sketch. Let's assume that
$$
\inf_{\mu(E)\neq 0}\mu(E) > 0. \qquad (*)
$$
Given $f \in L^q(\mu)$, let $E_n = \{ \vert f \vert^p > n \}$. Then
$$
\mu(E_n) \leq \frac{1}{n}\Vert f \Vert_p^p \to 0
$$
as $n \to \infty$. Therefore there is some $N$ for which $\vert f \vert \leq N^{1/p}$ a.e., hence $f \in L^q(\mu)$.

(This shows that $(*)$ implies $L^p(\mu) \subset L^q(\mu)$. One should also be able to show the opposite: $L^p(\mu) \subset L^q(\mu)$ implies $(*)$.)

Title: When are two $L^p$ spaces equal

Body:
Q) Consider a probability space i.e. $\mu(X)=1$. Under what conditions is $L^p(\mu)=L^q(\mu)$ for $0<p<q\leq\infty$?

I can see when $||f||_p=||f||q$ by looking at the fact the Holder's inequality is an equality if and only if $\alpha |f|^q=\beta$ a.e. for some $\alpha,\beta$ both not zero which leads to the fact that $f$ is constant a.e.

But I'm not sure under what conditions the whole of $L^p$ space is equal to $L^q$?


问:
Show that subspace contains unique element with minimal $L^2$-norm.
Let $S = \{x \in L^2(0,1) : \int_0^1 x(t) dt = 1\}$. I want to show that $S$ containts a unique element with minimal $L^2$-norm and I want to find this element.

I start of by finding a minimal element $x \in S$:

$$ 1 = \int_0^1 x(t) dt = \int_0^1 x(t)\cdot1\ dt= \langle\,x,1\rangle = |\langle\,x,1\rangle| \leq ||x||\cdot||1|| = ||x||$$

Solving the inequality above with equality gives us that a minimal element is given by $x = 1$.

How can I show that this $x$ is the unique minimal solution in $S$?

答: Condition for equality in C-S inequality : $ \langle f, g \rangle=\|f\| \|g||$ ($f,g \neq 0$) if and only if $f=cg$ for some positive number $c$. So if $x$ is any element of norm $1$ in your set the $x$must be a constant function and since its integral is $1$ the constant has to be $1$.


问:
How to show that the integral is convergent without actually evaluating it?

$$ \int _ 0 ^ { + \infty } \frac 1 { \sqrt { 1 + x ^ 8 } } \ \mathrm d x $$

I have a hint that says there is no elementary way to evaluate this integral. So I am supposed to bound this integrand by a simpler function and use the following fact without proof:

Let $ f : [a,+\infty) \to \mathbb R $ be a bounded and increasing function. Then $ \lim _ { x \to \infty } f ( x ) $ exists.

I specifically don't understand what it means to bound the integrand by a simpler function. Any help will be appreciated.

Body: Hint: $$ \frac{1} { \sqrt{1+ x^{8} } } \sim \frac{1} { x^4 }, x \to +\infty$$


问:
What are some reasonable things to prove about the Collatz Conjecture?
I am writing an undergraduate paper on the $3n+1$ problem, and I am looking for some theorems related to the problem that would be reasonable for someone with my mathematical background to prove. I'm near the end of my undergraduate studies. I have had a pretty basic introduction to Number Theory and Group Theory. I have also taken a few introductory Computer Science courses. Those are the only courses that I've taken that seem to be relevant to the problem but otherwise my math classes have been: the usually calculus sequence, real analysis, introductory topology, linear algebra.

A good answer to my question would do the following:

Either state one or more non-trivial theorems related to the Collatz conjecture, or reference sources that would contain this kind of information.

Would NOT state any proofs unless a proof or partial proof would be essential for someone with my requisites if they are to be able to provide a full proof.

Theorems that I have already proved:

If $L(n)$ is a function that counts the number of mappings in the Collatz function for $n$ to get to 1, then if $n = 8k+4$, $L(n) = L(n+1)$.

If $n = 128k+28$ then $L(n) = L(n+2)$.

答:
The starting point for the literature is the survey by Jeff Lagarias ten years ago:

math/0309224 The 3x+1 problem: An annotated bibliography (1963--1999) (sorted by author). Jeffrey C. Lagarias. math.NT (math.DS).
https://arxiv.org/abs/math/0309224

Other papers from an arxiv Front search for Collatz and 3x+1:

math/0608208 The 3x+1 Problem: An Annotated Bibliography, II (2000-2009). Jeffrey C. Lagarias. math.NT (math.DS) https://arxiv.org/abs/math/0608208

arXiv:0910.1944 Stochastic Models for the 3x+1 and 5x+1 Problems. Alex V. Kontorovich, Jeffrey C. Lagarias. in: The Ultimate Challenge: The 3x+1 problem, Amer. Math. Soc.: Providence 2010, pp. 131--188. math.NT. https://arxiv.org/abs/0910.1944

math/0509175 Benford's law for the $3x+1$ function. Jeffrey C. Lagarias, K. Soundararajan. J. London Math. Soc. 74 (2006), 289--303. math.NT (math.PR). https://arxiv.org/abs/math/0509175

math/0412003 Benford's Law, Values of L-functions and the 3x+1 Problem. Alex V. Kontorovich, Steven J. Miller. Acta Arithmetica 120 (2005), no. 3, 269-297. math.NT (math.PR).  https://arxiv.org/abs/math/0412003

and

https://arxiv.org/abs/math/0411140

https://arxiv.org/abs/math/0205002

https://arxiv.org/abs/math/0201102

https://arxiv.org/abs/math/0204170

There are many un-serious papers on this subject at the same site, I selected the above from papers that I read or authors whose other papers I have seen.

There is also a celebrated theorem by John H Conway in which he shows that the natural generalization of the Collatz problem, where different linear functions are applied to $n$ in several arithmetic progressions (partitioning all integers), can simulate a computer.  A consequence is that it is undecidable to determine if such an iteration loops when started from $1$.


问:
Struggling to describe the language that this pushdown automaton represents
Given this pushdown automaton :

I am trying to describe the language that this represents.

My analyze yet:

the numbers of a at the start will need to be the same that as the end.

Also, the starting letter if there's no a, will be one b.

Also, the last letter before the ending a's will be a b

Now I struggle with the middle loop ( state 3 and 4 ) I am not sure on how to analyze and represent it.

For now i have something that looks like this:

$a^n b(b^*aa^*bb^*)^*a^n :n \in\mathbb{N}$

But I am pretty sure it's not good since I think I need to have a relaiton between the exponent ( just like the $a^n$) the other exponent should also be represented by a letter but I can't seem to understand the pattern and relations here.
Just to give you an example i need something that could look like this :
$$
L=\left\{a^{m} b^{m} c^{n}: m, n \in \mathbb{N} \wedge m \geq n\right\}
$$

If this can help I figured some words that can be accepted : babb , aabaaababaa

Could anyone help me figure this out and how to approach it.

Thanks a lot.

答:
The language L is , L = { a^n b (a* b)*  a^n , n ≥ 0 }

We can see how the number of as at begging and end of the string must match from the transitions a,ε,c and a,c,ε , this is why we have a^n at begging and end

After the first a^n , the transition b,ε,ε shows that there comes a b

Now we arrive arrive at state 3 , we consider states 3 and 4 , once you enter 3 you are left with 2 options :

go from 3 to 3 , which is b , the same as a^0 b

go from 3 to 4 , loop on 4 if you want , then go back to 3 , which is a^+ b

now  you can do 1 or 2 , hopefully you can see that 1 U 2 is a^0 b U a^+ b =  a* b (this is why we wrote b as a^0 b) , and this can be repeated as many times as we want since after 1 or 2 we return back to 3 so we arrive at (a* b)*


答:
Start by noting that the existence of the proposed integral is equivalent to the existence of the integral $\int_1^{+\infty}\frac{1}{\sqrt{1+x^8}} dx$. Then, since
$$
\frac{1}{\sqrt{1+x^8}} \leq \frac{1}{\sqrt{x^8}} = \frac{1}{x^4},
$$

the convergence of $\int_1^{+\infty} \frac{1}{x^4} dx$ implies the convergence of our integral.

Title: How to show that the integral is convergent without actually evaluating it?

Body:
$$ \int _ 0 ^ { + \infty } \frac 1 { \sqrt { 1 + x ^ 8 } } \ \mathrm d x $$

I have a hint that says there is no elementary way to evaluate this integral. So I am supposed to bound this integrand by a simpler function and use the following fact without proof:

Let $ f : [a,+\infty) \to \mathbb R $ be a bounded and increasing function. Then $ \lim _ { x \to \infty } f ( x ) $ exists.

I specifically don't understand what it means to bound the integrand by a simpler function. Any help will be appreciated.


问:
Traces on a von Neumann algebra
Let $M$ be a von Neumann algebra. In Takesaki's book, the trace $\tau$ on $M$ is defined as follows:

$\tau$ is a function defined on the positive cone $M_{+}$ with values in $[0, +\infty]$   satisfying some conditions.

My question: Can we define $\tau$ on $M$?

When we define a tracial state $\phi$ on a $C^*$–algebra $A$, the domain of $\phi$ is $A$, rather than $A^+$.

答:
Note the second square bracket in $[0,\infty]$. What Takesaki is defining is a tracial weight. Many von Neumann algebras have no tracial states, but do have faithful semifinite tracial weights. These precisely those with no type III central summand, and where the identity is infinite (i.e., they have at least a component of type I$_\infty$ of II$_\infty$).

The typical example is the trace on $B(H)$ with $H$ infinite-dimensional.

Title: Traces on a von Neumann algebra


问:
Non-trivial connected semi-simple compact Lie group
This is a question about some terminology I'm not familiar with and was hoping to clarify.

I've seen it written that every "non-trivial connected semi-simple compact Lie group" contains a non-abelian Lie subgroup.

Question: But isn't $S^1$ such a group and yet abelian?

(E.g. This would make sense if "non-trivial" here refers to non-triviality of the Lie algebra; is this the standard usage?)

答:
As mentioned in the comments, $S^1$ is not semisimple. Unlike the case of groups, for Lie algebras the convention is that "simple" means "simple nonabelian."

(In my opinion this is such a useful convention it should be imported back into group theory; it would be very useful to have one word that means "simple nonabelian" in group theory also and it's annoying that there isn't one. Simple abelian and simple nonabelian groups behave very, very differently from each other, and this difference is so pronounced in the Lie case that terminology specific to it was unavoidable.)


答:
For $x > 1$,
\begin{align*}
  0 < x^8 &< x^8 + 1 < \infty  \\
  0 < x^4 = \sqrt{x^8} &< \sqrt{x^8 + 1} < \infty  \\
  \infty > \frac{1}{x^4} &> \frac{1}{\sqrt{x^8+1}} > 0  \text{.}
\end{align*}
So, $$  0 = \int_1^\infty 0 \,\mathrm{d}x < \int_1^\infty \frac{1}{\sqrt{x^8 + 1}} \,\mathrm{d}x < \int_1^\infty \frac{1}{x^4} \,\mathrm{d}x  \text{.}  $$
The right-hand $p$-integral is convergent, so $\displaystyle \int_1^\infty \frac{1}{\sqrt{x^8 + 1}} \,\mathrm{d}x$ is convergent.

For $0 \leq x \leq 1$, \begin{align*}
    1 &\leq x^8+1 \leq 2  \\
    1 = \sqrt{1} &\leq \sqrt{x^8+1} \leq \sqrt{2}  \\
    1 = \frac{1}{1} &\geq \frac{1}{\sqrt{x^8+1}} \geq \frac{1}{\sqrt{2}}  \text{.}  \\
\end{align*}
So,
$$  1 = \int_0^1 1 \,\mathrm{d}x \geq \int_0^1 \frac{1}{\sqrt{x^8+1}} \,\mathrm{d}x \geq \int_0^1 \frac{1}{\sqrt{2}} \,\mathrm{d}x = \frac{1}{\sqrt{2}}  \text{.}  $$
This shows the integral $\displaystyle \int_0^1 \frac{1}{\sqrt{x^8+1}} \,\mathrm{d}x $ is convergent.

Therefore, $\displaystyle \int_0^\infty \frac{1}{\sqrt{x^8+1}} \,\mathrm{d}x$ is convergent.


问:
Calculating results of a Coin Flip with dynamic odds based on the numbers of flips so far
Suppose you have a coin which instead of having 50:50 chance of Heads v Tails you have a A:B chance. At the start A is 0 and B is 100 so there is no way to land on heads, but each time you flip it, after each flip, A increases by 1 and B decreases by 1, so that the odds go:

```<code>0:100 <-first flip

1:99 <-second flip

2:98 <-third flip

3:97 <-fourth flip
</code>```

How many flips will you have to do to land on heads?

I built a simple computer program to simulate this a few million times and I know the answer is around 14.2 flips, but I'm curious how you could actually solve this just with math, and ideally have a function that even lets you adjust variables such as how the odds adjust over each flip (suppose it increases by 2 points instead of 1 each flip, for example).

I've observed that adding the A value (of the A:B) odds for the first 14 flips (that is 0+1+2+...+13) gets you a number under 100 while adding the A value of 1 more flip gets you a total sum above 100, so this seems to indicate that the solution can be derived somewhere within this pattern... Not sure where it possibly goes from there though

答:
Let $X_i$ be the outcome of the $i$'th flip (zero-indexed), with $p_i:=P(X_i=1)=i/100$ and $P(X_i=0)=(100-i)/100$

Let $Y$ denote the random variable for when you see the first head. Thus for $n\in[0,99]$:

$$P(Y>n)=\prod_{k=0}^{n}(1-p_i)=\frac{100!}{(100-n-1)!100^{n+1}},$$

and $P(Y>100)=0$. Then:

$$E[Y]=\sum_{n=0}^{100}P(Y>n)$$

which numerically gives: $\approx 12.2$.

If you want to enumerate flips with not zero-indexing, then the answer is around 13.2 flips. You got 14.2 flips, so I wonder if there's a silly error above on my part.


问:
Probability of an RNG?
For fun (relatively speaking), I'd like to do a little research on probabilities related to a pseudo-random number generator.

Specifically, how would I get started figuring out "what is the probability that a prng will generate the same random number twice?"

This is part of an investigation into a larger issue, but I'm not sure where to begin (or if my question is even relevant).

Edited to add:

After reading the comments and answers I see that my question isn't clear. As an example: the rng outputs a 5 What is the probability that another 5 will be the next output? How many "rolls" of the RNG must I make before I see another 5?

If we assume a good rng has uniform distribution, then is it 1 out of (2^32) (assuming 32bit integers)?

For a larger example, lets say I use the RNG to generate numbers that are used as an index into an array of characters. The array is 25 numbers long, so the output of the RNG is mod 25 I will generate a sequence of 5 characters.

How would I determine the probability of generating the same 5 character sequence?

If we assume that we have a uniform distribution of output numbers from the RNG, then the RNG's output shouldn't factor into the probability of generating the same character sequence.

However, is my assumption valid?

答: All pseudorandom number generators are periodic; you can thus expect any number you've seen to appear again if you do Monte Carlo long enough. What you really want to ask, I think, is how long their periods are. To list the ones I'm most familiar with, linear congruential generators (often used in calculators) typically have (very!) short periods (though two or more of these can be combined to produce a generator with a slightly longer period), the Marsaglia-Zaman generator has a period of $\approx 5\times 10^{171}$, and the now-popular Mersenne twister by Matsumoto and Nishimura has a period of $2^{19937} - 1\approx4\times 10^{6001}$. One necessary condition for a quality PRNG is a long period length (but this isn't a sufficient condition, as it should also pass a bunch of stringent statistical tests). Have a look at these two articles for more information.

Title: Probability of an RNG?


答:
The author probably means that compactness is inherent property of subspace: you don't need the space in which it is contained to deduce its compactness.

In more rigorous terms. Let $ (X, \, \tau_X)$ be a topological space. Consider its compact subspace $ K \subset X $. Then for any $ Y $ such that $ K \subset Y \subset X $ with topology $ \tau_Y $ induced from $ X $, $K$ as subspace of $ Y $ is also compact.

In particularly, when $ Y = K $ space $K$ is compact in $K$ itself. What does it mean? Consider a collection of open sets $\{U_{\alpha}\}$ with $\forall \alpha:\: U_{\alpha} \subset K $ that covers $K \subset \bigcup_{\alpha}U_{\alpha} $, or as everything is contained in $K$ we have $K = \bigcup_{\alpha} U_{\alpha}$. Then we have a finite subcover: $K = \bigcup_{i=1}^{N} U_i $.

It is easily proved by using the definition of subspace topology. For $ Y \subset X $ $\tau_Y$ is defined by the map $ i\ :\ Y \to X $. The set $U \subset Y $ is open, iff $\exists V \subset X $ such that $ U = i^{-1}(V)$. In other words $i$ is an open map (it maps open sets to open sets). Thus we can push any open cover $\{U_{\alpha}\}$ in $Y $ to open cover $\{f(U_{\alpha})\}$ in $X$. Therefore, because we consider $K$ compact in $X$, $\{f(U_{\alpha})\}$ has a finite subcover $\{f(U_i)\}$, which we pull back to $Y$ and get a finite subcover $\{U_i\}$ in $Y$. The statement is proved.

Title: To what extent is compactness inherited by subspaces?

Body:
In Valter Moretti's "Spectral Theory and Quantum Mechanics," Remark 4.2 (1), he claims:

Compactness is hereditary, in the sense that it is passed on to induced topologies.

I am not sure exactly what he means by this. In particular, one might interpret this in the most strong/general sense possible:

Let $(X,\mathcal T_X)$ be a topological space, and $(A,\mathcal T_A$) be a subspace, i.e., $A \subseteq X$ and $\mathcal T_A$ is the induced topology from $\mathcal T_X$. Let $K⊆ X$ be compact. Then $K\cap A$ is compact in $(A,\mathcal T_A)$.

Is this true? Please explain why or why not.


问:
In what condition $\operatorname{Hom}(E,F)$ is of finite dimension?
Let $X$ be a nice (say, proper and smooth) variety over $\mathbb C$, and $E,F$ be coherent sheaves on $X$. Is it true that $\operatorname{Hom}(E,F)$ is always a finite-dimensional vecter space over $\mathbb C$?

If $E,F$ are vector bundles, this follows from $\operatorname{Hom}(E,F)=H^0(E^\vee \otimes F)$. Does this also work for $E,F\in \operatorname{Coh}(X)$?

答: If $E$ and $F$ are coherent sheaves, then so is the sheaf Hom $\mathcal{Hom}(E,F)$.  Since $\operatorname{Hom}(E,F)$ is just the global sections of $\mathcal{Hom}(E,F)$, this implies $\operatorname{Hom}(E,F)$ is finite-dimensional.


问:
Set of polynomials being an ideal

Are all polynomials {$P \in\Bbb Q(X)\mid P(1)=P(3)=0$} an ideal? If so find the monic generator.

$\Bbb Q$ is the rationals here and I know that the above set is an ideal by writing out multiplication of an element from Q(x) and an element from the set which shows me when x=1 or x=3, the multiplication result is still 0., but how do i figure out the generator?  I do not know where to start.

答:
Hint:

Do some polynomial arithmetic:

$$ P(\alpha)=0\iff P(X)\text{ is divisible by }\enspace X-\alpha. $$

Also, $\mathbf Q[X]$, like all polynomial rings over a field, is a P.I.D. so you can apply Euclid's lemma, $X-\alpha$ being irreducible.

Title: Set of polynomials being an ideal


问:
$\forall xA\Rightarrow A[t/x]$ and rule of left universal quantification in sequent calculus
Consider the rule of left universal quantification in the sequent calculus:
$$\dfrac{\Gamma,A[t/x]\vdash \Delta}{\Gamma,\forall xA\vdash\Delta}.$$
This can be used to give the following proof:
$$\dfrac{A[t/x]\vdash A[t/x]}{\dfrac{\forall x A\vdash A[t/x]}{\vdash (\forall x)A\Rightarrow A[t/x]}}$$
What is the relation between $\vdash(\forall x)A\Rightarrow A[t/x]$ and left universal quantification rule? Obviously one is a sequent and the other a deductive rule. They aren't saying the same thing, right?

答:
Obviously one is a sequent and the other a deductive rule.

Yes, indeed, and that is the distinction.

The sequent, $~\vdash (\forall x~A)\to A[t\backslash x]~$, is the claim that the implication is provable to be a tautology.

The Left Universal Rule is one among the rules of inference used to do so.

$$\dfrac{\dfrac{\dfrac{}{A[t/x]\vdash A[t/x]\qquad\;}{\small\textsf{Id}}}{\forall x A\vdash A[t/x]\qquad}{\small{\forall}\textsf{L}}}{\qquad\quad\,\vdash(\forall x)A\to A[t/x]}{\small{\to}\textsf{R}}$$


问:
Quotient of amalgamated free product by normal subgroup
I was given the following question in my homework:

Given $Γ =G_1∗_HG_2$, suppose that $H$ is a normal subgroup in both
$G_1$ and $G_2$.

(a)  Show that $H$ is then a normal subgroup in $Γ$.

(b)  Show that the quotient group $Γ/H$ is canonically isomorphic to
the free product $(G_1/H)∗(G_2/H)$.

I've already submitted my work, but I've noticed that the proof I gave for (b) was wrong. I tried to fix it today (just for the sake of learning) but didn't manage to get a complete solution.
My attempt to solve it is as follows (suppose I've solved A correctly):

Using the universal property of $Γ$, and the natural homomorphisms $\phi_1, \phi_2$ from $G_1$ and $G_2$ to $(G_1/H)∗(G_2/H)$, (through the quotients), we get a unique homomorphoism $\theta:Γ\to (G_1/H)∗(G_2/H)$, where all the diagram commutes.

It's easy to see that $H \subseteq Ker(\theta)$ because for each $h \in H \subseteq G_1$ we get $\phi_1(h) = e$, so using the diagram, we get that $\theta(h) = e$ for $h$ as an element of $Γ$.

I'm also pretty sure that $\theta$ is surjective, because $(G_1/H)∗(G_2/H)$ is generated by $Im\phi_1 \cup Im\phi_2$, so using the fact that the diagram commutes, it must be generated by elements that are in the image of $\theta$.

To prove the claim, the only thing I have missing with this approach is to prove that $Ker(\theta)\subseteq H$, and I'll get the desired isomorphism, that $Γ/H \cong (G_1/H)∗(G_2/H)$.

My questions are:

Are there any mistakes or nonsense (wrong understanding of definition, etc..) in what I wrote so far?

Is that a good approach for this question? Is trying to build an isomorphism by hand an easier approach? How would you try to solve it?

If it's indeed a good approach - how do I proof that $Ker(\theta)\subseteq H$?

答:
No, it's all correct.

Usually, to prove an isomorphism of the form $A/N\cong B$, the first isomorphism theorem seems indeed a good approach.

Assume $w=x_1y_1x_2y_2\dots x_ny_n\in \ker\theta\subseteq G_1 *_H G_2,\ x_i\in G_1,\,y_i\in G_2$. We want to prove $w\in H$.

If any letter $x_i$ or $y_i$ is in $H$, then we can move it to the beginning of $w$, perhaps replacing the letters behind by other elements of the same $G_1$ or $G_2$, then we can ignore it.

So, we are basically left with the case to consider when $\forall i: x_i,y_i\notin H$.

But then, $\theta(w)=[x_1][y_1]\dots[x_n][y_n]$, as being an element of the free product, has no chance to have any cancellable substring, so it is definitely not the empty word.

Nevertheless, you might find it easier to prove that $\Gamma/H$ satisfies the universal property of being the free product of $G_1/H$ and $G_2/H$.


问:
Proof using formal definition of limit that $ \lim_{x\to0} (x^2\sin(x^2+2x)+2)=2 $
I'm having a hard time trying to understand how to prove the following, using the formal definition of limits:

$$
\lim_{x\to0} (x^2\sin(x^2+2x)+2)=2
$$

I already did the following the steps but I'm not sure if my thinking is correct:

$$
|x^2\sin(x^2+2x)+2-2|<\epsilon \implies  |x^2\sin(x^2+2x)|<\epsilon \implies |x^2||\sin(x^2+2x)|<\epsilon
$$

and assuming that $x\in{R}$ then $x^2|\sin(x^2+2x)|<\epsilon$

Now I also know that $-1<\sin(x)<1$ however I'm struggling to understand how to best apply this to the proof.

I'm not looking for a full solution to the problem, rather I'm interested in confirming if my thinking up until now is correct and maybe some hints to unblock the next steps.

答:
You are almost there! Since the sine function is bounded by one and minus one, we have that
\begin{align*}
|x^{2}\sin(x^{2}+2x) + 2 - 2| = |x^{2}\sin(x^{2}+2)| \leq |x^{2}|\leq \varepsilon \Rightarrow |x|\leq\sqrt{\varepsilon}
\end{align*}

Thus we conclude that for every $\varepsilon > 0$, there corresponds a $\delta = \sqrt{\varepsilon} > 0$ such that
\begin{align*}
|x - 0| \leq \delta  = \sqrt{\varepsilon} \Rightarrow |x^{2}| \leq \varepsilon & \Rightarrow |x^{2}\sin(x^{2} + 2x)| \leq |\sin(x^{2}+2x)|\varepsilon \leq \varepsilon
\end{align*}
and we are done.

Hopefully this helps!


问:
How are the ranges of these cdf determined?

In the solution, it says $a > 0$ is $P(aX+b\leq y)$ and $a < 0$ is $P(aX+b\geq y)$. Can I get an explanation for why that is, please?

答: It is because, when you divide by a, the inequality will change sign, so you get for a<0 : $$P(aX+b\leq y)=P\left(X\geq\frac{y-b}{a}\right)=1-F_X\left(\frac{y-b}{a}\right) $$


Body:
Here is a generalization:
$$
\begin{align}
\sum_{k=m}^{n-1}\binom{k}{m}2^{n-(k+1)}
&=\sum_{k=m}^{n-1}\binom{k}{m}\left(2^{n-k}-2^{n-(k+1)}\right)\tag{1a}\\
&=\sum_{k=m}^{n-1}\binom{k}{m}2^{n-k}-\sum_{k=m+1}^n\binom{k-1}{m}2^{n-k}\tag{1b}\\
&=\sum_{k=m}^{n-1}\binom{k}{m}2^{n-k}-\sum_{k=m}^{n-1}\binom{k-1}{m}2^{n-k}-\binom{n-1}{m}+2^n[m=0]\tag{1c}\\
&=\sum_{k=m}^{n-1}\binom{k-1}{m-1}2^{n-k}-\binom{n-1}{m}+2^n[m=0]\tag{1d}\\
&=2\sum_{k=m-1}^{n-2}\binom{k}{m-1}2^{(n-1)-(k+1)}-\binom{n-1}{m}+2^n[m=0]\tag{1e}
\end{align}
$$
Explanation:

$\text{(1a)}$: $2^{n-(k+1)}=2^{n-k}-2^{n-(k+1)}$

$\text{(1b)}$: substitute $k\mapsto k-1$ in the right-hand sum

$\text{(1c)}$: handle the edge terms from the right-hand sum

$\text{(1d)}$: $\binom{k-1}{m-1}=\binom{k}{m}-\binom{k-1}{m}$

$\text{(1e)}$: substitute $k\mapsto k+1$ and pull a factor of $2$ out front

$\phantom{\text{(1e):}}$ so that the sum matches with $(n,m)\mapsto(n-1,m-1)$

Applying $(1)$ twice gives
$$
\begin{align}
\sum_{k=1}^{n-1}\binom{k}{1}2^{n-(k+1)}
&=2\color{#C00}{\sum_{k=0}^{n-2}\binom{k}{0}2^{(n-1)-(k+1)}}-(n-1)\tag{2a}\\
&=2\color{#C00}{\left(2^{n-1}-1\right)}-(n-1)\tag{2b}\\[8pt]
&=2^n-n-1\tag{2c}
\end{align}
$$
Explanation:

$\text{(2a)}$: apply $\text{(1e)}$

$\text{(2b)}$: apply $\text{(1e)}$ to the red part

$\text{(2c)}$: simplify


问:
Is $k[x,y]_{(x,y)}/(x^2-y^3)$ regular local?
Let $k$ be field,  and $A:=k[x,y]_{(x,y)}/(x^2-y^3). $

Now, $A$ is 1-dimension Noetherian local normal domain.

In general, 1-dimension Noetherian local domain is DVR iff normal.

So I think $A$ is DVR, in particular regular local.

What is my mistake?

答:
The element $T = x/y$ in the fraction field of $A$ satisfies
$$
T^3 - x = 0,
$$
but $T$ is not in $A$. Therefore $A$ is not integrally closed in its fraction field and is not normal.


问:
Show whether or not the Lebesgue Theorem applies. Does $\int \int_Af(x,y)dxdy$ exist?
Let $A = (0, 1) \times (0, 1)$ and $f : A \to \mathbb{R}$ where $$f(x, y) = \begin{cases}
\frac{1}{x−y} &\text{ if } x \neq y \\
            x &\text{if } x = y
\end{cases}$$

I have this piecewise function and am trying to answer the question above but am not sure how to .

答: If the integral exists, then we can apply Fubini theorem, that is integrate on a line (x or y = constant).  But the line integral is notoriously non existent, so the integral does not exist.</p>


答:
Assuming you are trying to find a $\delta > 0$ such that $|x - 0| < \delta$ implies the inequality you are working on.  So you are trying to find an upper bound on $x$ in terms of $\varepsilon$.

You have more than $-1 \leq \sin x \leq 1$.  You have $-1 \leq \sin(x^2 + 2x) \leq 1$.  So $0 \leq |\sin(x^2 +2x)| < 1$.  (The left-hand side of this is true for all moduli.  The right-hand side comes from the previous inequality.)  So you have
$$  x^2 [\text{some number in $[0,1]$}] < \varepsilon  \text{.}  $$
The biggest the left-hand side can be is $x^2$ and this gives a bound on $|x|$ in terms of $\varepsilon$...

Title: Proof using formal definition of limit that $ \lim_{x\to0} (x^2\sin(x^2+2x)+2)=2 $

Body:
I'm having a hard time trying to understand how to prove the following, using the formal definition of limits:

$$
\lim_{x\to0} (x^2\sin(x^2+2x)+2)=2
$$

I already did the following the steps but I'm not sure if my thinking is correct:

$$
|x^2\sin(x^2+2x)+2-2|<\epsilon \implies  |x^2\sin(x^2+2x)|<\epsilon \implies |x^2||\sin(x^2+2x)|<\epsilon
$$

and assuming that $x\in{R}$ then $x^2|\sin(x^2+2x)|<\epsilon$

Now I also know that $-1<\sin(x)<1$ however I'm struggling to understand how to best apply this to the proof.

I'm not looking for a full solution to the problem, rather I'm interested in confirming if my thinking up until now is correct and maybe some hints to unblock the next steps.


问:
$\det A = a_{1 , 1}A_{1 , 1} - a_{1 , 2}A_{1 , 2} + \cdots + (-1)^{n+1} a_{1 , n}A_{1 , n}$.
$A = (a_{ij})$ is a square matrix  of order $n$ . It's determinant will be $a_{1 , 1}A_{1 , 1} - a_{1 , 2}A_{1 , 2} + \cdots + (-1)^{n+1} a_{1 , n}A_{1 , n}$. Here $A_{i , j}$ is the co-factor of $a_{i , j}$.

The definition of determinant is following.

I was trying to prove it only from the definition of determinant . I was not allowed to use any other result than the definition of determinant.

But I am not being able to do that. Can anyone please help me to do this ?

答:
Given two sets $X, Y$, let $B(X,Y)$ denote the (possibly empty) set of all bijections $X\to Y$. For $M_n = \{1,\ldots,n\}$ we let $\mathfrak S_n = B(M_n,M_n)$ denote the set of all permutations of $M_n$. For $k \in M_n$ let $M_n^k = M_n \setminus \{k\}$ and let $b_k : M_{n-1} \to M_n^k$ denote the unique order preserving bijection. Explicitly $b_k(i) = i$ for $i < k$ and $b_k(i) = i+1$ for $i \ge k$.

We have $\det A = \sum_{\sigma \in \mathfrak S_n} \left(\operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)} \right)$.

Let $\bar A_{1,k}$ denote the $(n-1) \times (n-1)$-matrix obtained from $A$ by deleting the first row and the $k$-th column. Then $A_{1,k} = \det \bar A_{1,k}$. We have $\bar A_{1,k} = (\bar a_{i,j})$ with $\bar a_{i,j} = a_{b_1(i),b_k(j)}$.

Define $\mathfrak S_n^{1,k} =\{ \sigma \in \mathfrak S_n \mid \sigma(1) = k \}$. Then
$$\det A = \sum_{k=1}^n a_{1,k}\sum_{\sigma \in \mathfrak S_n^{1,k}} \left(\operatorname{sgn}(\sigma) \prod_{i=2}^n a_{i,\sigma(i)} \right) .$$
There is an obvious bijection $c_{1,k} : B(M_n^1, M_n^k) \to \mathfrak S_n^{1,k}$. Thus
$$f_{1,k} : \mathfrak S_{n-1} \to \mathfrak S_n^{1,k}, f_{1,k}(\tau) = c_{1,k}(b_k \tau b_1^{-1})$$
is a bijection. Explicitly $\bar \tau = f_{1,k}(\tau)$ is given by
$$\bar \tau(i) = \begin{cases} k & i = 1 \\ b_k(\tau(i-1)) & i > 1 \end{cases}$$

For a permutation $\omega \in \mathfrak S_m$ we have $\operatorname{sgn}(\omega) = (-1)^{N(\omega)}$, where $N(\omega)$ is the number of inversions in $\omega$, i.e. the number of elements of the set $I(\omega) = \{ (i,j) \mid i,j \in \{1,\ldots,m\}, i < j , \omega(i) > \omega(j) \}$.

Let us count the pairs $(i,j)$,  $1 \le i < j \le n$, which belong to $I(\bar \tau)$.

Case 1. $i = 1$.

Then the inversion condition $k = \bar \tau(1) > \bar \tau(j)$ is satisfied iff $\tau(j-1) < k$. Among $j =2,\ldots,n$ there are exactly $k-1$ numbers such that $\tau(j-1) < k$ (since  $\tau$ is a bijection on $M_{n-1}$).

Case 2. $i > 1$.

We have $\bar \tau(i) > \bar \tau(j)$ iff $b_k(\tau(i-1)) > b_k(\tau(j-1))$. Since $b_k$ is order preserving, i.e. strictly increasing, this is equivalent to $\tau(i-1) > \tau(j-1)$. This shows that the number of inversions of $\bar \tau$ which have the form $(i,j)$ with $i > 1$ is exactly $N(\tau)$.

Cases 1. and 2. prove $N(\bar \tau) = N(\tau) + k - 1$. Hence
$$\det A = \sum_{k=1}^n a_{1,k}\sum_{\tau \in \mathfrak S_{n-1}} \left(\operatorname{sgn}(\bar \tau) \prod_{i=2}^n a_{i,\bar \tau(i)} \right) = \sum_{k=1}^n a_{1,k}\sum_{\tau \in \mathfrak S_{n-1}} \left((-1)^{k-1}\operatorname{sgn}(\tau) \prod_{i=2}^n a_{i,\bar \tau(i)} \right) \\ = \sum_{k=1}^n (-1)^{k-1}a_{1,k}\sum_{\tau \in \mathfrak S_{n-1}} \left(\operatorname{sgn}(\tau) \prod_{i=1}^{n-1} a_{i+1,\bar \tau(i+1)} \right).$$
But $i+1 = b_1(i)$ and $\bar \tau(i+1) = b_k(\tau(i))$ and therefore
$$\det A = \sum_{k=1}^n (-1)^{k-1}a_{1,k} \det \bar A_{1,k} = \sum_{k=1}^n (-1)^{k-1}a_{1,k} A_{1,k} .$$


答:
Let $Y=aX+b$, where $a,b\in\mathbb{R}$ with $a\ne 0$.

By definition,
$$
F_Y(y)=P(Y\le y)=P(aX+b\le y)
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;
$$
hence if $a > 0$ we get
\begin{align*}
F_Y(y)
&=\,
P(aX+b\le y)
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;
\\[4pt]
&=\,
P(aX\le y-b)
\\[4pt]
&=\,
P\Bigl(X\le \frac{y-b}{a}\Bigr)
\\[4pt]
&=\,
F_X\Bigl(\frac{y-b}{a}\Bigr)
\\[4pt]
\end{align*}
and if $a < 0$, correcting a typo in the referenced solution, we get
\begin{align*}
F_Y(y)
&=\,
P(aX+b\le y)
\\[4pt]
&=\,
1-P(aX+b\ge y)&&\text{[assuming $P(aX+b=y)=0$]}
\\[4pt]
&=\,
1-P(aX\ge y-b)
\\[4pt]
&=\,
1-P\Bigl(X\le \frac{y-b}{a}\Bigr)&&\text{[since $a < 0$, dividing by $a$}\\[-4pt]
&&&\;\text{reverses the direction of the inequality]}
\\[4pt]
&=\,
1-F_X\Bigl(\frac{y-b}{a}\Bigr)
\\[4pt]
\end{align*}

Title: How are the ranges of these cdf determined?

Body: In the solution, it says $a > 0$ is $P(aX+b\leq y)$ and $a < 0$ is $P(aX+b\geq y)$. Can I get an explanation for why that is, please?


问:
Proving two spanning set are equal
I'm trying to understand how I should approach the following question;

Prove the following subspaces of $\mathbb{R}^3$ are equal. $$U = \operatorname{span}\{(1,2,3),(−1,2,9)\}$$ and $$W = \{(x,y,z) \in \mathbb{R}^3 \mid z−3y+3x = 0 \}.$$

I understand that proving this involves proving a set inclusion in both directions, however its not very clear to how the elements/vectors of $W$ show look given the formula to which they must satisfy.

Your help would be very much appreciated.
Thanks

答:
It is clear that both $U$ and $W$ are each dimension $2$.

($W$ is dimension $2$ because it can be written as $span\{(1,0,-3),(0,1,3)\}$)

Since they are the same dimension and both subsets of the same parent space of $\Bbb R^3$, it suffices to merely show inclusion in one direction instead of in both directions.  After all, if you have an $n$ dimensional space, the only $n$-dimensional subspace of it is the original space itself.

So... checking inclusion in the one direction, it suffices to check whether $(1,2,3)$ satisfy $z-3y+3x=0$ and whether $(-1,2,9)$ satisfy $z-3y+3x=0$.

Title: Proving two spanning set are equal


问:
Solve the equation $y=\frac{ax + b}{cx + d}$ for $x$, noting where the solution assumes $ad-bc\ne 0$

These questions are from Stillwells The four Pillars of Geometry and in the context of linear fractional functions in projective geometry.

Solving for $x$ is straightforward where I get $x= \frac{b-yd}{yc-a}$ by some algebraic manipulation. But I can't spot where my manipulations assumed $ad-bc\ne 0$. What am I missing?

I have a hunch that another way to approach this could connected to the determinant of a matrix being zero and the impact that can have on whether or not an equation can be solved, but i'm not sure about that. I'm also looking to understand a bit more intuitively why $ad-bc\ne 0$ in general for linear fractional functions.

EDIT: Excerpt from the text regarding division by zero

" The linear fractional functions $f(x)= \frac{ax+b}{cx+d}$ we have used to describe projective mappings of lines are actually defective is the variable $x$ only runs through the set $\mathbb{R}$ of real numbers. For example, the function $f(x)=\frac{1}{x}$ we used to map points of the line L1 onto points of the line L2 does not in fact map all points. It cannot set the point $x=0$ anywhere because $\frac{1}{0}$ is undefined; not can it sent any point to $y=0$ because $0 \ne \frac{1}{x}$ for any real x. This defect is neatly fixed by extending the function $f(x)=\frac{1}{x}$ to an new object $x=\infty$ and declaring that $\frac{1}{\infty}=0$ and $\frac{1}{0}=\infty$. The new object $\infty$ is none other than the point at infinity of the line L1, which is supposed to map to the point $0$ on L2. Likewise, id $\frac{1}{0}= \infty$, the point 0 on L1 is sent to the point $\infty$ on L2, as it should be."

答:
You don't show your work, so it is impossible to spot where you used the assumption.

From the givens, $cx+d \neq 0$, so we can multply both sides of the equation by $cx+d$ without destroying all information in the equation.
$$  y (cx+d) = ax+b  $$
Move $x$-containing terms left and others right.
$$  c x y - ax = b - yd  $$
Un-distribute $x$s.
$$  (c y - a)x = b - yd  $$
Now, assuming $cy - a \neq 0$,
$$  x = \frac{b - yd}{cy-a}  \text{.}  $$
However, \begin{align*}
cy - a &\neq 0  \\
c \frac{ax+b}{cx+d} - a \frac{cx+d}{cx+d} &\neq 0  \\
\frac{acx+bc - acx - ad}{cx+d} &\neq 0 \\
\frac{bc - ad}{cx+d} &\neq 0 \text{.}
\end{align*}
We already know $cx+d \neq 0$, so this is defined and is satisfied as long as the numerator on the left-hand side is nonzero, that is, when $bc-ad \neq 0$.  Multiplying through by $-1$, when $ad - bc \neq 0$.

So you discover $ad - bc \neq 0$ is equivalent to $cy - a \neq 0$, which is necessary for inverting the relation between $x$ and $y$.  So, this map isn't a bijection unless $ad -bc \neq 0$.


问:
How to show that the integral is convergent without actually evaluating it?

$$ \int _ 0 ^ { + \infty } \frac 1 { \sqrt { 1 + x ^ 8 } } \ \mathrm d x $$

I have a hint that says there is no elementary way to evaluate this integral. So I am supposed to bound this integrand by a simpler function and use the following fact without proof:

Let $ f : [a,+\infty) \to \mathbb R $ be a bounded and increasing function. Then $ \lim _ { x \to \infty } f ( x ) $ exists.

I specifically don't understand what it means to bound the integrand by a simpler function. Any help will be appreciated.

答:
Just to take a slightly different approach, if we can establish that

$${1\over\sqrt{1+x^8}}\le{2\over1+x^2}$$

for all $x$, then we have

$$\int_0^\infty{dx\over\sqrt{1+x^8}}\le2\int_0^\infty{dx\over1+x^2}=\pi$$

To establish the key inequality, note that

$${1\over\sqrt{1+x^8}}\le{2\over1+x^2}\iff1+2x^2+x^4\le4+4x^8\iff 4x^8-x^4-2x^2+3\ge0$$

and

$$4x^8-x^4-2x^2+3\ge
\begin{cases}
0-1-2+3=0\quad&\text{if }|x|\le1\\
4x^8-x^8-2x^8+3=x^8+3\ge0\quad&\text{if }|x|\ge1
\end{cases}$$

The only advantage (if any) of this approach is that it uses a single formula for the bounding function, rather than one formula for $x\le1$ and another for $x\ge1$. But doing so, of course, comes at the expense of a slightly tricky verification of a polynomial inequality.


问:
Integrate the given function around the unit circle using Cauchy's

Integrate the given function around the unit circle

$\frac{z^3}{2z-i}$

I'm a bit confused here. So since the function is undefined at $z = \frac{1}{2}$, what does this imply? It implies that the function is not analytic everywhere does the integral does not equal 0 and we have to evaluate this by hand right? So...

Since the unit circle can be expressed as $z(t) = e^{it}$ and $\frac{dz}{dt} = ie^{it}$

we have:

$$f(z) = \frac{e^{3it}}{2e^{it} - i}$$

So the integral is:

$$\int_0^{2\pi} \frac{e^{3it}}{2e^{it} - i} \cdot ie^{it} $$

Is this the right setup? Where do I go from here?

答:
I am assuming from your question that you do not know the Cauchy integral formula. Here is a simple approach: $f(z)=\frac{z^3}{2(z-\frac{i}{2})}=c_2(z-\frac{i}{2})^2+c_1(z-\frac{i}{2})+c_0+\frac{c_{-1}}{z-\frac{i}{2}}.$ Multiplying though by $2(z-\frac{i}{2})$, expanding and equating coefficients, you get $c_{-1}=-\frac{i}{16}.$ The other constants do not matter because when you integrate around your curve, the only non-zero integral is $\int_{\gamma}\frac{c_{-1}}{z-\frac{i}{2}}dz$ and this is equal to $2\pi i c_{-1}$ by direct calculation. Thus, your integral is $\frac{\pi}{8}.$

What we did above amounts to realizing that $f$ has a pole of order $1$ at $z=\frac{i}{2}$ so its Laurent expansion about $z=\frac{i}{2}$ is $f(z)=P(z-\frac{i}{2})+\frac{c_{-1}}{z-\frac{i}{2}}$ where $P$ is a polynomial. There are formulas for calculating $c_{-1}$, using the Residue Theorem.

Remark: a useful fact is that, somewhat loosely speaking, if two curves are path homotopic, then their integrals are the same. So, if we set $\gamma': z=\frac{i}{2}+e^{it}$ then $\gamma\sim \gamma'$ and $\int_{\gamma'}f(z)dz=i\int_0^{2\pi}e^{it}\frac{(\frac{i}{2}+e^{it})^3}{e^{it}}dt=i\int_0^{2\pi}(\frac{i}{2}+e^{it})^3dt=\frac{\pi}{8}.$


问:
Function equation problem: $ f ( x y ) = f ( x ) ^ { y ^ \beta } f ( y ) ^ { x ^ \beta } $

Solve function equation
$$
f ( x y ) = f ( x ) ^ { y ^ \beta } f ( y ) ^ { x ^ \beta } \text ,
$$
where $ \beta $ is real number, $ f ( x ) $ is continuous single valued function.

Obviously, $ f ( x ) = 1 $ satisfies it, but are there other solutions?

答:
First of all, note that there are ambiguities in the statement of the problem that need to be resolved before starting to solve it. The most important one is that the domain and codomain of $ f $ should be specified. Another one is that it should be specified for which elements of the domain like $ x $ and $ y $ we have the functional equation
$$ f ( x y ) = f ( x ) ^ { y ^ \beta } f ( y ) ^ { x ^ \beta } \text . \tag 0 \label 0 $$
Note for example that if $ \beta < 0 $, $ 0 ^ \beta $ has no meaning, if $ \beta $ is irrational, $ ( - 1 ) ^ \beta $ is undefined, and exponentiation of complex numbers (to the power of a non-integer) is not a well-defined single-valued function.

Let's assume that $ f : \mathbb R ^ + \to \mathbb R ^ { 0 + } $ and \eqref{0} hold for all $ x , y \in \mathbb R ^ + $. Under this specification, we can show that the only solutions are of the form $ f ( x ) = 0 $ and $ f ( x ) = \exp \big( x ^ \beta A ( \log x ) \big) $ for some additive function $ A : \mathbb R \to \mathbb R $. It's straightforward to check that these are indeed solutions. We prove that they are the only ones.

Define $ g : \mathbb R ^ + \to \mathbb R ^ { 0 + } $ with $ g ( x ) = f ( x ) ^ { x ^ { - \beta } } $. Then, by \eqref{0} we have
$$ g ( x y ) = g ( x ) g ( y ) \tag 1 \label 1 $$
for all $ x , y \in \mathbb R ^ + $. Lettin $ x = y = 1 $ in \eqref{1} we can see that $ g ( 1 ) \in \{ 0 , 1 \} $. If $ g ( 1 ) = 0 $, then putting $ y = 1 $ in \eqref{1} shows that $ g $ is constantly zero, and thus $ f $ too is the constant zero function, which is one of the mentioned solutions. So, from now on, assume that $ g ( 1 ) = 1 $. Letting $ y = \frac 1 x $ in \eqref{1} we have
$$ g ( x ) g \left( \frac 1 x \right) = 1 \text , $$
which in particular shows that $ g ( x ) > 0 $ for all $ x \in \mathbb R ^ + $. This lets us define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = \log \big( g ( \exp x ) \big) $, and \eqref{1} will show that $ A $ satisfies Cauchy's functional equation. Thus we have
$$ f ( x ) = g ( x ) ^ { x ^ \beta } = \Big( \exp \big( A ( \log x ) \big) \Big) ^ { x ^ \beta } = \exp \big( x ^ \beta A ( \log x ) \big) \text , $$
which gives the other mentioned set of solutions.

It's well-known that assuming the axiom of choice, one can show that wild additive functions exists, using Hamel bases of $ \mathbb R $ (considered as a vector space over the field $ \mathbb Q $). On the other hand, assuming further regularities like continuity, local boundedness or even Lebesgue measurability, an additive function $ A : \mathbb R \to \mathbb R $ must be linear; i.e. there must be a constant $ a \in \mathbb R $ such that $ A ( x ) = a x $ for all $ x \in \mathbb R $. See this post for more information. In the case of our problem, assuming regularities on $ f $ forces regularities on $ A $, and thus regular nonzero solutions to \eqref{0} must be of the form $ f ( x ) = \exp \left( a x ^ \beta \log x \right) $, or equivalently
$$ f ( x ) = x ^ { a x ^ \beta } $$
for some constant $ a \in \mathbb R $. The case of $ f ( x ) $ being constantly equal to $ 1 $ that you've found by yourself corresponds to the case $ a = 0 $.


问:
Trying to figure out what is wrong in this way of thinking about proofs - Proof by contadiction
This may be super random, but I will use a question from calculus, but this is pure logic question.

I am trying to understand what is wrong with this way of thinking about proofs using assuming the opposite and then getting a contradiction.

Given this task: Prove that

$$ \limsup (a_n + b_n) \leq \limsup a_n + \limsup b_n \tag{1}$$

Why can't I say:

Assuming that $$\limsup (a_n + b_n) > \limsup a_n + \limsup b_n \tag{2}$$

And then giving a counter-example - thus proving (1).

This probably does not tell me right away that $\text{(1)}$ is correct, but logically I don't find it any different than other "assuming that P(x) is wrong then we get to a contradiction thus P(x) must be right because of the law of excluded middle (tertium non datur) "

Where do I get things wrong here?

答:
The issue is quantifiers.

The original statement has an implicit universal quantifier: it would be better phrased as "Every pair of sequences $(a_i),(b_i)$ has property [stuff]." The negation of this involves an existential quantifier: "There is a pair of sequences $(a_i),(b_i)$ without property [stuff]." When we negate, universals turn into existentials (and vice versa).

So in the OP, you've successfully proved that your statement $(2)$ (which also carries an implicit universal quantifier) is false. But your statement $(2)$ is not the negation of statement $(1)$ - it's a much stronger statement, so disproving it doesn't finish your problem.

This is something natural language unfortunately makes rather annoying: in my experience, the biggest confusion students have with proving complicated expressions is keeping track of quantifier structure. The OP gives an example of how the type of quantifier (universal or existential) can be obscured by the natural language phrasing, but there are other subtleties to keep an eye out for, especially order of quantifiers (to see that order matters, consider "For all $x$ there is some $y$ such that $x<y$" versus "There is some $y$ such that for all $x$ we have $x<y$").


问:
Gradient in Polar Coordinates ($\mathbb{R}^n$)
I am told that $\textbf{x} = r\omega$ where $r = |x|$ and $\omega \in S^{n-1}$
, then $(\textbf{x} \cdot \nabla)f = r\partial_r f$. Why is this so?

答:
If $n=2$, you can see this with the usual polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$. By the ordinary chain rule:

$$\frac{\partial}{\partial r}f(r\cos(\theta),r\sin(\theta)) = \cos(\theta) \frac{\partial f}{\partial x} + \sin(\theta) \frac{\partial f}{\partial y} $$

Multyply both sides by $r$, and you get $r \, \partial_r f = x f_x + y f_y = \mathbf{x} \cdot \nabla f$.

I'm sure the case for $n>2$ is similar, just using higher-dimensional spherical coordinates.