问:
Equation rearranging - what are $x$ and $y$ in terms of $u$ and $v$?
I am having trouble rearranging equations $u$ and $v$ into $x$ and $y$ in terms of $u$ and $v$.

The equations are: $u = e^x - y$ and $v = y^2 +4e^{-2x}$.

I wish to find $x = x(u,v)$ and $y = y(u,v)$ from these two equations.

So far I have rearranged $v$ in terms of $x$ for $x = \frac{1}{2}\ln(\frac{4}{v-y^2})$ and simplified substituting this into $u = e^x - y$ for:

$$ y^4+2uy^3+u^2y^2-y^2v-2uvy-u^2v +4 = 0$$

However, I am just at a loss on how to solve this quartic for $y$. What am I doing wrong? Is there a better method for this?

Body:
Be careful what you wish for.  Here's the solution for just $y$, produced by Mathematica:

$$y = -\frac{1}{2} \sqrt{\frac{1}{3} \left(u^2-v\right)+\frac{\sqrt[3]{2
   \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4
   v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}+\frac{\sqrt[3]{2}
   \left(u^4-2 u^2 v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288
   \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2
   v-v^4+32 v^2-256}}}+v}-\frac{1}{2} \sqrt{\frac{1}{3}
   \left(v-u^2\right)+u^2-\frac{\sqrt[3]{2 \left(u^2-v\right)^3-288
   \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2
   v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}-\frac{\sqrt[3]{2} \left(u^4-2 u^2
   v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24
   \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}-\frac{8
   u^3-8 u \left(u^2-v\right)}{4 \sqrt{\frac{1}{3} \left(u^2-v\right)+\frac{\sqrt[3]{2
   \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4
   v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}+\frac{\sqrt[3]{2}
   \left(u^4-2 u^2 v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288
   \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2
   v-v^4+32 v^2-256}}}+v}}+v}-\frac{u}{2}$$


问:
Show that from following equivalence we get homeomorphism
I have given X and Y topological spaces and f is a bijective function. I have to show that from B $\subseteq X$ : f(B) closed in Y $\iff $ B is closed in X it follows that f an Homeomorphism is.

I am struggling with this and I really would appreciate your help

答:
Well, by definition, a function $g : X \to Y$ is continuous if for any open set $U \subseteq Y$, the set $g^{-1}(U)$ is open in X.

Now, for $f$ to be a homeomorphism, you want both $f$ and $f^{-1}$ to be continuous... can you see where this is going? :)

ContentLicense: CC BY-SA 4.0


问:
Find $\angle ADB$ where $\angle BAC = 18^\circ$, $\angle ABC = 12^\circ$ and $AB = CD$.
Let $\triangle ABD$ be a triangle and $C$ be a point on $BD$ such that $ {AB} ={CD}$, $\angle BAC = 18^\circ$ and $\angle ABC = 12^\circ$. The question is to find $x$, where $\angle ADC = x^\circ$.

I think this is a pretty classic problem but I am not able to find a solution using the techniques which were used to solve a similar kind of problem (only the angles were different and they were $30^\circ$ and $24^\circ$ respectively). The idea was to create an equilateral triangle by drawing a perpendicular line segment from $B$ to $AC$ with the same length as $AB$. However, that technique does not work here and I cannot find another way. Any help is appreciated (I forgot to add this but an answer which uses "geometrical" methods like drawing line segments or forming new triangles etc. would be great), thank you. The figure is as follows:

答:
Use law of sines. In $\triangle ABC$:
$$\frac{\sin(180^\circ-12^\circ-18^\circ)}{AB}=\frac{\sin 12^\circ}{AC}$$
In $\triangle ACD$:
$$\frac{\sin(180^\circ-12^\circ-18^\circ-x)}{AB}=\frac{\sin x}{AC}$$ In the last equation I've used $CD=AB$.
Therefore, take $AB/AC$ from both equations:
$$\frac{\sin 150^\circ}{\sin 12^\circ}=\frac{\sin (150^\circ-x)}{\sin x}$$
Use $$\sin(150^\circ-x)=\sin 150^\circ\cos x-\cos 150^\circ\sin x$$
to get a simple equation for $\tan x$.


问:
X = a/ (1+a) . Would the value of x increase or decrease if “a” increases?
It’s pretty easy to find the answer after putting in random values but I can’t wrap my head around with my below explanation which is in direct contradiction to the result.

If the numerator increases the x value goes up and the value of x goes down if the denominator increases. Now if “a” increases the net  increase of the denominator would be more than the net increase of numerator as the denominator has “+1” in it. Therefore, since the denominator is more than that of numerator shouldn’t the value of x decrease with increase of “a”?

答: Note that $X = \frac{a}{1 + a} = 1 - \frac{1}{1 + a}$. Then, since $\frac{1}{1 + a}$ is a decreasing function, $X$ increases as $a$ increases. $\blacksquare$


问:
Is this relation symmetric? $R={(1,1),(0,0)}\ $
R={(1,1),(0,0)}

A relation is symmetric if xRy and yRx but can it be x=y?

答:
It symmetric.

If $\color{green}0R\color{red}0$ then $\color{red}0R\color{green}0$.

And if $\color{green}1R\color{red}1$ then $\color{red}1R\color{green}1$

And as those are the only two possible cases of $\color{green}x R\color{red}y$ possible, If we ever have $\color{green}x R\color{red}y$ it will always be the case we also have $\color{red}y R\color{green}x$.

Please let me know if you hare red/green colorblind.  If so consider this explanation:

If $\color{purple}{\large{\text{ZERO}}}R\color{orange}{\small{\text{zero}}}$ then $\color{orange}{\small{\text{zero}}}R\color{purple}{\large{\text{ZERO}}}$.

And if $\color{purple}{\large{\text{ONE}}}R\color{orange}{\small{\text{one}}}$ then $\color{orange}{\small{\text{one}}}R\color{purple}{\large{\text{ONE}}}$

And as those are the only two possible cases of $\color{purple}{\large{\text{X}}}R\color{orange}{\small{\text{y}}}$ possible, If we ever have $\color{purple}{\large{\text{X}}}R\color{orange}{\small{\text{y}}}$ it will always be the case we also have $\color{orange}{\small{\text{y}}}R\color{purple}{\large{\text{X}}}$.


问:
Upper bound on number of eigenvalues of an endomorphism
I am reading lecture notes on abstract algebra and came across a proof that doesn’t appear to be valid. The main theorem states that SpecT$\leq$DimV. It then says that this can be proven by showing that eigenvectors with distinct eigenvalues are linearly independent. This would be acceptable, however it requires that the number of eigenvalues of a given transformation(in this case an endomorphism) be finite. Is that the case or is there something I’m misunderstanding?

答:
It then says that this can be proven by showing that eigenvectors with distinct eigenvalues are linearly independent. This would be acceptable, however it requires that the number of eigenvalues of a given transformation(in this case an endomorphism) be finite.

Actually, once you know that eigenvectors with distinct eigenvalues are linearly independent, it follows immediately that there must be at most $\dim V$ eigenvalues, without any need to appeal to the existence of the characteristic polynomial. This is the content of the theorem! The argument is simple: if there are $n$ distinct eigenvalues, then we can find $n$ linearly independent eigenvectors, so $n \le \dim V$.

The existence of the characteristic polynomial proves the stronger fact that there are still at most $\dim V$ eigenvalues when counted with (algebraic) multiplicity, but you also don't need the existence of the characteristic polynomial to prove this if you define the algebraic multiplicity as the dimension of the generalized eigenspace $\dim \ker((T - \lambda)^{\dim V})$. The argument is a bit more complicated but similar: you show that the generalized eigenspaces are "linearly independent" in the sense that the natural map from their direct sum to $V$ is injective (in fact it's an isomorphism but that isn't needed here).


问:
Combination problem - picking a basketball team with restrictions
In how many ways can a basketball coach form a team of 2 centers, 2 guards and 1 forward out of a total of 10 basketball players, if
3 players can only play center,
3 only guard position,
1 only forward position,
2 players guard or forward,
1 center or forward?

The solution is as follows: ${5\choose 2}\cdot{4\choose 2}+{5\choose 2}\cdot{3\choose 2}+2\cdot{4\choose 2}\cdot{4\choose 2}-{3\choose 1}\cdot{4\choose 2}=144$

I understand ${5\choose 2}\cdot{4\choose 2}$ is for the case when we picked both "OR" cases as not forwards, but I don't get how we got the rest.

答:
First, pick who the forward is.  Break into cases based on which category the forward came from, be it the only forward category, the guard or forward category or the center or forward category.

In the event that your forward you picked was the person who could only play forward, you still have five people who can play guard (the 3 only guard plus the 2 guard or forward) of which we want to pick two of them and then you want to pick two people from the four you still have who can play center (the 3 only center plus the 1 center or forward).  This accounts for $\binom{5}{2}\binom{4}{2}$ different configurations.

The other terms in the above is what happens if the forward you selected for your team came from one of the other categories.  Supposing they came from the guard or forward category, that gives two options for who it was.  Next, pick who the two centers are and who the two guards are.  This is the $2\cdot \binom{4}{2}\binom{4}{2}$ term.

Similarly if the forward you select for your team came from the center or forward category, that leads to the $\binom{5}{2}\binom{3}{2}$ term.

I would have stopped here personally.  I would have expected the answer to have been $\binom{5}{2}\binom{4}{2}+\binom{5}{2}\binom{3}{2}+2\binom{4}{2}\binom{4}{2}$, having considered a team as being comprised of members who have clearly defined roles.

The $-\binom{3}{1}\binom{4}{2}$ term that appears seems to be trying to correct for something with some unwritten assumption opposite to what I had assumed, so let us explore that in greater detail to see if we can figure out what was intended...

Let the players be denoted by the digits $0,1,2,\dots,9$.  Let the only center players be denoted as $OC = \{0,1,2\}$, the only guard as $OG=\{3,4,5\}$, the only forward as $OF=\{6\}$, the guard or forward as $GF=\{7,8\}$ and the center or forward as $CF=\{9\}$

Let us denote a team using the digits and a subscript detailing what position they are on the team to play as.

I would have considered the team $0_C,1_C,3_G,7_G,8_F$ to be a different team than $0_C,1_C,3_G,7_F,8_G$.  It is possible that the author of the problem had intended these to be treated both as $0,1,3,7,8$ and are indistinguishable.  If that is the case, then let us consider which cases we overcounted.  Overcounting will only have occurred if it is unclear who was playing what position.

It is clear that if we have player $6$ on our team (the person who can play only forward) that he must be playing forward, so that case did not lead to any overcounting.  If player $6$ was not in our team and it was instead player $9$ who was playing forward, it could not be confused with another team arrangement because switching $9$ to playing center instead there isn't a center to swap back with them to keep the team balance correct and it would have changed it to being $3$ centers.

As such, the only situation we might have overcounted then was if both players $7,8$ were on the team with one of them playing forward.  We will have counted it once with $7$ as the forward and $8$ as a guard and again with $8$ as the forward and $7$ as a guard but otherwise the same teammembers.  To count how many times this happened, we know $7$ and $8$ are both on the team.  We still need to pick one guard and two centers, which can be done in $\binom{3}{1}\binom{4}{2}$ ways.

Since we had counted these twice, subtracting $\binom{3}{1}\binom{4}{2}$ has the effect of making it so that overall they will have been counted exactly once, bringing the final total to:

$$\binom{5}{2}\binom{4}{2}+\binom{5}{2}\binom{3}{2}+2\binom{4}{2}\binom{4}{2}-\binom{3}{1}\binom{4}{2}$$


问:
Can you use modulus to make 0 > 2?
I wanted to create a rock-paper-scissors game that didn't use a lot of conditionals, and I was wondering if there were any mathematical way of representing the cycle of rock-paper-scissors. So Rock beats Paper beats Scissors beats Rock, or Rock > Paper > Scissors > Rock. If you assigned numbers to these, it wouldn't work: 0 > 1 > 2 > 0. But I was wondering if there were some way you could use modulus function to make this work? I understand that if you had a series of integers and modded them by 3 you would get a repeating cycle of 0, 1, 2, e.g. 60 % 3 = 0; 61 % 3 = 1; 62 % 3 = 2, 63 % 3 = 0, etc.

Do you think there could be any way to make this programmable using mods? I'm interested to know if there's an algorithm or some way to use this to create a cycle where 0>1>2>0. Just a warning, though, I don't study maths or know a lot about complex maths, so if you know of a way to do this, I'd really appreciate it if you tried not to overwhelm me with maths terms!

答: Codify Rock by 0, Scissors by 1, and Paper by 2. Notice that, modulo $3$, option $k$ beats option $k+1$ and is beaten by option $k-1$. So, if $x$ and $y$ are the choices of player 1 and 2, respectively, then compute $k=x-y$ modulo $3$. Then player 2 wins iff $k=1$, player 1 wins iff $k=-1$, and they tie iff $k=0$. This easily generalizes to cycles of any length. Thus, after correctly codifying the options by numbers (actually elements of the group $\mathbb Z_n$), all you need to do is compute the difference, compute modulo $n$, check whether the result is $0$ (resulting in a tie), or is it in the range 1 to floor(n/2) (player 2 wins), or else player 1 wins (some care needed according to the parity of $n$ and the rules of the game). </p>

ContentLicense: CC BY-SA 3.0

Title: Can you use modulus to make 0 > 2?


问:
How to prove something with an inequality?
Given $y,z \in\Bbb Z$ and $x \in\Bbb R\setminus\Bbb Z$, I have to prove that, for a value of $x, y = z$ for $y$ and all $z$, given that $x<y<x+1$ and $x<z<x+1$.

答:
$x < z< x+1$ and $ x< y <x+1$ so $ -x > -y >-x-1$

$-1 < z -y < 1$ and $\mathbb{Z}$ is closed under addition and multiplication

So $z-y$ is integer between $-1$ and $1$ and so $z-y=0 $ which implies that

$$ z=y $$


问:
Let $G$ be a group, $H, K \le G$. Under what conditions does $G/H \cong G/K \Rightarrow H \cong K$?

Let $G$ be a group, $H, K \le G$. Under what conditions does $G/H \cong G/K \Rightarrow H \cong K$?

I believe there was a condition that could be simply stated under which it was true. Maybe having to do something with split exact sequences?

答:
I'm not aware of any useful sufficient conditions other than extremely restrictive ones like that $G$ is finite elementary abelian (equivalently, a finite-dimensional $\mathbb{F}_p$-vector space) or cyclic (in which case a subgroup is determined by its index). Consider the following very simple counterexample: $G = C_2 \times C_4$ has two non-isomorphic subgroups $H = C_2 \times C_2, K = 1 \times C_4$ both of whom have quotient $C_2$. This is the smallest finite abelian group which is not elementary abelian or cyclic.

$G$ is also a counterexample to the analogous statement about whether $H \cong K$ implies $G/H \cong G/K$: this time if we take $H = C_2 \times 1$ and $K = 1 \times C_2$, the two subgroups are isomorphic but their quotients are $C_4$ and $C_2 \times C_2$ respectively. Actually this counterexample is Pontryagin dual to the previous one.


答:
With $z:=e^x$,

$$u=z-y,\\(v-y^2)z^2=4.$$

We can eliminate $z$, to get

$$(v-y^2)(u+y)^2=4$$

or $y$, to get

$$(v-(z-u)^2)z^2=4.$$

None of these quartic equations seem to have a simple solution.

Title: Equation rearranging - what are $x$ and $y$ in terms of $u$ and $v$?

Body:
I am having trouble rearranging equations $u$ and $v$ into $x$ and $y$ in terms of $u$ and $v$.

The equations are: $u = e^x - y$ and $v = y^2 +4e^{-2x}$.

I wish to find $x = x(u,v)$ and $y = y(u,v)$ from these two equations.

So far I have rearranged $v$ in terms of $x$ for $x = \frac{1}{2}\ln(\frac{4}{v-y^2})$ and simplified substituting this into $u = e^x - y$ for:

$$ y^4+2uy^3+u^2y^2-y^2v-2uvy-u^2v +4 = 0$$

However, I am just at a loss on how to solve this quartic for $y$. What am I doing wrong? Is there a better method for this?


问:
Is $\int^a\frac{\sqrt{x^3-a^3}}{\sqrt{x-a}}dx $ equal to $0$?
Obviously this is not a nice integral for us to just analytically compute.

My question is whether this integral is $0$ or whether it is undefined and how we would should what it is? Of course, if $x=a$ then the integrand is undefined and if $x<a$ then the integrand is not real. I'm not sure if this has any relevance to this problem?

答: In case anyone's interested, the unique antiderivative that vanishes at $x=a$ is$$\begin{align}\int_a^x\sqrt{t^2+at+a^2}dt&=\tfrac34a^2\int_{\pi/3}^{\arctan\tfrac{1+2x/a}{\sqrt{3}}}\sec^3udu\\&=\tfrac38a^2[\sec u\tan u+\ln|\sec u+\tan u|]_{\pi/3}^{\arctan\tfrac{1+2x/a}{\sqrt{3}}}\\&=\tfrac38a^2\left(\tfrac{1+2x/a}{\sqrt{3}}\sqrt{1+\tfrac{(1+2x/a)^2}{3}}-2\sqrt{3}+\ln\left|\tfrac{\sqrt{1+\tfrac{(1+2x/a)^2}{3}}+\tfrac{1+2x/a}{\sqrt{3}}}{2+\sqrt{3}}\right|\right)\end{align}$$(using $t=\tfrac12a(\sqrt{3}\tan u-1)$ and this). The most general antiderivative is the above $+C$, equal to $C$ at $x=a$.


问:
Vacuous Quantification
In this formula, a vacuous quantification can be removed:

$ \exists x \forall x (P(x) \rightarrow Q(x)) \equiv \forall x (P(x) \rightarrow Q(x)) $

What about a formula like this?:

$ \forall x (P(x) \rightarrow \exists x Q(x)) $

Is this also a vacuous quantification or is it not a well formed formula?

Can it be rewritten in a different form like for the first formula?

答:
The formula $\exists x \forall x (P(x) \to Q(x))$ is logically equivalent to $\forall x (P(x) \to Q(x))$$-$and hence the former can be rewritten to the latter without changing its meaning$-$because $\exists x$ does not bind any occurrence of $x$ in $\forall x (P(x) \to Q(x))$ and so this $\exists x$ is superfluous (technically, it is often said dummy).

On the contrary, in the formula $\forall x (P(x) \to \exists x Q(x))$, the quantifier $\forall x $ is not dummy (or "vacuous" according to your terminology), because it binds the occurrence of $x$ in $P(x)$. Therefore, $\forall x$ cannot be removed. A similar discourse holds for $\exists x$ (with respect to the occurrence of $x$ in $Q(x)$).

But you can rewrite the formula $\forall x (P(x) \to \exists x Q(x))$ as $\forall x (P(x) \to \exists y Q(y))$: they are exactly the same formula, since formulas are identified up to renaming of bound variables. The advantage of the second version of the formula is that you can now move the existential quantifier in front of the implication, without changing the meaning of the formula. Indeed, the formula $\forall x (P(x) \to \exists y Q(y))$ is logically equivalent to
$$\forall x \exists y (P(x) \to  Q(y))$$
The latter is a formula in prenex normal form, which means that all the quantifiers are at the beginning of the formula. A well-known theorem in classical logic says that every formula is logically equivalent to a formula in prenex normal form.

Title: Vacuous Quantification


答:
Quantification is vacuous when it binds no variables at all, either because there are no occurrences of the binding variable or because it gets entirely overridden by another quantifier. The latter is the case with the first formula, where $\exists x$ is immediately followed by $\forall x$.

In the second formula, hower, $\forall x$ does have an active binding, namely the occurrence of $x$ in $P(x)$. It is only the occurrence in $Q(x)$ that gets overwritten by $\exists x$. $\forall x$ can hence not be removed.

In most definitions of FOL syntax, vacuous quantification, binding overriding and identically named binding across independent subformulas (such as $(\exists x P(x)) \land (\forall x Q(x))$) is permitted.

If you want to make the formula clearer by avoiding identical variable names for different bindings, you can perform bound renaming and e.g. write $\forall x (P(x) \to \exists y P(y))$. This formula will be logically equivalent to the original.

Title: Vacuous Quantification

Body:
In this formula, a vacuous quantification can be removed:

$ \exists x \forall x (P(x) \rightarrow Q(x)) \equiv \forall x (P(x) \rightarrow Q(x)) $

What about a formula like this?:

$ \forall x (P(x) \rightarrow \exists x Q(x)) $

Is this also a vacuous quantification or is it not a well formed formula?

Can it be rewritten in a different form like for the first formula?


问:
Multilevel integer programming
Recently, I have been dealing with an integer program that involves finding a $4$-level integral solution

$$x \in \{-2, -1, 1, 2\}$$

or even $x \in \{\pm n, \pm n-1, \dots, \pm 1\}$ which is multilevel integers solution. The cost function is quadratic and $x$ is the only optimization variable.

I search the network but I could not find useful information about this kind of problem. So is there any reference (algorithm) for solving this kind of problem? Could anyone provide some suggestions? Thanks in advance!

Add constraint:

$x \in \{1, -1, +1i, -1i\}$ or with even more levels with unitary amplitude $|x| = 1$.

答: One way to restrict $x$ to a discrete set $\lbrace a_1,\dots,a_n\rbrace$ is to create $n$ binary variables $z_1,\dots,z_n$ and add the constraints $$x=\sum_{i=1}^n a_i z_i$$ and $$\sum_{i=1}^n z_i =1.$$

Title: Multilevel integer programming


问:
Expressing the solution of integral using Special functions Gamma or/and Beta functions?
$$\int_{0}^{\infty}(1+x)^{2020}*\frac{1+x+x^2+...+x^{2020}}{2020^{{x}^{2020}}}dx$$

How can this integral be solved?

Body:
$$=\int_{0}^{\infty }{\frac{{{(1+x)}^{2020}}}{{{2020}^{{{x}^{2020}}}}}}\left( \sum\nolimits_{n=0}^{2020}{{{x}^{n}}} \right)dx$$

$$=\sum\nolimits_{n=0}^{2020}{\int_{0}^{\infty }{\frac{{{(1+x)}^{2020}}}{{{2020}^{{{x}^{2020}}}}}{{x}^{n}}dx}}$$

$$=\sum\nolimits_{n=0}^{2020}{\sum\nolimits_{k=0}^{2020}{\left( \begin{align}
  & 2020 \\  & \ \ k \\ \end{align} \right)\int_{0}^{\infty }{\frac{{{x}^{k+n}}}{{{2020}^{{{x}^{2020}}}}}dx}}}$$

$$=\frac{1}{2020}\sum\nolimits_{n=0}^{2020}{\sum\nolimits_{k=0}^{2020}{\left( \begin{align}
  & 2020 \\ & \ \ k \\ \end{align} \right)}}\frac{\Gamma \left( \frac{n+k+1}{2020} \right)}{{{\left( \ln 2020 \right)}^{\frac{n+k+1}{2020}}}}$$
for the integral in the third line:
$$=\int_{0}^{\infty }{{{x}^{k+n}}{{2020}^{-{{x}^{2020}}}}dx}=\int_{0}^{\infty }{{{x}^{k+n}}{{e}^{-{{x}^{2020}}\ln 2020}}dx}$$
use $$u=\ln 2020\ {{x}^{2020}}$$ or $$x={{\left( \frac{u}{\ln 2020} \right)}^{\frac{1}{2020}}}$$


问:
Does the function $h:\mathbb{Z}→\mathbb{Z}$ $h(x) = (x - 2)^3 - 3$ have an inverse, if so, how to calculate it?
There´s a theorem which says that $h(x)$ is bijective $\iff$ it´s invertible so that $h^{-1}(y)=x \iff y=h(x)$

So should I prove that it exists?

In any case h(x) is injective since $x_1,x_2 \in \mathbb{Z}$ so that $h(x_1)=h(x_2)\Rightarrow (x_1-2)^3-3=(x_2-2)^3-3\Rightarrow x_1=x_2$

But I´m thinking it´s not onto, the image of h should be $\mathbb{Z}$

Let $x=\sqrt[3]{y+3}+2$, so $(\sqrt[3]{y+3}+2-2)^3-3= y$, what about $0 \in \mathbb{Z}$?

$x=\sqrt[3]{0+3}+2\Rightarrow x=\sqrt[3]{3}+2$ which is not in $\mathbb{Z}$

So it shouldn´t have an inverse function, shoud it?

$h \circ h^{-1}= h(\sqrt[3]{x+3}+2)=(\sqrt[3]{x+3}+2)^3-3=x$

$h^-1 \circ h=h^{-1}((x - 2)^3 - 3)=\sqrt[3]{(x - 2)^3 - 3+3}+2=x$

So apparently it should be that one, Does it have or not?

答:
To show that the function is not surjective, it suffices to show that there is some integer $y$ such that
$$(x-2)^3-3=y,$$
has no integer solution, or equivalently, such that $y+3$ is not a perfect cube.


答:
Alternative approach:

$x = \frac{a}{1 + a} = \frac{1}{\frac{1}{a} + 1}.$

As $a$ increases, the first term in the denominator decreases.

Therefore, as $a$ increases, the denominator decreases.

Therefore, as $a$ increases, $x$ increases.

Title: X = a/ (1+a) . Would the value of x increase or decrease if “a” increases?

Body:
It’s pretty easy to find the answer after putting in random values but I can’t wrap my head around with my below explanation which is in direct contradiction to the result.

If the numerator increases the x value goes up and the value of x goes down if the denominator increases. Now if “a” increases the net  increase of the denominator would be more than the net increase of numerator as the denominator has “+1” in it. Therefore, since the denominator is more than that of numerator shouldn’t the value of x decrease with increase of “a”?


问:
A nonempty set is countable iff it is the image of a function whose domain is a nonempty countable set
A nonempty set is countable iff it is the image of a function whose domain is a nonempty countable set

Attempt:

Let $A$ be non empty and countable. If $A$ is countably infinite, we can find a bijection $f: \mathbb{N} \to A$. If I understand the problem correctly, I ${\bf need}$ to build a function $g: B \to A $ where $B$ is countable and $g(B) = A$. But, I have no idea how to proceed ? Can someone lead into the right direction?

The converse I believe is more manageable. Suppose there is some map whose domain is countable set then the image is also countable either finite or infinite... qed

I am still stuck on $\implies$ direction.

答:
I might be doing something circular, but:

I think you can just pick $B = \mathbb N$ and $g=f$ for $A$ countably infinite.

For $A$ finite of cardinality $n \ge 1$, choose $B=\{1, 2, ..., n\}$ and $g$ is any bijection or inj/surj between $A$ and $B$.


问:
Second degree linear derivate with exponential function, which general solution?
Suppose $y''-4y=xe^{2x}$.

Solution of the homogenous equation is $y_H(x) = C_{1} e^{2x} + C_{2} e^{-2}$, after solving the characteristic equation with the guess $y=e^{rx}$.

Now my instructor insists that the one solution is either

$y(x) = Ax^{2} e^{2x} + Bxe^{2x}$

or

$y(x) = Ax^{2}e^{2x} + Bxe^{2x}+Ce^{2x}$

...and then I get the solution $y(x) = y_{H}(x) +\frac{1}{8}x^{2}e^{2x}-\frac{1}{16}xe^{2x}$

but I cannot understand the choices. Why do just the forms work? I have tried many different forms there and get lost/wasted a lot time -- and my instructor just says that it is a "guess" and smiles. How can I find the one solution without guessing? For one solution, things such as $y(x)=e^{2x}$ or $y(x)=xe^{2x}$ work so I am a bit lost why the more complex form is "the solution" and why it provides all solutions.

For example, the case $y=e^{2x}$. We get $y'=2e^{2x}$ and $y''=4e^{2x}$ so

$y''-4y = 4e^{2x} - 4 e^{2x} = 0$

so $RHS = xe^{2x} = 0$ so $x=0$, a solution?! If not why? Ok it is naive, it does not contain all cases but how can I know that some form contains all?

答:
I believe this can be done without guesswork.

One way to get this mechanically is to use the method of Laplace Transforms.

Of course, I haven't tried out it out myself, but given that

$$\mathcal{L}[t^n e^{at}] = \frac{n!}{(s-a)^{n+1}}$$

I believe the approach will work for your problem.


问:
Show that ratio of positive functions is decreasing/increasing
Good evening, I wonder how to show formally that this function is increasing or decreasing. I have tried anything I know of, but I was not able to procede.

Does anyone has an idea ?

$$
\frac{x^2a(a-1)}{(b + ax)^2}
$$
where $a \geq 2$,  $b \geq 1$ and $x > 0$

答:
You can show this function is increasing, without calculus:

A little algebraic manipulation shows
$$\frac x{b+ax}=\frac1a - \frac ba\left(\frac1 {b+ax}\right).$$
Now reason as follows, keeping in mind that $a\ge 2$ and $b\ge 1$: For $x>0$, the function $x\mapsto b+ax$ is increasing and positive. So its reciprocal is decreasing and positive. Multiplying by $-\frac ba$ makes it increasing, and it remains increasing after adding $\frac1a$ . Conclude that the function
$$
x\mapsto\frac x{b+ax}
$$
is increasing. It's clearly positive for $x>0$. So squaring it keeps it positive and increasing, and it remains so after multiplying by $a(a-1)$.


问:
Why is Sorgenfrey Line normal?
I'm currently learning about normal spaces and I read the proof that the Sorgenfrey line is normal. The main idea was that for every two sets we can find an open set for each point such that the intersection of all the sets is empty, here is a more detailed proof. My question is - how can we separate the set Q of all rational numbers in R from the set P of all irrational numbers in R? Both Q and P are dense in R so shouldn't it be impossible to separate them from each other?</p>

答:
Normality only demands that we be able to separate arbitrary disjoint closed sets - since $\mathbb{Q}$ is neither closed nor open in the Sorgenfrey line (or in the usual topology on $\mathbb{R}$ for that matter), our inability to separate the rationals and the irrationals by closed neighborhoods is irrelevant.

It may not be obvious at first that $\mathbb{Q}$ is neither closed nor open in the Sorgenfrey topology. If that's the case, this is a good exercise. HINT: can you show that neither the rationals nor the irrationals contain an interval of the form $[a,b)$ with $a<b$? If so, do you see why this means that neither the rationals nor the irrationals are open in the Sorgenfrey sense?

Title: Why is Sorgenfrey Line normal?


问:
Numeric calculation of infinite Fourier integral in 2D
Consider a 2D function $f(x,y)$ on $\mathbb{R}^2$, which is finite and decays on some finite interval. I don't have a nice analytical/closed-form expression for $f(x,y)$, but can evaluate it at any $(x,y)$, and have thus computed it numerically in a discrete lattice of size $N_x$ and $N_y$, with finite limits $x \in [-a, a]$ and $y \in [-b, b]$. Here, the limits are chosen large enough such that $f(x,y) \to 0$ at the borders.

Question: How can I numerically evaluate the Fourier integral $$F(k_x,k_y) = \int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy e^{-i\mathbf{k}\cdot\mathbf{R}}f(x,y)$$
on a discrete lattice for $k_x$ and $k_y$?

Although these are continuous functions, I guess I could potentially express the answer in terms of discrete FT or FFT, which have many convenient implementations in most computer languages. I think I have to include an additional phase factor, but I don't know how to do this (at least not in 2D). Whatever I try to search for online is hidden by tons and tons of search results which seem irrelevant to me.

My approach so far has been to simply just discretize the integrals (using e.g. trapezoidal method or Simpson method). This is on the right level of complexity that I'm after, but feels crude and potentially error-prone.

答:
The given two-dimensional integration problem can be solved by multiple using of one-dimensional integral Fourier transforms (FT).

To calculate the FT, Filon's integration formula (M.Abramowitz - I.Stegun, p.890) is intended. Taking in account zero boundary conditions and the complexity of the third derivatives estimation, looks convenient using of detalized lattice and the simplified integration expression in the form of

$$\int\limits_{x_0}^{x_{2n}} f(x)\, e^{-itx}\,\text dx \approx h\,\big(\beta(t\,h)F_{2n}(t)+\gamma(t\,h)F_{2n-1}(t)\big),\tag1$$
where
$$\beta(\theta) = 2\left(\dfrac{1+\cos^2\theta}{\theta^2} - \dfrac{\sin 2\theta}{\theta^3}\right),\quad
\gamma(\theta) = 4\left(\dfrac{\sin \theta}{\theta^3} -\dfrac{\cos\theta}{\theta^2}\right),\tag2$$
$$F_{2n}(t) = \sum\limits_{k=0}^{n-1} f_{2k}\,e^{-2ikt},\quad
F_{2n-1}(t) = e^{-it}\sum\limits_{k=0}^{n-1} f_{2k+1}\, e^{-2ikt}.\tag3$$
Since formulas $(3)$ can be calculated via the FFT algorithm, then resulting calculation complexity of the proposed algorithm looks not big.


问:
Boolean functions satisfying a condition
What is the total number of Boolean functions $f :\{0,1\}^n \rightarrow \{1,-1\}$ such that

\begin{equation}\tag{1}\label{e}
\sum_xf(x)f(x+y) = 0
\end{equation}

for all $0\neq y$?

Is there some closed form construction to obtain them all? Or even one specific example if it exists at all? I tried to use inverse Fourier transform but no luck.

For $n = 2$ one such example is $f(00) = f(10) = f(01) = 1$ and $f(11) = -1$. But for larger $n$ it seems very difficult.

Note that $f$ cannot be additive.

EDIT: Alex below claims that no such functions exist for $n >2$. However, for $n =4$, I was able to construct (very ad-hoc) the following

Columns 1 through 14

```<code> 0     0     0     0     0     0     0     0     1     1     1     1     1     1
 0     0     0     0     1     1     1     1     0     0     0     0     1     1
 0     0     1     1     0     0     1     1     0     0     1     1     0     0
 0     1     0     1     0     1     0     1     0     1     0     1     0     1
 1    -1     1     1     1     1    -1     1     1     1     1    -1     1    -1
</code>```

Columns 15 through 16

```<code> 1     1
 1     1
 1     1
 0     1
-1    -1
</code>```

One way to to think of this is by considering the function $f_y: x \mapsto f(x)f(x+y)$ and impose that $f_y$ is additive for all $y$. So the question now becomes:

Question: Does there exist $\hat{y}$ such that $f_y = \chi_{\hat{y}}: x\mapsto (-1)^{\hat{y}x^T}$.

Then \eqref{e} becomes a simple consequence of characters.

答:
Following Peter Košinár's comment, I modified my Z3 model to use bit-wise exclusive or for $x+y$ rather than addition modulo $2$:

```<code>from z3 import *

n = 4
noOfMinterms = 2 << (n-1)

print("n = %s" % n)
print("minterm rows = %s" % noOfMinterms)

Minterms = [i for i in range(noOfMinterms)]

#  define truth table with one Bool variable for every row
table = []

for i in Minterms:
    table += [Bool("b" + str(i))]

def f(x):
    return If(table[x], 1, -1)

s = Solver()

for y in range(1, noOfMinterms):
    s.add(Sum([Product(f(x), f(x ^ y)) for x in Minterms]) == 0)

solutions = 0

while s.check() == sat:
    m = s.model()
    solutions += 1
    print(str(solutions) + ": ", [" 1 " if is_true(m[table[i]]) else "-1 " for i in Minterms])
    #  prevent the solution to be found again
    s.add(Or([table[i] != m[table[i]] for i in Minterms]))

if solutions == 0 :
    print("No solutions found. Sorry")
else:
    print("Solutions: %s" % solutions)
</code>```

The numbers of solutions for small $n$:

```<code>n   # Solutions
2      8
3      0
4      896
5      ?
</code>```

The modified MiniZinc model:

```<code>include "globals.mzn";

int: n = 4;
set of int: Bits = 0 .. n;

%  pre-calculate powers of 2: 1, 2, 4, ...
array[Bits] of int: twopow = array1d(Bits, [pow(2, i) | i in Bits]);

int: noOfMinterms = twopow[n];
set of int: Minterms = 0 .. noOfMinterms - 1;

array[Minterms] of var bool: X; %  true = 1; false = -1

constraint
forall(y in Minterms where y != 0) (
   sum([f(x) * f(bitxor(x, y)) | x in Minterms]) == 0
);

%  function to calculate the biwise xor of two ints
%  cf. https://en.wikipedia.org/wiki/Bitwise_operation#XOR
function int: bitxor(Minterms: x, Minterms: y) =
  sum([twopow[i] * (((x div twopow[i]) + (y div twopow[i])) mod 2) | i in Bits]);

function var int: f(int: x) = if X[x] then 1 else -1 endif;

function int: f2(Minterms: x) = if fix(X[x]) then 1 else -1 endif;

solve satisfy;

output ["\(n): "] ++
       [show_int(3, f2(i)) | i in Minterms];

</code>```

The modified check code in C# which confirms the example solution from the question:

```<code>int n = 4;
int noOfMinterms = (2 << (n - 1));
int[] f = { 1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, -1, 1, -1, -1, -1 };

for (int y = 1; y < noOfMinterms; y++)
{
    int sum = 0;

    for (int x = 0; x < noOfMinterms; x++)
    {
        sum += f[x] * f[(x ^ y) % noOfMinterms];
    }

    Console.WriteLine($"{y} {sum}");
}
</code>```

Body:
Following Peter Košinár's comment, I modified my Z3 model to use bit-wise exclusive or for $x+y$ rather than addition modulo $2$:

from z3 import *

n = 4
noOfMinterms = 2 << (n-1)

print("n = %s" % n)
print("minterm rows = %s" % noOfMinterms)

Minterms = [i for i in range(noOfMinterms)]

#  define truth table with one Bool variable for every row
table = []

for i in Minterms:
    table += [Bool("b" + str(i))]

def f(x):
    return If(table[x], 1, -1)

s = Solver()

for y in range(1, noOfMinterms):
    s.add(Sum([Product(f(x), f(x ^ y)) for x in Minterms]) == 0)

solutions = 0

while s.check() == sat:
    m = s.model()
    solutions += 1
    print(str(solutions) + ": ", [" 1 " if is_true(m[table[i]]) else "-1 " for i in Minterms])
    #  prevent the solution to be found again
    s.add(Or([table[i] != m[table[i]] for i in Minterms]))

if solutions == 0 :
    print("No solutions found. Sorry")
else:
    print("Solutions: %s" % solutions)

The numbers of solutions for small $n$:

n   # Solutions
2      8
3      0
4      896
5      ?

The modified MiniZinc model:

include "globals.mzn";

int: n = 4;
set of int: Bits = 0 .. n;

%  pre-calculate powers of 2: 1, 2, 4, ...
array[Bits] of int: twopow = array1d(Bits, [pow(2, i) | i in Bits]);

int: noOfMinterms = twopow[n];
set of int: Minterms = 0 .. noOfMinterms - 1;

array[Minterms] of var bool: X; %  true = 1; false = -1

constraint
forall(y in Minterms where y != 0) (
   sum([f(x) * f(bitxor(x, y)) | x in Minterms]) == 0
);

%  function to calculate the biwise xor of two ints
%  cf. https://en.wikipedia.org/wiki/Bitwise_operation#XOR
function int: bitxor(Minterms: x, Minterms: y) =
  sum([twopow[i] * (((x div twopow[i]) + (y div twopow[i])) mod 2) | i in Bits]);

function var int: f(int: x) = if X[x] then 1 else -1 endif;

function int: f2(Minterms: x) = if fix(X[x]) then 1 else -1 endif;

solve satisfy;

output ["\(n): "] ++
       [show_int(3, f2(i)) | i in Minterms];

The modified check code in C# which confirms the example solution from the question:

int n = 4;
int noOfMinterms = (2 << (n - 1));
int[] f = { 1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, -1, 1, -1, -1, -1 };

for (int y = 1; y < noOfMinterms; y++)
{
    int sum = 0;

    for (int x = 0; x < noOfMinterms; x++)
    {
        sum += f[x] * f[(x ^ y) % noOfMinterms];
    }

    Console.WriteLine($"{y} {sum}");
}


问:
Problem on entire function of finite order [Stein, Chapter 5, Problem 4]
I am working on the part (a) of Stein Complex Analysis, Chapter 5 Problem 4, which states as follows:

Let $F(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be an entire function of finite order.  Suppose $|F(z)|\leq Ae^{a|z|^{\rho}}$. Then show that $$\limsup_{n\rightarrow\infty}|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}<\infty.$$

Stein also gave hints, so I followed the hint.

First Step: Use Cauchy inequality to bound the coefficient.

Okay this one is easy: Let $R>0$, as $F(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ is entire, it must be holomorphic in $|z|<R$, so for any $0<r<R$ and $n\in\mathbb{N}$, we have $$|a_{n}|\leq \dfrac{\sup_{|z|=r}|F(z)|}{r^{n}}\leq \dfrac{Ae^{ar^{\rho}}}{r^{n}},$$ where the second inequality was obtained because the hypothesis that $F|(z)|\leq Ae^{a|z|^{\rho}}$ for any $z$ implies that the $\sup$ over all $|z|=r$ is the numerator of the quotient on the RHS.

Second Step: Prove the fact that the function $u^{-n}e^{u^{\rho}}$, $0<u<\rho$ attains its minimum value $e^{n/\rho}(\rho/n)^{n/\rho}$ at $u=n^{1/\rho}/\rho^{1/\rho}$. Then  choose $r$ in terms of $n$ to achieves this minimum.

I got stuck in this step. Firstly, Consider the function $f(x)$ defined by $f(x):=x^{-n}e^{x^{\rho}}$ for $0<x<\rho$. Its derivative is $$f'(x)=e^{x^{\rho}}x^{-n-1}(\rho x^{\rho}-n),$$ which means that the root for $f'(x)=0$ is $x_{0}=\rho^{-\frac{1}{\rho}}n^{\frac{1}{\rho}}.$ But how do you make sure $x_{0}\in (0,\rho)$? It is varying with $n$ right? Think about if we compare $x_{0}^{\rho}$ with $\rho^{\rho}$, we then compare $\rho^{-1}n$ with $\rho^{\rho}$, so we actually compare $n$ with $\rho^{\rho+1}$. Yes, at first $n$ may be small, but when $n$ goes higher it will eventually bigger than $\rho^{\rho+1}$. How can I conclude that this one is even in the interval $0<x_{0}<\rho$?

Also, for similar reason I do not know how to argue $f(x_{0})$ is the minimum in this interval since I don't know how to compare $f(x_{0})$ to $f(\rho)$. $f(x_{0})$ is varying with $n$.

Then I want to just skip this step for now and to see what's gonna happen. But Stein said it attains minimum right? I think we need a maximum to further the bound right?

I am really confused about this hint.. Any idea?

答:
While the question is about the hint and is addressed in the comments, I would suggest doing this in a different way by taking logarithms.

The problem reduces to showing $\frac{\log |a_n|}{n}+\frac{\log n}{\rho} <M$ for some positive $M$ uniformly in $n$ where the inequality is assumed satisfied automatically for $a_n=0$ as then $\log |a_n| =-\infty$

By Cauchy one knows that $\frac{\log |a_n|}{n} \leq C/n+\frac{ar^{\rho}}{n}-\log r$ for any positive $r$ so one needs to find some $r$ (depending on $n$) for which $\frac{ar^{\rho}}{n}-\log r+\frac{\log n}{\rho} <M$ uniformly in $n$

But now if we take $\log r =\frac{\log n}{\rho}$ (as that term is clearly going to infinity with $n$ so needs getting rid of) we notice that $r^{\rho}=e^{\rho \log r}=n$ so $\frac{ar^{\rho}}{n}=a$ constant and we are done!


问:
Why do fixed point theorems appear all over mathematics?
For example, the Banach fixed-point theorem is applied in the proof of the Picard–Lindelöf theorem about the uniqueness of solutions of ordinary differential equations and the Lefschetz fixed-point theorem (or a modification of it) is used in the proof or in the context of the Weil conjectures. There are so many more examples.

What is so special about the equation $f(x)=x$?

答: One thing that immediately comes to mind: it is by far the simplest equation one can write that's 'universal' (and not tautological). All one needs is some domain $D$ with an equality relation on it and a function $D\mapsto D$. $D$ can be discrete; it can be $\mathbb{R}$; it can be infinite-dimensional; it can have extra structure or no structure at all. We don't need to be able to combine elements of $D$ in any meaningful ways, we don't need to compare in any way other than equality.


问:
How were connections between different sorts of mathematical structures discovered?
The fundamental theorem of Galois theorem states that for a finite Galois field extension $E/F$, there is a one-to-one correspondence between the fields $K$ with $F\subset K\subset E$ and the subgroups of the Galois group $Gal(E/F)$ of the field extension $E/F$. I think this correspondence can be considered as an equivalence of categories if we consider the category of intermediate fields together with field homomorphisms fixing $E$ and the category of subgroups together with morphisms of corresponding $Gal(E/F)$-sets. (The details won't matter for my question.)

The Stone duality states that the category of Boolean rings is equivalent to the dual category of the category of zero-dimensional compact Hausdorff spaces.

There are many more one-to-one correspondences between different sorts of mathematical objects (Pontrjagin duality, Gelfand duality, ...).

My question isn't a precise mathematical one, but rather a soft-question: How on earth were such crazy correspondences discovered? Concerning the first example, I agree that both groups and fields are natural concepts to study, but I would never ever get the idea to conjecture that there is a correspondence between such-and-such groups and such-and-such fields, because fields and groups are rather different animals. Same with the second example. Rings are one thing, coming from algebra. Hausdorff spaces are quite another thing, coming from topology. How to even conjecture that there is a connection between the two?

I would be interested in rudimentary historic answers, but to me getting any clue on how one could have stumbled across these things matters more than historic correctness.

答:
A) The birth of the concept of group characters was somewhat "accidental".
They were discovered through the factorization of ... determinants of (Cayley) group tables! Dedekind has played there a fundamental role. I wrote some time ago an answer about this issue here.
See as well this very good reference.

$B_1$) The birth of Fourier Transform (and more generally other functional transforms) in the 1930s.... It wasn't evident to discover this kind of correspondence between multiplication and convolution just because the convolution operation hadn't received a full status ; one could even say that convolution has been co-discovered with Fourier Transform.

$B_2$) The birth of Discrete Fourier Transform (DFT), "co-discovered" by Numerical Analysts Cooley and Tuckey in the 1960s with an efficient algorithm for its computation, the Fast Fourier Transform (FFT), whose presence afterwards was found in Gauss works... It is a curious fact that many people, still now, are using FFT transform instead of DFT.


问:
Euclidean Norm Calculation
If ||x||$_2$ = 3, ||4x-5y||$_2$ = 14 and ||5x+2y||$_2$ = 8 find ||y||$_2$

I had attempted to write at the vectors in explicit form and in doing so arrived at a point where I would add (||4x-5y||$_2$)$^2$ = (14)$^2$ and (2)(||5x+2y||$_2$)$^2$= 8$^2$(2) to remove <x,y>. From there I would subtract out 66||x||$_2$ by adding on (-66)(||x||$_2$)$^2$ = (-66)(9). Now I get a negative answer of -270/33. Is it possible to get a negative answer? I was under the impression that Euclidean norms acted like absolute value signs. Any advice and explanation would be appreciated.

答: So, the answer is that there is no solution as seen by the fact that we get a negative solution which is not possible</p>

Title: Euclidean Norm Calculation


问:
Is the given complex series convergent or divergent? Give a reason. Show work.
$$\sum_{n=0}^{\infty} \frac{i^n}{n^2 - i}$$

I'm told to compare this to $\sum \frac{1}{n^2}$
but not sure what this means

I know the magnitude of the denominator is $\left | n^4 + 1\right |$ but not sure how this helps me. If I take the magnitude and compare it to $\frac{1}{n^2}$, how does this help me? Since $\frac{i^n}{n^4 + 1}$ is smaller than $\frac{1}{n^2}$ I guess our series converges too?

When we take the magnitude like this... why does this work? What are we doing when we take the magnitude of the denominator?

答:
We have

$$ \sum_{n=0}^\infty\left| \frac{i^n}{n^2-i}\right|= \sum_{n=0}^\infty \frac{|i^n|}{|n^2-i|}= \sum_{n=0}^\infty \frac{1}{|n^2-i|}$$

But we know that

$$|n^2-i|\geq ||n^2|-|i||=n^2-1$$

is true for $n\geq 2$ (this is the reverse triangle inequality). Then the sum becomes

$$\leq \frac{1}{|0-i|}+\frac{1}{|1-i|}+\sum_{n=2}^\infty \frac{1}{n^2-1}<1+\frac{1}{\sqrt{2}}+\sum_{n=2}^\infty \frac{1}{n^2-n}=2+\frac{1}{\sqrt{2}}$$

and therefore the original sum is absolutely convergent.


问:
What is the geometric interpretation of a unit circle having a line integral of zero along a curve?
Let C be an arbitrary arc of the unit circle. Give a geometrical interpretation for ∫C∇f·dr= 0 where f(x,y)=x^2+y^2.</p>

答:
The integral of the scalar field $f$ with respect to arc length on the curve is

$$ \int_C f(x,y)\; ds $$

This is never $0$ unless the curve reduces to a single point, because $x^2 + y^2 > 0$ except at the origin.

But I suspect that's not the integral you're thinking of.

Please clarify.

EDIT: OK, so now you want
$$ \int_C (\nabla f) \cdot dr$$
For any continuously differentiable function $f$, and any curve that starts at a point $a$ and ends at a point $b$, that is $f(b) - f(a)$.  So in this case all you need for that to be $0$ is that $f$ is the same at the starting and ending points of the curve, i.e. both are on the same circle centred at the origin.


问:
basis for null space of matrix in a certain field

Let $G=(V,E)$ be a connected graph. Let $T$ be a spanning tree of $G$. Fix a $2$-element field $\mathbb{F}$ (the elements are $\{0,1\}$). Let $S$ be an incidence matrix of $G$ in $\mathbb{F}$. Find a basis $A$ for the null space of $S$ so that for each edge $e\in E\backslash E(T),$ there is exactly one vector $w \in A$ for which $\mathbb{1}_w(e) = 1$ (here $\mathbb{1}_w(e)$ is the function that is $1$ if the entry corresponding to edge $e$ of vector $w$ is $1$ and $0$ otherwise). What are the sets of edges of $G$ corresponding to elements of $A$?

I know that the null space of $S$ has dimension $|E | - |E(T)|.$ However, I'm not sure how to find a basis for the null space. It might be useful to figure out what the left null space and row space of $S$ are, but again I'm not sure how to do that. I also think that a submatrix $F$ of $B$'s columns has linearly dependent columns iff $F$ contains a cycle. I'm not sure if an inductive argument might be useful, seeing as induction seems to be used quite often for proofs in combinatorics.

答:
Note of topological bias: My viewpoint is entirely motivated by cellular homology.

You are right that the null space has dimension $|E|-|E(T)|$ (this should give a hint that your basis has something to do with the edges $E\setminus E(T)$). You are also right that this has to do with cycles. First, let me redefine things a tad to work in a setting free of unnecessary coordinates.

(Note that my graphs can have loops and multiple edges and cycles are allowed to repeat edges. Also, my graphs can be infinite and the same sort of proof will go through).

Take vector spaces $\mathbb{F} E$ and $\mathbb{F}V$ corresponding to finite "sums" of edges and vertices. The incidence "matrix" is then a linear map $S:\mathbb{F}E\rightarrow \mathbb{F}V$ defined on generators by: if $e\in E$ has endpoints $u,v\in V$, then $S(e)=u+v.$ The goal is to understand the kernel of this map.

Definition. Call $y\in \mathbb{F}E$ a cycle if $y=e_1+\dots+e_j$, where $(e_1,\dots,e_j)$ are the edges of a cycle.

Lemma. An element $x\in \mathbb{F}E$ is in $\ker S$ if and only if $x$ is a sum of cycles.

Proof. You can check that cycles are in $\ker S$, using the fact that $2=0$ in $\mathbb{F}$. From this, we get that sums of cycles are also in $\ker S$. Conversely, suppose $x=e_1+\dots+e_j\in\ker S$, for some $e_1,\dots,e_j\in E$. We will show that some of these edges form a cycle. Then we can subtract off that cycle to get a sum of fewer edges, still in $\ker S$. Continuing this process inductively, we can decompose $x$ into a sum of cycles (with no edges occuring more than once). Let $S(e_1)=u+v$. If $u=v$, then $e_1$ is a loop and we're done. Otherwise, there must be some $e_i$ (where $i\neq 1$) with $S(e_i)=v+w$ to cancel out the $v$ coming from $e_1$. If $u=w$, then $(e_1,e_i)$ is a cycle and we're done. Otherwise, we can cancel $w$, etc. Eventually, we must indeed get back to $u$ and thus form a cycle. For otherwise, we will run out of edges without cancelling the $u$ coming from $e_1$, contradicting the fact that $S(x)=0. \ \square$

So now that we've characterized $\ker S$, how can we find an appropriate basis? This requires some facts that I will state without proof. I am going more off of algebraic topology intuition, but I will provide hints that can hopefully help you find inductive proofs form them.

If $(e_1,\dots,e_j)$ form a cycle in $T$, then $e_1+\dots+e_j=0\in \mathbb{F}E$. (Hint: The graph spanned by the edges $\{e_i:i=1,\dots,j\}$ is a tree. Why? At a leaf of this tree, the cycle "turns around," so the same edge is repeated twice in a row. Thus, this edge sums with itself to give $0$ in $\mathbb{F}E$ and the remaining edges form a smaller cycle. How does this facilitate a proof by induction?)

If $e\in E\setminus E(T)$, then there is a cycle $(e,e_1,\dots,e_j)$, where $e_1,\dots,e_j\in E(T)$. (Hint: Use the fact that $T$ is a spanning tree.)

If $x\in \ker S$, then we can write $x=y_0+y_1+\dots+y_k$, where each $y_i$ is a cycle with exactly one edge in $E\setminus E(T)$ and these edges outside of $T$ do not repeat. (Hint: Define a map $\ell:\mathbb{F}E\rightarrow \mathbb N$ as follows: if $x=f_1+\dots+f_m$, where $f_1,\dots,f_m\in E$ are distinct edges, then $\ell(x)$ is the number of $f_i$'s in $E\setminus E(T)$. You can prove (3) by induction on $\ell(x)$. The base case uses (1) and the lemma, where we need to show that $x=0$ and take an empty sum of $y_i$'s. For the inductive step, pick an $e_i\in E\setminus E(T)$ and find a cycle $y$ as in (2). Then we have $\ell(x+y)=\ell(x)-1$ and $x+y\in \ker S$ because $y$ is a cycle.)

For your basis, now take $\{e+e_1+\dots+e_j:e\in E\setminus E(T)\}$, where the cycles $(e,e_1,\dots,e_j)$ come from (2). Use (1) and (3), along with the lemma, to prove that this is indeed basis. By construction, each edge in $E\setminus E(T)$ occurs once in the basis, and each basis element has exactly one edge from $E\setminus E(T)$. This is your desired condition.

Final note: My solution is a little sloppy, in that it doesn't necessarily produce the canonical result: the fundamental cycles that Matt mentioned. That is because I am allowing cycles to have repeating edges. To fix this, you can refine (2) to produce cycles without repeating edges, simply by noting that any two vertices in $G$ are connected by a unique path in $T$ (with no repeating edges). This fact also allows for different proof of (1), which I think is quite nice. See if you can spot it.


问:
The expected number of heads
Given $10$ fair coins:

In the first round, we toss each coin once which gives us a combination of heads and tails.

In the second round, we only toss those coins that landed on the tail in the first round.

What is the expected number of heads after this experiment $?$

Intuition tells me is $5 + 2.5 = 7.5$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\overbrace{\color{#f44}{\sum_{k = 0}^{n}{n \choose k}\pars{1 \over 2}^{k}
\pars{1 \over 2}^{n - k}}}
^{\substack{\ds{First\ Round:}\\[1mm] \ds{\color{#f44}{k}\ \mbox{heads}}}}\,\,\,
\overbrace{\color{#44f}{\sum_{j = 0}^{n - k}{n - k \choose j}\pars{1 \over 2}^{j}
\pars{1 \over 2}^{n - k - j}}}
^{\substack{\ds{Second\ Round:} \\[1mm] \ds{\color{#44f}{j}\ \mbox{heads}}}}
\\[2mm] &\ \times\pars{k + j}
\\[5mm] = &\
{1 \over 2^{2n}}\sum_{k = 0}^{n}\sum_{j = 0}^{n - k}{n \choose k}
{n - k \choose j}2^{k}\pars{k + j} = \bbx{{3 \over 4}\,n}
\\[5mm] &\ \stackrel{\ds{n\ =\ 10}}{\ds{\implies}}\quad\bbx{7.5} \\ &
\end{align}

Title: The expected number of heads


答: Hint: By the limit comparison test, since $\lim_{n\to\infty}1/n^2/(1/(n^2-i))=1$, the series converges absolutely.

Title: Is the given complex series convergent or divergent? Give a reason. Show work.

Body:
$$\sum_{n=0}^{\infty} \frac{i^n}{n^2 - i}$$

I'm told to compare this to $\sum \frac{1}{n^2}$
but not sure what this means

I know the magnitude of the denominator is $\left | n^4 + 1\right |$ but not sure how this helps me. If I take the magnitude and compare it to $\frac{1}{n^2}$, how does this help me? Since $\frac{i^n}{n^4 + 1}$ is smaller than $\frac{1}{n^2}$ I guess our series converges too?

When we take the magnitude like this... why does this work? What are we doing when we take the magnitude of the denominator?


答:
Using asymptotic analysis, this series is absolutely convergent since
$$\biggl|\frac{i^n}{n^2-i}\biggl|=\frac 1{|n^2-i|}=\frac1{\sqrt{n^4+1}}\sim_\infty\frac1{\sqrt{n^4}}=\frac1{n^2}.$$


问:
How many patients will this doctor typically see in a $6$-hour surgery?
The amount of time patients spend in a doctor's office, $X$, is an exponential random variable with a mean of $10$ minutes.

$(a)$ Find the probability a patient spends less than $5$ minutes with the doctor.

$(b)$ Find the probability a patient spends between $7$ and $10$ minutes with the doctor.

$(c)$ How many patients will this doctor typically see in a $6$-hour surgery?

$(d)$ Another doctor spends $10$ minutes or more with half of his patients. How many fewer patients does he see in his $6$-hour surgery?

Ans: $(a)$ $X\sim \exp(\lambda)$ with $\lambda=\frac{1}{10}$, So, to find the probability that someone spends less than 5 minutes with a doctor,

$P(X< 5)= \int_{0}^{5}\lambda \exp(-\lambda x) dx$.

$(b)$ $P(7<X< 10)= \int_{7}^{10}\lambda \exp(-\lambda x) dx$.

$(c)$ No idea, could anyone please help.

$(d)$ No idea, could anyone please help.

Thanks!

答:
$(c)$ If the time spent per patient is an exponential random variable, then the number of patients seen in a fixed time interval is a Poisson random variable with rate proportional to the length of that time interval.

Specifically, since $6$ hours is equal to $360$ minutes, the random number of patients seen in this time interval is Poisson distributed with rate $36$ patients per $6$ hours, and the expected number of such patients seen is also $36$ patients.

$(d)$ let $Y$ be the random time spent per patient for the second doctor.  You are given $$\Pr[Y \ge 10] = 1/2.$$  If $Y$ is also exponentially distributed, this means $$1/2 = \Pr[Y \ge 10] = e^{-10 \lambda},$$ and solving for $\lambda$ in this equation gives you the corresponding rate parameter for this doctor's time spent.  The average number of patients seen for this doctor is $1/\lambda$.


问:
Question about special orthogonal Lie group construction
Working through homework and I run into this problem:

Suppose the Lie group $SO^{+}(2,2)$ is presented as the group of all transformations in its associated space. How do you determine whether a given 4x4 transformation matrix is an element of the group?

I'm slightly familiar with the $SO^{+}(1,3)$ Lie group, but have almost no experience with group theory as a whole. So far, I seem to remember the conditions require orthogonality between coordinates, preservation of the orientation (of subgroups?), and elements with a determinant of 1, but still feel like I'm missing something. I expect given a matrix representation of the form:

$\left( \begin{array}{ccc}
a & b & 0 & 0\\
c & d & 0 & 0\\
0 & 0 & e & f\\
0 & 0 & g & h\end{array} \right) $

Do I need to ensure the determinants of the components

$\left( \begin{array}{ccc}
a & b \\
c & d \\
\end{array} \right) $
and $\left( \begin{array}{ccc}
e & f \\
g & h \\
\end{array} \right) $

are related  in a certain way to prevent orientation shifts (such as reflections), or do I only need to pay attention to the total determinant of a transformation?

答: For $SO(2,2)$, You should check that it preserves the quadratic form $-x^2-y^2+z^2+w^2$, and that the matrix has determinant one. I guess the $+$ means that the transformation preserves both orientations, meaning that in you example both smaller determinants have to be equal to one.</p>


问:
limit superior of the sequence $|\frac{p_{n+1}}{n+1}-\frac{p_n}{n}|$ where $p_k$ is the $k$-th prime.
What is the best known estimatio for $\limsup |\frac{p_{n+1}}{n+1}-\frac{p_n}{n}|$ ?

I know about prime gaps, Prime Number Theorem etc. famous results, but this is something I don't know how to search info about. Does this problem have any name?

答:
A result from PNT is that there is prime number in the interval $ [x , x+ \frac{x}{25\ln^2 x}]$ for all $x \geq 396738$.

So we have that $ p_n < p_{n+1} < p_n + \frac{p_n}{25\ln^2 p_n} \leq p_n + \frac{n}{\ln n}$

Since $ n (\ln n+ \ln \ln n-1) < p_n < n(\ln n+\ln \ln n)$ for all $n \geq 6$

So $|\frac{p_{n+1}}{n+1}-\frac{p_n}{n}| \leq \frac{p_{n+1}-p_n}{n+1} +p_n |\frac{1}{n+1}-\frac{1}{n}| \leq \frac{\frac{n}{\ln n}}{n}+ \frac{p_n}{n^2} \leq \frac{1}{\ln n}+ \frac{2 n \ln n}{n^2}  = \frac{1}{\ln n}+\frac{2\ln n}{n} \to 0$ as $n \to \infty$

Note : all the inequalities and lemmas derived from PNT are found in Dusart's paper


问:
Show that AX is the median of a triangle
Let $\Delta ABC$ be a triangle. Choose $P\in BC$ such that $BC = 3 \cdot P C$ and let $X$ be the intersection point of the lines $AP$ and $CM_c$. Prove that the line $AX$ is the median of $\Delta AM_cC$. Here $M_c$ denotes the base point of the median starting from C. Picture:

Can someone tell me what connections or theorems I am supposed to use to prove this? I can't see any similar triangles and I tried to just make some equations from the given condition $BC=3PC$ but I didn't find anything useful.

答:
The best way to cover this and tons of similar problems is to take a look at the theorems of Ceva and Menelaus:

Ceva's Theorem

Menelaus' Theorem

Note that in most sources, proportions of segments built on the same line are considered to be signed.

Here is an application of Menelaus' Theorem for the case of $\Delta CBM_c$ with respect to the "secant line" AXP. We have the relation:
$$
1
=
\underbrace{\frac{PC}{PB}}_{=-1/2} \cdot
\underbrace{\frac{AB}{AM_c}}_{=2} \cdot
\frac{XM_c}{XC} \ .
$$
This gives the signed value $-1$ for the third proportion. This means $X$ is the mid point of $CM_c$.


问:
Parabolic subalgebra conjugate under Weyl group
Let $G$ be a semisimple (or reductive (using the definition of Knapp)) Lie group. Then any two minimal parabolic subalgebras resp. subgroups are conjugated under $\operatorname{Ad}(K)$ resp. $K$ ($K$ being maximally compact in $G$).

My question is: Is it sufficient to take $k\in K$ in the normalizer $N_K(\mathfrak a)$ of $\mathfrak a$ in $K$ (resp. the Weyl group $N_K(\mathfrak a)/Z_K(\mathfrak a)$), so that any two conjugated parabolics differ only be another choice of positive system in $\mathfrak a$? Here, $\mathfrak a$ denotes a maximal abelian subspace of $\mathfrak p$, the $-1$-eigenspace of the Cartan decomposition of $\mathfrak g=\mathfrak k\oplus\mathfrak p$.

答:
It is not sufficient.

Example:

$G= SL_2(\mathbb R)$, $\mathfrak g = \mathfrak{sl}_2(\mathbb R)$,

$\mathfrak k = \mathfrak{so}_2(\mathbb R) = \{ \pmatrix{0&\theta \\ -\theta & 0} \vert \theta \in \mathbb R \}$, $\mathfrak p = \{ \pmatrix{a & c \\ c & -a} \vert \; a,c \in \mathbb R \}$,

$K= SO_2(\mathbb R) = \{ R_\theta := \pmatrix{\cos(\theta) & \sin(\theta) \\ -\sin(\theta) &\cos(\theta) } \vert \;\theta \in \mathbb R \}$.

Now the first problem is that there are various choices for maximal abelian subspaces inside $\mathfrak p$ -- actually, each one-dimensional subspace of $\mathfrak p$ is one. The standard choice is

$$\mathfrak a := \{ \pmatrix{a & 0 \\ 0 & -a} \vert \; a \in \mathbb R \},$$

but

$$\mathfrak c := \{ \pmatrix{0 & c \\ c & 0} \vert \; c \in \mathbb R \}$$

theoretically is not better or worse. Now a well-known minimal parabolic (= Borel) subalgebra of $\mathfrak g$ is

$$\mathfrak b_a := \{ \pmatrix{a & b \\ 0 & -a} \vert \; a, b \in \mathbb R \}$$

which contains $\mathfrak a$; but another Borel subalgebra is

$$\mathfrak b_c := \{ \pmatrix{r & c+r \\ c-r & -r} \vert \; c,r \in \mathbb R \}$$

which contains $\mathfrak c$. And you can check that $\mathfrak b_a$ and $\mathfrak b_c$ are indeed conjugate via $K$ as predicted by the result quoted in your first paragraph, namely, with $R_{\pi/4} = \dfrac12 \sqrt{2} \cdot \pmatrix{1 & 1 \\ -1 & 1} \in K $ we have

$$ R_{\pi/4} \cdot \mathfrak b_a \cdot R_{\pi/4}^{-1} = \mathfrak b_c$$
including
$$ R_{\pi/4} \cdot \mathfrak a \cdot R_{\pi/4}^{-1} = \mathfrak c$$

i.e. $R_{\pi/4} \not \in N_K(\mathfrak a)$, and of course nothing in $N_K(\mathfrak a)$ could possibly conjugate $\mathfrak b_a$ and $\mathfrak b_c$ (as $\mathfrak a \cap \mathfrak b_c = \{0\}$).

Indeed $N_K(\mathfrak a)$ contains just four elements, $ \pm \pmatrix{1 & 0 \\ 0 & 1}$ and $\pm \pmatrix{0 & 1 \\ -1 & 0}$; the first two centralize $\mathfrak a$, so the Weyl group here indeed just has two elements, the non-trivial one being represented by $R_{\pi/2} = \pmatrix{0 & 1 \\ -1 & 0}$, and this conjugates $\mathfrak b_a$ and

$$\mathfrak b_a^- := \{ \pmatrix{a & 0 \\ b & -a} \vert \; a, b \in \mathbb R \}$$

which seem to be the only two minimal parabolics which contain $\mathfrak a$, and these two really correspond just to the other choice of the one positive root in the root system to $\mathfrak a$. But since you asked about general minimal parabolics ...


问:
Numbers $n$ with $n,n+2$ coprime to $p_k\#$ on $[1, p_k\#]$
In the context of sieving for twin primes ($p\#$ is the primorial function) the following seems true.

The number of $n$ such that $n, (n+2)$ are coprime to $p_k\#$ for $n=1,2,...,~p_k\#$ is $\prod_{i=2}^k (p_i-2).$

Calculation of the first 7 products and a bit of luck in OEIS suggest this is true but it does not seem so easy to prove. I notice that $n+2$ is a simple permutation of $GCD(n,p_k\#)$ and so maybe using characters? Partly I am posting this because I haven't seen it before so someone else might find it interesting. It does not lead (directly) to a useful sieve because $p_k\#$ grows too fast but probably it has been considered somewhere.

答:
I'm supposed to answer the question here. So I will mention that the proof is not hard, it can be reached by basic reasoning around the symmetry and divisibility of the numbers, without explicitly using the remainder theorem. Hint: what happens when expanding from $k$ to $k+1$? You're basically creating $p_{k+1}$ duplicates of the previous configuration; some of the duplicates are however divided by $p_{k+1}$. Since another and shorter proof has already been given, I do not go into more detail on this.

What I really want however, is some commentary on this entire concept.

I'm glad I found out, by some searching, that somebody else had found this formula, that I have also been playing with. I will create a linked question.


问:
Number Theory: Fibonacci Number Show that integers $n>1$, $f_{n+1}f_{n-1}={f_n}^2+{-1}^n$.
Help. I need help with a practice problem I am doing for while studying for my exam. I started working on the problem. I used mathematical induction.
We are given: F_(n+1)=F_n +F_(n+1) and F_1=F_2=1
Step 1: I prove the base case: n=1

f_1+1 f_1+1 = (f_1)^2 + (-1)^1
1(0)=1^2 + (-1)^1
0=0 Check, the base case.

Assume it is true for all k=n. That is (F_k+1)(F_k-1)=(F_k)^2 + (-1)^k.

Now, for k=n+1, we have

F_(n+1)+1 F_(n+1)-1 = (f_n+1)^2 + (-1)^n+1 =>

F_n+1F_n = (F_n+1)^2 + (-1)^(n+1)

LHS: F_(n+2)F_n = (F_(n+1) + F_n)Fn

= F_n+1 (F_n) + F_n(F_n)

Idk where to go know or if my work is incorrect.

答:
Using what you did so far -- which seems correct -- you wish to show that

$F_{n+1}^2+(-1)^{n+1}=F_{n+1}F_n+F_nF_n$.

Since $-(-1)^{n+1}=(-1)^n$, that is $F_{n+1}(F_{n+1}-F_n)=F_n^2+(-1)^n,$

which holds because of the inductive hypothesis $F_{n+1}F_{n-1}=F_n^2+(-1)^n$

and $F_{n+1}-F_n=F_{n-1}$.


问:
Evaluation of Gaussian integral $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx$
How to prove
 $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$</p>

答:
The following argument, similar to Bryan Yock's, is a Feynman parameter trick I invented in Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick

Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.

Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note

$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$

Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain

$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$


问:
A letter sequence problem
If I have an $8$ letter string of the letters $C$ and $N$, how many ways can I arrange the letters so that no two N's are adjacent to each other?

答:
Here we derive a recurrence relation. We consider words of length $n$ built from an alphabet $\{C,N\}$ and denote the number of valid words with $a_n$.

We obtain a recurrence relation for $a_{n+1}$ by partitioning valid words of length $n+1$ into two disjoint sets.

Termination with $C$: A word of length $n+1$ may end with the letter $C$. In this case we have $a_n$ possibilities to fill the left-most $n$ positions with valid words.

Termination with $N$: If on the other hand the word terminates with $N$ it terminates in fact with $CN$, since $NN$ is not admissible. We have therefore $a_{n-1}$ possibilities to fill the $n-1$ left-most positions with valid words of length $n-1$.
We obtain the recurrence relation
\begin{align*}
\color{blue}{a_{n+1}=a_n+a_{n-1}\qquad\qquad n\geq 1}
\end{align*}
It is easy to determine the initial conditions, since the valid words of length $n=1$ are $\{C,N\}$ and $n=2$ are $\{CC,CN,NC\}$, so that $\color{blue}{a_1=2}$ and $\color{blue}{a_2=3}$.

We finally get Fibonacci-numbers $a_n$:
\begin{align*}
\begin{array}{r|rrrrrrrr}
n&1&2&3&4&5&6&7&8\\
a_n&2&3&5&8&13&21&34&\color{blue}{55}\\
\end{array}
\end{align*}
and conclude there are $\color{blue}{a_8=55}$ valid words of length $8$.

Title: A letter sequence problem


答:
Suppose we have a dynamical system obtained by iterating a continuous mapping: $x(n+1) = f(x(n))$.  We are typically interested in what happens in the limit as $n \to \infty$.
Perhaps the simplest behaviour we might hope for is that
$x(n)$ goes to some limit $L$ as $n \to \infty$.   Taking $n \to \infty$ in the equation $x(n+1) = f(x(n))$ and using continuity of $f$ gives us $L = f(L)$, so $L$ should be a
fixed point of $f$.

Of course an equation $g(x) = 0$ can be transformed to a fixed-point equation $f(x)=x$, e.g. by taking $f(x) = g(x) + x$.  What can be nontrivial, though, is the use of the dynamical system $x(n+1) = f(x(n))$ to iteratively solve the original equation.

This doesn't explain every appearance of fixed points, but it is an important motivation for some of them.

Title: Why do fixed point theorems appear all over mathematics?

Body:
For example, the Banach fixed-point theorem is applied in the proof of the Picard–Lindelöf theorem about the uniqueness of solutions of ordinary differential equations and the Lefschetz fixed-point theorem (or a modification of it) is used in the proof or in the context of the Weil conjectures. There are so many more examples.

What is so special about the equation $f(x)=x$?


问:
Proving that if $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}\Rightarrow\operatorname{diam}(K_{n})\rightarrow 0$
let $(X,d)$ be a compact metric space if $K_{n}={}$closed and $K_{n+1}\subseteq K_{n} $

and $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$ show that $\operatorname{diam}(K_{n})\rightarrow 0$

I know there exists a different proof,  but my first thought was to prove this with subsequences which is probably wrong, I wonder whether  there is any way we can prove this with subsequences?

my attempt:

I think we can approach $\operatorname{diam}(K_{n})$ with subsequences $d(x_{k_{n}},y_{k_{n}})\rightarrow \operatorname{diam}(K_{n})$

know  $x,y\epsilon K_{n}$ such that $x_{k_{n}} \rightarrow x$ and $y_{k_{n}} \rightarrow y$

$d(x_{k_{n}},y_{k_{n}})\leq d(x_{k_{n}},x)+d(x,y)+d(y,y_{k_{n}})$ and since $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$

and $n \to \infty $  $x=y$ $\Rightarrow $

$\operatorname{diam}(K_{n})\leq 0$

I think I know where the mistake is. I just need to fully understand why this can't work ,

if there is a way to prove this sequences, I would appreciate if someone could write it below.

答:
Let $\epsilon > 0$.  Then $Q = X - B_\epsilon(x)$ is a compact set.  The sets $X - K_n$ cover $Q$ so they ahve a finite subcover; in this case we have for some
$n\in\mathbb{N}$ $X - B_\epsilon(x) \subseteq X - K_n$.  Hence $K_n \subseteq B_\epsilon(x)$.


答:
Here is a proof.

For all $\varepsilon >0$, call
$$H_n= K_n \setminus B(x, \varepsilon )$$
where $B(x, \varepsilon )$ is the open ball of radius $\varepsilon$ centered at $x$.

Clearly $H_n$ is compact and $H_{n+1} \subseteq H_n$.

Note that
$$\bigcap_n H_n = \bigcap_n K_n \setminus B(x, \varepsilon ) = \left\{ x \right\} \setminus B(x, \varepsilon ) = \emptyset $$

This means that one of the $H_n$ is empty (if all were not-empty, then their intersection would be not-empty). Say $H_N$ is the first one to be empty (and so are also $H_{N+1}, H_{N+2}, ...$)

In other words $K_N \subseteq B(x, \varepsilon )$. In particular $$\mathrm{diam} \ K_N \le \mathrm{diam} \ B(x, \varepsilon ) = 2 \varepsilon$$
Hence, for all $\varepsilon >0$ there exists $N >0$ such that for all $n \ge N$
$$\mathrm{diam} \ K_n \le 2 \varepsilon$$
which is by definition
$$\lim_{n \to \infty} \mathrm{diam} \ K_n =0$$

Title: Proving that if $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}\Rightarrow\operatorname{diam}(K_{n})\rightarrow 0$

Body:
let $(X,d)$ be a compact metric space if $K_{n}={}$closed and $K_{n+1}\subseteq K_{n} $

and $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$ show that $\operatorname{diam}(K_{n})\rightarrow 0$

I know there exists a different proof,  but my first thought was to prove this with subsequences which is probably wrong, I wonder whether  there is any way we can prove this with subsequences?

my attempt:

I think we can approach $\operatorname{diam}(K_{n})$ with subsequences $d(x_{k_{n}},y_{k_{n}})\rightarrow \operatorname{diam}(K_{n})$

know  $x,y\epsilon K_{n}$ such that $x_{k_{n}} \rightarrow x$ and $y_{k_{n}} \rightarrow y$

$d(x_{k_{n}},y_{k_{n}})\leq d(x_{k_{n}},x)+d(x,y)+d(y,y_{k_{n}})$ and since $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$

and $n \to \infty $  $x=y$ $\Rightarrow $

$\operatorname{diam}(K_{n})\leq 0$

I think I know where the mistake is. I just need to fully understand why this can't work ,

if there is a way to prove this sequences, I would appreciate if someone could write it below.


问:
Convert the following equation to a first-order vector differential equation.
How can I write the following equation as a first-order vector differential equation?

\begin{align} m \frac{\mathrm{d}^2 x}{\mathrm{d}^2 t}  + 2 \gamma m
 \frac{\mathrm{d} x}{\mathrm{d} t} + kx = 0 . \end{align}

This is a question originally from Kurt Jacobs, question 3 in chapter 2.

答:
Introduce a new variable $y(t) = \dot x(t)$. Then

$$\begin{align*}\dot x(t) &= \hspace{4.8em} y(t) \\ \dot y(t) &= -\frac k m x(t)  -2\gamma y(t) \end{align*}$$

Then if we take $v=\begin{bmatrix}x \\ y\end{bmatrix}$ and $A = \begin{bmatrix} 0 & 1 \\ -\frac k m & -2\gamma\end{bmatrix}$, we have the following vector differential equation:

$$\frac{\mathrm d}{\mathrm d t} v(t) = A v(t).$$


答:
What does it mean to be in the null space?

We are working in the field GF(2), so multiplying $S$ by a vector $w$ means that we can interpret $w$ as being a set of edges, and $Sw$ is the list of vertices that are incident to an odd number of edges in $w$.  So if $Sw=0$, then all vertices are incident to an even number of edges in $w$.  Recalling Eulerian graphs, this means that $w$ can be expressed as a union of cycles.

The basis $A$ for the null space, and edges outside the spanning tree

The problem asks for each edge in $E\,\backslash E(T)$ (each edge outside the spanning tree) to occur in exactly one vector of $A$.
This defines a mapping from edges outside the spanning tree to vectors in $A$.

This mapping must be surjective (onto), because every vector in $A$, being a union of cycles, contains at least one edge outside the spanning tree.
And as you observed, the null space has dimension $|E|-|E(T)|$, so the number of edges outside the spanning tree is the same as the number of basis vectors in $A$.

The only way to have a surjective mapping to a set of the same finite size is if the mapping is a bijection.
So every vector of $A$ contains exactly one edge outside the spanning tree.

The elements of $A$

The only way for a union of cycles to contain exactly one edge outside the spanning tree is for the "union" to be a single cycle, having all edges on the spanning tree except one.
Indeed, given any edge not on the spanning tree, there is a unique path in the spanning tree between the two ends of that edge.
This yields the unique cycle with this property.

So the elements of $A$ are the fundamental cycles,
and $A$ is the fundamental cycle basis.

Title: basis for null space of matrix in a certain field

Body:
Let $G=(V,E)$ be a connected graph. Let $T$ be a spanning tree of $G$. Fix a $2$-element field $\mathbb{F}$ (the elements are $\{0,1\}$). Let $S$ be an incidence matrix of $G$ in $\mathbb{F}$. Find a basis $A$ for the null space of $S$ so that for each edge $e\in E\backslash E(T),$ there is exactly one vector $w \in A$ for which $\mathbb{1}_w(e) = 1$ (here $\mathbb{1}_w(e)$ is the function that is $1$ if the entry corresponding to edge $e$ of vector $w$ is $1$ and $0$ otherwise). What are the sets of edges of $G$ corresponding to elements of $A$?

I know that the null space of $S$ has dimension $|E | - |E(T)|.$ However, I'm not sure how to find a basis for the null space. It might be useful to figure out what the left null space and row space of $S$ are, but again I'm not sure how to do that. I also think that a submatrix $F$ of $B$'s columns has linearly dependent columns iff $F$ contains a cycle. I'm not sure if an inductive argument might be useful, seeing as induction seems to be used quite often for proofs in combinatorics.


答:
Your basic idea works; you just have to be a little more careful.

Since $X$ is a compact metric space, you can show that each $K_n$ has finite diameter. Moreover, since the sets $K_n$ are compact, there are points $x_n,y_n\in K_n$ such that $d(x_n,y_n)=\operatorname{diam}K_n$. The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$, and the sequence $\langle y_{n_k}:k\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle y_{n_{k_i}}:i\in\Bbb Z^+\rangle$. Let $x_0=\lim_ix_{n_{k_i}}$ and $y_0=\lim_iy_{n_{k_i}}$.

Let $i\in\Bbb Z^+$; then $x_{n_{k_j}}\in K_{n_{k_i}}$ for each $j\ge i$, and $K_{n_{k_i}}$ is closed, so $x_0\in K_{n_{k_i}}$. Thus, $x_0\in\bigcap_{i\in\Bbb Z^+}K_{n_{k_i}}=\bigcap_{n\in\Bbb Z^+}K_n=\{x\}$, and hence $x_0=x$. Essentially the same argument shows that $y_0=x$. Thus, $\lim_ix_{n_{k_i}}=x=\lim_iy_{n_{k_i}}$.

Let $\epsilon>0$; there is an $i\in\Bbb Z^+$ such that $d(x_{n_{k_j}},x)<\epsilon$ and $d(y_{n_{k_j}},x)<\epsilon$ for all $j\ge i$, and therefore

$$\operatorname{diam}K_{n_{k_j}}=d(x_{n_{k_j}},y_{n_{k_j}})\le d(x_{n_{k_j}},x)+d(x,y_{n_{k_j}})<2\epsilon$$

for each $j\ge i$. This shows that $\langle\operatorname{diam}K_{n_{k_i}}:i\in\Bbb Z^+\rangle\to 0$ as $i\to\infty$.

To complete the proof, just notice that since $K_{n+1}\subseteq K_n$ for each $n\in\Bbb Z^+$, $\operatorname{diam}K_{n+1}\le\operatorname{diam}K_n$ for each $n\in\Bbb Z^+$; the sequence $\langle\operatorname{diam}K_n:n\in\Bbb Z^+\rangle$ is therefore monotonically non-increasing, so the fact that it has a subsequence converging to $0$ implies that it also converges to $0$.


问:
Linear Algebra - subspace proof
Let $A \in \mathbb{R}^{m\times{n}}$ and $b \in \mathbb{R}^{m\times{1}}$. Show that $Ax = b$ solution set

$R = \{x \in \mathbb{R}^{n\times{1}}$$:Ax = b\}$

is a $\mathbb{R}^{n\times{1}}$ subspace if and only if $b = 0$

Looking for hints how to start proving this (looking for a starting point, not the answer).

答:
Hint for "if": assume $b=0$, then consider $x,y \in R$ we want to show that $R$ is closed under addition and scalar multiplication. You can use the linearity property of $A$ to do this.

"only if": assume $R$ is closed under addition and scalar multiplication. Again use the linearity of $A$.

Title: Linear Algebra - subspace proof


问:
Integral of $\int(4-2x)^\frac{1}{3}dx$
I solved this integral then I did $\frac{d}{dx}$ of $F(x)$ and saw that it's not the same, so I did something wrong in my integration process.
$$\int(4-2x)^\frac{1}{3}dx$$

What I did is $$F(x) =\frac{(4-2x)^\frac{4}{3}}{-2\times\frac{4}{3}} + C$$
Thanks!

答:
Use the substitution, let $4-2x=u$, then $-2dx = du \implies  dx = \frac 1{-2}du$, thus the integral is
$$\int \frac{u^{\frac{1}{3}}}{-2}du$$</p>

Title: Integral of $\int(4-2x)^\frac{1}{3}dx$


问:
moment generating function uniquely determines distribution
Suppose that X and Y are random variables both taking possible values in {$0, 1, 2 ...$}.
Further, suppose that X and Y have the same mgf for all t around $0$:
Then it holds:
$$\sum_{j=0}^{\infty}e^{tj}f_X(j)-\sum_{j=0}^{\infty}e^{tj}f_Y(j)=0$$

$$\Rightarrow\sum_{j=0}^{\infty}e^{tj}[f_X(j)-f_Y(j)]=0$$
$$\Rightarrow \sum_{j=0}^{\infty}e^{tj}c_j=0 $$
with $c_j:=f_X(j)-f_Y(j)$.

Is there a way to justify that  $c_j=0$  for all $j \in \{0,1,...\}$

答:
The series $g(z) = \mathbb E[z^X] = \sum_{j=0}^\infty z^j f_X(j)$ converges
absolutely to an analytic function for $|z| < 1$, and the coefficients $f_X(j)$ can be obtainethbbd from the values of $g$ and its derivatives at $z=0$:

$$ f_X(j) = \frac{g^{(j)}(0)}{j!} $$

$g(z)$ for $z \ne 0$ can be obtained from the moment generating function:
$$ g(e^{-t}) = \mathbb E[e^{-tX}] = \sum_{j=0}^\infty e^{-tj} f_X(j)$$

So the moment generating function determines $g$, and that determines the distribution.


问:
What's the meaning of this relation statement?
$R =\{ (A,B) \ |\ A,B\ \epsilon \ P(\mathbb{N}), (A=B) \vee (A\cup\ \{ 1,2\} =B) \}$

What is the meaning of this relation? it means either of $A=B$ and $(A\cup\ { 1,2} =B)$ statements must be true?

hence it's a reflexive relation because the $A=B$ is true but the other not?

答:
Right, if $A$ and $B$ are two subsets of $\mathbb N$ the the set $A$ relates to $B$ if either $A = B$ or $A$ would equal $B$ if you add $1$ and $2$ to $A$.

So $\{1,5,7,8\}R \{1,5,7,8\}$ and $\{1,5,7,8\}R \{1,2,5,7,8\}$ but $\{1,2,5,7,8\} \not R \{1,5,7,8\}$ for example.

It's weird relation but it's certainly a well defined relation.

It's reflexive as for all $A\subset \mathbb N$ we always have $A=A$ so $A \mathbb R$.

It's certainly not symmetric.  If $\{1,2\}\subset B$ and $A = B\setminus\{1,2\}$ then $A\cup \{1,2\}=B$ so $ARB$ but $B\ne A$ and $B\cup\{1,2\} = B\ne A$ so $B\not R A$.  (I gave a counter example in paragraph 2, but the simplest would  be $\emptyset R\{1,2\}$ but $\{1,2\}\not R \emptyset$.)

I think it's anti-symmetric.  If $A R B$ and $A\ne B$, then $A\ne A\cup \{1,2\}=B$.  So $\{1,2\}\not \subset A$ but $\{1,2\}\subset B$ so $B\ne A$ and $B\cup\{1,2\}=B\ne A$ so $B\not A$.  So $ARB$ and $BRA$ if only if $A=B$.

And I think it is Transitive.

If $A R B$ and $B RC$ then

case 1: $A = B$ than $BRC  \implies ARC$.

case 2: $A\ne B$ then $A \cup \{1,2\}= B$.  And $\{1,2\}\subset B$ so $B=B\cup \{1,2\}$.  So $B RC$ means $B = C$ or $B\cup\{1,2\}=C$ but as $B= B\cup\{1,2\}$ it follows that $B=C$.  So $ARB\implies ARC$.

So either way $ARB, BRC\implies ARC$.


问:
Do Boolean rings always have a unit element?

Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)

BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

答:
Here is a positive result: Let $R$ be a commutative ring such that every element can be written as a sum of products of elements of $R$ (i.e. the multiplication map $R \otimes_{\mathbb{Z}}R \to R$ is surjective). If $R$ is finite, then $R$ is unital. In particular, every finite boolean ring is unital (which also can be proven directly, of course).

Proof: Consider the unitalization $R^+$ as an $R$-module. Then we have $R R = R$, hence by Nakayama there is some $e \in R$ with $(1-e) R=0$. But this means that $e$ is a unit of $R$.


问:
Why $\lim_{x\to 0^-}\left(1+\cfrac1x\right)^x=+\infty$?
According to plot of WolframAlpha, $\lim_{x\to 0^-}\left(1+\cfrac1x\right)^x=+\infty$

But I am confused because when I am calculating the limit I get $1$:

$\lim_{x\to 0^-}\left(1+\cfrac1x\right)^x=\lim_{x\to 0^-}e^{x\ln\left(1+\tfrac1x\right)}$

$\lim_{x\to 0^-} x\ln\left(1+\frac1x\right)=\lim_{x\to 0^-}\cfrac{\ln(1+\tfrac1x)}{\tfrac1x}$ Apply L'Hopital rule: $\lim_{x\to 0^-}\cfrac{\tfrac{x}{x+1}\times\tfrac{-1}{x^2}}{\tfrac{-1}{x^2}}=\lim_{x\to 0^-}\frac{x}{x+1}=0$

Therefore the original limit is equal to $1$. but according to plot of the function $\left(1+\frac1x\right)^x$ it should be $+\infty$.but why?

答: $\lim_{x \to 0^-}$ does not make sense here. For $-1 < x < 0$ we have $b_x = 1 + \frac{1}{x} < 0$ and thus $(1 + \frac{1}{x})^x$ is not defined. In Wolfram you see that $\lim_{x \to -1^-}(1 + \frac{1}{x})^x = \infty$.


问:
How do you use the gradient to find $n$?
$$z = \sqrt{1 - x^2 - y^2}$$

$$W = x^2 + y^2 + z^2 - 1$$$$\overrightarrow{N} = \nabla W   = \langle 2x,2y,2z \rangle$$

$$\overrightarrow{n} = \frac{\overrightarrow{N}}  {\|\overrightarrow{N}\|} = \frac{\langle x,y,z \rangle}{\sqrt{x^2 + y^2 + z^2}} = \langle x,y,z \rangle$$

My question is how did it come out to be ? And shouldn't you be plugging in $\langle 2x,2y,2z \rangle$ in the denominator for $x,y,z$ inside the sqrt, so shouldn't it be $\sqrt{(2x)^2 + (2y)^2 + (2z)^2}$? And then shouldn't it equal $$\left \langle \frac{x}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}, \frac{y}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}, \frac{z}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}} \right \rangle $$ and not just $\langle x,y,z \rangle$
?

The original question was evaluate integral of C of
$$$$$$\overrightarrow{F} *d \overrightarrow{r} $$ where $$$$$$\overrightarrow{F}(x,y,z) = <e^(-x^2)+sinx, e^(-y^2)-2yz, x^2+ e^(-z^2)=2z> $$ and C is the boundary of the surface $$z = \sqrt{1 - x^2 - y^2}$$ oriented Counterclockwise.

答: You should only "plug in $\langle 2x,2y,2z\rangle$ in the denominator" if you also use them in the numerator. Then you get $$\overrightarrow{n} = \frac{\overrightarrow{N}}  {\|\overrightarrow{N}\|} = \frac{\langle 2x,2y,2z \rangle}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}} = \frac{2\langle x,y,z \rangle}{2\sqrt{x^2 + y^2 + z^2}}$$ and a simple cancellation.</p>


问:
What is it called when an operator is composed with a second operator and its inverse ($H^{-1} \circ T \circ H$)? What are some examples?
A student of mine recently came up with a programming solution that is an instance of $H^{-1} \circ T \circ H$, where $T$ is a translation and $H$ and $H^{-1}$ are roughly an embedding/projection pair.  (The embedding and projection are not exact because in this example $T$'s domain and codomain are different.)

I'm pretty sure I've seen examples like this before, especially when $T$'s domain and codomain are equal.  Maybe in quantum mechanics?  But I have not been able to find any examples, and I have not been able to recall the name of this transformation, in which $H$ and its inverse are used to modify $T$.

Where can I find examples of this transformation?  And perhaps learn what the transformation is called?

答: This is called a conjugation of $T$ by $H$. It’s a general abstract algebra concept that works in any group. An example from linear algebra is matrix similarity, which can function as a change of basis.


问:
Lineal Transformation $T : V \rightarrow V$ injective implies that $T$ is surjective
Show that if $V$ is a vector space over a finite-dimensional field $K$ and $T: V → V $ is a linear transformation, then if $T$ is injective it implies that $T$ is surjective.

Hi, I have been stuck with this problem can someone help me please?

答:
Take a basis $v_1,\dots,v_{\dim V}$ for $V$. Then, as $T$ injective, the vectors $Tv_1,\dots,Tv_{\dim V}$ are linearly independent$^1$, and hence, a basis for $V$. Thus, for every $v \in V$ there is $a_1,\dots,a_{\dim V} \in K$ such that $v = a_1Tv_1 + \cdots + a_{\dim V}Tv_{\dim V} = T(a_1v_1 + \cdots + a_{\dim V}v_{\dim V})$, which means that $T$ is surjective.

$^1$ To prove this use the fact that $Tw = 0$ implies $w=0$.


问:
Asymptotic formula for Mean of Sum of Power of Divisors $\frac{ \sum_{i=1}^{n}\sigma_v(i)}{n}\sim\frac{n^v\zeta(v+1)}{v+1}$

Question:

Define the sum of $v$-powers of divisor $\sigma_v(n)=\sum_{d|n}d^v$ for $v \in \mathbb{R}$.

Prove that for all $v>0$,
$$\frac{ \sum_{i=1}^{n}\sigma_v(i)}{n}\sim\frac{n^v\zeta(v+1)}{v+1}$$
where $\zeta$ is the Riemann zeta function.

Attempts:

By using the convolution $\sigma_1=\mathbb{1}*\text{Id}$,
(where $\mathbb{1}(n)=1$, $\text{Id}(n)=n$)

we have $\displaystyle \sum_{i=1}^n\sigma_v(i)=\sum_{d=1}^{n}\sum_{k\leq \frac{n}{d}}k^v$.

When $v=1$, we know the sum $\displaystyle \sum_{k\leq \frac{n}{d}}k = \sum_{k=1}^{[\frac{n}{d}]}k=\frac{1}{2}([\frac{n}{d}]+1)[\frac{n}{d}]$,

then by bounding $\frac{n}{d}-1<[\frac{n}{d}]\leq\frac{n}{d}$,

we can show $\displaystyle \sum_{i=1}^n\sigma_1(i)=\frac{n^2}{2}\sum_{d=1}^n\frac{1}{d^2}+f(n)$ for some $|f(n)|<n\log{n}$ , when $n\geq 4$.

The "2" in $d^2$, $n^2$, $\frac{1}{2}$ comes from the rather easy formula of sum of positive integers.

However, when $v\neq 1$, the term becomes the sum of power of integers and I cannot easily find a formula for it.

The closest I have heard of is Faulhaber's_formula with Bernoulli numbers as coefficients, but the coefficients are hard to handle and the formula seems to be for integer power only.

Another way I think of is by inequality. Power-Mean Inequality looks usable, but it does not have the ratio $\displaystyle\frac{1}{v+1}$ so it does not seems to be the way.

The third way is possibly constructing the general case from the result of $\sigma_1$, but I don't know how to either.

Any help is very much appreciated.

答:
I don't think you'll do it this way, Euler-McLaurin (the extension of Faulhaber to non-integer $v$) gives a good approximation of $\sum_{k\le y} k^v$ but as a function of $\lfloor y\rfloor$ not $y$.

$$\sum_{n\le x}\sigma_v(n) n^{-v}= \sum_{n\le x}\sigma_{-v}(n)
=\sum_{k\le x}k^{-v}  \lfloor x/k\rfloor =
\sum_{k\le x}k^{-v}  (x/k+O(1))$$ $$= x (\zeta(v+1)+O(x^{-v}))+ O(1)+O(x^{1-v})$$

Then by partial summation (for $x$ integer) $$\sum_{n\le x}\sigma_v(n)=x^v (\sum_{n\le x}\sigma_v(n) n^{-v})+\sum_{m\le x-1} (\sum_{n\le m}\sigma_v(n) n^{-v})(m^v-(m+1)^v)$$
$$ = \ldots $$


问:
Proof that entropy of Markov Source is $H(S) = \sum_s P(s) H(s)$
Let $S$ be a markov source in which distribution $P$ doesn't depend on $P_0$ and which generates numbers from finite set $\Omega$. Proof that
$$H(S) = \sum_{s \in \text{process states}} P(s)H(s) $$
where
$$H(s) = \sum_{x\in\Omega} P(x| s) \log P(x | s)$$

My attempt

According to the definition of the entropy, for discrete random variable $X$ with possible occurrences $x_1, ..., x_n$ it is $H(X) = \sum_{i=1}^{n} P(x_i)\log P(x_i)$. So in my case, $X$ would be $S$ and $x_i, ... x_n$ would be $s_1, ..., s_n$. Then:
$$H(X) =H(S)= -\sum_{i=1}^n P(s_i) \log P(s_i) $$

So I want that $ - \log P(s_i) = H(s_i)$
But I don't know how to proceed with that further. I also looked for that in some books but I didn't find any proof.

答:
A random process $X_1,X_2,\cdots$ is associated with an entropy rate, defined as
\begin{eqnarray*}
\mathcal{H} &=& \lim_{n \rightarrow \infty} \frac{1}{n}H(X_n, X_{n-1}, \cdots, X_1)
\end{eqnarray*}
(when the limit exists). If the process is stationary, as it appears to be in your example, then you can use an equivalent definitions for entropy rate:
\begin{eqnarray*}
\mathcal{H} &=& \lim_{n \rightarrow \infty} H(X_n | X_{n-1}, \cdots, X_1)  \\
\end{eqnarray*}

The definition that makes sense to use in the case of a stationary Markov process is the second one. In your example, it appears that you are assuming a Markov source of order 1 (the current symbol depends on only one past symbol). In that case
\begin{eqnarray*}
H(X_n | X_{n-1}, \cdots, X_1)  = H(X_n | X_{n-1}) = H(X_1|X_0)
\end{eqnarray*}
where the last equality follows from the stationarity of the process. Then the equation for the entropy rate that you mentioned follows trivially, as
\begin{eqnarray*}
H(X_1|X_0) &=& \sum_{x_0} p(x_0) \sum_{x_1} p(x_1|x_0) \log \frac{1}{p(x_1|x_0)} \\
&=&\sum_{x_0} p(x_0) H(X_1 | X_0 = x_0)
\end{eqnarray*}

Now, you have one more additional condition in your problem which is that "$P$" does not depend on "$P_0$".  It isn't immediately clear what exactly you mean, but with high likelihood, you are trying to state that the stationary distribution of the Markov chain does not depend on the distribution of the initial state. This condition is not strictly needed to obtain the entropy formula above, however that condition is often very useful when one wants to study how the probability of long sequences is associated with the entropy.

If the stationary distribution does not depend on the initial conditions, the Markov chain is "ergodic". One of the consequences of ergodicity is that empirical averages converge to the expectation with probability 1, as the number of samples being averaged goes to infinity; specifically if $E[X_0] < \infty$, for a stationary and ergodic random process,

\begin{eqnarray*}
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_i f(X_i) = E[X_0]
\end{eqnarray*}

with probability 1. If you plug in $f = \log P_{X_1|X_0}$, then you obtain that

\begin{eqnarray*}
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_i -\log P_{X_1|X_0}(X_i | X_{i-1}) = E[-\log P(X_1|X_{0})] = H(X_1|X_0)
\end{eqnarray*}
with probability 1. So ergodicity makes entropy meaningful - it can be used to estimate the probability of long sequences of symbols, regardless of how the sequence starts.

For further references, take a look at the classical book Elements of Information Theory by Cover and Thomas.


问:
Determine the posterior density of $θ$ after doing $n$ coin tosses
I am given a that I toss a coin and the prior  $θ \stackrel{}{\sim} Uniform[0.4, 0.6]$

$\textbf{Note:}$ $\theta$ is the probability of getting a head on a single toss

If I toss the coin $n$ times and obtain $n$ heads, then what is the posterior density
of $θ$?

答:
Let's do this for one toss of the coin. Suppose that we toss this coin once to get the roll $X$ and it falls heads  (i.e. $X=H$).

Note that the probability of getting the head depends on the value of $\theta$, so
$$
\mathbb{P}[X=H]
 = \int_{0.4}^{0.6} \mathbb{P}[X=H,\theta=t]dt
 = \int_{0.4}^{0.6} \mathbb{P}[X=H|\theta=t] f_\theta(t) dt
 = \int_{0.4}^{0.6} \frac{t\ dt}{0.2} dt
 = 0.5,
$$
which intuitively makes a lot of sense since the uniform distribution is symmetric around $0.5$...

The new distribution would be given by
$$
f_\theta(t|X=H)
 = \frac{\mathbb{P}[X=H|\theta=t]}{\mathbb{P}[X=H]}
 = \ldots
$$

can you complete this and then generalize to $n$ rolls?

You can try reading up on Bayes' Theorem, especially in the section Making a prediction.


问:
Ring of Invariant
Let $G \subset SL_2(\mathbb{C})$ be a finite subgroup acting linearly on $\mathbb{C}[X, Y]$. Then it is claimed that the ring of invariants $\mathbb{C}[X, Y]^G$ is always a hypersurface. I am not able to see its proof by myself. Please help.</p>

答:
I don't think you can tell this a priori without actually computing a presentation of the invariants.

If you do that, then it is a matter of checking that you can generate each of those rings with three elements, and that these satisfy exactly one polynomial relation. Finding the invariants is easy in most cases, but hard for the binary icosahedral group. Everything can be done by computer, using Groebner bases; you can spot the relations by looking hard at generators (and harder for the binary icosahedral group...), and then checking that you have them all is a matter of using Molien's formula. ALl this was done before von Neumann was born, so it is quite possible to do by hand, of course!

The description of generators and relations is done in Klein's Lectures on the icosahedron, and in several more modern places. I think, for example, that this is done explicitly in Dolgashev's notes on McKay correspondence, which Google will find for you. See also https://mathoverflow.net/questions/16026/the-finite-subgroups-of-sl2-c/16030#16030 and the references listed there.

Title: Ring of Invariant


问:
What are some relationships between a matrix and its transpose?
All I can think of are that
If symmetric, they're equivalent

If A is orthogonal, then its transpose is equivalent to its inverse.

They have the same rank and determinant.

Is there any relationship between their images/kernels or even eigenvalues?

答:
There is a very strong connection between an $m\times n$ matrix $A$ and its transpose $A^T$, at least if the base field is $\def\R{\mathbb{R}}\R$.

Denote by $C(A)$ the column space, that is, the subspace of $\R^m$ generated by the columns of $A$. Denote by $N(A)$ the null space, that is, the set of vectors $v\in\R^n$ such that $Av=0$.

Associated to a matrix $A$ we have four subspaces:

$C(A)\subseteq\R^m$

$N(A)\subseteq\R^n$

$C(A^T)\subseteq\R^n$

$N(A^T)\subseteq\R^m$

Now the relations (of course, the second one is equivalent to the first)

$$
C(A)\oplus N(A^T)=\R^m,\qquad
C(A^T)\oplus N(A)=\R^n
$$

which means that every vector of $\R^m$ can be written in one and only one way as the sum of a vector in $C(A)$ and one in $N(A^T)$.

To see why, consider a vector $v\in C(A)\cap N(A^T)$; then $v=Ax$ for some $x\in\R^n$ because $v\in C(A)$; since also $v\in N(A^T)$ we get
$$
0=A^Tv=A^TAx
$$
from which we deduce
$$
0=x^T0=A^TAx=(Ax)^T(Ax),
$$
so that $Ax=0$ and therefore $v=Ax=0$. Thus $C(A)\cap N(A^T)=\{0\}$ and we only need to show that $C(A)+N(A^T)=\R^m$.

The relation between the rank of $A$ and of $A^T$ and the "rank-nullity theorem" tell then that
$$
\dim N(A^T)=m-\mathrm{rk}\,A^T=m-\mathrm{rk}\,A=m-\dim C(A),
$$
so that $\dim C(A)+\dim N(A^T)=m=\dim\R^m$ and hence $C(A)+N(A^T)=\R^m$.

The same holds for matrices over $\def\C{\mathbb{C}}\C$, but using the Hermitian transpose instead of the transpose.

This doesn't hold however for matrices over an arbitrary field, because the step “$(Ax)^T (Ax)=0$ implies $Ax=0$” cannot be done in general.


问:
Calculate average bill of a phone call
Let's assume that during a phone call we charge a fee every one minute. Time of phone calls is defined by exponential distribution with a parameter of 1. Please calculate the average length of phone call and the average cost of phone call ($1 per minute).

From the definition, it sounds like it's a problem to be solved by the Poisson distribution. How to estimate average 'ticks' during the phone call?

答: No, it is an exponential distribution of phone call lengths.  You get $1$ if the call lasts $0-1$ minute, $2$ if it lasts $1-2$ minutes and so on.  You can find the number of calls in each interval by subtracting the cumulative distribution function at those points.

Title: Calculate average bill of a phone call