问:
Stability of differential equations under nonlinear pertubations
I am trying to understand the proof of the following theorem :

Let $A$ be an $n\times n$ matrix such that it has only negative eigenvalues and let $f: \mathbb{R}\times \mathbb{R}^n$ continuous and locally Lipschits in $x$. If $f(t,0)=0$ for every $t\in \mathbb{R}$ and $\lim_{x\rightarrow 0}\frac{||f(t,x)||}{||x||}=0$ for any $t\in \mathbb{R}$ then the zero solution of the equation $x'=Ax+f(t,x)$ is asymptotically stable .

Now in the proof of this, the authors claim that given $x_0\in \mathbb{R}^n$ with $||x_0||<\delta$ the solution satisfies $||x(t)||<\delta $ for any $t\in [t_0,t_1]$ sufficiently small. And then they use that fact to prove that if $||x_0||<\frac{\delta}{M}$ then $||x(t)||<\delta$ using the following
\begin{align}
x(t)&=e^{A(t-t_0)}x_0+\int_{t_0}^t e^{A(t-s)}f(s,x(s))ds
\\
\|x(t)\|&\leq Me^{-\alpha(t-t_0)}\|x_0\|+\int_{t_0}^tMe^{-\alpha(t-s)}\epsilon \|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}\|x_0\|+\int_{t_0}^t M\epsilon e^{\alpha s}\|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}e^{M\epsilon (t-t_0)}
\\[.5em]
\|x(t)\|&\leq Me^{(-\alpha +M\epsilon) (t-t_0)}\|x_0\|
\end{align}
I am just a bit confused why we are doing this for $||x_0||<\frac{\delta}{M}$ since we know that if $||x_0||<\delta$ then $||x(t)||<\delta$. I thought this might be an attempt to enlarge the interval where we have that $||x(t)||<\delta$, to use the fact that $|f(s,x(s)|<\epsilon||x(t)||$ we need the fact that $||x(t)||<\delta$ so we stay in the same interval.
Can anyone help me figure out why we care that $||x_0||>\frac{\delta}{M}$? Thanks in advance.

答:
The constant $M$ is there as guard for the behavior of $e^{(A+αI)t}$ in the case of eigenvalues with "proper" multiplicities. Then the matrix exponential may also have polynomial terms, and as one can see in examples like $(1+2t) e^{-t}$ for $t\ge 0$, while the general behavior is monotone convergence to zero, initially this function is growing above its initial value of $1$. So with
$$
M=\sup_{t>0}\|e^{(A+αI)t}\|
$$
one gets
$$
\|e^{At}x\|\le Me^{-αt}\|x\|.
$$
One might get a more accurate upper bound $\|e^{At}x\|\le \min(1+Nt,M)e^{-αt}\|x\|$ which would be tight at $t=0$, but that complicates the ensuing computations unnecessarily.

For $u(t)=e^{αt}\|x(t)\|$ the inequality
$$
u(t)\le Mu(t_0)+\int_{t_0}^t Mϵ u(s)\,ds
$$
has the solution
$$
u(t_0)+\int_{t_0}^t ϵ u(s)\,ds\le u(t_0)e^{Mϵ(t-t_0)}
\\~\\
\implies
\|x(t)\|\le Me^{-(α-Mϵ)(t-t_0)}\|x(t_0)\|
$$

So the flow of the proof is that you select some $α$ so that the spectrum of $A$ has all real parts smaller $α$. Then with this $α$ the bound $M$ is determined. Now select some $0<ϵ<\frac{α}{M}$. This then determines via the properties of $f$ a $δ$ so that $\|f(s,x)\|\le ϵ\|x\|$ for all $\|x\|\le δ$. Then for $\|x_0\|\le\frac{δ}{M}$ the given estimates guarantee (in a slightly circular way) that the solution remains in the admissible region where all assumptions and constants above are valid. Additionally that gives the defining property of stability and asymptotic stability.


问:
Find the value of $a,b$ such that given limit exists

Let $f(x)$ be a function such that $\lim_{x\to 0} \frac{f(x)}{x} =1$. If $\lim_{x\to 0} \frac{x(1+a\sin x)-b\sin x}{(f(x))^3}$ is finite, then find $a,b$

Let $f(x)=\sin x$

$$\lim_{x\to 0} \frac{ \sin x+ a\sin ^2x -b\sin x}{\sin^3 x}$$
$$\implies \lim_{x\to 0} \frac{ \sin x(1-a) +a\sin ^2 x}{\sin^3 x}$$

Which leads to nowhere. If the assumption of $f(x)$ is wrong, then I don’t know what else to do. Can I get a hint?

答:
From @Claude's answer,
$$\frac{x(1+a\sin x)-b\sin x}{x^3}=\frac{1-b\frac{\sin x}{x}+a\sin x}{x^2}.$$

Now if $ a\neq0 $ or $b\neq1$ then clearly the limit doesn't exist .

But we didn't check that if $a=0,b=1$ then limit exists.

Hence suppose $a=0,b=1$ then $$L=\lim_{x\to0} \frac{1-\frac{\sin x}{x}}{x^2}$$

Now you can use Taylor's series or L'Hospital's rule, using former we get,
$$L=\lim_{x\to0} \frac{\frac{x^2}{3!}-\text{higher order terms}}{x^2}=\frac1 6$$

ContentLicense: CC BY-SA 4.0


问:
Why do projective bases need $n+2$ points?
I am a little bit confused about why we define a basis of an $n$-dimensional projective space $P(V)$ to be a set of $n+2$ points in general position, rather than a set of $n+1$ points in general position. For vector spaces a basis is supposed to provide (unique) coordinates. In the world of projective spaces we obtain coordinates by looking at representative vectors of points in general position: If $P(V)$ is an $n$-dimensional projective space, then any choice of representative vectors of $n+1$ points $P_1,\dots,P_{n+1}\in P(V)$ in general position forms a basis of $V$ which provides coordinates for $P(V)$ that are unique up to a common scaling factor. Intuitively, I would stop here and call $P_1,\dots,P_{n+1}$ a basis of $P(V)$.

Flying over my lecture notes the definition with $n+2$ points seems to be motivated by the following lemma.

Lemma. Let $P(V)$ be an $n$-dimensional projective space. For every set of $n+2$ points $P_1,\dots,P_{n+2}\in P(V)$ in general position there exist unique (up to a common scaling factor) representative vectors $v_1,\dots,v_{n+2}\in V$ such that $\sum_iv_i=0$.

However, I don't see what the significance of the $(n+2)$-th point is, or how the addition of the $(n+2)$-th point narrows down the family of possible $(n+1)$-point bases of $P(V)$ to some more desirable subfamily of bases of $P(V)$. After all, for any set of $n+1$ points $P_1,\dots,P_{n+1}\in P(V)$ in general position with representative vectors $v_1,\dots,v_{n+1}\in V$ I can simply create a/the $(n+2)$-th point $P_{n+2}:=\bigl[-\sum_{i=1}^{n+1}v_i\bigr]$.

Can someone explain to me why the $(n+2)$-point definition of 'basis' for projective spaces is more desirable than the more intuitive $(n+1)$-point definition?

答:
Reading further through my lecture notes I strongly suspect that we prefer bases with $n+2$ points over bases with $n+1$ points, because two sets of $n+2$ points in general position uniquely determine a projective transformation between the corresponding $n$-dimensional projective spaces:

Proposition. Let $P(V)$ and $P(W)$ be $n$-dimensional projective spaces. Any two sets of points $P_1,\dots,P_{n+2}\in P(V)$ and $Q_1,\dots,Q_{n+2}\in P(W)$ in general position uniquely determine a projective transformation $f\colon P(V)\to P(W)$ with $f(P_i)=Q_i$ for all $i=1,\dots,n+2$.

If we just take two sets of $n+1$ points, then any choice of representative vectors uniquely determine a vector space isomorphism $F\colon V\to W$ which uniquely determines a projective transformation $f\colon P(V)\to P(W)$. But this depends on the choice of representative vectors, whereas for $n+2$ points the resulting projective transformation only depends on the $n+2$ points and not their representative vectors.


问:
Series solution of a 2nd order ODE
Is the ODE </p>

<p>$(1-x^2)y''+y'+y=0$ solvable by simple power series (not Frobenius) method?
The reason I am asking this, is because if the eq were $(1-x^2)y''+xy'+y=0$, it could have been easy, since all the individual series could have been synched to $\sum A_nx^n$, i.e in terms of a single power series.
Is there any workaround to it, without using the generalised method, i.e the Frobenius method?</p>

答:
Let $x=t-1$ ,

Then $(1-(t-1)^2)y''+y'+y=0$

$(1-t^2+2t-1)y''+y'+y=0$

$(-t^2+2t)y''+y'+y=0$

$t(t-2)y''-y'-y=0$

Which is similar as the Gaussian hypergeometric ODE and can solve easily by Frobenius method

ContentLicense: CC BY-SA 3.0

Title: Series solution of a 2nd order ODE


问:
Let $p$ be a prime number. Prove that for $a \in\mathbb Z, 0 \le a \le p−1$: $C(a, p - 1) \equiv (-1)^a\pmod p$
$C(a, p-1) = \frac{(p-1)!}{a!(p-1-a)!}$

If $a$ is an odd number, then $C(a, p-1)  \equiv p - 1 \pmod p$

If $a$ is an even number, then $C(a, p-1)  \equiv 1 \pmod p$

I'm stuck and I don't know how to proceed, I would be grateful for any help.

答:
Hint:

Using Pascal's Identity

$$\binom{p-1}k\equiv-\binom{p-1}{k-1}\pmod p $$   as $p$ divides $\displaystyle\binom pk$ for $1\le k\le p-1$

$$\implies\binom{p-1}k\equiv(-1)^r\binom{p-1}{k-r}\pmod p$$  for $0\le r\le k$

Set $r=k$


问:
Existence of an inverse number to zero (null)? Existence a meta of number?
Sorry for the structure and maybe some mistakes in the post - I'm not a Mathematician and not a native English speaker, so I will use Google Translator + Grammarly for help.

Please, if you want and can - help me to find an answer to this sequence of questions:

Does an inverse number to zero (null) exist?
1.1) If yes - what is it?
1.2) Or/And - is the one solution or several?
1.3) Or/And - is there an equal or different weight of these solution/-s in mathematical society?

Does a meta of number exist?

My own naive thoughts (briefly and simply):

About 1 question:
if we can imagine that "zero" (null) is a mathematical concept of "nothing", so maybe exists a mathematical concept of "everything"? Maybe if I call "zero" (null) an "Absolute Emptiness", so maybe we should create "Absolute Fullness"?
Remark - I don't mean any sort of infinities, like infinity small values (maybe they called "infinitesimals") and their opposites - infinite big values (plus or minus, I suppose it doesn't matter in this context, and don't know their right definitions, sorry) - because these entities have values and they have some sort of scope of properties, let's call them "Relatives".

About 2 question:
if we can operate not just different values, but also their "Absolute Emptiness" (which for me looks confusing, because if I have the right understanding, we trying to operate by "nothing" with serious faces =))) - maybe there is an existing core property of a number, lets call it's "meta". This "meta" means the existence of some object, which we want to operate. Or maybe, it would be useful to separate operating of "Absolutes" (like only two objects - "zero" (null) and it's inverse) and "Relatives" which has a range from infinity small number (no matter of sign) to infinity big number (also not matter of sign)? If we do this separation, maybe we displace problems like division by zero or just will look at the problem from another angle - because it seems like when we dividing the non-zero value by zero (or do other 3 arithmetic operations) we work with 2 types of entities/properties ("Relative" and "Absolute") which maybe require a different approach, because for my understanding (of course with Dunning–Kruger effect xDD) there is no fundamental/crucial/disbalance problem to divide non-zero infinity small/big/mediocre value by non-zero infinity small/big/mediocre value (or do other 3 arithmetic operations, or other general operations), are you agree?

It would be so nice if you give me answers to those questions because I even don't how trivial these questions are.

P.S. If there are no one/multiple clear answers, maybe you can suggest a path/sequence of books (from zero to goal) which I must read/understand and that is will help me solo find answers to the questions above.

Thanks you in advance.

答:
Assume there was some number $\star$ that we add to the real numbers $\mathbb R$ with the property $0\cdot \star = \star \cdot 0 = 1$. From this it follows that for any number $x\in\mathbb R$ we have
$$
x = x\cdot 1 = x\cdot 0\cdot \star = 0\cdot \star =1.
$$
This is a contradiction of course, since there are real numbers that are not equal to $1$.

Hence, you can't add a multiplicative inverse of $0$ to the reals. If you attempt to, you end up with a structure where all elements are equal to $1$, the zero ring.


问:
Positive real number M such that for all n $|a_n|$ less than M
Click on the link to view

In this problem I have integrated the expression |f'(x) | to get $f(x)$<$x/2$ but after that how to proceed with the problem is clueless.Kindly help.

答:
Hint : Show by induction, using MVT, that for all $n \in \mathbb{N}$,
$$|a_{n+1}-a_n| \leq \frac{1}{2^n}|a_1-a_0|$$

Deduce that the sequence $(a_n)$ is Cauchy, and hence, is bounded.


Title: Positive real number M such that for all n $|a_n|$ less than M

Body:
Click on the link to view

In this problem I have integrated the expression |f'(x) | to get $f(x)$<$x/2$ but after that how to proceed with the problem is clueless.Kindly help.

Body: $|f(x)-f(0)| \leq \frac 1 2 |x|$ so $|f(x)| \leq c+\frac {|x|} 2$ where $c=|f(0)|$. So $|a_{n+1}| \leq c+\frac {|a_n|} 2$. This gives $|a_{n+1}| \leq c(1+\frac 1 2)+ \frac {|a_{n-1}|} {2^{2}}$ and so on. Finally we get $|a_{n+1}| \leq c(1+\frac  1 2+\frac 1 {2^{2}}+...+\frac 1 {2^{n-1}})+\frac {|a_1|} {2^{n}}$. Can you finish?


问:
How do we know which component belongs to which part in a separable differential equation
Take for instance dP/dt = kP

We get after separating: dP/P = kdt, but why shouldn't it be

dP/kP = dt instead, mathematically it doesn't make sense to say that k must belong absolutely to the right hand side of the equation.

答:
Who said $k$ has to be on the right hand side? It is a constant, so it can be on either side. The important point is that anything in terms of $P$ is on the side with $dP$ and everything in terms of $t$ is on the side with $dt$. Since a constant is independent of $P$ and $t$, it may go on either side.

Let's solve $\dfrac{dP}{dt} = kP$ both ways to illustrate why it doesn't matter.

First we separate the equation as
$$\frac{dP}{P} = k \, dt.$$
Integrating both sides, we get
$$\ln |P| = kt + C.$$
Exponentiating both sides, we get the solution
$$P = e^{kt + C} = Ae^{kt},$$
where $A$ is an arbitrary constant.

Now separate the equation as
$$\frac{dP}{kP} = dt.$$
Integrating both sides, we get
$$\frac{1}{k} \ln |P| = t + C.$$
Multiplying both side by $k$ results in
$$\ln |P| = kt + C',$$
where $C' = kC$. Since $C$ is an arbitrary constant, this is an irrelevant detail. Now exponentiating both sides gives
$$P = e^{kt + C'} = Ae^{kt},$$
where $A$ is an arbitrary constant.

So both methods give the same general solution. The reason for this is that we can factor out a constant from an integral, so the side that the constant is on doesn't affect the integration.

The reason every book will leave the $k$ on the right-hand side for this problem is because we want a solution for $P$; we'll just have to move $k$ back to the right-hand side after integrating if we start with it on the left-hand side with $P$.


问:
If $\sum_{n=1}^\infty a_n$ converges with $a_n \geq 0$, then $\sum_{n=1}^\infty n a_n^2$ also converges or diverges?
Given $\sum_{n=1}^\infty a_n$ converges, with $a_n \geq 0$ $\forall n \in \bf{N}$, I want to know whether $\sum_{n=1}^\infty n a_n^2$ also converges.

Given the condition that $S_n =\sum_{n=1}^\infty a_n$ converges with $a_n \geq 0$ $\forall n$, then $\sum_{k=1}^\infty a_n^2$ also converges (proved result).

In addition, it is not always the case that $\sum_{n=1}^\infty n a_n$ converges (counterexample: let $a_n = \frac{1}{n^2}$), although $\lim_{n \to \infty} n a_n$ converges.

I am trying to see if $\sum_{n=1}^\infty n a_n^2$ would converge given the previous information as well as the results that I have obtained above and given to me by theorems. I cannot seem to figure out any counterexamples (if at all). Note that the sum goes from $n = 1 \to \infty$, so no series of the form $a_n = \frac{1}{n^p(\ln{n})^q}$ cannot be used (as the case $n=1$ is left out).

I don't see a good comparison test nor other convergence tests to use, along with monotone-bounded or Cauchy series principles. I am also at a loss of certain theorems that may be handy.

答:
1. Note that the statement is true if $a_n$ is monotone decreasing. Indeed, if $a_n$ is monotone decreasing and $S = \sum_{k=1}^{\infty} a_k$ converges, then

$$ n a_n \leq \sum_{k=1}^{n} a_k \leq S, $$

and so,

$$ \sum_{n=1}^{\infty} n a_n^2 \leq \sum_{n=1}^{\infty} S a_n = S^2. $$

2. Without monotonicity, we have a counter example as in @Hagen von Eitzen's answer. As it seems that OP is not satisfied with a piecewise expression, here is a slight tweak of @Hagen von Eitzen's example:

$$ \bbox[10px,#ffd]{ a_n = \frac{\cos^{2n^2}(\pi n^{1/4})}{\sqrt{n}} } $$

The idea is that $\cos^{2n^2}(\pi n^{1/4})$ will take the value $1$ for $n = k^4$ but otherwise becomes almost zero, closely mimicking @Hagen von Eitzen's example.

To analyze the behavior of this sequence, for each $k \in \mathbb{N}_1 = \{ 1, 2, 3, \ldots \}$ we define

$$ I_k = \Bigl\{ n \in \mathbb{N}_1 : k-\tfrac{1}{2} < n^{1/4} \leq k+\tfrac{1}{2} \Bigr\}. $$

In the following answer, $c_1, c_2, \ldots$ are positive absolute constants whose values are not so important.

First, it is useful to note that there exist absolute constants $c_1, c_2 \in (0, \infty)$ such that

$$ c_1 k^4 \leq n \leq c_2 k^4 \tag{1} $$

for all $k \in \mathbb{N}_1$ and for all $n \in I_k$.

Next, fix $k \in \mathbb{N}_1$ and consider $n \in I_k$. Also, write $n = k^4 + r$. Then by the mean value theorem and $\text{(1)}$,

$$ \left| n^{1/4} - k \right|
= \left| (k^4 + r)^{1/4} - (k^4)^{1/4} \right|
\geq \frac{r}{4\max\{k^4+r,k^4\}^{3/4}}
\geq \frac{c_3 r}{k^3} \tag{2} $$

for $c_3 = \frac{1}{4\max\{c_2^{3/4},1\}}$. Now using the inequality $\cos(\pi x) \leq 1 - 4x^2$ for $\left| x \right| \leq \frac{1}{2}$,

$$ \left| \cos(\pi n^{1/4}) \right|
= \cos(\pi (n^{1/4} - k))
\leq 1 - 4(n^{1/4} - k)^2
\leq 1 - \frac{c_4 r^2}{k^3} $$

for $c_4 = 4c_3^2$. Then by using the inequality $1 - x \leq e^{-x}$, we get

$$ \cos^{2n^2}(\pi n^{1/4}) \leq \left( 1 - \frac{c_4 r^2}{k^3} \right)^{2n^2} \leq e^{-c_5 k r^2}. $$

for $c_5 = 2c_1c_4$. So it follows that

$$ a_n \leq \frac{c_6}{k^2} e^{-c_5 k r^2}. $$

for $c_6 = c_1^{-1/2}$.

Combining altogether, we have

\begin{align*}
S
= \sum_{n=1}^{\infty} a_n
= \sum_{k=1}^{\infty} \sum_{n \in I_k} a_n
\leq \sum_{k=1}^{\infty} \sum_{r=-\infty}^{\infty} \frac{c_6}{k^2} e^{-c_5 k r^2}
< \infty.
\end{align*}

On the other hand,

$$ \sum_{n=1}^{\infty} n a_n^2
\geq \sum_{k=1}^{\infty} k^4 a_{k^4}^2
= \sum_{k=1}^{\infty} 1
= \infty. $$


问:
Distance between 2 points without knowing coordinates
I have the following scenario depicted in the attached Figure. Assuming that there are no symmetries and any particular cases, assuming to know the distance $d_{12}$, the coordinates $x_1,y_1,x_2,y_2$ and the other distances depicted in the figure do you have any idea on how to compute the distance $dist$ between the 2 green points?

答:
Updated answer:

First find $\angle (a+b)$ using the cosine rule in triangle $ABC$, then find $\angle a$ using the cosine rule in triangle $ABD$.

Then you can find angle $b$. Now use the cosine rule to find the distance between the green points $C$ and $D$.

You will need the following formulas:

$$c^2 = a^2+b^2 - 2ab \cos C \Rightarrow \cos C = \frac{a^2+b^2-c^2}{2ab}$$

$a,b$ are sides in this formula, and $c$ is the opposite side to the angle you are trying to find. $C$ is the angle you are trying to find. It does not matter which sides you choose as $a$ and $b$, as long as they are not opposite to the given angle.

Call the point $(x_2, y_2)$ $A$, the two green points $C$ and $D$, and point $(x_1, y_1)$ $B$.

If $C, D$ lie on a circle of radius $3$ centered at $A$, then $C$, $D$ are guaranteed to be exactly $3$ units from $A$. Similarly, by construction, $B$ lies $3$ units away from $C$ and $5$ units away from $D$.

Imagine moving $C$ and $D$ in the diagram. Thus the length of $CD$ is not fixed and can clearly vary.


问:
Smallest number of factors to achieve every number up to (and including) x
Lets say there is a set of numbers, of size n elements. Any of the numbers can be multiplied together, but cannot be used more than once in the same combination. I want to be able to obtain every number, up to and including a particular number x (starting from 0), ONLY as a result of multiplying any of the numbers within the set together.

For example, if x = 20, I want to be able to get 1, 2, 3, ..., 19, 20, with each of those numbers being the result of any combination of the elements being multiplied together within the set.

My question is, how can I determine the smallest possible number of elements (n) within the set to achieve those 20 different combinations; and how can I determine what those numbers are?

答:
You will need to include $1$ as the only way to get it would be to have it already in the set.

Let's now say $p$ is a prime number less or equal than $x$. You will need to include $p$ because this is the only way to have it included in the set. If now $p^2\le x$, you will need to include $p^2$, because the only way to have $p^2$ is if it is included in the set. (You cannot have it as $p\cdot p$ because you have disallowed repetition.) However, now you've included $p^2$, if it happens that $p^3\le x$, you can make $p^3=p\cdot p^2$. However, if you don't include $p^3$, and if $p^4\le x$, you must include $p^4$ etc.

Needless to say, if we have such a set that enables us to see every prime power not greater than $x$ as a product, then every other composite number not greater than $x$ can also be seen as a product - simply break it up as a product of prime powers and make out those prime powers as products.

This suggests the following strategy:

Solve a simpler problem: given $n$, find the smallest subset of $\{1,2,\ldots,n\}$ so that every number from $1$ to $n$ is a sum of the different numbers from that subset. This is because, as we multiply prime powers, we add their exponents.

Use the solution of (1) above for every prime number $p\le x$, taking $n=\lfloor\log_p(x)\rfloor$ - the highest power of $p$ not larger than $x$.

At the end, don't forget to add number $1$ to the set.

So, for example, for $x=20$, you have primes $2,3,5,7,11,13,17,19$ and the highest powers of those not exceeding $20$ are $4,2,1,1,1,1,1,1$ respectively, so:

For prime $p=2$ we have seen we need to choose three numbers (either $2^1, 2^2, 2^3$ or $2^1,2^2,2^4$)

For prime $p=3$ we have to choose two numbers ($3^1, 3^2$)

For other primes, we have to choose them.

The set will end up being either $\{1,2,3,4,5,7,8,9,11,13,17,19\}$ or $\{1,2,3,4,5,7,9,11,13,16,17,19\}$

Of course, we have not solved the problem 1 above, but it looks plausible that, if you choose the numbers that are themselves powers of $2$ you will be able to make any number as a sum of those. (Let's say $n=15$ and you can choose numbers $1,2,4,8$ and write every number from $1$ to $15$ as a sum of those - using the binary representation.)

This can be shown to be the minimum. Note if $n$ is given, take the numbers $1,2,2^2,2^3,\ldots,2^k$ so that $2^{k+1}>n\ge 2^k$. There are $k+1$ of those numbers, and again recalling binary representation, you can make up any number up to not just $n$ but $2^{k+1}-1\ge n$ as a sum of some of those numbers. However, if you took fewer ($k$) numbers, you cannot possibly make up all the sums up to $n$ because the number of (nonempty) sums of $k$ numbers is up to $2^k-1<n$. As we have seen before, this may not be the only minimum ($n=4$ admits the minimum $\{1,2,3\}$ as well as $\{1,2,4\}$) but it is certainly one of the minima.

Summary Given $x$, find all prime numbers not bigger than $x$. For each such prime number $p$ find the highest degree $n=\lfloor\log_p(x)\rfloor$ such that $p^n\le x$. Take the set of numbers $1,2,2^2\ldots,2^k\le n\lt 2^{k+1}$ (i.e. $k=\lfloor\log_2(n)\rfloor$). Your set will now consist of:

The number $1$, and

All the sets $\{p, p^2, p^{2^2},\ldots, p^{2^k}\}$ for all those prime $p$'s


问:
Proving Proving $ \sqrt{a \cdot b} \le (a+b)/2$ with a and b in N*
We have variables $a$ and $b$ as natural numbers..

I tried using the reccurence but I got stuck proving:

$\sqrt{(a+1) \cdot (b+1)} \le (a+1+b+1)/2$

May someone help me with this?

答:
$ (\sqrt{a}-\sqrt{b})^2 \ge 0$ and $(\sqrt{a}-\sqrt{b})^2=a-2 \sqrt{ab}+b.$

Can you proceed ?


问:
$f(g(x))=x$ implies $f(x)=g^{-1}(x)$
Is it possible to find a necessary and sufficient condition to conclude when
$$f(g(x))=x \implies f(x)=g^{-1}(x) \wedge f^{-1}(x)=g(x),$$
if both functions are well defined?</p>

答: If either $f$ is injective or $g$ is surjective, then $f\circ g={\rm id}$ implies $\exists f^{-1},g^{-1}$ then necessarily $f^{-1}=g$ and $g^{-1}=f$.</p>

Title: $f(g(x))=x$ implies $f(x)=g^{-1}(x)$


答: Hint: If $f,g\colon X\to X$, then $f^{-1}=g$ if and only if $f\circ g=g\circ f=\text{Id}_X$

Title: $f(g(x))=x$ implies $f(x)=g^{-1}(x)$

Body:
Is it possible to find a necessary and sufficient condition to conclude when
$$f(g(x))=x \implies f(x)=g^{-1}(x) \wedge f^{-1}(x)=g(x),$$
if both functions are well defined?</p>


问:
Finding the key matrix of a 2x2 Hill Cipher
I'm trying to find the Hill Cipher key from the following given info:

(1,3)^T is encrypted as (-9, -2)^T, and (7, 2)^T is encrypted as (-2, 9)^T.

However I don't seem to end up with the correct answer as the encryption is invalid when I try to use the key. Below is my solution, did I make a mistake somewhere?

So

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix}\begin{bmatrix}1\\3 \end{bmatrix}= \begin{bmatrix}-9\\-2 \end{bmatrix}$$

and

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix}\begin{bmatrix}7\\2 \end{bmatrix}= \begin{bmatrix}-2\\9 \end{bmatrix}$$

or in one matrix notation:

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} \begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix}$$

which allows us to find the encryption matrix by

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}}^{-1}$$

The determinant of $\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}$ is $1\cdot 2 - 3\cdot 7 = 7 \pmod{26}$, so the inverse exists and equals

$$\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}$$

This allows us to compute the encryption matrix

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}}$$

$$
\begin{bmatrix}
-9*2+-2*-3=14 & -9*-7+-2*1=9\\
-2*2+9*-3=21 & -2*-7+9*1=23\end{bmatrix}
=
\begin{bmatrix} 14 & 9\\ 21 & 23\end{bmatrix}
$$

Then $$\begin{bmatrix} 14 & 9\\ 21 & 23\end{bmatrix}\begin{bmatrix}1\\3 \end{bmatrix}= \begin{bmatrix}15\\12 \end{bmatrix}$$ which does not match the given info as it should have been equal to

$$\begin{bmatrix}-9\\-2 \end{bmatrix}$$

答:
I fully agree with

$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}}^{-1}$$
and indeed the determinant of the right hand side matrix equals $$2- 21 \pmod{26} = -19 = 7 \pmod{26}$$ which is relatively prime to $26$ so has an inverse.

The general formula for an 2-by-2 inverse is:

$$A^{-1} = \begin{bmatrix}a & b\\c & d\end{bmatrix}^{-1} = \frac{1}{\det A}
\begin{bmatrix}d &-b\\-c & a\end{bmatrix}$$ so your inverse is wrong and all entries need to be multiplies by the invesre (modulo $26$) of the determinant $7$.

This inverse can be found by applying the extended Euclidean algorithm to $7$ and $26$ and we get the Bézout identity

$$1 = -11 \cdot 7 + 3\cdot 26$$

from which it follows that $$-11 \cdot 7 = 1 \pmod{26}$$

so that the inverse of $7$ is $-11 \equiv 15$.

So we multiply all elements of $$\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}$$ by $15$ to get the inverse matrix we're looking for (of course all modulo $26$)

and we get $$\begin{bmatrix} 4 & 18\\ 7 & 15\end{bmatrix}$$

and now you can do the multiplication from the first equation modulo 26:

$$\begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} \begin{bmatrix} 4 & 18\\ 7 & 15\end{bmatrix}$$ to find the encryption matrix $E$. I leave that final bit to you.

takeaway: division is multiplying by the inverse. The inverse is found by the extended Euclidean algorithm. For $n=26$ you could also find the table of inverses once (program or trial and error) and use them all the time.


问:
Frobenius series
Consider the differential equation

\begin{equation}
(1+x^3)y''+4xy'+y =0
\end{equation}

I need to find a lower bound on the radius of convergence for above equation at $x=0\ \&\ x=2$.

I wrote the solution in Frobenius form
\begin{equation}
y(x) =\sum_{k=0}^{\infty}a_k x^{r+k}
\end{equation}

Substituting back into the equation I got the following equations:

\begin{align}
a_0 r(r-1) &= 0\\
a_1 r(r+1 &= 0\\
a_2 (r+2)(r+1) +a_0(4r+1)  &=0\\
a_{k+3}(k+r+3)(k+r+2) + a_k (k+r-1)(k+r) + a_{k+1}(4k+4r+5) &=0 \quad \forall k\ge0
\end{align}

I try to use the fact that the necessary criterion for convergence of Frobenius series is

\begin{equation}
\left|\frac{xa_{k}}{a_{k-1}}\right|<1 \implies \left|x\right|<\left|\frac{a_{k-1}}{a_{k}}\right|
\end{equation}

For $r=0$, i get the recurrence relations as above. But can proceed on the bound.

Same thing for $r=1$, but I find only one parameter arbitrary instead of two even though this a second order ODE.

For $r=-1$, the solution is trivial. I'm stuck.

答:
Let $x=t-1$ ,

Then $(1+(t-1)^3)y''+4(t-1)y'+y=0$

$(1+t^3-3t^2+3t-1)y''+4(t-1)y'+y=0$

$(t^3-3t^2+3t)y''+4(t-1)y'+y=0$

$t(t^2-3t+3)y''+4(t-1)y'+y=0$

Not that
this belongs to a Heun's Equation and don't expect simple form of series solution.

Title: Frobenius series


问:
Projecting a surface segment of a cone onto a 2D plane?
Firstly, I'd like to apologise - I do not know the correct terms for what I am asking.

Assume that the top/bottom of the highlighted portion there is actually aligned with the base.

To help explain:
I need to wrap that section of the cone using a piece of paper. What shape (exactly) do I need to cut out from said paper so that it will wrap flawlessly?

答: If I've understood your problem correctly, I think this should help. This one's a right circular cone and its opened up paper version. Here we've taken an arbitrary curved surface of the form shown on the cone and visualized.


问:
If $n(A)=4, n(B)=5$, find number of functions A to B such that the range contains exactly 3 elements.
The domain is set A while the co domain is set B

For the given condition, two elements from the domain correspond to one element in the co domain, while the the other two get attached to separate elements.

This is one way I can think of fulfilling the required condition, I am not sure if it’s the only one.

As for the one given above, I don’t know how to count all of them, so perhaps there is some formula I am missing. Can I get a hint?

答:
Hints :

Read the question as

Find number of functions from $A=\{ 1,2,3,4 \}$ to $B=\{ 1,2,3,4,5 \}$ such that range has exactly three elements.

Define range $R$. There are $\binom{5}{3}$ ways to choose $3$ elements for $R$. Let $R := \{ a,b,c \}$

Find number of surjective functions from $A$ to $R$. Their cardinalities are small so you may take cases - $a$ is mapped twice rest once, $b$ is mapped twice rest once, etc.


问:
continuity of coefficients imply uniqueness in law of the solutions of an sde?

let's consider a simple sde of the form
$$dX_t=b(t,X_t)dt+\sigma(t,X_t)dW_t$$
for $t\leq T$ and $b$,$\sigma$ continuous in $(t,x)$, $t \geq 0, x \in \mathbb{R}$.

My question is: does uniqueness in law hold?

Thanks in advance

答:
Unfortunatly, no. The generic counterexample is
\begin{equation}
dX_{t}=|X_{t}|^{\alpha}dB_{t},\quad X_{0}=0
\end{equation}
with $\alpha\in(0,1/2)$. Then, $0$ gives one solution and another can be constructed by time-change. Take a look e.g. into the book 'Singular Stochastic Differential Equations' of Cherny et.al.


问:
Prove by induction $n!<4n^2+3$
How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?

Here´s my try:

The equation is valid for every $n\leq4$.

n=5: n!=5!=120>103

n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$

Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.

Any ideas how i could continue or whats wrong?

答:
Here is a variant without induction.

For $n\ge 5\ $ then $\, n!=n(n-1)\underbrace{(n-2)(n-3)}_{\ge 6}\underbrace{\cdots 1}_{\ge 1}\ge 6n(n-1)$

Let $f(n)=6n(n-1)-(4n^2+3)=2n^2-6n-3$

$f'(n)=4n-6$ so $f\ \nearrow$ for $n\ge \frac 32$ and since $f(5)=17>0$ then $n!>4n^2+3$ for $n\ge 5$

Title: Prove by induction $n!<4n^2+3$


问:
If $\alpha$ is a Pisot number, is $\alpha_i^k$ a conjugate of $\alpha^k$?
Let $\alpha>1$. $\alpha$ is a Pisot number of degree $d$ if $\alpha$ is an algebraic integer of degree $d$ and all its conjugates $\alpha_2,\dots,\alpha_d$ other than $\alpha$ all have modulus less than 1.

Wikipedia claims that if $\alpha$ is a pisot number and $k\in\mathbb{N}$, then $\alpha^k$ is also Pisot. My question is, how to show this? Will it imply that the conjugates of $\alpha^k$ are $\alpha_i^k$'s?

答: Yes, the conjugates of $\alpha^k$ are the $\alpha_i^k$. The conjugates of $\alpha$ are its images under the Galois group $\operatorname{Gal}(\overline{\mathbb Q}/ \mathbb Q)$. (Or under $\operatorname{Gal}(L/\mathbb Q)$ for any Galois extension $L/\mathbb Q$ containing $\alpha$.) If $\sigma$ is a field automorphism, $\sigma(\alpha^k) = \sigma(\alpha)^k$.


答: Since the distance from the apex to the base is constant. Assuming the cone is made of paper,  cutting it open along your yellow line will make a circular sector whose radius is the former distance. Your cross-sections will correspond to bands concentric with the sector's circumference.</p>

Title: Projecting a surface segment of a cone onto a 2D plane?

Body:
Firstly, I'd like to apologise - I do not know the correct terms for what I am asking.

Assume that the top/bottom of the highlighted portion there is actually aligned with the base.

To help explain:
I need to wrap that section of the cone using a piece of paper. What shape (exactly) do I need to cut out from said paper so that it will wrap flawlessly?


问:
Rotating angular velocities
I have a quadcopter with yaw, pitch, roll and their respective derivatives p, q and r. Although the orientation is already defined in the world frame, the derivatives are not. This may sound weird, so let me illustrate:

In (A), the drone is aligned with the world frame axes so that the yaw, pitch, and roll are all $0 °$. In (B), the drone has rotated about the z-axis so that its yaw is now $90 °$. Now imagine the drone starts tilting forward (i.e. pitching) from orientation B.

In drone frame, this corresponds with pitch movement with some nonzero p

In world frame, this should correspond with roll movement with some nonzero q as result

So given the orientation, how do I compute the p,q,r in the world frame? How do I rotate angular velocity?

答:
I have found the result, you do it in exactly the same way as you rotate positions. So

[p',q',r'] = [p,q,r] * R(yaw,pitch,roll)

Title: Rotating angular velocities


问:
Stability of differential equations under nonlinear pertubations
I am trying to understand the proof of the following theorem :

Let $A$ be an $n\times n$ matrix such that it has only negative eigenvalues and let $f: \mathbb{R}\times \mathbb{R}^n$ continuous and locally Lipschits in $x$. If $f(t,0)=0$ for every $t\in \mathbb{R}$ and $\lim_{x\rightarrow 0}\frac{||f(t,x)||}{||x||}=0$ for any $t\in \mathbb{R}$ then the zero solution of the equation $x'=Ax+f(t,x)$ is asymptotically stable .

Now in the proof of this, the authors claim that given $x_0\in \mathbb{R}^n$ with $||x_0||<\delta$ the solution satisfies $||x(t)||<\delta $ for any $t\in [t_0,t_1]$ sufficiently small. And then they use that fact to prove that if $||x_0||<\frac{\delta}{M}$ then $||x(t)||<\delta$ using the following
\begin{align}
x(t)&=e^{A(t-t_0)}x_0+\int_{t_0}^t e^{A(t-s)}f(s,x(s))ds
\\
\|x(t)\|&\leq Me^{-\alpha(t-t_0)}\|x_0\|+\int_{t_0}^tMe^{-\alpha(t-s)}\epsilon \|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}\|x_0\|+\int_{t_0}^t M\epsilon e^{\alpha s}\|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}e^{M\epsilon (t-t_0)}
\\[.5em]
\|x(t)\|&\leq Me^{(-\alpha +M\epsilon) (t-t_0)}\|x_0\|
\end{align}
I am just a bit confused why we are doing this for $||x_0||<\frac{\delta}{M}$ since we know that if $||x_0||<\delta$ then $||x(t)||<\delta$. I thought this might be an attempt to enlarge the interval where we have that $||x(t)||<\delta$, to use the fact that $|f(s,x(s)|<\epsilon||x(t)||$ we need the fact that $||x(t)||<\delta$ so we stay in the same interval.
Can anyone help me figure out why we care that $||x_0||>\frac{\delta}{M}$? Thanks in advance.

答:
The logic of this kind of continuation argument goes like this. What has been shown is that supposing $\|x_0\|<\frac\delta M$, if $\|x(t)\|<\delta$ for all $t$ in any interval $[t_0,T]$ with $T>t_0$, then $\|x(t)\|\le M\|x_0\|$ for all such $t$.

The number $M\|x_0\|$ is strictly less than $\delta$, so whatever $T$ you have, by continuity it can be made a bit larger. To be precise, let $T_*$ be the supremum of the set $S$ consisting of those $T\ge t_0$ such that $\|x(t)\|<\delta$ for all $t\in[t_0,T]$. Then it must be that $T_*=\infty$, since otherwise we infer $\|x(T_*)\|\le M\|x_0\|<\delta$ by continuity, so $T_*\in S$, and (again by continuity) there is some $T>T_*$ with $T\in S$.


问:
What does it mean to be divergence thorem applicable?
To be able to use divergence theorem, $S$ needs to be piecewise smooth curved, and $\vec{F}(x,y,z)$ needs to be continuously differentiable.

$$\int \int_{S} \vec{F} * \vec{n} dS$$ where

$$\vec{F}(x,y,z) = <yz\tan^{-1}(y^2 + z^2) - 2x, xze^{x-z} - 2y, \sin(x^2+y)+5z>$$
and S is the sphere of radius 1 centered at the origin.

How do I check if they are piecewise smooth curved without looking at the graph? And how do I check if $\vec{F}(x,y,z)$ is continuously differentiable?

Could someone show steps to checking?

答:
With regards to your question about why $F$ is differentiable.

It could depend on the development in a course or textbook. This makes the question more difficult. I may make explicit and implicit assumptions as I go. A lot of details may need to be filled in, but I hope you find it a decent outline of some main ideas that could occur in some particular approach to the problem.

I will assume that we know that functions such as

\begin{align}
\pi_1:\mathbb R^3&\to\mathbb R\\
(x,y,z)&\mapsto x
\end{align}

are continuously differentiable. For instance, consider the definition at http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions

I am going to replace the word continuously differentiable with morphism, and not worry if I am justified in this.

I will assume that the product of two morphisms is also a morphism.

I will assume that if $f$, $g$ and $h$ are morphisms from $X$ to $\mathbb R$, then the map

\begin{align}
F:X&\to\mathbb R^3\\
x&\mapsto(f(x),g(x),h(x))
\end{align}

is also a morphism.

Thus, you may note that each of
\begin{align}
f,g,h:\mathbb R^3&\to\mathbb R\\
(x,y,z)&\overset{f}\mapsto yz\tan^{−1}(y^2+z^2)−2x,\\
(x,y,z)&\overset{g}\mapsto xze^{x−z}−2y,\\
(x,y,z)&\overset{h}\mapsto \sin(x^2+y)+5z
\end{align}
is a morphism.

Finally, we note that $F=(f,g,h)$ is a morphism.


问:
Calculating the limit of the following series
I want to prove that $$\lim_{x\to1^-}(1-x)\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} = \frac14$$

So far I tried to manipulate the series for instance using $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} =  -\sum_{n=1}^{\infty}(-1)^{n}nx^n\sum_{m=0}^{\infty}\left(x^{2n}\right)^m$$ since $x < 1$. Interchanging the two sums (not sure if allowed) I obtained, assuming I did not make mistakes, the sum $$\sum_{m=0}^{\infty}\frac{x^{2m+1}}{(1+x^{2m+1})^2}$$
I am unable to continue from this point. Perhaps my work isn't actually useful at all. Can you help me?

答:
Your computation is very close to the answer. If we denote the sum by $S(x)$, then

$$ S(x)
= (1 - x) \sum_{m=0}^{\infty} \frac{x^{2m+1}}{(1+x^{2m+1})^2}
= \frac{1}{1+x} \sum_{m=0}^{\infty} \frac{x^{2m+1} - x^{2m+3}}{(1 + x^{2m+1})^2}.
$$

Now the idea is that $S(x)$ can be regarded as a Riemann sum for $\frac{1}{(1+t)^2}$ over $[0, 1]$. To make use of this idea, note that $t \mapsto \frac{1}{(1+t)^2}$ is decreasing for $t \geq 0$, and so

$$ x^2 \int_{x^{2m+1}}^{x^{2m-1}} \frac{\mathrm{d}t}{(1+t)^2}
\leq \frac{x^{2m+1} - x^{2m+3}}{(1 + x^{2m+1})^2}
\leq \int_{x^{2m+3}}^{x^{2m+1}} \frac{\mathrm{d}t}{(1+t)^2}. $$

Summing this over $m = 0, 1, 2, \dots$, we obtain

$$ \frac{x^2}{1+x} \int_{0}^{1/x} \frac{\mathrm{d}t}{(1+t)^2} \leq S(x) \leq \frac{1}{1+x} \int_{0}^{x} \frac{\mathrm{d}t}{(1+t)^2}. $$

Therefore, letting $x \to 1^-$ yields

$$ \lim_{x \to 1^-} S(x) = \frac{1}{2} \int_{0}^{1} \frac{\mathrm{d}t}{(1+t)^2} = \frac{1}{4}. $$


问:
Proof of inversion formula in Probability
Theorem : Let $X$ be a real random variable such that $\phi_X \in L^1$ i.e $\int_{\mathbb{R}} \vert \phi_X(t) \vert < \infty$, then $X$ has density $f_X(x) \in C_b(\mathbb{R})$ given by $$f_X(x) = \frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt$$

(Where $\phi_X$ denotes the characteristic function of $X$)

The proof articulates in two parts : The first where we assume we already know $X$ has a density $f$ and we prove the inequality above, the second where we don't know such $f$ exists and we lead back to the first case using the trick of taking $X+\epsilon N$ where $N \sim N(0,1)$.

The sketch of the first part of the proof, which is where my problems are is the following : We consider a $g \geq 0, g \in C_{b}(\mathbb{R})$, such that $g = 0$ outside a compact, i.e $g \in C_{c}(\mathbb{R})$ and we apply the isometry lemma to $f_X + g$ which leads us to (Using Fubini-Tonelli and the fact that $\frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt$ is real)

$$\int g(x)f_X(x)dx = \mathbb{E}[g(X)] = \int g(x)[\frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt]dx$$

Which concludes the proofs for $g$ assumed as mentioned at the beginning.
But how to extend the result to just continuos and bounded functions ?

During this next proof the following three observation are made :

$1) \frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt$ is continuos in $x$ (and I was able to prove it as a consequence of dominated convergence theorem, prooving continuity by sequences).

$2) \frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt \geq 0$

3)$\hspace{0.1cm} \exists \hspace{0.1cm} g_n \uparrow 1 $ with $g_n \in C_b(\mathbb{R})$

I was able to prove step $2)$ in the following way with an hint which I was unable to prove (so this should be the way to prove it). The proof goes like this : if it was strictly negative, by continuity (prooved in the first observation) exists an open set $U$ where the function is still strictly negative, and then I can find a $g > 0, g \in C_b(\mathbb{R})$ defined on $U$ and conclude by contradiction thanks to the fact that in this case we would have $0 > \frac{1}{2\pi}\int_{\mathbb{R}}\phi_X(t)e^{-itx}dt = E[g(X)] > 0$, thanks to $g$.

As far concerned the third point I don't know where to start, I thought it could be useful to use a simplified version of Urysohn Lemma but unsuccefully.
I was unable to prove the highlighted sentences, there are any way to explicit such functions ? Any direct proof, explicit or not, would be appreciated, and some reference of using Fourier trasfrom in Probability as well, those seemed nice tricks to know.

答:
You do not need to assume that $g$ has compact support in the first part, I believe.

You can use a trick very similar to what you propose in the beginning to show that
\begin{align}
\lim_{n\to\infty}f_{X + \frac{1}{n}N}(t) = \frac{1}{\sqrt{2\pi}}\int_{R}e^{-\text{i}ts}\varphi_{X}(s)\text{d}s = f(t)
\end{align}
where $N\sim N(0,1)$. Thus it follows, that $f$ is strictly positive and measurable, and we can therefore consider the measure $\nu$ with density $f$.

Now, if we can show that
\begin{align}
\int_{\mathbb{R}}\psi \text{d}\nu = \int_{\mathbb{R}}\psi \text{d}\mathbb{P}_{X}, \quad \psi \in C_{c}(\mathbb{R})
\end{align}
then $\nu = \mathbb{P}_{X}$, which would imply that $f = f_{X}$, as desired. But showing that the integrals are equal is simply using dominated convergence on the integrals wrt. $\mathbb{P}_{X+\frac{1}{n}N}$.


问:
Invertiblity of ST and TS for linear maps
Could somebody explain why for any given linear maps S and T over a finite dimensional vector space V, ST is invertible if and only if TS is? Why is it so important that V is finite dimensional for this to hold true?

Edit: I am looking for an approach without using matrices or determinants.

答:
If $V$ is finite- dimensional, we can use determinants:

$$\det(ST)=\det(S) \det(T) = \det(TS).$$

$ST$ is invertible $ \iff det(ST) \ne 0 \iff \det(TS) \ne 0 \iff TS $ is invertble.

Now let $W=\{(x_n): x_n \in \mathbb R, n \in \mathbb N\}.$

$W$ is a vector space with $\dim W= \infty$.

Let $T,S:W \to W$ be defined by

$$ T(x_1,x_2,....)=(0,x_1,x_2,...)$$

and

$$S(x_1,x_2,...)=(x_2,x_3,...).$$

It is your turn to show that $ST$ is invertible, but $TS$ is not invertible.


问:
Approximation for Beta distribution
I've heard that, when the parameters $\alpha$ and $\beta$ of the beta distribution approach infinity, the distribution becomes approximately normal. But I have only seen a proof of such fact for the case $\alpha = \beta$.

Is it true that beta distributions with large parameters are
approximately normal even when the parameters are not equal? If so,
how big have the parameters to be for the approximation to be good?

I am willing to use this (or any other approximation that simplifies the calculations) with values $\alpha \ge 100$ and $10^4\alpha \ge \beta \ge 10\alpha$.

答:
If I poperly remember, if you have a beta distribution with large values of parameters $(\alpha,\beta)$, it almost equivalent to a normal distibution with parameter
$$\frac{\alpha  }{\alpha +\beta} \sqrt{\frac{\alpha  \beta }{(\alpha +\beta )^2 (\alpha +\beta
   +1)}}$$ I think that this comes from Taylor expansions of the beta distribution probability density function.

Use a mean of $\frac\alpha{\alpha+\beta}$ and a variance of $\frac{\alpha\beta}{(\alpha+\beta)^{2} (1+\alpha+\beta)}$

Title: Approximation for Beta distribution


问:
Jacobian of non-square matrices
Let $m>n$ and $L\colon \mathbb{R}^m \to \mathbb{R}^n$ be a linear map(=matrix). The "$n$-dimensional Jacobian"
$$
J^n(L) = \sqrt{\det(LL^t)}
$$
is of geometric significance! Let's say it makes the co-area formula possible.

Suppose $C\colon \mathbb{R}^m \to \mathbb{R}^m$ is an invertible linear map (think of it as a change of coordinates). I am interested in the precise formula relating $J^n(L)$ and $J^n(L \circ C)$.

For simplicity assume $L$ is surjective, i.e. it has rank $n$. Denote by $C^{ker}$ the restriction of $C$ to the $(m-n)$-dimensional $(\ker L\circ C)$. So,  $C^{ker}\colon (\ker L\circ C) \to \ker L$ can be seen as a map $\mathbb{R}^{m-n} \to \mathbb{R}^{m-n}$. I think the answer should be:
$$
J^n(L) = \frac{|\det C^{ker}|}{|\det C|}\, \cdot J^n(L \circ C)\, ,
$$
Because for certain $C$ I do have a proof, albeit quite a tricky one.

Questions:

Is the claimed identity true?

If not, what is the right formula?

How does one prove it?

答:
This turned out a bit longer than expected and I'm not sure it answers your questions but it was helpful for me to think about this; so here we go:

Suppose $u\colon (M,g) \to (N,h)$ is either an immersion or submersion (so that its differential has full rank) where $g$ and $h$ are Riemannian metrics. We will denote coordinates on $M$ and $N$ by $(x^1,\ldots,x^m)$ and $(y^1,\ldots,y^n)$, respectively.
In these local coordinates we interpret $\mathrm{d}u_x$ as an $(n\times m)$-matrix.
In particular, its columns can be interpreted as (the coordinate representation of) the
push-forward by $u$ of the coordinate vector fields $\tfrac{\partial}{\partial x^1},\ldots,\tfrac{\partial}{\partial x^m}$, i.e. the $j$-th column of $\mathrm{d}u_x$ is (the coordinate representation of)
\begin{align*}
 u_{\#}\big(\tfrac{\partial}{\partial x^j}\big|_x\big)
  = \mathrm{d}u_x\big(\tfrac{\partial}{\partial x^j}\big|_x\big)
  = \tfrac{\partial u}{\partial x^j}\big|_x.
\end{align*}
Its rows can be interpreted as (the coordinate representation of) the pullback by $u$ of the coordinate covector fields $\mathrm{d}y^1,\ldots, \mathrm{d}y^n$, i.e. the $i$-th row of $\mathrm{d}u_x$ is (the coordinate representation of) the differential of the $i$-th coordinate component of $u$, or in other words the linear form which maps $v\in T_xM$ to
\begin{align*}
 u^{\#}(\mathrm{d}y^i)|_x(v)
 = \mathrm{d}y^i|_{u(x)}(\mathrm{d}u_x(v))
 = \mathrm{d}(y^i\circ u)_x(v)
 = \mathrm{d}u^i_x(v).
\end{align*}
Note that if $m<n$ then everything is independent of $g$ and we only need $g$ here to connect it to your formalism! Likewise, if $m>n$ then everything is independent of h!

The Jacobian that appears in the area formula is then
\begin{align*}
 J_m(u)(x) = \sqrt{\operatorname{det}(\mathrm{d}u_x^\ast \mathrm{d}u_x)}
\end{align*}
where $\mathrm{d}u_x^\ast\colon T_{u(x)}N \to T_xM$ is the adjoint of $\mathrm{d}u_x$ with respect to the inner products $g_x$ and $h_{u(x)}$.
The Jacobian in the coarea formula is
\begin{align*}
 J^n(u)(x) = \sqrt{\operatorname{det}(\mathrm{d}u_x \mathrm{d}u_x^\ast)}.
\end{align*}
In the case $m<n$ what this formula really says is that we need to take the determinant of the Gramian matrix of $(u^{\#}h)_x$ -- the pullback of $h$ by $u$ -- with respect to the coordinate basis above, i.e.
\begin{align*}
 J_m(u)(x)
  = \sqrt{\operatorname{det}\Big( (u^{\#}h)_x\Big(\tfrac{\partial}{\partial x^i}\big|_x, \tfrac{\partial}{\partial x^j}\big|_x\Big) \Big)}
  = \sqrt{\operatorname{det}\Big( h_{u(x)}\Big(\mathrm{d}u_x\big(\tfrac{\partial}{\partial x^i}\big|_x\big), \mathrm{d}u_x\big(\tfrac{\partial}{\partial x^j}\big|_x\big)\Big) \Big)}
\end{align*}
-- here we see that everything is independent of $g$.

In the case $m>n$ it says that we need to calculate the determinant of the Gramian matrix of the induced inner product $\bar{g}_x$ (given by the inverse matrix of $g_x$) on the cotangent space with respect to the basis of pullback covector fields $u^{\#}(\mathrm{d}y^i)$:
\begin{align*}
 J^n(u)(x)
  = \sqrt{\operatorname{det}\big( \bar{g}_x\big(u^{\#}(\mathrm{d}y^i)\big|_x, u^{\#}(\mathrm{d}y^j)\big|_x\big) \big)}
  = \sqrt{\operatorname{det}\big( \bar{g}_x\big(\mathrm{d}u^i_x, \mathrm{d}u^j_x\big) \big)}
\end{align*}
-- here we see that everything is independent of $h$.

Reformulating this again in terms of the exterior product, we can also write for $m<n$
\begin{align*}
 J_m(u)(x)
  = \Big| \mathrm{d}u_x\big(\tfrac{\partial}{\partial x^1}\big|_x\big) \wedge \cdots \wedge \mathrm{d}u_x\big(\tfrac{\partial}{\partial x^m}\big|_x\big) \Big|_{h_{u(x)},\bigwedge^m T_{u(x)}N }
\end{align*}
where the norm on the exterior product is the one induced by the inner product $h_{u(x)}$ (defined exactly as the determinant of its Gramian matrix above).
In this case, we can use the Plücker embedding to interpret the $m$-vector $\mathrm{d}u_x\big(\tfrac{\partial}{\partial x^1}\big|_x\big) \wedge \cdots \wedge \mathrm{d}u_x\big(\tfrac{\partial}{\partial x^m}\big|_x\big)$ as representing the tangent space $T_{u(x)}u(M)$ of the immersed manifold $u(M)$ as an element of the Grassmannian $\operatorname{Gr}_m(T_{u(x)} N)$. Its norm is then the infinitesimal area element at $u(x)$.

And similarly for the case $m>n$ (if we write $\mathrm{d} u^i_x = (\mathrm{d}u_x)^{\#}(\mathrm{d} y^i|_{u(x)})$)
\begin{align*}
 J^n(u)(x)
  = \Big| (\mathrm{d}u_x)^{\#}\mathrm{d} y^1|_{u(x)} \wedge \cdots \wedge (\mathrm{d}u_x)^{\#}\mathrm{d} y^n|_{u(x)} \Big|_{\bar{g}_{x},\bigwedge^n T^\prime_{x}M }.
\end{align*}
Again we can interpret the $n$-covector as an element in the (co-?)Grassmannian $\operatorname{Gr}_n(T^\prime_{x} M)$. Note the role reversal of $M$ and $N$ here. I'm a little less sure of its geometric interpretation in this case.

As one last step, note that if $m<n$ any linear map $A\colon V\to W$ induces a linear map $\bigwedge^m A\colon \bigwedge^m V \to\bigwedge^m W$ by
\begin{align*}
 (\textstyle\bigwedge^mA)(v_1\wedge\cdots\wedge v_m) = Av_1\wedge\cdots\wedge Av_m.
\end{align*}

This operation respects composition, i.e. $\bigwedge^m(A\circ B)=\bigwedge^mA\circ \bigwedge^mB$ (it is a covariant functor).
Furthermore, if $\operatorname{dim}(V)=m$ and $V=W$ then this gives a coordinate-free definition of the determinant of a matrix, i.e. if $A\colon V\to V$ then $\bigwedge^mA=\operatorname{det}(A)$. (See for example [Flanders, Differential forms with applications to the physical sciences]).
Similarly for $m>n$, we can formulate such operations on the dual spaces to get a similar operation for pullbacks by linear maps which reverses composition (it is a contravariant functor): if $A\colon V\to W$ then $(\bigwedge^n A)^{\#}\colon \bigwedge^n W^\prime \to \bigwedge^n V^\prime$ is defined by
\begin{align*}
 (\textstyle\bigwedge^n A)^{\#}(\omega_1\wedge\cdots\wedge \omega_n) = A^{\#}\omega^1 \wedge \cdots \wedge A^{\#}\omega^n
\end{align*}
and $(A^{\#}\omega)(v) = \omega(Av)$ for all $v\in V$.
Again, we can reformulate for $m<n$
\begin{align*}
 J_m(u)(x)
  = \Big| \big(\textstyle\bigwedge^m\mathrm{d}u_x\big)\big(\tfrac{\partial}{\partial x^1}\big|_x \wedge \cdots \wedge \tfrac{\partial}{\partial x^m}\big|_x\big) \Big|_{h_{u(x)},\textstyle\bigwedge^m T_{u(x)}N }
\end{align*}
and for $m>n$
\begin{align*}
 J^n(u)(x)
  = \Big| \big(\textstyle\bigwedge^m\mathrm{d}u_x\big)^{\#}
   \big(\mathrm{d} y^1|_{u(x)} \wedge \cdots \wedge \mathrm{d} y^n|_{u(x)} \big)
   \Big|_{\bar{g}_{x},\textstyle\bigwedge^n T^\prime_{x}M }
\end{align*}
Thus, finally we can compute how the Jacobian behaves under change of coordinates. Let $\tau\colon M\to M$ be a diffeomorphism. Then $\mathrm{d}\tau_x\colon T_xM \to T_{\tau(x)}M$ is a linear map between $m$-dimensional spaces but  -- as user1551 has pointed out -- not an endomorphism. Suppose $(x^1,\ldots, x^m)$ and $(\tilde{x}^1, \ldots, \tilde{x}^m)$ are local coordinates around $x$ and $\tau(x)$. Then we can define a determinant of $\mathrm{d}\tau_x$ as the determinant of the endomorphism $\mathrm{d}(\tilde{x} \circ\tau\circ x^{-1})_x\colon \mathbb{R}^m\to\mathbb{R}^m$ which is independent of the chosen local coordinates (take, for example, orthonormal frames). With this interpretation we have
$\bigwedge^m\mathrm{d}\tau_x = \operatorname{det}(\mathrm{d}\tau_x)$.
Then we have in the case $m<n$
\begin{align*}
 J_m(u\circ \tau)(x)
  &= \Big| \big(\textstyle\bigwedge^m\mathrm{d}u_{\tau(x)} \circ \textstyle\bigwedge^m\mathrm{d}\tau_x\big)\big(\tfrac{\partial}{\partial x^1}\big|_x \wedge \cdots \wedge \tfrac{\partial}{\partial x^m}\big|_x\big) \Big|_{h_{u(\tau(x))},\textstyle\bigwedge^m T_{u(\tau(x))}N } \\
  &= |\operatorname{det}(\mathrm{d}\tau_x)| \cdot \Big| \big(\textstyle\bigwedge^m\mathrm{d}u_{\tau(x)}\big)\big(\tfrac{\partial}{\partial x^1}\big|_{\tau(x)} \wedge \cdots \wedge \tfrac{\partial}{\partial x^m}\big|_{\tau(x)}\big) \Big|_{h_{u(\tau(x))},\textstyle\bigwedge^m T_{u(\tau(x))}N } \\
  &= |\operatorname{det}(\mathrm{d}\tau_x)| \cdot J_m(u)(\tau(x)).
\end{align*}
For the case $m>n$ we make the same calculation
\begin{align*}
 J^n(u\circ \tau)(x)
  &= \Big| \big(\textstyle\bigwedge^m\mathrm{d}(u \circ \tau)_x\big)^{\#}
   \big(\mathrm{d} y^1|_{u(\tau(x))} \wedge \cdots \wedge \mathrm{d} y^n|_{u(\tau(x))} \big)
   \Big|_{\bar{g}_{x},\textstyle\bigwedge^n T^\prime_{x}M } \\
  &= \Big| \big(\textstyle\bigwedge^m\mathrm{d}\tau_x\big)^{\#} \circ
    \big(\textstyle\bigwedge^m\mathrm{d}u_{\tau(x)}\big)^{\#}
   \big(\mathrm{d} y^1|_{u(\tau(x))} \wedge \cdots \wedge \mathrm{d} y^n|_{u(\tau(x))} \big)
   \Big|_{\bar{g}_{x},\textstyle\bigwedge^n T^\prime_{x}M } \\
  &= |\operatorname{det}(\mathrm{d}\tau_x)| \cdot J^n(u)(\tau(x)).
\end{align*}

Title: Jacobian of non-square matrices


问:
Is there really no analogue of the derivative product rule for integrals, or we just haven't found one yet?
For a product rule for integrals, I am not talking about integration by parts. That particular formula uses a integral of products inside the formula itself. The derivative product rule does not use a derivative of products inside the formula. I have never seen a book that proves that there is in fact no analogous formula for the integral of a product. Some books prove that there is no formula for the roots of a general 5th degree or higher polynomial function. Can someone prove that there is no such formula for the integral of a product? There is a formula for the integral of a sum, so maybe someone will discover that there is a formula for the integral of a product. I apologize if my notion of formula is not precise enough, but maybe there is a book that precisely defines what a formula is.</p>

答:
I thought there was already such a question here, but I did not find it.

As the comments say, there is no simple formula for $\int f g dx$ in terms of $\int f dx$ and $\int g dx$.  There are lots of ways to see this.

(A)
$$
\int x\;dx\quad\text{and}\quad \int\frac{1}{x^2}\;dx\quad
\text{are rational functions, but}\quad \int\frac{1}{x}\;dx\quad\text{is not} .
$$
(B)
$$
\int x e^{x^2}\;dx\quad\text{and}\quad \int\frac{1}{x}\;dx\quad
\text{are elementary functions, but}\quad\int e^{x^2}\;dx\quad\text{is not} .
$$
You come up with what "simple formula" means, then there should be an example
like these using that notion of "simple formula".


问:
Elements in a commutative ring ( Zero divisors , Unit)
This question was part of my algebra quiz and I was unable to solve 2 options out of 4. So, I am asking it here for help.

Let R be a finite non-zero commutative ring with unity. Then which of the sattements are true?

Any non zero element of ring is either a unit or a zero divisor.

There may exist an element which is neither a ring nor a zero divisor.

I studied algebra from Joseph Gallian.

Can you please tell how should I approach this particular problem as I am really confused if there exists an element which is neither unit nor zero divisor.

I can't think of any idea.

Thanks

答:
It's critical here that the ring is finite. Indeed, let $a \in R$ be nonzero. If $a$ is not a zero divisor then the map $R \longrightarrow R$ via $x \mapsto ax$ is injective. As $R$ is finite, it is therefore a bijection by the Pigeonhole Principle. Hence, there is some $x \in R$ such that $ax = 1$, i.e. $a$ is a unit.

Note that this is false in general for infinite rings. For instance, $2 \in \mathbb Z$ is not a zero divisor but $1/2$ is not an integer, so 2 is not a unit in.$\mathbb Z$.


问:
Prove that vector $[3,2]$ is tangent to the set $\{(x,y): x^2 - y^3 = 0\}$ at point $(1,1)$.
As in title.

Prove that vector $[3,2]$ is tangent to the set $\{(x,y): x^2 - y^3 = 0\}$ at point $(1,1)$.

I start from 'differentiating' condition defining the set: $2x - 3y^2 = 1$.

Then $2(3-1) - 3(2-1)^2 = 1$.

Is it right justification?

答:
The kernel of a function $f$ from $\mathbb{R}^n \to \mathbb{R}$ (called a "scalar function"), can be visualized as an embedded submanifold of dimension $n$ inside $(n+1)$-dimensionsal space. This $(n+1)$-dimensionsal space, here, is the direct product (combination, addition) of the input space (here $\mathbb{R}^n$) and the output space (here $\mathbb{R^1}$).

Another representation of a scalar function is as colored sets inside $\mathbb{R}^n$; where negative output values are mapped to blue (darker as you approach negative infinity), the kernel is mapped to white, and positive values to red (darker as you approach positive infinity). A certain shade of color corresponds precisely to what is called a level set, the solution to an equation of the form $f(u) = C$. Your kernel is specifically the level set for $C=0$. A level set is a manifold of dimension $n-1$ in this context.

Inside this representation in $\mathbb{R}^n$, the gradient at a given point, geometrically, is always perpendicular to the level set on which it starts. Proving that a certain vector is tangent to a level set is then simply a question of proving that its dot product with the gradient is null. That is the justification you are looking for. Why ? Because the tangent space at a point on a manifold is always just the orthogonal complementary of the $1$-dimensional subspace generated by the gradient (in this configuration of a $(n-1)$-dimensional manifold in $\mathbb{R}^n$).

So, now for some calculation, your gradient is: $\nabla f(x, y) = [2x \space \space -3y^2]$. This means $ \nabla f(1, 1) = [2 \space \space -3]$. And $<[2 \space \space -3], [3 \space \space 2]> = 0$, so indeed the vector $[3 \space \space 2]$ is tangent to the level set defined by the kernel.

Finally, some visuals:

Fig 1: General point of view of your manifold, the $(n+1)$-dimensional view. Your kernel is the curved black line. The grey plane is the input space; the vertical blue axis is the output space. The gradient at $[1 \space 1]$ is the black vector showing the direction of greatest increase on the blue manifold. The other vector is clearly tangent to the kernel, and perpendicular to the gradient.

Fig 2: Top-down view, projecting the output space to zero, to give you an idea of the $n$-dimensional visualization. Imagine parts below the plane are in blue, and parts above a red, the black curve is white, and the further away from the plan points are, the darker they get.


问:
About norm on polynomials
Is it possible to construct a norm on $\mathbb{C}[x,y]$ that satisfies the following condition: There exists a $C>0$ such that for any two polynomials $p,q$:

$||p(x,y) q(x,y)||<=C||p(x,y)||\cdot||q(x,y)||$

At first I thought this would follow from Cauchy-Swartz, but I'm talking about polynomial multiplication on the left side, so no inner product. If this is not possible, would it be possible to choose a norm such that $C$ that only depends on $q$?

答:
The sup/max-norm trivially fulfills this:
$$
\|p\| := \max_{|x|\le1} |p(x)|.
$$
This is a norm on these polynomials. Because for polynomials  $\|p\|=0$ implies $p=0$.
Then clearly for $x\in \mathbb C^2$ with $|x|\le 1$
$$
|p(x)q(x)|\le |p(x)| \cdot |q(x)| \le \|p\|\cdot \|q\|.
$$

Title: About norm on polynomials


问:
The role of Tangent lines in the method of Lagrange Multipliers
I have been studying the method of Lagrange multipliers for a machine learning course. From the following sources:

https://en.wikipedia.org/wiki/Lagrange_multiplier#Single_constraint

https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/lagrange-multipliers-and-constrained-optimization/v/meaning-of-lagrange-multiplier

https://www.byteofmath.com/the-math-behind-support-vector-machines/

I have been able to achieve a core understanding of the method. However this one source:

https://medium.com/@andrew.chamberlain/a-simple-explanation-of-why-lagrange-multipliers-works-253e2cdcbf74

makes it sound as though I can use comparisons between the slope of the tangent line of the constraint curve $g(x,y,z,..) = 0$ and the slope of the tangent line of the level curve $f(x,y,z..) = d$ at points where these curves intersect to determine which direction along the constraint curve I should move in order to reach a max or min of the surface $f(x,y,z,...)$ subject to the constraint curve $g(x,y,z,..) = 0$.

To quote the source:

"Imagine hiking from left to right on the constraint line. As we gain elevation, we walk through various level curves of f. ...~... .At each level curve, imagine checking its slope — that is, the slope of a tangent line to it — and comparing that to the slope on the constraint where we’re standing.

If our slope is greater than the level curve, we can reach a higher point on the hill if we keep moving right. If our slope is less than the level curve — say, toward the right where our constraint line is declining — we need to move backward to the left to reach a higher point."

Figure 1 in the Wikipedia source seems to give a counter example to this (if I am interpreting things correctly) hence my confusion. It would be a great help if this role of the tangent line in Lagrange Multipliers can be confirmed and the math behind it shown. Moreover, if this is confirmed, I would like clarification on whether the sign of the slopes matters (Which I would think that it does) or if its just the magnitude.

I realize my question is similar to and contains the same subject matter as these questions:

Lagrange multipliers tangency

Lagrange multiplier parallel slope intuition confusion?

but I'm going to post this question anyway

Thank you

答:
I am by no means qualified to answer your question. However, I was struggling with the same problem and would like to share what I found:

First of, let's get some intuition with the Lagrange formula, in the bivariate case.
We know that the function $g(x,y) = 0$
That is why we can rewrite $$f(x,y) = f(x,y) - \lambda 0 = f(x,y) - \lambda g(x,y)$$
This last equation is what we recognize as the Lagrange formula:
$$ \mathcal{L}(x,y) = f(x,y) - \lambda g(x,y) $$
As a result, maximizing $f(x,y)$ is no different than maximizing the Lagrangian
$$ D\mathcal{L}(x,y) = 0 \\ \Rightarrow Df(x,y) - \lambda Dg(x,y) = 0 \\ Df(x,y) = \lambda Dg(x,y) $$
We see that $Dg(x,y) $ is parallel to $Df(x,y)$. So, it is not the sign (or magnitude) of the derivative that matters but the direction.

Note that the Lagrange multiplier $\lambda$ is only introduced to generalize the formula: There are more solutions to $Df(A)= cDg(B)$ than $Df(a)=Dg(b)$ (seeing as $a \subset A$ and $b \subset B$).  Both equations maximize $f$ under some constraint(s), so it's intuitively better to find more answers for $\max f$.

As to your question about figure 1 on wikipedia, I have deciphered the following: the arrows are orthogonal to the directional derivatives. $d_2,d_3$ are arbitrary points where for values of $(x,y)$ the constraint is met. It shows us that, only when the curves' tangents are parallel (e.g. the tangents' orthogonal complements are parallel), the function $f(x,y)$ is maximized, with $g(x,y) = 0 $.

Hope this helps.


问:
Limit of ln summation
I am trying to find the limit $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=\left\lceil \sqrt{n}\right\rceil }^{n}\ln\left(\frac{n}{i}\right) $$ I can't  find a good way to handle it. I tried bounding it from below and from above, but the bounds I got to are not tight enough.

答:
This is essentially a Riemann sum. Noting that $f(x) = \log(1/x)$ is decreasing, we have

$$ \int_{\frac{i}{n}}^{\frac{i+1}{n}} f(x) \, \mathrm{d}x
\leq \frac{1}{n}f\left(\frac{i}{n}\right)
\leq \int_{\frac{i-1}{n}}^{\frac{i}{n}} f(x) \, \mathrm{d}x. $$

So, if we denote the sum by $S_n$, then

$$ \int_{\frac{\lceil\sqrt{n}\rceil}{n}}^{1+\frac{1}{n}} f(x) \, \mathrm{d}x
\leq S_n
\leq \int_{\frac{\lceil\sqrt{n}\rceil-1}{n}}^{1} f(x) \, \mathrm{d}x. $$

Since $f(x)$ is either improper Riemann-integrable or Lebesgue-integrable over $(0, 1]$, it follows from the squeezing theorem that

$$ \lim_{n\to\infty} S_n = \int_{0}^{1} f(x) \, \mathrm{d}x = 1. $$

Title: Limit of ln summation


问:
How to prove the property of convex function in higher dimension
Suppose $f:\mathbb{R}^n\mapsto \mathbb{R}$ is convex, could anyone tell me how to prove the following fact?

(1) If $f\in C^1$, then for any $u,v$

$$
f(v)\geqslant \langle\nabla f(u),v-u\rangle
$$

(2) If $f\in C^2$, then for any $u,v$
$$
\langle\nabla^2f(u)v,v\rangle\geqslant0
$$

答:
Useful fact: the restriction of a convex function to a line is a convex function.

Formally, for any point $a$ and any nonzero vector $w$ the function $\varphi(t) = f(a+tw)$ is convex on $\mathbb R$.

If $f$ is $C^1$, then so is $\varphi$. We have $\varphi(t)-\varphi(0)=\varphi'(\xi)\,t>\varphi'(0)\,t$ by the Mean Value Theorem and because $\varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.

If $f$ is $C^2$, then so is $\varphi$. Consequently, $\varphi''(0)\ge 0$ which translates into the inequality you want.


问:
Partial Differentiation $ \frac {log( \frac {S_0}{k}) + (r - \frac {\sigma^2}{2})T} {\sigma \sqrt{T}} $
I'm having difficulty in finding the $ \frac{\partial d1}{\partial \sigma}$ for the equation below.

I tried using quotient rule but I'm getting $ -\sqrt{T} -\frac {log( \frac {S_0}{k}) + (r - \frac {\sigma^2}{2})T} {\sigma^2 \sqrt{T}} $ for $ \frac{\partial d1}{\partial \sigma}$

$ d1 = \frac {log( \frac {S_0}{k}) + (r - \frac {\sigma^2}{2})T} {\sigma \sqrt{T}} $

答:
You can use the quotient rule as you tried already yourself.

So $g(x) = {log( \frac {S_0}{k}) + (r - \frac {\sigma^2}{2})T}$

and $h(x) =  \sigma\sqrt{T}$

and $g'(x) = \frac{\partial}{\partial \sigma} log( \frac {S_0}{k}) + \frac{\partial}{\partial \sigma}(r - \frac {\sigma^2}{2})T = -T\sigma $

and  $h'(x) = \frac{\partial}{\partial \sigma}  \sigma \times \sqrt{T} = \sqrt{T}$

and then it is just applying the rule:

$$ \frac{-T\sigma \times \sigma\sqrt(T) - \sqrt(T)\times[log( \frac {S_0}{k}) + (r - \frac {\sigma^2}{2})T]}{(\sigma\sqrt{T})^2} $$


问:
If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous?
Suppose $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous function. Define $f_n(x)=f(x+1/n)$ for all $n\in\mathbb{N}^+$. If $f_n(x)\to f(x)$ uniformly on $\mathbb{R}$, can we conclude that $f$ is actually uniformly continuous? If not, can you give a counterexample of $f$ to be not uniformly continuous but satisfies all above conditions?

I know that if $f_n(x)$'s are all uniformly continuous and $f_n(x)\to f(x)$ uniformly, then $f(x)$ must be uniformly continuous. However, here we do not assume $f(x)$ to be uniformly continuous, and $f_n(x)$ has specific structure related to $f(x)$, so at least you cannot directly conclude they are uniformly continuous. I guess the statement above is wrong but it's hard for me to find a counter-example. Can anyone help me?

答:
By assumption, for every $\epsilon >0$, there is an $n_0 \in \Bbb N$ such that for every $x \in \Bbb R$

$$n \geq n_0 \implies \left| f(x) - f \left(x + \frac{1}{n} \right)  \right| < \epsilon$$Now if you repeatedly substitute $x = \pm \frac{1}{n}$ in the above inequality, we can conclude that for any two rationals $a,b$ of the form(say $*$-type) $\frac{m}{n}$, where $m,n \in \Bbb Z, n \geq n_0$, we have

$$|f(a)-f(b)| < \epsilon$$

Now observe that the rationals of the form $m,n \in \Bbb Z, 0<n < n_0$ are nowhere dense in $\Bbb R$. Therefore rationals of $*$-type are dense in $\Bbb R$(we are using the fact that the rationals are dense in $\Bbb R$). Since $f$ is continuous, we obtain that for any two real numbers $x,y$

$$|f(x)-f(y)| \leq \epsilon$$

But since $\epsilon$ was arbitrary, we can conclude that $f$ must be a constant function which is trivially uniformly continuous.


问:
Ramification and algebraic closure
Let $k$ be a field and $f:C\rightarrow B$ a morphism of smooth curves. Let $C'$, $B'$ and $f'$ be the extensions of $C$, $B$, and $f$ to $\overline{k}$.

Let $c\in C$, $b=f(c)$, $c'\in C'$ above $c$ and $b'=f'(c')$.

Is it true that the ramification index $e(c/b)$ is the same as the ramification index $e(c'/b')$?

答:
As suggested by the author, I'm turning my comment into an answer.

It is true if $e(c/b)$ or $e(c′/b′)$ is equal to 1 and $k$ is perfect (thank you @Daniel Hast for detecting this), because then it's unramifiedness we're talking about, and it is detected by whether $\Omega^1$ vanishes. But $\Omega^1$ behaves nicely with base change (and it's a faithfully flat base change).

In general, I think it remains true if $k$ is perfect, because then the $C_{\overline{k}} \rightarrow C$ (same for $B$) correspond to unramified maps of valuation rings, thus $e(c/b)$ and $e(c′/b′)$ are both the ramification indices of $\mathcal{O}_{B,b}\rightarrow \mathcal{O}_{C′,c′}$ (using either $\mathcal{O}_{C,c}$ or $\mathcal{O}_{B′,b′}$ as intermediate rings).

So what I showed is that we can assume that $\overline{k}/k$ is purely inseparable. But then, this base change is a universal homeomorphism, and stranger things may happen (as shown in the answer linked by Daniel Hast again).

Title: Ramification and algebraic closure


问:
Show that $(R, +, 0)$ is generated by the closed interval $X = [0, 1]$
Show that $(R,+, 0)$ is generated by the closed interval $X = [0, 1]$

Show that $(Q,+,0)$ is generated by the set $X=\{(1/n)) :n≥1\}$

I can't find anything to help me with these questions.

答:
$(\Bbb R,+,0)$ is generated by any interval $(u,v)$ with $u<v$.

Indeed, with $r:=v-u>0$, we find $x\in\langle (u,v)\rangle$ for $|x|<r$ simply because $x=\frac{u+v+x}2-\frac{u+v-x}2$ where $u=\frac{u+v-r}2<\frac{u+v\pm x}2<\frac{u+r+v}2=v$.
Now for arbitrary $x\in\Bbb R$, let $n=\lfloor \frac{|x|}r\rfloor$. Then $y:=\frac1n x\in(-r,r)\subset \langle (u,v)\rangle$, and hence $x=\underbrace{y+\cdots +y}_n\in \langle (u,v)\rangle$.


问:
Solving $12n + 2n + 80 \times 2 \times 4.92 = 200n$ for $n$
Can someone please help me with this?

I'm studying chemistry for the medicine exam, and I haven't done math in a long time

$$12n + 2n + 80 \times 2 \times 4.92 = 200n$$

I need to find $n$.

If someone can please explain me how to do it, I would really appreciate it.

答:
You can transpose the left side terms having $n$ to the RHS to get $80 \times 2 \times 4.92 = 186n \implies 4.92 = \dfrac{186n}{160} = \dfrac{93n}{80} \implies 93n = 393.6 = 93n \implies n = \dfrac{393.6}{93} = \dfrac{131.2}{31}$

Hope you get it (as well as what you get from the division).

It's a simple problem, actually...


问:
Prove $h_{p\mu+(1-p)\nu}(T)=ph_\mu(T)+(1-p)h_\nu(T)$
Let $T:X\to X$ be a continuous map on a compact metric space. I'm trying to prove that for two $T$-invariant probability measures $\mu,\nu$ we have $h_{p\mu+(1-p)\nu}(T)=ph_\mu(T)+(1-p)h_\nu(T)$. Here $h_\mu(T)$ denotes the metric entropy of $T$, also known as the measure theoretic entropy.

I managed to show that we have $h_{p\mu+(1-p)\nu}(T)\geq ph_\mu(T)+(1-p)h_\nu(T)$ by defining $\Phi:[0,\infty)\to\mathbb R$ by $\Phi(x)=-x\log x$ for $x>0$ and $\Phi(0)=0$; then $\Phi$ is a concave function, hence by Jensen $$p\Phi(\mu(A))+(1-p)\Phi(\nu(A))\leq\Phi(p\mu(A)+(1-p)\nu(A))$$ for each measurable $A$. Applying this to finite measurable partitions $\alpha$ gives $H_{p\mu+(1-p)\nu}(\alpha)\geq pH_\mu(\alpha)+(1-p)H_\nu(\alpha)$. Next, we plug in $\bigvee_{i=0}^{n-1}T^{-i}\alpha$, divide by $n$ and letting $n\to\infty$ and obtain $h_{p\mu+(1-p)\nu}(\alpha,T)\geq ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T)$. Then we find partitions $\alpha_\mu,\alpha_\nu$ such that $h_\mu(\alpha_\mu,T)\geq h_\mu(T)-\frac{\epsilon}{p}$ and $h_\mu(\alpha_\nu,T)\geq h_\nu(T)-\frac{\epsilon}{1-p}$, yielding $$h_{p\mu+(1-p)\nu}(\alpha_\mu\vee\alpha_\nu,T)\geq ph_\mu(T)+(1-p)h_\nu(T)-\epsilon;$$ passing to the supremum over all measurable partition $\alpha$ on the left hand side gives the desired inequality.

For the reverse inequality, I have no idea what to do. Any help is much appreciated!

答:
I managed to prove the other inequality myself.

Again take $\Phi(x)=-x\log x$ and observe that $-\log x$ is decreasing. Therefore, for each measurable $A$ and $p\in(0,1)$, \begin{align*}&\phantom=\ \ \Phi(p\mu(A)+(1-p)\nu(A))\\&=-p\mu(A)\log\big(p\mu(A)+(1-p)\nu(A)\big)-(1-p)\nu(A)\log\big(p\mu(A)+(1-p)\nu(A)\big)\\&\leq -p\mu(A)\log\big(p\mu(A)\big)-(1-p)\log\big((1-p)\nu(A)\big)\\&=p\Phi(\mu(A))+(1-p)\Phi(\nu(A))-p\log(p)\mu(A)-(1-p)\log(1-p)\nu(A).\end{align*} We proceed as in the proof of the other inequality: sum over the sets in some finite measurable partition $\alpha$ to obtain $$H_{p\mu+(1-p)\nu}(\alpha)\leq pH_\mu(\alpha)+(1-p)H_\nu(\alpha)-p\log(p)-(1-p)\log(1-p).$$ Next, we plug in $\bigvee_{i=0}^{n-1}T^{-i}\alpha$, divide by $n$ and let $n\to\infty$ to obtain $$h_{p\mu+(1-p)\nu}(\alpha,T)\leq ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T).$$ Now pass over to the supremum over all measurable partitions $\alpha$, to obtain \begin{align*}h_{p\mu+(1-p)\nu}(T)&\leq\sup_\alpha\left\{ ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T)\right\}\\&\leq p\sup_\alpha h_\mu(\alpha,T)+(1-p)\sup_\beta h_\nu(\beta,T)\\&=ph_\mu(T)+(1-p)h_\nu(T),\end{align*} establishing the other inequality, hence finishing the proof.


问:
Why are topological spaces interesting to study?
In introductory real analysis, I dealt only with $\mathbb{R}^n$. Then I saw that limits can be defined in more abstract spaces than $\mathbb{R}^n$, namely the metric spaces. This abstraction seemed "natural"to me. Then, I knew the topological spaces. However, this time the abstraction did not seem natural/useful to me. Then one runs into problems of classifying spaces into normal/ first countable ... In my opinion, these resulted from the high  level of abstraction adopted by studying  topological spaces. When one uses a more general definition for a space, it is possible that the number of uninteresting objects increase. I guess this is what happened here, we use a very general definition for topological spaces, we get a lot of uninteresting spaces, then we go back and make classifications such as normal, Hausdorff,..

I was trying to justify to myself why are topological spaces are good to study. The best and only reason I can propose is that the category $Top$ is bicomplete.

Question 1: (Alternatives to $Top$)

If this is the only reason, can't there exist a "smaller" category   such that it contains all metric spaces and is bicomplete ?

Question 2: (History of topological spaces)

I mentioned that the abstraction from metric spaces to topological spaces does not seem very natural to me.  I suspect that historically, metric spaces were studied before topological spaces. If this is the case, I'd like to know what was the motivation/justification for this abstraction.

Question 3: (Applications of non-metric topology outside topology)

I mentioned earlier that  "we get a lot of uninteresting spaces". Perhaps I am wrong  (I hope I am wrong). I would value non-metric topological spaces more, if I see examples of theorems such that:

1) The theorems are in a branch of mathematics outside Topology

2) The theorems are proven with the aid of topology

3) The topological part about the proof of the theorem is about a non-metric space

Edit:  ${}$ non-artificial instances of non-metric spaces appearing in other branches of math are valuable as well.

Thank you

答:
These are further examples of answers to your third question, related to Funcional Analysis.

Direct Method of the Calculus of Variations: Let $M$ be a weakly closed subset of a reflexive Banach space $E$ and $F\colon M\to\mathbb{R}$ a weakly lower semicontinuous function such that there exists $a\in\mathbb{R}$ for which $F^{-1}(-\infty,a]$ is nonempty and norm-bounded. Then $F$ attains its infimum at some point of $M$.

The hypotheses might look a little complicated, but they are so mostly for the sake of generality.

Intuitively, minimization of functionals should already be interesting. Moreover, these types of problems appear when one is looking for the existence of solutions to PDEs. Sometimes it suffices to minimize some sort of "energy" functional on a Sobolev space (or some space of distributions), which satisfies conditions such as the ones above, in order to conclude that some problem (e.g. a Cauchy problem) has a solution. A very well-known application of this sort is Douglas' solution to the Plateau Problem.

The Direct Method of the Calculus of Variations is a direct application of the Banach-Alaoglu theorem.

Gelfand representation of commutative $C^*$-algebras: Every unital commutative C*-algebra is isomorphic to $C(K)$ for some compact Hausdorff space $K$.

This theorem again requires Banach-Alaoglu and Stone-Weierstrass for non-metrizable spaces.

This is one of the fundamental theorems in the theory of $C^*$-algebras, which allows one to construct a functional calculus and essentially treat self-adjoint elements as if they were functions. You can see more on any book on C*-algebras.

(Although if you only care about separable $C^*$-algebras you could get away with metrizable $K$. I'm not sure if the proof of Gelfand representation could be made significantly easier in this case, though.)


Title: Solving $12n + 2n + 80 \times 2 \times 4.92 = 200n$ for $n$

Body:
Can someone please help me with this?

I'm studying chemistry for the medicine exam, and I haven't done math in a long time

$$12n + 2n + 80 \times 2 \times 4.92 = 200n$$

I need to find $n$.

If someone can please explain me how to do it, I would really appreciate it.

Body:
$12n + 2n + 80 \times 2 \times 4.92 = 200n$

$\Rightarrow 14n + 160 \times 4.92 = 200n$

$\Rightarrow 14n + 787.2 = 200n$

$\Rightarrow 782.2 = 200n - 14n$

$\Rightarrow 782.2 = 186n$

$\Rightarrow 186n = 782.2$

$\Rightarrow n = \dfrac{782.2}{186}$

$\Rightarrow n = 4.23226$


问:
Distance between 2 points without knowing coordinates
I have the following scenario depicted in the attached Figure. Assuming that there are no symmetries and any particular cases, assuming to know the distance $d_{12}$, the coordinates $x_1,y_1,x_2,y_2$ and the other distances depicted in the figure do you have any idea on how to compute the distance $dist$ between the 2 green points?

答:
Let $\color{green}{O}$, $\color{green}{A}$, $\color{orange}{B}$, $\color{blue}{C}$ be the four points in figure
( $\color{green}{O}$ is the green dot at left, $\color{green}{A}$ the green dot at right, $\color{orange}{B}, \color{blue}{D}$ the dots with corresponding color). Let

$$\begin{cases}
a_1 &= |OA| = \verb/dist/\\
b_1 &= |OB| = 3\\
c_1 &= |OC| = 3\\
\end{cases}\quad\text{ and }\quad
\begin{cases}
a &= |BC| = d_{12}\\
b &= |AC| = 5\\
c &= |AB| = 3
\end{cases}
$$

Since these four points lie on a plane, the volume of the tetrahedron formed by these points is zero. In general, if we know the edge lengths of a tetrahedron, we can compute its volume $V$ by evaluating corresponding  Cayley-Menger determinant:

$$288V^2 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & a_1^2 & b_1^2 & c_1^2\\
1 & a_1^2 & 0 & c^2 & b^2\\
1 & b_1^2 & c^2 & 0 & a^2\\
1 & c_1^2 & b^2 & a^2 & 0
\end{matrix}\right|
$$
For the problem at hand, we have

$$V = 0 \iff
\left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & a_1^2 & 3^2 & 3^2\\
1 & a_1^2 & 0 & 3^2 & 5^2\\
1 & 3^2 & 3^2 & 0 & a^2\\
1 & 3^2 & 5^2 & a^2 & 0
\end{matrix}\right| = 0
$$
Throwing this to a CAS, you can simplify this to an quadratic equation in $y = a_1^2$:

$$y^2 + (a^2-52)y + \frac{2304}{a^2} = 0$$
This leads to two possible solution for $OA = \verb/dist/ {}^{\color{blue}{[1]}}$.

$$\verb/dist/ = a_1 = \sqrt{\frac{a(52-a^2) \pm \sqrt{a^6-104a^4+2704a^2-9216}}{2a}}$$

note

$\color{blue}{[1]}$ - For the configuration in figure, $O$ and $A$ lie on same side of $BC$, one should take the smaller root of $y$ (ie taking '-' sign in front of the inner square root) to get the right distance.


问:
Is the identity $a^n \equiv a^{n \mod m} \mod m$ true if m is a power of 10
I stumbled on this code, which finds the last digits of a tetration: https://blog.dreamshire.com/project-euler-188-solution/.
The code seems to imply that $a^n \equiv a^{n \mod m} \mod m$ and takes advantage of this property to avoid calulation very large numbers.
This implies that if you are working with powers (mod 10^m) you only need to care about the last m digits of the exponent.

I tried a few examples with pythons pow() and it seems to hold, but i cant find a proof.
*with m > 1.

答:
This is valid for $m=10^8$. Namely, if $\gcd(a,10)=1$ we have (Euler's theorem):

$a^{2^7}\equiv 1\pmod{2^8}$ as $\varphi(2^8)=2^7$

$a^{4\cdot 5^7}\equiv 1\pmod{5^8}$ as $\varphi(5^8)=4\cdot 5^7$

Now, because both $2^7$ and $4\cdot 5^7$ are factors of $m=10^8$ we have that $a^m\equiv 1\pmod{2^8}$ and $a^m\equiv 1\pmod{5^8}$, so $a^m\equiv 1\pmod m$.

From this your statement follows fairly simply, but note that this is only valid for this very special modulus $10^8$, though the same proof would apply to all other powers of $10$, excluding $10$ itself. This is certainly not valid for a lot of other modules, as you have seen from the comments.


答:
If we look for any arbitrary $f(x)$ and $g(x)$, there are only 2 combined derivatives that include the product of these function (and their individual derivatives): The quotient rule and the product rule. As its clear we are not dividing by $g^2(x)$, lets focus on the product rule:
$$\frac{d}{dx}\left(f(x)g(x)\right) = f'(x)g(x) + f(x)g'(x)$$

The (indefinite) integral represents the anti-derivative of a function, which can be described as: Find a function that if differentiated over $x$ yields the function inside the integral. If Im correctly, you want to find a way that allows us to compute a integral of the following shape:
$$\int f(x)g(x)dx$$
as a function without the integrated product of both terms. Sadly, if we look at the definition of the product rule, one might notice 2 things: First there are 2 products and both products contain both $f(x)$ and $g(x)$ or their derivatives.
Therefore you are always stuck with the following deduction:
$$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx$$
$$f(x)g(x)- \int f(x)g'(x)dx = \int f'(x)g(x)dx$$
Which equals to the integration by parts rule and does not remove the integral of the product in its equation.

Title: Is there really no analogue of the derivative product rule for integrals, or we just haven't found one yet?

Body: For a product rule for integrals, I am not talking about integration by parts. That particular formula uses a integral of products inside the formula itself. The derivative product rule does not use a derivative of products inside the formula. I have never seen a book that proves that there is in fact no analogous formula for the integral of a product. Some books prove that there is no formula for the roots of a general 5th degree or higher polynomial function. Can someone prove that there is no such formula for the integral of a product? There is a formula for the integral of a sum, so maybe someone will discover that there is a formula for the integral of a product. I apologize if my notion of formula is not precise enough, but maybe there is a book that precisely defines what a formula is.</p>


问:
relationship of polar unit vectors to rectangular
I'm looking at page 16 of Fleisch's Student's Guide to Vectors and Tensors.  The author is talking about the relationship between the unit vector in 2D rectangular vs polar coordinate systems.  They give these equations:

\begin{align}\hat{r} &= \cos(\theta)\hat{i} + \sin(\theta)\hat{j}\\
\hat{\theta} &= -\sin(\theta)\hat{i} + \cos(\theta)\hat{j}\end{align}

I'm just not getting it.  I understand how, in rectangular coordinates, $x = r \cos(\theta)$, but the unit vectors are just not computing.

答:
The (endpoint of the) vector $(\cos\theta,\,\sin\theta)$ is on the unit circle, exactly at angle $\theta$ (if angles are measured from the $x$-axis, towards the $y$-axis). So, in polar form, we can say $r=1$ and $\theta=\theta$.

The other one, $(-\sin\theta,\,\cos\theta)$ is its rotated version, by $+90^\circ$. So, this has $\hat r=1$ and $\hat\theta=\theta+90^\circ$.


问:
Invertiblity of ST and TS for linear maps
Could somebody explain why for any given linear maps S and T over a finite dimensional vector space V, ST is invertible if and only if TS is? Why is it so important that V is finite dimensional for this to hold true?

Edit: I am looking for an approach without using matrices or determinants.

答:
Assume $ST$ is invertible.  Then $\exists A$ such that $AST=I$ and $STA=I$.

Multiplying on the left by $T$ and on the right by $S$, respectively:

$TAST=T$ and $STAS=S$,

so multiplying the first equation on the right by $AS$ and letting $Q=TAS$, we see that $Q^2=Q$.  $Q$ is idempotent, and non-singular.  $SQ=S$ and $S$ must be of full rank, thus $Q$ is of full rank.

Thus (in the finite dimensional case), this implies $Q=I$, that is $TAS=I$.

Now we have that both $S$ and $T$ are invertible.  $S^{-1}=TA$ and $T^{-1}=AS.$

Thus, $TS$ is invertible:  $(TS)^{-1} = S^{-1} T^{-1}$.

The argument in the other direction is exactly the same (switching the roles of $T$ and $S$).


问:
How to rigourously define $G^{|A|}$? where G is a group and A being an infinite set.
It is clear to me that $G^n:=G\times G\times ...\times G$. Every member of $G^n$ looks like $(g_1,g_2,...g_n)$.

But how to rigorously define $G^{|A|}$, i.e how to explicitly write elements of $G^{|A|}$?

答:
$G^A$ is the set of functions $x:A\to G$, endowed with componentwise product and inverses. So to say, you gain access to the $a$-th component of the element $x$ by evaluating it, as a function, at $a$: id est $x_a:=x(a)$. More generally, you can define $\prod\limits_{i\in I}G_i$ as the set of functions $x:I\to\bigcup\limits_{i\in I}G_i$ such that $x(i)\in G_i$ for all $i\in I$. This works fine as a categorical product of groups.

When $G$ is a group, there is another (not the only other) useful notion, sometimes denoted as $G^{(A)}$, which is $$G^{(A)}:=\left\{x\in G^A\,:\, \operatorname{card}\{a\in A\,:\, x_a\ne e\}\text{ is finite}\right\}$$


问:
How should I set the integral route for this complex integral?
\begin{equation}
\int_0^{\infty} \dfrac{x^{\lambda -1}}{x^2+1} dx
=
\dfrac{\pi}{2 \sin{\dfrac{\pi \lambda}{2}}} \quad (0< \lambda < 2)
\end{equation}

I will prove this using complex integral, but I don't come up with how I should set the integral route.

I tried prove the equality using the integral route below and using residue theorem.

$
C_{R} : \text{semicircle in first and second quadrant whose radius is R }\\
C_1 : [-R, \epsilon]\\
C_{\epsilon} : \text{semicircle in first and second quadrant whose radius is $\epsilon$ }\\
C_2 : [\epsilon, R]
$

However, I failed to prove. It seems that it is impossible to calculate the integral with this integral route.

I would like you to teach me how I should set the integral route.

答:
I will use $a$ instead of $\lambda$, it is simpler to type. Since $a-1$ may take negative values, we may have some problems in $0$, so the contour from the OP avoiding $0$ is needed. The following is valid for
$$
a\ne 1\ .
$$
(For this particular value use an other argument or a posteriori the continuity in $a$.)

Let us denote the two contour pieces $C_1$ and $C_2$ from the OP by $C_1(\epsilon,R)$ and $C_2(\epsilon,R)$, also involving $\epsilon$ and $R$.

The expression $x^a$ is then rather defined as $\exp(a \log x)$ with a branch $\log$ for the logarithm defined to coincide with the real natural logarithm $\ln$ on $(0,\infty)$. So we have for $x>0$ the relation
$$(-x)^{a-1}=(-1)^{a-1}\cdot x^{a-1}=\exp(i\pi\; (a-1))\cdot x^{a-1}\ .$$

So let us compute the integral
$$
\begin{aligned}
J(a)
&:=
\int_0^{\infty}
\dfrac{x^{a -1}}{x^2+1} \;dx
\\
&=\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}}
\int_\epsilon^\infty
\dfrac{x^{a -1}}{x^2+1} \;dx
=\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}}
\int_{C_2(\epsilon,R)}
\dfrac{x^{a -1}}{x^2+1} \;dx\ .
\\[3mm]
&\qquad\text{We have:}
\\[3mm]
\int_{-\infty}^0
\dfrac{x^{a -1}}{x^2+1} \;dx
&
=\int_0^{\infty}
\dfrac{(-x)^{a -1}}{x^2+1} \;dx
=e^{i\pi \; a}\cdot
\int_0^{\infty}
\dfrac{(-x)^{a -1}}{x^2+1} \;dx
\\
&=
e^{i\pi\; a}J(a)\ .
\\[3mm]
\left|\int_{C(\epsilon)}
\dfrac{x^{a -1}}{x^2+1} \;dx
\right|
&\le
\int_{C(\epsilon)}
\dfrac{|x|^{a -1}}{|x^2+1|} \;dx
\le
\int_{C(\epsilon)}
\dfrac{\epsilon^{a -1}}{1-\epsilon^2} \;dx
=\pi\epsilon\cdot\dfrac{\epsilon^{a -1}}{1-\epsilon^2}
\\
&\to 0\qquad\text{ because $a>0$.}
\\[3mm]
\left|\int_{C(R)}
\dfrac{x^{a -1}}{x^2+1} \;dx
\right|
&\le
\int_{C(R)}
\dfrac{|x|^{a -1}}{|x^2+1|} \;dx
\le
\int_{C(\epsilon)}
\dfrac{R^{a -1}}{R^2-1} \;dx
=\pi R\cdot\dfrac{R^{a -1}}{R^2-1}
\\
&\to 0\qquad\text{ because $a<2$.}
\\[3mm]
&\qquad\text{This gives:}
\\[3mm]
(1+e^{i\pi\; (a-1)})\cdot J(a)
&=
\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}}
\int_{C_1(\epsilon,R)\ \sqcup\ C_2(\epsilon,R)}
\dfrac{x^{a -1}}{x^2+1} \;dx
\\
&=
\lim_{\substack{0<\epsilon<R\\\epsilon \to0\\R\to\infty}}
\int_{C_1(\epsilon,R)\ \sqcup \ C(\epsilon)\ \sqcup \ C_2(\epsilon,R)\ \sqcup \ C(R)}
\dfrac{x^{a -1}}{x^2+1} \;dx
\\
&=2\pi\cdot i\sum_{r\text{ Residue}}\operatorname{Res}_{x=r}\dfrac{x^{a -1}}{x^2+1}
\\
&=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\dfrac{x^{a -1}}{x^2+1}
\\
&=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\frac 1{2i}x^{a -1}
\left(\frac 1{x-i}-\frac 1{x+i}\right)
\\
&=2\pi\cdot i\cdot \operatorname{Res}_{x=i}\frac 1{2i}x^{a -1}\cdot \frac 1{x-i}
\\
&=\pi\cdot i^{a-1}=\pi\cdot e^{i\frac \pi 2(a-1)}\ .
\end{aligned}
$$
Now $1+e^{ib}=(1+\cos b)+i\sin b=2\cos ^2\frac b2+2i\sin\frac b2\cos \frac b2=2\cos\frac b2\cdot e^{ib/2}$.

This gives finally:
$$
J(a)
=
\frac
{\pi\cdot \color{blue}{e^{i\frac \pi 2(a-1)}}}
{2\cos\frac \pi 2(a-1)\cdot \color{blue}{e^{i\frac \pi 2(a-1)}}}
=
\frac
\pi
{2\cos\frac \pi 2(1-a)}
=
\frac
\pi
{2\sin\frac {\pi a} 2}
\ .
$$
$\square$


答:
I've never seen this before but I guess a reasonable way to define $G^\mathbb{R}$ would be to say that $f\in G^\mathbb{R}$ is a map

$$f:\mathbb{R}\longrightarrow G $$

And if I have $f,g\in G^\mathbb{R}$ then $(fg)[x]=f(x)g(x)$, this is a generalization of the $G^n$ case since you can think about elements of  $G^n$ in a similar way as maps from $\{1,...,n\}\rightarrow G$.

Title: How to rigourously define $G^{|A|}$? where G is a group and A being an infinite set.

Body:
It is clear to me that $G^n:=G\times G\times ...\times G$. Every member of $G^n$ looks like $(g_1,g_2,...g_n)$.

But how to rigorously define $G^{|A|}$, i.e how to explicitly write elements of $G^{|A|}$?


问:
Types of minimal ranks have unique extensions
Suppose $T$ is a $\aleph_0$ stable theory in a countable language. Fix some set $A$.

Let $S = \{ R[p]:p \in S(A), p$ is non-algebraic$\}$
Let $\alpha = min(S)$ and let $p \in S$ be such that $R[p] = \alpha$. Then show that for any $B$ such that $A \subseteq B$, and $q_1, q_2 \in S(B)$ non-algebraic such that $p_0 \subseteq q_1, q_2$, then we have $q_1 = q_2$.

Definition of R

$R$ here is a modified binary version of Morley rank. Given a type $p$, and an ordinal $\alpha$, $R[p] \geq \alpha + 1$ iff for every finite $p_0 \subseteq p$ there is $\varphi$ such that $R[p_0 \cup \{\varphi\}], R[p_0 \cup \{\neg \varphi\}] \geq \alpha$.

The base case and limit case are defined as you'd expect (base case: $\geq 0$ if consistent/realized, and limit case: $\geq \alpha$ iff $\geq \beta$ for every $\beta < \alpha$). I also supressed information relating to parameters and such to keep it from getting too wordy. Let me know if you'd like me to add details.

A few basic results that are basically the same as in Morley Rank:

$p \vdash q \implies R[p] \leq R[q]$

$R$ has finite character (i.e every type has a finite subtype with the same rank)

$R$ is invariant under automorphisms

Uniqueness: if $R[p] < \infty$, $p \subseteq q_1, q_2$ where $q_1, q_2$ are complete types over bigger domains,and $R[p] = R[q_1] = R[q_2]$, then $q_1 = q_2$.

Stability: $T$ is $\aleph_0$ stable iff $R[x = x] < \omega_1$ iff $R[x = x] < \infty$.

As you can see it's quite similar to Morley rank.

I've been working on it for a bit with no avail.

答:
You're probably having trouble proving this because it is not true as stated. Here's a counterexample.

Let $T$ be the theory of an equivalence relation $E$ with two infinite classes. Pick representatives $b_1$ and $b_2$ for the two classes. Let $A=\varnothing$, and let $B=\{b_1,b_2\}$.

Now there is a unique $1$-type $p$ in $S(A)$, which is of course the non-algebraic type of minimal rank (I think its rank is $\omega+1$), but $p$ has two distinct non-algebraic extensions to $S(B)$, isolated by $(xEb_1\wedge x\neq b_1)$ and $(xEb_2\wedge x\neq b_2)$ respectively.


问:
Corollary: Let X be a connected and completely regular space with more than one point, then $| X | \geq | \mathbb R |$.
Demonstration: Let $x, y  \in  X$ be different. Because $X$ is completely regular, there is $f: X  \to  [0,1]$ continuous such that $f (x) = 0$ and $f (y) = 1$. Since $X$ is connected, $f [X]$ is an interval in $\Bbb R$, but $1,0  \in  f [X]$, so $f [X] = [0,1]$.

I found this corollary, but I am unable to understand the proof. Can anyone help me with this? Thanks in advance.

答: HINT: The continuous image of a connected set is connected.


问:
linearly dependent linear tranformations
Let $E$ be a vector space over $k$ of finite dimension $n$. Let $f_1, \ldots, f_p \in \mathcal{L}(E)$ with $p \leq n$ and assume that the $f_i$ are simultaneously diagonalizable (put it simply, there exists a basis of $E$, where the matrices representing the $f_i$ are all diagonal).

Assume that for all $x \in E$, the vectors $f_1(x), \ldots, f_p(x)$ are linearly dependent. Then the vectors $f_1, \ldots, f_p$ are linearly dependent in $\mathcal{L}(E)$.

I have a complicated proof of that, and I'd like to know if there is a simple argument proving this result. Furthermore, I feel like the diagonalizability assumption could be withdrawn and one could only assume that for any $i,j$ we have $f_i \circ f_j = f_j \circ f_i$, but I have no clue on how to prove that last assertion.

答:
Counterexample for the case when the matrices are not simultaneously diagonalisable:

$$f_1:=\begin{pmatrix}1&1\\0&0\end{pmatrix},\qquad f_2:=\begin{pmatrix}1&2\\0&0\end{pmatrix}$$

Then $f_1,f_2$ are linearly independent, yet for any vector $x=(a,b)^t$, $$\tfrac{1}{a+b}f_1(x)+\tfrac{1}{a+2b}f_2(x)=0$$ (unless $a+b=0$ or $a+2b=0$; those cases correspond to eigenvectors of $f_1$, $f_2$ respectively, so $f_1(x)=0$ or $f_2(x)=0$, so $f_1(x)$, $f_2(x)$ are still dependent.)

Looking at the $2\times2$ case, commutativity of the matrices is not required for the property to hold; rather it is invertibility.

Suppose $f_1=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is invertible and $f_2=\begin{pmatrix}\alpha a&\beta b\\\alpha c&\beta d\end{pmatrix}$ are as in the question. The second matrix must be of this form since $f_1(\mathbf{i})$, $f_2(\mathbf{i})$ are linearly dependent, and similarly $f_1(\mathbf{j})$, $f_2(\mathbf{j})$. Furthermore, $f_1(x)$ and $f_2(x)$ are in the same direction, hence if $x=(u,v)^t$, then $$\lambda\begin{pmatrix}au+bv\\cu+dv\end{pmatrix}=\begin{pmatrix}\alpha au+\beta bv\\\alpha cu+\beta dv\end{pmatrix}$$ $$\therefore\ (au+bv)(\alpha cu+\beta dv)=(cu+dv)(\alpha au+\beta bv)$$ $$\therefore\ (\beta-\alpha)(ad-bc)uv=0$$ Hence $\alpha=\beta$, that is, $f_2=\alpha f_1$ as required.