问:
Integral of $f(x)f'(x)$
I found a problem that says:

Let $f$ be a real function with continuous derivative on $[a , b]$ such that $f(a)=f(b)=0$. Suppose $\int_a^b f(x)^2 dx =1$. Show that $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$.

The suggested solution was to do integration by parts using $u=f(x)^2$ and $dv=dx$ which leads to
$1=\int_a^b f(x)^2 dx = 2xf(x)^2 |_a^b - \int_a^b 2xf(x)f'(x) dx$. Then since $f(a)=f(b)=0$ that makes the first part go away and we get $1=- \int_a^b 2xf(x)f'(x) dx$ then we can make the conclusion.

But what am I missing? We showed that $\int_a^b xf(x)f'(x) dx = -\frac{1}{2}$ but the problem wanted to show $\int_a^b f(x)f'(x) dx = -\frac{1}{2}$. Was this just a typo?

答:
$$\int_a^bf(x)f'(x)dx=\int_{f(a)}^{f(b)}udu=\frac12\left[u^2\right]_{f(a)}^{f(b)}=\frac{f^2(b)-f^2(a)}{2}=0$$
because $f(a)=f(b)=0$ so the question is wrong

Title: Integral of $f(x)f'(x)$


问:
Any quicker way to find the residues here?
I have to integrate $\dfrac{1}{(z-4)(z^7-1)}$ w.r.t $z$ along $C : |z|=2$.

7 poles of the function lie inside $C$ and $z=4$ lies outside. I am trying to find the answer by applying residue theorem. i.e. $2\pi i \sum (\text{residues inside }C)$, but it is becoming too lengthy to handle. Any leads?

答:
We have
\begin{align}
\oint_{|z|=2}f(z)\,dz&=\int_0^{2\pi}f(2e^{it})\,{2ie^{it}}\,dt
=\int_0^{2\pi}f(2e^{-it})2ie^{-it}\,dt\\[0.3cm]
&=\int_0^{2\pi}\frac1{\frac14\,e^{2it}}\,f\Bigg(\frac1{\frac12\,e^{it}}\Bigg)\,\frac12ie^{it}\,dt
\\[0.3cm]
&=\oint_{|z|=\frac12}\frac1{z^2}\,f\Big(\frac1z\Big)\,dz.
\end{align}
Now your new function is
$$
\frac1{z^2}\,f\Big(\frac1z\Big)=\dfrac{1}{z^2\Big(\dfrac1z-4\Big)\Big(\frac1{z^7}-1\Big)}
=\frac{z^6}{4\Big(z-\dfrac14\Big)(z^7-1)}
$$
and you want to integrate on $|z|=\tfrac12$, so there is a single pole at $z=\tfrac14$. As the pole is simple, the integral is
$$
\frac{2\pi i (1/4)^6}{4((1/4)^7-1)}=\frac{2i\pi}{1-4^7}=-\frac{2i\pi}{16383}.
$$

ContentLicense: CC BY-SA 4.0


问:
Why does this trigonometric identity problem work?

Why do I multiply by 1 + sin (theta)? If I multiply on one side, shouldn't I multiply on the other side too? Sorry for the weird formatting. I'm still learning how to use MathJax.

答:
The proof above is making use of the identity
$$1-\sin^2x=\cos^2x$$
As we have a $\cos x$ in the numerator of the second given expression (on the right hand side of the equality) it make sense to try to create a $\cos x$ or a power of $\cos x$ in the numerator of our expression, which can easily be done as they have demonstrated.

Is that helpful?


问:
limit on incomplete gamma functions
I am trying to prove that
$$
\frac{x^{k}}{{\rm e}^{xk}\int_{x}^{\infty}
{\rm e}^{-tk}\,\,t^{k}\,{\rm d}t} =
ak + b + o\left(1\right)
$$
when $k\to \infty$ for any $x> 1$, or equivalently
$$
\int_{0}^{}\left({\rm e}^{-s}\,\,
\frac{x + s}{x}\right)^{k}\,{\rm d}s =
ck^{-1} + dk^{-2} + o\left(k^{-2}\,\right)
$$
where I can derive what should be the correct $a,b,c,d$.

Wolfram seems unable to expand or take limits when functions like $\Gamma(k + 1,kx)$ are involved.

Any idea $?$.

答:
We can start by factoring the $x$ out and breaking up the product:
$$f_n(x)=\frac{1}{x^n}\int_0^\infty e^{-ns}(x+s)^n\mathrm{d}s$$
Then assuming $n\in\Bbb{N}$ (doesn't matter too much if we're taking $n\to\infty$) we get
$$=\frac{1}{x^n}\int_0^\infty e^{-ns} \sum_{k=0}^n \mathrm{C}(n,k)x^{n-k}s^k~\mathrm{d}s=\sum_{k=0}^n \mathrm{C}(n,k)x^{-k}\int_0^\infty s^ke^{-ns}\mathrm{d}s$$

$$=\sum_{k=0}^n \mathrm{C}(n,k)x^{-k} ~\frac{k!}{n^{k+1}}$$
$$=(n-1)!\sum_{k=0}^n \frac{1}{(n-k)!}\cdot\frac{1}{(nx)^k}=(n-1)! \sum_{k=0}^n \frac{1}{k!}\cdot\frac{1}{(nx)^{n-k}}$$

$$=\frac{(n-1)!}{(nx)^{n}} \sum_{k=0}^n \frac{(nx)^k}{k!}\approx \frac{(n-1)!}{(nx)^n}\exp(nx)$$

$$f(x;n)\approx \frac{(n-1)!}{(nx)^n}\exp(nx)\tag{$n\gg 1$}$$
It has occurred to me that we can actually use Stirling's formula
$$(n-1)!=\Gamma(n)\approx \sqrt{\frac{2\pi}{n}}\frac{n^n}{e^n}$$
Meaning for large $n$,
$$f(x;n)\approx \frac{1}{x^n}\sqrt{\frac{2\pi}{n}}e^{n(x-1)}$$
Maybe this helps?

Title: limit on incomplete gamma functions


问:
How can I construct a variation problem whose solution is $f^{(3)}=0$?
For example, for the variation problem
$$
\min_{f\in H([a,b])} \int_a^b f''^2(x) \mathrm{d} x
$$

Based on Euler-Lagrange, we can obtain
$$
f^{(4)}=0
$$

Now I need to find a variation problem whose solution is
$$
f^{(3)}=0
$$

However, it seems really hard to construct such a variation problem. Something like $\min \int f' f'' \mathrm{d} x$ and $\min \int f f^{(3)} \mathrm{d} x$ cannot work.

Could anyone please help me with that?

答:
Any even number of differentiations is easy: $f^{(2n)}=0$ is the (higher) Euler-Lagrange equation for the functional $\int \! dx~|f^{(n)}(x)|^2$. Here $n\in\mathbb{N}_0$.

However OP wants to consider an odd number of differentiations: $f^{(2n+1)}~=~0$.

The remedies are standard:

If $f$ is a complex variable, then use the functional $\int \! dx~\overline{f(x)} f^{(2n+1)}(x)$.

If a Lagrange multiplier field $\lambda$ is allowed, then use the functional $\int \! dx~\lambda(x) f^{(2n+1)}(x)$.


问:
Projection of a Great Circle on another
Consider a great circle between $[lat_1, lon_1]$ and $[ lat_2, lon_2]$, on a perfectly spherical earth.

Consider a second one : between  $[lat_1, lon_1 +b]$ and $[ lat_2, lon_2+b]$. For a very small constant $b$, if the great circles themselves are small, they can be considered parallel to each other.

Both are segments of circles and thus does not go completely around the world. I want to project one great circle segment upon the other to find out if these segments "overlap" each other.

Overlap :

Overlap happens in between the two green dotted lines. If I move the second great circle by $-b$, then the "overlap" is the common sub-segment of both segments of great circles.

Edit after @michaelseifert 's comment :

The green lines should be normals dropped from one segment to another. However, I also do not have a proper algorithm / formula to do that. After the projection on the cylinder, I was painstakingly evaluating each endpoint to drop these normals. Even tho, it is done on a computer, I think there should be better ways of doing it.

Edit 2 :

Once I project the Great Circles on a cylinder (see below in attempt to solution), the arcs of latitude and longitude become straight lines. For very short lengths of great circle segments, I can approximate them as straight lines. Then I can drop normals.

The outcome will be the same, if I could draw the shortest cross track great circle segments between the two given great circle segments defined by $[lat_1, lon_1]$ , $[ lat_2, lon_2]$, and  $[lat_1, lon_1 +b]$ , $[ lat_2, lon_2+b]$ - from the points that delimit the "overlapping". I realize that overlapping and the normals are defined w.r.t. each other, and that bad. But the plan is to extract parallel GPS tracks, and trim them if one exceeds a certain length beyond overlapping.

Attempt to solution

I tried to convert the polar coordinates to cartesian, butthey are in 3D.

In order to reduce them to 2D i am projecting them on a cylinder like mercator projection and unwrapping it.

Problem with this approach

At higher latitudes the surface of the cylinder is very far away from the surface of the sphere leading to large errors.

Question

How can I project one great circle segment on another without first having to project them on a cylinder or other surface?

Thank you

答:
Comment... trying to understand your question..i.e,

By their definition, two geodesics/great circles can never be parallel. They intersect at two points.

For parallels at least one should be a small circle.

Every small circle is non-geodesic, that is, the normal to the surface at a given point P and normal to the small circle at P  makes the same angle between them.

I.e., the $\phi=5^{\circ}$ parallel circle on a sphere has this angle constant  all along the small circle length.

Two given points and sphere center determine a great circle. With respect to a great circle can define the relative latitude for a small parallel circle $\phi$ thus:

For a great circle Clairaut circle radius $  is a constant (blue). For a small circle it varies as shown (red). Using Liouville theorem of geodesic curvature we can establish that

$$\tan \phi= \dfrac{d (r \sin \psi)}{dz}=\dfrac{d (r_{Clairaut})}{dz}; $$
It vanishes for a great circle but is a constant for small circle, $\psi$ being angle between meridian and the small circle.

Title: Projection of a Great Circle on another


问:
How to find density function?
$X \sim N(1,4)$ and $Y = 3 - 5X$. How to find the density function of $Y$?

I tried first to find the distribution function of Y, but got stuck.

$$F(y) = P(Y <= y) = P(3 - 5X <= y) = P(X >= (-y + 3)/5).$$

Any help appreciated.

答: Hint: Try to calculate the mean and variance of $Y$. $Y$ also has a normal density, just its mean and variance are different from $X$.

ContentLicense: CC BY-SA 3.0

Title: How to find density function?


问:
Connecting the two definitions of $e$
There are two different ways I've been taught to understand the meaning of $e$.

$e$ can be defined as the total growth from continuously compounding interest in a single period. To make the growth rate the same as the dollar amount, consider I start with $\$1$. Then, $a_n$ represents the amount I have after compounding $n$ times.
\begin{align}
a_n&=1\cdot\left(1+\frac{1}{n}\right)^n\\[5pt]
e=a_{\max}&=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n
\end{align}
As we increase the number of times we compound the interest, the final amount (in the case that we start with $\$1$), approaches $e$.

All exponential functions increase at a rate proportional to their current value. For example, consider the derivative of $2^x$:
\begin{align}
\frac{d\left(2^x\right)}{dx}&=\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x2^h-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x\left(2^h-1\right)}{h}\\[5pt]
&=2^x\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}
\end{align}
From this we can see that the rate at which $2^x$ changes is proportional to its current value, with a proportionality constant of $$\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}.$$ For $2^x$, this constant happens to be $\ln(2)$ (why is this the case?). $e$ can be defined as the base where this proportionality constant is $1$; therefore, the rate $e^x$ grows at is exactly its current value.

My question is this: how can the same constant have two (apparently) separate definitions? Is there a way of looking at $e$ that will combine these two definitions, and make it obvious as to why they are true? Also, why is the proportionality constant of any exponential functions growth rate the natural log of its base? Is there a way to understand why this is the case intuitively?

答:
This Wikipedia article: Characterizations of the exponential function, shows six ways to define the exponential function $e^x$ (thus for the constant $e$) and a detailed discussion of equivalency of all the six definitions.

This happens a lot in mathematics. The same mathematical object may have many properties that can characterize the object. Once they are proved to be equivalent to the definition, they can all be regarded as a definition. More generally, there are equivalent definitions of mathematical structures.

For an intuitive understanding of the constant $e$ and exponential growth, you may read this article https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ from the "BetterExplained" website.

Title: Connecting the two definitions of $e$


问:
Understanding the proof of $\sim\!\!(\forall x)A(x)\!\iff\!(\exists x)\sim \!\! A(x)$
I need to understand the proof of $\sim\!\!(\forall x)A(x)\!\iff\!(\exists x)\sim \!\! A(x)$ in "A Transition to Advanced Mathematics: Edition 8" pg. 23 Theorem 1.3.1 a)

It states:

Let $U$ be any universe (universe of discourse, the set of values
being considered for $x$)

The sentence $\sim\!\!(\forall x)A(x)$ is true in U

iff the truth set of $A(x)$ is not the universe

iff the truth set of $\sim A(x)$ is nonempty

iff $(\exists x)\sim \!\! A(x)$ is true in U

I couldn't rationalize how "iff the truth set of $\sim A(x)$ is nonempty" led to " iff $(\exists x)\sim \!\! A(x)$ is true in U". Couldn't we also say "iff $(\forall x)\sim \!\! A(x)$ is true in U"?

Is there a better proof I can understand? So far I learned about negation, conjuctions, disjunctions, conditionals and biconditionals.

答:
$\lnot(∀x)A(x)$ means that in the "universe" $\text U$ not every object is an $A$.

In the "universe" $\mathbb N$ of natural numbers, not every number is even.

Thus, there is some object that is a not-$A$, i.e. $(∃x)\lnot A(x)$.

In $\mathbb N$ there are numbers that are not-even, i.e. odd.

And vice-versa.


问:
$E e^{B_s^2 e^{2B_s}}\mathbb{1}_{\{B_s\leq -1\}} < \infty$
I wan't to see that $E e^{2B_s e^{2B_s}}\mathbb{1}_{\{B_s\leq -1\}} < \infty$ when $B_s\sim N(0,s)$ (standard brownian motion). Any help, please?

答:
The expected value you are after is $$\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{-1}e^{-\frac{x^2}{2 s}}e^{e^{2x}x^2}dx=\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{-1}e^{-x^2\left(\frac{1}{2 s}-e^{2x}\right)}dx.$$
Now it is easy to show that for some $-\infty<x_0<-1$ we have that $\frac{1}{2 s}-e^{2x}>0$ for all $x<x_0$. And the result you are after follows.


答:
Take the second definition:

Consider:  $\lim_{h\to 0}\frac {b^h -1}h = \ln b = 1$. And solve for $b$.  That should be $e$ and we could define, supposedly, $e$ as the $b$ where that limit is $1$.

....

$\lim_{h\to 0}\frac {b^h -1}h = $

$\lim_{\frac 1n\to 0}\frac {b^{\frac 1n} -1}{\frac 1n} = $

$\lim_{n\to \infty} n(b^{\frac 1n} - 1)=1$

Now for any $k$ if $k(b_k^{\frac 1k} - 1)=1$ then $b_k= (1+ \frac 1k)^k$ and so if we wave our hands really fast and point out that "Look, it's Halley's comet" it should be reasonable that

If $\lim_{n\to \infty}n(b^{\frac 1n} -1) = 1$ then

$b = \lim_{n \to \infty}(1 + \frac 1n)^n$.

So the definitions are consistent.

Title: Connecting the two definitions of $e$

Body:
There are two different ways I've been taught to understand the meaning of $e$.

$e$ can be defined as the total growth from continuously compounding interest in a single period. To make the growth rate the same as the dollar amount, consider I start with $\$1$. Then, $a_n$ represents the amount I have after compounding $n$ times.
\begin{align}
a_n&=1\cdot\left(1+\frac{1}{n}\right)^n\\[5pt]
e=a_{\max}&=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n
\end{align}
As we increase the number of times we compound the interest, the final amount (in the case that we start with $\$1$), approaches $e$.

All exponential functions increase at a rate proportional to their current value. For example, consider the derivative of $2^x$:
\begin{align}
\frac{d\left(2^x\right)}{dx}&=\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x2^h-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x\left(2^h-1\right)}{h}\\[5pt]
&=2^x\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}
\end{align}
From this we can see that the rate at which $2^x$ changes is proportional to its current value, with a proportionality constant of $$\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}.$$ For $2^x$, this constant happens to be $\ln(2)$ (why is this the case?). $e$ can be defined as the base where this proportionality constant is $1$; therefore, the rate $e^x$ grows at is exactly its current value.

My question is this: how can the same constant have two (apparently) separate definitions? Is there a way of looking at $e$ that will combine these two definitions, and make it obvious as to why they are true? Also, why is the proportionality constant of any exponential functions growth rate the natural log of its base? Is there a way to understand why this is the case intuitively?


问:
Infinite Collection of Sets with These Properties
I passed by this problem in Abbott's "Understanding Analysis", page 11:

Produce an infinite collection of sets $A_1, A_2, A_3,...$ with the
property that every $A_i$ has an infinite number of elements, $A_i ∩ A_j = \phi$ for all $i \not= j$ , and
$\bigcup_{i=1}^{\infty} A_i = \mathbf N$

Finding the solution was not difficult conceptually (that is, if my attempt is correct), but I have no idea how to formulate it.
My idea is as follows:

If the question asked for two sets, an answer would be the sets of even and odd numbers. If they're $3$ sets, I would then give $A_1$ all the numbers starting from $1$ and increment $3$ to get $A_1 = \{1,4,7,...\}$, similarly, for $A_2$, starting from $2$ and incrementing by $3$, etc... For $n$ sets, $A_i = \{i, i+n,i+2n,...\}$.

However, for an infinite number of sets, formulating a more formal solution is confusing to me.

My other question is this: Is this the only way to solve this problem, and why?

答: Hint: use the Fundamental Theorem of Artithmetic and Euclid's proof that there are infinitely many primes. Good luck!</p>


问:
Deducing finiteness of the integrand from finiteness of integral
Suppose we know that:
$$
\int_{[-\pi, \pi]^d} \, g(k) \,  h^2(k) \, dk  \leq\int_{ [-\pi, \pi]^d} \, h(k) \, dk < \infty.
$$
where $h(k) = \sin(\frac{k_1}{2})^2 +  \sum_{i=2}^{d} \sin(k_i)^2.$
Assuming $d > 2$, is it possible from this to deduce that,
$$
\int_{[-\pi, \pi]^d} \, g(k)  < \infty?
$$

答:
The answer to your question is: no, at least for $d<5$.

Consider $g(x)=\frac{\int_{ [-\pi, \pi]^d} \, h(k) \, dk}{h^2(x)}$. Then clearly the desired conditions are satisfied. However, for $x\in\mathbb{R}^d$ such that $x\simeq 0$ one has
$$
h(x)\simeq |x|^2,
$$
since
$$
\sin y\simeq y, \quad \textrm{for}\quad y\sim 0,\, y\in\mathbb{R}.
$$
Consequently,
$$
g(x)\simeq |x|^{-4}, \quad \textrm{for}\quad x\sim 0,
$$
thus $g$ is not integrable for $d<5$.


问:
If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
I'm reading elementary number theory and trying to understand the following problem: If $x^2\equiv 1 \pmod{n}$, $n=pq$, $p$ and $q$ are odd primes and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of $n$.

EDIT: Andreas Caranti wrote an updated, corrected version of my problem, so I wrote the definition again.

答:
If $n=p\cdot q$ divides $x^2-1=(x+1)(x-1)$

If odd prime $r$ divides $x-1$ and $x+1,$ it will divide $(x+1)-(x-1)=2$ which is impossible $\implies r$ can divide exactly one of  $x-1,x+1$

Now let us check the possibilities

Case $1:$   $p$ divides $x-1$ and $q$ divides $x-1\implies (x-1)$ is divisible by lcm$(p,q)=pq=n\implies x\equiv1\pmod n$ which is not acceptable according to the given condition.

Similarly, Case $2:$ $p$ divides $x+1$ and $q$ divides $x+1$ is discarded

Case $3:$  $p$ divides $x-1$ and $q$ divides $x+1$

Then $x-1=p\cdot a,x+1=q\cdot b$ where $a,b$ are some integers

i.e., $q\cdot b-p\cdot a=2$ , we can always find such $a,b$ using Bézout's Lemma as $(p,q)=1$

If $q$ divides $a,$ the LHS will be divisible by $q$

$\implies 2$ must be divisible by $q$ which is impossible $\implies (q,a)=1$

Now, gcd$(x-1,n)=$gcd$(p\cdot a,p\cdot q)=p\cdot$gcd $(a,q)=q$ as gcd$(a,q)=1$

Case $4$  $p$ divides $x+1$ and $q$ divides $x-1$ can be handled similarly.


问:
How to explain this in propositional logic?
Suppose I am given the statement $a \rightarrow b$ (1), and I know $b$ is true. Can I also conclude $a$ is true from just the given information (1)?
My logic for trying to figure out this was —
Let me apply the ¬ operator to the given statement (1). Then, $¬ a \rightarrow ¬ b$ (2). ( I am not sure if this can be done.)
Then, if it were the case that $a$ is false, then $¬ a$ would be true. Which would also imply from (2) that $¬ b$ is true, $\implies b$ is false. This is a contradiction to the information that we are given that $b$ is true.
So, $a$ must be true.

What confuses me is that for the general truth table of $a \rightarrow b$, there is a particular case where $a=0, b=1$ and $a \rightarrow b$ is also $1$.

Edit: My confusion arose after seeing this being done in class. I had a statement I) $p \rightarrow q$ and II) $¬ q$. II) is true. From these two, the instructor concluded that $¬ p$ is true. I don’t quite understand how that happened.

答:
You're making some errors in your reasoning.

(i) From $a \Rightarrow b$, if b is true we cannot conclude that a is true. For example, if f is differentiable then f is continuous; but if f is continuous f doesn't need to be differentiable.

(ii) Let say $a \Rightarrow b$. Then, the implication is false only in the case that "a" is true but "b" is false. The case you say where $a=0,b=1$ corresponds to the case when the hypothesis is false, hence there's no implication so the statement is true. It might seem strange but try to reason why $\emptyset \subset A, \forall A$.

(iii) Finally, the negation of a statement is done as it follows: $a \Rightarrow b$ is equivalent to $\lnot b \Rightarrow \lnot a$. For example, the following definitions of inyective functions is equivalent: $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ or $x_1 \not = x_2 \Rightarrow f(x_1) \not = f(x_2)$.


问:
about valuation
Definition. Let $K$ be a field and let $G$ be a totally ordered abelian group. A valuation of $K$ with values in $G$ is a map $v: K\backslash \{0\} \rightarrow G$ such that for all $x, y \in K, x, y \neq 0,$ we have
$$
\begin{array}{l}
\text { (1) } v(x y)=v(x)+v(y) \\
\text { (2) } v(x+y) \geqslant \min (v(x), v(y)) .
\end{array}
$$

why $G$ must be abelian group in above definition ? if $G$ be any arbitrary group then definition is false ?

why $G$ must be totally ordered group in above definition ?

for 2) i think $G$ must be totally ordered group because we can understand  any element is  "positive" or "negative".(an element $g \in G$ is positive if $0<g$ and negative if $0>g$ )

is there a totally ordered abelian group $G$ such that every element $g \in G-\{0\}$ be negative ? or in every non trivial group we have at least one positive  element ?

also i want to find simplest example of valuation.

答:
The first valuation axiom says that $v$ is a group homomorphism from $K^\times$ (the group of nonzero elements of $K$ with multiplication) to $G$. Since $K$ is a field and one of the field axioms says that multiplication is commutative, that means that the image of $v$ is a commutative subgroup of $G$, regardless of whether $G$ is assumed to be commutative. So there's no point in allowing $G$ to be noncommutative, because we don't actually gain any extra generality that way.

Conceptually, the point of a valuation is that it gives a way of comparing the "size" or "multiplicity" of elements of a field. In order to do that, we have to be able to compare any two elements of the value group $G$. You could, of course, generalize this by allowing $G$ to only be partially ordered or something like that, but most of valuation theory probably wouldn't work with this generalization. (For example, you need the value group to be totally ordered for the valuation to induce a metric on $K$ and thus turn $K$ into a metric space.)

The definition of an ordered group requires the order relation to be translation-invariant, so if $g < 0$, then $0 = -g + g < -g + 0 = -g$. Thus the inverse operation in the group gives a bijective correspondence between negative elements and positive elements.

As for examples of valuations, here are the simplest ones that come to mind:

Let $p$ be a prime number. Define a valuation $v_p$ on the field of rational numbers $\mathbb{Q}$ as follows: for any nonzero $m \in \mathbb{Z}$, define $v_p(m) = k$, where $m = p^k d$ with $\gcd(d, p) = 1$. For any nonzero $m, n \in \mathbb{Z}$, define $v_p(m/n) = v_p(m) - v_p(n)$. So, for example, $v_3(18/5) = 2$, while $v_5(18/5) = -1$.

Let $K$ be a field, let $K(x)$ be the field of rational functions in one variable over $K$, and let $c \in K$ be arbitrary. Define a valuation $v_c$ on $K(x)$ as follows: for a polynomial $f \in K[x]$, define $v_c(f)$ to be the order of vanishing of $f$ at $c$, i.e., the multiplicity with which $(x - c)$ occurs in the factorization of $f$. For $f/g \in K(x)$, define $v_c(f/g) = v_c(f) - v_c(g)$; you can think of this as the "order of zeroes or poles". For example, $v_0(x^2/(x - 3) = 2$, while $v_{3}(x^2/(x - 3)) = -1$, and $v_2(x^2/(x - 3)) = 0$.

Title: about valuation


答:
No. This is known as the fallacy of affirming the consequent.

Applying negation to the implication yields $\neg(a \to b)$, not $\neg a \to \neg b$, and they are not equivalent.

I also don't see where you're trying to get by applying negation to the implication. If what you want to do is a proof by contradiction, you would have to assume that $a$ is false, then in combination with the other assumptions derive a contradiction from that assumption and conclude that $a$ must be true after all.

But there is no contradiction: $a \to b$ can be true when just $b$ is true and $a$ is false. There is nothing more to it than the fact that that's what the truth table states.

If you find this unintuitive, try to avoid translating "$\to$" in your mind as an English "if $\ldots$ then", and think of it as "(not $a$) or ($b$)" or "the only thing that must not happen is for $a$ to be true but $b$ false" instead, since that's the truth conditions $\to$ is intended to express.

There are many explanations of why these are the truth conditions; see for example In classical logic, why is $p \to q$ True if $p$ is False and $q$ is True?.

Re. your edit: It's different when $b$ is false. According to the truth table of implication, in order for $a \to b$ to be true it must be the case that either $a$ is false or $b$ is true, or in other words, what can not be the case is that $a$ is true but $b$ is false. So if $a \to b$ is true but $b$ is false, then $a$ must be false as well, since otherwise you'd get the line in the truth table in which $a \to b$ is false. In contrast, there are two possible lines in the truth table of $a \to b$ and $b$ being true, one where $a$ is true and one where $a$ is false, so you can not definitely conclude the truth of $a$ just from the truth of $a \to b$ and $b$.

Title: How to explain this in propositional logic?

Body:
Suppose I am given the statement $a \rightarrow b$ (1), and I know $b$ is true. Can I also conclude $a$ is true from just the given information (1)?
My logic for trying to figure out this was —
Let me apply the ¬ operator to the given statement (1). Then, $¬ a \rightarrow ¬ b$ (2). ( I am not sure if this can be done.)
Then, if it were the case that $a$ is false, then $¬ a$ would be true. Which would also imply from (2) that $¬ b$ is true, $\implies b$ is false. This is a contradiction to the information that we are given that $b$ is true.
So, $a$ must be true.

What confuses me is that for the general truth table of $a \rightarrow b$, there is a particular case where $a=0, b=1$ and $a \rightarrow b$ is also $1$.

Edit: My confusion arose after seeing this being done in class. I had a statement I) $p \rightarrow q$ and II) $¬ q$. II) is true. From these two, the instructor concluded that $¬ p$ is true. I don’t quite understand how that happened.


问:
Evaluate $\lim_{x\to \infty} \frac{\{x\} \ln x^n}{[x]}$
Here, $\{.\}$ represents fractional part function and $[.]$ represent integral part function

Can I get a hint to get me started, because I have no idea how to begin. Not the complete answer, just a hint.

答:
$\lim_{x\to \infty} \frac{\{x\} \ln x^n}{[x]}$

This thing looks scary, but it's surprisingly easy.

I'll assume $n>0.$

I'm not going to spoon-feed you the answer, and I'm going to go out on a limb here and guess that you're new to these two terms "fractional part" and "integral part".

I also haven't heard these terms before, but they're easy to understand.

The "fractional part" (also called the "decimal part") of $123.79$ is $0.79$.

The "integral part" ("integer part") of $123.79$ is $123$.

So for large $x$, the fractional part $\{x\}$ is between $0$ and $1$. But the integral part $[x]$ is large (a large integer).

Also, I'll wager that you're capable of simplifying $\ln(x^n)$.
$$$$
OP, get your spoon out. For every nonnegative integer $n$, we have all of the following:

For all $x>1, x^n\geq1,\ $ so $\ln x^n\ $ is positive.

Furthermore, for all $x,\ $ $0 \leq \{x\} < 1,\ $ and $\ 0< x-1 < [x].\ $ Therefore:

$$\forall x>1:\quad 0<\frac{\{x\} \ln x^n}{[x]} < \frac{1 \times \ln x^n}{x-1} = n\ \frac{\ln x}{x-1},\ $$ and

$$\lim_{x \to \infty} \left(n\ \frac{\ln x}{x-1}\right) = n \times \lim_{x \to \infty} \left(\frac{\ln x}{x-1}\right) \overset{\mathrm{(*)}}{=} n \times \lim_{x \to \infty} \left(\frac{\frac{1}{x}}{1}\right)\ = n \times \lim_{x \to \infty} \left(\frac{1}{x}\right)\ = n \times 0 = 0,$$

where L'hopital's rule was used at $(*).$

Therefore, by the squeeze theorem, $\ \lim_{x \to \infty} \left(\frac{\{x\} \ln x^n}{[x]}\right) = 0.$

$$$$
Finally, for negative integers $-n,\ (n >0),\  \large{\frac{\{x\} \ln x^{-n}}{[x]} =  -n \frac{\{x\} \ln x}{[x]}},\ $ and so $$\ \lim_{x \to \infty} \left(\frac{\{x\} \ln x^{-n}}{[x]}\right) = -n \times \lim_{x \to \infty} \left(\frac{\{x\} \ln x}{[x]}\right) = -n\times 0 = 0.$$


问:
Studying convergence of improper integral
I have a doubt on how to apply asymptotic comparison to this improper integral: $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$

I'll call $f(x)=\dfrac{4x}{4x^8 + 1}$, and notice it is non negative as $x \to +\infty$.

If I understand the idea correctly, the I need to find a function $g(x)$ (also non negative as $x \to +\infty$) such that:

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$ is finite and not zero: $\displaystyle\int_0^{+\infty} f(x) dx$ converges if and only if $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 0$ and $\displaystyle\int_0^{+\infty} g(x) dx$ converges: $\displaystyle\int_0^{+\infty}f(x)dx$ converges.

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = +\infty$ and $\displaystyle\int_0^{+\infty} f(x) dx$ converges: $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

If I choose $g(x) = \dfrac{1}{x^7}$, I have $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 1$, so the first case applies. I know that $\displaystyle \int_0^{+\infty}\dfrac{1}{x^7}dx$ diverges, so $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$ should diverge, but it does not..

What am I doing wrong?

答:
if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$ is finite and not zero: $\displaystyle\int_0^{+\infty} f(x) dx$ converges if and only if $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

This is false. The reason that $\int_0^\infty \frac1{x^7}dx$ diverges is not because of the behaviour as $x\to\infty$ but rather near $x=0$.

Try splitting up your integral into two intervals, say $(0,1)$ and $(1,\infty)$. In particular, note that $\int_1^\infty \frac1{x^7}dx$ converges.


问:
Is C an invertible matrix where CDCDC+C=0?
Its a hw question so i'd prefer a hint over a solution.
We know that $C$ and $D$ are 7x7 real matrices, and need to find out whether $C$ is invertible or not, based of of the equation: $CDCDC+C=0%$.
I've got to a point where $DC=-CD$

答: If $C$ is invertible then $C(DCDC+I)=O$ gives $(DC)^2=-I$ and $(\det{DC})^2=-1$.


问:
Range of $\frac{\cos\theta_1+\cdots+\cos\theta_{10}}{\sin\theta_1+\cdots+\sin\theta_{10}}$ given $\sin^2\theta_1+\cdots+\sin^2\theta_{10}=1$

Given that for $ \theta_i \in \left[0, \dfrac{\pi}{2}\right]$, where $1 \le i \le 10$ , $\sin^2\theta_1+\sin^2\theta_2+\cdots+\sin^2\theta_{10}=1$, find the minimum and maximum value of $$\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}}$$

I tried using Cauchy Schwarz inequality which just gives $\displaystyle \sum_{i=1}^{10} \sin\theta_i \le \sqrt{10}$ and $\displaystyle \sum_{i=1}^{10} \cos \theta_i \le \sqrt{90}$ because $\displaystyle \sum_{i=1}^{10}\cos^2\theta_i =9$ but I don't think it would be helpful here.

Maybe some inequality can be applied by writing it as $$\dfrac{\displaystyle \sum_{i=1}^{10} \sqrt{1-\sin^2\theta_i}}{\displaystyle \sum_{i=1}^{10} \sin \theta_i}$$

Any hints would be appreciated!

答:
The method suggested by @Albus Dumbledore can be tweaked slightly to prove that the upper bound of the given expression is $9$. Note that, since $\theta_i \in \left[0,\dfrac{\pi}{2}\right]$, $\cos(\theta_i) \geq 0$ and $\sin(\theta_i) \geq 0$. We will show that $\cos(\theta_i) \leq S_i$:

\begin{align}
\cos(\theta_i) \leq S_i & \iff \cos^2(\theta_i) \leq S_i^2 \\
& \iff \sum_{j \neq i} \sin^2(\theta_j) \leq \left( \sum_{j \neq i}  \sin(\theta_j) \right)^2
\end{align}
Which is obviously true.
Thus, $\dfrac{\cos \theta_1+\cos\theta_2+\cdots+\cos\theta_{10}}{\sin\theta_1+\sin\theta_2+\cdots+\sin\theta_{10}} \leq  \frac{S_1+S_2+S_3+\cdots+S_{10}}{\sin \theta_1+\sin \theta_2+\cdots+\sin \theta_{10}}=\boxed{9}$ . Equality holds when $\theta_1=\dfrac{\pi}{2}, \theta_2=...=\theta_n=0$, up to permutation.


问:
how to solve the following recurrence relation
Suppose I'm given the following recurrence relation, where $n, r, s$ are fixed

$$a_1 = ra_2+ sa_n$$
$$a_i = ra_{i+1}+sa_{i-1}, i=2,\dots n-1 $$
$$a_n = ra_1+sa_{n-1}$$

and we have $\sum_i^n a_i = 1$ and $r+s=1$.

I've to express the middle equation as $a_2-a_{k+1}=\sum_{i=2}^k = s(\sum_{i=2}^k a_{i-1}-a_{i+1}=a_1+a_2-a_k-a_{k+1}$ where I've used that the second relation is equivalent to

$$ a_i-a_{i+1}=s(a_{i-1}-a_{i+1}) $$

Apparently the solution should be $a_i=\frac{1}{n}$ but I don't see how I can get this from my attempts so far

答:
If $r=1$ or $r=0$ we have $a_1=a_2 = \cdots = a_n = \frac 1n$. From now on assume $r\neq 0, 1$. From $r+s=1$ we have
$$r(a_{i+1}-a_i) = (1-r)(a_i - a_{i-1}) \tag1$$

If $\exists j, a_{j+1}=a_j$, it's easy to see that $a_{i+1}-a_i=0, \forall i$.

If not, then $a_{i+1} \neq a_i, \forall i$. We multiple $(1)$ from $i=1$ to $n$ (we may assume $a_0=a_n, a_{n+1}=a_1$), and divide both sides by $\prod_{i=1}^n (a_{i+1}-a_i) \neq 0$, then $r^n = (1-r)^n \implies r=\frac 12$ (note that when $n$ is even we cannot have $r=-(1-r)$). But then $$a_{i+1}-a_i = d, \forall i \tag2$$

We add $(2)$ for all $i$ then $0 =\sum a_i - \sum a_i = nd=0, d=0, a_i=\frac 1n$.


答:
I think, you used some thing wrong. But it seems you have the idea.

Hint

$$\frac{4}{x^{-1}(4x^8+1)}$$ split the integral to $I_1$ and $I_2$ where $$I_1=\displaystyle\int_0^{1} \frac{4x}{4x^8 + 1}dx$$and $$I_2=\displaystyle\int_1^{+\infty} \frac{4x}{4x^8 + 1}dx$$ Now, $$I_1\leq \displaystyle\int_0^{+1} \frac{4}{x^{-1}}dx$$ and $$I_2\leq \displaystyle\int_1^{+\infty} \frac{4}{x^7}dx$$
I think, it is clear now. I hope that helps

Title: Studying convergence of improper integral

Body:
I have a doubt on how to apply asymptotic comparison to this improper integral: $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$

I'll call $f(x)=\dfrac{4x}{4x^8 + 1}$, and notice it is non negative as $x \to +\infty$.

If I understand the idea correctly, the I need to find a function $g(x)$ (also non negative as $x \to +\infty$) such that:

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$ is finite and not zero: $\displaystyle\int_0^{+\infty} f(x) dx$ converges if and only if $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 0$ and $\displaystyle\int_0^{+\infty} g(x) dx$ converges: $\displaystyle\int_0^{+\infty}f(x)dx$ converges.

if $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = +\infty$ and $\displaystyle\int_0^{+\infty} f(x) dx$ converges: $\displaystyle\int_0^{+\infty}g(x)dx$ converges.

If I choose $g(x) = \dfrac{1}{x^7}$, I have $\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 1$, so the first case applies. I know that $\displaystyle \int_0^{+\infty}\dfrac{1}{x^7}dx$ diverges, so $\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$ should diverge, but it does not..

What am I doing wrong?


问:
Boundedness of optimal input of scalar system
Context: Given is the scalar system
$$\dot{x}=-ax+u$$ with $x(0)=x_0$ and $|u(t)|\leq M$ for a constant $a$ and a positive constant $M$ along with cost criterion
$$J(u)=x(1)+\frac{1}{2}\int_{0}^{1}u^2(t)dt.$$

Tasks:

Determine the Hamiltonian for the optimal control problem and the differential equation for the costate $p$.

Determine the optimal input in the case that $M=\infty$.

Determine the optimal input in the case that $M$ is finite, distinguishing between the cases $a\geq0$ and $a<0$.

My attempt: For the first task, I found the Hamiltonian $$H(x,p,u)=p(-ax+u)+\frac{u^2}{2}$$ and that the differential equation for the costate is given by $$\dot{p}(t)=ap(t)$$ with the conditions $p(1)=1$, the solution of which is $$p(t)=e^{a(t-1)}.$$

For the second task, in minimising the Hamiltonian, we find that the optimal input is given by $$u(t)=-p(t)=-e^{a(t-1)}.$$

Question: From here my question is what effect the bound $|u(t)|\leq M$ has on this optimal input. Is it maybe that for particular values of finite $M$, the optimal input $u(t)$ is not given by $-e^{a(t-1)}$? Any help is appreciated.

答:
The optimal control in general when using Pontryagin's maximum principle can be found by minimizing the Hamiltonian with respect to the control variables. In your case this would be the constrained optimization problem

\begin{align}
\min_{u}\ & p(-ax+u)+\frac{u^2}{2} \\
\text{s.t.}\ & |u| \leq M
\end{align}

Since the cost function is convex and the decision variable is scalar the solution would either be equal to the unconstrained solution ($u(t)=-p(t)$) or on the closest boundary of the constraint if that solution violates that constraint. Therefore, the constrained solution can be written as

$$
u(t) = \left\{
\begin{array}{ll}
M & \text{if}\ \, p(t) < -M, \\
-M & \text{if}\ \, p(t) > M, \\
-p(t) & \text{otherwise}.
\end{array}
\right. \tag{1}
$$

Since the solution for $p(t)$ in this case is always positive one could also remove the expression for $p(t) < -M$ from $(1)$. Depending on the values of $M$ and $a$ only one or two expressions from $(1)$ will be used. For example for $M<1$ the expression from $p(t) > M$ will always be used, since $p(1)=1$, but for sufficiently large $a$ it would at least be the case that $p(0) \ngtr M$. It can be noted that the exponential expression for $p(t)$ is monotonicly increasing or decreasing, so there will be at most one switch of which expression is being used from the right hand side from $(1)$. Furthermore, in the limit of $M$ to infinity (and finite $a$) the constraint problem becomes equivalent to the unconstrained case.


问:
Why this iteration summation is wrong when I calculate it?
$V_n = 3^{n}$

I need to calculate its summation where

$S = (V_0)^{2} + (V_1)^{2} + ... + (V_{n-1})^2 $

Obviously, you can just put a new iteration $(W_n)$ where $W_n = (V_n)^{2}$ You find it $W_n = 9^{n}$ and then in the end you find

$S = (9^{n} - 1) / 8$

But when I tried putting it like this

$S = 3^{2(0)} + 3^{2(1)} + ... + 3^{2(n-1)}$

I find $S = (3^{2n-1} - 1) / 2$

And if you put $n = 1$ or $n = 0$ in both summations we found, you clearly don't find the same value which means the second one is probably wrong but I don't understand why. It should be the same.

答: No, $(3^{2n-1}-1)/2$ would be $3^0+3^1+\cdots+3^{2n-2}$, whereas $S$ lacks the odd-exponent terms. As @zwim notes, the even terms are $S$ and the odd ones are $3S$, and their sum is $4S$. Note $4=(3^2-1)/(3-1)$.


问:
The Basis for Product Toplogy
Hi i am reading Topology by Munkres and having 3 doubts under the section The product Topology at page no 86. There it is written that

The union of the two rectangles pictured in Figure is not a product of two sets so it cannot belong to $\mathcal{B}$ however it is open in $X\times Y$ .

My doubts are as following:

Why is it that the union of two rectangles is not a product of two sets?
I think what i am asking is why is this $(U_1\times V_1)\ \cup (U_2\times V_2) = (U_1\ \cup U_2)\times (V_1\ \cup V_2)$ not true?

How is that union  open in $X\times Y$ ?

Why is this $(U_1\times V_1)\ \cap (U_2\times V_2) = (U_1\cap U_2)\times (V_1\cap V_2)$ true? Is there a rule that applies to intersection but not to union?Or rather which law(like De Morgan's law) are we using?

For reference i am attaching the screenshot where i have highlighted the part.

答:
If you pick an element $u\in U_1\setminus U_2$ and $v\in V_2\setminus V_1$, it is clear that $(u,v)$ is in $(U_1\cup U_2)\times(V_1\cup V_2)$ but not in $(U_1\times V_1)\cup(U_2\times V_2)$.

Any union of sets of the basis of a topology is an open set, so this is a direct consequence of the definition of product topology.

A point $(u,v)$ is in $(U_1\times V_1)\ \cap (U_2\times V_2)$ iff it is in $U_1\times V_1$ and in $U_2\times V_2$, that is, if $u$ is in $U_1$ and in $U_2$ and $v$ is in $V_1$ and $V_2$. This is equivalent to saying that $u\in U_1\cap U_2$ and $v\in V_1\cap V_2$, which means that $(u,v)$ is in $(U_1\cap U_2)\times (V_1\cap V_2)$.

You can deduce all these laws regarding products, unions and intersections using just the definitions of these concepts and logic. The only logic laws that have been implicitly used in this reasoning are associativity and commutativity of conjunctions.

Title: The Basis for Product Toplogy


问:
How to find density function?
$X \sim N(1,4)$ and $Y = 3 - 5X$. How to find the density function of $Y$?

I tried first to find the distribution function of Y, but got stuck.

$$F(y) = P(Y <= y) = P(3 - 5X <= y) = P(X >= (-y + 3)/5).$$

Any help appreciated.

答:
Note that: $$P(X >= (-y + 3)/5)= 1 - P(X <= (-y + 3)/5) = 1 - \Phi((-y+3)/5,1,4)$$

where,

$\Phi(x,\mu,\sigma)$ stands for the cdf of a normal with mean $\mu$ and standard deviation $\sigma$.

Now differentiate both sides to get the pdf of $Y$ and show that it is a normal with the appropriate mean and variance.

Title: How to find density function?


问:
What is the best way to solve modular arithmetic equations such as $9x \equiv 33 \pmod{43}$?
What is the best way to solve equations like the following:

$9x \equiv 33 \pmod{43}$

The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.

Is there a more efficient way ?

Help would be greatly appreciated!

答:
$43$ is prime.  And even if it weren't, $\gcd(9,43)= 1$.

That means we know that $9^{-1}_{43}$ exists.  That is there is a number so that $9\cdot 9^{-1}_{43} \equiv 1 \pmod {43}$.  And we could calculate  (or guess) that $9^{-1}_{43}\equiv 24$ (because $9\cdot 24\equiv 1 \pmod {43}$) and we can solve $9x \equiv 33\pmod {43} \implies 24\cdot 9 x\equiv x \equiv 24\cdot 33\pmod {43}$.

But we don't have to.  If $\gcd(9,43) =1$ and if $9^{-1}$ exists, then in this case we are allows to know that division is acceptable, and we are, in this case, allowed to divide both sides by $3$.

$9x \equiv 33 \pmod {43}$

$3x\equiv 11 \pmod {43}$  (we wouldn't be allowed to do this if $\gcd(3,43)\ne 1$)

Now $3$ is such a small number that we know that one these:  $11, 11 + 43, 11+2*43$ is divisible by $3$.

As it turns out $3x \equiv 11 \equiv 11 + 43 \equiv 54\pmod {43}$.

And we can divide both sides by $3$ to get

$x \equiv 18\pmod {43}$.[1]

..........

Alternatively we can do baby steps.

$9x \equiv 33 \pmod {43} \implies$ there is a $k$ so that

$9x = 33 + 43k$.  Divide by $3$ and we get

$3x = 11 + 43\frac k3$.  That will be an integer so long as we choose a $k$ that is divisible by $3$.  Relable $m= \frac k3$

$3x = 11 + 43m$

$3x =11 + 43 + 43(m-1)$

$3x = 54 +43(m-1)$

$x = 18 + 43\frac {m-1}3$ which is an integer so long as we select $m-1$ so that $m-1$ is divisible by $3$.  (For example the easiest example is $m-1=0$, $m=1$, $k = 3$ an $9*18 = 33 + 3\cdot 43$.  But we don't need to find the actually $k$)

......

[1] Notice our earlier answer $24\cdot 33\equiv 24\cdot 32 + 24\equiv 48\cdot 16 + 24 \equiv 5\cdot 16 + 24\equiv 80 + 24 \equiv 86 + 18 \equiv 18\pmod {43}$.  So our answers are consistent.


问:
Why $[\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q}):\mathbb{Q}] = nm$?
As title says, I want to prove that $\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$ is degree $nm$ extension of $\mathbb{Q}$ when $p \neq q$ are distinct primes. By Eisenstein's criterion, $x^{n} - p$ is irreducible over $\mathbb{Q}$ and so $[\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}] = n$. However, I wonder if it is trivial that $x^{m} - q$ is irreducible over $\mathbb{Q}(\sqrt[n]{p})$, assuming the general form of Eisenstein's criterion over integral domain (see Wikipedia for the statement). I'm not sure if the polynomial satisfies assumptions of the Eisenstein's criterion.

Also, I tried some $p$-adic ($q$-adic will be more appropriate) strategy. If we assume that $\mathbb{Q}(\sqrt[n]{p})$ can be embedded in $\mathbb{Q}_{q}$, then the irreducibility of $x^{m} - q$ over $\mathbb{Q}(\sqrt[n]{p})$ follows from it over $\mathbb{Q}_{q}$, which is a result of Newton polygon. However, the assumption fails in general, and when we enlarge $\mathbb{Q}_q$ to its finite extension that contains $\sqrt[n]{p}$, the situation gets worse.

Another direction (which I believe that shouldn't be easy) is to prove that the polynomial
$$
f(x) = \prod_{0\leq i < n, 0 \leq j < m} (x - \sqrt[n]{p}\zeta_{n}^{i} - \sqrt[m]{q}\zeta_{m}^{j})
$$
is irreducible over $\mathbb{Q}$. Then the claim follows from the another proposition $$\mathbb{Q}(\sqrt[n]{p} + \sqrt[m]{q}) = \mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$$ (this is not trivial, but I have a proof by the light use of Galois theory.) According to this, this proposition implies our main claim when $n, m$ are coprime.

Can we use Eisenstein's criterion in this case? If not, is there a simple way to prove this?

Maybe related: Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?

答:
An elementary proof produced with the assistance of @KCd.

Let $\alpha = \sqrt[n]{p}$ and $\beta = \sqrt[m]{q}$ and
let $L=\mathbb{Q}(\alpha,\beta)$.

If $i\not\equiv 0\pmod n$ and $j\not\equiv 0\pmod m$, then some minimal power $k>1$ of $\gamma=\alpha ^i\beta ^j$ is a rational. Now let $\omega$ be a primitive $k$th root of unity and consider the polynomial $x^k-\gamma ^k$ with roots  $$\gamma, \gamma \omega ,\gamma \omega ^2, ...,\gamma \omega ^{k-1}.$$
If $x^k-\gamma ^k$ is not a minimal polynomial for $\gamma $ over $\mathbb{Q}$, there would be a polynomial over $\mathbb{Q}$ with roots a non-trivial subset of  $\gamma, \gamma \omega ,\gamma \omega ^2, ...,\gamma \omega ^{k-1}$ and therefore with a product  $\gamma ^x\omega ^y\in \mathbb{Q}$ where $0<x<k$. Then $\omega ^y$ is real and is therefore $\pm 1$ and so $\gamma ^x\in \mathbb{Q}$, a contradiction of minimality.

We conclude that $x^k-\gamma ^k$ is a minimal polynomial. Let $K=\mathbb{Q}(\gamma )$ and then

$$\mathrm{Tr}_{L/\mathbb{Q}}(\gamma )=[L:K]\mathrm{Tr}_{K/\mathbb{Q}}(\gamma )=0.$$
Now suppose that there are rationals $a_{ij}$ such that  $$\sum_{i=0}^{n-1} \sum_{j=0}^{m-1}a_{ij}\alpha ^i\beta ^j=0.$$
Then for any integers $u$ and $v$,  $$\sum_{i=0}^{n-1} \sum_{j=0}^{m-1}a_{ij}\alpha ^{i-u}\beta ^{j-v}=0.$$

Taking $\mathrm{Tr}_{L/\mathbb{Q}}$ we see that $a_{ij}=0$ if  $i\equiv u\pmod n$ and $j\equiv u\pmod m$ and therefore  all $a_{ij}=0$. Then the $\alpha ^i\beta ^j$ are linearly independent over $\mathbb{Q}$ and $[L:\mathbb{Q}]=mn$.


问:
Function satisfying both these limits
What is an example of a function $f(x)$ that satisfies both

$\lim_{x\rightarrow0} f(x)/x^2 = C$, any finite non-zero real number

and

$\lim_{x\rightarrow0} f(x)/x = 1$?

I'm struggling to find anything that works after trying lots of different functions.

答:
this is not possible, you can check that by dividing the second limit by the first one, and you get :

$$\lim_{x\to0}x=\frac{1}{C}=0 $$

Which is not possible

Title: Function satisfying both these limits


问:
Solving simple system of congruences
I have this example from wikipedia:

$$x \equiv 3 \pmod 4$$
$$x \equiv 4 \pmod 5$$

$$x = 4a + 3\\
4a + 3 \equiv 4 \pmod 5\\
4a \equiv 1 \equiv -4 \pmod 5\\
a \equiv -1 \pmod 5\\
x = 4(5b - 1) + 3 = 20b - 1
$$

But wikipedia shows $20b + 19$...what did I do wrong here?

Title: Solving simple system of congruences

Body:
Your answer is equivalent since $\ 19 \equiv -1\pmod{20}.\ $ Explicitly $\ 20b\!+\!19 = 20(b\!+\!1)\!-\!1.\,$

Simpler: $\, x \equiv -1\, $ mod $\,4,5\ \Rightarrow\ x\equiv -1\,$ mod $\,{\rm lcm}(4,5) = 20,\ $ i.e. $\ 4,5\mid x\!-\!1\,\Rightarrow\, 4\cdot 5\mid x\!-\!1.$


答:
The answer to all the questions are as follows:

Proof of this is here

Because it is the union of two of its basis element. As said by @J.W.Tanner

Proof of this is here

Title: The Basis for Product Toplogy

Body:
Hi i am reading Topology by Munkres and having 3 doubts under the section The product Topology at page no 86. There it is written that

The union of the two rectangles pictured in Figure is not a product of two sets so it cannot belong to $\mathcal{B}$ however it is open in $X\times Y$ .

My doubts are as following:

Why is it that the union of two rectangles is not a product of two sets?
I think what i am asking is why is this $(U_1\times V_1)\ \cup (U_2\times V_2) = (U_1\ \cup U_2)\times (V_1\ \cup V_2)$ not true?

How is that union  open in $X\times Y$ ?

Why is this $(U_1\times V_1)\ \cap (U_2\times V_2) = (U_1\cap U_2)\times (V_1\cap V_2)$ true? Is there a rule that applies to intersection but not to union?Or rather which law(like De Morgan's law) are we using?

For reference i am attaching the screenshot where i have highlighted the part.


问:
How to minimize the chance of making an error in algebraic calculations?
This is a question about solving problems under significant time constraints, for example under exam conditions. Even though this question might be seen as "subjective", it is seriously related to many people who study mathematics, so an answer will be very helpful.

The question is simple: what are some ways to reduce the amount of errors in algebraic calculations, for example integrals, derivatives, series expansions, and things like that?

This is important in pure maths, when we are, for example,  doing proofs in analysis/number theory. Needless to say, this is very very important for applied maths and physics.

Apparently, if one is really concentrated on finding the method to solve a problem, he/she becomes much more likely to make small mistakes along the way. Of course, such errors are not an issue if one has plenty of time to check, but it does become an issue when time is limited.

答:
Write down what you're thinking. Don't be afraid to write down intermediate steps, even if they may seem obvious. The smaller the steps you take, the easier it is to verify that they are correct.

Try to do as little as possible in your head. You can get distracted and you can forget things. Paper does not get distracted and also does not forget things. Your mind can play tricks on you and you may remember things incorrectly. But, if you write it down, you don't need to remember it, and therefore, you can't remember it incorrectly.

In short, take small steps and write all of them down.


问:
$a\mid b+c \implies a\mid b$ or $a\mid c$ ,which is the case when this statement is true?
I think that I've read this statement somewhere, but I am not sure if it's true or not.
If you have any idea, please post it here, and thank you.

I know there is a lot of counterexamples to this statement but I'm asking for a way (not an example) that lets this true for every $a,b,c \in \mathbb N$

答:
This statement is not true in general.

Counterexample: $6$ divides ($3+3$) but $6$ does not divide $3$.

An example for when it is true: $6$ divides $(6+6)$ and, of couse, $6$ divides $6$.


问:
Let $\vec{x} \in \mathbb{R}^n$ and define $U = \{A\vec{x} | A \in M_{mn}\}$. Show that $U=\mathbb{R}^m$ if $x \neq\vec{0}$.
Let $\vec{x} \in \mathbb{R}^n$ and define $U = \{A\vec{x} | A \in M_{mn}\}$. It is given that $U$ is a subspace of $\mathbb{R}^n$. Show that $U=\mathbb{R}^m$ if $x \neq\vec{0}$.

My Attempt

To show $U = \mathbb{R}^m$ you need to show each is a subset of another. Clearly if $U$ is a subspace of $\mathbb{R}^m$, $U \subseteq \mathbb{R}^m$.

Now show that $\mathbb{R}^m \subseteq U$:

Let $\vec{x} \in \mathbb{R}^m$, then I need a $\vec{y} \in \mathbb{R}^n$ such that $\vec{x} = A\vec{y}$.

$$A = (a_1, ... , a_n) \rightarrow x=Ay=(a_1y, ...,a_ny) = a_1y_1+....+a_ny_n$$

Not sure why $\vec{x}\neq \vec{0}$ though...could someone help point out the error in the proof?

答: Choose $b\in \mathbb{R}^m$ arbitrarily. Write $x=[x_1, \ldots, x_n]^T$. Without any loss of generality assume $x_1 \neq 0$. Let $A$ be the $m\times n$ matrix such whose first column is $\frac{1}{x_1}b$ and whose remaining $n-1$ columns are all $0$. Then $Ax=b\in U$. Since $b$ was arbitrary we have $U=\mathbb{R}^m$.


答: $2\mid 3+5$, but $2$ does neither divide $3$ nor $5$.

Title: $a\mid b+c \implies a\mid b$ or $a\mid c$ ,which is the case when this statement is true?

Body:
I think that I've read this statement somewhere, but I am not sure if it's true or not.
If you have any idea, please post it here, and thank you.

I know there is a lot of counterexamples to this statement but I'm asking for a way (not an example) that lets this true for every $a,b,c \in \mathbb N$


问:
Do graph homomorphisms preserve forests?
A graph homomorphisms $\phi:G\rightarrow H$ is an adjacency preserving map between graphs $G$ and $H$. In other words, if $u,v\in V(G)$ are adjacent in $G$, written $u\sim v$, then we must have $\phi(u)\sim \phi(v)$ in $H$.

Any graph $G$ is a matroid $M=(E,\mathcal{I})$, where the ground set $E$ is the set of edges $E(G)$ and the independent sets $I\in\mathcal{I}$ are the forests (disjoint collections of trees) $F\subset E(G)$. To give more intuition, the dependent sets in a graph (as a matroid) are its cycles.

A (possible) definition of a matroid homomorphism is a map $\varphi: M \rightarrow N$, between matroids $M$ and $N$ which preserves independence, in the sense that if $I$ is independent in $M$, then $\varphi(I)$ must be independent in $N$.

Note: in graph theory the term independent set is often used synonymously with the term coclique (or stable sets). In this context, I mean an independent set in the matroid sense i.e. a forest when referring to a graph.

Question: Let  $\phi:G\rightarrow H$ be an adjacency preserving map between graphs, is the image of each forest, $F$ in $G$, a forest $\phi(F)$ in $H$?

I think that one should consider non-simple graphs since loops are quite relevant in matroids.

Follow up question: Is there a study of forest preserving mappings between graphs? Or does this all get subsumed by graph minors?

答:
Not necessarily. Let $G$ be a graph on $\{0,1,2,\ldots, 19\}$ where there is an edge between $i$ and $j$ iff $|i-j|=1$ [no wraparound]. So 19 is adjacent to precisely 18, and 0 to precisely 1, and every other vertex has degree 2. This is a forest, a tree to be precise, actually a path on 20 vertices.

Then consider the mapping $\phi: G \mapsto \{0,1,2,3\}$ where $\phi(i) = i \mod 4$.
Then $\phi(G)$ is a 4-cycle.


问:
Complex Exponents
What does it mean to raise a number to a complex exponent, and why? A lot of the explanations that I've seen involve e, why is this?

I'm looking for an intuitive answer describing to how exponentiation works on a number plane (i.e. the complex number plane) rather than just a number line (the real numbers), where I understand exponentiation quite well. I'm not looking for a breakdown of Taylor series - I believe the proofs, I just don't grok them.

For example, if I was asking how sine worked, I'd want this, not this.

答:
Dear Nate, first, why is $e$ the preferred base for exponentials? Imagine that you have 1 dollar and your bank gives you 100% interest rate. After 1 year, you will have 2 dollars.

Now it offers you to add interests 100 times in a year but the interest is 1% at each moment. How much will you get? You will get
$$ (1+0.01)^{100} \approx 2.704 $$
What if they add you $1/N \times $  100% at $N$ moments of the year and you send $N$ to infinity? Well, you will have $e\approx 2.71828$ dollars after one year.

In fact, the general exponential - power with the base of $e$ - may be defined by this "repeated small interest" formula as
$$ e^X = \exp(X) = \lim_{N\to \infty} \left(1+\frac {X}{N}\right)^N $$
It only has this simple form if the base is $e$. A more general power may be defined as
$$ Y^X = \exp(X \ln Y) .$$
Here, $\ln$ is the natural logarithm so that $\exp\ln X = X$. If I replaced the base $e$ by another base such as $2$ or $10$, the "repeated small interest" formula above would have to contain $\ln 2$ or $\ln 10$ or other awkward factors at various places. It wouldn't be natural.

So instead of powers $Y^X$ and logarithms with general bases, you should think that in mathematics, only $\exp(X)$ and $\ln(X)$ are really needed, and all the other powers and logarithms may be expressed as composite functions. Also, $\exp(X)$ has the advantage that its derivative is exactly equal to the very same function $\exp(X)$. In particular, the derivative evaluated at $X=0$ is equal to one, very nice and simple. It would be $\ln(Y)$ if you used a different base $Y$ instead of $e$.

Now, what is the exponential of an imaginary exponent? Again, you may write
$$\exp(iX) = \lim_{N\to\infty} \left(1+\frac {iX}{N}\right)^N $$
You multiply $N$ copies of a number that is very close to one. What do you get?

Well, the multiplication by a complex number has the effect of magnifying (or reducing) the plane, and rotating it. In particular, the absolute value of the number $(1+iX/N)$ is essentially one, up to second-order corrections that disappear in the $N\to \infty$ limit. So in the limit, $(1+iX/N)$ is effectively a number whose absolute value equals one.

Multiplying by complex numbers whose absolute value is equal to one looks like a rotation of the complex plane. The angles are preserved - those are some things one should know about the complex numbers. Moreover, it's clear that multiplying by $(1+iX/N)$ is equivalent to the rotation by $X/N$ in radians. If you multiply the same factor $N$ times, you simply get a rotation by $X/N$ in radians.

So the $N$th power of $1+iX/N$, in the limit $N\to\infty$, is the number that you get by rotating $1$ in the counter-clockwise direction by the angle $X$ in radians. Clearly, the answer is
$$ \exp(iX) = \cos (X) + i \sin(X) $$
where the trigonometric functions have arguments in radians, of course. Once again, the mathematically natural unit of an angle is in radians for very similar reasons why the natural base of the powers or exponentials is $e$. Only in radians, it's true that the derivative of $\sin X$ equals $\cos X$ and many other things.

In fact, the previous formula makes it natural to say that $\cos X$ and $\sin X$ are not "independent" functions, either. They may be defined as
$$\cos (X) = \frac 12 ( e^{iX} + e^{-iX} ) $$
$$\sin(X)  = \frac{1}{2i} (e^{iX} - e^{-iX} ) $$
You may substitute the last two equations into the previous one or vice versa to check that everything is consistent.

Just to be sure, general complex numbers may also be exponentiated via $\exp(A+iB) = \exp(A)\exp(iB)$ where both factors are known.

Title: Complex Exponents


问:
initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t <3; t if t >3}
Solve the initial value problem:

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &t < 3 \\ t & t > 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution,
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$.

But, now where do I go next?

Thanks!

答:
Find two solutions for for $y'' + 4y = 0$ (Set $y(t) = e^{st}$ and solve for $s$, note that $s$ may be complex. If you don't want to use complex numbers, set $y(t)=e^{\sigma t}\sin(\omega t + \phi)$ instead).

Combine those solutions to find a solution of the IVP $y_1'' + 4y_1 = 0$ with boundary conditions $y_1'(0) = 0$ and $y_1(0)=0$.

Then find a solution for $y'' + 4y = t$. (Set $y(t) = At$ and solve for $A$)

Combine the two solutions from (1) and the one from (3) to find a solution fo the IVP $y_2'' + 4y_2 = t$ with initial conditions $y_2'(3) = y_1'(3)$ and $y_2(3) = y_1(3)$ for $t \geq 3$.

Now $$
    y(t) = \begin{cases}
      y_1(t) &\text{if $t \leq 3$} \\
      y_2(t) &\text{if $t > 3$}
    \end{cases}
  $$ is the solution you're looking for

Here's a plot of the solution as calculated by Mathematica


问:
On the number of extremum
Let $f$ be infinitely differentiable on $[-1,1]$, $f^{(n)}(-1)=f^{(n)}(1)=0,n=0,1,2,\cdots,$ and $f>0$ in $(-1,1)$. Prove that there is a positive integer $k$ such that $\dfrac{f(x)}{(1-x^2)^k}$ has at least three extreme points in $(- 1,1)$. I tried to use Taylor and $$f(x)=f(1)+f'(1)(x-1)+\dfrac{f''(1)}{2}(x-1)^2+\cdots$$
$$f(x)=f(-1)+f'(-1)(x+1)+\dfrac{f''(1)}{2}(x+1)^2+\cdots$$
But this leads to an expression of $0$. I'm a little confused...

答:
Since $f^{(0)}$ is the function $f$ itself one has $f(-1)=f(1)=0$

Put $F(x)=\dfrac{f(x)}{(1-x^2)^k}$. The derivative of $F$ should have at least three roots for some integer $k\gt0$.

Choose $k=2$. We have
$$\frac{dF}{dx}=\frac{f'(x)(1-x^2)-2xf(x)}{(1-x^2)^{3}}=0$$ so  $x=0\Rightarrow f'(0)=0$ and,the function $F$ has a minimum at zero.
On the other hand, since $f(-1)=0$ we have $F(-1)$ has an indeterminated form but
$$\lim_{x\to{-1}}F(x)=\frac{f'(-1)}{2x^3-2x}\\\lim_{x\to{-1}}F(x)=\frac{f''(-1)}{6x^2-2}=0$$Now by prolongation by continuity we can do $F(-1)=0$ and because of $F$ is continuous on the compact $[-1,0]$, by the extreme value theorem, has a maximum at the interior of $[-1,0]$ i.e. on the open interval $(-1,0)$.

Similarly happens on the interval $[0,1]$ with which we have in total three extremums at least as required.

Title: On the number of extremum


问:
The distribution of areas of a random triangle on the sphere - what are the second, third, etc. moments?
Suppose that we choose three points independently and uniformly at random on the surface of a unit sphere as the vertices of a triangle, and consider the area of this triangle. Call this random variable $X$.

The area of such a triangle is the sum of its angles minus $\pi$, so by linearity of expectation the expected value of $X$ is just $3q-\pi$, where $q$ is the expected value of one of the angles. But by symmetry we can fix one point to be at the north pole and the other to lie on the Prime Meridian, from which it is obvious that the angle distribution is uniform on $[0,\pi]$. Thus $\mathbb{E}[X]=\pi/2$, or one-eighth of the sphere's area.

However, because the angles are not independent (they cannot sum to less than $\pi$, for instance), we cannot use this sort of logic to easily infer the values of the second and third moments of this distribution.

After gathering some numerical data, it appears that $\mathbb{E}[X^2]=\frac{\pi^2}2$, but I am not sure how to prove this. (Note that this is equivalent to the statement that the standard deviation of $X$ is $\pi/2$, which may be easier to show?)

I also have $\mathbb{E}[X^3]\approx 20.36$, or that the third moment of $X$ is approximately $4.86$ (either can be inferred from the other, given lower-order moments - they should differ by $\pi^3/2$). I haven't found any particularly nice formulas that match either of these values, though I'm not sure about the last digit in either of these estimates.

In general, what is $\mathbb{E}[X^n]$ or $\mathbb{E}[(X-\frac{\pi}2)^n]$? If any of the values are open, has it been discussed in the literature? Is there a nice geometric argument for $\mathbb{E}[X^2]$?

Edit: Here is a histogram of the area distribution from $1000000$ samples. Interestingly, it seems not to decay to $0$ at the upper bound of $2\pi$.

答:
I'll confess some frustration with this answer, because I can quote various sources on this but not verify them readily for myself.

The major result to quote is the probability density for the area $x\in [0,2\pi]$ of a spherical triangle:

$$f_X(x)=-\frac{(x^2−4\pi x+3\pi^2−6)\cos(x)−6(x−2\pi)\sin(x)−2(x^2−4\pi x+3\pi^2+3)}{16\pi \cos(x/2)^4}.$$

This formula apparently goes back to at least 1867, appearing as problem 2370 in Mathematical Questions and Solutions from the “Educational Times” : It was proposed by M.W. Crofton and solved the pseudonymous "Exhumatus". This is available via Google Books here (p. 21-23). Finch and Jones 2010 review Exhumatus's approach; see also the discussion of Case 2 random spherical triangles in Philip 2015. Interestingly, there doesn't seem to be a direct derivation of the above formula from the trivariate density for the three angles (which is known.)

With this density in hand, one may use Mathematica to verify that the PDF is properly normalized and that

$$\mathbb E[X]=\frac{\pi}{2},\quad \mathbb E[X^2]=\frac{\pi^2}{2},\quad \mathbb{E}[(X-\pi/2)^2]=\frac{\pi^2}{4}$$ in agreement with the OP. Mathematica seems to struggle with finding closed-forms for the higher moments but does obtain

\begin{array}{ll}
\mathbb{E}[X^3]=\frac32 \pi^3-12\pi \ln 2\approx 20.3784,
&\mathbb{E}[(X-\pi/2)^3]=\pi^3-12\pi \ln 2,\\
\mathbb{E}[X^4]=4 \pi^4-24 \pi^2 \ln 2-\zeta(3) \approx 95.6281,
&\mathbb{E}[(X-\pi/2)^4]=\frac{25}{16}\pi^4-108\zeta(3)\\
\mathbb{E}[X^5]=10 \pi^5-120 \pi^3 \ln 2 \approx 481.167,
&\mathbb{E}[(X-\pi/2)^5]=\frac{13}{4}\pi^5 -90\pi^3 \ln 2+270\zeta(3)\\
\end{array}
and so forth. At this point computing the central moments becomes too laborious for my sessions of Wolfram Cloud, but the next few non-central moments are

\begin{align}
\mathbb{E}[X^6]&=24 \pi ^6-480 \pi^4 \ln 2-180 \pi^2 \zeta(3)+13500 \zeta(5)\\&\approx 2527.33 ,\\
\mathbb{E}[X^7]&=56 \pi^7-1680 \pi^5 \ln 2+1260 \pi^3\zeta(3)+47250 \pi \zeta(5)\\&\approx 13664.1 ,\\
\mathbb{E}[X^8]&=128 \pi^8-5376 \pi^6 \ln 2+12096 \pi^4 \zeta(3)+264600 \pi^2 \zeta(5)-1666980 \zeta(7)\\&\approx 75422.4,\\
\mathbb{E}[X^9]&=288 \pi^9-16128 \pi^7 \ln 2+66528 \pi^5 \zeta(3)+1020600 \pi^3 \zeta(5)-10001880 \pi \zeta(7)\\&\approx 422860.,\\
\mathbb{E}[X^{10}]&=640 \pi^{10}-46080 \pi^8 \ln 2+293760 \pi^6\zeta(3)+3175200 \pi^4 \zeta(5)-67869900 \pi^2 \zeta(7)+260253000 \zeta(9)
\\&\approx 2399821.43.
\end{align}
I'm not sure how much farther this can be pushed using Wolfram Cloud. The best route would be to determine some generating function like $\mathbb{E}[e^{tX}]$ or  some other convenient $\mathbb{E}[f(tX)]$, but I haven't succeeded in obtaining such.

As a last point of agreement with the OP, the density does not vanish at $x=2\pi$: $$f_X(2\pi)=-\frac{(-\pi^2-6)-2(-\pi^2+3)}{16\pi}=\frac{3}{4\pi}-\frac{\pi}{16}\approx 0.0424. $$ This seems in good agreement with the histogram obtained by the OP. My own intuitive explanation for this is that the area is nearly maximized occurs when the vertices lie near some common great circle, and there are "many" ways to form such triangles. Thus there's no reason for the probability density to vanish as $x\to 2\pi$.


答: Well, its true if both $b$ and $c$ are multiples of $a$. It is not true if, only one of the two ($b$ or $c$) is multiple of $a$.


问:
Calculate the following series --- need to understand why I have it incorrect
Okay I know I have the answer incorrect but not sure why...

Calculate the following series:

$$\sum_{j=0}^n 2^j \text{ for } n = 1,2,3,4$$

so I think it would be $2^1 = 2$ but the answer is $3$ and the rest are $7,9,32$... I do not understand how they got $3$. Any guidance will help.

Body:
The geometric series in general is $$\sum_{j=0}^n a^j = \frac{1-a^{n+1}}{1-a}.$$

For $a=2$, you get $$\sum_{j=0}^n 2^j = 2^{n+1}-1.$$

If your sum begins at $j=1$, $$\sum_{j=1}^n a^j = \frac{1-a^{n+1}}{1-a}-1=\frac{a-a^{n+1}}{1-a}.$$
If you use $a=2$, you get $$\sum_{j=1}^n 2^j =2^{n+1}-2.$$


答:
A related problem. In fact, you need to solve these two differential equations

$$ \begin{cases} y'' + 4y = 0 &t < 3 \\ y'' + 4y = t & t > 3\end{cases} .$$

Title: initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t <3; t if t >3}

Body:
Solve the initial value problem:

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &t < 3 \\ t & t > 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution,
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$.

But, now where do I go next?

Thanks!


问:
$\gcd(a,b)\gcd(x,y)\mid ax+by$
Suppose $a$ and $b$ are positive integers, and that $d=\gcd(a,b)$. Suppose we have found integers $x$ and $y$ such that $ax+by=d$. Prove that $x$ and $y$ are relatively prime.</p>

答:
Since, $d = (a, b)$
Let $a = dk_1, b = dk_2$
$\therefore ax + by = d \\ \Rightarrow dk_1x + dk_2y = d \\ \Rightarrow k_1x + k_2y = 1$

By Bezout's Theorem, x and y are relatively prime.

Title: $\gcd(a,b)\gcd(x,y)\mid ax+by$


问:
Please explain the meaning of N in this Venn Diagram.

Can someone explain what N means in this diagram? I am new to venn diagrams, and I was able to solve questions based on this Venn diagram without knowing what N means. But I am curious to know what it means.

答: I believe it is the number of elements, or the cardinality of $\xi$. I have seen this notated as $n(\xi)$ before, but never as a capital.


问:
Parallelogram and cosine rule
The parallelogram $ABCD$ is determined by $AB=8,AD=10$ and $\measuredangle BAD=60^\circ$. The perpendicular bisector $s_{BD}$ of $BD$ intersects $AD$ and $BC$ at $K$ and $M$, respectively. Find $BK$ and $KM$.

 $BK=7,KM=4\sqrt7$

First, I am trying to find $BK$. The cosine rule on triangle $ABK$ gives $$BK^2={AK}^2+AB^2-2.AK.AB.\cos60^\circ.$$ We don't know only the length of $AK$. How can I find it?

答:
Expanding on Magnum's comment.
As you pointed out $$BK^2=AK^2+AB^2-2\cdot AK\cdot AB\cdot\cos\measuredangle BAK$$ Note that $BK=DK$ so let $BK=DK=x$. Then $AK=10-x$. Plugging in gives $$x^2=(10-x)^2+64-2\cdot8\cdot(10-x)\cdot\dfrac12\Rightarrow x=7$$

Title: Parallelogram and cosine rule


答: See the earlier question $\operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm$, where my answer explains why this problem over all $m$ and $n$ is equivalent to the case $m=n$ (for all $n$) and you can replace 2 and 3 with a more general pair of numbers in the base field.

Title: Why $[\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q}):\mathbb{Q}] = nm$?

Body:
As title says, I want to prove that $\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$ is degree $nm$ extension of $\mathbb{Q}$ when $p \neq q$ are distinct primes. By Eisenstein's criterion, $x^{n} - p$ is irreducible over $\mathbb{Q}$ and so $[\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}] = n$. However, I wonder if it is trivial that $x^{m} - q$ is irreducible over $\mathbb{Q}(\sqrt[n]{p})$, assuming the general form of Eisenstein's criterion over integral domain (see Wikipedia for the statement). I'm not sure if the polynomial satisfies assumptions of the Eisenstein's criterion.

Also, I tried some $p$-adic ($q$-adic will be more appropriate) strategy. If we assume that $\mathbb{Q}(\sqrt[n]{p})$ can be embedded in $\mathbb{Q}_{q}$, then the irreducibility of $x^{m} - q$ over $\mathbb{Q}(\sqrt[n]{p})$ follows from it over $\mathbb{Q}_{q}$, which is a result of Newton polygon. However, the assumption fails in general, and when we enlarge $\mathbb{Q}_q$ to its finite extension that contains $\sqrt[n]{p}$, the situation gets worse.

Another direction (which I believe that shouldn't be easy) is to prove that the polynomial
$$
f(x) = \prod_{0\leq i < n, 0 \leq j < m} (x - \sqrt[n]{p}\zeta_{n}^{i} - \sqrt[m]{q}\zeta_{m}^{j})
$$
is irreducible over $\mathbb{Q}$. Then the claim follows from the another proposition $$\mathbb{Q}(\sqrt[n]{p} + \sqrt[m]{q}) = \mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$$ (this is not trivial, but I have a proof by the light use of Galois theory.) According to this, this proposition implies our main claim when $n, m$ are coprime.

Can we use Eisenstein's criterion in this case? If not, is there a simple way to prove this?

Maybe related: Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?


问:
Is it possible that a path $a$ is not homotopic to the constant path?
Is my argument correct?

Aim:

Let $X$ be a conected by paths set and $a:[0,1]\to X$ a path. So, $a\simeq c_a$, where $c_a(s)\equiv a(0)$ is the constant path based on $a(0)$.

I think this is trivial, because as $ X $ is connected, using a free homotopy $ H (a, t) = a_t $, where $ a_t (s) = a (ts) $ we deform $ a $ in order to obtain $ a \simeq c_a $.

Is it correct? Thank you so much.

edit $\simeq$ is the free homotopy

答:
It is correct. In fact, if $Y$ is any contractible space (as $I = [0,1]$) and $f  : Y \to X$ is any map to some space $X$ (no assumptions on $X$!), then $f$ is homotopic to a constant map. To see this, let $x_0 \in f(Y)$. Choose $y_0 \in Y$ such that $f(y_0) = x_0$. Since $Y$ is contractible, there exists a homotopy $H : Y \times I \to Y$ such that $H(y,0) = y$ and $H(y,1) = y_0$ for all $y \in Y$. Then $G = f \circ H : Y \times I \to X$ is a homotopy such that $G(y,0) = f(y)$ and $G(y,1) = x_0$ for all $y$.

Note that if we assume $X$ to be path connected, then $f$ is homotopic to each constant map $Y \to X$.

Let us emphasize that for $Y = I$ we cannot expect that the homotopy keeps fixed both endpoints $0, 1$ of $I$. This may confuse some readers because they misinterpret the word "homotopy" as "homotopy of paths" which by definition keeps endpoints fixed. However, here we deal with free homotopies. The best what can be achieved in general is that our homotopy keeps fixed a single point of $I$, for example $0$. Also note that that only closed paths have the chance that there exists a homotopy of paths to a constant path.


问:
Evaluate $\lim_{x\to \infty} \frac{\{x\} \ln x^n}{[x]}$
Here, $\{.\}$ represents fractional part function and $[.]$ represent integral part function

Can I get a hint to get me started, because I have no idea how to begin. Not the complete answer, just a hint.

答:
As others already stated, you have to reason on the meaning of the two functions. Assuming $n > 0$,

$$\{x\} \in [0, 1)$$
$$[x] \geq 1\ \text{for}\ x \geq 1 $$

Using the log property $\log(x^n) = n\log(x)$ your limit becomes

$$\lim_{x\to +\infty} \dfrac{n\{x\}\log(x)}{[x]}$$

Since $\text{max}\big(f(x) = \{x\}\Big) = 0.\bar{9}$ and the integer part is not bounded from above, (or more rigorously: $[x] = x - \{x\}$ hence $[x] \leq x - \{x\}$) the limit can be estimated as

$$\lim_{x\to +\infty} \dfrac{n\{x\}\log(x)}{[x]} \geq \lim_{x\to +\infty} \dfrac{0.\bar{9}n\log(x)}{x - \{x\}} \geq 0.\bar{9} n \lim_{x\to +\infty} \dfrac{\log(x)}{x-1} = 0$$

Where the very last result is obtained, for example, via De L'Hôpital rule.

Caveat

We all do know that $0.\bar{9} = 1$, but, well... you know... It's for the sake of an estimation.


答:
Generalization:

$a \equiv b(\mod n)$ if $b<n$, $a\equiv b-n (\mod n)$

The reason is quite simple, $a \equiv b-n (\mod n) \implies a+n \equiv b(\mod n)$

Title: Solving simple system of congruences

Body:
I have this example from wikipedia:

$$x \equiv 3 \pmod 4$$
$$x \equiv 4 \pmod 5$$

$$x = 4a + 3\\
4a + 3 \equiv 4 \pmod 5\\
4a \equiv 1 \equiv -4 \pmod 5\\
a \equiv -1 \pmod 5\\
x = 4(5b - 1) + 3 = 20b - 1
$$

But wikipedia shows $20b + 19$...what did I do wrong here?


问:
Integral of $1/x^2$ without power rule
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

答:
I thought this answer might be useful to you as it involves simple manipulations.

$$\begin{align} \int \dfrac {1}{x^2}dx=\int \dfrac {1}{x} \times \dfrac {1}{x}dx=\int x^{-1}d(\ln x)=\int e^{\ln x^{-1}}d(\ln x)=\int e^{- \ln x}d(\ln x)=- \int e^{- \ln x}d(- \ln x)=-e^{- \ln x}+C=-\dfrac {1}{x}+C.\end{align}$$

Title: Integral of $1/x^2$ without power rule


问:
Find $f$ such that it satisfies $\lim_{x\to\infty}\frac{f(x)}x=1$ and $\lim_{x\to0}\frac{f(x)}{x^2}=C$ for constant $C$
I need to find a function $f$ such that $$\lim_{x\to\infty}\frac{f(x)}x=1$$ and $$\lim_{x\to0}\frac{f(x)}{x^2}=C$$ for a constant $C$ that is not zero.
I’ve tried polynomials in $x$ but these seem not to work. I thought maybe using an exponential function might help, but I’m not sure what exactly to use?

答:
The rational functions
$$f_n(x)=\frac{x^{n+1}+Cx^2}{x^n+1}\quad(n\ge 2)$$
are solutions.


问:
Derivative of Dirac delta function
Is the relation of the Dirac delta function correct?

$$
\frac{\partial}{\partial x''}\delta(x''-x')
= -\frac{\partial}{\partial x'}\delta(x'-x'').\tag{1}
$$

If it is, how to derive the above relation?

答:
To prove
\begin{align}
\frac{\partial  \delta(x-x_0)}{\partial x}  = -\frac{\partial  \delta(x-x_0)}{\partial x'} . ~~~~~~~~~~~~~~~~~~~~ \text{(1)}
\end{align}
Proof: We know that
\begin{align}
\int f(x) \frac{\partial \delta(x-x_0) }{\partial x}= - f'(x_0)   ~~~~~~~~~~~~~~~~~~~~~ \text{(2)}
\end{align}
Now,
\begin{align}
& \int f(x) \frac{\partial \delta(x-x_0) }{\partial x_0}  dx  \\
= & \int f(x)  \left(  \frac{ \delta(x-(x_0+\epsilon) - \delta(x-x_0)    }{(x_0+\epsilon) - x_0} \right) dx   \\
= & \frac{  f(x_0 + \epsilon) -f(x_0 )  }{\epsilon} \\
= & f'(x_0),
\end{align}
which upon comparison with Eq. (2) immediately gives us Eq. (1).

Title: Derivative of Dirac delta function


问:
Please find the range of $V$ when $V = \frac {1} {X}$ & $0 <x<1 $
if $0 <x<1 $ then what is the range of $V$ when

$V = \frac {1} {X}$

i tried to compute it by :

when $x=0$ then $V = \frac {1} {X}= \frac {1} {0} =undefined$

when $x=1$ then $V = \frac {1} {X}= \frac {1} {1} =1$

so it comes    $\quad undefined <v<1 $

but the book, which i am reading , is showing the range $\quad v>1$

答:
I don't think you mean "limit", I think you mean "range".

When $x$ decreases from just below one, $1/x$ increases from just above one. E.g. work out $1/0.999$ on a calculator, and understand why it's like that.

As $x$ gets very close to zero, $1/x$ gets bigger and bigger, tending to $+\infty$. Therefore, $v$ takes on everything bigger than 1.


答:
What about
$$f(x)=\begin{cases}
x &\text{ if }|x|\geq 1 \\
Cx^2 &\text{ if }|x| < 1
\end{cases}\quad ?$$

Title: Find $f$ such that it satisfies $\lim_{x\to\infty}\frac{f(x)}x=1$ and $\lim_{x\to0}\frac{f(x)}{x^2}=C$ for constant $C$

Body:
I need to find a function $f$ such that $$\lim_{x\to\infty}\frac{f(x)}x=1$$ and $$\lim_{x\to0}\frac{f(x)}{x^2}=C$$ for a constant $C$ that is not zero.
I’ve tried polynomials in $x$ but these seem not to work. I thought maybe using an exponential function might help, but I’m not sure what exactly to use?


问:
Prove $\int_0^\infty e^{-x} x^k [L^k_n (x)]^2 \, dx=\frac{(n+k)!}{n!}$
How can I prove the normalization ratio of associated Laguerre polynomials: $$\int_0^\infty e^{-x} x^k [L^k_n (x)]^2 \, dx=\frac{(n+k)!}{n!}$$ using the generator function of Laguerre polynomial, $$\sum_{n=0}^\infty L_n^k (x) t^n=\frac{1}{(1-t)^{k+1}}\exp\left(\frac{-xt}{1-t}\right)?$$

答:
I think that the easiest (and well-known) way to do it is via Rodrigues' formula $$L_n^k(x)=\frac{x^{-k}e^x}{n!}\frac{d^n}{dx^n}(x^{n+k}e^{-x}).\label{rodrigues}\tag{*}$$

To obtain it from the generating function, one may use the Cauchy integral $$L_n^k(x)=\frac{1}{2\pi i}\oint_{(t=0)^+}\frac{1}{(1-t)^{k+1}}\exp\left(\frac{-xt}{1-t}\right)\frac{dt}{t^{n+1}}$$ (with the contour, say, a small circle around $t=0$), and substitute $t=1-x/z$ to get $$L_n^k(x)=\frac{x^{-k}e^x}{2\pi i}\oint_{(z=x)^+}\frac{z^{n+k}e^{-z}}{(z-x)^{n+1}}\,dz,$$ exactly the Cauchy integral for \eqref{rodrigues}. Now, returning to the question, we get
$$\int_0^\infty x^k e^{-x}\left(L_n^k(x)\right)^2 dx
=\frac{1}{n!}\int_0^\infty L_n^k(x)\frac{d^n}{dx^n}(x^{n+k}e^{-x})\,dx
\\=\frac{(-1)^m}{n!}\int_0^\infty\left(\frac{d^m}{dx^m}L_n^k(x)\right)\left(\frac{d^{n-m}}{dx^{n-m}}(x^{n+k}e^{-x})\right)dx$$
for $m\leqslant n$ (using induction on $m$ and integration by parts). Taking $m=n$ (recall that $L_n^k(x)$ is a polynomial of degree $n$ with the leading coefficient equal to $(-1)^n/n!$), we get the needed $$\int_0^\infty x^k e^{-x}\left(L_n^k(x)\right)^2 dx=\frac{1}{n!}\int_0^\infty x^{n+k}e^{-x}\,dx=\frac{(n+k)!}{n!}.$$ (The orthogonality $\int_0^\infty x^k e^{-x} L_m^k(x) L_n^k(x)\,dx=0$ for $m\neq n$ is shown the same way.)


问:
Necessary conditions for mixed concavity and convexity.
The questions is originally physical, but the problem is purely mathematical. We have a function

$$F(T,V,N) = A\left(\frac{N^5 T}{V^3}\right)^{1/2}\quad \text{for}\quad A,N,T,V\in\mathbb{R^+}, A\neq 0$$

which must be convexe w.r.t. $V$ and $N$, but concave w.r.t. $T$. Now we are asked to determine whether this conditions are satisfied by $F$. What is confusing me is the following: the corrector says that no, the conditions aren't satisfied because

$$\partial_N^2F\cdot\partial_V^2F - (\partial_V\partial_NF)^2 = -\frac{675}{16} A^2 N^3 T V^{-5} \leq 0$$

Why is this true? I'm getting no satisfying answer, and looking it up on internet hasn't given me any answer. My questions are, in particular, the following:

What are sufficient conditions for a function on a $n$-dimensional space to be concave w.r.t. to some variables and convexe w.r.t. to the rest? As far as I understand, if it is concave w.r.t. to $k$ variables, its Hessian matrix should have $k$ negative eigenvalues, and the opposite applies to the remaining $n-k$ variables which is convexe w.r.t. If a general answer can't be given, consider only 3-dimensional domains.

What are the necessary conditions that must be satisfied by the $2\times 2$ minors of the Hessian matrix of a function as described above (concave and convexe depending on the variable considered). Is the reasoning of my corrector really correct in the case of this exercise?

Thank you in advance.

答:
You should look at concave and convex separately. For (joint) convexity in $V$ and $N$, the Hessian of $F$ w.r.t. $V$ and $N$ needs to be positive semidefinite. You can also look at separate convexity (i.e., the diagonal of the Hessian), but joint convexity is a stronger property and guarantees that a local minimum is globally optimal. Separate convexity is hardly used. I do not have the full context, but it appears to refer to joint convexity.

The derivative of $F$ w.r.t. $V$ and $N$ is:
$$\begin{pmatrix} -\frac{3}{2} A N^{5/2} T^{1/2} V^{-5/2} \\ \frac{5}{2} A N^{3/2} T^{1/2} V^{-3/2} \end{pmatrix}$$
so the Hessian is:
$$\frac{15}{4} A \begin{pmatrix} N^{5/2} T^{1/2} V^{-7/2} & - N^{3/2} T^{1/2} V^{-5/2} \\
- N^{3/2} T^{1/2} V^{-5/2} & N^{1/2} T^{1/2} V^{-3/2} \end{pmatrix}$$

The determinant of the Hessian is $0$ (not $-(675/16) A^2 N^3 T V^{-5}$ as claimed), so the eigenvalues are $0$ and the trace. The trace is also positive (since $A$ is positive), so the function is convex.

As for necessary/sufficient: if the Hessian exists everywhere, a function is convex if and only if the Hessian is positive semidefinite.


问:
Element by element formulae for 3x3 matrix inversion
Given a 3 x 3 matrix:

$$
       A= \begin{bmatrix}
        a & b & c \\
        d & e & f \\
        g & h & i \\
        \end{bmatrix}
$$

Can $A^{-1}$ be shown as as a 3x3 matrix with each element in terms of $a,b,c,d,e,f,g,h$ and $i$. Showing  basic operators (only $+ - / *$)?

答:
$$A^{-1}=\frac{1}{\det(A)} adj(A)$$

$$\det(A)=aei+dhc+bfg-ceg-bdi-afh$$

$$adj(A)= \begin{bmatrix}
        ei-fh & ch-bi & bf-ce \\
        fg-di & ai-cg & dc-af \\
        dh-ge & bg-ah  & ae-bd \\
        \end{bmatrix}$$

_


问:
Proving,that if $a$ and $a \implies b$ are tautologies,then $b$ is a tautology.
How do I prove this?I know,that tautology is a statement that always is true.I'm pretty new to this type of logic and I got stuck.</p>

答: Use the fact that $(a\implies b) \iff ((\neg a) \vee b)$


问:
A possible upper bound on the fractional part of exponential functions

Does there exist a real number $a$ bigger than $1$ that's not a rational power of an integer such that $\displaystyle\lim_{n\to\infty}${$a^n$}$=0$ where values for $n$ are positive integers?

(P.S. $\{ x \} = x - \lfloor x \rfloor$ or namely the fractional part function)

I just saw this problem today. I've just started studying about the fractional part approximations so unfortunately I've made no progress solving this.

*I made a slight change in the problem to avoid any confusions about small values

答:
I don't have a positive answer to OP's question; but note that it's easy to find solutions to the seemingly similar question where you would like to have $\lim_{n\rightarrow\infty}\{a^n\}=1$.  For instance,
$$a_m = \frac{1}{2}\left(m + \sqrt{m^2-4}\right) \in (1,\infty)$$
has this property for any integer $m > 2$.  This works because the linear recurrence defined by $f_0=2,$ $f_1=m,$ and $f_{n+2}=mf_{n+1} - f_n$ can be explicitly solved to yield
$$
f_n = a_m^n+b_m^n,
$$
where $b_m=\frac{1}{2}\left(m - \sqrt{m^2-4}\right)\in(0,1)$.  Hence
$$
\{a_m^n\}=\{f_n - b^n_m\}=\{-b^n_m\}\rightarrow 1.
$$
It seems as if it should be possible, perhaps with a higher-range linear recurrence, to do the same thing with the opposite sign: we would need, say, $f_n=a^n-Bb^n+...$ for some $a>1$ and $B> 0$ and $b\in(0,1)$, and for the remaining terms to be real and decaying faster than $b^n$.  Some constraints are placed on this by the fact that the sum of the eigenvalues of an integer matrix is integral.


问:
Question on countability of family of holomorphic functions
Let $\mathcal{F} \subset \mathcal{H}(\mathbb{C})$ a family of holomorphic functions such that for every $z \in \mathbb{C}$ the set $\{f(z): f \in \mathcal{F}\}$ is countable, my question is if $\mathcal{F}$ is necessary countble.

I don't have any idea from where to start so any tips or help would be really appreciated.

答:
Apparently, the question you ask has name and surname: it is called Wetzel's problem. To that extent, we can mention the Wikipedia page, which has a brief historical overview of the problem, as well as the article by Erdös, which proves the following:

The conclusion of the problem (i.e., there are only countably many  analytic functions in the family) is equivalent to the continuum hypothesis not holding.

Therefore, there is no simple answer to your question. The article by Erdös contains a beautiful proof of such equivalence, which I recommend that anyone who likes mathematics checks.

On the other hand, if you replace "countable" by "finite" everywhere, there is a simple answer: yes, the family must be finite, given that $\{f(z) \colon f \in \mathcal{F}\}$ is finite for all $z \in \mathbb{C}.$ The proof, which is fairly simple, is as follows:

Suppose there is an infinite sequence $\{f_i\}_{i \ge 1} \subset \mathcal{F}$ of pairwise distinct analytic functions. Let first $B_{i,j} = \{z \in \mathbb{C} \colon f_i(z) = f_j(z)\}.$ As all functions are analytic and distinct pairwise, the set $B_{i,j}$ is at most countable. Thus, the set

$$B = \cup_{i,j \ge 1, i \neq j} B_{i,j}$$

Is also countable. Now let $A_N = \{z \in \mathbb{C} \colon |\{f(z) \colon f \in \mathcal{F}\}| = N\}.$ As $\cup_{N \ge 1} A_N = \mathbb{C},$ there is $M \in \mathbb{N}$ such that $A_M$ is uncountable.

On the other hand, as $A_M$ is uncountable and $B$ is countable, there is at least a point $z_0 \in A_M \cap B^{c}.$ But this is a contradiction, as $z_0 \in B^{c}$ implies that $f_i(z_0) \neq f_j(z_0), \forall i\neq j,$ and thus $z_0 \not \in A_M.$


问:
What is "induction on complexity of formula"
The fundamental theorem of ultraproducts

Given a language $L$ and a family of models $M_i$ of $L$ and ultrafilter $\mathcal U$ on $I$ and $\varphi$ a formula of $L$ then
$$ \prod_{i \in I}A_i/\mathcal U \models \varphi(s_0/_{\mathcal U}, \dots, s_n/_{\mathcal U}) \iff \{i \in I : \varphi (s_0(i), \dots, s_n (i))\}\in \mathcal U$$

Is proved by induction on the complexity of $\varphi$. I expect the proof to do atomic formula case first. Atomic formula is formula of form $\varphi = (t_1 = t_2)$ or of form $\varphi = R(t_1, \dots , t_n)$ where $t_i$ are term. But the proofs for example this prove something about terms not atomic formula. Can someone explain me please what is it we need to show if not atomic formula case first? Why terms?

答:
The construction of terms usually proceeds through the following recursive definition:

Any variable $x$ is a term;

Any constant $c$ is a term;

For an $n$-ary function symbol $f$ and terms $t_1,\ldots, t_n$, $f(t_1,\ldots,t_n)$ is a term.

(Some texts define constants as nullary function symbols, but this does not guarantee that this special case does not require special attention -- i.e. it is not trivial that the induction proof for function symbols of positive arity carries over to nullary function symbols.)

So in order to prove a statement $P(t)$ about terms $t$ by induction on the complexity, one needs to verify that each of these constructions is suitably well-behaved with regard to the formation rules.

That is, we need to verify that $P(x), P(c)$ for $x$ a variable, $c$ a constant, and that $P(t_1),\ldots, P(t_n)$ imply $P(f(t_1,\ldots,t_n))$.

Now why would we want to have such statements $P$? Well, a useful example of a statement $P$ about terms could be verifying your initial statement for the case $\varphi = (t_1 = t_2)$!

In conclusion, proving something about terms is entirely different from proving something for formulae. Therefore, they call for different approaches, and as such require a different basis for their "structural induction".


答: You can even prove by contradiction, if $b$ is not a tautology, then there must be a substitution such that $b$ is false, but then $a \implies b$ would not be a tautology because $(\top \implies \bot) \equiv \bot$ thus $b$ must be a tautology.

Title: Proving,that if $a$ and $a \implies b$ are tautologies,then $b$ is a tautology.

Body: How do I prove this?I know,that tautology is a statement that always is true.I'm pretty new to this type of logic and I got stuck.</p>


问:
Solve $z^4+2z^3+3z^2+2z+1 =0$
Solve $z^4+2z^3+3z^2+2z+1 =0$ with $z$: a complex variable.

Attempt at solving the problem:

We divide the polynom by $z^2$ and we get:

$z^2+2z+3+\dfrac{2}{z}+  \dfrac{1}{z^2}=0  $  $         $ We set $w=z+  \dfrac{1}{z}$

We now have $w^2+2w+5=0$

$\bigtriangleup = -16$

Let's find $\omega$ such that $\omega^2=-16$

We have $\omega=4i$

Therefore we have the 2 roots:
$w_    {1}=-1-2i$ and $ w_    {2}=-1+2i  $

The issue is: I don't know how to find z

答:
"I don't know how to find z"

Sure you do!

....

If you are correct in what you have done so far and you have

$z + \frac 1z = w$

And $w_1 = -1-2i$ and $w_2 = -1 + 2i$ then

you need to solve $z +\frac 1z = (-1-2i)$ or $z^2 +(1+2i)z + 1=0$

ANd $z + \frac 1z = (-1+2i)$ or $z^2 + (1-2i)z + 1 = 0$.

Both of which can be solved by quadratic equation to get the four solutions

$z = \frac {(-1\pm 2i) \pm\sqrt{(1\mp 2i)^2 -4}}{2}$

(And if you are worried about expressing $\sqrt{M}$ just convert to polar coordinates.)

That is.... if everything you've done is correct.

Title: Solve $z^4+2z^3+3z^2+2z+1 =0$


问:
Area of triangle formed by tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and coordinate axes
Using the Lagrange's multiplier we need to solve the following problem.

At what point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ does the tangent line to it form with the coordinate axes a triangle of smallest area ?

I am stuck at the application of Lagrange's multiplier.  Had it been some other way like coordinate geometry, I could have managed somehow. But here I have no idea how to proceed. Can someone help me out?

答:
General tangent line to ellipse at point $(p,q)$ has equation

$$\frac{px}{a^2}+\frac{qy}{b^2}=1;\;\frac{p^2}{a^2}+\frac{q^2}{b^2}=1$$

(see here, for instance.)

Suppose $p,q$ positive. (The problem is symmetric)

Triangle area is
$$A(p,q)=\frac{a^2b^2}{2pq};\;\frac{p^2}{a^2}+\frac{q^2}{b^2}=1$$

Define
$$f(p,q,\lambda)=\frac{a^2b^2}{2pq}+\lambda\left(\frac{p^2}{a^2}+\frac{q^2}{b^2}-1\right)$$
$$
\begin{cases}
\frac{2 \lambda  p}{a^2}-\frac{a^2 b^2}{2 p^2 q}=0\\
\frac{2 \lambda  q}{b^2}-\frac{a^2 b^2}{2 p q^2}=0\\
\frac{p^2}{a^2}+\frac{q^2}{b^2}=1\\
\end{cases}
$$

solution is $p=\frac{a}{\sqrt{2}},q= \frac{b}{\sqrt{2}},\lambda=ab$.

Minimum area is
$$A(p,q)=ab$$

Because of the symmetry the points are four
$$\left(\pm\frac{a}{\sqrt{2}},\pm \frac{b}{\sqrt{2}}\right)$$

Hope this helps


答:
Consider $x > 0$. It immediately follows that $V > 0$.

Consider $x < 1$ and substitute $x = 1/V$ in that inequality.

So, we get:

$$\frac{1}{V} < 1$$

Multiplying both sides by $V$, we get:

$$1 < V$$

Note that the inequality sign does not flip because $V > 0$. Or in other words:

$$V > 1$$

Title: Please find the range of $V$ when $V = \frac {1} {X}$ & $0 <x<1 $

Body:
if $0 <x<1 $ then what is the range of $V$ when

$V = \frac {1} {X}$

i tried to compute it by :

when $x=0$ then $V = \frac {1} {X}= \frac {1} {0} =undefined$

when $x=1$ then $V = \frac {1} {X}= \frac {1} {1} =1$

so it comes    $\quad undefined <v<1 $

but the book, which i am reading , is showing the range $\quad v>1$


问:
Any advice on how to develop intuition in math?
I've recently started to learn math online (studying high school algebra right now after being 7 years out of school),
and while I understand the formulas I'm learning I feel as though I'm not really getting much intuition about the subjects I learn, which for me is the most important thing about learning anything.
Last year I started to learn programming and pretty quickly I found that the best way for me was to do something "hands on" so to speak, like after learning a new concept to actually build a project using what I've just learned.
I want to do the same thing with algebra, doesn't necessarily means to code something but I have a hard time really grasping something until I actually do something with it other than some short practice questions

If you have a suggestion please post! Anything from an interesting article to maybe a project idea that will force me to use algebra or how you would study math if you had to start from scratch again

答:
I develop my mathematics intuition visually, seeing figures, studying graphs, and so on, more than analyzing equations.  This works better for some branches of math than others:  geometry, topology, knot theory, etc., rather than formal logic, category theory, etc.  Some branches of math can be presented in a formal, algebraic way and a visual, intuitive way, most notably group theory.

Einstein said he preferred to reason mathematically geometrically, spatially, visually, and only then turn to the formal symbol manipulation and proofs.  I and many others are the same way.

Personally, I learned calculus in high-school math class, but I understood calculus in high-school physics class, where the problems related to mass, motion, velocity, acceleration, moment of inertia, and problems that were so visual and intuitive.

There are several books that are great in this regard, my two favorites being Visual group theory by Nathan Carter and An illustrated theory of numbers by Martin Weissman.  You might like Proofs without words:  Exercises in visual thinking by Roger Nelson.

But some of the best approaches are from Grant Sanderson's Youtube channel, 3b1b, which you simply MUST watch.

Another technique, which need not rely on visualization, is to be explicitly aware of how you found the right approach to a problem.  Consider a tricky integral.  You try integration by parts, series expansion, substitution, and so on until you find that trigonometric substitution worked.  Once you've solved the integral, go back and ask yourself:  "What aspects of this integral meant that trigonometric substitution was the proper approach?"  In this way, you'll become more aware and gain intuition, for instance seeing the $\sqrt{1 - x^2}$ is a functional form that at the very least suggests trigonometric substitution.  Ponder that in order to build up your intuition.

Another suggestion that might help.  I once spent an afternoon with the late titan of theoretical physics John Wheeler.  He once said "Never calculate anything until you first know the answer."  Well, that doesn't always help, but sometimes it does:  think about a problem before you rush into solving it with known techniques.  Perhaps you can reason that the answer is $0$, or $\infty$, or some other special answer first.  Even reasoning that the answer has to be positive, or negative, or irrational helps you build intuition.  Once you solve the problem, go back and revisit and refine your intuitions in light of your initial thoughts.

As a general guide, though, practice graphing equations, sketching solids, and so on and visually understanding them first, even before delving into symbol manipulation.


问:
Segre embedding preserves locally closed sets
I want to prove that if $X \subseteq \mathbb{P}^n$ and $Y \subseteq \mathbb{P}^m$ are two locally closed sets, then $S(X \times Y)$ is locally closed in $\mathbb{P}^{(n+1)(m+1) -1}$, where $S(X \times Y)$ is the image of $X \times Y$ through Segre immersion.

If $X$ and $Y$ are locally closed, it means that they are open in their closures $\bar{X}$ and $\bar{Y}$. I was trying to prove that $S(X \times Y)$ is open in $S(\bar{X} \times \bar{Y})$ (which is closed because Segre immersion is closed). However, I got stuck here: I am not sure how to prove that $S(\bar{X} \times \bar{Y})$ is the closure of $S(X \times Y)$ and how to prove that $S(X \times Y)$ is open in its closure. Any hints (please do not show me solutions using schemes, I don’t know anything about those)?

答:
Here’s a sketch:

Write $X$ as the subset of points of $\overline{X}$ where some homogeneous polynomials $f_0,\ldots,f_k$ do not simultaneously vanish. Same for $Y$ with $g_0,\ldots,g_q$. We can assume that all the $f_i$ and $g_j$ have the same degree (up to taking powers). Then you can regard $f_ig_j$ as a homogeneous polynomial in the “bihomogeneous variables” considered in $S: \mathbb{P}^n \times \mathbb{P}^m \rightarrow \mathbb{P}^{(n+1)(m+1)-1}$.

Then $S(X \times Y)$ is the subset of $S(\overline{X} \times \overline{Y})$ where these polynomials do not simultaneously vanish, so is open in this set.