问:
The Method Of Frobenius

The ODE $xy'' + y = 0$ has a real degeneracy.  Use The Method Of
Frobenius to find a fundamental set of solutions.

Here is the procedure, as I understand it:

1)  Plug the guess $y = x^s \sum_{n = 0}^\infty a_n x^n$ into the ODE and do the algebra/calculus to separate out the indicial equation and the recurrence relation.

$xy'' + y = 0$

$x(x^s \sum_{n = 0}^\infty a_n x^n)'' + (x^s \sum_{n = 0}^\infty a_n x^n) = 0$

$\sum_{n = 0}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 0}^\infty a_n x^{n + s} = 0$

$\sum_{n = 0}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 1}^\infty a_{n - 1} x^{n + s - 1} = 0$

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + \sum_{n = 1}^\infty a_{n - 1} x^{n + s - 1} = 0$

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty (n + s)(n + s - 1) a_n x^{n + s - 1} + a_{n - 1} x^{n + s - 1} = 0$

$s(s - 1) a_0 x^{s - 1} + \sum_{n = 1}^\infty [(n + s)(n + s - 1) a_n + a_{n - 1}] x^{n + s - 1} = 0$

Indicial Equation:  $s(s - 1) = 0$

Recurrence Relation:  $(n + s)(n + s - 1) a_n + a_{n - 1} = 0$, where $n \geq 1$

2)  Solve the recurrence relation, treating $s$ as a constant, but without plugging in any indicial equation solutions.

$(n + s)(n + s - 1) a_n + a_{n - 1} = 0$

$a_n = -\frac{1}{(n + s)(n + s - 1)}a_{n - 1}$

$a_n = a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$

Here I would simplify using a formula I was given, $\Pi_{j = c}^d (j + k) = \frac{(d + k)!}{(c + k - 1)!}$.

$a_n = a_0 [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]$

$a_n = a_0 [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]$

$a_n = \frac{(-1)^n s! (s - 1)!}{(n + s)! (n + s - 1)!} a_0$

3)  To find one fundamental solution, plug the largest indicial equation solution $s_1$ and its associated recurrence relation solution into the guess, and make the particular choice $a_0 = 1$.

$s(s - 1) = 0$

$s = 0, 1$

$s_1 = 1$

$y = x^1 \sum_{n = 0}^\infty \frac{(-1)^n 1! (1 - 1)!}{(n + 1)! (n + 1 - 1)!} 1 x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!}\leftarrow$  FIRST FUNDAMENTAL SOLUTION

4)  To find another fundamental solution, plug the other indicial equation solution $s_2$ and its associated recurrence relation solution into $y = [$FIRST FUNDAMENTAL SOLUTION$]ln|x| +\ x^{s_2} \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - s_2)a_n)]_{s = s_2} x^n$, and make the particular choice $a_0 = 1$.

$s_2 = 0$

Here I use the earlier, less simplified expression for $a_n$ which is in $\Pi$ product notation, in anticipation of taking the logarithmic partial derivative.

$y = (\sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1}}{n! (n + 1)!})ln|x| + x^0 \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - 0) a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{\partial}{\partial s} (s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})]_{s = 0} x^n$

I make the substitution $b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$, appearing in the solution as $y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!}$
$+ \sum_{n = 0}^\infty [\frac{\partial}{\partial s} b_n]_{s = 0} x^n$.

$b_n = s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}$

$ln|b_n| = ln|s| + ln|a_0| + \sum_{m = 1}^n ln|\frac{1}{m + s}| + ln|\frac{1}{1 - m - s}|$

$\frac{\partial}{\partial s} ln|b_n| = \frac{\partial}{\partial s} ln|s| + \frac{\partial}{\partial s} ln|a_0| + \sum_{m = 1}^n \frac{\partial}{\partial s} ln|\frac{1}{m + s}| + \frac{\partial}{\partial s} ln|\frac{1}{1 - m - s}|$

I set $\frac{\partial}{\partial s} ln|a_0|$ to $0$ because the chapter stipulates that this procedure only seeks solutions where $a_0$ is nonzero and does not depend on $s$.

$\frac{\frac{\partial b_n}{\partial s}}{b_n} = \frac{1}{s} + 0 + \sum_{m = 1}^n -\frac{1}{m + s} + \frac{1}{1 - m - s}$

$\frac{\partial b_n}{\partial s} = \frac{b_n}{s} + b_n \sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)}$

$\frac{\partial b_n}{\partial s} = \frac{s a_0 \Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)}}{s} + s a_0 (\Pi_{m = 1}^n -\frac{1}{(m + s)(m + s - 1)})(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

Now I can simplify the $\Pi$ products using the above formula.  I also make the choice for $a_0$ here.

$\frac{\partial b_n}{\partial s} = \frac{s [\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}] [\frac{1}{\Pi_{m = 1}^n (m + s - 1)}]}{s} + s ([\Pi_{m = 1}^n -1] [\frac{1}{\Pi_{m = 1}^n (m + s)}][\frac{1}{\Pi_{m = 1}^n (m + s - 1)}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

$\frac{\partial b_n}{\partial s} = \frac{s [(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}]}{s} + s ([(-1)^n] [\frac{1}{\frac{(n + s)!}{s!}}] [\frac{1}{\frac{(n + s - 1)!}{(s - 1)!}}])(\sum_{m = 1}^n -\frac{2m + 2s - 1}{(m + s)(m + s - 1)})$

$\frac{\partial b_n}{\partial s} = \frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}$

Unwinding the substitution, I plug the expression for $\frac{\partial b_n}{\partial s}$ back into the fundamental solution.

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty [\frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}]_{s = 0} x^n$

$y = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} ln|x|}{n! (n + 1)!} + \sum_{n = 0}^\infty \frac{(-1)^n (-1)! x^n + (-1)^{n + 1} x^n \sum_{m = 1}^n \frac{2m - 1}{m^2 - m}}{n! (n - 1)!}$

And now we see that something is broken.  $(-1)!$ is undefined, and can't even be rescued by being generalized to the Gamma or Pi function.  What am I doing wrong, and how can I use this method to solve this ODE?

答:
The first mistake is with the book/material you are using, as $$y_2 = y_1\ln|x| + u$$ isn't a valid solution of the differential equation if $u = x^{s_2} \sum_{n = 0}^\infty [\frac{\partial}{\partial s} ((s - s_2)a_n)]_{s = s_2} x^n$ and $a_0 = 1$. Instead, you either want $a_0 = -1$ or $y_2 = y_1\ln|x| - u$. Not a rigorous proof, but you can see evidence to support this by looking at the constant term of $xy_2'' + y_2$, which must be $0$.

The main problem is the hole at $s = 0$. You ended up with $$\frac{\partial b_n}{\partial s} = \frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!} \tag{*}$$ and directly plugged in $s = 0$. But the function is undefined there because of the $(-1)!$ term. Instead, take the limit as $s \to 0$. The limit is $1$ for $n = 0$ and $$\frac{(-1)^{n+1}(1+2nH_{n-1})}{(n!)^2}$$  for $n \ge 1$, where $H_n$ is the harmonic number (found using Mathematica, proof at bottom). This means that the second solution is $$y_2 = \sum_{n = 0}^\infty \frac{(-1)^n x^{n + 1} \ln|x|}{n! (n + 1)!} - \left(1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(1+2nH_{n-1})}{(n!)^2} x^n\right)$$

Edit: Coming back to $(*)$: The problem is to evaluate $$\lim_{s \to 0}\frac{(-1)^n s! (s - 1)! + (-1)^{n + 1} s!^2 \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}}{(n + s)! (n + s - 1)!}$$ For $n = 0$, it is clearly $0$ since the sum is $0$ and then the numerator is the same as the denominator. For other $n$, we can factor it as $$\lim_{s \to 0}\frac{(-1)^ns!}{(n+s)!(n+s-1)!}\left( (s - 1)! - s! \sum_{m = 1}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}\right)$$ The factors converge separately, so moving the first term to the outside and splitting the sum makes it $$\frac{(-1)^n}{n!(n-1)!}\lim_{s \to 0}\left( (s - 1)! -s! \frac{2s +1}{(1 + s)s}- s! \sum_{m = 2}^n \frac{2m + 2s - 1}{(m + s)(m + s - 1)}\right)$$ which is equal to $$-\frac{(-1)^n}{n!(n-1)!}\lim_{s \to 0} \left(\frac{s!}{s+1}\right)-\frac{(-1)^n}{n!(n-1)!}\left( \sum_{m = 2}^n \frac{2m - 1}{m(m - 1)}\right)$$ which simplifies to $$\frac{(-1)^{n+1}}{(n!)^2}\left(n+n\sum_{m = 2}^n \left( \frac{1}{m}+\frac{1}{m-1} \right)\right) = \frac{(-1)^{n+1}}{(n!)^2}\left(1+2nH_{n-1}\right)$$

Edit 2: By the way, the solutions (found using Mathematica) are $$y_1 = \sqrt{x}\ J_1(2\sqrt{x}), y_2 = \sqrt{x}\ Y_1(2\sqrt{x})$$ where $J_1(z)$ is the Bessel function of the first kind, and $Y_1(z)$ is the Bessel function of the second kind.

Title: The Method Of Frobenius


问:
Why "some As are Bs" is $\exists x(A(x) \land B(x))$
Lets say we have an English sentence:

some $A$s are $B$s

We define some predicates:

$A(x)$ is "$x$ is $A$"

$B(x)$ is "$x$ is $B$"

Then to me , it seems like that there are two possible translations (point me if I am wrong)

$\exists x(A(x) \to B(x))$

$\exists x(A(x) \land B(x))$

In "Introduction to Mathematical Logic" by Elliot Mendelson , under exercise 2.7 , they have chosen the second translation.Does that mean the first translation is wrong.If it is wrong , can someone explain to me why it is wrong?

答:
If there are no $A$s at all in the world, then $A(x)$ is false for every $x$.
In particular, if we suppose that there is at least one possible value of $x$,
choose a value for $x$ and then $A(x)$ is false.

If $A(x)$ is false then $A(x) \rightarrow B(x)$ is true.

So in the world described above, $\exists x(A(x) \rightarrow B(x))$ is true,
but no $A$s are $B$s for the simple reason that there are no $A$s.

Another example, a little less extreme:
suppose there is an $x$ which is a $B$ but not an $A$.
Then this $x$ is a witness for the true statement
$\exists x(A(x) \rightarrow B(x))$.
And this statement will continue to be true even if there are no $A$s that also are $B$s.

Another way to look at it
(a heuristic, intuitive approach rather than a rigorous one):
if some $A$s are $B$s,
consider one of those $A$s which is a $B$:
that thing is also a $B$ which is an $A$.
So from "some $A$s are $B$s" it follows that "some $B$s are $A$s",
and vice versa.
This should lead us to expect a symmetric statement;
$\exists x(A(x) \land B(x))$ is such a statement, but
$\exists x(A(x) \rightarrow B(x))$ is not

ContentLicense: CC BY-SA 4.0


问:
number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$
I want to solve the following problem:

Find the number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$.

First of all, from Cayley's theorem, I know

Every group $G$ is isomorphic to a subgroup of the symmetric group $S_G$.

The element in $S_5$ with order $5$ is $(1,2,3,4,5)$ so my guess was $\mathbb{Z}_5 = < (1,2,3,4,5)>$

But having trouble understanding the number of groups.

Wikipedia states the total number of the subgroup is $6$, but how to count properly?

答:
Each of the copies of $\Bbb Z_5$ is generated by a cyclic permutation of $\{1,2,3,4,5\}$. For instance, one is generated by $p=(1,2,3,4,5)$ and has $p^2=(1,3,5,2,4)$, $p^3=(1,4,2,5,3)$, $p^4=(1,5,4,3,2)$, and $p^5=(1)(2)(3)(4)(5)$ as its other four elements.

There are $4!$ such permutations (why?), and each of these groups contains $4$ of them. The only element that two of these groups can have in common is the identity, $(1)(2)(3)(4)(5)$, so there must be $\frac{4!}4=3!=6$ of these groups.


问:
Largest prime that remains prime when any one of its digits is deleted
Recently I came across this beautiful question: Deleting any digit yields a prime... is there a name for this?

The question talks about primes that remain prime when any one of their digits is deleted (zeros allowed). The answers there claim that there is only a finite number of such primes. So naturally I started wondering what is the largest such prime?

According to A051362 the largest known prime with this property is 9977770001777. Can we find a larger one?

I searched for primes in the form $aa \ldots aabb \ldots bb$, where $a$ and $b$ are digits in $\{1,3,7,9\}$ and there are up to $n \leq 400$ of each one. But I couldn't find any solutions other than the trivial small ones.

答:
About $8$ years ago, not too long after the linked question was asked, Code Golf StackExchange took up the challenge. The winning answer found a $274$-digit solution:
$$\underbrace{444\dots 44}_{163} \underbrace{000 \dots 00}_{80} \underbrace{111\dots 11}_{31}$$
This outdoes my own searches considerably. The largest solution I've personally found is the following $84$-digit number, which is the second-largest solution found there:
$$\underbrace{444\dots 44}_{50} \underbrace{111 \dots 11}_9 \underbrace{333\dots 33}_{25}$$
It makes sense to test numbers with only a few blocks of consecutive repeated digits: then there's only a few digit deletions to check, so it's more likely that all digit deletions will remain prime. The search I did included:

All $2$-block numbers with at most $150$ digits per block. The largest solution I found of this form was $222\,223\,333\,333$, which is almost certainly the largest solution there is of this form.

All $3$-block numbers with at most $50$ digits per block. You can see that the solution above brushes up against this limit, but the second-biggest solution of this form I found only had $32$ digits total. The heuristic below tells us not to expect anything better than the $274$-digit solution.

All $4$-block numbers with at most $20$ digits per block; I did this before I found the $3$-block solution, or else I would've known not to bother. This gave me the second-largest solution I've found: an "inflation" of $3161$ with $19$ $3$'s followed by $12$ $1$'s followed by $10$ $6$'s followed by $5$ more $1$'s, for $46$ digits total.

Here's a rough probabilistic analysis of how likely such numbers are to have the property we want. There are $9^k \binom{n-1}{k-1}$ distinct numbers with $n$ digits divided into $k$ blocks of digits (like the above has $16$ digits divided into $4$ blocks). Very loosely, this is $9^k \cdot \frac{n^{k-1}}{(k-1)!}$. The probability that a random $n$-digit number is prime is roughly $\frac1{n \ln 10}$. Initially, I estimated that for an $n$-digit number and all $k$ possible digit-deletions to be prime, you pay a factor of $\frac1{(n \ln 10)^{k+1}}$. So we get a probability of $O\Big((\frac{9}{\ln 10})^k \cdot \frac1{(k-1)!} \cdot n^{-2}\Big)$ that there is any solution with $n$ digits and $k$ blocks. The part depending on $k$ is maximized when $k$ is around $4$.

However, looking at the numbers modulo small primes like $2$, $3$, or $5$, there's some correlation. Some of the correlation helps us (spot-checking primes using the last digit) and some hurts us on net (divisibility by $3$ enforces that either digits $1 \bmod 3$ or digits $2 \bmod 3$ must be missing from the number). It's hard to say what the best value of $k$ is, just that it should not grow with $n$.

I've also limited the number of blocks for practical considerations. Even if it turns out that $k=5$ is abstractly more promising than $k=4$, numbers with $5$ blocks will be at least an order of magnitude less likely to result in all the primes we want - there will just be more of them to compensate. So it will take longer to find such solutions.


问:
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$
For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

$\sin x \approx x$

$\cos x \approx 1-\frac{x^2}{2}$

$\tan x \approx x +\frac{x^3}{3}$

I am looking for a justification of these approximations without the use of series expansions.

By purely geometric considerations, it is easy to see that for small values of $x$, we have

$$ \sin x < x < \tan x.$$

Division by $\cos x$ and an application of the squeeze lemma yield

$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$

and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.

Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?

I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.

答:
Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}}=\frac16$

As shown in $(2)$ from this answer, since $\cos(x)$ is continuous, given any $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $0\lt|x|\le\delta$, we have
$$
1-\epsilon\le\cos(x)\le\frac{\sin(x)}x\le1\tag1
$$
Thus, assuming $0\lt|x|\le\delta$,
$$
\begin{align}
\frac{x-\sin(x)}{x^3}
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^n\sin\left(x/2^n\right)\right)\tag{2a}\\
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^{n+1}\sin\left(x/2^{n+1}\right)\cos\left(x/2^{n+1}\right)\right)\tag{2b}\\
&=\frac1{x^3}\sum_{n=0}^\infty2^{n+1}\sin\left(x/2^{n+1}\right)\left(1-\cos\left(x/2^{n+1}\right)\right)\tag{2c}\\
&=\frac1{x^3}\sum_{n=0}^\infty\frac{2^{n+1}\sin^3\left(x/2^{n+1}\right)}{1+\cos\left(x/2^{n+1}\right)}\tag{2d}\\
&=\frac12\sum_{n=0}^\infty\frac1{2^{2n+2}}\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2e}\\
&=\frac16\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2f}
\end{align}
$$
Explanation:

$\text{(2a)}$: apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and the telescoping sum

$\text{(2b)}$: $\sin(2x)=2\sin(x)\cos(x)$

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

$\text{(2d)}$: $\sin^2(x)=(1-\cos(x))(1+\cos(x))$

$\text{(2e)}$: apply $(1)$ with $[a,b]$ representing a number in that range

$\text{(2f)}$: evaluate the sum

Therefore, by $(2)$ and the Squeeze Theorem,
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3
$$

The Approximations

The limit $(3)$ gives the following approximation
$$
\sin(x)\approx x-\frac{x^3}6\tag4
$$
Since the Binomial Theorem gives $(1-x)^2=1-2x+x^2\approx1-2x$, we have $\sqrt{1-x}\approx1-\frac x2$.
$$
\begin{align}
\cos(x)
&=\sqrt{1-\sin^2(x)}\tag{5a}\\
&\approx\sqrt{1-x^2}\tag{5b}\\
&\approx1-\frac{x^2}2\tag{5c}
\end{align}
$$
Finally, since the Geometric Series gives $\frac1{1-x}=1+x+x^2+\dots\approx1+x$,
$$
\begin{align}
\tan(x)
&=\frac{\sin(x)}{\cos(x)}\tag{6a}\\
&\approx\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)\tag{6b}\\
&\approx x+\frac{x^3}3\tag{6c}
\end{align}
$$


问:
Problems relates to minimal polynomial and $AB, BA$
Let $A$ and $B$ be complex $3\times 3$ matrices. The minimal polynomial of $A$ is $x^3-1$ and the minimal polynomial of $B$ is $(x-1)^3$. At this time show that $AB\neq BA$

I know that the characteristic and minimal polynomials of $AB$ and $BA$ are the same at least one of $A$,$B$ is invertible. But in general, $AB$ and $BA$ do not have the same minimal polynomial. Am I right?

How do we answer this question? Any hints would be appreciated.

答:
The given information tells you that the eigenvalues of $A$ are distinct (in fact the complex third roots of unity), and that the only eigenvalue of $B$ is $1$ (not counting multiplicity here).

And let's assume for purposes of obtaining a contradiction that $AB = BA$. It is often easier to reason with commuting things.

Note that if $v$ is an eigenvector for $A$ corresponding to the eigenvalue $\lambda$, then $Bv$ is also an eigenvector for $A$ corresponding to the eigenvalue $\lambda$.  This is because from left multiplying both sides of $Av = \lambda v$ by $B$ (and pulling the scalar out on the right) we deduce $BAv = \lambda Bv$, and because $AB = BA$ we can deduce from this that $ABv = \lambda Bv$.  And the vector $Bv$ is nonzero because $v$ is nonzero, and we know that $0$ is not an eigenvalue of $B$.

Note also that because $A$ is $3 \times 3$ and has three distinct eigenvalues, the eigenspaces corresponding to each eigenvalue are one-dimensional (and span $\mathbb{R}^3$).  Pick eigenvectors $v_1, v_2, v_3$ corresponding to the distinct eigenvalues $a, b, c$ of $A$.

Fix any eigenvalue $\lambda$ of $A$, and let $v$ be a corresponding eigenvector.  Running through the above discussion we learn that $Bv$ is also an eigenvector of $A$ corresponding to the eigenvalue $\lambda$.  Since the eigenspace corresponding to $\lambda$ is one dimensional and since $v$ is a nonzero vector in it, $v$ spans this eigenspace.  Because $Bv$ is in there, there is therefore a scalar $\mu$ for which $Bv = \mu v$, i.e., $v$ is an eigenvector of $B$ corresponding to the eigenvalue $\mu$.  But the only eigenvalue of $B$ is $1$, so $\mu = 1$ and $Bv = v$.

Running through the previous paragraph with $\lambda$ taken to be each of the eigenvalues of $A$, respectively, with corresponding eigenvectors $v_1, v_2, v_3$ as $v$, respectively, we learn that $Bv_1 = v_1$, $Bv_2 = v_2$, and $Bv_3 = v_3$.  In other words, $B$ acts as the identity on a basis of $\mathbb{R}^3$ and hence $B = I$.

Note that I never used the minimal polynomials here, just eigenvalues (and we never used what the eigenvalues of $A$ were, other than distinct). You might be able to convince yourself that if $A$ is any $3 \times 3$ matrix with distinct eigenvalues and $B$ has only the eigenvalue $1$, and $AB = BA$, then the same argument just given would show that $B = I$ in that case.

But the minimal polynomial of $I$ is $x-1$, not $(x-1)^3$.  So there's your contradiction.

Generalizing this you may be able to convince yourself that if $A$ is $n \times n$ complex matrix with distinct eigenvalues (and hence diagonalizable), and $B$ is matrix commuting with $A$, then $B$ must be simultaneously diagonalizable with $A$ (and hence its minimal polynomial is not going to have repeated roots).  Or see e.g. Prove that simultaneously diagonalizable matrices commute and the answers/references cited therein for more.


问:
How many ways to fill 1xn board with tiles of 2 different sizes?
If I have a board of size 1xn and two different types of tiles, 1xa and 1xb, how many different ways can I fill the board?</p>

答:
You can do it by building a recurrence equation. If $T(n)$ denotes the number of ways to tile a board of size $1$ x $n$ then:

$T(n) = T(n - a) + T(n - b)$

where base conditions are:

$T(x) = 0$ if $x < a$ and $x < b$ and $ x != 0 $

$T(0) = 1 $


问:
How many permutations of the set n={1,2,3,3,3,4,4,5} are possible without any consecutive identical elements?
I understand that there are
$\frac{8!}{3!2!}$=3600 permutations possible without any restrictions in place. I follow this by attaching all like elements as such {1,2,{3,3,3},{4,4},5} where I can account for 5! permutations where I have three 3s and two 4s in consecutive positions. However, when it comes to accounting for the permutations that, for example, will have three consecutives 3s but not so for the 4s, I am at a loss. I wrote some simple code and came up with a set of 960 permutations as the acceptable answer. As the question implies, I am unsure of how to reach that through other means. My apologies if this is considered elementary and I appreciate any help whatsoever.

答:
Arrangements with $3's$ apart ($4's$ unrestricted), eg

$\uparrow2\uparrow 1\uparrow 4\uparrow 4\uparrow 5\uparrow \implies \frac{5!}{2!}\binom 6 3 = 1200$

Subtract arrangements with $3's$ apart but  $4's$ together

$\uparrow 1\uparrow44\uparrow2\uparrow5\uparrow \implies 4!\binom53 = 240$


问:
How do I prove $[0,3)$ is an open set in $([0,\infty),d)$ standard metric?
Question : How do I prove $[0,3)$ is an open set in $([0,\infty),d)$ standard metric?

Background and Approach : This is my first analysis class and I don't know if I'm on the right path. I'm thinking about proving if every point in the interval has a neighborhood contained in it, but the $[0,\infty)$ confuses me. I know that a set $X$ is open if given any $x \in X\; \exists \;r\in\mathbb{R^+}$ such that $B(x,r) \in X$.

答:
To prove $U = [0, 3)$ is an open set in $V = [0, \infty)$ with standard metric you need to find an set $X$ such that

$U = X \cap V $ and $X$ is open in the $\mathbb{R}$ w.r.t standard metric.

Taking $X = (-1, 3)$ would work in this case.


问:
How many ways can you pay $100 when the order in which you pay the coins matters?
Suppose there is a vending machine where you have to pay $100. You
have an unlimited amount of 1 dollar, 2 dollar and 5 dollar coins. How many ways
can you pay when the order in which you pay the coins does not matter?
Also, find the ways when the order matters.

Using generating functions, I solved the first part, i.e., we need to find the coefficient of $x^{100}$ in $$(1+x + x^2 + \dots)(1+x^2 +x^4+ \dots)(1+x^5+x^{10}+\dots).$$

However, I am stuck in the second part. How do we maintain the order while finding the number of ways?

答:
You could use the recursion
$$f(n)=f(n-1)+f(n-2)+f(n-5)$$


问:
How many ways can we arrange 3 boys and 3 girls in a circular table if all boys should sit beside one another?
This is what I've done so far:

I treat the groups of 3 people as 1 since the constraint is that all the 3 boys should sit beside one another. Since it's a group of 3 boys, and this would mean that the girls would do the same. We can treat each group as 1, then this means we will have 2 groups left. I applied the technique for the circular permutation for each groups, Then each group will be having 3 factorial.

$$(2-1)! \times (3!)^2 = 36$$

Is this correct?

答:
You are correct in that there are 36 possible permutations considering only who is next to whom and on which side. 6 permutations for the boys, 6 for the girls.

But if the chairs are numbered, the boy who sits to the right of a girl can be in six different chairs, so there are 216 possible seatings.


问:
How to evaluate this limit where $f(x)=x$ if $x\le 1$ and $1-x$ if $x>1$?
I am trying to evaluate $\displaystyle \lim_{x \to 1}f(x)$ where $f(x) =x$ if $x\le 1$ and $1-x$ if $x>1$. I graphed it and I am confused because it intersects at $(0.5, 0.5)$. I did my work and I said the limit as $x$ approaches $1$ from the left equals $x$ and the limit as $x$ approaches $1$ from the right equals $1-x$.

I am not sure if this is the correct answer.
Can someone guide me? The problem and work is posted below. Thank you.

答:
If $x<1$, then $f(x)=x$; therefore, $\lim_{x\to1^-}f(x)=1$. And if $x>1$, then $f(x)=1-x$; therefore, $\lim_{x\to1^+}f(x)=0$.

So, the limit $\lim_{x\to1}f(x)$ does not exist.


问:
Proving that cyclotomic polynomials have integer coefficients
I don't understand why Gauss's lemma is invoked in the proof in Dummit and Foote that $\Phi_n(x)$ (the $n$th cyclotomic polynomial) belongs to $\mathbb{Z}[x]$. I'm an analyst and I wanted to remind myself about cyclotomic polynomials. The following is the proof as I've written it, and because it's been a while since I've done algebra I want to know if there is something I'm taking for granted.

It is a fact that if $R$ is a unital commutative ring, $f \in R[x]$ is a monic polynomial and
$g \in R[x]$ is a polynomial, then there are $q,r \in R[x]$ with
$g = qf+r$,
$r=0$ or $\deg r < \deg f$.

First, $\Phi_1(x)=x-1 \in \mathbb{Z}[x]$. For $n>1$, assume that
$\Phi_d(x) \in \mathbb{Z}[x]$ for $1 \leq d < n$. Then let
$f = \prod_{d \mid n, d<n} \Phi_d$,
which by hypothesis belongs to $\mathbb{Z}[x]$; and since each $\Phi_d$ is monic, so is $f$.

On the one hand,
since $g(x)=x^n-1 \in \mathbb{Z}[x]$, there are $q,r \in \mathbb{Z}[x]$ with
$g = q f + r$ and
$r=0$ or $\deg r < \deg f$.
On the other hand, using $x^n-1= \prod_{d \mid n} \Phi_d(x)$ we have
$g = \Phi_n f \in \mathbb{C}[x]$.

Thus $\Phi_n f = qf+r \in \mathbb{C}[x]$,
so $r = f \cdot (\Phi_n - q) \in \mathbb{C}[x]$. If $\Phi_n \neq q$ then $\deg r = \deg f + \deg (\Phi_n-q) \geq \deg f$, contradicting that
$r=0$ or $\deg r < \deg f$. Therefore $\Phi_n = q \in \mathbb{C}[x]$, and because $q \in \mathbb{Z}[x]$ this means that
$\Phi_n \in \mathbb{Z}[x]$.

I don't see any tacit assumptions, like for example degree meaning two different things in $\mathbb{C}[x]$ and $\mathbb{Z}[x]$, but if I'm not assuming anything then I don't see why Gauss's lemma is being used in the proofs I've come across.

答:
This proof seems to be more elementary. We begin by contending that if $$(x^n -1) =({\sum}^p_{i=1}{a_i{x^i}})({\sum}^q_{j=1}{b_j{x^i}}),$$ where ${\sum}^p_{i=1}{a_i{x^i}}\in{\bf{Z}[x]},$ then every coefficient $b_j$, must be an integer. To prove this, we argue by backward induction. Since $a_p{b_q} = 1,$ and since $a_p\in{\bf Z},$ we conclude that $a_p=b_q = \pm{1}$ and $b_q$ is therefore an integer. Now, assume as our induction hypothesis, that $b_q,b_{q-1},\cdots b_{r+1}$ are all integers. Then, the coefficient of $x^{p+r}$ is $${a_p}{b_r}+{a_{p-1}}{b_{r+1}}+\cdots +{a_{(p+r)-q}}{b_q}=0.$$ Since ${a_p}=\pm1$, the above identity shows that $${b_r} = \pm({a_{p-1}}{b_{r+1}}+\cdots +{a_{(p+r)-q}}{b_q})\in{\bf{Z}}.$$ Thus, we conclude that every coefficient $b_j$, must be an integer by mathematical induction.
Now, we use the identity $$(x^n - 1) = {\prod}_{d|n}{\Phi}_d(x).$$ We claim that $\Phi_n(x)\in{\bf{Z}}[x]$ by induction on $n$. The assertion is clear for $n=1$ since $\Phi_1(x)=(x-1).$ Assume now, as our induction hypothesis, that $\Phi_d(x)\in{\bf{Z}}[x]$ for all positive integers $d<n$. Then, writing $$(x^n - 1) = {{\prod}_{d|n,d<n}}{\Phi}_d(x)\cdot{\Phi_n(x)},$$ since ${{\prod}_{d|n,d<n}}{\Phi}_d(x)\in {\bf{Z}}[x]$, it follows by our argument above that ${\Phi_n(x)}\in {\bf{Z}}[x].$

ContentLicense: CC BY-SA 3.0


答:
Since it is emphasized that the table is circular, it is reasonable to presume that rotational changes are not deemed to produce a different arrangement.

Your answer of $(3!)^2$ is correct:

Without Loss of Generality, the Head of the table (i.e. the 12 O'clock position in the circular table with 6 chairs) will be occupied by the middle boy (whoever that turns out to be).

Given that, with the boys seated at positions 10,12,2 and the girls seated at positions 4,6,8, each of the two groups (3 boys and 3 girls) can be permuted in $3!$ ways.

Title: How many ways can we arrange 3 boys and 3 girls in a circular table if all boys should sit beside one another?

Body:
This is what I've done so far:

I treat the groups of 3 people as 1 since the constraint is that all the 3 boys should sit beside one another. Since it's a group of 3 boys, and this would mean that the girls would do the same. We can treat each group as 1, then this means we will have 2 groups left. I applied the technique for the circular permutation for each groups, Then each group will be having 3 factorial.

$$(2-1)! \times (3!)^2 = 36$$

Is this correct?


问:
How to find the subgroups of $S_n$ which are isomorphic to $\mathbb{Z}_l$?
This post is somewhat generalization of my previous question, number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$

I want to generalize the previous post and want some procedure or known facts regarding this.
i.e.,

How to find the subgroups of $S_n$ which are isomorphic to $\mathbb{Z}_l$?

My naive idea is to express the explicit element of $S_n$ and see the cyclic order of $l$... But for me it seems it is not enough.

答:
Yes, every cyclic subgroup of $S_n$ of order $k$ (I'm using $k$ because I find $l$ difficult to read, too close to $1$ for my taste) is generated by an element of $S_n$ of order $k$ (and of course often the same subgroup can be generated by different elements, but you don't seem to be interested in counting subgroups in this question, only finding them - so that's another story - but the below may assist in counting them).

It helps to remember that the order of an element of $S_n$ can be determined from its cycle type - that is, the list of positive integers $(a_1, \dots, a_m)$ for which the element can be written as product of disjoint cycles of lengths $a_1, \dots, a_m$,  respectively.  In this event, fairly simple calculations can show you that the order of an element of cycle type $(a_1, \dots, a_m)$ is $\operatorname{lcm}(a_1, \dots, a_m)$.

So in terms of cycle types, you're looking for ones corresponding to tuples of positive integers $(a_1, \dots, a_m)$ with $\operatorname{lcm}(a_1, \dots, a_m) = k$.  And because the disjoint cycles that make up this cycle type have to fit in $S_n$ it is also required here that $a_1 + \dots + a_m = n$.  (There's no harm in thinking of fixed points of a permutation as cycles of length $1$, which is why I have equality here.)

Recall that tuples of positive integers summing to a positive integer $n$ are sometimes called partitions of $n$.  You're looking to see if there are any partitions of $n$, the $\operatorname{lcm}$ of which is $k$.  If there are any, then you just fill in cycles corresponding to that cycle type.  For many values of $n$ and $k$ computers would likely to be better at performing this computation than people.  I don't know of a way to simplify it, or a simpler way of approaching it.

For example if $n=10$ and $k=6$, I'm looking for partitions of $10$ with $\operatorname{lcm}(6)$.  Doublecheck, e.g. via your own reasoning or via http://oeis.org/wiki/10, that there are not so many such partitions, including e.g. $(3,2,1,1,1,1,1)$, $(3,2,2,2,1)$, and $(6,1,1,1,1)$ and others, but not so many.  If I've done this right that means the subgroups of $S_{10}$ of order $6$ must be generated permutations of these cycle types, e.g. those of the form $(abc)(de)$, or $(abc)(de)(fg)(hi)$, or $(abcdef)$, etc.  It's pretty easy to write down examples of these (and others), and from the above we know that those are all of them.

From there, maybe you can begin identifying when subgroups with these types of generators happen to be equal, and begin counting them.


答:
One of the ways to approach it -

First place $1,2, 5$ ($3!$ ways). Then,

i) Place two $4$'s in four places between $\{ \ 1 \ 2 \ 5 \ \}$. You then have five digits and hence six places in between where you can place $3$.

So number of arrangements $ = \displaystyle 3! {4 \choose 2} {6 \choose 3} = 720$.

ii) Place both $4$'s in one of the four places between $\{ \ 1 \ 2 \ 5 \ \}$ separated by a $3 \ (434$ being together). With $434$ being together, you now have five places and choose two places for remaining two $3$'s.

Number of arrangements $ = \displaystyle 3! {4 \choose 1} {5 \choose 2} = 240$.

So, total number of valid arrangements $ = 720 + 240 = 960$.

Title: How many permutations of the set n={1,2,3,3,3,4,4,5} are possible without any consecutive identical elements?

Body:
I understand that there are
$\frac{8!}{3!2!}$=3600 permutations possible without any restrictions in place. I follow this by attaching all like elements as such {1,2,{3,3,3},{4,4},5} where I can account for 5! permutations where I have three 3s and two 4s in consecutive positions. However, when it comes to accounting for the permutations that, for example, will have three consecutives 3s but not so for the 4s, I am at a loss. I wrote some simple code and came up with a set of 960 permutations as the acceptable answer. As the question implies, I am unsure of how to reach that through other means. My apologies if this is considered elementary and I appreciate any help whatsoever.


问:
Show that if $x\geq 0$ and $n$ is a positive integer, then $\sum_{k=0}^{n-1}\left\lfloor {x+\frac{k}{n}}\right\rfloor=\lfloor {nx}\rfloor$
I need help with this question, where $\lfloor x\rfloor$ means the floor function of $x$.

Show that if $x\geq 0$ and $n$ is a postive integer, then $$\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n} \right\rfloor =\lfloor nx\rfloor$$

答:
The main idea of the proof is splitting the $x$ into integer and fractional part. That is, $x=\lfloor x \rfloor+{x}$ and there exists exactly one $k' \in \{1,2, \dots n \}$.

$$\lfloor x\rfloor=\left \lfloor x+\frac{k'-1}{n}\right\rfloor\le x\left\lfloor x+\frac{k'}{n}\right\rfloor=\lfloor x\rfloor+1 $$

Subtracting $\lfloor x \rfloor$ from the equation would give:

$$\left\lfloor \{x\}+\frac{k'-1}{n}\right\rfloor\le \{x\} \left\lfloor \{x\}+\frac{k'}{n}\right\rfloor=1$$

$$1-\frac{k'}{n}\le \{x\} < 1-\frac{k'-1}{n}$$

Multiplying out $n$ throughout.

$$\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor
=\sum_{k=0}^{k'-1} \lfloor x\rfloor+\sum_{k=k'}^{n-1} (\lfloor x\rfloor+1)=n\, \lfloor x\rfloor+n-k'$$

$$=n\, \lfloor x\rfloor+\lfloor n\,\{x\}\rfloor=\left\lfloor n\, \lfloor x\rfloor+n\, \{x\} \right\rfloor=\lfloor nx\rfloor$$

Source: Hermite's Identity.


问:
How many ways (methods) there are to solve Cauchy-Euler equation?
I have seen at least three different ways to solve it. The first, by trial solution. The second, using a change of variable. The third, by Power series... Oh! And Frobenius too.... So I was wondering, should I just use the easiest?. For example, using trial solution instead of Frobenius. Thanks in advance.</p>

答:
If by "trial solution" you mean establishing the characteristic polynomial and getting the basis solutions from it, then that is the same as the Frobenius (and) power series method. As there the indicial condition for the leading power is the same as the characteristic polynomial, with the same consequences.

So in fact you only have 2 really different methods.


问:
Find all the possible numbers whose sum is x but ....
I want to find all the number whose sum is x but

but series of number start with a specific number and next number is always smaller then the number before it

For example I want 41 but first number in the series is
11
I want all the combination of number which start from 11 and succeeding number are smaller than previous number like
11 + 9 + 6 + 5 + 4 + 3 + 2 + 1 = 41

Numbers are 11 and less than 11 and each number occurs only one time, and sum of them is 41.
11+10+9+8+3 = 41

I want a method so I can effectively generate all such combinations.

答:
Hint

Since $\sum_{k=1}^{11} k = 66$, your question bijects to the (easier to manually count) question of :

How many distinct subsets of $\{1,2,\cdots,11\}$ are there such that the sum of all of the numbers in the subset exactly equals $(66-41) = 25?$

Given the simplicity of the alternative question, and given that no elegance presents itself to me, I would attack the alternative question manually.

Naturally, if anyone has a more elegant approach, I would consider that approach superior, regardless of the simplicity of the question being attacked.

Note

I reject the Stars and Bars approach because the number of (analogous) variables is not fixed.

Also, the actual query is not the number of solutions, but rather to obtain an enumeration of each possible solution.


问:
How to solve this (binomial?) series?
Im struggling with this equation. Could someone maybe explain how to get this solution?

$ \sum_{k=1}^{\infty} \binom{-k}{k-1} (p(p-1))^{k-1} = \frac{1}{2p-1}$

My first idea was to use the binomial series, but therefore the $-k$ would have to be fix, which it isn't in my case.

So what would be the best method to solve the equation?

Thank you in advance!

Title: How to solve this (binomial?) series?

Body: Hint: $\displaystyle\binom{-k}{k-1}=(-1)^{k-1}\binom{2k-2}{k-1}=2^{2(k-1)}\binom{-\frac12}{k-1}$


问:
A problem in understanding the Intermediate Value Theorem
So, IVT essentially says if a function $f$ is continuous over an interval $[a,b]$, then the function $f$ will take up all the values between $f(a)$ and $f(b)$ at least once at some point in the interval $[a,b]$.

Now, suppose $f$ is continuous between $[3,7]$ and $f(3) = 4$ and $f(7) = 25$

According to IVT, $f(3) = 4 < m < 25 = f(7)$, i.e., the function takes up all the values between $f(3)$ and $f(7)$ at least once in the interval. But, according to the below picture, clearly, the function within that interval takes up values different that do not satisfy the above inequality. Please help me fill the gap in my understanding of IVT here.

Please do correct me if I'm mistaken.

This doubt arose from the below KhanAcademy question

答: The intermediate value theorem tells you that, in that context, for each $y\in[4,25]$, there is some $x\in[3,4]$ such that $f(x)=y$. It does not tell you that if $x\in[3,4]$, then $f(x)\in[4,25]$. So, there is no contradiction if $f\bigl([3,4]\bigr)$ contains values outside $[4,25]$.


问:
Runge-Kutta identity for $y'=\lambda y$
An $r$-stage (explicit) RK method of order $r$, step size $h$, is applied to the scalar equation $y'=\lambda y$. Show that
$$y_{n+1}=\left(\sum_{i=0}^r\frac{1}{i!}(h\lambda)^i\right)y_n.$$
I have tried to write the method explicitly, as
$$y_{n+1}=y_n+h\sum_{i=1}^r b_ik_i,$$
where $k_1=\lambda y_n$, $k_2=\lambda y_n+\lambda^2 hc_2y_n$ etc, but it's all getting rather messy.

答:
You have to think more simply. Take the given data at face value. The exact solution with $y(0)=y_0$ is $y(t)=y_0e^{λt}$. Being of global order $r$ means for the step of size $h$ that
$$
y_1=y_0+h\Phi_f(0,y_0,h)=y(h)+O(h^{r+1})
$$
So now what is $e^{λh}$ truncated at the order $O(h^{r+1})$?

Then use that with $r$ stages you can also not get a higher polynomial degree.

Title: Runge-Kutta identity for $y'=\lambda y$


问:
Transposing the coefficients of a bivariate generating function
Consider a triangle of integers

$a_{1,0}$

$a_{2,1}, \, a_{2,0}$

$a_{3,2}, \, a_{3,1}, \, a_{3,0}$

$\quad \vdots$

and its "transpose"

$a_{1,0}$

$a_{2,0}, \, a_{2,1}$

$a_{3,0}, \, a_{3,1}, \, a_{3,2}$

$\quad \vdots$

Given a bivariate generating function $f(x,y)$ with series
$$
a_{1,0}*y + (a_{2,1}x + a_{2,0})y^2 + (a_{3,2}x^2 + a_{3,1}*x + a_{3,0})y^3 + \cdots
$$
what is the generating function which gives the coefficients of
the polynomials of $x$ as the transposed triangle,
i.e. how can $g(x,y)$ be obtained with series
$$
a_{1,0}y + (a_{2,1} + a_{2,0}x)y^2 + (a_{3,2} + a_{3,1}x + a_{3,0}x^2)y^3 + \cdots
$$

答: $g(x, y) = \frac{1}{x} f \left( \frac{1}{x}, xy \right)$. First you should know that there is a more basic trick here where you reverse the coefficients of a polynomial by substituting $x \mapsto \frac{1}{x}$ and clearing denominators, and after that it's not hard to tweak everything else to work out. </p>


问:
Is the number of enclosed areas always equal to number of intersections plus one?

For this shape, the number of intersections is 4, and the number of enclosed areas/regions is five.

For this shape, the number of intersections is 6, and the number of enclosed areas is 7.

It seems reasonable to conclude that number of enclosed areas equals the number of intersections plus one, but I was wondering if this is always the case.

答:
No. Consider $k$ non-intersecting concentric polygons. The number of intersections is zero, but the number of enclosed areas is $k$.

Other violations of the proposed rule occur in the degenerate case with more than two lines intersecting at a point as well as the wholly pathological case of overlapping sides when the number of intersections is infinite.


答:
You have the Diophantine equation $a + 2b + 5c = 100$, and you have  already found out the number of solutions.

If a,b,c, are distinct for each solution, the number of ways could be found out as $\dfrac{(a+b+c)!}{a!b!c!}$, but they may not always be distinct

You can't have $a = b = c$ (why ?) but you can have $a=b,\; a=c,\; b=c$

These can be ferreted out (they will not be many), and the usual correction for permutations with repeated elements be made.

Title: How many ways can you pay $100 when the order in which you pay the coins matters?

Body:
Suppose there is a vending machine where you have to pay $100. You
have an unlimited amount of 1 dollar, 2 dollar and 5 dollar coins. How many ways
can you pay when the order in which you pay the coins does not matter?
Also, find the ways when the order matters.

Using generating functions, I solved the first part, i.e., we need to find the coefficient of $x^{100}$ in $$(1+x + x^2 + \dots)(1+x^2 +x^4+ \dots)(1+x^5+x^{10}+\dots).$$

However, I am stuck in the second part. How do we maintain the order while finding the number of ways?


问:
Orthonormal functions as a combination of three complex functions
I have some problem to find three orthonormal functions in the interval $-1\le x\le 1$ as a linear combination of these three functions:
$$f_1(x)=1,f_2(x)=x\exp(i\pi x),f_3(x)=\exp(i\pi x)$$
Is it possible to use the Gram - Schmidt method?
Thanks</p>

答:
Using Gram-Schmidt process, as correctly pointed out in the comments, you will notice that, in order to get three orthogonal functions in $L^2$ with the inner product $\langle f_i,f_j\rangle=\int_{-1}^1dxf_i^*(x)f_j(x)$, it is enough to rewrite
$$
   f_2(x)=xe^{i\pi x}-\frac{i}{\pi}.
$$
Then you can properly normalize them.</p>


问:
Since we can fill a circle with infinitely-many copies of its radius, isn't $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$
In a circle, there are area and a radius

When we fill up an infinite radius in a circle it becomes its area.

In that case, doesn't it means that $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$

答: You are approximating the circle with the sum of the areas of triangles that fit inside. This is actually the first numerical way pi was calculated, done by Archimedes about 2300 years ago. Yes you can make the triangular slices thinner and thinner, in a process where you can fit in triangles without bound, however because they are getting thinner in proportion with the rate at which you are making more of them, the sum of the areas of the triangles actually stays finite.</p>


问:
order of "truncated" braid groups
Consider the braid group on n strands given in the usual Artin presentation. Then add extra relations: each Artin generator has order d. For example, if d=2, one recovers the symmetric group. I would like to know what the order of the group is for arbitrary n and d. Even knowing the name of such groups would be helpful, though, as my attempts to determine this by searching the literature have so far failed.</p>

答:
I thought I would add what is almost a complete answer which is outlined in the book

K. Murasugi & B. Kurpita, A Study of Braids, Kluwer Academic Publishers, 1999.

The following surprising theorem tells us when the truncated braid groups are finite, and the order of the groups when they are.

Theorem: Let $B_n(d)=B_n/\langle\sigma_i^d \rangle$. The group $B_n(d)$ is finite if and only if $d=2$ or $(n,d)$ is the type of one of the 5 platonic solids. For these cases, $$|B_n(d)|=\left(\frac{f(n,d)}{2}\right)^{n-1}n!$$ where $f(n,d)$ is the number of faces of the platonic solid of type $(n,d)$

The 5 platonic solids correspond to the pairs $(n,d)\in\{(3,3),(3,4),(4,3),(3,5),(5,3)\}$. This is equivalent to the pair $(n,d)$ being a solution to the inequality $$\frac{1}{n}+\frac{1}{d}>\frac{1}{2}.$$

For ease of calculation, we have the following table giving the number of faces of the corresponding platonic solids
$$\begin{array}{|r|l|}\hline (n,d)&f(n,d)\\\hline
(3,3)&4\\
(3,4)&8\\
(4,3)&6\\
(3,5)&20\\
(5,3)&12\\\hline
\end{array}$$
and so we can calculate the table of group orders
$$\begin{array}{|r|l|}\hline (n,d)&|B_n(d)|\\\hline
(3,3)&24\\
(3,4)&96\\
(4,3)&648\\
(3,5)&600\\
(5,3)&155520\\\hline
\end{array}$$

To me, this theorem and its application highlights one of the strangest links between two fairly weakly related areas of mathematics; finite groups arising from topological or combinatorial (pick you favourite description of the braid groups of the disk) considerations, and the geometric classification of regular solids in $\mathbb{R}^3$.

Title: order of "truncated" braid groups


问:
How to prove that the limit is 0?
I come across with this question:

Consider $\{a_n\}$ is a bounded sequence, $\lim_{n\rightarrow \infty} (a_n - 2a_{n+1}+a_{n+2}) = 0$, prove that
$$
\lim_{n\rightarrow \infty}(a_n - a_{n+1}) = 0
$$

I can only prove that $b_n = a_n - a_{n+1}$ is Cauchy, how can I conclude that the limit is $0$?

答:
Suppose $a_n -a_{n+1} \to l>0$. Then $a_n -a_{n+1} >\frac l 2$ for $n$ suffcienlty large, say for $n \geq n_0$. So $a_{n+1} <a_n-\frac l 2$ for $n \geq n_0$. Iterate this to get $a_{n+1} <a_{n_0} -\frac {(n+1-n_0) l} 2$ and conclude that $a_n \to -\infty$ as $n \to \infty$ contradicting the fact that $(a_n)$ is bounded.

Similarly, $a_n -a_{n+1} \to l<0$ leads to a contradiction. Hence, $a_n -a_{n+1} \to 0$.

Title: How to prove that the limit is 0?


问:
Showing that $\frac{(j+a)!(j-a)!}{a^2}$, for $a=1, \ldots, j$, reaches its minimum at $a\approx \sqrt{j}$
Let $j$ be a positive integer, and $a=1,\dots,j$. From numerical result, I get that $$\frac{(j+a)!(j-a)!}{a^2}$$ reaches its minimum at $a\approx\sqrt{j}$. But how to prove this?

More general form: $\alpha \neq 2$ case

答:
Consider $$f(a)=\frac{(j+a)!\,(j-a)!}{a^2}\qquad \text{and} \qquad g(a)=\log[f(a)]$$
$$g'(a)=\frac {f'(a)}{f(a)}$$ Simplify to show that
$$f'(a)=0 \implies H_{j+a}-H_{j-a}-\frac{2}{a}=0$$
Suppose that $j \gg a$ and use the asymptotics of harmonic numbers; it gives
$$H_{j+a}-H_{j-a}-\frac{2}{a}=-\frac{2}{a}+\frac{2 a}{j}-\frac{a}{j^2}+O\left(\frac{1}{j^3}\right)$$ If you use the expansion to $O\left(\frac{1}{j^2}\right)$, it gives your result. Using the next term would give, as better approximation
$$a \sim \frac{j\sqrt{2} }{\sqrt{2 j-1}}$$

Using more terms and series reversion,
$$a=\sqrt j \left(1+\frac{1}{12 j}+\frac{71}{1440 j^2}+\frac{2951}{34560 j^3}+O\left(\frac{1}{j^4}\right)\right)\tag 1$$

Trying for $j=16$, the above truncated series gives
$$a=\frac{28465051}{7077888}\approx 4.021687$$ while the solution given by Newton method is $4.021616$

Edit

Using $(1)$, we could easily show that
$$H_{j+a}-H_{j-a}-\frac{2}{a}=\frac{3463}{12096\,j^{7/2}}+O\left(\frac{1}{j^{9/2}}\right)$$ Knowing the possible form of $a$, let
$$a=\sqrt j \,\,\sum _{i=0}^3 \frac {b_i} { j^{i}}$$ This leads to
$$a=\sqrt j \left(1+\frac{1}{12 j}+\frac{71}{1440 j^2}+\frac{491}{5760j^3}+O\left(\frac{1}{j^4}\right)\right)\tag 2$$ This changed very little coefficient $b_3$

$$\frac{2951}{34560}-\frac{491}{5760}=\frac{1}{6912}$$ but this is sufficient to make
$$H_{j+a}-H_{j-a}-\frac{2}{a}=-\frac{4429051}{1036800\,j^{9/2}}+O\left(\frac{1}{j^{11/2}}\right)$$


问:
To find $k$ such that slanted line gets mapped to vertical line
Given, $T:R^2\to R^2$, a horizontal shear, whose standard matrix is $\begin{bmatrix}1 & k \\ 0 & 1 \end{bmatrix}$. It is asked to find $k$ if possible, so that line x=$\begin{bmatrix}2 \\ -3 \end{bmatrix} + t\begin{bmatrix}1  \\ 4 \end{bmatrix}$, gets mapped into a vertical line.

I think it is impossible to find such $k$. Because, horizontal shear will just shift the given line without changing it's slope, i.e. we will have some parallel line to the original one, so suck $k$ does not exist. Am I correct?

答: Map the matrix on to the line: $$\begin{bmatrix}1&k\\0&1\end{bmatrix}\begin{bmatrix}2+t\\-3+4t\end{bmatrix}=\begin{bmatrix}(2-3k)+(1+4k)t\\-3+4t\end{bmatrix}$$ This is a line passing through $(2-3k,-3)$ with direction $(1+4k,4)$. So to get a vertical line, choose $k=-1/4$.


问:
Number of codes with $5$ distinct characters using $5$ letters and $9$ numbers
The letters; $A, B, C, D, E$ and the number; $1, 2, 3, 4, 5, 6, 7, 8, 9$ are to be used to form codes with five distinct characters. Each code has at least one letter and has at least one number. How many codes can we form?

My attempt:

Choosing exactly one letter:
$^{5}P_{1} \times ^{9}P_{4}= 5 \times 3024 = 15120$ ways.

Choosing exactly two letters:
$^{5}P_{2} \times ^{9}P_{3}= 20 \times 504 = 10080$ ways.

Choosing exactly three letters:
$^{5}P_{3} \times ^{9}P_{2}= 60 \times 72 = 7320$ ways.

Choosing exactly four letters:
$^{5}P_{4} \times ^{9}P_{1}= 120 \times 9 = 1080$ ways.

So, we have a total of $15120+10080+7320+1080=33600$ different codes.

Please tell me if that is right or wrong. Thanks.

答:
The simplest approach is to think of all the codes you don't want and subtract these from the total number of codes built with all the 14 characters. The ones you don't want are those which are all numbers or are all letters.

This is just
$14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 - (9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)$.


问:
Show that α : R → R n (distance on R and R n being usual and Euclidean respectively) defined by α(x) = (x, 0, . . . , 0) is continuous
We have been given this question. I am new to topology so I'm having a hard time figuring how to solve this out. Can someone help?</p>

答: Hint: $$\| \alpha(x) - \alpha(y) \| \le |x-y|$$ for all $x,y \in \Bbb R$.


问:
Bilinear form respect to the representation $V/W$ of Lie algebra
I got a problem from the book An Introduction to Lie Groups and Lie Algebras written by Kirillov.

The exercise 5.1 says
$$\begin{array}{l} \text { (1) Let } V \text { be a representation of } \mathfrak{g} \text { and } W \subset V \text { be a subrepresentation. Then } B_{V}=B_{W}+B_{V / W} \text { , } \\ \text { where } B_{V} \text { is defined by }(5.14) .\end{array}$$

and 5.14 says $$\text { Let } V \text { be a representation of g and define a bilinear form on g by } B_{V}(x, y)=\operatorname{tr}_{V}(\rho(x) \rho(y)).$$

I'm not sure what $B_{V/W}$ is. For $\rho_{V/W}(x)$, it should be defined as $\rho_{V/W}(x)(v+W)=\rho_V(x)(v)+W$. However, what should $\operatorname{tr}_{V/W}$ be? I can't find any information about this trace. Is there any common definition for it? Thank you.

答: Your definition of $\rho_{V/W}$ is wrong.  $\rho_{V/W}$ is the representation of $\mathfrak{g}$ on $V/W$ so it is $\rho_{V/W}(x)(v+W)=\rho_V(x)(v)+W$ for all $x\in\mathfrak{g}$ and all $v\in V$.  You can take trace of linear maps $V/W\to V/W$ so that gives you $\operatorname{tr}_{V/W}(\rho_{V/W}(x)\rho_{V/W}(y))$ which of course you can drop some of the $V/W$ subscripts for brevity.


问:
Probability Problem on Gates in Subway station
A subway station in a metro city has 10 gates, five for entering into the subway station, and five for exiting the subway station. The number of gates observed in each direction is observed at a particular time of a day. Assume that each outcome of the sample space is equally likely.                                                                                4Marks
(a) What is the probability that at most one gate is open in each direction?
(b) What is the probability that at least one gate is open in direction?
(c) What is the probability that the number of gates open is the same in both
direction?
(d) What is the probability of the event that the total number of gates open is four?</p>

答:
Hints

There are 10 gates in total, and each outcome of the sample space is equally likely, I took the chance that a gate is open is 1/10!

No.  Each gate is open or closed.  Therefore, there are $2^{(10)}$ possibilities, each of which is equally likely.  As a consequence of identifying the sample space as having $(1024)$ equally likely possibilities, the chance of any individual gate being open is exactly $\frac{1}{2}$.

The easiest way to attack part $a$ is to consider the 5 entrance gates and 5 exit gates separately.  Examining only the entrance gates, the sample space is reduced to $(2^5) = 32.$  An examination of the corresponding row of Pascal's triangle indicates that the probability of there being at most 1 entrance gate open is

$$\frac{1 + 5}{1 + 5 + 10 + 10 + 5 + 1}.$$

More formally, if the chance of the success of an event is $p \in (0,1)$, and $q = (1-p)$, then the chance of exactly $k$ successes in $n$ bernoulli trials is

$$\binom{n}{k}p^k q^{(n-k)}.$$


问:
Surjective implies fixed point?
I'm in a discrete dynamical systems class and we have a function $f : [0, \infty) \longrightarrow [0, \infty)$ such that $f$ is surjective, I need a hint to prove that there exists $x_0 \in [0, \infty) \ \text{s.t.} \ f(x_0)=x_0$. I have tried to come up with counterexamples but I can't come up with any function.

Thanks in advance.

答: This is false. Let $f(x)=(x-2)^{2}$ if $x \neq 1, x \neq 4$. Let $f(1)=4$ and $f(4)=1$. Then you can check that $f$ is surjective but there is no $x$ such that $f(x)=x$.

Title: Surjective implies fixed point?


问:
Prove that if $G$ is a finite group with an even class number, then $G$ is of even order. Give an example that the converse fails.

Prove that if $G$ is a finite group with an even class number (the number of conjugacy classes of a group $G$ is called its class number) then $G$ is of even order. Give an example that the converse fails.

My attempt: If $n$ is even $$|G|=\sum_{i=1}^{n} |{\rm orb}(g_i)|$$
and for all $i$, $|{\rm orb}(g_i)|$ divides $|G|$. Let $a,b_i$ be integers. Does

$$a=\sum_{i=1}^{n \text{ even } }b_i$$ with $b_i\mid a$ $\forall i$ imply that $a$ is even?

Actualization: By contradiction, if $a$ is odd, $b_i\mid a$ implies $b_i$ is odd. Because odd+odd=even then $a$ is even a contradiction.

What counterexample can be work?

答:
Here is a much simpler solution based on the class equation. Assume that $|G|$ is odd, then since $|Z(G)| \mid |G|$, also $|Z(G)|$ is odd and for $g \in G$ also $\#Cl_G(g)=|G:C_G(g)|$ is odd. The class equation gives
$$|G|=|Z(G)| + \sum_{g_i \notin Z(G)} \#Cl_G(g_i)$$
for certain non-central $g_i \in G$. Write $k(G)$ for the number of conjugacy classes. Taking the equation mod $2$ at both sides yields
$$1 \equiv 1 +(k(G)-|Z(G)|) \text { mod 2},$$
hence $k(G) \equiv |Z(G)| \equiv 1 \text { mod } 2$. So if $|G|$ is odd then $k(G)$ is odd. A counterexample to the reverse is $G=S_3$, which has $ k(G)=3$.

Note One can even prove the following congruence (and this is a theorem of Burnside which he proved with the help of character theory, but also can be shown without (even improvements and variations, see here.))

Theorem If $|G|$ is odd, then $|G| \equiv k(G) \text { mod } 16$.


问:
What is the other way to write this triple integral?
I need to find all $5$ ways to write this integral and so far I have found $3$ of the $5$ but I need some help to find the last $2$ which are $dy$ $dz$ $dx$ and $dy$ $dx$ $dz$. I assume one of them might require you to add $2$ triple integrals  together. Anyways when evaluated the answer needs to come out to about $12.669$ which can be checked via Desmos. Can anyone help?

$$\int_0^2\int_{e^y}^{e^2}\int_0^{e^\left(\frac{y}{2}\right)} f(x,y,z) \ dz \ dx \ dy$$

答:
Please see this in a $3D$ online grapher tool or do rough $2D$ sketches to visualize better.

If you are going $dy$ first, you need to split the integral into two. For the order of integral $dy \ dz \ dx$, we make following observations.

At $y = 0, 0 \leq z \leq 1$. Also $y$ is not a function of $z$ for $0 \leq z \leq 1$.

a) For $0 \leq z \leq 1$, the lower limit of $y$ is $0$ and the upper limit of $y$ comes from the curve $x = e^y \implies y = \ln x$.

We already know that upper bound of $x$ is $e^2$. For lower bound, we note that at $y = 0, x = e^y = 1$.

b) For $z \geq 1$, the lower limit of $y$ comes from the curve $z = e^{y/2} \implies y = 2 \ln z$. Upper limit continues to be $y = \ln x$.

Also the upper bound of $z$ comes from intersection of curves $x = e^y, z = e^{y/2} \implies z = \sqrt x$.

So the integral becomes,

$\displaystyle \int_1^{e^2}\int_{0}^{1}\int_{0}^{\ln(x)}  f(x, y, z) \ dy  \ dz \ dx +
\int_1^{e^2}\int_{1}^{\sqrt x}\int_{2 \ln(z)}^{\ln(x)} f(x, y, z) \  dy \ dz \ dx$

Now can you try and do it for the order $dy \ dx \ dz$? Let me know if you get stuck.


答:
You could just as easily claim that, if a square of side length $L$ is positioned with its sides horizontal and vertical, infinitely many line segments parallel to the leftmost side can fit in the square, and therefore its area is infinite rather than $L^2$. The problem in both arguments is line segments have zero area, and infinitely many zeroes don't have an infinite sum. Having said that, we also, of course, can't say the area is $0$ either. So what's going on here?

The reason a line segment has zero area is because every positive number is an upper bound on its area, because it can be made part of an arbitrarily thin rectangle. If the square is divided into $n$ strips of width $L/n$, each has area $L^2/n$, and the total is $L^2$. Admittedly, "using a rectangle's area to compute a square's area" may sound circular; but the area axioms include a rectangle's area, as well as the "subset has at most a superset's area" principle I used to bound a line segment's area.

Now let's divide a circle into $n$ sectors with $n$ equally spaced radii, so each sector subtends $2\pi/n$ radians. Each sector's vertices are those of a triangle that's a subset of the sector, so the sector has at least the triangle's area, $\tfrac12 r^2\sin(2\pi/n)$. So for all positive integers $n$, the circle's area is at least $\tfrac12nr^2\sin(2\pi/n)$. But this doesn't grow in proportion to $n$, the way an "it contains infinitely many radii so has infinite area" intuition may expect. This approaches $\pi r^2$ as $n\to\infty$. Or you can use the OP's diagram's "rearrange them into an approximate rectangle argument, thereby avoiding the above trigonometric limit.

Title: Since we can fill a circle with infinitely-many copies of its radius, isn't $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$

Body:
In a circle, there are area and a radius

When we fill up an infinite radius in a circle it becomes its area.

In that case, doesn't it means that $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$


问:
Solve equation $2^{x} - 2^{-x} = 2*\cos(x/5)$
I'm trying to solve it by log two parts:

$$
\log_2(2^{x}/2^{-x}) = \log_2(2*\cos(x/5))
$$
Simplifying:
$$
2x - 1 = \log_2(\cos(x/5))
$$

And I'm stuck here, I think we could solve it by investigating functions(linear and logarithmic) and trying to find border value. Could you give the right direction?

答:
Looks like there's a typo in the textbook, the left part must be $2^x + 2^{-x}$. In this case, I found a solution:

Since $2^x + \frac{1}{2^{x}}\geq 2$ and $2\cos{\frac{x}{5}} \leq 2$,then

\begin{cases}
2^x + 2^{-x}=2 \\
2\cos{\frac{x}{5}}=2 \\
\end{cases}

After solving this simple system, the answer would be $ x = 0 $.


答:
Let' assume you want solution to $f:\mathbb{R}\mapsto\mathbb{R}$ that satisfies this equation $$\forall x,y\in\mathbb{R},f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)\tag{1}$$

Let's plug in $(x,y)=(0,0)$ in the equation $(1)$,
$$f(0)^2=f(0)\Rightarrow f(0)\in\{0,1\}$$

Case-1 : $f(0)=0$

For $x>0, f(x)=f\left(\sqrt{x^2+0^2}\right)=f(x)f(0)=0$.

For $x<0, f(x)^2=f(x)f(x)=f\left(\sqrt{x^2+x^2}\right)=f\left(\sqrt{2}|x|\right)=0\Rightarrow f(x)=0$.

So, if $f(0)=0$, then $\forall x\in\mathbb{R}, f(x)=0$

Case-2 : $f(0)=1$

In this case, we claim that this function must be even.

Proof :

$f(x)=f(x)f(0)=f\left(\sqrt{x^2+0^2}\right)=f\left(\sqrt{(-x)^2+0^2}\right)=f(-x)f(0)=f(-x)\tag{2}$

Now, we will consider two assumptions under this case.

Assumption-A :

$\exists a>0, f(a)=0$.

Then, $\forall x\geq a$, $$f(x)=f\left(\sqrt{a^2+\left(\sqrt{x^2-a^2}\right)^2}\right)=f(a)f\left(\sqrt{x^2-a^2}\right)=0$$

Assumption-B :

$\exists b>0, f(b)>0$

Then, $$f\left(\sqrt2b\right)=f\left(\sqrt{b^2+b^2}\right)=f(b)^2>0$$
Proceeding this way, we can find arbitrarily large real number $x$ for which $f(x)>0$. Which contradicts with Assumption-A that says after reaching a bound, we can't find and real number $x$ for which $f(x)>0$.

So, we conclude that $a$ and $b$ can't exist simultaneously. If we assume the existence of $a$, We get the solution : $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
We can check and indeed, this is a valid solution.

Now we assume the existence of $b$, that is, $\exists b>0, f(b)>0$. Since this assumption excludes the possibility of the existence of real number $a$ for which $f(a)=0$, we only have two options : either $f(x)>0$ or $f(x)<0$ for any real number $x$. We claim that $\forall x, f(x)>0$.

Proof :
$$f(x)=f\left(|x|\right)=f\left(\sqrt{\left(\frac{x}{\sqrt2}\right)^2+\left(\frac{x}{\sqrt2}\right)^2}\right)=\left(f\left(\frac{x}{\sqrt2}\right)\right)^2>0$$
So, $\forall x, f(x)>0$.

Now, we will plug in $(\sqrt{x},\sqrt{y})$ where $x,y\geq0$ in equation $(1)$,
$$f\left(\sqrt{x}\right)f\left(\sqrt{y}\right)=f\left(\sqrt{x+y}\right)\\\Leftrightarrow \ln\left(f\left(\sqrt{x}\right)\right)+\ln\left(f\left(\sqrt{y}\right)\right)=\ln\left(f\left(\sqrt{x+y}\right)\right)$$

Let's define $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ such that $\phi(x)=\ln\left(f\left(\sqrt{x}\right)\right)$. Note that $\phi(x)+\phi(y)=\phi(x+y)$ which satisfies the Cauchy's functional equation. So, $\phi$ is a solution to Cauchy's functional equation. The most trivial solution is $\forall x\geq0,\phi(x)=kx$ where $k\in\mathbb{R}$. There are non trivial solutions which are highly pathological functions.

So, analyzing all the cases we got this solutions :

$\forall x, f(x)=0$

$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$

$\forall x, f(x)=e^{\phi(x^2)}$ where $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ and $\phi$ satisfies Cauchy's functional equation.

Title: If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$

Body:
As we know, for the $$f(x)f(y)=f(x+y)$$  $f(x)=\mathrm e^{\alpha x}$ is a solution.

What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?

I tried to find $f(x)$.
See my attempts below to find $f(x)$.

$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$

$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$

$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$

$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$

$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$

if we use binom expansion for $(1+(x/y)^2)^{m}$

$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$

Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$

$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\  \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\  \\
&  + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\  \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$

If we equal for all $x^n$ terms in both sides

we can see $a_{2n-1}=0$, but to find  $a_{2n}$  seems hard for me.
Any idea to find $a_{2n}$

Thanks in advice.


问:
How to properly write down $\frac{\partial}{\partial xy } f(xy, x^2+3y-2)$
I am wanting to take the derivative with respect to $x$ of $f(xy, x^2+3y-2)$, so, I use the chain rule to find
$$
\frac{\partial}{\partial x} f(xy, x^2-2) = y \frac{\partial}{\partial xy}f(xy, x^2-2) + 2x\frac{\partial}{\partial (x^2-2)}f(xy, x^2-2)
$$
With which I would then continue
$$
\frac{\partial}{\partial xy} f(xy, x^2-2) = \frac{1}{x} \frac{\partial}{\partial y} f(xy, x^2-2)
$$
However, when differentiating w.r.t $xy$, I am not communicating to the reader that $x^2-2$ is expected to remain a constant, and it would seem to them that I should rewrite $x^2-2$ as $\left(\frac{xy}{y}\right)^2-2$.

How do I use mathematical notation to communicate to another mathematician that $x^2-2$ remains constant when differentiating with respect to $xy$?

答:
You start out wanting $\frac{\partial}{\partial x}f(xy,x^2-2)$.

There is this function $f$ of two variables that I might call $u$ and $v$: $f=f(u,v)$. Here, $u$ is composed with $xy$ and $v$ with $x^2-2$. The chain rule says:

$$\begin{align}
\frac{\partial}{\partial x}f(xy,x^2-2)
&=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\cdot\frac{\partial v}{\partial x}\\
&=\frac{\partial f}{\partial u}\cdot y+\frac{\partial f}{\partial v}\cdot2x\\
\end{align}$$

And that's where I'd leave it until you know more about the nature of $f(u,v)$. You have $u$ and $v$, independent variables. No one will think they have a relation. Later, if you have some formula for $f(u,v)$ and you can actually write a formula for $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$, you could substitute away the $u$s and $v$s with $xy$ and $x^2-2$. Or you could write it as

$$\left.\left(y\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v}\right)\right\rvert_{u=xy,v=x^2-2}$$


答:
All you have to do is to make the following simple modifications

Choosing exactly one letter:
$^{5}C_{1} \times ^{9}C_{4} \times 5! = $

Choosing exactly two letters:
$^{5}C_{2} \times ^{9}C_{3} \times 5! = $

Choosing exactly three letters:
$^{5}C_{3} \times ^{9}C_{2} \times 5! = $

Choosing exactly four letters:
$^{5}C_{4} \times ^{9}C_{1} \times 5! = $

Compute and add up.

Title: Number of codes with $5$ distinct characters using $5$ letters and $9$ numbers

Body:
The letters; $A, B, C, D, E$ and the number; $1, 2, 3, 4, 5, 6, 7, 8, 9$ are to be used to form codes with five distinct characters. Each code has at least one letter and has at least one number. How many codes can we form?

My attempt:

Choosing exactly one letter:
$^{5}P_{1} \times ^{9}P_{4}= 5 \times 3024 = 15120$ ways.

Choosing exactly two letters:
$^{5}P_{2} \times ^{9}P_{3}= 20 \times 504 = 10080$ ways.

Choosing exactly three letters:
$^{5}P_{3} \times ^{9}P_{2}= 60 \times 72 = 7320$ ways.

Choosing exactly four letters:
$^{5}P_{4} \times ^{9}P_{1}= 120 \times 9 = 1080$ ways.

So, we have a total of $15120+10080+7320+1080=33600$ different codes.

Please tell me if that is right or wrong. Thanks.


问:
How can we prove for a tychonoff space $X$ and a metric space $(Y, d)$, that a map $\phi : Y \rightarrow C(X, Y)$, is an embedding.
How can we prove for a tychonoff space $X$ and a metric space $(Y, d)$, that a map $\phi : Y \rightarrow C(X, Y)$, $\phi(y) = f$, $f$ is a constant map in $C(X, Y)$ is an embedding. Where $C(X, Y)$ is equipped with uniform topology.

答:
The uniform topology on $C(X,Y)$ has as a base all sets of the form $B(f,r)$ where $f \in C(X,Y)$ and $r>0$ and $$B(f,r) = \{g \in C(X,Y): \forall x \in X: d(f(x),g(x)) < r\}$$

Now $\phi$ which maps $y \in Y$ to $\phi(y) \in C(X,Y)$ defined by $\phi(y)(x)=y$ for all $x \in X$, is continuous:

Suppose $y_n \to y_0$ in $(Y,d)$ (it's enough to show sequential continuity, as $Y$ is a first countable space), then it's clear that $\phi(y_n) \to \phi(y_0)$: if $B(\phi(y_0), r)$ is a basic neighbourhood of $\phi(y_0)$, then there is some $N$ such that $n \ge N \to d(y_n, y_0) < r$ by convergence and then also $n \ge N \to \phi(y_n) \in B(\phi(y_0), r)$ and we're done.

$\phi$ is clearly 1-1 and $\phi$ is an embedding as for all $y \in Y$ and $r>0$ we have

$$\phi[ B_d(y,r)] = B(\phi(y), r) \cap \phi[Y]$$

and so $\phi$ maps open sets to relatively open subsets in $\phi[Y]$.


答:
You do not fill the circle with copies of its radii (which will only have linear length and not width and so will not have any area).  You fill the circle with tiny triangle slivers whose altitudes are nearly the same of the height and whose bases are really really small.

And you don't fill the circle with infinitely many.  You fill the triangle with a lot of them but always so finite number even though it could be very large. And you reason what the area becomes when you take more and more of them.

And we don't fill the circle with them we almost fill the circle with them and we estimate.

SO for example if we fill in with $n$, a huge number,  of triangle slivers, the height of on of this slivers will be nearly equal to the radius $r$ (just a teeny bit less) and base of these triangles, since there are $n$ of them and they are fitted along the circumference of the circle will have a $b$ very nearly equal to the the circumference divided by the number of triangles, that is $\frac Cn$.

So a triangle wedge has base = $b= \frac Cn$ and has a height $h = r$. So the area of the wedge is $\frac 12\cdot \frac Cn \cdot r$.  And there are $n$ of them so the total area is $\require{cancel} n\cdot \frac 12 \cdot \frac Cn \cdot r = \cancel n\cdot \frac 12 \cdot \frac C{\cancel n}\cdot r = \frac 12 C\cdot r$. (Approximately.  The larger $n$ is the closer this will be to the value and in a circle it will be reached.)

And $\frac {Area}{Radius} = \frac {\frac 12 Cr}{r} = \frac 12 C$.

Which considering $C = \pi \cdot diameter = 2\pi r$ is exactly what we would hope for.

Likewise the formula for the area is what we are used to if we consider $Area = \frac 12 C \cdot r = \frac 12 (\pi \cdot diameter)\cdot r= \frac 12(\pi \cdot 2r)\cdot r = \pi r^2$


问:
Normal distribution of a random variable proof
Suppose that $X$ is a random variable that follows a chi-square distribution with df $= 1$. Now suppose $W$ is another random variable such that its value is dependent on $X$. In particular a value of $X$ is sampled and a fair coin is flipped. If the coin lands on "tails", then $W = -\sqrt{X}$. Otherwise if "heads", then $W = \sqrt{X}$. Prove that $W$ is a standard normal variable.

Hint: Consider two cases for $w$.

My attempt:

When $w \geq 0$:

$F_W(w) = P(W \leq w) = P(\sqrt{X} \leq w) = P(X \leq w^2) = 2P(X \leq w^2) - 1$

We have that $F_W(w) = 2\int_{0}^{w^2} \frac{1}{\sqrt{2\pi xe^x}} \,dx - 1$

So, $f_W(w) = \frac{d}{dw}(2\int_{0}^{w^2} \frac{1}{\sqrt{2\pi xe^x}} \,dx - 1) = 4(\frac{e^{\frac{-w^2}{2}}}{\sqrt{2\pi}})$

When $w < 0$:

$F_W(w) = P(W \leq w) = P(-\sqrt{X} \leq w) = P(\sqrt{X} \geq w) = 1 - P(\sqrt{X} \leq w) = -2P(X \leq w^2)$

We have that $F_W(w) = -2\int_{0}^{w^2} \frac{1}{\sqrt{2\pi xe^x}} \,dx$

So, $f_W(w) = \frac{d}{dw}(-2\int_{0}^{w^2} \frac{1}{\sqrt{2\pi xe^x}} \,dx) = -4(\frac{e^{\frac{-w^2}{2}}}{\sqrt{2\pi}})$

So $W$ follows a standard normal distribution.

QED.

Is this proof correct? Did I make a mistake (especially in case 2? Is $P(-\sqrt{X} \leq w) = P(\sqrt{X} \leq w)$?). Any assistance is much appreciated.

答:
$w > 0$:
$$ P(W \le w) = P(W = - \sqrt{X}, W \le w) + P(W = \sqrt{X}, W \le w) = P(W = -\sqrt{X}) + \frac{1}{2}P(\sqrt{x} \le w) = \frac12 + \frac{1}{2}F_X(w^2), $$
Differentiating gives you the right solution in this case.
Then for $w \le 0$,
$$ P(W \le w) = P(W = -\sqrt{X}, - \sqrt{X} \le w) = \frac{1}{2} P(X > w^2), $$
and once again, differentiating gives you the answer.


问:
Units in Tensor Algebra
Let $M$ be an $R$-module, and let $T(M) = \bigoplus_{k \geq 0 } T^k(M)$ denote the tensor algebra of $M$.

By definition $T^0(M)=R$ . I am looking for a simple explanation to this problem.

I am looking for units in $T(M)$. I can figure out that the units in $R$ are also units in $T(M)$ but are there any other units in $T(M)$. Can anyone give few examples where units in $T(M)$ which are not contained in $T^0(M)$?

I am studying about tensor algebra and I am wondering about the existence of units in $T(M)$. Any such example would help me in finding how elements of $T(M)$ behave. I am in a misconception that elements of $T(M)$ of degree $m$ and $n$ will multiply and give me an element whose component is in $T^{m+n}(M)$ and hence it cannot be an unit.

答:
It's definitely true that if $a\in T^n(M)$ and $b\in T^m(M)$, then $ab\in T^{n+m}(M)$. But that doesn't preclude additional units other than those in $T^0$, it just shows that no element in $T^m$ has an inverse in $T^n$ for $m>0$.

Here's one way to produce a unit of $T(M)$: let $R=\Bbb Z$ and $M=\Bbb Q/\Bbb Z$. Then $M\otimes_R M=0$, so $T(M)=\Bbb Z\oplus \Bbb Q/\Bbb Z$ with the multiplication defined by $(a,b)(c,d)\mapsto (ac,ad+bc)$, so the units are precisely those elements of the form $(\pm 1, x)$ with inverse $(\pm 1, -x)$ for $x\in\Bbb Q/\Bbb Z$.

Title: Units in Tensor Algebra


问:
How do I prove $[0,3)$ is an open set in $([0,\infty),d)$ standard metric?
Question : How do I prove $[0,3)$ is an open set in $([0,\infty),d)$ standard metric?

Background and Approach : This is my first analysis class and I don't know if I'm on the right path. I'm thinking about proving if every point in the interval has a neighborhood contained in it, but the $[0,\infty)$ confuses me. I know that a set $X$ is open if given any $x \in X\; \exists \;r\in\mathbb{R^+}$ such that $B(x,r) \in X$.

答:
As mentioned in the last comment, you need to find an $r$ such that $B(x,r)\subset [0,3)$, whenever $x\in [0,3).$

Hint: Instead of $x$, you can write $0\le x=3-\epsilon<3,$ where $0<\epsilon\le 3.$ Namely, you can express every element in $[0,3)$ as $3-\epsilon$, for some $\epsilon\in(0,3].$

 Set $$r=\frac{\epsilon}{2}=\frac{3-x}{2},$$ for any $x\in[0,3)$.

(Anything less than $\epsilon$ should work.)


问:
Given a plane, is there a vector that points towards the highest dz when moving by dx and dy? What is that vector called?
The question came to my mind when trying to explain snowboarding up a ramp. When you don't use an edge and the snowboard is flat, it is only stable if your momentum is going straight up the ramp. You are likely to fall if your velocity is at an angle. So I was trying to explain which vector of the ramp you need to align to, but I struggled to define it.

I know that a plane can be defined with a vector that's perpendicular to it, but there are many vectors that are parallel to it. If the Z axis is gravity, I think "going straight up" means maximizing dz when stepping dx dy. I think there is only one such vector that is parallel to the plane. Is this a well defined vector? Does it have a name? I also suspect that given such a vector, you could derive its origin intersecting plane.

答:
Suppose $Z$ is a unit vector in the vertical direction, and $N$ is a unit vector that’s normal to the plane of the hill. Then the cross product $H = N \times Z$ is the horizontal direction on the slope. In other words, if you go in the direction of $H$, you will be neither gaining nor losing altitude. The vector $N \times H$ gives you the steepest downhill direction. I think that’s what you’re asking about.

This direction is often called the “direction of steepest descent”. This terminology comes from optimization theory, where sometimes the best way to get to a minimum is to go downhill as fast as you can. The term “gradient descent” is also used.


问:
Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$
(P1) Let $a, b, c$ be real numbers. Prove that $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge 0.$

This inequality is verified by Mathematica. I am particularly interested in (simple) SOS solutions (also for (P2) and (P3) below),
however, any comments and solutions are welcome. I also gave a SOS solution for (P2) below, but it is complicated.

Some relevant questions:

(P2) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a+b+c)^6}{27}.$

Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$

(P3) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a^2+b^2+c^2+ab+bc+ca)^3}{8}.$

Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$

Body:
My P2 SOS is ugly, the coefficients is too large, I will try to improve my tool:

We have:

$$27\left[4\left(a^6+b^6+c^6\right)+5\left(a^5b+b^5c+c^5a\right)\right]-(a+b+c)^6$$

$$={\frac {1}{418495910519665315875436}}{c}^{2} \left( 1877899493539b +2231128254114c \right) ^{2} \left( b-c \right) ^{2}$$

$$+{\frac {1}{1227564566709386794764}}\left( b-c \right) ^{2} \left( 111426599762{b}^{2}+57572325931bc-66554368268{c}^{2} \right) ^{2 }$$

$$+{\frac {1}{4982400602455205424}}\left( 11016800022a{c}^{2}- 8322034448{b}^{3}-16516031087{b}^{2}c+953764263b{c}^{2}+ 12867501250{c}^{3} \right) ^{2}$$

$$+{\frac {1}{110170254068928}}\left( 41114072abc-13516246\,a{c}^{2} -63487932{b}^{3}-24871895{b}^{2}c+68571471b{c}^{2}-7809470{c}^ {3} \right) ^{2}$$

$$+{\frac {1}{92668107698496}} \left( 58951728a{b}^{2}+20957464abc- 61903346a{c}^{2}+23285292{b}^{3}+20567219{b}^{2}c-22826379b{c} ^{2}-39031978{c}^{3} \right) ^{2}$$

$$+{\frac {1}{145642643664}}\left( 3143864{a}^{2}c-684660a{b}^{2}+ 1346196abc-252458\,a{c}^{2}-539136{b}^{3}-876131{b}^{2}c+54651b{c}^{2}-2192326{c}^{3} \right) ^{2}$$

$$+{\frac {1}{9913764}}\left( 23163{a}^{2}b+6338{a}^{2}c+11364a{ b}^{2}+3372abc-5897a{c}^{2}-19512{b}^{3}-9398{b}^{2}c-2028b{ c}^{2}-7402{c}^{3} \right) ^{2}$$

$$+{\frac {1}{428}} \left( 214{a}^{3}+129{a}^{2}b-6{a}^{2}c-108a{b}^{2}-56abc-87a{c}^{2}-8{b}^{3}-30{b}^{2}c-44b{c}^{2}-4 {c}^{3} \right) ^{2}+{\frac {5083151117301}{1877899493539}}{c}^{4} \left( b-c \right) ^{2}\ge 0,$$

which give us qed./

See also here.


问:
Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>

答:
The different is that in $x^n$ (or more generally, $x^\alpha$) it's the base which varies and the exponent stays fixed, while in $e^x$ (or more generally $\alpha^x$) it's the exponent which varies while the base stays fixed.

Let's look at a generalization of this. Assume that both can vary, i.e. we look at $$
  f(x) = b(x)^{p(x)} \text{.}
$$
for some functions $b$ (for base) and $p$ (for power, since e for exponent is taken). Let's also assume that $b(x) > 0$, to avoid having to deal with negative basis. By writing this as $$
  f(x) = h(b(x),p(x)) \quad\text{where}\quad h(u,v) = u^v = e^{v\log u}
$$
the structure of this becomes more explicit. Now, if we do know that $\frac{de^x}{dx} = e^x$, by extensive use of the chain rule we can find $$\begin{aligned}
  h_u(u,v) &= \frac{\partial h}{\partial u} = \frac{v}{u}e^{v\log u}\\
  h_v(u,v) &= \frac{\partial h}{\partial v} = e^{v\log u}\log u\\
  f'(x) &= h_u(b(x),p(x))b'(x) + h_v(b(x),p(x))p'(x) \\
     &= b(x)^{p(x)} \left(b'(x)\frac{p(x)}{b(x)} + p'(x)\log b(x) \right) \\
     &= b(x)^{p(x)-1} \left(b'(x)p(x) + p'(x)b(x)\log b(x) \right)
\end{aligned}$$

Now let's apply this to $x^n$, i.e. we set $b(x)=x$ and $p(x)=n$, and get $$
  f'(x) = x^{n-1}\left(n + 0\cdot x\log x\right) = nx^{n-1} \text{.}
$$
So we got the rule for powers of $x$ from the rule that $\left(e^x\right)' = e^x$, plus the chain rule. We can also set $b(x)=e$ and $p(x)=x$ and get back that $$
  f'(x) = e^{x-1}\left(0\cdot x + 1\cdot e\log e\right) = e^{x-1}\left(e\right) = e^x \text{,}
$$
but of course we expected that since we used that rule to derive our formula.


问:
Projection on modified L2-Ball
Suppose we are in $\mathbb{R}^n$ and we are given some integer $k$ such that $1 \leq k \leq n$. Let $\Vert x \Vert_0 = \#\{i\ |x_i \neq 0\}$ be the $L_0$-norm counting the number of non-zero entries of a vector (actually not really a norm).

Let $P = conv\{x\ |\ \Vert x \Vert_2 = 1, \Vert x \Vert_0 \leq k \}$ be the convex hull of all points on the unit ball that have at most $k$ non-zero entries. This can be seen as as the convex hull of all unit spheres/disks of dimension at most $k$ in $\mathbb{R}^n$.

Suppose now I have some arbitrary $x \in \mathbb{R}^n$ given. How could I determine whether $x$ lies in $P$ (apart from the obvious case when $\Vert x \Vert_0 \leq k$)? If $x$ doesn't lie in $P$, how can I find its projection onto $P$, i.e. the closest point w.r.t. the euclidean norm?

Any help is greatly appreciated!
Thanks

答: Found an answer to this myself: Apparently $P$ defines a ball w.r.t. the so-called $k$-Support-Norm, on which there's quite a bit of literature, see e.g. the original paper.

Title: Projection on modified L2-Ball


问:
Find a retraction $r:\mathbb{R} \to [0,1].$

Find a retraction $r:\mathbb{R} \to [0,1].$

I was trying to find examples on retractions, but couldn't find many online. How are these retractions constructed? It seems to be related to the identity map, but I'm not entirely sure what's the connection?

答:
$$r(t) = \begin{cases} 0 & t \le 0\\
                      t & t \in [0,1]\\
                      1 & t \ge 1\\
        \end{cases}$$

is a perfectly fine retraction....Or if you prefer formulae: $r(t)=\max(\min(t,1),0)$.


问:
This is sufficient to prove by induction that 2n < n! for n >= 4?
I start with base saying that this is valid for one by simple algebraic calculus:

This is my base for $ n = 4$
$$ 2 \cdot 4  \leq 4! \Leftrightarrow 8 \leq 24  $$

In my hypotheses just say that exists a number $k$ that satisfies the expression for some $k\geq 4$:
$$ 2k < k! $$

And my inductive step I multiply both sides from hypotheses by $(k+1)$ as follows:
$$ (k+1) \cdot 2k < k! \cdot (k+1) $$
$$ \Leftrightarrow (k+1) \cdot 2k < (k+1)! $$

So by hypotheses we know that $k\geq 4$ soon can afirms that $2k(k+1) > 2k$. Given that $(k+1)! > 2k(k+1)$ and $2k(k+1) > 2k$ is proved by induction that $2n < n!$

I wrong anything or is insufficient to prove the enunciated?

答:
It's often better to prove it in one line:

$2(k+1) = 2k + 2 < k!+2 < k!\cdot (k+1) = (k+1)!$,

where the first inequality is by induction hypothesis and the second inequality holds for $k\geq 4$ anyway (for completeness, it could also be proved by induction).


Title: Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$

Body:
(P1) Let $a, b, c$ be real numbers. Prove that $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge 0.$

This inequality is verified by Mathematica. I am particularly interested in (simple) SOS solutions (also for (P2) and (P3) below),
however, any comments and solutions are welcome. I also gave a SOS solution for (P2) below, but it is complicated.

Some relevant questions:

(P2) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a+b+c)^6}{27}.$

Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$

(P3) Let $a, b, c$ be real numbers. Prove that $4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) \ge \frac{(a^2+b^2+c^2+ab+bc+ca)^3}{8}.$

Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$

Body:
Since the exact SOS for the stronger problem of P1 is very very big, can not express in $\LaTeX;$

So here I only give the approximate SOS for
$$40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge \dfrac{31}{81}(a+b+c)^6$$

Have

$$40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) - \dfrac{31}{81}(a+b+c)^6$$

$$\approx 39.62\, \left( {a}^{3}+ 0.6399\,{a}^{2}b- 0.02898\,{a}^{2}c- 0.5041\,
a{b}^{2}- 0.3277\,abc- 0.4398\,a{c}^{2}+ 0.04222\,{b}^{3}- 0.08519\,{b
}^{2}c- 0.2490\,b{c}^{2}- 0.04745\,{c}^{3} \right) ^{2}$$
$$+ 25.53\,
 \left( {a}^{2}c- 0.2476a{b}^{2}+ 0.4192abc- 0.04827a{c}^{2}-
 0.1957\,{b}^{3}- 0.3349{b}^{2}c+ 0.03937b{c}^{2}- 0.6321{c}^{3}
 \right) ^{2}+ 17.98\left( {a}^{2}b+ 0.4437{a}^{2}c+ 0.4049a{b}
^{2}+ 0.1510\,abc- 0.1529\,a{c}^{2}- 0.8971\,{b}^{3}- 0.3941\,{b}^{2}c
- 0.08513\,b{c}^{2}- 0.4704\,{c}^{3} \right) ^{2}+ 9.789\, \left( a{b}
^{2}+ 0.5075\,abc- 1.492\,a{c}^{2}+ 0.5095\,{b}^{3}+ 0.7374\,{b}^{2}c-
 0.2611\,b{c}^{2}- 1.001\,{c}^{3} \right) ^{2}+ 6.027\, \left( abc+
 0.03308\,a{c}^{2}- 1.885\,{b}^{3}- 1.121\,{b}^{2}c+ 1.783\,b{c}^{2}+
 0.1908\,{c}^{3} \right) ^{2}+ 3.993\, \left( a{c}^{2}- 0.1356\,{b}^{3
}- 2.018\,{b}^{2}c- 0.7975\,b{c}^{2}+ 1.951\,{c}^{3} \right) ^{2}+
 0.06508\, \left( {b}^{3}- 0.6985\,{b}^{2}c- 1.349\,b{c}^{2}+ 1.047\,{
c}^{3} \right) ^{2}+ 0.03708\, \left( {b}^{2}c- 0.03128\,b{c}^{2}-
 0.9687\,{c}^{3} \right) ^{2}+ 0.005711\, \left( b{c}^{2}- 1.0\,{c}^{3
} \right) ^{2}\ge 0,$$

which is obvious.

If you want to see the exact SOS, please visit Github.


问:
Uniquely $3$-edge-colourable $3$-regular graph with $\chi'(G) = 3$ has exactly $3$ Hamiltonian cycles?
As the title says, I am trying to show that a uniquely $3$-edge-colourable $3$-regular graph $G$ with edge-chromatic number $3$ has exactly $3$ Hamiltonian cycles.

I've managed to show the existence of three Hamiltonian cycles as follows: pick any $3$-colouring of the edges of $G$, then deleting the edges of a single colour, the resulting graph $G_1$ is $2$-regular, and hence its connected components are cycles. If there is more than one component, then swapping the colours in a single component of $G_1$ induces a different $3$-colouring of $G$, which is a contradiction; hence $G_1$ is a single cycle, and it is a Hamiltonian cycle on $G$.

I'm having trouble proving that the three Hamiltonian cycles I've found are unique, though. Let $C$ be a Hamiltonian cycle on $G$; I was thinking of trying to $2$-colour $C$ and then extend that to a $3$-colouring of $G$. If that can be done, then I think that by uniqueness of $3$-colourings, it follows that $C$ is one of the Hamiltonian cycles I've found earlier, but I'm having trouble constructing the $2$-colouring of $C$.

Any help is appreciated!

答: Removing a Hamiltonian cycle leaves a $1$-regular graph, which can be coloured with one colour. And the cycle itself can be coloured by $2$ colours, assigning the colours to alternate edges in the cycle. By uniqueness of the colouring, this cycle must have been obtained through the process you have described earlier.</p>


问:
Finding $\tan(x+90^\circ)$ when $\sin x = k$ and $x$ is obtuse. I get the reciprocal of the correct answer.

If $\sin x = k$ and $x$ is obtuse, find an expression for $\tan(x+90^circ)$.

So $$\tan(x+90^\circ)=\frac{\sin(x+90^\circ)}{\cos(x+90^\circ)}$$
and after drawing a triangle, I obtain
$${\sin(x+90^\circ)} = -k \qquad\text{and}\qquad\cos(x+90^\circ)= -\sqrt{1-k^2}$$ because $x+90^\circ$ is in Quadrant 3 where both $\sin$ and $\cos$ are negative.

So $$\tan(x+90) = \frac{k}{\sqrt{1-k^2}}$$

However, the answer is the reciprocal of this. Why is that so?

答:
You should have identites that $\sin(x + 90) = \cos x$ and not $\sin(x+90) \ne -\sin x$. And that $\cos(x+ 90) =-\sin x$ and not $\cos (x+90) \ne \cos x$.

You can prove it by the $\sin(x+90) = \sin x\cos 90 + \cos x\sin 90$ and $\cos(x+90) = \sin x\cos 90 - \cos x \sin 90$.

But you should draw circles on graph paper and rotate it a quarter turn to convince yourself.


答:
I think the proof of the first part by contradiction is easier than anything yet proposed.

Since the size of each conjugacy class is a divisor of the order of the group, if the group has odd order then each class must have odd size (the group order has no even divisors). Since an even number of sets of odd size would have an even number of elements, and the group has odd order, there cannot be an even number of classes.

Therefore, if the number of classes is even, the group cannot have odd order, so in this case the order must be even.

Title: Prove that if $G$ is a finite group with an even class number, then $G$ is of even order. Give an example that the converse fails.

Body:
Prove that if $G$ is a finite group with an even class number (the number of conjugacy classes of a group $G$ is called its class number) then $G$ is of even order. Give an example that the converse fails.

My attempt: If $n$ is even $$|G|=\sum_{i=1}^{n} |{\rm orb}(g_i)|$$
and for all $i$, $|{\rm orb}(g_i)|$ divides $|G|$. Let $a,b_i$ be integers. Does

$$a=\sum_{i=1}^{n \text{ even } }b_i$$ with $b_i\mid a$ $\forall i$ imply that $a$ is even?

Actualization: By contradiction, if $a$ is odd, $b_i\mid a$ implies $b_i$ is odd. Because odd+odd=even then $a$ is even a contradiction.

What counterexample can be work?


答:
A simpler proof:

For all naturals, $$2<(n-1)!\implies 2n< n!$$
and it suffices to prove

$$n\ge3\implies n!>2.$$

Very easily,

$$3!>2$$ and

$$n!>2\implies (n+1)n!>2.$$

Title: This is sufficient to prove by induction that 2n < n! for n >= 4?

Body:
I start with base saying that this is valid for one by simple algebraic calculus:

This is my base for $ n = 4$
$$ 2 \cdot 4  \leq 4! \Leftrightarrow 8 \leq 24  $$

In my hypotheses just say that exists a number $k$ that satisfies the expression for some $k\geq 4$:
$$ 2k < k! $$

And my inductive step I multiply both sides from hypotheses by $(k+1)$ as follows:
$$ (k+1) \cdot 2k < k! \cdot (k+1) $$
$$ \Leftrightarrow (k+1) \cdot 2k < (k+1)! $$

So by hypotheses we know that $k\geq 4$ soon can afirms that $2k(k+1) > 2k$. Given that $(k+1)! > 2k(k+1)$ and $2k(k+1) > 2k$ is proved by induction that $2n < n!$

I wrong anything or is insufficient to prove the enunciated?


问:
Find all the possible numbers whose sum is x but ....
I want to find all the number whose sum is x but

but series of number start with a specific number and next number is always smaller then the number before it

For example I want 41 but first number in the series is
11
I want all the combination of number which start from 11 and succeeding number are smaller than previous number like
11 + 9 + 6 + 5 + 4 + 3 + 2 + 1 = 41

Numbers are 11 and less than 11 and each number occurs only one time, and sum of them is 41.
11+10+9+8+3 = 41

I want a method so I can effectively generate all such combinations.

答:
Hint: if we remove the condition that the numbers are in a decreasing order, than we left with searching for all the partitions of an integer - we can sort the values at the end.

The referenced Wikipedia article discusses variations of this problem, like if you want strictly smaller values than you want distinct parts, and if you have an initial value, than you want restricted part size. At the bottom there is a link that says how to generate these values, I did not check it out and I also never actually generated such values, but it should serve as a starting point.


问:
A Class of Diophantine Equations in Three Variables
This question is inspired by the problem https://projecteuler.net/problem=748

Consider the Diophantine equation
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{k}{z^2}$$
$k$ is a squarefree number. $A_k(n)$ is the number of solutions of the equation such that $1 \leq x+y+z \leq n$, $x \leq y$ and $\gcd(x,y,z)=1$. This equation has infinite solutions for $k=1$ and $k>1$ that can be expressed as sum of two perfect squares.

Let $$A_k=\lim_{n \to \infty}\frac{A_k(n)}{\sqrt{n}}$$

\begin{array}{|c|c|c|c|c|}
\hline
k& A_k\left(10^{12}\right)& A_k\left(10^{14}\right)& A_k\left(10^{16}\right)& A_k\left(10^{18}\right)& A_k \\ \hline
1& 127803& 1277995& 12779996& 127799963& 0.12779996...\\ \hline
2& 103698& 1037011& 10369954& 103699534& 0.1036995...\\ \hline
5& 129104& 1291096& 12911049& 129110713& 0.129110...\\ \hline
10& 90010& 900113& 9000661& 90006202& 0.0900062...\\ \hline
13& 103886& 1038829& 10388560& 103885465& 0.103885...\\ \hline
17& 86751& 867550& 8675250& 86752373& 0.086752...\\ \hline
\end{array}

From these data, it seems that these limits converge. I wonder if it is possible to write them in terms of known constants or some sort of infinite series. For Pythagorean triples, there are about $\frac{n}{2\pi}$ triples with hypotenuse $\leq n$.

The Python code to calculate $A_1(n)$:

```<code>def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 0
for a in range(1, 22000):
   a2 = a * a
   for b in range(1, a):
       if (a + b) % 2 == 1 and gcd(a, b) == 1:
           b2 = b * b
           x = 2 * a * b * (a2 + b2)
           y = a2 * a2 - b2 * b2
           z = 2 * a * b * (a2 - b2)
           if x + y + z > N:
               continue
           cnt += 1
print(cnt)
</code>```

The Python code to calculate $A_2(n)$:

```<code>def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 1     # (1, 1, 1)
for a in range(1, 22000):
   a2 = a * a
   a4 = a2 * a2
   for b in range(1, a):
       if gcd(a, b) == 1:
           b2 = b * b
           b4 = b2 * b2
           x = 2 * a * b * (a2 + b2) - (a4 - b4)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = 6 * a2 * b2 - (a4 + b4)
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
           x = (a4 - b4) - 2 * a * b * (a2 + b2)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = (a4 + b4) - 6 * a2 * b2
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
print(cnt)
</code>```

答:
To solve the Diophantine equation.

$$ \frac{ 1 }{ x^2 } +\frac{ 1 }{ y^2 } = \frac{ q }{ z^2 } $$

It is necessary to use the solution of the following equation.

$$a^2+b^2=qc^2$$

There are solutions when the coefficient can be represented as the sum of squares. $q=t^2+k^2$

$$a=-tp^2+2kps+ts^2$$

$$b=kp^2+2tps-ks^2$$

$$c=p^2+s^2$$

And the solution itself can be presented in this form.

$x=ac$

$y=bc$

$z=ab$


问:
Null and alternative hypotheses : Car model fuel usage
Some people assume that a specific car model does at least $\mu_0=120$ km with $1$ Lt of gasoline (petrol).

$10$ independent tests give the following results: $$104, \ 96, \ 80, \ 100, \ 108, \ 100, \ 112, \ 120, \ 130, \ 132$$

(a) Give the Null Hypothesis $H_0$ and the alternative Hypothesis $H_1$, for the test of that assumption.

(b) Give the statistic function of that test.

(c) late the p-value of the test.

(d) In what confidence level can the assumption be rejected?

$$$$

For (a) is the null hypothesis $H_0: \ m_0=120$ and the alternative $H_1: \ \mu_1\neq 120$ ?

答:
You have to assume normality

The Hypothesis you stated is correct

the test statistics is the following

$$t=\frac{\overline{X}_{10}-120}{S}\sqrt{10}\sim \mathcal{T}_{9}$$

say $t$ follows a Student T distribution with 9 d.o.f.

$S^2$ is the unbiased sample variance (to be calculated with the given data)

I think you can conclude by yourself

To calulate exactly the pvalue you need a calculator. It results to me $p_{\text{value}}\approx 4.41\%$

You can reject $H_0$ for any significance level less than $p$


问:
higher weak derivatives of the hat function
The hat function is defined as:
$$
\phi(x) = (1-|x|)_{+}
$$
where $(y)_{+} = \max\{y, 0\}$. I know that the hat function is a member of the Sobolev space $W^{1,p}(-1, 1)$ for $p \in [1, \infty]$. The question is: $\phi(x) \in W^{k, p}(-1, 1)$? for $k > 1$ and $ p \in [1, \infty]$.

答:
The key word is Sobolev Embedding. In fact in dimension $ 1 $, it implies that if
$$ k-\frac{1}{p} > 1 $$
then $ W^{k,p}(-1,1) \subset C^1 $.

The answer to your question is than $ \phi $ is in $ W^{k,p} $ if

$$ k-\frac{1}{p} < 1. $$
If $ k \in \mathbb{N} $, we proved for $ k \in \mathbb{R} $ you need an extra argument.


答:
Fixe that $2k(k+1) > 2(k+1)$ and not $2k(k+1) > 2k$ and your proof is fine.

Alternatively it might be simpler to go left to right rather than right to left

$2k < k!$

$2k + 2 < k! + 2 < k! + 2\cdot 3 < k! + 2\cdot 3\cdot 4 < ...... < k! + k! =$

$2k! < (k+1)k! = (k+1)!$.

....

You know.... as far as inequalities go... this aint even close.


问:
Are alternating and anticommutative forms equivalent?
A form is defined to be alternating iff having two equal arguments means it is equal to 0. It is defined to be anticommutative iff permuting its arguments means multiplying by the sign of the permutation.

Unless my definitions are wrong, I'm pretty sure the two are equivalent, right? Also, if it is the case, why do we use two different names for the same concept?

答:
Expanding upon the comment, alternating always implies anticommutative, and the converse is true provided that $2$ is invertible in your ground ring (e.g. field of characteristic $0$ or $p>2$).

Alternating $\implies$ Anticommutative

Suppose that $\def\o{\omega}\def\e{\eta}\def\w{\wedge}\o \w \o = 0$ for all forms $\o$. Then for any forms $\o, \e$,
$$
\o \w \e + \e \w \o = (\o + \e) \w (\o + \e) - \o \w \o - \e \w \e = 0.
$$

Anticommutative $\implies$ Alternating provided that $2^{-1}$ exists

Suppose that $\o \w \e + \e \w \o = 0$ for all forms $\o, \e$. Then for any form $\o$,
$$
2(\o \w \o) = \o \w \o + \o \w \o = 0.
$$
If $2$ is invertible in the ground ring, then by multiplying by $2^{-1}$, we have
$$
\o \w \o = 0.
$$