问:
How to solve such integration?
i am trying to solve the following integration but not reaching the final step as the integration results in infinity. $\int_0^\infty y^{2m+1}e^{-C_{2}(x+\rho)y}\text{d}y$. Where $m,C_2,x$ and $\rho$ are constants. Any help in this regard is highly appreciated.

答:
To flesh out Kavi's hint, assuming that $C_2(x + \rho)$ is positive (if it's negative, the integral doesn't converge) we can make the substitution $t = C_2(x + \rho)y$ gives $dy = \frac{1}{C_2(x + \rho)} dt$ and the integral becomes
$$\int_0^\infty \left( \frac{t}{C_2(x + \rho)}\right)^{2m + 1} e^{-t} \left(\frac{1}{C_2(x + \rho)}dt\right) = \left(\frac{1}{C_2(x + \rho)}\right)^{2 m + 2} \Gamma(2m + 2)$$
(In your final answer, you might want to remember that $\Gamma(2m + 2) = (2m + 1)!$)

Title: How to solve such integration?


问:
Advocating base 12 number system
I had a calculus professor who suggested we should be using base 12 number system. What are the advantages of using such a system?   </p>

答: As I see it, there are two advantages.  First, it's not too different from base 10, so it comes fairly naturally.  Second, 12 has many divisors, so $1/2, 1/3, 1/4, 1/6, 1/8, 1/9, 1/12,\ldots$ would all be terminating decimals.</p>

ContentLicense: CC BY-SA 3.0

Title: Advocating base 12 number system


问:
Find Maclaurin series and radius of convergence. Compare with geometric series?
What is the MacLaurin series and radius of convergence of: $$\frac{1}{(2+z^4)}$$

What can I compare this to? Can I compare it to a geometric series? What is the radius of convergence?

Since $$\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z  + z^2 + ...$$

Does this help here? Can I transform the original function to:

$$\frac{1}{1+\frac{z^4}{4}} =\frac{1}{1-\frac{-z^4}{4}} = \sum_{n=0}^{\infty} (\frac{-z^4}{4})^n = 1 - \frac{z^4}{4}  + \frac{z^8}{4} + ... $$

Is this valid? What's the radius of convergence of a geometric series? It's when |z| < 1 right? Does that apply here too?

答:
What you have done is correct except for the last equality. The terms of the series are alternately positive and negative,.

The series converges if $|\frac {-z^{4}} 4| <1$ or $|z|^{2} <2$ of $|z| <\sqrt2 $ and diverges if $|z| >\sqrt 2$ so the radius of convergence is $\sqrt 2$.

ContentLicense: CC BY-SA 4.0


问:
is there any difference between these three logarithms?
is there any difference between $\log(n)^{\log(n)}$ vs $(\log n)^{\log n}$ vs $\log n^{\log n} $ from asymptotic growth rate? maybe all the same. I doubt about notation.

maybe this is very basic or crazy question, but I need to know.

答: One must be careful, there is a crucial difference between $$(\log n)^{\log n}$$ and $$\log n^{\log n}$$ The exponent rule for the logarithm can be applied to this last one to obtain: $$\log n^{\log n}= \log n \log n=(\log n)^2 \neq (\log n)^{\log n}$$


问:
The Product of Consecutive Integers is Never a Power: Lemma 2 (Research Paper Study)
P. Erdos and J. L. Selfridge proved in the  paper THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER (click here), that the equation $(n + 1) \cdots(n + k)=x^l \cdots (1)$ has no solution in integers with $k > 2, l > 2, n > 0$. There is a lemma $2$ on page 294, 295 -

LEMMA 2. By deleting a suitably chosen subset of $\pi(k-1)$ of the numbers
$a_i (1 \leq i \leq k)$, we have
$$a_{i_l}...a_{i_k'},|(k -1)!\cdots  (9)$$
where $k’=k -\pi(k-1)$.

For each prime $p < k-1 $ we omit an $a_m$ for which $ n + m$ is divisible by $p$
to the highest power. If $1 \leq i \leq k$ and $i \neq m$, the power of $p$ dividing $n +i$ is the same as the power of $p$ dividing $i-m$. Thus $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)!$.

I understand  $p | i-m$ but how does  $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)! \:\:$ ?

Edit:

Here, $n + i= a_ix^l$, where $a_i$ is $l^{th}$-power free and all its prime factors are less than $k$ (supposing Theorem 2 is false, see page 293 of the  paper (click here) for the definition).

答:
$i-m$ is divisible by $p$,  just like $(i-m)$,  $p$ divides $(m-i)$, from
$(n+m)-(n+i) \equiv 0 \pmod p \implies m-i \equiv 0 \pmod p$, look at the definition in pdf file.

See that $(i-m)$ ,   $(m-i)$ act as a  a factor of $(k-m)!(m-1)!$.

it is easy to find that $(k-m)!(m-1)!| (k-1)!$.

That is why  $p^{\alpha}$ divides $a_{i_{1}} \cdots a_{i_{k'}}$, gives $p^{\alpha}$ divides $(k-m)!(m-1)!$and , $p^{\alpha}$ divides $(k-1)!$.


问:
Find MacLaurin series and radius of $cos^2 \frac{1}{2}z$
$$cos^2 \frac{1}{2}z$$

So I know that $$\cos{z} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} -+ ...$$ and I have also seen that using taylor series formula that:

$$cos^2{z} = 1 - z^2 + \frac{z^4}{3} - \frac{2}{45}z^6$$

So can I just replace all z's with $\frac{1}{2}z$ and be done? Is that the maclaurin series here? What is the radius of convergence and why? I'm not sure how to calculate that...

Body: $$\cos^2(z/2)=\frac{1}{2}(1+\cos z)=\frac{1}{2}(1+1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+....)$$


答:
Yes there is. Simple thumb rule: Power on the operand comes to side.

Suppose there is a logarithm exprerssion $\log_a{x}, a\neq1, x\gt0$. I am going to refer $x$ as the operand.Now, if an exponent $y$ is present on the operand $x$, then:

$$\log_a{(x^{y})} = y\log_a{x}$$

Proof:
Here I am considering $y\ge0$ (since I am assuming growth rates require only positive integral domain). A similar proof is present for negative y too.
$$\log_a{(x^{y})} = \log_a{(x\times x \times x\times x...y\ times)} = \log_a{x}+\log_a{x}+\log_a{x}...y \ times = y\log_a{x}$$

The other one is simply

$$(\log_a{x})^y = \log_a{x}\times\log_a{x}\times\log_a{x}...y \ times$$

EDIT:
$\log{(n)}^{\log{(n)}}$ is a little ambiguous. If up to me, I would consider it as an exponent on the operand (just because parenthesis are there about n)

EDIT 2:
To put it simply, one growth is polynomial$\times$logarithmic and other logarithmic exponentiated to logarithmic. To find the actual difference, you could find the ratio of their limits as n$\to\infty$. Then you can compare their relative growth rates.

Title: is there any difference between these three logarithms?

Body:
is there any difference between $\log(n)^{\log(n)}$ vs $(\log n)^{\log n}$ vs $\log n^{\log n} $ from asymptotic growth rate? maybe all the same. I doubt about notation.

maybe this is very basic or crazy question, but I need to know.


问:
Question on the proof of Subadditive Ergodic Theorem in Durrett's textbook.
This came up in Durrett's proof of the subadditive ergodic theorem.

Let $X_{m,n}$ satisfy the assumptions of subadditive ergodic theorem, which reads

$X_{0,m}+X_{m,n} \geq X_{0,n}$

$(X_{nk, (n+1)k})_{n\geq 1}$ is a stationary sequence for eack $k$.

The distribution of $(X_{m,m+k})_{k\geq 1}$ does not depend on $m$.

$ E X_{0,1}^+ <\infty$ for each $n$, $EX_{0,n} \geq \gamma_0 n$ where $\gamma_0 >-\infty$.

In STEP 3 of his proof, he defines $Z:= \epsilon + (\underline{X} \vee -M)$ where $\underline{X}:= \liminf_{n \to \infty} X_{0,n} /n$. He first shows that $\underline{X} = \underline{X}_{m} := \liminf_{n \to \infty} X_{m,m+n} /n$ almost surely for any $m$.
Then he says $$Y_{m,n} := X_{m,n}-(n-m)Z$$ also satisfies the assumptions of the subadditive ergodic theorem since $Z_{m,n}=-(n-m)Z$ trivially does satisfy the assumptions.

My question is : why does $\mathbf {Y_{m,n}}$ satisfy the assumtions?
I see that conditions 1 and 4 are trivial to check, but I don't get how the conditions 2 and 3 are satisfied. Sum of two stationary sequence are not in general stationary.

Any help is appreciated.

答:
Note that $Y_{nk, (n+1)k}=X_{nk, (n+1)k}-kZ.$ Since $(X_{nk, (n+1)k})_{n\ge 1}$ is staionary by assumption, it follows that $Y_{nk, (n+1)k}$ is also stationary. One can write out the details using the fact that $Z$ is independent of any finite collection of random variables $X_{m, n}.$

is  actually very easy. Just write out the definition and observe that $(Y_{m,m+k})_{k\ge 1}=(X_{m, m+k}-kZ)_{k\ge 1}$ which clearly does not depend on $m.$


问:
Parametrize $\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$ by its arc length
I want to parametrize this curve $\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$, $t\in[-1,0)$ by its arc length.
Now $\|\psi'(t)\|=\sqrt{1+\frac{1}{t^2}}$ is $\ne0, \forall t\in[-1,0)$, so this curve is regular.

Now the arc length of this curve is $s(t)=\int_{-1}^{t}\|\psi'(u)\|\,du$ and now I should solve for $t$ as $t=t(s)$, but the integral's solution it's very difficult and I can't find any way to solve for t.

What am I doing wrong?

答:
As @Michael Hoppe already commented, you cannot solve for $t$ in the equation
$$s=\tanh ^{-1}\left(\sqrt{t^2+1}\right)-\sqrt{t^2+1}$$ but you can have approximations.

For convenience, let $x=\sqrt{t^2+1}$ to make the problem
$$s=\tanh ^{-1}(x)-x=\sum_{n=1}^\infty \frac {x^{2n+1}}{2n+1}$$ If we truncate to some order, we have the Taylor series that we can inverse to obtain for example
$$x=y-\frac{y^3}{5}+\frac{3 y^5}{175}+\frac{2 y^7}{1575}-\frac{16
   y^9}{202125}-\frac{362 y^{11}}{9384375}-\frac{49711
   y^{13}}{12415528125}+O\left(y^{15}\right)$$ where $y=\sqrt[3]{3s}$.

Edit

In fact, there is a formal analytical solution.

Define $k=\tanh(s)$ to make
$$k=-\tanh \left(x-\tanh ^{-1}(x)\right)=\frac{x \cosh (x)-\sinh (x)}{\cosh (x)-x \sinh (x)}$$ which leads to
$$e^{-2x}=\frac{(k+1) }{(k-1) }\frac{ (x-1)}{ (x+1)}$$
the solution of which being given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).


问:
Finding the maximum of a function
I want to find the maximum of the following function,

$$
f(x, y) = e^{m e^{-x}+n e^{-y}-x-y}(mrxe^y+nsye^x+mn(r+s)xy), 0 \le x, y \le 1
$$

where $r, s, m,$ and $n$ are positive constants. At the maximum point we have the following equations,

$$
\begin{cases}
\frac{\partial{f(x, y)}}{\partial{x}} = 0 \\
\frac{\partial{f(x, y)}}{\partial{y}} = 0
\end{cases}
$$

Simplifying the above equations we will get,

$$
\begin{cases}
x = -W_0\left(\frac
{s(ny+ye^y-e^y)}
{m(nry+nsy+rye^y+sye^y-se^y)}\right) \\

y = -W_0\left(\frac
{r(mx+xe^x-e^x)}
{n(mrx+msx+rxe^x+sxe^x-re^x)}\right)
\end{cases}
$$

Where $W_0(.)$ is the upper branch of the Lambert W function. We know that $W_0(.)$ cannot be expressed in terms of elementary functions. Does it imply that the system of equations cannot be solved analytically? If not, how can I solve it analytically?

答: Hint: One possible solution is using complex analysis, by substituting $z=x+iy$ (and then $x=\frac{z+\overline{z}}{2}$ and $y=\frac{z-\overline{z}}{2}$), you can get a simpler form (using, for example, mathematica). Hopefully, the function would be analytic in $\{z|0\leq\Re(z),\Im(z)\leq 1\}$. Then the maximum would be on the boundary, which is relatively easy to calculate (you get a function of one variable).

Title: Finding the maximum of a function


问:
Probability question, choosing a random integer number from [0, a]
Every time choosing a random integer number from [0, a], what is the probability of after k times, the sum of all chosen numbers is greater or equal to b.</p>

答:
Hint:

In this query
compare the methods used in the answers of user2661923
[me : stars and bars + Inclusion-Exclusion]
and Yip Jung Hon [generating polynomials].

I think that you should specify in your query whether $a$ and $b$ are integers.  Note that you have to count the polynomial solutions for each possible value of $b$.  Presumably, the possible sums will range from $0$ through $a \times k$ inclusive.

Each collection of random integers is equally likely, as is each non-negative integer solution $x_1 + x_2 + \cdots + x_k = b.$

The math here is (in my opinion) extremely ugly.


问:
Balkan MO :Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.
Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.

This problem was posted 2 times in MSE, but was never solved. Can someone help me?

I am supposed to use orders and mods

答:
Answer with hidden hints. We assume that $3p^{q-1} +1$ divides $11^p+17^p$ for some primes $p$ and $q$.

If $p=2$, it is not possible. (Brute force)

 For $p=2$ we have $11^2+17^2 = 410$ whose factors are $1,2,5,10,41,82,205,410$. If $3 \times 2^{q-1}+1$ is a factor then that factor is of the form $3k+1$ i.e. that factor can only be $1,10,82$ or $204$, but in each case $k \neq 2^{q-1}$ for $q$ prime.

Now for $p>2$, $ 8 \nmid 11^p+17^p$.

 Since $p$ is odd, $$11^p + 17^p \equiv 3^p+1 \equiv 3+1\equiv 4 \mod 8$$

Let $r$ be an odd prime divisor of $3p^{q-1}+1$. Then $r \notin \{3,11,17\}$.

 Of course $r \neq 3$,otherwise it cannot divide a number of the form $3k+1$. But $3p^{q-1}+1$ divides $11^p+17^p$ so if $r$ divides the RHS then it cannot divide either of $11^p$ or $17^p$ as it would have to divide the other too, but both are coprime. Consequently $r \neq 11,17$.

There is a $b$ such that $17b \equiv 1 \mod r$ , by Bezout's theorem. Show that $ord_r(11b) \in \{2,2p\}$.

 If $17b \equiv 1 \mod r$ then $b^p(11^p+17^p) \equiv (11b)^p +1\mod r$, but $r$ divides $11^p+17^p$ so $(11b)^p + 1$ is a multiple of $r$, and therefore $(11b)^{2p}-1$ is a multiple of $r$. Therefore $ord_r(11b)$ divides $2p$, but is not equal to $1$ obviously.

Let $ord_r(11b) = 2$. Then $r=7$.

 If so then $r$ divides $(11b)^2-1$, but then it also divides $(11b)^2 - (17b)^2$, and because it is coprime to $b$ is a divisor of $11^2-17^2 = -168$. Since $r \neq 3$ we must have $r=7$.

If $ord_r(11b) = 2p$ then $2p$ divides $r-1$, by Fermat's little theorem. Hence we may write the prime decomposition $$
3p^{q-1}+1 = 2^{\alpha}7^{\beta}p_1^{\gamma_1}...p_k^{\gamma_k}
$$ where $p_i \neq 2,7$ are prime factors with $p_i \equiv 1 \mod 2p$.

We have $\beta \leq 1$, by observing that $\frac{11^p+17^p}{28}$ is not a multiple of $7$. (Hint : $11+17= 28$, so that fraction is definitely a natural number, but you can say more) Thus $\beta \leq 1$.

 The standard factorization $\frac{a^p-b^p}{a-b} = \sum_{k=1}^{n-1} a^kb^{n-k}$ applies with $a=11,b=-17$ and we get using $11 \equiv 4$ and $17 \equiv -4$ mod $7$, that $\frac{11^p+17^p}{28} \equiv p4^{p-1}$ mod $7$, and $p \neq 7$(why?)

If $q=2$, then the problem has no solutions.

 If so then $3p+1 = 2^{\alpha}7^{\beta}p_1^{\gamma_1}...p_k^{\gamma_k}$, but each of the $p_i$ is at least $2p+1$ which is more than half $3p+1$. Therefore no $p_i$ exist. By the existing bounds on $\alpha,\beta$ there are only the candidates $3p+1 = 2,4,14,28$, none of which work.

If $q>2$, then $\alpha = 2$ and $p=3$.

 If $q>2$, then $p^{q-1} \equiv 1 \mod 4$ since $q-1$ is even, hence $3p^{q-1} +1$ is a multiple of $4$, so $\alpha = 2$. Finally, $2^{\alpha}7^{\beta} p_1^{\gamma_1}...p_k^{\gamma_k}$ is congruent to $4$ or $28$ modulo $p$, because the $p_i$ are all congruent to $1$ modulo $p$. But $3p^{q-1}+1$ is congruent to $1$ modulo $p$. As these are equal, we get that either $1-4$ or $1-28$ is a multiple of $p$. Either way $p=3$.

See that $q=3$ must happen if $p=3$.

 Well, we have $11^3+17^3= 6244 = 2^2 \times 7 \times 223$, so if $3 \times 3^{q-1} +1$ is a multiple of $6244$, then this is the same as saying $3^q+1$ is a multiple of $6244$, and therefore $q=3$ is the only option since $q=1,2,4,...,7$ don't work and $q=8$ is too large.

Thus $p=q=3$ is the only solution.


问:
How to find details about function in $\mathbb{R}^2$ if $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in \mathbb{R}^2$
I am trying assignment questions of real analysis and I was unable to solve this question. In fact, I don't have any idea on how to start.

So, I am looking for help here.

Let $f$  be a function on $\mathbb{R}^2$ such that $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in  \mathbb{R}^2$. Now which of the following options are true:

1 $f$ is constant on all lines parallel to the line $x=-y$.

2 $f(x,y)=0 \quad\forall (x,y) \in \mathbb{R}^2$.

3 $f(x,y)=f(-y,x) \quad\forall (x,y) \in \mathbb{R}^2$.

4 $f(x,y)-f(y,x)=(x-y) \frac {df}{dx}(x*,y*) + (y-x) \frac {df}{dy} (x*,y*) \in \mathbb{R}^2$.

I am unable to get any idea on how to use the hypothesis $\frac{df} {dx}= \frac{df}{dy}$ to deduce or contradict  any of the options.

I studied multivariable calculus from Tom M Apostol in case it is of any help.

Please help me. Kindly just give outlines not complete solution.

答: The gradient of $f$ is parallel to $\vec i + \vec j$, so level curves are orthogonal to that constant vector. That makes lines $x+y=c$ the level curves, so $f(x,y) = g(x+y)$ for any differentiable function $g$.


问:
prove the identity $\sum_{k=0}^n {s+k \choose k}{n-k \choose m} = {s+n+1 \choose s+m+1} $
I am not able to understand how to apply the vandemonde identity to prove this. Please help me proving this.</p>

答:
Some progress:

This sum will vanish if $m>n$.
$$S=\sum_{k=0}^{n} {s+k\choose k} {n-k \choose m}$$
$$S=[x^m]\sum_{k=0}^{n} {s+k\choose k} (1+x)^{n-k}$$
next use the Binomial series with negative index as
$$\sum_{k=0}^{n} {s+k \choose k} z^k=(1-z)^{-s-1}-z^{n+1} {n+s+1 \choose n+1} ~_2F_1[n+s+2,1;2+n;z]~~~~~(2)$$
Then,
$$S=[x^m] (1+x)^nx^{-s-1}(1+x)^{s+1}\implies S=[x^{m+s+1}] (1+x)^{n+s+1}$$
$$\implies S={n+s+1 \choose m+s+1}.$$
Here $[x^j] f(x)$ means coefficient of $x^{j}$ in $f(x).$

Note:
One is welcome to point out why second term in (2) would not matter.


问:
Computing the limit $\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}$
For every $z\in\mathbb C\setminus\{0\}$ we define $\operatorname{Arg}(z)$ to be the principal argument of $z$, that is $\operatorname{Arg}(z)=\theta$, where $z=re^{i\theta }$ and $\theta\in(-\pi,\pi]$. Does the following limit exists?

$$
\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}
$$
I guess that if it exists, then it must be $0$ since if $z\to0$ along the real line, then the limit is $0$.

答:
Given
$$f(z)=u(x,y)+iv(x,y)$$
$$z_0=x_0+iy_0$$
$$w_0=u_0+iv_0$$
Then
$$\lim_{z\to z_0}f(z)=w_0$$
If and only if
$$\lim_{(x,y)\to (x_0,y_0)}u(x,y)=u_0$$
And
$$\lim_{(x,y)\to (x_0,y_0)}v(x,y)=v_0$$

Let $z=x+iy$
$$\lim_{(x,y)\to (0,0)}\frac{x}{x^2+y^2}\arctan\left(\frac{y}{1+x}\right)$$
$$=\cos\phi\lim_{r\to 0^+}\frac1r\arctan\left(\frac{r\sin\phi}{1+r\cos\phi}\right)$$
$$=\sin\phi\cos\phi\lim_{r\to 0^+}\frac{1}{r^2+2r\cos\phi+1}$$
$$=\sin\phi\cos\phi$$
Since this limit is clearly dependent on $\phi$, we can conclude that
$$\lim_{(x,y)\to (0,0)}\frac{x}{x^2+y^2}\arctan\left(\frac{y}{1+x}\right)$$
$$\mbox{does not exist}$$
Therefore
$$\lim_{z\to 0}\frac{1}{z}\operatorname{Arg}\left(1+z\right)$$
$$\mbox{does not exist}$$


问:
Find the distribution of $Z = X/(X + Y )$
Let $X$ and $Y$ be independent random variables with the same geometric distribution $\left(P(X=x|p)=p(1-p)^{x-1}\right)$.

Find the distribution of $Z = X/(X + Y )$, where we define $Z = 0$  if  $X + Y = 0$.

Hello, I need help with this question I start by letting x/(x+y)=r/s where r and s are positive integers and s>r but I don't know where to go from here. can someone help me?

答:
Suppose that $a/a+b$ is in lowest terms, $a<b$.  Then $X/(X+Y)=a/(a+b)$ if there is some integer $k$ such that $X=ka$ and $Y=kb$.  So

$$P\left(X/(X+Y)=a/(a+b)\right)=\sum_{k=1}^{\infty} P(X=ka)P(Y=kb)\\=\sum_k p^2(1-p)^{k(a+b)-2} \\= \frac{p^2(1-p)^{a+b-2}}{1-(1-p)^{a+b}}.$$

This interestingly enough doesn't depend on $a$, only on $a+b$.


问:
There are eight coins in a row, all showing heads.
There are eight coins in a row, all showing heads. In each move, we can flip over two adjacent coins provided that they are both showing heads or both showing tails. How many different patterns of heads and tails can we obtain after a number of moves?

My solution: We can replace each head with the number '1', and each tail with the number '0', we will then end up with a binary sequence with a starting decimal value of 255. Each move will add or subtract $3*2^n$, $(6 \geq n \geq 0)$, thus there will be $255 / 3 + 1 = 86 $ solutions.

There is already a solution on StackExchange here: How many different patterns of heads and tails can we obtain after a number of moves?

Although I understand the accepted solution for the above link, I still don't understand why my solution is wrong, can anyone help continue my solution or give me a hint?

答: You are suggesting that you can reach any multiple of 3. But you can't reach, for instance, $21$, that is $00010101$.


问:
Postitive definiteness of the Kronecker product of two positive definite matrices
Let $A$ and $B$ both be positive definite matrices. How do I
show that their Kronecker product is also positive definite?

I know we can use the fact that the eigenvalues of the Kronecker
product is $\lambda_A+\lambda_B$ which are all positive. But I want
to use a different approach here.

Thank you.

答:
Here is another approach: Let $A,B\geq0$, so there exists positive square roots $\sqrt{A}, \sqrt{B}$. Thus, we have (essentially by definition of the tensor/kronecker product of operators/matrices):

$$ A \otimes B = \sqrt{A} \otimes \sqrt{B} \cdot \sqrt{A} \otimes \sqrt{B}$$

But $\sqrt{A} \otimes \sqrt{B}$ is a self adjoint matrix, so it's square must be positive.


问:
An explicit algorithm to test whether a string of symbols is or is not well-formed in this language.
Consider the set of five symbols, $\{ (, ), p, ', *\}$. A variable is defined as $p$ followed by zero or more $'$s. A wff is inductively defined as follows: $1$. Every variable is a wff. $2$. If $x$ and $y$ are formulas, so is $(x*y)$. $3$. Nothing else is a formula. This is really just a simplified version of a propositional logic language. My problem, though, is with clause 3 of the definition. It is easy to prove that a particular sequence of symbols is a wff. How, though, does one actually prove that a particular sequence, like $p**pp$, is NOT a wff? More precisely, and this is my real question, is there an explicit algorithm that takes a string and returns Yes if it is a wff, and No if it is not? And can someone describe the algorithm?

答:
You're asking a particular question:

How… does one actually prove that a particular sequence, like p∗∗pp, is NOT a wff

and a general question:

is there an explicit algorithm that takes a string and [ decides if it is a wff ]

The particular question has a shorter answer.  The rules you gave require that a wff must begin with p (rule 1) or with ( (rule 2).  The string p**pp begins with p so if it is a wff it must be a variable.  But a variable is a p followed by zero or more 's, and p**pp is not that, so it is not a variable.  So by rule 3, it is not a wff.

The more general question has a very interesting answer.

There are many explicit algorithms that do what you want.  They come under the general heading of “parsing algorithms”.  A parsing algorithm takes an unstructured sequence of symbols and decides if it can be understood as a meaningful utterance in a certain language. In your example the language is the syntax of wffs. Parsing is an interesting topic, and it is impossible to give a complete answer here.  (I highly recommend Parsing Techniques: A Practical Guide by Grune and Jacobs, available for free download from the author's web site.)

I can outline one example an algorithm that parses the language you described.  This uses the method of recursive descent parsing:

You imagine that the parser will scan over the symbols of the input from left to right.

You write a function that recognizes a variable.  This is easy: if the next unscanned symbol is p the function will consume the p and then consume any following ' symbols, and yield success; otherwise it yields failure, which indicates that the unscanned part of the input does not look like a variable

You write a function that recognizes if the input begins with a wff, and if so scans the entire wff:

It first uses the variable-recognizing function to see if it can recognize a variable at the beginning of the unscanned input.  If so, it yields success, because a variable is a wff.

Otherwise, the function looks to see if the next unscanned symbol is a (.  If not, it can fail immediately because every wff that is not a variable begins with a (

If the upcoming symbol is a (, then it consumes the (.  Then it calls itself recursively scan the following wff.  If this recursive call fails, the failure is propagated back up, because what we have is ( followed by a non-wff.

But if the ( is followed by a valid wff, the function then looks to see if the following symbol is *.  If not, it fails immediately.

But if there is a *, the function consumes it and makes a second recursive call, looking for a second wff.

If this second recursive call also succeeds then the function checks to see if the upcoming symbol is ), and reports success if it is.


问:
Markov Chain with 2 machines and 2 repair men
If we have 2 machines and 2 repair men. The machines each break down at different rates $\alpha_a$ for machine A and $\alpha_b$ for machine B. If we assume that the repairmen have different abilities, they each repair with exponential random times with parameters $\beta_1$ and $\beta_2$ respectively(naturally, this rate does not depend on which machine they repair).

Define a Markov chain, specify the state space and the transition matrix. Also determine the communicating classes.

My attempt: I am unsure of how to proceed for this case. Should it been machine at the same rate, and the repairmen with same abilities, I know how to solve this question. But now that the rates have been split, I am unsure how to construct a state space.

答:
Guide:

One possibility,

No broken machine. $1$ state.

Machine $i_1$ is broken and it is being fixed by repairman $j_1$, machine $i_2$ is fine. $4$ such states.

Machine $i_1$ is broken and it is being fixed by repairman $j_1$, machine $i_2$ is being fixed by repairman $j_2$. $2$ states.

With that, try to write down the transition rate.


问:
Radius of convergence of the Bernoulli polynomial generating function power series.
The generating function of the Bernoulli Polynomials is: $$\frac{te^{xt}}{e^t-1}=\sum_{k=0}^\infty B_k(x)\frac{t^k}{k!}.$$

Would it be right to say that the radius of convergence of this power series is $2\pi$ ? I'm not sure since the power series above is in fact a double series: $$\sum_{k=0}^\infty\left(\sum_{j=0}^k {k\choose j}B_{k-j} x^j\right)\frac{t^k}{k!}.$$

What if I were to choose a fixed value for $x$? Would the radius be $2\pi$ then, even for the double power series?

答: For every fixed $x=c$, the radius of convergence of the power series is $2\pi$. This is because $$\frac{ze^{cz}}{e^z-1}$$ is analytic everywhere except at $z=i2\pi n, n=\pm 1,\pm 2,\cdots$ (not at $0$ though.) The disk $B(0,2\pi)$ is the smallest one centered at $0$ that contains a singularity on its boundary, so the radius of convergence is $2\pi$.</p>


问:
Proving the existence of the third-order derivatives
The statement is as follows:

Let $f ∶ R → R$ be a function such that $f′′′$ exists. Suppose a < b are real numbers such
that f(a) = f(b) = f′(a) = f′(b) = 0.
Then there exists c ∈ (a, b) such that f′′′(c) = 0.

** I am trying to formulate a proof in this way:
$f'''(c)$ = 0 indicating $\frac{f''(b)-f''(a)} {b-a}$ = 0 for some a,b

and it follows that $f''(b) = f''(a)$

Apply the Mean Value Theorem again we needs to prove $ \frac{f'(x)-f'(a)}{x-a}$ equals $\frac{f'(x)-f'(b)}{x-b}$

And I had a feeling that I use the condition  f(a) = f(b) = f′(a) = f′(b) = 0, but how?

答: If $f(a)=f(b)=0$, by Rolle's theorem there is $c_1\in (a,b)$ with $f'(c_1)=0$. Same argument gives $c_2\in (a,c_1)$ and $c_3\in (c_1,b)$ with $f''(c_2)=f''(c_3)=0$ (because $f'(a)=f'(c_1)=f'(b)=0$). Repeat again and get $c\in (c_2,c_3)$ with $f'''(c)=0$.


问:
Show $N(0, a^2) \cdot c = N(0, a^2c^2)$
Let $\xi$ is a continuous random variable such that $\xi$ ~ $N(a,b)$ normal distribution. How to show that
$$N(0, a^2) \cdot c = N(0, a^2c^2)$$

I found $$\mathbb E\xi_{N(0, a^2)} = \mathbb E\xi_{N(0, a^2c^2)} = 1$$
and $$D\xi_{N(0,a^2)} = a^4, D\xi_{N(0,a^2c^2)} = c^4a^4$$

What does this show? That is, it is clear that the coefficient $c$ must be introduced into the distribution to the second degree, but this is not visible here. How else can you prove equality?

答: If $\xi \sim N(0, a^{2})$ then $c\xi$ has normal distribution with mean $0$ and variance $E(c^{2}\xi^{2})=c^{2} var(\xi)=c^{2}a^{2}$. Hence $c \xi \sim N(0, c^{2}a^{2})$.

Title: Show $N(0, a^2) \cdot c = N(0, a^2c^2)$


问:
How to solve the eigenvector equation using an inverse?
For a Matrix

$$A=\begin{bmatrix}-2 & 1\\6 & 3\end{bmatrix}$$

The eigenvalues are

$$(A-\lambda I)x=0$$ $$det(A- \lambda I)= (-2-\lambda)(3-\lambda) - 6=0$$ $$\lambda^2 - \lambda -12=0$$ $$\lambda=-3,4$$

The eigenvectors should then be found by using the equation $$Ax=\lambda x$$
I have tried solving this by multiplying both sides by the inverse of $A$ $$A^{-1}Ax=A^{-1}\lambda x$$ However that just leads me to
$$x=A^{-1}\lambda x$$
I am unsure how to proceed in finding the eigenvector $x$ by this method, is this even possible or am I missing something crucial?

答: For $\lambda = 4$, the equation $(A-\lambda I)x = 0$ becomes $$\left(\begin{bmatrix}-2&1\\6&3\end{bmatrix} - \begin{bmatrix}4&0\\0&4\end{bmatrix}\right)\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$ $$\begin{bmatrix}-6&1\\6&-1\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$ $$ \begin{bmatrix}-6x_1+x_2\\6x_1-x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}.$$ Can you find a solution to this equation? Finding the eigenvector for $\lambda = -3$ is similar.


问:
Advocating base 12 number system
I had a calculus professor who suggested we should be using base 12 number system. What are the advantages of using such a system?   </p>

答: Well, in the base $10$ number system it is as easy to multiply by 5 as to divide by 2 (the answers differ by 0 at the end). And it is as easy to multiply by 2, 4, 8... as to divide by 5, 25, 125..., by the same reason. So in the base $12$ number system it is as hard to divide by 2,3,4,6,12,4,9,16... as to multiply by 6,4,3,2,1,36,16,9..., and, therefore, one can trade division to multiplication in more instances... </p>

Title: Advocating base 12 number system


问:
A question based on metric space $d(x_i , y_i) $= $\sum_{i\geq 1 }|x_i -y_i| 2^{-i}$
This question was asked in my analysis quiz and I was unable to solve some options although I tried at home also.

Question: Let X={$(x_i)_{i\geq 1}: x_i \in$ {0,1} for all$ i\geq 1 $} with metric $d(x_i , y_i)$ = $\sum_{i\geq 1 }|x_i -y_i| 2^{-i}$. Let $f: X \to [0,1]$ be the function defined by $f(x_i)_{i\geq 1} = \sum_{i\geq 1}x_i 2^{-i}$. Which of the following options are correct:

1 f is continuous.

2 f is onto

3 f is 1-1

4 f is open

I have proved 1. But I don't know how to approach 2, 4 . I proved 3 right . BUt answer doesn't have it as correct.

I hav egiven a lot of time on it. I request you to tell me 2,3,4.

Waiting for your help!!

答:
Note that we can prove that $d$ is in fact a metric.

(1) In a metric
space $(X,d)$, the metric $d$ is continuous so that $f(x)=d(o,x)$ is
continuous where $o=(0,0,0,\cdots)$.

(2) For any $y\in [0,1]$, then we have $y= \sum_{i=1}^\infty
\frac{\varepsilon_i}{2^i},\ \varepsilon_i\in \{0,1\}$ so that
$f(\varepsilon)=y$ where $\varepsilon=(\varepsilon_i)$, which proves
that $f$ is onto

(3) When $x=(1,0,0,0,\cdots ),\ y=(0,1,1,1,\cdots)$, then $f(x)=f(y)=
\frac{1}{2}$ so that not $1$-$1$.

(4) Consider a open ball $B=\{ x| d(x,y)< \frac{1}{2}\}$ where $y=
(1,0,0,0,\cdots )$

Note that $B$ is exactly the set of all points of type
$x=(1,x_2,x_3,x_4,\cdots )$ s.t. some $x_i$ is equal to $0$

Hence $f(B) = [\frac{1}{2},1)$ is not open so that $f$ is not open


问:
Check if a curve is contained in a plane

Let $$\gamma :\mathbb{R}\rightarrow\mathbb{R}^{3},$$
$$\gamma(t)=(t,t^{2},\ln(1+t^{2})).$$
be a parametrization of a curve. Establish if $\gamma$ is contained in a plane.

I have a problem here. First of all, I know that a curve $\alpha$ with curvature $k(t)\neq 0, \ \forall t\in Dom(\alpha)$ is contained in a plane if and only if $B(t)=\frac{\alpha '(t)\wedge\alpha ''(t)}{\mid\mid\alpha '(t)\wedge\alpha ''(t)\mid\mid }$ is constant. This curve seems to respect this condition but it's easy to see that $\gamma '(t)\mid_{t=0}=\underline{0}$ so in the origin it doesn't make sense to speak about curvature. I though to proceed in this way:
I consider two curves $\gamma_{1}:(-\infty,0)\rightarrow\mathbb{R}^{3}$ and $\gamma_{2}:(0,+\infty)\rightarrow\mathbb{R}^{3}$. I see that these two curves are contained on a plane so I check if the planes are the same. Is it the right way to proceed?

答: If this curve were contained in a plane, that plane would have to pass through $\gamma(0)=(0,0,0)$ and be spanned by $\gamma'(0)=(1,0,0)$ and $\gamma''(0)=(0,2,2)$. So the plane equation (again, if this were the case) would be $y=z$. But $t^2$ is not equal to $\ln(1+t^2)$ for all $t$, so here's your contradiction.


答:
Here is  a better way: If the curve lines in a plane and $(a,b,c)$ is a unit normal vector to the plane then $at+bt^{2}+c \ln (1+t^{2})=0$ for all $t$. To get a contradiction easily by differentiating twice and setting $t=0$.

[First differentiation gives $a=0$ and second differentiation gives $b+c=0$ But then you get $t^{2}=-\ln (1+t^{2})$ for all $t$ which is false when $t=1$].

Title: Check if a curve is contained in a plane

Body:
Let $$\gamma :\mathbb{R}\rightarrow\mathbb{R}^{3},$$
$$\gamma(t)=(t,t^{2},\ln(1+t^{2})).$$
be a parametrization of a curve. Establish if $\gamma$ is contained in a plane.

I have a problem here. First of all, I know that a curve $\alpha$ with curvature $k(t)\neq 0, \ \forall t\in Dom(\alpha)$ is contained in a plane if and only if $B(t)=\frac{\alpha '(t)\wedge\alpha ''(t)}{\mid\mid\alpha '(t)\wedge\alpha ''(t)\mid\mid }$ is constant. This curve seems to respect this condition but it's easy to see that $\gamma '(t)\mid_{t=0}=\underline{0}$ so in the origin it doesn't make sense to speak about curvature. I though to proceed in this way:
I consider two curves $\gamma_{1}:(-\infty,0)\rightarrow\mathbb{R}^{3}$ and $\gamma_{2}:(0,+\infty)\rightarrow\mathbb{R}^{3}$. I see that these two curves are contained on a plane so I check if the planes are the same. Is it the right way to proceed?


问:
Are complicated systems more likely to be chaotic?
Though the simple pendulum isn't chaotic, the double pendulum is. Is it true in general that more complicated differential equations are more likely to be chaotic than simple differential equations? There are many ways differential equations can be complicated, but these are some of the criteria I can think of:

No analytical solution

Nonlinear

Coupled

More variables

I know that there are complicated differential equations which aren't chaotic, but would you have more luck finding chaos in a simple or complicated differential equation?

答:
Yes.

First of all, for continuous-time systems, you need at least the following ingredients for chaos:

non-linearity

three dimensions (you need one positive, one zero and one negative Lyapunov exponent)

These conditions already imply a certain complexity – but not all systems fulfilling them are chaotic. One reason for this is that you may have strongly coupled dimensions or weak non-linearities. Thus, if you increase the number of dimensions or strength of non-linearities, you increase your chances to find chaos.

Moreover, consider a complex system consisting of many coupled subsystems, with some only driving.
Here, it take only a single chaotic subsystem to render the entire dynamics chaotic.

This also holds empirically: Complex systems show larger chaotic bands, i.e., parameter regimes in which they are chaotic.
For example, consider the following figure from this paper (which I co-authored):

Systems B and A are coupled oscillator networks designed in the same way, except that System B has more components or dimensions, respectively.
Whenever you see a vertical band in the diagram, you have chaos for that value of the control parameter $a$.
As you can see, the chaotic regions are considerably larger for the more complex system.

No analytical solution

The solutions that are considered analytical are not chaotic.
So this a also requirement for chaos, if you so wish.


问:
If Two Groups are both NONAbelian And Have same No. Of elements of each order .Are they isomorphic
I was trying to prove that S3×Z2 is isomorphic to D6 .
Well there are a lot of Methods to Do so.

But I was Queried that I We know If two ABELLEAN Groups have same no. Of elements of each order then they are ISOMORPHIC

Also there exist a Abellean and a NONAbelian group satisfying the property but still non ISOMORPHIC.

So my doubt Is -- Is it true If both are NONAbelian And Have same no. Of elements of each order then they are ISOMORPHIC

答: No; there are many more groups than there are possible sets of orders. In this MO answer you can find a counting argument that makes this precise and in this MO answer you can find an explicit minimal counterexample of order $16$ (although I haven't checked to see if you can find a counterexample where both groups are nonabelian).


问:
Balkan MO :Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.
Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.

This problem was posted 2 times in MSE, but was never solved. Can someone help me?

I am supposed to use orders and mods

答:
Another approach:

Due to Euler's criterion:

$A=3^{\frac{7-1}2=3}+1 \equiv 0 \ mod (7)$

$3\times 3^{q-1}+1\equiv 0 \ mod (7)$

These two relations give $p=q=3$

In this case $A=28$.

We check this with RHS, i.e $B=11^p+17^q$:

$11\equiv 4 \mod (7)$ , $\rightarrow 11^p \equiv 4^p \mod (7)=7a+4^p$

$17 \equiv 3 \ mod (7)$, $\rightarrow 17^q\equiv 3^q \ mod (7)=7b+3^q$

$17^q$ and $3^q$ are odd so $7b$ is even, therefore:

$17^q=14 b_1+3^q$

⇒ $11^p+13^q=7a+7b+4^p+3^q$

$11^p+17^q$ and $4^p+3^q$ are odd so $7a+7b$ must be even; $7b=14b_1$ is even, so $7a$ must also be even, let $7a=14a_1$, so we have:

$11^p+17^q=14(a_1+b_1)+4^p+3^q$

$4^p+3^q=(7-3)^p+3^q= 7t -3^p+3^q$

We need $11^p+11^q\equiv (4^p+3^q)\equiv 0 \ mod (7)$

This is only possible if $p=q$ such that:

$4^p+3^q=(7-3)^p+3^q= 7t +(-3^p+3^q=0)=7t$

Now if $p=q=3$ then $14a_1+14b_1=28 k$, because:

$11^3+17^3=223\times 28$

Hence $A=28\big|B$

That is $p=q=3$ can be a solution.


问:
Partitioning $\{1,2,...,1998\}$ such that $|a_i-b_i|$ is $1$ or $6$

Suppose that the set $\{1,2,3,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}(1\le i \le 999)$ so that for all $i, \mid a_i-b_i\mid$ equals $1$ or $6$. Then the sum $\mid a_1-b_1\mid +\mid a_2-b_2\mid+\cdots +\mid a_{999}-b_{999}\mid$ = ?

I found this problem on AoPS. I found out that the answer is 999 by partitioning the set into $\{1,2\},\{3,4\}, ... , \{1997, 1998\}$, but I am not sure about the proper way of solving it

答:
(Previous incorrect answer removed.)

As $1 \equiv 6 \pmod 5$, $|a_i-b_i| \equiv 999\cdot1 \equiv 4 \pmod 5$.

For minimum sum, take pairs of consecutive numbers so that each $|a_i-b_i|=1$ and $\sum |a_i-b_i|=999$.

Next rearrange groups of $6$ consecutive pairs as in @Hagen von Eitzen's answer, to increase sum by $30$ per group.

As sum only increases by $30$, every possible sum ends in digit $9$.

Note as $1998$ is only multiple of $6$ and not $12$, not all $|a_i-b_i|$ can be $6$. For maximum sum, take distance-$6$ pairs upto largest multiple of $12$ which is $1992$, as $\{(1,7)(2,8),\ldots,(1991,1992)\}$ followed by $\{(1993,1994),(1995,1996),(1997,1998)\}$ to yield $$ \dfrac{1992}{2}\cdot6+3=5979$$


问:
How to determine asymptotics for $\sum_{ab^2 < x} ab^2$ by summing separately over $a$ and $b$
I do not understand how to get asymptotics for the double sum
$$\sum_{ab^2 < x} ab^2$$

If I sum over $a$ first, I get
$$\sum_{b^2<x} b^2 \sum_{a < x/b^2} a = \frac12 \sum_{b^2<x} b^2 \frac{x^2}{b^4} \sim \frac{\zeta(2)}{2} x^2.$$

If I sum over $b$ first, I get
$$\sum_{a<x} a \sum_{b < \sqrt{x/a}} b^2 = \frac13 \sum_{a<x} a \left( \frac{x}{a} \right)^{3/2} \sim \frac{2}{3} x^2. $$

What is happening. Am I doing something forbidden in switching summations this way? I suppose it is a possible divergence problem, so I do I get precise asymptotics for this sum?

答:
TLDR: Your second method uses two asymptotics. The asymptotic approximation for the inner summation throws out "lower order terms" which end up mattering for outer summation.

For integers $N \ge 0$, we have $$\displaystyle\sum_{k = 1}^{N}k^2 = \dfrac{N(N+1)(2N+1)}{6}.$$ Combining this with the inequality $y-1 \le \lceil y \rceil-1 \le y$, we get $$\dfrac{(y-1)y(2y-1)}{6} \le \displaystyle\sum_{k = 1}^{\lceil y \rceil-1}k^2 \le \dfrac{y(y+1)(2y+1)}{6},$$ i.e.
$$\dfrac{1}{3}y^3-\dfrac{1}{2}y^2+\dfrac{1}{6}y \le \displaystyle\sum_{k < y}k^2 \le \dfrac{1}{3}y^3+\dfrac{1}{2}y^2+\dfrac{1}{6}y$$ for any real number $y \ge 1$. Applying this yields
$$\dfrac{x^{3/2}}{3a^{3/2}}-\dfrac{x}{2a}+\dfrac{x^{1/2}}{6a^{1/2}} \le \sum_{b < \sqrt{x/a}}b^2 \le \dfrac{x^{3/2}}{3a^{3/2}}+\dfrac{x}{2a}+\dfrac{x^{1/2}}{6a^{1/2}},$$ and thus, $$\sum_{a < x}\left[\dfrac{x^{3/2}}{3a^{1/2}}-\dfrac{x}{2}+\dfrac{x^{1/2}a^{1/2}}{6}\right] \le \sum_{a < x}a\sum_{b < \sqrt{x/a}}b^2 \le \sum_{a < x}\left[\dfrac{x^{3/2}}{3a^{1/2}}+\dfrac{x}{2}+\dfrac{x^{1/2}a^{1/2}}{6}\right].$$

However, note that $\displaystyle\sum_{a < x}\dfrac{x^{3/2}}{3a^{1/2}} \sim \dfrac{2}{3}x^2$, $\displaystyle\sum_{a < x}\dfrac{x}{2} \sim \dfrac{1}{2}x^2$, and $\displaystyle\sum_{a < x}\dfrac{x^{1/2}a^{1/2}}{6} \sim \dfrac{1}{9}x^2$. So the lower order terms that the approximation $\displaystyle\sum_{k < y}k^2 \sim \dfrac{1}{3}y^3$ drops end up contributing a significant amount to the double sum. Hence, all you know from the second method is that the double sum is asymptotically bounded between $\dfrac{5}{18}x^2$ and $\dfrac{23}{18}x^2$.

We can analyze the first method in the same manner. Using $$\dfrac{1}{2}y^2-\dfrac{1}{2}y \le \sum_{k < y}k \le \dfrac{1}{2}y^2+\dfrac{1}{2}y,$$ we get $$\dfrac{x^2}{2b^4}-\dfrac{x}{2b^2} \le \sum_{a < x/b^2}a \le \dfrac{x^2}{2b^4}+\dfrac{x}{2b^2},$$ and thus, $$\sum_{b < \sqrt{x}}\left[\dfrac{x^2}{2b^2}-\dfrac{x}{2}\right] \le \sum_{b < \sqrt{x}}b^2\sum_{a < x/b^2}a \le \sum_{b < \sqrt{x}}\left[\dfrac{x^2}{2b^2}+\dfrac{x}{2}\right].$$ As you already noted, $\displaystyle\sum_{b < \sqrt{x}}\dfrac{x^2}{2b^2} \sim \dfrac{\zeta(2)}{2}x^2$. The other term scales like $\displaystyle\sum_{b < \sqrt{x}}\dfrac{x}{2} \sim \dfrac{1}{2}x^{3/2} = o(x^2)$. Hence, the double sum does indeed scale like $\dfrac{\zeta(2)}{2}x^2$.


问:
Continuity of multivariable functions along axes
I was given the following problem:

Let $f:D(\subseteq R2) \to R$ be a function and $(a, b) \in D$. If $f(x, y)$ is continuous at $(a, b)$, then show that the functions $f(x, b)$ and $f(a, y)$ are continuous at $x = a$ and at $y = b$ respectively.

This fact seems obvious so it's hard to figure out how to prove it rigorously. My attempt:

$f(x, y)$ is continuous at $(a,b)$ implies that it is continuous in a neighborhood around (a, b). That is, $\forall \epsilon > 0$, there exists $\delta$ such that $|f(x, y) - f(a,b)| < \epsilon$ whenever $|(x,y) - (a,b)| < \delta$.

Since the one-dimensional neighbourhood $(x,b)$ such that $|(x,b) -(a,b)| < \delta$ is situated within the above neighbourhood, this implies $f(x, b)$ is continuous at $x=a$. Similarly for the other case.

Is this a rigorous proof?

答: Consider the function $g: \Bbb R \to \Bbb R^2$ defined by $g(y) = (a,y)$. Then you can easily verify that $g$ is a continuous function. Now observe that $f(a,y) = f \circ g$ which is a composition of two consinuous function and therefore is continuous. Similarly, $f(x,b)$ can be shown to be a continuous real function of one real variable.


问:
Find Taylor series and radius of convergence of $\frac{1}{1-z}, z_0 = i$
I cannot find a comparable series. What am I supposed to compare this to?

$$\frac{1}{1-z}, z_0 = i$$

So I guess this is the geometric series but $z_o = i$. So what's the Taylor series then? Is it just:

$$ \frac{1}{1-z} = 1 + x + x^2 ... = \sum_{n=0}^{\infty} x^n$$
z
I think not... but how do I incorporate the $z_0$


问:
Lower bound for the $f(x)=\frac{1}{2}x^TAx + c^Tx + b$ when $A$ is PSD
Let $A \in \mathbb{R}^{n \times n}$ be a positive semi-definite matrix. Let
$$
f(x)=\frac{1}{2}x^TAx + c^Tx + b$$

It is possible to show that when $f$ is bounded below, and $c$ is in the range of $A$, then $f$ has a global minimizer.

Suppose only the following assumptions hold:

$A$ is positive semi-definite matrix

$c$ is in the range of $A$

Would it be possible to show that $f$ is bounded below? If so, find that lower bound.

答:
Let $x^*$ be such that $Ax^*=-c$. (You are missing a negative sign) Then
$$
f(x) - f(x^*)= (x^*)^TA(x-x^*) + \frac12(x-x^*)^TA(x-x^*) + c^T(x-x^*)
= \frac12(x-x^*)^TA(x-x^*) \ge0,
$$
so $f(x^*)$ is the lower bound.


答:
Hint: $$\frac{1}{1-z}= \frac{1}{1-i} \cdot \frac{1}{1-\frac{z-i}{1-i}}.$$

Geometric series !

Title: Find Taylor series and radius of convergence of $\frac{1}{1-z}, z_0 = i$

Body:
I cannot find a comparable series. What am I supposed to compare this to?

$$\frac{1}{1-z}, z_0 = i$$

So I guess this is the geometric series but $z_o = i$. So what's the Taylor series then? Is it just:

$$ \frac{1}{1-z} = 1 + x + x^2 ... = \sum_{n=0}^{\infty} x^n$$
z
I think not... but how do I incorporate the $z_0$


问:
Convolution with a convex function
Let $f$ a convex function, and $j$ a $C^\infty$-function with compact support in $(-\infty,0]$ such that $\int j(s) ds = 1$ and set

$$ f_n(x) = n \int_{-\infty}^0 f(x+y)j(ny)dy $$

Prove that $f_n$ is continuous, convex, converges to $f$ pointwise and $f'_n$ increases to $f'_-$.

Im currently reading Continuous Martingales and Brownian Motion and they use this lemma. The fact that $f_n$ is convex is easy from definition. Im having trouble seeing the rest of the properties, since $f$ isn't necessarily integrable, so i can't use simple mollifier - convolution techniques to use dominated convergence or other teorems (i can't even see it's differentiable).

Any help could be grateful.

答:
Hints for the  proof when $j \geq 0$: $f_n(x)=\int_{-N}^{0} f(x+\frac t n) j(t)dt$ (by a simple change of variable) where $N$ is such that the support of $j$ is contained in $[-N, 0]$. It is quite easy to conclude that $f_n$ is continuous and converges pointwise to $f$.[Remember that $f$ is continuous].

It is well known that $f(x)=f(0)-\int_x ^{0}f_{-}'(t))dt$. Use Fubini's Theorem to conclude that $f_n'(x)=-\int_{-N}^{0} f_{-}'(x+\frac t n)dt$. Since $f'_{-}$ is increasing it follows that $f_n'$ increases to $f'_{-}$.

Title: Convolution with a convex function


问:
How to prove that $ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1$ holds?
Let $p_1, p_2>1$ and $a, b\in\mathbb{R}^+$ such that $a+b\geq2$. Could anyone please help me to prove that
$$ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1?$$
Here $p=\min(p_1, p_2)$. I am trying by using the inequality
$$(a+b)^p\leq 2^{p-1}(a^p + b^p),$$
but I am not able to obtain the desired result. Could anyone please help me or give me a hint?

Thank you in advance!

答:
By assumption $a+b\ge2$. So at least one of these two numbers is larger than $1$. Assume $a\ge1$. Then $a^p\le a^{p_1}$. If $b\ge1$ then $b^p\le b^{p_2}$, if $b\in (0,1)$ then $b^p \le 1$.
This proves
$$
a^p + b^p \le a^{p_1} + b^{p_2} + 1.
$$


答:
You found the partition with the minimal possible such sum.
But whenever you have six consecutive distance-$1$ pairs $\{a,a+1\},\{a+2,a+3\},\ldots,\{a+10,a+11\}$, you can replace them with distance-$6$ pairs $\{a,a+6\},\{a+1,a+7\},\ldots, \{a+5,a+11\}$. This increases the distance sum  for these from $6$ to $36$. You can repeat this process for several groups of consecutive $12$ numbers.

Title: Partitioning $\{1,2,...,1998\}$ such that $|a_i-b_i|$ is $1$ or $6$

Body:
Suppose that the set $\{1,2,3,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}(1\le i \le 999)$ so that for all $i, \mid a_i-b_i\mid$ equals $1$ or $6$. Then the sum $\mid a_1-b_1\mid +\mid a_2-b_2\mid+\cdots +\mid a_{999}-b_{999}\mid$ = ?

I found this problem on AoPS. I found out that the answer is 999 by partitioning the set into $\{1,2\},\{3,4\}, ... , \{1997, 1998\}$, but I am not sure about the proper way of solving it


问:
Why can I not solve this simple simultaneous equation?
$$\frac{B}{A}=\frac{C}{2B}=\frac{A}{4C}$$

Obviously, I assume $A \neq 0, B \neq 0, C \neq 0$.

From the above expression I chose three equations:

$$\frac{B}{A} = \frac{C}{2B}$$
$$\frac{C}{2B} = \frac{A}{4C}$$
$$\frac{B}{A}= \frac{A}{4C}$$

but cannot go further; if I solve them, it leads to an absurd identity (like $2B^2 = 2B^2$, for example) not a useful result.

But why?

答:
Although this looks like three equations in three unknowns, in fact there are only two equations. The third one is just a restatement of the other two. This means that you're not going to be able to find independent values for all three unknowns. Given that fact, decide on which two equations you're going to concentrate (it doesn't matter) and decide which variable you are going to keep as an unknown - then find the other two variables in terms of that variable.

Suppose you choose

$$\frac{B}{A} = \frac{C}{2B}$$
$$\frac{C}{2B} = \frac{A}{4C}$$

and decide to make $A$ your unknown.

Then your equations become

${2B^2}={AC}$ and $2C^2=AB$

Squaring the first gives $4B^4=A^2C^2$

Multiply by 2 gives $8B^4=2A^2C^2$

But we know that $2C^2=AB$ so $8B^4=A^2 \times AB$

$8B^3=A^3$

$B=\frac 12 A$

$\frac {\frac 12 A}{A}=\frac{C}{A}$

$C=\frac 12 A$


答:
From the equation $\dfrac BA = \dfrac C{2B} = \dfrac A{4C}$, we see that these are only the ratios between $A, B$ and $C$. There is no way to find their exact values. For example, if each of $A,B$ and $C$ are doubled, the equation still holds.

However we can still find the exact ratios between them. Observe that:

$$\left(\frac BA\right)^3 = \frac BA \cdot \frac C{2B} \cdot \frac {A}{4C}= \frac18$$

Hence we have $\dfrac BA = \dfrac12 = \dfrac C{2B} = \dfrac A{4C}$. This shows that $A:B:C = 2:1:1$.

Title: Why can I not solve this simple simultaneous equation?

Body:
$$\frac{B}{A}=\frac{C}{2B}=\frac{A}{4C}$$

Obviously, I assume $A \neq 0, B \neq 0, C \neq 0$.

From the above expression I chose three equations:

$$\frac{B}{A} = \frac{C}{2B}$$
$$\frac{C}{2B} = \frac{A}{4C}$$
$$\frac{B}{A}= \frac{A}{4C}$$

but cannot go further; if I solve them, it leads to an absurd identity (like $2B^2 = 2B^2$, for example) not a useful result.

But why?


问:
How to visualize quotient manifold theorem
The quotient manifold says that if a Lie group $G$ acts smoothly, freely and properly on a smooth manifold, then the quotient space is again a smooth manifold with natural topology.

All of the proofs seem complicated and I could not get much insight on the proofs. Though I can check everything. But it does not help much to understand what's going on. Can anyone tell me some natural route to the proof?

答:
Let me assume that $G$ is compact (otherwise proper action takes care of this).

Consider $x\in M$. Because $G$ acts freely the orbit $Gx$ is diffeomorphic to $G$. So all the orbits are copies of $G$. How to parametrize them? Well the tangent space at $x$ splits as the tangent space to the orbits $T_x Gx$ and the normal space $N_x=T_xM/T_x Gx$. Now this normal space parametrizes the orbits closeby the orbit $Gx$. Note that this a euclidean space and that that $\dim N_x=\dim M-\dim G$. Hence it is not too surprising that these can be made into charts of a quotient space, and that the quotient is manifold. The dimension of this quotient manifold is thus $\dim M-\dim G$.

Of course there is a lot to check, but this is the basic idea.


问:
Simplifying $\frac1{a(a-b)(a-c)}+\frac1{b(b-c)(b-a)}+\frac1{c(c-a)(c-b)}$ without opening brackets using interpolation theorems
I have such expression:
$$\frac{1}{a(a - b)(a-c)} + \frac{1}{b(b-c)(b - a)} + \frac{1}{c(c - a)(c-b)}$$

I need to use interpolation in order to simplify it and I basically have no idea how.

答:
With $P(x)=x(x-a)(x-b)(x-c)$, you can use a decomposition of $1/P(x)$ in simple fractions :
$$\frac{1}{P(x)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x}$$
where $A,B,C$ are the terms in your expression (for example, to find $A$, you multiply by $x-a$, simplify and take $x=a$), and $D=-1/abc$. Then multiply by $x$ and let $x\to \infty$ and we are done.


问:
Predicate logic from formula to English
I study for an exam and I have this exercise that I've practiced

Let: Monkey(x) = "x is a monkey"

Likes(x, y) = "x likes y"

∃x (Monkey(x) ∧ ∀y(Monkey(y) ∧ x ̸= y ⇒ Likes(y, x)))

My response:

There exists someone to like all monkeys, but not all monkeys like every one.

I could use some feedback, please.

答:
Let's write

A=Monkey(x)

B=∀y(Monkey(y) ∧ x ̸= y ⇒ Likes(y, x))

We look at "∃x(A and B)". So there exists x such that A and B both holds.

By A, x is a monkey.

By B, we know the following fact: "if y is a monkey and y is not x, then y likes x".

As a conclusion, we obtain that:
"there is a monkey (let's call him x) such that every other monkey likes him (i.e. any monkey that is not x likes x) . "

Title: Predicate logic from formula to English


答:
Propositional logic and its variable cousain, the predicate logic is not able to model all predicates in natural language, including that of English.

This is one motivation for higher order logic. And in fact, this is a form of set theory with one inaccessible cardinal. But as a language, it is more expressible.

Title: Predicate logic from formula to English

Body:
I study for an exam and I have this exercise that I've practiced

Let: Monkey(x) = "x is a monkey"

Likes(x, y) = "x likes y"

∃x (Monkey(x) ∧ ∀y(Monkey(y) ∧ x ̸= y ⇒ Likes(y, x)))

My response:

There exists someone to like all monkeys, but not all monkeys like every one.

I could use some feedback, please.


问:
Will differentiation always give us the maxima and minima
If a given a curve with a maxima and a minima will differentiation always give us the maxima and minima values ?

Is it possible that the curve will have the critical points and derivatives won't give us those values ?

Sorry if this seems to simple ,I am a newbie at calculus.

Edit : my question does not seem to be clear ," if a function has a maxima and minima will differentiation ALWAYS give us the critical points .

答:
The necessary condition for the existence of a local max or min at $x=a$ is that either $f'(a)$ does not exist or it it exists then $f'(a)=0$. Even if $f'(a)=0$ exists you may not get the explicit value of $a$.

For example $f(x)=x^4-x^3+2x+1 \implies f'(x)=4x^3-3x^2+2,$ but $4x^3+3x^2-1=0$ is not solvable by hand however its real root is at $x=-0.607007...$ which is obtainable numerically. In principle, $f(x)$ has one local min at $x=-0.607007...$

In books and in exams when you are asked to find max/min of $f(x)$, the Authors and Examiners see to it that $f'(x)=0$ is solvable by hand.


问:
Representations of Primitive roots
Start by letting $p$ be an odd prime and let $\Pi_{i=1}^{k}$ $q_{i}^{\alpha_i}$ be the canonical factorization of $p-1$.
The goal is to show that every primitive root $g$ (mod $p$) has a representation in the form

$g\equiv \Pi_{i=1}^{k}$ $a_i$ (mod $p$), with $ord_{p}$ $a_i=q_{i}^{\alpha_i}$.

I was given two hints to this, each being a different method of approach.

Count the residues of the form $\Pi_{i=1}^{k}$ $a_i$ (mod $p$), with $ord_p$$a_i=q_i^{\alpha_i}$.

Let $Q_i=(p-1)/q_{i}^{\alpha_i}$ and pick $r_i$ satisfying $r_iQ_i \equiv 1$ (mod $q_i^{\alpha_i}$).  Then examine $g^{\Sigma_{i} r_iQ_i}$ (mod $p$).

The first one has to do with the idea of using a one-to-one correspondence and counting, whereas, the second one uses the idea of "divide and conquer" to solve this.  I cannot figure out either of them, especially when considering hint 1.  How do we even know that there exists residues of order $q_i^{\alpha_i}$ in the residue class of $p$?  Any light to help me understand?

答:
How do we know there are units of order $q_i^{\alpha_i}$ in the integers mod $p$? Why, hint #2 tells you how to explicitly construct them: given $Q_i=(p-1)q_i^{-\alpha_i}$ and $r_iQ_i\equiv1\bmod q_i^{\alpha_i}$, try to compute the order of $a_i=g^{r_{\large i} Q_{\large i}}$. Observe that $a_i^{q_{\large i}^{\alpha_{\large i}-1}}=g^{r_{\large i}(p-1)/q_{\large i}}\ne1$ since $q_i\nmid r_i$ but $a_i^{q_{\large i}^{\alpha_{\large i}}}=g^{r_{\large i}(p-1)}=1$.

Now use Sun-Ze (aka Chinese Remainder Theorem) to see that

$$r_iQ_i\equiv1 \bmod q_i^{\alpha_i}\implies \prod a_i=g^{\sum r_{\large i}Q_{\large i}}\equiv g.$$

Here is a structure-theoretic approach relying on basic finite abelian group theory.

The group of units mod $p$ is cyclic of order $p-1$. By SZ,

$$\frac{\bf Z}{(p-1){\bf Z}}\cong \frac{\bf Z}{q_1^{\alpha_1}{\bf Z}}\times\cdots\times\frac{\bf Z}{q_k^{\alpha_k}{\bf Z}}.$$

Group theory exercise: if $n=w_1\cdots w_k$ with $w_i$ pairwise coprime, then $g\in{\bf Z}/n{\bf Z}$ is a generator if and only if each $g_i$ is a generator for ${\bf Z}/w_i{\bf Z}$, where $g\leftrightarrow(g_1,\cdots,g_k)$ in accordance with the isomorphism ${\bf Z}/n{\bf Z}\cong{\bf Z}/w_1{\bf Z}\times\cdots\times{\bf Z}/w_k{\bf Z}$.

Now, $(g_1,\cdots,g_k)=(g_1,0,\cdots,0)(0,g_2,0,\cdots)\cdots(0,\cdots,0,g_k)$ is a decomposition of $g$ into a product of elements of orders $w_i$ as $i$ varies $1$ through $k$. So pull these $(0,\cdots,0,g_i,0,\cdots,0)$ elements back to the cyclic group ${\bf Z}/(p-1){\bf Z}$ and then back to the multiplicative group $({\bf Z}/p{\bf Z})^\times$.

The hints you have point toward an explicit construction of the elements that implicitly mirrors this structure-theoretic approach.

Title: Representations of Primitive roots


问:
Imaginary part of expression involving complete elliptic integral
Suppose an expression
$$
\tag 1 G(z) = \frac{1}{\sqrt{(z-1)^{3}(z+3)}}\text{K}\left( \sqrt{\frac{16z}{(z+3)(z-1)^{3}}}\right),
$$
where
$$
K(x) \equiv \int \limits_{0}^{\frac{\pi}{2}}\frac{dy}{\sqrt{1 - x^2\sin^{2}(y)}}
$$
is complete elliptic integral of the first kind.

My task: I need to compute the imaginary part of $(1)$ for $0<z<3$.

What's the main challenge of the task? The region $0<z<3$ is divided on two sub-regions $0<z<1$ and $1<z<3$ because of a singularity in the factor $\frac{1}{\sqrt{(z-1)^{3}}}$, and it seems that I need to compute imaginary parts for these regions separately. Let's discuss behavior of the imaginary part of $(1)$ for these two regions.

In the region $0 < z < 1$ the prefactor in the front of $K(...)$ is purely imaginary, while $K(...)$ is real. In the region $1 < z < 3$ the prefactor is real, but the elliptic integral contains imaginary part, since its argument is real and larger than one. Therefore, naively the functional form of imaginary parts in two regions $0<z<1$ and $1<z<3$ must be different (at least I think so): they come from different functions.

It's possible to compute the imaginary parts in two regions analytically by using functional identities involving elliptic integrals. Surprisingly it turns out (after using one of identities) that functional dependences of imaginary parts in different regions are the same:
$$
\text{Im }G(z) = -\frac{1}{2}\frac{1}{2\sqrt{z} + \frac{1}{2}\sqrt{(3-z)(z+1)^3}}\text{K}\left( \frac{2\left(z(3-z)(z+1)^{3}\right)^{\frac{1}{4}}}{2\sqrt{z}+\frac{1}{2}\sqrt{(3-z)(z+1)^{3}}}\right), \quad 0 < z < 3
$$
I can show this step-by-step, by evaluating the imaginary parts in the region $0<z<1$ and in the region $1<z<3$ and applying definite elliptic integrals functional identity to obtained relations.

But is it possible to calculate the imaginary part directly from $(1)$, without intermediate step of dividing the region $0<z<3$ onto $0<z<1$ and $1<z<3$?

An edit

One can think about the problem in another way. Precisely, after few manipulations the expression $(1)$ can be converted to the form
$$
G(z) = \frac{i}{\sqrt{(3-z)(z+1)^{3}}}\text{K}\left( \sqrt{\frac{16z}{(3-z)(z+1)^{3}}}\right)
$$
Now the prefactor is always purely imaginary, and
$$
\text{Im }G(z) =  \frac{1}{\sqrt{(3-z)(z+1)^{3}}}\text{Re }\text{K}\left( \sqrt{\frac{16z}{(3-z)(z+1)^{3}}}\right)
$$
How to derive $\text{Re}[\text{K}(...)]$ without dividing the region onto two regions? It's obvious that the imaginary part is continuous function of $(...)$ except the singularity $z = 1$, where $(...) = 1$.

答:
All elliptic integrals in this answer follow Mathematica/mpmath convention; their last argument is $m$, the parameter.

In your original expression
$$G(z)=\frac1{\sqrt{(z-1)^3(z+3)}}K\left(\frac{16z}{(z-1)^3(z+3)}\right)$$
the parameter of $K$ is in $(1,\infty)$ for $z\in(0,3)$. Now apply the reciprocal modulus transformation and decompose the resulting incomplete integral of the form $F\left(\sin^{-1}\frac1{\sqrt m},m\right)$ into real and imaginary parts (Byrd and Friedman 162.02, DLMF 19.7.3):
$$K(m)=\frac1{\sqrt m}\left(K\left(\frac1m\right)-iK\left(1-\frac1m\right)\right)$$
This gives
$$\operatorname{Im}(G(z))=-\frac1{4\sqrt z}\operatorname{Re}\left(K\left(\frac{(3-z)(z+1)^3}{16z}\right)\right)$$
The parameter in this new expression is in $(0,\infty)$ for $z\in(0,3)$. Apply an ascending Gauss transformation (B&F 164.02):
$$\operatorname{Re}(K(m))=\frac{1+h}2K(1-h^2),\quad h=\frac{1-\sqrt m}{1+\sqrt m}$$
For the specific $m$ in the expression for $\operatorname{Im}(G(z))$ above
$$h^*=\frac{1-\sqrt{\frac{(3-z)(z+1)^3}{16z}}}{1+\sqrt{\frac{(3-z)(z+1)^3}{16z}}}=
\frac{4\sqrt z-\sqrt{(3-z)(z+1)^3}}{4\sqrt z+\sqrt{(3-z)(z+1)^3}}$$
Returning to the actual problem at hand, we get
$$\operatorname{Im}(G(z))=-\frac{1+h^*}{8\sqrt z}K(1-(h^*)^2)$$
with the parameter now in $(0,1)$, so we now have a valid and fully real expression for $\operatorname{Im}(G(z))$. Simplifying gives the result you found:
$$\boxed{\operatorname{Im}(G(z))=-\frac1{4\sqrt z+\sqrt{(3-z)(z+1)^3}}K\left(\frac{16\sqrt{z(3-z)(z+1)^3}}{(4\sqrt z+\sqrt{(3-z)(z+1)^3})^2}\right)}$$
At no point in this answer did I split the domain of $z$ into parts.


问:
Prove function $f:[a, b] \rightarrow \mathbb{R}$ is Riemann integrable if and only if $f$ is bounded and almost everywhere continuous.
I wonder how can we prove the backward direction (function $f:[a, b] \rightarrow \mathbb{R}$ is Riemann integrable, then $f$ is bounded and almost everywhere continuous.) without using the measure theory.

My attempt:

If $f\in R(\alpha)$ on [a,b], there exists a partition $U(P,f,\alpha)-L(P,f,\alpha)<\epsilon$. Consider an arbitrary partition $P: {a=x_0<x_1<\ldots<x_n=b}$.

By theorem 6.7(Rudin), If $s,t$ are arbitrary points in $[x_{i-1},x_i]$, then $\sum_{i=1}^n|f(s)-f(t)|\Delta\alpha_i<\epsilon$.

Since $U(P,f,\alpha)-L(P,f,\alpha)=\sum_{i=1}^n(M_i-m_i)\Delta\alpha_i<\epsilon$ by definition and $f$ is bounded, $M_i-m_i\leq\eta$ where $\eta>0$. Thus, given $\eta > 0,\exists \delta>0$  s.t. when $|s-t|<\delta$, $|f(s)-f(t)|<\eta.$ Thus, $f$ is continuous on each $[x_{i-1},x_i]$.

I felt the part "$M_i-m_i\leq\epsilon$ where $\epsilon>0$. Thus, given $\epsilon > 0,\exists \delta>0$" weird and not sure if what I'm doing is correct. Any help appreciated!

Edit: --------------------------

The original problem is : Let $a_{1}, a_{2}, \ldots$ be a strictly increasing sequence in $(a, b],$ and let $p_{n}>0$ be such that $\sum_{n=1}^{\infty} p_{n}=1$ Define $\alpha:[a, b] \rightarrow \mathbb{R}$ as follows: $\alpha(x)=0$ if $a \leq x<a_{1}$
$$
\alpha(x)=\sum_{k=1}^{n} p_{k} \text { if } a_{n} \leq x<a_{n+1}
$$
and $\alpha(x)=1$ if $\sup _{n} a_{n} \leq x \leq b$

I need to Show that a bounded function $f:[a, b] \rightarrow \mathbb{R}$ is Riemann-Stieltjes integrable with respect to $\alpha$ if and only if $f\left(a_{n}^{-}\right)=\lim _{x \rightarrow a_{n}^{-}} f(x)$ exists and $f\left(a_{n}^{-}\right)=f\left(a_{n}\right)$ for all $n,$ and that in this case
$$
\int_{a}^{b} f d \alpha=\sum_{n=1}^{\infty} f\left(a_{n}\right) p_{n}
$$

I'm stuck on proving "$f:[a, b] \rightarrow \mathbb{R}$ is Riemann-Stieltjes integrable, then it's left continuous" .

答:
This is an expansion of my comment.

Let's prove this lemma first (taken from Apostol's Mathematical Analysis, theorem 7.29, page 160).

Lemma: Assume $\alpha:[a, b] \to\mathbb {R} $ is increasing on $[a, b] $ and $c\in(a, b) $. Let both the functions $f:[a, b] \to\mathbb{R} $ and $\alpha$ be discontinuous from left at $c$. Then the Riemann-Stieltjes integral $\int_a^b f\, d\alpha$ does not exist.

By the given conditions there is an $\epsilon >0$ such that every neighborhood of type $[c-\delta, c] $ contains points $x, y$ such that $$|f(x) - f(c) |\geq\epsilon, \alpha(c) - \alpha(y) \geq \epsilon $$

Now given any partition $P$ of $[a, b] $ with norm however small we can choose another partition $P'= P\cup \{c, y\} $ whose norm does not exceed that of $P$ and the interval $[y, c] $ contribution to the difference $U(P', f, \alpha) - L(P', f, \alpha) $ is not less than $$\epsilon(\alpha(c) - \alpha(y)) \geq \epsilon^2$$ It follows that the upper and lower Darboux sums can't tend to same limit and hence the integral does not exist.

The function $\alpha $ in your question is increasing and the lemma above applies to it. It is discontinuous from left at each $a_n$ and hence in order for the integral $\int_a^b f\, d\alpha $ to exist the function $f$ must be continuous from left at each $a_n$.

The value of the integral can be found by letting $c=\sup a_n$ and noting that $$\int_{a} ^{a_n} f\, d\alpha=\sum_{k=1}^{n-1}f(a_k)p_k$$ (because $\alpha$ is a step function on $[a, a_n] $) and the integral on $[c, b] $ is $0$. The integral thus equals $$\int_{a} ^{a_n} f\, d\alpha+\int_{a_n} ^{c} f\, d\alpha$$ The first term tends to $\sum_{k=1}^{\infty} f(a_k) p_k$ and second term tends to $0$.

Note: In case anyone is worried about convergence of $\sum f(a_n) p_n$, it is absolutely convergent as $f$ is bounded $p_n>0$ and $\sum p_n=1$.


问:
Mean of unormalized Gamma distribution
If $X$ follows a gamma distribution with parameters $\alpha$ and $\beta$ then:

$ p(x|\alpha,\beta) = \frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}$
with $\Gamma(\alpha) = (\alpha - 1)!$

Then, $mean(x) = \frac{\alpha}{\beta}$

But, if for $x$ we have this probability density function instead:

$
p(x|\alpha,\beta) = K \times \beta^{\alpha}x^{\alpha-1}e^{-\beta x}
$ with no $\Gamma(\alpha)$ and $K$ a constant value

What would be the $mean$ value of $x$ in this case ?

Best regards

Aymeric

答:
Simply your density (that is a posterior density) is not exactly a density. The correct one is

$$p(\mathbf{x}|a,b)\propto b^a x^{a-1}e^{-xb}$$

The constant must be calculated in order to let $p(\mathbf{x}|a,b)$ be a nice density.

This constant $K$ can be calculated using Bayes Theorem (integrating the denominator of Bayes Theorem ) or simply recognizing in your posterior the kernel of a Gamma distribution, thus

$$K=\frac{1}{\Gamma(a)}$$

and your expectation is exactly what you showed, $\frac{a}{b}$

Title: Mean of unormalized Gamma distribution


问:
What happens if two functions have the same derivative on an interval?
Suppose that $ f $ and $ g $ are differentiable functions on the interval $ [a, b] $ such that $ f^\prime (x) = g^\prime (x) $ for all $ x \in [a, b]$.

How can I prove that " f = g modulates a constant ", that is, that there is a constant $ k $ such that $ f (x) = f (x) + k $ for all $ x \in [a, b] $. Considering the function $ f - g $ and applying the mean value theorem.

I appreciate the help you can give me.

答:
Incase you want it without using the mean value theorem:

let $h(x) = f(x) - g(x)$

Differentiating both sides, $h^\prime(x) = f^\prime(x) - g^\prime(x)$

Since $f^\prime(x) = g^\prime(x)$, we can say $h^\prime(x) = 0$

Therefore, $h(x) = \int h^\prime(x)dx = $ a constant ($k$).

So $f(x) - g(x) = k$ and $f(x) = g(x) + k$


答:
Note that $$\int_{0}^{\infty} \frac{\beta^{\alpha}e^{-\beta x}}{\Gamma(\alpha)} x^{\alpha-1} dx=1$$ If you use
$\int_{0}^{\infty} x^{m} e^{-kx}dx=k^{-m-1} \Gamma(m+1).$
Next the expatation value of $x$ is
$$E(x)=\int_{0}^{\infty} \frac{\beta^{\alpha}e^{-\beta x}}{\Gamma(\alpha)} x^{\alpha} dx=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\beta^{-\alpha-1} \Gamma(\alpha+1)=\frac{\alpha}{\beta}.$$

Title: Mean of unormalized Gamma distribution

Body:
If $X$ follows a gamma distribution with parameters $\alpha$ and $\beta$ then:

$ p(x|\alpha,\beta) = \frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}$
with $\Gamma(\alpha) = (\alpha - 1)!$

Then, $mean(x) = \frac{\alpha}{\beta}$

But, if for $x$ we have this probability density function instead:

$
p(x|\alpha,\beta) = K \times \beta^{\alpha}x^{\alpha-1}e^{-\beta x}
$ with no $\Gamma(\alpha)$ and $K$ a constant value

What would be the $mean$ value of $x$ in this case ?

Best regards

Aymeric


答:
You have a misconception about probability densities.  If $f_X(x)$ is a density for some real-valued random variable $X$, then we require $$\int_{x = -\infty}^\infty f_X(x) \, dx = 1.$$  Equivalently, if we wish to avoid using indicator functions, we can write the above as an integral over the support $\Omega$ of $X$:  $$\int_{x \in \Omega} f_X(x) \, dx = 1.$$  In the case of a gamma distribution, the support is $X \in [0, \infty)$.

As such, you are not free to choose any value of $K$ in the function $K \beta^\alpha x^{\alpha-1} e^{-\beta x}$.  You are constrained because when integrated on $[0,\infty)$, the result must be $1$, otherwise it is not a density.  This forces $K = 1/\Gamma(\alpha)$.

In fact, if we remove all of the factors that are independent of $x$, i.e. we consider the kernel of the gamma density $$x^{\alpha-1} e^{-\beta x},$$ the unique constant $K$ that makes this kernel a density function is precisely $$K = \frac{\beta^\alpha}{\Gamma(\alpha)}.$$

Title: Mean of unormalized Gamma distribution


问:
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

答:
From Stirling number's of the second kind,

$$ x^n = \sum_{k=0}^n S(n,k) \frac{x!}{(x-k)!}$$

Sum both sides over $x$

$$ \sum_{x=0}^j x^n = \sum_{k=0}^n \sum_{x=0}^j S(n,k) \frac{x!}{(x-k)!}$$

Or,

$$ \sum_{x=0}^j x^n = \sum_{k=0}^n k!  S(n,k)  \sum_{x=0}^j\binom{x}{k}$$

Now use hockey stick identty,

$$ \sum_{x=0}^j \binom{x}{k} =\binom{n+1}{i+1}$$

Hence,

$$ \sum_{x=0}^j x^n = \sum_{k=0}^n k!  S(n,k)  \binom{n+1}{i+1}$$

Q.E.D


问:
Does this ODE have an exact or well-established approximate analytical solution?
The equation looks like this:
$$\frac{\mathrm{d}y}{\mathrm{d}t} = A + B\sin\omega t - C y^n,$$
where $A$, $B$, $C$ are positive constants, and $n\ge1$ is an integer.  Actually I am mainly concerned with the $n=4$ case.

The $n=1$ case is trivial.  Otherwise the only method I can think of is kind of an iterative approximation, in which a "clean" expression seems not easy to obtain.

Just in case it is too "easy" for the experts, we may generalize it to the form
$$\frac{\mathrm{d}y}{\mathrm{d}t} = f(t) - g(y).$$

答:
If $\frac{dy}{dt} = F(t,y)$ has $F$ and $\partial_2F$ continuous on a rectangle then there exists a unique local solution at any particular point within the rectangle. This is in the basic texts. For $F(t,y) = f(t)-g(y)$ continuity requires continuity of $f$ and $g$ whereas continuity of the partial derivative of $F$ with respect to $y$ require continuity of $g'$. Thus continuity of $f$ and continuous differentiability of $g$ suffice to indicate the existence of a local solution. Of course, this is not an analytic expression just yet. That said, the proof of this theorem provides an iteratively generated approximation.

Alternatively, you could argue by Pfaff's theorem there exists an integrating factor which recasts the problem as an exact equation. So, you can rewrite $dy = [f(t)-g(y)]dt$ as
$Idy+I[f(t)-g(y)]dt=dG$ for some function $G$. Then the solution is simply $G(t,y)=k$ which locally provides functions which solve your given DEqn. Of course, the devil is in the detail of how to calculate $I$. The answer for most of us is magic.

Probably a better answer to your problem is to look at the Bernoulli problem. There a substitution is made which handles problems which look an awful lot like the one you state.


问:
Show that the polynomial ring $F[X]$ is not a free product in the category of associative algebra.
I want to show that

If $F$ is a field and $X$ is a set with $|X|>1$, then $F[X]$ is not a free object in the category of associative $F-$ algebras.

Now here is my attempt. Apparently, I got stuck so I am not sure how to proceed from what have. I never prove an object is not a free object.

Assume on the contrary that $F[X]$ is a free object. Let $A$ be an (noncommutative) associative $R-$algebra. Then since $F[X]$ is a free object, then there exists an associative algebra homomorphism $\gamma$ making the diagram commutates. Since the set function $i \colon X \to F(X)$ is unknown, it is hard for me to construct the set function $g \colon X \to A$ in order to get a contradiction. Also, I don't know many examples of noncommutative associative algebras $A$. I think it will be nice to construct a specific $A$. I think the idea is to contradict the idea that $A$ is noncommutative. But again, I got very stuck. I haven't thought about the second part of the questions yet. Any hint will be great. Thanks!

答:
So first of all, I want to clear up the definition a bit. You said:

Since the set function $i:X \longrightarrow F[X]$ is unknown, it is hard for me to construct the set function $g:X \longrightarrow A$ in order to get a contradiction.

The function $i$ is actually not unknown. The issue here is a slight abuse of notation in the problem statement. You are asked to show that $F[X]$ is not free on $X$ in this category, but this is not entirely accurate. Look back at your definition of a free object. It would really be more correct to say that $i: X \longrightarrow F$ is free, rather than to say $F$ is free. The map $i$ is part of the data here, as it does not make sense to say that an object in isolation is free. A more precise statement of the problem would be to show that $i: X \longrightarrow F[X]$ is not free, where $i(x) = x$.

To belabor this point a bit more, let's instead consider $F[X]$ in the category of commutative $F$-algebras. Here we do have that $i: X \longrightarrow F[X]$ as given is free. However, if I chose a different map $j: X \longrightarrow F[X]$ I could contradict this. For instance, I could have some $j(x_0) = 0$. Then I could consider a map $X \longrightarrow F$ sending $x_0 \mapsto 1$. By universal property, we would then need a map $F[X] \longrightarrow F$ sending $0 \mapsto 1$, which is impossible. The point of this example was to emphasize that we really do need to have the map $i$ as part of the initial data. We cannot simply take an object and ask if it's free.

Anyways, back to the problem at hand. Let $F\langle X\rangle$ denote the ring of noncommutative polynomials in $X$ with coefficients in $F$. That is, an element of this ring is a formal $F$-linear combination of terms like $x_1^{e_1} \cdots x_n^{e_n}$. We are not assuming that the $x_i$ commute. So in particular, take $x_0 \neq x_1$ in $X$, as we assumed $|X| > 1$. Then $x_0x_1 \neq x_1x_0$ in $F\langle X\rangle$.

Let's consider now the map $g: X \longrightarrow F\langle X\rangle$ via $g(x) = x$. Then if $F[X]$ (meaning $i: X \longrightarrow F[X]$) was free, we would have a map $\overline g: F[X] \longrightarrow F\langle X\rangle$ such that $\overline g \circ i = g$. Thus, $\overline{g}(x_0x_1) = x_0 x_1 \in F\langle X\rangle$. However, in $F[X]$ the variables commute so $x_1x_0=\overline{g}(x_1x_0) = \overline{g}(x_0x_1) = x_0 x_1$, a contradiction. Thus, $i: X \longrightarrow F[X]$ is not free on $X$ in this category.


问:
Using the four color theorem to check if a graph is Planar
I've just learned about the four color theorem. So my question is: "Given an undirected graph if I am able to color this undirected graph with 1,2,3 or at most colors 4 colors such that no two vertices with the same colors are adjacent, can I by the four color theorem conclude that the graph is a planar graph"?</p>

答: The four-colour theorem only stats that if a graph is planar then it can be coloured by at most four colours. It does not state the converse, and indeed $K_{3,3}$ is bipartite – can be coloured with only two colours – but is not planar.


问:
Show $N(0, a^2) \cdot c = N(0, a^2c^2)$
Let $\xi$ is a continuous random variable such that $\xi$ ~ $N(a,b)$ normal distribution. How to show that
$$N(0, a^2) \cdot c = N(0, a^2c^2)$$

I found $$\mathbb E\xi_{N(0, a^2)} = \mathbb E\xi_{N(0, a^2c^2)} = 1$$
and $$D\xi_{N(0,a^2)} = a^4, D\xi_{N(0,a^2c^2)} = c^4a^4$$

What does this show? That is, it is clear that the coefficient $c$ must be introduced into the distribution to the second degree, but this is not visible here. How else can you prove equality?

答:
Your choice of notation is clumsy, which is why you are confused.

it is sloppy and misleading to write something like $$N(0, a^2) c = N(0, a^2 c^2).$$  Instead, we should define random variables, and then perform computations on those random variables.  Specifically, suppose $$X \sim \operatorname{Normal}(\mu = 0, \sigma^2 = a^2),$$ that is to say, $X$ is normally distributed with mean $\mu = 0$ and variance $\sigma^2 = a^2$.  Thus $X$ has probability density $$f_X(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/(2\sigma^2)} = \frac{1}{\sqrt{2\pi} a} e^{-x^2/(2a^2)}, \quad -\infty < x < \infty,$$ and cumulative distribution function $$F_X(x) = \Pr[X \le x] = \Phi(x) = \int_{t = -\infty}^x f_X(t) \, dt = \int_{t = -\infty}^x \frac{1}{\sqrt{2\pi} a} e^{-t^2/(2a^2)} \, dt.$$

Now let $Y = cX$ with $c \ne 0$.  Then $$F_Y(y) = \Pr[Y \le y] = \Pr[cX \le y] = \Pr[X \le y/c] = F_X(y/c),$$ if $c > 0$, and if $c < 0$, we have $$F_Y(y) = \Pr[cX \le y] = \Pr[X \ge y/c] = 1 - F_X(y/c).$$  In both cases, differentiating with respect to $y$ yields $$f_Y(y) = \frac{d}{dy}\left[F_X(y/c)\right] = f_X(y/c) \cdot \frac{1}{c} = \frac{1}{\sqrt{2\pi} a c} e^{-(y/c)^2/(2a^2)} = \frac{1}{\sqrt{2\pi} (ac)} e^{-y^2/(2(ac)^2)}.$$  Thus $Y$ is normally distributed with mean $0$ and variance $(ac)^2$.

All of this is a special case of the more general location-scale transformation $$Y = c(X + d) \implies Y \sim \operatorname{Normal}(\mu + d, \sigma^2 c^2).$$  This is also the basis of standardizing normal random variables to have zero mean and unit variance, via the choice $d = -\mu$, $c = 1/\sigma$, thus if $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ in general, the transformation $$Y = \frac{X - \mu}{\sigma}$$ makes $Y$ a standard normal random variable with zero mean and unit variance.

Title: Show $N(0, a^2) \cdot c = N(0, a^2c^2)$


问:
How do find if (1,2) lies in between the acute or obtuse region of two lines .
The two lines are $\sqrt{3}x-y+5=0$ and $\sqrt{3}x+y-1=0$.

How will I find in which region the point lies: the acute region or the obtuse region?

I feel this can't be done if anyone could suggest a method that would be great

答:
It can be done, and quite easily!

Now for any point $P(x_1, y_1)$ , if I put it into the formula of the line $L: ax+by+c$, I get three outcomes

$ax_1 + by_1 + c < 0$ - the point is in the lower half space created by the line

$ax_1 + by_1 + c = 0$ - the point is on the line

$ax_1 + by_1 + c > 0$ - the point is in the upper half space created by the line

Now the primary angle between the two lines is $60^0$ (based on comparing slopes)

Therefore any point which satisfies the below is in the "obtuse" portion

$$(\sqrt3 x-y+5)(\sqrt3 +y-1) > 0$$


问:
Determine the domain, codomain and image of a linear map

Let $$A= \begin{bmatrix} 1 && 0 && 2 && 0 \\ 0 && 3 && 0 && 5 \\ 5 && 0 &&0 && -6 \end{bmatrix}$$ determine the domain, codomain and image of the linear map $L$ corresponding to $A$.

From $T(x) = Ax$ I get that $$\begin{bmatrix} 1 && 0 && 2 && 0 \\ 0 && 3 && 0 && 5 \\ 5 && 0 &&0 && -6 \end{bmatrix} \cdot \begin{bmatrix} x \\y\\z\\w \end{bmatrix} = \begin{bmatrix} x+2z+6w \\3y+5w \\5x-6w \end{bmatrix}$$

So $T(x,y,z,w) = (x+2z+6w, 3y+5w, 5x-6w)$

The domain seems to be $\mathbb{R}^4$, but how can I determine the codomain and image of a linear map? Also I would probably need to verity that $T(x,y,z,w)$ has the properties $T(u+v) = T(u) + T(v)$ and $T(\alpha u) = \alpha T(u)$ or is this true immediately from the matrix representation?

答:
We have $T: \mathbb R^4 \to \mathbb R^3$ and $T(x)=Ax.$ Yes, $T$ is linear.

It is easy to see that $A$ has rank $3.$  Thus $T( \mathbb R^4)= \mathbb R^3.$


问:
Composite Trapezoid Rule Error Function
The function $erf\left(t\right)\:=\:\frac{2}{\sqrt{\pi }}\int _0^t\:e^{-x^2}dx$ is called the error function in statistics. If we use the composite trapezoid rule to compute $erf\left(2\right)$, how many points are needed to ensure that the absolute error is not larger than $10^{-5}$?

I understand that the error with the composite trapezoid rule is given by $-\frac{1}{12}\left(b-a\right)h^2f''\left(z\right)$. Here, $a$ and $b$ would be the two endpoints of the integration and $h$ would be $\frac{a-b}{\:n}$, where $n$ is the number of subintervals. I understand I am supposed to figure out the value of $n$ for which the error formula above is below $10^{-5}$.

However, I am quite confused what the values of $a$, $b$, $h$, and $z$ are supposed to be in this specific case.

Any help would be greatly appreciated!

答:
You have $a=0$, $b=2$, $h=\frac{b-a}n=\frac2n$ and $z$ is an unknown midpoint. You need to take the worst possible case into account, that is, use $M=\max_{z\in[0,2]}|f''(z)|$. $n$ is the number of segments, so $n+1$ is the number of points.

For a first estimate you can also approximate the error as $\frac{|f'(b)-f'(a)|}{12n^2}$. The next term in the error expansion is of order $n^{-4}$.

Title: Composite Trapezoid Rule Error Function


问:
Complete elliptic integral $K(k) $ for $k>1$
I have always tried to work with elliptic integrals with modulus $k\in(0,1)$ to avoid the issues related to complex variables.

In what follows I have tried to link the integral of modulus greater than $1$ with those of modulus less than $1$.

Let $k>1$ and consider $$K(k) =\int_{0}^{1}\frac{dx}{\sqrt {(1-x^2)(1-k^2x^2)}} $$ Splitting the range of integration into $[0,1/k]$ and $[1/k,1]$ we get $$K(k) =\frac{1}{k}K\left(\frac{1}{k}\right)-i\int_{1/k}^{1}\frac{dx}{\sqrt{(1-x^2)(k^2x^2-1)}}\tag{1}$$ and let $x=1/\sqrt{1-k'^2y^2}$ then we have $$dx=\frac{k'^2y\,dy}{(1-k'^2y^2)^{3/2}}$$ and $$1-x^2=-\frac{k'^2y^2}{1-k'^2y^2}$$ and $$k^2x^2-1=-\frac {k'^2(1-y^2)}{1-k'^2y^2}$$ and hence we arrive at the relation $$K(k) =\frac{1}{k}\left\{K\left(\frac{1}{k}\right)-ikK\left(k' \right) \right\} $$ Here $k'=i\sqrt{k^2-1}$ is purely imaginary and with some manipulation one can show that $$K(k') =\frac{1}{k}K\left(\frac {\sqrt{k^2-1}}{k}\right)$$ and thus we arrive at $$K(k) =\frac{1}{k}\left\{K\left(\frac {1}{k}\right)-iK\left(\frac{\sqrt{k^2-1}}{k}\right)\right\}$$ where $k>1$. Replacing $k$ with $1/k$ we get the typical representation of the above formula as $$K(1/k)=k(K(k)-iK(k'))\tag{2}$$ where $0<k<1$.

My question is about the choice of $i$ in equation $(1)$. We could equally well have used $-i$ instead of $i$. How does one choose the correct sign of $i$?

The DLMF reference (equation 19.7.2) gives the rule that the sign of $i$ is opposite to that of imaginary part of $k^2$. But that does not help here as $k^2$ is real.

The derivation above was more to confirm that my calculations are correct and one does not need to get bogged down into them.

答:
$K$ has a branch cut on $[1,\infty)$:

(The above plot is actually that of $K(m)$, with the parameter, but $k>1\implies m>1$.)

So either sign of the imaginary part is equally valid. Most implementations of $K$, however, take the principal branch of the square root in the defining integral. This leads to the reciprocal of the principal square root of a negative number, a number of the form $\frac1{ix}=-i\cdot\frac1x$ where $x>0$. Therefore $K$ has negative imaginary part if $k>1$.


问:
Solution of $y''-x^2y=0$ comes out by series?
I tried to solve the differential equation $y''(x)-x^2y(x)=0$ by series solution, but but it doesn't seem right. My terms become zero when they depend on $a_2$ and $a_3$ and I can't find the general form of the recurrence relations. Does anyone know the solution?

With $y=\sum_0a_nx^n$ and $y''=\sum_2a_nn(n-1)x^{(n-2)}$

\begin{eqnarray}
\sum_2a_nn(n-1)x^{(n-2)}-\sum_0a_nx^{n+2}=0 \\
\sum_{-2}a_{n+4}(n+4)(n+3)x^{(n+2)}-\sum_0a_nx^{n+2}=0 \\
2a_2+6a_3x+\sum_{0}[a_{n+4}(n+4)(n+3)-a_n]x^{n+2}=0
\end{eqnarray}

then,

\begin{eqnarray}
a_{n+4}=\frac{a_n}{(n+4)(n+3)}
\end{eqnarray}

答:
Starting from your observation, we first note that $a_2 = a_3 = 0$. Also, if we write $n = 4q+r$ for some $q\geq 0$ and $r\in\{0,1,2,3,\}$, then for $q \geq 1$ we get

$$ a_{4q+r} = \frac{a_{4(q-1)+4}}{(4q+r)(4q+r-1)} = \frac{a_r}{\prod_{k=1}^{q} (4k+r)(4k+r-1)}, $$

where the empty product is regarded as $1$. So,

If $r = 0$, then
$$ a_{4q} = \frac{a_0}{\prod_{k=1}^{q} (4k)(4k-1)} = \frac{a_0}{16^q} \frac{\Gamma(\frac{3}{4})}{q!\Gamma(q+\frac{3}{4})} $$

If $r = 1$, then
$$ a_{4q+1} = \frac{a_r}{\prod_{k=1}^{q} (4k+1)(4k)} = \frac{a_1}{16^q} \frac{\Gamma(\frac{5}{4})}{q!\Gamma(q+\frac{5}{4})}. $$

If $r= 2$ or $r=3$, then we can show that $a_{4q+r} = 0$ for all $q \geq $ by the induction on $q$.

Therefore the general solution is given by

\begin{align*}
y(x)
&= a_0 \Gamma(\tfrac{3}{4}) \sum_{q=0}^{\infty} \frac{1}{q!\Gamma(q+\frac{3}{4})} \biggl( \frac{x}{2} \biggr)^{4q} + 2 a_1 \Gamma(\tfrac{5}{4}) \sum_{q=0}^{\infty} \frac{1}{q!\Gamma(q+\frac{5}{4})} \biggl( \frac{x}{2} \biggr)^{4q+1} \tag{*}
\end{align*}

Those two particular solutions appearing in $\text{(*)}$ can also be written in terms of the modified Bessel function of the first kind:

\begin{gather*}
\sum_{q=0}^{\infty} \frac{1}{q!\Gamma(q+\frac{3}{4})} \biggl( \frac{x}{2} \biggr)^{4q}
= \sqrt{\frac{x}{2}} I_{-\frac{1}{4}}\left(\frac{x^2}{2}\right), \\
\sum_{q=0}^{\infty} \frac{1}{q!\Gamma(q+\frac{5}{4})} \biggl( \frac{x}{2} \biggr)^{4q+1}
= \sqrt{\frac{x}{2}} I_{\frac{1}{4}}\left(\frac{x^2}{2}\right).
\end{gather*}


答:
Differentiation of a differentiable function $f$ always give you stationary points by looking for the points $x$ where $f'(x)=0$.

And amongst these will be inflexion points as well as local minimi and maxima and the global minimal or maximal; that is -  if it has them.

Title: Will differentiation always give us the maxima and minima

Body:
If a given a curve with a maxima and a minima will differentiation always give us the maxima and minima values ?

Is it possible that the curve will have the critical points and derivatives won't give us those values ?

Sorry if this seems to simple ,I am a newbie at calculus.

Edit : my question does not seem to be clear ," if a function has a maxima and minima will differentiation ALWAYS give us the critical points .


问:
showing $\mu(\lim\sup_{n\to\infty} E_n) = 0.$
Let $(X,\mathcal{M},\mu)$ be a measure space. Suppose $E_n\in \mathcal{M}$ such that $$\sum_{n=1}^\infty \mu(E_n) < \infty$$ show $\mu(\lim\sup_{n\to\infty} E_n) = 0.$

Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with $$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$$

Can I prove this by Fatou's Lemma? Thank you

答:
Proof of the first part:

The result you mention is the first Borel-Cantelli Lemma. First recall that
$$\limsup_n E_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k.$$
By continuity from above (check the conditions as an exercise) and sub-additivity, we have
$$\mu\left(\limsup_n E_n\right) = \lim_n\mu\left(\bigcup_{k=n}^\infty E_k\right)\leq \lim_n \left(\sum_{k=n}^\infty \mu(E_k)\right).$$
Note that last term is the tail of a convergent series, hence it converges to zero.

For the second part see Sam Wong's solution.


问:
Recursive expected value when ending state is up to the player
I am wanting to improve my understanding of solving expected value through recursion. I understand how to perform some problems (such as expected number to get X heads in a row).

However, I am lost when the ending condition is up to the player. I think someone going through an example like the one below would be very helpful.

For example, suppose a game where you flip 6 coins and the number of heads determines your payout. If you do not like your flip of 6 coins, then you can pay one dollar and reflip all 6 coins again. You can do this as many times as you like but once you choose to reflip, even if the new result is worse, you cannot go back to a previous flip. In other words, once you choose to reflip, you lose another dollar and “forget all flips before.” Once you decide to take the payout the game ends and you win # of heads on last set of 6 flips - #rerolls * cost to reroll.

Now I know that the expected value of heads is 3. So if I get something less than 3, I may be inclined to reroll, but I am pretty confident that this depends on the cost of a reroll. I can imagine that if it costed 5 dollars to reroll you would only play once.

How can I determine 1) expected value and 2) at what # of heads or less should I reroll?

This is just a specific example I have come across when practicing (I found no solution though). Any other resources for problems like these (with solutions) would be super helpful.

答:
Clearly, your strategy should be a cutoff strategy: reflip for less than $a$ heads, stop for $a$ heads or above. Let $x$ be the expected value of the game. Then the expected value should solve the following equation:

$$x=\tfrac{1}{6}\sum\limits_{i=1}^{a-1} (x-1) + \tfrac{1}{6}\sum\limits_{j=a}^6 j$$

Where the first part is lower results which require reflipping, the right side is the better results for which you start. You can write the $\Sigma$s explicitly, then solve for $x$ and get something of the sort $x=f(a)$. For each $a$ you can compute $x$ and choose the one that maximizes $x$.


问:
Does it exist a way to find one side of a triangle in polygon other than using similarity?
The problem is as follows:

Using the figure from below find $EB$. Assume $\triangle ABC$ is isosceles where $AB=AC$ and $\overline{BN} \parallel \overline{AC}$. Find $EB$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&6\,cm\\
2.&8\,cm\\
3.&10\,cm\\
4.&12\,cm\\
5.&11\,cm\\
\end{array}$

What I attempted to do is sumarized in the figure from below:

As it can be seen I redrawn the figure so it can describe better what it is mentioned in the passage of the problem.

From this part. I relied on triangle similarity.

This would meant:

$\triangle ABE \sim \triangle BED$

Hence:

$\frac{6}{x}=\frac{x}{24}$

Solving this:

$x^2=6\times 24$

Which makes $x=12\,cm$

This later answer does check with the official answer from my book. But since this question is placed before introducing the concept of triangle similarity I am wondering if it could be solved relying using fundamentals of bisector angles and perpendicular bisector segment concepts. Can it be solved only using those?.

As it can be noted, some information is unused which makes me to believe was intentionally put there to use the above mentioned concepts.

All and all, is my solution correct?. I'm not good in this topic thus I need assistance on it. Please, include a drawing in your answer as it could require some construction.

答: How about using $\sin \beta$ from two different triangles (ABE and DBE).


问:
terminology for the derivative of a trajectory in phase space
Suppose we have 2nd order ODE.  For example: $x'' + x = 0.$
We can view this as a first order ODE in two dimensions:

$x' = v$

$v' = -x$

What is the vector $(x', v') = (v, -x)$ called?

答:
When working with a system $\vec{x}'=f(\vec x)$, I call $\vec{x}'$ the velocity vector. And I am not alone (middle of first page). The idea is that $\vec x$ is the position vector of a point in the phase space, so the derivative naturally corresponds to velocity.

This may be awkward when you also think of $v$ as velocity of point at position $x$, as in your example. But you have to decide: do you want to think about second-order equation, or about the equivalent first-order system? If it's the latter, then you forget the original meaning of $x$ and $v$ and think of $(x,v)$ as coordinates of a point, with velocity $(x',v')$.


问:
$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$
I have come across the equation $$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$$. Of course we have $x \in (-\infty, -1) \cup(1, \infty)$. I noticed that $(|x| - \sqrt {x ^ 2 - 1}) ^ x = \frac{1} {(|x| + \sqrt {x ^ 2 - 1}) ^ x}$, but, still, I cannot find a way to write it more easier.
I tried proving the fact that there are 2 functions with different monotonies, but that is not true.

答:
Consider $A:=|x|+\sqrt{x^2-1}$; then we have, $$A^{-1}=\frac{1}{|x|+\sqrt{x^2-1}}=\frac{|x|-\sqrt{x^2-1}}{x^2-(x^2-1)}=|x|-\sqrt{x^2-1}$$

So, your equation is essentially,

$$A^x+A^{-x}=2(2x^2-1)$$

and note that we have $A+A^{-1}=2|x|$ and $A-A^{-1}=2\sqrt{x^2-1}$

Can you take it from here?


问:
Value of lim $\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$
How can I calculate this limit?

$$\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$$

I think it is a $0 \cdot \infty$ case, but I have no other ideas. I looked again at the properties of $\ln$, thinking that it might help me, but I still didn't figure it out.

Thanks.

答: Since $\displaystyle\lim_{x\to0}x\ln(x)=0$, $\displaystyle\lim_{n\to\infty}\frac xn\ln\left(\frac xn\right)=0$.


问:
Proof $(X×Y)-(A×B)$ is connected [Contradiction]
Let $A$ be a proper subset of $X$, and let $B$ is a proper subset of $Y$. If $X$ and $Y$ are connected, show that

$$(X×Y)-(A×B)$$

is connected.

I know there's already a question answering this problem, but as I was working on this problem, I did it from another way and wanted to make sure is right.

Proof Idea:

Suppose by contradiction that $H=(X×Y)-(A×B)$ is not connected. Therefore There exist a separation of H from two opens $U,V$.

$$(X×Y)-(A×B)=U\cup V = U_1\times V_1 \cup U_2 \times V_2$$

Where $U_1,U_2$ are open in $X$ and $V_1,V_2$ are open in Y. This way

$U_1\times V_1 \cap U_2 \times V_2 = (U_1 \cap U_2) \times (V_1 \cap V_2) = \emptyset$. So there are two cases. I will work only on the first case, the second is similar.

$U_1 \cap U_2 = \emptyset$, note that $U_1, U_2 \notin A$, as consequence $U_1 \cap U_2 \cap A = \emptyset$. Consider in particular $U_c=U_1 \cup U_2$. Observe that $U_c \cup A = X$, is a separation of $X$, but $X$ is connected. $\Rightarrow\!\Leftarrow$.

Let me know if there's any mistake, will help me prepare for my exam.

答:
An open set need not be of the form $U_1 \times V_1$, you can only say that $U = \cup_i (U_i \times V_i) - (A \times B)$, $U_i, V_i, i \in I$ are open in $X$, rep. $Y$. This doesn't help you for the proof, really.

A more direct appraoch is to write this difference $(X\times Y)- (A \times B)$ as a union of sets like $\{x\} \times Y, X \times \{y\}$, which are connected subspaces, and in such a way that there are intersections to ensure the connectedness of the union.


问:
Why the dual lagrange function is concave?
When we have a standard minimization problem expressed in the form

\begin{align}
&minimize  \,\, \, f_0(x)\\
&s.t. \, \, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f_i(x)\le0  \qquad i=1, \dots,m\\
&\, \, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, \, h_i(x)=0 \qquad i=1, \dots,p
\end{align}

then we can take account of the constraints by considering a weighted sum of them, considering lagrangian

$$L(x,\lambda,\nu) = f_0(x) + \sum_{i=1}^m\lambda_if_i(x) + \sum_{i=1}^p \nu_ih_i(x) $$

and also define the dual function $g$ as
$$ g(\lambda,\nu) := \inf_{x}L(x,\lambda,\nu)$$
Of course, when the lagrangian is unbounded below in $x$ the dual formulation takes the value $-\infty$.

What I don't understand is that since the dual function is the pointwise infimum of a family of affine functions of $(\lambda,\nu)$it is concave, even when the problem is not convex.

I understand that $g$ pointwise infimum of affine functions, (because they are basically linear functions), but I'm not relating this with the notion of always being concave.

Also my intuition brings me to consider that if $g$ is concave, then we end up with a maximization problem which will take us to the same optimal value of the original minimization problem, maybe under some additional conditions?

Thank you

答:
As mentioned, the function $L(x, \lambda, \nu)$ is affine with respect to $(\lambda, \nu)$. Therefore it is also concave. Hence, since $g(\lambda, \nu)$ is the minimum of concave functions, it is also concave. This is equivalent to the case of maximum of convex functions, which is convex.

Does this make sense? Or you still have a missing piece?


答:
Hint:

Notice that with exponent $2$,

$$(|x| + \sqrt {x ^ 2 - 1}) ^ 2 + (|x| - \sqrt {x ^ 2 - 1}) ^ 2= 2(2x^2 - 1),$$

so $x=2$ is a solution.

And as

$$(|x| + \sqrt {x ^ 2 - 1}) (|x| - \sqrt {x ^ 2 - 1})=1,$$ the function is even, so $-2$ is another solution.

With exponent $1$,

$$2|x|=2(2x^2-1)$$

and $|x|=1$ is also a solution.

Remains to show if they are the only ones. From the first identity, it is obvious that for $x>2$,

$$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x\\>(|x| + \sqrt {x ^ 2 - 1}) ^ 2 + (|x| - \sqrt {x ^ 2 - 1}) ^ 2= 2(2x^2 - 1).$$

Body:
I have come across the equation $$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$$. Of course we have $x \in (-\infty, -1) \cup(1, \infty)$. I noticed that $(|x| - \sqrt {x ^ 2 - 1}) ^ x = \frac{1} {(|x| + \sqrt {x ^ 2 - 1}) ^ x}$, but, still, I cannot find a way to write it more easier.
I tried proving the fact that there are 2 functions with different monotonies, but that is not true.


Title: $(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$

Body:
Hint:

Let $|x|=a\ge0, \sqrt{x^2-1}=b\implies a^2-b^2=1, a^2+b^2=?$

If $x\ge0, x=a$

$$(a+b)^a+(a-b)^a=2(a^2+b^2)=(a+b)^2+(a-b)^2$$

$$\implies (a+b)^a+\dfrac1{(a+b)^a}=(a+b)^2+\dfrac1{(a+b)^2}$$

If $p+\dfrac1p=q+\dfrac1q, p=q$ or $p=\dfrac1q$

What if $x<0?$


答:
Substitute $t:= \frac{n}{x}$, then the limit becomes
$$\lim_{t\to \infty}\frac{\ln\left(\frac{1}{t}\right)}{t} = -\lim_{t\to\infty}\frac{\ln(t)}{t}$$
Now, since $\ln(x) < x$, we have
$$0 \le \frac{\ln(t)}{t} = \frac{2\color{red}{\ln\left(\sqrt t\right)}}{\sqrt t \color{red}{\sqrt t}}\le \frac{2}{\sqrt t}\cdot\color{red}1$$ Therefore,$$
0 \le \frac{\ln(t)}{t} \le \frac{2}{\sqrt t}$$
Finally, by squeeze theorem, we have
$$\lim_{t\to\infty}\frac{\ln(t)}{t} = 0$$
which also means that
$$\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)= 0$$

Title: Value of lim $\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$

Body:
How can I calculate this limit?

$$\lim_{n\to\infty}\frac xn \ln\left(\frac{x}{n} \right)$$

I think it is a $0 \cdot \infty$ case, but I have no other ideas. I looked again at the properties of $\ln$, thinking that it might help me, but I still didn't figure it out.

Thanks.