问:
Lyapunov stability dynamic systems
$\dot{x}=-(x-1)\cdot(x-2)^2$

I want to find the stability with the 2 Lyapunov methods ( linearization and appropriate Lyapunov function).
I solved similar exercises with the first method of linearization but I don't understand how to solve this because is not a system of two equations. Is there a variable change involved?
And also I don't understand how to choose a suitable Lyapunov function. Thanks!

答:
The system comprises two equilibrium points $x=1$ and $x=2$

To study the first equilibrium point $x=1$,  consider the variable transformation $z=x-1$. Then the original dynamical system is rewritten as follows:

$$\dot{z}=-z(z-1)^2$$

Let $V(z)=\frac{1}{2}z^2$. Its time derivative:
$$\dot{V}(z)=z\dot{z}=-z^2(z-1)^2<0\;\;\forall\;\; |z|<1$$

then the equilibrium point $z=0$ (or equivalently $x=1$) is stable.

For the second equilibrium $x=2$, we can not construct a Lyapunov function because it is not stable.

Title: Lyapunov stability dynamic systems


问:
Find continuous $f$ such that $f(x+1)=f(x)+f(\frac{1}{x})$

Find all real continuous functions that verifies :
$$f(x+1)=f(x)+f\left(\frac{1}{x}\right)    \ \ \ \ \ \  (x\neq 0)    $$

I found this result $\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$ and I tried to study the behaviour of the function $g$ defined as $g(x)=\frac{x}{x-1}$ and compare it with $x$ in order to use fixed point theorem but it won't work.

I need a hint and thanks.

答:
Some results

Assume that $f$ is defined and continuous on all of $\mathbb{R}$ even if the equation $f(x+1) = f(x) + f(\frac{1}{x})$ is not defined for $x=0.$

From the given identity
$
f(x+1) = f(x) + f(\frac1x)
$
we get
$
f(\frac1x) = f(x+1) - f(x) \to f(1) - f(0)
$
as $x\to0$. Thus $\lim_{R\to\infty}f(R) = \lim_{R\to\infty}f(-R) = f(1) - f(0).$

Furthermore, we have
$$\begin{align}
f(2) &= f(1+1) = f(1) + f(\frac{1}{1}) = f(1) + f(1) \\
f(3) &= f(2+1) = f(2) + f(\frac{1}{2}) = f(1) + f(1) + f(\frac{1}{2}) \\
f(4) &= f(3+1) = f(3) + f(\frac{1}{3}) = f(1) + f(1) + f(\frac{1}{2}) + f(\frac{1}{3}) \\
\vdots \\
f(n) &= f(1) + \sum_{k=1}^{n-1} f(\frac{1}{k}) \\
\end{align}$$
Taking limits as $n\to\infty$ gives
$
\lim_{n\to\infty} f(n) = f(1) + \sum_{k=1}^{\infty} f(\frac{1}{k})
,
$
but since $\lim_{n\to\infty} f(n) = f(1) - f(0)$ this implies
$$
f(0) = -\sum_{k=1}^{\infty} f(\frac{1}{k})
.
$$
However, the convergence of the series implies that $|f(\frac1k)|\to 0$ as $k\to\infty.$ By continuity of $f$ this means that $f(0)=\lim_{k\to\infty}f(\frac1k)=0.$ And from that follows that
$
\sum_{k=1}^{\infty} f(\frac{1}{k}) = 0
$
and that
$
\lim_{R\to\infty}f(\pm R) = f(1)
.
$

We also get that
$$
f(0) = f((-1)+1) = f(-1) + f(\frac1{-1}) = 2 f(-1)
,
$$
i.e. $f(-1) = 0.$

ContentLicense: CC BY-SA 4.0


问:
Calculating conditional variance using two different methods

The stock prices of two companies at the end of any given year are modeled with random variables $X$ and $Y$ that follow a distribution with joint density function $f(x, y) = 2x$ for $0<x<1, x<y<x+1$ and $0$ otherwise. What is the conditional variance of $Y$ given $X=x$?

My attempt:

$$Var[Y|X=x] = E[Y^2|X=x] - (E[Y|X=x])^2$$

Now, $E[Y^2|X=x] = \int_0^2y^2\cdot f(y|X=x) \; dy$. Note that $f(y|X=x) = \dfrac{f(x,y)}{f_X(x)} = \dfrac{2x}{\int_x^{x+1}2x \; dy} = 1$. So, $E[Y^2|X=x] = \int_0^2y^2 \; dy = \dfrac{8}{3}$.

Next, $E[Y|X=x] = \int_0^2y\cdot f(y|X=x) \; dy = \int_0^2y\; dy = 2$.

So, $Var[Y|X=x] = \dfrac{8}{3} - 4 = \dfrac{-4}{3}$

which is not the correct answer. Where have I gone wrong?

Alternatively, when I recognize that $f(y|X=x) = 1 \implies W \sim Unif[0, 2]$ where $W = Y|X=x$, I can use the formula for the variance of a uniform distribution to calculate the conditional variance as follows:
$$Var[W] = \frac{(2-0)^2}{12} = \frac{1}{3}$$
which is not the correct answer either. Where have I gone wrong with this approach? Thanks!

答:
The main mistake is that the conditional density is actually $$f(y \mid X=x) = \begin{cases} 1 & x < y < x+1 \\ 0 & \text{otherwise}\end{cases}$$
It is very important to consider the support of the density.

In the first approach, you have

$$E[Y \mid X=x] = \int_x^{x+1} y \, dy = \frac{(x+1)^2 - x^2}{2} = x + \frac{1}{2}$$
and
$$E[Y^2 \mid X=x] = \int_x^{x+1} y^2 \, dy = \frac{(x+1)^3 - x^3}{3} = x^2 + x + \frac{1}{3}.$$

So the variance is
$$x^2 + x + \frac{1}{3} - (x+1/2)^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}.$$

In the second approach, you have $W \sim \text{Unif}[x, x+1]$, so the conditional variance is $\frac{((x+1)-x)^2}{12} = \frac{1}{12}$.


问:
Evaluating the series $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
Inspired by this answer, I'm trying to show that $$\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{24} - \frac{1}{8 \pi}$$ using the inverse Mellin transform.

But the answer I get is twice as much as it should be, and I don't understand why.

EDIT:

With Marko Riedel's help, I corrected the error in my evaluation.

Since
$$ \left\{ \mathcal{M} \ \frac{x}{e^{2\pi x}-1} \right\}(s) = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \, dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) $$ for $\operatorname{Re}(s) >1$,

we have
$$  \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s} \, ds , $$ where $c >1$.

Replacing $x$ with $n$ and summing both sides, we get $$ \begin{align} \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1} &= \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\, ds \\&= \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} f(s) \, ds. \end{align} $$

The integrand has simple poles at $s=-1, 0$, and $1$.

The fact that $\left|\Gamma(s)\right|$ decays exponentially fast to $0$ as $\text{Im} (s) \to \pm \infty$ allows us to shift the contour to the left.

I originally shifted the contour all the way to negative infinity.

But as Marko Riedel explains below, we want to shift the contour to the imaginary axis since the integrand is odd there.

Indeed, using the functional equation of the Riemann zeta function, we get $$ f(it) = \frac{it}{2 \pi} \sinh \left(\frac{\pi t }{2} \right) \operatorname{csch}(\pi t) \left|\zeta(1+it)\right|^{2}, \quad t \in \mathbb{R}.$$

Therefore,

$$ \int_{c-i \infty}^{c+i \infty} f(s) \, ds = 2 \pi i \ \text{Res}[f,1] + \pi i \ \text{Res}[f,0] ,$$

where

$$ \begin{align} \text{Res}[f,0] &= \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s) \\ &= 1\left(\frac{1}{2 \pi} \right)(1)\left(- \frac{1}{2} \right) \\ &=-\frac{1}{4 \pi} \end{align} $$

and

$$ \begin{align} \text{Res}[f,1] &= \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) \\&= 1\left(\frac{1}{4 \pi^{2}}\right)(1)\left(\frac{\pi^{2}}{6}\right) \\ &=\frac{1}{24} . \end{align} $$

The result then follows.

答:
There is another missing piece (in response to the comments) that must be added here, and that is the proof of the claim that $$\frac{1}{\pi^s}\Gamma(s)(s-1/4)\zeta(2s)$$ is odd and purely imaginary on the line $\Re(s) = 1/4.$ This is due to a deep connection to the functional equation of the Riemman zeta function. We will show that  $$\frac{1}{\pi^s}\Gamma(s)\zeta(2s)$$ is even and real when $\Re(s) = 1/4.$ The initial claim then follows.

We have from the proof of the functional equation of the Riemann zeta function that
$$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) =
-\frac{1}{2s(1-2s)} + \int_1^\infty (x^{s-1} + x^{-s-1/2})
\left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$
(The first term in the integral here is a straightforward Mellin transform while the second one requires more work.)

Putting $s=1/4+it$, we have that
$$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) =
-\frac{4}{1+16t^2} + \int_1^\infty (x^{-3/4+it} + x^{-3/4-it})
\left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$
But clearly $$x^{it} + x^{-it} = \exp(it\log x) + \exp(-it\log x) =
2\cos(t\log x)$$
is a real number  and even in $t$ as is the fractional term in front and the result follows.

ContentLicense: CC BY-SA 3.0


问:
The action of Galois group on the ideal class group, in a very special case.
Yesterday I asked a question and Franz Lemmermeyer answered my question in a comment. After that I came to another guess, but I can not prove or disprove it.

Let $L/\mathbb{Q}$ be a cyclic Galois extension of prime degree $r$. Suppose $p$ is a prime that is neither ramified nor inert. So we can conclude that it is split, and if $\mathfrak{p}$ is a prime above $p$, then $p\mathcal{O}_L=\prod\sigma(\mathfrak{p})=\mathfrak{p}_1 \cdots \mathfrak{p}_r$, where $\sigma$ runs over elements of Galois group, and all of $\sigma(\mathfrak{p})$'s are distinct. I am curious about some facts about the classes of $[\sigma(\mathfrak{p})]$'s, in the ideal class group. I do not know how they behave, but my observations evoke these two childish conjectures:

Assume that $\mathcal{Cl(O}_L) \cong \dfrac{\mathbb{Z}}{r\mathbb{Z}}$. Then $[\sigma(\mathfrak{p})]=[\mathfrak{p}]$, in the ideal class group.

I hope at least the first statement is correct.

Assume that $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{r\mathbb{Z}})^k$. Then $[\sigma(\mathfrak{p})]=[\mathfrak{p}]$, in the ideal class group.

In the case of this problem (Galois extension of prime degree), this relation is clearly satisfied for any ramified or inert prime, so we can eliminate the assumption of "splits completely" from these two statements.

What I can do:

If $r=2$, and $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{2\mathbb{Z}})^k$, then for any prime $p$ that splits completely ($p\mathcal{O}_L=\mathfrak{p}_1 \mathfrak{p}_2$), it is very easy to show that $[\mathfrak{p}_1]=[\mathfrak{p}_2]$.

[This is very very special and ridiculous]: Let $\alpha$ be a root of the polynomial $x^3-21x-35=0$, and let $L:=\mathbb{Q}(\alpha)$, then $r=3$, and $\mathcal{Cl(O}_L) \cong \dfrac{\mathbb{Z}}{3\mathbb{Z}}$, then for $p=5, 11$ (!!!), I can verify that $p$ splits completely ($p\mathcal{O}_L=\mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$), and $[\mathfrak{p}_1]=[\mathfrak{p}_2]=[\mathfrak{p}_3]$.

I strongly feel that even if these two statements are wrong, then the assumption of "$[\sigma(\mathfrak{p})]=[\mathfrak{p}]$" for "every prime $p$ (that splits completely)", implies that $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{r\mathbb{Z}})^k$.

答:
If $M$ is the Hilbert class field of $L$, $G = \mathrm{Gal}(M/\mathbf{Q})$ and $H = \mathrm{Gal}(M/L)$, then the conditions listed imply (by class field theory) that

$$G =  H \rtimes G/H = (\mathbf{Z}/r)^k \rtimes (\mathbf{Z}/r),$$

and moreover, the corresponding order $r$ automorphism $\varphi$ of $H = (\mathbf{Z}/r)^k$ may be identified with the action of a generator $\mathrm{Gal}(L/\mathbf{Q})$ on the class group of $L$. Here the fact that $G \rightarrow G/H$ splits follows from Minkowski: the extension $L/\mathbf{Q}$ must be ramified at some prime $q$, and then the inertia group $I_q \subset G$ has order $r$ and projects isomorphically onto $G/H$, giving the splitting.

Suppose the automorphism is non-trivial. Then there are certainly elements in $H$ which are not fixed by this automorphism. There is an isomorphism $\mathrm{Cl}(L) \rightarrow H$ given by the Artin map. By the Cebotarev density theorem, we deduce that there are infinitely many primes $p$ totally split in $K$ such that the Frobenius element associated to $\mathfrak{p} | p$ is not fixed by $\varphi$, and hence that $\mathfrak{p}$ inside the class group is not fixed by $\varphi$. Hence:

Your conclusion is true if and only if (under the assumptions) the Hilbert class field $M$ is abelian, and in particular $G = (\mathbf{Z}/r)^{k+1}$.

Note that without any assumption on the class group, the group $G$ is a semi-direct product (the splitting argument above applies). The condition that $\varphi$ is trivial then implies that

$$G = H \times G/H.$$

The ramification at any prime $q$ if non-trivial must be cyclic of order $r$. But now again by Minkowski this implies that $G = (\mathbf{Z}/r)^{k+1}$ and so:

The action of $\varphi$ is trivial if and only if $G = (\mathbf{Z}/r)^{k+1}$ assuming only that $L/\mathbf{Q}$  is cyclic of order $r$.

So now your first question is equivalent to asking if $H = (\mathbf{Z}/r)^k$ implies that $G = (\mathbf{Z}/r)^{k+1}$. There is no a priori reason to suppose this is true. However, as you noted, the result is true in your formulation by a quite easy argument for $r=2$. That means that $G$ must always be abelian in this case. Thus we should be a little careful. Note that one can write down all automorphisms of order $r$ on vector spaces over $\mathbf{F}_r$ - they have a Jordan decomposition where the block sizes are at most $r$. Hence we can even write $G$ explicitly as

$$G = \langle s,x_{i,j} | s^p, x^r_{ij}, [x_{i,j},x_{k,l}], [s, x_{i,j}] = x_{i,j-1} \rangle $$
where

$i$ ranges over $i = 1,\ldots, n$ for some $n$,

$j = 1, \ldots, m(i)$, where $m(i) \le r$,

by convention $x_{i,0} = e$.

$\sum_{i=1}^{n} m(i) = k$.

$\varphi(x_{i,j}) = s x_{i,j} s^{-1} = x_{i,j} x_{i,j-1}$.

Exercise: If $x \in H$, the element $sx$ has order $p$ if and only if $x$ lies in the subgroup generated by $x_{i,j}$ for $j \le \max\{m(i),r-1\}$.

This follows easily enough from the identity (true by induction, multiply by $sm$ on the right)

$$(sx)^m = \varphi(x) \varphi^2(x) \ldots \varphi^m(x) s^m x,$$

and hence

$$(sx)^r = x \varphi(x) \ldots \varphi^{r-1}(x) = (I + \varphi + \ldots + \varphi^{r-1})(x) = (\varphi - 1)^{r-1}(x),$$

where the polynomial identity $(1+X+\ldots X^{r-1}) = (X-1)^{r-1}$ follows by multiplying both sides by $(X-1)$ and then using Frobenius to see $(X-1)^r = X^r - 1$.

The ramification group $I_q$ at any prime $q$ is either trivial or provides a splitting $G/H \rightarrow G$. Each such group has an element which maps in $G/H$ to $s$. Hence the inertia groups $I_q$ are generated by elements of order $r$ of the form $ s x$. By the computation above, they are thus given by $s x $ where $x$  is a product of elements of the form $x_{i,j}$ for $j < r$. It follows that the images of the inertia groups $I_q$ for all $q$  lies inside the group generated by $s x_{i,j}$ for all $i$ and all $j < r$. Suppose that $m(i) = r$ for some $i$. Then the quotient obtained by setting $s = 0$ and $x_{i,j} = 0$ for all $i$ and $j$ except for $x_{i,m(i)}$ gives a quotient group $(\mathbf{Z}/r)$ where by this analysis the inertia groups are all trivial. But that means the corresponding extension is everywhere unramified which contradicts Minkowski. To summarize:

The only $\varphi$ which can arise must have Jordan blocks of size at most $r-1$, not simply $r$ as follows from group theory.

If $r = 2$ this means the Jordan blocks all have size $1$ and $\varphi$ is trivial and $G = (\mathbf{Z}/2)^{k+1}$ is abelian. However, if $r = 3$, there is no a priori obstruction, however, to

$$\mathrm{Gal}(M/\mathbf{Q}) = \langle x, y , s | x^3, y^3, [x,y], s^3, [s,x], [s,y]=x \rangle$$

occurring as a $G$. Here $G$ is the unique non-abelian group of order $27$ all of whose elements have order $3$ (the $3$-Sylow of $\mathrm{GL}_3(\mathbf{F}_3)$. (With the notation above, $x = x_{1,1}$ and $y = x_{1,2}$ and $m(1) = 2$.) The abelianization of $G$ is $(\mathbf{Z}/3)^2$, so the corresponding cyclic extension (by genus theory, omitted) should be ramified at exactly two primes which are either $3$ or $1 \bmod 3$.

Hence this suggests looking exactly for the following:

look for cyclic degree $3$ fields ramified at two primes $p$ and $q$ whose class group is $9$.

Now you can go to this website:

https://hobbes.la.asu.edu/NFDB/

And impose the following restrictions

degree $3$ extensions

discriminant in the range 1..1000000 (can be increased)

Galois group is a p-group

$h = 9$

$2$ ramified primes.

This returns (for example) the extension $x^3 - 219 x - 1241$ with class group $(\mathbf{Z}/3)^2$ (as given by the website), and thus this extension will lead to a counter-example to your guess.

To be completely explicit, the ring of integers is $\mathbf{Z}[x]$, and

$$11 = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3 = (x-1,11)(x-4,11)(x-6,11).$$

Now one can check by one's favourite method that these elements are distinct in the class group.


问:
Give a reason why $\mathbb Z_{64}\times \mathbb Z_4$ and $\mathbb Z_{64}\times \mathbb Z_{2}\times \mathbb Z_{2}$ are not isomorphic.
The question is about classifying groups of order $256$ with at least one element of order $64$, and justify why the elements of the list are non-isomorphic.
I'm done except for showing that
$$
  \mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_{2}\text{ and }
  \mathbb Z_{64}\times \mathbb Z_{4}
$$
are not isomorphic.
Of course they are not, that's the content of the classification theorem, but the question begs a direct approach.

All similar questions I found on MSE differentiate groups of this kind from order elements, e.g.  $\mathbb Z_2\times \mathbb Z_2$ and $\mathbb Z_4$ are not isomorphic as only the latter contains an element of order 4.
We can't apply that here.

My solution:
the order of a an element of a product is the minimum common multiple of the non-zero orders.
Thus an element of $\mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_2$ has order 2 iff it's nonzero and its $\mathbb Z_{64}$ factor is $0$ or $32$ (the elements of order 2 in $\mathbb Z_{64}$).
We get a total of 7 elements of order 2,
$$
  (32,0,0),(0,1,0) ,(0,0,1)  ,(0,1,1)
  ,(32,1,0),(32,0,1)  \text{ and }(32,1,1).
$$
On the other hand, the same reasoning shows that there are only 3 elements of order 2 in $\mathbb Z_{64}\times\mathbb Z_{4}$, namely $$(32,0), (0,2) \text{ and }(32,2).$$

Is this correct? I think so, but I spent more time in this question than I'd like to admit, that's why I'm posting it. Any other approaches?

答: Draw the lattice of subgroups, maybe. The lattice of the Klein four group is not a total order, but that of  the cyclic group of order four is. Then note how doing $G\times -$ changes the lattice.


答:
That was my first thought of a way to do it.

Depending on where you are in your learning you might be expected to show that if the first component is other than $0$ or $32$ the order is greater than $2$ (and similarly for the second component in the $\mathbb Z_4$ case). But the reasoning is completely sound.

You might also identify a subgroup $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ in one case rather than the other. Orders of elements and subgroups are two potentially distinguishing features to look out for.

Title: Give a reason why $\mathbb Z_{64}\times \mathbb Z_4$ and $\mathbb Z_{64}\times \mathbb Z_{2}\times \mathbb Z_{2}$ are not isomorphic.

Body:
The question is about classifying groups of order $256$ with at least one element of order $64$, and justify why the elements of the list are non-isomorphic.
I'm done except for showing that
$$
  \mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_{2}\text{ and }
  \mathbb Z_{64}\times \mathbb Z_{4}
$$
are not isomorphic.
Of course they are not, that's the content of the classification theorem, but the question begs a direct approach.

All similar questions I found on MSE differentiate groups of this kind from order elements, e.g.  $\mathbb Z_2\times \mathbb Z_2$ and $\mathbb Z_4$ are not isomorphic as only the latter contains an element of order 4.
We can't apply that here.

My solution:
the order of a an element of a product is the minimum common multiple of the non-zero orders.
Thus an element of $\mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_2$ has order 2 iff it's nonzero and its $\mathbb Z_{64}$ factor is $0$ or $32$ (the elements of order 2 in $\mathbb Z_{64}$).
We get a total of 7 elements of order 2,
$$
  (32,0,0),(0,1,0) ,(0,0,1)  ,(0,1,1)
  ,(32,1,0),(32,0,1)  \text{ and }(32,1,1).
$$
On the other hand, the same reasoning shows that there are only 3 elements of order 2 in $\mathbb Z_{64}\times\mathbb Z_{4}$, namely $$(32,0), (0,2) \text{ and }(32,2).$$

Is this correct? I think so, but I spent more time in this question than I'd like to admit, that's why I'm posting it. Any other approaches?


问:
Showing that the sequence is bounded without finding its limit
I am given a sequence ${x_n}=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ and I have to show that it's bounded.

I tried doing this: $$1\leq 1+\frac{1}{2}+...+\frac{1}{n}\leq n$$
$$1-\ln n\leq 1+\frac{1}{2}+...+\frac{1}{n}-\ln n\leq n-\ln n$$
$$1-\ln n\leq{x_n}\leq n-\ln n$$
But I don't think I am allowed to do this since the boundaries depend on n.
Is there maybe another way to solve this? and could you please help me?
Thanks a lot!
I can't use integrals, derivatives or limits

答:
You have too many limitations "no integrals, derivatives or limits". I will solve this problem using the elementary inequality $$x-\dfrac{x^2}{2}\lt \ln(1+x)\lt x$$ for all $x\ge 1.$ In case you haven't seen it before, lets prove it applying the mean value theorem for the function $f(t)=\ln(1+t)$ on the interval $[0, x].$ According to MVT, there exists $c\in (0,x)$ such that $f(x) - f(0) = f'(c)(x-0),$ and also we have $\dfrac{1}{1 + x}\lt f'(c)=\dfrac{1}{1 + c} \lt 1$. Hence $$\dfrac{x}{1 + x} \lt \ln(1 + x) \lt x,\qquad \forall x\gt0.$$ Since $$x-\dfrac{x^2}{2}-\dfrac{x}{1 + x}=\dfrac{x^2(1-x)}{2(1+x)}\le0$$ for all $x\ge 1,$ we have the desire result. Now lets observe that
\begin{align}
x_n & = 1+\dfrac12+\cdots+\dfrac1n-\ln n \\
 & = \left(1+\ln 1-\ln 2\right) + \left(\frac{1}{2}+\ln 2-\ln 3\right) + ... + \left(\frac{1}{n}+\ln n-\ln (n+1)\right) +\ln\left(\dfrac{n+1}{n}\right)\\
 & = \sum_{k=1}^n\left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right) + \ln\left(1+\frac{1}{n}\right).
\end{align}

But according to the inequality that we proved before $$0\lt\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\lt\dfrac{1}{2k^2}.$$ Since the infinite series correspond to the right most term of this last inequality is convergent $$0\lt x_n\lt \dfrac12\sum_{k=1}^{\infty}\dfrac{1}{k^2}+\max_{k\in\mathbb{N}}\ln\left(1+\frac{1}{k}\right)\lt\infty.$$ In fact, $\dfrac{\pi^2}{12}+\ln 2$ is an explicit upper bound for your sequence. Furthermore, if $$y_n = 1+\dfrac12+\cdots+\dfrac1n-\ln (n+1)$$ then $x_n=y_{n-1}+\dfrac1n$ and therefore one sequence converge if and only if so does the other. Also both of them share the same limit. Now using the same argument for $y_n$ we can get rid of $\ln 2$ term and obtain a much closer approximation to the potential limit.


问:
Calculate Probability Using Random Variable?
I found the following question in my text-book:

A bus comes in a delay which is uniform between $0$ and $10$ minutes.
Daniel comes $3$ minutes before the bus is supposed to arrive the earliest.

Calculate the possibility that Daniel waits more than $6$ min.

My solution:

Let $X$ be a random variable that describes the delay of the bus, we know that $X\thicksim\operatorname{Uni}(0,10)$ and that for sure he will wait at least $3$ min so we calculate $P(X>3)$:
$$P(X>3)=\int_3^\infty f(x)\,dx.$$
Now we can replace $\infty$ with $10$ since $f(x)=0$ for $x>10$ (Probability density function) thus getting $7/10$.

In a specific day Daniel woke up late and arrived to the bus station in a delay of $4$ minutes.
What's the possibility he missed the bus?

Daniel will miss the bus if it arrived in the first $4$ minutes so:
At first I though this is the same as
$$\frac{P(x<4\mid x=4)}{P(x=4)}=\frac{0}{P(x=4)}=0.$$
But It's clearly wrong and doesn't make sense.

答:
You established $X$ to be a random variable that describes the time Daniel waits for the bus, which is not the best choice here, because it does not follow a uniform distribution between $0$ and $10$.

For example, in the original settings, Daniel comes $3$ minutes before the bus is supposed to arrive the earliest, so the probability for him waiting more than $6$ min should $P(X>6)$ instead of $P(X>3)$.

However, in the second question, he indeed came to station of $4$ min late, so what does the event $X=4$ mean? It means the bus came $4$ min later than Daniel, i.e., the bus came in totally $4+4=8$ min late. Of course this does not make sense at all. In fact, in your setting, the probability should read $P(X\geq 0)$. If Daniel did not miss the bus, then the waiting time for him should be non-negative.

You should set $X$ to be the time for such bus come in a delay, which ranges uniformly from $[0,10]$. Then the first one is $P(X>3)$ and the second one is $P(X<4)$. If $X\geq 4$, then the bus came later than Daniel, so he would not miss the bus.


答:
For 1) you need to calculate $P(X>6\mid X>3)$ since we know he will wait at least 3 minutes.

For 2) We can safely assume that the probability of him arriving 4 minutes late is independent of the arrival time of the bus, so we get $P(X<4\mid \text{4 min late})=P(X<4)$

Title: Calculate Probability Using Random Variable?

Body:
I found the following question in my text-book:

A bus comes in a delay which is uniform between $0$ and $10$ minutes.
Daniel comes $3$ minutes before the bus is supposed to arrive the earliest.

Calculate the possibility that Daniel waits more than $6$ min.

My solution:

Let $X$ be a random variable that describes the delay of the bus, we know that $X\thicksim\operatorname{Uni}(0,10)$ and that for sure he will wait at least $3$ min so we calculate $P(X>3)$:
$$P(X>3)=\int_3^\infty f(x)\,dx.$$
Now we can replace $\infty$ with $10$ since $f(x)=0$ for $x>10$ (Probability density function) thus getting $7/10$.

In a specific day Daniel woke up late and arrived to the bus station in a delay of $4$ minutes.
What's the possibility he missed the bus?

Daniel will miss the bus if it arrived in the first $4$ minutes so:
At first I though this is the same as
$$\frac{P(x<4\mid x=4)}{P(x=4)}=\frac{0}{P(x=4)}=0.$$
But It's clearly wrong and doesn't make sense.


问:
Prove $\prod_{r=1}^{(n-1)/2} 2\cos \frac{\pi r}{n}=1$
I would like to know how to prove that

$$\prod_{r=1}^{(n-1)/2} 2\cos\frac{\pi r}{n}=1~~~~~\text{where $n$ is odd}$$
I have tried induction and using trigonometric identities to attempt to transform the product into something more helpful, but nothing has worked.

It may be more helpful to think of the product as
$$\prod_{r=1}^{n} 2\cos\frac{\pi r}{2n+1}=1$$

Many thansk for your help.

答: @Somos's tip is useful. In the second parameterization, write the product as$$\frac{\prod_{r=1}^n\sin\frac{2\pi r}{2n+1}}{\prod_{r=1}^n\sin\frac{\pi r}{2n+1}},$$then note the numerator's sines have the same values as the denominator's, but in another order. In particular, replacing $2r$ with $2n+1-2r$ when $2r\ge n+1$ produces a permutation of $\{1,\,\cdots,\,n\}$. For example, if $n=2$ the numerator is$$\sin\frac{2\pi}{5}\sin\frac{4\pi}{5}=\sin\frac{2\pi}{5}\sin\frac{\pi}{5}.$$


问:
Proving $(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)$
I wish to verify that:
$$
(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)
$$
I wish to present my attempt: (For the $(\subseteq)$ implication)
\begin{align*}
&x\in(A\setminus B)\setminus (C\setminus  B) \\
\implies&x\in(A\setminus B)\wedge x\notin(C\setminus B) \\
\implies&[(x\in A)\wedge(x\notin B)]\wedge[(x\notin C)\wedge(x\in B)]
\end{align*}
Unfortunately, I am stuck at this process and I am unable to proceed. I am new to elements of set theory and I am currently practicing such practice problems.

答: Make implications two sided:$$\begin{align}x\in(A\setminus B)\setminus(C\setminus B)&\iff x\in A\setminus B\land x\notin C\setminus B\\&\iff x\in A\land x\notin B\land x\notin C\\&\iff x\in A\land x\notin B\cup C\\&\iff x\in A\setminus(B\cup C).\end{align}$$Alernatively, you could prove the tautology $((a\land\neg b)\land\neg(c\land\neg b))\iff a\land\neg(b\lor c)$ with a truth table.


答:
Minor problem with your work so far: suppose $x \not \in C \setminus B$.

If $x \in C \setminus B$, then $x \in C$ and $x \not \in B$.

The negation of this statement, to yield $x \not \in C \setminus B$, is $x \not \in C$ or $x \in B$. More generally:

$$\neg (P \land Q) = (\neg P) \lor (\neg Q)$$

So what you have, then, is that

$$x\in(A\setminus B)\setminus (C\setminus B) \implies [(x\in A)\land (x\notin B)]\land [(x\notin C)\lor (x\in B)]$$

There are two possibilities for the second statement. Take $x \in B$, though, and you'll have that $x \not \in B$ contradicts this. Thus, only the choice for $x \not \in C$ is valid:

$$x\in(A\setminus B)\setminus (C\setminus B) \implies (x\in A)\land (x\notin B)\land(x\notin C)$$

From the definition of $A \setminus (B \cup C)$, you should be able to use this to reach your desired conclusion.

Title: Proving $(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)$

Body:
I wish to verify that:
$$
(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)
$$
I wish to present my attempt: (For the $(\subseteq)$ implication)
\begin{align*}
&x\in(A\setminus B)\setminus (C\setminus  B) \\
\implies&x\in(A\setminus B)\wedge x\notin(C\setminus B) \\
\implies&[(x\in A)\wedge(x\notin B)]\wedge[(x\notin C)\wedge(x\in B)]
\end{align*}
Unfortunately, I am stuck at this process and I am unable to proceed. I am new to elements of set theory and I am currently practicing such practice problems.


问:
Pick's Theorem and Primes.
For $n$ prime, each of the remaining triangles contains exactly the same number of grid points.
This was proved here.

However, is the converse true i.e., triangles containing exactly the same number of grid points implies a prime? I tried (geometrically) finding a non-prime which had the same number of internal grid points, but the lattice became too noisy to accurately manage.

Thanks.

答:
Let $n > 1$ be a positive integer and for each $0 \le i < n$, let $\Delta_n(i)$ be the triangle with coordinates at $(0,0)$, $(i,n-i)$ and $(i+1, n-(i+1))$.

Euclidean geometry tells us that these triangles have the same area since they have the same base and height if you take the base as the $((i,n-i), (i+1,n-(i+1)))$ edge and the height as the orthogonal segment to $((0,n), (n,0))$ that drops from the origin, namely the segment $((0,0), (n/2,n/2))$. So by Pick's theorem, denoting by $B_n(i)$ and $I_n(i)$ the number of grid points on the boundary and interior of $\Delta_n(i)$ respectively, we see that the quantity $2B_n(i) + I_n(i)$ must be constant when $i$ varies from $0$ to $n-1$ because $2B_n(i) + I_n(i) = 2 \, \mathrm{Area}(\Delta_n(i)) + 2$.

You'll notice two exceptions to the rule that the number of interior points remains the same, namely for the triangles $\Delta_n(0)$ and $\Delta_n(n-1)$; the problem with those two is that they share $n+1$ points with the $x$-axis (resp. $y$-axis) and one point at $(1,n-1)$ (resp. $(n-1,1)$), and no interior point. It doesn't help us figure out if $n$ is prime per se, but it gives us the value of
$$
2B_n(i) + I_n(i) = 2(n+2) = 2n+4.
$$
Here is the case where $n=7$:

And here is the case where $n=6$:

I labelled the intersection points with the integer grid for visibility. So basically the only way that $\Delta_n(i)$ will have the same number of interior points when $i=1,\cdots,n-2$ is if it also has the same number of boundary points by Pick's theorem, since those triangles all have the same area and all have integer coordinate vertices.

But how is it possible that a line segment from $(0,0)$ to $(i,n-i)$ intersects points on the integer grid? This happens if there exists a rational number $\lambda \in ]0,1[$ such that $\lambda i$ and $\lambda(n-i)$ are both integers. Writing $\lambda = a/b$ with $a$ and $b$ coprime, this means in particular that $a<b$ and $b$ divides both $i$ and $n-i$. The existence of such a $b$ is necessary and sufficient to find an intersection point on the integer grid with the segment (we can just take $a=1$) and $b$ exists if and only if $\gcd(i,n-i) > 1$. But $\gcd(i,n-i) = \gcd(i,n)$, so basically each integer $i$ between $1$ and $n-1$ satisfying $g_n(i) \overset{def}= \gcd(i,n) > 1$ will have $g_n(i) - 1$ points of intersection with the integer grid (excluding $(i,n-i)$ and $(0,0)$), namely corresponding to the values of $\lambda$ equal to $\frac 1{g_n(i)}, \frac 2{g_n(i)}, \cdots, \frac{g_n(i)-1}{g_n(i)}$.

We now show that the number of interior points of $\Delta_n(i)$ for $i=1,\cdots,n-2$ are going to be the same if and only if $n$ is prime or $n=4$, because in the first case, $g_n(i) = 1$ for $i=1,\cdots,n-1$, and in the second case, $g_4(1)=1, g_4(2)=2$ and $g_4(3) = 1$, so $g_4(1)+g_4(2) = g_4(2) + g_4(3)$.

The number $n=4$ is the only number small enough not to be prime that works, because whenever $n > 4$ and $n$ has a prime divisor say $p > 2$, we have $g_n(1) = 1$ and $g_n(2) = 1$ or $2$, but $g_n(p) = p$ and $g_n(p+1) = 1$. If $n$ is a power of $2$ greater than $4$, we have $g_n(1) = 1$ and $g_n(2) = 2$ but $g_n(n/2) = n/2$ and $g_n(n/2+1) = 1$.

Hope that helps,

Title: Pick's Theorem and Primes.


问:
Solving characteristic equation with matrix coefficients.
Given the equation \begin{equation}\det(\lambda^2 I+ B \lambda +K )=0\end{equation} where $I,B,K\in \mathbb R^{m \times m}$. $B$ and $K$ are symmetric matrices with no zero eigenvalues and $B>0$. Let $n^+(A)$ and $n^-(A)$ denote the number of eigenvalues of $A$ with positive real parts and negative real parts respectively. Is the number of roots of the above equation with positive real parts equal to $n^-(B)+n^-(K)$ and the number of roots with negative real parts equal to $n^+(B)+n^+(K)$?

My trial was as follows: let $v \in \mathbb{ R^m}$ if $K$ is positive definite
then $$\det(\lambda^2 I+ B \lambda +K )=0 \implies \exists v: v^T(\lambda^2 I+ B \lambda +K)v=0 \implies \lambda^2 v^2+ v^TBv \lambda +v^TKv=0.$$
All the coefficients are positive, so real part of $\lambda$ must be negative. The same argument can be used when $K<0$, the remaining case is when $K$ is indefinite.

Edit

As Ben Grossmann showed the system can be seen as the characteristic equation of:
$$
J = \pmatrix{0&&I \\ -K && -B}.
$$
where $B>0$, $J$ cannot have an eigenvalue with purely imaginary part if it did then $ \exists v: v^H((bi)^2 I+ B (bi) +K)v=0 \implies v^HBvb=0 \iff b=0$

Let $n_0(A)$ denote the number of zero eigenvalues of A.

lemma 1: $n_0(K)=n_0(J)$

we observe that the reduced echelon form of $J$ is:
$$
J_{red} = \pmatrix{I&&0\\0 && K}.
$$
so  $n_0(K)=n_0(J)$.

Let $G(t)=K+t I$ and let
$$
J(t)=\pmatrix{0&&I\\-G(t) &&-B}
$$
observe that as $t\rightarrow \infty$,$G(t)>0 \implies n_-(J(t))=2m$ and as $t\rightarrow -\infty$ , $G(t)<0 \implies n_-(J(t))=n_+(J(t))=m$.
Let the eigenvalues of $J(t)$ be $\{\lambda_i(t)\}$.

As $t$ goes from $-\infty$ to $+\infty$, the eigenvalues
of $G(t)$ become zero $m$ times which is when they change signs. By lemma 1, the eigenvalues
of $J(t)$ must also become zero $m$ times and those times are exactly when eigenvalues of $G(t)$ are zero. It is not difficult to see that the eigenvalues $J(t)$ are zero exactly when one of the positive eigenvalues change signs to negative, if not then one of the positive eigenvalues of $J(t)$ won't change signs and would remain positive as $t \rightarrow +\infty$ which is a contradiction.

So every time an eigenvalue of $G(t)$ changes signs from negative to positive, an eigenvalue of $J(t)$ will change signs from positive to negative, so we have that $\forall t$ , $n_-(J(t))=m+n_+(G(t))=n_+(B)+n_+(G(t))$, taking $t=0$ yields $n_-(J)=n_+(B)+n_+(K)$ as required.

Is there something wrong with this proof?

答:
A partial (but non-trivial) answer: if $K$ is negative definite, the number of positive/negative roots can be determined as follows.

First, observe that $\det(\lambda^2 I + \lambda B + K) = \det(\lambda I - M)$, where
$$
M_1 = \pmatrix{-B & -K\\I & 0}.
$$
This holds regardless of the definiteness of $K$. If $K$ is negative definite, then there exists a unique positive definite square-root $P$ of $-K$ (i.e. $P^2 = -K$). We note that the above matrix $M_1$ is similar to
$$
M_2 = \pmatrix{I\\ & P} \pmatrix{-B&-K\\I & 0}\pmatrix{I \\ & P}^{-1} = \pmatrix{-B & P\\P & 0}.
$$
The above matrix is symmetric and thus has real eigenvalues. Furthermore, since $B$ is invertible, we have
$$
n_\pm(M_2) = n_{\pm}(-B) + n_{\pm}(C),
$$
where $C$ denotes the Schur complement
$$
C = M_2/(-B) = PB^{-1}P.
$$
Note that $n_{\pm}(PB^{-1}P) = n_{\pm}(B^{-1}) = n_{\pm}(B)$. With that, we can conclude that regardless of the definiteness of $B$, the number of roots with positive real part and negative real part are both equal to $m$.

For the case where $K$ is positive definite, there is a positive definite (and symmetric) $P$ for which $K = P^2$. Applying the same manipulation as before leaves us with the matrix
$$
M_2 = \pmatrix{-B & -P\\ P & 0}.
$$
We note that the symmetric part of this matrix,
$$
M_S = \frac 12(M_2 + M_2^T) = \pmatrix{-B&0\\0 & 0},
$$
is negative semidefinite whenever $B$ is positive definite. It follows that $M_2$ will only have eigenvalues with non-positive real part in the case that $B$ is positive definite.

Thoughts on the general case:

For the general case, with $B$ positive definite. Let $P$ be such that $PKP^T$ is of the form $\operatorname{diag}(I_{n_+},-I_{n_-})$.

$$
M_2 = \pmatrix{P\\& P^{-T}} \pmatrix{-B & -K\\ I & 0} \pmatrix{P \\ & P^{-T}}^{-1}
= \pmatrix{-PBP^{-1} & -PKP^T\\ (PP^T)^{-1} & 0}
$$


问:
Expectation calculation of hypergeometric distribution with sampling without replacement
Given a population of $n$ black and $m$ green balls, the expected number of black balls in a random sample of $r$ balls can be calculated as follows

We define a random variable $X_k$ taking values 1 or according as the $k$-th element in the sample is black or not. Then $P(X_k = 1) = \frac{n}{n+m}$ and the expected number of black balls in a random sample of $r$ balls is $\frac{nr}{n+m}$.  This is the answer given in Feller's Book (Vol 1). The reason given by Feller is "For reasons of symmetry".

But I am not convinced by this approach. How can we assume a constant probability $P(X_k = 1) = \frac{n}{n+m}$ ?  The sampling process is without replacement and probability of success changes in every trial. I don't understand what is meant by "For reasons of symmetry".

答:
The sequence $(X_1, \dotsb, X_r)$ is exchangeable. Indeed the joint distribution of the $X_i$ is only dependent on the number of $X_i$ that are equal to $1$.

To see this consider the case $r=3$. Then
$$
P(X_1=1, X_2=0, X_3=1)=\frac{n(m)(n-1)}{(n+m)(n+m-1)(n+m-2)}=P(X_2=1, X_3=0, X_1=1).
$$

by using the decomposition $P(X_1=1, X_2=0, X_3=1)=P(X_1=1)P(X_2=0\mid X_1=1)P(X_3=1\mid X_2=0, X_1=1)$ for the first equality and something similar for the scond. More simply we are considering the number of red and black balls still left after each draw and multiplying the appropriate probabilites.

In general for any binary tuple $(x_1, x_2, x_3)$ the probability $P(X_1=x_1, \dotsc, X_3=x_3)$ will be a fraction with the same denominator as above and a numerator of the form $n(n-1)\dotsb(n-a+1)\times m(m-1)\dotsb(m-u+1)$ where $a$ is the number of ones in the sequence $(x_1, \dotsc, x_3)$ and $u$ the number of zeroes. This probability is invariant to permuatations of the indices of the $X_i$. We can argue similarly in the general case for given $r\geq 0$.

In particular, this means that all the marginal distributions of the $X_i$ are equal to one another. For example
$$
P(X_2=1)=\sum P(X_2=1, X_1=a_1, \dotsc, X_r=a_r)=\sum P(X_1=1, X_2=a_1, \dotsc, X_r=a_r)=P(X_1=1)
$$
where in the second equality we use exchangeability, interchanging the indices of $X_1$ and $X_2$ and fixing all other indices. But of course
$$
P(X_1=1)=\frac{n}{n+m}.
$$
The number of black balls drawn in the r draws $X$ is then found by linearity of expecation. Indeed $X=\sum_{i=1}^r X_i$ whence
$$
EX=\sum_{i=1}^r EX_i=\sum_{i=1}^r P(X_i=1)=\frac{rn}{n+m}
$$


问:
Logic problem - what kind of logic is it?
I would be most gratefull, if someone could verify my solution to this problem.

```<code> 1) Of Aaron, Brian and Colin, only one man is smart.

 Aaron says truthfully:
    1. If I am not smart, I will not pass Physics.
    2. If I am smart, I will pass Chemistry.

 Brian says truthfully:
    3. If I am not smart, I will not pass Chemistry.
    4. If I am smart, I will pass Physics.

 Colin says truthfully:
    5. If I am not smart, I will not pass Physics.
    6. If I am smart, I will pass Physics.

 While
   I. The smart man is the only man to pass one particular subject.
   II. The smart man is also the only man to fail the other particular subject.

 Which one of the three men is smart? Why?
</code>```

I would say that it could have been any one of them, as the implications in every statement are not strong enough to disprove the statements I,II. But I'm not sure if my solution is enough, as I'm not sure what kind of logic it is.

答: You could just create a simple table and read the solution:


问:
Is Einstein notation universally applicable?
It gets surprisingly monotonous to write the sum symbol repeatedly, so I'm wondering if one can just swap in Einstein notation at will to avoid the monotony, assuming that the symbol that's being summed over appears exactly twice?

For example, the wiki definition of $\nabla$ uses the sum symbol:

$\nabla = \sum_{i=1}^n \vec e_i {\partial \over \partial x_i} $

Could one simply write

$\nabla = \vec e_i {\partial \over \partial x_i} $

instead, and if so, would it need to be prefaced with "Using Einstein summation notation... " or is even that necessary?

If this question is more appropriate for physics.se, I'll post it there. Math.se has more hits on the search "Einstein notation," so I thought I'd ask here first.

Thank you for any help.

答: My feeling is that this is used much more broadly than in physics and it will be read as intended by any audience that's learned Einstein notation, so I use it readily, and will only mention what it is if I suspect the audience might not know it. For example, in a math.se answer, if the questioner didn't use the notation themselves, I try not to assume they know it already. I still use it, but briefly name it (once per document ought to do, especially with the hyperlink I've included above, even though you didn't need an explanation).


问:
Indefinite integral $\int \frac {\mathrm d x} {p^2 + q^2 \cosh^2 a x}$
Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.584$ gives:

$$\int \dfrac {\mathrm d x} {p^2 + q^2 \cosh^2 a x} = \begin{cases} \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \left({\dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } }\right) + C \end{cases}$$

(As usual he glosses over negative argument to the $\ln$, but all you need to do is modulus it.)

He does not give the conditions under which either case holds. This is crucial.

First I multiplied top and bottom by $\operatorname {csch}^2 a x$, then used $\operatorname {csch}^2 a x = \coth^2 a x - 1$ for the bottom, then substituted $u = \coth a x$ to convert it into the form:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\frac {p^2} {p^2 + q^2} - u^2}$$

I can't see there is any other case than that $\dfrac {p^2} {p^2 + q^2}$ is positive, hence:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\left({\frac p {\sqrt {p^2 + q^2} } }\right)^2 - u^2}$$

This is a standard integral, giving:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} }\right| + C$$

which leads to:

$$\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C$$

And that all works fine, no worries.

BUT, as far as I can see, there is no way to get to the arctangent result. To get there, we need $u^2 + k^2$ for some constant $k$, and I can't see how to get there from the given integrand.

答:
The second result is wrong it would seem, at least if it's supposed to hold for all values of $p, q,a$. Letting $p=q=a=1$ and using my calculator to evaluate the integral between $0$ and $1$:

$$\int_0^1\frac{1}{1+\cosh^2 x}dx=0.4257352463...$$
But $$\frac{1}{\sqrt2}\arctan\frac{\tanh 1}{\sqrt2}-\frac{1}{\sqrt2}\arctan\frac{\tanh 0}{\sqrt2}=0.3493059377...$$


问:
Coin toss game expectation
Person $A$ and $B$ are going to play a coin toss game. There is an initial score $0$, and whenever a head/tail appears, the score $+1$/$-1$. Repeat the coin toss until one wins, that is, when the score reaches $+2$/$-2$, $A$/$B$ wins the game. There is also an initial stake \$$1$ for the game and person $A$ has the option to double the stake every time before a coin toss. When one person wins the game, the other player needs to pay the amount of dollars on the stake to the winner. The question is: if you are person $A$, what is your strategy and what is your highest expected payoff of the game?

I suspect that the stake should be doubled when you have a score of $0$ or $+1$ but if that is the case, the expectation will be infinity according to my calculation. Also, I ran a computer simulation on $10$ runs of $1000000$ games, but for the $10$ means I get, the mean does not converge to a single value. Could someone enlighten me what is the correct approach?

答:
For the case in which you only double at $x=1$, a recursive approach can be used to find the expected value too.
Using each two-toss sequence TT, TH, HT, and HH we get the following equation:

$e = \frac{1}{4}*-1 + \frac{1}{4}*e + \frac{1}{4}*2e + \frac{1}{4}*2 $.

Solving yields e = 1.

Title: Coin toss game expectation


问:
Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).

The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."

I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.

My attempt:

Assume to the contrary that solution of this equation is greater then or equal to 2.

Let $x=\frac 2p$.

Then we have

$ (\frac 2p)^5-2(\frac 2p)^3-3=0 $

$ \frac {2^5}{p^5} - \frac {2^4}{p^3} - 3 = 0$

We now consider two case: when $p=1$ and $p<1$.

when $p=1$:

$ 2^5 - 2^4 - 3 = 32 - 16 - 3 = 13$.

Since $x = 2$ is not a solution this is a contradiction.

When $p<1$:

If $p<1$, then we know $1/p>1.$ This implies $\frac {1}{p^{n+1}} > \frac {1}{p^n}.$

Since $\frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality

$ \frac 1{p^5}(2^5)- \frac 1{p^3} 2^4-3 > \frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds

Since for all $x > 2$ is not a solution this is a contradiction.

Thus, solution $x$ is less then 2.

答:
Here is another way of doing this:

If $x\gt 2$ then $x^5\gt 4x^3\left(=2x^3+2x^3\right) \gt 2x^3+16\gt 2x^3+3$

I have added this because it clicked into my head immediately as a way of estimating and thinking though different approaches gives you a range of tools to solve such problems.


问:
Solving for an unknown value in inequalities
Q: Find the values of  for which the quadratic expression
4^2 + 12 −  is always positive.

What I’ve tried:

The discriminant is > 0 therefore
b^2-4ac>0

a=4
b=12k
c=-k

This resulted in:
12k^2+16k>0

I am not sure where to go next to solve for the value of ‘k’, any help would be greatly appreciated, thanks!

答:
You can simplify a bit  the computations using the reduced discriminant:
$$\Delta'=(6k)^2+4k=4k(9k+1)$$
The quadratic polynomial $4x^2 + 12x − $ has a constant sign (that of the leading coefficient) if & only if $\Delta'<0$. As the leading coefficient of $\Delta'$ (as a quadratic polynomial in $k$) is positive, this means $k$ belongs to the interval of the roots, namely $\;\bigl(-\frac19,0\bigr)$.


问:
Is this proof of, that there is no limit of lim 1/x as x approaches 0, correct?
I just wanted to ask, if the following proof can be used to proof, that there is no l, such as $\lim\limits_{x \to 0} \dfrac{1}{x} = l$.

If we assume such l exists, then for every $\epsilon>0$ there is some $\delta>0$, such that for all $x$, if $\left|x \right|< \delta$, then  $\left| \dfrac{1}{x}-l\right|<\epsilon$.
Then $\left|\dfrac{1-xl}{x} \right|<\epsilon$, so  $\left|\dfrac{xl-1}{x} \right|<\epsilon$, then $\dfrac{\left|xl-1 \right|}{\left|x \right|}<\epsilon$, and finally $\left| xl-1 \right|<\epsilon\cdot\left| x \right|$.

For contradiction, we need to find $\left|x\cdot l -1 \right| \geq \epsilon\cdot \left|x \right| $.

If we choose $\epsilon =1$, then we have, for all $\delta$ that $\dfrac{1}{n}< \delta$, as $n\in N$, then we can assume $x=\dfrac{1}{n}$.

From this we have $\left|\dfrac{l}{n}-1 \right|=\left|xl-1 \right|<\epsilon\cdot \left| x \right|=\left| \dfrac{1}{n}\right|$, furthermore $\left|l-n \right|<1$, but there is certainly some combination of l,n for which $\left|x-l \right|\geq 1 = \epsilon$, which contradicts our assumption about limits.
In conclusion, there is no such l for which $\lim\limits_{x \to 0}\dfrac{1}{x}=l$ works.

答:
I don't know, but just for laughs, I'll show a proof I came up with.

Suppose there is an $l \in \mathbb{R}$ such that $\lim\limits_{x \to 0} \dfrac{1}{x} = l$.

Then there exists $\delta \in \mathbb{R}$ such that,

$\left| \dfrac{1}{x}-l\right|<0.1\ $ for all $x$ satisfying$\ 0 < \left|x \right|< \delta < \frac{1}{3}$.

Now for example, both $-\delta^2\ $ and $\ \delta^2$ are in the region from the previous line.

However, $\left| \dfrac{1}{\delta^2}-l\right|< 0.1 \implies l > 8.9$, whereas

$\left| -\dfrac{1}{\delta^2}-l\right|< 0.1 \implies l < -8.9,$ a contradiction.


问:
How does $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...
How can $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...? Why odd numbers?</p>

Body: $$1+3+5+\ldots+(2n-1)=\sum_{k=1}^n (2k-1)=\sum_{k=1}^n [k^2-(k-1)^2]=n^2-0^2=n^2=\sum_{k=1}^{n^2} 1$$


问:
Is a function differentiable at the end point of its domain?
I've been going through Spivak's calculus and in it spivak states that for a function $f$ to be continuous on a closed interval $[a,b]$ the function is said to be continuous at the end points if $\lim_\limits{x\to{a^+}}f(x)=f(a)$
And
$\lim_\limits{x\to{b^-}}f(x)=b$

However can the same reasoning apply to the derivative of a function i know that limits are based off taking limits from any direction within the domain of the function that is $x$ lies in the domain of $f$ so for example if I had the function
$f:[a,b]\longrightarrow \mathbb{R}$
when calculating the derivative at the end point for example at b since f is not defined for points greater than b is it true to say that the derivative of the function at b is the left handed derivative of b
That is
$$
f'(b)=\lim_{h\to 0^-}\frac{f(b+h)-f(b)}{h}
$$
Thanks in advance.

答:
Yes, you can do that. The idea is that this derivative enables you to make sense of what would happen if you were to extend $f$ beyond the interval $[a,b]$. However, you need to be careful when extending this new notion of derivative to the theorems you are familiar with. Just take $f : [0,1] \to \mathbb R$ defined by $f(x) = x$. It has a well-defined derivative on $[0,1]$ equal to $1$ everywhere, but your theorem that says that if $f$ takes its maximum at $x_0$, then $f'(x_0) = 0$, is not true anymore. The theorem becomes "if $f$ takes its maximum at $x_0$, then either $f'(x_0) = 0$ or $x_0$ is at the endpoints of your interval".

Hope that helps,


问:
Does the reduced column echelon form of a matrix $A$ have the same null space as $A$?
Row reduction never changes the null space of a matrix $A$. But can column reduction change the null space?

I do not think this is true, my attempt at a counterexample is

$$ A =
\begin{pmatrix}
1&0&1&0\\
0&1&0&1\\
\end{pmatrix}
$$

The null space of $A$ equals the span of $[1, 1, -1, -1]$. The column reduced form of $A$ is

$$
R=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
\end{pmatrix}
$$

and the null space of $R$ is the span of $[0,0,c_1, c_2]$ where $c_1, c_2 \in \mathbf{R}$.

So $N(A) \neq N(R)$. But I may have made a mistake as these concepts are still new to me.

As a side note, I suspect that for any invertible matrix, column reduction would not change the nullspace, but I'm not sure how to prove this.

答:
Note that composition of column operations on $A$ is the same as $A$ right-multiplied with an invertible matrix $T$. e.g.

$$\begin{align*}
A &= \pmatrix{1&0&1&0\\0&1&0&1}\\
R &= \underbrace{\pmatrix{1&0&1&0\\0&1&0&1}}_A
\underbrace{\pmatrix{1&0&-1&0\\0&1&0&-1\\0&0&1&0\\0&0&0&1}}_T\\
&= \pmatrix{1&0&0&0\\0&1&0&0}
\end{align*}$$

If $\mathbf v\in N(A)$, i.e. $A\mathbf v = \mathbf 0$, then consider $T^{-1}\mathbf v$,

$$R \left(T^{-1}\mathbf v\right) = ATT^{-1}\mathbf v = \mathbf 0$$

so $T^{-1}\mathbf v \in N(R)$.

And similarly the inverse: if $\mathbf v \notin N(A)$, then $R\left(T^{-1}\mathbf v\right) = ATT^{-1}\mathbf v \ne \mathbf 0$, i.e. $T^{-1}\mathbf v \notin N(R)$.

Then your side note about the null space of invertible matrices is obvious:

The null space of an invertible matrix contains only the zero vector $\mathbf 0$, so the null space of the matrix after column operations contains only $T^{-1}\mathbf 0 = \mathbf 0$.

For your counterexample, the null space of $A$ has vectors of the form

$$\mathbf v
= \pmatrix{c_1\\c_2\\-c_1\\-c_2}
= c_1\pmatrix{1\\0\\-1\\0} + c_2\pmatrix{0\\1\\0\\-1}
$$

Then the vectors in $N(R)$ can be computed as:

$$\begin{align*}
T^{-1} &= \pmatrix{1&0&1&0\\0&1&0&1\\0&0&1&0\\0&0&0&1}\\
T^{-1}\mathbf v
&= \pmatrix{c_1-c_1\\c_2-c_2\\-c_1\\-c_2} = \pmatrix{0\\0\\c_1'\\c_2'}\\
&= T^{-1}\left[c_1\pmatrix{1\\0\\-1\\0} + c_2\pmatrix{0\\1\\0\\-1}\right]\\
&= c_1\pmatrix{0\\0\\-1\\0} + c_2\pmatrix{0\\0\\0\\-1}\\
\end{align*}$$

This shows how $N(R)$ may be derived from $N(A)$ and $T^{-1}$:

$$\begin{align*}
N(A) &= \operatorname{span}([1,0,-1,0],[0,1,0,-1])\\
N(R) &= \operatorname{span}\left(T^{-1}[1,0,-1,0],T^{-1}[0,1,0,-1]\right)
\end{align*}$$

But while the null spaces change, what's preserved here is that a set of independent vectors in $N(A)$, after left-multiplied with $T^{-1}$, will still be a set of independent vectors in $N(R)$.


答:
I agree with Don’s comment this is confusing and unwelcome outside areas of physics where it’s common. I disagree that it makes a lot of sense.

If $a_i\vec{e}_i$ is a sum, how would you talk about an particular but arbitrary $a_i\vec{e}_i$ when you needed to? Maybe physicists never have to do that?

If you multiply two sums together that both use the index $i$, you have to change one of the letters if you aren’t explicit about the sum. This is likely confusing to a mathematics student. Why do you sometimes have to change letters when you multiply things, but at other times can’t?

Another problem with using this generally: mathematics often includes various different index sets in single expressions: $\sum_{i=0}^n\sum_{j=1}^{i-1} a_{ij}$. It’s not easy to see when you can and when you can’t write such expressions in Einstein notation.

There are a lot of confusing consequences of this notation outside a narrow context.

[Disclaimer: I was a physics major for a while, and this notation (among other things) led me to study mathematics.]

Title: Is Einstein notation universally applicable?

Body:
It gets surprisingly monotonous to write the sum symbol repeatedly, so I'm wondering if one can just swap in Einstein notation at will to avoid the monotony, assuming that the symbol that's being summed over appears exactly twice?

For example, the wiki definition of $\nabla$ uses the sum symbol:

$\nabla = \sum_{i=1}^n \vec e_i {\partial \over \partial x_i} $

Could one simply write

$\nabla = \vec e_i {\partial \over \partial x_i} $

instead, and if so, would it need to be prefaced with "Using Einstein summation notation... " or is even that necessary?

If this question is more appropriate for physics.se, I'll post it there. Math.se has more hits on the search "Einstein notation," so I thought I'd ask here first.

Thank you for any help.


答: You are correct, but you also don't need a provisional definitions when considering limits. The rigorous definition of a limit for a function $f:A\subseteq \mathbb{R}\to\mathbb{R}$ at a point $x_0$ only consider the elements of the domain. That is we call $L=\lim_{x\to x_0}f(x)$ if for every $\epsilon > 0$ there is a $\delta >0$ such that if $x\in A$ and $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$. That means you only have to consider how the function approaches for $x$ close $x_0$ in the domain, so in the boundary case, the limit and the right-handed limit are the same as the elements above or below these limit points are not in the domain. Some people may have issues with this definition as it would imply all functions defined on, let's say, the integers are continuous. But further down the line you will see this is consistent with the notion of the subspace topology of a given topological space.

Title: Is a function differentiable at the end point of its domain?

Body:
I've been going through Spivak's calculus and in it spivak states that for a function $f$ to be continuous on a closed interval $[a,b]$ the function is said to be continuous at the end points if $\lim_\limits{x\to{a^+}}f(x)=f(a)$
And
$\lim_\limits{x\to{b^-}}f(x)=b$

However can the same reasoning apply to the derivative of a function i know that limits are based off taking limits from any direction within the domain of the function that is $x$ lies in the domain of $f$ so for example if I had the function
$f:[a,b]\longrightarrow \mathbb{R}$
when calculating the derivative at the end point for example at b since f is not defined for points greater than b is it true to say that the derivative of the function at b is the left handed derivative of b
That is
$$
f'(b)=\lim_{h\to 0^-}\frac{f(b+h)-f(b)}{h}
$$
Thanks in advance.


问:
How can I solve an integral by using hypergeometric function of the second kind?
This is an article that I've been studing for a while. I came across this integral equation
$$\mathrm dm=\frac{\mathrm dr}{{r}^{z+1}\left[1-{\left(\dfrac{r_h}{r}\right)}^{n+z-1}\right]}$$
where $n$, $z$ and $r_h$ are some parameters and $r$ and $m$ are my variables. The general solution for $n+z>1$ is
$$\ m=\frac{-{r}^{-z}}{z} _2F_1\big[1,\frac{z}{n+z-1};1+\frac{z}{n+z-1};{(\frac{r_h}{r})}^{n+z-1}\big]. $$

I want to know how that integral leads to a hypergeometric function of the second kind.

答:
The integral can transformed to an expression that's easier to relate to ${_2}F_1$. First, by a change of variables $r\to x\equiv r_h/r$ and, second, by another change $x\to y\equiv x^z$, the primitive would be given by
$$\int \frac{\mathrm d r}{r^{z+1}\left[1-\left(\frac{r_h}{r}\right)^{n+z-1}\right]}=-\frac{1}{r_h^z}\int\frac{x^{z-1}\mathrm d x}{1-x^{n+z-1}}=-\frac{1}{z\,r_h^z}\int\frac{\mathrm dy}{1-y^\alpha},$$ where I took the freedom to define $\alpha=(z+n-1)/z$. The last integral is $$\int\frac{\mathrm d y}{1-y^\alpha}=y\cdot {{_2}F_1}\left(1,\frac1\alpha;1+\frac1\alpha;y^\alpha\right)$$ which, upon undoing of the changes of variables, gives you the result you show for $m$.

Now, if you're interested in how the last integral is related to ${_2}F_1$ you should probably dig into the definition of the hypergeometric function through a power series and see how to solve the integral to obtain that kind of result explicitly.

OwnerDisplayName: Q Tara

OwnerDisplayName: anonymous


问:
Monotone functions are Borel-measurable
I know that there are many similar questions to mine. But this is more a question concerning a certain "logic".

I want to show, that every monotone function $f: \mathbb{R} \rightarrow \mathbb{R}$ is Borel-measurable.
Via definition of Borel-measurability, it suffices to show, that $$\forall y \in \mathbb{R}: \{f \geq y\} \in \mathfrak{B}(\mathbb{R}). $$

Proof:

Let wlog $f$ be a non-increasing function and $y \in \mathbb{R}$ fix.

(1)  If $\{f \geq y \}  = \emptyset$ than $\{f \geq y \} \in \mathfrak{B}(\mathbb{R}) $

(2) If $\{f \geq y \}  \neq \emptyset $ define $x^* :=\sup\{x \in \mathbb{R}: f(x) \geq y \} $. Than we have:

$x^* = \infty \Rightarrow \{ f \geq y \} = \mathbb{R} \in \mathfrak{B}(\mathbb{R}) $

$x^* < \infty$:

$x^* \in \{ f \geq y \} \Rightarrow \{f \geq y \} = (- \infty, x^*] \in \mathfrak{B}(\mathbb{R}) $

$x^* \notin \{ f \geq y \} \Rightarrow \{f \geq y \} = (- \infty, x^*) \in \mathfrak{B}(\mathbb{R}) $

Since all these sets are in $\mathfrak{B}(\mathbb{R})$ and $y \in \mathbb{R}$ arbitrary $\Rightarrow$ $f$ is measurable.

I hope that this is so far okay. But I am not really happy with the last distinction: How is it even possible, that  $x^* \notin \{ f \geq y \}$, since $f$ is a non-increasing function? I have a hard time thinking about this trivial fact; some help would be awesome!

答: Your proof is fine. Let $f(x)=0$ for $x<0$ and $f(x)=-1$ if $x \geq 0$. Take $y=0$. Then $\{x: f(x) \geq y\}=(-\infty, 0)$ and $x^{*}=0$.

Title: Monotone functions are Borel-measurable


问:
If $(f_n):[0, 1] \to [0, 1]$ are continuous and converge to $f$ pointwise, must $f$ be Riemann Integrable?
I am trying to solve the following question

True or False? If $(f_n):[0, 1] \to [0, 1]$ is a sequence of continuous functions which converges to $f$ pointwise, then $f$ is Riemann integrable and $\int_0^1{f_n} \to \int_0^1f.$

With help from the comments I found this counterexample, but I am hoping there is a simpler one.

If we replace the Riemann integrals by Lebesgue integrals, then the result is true by the Dominated Convergence Theorem. This implies that if $f$ is Riemann Integrable, then indeed  $\int_0^1{f_n} \to \int_0^1f.$ So in looking for a counterexample, we should try to find one where $f$ is not Riemann integrable.

Thank you very much for any help.

答: The classic counterexample is the following : $\mathbf{1}_{\mathbb{Q}}(x)=\lim\limits_{n\rightarrow +\infty}\lim\limits_{m\rightarrow +\infty}\cos(n!\pi x)^{2m}$. Let $f_n(x)=\lim\limits_{m\rightarrow +\infty}\cos(n!\pi x)^{2m}$ (which exists because it is the limit of a positive decreasing sequence), then either there exists $n_0$ such that $f_{n_0}$ is not Riemann-integrable which forms a counterexample because $x\mapsto\cos(n! \pi x)^{2m}$ is Riemann-integrable for all $m$, either the $f_n$ are all Riemann-integrable, but since $\mathbf{1}_{\mathbb{Q}}$ is not Riemann-integrable and $\mathbf{1}_{\mathbb{Q}}(x)=\lim\limits_{n\rightarrow +\infty}f_n(x)$, then this is a counterexample.


答:
Note
$$ \frac12\ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }
= \frac12\ln {\frac {\frac {p \tanh a x }{\sqrt {p^2 + q^2} }  + 1}{\frac{p \tanh a x }{\sqrt {p^2 + q^2} }-1}}
=\tanh^{-1}\frac {p \tanh a x }{\sqrt {p^2 + q^2} }\\
$$

So, there is a typo in the second expression.

Title: Indefinite integral $\int \frac {\mathrm d x} {p^2 + q^2 \cosh^2 a x}$

Body:
Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.584$ gives:

$$\int \dfrac {\mathrm d x} {p^2 + q^2 \cosh^2 a x} = \begin{cases} \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \left({\dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } }\right) + C \end{cases}$$

(As usual he glosses over negative argument to the $\ln$, but all you need to do is modulus it.)

He does not give the conditions under which either case holds. This is crucial.

First I multiplied top and bottom by $\operatorname {csch}^2 a x$, then used $\operatorname {csch}^2 a x = \coth^2 a x - 1$ for the bottom, then substituted $u = \coth a x$ to convert it into the form:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\frac {p^2} {p^2 + q^2} - u^2}$$

I can't see there is any other case than that $\dfrac {p^2} {p^2 + q^2}$ is positive, hence:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\left({\frac p {\sqrt {p^2 + q^2} } }\right)^2 - u^2}$$

This is a standard integral, giving:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} }\right| + C$$

which leads to:

$$\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C$$

And that all works fine, no worries.

BUT, as far as I can see, there is no way to get to the arctangent result. To get there, we need $u^2 + k^2$ for some constant $k$, and I can't see how to get there from the given integrand.


问:
Separation of variables on a second order ODE
I am attempting to grasp the basics of separation of variables for a second order separable differential equation, and am failing to do so:

Given the equation:

$$ x=\frac{d^2y}{dx^2}$$

I know from calculus class intuitively that the solution is:

$$ y=\frac{x^3}{6}$$

But if I am to use separation of variables, why don't I get:

$$ \frac{y^2}{2}=\frac{x^3}{6} $$

Since I should have had to integrate the $y$ side twice as well.

答:
This comes from bad standard notation of the second derivative.  If you want to do algebraic manipulations, the second derivative of $\frac{dy}{dx}$ is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$.  This is found by using the quotient rule on the first derivative.

However, we can simplify things by not using either notation, and simply writing the second derivative as being the differential of the first derivative divided by $dx$:

$$y'' = \frac{d\left(\frac{dy}{dx}\right)}{dx}$$

This is a form that (a) shows where the differentiation occurs, (b) is algebraically manipulable, and (c) isn't as messy.

So, on your problem, you start out with

$$ x = \frac{d\left(\frac{dy}{dx}\right)}{dx} $$

Multiplying both side by $dx$ gives you:

$$ x\,dx = d\left(\frac{dy}{dx}\right)$$

We can integrate that:

$$ \int x\,dx = \int d\left(\frac{dy}{dx}\right) \\
\frac{x^2}{2} + C_1 = \frac{dy}{dx}$$
Note that on the right hand side, this works because the integral and differential are opposites, so they basically cancel out (obviously leaving a $+C$ if the integral is the last operation).

Now, we multiply both sides by $dx$ again.  This yields:

$$\frac{x^2}{2}\,dx + C_1\,dx = dy$$

Integrating yields:

$$\int \left(\frac{x^2}{2}\,dx  + C_1\,dx\right) = \int dy \\
\int \frac{x^2}{2}\,dx + \int C_1\, dx = \int dy \\
\frac{x^3}{6} + C_1x + C_2= y $$

That is the mechanics of how it works.


问:
Interior points and measure zero set.
I have a question about my proof. I assume $A$ has measure zero, and I want to prove that $A$ has no interior points. I assume by contradiction that $A$ does have an interior point $x$. So, there exists $\delta>0$ s.t there exists an open ball $B_x(\delta)\subseteq A$. Now since $A$ is a measure zero set, I know that for $\varepsilon = \frac{\text{Vol}(B_x(\delta))}{3}$ there exist cuboids $\{R_k\}_{k=1}^{\infty}$ s.t $A\subseteq \bigcup_{k=1}^{\infty}R_k$, and $\text{Vol}(\bigcup_{k=1}^{\infty}R_k)\leq\sum_{k=1}^{\infty}\text{Vol}(R_k)<\varepsilon$, but thats a contradiction to the fact that $B_x(\delta)\subseteq A\subseteq \bigcup_{k=1}^{\infty}R_k$ (we found a subset of a set whose volume is bigger than the volume of the set).
Is this proof correct?

答:
That proof is correct, but can be cleaned up slightly.

In my mind, it is easier to say that $B_\delta(x) \subseteq A$ for some $x$ and some $\delta$. Moreover, we know that $B_\delta(x)$ has positive measure. Then by monotonicity of measure, we have $0 < m(B_\delta(x)) \leq m(A)$, and so $A$ cannot be null.

Obviously your proof uses the same idea, but an appeal to open covers of small measure seems a bit excessive in this case. I'll emphasize again that your proof absolutely works, though!

I hope this helps ^_^

Title: Interior points and measure zero set.


问:
Finding the angle between $x-5=0$ and $2x+2y-7=0$
I have 2 line equations: $x-5=0$ and $2x+2y-7=0$

I believe the angle is $45^\circ$, but I don't know how to calculate the second equation when there's a $y$ as well.

Thanks!

答:
Here's a way without trig or vectors.

We can find the intersection of the two lines. $x-5=0$ means that $x=5$, so plugging that into the second equation gets us $y=-\frac{3}{2}$. So the point of intersection is $$(5, -\frac{3}{2}).$$

Refer to the Desmos graph here. (When solving individually you should always draw out the problem.) The red line is $2x+2y-7=0$ and the blue line is $x-5=0.$ The intersection point $(5, -\frac{3}{2})$ is marked as $X$.

Let's try drawing a line perpendicular to $x-5=0$. We can make it stop when it touches $2x+2y-7=0$. The inspiration for this is because we can try creating a right angle with $x-5=0$, a strictly vertical line. This can give us a right triangle, which is usually helpful on geometry problems. The line is the purple dashed one. It intersects the red line at $A$ and the blue line at $B$.

We now have a right triangle, and it looks to be isosceles, too! More specifically, it seems that $AB =BX$. (Can you prove this? Hint: $A$ has to be the same distance from $B$ as $X$ is.)

Now, a right isosceles triangle has angles of $45-45-90.$ Since we already have the $90^\circ$ angle, our answer must be $$\boxed{45^\circ.}$$

Note: We got lucky that the triangle is a right isosceles. If you don't know trigonometry already, I strongly advise you to learn the basics to solve problems like these in less than a minute. The purple line we drew would also help if we solved using trig functions.


问:
mixture problem
From Stewart, Precalculus, $5$th ed, p.$71$, q.$55$

The radiator in a car is filled with a solution of $60\%$ antifreeze and $40\%$ water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only $50\%$ antifreeze. If the capacity of the radiator is $ 3.6$L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

I don't think there is a need to set up a model, since the currently concentration of antifreeze is at $60\%$ and desired contraction is $50\%$, the difference is $10\%$ of the total capacity of the radiator.

$10\%\times3.6L = 0.36L$

$0.36$L to be drained from the original $60\%$ concentration of $\dfrac{2.16}{3.6}L$ and replaced with water would give $50\%$ concentration at $\dfrac{1.8}{3.6}L$

The answer at the back of the book says $0.6$L to be drained. I don't understand why this is so, if $0.6$L is drained the new concentration would be $43.333\%$.

答:
Let $x$ be the amount in litres of coolant replaced by water.

Amount of water originally in the coolant: $0.4 \times 3.6$L

Amount of water removed when $x$ amount of coolant is removed: $(0.4\times x)$ L

Note that the removal of $x$ amount of coolant would result in the removal of fluids according to their proportion.

Amount of water when $x$ amount of water is added: $\left(0.4 \times 3.6-(0.4\times x)+x\right)$L $=(1.44+0.6x)$L

Amount of antifreeze originally in the coolant: $0.6 \times 3.6$L

Amount of antifreeze removed when $x$ amount of coolant is removed: $(0.6\times x)$ L

Amount of antifreeze when $x$ amount of water is added: $\left(0.6 \times 3.6-(0.6\times x)\right)$L $=(2.16-0.6x)$L

For equal fluid proportions, the amount of fluids should be equal.

$\therefore 1.44+0.6x=2.16-0.6x$

or $x=0.6$L

Title: mixture problem


问:
Proving $(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)$
I wish to verify that:
$$
(A\setminus B)\setminus (C\setminus B)=A\setminus (B\cup C)
$$
I wish to present my attempt: (For the $(\subseteq)$ implication)
\begin{align*}
&x\in(A\setminus B)\setminus (C\setminus  B) \\
\implies&x\in(A\setminus B)\wedge x\notin(C\setminus B) \\
\implies&[(x\in A)\wedge(x\notin B)]\wedge[(x\notin C)\wedge(x\in B)]
\end{align*}
Unfortunately, I am stuck at this process and I am unable to proceed. I am new to elements of set theory and I am currently practicing such practice problems.

答:
I like to approach such type of problem through the definitions and properties involved.

Precisely, one has that

\begin{align*}
(A - B) - (C - B) & = (A\cap B^{c})\cap(C\cap B^{c})^{c}\\\\
& = (A\cap B^{c})\cap(C^{c}\cup B)\\\\
& = (A\cap B^{c}\cap C^{c})\cup(A\cap B^{c}\cap B)\\\\
& = (A\cap B^{c}\cap C^{c})\cup\varnothing\\\\
& = A\cap(B^{c}\cap C^{c})\\\\
& = A\cap(B\cup C)^{c}\\\\
& = A - (B\cup C)
\end{align*}


问:
Linearity of Expectation - Random Variables
If I have the equation:
 $E[X+E[Y]]=E[X]+E[E[Y]]$
 Using the formula: $E[X+Y]=E[X]+E[Y]$

But from other hand I can say that: $E[X+E[Y]]=E[X]+E[Y]$
 Using the formula: $E[aX+b]=aE[X]+b$ because $E[Y]$ is a number like $b$

So what is the true one?

答: Both are correct. $E[Y]$ is not random so $E[E[Y]]$ is same as $E[Y]$.


问:
Find all positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$
All I have got so far is that $a$ must divide $2002$... Could anyone share some ideas? I mean is there any method rather than trial and error?

答:
This isn't an answer, but it's an idea on how you could further narrow down your options

You can try taking (mod a) of either side to get $-1\equiv2001($mod $a)$ (I'm assuming that's how you got a divides 2002).

Taking (mod 3) of either side shows that a isn't congruent to 0 or 2 (mod 3), so it must be congruent to 1 (mod 3), and n must be even.

Finally you can take the (mod a+1) of either side to get $-1^{n+1}\equiv 2001($mod $a+1) \Rightarrow (a+1)|2002$ since n is even.

This reduces the possible values of a to 1 and 13, and a=1 obviously won't have any solutions.


问:
The general formula of the series: $22+21+20 + \cdots + 1 = 253$
Would the general formula of the series: $22+21+20 + \cdots +  1 = 253 $ expressed as a summation would be: $ X_i-\sum_{i=22}^n X_i-1 $? Or am I wrong?

I am sure this formula is basic but I could not find it.

This is a series from $22$ to $1$ but I try to figure out the general formula. If this formula have a standard name would be appreciate to know it. Thank you in advance.

Edit: Sorry I fixed the question

For example:

$$5+4+3+2+1 = 14$$
$$3+2+1 = 6$$

Would $ X_i-\sum_{i=22}^n X_i-1 $  be the general expression?

答: Sometimes called Gauß's sum, because of a story of him as a youth, we have that $1+2+3+\dots+n=\sum_{i=1}^ni=n(n+1)/2$.


答: I think you can prove this in a more straightforward way, using Cauchy's definition (proving that there's an unbounded limit - so it's not a real number) or better - you could use Heine's definition. If you choose Heine, just assume that there exists a limit, L that is a real number. Then, according to Heine, we can choose any sequence a(n) that approaches 0, as n grows to infinity, that we can substitute x with in the function ( substituting x approching to 0 with n approching infinity) and its limit will be L. Specifically, we can choose a(n) = 1/n, which will leave us with 1/(1/n)=n as n approaches to infinity - which is an unbounded sequence. This contradicts our initial assumption.</p>

Title: Is this proof of, that there is no limit of lim 1/x as x approaches 0, correct?

Body:
I just wanted to ask, if the following proof can be used to proof, that there is no l, such as $\lim\limits_{x \to 0} \dfrac{1}{x} = l$.

If we assume such l exists, then for every $\epsilon>0$ there is some $\delta>0$, such that for all $x$, if $\left|x \right|< \delta$, then  $\left| \dfrac{1}{x}-l\right|<\epsilon$.
Then $\left|\dfrac{1-xl}{x} \right|<\epsilon$, so  $\left|\dfrac{xl-1}{x} \right|<\epsilon$, then $\dfrac{\left|xl-1 \right|}{\left|x \right|}<\epsilon$, and finally $\left| xl-1 \right|<\epsilon\cdot\left| x \right|$.

For contradiction, we need to find $\left|x\cdot l -1 \right| \geq \epsilon\cdot \left|x \right| $.

If we choose $\epsilon =1$, then we have, for all $\delta$ that $\dfrac{1}{n}< \delta$, as $n\in N$, then we can assume $x=\dfrac{1}{n}$.

From this we have $\left|\dfrac{l}{n}-1 \right|=\left|xl-1 \right|<\epsilon\cdot \left| x \right|=\left| \dfrac{1}{n}\right|$, furthermore $\left|l-n \right|<1$, but there is certainly some combination of l,n for which $\left|x-l \right|\geq 1 = \epsilon$, which contradicts our assumption about limits.
In conclusion, there is no such l for which $\lim\limits_{x \to 0}\dfrac{1}{x}=l$ works.


答:
You could go with $$\sum_{n=1}^{22}(23-n)$$Note that each summand is positive, so the $-$ signs have to be kept under control. What you have written is definitely wrong.

Note that I have organised the index to count the number of terms ($22$), and then expressed the summands in terms of the index. Doing that will help you to keep things under control.

Title: The general formula of the series: $22+21+20 + \cdots + 1 = 253$

Body:
Would the general formula of the series: $22+21+20 + \cdots +  1 = 253 $ expressed as a summation would be: $ X_i-\sum_{i=22}^n X_i-1 $? Or am I wrong?

I am sure this formula is basic but I could not find it.

This is a series from $22$ to $1$ but I try to figure out the general formula. If this formula have a standard name would be appreciate to know it. Thank you in advance.

Edit: Sorry I fixed the question

For example:

$$5+4+3+2+1 = 14$$
$$3+2+1 = 6$$

Would $ X_i-\sum_{i=22}^n X_i-1 $  be the general expression?


问:
Prime ideal factorization in non-monogenic ring of integers ($K=\mathbb{Q}(\sqrt[4]{24})$)
Consider the number field $K=\mathbb{Q}(\sqrt[4]{24})$. Then one can show that $\mathcal{O}_{K}=\mathbb{Z}[\alpha,\beta]$ where $\alpha=\sqrt[4]{24}$ and $\beta=\frac{\alpha^{3}}{4}$. The corresponding minimal polynomials are given by
$f_{\alpha}=X^4-24$ and $f_{\beta}=X^4-54$.

Then how does one factor the prime ideals in such a non-monogenic ring of integers. For monogenic ring of integers it can be done quite easily by using the Kummer-Dedekind Theorem.

For instance how would one factorise $5\mathcal{O}_{K}$? If we view the corresponding ideal in the subrings $\mathbb{Z}[\alpha]$ and $\mathbb{Z}[\beta]$, then the prime is actually inert. So my first thoughts were that it will also be inert in $\mathcal{O}_{K}$. But after I used some sage code I found the following factorisation

$5\mathcal{O}_{K}=(5,\frac{\alpha^2}{2}+1)(5,\frac{\alpha^2}{2}-1)$.

Any help would be appreciated!

答:
The Kummer-Dedekind theorem does not require $\mathcal O_K = \mathbf Z[\alpha]$. If $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer with a minimal polynomial $f(x) \in \mathbf Z[x]$, then for every prime $p$ not dividing the index $[\mathcal O_K:\mathbf Z[\alpha]]$, the factorization of $p\mathcal O_K$ resembles that of $f(x) \bmod p$.  Even if you're not sure about the value of the actual index $[\mathcal O_K:\mathbf Z[\alpha]]$, that number is a factor of the very computable disciminant ${\rm disc}(f) = {\rm disc}(\mathbf Z[\alpha])$, so you can apply the theorem to all primes not dividing ${\rm disc}(f)$ while remaining unsure if it would be valid to apply it to some of the prime factors of ${\rm disc}(f)$ (settling that would require further study of the particular situation).

The reason the theorem is called "Kummer-Dedekind" is because of the role each one played, Kummer in the 1830s and Dedekind many decades later:

Kummer only treated $\mathcal O_K = \mathbf Z[\alpha]$, and that itself is being generous since he did not have a general concept of algebraic integer.  He only worked with rings of integers in cyclotomic fields, where the full ring of integers very fortunately has the form $\mathbf Z[\alpha]$ and thus Kummer could succeed without realizing the true difficulty of the general case (non-monogenic rings of integers).

Dedekind came up with the general notion of algebraic integers and  he proved that Kummer's theorem still works at all primes $p$ such that  $p \nmid [\mathcal O_K:\mathbf Z[\alpha]]$. At first he hoped that even if $p \mid [\mathcal O_K:\mathbf Z[\alpha]]$ he could find a $\beta$ (depending on $p$) such that $p \nmid [\mathcal O_K:\mathbf Z[\beta]]$, thus allowing him to factor each $p$ by turning that into a problem of factoring a polynomial mod $p$.  That hope was dashed when he discovered the first example of a non-monogenic ring of integers ($K = \mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$, for which $[\mathcal O_K:\mathbf Z[\alpha]] = 2$ and in fact $2 \mid [\mathcal O_K:\mathbf Z[\beta]]$  for all $\beta \in \mathcal O_K - \mathbf Z$).

Example. $K = \mathbf Q(\sqrt[4]{24})$.  The discriminant of $x^4 - 24$ is $-2^{17}3^3$, so for each prime $p$ other than $2$ and $3$, the way $(p)$ factors is like the way $x^4 - 24 \bmod p$ factors over $\mathbf F_p$.  What about $p = 2$ and $p = 3$? The polynomial $x^4 - 24$ is Eisenstein at $3$, so $3 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ and therefore the Kummer-Dedekind theorem says the factorization of $(3)$ resembles the factorization of $x^4 - 24 \equiv x^4 \bmod 3$, so $3$ is totally ramified in $\mathcal O_K$.  To treat $p = 2$, we turn instead to $x^4 - 54$, which also has a root $\sqrt[4]{54}$ that is in $\mathcal O_K$ and generates $K$.  Since $x^4 - 54$ is Eisenstein at $2$, we have $2 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$ and thus the Kummer-Dedekind theorem tells us that $(2)$ factors in the same way as $x^4 - 54 \equiv x^4 \bmod 54$. So $2$ is totally ramified in $\mathcal O_K$.

I am not sure why you are saying "$5$ is inert" in $\mathbf Z[\sqrt[4]{24}]$ or $\mathbf Z[\sqrt[4]{54}]$.  You must have made a typographical error when coding in Sage or you misinterpreted the Sage output. Modulo $5$,
$$
x^4 - 24 = x^4 - 4 = (x^2 + 2)(x^2-2)
$$
and
$$
x^4 - 54 = x^4 - 4 = (x^2+2)(x^2-2).
$$
These are consistent with each other (both having two irreducible factors of degree $2$), as they must be since $5$ does not divide either index $[\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ or $[\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$. Here we have an "accident" that they factor mod $5$ literally in the same way because $x^4 - 24 \equiv x^4 - 54 \bmod 5$.
In any case, the Kummer-Dedekind theorem can be used with either $x^4 - 24$ or $x^4 - 54$ to factor $(5)$ and they both tell us that $(5) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ are prime ideals with norm $25$:
$$
\mathfrak p = (5,\sqrt[4]{24}+2) = (5,\sqrt[4]{54}-2), \ \ \ \mathfrak q = (5,\sqrt[4]{24}-2) = (5,\sqrt[4]{54}+2),
$$
where I am using positive fourth roots of $24$ and $54$ with the standard real embedding so that $\sqrt[4]{24}\sqrt[4]{54}$ is $6$ on the nose (as opposed to being $-6$).
When we declare $\mathfrak p$ to be the prime ideal $(5,\sqrt[4]{24}+2)$, modulo $\mathfrak p$ we have $\sqrt[4]{24}\sqrt[4]{54} = 6 \equiv 1 \bmod \mathfrak p$, so from $\sqrt[4]{24} \equiv -2 \equiv 3 \bmod \mathfrak p$ we must have $\sqrt[4]{54} \equiv 2 \bmod \mathfrak p$ as opposed to $\sqrt[4]{54} \equiv 3 \bmod \mathfrak p$.

I don't really need real numbers or positivity, but you have to be careful with $\sqrt[4]{24}$ and $\sqrt[4]{54}$ in
an abstract field $\mathbf Q(\gamma)$ where $\gamma^4 = 24$ because this field has two fourth roots of $24$ and two fourth roots of $54$. If you use fourth roots of both numbers together then you need to relate them to each other in a definite way to avoid making miscalculations. Since  $6/\gamma$ is a fourth root of $54$, and its product with $\gamma$ is $6 \equiv 1 \bmod \mathfrak p$, we can say
$$
\mathfrak p = (5,\gamma+2) = (5,6/\gamma-2), \ \ \ \mathfrak q = (5,\gamma-2) = (5,6/\gamma+2).
$$
It would be incorrect to say $\mathfrak p = (5,\gamma+2) = (5,6/\gamma+2)$ because then modulo $\mathfrak p$ we'd have $\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$ as well as $6/\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$, so $\gamma \equiv 2 \bmod \mathfrak p$, and such congruences for $\gamma \bmod \mathfrak p$ (congruent to both $3$ and $2$) would be inconsistent.


问:
Number of paths from vertex A or B or C to G or H or I
I am trying to find in how many ways I can get from any of A,B,C to any of G,H,I in the graph shown bellow visiting exactly n vertices without repeating any vertex. I have tried creating an adjacency matrix and setting the n as the exponent then adding up all the values at the bottom left corner where A,B,C and G,H,I intersect but this is giving me different values to the ones I calculated manually so I don't know where I went wrong and what I should be doing to solve this problem effectively.

The overall aim of solving this problem is to find an equation that given the "side length" (number of vertices on one row) in a graph that has the same square structure as the one above and a the size of the paths returns how many different paths there is.

答:
The problem is small enough that you can do this dynamic programming. Let num(start, end, n, mask) denote the number of paths, where

start is the start vertex,

end is the end vertex,

n is the length of the path, and

mask is a subset of the $9$ vertices; the vertices in this subset are prohibited from being used in the path. For your problem, you want $\text{mask}=\varnothing$, but intermediate calculations will have nonempty masks.

Then it follows that

num(end, end, n, mask) = 1 if n = 0, 0 otherwise.

If start != end, then num(start, end, n, mask) is the sum of num(neighbor, end, n-1, mask U {start}), where neighbor ranges over the neighbors of start which are not in mask.

Really, this is just using depth first search to trace out all possible paths. Since you want paths from any vertex in the first row to any vertex in the last row, you need to add up nine values with all possible start and end vertices as inputs.


问:
Show that $ |\operatorname{det}(x, y)| \leq|x|^{s}|y|^{t}|x-y|^{r} $
Show that for any $x, y \in \mathbb{R}^{2}$ and for any $r, s, t \in[0,1]$ with $r+s+t=2$ we have
$$
|\operatorname{det}(x, y)| \leq|x|^{s}|y|^{t}|x-y|^{r}
$$
Here $(x, y)$ represents the $2 \times 2$ matrix with vectors $x$ and $y$ as columns.

Hint: If you replace $y$ by $y-x \ldots$.

I don't know how to proceed, I guessed that replacing $y$ by $y-x$ will come at the end because of the symmetricity of $|y|^{t}$ and $|x-y|^{r}$ but I am not seeing any point to start. It is supposed to be a real analysis/calculus question.

答:
Let $D = |\det(x, y)|$. The estimate
$$
D \le |x| |y|
$$
follows from the Cauchy-Schwarz inequality:
$$
|D| = |x_1 y_2 - x_2 y_1| \le |x_1|| y_2| + |x_2|| y_1| \le \sqrt{x_1^2+x_2^2} \sqrt{y_2^2+y_1^2} = |x| |y| \, .
$$
Substituting $x$ or $y$ by $x-y$ does not change the absolute value of the determinant, so that also
$$
D \le |x| |x-y| \, ,\\
D \le |x-y| |y| \, .
$$
Now we can exponentiate these three inequalities with suitable exponents and multiply them to get the desired result:
$$
 D = D^{1-r} D^{1-t} D^{1-s} \le \bigl( |x| |y|\bigr)^{1-r}  \bigl( |x| |x-y|\bigr)^{1-t} \bigl( |x-y| |y|\bigr)^{1-s} = |x|^s |y|^t |x-y|^r \, .
$$


问:
Are there equivalent terms of injection and surjection about domain (not codomain)?
I think the terms "injection" and "surjection" are related to the codomain of a function, because when one say "this function is injective/surjective", it constrains the image of the funtion not to overlap/leave some elements in the codomain respectively. Are there any dedicated terms to constrain about not the codomain but the domain of a function (or relation?) in the same way?

(Context: I want to use such terms if exist, in my own programming language's documentation, in order to indicate a hashtable lookup function should always returns a single value (without "option" or similar) corresponding to the argument, if the hashtable is properly typed. My language's type system has some abstraction between exhaustiveness of patterm-matching and hashtable lookup.)

答:
At least one output

When discussing relations in general, there is precedent for "cosurjective" or "co-surjective". John Carlos Baez uses these terms in a Twitter thread and Kyle Miller uses them in "The theory of relations". This is similar to the category theory convention of using "co-" to denote a reversal of domains and codomains.

Some other terms are "serial" and "left-total". For "left-total", Wikipedia cites Mathematical Foundations of Computational Engineering: A Handbook, Semantics of Sequential and Parallel Programs, and Modelling of Concurrent Systems: Structural and Semantical Methods in the High Level Petri Net Calculus.

In a context like a complex analysis textbook, a relation with this property may be called a "multivalued function", a "multifunction", a "many-valued function", or perhaps/arguably a "set-valued function". For example, a multivalued square root function would take any complex number as input, but would output a pair of square roots (except at $0$). These terms are likely only used when the emphasis is on the possibility of more than one output, and they might even be used by some when, say, not every complex input produces an output.

Another term, mentioned in the comments, is  "defined everywhere" (or "everywhere-defined"). See A function that is defined everywhere but has unknown values [closed] on MathOverflow or Function defined everywhere but continuous nowhere on MathSE for positive examples, and Oliver Knill's lecture notes on limits and What is the limit of a function which is not defined everywhere? on MathSE for negative examples. This phrasing is used most often in cases like real-valued Calculus where it's already assumed that there is at most one output and we want to talk about things using function notation/terminology. In these contexts especially, the adjective "total" or the phrase "total function" may be used.

At most one output

Similarly to "cosurjective", there is the same precedent for "coinjective" or "co-injective".

Some other terms are "functional", "right-unique", "right-definite", and "univalent". For "right-unique", Wikipedia cites the same sources as for "left-total". For "right-definite", it cites Reasoning on Spatial Semantic Integrity Constraints. For "univalent", it cites a sci.physics.research post by John Carlos Baez.

In computer science especially, I see these relations called "partial functions". There is even special notation when this term is used. Instead of $f:X\to Y$, something like $f:X\not\to Y$ is sometimes written.

Occasionally I see "well defined" used for these relations, as in Function defined everywhere but continuous nowhere and outlined on Wikipedia. But I would not recommend this usage because the phrase has other meanings/implications. Perhaps most commonly, "well defined" usually carries extra baggage of a definition based on representatives of an equivalence class, and it's "well defined" from the set of equivalence classes if it yields the same result for all members of the class. Wikipedia also mentions well-defined notation and a special meaning in the context of partial differential equations, if MathWorld is to be believed.


问:
Limits and Integration Recursion

I was able to obtain an answer of $k/(k+1)$ for part 1, by interchanging the limit and sum and then evaluating the limit of the integrand. However, for part (2), I'm stuck because using the same idea from part (1), I obtain an infinity - infinity indeterminate form. The substitution x = u^(n^-2) didn't help much either. Any help for parts (1) and (2) would be appreciated.

答: The above is my attempt at answering this part. I got something similar to what Sanjoy Kundu suggested.

Title: Limits and Integration Recursion


问:
Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?
Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?

Similarly, let $f :\mathbb{C}→ \mathbb{C}$ be a function such that $f^2$ and $f^3$ are analytic. Is $f$ analytic?

答:
In short: not necessarily in the real case, but yes in the complex case.

In the real case like in the complex one, $f$ is differentiable whenever it does not vanish and
$$
f'=\left(\frac{f^3}{f^2}\right)'=\frac{(f^3)'f^2-f^3(f^2)'}{f^4}.
$$

Real case: if $f$ vanishes, it no longer needs to be differentiable. For an alternative example to $f(x)=|x|$, observe that
$$
f(x)=x\sin\left(\frac{1}{x}\right)\quad \forall x\neq 0\qquad f(0)=0
$$
is not differentiable at $0$, while $f^2$ and $f^3$ are differentiable everywhere.

Complex case: the zeros of $f^2$ and those of $f$ coincide. If $f$ is constant equal to $0$, the result is clear. Otherwise, its zeros are isolated since $f^2$ is holomorphic non constant equal to $0$. On the open set which is the complement of these zeros, the argument above shows that $f$ is holomorphic. And $f^2$, whence $f$, is bounded near each zero. So these are removable singularities. Hence $f$ is holomorphic on its domain.

Edit: for a different argument, see this thread.

Note: if you replace the assumptions by $f^2$ (or $f^n$, $n\geq 2$, more generally) differentiable and $f$ continuous, you get the same conclusions in both cases. Indeed, where $f(x_0)\neq 0$, we have
$$
\frac{f(x)-f(x_0)}{x-x_0}=\frac{f^2(x)-f^2(x_0)}{x-x_0}\cdot\frac{1}{f(x)+f(x_0)}\longrightarrow \frac{(f^2)'(x_0)}{2f(x_0)}.
$$
But if you remove the continuity assumption on $f$, the complex case can fail as well. Just pick a random square root of $z$.


问:
How to calculate $|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle$ for $\mathfrak{su}(2)$?
At $\mathfrak{su}(2)$, tensor product $3 \otimes 3$ can be decomposed to
$3\otimes3=5\oplus3\oplus1$.

In this space, $|1,1\rangle \otimes |1,-1\rangle$ has weight $0$.

So we should be able to write down
$$|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle.$$

How to calculate $\alpha,\beta,\gamma$?

答:
Apply $\langle 2,0\mid$ on both sides so you get $\alpha$ on the right hand side (assuming standard normalization) and the inner product

$$
\langle 2,0 \mid 1,1 ; 1,-1 \rangle
$$

in the

$$
\langle J M \mid j_1 j_2 ; m_1 , m_2 \rangle
$$

notation. Now look at the Clebsch-Gordan  recurrence relations (the recursion @JyrkiLahtonen refers to).

This is the special case of $j_1=j_2=m_1=-m_2$ in the article. But check that you are using the same normalization convention as the article before using that to check your answer.


问:
A problem with uniform distribution?

In a game Alis and Daniel shoot arrows on a circular target with a
radius of R.

Alis throws arrows such that their distance from the center have
Continuous uniform distribution (0,R) - o,R are the parameters for
uniform distribution.

While Daniel throws arrows such that their distance from the center have
random Continuous uniform distribution (From the
target)

I was asked to calculate the expected value for the distance of the arrows for both participant.

For Alis it's R/2 but what about Daniel? I didn't understand what's given about him, what are the parameters for him...

I claimed: "For Alis it's R/2" Proof:

The expected value for random variable with uniform distribution (a,b) (with parameters a and b) is (a+b)/2

答:
HINT: Presumably it means that the probability that Daniel’s arrow lands in a given region of the target is proportional to the area of the region. The area of an annulus of mean radius $r$ and width $dr$ is $dA=2\pi r\,dr$, and the area of the target is $\pi R^2$, so the probability that it lands in that annulus is

$$\frac{dA}{\pi R^2}=\frac{2\pi r\,dr}{\pi R^2}=\frac2{R^2}r\,dr\,.$$

Alternatively, the cumulative distribution function $F(r)$ for $0\le r\le R$ is given by

$$F(r)=\frac{\pi r^2}{\pi R^2}=\frac{r^2}{R^2}\,.$$

Use whichever approach you like to get the probability density function, and then use that to calculated the expected value from the definition of expected value.

Title: A problem with uniform distribution?


问:
Understanding geometry of quotient spaces
Let $L$ be a one-dimensional subspace of $\mathbb{R}^2$, namely, a line, say, the set of $(x,y)$ with $y = x$.  and consider the quotient $\mathbb{R}^2/L$. The equivalence classes, which make up the elements of this quotient space, are exactly the set of ordered pairs in $\mathbb{R}^2$ whose difference is an element of the line $y = x$. The equivalence class of the point $(a,b)$, assuming $a \neq b$, ends up being the line through $(a,b)$ parallel to the line $y = x$.

This geometry makes sense to me. By simple number-crunching, I can also see that
$$\dim \mathbb{R}^2/L  = \dim \mathbb{R}^2 - \dim L = 2 - 1 = 1.$$
I am trying to understand why, geometrically, it makes intuitive sense for the dimension to be $1$ or why it makes sense to say there is "one degree of freedom." When I sketch the picture, it almost looks like I'm picking a line up and "scaling it" across the $\mathbb{R}^2$ plane, in which case the only degree of freedom is "vertical distance" of some kind. Is there a better way to think of this?

答: Since you quotient by a line, it can be convenient to choose your coordinates so that one axis, say the $y$-axis is coincident with $L$.  Then every point has coordinates in this coordinate system and the quotient collapses each point $(x,y) \mapsto x$.  That is, the quotient acts like the linear map $(x,y) \mapsto 1 \cdot x + 0 \cdot y$.  Now look at the nullspace of this map, which is one dimensional and think about the rank-nullity theorem.


问:
Christoffel Symbols: when did the bracket notation first appear?
An early notation to Christoffel symbols is:

\begin{equation}
\begin{Bmatrix}
  m  \\
  \hspace{1.5mm}b\hspace{5mm} c\hspace{1.5mm}
\end{Bmatrix} = \frac{1}{2}g^{ma}\{ \partial_{b}g_{ca} + \partial_{c}g_{ab} - \partial_{a}g_{bc}\} \tag{1}
\end{equation}

with these curly brackets.

Now, in Einstein's notebook and in the seminal paper of Einstein Field Equations we see a clear usage of this notation:

and

I'm not sure if was Einstein indeed the first one to use this curly notation (or the german publishers). So, It was Elwin Christoffel the one who introduced this notation? If no,  would like to know the one who first have used this notation.

答: In Christoffel's original paper (according to wikipedia) introducing Christoffel symbols, he uses this notation (although with square brackets).</p>


问:
$\sigma$-weak closure of $B(H) \otimes B(K)$.
Let $H$ and $K$ be Hilbert spaces. I'm  trying to prove that  $B(H) \otimes B(K)$ (= minimal tensor product of $C^*$-algebras) is $\sigma$-weakly dense (= weak$^*$-dense) in $B(H \otimes K)$.

My progress so far:

I managed to show:

$B(H) \otimes B(K)$ is strongly dense in $B(H \otimes K)$

$B(H) \otimes B(K)$ contains the compact operators on $H \otimes K$, hence it suffices to show that the compact operators are $\sigma$-weakly dense in $B(H \otimes K)$.

Any ideas how I can proceed?

答:
Yes, the compact operators on any Hilbert space $H$ form a $\sigma$-weakly dense subset.  This is because otherwise there would be a nonzero trace class operator $S$ such that
$\text{tr}(TS)=0$, for all compact operator $T$, and it is not hard to see that this is impossible.