问:
Proving Sup(A+B) = Sup(A)+Sup(B), I got a question on something
$A,B $ : Nonemtpty bounded sets on R

I'm proving on $\sup(A+B) \geq \sup (A)+\sup(B)$

prove by contradicting assumption: $\sup(A+B) < \sup(A) + \sup(B)$

Then $\sup(A+B) -\sup (B)< \sup (A)$

Then there exists $a \in  A$ such that  $\sup(A+B)-\sup(B)<a$

$\sup(A+B) -a <\sup(B)$...

(next steps are omitted)

Now, I have a question on this prove on existence of $a$

For exmaple, Let $\sup(A+B)=X, \sup(B)=Y$. Then $X-Y<\sup(A)$

For sure, $a\le\sup(A)$. Then, how can we know $X-Y<a?$

Or am I wrong on proving this?

答:
Yeah this is a good question to ask. The fact is that there is such an $a$ because the supremums $X$ and $Y$ are fixed, whereas $\sup(A)$ is a limit point (i.e. elements of $A$ get arbitrarily close to $\sup(A)$).

Let $\sup(A+B)=X, \sup(B)=Y$. Then $X-Y <\sup(A)$ is your assumption that you want to contradict.

Note that $\sup(A+B)$ and $\sup(B)$ are fixed. That is, $X-Y \in \mathbb{R}$ is fixed.  If $X-Y < \sup(A)$, (strict inequality), then there is a gap $r$ between them:
$$\sup(A) - (X-Y) = r > 0.$$

Recall that by definition, the supremum is a limit point of $A$: there exists a sequence $\{a_i\}_{i=1}^\infty \in A$ such that $a_i \rightarrow \sup (A)$. So there is an element of this sequence that is arbitrarily close to $\sup(A)$, i.e. some $a_k$ such that
$$\sup(A) - a_k < r.$$

Combine these two facts to see that $$ X-Y <a_k.$$


问:
How many numbers are with $5$ different digits in decreasing order?
I mean that $54321$ is a valid number, but $16755$ is not.

I was thinking that there are $6^5$ possible numbers because there are 6 possibilities for each position in the number, but I don't know exaclty.

答: You can write $10\cdot9\cdot8\cdot7\cdot6\cdot5$ distinct numbers made of $6$ digits. Divide this amount by $5!$ to only allow the decreasing permutations.

ContentLicense: CC BY-SA 4.0


问:
Exact functors and their naive universal $\delta$-functors
This is Exercise 2.1.2 of Weibel's Introduction to homological algebra, and it might actually be a very simple exercise.

For $F: \mathcal A \to \mathcal B$ an exact functor, I want to show that $T_0 = F$ and $T_n = 0$ for $n \neq 0$ defines a universal $\delta$-functor.

Since $F$ is exact, then $T$ is almost trivially a $\delta$-functor, since the boundary maps are zero.

Then to check universality, if $S : \mathcal A \to \mathcal B$ is another $\delta$-functor for which there is a natural transformation of $S_0$ to $T_0 = F$, and $S_n$ to $T_n = 0$ for $n \neq 0$, then surely this is unique almost by construction? I am not sure if there is some nuance or detail to the argument that I am missing here, as I am not super familiar with homological algebra arguments in full generality.

答:
It is trivial but worth expanding on. As you say, the boundary map $T_1(C)\to T_0(A)$ is trivial because $F=T_0$ is left exact. Now for any other $\delta$ functor $S_*$ you can define $S_n\to T_n=0$ to be the trivial natural transformation whenever $n>0$ (this is the only possible nat. trans. into the trivial functor!). You just need to see that when applied to a short exact sequence this gives a map between the long exact sequences for $S_*$ and $T_*$.

I think getting used to working with commutative diagrams is something that this exercise will help with if you write it out in full.


问:
proof reading and a little question and refine english about college-algebra

Welcome to edit my post to revise any mistakes, both English and Math, thanks.

Theorem 12

polyIC=integer polynomial

polyRC=rational polynomial

Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\text{...}+a_0$ is a polyIC, and $\frac{r}{s}$ is a rational root, where $(r,s)=1$.

Then we must have $s\left|a_n\right.,r\left|a_0\right.$. Specially, If the leading coefficient of $f(x)$ satisfies $a_n=1$,

then the rational roots of $f(x)$ are all integer roots, and they are factors of $a_0$.

Proof$\unicode{ff1a}$

Because $\frac{r}{s}$ is an rational root of $f(x)$. On rational number field $\left.\left(x-\frac{r}{s}\right)\right|f(x)$.

Thus $[\text{Mark1}]$ $(\text{sx}-r)|f(x)$.  For $(r,s)=1$, $s x-r$ is an primitive polynomial.

Conclude from above, $f(x)=(s x-r)\left(b_{n-1}x^{n-1}+\text{...}+b_0\right)$,where $b_{n-1},\text{...},b_0$ are all integers.

Compare the coefficients of two sides, that is $a_n=s b_{n-1},a_0=-r b_0$.

Hence $s\left|a_n\right.,r\left|a_0\right.$.

one little question

Does $[\text{Mark1}]$ so obvious?

$\left.\left(x-\frac{r}{s}\right)\right|f(x)$ $\Longrightarrow $ $s|f(x)$

of course, this is obvious in this example: f below, has one rational root-4/5$\unicode{ff0c}$so 5(x-4/5)$|$f(x)

$\pmb{f=25\left(x-\frac{4}{5}\right)(x-2)\left(x+\frac{4}{5}\right)\text{//}\text{Expand}}$

$32-16 x-50 x^2+25 x^3$

$\pmb{\text{Solve}[f==0,x]}$

$\left\{\left\{x\to -\frac{4}{5}\right\},\left\{x\to \frac{4}{5}\right\},\{x\to 2\}\right\}$

$\pmb{\text{Factor}[f]}$

$(-2+x) (-4+5 x) (4+5 x)$

$\pmb{5(x-4/5)\text{//}\text{Expand}}$

$-4+5 x$

答:
By the Factor Theorem $\, f(r/s)=0\,\Rightarrow\, f = (x-r/s)\, g= (sx-r)\, h,\ h = g/s.\ $ Therefore $\,(r,s)=1\,\Rightarrow\,sx\!-\!r\,$ primitive, thus $\,(sx\!-\!r)h = f\in \Bbb Z[x]\Rightarrow h\in\Bbb Z[x],\,$ by Gauss's Lemma.

Therefore, with $\,\ell f := $ lead coef of $\,f,\,$ we have $\, s\mid \ell f = \ell((sx\!-\!r)h) = s\, \ell (h),\,$ and $\,r\mid f(0) = -r\,h(0),\ $ since $\ \ell(h),h(0)\in\Bbb Z,\,$ by $\,h\in\Bbb Z[x].$

ContentLicense: CC BY-SA 3.0


问:
Dirac measure $\delta_x$ not absolutely continuous for almost every $x$ in $[0,1]$ with respect to the Lebesgue measure $\mu$ on $[0,1]$.

Show that the Dirac measure $\delta_x$ not absolutely continuous with respect to the Lebesgue measure $\mu$ on $[0,1]$ for almost every $x$ in $[0,1]$ .

I though about using $\mathbb{Q}$ as the example set. However, it lead to the following which doesn't show what I'm trying to show:

Clearly $\mu(\mathbb{Q})=0$. However, $\delta_x(\mathbb{Q})=1$ if $x\in\mathbb{Q}$ and $\delta_x(\mathbb{Q})=0$ if $x\not\in\mathbb{Q}$.

答: $\delta_x <<\mu$ fails for every $x$. This is because $\mu\{x\}=0$ and $\delta_x \{x\}=1$.


问:
Nature of $u_n(x)=\sqrt{a_1(x)+\sqrt{a_2(x)+\sqrt{\dots+\sqrt{a_n(x)}}}},a_n(x)=x^{2^n}$
In a question posed in this forum convergence of $u_n=\sqrt{a_1+\sqrt{a_2+\sqrt{\dots+\sqrt{a_n}}}}$
, the author proved that if $a_n=\lambda^{2^n}$ then $a_n$ tends to $\phi\lambda,\phi=\frac{1+\sqrt{5}}{2}$. Now, if we take $a_n(x)=x^{2^n}$, does the sequence converges to $\phi x$ and how to show that? Thank you in advance.

答:
I guess the OP wants to know why if $a_n=x^{2^n}$ the series $u_n$ conveges to $\phi x$, something that is not proven in the linked question but taken for granted.

To do this, consider the following steps. Take first $x=1$, so that $a_n=1$ for each $n$.

(1) When $x=1$, the recursive relation is true:

$u_0=1, u_{n+1}=\sqrt{1+u_n}$

Suppose a limit exists $l$. Than it will satisfy (why?) $l=\sqrt{1+l}$

(2) To prove the limit exists, consider visually the function $f(x)=\sqrt{1+x}$ for positive $x$ and its intersections with the line $y=x$. Can you plot the succession of points $u_n$ ? In order to formalize the intuition show that:

$0\le x\le\phi \rightarrow 0\le f(x)\le\phi $

This shows that the sequence is bounded. Since it is increasing, it has a limit.

(3) Now change to $x>1$. Let's call $u_n(x)$ the new $u_n$. Can you show that $u_n(x)=x u_n(1)$ at a fixed $n$ ? (check small $n$ to develop an intuition) Than just take the limit of this relation and exploit the result you already found for $x=1$.


问:
Finitely generated modules over $\mathbb{C}^2$ are projective.
In a practice exercise on algebra, I want to show that there is an equivalence between the category of all finitely generated $\mathbb{C}^2$-modules and the finitely generated projective $\mathbb{C}^2$-modules. But I can hardly believe it is true (there are so many modules....), and I actually have no clue how I would go about tackling this problem. A module $P$ is projective if every short exact sequence of modules
$$
0\to A\to B\to P\to 0
$$
is split, or, if $P$ is a direct summand of a free $\mathbb{C}^2$-module. This last description seems to be the one I should go for, so how can I prove that any $\mathbb{C}^2$-module is the direct summand of a free $\mathbb{C}^2$-module?

答:
Hint:

A product of fields is an absolutely flat and noetherian ring, and a finitely generated flat module over a noetherian ring is projective.


问:
solution to $\text{div}(f)=0$
What is the most compact representation of the general solution to
$$
\text{div}(f)=\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+\frac{\partial f_3}{\partial x_3}=0
$$
for a $f:\mathbb{R}^3\to\mathbb{R}$ that goes to zero exponentially as $|\vec{x}|\to\infty$.

My first approach was the Helmholtz theorem with a potential that satisfies the Laplace equation.
Edit: Using the Helmholtz theorem,  $f=-\nabla \Phi + \nabla\times f$,   one can write
$$
\nabla\cdot f = -\nabla^2 \Phi + \nabla\cdot \nabla\times f=-\Delta \Phi =0
$$

答:
The Poincar'e Lemma, as in Spivak's Calculus on Manifolds, includes
the fact that for smooth functions ( $x$ and $f(x)$ in $R^3$, domain
of $f$ a star-shaped set) you have
$$
 f(x) = {\rm curl\,}\Big(
 \int_0^1 f(tx)\times tx\,dt
                    \Big)
 +\int_0^1 t^2x{\rm \, div\,}f(tx)\,dt.
$$
(I have changed the notation to express it for vector fields; Spivak
has it for differential forms.)

When the divergence of $f$ is zero in a star-shaped set there are many
functions $g$ such that $f = {\rm curl\,} g$; the first integral
above is one of them.

It looks like we can't guarantee the exponential decay:
Suppose $|f(x)| \le ce^{-|x|}$. Then
$$
 \bigg|
  \int_0^1 f(tx)\times tx\,dt
 \bigg|
  \le
  \int_0^1 ce^{-t|x|}t|x|\,dt
$$
$$
 =\bigg[
 -c\big(t+\frac{1}{|x|}\big)e^{-t|x|}
 \bigg]_0^1
 =
 c\Big(-\big(1+\frac{1}{|x|}\big)e^{-|x|}+\frac{1}{|x|}
 \Big).
$$

Title: solution to $\text{div}(f)=0$


问:
Solving A Linear Congruential System
I need to find a prime p which makes:
$p\equiv {\pm1}\text{ (mod }8)$ and
$p\equiv {\pm1}\text{ (mod }12)$

How could I find such $p$? Is there any specified method I can use?
I'd be grateful if anyone could point me to the solution. Thanks in advance!

答:
Iff:$$q\equiv {a}\text{ (mod }lcm(n,m))$$ then $$q\equiv {a}\text{ (mod }n)$$ and: $$q\equiv {a}\text{ (mod }m)$$
so it suffices to solve $q\equiv {a}\text{ (mod }lcm(n,m))$ and then check the primality of $q$. So if we take: $$n=8,m=12,a=-1$$ we try to solve:
$$q\equiv {-1}\text{ (mod }lcm(8,12)=24)$$
which is equivalent to:
$$q\equiv {23}\text{ (mod }lcm(8,12)=24)$$
and we're in luck that $q=23$ is a simple solution; i.e it is a prime. If instead we take:
$$q\equiv {1}\text{ (mod }lcm(8,12)=24)$$ the solutions will be:
$$q=1,24+1, 2\times24+1,3\times24+1...$$ We can then test for primality; I use maple with the comand "$\text{isprime}(1+24k)$", flick through $k$ and get "true" for $k=3$, so: $$p=q=1+24\times3=73$$ is a prime that solves:
$$p\equiv {1}\text{ (mod }8)$$
$$p\equiv {1}\text{ (mod }12)$$</p>

Title: Solving A Linear Congruential System


问:
Derivative of $\sin(x)$: how to avoid circular reasoning?
Dad here, helping his kid out with math.

I get it that to determine the derivative of $\sin(x)$ you do:

$$\frac{\sin(x+h) - \sin(x)}h = \frac{\cos(x)\sin(h)}h \to \cos(x)$$

However I assume that $\sin(h) = h$ for $h\to 0$. This correct because the "angle" (not sure what the correct word is in English) of $\sin(x) $ is $1$ for $x=0$. But this is what I'm trying to prove in the first place. So a circular reference. How to solve this?

答:
It should rather be
$$\lim_{h \to 0}\frac{\sin(x+h) - \sin(x)}h =\lim_{h \to 0} \frac{2 \cos (x+h/2) \sin(h/2)}{h} =\cos x \lim_{h \to 0} \frac{\sin(h/2)}{h/2}=\lim_{h \to 0} \frac{h/2-h^3/48+O(h^5)}{h/2}\cos x=\cos x$$
Here we have used $\sin z=x-z^3/3!+h^5/5!+...$


问:
Tensor product of $\partial_i$ and $dx^j$
In general the tensor product is not commutative. But I have a doubt:

in defining the absolute derivative (or covariant derivative) of a vector field $X$, as $\nabla X\in \cal{T_{1}^{1}(\cal{M})}$ (tensor field of type $(1,1)$) I have read that it is defined by:
$$\nabla X=\xi_{;j}^{i}dx^j\otimes \partial_i$$ with $X=\xi^i\partial_i$.

But in defining  a vector field of type $(1,1)$ in coordinates I know that it can be written as
$$t=t_i^j\partial_i\otimes dx^j$$
So it seems that $\partial_i$ and $dx^j$ are shifted in the order in defining $\nabla X$ with respect to the general definition of a tensor field of type $(1,1)$.

How it is possible? Maybe it is true that $\partial_i\otimes dx^j=dx^j\otimes \partial_i$?

答:
Some authors define a tensor of type $(1,1)$ to be, in local coordinates:
$$ \label{1}
T = \sum_{i,j} T^i_j \partial_i \otimes \mathrm{d}x^j, \tag{1}
$$
while others prefer to define it the other way
$$ \label{2}
T = \sum_{i,j} T^i_j \mathrm{d}x^j\otimes \partial_i \tag{2}
$$
I personally prefer \eqref{2} because if given a vector field $X$, one can feel that in the writing $\left(\mathrm{d}x^j\otimes \partial_i\right)(X)$, it becomes $\mathrm{d}x^j(X) \partial_i$ and the left part is a coefficient $\times $ the vector field $\partial_i$. In this case, a $(1,1)$-tensor is a section of $T^*M\otimes TM$.

In case \eqref{1}, $(1,1)$-tensors are sections of $TM\otimes T^*M$. This looks a bit unnatural to me as I learned during my studies that we have a natural isomorphism $\mathrm{End}(E) \simeq E^*\otimes E$ (coefficient $\times$ vector), not the other way. Again, this is a personal taste.

One a choice is made, you have to try being consistent. Your first equation $\nabla X = \xi^i_j \mathrm{d}x^j \otimes \partial_i$ is suggesting that the convention is \eqref{2}.

Edit I did not notice that there was another answer (I took my time writing this up), and it seems we are basically saying the same thing. The other answer is a bit more accurate as it explains how to understand the notation $\alpha \otimes X \left(Y,\omega \right)$.


问:
Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?
Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?

If it exist, what is the function, and if it doesn't, what is the proof to that?

My intuition is that big O refers to the worst case, meaning worst x, so no constant should have a significant effect on the time complexity, but I'm not sure at all, and I'm struggling to write it formally.

答:
I got an idea, would love some verification:

I will prove that no such function exist and that for any function $f(x-k)=O(f(x))$.

To prove that, I will use proof by contradiction assuming that there is a function that for $f(x-k)\neq O(f(x))$ does hold.

There for $\lim \frac{f(x-k)}{f(x)}=\infty$

$\lim f(x)=\lim f(x-k)$

If so, then $\lim \frac{f(x-k)}{f(x)}=\frac{\lim f(x-k)}{\lim f(x)}=1$, in contradiction to our assumption that $f(x-k)\neq O(f(x))\Leftrightarrow \lim \frac{f(x-k)}{f(x)}=\infty$.

There for indeed $f(x-k)=O(f(x))\blacksquare$


问:
Consistency in Statistical Decision Theory
Let $(\mathcal X,\mathcal F,\mathcal P)$ be a statistical model with $\mathcal P = \{P_\theta : \theta\in\Theta\}$. A decision rule is a measurable function $\delta:(\mathcal X, \mathcal F)\rightarrow(\mathcal A,\mathcal G)$. Examples for such decision rules include parameter estimates, or hypothesis tests. This question is concerned with certain properties of decision rules.

Definition. A decision rule is said to be unbiased iff $$\mathbb E_{P_\theta}[L(\vartheta, \delta)]\geq\mathbb E_{P_\theta}[L(\theta, \delta)]$$ for each $\theta,\vartheta\in\Theta$.

Assuming a certain loss function $L:(\mathcal A\times\mathcal A, \mathcal G\otimes\mathcal G)\rightarrow(\mathbb R,\mathcal B)$, this yields the familar concepts for unbiasedness of an estimator and unbiasedness of a hypothesis test:

Theorem 1. Assume the $L_2$ loss $L(t,a) = \Vert t - a\Vert_2$, and $\tau:\Theta\rightarrow\mathcal A$. Then $\delta$ is unbiased iff $\mathbb E_{P_\theta}[\delta] = \tau(\theta)$ for all $\theta\in\Theta$, provided that $\mathbb E_{P_\theta}[\Vert\delta\Vert^2_2] < \infty$, and $\mathbb E_{P_\theta}[\delta]\in\tau(\Theta)$.

Theorem 2. Assume the $0$-$1$-loss $L(t,a) = \ell_0 a\mathsf 1_{\Theta_0}(t) + \ell_1(1-a)\mathsf 1_{\Theta_1}(t)$, where $\ell_0,\ell_1>0$, $\Theta = \Theta_1\cup\Theta_0$ with $\Theta_1\cap\Theta_0=\emptyset$, and $\mathsf 1$ is the usual indicator function. Then $\delta$ is unbiased iff $\mathbb E_{P_\theta}[\delta]\leq\alpha$ for all $\theta\in\Theta_0$, and $\mathbb E_{P_\theta}[\delta]\geq \alpha$ for all $\theta\in\Theta_1$, where $\alpha = \frac{\ell_1}{\ell_0+\ell_1}$.

Another important concept in testing and estimation theory is consistency:

A sequence of estimtors $\delta_n$ is said to be consistent (for $\tau(\theta)$) iff $P_\theta(\Vert\delta_n - \tau(\theta)\Vert\geq\epsilon)$ converges to zero for each $\theta\in\Theta$.

Similarily, a sequence of tests $\delta_n$ is said to be consistent iff $\sup_{n\in\mathbb N}\mathbb E_{P_\theta}[\delta_n]\leq\alpha$ for all $\theta\in\Theta_0$, and $\mathbb E_{P_\theta}[\delta_n]$ converges to $1$ for all $\theta\in\Theta$.

The concepts are clearly related: for estimators, consistency states that the risk converges to zero in probability. For tests it states that the type I error is bounded, while the probability of a type II error converges to zero. It is thus natural to think about generalizations in the decision theory setup stated above. However, I was not able to do so. Is it even possible?

答:
Definition: We say that a sequence of decisions $\delta_n$ in $\mathcal{A}$ is consistent for the tuple $(L,\tau(\theta),\mathcal{A}',\Theta')$ iff for all $\theta\in\Theta'\subset\Theta$, $\delta_n$ is a sequence in $\mathcal{A}'\subset\{(\mathcal{X},\mathcal{F})→(\mathcal{A},\mathcal{G})\}$ and:
$$
L(\tau(\theta),\delta_n)\rightarrow 0,\quad \text{in $P_\theta$-probability}
$$
Let's check that we can recover estimation and testing with this definition:

Estimation:
Take $L(\tau(\theta),\delta):=||\tau(\theta)-\delta||$, $\Theta' := \Theta$ and $\mathcal{A}':=\mathcal{A}$. Then you recover the given definition of consistency for estimators.

Testing:
Here, we take $L(\tau(\theta),\delta):=1-\delta$, $\Theta' := \Theta_1$
$$
\mathcal{A}':=\{\delta:(\mathcal{X},\mathcal{F})→(\mathcal{A},\mathcal{G}):\sup_{n\in\mathbb{N}}\mathbb{E}_{P_\vartheta}[\delta]\leq \alpha,\forall\vartheta\in\Theta_0\}
$$
What is left to show is that $1-\delta_n\rightarrow 0$ in $P_\theta$-probability is equivalent to $\mathbb{E}_{P_\theta}[\delta_n]\rightarrow1$, for all $\theta\in\Theta_1$.
Using dominated convergence:
$$\begin{align*}
\lim_{n\rightarrow\infty}\mathbb{E}_{P_\theta}[\delta_n]]&=
\lim_{n\rightarrow\infty}\int_0^\infty P_\theta[\delta_n>t]dt\\&=
\int_0^\infty \lim_{n\rightarrow\infty}P_\theta[\delta_n>t]dt\\&=
\int_0^\infty 1\{t<1\}dt=1
\end{align*}$$


问:
factorization of 2 numbers
I tried to factor the numbers $1013^{13}+331\#$ and $98!+76!+54!+32!+1$ with GMP-ECM
because the quadratic sieve takes very much time. Perhaps someone factors
one of those numbers for me. The second number has a prime factor of 25 digits,
namely $1005693665747024080598971$. For those who are not familiar with the
number theory notations: $!$ is the factorial $n!=1\cdot 2\cdot 3\cdot\ldots\cdot n$ and $n\#$ is
the product of primes up to $n$: $n\#=2\cdot 3\cdot 5\cdot 7\cdot\ldots\cdot n$, if $n$ is a prime.
And please do not ask why I am interested in these factorization. It is
just curiousity because in the net it is stated that such numbers are already
quite easy to factor. Useful programs are GMP-ECM or msieve downloadable from
the net.</p>

答:
The first one took about 60 hours of NFS-ing on a moderately slow machine with unoptimized factoring settings while the machine was being used for other purposes too:

```<code>1013^13 + 331# = 13660997293197200778560512676198053659658942317253 *
469117945090530664165653254310752426123381482366114189309329070591871468575459348391
</code>```

The other one took about 64 hours; the smallest factor being discovered by ECM pretty quickly and the rest of the time spent on NFS-ing (again, unoptimized, running in background, ...):

```<code>98!+76!+54!+32!+1 = 1005693665747024080598971 *
79146647927616255967923163095656379252005739066059813 *
118432315205629597504117998137752645490949757279339848413269947393226465090487
</code>```

Altogether, this supports the statement that numbers of this size are quite easy to factor with current hardware. The time could be cut down to about one day on a dedicated, reasonably modern workstation and further cuts would be possible with better tuning of the factoring settings.

Title: factorization of 2 numbers

Body:
The first one took about 60 hours of NFS-ing on a moderately slow machine with unoptimized factoring settings while the machine was being used for other purposes too:

1013^13 + 331# = 13660997293197200778560512676198053659658942317253 *
469117945090530664165653254310752426123381482366114189309329070591871468575459348391

The other one took about 64 hours; the smallest factor being discovered by ECM pretty quickly and the rest of the time spent on NFS-ing (again, unoptimized, running in background, ...):

98!+76!+54!+32!+1 = 1005693665747024080598971 *
79146647927616255967923163095656379252005739066059813 *
118432315205629597504117998137752645490949757279339848413269947393226465090487

Altogether, this supports the statement that numbers of this size are quite easy to factor with current hardware. The time could be cut down to about one day on a dedicated, reasonably modern workstation and further cuts would be possible with better tuning of the factoring settings.


问:
What's the diffrence between Secant method and False position method?
I've been studying numerical analysis and I ran into false position method and I can't figure out its diffrence  between secant method , because I also  heard that the convergence is guaranteed in false position method but not in secant method.
And also what is regula-falsi method ?
There was a table which said the rate of convergence for secant,bisection and regula-fasi method is respectively 1.618 , 0.5 and 1 .
But I even don't know how it's calculated ! Aren't they linear ?</p>

答:
False-position method is another name for regula falsi.

The difference to the secant method is the bracketing interval. Meaning that the new secant root is not computed from the last two secant roots, but from the last two where the function values have opposing signs.

Yes, bracketing interval methods ensure convergence, as they shrink the interval that is known---via intermediate-value theorem---to contain a root. Thus the sequences of interval end-points are monotonous, both converging, at least one of them converging to a root.

The secant method, if it converges to a simple root, has the golden ratio $\frac{\sqrt5+1}2=1.6180..$ as superlinear order of convergence.

Bisection, in only considering the length of the bracketing interval, has convergence order 1, that is, linear convergence, and convergence rate $0.5$ from the halving of the interval in every step.

Regula falsi in most cases does not converge in the sense of shrinking the interval towards zero length. In such a situation, one side of the interval will converge to the root linearly, with order 1. And most likely a very unfortunate convergence rate close to $1$.

That is why anti-stalling versions of the method exist. For instance the Illinois modification shrinks the interval systematically towards zero length and reduces the error by the third power every 3 steps in a quite periodic pattern, so has a superlinear order of convergence of $\sqrt[3]3= 1.4422..$. One might claim that this is still faster than the $\sqrt2=1.4142..$ per function evaluation of the Newton method.

See Accuracy of approximation using linear interpolation


答:
The fact that $\sin(h) \approx h$ for small $h$ is usually proven non-circularly (although circles are involved :D) by resorting to the following inequalities
$$\begin{split}
\sin (h) &\leqslant h \leqslant \tan (h),\\
\cos (h) &\geqslant \frac 1 2,
\end{split} \tag 1$$
both valid when $0 < h < \pi/3$. You can see they hold by drawing a unit circle; here's a helpful picture for the first one:

$\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}$

Indeed, when $0 < h < \pi/3$, we obtain, by rearranging terms in inequality $(1\text a)$ that
$$0 \leqslant \sin (h) - h \leqslant \tan (h) - \sin (h ). \tag2$$
However, by the definition of $\tan$ and the trigonometric identity $1-\cos(h) = 2\sin^2(h/2)$,
$$\tan (h) - \sin (h )= \frac{\sin (h)}{\cos (h)}(1-\cos (h)) = \frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right). $$
By employing inequality $(1\text a)$ again in the forms $\sin(h/2) \leqslant h/2$ and $\sin(h) \leqslant h$, and inequality $(1\text b)$ for the cosine at the bottom, we see that
$$\frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right) \leqslant \frac h {1/2} \cdot 2 \left(\frac h 2\right)^2 = h^3. \tag 3$$
Merging inequalities $(2)$ and $(3)$ entails that $0\leqslant \sin(h) - h \leqslant h^3$, that is (adding $h$ to all sides) $h \leqslant \sin(h) \leqslant h^3 + h$, that is (dividing everything by $h$) $$1 \leqslant \frac{\sin(h)}{h} \leqslant 1+ h^2,$$ which ultimately implies that
$$\lim_{h\to 0^+} \frac{\sin(h)}{h} = 1 $$
by the squeeze theorem (both $1$ and $1+h^2$ clearly tend to $1$ as $h\to 0^+$).

The complete limit (as $h\to 0$ from both sides) follows from the fact that $\sin(-h) = -\sin (h)$.

Title: Derivative of $\sin(x)$: how to avoid circular reasoning?

Body:
Dad here, helping his kid out with math.

I get it that to determine the derivative of $\sin(x)$ you do:

$$\frac{\sin(x+h) - \sin(x)}h = \frac{\cos(x)\sin(h)}h \to \cos(x)$$

However I assume that $\sin(h) = h$ for $h\to 0$. This correct because the "angle" (not sure what the correct word is in English) of $\sin(x) $ is $1$ for $x=0$. But this is what I'm trying to prove in the first place. So a circular reference. How to solve this?


问:
Prob. 13 (i), Chap. 2, in Royden's REAL ANALYSIS: The translate of an $F_\sigma$-set is also an $F_\sigma$-set
Here is Prob. 13 (i), Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Show that the translate of every $F_\sigma$-set is also $F_\sigma$.

My Attempt:

Suppose that $E \subset \mathbb{R}$ is an $F_\sigma$-set, and let $y$ be any real number. We show that the set $E+y$, defined by
$$
E + y := \{ x+y \, | \, x \in E \},
$$
is also an $F_\sigma$-set.

As $E$ is an $F_\sigma$-set, so there exists some countable collection $\left\{ E_k \right\}_{k = 1}^\infty$ of closed subsets of $\mathbb{R}$  such that
$$
E = \bigcup_{k=1}^\infty E_k. \tag{1}
$$
We show that
$$
E + y = \bigcup_{k=1}^\infty \left( E_k + y \right). \tag{2}
$$
Let $x \in E + y$. Then $x = e+y$ for some $e \in E$, by (1) above this $e \in E_k$ for some $k = 1, 2, 3, \ldots$, which implies that $x = e+y \in E_k + y$ for that same $k$, and hence $x \in \bigcup_{k=1}^\infty \left( E_k + y \right)$, which shows that
$$
E+y \subset  \bigcup_{k=1}^\infty \left( E_k + y \right).
$$
Conversely, let $x \in \bigcup_{k=1}^\infty \left( E_k + y \right)$. Then this $x \in E_k + y$ for some $k = 1, 2, 3, \ldots$, which implies that $x = e +y$ for some $e \in E_k$ for that same $k$ and that same $e \in \bigcup_{k=1}^\infty E_k = E$ also, which in turn implies that $x = e+y \in E + y$, thus showing that
$$
\bigcup_{k=1}^\infty \left( E_k + y \right) \subset E+y.
$$
Thus (2) above indeed holds.

Let us fix some $k = 1, 2, 3, \ldots$. We show that the set $E_k + y := \left\{ x+y \, | \, x \in E_k \right\}$ is also closed; for this we show that the corresponding set $\mathbb{R} \setminus \left( E_k + y \right)$ is open; let $x \in \mathbb{R} \setminus \left( E_k + y \right)$. Then $x \in \mathbb{R}$ and $x \not\in E_k + y$, which implies that $x \neq e + y$ for any element $e \in E_k$, that is, $x-y \not\in E_k$, which implies that $x-y \in \mathbb{R} \setminus E_k$, and as this last set is open, so there exists a nonempty, bounded open interval $(a, b)$ such that
$$
x - y \in (a, b) \subset \mathbb{R} \setminus E_k,
$$
which implies that
$$
x \in (a + y, b+ y) \subset \left( \mathbb{R} \setminus E_k \right) + y.  \tag{3}
$$
Lastly we show that
$$
\left( \mathbb{R} \setminus E_k \right) + y = \mathbb{R} \setminus \left( E_k + y \right). \tag{4}
$$
Let $u \in \left( \mathbb{R} \setminus E_k \right) + y$. Then by definition $u = v + y$ for some element $v \in \mathbb{R} \setminus E_k$, that is, $u-y \in \mathbb{R} \setminus E_k$, and so $u-y \in \mathbb{R}$ such that $u-y \not\in E_k$ and hence $u \not\in E_k+y$, but as $u \in \mathbb{R}$, so $u \in \mathbb{R} \setminus \left( E_k + y \right)$, thus showing that
$$
\left( \mathbb{R} \setminus E_k \right) + y \, \subset \, \mathbb{R} \setminus \left( E_k + y \right).
$$
Conversely, let $u \in \mathbb{R} \setminus \left( E_k + y \right)$. Then $u \in \mathbb{R}$ and $u \not\in E_k + y$, which implies that $u \neq e + y$ for any element $e \in E_k$, that is, $u-y \not\in E_k$, and thus $u-y \in \mathbb{R} \setminus E_k$, which in turn implies that $u = (u-y)+y \in \left( \mathbb{R} \setminus E_k \right) + y$, thus showing that
$$
 \mathbb{R} \setminus \left( E_k + y \right) \, \subset \, \left( \mathbb{R} \setminus E_k \right) + y.
$$
Thus (4) does indeed hold. Using (4) in (3) we can conclude that, for any element $x \in \mathbb{R} \setminus \left( E_k + y \right)$, there exists a non-empty, bounded open interval $(a +y, b+y)$ such that
$$
x \in (a+y, b+y) \subset \mathbb{R} \setminus \left( E_k + y \right),
$$
thus showing that the set $\mathbb{R} \setminus \left( E_k + y \right)$ is open in $\mathbb{R}$ and hence $E_k + y$ is closed.

As each set $E_k + y$, for $k = 1, 2, 3, \ldots$, is closed, so by (2) above we can conclude that the set $E + y$ is an $F_\sigma$-set.

Is my proof correct and clear in each and every step? Or, are there issues?

答: It suffices to note that $t_y:x \to x+y$ is a homeomorphism ( even isometry) for each $y$ and homeomorphisms are closed bijections and so preserve $F_\sigma$ sets: $E+y = t_y[\bigcup_n F_n] = \bigcup_n t_y[F_n]$. You overcomplicate it.


问:
Show that $\alpha$ divides the determinant of a particular matrix.
To illustrate my problem with a specific example, consider the matrix
$$D=\left[ \begin {array}{ccc} a_{{1}}+\alpha&a_{{2}}&a_{{3}}
\\ a_{{1}}&a_{{2}}+\alpha&a_{{3}}
\\ b_{{1}}&b_{{2}}&b_{{3}}\end {array} \right]
$$
then show that $\alpha$ divides $\det(D)$.

More generally, my problem is as follows. Given the $n\times n$ matrices
$$
A = \left(
\begin{array}{c}
\mathbf{a}^T\\
\vdots\\
\mathbf{a}^T\\
\mathbf{b}_1^T\\
\vdots\\
\mathbf{b}_k^T\\
\end{array}
\right)
\quad\mathrm{and}\quad C = \mathrm{diag}([\overbrace{\alpha,...,\alpha}^{\text{m times}},\overbrace{0,...,0}^{\text{k times}}])
$$
That is, $\mathbf{a}\in\mathbb{R}^n$ make the first $m$ rows of $A$ and $\mathbf{b}_i \in \mathbb{R}^n$ make up the last $k$ rows of $A$ while $C$ is the matrix with $\alpha$ on the first $m$ diagonal elements and $0$ elsewhere. Also the vectors $\mathbf{b}_i$ are linearly independent s.t. $A+C$ has full rank when $\alpha\ne 0$.

Then show that that $\alpha^{m-1}$ divides $\det(A+C)$. That is, show that $\det(A+C) = \alpha^{m-1} d$. I do not need to know $d$, I only need to prove that $\alpha^{m-1}$ is always a factor of $\det(A+C)$.

It suffices (for my purposes) to consider the case where $m=2$.

For my example above in $n=3$, Maple gives $\det(D)=\alpha\, \left( b_{{3}}\alpha+a_{{1}}b_{{3}}+b_{{3}}a_{{2}}-a_{{3}}b_{
{1}}-a_{{3}}b_{{2}} \right)
$

答:
Here are the steps needed.

(1) Add all columns into column 1.

(2) Subtract row 1 from each of rows 2-m.

At the stage the (1,1) entry is $\alpha +\sum a_i$, and all entries in rows 2-m are zero, apart from those on the diagonal which are $\alpha$.

(3) Evaluate by expanding on rows 2-m first, this will provide the factor $\alpha^{m-1}$ as required.


问:
Is any Newton equation an Euler-Lagrange equation?
Let $$ r'' = \mathrm{F}(r', r)$$

be Newton equation in one variable whith $\mathrm{F}$ locally Lipschitz.

Is there a function $\mathcal{L}(r',r)$ such that the Newton equation is in fact Euler-Lagrange equation $$ \frac{\partial \mathcal{L}}{\partial r} = \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial r'} \quad ?$$

I know that it is true for a conservative force field, but what about the general case?

答:
your first question is also known as the inverse problem for the Lagrangian mechanics.

Answering your question, there exists a necessary and sufficient condition in order that a second order ode $F(r'',r',r)=0$ be of Euler-Lagrange kind.
This condition, called after Helmholtz, is expressed by the system
$$\left\{\begin{gathered}
\frac{\partial F_i}{\partial r_j''}=\frac{\partial F_j}{\partial r_i''},\\
\frac{\partial F_i}{\partial r_j'}+\frac{\partial F_j}{\partial r_i'}=2\frac{d}{dt}\frac{\partial F_j}{\partial r_i''},\\
\frac{\partial F_i}{\partial r_j}-\frac{\partial F_j}{\partial r_i}=\frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial r_j'}-\frac{\partial F_j}{\partial r_i'}\right),
\end{gathered}\right.$$
where the Einstein summation convention is active.


问:
How is $-4/7$ the slope of $4x+7y=1$?

How is $-4/7$ the slope of $4x+7y=1$?

Can someone explain? I think they have divided the coefficients of $x$ and $y$.

答:
It is a bit of algebra, but essentially you have:

$4x+7y=1,$

$7y=-4x+1,$

$y=-4/7x+1/7$

the slope is always a coefficient of the term containing $x$.

Title: How is $-4/7$ the slope of $4x+7y=1$?


问:
Solving the recurrence relation $T(n)=3T\Big(\frac{n}{2}\Big)+n$

Use the expand, guess and verify approach to solve the following recurrence relation
$$\cases{T(1) = 1\\T(n) = 3T\Big(\frac{n}{2}\Big)+n & for $n > 1$}$$

I think it's defined for $n=2^m$, $m>1$, so here's my attempt: $$T(n)=3T\Big(\frac{n}{2}\Big)+n=T(n)=3^2T\Big(\frac{n}{2^2}\Big)+\frac{3}{2}+n=3^3T\Big(\frac{n}{2^3}\Big)+\frac{3^2}{2^2}n+\frac{3}{2}n+n$$

I guess that after $k$ steps, we get $$T(n)=3^kT\Big(\frac{n}{2^k}\Big)+n\sum_{i=0}^{k-1} \Big(\frac{3}{2}\Big)^i$$

Setting $k=\log n$, $$T(n)=3^{\log n}+n\sum_{i=0}^{\log (n)-1} \Big(\frac{3}{2}\Big)^i$$

I'm having a hard time proving the previous formula by induction. I tried this:

Base case: $$T(1)=3^{0}+1\sum_{i=0}^{0-1} \Big(\frac{3}{2}\Big)^i =(?) 1$$

Is that summation equal to zero? Those indices got me wondering. Also, assuming everything is correct so far, I got stuck when trying to write down the induction step.

Inductive step: If $T(k)$ is true, then $T(2k)$ must be true too. So, $$T(2k)=3T(k)+2k$$ $$T(2k)=3^{\log 2k}+3k\sum_{i=0}^{\log (k)-1} \Big(\frac{3}{2}\Big)^i + 2k$$ The last step doesn't seem to be correct. Any help, please? Thanks in advance.

答:
As

$$
T\left(2^{\log_2 n}\right) = 3T\left(2^{\log_2 \frac n2}\right)+n
$$

making $\mathcal{T}(\cdot) = T\left(2^{(\cdot)}\right)$ and $z = \log_2 n$ we follow with

$$
\mathcal{T}(z) = 3\mathcal{T}(z-1) + 2^z
$$

with solution

$$
\mathcal{T}(z) = 3^{z-1}c_0+2(3^z-2^z)
$$

and going backwards with $z=\log_2 n$ we obtain

$$
T(n) = 3^{\log_2 n-1}c_0 +2(n-3^{\log_2 n})
$$

and obeying the initial conditions

$$
T(n) = 3^{\log_2(2n)}-2n
$$


答: Hints:-  For a linear equation $ax+by=c$, slop is always $-\dfrac{a}{b}$.

Title: How is $-4/7$ the slope of $4x+7y=1$?

Body:
How is $-4/7$ the slope of $4x+7y=1$?

Can someone explain? I think they have divided the coefficients of $x$ and $y$.


问:
Suppose 7 people are sitting at a round table. If three people are selecting at random, find the probability that at least two of then are together.
First we find the probability that all three sit together. The string method can be used for this. So
$$P(3)=\frac{4! 3!}{6!} =\frac 15$$

Now for finding probability that two people are together and one of them of is separate. I selected any two people out of the $3$ selected people and string them. Then number of ways to do that should be
$\binom{3}{2} 2!$

Now if this string is at any one position, then only $4$ positions are left to seat the third person. So number of ways should be $\binom{4}{1} 4!$

But multiplying all this gives a value more than 6!, which shouldn’t happen. What am I doing wrong?

I also though about using complementary method, but I didn’t know how to apply it for the circular permutation. So can I get a hint on that too?

Thanks

答:
Your working and my answer is considering the question that if $7$ people are randomly seated on a round table, what is the probability that at least $2$ of $3$ given people seat together.   See my edit for the other answer.

In the second case, we can choose two out of three people from the first group and seat them together. Number of ways to do that is $\displaystyle \small \binom 32 2! \ $.

We then have the second group seat in $4!$ ways. That gives us $5$ places in between them (two being together and hence no place between them) to seat the last person. But the last person cannot be with the rest two from first group and therefore cannot occupy $2$ of those $5$ places (to the left of two from the first group seated together or to the right of them).

So toal number of ways to seat them in the second case $ = \displaystyle \small 2! \cdot \binom 32 \cdot 3 \cdot 4!$

Adding first and second case and dividing by $6!$ gives us desired probability of $ \ \displaystyle \small \frac{4}{5}$.

Alternatively, probability that none of the people from the first group of three are seated together,

$ = \displaystyle \small 3! \cdot {4 \choose 3} \cdot 3! \ / \ 6! =   \frac{1}{5}$

Explanation for numerator: we first seat second group in $3!$ ways and then we seat the first group in $3$ out of the $4$ places.

That leads to probability of $ \ \displaystyle \small \frac{4}{5}$ that at least two of them are seated together.

EDIT: If the interpretation is that of $7$ seated people on a round table, if we pick any $3$ randomly, what is the probability that at least two of them were seated together, the working will be as below.

Total number of ways to pick $3$ people $ = \displaystyle \small {7 \choose 3} = 35$.

case $1$: Exactly two are seated together. There are $7$ ways to pick two people seated together. As third person is not neighbor with the first two, he or she can be chosen from $3$ non-neighbors.

case $2$: All three are seated together. There are $7$ ways to pick $3$ people seated together.

So desired probability $ = \displaystyle \small \frac{3 \cdot 7 + 7}{35} = \frac{4}{5}$.


问:
Given the arc length of a function and the lower limit, how to find the upper limit?
Given any arbitrary function $f(x)$, its arc length in the interval $[a,b]$ is defined as:

$$
s=\int_a^b\sqrt{1+(\nabla f(x))^2}\,dx,
$$

i.e., we integrate the above expression from $a$ to $b$, using the first derivative of the function $f(x)$.

Question: if $a,s$ are known, how to get $b$? In other words, I already have the arc length, and I do know from where it was measured (the beginning of the interval) - how to get $b$? $f(x)$ and its derivative $f'(x)$ are not analytical expressions, rather, I am doing numerical computations.

答:
Suppose that we know the derivative but
$$I=\int \sqrt{1+[f'(x)]^2}\,dx$$ is not analytical. Suppose that $I$ is analytical but cannot be inversed. What can be done beside numerical calculations ?

But the is not bad in the first case since Newton will perfect. We want to find the zero of function
$$F(b)=s -\int_a^b \sqrt{1+[f'(x)]^2}\,dx$$
$$F'(b)=-\sqrt{1+[f'(b)]^2}$$
$$b_{k+1}=b_k+\frac{s -\int_a^{b_k} \sqrt{1+[f'(x)]^2}\,dx } {\sqrt{1+[f'(b_k)]^2} }$$ the worst initial estimate being
$$b_0=a+\frac s{\sqrt{1+[f'(a)]^2} }$$

Just for illustration, using $f(x)=\text{erf}(x)$, $a=-1$ and $s=123.456$
So
$$F(b)=123.456-\int_{-1}^b \sqrt{1+\frac{4 }{\pi }e^{-2 x^2}}\,dx$$  Being very lazy, I shall start the above $b_0$ and the solution is obtained in $2$ iterations for $50$ exact significant figures.


问:
Find ${51\choose 50} + {52\choose 49} + {53\choose 48} + … + {100\choose 1} + {101\choose 0}$
I know the answer is $F_{100}$, where $F_{100}$ is $100^{th}$ number in Fibonacci sequence. And it is, in general, true for any n, I checked it for some initial values. How can we prove this?

Body:
Denote by $s_n$ such series:
$$s_n = C_n^0 + C_{n-1}^1+C_{n-2}^2 + ...$$
Then $s_0 = 1, s_1 = 1$, so by induction:
$$s_{n}+s_{n+1} = 1 + (C_{n}^0 +C_{n}^1) + (C_{n-1}^1 +C_{n-1}^2) + ... = C_{n+2}^0 + C_{n+2}^1 + C_{n+2}^2 + ... = s_{n+2}$$
The last equality is the property of Pascal's triangle.


问:
What is the difference between multiplicative group of integers modulo n and a Galois Field
What is the difference between multiplicative group of integers modulo n and a Galois Field?

Is $\mathbb{Z}^*_p$ with p prime the same as $GF(p)$?
Or is it the same as $\mathbb{Z}/n\mathbb{Z}$?

Sorry for the short and simple question, but this bit of notation is confusing me, so a clear different should help me alot!

Thanks!

答: If $p$ is prime then the ring $\mathbb Z/p\mathbb Z$ is in fact a field and is "the" field with $p$ elements (as there is only one up to isomorphism). Note however that for prime powers $q=p^n$ with $n>1$, the ring $\mathbb Z/q\mathbb Z$ is not a field, hence is different from the field $GF(q)$ (or $\mathbb F_q$, depending on author).


答:
Just for your curiosity :

Using Young's inequality for product we get for $x>1$ :

$$f(x)=\left(\frac{(\frac{75}{7})^{x}}{x}+\frac{x-1}{x}7^{\frac{x}{x-1}}\right)>75$$

Now we have to show for some $x>1$ :

$$g(x)=\frac{f(x)}{49*7^{\frac{7}{32}}}<1$$

And now the miracle of the inequality is :

$$g\Big(\frac{71}{39}\Big)<1$$

And as we now $71=39+32$

So the inequality can be rewritten :

$$\frac{\ln(7)}{\ln(75)-\ln(7)}+1>\frac{71}{39}$$

The LHS is the equality case for the Young's inequality .

So now I suspect we can use Reductio ad absurdum to solve it .

Well some explanations :

If we suppose :

$$7^{71}<75^{32}$$

It's easy to remark that it's equivalent to :

$$\frac{75*39}{71}+\frac{32}{71}7^{\frac{71}{32}}>75$$

By the hypothesis wich make the reductio ad absurdum plausible we have :

$$\Big(\frac{75}{7}\Big)^{\frac{71}{39}}>75$$

So we have :

$$\frac{\Big(\frac{75}{7}\Big)^{\frac{71}{39}}39}{71}+\frac{32}{71}7^{\frac{71}{32}}>\frac{75*39}{71}+\frac{32}{71}7^{\frac{71}{32}}>75$$

Wich is the Young's inequality !

Now to get the end we need to show :

$$7^{\frac{71}{32}}>\frac{\Big(\frac{75}{7}\Big)^{\frac{71}{39}}39}{71}+\frac{32}{71}7^{\frac{71}{32}}$$

So we need to find an upper bound instead of a lower bound .

Any idea ?

Title: Prove $7^{71}>75^{32}$

Body:
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$

I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}

But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$


问:
Evaluating $ \int_{-\infty}^{\infty} \cosh(x+s)^{-2}\cosh(x)^{-2}dx$
So I have this function:

$$ \int_{-\infty}^{\infty} \cosh(x+s)^{-2}\cosh(x)^{-2}dx$$

And when I try to integrate it, I can obtain $0$. And, when I evaluate the limits, it also cancels to $0$. The solution I was given stated that the answer should be some form of:

$$\frac{\cosh(s)\cdot s}{\sinh(s)^3}- \frac{1}{\sinh(s)^2} $$

None of what I'm doing seems to get me the answer and as you can see, it's not like Mathematica even makes the output easy to parse. This is my Mathematica expression:

```<code>In[42]:= Integrate[Cosh[x + s]^-2*Cosh[x]^-2, x]

Out[42]= -2 Coth[s] Csch[s]^2 Log[Cosh[x]]+2Coth[s] Csch[s]^2 Log[Cosh[s + x]]-Csch[s]^2 Sech[s] Sech[s+x] Sinh[x]-Csch[s]^2Tanh[x]
</code>```

答:
Let $x= \frac{t-s}2$ to reexpress the integral as
$$I=\int_{-\infty}^{\infty} \frac1{\cosh^2(x+s)\cosh^2(x)}dx
= \int_{0}^{\infty} \frac4{(\cosh t+ \cosh s)^2}dt
$$
Note that

$$\left( \frac{\sinh t}{\cosh t+ \cosh s} \right)’
= \frac{\cosh s}{\cosh t+ \cosh s} - \frac{\sinh^2s}{(\cosh t+ \cosh s)^2}
$$
Integrate both sides
$$I =\frac{4 }{\sinh^2s}\left(-1+\cosh s \int_0^\infty \frac{1}{\cosh t+\cosh s}dt\right)
$$
where
$$\int_0^\infty \frac{1}{\cosh t+\cosh s}dt
= \frac{2\tanh^{-1}(\tanh\frac t2\tanh\frac s2)}{\sinh s}\bigg|_0^\infty
= \frac s{\sinh s}$$
Thus
$$I =\frac{4s\cosh s }{\sinh^3s}-\frac4{\sinh^2s}
$$


问:
When continuous random variable is in both the conditional and the conditioned
Let $A,B$ be continuous random variables. Let $E,G,K$ be events. Let $t \in image(B)$.

(I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf. If not, then assume joint pdf is well-defined whenever necessary.)

Question 1: Can we say $P(B \le A|B=t) = P(t \le A|B=t)$?

Question 2: If so, then how? If not then why?

What I've tried: I think we're doing something like $P(E|K)=P(E \cap K|K)$, and I get it when $P(G|K)$ is defined for $P(K)>0$. What is being done here when technically $P(K)=0$? I mean, of course, in the 1st place when we say like '$P(\cdot|B=t)$', this is notational, we're not really conditioning on the $P$-null event $\{B=t\}$. But I still don't get exactly what's being done here. Well...

As I understand, $$RHS = P(t \le A|B=t) := \frac{\int_{t \le A} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da} = \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da}$$

$$= \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{f_{B}(t)}= \int_{t}^{\infty} \frac{f_{A,B}(a,t)}{f_{B}(t)} da,$$ where/whence $f_{A|B=t}(a) := \frac{f_{A,B}(a,t)}{f_{B}(t)}$.

That's all I got.

Not really sure how to evaluate LHS. Perhaps LHS := RHS, i.e. LHS is defined as RHS?

Update: Wait I think I have an idea how to do LHS

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db$$

Here,

'$f_{(A,B)|B=t}(a,b)$' is I guess meant like $f_{(A,B)|T=t}(a,b)$ where $T$ is a random variable that is (not just almost surely but really) surely equal to $B$.

'$f_{(A,B)|T=t}(a,b)$' is meant like 2 random variables conditioned on a 3rd random variable. As I understand multivariate conditional joint distributions, we have this is $f_{(A,B)|T=t}(a,b) := \frac{f_{A,B,T}(a,b,t)}{f_T(t)}$ and then...

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db = \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{f_T(t)} da db$$

$$= \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} da db$$

$$= \frac{\int \int_{b \le a} f_{A,B,T}(a,b,t) da db}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} $$

And then I don't know. Apparently we might not be able to have a joint pdf if 2 of the random variables are surely equal (maybe even for almost surely):

4.1. joint distribution of x with..itself

4.2. https://stats.stackexchange.com/questions/47162/what-is-the-joint-probability-distribution-of-two-same-variables

答:
Here is an argument that uses no measure theory that you might like.  A measure theory argument that runs parallel to this one can be made using conditional expectation $E[1\{B \leq A\}|B]$ and by slightly modifying the claim.

Assume the density of $B$ is continuous at point $t$ and $f_B(t)>0$. Under this assumption, for an event $C$ that satisfies either $P[C]=0$ or $f_{B|C}(x)$ exists and is continuous at $x=t$, it can be shown the following limits exist and are equal:
$$ \lim_{\delta \rightarrow 0^+} P[C|B \in [t, t+\delta]]=\lim_{\delta \rightarrow 0^+} P[C|B \in [t-\delta, t]]$$
So under these assumptions we can define
$$ P[C|B=t] = \lim_{\delta \rightarrow 0^+} P[C|B \in [t, t+\delta]]$$

Now for any $\delta>0$ we have
$$ P[B \leq A|B \in [t, t+\delta]] \leq P[t \leq A | B \in [t, t+\delta]] $$

Taking $\delta \searrow 0$ and assuming we converge appropriately gives
$$ P[B \leq A | B=t] \leq P[t \leq A |B=t]$$

For the reverse inequality, observe that for any $\delta>0$ we have
$$ P[B\leq A | B \in [t-\delta, t]] \geq P[t \leq A | B\in [t-\delta, t]] $$
Taking limits and assuming the limits converge appropriately gives
$$  P[B \leq A | B =t] \geq P[t \leq A |B=t]$$


问:
Why $w_i=\pi/n$ Chebyshev–Gauss quadrature?
In Chebyshev–Gauss quadrature we have
$$\int _{-1}^{+1}{\frac {f(x)}{\sqrt {1-x^{2}}}}\,dx\approx \sum _{i=1}^{n}w_{i}f(x_{i})$$
where
$$x_{i}=\cos \left({\frac {2i-1}{2n}}\pi \right)$$
and the weight $w_i=\pi/n$. Why $w_i=\pi/n$?

答:
If we change variables by $x=\cos\theta$, then the integral  can be evaluated by dividing the interval into $n$ parts, using the midpoint rule on a circle, $$\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}dx=\int_0^\pi f(\cos\theta)\,d\theta\approx\sum_{i=1}^nf(x_i)\frac{\pi}{n}$$ where the $x_i=\cos\theta_i$ are as written in the question.

For a fuller explanation of the weights:

The Cheybshev polynomials $T_n$ are orthogonal, that is $\int_{-1}^1\frac{T_i(x)T_j(x)}{\sqrt{1-x^2}}dx=\int_0^\pi \cos(i\theta)\cos(j\theta)\,d\theta=0$ for $i\ne j$.

To approximate the integral by $\sum_iw_if(x_i)$, we may require that the integral is exact on the $T_j$, $j=0,\ldots,n-1$, that is $$\sum_{i=1}^nw_iT_j(x_i)=\int_{-1}^1\frac{T_j(x)}{\sqrt{1-x^2}}dx$$
This is essentially the matrix equation $$\begin{pmatrix}1&\cdots&1\\ \cos\theta_1&\cdots&\cos\theta_n\\\cos2\theta_1&\cdots&\cos2\theta_n\\\vdots\end{pmatrix}\begin{pmatrix}w_1\\\vdots \\w_n\end{pmatrix}=\begin{pmatrix}\pi\\0\\\vdots\\0\end{pmatrix}$$

One can invert this matrix, or else notice that $$\begin{pmatrix}1&\cdots&1\\ \cos\theta_1&\cdots&\cos\theta_n\\\cos2\theta_1&\cdots&\cos2\theta_n\\\vdots\end{pmatrix}\begin{pmatrix}1\\\vdots\\1\end{pmatrix}=\begin{pmatrix}n\\0\\\vdots\\0\end{pmatrix}$$ to conclude that $w_i=\pi/n$.


问:
Show that for a principal ideal domain $R$ with identity, $R/(p^n)$ is an indecomposable module, where $p$ is a prime element in $R$.

Show that for a principal ideal domain $R$ with identity, $Q=R/(p^n)$ is an indecomposable module over $R$ where $p$ is a prime element in $R$ and $n$ is a positive integer.

Actually, I was trying to show that we can never decompose $Q$ into non-trivial submodules, but I am getting stuck.

A small hint is very much required at this moment.
Thanks in advance.

答:
If $R$ is a PID and $p \in R$ is a non-zero element generating a prime ideal $I:=(p)$ it follows the quotient $R^*:=R/(p^n)$ is a local ring: If $J^* \subseteq R/(p^n)$ is a prime ideal, it follows we get an inclusion $(p^n) \subseteq J \subseteq R$ and hence $(p) \subseteq J$. This implies that $(p)=J$ since $(p)$ is a maximal ideal. Hence $J^*$ is the only maximal ideal in $R/(p^n)$ and $R^* \rightarrow R^*/J^*$ is the only simple quotient of $R^*$. If $R^* \cong R_1 \oplus R_2$ is a non-trivial direct sum decomposition as $R^*$-modules, we may choose simple quotients $S_i:=R_i/m_i$ and get a surjection

$$\phi:R^* \rightarrow S_1 \oplus S_2$$

and since $R^*/J^*$ is the only simple quotient of $R^*$ we get a contradiction: The inverse image $m_i^*$ of $m_i$ in $R^*$ is an ideal, and since $R^*/J^*$ is the only simple quotient of $R^*$ we must have $m^*_i=J^*$. Since the map $\phi$ is surjective we must have $m^*_1$ and $m^*_2$ to be relatively prime - a contradiction.


问:
Proof that $a(1) \equiv 0 \pmod{q} \Rightarrow (x-1)|a(x)$
I need to prove the following result:

Let $a\in(\mathbb{Z}/q\mathbb{Z})[X]$ with $q$ prime. Then $$a(1)\equiv 0 \pmod{q} \;\Rightarrow\; (X-1)\,|\,a(X) \;\text{ in }\; (\mathbb{Z}/q\mathbb{Z})[X]$$

So far I have only managed to prove the converse. Any hint toward the right direction is therefore appreciated.

Body:
Let $a(X)= \sum a_k X^k$ then, since $a(1)=\sum a_k = 0 $ :

$a(X)= a(X)-a(1)=\sum a_k(X^k-1)=\sum a_k(X-1)(X^{k-1}+...+1)=\\(X-1)\times(\sum a_k(X^{k-1}+...+1) )$

Therfore $X-1 | a(X)$


问:
Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?
Is there a function $f(x)$ and a constant $k$ so that $f(x-k) \neq O(f(x))$?

If it exist, what is the function, and if it doesn't, what is the proof to that?

My intuition is that big O refers to the worst case, meaning worst x, so no constant should have a significant effect on the time complexity, but I'm not sure at all, and I'm struggling to write it formally.

答:
As Christian Blatter commented, let $f(x)=e^{-x^2}$. Then
$$
\begin{align}
f(x-k)
&=e^{-(x-k)^2}\\
&=e^{2kx-k^2}e^{-x^2}\\
&=e^{2kx-k^2}f(x)
\end{align}
$$
That is, for $k\gt0$, $\frac{f(x-k)}{f(x)}$ grows without bound as $x\to\infty$.

This means that $f(x-k)\not\in O(f(x))$.


问:
Proving Sup(A+B) = Sup(A)+Sup(B), I got a question on something
$A,B $ : Nonemtpty bounded sets on R

I'm proving on $\sup(A+B) \geq \sup (A)+\sup(B)$

prove by contradicting assumption: $\sup(A+B) < \sup(A) + \sup(B)$

Then $\sup(A+B) -\sup (B)< \sup (A)$

Then there exists $a \in  A$ such that  $\sup(A+B)-\sup(B)<a$

$\sup(A+B) -a <\sup(B)$...

(next steps are omitted)

Now, I have a question on this prove on existence of $a$

For exmaple, Let $\sup(A+B)=X, \sup(B)=Y$. Then $X-Y<\sup(A)$

For sure, $a\le\sup(A)$. Then, how can we know $X-Y<a?$

Or am I wrong on proving this?

答:
This is very simple using the definition of supremum. The actual definition, which says nothing about subsequences and limit points.

By definition $\sup(S)$ is the least upper bound of $S$. Which means two things: (i) $\sup(S)$ is an upper bound for $S$ and (ii) if $\alpha<\sup(S)$ then $\alpha$ is not an upper bound for $S$.

What you want ask about is this:

If $\alpha<\sup(S)$ there exists $a\in S$ with $\alpha<a$.

Proof: Since $\alpha<\sup(S)$, $\alpha$ is not an upper bound for $S$, which in turn means precisely that there exists $a\in S$ with $a>\alpha$.


答:
There’s another neat way to think about slopes. Suppose $x$ is increased by $1$ unit. How should $y$ change so that that the resulting point is still on the line, i.e. the equation remains unchanged?
The new LHS is $$4(x+1) +7y = 4x+7y +4 $$ If $y$ is decreased by $\frac 47$, the change in $x$ cancels out: $$4x+7(y-\frac 47) +4 = 4x +7y -4+4 =4x+7y $$

The slope is therefore $-\frac 47$.

Title: How is $-4/7$ the slope of $4x+7y=1$?


问:
Differentiation of a stochastic process by Ito's formula

Let $W_t$ denote a Brownian Motion, use ito's formula to compute the
differential of the following stochastic process:
$$X_t=e^{t/2}\cos(W_t)$$

Ito's formula is supposed to be used when we know a diffusion differential equation like $dx=\mu dt+\sigma dW_t$, then we can apply it to solve a $dG(x,t)$ by:
$$dG=\left(\frac{\partial G}{\partial t}+\frac{\partial G}{\partial x}\mu+\frac{1}{2}\frac{\partial^2G}{\partial x^2}\sigma^2\right)dt+\frac{1}{2}\frac{\partial G}{\partial x}\sigma dW_t.$$
But for this question, why we can solve the $dX_t$ directly?

Also, the answer to this question suggests that we should use the following Ito's formula to solve $dX_t$:
$$dX(t,W_t)=\frac{\partial X(t,W_t)}{\partial t}dt+\frac{\partial X(t,W_t)}{\partial W}dW_t+\frac{1}{2}\frac{\partial ^2 X(t,W_t)}{\partial W^2}dt$$

What's the logic behind this Ito formula?

答:
The usual form of It's formula you note is applied like chain rule, when you need to find $dG(x,t)$ but you only know the dynamics of $dx$. The "regular-Calculus" analog of this is when you know $x(t)$ and $x'(t)$ but need to find $\frac{d}{dx} g(x(t))$.

In the specific case you bring, you definitely could differentiate directly, as your second formula suggests.

Not sure what else you are asking, please post comments or update the question.


问:
Finding an Awesome method to Prove a fomular $\sum_{j=0}^n(x_j-x)^kl_j(x)\equiv 0$ for $l_j(x)$ , the Lagrange Interplotion Basis
Prove a fomular $\sum_{j=0}^n(x_j-x)^kl_j(x)\equiv 0$ for $l_j(x)$ , the Lagrange Interplotion Basis
$$
l_{j}(x)=\frac{\left(x-x_{0}\right) \cdots\left(x-x_{j-1}\right)\left(x-x_{j+1}\right) \cdots\left(x-x_{n}\right)}{\left(x_{j}-x_{0}\right) \cdots\left(x_{j}-x_{j-1}\right)\left(x_{j}-x_{j+1}\right) \cdots\left(x_{j}-x_{n}\right)}
$$
$x_j$ are different points $(j=0,1,2,\cdots,n)$

My works:
I have known the fomular
$$
\sum_{j=0}^{n} x_{j}^{k} l_{j}(x) \equiv x^{k}, \quad k=0,1, \cdots, n
$$
I want to prove $\sum_{j=0}^n(x_j-x)^kl_j(x)\equiv 0$

I use the above fomular to get the following for $k = 1,\cdots,n$
$$
\begin{aligned}
& \sum_{j=0}^{n} \sum_{l=0}^{k}\left(\begin{array}{l}
k \\
l
\end{array}\right) x_{j}^{l}(-x)^{k-l} l_{j}(x) \\
=& \sum_{l=0}^{k}\left(\begin{array}{l}
k \\
l
\end{array}\right)(-x)^{k-l} \sum_{j=0}^{n} x_{j}^{l} l_{j}(x) \\
=& \sum_{l=0}^{k}\left(\begin{array}{l}
k \\
l
\end{array}\right)(-x)^{k-l} x^{l} \\
=&\,(x-x)^{k}\\
\equiv & \,0
\end{aligned}
$$
I wonder is there a method doesn't use the above fomular, or can inspire me from a different opinion. Thanks!!!

答:
My way:

If $f(x)$ is a $k$-order polynomial with $0\le k\le n$, it holds that
$$ \sum_{j=0}^n f(x_j)l_j(x) \equiv f(x). $$
So, for every $t\in\mathbb{R}$, we can choose $f_t(x)=(x-t)^k$, then the equality
$$ \sum_{j=0}^n (x_j-t)^k l_j(x) = (x-t)^k $$
holds for every $t$ and $x$.

If $k=1,2,\cdots,n$, let $t=x$, then
$$ \sum_{j=0}^n (x_j-x)^k l_j(x) \equiv 0 $$
follows.


问:
How is $\exists$ formulated in a FOL proof system?
$\exists x \phi(x)$ in mathematics  means there is something $c$, s.t. $\phi(c)$ holds.
What is the formulation of that same idea as a inference rule, an axiom,  or something else in a proof system in FOL?

Particularly what is the elimination rule for $\exists$? (It is not correct that if $\Phi \vdash \exists x \phi$, there exists a term $t$ so  $\Phi \vdash \phi[t/x]$.)

What is the similar one for $\forall$?

答:
In the Natural Deduction proof system we have a pair of inference rules for the existential (as well as the universal) quantifier:

$(\exists \text I) \ \varphi [x/a] \vdash \exists x \varphi$

$(\exists \text E) \ \text { if } \Gamma \vdash \exists x \varphi \text { and } \Delta, \varphi [x/a] \vdash \psi, \text { then } \Gamma, \Delta \vdash \psi$.

In an Hilbert-style proof system we have at least one axiom for it, unless it is defined in terms of the universal one.

See e.g. J. Shoenfield, Mathematical Logic for the axiom

$A[x/a] \to \exists x A$

and the rule:

$\dfrac {A \to B}{\exists x A \to B}$, provided that $x$ is not free in $B$.

In Enderton's system (see page 78) the existential quantifier is an abbreviation: $∃x α$ abbreviates $(¬∀x(¬α))$.


问:
Context Free Grammar Equal Lengths
Consider the language L = {x#y is in {0,1}* where |x| = |y|}

Would this CFG be a sufficient definition of the language L?

S->0S0 | 0S1 | 1S0 | 1S1 | #

Thanks.

答:
No, from Definition of this language, we can generate '#0#1#' and in your grammar, we cant generate this.

First way: Change definition of language to:

$$L=\{X\#Y \mid X,Y\in\{0,1\}^* \& |X|=|Y| \}$$

Second way: Change grammar

$$S\rightarrow0S0|1S1|0S1|1S0|0S\#|\#S0|1S\#|\#S1$$

Of course i think your goal is First Way

Title: Context Free Grammar Equal Lengths


答:
There are different ways to introduce $\exists$ in a FOL proof system.

I think the most common approach is that $\exists$ is defined via $\forall$:

$$\exists x \phi(x) \stackrel{\text{def}}\iff \lnot\forall x \lnot \phi(x).$$

And $\forall$ is introduced axiomatically before that.

Title: How is $\exists$ formulated in a FOL proof system?

Body:
$\exists x \phi(x)$ in mathematics  means there is something $c$, s.t. $\phi(c)$ holds.
What is the formulation of that same idea as a inference rule, an axiom,  or something else in a proof system in FOL?

Particularly what is the elimination rule for $\exists$? (It is not correct that if $\Phi \vdash \exists x \phi$, there exists a term $t$ so  $\Phi \vdash \phi[t/x]$.)

What is the similar one for $\forall$?


答:
I think you have a slight misunderstanding of Itô's lemma. What it says (in a simplified version) is that for any drift-diffusion process $(Y_t)_t$ which is solution of $dY_t = \mu dt +\sigma dW_t $, and any twice-differentiable function $f(t,x)$, it holds that the process $(f(t,Y_t))_t$ is a solution of the SDE :
$$df(t,Y_t) =  \frac{\partial f}{\partial t}(t,Y_t)dt+\frac{\partial f}{\partial x}(t,Y_t)dW_t+\frac{1}{2}\frac{\partial ^2 f}{\partial x^2}(t,Y_t)\cdot\sigma^2dt$$

In your case, the process $Y_t := W_t$ is indeed a drift-diffusion process and $f:(t,x)\mapsto e^{t/2}\cos(x)$ is twice-differentiable w.r.t. its arguments so Itô's lemma applies.

Title: Differentiation of a stochastic process by Ito's formula

Body:
Let $W_t$ denote a Brownian Motion, use ito's formula to compute the
differential of the following stochastic process:
$$X_t=e^{t/2}\cos(W_t)$$

Ito's formula is supposed to be used when we know a diffusion differential equation like $dx=\mu dt+\sigma dW_t$, then we can apply it to solve a $dG(x,t)$ by:
$$dG=\left(\frac{\partial G}{\partial t}+\frac{\partial G}{\partial x}\mu+\frac{1}{2}\frac{\partial^2G}{\partial x^2}\sigma^2\right)dt+\frac{1}{2}\frac{\partial G}{\partial x}\sigma dW_t.$$
But for this question, why we can solve the $dX_t$ directly?

Also, the answer to this question suggests that we should use the following Ito's formula to solve $dX_t$:
$$dX(t,W_t)=\frac{\partial X(t,W_t)}{\partial t}dt+\frac{\partial X(t,W_t)}{\partial W}dW_t+\frac{1}{2}\frac{\partial ^2 X(t,W_t)}{\partial W^2}dt$$

What's the logic behind this Ito formula?


问:
How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have?
I got stuck on this question:

How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? (including complex roots and roots with multiplicity)

My attempt at a solution:

First we can write the equation as:

$$(1+z^3)^8 - (1 + z^4)^6 = 0$$
We can then factor $(1+z^3)$

$$(z+1)^8(z^2-z+1)^8 - (1 + z^4)^6 = 0$$

But I am stuck here, the expression $(1+z^4)$ cannot easily be factored, but it should have 4 complex roots. I think there should be a simpler way to solve this question...

Body:
$(z^3+1)^8 = (z^4+1)^6$

$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 = z^{24} + 6z^{20} + 15z^{16} + 20z^{12} + 15z^8 + 6z^4 + 1$

$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 - z^{24} - 6z^{20} - 15z^{16} - 20z^{12} - 15z^8 - 6z^4 - 1 = 0$

$z^3(8z^{18} - 6z^{17} + 28z^{15} - 15z^{13} + 56z^{12} + 50z^9 + 56z^6 - 15z^5 + 28z^3 - 6z + 8) = 0$

$z^3(4z^6 - 3z^5 + 6z^3 - 3z + 4)(2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2)$

Cannot be factored further, but we can try to solve it from there:

$z^3 = 0 \Rightarrow \sqrt[3]{z^3} = \sqrt[3]{0} \Rightarrow z = 0$

$4z^6 - 3z^5 + 6z^3 - 3z + 4 = 0$ has 6 complex roots (three pairs of complex conjugates)

$2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2 = 0$ has 12 complex roots (six pairs of complex conjugates)

The equation has 19 roots.


问:
Module of differentials of an extension of Dedekind domains is cyclic
Let  be a Dedekind domain with fraction field , | a finite separable field extension and  the integral closure of  in . Assume that all the residue field extensions are separable.

In the proof of Serre's "Local Fields", chapter III, Proposition 14 (page 59), it is stated that in order to show that the module of differentials $Ω_{|}$ is cyclic we can assume that  is local and complete. (See  this question.)

I understand the reduction to the local case, but I don't understand the reduction to the complete case. I tried to use that the canonical map from $A_p$ to its $p$-adic completion is faithfully flat but I didn't arrive to the desired conclusion. In particular, I don't understand why, if $Ω_{'/'}$ is cyclic with with annihilator the different ideal of B' over A' [where A' and B' denote the $p$-adic completion of the DVRs A and B respectively], then $Ω_{/}$ is cyclic with annihilator the different ideal of B over A.

Can you give me some details? Thank you very much.

答:
It's easy to say "this just follows from faithful flatness", but I think the actual argument is a bit more subtle. So I'll try to be really precise.

Let $\mathfrak P\subset B$ be a prime ideal and $\mathfrak p\subset A$ its preimage. As argued in the question you linked, it suffices to show that $(\Omega_{B/A})_{\mathfrak P}$ is cyclic with annihilator the different $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$. Note that $(\Omega_{B/A})_{\mathfrak P}=\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$; see [Stacks Project, Tag 00RT]. Let $\widehat{B}_{\mathfrak P}$ and $\widehat{A}_{\mathfrak p}$ denote the completions of the DVRs $B_{\mathfrak P}$ and $A_{\mathfrak p}$. The key observation is:

Observation. $\widehat{B}_{\mathfrak P}=\widehat{A}_{\mathfrak p}[x]$ is generated by a single element. Moreover, we may assume that $x$ is already contained in $B_{\mathfrak P}$ .

Proof. The first part is well-known and appears for example in Serre's book, Chapter II Proposition 12, or in Neukirch's "Algebraic Number Theory", Chapter II Lemma (10.4). The second part isn't stated explicitly, but it becomes apparent from the proof given in Neukirch's book (and maybe also from Serre's proof, but I'm not really familiar with Serre's book). $\square$

Now we wish to show that $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$ is generated by $\mathrm{d} x$ and that the annihilator is $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$, i.e., that the sequence
$$0\rightarrow \mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\rightarrow B_{\mathfrak P}\xrightarrow{\mathrm{d}x}\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\rightarrow 0 $$
is exact (where the second arrow is the inclusion of the ideal $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$ and the third arrow is the map sending $b\in B_{\mathfrak P}$ to $b\mathrm{d}x\in \Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$). Since $\widehat{B}_{\mathfrak P}$ is faithfully flat over $B_{\mathfrak{P}}$, it suffices to show that $$0\rightarrow \mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}\rightarrow \widehat{B}_{\mathfrak P}\xrightarrow{\mathrm{d}x}\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}\rightarrow 0 $$
is exact. We already know that $\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$ is generated by $\mathrm{d}x$ with annihilator $\mathfrak D_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$. Moreover,  Chapter III Proposition 10 in Serre's book shows that $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\mathfrak D_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$. Hence we are done once we show the following lemma:

Lemma. $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$.

Proof. Since $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$ is a finitely generated $B_{\mathfrak P}$-module, we get that
$$ \Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\lim_{n\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}B_{\mathfrak P}/\mathfrak P^n\right)$$ is its $\mathfrak P$-adic completion. Since $B_{\mathfrak P}$ and $A_{\mathfrak p}$ are DVRs, there is an integer $e\geq 1$ such that $\mathfrak P^e=\mathfrak pB_{\mathfrak P}$. Thus
$$ \lim_{n\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}B_{\mathfrak P}/\mathfrak P^n\right)=\lim_{m\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{A_{\mathfrak p}}A_{\mathfrak p}/\mathfrak p^m\right)=\lim_{n\geq 0}\Omega_{(B_{\mathfrak P}/\mathfrak p^mB_{\mathfrak P})/(A_{\mathfrak p}/\mathfrak p^m)}\,,$$
where the second equality follows from the base change property for Kähler differentials. Similarly, $\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$ is a finitely generated module over the complete DVR $\widehat{B}_{\mathfrak P}$, hence it also is $\mathfrak P$-adically complete. Thus
$$ \Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\lim_{n\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{B}_{\mathfrak P}}\widehat{B}_{\mathfrak P}/\mathfrak P^n\right)\,.$$
With the same trick as above, we can rewrite this as
$$ \lim_{n\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{B}_{\mathfrak P}}\widehat{B}_{\mathfrak P}/\mathfrak P^n\right)=\lim_{m\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{A}_{\mathfrak p}}\widehat{A}_{\mathfrak p}/\mathfrak p^m\right)=\lim_{n\geq 0}\Omega_{(\widehat{B}_{\mathfrak P}/\mathfrak p^m\widehat{B}_{\mathfrak P})/(\widehat{A}_{\mathfrak p}/\mathfrak p^m)}\,.$$
Now observe $\widehat{B}_{\mathfrak P}/\mathfrak p^m\widehat{B}_{\mathfrak P}=B_{\mathfrak P}/\mathfrak p^mB_{\mathfrak P}$ and $\widehat{A}_{\mathfrak p}/\mathfrak p^m=A_{\mathfrak p}/\mathfrak p^m$ and the lemma follows. $\square$


问:
PSD Matrix Inequality proof
Is it true that for any two matrices $A \preceq B \iff C^T A C \preceq C^T B C$ for any conformable matrix $C$?

This related question proves
($\Rightarrow$) We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x \ge 0$ for any conformable vector $x$ so that $C^T(B-A)C \ge 0.$

What about the other direction?

答:
The converse is clearly false. Consider $C=0$ for instance.

For another counterexample we have $\pmatrix{1&0}\pmatrix{1&0\\ 0&-1}\pmatrix{1\\ 0}\succeq0$ but $\pmatrix{1&0\\ 0&-1}$ is indefinite.

Title: PSD Matrix Inequality proof


问:
Can I conclude from $\sum_{n=1}^\infty u_n$ being convergent that $\sum_{n=1}^\infty (-1)^n\frac{u_n}{n}$ is convergent?
So the question is the following statement correct?
$$
\sum_{n=1}^\infty u_n < \infty \implies
\sum_{n=1}^\infty (-1)^n\frac{u_n}{n} < \infty.
$$
I think the answer is not, however I cannot find a counterexample or prove it otherwise.

答: Hint: What if $u_n=\frac{(-1)^n}{\log(n+1)}$?


问:
Prove that if $\langle A_1, ..., A_k \rangle = M_{m\times n}(\mathbb{F})$, then $ \langle A_1^t, ..., A_k^t \rangle = M_{n\times m}(\mathbb{F})$.
I need to prove that $$\langle A_1, ..., A_k \rangle = M_{m\times n}(\mathbb{F}) \Longrightarrow \langle A_1^t, ..., A_k^t \rangle = M_{n\times m}(\mathbb{F})$$ By definition, $S=\{A_1, ..., A_k\}$ is subset of the minor vector subspace $W$ where $W=\cap \{W_i\}_{i\in I}$ and $W_i \subseteq V$. We say that $M_{m\times n}(\mathbb{F}) = \langle S \rangle$ is the vector space generated by $S$.

Note: By definition of $W$, $$W \subseteq \{W_i\}_{i\in I} \Longrightarrow S \subseteq \{W_i\}_{i\in I}$$

Edit. I'm not sure this is true, but I'm sure that the one for $n \times n$ is true...

答:
Your solution for the square case can be extended to the general case.

Let $A \in M_{n \times m}(\Bbb F)$, then $A^T \in M_{m \times n}(\Bbb F)$. Because $\langle A_1,\dots,A_k\rangle = M_{m \times n}(\Bbb F)$, there exist coefficients $a_i$ (for $i = 1,\dots,k$) such that $a_1 A_1 + \cdots + a_k A_k = A^T$. By the linearity of the transpose, we have $a_1 A_1^T + \cdots + a_k A_k^T = A^{TT} = A$. Thus, the arbitrary matrix $A \in M_{n \times m}(\Bbb F)$ is an element of $\langle A_1^T,\dots,A_k^T \rangle$. That is, we have $\langle A_1^T, \dots, A_k^T \rangle = M_{n\times m}(\Bbb F)$, which is what we wanted.


问:
Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$
How to simplify the expression:

$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$

If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.

Thank you.

答:
Put
$$x_0:=0,\quad x_1:=\sqrt{5},\quad x_2:=\sqrt{5+2\sqrt{5}},\quad x_3:=\sqrt{5+2\sqrt{5+2\sqrt{5}}}\ ,$$
and so on, which amounts to
$$x_0:=0,\qquad x_{n+1}:=\sqrt{5+2x_n}\quad(n\geq0)\ .$$
Then
$$x_{n+1}-x_n=\sqrt{5+2x_n}-\sqrt{5+2x_{n-1}}={2(x_n-x_{n-1}) \over \sqrt{5+2x_n}+\sqrt{5+2x_{n-1}}}\ .$$
As $x_1-x_0>0$ this shows that the sequence $(x_n)_{n\geq0}$ is momotonically increasing.

Furthermore $0\leq x_0<4$, and for any $n\geq0$ the statement $0\leq x_n< 4$  implies $$0\leq x_{n+1}<\sqrt{5+2\cdot 4}<4\ .$$ This shows that our sequence is as well bounded, so it has a limit $\xi\in[0,4]\ $. This limit satisfies the equation $x=\sqrt{5+2x}$ and  is therefore given by $\xi=1+\sqrt{6}\doteq 3.45$.


问:
Find point for geodesic on two-sheeted hyperboloid
The two-sheeted hyperboloid $p$ is defined by $x^2 + y^2 - z^2 = -1$

We are given two points, $a = (x_a,y_a,z_a)$ and $b = (x_b,y_b,z_b)$.

$a$, $b$ and $c$ defines the plane $q$.

How can we choose a point $c$ so that the intersection between $p$ and $q$ is a geodesic on the surface of the hyperbola $p$.

答: The intersection of any plane through the origin with the hyperboloid is a geodesic. All there has to be done, is to make sure that the three points $a$, $b$ and $c$ describe a plane through the origin. If $a$ and $b$ are not parallel, it is easiest to choose $c=(0,0,0)$.


问:
GAP checking for symmetric group
In GAP one can use
StructureDescription(G) to find out that $G$ is isomorphic to (say) $S_{120}.$ ($G$ might be defined by a collection of permutation of $140$ elements). One can also ask IsSymmetricGroup(G) which is a lot faster but just returns a boolean, so if the answer is true, we don't know if it is isomorphic to $S_{120}$ or $S_{7}.$ Is there some trick to get the degree of the permutation rep? StructureDescription is much slower.

答:
As noted, StructureDescription (which is a toy routine for small examples) will perform horribly badly on this question. But there are two special functions for this task:

IsNaturalSymmetricGroup takes a permutation group and checks whether it is the the symmetric group on its MovedPoints. (That is, it returns true on the group $\langle (4,5,6),(4,5)\rangle$.) The algorithm used will be much faster than even computing the order and should easily work for degrees into the 10000s.
The degree is then given by NrMovedPoints.

IsSymmetricGroup in contrast tests whether a group (of arbitrary representation) is isomorphic to a symmetric group. If so, SymmetricDegree will
return the degree of this symmetric group. The algorithm calculates the order and then a composition series and tests (with special treatment for $n=6$) that the group has a normal subgroup of index 2 isomorphic to $A_n$ (which, apart from $A_8$ can be identified from its order once we know it is simple) on which the group induces an outer automorphism.

Title: GAP checking for symmetric group


答:
It's not true generally, but perhaps could be made true under certain conditions. For example, if all $u_n > 0$, then
$$
(-1)^n \frac{u_n}{n}<\frac{u_n}{n} < u_n, \quad \forall n\ge 2,
$$
so your statement  would hold.

Additionally, this suggests a good place to start looking for counter-examples if you want to claim it always holds.

Title: Can I conclude from $\sum_{n=1}^\infty u_n$ being convergent that $\sum_{n=1}^\infty (-1)^n\frac{u_n}{n}$ is convergent?

Body:
So the question is the following statement correct?
$$
\sum_{n=1}^\infty u_n < \infty \implies
\sum_{n=1}^\infty (-1)^n\frac{u_n}{n} < \infty.
$$
I think the answer is not, however I cannot find a counterexample or prove it otherwise.


问:
Existance and uniqueness for diverging and converging integrals
Let $f\in C[a,b]$ and $y_0\in (a,b)$ such that $f(y_0)=0$ and $f(y) \neq 0$ for any $y\in [a,b]$,$y\neq y_0$. Let $x_0\in\mathbb{R}$. Show that if both $\int_{a}^{y_0}\frac{du}{fu},\int_{y_0}^{b}\frac{du}{f(u)}$ are divergent, then $(x_0,y_0)$ is a point of existence and local uniqueness of $y'=f(y)$. If at least one of the two integrals converge, then $y_0$ is a singular solution for $y'=f(y)$.

My approach was to choose a integral, $F(y)=\int_{y_0}^{y}\frac{du}{f(u)}$,$y\in\mathbb{R}$. For a solution $\phi:I\rightarrow\mathbb{R}$ I obtain that $F(\phi(x))=x+c$ and, more, $F(\phi(x))=x-x_0$ and so $I\subset I+x_0$. But the limits of $F$ as $y\rightarrow a$ is negative infinity and for y towards $b$ positive infinity, so I can argue that $I=\mathbb{R}$. I don`t know how to end it from here.

答:
The singularity is at $y_0$, thus you get that
$$
\lim_{y\to y_0}\int_a^y\frac{du}{f(u)}=\pm\infty
$$
the sign depending on the sign of $f$ on $[a,y_0)$. The sign is certain as $y_0$ is the only root of $f$ on $[a,b]$.

Now continue as proposed, with $F$ you get $F(y(t))-F(a)=t-t_a$, and as $F$ is a monotonic function, it has an inverse. If $f$ is positive on $[a,y_0)$, then $F$ is growing towards infinity, $F(y(t))=F(a)+t-t_a$ is solvable for $a<y(t)<y_0$ for any $t>t_a$. So at no time can the level $y_0$ be reached.

In the case where that integral is finite, one gets that at the finite time $t=t_a+F(y_0)-F(a)$ the line $y=y_0$ is reached, so that there is no uniqueness at that line.

The situation on the other side $(y_0,b)$ is similar for negative $f$. With a switched sign in $f$, one has to consider the behavior of the function for falling $t$ values.


问:
Contangent function (approximate)
Approximate value of the  cotangent function we can find by the following formula, where $x$ represents the value of the semicircle minor arc angle

$ \sqrt [90]{\frac{{\pi}}{4\sin\left(\frac{{\pi}}{4}\right)}}^\left(2x-90\right)\cdot \left(\frac{90}{x}-1\right)\cdot \sqrt{2}+1$

For example if $x$= 60 we will get 1.73229..or for example $x$=30 we will get 3.7311 There is formula similare to mine, in literature

答:
More than $\large\color{red}{1,400}$ years ago were made the following approximations
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$
So, for $0 \lt x \leq \frac \pi 2$, it could have been possible to write
$$\cot(x) \sim  \frac{\left(5 \pi ^2-4 (\pi -x) x\right) \left(\pi ^2-4 x^2\right)}{16 (\pi -x) x
   \left(x^2+\pi ^2\right)}$$ Making $x=\frac \pi n$, this would give the simple
$$\cot \left(\frac{\pi }{n}\right)\sim \frac{\left(n^2-4\right) \left(5 n^2-4 n+4\right)}{16 (n-1) \left(n^2+1\right)}$$

By the way, your values are not correct.

Why don't you try my simpler
$$\cot(x)\sim \frac{15 \left(x^4-28 x^2+63\right)}{x \left(x^4-105 x^2+945\right)}$$

Title: Contangent function (approximate)


问:
Complex integral of $\lvert z-i\rvert$ along a line
I was given the following question in a small exam, which i had to answer within 5 minutes: Let $a,b \in \mathbb{C}$. Let $\gamma$ be the parameterization of the line between $a%$ and $b$. Compute:
$$ \int_\gamma \left\lvert z-\mathcal{i} \right\rvert ^2\mathbb{d}z$$

I tried to calculate the integral with cartesian coordinates and the parameterization $\gamma :t\mapsto a+t(b-a)$, using the definition of a line integral:
$$=\int_0^1|a+t(b-a)-\mathcal{i}|^2(b-a)\mathbb{d}t \ =\int_0^1\Bigl(\bigl(\Re(a)+t \left(\Re(b)-\Re(a)\right)\bigl)^2+ \bigl(\Im(a)+t\left(\Im(b)-\Im(a)\right) - 1\bigl)^2\Bigl)(b-a)$$

Which evaluates to

$$\frac{1}{3}(b-a)\bigl(\lvert a\rvert^2+ \lvert b\rvert^2+ \Re(a)\Re(b)+\Im(a)\Im(b)\bigl) + (a-b)\bigl(\Im(a)+\Im(b)-1\bigl)$$

While it is not hard to calculate this by multiplying and integrating the resulting polynomial, it is rather laborious and it took me far longer that 5 minutes. This leads me to believe that there is a more efficient way to do this. The only similar integrals that I was able to find on here were along a circle, where polar coordinates obviously facilitate the computation. Since the integrand is not holomorphic, I am not able to use Cauchy's integral formula either.

Is there a faster way to compute this, or am I just too slow using the method that I described?

答:
Its easier to compute a general case then check parameters:
$$\int_0^1 |A+tB|^2 Cdt = C\int_0^1|A|^2+2\Re \big(tA\overline B\big) + t^2|B|^2dt = C(|A|^2+\Re (A\overline B) + \frac13|B|^2)$$
now your integral is
$$ A=a-i, \quad B=b-a,\quad C=|b-a|$$
In a time-crunch situation, I would be happy to submit my answer as

$$\int_0^1 |A+tB|^2 Cdt =|b-a|\left(|a-i|^2 + \Re\big((a-i)\overline{(b-a)}\big) + \frac13|b-a|^2\right).$$


问:
Given $ A $ find $ B $ so that $ [ A \: B ] [ A \: B ] ^ { \mathrm T } $ is the identity matrix
Given $ A $ ($ 4 \times 2 $ matrix) I want to find $ B $ ($ 4 \times 2 $ matrix) such that when you append both in a single $ 4 \times 4 $ matrix $ [ A \: B ] $ the following relationship will hold:
$$
[ A \: B ] [ A \: B ] ^ { \mathrm T } = I
$$

Edit context:

I am going through a paper on unambiguous state discrimination where a process has been outlined for finding optimal discrimination matrices. This is the paper https://ieeexplore.ieee.org/document/4557236. In section (3) the paper uses the matrix I have called $ B $ but does not however tell how to get the matrix. $ A $ is made up of the null space vectors of a rank $ 2 $ matrix in a $ 4 $-dimensional Hilbert space. $ A $ is also real.

答:
I cannot access the paper you linked. However, I suspect that "null space vectors" refer to an orthonormal basis for the nullspace.

If this is the case, then the columns of $A$ are orthonormal, and it follows that the matrix $M = [A\ \ B]$ will satisfy $MM^T = I$ if and only if $M^TM = I$, which holds if and only all columns of $M$ are orthonormal. Thus, a satisfactory matrix $B$ can be build from the nullspace vectors of $A^T$.


问:
Calculating the integral of a function multiplied by two delta functions
I would like to calculate $\int_{a}^{d}\delta(x-b)\delta(x-c)f(x)dx$, where $a \le b \le d$ and $a \le c \le d$.

My imediate thought is to integrate by parts to obtain
$$\int_{a}^{d}\delta(x-b)\delta(x-c)f(x)dx = [\delta(b-c)f(x)]_{a}^{d} - \int_{a}^{d}\delta(b-c)f'(x)dx$$
$$= \delta(b-c)([f(x)]_{a}^{d} - [f(x)]_{a}^{d}) = 0$$

However, this does not seem correct. Any help on the matter would be appreciated.

答:
So I believe the answer can be found using test functions to represent the $\delta$-function. I obtained the general form

$$ \int_a^d \delta(x - b) \delta (x - c) f(x) dx = \delta(b - c) f(c), $$

where $f$ is evaluated at $c$ w.l.o.g as the integral is 0 when $b \ne c$.


问:
A set theory problem from Bruckner Analysis
Following is from Bruckner's book Analysis (Theorem 5.29 (Radon–Nikodym)) :

Fix $n\in\mathbb{N}$. Let
$$f_n(x)=\max_{i\leq n}g_i(x).$$
For $M\in\mathcal{M}$, $M\subset E$, let $B_0=\varnothing$ and, for $i\leqslant n$, inductively define
$$B_i=\bigl(\{x:f_n(x)=g_i(x)\}\cap M\bigr)\setminus B_{i-1}.$$
Then $\color{blue}{M=\bigcup_{i=1}^nB_i.}\ \color{red}{\text{This is a disjoint union.}}$ Thus
$$\int_Mf_n\,d\mu=\sum_{i=1}^n\int_{B_i}g_i\,d\mu=\sum_{i=1}^n\sum_{k=1}^\infty\int_{B_i\cap E_i^k}g_i\,d\mu.\tag{16}$$

I long thought for both the claims (highlighted in blue and red) and I couldn't prove them. Specially the sets being disjoint seems to be wrong by van diagram.  Please help!

PS  I think they are just elementary set theory questions, no need to know about $g_i$.

答: The first and more obvious statement is that $\bigcup_{i=1}^n B_i=M$. Each $B_i\subseteq M$ by definition and since there is a finite number of $g_i$ there is for each $x\in M$ $\max\{g_1(x)\cdots g_n(x)\}=f_n(x)$ for the least $i\in \{1,2\cdots n\}$ such that $g_i(x)=\max\{g_1(x)\cdots g_n(x)\}$ we have $x\in B_i$ since $g_i(x)=f_n(x)$ but we also have that since $i$ is the least $g_{i-1}\neq f_n(x)$ so $x\notin B_{i-1}$. For the statement unless we know something else about the $g_i$s or what $M$ is this is false. Here is a very simple counter example $g_1=1, g_2=0, g_3=1$ as constant funcitons. Then $f_3=1$ as well. We have then $B_1=M$ $B_2=\emptyset$ and finally $B_3=M\setminus B_2=M$. These are clearly non disjoint. I do suspect that the author meant to subtract the union of all $B_j$ with $j<i$


问:
When continuous random variable is in both the conditional and the conditioned
Let $A,B$ be continuous random variables. Let $E,G,K$ be events. Let $t \in image(B)$.

(I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf. If not, then assume joint pdf is well-defined whenever necessary.)

Question 1: Can we say $P(B \le A|B=t) = P(t \le A|B=t)$?

Question 2: If so, then how? If not then why?

What I've tried: I think we're doing something like $P(E|K)=P(E \cap K|K)$, and I get it when $P(G|K)$ is defined for $P(K)>0$. What is being done here when technically $P(K)=0$? I mean, of course, in the 1st place when we say like '$P(\cdot|B=t)$', this is notational, we're not really conditioning on the $P$-null event $\{B=t\}$. But I still don't get exactly what's being done here. Well...

As I understand, $$RHS = P(t \le A|B=t) := \frac{\int_{t \le A} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da} = \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da}$$

$$= \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{f_{B}(t)}= \int_{t}^{\infty} \frac{f_{A,B}(a,t)}{f_{B}(t)} da,$$ where/whence $f_{A|B=t}(a) := \frac{f_{A,B}(a,t)}{f_{B}(t)}$.

That's all I got.

Not really sure how to evaluate LHS. Perhaps LHS := RHS, i.e. LHS is defined as RHS?

Update: Wait I think I have an idea how to do LHS

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db$$

Here,

'$f_{(A,B)|B=t}(a,b)$' is I guess meant like $f_{(A,B)|T=t}(a,b)$ where $T$ is a random variable that is (not just almost surely but really) surely equal to $B$.

'$f_{(A,B)|T=t}(a,b)$' is meant like 2 random variables conditioned on a 3rd random variable. As I understand multivariate conditional joint distributions, we have this is $f_{(A,B)|T=t}(a,b) := \frac{f_{A,B,T}(a,b,t)}{f_T(t)}$ and then...

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db = \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{f_T(t)} da db$$

$$= \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} da db$$

$$= \frac{\int \int_{b \le a} f_{A,B,T}(a,b,t) da db}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} $$

And then I don't know. Apparently we might not be able to have a joint pdf if 2 of the random variables are surely equal (maybe even for almost surely):

4.1. joint distribution of x with..itself

4.2. https://stats.stackexchange.com/questions/47162/what-is-the-joint-probability-distribution-of-two-same-variables

答:
Here is a measure theory argument that runs parallel to my non-measure theory argument.

Let $A, B$ be random variables. Suppose $B$ has a density $f_B(x)$ that is continuous at the point $x=t$, and $f_B(t)>0$.   Fix $t \in \mathbb{R}$.  Let $h:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ be bounded and measurable functions that define versions of the following conditional expectations:
\begin{align}
&h(B) = E[1\{B \leq A\}|B]\\
&g(B) = E[1\{t \leq A\}|B]
\end{align}

Claim:   If $h(x)$ and $g(x)$ are continuous at $x=t$ then $h(t)=g(t)$.

Proof: Fix $\delta>0$. Observe that
$$ 1\{B \leq A\}1\{B \in [t, t+\delta]\} \leq 1\{t \leq A\}1\{B \in [t, t+\delta]\}$$
Thus
$$ E[1\{B \leq A\}1\{B \in [t, t+\delta]\}] \leq E [1\{t \leq A\}1\{B \in [t, t+\delta]\}]$$
and so, by the definition of a conditional expectation:
$$ E[h(B)1\{B \in [t, t+\delta]\}] \leq E[g(B)1\{B \in [t, t+\delta]\}]$$
Thus
$$ \int_{t}^{t+\delta} h(x)f_B(x)dx \leq \int_t^{t+\delta}g(x)f_B(x)dx$$
This holds for all $\delta>0$ and since $h, g$ are both continuous at $x=t$, and $f_B(x)$ is continuous at $x=t$ and $f_B(t)>0$, we get
$$ h(t)\leq g(t)$$
A similar argument shows the reverse inequality.


问:
The integral of $\mathrm{e}^{-\frac{1}{\sin^2 x}}$
I have tried playing around an integral
$$
\int \mathrm{e}^{-\frac{1}{\sin^2 x}} \,\mathrm{d}x
.
$$
I know this integral does not have a closed form (Wolfram Alpha and many more resources!), but I appreciate an approximation that holds as close as possible to this integral. The Taylor series is not accurate. It can be assumed that the bounds of the integral are $0$ and $x$, where $0<x<\frac{\pi}{2}$.

Appendix

Thanks to the @robjohn's useful answer, I have simulated his three approaches (in the original order presented) for $0<x<\frac{\pi}{2}$ in the following figure:

The blue solid line denotes the integral which should be approximated and the other curves are three closed forms for that. Unfortunately, there is a bad divergence between the approximations and the integral after about $x=0.9$.

答:
A First Approach

Since
$$
\frac1{\sin^2(t)}=\frac1{t^2}+\frac13+O\!\left(t^2\right)\tag1
$$
we have
$$
\begin{align}
\int_0^xe^{{}-\csc^2(t)}\,\mathrm{d}t
&=e^{-1/3+O(x^2)}\int_0^xe^{-1/t^2}\,\mathrm{d}t\tag{2a}\\
&=e^{-1/3+O(x^2)}\int_{1/x}^\infty\frac{e^{-t^2}}{t^2}\,\mathrm{d}t\tag{2b}\\
&=\frac12e^{-1/3+O(x^2)}\int_{1/x^2}^\infty\frac{e^{-t}}{t^{3/2}}\,\mathrm{d}t\tag{2c}\\
&=\frac12e^{-1/3+O(x^2)}\left(x^3+O\!\left(x^5\right)\right)e^{-1/x^2}\tag{2d}\\
&=\frac12e^{-1/3}x^3e^{-1/x^2}\left(1+O\!\left(x^2\right)\right)\tag{2e}
\end{align}
$$
Explanation:

$\text{(2a)}$: apply $(1)$

$\text{(2b)}$: substitute $t\mapsto1/t$

$\text{(2c)}$: substitute $t\mapsto t^{1/2}$

$\text{(2d)}$: integrate by parts

$\text{(2e)}$: collect the error terms

Verification of Fit

This shows that the error seems to be within a factor of $1+O\!\left(x^2\right)$.

A Better Approach
$$\newcommand{\arccsc}{\operatorname{arccsc}}
\begin{align}
&\int_0^xe^{{}-\csc^2(t)}\,\mathrm{d}t\tag{3a}\\
&=\int_0^{\sin(x)}e^{-1/t^2}\,\mathrm{d}\arcsin(t)\tag{3b}\\
&=\int_0^{\sin(x)}e^{-1/t^2}\left(1+\frac{t^2}2+\frac{3t^4}8+\frac{5t^6}{16}+O\!\left(t^8\right)\right)\mathrm{d}t\tag{3c}\\
&=\int_{\csc^2(x)}^\infty e^{-t}\left(\frac{t^{-\frac32}}2+\frac{t^{-\frac52}}4+\frac{3t^{-\frac72}}{16}+\frac{5t^{-\frac92}}{32}+O\!\left(t^{-\frac{11}2}\right)\right)\mathrm{d}t\tag{3d}\\
&=\left.e^{-t}\left(-\frac{t^{-\frac32}}2+\frac{t^{-\frac52}}2-\frac{23t^{-\frac72}}{16}+\frac{39t^{-\frac92}}8+O\!\left(t^{-\frac{11}2}\right)\right)\right|_{\csc^2(x)}^\infty\tag{3e}\\
&=e^{{}-\csc^2(x)}\left(\frac{\sin^3(x)}2-\frac{\sin^5(x)}2+\frac{23\sin^7(x)}{16}-\frac{39\sin^9(x)}8+O\!\left(x^{11}\right)\right)\tag{3f}
\end{align}
$$
Explanation:

$\text{(3b)}$: substitute $t\mapsto\arcsin(t)$

$\text{(3c)}$: use the Taylor Series for $\arcsin(t)$

$\text{(3d)}$: substitute $t\mapsto t^{-\frac12}$

$\text{(3e)}$: integrate by parts repeatedly

$\text{(3f)}$: evaluate

Instead of a relative error of $1+O\!\left(x^2\right)$ this formula has a relative error of $1+O\!\left(x^8\right)$.

Extending the Formula

We can easily extend $(3)$ to higher order. The formula with the next few terms is
$$
\begin{align}
e^{{}-\csc^2(x)}&\left(\frac{\sin^3(x)}2-\frac{\sin^5(x)}2+\frac{23\sin^7(x)}{16}-\frac{39\sin^9(x)}8\right.\\
&\left.+\frac{5651\sin^{11}(x)}{256}-\frac{31049\sin^{13}(x)}{256}+\frac{1614779\sin^{15}(x)}{2048}+O\!\left(x^{17}\right)\right)\tag4
\end{align}
$$
which has a relative error of $1+O\!\left(x^{14}\right)$.

Starting at the Other End

Since $e^{{}-\csc^2(x)}$ is not analytic at $x=0$, the expansion above is asymptotic and does not converge; it is not good for $x$ far from $x=0$. To compute the integral near $x=\frac\pi2$, we first compute the integral over the entire interval $\left[0,\frac\pi2\right]$:
$$\newcommand{\erfc}{\operatorname{erfc}}
\begin{align}
\int_0^{\pi/2}e^{{}-\csc^2(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}e^{{}-\sec^2(x)}\,\mathrm{d}x\tag{5a}\\
&=\int_0^\infty e^{-1-x^2}\frac{\mathrm{d}x}{1+x^2}\tag{5b}\\
&=\int_0^\infty\int_0^\infty e^{-t(1+x^2)}e^{-1-x^2}\,\mathrm{d}t\,\mathrm{d}x\tag{5c}\\
&=\frac{\sqrt\pi}2\int_0^\infty\frac{e^{-1-t}}{\sqrt{1+t}}\,\mathrm{d}t\tag{5d}\\
&=\sqrt\pi\int_1^\infty e^{-t^2}\,\mathrm{d}t\tag{5e}\\[3pt]
&=\frac\pi2\erfc(1)\tag{5f}
\end{align}
$$
Explanation:

$\text{(5a)}$: substitute $x\mapsto\frac\pi2-x$

$\text{(5b)}$: substitute $x\mapsto\arctan(x)$

$\text{(5c)}$: $\int_0^\infty e^{-t(1+x^2)}\,\mathrm{d}t=\frac1{1+x^2}$

$\text{(5d)}$: integrate in $x$

$\text{(5e)}$: substitute $t\mapsto t^2-1$

$\text{(5f)}$: $\int_a^\infty e^{-t^2}\,\mathrm{d}t=\frac{\sqrt\pi}2\erfc(a)$

We will also use the following integral, which can be evaluated by repeated integration by parts
$$
\begin{align}
\int_0^{\tan^2(x)}e^{-t}t^{k-1/2}\,\mathrm{d}t
&=e^{{}-\tan^2(x)}\frac{\tan^{2k+1}(x)}{k+1/2}+\frac1{k+1/2}\int_0^{\tan^2(x)}e^{-t}t^{k+1/2}\,\mathrm{d}t\tag{6a}\\
&=e^{{}-\tan^2(x)}\frac{(2k)!}{4^kk!}\sum_{j=k}^\infty\frac{4^{j+1}(j+1)!}{(2j+2)!}\tan^{2j+1}(x)\tag{6b}
\end{align}
$$
Thus, we get
$$\newcommand{\erfc}{\operatorname{erfc}}
\begin{align}
\int_0^xe^{{}-\sec^2(t)}\,\mathrm{d}t
&=\int_0^{\tan(x)}e^{-1-t^2}\frac{\mathrm{d}t}{1+t^2}\tag{7a}\\
&=\frac12\int_0^{\tan^2(x)}e^{-1-t}\sum_{k=0}^\infty(-1)^kt^{k-1/2}\mathrm{d}t\tag{7b}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}2\int_0^{\tan^2(x)}e^{-1-t}t^{k-1/2}\mathrm{d}t\tag{7c}\\
&=\frac{e^{{}-\sec^2(x)}}{2}\sum_{k=0}^\infty(-1)^k\frac{(2k)!}{4^kk!}\sum_{j=k}^\infty\frac{4^{j+1}(j+1)!}{(2j+2)!}\tan^{2j+1}(x)\tag{7d}\\
&=\frac{e^{{}-\sec^2(x)}}{2}\sum_{j=0}^\infty\sum_{k=0}^j(-1)^k\frac{(2k)!}{4^kk!}\frac{4^{j+1}(j+1)!}{(2j+2)!}\tan^{2j+1}(x)\tag{7e}\\
&=e^{{}-\sec^2(x)}\left(\tan(x)+\frac13\tan^3(x)+\frac13\tan^5(x)-\frac1{21}\tan^7(x)\right.\\
&\phantom{=e^{{}-\sec^2(x)}}\left.+\frac{19}{189}\tan^9(x)-\frac{151}{2079}\tan^{11}(x)+O\!\left(x^{13}\right)\right)\tag{7f}
\end{align}
$$
Explanation:

$\text{(7a)}$: substitute $t\mapsto\arctan(t)$

$\text{(7b)}$: substitute $t\mapsto t^{1/2}$ and expand $\frac{t^{-1/2}}{1+t}$

$\text{(7c)}$: exchange order of summation and integration

$\text{(7d)}$: apply $(6)$

$\text{(7e)}$: change the order of summation

$\text{(7f)}$: generate the coefficients

Putting the Two Ends Together

Since the formula in $(3)$ is designed to be very accurate near $x=0$, and the extra terms do not behave well for larger $x$, we will only use the first term for $x\in\left[0,\frac\pi4\right]$, which actually matches better than the formula in $(2)$.

For $x\in\left[\frac\pi4,\frac\pi2\right]$, we will use $(5)$ and $(7)$.
$$
f(x)=\left\{\begin{array}{}
e^{{}-\csc^2(x)}\frac{\sin^3(x)}2&\text{if }x\in\left[0,\frac\pi4\right]\\
\frac\pi2\erfc(1)-e^{{}-\csc^2(x)}p(\cot(x))&\text{if }x\in\left[\frac\pi4,\frac\pi2\right]
\end{array}\right.\tag{8a}
$$
where
$$
p(x)=x+\frac13x^3+\frac13x^5-\frac1{21}x^7+\frac{19}{189}x^9-\frac{151}{2079}x^{11}\tag{8b}
$$


问:
How to prove composition of two funtions of type $min(c,max (b,x+a))$ is of the same type
Suppose i have functions $f(x)=min(c_1,max(b_1,x+a_1))$ and $g(x)=min(c_2,max(b_2,x+a_2))$ , how to prove that $g(f(x))=min(c,max(b,x+a))$ . Here except $x$ all other values are constants.

Motivations of the problem : I was reading editorial of some problem here but i was not able to understand how 3rd step came from 2nd in the proof.

Body:
we have

$$f(x)=min(c_1,max(b_1,x+a_1))$$
$$g(x)=min(c_2,max(b_2,x+a_2))$$

Then it's something like

$$g(f(x))=min(c_2,max(b_2,min(c_1,max(b_1,x+a_1))+a_2))$$

$$=min(min(max(c_1,b_2),c_2),max(max(min(b_1,c_1),b_2),x+a_1+a_2^*))$$


问:
Prove that $\Vdash_{Coll(\omega_1, \kappa)} \Diamond$ where $\kappa$ is the cardinal $2^{\aleph_0}$
I'd like to prove that $\Vdash_{Coll(\omega_1, \kappa)} \Diamond$ where $\kappa$ is the cardinal $2^{\aleph_0}$

My attempt:

Recall that $Coll(\omega_1, \kappa)$ consists of functions $\omega_1 \rightarrow \kappa$ with domain of size $<\omega_1$.

Given an $M$-generic $G$ and corresponding $M[G]$, we first start with a surjection $g:\omega_1 \rightarrow \kappa$ where $g = \cup G \in M[G]$. We may also use basic cardinal arithmetic to find that, in $M[G]$, there is a bijection $h:\kappa \rightarrow [\omega_1]^{<\omega}$.

We define $A_{\alpha} = h(g(\alpha)) \cap \alpha$ for every $\alpha < \omega_1$. I'm trying to show that $\langle A_{\alpha}: \alpha < \omega_1 \rangle$ is our $\Diamond$ sequence. But I'm struggling to make this part work.

Would appreciate the help!

答:
(There was an error in my previous answer: my original attempt is reducing our goal to $\mathsf{Add}(\omega_1,1)$ forces $\diamondsuit$, by showing every $V[G]$ for a $\operatorname{Col}(\omega_1,\kappa)$-generic filter contains a $\operatorname{Add}(\omega,1)$-generic filter $H$. The flaw is that there is no reason to believe $V[G]$ satisfies $\diamondsuit$ even if $V[H]$ does. However, a slight modification still works.)

Let me work over $V$ instead of $M$.
Observe the following facts:

If $G$ is
a $\operatorname{Col}(\omega_1,\kappa)$-generic filter
over $V$,
then $V[G]$ contains
$\operatorname{Add}(\omega_1,1)$-generic filter
over $V$. In fact, the function
$\varphi: \operatorname{Col}(\omega_1,\kappa)=\operatorname{Fn}(\omega_1,\kappa,\omega_1)\to
\operatorname{Add}(\omega_1,1)=\operatorname{Fn}(\omega_1,2,\omega_1)$
given by $\varphi(p)(\alpha) := p(\alpha)\mod 2$,
that is,
$$\varphi(p)(\alpha)=\begin{cases}0 & \text{if $p(\alpha)$ is of the form $\xi+2n$ for some $n\in\omega$,}\\ 1 & \text{otherwise.}\end{cases}$$
satisfies the following: if $D\subseteq\operatorname{Add}(\omega_1,1)$ is dense,
then $\varphi^{-1}[D]$ is also a dense subset of
$\operatorname{Col}(\omega_1,\kappa)$.
Hence $\varphi^"[G]$ generates a generic filter of $\operatorname{Add}(\omega_1,1)$.

$\operatorname{Add}(\omega_1,1)$ forces $\diamondsuit$.

Hence we will mimic the proof of $\Vdash_{\operatorname{Add}(\omega_1,1)}\diamondsuit$. I borrowed the following proof from Kunen's new forcing textbook.

Let us replace $\operatorname{Col}(\omega_1,\kappa)$ to $\operatorname{Col}(I,\kappa)$, where
$$I=\{(\xi,\alpha)\mid \xi<\alpha<\omega_1\}.$$
For a generic filter $G$, let $f_\alpha(\xi):=(\bigcup G)(\xi,\alpha)\mod 2$, that is, $f_\alpha(\xi)=\varphi\circ (\bigcup G)(\alpha,\xi)$. Then $f_\alpha:\alpha\to 2$ is a characteristic function of some $A_\alpha\subseteq \alpha$. We claim that $\langle A_\alpha\mid\alpha<\omega_1\rangle$ is a $\diamondsuit$-sequence.

Work over $V[G]$, fix $h:\omega_1\to 2$ and a club $C\subseteq\omega_1$. Take a name $\dot{h}$, $\dot{C}$ and a condition $p_0$ such that
$$p_0\Vdash \dot{h}:\omega_1\to 2\text{ and $\dot{C}$ is a club.}$$

Now we forget about $V[G]$ and work over $V$. For each condition $p$, let $\operatorname{supp}p$ be the least ordinal $\beta$ such that $\operatorname{dom}p\subseteq \{(\xi,\alpha)\mid \xi<\alpha<\beta\}$.
Now define $p_n$, $\beta_n$, $\delta_n$, and $k_n$ such that

$p_0\ge p_1\ge\cdots$, $\beta_n=\operatorname{supp}p_n$,

$\beta_0<\delta_0<\beta_1<\delta_1<\cdots$, and

$p_{n+1}\Vdash \dot{h}\upharpoonright \beta_n = k_n,\> \delta_n\in\dot{C}$.

(Here we need that $\operatorname{Col}(\omega_1,\kappa)$ does not add any sequence of countable length, so for each ordinal $\beta$ and a condition $q$, we can find $r\le q$ and $k$ such that $r\Vdash \dot{h}\upharpoonright\beta=k$.)

Now $p=\bigcup_n p_n$, $\gamma=\bigcup_n\beta_n=\bigcup_n\delta_n$, and $k=\bigcup_n k_n$. Then $p$ forces $\gamma\in\dot{C}$ and $\dot{h}\upharpoonright\gamma=k$.
Moreover, $\operatorname{supp}p=\gamma$. Now extend $p$ to $s$ by letting
$$s(\xi,\gamma):= k(\xi).$$

Then $s\Vdash \dot{h}\upharpoonright\gamma=\dot{f}_\gamma$. Hence we proved the following:

For arbitrary $p$, $\dot{h}$, and $\dot{C}$ such that $p\Vdash \dot{h}:\omega_1\to 2\land \dot{C}\subseteq\omega_1\text{ is club}$, there is $s\le p$ and $\gamma<\omega_1$ such that $s\Vdash \dot{h}\upharpoonright \gamma=\dot{f}_\gamma$.

Fix $\dot{h}$ and $\dot{C}$, the set of all such $s$ forms a dense set. Hence $V[G]$ thinks $\langle f_\alpha\mid\alpha<\omega_1\rangle$ is a $\diamondsuit$-sequence.