问:
Measurable random variables constant on generators of sigma algebra?
Let $_1, … , _$ be a partition of $$ with $ℙ[_] > 0$ for all $$, and
let $G ≔ (_1, … , _)$. Now if  is a random variable that is $G$-measurable it means I thing that the preimage of each borel set is in $G$ right? But why is the random variable $X$ constant on each $E_1,...,E_n$? If I suppose that on for instance that on $E_j$ X takes two values $a$ and $b$ the I know $X^{-1}(a)$ and $X^{-1}(b)$ must be in $G ≔ (_1, … , _)$ but how to continue?

Edit

I think I get it now. Consider $E_j$ and let $\omega\in E_j$. Then say $X(\omega)=a$. Now $X^{-1}(a)\cap E_j$ is a non-empty measurable subset of  $E_j$ But. Since $E_j$ is the only non-empty measurable subset of $E_j$ then $X^{-1}(a)\cap E_j=E_j$ meaning $X|_{E_j}=a$

答:
To make thing more clear, let $\mathcal G=\sigma (E_1,E_2)=\{X,E_1, E_2,\varnothing \}.$

As you can see if there are $$X(\omega _1)=:a\neq b:=X(\omega _2),$$ where $\omega _1,\omega _2\in E_1$, then $X^{-1}(a)\subsetneq E_1$, and thus $X$ is not measurable.


问:
Why does $P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$ hold?
Let $X, X_n$ be $\mathbb{R}^d$-valued random variables.
The solution for an exercise uses the following step without further comment:
$$P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$
As far as I know, I should however be able to rewrite $\limsup$ on the left as follows, which seems a bit of a jump:
$$P(\limsup_n |X_n-X|>0) = P(|X_n-X|>0 \text{ i.o.}) \le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$
Can somebody please explain why this step is allowed?

答:
$\{\limsup_n |X_n - X| > 0\} \subseteq \{|X_n - X| > 0 \text{ i.o.}\}$, but equality does not hold.

Showing the inclusion: if the limsup is positive, then there is a subsequence $(|X_{n_k} - X|)_k$ that converges to a positive number. Thus, $|X_{n_k} - X| > 0$ i.o. and thus $|X_n - X| > 0$ i.o.

Showing the reverse inclusion need not hold: If $|X_n - X| = \frac{1}{n}$, then $|X_n - X| > 0$ always, but the lim sup is zero. Note that this is also a counterexample to the false claim $\{|X_n - X| > 0 \text{ i.o.}\} \subseteq \{|X_n - X| > 1/k \text{ i.o. for some $k$}\}$

To show $\{\limsup_n |X_n - X| > 0\} \subseteq \{|X_n - X| > 1/k \text{ i.o. for some $k$}\}$, modify the argument in first bullet point slightly.

ContentLicense: CC BY-SA 4.0


答: As stated, this is wrong unless you add some more conditions. Consider $X \not=0$ a.s. and consider the random variables $X_1 = 0$ and $X_n = X$ for $n\geq 2$. Then $\mathbb{P}(\limsup_n |X-X_n| > 0) = \mathbb{P}(|X-X_1|>0)=\mathbb{P}(|X|>0)=1$. But if $n\geq 2$, then $|X-X_n| = 0$ a.s. and for any $k$ we have that $\mathbb{P}(|X-X_n|>\frac{1}{k}) = 0$. So your statement is wrong because this fact holds for all $k$.

Title: Why does $P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$ hold?

Body:
Let $X, X_n$ be $\mathbb{R}^d$-valued random variables.
The solution for an exercise uses the following step without further comment:
$$P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$
As far as I know, I should however be able to rewrite $\limsup$ on the left as follows, which seems a bit of a jump:
$$P(\limsup_n |X_n-X|>0) = P(|X_n-X|>0 \text{ i.o.}) \le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$
Can somebody please explain why this step is allowed?


问:
A distributor of TVs determines that 1% of them are faulty. He sell them in bulk of 200 units and guarantees that 98% of the TVs won't be faulty.
Find the probability that any particular 200 unit batch violates the guarantee

exactly, using a binomial distribution

using the Poisson approximation

答: Sketch The number of faulty TVs is $X\sim B(200,\,0.01)\approx\operatorname{Poisson}(2)$. The guarantee is honoured iff $X\le4$, with probability $\sum_{k=0}^4\binom{200}{k}0.01^k0.99^{200-k}\approx e^{-2}\sum_{k=0}^4\frac{2^k}{k!}$. Once either expression is calculated by summing just $5$ terms, subtraction from $1$ gives the probability of guarantee violation. By my calculation, the exact result $0.05175$ admits Poisson approximation $0.05265$.


问:
How to go about solving this question on positive semi-definite matrices?
How to go about solving this question? I'm stuck and can't proceed. Appreciate any help.

答:
When both $A$ and $B$ are PSD, $(B^T)^{1/2}A(B^T)^{1/2}$ is also PSD. Hence
$$
\sum_{i,j}a_{ij}b_{ij}=\operatorname{tr}(AB^T)=\operatorname{tr}\left((B^T)^{1/2}A(B^T)^{1/2}\right)\ge0.
$$
Conversely, if $A$ is not PSD, then there exists a vector $v$ such that $v^\ast Av$ is not real nonnegative. Thus $\operatorname{tr}(AB^T)=v^\ast Av$ is not real nonnegative for the PSD matrix $B=(vv^\ast)^T$.


问:
Tangent space is independent of the coordiante function. Why $d\phi_0=d\psi_0\circ dh_0$ implies $Im(d\phi)\subseteq Im(d\psi)$
Tangent space is independent of the coordiante function.

Let $X$ be a smooth n-manifold and $\phi: U\to X$ be a local parametrization at $x$ ($\phi(0)=x$). Define tangent space as $T_x(X)=d\phi_0(\mathbb R^n)$

We claim tangent space is independent of the local parametrization so we take another local parametrization $\psi: V\to X$ and we shrink $U,V$ such that $\phi(U)=\psi(V)$ namely we take a diffeomorphism $h=\psi^{-1}\circ \phi:U\to V$

Such that $\phi=\psi\circ h$ and differentiating it yields:$$d\phi_0=d\psi_0\circ dh_0$$

Question: Why last equation would imply image of $d\psi$ consists image of $d\phi$?

答:
To spell it out, if $x\in\operatorname{im}(d\phi_0)$, then $x=d\phi_0(y)$ for some $y$, and so $x=(d\psi_0\circ dh_0)(y)=(d\psi_0)(dh_0(y))$ is in $\operatorname{im}(d\psi_0)$. That is, $\operatorname{im}(d\phi_0)\subset\operatorname{im}(d\psi_0)$.

Also, you may want to check out J. Milnor's Topology from the differentiable viewpoint for what I find to be a clear exposition.


问:
Eigenvalues of Symmetric/Hermitian Matrices
I am new to stack exchange so please excuse me as I write in JAX I have a question on linear algebra that goes like this :  If $A$ is an $n\times n$ real symmetric matrix or complex Hermitian matrix, then its eigenvalues are real.

I want to prove this theorem but I hope someone can this proof and correct me if I am wrong : We have that By the Schur’s decomposition, $A=QRQ^{*}$ however $A=A^{*}=(QRQ^{*})^{*}=QR^{*}Q^{*}$ where we have that $R$ an upper triangular matrix with eigenvalues of $A$ on its diagonal entries, so $R^{*}$ is a lower triangular matrix thus $A-A^{*}=Q(R-R^{*})Q^{*}=0\implies R-R^{*}=0\implies r_{i,i}=\overline{r_{i,i}}$. Therefore, $R$ is a diagonal matrix and its diagonal entries
are real. I would like to thank whoever checks my proof. Best regards!

答:
Yes, yes, the proof given by our OP Alexandre von Maylen looks fine.  However, it leans heavily on the Schur decomposition which is itself non-trivial. A simple proof may be had by proceeding directly from first principles and definitions, to wit:

We are given

$Ax = \mu x \tag 1$

and

$A^\dagger = A; \tag 2$

we may assume that

$\langle x, x \rangle = 1, \tag 3$

with $\langle \cdot, \cdot \rangle$ the standard  Hermitian inner product on $\Bbb C^n$; thus we have

$\mu = \mu \langle x, x \rangle = \langle x, \mu x \rangle$
$= \langle x, Ax \rangle = \langle A^\dagger x, x \rangle = \langle Ax, x\rangle$
$= \overline{\langle x, Ax \rangle} = \overline{\langle x, \mu x \rangle} = \overline{\mu \langle x, x\rangle}= \bar \mu. \tag 4$

Note that

$\langle x, Ax \rangle = \langle A^\dagger x, x \rangle \tag 5$

is essentially the definition of $A^\dagger$, which coincides with $A^T$ since $A$ is specified to be a real $n \times n$ matrix.


问:
Questions about foundations of mathematics
It seems to me that in trying to create a foundation of mathematics, mathematicians are trying to create a formal system that models the language used in what I view as common-sense, real mathematics. Whatever the number $1$ is, I doubt it's $\{\{\}\}$, no? The set containing the empty set is merely a representation of $1$ in this formalism. So then is a foundation of mathematics merely a formalism used to model 'real' mathematics? If not, what is it?

I thought I was in on a little-known secret when I was (perhaps not intentionally) lead to believe that $1$ is the same as $\{\{\}\}$, but now I see how preposterous a claim that is, especially considering that the empty set is probably but a linguistic construct.

If the answer to my first question is 'yes', then presumably the reason for the foundation is to check it for self-consistency, for if the foundation is self-consistent, the thought process probably goes, then 'real' mathematics is self-consistent, yes? How can one know that the foundation (always) parallels 'real' mathematics? How do we know that the two discourses are isomorphic? If we can't know then why bother with a foundation in the first place? If we can know then presumably it's due to some metalanguage, but then... is the metalanguage self-consistent? How do we know? Isn't this just kicking the can down the road?

答:
Set theory starts with axioms about set existence and set construction. Naive Cantor set theory, Zermelo-Fraenkel (ZF),and Quine's New Foundations (NF) are examples of set theory.
In any case, the set theory is supposed to acommodate numbers, starting with the natural numbers, that have to abide to the Peano axioms, briefly stating that there is a starting point $0$, and every number has a successor that is not equal to a number already in the list, effectively creating an infinite list that, when the set theory allows pushing all these into one new set, supplies us with our first infinite set.

In any set theory, the challenge then is to find sets that represent these natural numbers. After that, integers can be defined as classes of a specific equivalence relation on natural numbers, rationals likewise with another equivalence relation on pairs of integers, and reals as equivalence classess of (infinite) fundamental sequences of rationals. All mathematics can then be built on these number foundations plus symbolic logic and proof theory. Obviously, not everything is fixed then (axiom of choice usually being independent), nor complete (in the sense that not every true theorem has a proof, see Godel).

In ZF and NF, $0$ is defined as the empty set (existing due to an axiom in ZF, and as all elements satisfying the stratified formula $x \ne x$ in NF), and in ZF the successor of a natural number $n$ is defined as the set $n \cup \{n\}$. In ZF and NF, the Frege-Russel cardinal representing $1$ is the set of all sets containing 1 element. In NF it is possible to construct the successor of $n$ just as done in ZF, although not directly by using the formula $x=n \lor x \in n$, as this is not stratified. But then, the $\textit{set of naturals}$ $\mathbb{N}$ or $\omega$ cannot be constructed, and hence in NF a different successor construction is required (see for example my dissertation on NF in https://eprints.illc.uva.nl/574/1/X-1989-02.text.pdf, including Rosser's definition of $x+y=\{z \cup w | z \in x \land z \in w \land z \cap w = \{\}\}$).

Hence, this means that not $1=\{\{\}\}$ always - in fact it is not, in NF - it is just a matter of $\textit{defining}$ natural numbers as sets that conform to the Peano axioms. Other sets could do just as fine.


问:
After many years, my Grandad told me his 'magic trick' isn't magic... but why does it work so often?
Take a standard pack of playing card, no jokers and no magic shop gimmicks. The 'Magician' invites you to name 2 cards, but without their suit; eg "two" and "queen" (ie but not "two of hearts" and "queen of spades") and says that he can make them 'stick together'.

From the pack face down, he turns over the first card off the top and then second... and third... and keeps going.  Almost every time (and that's the point of the question!) - almost every time, long before he's got to over the last card, he'll have dealt a "two" immediately followed by a "queen" or vice-versa. There they are, 'stuck together'.

The audience is impressed and thinks there is sleight of hand but no, he hands them the pack. They shuffle it and the 'magician' names 2 cards which they turn over themselves. It works! Someone says that the magician must have already known of 2 cards that would come up; so the magician asks the volunteer to choose anyone else to name 2 cards and they deal the cards out themselves a second time. It works again and the magician isn't even involved!

Sometimes, if it doesn't work,  the magician blames the 'novice' magician and berates them for not concentrating. Worst case, the 2 cards are normally only separated from being "stuck together" by one stray card. "Try again!" says the magician - and it works!  This delights children and confounds adults!

So, clearly down to probability and not magic but it almost always works.  Question is... what is the probability of it working on any given 'deal'?

Thanks
Mike

答:
That’s extraordinary! Thank you. I wouldn’t have known where to start! So - half the time and, with a little patter to explain how the cards came super close every other turn in 4, that is perhaps why it seems so successful. I don’t imagine that most people intuitively think it would be that likely - and why it is so easy to amaze and believe some mystical forces are at work! Of course, the mathematician is is the real magician here - but thank you for sharing the secret.

Mike


问:
Laurent Series Expansion
Give the Laurent series development of the function $f(z)=\frac{1}{z(z-1)(z-2)}$ in the three rings $A_1=\{z:|z|<1\}$, $A_2=\{z:1<|z|<2\}$ and $A_3=\{z:2<|z|\}$.

I have gotten the partial fraction form $f(z)=\frac{1}{2z}-\frac{1}{z-1}+\frac{1}{z-2}$

For $A_1$, I think it's the standard Taylor expansion about 0 and I can do this for the second and third terms:

$-\frac{1}{z-1}=(1-z)^{-1}=\sum^\infty_{n=0}z^n$ and $\frac{1}{2(z-2)}=\frac{1}{4}\sum^\infty_{n=0}(\frac{z}{2})^n$. How do I get the Taylor expansion of the first term?

I foresee similar problems if I work out the expansion for $A_2$ and $A_3$ as well, so a general method is appreciated. Thanks!

答:
First of all, the partial fraction decomposition is
$$
f(z) = \frac{1}{2z} - \frac{1}{z - 1} + \frac{1}{2(z - 2)}.
$$
The Laurent series converges on an annulus, so $A_1$ must not contain the origin:
$$
A_1 = \left\{z : 0 < |z| < 1 \right\}.
$$
Then, you can just use the usual Taylor expansion around $z = 0$:
$$
\begin{align}
f(z) &= \frac{1}{2z} + \frac{1}{1 - z} - \frac{1}{4(1 - \tfrac{z}{2})} \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty z^n - \tfrac{1}{4} \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \sum_{n=-1}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \tfrac{1}{2}z^{-1} + \tfrac{3}{4} + \tfrac{7}{8}z + \tfrac{15}{16}z^2 + \cdots
\end{align}
$$

There is actually an easier way to expand the series (around $z = 0$) which generalizes nicely for the other regions:
$$
f(z) = \frac{1}{z} \cdot \frac{1}{(z - 1)(z - 2)} = \frac{1}{z} \cdot \left( \frac{1}{1 - z} - \frac{1}{2 - z} \right).
$$
Each of these is a geometric series that can be expanded inside or outside its radius of convergence.
$$
\frac{1}{1 - z} = \begin{cases}
\phantom{-}\sum_{n = 0}^\infty z^n, &|z| < 1, \\
-\sum_{n = 1}^\infty \frac{1}{z^n}, &|z| > 1.
\end{cases}
$$
Similarly,
$$
\frac{1}{2 - z} = \frac{1}{2(1 - \tfrac{z}{2})}
= \begin{cases}
\phantom{-}\frac{1}{2}\sum_{n = 0}^\infty \left(\frac{z}{2}\right)^n
= \frac{1}{2}\sum_{n = 0}^\infty \frac{z^n}{2^n}
= \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}}, &|z| < 2, \\
-\frac{1}{2}\sum_{n = 1}^\infty \frac{1}{\left(\frac{z}{2}\right)^n}
= -\frac{1}{2}\sum_{n = 1}^\infty \frac{2^n}{z^n}
= -\sum_{n = 1}^\infty \frac{2^{n - 1}}{z^n}, &|z| > 2.
\end{cases}
$$
Now, for each of the three regions, choose the appropriate expansion, and combine them.  For instance, on $A_2$, where $1 < |z| < 2$,
$$
\begin{align}
f(z) &= \frac{1}{z} \left(-\sum_{n = 1}^\infty \frac{1}{z^n}
- \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}} \right) \\
&= -\sum_{n = 1}^\infty \frac{1}{z^{n + 1}}
- \sum_{n = 0}^\infty \frac{z^{n - 1}}{2^{n + 1}}.
\end{align}
$$

ContentLicense: CC BY-SA 3.0

Title: Laurent Series Expansion


问:
How to prove $x^3$ is strictly increasing
I am trying to use $f(x)=x^3$ as a counterexample to the following statement.

If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.

But how can I show that $f(x)=x^3$ is strictly increasing?

答:
Suppose $a<b$. We want to prove that $a^3 < b^3$.

Well, $a0$ such that $b = a+\delta$. So:

$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.

Now we consider five different cases:

$\ a<-\delta$

$\ a = -\delta$

$\ -\delta<a<0$

$\ a=0$

$\ a > 0$.

In cases $1, 2, 4$ and $5,$ it is fairly easy to see that  $3a\delta(a+\delta)+\delta^3 > 0,\ $ so all that remains is case $3$:

$-\delta<a<0 \implies \exists k$ with $0<k<1,\ $ such that $a = -k\delta.\ $ Then, $\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$, and using the fact that $k(1-k) \leq \frac{1}{4},\ $ we see that $-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $ and so $\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$.

Body:
Suppose $a<b$. We want to prove that $a^3 < b^3$.

Well, $a<b, \ \therefore \exists \delta>0$ such that $b = a+\delta$. So:

$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.

Now we consider five different cases:

$\ a<-\delta$

$\ a = -\delta$

$\ -\delta<a<0$

$\ a=0$

$\ a > 0$.

In cases $1, 2, 4$ and $5,$ it is fairly easy to see that  $3a\delta(a+\delta)+\delta^3 > 0,\ $ so all that remains is case $3$:

$-\delta<a<0 \implies \exists k$ with $0<k<1,\ $ such that $a = -k\delta.\ $ Then, $\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$, and using the fact that $k(1-k) \leq \frac{1}{4},\ $ we see that $-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $ and so $\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$.


问:
Problem on Sum of Digits of a Number
I was given this problem to Prove if $Q(n)=Q(2n)$ implies $9∣n$. Where $Q(n)$ be the sum of digits of $n$. I tried to use the fact that sum of digits is congruent to the number modulo $9$ but I had no luck. Any help is appreciated.

答:
By your fact:

$$Q(n)\equiv n \pmod 9,\quad Q(2n)\equiv 2n \pmod 9$$

Hence:

$$Q(n) = Q(2n) \implies n\equiv 2n \pmod 9 \implies n \equiv 0 \pmod 9$$

thus $9 \mid n$.

Title: Problem on Sum of Digits of a Number


问:
How to get the conditional probability of $X|Y\sim Y N(0,\sigma_1^2)+(1-Y) N(0,\sigma_2^2)$?
If we have a random variable $Y\sim Ber(p)$ and
$$X|Y\sim Y N(0,\sigma_1^2)+(1-Y)N(0,\sigma_2^2)$$

How to get $f(x|y)$ proportion to the product of two normal distribution?

Is it just the sum of two normal distributions?
The expectation of $X|Y$ is $\mathbb{E}(X|Y)=0$. Do we have $ Var(X|Y)=Y^2\sigma_1^2+(1-Y)^2\sigma_2^2$

答:
The question is not 100% clear so I will get everything in one place.

Conditional distribution of $X|Y$ is normal whose parameter depends on $Y$, with density function
$$
f_{X|Y}(x|y) = y \frac{1}{\sqrt{2\pi}\sigma_1}\exp\left(-\frac{x^2}{2\sigma_1^2}\right) + (1-y) \frac{1}{\sqrt{2\pi}\sigma_2}\exp\left(-\frac{x^2}{2\sigma_2^2}\right), x \in (-\infty, \infty), y \in \{0,1\}
$$

Marginal distribution of $X$ is a mixture distribution with density
$$f_X(x) = p \frac{1}{\sqrt{2\pi}\sigma_1}\exp\left(-\frac{x^2}{2\sigma_1^2}\right) + (1-p) \frac{1}{\sqrt{2\pi}\sigma_2}\exp\left(-\frac{x^2}{2\sigma_2^2}\right), x \in (-\infty, \infty)$$

Conditional variance of $X|Y$is a function of $Y$:
$$
\text{var}(X|Y) = Y\sigma_1^2 + (1-Y) \sigma_2^2
$$
Marginal variance of $X$ is a number:
$$
\text{var}(X) = p\sigma_1^2 + (1-p) \sigma_2^2.
$$


问:
Solve for $f$ in an ODE involving two points: $f'(x)-f'(y)\geq g(x,y)$ where $g$ is a known function.

Prove that, for any $g$, there exists $f$ satisfying an ordinary differential inequity: $$f'(x)-f'(y)\geq g(x,y)$$ Where $g$ is a known function and $g(x,y)=-g(y,x)$.

$f$ is a real function. $x,y$ are real numbers. $g$ is a differentiable real function with two arguments. $g$ is also a monotonic function: increasing with the first argument and decreasing with the second argument. $g$ is defined on connected, convex, bounded, closed sets $C^2$. $f$ is defined on $C$.

Motivation: The ODE involving one point like $f^{(n)}(x)=g(x)$ is well studied. But the simplest ODE with two points: $f'(x)-f'(y)= g(x,y)$ seems never be studied?!

My Try: First try to solve the equity case, ODE: $f'(x)-f'(y)= g(x,y)$. But this case seems not too complicated. With some works you could see $g$ has to satisfy that there must exists $h$ such that:

$$g(x,y)=h(x)-h(y).$$

And $f$ is just the integration of $h$. So, for any $g$, $f$ can be non-existence in this ODE.

The true hard part is to solve the inequity.

Another approach is to treat it as a functional equation: $F(x)-F(y)\geq g(x,y)$.

The third approach would be transform it like: $\int_y^x f''(z)dz\geq g(x,y)$,

Then $\int_y^{y+\epsilon} f''(z)dz\geq g(y+\epsilon,y)$

Then $\lim_{\epsilon\to 0}\epsilon f''(y)\geq g(y+\epsilon,y)$

Since $g(y,y)=0$,

Then $$\lim_{\epsilon\to 0}\frac{g(y+\epsilon,y)}{\epsilon}=g_1(y,y)$$

So $f''(y)\geq g_1(y,y)$.

In fact, as $\int_y^xf''(z)dz\geq g(x,y)-g(y,y)=\int_y^xg_1(z,y)dz$.

So for all $y$ and all $z>y$, $f''(z)\geq g_1(z,y)$.

Let the upper bound $\sup\{g_1(z,y)||y\in C\}=U(z)$. The upper exists since $C$ is compact.

Then, let $f''(z)=U(z)$ and solve for $f$.

It seems solves the original ODIE?

答:
I'm afraid there are too many things mixed together, here. There are functional equations (i.e. equations where the unknown is a function), differential and integral equations would be examples important for applications, while things like $$f(x-y)\,f(x+y)=f(x)^2-f(y)^2$$ are rather exotic, reserved for contest questions and their likes.

In your case, $$f'(x)-f'(y)=g(x,y)\tag{F'}$$ would rather be a general functional equation, essentially, it is $$h(x)-h(y)=g(x,y)\tag{H}$$ with the additional requirement that $h$ be a derivative (meaning it satisfies the intermediate value theorem, so we're not just speaking of integration, here).

Unfortunately, your assumptions aren't very clearly stated, it starts with "any $g$" (that wouldn't work, obviously), soon followed by $$g(x,y)=−g(y,x)\tag{G}$$ and later by
"$g$ is a differentiable real function with two arguments. $g$ is also a monotonic function: increasing with the first argument and decreasing with the second argument" (why?).

Now you'll say "wait, I want a functional inequality:" $$h(x)-h(y)\le g(x,y)\tag{H'}.$$ but as a matter of fact, your assumption (G) reduces that to (H): exchanging $x$ and $y$ gives $$h(y)-h(x)\le g(y,x),$$ i.e.
$$h(x)-h(y)\ge -g(y,x)=g(x,y).$$

So we're left with (H). Under which assumptions is that solvable? Obviously, it would be necessary that $$g(x,y)+g(y,z)=h(x)-h(y)+h(y)-h(z)=h(x)-h(z)=g(x,z),$$ i.e. $$g(x,y)+g(y,z)=g(x,z)\tag{G'}.$$ That's also sufficient, because it implies $$g(x,y)=g(x,z)-g(y,z)=g(x,z_0)-g(y,z_0)=h(x)-h(y)$$ for any constant $z_0\in C$.

Speaking of $C$: you assume $C^2$ ($=C\times C$, I hope) to be "connected, convex, bounded, closed", but trust me, "convex" implies "connected", so $C$ must be convex, and since later in your post, we learn that $C$ is assumed to be compact, you could as well say $C=[a,b]$ from the very beginning.

In a word: for the solvability of your inequality/equation, you need (G'), together with some regularity assumption so that $h(x)=g(x,z_0)$ can be a derivative, right?

BTW, while (G') seems to be explicit in terms of $g$ (unlike "there exists a function $h$ so that..."), watching somebody checking this for, say $g(x,y)=\sin(x+y)\,\sin(x-y)$ would be... erm... entertaining, most likely. =D


问:
Convergence of $\int_0^\infty \sin(t)/t^\gamma \mathrm{d}t$

For what values of $\gamma\geq 0$ does the improper integral $$\int_0^\infty \frac{\sin(t)}{t^\gamma} \mathrm{d}t$$ converge?

In order to avoid two "critical points" $0$ and $+\infty$ I've thought that it would be easier to test the convergence of the sum (is this coherent?):
$$
\int_0^1 \frac{\sin(t)}{t^\gamma}\mathrm{d}t +
\int_1^\infty \frac{\sin(t)}{t^\gamma}\mathrm{d}t.
$$
For the second integral, it converges if $\gamma > 1$ (comparision) and also converges if $0 <\gamma \leq 1$. I'm stuck on proving the last part and the fact that the first integral converges for $\gamma < 2$. Any help would be appreciated. Thanks in advance.

PD: I've checked the answers for this question but I would not like to solve this integral using $(n\pi,(n+1)\pi)$ intervals.

答:
I was not going to answer, but the previous answers left me a bit anxious for $t$ near $\infty$.

Integrate by parts to get
$$
\int_1^\infty\frac{\sin(t)}{t^\gamma}\,\mathrm{d}t
=\left.\frac{-\cos(t)}{t^\gamma}\right]_1^\infty
-\gamma\int_1^\infty\frac{\cos(t)}{t^{\gamma+1}}\,\mathrm{d}t
$$
and both converge at $\infty$ when $\gamma\gt0$.

Of course, as the previous answers have said
$$
\int_0^1\frac{\sin(t)}{t^\gamma}\,\mathrm{d}t
$$
converges when $\gamma\lt2$ by comparison with $\dfrac{t}{t^\gamma}=\dfrac1{t^{\gamma-1}}$.

This shows that the interval of convergence is $(0,2)$.


答: $P(\limsup_{n\to\infty}|X_n−X|>0) = P(\inf_{n\to\infty} \sup_{m\ge n} |X_m-X|>0) \le P(\sup_{n\to\infty} |X_n -X|>0) =  P(\sup_{n\to\infty}|X_n−X|> \varepsilon ,\varepsilon> 0)\le P(\sup_{n\to\infty}|X_n−X|> 1/k ,1/k<\varepsilon) \le P(|X_n-X| > 1/k,i.o.,\text{ for some very large }k).$


问:
Funtion $y=x^{x^x}$. Find $y'$
Let $f(x)=x^x, x>0$.

Then $f(x)=e^{x\ln x}$ and $f'(x)=e^{x\ln x}(\ln x+1)$.

What about $g(x)=x^{x^x}$? How find $g'(x)$?

答:
Apply $\ln$ to both sides:$$y=x^{x^x} \\ \ln y = x^x\ln x $$ Differentiate both sides with respect to $x$:$$ \frac{y'}{y} = {(x^x)'}\ln x + x^x\frac{1}{x} \\ y' = x^{x^x}(\color{red}{(x^x)'}\ln x + x^x\frac{1}{x})$$
Then, plug in what you said you already $\color{red}{\text{know}}$.

Title: Funtion $y=x^{x^x}$. Find $y'$


答: It depends on how the question is asked. Answer changes for $(x^x)^x$ and $x^{(x^x)}$.

Title: Funtion $y=x^{x^x}$. Find $y'$

Body:
Let $f(x)=x^x, x>0$.

Then $f(x)=e^{x\ln x}$ and $f'(x)=e^{x\ln x}(\ln x+1)$.

What about $g(x)=x^{x^x}$? How find $g'(x)$?


答:
$$g(x)=e^{f(x) \ln x}.$$

$$g^\prime(x) = \left(\frac{f(x)}{x} + f^\prime(x) \ln x\right) g(x).$$

You have shown that $$f^\prime = f(x)\cdot (\ln x +1),$$ so

$$g^\prime(x) = f(x)\cdot g(x)\cdot \left( \frac{1}{x} + \ln x \cdot( \ln x +1 )\right)$$

$$g^\prime(x) = x^x \cdot x^{x^x} \cdot \left( \frac{1}{x} + \ln x \cdot( \ln x +1 )\right)$$

Title: Funtion $y=x^{x^x}$. Find $y'$


问:
Integral test of convergence for $X \in{\displaystyle L^{2}}$
I'm reading a script where it says that for $X\in{\displaystyle L^{2}}$ we have
$$\mathbb{E}[X^2]=\int_0^{\infty}\mathbb{P}(X^2\geq s)\text{d}s<\infty$$
and from this we can conclude with the integral test of convergence that
$$\mathbb{P}(X^2\geq s)=\mathcal{o}\left(\frac{1}{sln(s)}\right)$$
The formula with the expectation is clear to me but I don't know how to arrive at the statement regarding $\mathcal{o}\left((sln(s))^{-1}\right)$. Or more generally I'm guessing it would then hold that if we have decreasing functions $f,g\geq 0$ with $\int_0^{\infty}f(s)\text{d}s<\infty$ and $\int_0^{\infty}g(s)\text{d}s =\infty$ then $\frac{f(x)}{g(x)}\to 0$ as $x\to\infty$. But I don't know for certain if this is true and how I would prove it.

答:
I expect the statement regarding $\mathcal{o}\left((s\ln(s))^{-1}\right)$ can  fail because its counterpart for functions is fail. For instance, for each $x\ge 0$ let $f(x)=(s(x) \ln(s(x)))^{-1}$, where $s(x)$ is the smallest number $s$ from the sequence $(\exp(s^2))_{s=0}^\infty$ such that $s(x)\ge x$. Then $f(x)\not\in \mathcal{o}\left((s ln(s))^{-1}\right),$ but
$$\int_0^\infty f(x)dx=\int_0^1 f(x)dx+\sum_{s=0}^\infty \int_{\exp(s^2)}^{ \exp((s+1)^2)} f(x) dx\le$$
$$1+\sum_{s=0}^\infty \int_{\exp(s^2)}^{ \exp((s+1)^2)} (\exp((s+1)^2\ln(\exp((s+1)^2))^{-1}  dx\le$$
$$1+\sum_{s=0}^\infty \exp((s+1)^2) \exp (s+1)^{-2} (s+1)^{-2}=$$
$$1+\sum_{s=0}^\infty (s+1)^{-2}=1+\frac {\pi^2}6.$$


问:
Logarithm in $2$-adic numbers.

How can we find the first $10$ digits of $\dfrac{\log 5}{\log 13}$ in the $2$-adic numbers?
I mean how can we write down $\{ a_i\}_{i=0}^{10}$'s explicitly,
in $\dfrac{\log 5}{\log 13}=:\sum_{i=0}^{\infty} a_i 2^i$.

Motivation for asking this question:

Let's consider the set of prime numbers that are congruent to $5$ modulo $8$:

$$\{5, 13; 29, 37; 53, 61; 101, 109; 149, 157; 173, 181; 197, \cdots\}.$$

For any prime $p$ in this set we have $v_2(p-1)=2$.
From now on we assume that $2 \leq n$.
With a little work,
we can show that $\displaystyle{{\rm order}_{2^n}p = 2^{n-2}}$.
This implies that $p$ is a generator for the index $2$ subgroup
$$H:=\{a\in \dfrac{\mathbb{Z}}{2^n\mathbb{Z}} \mid a \stackrel{4}{\equiv} 1\},$$
of $\dfrac{\mathbb{Z}}{2^n\mathbb{Z}}$.
Considering these facts,
we can find that $\overline{\langle p\rangle}=1+4\mathbb{Z}_2$,
in the $2$-adic topology.

Now let $p_1=5$ and $p_2=13$. I am not sure, but I think we can conclude that $5=13^z$, for some $z \in \mathbb{Z}_2$.

I can not see why should we have $\dfrac{\log p}{\log q} \in \mathbb{Z}_2$,
for arbitrary prime numbers $p,q \stackrel{8}{\equiv} 5$.

答:
You can do it by hand, but it is much easier using a calculator (just as you do calculations of real numbers).

Paste the following into this site and press "Evaluate" to see the result.

```<code>F= Zp(2)
print(F)
print(log(F(5))/log(F(13)))
</code>```

Output:

```<code>2-adic Ring with capped relative precision 20
1 + 2 + 2^2 + 2^3 + 2^8 + 2^9 + 2^10 + 2^13 + 2^15 + 2^17 + O(2^18)
</code>```

Because $\log$ is an isometry from $1 + 4\Bbb Z_2$ to $4\Bbb Z_2$ (which is an easy exercise). Thus for every $p \equiv 5\mod 8$, we have $$v_2(\log(p)) = v_2(\log(p) - \log(1)) = v_2(p - 1) = 2,$$ and hence the quotient $\log(p)/\log(q)$ has $2$-adic valuation $0$.

Title: Logarithm in $2$-adic numbers.

Body:
You can do it by hand, but it is much easier using a calculator (just as you do calculations of real numbers).

Paste the following into this site and press "Evaluate" to see the result.

F= Zp(2)
print(F)
print(log(F(5))/log(F(13)))

Output:

2-adic Ring with capped relative precision 20
1 + 2 + 2^2 + 2^3 + 2^8 + 2^9 + 2^10 + 2^13 + 2^15 + 2^17 + O(2^18)

Because $\log$ is an isometry from $1 + 4\Bbb Z_2$ to $4\Bbb Z_2$ (which is an easy exercise). Thus for every $p \equiv 5\mod 8$, we have $$v_2(\log(p)) = v_2(\log(p) - \log(1)) = v_2(p - 1) = 2,$$ and hence the quotient $\log(p)/\log(q)$ has $2$-adic valuation $0$.


问:
Morphological Closing. Formula understanding
I am trying to do use the morphological closing operation on a greyscaled image. As I found out closing is a combination of dilation and erosion.
Now I am trying to understand the formulas of dilation and erosion.
The formula of dilation is
$$
\left( A \bigoplus X \right)\left( x,y \right) = max \left\{ A \left( x+s,y+t \right)+ X\left(s,t \right) \right\}
$$
I dont unterstand what s and t should be? Is it a max over s and t? And s and t are coordinates of image pixels? But if so, is it a max over the whole image or just over the neighborhood of (x,y)?

答:
The notation is incomplete. Let $D$ the domain of $X$, then

$$\left(A\bigoplus X\right)(x,y)=\max_{(s,t)\in D}\{ A(x+s,y+t)+X(s,t)\}.$$

This function is defined over all pixels of $A$, but some special convention must be specified for coordinates outside the image.


问:
Rotation of a vector and general value of angle according to given conditions: how do I find the correct answer among the two?
My question states:

For some non-zero vector $\overrightarrow{V}$, if the some of $\overrightarrow{V}$ and the vector obtained by rotating it by $\angle2\alpha$ equals the vector obtained from $\overrightarrow{V}$ by rotating it by an angle of $\angle \alpha$, then the general value of $\angle \alpha$ is?

Here's how I tried it:

Clearly, we can write from the given conditions that, $\overrightarrow{V} + \overrightarrow{V}_{2\alpha}  = \overrightarrow{V}_{\alpha}$

where, $\overrightarrow{V}_\alpha = \overrightarrow{V}$ rotated by angle $\angle \alpha$

$\overrightarrow{V}_{2\alpha} = \overrightarrow{V}$ rotated by angle $\angle 2\alpha$

Also, the magnitudes of each of them are equal, and this yields:

$|\overrightarrow{V} + \overrightarrow{V}_{2\alpha}|^2  = |\overrightarrow{V}_{\alpha}|^2$

From here, I get that $\cos 2\alpha = -\frac{1}{2} \implies 2\alpha = 2n\pi \pm \frac{2\pi}{3}\implies \alpha = n\pi \pm \frac{\pi}{3}$

But, if I used the fact that  $|\overrightarrow{V}_{2\alpha}|^2  = |\overrightarrow{V}_{\alpha} - \overrightarrow{V} |^2$

Then I seem to get that $\cos \alpha = \frac{\pi}{3} \implies \alpha = 2n\pi \pm \frac{\pi}{3}$

Why do I get some more solutions in my approach, and how to understand which one is the correct solution? There must be a way to check right? (The book states that the 2nd answer is correct). What was the error of my approach?

答:
In the first approach, you are finding solutions for

$$ \vec V + \vec V_{2\alpha} = \vec V_\alpha  \tag1$$

and also solutions for

$$ \vec V + \vec V_{2\alpha} = -\vec V_\alpha.  \tag2$$

Both Equations $(1)$ and $(2)$ are satisfied by vectors that also satisfy
$$ \lVert\vec V + \vec V_{2\alpha}\rVert^2 = \lVert\vec V_\alpha\rVert^2 $$
(just take the square of the magnitude on each side),
but only Equation $(1)$ is consistent with the requirements of the question.
The solutions of Equation $(2)$ are the incorrect angles
$2n\pi \pm \frac{2\pi}3.$

The second approach can also be written
$$ \lVert\vec V_\alpha - \vec V\rVert^2 = \lVert\vec V\rVert^2
 = \lVert\vec V_\alpha\rVert^2  \tag3$$
since $\lVert\vec V_{2\alpha}\rVert^2 = \lVert\vec V\rVert^2.$
Then since $\vec V,$ $\vec V_\alpha,$ and $\vec V_\alpha - \vec V$
are three sides of a triangle, it is clear that Equation $(3)$ describes an equilateral triangle, one of whose angles is the angle between
$\vec V$ and $\vec V_\alpha,$
which can only be accomplished by rotations through angles
$2n\pi \pm \frac\pi3.$

If you only thought of the first approach and don't think of the second one in time, you can still use the first approach, but keep this in mind:

Beware of squaring both sides of an equation. Always remember that while it is true that $A=B$ implies $A^2=B^2,$ it is not true that $A^2=B^2$ implies $A=B.$ Instead, $A^2=B^2$ implies $A=\pm B,$ so any solutions  you get from $A^2=B^2$ have to be carefully checked to see which ones are the true $A=B$ solutions and which are the false $A=-B$ solutions.

In this particular problem, a vector diagram for $\pm\frac\pi3$ covers all the even $n$ in $n\pi\pm\frac\pi3$ and a diagram for $\pi\pm\frac\pi3$ covers all the odd $n$. All other values of $n\pi\pm\frac\pi3$ can be obtained by adding some multiple of $2\pi$ to one of these examples.
In the second diagram, in fact, it may be easier to construct the angles if you construct the angle $-\pi+\frac\pi3$ rather than $-\pi+\frac\pi3$ for the $+$ case, that is, add $-2\pi$ to that case, while keeping the angle $\pi-\frac\pi3$ for the $-$ case, so that one angle is clockwise and the other counterclockwise.
You might even draw three or four vector diagrams if that's what it takes to cover all the cases without getting confused.

If you examine the case $\alpha = \frac\pi3$ in a vector diagram in which you have constructed $\vec V_\alpha$ and $\vec V_{2\alpha}$ relative to an arbitrary $\vec V,$ it should be clear that this $\alpha$ is a correct answer. Likewise with the case $\alpha = -\frac\pi3$.
But with $\alpha = \pi-\frac\pi3 = \frac{2\pi}3,$ after drawing $\vec V_\alpha$ and $\vec V_{2\alpha}$ relative to an arbitrary $\vec V$ it should be clear that $\vec V + \vec V_{2\alpha}$ points in the wrong direction and cannot be equal to $\vec V_\alpha$.

So you can see graphically that $n\pi\pm\frac\pi3$ gives correct answers only in the cases where $n$ is even, and you can correct the answer by writing $2n$ instead of $n$.


问:
Integration involving roots
$$\int\frac{dx}{(1+x^\frac{1}{4})x^\frac{1}{2}}$$
This is my work:
$$u^4=x$$
$$4u^3=dx$$
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u^3du}{(1+u)u^2}=-4(1+x^\frac{1}{4})^{-1}+2(1+x^\frac{1}{4})^{-2}+C$$
But wolframAlpha gives quite a different answer. So, have I done a mistake? Or somehow both answers are correct?</p>

答:
You substitutions are just fine. First, we cancel the common factor of $u^2$ in the numerator and denominator.
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u\,du}{1+u}$$

Dividing the numerator by the denominator gives us $$\int\frac{4u}{1+u}\,du = \int \left(4 - \frac 4{1+u} \right) \,du = 4\int \,du - 4\int \frac {du}{1+u}\tag{1}$$

Now, I'm not entirely sure how you obtained your integrated your terms, but I can only guess that you mixed up differentiating and integrating terms: subtracting 1 from the exponent of $(1+u)^0$ and dividing $4$ by $-1$, and subtracting 1 from the exponent of $(1+u)^{-1}$ and dividing $-4$ by $-2$.

First, when integrating a constant $\alpha$, $\int \alpha \,du$, we obtain the variable with respect to which we are integrating, and in this case, that's simply the variable $u$ (the $\,du\,$ indicates which variable). So we can think of $$\int \alpha\,du = \int \alpha u^0\,du = \alpha u^1 + C = \alpha u + C$$ Second, when integrating, we add (increment) exponents, and divide by the new exponent, except when the exponent is $-1$ (and in your second integral, we have $(1 + u)^{-1}$ as the integrand. Recall, $$\int \frac{f'(u)}{f(u)}\,du = \ln|f(u)| + C.$$

So back to $(1)$. Integrating, we get $$4\int \,du - 4\int \frac {du}{1+u} = 4u - 4\ln|(1+u)| + C = 4x^{1/4} - 4 \ln\left(1 + x^{1/4}\right) + C$$

Title: Integration involving roots


问:
$\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}\ln\left(\frac{m+k}{n}\right)=\int_{0}^{1}\ln x\, dx$
I am trying to proof:
$$\lim_{n\to \infty}\sum_{k=0}^{n-1}\frac{1}{n}\left(\ln\left(\frac{m}{n}+\frac{k}{n}\right)\right)=\int_{0}^{1}\ln(x)dx \hskip50pt (1)$$

where $m\in\mathbb{R}$ and $m$ positive.

I got inspiration in this post: Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ and the answer Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ but I am not sure about how to deal with the $\frac{m}{n}$ term since my partition $\{0 +\frac{m}{n},\frac{1}{n}+\frac{m}{n},\frac{2}{n}+\frac{m}{n},\dots,\frac{(n-1)}{n},1+\frac{m}{n}\}$ isn`t from $0$ to $1$.

Maybe (1) is false? Am I missing something?

Any hint is appreciated.

答:
Since $\ln x$ is strictly increasing, then
$$
\frac{1}{n}\ln\left(\frac{k}{n}\right)<\int_{k/n}^{(k+1)/n}\ln x\,dx
<\frac{1}{n}\ln\left(\frac{k+1}{n}\right)
$$
and hence
$$
\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k+1}{n}\right)
<
\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\int_{k/n}^{(k+1)/n}\ln x\,dx
<
\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k}{n}\right)
$$
But
$$
\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k}{n}\right)
=\frac{1}{n}\ln\left(\frac{m+k}{k}\right)< \frac{1}{n}\ln\left(1+\frac{m}{k}\right)<\frac{m}{kn},
$$
while
$$
\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k+1}{n}\right)=\frac{1}{n}\ln\left(\frac{m+k}{k+1}\right)=
\frac{1}{n}\ln\left(1+\frac{m-1}{k+1}\right)
$$
If $m\ge 1$, then $\ln\left(1+\frac{m-1}{k+1}\right)\ge 0$. If $m\in(0,1)$, then
$$
\ln\left(1+\frac{m-1}{k+1}\right)=\frac{1}{k+1}\ln\left(1+\frac{m-1}{k+1}\right)^{k+1}\ge \frac{1}{k+1}\ln\left(1+(k+1)\frac{m-1}{k+1}\right)=\frac{\ln m}{k+1}.
$$
Thus
$$
\frac{1}{(k+1)n}\min\{0,\ln m\}<\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\int_{k/n}^{(k+1)/n}\ln x\,dx<\frac{m}{kn}
$$
and therefore
$$
\frac{\min\{0,\ln m\}}{n}(\ln(n+1)-\ln 2) <
\frac{\min\{0,\ln m\}}{n}\sum_{k=1}^{n-1}\frac{1}{(k+1)}<\sum_{k=1}^{n-1}\frac{1}{n}\ln\left(\frac{m}{n}+\frac{k}{n}\right)
-\int_{1/n}^1\ln x\,dx<\sum_{k=1}^{n-1}\frac{m}{kn}
=\frac{m}{n}\sum_{k=1}^{n-1}\frac{1}{k}
<\frac{m}{n}(1+\ln n)
$$
since $1+\ln n>\sum_{k=1}^{n-1}\frac{1}{k}$.


问:
Simplification of integral $\int \frac{\sqrt{9-w^2}}{w^2} dw$
I am tasked to find the integral $$\int \frac{\sqrt{9-w^2}}{w^2} dw$$ To solve it, I use the substitution of $w = 3\sin u$, $\frac{dw}{du} = 3\cos u$.

Upon further integration, I arrive at an answer of $-\cot(u) - u + C$, where $C$ is the integration constant. Upon substitution of $u$, I get $$-\cot(\sin^{-1}\frac{w}{3}) - \sin^{-1}\frac{w}{3} + C$$ However, the model answer returns me $$\frac{-\sqrt{9-w^2}}{w^2} - \sin^{-1}\frac{w}{3} + C$$

I'm puzzled how $-\cot(\sin^{-1}\frac{w}{3})$ will end up as $\frac{-\sqrt{9-w^2}}{w^2}$. Is there something I'm missing out on? How do I just replace it with the original expression? Thank you very much!

答:
You wish to simplify:

$$-\cot(\arcsin(\frac{w}{3})) - \arcsin(\frac{w}{3}) + C$$

$$-\frac{\cos(\arcsin(\frac{w}{3})}{\sin(\arcsin(\frac{w}{3})} -\arcsin(\frac{w}{3}) +C$$

Note that $\cos(\arcsin(x)) = \sqrt{1 - \sin^{2}(\arcsin(x))} = \sqrt{1 -x^{2}}$ for $x$ in $\arcsin$'s domain:

$$-\frac{\sqrt{1 - \frac{w^{2}}{9}}}{\frac{w}{3}} - \arcsin(\frac{w}{3}) + C$$

$$-\frac{\sqrt{1 - \frac{w^{2}}{9}}}{\frac{w}{3}}\cdot\frac{3}{3} - \arcsin(\frac{w}{3}) + C$$

$$\boxed{-\frac{\sqrt{9 - w^{2}}}{w} - \arcsin(\frac{w}{3}) + C}$$


问:
Uniqueness of values of automorphisms on a primitive root
Let $L$ be a field and let $\zeta \in L$ be a primitive root of $n$-th degree. Let $\alpha$ and $\beta$ be automorphisms belonging to $Aut(L)$. Is it possible that $\alpha(\zeta) \ne \beta(\zeta)$? It strongly seems to me that the equality $\alpha(\zeta) = \beta(\zeta)$ must hold. Here is my reasoning behind this statement:

as $\alpha$ is automomorphism, it must be that $\alpha(1) = 1$. Since $\zeta$ is a primitive root of $n$-th degree, we have that $\zeta^n = 1$. Hence, we have that $\alpha(\zeta^n) = 1$. Similarly $\beta(\zeta^n) = 1$. So we have $$\alpha(\zeta^n) = \beta(\zeta^n)$$ Thus: $$(\alpha(\zeta^n))^{1-n} = (\beta(\zeta^n))^{1-n}$$ and finally we obtain: $$\alpha(\zeta) = \beta(\zeta)$$

答:
The following answer is thanks to Roland and Daniel Hast, I write it here for documentation purposes. The reasoning that I presented is seriously flawed. First of all, there is a mistake  because $(\alpha(\zeta)^n)^{1-n} \neq \alpha(\zeta)$, but $(\alpha(\zeta^n)^{1-n} = (\alpha(\zeta^{n \cdot 1-n \cdot n})) = (\alpha(\zeta^{n-n^2}))$. Second thing is that the idea itself is wrong. While it may seem for a while that we could rescue the situation by writing that $$\alpha(\zeta^n) = \beta(\zeta^n)$$ implies $$(\alpha(\zeta^n))^{1/n} = (\beta(\zeta^n))^{1/n}$$ and hence $$\alpha(\zeta) = \beta(\zeta),$$ that is plainly wrong. It is due to the fact that a given field element can have many (possibly different) roots which makes it impossible to deduce the above equality. As a concrete example let us take the field $C$ - we have $(i^2)^{\frac{1}{2}} = 1^\frac{1}{2}= \{1, -1\}$ but if we (incorrectly) try $(i^2)^\frac{1}{2} = i^{2\cdot \frac{1}{2}}$ then we get farther $= i^1 = i$.

The formula $(a^b)^\frac{1}{n} = a^\frac{b}{n}$ that works in $R$ if some common conventions are preserved does not work in fields in general.


问:
How do I find the limits to this double integral and convert into polar?
With $M = \{(x, y) ∈ R^2: 1 ≤ x^2 − y^2 ≤ 9, 2 ≤ xy ≤ 4,     x, y ≥ 0\}$, calculate
$$\int_M (x^2+y^2) \mathrm dV$$

What I´ve managed so far is to convert $x$ and $y$ into polar coordinates with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and got
$$\iint \left[(r\cos(\theta))^2+(r\sin(\theta))^2\right] r\mathrm dr\mathrm d\theta$$

is it right to use polar coordinates? if so, how do I set the limits of this double integral? Thx

答:
Double-check all my arithmetic. Even if I made no mistakes, it'll be instructive.

Since $M$ is defined by $1\le r^2\cos2\theta\le9,\,4\le r^2\sin2\theta\le8,\,0\le\theta\le\pi/2$, the integration range is$$0\le\theta\le\pi/4,\,\max\{\sec2\theta,\,4\csc2\theta\}\le r^2\le\min\{9\sec2\theta,\,8\csc2\theta\}.$$The constraint $\theta\le\pi/4$ follows from $\cos2\theta\ge0$. No, wait, we need to restrict $\theta$ further. Since $\sec2\theta\le8\csc2\theta$, $\theta\le\tfrac12\arctan8$. Similarly, $\theta\ge\tfrac12\arctan\tfrac49$. On this, we integrate $r^3drd\theta$, giving$$\begin{align}I&=\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan8}d\theta\int_{\sqrt{\max\{\sec2\theta,\,4\csc2\theta\}}}^{\sqrt{\min\{9\sec2\theta,\,8\csc2\theta\}}}r^3dr\\&=\tfrac14\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan8}d\theta\left[(\min\{9\sec2\theta,\,8\csc2\theta\})^2-(\max\{\sec2\theta,\,4\csc2\theta\})^2\right]\\&=\frac14\left(\int_{\tfrac12\arctan\tfrac49}^{\tfrac12\arctan\tfrac89}48\csc^22\theta d\theta+\int_{\tfrac12\arctan\tfrac89}^{\tfrac12\arctan4}(81\sec^22\theta-16\csc^22\theta)d\theta+\int_{\tfrac12\arctan4}^{\tfrac12\arctan8}80\sec^22\theta d\theta\right).\end{align}$$Using$$\begin{align}\int_{\tfrac12\arctan a}^{\tfrac12\arctan b}\csc^22\theta d\theta&=\left[-\frac12\cot2\theta\right]_{\tfrac12\arctan a}^{\tfrac12\arctan b}\\&=\frac{b-a}{2ab},\\\int_{\tfrac12\arctan a}^{\tfrac12\arctan b}\sec^22\theta d\theta&=\left[\frac12\tan2\theta\right]_{\tfrac12\arctan a}^{\tfrac12\arctan b}\\&=\frac{b-a}{2},\end{align}$$we can finish:$$4I=48\frac{\tfrac89-\tfrac49}{2\tfrac89\tfrac49}+81\frac{4-\tfrac89}{2}-16\frac{4-\tfrac89}{2\cdot4\cdot\tfrac89}+80\frac{8-4}{2}=27+126-7+160=306,$$making the original integral $76\tfrac12$.


答:
Let $\theta=\sin^{-1}\frac{w}3$.  Then $\sin\theta=\frac{w}3$, which means $\cot\theta=\frac{\sqrt{9-w^2}}w$.

Draw a right triangle where $w$ is the opposite side and $3$ is the hypotenuse, since $\sin\theta=\frac{w}3$.  You will find by the Pythagorean theorem that the adjacent side is $\sqrt{9-w^2}$, which gives the above $\cot\theta$.

Title: Simplification of integral $\int \frac{\sqrt{9-w^2}}{w^2} dw$

Body:
I am tasked to find the integral $$\int \frac{\sqrt{9-w^2}}{w^2} dw$$ To solve it, I use the substitution of $w = 3\sin u$, $\frac{dw}{du} = 3\cos u$.

Upon further integration, I arrive at an answer of $-\cot(u) - u + C$, where $C$ is the integration constant. Upon substitution of $u$, I get $$-\cot(\sin^{-1}\frac{w}{3}) - \sin^{-1}\frac{w}{3} + C$$ However, the model answer returns me $$\frac{-\sqrt{9-w^2}}{w^2} - \sin^{-1}\frac{w}{3} + C$$

I'm puzzled how $-\cot(\sin^{-1}\frac{w}{3})$ will end up as $\frac{-\sqrt{9-w^2}}{w^2}$. Is there something I'm missing out on? How do I just replace it with the original expression? Thank you very much!


问:
$\lim_{x\to 0}\frac{1}{x}−\frac{2}{e^{2x}−1}$ doesn't get solved as expected
A question says $f:\Bbb R\setminus\{0\} \to \Bbb R$, $f(x)=\dfrac{1}{x}−\dfrac{2}{e^{2x}−1}$. The limit as the function approaches $x=0$ is logically $0$, as in;

$\lim \limits_{x \to 0}\dfrac{e^{2x}-1}{2x} = 1$ This function then becomes $\dfrac{1}{x}−\dfrac{2}{2x} =0$

But when I graphed it online, the function approached $1$.

答:
Rearranging the equation and applying L'Hopital rule twice gives the correct answer.
$$\lim \limits_{x \to 0} \frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
$$=\lim \limits_{x \to 0} \frac{2e^{2x}-2}{(e^{2x}-1)+2xe^{2x}}$$
$$=\lim \limits_{x \to 0} \frac{4e^{2x}}{2e^{2x} + 2e^{2x} + 4xe^{2x}}$$
$$=1$$


问:
Using the Intermediate Value Theorem and Rolle's theorem to determine number of roots

Use the Intermediate Value Theorem and Rolle's Theorem to show the that the polynomial $$p(x) =  x^{5} + x^{3} + 7x - 2$$ has a unique real root.

Can someone please give some hints on how to do this question. Thanks in advance.

答: First, $p\,'(x)=5x^4+3x^2+7$; this is a quadratic in the variable $x^2$, so you can find all of its integer roots. Rolle’s theorem says that if $p(a)=0$ and $p(b)=0$, where $a<b$, then there is an $x$ between $a$ and $b$ such that $p\,'(x)=0$. Thus, if $p(x)$ has more than one real root, $p\,'(x)$ has at least one real root. Finally, remember that if $p(a)<0$ and $p(b)>0$, then the intermediate value theorem insures that there is an $x$ between $a$ and $b$ such that $p(x)=0$. That’s everything that you need to use for the problem.</p>


问:
Prove that $G=\{f\in L_p(\mu): \int_\mathbb{R} f(t) dt = 0\}$ is a dense subset of $L_p(\mu)$, where $\mu$ is the lebesgue measure on $\mathbb{R}$.
Here $1<p<\infty$, prove that $G=\{f\in L_p(\mu): \int_\mathbb{R} f(t) dt = 0\}$  is a dense subset of $L_p(\mu)$, where $\mu$ is the Lebesgue measure on $\mathbb{R}$.

How to prive this? I have no idea, can anyone give me a hint?

答:
Consider the linear operator $$\Lambda: (L^p \cap L^1,\|\cdot\|_p) \to (\mathbb{R},|\cdot|), f \mapsto \int f \, d\mu.$$

Since $L^p \cap L^1$ is dense in $L^p$ (with respect to $\|\cdot\|_p)$, we are done if we can show that the kernel of $\Lambda$ is dense in $(L^p \cap L^1,\|\cdot\|_p)$. To this end, we use the following general statement, which can be, for instance, found in the Functional Analysis book by Rudin (Theorem 1.18)

Let $X$ be a normed space and $\Lambda:X \to \mathbb{R}$ a non-zero linear functional. $\Lambda$ is not continuous if and only if its kernel $$\text{ker}(\Lambda) = \{x \in X\::\: \Lambda(x)=0\}$$ is dense in $X$.

Consequently, it is enough to show that $\Lambda$ is not continuous. Take

$$f_n(x) := \frac{1}{n^{1/p}} 1_{[0,n]}(x),$$

then $(f_n)_{n \in \mathbb{N}} \subseteq L^1 \cap L^p$ and the sequence is $L^p$-bounded since

$$\|f_n\|_p = n \frac{1}{n}=1, \qquad n \in \mathbb{N}.$$

On the other hand,

$$\Lambda(f_n) = \frac{1}{n^{1/p}} n \xrightarrow[p>1]{n \to \infty} \infty.$$

This shows that the operator $\Lambda$ is not bounded, hence not continuous.


答:
Hints:

$1).\ $ There is a compactly-supported continuous $g$ such that $\|f-g\|<\epsilon/2.$

$2).\ $ Define $g_n(x)=g(x)-\left(\frac{\int g}{2n}\right)\cdot1_{[-n,n]}(x).$

$3).\ $ Prove that $g_n$ is compactly-supported and $\int g_n=0.$

$4).\ $ Prove that $\|f-g_n\|<\epsilon$  if $n$ is large enough.

Title: Prove that $G=\{f\in L_p(\mu): \int_\mathbb{R} f(t) dt = 0\}$ is a dense subset of $L_p(\mu)$, where $\mu$ is the lebesgue measure on $\mathbb{R}$.

Body:
Here $1<p<\infty$, prove that $G=\{f\in L_p(\mu): \int_\mathbb{R} f(t) dt = 0\}$  is a dense subset of $L_p(\mu)$, where $\mu$ is the Lebesgue measure on $\mathbb{R}$.

How to prive this? I have no idea, can anyone give me a hint?


Body:
Note

$$\cot(\sin^{-1}\frac{w}{3})=
\cot(\csc^{-1}\frac{3}{w})
=\sqrt{\csc^2(\csc^{-1}\frac 3w)-1}\\
= \sqrt{(\frac 3w)^2-1}= \frac{\sqrt{9 - w^{2}}}{w}
$$


答:
$$\cot(\sin^{-1}(w/3))=\cot \theta , \sin \theta =w/3$$
So $$\cot \theta= \frac{\sqrt{9-w^2}}{w}.$$
Eventually both answers are the same.


问:
Let V be a vector space over $Z_3$ and dim(V)=3 ,then V has how many elements
The following question is part of a mock test which I am trying to solve and I need help in this question.

Let V be a vector space over $Z_3$ and dim(V)=3 ,then V has how many elements?

I used formula $(3^3 -1) (3^3 -3) (3^3 -3^2)$ but answer is given 125 elements.

I am asking here because I am not good in this type of questions so that you can tell me whether I am wrong and what is the right approach.

thnaks!!

答: The vector space will be isomorphic to $\mathbb Z_3^3$, which has $\vert\mathbb Z_3\vert^3=3^3=27$ elements


问:
Specify the inverse of the matrix $g_{ij}=\langle v_i,v_j\rangle$, where $(v_1,...,v_n)$ is a basis
Let $V$ be an inner product space with a basis $(v_1,...,v_n)$.

I know that the Gram matrix
\begin{equation}
G=\begin{pmatrix}
\langle v_1,v_1\rangle&\cdots&\langle v_1,v_n\rangle\\
\vdots&&\vdots\\
\langle v_n,v_1\rangle&\cdots&\langle v_n,v_n\rangle
\end{pmatrix}=\begin{pmatrix}
g_{11}&\cdots&g_{1n}\\
\vdots&&\vdots\\
g_{n1}&\cdots&g_{nn}
\end{pmatrix}
\end{equation}
is invertible, but is it possible to explicitly specify the inverse (maybe involving the dual basis)?

Motivation: For those who have heard about the metric tensor: I'd like to have a formula for $g^{ij}$ (which boils down to the question above).

答:
First of all, we know that
\begin{align*}
    \phi\colon V&\to V^*\\
    v&\mapsto\langle v|\,\cdot\,\rangle
\end{align*}
is an isomorphism. We can construct an inner product on $V^*$ by defining
\begin{equation}
\langle\omega,\eta\rangle=\left\langle\phi\,^{-1}(\omega),\phi\,^{-1}(\eta)\right\rangle
\end{equation}
for $\omega,\eta\in V^{*}$.
With this definition, you can easily prove that
\begin{equation}
\begin{pmatrix}
g^{11}&\cdots&g^{1n}\\
\vdots&&\vdots\\
g^{n1}&\cdots&g^{nn}
\end{pmatrix}:=
\begin{pmatrix}
\langle v^1,v^1\rangle&\cdots&\langle v^1,v^n\rangle\\
\vdots&&\vdots\\
\langle v^n,v^1\rangle&\cdots&\langle v^n,v^n\rangle
\end{pmatrix}
\end{equation}
is the inverse of $G$.


问:
Questions about Zorn's lemma
Zorn's lemma states that: Suppose a partially ordered set P has the property that every chain in P has an upper bound in P. Then the set P contains at least one maximal element.

Suppose $P$ satisfies the assumption to apply the Zorn's lemma. Does it then follow that given any $x \in P$ there exists a maximal element $m \in P$ such that $x \leq m$?

ps also by chain I thought it was of the form $x_1 \leq x_2 \leq x_3...$
Is this correct? Thank you

答:
YES.

Set $\,Y=\{y\in X: x\le y\}$. Then $Y$ inherits the partial order of $X$, and every chain $S\subset Y$, is also a chain of $X$, and hence possesses an upper bound $y_M$ in $X$. Clearly $x\le y_M$ and hence $y_M\in Y$. Therefore one may apply Zorn's Lemma in $Y$ as well.

Any maximal element $m$ of $Y$ satisfies $x\le m$.

Title: Questions about Zorn's lemma


问:
Showing that $ (v \cdot \nabla) v = (\nabla \times v) \times v + \frac{1}{2} \nabla (v \cdot v)$ (Fluid Mechanics , Feynman)
Right under equation 40.7 in the feynman lecture (here), this identity is written:

$$ (v \cdot \nabla) v = (\nabla \times v) \times v + \frac{1}{2} \nabla (v \cdot v)$$

I seek a proof for this identity/ an intuitive proof for why it is true. I'm not sure how I'd even start the derivation but I think this identity is the same as the one under the 'special sections' part of this wiki page

An attempt:

By the vector triple product identity

$$ a \times b \times c = (b ) c \cdot a - ( c ) b \cdot a$$

Now this gave me zero when applied to $ \nabla \times v \times v$ and that doesn't look right..

Hints would also be appreciated :)

答:
In the equation you have, in the $(\nabla \times \ \textbf{v}) \times \textbf{v}$ term, there are parantheses around the first two terms, so it's not necessarily zero since first you do $\nabla \times \ \textbf{v}$, which is perpendicular to $\textbf{v}$, and then you cross it with $\textbf{v}$.

Starting with the vector triple product seems like a good approach. You can always write it out in tensor notation with the Levi-Civita symbol to get the $n^{th}$ component:
$$\left((\partial_i v_j) \epsilon_{ijk}\right) v_m \epsilon_{kmn},$$
where $\left((\partial_i v_j) \epsilon_{ijk}\right)$ is the $k^{th}$ component of $(\nabla \times \ \textbf{v})$.

Since $\epsilon_{kmn} = \epsilon_{mnk}$ and $\epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$, the $n^{th}$ component is equal to
$$ \begin{eqnarray}(\partial_i v_j)  v_m \epsilon_{ijk}\epsilon_{mnk}\\
 &=&(\partial_i v_j) v_m (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})\\
&=&(\partial_i v_j) v_m \delta_{im}\delta_{jn}-(\partial_i v_j) v_m\delta_{in}\delta_{jm}\\
&=& v_m \partial_m   v_n -\partial_n v_m v_m.
\end{eqnarray}$$
Now, you can use what @PAM1499 noted (the chain rule) to write $\partial_n v_m v_m = \frac{1}{2} \partial_n (v_m v_m)$.

Putting everything back into vector form, you get:
$$(\nabla \times \ \textbf{v}) \times \textbf{v} = (\textbf{v} \cdot \nabla ) \textbf{v}- \frac{1}{2} \nabla(\textbf{v} \cdot \textbf{v}).$$
and all you need to do is rearrange the terms to get the identity.


问:
Can we define a derivative on the $p$-adic numbers?
Or on any complete valuated field, really. This question came to me when I was thinking about real and complex differentiability, which we can define in a very similar manner: Let $\mathbb F\in\{\mathbb R,\mathbb C\}$, and $V,W$ normed, finite dimensional vector spaces over $\mathbb F$. Also let $f:V\to W$. We call $f$ $\mathbb F$-differentiable at $x_0\in V$ if there is an $\mathbb F$-linear map $\mathrm Df_{x_0}:V\to W$ such that
$$\lim_{x\to x_0}\frac{\Vert f(x)-f(x_0)-\mathrm Df_{x_0}(x-x_0)\Vert}{\Vert x-x_0\Vert}=0.$$
This gives us a very natural way to talk about the real or complex differentiability of a function $\mathbb C\to\mathbb C$, for instance. Just consider $\mathbb C$ a two-dimensional vector space over $\mathbb R$ or a one-dimensional vector space over $\mathbb C$.

I'm wondering wether this can be generalized in an interesting way. This is my idea: We would need a generalization of a norm to general fields, and of course we would still need to work in complete spaces. Norms can be generalized using a valuation: Let $\mathbb F$ be a field. A valuation is a map $\vert\cdot\vert:\mathbb F\to\mathbb R$ satisfying the following for all $a,b\in\mathbb F$:

$\vert a\vert\geq0$ and $\vert a\vert=0$ iff $a=0$.

$\vert ab\vert=\vert a\vert\vert b\vert$.

$\vert a+b\vert\leq\vert a\vert+\vert b\vert$.

Now let $V$ be a vector space over $\mathbb F$. A norm on $V$ is a map $\Vert \cdot\Vert:V\to\mathbb R$ satisfying the following for all $v,w\in V,~a\in \mathbb F$:

$\Vert v\Vert\geq0$ and $\Vert v\Vert=0$ iff $v=0$.

$\Vert av\Vert=\vert a\vert\Vert v\Vert$.

$\Vert v+w\Vert\leq\Vert v\Vert+\Vert w\Vert$.

A norm on $V$ induces a metric $\Vert x-y\Vert$. We can now consider finite-dimensional, normed vector spaces $V,W$ over $F$ which are complete with respect to the induced metric, and are free to define: A function $f:V\to W$ is $\mathbb F$-differentiable in $x_0\in V$ if there exists an $\mathbb F$-linear map $\mathrm Df_{x_0}:V\to W$ such that
$$\lim_{x\to x_0}\frac{\Vert f(x)-f(x_0)-\mathrm Df_{x_0}(x-x_0)\Vert}{\Vert x-x_0\Vert}=0.$$

Now the question is wether such a definition is actually interesting. For instance, finite fields and finite-dimensional vector spaces over them will be discrete, so while they are complete, evaluating the above limit is still not possible. But I imagine infinite fields, like the $p$-adic numbers, complex $p$-adic numbers, or function fields, could make the definition a sensible one. Has such a thing been done before? Are there any obstacles to the well-definedness? Or something that would make the theory of such derivatives uninteresting?

答:
There is a short part on derivatives in $\mathbb{Q}_p$ in Fernando Q.Gouvêa's "$p$-adic numbers".

There is also no reason to restrict to finite-dimensional $V,W$. You might be interested in the notion of Fréchet derivatives.


问:
Uniformly convergent sequence of holomorphic function in every compact subset converges to holomorphic function

$\{f_n\}^\infty_{n=1}$ is a sequence of holomorphic functions that
converges uniformly to a function $f$ in every compact subset of
$\Omega$, then $f$ is holomorphic in $\Omega$.

We let $D$ be any disc whose closure is contained in $\Omega$. Then for any triangle $T$ contained in $D$, by Goursat's theorem, we have $\int _T f_n(z)dz=0$. It then asserts that

$$\int_T f_n(z)dz\to \int_T f(z)dz\text{ as }z\to \infty$$

in the closure of $D$, because of the uniform convergence of $f_n$. This seems a basic question, but can anybody please elaborate what is happening here?

答:
The idea of the proof is that according to Morera's theorem, if the integral of a continuous function $f$ along the boundary of all triangles contained in the domain vanishes, then $f$ is holomorphic. So it has to be proved that a) $f$ is continuous, and b) $\oint_T f(z)\mathrm dz=0$ for all triangles $T$ in $D$. Because then $f$ is holomorphic on $D$.

For a), we can use the usual result that uniform convergence preserves continuity. So since $f_n$ are all continuous and converge to $f$ uniformly on $\overline D$ (since that's a compact set), $f$ is also continuous.

For b), they use the fact that if $f_n\to f$ uniformly on the contour along which we integrate (which they do, since the contour is compact), then $\int f_n\mathrm dz\to\int f\mathrm dz$. Since the integrals over $f_n$ along triangles are all $0$ according to Cauchy-Goursat, the sequence converges to $0$, so we have
$$\oint_T f(z)\mathrm dz=0$$
as desired.


答:
If
$$
v_i=\sum_{j=1}^n a_{ij}e_j
$$
with the $e_j$'s forming an orthonormal basis,
$$
\langle v_i,v_j\rangle=\sum_{k=1}^n a_{ik} a_{jk}
$$
and hence
$$
U= \big(\langle v_i,v_j\rangle\big)_{i,j=1}^n=AA^T,
$$
where $A=(a_{ij})_{i,j=1}^n$.

So $U^{-1}=A^{-T}A^{-1}$.

Title: Specify the inverse of the matrix $g_{ij}=\langle v_i,v_j\rangle$, where $(v_1,...,v_n)$ is a basis

Body:
Let $V$ be an inner product space with a basis $(v_1,...,v_n)$.

I know that the Gram matrix
\begin{equation}
G=\begin{pmatrix}
\langle v_1,v_1\rangle&\cdots&\langle v_1,v_n\rangle\\
\vdots&&\vdots\\
\langle v_n,v_1\rangle&\cdots&\langle v_n,v_n\rangle
\end{pmatrix}=\begin{pmatrix}
g_{11}&\cdots&g_{1n}\\
\vdots&&\vdots\\
g_{n1}&\cdots&g_{nn}
\end{pmatrix}
\end{equation}
is invertible, but is it possible to explicitly specify the inverse (maybe involving the dual basis)?

Motivation: For those who have heard about the metric tensor: I'd like to have a formula for $g^{ij}$ (which boils down to the question above).


问:
Cellular Automata on the Collatz Conjecture
I have a Cellular Automaton that of any initial integer (initial condition of the automaton) generates states of Collatz sequences. The neighbourhood of the automaton is shaped like an L-tetromino (including two time-states). The (quiescent) background is represented as $0$ (the state zero). The other $1$-state (on-state) generates active border on the left side during the generation of the automaton. The border is the interesting part because that is where the "currect" generation either increases (is odd), are still (is odd) or decreases (is even). My earlier post on the matter was closed due to my question being unclear.

I have found out that if I can write a paper about this (or get somebody to help me writing it - helping me with terminology and such), we could represent the ($3n+1$) Collatz-problem with an Cellular Automaton (or Adaptive Cellular Automaton), and someone could approach the problem from that standpoint; like for example, Matthew Cook proved that Rule 110 was Turing Complete (i.e universal); and an undergraduate from UK proved that a $2,3$ Turing machine was universal. I have a feeling that we might perhaps find out if the conjecture is true, false or unprovable by solving some properties of this automaton.

Since now the Collatz Conjecture is represented as a Cellular Automaton, I think we can make one or more conjectures; one that either indirectly proves or disproves the conjecture, or one that gives us some information about its behaviour or if its undecidable. I have not been able to reach out to the mathematics community with my Cellular Automaton on the Collatz conjecture, and I don't know if it allready exists (that is; if someone else have allready done it), I have not yet found a paper that looks similiar to my representation (on an binary integer lattice using L-tetromino shaped neighbourhood).

On the automaton: I think convergence is difficult to prove in a Cellular Automaton, but it might be possible depending on the neighbourhood, rule and CA-setup perhaps. I think maybe proving that the paths that the neighbourhood takes (Transition Function) reaches some boundary (or not) will prove that the Conjecture is true or not. It's hard for me to actually describe what I mean about this. Please, if anyone have tried this before? I would like to hear more about it. Anyone how knows about information, or a paper on this specifically I would like to know.

My question is; does there allready exists a conjecture on the Collatz in the form of an Cellular Automaton, and such that if something about it can be proven true or false - it will also prove if the original Conjecture is true or false?

Thanks

Example of space-time diagram of the Cellular Automaton showing odd numbers only (leaving out the shift right rule):

答:
The Collatz problem can be mapped even more simply to a cellular automaton using base 6, because division by 2 in base 6 is extremely similar to multiplication by 3, and there is no problem of cascading carries.  It is fairly easy to create a 7-state automaton whose neighborhood is simply two adjacent cells in the previous generation.  The 7 states correspond to the 6 digits of base 6, and a null state to mark the edges of the computation.  Someone coded a base 6 cellular automaton representation of the Collatz conjecture on wolfram.com in 2011:
https://demonstrations.wolfram.com/CollatzProblemAsACellularAutomaton/</p>


问:
determine $\iiint zdv$ using spherical coordinates
Let $D$ be the region enclosed by the surface $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$

Determine:

$$\iiint zdv$$

using Spherical coordinates.

I'm trying to solve the above but I'm unsure how to go about it.

My attempt has been:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} zdv$$

which I then integrate the first layer leaving me with:
$$\int_0^{2\pi}\int_0^{2} 2-x^2-y^2dv$$

Should I have subbed in the spherical coordinates right at the start and integrated it as:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}p\cos\phi p^2\sin\phi dp d\theta d\phi$$

Based on a users input, would I have to first rewrite $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$

In terms of spherical coordinates?

答:
Combine $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$ to get $z=-\sqrt2$, which corresponds to the boundary $\cos\theta = \frac zr =-\frac{\sqrt2}2$, or $\theta =\frac{3\pi}4$, in spherical coordinates.

Then, the integral is
$$\int_V zdv = 2\pi \int_{\frac{3\pi}4}^\pi\int_0^{2}(r\cos\theta )\>r^2\sin\theta drd\theta=- 2\pi
$$


问:
How can you calculate $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$without using L'Hospital's law?
I'm trying to calculate this limit with substitution: $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$

I started by multiplying with $\frac{\root{n} \of {x}}{\root{n} \of {x}}$ $\to$ $\lim _{x\rightarrow 1}\; \frac{\root {mn} \of {x^{n+m}}-\root{n} \of {x}}{\root {n} \of {x^2}-\root{n} \of {x}}$

I then substituted $\root{mn}\of{x}$ with $t$, so I got $x=t^{mn}$ with $\lim _{x\rightarrow 1}\;\root{mn}\of{x}=1$

$\lim _{t\rightarrow 1}\;\frac{t^{n+m}-\root{n} \of {t^{mn}}}{\root{n} \of {t^{2mn}}-\root{n} \of {t^{mn}}}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^{n+m}-t^m}{t^{2n}-t^m}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^m(t^n-1)}{t^m(t^m-1)}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$

$\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{(t^{n-1}+t^{n-2}+...+t^1+t^0)}{(t^{m-1}+t^{m-2}+...+t^1+t^0)}$

So if we put $1$ in we will get $\frac{n*1}{m*1}$ which is $\frac{n}{m}$

Is this right?

Thanks

Body:
Binomial expension

$\lim_{x\to1}\frac{x^{1/m}-1}{x^{1/n}-1}$= $ \lim_{h\to 0} \frac{(1+h)^{1/m}-1}{(1+h)^{1/n}-1}$=
$ \lim_{h\to 0} \frac{1+\frac{h}{m}+
O(h^2)-1}{1+\frac{h}{n}+O(h^2)-1}$

= $ \lim_{h\to 0} \frac{\frac{h}{m}+
O(h^2)}{\frac{h}{n}+O(h^2)}$
=  $ \lim_{h\to 0} \frac{h(\frac{1}{m}+
O(h))}{h (\frac{1}{n}+O(h))}= \frac{n}{m}$

$O(h^2)$=  other terms contains $h^2$ and higher powers


问:
Finding maximum possible value

Let $ x $, $ y $ and $ z $ be non-negative real numbers, $ y \ne 0 $, and let $ 2 x + \frac 3 y + \frac z 6 = 24 $. Determine the maximum possible value of
$$ \sqrt [ 3 ] { \frac { x z } y } \text . $$

In class, we recently learned about inequalities between the arithmetic mean, geometric mean, harmonic mean, and quadratic mean. I started by stating that the given equation is less than or equal to $ \frac { x + z + \frac 1 y } 3 $. However, I don't know where to go from here. How can I relate the two equations we are given to finish this problem?

答:
Direct resolution:

We have $$\frac{18}y=144-12x-z$$

and we can equivalently maximize

$$xz\left(144-12x-z\right).$$

By canceling the gradient,

$$\begin{cases}144z-24xz-z^2=0,\\144x-12x^2-2xz=0.\end{cases}$$

Both equations can be factored and the largest solution is obtained with the nonzero solutions.

Title: Finding maximum possible value


问:
How to show that any two isomorphic groups have the same number of elements of order $n$ for each $n \in \mathbb{N} $
If $f :G \rightarrow  H$ is an isomorphism such that $|f(x) | = |x|$ for all $x \in G$. Prove that any
two isomorphic groups have the same number of elements of order $n$ for each $n  \in \mathbb{N} $

My attempt: If $|x|=n$ ,then we have $f(x)^n=f(x^n)=1$
But here im confusing that how to  show that any
two isomorphic groups have the same number of elements

答:
If the groups are isomorphic then there is a bijective homomorphism between them.  The bijective part means they have the same number of elements.

They also have the same number of elements of order $n$:  say $x$ has order $n$, then $x^n=e\implies f(x)^n=f(x^n)=f(e)=e\implies $ the order of $f(x)$ divides $n$.  Similarly for $f^{-1}$.  So that the order of $x$ divides the order of $f(x)$.  Thus $x$ and $f(x)$ have the same order.


问:
Additional assumptions for the robust optimization paradigm
Please, see here at page 3: could you help to understand the proofs from E.1 to E.4? In particular:

why, by adding the constraint $\operatorname{c^Tx}-t\leq 0$, did the objective function become certain?

What's the meaning of uncertain right-side side of a constraint? I don't understand the rule of $x_{n+1}=-1$.

Then, for E.3 and E.4, total darkness.

Thanks in advance.

答:
E.1 is the usual linearization of a min-max problem, replacing the max with an explicit variable $t$ and imposing a constraint on $t$.  The uncertainty of $c$ has now moved from the objective to a constraint.

For E.2, suppose you have some constraint $\sum_{j=1}^n a_{i,j} x_j \le d_i$, where $d_i$ is uncertain.  You can introduce a new variable $x_{n+1}$ and rewrite the constraint as two constraints:
\begin{align}
\sum_{j=1}^n a_{i,j} x_j + d_i x_{n+1} &\le 0 \\
x_{n+1} &= -1
\end{align}
The uncertainty of $d_i$ has now moved from the right-hand side to the left-hand side (a constraint coefficient).


问:
Regular expressions, is it always true that (r+s)*=r*+s*?
I'm really confused about this, can some one please help me understand this better.

If r and s are regular expressions then is it always true that (r+s)=r+s*?

Are r and s sets and does the plus mean the union?

答: Let $R$ be the set of words described by the regular expression $r$, and let $S$ be the set of words described by the regular expression $s$. Then the regular expression $(r+s)^*$ describes the words in the set $(R\cup S)^*$, and $r^*+s^*$ describes the words in the set $R^*\cup S^*$. Thus, the question boils down to asking whether it’s necessarily true that $(R\cup S)^*=R^*\cup S^*$. (Looking at some very simple regular expressions should give you the answer pretty easily.)</p>


问:
What is the Fourier transform of $f(t)=1$ or simply a constant?
Wolfram alpha gives the answer to be

$$F(\omega)=\sqrt{2\pi}\delta(\omega)$$

Does that mean that the function is valued $\sqrt{2\pi}$ at all points in the frequency domain? I think this is reasonable because such function i.e. $f(t)=1$ in the time domain would be sum of all the harmonics of a sinusoid and hence would contain all the frequencies. Maybe no, the function isn't varying at all and hence the frequency is $0$. But then the Fourier transform should have been $\delta(0)$ instead of $\delta(\omega)$.

Someone please shed some light on this!

答:
You can derive the answer very easily with the general formula for the fourier series of a complex exponential:

$\mathcal F(e^{jw_0 t}) = 2\pi \delta(w-w_0)$

This identity is very intuitive: Since a complex exponential only has one frequency ($w_0$), its fourier transform only has one pulse at that frequency[1].

Set $w_0 = 0$ and you get:

$\mathcal F(e^{0}) = \mathcal F(1) = 2\pi \delta(w-0) = 2\pi \delta(w)$

[1] Here is a proof: https://staff.fnwi.uva.nl/r.vandenboomgaard/SignalProcessing/FrequencyDomain/CTNP.html#complex-exponential


问:
linear transformation of the classical projective plane taking a circle to a hyperbola
Consider a circle with equation $x^2+y^2=1$ over $\mathbb{R}$ and hyperbola $xy=1$. Then with homogenenous coordinates we have that the circle has equation $X^2-Y^2-Z^2=0$ which gives the matrix
$$A=\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}.$$

For the hyperbola we have the equation $X^2-XY=1$ so we homogenize the matrix to look like
$$B=\begin{bmatrix}2 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0\end{bmatrix}.$$

I would like to map the circle to the hyperbola with a matrix $M$ such that $M^{-1}AM^{-T}=B$ where $-T$ denotes the inverse transpose of $M$. The problem is I am not sure how to get $M$. If we take the base points $(1:0:0),(0:1:0),(0:0:1)$ , only two of these belong to the circle over $\mathbb{R}$ and the circle has no points at infinity. Any hint is appreciated.

答:
First of all, the equation of the circle is $X^2+Y^2-Z^2=0$ correctly, so $$A=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}.$$
Also, the equation of the hyperbola is $$XY-Z^2=0,$$ so $$B=\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -2\end{bmatrix}.$$
Now we will find the projective transformation that sends the hyperbola to the circle. We will use that a conic is determined by its three points and the tangents at two of these points. So we will send three points and two corresponding tangents of the hyperbola to points and tangents of the circle.

First, $[1,0,0]$ is on the hyperbola, let its image be $[1,0,1]$.

$[0,1,0]$ is also on the hyperbola, let its image be $[0,1,1]$.

The tangents of the hyperbola at these points are the $x$-axis and $y$-axis, their intersection is $[0,0,1]$. Let its image be $[1,1,1]$, because the tangents of the circle at $(1,0)$ and $(0,1)$ meet at $(1,1)$.

Finally, $[1,1,1]$ is on the hyperbola, let its image be an arbitrary point of the circle, for example $[-1,0,1]$.

We use that in the coloumns of the matrix of a linear transformation we find the images of the base vectors. Thus our desired matrix is of form $$M=\begin{bmatrix}a & 0 & c \\ 0 & b & c \\ a & b & c\end{bmatrix}.$$ (For example, the first coloumn is $[a,0,a]$, because the image of $[1,0,0]$ is $[1,0,1]$, which is the same point as $[a,0,a]$ for any $a\neq 0$.)

Since the image of $[1,1,1]$ is $[-1,0,1]$, we get the equation $$\begin{bmatrix}a & 0 & c \\ 0 & b & c \\ a & b & c\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}=\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}.$$ Solving this equation we get $a=1$, $b=2$, $c=-2$. Thus the matrix of the projective transformation that sends the hyperbola to the circle is
$$M=\begin{bmatrix}1 & 0 & -2 \\ 0 & 2 & -2 \\ 1 & 2 & -2\end{bmatrix}.$$
You can check that if $N$ is the inverse of this matrix, then $NAN^T=B$ (up to some factor).


问:
Combinatorial design summation identities.
Let $\mathcal{D}$ be a $(v,k,\lambda)$-design and $B\in \mathcal{D}$ an arbitary block. Let $y_i$ be the number of block which have exactly $i$ common elements with $B$, $(i < k)$. Prove that:
$$\sum_{i=1}^{k-1}iy_i=k(r-1)\;\;\text{and}\;\; \sum_{i=1}^{k-1}{i\choose 2}y_i={k\choose 2}(\lambda-1)$$
I'm thinking basic combinatorial argumentation will suffice. In the first equations right-hand side since $r$ is the number of blocks containing a given point we just have to take the block $B$ out of the equation and we have $r-1$ and then multiply by $k$ which is the number of elements in each block. But I don't know how do we argue that we are calculating the same thing on the left-hand side. I have no idea why the second equation is as it is.

答:
For the first one, consider the following set $$A=\{(x,C):x\in B\cap C,C\neq B\},$$
on one side, we know what every $x\in B$ is in other $r-1$ blocks so $|A|=k(r-1)$ as you have noticed. On the other hand we can split $$A=\bigcup _{i=1}^{k-1} A_i,$$
where $A_i=\{(x,C)\in A: |C\cap B|=i\},$ notice that there are $|C\cap B|\cdot y_i=i\cdot y_i$ elements in each set and they are disjoint. This is the LHS

For the second one, remember that $\lambda$ is the number of blocks in which each pair of distinct elements appear. So the right hand side is pick two elements of $B$ in $\binom{k}{2}$ and record also the sets in which these appear(except $B$ itself). Do a split like the one before and you should get your answer.


Title: Finding maximum possible value

Body:
Let $ x $, $ y $ and $ z $ be non-negative real numbers, $ y \ne 0 $, and let $ 2 x + \frac 3 y + \frac z 6 = 24 $. Determine the maximum possible value of
$$ \sqrt [ 3 ] { \frac { x z } y } \text . $$

In class, we recently learned about inequalities between the arithmetic mean, geometric mean, harmonic mean, and quadratic mean. I started by stating that the given equation is less than or equal to $ \frac { x + z + \frac 1 y } 3 $. However, I don't know where to go from here. How can I relate the two equations we are given to finish this problem?

Body: Hint:by AM-GM$$8=\frac{2x+\frac{3}{y}+\frac{z}{6}}{3}\ge \sqrt[3]{2x.\frac{3}{y}.\frac{z}{6}}=\sqrt[3]{\frac{xz}{y}}$$


问:
Are these functions in $L^{1}(\mathbb{R}^{d})$?
Are the functions $f:\mathbb{R}^{d}\to\mathbb{R}$ with

$$f(x):=\frac{1}{(1+\vert x\vert)^{n}}$$

in $L^{1}(\mathbb{R}^{d})$ if $n>d$?

These functions are obvious measurable, so we only need to show

$$\Vert f\Vert_{L^{1}}=\int_{\mathbb{R}^{d}}\vert f(x)\vert\,\mathrm{d}x<\infty$$

My first try was to bound $\frac{1}{(1+\vert x\vert)^{n}}\leq \frac{1}{\vert x\vert^{n}}$, but this obviously not works, because this bound diverges for $\vert x\vert=0$.

Just by looking at the functions it is quite clear that the integral is finite, because it is everywhere finite and we have $f(x)\to 0$ for $\vert x\vert\to\infty$, but I am not sure how to prove it rigorously...

答:
Note that, polar coordinates in $\mathbb R^d$:
$$
\int_{\mathbb R^d}f(x)\,dx=\int_0^\infty \left(\int_{S^{d-1}}f(rw)dw\right)\,r^{d-1}\,dr
$$
where $S^{d-1}$ is the unit sphere in $\mathbb R^d$.

So
$$
\int_{\mathbb R^d}\frac{dx}{(1+|x|^n)}=\int_0^\infty \left(\int_{S^{d-1}}\frac{dw}{(1+r^n)}\right)\,r^{d-1}\,dr
=\omega_{d-1}\int_0^\infty \frac{r^{d-1}\,dr}{(1+r^n)}<\infty,
$$
if $d<n$, where $\omega_{d-1}$ is the $(d-1)-$dimensional area of $S^{d-1}$.


Title: How can you calculate $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$without using L'Hospital's law?

Body:
I'm trying to calculate this limit with substitution: $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$

I started by multiplying with $\frac{\root{n} \of {x}}{\root{n} \of {x}}$ $\to$ $\lim _{x\rightarrow 1}\; \frac{\root {mn} \of {x^{n+m}}-\root{n} \of {x}}{\root {n} \of {x^2}-\root{n} \of {x}}$

I then substituted $\root{mn}\of{x}$ with $t$, so I got $x=t^{mn}$ with $\lim _{x\rightarrow 1}\;\root{mn}\of{x}=1$

$\lim _{t\rightarrow 1}\;\frac{t^{n+m}-\root{n} \of {t^{mn}}}{\root{n} \of {t^{2mn}}-\root{n} \of {t^{mn}}}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^{n+m}-t^m}{t^{2n}-t^m}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^m(t^n-1)}{t^m(t^m-1)}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$

$\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{(t^{n-1}+t^{n-2}+...+t^1+t^0)}{(t^{m-1}+t^{m-2}+...+t^1+t^0)}$

So if we put $1$ in we will get $\frac{n*1}{m*1}$ which is $\frac{n}{m}$

Is this right?

Thanks

Body:
I suggest you to use the identity
$$
t^N-1=(t-1)\sum_{k=0}^{N-1}t^{k}
$$
Hence, by setting $x=t^{mn}$ we obtain that $t\to1$ as $x\to1$. Therefore
$$
\lim_{x\to1}\frac{\sqrt[m]{x}-1}{\sqrt[n]{x}-1}
=\lim_{t\to1}\frac{\sqrt[m]{t^{mn}}-1}{\sqrt[n]{t^{mn}}-1}
=\lim_{t\to1}\frac{t^n-1}{t^m-1}\\
=\lim_{t\to1}\frac{\sum_{k=0}^{n-1}t^{k}}{\sum_{k=0}^{m-1}t^{k}}
=\frac{\sum_{k=0}^{n-1}1}{\sum_{k=0}^{m-1}1}
=\frac{n}{m}
$$


答:
$r$ and $s$ are regular expressions, which represent regular sets.  If $r$ and $s$ are regular expressions that represent sets $R$ and $S$, then the regular expression $r+s$ denotes the set $R\cup S$.

If $r$ is a regular expression denoting the set $R$, then $$r^*$$ is a shorthand for $$\epsilon + r + rr + rrr + \cdots .$$

Title: Regular expressions, is it always true that (r+s)*=r*+s*?

Body:
I'm really confused about this, can some one please help me understand this better.

If r and s are regular expressions then is it always true that (r+s)=r+s*?

Are r and s sets and does the plus mean the union?


答:
A much simpler approach.  Let $y=x-1$.  Then the expression becomes $\frac{\sqrt[m]{y-1}-1}{ \sqrt[n]{y-1}-1}=\frac{1+\frac{y}{m}+...-1}{1+\frac{y}{n}+...-1}\to \frac{n}{m}$ as $y\to 0$.

Note that $m$ and $n$ do not have to be integers.


问:
Defining A Linear Approximation
Let $\mathbb{K}$ represent either $\mathbb{R}$ or $\mathbb{C}$ and let $E$ denote a normed vector space. Then, a map $f:X \subset \mathbb{K} \rightarrow E$ is said to be approximately linear  at a point $a \in X$ if if there exists an affine function $g:\mathbb{K} \rightarrow E$ such that $f(a) = g(a)$ and

$$\lim_{x \to a} \frac{\|f(x)-g(x)\|}{|x-a|} = 0$$

My question is, is the presence of the norm and absolute value in the above definition necessary? The expression $x - a$ is just a scalar in $\mathbb{K}$ and $f(x) - g(x)$ is just a linear combination of vectors and so the expression

$$\lim_{x \to a} \frac{f(x)-g(x)}{x-a} $$

is meaningful

In fact, would it not be the case that the normed quotient approaches the number $0$ if and only if the unnormed quotient approaches the vector $0_E$?

答: Yes. Since $\|\lambda e\|=|\lambda|\cdot\|e\|$ for any scalar $\lambda$ and any $e\in E$, and if $F\colon \mathbb{K}\to E$ then $\lim\limits_{x\to a}\;F(x) = 0_E\iff \lim\limits_{x\to a} \|F(x)\|=0$ by the definition of limits in a normed space, the two conditions are equivalent.</p>

Title: Defining A Linear Approximation


问:
multivariable limit question
Is this an acceptable solution?
$$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(2(x^2+y^2))}{x^2+y^2}$$
$$t=x^2+y^2$$
So $t\rightarrow0$. Now I change the limit to:
$$\lim_{t\rightarrow0}\frac{\sin(2t)}{t}=2$$
This solution makes sense to me, but I am still having doubts.</p>

答: Your proof is fine. In fact, it's very similar to the common technique of switching to polar coordinates. In that technique, one puts $x = r\cos\theta$ and $y = r\sin\theta$ and evaluates the limit as $r \to 0$. This is equivalent to putting $r^2 = x^2 + y^2$. You've put $r = x^2 + y^2$ instead. Check out this question for more information on this technique and others.</p>

Title: multivariable limit question


问:
Proof of "Induction proof method"
So I have been proving various logical statements using induction method (like structural induction , strong induction , weak induction etc ).I was wondering If there is a proof of this "Induction proof method" . So far , I came to this ,

Induction $\rightarrow$ Well ordering principal $\rightarrow$ Axiom of choice $\rightarrow$ ZFC $\rightarrow$ First-order logic theory

So now I wonder , Is there a way to prove (or show equivalence of) this method of proof using just Logic and no Set theory.Also point out if there is a flaw in my reasoning

Edit:It seems like structural induction doesn't do induction over numbers of any kind , it does in on structures .So I can't use peanos axioms to formulate it .I need ZFC .But ZFC is just a kind of first order logic.So structural induction comes from this particular first order logic . But there are some General theorems (which probably don't necessarily belong to ZFC) in Propositional Calculus which I have to prove using structural Induction .But Structural Induction can only be used inside ZFC , not outside of it.I am confused.In a simpler way , The following general theorem I will show at the end of my question is outside of set theory . And  I need structural induction to prove it . But structural Induction can only prove things inside set theory. Because Structural Induction is a axiom of Axiomatic set theory.

I will give just a example of one of these general theorem.

"Assume $A$$1$ $\equiv$  $A$$2$ . Show that for any formula $C$-containing $A$$1$ as a part , if we replace one of more occurences of the part $A$$1$ by $A$$2$ , then the resulting formula is logically equivalent to $C$."

答:
Mathematical induction is one of the Peano axioms, to which every definition of the natural numbers and the set of natural numbers, in every set theory, has to abide.

After defining natural numbers in a set theory, and after constructing the natural numbers and the set of natural numbers from this, the Peano axioms need to be proved within the set theory, making only use of set theory and symbolic logic.
It all takes place $\textit{within}$ a set theory, and hence one cannot do it $\textit{outside}$ a set theory.

The proof of the induction schema reduces to a simple set-theoretical argument; of course this may differ according to the basics of the set theory used. For example, in Quine's New Foundations (NF) the induction schema does not generally hold - as may be expected, it only holds for stratified formulae, or, more general, for formulae $\phi$ for which $\{x|\phi\}$ exists.

See my dissertation on NF in https://eprints.illc.uva.nl/574/1/X-1989-02.text.pdf .

Title: Proof of "Induction proof method"


答:
Under an isomorphism $f$, we not only have $x^n=1\implies f(x)^n=1$, but also the reverse direction, i.e., we have $x^n=1\iff f(x)^n=1$. Thus we also have
$$ \begin{align}|x|=n&\iff x^n=1\land \forall  k<n\colon x^k\ne 1\\&\iff  f(x)^n=1\land \forall k<n\colon f(x)^k\ne 1\\&\iff |f(x)|=n.\end{align}$$
Hence $f$ induces a bijection between the set of elements of $G$ of order $n$ and the set of elements of $H$ of order $n$.

More generally, an isomorphism bijects corresponding subsets of $G$ and $H$ whenever these subsets can be described in terms of any predicate in the language of first-order logic with equality extended by the group operation. Or in simpler words: isomorphic groups are "the same" for all "group-internal" intents and purposes.

Title: How to show that any two isomorphic groups have the same number of elements of order $n$ for each $n \in \mathbb{N} $

Body:
If $f :G \rightarrow  H$ is an isomorphism such that $|f(x) | = |x|$ for all $x \in G$. Prove that any
two isomorphic groups have the same number of elements of order $n$ for each $n  \in \mathbb{N} $

My attempt: If $|x|=n$ ,then we have $f(x)^n=f(x^n)=1$
But here im confusing that how to  show that any
two isomorphic groups have the same number of elements


问:
If $xy=z^2+1$ for natural $x$, $y$, $z$, prove that $x=a^2+b^2$, $y=c^2+d^2$, $z=ac+bd$ for some integers $a$, $b$, $c$, $d$
A contest-math number theory problem

If $x,y,z \in \mathbb{N}$ and $xy=z^2+1$, prove that there exist integers $a$, $b$, $c$, $d$ such that $x=a^2+b^2$, $y=c^2+d^2$, $z=ac+bd$.

After working for 20 hours, I learned that there is a solution using complex numbers.  But I have no idea about the solution.
I don't know what will come out of this.

$$xy=(z-i)(z+i)$$

I guess if I could prove that $x = (a-bi)(a+bi)$ and that $y=(c-di)(c+di)$ then maybe I would get a result.

But I don't know how to relate this to the fact that $a, b, c, d$ are integers.

Question 1: (Main)

Can we find a solution using pure number theory without using a complex numbers?

Question 2:

How can a solution be made using complex numbers?

I would love to get a detailed answer.  Thank you!

答:
The method below does not require knowing all the primes that can be expressed as the sum of two squares.

We have determinant $1$ in
$$
A =
\left(
\begin{array}{cc}
x & z \\
z & y
\end{array}
\right)
$$
As the entries and trace are positive, it is positive definite.
Gauss reduction is the finite process of multiplying successively in the form $A \mapsto P^T A P,$ with the choice of
$$
P =
\left(
\begin{array}{cc}
1 & n \\
0 & 1
\end{array}
\right)
$$
or
$$
P =
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
until the final output is the identity matrix. This comes from the inequalities of Gauss reduction. So $I = P_j^T A P_j $ for an integer matrix $P_j$ of determinant $1.$   By taking $Q = P_j^{-1}$ we get
$$ A = Q^T Q  $$

$$
Q =
\left(
\begin{array}{cc}
a & c \\
b & d
\end{array}
\right)
$$

NOTE: Gauss reduced means $\langle r,2s,t \rangle$
with $1 \leq r \leq t,$ then $-r < 2s \leq r,$ in the matrix below. A little fiddling with inequalities is needed.

$$
\left(
\begin{array}{cc}
r & s \\
s & t
\end{array}
\right)
$$


问:
How do I find the limits to this double integral and convert into polar?
With $M = \{(x, y) ∈ R^2: 1 ≤ x^2 − y^2 ≤ 9, 2 ≤ xy ≤ 4,     x, y ≥ 0\}$, calculate
$$\int_M (x^2+y^2) \mathrm dV$$

What I´ve managed so far is to convert $x$ and $y$ into polar coordinates with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and got
$$\iint \left[(r\cos(\theta))^2+(r\sin(\theta))^2\right] r\mathrm dr\mathrm d\theta$$

is it right to use polar coordinates? if so, how do I set the limits of this double integral? Thx

答: enter image description hereProps to J.G. I also figured it out, hope it is correct


问:
Let $R$ be a finite ring, and $S$ its subring. Then $|S|$ divides $|R|$
I'm supposed to answer if this statement is true or false. I am extremely confused... I don't even know where to start. Does anybody have any hints?</p>

答: Since a ring consists in an additive group with some extra structure, you can apply Lagrange's theorem for groups.</p>


问:
Conditional expectation on partition
Consider $X\in L^2(\Omega,\mathcal{A},P)$ with $\Omega = A_1 \cup A_2$, $A_1 \cap A_2 = \emptyset$, $P(A_i)>0$ and $\mathcal{F}=\sigma(A_1,A_2)$. How do I compute $E[X^2|\mathcal{F}]$?

I know that $E[X|\mathcal{F}]=\frac{E[X\mathbb{1}_{A_1}]}{P(A_1)}\mathbb{1}_{A_1}+\frac{E[X\mathbb{1}_{A_2}]}{P(A_2)}\mathbb{1}_{A_2}$.

Title: Conditional expectation on partition

Body:
Since we know $E[X|\mathcal{F}]=\frac{E[X\mathbb{1}_{A_1}]}{P(A_1)}\mathbb{1}_{A_1}+\frac{E[X\mathbb{1}_{A_2}]}{P(A_2)}\mathbb{1}_{A_2}$ by definition, then

$E[X^2|\mathcal{F}]=\frac{E[X^2\mathbb{1}_{A_1}]}{P(A_1)}\mathbb{1}_{A_1}+\frac{E[X^2\mathbb{1}_{A_2}]}{P(A_2)}\mathbb{1}_{A_2}$.


问:
Problem of minimum in physics.

It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.

Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.

I've called $x$ the distance of A from P.
The function is $$t(x)=\frac{x}{v_1}+\frac{\sqrt{(a-x)^2+b^2}}{v_2}$$
$$\frac{dt}{dx}=\frac{1}{v_1}+\frac{(a-x)*(-1)}{v_2* \sqrt{(a-x)^2+b^2}}=0 \Rightarrow v_2*\sqrt{(a-x)^2+b^2}=v_1*(a-x) \Rightarrow x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2)=0$$

The solutions of this equation are $x_{1/2}  =a \pm \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}  $
Putting $\frac{dt}{dx}>0$ it is verified for $0<x- \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and for $x>a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$.

So the minimum point is for $a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$
and  the minimum time is

$$t(a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b (v_1^2+v_2^2)}{v_1v_2 \sqrt{v_1^2-v_2^2}}+\frac{a}{v_1}$$

I'm not sure that I haven't done mistakes, because in the solution on the book it indicates
$a  - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}  $  as the solution of minimum.

and  the minimum time

$$t(a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}+\frac{a}{v_1}$$

but, in particular , it distinguishes the case of $\frac{v_2^2}{v_1^2}\ge \frac{a^2}{a^2+b^2}$
in which the minimum time is possible walking from A to B directly.

答:
Your calculations are mostly correct, you just have to interpret the results correctly.

You have a continuous function $t : \Bbb{R} \to \Bbb{R}$ and hence it must attain its minimum on the compact set $[0,a]$ in some point $x_{0} \in [0,a]$.

Moreover, $t$ is differentiable on $\langle 0,a\rangle$ and if it happens that $x_0 \in \langle 0,a\rangle$ then $x_0$ is a local extremum of $t$ so in particular $t'(x_0) = 0$. Assuming $v_1 > v_2$ (if $v_1 \le v_2$, then simply $x_0 = 0$), you correctly found that the possible zeroes of $t'$ are included in $\left\{a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+ \frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}$ (as @David K pointed out, the only actual zero of $t'$ is $a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}}$ but it doesn't matter here). Therefore, we conclude that $$x_0 \in \left\{0,a, a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}.$$
However, we are interested only in $x_0 \in [0,a]$. Notice that $a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}} > a$ so this option is certainly wrong.

Moreover, we have
$$a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a] \iff \frac{bv_2}{\sqrt{v_1^2-v_2^2}} \le a \iff \frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}.$$

Hence if $\frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}$, then $a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a]$ and this is indeed the minimum point. Namely,

$$t\left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}$$
which is smaller than both $t(0)$ and $t(a)$. We can use Cauchy-Schwarz:
$$\frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}\le \sqrt{a^2+b^2}\sqrt{\frac{1}{v_1^2}+\frac{v_1^2-v_2^2}{v_2^2}} = \frac{\sqrt{a^2+b^2}}{v_2} = t(0),$$
$$\frac{a}{v_1}+\frac{b}{v_2} \cdot \underbrace{\frac{\sqrt{v_1^2-v_2^2}}{v_1}}_{\le 1} \le \frac{a}{v_1}+\frac{b}{v_2} = t(a).$$

On the other hand, if $\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}}$, then this point is not in $[0,a]$ so it has to be simply $x_0 = 0$ or $x_0 = a$. It is not hard to see that
$$t(0) \le t(a) \iff \frac{\sqrt{a^2+b^2}}{v_2} \le \frac{a}{v_1}+\frac{b}{v_2} \iff \frac{v_2}{v_1} \ge \frac{\sqrt{a^2+b^2}-b}a$$
and the latter is true in our case since
$$\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}} \ge  \frac{\sqrt{a^2+b^2}-b}a.$$
So it has to be $x_0 = 0$.

Therefore:

$$x_0 = \begin{cases}
a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}, &\text{ if } \frac{v_2}{v_1} \in \left\langle 0, \frac{a}{\sqrt{a^2+b^2}}\right] \\
0, &\text{ if } \frac{v_2}{v_1} \in \left[\frac{a}{\sqrt{a^2+b^2}}, 1\right\rangle \\
\end{cases}$$

If $v_1 >> v_2$ then $x_0$ approaches $a$, namely
$$\lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{b}{\sqrt{\left(\frac{v_1}{v_2}\right)^2-1}}\right) = a$$
but it is always faster to go at least somewhat diagonally than straight left and then straight up.

On the other hand, if $v_1$ approaches $v_2$ then you may wonder why
$$\lim_{\frac{v_2}{v_1} \to 1} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = -\infty$$
and not $0$ but this expression stops being relevant as soon as $\frac{v_2}{v_1} \ge \frac{a}{\sqrt{a^2+b^2}}$ as is it is faster to simply go diagonally.

Title: Problem of minimum in physics.


问:
Find a recurrence relation for $a_{n,k}$ , the number of ways to order n doughnuts from k different types of doughnuts if
Find a recurrence relation for $a_{n,k}$ , the number of ways to order n doughnuts from k different types of doughnuts if two or four or six doughnuts must be chosen of each type.

I need help understanding the answer to this.

So I know the answer to this is

$a_{n,k}=a_{n-2,k-1} + a_{n-4,k-1}+a_{n-6,k-1}$

so I know that $a_{n-2,k-1}$ means you start by picking $2$ donuts of one type, then you pick the remaining $n-2$ donuts from the other $k-1$ types

$a_{n-4,k-1}$ means you start by picking $4$ donuts of one type, then you pick the remaining $n-4$ donuts from the other $k-1$ types

$a_{n-6,k-1}$ means you start by picking $6$ donuts of one type, then you pick the remaining $n-6$ donuts from the other $k-1$ types

But how does this satisfy the constraint that two or four or six doughnuts must be chosen of each type? That's what I'm confused by. Can't you just pick any amount of each type of donut from the other $k-1$ types, not just two or four or six.

If someone can explain this to me, I would really appreciate it.

答:
The definition of $a_{n,k}$ requires 2, 4, or 6 donuts of each of the $k$ types.  The recurrence relation is just splitting up the possibilities according to the number of donuts of type $k$.  Once you have decided whether that is 2, 4, or 6, you have a smaller problem with fewer donuts and only $k-1$ types of donuts.  The constraint on number of donuts of each type still applies to $a_{n-2,k-1}$, and $a_{n-4,k-1}$, and $a_{n-6,k-1}$.  If the constraints applied only to type $k$, the recurrence would instead be
\begin{align}
a_{n,k} &= b_{n-2,k-1} + b_{n-4,k-1} + b_{n-6,k-1} \\
b_{n,k-1} &= \sum_{j=1}^n b_{n-j,k-2}
\end{align}
where $b_{n,k}$ is the (unconstrained) number of ways to order $n$ donuts of $k$ types.

Note: as a sanity check, you should find that $a_{n,k}=0$ when $n$ is odd.


答:
Hint. Notice that every odd prime factor $p$ dividing $z^2+1$ satisfies $p\equiv 1\bmod 4$, since it is the only way that $-1$ is a quadratic residue (we will treat the case $2\mid z^2+1$ at the end). Besides, Fermat's theorem establishes that for every $p\equiv 1\bmod 4$ there exist integers $r,s$ such that $p=r^2+s^2$ (this can be proven, e.g., with Thue's Lemma). Thus, both $x,y$ are a product of odd numbers (and possibly powers of these odd numbers).

Yet, we also have from Brahmagupta's Identity that $$(r^2 + s^2)(t^2 + u^2) = (rt + su)^2 + (ru - st)^2 $$ You can, hence, take two odd primes - say $p_1$ and $p_2$ - , and write them as a sum of squares: $p_1\cdot p_2=(r^2+s^2)(t^2+u^2)=(rt+su)^2+(ru-st)^2$. Use this identity repeatedly. Can you finish?

For the case $2\mid z^2+1$, notice that $2=1^2+1^2$.

Now, we just need to prove that these $a,b,c,d$ satisfy $ad - bc=\pm1$. This looks harder using elementary techniques; I might try it later...

Title: If $xy=z^2+1$ for natural $x$, $y$, $z$, prove that $x=a^2+b^2$, $y=c^2+d^2$, $z=ac+bd$ for some integers $a$, $b$, $c$, $d$

Body:
A contest-math number theory problem

If $x,y,z \in \mathbb{N}$ and $xy=z^2+1$, prove that there exist integers $a$, $b$, $c$, $d$ such that $x=a^2+b^2$, $y=c^2+d^2$, $z=ac+bd$.

After working for 20 hours, I learned that there is a solution using complex numbers.  But I have no idea about the solution.
I don't know what will come out of this.

$$xy=(z-i)(z+i)$$

I guess if I could prove that $x = (a-bi)(a+bi)$ and that $y=(c-di)(c+di)$ then maybe I would get a result.

But I don't know how to relate this to the fact that $a, b, c, d$ are integers.

Question 1: (Main)

Can we find a solution using pure number theory without using a complex numbers?

Question 2:

How can a solution be made using complex numbers?

I would love to get a detailed answer.  Thank you!