问:
Proof that $\lim_{n \to \infty}\sqrt[n]{|a_n|}=1$.
Suppose that there is an arbitrary $a_n$ and that $\sum_{n=1}^{\infty}a_n$ is a conditionally convergent series. Then prove if $\lim_{n \to \infty}\sqrt[n]{|a_n|}$ exists, then $\lim_{n \to \infty}\sqrt[n]{|a_n|}=1$.

Intuition

This seems to be a difficult question, since it is about an arbitrary series. What information should be utilized? I think the phrase conditionally convergent is key here.

Attempted Solution (sort of)

An example of a conditionally convergent series would be $\frac{(-1)^n}{n}$, and we know that $(\frac{1}{n})^{\frac{1}{n}}$ exists and converges to 1. From this example I see distinctly that if the absolute value of the $n^{th}$ square root of a conditionally convergent series exists, then alternating series are possible examples. But how would I exactly go about proving that every $a_n$ will have the property $\lim_{n \to \infty}\sqrt[n]{|a_n|}=1$?

答:
Suppose that $\lim_{n\to\infty}\sqrt[n]{|a_n|}$ exists.

If it is greater than $1$, then $\sqrt[n]{|a_n|}>1$ if $n$ is large enogh, and therefore $|a_n|>1$ if $n$ is large enough. So, you don't have $\lim_{n\to\infty}a_n=0$, and therefore your series diverges.

And if it is smaller thatn $1$, the let $c\in\left(\sqrt[n]{|a_n|},1\right)$. Then $\sqrt[n]{|a_n|}<c$ is $n$ is large enough, and therefore $|a_n|<c^n$. So, your series converge absolutely.

So, $\lim_{n\to\infty}\sqrt[n]{|a_n|}=1$.


问:
topology- evaluation functions into products
If $X$ is a Tychonoff space, prove that the family of all continuous real-valued functions on $X$ distinguishes points</p>

答: Consider the definitions of a completely regular space and of a family of functions separating points.</p>

ContentLicense: CC BY-SA 3.0


问:
How big is the equilateral triangle inscribed in a triangle?
Given a scalene triangle $ABC$ and an inscribed equilateral triangle whose vertices lie on different sides of $\triangle ABC$, what is the maximal ratio of the area of the equilateral triangle to that of the original triangle?

I would expect an answer as a ratio of polynomials in sines and cosines of the angles of $\triangle ABC$. I got such an expression via a clunky unsymmetrical method, but it but it was so messy that I gave up trying to simplify it. However, a more intelligent method might well yield a formula of reasonable length.

答:
We start with the process of solving the problem of "inserting" an equilateral triangle in the given triangle. We select the side of the triangle in front the largest angle to create the locus the smaller angles of the equilateral triangle.

When you have (the drawing on the right) the two circles with centers $U$ (radius $r_1$) and $S$ (radius $r_2$), you select the side of the equilateral triangle $GH$, in the larger circle (center $S$ in the drawing), you stretch it to intersect the smaller circle (not shown) and then you draw the line through this intersection and the third vertex of the equilateral triangle intersecting the other circle.

The triangle you created is the smallest triangle subscribing the equilateral triangle and is similar to the original triangle. You scale up the equilateral triangle by the similarity factor to get the largest equilateral triangle.

Note that the largest equilateral triangle is sharing a side with the triangle in which it is inscribed and that the $60^\circ$ angle of the equilateral triangle is included in the largest angle of the original triangle.

This means that to draw the largest equilateral triangle you do not need to go through the lengthy process; you draw a $60^\circ$ angle on the larger side of the largest angle of the triangle and from that vertex to the intersection with the opposite side is the largest equilateral triangle.

Let me know if you have a question.

ContentLicense: CC BY-SA 4.0


问:
Second pair of matching birthdays
The "birthday problem" is well-known and well-studied.  There are many versions of it and many questions one might ask.  For example, "how many people do we need in a room to obtain at least a 50% chance that some pair shares a birthday?"  (Answer: 23)

Another is this: "Given $M$ bins, what is the expected number of balls I must toss uniformly at random into bins before some bin will contain 2 balls?"  (Answer: $\sqrt{M \pi/2} +2/3$)

Here is my question: what is the expected number of balls I must toss into $M$ bins to get two  collisions?  More precisely, how many expected balls must I toss to obtain the event "ball lands in occupied bin" twice?

I need an answer for very large $M$, so solutions including summations are not helpful.

Silly Observation:

The birthday problem predicts we need about 25 US Presidents for them to share a birthday.  It actually took 28 presidents to happen (Harding and Polk were both born on Nov 2).  We see from the answers below that after about 37 US Presidents we should have a 2nd collision.  However Obama is the 43rd and it still hasn't happened (nor would it have happened if McCain had won or Romney had won; nor will it happen if H. Clinton wins in 2016).

答:
Suppose there are $n$ people, and we want to allow $0$ or $1$ collisions only.

$0$ collisions is the birthday problem: $$\frac{M^{\underline{n}}}{M^n}$$

For 1 collision, we first choose which two people collide, ${n\choose 2}$, then the 2nd person must agree with the first $\frac{1}{M}$, then avoid collisions for the remaining people, getting $${n \choose 2}\frac{M^{\underline{n-1}}}{M^{n}}$$

Hence the desired answer is $$1-\frac{M^{\underline{n}}}{M^n}-{n \choose 2}\frac{M^{\underline{n-1}}}{M^{n}}$$
or
$$ 1-\frac{M^{\underline{n-1}}(M-n+1+{n\choose 2})}{M^n}$$

When $M=365$, the minimum $n$ to get at least a 50% chance of more than 1 collision is $n=36$.

Title: Second pair of matching birthdays


问:
necessary and sufficient conditions for weak convergence in $c_0$
First, some definitions:

A sequence $[x_n]$ in a normed linear space $X$ is said to converge weakly to an
element $x$ if for every $\phi \in X^*,\,\,  \phi(x_n) \rightarrow \phi (x),$ (where$X^*$ is the conjugate space of $X$.)

As for $c_0$, let $c_0$ denote the space of all real sequences that converge to zero, with the norm $\| x\| = \sup_n |x(n)|.$

Now, I need to show that in order for a sequence $[x_n]$ to converge weakly to an element $x$ in the space $c_0$, it is necessary and sufficient that the sequence be bounded and that (for each $i$) we have $\lim_{n \rightarrow \infty} x_n(i) =x(i).$

Here's my attempt at a proof:

First direction: assume that the sequence is weakly convergent, prove the boundedness and pointwise convergence:
We know that any weakly convergent sequence is bounded. Also, every continuous linear functional $\phi$ in $c_0$ can be written in the following form for some suitable point $\alpha$ in $l^1$. $$\phi(x) = \sum_{i=1}^{\infty } \alpha(i) x(i)$$
Similarly, $$\phi_n (x) = \sum_{i=1}^{\infty } \alpha^\prime (i) x_n(i).$$ Now, due to weak convergence, we have $lim_{n \rightarrow \infty} \phi_n(x) = \phi(x) $. We pick $\alpha(i) = \alpha ^ \prime (i) = e(i) $, where $e(i)$ is the $i$th basis for $c_0$. Thus, we can conclude $lim_{n \rightarrow \infty} x_n(i) = x(i).$

Am I right so far? And if so, I'd appreciate any hint/help on the inverse direction of the proof. Thanks.

答:
Even though this question is already answered,  let me offer another approach which works not only for $c_0$,  but also
for  $\ell ^p$, for
every $p\in (1, \infty )$, and  which also
highlights some important features of the problem.

Lemma.  Let $X$ be a Banach space and let $S\subseteq X^*$ be a total set (i.e., $S$ spans a dense subspace of $X$).   Then a
sequence $\{x_n\}_n\subseteq X$ converges to a given $x\in  X$ in the weak topology, if and only if

$\sup_{n\in {\mathbb N}}\|x_n\|<\infty $, and

$\displaystyle \lim_{n\to \infty }\sigma (x_n)=\sigma (x)$, for every $\sigma $ in $S$.

Proof.  Assuming weak convergence one gets (1) by Banach-Steinhauss and (2) is obvious.

Conversely, assuming (1) and
(2),  pick $\varphi $ in $X^*$,  and $\varepsilon >0$.   Next choose $\sigma $ in $\text{span}(S)$ such that $\|\sigma -\varphi \|<\varepsilon /3M$,  where $M$ is the
supremum in (2).

Observing that (2) implies that $\sigma (x_n)\to \sigma (x)$,   choose $n_0$ such that $|\sigma (x_n)-\sigma (x)|<\varepsilon /3$,  for all $n\geq n_0$.

For $n\geq n_0$ we then have
$$
  |\varphi (x_n)-\varphi (x)| \leq  $$$$ \leq
  |\varphi (x_n)-\sigma (x_n)| +  |\sigma (x_n)-\sigma (x)| + |\sigma (x)-\varphi (x)| \leq  $$$$ \leq
  \|\varphi -\sigma \Vert  \|x_n\| +  \frac\varepsilon 3 +   \|\varphi -\sigma \Vert  \|x\| \leq  $$$$ \leq
  \frac\varepsilon {3M} M +  \frac\varepsilon 3 +  \frac\varepsilon {3M} M = \varepsilon .
  \tag*{$\square$}
  $$

Back to the question,  let $X$ be either $c_0$ of $\ell ^p$,  with $p\in (1, \infty )$,  and let $S$ be the subset of $X^*$ consisting
of the functionals of the form
$$
  \sigma (x) = x(i),
  $$
where $i$ ranges in ${\mathbb N}$.  Then observe that condition (2) in the Lemma translates to
$$
  \lim_{n\to \infty }x_n(i)=x(i), \quad\forall i\in {\mathbb N}.
  $$

To conclude the proof of the statement made in the original post, it is enough to observe that $\text{span}(S)$ is dense
in $X^*$.

We should remark that the last point in the sentence above  fails for $\ell ^1$ and for $\ell ^\infty $, and this is why these values of $p$ must be  excluded.


问:
Clarification required for proof structure in Spivak's Calculus
There is a particular "proof structure" I routinely observe throughout Michael Spivak's Calculus that I hoped I could get some commentary on (I will provide the most recently encountered example shortly).

Such questions will be framed as "Prove $P$", and then Spivak's solutions proceed as follows:

Assume $P$ is true.

Algebraic manipulation

Arrive at a previously proven statement $Q$.

Claim $\square$

Here is one such example:

Prove that $\sqrt{(x_1+y_1)^2+(x_2+y_2)^2} \leq \sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}} + \sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$

The "$Q$" that we have in our back pocket is the Schwarz inequality, which reads as:

$$x_1y_1+x_2y_2 \leq \sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}}\sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$$

Here is Spivak's solution (verbatim):

This inequality is equivalent to the squared inequality, $$(x_1+y_1)^2+(x_2+y_2)^2 \leq (x_1^{\ \ \ 2}+x_2^{\ \ \ 2})+ (y_1^{\ \ \ 2}+y_2^{\ \ \ 2})+2\sqrt{x_1^{\ \ \ 2}+x_2^{\ \ \ 2}}\sqrt{y_1^{\ \ \ 2}+y_2^{\ \ \ 2}}$$ which is easily seen to be equivalent to the Schwarz inequality

My major confusion with this proof strategy is that I view it as a proof of the implication $P \rightarrow Q$. If I am keen on proving $P$, certainly I am trying to prove $Q \rightarrow P$, right?

Is Spivak glossing over the fact that in order to work backwards (i.e. to start from the Schwarz inequality and arrive at our initial $P$ statement), we must necessarily be in an algebraic field?

I ask this question because it appears as though one of the unspoken assumptions here by Spivak is that we can casually "reverse" all of the algebraic manipulations that allowed us to traverse from $P$ to $Q$, so that we can then go from $Q$ to $P$.

Is this the correct interpretation?

I believe another relevant assumed-to-be-true statement that must be in play for the reversed $Q \rightarrow P$ direction (at least for this particular example) is that $$x,y \gt 0 \land x^2 \geq y^2 \rightarrow x \geq y$$

答:
José Carlos Santos answered this specific example perfectly.

All that I'd add is that for real numbers $a$ and $b$,
$$a,b \gt 0 \land a^2 \geq b^2 \rightarrow a \geq b$$
comes from an earlier problem (3rd Edition, 1-5).

Results from past problems are employed throughout the text, both in the chapters themselves and in subsequent problems.

Spivak does slip up here and there. It's good you're keeping him honest.

In general, I think some of the book's arguments do sometimes rely on equivalence relations that aren't explicitly demonstrated, or even acknowledged.

Here are some other things the book uses without (I think) first providing adequate justification. Spivak likely (and fairly) assumes the reader is familiar with these things, but their tacit use is somewhat at odds with his "let's start from nothing" approach. Some of these are maybe not themselves used by the book, but are themselves used to prove others on the list:

Basic algebraic properties (ex: we can add the same number to both sides of an equation)

Relationships between angles generated by parallel lines cut by a transversal

Pythagorean Theorem

Triangle side/angle relationsip (bigger side, bigger angle)

Sum of two sides of triangle always greater than the third

Law of sines

Law of cosines

$\sqrt a$ (existence of roots is eventually proven, but not before many problems that involve terms like $\sqrt 2$ and $\sqrt{(x_1+y_1)^2+(x_2+y_2)^2}$

Several limits and a useful limit substitution trick (Chapter 5 "Limits")

There are likely more. My list of things Spivak leaves out probably leaves some things out.

Edit:
Spivak does acknowledge much of this, and is careful not to rely on geometry in his formal proofs. He works to define lines, curves, distance, trigonometric functions, and all in terms of pairs of real numbers.

It's just that some of these things are invoked a bit before they've been fully justified, especially in the examples and problems.


问:
Prove that: $\underset{x\to \infty }{\text{lim}}\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))} = \text{doesn't exist}$
Prove that:
$$\underset{x\to \infty }{\text{lim}}\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))} = \text{doesn't exist}$$

We were showing that the limit doesn't exist with L'hopital rule: We have done L'hopital 4 times and then seen that the zero in the denominator is not isolated (the epsilon neighbourhood of the zero contained another zero).

I understand that $e^{\sin{x}}$ is oscillating and bounded,however, I do not really understand why the $e^{\sin{x}}$ is "stronger" the $x$, since it is bounded and the limit is not one.

Another question is, how valid is a proof with L'hopital rulle used to proof that the limit doesn't exist.

答:
Let me point out two misconceptions:

I cannot say for sure, but I expect that the proof using de l'Hospital's rule you're mentioning is false. Simply because it's a theorem of the type if (...) then the limit exists, and not of the type if (...) the the limit doesn't exist.

You may consider this picky, but one shouldn't write
$$\lim_{x \to \infty} f(x) = \text{doesn't exist},$$
but rather
$$\lim_{x \to \infty} f(x) \quad \text{doesn't exist}.$$
The difference lies in the fact that $\lim_{x \to \infty} f(x)$ is a priori just a symbol. If the limit exists, we can assign a value to this symbol and write $\lim_{x \to \infty} f(x) = (\ldots)$. But when the limit doesn't exist, the symbol is not related to any number, and writing an equal sign next to it is misleading.

As for the problem itself, it's helpful to note that
$$
\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))}
= e^{-\sin x}
$$
for all $x$ for which the first expression makes sense. This reduces the problem to the study of $e^{-\sin x}$, or even $\sin x$.


问:
"Inflection " point of the sigmoid
I have a very short question, this might be the wrong place to post it so I do apologise.

In the standard sigmoid function, $x \mapsto \frac{1}{1+e^{-x}}$ we have a point that intuitively should be called its inflection, namely $0$ where the second derivative is $0$. However if I think back to my analysis an inflection point must be a root also of the first derivative.

Is there a technical term for the "inflection" , second derivative root , "point where the function starts going up faster" of the sigmoid function?

Again sorry for how trivial this might be for most of you.

I have attached an image in which the blue point is the one I mean.

答:
Edited:

Actually an inflection point is just a point where the second derivative changes sign irrespective of whether the first derivative is zero. The term "inflection point" does not capture the distinction between e.g. $x^3$ and $\frac{1}{1+e^{-x}}$ at $0$.

Title: "Inflection " point of the sigmoid


问:
Compute eigenvalues and eigenvectors problem
I really don't know how solve this problem:

Let $V$ be the space of real functions spanned by $\cos(x)$, $\cos(2x)$ and $\cos(3x)$. Let $T\in\mathcal{L}(V,V)$ con $T(\cos(x)) = 3\cos(x) + 2\cos(2x) - \cos(3x)$, $\;T(\cos(2x)) = 3\cos(2x) + \cos(3x)$ and $T(\cos(3x)) = \cos(3x)$. Find the eigenvalues and eigenvectors of $T$.

Please, I need help

Thanxs in advance.

答:
The matrix representation of your linear transformation can be written as:

$$ A:= \left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right) $$

Since the matrix is triangular the characteristic polynomial can be easily read off by looking at $ \det( xI-A)$. Can you proceed now with what you need to show?


问:
Limit of a continued product $\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $ as $n \to \infty$
Evaluate $\lim_{n\to \infty }\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $

I don't have any ideas on how to approach the problem. Hints or solutions are appreciated.

答:
If you factor the general term, you get:

$$\frac{r^2 \left(r^2+3\right)^2}{(r-2) (r+2)
   \left(r^2-2 r+4\right) \left(r^2+2 r+4\right)}$$

Notice that $r^2-2r + 4 = (r-1)^2 + 3,\ r^2+2r + 4 = (r+1)^2+3,$
So the product telescopes: and almost everything cancels. (the final answer is $\frac{72}7.)$


问:
Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$
I'm trying to solve the following sum for a project. Many thanks in advance for answering the question!

$$\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m+a)}{\Gamma(k+m+b)}(k+m)$$

where $m$ is an integer but $a$ and $b$ can be any real numbers.

答:
Using this result for hypergeometric functions of unity argument as well as the property of the gamma function $x\Gamma(x)=\Gamma(x+1)$ we find:
$$
\begin{align}
S
&=\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m
+a)}{\Gamma(k+m+b)}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(m
+a)_k}{(m+b)_k}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}(k+m)\\
&=m\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}
+\sum_{k=1}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}k\\
&=mF(1,m+a;m+b;1)
+\sum_{k=0}^{\infty}\frac{(1)_{k+1}(m
+a)_{k+1}}{(m+b)_{k+1}\,(k+1)!}(k+1)\\
&=m\frac{\Gamma(m+b)\Gamma(b-a-1)}{\Gamma(m+b-1)\Gamma(b-a)}
+\frac{m
+a}{m+b}\sum_{k=0}^{\infty}\frac{(2)_k(m
+a+1)_k}{(m+b+1)_k\,k!}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}F(2,m+a+1;m+b+1;1)\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}\frac{\Gamma(m+b+1)\Gamma(b-a-2)}{\Gamma(m+b-1)\Gamma(b-a)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=\frac{m+b-1}{b-a-1}\left(m
+\frac{m+a}{b-a-2}\right),
\end{align}
$$
which holds so long as $\Re(b-a-2)>0$. This result was numerically evaluated in Mathematica and compared with numerical values of the original sum $S$, which showed agreement.


问:
Detail in Massey's Classification Theorem for Compact Surfaces
So, I am going through Massey's Algebraic Topology: an Introduction, and I got stuck in a little detail in the proof of Theorem 5.1 of Chapter one (the Classification Theorem for Compact Surfaces).

In the very end of the fourth step, the author proves that edges of "the first kind" must appear separated by another pair of this kind (simbolically, it looks something like $c\dots d \dots c^{-1}\dots d^{-1}$). And then they argue the following:

To prove this assertion, assume that the edges labeled с are not
separated by any other pair of the first kind. Then our polygon has the
appearance indicated in Figure 1.20. Here A and В each designate a
whole sequence of edges. The important point is that any edge in A
must be identified with another edge in A, and similarly for B. No
edge in A is to be identified with an edge in B. But this contradicts the
fact that the initial and final vertices of either edge labeled "c" are to be
identified, in view of step number three.

Kinsey elaborates a little more on the argument in Topology of Surfaces, saying that if we had only pairs of the "second kind" between the edges labeled $c$ we would have at least two different vertices on the polygon, contradicting the previous step of eliminating all but one vertex. And then, they conclude that the only possibility that doesn't violate that last condition is to have pairs of "the first kind" the way I said above.

However, I can't even understand why that would imply that the vertices are different... Can someone help me in this argument?

答:
Your language lost some precision when you paraphrased from Kinsey regarding vertex identifications. So let me rewrite the implication you are asking about, using the more precise language of vertex identification as in Massey:

If each edge in $A$ is identified with some other edge in $A$ then for any vertices $v,w$ such that $v$ is identified with $w$, $v$ is in $A$ then $w$ is in $A$. Therefore, no vertex in $A$ is identified with a vertex in $B$.

As for why this statement is true, we'll use the fact that two different vertices $v \ne w$ of the polygon are identified if and only if they are forced to be identified by some sequence of edge identifications. More precisely, by definition $v$ and $w$ are identified if and only if there exists a chain of vertices
$$(*) \qquad v=v_0 \ne v_1 \ne ... \ne v_n=w
$$
such that for each $i$, the two vertices $v_{i-1}$ and $v_i$ are identified to each other by a single edge identification.

So, assuming that $v$ is in $A$, we prove that $w$ is in $A$ by induction on the length $n$ of the chain of vertices $(*)$.

For the base case of the induction where $n=1$, we just have to check on edge identification: if $v \ne w$ are two vertices of the polygon, if $v$ is an end point of one edge labelled $e$ (which I'll call "the first $e$ edge"), and if $w$ is the corresponding endpoint of the other $e$ edge to which $v$ is identified (which I'll call "the second $e$ edge"), and if $v$ is a vertex in $A$, then $w$ is also a vertex in $A$. There are two cases to consider:

The label $e$ is different from $c$. Because $v$ is in $A$, it follows that the first $e$ edge is in $A$.  Because any edge in $A$ is identified with another edge in $A$, it follows that the second $e$ edge is also in $A$. Therefore $w$ is in $A$.

The label $e$ is the same as $c$. So the first $e$ edge is one of the two $c$ edges, and $v$ is its final vertex. It follows that the second $e$ edge is the other $c$ edge, and $w$ is its final vertex. Therefore $w$ is in $A$.

For the inductive step, assume that statement $(*)$ is true for any chain of length less than $n$. For the chain of length $n$ as shown in $(*)$, we break it into the length $n-1$ chain $v=v_0,...,v_{n-1}$ followed by the length $1$ chain $v_{n-1},v_n=w$. Applying the inductive step to that length $n-1$ chain it follows that $v_{n-1}$ is in $B$. Then applying the inductive step to that length $1$ chain it follows that $w$ is in $B$.


问:
Integral inequality (Divergence theorem)
I'm trying to prove the following inequality:

$$2 \int_U |\nabla \phi|^2 dx \leq \int_U \phi^2 dx + \int_U |\Delta \phi|^2 dx$$

where $U \subset \mathbb{R}^n$ is bounded and open and $\phi \in C^\infty_c(U)$. I actually think I have managed to prove this just using one of Green's identities $$\int_UD \phi \cdot D \phi dx = -\int_U \phi \Delta \phi dx + \int_{\partial U}\frac{\partial \phi}{\partial \nu} \phi dx$$
(which comes from the divergence theorem), and then using the fact that $\phi = 0$ on $\partial U$. This then gives:

$$\int_U |\nabla \phi|^2 dx = -\int_U \phi \Delta \phi dx \leq \int_U |\phi| |\Delta \phi| dx \leq \int_U \frac{\phi^2}{2}+\frac{|\Delta \phi|^2}{2}$$ where the last inequality comes from Cauchy's inequality $ab \leq a^2/2 + b^2/2$.

Does this seem correct? The problem is that I have been given a hint which says to use the fact that $\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi$ and I'm not sure how to use this hint. Would this just lead to an alternative proof of the inequality?

答:
Divergence theorem is
$$
\int_{U} \nabla \cdot F\,dx = \int_{\partial U} F\cdot \nu\,dS.
$$
The hint you have is:
$$\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi.$$
So we let $F = \phi \nabla \phi$:
$$
\int_{U}(|\nabla \phi|^2 + \phi \Delta \phi)\,dx =  \int_{U} \nabla \cdot (\phi \nabla \phi)\,dx = \int_{\partial U} \phi \nabla \phi\cdot \nu\,dS = \int_{\partial U} \phi \frac{\partial \phi}{\partial \nu}\,dS .
$$
And here we have the Green's identity and the rest of the proof goes like you wrote. Also I want to point out that the integration on boundary is with respect to the surface measure $dS$, not the volume measure $dx$ anymore, you may wanna modify the Green's identity in your question a little bit. :)</p>


问:
Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs

Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs as $P_1P_2,P_2P_3....P_{m-1}P_m, P_mP_1$ and any player who wins two consecutive games (ie. k and (k+1)th game) wins the match. If the chance that the match is won at the rth game is $k$, then find relation between $k$ and $r$

I haven’t really understood the question or what it wants to do, so I don’t know how to begin. Can anyone explain the question and give a starter hint?

答:
A chart of what happens in the first few games would perhaps help supplement lulu's answer above:

Note that the outcome is split by the probabilities of the last two games, since that's how we know that the match finishes: when the "new" player wins one round and then - as the "old" player - wins the next round, shown as NO. Here the options for the new player winning are gathered into a single item by "forgetting" past history - so for example the ...ON and ...NN outcomes from game $3$ have a combined probability of $3/8$, which is then split into two $3/16$ probability outcomes of game $4$.


问:
System of three equations with $e$
I'm trying to solve the following system of equations:

$0.001213=A+B.e^{30C}$

$0.002817=A+B.e^{40C}$

$0.00793=A+B.e^{50C}$

I tried elimination method but I got no answer. What I did was multiplying the first equation with -1 and summing it.

Could you please help me on how to solve this?

Thanks

答:
COMMENT $$\begin{cases}a=A+B.e^{30C}\\b=A+B.e^{40C}\\c=A+B.e^{50C}\end{cases}\Rightarrow\frac{c-b}{b-a}=\frac{e^{50C}-e^{40C}}{e^{40C}-e^{30C}}=\frac{e^{20C}-e^{10C}}{e^{10C}-1}$$
Put $X=e^{10C}$ so you have a quadratic equation
$$(b-a)X^2-(c-a)X+(b-c)=0$$ from which $C=\dfrac{\ln(X)}{10}$, etc.

Title: System of three equations with $e$


Title: System of three equations with $e$

Body:
I'm trying to solve the following system of equations:

$0.001213=A+B.e^{30C}$

$0.002817=A+B.e^{40C}$

$0.00793=A+B.e^{50C}$

I tried elimination method but I got no answer. What I did was multiplying the first equation with -1 and summing it.

Could you please help me on how to solve this?

Thanks

Body:
Pay attention the pattern of $e^{30C},e^{40C},e^{50C}$ and notice 40-30=50-40, which gives us a hint  where to start.

$0.001213=A+B.e^{30C}$(I)

$0.002817=A+B.e^{40C}$ (II)

$0.00793=A+B.e^{50C}$  (III)

(II)-(I) $0.001604=B.(e^{40C}-e^{30C})$ =$B.e^{30C}(e^{10C}-1)$ (IV)

(III)-(II) $0.005133=B.(e^{50C}-e^{40C})$=$B.e^{40C}(e^{10C}-1)$ (V)

(V)/(IV) $\frac{5133}{1604}$=$\frac{e^{40C}}{e^{30C}}$=$e^{10C}$, $ln \frac{5133}{1604}$=10C $\to$
C= $\frac{ln \frac{5133}{1604}}{10}$

With C known, substract (I) and (II) to find B. With B,C known, use (I) to find A.


问:
Reference algorithm/formula for the distribution of the median of random variables?
The distribution of the mean of two random variables can be calculated using a convolution. I have a collection of $n$ independent random variables each with PDFs that are simple functions on $[0,1]$. I would like to know the exact distribution of the median of these variables. I understand there is a central limit theorem for the distribution of the sample median for i.i.d variables, but I don't have that assumption here. I also see that there's a way to get a formula for discrete random variables. Is there a reference for continuous random variables?

答:
Assuming the number of your random variables is $2n,$ the probability that the median is $x$ equals
$$m(x)=\sum_{\mbox{subsets I of size $n$}}\prod_{i \in I} F_i(x) \prod_{j\notin I}(1-F_i(x),$$ where $F_k$ is the CDF of the $k$-th variable. Needless to say, for $n$ large (as in, bigger than about 6), this is not super useful. If the variables are $i.i.d,$ this is a fairly civilized formula.


问:
Smallest subgroup of $A_{5}$ containing the permutation $(12)(13)$.
I need to find the smallest subgroup of $A_{5}$ that contains the permutation $(12)(13)$ and to also explain why.
I guess we can write this permutation as $(123)$ and I guess, according to a list of subgroups that I found, that the smallest would be $A_{3}$ in $A_{5}$.

If somebody could help me out with some reasoning or hints to understand this, I would appreciate it a lot.

It's important to say that this is my first ever algebra course, so it could be nice if the help would be kept simple and understandable for beginners.

Thanks in advance

答:
Evaluating $(12)(13)$ from right to left gives $(132)$, which has order three. By Lagrange's Theorem, then, the order of the smallest subgroup $H$ must be divisible by three. But, indeed, if we take $H=\langle (132)\rangle$ to be the (cyclic) subgroup of $A_5$ generated by $(132)$, we achieve the lowest such order.

You were right, too, in a sense, since $H$ is isomorphic (but not equal) to $\Bbb Z_3$ trivially, which, in turn, is isomorphic (but not equal) to $A_3$ by rotating the vertices of an equilateral triangle.


问:
Show that the boundary of each of the sets is contained in $b(A)\cup b(B)$ ..

Let $A$, $B$ be subsets of $\mathbb{R}^n$. Show that the boundary of each of the sets
  $$A\cap B,    \quad    A\setminus B,   \quad    A\cup B$$
  is contained in $b(A)\cup b(B)$. Hint: $b(A)=A^{-}\cap (\delta(A))^{-}$.

Case: $A\cap B$

Since $(A∩B)^{-}\subseteq A^{-}\cap B^{-}$ , it follows that
$$\begin{align*}
b(A\cap B) &=         (A\cap B)^{-} \cap (\delta(A\cap B))^-\\\\
           &\subseteq A^{-}\cap B^{-} \cap  (\delta(A)\cup \delta(B))^{-} \\\\
           &=         A^{-} \cap B^{-} \cap  (\delta(A)^{-}\cup \delta(B)^{-})\\\\
           &=         (B^{-}\cap b(A))\cup (A^{-}\cap b(B))\\\\
           &\subseteq b(A)\cup b(B).
\end{align*}$$

How can I prove case $A\setminus B$ and $A\cup B$?

答:
$A\setminus B$ is $A\cap \delta(B)$, and a set has the same boundary as its complement, so you can recycle your first result.

Similarly, $A\cup B\setminus B= \delta(\delta(A)\cap\delta(B))$, so you can reduce your workload and reuse your second result, too.


问:
Intuition for spherical coordinates of a highly symmetric point
Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought "oh, that point lies on the "dead center" of the sphere as far the first octant is concerned", so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions:
$$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$
$$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$
$$z(r,\theta)=r\cos(\theta)$$
I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?

答:
Direction cosines

$$l^2 + m^2+n^2= \cos^2 \alpha+ \cos^2 \beta+ \cos^2 \gamma=1; \text {if } l=m=n, \text{then }  l=m=n=\frac{1}{\sqrt 3}$$

The vector makes this angle to the coordinate axes including the $z$ axis. The parallel circle angle

$$ \theta=\pi/2 - \gamma = \pi/2 - \cos^{-1} \frac{1}{\sqrt3 }=\sin^{-1} \frac{1}{\sqrt3 }\approx 35.264^{\circ}; $$

Due to equal inclination to $x-, y-$ axes and projection on $x,y$ plane directly the azimuth (polar coordinate) angle $ \varphi= \dfrac{\pi}{4}.$


问:
Example of an embedding $F$ with a given property
Suppose $F:M\rightarrow N$ is an injective immersion between smooth manifolds. Is there an embedding $F$ such that there is a compact subspace $K$ of $N$ with $F^{-1}(K)$ is not compact?

答: Hint: Take $M, N$ both equal to the open interval $(0, 1)$ and try to find a counter-example.


问:
Unknown sequence with some values known
I know that $n$ and $m$ are independant of eachother, and $f(n,m)$ is symmetrical. I can see different sequences within it, but is there a definite function $f: \mathbb{N}^2 \to \mathbb{N}$ for all natural numbers, or does it have to be a composite function?

$$
\begin{array}{c|lcr}
n,m & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &\cdots \\
2 &   & 6 & 10 & 15 & 21 & 28 & 36     \\
3 &   &   & 20 & 35 & 56 & 84 & 120     \\
4 &   &   &    & 70 & 126 & 210 & 330&\\
5 &   &   &    &    & 252 & 462  & 792  &\\
6 &   &   &    &    &     & 924  & 1716  & \\
\end{array}
$$

答:
It looks like when $m \geq n$, the table entries are given by
$$
f(n,m) = \binom{m+1}{n},
$$
where $\binom nk$ denotes a binomial coefficient.

Title: Unknown sequence with some values known


问:
Why can two non-overlapping circles intersect in at most two points, while two non-overlapping ellipses can intersect at four?
When reading about why no Venn diagram for four sets can be formed by intersecting four circles, I found that the author claimed that any two distinct circles can intersect in at most two points, while any two distinct ellipses can intersect in at most four.

Why is this?  I can easily see from examples that it's intuitively obvious, but is there a geometric reason for it?

Thanks!

答:
It is because circles have a constant diameter.
What that means is that if you have two circles intersecting, they cannot cross each other completely without fully overlapping.
Whereas when using ellipses, there is a minor axis and a major axis. Which means that one side is shorter than the other, which means that one can completely cross the other. This produces 4 points of intersection, 2 on either side.</p>


问:
If $X^\ast $ is separable $\Longrightarrow$ $S_{X^\ast}$ is also separable
Let $X$ be a Banach space such that $X$* (Dual space of $X$) is separable

How can we prove that $S_{X^\ast}$ (Unit sphere of $X$*) is also separable

Any hints would be appreciated.

答:
Note: here we consider the question for $X^*$, and $S_{X^*}$, equipped with the topology induced by the norm. See this thread for the same question with the weak* topology.

This is a consequence of a much more general fact which has nothing to do with duals and Banach spaces.

Fact: if $(X,d)$ is a separable metric space, then for any subset $Y\subseteq X$, the subspace $(Y,d)$ is separable.

Note: as pointed out by Nate Eldredge, another approach is to observe that a subspace of a second countable space is obviously second countable. So it only remains to use or prove that a metric space is separable if and only it is second countable.

Proof: let $S=\{x_n\,;\,n\geq 1\}$ a dense subset of $X$. For every $n\geq 1$, consider the distance $d(x_n,Y)=\inf\{d(x_n,y)\,;\,y\in Y\}$. By definition of an infimum=greatest lower bound, for every $k\geq 1$, $d(x_n,Y)+\frac{1}{k}$ is not a lower bound of $\{d(x_n,y)\,;\,y\in Y\}$.  So there exists $y_n^k\in Y$ such that $d(x_n,y_n^k)< d(x_n,Y)+\frac{1}{k}$.

Since $T=\{y_n^k\,;\,n\geq 1, k\geq 1\}$ is a countable union of countable sets, it is countable. We will now show that it is dense in $Y$.

Take $y\in Y$ and $\epsilon>0$. By density of $S$ in $X$, there exists $n\geq 1$ such that $d(x_n,y)<\frac{\epsilon}{2}$. In particular, $d(x_n,Y)\leq d(x_n,y)<\frac{\epsilon}{2}$. So for $k$ large enough, precisely: $k>\left(\frac{\epsilon}{2}-d(x_n,Y)\right)^{-1}$, we have $d(x_n,Y)+\frac{1}{k}<\frac{\epsilon}{2}$. Then for this $n$ and such a $k$, we get
$$
d(y_n^k,y)\leq d(y_n^k,x_n)+d(x_n,y)<d(x_n,Y)+\frac{1}{k}+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
$$
So there are elements of $T$ arbitrarily close to $y$ for every $y\in Y$. That is $T$ is dense in $Y$. QED.

Remark: what really is about Banach spaces $X$ such that $X^*$ is separable is that $X$ is separable. And it follows that the unit sphere of $X$, $S_X$, is separable as well.


问:
Does entrywise bounded positive definite matrix also have bounded inverse?
Suppose $A(x)\colon \mathbb{R}\to\mathbb{R}^{n\times n}$ is a matrix-valued function.

Assume that for all $x\in\mathbb{R}$, $A(x)$ is positive definite, and $A(x)$ is uniformly entrywise bounded (i.e., absolute value of entry) by a constant $C$ independent of $x$.

Is it true that the matrix inversion $(A(x))^{-1}$ is also uniformly entrywise bounded? If not, do we have any counter-example?

答:
It is not true even in the case that $n=1$. Consider $A(x) = e^{-|x|}$.

The absolute values of the entries of $A^{-1}$ are bounded above by $1/\sigma_\min(A)$. If $A$ is positive definite and symmetric, then its singular values are equal to its eigenvalues.

To see that this is the case, note that $\sigma_\max(A^{-1}) = 1/\sigma_\min(A)$, and for all matrix $M$ and vectors $v$ we have $\|Mv\| \leq \sigma_\max(M) \cdot \|v\|$. If we use $e_j$ to denote the $j$th standard unit vector, then we have
$$
|A^{-1}_{ij}| = |e_i^T A^{-1}e_j| \leq \|e_i\| \cdot \|A^{-1}e_j\| \\
\leq \|e_i\| \cdot \sigma_\max(A^{-1}) \cdot \|e_j\|\\
= 1/\sigma_\min(A).
$$


问:
Show that $G$ satisfies Hall's condition

Show that $G$ satisfies Hall's condition

I knew the definition:

Let $G$ be a bipartite graph with partite sets $U$ and $W$, where $r=|U|\leq|W|$. Then $G$ satisfies Hall's condition if, for every $S\subseteq U$, $|S|\leq|N(S)|$.

But did I need to check it for all subsets manually to answer that question? Because I easily found a perfect match. And Hall condition implies the existence of a perfect match, but I don't know the reverse is true or not? Then how to deal with this question without huge amount of checking?

答:
And Hall's condition implies the existence of a perfect match, but I don't know the reverse is true or not?

Suppose that you found a perfect match but that Hall's condition were not satisfied. Then there would exist some $S \subseteq U$ for which $|S| > |N(S)|$. It would be impossible to match all the elements of $S$, a contradiction.

In other words, if you found a perfect match, then you are done.


问:
How is G derived from this formula?
I'm a bit stumped. I've stumbled upon these formulas for converting $RGB$ to $YC_BC_R$ color space...:

$$Y=a*R+b*G+c*B$$
$$C_B=\frac{B-Y}d$$
$$C_R=\frac{R-Y}e$$

...and back again...:

$$R =Y+e*C_R$$
$$G =Y-\Bigl(\frac{a*e}b\Bigr)*C_R-\Bigl(\frac{c*d}b\Bigr)*C_B$$
$$B =Y+d*C_B$$

...where $a$, $b$, $c$, $d$ and $e$ are constants. While I see how to get $C_R$ from the formula for $R$ and, similarly, $C_B$ from $B$, I don't really know how to rearrange formula for $Y$ so that it gives $G$ precisely as in the formula above.

Here's what I have so far:

$$Y=aR+bG+cB$$

I subtract $bG$ from both sides and replace $R$ and $B$ with $C_R$ and $C_B$...

$$Y-bG=aC_R+cC_B$$

...then I subtract $Y$...:

$$-bG=(aC_R+cC_B)-Y$$
...followed by dividing by $-b$...:

$$G=\frac{-Y+(aC_R+cC_B)}{-b}$$

...and I'm not really sure what to do next.

I probably shouldn't be playing around with such things while being bad at math, but my curiosity is killing me, and I'm more than willing to make a fool of myself if it means I'll get a few tips on how to proceed from here. Thank you in advance!

答:
Right, so step by step:

$$Y=aR+bG+cB\quad$$

subtract $bG$ from both sides and replace $R$ with $Y+e*C_R$ and $B$ with $Y+d*D_B$...

$$Y-bG=a*(Y+e*C_R)+c*(Y+d*C_B)$$

...expand the parentheses...

$$Y-bG=aY+aeC_R+cY+cdC_B$$

...subtract $Y$ from both sides...

$$-bG=(aY+aeC_R+cY+cdC_B)-Y$$

...then divide by $-b$...

$$G=\frac{(aY+aeC_R+cY+cdC_B)-Y}{-b}$$

...now we can rearrange stuff a bit...

$$G=\frac{-Y+aY+cY}{-b}+\frac{a*e}{-b}*C_R+\frac{c*d}{-b}*C_B$$

...now let's simplify the first term, keeping in mind that $a+b+c=1$. That means that $-b=-1+a+c$

$$G=\frac{-Y+aY+cY}{-1+a+c}+\frac{a*e}{-b}*C_R+\frac{c*d}{-b}*C_B$$

Let's factor out $Y$ in the numerator...

$$G=\frac{Y(-1+a+c)}{-1+a+c}+\frac{a*e}{-b}*C_R+\frac{c*d}{-b}*C_B$$

...which, after cancelling the common factor leaves us with:

$$G=Y+\frac{a*e}{-b}*C_R+\frac{c*d}{-b}*C_B$$

...and we're almost home, we just need to subtract instead of adding:

$$G=Y-\Bigl(\frac{a*e}{b}\Bigr)*C_R-\Bigl(\frac{c*d}{b}\Bigr)*C_B$$

...which takes us to the final form of the formula for calculating the green channel.

Thanks again for all the help @octave!

Title: How is G derived from this formula?


问:
Why the $\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$ is not equal to $0$?
I saw this question on the site. It asked to evaluate:

$$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x.$$

Although the answer is $e^{\tfrac{-2}{\pi}}$, I don't completely understand why the limit is not equal to zero. I think it should be zero because as $x\to\frac{\pi}2^-$, $\tan x\to+\infty$. Hence we have the limit of $(\frac2{\pi}\times{(\frac{\pi}2}^-))^{\infty}$. and it should be zero (for example the value of $0.99999^{1000000}$ is very close to zero.

答:
Note that you can not just plug the limit it, since the form you get is indeterminate: $1^{\infty}$. To show you another (very famous) example why this does not work:

Consider
$$\lim_{n\to\infty}(1+\frac{1}{n})^n.$$
You may recall that this limit is the defintion of $e\approx2.7$. But if we were to plug in the limit immediatly we'd get
$$(1+0)^\infty.$$
And $e\neq0$ (nor $1$ or $\infty$).

Same thing for
$$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x.$$
We also have $1^\infty$, but as you said correctly it's $e^{-2/\pi}$.

Addendum: This post might be helpful.


问:
Is there any other approach available without trig substitution?

Let $f\colon [0, 1] \to \mathbb{R}$ be a continuous function such that for any $x, y \in [0, 1]$, $xf(y) + yf(x) \le  1$. Find
maximum value of $\int_0^1f(x)\,dx$.

Method given was using $x= \sin\theta$, we get $$\int_0^1 f(x)\, dx = \int_0^{\pi/2} f(\sin\theta)\cos\theta\, d\theta \tag{I}$$and similarly by $x= \cos\theta$, we would get $$\int_0^1 f(x)\,dx = \int_0^{\pi/2} f(\cos\theta)\sin\theta \,d\theta \tag{II},$$ adding $(\text{I})$ and $(\text{II})$ we get max value of the required integral to be $\pi/4$. Any other approach because the idea is very tricky and one may not realize for this type of substitution.

答:
I don't have an alternative approach to suggest. The first thing that occurred to me, which was to integrate over the two-dimensional region $[0,1]^2$, doesn't work:
\begin{align}
1=\int_0^1\int_0^1 1\,dx\,dy\ge\int_0^1\int_0^1(yf(x)+xf(y))\,dx\,dy&=2\int_0^1y\,dy\int_0^1 f(x)\,dx\\
&=\left[y^2\right]_0^1\int_0^1f(x)\,dx\\
&=\int_0^1f(x)\,dx.
\end{align}
My upper bound is not as tight as yours: $1>\pi/4\approx0.785$.

What you have done is to cleverly bound the integral by restricting the integration to a single contour, namely the arc of the unit circle that lies in the first quadrant. Your bound is, in fact, the best possible, as the example $f(x)=\sqrt{1-x^2}$ shows. First confirm that your maximum is attained:
\begin{align}
\int_0^1\sqrt{1-x^2}\,dx&=\int_0^{\pi/2}\sqrt{1-\sin^2\theta}\cos\theta\,d\theta\\
&=\int_0^{\pi/2}\left(\frac{1}{2}\cos2\theta+\frac{1}{2}\right)\,d\theta\\
&=\frac{\pi}{4}.
\end{align}
To verify that $f$ satisfies the required condition, observe that the function
$$
g(x,y)=y\sqrt{1-x^2}+x\sqrt{1-y^2}
$$
achieves a maximum of $1$ for $(x,y)\in[0,1]^2$ all along your contour. You can check this by observing that the equations
\begin{align}
0&=\frac{\partial}{\partial x}\left[
y\sqrt{1-x^2}+x\sqrt{1-y^2}\right]=\sqrt{1-y^2}-\frac{xy}{\sqrt{1-x^2}}\\
0&=\frac{\partial}{\partial y}\left[
y\sqrt{1-x^2}+x\sqrt{1-y^2}\right]=\sqrt{1-x^2}-\frac{xy}{\sqrt{1-y^2}}
\end{align}
are satisfied when $x^2+y^2=1$. To see that it is a maximum that is achieved at this locus of critical points, switch to polar coordinates,
$$
h(r,\theta)=g(r\cos\theta,r\sin\theta),
$$
and compute
\begin{align}
\frac{\partial}{\partial r}h(r,\theta)=&\left(\sin\theta\sqrt{1-r^2\cos^2\theta}+\cos\theta\sqrt{1-r^2\sin^2\theta}\right)\\
&\times\left(1-\frac{r^2\sin\theta\cos\theta}{\sqrt{1-r^2\cos^2\theta}\sqrt{1-r^2\sin^2\theta}}\right),\\
\frac{\partial}{\partial\theta}h(r,\theta)=&r\cos\theta\sqrt{1-r^2\cos^2\theta}+\frac{r^3\sin^2\theta\cos\theta}{\sqrt{1-r^2\cos^2\theta}}\\
&-r\sin\theta\sqrt{1-r^2\sin^2\theta}-\frac{r^3\cos^2\theta\sin\theta}{\sqrt{1-r^2\sin^2\theta}}.
\end{align}
One can check that, for fixed $\theta$,  $\frac{\partial h(r,\theta)}{\partial r}$ is positive when $r<1$, zero when $r=1$, and negative when $r>1$, while, for $r$ fixed to $1$, $\frac{\partial h(r,\theta)}{\partial\theta}$ is zero, as  expected.


问:
Determining the constant $k$ of a density function
The diameter of a grain of sand in a sand roof, measured in mm, can be considered as a random variable X with probability density

$f(x) = k(x-x^2)$ for $0\leq x \leq 1$

$f(x) = 0$ otherwise

Decide the constant $k$ such that this becomes a legitimate probability density.

For $f(x)$ to be legitimate, it must satisfy the following:

$f(x)\geq 0$ $\forall$ $X$

$\int_{-\infty}^{\infty} f(x)dx$ = [the area under the graph of $f$]=1

Now, we can see that $f(x) \geq 0$ for all $k \geq 1$ because the numbers in the given interval $[0,1]$ yields a positive number for $(x-x^2)$ for all $x$. But how do I prove that $f(x)$satisfies the second demand?

答:
We can see that $\int_{-\infty}^{\infty}f(x)dx$ = $\int_{0}^1 k(x-x^2)dx $ (we are restricting the integral to the given domain)

We then get:

$=k\big[\frac{x^2}{2}-\frac{x^3}{3}\big]_0^1=k\big[\frac{1}{2}-\frac{1}{3}\big]=1\Rightarrow k=6.$


答:
The statement

As $x \to \frac{\pi}{2}^-$, $\arctan x \to +\infty$

is incorrect.  You are confusing $\tan x$ and $\arctan x$.

Second, your suggestion that the indeterminate form $1^\infty \to 0$ is also incorrect.  For instance,

$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e,$$ in which $1 + 1/x \to 1$ as $x \to \infty$.  However,

$$\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x = 1,$$ yet $1 + 1/x^2 \to 1$ also as $x \to \infty$.  And

$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x^2} = \infty.$$  All of these are indeterminate forms of type $1^\infty$ yet they have different limits, none of which are zero.

Title: Why the $\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$ is not equal to $0$?

Body:
I saw this question on the site. It asked to evaluate:

$$\lim_{x\to \infty}\left(\frac{2\arctan(x)}{\pi}\right)^x.$$

Although the answer is $e^{\tfrac{-2}{\pi}}$, I don't completely understand why the limit is not equal to zero. I think it should be zero because as $x\to\frac{\pi}2^-$, $\tan x\to+\infty$. Hence we have the limit of $(\frac2{\pi}\times{(\frac{\pi}2}^-))^{\infty}$. and it should be zero (for example the value of $0.99999^{1000000}$ is very close to zero.


答:
The result highly depends on the "speed" with which $\frac{2}{\pi} \arctan x$ tends to $1$ as $x$ tends to infinity . You cannot just come up with an example like $0.99999^{1000000}$ and expect that the value has any connection with the value of the limit... If that was the case, there would be no indeterminations: every limit of the type $1^{\infty}$ would be zero.

The appropriate approximation here would be

$$
0.9999993633802277^{1000000}\approx 0.529078 \approx e^{-2/\pi}
$$


答:
Write

$$\left(\frac{2\arctan x}\pi\right)^x=e^{x\log\frac{2\arctan x}x}$$

and now L'hospital

$$\lim_{x\to\infty}\frac{\log\frac{2\arctan x}\pi}{\frac1x}\stackrel{\text{L'H}\frac00}=\lim_{x\to\infty}\frac{\frac1\pi\frac2{1+x^2}\frac\pi{2\arctan x}}{-\frac1{x^2}}=\lim_{x\to\infty}-\frac{ x^2}{1+x^2}\cdot\frac1{\arctan x}=-1\cdot\frac1{\frac\pi2}=-\frac2\pi$$

and by continutiyt of the exponential function, the limit is $\;e^{-2/\pi}\;$


问:
If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?
If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?

I attempted to solve this question as follows:

$\cos{2x}=\cos^2{x}-\sin^2{x}$

$\implies \cos{x}(\cos^2{x}-\sin^2{x})=\frac{1}{4}$

$\cos^3{x}-\sin{x}(\cos{x}\sin{x})$

And I got stuck here, I did not know how to continue. I plugged in a few values for $x$ and worked out that $x=36^o$ is the solution. Could you please help me solve this question?

答:
Since $x \neq 0$, multiplying both sides by $4\sin x$ and using double angle formula, $\sin 2\theta = 2\sin \theta \cos \theta \,$ twice, it is obtained
$$\sin 4x = \sin x$$

whence $$4x+x=\pi \Rightarrow x=\pi/5=36^{\circ}$$


答: $\cos x\cdot \cos(2x) = \dfrac{1}{4} \implies 4\cos x\cdot (2\cos^2 x - 1) =1\implies 8\cos^3x-4\cos x-1=0$. Observe that $\cos x = -\frac{1}{2}$ is a solution. Can you finish it with factoring ?

Body:
If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?

I attempted to solve this question as follows:

$\cos{2x}=\cos^2{x}-\sin^2{x}$

$\implies \cos{x}(\cos^2{x}-\sin^2{x})=\frac{1}{4}$

$\cos^3{x}-\sin{x}(\cos{x}\sin{x})$

And I got stuck here, I did not know how to continue. I plugged in a few values for $x$ and worked out that $x=36^o$ is the solution. Could you please help me solve this question?


问:
Convergence of a sequence of norm
Let be (X,$||.||$) a normed space and $\{x_n\}$ a Cauchy sequence. Then $\{||x_n||\}$ is convergent.

My attemp: if $\{x_n\}$ is Cauchy then $\forall \epsilon >0$ exists $N>0: \forall n,m\ge N \quad||x_n-x_m||<\epsilon$.

Then $\big| ||x_n||-||x_m||\big |\le ||x_n-x_m||<\epsilon \quad \forall n,m\ge N$ so $\{||x_n||\}$ is Cauchy sequence.

But how can I prove is convergent?

答: You’re almost there! $\left\|x\right\|$ is a Cauchy sequence in $\mathbb{R}$. And since the reals are a complete metric space them every Cauchy sequence converges.

Title: Convergence of a sequence of norm


问:
Average of two i.i.d. random variables and their conditional expectations.
I would like to see some indications on how to approach this: Let $\xi$ and $\eta$ be independent, identically distributed random variables, with $E\xi$ defined (that is, $\xi$ is semi-integrable). Show that $E(\xi | \xi + \eta)=E(\eta | \xi + \eta ) = (\xi + \eta) / 2$ almost surely.</p>

答:
This seems intuitive. The first equality $\mathbb{E}[\xi|\xi + \eta] = \mathbb{E}[\eta|\xi + \eta]$ follows by symmetry in $\xi$ and $\eta$. Let $x = \mathbb{E}[\xi|\xi + \eta]$.

Note that

$$
2x = x + x = \mathbb{E}[\xi|\xi + \eta] + \mathbb{E}[\eta|\xi + \eta]
   = \mathbb{E}[\xi + \eta|\xi + \eta] = \xi + \eta.
$$
Therefore,
$$x = \frac{\xi+\eta}{2}$$
as desired.


问:
Prove that define topology in ${\mathbb{R}}^2$
I was studying topology and I have problems with this exercise

On the plane ${\mathbb{R}}^2$ consider the family $\tau$ consisting of the $\emptyset$ , ${\mathbb{R}}^2$ and all open discs $\{ x^2 + y^2 < r^2 \}, r > 0$. Prove that $\tau$ defines a topology and determine
the closure of the hyperbola $xy = 1$.

$A1$ For a definition $\emptyset$ and  ${\mathbb{R}}^2$ are in $\tau$.

$A2$ ${\mathbb{R}}^2 \cup$ all open discs $\{ x^2 + y^2 < r^2 \}, r > 0$ it results ${\mathbb{R}}^2$ and ${\mathbb{R}}^2 \in \tau$

$A3$ ${\mathbb{R}}^2 \cap $ all open discs $\{ x^2 + y^2 < r^2 \}, r > 0$ it results all open discs $\{ x^2 + y^2 < r^2 \}, r > 0$ and $\{ x^2 + y^2 < r^2 \}, r > 0 \in \tau$

I have problems determining  the closure of the hyperbola

Thanks

答:
The proof that $\tau$ is a topology is wrong. You were supposed to prove that an union of an arbitrary set of elements of $\tau$ belongs to $\tau$ and that the intersection of finitely many elements of $\tau$ belongs to $\tau$ too.

You know that the closed subsets of $(\Bbb R^2,\tau)$, other than $\emptyset$ and $\Bbb R^2$, are the sets of the form $\{(x,y)\in\Bbb R^2\mid x^2+y^2\geqslant r^2\}$. The smallest of these sets which contains the hyperbola is $\{(x,y)\in\Bbb R^2\mid x^2+y^2\geqslant2\}$. So, this is the closure of the hyperbola.


问:
How do I compute Lebesgue integrals of non-simple functions?
I know this is not the purpose of Lebesgue integration, but it would still be nice to know how to do this. For example, how do I compute the integral

$$\int_{[0,1]} \sin(x)\,\mu(\mathrm dx).$$

I know how to compute it the Riemann way—I just want to get practice on Lebesgue integration.
I know I have to make $\sin x$ a pointwise limit of simple functions which would be $$\lim_{n \rightarrow \infty} \frac{\left \lfloor{10^n\cdot\sin x}\right \rfloor}{10^n}.$$

答:
I take it that $\mu$ here is the Lebesgue measure on $\mathbb{R}$. Assuming such, you actually calculate your integral exactly how you'd calculate a Riemann integral and there equal. Reason being, what you have here is the Lebesgue integral of a bounded measurable function over a set of finite measure. As such, there is the following theorem,

Theorem: Let $f$ be a bounded function defined on the closed, bounded interval $[a, b]$. If $f$ is Riemann integrable over $[a, b]$, then it is Lebesgue integrable over $[a, b]$ and the two integrals are equal.


答:
$\cos x(2\cos^2x-1)=\frac 14=(-\frac 12)^2$

$\rightarrow\cos x=-\frac 12=\cos (\pi-\frac{\pi}3) $

$\rightarrow x=(2k+1)\pi-\frac{\pi}3$

Title: If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?


问:
How to calculate intermediate point?
I am developing an Android application, and I am trying to develop a chart with points, but I am struggling with some (probably) basic math.

So, the point is: I am drawing a price chart, and I know that the maximum price, being 50000€ is drawn at the 0px. At the other end, the minimum price, being 20000€ is drawn at the 800px. Now, let's suppose that I got a price between the minimum and maximum like 27500€. How can I calculate the px for that price knowing the information I gave earlier?

If someone could help me with this calculation, or at least point me in the right direction, I would really appreciate it!

答:
Begin by seeing you have 800 px for $(50,000-20,000) = 30,000  units$ so you have $\frac{800}{30000} \frac {p}{unit}$. Since 50,000 is your zero mark, it simply becomes $px = \frac{800}{30000}*(50,000-Price) $

So for Price = 27500, you have $px = \frac{8}{300}*(50,000-27,500) = 600px$

Title: How to calculate intermediate point?


问:
Prove that $G$ is an abelian group if $\{(g, g):g\in G\}$ is a normal subgroup.
Let $G$ be a group and let $D=\{(g, g):g\in G\}$. If $D$ is a normal subgroup of $G\times G$, prove that $G$ is an abelian group.

My attempt:

$D$ is a normal subgroup of $G\times G$.

$\implies(a, b)D=D(a, b)\ \forall(a, b)\in G\times G$

So for a given $(a, b)\in G$ and $(g, g)\in D$, $\exists(g', g')\in D$ such that

$(a, b)(g, g)=(g', g')(a, b)$

$(ag, bg)=(g'a, g'b)$

$ag=g'a$ and $bg=g'b$

I feel like I've used all of the information given but don't know how to conclude that $G$ is abelian. Any suggestions?

答:
Since D is normal in GxG
So using the property of cosets and for (g,g) in GxG
We have
=> D(g,g)=(g,g)D .
Let (d,d) in D ve arbitray where d is in G .
=>(d,d)(g,g)=(g,g)(d,d)
=> (dg,dg)=(gd,gd) .
Now comparing corresponding entries we have
dg=gd and  g,d both are in G .
Hence G is abelian</p>


问:
Bounding the uniform deviation of the empirical risk from the risk over a finite function class.
I am having difficulty interpreting the following theorem from here as a probability statement:

Theorem.

For all $\delta$ such that $0 < \delta < 1/2$, with proability at least $1 - \delta$ the following is true
$$\lvert \hat{R}_n(f) - {R}(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \quad \forall f \in \mathcal{F} \tag{4.3}$$

Where $\hat{R}_n(f) = n^{-1} \sum^n_{i=1} \mathbb{I}(Y_i \neq f(X_i))$ is the empirical risk under a 0/1 loss function from $n$ training data points $(X_i, Y_i)$ drawn from an unknown distribution $p^*$, and $R(f) = \mathbb{E}_{(X, Y) \sim p^*}[\mathbb{I}(Y \neq f(X))]$ is the risk. $f$ is a linear classifier in a function class $\mathcal{F}$ that is finite i.e. $\lvert \mathcal{F} \rvert < \infty$.

Query.

If I were to rewrite this as a probability statement, would the qualifier $\forall f \in \mathcal{F}$ go inside or outside the probability statement?

That is, should the above theorem read:

$$\mathbb{P} \left( \forall f \in \mathcal{F} \lvert \hat{R}_n(f) - R(f) \rvert   \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \tag{A}$$

Or should it read:

$$\forall f \in \mathcal{F} , \mathbb{P} \left( \lvert \hat{R}_n(f) - R(f) \rvert   \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \tag{B}$$

Further context - proof of the theorem.

I will now detail each step of the proof, which is fairly standard in the statistics and statistical learning theory literature, together with my understanding.

Step 1 - Bound the deviation of the empirical risk from the risk for a single function $f \in \mathcal{F}$ using a concentration inequality.

For a single function $f \in \mathcal{F}$, as the empirical risk $\hat{R}(f)$ is the sample average of i.i.d. Bernoulli rvs, and the risk $R(f)$ is the theoretical expectation, we have that

$$\mathbb{P} \left( \lvert \hat{R}(f) - R(f) \rvert \geq \epsilon \right) \leq 2\exp(-2n\epsilon^2)$$

Step 2 - Bound the uniform deviation using the union bound.

My understanding is that we require the empirical risk to be a good estimate of the risk in the sense of being 'uniformly close' to the risk over the entire function class $\mathcal{F}$ . And that Hoeffding's inequality by itself is insufficient, because for a particular $f \in \mathcal{F}$, there will be a set of training samples $S_f$ such that $\hat{R}(f) - R(f) \leq \epsilon$ with probability $\mathbb{P}(S_f) \geq 1 - 2\exp(-2n\epsilon^2)$, and the crux is that in general, this set $S_f$ will be different for different functions in $\mathcal{F}$.

We define the event $E_f = \{\lvert \hat{R}(f) - R(f) \rvert \geq \epsilon\}$ and as the function class $\mathcal{F}$ is finite, apply the union bound over $E_f$:

$$\begin{align}
\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon \right) &= \mathbb{P}\left( \bigcup_{f \in \mathcal{F}} \{ \lvert \hat{R}_n(f) - R(f) \rvert  \geq \epsilon\}  \right) \\
&\leq \sum_{f \in \mathcal{F}} \mathbb{P} \left(\lvert  \hat{R}_n(f) - R(f) \rvert \geq \epsilon\right) \\
&\leq \sum_{f \in \mathcal{F}} 2 \exp(-2n \epsilon^2) \\
&= 2 \lvert \mathcal{F} \rvert \exp(-2n \epsilon^2)
\end{align}$$

Rewriting in terms of $\delta$, that is taking $\delta := 2 \lvert \mathcal{F} \rvert \exp(-2n \epsilon^2)$ and rearranging for $\epsilon$ to get

$$\epsilon = \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}$$

Yields

$$\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta$$

Up to here, I am comfortable with the proof steps - and I read the above line as, "the probability that the maximum deviation between the empirical risk and risk over the entire function class $\mathcal{F}$ lies in $[-C_{\delta}, C_{\delta}]$ is at least $1 - \delta$.

My confusion.

However, where it gets unfamiliar is that the authors state that this proves $(4.3)$, but I am having difficulty trying to figure out whether this means that

$$\begin{align}
&\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta \\
& \implies \mathbb{P} \left( \forall f \in \mathcal{F} \lvert \hat{R}_n(f) - R(f) \rvert   \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta
\end{align}$$

or whether it means that

$$\begin{align}
&\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta \\
& \implies \forall f \in \mathcal{F} , \mathbb{P} \left( \lvert \hat{R}_n(f) - R(f) \rvert   \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \\
\end{align}$$

答:
The statement the authors intended is (A) for the following reasons.

Addressing the confusion.

Much of your reluctance to consider $(A)$ as the correct statement amounted to insufficient attentiveness concerning the following basic distinction between:

$$\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon \right) = \mathbb{P} \left( \bigcup_{f \in \mathcal{F}} \{\lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon \} \right)$$

and

$$\begin{align}
\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \epsilon \right) &= \mathbb{P} \left( \bigcap_{f \in \mathcal{F}} \{ \lvert \hat{R}_n(f) - R(f) \rvert \leq \epsilon \} \right) \\
&= \mathbb{P} \left( \forall f \in \mathcal{F}, \space \lvert \hat{R}_n(f) - R(f) \rvert \leq \epsilon  \right)
\end{align}$$

Consider the events within each statement.

In the first statement, the event that the "maximum absolute deviation between the empirical risk and the risk over the function class $\mathcal{F}$ exceeds $\epsilon$" is the same as the compound event that "there is at least one function $f$ whose absolute deviation between the empirical risk and risk exceeds $\epsilon$".

In the second statement, the event that the "maximum absolute deviation between the empirical risk and the risk over the function class $\mathcal{F}$ does not exceed $\epsilon$" is the same as the compound event that the "absolute deviation between the empirical risk and the risk does not exceed $\epsilon$ for all functions $f \in \mathcal{F}$.

Difference between statements $(A)$ and $(B)$.

The reason why the statement the authors intended is not statement $(B)$ goes back to the argument about the insufficiency of concentration inequalities for understanding generalisation error, and why we are instead interested in bounding uniform deviations. Statement $(B)$ is of the same flavour as a concentration inequality applied to every $f \in \mathcal{F}$, except it is much looser than the one derived using Hoeffding in step 1, due to the inclusion of the $\ln \lvert \mathcal{F} \rvert$ term:

$$ \space \mathbb{P} \left( \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta, \quad \forall f \in \mathcal{F}$$

Setting the looseness of the bound aside, this is qualitatively distinct from statement $(A)$.

Statement $(B)$ is saying that for every single function $f \in \mathcal{F}$, there exists a set of training samples associated with that function $f$, which we define as $S_f$, such that $\hat{R}_n(f) - R(f) \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}$, and that the probability of encountering this sample exceeds $1 - \delta$, that is, $P(S_f) \geq 1 - \delta$.

Even though for all $f \in \mathcal{F}$, the probability $P(S_f) \geq 1 -\delta$, this does not change the fact that in general, each training sample $S_f$ for which $\hat{R}_n(f) - R(f) \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}$ holds is different for different functions $f \in \mathcal{F}$. Hence what is required is something stronger than this.

What instead is required is the probability of a set of training samples $S$ for which every absolute deviation $\lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}$. Only statement $(A)$ captures this sense.

Key take away: The position of the qualifier $\forall f \in \mathcal{F}$ is of the utmost importance.

Other sources.

Statements equivalent to $(A)$, though not written explicitly as probability statements, and with some mild idiosyncratic differences in notation, can be found in:

Anthony, M., & Bartlett, P. (1999). Neural Network Learning: Theoretical Foundations. Cambridge: Cambridge University Press. p22 doi:10.1017/CBO9780511624216

Bousquet O., Boucheron S., Lugosi G. (2004) Introduction to Statistical Learning Theory. In: Bousquet O., von Luxburg U., Rätsch G. (eds) Advanced Lectures on Machine Learning. ML 2003. Lecture Notes in Computer Science, vol 3176. Springer, Berlin, Heidelberg. p179 https://doi.org/10.1007/978-3-540-28650-9_8


问:
A question on mean value inequality
It is known that mean value inequality is very useful. It is:

For any $0 \le a_i (i=1,2,\dots,n)$,
  $$
a_1 a_2\dots a_n\le (\frac{a_1+a_2+\dots + a_n}{n})^n \tag1
$$

My question is: how many ways by which the mean value inequality can be proved?

Thanks for your help.

答:
One method, which is often employed to prove AM-GM, is Cauchy induction.
Such proof is sketched in this answer, this answer, this answer, Wikipedia and in many other places.

More about Cauchy induction:

Cauchy induction at AoPS

Proof of AM-GM using this type of induction at Wikipedia. They call this technique forward-backward-induction.

One section of Pete L. Clark's notes on induction is devoted to this type of induction. He calls it upward-downward induction.

Perhaps also this question

Title: A question on mean value inequality


答:
In general, let the minimum price be $a$ and the maximum price be $b$.  Suppose the minimum price is to be plotted at coordinate $y_a$ and the maximum price plotted at coordinate $y_b$.

So in your case, $a = 20000$, $b = 50000$, $y_a = 800$, and $y_b = 0$.

Then suppose you have a third price, say $c$, satisfying $a < c < b$.  In your case, $c = 27500$.  You want to find the coordinate $y_c$ such that $c$ is plotted to scale.  Then we require $$y_c = \frac{c-a}{b-a} y_b + \frac{b-c}{b-a} y_a.$$  Substituting, you get $y_c = 600$.

Title: How to calculate intermediate point?

Body:
I am developing an Android application, and I am trying to develop a chart with points, but I am struggling with some (probably) basic math.

So, the point is: I am drawing a price chart, and I know that the maximum price, being 50000€ is drawn at the 0px. At the other end, the minimum price, being 20000€ is drawn at the 800px. Now, let's suppose that I got a price between the minimum and maximum like 27500€. How can I calculate the px for that price knowing the information I gave earlier?

If someone could help me with this calculation, or at least point me in the right direction, I would really appreciate it!


问:
Show that if $(x_k)$ and $(y_k)$ are sequences in $V$ such that $x_k \to x$ and $y_k \to y$, then $x_k + y_k \to x + y$.

Let $(V, \|\cdot\|)$ be a normed space. Show that if $(x_k)$ and $(y_k)$ are sequences in $V$ such that $x_k \to x$ and $y_k \to y$, then $x_k + y_k \to x + y$.

Let $U$ be a neighborhood of $x+y$ and $\varepsilon > 0$ s.t $B(x+y, \varepsilon) \subset U$. Since $x_k \to x$ there is a $n_0 > 0$ such that $x_k \in B(x, \varepsilon/2)$, whenever $k \ge n_0$. Similarly there exists a $m_0 > 0$ such that $y_k \in B(y, \varepsilon/2)$ whenever $k \ge m_0$. Now I’m not sure if I should pick $k_0 = \max(n_0, m_0)$ or $k_0 = \min(n_0, m_0)$ to make this work?

I have that $$\|(x_k + y_k) -(x+y) \| = \|(x_k - x) + (y_k - y) \| \le \|x_k - x \| + \|y_k - y \| \le \varepsilon$$ when $k \ge k_0$?

答:
You don't need to use open neighborhoods and stuff like that. Simply observe that
$$\|x_n + y_n -(x+y)\| = \|(x_n-x) + (y_n-y)\| \leq \|x_n-x\| + \|y_n-y\|$$
for all $n$. Letting $n \to \infty$, we obtain
$$\lim_n \|x_n + y_n-(x+y)\|= 0$$
which means that $\lim_n (x_n + y_n) = x+y$.

Nevertheless, your proof is correct if you consider the maximum. Why the maximum, you may ask? Well, if $k \ge k_0$, you have $k \geq n_0$ AND $k \geq m_0$ so both terms become smaller than $\epsilon/2$.  If you would have taken the minimum, you won't be able to garantuee that both terms become smaller than $\epsilon/2$, but rather only one of them.


问:
Non-locally compact subset of $\mathbb{R}^2$ and its compactification

Find a non-locally compact subset $A$ of $\mathbb{R}^2$ which has a compactification $K$ such that $K\setminus A$ is has the discrete topology.

I know that the Alexandroff compactification $X^*$ of $X$ has the property that $X^*\setminus X$ has the discrete topology. But I think this is rather irrelevant.

答:
I checked the definition of locally compact: every point in the set has a compact neighborhood.

Thinking of the case in $\mathbb{R}$, wouldn't the following work?
Let $A=[0,1]\setminus \{1/n : n\in\mathbb{N}\}$. It is not locally compact because every neighborhood of $0$ fails to be closed. The interval $[0,1]$ would be a compactification. But $[0,1]\setminus A=\{1/n : n\in\mathbb{N}\}$.

Then you can construct something very similar in $\mathbb{R}^2$. Work with a box instead of an interval.


问:
Taking the homotopy pullback always results in fibrations?
My question is quite simple, if we have a diagram $X \rightarrow Z \leftarrow Y$ in the category of topological sapces, then when taking the homotopy pullback we get a homotopy commutative diagram \begin{alignat}{9} P&\rightarrow\ &X \\ \downarrow & &\downarrow \\ Y &\rightarrow &Z \end{alignat}

My question is whether $P\rightarrow X$ and $P\rightarrow Y$ are always fibrations.

I think this is true because one equivalent way of obtaining a homotopy pullback is to first take for example $X \rightarrow Z$ and write it as a fibration $E \rightarrow Z$ ($E$ is of the same homotopy type as $X$), and then take the conventional pullback of $E \rightarrow Z \leftarrow Y$ \begin{alignat}{9} P&\rightarrow\ &E \\ \downarrow & &\downarrow \\ Y &\rightarrow &Z \end{alignat} then since $E \rightarrow Z$ is a fibration $P \rightarrow Y$ is also a fibration (common known propety of pullbacks). And one may substitute $Y\rightarrow Z$ by aa fibration, to conclude that $P\rightarrow X$ is a fibration.

So my question is if this implies that any choice of $P$, $P \rightarrow X$ and $P \rightarrow Y$ will be so that $P \rightarrow X$ and $P \rightarrow Y$ are fibrations. If this is not true, is it atleast true for the standard construction? Where one takes $$P=\{(x,\gamma,y)\in X\times Z^I\times Y\ |\ \gamma(0)=x\ \text{and}\ \gamma(1)=y)\} $$ and $P \rightarrow X$ and $P \rightarrow Y$ are simply the projections.

答:
The double mapping path space $P=P_{f,g}$ associated with a cospan
$$X\xrightarrow{f}Z\xleftarrow{g}Y$$
is obtained as the strict pullback of the cospan
$$X\times Y\xrightarrow{f\times g}Z\times Z\xleftarrow{e_{0,1}} Z^I$$
where $Z^I$ is the space of maps $I\rightarrow Z$ in the compact-open topology. Here $e_{0,1}$ is the fibration given by the endpoint evaluation $e_{0,1}(\ell)=(\ell(0),\ell(1))$.

Thus
$$P_{f,g}\cong \{(x,\ell,y)\in X\times Z^I\times Y\mid f(x)=\ell(0),\;g(y)=\ell(1)\}.$$
Since fibration are stable under pullback we obtain a fibration,
$$\pi:P_{f,g}\rightarrow X\times Y,\qquad (x,\ell,y)\mapsto (x,y)$$
as the pullback of $e_{0,1}$. Since a composition of two fibrations is another fibration, we obtain fibrations
$$\pi_X:P_{f,g}\rightarrow X,\qquad \pi_Y:P_{f,g}\rightarrow Y$$
as the composites of $\pi$ with the projections
$$X\leftarrow X\times Y\rightarrow Y.$$

Note that there is nothing to be said about the general case: the square
\begin{array}{ccc}
\ast &\rightarrow & \mathbb{R} \\
\downarrow & &\ \downarrow \\
\mathbb{R} &\rightarrow& \mathbb{R}
\end{array}
is a homotopy pullback.


问:
Homeomorphism over it's image takes open sets to open sets in respect to what set?
Let $M \subseteq \mathbb{R^n}$, and suppose $f: U \subseteq \mathbb{R^n} \rightarrow \mathbb{R^n}$ is an homeomorphism over it's image $f(U) \subseteq M$, where $U$ is an open set of $\mathbb{R^n}$.

Let $U_0 \subseteq U$ be another open set of $\mathbb{R^n}$. Is $f(U_0)$ an open set of $\mathbb{R^n}$? Is it open in $M$? And in $f(U)$?

I'm a little bit confused on where the image of open sets are open in this case. My judge is that it's open in $f(U)$, but i was hoping it's also open in $\mathbb{R^n}$. Any ideas? Thanks!

答:
For a given homeomorphism onto image $f:X\to Y$ we have that $f(V)$ is always open in the full image $f(X)$. However it doesn't have to be open in $Y$ in general, because the full image $f(X)$ is not necessarily an open subset of $Y$.

But in the special case $f:U\subseteq\mathbb{R}^n\to\mathbb{R}^n$ with $U$ open we have that the image (of any open subset) indeed is open by the invariance of domain which is a highly nontrivial theorem.


问:
Commutant of a set of operators in $B(H)$
Let $\mathcal{T} \subset B(T)$ be a set of operators.

Then we define the commutatnt $\mathcal{T}' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}\}$.

I'm trying to show $\mathcal{T}' = \mathcal{T}'''$.

Attempt:

By definition, we have that $\mathcal{T}'' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}' \}$, and $\mathcal{T}''' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}''\}$.

"$\subset$":

Let $S \in \mathcal{T}$. Then we have $ST=TS$ for all $T\in \mathcal{T}$.

Then I'm a bit confused on what to do next, because the $\mathcal{T}'''$ I wrote above doesn't really give me a clear idea of what an element in $\mathcal{T}'''$ looks like.

Any help will be appreciated!

Thank you.

答:
We make two trivial observations:

(1) $A \subseteq B \implies B' \subseteq A'.$

(2) $A \subseteq A''$.

Your claim then follows:

By $(2)$, $A \subseteq A''$. Hence, by $(1)$, $A'''= (A'')' \subseteq A'$.

On the other hand, another application of $(2)$ yields $A' \subseteq (A')''= A'''$.


问:
Complex function defined as an integral: is it analytic?
I'm trying to study for my final this Thursday and for some reason the problem that is giving me the most trouble is, according to the professor, the "easiest one". I'm just not seeing something here, it's been a long time since I've done anything in complex that looks like this. The problem is:

$$A = \left\{ \zeta: |\zeta|= \frac{1}{2}\right\}, B: = \left\{\omega : |\omega|< \frac{3}{4}\right\}. \text{ Define }
 f(w) : = \oint_A \frac{z(z+1)}{z^2 + 2z -w} dz.$$

I'm asked to show that $f$ is analytic in $B$ and to find $f'(0)$. I need to learn the idea behind this much more than I need an answer to this specific question, so hints or related questions are much appreciated.

Thank you fixed!

答:
Let's write $$f(w)=\oint_A \frac{g(z)}{h(z)-w}\,dz,\quad g(z)=z^2+z, \quad h(z)=z^2+2z$$
When $|w|<3/4$, we have $|2z|>|z^2-w|$ on $A$. By Rouche's theorem there is a unique root of $h(z)=w$ within the region $D$ bounded by $A$. Write it as $h^{-1}(w)$. Since $h'(z)=2z+2$ does not vanish in $D$, it follows that $h^{-1}(w)$ is a holomorphic function of $w\in B$.

By the residue theorem,
$$f(w)=2\pi i \frac{g'(h^{-1}(w))}{h'(h^{-1}(w))}\tag1$$ which is evidently a holomorphic function in $B$. Before calculating $f'$ it is helpful to plug in explicit $g,h$ and simplify:
$$f(w)=2\pi i\frac{2h^{-1}(w)+1}{2h^{-1}(w)+2}=2\pi i -\frac{\pi i}{h^{-1}(w)+1} \tag2$$
Now take the derivative using the chain rule and the inverse function theorem:
$$f'(w) = \frac{\pi i}{(h^{-1}(w)+1)^2} \frac{d(h^{-1}(w))}{dw} =\frac{\pi i}{(h^{-1}(w)+1)^2 h'(h^{-1}(w))} = \frac{\pi i/2}{(h^{-1}(w)+1)^3 } \tag3$$
Finally, plug in $w=0$, which yields $h^{-1}(0)=0$ and $f'(0) = \pi i/2$.


问:
Proving $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent
I need to prove that  $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent only by Cauchy's test.

Cauchy's test for sequences convergence :

The $\sum \limits _{n=1}^{\infty}a_n$ convergent $\iff$ $\forall \varepsilon >0$ $ \exists$ $ n_0\in \mathbb{N}$ such that $\forall $ $n_0 \le n$ , $\forall$ $p\in \mathbb{N}$ the following holds $\mid S_{n+p}-S_n\mid = \mid a_{n+1}+\dots+a_{n+p}\mid<\varepsilon$

My Attempt:

Let $\varepsilon>0$,$\ p\in\mathbb{N}$. We define $n_0=\frac{p}{\varepsilon}$, and since $-1\le \sin(x)\le 1$ therefore, for each $n\le n_0$: $$\mid \frac{\sin (n+1)}{n+1}+\frac{\sin (n+2)}{n+2}+...+\frac{\sin (n+p)}{n+p} \mid \ \le \ \mid \frac{\sin (n+1)}{n+1}\mid + \mid \frac{\sin (n+2)}{n+2}\mid+...+\mid \frac{\sin (n+p)}{n+p} \mid \le \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+p}\le \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}=\frac{p}{n}<\varepsilon$$

Is my proof correct? if not so how can I prove it only by Cauchy's test as I said earlier, without any further test - such as Dirichlet, Abel...

答:
Your proof is wrong because you have
$\sin(n)$ in the numerators
where you should have
$\sin(n+p)$.


答:
Since the sequence of simple functions you gave converges uniformly, you can simply evaluate the integrals of the simple functions, and then take the limit of that sequence.
$$\int_{[0,1]} \sin(x)\,\mathrm{d}(\mu x) = \lim_{n \rightarrow \infty} \int_{[0,1]} \frac{\left \lfloor{10^n\cdot\sin(x)}\right \rfloor}{10^n}\,\mathrm{d}(\mu x)$$

Since $\sin(x)$ is monotone on $[0,1]$, the integral of the simple functions can be expressed by the following sum:
$$\sum_{k=0}^{10^n-1} 10^{-n}k\, \mu\!\left([\arcsin\lfloor10^{-n}k\rfloor,\arcsin\lfloor10^{-n}(k+1)\rfloor)\right) + \mu(\{1\})$$

Title: How do I compute Lebesgue integrals of non-simple functions?

Body:
I know this is not the purpose of Lebesgue integration, but it would still be nice to know how to do this. For example, how do I compute the integral

$$\int_{[0,1]} \sin(x)\,\mu(\mathrm dx).$$

I know how to compute it the Riemann way—I just want to get practice on Lebesgue integration.
I know I have to make $\sin x$ a pointwise limit of simple functions which would be $$\lim_{n \rightarrow \infty} \frac{\left \lfloor{10^n\cdot\sin x}\right \rfloor}{10^n}.$$


问:
In how many ways can we pick three (3) balls so that we have a ball of each color?

We have a bin with 9 balls, 4 blue balls and 5 red balls. In how many ways can we pick three (3) balls so that we have a ball of each color?

We have a deck of 52 playing cards, 13 of each of the four suits. In how many ways can we pick five (5) cards so that there are at most two suits represented among the five cards? (As e.g. in three hearts and two clubs, but no spades and no diamonds.)

For the first question, I got 140 as my answer. I divided it into two cases: 1R 2B or 1B 2R, but i don't know if I did it correctly.

2((5C1)(4C2)) + 2((5C2)(4C1)) = 140

I am just getting started with this topic, can anyone help me with these questions, thanks.

答:
For the first one, there are $4$ blue balls (B) and $5$ red balls (R). For the given condition, you choose either ($1$ B, $2$ R) or ($2$ B, $1$ R). The question does not specify that order matters so you should not multiply by $2$.

$\displaystyle \small 4 \cdot {5 \choose 2} + 5 \cdot {4 \choose 2} = 70$

For the second one, divide it into two cases, you either have all $5$ cards from one suit or from two suits. So first choose the suit(s) from where we pick $5$ cards and then choose $5$ cards either from $13$ (one suit) or $26$ (two suits)

So it should be  $ \ \displaystyle \small 4 \cdot {13 \choose 5} + {4 \choose 2} \cdot \left[{26 \choose 5} - 2 \cdot {13 \choose 5}\right]$.

$\small \displaystyle {26 \choose 5} - 2 \cdot {13 \choose 5} \ $ is the number of ways in which we select $5$ cards such that it contains cards from both suits.


问:
Can I find the equation of an ellipse with these points?
How can I solve an ellipse with its major axis on the $x$-axis, given one focus, and two points on the ellipse, one of which I know to be on the major axis?</p>

答:
You can ignore the coordinate system to find the semi-minor axis. Choose one of the points on the ellipse $(x,y)$. You can find semi-minor axis by: $$\frac { { y }^{ 2 } }{ { b }^{ 2 } } =\frac { { x }^{ 2 } }{ { a }^{ 2 } } -1\\ b=ay\sqrt { \frac { 1 }{ { x }^{ 2 }-{ a }^{ 2 } }  } $$

Now imagine a coordinate system the center of ellipse is at the origin of that coordinate system. Find out how much the ellipse is rotated with this coordinate system;$$\theta =\arctan { \frac { { x }_{ f } }{ { y }_{ f } }  } $$ where $({ x }_{ f },{ y }_{ f })$ is the point of the foci which is already known. Since the ellipse is rotated, which can be understand as coordinate system is rotated by $\theta$. Moreover the ellipse may be shifted. I don't want to continue to not to give you wrong answer since I am not confident. But you can find more information at here and here


问:
"Negative" versus "Minus"
As a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and
not "minus $0.8$" to denote $-0.8$?

The so called "textbook answer" regarding this question reads:

A number and its opposite are called additive inverses of each other because their sum is zero, the identity element for addition. Thus, the numeral $-5$ can be read "negative five," "the opposite of five," or "the additive inverse of five."

This question involves two separate, but related issues; the first is discussed at an elementary level here. While the second, and more advanced, issue is discussed here. I also found this concerning use in elementary education.

I recently found an excellent historical/cultural perspective on What's so baffling about negative numbers? written by a Fields medalist.

答: In modern mathematics, there is not much of a difference between the unary minus and the binary minus anyway. '$a-b$' is most often regarded as an abbreviation of '$a+(-b)$', and so using different terminology seems superfluous.

Title: "Negative" versus "Minus"


问:
Proving there are only finitely many $n$-tuples of prime numbers such that $(p_{1}p_{2}...p_{n})|(p_{1}+k)(p_{2}+k)...(p_{n}+k)$
I have stumbled upon this claim: $\forall k\in \mathbb{N}$ there are only finitely many completely distinct $n$-tuples of prime numbers $p_{1},p_{2},...,p_{n}$ such that $(p_{1}+k)(p_{2}+k)...(p_{n}+k)$ is divisible by $p_{1}p_{2}...p_{n}$. There was no proof provided so I tried to have a go at this myself.

I tried expanding the terms but the results did not appear to help with the general case. I also attemped to split the problem into cases on the basis of parity of $n,k$ but without much success there either.

Is the right approach trying to directly express the form of such $n$-tuples and then enumerating or is there a neat method that allows us to enumerate right away?

I am grateful for any suggestions.

答:
Let $p_1,\ldots,p_n$ be such a tuple. Choose a sequence $q_0,q_1,\ldots\in\{p_1,\ldots,p_n\}$ such that $q_i|q_{i+1}+k$ for all $i$, and $q_0 = \max\{p_i\}$. Let $i_0$ be the smallest value of $i$ for which $q_{i}\neq q_{i+1}+k$. Write $\ell$ for the smallest prime not dividing $k$.

I claim $i_0< \ell$. If instead $i_0\geq\ell$, then since $k$ is coprime to $\ell$, one of the $\ell$ values $q_i=q_0-ik$ for $0\leq i\leq \ell-1$ would be divisible by $\ell$, hence equal to $\ell$, in which case $q_{i+1}=\ell-k\leq 1$ could not be prime.

Now, by assumption $q_{i_0} = q_0-i_0k$ is a proper divisor of $q_{i_0+1}+k$, so
$$
q_0 + k\geq q_{i_0+1} + k \geq 2(q_0-i_0k),
$$
which implies
$$
q_0\leq (2i_0+1)k <2\ell k.
$$
So we have a bound on $q_0=\max\{p_i\}$ depending only on $k$. This implies that for fixed $k$, there are only finitely-many tuples satisfying the given divisibility.


问:
Why doesn't the circle retract to a point?
OK, this appears to me like perhaps a dumb question. I am reading Allen Hatcher's Algebraic Topology. I've seen bits and pieces of further material here and there before, now I'm restarting from the beginning.

OK, visually I can see why, say, the 2-sphere $S^2$ can not deformation retract onto it's equator. Intuitively, we can't do this without tearing a hole (or rather two holes). Even with visualizing the 'cylinder' $S^2$ x I (well as a volume) for the function $F$ to $S^1$,

$F:S^2 \times I \rightarrow S^1$,

$F(x,t) = f_t(x)$ the family of homotopic maps. You can see that this mapping cylinder won't work in trying to identify points $F(x,1)$ to points in the circle.

The mapping cylinder for a map and spaces $f:X \rightarrow Y$, is the quotient space

$[(X \times I) \amalg Y]/((x,1) \sim f(x))$.

Or even $S^1$ to its equator, $S^0$ the union of two separate points: not "deformation-retractable". Again, looking at $S^1 \times I$ doesn't appear to have a cont. function to $\{x\} \cup \{y\}$ for distinct points $x,y$ in $S^1$. It doesn't seem to have any mapping cylinder as well.

Disclaimer: OBVIOUSLY a circle is NOT homotopic to a point. Just in case anyone gets any wrong ideas as to where I'm going with this. It's just a question on my part :) If anyone could guide me to the a better intuition or visualization to correct the error of what I'm seeing.

However, if we just pick one point $x_0$ from $S^1$, then the mapping cylinder looks just like a cone, with $F(x,0) = x$, $F(x_0, t) = x_0$ (which is a line going down the cone from the circle to the bottom apex), and $F(x,1) = x_0$. And it looks like the mapping cylinder shows how to continuously deform the circle to the point, even if the point is an element of the circle.

I am thinking this "homotopic picture" of what merely looks like, the circle def. retracting to a point, is somehow misleading me, in the sense that I am just missing something or looking at it wrong or etc...

Can anyone elucidate?

答:
First of all, let's clarify one thing.  A circle does retract onto a point, because a retract of a circle to a point on it is just a constant map $r : S^1 \to \{p\}$.  What you're really asking about is the fact that a circle doesn't deformation retract onto a point.

A deformation retract would be a homotopy $F : S^1 \times I \to S^1$ taking the circle to one of its points, so to deformation retract a circle to a point you'd need to retract it to a point on the circle, via a series of maps $F_t$ that map the circle to itself.  Your map seems to be shrinking the circle to a point, which doesn't work because you're moving the rest of the circle off of itself into the "empty space inside of it," which isn't allowed.

In other words, you're viewing the circle as being embedded in the plane, like you'd draw a circle on a piece of paper.  But, topologically, the points "inside the circle" don't exist -- there's only the circle itself.


问:
Evaluating $\sin(\pi x)$ correctly using complex exponentials
Clearly, one has $\sin(\pi x)\neq0$ for all $x\in\mathbb{R}\backslash\mathbb{Z}$. But, one also has $\sin(\pi x)=\frac{e^{\imath\pi x}-e^{-\imath\pi x}}{2\imath}=\frac{(e^{\imath\pi})^x-(e^{-\imath\pi})^x}{2\imath}=\frac{(\cos(\pi)+\imath\sin(\pi))^x-(\cos(-\pi)+\imath\sin(-\pi))^x}{2\imath}=\frac{(-1+\imath0)^x-(-1-\imath0)^x}{2\imath}=\frac{(-1)^x-(-1)^x}{2\imath}=0$ seemingly for all $x\in\mathbb{R}$ where $(\pm\imath)^2=-1$. With $x=0.5$, for example, one would then have $\sin(0.5\pi)=\frac{(-1)^{0.5}-(-1)^{0.5}}{2\imath}=\frac{\sqrt{\imath^2}-\sqrt{\imath^2}}{2\imath}=0$ rather than $\sin(0.5\pi)=\frac{\sqrt{-1} - \sqrt{-1}}{2\imath}=\frac{\sqrt{\imath^2} - \sqrt{(-\imath)^2}}{2\imath}=1$ (which is correct) while $\sin(0.5\pi)=\frac{\sqrt{-1} - \sqrt{-1}}{2\imath}=\frac{\sqrt{(-\imath)^2} - \sqrt{\imath^2}}{2\imath}=-1$. But how to decide when to take $-1=\imath^2$ or $-1=(-\imath)^2$, how do the sign of the arguments of $\cos$ and/or $\sin$ (in Euler's identity) relate to $(\pm\imath)^2=-1$ and how about $0\neq x\in\mathbb{C}$?

答:
This question states basically the exact same problem but in an easier way. As the helpful comments already wrote $e^{i\pi x}=(e^{i\pi})^x$ only if $x$ is an integer. This Wikipedia article about "De Moivre's formula" might be interesting.

De Moivre's formula does not hold for non-integer powers. The derivation of de Moivre's formula above involves a complex number raised to the integer power $n$. If a complex number is raised to a non-integer power, the result is multiple-valued.

This is because the complex logarithm is not a bijective funtion. An example: $e^{2i\pi}$ and $e^{4i\pi}$ are equal. Taking $\log$'s on both expressions yields $2\pi i=4\pi i$, which is false. See here for more detail.


问:
Calculating a Jordan Basis for a matrix A
I am trying to calculate a Jordan basis for the matrix A.
$$A=\begin{pmatrix}1&1\\-1&3\end{pmatrix}$$
$$\chi(A)=(\lambda-2)^2 \to \lambda =2 , \alpha =2$$
$$V(2)=ker(A-2I)= span\begin{pmatrix}1\\1\end{pmatrix}$$
In my understanding to find the other basis vector, one chooses $v=\begin{pmatrix}1\\1\end{pmatrix}$ and computes $Bv \ne 0 , B=A-2I$, which will give the other vector in the Jordan basis of A. However I get the zero matrix in this case, I also tried to compute $ker(A-2I)^2$ but $(A-2I)^2$ is also the zero matrix.

Help would be greatly appreciated.

答:
Pick a vector $w$ which is not in $\ker(A-2I)$, say $w = \pmatrix{0 \\ 1}$ and define $v = (A-2I)w = \pmatrix{1 \\ 1}$.

Then $v \in \ker(A-2I)$ so $$Av = 2v, \quad Aw = (A-2I)w +2w = v + 2w$$
which means that with respect to the basis $\{v,w\}$ the operator $A$ has matrix
$$\pmatrix{2 & 1 \\ 0 & 2}.$$


问:
Divisors in an abelian surface
How to compute the Néron-Severi group of the abelian surface $Y = \mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i]$. More generally, are there any result that compute the Néron-Severi group of product of curves?

Suppose that surface $Y$ has a divisor class $[F]$ such that $[F]^{2} > 0$. Why is $[F]$ or $-[F]$ ample?

Thanks.

答:
For your first question, the answer is based on your comments.

By Lefschetz (1,1) theorem, $rk(NS(Y))=\dim_\mathbb{Q}H^2(Y,\mathbb{Q})\cap H^{1,1}(Y)$, so for any abelian surface $Y$, we have $rk(NS(Y))\leq h^{1,1}(Y)=4$.

Claim: Suppose $Y=\mathbb{C}/\mathbb{Z}[i]\times \mathbb{C}/\mathbb{Z}[i]$, then $rk(NS(Y))=4$.

Proof: Let $(z,w)$ be the complex coordinate of $\mathbb{C}^2$, the universal cover of $Y$. Here $Y=\mathbb{C}^2/\Lambda$, $\Lambda$ is a free abelian group generated by $(1,0),(i,0),(0,1),(0,i)$.

Consider $\alpha=idz\wedge d\bar z$, $\beta=idw\wedge d\bar w$, $\gamma=i(dz\wedge d\bar w+dw\wedge d\bar z)$, $\delta=dz\wedge d\bar w-dw\wedge d\bar z$, so $\{\alpha,\beta,\gamma,\delta\}$ is a basis for the complex vector space $H^{1,1}(Y)$.

Moreover, it follows by simple calculation that $[\mathbb{C}/\mathbb{Z}[i]\times\{\ast\}]=\beta/2$, $[\{\ast\}\times\mathbb{C}/\mathbb{Z}[i]]=\alpha/2$, $[\{(z,z)\in Y\}]=-\gamma/2$, $[\{(z,iz)\in Y\}]=\delta/2$, so $\alpha,\beta,\gamma,\delta\in H^2(Y,\mathbb{Q})\cap H^{1,1}(Y)$ and they are linearly independent over $\mathbb{Q}$. QED

For your second question, since $K_Y=0$, by Riemann-Roch theorem
$$
h^0(nF)+h^0(-nF)=h^0(nF)+h^2(nF)
 \geq \frac{(F^2)}{2}n^2 +\chi(\mathcal{O}_Y).
$$

Therefore, for $n>>1$, either $nF$ or $-nF$ is effective. WLOG, we assume $D=nF$ is effective. Since $Y$ is abelian surface, every curve $C$ in $Y$ is movable (in the sense of algebraic equivalence), so for any two (possibly reducible) curves $C_1$ and $C_2$ we have $(C_1\cdot C_2)\geq 0$. In particular, for any curve $C$ we have $(D\cdot C)\geq 0$.

If $(D\cdot C)=0$ for some curve $C$, together with $(D^2)>0$, the Hodge index theorem implies $(C^2)<0$, we get a contradiction! Thus for any curve $C$, we have $(D\cdot C)>0$, together with $(D^2)>0$, the Nakai-Moishezon criterion tells us that $D=nF$ is ample, so $F$ is ample.

Remark: This argument for your second question applies for any abelian surface $Y$.

Title: Divisors in an abelian surface


问:
Finding least-norm solutions of a linear Diophantine equation
I am looking for an algorithm solving the following problem.

Let $\Vert x\Vert_{*} := \sqrt{x^\mathrm{T} D \,x}$, where $D$ is a diagonal matrix with positive diagonal entries.
Given $a\in \mathbb{N}^n$, $b \in \mathbb{N}$, find
$$\arg \min_{x \in \mathbb{N}^n} {\vert a^{\mathrm{T}} x - b\vert}$$
with minimal $\Vert x \Vert_{*}$.

Motivation

I am creating a tools website for tabletop role-playing game (TTRPG) players. In these games, players fight together against monsters, and one player (called "GM") has to ensure these fights are of appropriate difficulty. This is done using "experience points" (XP). XP is used to estimate the strength of monsters. Each monster awards a set amount of XP (tougher monsters award more XP) and each difficulty has a target XP number. When the players defeat the monsters they gain these XP, becoming stronger as a result. If the GM selects a few monsters, how many of each do we need to ensure a good battle?

For example, the GM selects goblins, hobgoblins, and bugbears (who award 1,2, and 4 XP) to be his monsters. As threshold they select 12 XP. We would have $a^T=[1,2,4], b=12.$ $x$ describes the number of monsters picked. It is clear that there are many ways to select monsters, but most of them are undesirable. We don't want a fight against just 12 goblins, or only 3 bugbears. My idea is giving each monster a somewhat arbitrary weight: $D=diag(1,6,12)$. With this we obtain the optimal $x^T=[4,2,1]$ (4 goblins, 2 hobgoblins and 1 bugbear), a well balanced encounter.
If the GM selects creatures such that we can't meet the threshold, we want to get as close as possible to it.

Why don't I use a greedy algorithm? Because I thought it would be cool and interesting not to.

答:
First,

$$\begin{array}{ll} \underset{\mathrm x \in \Bbb N^3}{\text{minimize}} & \left| \, {\rm a}^\top {\rm x} - b \,  \right|\end{array}$$

If the minimum is zero, then solve the following least-norm problem

$$\begin{array}{ll} \underset{\mathrm x \in \Bbb N^3}{\text{minimize}} & \| \mathrm x \|_*^2\\ \text{subject to} & \mathrm a^\top \mathrm x = b\end{array}$$

Using Python (NumPy + CVXPY),

```<code>import cvxpy as cp
import numpy as np

# parameters
a = np.array([1, 2, 4])
b = 12
D = np.diag([1, 6, 12])

# optimization variables
x = cp.Variable(3, integer=True)

# create and solve linear optimization problem over the integers
objective = cp.Minimize(cp.abs(a.T @ x - b))
prob      = cp.Problem(objective, [ x >= np.zeros(3) ])
solution  = prob.solve()

print("—————————————————————————————")
print("IP ——————————————————————————")
print("Minimum  = ", prob.value)
print("Solution = ",    x.value)

# create and solve quadratic optimization problem over the integers
objective2  = cp.Minimize(cp.quad_form(x, D))
constraints = [ a.T @ x - b ==  0,
                x >= np.zeros(3)  ]
prob2       = cp.Problem(objective2, constraints)
solution2   = prob2.solve(solver=cp.ECOS_BB)

print("—————————————————————————————")
print("MIQP ————————————————————————")
print("Minimum  = ", prob2.value)
print("Solution = ",     x.value)
</code>```

which outputs

```<code>—————————————————————————————
IP ——————————————————————————
Minimum  =  0.0
Solution =  [0. 0. 3.]
—————————————————————————————
MIQP ————————————————————————
Minimum  =  51.99999927150701
Solution =  [4. 2. 1.]
</code>```

Unfortunately, according to the CVXPY documentation, there are correctness issues with this solver.

Body:
First,

$$\begin{array}{ll} \underset{\mathrm x \in \Bbb N^3}{\text{minimize}} & \left| \, {\rm a}^\top {\rm x} - b \,  \right|\end{array}$$

If the minimum is zero, then solve the following least-norm problem

$$\begin{array}{ll} \underset{\mathrm x \in \Bbb N^3}{\text{minimize}} & \| \mathrm x \|_*^2\\ \text{subject to} & \mathrm a^\top \mathrm x = b\end{array}$$

Using Python (NumPy + CVXPY),

import cvxpy as cp
import numpy as np

# parameters
a = np.array([1, 2, 4])
b = 12
D = np.diag([1, 6, 12])

# optimization variables
x = cp.Variable(3, integer=True)

# create and solve linear optimization problem over the integers
objective = cp.Minimize(cp.abs(a.T @ x - b))
prob      = cp.Problem(objective, [ x >= np.zeros(3) ])
solution  = prob.solve()

print("—————————————————————————————")
print("IP ——————————————————————————")
print("Minimum  = ", prob.value)
print("Solution = ",    x.value)

# create and solve quadratic optimization problem over the integers
objective2  = cp.Minimize(cp.quad_form(x, D))
constraints = [ a.T @ x - b ==  0,
                x >= np.zeros(3)  ]
prob2       = cp.Problem(objective2, constraints)
solution2   = prob2.solve(solver=cp.ECOS_BB)

print("—————————————————————————————")
print("MIQP ————————————————————————")
print("Minimum  = ", prob2.value)
print("Solution = ",     x.value)

which outputs

—————————————————————————————
IP ——————————————————————————
Minimum  =  0.0
Solution =  [0. 0. 3.]
—————————————————————————————
MIQP ————————————————————————
Minimum  =  51.99999927150701
Solution =  [4. 2. 1.]

Unfortunately, according to the CVXPY documentation, there are correctness issues with this solver.


问:
Using $f(x+a) = e^{a\frac{d}{dx}}f(x)$
Using Taylor series for an infinitly differentiable function we've
$$f(x+a) = e^{a\frac{d}{dx}}f(x)$$

But if I choose $f(x)=e^x$ and $x=1,a=2$

I end up with $$e^{3}=e^{2 e}$$.

Can anyone please hint at my mistake. Thank you so much.

答: You've mistakenly replaced $\left[\exp\left(2\tfrac{d}{dx}\right)\right]e^x$ with $\exp\left(2\tfrac{d}{dx}e^x\right)$.

Title: Using $f(x+a) = e^{a\frac{d}{dx}}f(x)$


答:
Understand what an "operator" is, and how it acts.
$$
f(x) = e^x
\\
\left(\frac{d}{dx}\right)^k f(x) = \frac{d^k}{dx^k}\; f(x) = e^x
\\
e^{a\frac{d}{dx}} = \sum_{k=0}^\infty \frac{a^k}{k!}\left(\frac{d}{dx}\right)^k
$$
so that
\begin{align}
e^{a\frac{d}{dx}} f(x) &= \sum_{k=0}^\infty \frac{a^k}{k!}\left(\frac{d}{dx}\right)^k f(x)
= \sum_{k=0}^\infty \frac{a^k}{k!}e^x = e^x\sum_{k=0}^\infty \frac{a^k}{k!}
= e^x e^a = e^{x+a}
\end{align}

Title: Using $f(x+a) = e^{a\frac{d}{dx}}f(x)$

Body:
Using Taylor series for an infinitly differentiable function we've
$$f(x+a) = e^{a\frac{d}{dx}}f(x)$$

But if I choose $f(x)=e^x$ and $x=1,a=2$

I end up with $$e^{3}=e^{2 e}$$.

Can anyone please hint at my mistake. Thank you so much.


问:
Optimize quadratic form weighted Gram matrix
Let $\alpha\in [0,1]^K: \sum_{i=1}^K\alpha_i=1$, and assume we have $K$ vectors $\{x(1),...,x(K)\}$ such that $x(i)\in\mathbb{R}^d, \forall i \in [K]$.

Consider the Gram matrix $G(\alpha) = \sum_{i=1}^K \alpha_i x_i x_i^T$.

Now consider the optimisation problem for $y\in\mathbb{R}^d$

$$\sup_{\alpha\in[0,1]^K: \sum_{i=1}^K\alpha_i=1} y^\top G(\alpha)y. $$

Does this problem admit a unique solution, many solutions, no solutions?

I think it should have a unique solution since we have the maximum of $G(\alpha)$ that is a polynomial function of $\alpha$ over a compact set, but I am not sure about it.

How is it possible to solve it?

答:
Let $\Delta \subset \Bbb R^K$ denote the simplex of dimension $K-1$. This problem can be more simply rewritten as
$$
\sup_{\alpha \in \Delta} \sum_{i=1}^k \alpha_i(y^Tx_i)^2.
$$
It is easy to see that this maximum is attained with $\alpha_j = 1$ and $\alpha_i = 0$ for all $i \neq j$, where $j = \operatorname{argmax}_i (y^Tx_i)^2$. This is the unique point at which the maximum is obtained if and only if $j$ is the unique maximizing index.