问:
Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$.

Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.

For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le  1
\end{align}

For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.

My questions are:

Does someone have a hint or a proof?

Can someone give me some advice on how to handle such problems during an exam?

Thanks :)

答:
It's easy to notice that the function $f(x) =\prod_{k=0}^n |x-k|$ is a symmetric function at $x = \frac{n}{2}$ (in other words, $f(x) = f(n-x)$). So, it suffice to prove $f(x) \le \frac{(n-1)!}{2}$   in the interval $x \in [1, \frac{n}{2}]$.

For $x \in [p,p+1)$ with $p\in \{0,...,[\frac{n}{2}]-1\}$, we have

$$ f(x+1) = \prod_{0 \le k \le n}|(x+1)-k| = \frac{x+1}{|x-n|}\prod_{0 \le k \le n}|x-k| = \frac{x+1}{|x-n|} f(x)$$
As $x \in [p,p+1)$ with $p\in [0,...,([\frac{n}{2}]-1)]$, we have then $x < \frac{n-1}{2}$, so $\frac{x+1}{|x-n|} <1$ or $f(x+1) < f(x)$.

We deduce that for all $p\in \{0,...,[\frac{n}{2}]-1\}$
$$\max_{x \in [p+1,p+2)}f(x) = \max_{x \in [p,p+1)}f(x+1) < \max_{x \in [p,p+1)}f(x)$$
or
$$\max_{x \in [1,\frac{n}{2}]}f(x) = \max_{x \in [1,2)}f(x)$$
So, it suffices to prove $f(x) \le \frac{(n-1)!}{2}$   in the interval $x \in [1, 2)$.
In this interval, we have
$$f(x)  = \prod_{0 \le k \le n}|x-k| = x(x-1)(2-x) \prod_{3 \le k \le n}(k-x) = x(x-1)(4-2x) \frac{1}{2}\prod_{3 \le k \le n}(k-x)   \tag{1}$$
By AM-GM inequality,
$$x(x-1)(4-2x) \le \left(\frac{x + (x-1) + 2(2-x)}{3}  \right)^3 = 1 \tag{2}$$
and because $x \in [1, 2)$, then
$$\prod_{3 \le k \le n}(k-x) \le \prod_{3 \le k \le n}(k-1) = (n-1)! \tag{3}$$

From (2) and (3), we deduce that (1) holds true. Hence, for all $x\in [1,n-1]$
$$f(x) \le \frac{(n-1)!}{2}$$

Note: The equality can't occur because the equality in (2) occurs if $x = x-1 = 4-2x$, which is impossible.


问:
Joint Cummulative Distribution Identity
I understand how to prove the following using Venn diagrams (asked and answered in this question) but would also like to be able to show it using the general properties of distributions. The following comes from chapter 6 of Ross's First Course In Probability.

Let $P\{X\leq a \, \wedge Y\leq b\} = F(a,b)$ be the probability that the $X\leq a$ and $Y\leq b$. Prove the following claim:

$P\{a_1 \lt X \leq a_2 \,\, \wedge \,\, b_1 \lt Y \leq b_2 \} = F(a_2,b_2) + F(a_1,b_1) - F(a_1,b_2) - F(a_2,b_1)$

答:
The way to show this is to express $A\cup B$ as the equivalent union of disjoint sets $A \cup A^c B$.

We will use the following notation:

$$P(X\leq a) = F_X (a)$$

$$P(Y\leq b) = F_Y (b)$$

$$P(X\leq a \cap Y \leq b) = F(a,b)$$

We will also use the following identity in the below solution:

$$
\begin{array}{rcl}
P\{X > a \cap Y > b\} &=& 1 - P\{ X \leq a \cup Y \leq b\} \\
&=& 1 - P\{X \leq a\} - P\{Y \leq b\} + P\{ X \leq a \cap Y \leq b\} \\
&=& 1 - F_X (a) - F_Y (b) + F(a,b)
\end{array}
$$

We rewrite the original problem as

$$
\begin{array}{rcl}
P(a_1 < X \cap X \leq a_2 \cap b_1 < Y \cap Y \leq b_2) &=& 1 - P(X\leq a_1 \cup X>a_2 \cup Y \leq b_1 \cup Y > b_2 ) \\
&=& 1 -P[ ( X\leq a_1 \cup Y \leq b_1) \cup ( X>a_2 \cup Y > b_2 ) ]
\end{array}
$$

Using $P(A \cup B) = P(A) + P(A^cB)$ and $P(A\cup B)=P(A) + P(B) - P(AB)$, we have

$$
\\
1 - \left[ \left(F_X (a_1) + F_Y (b_1) - F(a_1,b_1) \right) + \left( P( (X > a_1 \cap Y > b_1 ) \cap (X > a_2 \cup Y > b_2 ) ) \right) \right]
\\
$$

Since $\{X > a_1\} \supset \{X > a_2\}$ and $\{Y>b_1\} \supset \{Y>b_2\}$, the last term in the above equation is equal to

$$
\\
\begin{array}{rcl}
P( (X>a_2 \cap Y>b_1) \cup (X>a_1 \cap Y>b_2) ) &=& P(X>a2 \cap Y>b1) + P(X>a_1 \cap Y > b_2)\\
&&  - P(X>a_2 \cap Y > b_2)
\end{array}
$$

Using the above identity for $P(X>a \cap Y>b)$ and combining terms yields:

$$
P(a_1 < X \leq a_2 \cap b_1 < Y \leq b_2 ) = F(a_1,b_1)+F(a_2,b_2) - F(a_1,b_2) - F(a_2,b_1)
$$

ContentLicense: CC BY-SA 3.0

Title: Joint Cummulative Distribution Identity


问:
Generating words in a finitely presented group in SAGE
I'm trying to get a list of all words of length $n$ (in the word metric sense) in some finitely presented group. I have tried some very naive enumerations but it is very slow. Is there an efficient way of doing this?

答:
It might be too naive, but you can iterate on the elements of the free group, then use these elements to compute a representative in the group. It works decently well on the example I picked from GAP documentation, but it is certainly not perfect. If you give a more concrete example of a group you're looking at this can certainly be optimized.</p>
```<code>m1 := [ [ Z(3)^0, Z(3)^0,   Z(3) ],
        [   Z(3), 0*Z(3),   Z(3) ],
        [ 0*Z(3),   Z(3), 0*Z(3) ] ];;
m2 := [ [   Z(3),   Z(3), Z(3)^0 ],
        [   Z(3), 0*Z(3),   Z(3) ],
        [ Z(3)^0, 0*Z(3),   Z(3) ] ];;
G := Group( m1, m2 );

hom:=EpimorphismFromFreeGroup(G:names:=["a", "b"]);;

PreImagesRepresentative( hom, m1*m2^(-1) );

iter := Iterator(G);
##test if we iterate directly on the matrix group, not very efficient

F2:=FreeGroup("a","b");

iterF2 := Iterator(F2);
##iterator on the free group, used to get the reduced words

toMatrixGp:=function(gen,w)
##computes the element in G via a->m1 and b->m2
    local temp;
    x:=gen[1]^0;
    for i in [1..NumberSyllables(w)] do

            temp:=(gen[GeneratorSyllable(w,i)]^(ExponentSyllable(w,i)));
        x:=x*temp;

    od;

    return x;
end;

elmtSet:=Set([]);
##set of all different elements
for i in [1..1000] do
    w:=UnderlyingElement(NextIterator(iterF2));
    g:=toMatrixGp(gen,w);
    elmt:=(PreImagesRepresentative( hom, g ));
    Print(elmt,"\n");
    AddSet(elmtSet,elmt);
od;
</code>```

ContentLicense: CC BY-SA 4.0

Body:
It might be too naive, but you can iterate on the elements of the free group, then use these elements to compute a representative in the group. It works decently well on the example I picked from GAP documentation, but it is certainly not perfect. If you give a more concrete example of a group you're looking at this can certainly be optimized.

m1 := [ [ Z(3)^0, Z(3)^0,   Z(3) ],
        [   Z(3), 0*Z(3),   Z(3) ],
        [ 0*Z(3),   Z(3), 0*Z(3) ] ];;
m2 := [ [   Z(3),   Z(3), Z(3)^0 ],
        [   Z(3), 0*Z(3),   Z(3) ],
        [ Z(3)^0, 0*Z(3),   Z(3) ] ];;
G := Group( m1, m2 );

hom:=EpimorphismFromFreeGroup(G:names:=["a", "b"]);;

PreImagesRepresentative( hom, m1*m2^(-1) );

iter := Iterator(G);
##test if we iterate directly on the matrix group, not very efficient

F2:=FreeGroup("a","b");

iterF2 := Iterator(F2);
##iterator on the free group, used to get the reduced words

toMatrixGp:=function(gen,w)
##computes the element in G via a->m1 and b->m2
    local temp;
    x:=gen[1]^0;
    for i in [1..NumberSyllables(w)] do

            temp:=(gen[GeneratorSyllable(w,i)]^(ExponentSyllable(w,i)));
        x:=x*temp;

    od;

    return x;
end;

elmtSet:=Set([]);
##set of all different elements
for i in [1..1000] do
    w:=UnderlyingElement(NextIterator(iterF2));
    g:=toMatrixGp(gen,w);
    elmt:=(PreImagesRepresentative( hom, g ));
    Print(elmt,"\n");
    AddSet(elmtSet,elmt);
od;


问:
Solving $y = x \tan\theta + \frac{g x^2}{2v^2\cos^2\theta} $ for $\theta$
Can anyone solve for $\theta$?

$$y = x \tan\theta + \frac{g x^2}{2v^2\cos^2\theta} $$

I've gone to like 3 teachers and I just need $\theta$ in terms of these other variables/constants, but I can't get it into few enough terms to be solvable.

答:
Observe that

$$\sec^2\theta=\frac1{\cos^2\theta}=1+\tan^2\theta$$

so

$$x\tan\theta+\frac{gx^2}{2v^2\cos^2\theta}=x\tan\theta+\frac{gx^2}{2v^2}\left(1+\tan^2\theta\right)\implies$$

our equation is

$$\frac{gx^2}{2v^2}\tan^2\theta+x\tan\theta+\frac{gx^2}{2v^2}-y=0$$

The above is a quadratic in $\;\tan\theta\;$ whose discriminant is

$$\Delta=x^2-4\frac{gx^2}{2v^2}\left(\frac{gx^2}{2v^2}-y\right)$$

and thus

$$\tan\theta_{1,2}=\frac{-x\pm\sqrt\Delta}{2\frac{gx^2}{2v^2}}$$

and then from here to obtain $\;\theta\;$ using $\;\arctan\;$...but still not very pretty.


问:
Canonical form of a PDE of order 2
We consider the following PDE:
$$\frac{\partial^2u}{\partial x^2}+2\frac{\partial^2u}{\partial x \partial y}+\frac{\partial^2u}{\partial y^2}+3\frac{\partial u}{\partial y}+4u=0$$

We have $B^2-AC=1^2-1\times 1=0$ then the PDE is Parabolic.

Could you please help me to find the canonical form of this PDE?

Thanks in advance

答:
If you choose the transformation $u(x,y) \mapsto u(\xi)$, with $\xi  = x+y$ given by one of the characteristics, then your PDE reduces to the ODE $u''(\xi)+ \frac{3}{4}u'(\xi) + 4 u(\xi) = 0$, whose solution is some combination of sine/cosine times an exponential. By reverting the transformation, one can show that $u(x+y)$ is indeed a solution of the original PDE.

For a more general integral, I think you need to specify another expression for the second characteristic $\eta = f(x,y)$ and then apply the transformation $u(x,y)\mapsto u(\xi,\eta)$. However, $f$ is somehow arbitrary and I haven't found a choice of $f$ so as to make the resulting PDE easy to solve.

I hope this helps.

Title: Canonical form of a PDE of order 2


问:
Proving that $f$ verifies $f(1−x) + f(x) = 1$
I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression

\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}

I would like to show that $f(1-x) + f(x) = 1$.

What I have done so far

With a little bit of algebra I have

\begin{align*}
f(x)
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
&& \text{using }\binom{n}{m} = \frac{n!}{m!(n-m)!}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p, j\neq-k}^{p} (x-j)\right)
&& \text{using the change of indices j=i-k}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p}^{p} (x-j)\right) \frac{1}{x+k}
&& \text{completing the product}
\\
&= \frac{1}{(2p-1)!} \left(\prod_{j=1-p}^{p} (x-j)\right) \left(\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}\right)
&& \text{factoring the product from the sum}
\end{align*}

Then I introduce the functions
\begin{align*}
F(x) &= (2p-1)! f(x)  = G(x) H(x)\\
G(x) &= \prod_{j=1-p}^{p} (x-j) \\
H(x) &= \sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\end{align*}

So the initial question reduces to proving that $F(1-x) + F(x) = (2p-1)!$

Next, playing with the indices of the product I have been able to show that $G(1-x) = G(x)$.

So that now I only need to show $G(x) \left( H(1-x) + H(x)\right) = (2p-1)!$

EDIT:

I have just been able to simplify $H(1-x) + H(x)$ as follows :

\begin{align*}
H(x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\\
&=\sum_{s=p}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p+k
\end{align*}

And

\begin{align*}
H(1-x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p+k}
\\
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p-1-k}
&& \text{using } \binom{2p-1}{p+k} = \binom{2p-1}{p-1-k}
\\
&=\sum_{s=0}^{p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p-1-k
\end{align*}

So finally

\begin{align*}
H(1-x) + H(x)
&=\sum_{s=0}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
\end{align*}

EDIT2:

I have just made the same work on $G(x)$ as follows

\begin{align*}
G(x)
&=\prod_{j=1-p}^{p} (x-j) \\
&=\prod_{s=0}^{2p-1} (x-p+s)
&& \text{using a change of index } s=p-i
\end{align*}

Now I think the constant $(2p-1)!$ in the relation $F(1-x)+F(x)=(2p-1)!$ comes from the binomial factor in $H(1-x) + H(x)$ but I don't see how to prove that...

Any lead ?

答:
I have actually found a way to solve the question from EDIT1 and EDIT2
So

\begin{align*}
F(1-x) + F(x)
&= G(x) \Big( H(1-x) + H(x) \Big) \\
&= \Bigg(\prod_{s=0}^{2p-1}(x-p+s)\Bigg)\Bigg(\sum_{k=0}^{2p-1}\frac{(-1)^{k}}{x-p+k}\binom{2p-1}{k}\Bigg)\\
&= \sum_{k=0}^{2p-1} (-1)^{k}\binom{2p-1}{k}\Bigg(\prod_{s=0, s \neq k}^{2p-1}(x-p+s)\Bigg)
\end{align*}

This the (2p-1)-th order finite difference scheme with step $h=1$ applied to the function

\begin{align*}
g(x) = \Bigg( \prod_{s=0, s \neq k}^{2p-1}(x-p) \Bigg) = (x-p)^{2p-1}
\end{align*}

So clearly this would mean that :

\begin{align*}
F(1-x) + F(x)
&= \Delta_{h}^{2p-1} (x-p)^{2p-1} \\
&= \frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1} \\
&= (2p-1)!
\end{align*}

Hot point to justify in the conclusion :

So we just need now to justify the passage from $\Delta_{h}^{2p-1}(x-p)^{2p-1}$ to $\frac{d^{2p-1}}{dx^{2p-1}} (x-p)^{2p-1}$.

Is there a property that says the nth-order finite difference scheme is exact on polynomials ?

Found a justification here :

https://hal.archives-ouvertes.fr/hal-01350976/file/on_the_link_between_finite_difference_and_derivative_of_polynomials_kolosov_petro_2017.pdf


问:
Average degree of convex hull vertices in a Delaunay triangulation
Let $P \subset \mathbb{R}^2$. The boundary of $DT(P)$, the Delaunay triangulation of the point set $P$, is  $conv(P)$. It is also known that the average degree of the vertices of $DT(P)$ is $\lt 6$. My question is there any known bound on the expected degree of the hull vertices of $DT(P)$? Is there an intuitive argument at least for this bound to also be $O(1)$?

Edit: Consider the two cases:

Case 1: $P$ is drawn uniformly and independently from the unit square.

Case 2: $P$ is drawn uniformly and independently from the unit disk.

Are there known results for point sets with other distributions? Any valuable reference on this questions?

答:
I have an intuitive argument for why the average degree of the hull vertices should still be $O(1)$ when the point set is drawn uniformly from a convex shape (like a unit square or unit disk).

The environment of a hull vertex is much "the same" as the internal vertices with respect to nearby other vertices, except that there are no vertices on one side of it. Stated another way, if I take some convex shape and generate points uniformly inside it, then slice it down the middle and throw away all the points on one side, the resulting point set may as well have come from a uniform distribution on the new shape, but some points that were previously internal vertices are now hull vertices, and the point distribution on one side is the same as it was before, when they were internal vertices. This means that hull vertices don't have to worry about neighbor points being any closer or farther away than normal, just that the distribution is only on one side.

Okay, so the points "look" the same at a hull vertex, except there are only half as many and they are all on one side. But what about the edges? Surely as a hull vertex, more edges will come to them than if they were deep in the middle. Well, we can estimate the edges at the hull to good approximation by considering the "sub-hull", which I define as $\operatorname{conv}(P-\operatorname{conv}(P))$, i.e. the points in the convex hull of the points that remain after you remove the true convex hull points. These are the "faraway" points that will form edges with the hull points simply because there are no closer points on the other side of the hull to connect to.

There are $O(\log n)$ hull points and $O(n)$ points, so $|P-\operatorname{conv}(P)|$ is also $O(n)$ and the sub-hull is also $O(\log n)$. Thus (I'm ignoring a lot of variables because I'm only after a big-$O$ estimate) there will be on average $O(1)$ connections from adjacent hull points and sub-hull points, and $O(1)$ from nearby points (this is the factor that internal vertices have, perhaps divided by $2$), for a total of $O(1)$ edges at a typical hull point.

Note: This is not even close to a mathematical proof, but it should give at least strong evidence to believe that hull vertices have $O(1)$ average degree.


问:
Transformation of functions help
$f:[0,4] \to\mathbb{R}$, $f(x)=6\sin(\pi x/2)$

Let $h:[-3,1] \to\mathbb{R}$, $h(x)=6\cos(\pi x/2)$
State a sequence of two transformations that takes the graph of $f$ to the graph of $h$.

I would love some help on this because I am confused about how to go from graph $f$ to graph $h$. For example, I can only see one transformation to go from $f$ to $h$ which is changing sin to cos but I can't see a second transformation. Thanks

答:
If you do a single transformation, you shift the graph of the function to the left by $1$. I can think of several options:

Split the shift into two separate shifts, say by $0.2$ and $0.8$

Multiply $f$ by $-1$ and you need a shift by $3$

Shift by $1$ and change $x$ to $-x$. The last transformation should not change $h$

Title: Transformation of functions help


问:
Why is the derivative of f equal to the sum of its partials along its components?
I was trying to understand the development of the solution in this answer, where $\overline{f(z)}f'(z)dz = (u - iv)(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial y})(dx + idy)$. The first term is pretty obvious, as it is just the conjugate. However, where does the equality

$$f'(z)dz = \left(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\right)(dx + idy)$$
come from?

I know that we can write the derivative of an analytic function $f$ as

$$f'(z) = \frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y} = \frac{\partial v}{\partial y} + i \frac{\partial v}{\partial x}$$

but the above term does not seem to follow algebraically. Where does this term come from?

答:
This comes from the Cauchy-Riemann equations, since $f$ is analytic:

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

so it follows from my second expression.


问:
Is my proof correct???
The following question was asked in INMO 2005:

"Find all functions $f:\Bbb{R} \rightarrow \Bbb{R} $ such that
$$f(x^2 + yf(z)) = xf(x) + zf(y)$$

for all $x,y,z \in \Bbb{R}$.

I am thinking of this as a solution:

"Put $x = y = 0 $ in the given equation, I get $$f(0) = zf(0)$$ for all $z$ in $R$ which means that $f(0) = 0$

Further take $y=0$ in the original equation to give me $$f(x^2)=xf(x)$$

Put $x = 0$ in the original equation to give $$f(yf(z)) = zf(y) *$$

In the above equation put $z=1$ to give $$f(yf(1)) = f(y)$$

Let $f(1) = c$ ,which gives me $f(cy) = f(y) **$
In the equation marked * put $y = 1$ to give $$f(f(z)) = zc$$

Taking $f$ on both sides gives $f(f(f(z))) = f(zc) = f(z)$ from **

Let $f(z) = x$ , say then I get $f(f(x)) = x$, but I know that $f(f(x)) = xc$ which means that $$c=1$$

Not put $z = 1$ and since $xf(x) = f(x^2)$ I get $$f(x^2 + y) = f(x^2) + f(y)$$ Now I replace x^2 by x to get $$f(x + y) = f(x) + f(y)$$ to give me the cauchy equation with solns $f(x) = 0$ for all x or $f(x) = x$(from f(1) = 1)

Now though I have given the solutions, there are two pints where I doubt my solution a little. They are when I replace$f(z)$ by some $x$. Is this step valid? And the other doubt is when I replace $x^2$ by $x$. Are these two steps valid and are there any errors in this solution?

Thanks in advance

答:
$$f\left ( x^2+yf\left ( z \right ) \right )=xf\left ( x \right )+zf\left ( y \right )$$
$$y=0\Rightarrow f\left ( x^2 \right )=xf\left ( x \right )$$
$$x=0\Rightarrow f\left ( yf\left ( z \right ) \right )=zf\left ( y \right )$$
$$y=1\Rightarrow f\left ( f\left ( z \right ) \right )=zf\left ( 1 \right ),\forall x\in \mathbb{R}$$
$$f\left ( x\left ( f\left ( x \right ) \right ) \right )=f\left ( f\left ( x^2 \right ) \right )=x^2f\left ( 1 \right )$$
$$y=x=z\Rightarrow f\left ( xf\left ( x \right ) \right )=xf\left ( x \right )$$
$$x^2f\left ( 1 \right )=xf\left ( x \right )\Rightarrow f\left ( x \right )=cx,x\neq 0$$
$$c^2yz=cyz,\forall y,z\in \mathbb{R}\Rightarrow c=\left \{ 0;1 \right \}$$
$$f\left ( x \right )=0 \wedge f\left ( x \right )=x$$

Title: Is my proof correct???


问:
difference between the metric tensor, dot product and metric
From my understanding the metric tensor is a function/tensor field that specifies a tensor at each point on a manifold, the tensor is used to define the inner product for the tangent space (in this sense the inner product can also be referred to as the dot product) and then from there we can naturally induce a norm and subsequently a metric for said tangent space at any given point on a manifold forming a metric vector space. Is this correct?</p>

答:
Strictly speaking, we should always refer to "metric tensor field" because it's actually a tensor field, although for convenience we simply refer to "metric tensor". You're right that the metric tensor field (usually denoted as $g$) assigns to each point $p$ of the manifold an inner product (which is of course a $(0,2)$ tensor) $g_p$ on the tangent space $T_pM$. After this, you're right: the inner product $g_p$ on $T_pM$ gives rise to a norm $\lVert \cdot\rVert_p$ on $T_pM$, which can in turn be used to define a distance function $d_p$ on $T_pM$. This is all just basic linear algebra/analysis/topology; the only difference being that now the vector spaces in question are the various tangent spaces to our manifold.

The only thing you should not confuse is the double usage of the word "metric". In geometry, the term "metric" refers to the metric tensor field $g$. The word "metric" in "metric spaces" is the distance function $d$. As long as you do not mistake one for the other (based on the way you wrote things it doesn't seem you're confused by this but I just wanted to emphasize it), your understanding is right.


问:
Probability distribution of expression containing random variables
Two independent random variables $A$ and $B$ share the same probability distribution
$$p_{A}(x) = p_{B}(x) = \frac{1}{2^{x+1}} $$
where $x$ is a non negative integer.

N is a random variable given by:
$$N = 4 + 2A + 3B$$

What is the distribution of N? Is there a general method for generating a new probability distribution from an expression containing random variables with given distributions?

答:
There is a closed form for this case, but I suspect it could be ugly involving the three cube-roots of $2$ (two of them complex) and specific to this example.

So instead for an example of the more general case, consider the probability generating function of $A$ and $B$: $\dfrac{1}{2-x}$, Then $2A$ has  probability generating function $\dfrac{1}{2-x^2}$ and $3B$ has  probability generating function $\dfrac{1}{2-x^3}$ and the constant $4$ has generating function $x^4$.

If they are independent then $N=4+2A+3B$ has  probability generating function $$\dfrac{x^4}{(2-x^2)(2-x^3)} = \dfrac{x^4}{4-2x^2-2x^3+x^5}$$ and expanding this gives $$\frac14x^4+\frac18x^6+\frac18x^7+\frac1{16}x^8+\frac1{16}x^9+\frac3{32}x^{10}+\frac1{32}x^{11}+\frac3{64}x^{12}+\cdots$$ where the coefficients of $x^n$ gives the probability $\mathbb P(N=n)$.

We can turn this into a recurrence with $$\mathbb P(N=n)=\tfrac12\mathbb P(N=n-2)+\tfrac12\mathbb P(N=n-3)-\tfrac14\mathbb P(N=n-5)$$ starting with $\mathbb P(N=0)=\mathbb P(N=1)=\mathbb P(N=2)=\mathbb P(N=3)=0$ and $\mathbb P(N=4)=\frac14$.


问:
Help with a theorem in Hungerford's book
I'm trying to understand a theorem of Hungerford's book

First, Hungerford defines a localization in the following manner:

After that, there is a theorem:

I have just a little doubt with this theorem, I didn't understand why $Q\subset P$ in the fourth line.

It's just a detail, but it's a crucial point in the core of the proof.

Thanks in advance

答: This is just part (i) which was proven before. Since there is a one-to-one correspondence between the set of prime ideals of $R$ which are contained in $P$ and the set of prime ideals of $R_P$, the maximal ideal $M$ has a preimage $Q$, i.e. $M=Q_P$ and this $Q$ is contained in $P$, in symbols $Q\subset P$.</p>


问:
Four dimensional field over complex numbers
A guy in Facebook claims he's come up with an algebraic field extension to the complex plane. He's defined the unit multiplications as $i^2=-1$, $j^2=i$ and $k^2=-i$. This implies that $ij=ji=k$, $ik=ki=-j$ and $jk=kj=-1$.

Furthermore complex numbers are seemingly a subset of this space, since each complex number $z = a + bi$ can be represented as $a + bi + 0j + 0k$.

So far I've checked that both addition and multiplication in this space are indeed associative and commutative, the addition and multiplication are distributive, and that $0$ and $1$ are the additive and multiplicative inverses respectively. I've yet to find an example of a non-inversible element.

Could someone help explain what's going on?

答:
You're looking at the extension $\Bbb{C}\subset\Bbb{C}[j,k]$ where $j$ and $k$ satisfy $j^2=i$ and $k^2=-i$. The key is that you are requiring this to be an extension of fields. Over a field, a polynomial of degree $d$ has at most $d$ roots. As $j$ and $k$ are zeros of $X^2\pm i\in\Bbb{C}[X]$, and we can find two roots of both quadratics in $\Bbb{C}$, it follows that $\Bbb{C}[j,k]$ is isomorphic to (a subfield of) $\Bbb{C}$.

More explicitly, we can find the polar forms of these roots from the fact that $\pm i=e^{\pm\tfrac{\pi}{2}i}$. It follows that
$$j=\pm e^{\tfrac{\pi}{4}i}\qquad\text{ and }\qquad k=\pm e^{-\tfrac{\pi}{4}i},$$
because these four complex numbers are precisely the roots of $X^2\pm i$.

Similarly we could find the Cartesian forms of these roots by solving $(a+bi)^2=\pm i$. This is equivalent to the simultaneous equations
$$a^2-b^2=0\qquad\text{ and }\qquad 2ab=\pm1.$$
The former implies that $b=\pm a$ and hence the latter reduces to $2a^2=\pm1$, or equivalently $a=\pm\sqrt{\pm\tfrac12}$, which we can express more concisely as $a=i^k\tfrac{1}{\sqrt{2}}$. Tracing our steps back then shows that
$$j=\pm\frac{1}{\sqrt{2}}(1+i)\qquad\text{ and }\qquad k=\pm\frac{1}{\sqrt{2}}(1-i).$$
Either way we see that $\Bbb{C}[j,k]$ is isomorphic to a subfield of $\Bbb{C}$, and hence isomorphic to $\Bbb{C}$ itself.


问:
Linear system of inequalities.
Given linear system:

$$4+\delta_1-\delta_2\ge 0 \\ 2-\delta_1+\delta_2 \ge 0 \\ 1+\delta_1-\delta_2+\delta_3 \ge 0$$

How from there it follows that $\delta_2=\delta_3 = 0$ and $\delta_1\neq0$?

答: It doesn't, try $$\delta_1=\delta_2=\delta_3=1.$$

Title: Linear system of inequalities.


问:
partition of unity and $W^{2,2}_{\mathrm{loc}}$
Let $K$ be a compact subset of $\mathbb{R}^n$. Choose a finite open cover $\{B_i\}_{i=1,\dots,k}$ so that $K\subset \bigcup_{i=1}^k B_i$. Take a partition of unity $\{\zeta_1,\dots,\zeta_k\}$ so that $\zeta_i \in C^\infty_c(B_i)$, $\mathrm{supp}\, \zeta _i \subset B_i$ for all $1\leq i\leq k$ and $\sum_{i=1}^k \zeta_i =1$ on $K$.

Here is my question. Suppose that $\zeta_i u \in W^{2,2}(B_i)$. Can we say that $u\in W^{2,2}(K)$? It seems to be wrong, but I cannot find exact counterexamples.

Thanks for your time.

答:
It is true that $u\in W^{2,2}(K)$ due to the simple fact that $u=\sum \zeta_i u$ since $\sum \zeta_i=1$ on $K$. This is different from being in $W^{2,2}_{loc}$ however; that would mean that for any compactly contained subset of $K$, say $V$, then the restriction of $u$ to $V$ lies in the Sobolev space.

An example of a function in $W^{2,2}_{loc}$ but not in $W^{2,2}$ can be seen by e.g. $f(x)=x^{-2}$ on $K=[0,1]$. This is not even in $L^2$, but $f$ and its derivatives are easily seen to be in $L^2_{loc}$ since any compactly contained subset is bounded away from 0.


问:
Finding the distance between a unit circle and a line analytically
Let $A = \{(x, y) \in \mathbb{R}^2\mid x^2 + y^2 = 1\}$ and $B = \{(x, y) \in \mathbb{R}^2 \mid y = 3 - x\}$. I am currently taking my analysis/metric spaces class and would like to find the distance between these two sets in a more "analytical" way. That is, using only the definition $\mathrm{dist}(A, B) = \inf \{d_{\mathbb{R}^2}(a, b) \mid a \in A, b \in B\}$. I know this might sound like heresy, but I would prefer for the proof to not contain any argumentation by pictures etc. Just by the tools of analysis. The issue is that I honestly do not even know where to start. Sure, $d_{\mathbb{R}^2}(a, b) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$ and points in $A$ and $B$ satisfy the properties of a circle and a line. But by using only the tools of analysis, how can you derive the distance between the sets?

I do know that this problem is trivial with the use of the formula $\frac{\left|ax_0 + by_0 + c\right|}{\sqrt{a^2 + b^2}} - r$, where $(x - x_0)^2 + (y - y_0)^2 = r^2$ is a circle and $ax + by + c  = 0$ is a line. But so far (through HS) I have only ever used these ready made tools and argumentation by e.g. geometry to find the answers. Therefore I do not know how to approach this problem in the "analysis" way.

答:
A point on the line has coordinates $P(t,3-t)$

The squared distance from the center of the circle $O(0,0)$ is
$$OP^2=f(t)=t^2+(3-t)^2=2t^2-6t+9$$
This is minimum when $f'(t)=0$ and $f''(t)>0$

$f'(t)=4t-6=0\to t=\frac32$

$f''(t)=4$ so $P^*\left(\frac{3}{2},\frac{3}{2}\right)$ is the point of the line closest to the circle.

$OP^*=\frac{3}{\sqrt{2}}$

So the distance between the line and the circle is $OP^*-r=\frac{3}{\sqrt{2}}-1$


问:
Finding covariance of this random variables
I could find the mean and variance but I found some difficulties in computing the covariance of the following random variables.

Here is the problem:

Suppose $Y_1,Y_2,\ldots$ are i.i.d RVs with $\mathbb{E}Y_i=\mu$ and Var$(Y_i)=\sigma^2\in(0,\infty)$. Set $X_k:=Y_k-3Y_{k+1}+Y_{k+2},k=1,2,\ldots$, and put $S_n:=X_1+\dots+X_n$. Compute Cov$(X_j,X_k)$ for $j\not=k$.

Here is my attempt:

We let $\tilde{Y_j}:=Y_j-\mu$ (so that $\mathbb{E}\tilde{Y_j}=0$), then we have Cov$(X_j,X_k)=\mathbb{E}(\tilde{Y_j}-3\tilde{Y_{j+1}}+\tilde{Y_{j+2}})(\tilde{Y_k}-3\tilde{Y_{k+1}}+\tilde{Y_{k+2}})$.

I know that we have to separate into cases of different values of $|j-k|$, but I am not sure how to proceed. I am a bit confused in manipulating random variables with indexes. Any helps on the detailed workings will be really appreciated.

Many many thanks!

答:
Start by expanding the two parenthesis under the expectation. Alternatively, save a bit of time by using the fact that covariance is multilinear:

$$\mbox{Cov}(aX+bY,cR+dS)=ac\mbox{Cov}(X,R)+ad\mbox{Cov}(X,S)+bc\mbox{Cov}(Y,R)+bd\mbox{Cov}(Y,S)$$

and that two independent random variables have zero covariance.


问:
Area between curves $r=5\sin(3\theta)$ and $r=8\sin(3\theta)$
So I was studying for a test and came across the following prompt:

"What is the total area between the polar curves $r=5\sin(3\theta)$ and $r=8\sin(3\theta)$?"

I guess I'm a bit confused about what the bounds of my integral might look like here. I understand that the setup for the area integral is as follows:
$$\frac{1}{2}\int_a^b(f(\theta)^2-g(\theta)^2)d\theta$$
but I'm a bit confused about how I'd go about finding my bounds for this integral. Any help would be appreciated!

答:
Note that each curve represents a set of three equally spaced pedals with each pedal subtending a polar angle of $\frac\pi3$.
Thus, the area is
$$\frac{3}{2}\int_0^{\frac\pi3} (f(\theta)^2-g(\theta)^2)d\theta
= \frac{3}{2}\int_0^{\frac\pi3} (64\sin^2(3\theta)-25\sin^2(3\theta))d\theta=\frac{39\pi}4
$$


答: Due to centrality of alignment and polar symmetry $ \theta \text{ varies between ( 0 ,} \pi/3) $ for both.

Title: Area between curves $r=5\sin(3\theta)$ and $r=8\sin(3\theta)$

Body:
So I was studying for a test and came across the following prompt:

"What is the total area between the polar curves $r=5\sin(3\theta)$ and $r=8\sin(3\theta)$?"

I guess I'm a bit confused about what the bounds of my integral might look like here. I understand that the setup for the area integral is as follows:
$$\frac{1}{2}\int_a^b(f(\theta)^2-g(\theta)^2)d\theta$$
but I'm a bit confused about how I'd go about finding my bounds for this integral. Any help would be appreciated!


问:
Simplify sum $\sum_{m_1,m_2,\ldots,m_k}{p \choose m_1\; m_2 \; \ldots \; m_k}\Pi_{i=1}^k (2m_i-1)!!$ for $p\to \infty$
Let $1 \le k \le p$ be integers. Define
$$
S_{p,k} := \sum_{m_1,m_2,\ldots,m_k}{p \choose m_1\; m_2 \; \ldots \; m_k}\Pi_{i=1}^k (2m_i-1)!!,
$$
where the sum if over all nonnegative $k$-partitions $(m_1,\ldots,m_k)$ of $p$, and
$n!!$ is the product of all integers between $n$ and $1$ inclusive, with the same parity as $n$, and $n!!=1$ if $n \le 0$.

Question. Does $S_{p,k}$ simplify (say via special functions) ?

Note. I'm fine with asymptotic formula for $S_{p,k}$ in the limit $p \to \infty$.

Edit: motivation of the question

Let $T_{n,d,p}:= \mathbb E[\mbox{Tr}(WW^\top)^p]=\mathbb E[(\sum_{i=1}^n\sum_{j=1}^jw_{ij}^2)^{p}]$ be the expectation of the $p$th power of the trace of the Wishart matrix $WW^\top$, shere $W$ is an $n \times p$ matrix with iid entries from $N(0, 1)$. Via a quick-and-dirty calculation, I it can be seen that $T_{n,d,p} = \sum_{k=1}^{\min(p,n)}{n \choose k}S_{p,k}$. This is because the general term in the sum $(\sum_{i=1}^n\sum_{j=1}^jw_{ij}^2)^{p}$ is $(w_{i_1,*}^2)^{m_k} \ldots  (w_{i_k,*}^2)^{m_1}$ for distinct $i_1,\ldots,i_k \in [n]$ and $k \le \min(n,p)$, and the expectation of this term is $\Pi_{i=1}^k(2m_i-1)!!$, which only depends on the numbers $m_1,\ldots,m_k$, and not on the specific $i$'s!

答:
Following Phicar's very insightful comment, it is clear that
$$
S_{p,k}=\mathbb E\Big[(Z_1^2+Z_2^2+\dots+Z_k^2)^p\Big]
$$
where $Z_1,\dots,Z_k$ are independent standard normal variables. That is, $S_{p,k}$ is the $p^{th}$ moment of a $\chi^2$ random with $k$ degrees of freedom! Once you know the moment generating function of such a random variable (see here for a derivation), you can read off the moments by expanding into a Taylor series. The result is that
$$
S_{p,k}=2^p\;\Gamma(p+\tfrac{k}2)/\Gamma(\tfrac{k}2)=(k+2p-2)(k+2p-4)\cdots(k+2)k
$$
As a check, when $p=2$, every composition is a permutation of $(2,0,\dots,0)$ or $(1,1,0,\dots,0)$. There are $k$ of the former, each contributing $\binom{2}{2,0,\dots,0}(2\cdot 2-1)!!=3$, and there are $\binom{k}2$ of the latter, each contributing $\binom{2}{1,1,0,\dots,0}1!!\cdot 1!!=2$. Therefore, $S_{3,k}=3k+2\binom{k}2$, which is indeed equal to $k(k+2)$.

Again, following Phicar's prompt, one can give a combinatorial proof as well. Both sides of the equation
$$
\sum_{m_1+\dots+m_k=p}\binom{p}{m_1,\dots,m_k}\prod_i (2m_i-1)!!=k(k+2)(k+4)\cdots(k+2(p-1))
$$
count the number of ways to take $p$ couples, assign each couple one of $k$ colors, and then partition the set of people in each color into pairs. That is, in the end, the $2p$ initial people will all be divided into pairs, and each pair will have a color, in such a way that two people who were initially in the same couple must end up with the same color.

This interpretation is a direct reading of the LHS; if you want $m_i$ people to have color number $i$ for each $i\in \{1,\dots,k\}$, then there are $\binom{p}{m_1,\dots,m_k}$ ways to assign the colors, and then $(2m_i-1)!!$ ways to form a perfect pairing within the $i^{th}$ color.

For the RHS, imagine assigning both the colors and pairings for each couple, going one couple at a time.

For the first couple, there are $k$ ways to choose their color. You then temporarily pair these two together.

For the second couple, there are $k+2$ options. For $k$ of these options, you pair off this couple with each other, and give them one of $k$ colors. For the other two options, they will trade partners with the previous couple, and take their color. This can be done in two ways; if the couples are $(A,B)$ and $(C,D)$, you can pair $C$ with either $A$ or $B$, and then $D$ gets the other person in $(A,B)$.

For each subsequent couple, the logic is the same. They can either pair together and get a new color in $k$ ways, or they can trade partners with any of the previously formed pairs. If we are assigning couple number $i$ ($1\le i\le p$), then there are $2(i-1)$ ways to do this, as one of the new couple members can "steal" the partner of any previous pair, and then the other couple member gets the member of the previous pair.

Therefore, there are $p$ stages, and $k+2(i-1)$ ways to ways to complete the $i^{th}$ stage, so in total there are $k(k+2)(k+4)\cdots(k+2(p-1))$ ways. QED.


问:
$f(x,y)=\langle y- \cos y, x \sin y\rangle$
$f(x,y)=\langle y-\cos y,x\sin y\rangle$

$C$ is the circle $(x-3)^2 + (y+4)^2 = 4$ orientated clockwise.

Relevant theorems:

Green's theorem (this is under the Green's theorem section of our book).

What I Have Tried

Using Green's theorem, I have come up with the double integral $\iint ( \sin y - 1) \,dA$.  Beyond here, I can't figure out what to do for my limits of integration.  I have come to the conclusion that polar might be the way to go, but am at a loss of how to proceed with that, since I have a circle whose center is not the origin.

If it matters, this is out of Stewart Calculus Edition 7.  It is chapter 16.4 question 13.  The answer is $4 \pi$.

答:
You're doing a double integral over $C$, so one option for your limits of integration will be:

$$\int_1^5 \int_{-4-\sqrt{4-(x-3)^2}}^{-4+\sqrt{4-(x-3)^2}} \dots dy dx$$

Explanation: The boundaries of the circle you're integrating over stretch from $1$ to $5$, since it's a circle of radius $2$ centered at $\langle 3, -4 \rangle$.

The $y$-coordinates of the boundary will vary with $x$: we have arrived at the bounds above by simply solving for $y$ from the equation of the circle.


问:
The average number of handshakes between people around a circle.

$5$ people are around a circle. At a moment and at the same time each
person extend their hand to one of their adjacent person. people who
extended their hand to each other, shaking their hands. In average
between different possibilities how many hand shakings are done?

$1)\frac32\quad\quad\quad\quad\quad2)1\quad\quad\quad\quad\quad3)\frac23\quad\quad\quad\quad\quad4)\frac54\quad\quad\quad\quad\quad5)\frac12$

First I noticed that for $5$ people we have $2^5=32$ different ways of extending hands (from this we can see third choice $\frac23$ can't be the answer). also maximum number of handshakes is $2$ and it happens in five different cases (at each case there is a person who is not shaking a hand). also minimum number of hand shaking is zero which happens in two cases (all persons extend their hand in a clockwise or counterclockwise directions). so I conclude there are $32-5-2=25$ cases with only one handshake hence the average is $\cfrac{2\times5+25\times1}{32}=\cfrac{35}{32}$. but I don't have this in the choices.

答:
Establish five random variables $X_i$ for $i = 1, \ldots, 5$ where
$$
X_i = \cases{\frac12 & if person $i$ shook hands with somebody\\
0 & otherwise}
$$
Then note that the quantity we're interested in is $X_1 + X_2 + X_3 + X_4 + X_5$. Or rather, we're after the expected value of this sum.

It is known that expectation is linear. Which is to say, we have
$$
E(X_1 + X_2 + X_3 + X_4 + X_5) = E(X_1) + E(X_2) + E(X_3) + E(X_4) + E(X_5)
$$
Each of these five expectations equals $\frac 1{4}$, as there is a $\frac12$ probability that the person they extend their hand to extends back at them.

Which is to say, the final answer is $\frac54$.

As for your approach, there are acutally 10 ways to have two handshakes, as the lone person can reach out in two directions. So you get
$$
\frac{2\times 10 + 20}{32} = \frac{40}{32} = \frac54
$$
Your approach might ultimately be faster, but it is, in my opinion, easier to miss something and get the wrong answer.


问:
$L^p$ convergence implies $\int_{\mathbb{R}^d \setminus E} |f_n|^p < \epsilon$ for all $n$.

Question. Suppose $f_n, f \in L^p(\mathbb{R}^d)$ such that $f_n \to f$ in $L^p$-norm. Is it necessarily true that for all $n \in \mathbb{N}$, and any $\epsilon > 0$, there exists a $E \subset \mathbb{R}^d$ with finite measure $\mu(E) < \infty$ such that $\int_{\mathbb{R}^d \setminus E} |f_n|^p < \epsilon$.

I believe this is true. However, I'm a little unsure of how to rigorously prove.

What I have: Since $E$ is measurable, then $F = \mathbb{R}^d \setminus E$ is also measurable. Then

$$\int_F |f_n|^p = \int_F |f_n - f + f|^p \leq \int_F |f_n - f|^p + \int_F |f|^p.$$

Since $f_n \to f$ in $L^p$-norm, for each $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that
$$\int_F |f_n - f|^p < \frac{\epsilon}{2},$$

whenever $n \geq N$. Further, $f$ is integrable. Meaning there exists some $\epsilon > 0$ such that
$$\int_F |f|^p < \frac{\epsilon}{2}.$$

Thus, for $n \geq N$,

$$\int_F |f_n|^p < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

But how does one proceed when $n < N$?

答:
Let us argue by contradiction. Suppose it does not hold. That means there is an $\varepsilon > 0$ such that for all finite measurable $E \subset \mathbb{R}^d$ there exists some $n\in\mathbb{N}$ so that $\int_{E^c} |f_n|^p > \varepsilon$.

Choose $\varepsilon > 0$. Since $f\in L^p$ we know there is some finite and measurable $F \subset \mathbb{R}^d$ such that

$$ \int_{F^c} |f|^p < \frac{\varepsilon}{2}. $$

Now we proceed as you do. Since $f_n\rightarrow f$ in $L^p$, there is some natural number $N$ such that for every $n\geq N$

$$ \int_{F^c} |f_n|^p \leq \int_{F^c} |f_n-f|^p + \int_{F^c} |f|^p < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon.  $$

Therefore we already know that our new claim is not true for any $n\geq N$. But the remaining $N-1$ functions are a finite amount, so we can simply define finite and measurable sets $E_n \subset \mathbb{R}^d$ such that

$$ \int_{E_n^c} |f_n|^p < \varepsilon $$

for all $1\leq n< N$. Since we can choose whatever $\varepsilon > 0$ and repeat the same argument the new claim is false and the answer to your question is a positive answer.


问:
Probability of complete pairings from permutation of 14 people
14 people put their coats in a coatroom and are given back their coats completely at random. What is the probability that 7 pairs of people have received their coats switched?

(For example, if there were only 4 people, Harry, Ron, Hermione, and Tom: Harry’s and Tom’s coats are switched and Ron’s and Hermione’s coats are switched.)

Answer:
There are 14! ways of assigning the coats, all equally likely. There are
${14 \choose 2,2,2,2,2,2,2}$ ways of dividing the people into 7 pairs.
the people into 7 distinct pairs, and therefore
${14 \choose 2,2,2,2,2,2,2} \dfrac{1}{7!}$. By the law of equally likely outcomes, the required probability is: $\dfrac{{14 \choose 2,2,2,2,2,2,2} \dfrac{1}{7!}}{14!} = \dfrac{1}{2^7 7!}$

I was wondering, since the $\dfrac{1}{2^7 7!}$ looks so simple, was there another way to think about the question that also arrived at the $\dfrac{1}{2^7 7!}$ ?

答:
"Is there another way to think about the question that also arrived at the $\frac{1}{2^77!}$?"

Yes.

You should be able to convince yourself that it doesn't matter what order people arrive to claim their coats... all that matters is who receives what coat.  Let's assume the people are able to be alphabetically ordered.

Let the first person go and claim a coat.  They don't receive their correct coat with probability $\dfrac{13}{14}$.  Supposing this happens, let them look at the nametag on their coat and tell that person it is their turn to grab a coat.  That person grabs the first person's coat with probability $\dfrac{1}{13}$.

Now, given that this has all happened so far, the next remaining person earliest in the alphabet who hasn't picked up their coat yet goes to grab a coat.  They don't get theirs with probability $\dfrac{11}{12}$ and they again call the person whose coat they did grab to go in for their turn to grab a coat which happens to be this third person's coat with probability $\dfrac{1}{11}$.

Continuing in this fashion, given each outcome has occurred in a way such that we result in pairs of people whose coats were swapped, (i.e. that the permutation of coats can be written in disjoint cyclic form consisting only of $2$-cycles with no stationary points) we have as a probability:

$$\dfrac{13}{14}\times\dfrac{1}{13}\times\dfrac{11}{12}\times\dfrac{1}{11}\times\cdots \times \dfrac{3}{4}\times\dfrac{1}{3}\times\dfrac{1}{2}\times\dfrac{1}{1}$$

or simplified

$$\dfrac{1}{14\times 12\times 10\times\cdots\times 2}=\dfrac{1}{14!!}=\dfrac{1}{2^7\cdot 7!}$$


答:
Defining $\delta_{12} = \delta_1-\delta_2$:

$$
\begin{matrix}
4+\delta_{12}\ge 0           && \rightarrow && \delta_{12}\ge -4\\
2-\delta_{12} \ge 0          && \rightarrow && \delta_{12} \le 2\\
1+\delta_{12}+\delta_3 \ge 0 && \rightarrow && \delta_3 \ge -\delta_{12} -1\\
\end{matrix}$$

So the solution domain in the $\delta_{12} \, \text{x} \, \delta_3$ plane is something like:

Now, focusing on the $\delta_{1} \, \text{x} \, \delta_{2}$ plane:

$$
\begin{matrix}
\delta_{12}\ge -4 && \rightarrow && \delta_{2} \le \delta_{1} + 4\\
\delta_{12} \le 2 && \rightarrow && \delta_{2} \ge \delta_{1} - 2\\
\end{matrix}$$

Title: Linear system of inequalities.

Body:
Given linear system:

$$4+\delta_1-\delta_2\ge 0 \\ 2-\delta_1+\delta_2 \ge 0 \\ 1+\delta_1-\delta_2+\delta_3 \ge 0$$

How from there it follows that $\delta_2=\delta_3 = 0$ and $\delta_1\neq0$?


答:
Edit: Apologies to @Raffaele

I took too long writing this and by the time I'd finished you'd already posted the answer. Leaving mine up in case it's useful, but please don't award me points that should go elsewhere!

Points on the line are given by $(x, f(x))$ where $f(x) = 3-x$.

If we have some other point $(a, b)$, we can find the distance from this point to the line by finding $x$ which minimizes

$$d(x) = \sqrt{(x-a)^2 + (3-x - b)^2}$$

We can simplify things by minimizing $[d(x)]^2$ instead:

$$[d(x)]^2 = (x-a)^2 + (3-x-b)^2$$

In the case of a line and a circle, we can find the point on the line that minimizes the distance to the center point of the circle. We can then use this to find the minimum distance.

A calculus solution is below.

Here we want to minimize

$$d(x) = \sqrt{x^2 + (3-x)^2}$$

The $x$ that minimizes this will also minimize the square of the distance, which we can call $D(x)$

$$D(x) = x^2 + (3-x)^2$$

Differentiating we have
$$D'(x) = 2x + 2(3-x)(-1)$$
$$D'(x) = 4x - 6$$

So we see there's a critical point at
$$0 = 4x - 6$$
$$x = 3/2$$

We can see $D' < 0$ for $x < 3/2$ and $D' > 0$ for $x > 3/2$, i.e. $3/2$ is indeed a minimum point for $D$.

Plugging this point back into $f(x)$ we find that the point on the line that's closest to the origin (the center of the circle) is $(3/2,3/2)$

The distance from the line to the origin is

$$d_{center} = \sqrt{(3/2)^2 + (3/2)^2} = \frac{3}{2}\sqrt{2}$$

The distance from the line to the circle is

$$d_{center} - 1 = \frac{3}{2}\sqrt{2} - 1 = \frac{3\sqrt{2}-2}{2}$$

Edit also, sorry of this isn't quite what you're looking for. I'm not super versed in formal analysis.

Title: Finding the distance between a unit circle and a line analytically

Body:
Let $A = \{(x, y) \in \mathbb{R}^2\mid x^2 + y^2 = 1\}$ and $B = \{(x, y) \in \mathbb{R}^2 \mid y = 3 - x\}$. I am currently taking my analysis/metric spaces class and would like to find the distance between these two sets in a more "analytical" way. That is, using only the definition $\mathrm{dist}(A, B) = \inf \{d_{\mathbb{R}^2}(a, b) \mid a \in A, b \in B\}$. I know this might sound like heresy, but I would prefer for the proof to not contain any argumentation by pictures etc. Just by the tools of analysis. The issue is that I honestly do not even know where to start. Sure, $d_{\mathbb{R}^2}(a, b) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$ and points in $A$ and $B$ satisfy the properties of a circle and a line. But by using only the tools of analysis, how can you derive the distance between the sets?

I do know that this problem is trivial with the use of the formula $\frac{\left|ax_0 + by_0 + c\right|}{\sqrt{a^2 + b^2}} - r$, where $(x - x_0)^2 + (y - y_0)^2 = r^2$ is a circle and $ax + by + c  = 0$ is a line. But so far (through HS) I have only ever used these ready made tools and argumentation by e.g. geometry to find the answers. Therefore I do not know how to approach this problem in the "analysis" way.


问:
find all primes $p$ such that $3^p-(p+2)^2$ is a prime
find all primes $p$ such that $3^p-(p+2)^2$ is also a prime.

My proof

set  $3^p-(p+2)^2=q$

case $1$:$p=2 \implies q = -7 \notin \mathbb N$. so there is no solution.

case $2$: $p \geq 3$

notice that $p=3 \implies q = 2$ so there is a solution.

claim: with $p\geq 4$ there is no solution.

Proof: we gonna show that $3^p-(p+2)^2$ is always even, and we know that for $p\geq3$ there is no even prime.

so :$$q = 3^p-(p+2)^2= 3^p-p^2-4p-4$$ $3^p$ and $p^2$is odd and $-4p-4$ is even so :

$$odd + odd+even+even=even+even=even$$
therefore there is no solution with $p\geq4$.

is this a valid proof?

答:
Hint: As is very often the case, you don't need calculus when dealing only with integers. In stead of differentiating, you can consider the differences between consecutive values of $3^k-(k+2)^2$.

The details:

For every nonnegative integer $k$ define $f(k):=3^k-(k+2)^2$. With a bit of algebra you'll find that the first difference of this sequence equals
\begin{eqnarray*}
f(k+1)-f(k)&=&
\big(3^{k+1}-((k+1)+2)^2\big)-\big(3^k-(k+2)^2\big)\\
&=&\big(3^{k+1}-3^k\big)-\big((k+3)^2-(k+2)^2\big)\\
&=&3^k\big(3-1\big)-\big(2k+5\big)\\
&=&2\cdot3^k-2k-5.
\end{eqnarray*}
Similarly, the second difference equals
\begin{eqnarray*}
\big(2\cdot3^{k+1}-2(k+1)-5\big)-\big(2\cdot3^k-2k-5\big)
&=&\big(2\cdot3^{k+1}-2\cdot3^k\big)-\big(2(k+1)-2k\big)\\
&=&2\cdot3^k\big(3-1\big)-2\\
&=&4\cdot3^k-2.
\end{eqnarray*}
Clearly the second difference is positive for all $k\geq0$. Then looking at the first difference, for $k=2$ we have
$$2\cdot3^2-2\cdot2-5=4>0,$$
and hence the first difference is positive for all $k\ge2$. It follows that for all $p>3$ we have
$$f(p)>f(3)=2,$$
and hence $f(p)$ is not prime for any prime number $p>3$, because $f(p)$ is even and greater than $2$. For $p=2$ we have $f(2)=-7$ which is of course not prime. So $p=3$ is the unique prime number for which $f(p)$ is prime.


答:
We can assume that $j> k\geq 1$. Using that the covariance is bi-linear we obtain the expression:
$$
\begin{align*}
\mathrm{Cov}(X_j,X_k)=&\mathrm{Cov}(Y_j,Y_k)-3\mathrm{Cov}(Y_j,Y_{k+1})+\mathrm{Cov}(Y_j,Y_{k+2})-3\mathrm{Cov}(Y_{j+1},Y_k)\\
+&9\mathrm{Cov}(Y_{j+1},Y_{k+1})-3\mathrm{Cov}(Y_{j+1},Y_{k+2})+\mathrm{Cov}(Y_{j+2},Y_k)\\
-&3\mathrm{Cov}(Y_{j+2},Y_{k+1})+\mathrm{Cov}(Y_{j+2},Y_{k+2}).
\end{align*}
$$
Now, use that $\mathrm{Cov}(Y_i,Y_j)=0$ whenever $i\neq j$ (due to independence). This allows you to deal with seperate cases:  $j=k+1$, $j=k+2$ and lastly $j>k+2$.</p>

Title: Finding covariance of this random variables

Body:
I could find the mean and variance but I found some difficulties in computing the covariance of the following random variables.

Here is the problem:

Suppose $Y_1,Y_2,\ldots$ are i.i.d RVs with $\mathbb{E}Y_i=\mu$ and Var$(Y_i)=\sigma^2\in(0,\infty)$. Set $X_k:=Y_k-3Y_{k+1}+Y_{k+2},k=1,2,\ldots$, and put $S_n:=X_1+\dots+X_n$. Compute Cov$(X_j,X_k)$ for $j\not=k$.

Here is my attempt:

We let $\tilde{Y_j}:=Y_j-\mu$ (so that $\mathbb{E}\tilde{Y_j}=0$), then we have Cov$(X_j,X_k)=\mathbb{E}(\tilde{Y_j}-3\tilde{Y_{j+1}}+\tilde{Y_{j+2}})(\tilde{Y_k}-3\tilde{Y_{k+1}}+\tilde{Y_{k+2}})$.

I know that we have to separate into cases of different values of $|j-k|$, but I am not sure how to proceed. I am a bit confused in manipulating random variables with indexes. Any helps on the detailed workings will be really appreciated.

Many many thanks!


问:
Convergence of an oscillating power series to a positive number
I have a series

$$ S =  \sum_{i=1}^{\infty} (-1)^{i-1} c_{i} x^{2i} = c_1 x^{2} - c_2 x^{4} + c_3 x^{6} - \dots$$

The coefficients $c_i$ are all positive numbers, and they do not depend on $x$.

$x$ is also assumed to be positive.

Are there are any conditions which may be met by the $c_i$ which ensures the series converges to a positive number?

For example, if I can bound the coefficients from beneath and above, and write the series as a difference or sum of two geometric series?

Body:
$$ S =  \sum_{i=1}^{\infty} (-1)^{i-1} c_{i} x^{2i} = c_1 x^{2} - c_2 x^{4} + \dots.$$

Let $a_i=(-1)^{i-1}c_ix^{2i}$,

By root test,

Let $\alpha=\lim_{i\to\infty}\text{sup}\sqrt[i]{|a_i|}.$

$\alpha=\lim_{i\to\infty}\text{sup}\sqrt[i]{|c_ix^{2i}|}$

and so on...


问:
Why log disappears when time step goes to limit

Hello I’m newbie of math.

And I couldn’t understand why $\log(1 + \mu\delta t)$ goes to $\mu\delta t$ when time step ($\delta t$) goes to zero above the picture.

Please let me know what reference should I study?

Thanks

答: Note that $\log'(1)=1$. In other words,$$\lim_{x\to0}\frac{\log(x+1)}x=1.$$Therefore, if $x$ is very close to $0$, then $\log(x+1)$ behaves as $x$.


答:
By L'hopital we have:
$$\lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}=\lim_{x\rightarrow 0}\frac{\frac{1}{1+x}}{1}=1$$

Title: Why log disappears when time step goes to limit

Body:
Hello I’m newbie of math.

And I couldn’t understand why $\log(1 + \mu\delta t)$ goes to $\mu\delta t$ when time step ($\delta t$) goes to zero above the picture.

Please let me know what reference should I study?

Thanks


问:
If $A$ is open, $\operatorname{Int}(\operatorname{Bdry}(A))=\emptyset$
I'm trying to prove that if $A\subset X$ is open then $\operatorname{int}(\operatorname{Bdry}(A))=\emptyset$. I have that $\operatorname{Bdry}(A) = \operatorname{Cl}(A)\cap (X-\operatorname{int}(A))=\operatorname{Cl}(A)\cap(X-A)$ because $A$ is open, but I'm stuck in that.

答: $x\in\operatorname{Bd}(A)$ implies $x\in\operatorname{Cl}(A)$ and every open neighbourhood $U$ of $x$ intersects with $A$. Now $\operatorname{Bd}(A)^C=(X-\operatorname{Cl}(A))\cup A$ and thus $U$ intersects with $\operatorname{Bd}(A)^C$, which means $x\not\in\operatorname{Int}(\operatorname{Bd}(A))$.


问:
How to show that $\int \lambda((A-x)\cap B) \, d\lambda = \lambda(A)\lambda(B)$?
Given measurable sets $A, B\subset\mathbb R$, how can we show that
$$\int_\mathbb R \lambda((A-x)\cap B) \, d\lambda = \lambda(A)\lambda(B)$$
holds?

I don't see how the integrand might be simplified. The Lebesgue-measure is invariant under translation, so it would be sufficient to show that $\lambda((A-x)\cap B)=\lambda(A-x) \chi_B$ somehow, but I don't know how that might be argued.

Is there an argument that allows that transformation? Or is there a better way entirely?

答:
One has $\lambda((A-x)\cap B) = \int 1_{(A-x) \cap B}(y) \,d\lambda(y)$ so that

\begin{align}
& \int \lambda((A-x)\cap B)\,d\lambda(x) \\[6pt]
= {} & \iint 1_{(A-x) \cap B}(y) \,d\lambda(y)\,d\lambda(x) \\[6pt]
= {} & \iint 1_{y \in (A-x) \cap B} \,d\lambda(y)\,d\lambda(x) \\[6pt]
= {} & \iint 1_{y \in B}1_{x  \in (A-y)} \,d\lambda(y)\,d\lambda(x) \\[6pt]
= {} & \iint 1_{y \in B}1_{x  \in (A-y)} \,d\lambda(x)\,d\lambda(y) \\[6pt]
= {} & \int 1_{y \in B}\lambda(A-y)\,d\lambda(y) \\[6pt]
= {} & \lambda(A)\int 1_{y \in B}\,d\lambda(y) \\[6pt]
= {} & \lambda(A)\lambda(B).
\end{align}


问:
Two linear maps, domain and codomain of both connected via a bijective linear map
Let $L_A : \Bbb{R^n}\to\Bbb{R^m}$, which has a matrix $A$. Let $L:V\to W$. Let $E$ be a basis of $V$ and $F$ a basis of $W$. Moreover we have bijective linear maps $P_E : V\to \Bbb{R^n}$ and $P_F : W\to \Bbb{R^m}$.

How can I show that $P_E(Ker (L))=Null(A)$? I was told to draw a diagram, which is a sort of four directional map, but I cant make any sense of it. The point is, these are both subspaces of $\Bbb{R^n}$ so we can prove they are equal by inclusion, or by showing they have the same basis. I don't think there is any other way to do this.

答:
The key piece of information the diagram provides is the commuting relationship $$L_A \circ P_E=P_F \circ L$$
First if $x\in P_E\Big(\ker(L)\Big) $ then $x=P_E (v)$ for some $v\in \ker(L)$ and $$Ax=L_A(x)=(L_A \circ P_E)(v)=(P_F\circ L)(v)=P_F(0)=0$$ This shows $x\in N(A)$. On the other hand, if $x\in N(A),$ and we take $v=P_E^{-1}(x)$ then $$L(v)=(L\circ P_E^{-1})(x)=(P_F^{-1}\circ L_A)(x)=P_F^{-1}\big(Ax\big)=P_F^{-1}(0)=0$$ This shows $v\in \ker(L)$ meaning $x=P_E(v)\in P_E\Big(\ker(L)\Big)$ and we have our result.


问:
Doubt in homeomorphism
A mapping is a homeomorphism if it is a continuous, bijective map such that it's inverse is also continuous. This is the definition of homeomorphism which was taught to me.
What does the following statement mean:"A point P is homeomorphic to an open disc in $\mathbb{R}^2$".

答: The statement says that there exists a continuous bijection from $\{p\}\to D\subseteq \Bbb{R}^2$ where $D$ is an open disk in $\Bbb{R}^2$ with continuous inverse. That is, there exists a homeomorphism $\phi:\{p\}\to D$. This is, of course, false. There is not even a bijection between $\{p\}$ and $D$. One of them has cardinality $1$, while the other has uncountable cardinality.

Title: Doubt in homeomorphism


问:
Why do I get $0$ in the following integral?
The solution to the following integral should be $-2$,
$$
\int_{\mathbb{R}} \exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  \frac{|v|}{(1 - u^2)}\,\mathrm{d}v = -2.
$$
However, I obtain $0$ (this looks to me just like the Gaussian integral.
$$
\left[\lim_{t \to-\infty} \int_{t}^0 - \frac{|v|}{(1 - u^2)}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}dv + \lim_{t\to\infty} \int_{0}^{t} - \frac{|v|}{(1 - u^2)}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}dv\right] = 0
$$
What am I doing wrong?

Following Comments

Since it is an even function we have
$$
\begin{align}
    \int_{-\infty}^{\infty} \exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  \frac{|v|}{(1 - u^2)}dv
&= 2\int_0^{\infty}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  \frac{|v|}{(1 - u^2)}dv  \\
&= 2\left[\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}\right]_0^{\infty} \\
&= 2\left[0 - 1\right] = -2.
\end{align}
$$
How does one solve it without using the fact that the integrand is an even function?

答:
For simplification purposes we will define $\tau=1-u^2$. We want to show that
$$\left[-\lim_{t \to-\infty} \int\limits_{t}^0  \frac{|v|}{\tau}\exp\left\{-\frac{v^2}{2\tau}\right\}\,\mathrm{d}v - \lim_{t\to\infty} \int\limits_{0}^{t}  \frac{|v|}{\tau}\exp\left\{-\frac{v^2}{2\tau}\right\}\,\mathrm{d}v\right] = -2.
$$
For the first integral substitute $v=-\nu.$ This yields
$$\lim_{t \to-\infty} \int\limits_{0}^{-t}  \frac{|-\nu|}{\tau}\exp\left\{-\frac{(-\nu)^2}{2\tau}\right\}\,\mathrm{d}\nu= \lim_{t \to\infty} \int\limits_{0}^{t}  \frac{\nu}{\tau}\exp\left\{-\frac{\nu^2}{2\tau}\right\}\,\mathrm{d}\nu.$$
Now $\nu$ is running over non-negative values only. That means we can omit the absolut value. Furthermore this integral is equivalent to our second integral. If we call them both $I$ our final solution will be $-I-I$.

Substitute $\mu=-\frac{\nu^2}{2\tau}.$ The differential is going to be $d\nu=-\frac{\tau}{\nu}d\mu.$ That leaves us with
$$I=-\lim_{t \to\infty} \int\limits_{0}^{-\frac{t^2}{2\tau}}  \exp\left\{\mu\right\}\,\mathrm{d}\mu=-\left[\lim_{t \to\infty}e^{-\frac{t^2}{2\tau}}-e^0\right]=1.$$
We know $I=1$. Our solution is $-I-I$, so it's your desired $-2$.


问:
How do one solve a nonlinear combinatoric problem?
I am an undergraduate CS student and I am struggling with a problem.

$Qx = b$ where $Q$ is a constant $m \times n$ matrix (with $m>n$), $x$ is a $n \times 1$ vector and $b$ is a $m\times 1$ vector.

I want to maximize the number of zeros in vector $b$.

subject to: $x(i)>0$ for $i=1,2, \ldots,n$

How do one tackle a problem like this ?

答:
For any given subset $S$ of $\{1\ldots m\}$, making $b_j = 0$ for all $j \in S$ corresponds to solving
$Q_S x = 0, x > 0$ where $Q_S$ is the submatrix of $Q$ corresponding to the rows in $S$.  By homogeneity, if there is a solution there will be one with all $x_j \ge 1$.
Linear programming software may be helpful.

Of course there may be a lot of subsets to consider.  Fortunately, you didn't ask
for an efficient solution.


问:
The average order of $\sigma(n)$.
I'm reading through Apostol's wonderful book "Introduction to Analytic Number Theory" but became confused when reading the statement of theorem 3.4. It states that
$$
\sum_{n\leq x}\sigma(n) = \frac{1}{2}\zeta(2)x^2 + O(x\log(x)).
$$

This would imply that the average order of $\sigma(n)$ is $\frac{1}{2}\zeta(2)n = \frac{\pi^2}{12}n < n$.
How can the average order of $\sigma(n)$ be less than $n$, when $\sigma(n) > n$ for all $n$?

Wikipedia claims that the average order of $\sigma(n)$ is $\zeta(2)n$ under "Examples" here

but further down the page, they claim that for any $\alpha > 0$ we have that
$$
\sum_{n\leq x}\sigma_\alpha(n) = \frac{\zeta(\alpha+1)}{\alpha+1}x^{\alpha+1} + O(x^\beta)
$$
where $\beta = \max(1,\alpha)$ which gives the same result as Apostol for $\alpha = 1$. What am I not understanding?

答:
The result in Apostol is correct.

The average is over all the integers up to $x$, so we should expect it to be smaller than $x$.

Title: The average order of $\sigma(n)$.


答: Indeed you have $\sigma(n)>n$ for all $n$, but for 'most' values of $n\leq x$ you have $\sigma(n)<x$. So it should come as no surprise that the average value of $\sigma(n)$ for all $n\leq x$ is less than $x$.

Title: The average order of $\sigma(n)$.

Body:
I'm reading through Apostol's wonderful book "Introduction to Analytic Number Theory" but became confused when reading the statement of theorem 3.4. It states that
$$
\sum_{n\leq x}\sigma(n) = \frac{1}{2}\zeta(2)x^2 + O(x\log(x)).
$$

This would imply that the average order of $\sigma(n)$ is $\frac{1}{2}\zeta(2)n = \frac{\pi^2}{12}n < n$.
How can the average order of $\sigma(n)$ be less than $n$, when $\sigma(n) > n$ for all $n$?

Wikipedia claims that the average order of $\sigma(n)$ is $\zeta(2)n$ under "Examples" here

but further down the page, they claim that for any $\alpha > 0$ we have that
$$
\sum_{n\leq x}\sigma_\alpha(n) = \frac{\zeta(\alpha+1)}{\alpha+1}x^{\alpha+1} + O(x^\beta)
$$
where $\beta = \max(1,\alpha)$ which gives the same result as Apostol for $\alpha = 1$. What am I not understanding?


问:
I am trying to prove the following theorem: Let $x$ and $y$ be real numbers. Show that if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne y^2$.
I am trying to prove the following theorem:

Let $x$ and $y$ be real numbers. Show that if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne  y^2$.

$Proof.$

The contrapositive of the statement is: If $x^2 =  y^2$, then $x=y$ or $x,y < 0$.

Each real numbers is negative,zero,or positive.
By taking the square root on both sides of the equation $x^2 =  y^2$, one obtains the equations

$x = y$ for each $x,y \geq 0$ or $-x = -y$ for each $x,y<0$.

By multiplying both sides of the equation $-x = -y$ by $-1$, one obtains the equation $x=y$

Combining all the cases: If $x^2 =  y^2$,then $x = y$, or $x,y<0$, or both $x,y<0$ and $x = y$.

Since each implications and its contrapositive are equivalent, given that $x$ and $y$ are real numbers, if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne  y^2$.

I need help verifying my proof.

The question is from the book:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.

答: Hint: If $x^2=y^2$, then $(x-y)(x+y)=0$ and so $x=y$ or $x=-y$.


问:
Location of roots of certain transcendental equations.
If $a$ and $b$ are two solutions of $\,e^x \cos x -1=0$, then how many solutions of the equation $ e^x \sin x-1=0$ lie between $a$ and $b$ ?  </p>

答:
One solution of $f(x)=e^x\cos x-1=0$ is $x=0$ and there is clearly no solution with $x<0$. Between any two solutions, there must be a root of $f'(x) =(-\sin x+\cos x) e^x=\sqrt 2\sin(x-\frac\pi4)e^x$. Thus the first positive root is at $x\ge \frac\pi4$, so that $f(x)=0$ implies $|\cos x|\le e^{-\pi/4}<\frac12$ and hence there is $k\in\mathbb Z$ with $|x-((k+\frac12)\pi|<\frac\pi6$. Also, we have $f'(0)=1$ and for positive roots of $f$, $|\cos x|<\frac12$ implies $|\sin x|>\frac12$, hence $f'(x)\ne 0$, i.e. all solutions of $f(x)=0$ have multiplicity one.

Similarly, we find for $g(x)=e^x\sin x-1$ that all solutions of $g(x)=0$ are positive and between two such solutions must be a point with $g'(x)=0$, i.e. $(\sin x+\cos x)e^x=\sqrt 2\sin(x+\frac\pi 4)e^x=0$, hence the second positive solution of $g(x)=0$ is $\ge\frac{3\pi}4$ so that for all except the first roots of $g(x)$ we have $|\sin x|\le e^{-3\pi/4}<\frac12$, hence $|x-k\pi|<\frac16\pi$ for some $k\in\mathbb Z$. The first root on the other hand, is clearly between $x=0$ ($g(x)=-1$) and $x=\frac\pi4$ ($g(x)=e^{\pi/4}>1$).

In short: Apart from the first roots, the roots of $f(x)$ and $g(x)$ are relatively close to the well-behaved roots of $\cos$ and $\sin$ so that we conclude that the number of roots of $g$ between two numbers $0<a<b$ with $f(a)=f(b)=0$ is obtained by rounding $\frac{b-a}\pi$ to the nearest integer. If $a=0<b$, the number is obtained by rounding $\frac b\pi+\frac12$ to the nearest integer.


答:
More generally, let $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$ be topological spaces. We say that $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$ are homeomorphic if there exists a homeomorphism $X\to Y$; that is, a bijection $\phi:X\to Y$ which is continuous (with respect to the topologies $\mathscr{T}_1$ and $\mathscr{T}_2$) and which has a continuous inverse $\phi^{-1}$. We often shorten this and say that $X$ and $Y$ are homeomorphic if the topologies are clear, and we often write $X\cong Y$ (or similar notations).

In your case, we are abbreviating the statement that $\big(\{x\},\mathscr{T}^\text{ind}_{\mathbb{R}^2}\big)\cong\big(\mathbb{R}^2,\mathscr{T}_{\mathbb{R}^2}\big)$ (already demonstrated by other answers to be false) as $\{x\}\cong\mathbb{R}^2$, where $\mathscr{T}_{\mathbb{R}^2}$ is the Euclidean topology on $\mathbb{R}^2$ and $\mathscr{T}^\text{ind}_{\mathbb{R}^2}$ is the induced topology on the point set $\{x\}\subseteq\mathbb{R}^2$.

Title: Doubt in homeomorphism

Body:
A mapping is a homeomorphism if it is a continuous, bijective map such that it's inverse is also continuous. This is the definition of homeomorphism which was taught to me.
What does the following statement mean:"A point P is homeomorphic to an open disc in $\mathbb{R}^2$".


问:
5 books in 3 bundles
5 distinct books can be tied in 3 indistinct bundles in how many ways, if order of book within a bundle does not matter?

I have counted in following way, and I saw another person counting in a different way, which also seem correct to me, but our results do not tally. Can anyone point out why?

My way :

$5$ books can be tied in $3$ bundles in two different ways 1 + 1 + 3 or 1+2+2.

$1+1+3\rightarrow \binom{5}{3} = 10$ ways

$1+2+2\rightarrow \binom{5}{1} \binom{4}{2}=5.6=30$ ways.

Total 10+30 = 40 ways.

Another person's way :

5 distinct books can be divided in 3 distinct bundles (empty allowed) in $3^5$ ways. Subtracting the cases where one bundle is empty  $\binom{3}{1} 2^5$, adding the cases where 2 bundles are empty $\binom{3}{1} 1^5$. Total $3^5-3.2^5+3=243-96+3=150$. Since the bundles are indistinguishable, all 3! permutations among the bundles are to be counted only once. So total number of ways = $150/6=25$

答:
Okay, if you allow an empty bundle, then you can have the following situations:

A:  $\{5,0,0\}$.  There is one way to do this.

B:  $\{4,1,0\}$.  There are five ways to do this (select the one that is in its own bundle.

C:  $\{3,2,0\}$.  There are $\pmatrix{5\\3}=\frac{5!}{3!2!}=10$ ways.

D:  $\{3,1,1\}$.  There are $\pmatrix{5\\3} = 10 $ ways.

E:  $\{2,2,1\}$.  There are $\frac{1}{2} \cdot \pmatrix{5\\2} \cdot \pmatrix{3\\2}=10\cdot 3 /2=15$ ways.

That is it.  Adding them up:  $1+5+10+10+15=41.$

The comments are correct!  We've got duplicates in E and need to divide by two.  (Bundle 1 and bundle 3 are exchangeable).

Title: 5 books in 3 bundles


答:
You have a mistake in your $1+2+2$ case.

You begin by choosing the one book that gets bundled by itself.  This can be accomplished in $5$ ways.

The next step, in presumably what you are thinking, is to "pick which two books go into the first 2-bundle" and then the remaining books go into the second 2-bundle.  This can be done in $\binom{4}{2}$ ways.

However... we can't tell after the fact which of the two 2-bundles was the "first" or the "second"!  You have double counted these outcomes.

You can correct this by recognizing that you counted each outcome twice and dividing by symmetry... giving $\binom{5}{1}\cdot \binom{4}{2}\cdot \frac{1}{2}=15$ ways for this case.

I dislike "division by symmetry arguments" and so instead prefer to think of this as after having chosen the book for the 1-bundle, among those four remaining books one has the smallest ISBN number (or other distinguishable feature).  We then choose which book gets paired with that book.  The remaining books get paired together.  Choosing which book gets paired with that book with the smallest ISBN number can be done in $3$ ways, giving a total of $5\cdot 3 = 15$ ways for this case.

Title: 5 books in 3 bundles

Body:
5 distinct books can be tied in 3 indistinct bundles in how many ways, if order of book within a bundle does not matter?

I have counted in following way, and I saw another person counting in a different way, which also seem correct to me, but our results do not tally. Can anyone point out why?

My way :

$5$ books can be tied in $3$ bundles in two different ways 1 + 1 + 3 or 1+2+2.

$1+1+3\rightarrow \binom{5}{3} = 10$ ways

$1+2+2\rightarrow \binom{5}{1} \binom{4}{2}=5.6=30$ ways.

Total 10+30 = 40 ways.

Another person's way :

5 distinct books can be divided in 3 distinct bundles (empty allowed) in $3^5$ ways. Subtracting the cases where one bundle is empty  $\binom{3}{1} 2^5$, adding the cases where 2 bundles are empty $\binom{3}{1} 1^5$. Total $3^5-3.2^5+3=243-96+3=150$. Since the bundles are indistinguishable, all 3! permutations among the bundles are to be counted only once. So total number of ways = $150/6=25$


问:
Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$
\begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}

Did I do the right way with those step above?

I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$

Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.

Please help me!!?

答:
You're on the right track. You can continue by dividing out the problematic factor $1-x$. What does long division of the numerator by $1-x$ give?

Hint:

 If $P(x)=(x-1)Q(x)$ then $P'(x)=Q(x)+(x-1)Q'(x)$, so $Q(1)=P'(1)$.

Details:

 Denote the polynomial in the numerator of the fraction by $P(x)$, so that$$P(x)=2017x^{2017}-x^{2016}-\ldots-x-1.$$By polynomial longdivision there exist polynomials $Q(x)$ and $R(x)$ with integer coefficients such that $P(x)=(x-1)Q(x)+R(x)$, where $\deg R(x)<\deg(x-1)$. That is to say $R(x)$ is constant, so $R(x)=R$ for some integer $R$. Plugging in $x=1$ shows that $$P(1)=(1-1)Q(1)+R(1)=R(1)=R,$$and it is not hard to see that $P(1)=0$. This means $P(x)=(x-1)Q(x)$ and so$$\lim\limits_{x\to 1}\dfrac{P(x)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\lim\limits_{x\to 1}\dfrac{Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}.$$Here the hint comes in handy; we have\begin{eqnarray*}Q(1)&=&P'(1)=2017^2-2016-\ldots-2-1\\&=&2017^2-\frac12\cdot2017\cdot2016=2017\cdot1009.\end{eqnarray*}It follows that$$\lim\limits_{x\to 1}\dfrac{-Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\frac{-2017\cdot1009}{2017\cdot2018}=-\frac12.$$


答: If $x\neq y$ WLOG we can suppose $y=x+a$ with $a>0$, then $y^2=x^2+2\underbrace {ax}_{\ge 0}+\underbrace{a^2}_{>0}>x^2$ so $x^2\neq y^2$.

Body:
I am trying to prove the following theorem:

Let $x$ and $y$ be real numbers. Show that if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne  y^2$.

$Proof.$

The contrapositive of the statement is: If $x^2 =  y^2$, then $x=y$ or $x,y < 0$.

Each real numbers is negative,zero,or positive.
By taking the square root on both sides of the equation $x^2 =  y^2$, one obtains the equations

$x = y$ for each $x,y \geq 0$ or $-x = -y$ for each $x,y<0$.

By multiplying both sides of the equation $-x = -y$ by $-1$, one obtains the equation $x=y$

Combining all the cases: If $x^2 =  y^2$,then $x = y$, or $x,y<0$, or both $x,y<0$ and $x = y$.

Since each implications and its contrapositive are equivalent, given that $x$ and $y$ are real numbers, if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne  y^2$.

I need help verifying my proof.

The question is from the book:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.


Title: Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$

Body:
\begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}

Did I do the right way with those step above?

I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$

Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.

Please help me!!?

Body:
Yes, your approach is correct and you are doing fine. Now note that
\begin{align*}
2017x^{2017} - x^{2016} - \dots - 1&=2017(1+t)^{2017}-\frac{(1+t)^{2017}-1}{t}\\
&=2017(1+2017t+o(t))-\frac{1+2017t+\binom{2017}{2}t^2+o(t^2)-1}{t}\\
&=\left(2017^2-\frac{2017\cdot 2016}{2}\right)t+o(t)=(2017\cdot 1009)t+o(t)
\end{align*}
where $t=x-1\to 0$.

Hence, according to your work
$$\lim_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
= \lim_{t\to 0}\dfrac{(2017\cdot 1009)t+o(t)}{-t(2017+o(1))(2018+o(1))}=-\frac{1}{2}.$$

Bonus exercise. More generally, show that for $n,m\in\mathbb{N}^+$,
$$\lim\limits_{x\to 1}\left(\dfrac{n}{1-x^{n}} -\dfrac{m}{1-x^{m}}\right)=\frac{n-m}{2}.$$


答:
Let's begin.

If $x,y≥0$ and $x≠y$ we have

$$x-y≠0$$ and $$x+y>0$$

Because, $x=y=0$ contradicts with $x≠y$.

Then, we get

$$(x-y)(x+y)≠0$$

$$\implies x^2-y^2≠0$$

$$\implies x^2≠y^2$$

Title: I am trying to prove the following theorem: Let $x$ and $y$ be real numbers. Show that if $x \ne y$ and $x,y \geq 0$, then $x^2 \ne y^2$.


问:
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ .
I am trying to find $x+y$ given that

$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$

It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side

$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$

and see if a term $x+y$ comes out, but it looks hopeless at this point.

答:
I want to add another perspective to this problem. Not from the point of view of algebraic varieties, see zyx's answer for that, but from the point of view of teaching a problem solving strategy to beginners. The solution (as seen from several posts) depends on a useful trick of clearing the square roots from the denominator, but IMVHO there are important steps before we reach that point.

Let's take a look at the problem. We are given a single equation tying the values of two variables $x$ and $y$ together. Fine - business as usual. The equation looks a bit scary because of the square roots and all that, but presumably such an equation defines a curve of some kind. But we are asked about the value of another function, a much simpler one, the sum of the two coordinates. Wait one red minute!? What is this? It sounds as if a single number is expected? May be a range of values? No, that wouldn't quite match the chosen wording. As it stands the question is a bit strange, for a priori there is little reason to suspect that we could say much at all about the value of $x+y$. But let's keep an open mind!

None of the elementary things seem to say much about the equation we were given. How could we even get started? It's time for my battle cry: I couldn't make sense out of it, so I calculated some examples!  Let's plug in some values for $x$, solve for $y$, and see how the land lies. A physicist would call this step "experimental" or "making observations" - the technique is somewhat underrated in math education (though professionals do it all the time). Let's try something simple. What if $x=0$? Then $x+\sqrt{x^2+3}=\sqrt3$, and we are to solve $y+\sqrt{y^2+3}=3/\sqrt3=\sqrt3.$ Hmm? A bit ugly. Let's try $x=1$. Then we get $x+\sqrt{x^2+3}=1+\sqrt{4}=3$, and we are to solve $y+\sqrt{y^2+3}=1$. This time the value on the r.h.s. was a clean integer, which is an improvement I guess, but no flashes of insight yet. Any other values of $x$ that would make $x+\sqrt{x^2+3}$ simple? $x=-1$ comes to mind given that again $\sqrt{x^2+3}$ will be equal to two. So this time $x+\sqrt{x^2+3}=-1+2=1$, and we need to solve $y+\sqrt{y^2+3}=1$.

Ok, to get any example points $(x,y)$ we need to solve equations of the form
$$
y+\sqrt{y^2+3}=a
$$
for some $a$. How do we do that? Well, this is something where we can apply things about equations that we have learned earlier. Keep the square root on one side, move the rest to the other, and square both sides. We need to keep track of the signs, but that's routine. So we get as a consequence
$$
y^2+3=\left(\sqrt{y^2+3}\right)^2=(a-y)^2=a^2-2ay+y^2.
$$
Let's see, the $y^2$ terms cancel, so we get $2ay=a^2-3$ and $y=(a^2-3)/(2a)$.
When $x=1$, we had $a=1$, so $y=(1^2-3)/(2\cdot1)=-1$. Similarly, when $x=-1$
we have $a=3$ and $y=(3^2-3)/6=1$. And when $x=0$, we have $a=\sqrt3$ and $y=(3-3)/2a=0$.

So we at long last have observed some data points: $(x,y)$ can be one of $(1,-1)$, $(0,0)$ or $(-1,1)$. There are countless other points, and you are advised to produce a few others, if this is not enough. What was the question again? The value of $x+y$? Hmm, at all these "observed" points the coordinates sum to zero. Coincidence? Possibly, we are math folks, so suspicion is second nature. But let's try this further (for lack of better ideas, really). If we really have $x+y=0$ always, then we should have $y=-x$. Does that fit? Let's check
$$
(x+\sqrt{x^2+3})(-x+\sqrt{(-x)^2+3})=(x+\sqrt{x^2+3})(-x+\sqrt{x^2+3})??
$$
Ahh, it's the good ol' sum of two things $\times$ their difference: $(a+b)(a-b)=a^2-b^2$ with $a=\sqrt{x^2+3}$ and $b=x$. We have $a^2-b^2=(x^2+3)-x^2=3$! Success! Success? We just showed that points on the line $x+y=0$ do satisfy the given equation, but the direction of the implication is wrong. We wanted the converse. Leaving that to you! Insert a devilish grin here, but also the hint that now you may peek at the other solutions. Also our calculations above showed that to each $x$ there is a single $a$ (show that you never need to divide by zero!) and to each $a$ there will be a single $y$. We already know which $y$ will work at least, so it looks like there are no others???

This is by no means the only approach, but it may take you the distance, when you are in unfamiliar territory. The successful answerers apparently drew from their personal databanks and tuned up pattern recognition heuristics. Of course, building up the said databanks takes time, and you need to work your way through scores of examples to develop and improve those pattern recognition skills.


答:
One way to define the natural logarithm is as an integral:
$$
\log x = \int_{1}^{x}\frac{1}{t} \, dt \, .
$$
When the logarithm is seen in this light, it is easy to argue geometrically that $\log (1+x) \approx x$:

The value of $\log(1+x)$ is given by the area of the blue region. An upper bound for this region is given by the rectangle with vertices $(1,0)$, $(1+x,0)$, $(1+x,1)$, and $(1,1)$. This rectangle has a width of $x$ and height of $1$, and hence has an area of $x$. As $x$ approaches $0$, the area of the blue region approaches that of the rectangle. Hence, $\log(1+x) \approx x$, with $x$ always being a slight overestimate for the value of $\log(1+x)$.


问:
Geometry find the angle
Good day, excuse me, I am looking at congruence themes in Geometry, and I am trying to solve this problem, but I am not sure how to proceed. The coloured numbers is what I have so far (although now looking at it, my angles could be wrong). Any help would be very much appreciated. Thanks!

答:
Join $AB$. Then $AM=MB \Rightarrow \triangle AMB$ is equilateral.

Hence in $\triangle BAD$, $\angle BAD=60+30=90$ and $AD=AB$. Thus $\angle ADB = \angle ABD=45$.

Therefore the exterior $\angle X=\angle ADB + \angle DAM=75 $

Title: Geometry find the angle


问:
A question about univalent functions
The following question was part of my complex analysis assignment and I was unable to solve it. Possibility of discussion of assignment is bleak so I am putting it here for help.

Suppose that f(z) is analytic at $z_0$ with $f'(z_0) \neq 0$ . Show that there exist neighbourhood U and V of $z_0$ and $f(z_0)$, respectively ,such that f(z) is a univalent mapping from U onto V.

I chose a ball $B(z_0 , \epsilon)$ , also in  some neighbourhood of $z_0$ , let $U$, $f$ is 1-1. I need help in showing that $V$ exists.

Can you please help with that?

答:
As user @MartinR noted in the comments to the question, once you have $U$, it is easy to find $V$: just take $V=f(U)$.
The difficult part is to find $U$ on which $f$ is univalent.

One way to prove the result is to use the inverse function theorem from calculus in two variables:
if one writes (identifying complex numbers with vectors in $\mathbb{R}^2$) $f(z)=u(z)+iv(z)=(u(x,y),v(x,y))\in\mathbb{R}^2$, $f'(z_0)\neq0$ implies
$$\text{Jac}\left(\begin{pmatrix}u(x,y)\\v(x,y)\end{pmatrix}\right)_{(x_0,y_0)}=\det\begin{pmatrix}
\partial_xu&\partial_yu \\
\partial_xv&\partial_yv
\end{pmatrix}=\partial_xu\partial_yv-\partial_xv\partial_yu\overset{C-R}{=}(\partial_xu)^2+(\partial_xv)^2=|f'(z_0)|^2\neq 0$$
Thus $f(z)=(u(x,y),v(x,y))$ is $1-1$ on a small enough disc centered at $z_0$. The result follows taking as $U$ such disk and $V=f(U)$.

Another way is by means of Rouche's theorem:
suppose wlog $z_0=0,f'(z_0)=1, f(z_0)=0$.  By continuity, there's a disk $D(0,r)$ on which $\frac32 |z|\ge|f|\ge \frac12|z|$.
Thus, for $w:|w|<\frac r3$, we have that $$|f(z)-f(w)-f(z)|=|f(w)|\le \frac32|w|<\frac r2\le\min_{\partial D}|f|$$ By Rouche, $f(z)-f(w)$ and $f(z)$ have the same number of zeros on $D$, $1$.
Thus
$f$ is $1-1$ on $D(0,\frac r3)$.

The result follows taking $U=D(0,\frac r3)$, $V=f(U)$.

Title: A question about univalent functions


问:
I need visual examples of topological concepts
I'm trying to understand basic concepts of topology, unfortunately I'm a very visual person, and as much documentation there is on how to come up with closed/open/clopen/etc. There are very few visual examples (using actual sets of integers, shapes, etc). So it's very hard for me to understand. Not everyone learns through text or mathematical definitions, so I figured this could help other people that have the same difficulties I have. Thanks for your help!

Can anyone get an example of a topology over a set.

I need one for a topology over a set, an open set, closed set, and clopen set. It needs to be something real. Like, say a set contained numbers { 1,2,3,4 }

It doesn't have to be that set, but it has to be a actual set of something (numbers, letters, geometric shapes), no labels (set X is the union of set Y and set Z), etc.

答:
Consider the set $X=\{1,2,3\}$.

We have the trivial topology, namely $T=\{\emptyset,\{1,2,3\}\}$.

We have the discrete topology, in which every singleton is open. This produces the topology to be $P(X)=\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$

We may have something in between declaring the nonempty open sets to be those containing $1$, and then we have $\{\emptyset,\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$.

Closed sets are those whose complement is open, and clopen sets are those which are open and closed as well.

Note that $X$ is always clopen relatively to $X$ (it might not be if we take a larger set and endow it with a different topology).

In the trivial topology we only have clopen sets, because we only require the minimum from the topology, while the set $\{1,2\}$ is neither open nor closed.

In the discrete topology every subset of $\{1,2,3\}$ is open, therefore every subset is closed. Since every subset is both open and closed, every subset is clopen.

In the last example note that all those that include $1$ are open, so those who do not include $1$ are closed. Since $1$ cannot be in both a set an its complement we have that there are no clopen sets except the empty set and $\{1,2,3\}$.

However it is important to remember that this is all relative and clopen, open and closed sets are only in this relationship with a specific topology and space.

Addendum: I feel that I should add a very basic explanation about topology.

Suppose we have an underlying set $X$. A topology on $X$ is a collection of subsets which includes the empty set as well $X$ itself. It is closed under unions and finite intersections.

The sets which are in the topology are called open, and their complements are called closed. A set which is both open and closed is usually called clopen.


问:
Adjoint operator, bijective
Let $A\in\mathcal{L}(X,Y)$, where $X,Y$ are normed vector spaces. Define the adjoint operator
$$\begin{array}{ll}
A^{\prime}\ : & Y^{\prime}\rightarrow X^{\prime},\\
 & G \mapsto A^{\prime}(G)\ =\ G\circ A.
\end{array}$$
Its easy to show that $A^{\prime}\in\mathcal{L}(Y^{\prime}, X^{\prime})$ and $\|A\| = \|A^{\prime}\|$.

Now, suppose that $A$ is bijective, then show that $A^{\prime}$ is also bijective.

First at all, I have proven that $[\mbox{Im}(A)]^{\circ} = \mbox{Ker}(A^{\prime})$, and then $\mbox{Ker}(A^{\prime}) = \{O\}$. Therefore $A^{\prime}$ is injective. My problem is that I don't know how I can show that $A^{\prime}$ is onto.

Please I need help.

Thanks in advance.

答:
This is not true in general. Let $1<p<q<\infty$. Consider bijective bounded linear operator
$$
I:(\ell_p,\Vert\cdot\Vert_p)\to(\ell_p,\Vert\cdot\Vert_q):x\mapsto x
$$
which is set theoretic identity. Its adjoint is
$$
I^*:(\ell_p,\Vert\cdot\Vert_q)^*\to(\ell_p,\Vert\cdot\Vert_p)^*:f\mapsto f\circ I
$$
Assume it is bijective. Since dual spaces are always complete, then $I^*$ is a bijection between Banach spaces. By open mapping theorem $I^*$ is an isomorphism.
Note that
$$
(\ell_p,\Vert\cdot\Vert_p)^*\cong(\ell_q,\Vert\cdot\Vert_q)
$$
$$
(\ell_p,\Vert\cdot\Vert_q)^*
\cong\left(\operatorname{completion}(\ell_p,\Vert\cdot\Vert_q)\right)^*
\cong(\ell_q,\Vert\cdot\Vert_q)^*
\cong(\ell_p,\Vert\cdot\Vert_p)
$$
so using that $I^*$ is an isomorphism we see that $(\ell_p,\Vert\cdot\Vert_p)$ and $(\ell_q,\Vert\cdot\Vert_q)$ are isomorphic via some operator $J$. But this is impossible, so we get a contradiction. Thus $I^*$ is not bijective, though  $I$ is bijective.

On the other hand, if you assume that $X$ and $Y$ are Banach spaces then you have an equivalence
$$
A\text{ is bijective}\Longleftrightarrow A'\text{ is bijective}
$$
For proof see this and this answers.

Title: Adjoint operator, bijective


问:
On convergence of WOT
Let $H$ Hilbert space. $\{A_n\}_{n\ge0}$ a sequence in $B(H)$ such that the sequence
$\{\langle A_nx,y\rangle\}_{n\ge0}$ converges for all $x,y\in H$.

Can we show that there exist $A\in B(H)$ such that
$$\lim_{n \to \infty}\langle A_nx,y\rangle=\langle Ax,y\rangle$$
for all $x,y\in H$.

I guessed one can use Uniform Bounded Principle but couldn't figure out a concrete argument.

This question is motivated from the fact (can be proved using Uniform Boundedness Principle)-
If $\{A_n\}_{n\ge0}$ a sequence in $B(H)$ such that the sequence
$\{ A_nx\}_{n\ge0}$ converges for all $x\in H$ then there exist $A\in B(H)$ such that
$$\lim_{n \to \infty}A_nx= Ax$$

Thanks in advance.

答:
Define $\sigma:H\times H\to\mathbb{C}$ by setting
$$\sigma(x,y):=\lim_{n\to\infty}\langle A_nx,y\rangle $$

This is a well defined map and one can easily see that this is linear in the first argument and conjugate linear in the second argument. Moreover, note that
$\{\|A_n\|\}_{n=1}^\infty\subset[0,\infty)$ is a bounded sequence: You can find a proof of this result (using the principle of uniform boundedness) here.

So now there exists $M>0$ such that $|\sigma(x,y)|\leq M\cdot\|x\|\cdot\|y\|$ for all $x,y\in H$. This proves that there exists a bounded operator $A\in B(H)$ so that $\sigma(x,y)=\langle Ax,y\rangle$, as you can deduce from this theorem.

Title: On convergence of WOT


问:
If $p,q,r$ are distinct primes such that $p<q<r$, then find all solutions to $p+q+r=pq$
I was solving another question, and was unable to solve the case written above. The question is as follows:

If $p,q,r$ are distinct primes such that
$$\frac{p^{3}+q^{3}+r^{3}}{p+q+r}=249$$then find all possible values of r

On using the identity $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-xz)$, one can reduce the above question to the following,

Find all distinct primes $p,q,r$ such that
$$p+q+r|pqr$$

We have four cases for the following:

$p+q+r=pqr$

$p+q+r=qr$

$p+q+r=pr$

$p+q+r=pq$

I concluded that no solutions exist for the first three cases by assuming W.L.O.G, $p<q<r$. However, I am unable to do the same for the last case.

I request somebody to please provide a solution for this case or an alternate solution for the main question itself. Any help is much appreciated. Thanks a lot :)

答:
As mentioned in the comments, there are many solutions to $p+q+r = pq$. In particular, this can be rearranged to $r + 1 = (p-1)(q-1)$, so it is not unlikely to generate a pair $(p,q)$ from a prime $r$. In light of this, we should return the the given relation directly. WLOG suppose $p < q < r$. Then

$$249 = \frac {p^3+q^3+r^3}{p+q+r} > \frac {r^3}{3r} = \frac {r^2}{3}$$

This gives an upper bound: $r < \sqrt{744} < 28$. Hence $p<q<r\le23$.

Initially I listed the remainder of the cubes of these primes modulo $83$ (since $249 = 3\times 83$) and determined that $(3,11,19)$ is the only solution. [Warning: not fun]

However I believe that using your criterion $p+q+r=pq$, or $r+1 = (p-1)(q-1)$ would actually be simpler.

First note that $p \ne 2$: $2 + q + r = 2q \implies 2 +r = q$, which contradicts $r > q$.

Then, notice that $p < 5$: else $p \ge 5, q \ge 7, r = (p-1)(q-1)-1 \ge 23$. This gives the only case $(5,7,23)$, where $\dfrac {p^3+q^3+r^3}{p+q+r} = 361$. Hence we can only have $p=3$.

Since $r+1 = 2(q-1) \le 24$, we just need to check $q=5,7,11$, which gives $r = 7,11,19$, and $\dfrac {p^3+q^3+r^3}{p+q+r} = 33, 81, 249$. Hence $(3,11,19)$ is the only solution. Removing the WLOG, $r$ can take on all these values, so the required sum of possible values of $r$ is $33$.


问:
Maximal element of a poset
I am currently studying Zorn's lemma and its my understanding that the definition of a maximal element is a primitive to the lemma.

From my understanding a maximal element $M$ of a partially ordered set  $(S,<)$  is an element $M \in S$ where, if $M \leq x $ for some $ x \in S$, then $x=m$.

This easilty illustrated when the set is {∅, {1}, {2}, {3}, {1, 2}}. Then the maximal elements are {3} and {1,2}.

But what if the partially ordered set i am considering is $(S,R)$.
So for example let S=P({a,b,c}) which is the power set of {a,b,c} and the relation $\subseteq$.
Now whats is the maximal element in this set?

Thank you in advance.

答: The only maximal element in the power set $S = P(\{a,b,c\})$ is $\{a,b,c\}$. Clearly $\{a,b,c\}$ is itself maximal and any other maximal subset $M$ of $\{a,b,c\}$ satisfies $M \subseteq \{a,b,c\} = x$ and thus by the maximality has $M = x= \{a,b,c\}$.

Title: Maximal element of a poset


问:
Show that $n^2 + m^2 = 2nm$ is an equivalent relation
How can I show this is an equivalence relation:

$nRm ⟺ n^2 + m^2  = 2nm$

I had no problem proving the first two steps (reflexivity and symetric). However, for the transitivity part, I can't get to show it. Anyone can help me?

Body: Hint: $n^2+m^2=2nm\iff(n-m)^2=0\iff n=m$.


问:
Prove or disprove: 'Island' of compact set in $\mathbb{R}$ is closed?
I've been trying to prove something, and this seems to be getting in the way:

Let $X \subset \mathbb{R}$ be compact. Then $X$ is bounded, and let $k \in (\inf X, \sup X)$ such that $k \notin X$. Then is the set $X' = \{x: x \in X, x>k \}$ closed?

Intuitively the answer appears to be a yes, since the only closed sets in $\mathbb{R}$ I've dealt with are unions of finite sets and closed intervals, and if this is the case then there doesn't appear to be any tangible way that $X'$ isn't either a union of closed intervals and finite sets.

I've tried proving that the complement of $X'$ is open and tried proving compactness using open covers, but have been unable to go anywhere. At the same time, splitting something like $[0,1]$ into $[0,a] \cup [a,1]$ and $[0,a] \cup (a,1]$ doesn't satisfy the condition, and so I haven't found a counterexample either.

Is there any proof/disproof for the above assertion?

答:
$k \not\in X$, so that
$$
X' = \{x: x \in X, x>k \} = \{x: x \in X, x \ge k \}
 = X \cap [k, \infty)
$$
is the intersection of two closed sets, and therefore closed.

It suffices that $X$ is a closed set. If $X$ is compact then $X'$ is a closed subset of $X$ and therefore compact as well.

The condition $k \in (\inf X, \sup X)$ is not needed for this argument.


问:
What happens when C is only partly within the vector field?
So I have the following question where $$\vec{G}=(x,y)/\sqrt{x^2+y^2-1}$$ is a vector field.
I have to calculate the line integral for $$\int_{\mathrm{C}}\vec{G}\, d\vec{r}\, $$ where $C$ is the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.

I know that $\vec{G}$ has the domain $x^2+y^2 \gt 1,$ (meaning that the circle with radius 1 is not part of the domain) and the ellipse has radius 2 with x-component and 3 for y-component.
I have calculated param $x=2cost$ and $y=3sint$ where $0\le t\le 2\pi$.
So does this mean that only a part of the ellipse is within $\vec{G}$?

How do I calculate the line integral in this case?

答:
The ellipse $C: x^2/4+y^2/9=1$ is well within the field $\vec{G}=\frac{x\hat i+y\hat j}{\sqrt{x^2+y^2-1}}$ which is defined for the exterior of circle $x^2+y^2=1$ so you can easily evaluate the line integral using Green's theorem(after making it simply connected using a two-way cut).

$\int_C\vec{G}.\vec{dr}=\int\int_R curl(\vec{G})dR$

where $R$ is the region bounded by ellipse on $xy-$plane.