问:
Tensor with an Exact Sequence of Free Module Still Exact?
Let $F\rightarrow G\rightarrow H$ be an exact sequence of free $R$-modules and $M$ be any $R$-module.

Is that true that $F\otimes_RM\rightarrow G\otimes_RM\rightarrow H\otimes_RM$ is exact?

If the answer is no, does that hold when all the modules above are finitely generated?

答:
Let $R=\mathbb{Z}$. Then
$$0\rightarrow\mathbb{Z}\xrightarrow{\times2}\mathbb{Z}$$
is an exact sequence of free $R$-modules, but the tensor product with $M=\mathbb{Z}/2\mathbb{Z}$ is not exact.


问:
Connected sums where the separating "2"-sphere bounds a punctured ball
I have a question about connected sums in 3-dimensional spaces in relation to normal surfaces.
I was reading the paper https://arxiv.org/abs/math/9712269 by Joel Hass, which at one point outlines a proof of "Knesers Theorem", which states that a triangulated $3$-dimensional manifold (without boundary) $M$ can be decomposed
non-trivially along $2$-spheres (in terms of the connected sum) up to a number of times bounded by a constant $k(M)$ dependent on $M$ and its triangulation.

For the proof, the notion of a normal surface was introduced, which is a surface embedded in $M$ which is a disjoint union of normal triangles or normal quadrilaterals, where the latter are intersecting a tetrahedron of $M$ in 3 or 4 edges and faces.

At Page 7 in the proof, the notion of a "punctured ball" is introduced, which is just defined as "a ball with some open balls removed". At Page 8 in the proof it is mentioned, that if we use a separating (normal) $2$-sphere $S$ to decompose $M$ into a connected sum and $S$ bounds a punctured ball, then one of the summands in the connected sum is trivial, meaning its homeomorphic to $S^3$. I can't get my head around the following things:

Do we have specific extra conditions regarding the definition of a punctured ball? For example that we only allow for finitely many open balls cut out or that they have to be completely enclosed in the main ball?

I can't really see why a $2$-sphere bounding a punctured ball leads to a $S^3$-summand. Is it a simple trick I don't see (maybe having to do with normal $2$-sphere) or is this maybe done by calculation of homology groups? I think I am missing something important here.

I appreciate any kind of help and ideas, thanks in advance for your time!

答:
Yes, there are finitely many and they have disjoint closures.

The summands (by definition of the connect sum) are gotten by gluing 3-balls to each 2-sphere in a connected region in the complement of the 2-spheres, This results in a  3-manifold with no boundary for each component.

If one of the summands is a punctured 3-ball then gluing in 3-balls to each boundary 2-sphere (ie filling back in the punctures) results in a 3-sphere, which is a trivial summand.

ContentLicense: CC BY-SA 4.0


问:
Definite integral of an exponential function
I are trying to solve the following definite integral
$$
\int_{0}^{\infty}\exp\left(-x+\frac{x^{1-\gamma}}{1-\gamma}\right) {\rm d}x
\quad\mbox{where}\ \gamma > 1\ \mbox{is a parameter}.
$$
My derivation is
\begin{align}
\int^\infty_0 e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x =& - \int^\infty_0 e^{\frac{x^{1-\gamma}}{1-\gamma}} d e^{-x} \\
=& - e^{\frac{x^{1-\gamma}}{1-\gamma}} \cdot e^{-x} |^\infty_{x=0} +
\int^\infty_0 e^{-x} d e^{\frac{x^{1-\gamma}}{1-\gamma}} \\
=& \int^\infty_0 x^{-\gamma} e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x \ .
\end{align}
I was stuck here. Is it possible to find an explicit solution for this integral？If yes, how can I proceed further? How about the case $0 < \gamma < 1$?

答:
This question makes me feeling younger (I need it) since, in my former research group, we studied the integrals for integer values of $\gamma$.

$$I_\gamma=\int^\infty_0 e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x$$

$$I_2=2 K_1(2)$$
$$I_3=\frac{1}{\sqrt{\pi }}G_{0,3}^{3,0}\left(\frac{1}{2^3}|
\begin{array}{c}
 0,\frac{1}{2},1
\end{array}
\right)$$
$$I_4=\frac{\sqrt{3}}{2 \pi }G_{0,4}^{4,0}\left(\frac{1}{3^4}|
\begin{array}{c}
 0,\frac{1}{3},\frac{2}{3},1
\end{array}
\right)$$
$$I_5=\frac{1}{\sqrt{2} \pi ^{3/2}}G_{0,5}^{5,0}\left(\frac{1}{4^5}|
\begin{array}{c}
 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},1
\end{array}
\right)$$
$$I_6=\frac{\sqrt{5}}{4 \pi ^2}G_{0,6}^{6,0}\left(\frac{1}{5^6}|
\begin{array}{c}
 0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1
\end{array}
\right)$$ and you can notice simple patterns in the Meijer G functions.

For the non-integer (but rational) values of $\gamma > 1$, I checked using a CAS and the results still express in terms of  Meijer G functions.

From a numerical point of view, for large values of $\gamma$, the product
$\gamma I_\gamma$ is quite close to linearity.


问:
Finding limits with substitution
$\lim_{x\to0+}(\sinh(x))^{1/x}$

I started by setting $y=\frac{\sinh(x)}{x}$ and taking the natural logarithm of both sides and trying to solve the limit for $ln(y)$ but I got stuck trying to solve $\lim_{x\to0+}\frac{\ln(\sinh(x))}{x}$. Any ideas? Thanks in advanced.

答:
$$\lim_{x\rightarrow 0^+}(\sinh(x))^{\frac{1}{x}}=\lim_{x\rightarrow 0^+}e^{\frac{1}{x}\ln(\sinh(x))}$$

$$\lim_{x\rightarrow 0^+}\frac{1}{x}=+\infty $$

$$\lim_{x\rightarrow 0^+}\sinh(\ln(x))=-\infty $$

so $$\lim_{x\rightarrow 0^+} e^{\frac{1}{x}\ln(\sinh(x))}=e^{-\infty}=0$$

ContentLicense: CC BY-SA 3.0

Title: Finding limits with substitution


问:
Is the Jackson integral of $e_q(x)$ , $e_q(x)$ itself?
Some info-

Jackson integral (q-analog of standard integration) simply defined as -$$\int f(x){\mathrm d}_qx=(1-q)x\sum_{k=0}^\infty q^kf(q^kx)$$

q-exponential defined as $$e_q(x)=\sum_{n=0}^\infty \frac{x^n}{[n]_q!} =
\sum_{n=0}^\infty \frac{x^n (1-q)^n}{(q;q)_n} =
\sum_{n=0}^\infty x^n\frac{(1-q)^n}{(1-q^n)(1-q^{n-1}) \cdots (1-q)}$$

My question - $$\text{is}\int e_q(x){\mathrm d}_qx = e_q(x) ?$$
My approach- $$\int e_q(x){\mathrm d}_qx=(1-q)x\sum_{k=0}^\infty q^ke_q(q^kx) \\ =(1-q)x\sum_{k=0}^\infty q^k\sum_{n=0}^\infty(q^kx)^n\frac{(1-q^k)^n}{(q^k;q^k)_n} \\ $$
How should I proceed?

答:
Write out $e_q(q^kx)$ explicitly and then interchange the summations:

$$ \begin{array}{ll} \displaystyle \int e_q(x)\,\mathrm{d}_qx & \displaystyle =(1-q)x\sum_{k=0}^\infty q^k e_q(q^kx) \\[5pt] & \displaystyle  =(1-q)x\sum_{k=0}^\infty q^k \sum_{n=0}^\infty \frac{(q^kx)^n}{[n]_q!} \\[5pt] & \displaystyle = (1-q)x\sum_{n=0}^\infty \frac{x^n}{[n]_q!} \sum_{k=0}^\infty (q^{n+1})^k \\[5pt] & \displaystyle =\sum_{n=0}^\infty \frac{x^{n+1}}{[n]_q!} \frac{1-q~~~~~}{1-q^{n+1}} \\[5pt] & \displaystyle = \sum_{n=0}^\infty \frac{x^{n+1}}{[n+1]_q!} \\[5pt] & = e_q(x)-1. \end{array} $$

In particular, $\displaystyle \int x^n\,\mathrm{d}_qx=\frac{x^{n+1}}{[n+1]_q} $.


问:
Proof related to Ramsey number
I am studying about Ramsey theorem and I came across this question as mentioned below:
Suppose $R(K_3; k)$ be such minimal n that in every edge colouring of $K_n$ with
k colors there is a monochromatic $K_3$ (that is such that all edges have the same color.). So how do I show that $R(K_3; k) > 2^k$

Can anyone please help me with this and thanks in advance.

答:
This is obviously true for $k=0$. Suppose you can edge-color $K_{2^k}$ with $k$ colors so that there is no monochromatic $K_3$. Partition vertices of $K_{2^{k+1}}$ into $2$ sets $U$ and $V$ of size $2^k$ and edge-color the two induced subgraphs $K_{2^k}$ on $U$ and $V$ with colors $1$ through $k$, then color all edges between a vertex in $U$ and a vertex in $V$ with color $k+1$. Check that this coloring has no monochromatic $K_3$.

To find an edge coloring without a monochromatic $K_3$ non-inductively, label the vertices of $K_{2^k}$ with binary strings of length $k$, then color an edge with color $i\in[1,k]$ if the leftmost bit in which the two endpoints' labels differ is the $i$th bit.

Title: Proof related to Ramsey number


问:
Rewriting a Binomial expansion and expanding the rewritten version after making a convenience substitution to get a totally different answer
I know that if I have a function $$f(x) = \frac{1}{x-1}$$ that if I perform a binomial expansion on I get the  result $$f(x)\approx -1 -x -x^2-x^3-x^4 ...$$

However if I rewrite the function $$f(x) = \frac{1}{x-1} = \frac{1}{x(1-\frac{1}{x})} = \bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$$

Then we make the expansion of $\bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$ by using the subsitution $\frac{1}{x} = z$

We get $$(1-z)^{-1} = 1+z+z^2+z^3+z^4... = 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+ \frac{1}{x^4}...$$
so that
$$x^{-1}(1-z)^{-1} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4}+ \frac{1}{x^5}...$$

This is completely different result!
Why is this the case?

答: The expansion of $\frac 1 {x-1}$ you are using is valid only for $|x|<1$. When you change $x$ to $\frac 1 x$ and use the expansion you are assumung that $|x|>1$. Since there is no number with $|x|<1$ and $|x| >1$ you cannot use both of them for the same $x$.


问:
how to find a data set of 5 values from a given variance?
the question is to create a data set of 5 values with variance=40 and mean=median, so how do we do that ?

I tried simplifying the variance formula:

40=(x^2)+(mean)^2/(5-1)

so if i assume mean as 10 then : 160 + 100=x^2

and as mean is assumed as 10:   total/5=10, so total=50

if the values are a,b,c,d,e , as the median=mean, c=10

so i assumed a data set to be: 1,4,10,15,20 as it gives a total of 50 but it doesn't satisfy the equation created from variance:

**(a^2)+(b^2)+(c^2)+(d^2)
+
(e^2)=260(160+100)

(a^2)+(b^2)+(100)+(d^2)+
(e^2)=260(160+100)

(
a^2)+(b^2)+(d^2)+(e^2)=260-100(160)**

now iam confused how to create the data set as the square of the individual data set doesn't meet the requirements for variance equation.as the data set should be in order d^2 and e^2 values should be greater than 100,but that will add more than 160.

can you please help me with this, iam trying to figure this out for long.

it will be a great help if i can get answers, thank you in advance

答:
in a very easy way, chose a symmetric set around zero with $E(X^2)=40$

Example:

$$\{x_1;x_2;x_3;x_4;x_5\}=\{-8;-6;0;6;8\}$$

In this case you have

$$\mathbb{E}[X]=\text{median}[X]=0$$

$$\mathbb{V}[X]=\mathbb{E}[X^2]=\frac{2\times 64+2\times 36}{5}=40$$

Of course if for "variance" you mean the Sample variance, say the one divided by $n-1$ the reasoning remains valid but you have to take into consideration the definition of variance you have to apply: in this second case, a nice sample example is the following:

$$\{x_1;x_2;x_3;x_4;x_5\}=\{-8;-4;0;4;8\}$$


问:
Find expression that represent dots quantity in terms of radius
For $n$ dots, such that $n=10$:

pick a pivot, such that $p=[8, 11]$, calculate the distance to each:

let
let $r_{min} = $ smallest radius distance, which in all cases $0$

let $r_{max} = $ largest radius distance, which in this case $16$

Perform a circular expansion originated from $[8, 10]$ from $t_0 \rightarrow t_1$, where $t_0 = 0$ &  $t_1 = 1$.

Both time and radius increment evenly, which

At $t_0$, $r_0 = r_{min} = 0$
 At $t_1$, $r_1 = r_{max} = 16$

Question:

let

quantity of dots be $n$

vertexes of dots be $[[x_1, y_1], [x_2, y_2] ..]$

pivot be $p = [x_p, y_p]$

circular expansion starts at $t_0 = r_0 = r_{min}= 0$

circular expansion end at $t_0 = 1$ & $r_1 = r_{max}$

Find $n(r)$, that $\int_{r_{min}}^{r_{max}} n(r) \,dr = n$

for the example above, here are some expected results:

$\int_{0}^{2} n(r) = 2$

$\int_{0}^{3} n(r) = 3$

$\int_{0}^{7} n(r) = 7$

$\int_{0}^{16} n(r) = 10$

The function should reflect in which segment of radius increment, there are relatively a large amount of dots getting covered, therefore, for any interval, for example, $n(6)$ the function should returns a decimal, perhaps approximately $6.7$

Thanks very much for reading to the end. I am not certain if I have spoken with enough clarity. It's a vague idea, I am not sure if it's tangible at all.
I will be so excited if someone give me a hand.

答:
We consider that the distances to the pivot point are ranked in increasing order, i.e.,  $r_k$ = distance to the $k$th closest point with $r_0=0$.

Your issue boils down to a linear interpolation between the following points :

$$(x,y)=(r_k, k)$$

(of course not in the same plane as the initial plane !).

Otherwise said, your function is affine on each interval $(r_k,r_{k+1})$ with:

$$\text{If} \ \ r_k \le x \le r_{k+1}, \ \ \ n(x)=\dfrac{(x-k)}{(r_{k+1}-r_k)}+k$$

Matlab program for the figure:

```<code>A=[...
10, 2, 6, 8, 7,11,16, 8,11,20
17,15,12,11, 8, 5, 5, 4, 2, 0];% coordinates
n=size(A,2);% number of points
p=4;% pivot rank;
for k=1:n
    r(k)=norm(A(:,k)-A(:,p));% distances
end;
r=sort(r);% distances sorting
for k=0:n-2;
    plot([r(k+1),r(k+2)],[k,k+1]);
end;
for k=0:n-1;
    plot(r(k+1),k,'or');
end;
</code>```

Body:
We consider that the distances to the pivot point are ranked in increasing order, i.e.,  $r_k$ = distance to the $k$th closest point with $r_0=0$.

Your issue boils down to a linear interpolation between the following points :

$$(x,y)=(r_k, k)$$

(of course not in the same plane as the initial plane !).

Otherwise said, your function is affine on each interval $(r_k,r_{k+1})$ with:

$$\text{If} \ \ r_k \le x \le r_{k+1}, \ \ \ n(x)=\dfrac{(x-k)}{(r_{k+1}-r_k)}+k$$

Matlab program for the figure:

A=[...
10, 2, 6, 8, 7,11,16, 8,11,20
17,15,12,11, 8, 5, 5, 4, 2, 0];% coordinates
n=size(A,2);% number of points
p=4;% pivot rank;
for k=1:n
    r(k)=norm(A(:,k)-A(:,p));% distances
end;
r=sort(r);% distances sorting
for k=0:n-2;
    plot([r(k+1),r(k+2)],[k,k+1]);
end;
for k=0:n-1;
    plot(r(k+1),k,'or');
end;


问:
Is there a solution to $\int \frac{x^7-2}{\ln x}dx$ without the $\text{Ei}(x)$ and $\text{Li}(x)$ integral?
I found this question in writing so perhaps I have made an error in the handwriting interpretation. I think it should involve a simple $u$-substitution but Wolfram Alpha gives an answer involving the $\text{Ei}(x)$ and $\text{Li}(x)$ integral. Is there a substitution if the exponent was to power $2$ instead of $7$?

答:
The primary definition of $\text{Li}(x)$ is
$$ \text{Li}(x) = \int_{0}^{x} \frac{dt}{\ln(t)} $$
so there is no way, in this case, to get away from $\text{Li}(x)$. The integral in question is seen to be:
\begin{align}
\int \frac{ x^{a} - b}{\ln(x)} \, dx &= \int \frac{e^{a \, \ln(x)}}{\ln(x)} \, dx - b \, \text{Li}(x) = \text{Ei}((a+1) \, \ln(x)) - b \, \text{Li}(x) + c_{0}.
\end{align}

If the limits of the integral, for example, are $(0,1)$ and $b=1$, then
\begin{align}
\int_{0}^{1} \frac{x^{a} - 1}{\ln(x)} \, dx &= [\text{Ei}((a+1) \, \ln(x)) - \text{Li}(x) ]_{0}^{1} = \ln(1+a)
\end{align}
For a different set of methods see this question.


问:
To find all integers n such that the given expression is a perfect square
The question is as follows

Determine all integers n such that $n^{4}-n^{2}+64$ is the square of an integer

Here is my approch

Let $n^{4}-n^{2}+64=k^{2}$. On multiplying both sides by 4, we get
$$4n^{4}-4n^{2}+256=4k^{2}$$
$$\Rightarrow(2n^{2}-1)^{2}+255=4k^{2}$$
$$\Rightarrow(2k+2n^{2}-1)(2k-2n^{2}+1)=255$$
The only ordered pair of factors possible are (3,85); (5,51); (15,17). We need not consider negative factors as we are squaring later.

Thus, we get $k=8, 14, 22$, which implies $k^{2}=64, 196, 484$ and finally $k^{2}-64=0, 132, 420$
$$\Rightarrow n^{4}-n^{2}=0, 132, 420$$
These 3 equations admit only $n=0, \pm1$ as integer solutions.

However, $n=\pm8$ also satisfies. Thus, my solution is incomplete or wrong

 Could anyone please point out the mistake in my solution or provide the necessary extension to fix it.
Thanks a lot for help!! :)

答:
Everything you did was correct, but you missed the factorization $255=1\cdot 255$. This gives $4k=1+255=256$ and so $k=64$, giving
$$n^4-n^2=k^2-64=64^2-64,$$
which gives that $n=\pm 8$ -- exactly what you missed.


问:
Obtain B to minimize $\|\boldsymbol{X}-\boldsymbol{B}\|_{\boldsymbol{F}}^{2}$ with constraint on the spectral norm of B to be ≤ 1
I stumbled upon this problem where I have to

I want to obtain B such that we Minimize $\|\boldsymbol{X}-\boldsymbol{B}\|_{\boldsymbol{F}}^{2}$ subject to the constraint
that the spectral norm of  is less than or equal to one.

I tried to proceed using the SVD decomposition and trying to involve some inequalities of norms. but with no result.
Any suggestions??

答:
According to Wikipedia, the Frobenius norm is equivalent to $\displaystyle \Vert X-B\Vert_F^2=\sum_{i=1}^{min(m,n)}\sigma_i^2(X-B)$

Write X in its svd form $X=\sum_i\lambda_iu_iv_i^T$

Write B as $\sum_ib_iu_iv_i^T$

Then $X-B=\sum_i(\lambda_i-b_i)u_iv_i^T$

Take $b_i=min\{\lambda_i, 1\}$

Then this achieves the minimum frobenius norm.

Disclaimer: not sure if this works.


答:
Deal with the spread first. If I have a symmetric set of $5$ values around zero, and $E(X^2)=40$, and sum of all squares to be $40\cdot 5=200$, then it would be safe to picture $2$ values (from either positive or negative side of the distribution) where their sum of squares is $100$. Casually pick $6$ and $8$ to be our candidates, then we have:

$${-8, -6, 0, 6, 8}$$

Finally, shift each value with reference to your mean, then we have:

$${2, 4, 10, 16, 18}$$

Title: how to find a data set of 5 values from a given variance?

Body:
the question is to create a data set of 5 values with variance=40 and mean=median, so how do we do that ?

I tried simplifying the variance formula:

40=(x^2)+(mean)^2/(5-1)

so if i assume mean as 10 then : 160 + 100=x^2

and as mean is assumed as 10:   total/5=10, so total=50

if the values are a,b,c,d,e , as the median=mean, c=10

so i assumed a data set to be: 1,4,10,15,20 as it gives a total of 50 but it doesn't satisfy the equation created from variance:

**(a^2)+(b^2)+(c^2)+(d^2)
+
(e^2)=260(160+100)

(a^2)+(b^2)+(100)+(d^2)+
(e^2)=260(160+100)

(
a^2)+(b^2)+(d^2)+(e^2)=260-100(160)**

now iam confused how to create the data set as the square of the individual data set doesn't meet the requirements for variance equation.as the data set should be in order d^2 and e^2 values should be greater than 100,but that will add more than 160.

can you please help me with this, iam trying to figure this out for long.

it will be a great help if i can get answers, thank you in advance


问:
Proving topologist's sine curve limit points.
I am not sure if this makes sense or not.

I was trying to prove that $\{0\} \times [-1,1]$ is a subset of the set of limit points of the topologist's sine curve. The topologist's sine curve being $S=\{(x,\sin\frac{1}{x})|x \in (0,1)\}$. I was wondering if this approach is ok. I tried to show if $y$ is fixed in $[-1,1]$, then there is a sequence of points $x_n \rightarrow 0$, such that any open ball around $(x_n,y)$ intersects $S$ for each positive epsilon, and natural number $N$. Here was my attempt:

Because of the rapid oscillations of $\sin(\frac{1}{x})$ near $0$, consider the points in $\mathbb{R}^2$ of the form $(0,y), y \in [-1,1]$.I will try to show there is a sequence $x_n$ of points converging to $0$, such that for a fixed $y \in[-1,1]$, and each point in the sequence, every neighborhood of $(x_n,y)$ is contained in an open ball, that intersects the set $S$. For each $n \in\mathbb{N}$ let $\epsilon_n>0$ be an arbitrary small positive number.By the archimedean property choose $N \in \mathbb{N}$  such that $n \geq N \implies \frac{1}{2\pi n}<\epsilon_{n}$. Then for each $\epsilon_n>0$,choose a point  $x_n \in [\frac{1}{2 \pi(n+1)},\frac{1}{2 \pi n})$, as a term in the sequence. Then  for $n \geq N$,  $B((x_n,y),\epsilon_n) \cap S \neq \varnothing$. Since $\text{lim}x_n=0,B((0,y),\epsilon) \cap S \neq \varnothing$ for any positive $\epsilon$.

答:
"Rapid oscillations of $\sin 1/x$ near $0$" is not a rigorous argument. How do you demonstrate $B_\epsilon((x_n,y))\cap S\ne\emptyset\forall\epsilon>0$?

Let $y\in[-1,1]$. Consider any basis element $B_\delta((0,y))$. Now, $0<x<\delta\implies1/\delta<1/x<\infty$ and $\sin\left(\left(\frac1\delta,\infty\right)\right)=[-1,1]$, implying that there exists $x_1\in(0,\delta)$ such that $\sin(1/x_1)=y$. So $(x_1,y)\in S\cap B_\delta((0,y))$ and $(x_1,y)\ne(0,y)$. This completes the proof.


问:
Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle

If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$

My attempt: $$\sum_{cyc} cos \,\frac{A}{2}=\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \frac{3 \sqrt{3}}{2}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. As the first inequality is obvious, it is enough to show that $$\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. We prove that $$\sqrt{\frac{s(s-a)}{bc}} \leq\frac{\sqrt{s(s-a)}}{a}$$ or equivalently $a^{2}\leq bc$. Similary $b^{2}\leq ac$ and $c^{2}\leq ab$ and adding the inequalities we get $(a-b)^{2}+(b-c)^{2}+ (c-a)^{2}\leq 0$.

答:
Let $a = y+z,\,b = z+x,\,c = x+y$ for $x,\,y,\,z>0,$ we have
$$\sum \frac{\sqrt{s(s-a)}}{a} = \sum \frac{\sqrt{\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)}}{a} = \sum \frac{\sqrt{x(x+y+z)}}{y+z}.$$
We will show that
$$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \geqslant \frac{3\sqrt 3}{2}.$$
Indeed, suppose $x+y+z=3$ then the inequality become
$$\sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{3}{2}.$$
We have
$$\left(\frac{\sqrt{x}}{3 - x}\right)^2 - \left(\frac{x}{2}\right)^2  = \frac{x(4 - x)(x-1)^2}{4(3-x)^2} \geqslant 0.$$
So
$$ \frac{\sqrt{x}}{3 - x} \geqslant \frac{x}{2},$$
or
$$ \sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{x+y+z}{2} = \frac32.$$
The proof is completed.


问:
Finding the self intersection point of two parametric equations.
Find the values of t where the graph of the parametric equations crosses itself:
$x = t^3-4t^2+t+7$ and $y = t^2-t$.

I am asking this problem on behalf of me and my Calculus 3 class. We understand the idea of $t$ being two different times, so making them say $s$ and $t$, then setting $x(s) = x(t)$ and $y(s) = y(t)$.

So we get:

$$x(t) = t^3-4t^2+t+7 = s^3-4s^2+s+7 = x(s)$$

$$y(t) = t^2-t = s^2-s = y(s)$$

However, past that we are completely lost. Our textbook goes over this type of problem for all of one paragraph, summarizing the process heavily. We asked our professor to go over this type of problem again but it still makes next to no sense. We also looked on YouTube and here on StackExchange for examples but they are either not the same type of problem, or the answers given are heavily summarized with no clear process. We are looking for a concrete process or set of ideas to follow in order to solve a problem like this.

Any help to this is greatly appreciated!

答:
First attack $y,$ the weakest link.  Substitute $s=t+u$ in $y$ to obtain $0=2tu+u^2-u.$ Cancel $u$ to obtain $u=1-2t.$ Therefore, $s=1-t.$ If $t=s,$ then $t={1\over 2},$ a case we must reject.
Put $s =1-t$ in $x$ to obtain
$$f(t)=2t^3-3t^2-3t+2=0.$$
The rejected case tells us that $2t-1$ must divide $f(t).$ In fact, $$f(t)=(2t-1)(t^2-t-2)=(2t-1)(t+1)(t-2).$$ Thus, the curve crosses itself at $(x,y)=(1,2)$ when $t$ is $-1$ and $2.$


问:
Can anyone explain this to me? (Epsilon and Delta)
If I have the statement $\lim_{x\to 0}f(x)=0$ and
$$
f(x)=\begin{cases}
x,&x>0\\
x-0.3,&x<0
\end{cases}
$$
how do I prove that the statement is false?

Is this correct?
$[f(x)-0]$ < e for every $[x-0]<d$

So if we choose e= 0.3 then it is $[f(x)]<0.3$ For every $[x]<d$

$-0.3<f(x)<0.3$ for x<d
But where do i go next?

答: The notation $\lim_{x\to 0^+}f(x)$ means the limit of $f(x)$ as $x$ approaches $0$ through positive values. Precisely, $$L^+=\lim_{x\to 0^+}f(x) \iff$$ $$\iff  \forall e>0\;\exists d>0\;\forall x\in (0,d)\;(|f(x)-L^+|<e).$$ Similarly we define $$L^-=\lim_{x\to 0^-}f(x) \iff$$ $$ \iff \forall e>0\;\exists d>0\;\forall x\in (-d,0)\;(|f(x)-L^-|<e).$$ Meanwhile we have $$L=\lim_{x\to 0}f(x) \iff$$ $$\iff  \forall e>0\;\exists d>0\;\forall x\in (0,d)\cup (-d,0)\;(|f(x)-L|<e).$$ Show that $$L=\lim_{x\to 0}f(x) \iff \,\lim_{x\to 0^+}f(x)=L=\lim_{x\to 0^-}f(x)\,.$$ For the $f$ of your Q, do $\lim_{x\to 0^+}f(x)$ and $\lim_{x\to 0^-}f(x)\;$ exist, and, if so, what are they?


问:
Why the probability of single element event is "zero" on continuous interval?
Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.

cool thanks!

答:
Assume $\xi$ is a random variable and $p(x)$ is the density function of $\xi$.

For all $t\in R$, we have $P(\xi =t)\leq P(t\leq \xi\leq t+h)=\int_{t}^{t+h}p(x)dx$ for any $h>0$.

Therefore $0\leq P(\xi =t)\leq \lim\limits_{h\to 0^+}\int_{t}^{t+h}p(x)dx=0$, so $P(\xi =t)=0$.


问:
By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$
Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$

To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.

Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity

答:
hint

if $f'(z) = 0$ for all $z$ then $f(z) = $ a constant .

this means that for all $z, f(z) = f(0).$

let $f(z) = [s(z)]^2 + [c(z)]^2.$

Addendum

Per request of OP

determine $s(z)$ and $c(z)$ completely.

Actually, the necessary analysis has already been provided by the comments of leslie townes.  All I need to do is fill in all of the ground work.

First of all, I completely agree with the response of Hagen von Eitzen, so following
through with the first part of my answer (before the Addendum) is the place to start.

$f'(z) = s(z)s'(z) + c(z)c'(z) = s(z)[c(z)] + c(z)[-s(z)] = 0.$

This establishes that for all $z, f(z) = f(0) = 1,$ which completely resolves
the question in the title of the query.

Next, since $s'(z) = c(z)$ and $c'(z) = -s(z)$, you have that

$s''(z) = c'(z) = -s(z)$ and that $c''(z) = -s'(z) = -c(z).$

Therefore,

$$\sum_{n=2}^\infty (n)(n-1)[a_n]z^{(n-2)} = s''(z) = -s(z) =
\sum_{n=0}^\infty -a_nz^n = \sum_{n=2}^\infty -a_{(n-2)}z^{n-2}.$$

As a result, you have that for each $n \in \{2,3,4,\cdots\}$

$(n)(n-1)a_n = -a_{(n-2)}.$

By identical analysis, for each $n \in \{2,3,4,\cdots\}$

$(n)(n-1)b_n = -b_{(n-2)}.$

All that remains is to compute $a_0, a_1, b_0, b_1.$

Since $a_0 = s(0) = 0$ and $b_0 = c(0) = 1$,

it is immediate that $a_0 = 0$ and $b_0 = 1.$

Also, remember that for all $z$, you have that $s'(z) = c(z)$ and $c'(z) = -s(z).$

Therefore, since $a_1 = s'(0) = c(0) = 1$, and $b_1 = c'(0) = -s(0) = 0$,

it is immediate that $a_1 = 1$ and $b_1 = 0.$


问:
The converse problem about duality of $L^p$

Let $Ω$ be a bounded open subset in $R^n$ and $f$ be measureable with respect to Lebesgue measure on $Ω$. If there exist $M＞0$ s.t.$$\left|\int fg\,\mathrm{d}x\right|≦M\left(\int |g|^q\mathrm{d}x\right)^\frac{1}{q}$$ for all continuous function $g$ on $Ω$ (the integral is on $Ω$), where $1＜q＜+∞$, $\frac{1}{p}+\frac{1}{q}=1$, then $f \in L^p(Ω)$ and $\|f\|_{L^p(Ω)}≦M$.

I find some references: the conclusion is true if $g$ can be substituted by all simple functions.

But how can we do this in the continuous case? Should we choose simple functions that approximate the continuous $g$ by using Lusin's theorem  (but I have some problem in seeing how to do this)? Could someone show me the details? Thank you!

My doubt: even though we can use a continuous $g$ to approximate simple functions in $L^1$ (i.e. in the left side integral), we need that to prove they are close enough also in $L^q$ (i.e. in the right side integral).

答:
Abstract: In this article, the norm in $L^p(\Omega)$ space could be descriped by all continuous functions in theorem 1 on $ \Omega $, where  $ \Omega $ is a bounded open set in $\mathbb{R} ^n$, $1<p<\infty$. This could be regard as the converse problem for H{"o}lder inequality.

Theorem 1. $ \Omega $ is a bounded open set in $\mathbb{R} ^n$, $1<q<\infty$, $f$ is Lebesgue measurable on $ \Omega $. If there exist a $M>0$ s.t.
\begin{equation}
 |\int_{\Omega} fgdx|\leq M(\int_{\Omega} |g|^qdx)^\frac{1}{q}
\end{equation}
then, $f\in L^p(\Omega)$,  $||f||_p \leq M$, where $\frac{1}{p}+\frac{1}{q}=1$ and $||\;||_p$ denote the norm in $L^p(\Omega)$

Proof:

Claim 1. if the continuous funtion  could be sbustituted by all integrable simple functions in (1), then the results hold . We can get it from reference 1.

Claim 2. $|x^p-y^p|\leq p|x-y|(x^{p-1}+y^{p-1})$ when $1<p<\infty$, Real and Complex Analysis Rudin's book in chapter 3, problem 24.

We only need to show
\begin{equation}
  |\int_{\Omega} fhdx|\leq M(\int_{\Omega} |h|^qdx)^\frac{1}{q}
 \end{equation}
for all integrable simple function $h$ by claim 1. Thus we only need to prove the inequality (2).

case 1.   $\int_{\Omega} |f|=0$, (2) is trivial.

case 2.   $\int_{\Omega} |f|\neq0 $
for any $\epsilon >0$, we choose a bounded continuous function $g$.(Choose a real number $G$ s.t. $h,g$ are bounded by $G$) s.t.$\int_{\Omega}|h-g|^qdx<2\epsilon$

Claim 3. $ \int _{\Omega}|f(h-g)|dx<\epsilon$

We have $\int_{\Omega}|h^p-g^p|dx<4p\epsilon G^(p-1)$ by Claim 2.
Now we have
$$|\int_{\Omega} fhdx|\leq M(\epsilon)(\int_{\Omega} |h|^qdx)^\frac{1}{q} $$
and $M(\epsilon)$ is bigger but close enough to $M$ by claim 1.,2.and 3.
Let $\epsilon $ converges to $0$. We  have proved the inequality (2) and complete the proof.

Proof of claim 1.
(I translate the reference into English):

We choose compact supported nonnegative simple measurables function sequence ${\phi_k(x)}$ ($0\leq\phi_1(x)\leq\phi_2\leq...\leq \infty$) s.t. $$ \lim_{k\rightarrow\infty}\phi_k(x)=|f|^p,x\in \Omega$$
Let $\psi_k(x)=(\phi_k(x))^\frac{1}{q} \text{sign} f(x)$,
$$||\psi_k||_q=(\int_{\Omega}\psi_k(x)dx)^{\frac{1}{q}}$$
$$0\leq\psi_k(x)=(\psi_k(x))^{\frac{1}{q}}(\psi_k(x))^{\frac{1}{p}}\leq(\psi_k(x))^{\frac{1}{q}}|f(x)|=\psi_k(x)f(x)$$
So, we have
$$\int_{\Omega}\phi_k(x)dx\leq\int_{\Omega}\psi_k(x)f(x)\leq M||\psi_k(x)||_q$$.
Thus
$$\int_{\Omega}\phi_k(x)dx\leq M^p $$.
After taking the limit, we have
$$\int_{\Omega}|f(x)|^pdx\leq M^p $$.

Proof of claim 2.

$f$ is integrable, then there exist $\delta>0$ s.t. $\int_{E} |f|<\frac{1}{2G}\epsilon$
Then there exist closed $E_{\delta}\subset \Omega$ and $m(\Omega \backslash E_{\delta})<\delta$. The integral on $E_{\delta})$ is dominated by  H{"o}lider inequality.


问:
Probability of throwing darts when likelyhood of hitting one side is more than the other
Currently learning probability from MIT OCW 6.041. I want to understand the following regarding Problem #5 from the PSET1(read the question here at MIT OCW). I am unable to understand part c of the given problem.

We need to calculate the probability of John hitting 50 point mark on the dartboard. I understand that we are looking for values of $\alpha$ and $\beta$ which are the proportionality constants, I want to understand why the probability of throwing the dart in the right half of the board is $\frac{2}{3}$

The question being asked above(in bold) is being referenced from the answer given for the Problem #5 referenced from the solution of the PSET1-Solution from the MIT OCW site.

The  first  term  in  the  sum  is  proportional  to  the  area  of  the  right  half  of  the  inner  disk and  is  equal  to απR2/2 = απ/2, where α is  a  constant  to  be  determined. Similarly,  the  probability  of  him  throwing  in  the  left  half  of  the  board  and  scoring  50 points  is  equal  to βπ/2, where β is  a  constant (not  necessarily  equal  to α). In  order  to  determine α and β,  let  us  compute  the  probability  of  throwing  the  dart  in the  right  half  of  the  board.  This  probability  is  equal  to απR2/2 = απ102/2 = α50π.
Since  that  probability  is  equal  to  2/3, α = 1/(75π). In  a  similar  fashion, β can  be determined  to  be  1/(150π).   Consequently,  the  total  probability  is  equal  to  1/150  + 1/300  = 0.01

答:
Since John is twice as likely to hit the right part of the board, for every one dart that he lands on the left half, he would have landed two darts on the right half.

Thus out of every 3 darts thrown, two land on the right half and one lands on the left half. Thus $\alpha = \frac{2}{3}$ and $\beta = \frac{1}{3}$.


问:
How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
In showing that $(x_i\rightharpoonup x)\not\Rightarrow(x_i\to x)$ or similar noncorallaries, one frequently uses the counterexample
$$
  (u_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad u_i = (\underbrace{0,\ldots,0}_{i-1},1,0,\ldots)
$$
This, if I can trust my professor, is a weak convergent sequence in $\ell^2$, because it is bounded (I apparently missed that bit, this was the cause of the confusion) and for each $\varphi_j=(0,\ldots,1,0,\ldots)$, there exists an $N\in\mathbb{N}$, namely $N=j$, such that for every $i>N$,
$$
  \langle u_i, \varphi_j \rangle_{\ell^2} = 0 + \ldots + 0\cdot1 + 0 + \ldots + 1\cdot 0 + 0\ldots = 0 < \varepsilon
$$
for all $\varepsilon>0$. Because the $\varphi_j$ form a complete orthonormal system in $\ell^2$, this is sufficient for $(u_i)_i$ to be weakly convergent.

The problem is now: consider the sequence
$$
  (v_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad v_i = (\underbrace{0,\ldots,0}_{i-1},i,0,\ldots).
$$
You could show that this is weakly convergent in exactly the same way as I just did for $(u_i)_i$, but on the other hand, $(v_i)_i$ is obviously an unbounded sequence and according to the principle of uniform boundedness (which I don't quite understand) every weak convergent sequence is bounded.

So what's wrong here?

Well, as the answers pointed out that it was the requirement of $(u_i)$ being bounded which was necessary for using just the $\varphi_j$ to show weak convergence.

答:
Dear leftaroundabout, your $v_i$ sequence is not weakly convergent to zero. Take its inner product with
$$ x = (1/1, 1/2, 1/3, 1/4, 1/5, \dots ) $$
to see that this inner product of $x$ with $v_i$ is actually equal to one in the infinite $i$ limit (and not zero as required by the weak convergence to zero). Note that $x$ is $l^2$ summable, because $\sum 1/n^2 = \pi^2/6$ is convergent, i.e. that it is a vector in the Hilbert space.</p>


答: For the desired equation, you do not need the power series, but only the given relations between derivatives, $s'=c$ and $c'=-s$ (plus that the equation holds at $z=0$)

Body:
Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$

To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.

Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity


问:
Existence of $f \in L^2(\Bbb R^n)$ with $f=g_1$ on $E$ and $\mathscr{F}(f)=g_2$ on $F$
Question:

Suppose $E,F$ subsets of $\Bbb R^n$ have finite measure. Show that for any $g_1,g_2 \in L^2(\Bbb R^n)$ there exists $f \in L^2(\Bbb R^n)$ with
$f=g_1$ on $E$ and $\mathscr{F}(f)=g_2$ on $F$,where $\mathscr{F}(f)$ is the Fourier transform of $f$.

My attempt:

This is a problem from Schlag's book Classical and multilinear harmonic analysis, chapter 10 problem 10.3. I try to prove it by contradiction and Amrein-Berthier thm but failed.

Any solution or hint is highly appreciated.

A-B thm: Let E,F be sets of finite measure in $\Bbb R^n$.Then $ \left\|f\right\|_{L^2(\Bbb R^n)} \leq C( \left\|f\right\|_{L^2(E^c)}+ \left\|{F}(f)\right\|_{L^2(F^c)} )$ for some $C=C(E,F,n)$

答:
Let $V$ be the closed subspace of $L^2(\mathbb{R}^n)$ consisting of those functions that vanish almost everywhere on $E$, let $W \subseteq L^2$ be the closed subspace of functions whose Fourier transforms vanish almost everywhere on $F$, and let $V^{\perp}$ and $W^{\perp}$ be their orthogonal complements.

It suffices to show that there exists $f \in L^2$ with $f = g_1$ on $E$ and $\mathscr{F}(f) = 0$ on $F$. Or, put another way, we'd like to show that an arbitrary coset of $V$ intersects $W$.

Now, if we knew that $V + W = L^2$, the result would follow from the (purely algebraic) isomorphism $(V + W) / V \cong W / (V \cap W)$. To see this, note that the natural isomorphism is given by $h + (V \cap W) \mapsto h + V$. So given an arbitrary coset $g + V$ ($g \in L^2 = V + W$), we could find a $\tilde{g} \in W$ so that $\tilde{g} + V = g + V$. In particular, $\tilde{g} \in g + V$, so this function demonstrates that the coset $g + V$ intersects $W$.

To finish, we show that $V + W = L^2$. We rely on the fact that the $L^2 \to L^2$ operator norm of $P_{V^{\perp}} P_{W^{\perp}}$ is strictly less than $1$, where $P_{\cdot}$ represents the orthogonal projection onto the relevant subspace. (This norm statement is connected to the Amrein-Berthier theorem, and follows from Theorem 10.4 and Lemma 10.5 in Muscalu and Schlag's book.) Using the fact that $I = P_W + P_V P_{W^{\perp}} + P_{V^{\perp}} P_{W^{\perp}}$, we can write an arbitrary $f \in L^2$ as $f = g_1 + h_1 + P_{V^{\perp}} P_{W^{\perp}} f$, where $g_1 \in W$ and $h_1 \in V$. Iterating this expansion on the remainder term and using the norm bound, we obtain convergent series $\sum_j g_j$ in $W$ and $\sum_j h_j$ in $V$ satisfying $f = (\sum_j g_j) + (\sum_j h_j)$. This completes the proof.


问:
How would I evaluate this integral?
I have an integral
$$\int\sqrt{b^2-\frac{x^4}{a^4}} dx \quad a, b\in
\mathbb{R}$$

And I have no idea how to solve it. The $x^4$ term is stumping me. I've tried the standard $u$ and trig substitution methods but to no avail. Any suggestions?

答:
First of all, with you numbers, the integration bounds are exactly
$$t_\pm=\pm \frac 3 5 \sqrt{\frac{31}{5}} \sim \pm 1.493987952$$ that is to say the limit of definition of the integrand.

$$I=\int\sqrt{b^2-\frac{x^4}{a^4}}\, dx$$

If you do not want to face nasty elliptic integrals, assuming $a>0$ and $b >0$, use
$$\sqrt{b^2-\frac{x^4}{a^4}}=\sum_{n=0}^\infty (-1)^n a^{-4 n} b^{1-2 n} \binom{\frac{1}{2}}{n}x^{4n}$$ to make after integration
$$I=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\frac {x^{4n+1}}{a^{4n}b^{2n-1}(4n+1) }$$

Now, for the integral
$$J=\int_{-a \sqrt{b} }^{+a \sqrt{b} }\sqrt{b^2-\frac{x^4}{a^4}}\, dx=2 a b^{3/2}\sum_{n=0}^\infty (-1)^n \frac{\binom{\frac{1}{2}}{n}}{4 n+1}$$
$$\color{blue}{J= a\, b^{\frac 32}\,\sqrt{\pi }\,\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}}$$

More generally
$$K=\int_{-ka \sqrt{b} }^{+ka \sqrt{b} }\sqrt{b^2-\frac{x^4}{a^4}}\, dx=2 a b^{3/2}\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\frac{k^{4n+1}}{4 n+1}$$
$$\color{blue}{K=2\, k\, a\, b^{\frac 32}\,\, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};k^4\right)}$$

Title: How would I evaluate this integral?


答:
Intuition only approach (because I'm ignorant of the corresponding math).

You have the interval $[0,1]$ and are asked to choose a number at random from this interval.

$E$ is the event that you choose the number $(1/2)$ and you are asked to compute
$p(E).$

So, as a worshipper of Archimedes (or whoever), you reason as follows:

I can estimate $p(E)$ by assuming that the interval $[0,1]$ is broken into exactly $n$ sub intervals :

$[0,\frac{1}{n}), [\frac{1}{n},\frac{2}{n}), \cdots [\frac{n-1}{n},1].$

Then, the number $(1/2)$ is in only one of these subintervals.

Therefore, $p(E) < $ probability of choosing the corresponding subinterval, which is clearly $\frac{1}{n}$.

This analysis holds, even if you allow $n$ to grow unbounded (i.e. approach $\infty$).  In such a scenario, the chance of selecting the corresponding sub-interval, which is $\frac{1}{n}$ must approach $0$.

Title: Why the probability of single element event is "zero" on continuous interval?

Body:
Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.

cool thanks!


问:
Prove that if $E[X_1^p]<\infty$, then $\frac{\max_{1\le i\le n} X_i}{n^{1/p}} \rightarrow 0$ in probability where $\{X_n\}$ is i.i.d and non-negative
Suppose $\{X_n\}_{n\geq 1}$ are iid and non negative. Define $M_n=\max \limits_{i=1,\ldots,n}\{X_i\}.$

Prove if $E[X_1^p]<\infty$, then $\frac{M_n}{n^{1/p}}\rightarrow 0$ in probability.

As a previous result I have that $P[M_n>x]\leq nP[X_1>x].$ Is this result useful to prove this convergence? I would appreciate any hint.

答:
You can use the result $P[M_n>x]\leq nP[X_1>x]$. Observe that

$$P[M_n/n^{1/p}>\epsilon]=P[M_n>\epsilon n^{1/p}]
\le nP[X>\epsilon n^{1/p}]=nP[X^p>\epsilon^pn].$$ Now use the inequality $P[Y>a]\le E\left[\frac YaI(Y>a)\right]$ to conclude $$nP[X^p>\epsilon^pn]\le E\left[\frac{X^p}{\epsilon^p}I(X^p>\epsilon^pn)\right],$$ and finally argue that the RHS tends to $0$ as $n\to\infty$.


问:
Evaluating the chi-square integral
The question is to find that value of x for which F(x)=0.3, where X follows Chi-square distribution with 13 df.

∫_0^x▒〖(e^((-x)/2)   x^(n/2-1))/(2^(n/2)   ⌈n/2┤ )=0.3〗

∫_0^x▒〖e^((-x)/2)  x^(13/2-1 )=0.3*  2^(13/2)* ⌈13/2┤  〗

I cannot move further. Can someone help?

答:
Usually this is solved using the chi-squared tables

$$x\approx 9.93$$

Usually Paper Tables do not show $F(0.3)$. In a standard chi-square table you usually find $F(0.25)$ and $F(0.5)$ but with a simple gaussian approx you get

$$u_{0.3}=\frac{1}{2}\Bigg[z_{0.3}+\sqrt{2\times 13-1}\Bigg]^2=\frac{1}{2}\Big[-0.53+5\Big]^2\approx 9.99$$

Which is very close to the exact result of $9.93$ that I got with a calculator

Title: Evaluating the chi-square integral


问:
Find the condition in which $\angle ADB = 3\angle BAC$
See below in acute triangle $ABC$, $D$ is on $AC$ such that $AD=BC$. $CF$ is the angle bisector of $\angle ACB$. $DE \parallel CF$. $E$ is on $AB$. $AE = CD$.

Prove that $\angle ABC = 2 \angle BAC$. In addition, find the condition in which $\angle ADB = 3 \angle BAC$

I struggle to find ways to use two conditions $AD = BC$ and $AE = CD$

答:
Here I prove that $\widehat{ABC} = 2\widehat{BAC}$ . Please note that the figure below is not precise.

From $B$ draw $BM$ perpendicular to $CF$ and extend it to meet $AC$ at point $N$ . Because $CM$ is the bisector of angle $\widehat C$ and perpendicular to $BN$ , triangle $CBN$ is isosceles and $BC = NC$ . The problems states that $BC = AD$ . Therefore $AD = NC$ , which in turn leads to
$$AN = DC \qquad(1)$$

From $E$ draw a line parallel to $AC$ and let it meet $CF$ at point $O$. The quadrilateral $CDEO$ is a parallelogram because it has two pairs of parallel sides. We have $DC = EO$ . But the problem states that $DC = AE$ , therefore
$$AE = EO = DC \qquad(2)$$
From (1) and (2) we have $AN = EO$ . The quadrilateral $AEON$ has a pair of parallel sides of equal length ($AN$ and $EO$) therefore it is a parallelogram. But from (2) we know that three sides of this parallelogram are equal, so it must be a diamond, and
$$AE = EO = AN = NO \qquad(3)$$
Moreover, because $AEON$ is a diamond, the diagonal $AO$ bisects angle $\widehat A$. Now, $O$ is the meeting point of bisectors of $\widehat A$ and $\widehat C$ , therefore $BO$ is the bisector of $\widehat B$ .

Next, consider the triangle $OBN$ . Because $OM$ bisects $BN$ and is perpendicular to it, the triangle is isosceles and
$$BO = NO \qquad(4)$$
From (3) and (4) we have $BO = EO$ , therefore the triangle $OBE$ is also isosceles and $\widehat{OBE} = \widehat{OEB}$ . But because $EO$ and $AC$ are parallel, $\widehat{OEB} = \widehat{BAC}$ . Therefore
$$\widehat{OBE} = \frac12 \widehat{ABC} = \widehat{BAC}$$

Update:
Here I prove that the statement $\widehat{ADB} = 3\widehat{BAC}$ which was initially written in the problem is not valid in general, and therefore cannot be proven. In fact, I demonstrate that the above statement is valid only for a particular class of similar triangles, and I specify the three angles of triangles of that class.

We previously saw that given the conditions of the problem, we necessarily have $\widehat{ABC} = 2 \widehat{BAC}$ . Therefore in triangle $ABC$ :
$$\widehat{ACB} = 180^o - 3 \widehat{BAC} $$
Now let us assume that the construction is such that $\widehat{ADB} = 3\widehat{BAC}$ . This means that
$$\widehat{BDC} = 180^o - \widehat{ADB} = 180^o - 3 \widehat{BAC}$$
which, in turn means that the triangle $BDC$ must be isosceles for this condition to hold. So
$$BC = BD$$
Recall that the problem states that
$$BC = AD$$
Therefore:
$$BD = AD$$
which means the triangle $ADB$ must also be isosceles. Therefore
$$\widehat{ABD} = \widehat{BAD} $$
By writing the sum of angles in triangle $ABD$ we are able to find the angle $\widehat{BAC}$ :
$$\widehat{ABD} + \widehat{BAD} + \widehat{ADB} = 5\widehat{BAD} = 180^o$$
$$\widehat{BAD} = \widehat{BAC} = \frac{180^o}{5} = 36^o $$
So, given the conditions of the problem, the statement $\widehat{ADB} = 3 \widehat{BAC}$ is valid only if angle $\widehat A$ in triangle $ABC$ be $36^o$ . Hence that statement is not valid for other triangles. It follows that in the same triangle $ABC$, angle $\widehat B$ will be $2 \times 36^o = 72^o$ and $\widehat C$ will be $180^o - (36^o + 72^o) = 72^o$. So the triangle $ABC$ will be isosceles. Finally, note also that in this case $\widehat{ABD} = \widehat{DBC} = 36^o$ , which means $BD$ bisects angle $\widehat B$ in triangle $ABC$ , so $BD$ passes through point $O$ and in the above figure, $D$ falls on $G$ .


问:
On a question chosen at random, what is the probability that the student answers it correctly?
I'm really confused about this question. I appreciate your help.

A student takes a multiple choice exam where each question has five possible answers. He answers correctly if he knows the answer, otherwise he guesses at random. Suppose he knows the answer to $70$% of the questions.

Question 1

On a question chosen at random, what is the probability the student answers it correctly?

We know $P(\text{know the answer}) = 0.7$. On a question chosen at random, he either knows the answer or he doesn't. If he does, then $P(\text{answer correctly}|\text{know correct answer})= 1$. If he doesn't, then $P(\text{answer correctly}|\text{doesn't know correct answer})= 0.7(0.3)^4$. I don't know how to continue the answer.

Question 2

Given that he did answer correctly, what is the probability that he actually knew the correct answer?

All I can think of is the following conditional probability.

$\begin{align}
P(\text{know correct answer}|\text{answer correctly}) & = P(\text{know correct answer}|\text{answer correctly}) \\
& =\frac{P(\text{know correct answer and answer correctly})}{P(\text{answer correctly)}}. \\
\end{align}$

答:
For the first one, drawing a probability tree might help.

At first, you have the two branches of whether he knows the answer (0.7) and he doesn't know the answer (0.3).

If he knows the answer, then he answers correctly (1), so that, the probability that he knows and answer AND answers correctly is (0.7*1 =) 0.7.

If he doesn't know the answer, he has a probability of (1/5 =) 0.2 of getting the right answer, so that the probability that he doesn't know the answer AND answers correctly becomes: (0.3*0.2 =) 0.06

The rest should be a little easier.


问:
How to complete this proof process of Arzelà-Ascoli theorem?

Theorem. Let $X$ be a compact space and $Y$ be a metric space. A set $\mathscr F \subseteq C(X,Y)$ is precompact if and only if it is
pointwise precompact and equi-continuous.

Definitions. A subset $\mathscr F \subseteq C(X,Y)$  is called:

equi-continuous if for every $x ∈ X$ and every $ε > 0$, there exists an
open neighborhood $U \subseteq X$ of $x$ such that $d_Y(f(x), f(x')) < ε$ for all $ x'∈ U$ and all $f ∈ \mathscr F$.

precompact  if the closure of $\mathscr F$ is compact.

pointwise precompact if for each $x∈X$, $\mathscr F(x) = \{f(x): f∈\mathscr F \}$ is precompact.

Lemma. Let $(X,d)$ be a metric space and let $K \subseteq X$. Then every sequence in $K$ has a Cauchy subsequence if and only if $K $is totally bounded.

I know how to prove it from left, the way is similar with $X$ is compact metric space. But from right: auth give me some hint.
The key is is to prove every sequence has Cauchy subsequence. By the lemma, we only need to prove $\mathscr F$ is totally bounded. He gives me two hint:

The set $F = \{f(x) :  x ∈ X, f ∈ \mathscr F\} \subseteq Y$ is precompact so  is totally bounded.

Let $ε > 0$. Cover $F$ by finitely many open balls $V_1,\dots,V_n$ of radius $ε/3$ and cover $X$ by finitely many open sets $U_1,\dots,U_m$ such that
$$\sup_{x,x' ∈ U_i} \sup_{f ∈ \mathscr F} d_Y(f(x),f(x'))＜\frac ε3.$$
Let $\alpha $ be a any map from $\{1,\dots,m\}$ to $\{1,\dots,n\}$.
Define $$\mathscr{F}_α = \{f ∈ \mathscr F : f(U_i) ∩ V_{\alpha(i)} ≠ \varnothing \} .$$  Let $A$ be
the set of all $α$ such that $\mathscr F_α ≠ \varnothing$. Prove that $\mathscr F = \bigcup_{α∈A} \mathscr F_α$.

I can following the hint 2 to achieve it, but I don't how to prove hint 1.

My try: Originally, I choose  $f_n(x_n)$ be sequence of $F$, I use diagonal method to find a subsequence $g_n$ of $f_n$ such that $f_{n_k}$ converges at each $x_i$, and I want to prove that $f_{n_k}(x_{n_k})$ converges to subsequence of $f_n(x_n)$. However I think this way is not ture now.

答:
It's not said that $Y$ is a complete metric space, but we can work in the completion $\overline{Y}$.

If I'm reading this right, we want to prove

If $\mathscr F \subseteq C(X,Y)$ is equicontinuous and pointwise precompact, then $\mathscr F$ is precompact.

Regarding Hint $\# 1$. Fix $f_n(x_n)\in F, n\in\mathbb N$. Without loss of generality, we may assume $x_n$ to be convergent due to compactness of $X$.

By assumption $\{f_n(x_i) \mid n\in\mathbb N\}$ is precompact for every $i\in\mathbb N$. Apply diagonal argument. Put $\lim\limits_{n\in N_1} f_{n}(x_1)  := f(x_1)\in\overline{Y}$ for some subsequence $N_1\subseteq \mathbb N$. Then, again by precompactness, there exists $N_2\subseteq N_1$ s.t $\lim\limits _{n\in N_2} f_n(x_2) := f(x_2)$ etc. Now define $N\subseteq \mathbb N$ such that the $k$-th component is the $k$-th
component of $N_k$. Then $\lim \limits _{n\in N} f_n(x_i) = f(x_i)$ for every $i\in\mathbb N$.

Now we have $f_{\ell_n}(x_i)\to f(x_i)$ for every $i\in\mathbb N$. Since $x_n\to x$, we also have $x_{\ell_n} \to x$. It suffices to show $f_{\ell_n}(x_{\ell _n})$ is Cauchy.

To ease notation, identify $f_{\ell_n}(x_{\ell _n})$ with $f_n(x_n), n\in\mathbb N$. By triangle inequality
$$d(f_n(x_n), f_m(x_m)) \leqslant d(f_n(x_n), f(x_n)) + d(f_m(x_m), f(x_m)) + d(f(x_m),f(x_n)). $$
The first two terms can be made small due to $f_n(x_i)\to f(x_i)$.

To make the third term small eventually, it suffices to show that $f(x_i)$ is Cauchy. We have $f(x_i) = \lim f_n(x_i)$. Due to continuity, the metric respects limits, so
$$d(f(x_i),f(x_j)) = \lim d(f_n(x_i),f_n(x_j)) \leqslant \lim d(f_n(x_i), f_n(x)) + d(f_n(x_j),f_n(x)). $$
The term $d(f_n(x_i), f_n(x))$ is small eventually due to equicontinuity of $\mathscr F$:

Firstly,  $d(f_n(x_i),f_n(x))$ can be made small around $x$ by definition. But $x_i\to x$ as well, so $x_i\in U_x$ eventually.


问:
Hi, so I find olympiad geometry really difficult, ACM12.
I tried EGMO (Euclidean Geometry for Mathematical Olympiads) but I end up solving every problem using the same technique, bashing. Are there any ways to improve at synth geometry? Any YouTube channels which could help or something? I know so many formulae, even weird ones like Lagrange's identity and stuff like that yet when I see a solution I think: How could I have come up with this?</p>

答:
The best way to learn how to solve problems is to solve problems. You don't learn how to draw just by looking at paintings. You don't learn how to play the piano just by memorizing notes. You have to actually engage in the activity along with learning about the activity to truly master a skill. You also need to put in the time and effort as well. Here are some problem solving resources I use to practice my math skills.

https://artofproblemsolving.com/store Art of Problem Solving Books

https://artofproblemsolving.com/alcumus/problem Practice system that gives you lots of problems of various topics to solve.

Hope that helps.


问:
Linear recurrence for a sequence
I am new to linear recurrence theory. I want to find a linear recurrence for the sequence $f_{n}$ where $f_{n}$ is the number of subsets $G'\subset G(P_{n})$ where each pair of vertices in $G'$ are neither adjacent nor have a common neighbour.

Here $P_{n}$ means a chordless path with $n$ vertices and $G(P_{n})$ is the set of vertices $\{G_{1},...,G_{n}\}$. I don't know how to proceed, any kind of hints would be appreciated. I think I need to use the adjacency matrix here.

答:
If I'm understanding the problem correctly, it's basically a subset of a path where the chosen vertices have to be at least distance 3 from each other.

Given a path of length $n$, we can iterate on the first vertex to be present in our subset. We can choose the first vertex in the subset to be:

The first vertex in the path (after which, we have $f_{n-3}$ ways to choose the remaining)

The second vertex in the path (after which, we have $f_{n-4}$ ways to choose the remaining)

The third vertex in the path (after which, we have $f_{n-5}$ ways to choose the remaining)

None of the first three vertices (after which, we have $f_{n-3}$ ways to choose the remaining)

Note that these 4 cases are disjoint and also cover all possibilities. Thus,

$f_n = f_{n-3} + f_{n-4} + f_{n-5} + f_{n-3}$

Title: Linear recurrence for a sequence


问:
Using the Taylor expansion to get the solution of a differential equiation
I've got stuck with this differential equation. I've to solve the problem using the exponential expansion.

$$
y'(x) =
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
y
\text{ }\text{ };
y(1) =
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$

Let's get $A^n$:

$$
A^2 =
\begin{pmatrix}
2 & 2 \\
2 & 2 \\
\end{pmatrix}
\text{ }\text{ }
A^3 =
\begin{pmatrix}
4 & 4 \\
4 & 4 \\
\end{pmatrix}
\text{ }\text{ }
...
\text{ }\text{ }
A^n =
2^{n-1}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
$$

We can do the exponential expasion like:

$$
e^{Ax} =
\sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} =
\sum_{n=0}^{\infty} 2^{n-1}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix} \frac{x^n}{n!} =
\frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
\sum_{n=0}^{\infty} \frac{(2x)^n}{n!} =
\frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}e^{2x}
$$

So that,  I think the solution should be:

$$
y(x) = \frac {C}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}e^{2x}
$$

But I think this is wrong because it looks like it's impossible to get the particular solution:

$$
y(1) = \frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
C_1 \\
C_2 \\
\end{pmatrix}
e^2 =
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$

As you can see, this is not possible to get a solution from this equation.

Does anyone know where I made a mistake?
Thanks in advance.

答:
The formula
$$
A^n =
2^{n-1}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
$$
is not correct for $n=0$.
$$
e^{Ax} =
\sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} =
I+\sum_{n=1}^{\infty} 2^{n-1}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix} \frac{x^n}{n!} = I+
\frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
\sum_{n=1}^{\infty} \frac{(2x)^n}{n!}
$$
$$
= I+\frac12\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}+
\frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
\sum_{n=1}^{\infty} \frac{(2x)^n}{n!} -\frac12\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
$$
$$
=
I+\frac {1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}e^{2x}-\frac12\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}
$$


问:
Question related to Vector Space
A span, of a set of vector, is the smallest vector subspace containing it.  Give example of Vector space, such that the subspace containing the set of vectors, which is greater than span, is infinite infinite in  number.</p>

答: Can you clarify your question? Given a set $X$ of vectors, we always have that $X\subset span(X)$. You can have a countable set of vectors such that the span is uncountable, the $\operatorname{span}({(1,0),(0,1)})=\mathbb{R}^2$ for instance.

Title: Question related to Vector Space


问:
To find all integers n such that the given expression is a perfect square
The question is as follows

Determine all integers n such that $n^{4}-n^{2}+64$ is the square of an integer

Here is my approch

Let $n^{4}-n^{2}+64=k^{2}$. On multiplying both sides by 4, we get
$$4n^{4}-4n^{2}+256=4k^{2}$$
$$\Rightarrow(2n^{2}-1)^{2}+255=4k^{2}$$
$$\Rightarrow(2k+2n^{2}-1)(2k-2n^{2}+1)=255$$
The only ordered pair of factors possible are (3,85); (5,51); (15,17). We need not consider negative factors as we are squaring later.

Thus, we get $k=8, 14, 22$, which implies $k^{2}=64, 196, 484$ and finally $k^{2}-64=0, 132, 420$
$$\Rightarrow n^{4}-n^{2}=0, 132, 420$$
These 3 equations admit only $n=0, \pm1$ as integer solutions.

However, $n=\pm8$ also satisfies. Thus, my solution is incomplete or wrong

 Could anyone please point out the mistake in my solution or provide the necessary extension to fix it.
Thanks a lot for help!! :)

答:
Another approach:

$(n^2-n)(n^2+n)=(k-8)(k+8)\space\space\space\space\space (1)$

Since $n^2+n>n^2-n$ and $k+8>k-8$ we cnstruct following system of equations:

$\begin{cases} n^2-n=k-8\\n^2+n=k+8\end{cases}$

Sum these relations you get:

$2n^2=2k$ $\rightarrow$ $n=\pm k$

subtracting gives:

$2n=16$ $\rightarrow$ $n=8$

For $k^2$ also $-8$ can be solution.If we put $\pm 8$ in (1) RHS beomes zero and we get :

$n= 0, 1, -1$


问:
What does it mean for a finite product topology to be a T1 space?
If we define T1 space to be where all singletons are closed, then for a finite product topology, does T1 mean that all possible product of singletons are closed?</p>

答:
Yes, if $X=\prod_{i=1}^n X_i$ is a finite product of $T_1$ spaces, then $\{(x_1,\ldots,x_n)\}= \{x_1\} \times \ldots \{x_n\}$ is closed as a product of closed singletons in the factor spaces.

So yes, a (finite, in fact arbitrary) product of $T_1$ spaces is again $T_1$.

And if $X$ is $T_1$, then so is every $X_i$, $i=1, \ldots, n$ as these are open continuous images of $X$.


问:
In a discrete metric space $X$ no proper subset of $X$ can be dense in $X$
While reading a book on Introductory Functional analysis, I saw the statement that,

In a discrete metric space $X$ no proper subset of $X$ can be dense
in $X$

So i thought to prove it. But before going to prove it, i was thinking to verify this statement first!

Verification of statement: So i consider the set $X=[0,5]$ equipped with discrete metric. Let $Y=(0,5)$ clearly $Y$ is proper subset of $X$. So according to above statement, $Y$ can't be dense in $X$. So that, we must have $\bar Y≠X$, where $\bar Y$ denotes closure of $Y$ in $X$. So i think we need to show there exists some point $a\in X$ which is neither a point of $Y$ nor a limit point of $Y$.(Am, I correct?) I claimed that, such a point is $0\in X$. Because, $0\not\in Y$ and there exists a neighborhood $B(0,1/2)=\{x\in X: d(x,0)<1/2\}=\{0\}$
Such that, $B(0,1/2)$ contains no points of $Y$ different from $0$ itself. Hence $0$ can't be limit point of $Y$ (Am i correct?)
Hence, we have $\bar Y≠X$ and so that, $Y$ is not dense in $X$.

Is my verification is correct?

Proof of statement let $Y$ be any proper subset of $X$. Then to show $Y$ is not dense in $X$, that is, to show $\bar Y≠X$. Let $a\in X$ such that, $a\not\in Y$. Then, it is sufficient to show that, $a$ is not a limit point of $Y$. Consider a neighborhood $B(a,1/2)=\{x\in X: d(x,a)<1/2\}=\{a\}$ thus there exists a neighborhood of $a$ which contains no point of $Y$ other than $a$ itself. Hence $a$ can't be a limit point of $Y$. Hence $\bar Y≠X$. As required.

Is my proof is correct?

答: Yes, your proof is correct. But there is another way to look at it too: If $X$ is discrete, then if $Y$ is a proper subset, $X\setminus Y$ is open (as all subsets are open in a discrete space!) so $Y$, its complement, is closed. So $\overline{Y}=Y \neq X$ is $Y$ is not dense.


问:
Continuity of $d(x,A)$
I am doing a head-check here. I keep seeing this theorem quoted as requiring $A$ to be closed (as in Is the distance function in a metric space (uniformly) continuous?), but I don't think that it is needed.

Theorem. Let $(X,d)$ be a metric space and $\emptyset \neq A\subseteq X$. Then $x \longmapsto d(x,A)$ is Lipschitz continuous.

Proof. Fix $x,y \in X$. We note that for any $a \in A$ we have $d(x,a) \leq d(x,y) + d(y,a)$.
Taking the infimum over $a \in A$ then gives $d(x,A) \leq d(x,y) + d(y,A)$. It follows quickly that
$$
|d(x,A)-d(y,A)| \leq d(x,y)
$$
and the claim is proved.

Have I accidentally used closedness of $A$ somewhere in my proof? I don't think it is necessary. Maybe the reason that it is usually quoted with $A$ closed is so that there is an $a \in A$ with $d(x,A) = d(x,a)$? (As @Martin points out, this also requires some compactness assumption on the unit ball, e.g. it holds if $X=\mathbb{R}^n$ with Euclidean metric)

答:
Summarizing the comments and expanding slightly:

Yes, your argument is perfectly okay as it stands and you didn't make any hidden use of closedness of $A$ anywhere, only non-emptiness of $A$ is used to ensure that $d(\cdot,A)$ is a real-valued function.

A good additional exercise would be to show that $1$ is the best Lipschitz constant (unless $A$ is dense in $X$ where $d(\cdot,A) = 0$).

Why people often state this result for closed $A$ only is unclear to me, but I agree that it is quite common to make this unnecessary assumption. See for example this question or this question for further examples. The answers point out more or less explicitly that this is not needed.

It could be that the reason is related to the fact you mention---that the infimum is a minimum for closed subsets of $\mathbb{R}^n$, or, more generally, in metric spaces for which closed balls are compact. Another possible "reason" could be that $\overline{A} = \{x \in X \mid d(x,A) = 0\}$, which shows that a closed set $A$ can be recovered from the zero-set of $d(\cdot,A)$.

Title: Continuity of $d(x,A)$


问:
Direct sum of Null space and its complement
My textbook has this problem.

Suppose $T:V\rightarrow F$. Suppose $u\in V$ is not in null($T$). Prove that $$V=Null(T) \bigoplus \{au: a\in F\}$$

Now, if the set $\{au: a\in F\}$ is a subspace of $V$, I can prove it using definition of direct sum, as below. But I think $\{au: a\in F\}$ is not a subspace of V. Can someone point out where I am wrong in my argument/understanding?

Proof (assuming the set $\{au: a\in F\}$ is a subspace of $V$):

Definition of direct sum : If $U,W$ are subspaces of $V$, and $U+W=V, U\cap W=\{0\}$, then $U\bigoplus W=V$.

So, I can prove Null$(T)\cap \{au: a\in F\}=\{0\}$, using the fact $T(au)=0\Rightarrow a=0.$, and since $\{au: a\in F\}$ is the union of $\{0\}$ and $u\notin Null(T), V=Null(T)+{au: a\in F\}$.

Why I think $\{au: a\in F\}$ is NOT a subspace of $V$:

Counter example:

Let $V$ is $R^3$. $T$ is a linear map from $R^3$ to $R$, $T:(x,y,z)\rightarrow x$ i.e. $T$ maps $3-D$ space to x-axis. Here null space of $T$ is the $y-z$ plane, $x=0$. $\{au: a\in F\}$ is the set $x\neq 0$. Now consider two elements $(6x + y)$ and $(-6x + y)$, $x,y \neq 0$. $(6x + y),(-6x + y)\in \{au: a\in F\}$, but $(6x + y)+(-6x + y)=2y \notin \{au: a\in F\}$,  $$(6x + y)+(-6x + y)=2y\in Null(T)$.

So   $\{au: a\in F\}$ is not a subspace. Now, since the definition of direct sum specifies that the elements should be subspaces, then I can't say $V=Null(T) \bigoplus \{au: a\in F\}$

答: You have to fix one vector $u$ not in the kernel of $T$ and consider $\{au:a \in F\}$. In your example you can take $(1,0,0)$ for $u$. In that case  $\{au:a \in F\}=\{(a,0,0): a \in F\}$ which is a subspace.


问:
Finding the self intersection point of two parametric equations.
Find the values of t where the graph of the parametric equations crosses itself:
$x = t^3-4t^2+t+7$ and $y = t^2-t$.

I am asking this problem on behalf of me and my Calculus 3 class. We understand the idea of $t$ being two different times, so making them say $s$ and $t$, then setting $x(s) = x(t)$ and $y(s) = y(t)$.

So we get:

$$x(t) = t^3-4t^2+t+7 = s^3-4s^2+s+7 = x(s)$$

$$y(t) = t^2-t = s^2-s = y(s)$$

However, past that we are completely lost. Our textbook goes over this type of problem for all of one paragraph, summarizing the process heavily. We asked our professor to go over this type of problem again but it still makes next to no sense. We also looked on YouTube and here on StackExchange for examples but they are either not the same type of problem, or the answers given are heavily summarized with no clear process. We are looking for a concrete process or set of ideas to follow in order to solve a problem like this.

Any help to this is greatly appreciated!

答:
After we know that $(1,2)$ is a double point, it's even clearer if we notice that
\begin{eqnarray}
x-1 &=& (t-3)(t-2)(t+1) \\
y-2 &=& (t-2)(t+1)
\end{eqnarray}
a parametrization of the cubic $-35 + 12 x - x^2 + 33 y - 5 x y - 10 y^2 + y^3=0$ with a singular point $(1,2)$.


问:
Confidence Interval, why is this unusual?
A Normal population has a known mean of $\mu = 50$ and a known variance of
$\sigma = \sqrt 2$. A random sample of size $n = 16$ is selected from this population and find that the sample mean is $x = 52$. This is an out-of-the-box sample result. common Why? Answer using a $95\%$ confidence interval.

I got the interval $(51.30, 52.69)$...
But what should I do after?

答:
Let X denote the distribution of the existing population.

$X \sim N(50, 2)$

$X_1 + X_2 + ... + X_{16} \sim N(16*50, 16 * 2)$

$X_1 + X_2 + ... + X_{16} \sim N(800, 32)$

Let S denote the distribution of the sample.

$S =\frac{X_1 + X_2 + ... + X_{16}}{16} \sim N(800/16, (1/16)^2 * 32) $

$S = \frac{X_1 + X_2 + ... + X_{16}}{16} \sim N(50, 1/8) $

So, our 95% interval (2 standard deviations) is $(50-2*\sqrt{1/8},50+2*\sqrt{1/8}) = (49.3, 50.7)$

So if we did this experiment, 95% of the time, our sample mean would lie between 49.3 and 50.7. So it could be said that finding the sample mean to be 52 is a bit unusual.


问:
Completion of a function space with sup metric
Let $X$ be a non-empty set and let $(Y , d)$ be a metric space. Let $(\mathcal B(X,Y), ρ)$ be the
set of all bounded functions from $X$ to $(Y,d)$ endowed with the sup-metric $ρ$ defined as follows.

$$\forall f,g \in \mathcal B(X , Y) \quadρ(f,g) := \sup\{ d(f(x),g(x)) : x \in X \}.$$
I know that the metric space $(\mathcal B(X,Y), ρ)$ is complete if and only if the metric space $(Y,d)$ is complete.

Now, assume that $(Y,d)$ is not complete and let $(Z,d)$ be a completion of $(Y, d)$.

Let $(\mathcal B(X,Z), ρ)$ be the set of all bounded functions from $X$ to $(Z,d)$ endowed with the sup-metric. Then how to prove that the metric space $(\mathcal B(X,Z), ρ)$ is a completion of the metric space $(\mathcal B(X,Y), ρ)$?

I assume that it is true and if not then can we get a counterexample. If $X$ is finite then it is easy for me to prove it, but I am struggling if $X$ is infinite.

Thanks in advance for your help.

答:
The function $T$ that sends $f \in B(X,Y)$, so $f:X \to Y$, to $T(f)= e \circ f: X \to Z$ in $B(X,Z)$, where $e: (Y,d_Y) \to (Z,d_Z)$ is the distance preserving embedding of $Y$ into its completion, is also distance preserving, as is easy to check.

It's also easy to check that the image of $T$ is dense in $B(X,Z)$, based on the fact that $e[Y]$ is dense in $Z$.

It follows that $T$ witnesses that $B(X,Z)$ is a (or the, up to isometry) completion of $B(X,Y)$, by standard facts from the theory of completions.


问:
Reducibility of a rational polynomial with two identically complex root

Let $n\geq 2$, $f(x)$ is a rational polynomial with degree $n$, and $x_1,\cdots,x_n$ are the complex roots of $f(x)$. If exists $1\leq i<j\leq n, x_i=x_j$, show that $f(x)$ is reducible in $\Bbb Q$.

My attempt: Argue by contradition. We assume $$f(x)=(x-x_i)^2(x-\bar x_i)^2g(x)$$ for some $x_i\in\Bbb C$ and $g(x)$ is a real polynomial. Then $f$ is reducible in $\Bbb R$.

What to do next?

答:
Hint: consider $f'$ and $\mathbf{gcd}_{\Bbb R[X]}(f,f')$. Now, both $f$ and $f'$ are in $\Bbb Q[X]$, so what does that tell you about $\mathbf{gcd}_{\Bbb Q[X]}(f,f')$?

Remark. Recall that the derivative of a polynomial $f = a_0+a_1x+ \ldots a_n x^n$ is
$$
f' := a_1+2a_2x + \ldots + na_n x^{n-1}
$$

Here's a general fact about derivatives and multiple roots over $\Bbb Q$. Everything that follows can be made more general, but let's stick with polynomials over $\Bbb Q$ for a moment.

Note that "algebraic derivatives" as defined above still satisfy the usual rules $(f+g)' = f'+g'$ and $(fg)' = f'g'$. You can prove this inspecting the coefficients at both sides. Here is a characterization of double roots in terms of derivatives:

Lemma. The following are equivalent,

$\alpha \in \Bbb C$ is a double root of $f$.

$\alpha$ is a root of $f'$ and $f$.

Proof. If $(X-\alpha)^2$ divides $f$ in $\Bbb C$, then $f = (X-\alpha)^2 g$ and taking derivatives we see that $$f' = (X-\alpha)^2 g' + 2(X-\alpha)g$$ is divisible by $X-\alpha$, hence $f'(\alpha) = 0$. Reciprocally, if $f'(\alpha) = 0$ dividing $f$ by $(X-\alpha)^2$ we get

$$
f = (X-\alpha)^2 q + r \tag{1}
$$

with $r = 0$ or $\deg r \leq 1$. and taking derivatives,

$$
f' = (X-\alpha)^2q' + 2(X-\alpha)q + r'
$$
hence evaluating at $\alpha$ yields $0 = f'(\alpha) = r'(\alpha)$. By degree considerations on $r$ it must be $r' = 0$, hence $r$ is constant. Evaluating at $(1)$ gives then $0 = f(\alpha) = r(\alpha)$ and thus $r = 0$, as desired. $\square$

Now, a double root $\alpha$ of $f$ is a common root of $f'$ and $f$, hence $X-\alpha \mid \mathbf{gdc}_{\Bbb C[X]}(f,f')$.

Note that $f$ and $f'$ are polynomials in $\Bbb Q[X]$ and $\Bbb C[X]$ and in both rings one can consider their greatest common divisor. If a polynomial divides both $f$ and $f'$ in $\Bbb Q[X]$, so it does over $\Bbb C[X]$. In particular $\mathbf{gdc}_{\Bbb Q[X]}(f,f') \mid f,f'$ in $\Bbb C[X]$ and so $\mathbf{gdc}_{\Bbb Q[X]}(f,f') \mid \mathbf{gdc}_{\Bbb C[X]}(f,f')$.

However, since there exists $g,h \in \Bbb Q[X]$ for which $\mathbf{gdc}_{\Bbb Q[X]}(f,f') = fg+f'h$ and $\mathbf{gdc}_{\Bbb C[X]}(f,f')$ divides both factors, we get the other division, hence

$$
\mathbf{gdc}_{\Bbb C[X]}(f,f') = \mathbf{gdc}_{\Bbb Q[X]}(f,f') =: (f,f').
$$

In particular, the former was already in $\Bbb Q [X]$. By the lemma, the polynomial $f$ has no double roots if and only if $(f,f') = 1$. In particular $f$ has no double roots if it is irreducible (here we use that $f'\neq 0$, essentially the same argument can be made in any field of characteristic zero). The contrapositive is what you needed.


答:
If $X\sim N(50;2)$ then $\overline{X}_{16}\sim N(50;1/8)$

This is the graph of the normal density...

as you can see... 52 is a very unusual value for $\overline{X}_{16}$

Title: Confidence Interval, why is this unusual?

Body:
A Normal population has a known mean of $\mu = 50$ and a known variance of
$\sigma = \sqrt 2$. A random sample of size $n = 16$ is selected from this population and find that the sample mean is $x = 52$. This is an out-of-the-box sample result. common Why? Answer using a $95\%$ confidence interval.

I got the interval $(51.30, 52.69)$...
But what should I do after?


问:
Sum of powers of a Matrix and the inverse
$A$ is a matrix, then for some $p$:

if $A^0+A^1+A^2+ \cdots +A^p = 0 $ then $A$ has an inverse

I have tried to prove this using:

$\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$

$\det(A^n) = \det(A)^n$

$\det(AB) = \det(A)\cdot\det(B)$

It shouldn't be difficult as it is a problem for first year in college, but I don't get any useful to prove it, can anybody help?

Thanks in advance.

答: Look directly for an inverse: Note that $A(-1(I+A+\cdots+A^{p-1})) = I$.


问:
Convergence of: $\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$
Convergence of:

$$\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$$

Tried ratio test but somehow it doesn't work for me.

答:
Using
$$\left( \frac{n}{e}\right)^n \lt n! $$
we have
$$\frac{3^n \cdot n^n}{(n!)^2} \lt \frac{3^n e^{2n}}{n^n}$$
Then by root test ..


问:
Proving topologist's sine curve limit points.
I am not sure if this makes sense or not.

I was trying to prove that $\{0\} \times [-1,1]$ is a subset of the set of limit points of the topologist's sine curve. The topologist's sine curve being $S=\{(x,\sin\frac{1}{x})|x \in (0,1)\}$. I was wondering if this approach is ok. I tried to show if $y$ is fixed in $[-1,1]$, then there is a sequence of points $x_n \rightarrow 0$, such that any open ball around $(x_n,y)$ intersects $S$ for each positive epsilon, and natural number $N$. Here was my attempt:

Because of the rapid oscillations of $\sin(\frac{1}{x})$ near $0$, consider the points in $\mathbb{R}^2$ of the form $(0,y), y \in [-1,1]$.I will try to show there is a sequence $x_n$ of points converging to $0$, such that for a fixed $y \in[-1,1]$, and each point in the sequence, every neighborhood of $(x_n,y)$ is contained in an open ball, that intersects the set $S$. For each $n \in\mathbb{N}$ let $\epsilon_n>0$ be an arbitrary small positive number.By the archimedean property choose $N \in \mathbb{N}$  such that $n \geq N \implies \frac{1}{2\pi n}<\epsilon_{n}$. Then for each $\epsilon_n>0$,choose a point  $x_n \in [\frac{1}{2 \pi(n+1)},\frac{1}{2 \pi n})$, as a term in the sequence. Then  for $n \geq N$,  $B((x_n,y),\epsilon_n) \cap S \neq \varnothing$. Since $\text{lim}x_n=0,B((0,y),\epsilon) \cap S \neq \varnothing$ for any positive $\epsilon$.

答:
Let $(0,p) \in \{0\} \times [-1,1]$. Find some $\alpha_0 \in [0, 2\pi]$ such that $\sin(\alpha_0)=p$. Note that this means that $\sin(\alpha_0 + 2n\pi) = p$ for all $n \in \Bbb N^+$ as well, by periodicity.

Then $x_n = (\frac{1}{\alpha_0 + 2n\pi} ,p)$ is a sequence that lies on $\mathcal{S}$ and converges to $(0,p)$. So $(0,p) \in \overline{\mathcal{S}}$

No need for vagueness about "rapid oscillations".


问:
Prime of the form $n^2-2m^2$
I need to prove that a prime $p$ is of the form $n^2-2m^2$ iff $p=2$ or $p=8k\pm 1$. So first I tried figuring out, for which $p$ does $n^2-2m^2=0$ has a solution if $\mathbb{F}_p$, or similarly all primes $p$ s.t there's a solution to the equation $(\frac{n}{m})^2=2$.

So now I know that if $\mathbb{F}_p$ has a square root for $2$, then $p$ divides a number of the form $n^2-2m^2$, but how can I prove that $p$ is equal to it? I tried looking at $\mathbb{Z}[\sqrt2]$ but that didn't get me so far.

Any help would be apperciated.

答:
The number field $\Bbb Q[\sqrt 2]$ has class number $1$.

This means that the ring $\mathcal O = \Bbb Z[\sqrt 2]$ is a principal ideal domain.

If $p = 8k \pm 1$ is a prime, then we know that the ideal $p\mathcal O$ splits into a product $p\mathcal O = \mathfrak p \mathfrak q$, with both $\mathfrak p, \mathfrak q$ not equal to $\mathcal O$ (this is essentially the fact that $2$ has two different square roots in $\Bbb F_p$). This implies that the norms of $\mathfrak p, \mathfrak q$ are both equal to $p\Bbb Z$.

Since $\mathfrak p$ is principal, we may write $\mathfrak p = x\mathcal O$ for some $x \in \mathcal O$. We can write $x = n + m\sqrt 2$ with $n, m \in \Bbb Z$.

Since the norm of the ideal $x\mathcal O$ is equal to $p\Bbb Z$, we know that $N(x) = \pm p$, where $N$ denotes the norm map from $\mathcal O$ to $\Bbb Z$.

If $N(x)$ is equal to $p$, then we are done, as we get $n^2 - 2m^2 = p$.

If $N(x)$ is equal to $-p$, then we have $N(\epsilon x) = p$, where $\epsilon$ is the unit $1 + \sqrt 2$. By writing $\epsilon x = u + v\sqrt 2$, we get $u^2 - 2v^2 = p$.

Title: Prime of the form $n^2-2m^2$


问:
A ring is Jacobson iff $Spec_{max}(R)$ is dense in any closed topological subspace of $Spec(R)$
I am trying to prove that $R$ is a Jacobson ring iff for any $Y \subseteq Spec(R)$ closed in the Zariski topology, one has that the closure of $Spec_{max}(R) \cap Y$ is $Y$ itself.

I denote for any subset $S \subseteq R$,  $V(S) := \left\{ P \in Spec(R): S \subseteq P \right\}$. In this notation $Y$ is closed iff $Y = V(S)$ for some $S \subseteq R$ by definition.

Firstly, to argue from left to right, I took $Z = V(T)$ and $Y = V(S)$ closed such that $Spec_{max}(R) \cap Y \subseteq Z \cap Y$. To show the implication, I would have to show that $Y \subseteq Z \cap Y$. Since $Spec_{max}(R) \cap Y \subseteq Z \cap Y$, we have that

\begin{equation}\label{lab} \bigcap\limits_{P \in Z \cap Y} P \subseteq \bigcap\limits_{P \in Spec_{max}(R) \cap Y} P \end{equation}

The RHS of the inclusion is just (denoting by $(\cdot)$ the ideal generated by $\cdot$ in $R$):

$$\bigcap\limits_{ \substack{\mathfrak{m} \in Spec_{max}(R) \\ (S) \subseteq \mathfrak{m} } } \mathfrak{m} \ ,$$

while the LHS of the inclusion is the intersection of primes in $V(S \cup T)$, which is just $\sqrt{(S \cup T)}$. By hypothesis $R$ is Jacobson, so:

$$\sqrt{(S \cup T)} = \bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ (S \cup T) \subseteq \mathfrak{m} }} \mathfrak{m} \ .$$

So
$$
\bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ (S \cup T) \subseteq \mathfrak{m} }} \mathfrak{m}
\subseteq
\bigcap\limits_{ \substack{\mathfrak{m} \in Spec_{max}(R) \\ (S) \subseteq \mathfrak{m} } } \mathfrak{m}
$$

The intersection on the right is over the same terms as the one on the left and possibly more, so it contains the one on the left. Hence the two intersections must be equal, so using again the hypothesis for $\sqrt{(S)}$, we get that $\sqrt{(S)} = \sqrt{(T \cup S)}$. Now an easy check shows that $V(\sqrt{I})=V(I)$ for any ideal $I$, hence we get that $Z \cap Y = V((T \cup S)) = V(\sqrt{(T \cup S)}) = V(\sqrt{(S)})= V((S)) = Y$ as desired. Is this correct? Is there a less "messy" proof (using only the very elementary properties of the Zariski topology, which are the only ones with which I'm familiar now)?

As for the second implication, I am a bit stuck: for $I$ an ideal in $R$ I have to show that it has the Jacobson property if $Spec_{max}(R) \cap V(I)$ is dense in $V(I)$. I think the problem is I can't figure out a way to take advantage of the density property. I tried deriving something from the density property expressed as:

$$
\bigcap\limits_{\substack{I \subseteq T \\ Spec_{max}(R) \cap V(I) \subseteq V(T)}} V(T) = V(I) = V(\sqrt{I}) \,
$$

but this expression doesn't seem to help. So how is this implication proven?

答:
I figured out the other implication too, it was almost there but I didn't realise it: if $Spec_{max}(R)\cap V(I)$ is dense in $V(I)$:

$$
\bigcap\limits_{\substack{I \subseteq T \\ Spec_{max}(R) \cap V(I) \subseteq V(T)}} V(T) = V(I) \,
$$

The intersection on the LHS is necessarily contained in $Spec_{max}(R)\cap V(I)$, so we get that $V(I) \subseteq Spec_{max}(R) \cap V(I)$, so going to intersection of the primes in each set (which reverses inclusion) we get

$$
\bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ I \subseteq \mathfrak{m} }} \subseteq \sqrt{I}
$$

Since we always have the other inclusion, we get equality.


问:
To check whether the set $S$ is compact.
The set $S=\{(x, y):x=0,|y|<=1\}\cup \{(x, y):0<x \le 1,y=sin(\frac{1}{x})\}$.We need to check whether $S$ is compact.

1)
$S_1 = \{(x, y):x=0,|y|<=1\} $.Now, $S_1$ is closed and bounded subset of $\mathbb{R}^2$ ,hence it is compact.

2)$S_2= \{(x, y):0<x \le 1,y=sin(\frac{1}{x})\}$.

Now,I know this theorem that

if $E$ is a compact set then $f$ is continuous on $E$ iff the graph of $f$ is compact.

We know that value of $x$ comes from the interval $[0,1]$

However we don't have the function $sin(\frac{1}{x})$ continuous in the interval [0,1] even if we define $sin(\frac{1}{x})=0$.So if graph ($S_2 \cup {(0,0)}$)becomes compact then sin($\frac{1}{x}$) becomes continuous on the interval $[0,1]$,which is a contradiction.

Hence my intuition is that $S$ is not compact. Is this OK?

答: The set $S_2$ is also compact. Note that$$S_2=\overline{\left\{\left(x,\sin\left(\frac1x\right)\right)\,\middle|\,x\in(0,1]\right\}}.$$So, $S_2$ is closed. Since it is also bounded (it is a subset of $[0,1]\times[-1,1]$), it is compact.


问:
Condition on a parameter for a certain matrix to be idempotent
A matrix $A$ is called idempotent if $A^2=A.$

If $$A=α\begin{pmatrix}-\frac{3}{2}&1\\ -3&2\end{pmatrix}$$ is idempotent, find α.

How can I solve this question? Do I just need to insert α into the matrix and equal to zero? I need help regarding this question. I try to solve it for hours but I cannot understand it

答:
$A=a\begin{pmatrix}-1.5& 1\\ -3 &2\end{pmatrix}$. $A^2=a^2\begin{pmatrix}-0.75& 0.5\\ -1.5 &1\end{pmatrix}$

As $A^2=A$, $2a=a^2\implies a(2-a)=0\implies a=0$ or $a=2$.

I am not sure where you had problem (as it is straight-forward), so I am writing the answer explicitly.


问:
weak topology on a topology vector space
$X$ is a vector space. $X'$ is a family of linear functionals on $X$ and separate point on $X$ (for any point $x\neq y$ in $X$, there is a linear functional $L$ in the family $X'$ s.t. $Lx\neq Ly$) . The weak topology on $X$ is the weakest topology making all the element in $X'$ continuous. And $X^*$(denote the dual space of $X$) is the span of $X'$.

Now for any linear functional $L$ on $X$, the function ,$$p_L:X\rightarrow \mathbb{R},p_L(x)=|Lx|$$
is a seminorm on $X$, $X'$ separates the point implies that the family of semi-norms is a separating family.
My teacher told me that the topology generated by $X'$ is exactly the same the topology generated by the family of semi-norms(why?) In other words, why the weak topology has a local base $\mathcal{B}_0=\{V_{L_1,…,L_n;r_1,…r_n}|L_1,…,L_n\in X',r_1,…,r_n＞0\}$ , where $V_{L_1,…,L_n;r_1,…r_n}=\{x\in X:|L_ix|\leq r_i, 1\leq i\leq n\}$.

My second question: why $x_n\rightarrow x$ in the weak topology iff $L(x_n)\rightarrow L(x)$ for all $L\in X^*$($X^*$ is the dual space of $X$).

答:
Let $X$ be  a set and $(f_i)$ be a family of scalar valued functions on $X$. Let $\tau$ be the weak topology generated by this family. If $x_n \to x$ then $f_i(x_n) \to f_i(x)$ for each $i$ because $f_i$ is continuous for $\tau$. This proves one way of your result since,for any finite linear combination $g$ of $f_i$'s also $g(x_n) \to g(x)$.

Conversely, suppose  $f_i(x_n) \to f_i(x)$ for each $i$. Consider any basic neighborhood of $x$ in $\tau$: $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$. Since $f_{i_j}(x_n) \to f_{i_j}(x)$ for each $j \leq N$ it follows that the inequalities $|f_{i_j}(x_n)-f_{i_j}(x)| <r_i: 1 \leq j \leq N$ hold for $n$ suffcientyl large. Hence $x_n$ lie in the basic neighborhood for $n$ suffcientyl large.

HOWEVER there is a flaw in your approach. Use of sequences here is fine in metrizable case but not in the case of general topological vector spaces. For  a proper proof you have to use nets, or work completely with neighborhoods.

EDIT: Weak topology $\tau$ generated by $(f_i)$: Consider sets of the type $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$. Form all possible unions of these sets. You can check that this gives a  topology $\tau'$ on $X$. It is clear that $\tau' \subseteq \tau$ because each of the sets $\{y: |f_{i_j}(y)-f_{i_j}(x)| <r_i: 1 \leq j \leq N\}$ is open in $\tau$.
It is  also easy to check  that each $f_i$ is continuous for $tau'$. By definition of $\tau$ it follows that $\tau \subseteq \tau'$.


问:
Visualizing infinitesimals
I am having problems visualizing infinitesimals. Like in a curve, when we integrate we can adjust in it a perfect infinitely thin rectangle. But how do I visualize it? Every time I try to visualize it, I always notice some gap area or error in the curve. Also in differentiation, $ dx $ is supposed to an infinitely close step. But every time I draw a line, I can cut it into smaller line and that becomes $ dx $. So I am facing bit problema visualizing infinitesimals.

答: Just like $x$ has no fixed value (it's variabel), so is $dx$. So you shouldn't be irritated by the fact that, even if you manage to visualise $dx$, there is still a "smaller $dx$". The way I visualise $dx$ is by contrasting it to $\Delta x$ with a looking glass that zooms in "infinitely deep". Here is a sketch:

Title: Visualizing infinitesimals


问:
Which does "themselves" qualify, "proofs" or "three theorems" in this sentence? (Michael Spivak "Calculus 3rd Edition")
I am reading "Calculus 3rd Edition" by Michael Spivak.

There is the following sentence in this book:

The proofs of the three theorems themselves will not be given until the next chapter, for reasons which are explained at the end of this chapter.

Which does "themselves" qualify, "proofs" or "three theorems"?

Which is correct?:

"The proofs themselves will not be given until the next chapter"

or

"The proofs of (the three theorems themselves) will not be given until the next chapter"

答:
This is more of an English question than a math question. Regardless of what "themselves" point to, it is clear (at least to me) that what comes in the next chapter is the proof of the three theorems.

That being said, I think it makes more sense to let "themselves" qualify the theorems. I don't know this book and the context here. However, my guess is that there are these three theorems which are stated and used prolifically in other proofs in this chapter. But these three theorems themselves will have to be proven later.

As a test, consider the hypothetical that there is a single theorem with multiple proofs. I personally feel that "The proofs of the theorem itself will not be given until next chapter." reads better than "The proofs of the theorem themselves will not be given until next chapter." At least with my guess at what the context is like, detailed in the previous paragraph.


答: It concerns “proofs”; this chapter contains the statements of the three theorems.</p>

Title: Which does "themselves" qualify, "proofs" or "three theorems" in this sentence? (Michael Spivak "Calculus 3rd Edition")

Body:
I am reading "Calculus 3rd Edition" by Michael Spivak.

There is the following sentence in this book:

The proofs of the three theorems themselves will not be given until the next chapter, for reasons which are explained at the end of this chapter.

Which does "themselves" qualify, "proofs" or "three theorems"?

Which is correct?:

"The proofs themselves will not be given until the next chapter"

or

"The proofs of (the three theorems themselves) will not be given until the next chapter"


问:
A classical example of a sequence satisfy the strict inclusions of sequences
There is no doubt that clear examples consolidate the understanding of concepts being learnt. I am new to Analysis of sequences, I learnt theoretically that suppose in the sequence space, $s = \{ x: \mathbb{N} \to \mathbb{R}\}$, we have that

$$\ell^r, \ell^s = \{x: \mathbb{N} \,\to\, \mathbb{R} \Big| \ \sum\limits_{i=1}^\infty |x|^r < \infty \}, \quad  \ell^\infty = \{x: \mathbb{N} \,\to\, \mathbb{R} \big| \  x =(x_i) < \infty \} \\ \mathbf{c} = \{x: \mathbb{N} \,\to\, \mathbb{R} \Big| \  x = (x_i) \ \text{is a convergent sequence} \}, \qquad \mathbf{c}_0 = \{x: \mathbb{N} \,\to\, \mathbb{R} \big| \  x_i \,\to\, 0 \ \text{ as } i \,\to\, \infty \}$$ then this inequality:
$$ \ell^r \subsetneq  \ell^s \subsetneq \mathbf{c}_0 \subsetneq \mathbf{c} \subsetneq \ell^\infty  \subsetneq \mathbf{s} \quad  \text{ for } 1 \le r<s<\infty \qquad (1). $$
is strictly satisfied.

After going through https://math.stackexchange.com/ to try and get more knowledge, I read from How do you show monotonicity of the $\ell^p$ norms?,  Inequality between $\ell^p$-norms, An example of a sequence which satisfies a number of conditions, and here What sequences could satisfy these requirements? just to catch a glimpse of an example but there wasn't a clear cut example.

Assuming without admitting that the inequality was just an improper subset $"\subseteq"$, then the sequence would have been as easy as $(1, 0, 1, 0, \ldots)$. Now my task is finding that example satisfying (1). Benevolent contributors, I rely on you for assistance.

答:
$\ell^r\subsetneq \ell^s$ for $r<s$:

We will use the fact that for $\alpha\in \mathbb{R}$, $$\sum_{n=1}^\infty n^{-\alpha}$$ converges if and only if $\alpha >1$ (which can be shown by the integral test). Pick $1/s<t<1/r$. Then $x(n)=n^{-t}$ is in $\ell^s$ but not $\ell^r$.

$\ell^s\subsetneq \mathbf{c}_0$ for any $s$:

We can construct a sequence that works for all $s$. Consider the sequence $x(n)=1/\ln n$ (for $n>1$, it doesn't matter what happens at $n=1$, since we only need the limiting behaviour). To prove this, it suffices to show that for sufficiently large $n$, $$\ln n<n^{1/s},\tag{*}$$for then we use a comparison with the harmonic series to show that
$$\sum_{n=1}^\infty \frac{1}{(\ln n)^s}$$
diverges. To prove (*), we compute the limit
$$\lim_{t\to\infty}\frac {\ln t}{t^{1/s}}=\lim_{t\to\infty}\frac{1/t}{(1/s)t^{1/s-1}}=\lim_{t\to\infty}\frac{s}{t^{1/s}}=0,$$
where the first equality follows from l'hopital's rule.

The strictness of the other inclusions are relatively straightforward. For $\mathbf{c}_0\subsetneq \mathbf{c}$, pick your favourite sequence converging to $0$ and then adding some non-zero constant will produce an example. For $\mathbf{c}\subsetneq \ell^\infty$, examples include $x(n)=(-1)^n$.


问:
On a question chosen at random, what is the probability that the student answers it correctly?
I'm really confused about this question. I appreciate your help.

A student takes a multiple choice exam where each question has five possible answers. He answers correctly if he knows the answer, otherwise he guesses at random. Suppose he knows the answer to $70$% of the questions.

Question 1

On a question chosen at random, what is the probability the student answers it correctly?

We know $P(\text{know the answer}) = 0.7$. On a question chosen at random, he either knows the answer or he doesn't. If he does, then $P(\text{answer correctly}|\text{know correct answer})= 1$. If he doesn't, then $P(\text{answer correctly}|\text{doesn't know correct answer})= 0.7(0.3)^4$. I don't know how to continue the answer.

Question 2

Given that he did answer correctly, what is the probability that he actually knew the correct answer?

All I can think of is the following conditional probability.

$\begin{align}
P(\text{know correct answer}|\text{answer correctly}) & = P(\text{know correct answer}|\text{answer correctly}) \\
& =\frac{P(\text{know correct answer and answer correctly})}{P(\text{answer correctly)}}. \\
\end{align}$

答:
Consider a typical run of 100 questions.

How many will he know the answer for?  Obviously, he gets all of these right.

How many are left to guess about?  How many of these will he get right?

How many correct all together from the entire 100 question run?

Your logic for the second question is correct, given the correct numbers to insert...


问:
Norm determines lower bound for invertible elements
Let $A$ be some unital Banach *-algebra. M. Takesaki, Theory of Operator Algebras, vol. 1. Springer, 1979. claims that if $|\lambda|>r(x):=\lim\inf_{n\rightarrow\infty}\|x^n\|^{1/n}$ then $\sum_{n=0}^\infty x^n/\lambda^{n+1}$ converges. I don't see why this is the case. This is used to claim that $x-\lambda$ is invertible. I would appreciate if someone could help me pin point why the sum converges.

As an alternative, the way that W. Rudin, Functional Analysis, 2nd ed. McGraw-Hill, Inc., 1991. does this is by showing that if $x-\lambda$ is not invertible then $|\lambda|\leq \inf_{n\geq 1}\|x^n\|^{1/n}\leq r(x)$. For this he notes that $x^n-\lambda^n=(x-\lambda)(x^{n-1}+x^{n-2}\lambda+\cdots\lambda^{n-1})$. He claims that from this decomposition it is clear that $x^n-\lambda^n$ can't be invertible. Once again, I don't see why this is true, for a product containing a non-invertible element can still be invertible.

答:
Mark solved the second half in the comments. As far as the first half is concerned,  I saw the answer in O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics, Second., vol. 1. Springer, 1987. The idea is that $|\lambda|>r(x)$ implies that there is an $n$ such that $|\lambda^n|>\|x\|^n$. Now, due to the Euclidean algorithm, one can write
$$\sum_{n=0}^\infty\frac{x^n}{\lambda^{n+1}}=\frac{1}{\lambda}\sum_{p=0}^\infty\sum_{q=0}^{n-1}\left(\frac{x}{\lambda}\right)^{np+q}=\frac{1}{\lambda}\sum_{q=0}^{n-1}\left(\frac{x}{\lambda}\right)^{q}\sum_{p=0}^{\infty}\left(\frac{x^n}{\lambda^n}\right)^{p}.$$
In this last form, convergence is clear.


问:
Notation for multilinear maps - is there a good reason why it is not common to write $\text{ML}(V_1\times\ldots\times V_n,W)$?
Since we write $\text{L}(V,W)$ for the set of linear maps from $V$ to $W$, it seems to suggest itself to denote multilinear maps from $V_1\times\ldots\times V_k$ to $W$ by $\text{ML}(V_1\times\ldots\times V_k,W)$. Is this notation already used for something else or is there some other reason not to use it?

Edit: I know that it is common to use the notation $\text{L}(V_1,\ldots,V_n,W)$ for $n$-linear maps from $V_1\times\ldots\times V_n$ to $W$, but authors who use this notation (e.g. John Lee in his introduction to manifolds) constantly have to use underbraces, e.g.
\begin{equation}
\text{L}(\underbrace{U,\ldots,U}_{m\text{ copies}},\underbrace{V,\ldots,V}_{n\text{ copies}},W).
\end{equation}
I was searching for a way to avoid this, especially since this can't be used in inline equations. My idea was to write $\text{ML}(U^m\times V^n,W)$ - more generally, $\text{ML}(V_1^{n_1}\times\ldots\times V_k^{n_k},W)$.

答:
I just realized that the notation is ambiguous: It's not clear whether the elements of $\text{ML}(V_1^{n_1}\times\ldots\times V_k^{n_k},W)$ are $k$-linear or $(n_1+\ldots+n_k)$-linear (consider for example bilinear maps from $\mathbf{R}^n\times\mathbf{R}^n$ to $\mathbf{R}$).

I guess we'll have to stick to the common notation $L(V_1,\ldots,V_n,W)$. One idea would be to define
\begin{equation}
L(V_1\colon n_1,\ldots,V_k\colon n_k,W):=L(\underbrace{V_1,\ldots,V_1}_{n_1\text{ copies}},\ldots,\underbrace{V_k,\ldots,V_k}_{n_k\text{ copies}},W)
\end{equation}
to avoid having to use underbraces all the time.


答: Even though the two other answers are contradictory, both are good because the formulation in the book is indeed ambiguous when taken out of context. I tend to think it refers to "proofs" because "themselves" seems to me like it indicates a disjunction from something stated earlier. It may be that the author writes a short intuitive explanation for the theorems before this statement and so aknowledges the reader that the proofs themselves will be given later. It's also possible that the author didn't state the theorems yet and is just about to. In this case, they first tell the reader that the proofs of the * theorems themselves * will not be given together with them.</p>


问:
Self-intersection number of a loop on a surface
Let $M$ be an orientable surface$($without boundary$)$, possibly non-compact. Let $\alpha\in \pi_1(M)$ be a primitive element i.e. $\alpha$ is not a proper power of some other element.  Let $f:\Bbb S^1\to M$ be any loop representing $\alpha$. Consider the lifting problem

Here, $M_\alpha$ be the cover corresponding to the subgroup $\langle\alpha\rangle$ of $\pi_1(M)$ and $\widetilde M$ be the universal cover of $M$. Also, all unleveled maps are covering maps.

Note that $\pi_1(M)$ acts on $\widetilde M$ via covering transformations, and $M_\alpha$ be an open annulus$($any open connected surface with a finitely generated fundamental group is homeomorphic to the interior of a closed surface$)$.

$\textbf{Problem 1:}$ Consider two sets $$\mathscr A_f:=\big\{g\in
 \pi_1(M):\text{ the map } \Bbb R\xrightarrow{g\cdot \ell}\widetilde
 M\to M_\alpha\text{ runs from one end to other of the open annulus
 }M_\alpha\big\},$$ $$\mathscr B_f:=\big\{(z,w)\in \Bbb S^1\times\Bbb
 S^1:z\not= w\text{ and   }f(z)=f(w)\big\}$$ Is the cardinality of
$\mathscr B$ two-times the cardinality of $\mathscr A$, i.e. $|\mathscr A_f|=\frac{1}{2}|\mathscr B_f|$?

$\textbf{Problem 2:}$ Give a Riemannian metric on $M$ such that
$\alpha$ can be represented by a shortest loop $f^\#:\Bbb S^1\to M$,
i.e. $f^\#$ has the minimum length in its free homotopy class. Is
$|\mathscr A_f|\geq |\mathscr A_{f^\#}|$?

答:
For problem one, the answer is "no" as there may be "bigons".  The pair of intersections forming the bigon contribute to $B_f$ but not to $A_f$.

For problem two, the answer is "yes" as the shortest representative has no bigons.

A version of this is discussed in the "Primer on mapping class groups" by Farb and Margalit under the name "the bigon criterion".


答:
Semantically it's surely referred to "proofs" (in the sense that it means that the proofs themselves will be et cetera). I would say grammatically as well, but I'm not British: perhaps the principle of proximity is stronger and it turns out to be formally referred to "theorems". You should ask on English.SE or something which ones are best among:

the doors of the wardrobe themselves, I will buy separately

the doors of the wardrobe itself, I will buy separately

the doors themselves of the wardrobe, I will buy separately

the stairs in the house themselves were made by my grandfather

the stairs in the house itself were made by my grandfather

the stairs themselves in the house were made by my grandfather


问:
$S$ and $T$ are idempotent linear operators in $V$ which is a vector space in $\mathbb{C}$. Prove that if $S+T$ is idempotent then $ST=TS=0$.
If $S+T$ is idempotent, then
$$(S+T)^2=S+T$$
$$\implies S^2+ST+TS+T^2=S+T$$
$$\implies S+T+ST+TS=S+T$$
$$\implies ST+TS=0$$

Now, from here how do I show that $ST=TS=0$?

答: Essentially equivalent answer but more elegant solution: $ST = SST = -STS = TSS = TS$, so $2ST = ST+TS = 0$ and hence $ST = 0$. Note that you do not need the field $F$ underlying the vector space to be $ℂ$, but you do need $\text{char}(F) ≠ 2$ otherwise you cannot divide by $2$.