问:
"Topological disjoint union" vs "disjoint union"
I want to understand the definition of topological disjoint union. Let $X$ be a topological space such that we can write $X=\bigsqcup X_i$. We say that $X$ is the "topological disjoint union" of the $X_i$ if each $X_i$ is an open subset of $X$, but if the $X_i$ are not necesserily open subsets of $X$ we say that $X$ is a the "disjoint union" of the $X_i$ only as sets (without the topological adjective) which means that $X =\cup X_i$ is the union of the $X_i$ such that all the $X_i$ are pairwise disjoint as sets. Is my understanding correct? thank you for your help!

答:
Yes, you are correct. But sometimes the adjective “topological” is implied as you are working in the category of topological spaces. One can make the distinction also by saying “disjoint union of subspaces” vs. “disjoint union of subsets”. It really depends on the context how careful one has to be to distinguish the two situations.

I would also like to add that there two versions of the sum / disjoint union. One of them is a construction – you start with any family of sets/spaces, and then you construct their sum. For this you often have to replace the original spaces by isomorphic copies that are disjoint. The other one is a property – you already have the whole space and a family of its subspaces, and you ask whether the whole space is the sum of those subspaces. Of course the two variants are closely connected – the latter propery holds if and only if the given space is canonically isomorphic to the constructed one. Also note that this phenomenon is not special to topological spaces and sets, it happens for example with the direct sum of modules / vector spaces.


问:
Evaluating a Gaussian Integral
How to prove that
$$\int_{\mathbb{R}^N}e^{-\langle Ax,x\rangle}\operatorname{dm}(x)=\left(\frac{\pi^N}{\det A}\right)^{\frac{1}{2}}$$

Where $A:\mathbb{R}^{N}\to\mathbb{R}^{N}$ is a symmetric positive-definite linear transformation.

My approach: Given a $x\in\mathbb{R}$, one have
$$\left(\int_{\left[-r,r\right]} e^{-x^2}\right)^{N}=\prod_{j=1}^{N}\left(\int_{\left[-r,r\right]}e^{-y_{j}^{2}}\operatorname{dm}(y_{j})\right)$$
So, by Fubinni's therem

$$\left(\int_{\left[-r,r\right]} e^{-x^2}\right)^{N}=\int_{\left[-r,r\right]^{N}}e^{-(y_{1}^{2}+\dots+y_{N}^{2})}\operatorname{dm}(y_1,\dots,y_N)=\int_{\left[-r,r \right]^N}e^{-||y||^2}\operatorname{dm}(y)$$

And, since $A$ is symmetric and positive-definite, there exist a $B:\mathbb{R}^N\to\mathbb{R}^N$ invertible such that $B^{-1}AB$ is diagonal. But I got stuck and can't find a change of variables that work.

答:
Since $A$ is symmetric positive definite, we have the Cholesky decomposition, i.e., $A = LL^T$. Hence, we have
$$\langle Ax,x \rangle = x^TL^TLx$$
Now make the transformation $y = Lx$. We then get
$$\int \exp(-\langle Ax,x \rangle) dx = \int \exp \left(-y^Ty \right) \dfrac{dy}{\det(L)} = \dfrac1{\sqrt{\det(A)}} \int\exp(-y^Ty) dy = \dfrac{\pi^{N/2}}{\sqrt{\det(A)}}$$</p>

ContentLicense: CC BY-SA 3.0

Title: Evaluating a Gaussian Integral


问:
Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$
I need to compute a limit:

$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.

$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$

I've calculated several values of this function, and it seems to have a limit of $1$.

答:
For small positive angles, the inequalities ${3\over4}\theta\le\sin\theta\le\theta$ apply. (The fraction $3/4$ is somewhat arbitrary; any fraction less than $1$ will do.)  Letting $\theta=\sqrt x$ and noting that $-1\le\sin(1/x)\le1$ for all $x\not=0$, we have

$${3\over2}\sqrt x-\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le2\sqrt x+\sqrt x$$

which simplifies to

$${1\over2}\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le3\sqrt x$$

If we take for granted the limit $x^x\to1$ as $x\to0$ (easily proved with L'Hopital applied to $\ln x\over1/x$), we have $\lim_{x\to0^+}(\sqrt x/2)^x=\lim_{x\to0^+}(3\sqrt x)^x=1$ so the Squeeze Theorem tells us

$$\lim_{x\to0^+}(\sin\sqrt x+\sqrt x\sin(1/x))^x=1$$

ContentLicense: CC BY-SA 4.0


问:
Two proof by ϵ−δ definition
$$(1.)\text{ }f(x) =
\begin{cases}
sin(x), &\text{ }x\in \mathbb{Q}\\
cos(x), &\text{ }x\notin \Bbb{Q}
\end{cases}
,\text{ prove that f is continuous at x = }\cfrac{\pi}{4}\text{ }by\text{ }\epsilon-\delta \text{ } \text{definition}$$

$$(2.)\text{ }f(x) =
\begin{cases}
x^2, &\text{ }x\in \mathbb{Q}\\
x^3, &\text{ }x\notin \Bbb{Q}
\end{cases}
,\text{ prove that f is differentiable at x = 0}\text{ }by\text{ }\epsilon-\delta \text{ }  \text{definition}$$

I know I have to prove $\lim_{x\to \cfrac{\pi}{4}}$ f(x) = f($\cfrac{\pi}{4}$) in problem 1 and $\lim_{h\to 0} \cfrac{f(x+h)-f(x)}{h}$ exists in problem 2. But I don't know how to use ϵ−δ definition in these problems. Also, what should I do in such problems that the value of f(x) is given based on if the value of x is rational number or not. Any suggestion or hint is appreciated. Thanks a lot!

答:
CLAIM: Let $h(x)$ and $g(x)$ be continuous functions at a point $a\in\Bbb R$ such that $h(a)=g(a)$. The the function $$f(x):=\begin{cases}h(x)& x\in\Bbb Q\\ g(x)&x\notin\Bbb Q\end{cases}$$ is continuous at $a$.

PROOF:

Observe that no matter if $a$ is rational or not, $f(a)=g(a)=h(a)$ is uniquely defined.

Fix $\varepsilon>0$.

AS $g(x)$ is continuous in $a$, there is $\delta_g>0$ such that $|x-a|<\delta_g$ then $|g(x)-g(a)|<\varepsilon$.

AS $h(x)$ is continuous in $a$, there is $\delta_h>0$ such that $|x-a|<\delta_g$ then $|h(x)-h(a)|<\varepsilon$.

Take $\delta:=\min(\delta_g,\delta_h)>0$ and observe that $\delta<\delta_g$ and $\delta<\delta_h$.

Take $x\in\Bbb R$ such that $|x-a|<\delta$

If $x\in\Bbb Q$, then $|f(x)-f(a)|=|h(x)-h(a)|<\varepsilon$, since $|x-a|<\delta<\delta_h$.

If $x\notin\Bbb Q$, then $|f(x)-f(a)|=|g(x)-g(a)|<\varepsilon$, since $|x-a|<\delta<\delta_g$.

So, in both cases $|f(x)-f(a)|<\varepsilon$ and $f$ is continuous at $a$.

You can apply this same $\varepsilon-\delta$ argument to both problems:

(1) $h(x)=\sin x$, $g(x)=\cos x$ and $a=\pi/4$.

(2) $h(x)=x^2$, $g(x)=x^3$ and $x=0$ for continuity and $h(x)=2x$, $g(x)=3x^2$ $a=0$ for derivability.

Title: Two proof by ϵ−δ definition


问:
What is $\int_\Omega 1_{X^{-1}(A)}dP=\int_\mathbb R 1_AdP_X$ intuitively and why does it make sense?

What I don't get is:

why does this make sense, we are integrating in two different measured spaces

is this equality important? Why do we need it?

EDIT : I'm talking about the integral equality and not the indicator function.

P.S: I'm not looking for the proof, just the intuition.

答:
Yes, the equality is very important. A priori it makes sense since, being integrals, the left- and the right-hand side are both real numbers. For an explanation why this holds as you wrote it down, recall that the integral over an indicator function is (by definition) the measure of the underlying set i.e. the left-hand side

$$ \int_{\Omega} 1_{X^{-1}(A)}(\omega) d\mathbb{P}(\omega) = \mathbb{P}(X^{-1}(A)) $$

and on the right-hand side:

$$ \int_{\mathbb{R}} 1_{A}(x) d\mathbb{P}_X(x) = \mathbb{P}_X(A) $$

The reason for why this is true is precisely the definition of the probability measure $\mathbb{P}_X$, the distribution of $X$.

So in the case of an indicator function it is clear that this holds, but it also holds for arbitrary (integrable) functions i.e. for any integrable function $f: \mathbb{R} \rightarrow \mathbb{R}$ it is true that

$$ \int_{\Omega} f(X(\omega)) d\mathbb{P}(\omega) =  \int_{\mathbb{R}} f(x) d\mathbb{P}_X(x)$$

and that's the important part.

Usually one has no (or only little) information about $\mathbb{P}$, let alone $\Omega$) but there is information about $\mathbb{P}_X$. In many cases this measure even has a density. So what we are doing is transformaing some abstract,inaccesible into a very concrete, computable one.

Example: Consider a normally distributed random variable $X$ with mean $\mu = 400$ and standard deviation $\sigma = 100$, and we want to compute the percentage that $X$ is below $300$. Now that would mean to evaluate the integral

$$ \int_{\Omega} 1_{\{\omega \in \Omega: X(\omega) < 300\}} (\omega) d \mathbb{P}(\omega)$$

How can one possibly evaluate that? Apply the transformation above to obtain

$$ \int_{-\infty}^{300} d\mathbb{P}_X(x) = \int_{-\infty}^{300} \frac{1}{\sqrt{2 \pi 100^2}} \exp(- \frac{(x-300)^2}{2 \cdot 100^2}) dx $$

where the last integrand is just the density of the normal distribution.


问:
Approximate increase in circumference of circle giving increase in area of 3%
Find the percentage increase in the circumference of a circle, which will result in an increase in the area of the circle of approximately 3%.

I have assumed the percentage change in circumference is the same as the percentage change in r, since circumference is $2\pi r$.

I have said $A = \pi r^2, \frac{dA}{dr} = 2\pi r$

$\frac{\delta A}{\delta r} \approx \frac{dA}{dr}$

where $\delta$ refers to "a small increase".

$\delta A = \frac{dA}{dr}.dr$

$\frac{3A}{100} \approx 2\pi r.\delta r$

But I cannot get this to equal the answer in the book, which is 1.5%.

答:
$$A=\pi r^2 \implies dA=2\pi r dr \implies  \frac{dA}{A}\%=2\frac{dr}{r}\% \implies \frac{dr}{r}\%=3/2$$

Npw $$P=2\pi r \implies \frac{dP}{P}\%=\frac{dr}{r}\% \implies \frac{dP}{P}\%=\frac{3}{2}$$


问:
Squared linear differential operator meaning
What does the linear differential operator, $Q$, defined as

$Q = (1 + \frac{1}{\alpha}\frac{d}{dt})^2$

mean exactly? That is, in the context of ODEs.

Does it mean the following?

$Q = 1 + \frac{2}{\alpha}\frac{d}{dt} + \frac{1}{\alpha^2}\frac{d^2}{dt^2}$

or this?

$Q = 1 + \frac{2}{\alpha}\frac{d}{dt} + \frac{1}{\alpha^2}(\frac{d}{dt})^2$

答:
It means applying the operator twice. Remember, $1$ and $\frac{1}{\alpha}$ are not simply numbers, they are operators, so you can't just multiply them that way (at least, usually you can't). Here it works out since all the operators commute.

To your question, that just looks like different ways of writing the same thing, since $\left(\frac{d}{dt}\right)^2$ is simply differentiating twice with respect to $t$, as is $\left(\frac{d^2}{dt^2}\right)$. Again, it's just notational conventions.


问:
limit and integral in the Lp space with Schwartz class.
Can we actually interchange limit and integral for $f \in S(\mathbb{R})$. Or it's just the limit and integral happen both equal to 0?

An exercise:
Show that we can interchange the limit and integral:
$\lim_{y\rightarrow0} \int_{\mathbb{R}} |f(x-y)-f(x)|^p dx = \int_{\mathbb{R}} \lim_{y\rightarrow0} |f(x-y)-f(x)|^p dx = 0,$ for $f \in S(\mathbb{R})$.

This an exercise from a book, I showed they separately equal to 0 and therefore forms the equal sign

Body:
For $|y|\le 1$ and $p > 1/(2n)$

$$\sup_x\ (1+x^{2n}) |f(x)-f(x-y)|\le \sup_x \int_{-y}^0 (1+x^{2n})|f'(x+t)|dt $$ $$\le \sup_x \int_{-y}^0 (1+(x+t+1)^{2n}) |f'(x+t)|dt$$ $$ \le \int_{-y}^0 \sup_x (1+(x+t+1)^{2n}) |f'(x+t)|dt = |y| A$$

where $A = \sup_x (1+(x+1)^{2n}) |f'(x)|$ is finite because $f'$ is Schwartz.

Thus

$$\int_{\mathbb{R}} |f(x-y)-f(x)|^p dx\le \int_{\mathbb{R}} \frac{A|y|}{(1+x^{2n})^p} dx$$


问:
Sum of the series $S_k$
Suppose $S_k$ be the sum of an infinite geometric series with first term $\frac{k-1}{k!}$ and common ratio $1/k$  for $k = 1,2...$.

What's the value of $\sum_1^\infty S_k $?

I get $S_1 =0$, $S_2= 1$ and $\sum_1^\infty S_k = 1-e$. However the answer options don't have this choice. Am I missing something or is there an error in the question?

Title: Sum of the series $S_k$

Body:
You have $S_1=0$ and for all $k\geq 2$,
$$S_k = \frac{k-1}{k!} \times \frac{1}{1- \frac{1}{k}} = \frac{1}{(k-1)!}$$

so $$\sum_{k=1}^{+\infty} S_k =\sum_{k=2}^{+\infty} \frac{1}{(k-1)!}  = e-1$$


问:
Biggest circle under a polynomial curve
What is the biggest circle that is contained in the region bounded by the graph of the polynomial $f(x) = x(1-x)(2x+1)$ and the x-axis interval $[0, 1]$?

(Here's the thing in Desmos)

Here's what I have tried

Let's denote the center of the circle by $(c_x, c_y)$ and the points of tangency by $(x_k, y_k)$, $k=1,2$. From distance to the x-axis, we know that the radius of the circle is then $c_y$. We can express the tangency by the dot product of the vectors $(x_k-c_x, y_k-c_y)$ and $(1, f'(x_k))$ being zero. So we get the group of equations (all for $k=1,2$)

$$\begin{cases}
(x_k-c_x)^2 + (y_k-c_y)^2 &=& c_y^2 \\
x_k-c_x + (y_k-c_y)f'(x_k) &=& 0 \\
f(x_k) &=& y_k
\end{cases}
$$

Then I tried to solve this group by using Gröbner basis and elimination ideals. I wrote this SageMath code:

```<code>f(x) = x*(1-x)(2*x+1)
fd = f.derivative()

R.<x1, y1, x2, y2, cx, cy> = PolynomialRing(QQ, order='lex')
polys1 = [(cx-xk)**2+(cy-yk)**2 - cy**2 for (xk, yk) in [(x1,y1), (x2, y2)]]
polys2 = [(xk-cx)+(yk-cy)*fd(xk)  for (xk, yk) in [(x1,y1), (x2, y2)]]
polys3 = [f(xk)-yk for (xk, yk) in [(x1,y1), (x2, y2)]]
I = R.ideal(polys1+polys2+polys3)
gb = I.groebner_basis()
for poly in gb:
    print (poly)
print (I.dimension())
</code>```

It gave me a long list of polynomials, the last one being

$$c_x^6 - 2c_x^4c_y^2 + \frac{5}{4}c_x^4c_y + \frac{1}{64}c_x^4 + c_x^2c_y^4 + \frac{3}{4}c_x^2c_y^3 + \frac{3}{16}c_x^2c_y^2 + \frac{1}{64}c_x^2c_y$$

and said the dimension is $1$. I now realize the reason we don't get dimension $0$ has probably to do with the fact that the points $(x_1, y_1)$ and $(x_2, y_2)$ could be the same point(?) So for a solution, the point $(c_x, c_y)$ should be a zero of that polynomial? But is even that correct, since the points could be collapsed and the other side go over the curve??

How to solve this even numerically? I tried with Sage by minimizing the square sum of all those polynomials that should be zero. I also tried adding constraints for the variables to be near the solution that can be seen from the picture but that didn't work either.

答:
I now managed to solve it at least numerically with Sage. I first find the closest point on the curve to a given point $(c_x, c_y)$. This can be done by finding the minimum of

$$(x-c_x)^2 + (f(x)-c_y)^2$$

by finding roots of its derivative (equivalent to the "tangency equation")

$$12x^5-10x^4-6x^3+(3+6c_y)x^2+(2-2c_y)x-c_x-c_y.$$

This allows to write the constraint for the radius.

Here's another Desmos link to test it out.

Take the radius $r$ to be another variable, having just $c_x$ and $c_y$ didn't work for some reason (the radius constraint was violated in the answer).

```<code>f(x) = -2.0*x**3 + x**2 + x
fd = f.derivative()

def find_closest_on_curve(cx, cy):
    bestP = None
    bestD = None
    #for r in (x-cx+(f-cy)*fd).roots(ring=CDF): #gives bug!!?!?!?
    for r in numpy.roots([12, -10, -6, 3+6*cy, 2-2*cy, -cx-cy]):
        #print (r)
        imag = r.imag
        real = r.real
        if abs(imag)<0.000001 and 0<=real and real<=1:
            xVal = real
            yVal = f(xVal)
            d = (xVal-cx)**2+(yVal-cy)**2
            if bestP==None or d<bestD:
                bestP = (xVal, yVal)
                bestD = d
    return bestP

def radius_constraint(p):
    cx, cy, r = p
    closeP = find_closest_on_curve(cx, cy)
    if not closeP: return -1
    d = (cx-closeP[0])**2 + (cy-closeP[1])**2
    return min(d, cy**2)-r**2

constraints = ([
    lambda p: p[0]-0.3,
    lambda p: 0.7-p[0],
    lambda p: p[1]-0.1,
    lambda p: 0.4-p[1],
    lambda p: p[2],
    radius_constraint
])

startP = (0.6, 0.2, 0.1)
#print (radius_constraint(startP))
sol = minimize_constrained(lambda p: -p[2], constraints, startP)
print (sol)
#print (radius_constraint(sol))

g = Graphics()
g += plot(f, 0, 1)
nn=20
for (cx, cy, r) in [sol]:
    solP = find_closest_on_curve(cx, cy)
    #print (solP)
    if not solP: continue
    x1,y1 = solP
    g += points([(cx, cy), (x1, y1)])
    g += line([(cx, cy), (x1, y1)])
    g += circle((cx, cy), r, color="green")

g.show(aspect_ratio=1)
</code>```

It gives the solution $(c_x, c_y, r) = (0.5965, 0.2577, 0.2577)$

Title: Biggest circle under a polynomial curve

Body:
I now managed to solve it at least numerically with Sage. I first find the closest point on the curve to a given point $(c_x, c_y)$. This can be done by finding the minimum of

$$(x-c_x)^2 + (f(x)-c_y)^2$$

by finding roots of its derivative (equivalent to the "tangency equation")

$$12x^5-10x^4-6x^3+(3+6c_y)x^2+(2-2c_y)x-c_x-c_y.$$

This allows to write the constraint for the radius.

Here's another Desmos link to test it out.

Take the radius $r$ to be another variable, having just $c_x$ and $c_y$ didn't work for some reason (the radius constraint was violated in the answer).

f(x) = -2.0*x**3 + x**2 + x
fd = f.derivative()

def find_closest_on_curve(cx, cy):
    bestP = None
    bestD = None
    #for r in (x-cx+(f-cy)*fd).roots(ring=CDF): #gives bug!!?!?!?
    for r in numpy.roots([12, -10, -6, 3+6*cy, 2-2*cy, -cx-cy]):
        #print (r)
        imag = r.imag
        real = r.real
        if abs(imag)<0.000001 and 0<=real and real<=1:
            xVal = real
            yVal = f(xVal)
            d = (xVal-cx)**2+(yVal-cy)**2
            if bestP==None or d<bestD:
                bestP = (xVal, yVal)
                bestD = d
    return bestP

def radius_constraint(p):
    cx, cy, r = p
    closeP = find_closest_on_curve(cx, cy)
    if not closeP: return -1
    d = (cx-closeP[0])**2 + (cy-closeP[1])**2
    return min(d, cy**2)-r**2

constraints = ([
    lambda p: p[0]-0.3,
    lambda p: 0.7-p[0],
    lambda p: p[1]-0.1,
    lambda p: 0.4-p[1],
    lambda p: p[2],
    radius_constraint
])

startP = (0.6, 0.2, 0.1)
#print (radius_constraint(startP))
sol = minimize_constrained(lambda p: -p[2], constraints, startP)
print (sol)
#print (radius_constraint(sol))

g = Graphics()
g += plot(f, 0, 1)
nn=20
for (cx, cy, r) in [sol]:
    solP = find_closest_on_curve(cx, cy)
    #print (solP)
    if not solP: continue
    x1,y1 = solP
    g += points([(cx, cy), (x1, y1)])
    g += line([(cx, cy), (x1, y1)])
    g += circle((cx, cy), r, color="green")

g.show(aspect_ratio=1)

It gives the solution $(c_x, c_y, r) = (0.5965, 0.2577, 0.2577)$


答:
Using the OP's notation, in particular $\delta A/\delta r\approx dA/dr=2\pi r$, we have
$${\delta A\over A}={2\pi r\delta r\over\pi r^2}=2{\delta r\over r}$$

so if $\delta A/A\approx3\%$, then $\delta r/r\approx1.5\%$. (And since $C=2\pi r$, we have $\delta C=2\pi\delta r$, hence $\delta C/C=\delta r/r=1.5\%$ as well, as the OP observed.)

Title: Approximate increase in circumference of circle giving increase in area of 3%

Body:
Find the percentage increase in the circumference of a circle, which will result in an increase in the area of the circle of approximately 3%.

I have assumed the percentage change in circumference is the same as the percentage change in r, since circumference is $2\pi r$.

I have said $A = \pi r^2, \frac{dA}{dr} = 2\pi r$

$\frac{\delta A}{\delta r} \approx \frac{dA}{dr}$

where $\delta$ refers to "a small increase".

$\delta A = \frac{dA}{dr}.dr$

$\frac{3A}{100} \approx 2\pi r.\delta r$

But I cannot get this to equal the answer in the book, which is 1.5%.


问:
Prove that it is a dense space in $H$ indexed to $\mathbb Z$.
Let $(e_n)_{n\in \mathbb Z}$ be an orthonormal basis in $H$ indexed in $\mathbb Z$. We consider the sequence $u_n=e_{-n}+ne_n$,
$M =\overline{\operatorname{span}\left\lbrace u_n: n ≥ 1\right\rbrace}\,$ and
$\,F =\overline{\operatorname{span}\left\lbrace e_n: n ≥ 0 \right\rbrace}\,$.

I would like to prove that $F + M$ is dense in $H$, and that $F + M  \subsetneq H$.

I have tried to do the analysis of this problem with the vector $\sum_{n=1}^{\infty}\frac{1}{n}e_{-n}$

but I don't know how to proceed.
I would appreciate someone to help me.

答:
For showing that $F+M$ is dense you only have to show that any vector orthogonal to both $F$ and $M$ is $0$ and this is easy.

For the second part your idea is good. Let $x=\sum\limits_{n=1}^{\infty} \frac 1 n e_{-n}$. Observe that $e_{-n}+ne_n, n \geq 1$ are orthogonal.  This implies that any element of $M$ is of then form $\sum\limits_{n=1}^{\infty} a_n u_n$. Of course, any element of $F$ is of the form $\sum\limits_{n=0}^{\infty} b_n e_{n}$. If $x \in F+M$ then we get $\sum\limits_{n=1}^{\infty} \frac 1 n e_{-n} =\sum\limits_{n=1}^{\infty} a_n u_n+\sum\limits_{n=0}^{\infty} b_n e_{n}$. But then $a_n=\frac 1 n$ for all $n \geq 1$ and $na_n+b_n=0$ for all $n \geq 1$. But this gives $b_n=-na_n=-1$ a contradiction to $\sum b_n^{2} <\infty$.


答:
For $x\in\left(0,\frac\pi2\right]$, the concavity of $\sin(x)$ says
$$
\frac2\pi\le\frac{\sin(x)}x\le1
$$
Therefore,
$$
\underbrace{\left(\frac4\pi\sqrt{x}-\sqrt{x}\right)^x}_{\left(\frac4\pi-1\right)^x\sqrt{x^x}}\le\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x\le\underbrace{\left(2\sqrt{x}+\sqrt{x}\right)^x}_{3^x\sqrt{x^x}}
$$
The Squeeze Theorem says
$$
\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x=1
$$

Title: Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$

Body:
I need to compute a limit:

$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.

$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$

I've calculated several values of this function, and it seems to have a limit of $1$.


问:
Find maximum of the function $f(x)=log_{2}^{4}x+12log_{2}^{2}x \cdot log_{2}(\frac{8}{x}) $ on interval $\left [ 1,64 \right ] $
In solution it writes that we can write this function in this form???

$$f(x)=(log_{2}^{2}x-6log_{2}x)^{2} $$

And that quadratic function in brackets has minimum -9 for x=8

Anyway can someone explain me this?

答: Let $\ell:=\log_2x$ so $f-\ell^4+12\ell^2(3-\ell)=\ell^4-12\ell^3+36\ell^2=(6\ell-\ell^2)^2$, as claimed. On the given range, $6\ell-\ell^2\ge0$, so its maximum at $\ell=3$ gives $f=81$.


问:
Convolution of f with itself
I was trying the following exercise:

(a) Find the fourier transform of the unit rectangular distribution of

\begin{equation}
  f(t) =
    \begin{cases}
      1, \ |t|<1 \\
      0 & \text{otherwise}
    \end{cases}
\end{equation}

which I found to be $\frac{2sin(w)}{w\sqrt{2\pi}}$

(b) Determine the convolution of f(t) with itself:

I have learned convolutions recently and I'm confuse on how to calculate:

$(f*f)(t)=\frac{1}{\sqrt2\pi}\int^{\infty}_{-\infty}f(u)f(t-u)du$

can someone explain me with detail the range changes of the integral?

Thanks!

答:
$$ \int_{-\infty}^\infty f(u)f(t-u)du$$

the only way for the integrand not to vanish is $|u|<1$ and $|t-u|<1$, so you need to have $$u\in (-1,1)\cap (-1+t,1+t)$$

Check for which $t$ the intersection is not empty, and compute the integral inside the intersection (since $f=1$ where it doesn't vanish, it's just the length of the intersection).

Title: Convolution of f with itself


答: $(f*f)(t)=\int_{(-1,1)\cap (t-1,t+1)} du= m((-1,1)\cap (t-1,t+1))$ where $m$ is Lebesgue measure. Consider then  case $t \leq -2, -2<t\leq 0, 0 <t\leq 2$ and $t >2$ to evaluate this.

Title: Convolution of f with itself

Body:
I was trying the following exercise:

(a) Find the fourier transform of the unit rectangular distribution of

\begin{equation}
  f(t) =
    \begin{cases}
      1, \ |t|<1 \\
      0 & \text{otherwise}
    \end{cases}
\end{equation}

which I found to be $\frac{2sin(w)}{w\sqrt{2\pi}}$

(b) Determine the convolution of f(t) with itself:

I have learned convolutions recently and I'm confuse on how to calculate:

$(f*f)(t)=\frac{1}{\sqrt2\pi}\int^{\infty}_{-\infty}f(u)f(t-u)du$

can someone explain me with detail the range changes of the integral?

Thanks!


问:
An open cover that is not locally finite

I could not understand why example $13.4$ is not locally finite.

Can you give me an explanation please.

答:
An open cover $\mathscr{U}$ of a space $X$ is point-finite if $\{U\in\mathscr{U}:x\in U\}$ is finite for each $x\in X$: each point of $X$ belongs to only finitely many members of $\mathscr{U}$. Every locally-finite cover of a space is automatically point-finite as well. And this open cover of $\Bbb R$ is not even point-finite. In fact, it fails in the worst way to be point-finite: each $x\in\Bbb R$ belongs to infinitely many members of the cover.

To see this, let $x\in\Bbb R$. For each $n\in\Bbb Z^+$ there is a rational number $q_n$ such that $\left|x-q_n\right|<\frac1n$, and we may further assume that $q_n\notin\{q_k:k<n\}$, since there are infinitely many rational numbers within $\frac1n$ of $x$. But then $x\in U_{q_n,n}$ for each $n\in\Bbb Z^+$, and the sets $U_{q_n,n}$ are all distinct, so $x$ belongs to infinitely many members of the cover.

(This is a real distinction, by the way: there are point-finite covers that are not locally finite.)


问:
How to solve $4p^3-4p-1=0$
Could anyone please help and clarify how this particular cubic equation

$$4p^3-4p-1=0$$

could be solved?

I am familiar with the approach of solving quadratic equations through quadratic formula and discriminant, but I am a bit confused about how to approach above cubic equation.

答:
$4p^3-4p-1=0$

$p^3-p-\frac 14=0$

Substitute $p = q + \frac 1{3q}$, expand the above.

You will get an equation of the form $q^3 + \frac a{q^3} + b = 0$, which can be transformed into a quadratic by substituting $q^3 = r$ and multiplying throughout by $r  \ (\ne 0)$. Solve for $r$, retaining any complex roots and reverse all the substitutions.

More explanation: what you presented is essentially a "depressed cubic" as it lacks a square term. Any general cubic can be transformed into a depressed cubic via a linear substitution, following which it can be solved by further substitutions and transformations as above. It's just tedious and not commonly done.

(Also, I presented the technique as applied in your particular case. The general method can be elucidated by using symbolic variables in place of the coefficients, but that would make my answer longer. I am working on the assumption that you just want to know how to solve this particular cubic).

Title: How to solve $4p^3-4p-1=0$


问:
How to prove fg doesn't equal to gf in permutation?
1.Let f ∈ S3 be a permutation which is not the identity. Prove that there exists g ∈ S3
such that fg ≠ gf.

2.Let n ≥ 3, and let f ∈ Sn be a permutation which is not the identity. Prove that
there exists g ∈ Sn such that fg ≠ gf.

Hey guys I am completely stuck in these two questions I'm really not sure where to start with solving it, can anyone please give me a hint? Any helps would be appreciated!

Update: Thanks everyone for giving me hint and helping me with that question. I have attempted to work on the first question and here is what I've got:
Since f ∈ S3 and it is not identity. This can conclude that there would be 5 permutations in f and 6 permutations for g as g ∈ S3.

We then test the permutations by multiple one f permutation to all g permutations. For example we take f = (321) and we multiple it with all g permutations available.
We would then be able to prove fg ≠ gf in most sets except the one where g is identity and the one that g has the same permutation as f.

Then, we can take this set to conclude that fg ≠ gf in f ∈ S3 and f is not identity and g ∈ S3 except when g is identity and the one that g has the same permutation as f.

Would this be correct at all?

答:
Write $f$ as a bijective function from $\{1,2,...,n\}$ to itself. Since $f$ is not the identity, we have at least one $k \in \{1,2,...,n\}$ such that $f(k)\ne k$. Let $g \in S_n$ such that $g(k)=k$ and $g(f(k)) \ne f(k)$ (note that we need $n\ge 3$ here. Why?).

Now we have

$f(g(k))=f(k) \ne g(f(k))$

so $f$ and $g$ don't commute.


问:
Example of non-abelian group with element have all order p prime?
I am searching for an example of a non-abelian group $G$, s.t. all elements in $G$ have order $p$, where $p$ is a prime number.
For $p=2$ such an example does not exist.

How did I come up with such a question? If $G$ is abelian with the property above, we can define a $\mathbb{F}_p$-vector space structure on $G$. And my proof used that $G$ is a $\mathbb{Z}$-module. So the above question arose naturally.

答:
An example is the following group, for an odd prime $p>2$:

$$G= \langle x, y, z  |   x^p=1,  y^p=1,  z^p=1,  [x,z]=1,  [y,z]=1,  [x,y]=z^{-1} \rangle$$
is non-abelian of order $p^3$, and all its non-trivial elements have order $p$.
The group is exactly the Heisenberg group over $\mathbb{F}_p$.

More generally, this question is related to the restricted Burnside problem:

Proposition: A finite group $G$ has the property that all non-trivial elements have the same order $p$ if and only if $p$ is prime and $G\ne 1$ is a quotient of $B_0(m,p)$ for some $m.$


问:
Monotonicity of Convex/Concave Function's Gradient Function in High Dimension
For a concave function $f(\mathbf{x}):[0,\infty)^n\rightarrow[0,\infty)$ in high dimension, e.g. $f(\mathbf{x}):[0,\infty)^2\rightarrow[0,\infty)$ when $f(\mathbf{x})=\sqrt{x_1x_2}$. Let $\mathbf{x}$ be element-wise less than or equal to $\mathbf{y}$, I am wondering the relationship between $\nabla f(\mathbf{x}) \cdot \mathbf{x}$ and $\nabla f(\mathbf{y}) \cdot \mathbf{x}$. I thought it would be $\nabla f(\mathbf{x}) \cdot \mathbf{x} \geq \nabla f(\mathbf{y}) \cdot \mathbf{x}$. Appraently, in one dimension we'll have $\nabla f(x) \cdot x\geq \nabla f(y)\cdot x$ when $x\leq y$ (because $\nabla f(x) \geq \nabla f(y)$ and $x\geq 0$). However, it turns out this is not true in high-dimension when $\mathbf{x}$ is element-wise less than or equal to $\mathbf{y}$.

Consider, for example when $\mathbf{x}=[4, 1]$ and $\mathbf{y}=[9,9]$ where $\mathbf{x}$ is element-wise less than or equal to $\mathbf{y}$. And we'll have $\nabla f(\mathbf{x}) =[\frac{1}{4},1]$ and $\nabla f(\mathbf{y}) =[\frac{1}{2},\frac{1}{2}]$, then we have $\nabla f(\mathbf{x}) \cdot \mathbf{x} = 2 <  \nabla f(\mathbf{y}) \cdot \mathbf{x} = \frac{5}{2}$ which is very surprising to me.

So I'm wondering an even more specific case when $\mathbf{x}$ is not only element-wise less than or equal to $\mathbf{y}$ but also proportional, in other words, let $\mathbf{x} = \alpha \cdot \mathbf{y}$ where $\alpha \in [0,1]$. In this case, is it true  that $\nabla f(\mathbf{x}) \cdot \mathbf{x} \geq \nabla f(\mathbf{y}) \cdot \mathbf{x}$? The previous counter-example doesn't work, but I can't prove this is true as well. I also tried several other examples, but none of them turned out to be a successful counter-example. In this case, my intuition is that $\mathbf{x}$ and $\mathbf{y}$ are in the same direction, any ideas on how to prove $\nabla f(\mathbf{x}) \cdot \mathbf{x} \geq \nabla f(\mathbf{y}) \cdot \mathbf{x}$ is true or a counter-example for this is not true will be super helpful.

答:
If $f$ is convex, then $\nabla f$ is monotone, in the sense that
$$
(\nabla f(x) - \nabla f(y) ) ^T(x-y) \ge 0
$$
for all $x,y$. Maybe this can be useful for your computations.


问:
Isolate 3 variables in the compound interest formula
I'm trying to create three new formulas from a compound interest formula: I'm trying to respectively isolate M, r, and P. The original formula comes from https://www.thecalculatorsite.com/articles/finance/compound-interest-formula.php#regular-contributions, and it's the standard formula for finding future compound interest with an initial investment.

$$ A = P\left(1 + \frac{r}{n}\right)^ {n t} + M\biggl(\frac{\left(1 + \frac{r}{n}\right)^ {n t} - 1}{\frac{r}{n}}\biggr)$$

These are what the variables stand for:

A = the future value of the investment/loan, including interest

P = the principal investment amount

M = the monthly payment

r = the annual interest rate

n = the number of times that interest is compounded per unit t

t = the number of years the money is invested or borrowed for

I contacted the site owner and he wasn't able to help with this, since he's not on the mathematical side of things, he just builds calculators on the back of formulas that already exist. I've also been trying to achieve this in Wolfram Alphra with no luck:

```<code>solve P * (1 + r / n) ^ (n * t) + PMT * (((1 + r / n) ^ (n * t) - 1) / (r * n)) = A for r
</code>```

Any help here would be appreciated! Thanks. It's been too many years since my college math.

答:
To make notations simpler, let $\frac r n=x$ and $m=nt$ to make the equation
$$A=P (x+1)^m+M\frac{ (x+1)^m-1}{x}$$ and use the first iteration of Newton or higher order iterative methods (let $k$ be the order of the method).

This would give as estimates
$$x_{(2)}=\frac 2 m\,\frac{ A-P-m M}{ (m-1) M+2 P}$$
$$x_{(3)}=\frac {6 ((m-1) M+2 P) (A-P-m M) } {\alpha A +\beta (m+1) }$$
$$\alpha=2 (m-1) ((m-2) M+3 P)$$
$$\beta=(m-1) m M^2+4 (m-1) M P+6 P^2$$

Trying with $A=3000$, $P=1000$, $M=10$, $n=1$, $t=20$ these approximations would give
$$x_{(2)}=\frac{6}{73}\approx 0.08219 \qquad \qquad x_{(3)}=\frac{146}{3119}\approx 0.04681$$ while the exact solution is $0.05030$.

The next one (the formula is not reported here since a bit too long but available on request) woulf give
$$x_{(4)}=\frac{56142}{1119719}\approx 0.05014$$

If you want easier to implement formulae, consider that you look for the zero of function
$$f(x)=P (x+1)^m+M\frac{ (x+1)^m-1}{x}-A$$ and use the formulae given here with
$$f^{(0)}(0)=nM+P-A$$
$$ f^{(1)}(0)=\frac{1}{2} m ((m-1) M+2 P)$$
$$f^{(2)}(0)=\frac{1}{3} (m-1) m ((m-2) M+3 P)$$
$$f^{(3)}(0)=\frac{1}{4} (m-2) (m-1) m ((m-3) M+4 P)$$
$$f^{(4)}(0)=\frac{1}{5} (m-3) (m-2) (m-1) m ((m-4) M+5 P)$$
$$f^{(5)}(0)=\frac{1}{6} (m-4) (m-3) (m-2) (m-1) m ((m-5) M+6 P)$$ where you can notice the nice patterns.

With this, compute the coefficients $a_k^n$ of the linked paper and you can easily obtain "quintic, sextic, septic, octic ... iterations".


问:
Probability of $w<X$ where $X$ is a random variable equals the expectation of the probability of $w<X$ where $X$ is a constant
Why is it, that if $w \sim Exp(\lambda)$ and $X$ is a random variable, then

$$ P(w<X)=E(1-e^{-\lambda X})\text{ ?} $$

I know that if $t$ is a fixed value, then
$$ P(w<t) = \int _0^t \lambda e^{-\lambda s}ds = 1-e^{-\lambda t} \text{.}$$

Why is the probability of $w<X$, where $X$ is a variable, the expected value of the probability of $w<X$, where $X$ is a constant?

答:
Certainly false without independence of $X$ and $w$. Take $X=w$ for  a counterexample.

Under independence  this follows by Fubini's Theorem . You are just computing the intergral of $I_{w <X}$ w.r.t. the joint distribution of $w$ and $X$ ( which is  a product measure) and integrating w.r.t. $F_w$ and then w.r.t. $F_X$ gives this result.


问:
Unit ball in $L_2[0,1]$ is not relatively compact in $L_1[0,1]$
I want to prove that the following set
$$
B = \{f\in L_2[0,1]: \int_{0}^{1}|f|^2d\mu\leq 1\}
$$
is not relatively compact in $L_1[0,1]$.

I know the general ctiterion of relatively compactness in $L_p[a,b]$. A set $T \subset L_p[a,b]$ is relatively compact iff it is bounded and for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $h$ with the property $|h| < \delta$ we have
$$
\left(\int_a^b|f(x+h) - f(x)|^pdx\right)^{\frac{1}{p}} < \varepsilon
$$

I don't know how to use this criterion here. Maybe there are some other approaches.

答:
The unit ball in $L^2([0,1])$ is not relatively compact in $L^p([0,1])$ for any $p\in [1,2]$. Fix $p\in [1,2]$ and define for $n\in \mathbb{N}$ the set $A_n = \bigcup_{j=0}^{n-1} [2j/(2n), (2j+1)/(2n)]$ (i.e. we divide the unit interval into $2n$ pieces of equal length and kick out every second one) and the function $f_n = 1_{A_n}$. Then we have
$$ \int_0^1 \vert f_n(x)\vert^2 dx = \int_0^1 f_n(x) dx = 1/2 \leq 1. $$
On the other hand we have for every $n\in \mathbb{N}$ and every $p\in [1,2]$
$$ \int_0^1 \vert f_n(x+1/(2n)) - f_n(x) \vert^p dx = 1. $$
Thus, for $\varepsilon = 2/3$ the criterion fails and you arrive at the desired conclusion.


问:
Help understanding the meaning of this definition of a random variable?
In paper I am currently reading, the following passage appears:

Given a prior random variable $X:\Omega \rightarrow \mathbb{R}^N$, we
denote the probability density function (PDF) by $\pi_{X}(x),x \in \mathbb{R}^N$.

and later on:

...assume that an ensemble $x_i \in \mathbb{R}^N,i=1,...,M$ is given,
which mathematically are considered as realizations
$X_i(ω), ω \in \Omega$, of $M$ independent (or
dependent) random variables $X:\Omega \rightarrow \mathbb{R}^N$ with
law $\pi_X$.

I don't quite understand the interpretation of the random variable $X$ here. My intuition so far was that the random variable $X$ is, well, a random variable which can take on specific values $x$ in $\mathbb{R}^N$ with probability (density) $\pi_X(x)$. Is this interpretation correct (or complete)? What does the interpretation of $X$ as a map from $\Omega$ to $\mathbb{R}^N$ represent?

Unfortunately, the paper does not explain what $\Omega$ or $ω$ is, nor are any of these variables used later. Since it only plays a role in the definition, I assume it is something seemingly obvious enough that the authors did not bother to define it. Do you have any idea what the authors want to show here?

答:
To turn my comments into an answer:

$\Omega$  is the (non-empty) set of all possible outcomes $\omega$. $X(\omega)$ is the value (in $\Bbb{R}$) of the random variable $X$ when scenario $\omega$ occurs, therefore $X$ is a function of $\Omega$ to the real numbers.

$X$ is a function that assigns a real value to an outcome. For example, $X$ could be the result of a coin flip, $\omega_1$ is the event of "heads", $\omega_2$ is the event of "tails". Define $$X=\begin{cases}0,\quad\text{if heads}\\1,\quad\text{if tails}\end{cases}$$Then $X(\omega_1)=0$ and $X(\omega_2)=1$. In this case $\Omega=\{\omega_1,\omega_2\}\stackrel{\frown}{=} \{\text{heads, tails}\}$.


问:
$X\subsetneqq Y$ but $X^\star=Y^\star$
Are there $X,Y$ real Banach spaces, such that $X\subsetneqq Y$ (strictly contained) and $X^\star=Y^\star$, where $\star$ denotes the topological dual?

This property is not true for Hilbert spaces, nor even for $L^p$ spaces, so I was thinking to try some function space: continuous functions or bounded...

Any help is appreciated.

Thank you

答:
It seems that you are asking whether $X^*$ and $Y^*$ can be isomorphic (there exists a (bi)continuous isomorphism between them) when $X$ is a proper closed subspace of a Banach space $Y$.

Yes, this can happen. Even with Hilbert spaces. I would say especially with Hilbert spaces. It suffices that $X$ and $Y$ be isomorphic. And that's necessary if they are reflexive, like in the Hilbert case.

E.g.

a) $X=\ell^2(\mathbb{N})\subsetneq \ell^2(\mathbb{Z})=Y$ has $X^*\simeq Y^*\simeq \ell^2(\mathbb{N})$ isometrically.

b) More generally, for an infinite dimensional separable Hilbert space $H$ and any proper closed infinite dimensional subspace $K\subsetneq H$, we have $K^*\simeq H^*\simeq \ell^2(\mathbb{N})$ isometrically.

c) In $\ell^\infty(\mathbb{N})$, we have $c_0\subsetneq c$ and $c_0^*\simeq c^*\simeq \ell^1$ isometrically where $c$ (resp. $c_0$) is the subspace of sequences that converge (resp. to $0$).

d) And here is a somehow silly example: take $X=C([0,1])\subsetneq C([0,2]=Y$ with the obvious embedding which extends a function continuously by a constant on $[1,2]$. Then $X^*\simeq Y^*$ isometrically as follows from Riesz representation theorem. Or much easier, from the fact $X$ and $Y$ are obviously isometric.

Title: $X\subsetneqq Y$ but $X^\star=Y^\star$


答:
It is quite simple to follow the steps given here.

The key point is to first compute $\Delta$; in your case it is $592$ so we know that there are three ral roots. Now, we know that we have three real roots. For this case, the trigonometric method gives nice formulae.

Using the method, the roots are given by
$$x_k=\frac{2 }{\sqrt{3}}\cos \left(\frac{1}{3} \left(2 \pi  k-\cos ^{-1}\left(\frac{3
   \sqrt{3}}{8}\right)\right)\right)\qquad \text{for} \qquad k=0,1,2$$

Title: How to solve $4p^3-4p-1=0$

Body:
Could anyone please help and clarify how this particular cubic equation

$$4p^3-4p-1=0$$

could be solved?

I am familiar with the approach of solving quadratic equations through quadratic formula and discriminant, but I am a bit confused about how to approach above cubic equation.


问:
What is the difference between $:=$ and $:\iff$?
What is the difference between
$:=$ and $:\iff$? Which one is used when (and how ist the latter called)?

I've seen the following: $$ X \subset Y :\iff \forall x \in X: x \in Y$$

Why cannot it be written with like this/What difference does this formulation make?: $$ X \subset Y := \forall x \in X: x \in Y$$

Thank you!

EDIT I'm asking specifically about the $:\iff$ Operator and not about the $\iff$ without the colon.

答: The mysterious "$:\iff$" symbol is, disappointingly, the result of a typo in the first expression. It should say $$ X\subset Y \iff \forall x \in X : x \in Y.$$ We could, if we wanted to, replace $\iff$ with $:=$ in the above expression, but this wouldn't be standard notation, as $:=$ is mostly reserved for direct definitions, e.g. $x:=10$, and not for more complex mathematical statements.


问:
Drawing balls from a box
We have a box with $n+1$ white balls and $k(n+1)$ black balls. One white ball and $t$ black balls are signed with $1$ and the rest (so $(n+1)+k(n+1)-t-1$) with $0$. We draw one ball from this box and denote:

$A -$ drawn ball is white

$B - $ drawn ball has $0$ on it

$A, B$ are independent. How many black balls have $1$?

My solution:

$A \perp B \rightarrow \Bbb P(A \cap B)=\Bbb P(A)\cdot \Bbb P(B)$

$\Bbb P(A\cup B) =\Bbb P(A)+\Bbb P(B) -\Bbb P(A)\cdot \Bbb P(B)$

$\Bbb P(A)= \frac{n+1}{(n+1)+k(n+1)}=\frac{1}{k+1}$

$\Bbb P(B)= \frac{(n+1)+k(n+1)-(t+1)}{(n+1)+k(n+1)}=1-\frac{t+1}{(k+1)(n+1)}$

$\Bbb P(A\cap B)=\frac{1}{k+1}\cdot (1-\frac{t+1}{(k+1)(n+1)})$

$\Bbb P(A\cup B)=1-\frac{t}{(n+1)(k+1)}$ as an union of these events comprises White$1$, White$0$, Black$0$.

$1-\frac{t}{(n+1)(k+1)}=\frac{1}{k+1} + 1-\frac{t+1}{(k+1)(n+1)}-\frac{1}{k+1}\cdot (1-\frac{t+1}{(k+1)(n+1)}) \rightarrow t=...$

Am I right?

答:
If $A$ and $B$ are independent then $A$ and $B^c$ are independent and $\mathbb P(A \cap B^c)= \mathbb P(A)\mathbb P(B^c)$

so $\frac{1}{(n+1)+k(n+1)} = \frac{n+1}{(n+1)+k(n+1)} \frac{t+1}{(n+1)+k(n+1)}$ i.e.  with cancellation $1=\frac{t+1}{1+k}$ and so $t=k$

Title: Drawing balls from a box


问:
What is $\int_\Omega 1_{X^{-1}(A)}dP=\int_\mathbb R 1_AdP_X$ intuitively and why does it make sense?

What I don't get is:

why does this make sense, we are integrating in two different measured spaces

is this equality important? Why do we need it?

EDIT : I'm talking about the integral equality and not the indicator function.

P.S: I'm not looking for the proof, just the intuition.

答:
$$(*)\int_\Omega1_{X^{-1}(A)}dP=\int_{\mathbb{R}}1_AdP_X$$
Why do we need it? Well, exactly because thanks to this, we can integrate in two different spaces. As to why it makes sense, there are whole books/courses carefully developing this theory, so all I can do here is short sketch of an idea. Basic idea is, that measurable function induces pushforward measure on the codomain.

Lets fix any probability space $(\Omega, \Sigma, P)$ and real line with Lebesgue measure over borel sets $(\mathbb R, \mathcal B (\mathbb R), \mathcal L)$, and measurable function $X:\Omega\to\mathbb R$. Then the distribution $P_X$ of $X$ is defined by
$$(**)P_X:\mathcal B (\mathbb R)\ni A\mapsto P(X^{-1}(A))\in [0,1]$$

Such defined $P_X$ is probabilistic measure on $\mathbb R$. In plain english $P(X^{-1}(A))$ means "probability, that outcome of $X$ has value in $A$". Standard abuse of notation is to use $P(X\in A)$ in place of $P(X^{-1}(A))$.

Because every probability space is measure space, we can develop Lebesgue integral in it. Long story short, characteristic function of a set is first step in the construction of the integral. Knowing the definition of integral over characteristic function:
$$(***.1)\ \int_\Omega1_{X^{-1}(A)}dP=P(X^{-1}(A))$$
$$(***.2)\ \int_{\mathbb{R}}1_AdP_X=P_X(A)$$

then the equation in your question (*) is indeed immediate consequence of the definitions of integral (***) and distribution (**), and may look confusingly trivial. Here is at last a turning point which leads from this abstract considerations of measure theory to actual probability theory: what are properties of the distribution $P_X$? Without any further informations about $P_X$ we can look at cumulative distribution function:
$$CDF_X:\mathbb R\ni t\mapsto P_X((-\inf,t])\in [0,1]$$

If $P_X$ is absolutely continuous with respect to $\mathcal L$ - standard measure in $\mathbb R$, then by Radon-Nikodym theorem there exists function $f_{X}:\mathbb R\to \mathbb R$, such that:
$$\int_{\mathbb R}1_AdP_X=\int_{\mathbb R}1_Af_Xd\mathcal L$$
$f_X$ is called probability density function. Now if we can find $f_X$, we can also do all sort of rudimentary calculus. In particular:
$$\int_\mathbb R ^t f_Xd\mathcal L=CDF_X(t)$$

On the other hand, if $P_X$ is not absolutely continuous, we can examine it more closely by Lebesuge's decomposition theorem. But among others, there is well known discrete measure. If probability is discrete, all integral symbols can be simplified to sums in finite case, and series in countably infinite. Such expressions are meaningful, because there are at most countably infinite real values $t$ such, that $P_X(t)\neq 0$. For example cumulative distribution function takes form:
$$CDF_X(t)=\int_\mathbb R 1_{(-\infty, t]}dP_X=\sum_{u\in(-\infty, t]}P_X(u)$$
and such discrete measure $P_X$ is called probability mass function.


问:
Alternative ways to say "if and only if"?
There are some scenarios about which I would like to get some confirmation:

when defining a concept A,

We call A, if ... [definition of
  concept A]

Does "if" here mean equivalence
instead of just sufficiency? Is it
incorrect to replace "if" with "if
and only if"?

For precisely what condition is
   B satisfied?

Does "precisely" mean asking for
necessary and sufficient condition
for B?

Some other ways to say "if and only
if" / "necessary and sufficient"?

Some references that summarize some standard terminologies in Mathematics  such as this one?

Thanks and regards!

答:
3) Necessary and sufficient condition.

Addapted from Wikipedia: "$p$ if and only if $q$" means that $q$ is a necessary and sufficient condition of $p$.    It is the same as "$p$ if $q$" ($q$ implies $p$) and "$p$ only when $q$" (not $q$ implies not $p$).

From Wikitionary: "if and only if" is equivalent to; implies and is implied by; is true and false in the same cases as.


问:
Integer solution of $\tfrac ax+\tfrac by=\tfrac cz$ for pairwise coprime positive integers $a$, $b$ and $c$.
Given positive pairwise coprime integers $a,b,c$, how to find positive coprime integer solutions $(x,y,z)$ of
\begin{equation*}
\frac{a}{x}+\frac{b}{y}=\frac{c}{z}?
\end{equation*}
If there is no general formula or general way to solve it, then is there anything we can say about the positive integer solution set, for example it is easy to see that each $x\in\mathbb{N}$ corresponds to finitely many $y$ and $z$ values but this is not very meaningful.

For example, I calculated $(a,b,c)=(2,5,3)$: there are coprime solutions $(1,5,1)$, $(3,6,2)$, $(3,60,4)$, $(4,2,1)$, $(4,10,3)$, $(4,50,5)$ and so on.

答:
Claim: Let $a$, $b$ and $c$ be pairwise coprime integers, and $x$, $y$ and $z$ positive integers satisfying
$$\frac ax+\frac by=\frac cz.\tag{1}$$
Then there exist coprime positive integers $r$ and $s$ and positive divisors $a'$, $b'$ and $c'$ of $\gcd(a,r)$, $\gcd(b,s)$ and $\gcd(c,r+s)$, respectively, such that
$$(x,y,z)=\left(d\frac{a}{a'}\frac{s}{b'}\frac{r+s}{c'},d\frac{r}{a'}\frac{b}{b'}\frac{r+s}{c'},d\frac{r}{a'}\frac{s}{b'}\frac{c}{c'}\right).$$
for some positive integer $d$. Note that all nine fractions are in fact integers.

Conversely, every such triplet is a solution to $(1)$.

Proof. First a few reductions of the claim. If $(x,y,z)$ is a solution to $(1)$, then $(dx,dy,dz)$ is another solution for any positive integer $d$. So we obtain all solutions to $(1)$ by taking all positive integer multiples of solutions to $(1)$ with $\gcd(x,y,z)=1$.

Next note that if $(x,y,z)$ is solution to $(1)$ with $\gcd(x,y,z)=1$ and we let $e=\gcd(a,x)$, $f=\gcd(b,y)$ and $g=\gcd(c,z)$, so that
$$(a,b,c)=(ea',fb',gc')\qquad\text{ and }\qquad(x,y,z)=(ex',fy',gz'),$$
for some pairwise coprime positive integers $a'$, $b'$, $c'$, and positive integers $x'$, $y'$ and $z'$ with
$$\frac{a'}{x'}+\frac{b'}{y'}=\frac{c'}{z'},\tag{2}$$
and $\gcd(x',y',z')=1$ and moreover $\gcd(a',x')=\gcd(b',y')=\gcd(c',z')=1$.
So we obtain all solutions to $(1)$ with $\gcd(x,y,z)=1$ by appropriately scaling all solutions to $(2)$ with $\gcd(a',x)=\gcd(b',y)=\gcd(c',z)=1$, where $a'$, $b'$ and $c'$ range over the positive divisors of $a$, $b$ and $c$, respectively. Here "appropriately scaling" means taking $(x,y,z)=(\tfrac{a}{a'}x',\tfrac{b}{b'}y',\tfrac{c}{c'}z')$ for a solution $(x',y',z')$ to $(2)$.

So suppose $(x',y',z')$ is a solution to $(2)$ with $\gcd(a',x')=\gcd(b',y')=\gcd(c',z')=1$ and $\gcd(x',y',z')=1$, where $a'$, $b'$ and $c'$ are positive divisors of $a$, $b$ and $c$ respectively. Then clearing denominators shows that
$$y'z'a'+x'z'b'=x'y'c',\tag{3}$$
and hence that $x'\mid y'z'$ and $y'\mid x'z'$ and $z'\mid x'y'$. Because $\gcd(x',y',z')=1$ it follows that there exists a triplet of pairwise coprime positive integers $(u,v,w)$ such that
$$(x',y',z')=(vw,uw,uv),\tag{4}$$
and plugging this back into $(3)$ shows that
$$ua'+vb'=wc'.\tag{5}$$
In particular we see that for the trivial divisors $a'=b'=c'=1$ of $a$, $b$ and $c$ we have $u+v=w$, corresponding to the solutions of the form
$$(x,y,z)=(kav(u+v),kbu(u+v),kcuv),$$
where $k$ is any positive integer. More generally we can parametrize all solutions to $(5)$, for all positive divisors $a'$, $b'$ and $c'$ of $a$, $b$ and $c$ respectively, by first choosing $ua'$ and $vb'$ and then choosing appropriate $a'$, $b'$ and $c'$:

First note that $ua'$ and $vb'$ are coprime, because
$$\gcd(u,v)=\gcd(a',b')=1
\qquad\text{ and }\qquad
\gcd(a',x')=\gcd(b',y')=1,$$
where $x'=vw$ and $y'=uw$. So let $r$ and $s$ be coprime positive integers, and let $a'$, $b'$ and $c'$ be positive divisors of $\gcd(r,a)$, $\gcd(s,b)$ and $\gcd(r+s,c)$, respectively. Then $(u,v,w)=(\tfrac{r}{a'},\tfrac{s}{b'},\tfrac{r+s}{c'})$ is a solution to $(5)$. The corresponding solution to $(2)$ is then
$$(x',y',z')=(vw,uw,uv)=\left(\frac{s(r+s)}{b'c'},\frac{r(r+s)}{a'b'},\frac{rs}{a'b'}\right),$$
and the corresponding solutions to $(1)$ are then all positive integer multiples of
\begin{eqnarray*}
(x,y,z)&=&\left(\frac{a}{a'}x',\frac{b}{b'}y',\frac{c}{c'}z'\right)
=\left(\frac{as(r+s)}{a'b'c'},\frac{br(r+s)}{a'b'c'},\frac{crs}{a'b'c'}\right)\\
&=&\left(\frac{a}{a'}\frac{s}{b'}\frac{r+s}{c'},\frac{r}{a'}\frac{b}{b'}\frac{r+s}{c'},\frac{r}{a'}\frac{s}{b'}\frac{c}{c'}\right).
\end{eqnarray*}
Note that in this last expression, all nine fractions are in fact integers.

For the converse it is easy to verify that such a triplet indeed satisfies the equation.


问:
Is there other method to map $G$ to $D$ using holomorphic functions?
I'm trying to map G to D using holomorphic function where
$$
G = \{ z\in\mathbb{C}: Re(z) + 1 < Imz < Re(z) + 3 \}
$$
$$
D = \{ w \in \mathbb{C}: Im(w) < 0, w(2i) = -2i \}
$$
Here what i did.

$w_{1} = z - 1$

$w_{2} = w_{1}e^{-i\frac{\pi}{4}}$

$w_{3} = \sqrt2\pi w_{2}$

$w_{4} = e^{w_{3}}$

$w_{5} = e^{i\pi} w_{4}$

$$
w = e^{i\pi}e^{w_{3}} = e^{\pi + \sqrt2\pi w_{2}} = e^{\pi + \sqrt2\pi w_{1}e^{-i\frac{\pi}{4}}} = \text{notice that } e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt2}(1 - i) = e^{\pi((1-i)(z-1)+i)}
$$
but $w(2i) \not = -2i$

I'm looking for other solution or a place where I did mistake

答:
You meant $D = \{ w \in \mathbb{C}: Im(w) < 0\}$ and $f$ is a biholomorphic map $G\to D$ such that $f(2i)=-2i$.

Construct a biholomorphic map $\phi_1$ from $D$ to the unit disk,

a biholomorphic map $\phi_2$ from $G$ to $|\Im(z)|<\pi$,

a biholomorphic map $\phi_3$ from $|\Im(z)|<\pi$ to the unit disk,

then look at $f=\phi_4\circ \phi_1^{-1}\circ \phi_3 \circ \phi_2$ where $\phi_4\in Aut(D)=PSL_2(\Bbb{R})$ which acts transitively on $D$.

Note that from the stabilizer of $-i$ which is $PSO_2(\Bbb{R})$ we know the stabilizer of $-2i$ and hence all the $f$ satisfying the hypothesis.


问:
Fourier coefficient equal DFT coefficient for trigonometric functions
The Fourier coefficient for $f:[0,2\pi]\to\mathbb{R}$ are given by:

$c_k=\frac{1}{2\pi}\int_0^{2\pi}f(t)\exp(-ikt)dt,k\in \mathbb{Z}$

The DFT coefficient are:

$a_{k,N}=\frac{1}{N}\sum_{j=0}^{N-1}f(j\frac{2\pi}{N})\exp(-ijk\frac{2\pi}{N}), k=0,...,N-1$

prove that for a trigonometric function $f=\sum_{l=-n}^nc_l\exp(ilx)$ you get that $a_{k,N}=c_k$ for $k=-n,...,n$ where $a_{-k,N}=a_{N-k,N}$ white the rest of $a_{k,N}$ are zero.

My approach: I simply put $f$ in $a_{k,N}$:

$a_{k,N}(f)=\frac{1}{N}\sum_{j=0}^{N-1}\sum_{l=-n}^nc_l\exp(il\cdot j\frac{2\pi}{N})\exp(-ijk\frac{2\pi}{N})=\frac{1}{N}\sum_{j=0}^{N-1}\sum_{l=-n}^nc_l\exp(ij\frac{2\pi}{N}(l-k))=\frac{1}{N}\sum_{l=-n}^nc_l\sum_{j=0}^{N-1}\exp(ij\frac{2\pi}{N}(l-k))=\frac{1}{N}\sum_{l=-n}^nc_l\frac{1-exp(i\frac{2\pi}{N}\cdot N(l-k))}{1-\exp(i\frac{2\pi}{N}(l-k))}=0$

because $\exp(2\pi i(l-k))=1$ for all $l-k\in \mathbb{Z}$

Please help

答: I solved it. Needed to notice that for l=k I divide by zero, so I need to sum it differently.</p>


问:
Biggest circle under a polynomial curve
What is the biggest circle that is contained in the region bounded by the graph of the polynomial $f(x) = x(1-x)(2x+1)$ and the x-axis interval $[0, 1]$?

(Here's the thing in Desmos)

Here's what I have tried

Let's denote the center of the circle by $(c_x, c_y)$ and the points of tangency by $(x_k, y_k)$, $k=1,2$. From distance to the x-axis, we know that the radius of the circle is then $c_y$. We can express the tangency by the dot product of the vectors $(x_k-c_x, y_k-c_y)$ and $(1, f'(x_k))$ being zero. So we get the group of equations (all for $k=1,2$)

$$\begin{cases}
(x_k-c_x)^2 + (y_k-c_y)^2 &=& c_y^2 \\
x_k-c_x + (y_k-c_y)f'(x_k) &=& 0 \\
f(x_k) &=& y_k
\end{cases}
$$

Then I tried to solve this group by using Gröbner basis and elimination ideals. I wrote this SageMath code:

```<code>f(x) = x*(1-x)(2*x+1)
fd = f.derivative()

R.<x1, y1, x2, y2, cx, cy> = PolynomialRing(QQ, order='lex')
polys1 = [(cx-xk)**2+(cy-yk)**2 - cy**2 for (xk, yk) in [(x1,y1), (x2, y2)]]
polys2 = [(xk-cx)+(yk-cy)*fd(xk)  for (xk, yk) in [(x1,y1), (x2, y2)]]
polys3 = [f(xk)-yk for (xk, yk) in [(x1,y1), (x2, y2)]]
I = R.ideal(polys1+polys2+polys3)
gb = I.groebner_basis()
for poly in gb:
    print (poly)
print (I.dimension())
</code>```

It gave me a long list of polynomials, the last one being

$$c_x^6 - 2c_x^4c_y^2 + \frac{5}{4}c_x^4c_y + \frac{1}{64}c_x^4 + c_x^2c_y^4 + \frac{3}{4}c_x^2c_y^3 + \frac{3}{16}c_x^2c_y^2 + \frac{1}{64}c_x^2c_y$$

and said the dimension is $1$. I now realize the reason we don't get dimension $0$ has probably to do with the fact that the points $(x_1, y_1)$ and $(x_2, y_2)$ could be the same point(?) So for a solution, the point $(c_x, c_y)$ should be a zero of that polynomial? But is even that correct, since the points could be collapsed and the other side go over the curve??

How to solve this even numerically? I tried with Sage by minimizing the square sum of all those polynomials that should be zero. I also tried adding constraints for the variables to be near the solution that can be seen from the picture but that didn't work either.

答:
We have

$$
\cases{
y = x(1-x)(2x+1)\\
(x-x_0)^2+(y-y_0)^2=y_0^2
}
$$

substituting the first into the second we have

$$
p(x)=(x-x_0)^2+(x(1-x)(2x+1)-y_0)^2-y_0^2 = 0
$$

This polynomial at tangency should have the following structure

$$
p(x) = c(x-x_1)^2(x-x_2)^2((x-a)^2+b^2)\ \ \ \ (\text{Why?})
$$

now comparing coefficients we have

$$
0=\eta = \left\{
\begin{array}{l}
x_0^2-c x_1^2 x_2^2 \left(a^2+b^2\right) \\
 2 c x_1 x_2 \left(x_2 \left(a^2+b^2\right)+x_1 \left(a^2-a x_2+b^2\right)\right)-2 x_0-2 y_0 \\
2 -c x_1^2 \left(a^2-4 a x_2+b^2+x_2^2\right)-4 c x_2 x_1 \left(a^2-a x_2+b^2\right)-c x_2^2 \left(a^2+b^2\right)-2 y_0 \\
 2 \left(c x_1 \left(a^2-4 a x_2+b^2+x_2^2\right)+c x_2 \left(a^2-a x_2+b^2\right)+c x_1^2 \left(x_2-a\right)+2 y_0+1\right) \\
 c \left(a^2+b^2\right)+c \left(4 x_1 \left(x_2-a\right)+x_2 \left(x_2-4 a\right)+x_1^2\right)+3 \\
 2 c \left(x_1+x_2\right)-2 (a c+2) \\
 4-c \\
\end{array}
\right.
$$

and we finish by solving this nonlinear system for $x_0,y_0,x_1,x_2,a,b,c$ using a minimization procedure such as

$$
\min_{x_0,y_0,x_1,x_2,a,b,c} \eta\cdot\eta\ \ \ \text{s. t.}\ \ \ \{x_0 > 0,y_0 >0, x_1 > 0, x_2 > 0\}
$$

Follows a MATHEMATICA script which materializes those computations

```<code>Clear[g1, g2]
g1[x_, y_] := y - x (1 - x) (2 x + 1)
g2[x_, y_] := (x - x0)^2 + (y - y0)^2 - y0^2
dif = g2[x, x (1 - x) (2 x + 1)] - c (x - x1)^2 (x - x2)^2 ((x + a)^2 + b^2)
eta = CoefficientList[dif, x]

sol = NMinimize[{eta.eta, x0 > 0, y0 > 0,x1 > 0, x2 > 0}, {x0, y0, x1, x2, a, b, c }]

g2x = g2[x, y] /. sol[[2]]
gr1 = ContourPlot[g2x == 0, {x, 0, 1}, {y, 0, 0.6}, ContourStyle -> Red];
gr2 = ContourPlot[g1[x, y] == 0, {x, 0, 1}, {y, 0, 0.6}];
gr3 = Plot[0, {x, 0, 1}]
Show[gr1, gr2, gr3, AxesOrigin -> {0, 0}, AspectRatio -> 0.6]
</code>```

Title: Biggest circle under a polynomial curve

Body:
We have

$$
\cases{
y = x(1-x)(2x+1)\\
(x-x_0)^2+(y-y_0)^2=y_0^2
}
$$

substituting the first into the second we have

$$
p(x)=(x-x_0)^2+(x(1-x)(2x+1)-y_0)^2-y_0^2 = 0
$$

This polynomial at tangency should have the following structure

$$
p(x) = c(x-x_1)^2(x-x_2)^2((x-a)^2+b^2)\ \ \ \ (\text{Why?})
$$

now comparing coefficients we have

$$
0=\eta = \left\{
\begin{array}{l}
x_0^2-c x_1^2 x_2^2 \left(a^2+b^2\right) \\
 2 c x_1 x_2 \left(x_2 \left(a^2+b^2\right)+x_1 \left(a^2-a x_2+b^2\right)\right)-2 x_0-2 y_0 \\
2 -c x_1^2 \left(a^2-4 a x_2+b^2+x_2^2\right)-4 c x_2 x_1 \left(a^2-a x_2+b^2\right)-c x_2^2 \left(a^2+b^2\right)-2 y_0 \\
 2 \left(c x_1 \left(a^2-4 a x_2+b^2+x_2^2\right)+c x_2 \left(a^2-a x_2+b^2\right)+c x_1^2 \left(x_2-a\right)+2 y_0+1\right) \\
 c \left(a^2+b^2\right)+c \left(4 x_1 \left(x_2-a\right)+x_2 \left(x_2-4 a\right)+x_1^2\right)+3 \\
 2 c \left(x_1+x_2\right)-2 (a c+2) \\
 4-c \\
\end{array}
\right.
$$

and we finish by solving this nonlinear system for $x_0,y_0,x_1,x_2,a,b,c$ using a minimization procedure such as

$$
\min_{x_0,y_0,x_1,x_2,a,b,c} \eta\cdot\eta\ \ \ \text{s. t.}\ \ \ \{x_0 > 0,y_0 >0, x_1 > 0, x_2 > 0\}
$$

Follows a MATHEMATICA script which materializes those computations

Clear[g1, g2]
g1[x_, y_] := y - x (1 - x) (2 x + 1)
g2[x_, y_] := (x - x0)^2 + (y - y0)^2 - y0^2
dif = g2[x, x (1 - x) (2 x + 1)] - c (x - x1)^2 (x - x2)^2 ((x + a)^2 + b^2)
eta = CoefficientList[dif, x]

sol = NMinimize[{eta.eta, x0 > 0, y0 > 0,x1 > 0, x2 > 0}, {x0, y0, x1, x2, a, b, c }]

g2x = g2[x, y] /. sol[[2]]
gr1 = ContourPlot[g2x == 0, {x, 0, 1}, {y, 0, 0.6}, ContourStyle -> Red];
gr2 = ContourPlot[g1[x, y] == 0, {x, 0, 1}, {y, 0, 0.6}];
gr3 = Plot[0, {x, 0, 1}]
Show[gr1, gr2, gr3, AxesOrigin -> {0, 0}, AspectRatio -> 0.6]


问:
What is an overlap?

I want to ask what an overlap is.

My teacher said that for example $1$: Everything is an overlap hence it is not locally finite.

For example $2$, it doesnt overlap.

Please teach me these two examples.

Thanks.

答:
There are two families of sets mentioned on that page.

One is $\left\{\left(-\frac1n,\frac1n\right):n\in\Bbb Z^+\right\}$; it’s an open cover of $(-1,1)$, and it is not even point-finite, since $0$ belongs to all of the sets. The sets $U_p$ mentioned just below it do not seem to have anything to do with it, however, or with anything else on the page.

The other is $\mathscr{F}=\{[n,n+1]:n\in\Bbb Z^+\}$; it’s a closed cover of $[1,\to)$ (or if you prefer, $[1,\infty)$), and it is locally finite. To see this, suppose that $x\ge 1$. If $x$ is not an integer, then there is a positive integer $n$ such that $n<x<n+1$; let $U_x=(n,n+1)$. Then $U_x$ is an open neighborhood of $x$, and the only member of $\mathscr{F}$ that intersects $U_x$ is $[n,n+1]$: if $m\in\Bbb Z^+$, and $m\ne n$, then $[m,m+1]\cap(n,n+1)=\varnothing$, because either $m<n$, in which case $m+1\le n$, or $n<m$, in which case $n+1\le m$. If $x$ is an integer, let $U_x=(x-1,x+1)$. Suppose that $n\in\Bbb Z^+$ and $U_x\cap[n,n+1]\ne\varnothing$; then either $x=n$, or $x=n+1$, since $x$ is the only integer in $U_x$. In either case, then, $U_x$ intersects at most two members of $\mathscr{F}$. Thus, every $x\ge 1$ has an open neighborhood that intersects only finitely many members of $\mathscr{F}$, which by definition means that $\mathscr{F}$ is locally finite.

Finally, it appears that your teacher means that sets $A$ and $B$ overlap if $A\cap B$ contains an open interval, i.e., is more than just one point. With this definition $(0,2]$ and $[1,3)$ overlap, because their intersection, $[1,2]$, contains the open interval $(1,2)$, but $[0,1]$ and $[1,2]$ do not overlap, because $\{1\}$, their intersection, does not contain an open interval.

Title: What is an overlap?


问:
Help needed in solving a differential equation
Please help me in solving:
$$a^2z+\frac{\partial^2z}{\partial x^2}-\frac{\partial^2 z}{\partial y^2}=0$$
($a$ is a constant)

I plugged this in Wolfram Alpha and it outputs that this is a second order linear partial differential equation but unfortunately, I have never solved these type of equations before. I came across this equation while solving some other problem. I tried to look up the methods to solve such equations but they seem too advanced to me. :( Can somebody please please solve this? I am not even sure if it is possible to solve this equation.

Any help is appreciated. Thanks!

答:
Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,

Let $\begin{cases}p=\dfrac{x+y}{2}\\q=\dfrac{x-y}{2}\end{cases}$ ,

Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)$

$\dfrac{\partial^2z}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial x}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$

$\dfrac{\partial^2z}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial y}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$

$\therefore a^2z+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=0$

$a^2z+\dfrac{\partial^2z}{\partial p\partial q}=0$

$\dfrac{\partial^2z}{\partial p\partial q}=-a^2z$

Similar to Second order hyperbolic PDE,

The general solution is $z(p,q)=\int_0^pf(s)J_0\left(\dfrac{2\sqrt{q(p-s)}}{a}\right)~ds+\int_0^qg(s)J_0\left(\dfrac{2\sqrt{p(q-s)}}{a}\right)~ds$

$z(x,y)=\int_0^\frac{x+y}{2}f(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x-y}{2}\left(\dfrac{x+y}{2}-s\right)}\right)~ds+\int_0^\frac{x-y}{2}g(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x+y}{2}\left(\dfrac{x-y}{2}-s\right)}\right)~ds$

$z(x,y)=\int_0^{x+y}F(s)J_0\left(\dfrac{\sqrt{(x-y)(x+y-s)}}{a}\right)~ds+\int_0^{x-y}G(s)J_0\left(\dfrac{\sqrt{(x+y)(x-y-s)}}{a}\right)~ds$


问:
How to solve $4p^3-4p-1=0$
Could anyone please help and clarify how this particular cubic equation

$$4p^3-4p-1=0$$

could be solved?

I am familiar with the approach of solving quadratic equations through quadratic formula and discriminant, but I am a bit confused about how to approach above cubic equation.

答:
Substitute $p=\frac2{\sqrt3}\cos\theta$ into $4p^3-4p-1=0 $ to get

$$4\cos^3\theta - 3\cos\theta =\cos3\theta=\frac{3\sqrt{3}}8$$
which leads to $\theta = \frac13 \cos^{-1}\frac{3\sqrt{3}}8+\frac{2\pi k}3$ and the solutions

$$p= \frac{2 }{\sqrt{3}}\cos \left(\frac13 \cos^{-1}\frac{3\sqrt{3}}8 +\frac{2\pi k}3 \right), \>\>\>k=0,1,2$$

Title: How to solve $4p^3-4p-1=0$


问:
Show when $x \notin R(A)$, then $x \in N(A^T)$?
Let $A \in \mathbb{R}^{m \times n}$. Can we say when $x \in \mathbb{R}^n$ is not in range of $A$, then it is in the null space of $A$?

My try:

We know that $R(A)$ is a subspace in $\mathbb{R}^{m}$ whose dimension is $r \leq m$ and its complement is $N(A^T)$, i.e., $R(A) \bigoplus N(A^T) = \mathbb{R}^{m}$. Therefore, when $x$ is not in $\mathbb{R}^{m}$, it would be in the other one.

Am I right? If so, can you please provide me with other proofs and explanations?

答:
I think your question boils down to the following statement.
Let $V$ a vector space and $W_1$ and $W_2$ be two subspaces of $V$. Given:
\begin{equation}
V = W_1 \oplus W_2
\end{equation} For some $v \in V$, if $ v \notin W_1$ does that say anything about $v$ being in $W_2$. The answer is no.

Let us look at one consequence of the direct sum decomposition above. It tells us there exist bases of $W_1$ and $W_2$, let them be $\mathcal{B}_1$ and $\mathcal{B}_2$, such that:
\begin{equation}
\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2
\end{equation} where $\mathcal{B}$ is a basis for $V$.
Now if an element cannot be written as a linear combination of elements of $\mathcal{B}_1$, does that mean it can be written a linear combination of elements of $\mathcal{B}_2$?

For a simple counter example, take
\begin{equation}
\mathcal{B}_1 = \{b_1,b_2\} \text{ and } \mathcal{B}_2 = \{b_3\}
\end{equation}
Let $v=b_1+b_2+b_3$. As evident, $v \notin W_1$ and $v \notin W_2$.


问:
Question on tower property
in a book of mine it says that for sub $\sigma$-algebras $\mathbb D \subset \mathbb E$
$$
\mathrm E [X \mid \mathbb D]
=
\mathrm E [
\mathrm E [ X \mid \mathbb E ]
\mid \mathbb D ]
=
\mathrm E [
\mathrm E [ X \mid \mathbb D]
\mid \mathbb E ]
$$
The first equality I recognize as the tower property, but the later I cannot make sense of.
If $\mathbb E \not = \mathbb D$, then I can find $E \in \mathbb E$ subject to $E \not \in \mathbb D$ where it should hold that
$$
\int_E
\mathrm E [
\mathrm E [ X \mid \mathbb D]
\mid \mathbb E ]
dP
=
\int_E
E[X \mid \mathbb D] dP
$$
but $E[X \mid \mathbb D]$ is only $\mathbb D$-measurable. Can anyone help me out?

答:
Recall that if $X$ is $\mathbb F-$measurable, then $$\mathbb E[X\mid\mathbb F]=X.$$

Since $\mathbb E[X\mid \mathbb D]$ is $\mathbb E-$measurable,

$$\mathbb E[\mathbb E[X\mid \mathbb D]\mid \mathbb E]=\mathbb E[X\mid \mathbb D].$$

Title: Question on tower property


问:
How to find the inverse Laplace Transforms?
How do you find the inverse Laplace Transform of the following,

$$\frac{2 (s^2+4 s+5)^2+40}{(s^2+4 s+5)^2}$$

Separating them into complex coefficients is to long.

答:
Let $F(s)$ denote the fraction in the post, hence $F(s)=2+40\frac1{(s^2+4s+5)^2}$. The $2$ part of $F(s)$ is the Laplace transform of twice the Dirac measure at $0$. The fraction $\frac1{s^2+4s+5}$ is a linear combination of $\frac1{s+2\pm\mathrm i}$ hence it is the Laplace transform of a linear combination of the functions $t\mapsto\exp(-(2\pm\mathrm i)t)$ on $t\geqslant0$, namely, the Laplace transform of $f$ where $f(t)=\mathrm e^{-2t}\sin(t)$ for every $t\geqslant0$. Thus, the fraction $\frac1{(s^2+4s+5)^2}$ is the Laplace transform of $f\ast f$, which might (or might not, check for yourself) be such that $(f\ast f)(t)=\frac12\mathrm e^{-2t}(\sin(t)-t\cos(t))$ for every $t\geqslant0$.

Finally, the function $F$ is the Laplace transform of the measure $\nu=2\delta_0+40\mu$ where $\mu$ has density $f\ast f$ with respect to the Lebesgue measure on $t\geqslant0$. Note that $F$ is the Laplace transform of no function, only a measure can do, and one has
$$
F(s)=\int_0^{+\infty}\mathrm e^{-st}\mathrm d\nu(t).
$$


问:
If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim\limits_{n\to\infty}{n a_n} = 0$
I'm studying for qualifying exams and ran into this problem.

Show that if $\{a_n\}$ is a nonincreasing sequence of positive real
numbers such that $\sum_n a_n$ converges, then $\lim\limits_{n \rightarrow \infty} n a_n = 0$.

Using the definition of the limit, this is equivalent to showing

\begin{equation}
\forall \varepsilon > 0, \; \exists n_0\;\text{such that}\;
|n a_n| < \varepsilon,\; \forall n > n_0
\end{equation}

or

\begin{equation}
\forall \varepsilon > 0, \; \exists n_0\;\text{such that}\;
a_n < \frac{\varepsilon}{n},\; \forall n > n_0
\end{equation}

Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!

答:
There is also quite nice proof using Stolz-Cesaro Theorem:
$\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}a_{n+1}=0$ because of series convergence.

$$
\lim_{n\rightarrow\infty}\sum_{k=1}^n a_k=g
$$
Using Stolz-Cesaro Theorem:
$$
g=\lim_{n\rightarrow\infty}\frac{n \sum_{k=1}^n a_k}{n} = \left[\frac{\infty}{\infty} \right] = \lim_{n\rightarrow\infty}\frac{(n+1) \sum_{k=1}^{n+1} a_k-n\sum_{k=1}^n a_k}{n+1-n} = \lim_{n\rightarrow\infty}(\sum_{k=1}^n a_k + (n+1)a_{n+1})
$$
$$
g=g+\lim_{n\rightarrow\infty}a_{n+1}+\lim_{n\rightarrow\infty}na_{n+1}
$$
$$
\lim_{n\rightarrow\infty}na_{n+1} = \lim_{n\rightarrow\infty}na_{n}=0
$$
$\square$


问:
Help understanding the meaning of this definition of a random variable?
In paper I am currently reading, the following passage appears:

Given a prior random variable $X:\Omega \rightarrow \mathbb{R}^N$, we
denote the probability density function (PDF) by $\pi_{X}(x),x \in \mathbb{R}^N$.

and later on:

...assume that an ensemble $x_i \in \mathbb{R}^N,i=1,...,M$ is given,
which mathematically are considered as realizations
$X_i(ω), ω \in \Omega$, of $M$ independent (or
dependent) random variables $X:\Omega \rightarrow \mathbb{R}^N$ with
law $\pi_X$.

I don't quite understand the interpretation of the random variable $X$ here. My intuition so far was that the random variable $X$ is, well, a random variable which can take on specific values $x$ in $\mathbb{R}^N$ with probability (density) $\pi_X(x)$. Is this interpretation correct (or complete)? What does the interpretation of $X$ as a map from $\Omega$ to $\mathbb{R}^N$ represent?

Unfortunately, the paper does not explain what $\Omega$ or $ω$ is, nor are any of these variables used later. Since it only plays a role in the definition, I assume it is something seemingly obvious enough that the authors did not bother to define it. Do you have any idea what the authors want to show here?

答:
In probability, it's conventional to use the Greek letter $ω$ (omega) to denote an individual outcome in the set $Ω$ (Omega) of all possible outcomes, called the sample space. $\Omega$ itself is directly equipped with a $σ$-algebra, $\mathcal{F}$, of subsets called events, and a probability measure $P$ on $\mathcal{F}$ that is $σ$-additive and satisfies $P(Ω) = 1$, $P(∅) = 0$. Using this formalism, a random variable $X$ is conventionally defined as a measurable function $X: \Omega \to S$ for some other measure space $S$ (often, as here, $\Bbb{R}^n$).

While in many cases it's possible to let $\Omega = \Bbb{R}^n$ and define $P$ on subsets of $\Bbb{R}^n$ directly, sometimes this misses relationships between certain random variables that would be clunky to formalize alternatively. For example, suppose I'm playing the St. Petersburg game with a fair coin. My sample space $\Omega$ looks like this: $$\Omega = \{ T, HT, HHT, HHHT, ..., H^k T, ... \}$$ where $H^k T$ denotes a string of $k \geq 0$ heads, followed by a tail.

Two natural random variables to define on this sample space are the stopping time $N: \Omega \to \Bbb{R}$ given by $$N(H^k T) := k+1, k = 0, 1, 2, 3, ...$$ and the winnings $W: \Omega \to \Bbb{R}$ given by $$W(H^k T) := 2^{k+1}, k = 0, 1, 2, 3, ...$$ Clearly we should have $W = 2^N$, but if I consider $W$ and $N$ as functions from $\Bbb{R} \to \Bbb{R}$ (i.e. their sample space is $\Bbb{R}$), I have 2 different and incompatible probability mass functions $p_N$ and $p_W$ on the domain $\Bbb{R}$, and it's hard to make sense of the equation $W = 2^N$. If I let my sample space instead be the original sequence of coin flips in the game itself, this difficulty vanishes, as both $W$ and $N$ are functions of an individual sequence of coin flips and it's easy to formalize their relationship properly.


问:
Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).
Question:

Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).

What I did:
I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this Proving limit doesn't exist using the $\epsilon$-$\delta$ definition
which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked.
Anyway I'd really love someone to verify/correct this solution before I hand it in.
Thanks a lot.

We assume there exists a limit, therefore:
$\forall \varepsilon>0 \exists M>0: x>M \Rightarrow |f(x)-L|<\varepsilon$
and specifically, if we pick some $x_1,x_2>M$: $|f(x_1)-L|<\frac \varepsilon2,
|f(x_2)-L|<\frac \varepsilon2$

Also, this applies for all $\varepsilon$, and specifically for  $\varepsilon=0.5$

(1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L |  + |f(x_2)-L| \le \frac \varepsilon2 +  \frac \varepsilon2 = \varepsilon$

This applies for all $x>M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} > M , x_2=\pi \lfloor M+1 \rfloor$

We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 & \lfloor M+1 \rfloor   odd  \\ 0 &\lfloor M+1 \rfloor  even\end {cases} $

So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.

答:
Solution by means of sequences:

$$x_n:=\pi n\implies x_n\sin x_n=0\xrightarrow[n\to\infty]{}0$$

$$y_n:=\frac{(4n-3)\pi}2\implies y_n\sin y_n=\frac{(4n-3)\pi}2\xrightarrow[n\to\infty]{}\infty$$


问:
Urn experiment: pair-wise drawing of colored balls without replacement
Given an urn with colored balls from which we draw balls in a pair-wise manner (without replacement). If we draw $k$ pairs, what is the expected number of pairs with the same color?

答:
Say there are $n$ colors and that there are $B_i$ balls of color $i$.  Let $B=\sum B_i$ be the total number of balls.  Then the probability that both of the balls in a given pair have color $i$ is $\frac {B_i(B_i-1)}{B(B-1)}$  from which it follows that the probability that a given pair is monochromatic is $$\sum_{i=1}^n\frac {B_i(B_i-1)}{B(B-1)}$$  As you are choosing $k$ pairs, Linearity of Expectation tells us that the expected number of single color pairs is therefore $$\boxed {k\times \sum_{i=1}^n\frac {B_i(B_i-1)}{B(B-1)}}$$

Sanity checks:  If there is only one color then $B_1=B$ so this is just $k$, as it should be.  If there are two colors with equal probability and infinitely many of each then the answer should be $\frac k2$ since each pair independently has probability $\frac 12$ of being single colored.  Our formula would give $$2k\lim_{b\to \infty}\frac {b(b-1)}{2b(2b-1)}=2k\times \frac 14=\frac k2$$ as desired.  Of course, it is easy to work out small cases by hand, or to simulate larger cases, and to thereby get additional numerical support for the formula.


问:
Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively.

Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively. If $AO = 30, FO = 20, BO = 60, DO = 10, CO = 20$, find $EO$.

What I Tried: Here is a picture in Geogebra :-

The statement of the problem is really simple, every length surrounding point $O$ is given except $EO$ , which we have to find. My question is, how?

I am only given the lengths, which I don't even know how to use them. They neither help in finding lengths of other unknown lengths, nor help in angle-chasing, and so on. I don't find any similar triangles where I could have used the lengths. The only thing which I find interesting is $FO = CO$. Maybe we can use them in the areas of the triangles considering them as bases, but I can't see to find a way out there. Also that dosen't make the point $O$ special, does it, or am I missing something?

Can anyone help me or give me some hints? Thank You.

答:
$$\frac{\triangle OBC} {\triangle ABC} =\frac{OD} {AD} , \frac{\triangle AOC} {\triangle ABC} = \frac{OE} {BE} ,
\frac{\triangle AOB} {\triangle ABC} =\frac{OF} {FC}$$
$$\frac{\triangle OBC} {\triangle ABC} + \frac{\triangle AOC} {\triangle ABC} +\frac{\triangle AOB} {\triangle ABC} = \frac{OD} {AD} +\frac{OE} {BE} +\frac{OF} {FC} =1$$


问:
Constant solution of Euler-Cauchy equation
I want to prove that if $u \in C^2([-1, 1])$ solves
$$
x^2 u''(x) + 2xu'(x) = 0
$$
on $[-1, 1]$, then $u$ is constant. Is this possible without using arguments from ODE theory, i.e. Picard-Lindelöf's theorem or explicitly calculating solutions?

So far I tried to partially integrate
$$
0 = \int^1_{-1} x^2u''(x)+2xu'(x) ~\mathrm{d}x
$$
which did not help me that much...

答:
Method 1

Set $v(x)=u'(x)$. Then if $a\neq 0$ (suppose WLOG that $a>0$) is s.t. $v(a)\neq 0$ (suppose WLOG that $v(a)>0$), there is $\delta >0$ s.t. $v(x)>0$ for all $x\in (a-\delta ,a+\delta )$ where $a-\delta >0$. Then, for all $x\in (a-\delta ,a+\delta )$,

\begin{align*}
x^2u''(x)+2xu'(x)=0&\iff x^2v'(x)+2xv(x)=0\\
&\iff \frac{v'(x)}{v(x)}=-\frac{2}{x}\\
&\iff \ln(v(x))=\ln\left(\frac{1}{x^2}\right)+C\\
&\iff v(x)=\frac{D}{x^2}.
\end{align*}

Therefore, $u(x)=C+\frac{D}{x},$ where $C$ and $D$ are constant. You can prolonge $(a-\delta ,a+\delta )$ to $(0,\infty )$ if $a>0$ and $(-\infty ,0)$ if $a<0$.

Method 2

Suppose $x\neq 0$ (suppose WLOG $x>0$). Then, using an integrating factor yields
\begin{align*}
x^2u''(x)+2xu'(x)=0&\iff u''(x)+\frac{2}{x}u'(x)=0\\
&\iff u''(x)e^{2\ln(x)}+\frac{2}{x}u'(x)e^{\frac{2}{x}}=0\\
&\iff \left(u'(x)e^{2\ln(x)}\right)'=0\\
&\iff u'(x)e^{2\ln(x)}=C\\
&\iff u'(x)=Ce^{-2\ln(x)}\\
&\iff u'(x)=\frac{C}{x^2}\\
&\iff u(x)=D+\frac{C}{x}.
\end{align*}


问:
Maximized Likelihood of mu and sigma of a multivariate normal distribution
Given a p-dimensional multivariate normal distribution, I want to derive the maximized likelihood:

$$L(\hat{\mu}, \hat{\Sigma} ; X_{1}, ... , X_{n}) = (2\pi)^{-np/2}\left | \Sigma \right |^{-n/2}e^{-np/2}
$$

Start with MLE first step, the product of n p-dimensional normal PDFs:

$$ L(\hat{\mu}, \hat{\Sigma} ; X_{1}, ... , X_{n}) = \prod_{i=1}^{n}(2\pi)^{-p/2}\left | \Sigma \right |^{-1/2}e^{-1/2(\vec{x}_{i}-\vec{\mu})^{T}\Sigma^{-1}(\vec{x}_{i}-\vec{\mu})}
$$

Simplify the product:

$$ L(\hat{\mu}, \hat{\Sigma} ; X_{1}, ... , X_{n}) = (2\pi)^{-np/2}\left | \Sigma \right |^{-n/2}e^{-1/2 \Sigma_{i=1}^{n}(\vec{x}_{i}-\vec{\mu})^{T}\Sigma^{-1}(\vec{x}_{i}-\vec{\mu})} $$

And then I cannot seem to connect the term in the exponential with np in the "solution". Can someone give me a hint?

答:
Solution

The trick is in a decomposition of the sum in the exponential with the theorem:

For an n by 1 vector $\vec{a}$ and an n by n symmetrical matrix A, it follows that $$\vec{a}^{T}A\vec{a} = trace(\vec{a}A\vec{a}^{T}) = trace(A\vec{a}\vec{a}^{T})$$
Hence $$\Sigma_{i=1}^{n}(\vec{x}_{i}-\vec{\mu})^{T}\Sigma^{-1}(\vec{x}_{i}-\vec{\mu}) = trace(\Sigma^{-1}\Sigma_{i=1}^{n}(\vec{x}_{i}-\vec{\mu})(\vec{x}_{i}-\vec{\mu})^{T})$$

Now an assumption follows that the variance matrix $\Sigma$ can be estimated by:
$$\Sigma = \frac{1}{n}\Sigma_{i=1}^{n}(\vec{x}_{i}-\vec{\mu})(\vec{x}_{i}-\vec{\mu})^{T}$$

Therefore

$$ trace(\Sigma^{-1}\Sigma_{i=1}^{n}(\vec{x}_{i}-\vec{\mu})(\vec{x}_{i}-\vec{\mu})^{T}) = n trace(\Sigma^{-1}\Sigma) = ntrace(I) = np$$


答:
First, you should convince yourself with a picture:

Second, you need to show that for any $L$, that there exists an $\epsilon>0$ such that for any $N$, there exists some $x\ge N$ such that $|L-x \sin x| \ge \epsilon$.

In this example, one easy way to show this is to let $x_n = (\frac{1}{2}+2n)\pi$, and note that $x_n \sin x_n = x_n$. So pick any $L$, and let $\epsilon=1$. Now choose $N$ and select $k$ such that $x_k > \min(L+1, N)$. Then $|L-x_k \sin x_k| > 1$.

Body:
Question:

Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).

What I did:
I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this Proving limit doesn't exist using the $\epsilon$-$\delta$ definition
which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked.
Anyway I'd really love someone to verify/correct this solution before I hand it in.
Thanks a lot.

We assume there exists a limit, therefore:
$\forall \varepsilon>0 \exists M>0: x>M \Rightarrow |f(x)-L|<\varepsilon$
and specifically, if we pick some $x_1,x_2>M$: $|f(x_1)-L|<\frac \varepsilon2,
|f(x_2)-L|<\frac \varepsilon2$

Also, this applies for all $\varepsilon$, and specifically for  $\varepsilon=0.5$

(1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L |  + |f(x_2)-L| \le \frac \varepsilon2 +  \frac \varepsilon2 = \varepsilon$

This applies for all $x>M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} > M , x_2=\pi \lfloor M+1 \rfloor$

We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 & \lfloor M+1 \rfloor   odd  \\ 0 &\lfloor M+1 \rfloor  even\end {cases} $

So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.


问:
When does the boundary of a ball not equal to sphere?
Let $(M, d)$ be a metric space, $a \in M, r > 0$, and the boundary of $B$r$(a)$ be non-empty. Under what conditions of this metric space, the boundary of $B$r$(a)$ does not equal to the sphere {x ∈ M : d (x, a) = r} ? In other words, how do we find such a metric space? I would appreciate any concrete examples.

答: Take $M=(-\infty, 0]\cup\{1\}\subset\mathbb R$ with ordinary metric on $\mathbb R$. Then, $B_1(0)=(-1,0]$ and $\partial B_1(0)=\{-1\}$, but $S_1(0)=\{-1, 1\}$.


答: Consider $\Bbb{R}\times\Bbb{Z}$ under the metric it inherits as a subset of the Euclidean plane. The boundary of the open ball $B_1((0, 0))$ comprises the points $(\pm 1, 0)$, but the sphere of radius $1$ around $(0, 0)$ also includes the points $(0, \pm1)$.

Title: When does the boundary of a ball not equal to sphere?

Body: Let $(M, d)$ be a metric space, $a \in M, r > 0$, and the boundary of $B$r$(a)$ be non-empty. Under what conditions of this metric space, the boundary of $B$r$(a)$ does not equal to the sphere {x ∈ M : d (x, a) = r} ? In other words, how do we find such a metric space? I would appreciate any concrete examples.


问:
A simpler argument to prove n^4 + 4^n is composite for n ending in 5?
The expression $n^4 + \,4^n$ is known to be prime only for $n = 1$ corresponding to the prime $5$. The usual proof makes use of the Sophie Germain identity $$a^4 + \,4b^4 ≡ \left\{(a - b)^2 +\, b^2\right\}\left\{(a + b)^2 +\, b^2\right\}$$ since $n^4 + \,4^n$ is odd only for odd $n$ in which case $4^n$ has the form $4\left(2^k\right)^4$ and then both factors $\left(n ± 2^k\right)^2 + \,4^k > 1$ for $ k > 0 $. We observe, on the other hand, that for all $n$ ending in $1, 3, 7, 9$ expression ends in $5$ and obviously cannot be prime. I was wondering if a similar line of thought could dispose of the case when $n$ ends in $5$, but here I can't proceed further beyond the fact that they generate numbers ending in $49$ and primes do end in $49$ and hence doesn't preclude the existence of a prime. Can we somehow obviate recourse to Sophie Germain identity (or the like) to prove the expression is composite also in the case $n$ ending in $5$ ?

答:
In general for non negative $a,b$ $$a^2+b^2 = (a -\sqrt{2ab}+b)(a +\sqrt{2ab}+b)$$

so you can say $$n^4 + 4^n = (n^2 - 2^{\frac{(n+1)}2}n +2^n)(n^2 + 2^{\frac{(n+1)}2}n +2^n)$$

which is the product of two integers when $n$ is odd,

and both those integers are greater than $1$ when $n\ge 3$.

Numbers ending in $5$ are odd and greater than $3$.


问:
Count of subsets of size r disjoint with a subset of size r
I apologize if I do not use 100% correct terminology:

Imagine I have a set $S$ with $n$ elements. S = {1,2,3,4,5}. If we want to calculate how many subsets of $S$ there are of size 2, then we must take $n$ choose 2. In this case we have the following 10 combinations:

{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}

Now choosing one of these combinations, I would like to calculate how many other subsets are disjoint with it.

For example: Choose subset {1,2}. It is disjoint with 3 subsets: {3,4},{3,5},{4,5}

Is there a formula based on $n$ and size of the subset $r$ (in this example $r=2$) where I can calculate this number of disjoint subsets (in this example=3)?

答: Yes. Take out the $r$ objects from the $n$ and you get $\binom{n-r}{r}.$ That if you care about the size of the disjoint set to be also $r$. If you want any size set it would be $2^{n-r}$.


问:
ode; zero input response: what does q(D)y = 0 mean?
my textbook on ordinary differential equations has a chapter on system modeling, and I 'm not sure if I understand what is meant by

The zero-input response is the solution to  $q(\mathbf{D})y=0$,
$y(0)=a_0$, $y'(0)=a_1$, ...

One of the examples given of how to determine the zero-input respnse from a system modeled by q(D) is

$q(\mathbf{D}) = (\mathbf{D} + 1)^2 + 4$,  $y(0)=1$,  $y'(0)=3$

I can see how the answer

$y=e^{-t}(\cos(2t) + 2\sin(2t))$

is derived by searching for the characteristic modes from the standard basis, but then I recalculated the linear equation with the derivative back from this answer, and I expected the result to be a constant 0.

I tried this in matlab:

```<code>syms x

y = exp(-x) * (cos(2*x) + 2*sin(2*x));

qD = (diff(y) + y)^2 + 4*y;

diffy = diff(y);

hold on

fplot(y, [0 5])
fplot(qD, [0 5])
fplot(diffy, [0 5])

%fplot(hn)

legend ('y', 'qD', 'diffy')
</code>```

but I got anything but a constant 0 for $q(\mathbf D)$.

so is my matlab formula for $q(\mathbf D)$ wrong? or what else is meant by $q(\mathbf{D})=0$?

regards, Danny.

答:
Apparently you are using $\mathbf{D}$ as the derivative operator. So an ODE of a dynamic system can be written as:
$$q(\mathbf{D})y(t)=u(t)$$ being $q(\mathbf{D})$ a polynomial of $\mathbf{D}$ and $u(t)$ the input of the system.

The zero input response of a system is the response taking into account only its internal dynamics. So $y(t)$ is the solution of the equation $q(\mathbf{D})y(t)=0$.

The function you calculated in matlab is already the application of the operator $\mathbf{D}$, as you are using the diff functions, so it is expected to result 0.


问:
Weird definition for convergent sequences
if exists $N\in\mathbb N$ that for all $\varepsilon>0$ so that for all $n>N \,\,$ $|a_n-L|<\varepsilon \Longrightarrow \underset{n\rightarrow\infty}{\lim}a_{n}=L$

Any ideas on how to prove this definition is equal to the conventional definition of a the limit of sequence?

答:
Hint

Notice that $$\exists N\in\mathbb N: \forall \varepsilon >0, \forall n\in\mathbb N, n\geq N\implies |a_n-L|<\varepsilon $$
means that $a_n=L$ for all $n\geq N$.


答: The implication is certainly true, but the given condition is not equivalent to the definition of limit. For that condition to hold, the terms of $a_n$ are all equal to $L$ from a certain order on.

Title: Weird definition for convergent sequences

Body:
if exists $N\in\mathbb N$ that for all $\varepsilon>0$ so that for all $n>N \,\,$ $|a_n-L|<\varepsilon \Longrightarrow \underset{n\rightarrow\infty}{\lim}a_{n}=L$

Any ideas on how to prove this definition is equal to the conventional definition of a the limit of sequence?


问:
Rouche's theorem for two functions that have the same number of roots
I hope this is not too long. Thanks in advance!

Edit: I edited it for a great deal, most of the information was unnecessary.

Let us define a function $h(z) = f(z) + g(z)$. We know that $f(z)$ has $K$ roots for $z$ within the complex unit circle and $g(z)$ also has $K$ roots for $z$ in the complex unit circle.

Rouche's theorem states that if $|f(z)| > |g(z)|, ~ |z| = 1$, $h(z)$ has the same number of roots in the complex unit disc that $f(z)$ has. In my case, this inequality does not always hold, it depends on other coefficients.

My question is, does it matter that $|f(z)|$ can be greater, equal, or smaller than $|g(z)|$? My gut feeling tells me that this should not change the number of roots $h(z)$ has, namely $K$.

答:
It absolutely matters. Anything can happen without that inequality.

For example, you might have $g=-f$, in which case $h$ is identically zero! Or, you could have $f(z) = 2z$ and $g(z) = 1-2z$, so that both $f$ and $g$ have one root inside the unit circle but $h(z) = 1$ has no roots.

On general linear algebra grounds, if you have any polynomial $h(z)$ of degree $2K$ whatsoever, then you should be able to write $h(z) = f(z) + g(z)$ where $f$ and $g$ are polynomials of degree $2K$, each of which has $K$ zeros at any points you want to specify in advance. (I admit that this construction doesn't rule out additional zeros of $f$ and $g$, but it might be illuminating nonetheless.)


问:
When is full SVD used, if at all?
Whenever I see SVD, it's first introduced in a full form (i.e. for $X_{n \times m}$, where $n > m$, matrix $U$ has $n$ columns) and then truncated (so $U$ has $m$ columns). It's always also stated that we "almost never" use full SVD. So when would we use it, apart from theory (e.g. having orthgonal $U$ for proofs)? Or even in theory, where would we use full SVD instead of reduced?

答: The only setting in which the full SVD would be preferred is in the "theoretical" setting. In this setting, the full SVD is indeed strongly preferred. For example, if you look up proofs of the EYM theorem, pretty much all of them use the full SVD because it is much easier to work with an orthogonal $U$ that with a $U$ for which we only have $U^TU = I$ (for the $n>m$ case)

Title: When is full SVD used, if at all?


问:
Conjugacy classes of a nonabelian group of order $p^4$
Apologies in advance for the long text! I will explain what I have done so far, and then I will present my question.

Consider a nonabelian group of order $p^4$, where $p$ is a prime. Then, the center $Z(G)$ cannot have order $p^3$ or $p^4$, since both these cases would lead to the contradiction that $G$ is abelian. So we consider:

Case I: Order of $Z(G)$ is $p^2$     i.e.,$|Z(G)|=p^2$

The class equation is given by $|G|=|Z(G)|+\Sigma|G:C_G(g_i)|$, where the $g_i$ are representatives of the distinct conjugacy classes of G not contained in the center $Z(G).$ I deduced that for each $g\notin Z(G),$ $|G:C_G(g)|=p.$ So the class equation becomes $p^4=p^2+kp,$ where $k$ is the number of distinct conjugacy classes not contained in the center. This gives $k=p^3-p$ and hence the number of conjugacy classes of $G$ is $p^3+p^2-p$

Case II: $|Z(G)|=p$

I was able to prove that for any $g\notin Z(G),|G:C_G(g)|=p$ or $p^2.$ That means that, a conjugacy class not contained in the center can have size $p$ or $p^2.$

My question is: Is there any way to determine how many conjugacy classes of size $p$ is there, and how many conjugacy classes of size $p^2$ is there?

I would really appreciate any help regarding this question. Thank you!

答:
Here is a quick summary of what is happening - I have to admit that I did some computer calculations first and worked out what was happening.

It is not hard to see that a group $G$ of order $p^4$ with class two has centre of order $p^2$, so if $|Z(G)|=p$ then $G$ has class three and $|[G,G]|=p^2$.

Since the order of the automorphism group of a group of order $p^2$ is not divisible by $p^2$, we must have $|C_G([G,G])| = p^3$.

Then the elements in $C_G([G,G]) \setminus Z(G)$ have centralizers of order $p^3$ and classes of order $p$. There are $p^3-p$ such elements, and the remaining $p^4-p^3$ elements have centralizers of order $p^2$ and lie in classes of size $p^2$.

To see that a group of class $2$ has centre of order $p^2$, suppose not and let $G$ be a counterexample. Then $|Z(G)|=p$ and $G/Z(G)$ is abelian of order $p^3$. Then the commutator map $(g,h) \to [g,h]$ induces a bilinear map $G/Z(G) \times G/Z(G) \to Z(G)$.

Since $G/Z(G)$ cannot be cyclic, it is isomorphic to $C_p \times C_p \times C_p$ or to $C_p \times C_{p^2}$.

In the first case, let $a,b,c$ be inverse images in $G$ of generators of the three copies of $C_p$, and let $Z(G) = \langle z \rangle$.

Suppose that $[a,b]=z^i$, $[a,c]=z^j$, $[b,c]=z^k$ with $0 \le i,j,k < p$. We cannot have $i=j=k=0$, so suppose that $i \ne 0$. Then $[a,b^tc] = z^{it+j}$ and so choosing $t = -j/i\bmod p$ gives $[a,b^tc] = 1$. Similarly we can find $s$ with $[b,a^sc] = 1$, and then $a^sb^tc \in Z(G)$, contradiction.

The case $G/Z(G) = C_p \times C_{p^2}$ is easier, because the $p$-th power of the inverse image in $G$ of a generator of the $C_{p^2}$ factor is in $Z(G)$.


答:
Clearly the equation can be written as
$$ (x^2u'(x))'=0 $$
by using product-rule. So
$$ x^2u'(x)=C $$
which is easy to solve.

Title: Constant solution of Euler-Cauchy equation

Body:
I want to prove that if $u \in C^2([-1, 1])$ solves
$$
x^2 u''(x) + 2xu'(x) = 0
$$
on $[-1, 1]$, then $u$ is constant. Is this possible without using arguments from ODE theory, i.e. Picard-Lindelöf's theorem or explicitly calculating solutions?

So far I tried to partially integrate
$$
0 = \int^1_{-1} x^2u''(x)+2xu'(x) ~\mathrm{d}x
$$
which did not help me that much...


问:
Prove that $c^3f(c)+c f(c)\ge1$
Let $f:[0,1] \rightarrow \mathbb{R}$ a continuous function that satisfies $\int_0^1 2 x f(x) dx >\pi/2$. I need to prove there is

$c\in(0,1)$ such that $c^3f(c)+c f(c)\ge1$.

My guess is the mean value theorem. Am I on the right track?

答:
Since
$$\int_0^1\big(xf(x)-\frac{1}{1+x^2}\big)dx=\int_0^1xf(x)dx-\frac{\pi}{4}>0,$$
there exists $c\in(0,1)$, such that $cf(c)-\frac{1}{1+c^2}>0$. The conclusion follows.

Remark: It suffices to assume that $f$ is (Riemann or Lebesgue) integrable on $[0,1]$. Continuity is not necessary.

Title: Prove that $c^3f(c)+c f(c)\ge1$


问:
Order of an element in $GF(2^9)$
Let $a$ be an element in $GF(2^9)$ which satisfies $a^{9}+a^{8}=-1$.  If $H=\langle{a\rangle}$, then what is the order of $H$ in $GF(2^9)$. I have no idea how to start. $a^{9}+a^{8}=-1$, tried taking powers of this expression, to see if it gives anything, but I am lost. A brief solution would be of great help.

答:
This field is of characteristic $2$, so $(x+y)^2=x^2+y^2$ for all $x,y\in GF(2^9)$, and $-1=1$. Thus:

$$a^9=a^8+1=(a+1)^8$$

and

$$a^8(a+1)=1$$

In other words, $a^9=b^8$ and $a^8=b^{-1}$ where $b=a+1$. Thus, $a^9\times (a^8)^8=a^{73}=1$. As $73$ is prime, the order of $a$ must be $73$.

Title: Order of an element in $GF(2^9)$