问:
How to integrate over all space in polar coordinates?
I'm reading a paper (download here) which uses a polar coordinates to reach a result, but I can't understand how it uses polar coordinates at all. I try to explain all steps using photos of paper.

The system is a helium atom including two electrons and one nucleus named 1 , 2 and 0 respectively. There are three distances in this system: regarding nucleus located in the origin, $r_1$: distance of electron 1 from the origin, $r_2$ distance of electron 2 from the origin, $r_{12}$ distance between two electrons (you can recognize these in the second image easily)

First the paper defines a function called distribution function $f(r_{12})$ and then normalizes it as you can see in the following image:

After that it defines wave function of the system and normalizes it too. Then it equates them (because integration over both lead to 1), First question arises here: Why it uses $f(r_{12}) dr_{12}$ in equation 1, I mean why the integral sign is absent?!

Then It makes $d\tau_1d\tau_2$ equal to all distances and angels which is absolutely unclear to me. See the following image:

On one hand it says we are using polar coordinates. We know there is only one angle in this coordinate while here there are three angels $\theta, \phi$ and $\chi$ ! On the other hand we know integration over all space is $r^2 \sin[\theta] \;dr\;d\theta \;d\phi$ in spherical coordinates and $r dr d\theta$ in polar coordinates, so I can't understand how it reaches to such an equivalent for $d\tau_1d\tau_2$. It doesn't match what I know about integrating over all space. I will be so grateful if someone can help me to answer my questions. Thanks in advance,

答:
To answer your first question,

Why is the integral sign absent from the expression $f(r_{12}) d r_{12}$

I will just point out that right below that expression, the paper adds that

the integration is taken over all positions of the two electrons such that the interelectronic distances lies between $r_{12}$ and $r_{12} + d r_{12}$.

Thus the right hand side should be interpreted as an integral over a region that satisfies this description (i.e. a thin spherical shell at a distance of $r_{12}$ from particle $1$). The authors resorted to this verbal description probably because it is a bit cumbersome to explicitly write down the domain of the integral, which could be described as $r_{12}<|\boldsymbol{\tau}_1 - \boldsymbol{\tau}_2| < r_{12} + d r_{12}$. The differential $d r_{12}$ is the reason why there is a differential on the left hand side. (Note that I have made the variables $\boldsymbol{\tau}_1$ and $\boldsymbol{\tau}_2$ bold, because I want to make it clear that these are short hands for two volume integrals. For example $d\boldsymbol{\tau}_1$ can stand for $d x_1 d y_1 dz_1$ in Cartesian coordinates or $r_1^2 \sin\theta_1 dr d\theta d\phi$ in spherical coordinates.)

As for your second question concerning the coordinates, I would say that it is indeed a little different from the conventional polar (spherical) coordinates, but presumably this is to exploit the symmetries in the problem or some special properties of the integrand. Let me walk through the transformation first, then address some of the specific concerns you raised.

First, just as a sanity check, let's count the number of variables on the two sides of the transformation
$$
d\boldsymbol{\tau}_1 d\boldsymbol{\tau}_2 = r_1 r_2 r_{12} \sin\theta_1 d r_1 d r_2 d r_{12} d \theta_1 d \phi_1 d \chi \ .
$$
As I noted above, the left hand side is two 3D volume forms, so we are integrating over six variables. The right hand side also has six variables. Three of these are the usual spherical coordinates for particle $1$, namely $r_1$, $\theta_1$, and $\phi_1$. The other three require some explaining, but essentially they specify the position of the second particle. In any case we can see now that at least the number of variables match up.

To get to these three variables that describe the position of particle $2$, we can start from a set of spherical coordinates that takes $\boldsymbol{\tau}_1$ as the $z$-axis. In this "relative spherical coordinate system", we use the coordinates $r_2$, $\theta_{12}$, and $\chi$. The first is of course the usual radial coordinate. The second is the angle between the directions $\boldsymbol{\tau}_1$ and $\boldsymbol{\tau_2}$. The third is the angle of rotation from the $01z$ plane to the $012$ plane. To clarify, $01z$ is the plane that contains the $z$ axis and $\boldsymbol{\tau}_1$, and $012$ is the plane that contains $\boldsymbol{\tau}_1$ and $\boldsymbol{\tau}_2$. (See Figure 1 in the paper. You should think of point $2$ as hovering above the page and the triangle $012$ as folding away from the page, hinged along the $01$ line segment.) The volume form in these coordinates is just the ordinary spherical form $r_2^2 \sin\theta_{12} d r_2 d \theta_{12} d \chi$.

Next we want to get rid of the coordinate $\theta_{12}$ and use $r_{12}$ instead. The coordinate transform is given by the law of cosines,
$$
r_{12} = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos\theta_{12}}\ .
$$
We also need the Jacobian factor of this substition from $(r_2, \theta_{12}, \chi)$ to $(r_2, r_{12}, \chi)$, which turns out to be $r_1 r_2 \sin\theta_{12} / r_{12}$.
This means that the volume form transforms as
$$
r_2^2 \sin\theta_{12} d r_2 d \theta_{12} d\chi = \frac{r_2^2 \sin\theta_{12} d r_2 d r_{12} d\chi}{r_1 r_2 \sin\theta_{12}/r_{12}} = \frac{r_2 r_{12}}{r_1} d r_2 d r_{12} d\chi\ .
$$
Putting this together with the first part of the volume form $r_1^2 \sin\theta_1 dr_2 d\theta_1 d \phi_1$, we get the desired $r_1 r_2 r_{12} d r_1 d r_2 d r_{12} d\theta_1 d \phi_1 d\chi$.

Finally let me address the problem with terminology that might have tripped you up in the first place. It seems that you felt uneasy because the phrase "polar angles" seems to suggest the usage of polar coordinates, which is a two dimensional coordinate system that only contains one angular variable. However, people often use "polar coordinates" to mean "spherical coordinates" when the context makes it clear that the space in question is three-dimensional. The polar angle $\theta$ and azimuthal angle $\phi$ are then referred to collectively as the "polar angles".


问:
Difficult interpreting Real Analysis question
Sorry the vague title. I really have no idea how to classify this question. Could I get some help interrupting what the question is asking me to prove and hints how to solve?

Suppose that $f:R^n\to R$ is continuous. If there are $x\in R^n$ and $C\in R$ such that $f(x)<C$, then prove that there is $r>0$ such that for all $y\in B_r(x), f(y)<C$

I truly have absolutely no idea what this question is asking me to do.

答:
Continuity tells us that if you barely move around a point then the function doesn't change that much. The function at a point less than C is telling you that there is some "space" between $f(x)$ and C, there is in fact $\alpha: \quad f(x)<\alpha<C$.

Then provided you stay close enough to x, the function doesn't vary that much and is still below C.

Formally, given $\varepsilon =\frac{C-f(x)}{2}$.

By continuity, $\exists \ r>0: ||x-y||<r \implies |f(x)-f(y)| < \varepsilon$.
$$f(y)<\varepsilon+f(x) \leq \frac{C-f(x)}{2}+f(x) < C \quad \quad \forall \ y\in B_r(x)$$

ContentLicense: CC BY-SA 4.0


问:
Retracts preserving De Rham cohomology
I have to show an example of a retract $S$ from a $M$ manifold such that $S$ and $M$ doesn't have the same De Rham cohomology. Because of homotopy, I'm looking for a retract that is not a deformation retract but I don't have a clear intuition of how can I choose it. Thanks in advance

答: Hint: Every non-empty space retracts to a point

Title: Retracts preserving De Rham cohomology


问:
Parameterization of a curve (line & circle)
Let the curve C be the boundary of the region bounded by $y=2$ and $x^2+y^2=9$.

a) How would C look like?

(I think it would be circle segment?)

b) Which of the following characteristics does C have?

smooth

piece wise smooth

Jordan curve

closed

(I think it is closed and smooth and also a Jordan curve?) But how can I prove this mathematically?

c) I need to find a parameterization of the curve, and here I do not know how to start, do I parameterize the line and the circle arc separately?

the intersection points would be $P(\sqrt{5} | 2), Q(-\sqrt{5} | 2)$

答:
For the parametrization, if it is the upper sliver, you can solve the quadratic equation for $y$:

$y=\pm\sqrt{9-x^2}$ and since we’re working in $y>0$, simply $y=\sqrt{9-x^2}.$

Then letting $x=t$ we have the parametrization $(t,\sqrt{9-t^2})$ for $-\sqrt{5}\leq t\leq\sqrt{5}$.

Your line is $y=2$ thus the parametrization $(t,2)$ for $-\sqrt{5}\leq t\leq\sqrt{5}$.

These are easy ways to think of parameterizations at first, you can work your way to different methods such as polar coordinates.


问:
Fourier analysis question, orthonormal basis.
I need some help with this exercise:

Given $A>0$, let $L_{A}^2(\mathbb{R})$ the subspace of $L^2(\mathbb{R})$ of the functions $f$ that satisfy $\hat{f}=\chi_{[\frac{-A}{2},\frac{A}{2}]}\hat{f}$. Prove that the set:
$\left\{\sqrt{A}\frac{\sin(\pi(At-k))}{\pi(At-k)}/ k\in{\mathbb{Z}}\right\}$
is an orthonormal basis of $L^2_A(\mathbb{R})$.

This is an exercise of a Fourier series course, and I really don't know how to lead with it...
Thanks a lot for any help.

答:
Expanding $\sin(2\pi x)$ yields
$$
\int_{-\infty}^\infty\frac{\sin(2\pi x)}{\pi x}e^{-2\pi ix\xi}\,\mathrm{d}x
=\int_{-\infty}^\infty\frac{e^{2\pi ix(1-\xi)}-e^{-2\pi ix(1+\xi)}}{2\pi ix}\,\mathrm{d}x\tag{1}
$$
Using the contour, $\gamma$, from $-\infty-i\epsilon$ to $+\infty-i\epsilon$, then circling back counterclockwise through the upper half-plane for positive $\alpha$, or clockwise through the lower half-plane for negative $\alpha$, we  get
$$
\int_\gamma\frac{e^{2\pi ix\alpha}}{2\pi ix}\,\mathrm{d}x
=\left\{\begin{array}{cl}
1&\text{if }\alpha\ge0\\
0&\text{if }\alpha\lt0
\end{array}\right.\tag{2}
$$
Thus, using $(1)$ and $(2)$, we have
$$
\int_{-\infty}^\infty\frac{\sin(2\pi x)}{\pi x}e^{-2\pi ix\xi}\,\mathrm{d}x
=\left\{\begin{array}{cl}
1&\text{if }-1\le\xi\le1\\
0&\text{otherwise}
\end{array}\right.\tag{3}
$$
Usual scaling of the Fourier Transform yields
$$
\sqrt{A}\int_{-\infty}^\infty\frac{\sin(\pi Ax)}{\pi Ax}e^{-2\pi ix\xi}\,\mathrm{d}x
=\left\{\begin{array}{cl}
\frac1{\sqrt{A}}&\text{if }-\frac A2\le\xi\le\frac A2\\
0&\text{otherwise}
\end{array}\right.\tag{4}
$$
Usual translation of the Fourier Transform yields
$$
\sqrt{A}\int_{-\infty}^\infty\frac{\sin(\pi(Ax-k))}{\pi(Ax-k)}e^{-2\pi ix\xi}\,\mathrm{d}x
=\left\{\begin{array}{cl}
\frac1{\sqrt{A}}e^{-2\pi i\xi k/A}&\text{if }-\frac A2\le\xi\le\frac A2\\
0&\text{otherwise}
\end{array}\right.\tag{5}
$$
By the standard $L^2$ duality of the Fourier Transform, since the functions on the right hand side of $(5)$ are orthonormal basis elements of $L^2\left[-\frac A2,\frac A2\right]$, so are $\sqrt{A}\frac{\sin(\pi(Ax-k))}{\pi(Ax-k)}$ an orthonormal basis of $L_A^2(\mathbb{R})$.</p>

ContentLicense: CC BY-SA 3.0


问:
Simplifying trig expression
I was working through some trig exercises when I stumbled upon the following problem:

Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.

I started out by expanding it such that
$$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$
which simplifies to:
$$ \cos^2 A \cos^2 B- \sin^2 A \sin^2 B .$$
However, I don't know how to proceed from here. Does anyone have any suggestions on how to continue.

答:
Here is a detailed answer.
Let's rock!
$$
\require{cancel}\begin{align}
\cos\left(A-B\right)\cdot\cos\left(A+B\right)&=\left(\cos A\cos B-\sin A\sin B\right)\left(\cos A\cos B+\sin A\sin B\right)\\
&=\cos^2A\cos^2B-\sin^2A\sin^2B\\
&=\cos^2A\left(1-\sin^2B\right)-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cos^2A-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cancelto{1}{\left(\cos^2A+\sin^2A\right)}\\
&=\cos^2A-\sin^2B
\end{align}
$$
I hope this helps.

Title: Simplifying trig expression


问:
Is there a cubic spline interpolation method for curves in $\mathbb R^2$
We know that given sample points, $(x_i, f(x_i)),\ i=0,\cdots,n$, where $a=x_0<x_1<\cdots<x_n=b$, we can do a cubic spline interpolation to find a $C^2$ function on $[a,b]$ that is piecewise a cubic polynomial in each interval $[x_{i+1},x_i]$.

I wonder if we can extend this to curves in $\mathbb R^2$. Say, we have a smooth curve $C$ in $\mathbb R^2$. If we are given sample points $(x_i,y_i),\ i=0,\cdots,n$ on $C$, is it possible to find a piecewise "cubic" curve to interpolate $C$?

The reason I used quotes for "cubic" above is that there are possibly more than one way to define what is "cubic" for a curve. I can think of three ways:

Think of $y$ as a function of $x$, and do the usual interpolation. But this may not be suitable for curves.

Try to interpolation using $x(t)=At^3+Bt^2+Ct+D,\ x(t)=A't^3+B't^2+C't+D'$. But this has too many parameters. Also, the possible reparametrizations of $t$ is also a problem. I guess we can put some additional constrains such as $(x'(t))^2+(y'(t))^2=1$. But I am not sure this is suitable.

So is there currently any such effort?

答:
A smooth curve C in $R^2$ might have multiple $y$ values given the same $x$, so it is not suitable to use the explicit representation $y=f(x)$. Typically,  parametric representation $C(t)=(x(t),y(t))$ is used.

You can easily extend the cubic spline interpolation to interpolate a set of points $(x_i, y_i)$ as long as you know the order of these points. Knowing the order of these points will allow you to assign parameter to each point, a process known as "parametrization". Chord-length parametrization and centripetal parametrization are two typical methods. Once you know the $t$ value for each point, the cubic spline interpolation for $(x_i, y_i)$ will be essentially the same as doing it twice: one for $(t_i, x_i)$ and another for $(t_i, y_i)$.

You can also extend this to interpolate a set of 3d points $(x_i, y_i, z_i)$ sampled from a 3d curve $C(t)=(x(t),y(t), z(t))$.


问:
Intuitition for the vector Laplacian and the vector Laplace's equation
I have an intuitive understanding of the gradient, divergence, curl, and Laplacian operators in multivariable calculus, but not of the vector Laplacian. Is there a visualizable intuition for the meaning of the Laplacian of a vector field on a Riemannian manifold?

And a closely related question: is there an intuitive explanation for what's "special" about solutions to the vector Laplace equation?

答:
I found an answer. The property discussed at https://math.stackexchange.com/a/50285/268333 generalizes easily to the vector case for vector fields on $\mathbb{R}^d$:

$$
\nabla^2\, {\bf F}({\bf x}_0) = 2d \lim_{r \to 0^+} \frac{\langle {\bf F} \rangle_r - {\bf F}({\bf x}_0)}{r^2} ,
$$

where $\langle {\bf F} \rangle_r$ denotes the average value of the vector field over the small sphere of radius $r$ centered at ${\bf x}_0$. This means that the vector Laplacian of a vector field at a point is proportional to the leading-order amount by which the average value of the vector field near the point differs from its value at the point.


问:
Prove $1^3 + 2^3 + \ldots n^3 = \frac{n^2(n + 1)^2}{4}$ with mathematical induction
Hello guys I'm trying to prove this equation with mathematical induction method.

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D&space;+&space;2%5E%7B3%7D&space;+&space;...&space;+&space;n%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dn%5E%7B2%7D(n&space;+&space;1)%5E%7B2%7D" title="1^{3} + 2^{3} + ... + n^{3} = \frac{1}{4}n^{2}(n + 1)^{2}" />
```

So for n = 1 I know that :

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7D1%5E%7B2%7D(1&space;+&space;1)%5E%7B2%7D" title="1^{3} = \frac{1}{4}1^{2}(1 + 1)^{2}" />
```

1 = 1

Now I know that it works for some k

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} = \frac{1}{4}k^{2}(k + 1)^{2}" />
```

I wanna prove it for k + 1:

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;+&space;(k+1%5E3)&space;=&space;%5Cfrac%7B1%7D%7B4%7D(k+1%5E%7B2%7D)(k&space;+&space;2)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} + (k+1^3) = \frac{1}{4}(k+1^{2})(k + 2)^{2}" />
```

Since :```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} = \frac{1}{4}k^{2}(k + 1)^{2}" />
```

I can rewrite it as :

```
<img src="https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D&space;+&space;(k&space;+&space;1)%5E%7B3%7D" title="\frac{1}{4}k^{2}(k + 1)^{2} + (k + 1)^{3}" />
```

So I'm stuck here.. I don't know what should I do.. Should I first square (k + 1)^2 or rewrite (k + 1)^3 ??'
Either way I get weird result and don't know how to proceed with this next step to prove mathematical
induction. I mean I can multiple k^2 with the first parenthesis but I think I should leave it that way and do smth other than that. I'll appreciate any help. Thanks :)

```
<img src="https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k%5E%7B2%7D&space;+&space;2k&space;+&space;1)&space;+&space;(k%5E%7B3%7D&space;+&space;3k%5E%7B2%7D&space;+&space;3k&space;+&space;1)" title="\frac{1}{4}k^{2}(k^{2} + 2k + 1) + (k^{3} + 3k^{2} + 3k + 1)" />
```

Body:
You are almost there! Notice that
\begin{align*}
\frac{k^{2}(k+1)^{2}}{4} + (k+1)^{3} & = \frac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4}\\\\
& = \frac{(k^{2}+4(k+1))(k+1)^{2}}{4}\\\\
& = \frac{(k^{2} + 4k + 4)(k+1)^{2}}{4}\\\\
& = \frac{(k+2)^{2}(k+1)^{2}}{4}\\\\
& = \frac{(k+1)^{2}(k+2)^{2}}{4}\\\\
& = \frac{(k+1)^{2}((k+1)+1)^{2}}{4}
\end{align*}
and we are done.

Hopefully this helps!


问:
Line Integral: why does it require |r'(t)|?
I'm a confused by the following definition of the line integral from Wiki:

For some scalar field $f : U \subset R^n \to R$ the line integral
along some piecewise curve $C \subset U$ is defined as:   $\int_C f(r) \ ds = \int_a^b f(r(t)) | r'(t) | dt$

where $r : [a,b] \to C$ is an arbitrary bijective parametrization
of the curve $C$ such that $r(a)$ and $r(b)$ give endpoints of $C$.

QUESTIONS

On the LHS $r$ is used.  Is it still defined as a bijective parametrization?

On the RHS why is the Euclidean norm $|r'(t)|$ required in the equation?  Intuitively, it seems the line integral could just be calculated $\int_a^b f(r(t)) dt$?

答:
A stupid example of why you need the $|r'(t)|$ factor.

Suppose you have your parametrization of $x$ and $y$ in terms of some parameter t.  We could replace $t$ with $2t$ and it would still describe the same curve.  You would be moving along the curve at twice the speed, but it is the same curve.  As a result of moving along that curve at twice the speed, the interval of integration will be cut in half though.  The $|r'(t)|$ will correct for such shenanigans.

$\int_a^b f(r(t))|r'(t)|\ dt = \int_{\frac {a}{2}}^{\frac {b}{2}} f(r(2t))|r'(2t)|\ dt$

We want to analyze the integrand such that ever piece of the curve of equal length has equal contribution regardless of the "speed" we are moving across those lengths.

The $|r'(t)|$ factor gives us a consistent integral for a path regardless of the parameterization of that path.


问:
Find the number of quadratic polynomials ax² + bx + c

Find the number of quadratic
  polynomials ax² + bx + c such that:

a)
  a, b, c are distinct.

b) a, b, c ε {1,
   2, 3, …2008}

c) x + 1 divides ax² + bx
  + c

Reasoning->b=a+c  b ranges from 3 to
  2008 x1+x2=n has n-1C1 solns. Using th
  elogic we get 2+3+..2007 solutions. A
  total of 2015027
  solutions.here,condition 1 comes into
  picture telling us that a,b,c are
  distinct. Therefore the even values of
  b ie.4,6,8,..2008 will have one
  solution less ie.from the given
  solution subtract 1003 thus giving us
  20104024

答: Hint:  try polynomial long division to find a condition on $a,b,c$.  Then it is a combinatoric question how many sets of that kind there are.</p>


问:
Probability and random selection question
Two integers are to be selected from random and without replacement from the set of the first $25$ positive integers. What's the probability that product of the two integers will be even?

I'm stuck on this question for calculation, I know there are total $25\choose{2}$ ways to randomly select two numbers. Within $25$ integers we have $13$ odd integers and $12$ even integers. So maybe we have combination of one odd and one even, or two even.

But not sure how to calculate the probability when including selection.

Any response is helpful, thanks.

答:
There are 13 odd and 12 even numbers.

Probability of drawing a odd first number is $\frac{13}{25}$ and odd second number is $\frac{12}{24}$ giving probability that product is odd = $\frac{13}{25} * \frac{12}{24}$

In any other case the product will always be even.  So probability that product is even = $1 - \frac{13}{25} * \frac{12}{24} = \frac{156}{600}$

Another ways is to calculate the probabilities of all selections resulting into an even product and sum them up.  which will be $\frac{12}{25} * \frac{24}{24} + \frac{13}{25} * \frac{12}{24} = \frac{156}{600}$.


问:
$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$

Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
 1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
 {\rm d}t=\dfrac{\pi }{\sqrt{2}}$.

We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?

答:
I'm somehow not really satisfied with all those great solutions above. To me, they didn't show quite well the underlying reason that makes that convergence occur. So I'll give another solution by trying to generalize the very first problem.

I'll start by stating something trivial.

Lemma 1:
 (Convergence of measures)

Let $\mu_1, \mu_2,...$ be a sequence of measures on $[0,1]$ such that $$ \lim_{ n \rightarrow \infty} \mu_n[0,1-\epsilon) = 0 $$
for all $\epsilon>0$, and $$ \lim_n \mu_n [0,1]=C$$
for some $C>0$, then for all continuous function $g$ on $[0,1]$ , we have:
$$ \int g d\mu_n \longrightarrow Cg(1)$$
$\square$

Now, let's go back to the initial question.

Main solution

I'll define:

$g(t):= \frac{ \sqrt{2t} f(t)^{3/2}}{f'(t)}$

$d\mu_x(t)= \mathbb{1}_{[0,x]}(t)\frac{1}{ \sqrt{\left(1- h_x(t)\right)h_x(t)}} h_x'(t)dt $, where $h_x(t)= \frac{f(t)}{f(x)}$

So we see that:

The integral whose limit we want to calculate is equal to $ \int g d\mu_x$

$g$ is continuous on $[0,1]$ and  $g(1)=\frac{\sqrt{2}}{2}$

Note that $h_x$ is increasing in $t$, thus by changing variables, we can show that $$\mu_x[0,1] = \int_{[0,1]} \frac{1}{\sqrt{u(1-u)}} du= \pi$$

Also by changing variables and noting that $\lim_{x \rightarrow 1^-} f(x)=+\infty$, we can show that:
$$\lim_{x \rightarrow 1^{-}}  \mu_x[0,\alpha] =0$$ for all $\alpha \in (0,1)$

Hence by applying the very first lemma, we have the final result

Comment:

TL;DR: the main idea is just to find the right $g$ and $\mu_x$ to compactify our integral.


Title: Prove $1^3 + 2^3 + \ldots n^3 = \frac{n^2(n + 1)^2}{4}$ with mathematical induction

Body:
Hello guys I'm trying to prove this equation with mathematical induction method.

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D&space;+&space;2%5E%7B3%7D&space;+&space;...&space;+&space;n%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dn%5E%7B2%7D(n&space;+&space;1)%5E%7B2%7D" title="1^{3} + 2^{3} + ... + n^{3} = \frac{1}{4}n^{2}(n + 1)^{2}" />
```

So for n = 1 I know that :

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7D1%5E%7B2%7D(1&space;+&space;1)%5E%7B2%7D" title="1^{3} = \frac{1}{4}1^{2}(1 + 1)^{2}" />
```

1 = 1

Now I know that it works for some k

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} = \frac{1}{4}k^{2}(k + 1)^{2}" />
```

I wanna prove it for k + 1:

```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;+&space;(k+1%5E3)&space;=&space;%5Cfrac%7B1%7D%7B4%7D(k+1%5E%7B2%7D)(k&space;+&space;2)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} + (k+1^3) = \frac{1}{4}(k+1^{2})(k + 2)^{2}" />
```

Since :```
<img src="https://latex.codecogs.com/gif.latex?1%5E%7B3%7D+&space;2%5E%7B3%7D&space;+&space;....&space;+&space;k%5E%7B3%7D&space;=&space;%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D" title="1^{3}+ 2^{3} + .... + k^{3} = \frac{1}{4}k^{2}(k + 1)^{2}" />
```

I can rewrite it as :

```
<img src="https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k&space;+&space;1)%5E%7B2%7D&space;+&space;(k&space;+&space;1)%5E%7B3%7D" title="\frac{1}{4}k^{2}(k + 1)^{2} + (k + 1)^{3}" />
```

So I'm stuck here.. I don't know what should I do.. Should I first square (k + 1)^2 or rewrite (k + 1)^3 ??'
Either way I get weird result and don't know how to proceed with this next step to prove mathematical
induction. I mean I can multiple k^2 with the first parenthesis but I think I should leave it that way and do smth other than that. I'll appreciate any help. Thanks :)

```
<img src="https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B4%7Dk%5E%7B2%7D(k%5E%7B2%7D&space;+&space;2k&space;+&space;1)&space;+&space;(k%5E%7B3%7D&space;+&space;3k%5E%7B2%7D&space;+&space;3k&space;+&space;1)" title="\frac{1}{4}k^{2}(k^{2} + 2k + 1) + (k^{3} + 3k^{2} + 3k + 1)" />
```

Body:
Note:  It's enough to prove $\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4 = n^3$.

That is to say, if we assume Equation 1: $1^3 + 2^3 + ...... + (n-1)^3=  \frac {(n-1)^2n^2}4$

And we want to prove Equation 2: $1^3 + 2^3 +.... + (n-1)^3 + n^3 = \frac {n^2(n+1)^2}4$

That is a matter of proving by subtraction Equation 2 for Equation 1, that

$n^3= \frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4$

So... Lets do that.

$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$

$\frac {n^2}4[ (n+1)^2 - (n-1)^2] =$

$\frac {n^2}4[\{(n+1)+(n-1)\}\{(n+1)-(n-1)\}]=$

$\frac {n^2}4[\{2n\}\{2\}]=$

$\frac {n^2}4\cdot 4n = n^3$.

And ... that's that.

....

If somehow that doesn't feel right..... you can expand...

$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$

$\frac {n^2 (n^2 + 2n + 1)}4 -\frac {n^2(n^2 -2n+1)}4 = $

... and, yeah, yeah, we get it.


问:
Sub Rings of the Real Field $\mathbb{R}$
Are there proper sub-rings of $\mathbb{R}$ which are not fields , which contain $\mathbb{Q}$? Towards answering this, I am aware of the result that proper sub rings of $\mathbb{Q}$, properly containing $\mathbb{Z}$ are principal ideal domains. My hunch is that proper sub-rings of $\mathbb{R}$ properly containing $\mathbb{Q}$ should be a field. Can anyone help me please with the proof or with a counter example?

答:
You already have a correct answer for your question. But let me add few more words to it. If you want to construct a sub-ring of $\mathbb{R}$ that properly contains $\mathbb{Q},$ the simplest thing we can do is adjoin an irrational number $\zeta$ to $\mathbb{Q}$ and close the resulting set with rational linear combinations and products. But it turns out that if $\zeta$ is an algebraic number then $\mathbb{Q}[\zeta]$ is a field. Why? Because any algebraic number has a unique monic minimal polynomial with rational coefficients, say, of degree $n$. Then $\{1,\zeta,\cdots, \zeta^{n-1}\}$ form a basis for $\mathbb{Q}[\zeta]$ as a $\mathbb{Q}$-vector space. Moreover for any $a_0, a_1,\cdots a_{n-1}\in\mathbb{Q}$ (not all zero), you can find the (unique) multiplicative inverse of $a_0+a_1\zeta+\cdots+a_{n-1}\zeta^{n-1}$ by solving $$(a_0+a_1\zeta+\cdots+a_{n-1}\zeta^{n-1})(b_0+b_1\zeta+\cdots+b_{n-1}\zeta^{n-1})=1$$ for $b_0, b_1,\cdots b_{n-1}\in\mathbb{Q}.$ The well know examples for this construction are quadratic number fields, cyclotomic fields and generally splitting fields of rational polynomials.

So, we cannot construct a pure ring by adjoining one (or more) algebraic numbers. On the other hand, if we adjoin a transcendental number $\xi,$ the resulting set $\mathbb{Q}[\xi]$ is a ring (in fact, it is a Euclidean domain), but not a field. In particular, there is no polynomial $P$ over $\mathbb{Q}$ such that $\xi P(\xi)=1.$ In the same way, for any sub-field $F$ (other than rationals) of $\mathbb{R}$ you can find a ring $R$ that properly contains $F.$ Also, You can modify this idea to obtain some other rings, for example the $\mathbb{Q}$-vector space spanned by $\{1, \pi^2, \pi^3, \pi^4,\cdots \}$ (this is not even a GCD domain).

However, there are other more obscure rings of $\mathbb{R}$ that contains rationals. For an example see here.

I think, we can (non-canonically) embedded any ring of characteristic $0$ with cardinality less than $\mathfrak{c}$ as a proper sub-ring of $\mathbb{R}.$


问:
Inclusions between $H^1_0(\Omega) \cap H^2(\Omega)$ and $H^2_0(\Omega)$
Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. I would like to understand something about the relationships between these two spaces: $$H^1_0(\Omega) \cap H^2(\Omega)\quad \text{and}\quad H^2_0(\Omega).$$

Is there any inclusion? If there is not, could you exhibit a counterexample?
Thank you!

答: Using the definitions, you should be able to prove that $H^2_0(\Omega) \subset H_0^1(\Omega) \cap H^2(\Omega)$. The reverse inclusion does not hold; as a counterexample, try picking a really nice $\Omega$ (such as a disc) and writing down a well-behaved function that vanishes at the boundary, but whose first derivatives do not. Then such a function will be in $H_0^1(\Omega) \cap H^2(\Omega)$ but not $H_0^2(\Omega)$.</p>


问:
Clarification of inner and outer approximations of a set $E \in \mathbb{R}^n$ using dyadic cubes
In Folland's Real Analysis, he lets $Q_k$ be the collection of cubes whose side length is $2^{-k}$ where $k \in \mathbb{Z}$. Then for a set $E \in \mathbb{R}^n$ he lets $A_k(E) = \bigcup \{Q  \in Q_k : Q \subset E\}$ be the inner approximation of E and $B_k(E) = \bigcup \{Q  \in Q_k : Q \cap E \neq \emptyset\}$ the outer approximation of E. What I don't understand is he says "the sets $A_k$ increase with $k$ while the sets $B_k$ decrease", but I cannot see how. Would someone be able to explain how this holds?

答:
Each time you increase $k$, each "cube" is subdivided into $2^n$ cubes. In $A_{k+1}$, each old cube of $A_k$ remains, and new smaller cubes appear to fill the gaps. In $B_{k+1}$, only the old cubes remain, and some of the subdivisions will be lots.

If you don't see the above, just draw a disk in $\mathbb R^2$, and draw the squares, and their subdivisions.


问:
if $X \subseteq Y $ are flats s.t. $r(Y) = r(X)-1$ then $\exists$ hyperplane $H$ s.t. $Y = H \cap X$.
Let $X$ and $Y$ be flats of a matroid $M$ such that $Y \subseteq X$ and $r(Y) = r(X)-1$. How can i prove that $M$ has a hyperplane $H$ such that $Y = H \cap X$ ?

答: You can do this just as you would for a representable matroid. Take a basis for $Y$, then add an element from $X$ to get a basis for $X$. Extend this to a basis for the entire ground set, then delete the single element of $X$ to get a basis for your hyperplane.


问:
Compound Interest Formula for monthly deposits
I am currently developing a savings calculator to some set requirements, such as using a compounding interest formula. My calculator currently is annual interest only, and compounds only annually as well.

I am currently using this formula: FV = PV * (1+rate) ^ n (in years)
My calculator is fine when using this formula.

enter image description here

However, I am required to make it that the calculator can take a monthly deposit and calculate the future value.

I searched online for how to do this and found this formula: FV = (previous formula) + Pmt x (((1 + r) ^ n) – 1) ÷ r)

I saw that I needed to convert my annual interest rate (r) to monthly interest rate, so my annual rate would be: r/12.
I also saw that my time (years) needed to be converted to months, so n*12.

Okay. But wouldn't this mean I am essentially compounding monthly, not annually, as I am dividing my rate by 12?

As you can tell I am a tad confused. I just want a formula for monthly contributions that'll compound annually, so I can do:

FV = (PV * (1+rate) ^ n (in years)) + (Pmt x (((1 + r) ^ n) – 1) ÷ r))

If anyone can shine some light on this that'll be great, thank you.

答:
Sorry nobody has replied!

I'm not entirely sure that this formulation is correct - but I'll give it a shot. You can conceptualize the both terms of the compound interest formula (principal appreciation & contribution appreciation) as such:

Future = Amount * Factor

So here we need to convert the what I'd call the "actual" interest rate between periods.

In the interest formula with principal only the Factor is expressed as (1+r)^n. So to convert we need to make an equivalency between the factor for two different periods/rates.

(1+r)^n = (1+i)^t (here i and t are the alternate rate and period)

So lets solve this for i to get our equivalent interest rate in terms of our original rate.

i = e^((n/t)*ln(1+r)) - 1

We can now use i and t in place of r and n in the "contribution" part of the formula you have posted. Again, I'm not in finance, so this may not be the proper way! Just the way I've conceptualized the solution.


问:
How to find the eigenvalues of this matrix in a simple way?
Can anyone help how to find the eigenvalues of the following matrix in a simple way? I expand the characteristic polynomial being,
$$
    \lambda(\lambda-3)(\lambda - 2k) = 0
$$
and get the answer but intuition is that there must be a simple way to find it.
$$
    \begin{bmatrix} 1 + k & 1 & 1 - k \\ 1 & 1 & 1 \\ 1 - k & 1 & 1 + k \\ \end{bmatrix}
$$

答:
All three rows sum to 3.  That means $(1,1,1)$ is an eigenvector with eigenvalue 3.

$(1,0,-1)$ looks to also be an eigenvector.  The eigevalue for this one is $2k.$

And the sum of the eigenvalues equals the trace of the matrix.  (And the product equals the determinant, but that is what we are trying to avoid calculating.)

$0$ is the 3rd eigenvalue.  We could have figured this out as well by the fact that the rows are linearly dependent.

This method is not, in general, a good way to find eigenvalues, but works in special situations.


答:
Alternative approach that yields identical math as TKA's answer.

$$1 - \frac{N\text{(umerator)}}{D\text{(enominator)}}$$

where $N =$ total # ways of selecting 2 odd numbers and

$D =$ total # ways of selecting any 2 numbers.

Answer is

$$1 ~-~ \frac{\binom{13}{2}}{\binom{25}{2}}.$$

Title: Probability and random selection question

Body:
Two integers are to be selected from random and without replacement from the set of the first $25$ positive integers. What's the probability that product of the two integers will be even?

I'm stuck on this question for calculation, I know there are total $25\choose{2}$ ways to randomly select two numbers. Within $25$ integers we have $13$ odd integers and $12$ even integers. So maybe we have combination of one odd and one even, or two even.

But not sure how to calculate the probability when including selection.

Any response is helpful, thanks.


问:
Proving that two paths in $\mathbb{R}^{2}-\{(0,0)\}$ are homotopic.
Let $f$ and $g$ be paths in $\mathbb{R}^{2}-\{(0,0)\}$. Show that $f$ is homotopic to $g$.

By the definition of a path, we have $f(x):[0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$ and $g(x):[0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$. We want to find a homotopy $F$ such that $F:[0,1]\times [0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$, and $F(x,0)=f(x)$ ; $F(x,1)=g(x)$. I suspect that the homotopy $F(x,t)=(1-t)f(x)+tg(x)$ won't work because the book gives this hint:

Hint: Show that every path is homotopic to the constant path that sends the entire interval to the path's starting point. Then show that two constant paths are homotopic using the fact that $\mathbb{R}^{2}-\{(0,0)\}$ is path connected.

So my question is, why do we have to do this work when the homotopy I have defined seems to work?

答:
Since $\mathbb{R}^2 - (0, 0)$ is path connected, there exists a continuous path $h: [0, 1] \rightarrow \mathbb{R} - (0, 0)$ such that $h(0) = f(0)$ and $h(1) = g(0)$.
Define $F:[0, 1]\times [0, 1] \rightarrow \mathbb{R}^{2}-\{(0,0)\}$ as

$$
F(x, t) =
\begin{cases}
f((1-3t)x),  & t \in [0, \frac{1}{3}] \\
h(3t-1),  & t \in [\frac{1}{3}, \frac{2}{3}] \\
g((3t-2)x), & t \in [\frac{2}{3}, 1]
\end{cases}
$$

where the three cases are continuous and agree at the two overlapping points $t=\frac{1}{3}$ and $t=\frac{2}{3}$. Therefore, $F$ is continuous. Moreover, $F(x, 0) = f(x)$ and $F(x, 1) = g(x)$ for all $x \in \mathbb{R}^2 - (0, 0)$. Thus, $F$ is a desired homotopy between $f$ and $g$.

Note that we cannot define $F$ using $F(x,t)=(1-t)f(x)+tg(x)$ since there may exist $x$ and $t$ such that $F(x, t) = (0, 0)$. Above, we avoid this problem since all three functions $f$, $g$ and $h$ map into $\mathbb{R}^2 - (0, 0)$.


答:
The factor makes the integral well-defined, meaning that it doesn’t depend on the choice of parametrization of the curve.

See the below answer of @Doug M for an example.

Suppose you have your parametrization of $x$ and $y$ in terms of some parameter t.  We could replace $t$ with $2t$ and it would still describe the same curve.  You would be moving along the curve at twice the speed, but it is the same curve.  As a result of moving along that curve at twice the speed, the interval of integration will be cut in half though.  The $|r'(t)|$ will correct for such shenanigans.

$\int_a^b f(r(t))|r'(t)|\ dt = \int_{\frac {a}{2}}^{\frac {b}{2}}
> f(r(2t))|r'(2t)|\ dt$

We want to analyze the integrand such that ever piece of the curve of
equal length has equal contribution regardless of the "speed" we are
moving across those lengths.

The $|r'(t)|$ factor gives us a consistent integral for a path
regardless of the parameterization of that path.

Title: Line Integral: why does it require |r'(t)|?

Body:
I'm a confused by the following definition of the line integral from Wiki:

For some scalar field $f : U \subset R^n \to R$ the line integral
along some piecewise curve $C \subset U$ is defined as:   $\int_C f(r) \ ds = \int_a^b f(r(t)) | r'(t) | dt$

where $r : [a,b] \to C$ is an arbitrary bijective parametrization
of the curve $C$ such that $r(a)$ and $r(b)$ give endpoints of $C$.

QUESTIONS

On the LHS $r$ is used.  Is it still defined as a bijective parametrization?

On the RHS why is the Euclidean norm $|r'(t)|$ required in the equation?  Intuitively, it seems the line integral could just be calculated $\int_a^b f(r(t)) dt$?


问:
All open coverage, admits an enumerable sub-coverage
Prove that if $X \subseteq \mathbb{R^n}$, then all open coverage of $X$, admits an enumerable sub-coverage.

If {$C_{n}$} $_{n \in I}$ is a cover of $X$, we say that it is open, if $C_{n}$ is open for all $n \in I$.

If {$C_{n}$} $_{n \in I}$ is a coverage of $X$, we say that the coverage is enumerable, if $I$ is enumerable.

答: HINT: Let $\mathscr{B}$ be a countable base for $\Bbb R^n$. Let $\mathscr{U}$ be any relatively open cover of $X$. For each $x\in X$ there are a $U_x\in\mathscr{U}$ and a $B_x\in\mathscr{B}$ such that $x\in B_x\cap X\subseteq U_x$. Let $\mathscr{B}_X=\{B_x:x\in X\}$. $\mathscr{B}_X\subseteq\mathscr{B}$, so $\mathscr{B}_X$ is countable, and clearly $X\subseteq\bigcup\mathscr{B}_X$. For each $B\in\mathscr{B}_X$ there is a $U_B\in\mathscr{U}$ such that $B\cap X\subseteq U_B$; why? What can you say about $\{U_B:B\in\mathscr{B}_X\}$?


答:
$\Bbb R^n$ is second countable:  just consider the open balls with rational radii and centered at points with rational coordinates.  These form a basis for the standard topology.

It now follows easily that $\Bbb R^n$ has the Lindelof property (which is another name for the property you are referring to).

Title: All open coverage, admits an enumerable sub-coverage

Body:
Prove that if $X \subseteq \mathbb{R^n}$, then all open coverage of $X$, admits an enumerable sub-coverage.

If {$C_{n}$} $_{n \in I}$ is a cover of $X$, we say that it is open, if $C_{n}$ is open for all $n \in I$.

If {$C_{n}$} $_{n \in I}$ is a coverage of $X$, we say that the coverage is enumerable, if $I$ is enumerable.


问:
Conjecture: any factorial is a sum of two Jordan-Polya numbers.
A Jordan-Polya number A001013 is a product of factorials
$\prod^m_{k=1}a_k!$. For $n=2,\dots,12$ there are Jordan-Polya numbers $a,b$ such that $n!=a+b$.

```<code>2!=1+1
3!=2+4
4!=8+16
5!=48+72
6!=288+432
etc
</code>```

Is this a coinsidence?

答:
Writing a quick script to search for solutions, $n=15$ does not appear to work.

All other $n\le 21$ do; I may extend my program to search further than this by generating Jordan-Polya numbers beyond those in the OEIS b-file. Note that as Ty Jensen mentions in a comment, one obtains solutions for every number which is a sum of two Jordan-Polya numbers, so possible candidates for failure are $n=11,15,19,21,23,27,29,31,35,39,\ldots$.

Edit: After generating the $91,802$ Jordan-Polya numbers up to $30!$, I have confirmed that the values of $n$ up to $30$ for which $n!$ is not expressible as a sum of two Jordan-Polya numbers are $n=15,23,27,29$. Curiously, every other value of $n!$ has at least two such expressions (other than $n=2,3$).


问:
Explicit bijection between the set of binary sequences and $\mathbb{R}$?
I know there is a standard proof (Cantor’s diagonalization argument) to show that the set of infinite binary sequences, call it $\Omega := \{0,1\}^\mathbb{N}$, is uncountable.

However, I would like to describe an explicit bijection between $\mathbb{R}$ and $\Omega$ to show that they are equipotent. Is there a direct way to do this? Or is my best bet to find a bijection between $\Omega$ and $(0,1)$ (or something) and compose it with a bijection from $(0,1)$ to $\mathbb{R}$?

On that note, is there a "standard" bijection between $\Omega$ and $(0,1)$?

Many thanks.

答:
There's not really a 'standard' one but here's a simple construction (from $\Omega$ to $(0,1)$; bijecting $(0,1)$ and $\mathbb{R}$ is easy as you can just take something continuous like $f(x) = \frac{2x-1}{x(1-x)}$):

Let $A$ be the set of dyadic rational numbers in $(0,1)$, and let $B$ be the set of binary sequences that have only finitely many $0$'s or have only finitely many $1$'s. The binary expansion gives a bijection from $\Omega\setminus B$ to $(0,1)\setminus A$. Both $A$ and $B$ are countable so it should be straightforward to define a bijection between them, for example, we can enumerate them in parallel like so
\begin{eqnarray}
n &\hspace{25pt} A \hspace{25pt}& B\\
1 & \frac12& 1\color{gray}{00000...}\\
2 & \frac14& 0\color{gray}{11111...}\\
3 & \frac34& 01\color{gray}{0000...}\\
4 & \frac18& 10\color{gray}{1111...}\\
5 & \frac38& 11\color{gray}{0000...}\\
6 & \frac58& 00\color{gray}{1111...}\\
7 & \frac78& 001\color{gray}{000...}\\
8 & \frac1{16}& 110\color{gray}{111...}\\
9 & \frac3{16}& 011\color{gray}{000...}\\
10& \frac5{16}& 100\color{gray}{111...}\\
11&\frac{7}{16}& 101\color{gray}{000...}\\
12&\frac{9}{16}& 010\color{gray}{111...}\\
13&\frac{11}{16}& 111\color{gray}{000...}\\
14&\frac{13}{16}& 000\color{gray}{111...}\\
\vdots&\vdots&\vdots
\end{eqnarray}
Hopefully the pattern is clear at this point.


问:
General proof that the minimum value of $x+y$ from $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$ is $4+11+3+8=26$
I found this problem and the solution on Twitter (translated).

$x$ and $y$ is a (positive) integer that satisfies $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$. If $y$ is the smallest number, what is $x+y$?

The given solution is $\frac{x}{y} = \frac{4+3}{11+8} = \frac{7}{19}$. Therefore, $x+y=7+19=26$.

Is the solution correct? If it is, why is the solution correct? What is the proof that it is correct? And if possible, what is the general formula?

Generally, the solution will be, given $a,b,c,d \in \mathbb{N}$ where $\frac{a}{b} < \frac {c}{d}$, $a$ and $b$ are coprimes, $c$ and $d$ are coprimes, and $a \neq b \neq c \neq d$, the minimum value of $x+y$ where $x,y \in \mathbb{N}$ and $\frac{a}{b} < \frac{x}{y} < \frac{c}{d}$ is $a+b+c+d$.

Edit: it seems the bounds are also unclear. What are the correct bounds for the general statement?

What is the proof that it's right or wrong? Thank you.

Edit: If it's possible, please explain it with only 10th-grade math or lower.

答:
If I were starting out from scratch here, I'd note that:

$\frac{a}{b}<\frac{x}{y}$ means $bx-ay>0$. Since we are working with integers, $bx-ay\geq1$.

$\frac{x}{y}<\frac{c}{d}$ means $-dx+cy>0$. Since we are working with integers, $-dx+cy\geq1$.

So we have the system
$$\left\{
\begin{aligned}
bx-ay&\geq1\\
-dx+cy&\geq1
\end{aligned}
\right.$$

So it is a linear programming question. It means
$$\begin{bmatrix}b&-a\\-d&c\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\in R$$
where $R$ is the closed infinite rectangular region that has $(1,1)$ as a lower left corner.

So applying the matrix inverse,$$\begin{bmatrix}x\\y\end{bmatrix}\in \frac{1}{bc-ad}\begin{bmatrix}c&a\\d&b\end{bmatrix}R$$

Now $R=\{(u,v)|u,v\geq1\}$. So $$\begin{bmatrix}x\\y\end{bmatrix}\in \left\{\frac{1}{bc-ad}\begin{bmatrix}cu+av\\du+bv\end{bmatrix}\middle| u,v\geq1\right\}$$

Each of $a,b,c,d,u,v,bc-ad$ is positive. So the smallest that the second coordinate of the above could be is when $u$ and $v$ are at their smallest, $1$. So you will have the smallest possible value for $y$ when
$$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{bc-ad}\begin{bmatrix}c+a\\d+b\end{bmatrix}$$

BUT we still require that $x$ and $y$ be integers. One way to guarantee this is to require that $bc-ad=1$. Then $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}c+a\\d+b\end{bmatrix}$. And the smallest $y$ is acieved when $y=d+b$ and $x=c+a$.

So it is sufficient for $bc-ad=1$ to make it such that the minimal solution to the inequality $\frac{a}{b}<\frac{x}{y}<\frac{c}{d}$ is $y=d+b$, $x=c+a$.

For example, with $\frac{4}{11}<\frac{x}{y}<\frac{3}{8}$, the minimal solution is $y=19$, $x=7$.

Furthermore if $bc-ad\neq1$ but it happens to divide each of $d+b$ and $c+a$, that would make the minimal solution be $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$.

For example, with $\frac{3}{7}<\frac{x}{y}<\frac{5}{11}$, the minimal solution is $y=\frac{11+7}{2}=9$, $x=\frac{5+3}{2}=4$. (The fraction is $\frac{4}{9}$.)

If $bc-ad$ does not divide either $d+b$ or $c+a$, then $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$ give the minimal real solutions to the inequality, but at least one of them is not an integer. Instead, we would have to look for values of $(u,v)$ not both equal to $1$ such that $bc-ad$ divides both $du+bv$ and $cu+av$. And it's not immediately clear which values of $u$ and $v$ would lead to the minimal solution.


问:
Solving a recurrence relation, $a_n = \sqrt{n(n+1)}a_{n-1} + n!(n+1)^{3/2}$
I'm trying to solve the following recurrence relation, but I have a problem with the factorial part. I would like to evaluate its particular solution.
I would like also to suggest a textbook for solving these kinds of questions. We have:

$$a_n = \sqrt{n(n+1)}a_{n-1} + n!(n+1)^{\frac{3}{2}}$$

答:
OK, here's my solution. Note you didn't give any boundary values. Your equation can be rewritten as
$$
\frac{a_n}{n!}=\frac{a_{n-1}\sqrt{n(n+1)}}{n!}+(n+1)^{\frac{3}{2}}
$$
Denote $b_n=\frac{a_n}{n!}$:
$$
b_n=b_{n-1}\sqrt{\frac{n+1}{n}}+(n+1)^{\frac{3}{2}}\\
\frac{b_n}{\sqrt{n+1}}=\frac{b_{n-1}}{\sqrt{n}}+(n+1)^3
$$
Denote $s_n=\frac{b_{n}}{\sqrt{n+1}}$:
$$
s_n=s_{n-1}+(n+1)^3=\ldots=s_0+S_n
$$
here
$$
S_n=\sum_{k=1}^{n+1}k^3=O(n^4)
$$
I don't give the calculation of this expression, as you can easily find it by methods such as perturbation, induction, etc. Hence your expression is
$$
a_n=n!\sqrt{n+1}(a_0+S_n)
$$</p>


问:
Images of homotopic paths are homeomorphic?
Let $X$ be a space and $f,g:I\to X$ be two paths that are homotopic to each other. Denote $\tilde{f}$ and $\tilde{g}$ to be the images of $f,g$ in $X$. Is it true that $\tilde{f}$ is homeomorphic to $\tilde{g}$ under their subspace topologies induced from $X$? Intuition suggests so, but I'm not sure of how to prove this. Thanks!

答: This is not true. For instance, let $X = \mathbb R$. Consider $f,g:I\longrightarrow \mathbb R$ via $f(x)=x$ ang $g(x)=0$. Then $f$ and $g$ are homotopic via $H(x,t)= tx$. However, the image of $g$ is a single point while the image of $f$ is $[0,1]$. These are certainly not homeomorphic.


问:
Partial derivatives of multivariable parametric function
If
$f\left( \vec{v} \left( s,t,r\right)  \right)  =p\left( x,y,z\right)  \vec{i} +q\left( x,y,z\right)  \vec{j} +u\left( x,y,z\right)  \vec{k}$

$x=x\left( s,t,r\right)$

$y=y\left( s,t,r\right)$

$z=z\left( s,t,r\right)$

We have a multivariable vector-valued function (takes inputs as vectors too) where:

The input space (parameters) is a 3D in $\left( s,t,r\right)$

The intermediate space is a 3D in $\left( x,y,z\right)$

The output space is a 3D in $\left( p,q,u\right)$

I wanna get the derivative of f ... I thought of the following as the
Partial derivative of f() w.r.t (s)

$ \frac{\partial f}{\partial s} =  \left[ \begin{matrix}\frac{\partial p}{\partial x} &\frac{\partial p}{\partial y} &\frac{\partial p}{\partial z} \\ \frac{\partial q}{\partial x} &\frac{\partial q}{\partial y} &\frac{\partial q}{\partial z} \\ \frac{\partial u}{\partial x} &\frac{\partial u}{\partial y} &\frac{\partial u}{\partial z} \end{matrix} \right]\  .\left[ \begin{matrix}\frac{\partial x}{\partial s} \\ \frac{\partial y}{\partial s} \\ \frac{\partial z}{\partial s} \end{matrix} \right]  $

3x3 matrix multiplied by a vector (3x1) and the  result is a 3 element vector (3x1)

$\frac{\partial f}{\partial s} =\left[ \begin{matrix}\frac{\partial p}{\partial x} .\frac{\partial x}{\partial s} +\frac{\partial p}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial p}{\partial z} .\frac{\partial z}{\partial s} \\ \frac{\partial q}{\partial x} \cdot \frac{\partial x}{\partial s} +\frac{\partial q}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial q}{\partial z} .\frac{\partial z}{\partial s} \\ \frac{\partial u}{\partial x} .\frac{\partial x}{\partial s} +\frac{\partial u}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial u}{\partial z} .\frac{\partial z}{\partial s} \end{matrix} \right]  $

Is that correct?
And can we similarly find $ \frac{\partial f}{\partial t}$ and $ \frac{\partial f}{\partial r}$ ?

Is there a compact form for the rule?

Can we use the following notation?

$\frac{\partial f}{\partial s} =\triangledown f.\frac{\partial }{\partial s} \left[ \begin{matrix}x\\ y\\ z\end{matrix} \right]  =\left[ \begin{matrix}p\\ q\\ u\end{matrix} \right]  \left[ \begin{matrix}\frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \end{matrix} \right]  .\frac{\partial }{\partial s} \left[ \begin{matrix}x\\ y\\ z\end{matrix} \right] $

As in simpler forms:

if

$f\left( \vec{v} \left( t\right)  \right)  =f\left( x,y\right)$

$\vec{v} \left( t\right)  =\left[ \begin{matrix}x\left( t\right)  \\ y\left( t\right)  \end{matrix} \right]  $

Then

$\frac{d}{dt} f\left( \vec{v} \left( t\right)  \right)  =\triangledown f.\frac{d}{dt} \vec{v} \left( t\right) = \left[ \begin{matrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{matrix} \right]  .\left[ \begin{matrix}\frac{dx}{dt} \\ \frac{dy}{dt} \end{matrix} \right] =\frac{\partial f}{\partial x} .\frac{dx}{dt} +\frac{\partial f}{\partial y} .\frac{dy}{dt}   $

答:
Yes, this is just the general form of the "chain rule". It can be written in a slightly more compact notation. You can put all the derivatives into one Jacobian matrix:

$$ D(f \circ \vec{v}) = Df \cdot D \vec{v} =
 \begin{pmatrix} \frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} & \frac{\partial p}{\partial z} \\
\frac{\partial q}{\partial x} & \frac{\partial q}{\partial y} & \frac{\partial q}{\partial z} \\
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}
\end{pmatrix}
\begin {pmatrix}
\frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} & \frac{\partial x}{\partial r} \\
\frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} & \frac{\partial y}{\partial r} \\
\frac{\partial z}{\partial s} & \frac{\partial z}{\partial t} & \frac{\partial z}{\partial r}
\end {pmatrix} =
\begin {pmatrix}
\frac{\partial p}{\partial s} & \frac{\partial p}{\partial t} & \frac{\partial p}{\partial r} \\
\frac{\partial q}{\partial s} & \frac{\partial q}{\partial t} & \frac{\partial q}{\partial r} \\
\frac{\partial u}{\partial s} & \frac{\partial u}{\partial t} & \frac{\partial u}{\partial r}
\end {pmatrix}
$$

What you wrote is just looking at one column of $D \vec{v}$ at a time.


问:
Show that singularities of a specific complex function are poles.
Consider the function $g:\mathbb{C}\longrightarrow\mathbb{C}$ defined by $$g(z):=\dfrac{1}{2}\Bigg(\dfrac{1}{1-e^{z(i-1)}}+\dfrac{1}{1-e^{z(-1-i)}}\Bigg)=\dfrac{1}{2}\Bigg(\dfrac{1-e^{z(-1-i)}+1-e^{z(i-1)}}{(1-e^{z(i-1)})(1-e^{z(-1-i)})}\Bigg).$$

It has singularities at $$z_{k}=(1-i)k\pi\ \ \text{and}\ \ \omega_{k}=(1+i)k\pi\ \ \text{for}\ \ k\in\mathbb{Z}.$$

I want to classify these singularities. What I can tell so far is the following:

Note that for $k=0$, we have a double zeros in the denominator, but we have a simple zero in the numerator, so even though we have double zeros, the singularity for $k=0$, is still of order 1. And clearly there is no way to remove it.

For all other $k$, we can only have one component of the product in the denominator to be $0$ at one time, and $z_{k}$ and $\omega_{k}$ for each $k$, cannot individually make the numerator $0$, and thus for all other $z_{k}$ and $\omega_{k}$, they are simple singularities that cannot be removed.

I want to say more than these. If they are not removable, are they poles or essential singularities? I can prove neither.

So I expect that they are poles, but I am having a hard to to write $g(z)$ in the form of $$g(z)=(z-z_{k})^{-1}h(z)$$ for some $h(z)$ holomorphic non-vanishing in a neighborhood of $z_{k}$. This is difficult since we don't have things like $z^{4}+1$ so that we can factorize.

For $z=0$, for sure I can use power series expansion of $e^{z(i-1)}$ and $e^{z(-1-i)}$ around $0$, and conclude what I want. For all other $z_{k}$, it seems really time-consuming to do the power series around them.

I don't have any idea about how to prove them are essential either..(I don't believe they are essential actually)

Any idea? Thank you so much!

答:
For $z_{k}\neq 0$, we use L'Hopital's Rule to get
\begin{align*}
\lim_{z\rightarrow z_{k}}(z-z_{k})g(z)&=\dfrac{1}{2}\lim_{z\rightarrow z_{k}}\dfrac{(z-z_{k})(2-e^{z(-1-i)}-e^{z(i-1)})}{(1-e^{z(i-1)})(1-e^{z(-1-i)})}\\
&=\dfrac{1}{2}\lim_{z\rightarrow z_{k}}\dfrac{(2-e^{z(-1-i)}-e^{z(i-1)})+(z-z_{k})((1+i)e^{z(-1-i)}-(i-1)e^{z(i-1)})}{-(i-1)e^{z(i-1)}(1-e^{z(-1-i)})+(1-e^{z(i-1)})(1+i)}\\
&=\dfrac{1}{2}\dfrac{2-e^{z_{k}(-1-i)}}{(1+i)}<\infty.
\end{align*}

A similary computation shows that $$\lim_{z\rightarrow\omega_{k}}(z-z_{k})g(z)<\infty,$$ so all $z_{k},\omega_{k}\neq 0$ are simple poles.

A similar computation shows that $$\lim_{z\rightarrow z_{k}}(z-z_{k})g(z)=\dfrac{1}{(1+i)},$$ similar for $\omega_{k}$.

So all the $z_{k}$ and $\omega_{k}$ are simple poles.


问:
Weak Convergence - similar to continuous mapping theorem
I recently revisited the following textbook exercise (Probability and Measures by Billingsley, Problem 25.8 in the third edition):

Let $X_n$ and $X$ be real-valued random variables and $h_n, h: \mathbb{R} \to \mathbb{R}$ be Borel functions. Let $E$ be the set of $x \in \mathbb{R}$ such that $h_n(x_n) \to h(x)$ fails for some sequence $x_n \to x$. Suppose that $X_n$ converges to $X$ weakly (a.k.a. in distribution), denoted by $X_n \Rightarrow X$, and that $E$ is measurable with $P(X\in E) = 0$. Show that $h_n(X_n) \Rightarrow h(X)$.

This exercise can be solved quite easily using Skorohad's theorem: on some probability space, there exist real-valued random variables $Y_n$ and $Y$ with the same distribution as $X_n$ and $X$ satisfying $Y(\omega) \to Y(\omega)$ pointwise.

But now I want to prove it without using Skorohad's theorem. I wonder if I can do so by using the common characterization in Portmanteau theorem: $\limsup_n \mu(C) \to \mu(C)$ for any closed set $C$ if and only if $\mu_n \Rightarrow \mu$.

My efforts:

Let $\mu_n:= P \circ X_{n}^{-1}$ and $\mu:= P \circ X^{-1}$ denote the probability measures induced by $X_n$ and $X$. Also, we let $\nu_n := P \circ X_n^{-1} \circ h_n^{-1}$ and $\nu:= P \circ X^{-1} \circ h^{-1}$ denote the probability measures induced by $h_n(X_n)$ and $h(X)$. The goal is to show $\nu_n \Rightarrow \nu$.

Let $C$ be an arbitrary closed subset of $\mathbb{R}$. Then we have
$$
  \limsup_n \nu_n(C) = \limsup_n \mu_n (h_n^{-1}(C) ) \leq \limsup_n \mu_n( \overline{ h_n^{-1}(C)} ),
 $$
where $\overline{ h_n^{-1}(C)}$ denote the closure of $h_n^{-1}(C)$. I would like to use Portmanteau theorem on $\mu_n \Rightarrow \mu$, but the set $\overline{ h_n^{-1}(C)}$ also depends on $n$. Is there a way to bypass this issue?

答:
Here is a proof using Portmanteau. It is adapted from Billingsley's Convergence of Probability Measures.

Claim: Let $G$ be an open set. Then $$h^{-1}(G) = E \cup \bigcup_k \operatorname{int}(T_k)$$ where $\operatorname{int}(A)$ denotes the interior of set $A$ and we define
$$T_k:=\bigcap_{i\ge k}\left\{ x:h_i(x)\in G\right\}.\tag1$$

Proof: Suppose $x\in h^{-1}(G)$, i.e., $h(x)\in G$. Since $G$ is open, there exists $\epsilon>0$ such that $$\text{$y\in G$ whenever $|y-h(x)|<\epsilon$}.\tag2$$ Suppose also that $x$ is not in $E$. By definition of $E$, this means
$$
\text{$\exists k\exists\delta>0$ such that $|h_i(x')-h(x)|<\epsilon$ whenever $i\ge k$ and $|x'-x|<\delta$.}\tag3
$$
By (2), this implies
$$
\text{$\exists k\exists\delta>0$ such that $h_i(x')\in G$ whenever $i\ge k$ and $|x'-x|<\delta$.}
$$ Equivalently,
$$
\text{$\exists k\exists\delta>0$ such that $x'\in T_k$ whenever $|x'-x|<\delta$,}$$ which is to say there exists $k$ such that $x$ is an interior point of $T_k$.

To finish the proof, let $G$ be open. We show $\liminf P(h_n(X_n)\in G)\ge P(h(X)\in G)$. By the Claim,
$$P(h(X)\in G)\le P(X\in E) + P\left(\bigcup_k \left\{X\in \operatorname{int} (T_k)\right\}\right).$$
The first term on the RHS is zero, while $$P\left(\bigcup_k \left\{X\in \operatorname{int} (T_k)\right\}\right)=\lim_k P(X\in \operatorname{int}(T_k))\tag a$$ by continuity of $P$ from below (note the $\operatorname{int}(T_k)$ are increasing). Let $\epsilon>0$. By (a), there exists a (large) $k$ for which
$$P(h(X)\in G)\le P(X\in \operatorname{int}(T_k)) +\epsilon.\tag b$$
Portmanteau implies $$P(X\in \operatorname{int}(T_k))\le\liminf _nP(X_n\in\operatorname{int}(T_k)).\tag c$$
But $\operatorname{int}(T_k)\subset h_n^{-1}(G)$ for all large $n$, by definition (1) of $T_k$. Putting it all together we've shown
$$P(h(X)\in G)\le \liminf P(h_n(X_n)\in G)+\epsilon.$$ Since $\epsilon>0$ is arbitrary, this completes the proof.


问:
Question about VC-dimension
Because I'm not used to the theorem of VC-dimension, I don't know why these statements hold.

Why axis parallel square can shatter a set of three points?

and can't shatter any set of four points?

答:
If you have $3$ points that form triangle, it should be clear that you can enclose every point with an square, you just have to make it as small as you need and it is also clear that there must happen that $x_1\leq x_2<x_3$ with either $x_1<x_2$ or $x_2<x_3$(because they are collinear) so you can separate two points.

The problem with $4$ points is that if they are collinear, and you order them somehow in that line say by $x_1<x_2<x_3<x_4,$ then is not possible to have $x_1,x_3$ separated from $x_2,x_4.$ Because $x_2$ has to lie in any square containing $x_1,x_3.$

Title: Question about VC-dimension


问:
In which order should I read Pete Clark's general topology or what is the prerequisite for Pete Clark's general topology?
Recently,I start to read Prof. Pete Clark's lecture notes on general topology,and first I read the appendix,however I find it very hard for me to learn. There are some objects with no definition in the appendix.In fact,I am a self learner,I have only learned calculus and linear algebra,and know some basic real analysis objects,such as countable sets,uncountable sets,mappings,etc.Many people tell me that if you have knowledge of calculus,you can start to learn general topology.However,for now,I think that Pete Clark's general topology is hard for me.</p>

答:
As I write this answer now (December 2020) I am coming up on the end of the second time I have taught general topology; the first was in Spring 2015.  When I taught the course the first time I already had my notes on convergence (i.e., nets and filters) but that was it --
in the first half of the course I was largely following Kaplansky's Set Theory and Metric Spaces and in the second half of the course I was trying to cover as much of what I think of as the "standard stuff" (my notions for what this is are probably formed from Munkres's Topology more than anything else) on connectedness, compactness, separation axioms and so forth before getting to the big theorems at the end: Tietze Extension Theorem/Urysohn's Lemma, Tychonoff's Theorem and applications to Urysohn's Embedding and Metrization Theorems.  Much of the rest of the textbook came from my own notes on these lectures.  Then I decided that I wanted to enrich the text by including lots of things on metric spaces, so over the years I went back to add more material there.

The next thing to say is that I regard my notes as being very much incomplete as a general topology text.  There is on the other hand enough there to read through, and in this run through of the semester course I covered perhaps 2/3 of the material in the text.

In the course of teaching the 2020 course I had the occasion to add to the notes a bit: the latest copy is available here.  From my perspective the most significant addition is a discussion of Arrow's Impossibility Theorem and its infinite (non?-)analogue using ultrafilters.  What can I say -- at the end of 2020 this material felt timely.  The introduction (which, appropriately, is also unfinished) contains some indications of what I wish would be there -- for me personally the lack of discussion of the Stone-Cech compactification is the most outstanding.  Also, in the (at long last!) published version of my instructor's guide to real induction I have a fairly systematic discussion of order topologies, and for some reason most of this does not appear in the text yet.  When I taught the course this time around I did briefly introduce the topology on an ordered space before going on to metric spaces and I think this might be a good idea but my thoughts about the pedagogical merits of this approach are still a little fuzzy.

And now I will address your question directly.  The course that these notes accompany is an undergraduate course, but it is the most advanced undergraduate course that I have ever taught (at UGA or elsewhere).  This course is for students in transition from a study of undergraduate level mathematics to a study of graduate level mathematics.  It is in order to facilitate this transition that I start with the topology of the real line (Real Induction is, more than anything else, an excuse to begin by thinking deeply about intervals on the real line, whereas a more standard approach to this would seem "unfresh" to many students and maybe not get their full attention) and then transition to and linger for quite a while on metric spaces, which again the majority of my students have seen in a previous course, but not in much depth.

So you get from this that the majority of my students -- and I think even more of the students who really engage with the course material -- certainly do have a background beyond calculus.  All of the students have had a course on sequences and series of real numbers and of real functions, and the majority have had an undergraduate course in real analysis at approximately the level of (some initial segment of) Rudin's Principles of Mathematical Analysis.  Also students are expected to have prior exposure to reading, understanding and writing proofs (there is a prior course on precisely this at UGA) and the course is "fully theoretical" in the sense that the material is definitions, theorems and proofs and whatever else helps to motivate and aid understanding of them, and most of the exercises are proofs.

In terms of self study, I would say that a student who just finished an undergraduate math major (or equivalent content) but hasn't had a general topology course yet would be at the right level to read these notes.

Finally, in terms of the set theory: yes, I agree with Brian M. Scott.  No one puts something in an appendix that they want you to read first!  I think the title "Very Basic Set Theory" is accurate in the sense that set theory is an entire mathematical field and the material here is something like the first 3-4 weeks of material in that field.  On the other hand, nowadays very few working mathematicians know more set theory than this, and a lot of us can get away with knowing even less.  (But it is nice to know these 3-4 weeks.  I have a paper on the border of commutative algebra and elliptic curve theory where the main result is proven using transfinite induction.  I was so excited to prove a real theorem using transfinite induction that I bounded giddily around my department for a while until I ran into a topologist colleague of mine and told them the news.  Their reply: "That's great!  [pause] What's transfinite induction?")

I forgot to mention this before, but the set theory notes were also written much earlier and not for this particular course -- I just figured it couldn't hurt to lump them in.  In fact the basic narrative of the course notes requires only the trichotomy finite / countably infinite / uncountably infinite.  Cardinal numbers appear only in corners, and ordinal numbers -- which do have applications in topology -- appear as pointers to those who care but are not really developed.  (This might change a bit: in an exam that I wrote today there is an optional problem introducing the long line.  If I write up solutions to that problem, I might as well stick it in the text!)

I hope these remarks are helpful.  By the way, anyone is more than welcome to contact me with questions or comments about the notes.  Unfortunately I mostly have time to make changes when I am teaching the corresponding course -- I believe that someone wrote me years ago with pages and pages of (good!) comments, and I implemented the first 20 or so of them before I had to get back to what I was doing at the time.  But I would be very interested to hear about what people like or don't like or want to see in the notes.  After all these years I am still really excited that if I post a couple hundred pages of mathematics on the internet, some people will actually read it!


问:
Critical points in multivariable calc
Find the critical points of

$z = x^{3} + 3xy^{2} - 3x^{2} - 3y^{2} + 7$

I understand if it was $f(x,y)$ but this z is really throwing me off..

I could take the partial derivs of x and y, but if I take the partial of z I get 0=0?

EDIT after comment. Ok, so I will take the partial derivatives.

$Fx = 3x^{2} + 3y^{2} - 6x$

|

$Fy = 6xy - 6y$

Where would I go from here in obtaining the critical points.. solve for x and y @ 0?

答:
Don't let the $z$ confuse you. It's in the same spirit as denoting $y=f(x)$ for some function of one variable. In single variable calculus, if we have some polynomial like
$$
y=f(x)= x^3 - 2x + 3
$$
then we can either write
$$
f'(x) = 3x^2 - 2
$$
or
$$
\frac{dy}{dx} = 3x^2 - 2.
$$
Likewise, if we want to take the partial derivatives of $z=f(x, y) = x^3 + 3xy^2 - 3x^2 - 3y^2 +7$ with respect to $x$ and $y$, we can write
$$
f_x(x, y) = \frac{\partial z}{\partial x} = 3x^2 + 3y^2 - 6x
$$
$$
f_y(x, y) = \frac{\partial z}{\partial y} = -6y + 6xy.
$$
To find the critical points of $f$, simply find the points $(x, y)$ where $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0$.</p>

Title: Critical points in multivariable calc


问:
Prove that there is a unit circle intersecting all n lines
Suppose there are n lines in the plane such that for any three of them there is a
unit circle intersecting each of them. Prove that there is a unit circle intersecting all n lines.

How I have imagined them pictorially is that there will be some straight lines passing through the circle and some forming tangents. At first, I thought its an example of Helly's theorem but I read about Jung's theorem. And I thought maybe its an example of it. Moreover, to apply Helly's theorem we need to have convex sets which are nowhere mentioned explicitly in the question. This is what I thought. And I found the answer of it here prove that all line can be intersected with a unit circle where someone has asked the similar question and someone pointed out for n=2 it is false for parallel lines.

And I am still not able to understand it. I will be thankful if someone tells me.

答:
The second part is a separate proof from Helly's theorem. It looks like Dave wrote this proof out first, then later realized Helly's theorem was applicable at the indicated point and added the edit.

For his own idea, Dave needs a pair of lines that are not parallel to each other. But of course, this isn't guaranteed, so he breaks the proof into two cases: Either

All $n$ lines are parallel to each other, or

There are at least two lines that intersect.

Then he examines the first case: If all the lines are parallel, then they have to be in order, with two outside lines and all the others inbetween. If you take the two outside lines and pick any of the middle lines (this is where the proof fails for the $n = 2$ case, as mentioned in the comments), then you have a triple of lines, and per the hypothesis, there is some circle intersecting all three. But the only way the circle can intersect both outside lines is to pass through all the lines in the middle. So it intersects every line. The parallel case is proved.

Next he proves the second case: Two of the lines $\ell_1$ and $\ell_2$ intersect. The cylinders of radius $1$ about them, $C_1$ and $C_2$ therefore intersect to form a parallelogram (actually, a rhombus). This is a closed polygon. Number the cylinders about the remaining lines as $3$ and on in some arbitrary order.

Now we note a couple things about the Sets $P_k = \bigcap_{i=1}^k C_i$, for any $k \ge 2$. Because the intersection of convex sets is also convex, each $P_k$ is convex. Further, if $P_{k-1}$ is a polygon and $P_k \ne \varnothing$, then $P_k = C_k \cap P_{k-1}$ is also a polygon. This is because the intersection just cuts off from $P_{k-1}$ the portions that extend beyond $C_k$'s two boundary lines. And since $P_2$ is a polygon, by induction, so are all the $P_k$ for $k > 2$.

Now suppose that $\bigcap_{k=1}^n C_k = \varnothing$. Then there has to be some largest $p < n$ such that $P_p = \bigcap_{k=1}^p C_k \ne \varnothing$. Now since $p$ is the largest possible, $P_p\cap C_{p+1} = \varnothing$, and in particular $P_p$ does not intersect the line $\ell_{p+1}$. Dave uses an optimization argument to prove the intuitively obvious fact that the closest point of the solid polygon $P_p$ to $\ell_{p+1}$ will be some vertex $v$ of $P_p$. Label the two boundary segments meeting at $v$ by $b_i$ and $b_j$. Because of how the polygon $P_p$ is formed, these segments are parts of the boundaries of two cylinders $C_i, C_j$ for some $i, j \le p$. Now $b_i$ and $b_j$ are at best parallel to $\ell_{p+1}$, or else they slant away from $\ell_{p+1}$ as you move away from $v$. Otherwise $v$ would not be the closest point of $P_p$ to $\ell_{p+1}$.

But the intersection $C_i\cap C_j$ is a rhombus lying on the same side of $b_i$ and $b_j$ as its subset $P_p$. Which means that $v$ is also the closest point of $C_i\cap C_j$ to $\ell_{p+1}$. But $v \notin C_{p+1}$, so it is a distance of more than $1$ away from $\ell_{p+1}$. Thus the same must be true of every point of $C_i\cap C_j$ to $\ell_{p+1}$. In particular, $C_i\cap C_j \cap C_{p+1} = \varnothing$, which means there can be no circle intersecting all three of $\ell_i, \ell_j, \ell_{p+1}$, a contradiction.


问:
Proving infinite series involving the zeta-function
How do I prove that
$$\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n(2n+1)} = \ln\frac{2\pi}{e}$$
and $$\sum_{n=1}^∞ \frac{\zeta(2n)}{n(2n+1)4^n} = \ln\frac{\pi}{e}$$

I tried to use bernoulli numbers but failed. Could someone please help? Thanks a lot

答:
Let, for $\displaystyle x\in [0,1), \varphi(x)=\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{n(2n+1)}x^{2n+1}$, then
$$ \begin{aligned} \varphi''(x)&=2\sum_{n=1}^{+\infty}\zeta(2n)x^{2n-1} \\
&=\frac{2}{x}\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty}\frac{x^{2n}}{k^{2n}} \\
&=\frac{2}{x}\sum_{k=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{x^{2n}}{k^{2n}} \\
&=\frac{2}{x}\sum_{k=1}^{+\infty}\frac{\frac{x^2}{k^2}}{1-\frac{x^2}{k^2}} \\
&=2x\sum_{k=1}^{+\infty}\frac{1}{k^2-x^2} \\
&=\frac{1}{x}-\pi\cot(\pi x)
\end{aligned} $$
But $\displaystyle \int \cot(\pi x)dx=\frac{\log\sin(\pi x)}{\pi}+c$ thus
$$ \varphi'(x)=\log(x)-\log(\sin(\pi x))+c $$
Since $\varphi'(0)=0$ and $\log(\sin(\pi x))=\log(\pi x+o(x))=\log(\pi x)+\log(1+o(1))=\log(\pi x)+o(1)$, we have $c=\log\pi$ and thus, using $\varphi(0)=0$, we get
$$ \varphi(x)=x\log(x)+\log\left(\frac{\pi}{e}\right)x-\int_0^x\log(\sin(\pi x))dx $$
It only remains to show that
$$ \int_0^1\log(\sin(\pi x))dx=-\log 2 \text{ and } \int_0^{\frac{1}{2}}\log(\sin(\pi x))dx=-\frac{\log 2}{2} $$
However using the identity $\sin(\pi-x)=\sin(x)$, the substitution $x\mapsto 1-x$ gives that
$$ \int_{\frac{1}{2}}^1\log(\sin(\pi x))dx=\int_0^{\frac{1}{2}}\log(\sin(\pi x))dx $$
Let $\displaystyle I=\int_0^{\frac{1}{2}}\log(\sin(\pi x))dx$, then because of what said above, $\displaystyle \int_0^1\log(\sin(\pi x))dx=2I$, thus it only remains to show that $I=-\frac{\log 2}{2}$. Let $\displaystyle J=\int_0^{\frac{1}{2}}\log(\cos(\pi x))dx$, the substitution $x\mapsto\frac{1}{2}-x$ leads to $I=J$, thus
$$ 2I=I+J=\int_0^{\frac{1}{2}}\log(\sin(\pi x)\cos(\pi x))dx=\int_0^{\frac{1}{2}}\log(\sin(2\pi x))dx-\frac{\log 2}{2} $$
Substituting $x\mapsto 2x$ gives
$$ \int_0^{\frac{1}{2}}\log(\sin(2\pi x))dx=\frac{1}{2}\int_0^1 \log(\sin(\pi x))dx=I $$
because of what said above. Thus $2I=I-\frac{\log 2}{2}$ and $I=-\frac{\log 2}{2}$. This ends the proof.


问:
Property for 4 consecutive positive integers?

I derive the answer G by randomly choosing four integers multiple times, which is a bit time-consuming for ACT Math. Could anyone offer me a quicker method to cope with it?

答:
I is false since $24, 25, 26, 27$ are all not prime. To see this, I simply went through the primes in my head $2,3,5,7,11,13,17,19,23,29,\ldots$ until I found two which were more than four apart.

III is false since none of $4,5,6,7$ is a factor of the others. To see this, I simply noted that once $n$ gets "large enough", $n-3$, $n-2$, $n-1$, and $n$ are all bigger than $\tfrac{n}{2}$, which means none of them are small enough to be a factor of the others.

II is true since two of the four consecutive numbers must be even, and therefore have $2$ as a common factor.

Hence, the answer is G.


问:
What are the proper ways to measure the distance between two adjacency matrices of two graphs
Suppose I have two undirected unweighted graphs $G_1$ and $G_2$, which have the same set of nodes. Let $A_1$ and $A_2$ be their (binary) adjacency matrices, respectively.

I wonder what are the proper ways to quantitatively measure the similarity (or distance) between $G_1$ and $G_2$ (or specifically, $A_1$ and $A_2$).

For example, we can reshape $A_1$ and $A_2$ as vectors and measure the $p$-norm distance. I feel that there should be a correspondence between a metric and a physical meaning (e.g., how many edges we need to move to transform $G_1$ to $G_2$).

I wonder if there is some serious discussion (like papers) on measuring distances between graphs in graph theory literature.

答:
Here's a brainstorm: in biological sequence analysis, there is a function called the "edit distance" that solves an analogous problem: e.g., how far apart are any 2 finite-length DNA sequences? This is defined in terms of the minimum number of "edits" required to convert one sequence into the other, using insertions and deletions.

Maybe we could define a type of "edit distance" for graphs, where we allow node and edge insertion/deletion operations, and count the minimum number of "graph edits" required to convert one graph to another?


问:
Is there a rigorous definition of a Young tableau?
In all combinatorics and algebra texts that I have seen so far, the notion of a "Young tableau" is defined in a somewhat informal fashion. The most common approach is stating that a Young tableaux is a "filling" of a Young diagram (or Ferrers diagram, in some texts) with symbols from some (usually totally ordered) alphabet.

In the specific case of standard Young tableaux, the alphabet is the set of numbers from $1$ to $n$, and the entries should increase along the lines and columns.

My question is: is there a way of making this definition rigorous? Specifically, I have a problem with the word "filling" in the above definition. What exactly does that mean, in a formalized way?

(I do realize that a rigorous definition os a Young tableau is not really necessary to work with that notion; I really have no problem following and accepting most standard proofs of theorems involving Young tableaux. I was just curious to know if there is such a formalization.)

答:
Frankly, it's as formal as it needs to be. But if you're just curious for curiosity sake, it is not too difficult to create an ad hoc formal definition to capture our conceptual understanding.

For naturals $n\ge1$, use the standard notation $[n]=\{1,\cdots,n\}$. Encode a Ferrers diagram by an integer partition $(\lambda_1,\cdots,\lambda_k)$ (i.e. a nonincreasing sequence of positive integers). Define a semi-standard tableau to be a function $f$ from $\bigsqcup[\lambda_i]:=\bigcup\big(\{i\}\times[\lambda_i]\big)$ into an ordered set $A$ (where $A$ is usually something like $[m]$ for some fixed $m$, or $[n]$ with $n:=\lambda_1+\cdots+\lambda_k$, or  $\bf N$) such that

Strictly increasing columns: $i<j, c\le \lambda_j\implies f(i,c)<f(j,c)$

Weakly increasing rows: $a\le b\le \lambda_i\implies f(i,a)\le f(i,b)$

This is implicitly the definition used in e.g. the article The Littlewood-Richardson rule, and related combinatorics.


问:
How does one determine what group homomorphisms are injective and which ones are surjective?
I have been tasked to solve both parts of this question.

(a): How many group homomorphisms $\phi :\mathbb{Z}_{45} \to \mathbb{Z}_{27}$ are there?

(b): How many of these $\phi$ are injective and how many are surjective?

For part (a): I recall, that in general, the # of group homomorphisms from $Z_m\to Z_n$ can be found by the $\gcd(m,n)$. So here, I got:
$\gcd(45,27) = 9$.

For part (b) though, I'm actually entirely lost on how to determine which of these $\phi$ are injective and surjective. Is there a general way to find these? and if so, how would one do that with this example?

Thanks in advance. Any help, advice, or direction would be greatly appreciated.

答:
Recall that $$\mathbb{Z}_{45}\cong\mathbb{Z}_{5}\times\mathbb{Z}_{9}$$ via the map $1\mapsto(1,1).$ Therefore it is enough to determine homomorphisms between $\varphi: \mathbb{Z}_{5}\times\mathbb{Z}_{9}\to\mathbb{Z}_{27}.$ Note that $$\varphi((a,b))=\varphi((a,0)+(0,b))=\varphi((a,0))+\varphi((0,b)),\qquad a\in\mathbb{Z}_5, b\in\mathbb{Z}_9.$$ Now, since $\mathbb{Z}_{27}$ has no element whose order is $5,$ under any homomorphism the first factor of $\mathbb{Z}_5$ maps to $0.$ This in fact shows that there are no injective or surjective homomorphism between them.

Then, if you determine the number of ways to map the other factor of $\mathbb{Z}_{9}$ in to $\mathbb{Z}_{27}$  we are done. The image of any homomorphism in to $\mathbb{Z}_{27}$ is a (unique) subgroup of order $1, 3, 9$ or $27.$   By looking at image of the generator we can see that there are three different such maps. Explicitly they are $$(0,1)\mapsto 0\qquad (0,1)\mapsto3,\qquad (0,1)\mapsto9.$$

Added:

Now that I'm thinking, we can directly establishes all group homomorphisms $\psi: \mathbb{Z}_{45}\to\mathbb{Z}_{27}$ by simply observing that $$0=\psi(45)=45\psi(1)\equiv18\psi(1)(\text{mod}27).$$ Then we have only three solutions $\psi(1)=0, 3,$ and $9$ in $\mathbb{Z}_{27}.$ Each of these choices give a group homomorphism.


问:
Why is $dy/dx = y\cos(x)$ a non-linear differential equation?
I thought it would be linear because it can be made in the form $$dy/dx + P(x)y = Q(x).$$
$$dy/dx = y\cos(x),$$
$$dy/dx - y\cos(x) = 0,$$
with $P(x) = \cos(x)$ and $Q(x) = 0$.

Where am I wrong?

Thank you.

答: It is a linear, first order differential equation of the form $\displaystyle y' + y(-\cos x) = 0$ and it is easily solvable by multiplying throughout by the integrating factor $\displaystyle e^{\int (-\cos x)dx}$. Can you proceed?


问:
A set of circles in $\mathbb{R}^2$ is called E. Is $f∶ E → \mathbb{R}^3$, defined as $f(\mathcal{K}) = (x, y, r)$ an surjection / injection?
A set of all open circles (circles without elements lying on the border) in $\mathbb{R}^2$ is called E. A function $f∶ E → \mathbb{R}^3$ is defined as $f(\mathcal{K}) = (x, y, r)$, where $(x, y)$ is the center of the circle $\mathcal{K}$, and $r$ is the radius of the circle $\mathcal{K}$.

Is $f$ "onto" $\mathbb{R}^3$ (is that an surjection)? If not, determine its set of values $R_f$.

Is $f$ an injection? If not, point out a counter-example. If yes, determine $f^{-1}$

Now, I think that $f$ is "onto" $\mathbb{R}^3$. I can't think of any point in $\mathbb{R}^3$ that could be excluded from being covered by those circles.

As for the second one, I am sure that this is an injection. The only possibility to get the same circle is to enter the same information into the function.

However, I don't know how to express that more formally. Am I right in my way of thinking? Could somebody please help me structure my thoughts?

答: $f$ ia not onto. The radius of a ball must be positive, so in particular $(0,0,-1)$ is not in the image. The image of $f$ will be $\mathbb R^2 \times (0, \infty)$. It is, however, injective. Indeed, let $A, B \in E$ such that $f(A)=f(B)$. Let $p$ denote their common center and $r$ their common radius. Then $q \in A$ iff $|p-q| < r$ iff $q \in B$. Hence, $A=B$. Because the map is not onto, the inverse can only be defined from the image $\mathbb R^2 \times (0, \infty)$. For $(p,r) \in \mathbb R^2 \times (0, \infty)$, we define $f^{-1}(p,r)=\{q\in \mathbb R^2 : |p-q| < r\}$, i.e. the open ball of radius $r$ centered at $p$.


问:
Barycentric Coordinates are Continuous Functions
While studying some basic homology stuff, I encountered the idea of Barycentric Coordinates. I am using Elements of Algebraic Topology by James Munkres. On p.g. 4 of his book, Munkres roughly states that barycentric coordinates are continuous functions ($$the\ barycentric\ coordinates\ t_i(x) \ of \ x $$ with respect to $$a_0,..a_n\ are\ continuous\ functions\ of\ x$$). I do not understand how this is working. Some enlightenment on this would be highly appreciated. Thank you.

答:
Notation and definitions will be that of Munkres' Elements of Algebraic Topology.

Throughout, let $x, y$ be points in the $n$-plane spanned by the geometrically independent set $a = \{a_0, \dots, a_n\} \subset \Bbb{R}^N$.

Property (1) on page 5 reads:

The barycentric coordinates $t_i(x)$ of $x$ with respect to $a$ are continuous functions of $x$.

It is already shown in an earlier exercise that indeed the values $t_i(x), i =0..n$ are uniquely determined by $x$, so that we indeed have functions of $x$.

Define the $n$-plane $P$ to be the set of points spanned by the vectors in $a$ or all $z = \sum_{i=0}^n \lambda_i a_i$ such that $\lambda_i \in \Bbb{R}$.  This is a superset of the simplex $\sigma$ which is defined by having the extra condition that $\sum_{i=0}^n t_i = 1$.

Thus it suffices to prove continuity of $t_i : P \to \Bbb{R}$ and then continuity on $\sigma$ follows via restriction of a continuous map which is then in turn continuous.

Which definition of continuity of a function $P \to \Bbb{R}$ should we use, where $P \subset \Bbb{R}^N$ is a vector subspace?

This is the difficult part to answer, since once we have a suitable definition to use, it's just a matter of expanding the definition.

First of all, are the $t_i(x)$ linear functions?  Since then we automatically get continuity.

We know that if $x - y = \sum_{i=0}^n (t_i(x) - t_i(y)) a_i$.  But such coefficients are uniquely determined by $x-y$ and called $t_i(x-y)$.  Thus we have $t_i(x-y) = t_i(x) - t_i(y)$.  That is equivalent to $t_i(x+y) = t_i(x) + t_i(y)$ for all $x,y \in \Bbb{R}^N$.

And $\alpha x = \sum_{i=0}^n \alpha t_i a_i$.   But we know that $\alpha x$ uniquely determines its $t_i$ so that $t_i(\alpha x) = \alpha t_i(x)$ for all $\alpha \in \Bbb{R}$.

Thus we've proven that $t_i$'s are linear maps of $x \in \Bbb{R}^N$.  Every linear map on a finite dimensional vector space is continuous.  See further answers on MSE for proof.

$\blacksquare$


问:
If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$
If $f$ is a function differentiable at $a$ find:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$

I figure that the answer is $\infty$, but I a torn on whether I am correct. Any idea whether I am correct or if I have an issue.

My work is the following:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-(f(a-h^2)-f(a))}{h^2}$ = $\underset{h\rightarrow 0}{\lim} \frac{-1}{h}\frac{f(a-h^2)-f(a)}{h}$= $\infty \cdot f{'}(a)= \infty.$

edit:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-h}{-h}\frac{f(a-h^2)-f(a)}{h}$= $\underset{h\rightarrow 0}{\lim} {-h}\frac{f(a-h^2)-f(a)}{-h^2}$= $0 \cdot f{'}(a)= 0.$

答:
$$ \frac{f(a+h^2)-f(a)}{h}=h\frac{f(a+h^2)-f(a)}{h^2}\underset{h\rightarrow 0}{\longrightarrow}0 $$
Because $\lim\limits_{h\rightarrow 0}\frac{f(a+h^2)-f(a)}{h^2}=f'(a)$.


问:
Contour method to show that $\int_0^\infty\frac{x-\sin x}{x^3} \, dx=\frac\pi4$

Show that $$\int_{0}^{\infty} \frac{x - \sin(x)}{x^3} \, dx = \frac{\pi}{4}$$

My attempt is as follows: Let $$f(z) = \frac{z - i e^{iz}}{z^3}$$ and consider the contour on $[\epsilon, R]$ followed by a semicircular arc in the counter clockwise direction, then on $[-R, -\epsilon]$, then the semicircular clockwise contour avoiding the origin. We have, then, that

$$0 = \int_{\Gamma} f(z) dz = \int_{[\epsilon, R]} f(t) dt + \int_{C_R}f(Re^{it})Rie^{it}dt + \int_{[-R, -\epsilon]}f(t)dt + \int_{C_{\epsilon}}f(\epsilon e^{-it})\epsilon i e^{-it} dt$$

Then the first and third integrals ( $I_1$ and $I_3$) combine so that

$$I_1 + I_3 = 2\int_{\epsilon}^R \frac{t - \sin{t}}{t^3}\,dt$$

Further,

$$|I_{C_R}| \leq \int_0^\pi \left|\frac{Re^{it} - ie^{-R\sin{t}}e^{iRcos{t}}}{R^2 e^{2it}} \right|dt \rightarrow 0 \text{ as } R\rightarrow \infty$$

(I've omitted the details, it isn't too bad to bound)

However, I'm having trouble computing the limit

$$\lim_{\epsilon \rightarrow 0}\int_{C_{\epsilon}} f(\epsilon e^{-it})\epsilon i e^{-it} dt$$

No matter which way I look at it, it seems like this limit does not exist. Perhaps I'm seeing something wrong or have I chosen a bad $f(z)$?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{0 < \epsilon < R}$, I'll perform an integration along a quarter circle -of radius $\ds{R}$- in the complex plane first quadrant which include an indent -an arc of radius $\ds{\epsilon}$- about the origin of coordinates. Namely,
\begin{align}
&\int_{\epsilon}^{R}{\expo{\ic x} - 1 - \ic x \over x^{3}}\,\dd x
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
\left.-\int_{0}^{\pi/2}{\expo{\ic z} - 1 - \ic z \over z^{3}}\,\dd z
\,\right\vert_{z\ =\ R\exp\pars{\ic\theta}}
\\[2mm] - &\
\int_{R}^{\epsilon}{\expo{-y} - 1 + y \over -\ic y^{3}}\,\,\ic\,\dd y -
\int_{\pi/2}^{0}{-\epsilon^{2}\expo{2\ic\theta}/2 \over \epsilon^{3}\expo{3\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta
\end{align}
The RHS first integral vanishes out as $\ds{R \to \infty}$. Then,
\begin{align}
&\int_{\epsilon}^{\infty}{\expo{\ic x} - 1 - \ic x \over x^{3}}\,\dd x
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
-\int_{\epsilon}^{\infty}{\expo{-y} - 1 + y \over y^{3}}\,\dd y -
{1 \over 2}\ic\int_{0}^{\pi/2}\dd\theta
\\[5mm]
&\int_{0}^{\infty}{\sin\pars{x} - x \over x^{3}}\,\dd x =
- {\pi \over 4}\quad\mrm{as}\ \epsilon \to 0^{+}
\\[5mm] &
\bbx{\int_{0}^{\infty}{x - \sin\pars{x} \over x^{3}}\,\dd x =
{\pi \over 4}} \\ &
\end{align}


问:
Compute $\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x} dx$
Compute the integral
$$I_n=\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x} dx$$
where $n$ is any integer.

Using trig identities to evaluate special cases I found answers for $n=0$ and $n=1$.

For $n=0$
$$I_0=\frac{2\pi}{ a \sqrt{a^2 +1}}$$

For $n=1$
$$I_1=2\pi\frac{ 2a( \sqrt{a^2 +1} - a ) -1 }{ a \sqrt{a^2 +1}}$$

How do I find a generic answer for any integer n?

答:
Assume $n\ge0$, $ a>0$ (wlog) and let $t=2x$. Then

$$I_n=\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x } dx
=2\int_0^{2\pi} \frac{\cos(nt )}{{2a^2+1+\cos t}}dt
$$
Apply the Fourier series below and observe that only the term of $k=n$ survives the integration due to $\cos(nt ) $ in the numerator (the constant term does in the case of $n=0$)
$$\frac{1}{{2a^2+1+\cos t}}=
\frac1{2a\sqrt{1+a^2}}\left(1+2\sum_{k=1}^\infty (-1)^k (\sqrt{1+a^2}-a)^{2k}\cos (kx)\right)
$$
As a result, the integral yields
$$I_n= 2\pi (-1)^n \frac{(\sqrt{1+a^2}-a)^{2n}}{a\sqrt{1+a^2}}
$$


问:
Is sequence $x_n=\sum_{i=1}^n \sin(i)$ a Cauchy Sequance?
Is $x_n=\sin(1)+\sin(2)+...+\sin(n)=\sum_{i=1}^n \sin(i)$ a Cauchy sequence?

By defenition $|x_{p+m}-x_p|=|sin(p+1)+...+sin(p+m)|=|cos(p+1/2)-cos(p+m+1/2)|/sin(1/2)$

答:
If $x_n$ is Cauchy then there exists $\epsilon > 0$ and $N$ such that for $m,n > N$ then $|x_n - x_m| < \epsilon$.

But w.l.g. if $n > m$ then $|x_n - x_m| = \sin (m+1) + ... + \sin (n) = \sum_{i=m+1}^n \sin (i) $ which diverges as $n >> m$.


答:
Since $f$ is continuous, $f(a-h^2) - f(a) \rightarrow 0$ as $h \rightarrow 0$, so we have a limit of the indeterminate form $\left[ \frac{0}{0} \right]$.  Applying L'Hopital's rule, we obtain \begin{align*}
\lim_{h \rightarrow 0} \frac{f(a-h^2) - f(a)}{h}
    &\overset{L'H}= \lim_{h \rightarrow 0} -2 h f'(a - h^2)  \\
    &= -2 \cdot 0 \cdot f'(a)  \\
    &= 0  \text{,}
\end{align*}
where in the last, we have used that differentiability of $f$ at $a$ implies continuity on an open set containing $a$.

Body:
If $f$ is a function differentiable at $a$ find:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$

I figure that the answer is $\infty$, but I a torn on whether I am correct. Any idea whether I am correct or if I have an issue.

My work is the following:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-(f(a-h^2)-f(a))}{h^2}$ = $\underset{h\rightarrow 0}{\lim} \frac{-1}{h}\frac{f(a-h^2)-f(a)}{h}$= $\infty \cdot f{'}(a)= \infty.$

edit:
$\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$ = $\underset{h\rightarrow 0}{\lim} \frac{-h}{-h}\frac{f(a-h^2)-f(a)}{h}$= $\underset{h\rightarrow 0}{\lim} {-h}\frac{f(a-h^2)-f(a)}{-h^2}$= $0 \cdot f{'}(a)= 0.$


问:
Proving $\frac{1}{2}(5x+4),\;2 < x,,\;\text{isPrime}(n)\Rightarrow n = 10k+7$
How is it possible to establish proof for the following statement?

$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$

Where $n,x,k$ are $\text{integers}$.

To be more verbose:

I conjecture that;

If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$

How could one prove this?

答:
What is $x$? I assume that it is some natural number: then the conjecture is false. If you let $x=0$, then $n=2$, but $n\equiv 2\text{ (mod }10)\not\equiv 7\text{ (mod 10})$.

For $n$ to be an integer, we must have $x$ even, and since $n\ne 2$, we must have $n$ odd. So in particular, we cannot have $5x+4$ even. This forces $x=4k+2$, hence
$$
\frac{1}{2}(5x+4)=\frac{1}{2}(5(4k+2)+4)=\frac{1}{2}(20k+14)=10k+7
$$


答:
Your answer is correct. You can solve the problem by using L'hospital rule:
$\displaystyle \lim_{h \to 0} \dfrac{f(a-h^2) - f(a)}{h}= \displaystyle \lim_{h \to 0} \dfrac{\frac{df(a-h^2)}{dh}}{1}= \displaystyle \lim_{h \to 0} (-2h\cdot f'(a))=0. $

Title: If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a-h^2)-f(a)}{h}$


问:
Compactness preserved in subspace
Suppose $A,B$ are topological spaces, $B $ is a subspace of $A $ and  $X\subseteq B $.

If $X $ is compact in $ A$ then is $X $ compact in B?

If an open covering of $X$ in $B$ is in fact an open covering in $A$ then by compactness of $A$ we would have a finite sub covering.

答: Yes, and you can prove it directly from the definitions.  You are assuming we have a chain of subspaces $X \subseteq B \subseteq A$.  If $\{O_\beta \mid \beta \in I\}$ is an open cover of $X$ in $B$, then $O_\beta = B \cap U_\beta$ where $U_\beta$ is open in $A$.  Then $\{U_\beta \mid \beta \in I\}$ is an open cover of $X$ in $A$, so this would reduce finitely by compactness.  Now use the same finite number of $U_\beta$'s to create a finite number of $O_\beta$'s derived from the original cover.

Title: Compactness preserved in subspace


问:
What is wrong with this way of solving trig equations?
Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:

$\cos^3 \theta$ = $\cos \theta $

$3\tan^3 \alpha = \tan \alpha$

on the interval $[0; 2 \pi]$.

If I try to solve, for instance, the first equation like this:

$$\cos^3 \theta = \cos \theta $$

$$\cos^2 \theta * \cos \theta = \cos \theta$$

$$\cos^2 \theta  = \cos \theta  \div \cos \theta $$

$$\cos^2 = 1$$

$$\cos \theta = \pm \sqrt {1}$$

$$\cos \theta = \pm 1$$

I end up getting only:

$\theta = 0 ; \pi ; 2 \pi$

But I know that $\pi /2$ and $3\pi /2$ would also make the equation true since $\cos^3 (\pi /2) = cos (\pi /2) = 0$ and $\cos^3 ( 3 \pi /2) = cos (3 \pi /2) = 0$.

The same problem arises when I try to solve the second equation:

$$3\tan^3 \alpha = \tan \alpha$$
$$3\tan^2 \alpha * \tan \alpha = \tan \alpha$$
$$3\tan^2 \alpha = \tan \alpha \div \tan \alpha$$
$$3\tan^2 \alpha =1$$
$$\tan^2 \alpha =1/3$$
$$\tan \alpha =\pm \sqrt{1/3}$$
$$\tan \alpha =\pm 1/\sqrt{3}$$

The values of $\alpha$ that make the tangent equal to $\pm 1/\sqrt{3}$ between 0 and $2 \pi$ are only these:

$ \alpha = \pi / 6 ; 5 \pi /6 ; 7 \pi / 6 ; 11 \pi /6$

However,  I should also find $0$ and $\pi$ because $3\tan^3 (0) = \tan (0) = 3 tan^3 (\pi) = tan ( \pi ) = 0 $.

Something similar happens when I'm looking for the local minima and maxima of this function:
$$ f(x) = \sin^2(x) + \cos(x)$$

on the interval $[0; 2 \pi]$

$f'(x) = 2 \sin(x) \cos(x) - \sin(x)$

$ 0 = 2 \sin(x) \cos(x) - \sin(x)$

$ \sin(x) = 2 \sin(x) \cos(x)$

$ \sin(x) / \sin(x) = 2 \cos(x)$

$1 = 2 \cos(x)$

$1/2 =  \cos(x)$

$x = \pi / 3 ; 5 \pi /3$

And again, plugging $x = 0$ or $x= \pi$ or $x = 2 \pi$ also make the derivative 0.

I've noticed that in all of these 3 cases I have the same trigonometric function on both sides of the equation and I'm dividing both sides by that function. This is making one side of the equation equal to 1 at some point. What is wrong or incomplete with this method? Why am I missing some results when I do this?

答:
Clearly, when $\cos \theta = 0$, then $\cos^3 \theta = \cos \theta$.

So, in the interval $[0, 2\pi]$, we get $\theta \in \left\{\frac 12 \pi, \frac 32 \pi \right\}$

If $\cos \theta \ne 0$, then we can divide both sides by $\cos \theta$...


答:
I) should be rejected outright.  It should be well known that we can have a gap in primes of more than $4$.  There is the excercise that $n! + 2$ through $n! + n$ will be $n-1$ consecutive digits with no primes.  But even if you don't think of that the statement should just seem absurd.  After all no even number over $2$ is prime so that would mean every other odd number is prime and surely that can't be true!  $25$ and $27$ are not prime so $24,25,26,27$ are a counter example.

II) is clear when you realize $2$ is prime and two of the numbers must be even.  (Otherwise $3$ being a common factor of the first and last is the only other possible option but its not guarenteed.)

III) is clearly false.  If two numbers have a common factor, $k$ the must be a multiple of $k$ apart.  As four consecutive numbers are at most $3$ apart $1,2,3$ are the only possible factors they can have in common and if these numbers are larger than $3$ this statement is false.

So II is the only correct one.

Title: Property for 4 consecutive positive integers?

Body: I derive the answer G by randomly choosing four integers multiple times, which is a bit time-consuming for ACT Math. Could anyone offer me a quicker method to cope with it?


答: Let $\bigcup U_{i \in I} \supset X$ be a finite covering of $X$ in $A$.  Since $B$ is a subspace of $A$ then $\tau_B = \{B \bigcap U | U \in \tau\}$.  This proves that there is a finite covering of $X$ in $B$.

Title: Compactness preserved in subspace

Body:
Suppose $A,B$ are topological spaces, $B $ is a subspace of $A $ and  $X\subseteq B $.

If $X $ is compact in $ A$ then is $X $ compact in B?

If an open covering of $X$ in $B$ is in fact an open covering in $A$ then by compactness of $A$ we would have a finite sub covering.