问:
Arrangements of affine hyperplanes
Fix $n>0$ and $X\subseteq\mathbb{R}^n$. Call a function $f:X\longrightarrow \mathbb{R}$ linear if it is of the form
$$
f(\bar{x})=a_1x_1+\ldots+a_nx_n+b
$$
for some $a_i,b\in\mathbb{R}$.

Now suppose we have linear functions $f_1,\ldots,f_t$ and $g_1,\ldots,g_t$ with the following property:

For all $i\in\{1,\ldots,t\}$ and for all $\bar{x}\in X$,

$$
\textrm{there are }j,k\in\{1,\ldots,t\}\textrm{ such that }f_i(\bar{x})\leq g_j(\bar{x})\textrm{ and }f_k(\bar{x})\leq g_i(\bar{x}).
$$

Is it true that there must be some $i,j\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X$?

答:
I had an idea to approach this differently, and after fighting it, took a simpler case where each $f$ or $g$ has only one element that is non zero, in the hope that this is still useful and suggests a path to the general case.

For the case when each $f$ or $g$ has only one element that is non zero (i.e. the planes are 'axis' aligned):

Given an index $p$, up to $n$, we can construct a vector $\bar v$ in $X$ with a 1 in the $p$ position and otherwise $0$s, and following the direction of $\bar v$ there is some point beyond which a function $f$` having the largest negative value in position p becomes (and remains) the minimum function $f$.

This gives us a lower bound for components in the $p$ position for all functions in $g$ - since any $g$ that has a lower value in the $p$ position will eventually fall below and remain below $f$` in the direction of $\bar v$, and thus break an assumption.

Heading in the reversed direction of $\bar v$, we can find a point at which the same $f \prime$ (or some parallel $f\prime\prime$ above it with equal largest negative component in position $p$) becomes and remains the maximum $f$ function. Since $f \prime$ and $f \prime\prime$ share the component in position p we are interested in, we will take the case where $f \prime$ is the upper function, without loss of generality.

Since we know there must be at least one function $g \prime$ which is eventually above $f \prime$ and remains above $f \prime$ in this direction by assumption (since there are a finite number of $g$, eventually at least one $g \prime$ must remain above $f \prime$ to infinity), we know this $g \prime$ must have a value in the p position of not more than that of that of $f \prime$, since otherwise it will eventually fall below $f \prime$.

So now we have a maximum bound for the position p for $g \prime$, and since it is equal to the minimum bound we have above for all g in this position, we have an equality for this position p in (at least one) $g \prime$.

The conclusions then follows since for each position $p$, we determine that the function $f \prime$ which has the minimum value of all $f$ in this position, has some $g \prime$ parallel and $\ge$ to it (possibly also parallel and $\ge$ to some parallel $f \prime\prime$ above it).

Note: I think it can be argued similarly for each position $p$ from both the maximum positive and minimum negative values in $f$, since both will force this situation for some $g$.

I think the underlying idea also works for the general case, but requires taking combinations of multiple positions and allowing for $g$ that are parallel to intersections of planes, and I've not managed to find a clean (and reasonably short) way to attack that yet.

Title: Arrangements of affine hyperplanes


问:
If $f, g$ are smooth functions on $I$, $g$ has no flat point, and $f/g$ has a continuous extension to $I$, is this extension smooth?
Let $f,g$ be smooth functions $I \to \mathbb{R}$. Suppose that the following two conditions are met:

For all $t \in I$, there exists $m \in \mathbb{N}$ such that the $m$-th derivative $g^{(m)}$ is nonzero at $t$. (This implies, in particular, that every zero of $g$ is isolated.)

The function $f/g$ admits a continuous extension to $I$. (This happens when the order of every zero of $g$ is smaller than or equal to the one of $f$.)

Does it then follow that such extension is smooth?

答:
Yes.

Claim: If $g(x_0)=0$, then there is a smooth $\tilde{g}\colon I\to\mathbb{R}$ satisfying $g(x)=(x-x_0)\tilde{g}(x)$ for all $x$.

Proof: The only point need checking is $x_0$.  For any $n\in\mathbb{N}$, Taylor expand $g$ about $x_0$ up to $(x-x_0)^{n+1}$ with remainder term, and divide by $x-x_0$ to get Taylor expansion of $\tilde{g}$ up to degree $n$ with remainder.  QED.

So $f(x)=(x-x_0)^m F(x)$, $g(x)=(x-x_0)^n G(x)$ near $x_0$, $F,G$ smooth and $m\geq n$.  Cancel the $(x-x_0)$s, and you have product and quotient of smooth functions $(f/g)(x)=(x-x_0)^{m-n}F(x)/G(x)$ near $x_0$.

ContentLicense: CC BY-SA 4.0


问:
Boundary condition for derivative of trigonometric behaving function
I want to find a trigometric function for which:

$u(0)=u(1)=0$

and

$\frac{\partial u}{\partial y}(y=0)=\frac{\partial u}{\partial y}(y=1)=0$.

Lets look at e.g. $u=C1 cos(a y+\phi_0)+C2 sin(a y +\phi_1)$.

It is possible to choose a, $\phi_0, \phi_1$, C1 and C2 such that u(0)=u(1)=0, however I can only find the trivial solution in which u=0, to also solve the boundary condtions for the derivatives.

The function u itself should behave like a trigonometric function, but does not necessarily have to be one. So you can also modify the shown function itself. I feel like I am overlooking something very obvious right now, I would appreciate any kind of help.

答:
What you believe to be a combination of trigonometric functions with four independent parameters only has two in fact. Because

$$c_c\cos(ay+\phi_c)+c_s\sin(ay+\phi_s)\\=(c_c\cos\phi_c+c_s\cos\phi_s)\cos(ay)+(-c_c\sin\phi_c+c_s\sin\phi_s)\sin(ay)\\=p\cos(ay)+q\sin(ay)=c\cos(ay+\phi).$$

This is a general sinusoid, which cannot enjoy the requested properties.

A possible answer is

$$u(y)=2\sin^2(\pi y)=1-\cos(2\pi y),$$ if that is "trigonometric" enough.


问:
Solve transcendental equation : $\tanh(\alpha x) =\arctan(x)$, at $0<\alpha-1 \ll 1$ and at $\alpha\gg 1$?
To solve this transcendental equations approximately :

Preivous: $\tanh(\alpha x) =\arctan(x)$, at $0<\alpha-1 \leq 1$ and at $\alpha\geq 1$.

Edit: $\tanh(\alpha x) =\arctan(x)$, at $0<\alpha-1 \ll 1$ and at $\alpha\gg 1$.

I need hint on how to start? I'm thinking to use Newton's Method of Iteration but don't know how?

Body:
We can have an extremely good estimate of the zero of function
$$f(x)=\tanh (a x)-\tan ^{-1}(x)$$
making one iteration of Halley's mathod starting with $x_0=k$.

We have
$$f(k)=\tanh (a k)-\tan ^{-1}(k)$$
$$f'(k)=a\, \text{sech}^2(a k)-\frac{1}{k^2+1}$$
$$f''(k)=\frac{2 k}{\left(k^2+1\right)^2}-2 a^2 \tanh (a k)\, \text{sech}^2(a k)$$
$$x_1=k+\frac{2\, f(k)\, f'(k)}{f(k) \,f''(k)-2 f'(k)^2}$$

After @Lutz Lehmann's comment, choose $k=\tan(1)$.

Some results

$$\left(
\begin{array}{ccc}
 a & \text{estimation }& \text{solution} \\
 2 & 1.5432449040615292748 &  1.5432446279235773240 \\
 3 & 1.5568066556972442438 &  1.5568066556883083119 \\
 4 & 1.5573811263977578542 &  1.5573811263977572906 \\
 5 & 1.5574065442653326711 &  1.5574065442653326711
\end{array}
\right)$$

For small values of $a$
$$\left(
\begin{array}{ccc}
 a & \text{estimation }& \text{solution} \\
 1.1 & 1.1525726521622667697 &  1.1296349644856700163 \\
 1.2 & 1.3059331878896915304 &  1.3018024399310428716 \\
 1.3 & 1.3927625406966437033 &  1.3917949581555723633 \\
 1.4 & 1.4460683663131917454 &  1.4458089619371594747 \\
 1.5 & 1.4804871094885072890 &  1.4804115807485915832 \\
 1.6 & 1.5034696046136527262 &  1.5034463799965303017 \\
 1.7 & 1.5191789474986005266 &  1.5191715316939800960 \\
 1.8 & 1.5300991535713895614 &  1.5300967212675288608 \\
 1.9 & 1.5377849843609131701 &  1.5377841708190835133 \\
 2.0 & 1.5432449040615292748 &  1.5432446279235773240
\end{array}
\right)$$

But, for small values of $a$, a Taylor expansion followed by a series reversion leads to
$$x=1+t+\frac{ \left(4 a^2 \tanh ^3(a)-4 a^2 \tanh (a)+1\right)}{2 \left(-2 a+2 a
   \tanh ^2(a)+1\right)}t^2+O\left(t^3\right)$$ where
$t=\frac{\pi -4 \tanh (a)}{4 a \text{sech}^2(a)-2}$


问:
Inverse Laplace with formula table
In my book I have to do the inverse Laplace transform of:
$$\displaystyle \frac{4 + 15 s}{s(2+5s)} = \frac{15}{5} \frac{\left(s + \frac{4}{15}\right)}{s\left(s + \frac{2}{5}\right)}$$

With a partial fraction division, I can find that one fraction is $\displaystyle\frac{2}{s}$ and the second $\displaystyle\frac{5}{5s+2}$ and I can find $e^{-\frac{2}{5}t} + 2$.

But the book states that I should get it directly by looking at the summary of usual Laplace transforms? I checked the formulas and did not find what I should be able to see.
Can someone help me to navigate the classical formula table given that equation?

Title: Inverse Laplace with formula table

Body:
Here. Formula  $14.$

Write $\frac{15 s+4}{s (5 s+2)}$
as $$\frac{15}{5} \frac{\left(s + \frac{4}{15}\right)}{s\left(s + \frac{2}{5}\right)}$$
$\alpha=\frac{4}{15};\;a=0;\;b=\frac{2}{5}$
$$3\cdot \frac{1}{\frac{2}{5}} \left[\frac{4}{15}e^0-\left(\frac{4}{15}-\frac{2}{5}\right)e^{-2/5t}\right]=\frac{15}{2}\left(\frac{4}{15}+\frac{2}{15}e^{-2/5t}\right)=2+e^{-2/5t}$$

Tbh I prefer your solution :)


问:
Expansion of $(\sinh(x))^{\frac15}$ around 0 for x > 0
I'm aware of the series expansion of the hyperbolic functions, but how does one expand a fractional power of sinus hyperbolicus, i.e. e.g. $(\sinh(x))^{\frac15}$?

Body:
If you want a series expansion with integer power, let $x=t^5$
$$y=\sqrt[5]{\sinh \left(t^5\right)}\implies \log(y)=\frac 15 \log \left(\sinh \left(t^5\right)\right)$$ Now, compose series (one piece at the time)
$$\sinh \left(t^5\right)=t^5+\frac{t^{15}}{6}+\frac{t^{25}}{120}+\frac{t^{35}}{5040}+\frac{t^{45}}{362880}+O\left(t^{55}\right)$$
$$\log \left(\sinh \left(t^5\right)\right)=5 \log
   (t)+\frac{t^{10}}{6}-\frac{t^{20}}{180}+\frac{t^{30}}{2835}-\frac{t^{40}}{37800}+
   O\left(t^{50}\right)$$
$$\log(y)=\log
   (t)+\frac{t^{10}}{30}-\frac{t^{20}}{900}+\frac{t^{30}}{14175}-\frac{t^{40}}{18900
   0}+O\left(t^{50}\right)$$
$$y=e^{\log(y)}=t+\frac{t^{11}}{30}-\frac{t^{21}}{1800}+\frac{t^{31}}{25200}-\frac{131
   t^{41}}{45360000}+O\left(t^{51}\right)$$


问:
What's the property that same inputs to functions result in the same output?
Let $f:X \to Y$ a function.
Then, $$\forall x_1, x_2 \in X.~x_1=x_2 \implies f(x_1) = f(x_2)$$

What would this property be called?

答:
Informally, a function is a rule that assings a number to a specific output. Suppose that $x_1$ and $x_2$ were the same number, but produced a different output. Then, as Arun has mentioned in the comments, we wouldn't even be dealing with a function. For instance, consider the square root function $S:x\mapsto \sqrt{x}$. In order for this function to be well-defined, we have to designate $\sqrt{x}$ as the nonnegative square root of $x$. Otherwise, $S$ would output not a single number, but a set of numbers*:
$$
S(x^2)=\{x,-x \} \, .
$$
The formal definition of a function $f$ is that it is a binary relation between two sets $X$ and $Y$. The graph of a function $G$  refers to the following set
$$
G(f)=\{(x,f(x)) \mid x \in X\} \, .
$$
In order for a function to be well-defined, we require that if $(x,y)$ and $(x,z)$ are members of $G$, then $y$ must be equal to $z$. Hence, all functions must have the property that the same input produces the same output, otherwise we wouldn't be dealing with a function.

*In complex analysis, sometimes we relax the rules a little, and allow a function to have multiple outputs. For instance, we would allow $S$ to associate a number with its two square roots:
$$
S(-1)=\{i,-i\}
$$
This is called a 'multifunction'. It's worth noting, however, that in the strict interpretation of what a function is, a 'multifunction' is not a function at all.


问:
Calculate $E[X_t]$ and Var$(X_t)$
Let $(X_t)_{t \ge 0}$ be the process given by $X_t= e^{-B_t-t/2}$ for all $t\ge 0$.

Calculate $E[X_t]$ and $Var(X_t)$.

For the expectation, one would have

$E[X_t]=E[e^{-B_t}]e^{-t/2}=1$ due to the moment generating function?

My other idea was to consider the Ito form $dX_t= -X_t dB_t $ . So one would get $X_t-1= \int_0^t -X_ss dB_s$. Following that logic one has: $$E[X_t]=E[\int_0^t-X_sdB_s]+1= 1$$

For $Var(X_t)$, its just  $Var(X_t)=E[X_t^2]-1=e^{-t}E[e^{-2B_t}]-1=e^t-1$ ?

答:
There is no need to use Itô's formula here, as everything can be done using the mgf as you mentioned. Your calculation for $\mathbb{E}[X_t]$ is correct. Then, for $Z
 \sim N(0,1)$:
\begin{equation}
\mathbb{E}[X_{t}^{2}] = e^{-t} \mathbb{E}\left[ e^{2 \sqrt{t}Z }\right] = e^{-t}e^{2t} = e^t.
\end{equation}
So $\text{Var}[X_t] = \mathbb{E}[X_{t}^{2}] - \mathbb{E}[X_t] = e^t -  1$ as you mentioned.

It's usually good practice to avoid using Itô's formula when more elementary tools are available. The reason for this is that local martingale terms in the semimartingale decomposition may not be true martingales, and  the most common reason for this is lack of integrability. Hence, sometimes the expectation
\begin{equation}
\mathbb{E}\left[ \int_{0}^{t} X_s dB_s \right]
\end{equation}
may  not exist. In  this case you are fine, since you can check that the stochastic exponential is a true martingale, for example as presented here, so you can readily apply Itô's formula and the Itô isometry. But in more general cases, it is something worth worrying about.

Title: Calculate $E[X_t]$ and Var$(X_t)$


问:
Determining the convex hull of the union of two polyhedra
I'm doing an introductory course to linear programming and I'm working through some exercises to prepare for the final exam, I'm stuck on an exercise and I would really appreciate a hint:

Let $a\in \mathbb{R}^n $ where $\Vert a\Vert = \sqrt{a^T a}=1\,$ and let$\,\,b_1 , b_2 \in \mathbb{R}$ with $b_1 < b_2$. Define $H_1=\{ x\in \mathbb{R}^n : a^Tx=b_1\}$ and $H_2=\{ x\in \mathbb{R}^n : a^Tx=b_2\}$

Determine the convex hull of $H_1 \cup H_2$

The first thing I would do here is to define $C=\{ x\in \mathbb{R}^n : b_1 \leq a^Tx \leq b_2\}$, which is a polyhedron and therefore convex, and so $conv( H_1 \cup H_2) \subset C$.

$conv(S)$ here means the intersection of all convex sets that contain $S$.

I just need to show that $C \subset conv( H_1 \cup H_2)$, but I don't really know where to start.

Thanks in advance!

答: Hint: $H_1$ and $H_2$ are hyperplanes, so what happens if you take any old halfspace $H'= \{x \mid (a')^T x \leq c \}$ for $a\neq a'$? Does $H'$ contain $H_1$ or $H_2$? Why or why not?</p>

ContentLicense: CC BY-SA 3.0


问:
Solving $z^4 + 4i\bar{z} = 0$
How to solve $z^4 + 4i\bar{z} = 0$ (efficiently)?

I managed to compute the radius of $z$:

Denote $z = rcis(\theta)$

Rearranged the equation to $z^4 = -4i\bar{z}$

Taken absolute value on both sides $r^4 = 16r$

And found $r=0$ or $r = \sqrt[3]{16}$

How can we calculate $\theta$ from here?

Or is there a better way to solve this?

答:
Actually, if $z^4=-4i\overline z$, then $r^4=4r$, and therefore $r=\sqrt[3]4$.

If $r=0$, the $z=0$, which is a solution.

And if $r=\sqrt[3]4$, then, if $z=re^{i\theta}$, we have $z^4=4\sqrt[3]4e^{4i\theta}$ and $-4i\overline z=4ie^{-\pi i/2-i\theta}$. So,$$4\theta=-\frac\pi2-\theta+2k\pi,$$for some $k\in\Bbb Z$. Can you take it from here?

Title: Solving $z^4 + 4i\bar{z} = 0$


问:
A question on limit of ratio of exponentials?
I am quite rusty on calculus, when I plot this $f(x) = \frac{2^{x^2}}{L^{x(x-1)/6}}$ on desmos, It seems to me both intuitively and from the plot that the $\lim_{x\rightarrow \infty f(x)}$ will shift between $0$ and $\infty$ depending on $L$. For example check $L=20$ and $L=100$.

I would like to evaluate $L$ for which this transition would happen, I tried with L'Hopital, but I don't think it would work as th exponentials will remain after differentiating too.

答:
Taking the logarithm of your function we get

$\log(2)x^2-\frac{\log(L)}{6}(x^2+x)$

From this we see (the log) of $f$ goes to -infinity if $\log(2)<\frac{\log(L)}{6}$, or in other words, if $L>e^{6\log(2)}=64$. Therefore, $f$ goes to $0$ when $L\ge 64$, and infinity otherwise.


答:
Let $M=L^{1/6}$ and $N=\frac{2}{M}$. Then

$$f(x) = \frac{2^{x^2}}{M^{x^2-x}}  = \frac{2^{x^2}}{M^{x^2}M^{-x}}
=
\left(\frac{2}{M}\right)^{x^2} M^x
=
N^{x^2}M^x
=
e^{x^2\ln(N)+x\ln(M)}$$

Therefore, we only need to consider the limit of the function $g(x)=x^2\ln(N)+x\ln(M)$:

$\ln(N)>0$ (equivalent to $N>1 \iff 2>M \iff L<2^6=64$), then $\lim_{x\to\infty}g(x)=\infty$, and $\lim_{x\to\infty}f(x)=\infty$.

if $\ln(N)<0$ (equivalent to $L>2^6=64$), then $\lim_{x\to\infty}g(x)=-\infty$, and $\lim_{x\to\infty}f(x)=0$

if $\ln(N)=0$ (equivalent to $L=64$), then $M=2$ so $\ln(M)>0$. It follows that $\lim_{x\to\infty}g(x)=\infty$, and $\lim_{x\to\infty}f(x)=\infty$.

Title: A question on limit of ratio of exponentials?

Body:
I am quite rusty on calculus, when I plot this $f(x) = \frac{2^{x^2}}{L^{x(x-1)/6}}$ on desmos, It seems to me both intuitively and from the plot that the $\lim_{x\rightarrow \infty f(x)}$ will shift between $0$ and $\infty$ depending on $L$. For example check $L=20$ and $L=100$.

I would like to evaluate $L$ for which this transition would happen, I tried with L'Hopital, but I don't think it would work as th exponentials will remain after differentiating too.


问:
A question about floating point representation
This question is simple but I don't know the exact answer: My Professors' lecture notes say $(0.01)_2 = 0.1 \times 2^{-1}$ and he insists that all numbers are written in base 2. However, I think it's wrong and I interpret this expression in this way: If we convert $(0.1)_2$ to its corresponding number in base 10 and multiply it by $2^{-1}$ then we get the value of $(0.01)_2$ in base 10 or equivalently if we convert $2^{-1}$ to its corresponding number in base 2 and multiply it by $(0.1)_2$ we get $(0.01)_2$. Am I right? Thanks.

答:
I intepret this as a question about notation, and your professor is right. $(0.01)_2$ or sometimes $0.01_2$ are both common notations for binary expansions, so by definition

$$(0.01)_2 = 2^{-2},$$

Edit: Notice that according to your transcription, your professor’s notation is sloppy. He should indeed write

$$(0.01)_2 = (0.1)_2 \cdot 2^{-1}.$$

More generally for any $\beta > 1$. We have a $\beta$-expansion of a number:

$$(d_m \ldots d_1. d_{-1} \ldots d_{-n})_{\beta} = d_m \beta^m + \cdots d_1 \beta + d_{-1} \beta^{-1} + \cdots d_{-n} \beta^{-n}$$


问:
If $f_n\rightarrow f$ in $L^2$ and $\lVert f_n\rVert_\infty\le C$ then why is $f\in L^\infty$?
This exercise from the $L^p$-theory section of my measure theory course is giving me some trouble.

Let $(f_n)_n\subset L^2\cap L^\infty$ be a sequence such that $f_n\rightarrow f\in L^2$ in $L^2$ and $\lVert f_n\rVert_\infty\le C$ for all $n$. I want to show that

$f\in L^\infty$ and

$f_n\rightarrow f$ in $L^p$ for any $2\le p<\infty$

My most promising attempt at 1. was
$$\lVert f\rVert_\infty=\lVert f-f_n+f_n\rVert_\infty\le \lVert f-f_n\rVert_\infty+C,$$
but how can we bound $\lVert f-f_n\rVert_\infty$?

I'm having similar problems with 2., except of course there we want to bound $\lVert f-f_n\rVert_p$.

答:
Hints :

Use the fact that because $f_n \rightarrow f$ in $L^2$, then there exists a subsequence $f_{\varphi(n)}$ that converge pointwise almost everywhere to $f$.

Use the fact that

$$|f_n-f|^p \leq\left\{
    \begin{array}{ll}
        |f_n-f|^2  & \text{if } |f_n-f|\leq 1\\
        (2C)^p  & \text{if } |f_n-f|^2\geq 1
    \end{array}
\right. $$


问:
Which book to use in conjunction with Munkres' TOPOLOGY, 2nd edition?
Although Topology by James R. Munkres, 2nd edition, is a fairly easy read in itself, I would still like to know if there's any text (or set of notes available online) that is a particularly good choice to serve as an aid to Munkres' book, in case one gets stuck in some place in Munkres or in case one need to suggest some supporting text to one's pupils.

I know that there's a website where solutions to some of Munkres' exercises are also available.

Is the book Introduction to Topology and Modern Analysis by Georg F. Simmons a good choice for this same purpose?

Or, is Introduction to Topology Pure and Applied by Colin Adams a good companion to Munkres?

And, what about the General Topology text in the Schaum's Series?

P.S.:

Thank you so much Math SE community! But I also wanted to ask the following:

Which book(s) are there, if any, that support Topology by James R. Munkres, 2nd edition, in the sense that they cover the same material as does Munkres; prove the same theorems as are proved in Munkres, but filling in the details omitted by Munkres; use the same definitions as used by Munkres; include as solved examples some, most, or all of Munkres' exercise problems?

Of course, one cannot expect a text to fulfill all the above requirements, but which one(s) do(es) this the best?

答:
I would say Topology Without Tears by Sidney A. Morris. The pdf is freely available at:

https://www.topologywithouttears.net/

There is also a playlist of videos from the same author for some supplementary material.


问:
Probability of exactly k runs of wins
How to calculate probability of exactly k run of wins out of exactly m wins and n losses? A run of win means 1 or more successive wins.

My try: Divide the m+n matches to portions

$L_1W_1L_2W_2L_3W_3....L_kW_kL_{k+1}$

here $L_i$ denotes a run of losses, i.e. one or more consecutive loss(es). Similarly $W_i$ denotes a run of wins.

For exactly k runs of wins, $L_2,..L_k$ should have atleast $1$ loss, but $L_1, L_{k+1}$ can have $0$ loss as well. All $W_i$s should have atleast one win.

Now, number of ways to get all different $L_i$s satisfying this criteria, is same as arranging $k$ sticks between $n$ stars, so that none of the sticks are adjacent. This is same as putting (n-k+1) balls into k-1 boxes, with empty boxes allowed= $(k-1)^{n-k+1}$. But each loss is indistinguishable from one another, so total number of ways of forming $L_1,...,L_{k+1}$ is $\frac{(k-1)^{n-k+1}}{n!}$.

Similarly, number of ways to get all different $W_i$s satisfying the criteria, is same as putting m indistinguishable balls in to k boxes, with empty not allowed. Which is same as putting (m-k) indistinguishable balls into k boxes, with empty boxes allowed = $\frac{k^{m-k}}{m!}$

So, total number of ways = $\frac{(k-1)^{n-k+1}.k^{m-k}}{m!n!}$

Probability of exactly k runs of wins = $\frac{(k-1)^{n-k+1}.k^{m-k}}{m!n!(m+n)!}$

But I am not feeling confident with this logic. It will be good if anyone can confirm if it is correct, or contradict if there is any loophole.

答:
There are ${m+n \choose m}$ ways in total of ordering the equally likely outcomes given $m$ wins and $n$ losses.

Easy bits first:

$\mathbb P(K=0 \mid 0,n)=1$ since there are no wins.

$\mathbb P(K=1 \mid m,0)=1$ if $m \ge 1$ since there are only wins, so one run.

$\mathbb P(K=0 \mid m,n)=0$ if $m \ge 1$ since there is at least one winning run.

$\mathbb P(K=k \mid m,n)=0$ if $m \lt k$ or $n \lt k+1$ as $k$ winning runs impossible.

Otherwise with $m,n\ge1$ and $k\ge 2$, then of these:

${m-1 \choose k-1}{n-1 \choose k-2}$ start with a win, end with a win, have $k$ winning runs and $k-1$ losing runs

${m-1 \choose k-1}{n-1 \choose k-1}$ start with a win, end with a loss, have $k$ winning runs and $k$ losing runs

${m-1 \choose k-1}{n-1 \choose k-1}$ start with a loss, end with a win, have $k$ winning runs and $k$ losing runs

${m-1 \choose k-1}{n-1 \choose k}$ start with a loss, end with a loss, have $k$ winning runs and $k+1$ losing runs

So ${m-1 \choose k-1}\left({n-1 \choose k-2}+{n-1 \choose k-1}+{n-1 \choose k-1}+{n-1 \choose k} \right) ={m-1 \choose k-1}{n+1\choose k}$ have $k$ winning runs

making the probability  $$\mathbb P(K=k \mid m,n)=\dfrac{{m-1 \choose k-1}{n+1\choose k}}{{m+n \choose m}}$$

Title: Probability of exactly k runs of wins


答:
Note that $z=0$ is a solution. We can then proceed with the assumption that $z=re^{i\theta}$ is nonzero.

Your equation becomes $r^4e^{4i\theta}+4ire^{-i\theta}=0$ and therefore
$$
r^3e^{5i\theta}=-4i
$$
By uniqueness of the polar representation, you obtain
$$
r^3=4,\qquad e^{5i\theta}=e^{-i\pi/2}
$$
and so
$$
r=\sqrt[3]{4},\qquad \theta=-\frac{\pi}{10}+\frac{2k\pi}{5},\ k=1,2,3,4,5
$$
if you want $\theta$ in the $[0,2\pi)$ interval.

Title: Solving $z^4 + 4i\bar{z} = 0$

Body:
How to solve $z^4 + 4i\bar{z} = 0$ (efficiently)?

I managed to compute the radius of $z$:

Denote $z = rcis(\theta)$

Rearranged the equation to $z^4 = -4i\bar{z}$

Taken absolute value on both sides $r^4 = 16r$

And found $r=0$ or $r = \sqrt[3]{16}$

How can we calculate $\theta$ from here?

Or is there a better way to solve this?


问:
Prove that the implicit function is defined everywhere
I'm trying to prove this problem:

Let $f\colon \mathbb{R}^2 \to \mathbb{R}$ from $C^k$ class $(k \geq 1)$. Suppose there is a point $(x_0, y_0) \in \mathbb{R}^2$ and a constant M such as $f(x_0,y_0) = 0, \frac{\partial f}{\partial y}(x,y) \neq 0$ and $\left|\frac{\partial f}{\partial x}(x,y) \bigg/ \frac{\partial f}{\partial y}(x,y) \right| \leq M$ for all $(x,y) \in \mathbb{R}^2$. Prove that, for all $x \in \mathbb{R}$, there is an unique $y = \xi(x) \in \mathbb{R}$ such as $f(x,\xi(x)) = 0$ and the function $\xi: \mathbb{R} \to \mathbb{R}$ is $C^k$.

For the most part this problem can be proved using the implicit function theorem. However, the implicit function theorem can only guarantee an open interval around $x_0$, say I, within which the function $\xi(x)$ is defined and its unique. The problem asks to prove that this function is defined everywhere, from the fact that there is a single point $(x_0,y_0)$ there the function $f$ is constant equals to $0$.

From what I understand (and please correct me if I'm wrong), because we have added the condition
$\left|\frac{\partial f}{\partial x}(x,y) \bigg/ \frac{\partial f}{\partial y}(x,y) \right| \leq M$, the function $\xi(x)$ is uniformly continuous (actually it's Lipchitz, but we don't need that). Being uniformly continuous we know that this function is not "exploding" anywhere for each limited subinterval within $I$ (so it doesn't have asymptotes in at $x$ so to speak). However I was not able to formally prove this last part. Maybe I can use connectivity of $\mathbb{R}$, but still not clear.

Can anyone help me understanding how to prove this last fact assuming that we known $\xi(x)$ is unique and defined within an open interval $I$ around $x_0$ (as implicit function theorem states).

Thanks

答:
Assume $\xi\colon\operatorname{dom}\xi\subseteq\mathbb{R}\to\mathbb{R}$ is maximally defined.

By IFT, if $\xi$ is defined at $x$, it is defined on an open neighbourhood of $x$.  In other words $\operatorname{dom}\xi$ is open.

On the other hand, since we have $\xi$ uniformly continuous and $\mathbb{R}$ complete, we can extend $\xi$ to a continuous $\overline{\xi}\colon\overline{\operatorname{dom}\xi}\to\mathbb{R}$.  Since $f$ is continuous, we would still have $f(x,\overline{\xi}(x))=0$ for all $x$.  So maximality of $\xi$ gives $\xi=\overline{\xi}$.

So $\operatorname{dom}\xi$ is both open and closed, hence connectedness of $\mathbb{R}$ and $\operatorname{dom}\xi\neq\varnothing$ (since $\xi(x_0)=y_0$) gives $\operatorname{dom}\xi=\mathbb{R}$.


问:
What's the answer to this shaded area question?

I've been thinking about this for quite long, but I can't find an answer. Problem is, I lost the given answer some time ago.

答:
Hint for A:

You can move the inner segments out of the central disk and form rhombi. Then the black area is six times that of a third of a disk.


答:
By taking the modulus we get $|z|^4=4|z|\iff |z|=0\text{ or }|z|=\sqrt[3]{4}$

Let set aside the trivial solution $z=0$ then by multiplying by $z$ we get

$z^5=-4iz\bar z=-4i|z|^2=-8i\sqrt[3]{2}=-i\,2^\frac{10}{3}$

Therefore $z=\Big(-i\, 2^{\frac 23}\Big)\omega_k\ $ for $\, k=0,1,2,3,4\ $ and $\omega^5=1$ is a fifth root of unity.

Title: Solving $z^4 + 4i\bar{z} = 0$


问:
Is there an algorithm to compute the spanning trees from an specific node in digraphs?
Given a strongly connected digraph $\mathcal{D}=(\mathcal{V},\mathcal{E})$, where $\mathcal{V}={1,\cdots,N}$ is the set of vertices and $\mathcal{E}\subseteq\mathcal{V}\times\mathcal{V}$ is the set of directed edges, denote as $\mathbb{T}_v$ the set of all spanning trees rooted at vertex $v\in\mathcal{V}$. Is there a distributed algorithm to compute $\mathbb{T}_v$?

Since $\mathcal{D}$ is strongly connected, then $\mathbb{T}_v$ is nonempty. Also, from the Caley's Formula, $|\mathbb{T}_v|\leq N^{N-2}$, which corresponds to fully connected digraph. I need this information being available on each node to complete a distributed controller design for multi-agent systems.

答:
There is a non-distributed algorithm to do this kind of things, which computes determinants in a version of the adjacency matrix, see https://en.wikipedia.org/wiki/Kirchhoff%27s_theorem#Kirchhoff's_theorem_for_multigraphs

To get the number of spanning trees, you just compute a integer-valued determinant. If you want an enumeration of the spanning trees, you replace edges by indeterminates and compute the determinant as a polynomial : each monomial will be a spanning tree.


答:
Use that the set $\{f\in L^2: \ |f|\le M \ a.e.\}$ is closed (use pointwise convergent subsequences).

Follows from $\|f\|_{L^p}^p \le \|f\|_{L^2}^{2} \|f\|_{L^\infty}^{p-2}$:
$$
\int |f|^p dx \le \|f\|_{L^\infty}^{p-2} \int |f|^2 dx.
$$

Title: If $f_n\rightarrow f$ in $L^2$ and $\lVert f_n\rVert_\infty\le C$ then why is $f\in L^\infty$?

Body:
This exercise from the $L^p$-theory section of my measure theory course is giving me some trouble.

Let $(f_n)_n\subset L^2\cap L^\infty$ be a sequence such that $f_n\rightarrow f\in L^2$ in $L^2$ and $\lVert f_n\rVert_\infty\le C$ for all $n$. I want to show that

$f\in L^\infty$ and

$f_n\rightarrow f$ in $L^p$ for any $2\le p<\infty$

My most promising attempt at 1. was
$$\lVert f\rVert_\infty=\lVert f-f_n+f_n\rVert_\infty\le \lVert f-f_n\rVert_\infty+C,$$
but how can we bound $\lVert f-f_n\rVert_\infty$?

I'm having similar problems with 2., except of course there we want to bound $\lVert f-f_n\rVert_p$.


问:
Star-shaped domain and approximation by smooth functions
Let $1 \leq p  <  \infty$ and let $\Omega \subset \mathbb{R}^n$ be a domain. Suppose $\Omega$ is star-shaped in the sense  that there exists $x_0 \in \Omega$, such that for every $x \in \Omega$, the line segment connecting $x$ and $x_0$ stays in $\Omega$. Let $u \in W^{1,p}(\Omega)$, and  for  $\lambda >  0$ define $\Omega^{\lambda} = \{x : x/ \lambda \in \Omega \}$ and $u^{\lambda}(x) =u(x/ \lambda)$  for  any $x \in \Omega^{\lambda}$.

Using these definitions, one may easily show that $u^{\lambda} \in W^{1,p}(\Omega^{\lambda})$. Next, I aim to show that $C^{\infty}(\bar{\Omega)}$ is dense in $W^{1,p}(\Omega)$, by showing $u$ can be approximated by functions in $C^{\infty}(\bar{\Omega)}$. The idea is to suitably mollify $u^{\lambda}$  with $\lambda$ close to $1$ in order to subsequently approximate $u$. However, I do not know how to proceed and even if I did, how this would lead to the solution. What confuses me is that the introduction of a mollifier would introduce another variable whose limit we have to take, and I also do not see how this approach would lead to working with $\bar{\Omega}$.

To elaborate, if we mollify with a smooth function $\rho \in C_{c}^{\infty}(B_1(0))$ with compact support in the unit ball, we'd introduce a new convolution $u_{\lambda}^{\epsilon} := \rho_{\epsilon} * u_{\lambda}$, then this is only well-defined on the set:
\begin{equation}
\Omega_{\epsilon}^{\lambda} = \{ y \in \Omega_{\lambda} | B_{\epsilon}(y) \in \Omega^{\lambda}\}.
\end{equation}
So although these approximations are infinitely smooth and can be shown to converge in $L^{p}_{loc}(\Omega^{\lambda})$ to $u^{\lambda}$ as $\epsilon \to 0$ for $\lambda$ fixed, I don't see how to handle the limit $\lambda \to 1$ and how one would get the closure of $\Omega$ involved.

I am not necessarily looking for a full solution - any hints or intuition would be very appreciated as well. Thanks!

答:
I just remembered that this question was never answered and that I was working with a wrong formulation of the problem. The proper definition of star-like here is  that the segment $(xx_0]$ is in $\Omega$, for any $x \in \overline{\Omega}$ (the original formulation of star-like is a bit too constraining). Then a sketch of a solution would go as follows:

WLOG suppose that $x_0 =0$, so the domain is star-like about the origin. By continuity of the dilation in $L^p$, we can obtain via routine calculations:
\begin{equation}
||u - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as $\lambda \to 1$.}
\end{equation}

The next idea is to observe that the star-shapedness means that $\overline{\Omega} \subset \Omega^{\lambda}$ is a compact subset for $\lambda >1$, then for all $\epsilon \ll 1$ sufficiently small, we can take $u^{\lambda} * \rho_{\epsilon} \in C^{\infty}(\overline{\Omega})$. Moreover:
\begin{equation}
||u^{\lambda}* \rho_{\epsilon} - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as $\epsilon \to 0$.}
\end{equation}

So we're pretty close to finished, but the issue is that we need to rigorously reconcile the two limits of $\epsilon$ and $\lambda$. So given a fixed $\delta > 0$, we take $\lambda_{\delta}$ close to $1$ so that:
\begin{equation}
||u^{\lambda_{\delta}} - u||_{W^{1,p}(\Omega)} < \frac{\delta}{2}.
\end{equation}
Then, select $\epsilon = \epsilon(\delta)$ sufficiently small, so that:
\begin{equation}
||u^{\lambda_{\delta}} * \rho_{\epsilon} - u^{\lambda_{\delta}}||_{W^{1,p}(\Omega)} < \frac{\delta}{2},
\end{equation}
and so that $u^{\lambda_{\delta}} * \rho_{\epsilon} \in C^{\infty}(\overline{\Omega})$. So by the triangle inequality,
\begin{equation}
||u^{\lambda_{\delta}} * \rho_{\epsilon} - u||_{W^{1,p}(\Omega)} < \delta,
\end{equation}
so $C^{\infty} (\overline{\Omega})$ is dense in $W^{1,p}(\Omega)$.


答:
Hint for B:

$6 \times 2=12$ equilateral triangles plus $6 \times 2=12$ circle segments.  Each segment is a sixth of a circle minus an equilateral triangle.

So the same area as A, since a circle is six equilateral triangles plus six segments.

Title: What's the answer to this shaded area question?

Body: I've been thinking about this for quite long, but I can't find an answer. Problem is, I lost the given answer some time ago.


问:
Slope-ratio method - Calculation of average of observations vs. individual.
I am using Slope ratio method for estimation of relative potency in a drug. The rows in matrices correspond to number of parallels and columns are doses.

I have a two methods for calculating the slope ratio: Case 1 and Case 2.

Case 1

Let,
$$
  X =   \begin{bmatrix}
    a_{11} & a_{12} & a_{13} \\
     a_{22} & a_{22} & a_{23}\\
  a_{31} & a_{32} & a_{33}
    \end{bmatrix}  , \quad  T_1 =   \begin{bmatrix}
    t_{11} & t_{12} & t_{13} \\
     t_{22} & t_{22} & t_{23}\\
  t_{31} & t_{32} & t_{33}
    \end{bmatrix} ,  \quad   T_2 =   \begin{bmatrix}
    y_{11} & y_{12} & y_{13} \\
     y_{22} & y_{22} & y_{23}\\
  y_{31} & y_{32} & y_{33}
    \end{bmatrix},  \quad T_3 =   \begin{bmatrix}
    z_{11} & z_{12} & z_{13} \\
     z_{22} & z_{22} & z_{23}\\
  z_{31} & z_{32} & z_{33}
    \end{bmatrix}
$$

let $\bar{x_i}$ be the average of columns of $X$,  $\bar{t}$ of $T_1$,  $\bar{y}$ of $T_2$ and $\bar{z}$ of $T_3$.

Define,
$$
S = \begin{bmatrix}
s_1 \\ s_2 \\ s_3
\end{bmatrix}, \quad K  = \begin{bmatrix}
k_1 \\ k_2 \\ k_3
\end{bmatrix}
$$

Regression:
$$
\left.
\begin{align}
\bar{x}' &= \alpha + \beta_S S, & &\\
\bar{t}' &= \alpha_1 + \beta_{t} K,&\lambda_1 &= \frac{\beta_{t}}{\beta_S}  \\
\bar{y}' &= \alpha_2 + \beta_{y} K,& \lambda_2 &= \frac{\beta_{y}}{\beta_S} \\
\bar{z}' &= \alpha_3 + \beta_{z} K & \lambda_3 &= \frac{\beta_{z}}{\beta_S}
\end{align}
 \quad \right\rbrace \quad \lambda = \frac{\lambda_1 + \lambda_2 + \lambda_3}{3}
$$

Case 2

Averages of columns are not taken. Say,
$$
X = \begin{bmatrix}
x_1 \\ x_2 \\ x_3
\end{bmatrix}
,\quad \text{where}: x_1 = \{a_{11}, a_{12}, a_{13}\}, \quad x_2= \{a_{21}, a_{22}, a_{23}\} \dots
$$
Similarly for rows of $T_1$, $T_2$, and $T_3$. So that, e.g. $t_1 = \{t_{11}, t_{12}, t_{13}\}, t_2=\{t_{21}, t_{22}, t_{23}\} \quad \dots$

Now the slope ratios are:
$$
\begin{align}
x_1' &= \alpha_1 + \beta_{x_1} S, & t_1' &= \alpha_4 + \beta_{t_1} K,\quad t_2' = \dots\quad, & \lambda_1 &=\frac{(\beta_{t_1} + \beta_{t_2} + \beta_{t_3})/3}{\beta_{x_1}} \\
x_2' &= \alpha_2 + \beta_{x_2} S, & y_1' &= \alpha_5 + \beta_{y_1} K,\quad y_2' = \dots \quad,& \lambda_2 &=\frac{(\beta_{y_1} + \beta_{y_2} + \beta_{y_3})/3}{\beta_{x_2}} \\
x_3' &= \alpha_3 + \beta_{x_3} S, & z_1' &= \alpha_6 + \beta_{z_1} K,\quad z_2' = \dots\quad, & \lambda_3 &=\frac{(\beta_{z_1} + \beta_{z_2} + \beta_{z_3})/3}{\beta_{x_3}}
\end{align}
$$
Then, the overall slope ratio is $\bar{\lambda}$.

For those wondering why I am complicating the calculations, the regression coefficients for individual tests, i.e. $T_i$ must be calculated and then taken average of.

Questions

Intuitively, I believe that Case 1 is not representative of the sample population, but I am not sure. How can I prove it? Or are both methods the same?

答: My question was regarding bioassays. I needed to find parallelism between $S$ and $T_i$. Both cases given in the question were wrong. The correct way to do this is described in European Pharmacopoeia, chapter "5.3 Statistical Analysis of Results of Biological Assays and Tests" and USP <1210> " Statistical Tools for Procedure Validation". For parallelism test can be conducted by the equivalence test (TOST procedure).


问:
Flip a (loaded) coin and get a head only on the last two tosses
So on this loaded coin, the probability of a head is 0.4 and a tail is 0.6, and the coin is flipped 4 times. How would I go about calculating the probability of only getting a head on flips 3 and 4?</p>

答:
HINT
Multiply the probability of a tail on flip 1 by the probability of a tail on flip 2 by ....

 $0.6^2\times0.4^2=0.0576$


问:
Given $H\triangleleft G, K\le G\ $, $\;K\nsubseteq H,\;$ and $(G:H) = p$, $\;p$ prime, prove $HK=G$
Suppose that $H\triangleleft G, K\le G\ $ and $K\nsubseteq H$. How we can prove that $HK=G?$

Also $(G:H)=p$ where p is prime.

答:
$K\nsubseteq H$, so there is a non-trivial element $k$ of $K$ such that $kH$
non-trivial in $G/H$. Since $(G:H)=p$, $G/H$ is cyclic, so this $kH$
generates $G/H$. So any element in $G$ can be written in the form $%
hk^{\alpha }$ (since $H$ normal in $G$) where $0\leq {\alpha }\leq p-1$ and $h\in H$.</p>


问:
Is a bounded function in $H^1(\Omega)$ also continuous for a Lipschitz domain $\Omega \subset \mathbb R^2$?
Let $\Omega \subset \mathbb R^2$ be a bounded Lipschitz domain (lets say $\Omega = (0,1)^2$). Let $u \colon \Omega \longrightarrow \mathbb R$ be a function in the Sobolev space $H^1(\Omega)$, that is $u$ is weakly differentiable and $u, |\nabla u| \in L^2(\Omega)$. Additionally, assume $0 \leq u \leq 1$.

Can I conclude, that $u$ is continuous?

I know that merely $u \in H^1(\Omega)$ does not provide $u \in C^0(\Omega)$ e.g. $u(x) = \log \log |x|$. But the continuity only fails to hold because of the singularity in $x = 0$, which is removed when I assume $u$ to be bounded.

Because the dimension of $\Omega$ and the integrability and differentiability of $H^1$ are critical, I was not able to find a result which is connected to imbedding theorems.

I would be glad if someone could clarify my question with literature, or a counterexample.

Thanks in advance.

答:
Counterexample:  $f(x)=\sin^2 \log\log \frac1{|x|}\in H^1\cap L^\infty$, $0\le f \le 1$, and is not continuous at $0\in \Omega:= B_{1/2}(0)$.

This is because as $x\to0$, $\log\log\frac1{|x|}\to\infty$, so $\sin^2\log\log\frac1{|x|}$ takes all values in $[0,1]$ on any neighbourhood of $0$.

That $f\in H^1$ is a consequence of Exercise 17 of Evans Chapter 5 (2nd ed.), or by directly computing the weak derivative; set $\operatorname{LL}(x)=\log\log\frac1{|x|}$. Then the $\alpha$th-derivative everywhere except 0, $|\alpha|=1$,  is
$$ g_\alpha:= 2(\sin \operatorname{LL} \cos \operatorname{LL} ) D^\alpha \operatorname{LL} \in L^2$$
and for  $\phi\in C_c^\infty(\Omega)$,
$$ \int_{\epsilon<|x|<1/2} f D^\alpha \phi = \int_{|x|=\epsilon} f\phi n^\alpha d\sigma(x) -\int_{\epsilon<|x|<1/2} g_\alpha \phi$$
sending $\epsilon\to 0$ shows $D^\alpha f = g_\alpha$ in the sense of weak derivatives.


问:
Limit of $(x^3 - 3x^2 + 2x)$ as $x \to 1$ using epsilon delta definition
I need to prove the following limit using epsilon delta method. I have come up this on my own just to practice skills.

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

So, I need to come up with some $\delta >0$ such that given $0 < \lvert x - 1 \rvert <\delta$, I need to prove that

$$ |(x^3 - 3x^2 + 2x) - 0| < \varepsilon \tag{1}$$

I can simplify eq $1$ as follows.

$$ |x| |x-1||x-2| < \varepsilon $$

When we have a non linear term, then we restrict $x$ to some distance from $1$. So, suppose we have
$ 0 < |x-1| < 10$. This leads to $-10 < x-1 < 10$. Doing some algebra, we can get inequalities

$$ 0 \leqslant |x| < 11 \tag{2}$$

$$ 0 \leqslant |x-2| < 11 \tag{3}$$

From eq 2 and 3, it follows that

$$ 0 \leqslant |x||x-2| < 121 $$

And since $0 < |x-1| $, it follows that

$$ |x||x-1||x-2| < 121 |x-1| \tag{4}$$

To show that $|x||x-1||x-2| < \varepsilon$ , its sufficient to show that $ 121 |x-1| < \varepsilon$. And this will happen if $ |x-1| < \frac{\varepsilon}{121} $. So, this is another restriction on $|x-1|$. So, we can let

$$\delta = \text{min}\left(10, \frac{\varepsilon}{121}\right) $$

So, we can let this be our choice of $\delta$. Its clear that $\delta > 0$. So, now the official proof will follow.

Let $\varepsilon > 0$ be some arbitrary real. Suppose $0 < |x-1| < \delta$. With the choice of $\delta$ we have done, this means that $ 0 < |x-1| < 10$. As demonstrated above, equation 4 follows from this inequality

$$ |x||x-1||x-2| < 121 |x-1| $$

But now, we also have $ |x-1| < \frac{\varepsilon}{121} $. It means that $ 121 |x-1| < \varepsilon $. Using this, we get that

$$ |x||x-1||x-2| < \varepsilon  $$

which can be rewritten as

$$ | (x^3 - 3x^2 + 2x) - 0 | < \varepsilon  $$

Since, $\varepsilon > 0$ was arbitrary to begin with, its proven that

$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$

Is the proof correct ?

答:
This proof is fine, but manipulating the bounded part can sometimes prove more challenging.

Here it is easy since it is only $|x||x-2|$, but often you will get benefits while shifting the limit towards $0$ by setting $x=x_0+u$ with $u\to 0$.

The main benefit is that limits in zero are part of your mathematical toolbox, also $|u|<\delta$ is easy to manipulate compared to $|x-x_0|<\delta$.

For instance here $|x(x-1)(x-2)-0|=\overbrace{|u|}^{<\epsilon}\overbrace{|1-u^2|}^{\le 1}<\epsilon$

Granted we take $\delta=\min(1,\epsilon)$ so that $\begin{cases}|u|<\delta<\epsilon\\0\le u^2<\delta^2<1^2\implies|1-u^2|\le 1\end{cases}$

The inequalities did appear more naturally than when you calculated the $10$ and $\frac 1{121}$ factors.

Another advantage you will encounter in your epsilon-delta proofs is how it is easy to bound polynomials.

For instance imagine that in the problem instead of $|1-u^2|$ you had $|1-u^2+7u^3-2u^4|$.

Just proceed using the fact that for $|u|<1$ then $\cdots<|u|^4<|u|^3<|u|^2<|u|<1$.

The triangle inequality gives you $$|1-u^2+7u^3-2u^4|<|1|+|u^2|+7|u^3|+2|u^4|<1+1+7+2=11$$

Conclude by taking $\delta=\min(1,\frac{\epsilon}{11})$.


问:
Not able to calculate this expectation
If $h_{1}, h_{2}, \dots, h_{N}$ are zero mean i.i.d Gaussian random variables with variance $\sigma_{h}^{2}$, then how do I calculate the value of

$$\mathbb{E}\left[\left(h_{1}^{2} + h_{2}^{2} + \dots, h_{N}^{2} \right)^{2}\right]$$

The only clue I have is that

$$\mathbb{E}\left[h_{i}^{2}\right] = \sigma_{h}^{2}$$

I do not know how to compute $\mathbb{E}\left[h_{i}^{2} h_{j}^{2}\right]$ (for all $i \neq j$). Please help.

答:
$$Y=h_1^2+\dots + h_N^2 =\sigma^2\left[\left(\frac{h_1}{\sigma}\right)^2+\dots+\left(\frac{h_N}{\sigma}\right)^2 \right]$$

It is self evident that

$$Y\sim Gamma\left(\frac{N}{2};\frac{1}{2\sigma^2}\right)$$

Thus its second moment is

$$\mathbb{E}[Y^2]=\mathbb{V}[Y]+\mathbb{E}^2[Y]=\frac{N}{2}4\sigma^4+\left(\frac{N}{2}2\sigma^2\right)^2=(N+2)N\sigma^4$$

Title: Not able to calculate this expectation


问:
$ \mathbb{E} $ is a linear subspace of $ \mathbb{R^n}$. What $ \langle x, y\rangle \leq0 \implies \langle x, y\rangle= 0 $ exactly means?
$E$ is a subspace of $\Bbb R^n $. Let $ x \in \mathbb{R}^n $. for every $ y \in E $, we have : $$  \langle x, y\rangle \leq 0 \implies \langle x, y\rangle = 0 $$ We also have the same for $\geq$ from what I understood.

I understand the proof behind it; you just need to say that $E$ is a linear subspace, so: $ y \in E \implies -y \in E $ and you find that the dot product is null. Here is my question:

Does it mean that the dot product of $x$ and $y$, $y$ being in a linear subspace, is always null? And why is it suppose to be like this ? I don't understand the geometrical intuitions behind it.

Please, if my question is ambiguous or unclear, let me know !
Thank you very much :)

答: Fix $y\in E$. Then $\langle x, y \rangle=-\langle x, -y \rangle$ (1). So either $\langle x, y\rangle$ or $\langle x, -y\rangle$ is negative. By assumption it follows that one of them is $0$. (1) then implies that $\langle x, y \rangle=0$. So $E$ is orrhogonal to $x$.


问:
How to determine the function of a data list?
I'll be the first to admit, I am not a math wiz. That said, please forgive my ignorance. I need to write a script for work that will take a given temperature/pressure and output the other (pressure/temperature). How do I find the function of the data?</p>

答: First you should plot the data to see what it looks like.  The ideal gas law suggests that if you have a closed volume and fixed amount of gas, the pressure will be proportional to the temperature.  Excel will run a linear regression to give a straight line fit through the data and give you the equation of the line, so you could try that first.  Excel will also do a higher order polynomial fit if the errors in the straight line fit are too large.</p>


问:
Prove that: If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$
Let $f$ be differentiable in $\mathbb R$, with $f'$ differentiable and $f'(x) \geq 0,f''(x) \geq 0$ for every $x \in \mathbb R$.

How do I rigorously prove that (using Lagranges Mean Value Theorem):

If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$

I know that if f is monotonic increasing and $f'>m>0$ for $x>a$ (where $f'(a)=m>0$) then $\lim_{x \to \infty} f(x) > \lim_{x \to \infty} mx = \infty$

But I want something more rigorouse.

Thanks for any idea or direction

答: For $x >a$ there exists $t$ in $(a,x)$ with $f(x)-f(a)=(x-a)f'(t)$ so $f(x)-f(a) \geq (x-a)f'(a) \to \infty$.  [I have used the fact that $f'$ is increasing since its derivative is non-negative].


问:
What is the inverse Laplace transform of $\sqrt{1+a/s}$?
My question is that, what is the inverse Laplace transform of $\sqrt{1+a/s}$  where $a \gt 0$?

I tried to find solution in the Integral and Series books, but failed.

答:
The "solution from a book" uses modified Bessel functions: $$\mathcal{L}^{-1}\left[\sqrt{1+2a/s}-1\right](t)=ae^{-at}\big(I_0(at)+I_1(at)\big)$$ (easily obtained from the LT of $I_\alpha$; should I find a reference to the latter?..).

Knowing the ILT of $1$ (the first entry of this table), you're done.


问:
How to inverse the function $\frac13e^x+\ln(3x+1)$
What would be the strategy and final result of inverting the function

$$f(x) = \frac{1}{3}e^x+\ln(3x+1)$$

答:
Almost impossible to find a nice formula, but simple to get a graph.

Once you have graphed $y=f(x)$ you simply draw the symmetric wrt the line $y=x$.

Hope this can be useful


答:
With your modification, I give you a hint. By the index formula
$$p=(G:H)=(G:HK)(HK:H)$$
Since $p$ is prime, what can you say about the possibilities? Which one is impossible under the given assumptions?</p>

Title: Given $H\triangleleft G, K\le G\ $, $\;K\nsubseteq H,\;$ and $(G:H) = p$, $\;p$ prime, prove $HK=G$

Body:
Suppose that $H\triangleleft G, K\le G\ $ and $K\nsubseteq H$. How we can prove that $HK=G?$

Also $(G:H)=p$ where p is prime.


问:
Proving that there is some $x$ such that $\gcd(a + bx, c) = 1$
Let $\gcd(a,b) = 1$ and $c>0$. Pove that there is an integer x such that $\gcd(a + bx, c) = 1$.

This question is from Section 2.3 (Covers the Chinese Remainder Theorem) Problem 15 of An Introduction to the Theory of Numbers by Niven and Zuckerman.

I didn't use the Chinese Remainder Theorem in my attempt, but I suspect that it should, since $\gcd(a,b) = 1$. In fact I didn't even use that $\gcd(a,b)=1$, either.

Attempt: Instead of finding $x$, we find what $x$ cannot be. That is, $y$ such that $\gcd(a+by,c) \neq 1$.
Reduce $c$ into its prime factors, so that $c = c_{1}^{i_{1}}c_{2}^{i_{2}}c_{3}^{i_{3}}...c_{n}^{i_{n}}$. Then we can solve, for each $1 \leq m \leq n$,

$a + by \equiv _{c_{m}}0 \Leftrightarrow by \equiv _{c_{m}} -a$

The point is that if $a + by$ and $c$ have a GCD other than $1$, then one prime factor must be in common.

$by \equiv _{c_{m}} -a$ has solutions when $gcd(b,c_{m})$ divides $-a$ and has no solution otherwise. Suppose there are no solutions for all $m$. Then we are done, for there are no restrictions on $x$. Suppose for some or all $m$, there are solutions. Then $by \equiv _{c_{m}} -a$, has solutions $y \equiv _{c_{m}} -(\frac{a}{g})y_{0}+t(\frac{c_{m}}{g})$. Where $g = \gcd(b,c_{m})$, $t = 0,1,..,g - 1$, and $y_{0}$ is a solution to $(\frac{b}{g})y \equiv _{c_{m}}1$. But the set of all these solutions cannot equal $\mathbb{Z}$, because there are infinite primes and finite $c_{m}$. So there must be some integer $x$ such that $gcd(a + bx, c) = 1$.

$\blacksquare$

I don't like this proof, because it's nonconstructive. Is there a way to use the CRT or some other basic number theorem?

答: This is almost certainly not what the problem in Niven and Zuckerman is looking for, but if you have Dirichlet's Theorem at your disposal, then from the assumption $\gcd(a,b)=1$ you can conclude there are infinitely many prime numbers of the form $a+bx$, at which point you can just take one that's greater than $c$.


问:
A/B Test: t-test vs. chi-squared test
I am working on a marketing campaign analysis between treatment-A and treatment-B. The goal is to find which treatment yields more clicks. Say, I have one month of experiment duration and at the end I want to determine if one treatment is statistically better than the other. I can formulate the problem in two ways:

t-test: I find the average daily click rates of A and B, and then perform a t-test.

chi2 test: I find the total clicks of these A/B treatments at the end, and perform a chi2 test.

Can someone comment on the differences of these two approaches?

答:
I'm by no means a statistician, so what I write below is just some common sense advice.

The Student test is the canonical way to go for the mean comparison, but it relies on having a few samples from each of the two distributions. If you have a long streak of clicks, then you are welcome to split it into several subsets but you need to be reasonably sure that the distribution is the same and approximately normal within each subset. The splitting into days is very questionable because it is hard to justify that the people clicking habits do not change from day to day. As to myself, when I come home after a work day and go to the web, I definitely do not want any stupid ads, but on weekends I may be more benevolent if I'm in a really good mood. The Student test is robust in the sense that if you do any splitting (of course, decided upon in advance and without looking at the data) and it shows statistical significance, then it is, indeed, there. But if you split into subgroups with different underlying distributions, then your empirical variance may be a noticeable overestimate of the true variance and you lose sensitivity, so if the change in the clicking means from day to day is comparable to or greater than the difference in overall means between design A and design B, with daily averages you'll detect nothing even if the difference is there. Approximately normal is usually not a problem as long as you have not too few members in each subset.

So, if you have a timestamp on, say, users' entering the page, I would definitely go for Student, but I'd split differently. I would partition the whole observation period into 10 minute intervals and within each interval assign 1 minute to each group (I would even rotate or randomize that assignment from one 10 minute interval to another). That will give me more groups to estimate the variance and I will be way more sure that the underlying distributions for different groups are the same. By the way, I would also do it to compare the daily means in this way, and, if I see that there, indeed, exists a significant difference between them, request that the observations be complemented by two missing days of the week before doing any analysis because the natural human life cycle is a full week, not 5 days (just make sure that you don't hit holidays, etc.)

What to do if you don't have the timestamps but only the daily totals? Then, as I said, you may still perform Student and if it shows the difference, it will be there, but if it doesn't, then you may just suffer from the effect I described. So in this case I would go for a quick and dirty version of $\chi^2$.

Note that the original Pearson $\chi^2$ test is designed for a different situation that you are in (just look up the assumptions and the setup on the Wikipedia page or in a textbook), so I doubt you can run it the way you intend with a meaningful result. However, we can get something similar from the first principles. Suppose that $A,B$ are the total numbers of users reached and $X,Y$ are the corresponding numbers of users who clicked. You want to know if the difference has occurred by pure chance or not. If you accept that all users and design effects are the same and there is some underlying probability $p$ to click, then the distribution of the vector $(\frac{X-Ap}{\sqrt{Ap(1-p)}},\frac{Y-Bp}{\sqrt{Bp(1-p)}})$ should be approximately standard 2D normal and, in particular,
$$
\frac{(X-Ap)^2}{Ap(1-p)}+\frac{(Y-Bp)^2}{Bp(1-p)}
$$

would have the $\chi^2$ distribution with 2 degrees of freedom. Since you don't know $p$, you have to take the infimum over it, so your actual statistics is
$$
\inf_{p\in(0,1)}\left[\frac{(X-Ap)^2}{Ap(1-p)}+\frac{(Y-Bp)^2}{Bp(1-p)}\right]
$$

This is not exactly $\chi^2$ but it is concentrated better than $\chi^2$, so you still can use the $\chi^2$ table for it. When $A,B$ are not to small and the probabilities to click are not too close to $0$ or $1$, this approximation is fairly good. This is robust too (in the above sense) and will also suffer from the loss of sensitivity if the probability to click depends significantly on the day or on the user group, but it is still better than Student with a bad partition (though worse than the Student with a good one).

Just my two cents. Comments from people who know better are invited :-) (Questions from people who know worse too).

Title: A/B Test: t-test vs. chi-squared test


问:
if two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?
If two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?

If $p$ and $q$ are two strongly equivalent metrics (in the sense that $ A p(x,y) \leq q(x,y) \leq B p(x,y)$ for some $A,B>0$) on some set $X$, then $(X,p)$ is complete if and only if $(X,q)$ is complete. My question is whether the converse is true as well, i.e. given two metrics $p$ and $q$ on $X$ such that $(X,p)$ and $(X,q)$ are complete, does it follow that $p$ and $q$ are strongly equivalent or at least equivalent (generating the same topology)?

If the answer is no, does the situation change in the context of Banach spaces, with metrics replaced by equivalent norms on some (of course infinite-dimensional) vector space?

In case the above converse is true, I guess one should argue by contradiction, assuming that the two metrics do not generate the same topology. Even in the case where the topology generated by one topology is strictly stronger than the other one, it is not clear to me how to find a sequence $(x_n)$ converging in the weaker metric (say $p$) but diverging in the stronger one.

答: That is not true. Let $f\colon\Bbb R\longrightarrow\Bbb R$ be any bijection. Let $p$ be the usual distance in $\Bbb R$ and define $q(x,y)=\bigl|f(x)-f(y)\bigr|$. Then both $(\Bbb R,p)$ and $(\Bbb R,q)$ are complete, but, in general, $p$ and $q$ are not equivalent.


问:
To get a recursive relation for this type of problem
Twelve square sheets are placed in a row, and each can be painted with one of the four colors Y,R,O,B. Find the number of ways this can be done so that each block of five adjacent sheets contains at least one sheet of each color . What i tried is that for lower cases of sheets i tried to see the pattern so as to motivate for a recursive relation but going from 7 to 8 case and higher ones needed a recursive relation which i am not able to generate . As lower cases (<8) dont need any recursion it can be easily solved.</p>

答:
Let $n  \geqslant  5$ be the number of colored sheets, numbered from $1$ to $n$. We extend the system by adding a new sheet in position $0$ and shifting the sheets to the right.

For any coloring $c$, let the repetition index $r \in  \left\{2 , 3 , 4 , 5\right\}$ be
the index of the first sheet with a repeated color. If $r = 5$, it means that the $4$ first sheets have different colors. One can add a new sheet of any of the four colors, it gives $4$ ways to extend the coloring, which respective repetition indexes are $2 , 3 , 4 , 5$.

If $r < 5$, it means that there are only $3$ colors among the
$4$ first sheets and the only way to extend the system is to add a new
sheet which color is the missing color among the $3$. The result is
a coloring which repetition index is $r+1$.

Let ${X}^{\left(n\right)} = {\left({X}_{2}^{\left(n\right)} , {X}_{3}^{\left(n\right)} , {X}_{4}^{\left(n\right)} , {X}_{5}^{\left(n\right)}\right)}^{T}$
be the numbers of coloring with repetition indexes $\left(2 , 3 , 4 , 5\right)$ for
$n$ sheets. The previous reasoning shows that

\begin{equation}
{X}^{\left(n+1\right)} = A {X}^{\left(n\right)} = \left[\begin{array}{cccc}0&0&0&1\\
1&0&0&1\\
0&1&0&1\\
0&0&1&1
\end{array}\right] {X}^{\left(n\right)}
\end{equation}

In particular, the number ${C}_{n}$ of colorings of $n$ sheets is
${C}_{n} = U {X}^{\left(n\right)}$ where $U = \left(1 , 1 , 1 , 1\right)$. It follows that

\begin{equation}
{C}_{n+k} = U {A}^{k} {X}^{\left(n\right)}
\end{equation}

We claim that $U {A}^{k} = \left({u}_{k-3} , {u}_{k-2} , {u}_{k-1} , {u}_{k}\right)$ where the real sequence ${u}_{k}$ satisfies the recurrence relation

\begin{equation}
\boxed{{u}_{k+1} = {u}_{k}+{u}_{k-1}+{u}_{k-2}+{u}_{k-3}}
\end{equation}

Indeed, this is an immediate consequence of the relation
$U {A}^{k+1} = U {A}^{k} A$. The initial values of $u$ are

\begin{equation}
{u}_{{-3}} = {u}_{{-2}} = {u}_{{-1}} = {u}_{0} = 1
\end{equation}

It follows that

\begin{equation}
{C}_{n+k} = {u}_{k-3} {X}_{2}^{\left(n\right)}+{u}_{k-2} {X}_{3}^{\left(n\right)}+{u}_{k-1} {X}_{4}^{\left(n\right)}+{u}_{k} {X}_{5}^{\left(n\right)}
\end{equation}

For $n = 5$, one has ${X}^{\left(5\right)} = 4 ! \left(1 , 2 , 3 , 4\right)$, hence the formula

\begin{equation}
\boxed{{C}_{5+k} = {24}\times{\left(4 {u}_{k}+3 {u}_{k-1}+2 {u}_{k-2}+{u}_{k-3}\right)}}
\end{equation}

In particular ${C}_{12} = {24}\times{1129} = 27096$

Remark: note that $A$ has the form of a Companion matrix


答:
Hints:

Show that $(G:H)=p$ implies $H$ is maximal.

Show that $H\trianglelefteq G$ and $K\not\subseteq H$ imply $H\neq HK$.


答:
Let $c=c_1c_2$ where every prime divisor of $c_1$ is a prime divisor of $a$ and $c_2$ is coprime with $a$.

Then $a+bc_2$ and $c$ are coprime.

Title: Proving that there is some $x$ such that $\gcd(a + bx, c) = 1$

Body:
Let $\gcd(a,b) = 1$ and $c>0$. Pove that there is an integer x such that $\gcd(a + bx, c) = 1$.

This question is from Section 2.3 (Covers the Chinese Remainder Theorem) Problem 15 of An Introduction to the Theory of Numbers by Niven and Zuckerman.

I didn't use the Chinese Remainder Theorem in my attempt, but I suspect that it should, since $\gcd(a,b) = 1$. In fact I didn't even use that $\gcd(a,b)=1$, either.

Attempt: Instead of finding $x$, we find what $x$ cannot be. That is, $y$ such that $\gcd(a+by,c) \neq 1$.
Reduce $c$ into its prime factors, so that $c = c_{1}^{i_{1}}c_{2}^{i_{2}}c_{3}^{i_{3}}...c_{n}^{i_{n}}$. Then we can solve, for each $1 \leq m \leq n$,

$a + by \equiv _{c_{m}}0 \Leftrightarrow by \equiv _{c_{m}} -a$

The point is that if $a + by$ and $c$ have a GCD other than $1$, then one prime factor must be in common.

$by \equiv _{c_{m}} -a$ has solutions when $gcd(b,c_{m})$ divides $-a$ and has no solution otherwise. Suppose there are no solutions for all $m$. Then we are done, for there are no restrictions on $x$. Suppose for some or all $m$, there are solutions. Then $by \equiv _{c_{m}} -a$, has solutions $y \equiv _{c_{m}} -(\frac{a}{g})y_{0}+t(\frac{c_{m}}{g})$. Where $g = \gcd(b,c_{m})$, $t = 0,1,..,g - 1$, and $y_{0}$ is a solution to $(\frac{b}{g})y \equiv _{c_{m}}1$. But the set of all these solutions cannot equal $\mathbb{Z}$, because there are infinite primes and finite $c_{m}$. So there must be some integer $x$ such that $gcd(a + bx, c) = 1$.

$\blacksquare$

I don't like this proof, because it's nonconstructive. Is there a way to use the CRT or some other basic number theorem?


问:
Uniform convergence to swap summation and integration
I am integrating a meromorphic function $f(z)$ with pole $z-z_0$, $\int_{|z-z_0|=r}f(z)dz$. Expanding this as a Laurent series we get $\int_{|z-z_0|=r}\sum_{k=-\infty}^{\infty}a_k (z-z_0)^kdz$. As, $\sum_{k=-\infty}^{\infty}a_k (z-z_0)^k$ is uniformly convergent, we can re-write the integral as $\sum_{k=-\infty}^{\infty}a_k \int_{|z-z_0|=r} (z-z_0)^kdz$. Can anyone show me the formal steps, as how this integral and sum are getting swapped using uniform convergence?

答:
If $(f_n)$ converges uniformly to $f$, then you have for every finite-length path $\gamma$, that
$$\left|\int_{\gamma} f_n(z) dz - \int_{\gamma} f(z) dz \right| =\left|\int_{\gamma} f_n(z) -f(z) dz \right| \leq \int_{\gamma} \left| f_n(z) -f(z) \right| dz \leq ||f_n-f||_{\infty} L(\gamma)$$

where $L(\gamma)$ denotes the length of the path $\gamma$.

Because $||f_n-f||_{\infty}$ tends to $0$, you deduce that
$$\lim_{n \rightarrow +\infty} \int_{\gamma} f_n(z) dz = \int_{\gamma} f(z) dz$$

i.e.

$$\lim_{n \rightarrow +\infty} \int_{\gamma} f_n(z) dz = \int_{\gamma} \lim_{n \rightarrow +\infty} f_n(z) dz$$


问:
Prove $\log x!$ is $\Omega (xlogx)$
Find a positive real number $C$ and a nonnegative real number $x_o$ such that

$Cx$$\log x$ $\leq$ $\log x!$ for all real numbers $x > x_o$.

I tried to expand $\log x!$ into $\log 1 + \log2 +\log3 +....\log x$. But how do I choose $C$ and $x_o$ so the above inequality hold?

Any hints would be appreciated.

答:
Note that
$$e^n = \sum_{k=1}^{\infty} \dfrac{n^k}{k!} \geq \dfrac{n^n}{n!} \implies \log(n!) \geq n \log(n) - n$$
Also,
$$\log(n!) = \sum_{k=1}^n \log(k) \leq \sum_{k=1}^n \log(n) = n \log(n)$$
We hence have
$$n \log(n)-n \leq \log(n!) \leq n \log(n)$$</p>

Title: Prove $\log x!$ is $\Omega (xlogx)$


问:
Is the limit of a squence of p-adic integer also a p-adic integer?
Given a squence of p-adic integers $\{x_n\}$, and assume $\lim_{n\to\infty}x_n\in\mathbb Q_p$ exists in the sense of p-adic metric. Do we know the limit is also a p-adic integer, i.e. $\lim_{n\to\infty}x_n\in\mathbb Z_p$?

I was thinking about decimal expansion of some square root. Say $\sqrt2$ in $\mathbb Z_7$, we can approximate the square root with decimal expansion $3+1\cdot7+2\cdot7^2\cdots$, which gives a sequence of p-adic integers. Then if the limit exists, does it mean $\sqrt2\in\mathbb Z_7$? I am fairly new to p-adic numbers so any help is appreciated!

答:
Yes, $\mathbb{Z}_p$ is a closed subset (it is the closed ball of radius $1$).

To answer your second question, this does mean that $\sqrt{2} \in \mathbb{Z}_7$. Another way to see this is that $\mathbb{Z}_p$ is a PID, hence a UFD, and so integrally closed in its field of fractions $\mathbb{Q}_p$. The equation for $\sqrt{2}$ is $x^2 - 2 \in \mathbb{Z}_p[x]$.


答:
Also fails for Banach spaces.  Indeed, the non-isomorphic Banach spaces $l^1$ and $l^2$ are isomorphic as vector spaces; use that isomorphism to get two non-equivalent complete norms on the same vector space.

Remark

To get strong equivalence of two complete norms on one vector space, good tools are; closed graph theorem, open mapping theorem, Banach-Steinhaus theorem, and so on.

Title: if two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?

Body:
If two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?

If $p$ and $q$ are two strongly equivalent metrics (in the sense that $ A p(x,y) \leq q(x,y) \leq B p(x,y)$ for some $A,B>0$) on some set $X$, then $(X,p)$ is complete if and only if $(X,q)$ is complete. My question is whether the converse is true as well, i.e. given two metrics $p$ and $q$ on $X$ such that $(X,p)$ and $(X,q)$ are complete, does it follow that $p$ and $q$ are strongly equivalent or at least equivalent (generating the same topology)?

If the answer is no, does the situation change in the context of Banach spaces, with metrics replaced by equivalent norms on some (of course infinite-dimensional) vector space?

In case the above converse is true, I guess one should argue by contradiction, assuming that the two metrics do not generate the same topology. Even in the case where the topology generated by one topology is strictly stronger than the other one, it is not clear to me how to find a sequence $(x_n)$ converging in the weaker metric (say $p$) but diverging in the stronger one.


答:
This is true in finite dimensional Banach spaces (since all norms on a given finite dimensional vector spaces are equivalent) but certainly not the case in the infinite dimensional case. In some sense, this is precisely the reason why topology plays such an important role in the general theory of Banach spaces.

Take any two significantly different Banach spaces, say $\ell_p$ and $\ell_q$ for $p\ne q$. The underlying sets are different, but they have the same cardinality. So, you can transport the structure from each Banach space to some fixed common set $X$. You now have two non-equivalent norms on $X$ inducing each a complete metric.


答:
The fact that $f$ tends to $\infty$ is rigorously written as follows : $\lim_{x \to \infty} f(x)= \infty  \iff \forall N > 0 \space \exists \space  M(N) > 0 \space |\space  x>M \Rightarrow  f(x) > N$

Lagrange Mean Value Theorem ensures that in the conditions specified by the question it's $\forall \space a,b \in \mathbb{R}, a 0 $

then choosing as $N=f(a)$, since both $f'(c)>0$ and $(b-a)>0$, it's

$f(b)=f'(c)(b-a)+f(a)> f(a)=N $

and so picking $a<\frac{b+a}{2}:=M(N)<b$ it's true that $b>M \Rightarrow f(b)>N$. Then for the condition below it's

$\lim_{x \to \infty} f(x)= \infty $

Title: Prove that: If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$

Body:
Let $f$ be differentiable in $\mathbb R$, with $f'$ differentiable and $f'(x) \geq 0,f''(x) \geq 0$ for every $x \in \mathbb R$.

How do I rigorously prove that (using Lagranges Mean Value Theorem):

If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$

I know that if f is monotonic increasing and $f'>m>0$ for $x>a$ (where $f'(a)=m>0$) then $\lim_{x \to \infty} f(x) > \lim_{x \to \infty} mx = \infty$

But I want something more rigorouse.

Thanks for any idea or direction

Body:
The fact that $f$ tends to $\infty$ is rigorously written as follows : $\lim_{x \to \infty} f(x)= \infty  \iff \forall N > 0 \space \exists \space  M(N) > 0 \space |\space  x>M \Rightarrow  f(x) > N$

Lagrange Mean Value Theorem ensures that in the conditions specified by the question it's $\forall \space a,b \in \mathbb{R}, a<b, \space \exists \space c \in (a,b) \space | \space f'(c)= \frac{f(b)-f(a)}{b-a} > 0 $

then choosing as $N=f(a)$, since both $f'(c)>0$ and $(b-a)>0$, it's

$f(b)=f'(c)(b-a)+f(a)> f(a)=N $

and so picking $a<\frac{b+a}{2}:=M(N)<b$ it's true that $b>M \Rightarrow f(b)>N$. Then for the condition below it's

$\lim_{x \to \infty} f(x)= \infty $


问:
elements of sets, subsets relations
give an example (if possible) such that:

a)$x\in y$ and $x\not\subseteq y$

here I take $x=\{1,2\}$ and $y=\mathcal{P(x)}=\{\emptyset, \{1\},\{2\},\{1,2\}\}$ then $x\in y$ but $x\not\subseteq y$ as e.q $1\not\in y$

b)$x\subseteq y$ and $x\not\in y$ here I cannot find any counter example...By definition of subset any element in x must be in y...
is it true?

c)$x\in y$ and $x\subseteq y$

Here I take $x=\emptyset$ and $y=\mathcal{P(x)}=\{\emptyset, \{\emptyset\}\}$ then $x\in y$ and $x\subseteq y$

could someone please check?

答:
Your answers for (a) and (c) are fine. For (b), remember that $\varnothing\subseteq y$ for any set $y$ whatsoever, so take $x=\varnothing$ and let $y$ be any set that does not have $\varnothing$ as an element, e.g., $\{\{\varnothing\}\}$.

Added: For (b) you might also consider the fact that a finite set $y$ has $2^{|y|}$ subsets and $|y|$ elements, and $n<2^n$ for all $n\in\Bbb N$. Thus, every finite $y$ will give you an example: it has more subsets than elements, so at least one of those subsets cannot be an element. In fact you can always take $x=y$: $y\subseteq y$, but $y\notin y$. If you’re working in $\mathsf{ZF}$ this is true for all $y$ as a consequence of the axiom of regularity (also called foundation), but you can check it explicitly with any simple finite example that you might construct, like your answers to (a) and (c).

Title: elements of sets, subsets relations


问:
Prove that a graph with $100$ vertices with each edge having degree at most $4$ contains an empty graph of order $20$.

Prove that for every undirected graph $G$ of 100 vertices, if every
  vertex has degree at most $4$, then there exists a subset $S$ of 20
  vertices such that no two vertices in $S$ are neighbors of one
  another.

I found this problem in the book Introduction to Theoretical Computer Science by Boaz Barak.

Here's my work:

We denote $e(G)$ to be the set of edges in a graph $G$ and $v(G)$ to be the set of vertices. Suppose for contradiction that there exists a graph $G$ such that $|v(G)| = 100$ and for all subsets of $v(G')\subset v(G)$ with $|v(G')|=20$ there exists a nonempty subgraph $G'$ with vertex set $v(G')$.

Since every vertex has degree at most $4$, we have $|E|\leq \frac{4\times 100}{2} = 200$. Then

$$\sum_{\substack{G'\subset G \\ |v(G')| = 20}}|e(G')| \geq \binom{100}{20}.$$
In summing the number of edges over all subgraphs of $G$ of size 20, it is also equivalent to counting each edge $\binom{98}{18}$ times. Then since
$$\sum_{e\in e(G)}\binom{98}{18} = \binom{98}{18}|e(G)|=\sum_{\substack{G'\subset G \\ |v(G')| = 20}}|e(G')|,$$
it follows that $\binom{98}{18} 200 \geq \binom{100}{20}$.

I'm not sure how to get a contradiction from here.

Thanks in advance for any help.

答: You can use Caroi-Wei theorem. In graph $G$ there is an independent subgraph $H$ with at least $$\alpha (G)=\sum _{i=1}^{100}{1\over d_i+1} \geq 100\cdot {1\over 5} = 20$$ vertices


问:
Why care about ultrafilters (in operator algebras)?
I recently learned about filters and ultrafilters and the notion of convergence with respect to an ultrafilter in a topological space. I have to say, at first I could not appreciate their potential. I think the two central theorems are the following:

Proposition: Let $X$ be a set. Then any $S\subset\mathcal{P}(X)$ having the finite intersection property is included in some ultrafilter.

Proposition: Let $X$ be a topological space. Then $X$ is compact if-f every ultrafilter converges to at least one point in $X$. Also, $X$ is Hausdorff if-f every ultrafilter converges at most to one point in $X$.

I realized that ultrafilters are a powerful tool when I saw a very simple proof of Tychonoff's theorem using ultrafilters. I always thought that Tychonoff's theorem had a difficult proof and I was amazed to see such a simple proof using ultrafilters.

I would like to ask the following question: how are ultrafilters used in operator algebras? What are some interesting/ standard applications they have in this field? What books or papers would you recommend to someone that wants to understand the use of ultrafilters in operator algebras?

Bonus question: What about topology, in general? To what cause would someone use ultrafilters besides proving that a space is compact or Hausdorff? Are there any interesting constructions based on ultrafilters?

答:
The use of ultrafilters in topology is extremely well established. Bourbaki would be a standard reference. Filters in general are fundamental tools in set theory and analysis, albeit not so common in the standard pedagogical route. One can even construct the real numbers in terms of filters of rational numbers.

As you point out, ultrafilters are a convenient tool when proving various results. It is a way avoid the cumbersome details of sequences and nets (when sequences are not enough). The set-theoretic formalism of filters encapsulates things quite nicely. However, of course, ultrafilters are highly non-constructive. You will never see one that is not principal. So, their use is only theoretical.

Ultrafilters are very cool, no doubt. However, one should be aware of other cool options to prove results, e.g., Tychonoff's theorem, without ultrafilters. For instance, the characterisation of compactness in terms of the closedness of the projection mapping with respect to all spaces leads to a beautiful proof of Tychonoff's theorem [Tychonoff’s theorem in a category, Proc. Amer. Math. Soc.]. That proof is applicable in a much wider context than the ultrafilter proof since it is not set-theoretic based.

I hope this somewhat answers your question.


问:
Independence of Function of Random Vectors
For $i=1,2$, let $X_i(k)$ be positive random variables which are i.i.d. in $k = 0,1,2 \dots.$. Now we form two random vectors:
$$Y:=[X_1(0)X_1(1) \;\; X_2(0)X_2(1)]^T$$
and
$$Z=[X_1(3) \;\; X_2(3)]^T$$

I wonder if $Y$ and $Z$ are independent random vectors?

My guess is not. While we have component-wise independence (in this case, we have $X_1(0)X_1(1)$ is independent of $X_1(3)$ and $X_2(0)X_2(1)$ is independent of $X_2(3)$), it does not imply mutual independence in general. Is my thinking correct? Any comment/suggestion is appreciated.

答: Your thinking is right. For any given i.i.d sequence $(X_1(k): k \geq 0)$ we can construct an i.i.d sequence  $(X_2(k): k \geq 0)$ such that $X_1(0)X_1(1)=X_2(3)$. In this case then first component of $Y$ is same as the second component of $Z$ so $Y$ and $Z$ are not independent as long as $X_1(0)X_1(1)$ is not almost surely constant.


问:
Relation between side and radius of circle without any trigonometric functions
We have an equilateral triangle with 3 circles of equal radii inscribed in it. How do we find a relation between the side length "a" of the triangle and radius "r" of the circles? With trigonometric knowledge, mainly 30-60-90 triangles, the question is pretty straightforward, but how does one go about it purely from a geometric viewpoint. I got my answer as a = 2r(1 + $\sqrt{3}$).
But have no clue about how to approach the problem without trig.

答:
Using $30°-60°-90°$ triangles:

$$2\sqrt 3r+2r=a$$

$\hspace{3cm}$

"Purely Geometric"

Since the given figure is an equilateral triangle, you will need to use the $60°$ angles somehow. Let's assume you don't know any trigonometry. Let's prove the side length ratios of $30-60-90$ triangles purely geometrically.

Suppose we have a $30-60-90$ triangle with only $HC$ equal to $k$ (in the below figure). When you stick a copy of this triangle next to the original you form an equilateral triangle (all angles are $60°$). Therefore, we know $AC=2k$ and we can apply Pythagorean Theorem in $AHC$ to obtain $AH= r \sqrt3$.

$\hspace{3cm}$


答: $(a)$ and $(c)$ are alright. For $(b)$ let $x=\{1,2\}$ and $y=\{1,2,3\}$, Then $x\subseteq y$ but $x\notin y$. In contrast an example for $x\subseteq y$ but $x\in y$ would be when $x=\{1,2\}$ and $y=\{1,2,\{1,2\}\}$.</p>

Title: elements of sets, subsets relations

Body:
give an example (if possible) such that:

a)$x\in y$ and $x\not\subseteq y$

here I take $x=\{1,2\}$ and $y=\mathcal{P(x)}=\{\emptyset, \{1\},\{2\},\{1,2\}\}$ then $x\in y$ but $x\not\subseteq y$ as e.q $1\not\in y$

b)$x\subseteq y$ and $x\not\in y$ here I cannot find any counter example...By definition of subset any element in x must be in y...
is it true?

c)$x\in y$ and $x\subseteq y$

Here I take $x=\emptyset$ and $y=\mathcal{P(x)}=\{\emptyset, \{\emptyset\}\}$ then $x\in y$ and $x\subseteq y$

could someone please check?


问:
Finding a base of matrices of rank 1
I came up with an interesting question from linear algebra that I couldn't resolve. I am interested to find out an algorithm, a characterization or perhaps a proof that only one case is possible. Of course, if some other branch of mathematics, more specialized than linear algebra, is suited better for this problem, it will useful to tell what is it.

The statement of the problem:

Let $n>1$ be a natural number. Let $\mathbb{F}$ be some algebraically closed field. Let $M_1,M_2,\dots,M_n$ be $n$ matrices over $\mathbb{F}$ of size $n\times n$ that are linearly independent, meaning that the linear closure $L=\mathcal{L}(M_1,\dots,M_n)$ (linear space span over this set of matrices) has dimension $n$. What can we say about dimension $m$ of linear subspace that is generated by matrices of rank $1$?  Namely, the dimension of $\mathcal{L}\{M\in L\mid\text{rank}M=1\}$?

I believe that in the case of $n=2$, this rank has to be always $2$. However, I couldn't prove it. Also, for other values of $n$, does it still hold that $m=n$? What are all the possible pairs $(m,n)$? Is there perhaps an algorithm or method to calculate that dimension $m$ given the matrices $M_1,M_2,\dots,M_n$?

答:
That obviously depends on the $n$ $n\times n$ matrices $M_1,\dots,M_n$ you pick and the field $\mathbb{F}$.

For example, if $n=2$ and you pick
$$M_1=\begin{pmatrix}1\\&1\end{pmatrix},M_2=\begin{pmatrix}&-1\\1&\end{pmatrix}$$
with $-1\notin(\mathbb{F}^\times)^2$ (e.g., $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=GF(p)$ with $p\equiv 3\pmod{4}$), then there are no rank 1 matrices in $L$.  But if you pick $M_1$ to be a rank-1 matrix, then it is obviously are going to be in $L$.

Title: Finding a base of matrices of rank 1


问:
How to calculate $\int_0^\infty \frac{1-\cos(x)}{x^2e^x}dx$ (it beats Maxima)?
This is an integration theory practice exam question: Show that
$$\int_0^\infty \frac{1-\cos(x)}{x^2e^x} d\lambda (x)= \frac\pi4-\frac12\ln(2).$$
Hint: You may use without proof that
$$\frac \pi 4=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}, \quad \ln(2)=\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \quad\text{and}\quad \int_0^\infty x^ne^{-x}dx = n!.$$

I thought it might be a good idea to use the exponential identity of $\cos(x)$, but that seemed to make the situation worse because it introduces complex numbers. Then looking at the hint I thought it might be worth it to insert the power series of $e$ and $\cos$ directly, but (because of the $1-$ term) those power series were "incompatible", nothing seemed to simplify which is always a bad sign.

The first part of this question showed that term by term integration would be justified.

Am I missing some steps before introducing the power series? Or can we simplify the power series expressions somehow?

答: Let $$I(a)=\int^\infty_0\frac{1-\cos ax}{x^2}\,e^{-x}\,dx.$$ Then, $$I'(a)=\int^\infty_0\frac{\sin ax}{x}\,e^{-x}\,dx$$ and $$I''(a)=\int^\infty_0\cos ax\,e^{-x}\,dx=\frac1{1+a^2}.$$ Differentiating under the integral sign is justified, because the resulting integrals are absolutely convergent. The last integral can be evaluated, integrating by parts twice. Since $I'(0)=I(0)=0,$ we have $$I'(a)=\int^a_0 \frac1{1+t^2}\,dt=\arctan a,$$ and $$I(a)=\int^a_0 \arctan x\,dx=a\,\arctan a -\frac12\,\ln(1+a^2),$$ so the original integral is $$I(1)=\arctan 1-\frac12\,\ln 2=\frac\pi4-\frac12\,\ln 2.$$


问:
A system with a hidden mistake
Consider the square $ABCD$ such that $l$ is the length of one side and $x$ is the lenght of the segment $OA$, where $O$ is the center of the square. If the perimeter of the triangle $AOB$ is $24$ and the perimeter of the triangle $ACD$ is $34$, find the area of the square $ABCD$.

So, let us write mathematically the given hypothesis. We have that
$$
2x+l=24,\qquad 2x+2l=34.
$$
The system has a unique solution $(x,l)=(7,10)$. But if $l=10$, by Pitagora's theorem, then $x=5\sqrt{2}$. So, I don't see where is the problem. Why we get two, even three, different results when come back to geometric configuration?

答:
The second information is unnecessary (and causes problem). We know that the length of the diagonal of the square is $l\sqrt2$ and that the diagonals bisect each other. Only using the fact that the perimeter of  $\bigtriangleup AOB$ is $24$:
$$l+\frac{l\sqrt2}{2}+\frac{l\sqrt2}{2}=24$$
which gives: $$l = 24 \;(\sqrt2 - 1)$$

Title: A system with a hidden mistake


答: Use the Taylor series for the cosine, so your integral is$$\sum_{n\ge0}\tfrac{(-1)^n}{(2n+2)!}\int_0^\infty x^{2n}e^{-x}d\lambda(x)=\sum_n\tfrac{(-1)^n}{(2n+1)(2n+2)}=\sum_n\left(\tfrac{(-1)^n}{2n+1}+\tfrac{(-1)^{n+1}}{2n+2}\right)=\tfrac{\pi}{4}-\tfrac12\ln2,$$where the linear combination of two alternating series is legal because the resulting alternating series still converges by the alternating series test.

Title: How to calculate $\int_0^\infty \frac{1-\cos(x)}{x^2e^x}dx$ (it beats Maxima)?

Body:
This is an integration theory practice exam question: Show that
$$\int_0^\infty \frac{1-\cos(x)}{x^2e^x} d\lambda (x)= \frac\pi4-\frac12\ln(2).$$
Hint: You may use without proof that
$$\frac \pi 4=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}, \quad \ln(2)=\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \quad\text{and}\quad \int_0^\infty x^ne^{-x}dx = n!.$$

I thought it might be a good idea to use the exponential identity of $\cos(x)$, but that seemed to make the situation worse because it introduces complex numbers. Then looking at the hint I thought it might be worth it to insert the power series of $e$ and $\cos$ directly, but (because of the $1-$ term) those power series were "incompatible", nothing seemed to simplify which is always a bad sign.

The first part of this question showed that term by term integration would be justified.

Am I missing some steps before introducing the power series? Or can we simplify the power series expressions somehow?


问:
We are labelling some subsets from set $\{1,2...,100\}$. Prove that...

Let $S = \{1, 2, 3, ..., 100\}$ and $P$ is the collection of all subset $T$ of $S$ that have $49$ elements, or in other words: $$P = \{T \subset S : |T| = 49\}.$$ Every element of $P$ is labelled by the element of $S$ randomly (the labels may be the same). Show that there exist subset $M$ of $S$ that has $50$ members such that for every $x \in M$, the label of $M -\{x\}$ is not equal to $x$

This problem, from USAMO 1988 and Indonesia IMO TST 2008, seems to me very easy so I need proof verification.

Edit: The following ''proof'' is wrong as Especially Lime pointed out.

Proof: Suppose it is not true. Then there exists such labelling that for each $M$ there exists $m\in M$ such that labeling of $M -\{m\}$ is $m$. We make a bipartite graph, one side with subsets $M$ of $S$ with power $50$ and the other is set $S$ it self. We connect $M$ with $s\in S$ iff labelling $M -\{s\}$ is $s$.

By assumption, every $M$ has degree at least $1$, and every element $s$ in $S$ has degree at most ${99\choose 49}$, so we have $${100\choose 50}\cdot 1\leq 100\cdot {99\choose 49}\implies 51\leq 50,$$
a contradiction.

答:
Your final simplification is incorrect; $\binom{100}{50}\cdot 1=\frac{100}{50}\cdot\binom{99}{49}$, so the penultimate inequality is true.

This method couldn't possibly be sufficient since you haven't yet used the fact that each set only has one label. (Your solution so far would still apply if multiple labels are allowed and you have an edge if any one of the labels of $M-s$ is $s$, when the result clearly isn't true.)


问:
Given $H\triangleleft G, K\le G\ $, $\;K\nsubseteq H,\;$ and $(G:H) = p$, $\;p$ prime, prove $HK=G$
Suppose that $H\triangleleft G, K\le G\ $ and $K\nsubseteq H$. How we can prove that $HK=G?$

Also $(G:H)=p$ where p is prime.

答:
By this formula:
$$
[G : H] = [G : HK][HK : H]
$$

Since $[G : H] = p$ is prime, one of $[HK : H]$ and $[H : HK]$ must be $1$. It cannot be $[HK : H]$ because $K \not \subset H$ so $HK \ne H$. This forces $[G : HK] = 1$ and hence $G = HK$.


问:
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots

Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.

But I wonder that there is some logical ways we can find intervals that contain roots without guessing?

答:
We may apply Sturm's theorem https://en.wikipedia.org/wiki/Sturm%27s_theorem.

Let $\mathrm{rem}(p(x),q(x))$ denote the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.

The Sturm sequence is given by:

$p_0(x) = x^5-5x^3+4x-1$

$p_1(x) = p_0'(x) = 5x^4 - 15x^2 + 4$

$p_2(x) = -\mathrm{rem}(p_0, p_1) = 2x^3 - \frac{16}{5}x + 1$

$p_3(x) = -\mathrm{rem}(p_1, p_2) = 7x^2 + \frac{5}{2}x - 4$

$p_4(x) = -\mathrm{rem}(p2, p_3) = \frac{883}{490}x - \frac{29}{49}$

$p_5(x) = -\mathrm{rem}(p3, p4) = 1889881/779689$

As this is a constant, this finishes the computation of the Sturm sequence.

We have $V(-\infty) - V(+\infty) = 5 - 0 = 5$.

So there are five real roots.


问:
Countable set of finite sets is countable (without the axiom of choice).
So i have the infinite family of sets:
$$\{A_0, A_1, A_2 \dots\}$$
and $\forall i \in \omega,\; A_i$ is finite. I need to show that $\cup A_i$ is countable  ($\exists f: \cup A_i \rightarrow \omega $ and f is bijection).
I seem to understand it intuitively, but I can't prove it. I also don't have to use the axiom of choice. Any tips ?

答:
Proof with axiom of choice :

For every $i \geq 0$, let $\varphi_i : A_i \rightarrow \mathbb{N}$ be an injective map. Define
$\varphi : \bigcup_{i \geq 0} A_i \rightarrow \mathbb{N}^2$ by
$$\varphi(x)=(i_0, \varphi_{i_0}(x))$$

where $i_0$ is any integer such that $x \in A_{i_0}$. Then it is easy to see that $\varphi$ is injective.

It remains to compose $\varphi$ with an injective map $\mathbb{N}^2 \rightarrow \mathbb{N}$ (which is easy to construct) to get an injection from $\bigcup_{i \geq 0} A_i $ to $\mathbb{N}$, which let you conclude that $\bigcup_{i \geq 0} A_i $ is countable.


问:
Does Borda count satisfies the participation criterion?
The Borda count, or average rank method, is said to satisfies the participation criterion.

This means that this ranking method is free from the "no show paradox".

Do you know any proof of this claim? Can you please link to a reference?

答:
The proof is fairly simple, so I don't know of any work that spells it out (though there might well be).

To satisfy the Participation Criterion, The following situation should be impossible:

Voter Anne prefers candidate $A$ over $B$.

With all votes tallied, $B$ wins the election.

Had Anne not voted, $A$ would have won.

For Borda count, the candidate with the most "points" is elected, where the points of a candidate are simply the sum of the points they get per ballot, in which a better place on the ballot gives more points. This means that with all votes tallied, $B$ has more points, but with Anne's ballot removed, $A$ has more points. This necessarily means that she assigned more points to $B$ than $A$, which is in contradiction with Anne preferring candidate $A$ over $B$.


问:
Complex inequalities with fractions
The circle of radius $r$ in the xy-plane centred at $(a, b) \in \Bbb{R}^2$
is described by the equation $(x − a)^2 +(y − b)^2 = r^2$

Let $z = x + iy \in \Bbb C$.
Find the radius and the centre of the circle described on the Argand diagram given by the equation $\displaystyle \left\vert \frac{z+i}{z-1}\right\vert =2$.

I have many similar questions without fractions however i dont know what to do with the fraction should I do the division with $z=a+bi$ or not any help is apricated

答:
Note that the equation $$\left|\dfrac{z-a}{z-b}{}\right|=c$$ where $c$ is a constant real number represents a circle known as Apollonius Circle. It is the locus of points whose ratio of distances from two fixed points $a$ and $b$ is a constant $c$.

Now, if we consider the question given, the two fixed points from where the distances are measured (the focii) are $A(z=-i)$ and $B(z=1)$.

The distance between the points $z=-i$ and $z=1$ is simply $\sqrt{2}$.

Therefore if we if we call the point $C$ where circle cuts the line joining $A$ and $B$, we know that $\dfrac{AC}{BC}=2$ and $AC+BC=\sqrt{2}$ so we get $AC=\dfrac{2\sqrt{2}}{3}$ and $BC=\dfrac{\sqrt{2}}{3}$. Now, also since $\dfrac{AD}{BD}=2$, we get $BD=\sqrt{2}$ which means diameter is $BD+BC=\sqrt{2}+\dfrac{\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{3}.$ Hence radius is $\dfrac{2\sqrt{2}}{3}$.

For coordinates of $B$, we can write the equation of the line joining them as $y=x-1$ in the $xy$ plane. So coordinates of $B$ are $\left(\dfrac{2}{3}+\dfrac{2\sqrt{2}}{3}\cos(45^{\circ}),\dfrac{-1}{3}+\dfrac{2\sqrt{2}}{3}\sin(45^{\circ})\right)\equiv \left(\dfrac{4}{3},\dfrac{1}{3}\right)$

Title: Complex inequalities with fractions


答:
Call $z=x+iy$. Then you have
$$\left| \frac{x+i y+i}{x+i y-1}\right|=2$$
that is
$$\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{(x-1)^2+y^2}}=2$$
square both sides and simplify
$$3 x^2+3 y^2-8 x-2 y+3=0 \\
x^2 + y^2 -\frac8{3}x - \frac2{3}y + 1 = 0 \\
x^2 - 2 \cdot \frac4{3}x + \left(\frac4{3}\right)^2 - \left(\frac4{3}\right)^2 + y^2 - 2 \cdot \frac1{3}y + \left(\frac1{3}\right)^2 - \left(\frac1{3}\right)^2 + 1 = 0 \\
\left(x-\frac4{3}\right)^2 + \left(y - \frac1{3}\right)^2 = \frac8{9} = \left(\frac{2\sqrt{2}}3\right)^2
$$

circle with center $\left(\frac43,\frac13\right)$ and radius $r=\frac{2 \sqrt{2}}{3}$

Title: Complex inequalities with fractions

Body:
The circle of radius $r$ in the xy-plane centred at $(a, b) \in \Bbb{R}^2$
is described by the equation $(x − a)^2 +(y − b)^2 = r^2$

Let $z = x + iy \in \Bbb C$.
Find the radius and the centre of the circle described on the Argand diagram given by the equation $\displaystyle \left\vert \frac{z+i}{z-1}\right\vert =2$.

I have many similar questions without fractions however i dont know what to do with the fraction should I do the division with $z=a+bi$ or not any help is apricated


问:
Find the Laurent expansion of $(1-z)e^{1/z}$ - When can we use Taylor series to find Laurent series?
I'm currently taking a course in mathemathical tools, where we are covering complex analysis (Note that this course is not very rigorous and we cover complex analysis in only 4 lectures). The Laurent series have been introduced as:

Laurent series$$
\begin{array}{l}
f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n} \\
a_{n}=\frac{1}{2 \pi i} \oint_{C} \frac{f\left(z^{\prime}\right) d z^{\prime}}{\left(z^{\prime}-z_{0}\right)^{n+1}}
\end{array}
$$

However, I noticed that the above integral isn't used in the solutions to any of the problems regarding laurent series. So I'm trying to understand why we can avoid using it.
What I understood so far, is that the difference between the Taylor and Laurent series, is that the Laurent series also contains negative powers. Where it's not possible to expand a Taylor series around a point $z_0$ where f is not analytically, the opposite applies for the Laurent series.
In the case we expand f around a point where f is analytically, the Laurent series and Taylor expansion will be the same.
If that is the correct understanding, please help me understand the solution to the following problem:

Find the Laurent series for $f(z)=(1-z)e^{1/z}$ about $z=0$

What I thought should be my approach is using the formula above solving the contour integral, since $f(z)$ is non-analytically at $z=0$. However, the solution uses directly that $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$. So here comes my first question:

Why is the Laurent series of $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$.

It seems to me, that they have just substituted $z \rightarrow 1/z$ in the Taylor series for $e^z$. But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist? And if that is not the Taylor series, how do we know it's the Laurent series?

After this they multiply the two expressions together:
$(z-1) e^{1 / z}=(z-1) \sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=z-\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right) \frac{z^{-n}}{n !}$

Is that because $z-1$ is a polynomial and thus is analytically around $z=0$ and is it's own Taylor series and Laurent series. So we can just find the Laurent series of each factor and multiply them together to find the final Laurent series?

Fx: Consider $h(z)=f(z)g(z)$, if we want to find the Laurent series around $z=z_0$. If the Laurent series of f(z) around $z_0$ is $\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}$ and of $g(z)$ is $\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n}$. Is the Laurent series of $h(z)$ then:
$h(z)=(\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}) (\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n})$?

答:
Since, for each $z\in\Bbb C\setminus\{0\}$,$$e^{1/z}=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots,$$you have (again, for each $z\in\Bbb C\setminus\{0\}$),\begin{align}(1-z)e^{1/z}&=-z+(1-1)+\left(1-\frac1{2!}\right)\frac1z+\left(\frac1{2!}-\frac1{3!}\right)\frac1{z^2}+\left(\frac1{3!}-\frac1{4!}\right)\frac1{z^4}+\cdots\\&=-z+\sum_{n=0}^\infty\left(\frac1{n!}-\frac1{(n+1)!}\right)\frac1{z^n}.\end{align}
Concerning your final question: $\displaystyle\left(\sum_{n=-\infty}^\infty a_n\left(z-z_{0}\right)^{n}\right)\left(\sum_{n=-\infty}^\infty b_n\left(z-z_{0}\right)^{n}\right)$ is not a Laurent series.


问:
Index of twisted Dirac operator
I do not understand a step in the proof of the Lemma 11.4.1 in the book "Analytic K-Homology" by Higson, Roe.

Let $S$ be a Dirac bundle over a closed manifold $M$ and $D$ the corresponding Dirac operator. Let $E$ be a Hermitian vector bundle over $M$. Then $S \otimes E$ is again a Dirac bundle, so let $D_E$ be the corresponding twisted Dirac operator.

Let $P \colon M \to M_n(\mathbb{C})$ be the projection-valued function determining $E$. Let $D_n$ be the operator $1 \otimes D$ on $H_n := \mathbb{C}^n \otimes L^2(S)$. Let $\chi$ be a normalizing function.

Then it is claimed that $\operatorname{Index}(\chi(D_E))$ equals the index of the operator $\chi(PD_nP)$ on the space $PH_n$.

What bothers me is that the operator  $PD_nP$ does not depend on the chosen metric / connection on $E$, whereas the operator $D_E$ does depend on it. And it is never mentioned in the book that the index of the twisted operator $D_E$ is independent of this choices.

答:
Actually, the datum $P : M \to M_n(\mathbb{C})$ defines both a metric $(\cdot,\cdot)^P$ and a connection $\nabla^P$ on $E$:

If $P\xi$, $P\eta \in P(C^\infty(M) \otimes \mathbb{C}^n) = \Gamma(E)$, then $$(\xi,\eta)^P_x := \langle P(x)\xi(x), P(x)\eta(x)\rangle$$ for $\langle \cdot,\cdot \rangle$ the inner product on $\mathbb{C}^n$.

If $P\xi \in P(C^\infty(M) \otimes \mathbb{C}^n) = \Gamma(E)$, then $$\nabla^P P\xi := (1_{\Omega^1(M)} \otimes P) d P\xi,$$ where $d$ is the exterior derivative on $C^\infty(M)$ acting on $C^\infty(M) \otimes \mathbb{C}^n$ as $d \otimes 1_n$.

One then has that $PD_n P$ is, in fact, the twisted Dirac operator of the twisted spinor bundle $S \otimes E$, for $E$ endowed instead with the metric $(\cdot,\cdot)^P$ and connection $\nabla^P$. In particular, the fact that $\operatorname{Ind}(D_E) = \operatorname{Ind}(PD_n P)$ shows that $\operatorname{Ind}(D_E)$ is independent of the choice of metric and connection on $E$.

If you want another perspective on this issue, take a look at Chapter 11 of Roe's Elliptic operators, topology and asymptotic methods, which, in general, offers a very nice, concretely global-analytic complement to Higson--Roe.

So, one can define a continuous family of Dirac operators on a vector bundle $F \to M$ (e.g., your $F = S \otimes E$) as a one-parameter family $(g_t;c_t,(\cdot,\cdot)_t,\nabla^t)$, where for each $t \in [0,1]$,

$g_t$ is a Riemannian metric on $M$;

$c_t$ is a Clifford action, $(\cdot,\cdot)_t$ a Hermitian metric, and $\nabla^t$ a Clifford connection, making $F$ into a Dirac bundle over $(M,g_t)$;

for each $t$, then, let $D_t$ be the Dirac operator of the Dirac bundle $(F,(\cdot,\cdot)_t,c_t,\nabla^t) \to (M,g_t)$. One can then show that the $D_t$ satisfy elliptic estimates uniformly in $t$, and hence prove by functional-analytic arguments [Roe, Prop. 11.13] that $t \mapsto \operatorname{Ind}(D_t)$ must therefore be constant. In your particular context, $g$ and $c$ are constant, particularly since the spinor bundle $S$ is fixed, but given your vector bundle $E$, you'd need to construct $(\cdot,\cdot)_t$ and $\nabla^t$ to appy Roe's proposition. Then main ideas are these:

If $(\cdot,\cdot)_0$ and $(\cdot,\cdot)_1$ are two smooth metrics on $E$, then there exists a fibre-wise positive-definite (with respect to $(\cdot,\cdot)_0$) $T \in \operatorname{End}(E)$ such that $(\xi,\eta)_1 = (\xi,T\eta)_0$ for all $\xi,\eta \in \Gamma(E)$. Since $T$ is fibrewise positive-definite with respect to $(\cdot,\cdot)_0$, you can define $\log T \in \operatorname{End}(E)$ with $T_x = \exp((\log T)_x)$ for each $x$; then $(\cdot,\cdot)_t$ can be defined by $$(\xi(x),\eta(x))_{t,x} := (\xi(x),\exp(t(\log T)_x)\eta(x))_{0,x}.$$

If the metric $(\cdot,\cdot)$ is fixed, and $\nabla$ and $\nabla^\prime$ are two metric connections on $E$, then $M := \nabla^\prime - \nabla$ is a skew-adjoint $\operatorname{End}(E)$-valued $1$-form, and hence you can take $\nabla^t := \nabla + t M$.

These ingredients then allow you to define the corresponding continuous family of Dirac operators on $S \otimes E$.

Title: Index of twisted Dirac operator


问:
Does the sequence $\frac{1}{1-n}$ converge?
New to analysis here. I recognise this sequence converges to $0$, but surely it's unbounded since it's undefined at $n = 1$, which is a contradiction. Could someone just point out the problem in my thought process?

Thanks

答: As a sequence, it's not unbounded. It's just not defined when $n=1$. This sequence could reasonably be defined for $n\ge 2$. And so yes it is indeed convergent to $0$.