问:
Hypothesis testing in elections
The following is a problem I am working on for my stats class regarding hypothesis testing:

I have a random sample $X_1,...,X_n$ of voters who either voted for Candidate A ($X_1=1$) or for Candidate B ($X_i=0$), with probability of voting for A denoted as $p$.

I want to test the hypothesis that A will not lose the election: $H_0:p\geq 0.5$ vs. $H_1:p<0.5$. I set the null parameter space $\Theta_0=[0.5,1]$ and the power test to $\beta_\psi(p)=\mathbb{P}_p(T\leq c)=\sum_{k=0}^c{n\choose k}p^k(1-p)^{n-k}$, observing that for $c\in\{0,...,n-1\}$, the power test is strictly decreasing in $p$ and that size of $\psi$ is thus $\beta_\psi(0.5)$.

Now, I want to find $c$ so that the power test is at level 5% with the largest possible power, at sample size $n=100$. Is it correct for me to assume that I must find the largest $c$ such that $\beta_\psi(0.5)\leq 0.05$? I do intuit that this is because of the null hypothesis $H_0:p\geq 0.5$ and $c$ being in the "same direction" as $p$ (number of A voters in the sample is used to "estimate" the probability of voting A, which we want to increase to the minimal degree to satisfy the level-$\alpha$ test; the "increasing" being to control type I error and the "to the minimal degree" being to minimize type II error).

Could somebody please verify this? If there are any mistakes or a more rigorous explanation in what I said, please let me know.

答:
Assume polling is random, that subjects of the poll are registered voters, will vote, have decided their preference, tell the truth, and will not change their preference before the election.

Let $Y$ be the number out of $n = 100$ who will vote for A. Under $H_0: p = 0.5,$ we have $Y \sim\mathsf{Binom}(n=100, p=.5).$
We will reject $H_0$ against $H_a: p < .5$ for small observed
$Y.$ To test at the 5% level, we need to find the critical
value $c = 41$ such that $\alpha = P(Y \le c\,|\,p=.5) < 0.05,$ but as near to $0.05$ as possible; with $\alpha = 0.443.$
[You are correct that this $\alpha$ and $c$ give the best power.]

Computations in R:

```<code>qbinom(.05, 100, .5)
[1] 42
pbinom(42, 100, .5)
[1] 0.06660531
pbinom(41, 100, .5)
[1] 0.04431304
</code>```

I do not understand your notation for the power of the test. So I will define and use my own.

The power of a test at level $\alpha = 0.0443$ against alternative $p_a < .5$ is
$\pi(p_a) = P(Y \le 41 \,|\, p_a).$ We can plot a power curve
for various alternative values $p_1,$ as follows:

```<code>p.a = seq(0, .5, by=.01)
pwr = pbinom(41, 100, p.a)
plot(p.a, pwr, type="l", ylim=c(0,1), lwd=2, main="Power Curve")
 abline(h = c(.05, 1), col="green2")
 abline(v = c(0,.5), col="green2")
</code>```

It seems that we are almost sure to reject $H_0$ if the true
proportion in favor of A is less than about $p_a = 0.27.$

However, if only 40% favor A, we have only about a $60\%$ chance
of detecting that A will lose the election. In serious election
polling it takes a poll with about $n = 2500$ subjects to be
95% sure that the margin of sampling error is smaller than $\pm 2\%.$

I used exact binomial probabilities from R throughout. However, with $n = 100,$ you
could get reasonably precise values for the probabilities by using normal approximations to binomial distributions for $p > 0.05.$

Below is output from the 'power and sample size' procedure using
a recent release of Minitab statistical software. (The dotted
power curve is for $n=2500.)$

```<code>Power and Sample Size
Test for One Proportion

Testing p = 0.5 (versus < 0.5)
α = 0.05

              Sample
Comparison p    Size    Power
         0.1     100  1.00000
         0.1    2500  1.00000
         0.2     100  1.00000
         0.2    2500  1.00000
         0.3     100  0.99491
         0.3    2500  1.00000
         0.4     100  0.64150
         0.4    2500  1.00000
</code>```

Title: Hypothesis testing in elections

Body:
Assume polling is random, that subjects of the poll are registered voters, will vote, have decided their preference, tell the truth, and will not change their preference before the election.

Let $Y$ be the number out of $n = 100$ who will vote for A. Under $H_0: p = 0.5,$ we have $Y \sim\mathsf{Binom}(n=100, p=.5).$
We will reject $H_0$ against $H_a: p < .5$ for small observed
$Y.$ To test at the 5% level, we need to find the critical
value $c = 41$ such that $\alpha = P(Y \le c\,|\,p=.5) < 0.05,$ but as near to $0.05$ as possible; with $\alpha = 0.443.$
[You are correct that this $\alpha$ and $c$ give the best power.]

Computations in R:

qbinom(.05, 100, .5)
[1] 42
pbinom(42, 100, .5)
[1] 0.06660531
pbinom(41, 100, .5)
[1] 0.04431304

I do not understand your notation for the power of the test. So I will define and use my own.

The power of a test at level $\alpha = 0.0443$ against alternative $p_a < .5$ is
$\pi(p_a) = P(Y \le 41 \,|\, p_a).$ We can plot a power curve
for various alternative values $p_1,$ as follows:

p.a = seq(0, .5, by=.01)
pwr = pbinom(41, 100, p.a)
plot(p.a, pwr, type="l", ylim=c(0,1), lwd=2, main="Power Curve")
 abline(h = c(.05, 1), col="green2")
 abline(v = c(0,.5), col="green2")

It seems that we are almost sure to reject $H_0$ if the true
proportion in favor of A is less than about $p_a = 0.27.$

However, if only 40% favor A, we have only about a $60\%$ chance
of detecting that A will lose the election. In serious election
polling it takes a poll with about $n = 2500$ subjects to be
95% sure that the margin of sampling error is smaller than $\pm 2\%.$

I used exact binomial probabilities from R throughout. However, with $n = 100,$ you
could get reasonably precise values for the probabilities by using normal approximations to binomial distributions for $p > 0.05.$

Below is output from the 'power and sample size' procedure using
a recent release of Minitab statistical software. (The dotted
power curve is for $n=2500.)$

Power and Sample Size
Test for One Proportion

Testing p = 0.5 (versus < 0.5)
α = 0.05

              Sample
Comparison p    Size    Power
         0.1     100  1.00000
         0.1    2500  1.00000
         0.2     100  1.00000
         0.2    2500  1.00000
         0.3     100  0.99491
         0.3    2500  1.00000
         0.4     100  0.64150
         0.4    2500  1.00000


问:
Image of Sphere is a submanifold of $R^6$
I was trying to prove that $\alpha(S^2)$ is a submanifold of $\mathbb{R}^6$.

$\alpha(x,y,z) = (x^2, y^2, z^2, \sqrt2 xy, \sqrt2 yz, \sqrt2 zx)$

$\textbf{Attempt}$

I started by proving that  $\alpha(S^2)$ is contained in $S^5$, then I noticed that $S^5$ is a submanifold of $\mathbb{R}^6$ in this point can i conclude that $\alpha(S^2)$ is also a submanifold of $\mathbb{R}^6$?

If my approach is wrong, I would love some hints!

Thank you.

答: Hint: Verify that $\alpha$ is an immersion and then look at Homeomorphism from $P^2\mathbb{R}$ onto the image of $\mathbb{S}^2$ through the Veronese map .

ContentLicense: CC BY-SA 4.0


问:
$Q(x)=f(x^3)+xg(x^3)$ and $Q(x)$ is divisible by $x^2+x+1$
Let $Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$ and $Q(x)$ is divisible by $x^2+x+1$, then find the value of $Q(1)$

My approach is as follow

$Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$

$Q\left( x \right) = \phi \left( x \right)\left( {{x^2} + x + 1} \right)$

$Q\left( \omega  \right) = \phi \left( \omega  \right)\left( {{\omega ^2} + \omega  + 1} \right) = 0$

$Q\left( {{\omega ^2}} \right) = \phi \left( \omega  \right)\left( {{\omega ^4} + {\omega ^2} + 1} \right) = 0$

$Q\left( \omega  \right) = f\left( {{\omega ^3}} \right) + \omega g\left( {{\omega ^3}} \right) = f\left( 1 \right) + \omega g\left( 1 \right) = 0$

$Q\left( {{\omega ^2}} \right) = f\left( {{\omega ^6}} \right) + {\omega ^2}g\left( {{\omega ^6}} \right) = f\left( 1 \right) + {\omega ^2}g\left( 1 \right) = 0$

Adding we get $2f(1)-g(1)=0$

$Q\left( 1 \right) = f\left( 1 \right) + g\left( 1 \right)$

Not able to proceed from here

答:
Hint: if $x^2+x+1|Q(x)$, then
$$x^3-1\ \big| (x-1)Q(x)=(x^2-x)g(x^3)+(x-1)f(x^3).$$
See if you can use this, combined with the property that
$$f(x)-g(x)\ \big|\ h(f(x))-h(g(x))$$
for any polynomials $f,g,h$, to find $f(1)$ and $g(1)$.


问:
Find the natural number solution to the equation $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=1$
What I can do is, $x\gt 1, y \gt 2$ and $z \gt 3$. One of the solution I found is $x=3, y=6, z=9$. Is it the only solution?

答:
There are exactly 32 solutions to the equation $$\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=1;x,y,z\in\mathbb{N}$$The solutions in the form of $(x,y,z)$ are: $(2,5,30),(2,6,18),(2,7,14),(2,8,12),(2,10,10),(2,12,9),(2,16,8),(2,28,7),(3,4,18),(3,6,9),(3,12,6),(3,30,5),(4,3,12),(4,4,12),(4,8,6),(5,4,10),(5,40,4),(5,10,5),(6,3,18),(6,4,9),(6,6,6),(6,24,4),(8,4,8),(8,16,4),(10,5,6),(12,3,12),(12,12,4),(14,4,7),(15,6,5),(20,10,4),(30,3,10),(39,9,4)$

To get these solutions, I checked every natural number pair of $(x,y)$ up to $(400,400)$. We only need the two variables because we can solve for the last one. Solving for $z$, we get $$z=\frac{3xy}{xy-y-2x}$$ We know that $z$ is a natural number, so this is essentially asking the same thing as "When is $z=\frac{3xy}{xy-y-2x}$ a natural number?"

Either those are the only solutions, or the next solution just has really big numbers. To prove the former case, we know that if there is another solution, then either $x$ or $y$ must be greater than $39$ and $40$ respectively since those are the largest values for $x$ and $y$ in the set I've provided.

We'll start with your original problem and manipulate it to get this: $$\frac{2}{y}+\frac{3}{z}=1-\frac{1}{x}$$Since $x\gt39$,$$1-\frac{1}{x}\gt\frac{38}{39}\Rightarrow\frac{2}{y}+\frac{3}{z}\gt\frac{38}{39}$$So, now we know that if there is a solution to your original problem with $x\gt39$, then that second inequality must also be true. There are fourteen integer pairs that fit that description if $y\gt2$ and $z\gt3$ (you can graph this on Desmos to see). Checking all of them, we see that none of them are solutions to your original problem. That must mean that our assumption of $x\gt39$ was false.

This also works if you replace $y$ instead of $x$ which gets you $$\frac{1}{x}+\frac{3}{z}\gt\frac{39}{40}$$resulting in five possible integer pairs given $x\gt1$ and $z\gt3$. Again, we find that none of them are solutions to your problem. We started by assuming that if there is another solution, then either $x\gt39$ or $y\gt40$. We proved that neither of those cases are true. So that means there are no other solutions.


问:
Find matrix square roots
I recently came across this exercise where you are supposed to find the set of all square roots of the following matrices:

$\begin{pmatrix} 0 & 0 \\ 1 & 0  \end{pmatrix}$

$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

First, I'll explain the easy way I did it: the first is pretty obvious that there are no solutions (and a well known example). For the second one, its Jordan submatrix for the eigenvalue $0$ is just the $2\times 2$ block above which can't be factored on its own.

I am unsure about the third one, since I don't know how to prove that there aren't any other solutions than the two obvious ones which are $\begin{pmatrix} 0 & ±1 & 0 \\ 0 & 0 & ±1 \\ 0 & 0 & 0 \end{pmatrix}$.

So, my $3$ questions are: are there any other square roots for the third example, if not why? how to do this just using simpler linear algebra (this was a beginning of the course exercise), without just brute forcing the system of equations? Are there any good results about the number of solutions of the equation $X^2=A$ for singular matrices (or even non-diagonalisable)?

Thanks

答:
A $3 \times 3$ matrix commutes with $A = \pmatrix{0 & 0 & 1\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$ iff it is upper triangular with equal $(1,1)$ and $(3,3)$ entries.  If it's a square root of $A$ the eigenvalues (which are the diagonal entries) must be $0$.  And then the requirement for the square of $B = \pmatrix{0 & b_{12} & b_{13}\cr 0 & 0 & b_{23}\cr 0 & 0 & 0\cr}$ to be $A$ is $b_{12} b_{23} = 1$.

EDIT:
As for $X^3 = A$, that can have infinitely many solutions as well, if the matrices are at least $4 \times 4$.  Thus
$$\pmatrix{0 & x_{12} & x_{13} & x_{14}\cr 0 & 0 & x_{23} & x_{24}\cr
0 & 0 & 0 & x_{34}\cr 0 & 0 & 0 & 0\cr}^3 = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}$$ iff $x_{12} x_{23} x_{34} = 1$.

Title: Find matrix square roots


问:
How useless can the Mayer-Vietoris sequence be in general?
In an algebraic topology course I'm taking we are often asked to compute the homology groups of a space $X = A \cup B$ using the Mayer-Vietoris sequence, and it happens in all of the examples I've seen so far that it's possible to do this without knowing anything about the connecting homomorphisms $\partial_{\ast}$ (say on the level of chains); we only end up needing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ and possibly some of the inclusion maps. </p>

<p>My guess is that this is not a typical situation; is there a relatively simple example of a nice space $X$ and nice subspaces $A, B$ such that knowing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ is not enough to compute $H_{\ast}(X)$ without knowing the specific form of the connecting homomorphisms? (For maximal relevance to the course $X, A, B, A \cap B$ should be finite simplicial complexes.) </p>

答:
Given a long exact sequence

$$ \cdots \to C_{i+1} \to A_i \to B_i \to C_i \to A_{i-1} \to \cdots $$

let the map $A_i \to B_i$ be denoted $f_i$.  Then you have that $C_i$ is an extension

$$ 0 \to coker(f_i) \to C_i \to ker(f_{i-1}) \to 0$$

so up to that extension problem, the maps $f_i$ always determine the $C_i$ groups.  So if you want a situation where the group $C_i$ is ambiguous, you could have $ker(f_{i-1}) = \mathbb Z_2$ and $coker(f_i) = \mathbb Z$, that way $C_i$ could be either $\mathbb Z$ or $\mathbb Z \oplus \mathbb Z_2$.

Regardless, the connecting map $\partial_i : C_i \to A_{i-1}$ is determined by this extension problem, and it's easy enough to cook up examples either-way.

So I'm a little confused as to the nature of your question. I guess what I'm saying is that you are in the typical situation, and Grigory's example is also typical in that it's the inclusion map that makes the differences between his examples.

Regarding how useful/useless the MVS is for a typical problem, it really depends on how easily-expressible your space is as a union of spaces you understand (and their intersections).  If your space doesn't fit that profile, you've got potentially a lot of work to do.  The Serre Spectral Sequence of a Fibration is in a sense something of a souped-up Mayer-Vietoris sequence, and there are plenty of papers where people are happy just computing the $E_3$-page, or determining which page the SS collapses on, or computing a differential.  These extension issues tend to be very thorny and consume much literature.

ContentLicense: CC BY-SA 3.0


问:
How do you prove that there are infinitely many primes of the form $5 + 6n$?
There should be infinitely many primes of the form $5+6n$. How do you prove it? The same should be true for $7+6n$.</p>

答:
Primes of the form $6n+5$ is particularly easy:

Suppose that there are finitely many primes of the form $6n+5$, namely
$p_1,\cdots, p_n$.

Consider
$p^*=6p_1\cdots p_n -1$.

Note that any odd prime other than 3, is of the form $6n+1$ or $6n+5$.

Thus, prime divisors of $p^*$ are either of the form $6n+1$ or $6n+5$.

The prime divisors of $p^*$ should have at least one prime divisor of the form $6n+5$.

This is a contradiction.

For primes of the form $6n+1$, use the following:

"Existence of $x$ in $x^2-x+1\equiv 0\textrm{ mod }p$ $\Longleftrightarrow$ $p$ is of the form $6n+1$."

Suppose there are only finitely many $6n+1$ primes, namely $p_1,\cdots, p_n$,

Then consider $p^*=(p_1\cdots p_n)^2-(p_1\cdots p_n) + 1$.

Prime divisor of $p^*$ should be of the form $6n+1$ according to the above equivalence.

This is a contradiction.


问:
What is the sum of all elements in a symmetric group applied to a function of a single label?
Let $S_n$ be the symmetric group over integers $[1,n]$. Let us also define an arbitrary function $f[i]$ of integer $i$.

I want to calculate
$$\sum_{\sigma\in S_n}f[\sigma(i)].$$

Intuitively, I would argue that the images $\sigma(i)$ should be evenly distributed across the group; therefore,
$$\sum_{k=1}^{n!}f[\sigma_k(i)]=(n-1)!\sum_{j=1}^n f[j].$$

However, that reasoning doesn't sound very formal. Is my assumption correct? How can this result be more soundly justified?

答:
Yes, your reasoning is correct. You can first group the terms by $\sigma(i)$, which it sounds like is what you've been doing.
$$\sum_{\sigma\in S_n}f[\sigma(i)]=\sum_{j=1}^nf[j]\left|\big\{\sigma\in S_n\ \colon \sigma(i)=j\big\}\right|.$$
So, all you need to show is that there are $(n-1)!$ permutations in $S_n$ for which $\sigma(i)=j$, given fixed $i$ and $j$. There are a few ways to do this -- you can explicitly count them (by repeatedly fixing values of $\sigma$). Alternatively, if you define
$$T_j=\big\{\sigma\in S_n\ \colon \sigma(i)=j\big\},$$
and biject $T_j$ with $T_1$ for each $j$ (see if you can figure out such a bijection). From here, the $n$ sets $T_j$ are disjoint, of the same size, and comprise all of $S_n$, so they are each of size $|S_n|/n=(n-1)!$, as desired.


问:
Eigenspaces and subspaces
If $\lambda$ is an eigenvalue for an $n \times n$ matrix $A$, would the associated eigenspace be a subspace of $\mathbb{R}^n$?

答:
Yes.  The eigenspace of $\lambda$ is
$$
E_\lambda:=\{v\in \mathbb{R}^n~|~Av=\lambda v\}
$$
Lets check that this is a subspace.  First $0\in E_\lambda$ since $A0=\lambda 0=0$.  Next, if $v_1,v_2\in E_\lambda$ and $r\in \mathbb{R}$, then
$$
A(rv_1)=rAv_1=r(\lambda v_1)=\lambda (rv_1)
$$
and
$$
A(v_1+v_2)=Av_1+Av_2=\lambda v_1+\lambda v_2=\lambda (v_1+v_2).
$$
Hence, $rv_1\in E_\lambda$ and $v_1+v_2\in E_\lambda$ which proves that $E_\lambda$ is a subspace of $\mathbb{R}^n$.

Title: Eigenspaces and subspaces


问:
Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
On the interval [0,1].  Define $f(x)=x\sin(1/x)$ for $x\in(0,1]$ and $f(0)=0$.  I didn't work out the exact details but I'm pretty sure that then $$\Big |\int_0^xf'(t)dt\Big |=\infty,$$ due to a process similar to something of the form $1-2+3-4+5-...$ , as one approaches zero from above.

However according to the measure-theoretic definition of absolute continuity, there should in fact be some set of measure zero $E\in[0,1]$ such that $$\Big |\int_Ef'd\mu\Big | > 0.$$

I wasn't under the impression that this was even possible.

Edit:  Maybe I wasn't clear about what my question is.  What I want is a proof (constructive or not) that there exists a set of measure zero $E$ such that $\Big |\int_Ef'd\mu\Big | > 0.$  Or if that's not possible then for someone to explain to me what my misconception is concerning the measure theoretic definition of absolute continuity:

For $v(E)=\int_Efd\mu$.

If $\mu(E)=0$ then $v(E)=0$.

link to definition

definition can also be found in Royden's Real Analysis

答:
Note that we are taking as given the fact that $f$ is not absolutely continuous.

I'm going to answer my own question and in the process pose another question:

If a measure $v$ isn't absolutely continuous with respect to $\mu$, then it doesn't have a representation of the form $$v(E)=\int_Egd\mu.$$

Thus given $$f(x) = \left\{\begin{matrix}
x\sin(1/x) &\;\;\;\;\;\;\;\;x\in[-1,1]\backslash\{0\} \\
 0&x=0
\end{matrix}\right.$$

Then the measure $v$ on $[-1,1]$ induced by $f$, that is to say $v(E)=\mu(\;\{x\in[-1,1]:f(x)\in E\}\;)$, isn't absolutely continuous on $[-1,1]$ with respect to the Lebesgue measure, this means it can't be written in the integral form above.

However it still holds true that absolute continuity of a measure $v$ with respect to another measure $\mu$ (in this case $\mu$ being Lebesgue measure), is equivalent to the property that $$\mu(E)=0\Rightarrow v(E)=0.$$

**Provided $v$ and $\mu$ are $\sigma$-finite, which they are.

Thus we are forced to conclude that there exists some $E\in[-1,1]$ such that $\mu(E)=0$, but where the Lebesgue measure of the set which maps to $E$ under $f$ is not zero; which honestly doesn't seem possible.  The only way it seems possible is if some weird uncountable set of measure zero (such as the Cantor set), happens to get mapped to a set of positive measure.  Can anyone find this set?


问:
Integration by parts? Or does not
It is known that the following integrals can be solved like this:

$$I_1=\int{xe^xdx}=xe^x-e^x+C_1$$
$$I_2=\int{(xe^x)^2dx}=\frac 1 2x^2e^{2x}-\frac 1 2 xe^{2x}+\frac 1 4e^{2x}+C_2$$

What about $\int{(xe^x)^\sqrt2dx}$ or $\int{(xe^x)^edx}$?

答:
Let,
$$
I_a=\int (xe^x)^a\,\mathrm dx.
$$
If $a$ is a nonnegative integer we may evaluate $I_a$ in closed-form using repreated integration by parts. If, however, $a$ is not a nonnegative integer integration by parts no longer works and we instead perform the substitution $u=-ax$ to write
$$
\begin{align}
I_a
&=(-a)^{-a-1}\int u^ae^{-u}\,\mathrm du\\
&=(-a)^{-a-1}\int_0^u t^ae^{-t}\,\mathrm dt\\
&=(-a)^{-a-1}\gamma(a+1,u)+C\\
&=(-a)^{-a-1}\gamma(a+1,-a x)+C,
\end{align}
$$
where $\gamma(s,z)$ is the lower incomplete gamma function. Indeed, if $a=n=0,1,2,\dots$  we may use the specialized value found here to write
$$
I_n=(-n)^{-n-1} n!\left(1-e^{ax}\sum_{k=0}^n\frac{(-ax)^k}{k!}\right)+C,
$$
which is the solution you would obtain via integration by parts.

Title: Integration by parts? Or does not


问:
Expectation and Covariance of Random Variable with Mixture Distribution
A random vector $\mathbf{X}$ has a $p$-variate normal mixture distribution when with probability $q$ it is drawn from $f_1(\mathbf{X}) \sim N_m(\mathbf{\mu}_1, \mathbf{\Sigma}_1)$ and with probability $1-q$ from $f_2(\mathbf{X}) \sim N_m(\mathbf{\mu}_2, \mathbf{\Sigma}_2)$. It has probability density function $f(\mathbf{X})=qf_1(\mathbf{X})+(1-q)f_2(\mathbf{X})$

(a) How many parameters are in this distribution?

(b) Find $E(\mathbf{X})$ and $\text{Covar}(\mathbf{X})$

$\mathbf{Proof}$

$\mu_1,\mu_2$ trivially both have $m$ components. For $\mathbf{\Sigma_1}, \mathbf{\Sigma_2}$ we need only count the upper triangle since they are each symmetric about the diagonal. The diagonal has $m$ components, the adjacent upper diagonal has $m-1$ components, and if you continue this counting process we have that each covariance matrix has $m+m-1+m-2 +\dots + 1=\frac{m^2-m}{2}+m$. So all together we have:

$$m+m+\Bigl(\frac{m^2-m}{2}+m\Bigr)+\Bigl(\frac{m^2-m}{2}+m\Bigr)=2m+m^2-m+2m=3m+m^2$$

parameters in this distribution.

$$E(\mathbf{X})=E(qf_1(\mathbf{X})+(1-q)f_2(\mathbf{X}))=E(qf_1(\mathbf{X}))+E((1-q)f_2(\mathbf{X}))=qE(f_1(\mathbf{X}))+(1-q)E(f_2(\mathbf{X}))=q\mathbf{\mu_1}+(1-q)\mathbf{\mu}_2$$

I feel fairly confident about the number of parameters piece of this problem but am unsure whether or not I have communicated the expectation operations correctly. How do you interpret $E(f_1(\mathbf{X}))$ if not the way I already have it? I could also use help understanding how to calculate the covariance. Is it correct to say that $\text{Covar}(f_1(\mathbf{X}))=\mathbf{\Sigma}_1$?

答:
Writing things like "$E[f_1(X)]$" is a bit problematic; you aren't ever really plugging in the random variable into the density function. But your computation leads to the right answer, even if the notation is not standard.

Let $Z$ be the random variable satisfying $P(Z=1)=q$ and $P(Z=2)=1-q$. Then the conditional distribution of $X$ given $Z=1$ is $N_m(\mu_1, \Sigma_1)$, and the conditional distribution of $X$ given $Z=2$ is $N_m(\mu_2, \Sigma_2)$.

Then,
$E[X] = E[E[X \mid Z]] = E[X \mid Z=1] P(Z=1) + E[X \mid Z=2] P(Z=2) = q\mu_1 + (1-q) \mu_2$ which is what you got.

The law of total covariance applies to each entry of the matrix $\text{Cov}(X)$, and ultimately becomes
$$\text{Cov}(X) = E[\text{Cov}(X \mid Z)] + \text{Cov}(E[X \mid Z]).$$
The first term is $q\Sigma_1 + (1-q) \Sigma_2$ by a computation similar to the expectation computation above. The other term is
$$\text{Cov}(E[X \mid Z]) = E[E[X \mid Z]E[X \mid Z]^\top] - E[E[X \mid Z]]E[E[X \mid Z]]^\top =  q \mu_1 \mu_1^\top + (1-q) \mu_2 \mu_2^\top - (q\mu_1 + (1-q) \mu_2)(q\mu_1 + (1-q) \mu_2)^\top$$
where I am just using the formula $\text{Cov}(Y) = E[YY^\top] - E[Y]E[Y]^\top$ with $Y=E[X \mid Z]$.


问:
Question about the Fundamental Group as a notion
Can the fundamental group of some surface have more than one path homotopy class? Maybe this is a dumb question, but I'm really struggling with this</p>

答: Yes. The fundmental group of a torus is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$ so has infinitely many homotopy classes.


问:
Test for convergence of a series
How to test the convergence of the series $\sum_{n=1}^{\infty}(n!)^{\frac{1}{n}}$?

I think ratio test or comparison test is doable. But for ratio test, I cannot calculate the limit value.

答: Hint: $n! > 2^n \implies a_n = (n!)^{\frac{1}{n}}> 2 \implies a_n $ doesn't converge to $0$, hence your series not convergent. Hope it's clear now...

Title: Test for convergence of a series


问:
Does every $L^1_{\text{loc}}$-function have a signed measure as a distributional derivative?
Edit. To clear up the confusion that I caused, I will define a signed measure here. The literature sometimes calls it "extended signed measure":

Definition. A signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ is a function $$\mu:\text{Borel sets}\to\mathbb R\cup\{\infty\}$$
or a function
$$\mu:\text{Borel sets}\to\mathbb R\cup\{-\infty\}$$
such that

$\mu(\emptyset)=0$,

for any disjoint Borel sets $A_1, A_2, A_3, \dots$, we have
$$\mu\left(\bigcup_{n\in\mathbb N} A_n\right)=\sum_{n\in\mathbb N} \mu(A_n),$$
with the convention that $\infty+\text{anything}=\infty$ and $-\infty+\text{anything}=-\infty$. Note that $\infty-\infty$ can never occur, since $\{-\infty, \infty\}\subset\operatorname{Image}\mu$ is impossible by definition.

Back to the question. Let $f\in L^1_{\text{loc}}(\mathbb R)$, i.e. $f$ is a locally absolutely integrable function. It is well-known that the distributional derivative of $f$ doesn't have to be expressible as a $L_{\text{loc}}^1$ function again. For example, if $f$ is the characteristic function of $[0,\infty[$ (or the characteristic function of $]0,\infty[$, for that matter), then its distributional derivative corresponds to the Dirac measure $\delta_0$, which has no $L^1_{\text{loc}}$-density with respect to the Lebesgue measure.

Similarly, if we have a measure on $\mathbb R$, its distributional derivative need not be a measure again. Continuing the above example, the distributional derivative of $\delta_0$ is given by the bounded linear operator

\begin{split}\delta_0': \mathcal C_{\text c}^\infty(\mathbb R) &\to \mathbb R \\ \phi&\mapsto-\phi'(0),\end{split}

which is not expressible as a measure on $\mathbb R$.

My question: Does every $L_{\text{loc}}^1$-function have a distributional derivative that can be expressed as a signed measure? More explicitly, if $f\in L_{\text{loc}}^1(\mathbb R)$, does there always exist a signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ such that

\begin{equation}\tag{*}\label{*}\bbox[15px,border:1px groove navy]{\int_{\mathbb R}\phi\,\mathrm d\mu = -\int_{\mathbb R}\phi'(t)\cdot f(t)\,\mathrm dt}\end{equation}

for every $\phi\in\mathcal C_{\text c}^\infty(\mathbb R)$ ? Note: In particular, I demand that $\int_{\mathbb R}\phi\,\mathrm d\mu$ is well-defined for every $\mathcal C_{\text c}^\infty(\mathbb R)$ (which, since $\mu$ is signed, can be actually quite a messy affair.)

答:
Consider \begin{split}f:\mathbb R&\to\mathbb R, \\ x&\mapsto \begin{cases}\sin\left(\frac1x\right)&\text{if }x\in (0,1) \\ 0 & \text{else}\end{cases}.\end{split}
We have $f\in L^1(\mathbb R)\cap L^\infty(\mathbb R)$.

Consider the set $$A=\{x\in (0,1), \cos(1/x)>0\}.$$ Then $$\int_A\,\mathrm df=\infty,\qquad \int_{(0,1)\setminus A}\,\mathrm df=-\infty,$$ so $\mathrm df$ is not a signed measure, $\int_{(0,1)} \,\mathrm df$ is not well-defined.

Even assuming that $f$ is continuous doesn't help, see $x(1-x)\sin(1/x^2) 1_{x\in (0,1)}$.


问:
Why does $\frac{d}{d\nu}\ln\Gamma\left(\frac{\nu}{2}\right)$ simplify to digamma function?
$$-(\nu+1)\frac{d}{d\nu}\left[\ln\Gamma\left(\frac{\nu}{2}\right)-\ln\Gamma\left(\frac{\nu+1}{2}\right)\right]=\frac{\nu+1}{2}\left[\psi\left(\frac{\nu+1}{2}\right)-\psi\left(\frac{\nu}{2}\right)\right]$$

How to derive the above equality? I don't understand how taking the derivative of the log gamma function results in the digamma functions and other resulting side-effects. What steps occurred in between?

答:
For the sake of answering the question we have from the definition of the polygamma function and the chain rule:
$$
(\nu+1)\partial_\nu^{n+1}(\log\Gamma(\tfrac{\nu+1}{2})-\log\Gamma(\tfrac{\nu}{2}))=(\nu+1)2^{-n-1}(\psi^{(n)}(\tfrac{\nu+1}{2})-\psi^{(n)}(\tfrac{\nu}{2})),
$$
for which your problem is given by the special case $n=0$.


问:
How to disprove Russell's paradox in ZF?
How to avoid a Russell-style paradox in ZF?

The Zermelo-Fraenkel system, by not using a theory of types, does not prevent us from posing Russell's paradoxical scheme. Only when the argument is replaced by the set being defined, the resulting statement is false, not paradoxical.

Set R in Frege's system:

By the axiom of unrestricted comprehension:

∃y ∀x ( x ∈ y ) ⟺ F(x)

... where F (x) is any well-formed formula of the system. Replacing the formula, and the existential by the name of the defined set:

∀x ( x ∈ R ) ⟺ ( x ∉ x )

Instantiating the variable x with R:

( R ∈ R ) ⟺ ( R ∉ R )

Set R in the ZF system:

By the axiom of specification (also called separation):

∀z ∃y ∀x [ ( x ∈ y ) ⇔  ( x ∈ z  ) ∧ F(x) ]

Replacing the schema of the formula with a concrete formula, and the existential one with the name of the defined set:

|1|:    ∀z ∀x [ ( x ∈ R ) ⟺  ( x ∈ z  ) ∧  ( x ∉ x ) ]

We cannot replace x by R, because there is no set z of which it is a member, therefore the second member of the equivalence does not hold.

The ZF system also allows defining the R’s complement independently, that is, without using the set R in its definition, but simply as the set M of all self-belonging sets, those that are members of themselves.

```<code>∀z ∀x [ ( x ∈ M ) ⟺  ( x ∈ z  ) ∧  ( x ∈ x ) ]
</code>```

Then you can build the complement of M, which is the complement of the complement of R, without in principle its double complement condition being evident:

```<code>∀z ∀x [ ( x ∈ MC ) ⟺  ( x ∈ z  ) ∧  ( x ∉ M ) ]
</code>```

Then instantiating in the formula |1|:

∀z [ ( MC ∈ R ) ⟺  ( MC ∈ z  ) ∧  ( MC ∉ MC ) ]

But MC is the complement of the complement of R, so that it can be shown that:

```<code>( MC ⊆  R ∧ R ⊆ MC ) ⟹ ( R ≡ MC )
</code>```

That is, MC and R are two names for the same set; thus:

```<code>∀z [ ( R ∈ R ) ⟺ ( R ∈ z  ) ∧  ( R ∉ R ) ]
</code>```

If we assume that R ∈ R, then:

( R ∈ R )  ⟹  ∀z ( R ∈ z  ) ∧  ( R ∉ R )
∀z ( R ∈ z  ) ∧  ( R ∉ R )                            [ by Modus Ponens ]
( R ∉ R )                                                           [ conjunction elimination ]

If we assume that R ∉ R, then:

```<code>∀z ( R ∈ z  ) ∧  ( R ∉ R )  ⟹  ( R ∈ R )
</code>```

If we suppose that R is a member of all sets it is easily followed that it is a member of itself, by Modus Ponens.

If we assume that R is not a member of some set, R would not be defined, contrary to what we had assumed in the formula |1|.

For all of which:

( R ∈ R  ) ⟺  ( R ∉ R )

I assume there will be a way to disprove this reasoning, but at the moment I can't get a glimpse of it.

[

答:
This is a nice example of the subtlety of existential instantiation (replacing an existentially-quantified variable with a constant symbol). Specifically:

You cannot instantiate while bound by a universal quantifier.

To see what I mean, consider the following. The sentence $$\forall u\exists v(u<v)$$ is obviously true in $\mathbb{N}$. However, if we try to instantiate $v$, we get $$\forall u(u<C)$$ (here "$C$" is our new constant symbol) which is clearly false in $\mathbb{N}$ no matter how we interpret $C$.

In your argument, this is exactly what you do e.g. when you write

Replacing [...] the existential one with the name of the defined set:

the sentence "$\forall z\color{red}{\exists y}\forall x[x\in \color{red}{y}\leftrightarrow (x\in z\wedge x\not\in x)]$" is provable from separation but the sentence "$\forall z\forall x[x\in \color{red}{R}\leftrightarrow (x\in z\wedge x\not\in x)]$" is completely unjustified, and indeed $\mathsf{ZF}$ disproves the existence of an $R$ having that property.

There is a way to correctly instantiate inside a universal quantifier: introduce a function symbol taking as input the variables corresponding to universal quantifiers outside the relevant existential quantifier. Thinking of constant symbols as $0$-ary function symbols, this generalizes the version of existential instantiation we're used to in the case where there are no "outside universals."

This transformation is actually extremely useful - it's Skolemization. But that doesn't lead to a contradiction here: we just wind up building the function $z\mapsto \{x\in z: x\not\in x\}$, which is in fact $z\mapsto z$ by regularity.


问:
What's the maximum and minimum of $\sin Z$ and $\cos Z$ where $Z$ is a complex number?
I see that a lot of people try solving something  like $\sin Z=2$. So that breaks the the maximum of $\sin x$. Now I wonder if there is a maximum and minimum of $\sin Z$ and $\cos Z$ if $Z$ is a complex number.

答:
By Liouville's theorem, a bounded entire function must be constant. Since sine and cosine are entire and obviously non-constant, they must be unbounded. Hence they have no largest value. Note: by "largest" we mean largest modulus.

References: https://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)
and https://en.wikipedia.org/wiki/Entire_function.

EDIT: They still have a "smallest" value, namely zero (this is not true for all non-constant entire functions, as noted in the comments below).


问:
How can I find the limit of this fraction? ($0/0$ type)
Suppose I have two complex numbers $g_0$ and $g_1$, when $g_0\rightarrow1$ and $g_1\rightarrow0$, how can I evaluate
$$
\sqrt\frac{g_1^2}{1-g_0^2}
$$
Intuitively I think that should be infinity, but I'm not pretty sure how can I justify that? Thanks!!

答:
I want to say  again that your question is not well defined as it is stated now. We can modify $g_0$ and $g_1$ so that we have infinitely many solutions. If I'm not mistaken, you want to calculate the double limit
$$
\lim_{(g_0,g_1)\to(1,0)}\frac{g_1}{\sqrt{1-g_0^2}}.
$$

Let $S$ be an open set containing $(x_0,y_0)$ , and let $f$  be a function of two variables defined on $S$ , except possibly at  $(x_0,y_0)$ . The limit of  $f(x,y)$  as  $(x,y)$  approaches  $(x_0,y_0)$  is  $L$ , denoted
$$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L,$$
means that given any $\varepsilon>0$, there exists $\delta>0$  such that for all  $(x,y)≠(x_0,y_0)$ , if $(x,y)$  is in the open disk centered at $(x_0,y_0)$  with radius $\delta$ , then  $|f(x,y)−L|<\varepsilon$.

To show our limit is $0$, let $\varepsilon$ be given. We want to find $\delta>0$ such that if $\sqrt{(g_0-1)^2+(g_1-0)^2}<\delta$, then $$\left|\frac{g_1}{\sqrt{1-g_0^2}}-0\right|<\varepsilon.$$

Observe that $g_1<\delta$. Let $1/\sqrt{1-g_0^2}$ be bounded by a finite constant $\kappa$. (Don't forget that $g_0\neq1$ for this proof.) Choose $\delta<\varepsilon/\kappa$.

$$
\begin{align}
\left|\frac{g_1}{\sqrt{1-g_0^2}}\right|<{\delta}{\kappa}<\frac{\varepsilon\kappa}{\kappa}=\varepsilon.
\end{align}
$$
And this concludes the proof
$$\lim_{(g_0,g_1)\to(1,0)}\frac{g_1}{\sqrt{1-g_0^2}}=0.$$

Addendum: Looking at another interpretation of your limit would be way easier:
$$
\lim_{g_0\to1}\lim_{g_1\to0}\frac{g_1}{\sqrt{1-g_0^2}}=\lim_{g_0\to1}\frac{0}{\sqrt{1-g_0^2}}=0.
$$
It is also equal to zero but it is not the same as the first limit we calculated above!


问:
Elementary embedding from hereditarily finite to hereditarily countable sets
Let $(H(\omega),\in\vert_{H(\omega)})$ be the structure of hereditarily finite sets and $(H(\omega_1),\in\vert_{H(\omega_1)})$ be the structure of hereditarily countable sets. Does there exist an elementary embedding from $(H(\omega),\in\vert_{H(\omega)})$ into $(H(\omega_1),\in\vert_{H(\omega_1)})$ with respect to the language of set-theory, I.e. that the first order language with one binary predicate $\in$.

答:
They aren't even elementary equivalent (which is a stronger fact than the mere non-existence of an elementary embedding). HINT: only $H(\omega_1)$ has elements which are infinite.

In more detail:

 $H(\omega_1)$ satisfies "There is an $x$ with a non-surjective injection to itself" - that is, "There is a Dedekind-infinite set." For example, take $x=\omega$. Keep in mind that every function between two hereditarily countable sets is itself a hereditarily countable set, so $H(\omega_1)$ sees all the basic combinatorics of hereditarily countable sets that $V$ does. By contrast of course $H(\omega)$ does not satisfy that sentence.

A more interesting problem is to show that $H(\omega_1)\not\equiv H(\omega_2)$. Here we need to think about cardinalities:

 Only $H(\omega_2)$ satisfies "There are two Dedekind-infinite sets of different cardinalities."

And this extends to showing that $H(\omega_m)\not\equiv H(\omega_n)$ for all distinct finite $m,n$. In fact, there's a stronger phenomenon occurring here. Say that an ordinal $\alpha$ is $H$-isolated iff there is no $\beta\not=\alpha$ with $H(\omega_\alpha)\equiv H(\omega_\beta)$. The arguments above really show that every finite ordinal is isolated, and with a bit of thought we can push this much further (think about counting limit ordinals, counting limits-of-limit ordinals, etc.). Indeed, it's not hard to show that the smallest non-isolated ordinal must be rather large (of course by a counting argument it is guaranteed to be $<(2^\omega)^+$).


问:
Solving modular system involving binomial coefficients. Does there even exist a solution?
Consider the following system:

$$\begin{align*}
\sum\limits_{k =0\space (even)}^{n}{n \choose k} \sqrt{7}^{k} &= 0 \pmod{u}\\
\sum\limits_{k=0 \space (odd)}^{n}{n \choose k} \sqrt{7}^{k-1} &= 1\pmod{u}\end{align*}$$

For a given $u$, how can I determine whether this system has a solution? This means:

Does the system even have a solution?

How can one determine a $n$ for what it has a solution?

答:
Multiply the second equation by $\sqrt{7},$ and add the two equations together, to get

$$(1+\sqrt{7})^n \equiv (1+\sqrt{7}) \mod u.$$, so
$$(1+\sqrt{7})^{n-1}\equiv 1 \mod u.$$

So, if $7$ is a quadratic residue mod $u$ and $n-1$ is not divisible by $\phi(u),$ there is no solution.


答:
No, if you are looking at the derivative of a map between two vector spaces, it has to be a linear map. In this context, the derivative of $f$ at $x$ is an endomorphism $Df(x):\mathbb R^n\rightarrow \mathbb R^n$.

So when you give $Df=3\|x\|^2$, it can't be a good answer for $n\geq 2$. However if $n=1$, I agree.

To answer question 1) and 2), you can easily compute $Df(x)$ for any $x\in \mathbb R^n$ : since $f(x)=x\|x\|^2=\langle x,x\rangle x$, $$f(x+h)-f(x)=\langle x+h,x+h\rangle (x+h)-\langle x,x\rangle x= \cdots$$

For question 3), assume that $f(x)=f(y)$ and try to prove that $x=y$ i.e. $f$ is injective.

Then by question 3), if $f$ is injective, what else does $f$ need to be invertible ?

Body:
$f:\mathbb{R}^n\to\mathbb{R}^n,f(x)=x\|x\|^2$ Then

$1.$ $(Df)(0)=0$

$2.$ $(Df)(x)=0\forall x$

$3.$ $f$ is one one

$4$. $f$ has an inverse.

let, $x=(x_1,x_2,x_3,\dots,x_n)$, $\|x\|^2=(x_1^2+\dots,+x_n^2)$
,Then $Df=3\|x\|^2$ am I right?  Then $1$ is true, $2,3,4$ are false.


问:
Where should a water pump be placed to minimize the length of the pipe?
The problem is as follows:

Two buddies Reggie and Mathias have their homes nearby a fresh water river as indicated in the picture from below. Assuming they have only one available water pump which they can use. Where must be located this pump so that the pipes length is the minimum to provide water to both homes?.

The choices given in my workbook are as follows:

$\begin{array}{ll}
1.&\textrm{at 25 m from A}\\
2.&\textrm{at 25.71 m from A}\\
3.&\textrm{at 30 m from B}\\
4.&\textrm{at 26 m from B}\\
\end{array}$

I'm totally lost in this geometry problem. Can someone help me here?. I believe that the catch here is to make a right triangle or some sort of isosceles but I'm not sure of the rationale or most importantly the reason why?. Therefore I'm asking for assistance.

I would really like to provide more effort into this, but I'm stuck. I don't know what else can I do?.

Can you please include some drawing in your answer as a justification. I really beg someone include a drawing becuase I can't spot where should this pump be located.

Please help because I can't figure this out.

答: Imagine that line segment AB is a mirror and that Mathias shines a light from his house to Reggie's house by reflecting it off the mirror.  We know from physics that the light ray will take the shortest path, and that the angle of incidence is equal to the angle of reflection.  That is, if Reggie is at point $R$ and Matthias is at point $M$, then in order for the light to go from $M$ to $R$, it must strike the mirror at a point $X$ such that $\angle MXA  = \angle RXB$.  But then triangles $MXA$ and $RXB$ are similar, so $X$ must divide segment $AB$ in the ratio $3:4$.  That is, $X$ is $\frac{180}7$ meters from $A$ and the correct answer is $2$.


问:
Better way to write the sum of multiple terms with cross-correlation
how can I write this expression using summation?

$y^2=x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2x_1x_2 + 2x_1x_3 + 2x_1x_4 + 2x_2x_3 + 2x_2x_4 + 2x_3x_4$

Body: Consider: $$y^2=\sum_{i=1}^4x_i^2+\sum_{1\leq i<j\leq 4}2x_ix_j=\left(\sum_{i=1}^4x_i\right)^2.$$


问:
Finite Axiomatizability
In the situation I have an infinite set of closed wfs $\mathfrak{B}=\{\mathfrak{B}_{1}, \mathfrak{B}_{2}, \mathfrak{B}_{3}, ...\}$. It is also the case that no $\mathfrak{B}_{k}$ can be proven with the conjunction of all $\mathfrak{B}_{j}$ where $j < k$.

$T$ is the theory of all sentences that can be proven from $\mathfrak{B}$, and so prove that the theory $T$ is not finitely axiomatizable.

Naively, I understand this to be because there can be no model for every finite subset of $\mathfrak{B}$, so there is no model for $\mathfrak{B}$. In other words, not every theorem  of $\mathfrak{B}$ holds as $\mathfrak{B}_{1},\mathfrak{B}_{2},\mathfrak{B}_{3}, ..., \mathfrak{B}_{j} \vdash \mathfrak{B}_{k}$ does not hold. Is this along the right lines?

答: This is an application of compactness. The single sentence $\mathfrak A$ axiomatizing $T$ must be a logical consequence of $\{\mathfrak B_1,\mathfrak B_2,\ldots\}$ and thus of finitely many $\mathfrak B_1,\ldots,\mathfrak B_n$. But it also must imply $\mathfrak B_{n+1},$ since that is a consequence of $\{\mathfrak B_1,\mathfrak B_2,\ldots\}.$ So then $\mathfrak B_{n+1}$ is a logical consequence of $\{\mathfrak B_1,\ldots,\mathfrak B_n\},$ contrary to assumption.

Title: Finite Axiomatizability


问:
constrained optimization in calculus of variations
Consider the problem
\begin{align*}
\max_{X\in\mathcal{F}} \int_{0}^1 X(t) \phi(t)dt \\
\int_{0}^1 X(t) \psi(t) dt= 0
\end{align*}
where $\mathcal{F}$ is a  "well-behaving" set of functions. (In my application, $\mathcal{F}$ is defined as the set of CDFs  that are second-order stochastically dominated by some given CDF $F_0$.) I am looking for a formal reference/argument that relates this problem to the one obtained from writing the Lagrangian
$$\max_{X\in\mathcal{F}} \int_{0}^1 X(t) (\phi(t)-\lambda \psi(t))dt, $$
and in particular,  formally showing that the optimal solution in the first problem corresponds to a local maxima of the second problem. Any ideas/references would be helpful!

答:
The argument is really not any different from optimizing over $\mathbb R^n$. I will prove that the statement you want holds assuming a form of Slater's condition: $\mathcal F$ is convex, and $0$ is an interior point of $\left\{\int_{0}^1 X(t) \psi(t) dt : X \in \mathcal F\right\}$.

For convenience, let
\begin{align}
f(X) &= \int_{0}^1 X(t) \phi(t)dt \\
g(X) &= \int_{0}^1 X(t) \psi(t) dt \\
h(\lambda) &= \sup_{X \in \mathcal F} \Big\{f(X) - \lambda g(X)\Big\}
\end{align}
Let $M(z)$ be the maximum value (or maybe the supremum) of $f(X)$ over all $X \in \mathcal F$ such that $g(X) =z$. Its domain is the set $\{g(X) : X \in \mathcal F\}$. We allow $M(z) = +\infty$ when things work out that way; nothing special happens in this case. We have $M(0) = f(X^*)$, where $X^*$ is the optimal solution to our problem.

Assuming $\mathcal F$ is a convex domain, $M$ is a concave function of $z$. To see this, first suppose for simplicity that $M(z_1) = f(X_1)$ and $M(z_2) = f(X_2)$ with $g(X_i) = z_i$. Then for any $t \in [0,1]$, if $z = tz_1 + (1-t)z_2$, let $X = tX_1 + (1-t)X_2$; we get $g(X) = z$, so
$$
 M(z) \ge f(X) = t f(X_1) + (1-t) f(X_2) = t M(z_1) + (1-t) M(z_2)
$$
proving that $M$ is concave. When $M(z_i)$ can be approached by not reached by any $X$ with $g(X) = z_i$, we consider a sequence of $X$'s approaching it and the same thing falls out eventually.

By concavity of $M$, there is a slope $\lambda^*$ such that $M(z) \le M(0) + \lambda^* z = f(X^*) + \lambda^* z$... assuming $0$ is an interior point of the domain of $M$. To see that this condition is necessary, imagine a case where $M(z) = \sqrt{z}$; then the tangent line at $0$ is vertical. We can set $\lambda^* = M'(0)$ when that derivative exists; if $0$ is a corner point of $M$, there are many valid choices of $\lambda^*$.

For any $X \in \mathcal F$, we have $M(g(X)) \ge f(X)$, because $X$ is feasible for the optimization problem defining $M(g(X))$. Therefore $f(X) \le f(X^*) + \lambda^* g(X)$, or $f(X^*) \ge f(X) - \lambda^* g(X)$.

Since $g(X^*) = 0$, we have $f(X^*) - \lambda^* g(X^*) \ge f(X) - \lambda^* g(X)$. In other words:

$f(X) - \lambda^* g(X)$ is maximized at $X^*$.

$f(X^*) = h(\lambda^*)$, and it is easy to see that for all $x \in \mathcal F$ satisfying $g(X) = 0$ and for all $\lambda \in \mathbb R$, $f(X) \le h(\lambda)$. Therefore strong duality holds between the original problem and the problem of minimizing $h(\lambda)$ over $\lambda \in \mathbb R$, and $\lambda^*$ is the optimal dual solution.


问:
Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions
Show that lower semicontinuous function $f:X\rightarrow [0,1]$ on metrizable X is the supremum of an increasing sequence of continuous functions.

My attempt: I don't know how to approximate $f(x)$ to within $[f(x)-1/n,f(x)]$ by $h_n(x)$ which is a linear combination of characteristic function of open sets, using lower semicontinuity of $f$.

Can anyone help?

答:
This is a problem from Willard's general topology textbook. I shall provide an outline of the proof and leave the details to the reader.

The function $f(x)$ may be approximated to within $[f(x)-1/n, f(x)]$ by the function $$h_{n} = \frac{1}{n}\sum_{k = 1}^{n-1}\chi_{f^{-1}\left(\left(\frac{k}{n}, 1\right]\right)}$$ where $\chi_{A}$ denotes the characteristic function of a subset $A$ of $X$. Note that $h_{n}$ is a linear combination of characteristic functions of open sets.

The characteristic function of any open set $A$ in a metric space $(X, \rho)$ can be written as the supremum of an increasing sequence of continuous functions. To prove this, note that every open set in a metric space is $F_{\sigma}$. In particular, if $A$ is an open set then $A$ can be written as the union of an increasing sequence $A_{1}\subseteq A_{2}\subseteq\cdots$ of closed sets where $$A_{n} = \left\{y\in X: \rho(y, X-A)\geq \frac{1}{n}\right\}.$$ For each positive integer $n$ define $g_{n}:X\rightarrow\mathbb{R}$ by $g_{n}(x) = \min\{n\rho(x, X-A), 1\}$ and verify that $\chi_{A} = \sup_{n}g_{n}$.

For each $1\leq k\leq n-1$ let $$\frac{1}{n}\chi_{f^{-1}\left(\left(\frac{k}{n}, 1\right]\right)} = \sup_{m}g_{k,m}$$ for an increasing sequence of continuous functions $\{g_{k, m}\}_{m\in\mathbb{Z}^{+}}$. One can verify that $h_{n} = \sup_{m} h_{n, m}$ where $h_{n, m} = \sum_{k = 1}^{n-1}g_{k,m}$.

Finally, one can show that $$f = \sup_{(n, m)\in\mathbb{Z}^{+}\times\mathbb{Z}^{+}}h_{n, m}.$$ Using any bijection between $\mathbb{Z}^{+}\times\mathbb{Z}^{+}$ and $\mathbb{Z}^{+}$, rearrange the functions $\{h_{n,m}\}$ into a sequence $\{\tilde{k}_{n}\}$ and let $k_{n} = \max\{\tilde{k}_{1},\dots,\tilde{k}_{n}\}$. Then $\{k_{n}\}$ is the required sequence of increasing continuous functions such that $f = \sup_{n}k_{n}$.


问:
Algorithms for indefinite symmetric eigenvalue problems
I want to learn more about algorithms which deal with the following problems:

Given a symmetric positive semi-definite matrices $A, B \in \mathbb{R}^{d \times d}$, with $\lambda_{min}(A) = 0$ find the minimum eigenvalue of the matrix $C = A - B$. This can be framed as minimizing the Rayleigh quotient:
$$ \rho(v) = \min_v \frac{v^T C v}{v^T v} $$
Note that the matrix $C$ is actually indefinite. There are quite a few algorithms for solving this kind of problems when $C$ is PSD (like for instance LOBPCG). However, I can't seem to find much literature on how to solve this for a general $C$ or if there is anyway to exploit the structure of the problem.

答:
I don't know about really exploiting the structure, but the following seems to work fine:

Compute the maximal eigenvalue $\lambda_B$ of $B$ (this can be done fast by iterative methods).

Note that the minimal eigenvalue of $C=A-B$ is at least $-\lambda B.$ So, take $C + 2\lambda_B I$. It will be positive definite, so you can use your favorite algorithm for positive definite matrices (conjugate gradient?) At the end, subtract $2\lambda_B$ from the result.


答:
Your $f$ maps $(x_1,\dots,x_n)$ to $$((x\cdot x)x_1,\dots,(x\cdot x)x_n)$$
Let $f_k(x)=(x\cdot x)x_k$. Then $$\nabla f_k(x)=\nabla (x\cdot x)x_k+(x\cdot x)\nabla (x_k)$$

$$\nabla f_k(x)=2x_k x+(x\cdot x)e_{k}$$

Thus $$M_{(Df)(x)}=2 \begin{pmatrix} x_1x_1&x_1x_2&\cdots&x_1x_n\\x_2x_1&x_2x_2&\cdots&x_2x_n\\ \vdots &\vdots&\ddots&\vdots\\x_nx_1&x_nx_2&\cdots&x_nx_n\end{pmatrix}+\Vert x\Vert^2{\bf Id}$$

Note that your functions is of the form $f(x)=\lambda(x)\operatorname{id}(x)$ with $\lambda:\Bbb R^n\to \Bbb R$. We know that $\lambda(x)$ is not injective, but we know that $\operatorname{id}(x)$ is. Note $f$ is injective on every $\{\Vert x\Vert =r\}$, so we must look at $x,y$ with $\Vert x\Vert \neq \Vert y \Vert$, assume $\Vert x\Vert \geq \Vert y \Vert$. Since the norm is always non-negative, $f$ may only fail to be injective when each coordinate of $x$ has the same sign as the corresponding coordinate of $y$. Could you look at what happens in said case? Aim to show $x\neq y\implies f(x)\neq f(y)$.

Title: $f:\mathbb{R}^n\to\mathbb{R}^n,f(x)=x\|x\|^2$ Then


问:
Variance Definition: question on expansion
Wiki defines variance as follows.  I have a question on the expansion which I will point out.

$$
Var(X) = E[(X - E[X])^2] = E[X^2 - 2XE[X] + E[X]^2]
$$

I'm assuming $E[X - Y] = E[X] - E[Y]$ applies next.  Is this true?  Also, does anyone know how $-2E[X]E[X]$ is derived in the next term?

$$
Var(X) = E[X^2] - 2E[X]E[X] + E[X]^2 = E[X^2] - E[X]^2
$$

答:
$\def\E{\operatorname{\mathsf E}}$When $c$ is a constant $\E(c X)=c\E(X)$.  Well, $\E(X)$ is a constant so $\E\bigl(\E(X)X\bigr)=\E(X)\E\bigl(X\bigr)$.

Similarly $\E(\E(X)^2)=\E(X^2)$ because $\E(X^2)$ is a constant and of course when $c$ is a constant, then $\E(c)=c$.

$$\begin{align}
\E\big((X-\E(X))^2\big)
 &= \E\bigl(X^2-2\E(X)X+\E(X)^2\bigr)
\\ &=\E(X^2)-\E\bigl(2\E(X)X\bigr)+\E(\E(X)^2)
\\ &=\E(X^2)-2\E(X)\E(X)+\E(X)^2
\\ &=\E(X^2)-2\E(X)^2+\E(X)^2
\\ &=\E(X^2)-\E(X)^2
\end{align}$$


答:
Let AB be the perpendicular bisector of MM' and RR'.
MR',M'R and AB meet at P.  Here is where you should build the water pump.

For any other point along the bank, for example X,
MX+XR =M'X+XR>M'R=MP+PR.

Set $\angle R=\angle M=\theta$,

30 $\tan\theta+40\tan \theta=60$, $\tan \theta=\frac{6}{7}$,

$\theta = 40.6^\circ$,   MP= $30\tan 40.6^\circ$=25.71m$

Title: Where should a water pump be placed to minimize the length of the pipe?

Body:
The problem is as follows:

Two buddies Reggie and Mathias have their homes nearby a fresh water river as indicated in the picture from below. Assuming they have only one available water pump which they can use. Where must be located this pump so that the pipes length is the minimum to provide water to both homes?.

The choices given in my workbook are as follows:

$\begin{array}{ll}
1.&\textrm{at 25 m from A}\\
2.&\textrm{at 25.71 m from A}\\
3.&\textrm{at 30 m from B}\\
4.&\textrm{at 26 m from B}\\
\end{array}$

I'm totally lost in this geometry problem. Can someone help me here?. I believe that the catch here is to make a right triangle or some sort of isosceles but I'm not sure of the rationale or most importantly the reason why?. Therefore I'm asking for assistance.

I would really like to provide more effort into this, but I'm stuck. I don't know what else can I do?.

Can you please include some drawing in your answer as a justification. I really beg someone include a drawing becuase I can't spot where should this pump be located.

Please help because I can't figure this out.


问:
An ellipse inside a heptagon
Can you provide a proof for the following claim:

Claim. Construct a convex heptagon circumscribed about an ellipse. Intersection points of its non-adjacent longer diagonals lie on a common ellipse.

GeoGebra applet that demonstrates this claim can be found here.

I have noticed that principal diagonals of the hexagon defined by the points: $,FG \cap ED,,,,$ concur at the point $P$. Similar is true for the points $Q,R,S,M,N,O$ as well . So I guess we should apply Brianchon's theorem somehow.

答:
The comments mention the Heptagon Theorem:

Let $H$ be a heptagon with seven vertices given in cyclic order
inscribed in a conic. Then the Pascal lines of the seven hexagons
obtained by omitting each vertex of $H$ in turn and keeping the
remaining vertices in the same cyclic order are the sides of a
heptagon $I$ which circumscribes a conic.

Moreover, the Brianchon points of the seven hexagons obtained by
omitting the sides of $I$ one at a time and keeping the remaining
sides in the natural cyclic order are the vertices of the original
heptagon.

Since this is a theorem about points, lines, conics, and incidence, it is a theorem of projective geometry and can be dualized.  The dual of the first paragraph is

Let $H$ be a heptagon with seven sides given in cyclic order
which circumscribes a conic. Then the Brianchon points of the seven hexagons
obtained by omitting each side of $H$ in turn and keeping the
remaining sides in the same cyclic order are the vertices of a
heptagon $I$ inscribed in a conic.

As noted in the OP, the Brianchon points of the seven hexagons are the pairwise intersections of consecutive long diagonals (by which we mean their endpoints are in cyclic order, such as $AD,BE$).  So the dualized first paragraph proves the OP's claim.

The Heptagon theorem is stated and proven in Evelyn et al, The Seven Circles Theorem and Other New Theorems (if you can't get access to the book, the proof is shown below).  As done above, the Heptagon theorem can be dualized, or the proof of the Heptagon Theorem can be dualized, which is a lot more work.

It's worth pointing out, if it isn't already obvious, that the Heptagon Theorem contains dual constructions.  Starting with an inscribed heptagon one uses Pascal lines to construct a circumscribed heptagon.  One can then reverse the construction using Brianchon points to construct an inscribed heptagon.

The picture given in the Mathworld article uses a doubly winding blue heptagon to obtain a smaller red heptagon.  But, if you start with a singly winding heptagon, you end up with a red heptagon typically much larger than the blue heptagon.  For example the Pascal lines of the orange heptagon $MNOPQRS$ in the OP are the sidelines of the blue heptagon $ABCDEFG$.

Below is a copy of the proof of the Hexagon Theorem.

Title: An ellipse inside a heptagon


问:
Homology groups and embeddings
I'm working on the following problem:

Let $A$ be the image of an embedding $g: S^{1} \times D^{2} \rightarrow S^{3}$ . If $Y= S^{3} - $Int$(A)$
be the complement of its interior. Compute the homology groups $H_{*}(Y; \mathbb{Z}$) and $H_{*}(Y, \partial Y ; \mathbb{Z} $).

Since I'm a beginner in homology theory I have only learned to compute homology groups for very simple examples. Can someone help me go through this problem?

答:
For the first one, use the Mayer-Vietoris sequence for homology, the relevant piece of which is

$$H_2(S^3) \longrightarrow H_1(T^2) \longrightarrow H_1(A) \oplus H_1(Y) \longrightarrow H_1(S^3).$$ The first and the last groups are trivial, and $H_1(T^2) \simeq \mathbb{Z}^2,$ so $H_1(Y) \simeq \mathbb{Z}.$

Your second question can be done similarly, but I leave it to you.

Title: Homology groups and embeddings


问:
Finding a direct basis for tangent space of piece with boundary of an oriented manifold.
I have the following definition (from Hubbard's vector calculus book) for an oriented boundary of piece with boundary of an oriented manifold:

Let $M$ be a $k$ dimensional manifold oriented by $\Omega$ and $P$ a piece with boundary of $M$. Let $x$ be a point of the smooth boundary $\partial^{ \ S}_MP$ and let $\vec{V}_{\text{out}}\in T_xM$ be an outward pointing bector. Then the function $\Omega^\partial : \mathcal{B}(T_x\partial P)\to\left\{+1,-1\right\}$ given by
  $$
\Omega_x^\partial(\vec{v}_1,...,\vec{v}_{k-1}) = \Omega_x(\vec{V}_{\text{out}},\vec{v}_1,...,\vec{v}_{k-1})
$$
  defines an orientation on the smooth boundary $\partial_M^{ \ S}P,$ where $\vec{v}_1,...,\vec{v}_{k-1}$ is an ordered basis of $T_x\partial_M^{ \ S}P$.

I'm working on a problem that asks me to find a basis for the $T_x\partial P$ that is direct to a certain orientation (given by an elementary 3-form). My question is this:

When I choose a basis for $T_x\partial P$, does this basis also need to lie in $T_xM$? Also, are there restrictions to how I should choose $\vec{V}_{\text{out}}$? In other words, does $\vec{V}_{\text{out}}$ need only lie in $T_xM$ and not in $T_x\partial P$?

答: Yes, $T_x\partial P$ is a hyperplane in $T_xM$. $\vec V_{\text{out}}$ is by definition the outward-pointing normal to $\partial P$. This means that if $\vec v_1,\dots,\vec v_{k-1}$ are chosen as a basis for $T_x\partial P$, then $\vec V_{\text{out}},\vec v_1,\dots,\vec v_{k-1}$ will give you a basis for $T_xM$. The whole point of this orientation stuff is that when you pick $\vec v_1,\dots,\vec v_{k-1}$ so that $\vec V_{\text{out}},\vec v_1,\dots,\vec v_{k-1}$ gives you a positively-oriented basis for $T_xM$, then you win: You have achieved a positively-oriented basis for $T_x\partial P$.</p>


问:
Roots of a integral having upper bound as variable

We are given above the graph of $f(x) = x \sin x$. Define $G(x) = \int\limits_0^x f(t) dt $. It is clear that since $G( \pi ) > 0$ and $G(2\pi) < 0$, then we have a root $c$ in $[\pi, 2 \pi ]$. Similarly, there is a root in the interval $[2 \pi, 3 \pi]$, same argument.

Here is the subtle part: Notice that $G(0) = \int_0^0 f = 0 $. Do we consider $x=0$ to be a root of $G(x)$? is it by convention or there any underlying issues to consider here?

Im uncomfortable about it since we do not know what happes as $x \to 0^-$. Any thoughts?

答:
Yes, $F(0) = 0$ is a root.  You don't need to look at the limit because the function $f(x) = x \sin x$ is well-defined for all real numbers, even negative ones, and at zero; moreover, the way $F$ is defined as a definite integral from $0$ to $x$, it should be clear that $F(0) = 0$.

In fact, we can compute an exact antiderivative:  with the choice $$u = x, \quad du = dx, \\ dv = \sin x \, dx, \quad v = -\cos x,$$ we obtain via integration by parts
$$\int x \sin x \, dx = \int u \, dv = uv - \int v \, du = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C.$$
It follows that $$F(x) = -x \cos x + \sin x$$ for all real $x$, and $F(0) = 0$.  We have roots of $F$ whenever $x = \tan x$; this equation has no elementary closed-form solution, but it does tell us that, since $\tan x$ is periodic with period $\pi$ and is monotone increasing on each open interval $(-\pi/2 + \pi k, \pi/2 + \pi k)$ for integers $k$, that there is a unique root of $F$ in each such interval.


问:
Is $\sum_{n=2}^{\infty} (-1)^n \log (\frac{n+1}{n})$ convergent?
Determine what kinds of convergent it is: absolute convergent, conditonal convergent, or divergent. $\sum_{n=2}^{\infty} (-1)^n \log (\frac{n+1}{n})$. well to tell that maybe it's first to seperate log? $\log (\frac{n+1}{n})$ is $\log (n+1)- \log(n)$. Then the absolute value of it is $\log (n+1)- \log(n)$. Then I notify it's best using integral test. Then for the not absolute value, I use the alternate series test, because it's alternate series with $(-1)^n$ and it's going to 0 too.

答:
Define $\sum_{n=1}^{\infty}(-1)^na_n,$ where $a_n>0$. The alternating series test then says: if $|a_{n}|$ decreases monotonically and $\lim _{n\to \infty }a_{n}=0$ then the alternating series converges.

Set $a_n=\log(\frac{n+1}{n})=\log(1+\frac{1}{n})$ is monotonic decreasing for $a_n>0$ and it's limit
$$\lim_{n\to\infty}\log(1+1/n)=0$$
is equal to zero. By the Leibniz criterion this series converges. Obviously the limit is not absolutely convergent.


问:
Find a value of $\;\lim\limits_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$

I'm going to give you an extremal problem.

Given $a_{n}:=\left ( \text{the solution for the equation}\,x^{n}= \cos x,\quad x> 0 \right ),$ find a value of
$$\lim_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$$
Source: Fujino_Yusui

If you imagine the graph of $x^{n},$ it should intersect at the point where it increases rapidly by $1,$ which is about where $x^{n}= \cos 1\,(y$ is increasing rapidly, so if you look at $y,$ you should be able to approximate $x$ well enough$).\,n\left ( 1- \cos^{\frac{1}{n}}1 \right )$ looks good, and $\frac{1}{n}= h$ is the derivative of $h\rightarrow 0^{+}.$ I guess $-\ln\cos 1$ by definition.

答:
Here is an easy solution.

$x^n=\cos x$,  as $a_n$ satisfies the equation, $\left(a_n\right)^n=\cos \left(a_n\right)$

Taking log on both sides,

$n.\log \left(a_n\right)=\log \left(\cos \left(a_n\right)\right)$

$\therefore n=\frac{\log \left(\cos \left(a_n\right)\right)}{\log \left(a_n\right)}$

Now as n tends to infinity, the rhs must also tend to infinity. As cos lies between 0-1, only alternative is that $a_n$ tends to 1.  ($a_n=\pi/2$ wil never satisfy the original equation)

$n\rightarrow \infty \therefore a_n\rightarrow 1$

Now we substitute this value of n in our limit,

$\lim _{n\rightarrow \infty }\left(n\left(1-a_n\right)\right)=\lim _{a_n\rightarrow 1}\left(\frac{\log \left(\cos \left(a_n\right)\right)}{\log \left(a_n\right)}\left(1-a_n\right)\right)$

As it is now a 0/0 form, we use the l's hospital, and get the required answer as $a_\infty=-\ln\cos 1$


问:
Curious basic Euclidean geometry relation
I was plotting some equations and I got with the curious relation

If we build the triange

Such that it follows the following relation:
$$AD=a$$
$$DB=b$$
$$AC=ak$$
$$CB=bk$$

Then when we vary $k$ from the smallest possible value to the greatest possible one according to triangle inequality, the point C plots a circle around the midpoint of the maximum and minimum solutions, with it's diameter being the segment between them.

In a more formal way

If $C$ is the center of the hypotetic circle, $$\frac{AD}{DB}=\frac{AE}{EB}$$
and according to triangle inequality
$$\frac{AD}{DB}=\frac{FA}{FB}$$
Prove that
$$CD=CE$$
That is, prove that the point E in fact lies in the hypotetic circle.

This came out of curiosity when i plotted the point C as k varies, I was expecting a more weird shape but I got with a circle!.I have empirically verified it in geogebra and such, but i have not been able to prove it.The closest i have gotten is that $CA=\frac{AD^2}{DB-AD}$, imposing that $DB>AD$. If they are equal, a division by cero occurs, but if $DB<AD$, the circles moves to the other side.

答:
Here's a brute-force coordinate approach ...

Edit. My point names aren't consistent with OP's (which vary between his diagrams anyway :). In case of edits to the question, I'll explain the roles of my points here.

We take point $D$ on segment $AB$, with $a := |AD|$ and $b := |BD|$. A variable point $P$ is defined by $|AP|=k |AD|$ and $|BP|=k|BD|$. Note that, if $a=b$ the locus of $P$ is the perpendicular bisector of $AB$ (which happens to be the limiting case of circles), so we take $a > b > 0$.

Coordinatizing with points $A( -a, 0 )$, $B( b, 0 )$, $D( 0, 0 )$, $P(x,y)$, we have
$$\begin{align}
|AP| &= k |AD|: \qquad ( x + a )^2 + y^2 = a^2 k^2 \\
|BP| &= k |BD|: \qquad ( x - b )^2 + y^2 = b^2 k^2
\end{align}$$

Equating $k^2$ obtained from both equations gives
$$\frac{(x+a)^2 + y^2}{a^2}=k^2=\frac{(x-b)^2+y^2}{b^2}$$
so that
$$b^2 (x+a)^2 + b^2 y^2 = a^2 ( x-b)^2 + a^2 y^2$$

$$( a^2 - b^2 )x^2 - 2 x a b ( a + b ) + ( a^2 - b^2 ) y^2 = 0$$

$$( a - b ) x^2 - 2 x a b + ( a - b ) y^2 = 0$$

Since the (non-zero!) coefficients of $x^2$ and $y^2$ match (and there's no $xy$ term), we definitely have a circle; without a constant term, we see that the circle passes through the origin (point $D$). For more information, we put the equation into standard form:

$$\left( x - \frac{a b}{a-b} \right)^2 + y^2 = \left(\frac{ab}{a-b}\right)^2$$

So, the circle has center $C = \left(\frac{ab}{a-b},0\right)$ and radius $|CD| = \frac{ab}{a-b}$, and we can verify your conjecture:

$$|CA| = \frac{ab}{a-b}+a = \frac{a^2}{a-b} \qquad |CB| = \frac{ab}{a-b}-b=\frac{b^2}{a-b}$$


问:
How to count partitioned subsets of a binomial coefficient
I have been reading Dr. Carl Wagner's book Basic Combinatorics, and I cannot wrap my head around a particular theorem. Someone else has asked a very similar question and accepted an answer, but I'm afraid I still don't understand.

As stated, the theorem runs thus:

For all $ n, \; k \in \mathbb{N} $
$ \displaystyle \sum_{j = 0}^{n} \binom{j}{k} = \displaystyle \sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1} $

There follows a proof of the $\binom{n + 1}{k + 1} $ bit using some handy substitutions. What confuses me is the next part:

#2. Clearly $ \binom{n+1}{k+1} $ counts the class of all (k + 1)-element subsets of [n + 1]. But this class of k+1 subsets may be partitioned into subclasses corresponding to j = k, k + 1, . . . , n as follows.

The subclass of subsets with largest element equal to k + 1 is counted by $ \binom{k}{k} $ (why?), the subclass of subsets with largest element equal to k + 2 is counted by $ \binom{k+1}{k} $ (why?), . . ., and the subclass of subsets with largest element equal to n + 1 is counted by $ \binom{n}{k} $ (why?).

I don't understand why at all! Indeed, it makes no sense to me.

It seems to me that the number of subsets of [n+1] with largest element k+1 would be $\binom{k+1}{k+1}$. Sure, that's the same as $ \binom{k}{k} $  since they both equal one, but that just seems to be a coincidence. We're choosing k+1 numbers from a set that runs from 1 to k+1. There can, naturally, be only one set that fits that bill. Yet to use $ \binom{k}{k} $ to represent that one set seems confusing to me, and every example that follows makes less sense.

What am I missing?

答:
Before I posted this question, I wrote to the author of the text, and to my great surprise, he wrote back! Though RobPratt's answer was enough to get me there, I thought I should share the author's answer here.

Text Author's Response

Perhaps a concrete numerical example will help. Take n=6 and k=3. We want to explain why

$\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} = \binom{7}{3}$

The right-hand side, $\binom{7}{4}$, counts all of  the 4-element subsets of the 7-element set [7] = {1,...,7}.

So does the left-hand side, in four disjoint, exhaustive categories (in fancier language, this is called a partition of the set of all 4-element subsets of [7] into four blocks), namely:

(1) the category of all 4-element subsets of [7] with largest element equal to 4. Since it has already been determined that 4 will be an element of any such subset, we need to choose 3 more elements, different from, and smaller than 4, to create a 4-element subset of [7] with largest element equal to 4. So we need to choose 3 elements from the set [3], which can be done in $\binom{3}{3} = 1$ way. And, indeed, there is just one 4-element subset of [7] with largest element equal to 4, namely, {1,2,3,4}

(2) the category of all 4-element subsets of [7] with largest element equal to 5. Since it has already been determined that 5 will be an element of any such subset, we need to choose 3 more elements, different from and smaller than 5, to create a 4-element subset of [7] with largest element equal to 5. So we need to choose 3 elements from the set [4], which can be done in $\binom{4}{3} = 4$  ways. And, indeed, there are four
4-element subsets of [7] with largest element equal to 5, namely, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, and [2,3,4,5}.

If you'd like, carry this out for the remaining categories

(3) the category of all 4-element subsets of [7] with largest element equal to 6.

(4) the category of all 4-element subsets of [7] with largest element equal to 7

and email me the results. And include an explanation of why, among all (k+1)-element subsets of [n+1], if j is an integer in the set {k+1, k+2, ..., n+1}, there are (j-1)C(k) subsets with largest element equal to j.

My Response to the proposed exercise

(3) the category of all 4-element subsets of [7] with largest element equal to 6.

This, as predicted, had $\binom{5}{3}$ subsets:
{1, 2, 3, 6}, {1, 2, 4, 6} , {1, 2, 5, 6} , {1, 3, 4, 6}, {1, 3, 5, 6}, {1, 4, 5, 6}, {2, 3, 4, 6}, {2, 3, 5, 6}, {2, 4, 5, 6}, {3, 4, 5, 6},

(4) the category of all 4-element subsets of [7] with largest element equal to 7

Likewise,  $\binom{6}{3}$ = 20 subsets.
{1, 2, 3, 7}, {1, 2, 4, 7} , {1, 2, 5, 7} , {1, 2, 6, 7} , {1, 3, 4, 7}, {1, 3, 5, 7}, {1, 3, 6, 7}, {1, 4, 5, 7}, {1, 4, 6, 7}, {1, 5, 6, 7}, {2, 3, 4, 7}, {2, 3, 5, 7}, {2, 3, 6, 7}, {2, 4, 5, 7}, {2, 4, 6, 7}, {2, 5, 6, 7}, {3, 4, 5, 7}, {3, 4, 6, 7}, {3, 5, 6, 7}, {4, 5, 6, 7}

And include an explanation of why, among all (k+1)-element subsets of [n+1], if j is an integer in the set {k+1, k+2, ..., n+1}, there are  $\binom{j-1}{k}$ subsets with largest element equal to j.

The reason that the number of elements chosen from (let's call them "the candidates") is j-1 is because setting the largest element equal to j effectively removes a choice to be made. It's a forgone conclusion that j will be in the set, and that any elements larger than j will be excluded from candidacy. That naturally leaves only j-1 candidates remaining. The reason that the number of "open seats" is equal to k, even though the set size is k+1, is that we have already decided how one of the k+1 seats will be filled, leaving only k open seats remaining.


问:
"Obvious" integral inequality for radially decreasing function
I would like to show the following inequality in $\mathbb{R}^2$:

$$
\iint_{B\left( x_0, R \right)} \frac{1}{\lVert x \rVert} \,\mathrm{d}A
\leq \iint_{B\left( 0,R \right)} \frac{1}{\lVert x \rVert} \,\mathrm{d}A.
$$

Here, $\mathrm{d}A$ denotes the usual Lebesgue measure on $\mathbb{R}^2$. Intuitively, this should be true since points closer to the origin should on average give larger values, but I'm having a lot of trouble proving it rigorously. Any guidance would be appreciated.

答:
Let $A=B(x_0,R)\cap B(0,R)$ with $A_1=B(x_0,R)-A$ and $A_0=B(0,R)-A$.  Note that $A_1$ and $A_0$ have the same area.

$\int_{A_1} \frac{1}{||x||}dx \le \int_{A_0} \frac{1}{||x||}dx$ since $||x||\ge R$ for all points in $A_1$ while $||x||\le R$ for all points in $A_0$.

Add $\int_A \frac{1}{||x||}dx $ to both and get the desired inequality.


问:
Determine the value of K so the function have only one point of one horizontal point of tangency

Consider the polynomial function  define by

$ f(x) = x^3 + x^2 + kx - 1 $

Determine the value of k so f has only on point of horizontal
tangency.

So I know that $   f^{'}(x) = 3x^2 + 2x + k  $
And I should consider  $   f^{'}(x) = 0 $ but how I find a k that makes the $   f^{'}(x) = 3x^2 + 2x + k  $   have only one root?

I also thought that if $ k = -x^2 $ then the function would have only one point of horizontal tangency. But this does not works because this not equal the answer.  ( Though I think it shouldwork too )

Btw the answer is  $ \frac{1}{3} $ And I still can't figure why.

答: For $f(x) = x^3 + x^2 + kx - 1$ , when $\frac{df(x)}{dx}= 3x^2+2x+k$=0 has only one solution or its discriminant $\Delta = b^2-4ac=0$, you will  have only one horizontal tangent line. Now, a=3,b=2,c=k, solve  $2^2$-4 x 3 k=0, k=1/3.


问:
Question about basis change/transformation
Let

$B=\left[ \begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix},   \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix},  \begin{pmatrix} 0 \\ 0\\ 1   \end{pmatrix}\right]$

$B_1=\left[ \begin{pmatrix} 1 \\ -2 \\ 0  \end{pmatrix},   \begin{pmatrix} 1\\ 0 \\ -2\end{pmatrix},  \begin{pmatrix} 2 \\ 5\\ 6   \end{pmatrix}\right]$

and the following map

$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $f\left[\begin{pmatrix} \alpha \\ \beta \\ \gamma  \end{pmatrix}\right]=\begin{pmatrix} \alpha+\gamma   \\ 2\alpha+2\beta+\gamma  \\ 4\alpha+2\beta+\gamma  \end{pmatrix}$

I need to determine $\left[f\right]_{B,B}$ and $T=[id]_{B,B_1}$

So I have an idea how to start with  $\left[f\right]_{B,B}$

I need to take the first column vector and

$f\begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 4  \end{pmatrix}$
and then I need to disassemble the resulting vector  $\begin{pmatrix} 1 \\ 2 \\ 4  \end{pmatrix}$ into = $1 \cdot\begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix}+2\cdot \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}+4 \cdot \begin{pmatrix} 0 \\ 0\\ 1   \end{pmatrix}$.

and the same also with the second and third column vector. Is this correct?

In this case the breakdown into the basis vectors was very easy, but what if it's not the standard basis? How can I see how to disassemble in such cases? Can anyone recommend me any good .pdf with examples provided where I can read about this topic?

my second question how to determine $T=[id]_{B,B_1}$?

Do I need to invert something here?

答:
Yes, you are correct. In general, given a linear transformation from $V$ to $W$ vector spaces $f:V \to W$, $\alpha=\{v_1, \cdots, v_n\}$ basis of $V$ and $\beta=\{w_1, \cdots, w_n\}$ basis of $W$ to construct $[f]_{\alpha,\beta}$ you proceed like this: The $i$-th column of the matrix $[f]_{\alpha,\beta}$ is given by the coordinates of $f(v_i)$ with respect to the basis $\beta$. That is the $i$-th column of $[f]_{\alpha,\beta}$
$$
\begin{bmatrix}
    c_1 \\
    c_2\\
    \cdot\\
 \cdot\\
 \cdot\\
    c_m
\end{bmatrix}
$$
is such that $f(v_i)=c_1 w_1+ c_2 w_2 + \cdots + c_m w_m$. As you noticed this is not always so simple as it was in this case with the canonical basis of $\mathbb{R}^3$. But in your case with $V=W=\mathbb{R}^3$ and $\alpha=\beta=\{u_,u_2,u_3\}$ we must find $[f]_{\alpha,\alpha}$. Then we need to decompose $f(u_i)$ with respect to the basis $\{u_1,u_2,u_3\}$. Hence we must find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ such that $f(u_i)=a_{1,i} u_1 + a_{2,i} u_2 + a_{3,i} u_3$. Now denoting the vectors as
$$
f(u_i)=\begin{pmatrix}
    y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}
, u_i=\begin{pmatrix}
    x_{1,i}\\
    x_{2,i}\\
   x_{3,i}
\end{pmatrix}
$$
we have that
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    a_{1,i}x_{1,1}\\
    a_{1,i}x_{2,1}\\
   a_{1,i}x_{3,1}
\end{pmatrix}+\begin{pmatrix}
    a_{2,i}x_{1,2}\\
    a_{2,i}x_{2,2}\\
   a_{2,i}x_{3,2}
\end{pmatrix}+\begin{pmatrix}
    a_{3,i}x_{1,3}\\
    a_{3,i}x_{2,3}\\
   a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    a_{1,i}x_{1,1}+a_{2,i}x_{1,2}+a_{3,i}x_{1,3}\\
    a_{1,i}x_{2,1}+a_{2,i}x_{2,2}+a_{3,i}x_{2,3}\\
   a_{1,i}x_{3,1}+a_{2,i}x_{3,2}+a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    x_{1,1} && x_{1,2} && x_{1,3}\\
    x_{2,1}  && x_{2,2} && x_{2,3}\\
   x_{3,1} && x_{3,2} && x_{3,3}
\end{pmatrix} \begin{pmatrix}
    a_{1,i}\\
    a_{2,i}\\
   a_{3,i}
\end{pmatrix}.
$$
And in order to find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ you solve the $i$-th linear system. Now $[id]_{B,B_1}$ is the representation of the identity transformation $id:\mathbb{R}^3 \to \mathbb{R}^3$ relative with the basis $B$ and $B_1$. The first column of the matrix is the coordinates of the identity applied to $(1, 0, 0)$ with respect to the basis $B_1$. That is we must find $a, b$ and $c$ such that
$$
\begin{pmatrix}
    1\\
    0\\
   0
\end{pmatrix}=id \begin{pmatrix}
    1\\
    0\\
   0
\end{pmatrix}=a\begin{pmatrix}
    1\\
    -2\\
   0
\end{pmatrix}+b\begin{pmatrix}
    1\\
    0\\
   -2
\end{pmatrix}+c\begin{pmatrix}
    2\\
    5\\
   6
\end{pmatrix}
$$
$$
\begin{pmatrix}
  1\\
    0\\
    0
\end{pmatrix}=\begin{pmatrix}
    1 && 1 && 2\\
    -2  && 0 && 5\\
   0 && -2 && 6
\end{pmatrix} \begin{pmatrix}
    a\\
    b\\
   c
\end{pmatrix}.
$$
Then you can apply the same procedure to $(0,1,0)$ and $(0,0,1)$ in order to find the second and third columns of $[id]_{B,B_1}$. Another way to calculate $[id]_{B,B_1}$ is to find $[id]_{B_1,B}$ and since $B$ is the canonical basis
$$
[id]_{B_1,B}=\begin{pmatrix}
    1 && 1 && 2\\
    -2  && 0 && 5\\
   0 && -2 && 6
\end{pmatrix}.
$$
And then you can use the fact that $[id]_{B,B_1}=([id]_{B_1,B})^{-1}$ and calculate the inverse matrix.


问:
I do not understand this particular step in a proof using the Well Ordering Principle
Below is a proof using the Well Ordering Principle. I get lost starting at $(13)$...

$$
\begin{aligned}
P(c): &:=c^{3} \leq 3^{c} \\
& \equiv c^{3} \leq 3(c-1)^{3} \\
& \equiv c \leq \sqrt[3]{3} \times(c-1)
\end{aligned}
$$

I don't understand how we get $c^{3} \leq 3(c-1)^{3}$ from $c^{3} \leq 3^{c}$.. why does the righthand side of the inequality change like that?

答:
You want to show that
$c^3 \le 3^c$
by induction.

This means that you want to show that
$(c-1)^3 \le 3^{c-1}$
implies that
$c^3 \le 3^c$.

By the induction hypothesis,
$3^c = 3\cdot 3^{c-1}
\ge 3(c-1)^3
$.

Therefore,
if you can show that
$3(c-1)^3
\ge c^3
$
then
$3^c \ge c^3
$.

Now we can proceed as above.
Taking cube roots,
$3(c-1)^3
\ge c^3
$
becomes
$3^{1/3}(c-1)
\ge c
$
or
$c(3^{1/3}-1)
\ge 3^{1/3}
$
or
$c
\ge \dfrac{3^{1/3}}{3^{1/3}-1}
= \dfrac{1}{1-3^{-1/3}}
\approx 3.26...
$
so the induction will work for
$c \ge 4$.

Since
$4^3 (64) < 3^4 (81)
$,
$c^3 \le 3^c$
for $c \ge 4$.
(With induction,
you always have to check
the base case.)


问:
Find orthonormal vectors spanning column space:
Question is out of Strang's Intro Linear Algebra:

(a) Find orthonormal vectors $ q_1, q_2, q_3 $ such that $ q_1, q_2 $ span the column space of:
$$
A=\left[\begin{array}{rr}1 & 1 \\ 2 & -1 \\ -2 & 4\end{array}\right]
$$
(b) Which of the four fundamental subspaces contains $ q_3$?

In finding $ q_1, q_2 $ we solve the problem through Gram-Schmidt.  I set $q_1$ as is, except for normalizing and dividing by length, and set that first column of A to $q_1$.  To find $q_2$, I subtract from the column 2, its own projection onto column 1.  After normalizing I get:
$$
\boldsymbol{q}_1 = \frac{1}{3}(1,2,-2) \text { and } \boldsymbol{q}_{2}=\frac{1}{3}(2,1,2)
$$

I'm unclear how to generally find $ q_3$. Is there a formula or method for this?  The answer is:
$$
\boldsymbol{q}_{3}=\frac{1}{3}(2,-2,-1)
$$

Beyond that, the answer to (b) is that $q_3$ lies in the nullspace of A transpose.  I do not understand this.  Isn't $q_3$ supposed to be in the column space of A?

答:
Good, you have already found $q_1, q_2$ by Gram-Schmidt.

Now, we want $q_3$ to be a unit vector orthogonal to $q_1$ and $q_2$. For some vector $q_3 = (x,y,z)$ to be orthogonal to $q_1$ and $q_2$ means $(x,y,z) \cdot q_1 = 0$ and $(x,y,z) \cdot q_2 = 0$. This determines two equations in the three variables $x, y, z$. In fact, it is a homogeneous system, and hence always has a solution (in fact, infinitely many), which can be found by row-reduction. Then, divide by the length to get a unit vector. I leave all this to you to do.

$q_3$ is not in the column space of $A$: The column space of $A$ equals the span of $q_1$ and $q_2$ (recall that Gram-Schmidt does not change the span of the original vectors), and $q_3$ is linearly independent from them (recall that nonzero orthogonal vectors are linearly independent). As you may know, the nullspace of $A^T$ is the orthogonal complement of the column space of $A$. Thus $q_3$ is in the nullspace of $A^T$.


答: The proof is in the reverse order of what it should be.  They start from $P(c)$, which is what they want to prove and work to something we know.  It should go the other way.  We start with the claim that for $c \ge 4$ (they say $n$ but we are not talking about $n$ here) we have $c \le \sqrt[3]3(c-1)$, which is true but asserted without proof.  We should say that for $c \ge 4$ we have $\frac c{c-1}\le \frac 43 \le \sqrt[3]3$, so $c \le \sqrt[3]3(c-1)$.  We then go upward to $(13)$, which is what we want to prove.

Title: I do not understand this particular step in a proof using the Well Ordering Principle

Body:
Below is a proof using the Well Ordering Principle. I get lost starting at $(13)$...

$$
\begin{aligned}
P(c): &:=c^{3} \leq 3^{c} \\
& \equiv c^{3} \leq 3(c-1)^{3} \\
& \equiv c \leq \sqrt[3]{3} \times(c-1)
\end{aligned}
$$

I don't understand how we get $c^{3} \leq 3(c-1)^{3}$ from $c^{3} \leq 3^{c}$.. why does the righthand side of the inequality change like that?


答:
The whole thing is a rather appalling mess; I might have given it as much as $7$ points out of $10$ if it had come from a weak student. As Ross Millikan points out in the comments, $P(n)$ should be simply $n\le 3^{n/3}$; the statement given as $P(n)$ is not in fact a function of $n$ at all.

$C$ is apparently supposed to be the set of non-negative integers for which the proposition fails, the goal being to show that $C$ must be empty. What is actually ‘defined’ is something called $C(n)$ that apparently depends on $n$ and yet is defined as $\{n\in\Bbb N:n\ne 3^{n/3}\}$, something that does not depend on $n$ and is not what is wanted for $C$ anyway, since we’re not trying to prove that $n=3^{n/3}$ for all $n\in\Bbb N$.

Presumably $c$ is supposed to be the least element of $C$, not ‘the lesser’ element of $C$.

The two lines immediately below $(13)$ do not follow from $(13)$. The argument is probably intended to be that $c\le\sqrt[3]3(c-1)$, since $c-1\ge 4$, so $c^3\le 3(c-1)^3$, and $P(c)$ now follows from $(12)$.


答:
Continuing where you left off:

We know that the horizontal tangent line is found when $$f'(x) = 3x^2 + 2x + k =0$$

So we need to find the value of $k$ which only gives us $1$ root for this polynomial. Let's us use the quadratic formula to solve for the roots.

$$x=\frac{-2\pm \sqrt{2^2-4(3)(k)}}{2(3)}\\
x=\frac{-2\pm \sqrt{4-4(3)(k)}}{6}\\
x=\frac{-2\pm \sqrt{4(1-3k)}}{6}\\
x=\frac{-2\pm 2\sqrt{1-3k}}{6}\\
x=\frac{-1\pm\sqrt{1-3k}}{3}$$

Now, the only way that $x$ can have one solution is if $$\sqrt{1-3k}=0\\1-3k=0\\ 3k=1\\ \therefore k=\frac{1}{3}$$

Plugging $k$ back in to solve for our single root of $x$, $$x=\frac{-1\pm\sqrt{1-3\left(\frac{1}{3}\right)}}{3}\\
x=\frac{-1\pm\sqrt{1-1}}{3}\\
x=\frac{-1}{3}$$

The only way $x$ can have a single root is if $k=\frac{1}{3}$.

Title: Determine the value of K so the function have only one point of one horizontal point of tangency

Body:
Consider the polynomial function  define by

$ f(x) = x^3 + x^2 + kx - 1 $

Determine the value of k so f has only on point of horizontal
tangency.

So I know that $   f^{'}(x) = 3x^2 + 2x + k  $
And I should consider  $   f^{'}(x) = 0 $ but how I find a k that makes the $   f^{'}(x) = 3x^2 + 2x + k  $   have only one root?

I also thought that if $ k = -x^2 $ then the function would have only one point of horizontal tangency. But this does not works because this not equal the answer.  ( Though I think it shouldwork too )

Btw the answer is  $ \frac{1}{3} $ And I still can't figure why.


问:
Proof that $1-\frac12+\frac13-\frac14+\dots=\ln2$
Proof that $\displaystyle 1-\frac12+\frac13-\frac14+\dots=\ln2$.

I have read this answer but I don't understand the fifth line.
\begin{align*}
S_{2n}&=1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}
\\&=\left(1+\frac13+\dots+\frac1{2n-1}\right)-\left(\frac12+\frac14+\dots+\frac1{2n}\right)
\\&=\left(1+\frac12+\frac13+\dots+\frac1{2n}\right)-2\left(\frac12+\frac14+\dots+\frac1{2n}\right)
\\&=\left(1+\frac12+\frac13+\dots+\frac1{2n}\right)-\left(1+\frac12+\frac13+\dots+\frac1n\right)
\\\color{gray}{[\text{ why? }]}\quad&=[\ln 2n+\gamma+o(1)]-[\ln n+\gamma+o(1)]
\\&=\ln2+o(1)
\end{align*}
when $n\to\infty$, with $\gamma=\displaystyle\lim_{x\to\infty}\left(1+\frac12+\frac13+\dots+\frac1n-\ln n\right)$.

Otherway, $S_{2n+1}=S_n+\displaystyle\frac{1}{2n+1} \implies \lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}S_{2n}=\ln2$

答:
This proof uses the fact that $\lim_{n\to\infty} \left ( \left(\sum_{k=1}^n \frac{1}{k}\right) - \ln n\right )$ converges to a particular value, $\gamma\approx0.577...$ By using the definition of a limit, we can write:

$$
\gamma = \left(\sum_{k=1}^n \frac{1}{k}\right) - \ln n + \epsilon
$$

Where $\epsilon$ can be made arbitrarily small by making $n$ larger and larger. (Values that behave this way can also be denoted by "$o(1)$")


问:
Derivative of $f(x) = \sqrt[3]{(x-1)(x-2021)}|\sin(\pi x)|$ at $x = 1$
So I got this really strange contradiction that I couldn't explain to myself and I'd really love some clarification on the pitfalls that I'm potentially facing.

Let $$f(x) = \sqrt[3]{(x-1)(x-2021)}|\sin(\pi x)|$$

The very strange thing that's happening is that when I try to evaluate $f'(1)$ with the direct definition as:
$$f'(1) = \lim_{x\to 1}\frac{f(x) - f(1)}{x - 1} = \lim_{x\to 1}(x-1)^{-\frac{2}{3}}(x-2021)^\frac{1}{3}|\sin(\pi x)| = 0$$

But when I try to explicitly find $f'(x)$ as an expression for $x$:
$$f'(x) = \frac{(4x-4044)|\sin(\pi x)|^2 + 3\pi(x-2021)(x-1)\sin(2\pi x)}{6\left((x-1)(x-2021)\right)^\frac{2}{3}|\sin(\pi x)|}$$
and then plug in $x = 1$, I clearly get a $0$ in the denominator, hence $f'(1)$ is undefined.

There's obviously something smelly, but I find this one to be a really interesting case that I couldn't explain to myself. Any advice is welcome and appreciated! Thanks in advance!

Body:
$\lim_{x \to 1} (x-1)^{-2/3}
=\lim_{x \to 1} \dfrac1{(x-1)^{2/3}}
\to \infty$
since
$\lim_{x \to 1}(x-1)^{2/3}
=0
$.


问:
Can a function $f: \Bbb R \to \Bbb R$ have finitely many local maxima, but infinitely many local minima ? What happens if $f$ is continuous?
Can a function $f: \Bbb R \to \Bbb R$ have finitely many local maxima, but infinitely many local minima ? What happens if $f$ is continuous ? I am aware of the fact that for a continuous function, between two local maxima, there is a local mimimum

答: If $f$ is allowed to be discontinuous, we can just take $f(x) = x$ when $x$ is not an integer, and $f(x) = x-1$ when $x$ is an integer. Infinitely many local minima, and zero local maxima.


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
Here's my non-mathematical answer:

It works for elements $1, 2, 4$ and $8$ of your series (whose values are $1, 2, 8$ and $128$): $1 \times 100 + 2 \times 10 + 8 = 128$

If you start anywhere else (say, at term $16$, whose value is $\,\,32768$), then you can just divide all those terms by $32768$ to have the same expression as in step 1.

So I think any expression involving a series of terms in the $2^n$ series would work.


答: Not if $f$ is allowed to be discontinuous. Consider $|\sec(x)|$.

Title: Can a function $f: \Bbb R \to \Bbb R$ have finitely many local maxima, but infinitely many local minima ? What happens if $f$ is continuous?

Body: Can a function $f: \Bbb R \to \Bbb R$ have finitely many local maxima, but infinitely many local minima ? What happens if $f$ is continuous ? I am aware of the fact that for a continuous function, between two local maxima, there is a local mimimum


问:
Which Slope is larger?
If a line has $\tan(120^\circ)$ as gradient or slope and another line has $\tan(130^\circ) $as slope, which slope is larger? I mean numerically $\tan(120^\circ)=-1.7320$ and $\tan(130^\circ)=-1.1917$. Since $-1.1917>-1.7320$ so this slope seems grater to me. But the steepness of slope $\tan(120^\circ)$ is higher. So which slope is larger?

答: In mathematics, we will investiget the slope, or gradient, or the angle of inclination of a line. There is no definition on the term "steepness".</p>

Title: Which Slope is larger?


问:
How can I find the limit of this fraction? ($0/0$ type)
Suppose I have two complex numbers $g_0$ and $g_1$, when $g_0\rightarrow1$ and $g_1\rightarrow0$, how can I evaluate
$$
\sqrt\frac{g_1^2}{1-g_0^2}
$$
Intuitively I think that should be infinity, but I'm not pretty sure how can I justify that? Thanks!!

答:
Let
$g_1 = a$
and
$g_0 = 1-b$
so
$\sqrt\frac{g_1^2}{1-g_0^2}
=\sqrt\frac{a^2}{1-(1-b)^2}
=\sqrt\frac{a^2}{b(2-b)}
$
where
$a, b \to 0$.

If
$a=b$ this is
$\sqrt\frac{b^2}{b(2-b)}
=\sqrt\frac{b}{(2-b)}
\to 0
$.

If $a = b^{1/4}$ this is
$\sqrt\frac{b^{1/2}}{b(2-b)}
=\sqrt\frac{1}{b^{1/2}(2-b)}
\to \infty
$.

If $a = cb^{1/2}$ this is
$\sqrt\frac{c^2b}{b(2-b)}
=\sqrt\frac{c^2}{(2-b)}
\to \dfrac{c^2}{2}
$
so this can approach any positive value.


问:
How to find $AM$ which is a distance between two points in a mountain when looked through a binoculars?
The problem is as follows:

The figure from below shows a cartoon which represents the view of a pair of mountains when looked through a binoculars. Assuming that according to the inner ruler of the binoculars you can spot $BM=MN=4\,cm$ and $NC=6\,cm$, the inner goniometer indicates that $\angle ANC=\angle AMB=90^\circ$ and $\angle  BAM=\angle MAC$. Also it is known $A,\,M,\,N$ are colinear. With these information find the actual distance between $A$ and $M$. Assuming the scale is $\frac{1}{1000}$. Take into account that $\triangle ABM$ and $\triangle ANC$ are triangles.

The choices given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{40 m}\\
2.&\textrm{50 m}\\
3.&\textrm{60 m}\\
4.&\textrm{80 m}\\
\end{array}$

I don't know exactly what to do in this problem. Can someone help me here?. I'm not sure if should I use the given relations in a right triangle or what?. It would be nice if someone could help me with how to use the drawing given to assign the angles and find the length that it is requested which is $AM$. But just I don't know should congruence be used here?. Can someone help me with this?. Or would it be similarity of triangles?.

Help please! I'm lost.

答:
So, first.
The 2 triangles are  similar because they share two angles in common,  so we can make the following equation:

$ \frac{BM}{AM} = \frac{NC}{AN} $

And we know that $ AN = 4 + AM $ so:

$ \frac{4}{AM} = \frac{6}{4+AM} $

Then solving this equation we get:

$ AM = 8cm $

$ AM = 8 * 1000 $ So as 1 centimeter equals 1000 centimeters

$ AM = 8000 cm $

$ AM = 80 m $

P.S You asked for help about assigning the angles, idk what do you really mean, but if you need help to know where the angles are, they're always in the point of the middle, between the two segments ( first point and middle point, middle point and last point ).
So for example, the angle ABC is in point B, and it's between segment AB and segment BC.

Hope that this helps you!


问:
Inequalities with multiple variables : Find out the range of another variable
Given that $-1 \le v \le1,-2 \le u \le -0.5 \ and -2 \le z \le-0.5 \ and \ w=\frac{vz}{u} ,$ then what is the range of values for $w$?

I solved this question by trial and error method i.e. by plugging in the values of $v,u \ and \ z$ in the equation $w=\frac{vz}{u}$ to find the maxima and minima of the function. But it feels like a tedious method and not a reliable one.

Can someone please help me with a better solution or approach for this question?

答:
It is clear that the range would be symmetric about $0$ due to the range of $v$.

Now, let's consider the largest value that $\left| \frac{z}{u} \right|$ can take. Let's make $|z|$ as large as possible and $|u|$ as small as possible, the maximum value is $\frac{2}{0.5}=4$.

Hence the range of $w$ is $[-4,4]$.


问:
Find the area of an equilateral triangle inscribed in a unit square
If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.

I attempted to say, I state that $x=MD$

then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?

答:
Due to symmetry about the diagonal $AC$, $\angle NCB =( \angle DCB - \angle MCN) /2=15^{\circ}$

Hence $a=\sec 15^{\circ}$ and area of equilateral triangle is
$$\frac{\sqrt{3}}{4}\cdot \sec^2 15^{\circ}=2\sqrt{3}-3$$


问:
How can a parallel edge be identified?
An edge will have the same vertices as another edge that it is parallel to, so how can it be uniquely described? </p>

答:
I would prefer to use a variant of the second definition offered by Zev. A multigraph has a set of vertices $V$, a set of edges $E$. The connection between vertices and edges can be described by a relation on $V\times E$, such that
each edge is incident with one or two vertices (just one if you're not inclined to allow loops). Or it can be described by a function that assigns to each edge either one vertex or an unordered pair of vertices.

One place where multigraphs force themselves on us is when we consider duals of plane graphs. So we need to work with embeddings of multigraphs and in this situation the multiset approach would be a little awkward. My suspicion is that working graph theorists would not usually use the multiset definition (but I have not carried out a survey).

Title: How can a parallel edge be identified?


问:
What's the maximum and minimum of $\sin Z$ and $\cos Z$ where $Z$ is a complex number?
I see that a lot of people try solving something  like $\sin Z=2$. So that breaks the the maximum of $\sin x$. Now I wonder if there is a maximum and minimum of $\sin Z$ and $\cos Z$ if $Z$ is a complex number.

答:
I'll be elementary here.

Since,
for all complex $x$,
$e^{ix}
=\cos(x)+i\sin(x)
$
and
$e^{-ix}
=\cos(x)-i\sin(x)
$,
$\sin(x)
=\dfrac{e^{ix}-e^{-ix}}{2i}
=-i\dfrac{e^{ix}-e^{-ix}}{2}
=-i\sinh(ix)
$.

This holds for all complex $x$.

Letting $x = -iy$,
so $ix = y$,
$\sin(-iy)
=-i\sinh(y)
$.

Letting
$y \to \infty$,
$\sinh(y) \to \infty$
so
$|\sin(iy)|
\to \infty$.


问:
One sided ideal in finite dimensional $\mathrm{C}^*$-algebras
Let $\mathcal{A}$ be a finite dimensional $\mathrm{C}^*$-algebra and $\mathcal{A}p$ a one-sided, non-self-adjoint left ideal. That is there exists $f_1,f_2\in\mathcal{A}$ such that
$$f_1pf_2\not\in\mathcal{A}p.$$

Can the "not a right ideal"ness of $\mathcal{A}p$ be exhibited by a projection $q=f_1=f_2$?

答:
Lemma.  If $p$ and $f$ are projections in a C*-algebra such that $fpf$ is idempotent, then $p$ and $f$ commute.

Proof. We have
$$
  (pf-fp)^3=
  (pfpf - fpf - pfp + fpfp)(pf-fp) = $$$$ =
  pfpfpf - fpfpf - pfpf + fpfpf  - pfpfp + fpfp + pfpfp - fpfpfp =  $$$$ =
  pfpf - fpf - pfpf + fpf  - pfpfp + fpfp + pfpfp - fpfp = 0.
  $$
Since $pf-fp$ is skew-adjoint,  it follows that $pf-fp=0$, so $p$ and $f$ commute. $\qquad\square$

Proposition.
Let $A$ be a C*-algebra generated by projections and let $p$ be a fixed projection in $A$.  If
$$
  fpf\in Ap,
  $$
for every
projection $f$, then $p$ is central (and hence $Ap$ is a two sided ideal).

Proof.
If $fpf \in  Ap$,  then $fpf = fpfp$.  Multiplying on the right by $f$ gives  $fpf = fpfpf$, which means that $fpf$ is idempotent,  and hence $f$ and $p$
commute by the Lemma.  Since $f$ is an arbitrary projection in  $A$, and $A$ is generated by projections, then $p$ is central. $\qquad\square$


答:
What you described in your comment is essentially all you need to do for a solution.  If you want something a bit more formal, you could write, for instance:

Case I: $ \quad   0 \ \le \ v \ \le \ 1 $
$$ 0  \ \le v \ \le \ 1 \ \ \Rightarrow \ \ 0 \ \ge \ vz  \ge \  -2 \ \ \Rightarrow \ \ 0 \ \le \ \frac{vz}{u} \ \le \  \frac{-2}{-0.5} \ = \ +4 \ \ ;  $$

Case II:  $ \quad   -1 \ \le \ v \ \le \ 0 $
$$ -1  \ \le v \ \le \ 0 \ \ \Rightarrow \ \ 0 \ \le \ vz  \le \  +2 \ \ \Rightarrow \ \ 0 \ \ge \ \frac{vz}{u} \ \ge \  \frac{+2}{-0.5} \ = \ -4 \ \ .  $$

Hence, $ \ -4 \ \le \ w \ \le \ 4 \ \ . $

Title: Inequalities with multiple variables : Find out the range of another variable

Body:
Given that $-1 \le v \le1,-2 \le u \le -0.5 \ and -2 \le z \le-0.5 \ and \ w=\frac{vz}{u} ,$ then what is the range of values for $w$?

I solved this question by trial and error method i.e. by plugging in the values of $v,u \ and \ z$ in the equation $w=\frac{vz}{u}$ to find the maxima and minima of the function. But it feels like a tedious method and not a reliable one.

Can someone please help me with a better solution or approach for this question?