问:
Follow-up regarding disproof of modified conjecture found in MathOverflow
I was looking through the OEIS, and I found the following conjecture on sequence $\text{A}007097$:

If $a_n$ is a sequence for which $a_1=2$ and $a_{n+1}$ is the $a_n$-th prime, $$\prod_{i=1}^{n}\log(a_i)\sim a_n$$

I then saw that this conjecture had been disproved on MathOverflow by GH from MO in the comments on this question, but with a substituted possible replacement conjecture. The solver claimed the replacement was probably false as well, but that a proof would be considerably more difficult. So my question is, is the replacement conjecture true? Or more specifically, determine the veracity of the following:

If $a_n$ is a sequence for which $a_1=2$ and $a_{n+1}$ is the $a_n$-th prime, $$\prod_{i=1}^{n}\log(a_i)\sim a_{n+1}$$

My own math skills are decidedly insufficient to determine this question one way or the other, but I'm really interested in the problem and would be curious to see a resolution. Any help would be appreciated. Thanks!

Edit: I've learned that the person who "disproved" the original conjecture on MathOverflow has since then realized that they made a mistake in their disproof, so the original conjecture may still be valid. Maybe a fun challenge for someone else.

答:
Let, for each $n \geq 1$, $p_n$ be the $n$-th prime number, and $1+r_n=\frac{p_n}{n\ln{n}}$, so that $r_n \sim \frac{\ln{\ln{n}}}{\ln{n}}$ by refinements of the prime number theorem.

Let $\delta_n=r_{a_n}$, so that $\ln{a_n}=\frac{a_{n+1}}{a_n(1+\delta_n)}$.

Taking products, it follows $a_{n+1}=2\prod_{k=1}^n{(1+\delta_k)}\prod_{k=1}^n{\ln{a_k}}$.

Now we’ll show that $\prod_{k=1}^n{(1+\delta_k)}$ diverges. It’s enough to show that $\sum_{k \geq 1}{\frac{\ln{\ln{a_k}}}{\ln{a_k}}}$ diverges.

Let $b_k=\ln{a_k}$, clearly $b_k$ grows to infinity and by the prime number theorem $b_{k+1}-b_k=\ln{b_k}+o(1)$. We want to show that $\sum_{k \geq 1}{\frac{\ln{b_k}}{b_k}}$ diverges to infinity. But as $\frac{\ln{b_k}}{b_k}\sim \frac{b_{k+1}-b_k}{b_k} \sim \ln{\frac{b_{k+1}}{b_k}}$, this is clearly divergent.

By doing everything explicitly, we can even write $a_{n+1}=2\left(\prod_{k=1}^n{\ln{a_k}}\right)(b_{n+1}/b_1)^{1+o(1)}$.


问:
Estimate for integral with integrand $e^{(\log t)^{\alpha}}$.
Let $\alpha>1$. Is there an estimate for large $r$ (where large means that there is $r_{0}>0$ such that for all $r\geq r_{0}$ the estimate holds) for the following (positive) function $$\Phi(r)=\int_{1}^{r}{e^{(\log t)^{\alpha}}dt},$$ in the form that there is a (positive) function $\Psi(r)$ such that $$\Psi(r)\simeq \Phi(r),$$ where $"\simeq"$ means that there are constants $c_{1}, c_{2}>0$ such that $$c_{1}\leq\frac{\Psi(r)}{\Phi(r)}\leq c_{2}.$$

Thanks in advance!

答:
ok how about:
$$x=\log(t)\Rightarrow dt=tdx=e^xdx$$
and so:
$$e^{(\log t)^\alpha}dt=e^{x^\alpha}e^xdx$$
and your integral becomes:
$$\Phi(r)=\int_0^{\ln(r)}e^{x^\alpha}e^xdx$$
$$e^{x^\alpha}=\sum_{k=0}^\infty\frac{x^{\alpha k}}{k!}$$
$$\Phi(r)=\int_0^{\ln r}\sum_{k=0}^\infty\frac{x^{\alpha k}e^x}{k!}dx=\sum_{k=0}^\infty\frac{1}{k!}\int_0^{\ln r}x^{ak}e^xdx$$
$$u=-x\Rightarrow dx=-du$$
$$\Phi(r)=\sum_{k=0}^\infty\frac{(-1)^{\alpha k+1}}{k!}\int_0^{\ln(1/r)}u^{\alpha k}e^{-u}du$$
Which can be expressed in terms of a sum of the incomplete gamma function.

ContentLicense: CC BY-SA 4.0


问:
Confused: using Taylor series to find derivative
TL;DR: read bolded parts

Lets say I have f(x) = sin(x^2) and I want the f''''''(0) (6th derivative). Using taylor series, this is really simple. We plug in x^2 into the taylor polynomial of sin(x), and get this:

Then the 6th derivative is 1/3! * 6! = 120.

I am confused because taylor series seems really unrelated; there should be an equally easy way to do this just with derivatives and chain rule (no detour to taylor series). But when I bash it out, I don't get a simple solution.

(120 on the last line, typo)

Why does taylor series come up in finding derivatives?

答:
Ok, perhaps there is another  way!

$$ f= \sin(x^2) $$

Rewrite this as:

$$ f= \sin(u)$$

where $ u =x^2$, with  $ \frac{du}{dx} = 2x $ , $ \frac{d^2 u}{dx^2} = 2 $ and$ \frac{d^3 u}{dx^3} = 0$

then:

$$ \frac{df}{dx}= g_1 = \cos u \frac{du}{dx}$$

At this point, we can apply the leibniz rule for derivatives (I wrote an article on this ),

$$ \frac{d^5 g_1}{dx^5} = \sum_{i=0}^5 \binom{5}{i} \frac{d^{i+1} u}{dx^{i+1}} \frac{d^{5-i} \cos u}{d^{5-i} x}$$

Now, the important thing to notice that the sum is zero except for $ i=1$ at x=0, (can you figure out why?). Hence,

$$ \frac{d^5 g_1}{dx^5} = f= \binom{5}{1} (2) \frac{d^4}{dx^4} \cos u |_{0}$$

Now rinse and repeat:

$$ g_2 = \frac{d^4}{dx^4} \cos u  =-\frac{d^3}{dx^3} ( \frac{du}{dx} \sin u ) =- \sum_{i=0}^3 \binom{3}{i} \frac{d^{u+i} u}{dx^{u+i}} \frac{d^{3-i} \sin u }{d^{3-i}x}=-\binom{3}{1}(2) \frac{d^2}{dx^2} \sin(u)$$

And again,

$$ g_3 = -\frac{d^2}{dx^2} \sin(u) = -\frac{d}{dx} ( \cos(u) \frac{du}{dx}) = 2\cos(u) \binom{1}{1}+ \text{stuff going to zero}$$

Piecing everything together,

$$ f= \binom{5}{1} 2 \binom{3}{1} 2 \binom{1}{1} 2 \cos(u)$$

Evaluating 'officially' at $x=0$

$$ f(0)=120$$


问:
How do we know that the Sine function has no Non-Real Roots?
In this question and answers (How was Euler able to create an infinite product for sinc by using its roots?) we use the fact that the real roots of $f(x)=\sin x$ occur when $x$ is an integer multiple of $\pi$ to obtain an infinite product for $\sin x$ in terms of of its factors.

My question is, how do we know that $f(x)=\sin x$ has no non-real roots?

Thanks in advance.

答:
$$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$$
Note that $\cosh y$ is never zero for $y$ a real number. Thus, if $\sin(x+iy)=0$, $\sin x=0$ (the real and imaginary parts must be identically zero). But this forces $\cos x=\pm1$, which in turn forces $\sinh y=0$ and thus $y=0$. All the roots must be real.


问:
What are the list of math books one must complete before entering into a Ph.D.?
I am interested to know what books one should complete before entering their Ph.D. in mathematics? What topics I should have a strong grip on in order to feel eligible for entering into a Ph.D.

One of my seniors said I must complete Baby Rudin, Munkres's Topology, Conway's Complex Analysis.

I want a complete list of such topics and books. This will help me prepare for my PhD.

答: Just look at what ppl who got a Ph.D. got to read and found to be insightful instead of asking this. The math sorcerer has book recommendations like this.</p>


问:
How does knowing the norm of an ideal tell us about its factors?
In the ring $\mathbb{Z}[\sqrt{223}]$ we have that the ideal $\mathfrak{p}_3=(3,\sqrt{223}+1)$ is a prime ideal lying over 3, and that the ideal $\mathfrak{p}_{11}=(11,\sqrt{223}+5)$ is a prime ideal lying over 11. If we find an element $16+\sqrt{223}$ with norm 33, this is supposed to imply that $(16+\sqrt{223})=\mathfrak{p}_3\mathfrak{p}_{11}$ But I am having trouble seeing why this is the case.

答:
Let $K$ be a number field. There are a couple of facts coming in handy here:

Any ideal of $\mathcal{O}_K$ has a unique factorization into positive powers of prime ideals.

The ideal norm is multiplicative.

If an ideal is integral, then its norm is in $\mathbb{Z}$.

The norm of a prime $\mathfrak{p}$ over $p\in \mathbb{Z}$ is always a power of $p$.

The norm of the ideal $x\mathcal{O}_K$ is the same as the norm of $x$.

Then we just have to put these things together... An ideal of norm $33$ has to be the product of some prime lying over $3$ and one lying over $11$. Now you just have to make sure that these ideals over $3$ and over $11$ are the right ones.

EDIT: As Jyrki Lahtonen correctly remarked, we are not done here and the computations are not as easy as I hoped:

Be careful here since both $3$ and $11$ split, so there are $4$ ideals with norm $33$ and we have to be sure that $\mathfrak{p}_3\mathfrak{p}_5$ is the ideal we are looking for and not one of the other three.
Let's calculate (by multiplying generators):
$$ \mathfrak{p}_3\mathfrak{p}_5 = (33, 11\sqrt{223}+11,3\sqrt{223}+15,228+6\sqrt{223})$$

A quick computation yields that
$$ 2\cdot (11\sqrt{223}+11)-7\cdot (3\sqrt{223}+15)+3\cdot 33 = 16+\sqrt{223}$$

Thus $(16+\sqrt{223})\subset \mathfrak{p}_3\mathfrak{p}_5$ meaning that $\mathfrak{p}_3$ and $\mathfrak{p}_5$ divide $(16+\sqrt{223})$, so by the first paragraph we have equality.


问:
Range of analytic function
Let $f:U \to \mathbb{C}$ be an analytic function, where $U$ is an open set containing the closed unit disc. Suppose that $|f(z)| \leq 1$, for all $|z|\leq 1$ and $f$ maps infinitely many points of $S^1$ to $S^1$. Can we characterize such functions? (Here $S^1$ means the unit circle).

Note that finite Blaschke products and unimodular multiple of monomials and the constant unimodular function are already in this class. Will there be more?

Thanks in advance.

答: Write $f=u+iv.$ Then $w(t) = u(e^{it})^2 + v(e^{it})^2$ is a real analytic function on $\mathbb R.$ We are given that $w(t)=1$ for infinitely many $t\in [0,2\pi].$ Thus $w=1$ on set with limit point in $[0,2\pi].$ By the identity principle for real analytic functions, $w\equiv 1.$ This implies $|f(e^{it})|\equiv 1.$ A well known exercise then shows $f=cB_1\cdots B_n,$ where $c$ is a constant of modulus $1,$ and the $B_k$ are Blaschke factors.

Title: Range of analytic function


问:
Calculate the following limit:
Calculate $\;\lim\limits_{x\to\infty} \left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]$.

It is a $\frac00$ case of indetermination if we rewrite as $\lim_{x\to\infty} \frac{((1+\frac1x)^x-e\ln(1+\frac1x)^x)}{\frac{1}{x^2}}$, since $\lim_{x\to\infty} \frac{1}{x^2} = 0$, $\lim_{x\to\infty} (1+\frac1x)^x = e$ and $\lim_{x\to\infty} \ln(1+\frac1x)^x = 1$.

I think that it is the type that has a solution without l'Hospital's rule, but it's quite difficult to find, so l'Hospital still remains the best try to me. I tried using it with different rewrites, but it seems that it needs to be used multiple times, and the expression gets harder and harder to calculate, so I assume that some other limit must be applied first to make the expression nicer.

Also, I futilely tried to use the following known limits by changing $x$ into $y = \frac1x$ if needed (and adding and substracting $ex^2$ in the main parenthesis and trying to use the last 2 limits), but maybe it can help you: $\lim_{x\to0} \frac{a^x-1}{x} = \ln a$, $\lim_{x\to0} \frac{\ln(1+x)}{x} = 1$, $\lim_{x\to0} \frac{(1+x)^r-1}{x} = r$, $\lim_{x\to0} \frac{(1+x)^\frac1x-e}{x} = -\frac{e}{2}$, $\lim_{x\to\infty} (x-x^2ln(1+\frac1x)) = \frac12$.

Can you help me with this problem?

答:
We can use the following asymptotic relations

$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x}+\frac{11e}{24x^2}+O\left(\frac{1}{x^3}\right)$$

$$\ln\left(1+\frac{1}{x}\right) = \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+O\left(\frac{1}{x^4}\right)$$

to get that

$$x^2\left(1+\frac{1}{x}\right)^x-ex^3\ln\left(1+\frac{1}{x}\right) \sim x^2\left(e - \frac{e}{2x}+\frac{11e}{24x^2} - e+\frac{e}{2x}-\frac{e}{3x^2}\right) = \frac{e}{8}$$

Title: Calculate the following limit:


问:
implication of $\sigma$-finite measure
For $(X, M, \mu)$, if $\mu$ is $\sigma$-finite, then does it imply that $\exists E_1 \subset E_2 \subset \cdots \text{ with } \cup_i E_i = X$ and $\mu(E_i) < \infty$?

In my opinion, the statement isn't right. It's equivalent to saying it's finite measure? Since with this statement, by continuity from below, $\mu(X) = \mu(\cup_i E_i) = \lim_i \mu(E_i) < \infty$

This is actually part of one solution given by our prof. similar to this for the problem but instead of assuming $E_i$ being partitions.

答:
Consider on $\mathbb{R}$ with Lebesgue measure the sets $E_i=[-i,i]$.

The crux is that each individual set $E_i$ having finite measure does not mean that the limit of the measures is finite. This has little to do with measure theory, it's about real-valued sequences and their limits.

Title: implication of $\sigma$-finite measure


问:
Representations of Lie Groups in Arbitrary Co-ordinates
Seeing that a Lie group is a manifold, one should be able to work in arbitrary co-ordinates on that manifold. However I have not seen any discussion of how representations of a Lie group and it's algebra are affected by such co-ordinate transformations.

For example, consider the 1-dimensional translation group, with representations of group element $x$ denoted by $U[x]$. An example of such a representation is:
\begin{equation}
U[x]\vert p\rangle=e^{-ipx}\vert p\rangle
\end{equation}
Now I should be able to change co-ordinates on the group manifold by:
\begin{equation}
\xi=\tanh(x)
\end{equation}
So that the action of group element $\xi$ (which is just group element $x$ in another co-ordinate system) is:
\begin{equation}
U[\xi]\vert p\rangle=e^{-i p\tanh^{-1}(\xi)}\vert p\rangle
\end{equation}
And this still forms a representation because one can quickly check that:
$U[\xi_a]U[\xi_b]=U[\xi_a\cdot \xi_b]$.

Questions: (1) Is it true that one can work in any co-ordinates on the group manifold for a representation of a Lie group and it's Lie algebra?
(2) If one does so, then group elements are typically no longer the exponential of Lie algebra elements?
(3) Has anything useful ever been learnt by working in non-exponential co-ordinates? (4) Is the following statement is correct: there is a preferred class of co-ordinates (canonical co-ordinates) on a Lie group because the action of a group element on the manifold generates a diffeomorphism so one might as well use the orbits generated by such a diffeomorphism as co-ordinates?

答:
Yes, you can work in arbitrary coordinates, but often you don't need to work in coordinates at all.

Group elements near the identity are always the exponential of elements of the Lie algebra; this is a coordinate-invariant fact.

I don't know. Again, often you don't need to work in coordinates at all. Off the top of my head I'm not aware of any results in Lie theory proven using any coordinates other than possibly the local coordinates given by the exponential map or possibly the local coordinates given by the exponential map with respect to a bi-invariant Riemannian metric.

I don't know what you mean by this.


答: Yes it does imply this. For example just take $\mathbb R$ with Lebesgue measure, and set $E_n=(-n,n)$. Your error is in stating that $\lim_{n\to\infty} \mu(E_n)<\infty$, as that is clearly not the case. If each $\mu(E_n)$ is larger than the last why do you think the limit must be finite? This is just a sequence of real numbers. You don't think that $\lim_{n\to\infty} n^2<\infty$.

Title: implication of $\sigma$-finite measure

Body:
For $(X, M, \mu)$, if $\mu$ is $\sigma$-finite, then does it imply that $\exists E_1 \subset E_2 \subset \cdots \text{ with } \cup_i E_i = X$ and $\mu(E_i) < \infty$?

In my opinion, the statement isn't right. It's equivalent to saying it's finite measure? Since with this statement, by continuity from below, $\mu(X) = \mu(\cup_i E_i) = \lim_i \mu(E_i) < \infty$

This is actually part of one solution given by our prof. similar to this for the problem but instead of assuming $E_i$ being partitions.


问:
Writing $\sqrt[4]{16}$ in polar form and acquiring a solution that doesn't exist
$$\sqrt[4]{16} = ?$$

$$16 = |16|(\cos(0)+i\sin(\pi))$$
the equality above holds.

$$\sqrt[4]{16} = |\sqrt[4]{16}|(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4})) $$
using De Moivre"s formula we get:
$$ \sqrt[4]{16} = 2(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4}))$$
$$\text{for k = 0, we get one of the correct solutions,  but for k = 1 we get $-i\sqrt{2}$ which is not a solution.}$$

Why does this kind of approach not work?

答:
The proper way to express a complex number in polar coordinates is $z = |z| (\cos(\mathrm{arg \ z}) + i\sin (\mathrm{arg \ z}))$. Crucially, note that both the cosine and sine are of the same angle, the argument of the complex number.

What you did is wrong because you have $0$ as one angle and $\pi$ as the other. You cannot apply De Moivre's to that expression because that is not a proper polar form.

Instead, start with $16 = 16 (\cos 2k\pi + i \sin 2k\pi)$.


问:
Can a Accumulation Point be an Eigenvalue?
I have a discrete (separable) infinite dimensional Hilbert Space with a compact operator defined on it. So 0 is an accumulation point (some theorem says so). Can 0 also be an eigenvalue? And how would I prove it is/isnt?

Thanks

答:
The consensus, expressed in comments, is that $0$ can indeed be an eigenvalue: for example, the zero operator has $0$ as the only eigenvalue. For a less trivial example, consider the multiplication operator on $\ell_2$ with  entries $0,1/2,1/3,1/4,\dots$, that is,
$$(x_n)\mapsto (0,x_2/2,x_3/3,x_4/4\dots)$$
However, I advise some caution in the interpretation of

So 0 is an accumulation point (some theorem says so).

You should understand  accumulation point as the limit of the sequence of eigenvalues, not a limit point of the set of eigenvalues. For example, orthogonal projection onto a line has the spectrum $\{0,1\}$. Looking at it as a set, we would not say that $\{0\}$ is an accumulation point. But it is the limit of the sequence of eigenvalues, which is $1,0,0,0,0,\dots$.

And finally, you may want to include something about normality or self-adjointness of the operators in question.

ContentLicense: CC BY-SA 3.0


问:
Find a mistake in summing a series
I was doing one task and i can't find where I'm wrong. I have to find sum $S$ of given series. The answer is  $\frac{1}{8}$.
$$S = \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2(2n+1)^2} $$
First $\frac{n}{(2n-1)^2(2n+1)^2} = \frac{1}{2}(\frac{1}{(n - \frac{1}{2})^2}-\frac{1}{(n + \frac{1}{2})^2}) $
From this $ S_N = \frac{1}{2}(4-\frac{1}{(N+\frac{1}{2})^2}) $ and $ S = \lim_{N\to\infty}S_N = 2$
Where did I make a mistake?

Title: Find a mistake in summing a series

Body:
Partial fractions $$\frac{n}{(2 n-1)^2 (2 n+1)^2}=\frac{1}{8 (2 n-1)^2}-\frac{1}{8 (2 n+1)^2}$$
telescoping series
$$
\begin{array}{rl}
 s_1= & \frac{1}{8}-\frac{1}{72} \\
 s_2= & \frac{1}{8}-\frac{1}{72}+\frac{1}{72}-\frac{1}{200}= \frac{1}{8}-\frac{1}{200} \\
 s_3= &  \frac{1}{8}-\frac{1}{200}+\frac{1}{200}-\frac{1}{392}=\frac{1}{8}-\frac{1}{392}\\
\ldots & \ldots\\
 s_n= & \frac{1}{8}-\frac{1}{8 (2 n+1)^2}\\
\end{array}
$$
as $n\to\infty$, $s_n\to \frac18$


问:
Determine the convergence and divergence of $\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}dxdy}}$
Determine whether $$\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}\mathbb dx \mathbb dy}}$$
is convergent or divergent.

It looks like that it needs to be scaled by inequalities, but I failed. Can anyone help?

答:
$$I=\int_0^1\int_0^1\frac{1}{x^{1-y}(1-x)^y}dxdy=\int_0^1\int_0^1x^{y-1}(1-x)^{-y}dxdy$$
now using the definition of the Beta function we can say:
$$I=\int_0^1B(y,1-y)dy$$
Now we also have that:
$$B(y,1-y)=\frac{\pi}{\sin(\pi y)}\,\,\,\,y\notin\mathbb{Z}$$
so your integral becomes:
$$I=\pi\int_0^1\csc(\pi y)dy$$
$z=\pi y\Rightarrow dy=\frac{dz}\pi$
$$I=\int_0^\pi\csc(z)dz$$
Which you can see is divergent due to the values of $\csc(z)$ at $z=0,\pi$


答: Because starting at $\sin (\pi)$ as opposed to $\sin(0)$, you’re basically multiplying the imaginary part with $-1$.

Title: Writing $\sqrt[4]{16}$ in polar form and acquiring a solution that doesn't exist

Body:
$$\sqrt[4]{16} = ?$$

$$16 = |16|(\cos(0)+i\sin(\pi))$$
the equality above holds.

$$\sqrt[4]{16} = |\sqrt[4]{16}|(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4})) $$
using De Moivre"s formula we get:
$$ \sqrt[4]{16} = 2(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4}))$$
$$\text{for k = 0, we get one of the correct solutions,  but for k = 1 we get $-i\sqrt{2}$ which is not a solution.}$$

Why does this kind of approach not work?


问:
what is the relation of smooth compact supported funtions and real analytic function?
What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.

答:
I am trying to prove that if some test function $f$ in $\Bbb R^n$ is real analytic then it must be equivalent to zero function. Please correct me if there is any problem.

Proof:

Let us define a test function,

$ f_\epsilon (x)=
\begin{cases}
\frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}},  &\text{for |x|< $\epsilon$ } \\
0, & \text{everywhere else}  \\
\end{cases} $

$\color{red}{\text{Edited in: This was a false start}}$. You should work with a general test function, not  a particular one. That is, simply say: let $f:\mathbb R^n\to \mathbb R$ be a test function. By definition of a test function, $f$ has compact support. The goal is to show that if $f$ is real-analytic, then the support is  empty.

We know, this function has derivatives of all orders and also has a compact support.

Now, Let us suppose that it is also a real Analytic function.

If this function is a real analytic then it must has a Taylor series expansion about any point $x_0$ in its domain which will converge to $f_\epsilon (x)$ for a nbdd. of $x_0$.

Now in particular take Taylor expansion about $\pm\epsilon$

$f_\epsilon (x) \approx  0 + 0x + 0x^2 + . . .$

The Taylor series expansion does in fact converge to the function $f_\epsilon^˜ (x) = 0$.

We can see that $f_\epsilon^˜ (x)$ and $f_\epsilon (x)$ are two different functions for |x|< $\epsilon$.
But they must be equal since $f_\epsilon (x)$ is real analytic, which is only possible when $f_\epsilon (x) \equiv 0$.

Is it enough to prove it. Personally I don't feel that its sufficient...

If I need to elaborate it by writing its Taylor series then calculations of derivatives is again a problem for me. I don't know how to write derivatives of $f_\epsilon (x)$. And in above proof, the way I approximate its Taylor series at $\epsilon$ looks quite tricky... isn't it? Please give me your suggestions and feel free to edit this proof.

Thanks you so much.


问:
Boundary of a connected sum: is it connected?
Let $M^3$ and $N^3$ be compact orientable manifolds with connected boundary. When we perform the connected sum operation, will the resulting manifold $X = M \#N$ have a connected or disconnected boundary?

For instance, if $M$ and $N$ are two solid tori, their connected sum will be a genus $2$ handlebody, and hence have a connected boundary, or will it be another $3$-manifold whose boundary has $2$ connected components?

答:
The trouble is that for manifolds with connected boundary $M_1, M_2$ there are two notions of connected sum:

The ordinary connected sum, denoted $M=M_1\# M_2$ where you take out open balls $B_1, B_2$ from the interiors of $M_1, M_2$ and glue $M_i-B_i$ via a homeomorphism of the new boundary spheres $\partial B_1, \partial B_2$.  Then, clearly, $M$ has two boundary components, homeomorphic to $\partial M_i, i=1, 2$.

The boundary connected sum, denoted $N=M_1\#_{\partial} M_2$, where you pick two closed (tame) disks $D_i\subset \partial M_i$ and  glue $M_1, M_2$ via a homeomorphism $D_1\to D_2$. Then $N$ will have connected boundary, homeomorphic to $\partial M_1 \# \partial M_2$.

(In both cases, there are some issues with the choice of gluing homeomorphisms, I will ignore these. Separately, there are issues with the category in which everything is done: TOP, PL or DIFF, in dimension 3 this does not matter.)

Thus, the answer to your question is: It depends on your notion of connected sum.


问:
Does a function from the empty set to the empty set exist?
I know that every function from the empty set to any other set is the empty function.

I also know that there is no function to the empty set from any other set.

Now, what if both the domain and codomain are empty? Would such a function exist?

答:
I know that every function from the empty set to any other set is the empty function.

I also know that there is no function to the empty set from any other set.

From your first statement follows that the function from the empty set to the empty set is the empty function. From your second statement follows the that there is not function from the empty set to the empty set. So I think at least owne of these statements you "know" is wrong.

A function $f$ from $A$ to $B$ is a subset of $A\times B$ such that

$$\forall x\in A \;\exists y \in B: (x,y)\in f\tag 1$$
and further
$$\forall x\in A \; \forall y \in B\;\forall z \in B: (x,y)\in f \land (x,z) \in f \implies y=z \tag 2$$

The product $A\times B =\{(x,y)\mid x\in A \land y\in B\}$ is empty if $A$ or $B$ is empty. So $\emptyset \times \emptyset=\emptyset$ and the only subset of $\emptyset$ is $\emptyset$. And so the only possible candidate for a function $f:A\mapsto B$ is $\emptyset$. Becuase $(1)$ and $(2)$ is true for $\emptyset$ the set $\emptyset$ is a function from $\emptyset \mapsto \emptyset$ and is is the only one.


问:
Choosing a number from infinite numbers
Imagine infinite set of, let's say, natural numbers. I choose one of the infinite numbers randomly. Let's call that number n. If I choose another number too, can it be the same number (n), theoretically?

答:
As mentioned in the comments, there is no way that each natural can have the same probability. See this answer to "Can you pick a random natural number?..., for instance.

However, we can still pick a number randomly with unequal probabilities. For instance, suppose I choose a number by flipping a coin, and the number is the number of heads before the first tail. (If you want to include a case of flipping heads forever, let's say that also counts as the number $0$.)

I just got the following sequences of flips: $HHT,T,HT,T$. Which means that the number $0$ was randomly selected twice.

Title: Choosing a number from infinite numbers


问:
Characteristic function of sum with random amount of terms.

Let $X_1, X_2, ...$ be iid random variables with characteristic function $\phi$ and let $N$ be Poisson distributed with parameter $\lambda$. Define $Y = \sum_{n = 1}^N X_n$. Express the characteristic function $\psi$ of $Y$ in terms of $\lambda$ and $\phi$.

I am having some trouble with this exercise. We know that the characteristic function of $X_1 + ... + X_n$ is $\phi^n$, but the amount of terms in $Y$ is not fixed, so we cannot use this directly. By definition we have

$$\psi(u) = \sum_{k = 0}^\infty\left( e^{iuk} \mathbb{P}\left(\sum_{n = 1}^\infty X_n 1_{\{n \leq N\}} = k)\right)\right)$$

but I do not know how to compute the probability in there. Any hints?

答:
This is going to be a mere reformulation of your comment, possibly with more rigourous notations. First

$$\mathop{\mathsf{E}}e^{itY}=\mathsf{E}\left[\mathsf{E}\left[e^{it\sum_{n=1}^N X_n}\mid N\right] \right]=\mathsf{E}\left[\prod_{i=1}^N \mathsf{E}\left[e^{it X_i}\mid N\right]\right]$$

where we used the independence of $(X_i)$ and $N$ (this is not mentioned in your statement, but I assume this is the case). For the same reasons

$$\mathop{\mathsf{E}}e^{itY}=\mathsf{E}\left[\prod_{i=1}^N \mathsf{E}\left[e^{it X_i}\right]\right]=\mathsf{E}\left[\phi(t)^N\right]=\sum_{k\ge 0} \frac{\phi(t)^k \lambda^k}{k!}e^{-\lambda}=e^{\lambda(\phi(t)-1)}.$$


问:
Nilpotent groups can be constructed by means of abelian groups.
I am studying A course in theory of groups by Robinson. When defining the central extension of groups, the author says that

Every nilpotent group can be constructed from abelian groups by means of a sequence of central extensions.

What do I understand by this line is that for every nilpotent group $G$ there exists groups $H$ and $K$ such that $H\subseteq Z(K)$ and the exact sequence.
$$\{1\}\to H \to K\to G\to \{1\}.$$
But I can not prove this, please help me to do this.

答:
I think it may take more than one central extension to get $G$, as indicated by the language "...by means of a sequence of..".

Also the nilpotency class measures how many central extensions are necessary.


问:
Find all group of order $20$ which is a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$
Problem: Find all group of order $20$ that are a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$.

My attempt: We knew that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$, so we have $C_4 \cong \mathbb{Z}_4$, $C_5 \cong \mathbb{Z}_5$ and $|C_4| = |\mathbb{Z}_4| = 4$, $|C_5| = |\mathbb{Z}_5| = 5$. Then we find a semidirect product of $\mathbb{Z}_4$ by $\mathbb{Z}_5$. To find this, we define an automorphism $\theta \colon \mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$, while $\mathrm{Aut}(\mathbb{Z}_n) = \{\sigma_k \mid k \in \mathbb{Z}, (k,n) = 1\}$, so $\mathrm{Aut}(\mathbb{Z}_4) = \{\sigma_1, \sigma_3\}$ and $|\mathrm{Aut}(\mathbb{Z}_4)| = 2$.

My professor told that there is only the trivial homomorphism $\mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$ since $(5,2) = 1$.

So my question:

Why from $(5,2) = 1$ can we conclude that there is only one trivial homomorphism, and for the trivial homomorphism, how do I define it?

答:
$$
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\Im}{Im}
\DeclareMathOperator{\Aut}{Aut}
$$

Also, there's another way to prove $f$ is trivial. Consider corollary from homomorphism theorem:
$$f: G → H, \quad\textrm{where $f$ is a homomorphism}$$
$$|G| = |\Im f|\cdot|\Ker f| $$ So in our case $5 = |\Im f| |\Ker f|$. We know that $|\Im f| <= |\Aut(Z_4)| = |Z_2| = 2$. Thus, the only possible way is that $|\Ker f| = 5 $ and $|\Im f| = 1$, which concludes $f$ is trivial homomorphism.


问:
Computing Chern Classes of Tautological Line Bundles
I cannot find any references with how to handle this tautological line bundle $V \rightarrow P(\mathbb{H}^2)$ where $\mathbb{H}$ are the Hamilton quaternions.

I know it is a complex rank 2 vector bundle, so the total chern class is $1 + e(V_\mathbb{R})$ and the top chern class is $c_2(V) = e(V_{\mathbb{R}})$, so $c_0(V) = 1, c_1(V) = 0$ but I want to know what $c_2(V)$ is.

Does anyone have an idea for approaching this?

I think it I should use the splitting principle. From Bott and Tu, we can use the complex projectivization of V as a splitting manifold. And we know this is diffeomorphic to $P(\mathbb{C}^4)$

答:
I'm not really $100\%$ sure my solution is correct, also, there remains an indeterminacy in the end result: I don't quite identify the class you are looking for in $P^1\Bbb H\simeq \Bbb S^4$, only up to sign.

Let $Q$ be the projectivisation of $p:V\rightarrow P^1\Bbb H$. As you note, $Q$ is isomorphic to $P^3\Bbb C$. To be precise, we consider $\Bbb C\subset \Bbb H$ through $1,i\mapsto 1,i$ respectively. Thanks to the standard left $\Bbb H$-vector space structure on $\Bbb H^2$, $\Bbb H^2$ can be seen as a complex vector space  isomorphic $\Bbb C^4$. Now by definition of $Q$ we get a homeomorphism
$$\begin{array}{rcl}Q=\coprod_{S\subset\Bbb H^2}P(S)&\longleftrightarrow &P^3\Bbb C\\ l&\mapsto&l\\l\in P(\Bbb H\cdot l)&\gets&l\end{array}$$
The direct sum is taken over all quaternionic (left) lines $S$, and for every such line $S$, $P(S)$ is the complex projective space on $S$.

The pullback bundle of $V$ over $Q$ splits into two complex line bundles $L\oplus L'$ ver $Q$, with the fiber of $L$ over $l$ being $l$ itself, so that (modulo the above isomorphism), $$L\text{ is isomorphic to the tautological line bundle over }P^3\Bbb C$$

@Matt E gave a convincing argument for the vanishing of the first Chern class of $V$. Computing the total Chern class of the pullback bundle gives $$\pi^*(1+c_2(V))=c(L\oplus L')=c(L)c(L')=1+c_1(L)+c_1(L')+c_1(L)c_1(L')$$ The class $c_1(L)+c_1(L')$ vanishes, and it follows (I believe) that $L'\simeq L^*$, since both are line bundles, and thus completely caracterized by their first Chern class. If $c$ is the standard degree $2$ generator of $H^3(Q)=H^2(P^3(\Bbb C))$ (i.e. the first Chern class of the tautological line bundle over $P^3(\Bbb C)$), then $$\pi^*(c_2(V))=-c^2.$$

It remains to understand $\pi$. The canonical map $Q\to P^1\Bbb H$ is a fibration. Actually, it is obtained from the Hopf fibration $\Bbb S^3\hookrightarrow\Bbb S^7\hookrightarrow\Bbb S^4$ by quotienting out the action of $\Bbb S^1$).
$$\begin{array}{rc}\Bbb S^2\simeq P^1\Bbb C\hookrightarrow & P^3(\Bbb C)\\&\downarrow\\ &\Bbb S^4\end{array}$$
The associated spectral sequence collapses at rank $2$ because of how the nonzero nodes are placed, and this tells us that the cohomology of $\Bbb S^4$ in degree $4$ is isomorphic to that of $P^3\Bbb C$ in degree $4$ through $\pi^*$. Since $\pi^*(c_2(V))=-c^2$ is a generator in degree $4$, we necessarily have $c_2(V)=$ one of the two generators of $H^4(\Bbb S^4)$ . I don't know which one this is.


问:
Should I use induction in this proof on leading coefficient ideals?
Here is the question I want to answer letter $(c)$ of it:

Let $R$ be a commutative ring, and let $R[x] = R^{[1]}.$ Given an ideal $I \subset R[x]$ and $k \in \mathbb N,$ let $L_k(I) \subset R$ consists of $0$ and all leading coefficients of degree$-k$ polynomials of $I.$ Given ideals $I \subset J$ of $R[x],$ show the following.

$(a)$ $L_k(I)$ is an ideal of $R$ for each $k \in \mathbb N.$

$(b)$ $L_k(I) \subset L_k(J)$ for each $k\in \mathbb N.$

$(c)$ $L_k(I) \subset L_{k + 1}(I)$ for each $k\in \mathbb N.$

Here is my trial for letter $(c)$:

Let $x_{0} \in L_k(I)$ and we want to show that $x_{0} \in L_{k + 1}(I)$ for each $k\in \mathbb N.$

Now, since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of some degree$-k$ polynomial of $I,$ name it $p(x).$ Now, since $I$ is an ideal of $R[x],$ then $xp(x) \in I$ as $x \in R[x].$ If $p(x) = x_{0}x^k + a_{k-1}x^{k-1} + \dots + a_1x +a_0,$ then $xp(x) = x_0 x^{k+1} + a_{k-1}x^{k} + \dots + a_1x^2 +a_0x + 0,$ i.e. $xp(x)$ is a polynomial of degree $k+1$ with leading coefficient $x_0,$ which by definition of $L_{k + 1}(I)$ means that  $x_{0} \in L_{k + 1}(I).$ Since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$

Is my trial correct? or should I use induction in the proof? could anyone clarify this to me please?

EDIT:Also, here are my trials for the solutions of $(a)$ and $(b)$ and I have the same question: Should I use induction in any of the following solutions:

$(a)$

Given an ideal $I \subset R[x]$ and $k \in \mathbb N,$ Define $L_k(I) \subset R$ as consisting of $0$ and all leading coefficients of degree$-k$ polynomials of $I.$ We want to prove that $L_k(I)$ is an ideal of $R$ for each $k \in \mathbb N.$

\textbf{(1) Showing that $L_k(I)$ is a subgroup of $(R, +)$ for each $k \in \mathbb N.$}

$0_R$ is in $L_k(I)$ by its definition.

Now, let $x_{0}, y_{0} \in L_k(I),$ we want to show that $x_{0} - y_{0} \in L_k(I).$

Since $x_{0}, y_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $p(x)$ and also, $ y_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $q(x).$ Now, since $I$ is an ideal, then the polynomial $p(x) - q(x) \in I$ which is a degree$-k$ polynomial of $I$ with leading coefficient $x_{0} - y_{0}.$ Now, by the definition of  $L_k(I),$ we have that $x_{0} - y_{0} \in L_k(I)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$

So the previous 3 paragraphs implies that $L_k(I)$ is an abelian subgroup of $(R, +)$ as $R$ is commutative.

\textbf{(2) Showing that for $x_{0} \in L_k(I),$ and $r \in R,$ we have that $rx_{0} \in L_k(I),$ for each $k \in \mathbb N.$}

Let $x_{0} \in L_k(I),$ and $r \in R.$ Since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of a degree$-k$ polynomial of $I,$ name it $p(x).$ Now, since $I$ is an ideal, then the polynomial $rp(x) \in I$ which is a degree$-k$ polynomial of $I$ with leading coefficient $rx_{0}.$ Now, by the definition of  $L_k(I),$ we have that $rx_{0} \in L_k(I)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$

So, from $(1)$ and $(2),$ we get that $L_k(I)$ is an ideal of $R.$

$(b)$

Assume that $I,J$ are ideals of $R[x]$ such that $I \subset J.$

First, $0_{R}$ is in both $L_k(I)$ and $L_k(J)$ for each $k\in \mathbb N$ by their definitions.

Second, assume that $x_{0} \in L_k(I)$ and we want to show that $x_{0} \in L_k(J)$ for each $k\in \mathbb N.$

Now, since $x_{0} \in L_k(I),$ then $ x_{0}$ is a leading coefficient of some degree$-k$ polynomial of $I,$ name it $p(x).$ But, by the given, we have that $I \subset J,$ then $p(x)$ is also a degree$-k$ polynomial in $J$ that has a leading coefficient $ x_{0}.$ So, by definition of $L_k(J)$ we have that $x_{0} \in L_k(J)$ and since $k$ was arbitrary then the required is proved for every $k \in \mathbb N.$

答:
I can't imagine that induction is necessary here. We must simply fix a non-negative integer $k,$ and show that each of the claims holds for that $k.$ Considering that $k$ was chosen arbitrarily, each of the claims will hold for each non-negative integer $k,$ and our proof is complete.

I notice that there is a more succinct way of completing these proofs. Particularly, for any commutative ring $R,$ a nonempty subset $I$ of $R$ is an ideal of $R$ if and only if for any elements $i, j \in I,$ we have that (1.) $i - j \in I$ and (2.) $ri \in I$ for any element $r \in R.$ By the one-step subgroup test, the first shows that $(I, +)$ is a subgroup of $(R, +);$ the second shows that $I$ is closed under multiplication by elements of $R.$ Combined, these say that $I$ is an ideal of $R.$

Bearing this in mind, the proof of point (a.) should go as follows.

Proof. Given any non-negative integer $k,$ observe that $L_k(I)$ is nonempty by hypothesis that $0_R$ is in $L_k(I).$ Given any two elements $r, s \in L_k(I),$ therefore, there exist polynomial $p_r(x)$ and $p_s(x)$ of degree $k$ in $I$ whose leading coefficients are $r$ and $s,$ respectively. Consequently, the leading coefficient of the degree $k$ polynomial $p_r(x) - p_s(x)$ is $r - s,$ hence we have that $r - s \in L_k(I).$ Further, for any element $t$ of $R,$ we have that $tr = 0_R$ or $tr \neq 0_R.$ For the case of the former, it follows that $tr = 0_R$ is an element of $L_k(I),$ and for the latter, we have that $tp_r(x)$ is a polynomial of degree $k$ with leading coefficient $tr.$ Either way, it follows that $L_k(I)$ is closed under multiplication by elements of $R.$ We conclude that $L_k(I)$ is an ideal of $R.$ Considering that $k$ is arbitrary, $L_k(I)$ is an ideal of $R$ for all non-negative integers $k.$ QED.


问:
How to show that $\max(a,b)^2+\max(a,b)+a-b$ is bijective?
How can I show that $f:\mathbb{N}^2 \to \mathbb{N}$ given by $$f(a,b)=\max(a,b)^2+\max(a,b)+a-b$$ is bijective?
How could I show it to be injective and surjective? Or does it suffice to give a explicit inverse function? If so, what is the process of inversing this function?

答:
Hint: We have
$
f(a,b) = (a+1)^2-(b+1)
$ when $a\ge b$
and
$
f(a,b) = b^2+a
$ when $a\le b$.

Surjectiveness comes from considering whether $\lfloor \sqrt n \rfloor = \lfloor \sqrt n +0.5 \rfloor$. See OEIS/A063656 and OEIS/A063657. This also provides a recipe for the inverse.


问:
Show that the following series is always convergent.
I need to show that the following series is always convergent:
$$\left|\left|v\right|\right|=\sum_{k=0}^{\infty}\frac{\left|\mathcal{A}^kv\right|}{R^k},$$

where $\mathcal{A}$ is a linear operator and $R>\rho\left(\mathcal{A}\right)=\underset{i}{max}|\lambda_i|$. The R.H.S. of the inequality is the spectral radius of the linear operator $\mathcal{A}$.

I know that i need to show that there exists some $C$ such that $\left|\left|\mathcal{A}\right|\right|_2\leq Cr^n$. Thereafter i propably need to show that the sum is bounded by some series involving the above mentioned $C$.

But i cant seem to do so. How do i go about this?

答:
Take $r$ so $\rho(A) < r < R$.
By Gelfand's spectral radius formula,
$\|A^k\|^{1/k} < r$ for sufficiently large $k$, where $\|\cdot\|$ is the operator norm induced by the norm $|\cdot |$, and then $|A^k v|/R^k \le \|A^k\| |v|/R^k <
(r/R)^k |v|$ so the series converges.


问:
Evaluating $\sum_{k=0}^{2020} \cos( \frac{2πk}{2021}$)
I am trying to come up with a solution to  $$\sum_{k=0}^{2020}\cos\left(\frac{2\pi k}{2021}\right) $$ so far I have proceeded as to acknowledge the cosine is just the real part of the Euler's formula form: $$ \sum_{k=0}^{2020}\cos\left(\frac{2\pi k}{2021}\right)=\textrm{Re}\left(\sum_{k=0}^{2020}e^{2\pi ik/2021}\right)
 $$ and so, rephrasing since it is basically a geometrical series (for all k>1): $$\sum_{k=0}^{2020}e^{2\pi ik/2021}=\frac{1-e^{2021(2\pi i)/2021}}{1-e^{2\pi ik/2021}}
 $$ now, simplifying $$ {e^{2021(2\pi i)/2021}} = {e^{2\pi i}}  $$ and using, that $$e^{2\pi i}=\cos\left(2\pi\right)+i\sin\left(2\pi\right)=1 $$ the result will be $$\sum_{k=0}^{2020}e^{2\pi ik/2021}=\frac{1-1}{1-e^{2\pi ik/2021}} = \frac{0}{1-e^{2\pi ik/2021}}= 0  $$

Did I happen to make a mistake anywhere, or should this be correct? I have seen other people solve this problem with the formula for sine and cosine sums in arithmetic progression, however, I am not allowed to use that formula since it was never introduced in our course.

Any help would be very appreciated :)

答:
Seems correct to me. Here is another way to see it :

$ \left\lbrace z_{k}=\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}\mid k\in\left[\!\left[0,n-1\right]\!\right]\right\rbrace $ is the set of all roots of the polynomial $ P_{n}=X^{n}-1=\sum\limits_{k=0}^{n}{a_{k}X^{k}} $, thus their sum will be exactly $ \sum\limits_{k=0}^{n-1}{z_{k}}=-\frac{a_{n-1}}{a_{n}}=0 \cdot $


问:
let Y, YZ $\in$ $\mathcal{L}^1$ and let Z be $\mathcal{G}$ measurable then show E(YZ|$\mathcal{G}$) = ZE(Y|$\mathcal{G}$)
let Y, YZ $\in$ $\mathcal{L}^1$ and let Z be $\mathcal{G}$ measurable then show E(YZ|$\mathcal{G}$) = ZE(Y|$\mathcal{G}$)

I know we want to start with simple functions then move to non negative and integrable functions
and for each show $\int_{\Lambda} ZE(Y |\mathcal{G}$)dP = $\int_{\Lambda} YdP$

For the simple case we have Z = $1_A A \in \mathcal{G}$ then

$\int_{\Lambda} ZE(Y |\mathcal{G}$)dP = $\int_{\Lambda \cap A} E(Y |\mathcal{G}$)dP = $\int_{\Lambda \cap A} Y $dP

since $\Lambda \cap A \in \mathcal{G}$

=$\int_{\Lambda} Y1_A $dP = $\int_{\Lambda} YZ $dP

Is this the right path? I am struggling a little bit with the Non negative and integrable functions.

答:
The conditional expectation $E(YZ | \mathcal{G})$ is characterized by the two properties

$E(YZ|\mathcal{G})$ is $\mathcal{G}$-measurable, and

$\int E(YZ|\mathcal{G}) \cdot \varphi \,d\mathbb{P} = \int YZ\varphi \,d\mathbb{P}$ whenever $\varphi$ is $\mathcal{G}$-measurable.

It's clear that $Z \cdot E(Y|\mathcal{G})$ is $\mathcal{G}$-measurable because both $Z$ and $E(Y|\mathcal{G})$ are.  Therefore it's enough to show that $Z E(Y|\mathcal{G})$ also satisfies property $2$ above.

Let $\varphi$ be any $\mathcal{G}$-measurable function.  Then $Z\varphi$ is also $\mathcal{G}$-measurable, so $\int E(Y|\mathcal{G}) Z\varphi \,d\mathbb{P} = \int YZ\varphi \,d\mathbb{P}$ by the definition of $E(Y|\mathcal{G})$.  Therefore the random variable $Z E(Y|\mathcal{G})$ satisfies both properties 1 and 2, so it must be a version of $E(YZ|\mathcal{G})$.


问:
Probability that we choose a two headed coin
We have a $501$ coins on the table, and assume that they have all been flipped onto that table (i.e., there is a mix of heads and tails). This also includes a two-headed coing.

Now if we pick up $1$ coin and its heads, what is the probability it is also the two-headed coin?

I've seen questions similar to this, but not the exact same. I did this out but want to hear your thoughts, before I explain how I arrived at the probability.

答:
We will use the powerful Bayes theorem.
\begin{align}
\mathbb{P}(\text{Two sided coin }\vert \text{ we have a head}) & = \dfrac{\mathbb{P}(\text{we have a head }\vert \text{ Two sided coin}) \mathbb{P}(\text{Two sided coin})}{\mathbb{P}(\text{we have a head})}\\
&= \dfrac{1 \times \dfrac1{501}}{\dfrac12 \cdot \dfrac{500}{501} + \dfrac1{501}} = \dfrac1{251}
\end{align}</p>


问:
Singular solution of the differential equation $(y')^2-3xy'+y^2=0$.
To  find singular solution of the differential equation $$(y')^2-3xy'+y^2=0$$ I find its $p $-discriminant  as $$y=\pm \frac {3}{2}x $$ but none of the factor satisfies differential equation.  How to find its  singular solution? Thank you .

答:
Your equation has one clear singular solution which is $y(x) = 0$. It may also have other singular solutions, but if it does these are not so easy to find. You are correct to use the $p$-discriminant method, but in general solving via this method doesn't give singular solutions, only the locus of points in the curve $f(x,y,p) = 0$ which have constant $p$ (in your case, $f(x,y,p) := p^2-3xp+y^2$). So you shouldn't expect the $p-$discriminant that you found to solve the equation.

Based on my understanding from here, if you want to construct all singular solutions it's necessary to find both the $p$- and $c$-discriminants of your equation. This would require finding the general solution of your equation $\phi(x,y,c)$ in terms of integration constant $c$ to be able to find the $c$-discriminant. I spent a while trying to solve your equation and I couldn't find a general solution, and I also put it into Mathematica which doesn't know how to solve it either. I was surprised Mathematica didn't know, I guess the equation itself looks deceptively simple.

Here is my calculation of the $p-$discriminant, based on the analysis here. First $$\frac{\partial f}{\partial p} = 2p - 3x=0.$$ Then $$2 f - p \frac{\partial f}{\partial p} = 2 y^2 - 3 x p  = 0\qquad\implies\qquad p = \frac{2 y^2}{3x}.$$Now substitute back into $f(x,y,p) = 0$ to find that $$\psi(x,y):=y^2 \left(\frac{4 y^2}{9 x^2}-1\right) = 0.$$ I believe that this $\psi(x,y)$ is the $p$-discriminant. I think importantly, this method avoids dividing by $y$ in coming to the solution, and we see that $y=0$ turns out to be a singular solution here, so we don't want to divide by $y$ while solving the system to arrive at the form for $\psi(x,y)$.

Then setting $\psi(x,y)$ to zero gives two solutions to the $p$-discriminant equation; the first one is $y(x) = 0$, and the second is $y(x)^2=\frac{9}{4}x^2$. This gives $|y(x)| = \pm \frac{3}{2} |x|$, which is almost what you have. But as I said, I think it's fine that this doesn't solve $f(x,y,y') = 0$.

Out of interest, where does this equation come from? Is it from some research problem, or from a course on differential equations?


问:
Maps in the long exact sequence of relative homology groups
In Hatcher page 115: Elements of $H_n(X,A)$ are represented by $n$-chains $\alpha \in C_n(X)$ such that $\partial \alpha \in C_{n-1}(A) \subset C_{n-1}(X)$. We think of an element of $H_n(X,A)$ as being an $n$-thing in $X$ whose boundary (an $(n-1)$-thing) lies in $A$.

Since the relative homology groups fit into the long exact sequence
$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A)  \to H_{n-1}(X) \to H_{n-1}(X,A) \to \cdots,$$

My questons:

Are these maps surjectiveness or injectiveness, or
non-surjectiveness or non-injectiveness? We can clarify one by one?

How to be precise about the maps given above? I meant there are precise relations of
$$ker f_{j+1} = im  f_{j}
$$
for every map? even the dimensionality jumps?

答:
Yes, at each object in a long exact sequence the kernel of the map on the right is equal to the image of the map on the left. This is the definition of exactness.

None of the involved maps have to be injective or surjective a priori. However, the definition of exactness immediately shows that a map $f\colon Z\to W$ in an exact sequence …

… is injective if and only if the map to its left is zero: $$\cdots\xrightarrow{0} Z\xrightarrow{f} W\to \cdots$$

… is surjective if and only if the map to its right is zero: $$\cdots \to Z \xrightarrow{f} W\xrightarrow{0} \cdots$$

This is because injectivity is equivalent to $\operatorname{ker}(f)=0$ which then means that the map on the left has to have $0$ as its image, so it's the zero map. Surjectivity is equivalent to $\operatorname{im}(f)=W$ which means that the map on the right has to have $W$ as its kernel, so it's the zero map.

Note that for a map $P\xrightarrow{g} Q$ in the sequence it is sufficient that $P=0$ or $Q=0$ to have $g=0$. So when an object is $0$, the map two to the right is injective and the map two to the left is surjective:
$$
\cdots\xrightarrow{\text{surj.}} * \xrightarrow{0} 0 \xrightarrow{0} * \xrightarrow{\text{inj.}} \cdots
$$


问:
Solving a recurrence with circles
Consider n circles in the plane such that each intersects every other, and no three circles met at a point. I need to find a recurrence for the number of regions formed then solve the recurrence.

I started by diving the plane into rn regions. I then added a new circle and tried figuring out how many new regions that added to get an idea but struggle with the concept.

答:
Hint: Suppose you have an arrangement of $n$ circles, and you add a new circle, $C$. For each intersection point of $C$ with a previous circle, one of the previous regions will be split in two (if this is unclear, look at some small examples). Therefore, if $r_n$ is the number of regions, you have
$$
r_{n+1}=r_n+\text{# of intersections of $C$ with previous circles}.
$$
All that is left is to count the number of intersections.

Title: Solving a recurrence with circles


答:
It is helpful to consider the following two cases. When $a\leqslant b$, we have $f(a,b)=b^2+a$, and when $b\leqslant a$ we have $f(a,b)=a^2+2a-b$.

So, here is a way of constructing the inverse function $g:\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}$. Let $x\in\mathbb{N}$, and let $y:=\lfloor{\sqrt{x}}\rfloor\in\mathbb{N}$ be maximal such that $y^2\leqslant x$; we want to find $(a,b)\in\mathbb{N}\times\mathbb{N}$ such that $f(a,b)=x$. There are two cases to consider; either $x-y^2\leqslant y$, or $x-y^2>y$. In the first case, let $\fbox{$b=y$}$ and $\fbox{$a=x-y^2$}$. Then $a\leqslant b$ and so $$f(a,b)=b^2+a=y^2+(x-y^2)=x,$$ as desired.

In the second case, let $\fbox{$a=y$}$. We want to find a natural number $b\leqslant a$ such that $f(a,b)=x$. By the formula above, this amounts to finding $b\leqslant a$ such that $y^2+2y-b=x$, so we can just let $\fbox{$b=y^2+2y-x$}$. We need to show that $b\geqslant 0$ and that $b\leqslant a$.

For the first claim, suppose for contradiction that $b<0$. Then $y^2+2y-x<0$, so $y^2+2y<x$, so $(y+1)^2=y^2+2y+1\leqslant x$, contradicting maximality of $y$. For the second claim, suppose that $b>a$, ie that $y^2+2y-x> y$. Then we would have $y^2-x>-y$, and so $x-y^2<y$, a contradiction.

Thus, in sum, we define a function $g:\mathbb{N}\rightarrow\mathbb{N}^2$ by $$x\mapsto \begin{cases} (x-y^2,y) & x-y^2\leqslant y \\ (y,y^2+2y-x) & x-y^2>y\end{cases},$$ where $y=\lfloor{\sqrt{x}}\rfloor$ is maximal such that $y^2\leqslant x$.

The paragraphs above show that $f\circ g=\operatorname{id}_\mathbb{N}$, so we just need to show that $g\circ f=\operatorname{id}_{\mathbb{N}\times\mathbb{N}}$. To see this, let $(a,b)\in\mathbb{N}\times\mathbb{N}$. We divide into two cases. First suppose $a\leqslant b$, and let $x:=f(a,b)=b^2+a$. To find $g(x)$, we first need to find $y$ maximal such that $y^2\leqslant x$; we claim that $y=b$. Clearly $b^2\leqslant x$, so to show this claim it suffices to show that $(b+1)^2>x$. We have $(b+1)^2-x=2b+1-a\geqslant b+1>0$, as desired, since $a\leqslant b$. Thus indeed $y=b$. Now, $x-y^2=a\leqslant b=y$, so we are in the first case of the definition of $g$, and thus $g(f(a,b))=g(x)=(x-y^2,y)=(a,b)$, as desired.

Hopefully you should now be able to complete the proof that $g(f(a,b))=(a,b)$ when $a>b$, which is very similar. Once this is done we have proven that $f$ and $g$ are mutual inverses, so that $f$ is indeed a bijection.

Title: How to show that $\max(a,b)^2+\max(a,b)+a-b$ is bijective?

Body:
How can I show that $f:\mathbb{N}^2 \to \mathbb{N}$ given by $$f(a,b)=\max(a,b)^2+\max(a,b)+a-b$$ is bijective?
How could I show it to be injective and surjective? Or does it suffice to give a explicit inverse function? If so, what is the process of inversing this function?


问:
Additional question on Entire function bounded by a polynomial is a polynomial
I was reading this question. Here we are only considering $\frac{f(z)}{z^n}$ to be bounded for large $|z|$, i.e, $\exists R>0 $ such that  $\frac{f(z)}{z^n}$ is bounded for $|z|\geq R$.
I was able to understand how we solved this question but I have an additional question.

What will happen if $\frac{f(z)}{z^n}$ is bounded on whole of
$\mathbb{C}$?

答: Then $f(z)=kz^n$ for some constant $k$. That's because $0$ is a removable singularity of $\frac{f(z)}{z^n}$ and therefore, near $0$, $\frac{f(z)}{z^n}$ can be written as $a_0+a_1z+a_2z^2+\cdots$. But then, by Liouville's theorem, $a_0+a_1z+a_2z^2+\cdots=k$, for some constant $k$, and therefore $f(z)=kz^n$.


问:
Find a sequence of numbers such that random variable converges
The problem

There are independant random variables $\xi_1, \xi_2, ...$ with the same distribution $U_{0,4}$

Let us $S_n = \xi_1 + ... + \xi_n$

Find a sequence of $c_n$ such that $P(S_n < c_n) \rightarrow 0.4$

My Attempt

$\begin{array}{l}
P\left(S_{n}<c_{k}\right)=1-P\left(S_{n} \geq c_{k}\right) \\
P\left(S_{n} \geq c_{n}\right)=P\left(\frac{S_n}{n} \geq \frac{c_{k}}{n}\right) \leq {\text{Markov inequality}} \leq \frac{E S_{n}}{c_{k}}=\frac{n \mathbb{E}_{1}}{c_{k}} \rightarrow 0.6
\end{array}$

Find a $c_n$:

$\lim _{n \rightarrow \infty} \frac{4n}{2 c_n}=0.6 \Rightarrow c_{n}=\frac{4n}{1.2}$

$P\left(s_{n}<c_{k}\right)=1-P\left(S_{n} \geq C_{k}\right) \rightarrow 1 - 0.6 = 0.4$

What do I need

I am not sure about correctness of my solution. I would like to see my mistakes and figure out how to solve this problem correctly.

答:
Using CLT, we know that $$P\left(\frac{S_n-2n}{\sqrt{n}}< x\right)\to P(Z< x),$$
as $n\to \infty,$ where $Z$ is a standard normal random variable. It is a matter of re-writing the above expression to obtain
$$P(S_n<2n+x\sqrt{n})\to P(Z<x).$$

This solves your problem. Choose $x_{0.4}$ such that $P(Z<x_{0.4})=0.4$ and then $c_n=2n+x_{0.4}\sqrt{n}$ works. More generally, for any $r\in (0, 1)$ you can find $x_r$ such that $P(Z<x_r)=r$ and it would give you that
$P(S_n<c_n^r)\to r,$ where $c_n^r:=2n+x_r\sqrt{n}.$


答: Perhaps this figure will help:

Title: Solving a recurrence with circles

Body:
Consider n circles in the plane such that each intersects every other, and no three circles met at a point. I need to find a recurrence for the number of regions formed then solve the recurrence.

I started by diving the plane into rn regions. I then added a new circle and tried figuring out how many new regions that added to get an idea but struggle with the concept.

Body: Perhaps this figure will help:


问:
$P(N(2)=1, N(3)-N(1)=1)$
Suppose we have a Poisson process $\{N(t),t\geq 0\}$ with intensity $\lambda=1/2$.

I want to calculate $P(N(2)=1\cap N(3)-N(1)=1)$. I know that $N(3)-N(1)$ and $N(2)$ are $Po(1)$ but they are not necessarily independent. So how can I calculate what event $N(2)=1\cap N(3)-N(1)=1$ represents?

答: If $N(t)$ denotes the number of points in interval $(0,t]$, then you can rewrite your original event in terms of independent events. Break up the interval $(0,3]$ into the three disjoint intervals $(0,1]$, $(1,2]$, and $(2,3]$ and consider what possibilities are equivalent to your original event. For example, one possibility is that there is one point in $(0,1]$, zero in $(1,2]$, and one point in $(2,3]$.

Title: $P(N(2)=1, N(3)-N(1)=1)$


问:
Does an isometric transformation preserve the Lebesgue outer measure?
Let $T:\mathbb{R}\rightarrow\mathbb{R}$ be an isometric function i.e.
$$|x-y|=|T(x)-T(y)|$$
for all $x,y\in \mathbb{R}$

Consider now Lebesgue outer measure:
$$\mu^*(A)  := \inf\{\sum _{i\in I}|P_i|:A\subseteq \bigcup_{i\in I} P_i\}$$
Wherer $I$ is countable set and $P_i$ is an open interval.

Does equality $\mu^{*}(A) = \mu^{*}(T(A))$ holds?

I am only able to see that for open intervals we have $|P|=|T(P)|$. I have an intuition that this is somehow sufficient to solve the problem.

Thank you in advance for help.

答:
There is a theorem that any isometry of $\mathbb R^n$ is affine. Hence, you actually have $T(x)=x+c$ or $T(x)=-x+c$, and your proof becomes much simpler:

Now you can use that actually $T(P)$ is an open interval whenever $P$ is, and conversely if $Q$ is an open interval then so is $P=T^{-1}(Q)$ with $T(P)=Q$. Depending on your definition of outer measure the claim is now trivial.


问:
Proving limit of quadratic using formal definition without explicitly given f(x)
How would you prove the following statement?:

Assume that $\lim_{x\to1}{f(x)=6}$.
Prove using the epsilon-delta definition that $\lim_{x\to1}[5f(x)^2+ 3f(x)] = 198$.

I set up the following:

We have $\forall \epsilon_1$ in the reals, $\exists \delta_1$ such that $0<|x-1|<_1 ⇒ |f(x)-6|<\epsilon_1$

We want a $\delta$ such that $|5f(x)^2 + 3f(x) - 198|<\epsilon$

This becomes $|5f(x)+33| |f(x)−6| <\epsilon$, then, we have $|f(x)−6| <\frac{\epsilon}{|5f(x)+33|}$

Now, I am not sure how to continue and how to implement the $\delta = \operatorname{min}\{..., ...\} $thing

答:
HINT:

If $|f(x)-6|<\varepsilon_1$ then $|f(x)|\le |f(x)-6|+6\le 6+\varepsilon_1$.

And, in particular, if you assume that $\varepsilon_1<1$, then $$|5f(x)+33|\le 5|f(x)|+33\le 5(6+\varepsilon_1)+33 < 68.$$


问:
Help with Taylor polynomial
I need to find the Taylor polynomial of order $2n$ of the function
$$f(x,y)=\frac{1}{1+xy}$$ on  $(x_0,y_0)=(0,0)$.
Can anyone give me a hand please?

Sorry, i make a mistake, their should say "polynomial" instead of "series, i have corrected now.

答:
A related technique. Here is the general formula for the Taylor series of a function in two variables around the point $(x,y)=(0,0)$

$$ f(x,y)=\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)|_{x=0,y=0} . $$

So, to get the order $2n$, you can consider the truncated series

$$ f(x,y)=\sum_{k=0}^{n}\sum_{m=0}^{n}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)|_{x=0,y=0}. $$

You may want to show that

$$ \frac{\partial^{m+k}}{\partial x^m \partial y^k }\frac{1}{1-xy}\Big|_{x=0,y=0}=0\,\quad \mathrm{ when }\quad  m\neq n. $$

Otherwise, use Jonathan's technique to get the required series.

Title: Help with Taylor polynomial


问:
On an identity that appears in Apéry's proof of the irrationality of $\zeta(3)$
In proving the irrationality of $\displaystyle \zeta(3)=\sum_{n\geq1}\frac{1}{n^3}$, Apéry introduced the accelerated series

$$
\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}.
$$

A crucial ingredient of his proof of the above identity is the well-known equality
$$
\sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)}=\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}}.
$$
(note that an excellent overview of his proof can be found in Van der Poorten's "A Proof that Euler Missed...".)

I was trying to verify the above equality numerically, in Wolfram Mathematica. However, when asking to evaluate the LHS, Mathematica surprisingly gave:

```<code>1/n^2
</code>```

This is obviously false, as quick numerical verification reveals.

This was the simple Mathematica instruction I ran:

```<code>Clear[n]
N[Sum[((-1)^(k - 1) ((k - 1)!)^2)/(Product[(n^2 - j^2), {j, 1, k}]), {k, 1, n - 1}]]
</code>```

Note that Wolfram Alpha gives the same incorrect evaluation for the same expression, as can be seen here.

So what am I doing wrong? Clearly, the instruction itself is correct (click the Alpha link to see this). So is the Wolfram engine making an elementary mistake, or am I doing so?

答:
You already seem to have indication from Mathematica SE site that this might be a bug, here is I think an interesting observation (too long for a comment).

I reproduced the same issue in another CAS system - Maple 2020. Specifically

simplify(sum((-1)^(k-1)*(k-1)!^2/product(n^2-i^2,i=1..k),k=1..n-1))

gives the result $\frac{1}{n^2}$, which is of course incorrect for positive integer $n$. However it seems the tool somehow makes a sense of the input for non-integral $n$ and keeps going with that...

If we look at what is the intermediate result (before passing it to simplify), it is an expression in terms of gamma function, specifically:

$$
\frac{(1-n)(1+n)\Gamma(1-n)\Gamma(1+n)}{n^2\Gamma(2-n)\Gamma(2+n)}\tag{*}.
$$

However for integer $n \geq 1$ it is not hard to see that this expression is not even defined (specifically $\Gamma(1-n)$ and $\Gamma(2-n)$ are not defined because gamma function is not defined for negative integers). So somehow Maple simplified the sum to a result that is more generic, but is not defined for integral values, which paradoxically are the only ones for which the original sum was defined....

Then of course the final simplifcation is done in general without any assumption on $n$, and so for all $n$ for which the $(*)$ is defined, it simplifies to $\frac{1}{n^2}$.

One way around this is to add additional assumptions, in this case we want $n$ to be integer, and indeed if we add assume(n,integer) before the command, we get...

Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

Well it is not what we hoped for, but at least it is giving us an error instead of incorrect result (clearly it is now aware of the limitations of methods in the given domain).

I can only assume it is a similar case in Mathematica...


答:
You can take cue from Venn diagram. Take this for $3$ circles as an example.

Every circle is getting divided into $3$ regions but only one region of each circle is not shared. So that gives you $3$ regions that are not shared.

Then the next $3$ are shared between each of the two circles.

The last one is shared among all of them and that is only one.

So while there are $3$ regions for each circle, there are only $7$ different regions in total.

Similarly for $4$ circles, you get $13$ different regions.

$a(1) = 1$

$a(2) = a(1) + 2 \times 1 = 3$

$a(3) = a(2) + 2 \times 2 = 7$

$a(4) = a(3) + 2 \times 3 = 13$

$a(n+1) = a(n) + 2n$

It translates to $n(n-1) + 1$ regions for $n$ intersecting circles.

Title: Solving a recurrence with circles


问:
Minimize $\{tr^Trt^T\}_{ii}$ subject to $r^Tt=1$
I want to solve the following optimization problem over $r,t\in\mathbb{R}^{n\times 1}$:

\begin{array}{ll} \text{minimize} & \{tr^Trt^T\}_{ii}\\\quad r,t\\ \text{subject to} & r^Tt=1\end{array}

My attempt: since $r^Tr$ is scalar and $\{tt^T\}_{ii}=t^2_{i}$ , we can rewrite above problem as:

\begin{array}{ll} \text{minimize} & r^Trt^2_{i}\\\quad r,t\\ \text{subject to} & r^Tt=1\end{array}

Can I replace $r^T$ by its pseudo-inverse $r^T=(t^Tt)^{-1}t^T$? Then

\begin{array}{ll} \text{minimize} & \frac{t^2_{i}}{t^2_{1}+t^2_{2}+\cdots+t^2_{n}}=0?\\\quad t\end{array}

What I want is to minimize each diagonal element of $\{tr^Trt^T\}_{ii}$ as much as possible., subject to $r^Tt=1$, we can assume that diagonal elements of $t$ are ordered in nondecreasing order.

答: In vector terms, minimize $r^2t^2$ subject to $r\cdot t=1$; by Cauchy-Schwarz, this happens with value $1$ when $r=t$ is a unit vector.


答: Note that\begin{align}5f^2(x)+3f(x)-198&=5f^2(x)-180+3f(x)-18\\&=5\bigl(f^2(x)-36\bigr)+3\bigl(f(x)-6\bigr)\\&=5\bigl(f(x)+6\bigr)\bigl(f(x)-6\bigr)+3\bigl(f(x)-6\bigr)\\&=\bigl(5f(x)+33\bigr)\bigl(f(x)-6\bigr).\end{align}So, take $\delta_0>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<1.$$Then, if $|x-1|<\delta_0$, you have $5<f(x)<7$. But\begin{align}5<f(x)<7&\iff25<5f(x)<35\\&\iff58<5f(x)+33<68\\&\implies\bigl|5f(x)+33\bigr|<68.\end{align}Now, given $\varepsilon>0$, take $\delta_1>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<\frac\varepsilon{68}$$and then$$|x-1|<\min\{\delta_0,\delta_1\}\implies\bigl|5f^2(x)+3f(x)-108\bigr|<\varepsilon.$$

Title: Proving limit of quadratic using formal definition without explicitly given f(x)

Body:
How would you prove the following statement?:

Assume that $\lim_{x\to1}{f(x)=6}$.
Prove using the epsilon-delta definition that $\lim_{x\to1}[5f(x)^2+ 3f(x)] = 198$.

I set up the following:

We have $\forall \epsilon_1$ in the reals, $\exists \delta_1$ such that $0<|x-1|<_1 ⇒ |f(x)-6|<\epsilon_1$

We want a $\delta$ such that $|5f(x)^2 + 3f(x) - 198|<\epsilon$

This becomes $|5f(x)+33| |f(x)−6| <\epsilon$, then, we have $|f(x)−6| <\frac{\epsilon}{|5f(x)+33|}$

Now, I am not sure how to continue and how to implement the $\delta = \operatorname{min}\{..., ...\} $thing


问:
How to generate an oscillating integer sequence without trig functions?
I would like a function of t, where t is an integer increasing over time, that an average person could compute in their head that results in a sequential, oscillating sequence of integers. For example:

0, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3...

I have a function that almost computes what I want using sine (it doubles up on the peaks and valleys), but given a value for t, most humans cannot compute sine in their head.

x = floor(4*sin(t/4 + 5.5) + 1

I've been playing around with mod operation. Mod nicely builds a cycle, obviously, but I haven't been able to figure out a modification to avoid the discontinuity -- basically, I want to reverse every odd sequence of the mod cycle.

Is there a function that satisfies my requirements? Duplicating the peaks and valleys is acceptable, but I'd prefer to avoid that. Bonus points if you can express the function in terms of min and max, the top and bottom integers of the cycle!

答:
Assuming $t$ starts at $0$ (to start $t$ at $1$ instead, just replace $t$ with $t - 1$ below), then for any positive integer $n$, the cycles from $0$ to $n$ and back to $1$ before starting again at $0$ can be obtained using

$$f(t) = t\left(-1\right)^{\left\lfloor t/n \right\rfloor} \bmod{2n} \tag{1}\label{eq1A}$$

where $\bmod{2n}$ means the remainder $0 \le r \le 2n - 1$ when the result is divided by $2n$.

Your specific example sequence has $n = 7$, so it can be derived from

$$f(t) = t\left(-1\right)^{\left\lfloor t/7 \right\rfloor} \bmod{14} \tag{2}\label{eq2A}$$

Note \eqref{eq1A} works for $0 \le t \le n - 1$ because $\left\lfloor\frac{t}{n}\right\rfloor = 0$, giving $(-1)^{0} = 1$ so $f(t) = t$. For $n \le t \le 2n - 1$, we have $\left\lfloor\frac{t}{n}\right\rfloor = 1$, giving $(-1)^{1} = -1$ so $f(t) = 2n - t$, e.g., $f(n) = n, f(n + 1) = n - 1, \; \ldots \; , f(2n - 1) = 1$. With \eqref{eq2A}, this gives $f(7) = 7, f(8) = 6, \; \ldots \; , f(13) = 1$.

For $2n \le t \le 3n - 1$, then $\left\lfloor\frac{t}{n}\right\rfloor = 2$, giving $(-1)^2 = 1$, so since $t \bmod{2n} = t - 2n$, then $f(t) = t - 2n$ with $f(2n) = 0, f(2n + 1) = 1, \; \ldots \; , f(3n - 1) = n - 1$, i.e., the cycle is now repeating. Continuing with increasing $t$ gives the rest of the sequence you are asking for.


答:
You are right! In fact, as a general principle (which can be made precise) you can think that any operation that does not change the measure of open intervals, would not change the measure or outer measure of any set.

For your particular problem, you already know that $|P|=|T(P)|$ for every open interval $P.$ Another little fact that you might want to prove is that if $P$ is an open interval, then $T(P)$ is also an open interval.

Now fix $A\subseteq \mathbb{R}$ such that $\mu^*(A)<\infty.$ This means that for any $\epsilon>0,$ you have a collection of open intervals $(P_{i})_{i\in I}$ such that $A\subseteq \bigcup_{i\in I} P_i$ and $$\sum_{i} |P_i|\ge \mu^*(A)\ge \sum_{i} |P_i|-\epsilon.$$
Argue that $T(A)\subseteq \bigcup_{i\in I}T(P_i).$ Using the fact that $T(P_i)$ is open intevral, it follows that
$$\mu^*(T(A))\le \sum_{i\in I}|T(P_i)|=\sum_{i\in I} |P_i|\le \mu^*(A)+\epsilon.$$

Since $\epsilon$ was arbitrary, it follows that $\mu^*(T(A))\le \mu^*(A).$

Now use the fact that if $T$ is an isometry then $T^{-1}$ can be defined and it is also an isometry. Using this and repeating the above argument exactly one obtain $\mu^*(A)\le \mu^*(T(A)).$

Title: Does an isometric transformation preserve the Lebesgue outer measure?

Body:
Let $T:\mathbb{R}\rightarrow\mathbb{R}$ be an isometric function i.e.
$$|x-y|=|T(x)-T(y)|$$
for all $x,y\in \mathbb{R}$

Consider now Lebesgue outer measure:
$$\mu^*(A)  := \inf\{\sum _{i\in I}|P_i|:A\subseteq \bigcup_{i\in I} P_i\}$$
Wherer $I$ is countable set and $P_i$ is an open interval.

Does equality $\mu^{*}(A) = \mu^{*}(T(A))$ holds?

I am only able to see that for open intervals we have $|P|=|T(P)|$. I have an intuition that this is somehow sufficient to solve the problem.

Thank you in advance for help.


问:
Implicit Differentiation Help
I have to use implicit differentiation to find $\frac{dy}{dx}$ given:

$$x^2 \cos(y) + \sin(2y) = xy$$

I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.

Thank you everyone.

Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that

答:
The idea is that $y$ is a function of $x$.  Therefore, to differentiate terms like $f(y)$, one uses the chain rule:

$$
\frac{d}{dx}(f(y)) = f'(y) \frac{dy}{dx}.
$$

One also must use the product rule to differentiate the product of two or more terms (things like $x^2 y$ or $x^5 \sin(y)$ or $x \sin(xy^2)$, etc.).

Along with the comments and links above, this should be more than enough to get you on your way.

Title: Implicit Differentiation Help


问:
Find max and min of $IJ + FE + GH$
Let $D \in \triangle ABC$. Passing through D, contruct$\, FE \parallel AB, IJ \parallel AC, GH \parallel BC$. Find max and min of IJ + FE + GH

Can this problem be solved by AM-GM ? I tried $IJ + FE + GH \ge 3\sqrt[3]{\mid IJ\mid \cdot \mid FE\mid \cdot \mid GH\mid}$
and using $$\frac {FE}{AB} = \frac {CE}{CA}\implies FE = \frac {CE}{CA}\cdot AB\ ...$$
But still stuck

答:
Claim: $$\frac{IJ}{AC} + \frac{ EF} { AB} + \frac{ GH} { BC} = 2. $$

Corollary:

$$ AB + BC + CA - \max(AB, BC, CA) \leq EF + GH + IJ \leq AB + BC + CA - \min(AB, BC, CA)$$

Proof of Corollary

$x, y, z$ are real numbers from 0 to 1 subject to $x+y+z$ = 2. $a,b,c$ are non-negative values. Then, the minimum and maximum value of $ax+by+cz$ is $a+b+c-\max(a,b,c)$ and $a+b+c-\min(a,b,c)$ respectively.

Apply this to $x = \frac{IJ}{AC}, y = \frac{EF}{AB}, z = \frac{GH}{BC}$, $a=AC, b=AB, c=BC$.

Proof of claim: Use similar triangles to show that $\frac{IJ}{AC} = \frac{IJ+BI+NJ} {AB+BC+CA}$. Combine fractions and show that the numerator is twice the denominator.

Hint: $IJ = ID+DJ = AE+ HC$.

This arose from considering the equilateral triangle, and realizing that it was a constant 2 throughout. The extremes in the scalene case is obtained by setting $D$ at an extreme point, i.e. a vertex.

Title: Find max and min of $IJ + FE + GH$


问:
Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

So far I only found that I can find the angle at $C$ ($\gamma$) by constructing the right triangle with leg $h_c$ and hypotenuse $b$. I played a bit with the angles, but I can't really see how to use the length of the angle bisector, since there aren't many theorems helping me.

I feel like I have to use the inscribed angle theorem and find $B$ as the intersection of 2 loci, but I only found that it lies on side $c$

Any help would be appreciated. Thanks.

答:
COMMENT.-Where did you get this problem from? It is very difficult. Analytically you can solve it if you calculate the value of $x$ which gives you the position of the vertex $C = (x, h)$, but $x$ is a solution of the equation

$$k^2(\sqrt{x^2+h^2}+x+b\cos\alpha)^2=(x^2+h^2)(x+b\cos\alpha)(\sqrt{x^2+h^2}+2b)$$ where $k,h,b,\alpha$ are data of your problem (the angle $\alpha$ is determined by the length of the side $b$, $h$ is your altitude and $k$ is the length of the angle bisector).

This equation is of degree eight!


问:
Nilpotent groups can be constructed by means of abelian groups.
I am studying A course in theory of groups by Robinson. When defining the central extension of groups, the author says that

Every nilpotent group can be constructed from abelian groups by means of a sequence of central extensions.

What do I understand by this line is that for every nilpotent group $G$ there exists groups $H$ and $K$ such that $H\subseteq Z(K)$ and the exact sequence.
$$\{1\}\to H \to K\to G\to \{1\}.$$
But I can not prove this, please help me to do this.

答:
A nilpotent group of class $0$ is the trivial group.

A nilpotent group of class $1$ is a nontrivial abelian group. In particular, $G/Z(G)$ is nilpotent of class $0$ (trivial).

A nilpotent group of class $2$ is a group $G$ such that $G/Z(G)$ is nilpotent of class $1$ (that is, nontrivial and abelian).
In particular, you have an exact sequence
$$ 1 \to Z(G)\to G\to G/Z(G)\to 1$$
with $G/Z(G)$ nontrivial abelian; or, in your notation, nontrivial abelian groups $H$ and $K$ such that
$$1 \to H\to G\to K\to 1$$
where $H\to G$ sends $H$ into $Z(G)$.

A nilpotent group of class $3$ is a group $G$ such that $G/Z(G)$ is nilpotent of class $2$. Thus, there exists an exact sequence
$$1 \to H \to G \to K\to 1$$
with $H$ nontrivial abelian, where $H\to G$ sends $H$ into $Z(G)$, and $K$ is nilpotent of class $2$.

In general, $G$ is nilpotent of class $n+1$ if and only if $G/Z(G)$ is nilpotent of class $n$; so there is an exact sequence
$$1 \to H \to G \to K\to 1$$
with $H$ nontrivial abelian, the map $H\to G$ sends $H$ into $Z(G)$, and $K$ is nilpotent of class $n$.

In general, if $G$ is a group and we have an exact sequence
$$1 \to H \stackrel{f}{\to} G\to K\to 1$$
where $H$ is abelian, $f(H)\subseteq Z(G)$, and $K$ is nilpotent of class at most $n$, then $G$ is nilpotent of class at most $n+1$.


问:
(Almost) fill a square with non-overlapping circles
How do I put a finite number of non-overlapping circles into a square, so that their total area approaches the area of the square as the number of the circles approaches infinity? Their radius can vary.

This looks obvious to me, but I am unable to give it a formal proof. I tried induction, but I am unable to track the area of the smaller-and-smaller circles when I keep putting circles into the square.

答: What you are describing sounds like an Apollonian gasket for the square.  Obviously, one would begin with a single circle inscribed in the square; then the four corner shapes can be filled recursively in a symmetric fashion.  The centers and radii of the circles can be calculated using the complex Descartes' circle theorem.  This is indeed a space-filling packing.</p>


问:
Find all ordered pairs (m.n) of natural numbers that satisfy the equation
Find all ordered pairs $(m, n)$ of natural numbers that satisfy the equation

$9^m +3^m-2 = 2 p^n$

Where $p$ is a prime number.

This is what I have tried. I wrote the equation as
$ (3^m+2)(3^m-1) = 2p^n$

$(3^m-1)$ is always even and $(3^m+2)$ is  always odd. So $p$ cannot be even i.e. $p$ cannot be $2$.

When $(3^m-1) = 2 \implies m = 1.$ Then $p=5$ and $n=1$. $(1,1)$ is one ordered pair.

I feel there are no other solutions but I don't know how to prove it. Can someone help?

答: If there is any other solution, where m is not equal to 1 then observe that both factors necessarily have gcd as $p^k$ where k is a positive integer. But this is impossible since their difference is 3. Since  you showed uniquness for m=1, you are done.


问:
If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.
If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.

I came across this statement as lemma 2.1 in Factoring polynomials over large finite fields, ER Berlekamp.

The author also mentions that this Lemma is discussed among the others in two other publications. However, I could not find adequate proofs there either. The books just state them as fact.

Is there something obvious here that I am missing that can be used to prove this statement?

答:
Show that the claim is true when $g$ is constant

Expand 1. to monomials

Show (e.g., using properties of binomial coefficients) that the claim is true for $g_1+g_2$ if it is true for $g_1$ and $g_2$.


问:
How to show that $\max(a,b)^2+\max(a,b)+a-b$ is bijective?
How can I show that $f:\mathbb{N}^2 \to \mathbb{N}$ given by $$f(a,b)=\max(a,b)^2+\max(a,b)+a-b$$ is bijective?
How could I show it to be injective and surjective? Or does it suffice to give a explicit inverse function? If so, what is the process of inversing this function?

答:
Having a look at how it is constructed geometrically is really helping understanding that it is a bijection (I will not prove it in details, but rather show you how to find the inverse):

First point is $f(0,0)=0$ then $f(0,1)=1$ then $f(1,1)=2$ then $f(1,0)=3$ , etc..., and then we start again with a new square of greater size. In the end, we map all points of $\mathbb N^2$ in a sequence (i.e. $\mathbb N$).

We can notice that the "jump" occurs at all points of abscissa $0$.

The corresponding index is $f(0,n)=n^2$

Therefore at index $n^2$ we start an horizontal walk of $n$ points where $a\nearrow$ then a vertical walk of $n$ points where $b\searrow$

Starting from a random integer $x$, we start by writing it $\, x=n^2+y\quad$ where $n=\lfloor\sqrt{x}\rfloor$

Our trajectory is $\, \underbrace{(0,n)}_{y=0}\to\underbrace{(1,n)}_{y=1}\to\cdots\underbrace{(n,n)}_{y=n}\to\underbrace{(n,n-1)}_{y=n+1}\to\cdots\to\underbrace{(n,0)}_{y=n+n}$

In term of $(a,b)$ this gives $\begin{cases}(y,n) & y\le n\\(n,n-(y-n)) & y\ge n\end{cases}\quad$ and we have our inverse.

We can eventually gather everything in a single expression

$\, f^{-1}(x)=(\min(y,n),2n-\max(y,n))$

Or fully expanded:

$$f^{-1}(x)=\bigg(\min(x-\lfloor\sqrt{x}\rfloor^2,\lfloor\sqrt{x}\rfloor)\quad,\quad2\lfloor\sqrt{x}\rfloor-\max(x-\lfloor\sqrt{x}\rfloor^2,\lfloor\sqrt{x}\rfloor)\bigg)$$


问:
Approximation of $f(x, y)$ by $\sum_{n=1}^N c_n \chi_{A_n}(x) \chi_{B_n}(y)$
I believe the following statement is true, but I cannot prove it.
I would be grateful for your advice.

Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be $\sigma$-finite measure spaces, and let $(X \times Y, \mathcal{A} \times \mathcal{B}, \mu \times \nu)$ be product measure space. Define
\begin{equation}
D := \left \{\sum_{n=1}^N c_n \chi_{A_n \times B_n} \mid N \in \mathbb{Z}_+ , c_n \in \mathbb{C}, A_n \in \mathcal{A} , B_n \in \mathcal{B}\right \}.
\end{equation}
Then, $D$ is dense in $L^p(X \times Y; \mathbb{C})$ for all $p \in [1, \infty)$.

答: Here's a sketch.  You already know that the set $D' := \big\{ \sum c_n \chi_{E_n} : E_n \text{ is $(\mathcal{A} \times \mathcal{B})$-measurable} \big\}$ is dense in $L^p(X \times Y)$.  Then just use the fact that the product $\sigma$-algebra $\mathcal{A} \times \mathcal{B}$ is generated by sets of the form $A \times B$ where $A \in \mathcal{A}$ and $B \in \mathcal{B}$.  This allows you to conclude that any $(\mathcal{A} \times \mathcal{B})$-measurable set can be approximated up to an arbitrarily small error by a finite union of sets of the form $A \times B$.  This then implies that $D$ is dense in $D'$.


问:
Using trigonometric substitution to integrate $\int\frac{x^3dx}{\sqrt{25-x^2}}$

I tried solving this calculus problem so many times but I never got an answer that at least looked similar to the choices in the image. Can someone help?
I don't understand where the Sine came from. This is the answer I get

$$\frac13(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C $$

Edits are appreciated

Body:
Integration by parts. Your result is equal to mine, BTW

$$\int \frac{x^3}{\sqrt{25-x^2}} \, dx=\int (-x^2)\,\frac{ -x}{\sqrt{25-x^2}}\,dx=$$
$$\int \frac{ -x}{\sqrt{25-x^2}}\,dx=\sqrt{25-x^2}+C$$
$$=-x^2\sqrt{25-x^2}-\int\left(-2x\sqrt{25-x^2}\right)\,dx=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3} \left(25-x^2\right)^{3/2}+C=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3}\sqrt{25-x^2}(25-x^2)+C=$$
$$=-\sqrt{25-x^2}\left(x^2+\frac{2}{3}(25-x^2)\right)+C=-\frac{1}{3} \sqrt{25-x^2} \left(x^2+50\right)+C$$


问:
Advanced combinatorial probability
There are $5$ rooms $R_1,R_2,R_3, R_4, R_5$ in a hallway. $17$ people may enter any of those $5$ rooms upon their wish. What is the probability that $5$ of them will enter $R_1$, $2$ will enter $R_2$, $4$ will enter $R_3$, $4$ will enter $R_4$  and  $2$  will enter $R_5$?

答:
Assume that the rooms are filled sequentially with the desired number of people from oldest to youngest, say, then

$Pr = (\frac 1{5})^5\cdot(\frac 1{5})^2\cdot(\frac1{5})^4\cdot(\frac 1{5})^4\cdot(\frac 1{5})^2 = (\frac1{5})^{17}$

But actually it can be filled in any order, hence use permutations with repetitions to get

$Pr = \left(\dfrac 1 5\right)^{17}\times\dfrac{17!}{ 5!2!4!4!2!}$

Title: Advanced combinatorial probability


问:
Missing solution after solving system of equations.
I have been learning linear algebra from book ,,Mathematics for Machine Learning", all seems pretty clear and straight-forward for now, but I have encountered one problem. There was a system of equastions to solve, and i think that there is one solution missing. Here are the solutions from the text book.
I noticed that matrix [0,-1,-1,2,1]^T also solves this equation, and it is not in solutions (and they clearly stated that general solution contains set of all possible solutions).

The solution with matrix they gave in solutions

The solution I think also fits this system of equations

I guess that there wouldn't be a mistake like this in a textbook, so maybe I am missing something after all, could someone help me with this one, and write why I am right/wrong?

Thanks for every reply!

答:
The solution from the textbook is correct.

The solution you mention is a solution of the homogeneous system with the same matrix, but not of the full system. I don't know why you expect it to appear as a solution of the non-homogeneous system.


问:
Show $5z^n=e^z$ has a finite number of zero in $\{a<\Im z < v\}$ and $\{a < \Re z < b \}$
Show that the number of roots $N$ of the function
$$
h(z)= f(z)-g(z)=5z^n - e^z, n\ge 1,
$$
is at most finite in any horizontal strip
$$
\alpha=\{z:a<\Im z<b\},
$$
and any vertical strip,
$$
\beta=\{z:a<\Re z<b\}.
$$

I'm not clear on the meaning of the question.

Should I prove that $N$ is finite on $\alpha$ and $\beta$ separately, or on $\alpha \cap \beta$.

In any case, I would use Rouche theorem, but has none of the contour is a circle
I don't see how to proceed.

Also, $|g|$ is not always less than $|f|$ on circle of arbitrary radius $a$ or $b$.

答:
There are two different problems here:

Show that $h$ has finitely many zeroes on $\alpha$

Show that $h$ has finitely many zeroes on $\beta$

On $\alpha\cap \beta$ the problem would be too easy: an entire function can have only finitely many zeroes on any bounded region, since the zeroes cannot accumulate.

The right approach is to rewrite $h(z)=0$  as the equation $$5z^n=e^z\tag1$$ and consider the size (absolute value) of both sides of (1). Namely, if (1) holds then we certainly have $$5|z|^n=e^{\mathrm{Re}\,z}\tag2$$

Problem 2 is a bit easier. Since $\mathrm{Re}\,z$ is bounded within the strip, the right side of (2) is bounded (by some constant $M$). Therefore, (2) can only hold when $5|z|^n\le M$. But this constrains $z$ to lie in some bounded region, and $h$ can have only finitely many zeroes on a bounded region (as mentioned before).

In problem 1, you should consider the left and right "tails" of the strip separately. If you go far enough to the left, you  will find $5|z|^n>e^{\mathrm{Re}\,z} $.  If you go far enough to the right, you  will find $5|z|^n<e^{\mathrm{Re}\,z} $.  This leaves only the middle part, where, again, $h$ can have only finitely many zeroes.


答:
Your vector $\begin{bmatrix}0\\-1\\-1\\2\\1\end{bmatrix}$ is a linear combination of their vectors $(-1) \begin{bmatrix}2\\1\\0\\0\\0\end{bmatrix} + (1)\begin{bmatrix}2\\0\\-1\\2\\1\end{bmatrix}$, so it also solves the associated homogeneous equation, but is already included in their solution for $\lambda_1 = -1$ and $\lambda_2 = 1$.

As noted in Martin's answer, the solution to the original equation would require you to add the particular solution to the solution to the associated homogeneous equation.

Title: Missing solution after solving system of equations.

Body:
I have been learning linear algebra from book ,,Mathematics for Machine Learning", all seems pretty clear and straight-forward for now, but I have encountered one problem. There was a system of equastions to solve, and i think that there is one solution missing. Here are the solutions from the text book.
I noticed that matrix [0,-1,-1,2,1]^T also solves this equation, and it is not in solutions (and they clearly stated that general solution contains set of all possible solutions).

The solution with matrix they gave in solutions

The solution I think also fits this system of equations

I guess that there wouldn't be a mistake like this in a textbook, so maybe I am missing something after all, could someone help me with this one, and write why I am right/wrong?

Thanks for every reply!


问:
Determine the convergence and divergence of $\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}dxdy}}$
Determine whether $$\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}\mathbb dx \mathbb dy}}$$
is convergent or divergent.

It looks like that it needs to be scaled by inequalities, but I failed. Can anyone help?

答:
Here is an answer that avoid the use of beta function:

$$ I
= \int_{0}^{1} \frac{1}{x} \int_{0}^{1} \left(\frac{x}{1-x}\right)^y \, \mathrm{d}y \mathrm{d}x
= \int_{0}^{1} \frac{2x-1}{x(1-x)\log\bigl(\frac{x}{1-x}\bigr)} \, \mathrm{d}x. $$

Near $x = 0$, the integrand is asymptotically

$$ \frac{2x-1}{x(1-x)\log\bigl(\frac{x}{1-x}\bigr)} \sim -\frac{1}{x\log x}, $$

hence the integral diverges by the limit comparison test with

$$ - \int_{0}^{\frac{1}{2}} \frac{1}{x\log x} \, \mathrm{d}x = \biggl[ -\log(-\log x) \biggr]_{0}^{\frac{1}{2}} = \infty. $$


问:
Mistake in proof of sum of divisors function $\sigma(n)$
The proof derives the correct result, but I cannot see how the first equality is correct.

To begin we use the formula $\sigma(n)=\sum_{d\mid n}d$

This is the first step in the proof:

$$\sum_{1\leq n\leq x}\sigma(n)=\sum_{1\leq n\leq x}\sum_{d\mid n}n/d$$

Surely this first step is a simple substitution of our formula

Shouldn't it be $$\sum_{1\leq n\leq x}\sigma(n)=\sum_{1\leq n\leq x}\sum_{d\mid n}d$$

If not, why do we choose substitute in this way to sum over $n/d$?

答: I understand the confusion.  When looking at number theory calculations you have to think through each step and avoid the simple algebraic simplifications that one usually applies when manipulating sums.  Otherwise you see steps that may not at first make sense.  So in affect you read these expressions slowly.  Hope this helps.</p>


问:
Probability Q on rolling two 8-sided dice
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?

答:
This assumes you win if you get both $7$ and $11$ before the two $9$s.  Some people seem to be reading as if you win if you get one of these. I also assume the two $9$s do not need to be consecutive.

We can ignore the rolls that are not $7,9,11.$ The next roll in $7,9,11$ has probability $3/10,2/5,3/10,$ respectively.

Let the states of a Markov process bet $s_1,\dots,s_6$ with

$s_1$ be the state where no $7,9,11$ has occurred.

$s_2$ has gotten one $9$ and no $7,11.$

$s_3$ has gotten one of $7,11$ and no $9$

$s_4$ has gotten one of $7,11$ and one $9$

$s_5$ has gotten two $9$s first

$s_6$ has gotten $7,11$ first.

Then the Markov chain’s transition matrix is:

$$T=\begin{pmatrix}
0&2/5&3/5&0&0&0\\
0&0&0&3/5&2/5&0\\
0&0&3/10&2/5&0&3/10\\
0&0&0&3/10&2/5&3/10\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{pmatrix}$$

I used Wolfram Alpha to get that:

$$\lim_{n \to\infty}(1,0,0,0,0,0)T^n =\left(0,0,0,0,\frac{604}{1225},\frac{621}{1225}\right)$$

So the probability of winning is $\frac{621}{1225}\approx 0.507.$

You can avoid the Markov chain if you let $p_i$ be the probability of
winning starting at state $s_i.$ Then you get the equations:

$$\begin{align}p_1&=\frac{2}{5}p_2+\frac35p_3\\
p_2&=\frac35p_4+\frac25p_5\\
p_3&=\frac3{10}p_3+\frac25p_4+\frac3{10}p_6\\
p_4&=\frac3{10}p_4+\frac25p_5+\frac3{10}p_6\\
p_5&=0\\
p_6&=1
\end{align}$$

Solving for $p_1$ gives you your answer. This can be done by hand:

$\frac7{10}p_4=\frac3{10}$ give $p_4=\frac37.$

$\frac7{10}p_3=\frac25\cdot \frac37+\frac3{10}$ gives $p_3=\frac{33}{49}.$

$p_2=\frac35\cdot\frac37=\frac9{35}.$

$p_1=\frac25\cdot \frac9{35}+\frac35\frac{33}{49}=\frac{18}{5^2\cdot 7}+\frac{99}{5\cdot 7^2}=\frac{621}{1225}$

Getting the same results.


问:
Show that $\left|z+z\sqrt{1-\frac{1}{z^2}} \right| > 1$, for all $|z| < 1$ where $\mathrm{Im}(z) \ne 0$
I am having difficulty showing that

$$\left|z+z\sqrt{1-\frac{1}{z^2}} \right| > 1$$

for all $|z| < 1$ where $\mathrm{Im}(z) \ne 0$. Here, the square root is defined as
$$z = re^{i\theta} \text{ where } r > 0, \theta \in (-\pi, \pi] \text{ and } \sqrt{z} = \sqrt{r}e^{i\frac{\theta}{2}}.$$
I tried by writing $z = x+yi$ but I wasn't able to get much.
I would appreciate any hints.

答:
Hint:

Let $w_{1,2}=  z\pm z\sqrt{1-\frac{1}{z^2}}$ and note that

$$w^2-2zw+1=0$$

Then, $|w_1||w_2|=1$. The discriminant is of complex if $\mathrm{Im}(z) \ne 0$, ensuring $|w_1|\ne|w_2|$.


问:
Extreme values of $2\cos^2 \frac{x}{2}\cdot\sin^2 x$ without calculus

To find the extreme values of
$$2\cos^2 \frac{x}{2}\cdot\sin^2 x$$

I Have not learnt calculus yet but this is a question within our syllabus .which made me to think if this can be done with it .

I am not able approach this problem any help is appreciated.

Body:
Writing $x=2y$

$$\cos y\sin2y=2(1-\sin^2y)\sin y$$

$$\dfrac{2\sin^2y+1-\sin^2y+1-\sin^2y}3\ge\sqrt[3]{2\sin^2y(1-\sin^2y)^2}$$

$$\implies\left(\sqrt2\sin y(1-\sin^2y)\right)^{2/3}\le\dfrac23$$


问:
How to show "a set which is infinite" can not be finitely axiomatized?
I met a strange question which gives me a Language L which only has one unary Relation R.(Set R) My concern is how to prove that the theory which axiomatizes "R is infinite" can not be finite.

We know that the theory which axiomatizes "R is infinite" can be something like {R has at least 1 element, R has at least 2 elements... etc} which is infinite. But not sure if it can be done with finite sentences.

So generally how-to, I suppose, use compactness theorem to show that "R is infinite" can not be finitely axiomatized?

A previous relevant question is: The theory which axiomizes infinity.

答:
We can easily spell out "$|R|\le n$" for any fixed $n$, namely $$\exists x_1\ldots \exists x_n \forall y\colon y\in R\to (y=x_1\lor\cdots \lor y=x_n).$$
Hence we also can formulate $|R|> n$
assume we can spell out some $|R|<\infty$. Then the infinitely many sentences
$$ |R|<\infty,|R|> 1, |R|> 2,|R|> 3,\ldots $$
cannot have a model. By compactness, hence some finite subset has no model. But any such finite subset just amounts to $$n<|R|<\infty$$ for some $n$. Now exhibit a model where $R$ is finite but larger than $n$ to arrive at a contradiction.


问:
Expected stop time for biased gambler's ruin
Supposed you have \$10 at hand, and you play a game which costs \$2 each time. You will have $p=\frac{1}{3}$ to get \$5 back, and $q=\frac{2}{3}$ to get nothing. You would like to play the game as many as you can (which means whenever you have at least $2 at hand, you would play it). What is the expected time you could play?

Here is my attempts:

Let $E(X_t)$ be the expected money at hand after $t$. We also know $E(X_0) = 10$.

From the question, we know $E(X_{t+1}) = E(X_t) - \frac{1}{3}$, then we can have $E(X_{t+1} + \frac{t+1}{3}) = E(X_t + \frac{t}{3})$. Hence, we can have:
$$E(X_t + \frac{t}{3}) = E(X_0) = 10$$

Let $\tau$ be the stop time, then we could have:
$$E(X_\tau + \frac{\tau}{3}) = 10$$

So the expected stop time is:
$$E(\tau) = 3*(10 - E(X_\tau))$$

However, I don't know the value of $E(X_\tau)$ since when the game is over, we could possibly end with \$1 or \$0. And I couldn't get their corresponding probability.

I also try to simulate it for 10000000 times. It gives me around 28.41

答:
A lot of this comes down to what you know, as there are many moving parts underlying this.  Since this is tagged 'martingale' the solution I give leans heavily on martingales and basics of linear recurrences.

note:

there are essentially 2 different ways to represent the problem, one is you start at 10 dollars and terminate when bankroll is $\leq 1$.  The other is you start at 0 dollars and terminate when bankroll $\leq -9$.  There is a simple translation between one and the other so I'll freely use whichever is most convenient.

i.) let $S_n =X_1+...+ X_n$
Where the $X_i$ are iid taking values $\{-2,3\}$ with probabilities $\frac{2}{3}$ and $\frac{1}{3}$ respectively

(this is the start at 0 formulation where we want a stopping rule of terminate when bankroll $\leq -9$).

Let $A_n$ denote the event $S_n \gt -9$, then for $n$ large enough we have

$P\big(\text{threshold of -9 has not been crossed over n trials}\big)=P\big(A_1\cap ...\cap A_n\big)\leq P\big(A_n\big)\leq \rho^n $

for some $\rho\in(0,1)$

(Use Chernoff type bound or Azuma-Hoeffding)

This means that the tail of our stopping time random variable $T$ is bounded above by a geometric random variable.  (Or for some constant $c$ we have $T+c$ is stochasticaly smaller than some  geometric random variable.)

Thus $T$ is finite WP1 and has all moments.  In particular $E\big[T\big]\lt \infty$

ii.) Applying the Wald Equation we have

$E\big[S_T\big] = E\big[T\big]\cdot E\big[X\big]\implies E\big[T\big] = \frac{E\big[S_T\big]}{E\big[X\big]}$

now

$E\big[X\big] = \frac{2}{3}\cdot -2 + \frac{1}{3}3=-\frac{1}{3}$ and

$E\big[S_T\big]= p\cdot -9 + (1-p)\cdot -10$

so we need to solve for $p$, the probability that we stop after total net losses of exactly 9

iii.) Write this as a countable state Markov chain with states $\big\{0,1,2,....,\big\}$

(This is the start at 10 and terminate at 0 or 1 formulation)

States 0 and 1 are absorbing states, and states $\geq 2$ are transient, forming a single communicating class.  What (i) tells us is that states 0 and 1 form a persistent set, i.e. absorption happens WP1 in either state 0 or 1, no matter what the starting state is.  (Technically we only verified this for starting in state 10 but the transient states form a single communicating class so they all inherit this property.)

There is a transition matrix $P$ associated with this chain, and we want to find some $\mathbf x$ that satisfies $P\mathbf x = \mathbf x$

So, for state $k\geq 2$ we have a recurrence

$x_k = \frac{2}{3}x_{k-2}+\frac{1}{3}x_{k+3}$ or

$x_{k+3}=3x_k-2 x_{k-2}$

and we also have boundary conditions of $x_1:=1$ and $x_0:=0$.  (When there are only 2 absorbing states the standard technique is to give a reward of 1 for one of them and reward of 0 for the other to in effect convert them into a Bernouli.)

thus if we solve the recurrence we will have

$P\mathbf x = \mathbf x$  where
$x_0=0$ and $x_1 = 1$ and for probabilistic reasons we want to impose the additional constraint that each

$x_k\in [0,1]$

as a result we'll, given our start in state 10:

$\mathbf e_{10}^TP^n \mathbf x=\mathbf e_{10}^T\big(P^n \mathbf x\big) = \mathbf e_{10}^T \mathbf x = x_{10}$

but since $\lim_{n\to \infty}\mathbf e_{10}^TP^n = \mathbf z^T$ where $z_0 =1-p$ and $z_1=p$ and $z_i =0$ for $i\geq 2$, this means

$p =(1-p)\cdot 0 + p \cdot 1= \mathbf z^T\mathbf x=\lim_{n\to \infty}\big(\mathbf e_{10}^T P^n\big) \mathbf x  = \lim_{n\to \infty}\mathbf e_{10}^T P^n \mathbf x  = x_{10}$

in short $P\mathbf x = \mathbf x$ gives us a very useful martingale.

(if one is concerned about analytic subtleties relating to the above limits, there are many ways to add further justification.  A very simple exercise is to show that if state $j$ is absorbing, and $x_j:=1$, and in general $x_k\in[0,1]$ for some $\mathbf x = P\mathbf x$, then $p_{k,j}^{(n)}\leq x_k$ for all n.  After taking a limit of n with $j:=1$ and then repeating with $j:=0$, using this time $\mathbf x':=\mathbf 1-\mathbf x$ we see that these inequalities must in fact be met with equality, after taking the limit, in each case.)

iv.)
solve the recurrence however you like.  You can observe 5 distinct roots

https://www.wolframalpha.com/input/?i=x%5E5-3x%5E2%2B2

but $\lambda_1=1$ and $\lambda_2 \approx-0.76219$ are the only two with modulus $\leq 1$.  Note after factoring out the root of 1 the polynomial is $x^4+x^3+x^2-2x-2$ which in principle is solvable exactly.

since the linear recurrence has 5 distinct roots you'd normally take it to be of the form

$x_k = y_1 \cdot \lambda_1^k+y_2 \cdot \lambda_2^k+y_3 \cdot \lambda_3^k+y_4 \cdot \lambda_4^k+y_5 \cdot \lambda_5^k$

but there are 5 distinct $\lambda_i$ and only 2 boundary conditions.

heuristic: throw out all the $\big \vert\lambda_i\big\vert \gt 1$  because if we kept them then we couldn't ensure that $x_{k}\in[0,1]$ for all $k$ - the solution would get 'too big' in modulus.  Given the simplicity of the problem one could manipulate the symbols to rigorize this intuition, but a much better justification is to simply note

$x_k = y_1 \cdot \lambda_1^k+y_2 \cdot \lambda_2^k$

gives a particular solution that satisfies our constraints, and that particular solution must in fact unique for analytical reasons (i.e. calling on standard martingale or countable state markov chain limit theorems).

So solve

$ \begin{bmatrix}
1   & 1 \\
1 & \lambda_2 \\
\end{bmatrix} \begin{bmatrix}
y_1\\
y_2\\
\end{bmatrix}=\begin{bmatrix}
1   & 1 \\
\lambda_1 & \lambda_2 \\
\end{bmatrix} \begin{bmatrix}
y_1\\
y_2\\
\end{bmatrix} = \begin{bmatrix}
x_0\\
x_1\\
\end{bmatrix}=\begin{bmatrix}
0\\
1\\
\end{bmatrix}$

if you do this with the numeric approximation of $\lambda_2 \approx -0.76219$ then you get

$\begin{bmatrix}
y_1\\
y_2\\
\end{bmatrix} = \begin{bmatrix}
0.56748\\
-0.56748\\
\end{bmatrix}$

thus

$x_k=0.56748\cdot 1^k - 0.56748\cdot (-0.76219)^k=0.56748- 0.56748\cdot (-0.76219)^k$

which for $k:=10$ gives

$x_{10}=0.56748 - 0.56748\cdot (-0.76219)^{10}\approx 0.5299$

plugging this into (ii) gives  $E\big[T\big] \approx 28.410$