问:
How to prove $ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$
How can I prove that

$$ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$

The result is given here (result 48).

The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.

Some of my attempts include:

Multiplying known infinite products for $\frac{\pi}{2}$  and
$\frac{1}{e}$

$$ \frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots} $$
$$ \frac{1}{e} =  \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$

as well as others given in https://arxiv.org/pdf/1005.2712.pdf

Trying to find the partial products from ${n=1} $ to $k $, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}} $$

However, I have been unsuccessful producing partial products on my own.

Taking the $\ln$ of the product to try to find the partial sums;

Trying to transform the following integral into infinite product.

$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e} $$

From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.

Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.

Thank you kindly for your help and time.

答:
This infinite product could be computed (more or less) directly.

At first, let's compute infinite sum instead of infinite product:
$$P=\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}\,\,n}$$
$$S = \ln P = \sum_{n=1}^\infty (-1)^{n+1}n \ln\left(1+\frac2n\right) = \sum_{n=1}^\infty (-1)^{n+1}a_n$$
At first we need some regularization:
$$a_n=2-\frac2n+\frac{8}{3n^2}+\mathcal O(\frac{1}{n^3}) \nrightarrow 0$$
Nevertheless we could only consider partial sums of the $S$ with odd (or even) $n$; in other words, we'll assume that $\sum_{n=1}^\infty (-1)^{n+1} = 0$. (I know, I know, this is a not rigorous, but let me advance).

Ok, let's convert $a_n$ to power series:
$$a_n = \sum_{k=0}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}=2+\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k},$$
so we need to compute
$$S = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}$$
(look, we «cancel» 2 in $a_n$'s expansion). Let's change summation order:
$$S = \sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1} \left(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k}\right).$$
Series in the parentheses is alternating zeta-function:
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k} = \left(1-\frac{1}{2^{k-1}}\right)\zeta(k),$$
where $\zeta(k)$ is the Riemann zeta function ($k>1$). At $k=1$ we have
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\ln2.$$

So,
$$S = -2\ln2 + \sum_{k=2}^\infty (-1)^k\frac{2^{k+1}}{k+1}\left(1-\frac{1}{2^{k-1}}\right)\zeta(k)\\=-2\ln2+\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\zeta(k).$$
This series seems hopeless, but zeta function involves in many other series. We'll use this:
$$\Gamma(s)\zeta(s)=\int\limits_0^\infty \!\frac{x^{s-1}}{e^x-1}dx.$$

Hence
$$S + 2\ln2 = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{\Gamma(k)}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx \\=
\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{(k-1)!}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx,$$
or
$$T = S + 2\ln2 = \int\limits_0^\infty \!\frac{dx}{e^x-1} \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!}$$

The last series could be computed by standard methods via series for exponent and differentiation w.r.t. $x$:
$$\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!} = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{(k+1)!}kx^{k-1}\\=
\frac{3}{x^2}+\frac{1}{x^2}e^{-2x}+\frac{2}{x}e^{-2x}-\frac{4}{x^2}e^{-x}-\frac{4}{x}e^{-x}$$

Let's collect all these pieces together:
$$T = \int\limits_0^\infty \frac{e^{-2x}dx}{x^2(e^x-1)}(3e^{2x}-4(1+x)e^x+2x+1).$$
We have here terms with powers of $\mathrm{exp}$ and $x$. For me simplest and straightforward method for calculation of this integral is expansion of denominator:
$$\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=e^{-x}\sum_{m=0}^\infty e^{-mx} = \sum_{m=1}^\infty e^{-mx}.$$
So we have
$$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-2x}\sum_{m=1}^\infty e^{-mx},$$
or
$$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} \sum_{m=3}^\infty e^{-mx}.$$
Let's change summation and integration:
$$T = \sum_{m=3}^\infty  \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$

Now we need to compute
$$I_m = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$
Maple says
$$I_m = m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$
We could believe in it or could to get it:) At first,
$$\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} = \sum_{k=0}^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k,$$
and
$$\int\limits_0^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k e^{-mx} dx = \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$
(we used Euler's formula for $\Gamma(x)$). So,
$$I_m = \sum_{k=1}^\infty \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$
(term at $k=0$ vanishes). Now we can diff. w.r.t. $m$:
$$\frac{d^2 I_m}{dm^2}= \sum_{k=1}^\infty (12\cdot 2^k-4k-12) \frac{1}{m^{k+3}} = \frac{1}{m^3} \sum_{k=1}^\infty \frac{12\cdot 2^k}{m^k}-\frac{4k}{m^k}-\frac{12}{m^k}, $$
or
$$\frac{d^2 I_m}{dm^2}= \frac{1}{m^3} \left(\frac{24}{m - 2} - \frac{4m}{(m - 1)^2} - \frac{12}{m - 1}\right)$$
(while $m>2$). Integrate twice and we have indeed
$$I_m = (3m - 6)\ln(m - 2) + (-4m + 8)\ln(m - 1) + 2 + (m - 2)\ln(m)\\= m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$
We start summation from $m=3$; let $m=2+p$:
$$I_{p+2} = p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2$$

Now we have
$$T = \sum_{p=1}^\infty p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2 = \sum_{p=1}^\infty t_p.$$
We could find approximation for this sum by some methods, but I prefer more detailed calculation. At first, let's rearrange $t_p$:
$$t_p = [(p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p)] \\ +[2-2\ln(p + 2) + 4\ln(p + 1) ].$$
First sum could be found by kind of telescoping summation:
$$T^1_N = \sum_{p=1}^N (p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p) \\= -3(N + 1)\ln(N + 1) + (N + 2)\ln(N + 2) - 2\ln(2)$$
(put $f(x)=x\ln x$ and look at partial sums). Analogously for second term we have
$$T^2_N = \sum_{p=1}^N 2-2\ln(p + 2) + 4\ln(p + 1) \\= 2\sum_{p=1}^N \ln p + 2\ln(N + 1) - 2\ln(N + 2) + 2\ln(2) + 2N$$

Finally we have
$$T_N = 2\sum_{p=1}^N \ln p - (3N + 1)\ln(N + 1) + N\ln(N + 2) + 2N.$$
Asymptotic expansion for the first series is well-known and could be found from Euler—Maclaurin:
$$\sum_{p=1}^N \ln p = N(\ln(N) - 1) + \frac12\ln(2\pi) + \frac12\ln(N) + \frac{1}{12N}+\mathcal O \left(\frac{1}{N^2}\right),$$
and finally (really finally :)
$$T_N = \ln(2\pi)-1 - \frac{4}{3N}+\mathcal O \left(\frac{1}{N^2}\right).$$

So,
$$T = \ln(2\pi)-1,$$
$$S = T - 2\ln2 = \ln\frac{\pi}{2}-1 = \ln\frac{\pi}{2e},$$
and
$$P = \frac{\pi}{2e}.$$


问:
If $\beta$ is a primitive element in extension field GF$(2^m)$ does $\beta$ and $\beta^3$ are for all $m$ non-conjugate elements?
If $\beta$ is a primitive element in extension field GF$(2^m)$ does $\beta$ and $\beta^3$ are always non-conjugate elements?

In my textbook, I saw they assume they are, but I couldn't understand it for all $m$. It seems reasonable, but as I thought it seems reasonable that $\beta^3$ and  $\beta^5$ are non-conjugate elements, I found that for $m=3$, they are conjugate to each other.

答:
I assume that to you a primitive element is a generator of the multiplicative group. This is standard meaning of primitive in the context of finite fields.

The conjugates of $\beta$ in $GF(2^m)$ are
$$
\beta,\beta^2,\beta^4,\ldots,\beta^{2^{m-1}}.\qquad(*)
$$
Because $\beta$ is primitive the exponent $\ell$ identifies the power $\beta^\ell$ uniquely while it stays in the range $0\le \ell<2^m-1$.

The exponents in $(*)$ are all in this range, so $\beta^3$ is not a conjugate simply because $3$ is not in that list of exponents.

More generally, we need to look at the exponents modulo $2^m-1$. So $\beta^3$ and $\beta^5$ are conjugates if and only if
$$
3\cdot2^i\equiv 5\pmod{2^m-1}
$$
for some $i$. Indeed, $3\cdot2^2\equiv 5\pmod7$, $7=2^3-1$. Thus $\beta^3$ and $\beta^5$ are conjugates when $\beta$ is a primitive element of $GF(8)$.

With $i$ ranging over the interval $0\le i \le m-1$ we see that $3\cdot2^i$ "overflows" $2^m-1$ only when $i=m-1$ exactly. We have
$$
3\cdot2^{m-1}=2^m+2^{m-1}\equiv2^{m-1}+1\pmod{2^m-1}.
$$
So $\beta^3$ and $\beta^5$ can be conjugates only when $2^{m-1}+1=5$, a case we already handled.

ContentLicense: CC BY-SA 4.0


问:
Similar and congruent triangles
Let $ABC$ be a triangle and $h_a,h_b,h_c$ the length of the heights.

(a) Find similar triangles at which it holds that $ah_a=bh_b$.

(b) Use (a) to show : Is $A'B'C'$ a second triangle and $h_a,h_b,h_c$ are again the length of the heights of this triangle, so these triangles are congruent.

$$$$

For (a) do we have to give two specific triangles, i.e. give the negth of the sides? Or what exactly is asked?

答:
For part (a), yes, we have to find two similar triangles for which $ah_a=bh_b$ holds.

Draw $h_a$ and $h_b$ inside $\triangle ABC$. $h_a \cap BC \in D$. $h_b \cap CA \in E$. With simple angle chasing, $$\triangle ADC \sim \triangle BEC$$

Title: Similar and congruent triangles


问:
let $A\in \mathbb{R}^{n\times n} $...
vector $v\in \mathbb{R}^{n}$ is called a special vector for A, if $Av=v$, and is called untrivial special vector if $v\ne0$.

Prove that:
 $A$ has an untrivial special vector if and only if $A^t$ (transpose) has an untrivial special vector.

My attempt: I said that let $v$ be an untrivial special vector for $A$, then $Av=v$, then I did transpose on all equation sides and got $v^tA^t=v^t$.
 I really think that this is the same as $A^tv=v$ (tried a couple examples), but I can't somehow prove it. maybe because im wrong or because I just didn't succeed at it.

Any help is appreciated, thanks in advance.

答:
We have

$$\det(I-A)= \det((I-A)^t)= \det(I-A^t).$$

Then it follows that

$A$  has an untrivial special vector $ \iff \det(I-A)=0 \iff \det(I-A^t)=0 \iff A^t$ has an untrivial special vector

Title: let $A\in \mathbb{R}^{n\times n} $...


问:
Equivalent condition of the following group presentation to give a group of order $mn$
Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$
I have to prove that $G$ is of order $mn$ if and only if  $r^n = 1 \pmod{m}$. I was able to prove the $\leftarrow$ side, but I'm stuck in proving the $\rightarrow$ side.

答:
From $b^n = 1$ and $bab^{-1} = a^r$ it follows, that $a = b^{n}ab^{-n} = a^{r^n}$, which means $a^{r^n - 1} = 1$.

Because we also have $a^m = 1$ we can conclude that $a^{GCD(m, r^n - 1)} = 1$. And because it is not hard to see, that $a \mapsto a^r$ is an automorphism of $\langle a \rangle_{GCD(m, r^n - 1)}$ we can see that

$$\langle a, b | a^m = 1, b^n = 1, bab^{-1} = a^r\rangle \cong \langle a \rangle_{GCD(m, r^n - 1)} \rtimes_{a \mapsto a^r} \langle b \rangle_{n}$$

which is a group of order $nGCD(m, r^n - 1)$


问:
Circle Packing: Unsolved Problem in Geometry?
Graham and Sloane minimize the second moment of the centres of a number discs in order to maximize their compactness. They use computational geometry techniques to find the optimal packings for non-trivial $n$.

I have adopted this metric in my research. One of the reviewers of my paper insists that there must be a closed-form solution for the lower-bound of this second moment (i.e. the optimal packing) for general $n$, because the problem "seems simple".

I do not believe that this is so. Can you please confirm (or prove me wrong)? A reference confirming the unsolved status of this problem would be most welcome.

答: Slightly old, but the only proved (circle in circle) optimal packings are for $n \leq 20$ http://hydra.nat.uni-magdeburg.de/packing/cci/</p>

ContentLicense: CC BY-SA 3.0


问:
Isomorphism of Vector Spaces between Manifolds
Hi i am reading "An introduction to Manifolds" by Loring Tu. And there in one proof at the last line it is written that "Hence, $F_*$ and $G_*$ are isomorphisms." But they've not proved that the mappings $F_*$ and $G_*$ are one-one,onto and linear transformation. I am able to understand all the steps done there except the very last step where they've written "hence $F_*$ and $G_*$ are isomorphisms". I am attaching the screenshot where the step is highligted . They've shown:$$(G\circ F)_*=\mathbb 1_{{T_p}N}$$ and $$(F\circ G)_*=\mathbb 1_{T_{F(p)}M}$$ My problem is that how does this prove that the mappings are isomorphisms.Also i am attaching another screenshot where the definition of the differential $F_*$ is given .

答:
An isomorphism of vector spaces is by definition a linear transformation which is invertible (and whose inverse is also linear... but this is automatically true). What does it mean for a function $f:A\to B$ to be invertible? By definition this means there must exist a $g:B\to A$ such that $f\circ g = \text{id}_B$ and $g\circ f=\text{id}_A$ (and of course, if such a $g$ exists, then it is unique).

What the proof shows is that the linear transformations $F_{*,p}:T_pN\to T_{F(p)}M$ and $G_{*,F(p)}:T_{F(p)}M\to T_pN$ are inverse functions hence they're isomorphisms of the appropriate tangent spaces.


问:
Numerical method for functional equations similar to $2f(x) - f(x+2) - f(x-2) = 0$
I have some functional equations, that are of the form similar to something like
$$
2f(x) - f(x+2) - f(x-2) = 0 \text,
$$
which I need to find the solution to. I know that equations like these are generally hard to solve analytically (if even possible) so I was hoping to know if there were any numerical methods that could be used to solve functional equations of this form.

In general I'm looking to solve functional equations of the form
$$
\sum_{i=1}^n a_i(x) f\bigl(g_i(x)\bigr) = 0 \text.
$$

I know in some cases it is entirely possible to have there be multiple solutions, so that may throw a wrench in the works.

答:
What would the numerical method achieve in this case? If you try to get the solution on groups of points using the recursion, there is no guarantee that you will not be switching between different solutions.

If you restrict the domain to natural numbers and solve the corresponding difference equation, you'll see that every solution must satisfy
$$
f(n) = (c_1 n +c_2) (-1)^n + c_3 n + c_4, \quad n \in \mathbb{N}
$$

and the constants $c_i$ are computed from initial conditions.


问:
How to study for analysis?
I am currently a first year undergraduate majoring in mathematics. I'm taking an introductory analysis course and find it very hard compared to other math couses. I know that the topics covered in the course are really the basics of real analysis, such as properties of $\mathbb{R}$, sequences and series, limits, continuity, Riemann integral, etc. I work much harder in analysis than in other courses such as abstract algebra, and am spending a lot of time to memorize all the theorems and their proofs mentioned in class. However, when it comes to work out a problem in the book or in the assignment on my own, I'm stuck. My guess is that I never learned how to do math rigorously, and I always rely on my intuition, which proved usually accurate in the past.

The textbook we are using is "Introduction to Real Analysis" by Robert Bartle, 3rd ed., but I also downloaded and use some extra analysis notes from a few professors' webpages.

Could you please give me any advice on how to study analysis? I'm now really desperate :(

答:
What a beginning student, faced with the difficulties of Real Analysis, needs are examples, tons and tons of well written out and very detailed examples. - ( An example of the theorem is often more illuminating than the theorem itself, at least in the beginning. ) - The best source for examples is really your Analysis professor, but if you are self-studying, or working ahead, this book might help.

https://www.springer.com/gp/book/9783030358433

"Solving Problems in Mathematical Analysis" by  Radożycki, Tomasz. I believe this is the first book in a series of three.

Title: How to study for analysis?


问:
Expected value of sum of two dependent Binomial variables
Question: Assume we toss a coin in two trials. In the first trial we toss it n times. In the second trial we toss it as many as the number of tails observed in the first trial.

Calculate the expectation of total number of tails in both trails:

Solution: This is what I have tried but can't find any closed form for the answer when n is not given. I am not sure if I have taken the right approach:

$T_1$ = Number  of  tails in the 1st trial

$T_2$ =  Number of tails in the 2nd trial

T1 ~ Binomial(n,p)

T2 ~ Binomial(T1,p)

$E(T1 + T2) =
E(T1) + E(T2) = \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} E\left(T_{2} \mid T_{1}=i\right) P\left(T_{1}=i\right) =\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} i p \times\left(\begin{array}{l}n \\ i\end{array}\right) p^{i}(1-p)^{n-i}$

I couldn't find any closed form for the answer. Any advice on how to approach this?

答:
Let's consider an equivalent experiment. We have $n$ identical coins, numbered from $1$ to $n$. We toss each of the $n$ coins once; then each coin which came up tails is tossed one more time. Let $X_i$ be the number of times the $i^\text{th}$ coin comes up tails, and $X=\sum_{i=1}^nX_i$ the total number of tails obtained.
$$E(X)=\sum_{i=1}^nE(X_i)=\sum_{i=1}^n(p+p^2)=n(p+p^2)$$


问:
For x < 5 what is the greatest value of x
It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?</p>

答: The answer is $4$, assuming the domain of $x$ is $\Bbb Z$.</p>


问:
Jacobian matrix: Inequality between number of rows and columns
To find the Jacobian matrices, we have to take partial derivatives of the differential equations according to state variables. For example from Wikipedia,

In my system, i have 8 differential equations, but number of state variables is 9. So, Jacobian is not square matrix. As a result, i cannot use this matrix in state space modelling.

How can i fix my problem? Should i define new differential equation or change the state variables?

Thanks,

答:
I would guess that the 9th variable is not a state, but an external input, e.g., the control signal. Consider the example
$$\begin{aligned}\dot{x}_1 &= f_1(x_1,x_2,x_3),\\ \dot{x}_2 &= f_2(x_1,x_2,x_3).\end{aligned}$$
Then your linearization (I assume here that you linearize around the origin that is the equilibrium) is
$$
\begin{aligned}
 \begin{bmatrix}\dot{x}_1 \\ \dot{x}_2\end{bmatrix} = \begin{bmatrix} \frac{\partial}{\partial x_1}f_1 & \frac{\partial}{\partial x_2}f_1 \\ \frac{\partial}{\partial x_1}f_2 & \frac{\partial}{\partial x_2}f_2\end{bmatrix}  \begin{bmatrix}x_1 \\ x_2\end{bmatrix} +\begin{bmatrix}\frac{\partial}{\partial x_3}f_1 \\ \frac{\partial}{\partial x_3}f_2 \end{bmatrix} x_3
\end{aligned}
$$
I.e., the Jacobian is $2\times 3$, but the linearization state matrix is only $2 \times 2$ as it includes only those state for which you have the differential equations.


问:
Solving $\sqrt{a-\sqrt{a+x}} = x$
Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$

My approach:

Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.

答:
$$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
$$x^4-2ax^2-x+a^2-a=0$$
$$(x^2+x+1-a)(x^2-x-a)=0$$

You can factor the last expression by setting up $(x^2+px+q)(x^2+rx+s)=x^4-2ax^2-x+a^2-a$. And comparing the coefficient to find $p,q,r,s$.

Title: Solving $\sqrt{a-\sqrt{a+x}} = x$


问:
Characterizing Vector Spaces without (provable) Basis in ZF
Without the axiom of choice certain vector spaces cannot be proven to have a (hamel) basis in ZF alone and I am wondering whether there exists some criteria characterizing such spaces.

Here is what I know so far: Since "every vector space has a basis" implies AC (http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf) every model of $ZF+\neg AC$ contains vector spaces without a basis. There are also concrete examples of vector spaces which do not have a basis in certain models of ZF such as $\mathbb{R}$ over $\mathbb{Q}$ (https://people.clas.ufl.edu/zapletal/files/lru3.pdf). However I have not yet been able to find a source giving some sufficient condition $(*)$ and a theorem of the form:

For every field $K$ and $K$-vector-space $V$ sattisfying $(*)$ it is consistent with ZF to assume that $V$ has no basis over $K$.

I strongly suspect that one can construct (via symmetric forcing) a model of ZF where

For every field $K$ and $K$-vector-space $V$ such that $K^\mathbb{N}$ embeds into $V$, $V$ has no basis over $K$.

(here $K^\mathbb{N}$ denotes the space of all functions from $\mathbb{N}$ into $K$ with pointwise addition and multiplication)

$\bullet$ Does anyone know a source confirming this statement? (If true then I imagine someone must have already proven either this or some stronger result.)

$\bullet$ Is there maybe even a more general characterization of those spaces that can't be proven to have a basis in ZF and those that can? (e.g. the criterion I've given above does not seem to cover $\mathbb{R}$ over $\mathbb{Q}$, so there must be a more general one)

All I could find on the subject was

Howard, Paul; Tachtsis, Eleftherios, On vector spaces over specific fields without choice, Math. Log. Q. 59, No. 3, 128-146 (2013). ZBL1278.03082.

but it doesn't seem to answer my question (or maybe I missed something).

答:
To my knowledge there is no such condition. Certainly not at the generality you're asking for.

The closest we can get is "infinite dimensional Banach space" (well, to be honest we can go quite a lot more, but I don't remember the exact limitation there). The reason being that under certain assumptions (most notably $\sf ZF+DC+BP/LM$, where $\sf BP$ states that every set of reals has the Baire property and $\sf LM$ states that every set of reals is Lebesgue measurable) every linear operator from a complete Polish group into a normed group is continuous, and from that we can deduce the result for any Banach space. And of course, if there is an infinite dimensional vector space, we can arrange for a discontinuous functional.

You might want to think that if $K^\Bbb N$, which might not have a basis, embeds into a vector space then that space won't have a basis either. Unfortunately it is possible for a vector space to have a basis, while some of its subspaces don't have a basis. An example can be found in If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?.

In general, these sort of questions are very hard since we don't have precise enough tools to deal with arbitrary vector spaces over arbitrary fields. Hopefully with recent progress in the field of symmetric extensions we can try and improve the situation there, but frankly there is a long long way to go.


问:
Universal quadratic forms
A quadratic form is a polynomial $p(x_1,\dots,x_n)$ of the form $$ p(x_1,\dots,x_n)=\sum_{i \leq j}a_{ij}x_ix_j. $$
For example, $p_1(x,y,z,w)=x^2+y^2+z^2+w^2$ and $3x^2-5y^2$ are quadratic forms. I'm not sure this is relevant here, but for any such $p$ there is a symmetric matrix $A$ such that $p(\vec x)= (\vec x)^T A(\vec x)$, where we think of $\vec x$ as a column vector.

$p$ is positive definite iff $p(\vec x)>0$ whenever $\vec x\ne 0$.

The form is integer valued iff $p(\vec a)\in{\mathbb Z}$ whenever $\vec a\in {\mathbb Z}^n$.

For short, let's say "form" instead of "positive definite integer valued quadratic form".
A form is universal iff it represents every positive integer, i.e., for all $m>0$ there is $\vec a\in{\mathbb Z}^n$ with $p(\vec a)=m$. For example, $p_1$ above is universal.

I read that no form in 3 or fewer variables is universal, and I was hoping for a proof (or at least a reference if the proof is too long/convoluted for it to be written here). I was also wondering whether the requirement of positive-definiteness is relevant here, and whether this is only for forms $p$ where the matrix $A$ above has integer entries.

答: John Conway gives a proof (that no form in 3 or fewer variables is universal) in 2 paragraphs on page 142 of his book, The Sensual Quadratic Form. But the two paragraphs depend on $p$-adic squareclasses and other bits of theory developed elsewhere in the book. </p>

Title: Universal quadratic forms


问:
Prove or disprove: $\sum\limits_{i=1}^n i^2 = O(n^2) $
Prove or disprove:

$$\sum_{i=1}^n i^2 = O(n^2) $$

If we want to prove this, find some summation that we know the $ O(n)$ runtime for, and is $ O(n^2) $ or smaller.

Otherwise, we could disprove this by finding some summation that is less than this one, but has a $O(n)$ runtime that is greater than $ O(n^2)$.

No idea where to go from here though. Any assistance is appreciated.

答:
Did you know that $$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6\text{ ? }$$

More generally, $$f_p(n)=\sum_{k=1}^n k^p$$ is a polynomial of degree $p+1$, with leading term $$\frac{n^{p+1}}{p+1}$$ You can prove this by induction.


答:
$$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
Consider it in terms of a quadratic in $a$
$$a^2+a(-2x^2-1)+x^4-x=0$$
Now we see that $x^4-x=x(x^3-1)=x(x-1)(x^2+x+1)=(x^2-x)(x^2+x+1)$. And $(x^2+x+1)+(x^2-x)=2x^2+1$. By Vieta, its roots are $x^2-x$ and $x^2+x+1$. You could also do it using quadratic formula.

Title: Solving $\sqrt{a-\sqrt{a+x}} = x$

Body:
Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$

My approach:

Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.


问:
Finite sets and cardinality.
Let $$ I_N = \{n \in \mathbb N \mid 1 \leq n \leq N \}=\{1,2,3, \ldots, N \}.$$I want to show that $I_n \times I_m \sim  I_{nm}$.  For that, I defined $ \psi: I_n \times I_m \to I_ {nm} $ by $$ \psi (i, j) = i + n (j-1) $$ for all $ (i, j) \in I_n \times I_m $.  It is trivially apparent that $ \ psi $ is well defined.

Now let's show that $ \psi $ is injective, for that let's suppose that $ \psi (i, j) = \psi (i ', j') \Rightarrow i + n (j-1) = i '+ n (j'  -1) \Rightarrow n (j-j ') = i-i' $.  Since $ 1 \leq i, i '\leq n \Rightarrow | i-i' | <n \Rightarrow n (j-j ') $ is an element of the set $ \{- n + 1, -n + 2, \ldots, n-2, n-1 \} $.  On the other hand, $ n (j-j ') $ is a multiple of $ n $, that is, an element of the set of numbers $ \{0, \pm n, \pm 2n, \ldots \}. $ The only one of  these numbers that satisfy that $ | i-i '| <n $ is $ 0 $.  Which implies that $ i'-i = n (j-j ') = 0 $ and from there it is easily obtained that $ i = i' $ and $ j = j '$.  Therefore $ \psi $ is injective.

To show that it is surjective.  Let $ k \in I_{nm} $ and write $ k-1 = r + ns $ with $ r, s \in \mathbb N $ and $ 0 \leq r <n $.  So for $ i = r + 1, j = s + 1 $ we have $ k = i + n (j-1) $, and also $ 1 \leq i \leq n \Rightarrow i \in I_n $.  Then $ s \geq 0 \Rightarrow j = s + 1 \geq 1 $ and $ i + n (j-1) = k \leq nm \Rightarrow n (j-1) \leq nm-i \leq nm-  1 \Rightarrow j-1 \leq m-1 / n \Rightarrow j-1 \leq m-1 \Rightarrow j \leq m \Rightarrow j \in I_m \Rightarrow (i, j) \in I_n \times I_m $.

That way $ \psi $ is bijective.  Knowing that, someone can tell me if that is correct and can explain to me how to get the inverse of said function.

答:
Firstly, you should write your function as $ \psi \left(\ (i, j)\ \right) = i + n (j-1) $, not $ \psi (i, j) = i + n (j-1) $.

Next, I think the first part of your proof is actually quite good. The second part is honestly hard for me to parse, so I'll let someone else have that job. Obviously your function is bijective.

The inverse function then is surely $ \psi^{-1} :\ k \to \left(\ k\ \%\ n,  \left(k-1\right)/n\ + 1 \right), $ where $/$ means the quotient (e.g. $7 / 3 = 2$ ), and $\%$ means the remainder under division e.g. $7\ \%\ 3 = 1.$

In fact, it is more natural for me to prove that $\gamma: I_{nm}\to I_n \times I_m$ defined by: $$\gamma(k) = \left(\ k\ \%\ n,  \left(k-1\right)/n\ + 1 \right)$$

is bijective [which is fine since you should already know that  $I_{nm} \sim I_n \times I_m \iff I_n \times I_m \sim  I_{nm}].$

Proof that $\gamma$ is injective: Let $(a,b) \in I_n \times I_m.$ Suppose $\exists k \in I_{n,m},k' \in I_{n,m}$ such that $\gamma(k) = (a,b)$ and $\gamma(k') = (a,b).$ This implies: $$(k\ \%n, (k-1)/n + 1) = (a,b) = (k'\%n, (k'-1)/n + 1) \implies k\ \%n = k'\ \%n$$ $$\implies \exists r \in \mathbb{N} \text{ such that } k = k' + rn \quad (1), \text{ and also}$$ $$\implies (k-1)/n = (k'-1)/n$$ $$ \implies |k-k'| \leq n-1 \quad (2).$$

$(1)$ and $(2)$ together imply that $r = 0,$ and so $(1)$ reduces to $k=k',$ and we have injectivity.

Proof that $\gamma$ is surjective:

Let $(c,d) \in I_n \times I_m,\ $ and define $\ C = \{c + (e-1)n: e \in \mathbb{N} \}.$

Then there are only $n$ consecutive integers $k_i$ such that $(k-1)/n = d-1,\ $ namely

$$k_i \in D = \{ n(d-1) + 1, n(d-1) + 2, ..., n(d-1) + n = nd \}.$$

Clearly $D \cap C$ has only one element, $q\ $ say. Moreover, $q \in I_{nm}$ because $q \in D$ and $\max (D) = nm$, and $\min (D) = 1.$ Finally, $q$ satisfies:

$$\gamma(q) = \left(c,d\right).$$

This shows that $\gamma$ is surjective.

Title: Finite sets and cardinality.


Title: Solving $\sqrt{a-\sqrt{a+x}} = x$

Body:
Let $\sqrt{a+x}=y\ge0\implies x=y^2-a$

$$y^2-a=\sqrt{a-y}\ge0\implies  y^2\ge a$$

$$(y^2-a)^2=a-y\iff a^2-(2y^2+1)a+y^4-y=0$$

$$a=\dfrac{2y^2+1\pm\sqrt{(2y^2+1)^2-4(y^4-y)}}2=\dfrac{2y^2+1\pm(2y+1)}2$$


问:
Non-linear optimization: Levenberg-Marquardt gives different results using different forms of the same function
Background:

I'm using non-linear least squares to solve indoor trilateration problem using BLE (bluetooth low energy) beacons. My starting point is this: https://nrr.mit.edu/sites/default/files/documents/Lab_11_Localization.html B part of the document.

(I've been using this JAVA lib for trilateration ( https://github.com/lemmingapex/Trilateration ) which works fine but needed to switch to c++ and I couldn't find a similar lib. So I experimented with google's ceres solver and eigen's levenberg-marquardt lib)

Problem/question:

$(x*,y*) = argmin \sum_{i=0}^n [ri^2-(x-xi)^2 - (y-yi)^2$

For these non-linear optimization libs, you have to define the nonsquared function.

Using this function for L-M algorithm:
$\sqrt{(x-xi)^2 + (y-yi)^2} - ri$

c++ code:

```<code>fvec[i] = sqrt(pow(matrix(i, 0) - b[0], 2) + pow(matrix(i, 1) - b[1], 2)) - matrix(i, 2);
</code>```

I get different results when using this:
$1 - \frac{\sqrt{(x-xi)^2 + (y-yi)^2}}{ri}$

c++ code:

```<code>fvec[i] = 1 - sqrt(pow(matrix(i, 0) - b[0], 2) + pow(matrix(i, 1) - b[1], 2)) / matrix(i, 2);
</code>```

The latter gives much better results. Shouldn't this function defined in different forms give the exact same result when using L-M algorithm to find minimum? What am I missing?

By the way for predefined problems (in the JAVA github lib tests) both give correct results, but the considerable difference happens when I'm testing with reallife data (many beacons).

Notes:

First defined function (fvec) returns RelativeReductionTooSmall and second (which gives good results) returns with RelativeErrorTooSmall. So I guess the stopping condition may not be set properly. But this shouldn't have an effect on my problem.

I'm new to optimization. I understand how gradient-descent works, but didn't have the time to dig deeper into L-M algorithm. I know it "balances"/switches between gradient descent and Newton-Gauss approach. But my gut says I'm missing something more basic.

I'm treating this function as an eqution (hence the above mentioned 2 forms):
$0 = 1 - \frac{\sqrt{(x-xi)^2 + (y-yi)^2}}{ri}$ but maybe I can't treat the left side just as 0? Why not?

答:
Q: "Shouldn't this function defined in different forms give the exact same result when using L-M algorithm to find minimum?"

A: Both forms of the function should work correctly. I tested both forms using java lib 1 2 and it's giving good results.
So theoretically my assumption was correct. The form of the function to be minimized doesn't really matter(*). The problem most likely lies in the c++ libs that I tried out (eigen, ceres). (Or I'm not using those correctly.)

*Both forms don't give the exact same results (only very close) but I couldn't find out the reasons for that.

Followup question on stackoverflow if anyone is interested: https://stackoverflow.com/questions/65599497/what-is-this-strange-difference-between-c-eigen-levenberg-marquardt-and-java-l


问:
Ex.1.1 9 from topology without tears

this question says only three of the below are topology.

So,according to me,$\tau_2$ is topology it is satisfying all three definition for topology.

And I believe $\tau$ consisting of $\mathbb{R}$,$\phi$ and  every interval $[-r,r]$,for $r$ any positive real number is also a topology.So $\tau_9$ is union of both $\tau$ and $\tau_2$ with $\tau_2 \subset \tau$ so is also a topology.

Similarly $\tau_0$ consisting of $\mathbb{R},\phi$ and every interval $[-n,n]$,for $n$ any positive integer is also topology.So $\tau_{10}$ is union of both the topology with one containing the other so is also a topology.

so i think $\tau_2$,$\tau_9$ and $\tau_{10}$ are the required topology,am i correct?

And I also want to know whether $\tau_7$ and $\tau_8$ are topology or not with justification.

答:
$\tau_1$ is not a topology since $(a,b)\cup(c,d) \notin \tau_1$ where $a<b<c<d$.

$\tau_2$ is a topology ($\checkmark$)  since  $$\bigcup_{i\in I}(-r_i,r_i) =\begin{cases} (-R,R) & I \text{ is finite} \land R = \max\{r_i \mid i \in I\} \\ \mathbb R & I \text{ is not finite} \end{cases}$$  and $$\bigcap_{i\in J}(-r_i,r_i) = (-r,r)$$ where $r = \min\{r_i \mid i \in J\}$ ($J$ is finite).

$\tau_3$ is not a topology since for some irrational $a$ we have
$$\bigcup_{n=1}^\infty(-r_n,r_n) = (-a,a)\notin \tau_3$$ where $\lim_{n\to\infty}r_n = a$.

$\tau_4$ is not a topology since $$\bigcup_{n=1}^\infty\left[-r_n+\frac 1n,r_n-\frac 1n\right] = (-r,r)\notin \tau_4$$

$\tau_5$ is not a topology for a similar reason as 3.

$\tau_6$ is not a topology for a similar reason as 4 (note that if $a$ is irrational then so is $a + \frac{1}{n}$ for any $n\in \mathbb N$).

$\tau_7$ is not a topology for a similar reason as 4.

$\tau_8$ is not a topology for a similar reason as 4.

$\tau_9$ is a topology ($\checkmark$) since one cannot use the "trick" that was used in 4. (try to show it rigorously)

$\tau_{10}$ is a topology ($\checkmark$) since one cannot use the "trick" that was used in 4. (try to show it rigorously)

Title: Ex.1.1 9 from topology without tears


问:
Is there an analytic function $f=u+iv$ defined on the unit disk, such that $u_x=0$ but $f$ is not constant?
Using the following theorem,

$$\text{If }f\text{ is analytic on }D,\text{ and if }f'(z)=0\text{ for all }z\in D,\text{then }f\text{ is constant in }D $$

with Cauchy-Riemann equations has consequences:

$$\text{If }f=u+iv\text{ is analytic on }D\text{ with }u\text{ being constant, then }f\text{ is constant in }D $$

Which give me answer with "There is no such function" But when I see the solution I get the following statement:

The answer is yes, there is such a function! Take for instance the function $f(z)=y-ix$ defined on the unit disk, where $z=x+iy$. The associated real part and imaginary part to $f$ are: $u(x,y)=y$ and $v(x,y)=-x$. So in particular, $u_x=0$.
Furthermore, $u_x=0=v_y$ and $u_y=-1=-v_x$ (checking that the Cauchy-Riemann equations hold), and those partial derivatives of $u$ and $v$ are constant, thus clearly continuous in the unit disk. Therefore, $f$ is analytic in the unit disk. However, $f$ is not constant on the unit disk.

I couldn't connect my logic with the theorem. The solution seems to contradict the theorem, isn't it?

答: Your theorem applies to functions $f(z)=u(z)+iv(z)$ with $u(z)$ constant. The solution concerns a function $f(z)=u(z)+iv(z)$ with $u(z)=y$, where $z=x+iy$ which is nonconstant. So the theorem does not apply.


问:
General form of an isotropic fourth rank tensor
According to [1], a representation of the most general isotropic tensor of rank 4 is

$$ T_{ijkl} = \lambda \delta_{ij}\delta_{kl} + \mu(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) + \nu(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) $$

How is it derived?

I'm trying to understand the Navier-Stokes equation [2] and is crucial to understand that tensor representation.

答:
A general derivation can be found in 'On Isotropic Cartesian Tensors' by Hodge in 1961 or here which is based on the mentioned reference. Another 'derivation' based on the individual components is here

Since the derivation is well documented I only outline the necessary steps.
Below we follow the one given by Hodge.

You start from the definition of isotropic tensors and require invariance under infinitesimal rotations to arrive at the condition
$$
\epsilon_{mis}a_{sjkl}
+ \epsilon_{mjs}a_{iskl}
+ \epsilon_{mks}a_{ijsl}
+ \epsilon_{mls}a_{ijks} = 0
$$
Which you multiply by $\epsilon_{mit}, \epsilon_{mjt}, \epsilon_{mkt}, \epsilon_{mlt}$ and set $t=i, t=j, t=k, t=l$ respectively to obtain 4 equations.
This equations you 'simplify' by realizing that the 4th order isotropic tensors with two internal indices contracted are actually 2nd order isotropic tensors, which are proportional to the Kronecker delta, i.e. you can replace terms like $a_{iikl} = \lambda \delta_{kl}$. This results in equations like
$$
2 a_{ijkl} + a_{jikl} + a_{kjil} + a_{ljki} =
\lambda\delta_{ij}\delta{kl}
+ \mu\delta_{ik}\delta_{jl}
+ \nu\delta_{il}\delta_{jk} \\
\vdots
$$

From this 4 equations you can deduce relations like
$$
a_{ijkl} = a_{jilk} = a_{klij} = \ldots
$$
by subtracting the summed pairs from each other.
Using the relations you found in this way you can rewrite the former 4 equations to yield
$$
2 a_{ijkl} + a_{ijlk} + a_{ikjl} + a_{ilkj} =
\lambda\delta_{ij}\delta{kl}
+ \mu\delta_{ik}\delta_{jl}
+ \nu\delta_{il}\delta_{jk} \,.
$$
Here you obtain two more equations via cyclic permutation of $j,k,l$ while leaving $i$ untouched.
The resulting 3 equations are then summed followed by a symmetry argument to give
$$
a_{ijkl} + a_{iklj} + a_{iljk}
= a_{ijkl} + a_{ikjl} + a_{ilkj}
= \frac{1}{5} (\lambda + \mu + \nu) (
\delta_{ij}\delta{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}
)\,.
$$
Substituting this int he equation before while defining
$$
    \alpha = \frac{4 \lambda - \mu     - \nu}{10}, \qquad \qquad
    \beta  = \frac{4 \mu     - \lambda - \nu}{10}, \qquad \qquad
    \gamma = \frac{4 \nu     - \lambda - \mu}{10}
$$
You get the general form of a isotropic tensor of 4th rank
$$
a_{ijkl} = \alpha \delta_{ij}\delta_{kl}
+  \beta \delta_{ik}\delta_{jl}
+  \gamma \delta_{il}\delta_{jk}
$$
You can easily find a relation that yields the form you stated with the one given here.


问:
Homotoping non zero degree arbitrary maps to a covering map
Let $f:M\rightarrow N$ be a non zero degree smooth map between compact connected oriented manifolds of the same dimension. Must  $f$ be homotopic to a covering map ?

This seems too good to be true, So I m expecting counterexamples

答: For every $d \in \mathbb{Z}$, there is a map $f_d : S^2 \to S^2$ with $\deg f_d = d$. On the other hand, as $S^2$ is simply connected, the only connected covering is $\operatorname{id} : S^2 \to S^2$ which has degree $1$. As degree is a homotopy invariant, we see that the claim is false.


问:
Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$?
Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$? (EDITED: Both of these sets have anti lexicographic order, that is $(a,b)<(c,d)⟺b<d$ or $(b=d$ and $a<c)$).

Definition of similarity: Let $(A, <)$ and $(B,≺)$ be partially ordered sets. Then $f:A\rightarrow B$ is a similarity between $A$ and $B$ if $f$ is bijective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$ and $\forall m,n \in B, m≺n \Rightarrow f^{-1}(m)<f^{-1}(n)$. We write $A \simeq B$ if $A$ and $B$ are similar.

We usually check if the sets are (not) similar by checking for possible similarity invariants. (Let $A,B$ be posets. Property P is similarity invariant if $A \simeq B \Rightarrow (P(A) \iff P(B))$. That is, if $P(A)$ and $\lnot P(B)$, then $A \not \simeq B$.)

I have tried to find an invariant that one set has and the other doesn't, but haven't been able to find it. For example, they are both not separable (they both don't have a countable subset that's dense in them), they both aren't complete (there exists a nonempty subset that is bounded from above but doesn't have a supremum ( $\Bbb Z \times \{0\}$ is example for both), they both aren't locally finite, they are both unbounded on both sides, they both aren't dense in themselves.

On other hand, I can't find a similarity between them but I also don't know how to prove that there is no similarity between them.

(Also, if $(A, <)$ and $(B,≺)$ are both totally ordered (and in my case, they are), then $f:A\rightarrow B$ is similarity between $A$ and $B$ if $f$ is surjective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$.)

答:
First some terminology.

What you call a 'similarity' is usually called an isomorphism. A slightly easier way of formulating that property is $f: A \to B$ is an isomorphism if $f$ is bijective and for all $x, y \in A$ we have $x < y \Leftrightarrow f(x) < f(y)$.

A useful invariant will be the following.

An order $(L, <)$ is called a dense linear order if for any $x < y$ in $L$ there is $z \in L$ such that $x < z < y$.

A useful fact about dense linear orders is that every interval in them is again a dense linear order. More precisely, if $L$ is a dense linear order then for any $a < b$ in $L$ we have that $\{x \in L : a < x < b \}$ is a dense linear order.

In $\mathbb{Z} \times \mathbb{Q}$ there is a copy of $\mathbb{Q}$, namely $Q = \{(0, x) : x \in \mathbb{Q}\}$. Clearly this is a dense linear order.

I claim that there is no subset of $A \subseteq \mathbb{Z} \times \mathbb{N}$ that is a dense linear order. Suppose for a contradiction that there is such an $A$. Then for some $n \in \mathbb{N}$ we must have that $A \cap (\mathbb{Z} \times \{n\})$ contains at least two elements $a$ and $b$, because otherwise $A$ would be isomorphic to $\mathbb{N}$. Without loss of generality we assume $a < b$. The interval $\{ x \in A : a < x < b \}$ can then not be a dense linear order because it is a subset of (something isomorphic to) $\mathbb{Z}$, so we find our contradiction.

I will also leave my old answer, which was assuming that we order the sets with the usual lexicographical ordering (instead of the anti lexicographical ordering).

The order $\mathbb{Z} \times \mathbb{Q}$ a dense linear order, but $\mathbb{Z} \times \mathbb{N}$ is not. For example, there is no element strictly between $(0, 5)$ and $(0, 6)$ in $\mathbb{Z} \times \mathbb{N}$.


问:
If a die is thrown $5$ times, what are the odds of getting a $5$ AT LEAST $3$ times
The probability of getting a $5$ three times is $0.03215$ but as we are looking for a $5$ to be rolled at least $3$ times, does that mean we add up the probabilities for getting a $5$ four and five times as well?

答:
If $X$ is the number of 5s, then $X\sim B\left(5,\frac{1}{6}\right)$. The probability of getting at least 3 fives is $$\mathbb P(X=3)+\mathbb P(X=4)+\mathbb P(X=5)$$
meaning that you have to sum the probabilities for getting 3, 4 and 5 fives.


问:
Seating four people next to each other probability problem
The following problem is from Carol Ash's The Probability Tutoring Book

If $4$ people are assigned seats at random in a $7$ seat row, what is the probability that they are seated together?

My approach:

We can choose $4$ from $7$ seats: $_{7}C_4$ ways,

We can permute the people in $4!$ ways.

So, my answer was $\frac{4!}{_{7}C_4}$

However the answer in the book is $\frac{4}{_{7}C_4}$. did they miss the factorial? Or is this the correct answer?

Why is the numerator just $4$?

Is it because if I permute the numerator, I also have to permute the denominator? Can someone explain why this is logically so?

答:
You can notice that all 4 people must be seated in a row, so all the possible outcomes are : the seats from $1$ to $4$, from $2$ to $5$, from $3$ to $6$, and from $ 4$ to $7$.

Hence, there are 4 possible configuration, then they can be permuted $_{7}C_4$ ways, hence the total result is :

$$\frac{4}{_{7}C_4} $$


问:
MLE for $p$ in Geometric distribution from Exponential distribution (two methods, two results)
Let $Y_n$ given as $\mathrm{ceil}(X_n)$, where $\mathrm{ceil}(x):=$ the least integer greater than or equal to $x$ and $(X_n)$ is a sequence of iid random variables from $\mathrm{Exp}(\theta),~\theta>0$.

Then $Y_n\sim Geo(p)$, where $p=p(\theta):=1-e^{-\theta}$ and since the maximum likelihood estimator (mle) for $\theta$ is given as $\frac{1}{\overline{X}}$, the mle for $p(\theta)$ is
$1-e^{-\frac{1}{\overline{X}}}$.

If we compute directly the mle for $p$
using $\mathbb{P}(Y_1=y)=(1-p)^{y-1}p$ for $y\in \mathbb{N}$, we get that the maximum likelihood estimator for $p$ is given as $\frac{1}{\overline{Y}}=\frac{1}{\overline{\mathrm{ceil}(X)}} $, which is not the same as the previous result.

Is there some contradiction in these two results, or some fallacy?

Thank for the help.

答:
As Math-fun says, you are in effect using two different sets of information, one unrounded and the other rounded up, so you should not expect the same result.

For example,

if you see the data $0.1, 0.2, 0.3, 1.1, 2.1$ then $1-e^{-1/\overline{x}} \approx 0.732$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$.

if you see the data $0.7, 0.8, 0.9, 1.9, 2.9$ then $1-e^{-1/\overline{x}} \approx 0.501$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$ again since you round up to the same integers.

Here is some R code to simulate a thousand samples of size $10$.  From the chart below you can reasonably conclude:

the two maximum likelihood estimators for $p$ give slightly different estimates in practice, largely due to the particular sample data effects on the two expressions

the estimators are usually closer to each other than they are to the actual parameter used in the simulation ($0.3$ in this case)

the estimator using the ceiling function will only give discrete values; in retrospect this is obvious looking at the formula, as it can only be the sample size divided by an integer

.

```<code>set.seed(2021)
n <- 10
p <- 0.3
theta <- -log(1-p)
Xdat <- matrix(rexp(n*10^3, rate=theta), ncol=n)
MLE_p_exp <- 1 - exp(-1/rowMeans(Xdat))
MLE_p_geo <- 1 / rowMeans(ceiling(Xdat))
plot(MLE_p_exp, MLE_p_geo, xlim=c(0,0.8), ylim=c(0,0.8), pch=3)
abline(0,1, col="red")
</code>```

Body:
As Math-fun says, you are in effect using two different sets of information, one unrounded and the other rounded up, so you should not expect the same result.

For example,

if you see the data $0.1, 0.2, 0.3, 1.1, 2.1$ then $1-e^{-1/\overline{x}} \approx 0.732$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$.

if you see the data $0.7, 0.8, 0.9, 1.9, 2.9$ then $1-e^{-1/\overline{x}} \approx 0.501$ while $\frac{1}{\overline{\lceil x \rceil}}=\frac{5}{8}= 0.625$ again since you round up to the same integers.

Here is some R code to simulate a thousand samples of size $10$.  From the chart below you can reasonably conclude:

the two maximum likelihood estimators for $p$ give slightly different estimates in practice, largely due to the particular sample data effects on the two expressions

the estimators are usually closer to each other than they are to the actual parameter used in the simulation ($0.3$ in this case)

the estimator using the ceiling function will only give discrete values; in retrospect this is obvious looking at the formula, as it can only be the sample size divided by an integer

.

set.seed(2021)
n <- 10
p <- 0.3
theta <- -log(1-p)
Xdat <- matrix(rexp(n*10^3, rate=theta), ncol=n)
MLE_p_exp <- 1 - exp(-1/rowMeans(Xdat))
MLE_p_geo <- 1 / rowMeans(ceiling(Xdat))
plot(MLE_p_exp, MLE_p_geo, xlim=c(0,0.8), ylim=c(0,0.8), pch=3)
abline(0,1, col="red")


问:
First time passage decomposition for continuous time Markov chain
For discrete time finite Markov chain, the first passage time $T_j$ to visit state $j$, is determined from the recurrence equation:
$$
   p^{(n)}_{ij} = \sum_{k=0}^n f_{ij}^{(k)} p^{(n-k)}_{jj} =
      \sum_{k=0}^n \mathbb{P}(T_j = k ; X_0 = i) \mathbb{P}(X_{n-k} = j ; X_0 = j)
$$
where $p^{(n)}_{ij} = \mathbb{P}(X_n = j ; X_0 = i)$ and $f^{(k)}_{ij} = \mathbb{P}(T_j = k ; X_0 = i)$. This recurrence equation allows to find probability generating function for the first passage time distribution (exerices 1.5.3 of J.R. Norris's book on "Markov Chains", relevant chapter 1.5 is available from Norris's website).

It strikes me odd that no book I have seen discusses first passage time distribution for continuous time finite Markov chain.

Can anyone suggest a reference where this is discussed/worked-out, or hint me at how to do this for myself.

I suspect a similar integral equation would be set-up, and moment generating function for the first-passage-time distribution would satisfy some simple equation.

Thanks for reading.

答:
Concerning references, I've seen the problem of first passage time distributions treated in two books, unfortunately both are in German:

Nollau, V.: Semi-Markovsche Prozesse, H. Deutsch (1981).

Störmer, H.: Semi-Markoff-Prozesse mit endlich vielen Zuständen, Springer Lecture Notes in Operations Research and Math. Systems, Vol. 34 (1970).

As the titles imply, these references discuss the problem for semi-Markov processes, which of course includes the continuous time Markov chain case. For example, Nollau treats a more general problem: he provides a Markov renewal type equation, similar to the one posted by Ilya, for the probability $\Psi_{x,ABC}(t)$ of a transition from $A$ to $B$ not later than $t$, without leaving $C$, where $A$, $B$ and $C$ are subsets of states, $A\subseteq C$, and $x\in C$ is the initial state.


问:
Prime ideal and ring of integers
I am reading an introduction into algebraic number theory and struggling on the following sentence:

Let $M/F$ be an extension of number fields, and $\mathfrak p$ an prime ideal of $\mathcal{O}_F$. Then $\mathcal{O}_M \mathfrak p$ is an ideal in $\mathcal{O}_M$ which is no longer a prime ideal in general.

My question is: what does the notation $\mathcal{O}_M\mathfrak p$ mean? I think it means the product of the two ideals, but then why is $\mathfrak p$ also an ideal of $\mathcal{O}_M$? Why is $\mathcal{O}_M\mathfrak p$ even an ideal of $\mathcal{O}_M$?

Any help is appreciated.

答:
Question:"Why is $O_Mp$ even an ideal of $O_M$?"

Answer: If $\phi: A\rightarrow B$ is a map of commutative unital rings and $I$ is an ideal in $A$, you define $IB$ (or $BI$) to be the following ideal:

I1. $IB:=\{\sum_i b_ix_i$ with $b_i \in B, x_i:=\phi(y_i)$ with $y_i\in A\}.$

You may check that $IB\subseteq B$ is an abelian group with the property that  for
all $x\in IB, b\in B$ it follows $bx\in IB$. Hence $IB$ is an ideal in $B$.

As mentioned above: The standard example of a ring extension where the extension of a prime ideal is no longer a prime ideal is the extension

$\phi: A:=\mathbb{Z} \subseteq \mathbb{Z}[i]:=B$

let $I:=(2) \subseteq A$. The ideal $I$ is prime in $A$ but in $B$ there is a factorization $2=(1-i)(1+i)$, hence the ideal $IB$ equals $(1-i)(1+i)B$ which is no longer prime. If $p\neq 2$ is a prime with $p\equiv 1 (mod 4)$ there is an equality

$p=a^2+b^2=(a-ib)(a+ib)$ in $B$ for two integers $a,b\geq 1$

Hence the prime ideal $(p)B$ is no longer prime in $B$. This is Theorem 1.1.1 in Neukirch's book "Algebraic number theory". In Theorem 1.4 in the same book the prime elements in $B$ are classified, and a prime number $p \in A$
remains prime in $B$ iff $p\equiv 3 (mod 4)$. If $p \equiv 1(mod 4)$ with $p=a^2+b^2$ it follows the elements $a+ib, a-ib$ are prime elements in $B$,
hence

I1. $(p)B=(a+ib)(a-ib)B$ gives rise to the decomposition

I2. $(p)B=\mathfrak{p}_1\mathfrak{p}_2$

with $\mathfrak{p}_1=(a+ib)$ and $\mathfrak{p}_2=(a-ib)$. Here $\mathfrak{p}_i$ are prime ideals in $B$.

You get a canonical map of rings

I3. $A/(p) \rightarrow B/(p)B \cong B/\mathfrak{p}_1\mathfrak{p}_2 \cong B/\mathfrak{p}_1 \oplus B/\mathfrak{p}_2$

and since $\mathfrak{p}_i$ are non-zero prime ideals it follows they are maximal ideals. Hence the ring $B/(p)B$ is a direct sum of the fields $B/\mathfrak{p}_i$. Since $A/(p):=\mathbb{F}_p$ is the finite field with $p$
elements it follows there is an isomorphism

I4. $B/\mathfrak{p}_i \cong \mathbb{F}_{p^{r_i}}$

for some integer  $r_i \geq 2$. You may calculate the integer $r_i$ explicitly.

Title: Prime ideal and ring of integers


答:
Here is a graph representation of your problem

The result is the sum of the 3 yellow bars (the latest 2 are very low...but they exist)

Same solution if you calculate 1 minus the burgundy bars

Title: If a die is thrown $5$ times, what are the odds of getting a $5$ AT LEAST $3$ times

Body: The probability of getting a $5$ three times is $0.03215$ but as we are looking for a $5$ to be rolled at least $3$ times, does that mean we add up the probabilities for getting a $5$ four and five times as well?


问:
Show that $\int^b_a f^3(x)dx \le ( \int^b_a f(x)dx )^2$ when $f(a)=0$ and $0 \le f'(x) \le 1$
Assume that $f:[a,b] \to \mathbb R$ is a continuously differentiable function. We want to show that $\int^b_a f^3(x)dx \le ( \int^b_a f(x)dx )^2$ if we have $f(a)=0$ and $0 \le f'(x) \le 1$ for all $x \in [a,b]$.

My attempt:

I don't know if my approach is correct, but I started by changing the upper limits of the integrals to a variable $y$ so that I can differentiate and obtain $f'$. After changing the upper limits, I want to show

$$( \int^y_a f(x)dx )^2 - \int^y_a f^3(x)dx \ge 0$$

for $y \in [a,b]$. This holds for $y=a$. So I will now show that the derivative of LHS w.r.t. $y$ is positive. That is, I want to show

$$2f(y)(\int^y_a f(x)dx) - f^3(y) \ge 0$$

Again, this holds for $y=a$ since $f(a)=0$. And I take the derivative again and want to show that it is positive. The derivative is

$$2f'(y)(\int^y_a f(x)dx) +2f^2(y)-3f^2(y)f'(y)$$

I couldn't get rid of the integral and I am stuck here. Can you help me by proposing another approach or giving a solution?

答:
Let $$F(x)=\left(\int_{a}^{x}f(t)dt\right)^2-\int_{a}^{x}f^{3}(t)dt$$
We have $F(a)=0$,and$$F^{\prime}(x)=2f(x)\int_{a}^{x}f(t)dt-f^{3}(x)=2f(x)\left(\int_{a}^{x}f(t)dt-f^{2}(x)\right)=f(x)G(x)$$
$$f^{\prime}(x)\geqslant0\implies f(x)\geqslant f(a)=0$$
among $G(a)=0,G^{\prime}(x)=2f(x)(1-f^{\prime}(x))\geqslant0$.
Hence$$F(x)\nearrow\implies F(x)\geqslant F(a)\implies F(b)\geqslant0$$


问:
Murray V.N equivalence of projections in finite C* algebras.
In general, Murray von Neumann equivalence of projections is not the same as unitary equivalence. In fact in a unital infinite C* algebra 1 is Murray von Neumann equivalent to a proper subprojection p and so these are going to be murray von Neumann equivalent but certainly not unitarily equivalent. I was wondering if in a finite C* algebra these equivalence relations are the same? I certainly can't find a counterexample but I struggle to show in general that if p Murray von Neumann equivalent to q the 1-p is Murray von Neumann equivalent to 1-q when the C* algebra is finite.</p>

答:
Let $\mathbb K$ be either the field of real or complex numbers.
A $\mathbb K$-vector bundle $\eta $ over a compact Hausdorff space  $X$ is said to be stably trivial if there
are  integers $k, n\geq 0$, such that
$$
  \xi _k\oplus \eta  = \xi _n,
  \tag 1
  $$
where
$\xi _k$ and $\xi _n$ are the trivial bundles of rank $k$ and $n$, respectively.

In case $\mathbb K = \mathbb R$, an example is  the tangent bundle of the sphere $S^2$ which is not
trivial by the hairy sphere Theorem,  but is stably trivial since its direct sum with the (trivial) normal bundle is the
trivial bundle of rank 3.

Assuming we are in the situation of (1),  consider the algebra $A=M_n\big (C_{\mathbb K}(X)\big )$.  Then we may find
pairwise orthogonal
projections $p,q\in A$, corresponding to $\xi _k$ and $\eta $, respectively, such that $p+q=1_n$, where $1_n$  is the identity of $A$.

One then has that $p$ is Murray-von Neumann equivalent to $1_k$, but $1_n-p$ is not Murray-von Neumann equivalent to
$1_n-1_k=1_{n-k}$ because $1_n-p=q$, and $\eta $ is not the trivial bundle.


问:
Solution verification (volume)
In the following example I am getting that volume of body bounded by surfaces $x^2+y^2=8, x^2+y^2+4x=0, z=10-x^2-y^2, x+2z=3$ is negative, so I want to know where am I mistaking.
First those two cylinders will give me bounds for $x,y$. Since first one has center in $(0,0)$ and radius
$\sqrt{8}$ and second center in $(-2,0)$, radius $2$.
Now, after using cylindrical coordinates, first circle gives $r=2\sqrt{2}$ and the second $r=-4\cos\phi$. And then $-4\cos\phi\le r\le \sqrt{8}$ , $\frac{\pi}{2}\le \phi \le \frac{3\pi}{2}$(since $\cos\phi$ needs to be negative). And the I get volume $\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}d\phi\int\limits_{-4\cos\phi}^{\sqrt{8}} (10-r^2-\frac{3r\cos\phi}{2})rdr$

答:
For the intersection of both cylinders, $-4 \cos \phi = 2 \sqrt 2 \implies \phi = - \frac{1}{\sqrt2}$ and you get $\phi = \frac{3 \pi}{4}, \frac{5 \pi}{4}$. Now for $\frac{3 \pi}{4} \leq \phi \leq \frac{5\pi}{4}, \,$ you are bound by $r = 2\sqrt2$ and for $\frac{\pi}{2} \leq \phi \leq \frac{3\pi}{4}, \, \frac{5\pi}{4} \leq \phi \leq \frac{3\pi}{2}$, $r = -4 \cos \phi$.

If you go by this, the bound of $z \,$ will be $\, 10-r^2 \leq z \leq \frac{3-r\cos \theta}{2}$ but this is not correct. For example, if you check for $r = \sqrt 8$ and $\theta = \pi$, you realize your lower and upper bounds are not correct.

So to find the right limits, let's find intersection points.

i) Intersection of plane and paraboloid for $x \lt 0$,

$z = 10 - x^2 - y^2 = \frac{3-x}{2} \implies y = \pm \sqrt {\frac{17+x}{2} - x^2}$. Also substituting $y = 0, $ we get $\frac{1 - \sqrt{137}}{4} \leq x$.

ii) Common intersection of paraboloid and cylinder with plane,

$y^2 = 10 - z - x^2 = -4x-x^2 \implies z = 4x + 10 = \frac{3-x}{2}, x = -\frac{17}{9}, $

So the integral to find volume $(V = V_1 + V_2)$ is as below -

$V_1 = \displaystyle 2 \, \int_{(1 - \sqrt{137}) / 4}^{-17/9} \int_0^{\sqrt{\frac{x+17}{2} - x^2}} \int_{(3-x)/2}^{10-x^2-y^2} \, dz \, dy \, dx$

$V_2 = \displaystyle 2 \, \int_{-17/9}^0 \int_0^{\sqrt{-4x-x^2}} \int_{(3-x)/2}^{10-x^2-y^2} \, dz \, dy \, dx$

$\,$

Some additional details if you require -

To visualize how the plane cuts these solids, refer to the above diagram. We need to keep the cylinder $x^2 + y^2 = 8$ aside for now. I will come to that later.

At any given $z,\,$ think of the circle centered at the origin as cross section of the paraboloid. If you start at $z = 10, \,$ the plane cuts the cylinder $x^2 + y^2 + 4x = 0$ outside the paraboloid. Then as $z$ reduces, the radius of the paraboloid increases and so does the area of intersection of the cylinder and paraboloid. At some point the plane cuts the paraboloid within the intersection that you see in the diagram. Then as you reduce $z$ further, the plane cuts the cylinder within the intersection that you see in the diagram.

Now coming to the cylinder $x^2+y^2 = 8,$ which is only relevant for $z \leq 2$ as that is when the radius of the paraboloid is $ \geq \sqrt 8$ but for $z \leq 2, $ the plane cuts the other cylinder to the right of the intersection of both cylinders and so if we find that bound, we will automatically be within the intersection of both cylinders.

Title: Solution verification (volume)


问:
Endomorphism ring of supersingular elliptic curve
Let $E$ be a supersingular elliptic curve over $F_q$ where $q=p^n$, then $\operatorname{End}(E)$ is an order in quaterion algebra, hence a non-commutative ring.

Question: Is there an endomorphism $\psi$ of $E$ such that $\psi$ doesn't commute with Frobenius morphism $\phi_q: x\rightarrow x^q$?

Thanks in advance.

答:
Iff $\phi_q\not \in \Bbb{Z}$

(The largest subfields of $End(E)\otimes_\Bbb{Z}\Bbb{Q}$ are $2$-dimensional $\Bbb{Q}$-vector spaces and it is a division ring so any element not in $\Bbb{Q}[\phi_q]$ doesn't commute with $\phi_q$)

Try with $E/\Bbb{F}_9:y^2=x^3+x$ then $\phi_9=-3$ (and $\phi_3\approx \sqrt{-3}$ doesn't commute with $i(x,y)=(-x,iy)$)


问:
What is the correlation and covariance of a function of random variables?
If I have that $X$ and $Y$ are independent r.v with mean 0 and variance $\sigma^2$ and I have that

$Z=X\sin\alpha + Y\cos\alpha \quad$, what is then the covariance and the correlation between X and Z?

I know that $cov(X,Z)=E[X(X\sin\alpha + Y\cos\alpha)]$

I dont know now how to proceed with covariance and correlation. Could somebody help me please?

答:
You have :

$$cov(X,Z)=E[X(X\sin\alpha + Y\cos\alpha)]\\
=\sin\alpha\times E(X^2)+\cos\alpha\times E(XY)\\
=\sigma^2\sin\alpha+\cos\alpha\times cov(X,Y)\\
=\sigma^2\sin\alpha \text{ (because $X$ and $Y$ are independant)}$$

And you can also get the correlation :
$$R=\sin\alpha $$


问:
Proof of non-reflexivity of $l^\infty$ by finding a bounded sequence such that it contains no weakly convergent subsequence
I want to prove that $l^\infty$ is not reflexive, however I do not want to check why canonical inclusion of $l^\infty$ to $(l^{\infty})^{**}$ is not an isometric isomorphism. I know that if a normed space is reflexive then every bounded sequence of its elements has weakly convergent subsequence. In the case of $l^\infty$ which sequence should I take to break this property?

Thank you.

答:
Let $\phi$ be a Banach limit. Define $x_n$ as
$$
x_n = (1,0,1,0,\dots, 1,0,0, \dots) = \sum_{k=1}^n e_{2n-1}.
$$
Since only finitely many entries of $x_n$ are non-zero, $\phi(x_n) = 0$. The sequence $(x_n)$ converges weak-star in $l^\infty=(l^1)^*$ to the periodic sequence $x=(1,0,1,0,\dots)$. Now due to the properties of the Banach limit, $\phi(x)=\frac12 \ne \phi(0) = \lim_{n\to\infty} \phi(x_n)$. Hence $x_n \not\rightharpoonup x$ in $l^\infty$, and there can be no other weak limit, as the weak-star limit $x$ is uniquely determined.


答: What is your logic in using 4! ? If you choose 1 to 4, ..., 4 to 7, you can place the people in different permutations in the same consecutive seating. $$A-B-C-D$$ $$1-2-3-4$$ Or $$A-C-B-D$$ $$1-2-3-4$$ and so on but this is technically the same seating. $1-2-3-4$ or 2 to 5 and so on is applicable.

Title: Seating four people next to each other probability problem

Body:
The following problem is from Carol Ash's The Probability Tutoring Book

If $4$ people are assigned seats at random in a $7$ seat row, what is the probability that they are seated together?

My approach:

We can choose $4$ from $7$ seats: $_{7}C_4$ ways,

We can permute the people in $4!$ ways.

So, my answer was $\frac{4!}{_{7}C_4}$

However the answer in the book is $\frac{4}{_{7}C_4}$. did they miss the factorial? Or is this the correct answer?

Why is the numerator just $4$?

Is it because if I permute the numerator, I also have to permute the denominator? Can someone explain why this is logically so?


问:
Find a constant in two simultaneous equations with no solutions.
I am having trouble figuring out this question; could anyone explain it to me? Thanks a lot in advance!

The simultaneous equations do not have a solution. Find k.
$$2x-3ky=1$$
$$4x+(k+2)y=5$$

I am unable to find x or y as expected using substitution and cancellation, since there are no solutions. I do not know how to isolate k or find k in terms of x or in terms of y.

答: That system will have no solutions if and only if there is some number $\alpha\ne0$ such that$$4x+(k+2)y=\alpha(2x-3ky)$$and that $5\ne\alpha\times1=\alpha$. But$$4x+(k+2)y=\alpha(2x-3ky)\iff\alpha=2$$and, indeed, if $\alpha=2$, then $5\ne\alpha$. On the other hand\begin{align}4x+(k+2)y=2(2x-3ky)&\iff k+2=-6k\\&\iff k=-\frac27.\end{align}


答:
$2x-3ky=1\iff 4x-6ky=2$.

Now do you see there is no solution if $-6k=k+2$?

Now can you find $k$?

Title: Find a constant in two simultaneous equations with no solutions.

Body:
I am having trouble figuring out this question; could anyone explain it to me? Thanks a lot in advance!

The simultaneous equations do not have a solution. Find k.
$$2x-3ky=1$$
$$4x+(k+2)y=5$$

I am unable to find x or y as expected using substitution and cancellation, since there are no solutions. I do not know how to isolate k or find k in terms of x or in terms of y.


问:
Convergence of these series
$$\sum_{n=1}^\infty \frac{2nx^{n}}{(n+1)^{2}3^{n}} \tag{1}$$

$$\sum_{n=1}^\infty x^{n}\tan\frac{x}{4^{n}}\tag{2}$$

Is there any good article that describes an equivalents like if $$ \lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$$

(am I right about $(3)$?)

答:
The assertion
$$ \lim_{x\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}$$
is perhaps not precise enough for our needs. But it can be made precise, in a form useful for Ratio Test calculations.

We have
$$\lim_{t\to 0}\frac{\sin t}{t}=1.$$
We can replace $\sin\left(\frac{2\pi}{3^n}\right)$ by
$$\frac{\sin\left(\frac{2\pi}{3^n}\right)}{\frac{2\pi}{3^n}}\frac{2\pi}{3^n}.$$
The first term goes safely to $1$, while $\frac{2\pi}{3^n}$ "plays nicely" with the Ratio Test.

We can do precisely the same thing with $\tan\left(\frac{x}{4^n}\right)$ for $x\ne 0$.  Since
$$\lim_{t\to 0}\frac{\tan t}{t}=1,$$
use
$$\frac{\tan\left(\frac{x}{4^n}\right)}{\frac{x}{4^n}}\frac{x}{4^n}$$
in the Ratio Test limit calculation. The front part has limit $1$, so it ford not affect the Ratio Test limit.

Remark: More informally, we can simply replace $\tan(x/4^n)$ by $x/4^n$. However, that is likely to be considered too informal in a homework problem when an important  focus of the course is careful justification. Later, when one is doing "real" work, we can be more casual, for we know that formal detail could be supplied.

In answer to an earlier comment of yours, $\cos x$ is nearly $1$ if $x$ is close to $0$. But it can be useful to know that since the first $2$ terms of the power series of $\cos x$ are $1-\frac{x^2}{2}$, near $0$ the function $1-\cos x$ behaves like $\frac{x^2}{2}$.

Title: Convergence of these series


问:
Law of large numbers without independence and identical distributed assumption
I find the theorem below on https://math.stackexchange.com/a/1366549/533565. But I cannot find it at neither the textbook I have nor the published paper in google scholar. I want to cite this theorem in my project. Does anybody know where is the theorem from?

A theorem by Markov states that if a sequence of random variables $X_1, X_2, \ldots$ with finite variances fulfills one of conditions:

$\lim_{n \to \infty} \frac{\mathrm{Var} X_n}{n^2} = 0$;

$X_1, X_2, \ldots$ are independent and $\lim_{n \to \infty}\frac{1}{n^2}\sum_{i = 0}^n \mathrm{Var} X_i = 0$;

then the sequence $Y_n = \frac{1}{n}\sum_{i=1}^n (X_i - \mathsf{E} X_i)$ converges for $n \to \infty$ to $0$ in probability.

答:
This is false.  The first condition is insufficient for convergence.  Even if the first condition is augmented by assuming the $\{X_i\}_{i=1}^{\infty}$ variables are mutually independent, it is still insufficient. However, the second condition is sufficient.

Define $Y_n = \frac{1}{n}\sum_{i=1}^n (X_i-E[X_i])$.

Counter-example to first condition.

Let $\{X_n\}_{n=1}^{\infty}$ be mutually independent Gaussian random variables such that $X_n \sim N(0, n)$ for $n \in \{1, 2, 3, ...\}$.  Then the first condition holds because
$$ \frac{Var(X_n)}{n^2} = \frac{n}{n^2}\rightarrow 0$$
However, $Y_n$ is Gaussian with mean zero and variance $\frac{1}{n^2}\sum_{i=1}^n i$. That is,
$$Y_n \sim N\left(0, \frac{n+1}{2n}\right)$$
Thus, $Y_n$ does not converge to zero in probability. The limiting distribution of $Y_n$ is $N(0,1/2)$.

Proof of convergence under the second condition.

We just use the standard Markov/Chebyshev inequality.  Fix $\epsilon>0$.  Then
$$ P[|Y_n-0|\geq \epsilon] = P[Y_n^2 \geq \epsilon^2] \leq ...$$


问:
Automatically factor in price increases of 10% for every 100 items for bulk purchases
I have a problem that can be described like this:

we have an apple vendor who sells a rare sort of apple

In order to keep his apples desirable and rare he wants to limit the amount of apples in circulation (meaning the amount of apples he has sold)

but he also wants to make some decent money on his apples if they are not very popular currently

an apple takes 2 weeks to rot from the point of sale

In order to achieve this he develops the following scheme

The apple vendor keeps careful track of how many apples he sold and when they were sold

that way he knows how many apples have spoiled and how many edible ones are still in circulation (A is the amount of apples sold by the vendor that have not yet spoiled)

The vendor sells the apple for a base price plus 10% for all 100 apples in circulation according to this formula:

$price=B*(1.1^{\frac{A}{100}})$

This way the apples will remain affordable when they are not very popular but become pricey fast if theres a run on the market

Meaning apple price should increase with the baseprice (B), increase with the amount of apples in circulation (A) (ie first apple costs 1\$, second costs $1.1^{\frac{1}{100}}$\$, third costs $(1.1^{\frac{1}{100}})^{2}$\$ and 101th costs $(1.1^{\frac{1}{100}})^{100}= 1.1)$\$, assuming apples would be sold one at a time

To clarify: as long as apples don't rot, and if they are sold apple by apple the apples prices form a simple curve of the form $price=baseprice*(1.1^\frac{totalApplesSold}{100})$

Now heres the problem: Most people want to buy more than one apple. But with this formula, the apple vendor has to sell an apple, recalculate the price, then sell an apple again. Is there a way that I can modify this formula, so that I can just plug in the amount of (unspoiled) apples that have been sold (A), the amount of apples a customer wants to buy and then get the price for all the apples, which have become more expensive with each apple?

I know one way to solve this would be to take the derivative of the function. But I'm not sure if this would work since we are talking about individual apples, which get progressively more expensive with each individual one instead of gradually so I'm unsure if derivatives fit here.

Another one would treating the whole process as a series and using a riemann zeta formula, but I don't know enough about those to definitively say that they are the solution here.

答: I think your formula for the price is still wrong. It should be $P=B\cdot 1.1^{\frac{A}{100}}$. If customer buys $n$ apples, the total price can be calculated as (using geometric series)$$Total=B\sum_{i=1}^n1.1^{\frac{A-i}{100}}=B\cdot 1.1^{\frac{A}{100}}\sum_{i=1}^n(1.1^{\frac{-1}{100}})^i=B\cdot 1.1^{\frac{A}{100}}\cdot \frac{1-1.1^{\frac{-n}{100}}}{1-1.1^{\frac{-1}{100}}}$$ Of course, you still need to adjust number $A$ after each sale.


问:
Is (Inverse) Fourier Transform still a "sum" of sinusoids?
I'm trying to wrap my mind around the FT and specifically the IFT:

$$x(t) = \frac{1}{2T}\int_{-\infty}^{\infty}x(\omega)e^{i\omega t}d\omega
$$

I understand that they are generalization of the Fourier Series to non-periodic functions, where you take the period to be infinite, and the interval to be 0.

My question is - does it still hold that what you get is, sort-of, a "sum" of sinusoids? The $x(\omega)$ representing the frequency-domain, are the magnitudes of the different frequencies, but how do you get rid of the imaginary part $i$ when expanding $e^{i\omega t}$ using Euler's formula?

Specifically, is there a way to (numerically approximate) the rect function (time domain) by "sampling" of frequencies from the sinc function (frequency domain) and appropriate sinusoids? (To be honest, the sinc function itself is already somewhat similar to the rect, so I think some approximation is possible...)

答:
I think what you are looking for is the Nyquist–Shannon sampling theorem, a striking result of the last century I came to know in a seminar. It roughly says that, if the Fourier transform is zero outside an interval, then you can perfectly reconstrct the function with discrete samples, "summing up" sinc function. The right hypothesis to assume is that your signal has "limited bandwidth". Let me also remark that this is not too restrictive in digital signal processing: since human can sense from $20 Hz$ to $20.000Hz$, it does not make any difference if you cut the Fourier transform to zero outside this interval.

The explicit reconstruction is given by the Whittaker-Shannon interpolation formula:
$$x(t) = \sum_{n=-\infty}^{\infty} x(nT) \cdot  \text{sinc} \left ( \frac{t-nT}{T} \right ) $$

where $T$ is the sampling time you choose. There is also an explicit bound in how small you should choose $T$ in order to have a perfect approximation: if the frequencies are at most $B$, a sampling time of $T=1/2B$ will perfectly reconstruct the function.

Personally I was strike by the fact that a finite number of samples can give a perfect interpolation, as one intuitively thinks that the fidelity is proportional to how small the sampling time is.


问:
Variance of sum of two dependent Binomial random variables
Question: Suppose $X1 \sim Binomial(n,p)$ and $X2 \sim Binomial(X1,p)$. Find the variance of X1 + X2.

This is my approach:

$Var(X1 + X2) = Var(X1) + Var(X2) + Cov(X1,X2)$

$Var(X1) = np(1-p)$

$Var(X2) = E[Var(X2|X1)]+ Var(E[X2|X1]) = E[p(1-p)X1] + Var(pX1) = np^{2}(1-p^{2})$

$Cov(X1,X2) = E(X1X2) - E(X1)E(X2) = E(X1X2) - n^{2}p^{3}$

At this point I have no idea how to calculate $E(X1X2)$ and I am hoping that I got the other steps correctly. Anyone can help me with this?

答:
Firstly the formula for $\text{Var}(X_1+X_2)$ is given by $$\text{Var}(X_1+X_2)=\text{Var}(X_1)+\text{Var}(X_2)+2\text{Cov}(X_1,X_2)$$
We can find $\mathbb E(X_1X_2)$ by conditioning:
$$\mathbb E(X_1X_2|X_1)=X_1\mathbb E(X_2|X_1)=X_1\cdot X_1p=X_1^2p$$
By the law of total expectation,
$$\mathbb E(X_1X_2)=\mathbb E(\mathbb E(X_1X_2|X_1))=\mathbb E(X_1^2p)=p\mathbb E(X_1^2)$$
We find the second moment of $X_1$:
$$\mathbb E(X_1^2)=\text{Var}(X_1)+\mathbb E(X_1)^2$$
and the result follows.


问:
Consider the set $Y=[-1.1]$ a subspace of $\mathbb{R}$
Consider the set $Y=[-1,1]$ a subspace of $\mathbb{R}$.

Be $A=\{x| \frac{1}{2}<|x|<1\}$.

I know that $A$ is opened in the usual topology of the straight, because for every point in $A$ just take $\epsilon=\min\{|x|-\frac{1}{2}, 1-|x|\}$, then $B(x,\epsilon)\subset A$.

But how do I know $A$ is opened in $Y$ by the induced topology of the straight?

答:
Since $Y$ is closed in $ \mathbb R$, we have for a subset $B$ of $Y$:

$B$ is closed in $Y \iff B$ is closed in $\mathbb R.$

Now let $B$ the complement of $A$ in $Y$, hence $B= Y \cap (\mathbb R \setminus A).$

Show that

$$B=\{-1,1\} \cup [-1/2, 1/2].$$

Conclusion: $B$ is closed in $Y$.


问:
Is it possible to rearrange this equation to solve for x?
$$0 = -i+(iq-ix+1)\ln(1+i)-(1-d)^xln(1+i)-(1-d)^xln(1-d)+t (1-d)^xln(1+i)+t \ln(1-d)^x$$

Variables $i, q, d, t, x$

Wanting to rearrange equation for $x$, online calculators don't seem to like it.

答:
We assume here that $\ln(x)(y) = \ln(xy)$ to clear up ambiguity. Let $1 - d = d'$ to simplify things. First, we will use $\ln(xy) = \ln(x) + \ln(y)$:

$$0 = -i + (iq - ix + 1)\ln(1 + i) - \ln(1 + i) - \ln(d'^{x}) - \ln(d'^{x + 1}) + t\ln(1 + i) +t\ln(d'^{x}) + t\ln(d'^{x})$$

Using the closely related property that $\ln(x^{y}) = y\ln(x)$, and also grouping like terms:

$$0 = -i + (iq - ix + t)\ln(1 + i) + (2tx - 2x - 1)\ln(d')$$

Isolating terms with $x$ on one side:

$$i - (iq + it)\ln(1 + i) + \ln(d') = -ix\ln(1 + i) + (2tx - 2x)\ln(d')$$

$$i - (iq + it)\ln(1 + i) + \ln(d') = ((2t - 2)\ln(d') - i\ln(1 + i))x$$

Isolating $x$:

$$x = \frac{i - (iq + it)\ln(1 + i) + \ln(d')}{(2t - 2)\ln(d') - i\ln(1 + i)}$$

Substituting back $d' = 1 - d$:

$$\boxed{x = \frac{i - (iq + it)\ln(1 + i) + \ln(1-d)}{(2t - 1)\ln(1 -d) - i\ln(1 + i)}}$$

You could use the log rules to bring the coefficients into the logarithms and then combine them, but it doesn't really matter for purposes of calculation.

Let me know if I have made any mistakes.


问:
Be $B=\left\{x\mid \frac{1}{2}<|x|\le1\right\}$
Consider the set $Y=[-1.1]$ a subspace of $\mathbb{R}$.  Let $B=\left\{x\mid \frac{1}{2}<|x|\le1\right\}$.

I know that $B$ is not open in the straight, because we will have problem in the points $-1$ and $1$ because we can not get a ball centered on them entirely contained in $B$.

But because $B$ would be open in $Y$, I haven't been able to see that yet. I know that those opened in $Y$ would be the intersections of the common open of the straight with the $Y$. So how do I prove $B$ is open at $Y$?

Body: $B=Y\cap\underbrace{[(-2,-1/2)\cup(1/2,2)]}_\text{open in $\Bbb R$}$


问:
Finding the equivalence classes of a trigonometric relation
I have been asked to respond to the following:

Define a binary relation R on $\mathbb{R}$ as ${\{(x, y) \in
 \mathbb{R} \times \mathbb{R} \mid \sin(x) = \sin(y)\}}$. Prove that R
  is an equivalence relation. What are its equivalence classes?

Given that the relation R is based on equality, the first part of the question is rather simple:

Is R reflexive?

Let $a \in \mathbb{R}$, then $(a, a) \in R$ because $\sin(a) = \sin(a)$.

Is R symmetric?

Let $a, b \in \mathbb{R} \mid (a,b) \in R$. Then $(b,a) \in R$ by the symmetric property of equality.

Is R transitive?

Let $(a,b) \in R$; thus, $\sin(a) = \sin(b)$.

Let $(b,c) \in R$; thus, $\sin(b) = \sin(c)$.

Thus, $(a,c) \in R$, as $\sin(a) = \sin(c)$.

However, I am having trouble following a process to find the equivalence classes for R. As we have demonstrated that R is an equivalence relation, we know that we can decompose R into a series of equivalence classes such that, for any $x \in \mathbb{R}$, then $x \in [x]$, and that $[x] = \{ y \in \mathbb{R} \mid (x,y) \in R \}$ (or, more specifically, $[x] = \{ y \in \mathbb{R} \mid \sin(x) = \sin(y) \}$).

Examining the unit circle, we know that, for any value $x \in \mathbb{R}$, the value of $\sin(x)$ will be equivalent to that of any value $x_1 \in \mathbb{R}$ such that $x_1 = x * 2\pi k$, where $k \in \mathbb{Z}$ (after all, sin is a periodic function).

We can also see that $\sin(x) = \sin(\pi - x)$ for any x in the range $[0, \pi / 2]$; similarly, we know that $\sin(\pi + x)$ and $\sin(2\pi - x)$ are both equal to $-\sin(x)$. Along a single period of $\sin(\theta)$, there are exactly two values for $\theta$ (in that domain) for which $\sin(\theta)$ will be equal.

I don't know which of the above information is relevant to the task at hand, and I'm unsure about how to proceed. Any helpful explanations or clues would be appreciated!

EDIT:

I can begin to define some of the equivalence classes of R:

$[0] = \{ k\pi \mid k \in \mathbb{Z} \}$

$[1] = \{ \pi/2 + 2k\pi \mid k \in \mathbb{Z} \}$

$[-1] = \{ 3\pi/2 + 2k\pi \mid k \in \mathbb{Z} \}$

$[\pi/6] = \{ \pi/6 + 2k\pi, 5\pi/6 + 2k\pi \mid k \in \mathbb{Z} \}$

$[-\pi/6] = \{ -\pi/6 + 2k\pi, 7\pi/6 + 2k\pi \mid k \in \mathbb{Z} \}$

How can I generalize this to include all possible equivalence classes (accounting for all possible values of sin x)?

答: Each element $x \in \left [-\frac{\pi}{2}, \frac{\pi}{2} \right ]$ defines a unique equivalence class. An element $x$ is a member of a given equivalence class $\left [x_{i}  \right ]$ such that: $x \equiv x_{i} \pmod {2 \pi}$.


问:
Question about relations and equivalence classes.

Hi so we've started learning about relations and I'm completely lost..
I don't get how to represent the equivalence classes and I'm not even sure the way I've proved the equivalence relation is correct..

Question 1

a.

Note that if for every $a,b ∈ R$ if $a = b = 0 \ $ ➜ $a*b$ $\not>$ $0$, but in this case it's already part of the relation.

Also note that in order for this inequality to work, $a,b$ must be both negative, or both positive.

Reflexive:

Since $a$ obviously has the same sign as itself, $a * a > 0$.

Therefore for every $a ∈ R/(0) $ $ \ a R a \ $

Symmetric:

By associativity, if $a * b > 0 ➜ b * a > 0$.

Therefore for every $a,b ∈ R/(0) $ if $aRb ➜ bRa$.

Transitive:

Since $a,b,c$ must share the same sign, if $a * b > 0$ and $b * c > 0 ➜ a * c > 0$.

Therefore for every $a,b,c ∈ R/(0) $ if $aRb ∧ bRc ➜ aRc$.

Equivalence classes:

$\left\{R^{+},R^{-},\left\{0\right\}\right\}$

b.

Note that if $8 ∉ XΔY$ then either $8 ∈ X  ∧ 8 ∈ Y \ $ or $ \ 8 ∉ X  ∧ 8 ∉ X$.

Reflexive:

Since $XΔX$ is the $∅$, then it is obvious $8 ∉ XΔX$.

Therefore for every $X ∈ P(NxN)$ $XRX$.

Symmetric:

Δ is associative, therefore if $8 ∉ XΔY ➜ 8 ∉ YΔX $

Therefore for every $X,Y ∈ P(NxN) \ XRY➜YRX$.

Transitive:

Since $X,Y,Z$ must all contain or all not contain $8$ for the relation to work,

then if $8 ∉ XΔY ∧ 8 ∉ YΔZ ➜ 8 ∉ XΔZ$.

Therefore for every $X,Y,Z ∈ P(NxN) \ $ if $XRY ∧ YRZ ➜ XRZ$.

Equivalence class:

$\left\{\left\{X\ ∈\ \ P(N)\ :\ 8\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 8\ ∈\ X\ \right\}\right\}$

c.

Note that we're proven in class equality is both reflexive, symmetric and transitive.

Reflexive:
Obviously $X ∩ {1,2} = X ∩ {1,2}$.

Symmetric:

It is also obvious $X ∩ {1,2} = Y ∩ {1,2} ➜ Y ∩ {1,2} = X ∩ {1,2}$.

Transitive:

Again, obviously if $X ∩ {1,2} = Y ∩ {1,2} ∧ Y ∩ {1,2} = Z ∩ {1,2} \ ➜ \ X ∩ {1,2} = Z ∩ {1,2}$.

Equivalence classes:

$\left\{\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1\ ∉\ \ X\ \ ∧\ 2\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 2\ ∉\ \ X\ \ ∧\ 1\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∈\ \ X\ \right\}\right\}$

答:
Instead of 'describing the equivalence classes' it might be more appropriate to rather ask for a set $B$ and a surjective function $f:A\to B$ for each example such that the given equivalence relation on $A$ is just its 'kernel' $\{(a,a'):f(a)=f(a')\}$.

This set $B$ represents the best the equivalence classes.

For 1., the positive numbers are in relation with all each other, and so are the negative numbers, and there's also the $0$, so the function we're looking for is the signum function ${\rm sign}:\Bbb R\to\{-1,0,+1\}$.

For 2. and 3., observe that condition in 2. holds for sets $X,Y$ if and only if either both of them contains $8$ or neither of them, so that this can be translated to the same condition as for 3.:
$$X\sim Y\iff X\cap\{8\}=Y\cap\{8\}\,.$$
Generalizing this, for any set $U$ and any subset $A\subseteq U$, the relation $X\sim Y\iff X\cap A=Y\cap A$ is an equivalence relation on $P(U)$, and it is the kernel of the 'restriction' function
$$f:P(U)\to P(A)\quad X\mapsto X\cap A\,.$$

So, the equivalence classes of 2. are represented by elements of $P(\{8\})=\big\{\emptyset,\,\{8\}\big\}$ and similarly, those of 3. by $P(\{1,2\})=\big\{\emptyset,\,\{1\},\, \{2\},\, \{1,2\}\big\}$.


问:
Write the equation $f(x,y)=0$ that describes the set $S$
Let $A=(1,1)$, $B=(1, -2)$ and $C=(5,-5)$.

Write the equation $f(x,y)=0$ with $f\in \mathbb{R}[x,y]$ that describes the set $S$ of points that have same distance of $A$ as to $BC$.

$$$$

I have done the following:

Let $P=(x,y)$ be a point of $S$.

The it must be $d(P,A)=d(P,BC)$.

The line through $B$ and $C$ is $$y+2=\frac{-2+5}{1-5}(x-1) \Rightarrow y+2=-\frac{3}{4}(x-1) \Rightarrow 4y+8=-3x+3 \Rightarrow 3x+4y+5=0$$

From $d(P,A)=d(P,BC)$ we get $$\sqrt{(x-1)^2+(y-1)^2}=\frac{|3x+4y+5|}{\sqrt{3^2+4^2}}$$ right?

Do we get from that the desired equation?

Body:
We have

$$\sqrt{(x-1)^2+(y-1)^2}=\frac{|3x+4y+5|}{\sqrt{3^2+4^2}} \iff 25(x-1)^2+25(y-1)^2-(3x+4y+5)^2=0.$$


问:
Evaluating the integral $\int_{-\infty}^{\infty}\frac{1}{\sqrt{k-p}\sqrt{k+p}}\,\mathrm dp$
I was wondering about evaluating the following definite integral analytically:
\begin{equation}
\int_{-\infty}^{\infty}\frac{1}{\sqrt{k-p}\sqrt{k+p}}\,\mathrm dp
\end{equation}

Does someone know how to approach this?

答: Wolfram Alpha checked that this improper integral does not converge.</p>


问:
If the eigen values of an orthogonal matrix A are positive, then the determinant of A is 1 why??
can anyone say why the above statement is the way it is. Why determinant is 1?</p>

答:
Hint: Determinant of an orthogonal matrix is either $1$ or $-1.$

Also product of Eigen values of a matrix is equal to the determinant of that matrix.

Now, given that Eigen values of $A$ are positive, so its determinant is $1.$


问:
Eliminate $\theta$ and prove $x^2+y^2=1$
We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$

And want to prove $x^2+y^2=1$

My works:

I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:

$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.

答:
You are almost there. You already have $$2x\cos\theta=y\sin\theta+\sin\theta\cos\theta \tag1$$
and $$x\cos \theta = y \sin \theta \tag 2$$

Then $$x\cos \theta = y \sin \theta = 2x\cos \theta - y \sin \theta = \sin \theta \cos \theta$$
and you know $\sin \theta \ne 0, \cos \theta \ne 0$. Can you end it now?


问:
The nature of $\lim_{n\to\infty}\sum_{i=n/a}^{n}\frac{1}{i} = \ln a$ relation.
Actually there is a simple formula $$\lim_{n\to\infty}\sum_{i=n/a}^{n}\frac{1}{i} = \ln a$$
But not sure if I understand what is the nature of this relation. This is true for all $a>0$.
Just wanted to point this interesting formula as it shows another face of the natural logarithm.
Let me also say that $a$ can be any irational number.
The question I have is this, how does this work for non integer numbers?
$$\lim_{n\to\infty}\sum_{i=n/a}^{n}\frac{1}{i}=\lim_{n\to\infty}\big(\sum_{i=1}^{n}\frac{1}{i} - \sum_{i=1}^{n/a}\frac{1}{i}\big)$$ As the $n/a$ will have very similar values for many $a$ numbers because of ceiling.

答:
When you say $\sum\limits_{i=n/a}^{n}\frac{1}{i}$ you have to deal with the issue when $n$ is not a multiple of $a$.  You have said in comments you want to take the floor.

Gae. S. has raised the point of $0 < a < 1$ in which case $\lfloor n/a\rfloor$ is   $n$ when $n < \frac{a}{1-a}$, but when $n$ is larger then $\lfloor n/a\rfloor > n$  leaving you with an empty sum of $0$.

Otherwise you have in general $\sum\limits_{i=1}^{n}\frac{1}{i} = \log_e(n) +\gamma + O(\frac1n)$

so $\sum\limits_{i=\lfloor n/a\rfloor}^{n}\frac{1}{i} = \log_e(n) -\log_e(\lfloor n/a\rfloor-1)+O(\frac1n) = \log_e(n) -\log_e( n/a)+O(\frac1n)$  since $\frac{1}{\lfloor n/a\rfloor}$ is $O(\frac1n)$ and this means $$\lim\limits_{n \to \infty}\sum\limits_{i=\lfloor n/a\rfloor}^{n}\frac{1}{i} = \lim\limits_{n \to \infty}\left(\log_e(n) -\log_e( n/a)+ O(\tfrac1n)\right) = \log_e( a)$$

when $a \ge 1$, and $0$ when $0 < a < 1$


问:
For which $n$, $n=x+y+xy$ has no solution in natural numbers?
For which value of $n$, equation $n=x+y+xy$ has no solution in natural numbers ?

$a)n=100$

$b)n=105$

$c)n=110$

$d)n=115$

$e)n=120$

It was a problem from an Olympiad competition. to solve this I considered quadratic equation $m^2+Sm-P=0$ and if $x$ and $y$ are roots of the equation we should find $n=S+P$ where $n$ is a natural number. but I can't continue this idea to solve the problem.

答:
Hint :

Since $x+y+xy+1=n+1=(x+1)(y+1)$,

if $n+1$ is prime, there won't be solutions in $\mathbb{N}$.


问:
Count 0,1 codes with specific conditions
how many $0,1$ codes of size $n$ that each $1$ is between two $0$s?

$0100010010$ is acceptable but $010110$ is not.

i know it should be solved with some recursive function.

答:
Sketch:

Note that every good (non-empty) string ends in $0$ and that appending a $0$ or a $10$ to any good (non-empty) string results in a good string.

If the second to last character is a $0$ then deleting the final $0$ gets you a good string of length $n-1$.

If the second to last character is a $1$ then deleting the final $10$ gets you a good string of length $n-2$.

This results in a very familiar recursion.


问:
Calculate $\lim_{n\to\infty}\int_0^nf(\frac{x}{n})^ndx$.
Consider the following problem:

Suppose $f:[0,1] \to \mathbb{R}$ is a differentiable function with $f'(0)<0$, $f(0)=1$ and for all $x\in(0,1]:0<f(x)<1$. Calculate $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx.$$

My attempt so far:

Let $$f_n:x\mapsto\chi_{[0,n]}(x)\cdot f\left(\dfrac{x}{n}\right)^n,$$ where $\chi_{[0,n]}$ is the indicator function on $[0,n]$. We wish to do two things:

Show that $f_n\to h$ pointwise for some integrable function $h:\mathbb{R}^+\to\mathbb{R}$ and

show that $|f_n|\leq k$ for some positively integrable $k:[0,1]\to \mathbb{R}^+$ so that we can use the dominated convergence theorem to show that $$\lim_{n\to\infty}\int_{[0,\infty)}f_n(x) \,dx = \int_{[0,\infty)}h(x)dx.$$

To do this, we first construct the function $$g:[0,1]\to\mathbb{R}:y\mapsto\begin{cases}\dfrac{\ln f(y)}{y}& y\in(0,1],\\ f'(0) & y=0.\end{cases}$$
This is a continuous extention of $\ln f(y)/y$ since $$\lim_{y\to0}\dfrac{\ln f(y)}{y}=f'(0)$$
by l'Hôpital's rule. The idea is to show that  $$f\left(\dfrac{x}{n}\right)^n\leq e^{f'(0)x}$$ for all $x\in[0,n]$. To do this we can use the function $g$ as constructed above. Indeed, taking the natural log of this inequality gives us

$$n\ln f\left(\dfrac{x}{n}\right)\leq f'(0)x$$
which is equivalent to $$g(y)\leq f'(0),$$ for all $y\in[0,1]$. The problem is that I have trouble showing that this last inequality is true. Although I'm convinced it is. I've tried calculating $g'$ to see if it is negative everywhere, but have failed to do so.

Suppose it is true. Then, since $f'(0)<0$, the function $x\mapsto e^{f'(0)x}$ is integrable on $x>0$, so we have found a dominating function for the $f_n$.

Furthermore, we need to show that there is an integrable function $h$ to which the $f_n$ converge. I cannot see how to show this pointwise convergence.

Edit: I have found conclusive proofs for my two problems and have added them as an answer.

To conclude the problem, the limit is $$\lim_{n\to\infty}\int_0^nf\left(\frac{x}{n}\right)^ndx = \int_0^\infty e^{-|f'(0)|x}dx = \dfrac{1}{|f'(0)|}.$$

答:
For all $x\ge 0$: $$f\left(\dfrac{x}{n}\right)^n\to e^{f'(0)x}.$$

Proof:

For $x=0$, this is true since $f(0)=1$. Let $x>0$. We whish to show that $$\left|f\left(\dfrac{x}{n}\right)^n - e^{f'(0)x} \right|\to 0.$$
This is equivalent to $$\dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}}\to 1.$$ (Note that both denominator and enumerator are positive) Which on its turn is equivalent to $$d_n =\ln \dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}} \to 0.$$
We will show that this is indeed true. Note that $$d_n = n\ln f\left(\dfrac{x}{n}\right)-f'(0)x$$
and hence that (we took $x>0$) $$\dfrac{d_n}{x} = g\left(\dfrac{x}{n}\right)-f'(0).$$

We already know that $g(0) = f'(0)$, so then $d_n \to 0$.

For all $y\in[0,1]:$ $$g(y) \leq f'(0).$$

Proof:

Note that from what we know about $f$, it must be true that for all $y\in [0,1]: f(y)\leq f(0)=1$. Then it is also true that for all $y\in[0,1]:$ $$\ln f(y) \leq \ln f(0).$$ Now, it is also true that for all $y\in (0,1]:$ $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y}.$$ Now, note that $\ln f(0) = 0$, so for any $\varepsilon > 0$ we get that $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y-\varepsilon}.$$ Taking the limit $\varepsilon \to y$ makes the right hand side $g(0) = f'(0)$.


问:
Intersection of a pointed cone and a hyperplane is a polytope?
Let $C = \text{cone}(u_1,\dots,u_m)$ for some $u_1,\dots,u_m \in \mathbb{R}^d \setminus \{\textbf{0}\}$ be a finitely generated pointed cone. Let $H_0 := \{x \in \mathbb{R}^d: \langle a,x \rangle = 0\}$ be a hyperplane such that $C \subseteq H_0^\geq$ and $C \cap H_0 = \{\textbf{0}\}$. For some $\delta > 0$, let $H_\delta := \{x \in \mathbb{R}^d: \langle a,x \rangle = \delta \}$ and $C_\infty := C \cap H_\delta$.

I want to show that $C_\infty$ is a polytope. Since both $C$ and $H_\delta$ could be represented by a system of inequalities, so is $C_\infty$, which shows that it is a polyhedron. Now I just need to show that it is also bounded.

I was attempting a proof by contradiction, assuming that $C_\infty$ is not bounded, hence containing a ray $\{x + ty: t \geq 0\}$ for some $x \in C_\infty$ and $y \neq \textbf{0}$ (and in particular a point $q = x + ty, t > 0$ on this ray).  This should lead to $C \cap H_0$ contains more than just $\textbf{0}$, but I haven’t been able to show it so far. I know that $y$ as defined above should be in $H_0$, but that has not helped.

答:
How about this.

If $x\in H_\delta\cap C$, then $\epsilon x\in  H_{\epsilon\delta}\cap C$ for every $\epsilon>0$.

That is, if for some $y\in H_0\setminus\{0\}$ and all $t\ge 0$ holds $x+ty\in  H_\delta\cap C$, then also $\epsilon x  + ty\in H_{\epsilon\delta}\cap C$ for all $t\ge 0$ (the $\epsilon$ in the last term got absorbed into the $t$). Note that $C$ can be written as the union

$$C=\bigcup_{\epsilon \ge 0} (H_{\epsilon\delta}\cap C),$$

and so $\epsilon x + t y\in C$ for all $\epsilon > 0, t \ge 0$. But since $C$ is a  closed set, with $\epsilon\to 0$ we find $yt\in C$ for all $t\ge 0$. In particular, for $t>0$ and since $y\in H_0\setminus\{0\}$, we found $0\not= yt\in H_0\cap C$.

The crucial part is indeed that $C$ is closd, because the statement is wrong for open cones.


问:
Doubt on Double Covariant Derivatives
I simply want to calculate the Components of Riemann Tensor for a general connection from the definition given by:

\begin{equation}
    \mathrm{Riem}(V,W,U) := \nabla_{V}\nabla_{W}U - \nabla_{W}\nabla_{V}U - \nabla_{[V,W]}U
\end{equation}

But I'm struggling a little with the double covariant derivatives.

My calculation is given by:

$$\nabla_{V}\nabla_{W}U = \nabla_{V}(\nabla_{W}U):=\nabla_{V}(A) \tag{1}$$
In a local chart we have:
$$\nabla_{V}(A) = (V^{a}\nabla_{a}A^{c})\partial_{c} \tag{2}$$
But, note that $A^{c}$ are the components of $\nabla_{W}U$,
$$A^{c} = (\nabla_{W}U)^{c} = W^{b}\nabla_{b}U^{c} \tag{3}$$
Therefore,
$$\nabla_{V}\nabla_{W}U = (V^{a}\nabla_{a}(W^{b}\nabla_{b}U^{c}))\partial_{c} \tag{4}$$

So, my doubt is: the components of $\nabla_{V}\nabla_{W}U$ are, in fact, that of $(4)$?

答:
I find it pretty dangerous to ask for the component of $\nabla_V\nabla _W U$, while $$(U, V, W)\mapsto \nabla_V\nabla _W U$$

is not a tensor: It is not $C^\infty(M)$-linear in $U, W$. On the other hand, it makes sense to ask for the component of $\operatorname{Riem}$, since it is a tensor (some checking is needed).

Anyway, if we know that $\operatorname{Riem}$ is a tensor, then

$$ \operatorname{Riem} = \sum_{i,j,k,l=1}^n {R_{ijk}}^l dx^i \otimes dx^j \otimes dx^k \otimes \frac{\partial }{\partial x^l}, $$
where ${R_{ijk}}^l$ are the components of $\operatorname{Riem}$ given by

$$ \tag{1} \operatorname{Riem}\left( \frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j},\frac{\partial }{\partial x^k}\right) = \sum_{l=1}^n  {R_{ijk}}^l\frac{\partial }{\partial x^l}. $$

Calculating the left hand side of (1) is easy:

\begin{align}
\operatorname{Riem}\left( \frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j},\frac{\partial }{\partial x^k}\right) &= \nabla_{\partial_i }\nabla_{\partial_j} \frac{\partial}{\partial x^k} - \nabla_{\partial_j }\nabla_{\partial_i} \frac{\partial}{\partial x^k} \\
&= \sum_l \left( \nabla_{\partial_i } \left(\Gamma_{jk}^l \frac{\partial}{\partial x^l}\right)-\nabla_{\partial_j } \left(\Gamma_{ik}^l \frac{\partial}{\partial x^l}\right)\right) \\
&= \sum_l (\partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l) \frac{\partial}{\partial x^l} + \sum_m \left( \Gamma_{jk}^m \nabla_{\partial_i} \frac{\partial}{\partial x^m} - \Gamma_{ik}^m \nabla_{\partial_j} \frac{\partial}{\partial x^m} \right) \\
&=\sum_l \left(\partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l + \sum_m\left( \Gamma_{jk}^m \Gamma_{im}^l - \Gamma_{ik}^m \Gamma_{jm}^l\right) \right)\frac{\partial}{\partial x^l}
\end{align}

Thus
$${R_{ijk}}^l = \partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l + \sum_m\left( \Gamma_{jk}^m \Gamma_{im}^l - \Gamma_{ik}^m \Gamma_{jm}^l\right).$$

Title: Doubt on Double Covariant Derivatives


问:
$X_1,X_2$ distributed $U(0,1)$ and independent, find the distribution of $D=|X_2-X_1|$
I know it shouldn't be this hard but I am having trouble anyway.

I tried using $F_D(t)=\mathbb{P}(|X_2-X_1|\leq t)=\mathbb{P}(-t\leq X_1-X_2\leq t)$ and sketching the unit square $[0,1]^2$ but to no avail, I can't seem to find a formula that fits.

答:
If you to a drawing of the problem it will be very easy to find

$$F_D(d)=1-(1-d)^2$$

and derivating you get the density

$$f_D(d)=2(1-d)\cdot \mathbb{1}_{[0;1]}(d)$$

The CDF is the white area, in function of $d \in [0;1]$
To calculate it, it is easier to calculate the purple area $(1-d)^2$ and take the complement to 1


问:
Prove or disprove: $\sum\limits_{i=1}^n i^2 = O(n^2) $
Prove or disprove:

$$\sum_{i=1}^n i^2 = O(n^2) $$

If we want to prove this, find some summation that we know the $ O(n)$ runtime for, and is $ O(n^2) $ or smaller.

Otherwise, we could disprove this by finding some summation that is less than this one, but has a $O(n)$ runtime that is greater than $ O(n^2)$.

No idea where to go from here though. Any assistance is appreciated.

Body:
Hint:

$$
\sum_{i=1}^n i^2 \geq \int_0^n x^2\,dx.
$$


问:
Permutation problem of trading dolls
There are $n$ girls, numbered $1,2,...,n$. In the morning, each girl has a doll: Girl number $k$ has doll number $k$. Then, each pair of girls (in some order) traded their dolls.

My question is: When is it possible that at the end of the day, each girl will have the doll she started with?
 Of course that because of permutation pairity arguments, since there were $n \choose 2$ trades, we must have $n \choose 2$ even.
 Is it enough? Whenever we have $n \choose 2$ even, is it always possible to have any pair of girls trading their dolls such that at the end girl $i$ has doll number $i$?

答:
Consider any solution for $4k$ girls, such as Gerry Myerson's when $k=1$.

For $1\le i \le k$, replace each transposition $(ab)$ where $a=2i-1$ and $b=2i$ by
$(af)(ae)(ad)(ac)(ab)(bc)(bd)(be)(bf)$,where $c,d,e,f$ are four new girls and multiply by Gerry Myerson's solution for $c,d,e,f$.

This gives a solution for $4k+4$ girls. Note that a solution for $4k+1$ girls is obtained simply by replacing each transposition $(ab)$ where $a=2i-1$ and $b=2i$ by
$(ac)(ab)(bc)$.

Title: Permutation problem of trading dolls


答: What is the pink area in this unit square?

Title: $X_1,X_2$ distributed $U(0,1)$ and independent, find the distribution of $D=|X_2-X_1|$

Body:
I know it shouldn't be this hard but I am having trouble anyway.

I tried using $F_D(t)=\mathbb{P}(|X_2-X_1|\leq t)=\mathbb{P}(-t\leq X_1-X_2\leq t)$ and sketching the unit square $[0,1]^2$ but to no avail, I can't seem to find a formula that fits.