问:
A basic question on factorization
Is the following true? If not, can anyone add some reasonable assumptions to make it true?

Statement. Let $E/F$ be a field extension and let $\alpha \in E \setminus F$ be algebraic over $F$. If $f(x) \in F[x]$ is irreducible, then $f(x/\alpha) \in F[\alpha][x]$ is irreducible.

Attempt to prove the statement. Suppose for contradiction that $f(x/\alpha) = p(x)q(x)$ where $p(x), q(x) \in (F[\alpha])[x]$ and $\deg(p(x)), \deg(q(x)) > 0$. Over a splitting field $K$ that contains $\alpha$, we can write $f(x) = a(x - r_{1}) \cdots (x - r_{n})$ for some $a \in F^{*}$, so we have

$f(x/\alpha) = (a/\alpha^{n})(x - \alpha r_{1}) \cdots (x - \alpha r_{n})$.

Without loss of generality, we can write $p(x) = \gamma(x - \alpha r_{1}) \cdots (x - \alpha r_{j})$ and $q(x) = \delta(x - \alpha r_{j+1}) \cdots (x - \alpha r_{n})$ for some $\gamma, \delta \in F[\alpha]$ such that $\gamma\delta = a / \alpha^{n}$. Since $p(x), q(x) \in (F[\alpha])[x]$, we have

$p_{0}(x) := \delta p(x) = a(x/\alpha - r_{1}) \cdots (x/\alpha - r_{j}) \in (F[\alpha])[x];$
$q_{0}(x) := (\gamma/a) q(x) = (x/\alpha - r_{j+1}) \cdots (x/\alpha - r_{n}) \in (F[\alpha])[x].$

We have $f(x) = p_{0}(\alpha x)q_{0}(\alpha x)$, and I was wishing that $p_{0}(\alpha x), q_{0}(\alpha x)$ were polynomials over $F$, but there are no reasons to believe so.

Motivation. The motivation was to show $\Phi_{p}(x/\alpha) \in \mathbb{Q}[\alpha][x]$ is irreducible where $\alpha := 2^{1/p}$ and $\Phi_{p}(x) \in \mathbb{Q}[x]$ is the $p$th cyclotomic polynomial where $p$ is a prime number. I did attempt this in a different way but I am not sure how to conclude whether this is true.

A connection to the statement. Write $\zeta := e^{2\pi i/p}$, and we have $\mathbb{Q}[\zeta, \alpha]$ as a splitting field of $x^{p} - 2 \in \mathbb{Q}[x]$ of degree $(p-1)p$, which shows that $[\mathbb{Q}[\alpha\zeta, \alpha] : \mathbb{Q}[\alpha]] = p-1$, thus $\alpha \zeta$ must have the minimal polynomial $g(x) \in \mathbb{Q}[\alpha][x]$ whose degree is $p-1$.

If the previous statement is true, there is no choice but $g(x) = \alpha^{p}\Phi_{p}(x/\alpha)$, but like I said, I am not sure.

答:
I am a little regretful that I have not thought about this much before I asked this question. After talking to Gerry Myerson, I realized that the statement that I stated was very far from being true. Instead, we have the following.

Lemma. A polynomial $f \in F[x]$ is irreducible if and only if it is a minimal polynomial of some $\alpha$ in some extension $E/F$.

Proof. Suppose $f$ is irreducible. Then take $E = F[x]/(f)$. The converse is trivial. Q.E.D.

This seems like a very naive lemma, but I think the following variation looked very nontrivial to me before.

Corollary. A polynomial $f \in F[x]$ is irreducible if and only if there exists an extension $E/F$ and $\alpha \in E$ such that $f(\alpha) = 0$ and $\deg(f) = [E:F]$.

Proof. If we have an extension $E/F$ and $\alpha \in E$ such that $f(\alpha) = 0$, then the minimal polynomial of $\alpha$ must divide $f$ since it is the generator of the kernel of evaluation $(F[x], x) \rightarrow (F[\alpha], \alpha)$. The rest follows from Lemma. Q.E.D.

My previous question was as follows.

Question. Let $\alpha \in E$ is algebraic over $F$ where $E/F$ is an extension. How can we tell $f(x/\alpha)$ is irreducible over $F[\alpha]$ given $f(x)$ is irreducible over $F$?

Answer. By Corollary, the polynomial $f(x/\alpha)$ is irreducible over $F[\alpha]$ if and only if we have an extension $L/F[\alpha]$ and $\beta \in L$ such that $f(\beta/\alpha) = 0$ and $\deg(f) = [L:F[\alpha]]$.

Title: A basic question on factorization


问:
Is there a complex number $c$ such that $c^n$ goes to 1 as n (which is real) goes to infinity?
Is there a complex number $c$ such that $c^n$ goes to 1 as n goes to infinity?

What is my motivation for that? I learned some dynamical system (such as fibonacci series) are expressible in the form of matrices. In this case we can apply diagonalization and obtain a matrix of eigenvalues ($D$) for our model matrix $M = SDS^{-1}-$ Since it is straightforward to calculate powers of such a diagonal matrix, this gives us a lot info about this system. Lets think about a system which moves but starts to get slower and slower. This would happen when entries of the diagonal matrix, or lets say eigenvalues of the $M$, are such that when we take their power to infinity, they converge to 1. If eigenvalues are real this would only happen with $1$. But in this case $M$ would be the trivial identity matrix. So my question is if we can construct a matrix with nontrivial eigenvalues such that their power to infinity converges to 1, or formally,

$\lambda^n \to 1$ as $n \to \infty$

I appreciate any help thanks!

Edit: $\lambda$ has to be a complex number and $\neq 1$

答:
Suppose that $\lambda^n\to1$. Then $|\lambda|^n=|\lambda^n|\to1$. This forces $|\lambda|=1$.

So $\lambda=e^{i\theta}$. Now $$\tag1|e^{i(n+1)\theta)}-e^{in\theta}|=|e^{i\theta}-1|=2|1-\cos\theta|.$$ If $\{\lambda^n\}$ converges, then $\{e^{in\theta}\}$ is Cauchy. But the only way that $(1)$  can go to zero is that $\cos\theta=1$, so $\lambda=1$.

ContentLicense: CC BY-SA 4.0


问:
What is the equation for a line in a family ruling a rectangle, given the rectangle's dimensions and a point on the line inside the rectangle?
I have questions about the following ruling of rectangle $ABCD$ with top and bottom sides, $AB$ and $DC$, being collinear with $y=1$ and $y=−1$, respectively, and the midpoint of each lies on $x=0$. The ruling is defined as the sweep of a line through the rectangle as it follows this path: given a line collinear with side $AB$, rotate counterclockwise around vertex $B$ until it intersects with the opposite vertex $D$ (i.e., is collinear with diagonal $BD$), then rotate it clockwise around vertex $D$ until it is collinear with side $DC$.

There are an infinite number of families of lines that fit these criteria, one family per rectangle with a unique ratio of adjacent side lengths; for all families, it follows that: a) every point inside the rectangle lies on exactly one of the lines, so no two lines intersect, b) every line that is not a side of ABCD has a positive slope, and c) every line has a unique intercept.

I'm trying to find a general equation for lines in such a family, which takes as variables the ratio of one side of the rectangle to its adjacent side (to identify the family) and the xy coordinates of a point inside the rectangle, and returns the slope of the line in that family through the given point.

I suspect what I'm describing is equivalent to a rectangle cut out of the side of a single-sheet hyperboloid or some other quadric, but even so, I don't know how to go from the equation for the hyperboloid to the equation I want, other than the diagonal of the rectangle would be an asymptote of the hyperbola.

Am I right that this equation must exist, and, if so, what is it or how can I find it? Finally, is the equation associated with a particular quadric or other surface?

答:
Here's a re-statement of my re-statement of the problem (approved by OP in a comment, but with an orientation fix and an explicit vertical dimension for sanity-checking purposes):

Consider an axis-aligned, clockwise-oriented rectangle $\square ABCD$ with
$$A=(−r,s) \qquad B=(r,s) \qquad C=(r,-s) \qquad D=(−r,-s)$$
and consider a point $P=(p,q)\in \square ABCD$. We seek a function $f(p,q,r,s)$
such that
$$f(p,q,r,s) = \begin{cases} \text{equation of $\overleftrightarrow{PB}$}, & P\in\triangle ABD\\[4pt]
\text{equation of $\overleftrightarrow{PD}$}, & P\in\triangle CDB \end{cases} \tag{1}$$

The line equations are easy enough to generate:

$$\begin{align}
\overleftrightarrow{PB}:&\quad (s-q) x - (r-p) y = ps - qr \\
\overleftrightarrow{PD}:&\quad (s+q) x - (r+p) y = ps - qr
\end{align} \tag{2}$$

The cases are determined by whether $P$ lies above or below the diagonal $\overline{BD}$. The equation of that diagonal is $sx-ry=0$, so the condition corresponds to whether $(x,y)\to(p,q)$ makes $sx-ry$ negative (for "above") or positive (for "below"). With an explicit zero case, we have

$$f(p,q,r,s) = \begin{cases}
(s-q) x - (r-p) y = ps - qr, & ps < qr \\[4pt]
sx - ry = 0, & ps = qr \\[4pt]
(s+q) x - (r+p) y = ps - qr, & ps > qr \end{cases} \tag{3}$$

Defining $\lambda := \operatorname{sgn}(ps-qr)$, we can unify the above into

$$f(p,q,r,s) \quad\to\quad (s+\lambda q) x - (r+\lambda p) y = ps - qr \tag{$\star$}$$

or this symmetric-looking alternative

$$f(p,q,r,s) \quad\to\quad (s+\lambda q) (x-p) = (r+\lambda p) (y-q) \tag{$\star\star$}$$

Substituting $s=1$ for OP's particular needs is left as an exercise to the reader. :)

Here's the function in action (with $P$ not constrained to lie within the rectangle):


问:
Chi distribution approaches the normal distribution?
Wikipedia states here that the chi distribution $\chi _{k}$ with $k$ degrees of freedom converges to the standard normal distribution

${\displaystyle \lim _{k\to \infty }{\tfrac {\chi _{k}-\mu _{k}}{\sigma _{k}}}{\xrightarrow {d}}\ N(0,1)\,} $

I cannot find any references or equivalent statements on the internet. Is this statement true and, if yes, does someone know a reference or a proof?

答: The $x\ge0$ PDF $\tfrac{1}{2^{k/2-1}\Gamma(k/2)}x^{k-1}e^{-x^2/2}$ is asymptotic to $e^{-S}$ with$$S:=x^2/2-(k-1)\ln x+(k/2-1)\ln2+\ln\Gamma(k/2).$$As per Laplace's method, for large $k$ we can approximate the log-PDF as a quadratic, based on its first two derivatives. Since $S_x=x-\tfrac{k-1}{x}$ is $0$ at $x=\sqrt{k-1}$, and $S_{xx}=1+\tfrac{k-1}{x^2}>0$, this value of $x$ minimizes $S$ and maximizes the PDF. At this value of $x$,$$S=(k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)$$and $S_{xx}=2$, so in general$$S\approx (k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)+\left(x-\sqrt{k-1}\right)^2,$$and we've approximated the $\chi_k$ distribution as $N(\sqrt{k-1},\,\tfrac12)$. This Gaussian approximation can be restated as $\frac{\chi_k-\mu_k}{\sigma_k}$ being approximately Gaussian, with mean $0$ and variance $1$, and hence $\approx N(0,\,1)$.


问:
First ODEs Nonlinear Polynomial
I have the following system of nonlinear polynomial differential equations:

$\begin{cases}
\dfrac{dx(t)}{dt}=-3y(t)-3m(2p+x(t)^2-8y(t)^2) \\
\dfrac{dy(t)}{dt}=ex(t)+6mx(t)y(t) \\
\dfrac{dz(t)}{dt}=1+kp + \frac{3}{4}e - 6my(t)
\end{cases}$

where $m$, $e$, $p$ and $k$ are constant parameters.

I know that the third equation can not be separated and resolved later by quadrature since $z(t)$ is not present in the first two equations.

The biggest problem is certainly the strong non-linearity in the second equations.

Can anyone tell me how they can solve or the method to solve the first two equations.

Thanks in advance to everyone:)

答:
The first two equations fall into the class of projective Riccati equations, the analysis of which goes back to Lie and Vessiot.

Refer Robert L Anderson, "A Nonlinear Superposition Principle Admitted by Coupled Riccati Equations of the Projective Type", Letters in Mathematical Physics 4 (1980) pp. 1-7.

Title: First ODEs Nonlinear Polynomial


问:
If $u(z)$ is non-constant and harmomic in the plane, then show that $u(z)$ comes arbitrarily close to every real value
This question was asked in my complex analysis assignment and I am not able to solve it.

If $u(z)$ is non constant and harmonic in the plane, then show that $u(z)$ comes arbitrarily close to every real value.

Every harmonic function can be proved both real part and imaginary part of an aanalytic function. So, $u(z)$ itself would be analytic.
So function can't be bounded and can be  defined for all $z\in \mathbb{C}$. Now there is a theorem that an entire function comes in $\epsilon$ neighbourhood of every complex number (I forgot the theorem's name). So, I am done.

Am i right?

Is the proof correct?

答:
Your argument is correct and the  theorem you want to use is Picard's Theorem. But you to complete it by observing that if $u+iv$ attains some value in the disk of radius $\epsilon$ around a real number $a$ then $u$ takes a value in $(a-\epsilon, a+\epsilon)$.

Actually $u$ attains very real value (by continuity). To see this consider $e^{u+iv}$ and $e^{-(u+iv)}$ Using Liouville's Theorem show that $u$ can neither be bounded above nor be bounded below.


问:
Surface Area of Hypercylinder in d - dimensions
The volume of a hypercylinder in d-dimensions can be derived in a general way using the Cartesian product.

What is the volume of a hyper cylinder in d - dimension?

I want to find the surface area of a hypercylinder, is there an analogous way to calculate this?

答:
The analysis above is perfectly correct.

There is an ambiguity in the question regarding surface area with regards a 4-dimensional object.  This is equivalent to asking the circumference of a 3-dimensional object.  There is no single answer.  For example, the circumference of a cylinder depends both on its dimensions and where this circumference is taken.  It is the locus of points of the intersection of a plane with with the cylinder and has a range of values.  Even a sphere has a range of circumferences associated with is but we usually, unambiguously associate the question with the great circle of maximum circumference with it.  Unfortunately, we cannot do this with any other shape.  In all dimensions only the N-volume and N-surface area have unambiguous meanings.  This is probably why no additional formula was forthcoming for this question as the first was a complete as it could be.


问:
Power series of $f(x)=\sqrt{\frac{1+x}{1-x}}$
How do I find the power series form of $\,f(x)\,$:

$$\displaystyle f(x)=\sqrt{\frac{1+x}{1-x}}$$

I tried to multiply the fraction by $\,\dfrac{1+x}{1+x}\,$ but it didn't help...

答: Try writing it as $ f(x) = \sqrt{ 1- x^2 } \times 1/(1-x) $  . Then write $ 1/(1-x) $ as a geometric series and the other term can be expanded using binomial theorem. Hope that helps.</p>

ContentLicense: CC BY-SA 3.0


问:
Is $\bigoplus_{x \in X} \mathbb{Z}$ reflexive?
Basically the title. I am trying to figure out whether or not $Hom(Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z})) \cong \bigoplus_{x \in X} \mathbb{Z}$ with the isomorphism $i(y)(f)=f(y)$.

Here is my intuition as to why that might be true: $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z}) = \times_{x \in X} \mathbb{Z}$, since any "functional" in $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z})$ is determined by its values on each entry of $\bigoplus_{x \in X} \mathbb{Z}$ and any homomorphism $\mathbb{Z} \to \mathbb{Z}$ is just multiplication by some whole number, so there is a natural isomorphism from $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z}) \to \times_{x \in X} \mathbb{Z}$ (is this reasoning correct?)

So I am left with showing whether or not $Hom(\times_{x \in X} \mathbb{Z}, \mathbb{Z}) = \bigoplus_{x \in X} \mathbb{Z}$. Again, any "functional" would be determined by its' values on each entry, but it cant be nonzero on infinitely many entries since then we could find a corresponding element on which the functional would not be defined. My question is - is this reasoning fine? Turning it into a formal proof isn't hard, but something still feels "fishy" here.

答:
The first step of your argument is false: the Hom functor converts infinite direct sums into infinite products, so we have

$$\text{Hom}(\bigoplus_X \mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z}^X$$

which is much bigger than the infinite direct sum (e.g. if $X$ is countably infinite, the infinite product is uncountable while the infinite direct sum is countable). What this says explicitly is that every homomorphism $\oplus_X \mathbb{Z} \to \mathbb{Z}$ has the form

$$\bigoplus_{x \in X} n_x \mapsto \sum_{x \in X} c_x n_x$$

for some coefficients $c_x$ and that these coefficients are arbitrary, and in particular can all be nonzero. In the infinite direct sum only finitely many of them can be nonzero.

If $X$ is countably infinite, it is nevertheless still true that $\text{Hom}(\mathbb{Z}^X, \mathbb{Z}) \cong \bigoplus_X \mathbb{Z}$, so we still get reflexivity; this is very surprising, e.g. the corresponding statement with $\mathbb{Z}$ replaced by $\mathbb{Q}$ is false given the axiom of choice. This is due to Specker.

For larger $X$ the answer appears to be independent of ZFC.


问:
What is the observed threshold (p-value) of this two-tailed test?
Two teams of researchers have each developed a treatment against atypical pneumonia. The effectiveness of these two treatments is compared by administering them to two separate groups of patients chosen at random from the victims of this disease. Group 1 comprises 45 individuals while group 2 comprises 35. We observed 34 cures in the first group and 28 in the second group.

What is the observed threshold (p-value) of this two-tailed test?

I found 0.34 here but I am really not sure.

答:
This is a test to compare to proportions. Usually, the test uses a normal approximation for sufficiently large samples, which you have. There are
several formulas for the z statistic, some using a pooled estimate for
the common proportion (under $H_0)$ and some estimating proportions separately.

Here is output for the prop.test in R:

```<code>prop.test(c(34, 28), c(45,35), cor=F)

        2-sample test for equality of proportions
        without continuity correction

data:  c(34, 28) out of c(45, 35)
X-squared = 0.22302, df = 1, p-value = 0.6367
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.2270020  0.1381131
sample estimates:
   prop 1    prop 2
0.7555556 0.8000000
</code>```

The P-value is 0.6368 for this version of the test. Thus the
null hypothesis is not rejected at the 5% level of significance. [This
version of the test gives a chi-squared statistic, which is the square
of the relevant z-statistic.]

Here is the version of the test implemented in a recent release of Minitab.

```<code>Test and CI for Two Proportions

Sample   X   N  Sample p
1       34  45  0.755556
2       28  35  0.800000

Difference = p (1) - p (2)
Estimate for difference:  -0.0444444
95% CI for difference:  (-0.227002, 0.138113)
Test for difference = 0 (vs ≠ 0):  Z = -0.47  P-Value = 0.637

The P-value is about the same as from R.
</code>```

I suspect that your P-value 0.34 is for a one-sided test.
(If I do the appropriate one-sided test in R, I get P-value 0.317.)
Without knowing
the formula for your z-statistic, I can't say for sure how you got
your P-value.

Body:
This is a test to compare to proportions. Usually, the test uses a normal approximation for sufficiently large samples, which you have. There are
several formulas for the z statistic, some using a pooled estimate for
the common proportion (under $H_0)$ and some estimating proportions separately.

Here is output for the prop.test in R:

prop.test(c(34, 28), c(45,35), cor=F)

        2-sample test for equality of proportions
        without continuity correction

data:  c(34, 28) out of c(45, 35)
X-squared = 0.22302, df = 1, p-value = 0.6367
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.2270020  0.1381131
sample estimates:
   prop 1    prop 2
0.7555556 0.8000000

The P-value is 0.6368 for this version of the test. Thus the
null hypothesis is not rejected at the 5% level of significance. [This
version of the test gives a chi-squared statistic, which is the square
of the relevant z-statistic.]

Here is the version of the test implemented in a recent release of Minitab.

Test and CI for Two Proportions

Sample   X   N  Sample p
1       34  45  0.755556
2       28  35  0.800000

Difference = p (1) - p (2)
Estimate for difference:  -0.0444444
95% CI for difference:  (-0.227002, 0.138113)
Test for difference = 0 (vs ≠ 0):  Z = -0.47  P-Value = 0.637

The P-value is about the same as from R.

I suspect that your P-value 0.34 is for a one-sided test.
(If I do the appropriate one-sided test in R, I get P-value 0.317.)
Without knowing
the formula for your z-statistic, I can't say for sure how you got
your P-value.


问:
Is this a mistake $\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0?$
Can anyone show that,
$$\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0\tag1$$
Suppose $\ n\ $ and $\ r\ $ are integers.

I can't understand why it should be zero.

making a basic substitution:

$u=1-x^r$

$\mathrm dx=\frac{\mathrm du}{r(1-u)}$

$$\int_{0}^{1}\frac{u^{n/r}-n\cdot u^{1/r}(1-u)^{(n-1)/r}}{r(1-u)}\mathrm du$$

Body:
Use $\beta$integral and Gamma functions:https://en.wikipedia.org/wiki/Beta_function
Take $x=\sin^{2/r} t$ $$I_1=\int_{0}^{1} (1-x^r)^{n/r} dx=\frac{2}{r}\int_{0}^{\pi/2} \cos^{2n/r+1}t ~~\sin^{2/r-1}t~dt=\frac{1}{r}\frac{\Gamma(n/r+1)\Gamma(1/r)}{\Gamma(n/r+1/r+1)}$$
$$\implies I_1=\frac{n}{r^2}\frac{\Gamma(n/r)\Gamma(1/r)}{\Gamma(n/r+1/r+1)}~~~~(1)$$
$$I_2=\int_{0}^{1} nx^{n-1}(1-x^r)^{1/r} dx=\frac{2n}{r}\int_{0}^{\pi/2} \sin^{2n/r-1} t \cos ^{2/r+1} dt=\frac{n}{r}\frac{\Gamma(n/r) \Gamma(1/r+1)}{\Gamma(n/r+1/r+1)}$$
$$\implies I_2=\frac{n}{r^2}\frac{\Gamma(n/r)\Gamma(1/r)}{\Gamma(n/r+1/r+1)}~~~~(2).$$
Hence $$I_1=I_2 \implies I=I_1-I_2=0$$


问:
No closed form for the partial sum of ${n\choose k}$ for $k \le K$?
In Concrete Mathematics, the authors state that there is no closed form for
$$\sum_{k\le K}{n\choose k}.$$
This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the book).

How do they know this is true?

答:
There's a number of asymptotic approximations. One of the simpler (assume wlog $N<\frac{n}{2} \in \mathbb{Z}$) upper bounds is obtained by dividing the partial sum by the largest term, $\binom{n}{N}$:
$$
\frac{S_{n,N}}{\binom{n}{N}} = \frac{\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{N}}{\binom{n}{N}} = 1+ \frac{N}{n-N+1} + \ldots \frac{1}{\binom{n}{N}}
$$
Each term in this sum, $\frac{\binom{n}{k}}{\binom{n}{N}}$, after cancellations, reduces to
$$
a_{k,N,n} = \frac{(n-N)!}{(n-k)!}\cdot\frac{N!}{k!} \leq N^k
$$
Since $N\geq k$, the first term is upper-bounded by $1$. As as result, the upper bound is just
$$
S_{n,N} \leq \binom{n}{N}\sum_{k=0}^{N+1}N^k
$$
the latter being a simple Geometric sum that exists in closed form. If you want to reduce it further, you get something like (Geometric sum approximation + Stirling's formula)
$$
S_{n,N} \leq \bigg(n\bigg(1-\frac{1}{N}\bigg)\bigg)^{N+1}
$$


问:
Galois group is dihedral iff $\exists \beta_i\in F$
I was solving the following problem:

Let $char(F)\ne 2,3$ and $f(x) = x^4 +px^2 +qx+r$ be an irreducible separable polynomial in $F [x]$ with roots being $α_1, α_2, α_3, α_4$. Let $L = F(α_1,α_2,α_3,α_4) $ and $G = Gal(L/F) ≤ S_4$. Set $β_1 = α_1α_2 + α_3α_4, β_2 =α_1α_3+α_2α_4$ and $β_3 =α_1α_4+α_2α_3$. Show that there exists $i$ such that $β_i \in F$ if and only if $G≤D_8.$

Attempt: I first showed that $Gal(F(\beta_1,\beta_2,\beta_3)/F)\cong G/(G\cap V)$, where $V =\{id,(12)(34),(13)(24),(14)(23)\}$. Assume $\beta _1\in F$, then $[F(\beta_2,\beta_3):F]=\dfrac{|G|}{|G\cap V|}\ge \dfrac{|G|}{4}$. For contradiction, I supposed $G\gneq D_8\implies |G|\gneq 8, [F(\beta_2,\beta_3):F]\gneq2$, but this actually turned out to make sense to me, and I couldn't get a contradiction, which is imo weird. Is there anything I have missed? What did I do wrong?

答:
From your remark $Gal(F(\beta_1,\beta_2,\beta_3)/F)\cong G/(G\cap V)$  we see this group must be a subgroup of $S_3$.

If it contains an element of order $3$ then the $\beta_i$ are all conjugate and so none can lie in $F$.

If it does not contain an element of order $3$ then it is either trivial or cyclic of order $2$; in each case one $\beta_i$ lies in an orbit of size $1$ and so is in $F$.

The only subgroups of $S_4$ not containing an element of order $3$ are subgroups of $D_8$.


答: No. Take $a=c=y=z=0$ and $b=x=1$.

Title: If $a^2+b^2+c^2=1$ and $x^2+y^2+z^2=1$, can it be shown that $(ax)^2+(by)^2+(cz)^2=1$?

Body: If $a^2+b^2+c^2=1$ and $x^2+y^2+z^2=1$, can it be shown that $(ax)^2+(by)^2+(cz)^2=1$ for all real a, b, c, x, y, z?

Body: No. Take $a=c=y=z=0$ and $b=x=1$.


问:
How to generate a N*D random matrix with columns of unit length?
Is it possible to generate a N*D random matrix with columns of unit length?

If not, I also think it is possible of generating a N*D random matrix and, after that, normalizing it in order to have columns of unit length.

Is this last case, how is this possible mathematically?

答: If you want the column vectors to be uniformly distributed on the unit sphere, you may first generate an $N\times D$ rectangular array of standard normal pseudorandom numbers, and then normalise the length of each column vector.</p>


问:
Stoke's Theorem for an open cylinder
How do you use Stokes' Theorem to calculate the surface integral over a cylinder of $\nabla \times F$? Do you have to calculate the line integrals along the top and the bottom? If so, is this example done incorrectly? Should the top line integral also be calculated? I don't understand why they only calculate the line integral in the $xy$ plane.

Also, if I do need to calculate both integrals, do I need to orient them differently?

If you want the actual problem, it is:

Use Stokes' Theorem to evaluate
$\int\int_T curl(xz \vec{j})d\vec{S}$,
where T is the cylinder
$x^2 + y^2 = 9$ with $0\leq z \leq 2$, orientated with an outward pointing normal.

But don't worry too much about the computation, I'm struggling more with the concept. I'm also pretty sure I could just do the integral without Stokes', but it's in the section on Stokes' theorem, so I should probably learn how to do it that way.

答:
Yes, you do need to calculate both integrals and add them to get the answer.

However, you must be careful with the orientation of the curves.

Since the normal vector points outward($ \hat n = cos \phi \space \hat x \space + sin \phi \space \hat y$), the 2 curves $c_{top}$ and $c_{bot}$ that together make up the boundary of the cylinder, should inherit their orientation from the cylinder by applying the rule.

Rule:

Work for each curve seperately. Imagine yourself walking along the curve with your foot stepping on the curve and your head pointing as the normal vector indicates(i.e. radially outward). The orientation of the curve is chosen so that ,in your walk, you always have the surface to your left.

This image may help with the visualisation

Image: example for outward oriented open cylinder

This pdf helped me understand how to properly orient the curves of the boundary of an open cylinder, hope it helps someone else as well.

@page 7

http://www.owlnet.rice.edu/~fjones/chap13.pdf

I also found this relative topic.

Stokes' Theorem which boundaries to integrate

Title: Stoke's Theorem for an open cylinder


问:
Trying to calculate matches in a tournament using combinations, in games that have multiple players
I am trying to calculate opponents in matches for my app game.

In it I will have a league of a varied number (between 5-14) of players and I want to know, given the number of players in a league, what's the smallest number of games each person can play so they've played all other people in the league and equal number of times. I could obviously play out every combination, but I want to play less games if possible. Each match will have either 3 or 4 players in it (including yourself)

Some examples to help better explain...

I know that in a 9 player league, if each match has 3 opponents only, then each player can play only 8 games, resulting in playing each opponent twice. This is fewer than the full 84 possible combinations.

In an 8 player league with 4 people then it only needs 7 games per person playing each opponent 3 times.

I THINK I have worked out the maths on how to do get the number of matches and number of times to play each opponent, but I can't work out how to come up with the combinations of players in order to satisfy this.

I can add the maths I've worked out, but it would be quite long so I'll only do that if people need it. As that's not actually the bit I need solving. (though it's part of the solution)

I've tried both on pen and paper, and writing a program to do it. The program has had the most success, working on a variety of combinations of league and match size, but it doesn't work on all of them.

I'm at a loss right now, and looking for any sort of help.

答:
After @Saulspatz told me what to look for, (BIBD) I found a library in the R language that worked it out for me.

https://rdrr.io/cran/crossdes/man/find.BIB.html

@Saulspatz if you want to mark up a quick answer talking about balanced incomplete block designs I'll give you a tick


答:
Even in the reduced case with only two couples of variables $\begin{cases} c=z=0\\a=y=\cos(t)\\b=x=\sin(t)\end{cases}$

Then $(ax)^2+(by)^2+(cz)^2=2\big(\cos(t)\sin(t)\big)^2=\frac 12\sin(2t)^2\le \frac 12\neq 1$

So there are infinity of values for which the equation is not verified.


问:
How many primitive polynomials are there in $\mathbb{F}_p[x]$ where $p$ is prime?
I've read the post The number of primitive polynomials of degree $m$ over a finite field $GF(p^m)$, but I don't understand why there can't be common roots of distinct primitive polynomials of the same degree. Any help would be highly appreciated! Thank you in advance!</p>

答:
Primitive polynomials are irreducible

If two polynomials $f, g$ have roots in common, then $\gcd(f, g)\neq 1$

Couple these two facts together, and you get that two primitive polynomials with any roots in common must have all their roots in common.

Somewhat alternate reasoning (and the one hinted at in the linked post):

Given a primitive polynomial $f$ over $\Bbb F_p$, extending $\Bbb F_p$ by any of the roots of $f$ gives $\Bbb F_{p^m}$. Given another primitive polynomial $g$ of the same degree, any one of its roots generates the same (or at least an isomorphic) field.

If $f$ and $g$ have roots in common, then $h = \gcd(f, g)$ will also have those roots in common. And thus any root of $h$ will also generate $\Bbb F_{p^m}$. But that's a degree $m$ extension, so $h$ must have degree $m$. Thus $\gcd(f, g)$ has the same degree as $f$ and $g$, so $f$ and $g$ are (up to multiplication by a constant) the same polynomial.


问:
Let $2^x=3x$. Show that the equation has a solution in the range $(0, 1).$

Let $2^x=3x$. Show that the equation has a solution in the range $(0, 1).$

I was trying to apply just Bolzano's theorem here, but seems that that's not quite enough?

Denoting $f(x) = 2^x-3x$ and differentiating one has $f'(x) = 2^{x-1}x -3$ and then picking for example $x=\frac32$ I have $f'(\frac{3}{2})=\frac{3\cdot \:2^{\frac{1}{2}}}{2}-3 <0$ and $x=\frac{1}{2}$ I have $f'(\frac{1}{2})=\frac{\sqrt{2}}{4}-3<0$ which are both unfortunately negative...

Looking at the graph of the derivative it seems that it's negative everywhere in $(0,1).$ I guess I just have to use limit definition or something here?

答:
$$ f= 2^x - 3x$$

$$ f(1) = -1$$

$$ f(0) = 1$$

By intermediate value theorem ,we have the required statement. QED


问:
Derivative of Mean Squared Error
I'm studying with a book and I'm at the Linear Regression part. The author is showing that we have to calculate the derivative of each part of the equation that leads to the loss.

But he's using the MSE to calculate the loss and so, I tried to calculate the derivative of MSE:

the derivative of $ (y-p)^2 $ with respect to y (the target) is equal to $2(y-p)$ but  in the book it is written $-1*(2(y-p))$ which is simplified as $-2(y-p)$. Why do I have different values ? Where is this $-1$ coming from?

答:
Since you take derivative with respect to $p$, you first take the derivative of $(y−p)^2$ and then $−p$. Finally multiply them.

So first part is  $2(y−p)$ and the second part is simply $−1$.

Therefore the derivative is  $-1 \times2\times(y-p) =−2(y−p)$

Title: Derivative of Mean Squared Error


问:
Problem: Integrate using the Trigonometric Substitution method
Exercise about Trigonometric Substitution

Objective: Resolve using the Trigonometric Substitution method.

I've already tried solving the exercise by taking the following steps:

I take e^x as 2sen(t) and I substitute.

Take dx as 2cos(t)dt.

After some easy straight-foward steps, I came out with this:

The result has to be  and the integral of sen(t) is cos(t).

答:
In STEP 2: $\color{red}{e^x }dx=2\cos t\, dt$. so
$$\int\frac{e^x}{\sqrt{4-2e^{2x}}}\,dx =\int \frac{2\cos t}{2\cos t}\,dt=t+C=\arcsin\left(x/2\right)+C$$


问:
Doing modular division when denominator and modulus not coprime
So normally if you calculate $n/d \mod m$, you make sure $d$ and $m$ are coprime and then do $n[d]^{-1}\mod m$ , all $\mod m$. But what if $d$ and $m$ are not coprime? What do you do?</p>

答:
If $\gcd(m,d)=g$ and $g\mid n$, then you can perform the standard modular division on
$$
\left.\frac{n}{g}\middle/\frac{d}{g}\right.\left(\text{mod}\frac{m}{g}\right)\tag{1}
$$
Note that the division reduces the modulus, too.

The original equation
$$
dx\equiv n\pmod{m}\tag{2}
$$
is equivalent to
$$
dx+my=n\tag{3}
$$
To solve $(3)$, we need to divide through by $g$:
$$
\frac{d}{g}x+\frac{m}{g}y=\frac{n}{g}\tag{4}
$$
and $x$ in $(4)$ is given by $(1)$.

For example, suppose we know that
$$
12x\equiv9\pmod{15}
$$
we would solve
$$
4x\equiv3\pmod{5}
$$
and any solution would only be known mod $5$; that is,
$$
x\equiv2\pmod{5}
$$


问:
Definition of $C^r$ tensor fields in "Analysis on manifolds" by Munkres
The text defines a $k-$tensor field as a function assigning, to each point $x$ in an open subset $A$ of $\mathbb{R}^n$, a $k-$tensor defined on the tangent space $T_x(\mathbb{R}^n)$, which can be written in the form $$\omega(x)((x;\mathbf{v_1}),...,(x;\mathbf{v_k}))$$
Then it requires this function to be continuous as a function of $(x,v_1,...,v_k)$ and, if it is of class $C^r$, it says that $\omega$ is a tensor field of class $C^r$.

My problem is with this last statement.
For each $x \in A$, $\omega(x)$ is a tensor. I don't really know what it means for a tensor to be continuous or differentiable.

Does it mean that $\omega$ should be continuous for $x$, continuous for $v_1$, etc. separately?

答:
A manifold in the sense of Munkres is a subspace $M$ of some $\mathbb R^n$ having suitable properties (see the definition in §23). in §29 Munkres also introduces the tangent space $T_x \mathbb R^n$ at $x \in \mathbb  R^n$ as the set $\{x\} \times \mathbb R^n$ which gets the structure of a vector space via the canonical bijection $\phi_x : \mathbb R^n \to T_x \mathbb R^n, \phi_x(v) = (x,v)$.

A $k$-tensor field on an open subset $A \subset \mathbb{R}^n$ (Munkres restricts to this case and does not define tensor fields on arbitrary manifolds $M \subset \mathbb R^n$) is defined as a function $\omega$ assigning to each point $x \in A$ a $k$-tensor $\omega(x)$ defined on the tangent space $T_x(\mathbb{R}^n)$. Thus (as he explicitly says) $\omega(x)$ is a function $(T_x \mathbb R^n)^k \to \mathbb R$. We therefore get a function
$$\bar \omega : A \times  (\mathbb R^n)^k \to \mathbb R, \bar \omega (x,v_1,\ldots, v_k) =  \omega(x)(\phi_x(v_1),\ldots, \phi_x(v_k)) .$$
But of course $A \times  (\mathbb R^n)^k$ is an open subset of $\mathbb R^{(k+1)n}$, thus it makes sense to call $\bar \omega$ continuous or $C^r$.


问:
Square integrable integral by Brownian motion is martingale
Let $f \in C_b^2(\mathbb{R}_{+} \times \mathbb{R})$. Then my lecturer said that for a Brownian motion $B_t$
\begin{align}
X_t = \int_0^t f(s, B_s) \, dB_s
\end{align}
is a martingale, which should follow from the integral being square integrable, but I can't how to derive this from Ito's formula.

答:
The condition @d.k.o. provided in the comments implies that $E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=0.$
Then, using the linearity property of the integral, we have
$$E[X_t |\mathscr F_s]=E[X_s |\mathscr F_s]+E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=X_s.$$


问:
How does one improve their exam performance and avoid "stupid/careless" mistakes?
I have this problem in most Mathematics courses (in fact the only course I haven't had this problem in is Linear Algebra for some reason) but I study really hard, practice, and make sure to understand every concept deeply instead of just memorizing ways to solve certain types of problems but when doing an exam (note that I usually find the exam easy and familiar as it would be similar to exercises done in class and in the book) my performance is very bad usually, I do stupid/careless of mistakes due to answering with what's on top of my head instead of thinking well, I don't pay attention to certain things, and I have trouble proof-reading my exam well, I think that happens mostly due to my anxiety as I solve most things correctly during practice and they would be similar level and style, any advice on how to improve this and translate my actual knowledge into grades?</p>

答:
How does one improve their exam performance?

For improving exam performance I suggest you do practice problems but timing it. Timing is very important as it practices under time pressure which will help with performance gradually over time. Also do a lot of practices.

Avoid “stupid/careless” mistakes

I personally found this article very helpful about https://artofproblemsolving.com/news/articles/stop-making-stupid-mistakes
I will not write it here so you'll get the whole message.


问:
Algebra / Equation with 2 or more vaiables
A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds
respectively. What is the speed of the train?

I'm really confused if there are 2 or more variables. Can someone explain this in step by step? How x is removed until the end. Thanks.

Or a link on any tutorial or article regarding this. I don't know what to search, I only find simple examples with 1 variable. Thanks.

答:
Let the length of the train is $x\,\, m$ and speed of the train is $y \,\,m/s$
So you have $x=8y$ and $x+264=20y$.Hence your result is true.

If you have two variable then you must have two equation and from that you can solve the problem.


问:
$n$th derivative of a function that depends both explicitly and implicitly on a variable
I would like to have a closed-formula for the $n$th derivative of a function of the type $f(x,g(x))$, i.e. a function that depends both explicitly and implicitly on a (scalar) variable:
$$
\frac{d^n f}{d x^n} = ?
$$
It can be assumed that all partial derivatives up to order $n$ included exist.

I can calculate the first (or second, or third...) derivatives by hand by using the chain rule. For example:
$$
\frac{d f}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g}\frac{\partial g}{\partial x} = f^{(1,0)} + g'f^{(0,1)}.
$$
or
$$
\frac{d^2 f}{d x^2} =  f^{(2, 0)} +  2 g' f^{(1, 1)} + (g')^2 f^{(0, 2)} + g'' f^{(0, 1)}
$$

I however cannot see a pattern for a general formula. I am aware of the Faà di Bruno formula for calculating high order derivatives of a purely implicit function, but I am failing in seeing how to use it in combination with an explicit dependence of the function on the variable.

答:
The general higher derivative of a composition can be calculated by Faà di Bruno's formula (Higher Chain rule). Faà di Bruno's formula can be represented i.a. by Exponential Bell polynomials.

[Hardy 2006] generalizes Faà di Bruno's formula to the general higher partial derivatives of a multivariate function.

The answer to your question is given in [Bernardini/Natalini/Ricci 2005]:

"Abstract - We develop some extensions of the classical Bell polynomials, previously obtained, by introducing a further class of these polynomials called multidimensional Bell polynomials of higher order. They arise considering the derivatives of functions $f$ in several variables $\varphi^{(i)}$, $(i=1,2,...,m)$, where $\varphi^{(i)}$ are composite functions of different orders, i.e. $\varphi^{(i)}(t)=\phi^{(i,1)}(\phi^{(i,2)}(...(\phi^{(i,r_i)}(t)))$, $(i=1,2,...,m)$. We show that these new polynomials are always expressible in terms of the ordinary Bell polynomials, by means of suitable recurrence relations or formal multinomial expansions. Moreover,
we give a recurrence relation for their computation."


问:
closure of addition and multiplication in $\mathbb{Z}_n$
I need to understand what guarantess closure in $\mathbb{Z}_n$ under the operations

$[a]_n + [b]_n := [a + b]_n$

$[a]_n \cdot [b]_n := [a \cdot b]_n$

Different manuals state this:

It is clear that the rule $[a]_ + [b]_n := [a + b]_n$ yields an element of $\mathbb{Z}_n$, but the uniqueness of this result needs to be verified.

These properties are an imnmediate consequence of the definition of + and * in $\mathbb{Z}_n$.

I need to understand why it is so obvious (not for me) that closure is implied in the definition.

Previously this theorem was stated

If $a ≡ b \mod n,  c ≡ d \mod n$ Then $a + c ≡ b + d \mod n$ and $ac ≡ bd \mod n$

I think that $a ≡ b \mod n, c ≡ d \mod n$ implies $a + c ≡ b + d \mod n$ and this implies that $[a+c]$ is in $\mathbb{Z}_n$.

But the manual seems to suggest that closure is implied in the definition of the operations I given above. I'm not interested to prove that those are well-defined for the moment.

答: Since $a$ and $b$ are just integers, $a+b$ and $ab$ are also integers, and therefore $[a+b]_n$ and $[ab]_n$ are in $\mathbb{Z}_n$ by definition (if your definition of $\mathbb{Z}_n$ is the set of equivalence classes of integers modulo n).


答:
Making computational errors is inevitable, both within and without the context of an exam.

The simplest way to address these errors is to check your work:  after solving a problem, we go back and review the solution.  Therefore, one should take into consideration the additional time needed to perform this check during a timed examination.  When I say "check," I don't just mean re-reading your work.  I mean using specific strategies to ensure the result you obtained is correct, such as

substitution of the solution into the question to see if it is consistent

checking boundary cases

see if the result is the correct order of magnitude

try to use a different approach to the solution to see if you get the same result

go through your reasoning backwards

eliminate extraneous solutions

This is by no means an exhaustive list, and not all are applicable to every problem.  The point is, part of doing mathematics is not just about having the knowledge to answer the question; it is also about having the ability to review your work critically, in such a way that you can confirm the logical steps you took are correct.

In an exam setting, my preference would be to do this checking after most of the questions are answered.  If there are many remaining problems, checking may be of limited use.  Moreover, I am more likely to discover an error if I let some time pass between the initial write-up and the check.

Outside of what one can do during the exam, I would say the most effective approach to avoiding errors is through practice and experience.  Solving many problems, then going through and reviewing where you tend to make mistakes, is important in developing mathematical and computational proficiency.  This also involves fostering good habits with respect to problem solving.  Don't take shortcuts.  Write down every step.  Do not make an assertion or claim unless you are absolutely sure it is correct and you can justify it.  All too frequently, I see students write things that are completely wrong, because they skipped several steps and were too lazy to write things out.  Finally, do not blindly rely on formulas.  Many mistakes are the result of improper application of formulas, identities, or theorems due to over-reliance on memorized information rather than logical deduction.

The above is just a few suggestions from a much broader picture of how to approach problem solving effectively.

Title: How does one improve their exam performance and avoid "stupid/careless" mistakes?

Body: I have this problem in most Mathematics courses (in fact the only course I haven't had this problem in is Linear Algebra for some reason) but I study really hard, practice, and make sure to understand every concept deeply instead of just memorizing ways to solve certain types of problems but when doing an exam (note that I usually find the exam easy and familiar as it would be similar to exercises done in class and in the book) my performance is very bad usually, I do stupid/careless of mistakes due to answering with what's on top of my head instead of thinking well, I don't pay attention to certain things, and I have trouble proof-reading my exam well, I think that happens mostly due to my anxiety as I solve most things correctly during practice and they would be similar level and style, any advice on how to improve this and translate my actual knowledge into grades?</p>


问:
A question related to $f(n)=n^4+n^2+1$
If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 61. But obviously we can't evaluate this with brute force. The farthest I have reached is that $$n^4+n^2+1=n^2(n+1)+\frac{n+1}{n+1}=(n^2+\frac{1}{n+1})(n+1)$$ Any hints?

Title: A question related to $f(n)=n^4+n^2+1$

Body:
We note that $n^4+n^2+1 = (n^4+2n^2+1)-n^2 = (n^2+1)^2-n^2 = (n^2+n+1)(n^2-n+1)$

Writing the value of $(n^2+n+1) = g(n)$, the value of $g(n-1) = n^2-n+1$

So we get eg $f(n)=g(n)g(n-1)$

And  $\frac{f(3)f(5)f(7)...f(13)}{f(2)f(4)f(6)...f(12)}$

is equal to $\frac{g(2)g(3)g(4)...g(12)g(13)}{g(1)g(2)g(3)g(4)...g(12)} = \frac{g(13)}{g(1)}$

This evaluates to $(13*14+1)/(1*2+1)$ or $\frac{183}{3} = 61$


问:
Prove this set is closed.
I just need to prove that this set
$$E := \left\{ (x, \lambda x): x \in \mathbb{R}^{n+1} \setminus \{ 0 \}, \lambda \in  \mathbb{R} \setminus \{0\} \right\}$$
is closed in $(\mathbb{R}^{n+1} \setminus \{0\}) \times (\mathbb{R}^{n+1} \setminus \{0\})$ with the usual topology, but i cannot see the way... Just tried to used the definition or arguing by contradiction, but it does not seem to be easy.
Thanks in advance.

答:
Hint:

Show that

$$\hat E := \left\{ (x, \lambda x): x \in \mathbb{R}^{n+1}, \lambda \in  \mathbb{R} \right\} $$

is a closed subset of $\mathbb R^{2n+2}$, then use the definition of subspace topology.

Title: Prove this set is closed.


问:
How do you "regularize" infinite integrals?
This question was inspired by the post:
" Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds? " (and the follow-up discussion). Long story short, the OP was using some technique to "regularize" infinite integrals (and claims, for example, that $\int_2^{\infty}\frac{dx}{x}$ has a "regularized value" of $-\log 2$) - and then complaining that whatever regularization they were using did not play well with the rules for change of variable in an integral.

So... What is the "regularization" of an infinite integral? Are there multiple ways to do it, or do all of them boil down to the same thing whenever they "work"? Is there a preferred way of doing it? What generalizations of this notion have been used/studied/are practical? Is it a topic of active research? What are known unsolved problems related to it?

(Of course, I don't expect all those questions to be answered here, I am more after some sort of reference that would possibly be understandable with basic knowledge of real and complex analysis and Lebesgue integration, or perhaps slightly more than that. Ideally an article of the type "Everything you've always wanted to know about regularization but were afraid to ask...")

答:
Well.

First, you can regularize some integrals with the same techniques as you regularize divergent series. For instance, you can regularize some integrals using Cesaro summation. Basically it means finding the average of the integral as the integration limit goes to infinity.

Using these and related techniques one can find that

$$\int_0^\infty \sin x\, dx=1$$

and even

$$\int_0^\infty \tan x\,dx=\ln 2$$
(see here)

But these integrals go to some limit at least in a sense of mean value. There are other integrals that go to infinity, and they allow regularization as well.

For instance, we know that the harmonic series is regularized to Euler-Mascheroni constant (by Zeta regularization or Ramanujan):

$$\operatorname{reg}\sum_{k=1}^\infty \frac1k=\gamma$$

At the same time, the difference between the partial sum and integral goes to $\gamma$
as well (the blue area):

$$\gamma = \lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\int_1^n\frac1t dt\right)$$

This allows us to conclude that $$\operatorname{reg}\int_1^\infty\frac1t dt=\operatorname{reg}\sum_{k=1}^\infty \frac1{k}-\gamma=0$$ as regularization is always a linear operator.

In general, one can reduce regularizing an integral to regularizing the corresponding series:

$$\operatorname{reg}\int_0^\infty f(x)\,dx=\lim_{s\to0}\left( s \operatorname{reg} \sum_{k=1}^\infty f(sk)\right)$$


问:
All nested partial sums of a sequence tend to $0$. Is the sequence constant?
$S^0:\mathbb N\to \Bbb R$ is a function. For any $m\in \Bbb N$, we define

$$S^m:\Bbb N\to \Bbb R$$
$$S^m(n)=\sum_{k=1}^n S^{m-1}(k)$$

For each $m\in \Bbb N$, we have:
$$\lim_{n \to \infty}S^m(n)=0$$

Can we deduce $S^0\equiv 0?$

答:
Hmm, it might be that I did not understand the question correctly, but I find the problem interesting so I'll give it a try and please correct me if I'm wrong.

Here is, what I understand from the definitions of the $S_m$ with an example at $n=4$:
$$ \begin{eqnarray}
S^1(4) &=& S^0(1)+ S^0(2)+ S^0(3)+ S^0(4)  \\
 &=&    1 S^0(1)+ 1 S^0(2)+ 1 S^0(3)+ 1 S^0(4) \\
S^2(4) &=& S^1(1)+ S^1(2)+ S^1(3)+ S^1(4)  \\
 &=&    4 S^0(1)+ 3 S^0(2)+ 2 S^0(3)+ 1 S^0(4) \\
S^3(4) &=& S^2(1)+ S^2(2)+ S^2(3)+ S^2(4) \\
 &=& 10 S^0(1)+ 6 S^0(2)+ 3 S^0(3)+ 1 S^0(4) \\
\end{eqnarray} $$
and
$$ S^m(n) = \sum_{k=1}^n \binom{k+m}{m}S^0(k)$$ where also $ \lim _{n \to \infty}S^m(n) = 0$ is assumed. I hope I've got this correct so far.

Now in the example, if we subtract $S^1(4)$ from $S^2(4)$ we get $$S^2(4)-S^1(4) = S^2(3)$$ which is generalizable in the obvious way $$S^m(n)-S^{m-1}(n) = S^m(n-1) $$ or
$$S^m(n)-S^m(n-1) = S^{m-1}(n) $$
But as $S^m(n)$ goes to zero so does $S^m(n-1)$, and also $ \lim_{n \to \infty}S^m(n)-S^{m-1}(n) = 0 $

If that reasoning is correct, then we have also from $m=1$ only that $$ \lim_{n \to \infty}S^0(n) = 0 $$

Hmm, that's not the final solution I'm afraid. Did your question actually mean that $S^0(n)=0$ for all $n \in \mathbb N$ ?

[Update]: Ok, after that we can write things in matrix-notation. For shortness let's denote $S^0(n)$ as $a_n$ and $S^m(n)$ as $z_{m,n}$ and let us then rewrite the vector/matrix-product for the $n=4$ example:
$$
\begin{array} {lll}
    & * &\left[ \begin{array}  {rrrr} \quad 1  & \quad 4  & \quad 10  & \quad 20\\
                          1 & 3& 6  &10\\
                          1 & 2& 3  & 4\\
                          1 & 1& 1  & 1
        \end{array} \right]
   \\
  \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 \end{array} \right]
 & = &\left[ \begin{array} {rrrrr} \ \  z_{1,4} & \  z_{2,4} & \  z_{3,4} &  \ \  z_{4,4} \end{array} \right]
 \end{array}
$$
If this scheme is extended to infinite size ($ n \to \infty$) in the obvious way, the problem says that we expect that for all k we find $z_{k,n} \to 0$

To divide&conquer the problem, we can LR-factorize the coefficients matrix, let's call it $C_n$ with the index n for the size such that $C_n = L_n \cdot R_n$ with $L_n$ and $R_n$ triangular matrices. We have then the formal definition
$$ \begin{eqnarray} A_n \cdot C_n &=& Z_n \\ A_n \cdot (L_n \cdot R_n) &=& Z_n \\ \end{eqnarray}$$
We get then the interesting effect, that $L_n$ becomes the lower triangular Pascal-matrix and its entries are fixed for any size; so we can look at the left dot-product $ A_n \cdot L_n = B_n $ first and this looks then like

$$
\begin{array} {lll}
    & * &\left[ \begin{array}  {rrrr}
                    \quad 1  &   &  & \\
                          1 & 1&   &\\
                          1 & 2& 1  & \\
                          1 & \  3& \  3  &  \ 1 & \cdots\\
                      \vdots & \vdots & \vdots & \vdots & \ddots
        \end{array} \right]
   \\
  \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 & \cdots \end{array} \right]
 & = &\left[ \begin{array} {rrrrr} \quad  b_0 &  b_1 &  b_2 &  b_3 & \cdots \end{array} \right]
 \end{array}
$$
where if we understand the values $a_n$ as coefficients of a power series in x we have that the $b_0$ can expressed as a function $f(x)$ at $x=1$ by the power series  $$ f(x) = \sum_{k=0}^\infty a_{1+k} x^k  $$and the $b$'s of the following columns are then the derivatives such that
$$ b_k= x^k {f^{(k)}(x)\over k!} \qquad \qquad \text{ at } x=1 $$

Now the right matrix $R$ is upper triangular; and its entries change with the increasing $n$; we get with the $n$ kept indeterminate
$$ R = \left[ \begin{array} {ccccc}
 1 & n & \binom{n+1}{2} &\binom{n+2}{3} &\binom{n+3}{4}& \cdots \\
 .  & -1 & -n & -\binom{n+1}{2} &-\binom{n+2}{3} & \cdots \\
 . & .  & 1 & n & \binom{n+1}{2} & \cdots \\
 . & . & .  &- 1 & -n &  \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots &\ddots\\
\end{array} \right] $$
and we want, that the matrix-product of $B_n \cdot R_n = Z_n $ in the limit $ n \to \infty$ arrives at the nullvector $ \lim_{n \to \infty}  Z_n = [0,0,0,...]$.

From the first column of $R$ we see, that $b_0=f(1)=0$ is required to make $z_1=S^1(n)=0$.

From this seems(!) to follow, that for the second column to provide $z_2=0$ also $b_1=0$ is required and from this for all other columns too. However, because n goed to infinity it might be, that there is some more consideration of the limits required - but I'm not yet sure of this.

But given, that this considerations are all correct, then we see, that $f(x)$ must be a function, where all derivatives at $x=1$ become zero - and I know only one function, which is not constant and has still all derivatives at some (single (!)) point $x_0$ equal to zero, this is $g(x)=\exp(-{1 \over x^2})$      at $x_0=0$ .

Caution: if divergent expressions are involved this might look differently - I don't see this yet.

So, although this small study was a nice experience I still cannot answer this question to one side so far..., sorry...hmmm...


问:
Analyticity of this multivariate Laplace transform
Consider a $d$-dimensional random vector $\mathbf X$ with laplace transfrom $$L(\mathbf t) = \left(1 + \langle \mathbf \theta, \mathbf t \rangle\right)^{-\alpha}$$

Where $\mathbf \theta$ is a $d$-dimensional vector of non-negatives reals (some can be zero, not all), and $\alpha$ is a strictly positive real (that cannot be zero). The coresponding random vector writes as $\mathbf X = \left(\theta_1 Y,...,\theta_d Y\right)$ for $Y$ a random variable following a gamma distribution with shape $\alpha$ and unit scale, that is with laplace tranform $\left(1 + t\right)^{-\alpha}$. Therefore, $\mathbf X$ is a comonotonous vector, living on a straigth half-line in $\mathbb R^d$.

Question: Considering $L$ as a function from $\mathbb C^d$ to $\mathbb C$, what is the region of convergence of $L$ ? Where are it's singularities ?

答:
$\alpha >0$ and
$$\int_{\Bbb{R}^n} e^{-\langle t, x\rangle} d\mu_X(x)= \int_{ \Bbb{R}}e^{-\langle t, \theta y\rangle} d\mu_Y(y)=
\int_{[0,\infty)}e^{-\langle t, \theta \rangle y} \frac{y^{\alpha-1} e^{-y}}{\Gamma(\alpha)} dy
 $$ converges for $\Re(\langle \theta,t\rangle) > -1$ and diverges for $\Re(\langle \theta,t\rangle) \le -1$.

It extends analytically to $\{ t\in \Bbb{C}^n,\langle \theta,t\rangle\in U\}$ where $U$ is any simply connected open of $\Bbb{C}-\{-1\}$.

If $\alpha$ is an integer then it extends to a meromorphic function on $\Bbb{C}^n$ with only one pole of order $\alpha$ at the hypersurface $\langle \theta,t\rangle=-1$.


问:
Problem using Lagrange Multipliers in Utility function
I am to optimize utility given the utility function
$$ u(c,l):=c-\frac{\eta}{\eta+1}(24-l)^{\frac{\eta+1}{\eta}},$$

where $c$ represents consumption and $l$ represents leisure. The budget constraint given: $pc+wl=24w$; $p$ represents the price of consumption and $w$ represents hourly wage. I have found the optimal amount of leisure, but I can’t find the optimal amount of consumption for the life of me. I have found the optimal amount of leisure, $l^*$, as follows:

I have written it as a Lagrangian maximization problem:
$$ L(c,l,\lambda)=c-\frac{\eta}{\eta+1}(24-l)^{(\eta+1)/\eta} - \lambda(pc+wl-24w) $$

and found the following:
$$ \frac{\partial (U)}{\partial(l)}=-\lambda w+ (24-l)^{1/\eta}=0 $$

Solving for $\lambda$:
$$ \lambda = \frac{(24-l)^{1/\eta}}{w} $$
and
$$ \frac{\partial (U)}{\partial(c)}=-\lambda p+1=0 $$

Solving for $\lambda$:
$$ \lambda = 1/p$$

I then set $\lambda$ equal to $\lambda$ and solved the equation for $l$:
$$ \frac{(24-l)^{1/\eta}}{w} = 1/p $$
$$ l^*= 24-\left(\frac{w}{p}\right)^{\eta} $$

I have concluded that this is the optimal amount of leisure, however, I can’t find the optimal amount of consumption, as $c$ of course vanishes unless you find the partial derivative with respect to $\lambda$; this, however, of course doesn’t yield an equation containing $\lambda$, so I can’t isolate $\lambda$ and find $c^*$ how I found $l^*$. Any help would be much appreciated :)

答:
So far so good. Next you make the  derivative of  $\mathcal L(c,l,\lambda)$ w.r.t $\lambda$, which is just the constraint.

$$\frac{\partial\mathcal L(c,l,\lambda)}{\partial \lambda}=24w-pc-wl=0$$

In combination with $l^*= 24-\left(\frac{w}{p}\right)^{\eta}$ we obtain

$$24w-pc-w\cdot \left(24-\left(\frac{w}{p}\right)^{\eta} \right)=0$$

It remains to solve the equation for $c$. I don´t think that you need the next steps. So I hide them and you can compare them with your own steps at the end.

$$-pc+w\cdot \left(\frac{w}{p}\right)^{\eta} =0$$
$$c^*=\frac{w}p\cdot \left(\frac{w}{p}\right)^{\eta} $$
$$c^*= \left(\frac{w}{p}\right)^{\eta+1} $$


问:
solving for one variable in terms of others
A question from Steward's Precalculus textbook 5th, Pg 55,

the original formula is $$h=\frac{1}{2}gt^2+V_0t$$

the question asks to write the formula in terms of $t$, the answer is $$t=\dfrac{-V_0\pm\sqrt{v_0^2+2gh}}{g}$$

I don't know the steps on how to get there

Body:
$$h=\frac{1}{2}gt^2+V_0t$$
$$2h=gt^2+2V_0t$$
$$gt^2+2V_0t-2h=0$$

now use this formula for Quadratic equation:$ax^2+bx+c=0$;
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

so: $$t=\dfrac{-2V_0\pm\sqrt{4V_0^2+8gh}}{2g} $$
$$t=\dfrac{-V_0\pm\sqrt{V_0^2+2gh}}{g} $$


问:
How to use mean value theorem For harmonic functions to prove this
The following question was asked in my complex analysis assignment and I am confused about how this should be done.

Show that $\int_{0}^{\pi} \ln( \sin {\theta}) d{\theta} = -\pi \ln 2$ by applying the mean value theorem to $ \ln|1+z|$ for $|z|\leq r <1$ and then letting $r\to 1$.

If I use mean value theorem to $\ln(1+z)$,I get $ln|(1+z_0)| = \frac{1}{2\pi} \int_{0}^{2\pi}\ln|1+z_0+ r e^{i\theta}|d {\theta}$.

But how to change the $1+z_0 + re^{i \theta}$  in RHS into $\sin\theta$ if $r$ approaches $1$?

I am not able to manipulate into that.

答:
Set $z_0=0$, then
$$\begin{split}
0&=\frac 1 {2\pi} \int_0^{2\pi}\ln |1+re^{i\theta}|d\theta\\
&\rightarrow \frac 1 {2\pi} \int_0^{2\pi}\ln |1+e^{i\theta}|d\theta\\
&= \frac 1 {2\pi} \int_0^{2\pi}\ln\left |2\cos \frac \theta 2\right|d\theta\\
&= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln |\cos \theta |d\theta\\
&= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln (\sin \theta )d\theta\\
\end{split} $$


问:
Why is it called 'King's Property of Integration'?
Recently I learned about "King's Property" or "King's Rule" and I was wondering about its etymology? I understand that it's a basic change of variables but it has a name for a reason. I've been unable to find this reason online. I speculate that the person who first encountered it had the popular surname King.

In case terminology is different elsewhere, here's what I'm talking about:
$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$

答:
I have never heard this referred to as "King's Property" or "King's Rule".  And there is almost nothing (except this thread) in Google.  There is this video, maybe your answer is there. (Video can be viewed only by paid students)  Maybe the term "King's Property" is unique to that video.

The video is from an educational company (?) called Nucleon, located in Rajasthan, in northern India.  "Rajasthan" means "Land of Kings".

As pointed out in a comment, the "King's Property" has been mentioned in math.se before.  The oldest one seems to be Trigonometric definite integration .  That post is from July, 2017.  The post is by Tarun Raj Latiyan.  Wouldn't it be interesting if Tarun is from India and learned this terminology from the Nucleon video?


问:
How to calculate the divergence by the definition
I am trying to find the divergence of this field: $E = \frac{q(1-\sqrt{\alpha r})}{4 \pi \epsilon r^2} \hat r$

I already found the surface integral, that is, $\int EdS = \frac{q(1-\sqrt{\alpha r_{1}})}{\epsilon}$

I am having trouble to evaluate how to calculate the divergence by the definition now. (It has to be by definition

答:
Please note that the vector field is undefined at the origin.

Here is how you would find divergence in spherical coordinates -

$\displaystyle \vec{E} = \frac{q(1-\sqrt{\alpha r})}{4 \pi \epsilon r^2} \hat r$

$\nabla\cdot\vec{E} = \displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E) + 0 \,$ as  $\frac{\partial}{\partial \phi}, \frac{\partial}{\partial \theta} \,$ will be zero.

$\displaystyle  = \frac{q}{4 \pi r^2 \epsilon}\frac{\partial}{\partial r}(1-\sqrt{\alpha r}) = -\frac{q \sqrt \alpha}{8 \, \pi \, \epsilon \, r^{5/2}}$

If you want to do it in Cartesian coordinates -

As $\displaystyle \hat{r} = \frac{\vec{r}}{r} = \frac{(x, y, z)}{r}, $

$\displaystyle E_x = \frac{q}{4 \pi \epsilon}\frac{\partial}{\partial x} \Bigg(\frac{x}{(x^2+y^2+z^2)^{3/2}} - \frac{\sqrt \alpha \, x}{(x^2+y^2+z^2)^{5/4}}\Bigg)$

$E_y, E_z$ will be similar to $E_x$. So once you find $E_x$, you can find,

$\nabla \cdot \vec{E} = E_x + E_y + E_z \, $ and you will get the same value as above in spherical coordinates.


问:
Rate of convergence from asymptotic limit
I am working on an algorithm in which one parameter $a(t)$ verifies the following asymptotic limit: $f(a(t))=\mathcal{O}(1/t^{2})$. I am using it to assess an empirical rate of convergence of $f(a(t))$ as $\frac{f(a(t))}{f(a(10t))}\approx100$. It works more or less well, but I know that mathematically it is not formal. Do you have any suggestion on how to compute $\frac{f(a(t))}{f(a(10t))}$ from the asymptotic limit?

答:
$f(a(t))=O(1/t^2)$ means that $f(a(t))=\frac{g(t)}{t^2}$, $g$ being a bounded function. So
$$\frac{f(a(t))}{f(a(10t))} = \frac{\frac{g(t)}{t^2}}{\frac{g(10t)}{100t^2}} = 100\frac{g(t)}{g(10t)}$$
Problem is, you have to know both bounds for $g$ to make this accurate. If you know
$$(\forall t)\ 0<A\le g(t)\le B$$
which is quite common in algorithms theory, then
$$\frac{f(a(t))}{f(a(10t))} \le \frac{100B}{A}$$
But in general, you can not find any good bound (and this bound may not exist).


问:
Finding algebraic integer having a non-real conjugate $\beta$ with smaller absolute value and $\beta^n$ is real
I have a math model and I would like to find some algebraic parameters which shows periodic behavior when raising powers. The problem is:

Given two positive real numbers $L\geq\ell\geq1$, can we construct an algebraic integer $\alpha$ satisfying the following properties:

$\alpha$ is a real number and $\alpha\geq L$

$\alpha$ has an algebraic conjugate $\beta$ with $\ell\geq|\beta|\geq1$

$\beta$ is not a real number, but $\beta^n$ is a real number for some $n\in\mathbb{N}$.

So the last property is that $\arg\beta$ is a rational number times $\pi$, and the condition $L\geq\ell$ forbids examples like $\alpha = 13^{1/3}$.

Can we have an algorithm or examples for e.g. $(L, \ell) = (3, 2)$ and arbitrary $(L, \ell)$?

答:
If $\ell=1$ then $\beta$ is a root of unity so $\alpha=\pm 1$. Otherwise

$$\phi = \frac{1+\sqrt5}{2},\qquad \phi^* = \frac{1-\sqrt5}{2}=-\phi^{-1}$$
Take $k$ odd large enough and $m$ chosen such that
$$\phi^{k/2}> L, \qquad m\phi^{-k/2}\in (1,\ell) $$
Let $$\alpha = m\phi^{k/2}, \qquad \beta = i m\phi^{-k/2} = m(\phi^*)^{k/2}$$


问:
On convergence of problematic series.
Determine if the following series is converges or not

$$\sum_{n=2}^{\infty}\frac{1}{\ln^{\ln(\ln n)} (n)}$$

答:
Not too hard seeing that the following limit is infinite:

$$\lim_{n\to\infty}n^1\times\frac{1}{\ln^{\ln(\ln n)}(n)}=+\infty$$ So according to a result of the Comparison Test, the series diverges.

Title: On convergence of problematic series.


问:
two integrations with functions inside
I have thee next problem

If f is a continuous function that satisfies $$\int_3^4 f(t)\,dt=3$$
calculate $$\int_9^{16}\frac{f(\sqrt x)}{\sqrt x}\,dx$$

I don't even know how to start, does it talk about the first function being applied in the nominator of the second? so $3 =\sqrt x$?

答:
Substitute $$
\sqrt{x}=t,x=t^2
$$
And we get
$$
\int_9^{16}{\frac{f\left( \sqrt{x} \right)}{\sqrt{x}}dx}=\int_3^4{\frac{f\left( t \right)}{t}2tdt}=2\int_3^4{f\left( t \right) dt}=6
$$

Title: two integrations with functions inside


问:
Exact match of assets and liabilities.
Liabilities of $1$ each are due at then ends of periods $1$ and $2$. There are three securities available to produce asset income to cover these liabilities, as follows:

(i) A bond due at the end of period 1 with coupon at rate .01 per period, valued at a periodic yield of 14%

(ii) A bond due at the end of period 2 with coupon at rate .02 per period, valued at a periodic yield of 15%

(iii) A bond due at the end of period 2 with coupon at rate .20 per period, valued at a periodic yield of 14.95%

Determine the cost of the portfolio that exactly-matches asset income to liabilities due using

(a) bonds (i) and (ii) only.

(b) bonds (i) and (iii) only

(c) Show that the combination of securities in (b) minimizes the cost of all exact-marching portfolios made up of a combination of the three securities. Note that the minimum cost exact-matching portfolio does not use the highest yielding security in this case.

My attempt was to match the assets with liabilities for t=1 and t=2 here is how I did it:

$L_1$ at $t=1$ is $1(1.14^{-1})=0.877$

$L_2$ at $t=2$ is $1(1.15^{-2})=0.75614$

$X(1.01)(1.14^{-1})+Y(0.02)(1.15^{-1})=0.877$

$Y(1.02)(1.15^{-2})=0.75614$

Solving these leads to $X=0.97$ and $Y=0.98$

$PV=0.97(1.01)(1.14^{-1})+0.98(0.02(1.15^{-1})+1.02(1.15^{-2})=1.633$

Granted that review this afterward I realize I am just turning around and not actually solving the problem. Could anyone help me out and explain what are the proper steps and why?

答:
Myy solution is to solve for bond 2 first, because to reach the exact matching, we need to ensure that A=L at all time, and the 1st bond couldn't cover the liability at the end of period 2. Then substitute the result of bond 2 into the equation of period 1.
Hope this will be helpful.</p>

Title: Exact match of assets and liabilities.


问:
Finding the Taylor series of $f(z)=\frac{1}{z}$ around $z=z_{0}$
I was asked the following (homework) question:

For each (different but constant) $z_{0}\in G:=\{z\in\mathbb{C}:\,
 z\neq0$} find a power series $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$
  whose sum is equal to $f(z)$ on some subset of $G$. Please specify
  exactly on which subset this last claim holds.

Suggestion: Instead of calculating derivatives of f, try
  using geometric series in some suitable way.

What I did:

Denote $f(z)=\frac{1}{z}$ and note $G\subseteq\mathbb{C}$ is open.
For any $z_{0}\in G$ the maximal $R>0$ s.t $f\in H(D(z_{0},R))$
is clearly $R=|z_{0}|$.

By Taylor theorem we have it that $f$ have a power series in $E:=D(z_{0},R)$
(and this can not be expended beyond this point, as this would imply
that $f$ is holomorphic at $z=0$.

I am able to find $f^{(n)}(z_{0})$ and to solve the exercise this
way: I got that $$f^{(n)}(z)=\frac{(-1)^{n}n!}{z^{n+1}}$$ hence $$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z_{0}^{n+1}}(z-z_{0})^{n};\, z\in E$$

but I am not able to follow the suggestion, I tried to do a small
manipulation $$f(z)=\frac{1}{z-z_{0}+z_{0}}$$ and tried to work a bit
with that, but I wasn't able to bring it to the form $$\frac{1}{1-\text{expression}}$$
or something similar that uses a known power series .

Can someone please help me out with the solution that involves geometric
series ?

答:
$ f(z) $ does not have a Maclaurin series (a Taylor series centered at $ x = 0 $) because its derivative does not exist there. However, it does have a Laurent series there, namely $ \frac{1}{x} $ is the Laurent series as well as the function.

A Taylor series centered at $ x = x_0 $ only exists if the derivatives of $ f $ at $ x_0 $ exist as well.


问:
Is this a mistake $\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0?$
Can anyone show that,
$$\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0\tag1$$
Suppose $\ n\ $ and $\ r\ $ are integers.

I can't understand why it should be zero.

making a basic substitution:

$u=1-x^r$

$\mathrm dx=\frac{\mathrm du}{r(1-u)}$

$$\int_{0}^{1}\frac{u^{n/r}-n\cdot u^{1/r}(1-u)^{(n-1)/r}}{r(1-u)}\mathrm du$$

答:
Here is an elementary solution: Let $y = (1 - x^{r})^{1/r}$. In other words, $x^r + y^r = 1$. Then

\begin{align*}
\int_{x=0}^{x=1} (1 - x^r)^{n/r} \, \mathrm{d}x
&= \int_{x=0}^{x=1} y^n \, \mathrm{d}x
= \Bigl[ y^n x \Bigr]_{x=0}^{x=1} - \int_{x=0}^{x=1} x \, \mathrm{d}(y^n) \\
&= - \int_{x=0}^{x=1} n y^{n-1} x \, \mathrm{d}y.
\end{align*}

Now using that $x = (1 - y^r)^{1/r}$, and noting that $x = 0$ (resp. $x=1$) implies $y=1$ (res.p $y=0$) the above integral is equal to

\begin{align*}
- \int_{y=1}^{y=0} n y^{n-1} (1 - y^r)^{1/r} \, \mathrm{d}y
= \int_{y=0}^{y=1} n y^{n-1} (1 - y^r)^{1/r} \, \mathrm{d}y.
\end{align*}

Since the name of the dummy variable is irrelevant for the value of the integral, this tells that

$$ \int_{0}^{1} (1 - x^r)^{n/r} \, \mathrm{d}x = \int_{0}^{1} n x^{n-1} (1 - x^r)^{1/r} \, \mathrm{d}x $$

and therefore the integral in question is zero.

Addendum. The above computation is best understood if one is familiar with the concept of line integral. If one wants to stick to single-variable flavor, the above computation can be rephrases as follows:

Define $\varphi(x) = (1 - x^{r})^{1/r}$. Then it is straightforward to check that $\varphi(\varphi(x)) = x$, i.e., $\varphi$ is its own inverse. Then by substituting $y = \varphi(x)$, or equivalently, $x = \varphi(y)$,

\begin{align*}
\int_{0}^{1} \varphi(x)^n \, \mathrm{d}x
&= \int_{\varphi(0)}^{\varphi(1)} y^n \varphi'(y) \, \mathrm{d}y \\
&= \Bigl[ y^n \varphi(y) \Bigr]_{1}^{0} - \int_{1}^{0} n y^{n-1} \varphi(y) \, \mathrm{d}y \\
&= \int_{0}^{1} n y^{n-1} \varphi(y) \, \mathrm{d}y.
\end{align*}


问:
Is my proof to this number theory question valid?
I had the following question:

Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?

$P(x)$ has no rational root;

For every positive integer $n$, there exist an integer $m$ such that $n$ divides $P(m)$.

I created a proof showing that there was no polynomial satisfying both of these conditions:

Suppose that we have a nonzero polynomial with integer coefficients $P(x)=\sum_i c_i x^i$ without a rational root, and for all positive integer $n$, we have an integer $m_n$ such that $n|P(m_n)$. This would imply $P(m_n)\equiv0\pmod n\Rightarrow \sum_i c_i m_n^i\equiv0$. By Freshman's Dream we have $P(m_n+an)=\sum_i c_i(m_n+an)^i\equiv\sum_i c_im_n^i+c_ia^in^i\equiv\sum_ic_im_n^i=P(m_n)\equiv0\pmod n$ for some integer $a$. Therefore if $b\equiv m_n\pmod n$ then $P(b)\equiv P(m_n)\equiv0\pmod n$.

Now the above conditions and findings imply for all prime $p$, we have a number $m_p$ such that $p|P(m_p)$, and that if $b\equiv m_p\pmod p$ then $P(b)\equiv 0\pmod p$. Consider the set of the smallest $n$ primes $\{p_1,p_2,p_3,\cdots,p_n\}$. By Chinese Reamainder Theorem there exists an integer $b$ such that $b\equiv m_{p_1}\pmod{p_1},b\equiv m_{p_2}\pmod{p_2},b\equiv m_{p_3}\pmod{p_3},\cdots,b\equiv m_{p_n}\pmod{p_n}$ by Chinese Remainder Theorem. Then $p_1,p_2,p_3\cdots,p_n|P(b)\Rightarrow p_1p_2p_3\cdots p_n|P(b)$. As $n$ approaches infinity (as there are infinitely many primes), $p_1,p_2,p_3\cdots,p_n$ approaches infinity. Therefore either $P(b)=\infty$ or $P(b)=0$. Since for finite $b$ and integer coefficients $P(b)$ must be finite, then $P(b)=0$. However as $a$ is an integer, this implies $P$ has a rational root, a contradiction.

I'm not sure if my proof is correct, and my main concern is that I am incorrectly using Chinese Remainder Theorem since I am not sure if I can apply it to infinitely many divisors.

Is this proof correct, and if not, how do I solve this question?

EDIT: It appears not only is my proof incorrect (as$b$ does not converge) as Paul Sinclair has shown, but that according to Jaap Scherphuis there are examples of polynomials that satisfy the conditions. Therefore, my question now is how one can prove the existence of these polynomials while using elementary methods (as this is an IMO selection test problem).

答:
Consider the polynomial

$$
P(x) = (x^2 - 13)(x^2 - 17)(x^2 - 221)
$$

Clearly this has no rational solutions. We wish to show that the congruence $P(x) \equiv 0 \bmod{m}$ is solvable for all integers $m$. By the chinese remainder theorem, this is the same as showing $P(x) \equiv 0 \bmod{p^k}$ is solvable for all prime powers $p^k$.

Notice that if either $13$, $17$, or $221$ is a quadratic residue modulo $p^k$, then one of the quadratic terms in $P$ is a difference of squares modulo $p^k$ and factors into linear terms and we are done. We show this is always the case.

For odd $p \neq 13,17$, not all of $13$, $17$, $221$ can be quadratic non-residues modulo $p^k$, otherwise we would have the following contradiction

$$
-1 = \left( \frac{221}{p^k}\right) = \left( \frac{13}{p^k}\right)\left( \frac{17}{p^k}\right)=(-1) (-1)=1
$$

If $p = 13$ then 17 is a quadratic residue modulo $p^k$ because by quadratic reciprocity we have

$$
\begin{eqnarray}
\left( \frac{17}{13^k}\right)  &=& \left( \frac{13^k}{17}\right) (-1)^{\frac{13^k -1}{2} \frac{17 -1}{2}} \\
&=& \left( \frac{13}{17}\right)^k (-1)^{\frac{13^k -1}{2} \frac{17 -1}{2}} \\
&=& 1
\end{eqnarray}
$$

If $p=17$ a similar argument applies.

For $p=2$ we can use Hensel's lemma to show that for any odd $a$, $x^2 \equiv a \bmod{2^k}$ is solvable for all $k$ if and only if $a \equiv 1 \bmod{8}$. Therefore $17$ is a quadratic residue modulo $2^k$ for all $k$.


问:
Find all functions that satisfy the identity $f(x+5y)+f(x-5y)=2x^2+50y^2$.
My method goes like this.

Since I found that $2x^2+50y^2=2(x^2+25y^2)$ which can be written as a difference of squares in complex numbers. So I suppose $x+5y=a+bi=z$, so $x-5y=a-bi=\bar z$.

So basically, $f(z)+f(\bar z)=2z\bar z=z\bar z+z\bar z=z\bar z+\bar z\bar {\bar z}$.

So, $f(z)=z\bar z$. In reals, that is $f(x)=x^2$.

Is my method valid? I feel like getting this by luck. Could anyone introduce me a more general method?

答: Set $y = 0$ to obtain $f(x) = x^2$ for all $x\in\mathbb{R}$. Plugging this into the functional equation, we see that $f(x) = x^2$ is indeed a solution, and this is the only solution.


答: $$(A+B)^2+(A-B)^2=2(A^2+B^2)$$ is simpler in this regard.

Title: Find all functions that satisfy the identity $f(x+5y)+f(x-5y)=2x^2+50y^2$.

Body:
My method goes like this.

Since I found that $2x^2+50y^2=2(x^2+25y^2)$ which can be written as a difference of squares in complex numbers. So I suppose $x+5y=a+bi=z$, so $x-5y=a-bi=\bar z$.

So basically, $f(z)+f(\bar z)=2z\bar z=z\bar z+z\bar z=z\bar z+\bar z\bar {\bar z}$.

So, $f(z)=z\bar z$. In reals, that is $f(x)=x^2$.

Is my method valid? I feel like getting this by luck. Could anyone introduce me a more general method?


问:
Split short exact sequence contradiction
Am I doing it wrong or is this an exception to the split rules?

There is a short exact sequence
$$1 \longrightarrow \mathbb Z/_{2\mathbb Z} \hookrightarrow \mathbb Z/_{4\mathbb Z} \twoheadrightarrow \mathbb Z/_{2\mathbb Z} \longrightarrow 1$$
Using group representations
$\mathbb Z/_{2\mathbb Z} \simeq \left\{\begin{pmatrix}-1&0\\0&-1\end{pmatrix}, \begin{pmatrix}1&0\\0&1\end{pmatrix}\right\}$ and $\mathbb Z/_{4\mathbb Z} \simeq \left\{\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\begin{pmatrix}1&0\\0&1\end{pmatrix}, \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \begin{pmatrix}0&-1\\1&0\end{pmatrix}\right\}$

There cannot be any retraction but using the surjection $\rho : \begin{pmatrix}a&b\\c&d\end{pmatrix} \longmapsto ad + bc$
isn't there a section $\iota : \begin{pmatrix}-1&0\\0&-1\end{pmatrix} \longmapsto \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ ?

This would imply that we have a semi-direct product but I know this is not the case because $\mathbb Z/_{4\mathbb Z} \not\simeq \mathbb Z/_{2\mathbb Z} \rtimes \mathbb Z/_{2\mathbb Z} \simeq \mathbb Z/_{2\mathbb Z} \times \mathbb Z/_{2\mathbb Z}$...
What gives ?

答: Your $\iota$ is not a homomorphism. $\iota(-I)^2 \ne \iota((-I)^2)$.


问:
Why is this Isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Why is this table isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$?

Below is the additive composition table of four cosets of $\mathbb{Z}_2[x]/(x^2+x+1)$.

I don't understand why it's isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. The reasoning should be really simple but I can't find why exactly..

答:
From having all zeroes in the diagonal, every nonzero element has order 2. So, if it is a group, it is $\mathbb Z_2\times\mathbb Z_2$.

Now you can start to think who can be $(1,0)$, $(0,1)$, and $(1,1)$.


答:
Hint: map $(0,0) \mapsto 0$, $(1,0) \mapsto 1$, $(0,1) \mapsto x$, $(1,1) \mapsto 1+x$, can you show this is an isomorphism?

Another way to look at this: $x^2+x+1$ is irreducible over the field $\mathbb{F}_2$ and hence $\mathbb{F}_2[x]/(x^2+x+1) \cong \mathbb{F}_4$, which is a $\mathbb{F}_2$-vectorspace of dimension $2$. Hence the underlying abelian group structure is $\mathbb{Z}_2 \oplus \mathbb{Z}_2$.

Title: Why is this Isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Body:
Why is this table isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$?

Below is the additive composition table of four cosets of $\mathbb{Z}_2[x]/(x^2+x+1)$.

I don't understand why it's isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. The reasoning should be really simple but I can't find why exactly..


问:
Showing $\sum_{i=1}^n y_i = O_p(\sqrt{n})$ if $E[y_i] = 0$ and $\sum_{i=1}^n y_i = O_p(n)$ if $E[y_i] \neq 0$?
Let $y_1,\dots,y_n$ be a set of i.i.d random variables with mean $\mu$ and variance $\sigma^2$. Can it be shown that
$$
\begin{align}
\sum_{i=1}^n y_i = O_p(\sqrt{n}), & \quad \quad \text{if} \ E[y_i] = 0, \\
\sum_{i=1}^n y_i = O_p(n), & \quad \quad \text{if} \ E[y_i] \neq 0.
\end{align}
$$ using the definition of convergence in probability for Big $O_p$ notation?

I saw a simplified explanation for the results in page 4 of these notes but I'm wondering can it be done directly using the the definition of $O_p$, i.e. $X_n = O_p(n)$ if for all $\varepsilon > 0$, there exists $M,N>0$ such that $P(|X_n/n|>M) < \varepsilon$ for all $n > N$.

答:
The CLT says
$$
n^{1/2}\bigg(\frac{1}{n} \sum_{i=1}^n y_i - \mu\bigg) \stackrel{d}{\to} N(0,\sigma^2).
$$
Since the term on the left hand side converges in distribution it is bounded in probability:
$$
n^{1/2}\bigg(\frac{1}{n} \sum_{i=1}^n y_i - \mu\bigg) = O_p(1)
$$
which menas
$$
\begin{align}
\frac{1}{n} \sum_{i=1}^n y_i = \mu + O_p(n^{-1/2}) & \Longleftrightarrow \sum_{i=1}^n y_i = n\mu + O_p(n^{1/2}).
\end{align}
$$
Now if $E[y_i] = 0$ we get the rate $O_p(n^{1/2})$. Otherwise $\mu = O_p(1)$ since it is a 'constant random variable' and thus
$$
\begin{align}
\sum_{i=1}^n y_i = nO_p(1) + O_p(n^{1/2}) = O_p(n).
\end{align}
$$


答:
I think you were on the right track:

$$\frac1z=\frac1{z_0+(z-z_0)}=\frac1{z_0}\cdot\frac1{1+\frac{z-z_0}{z_0}}=\frac1z_0\left(1-\frac{z-z_0}{z_0}+\frac{(z-z_0)^2}{z_{0}^2}-\ldots\right)=$$

$$=\frac1z_0-\frac{z-z_0}{z_0^2}+\frac{(z-z_0)^2}{z_0^3}+\ldots+\frac{(-1)^n(z-z_0)^n}{z_0^{n+1}}+\ldots$$

As you can see, this is just what you got but only using the expansion of a geometric series, without the derivatives explicitly kicking in...The above is true whenever

$$\frac{|z-z_0|}{|z_0|}<1\iff |z-z_0|<|z_0|$$

Title: Finding the Taylor series of $f(z)=\frac{1}{z}$ around $z=z_{0}$

Body:
I was asked the following (homework) question:

For each (different but constant) $z_{0}\in G:=\{z\in\mathbb{C}:\,
 z\neq0$} find a power series $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$
  whose sum is equal to $f(z)$ on some subset of $G$. Please specify
  exactly on which subset this last claim holds.

Suggestion: Instead of calculating derivatives of f, try
  using geometric series in some suitable way.

What I did:

Denote $f(z)=\frac{1}{z}$ and note $G\subseteq\mathbb{C}$ is open.
For any $z_{0}\in G$ the maximal $R>0$ s.t $f\in H(D(z_{0},R))$
is clearly $R=|z_{0}|$.

By Taylor theorem we have it that $f$ have a power series in $E:=D(z_{0},R)$
(and this can not be expended beyond this point, as this would imply
that $f$ is holomorphic at $z=0$.

I am able to find $f^{(n)}(z_{0})$ and to solve the exercise this
way: I got that $$f^{(n)}(z)=\frac{(-1)^{n}n!}{z^{n+1}}$$ hence $$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z_{0}^{n+1}}(z-z_{0})^{n};\, z\in E$$

but I am not able to follow the suggestion, I tried to do a small
manipulation $$f(z)=\frac{1}{z-z_{0}+z_{0}}$$ and tried to work a bit
with that, but I wasn't able to bring it to the form $$\frac{1}{1-\text{expression}}$$
or something similar that uses a known power series .

Can someone please help me out with the solution that involves geometric
series ?


答: Use your Cayley table and the one of $\Bbb Z_2\times \Bbb Z_2$ to prove that $0\mapsto (0,0)$, $1\mapsto (1,0)$, $x\mapsto (0,1)$, $1+x\mapsto (1,1)$ is indeed an isomorphism.


问:
closure of addition and multiplication in $\mathbb{Z}_n$
I need to understand what guarantess closure in $\mathbb{Z}_n$ under the operations

$[a]_n + [b]_n := [a + b]_n$

$[a]_n \cdot [b]_n := [a \cdot b]_n$

Different manuals state this:

It is clear that the rule $[a]_ + [b]_n := [a + b]_n$ yields an element of $\mathbb{Z}_n$, but the uniqueness of this result needs to be verified.

These properties are an imnmediate consequence of the definition of + and * in $\mathbb{Z}_n$.

I need to understand why it is so obvious (not for me) that closure is implied in the definition.

Previously this theorem was stated

If $a ≡ b \mod n,  c ≡ d \mod n$ Then $a + c ≡ b + d \mod n$ and $ac ≡ bd \mod n$

I think that $a ≡ b \mod n, c ≡ d \mod n$ implies $a + c ≡ b + d \mod n$ and this implies that $[a+c]$ is in $\mathbb{Z}_n$.

But the manual seems to suggest that closure is implied in the definition of the operations I given above. I'm not interested to prove that those are well-defined for the moment.

答:
The key idea is that each equivalence class such as $\,[a]_n\,$ can be
specified by any of its elements. That is, $\,[a+kn]_n = [a]_n\,$ for
all integers $\,k.\,$ The definition $\,[a]_n + [b]_n = [a+b]_n\,$ for
addition of equivalence classes automatically implies closure, but the
problem is to prove that it is well defined. That is, do you get the
same result no matter which element you pick for both equivalence
classes? Yes, because $\, (a+jn) + (b+kn) = a+b+(j+k)n.\,$
A similar proof holds for multiplication of equivalence classes.


答: @Alan's answer has to be the slickest one possible, but what would you do if you didn't spot that trick? Defining $g(x):=f(x)-x^2$ gives $g(x+5y)+g(x-5y)=0$. Any two values can be written as $x\pm5y$ for some choice of $x,\,y$, so $g$ must always be $0$.


问:
Finding equation for $g^{-1}$ by $f^{-1}$
There are two function $f:\mathbb{Q} \rightarrow \mathbb{Q}\setminus \{-1\}, g:\mathbb{Q}\rightarrow\mathbb{Q}\setminus \{1\}$ such that for every $x\in\mathbb{Q}$,  $g(x)=\frac{f(x)}{f(x)+1}$.
Show that if $f$ is invertible, $g$ is invertible, and find an equation that depicts $g^{-1}$ by $f^{-1}$.

I proved that if $f$ is invertible then $g$ is invertible by showing that if $f$ is injective and surjective then $g$ is injective and surjective, but I don't know how to proceed with the second part of the question (finding the equation). Would appreciate your assistance. Thank you.

Body:
Suppose that $y=g(x)$. Then
$$
y=\frac{f(x)}{f(x)+1},
$$
and
$$
f(x)=\frac{y}{1-y}.
$$
Thus
$$
g^{-1}(y)=x=f^{-1}\Big(\frac y{1-y}\Big).
$$


问:
A question related to $f(n)=n^4+n^2+1$
If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 61. But obviously we can't evaluate this with brute force. The farthest I have reached is that $$n^4+n^2+1=n^2(n+1)+\frac{n+1}{n+1}=(n^2+\frac{1}{n+1})(n+1)$$ Any hints?

答:
By using Sophie German Identity

$$f(n)=n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$

$$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}=\frac{7\times 13\times 21 \times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157\times 183}{3\times 7\times 13 \times 21\times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157}=61$$

Title: A question related to $f(n)=n^4+n^2+1$


问:
Absolute convergence of $\sum_{n=1}^{\infty} a_nx^n$
I have to find an $a_n$ such that $\sum_{n=1}^{\infty}a_nx^n $ converges absolutely on $[-1,1]$. I have choosen $a_n = \frac {1}{n^2}$ and did the root test and see that $\lim_{n \to \infty} (|a_n|) =1$. Now I know that if the result of the root test was < 1 then it would converge $\forall |x| < 1 $. How can i show it it holds even on -1 and 1?

答: $|a_nx^{n}| \leq \frac  1 {n^{2}}$ so you can apply Comparison Test.


问:
Prove by induction $\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+2}{2n+2}$
How to prove this by induction?
$$\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+2}{2n+2}$$

答:
I think the problem should be a product on the left side:
$$
\prod_{k=1}^n\left(1-\dfrac{1}{(k+1)^2}\right)=\dfrac{n+2}{2n+2}
$$
rather than summation.

Now let's begin.
For $n=1$, it is obviously true.

Assume it is true for $n=p$. That is, $\displaystyle \prod _{k=1}^p \left(1-\frac{1}{(k+1)^2}\right) = \frac{p+2}{2 p+2}$ is already true. Now let's check $p+1$:
$$
\begin{aligned}
&\quad\prod_{k=1}^{p+1}{\left( 1-\frac{1}{\left( k+1 \right) ^2} \right)}
\\
&=\prod_{k=1}^p{\left( 1-\frac{1}{\left( k+1 \right) ^2} \right)}\times \frac{p^2+4p+3}{\left( p+2 \right) ^2}
\\
&=\frac{p+2}{2p+2}\times \frac{p^2+4p+3}{\left( p+2 \right) ^2}
\\
&=\frac{p+2}{2\left( p+1 \right)}\times \frac{\left( 1+p \right) \left( 3+p \right)}{\left( p+2 \right) ^2}
\\
&=\frac{p+3}{2p+4}
\\
&=\frac{\left( p+1 \right) +2}{2\left( p+1 \right) +2}
\end{aligned}
$$

Thus we finished the proof.


答:
Your manipulation was along correct lines. Start with $\frac{1}{z_0+z-z_0}$. If $z_0\ne 0$, we can rewrite this as
$$\frac{1}{z_0}\frac{1}{1-\frac{(-1)(z-z_0)}{z_0}},$$
and it's over, we can recognize the sum of a geometric progression.  This trick comes up fairly often. </p>


答:
We have the following:

Base case: $k=1$

We have: $1-\frac{1}{2^2}=\frac{1+2}{2+2}=\frac{3}{4}$ which is correct.

Assuming this is true for $n$ such that $\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+2}{2n+2}$ is true, we need to show it is true for $n+1$ such that  $\sum_{k=1}^{n+1}\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+3}{2n+4}$

We have $\sum_{k=1}^{n+1}\left(1-\tfrac{1}{(k+1)^2}\right)=\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)+ 1-\tfrac{1}{(n+2)^2}=\tfrac{n+2}{2n+2}+1-\tfrac{1}{(n+2)^2}\neq \tfrac{n+3}{2n+4}$

Which means your preposition is wrong and this can't be proven.

Title: Prove by induction $\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+2}{2n+2}$

Body:
How to prove this by induction?
$$\sum_{k=1}^n\left(1-\tfrac{1}{(k+1)^2}\right)=\tfrac{n+2}{2n+2}$$


问:
Solve a Linear System
Trying to solve the linear system below for $x,y$ and $z$ in terms of $a,b,c$ and $\zeta_3$:

$x+y+z=a$

$x+{\zeta_3}y+{\zeta_3}z=b$

$x+{\zeta_3}^2y+{\zeta_3}^2z=c$

where $\zeta_3=-1/2+-i(\frac{\sqrt 3}{2})$ i.e. ${\zeta_3}^3=1$

From a book by the Russian author Smirnoff (translated by Silverman) - the author says the answer is $x=(a+b+c)/3$; $y=(a+b{\zeta_3}^2+c{\zeta_3})/3$; $z=(a+b{\zeta_3}+c{\zeta_3}^2)/3$

Got part of the way with Gaussian elimination, but couldn’t figure out how to use the information about e to finish the solution.  Any thoughts?

答:
If you make a few steps of Gauss elimination you can get
$$\left[\begin{array}{ccc|c}
1 & 1 & 1 & a\\
1 & \zeta_3 & \zeta_3 & b\\
1 & \zeta_3^2 & \zeta_3^2 & c
\end{array}\right]\begin{array}{l}
\\
F_2\leftarrow F_2-F_1\\
F_3\leftarrow F_3-F_1\\
\end{array}
\left[\begin{array}{ccc|c}
1 & 1 & 1 & a\\
0 & \zeta_3-1 & \zeta_3-1 & b-a\\
0 & \zeta_3^2-1 & \zeta_3^2-1 & c-a
\end{array}\right]$$
$$\begin{array}{ll}
\ \\
\ \\
F_3\leftarrow F_3-(\zeta_3+1)F_2
\end{array}\left[\begin{array}{ccc|c}
1 & 1 & 1 & a\\
0 & \zeta_3-1 & \zeta_3-1 & b-a\\
0 & 0 & 0 & c-a -(\zeta_3+1)(b-a)
\end{array}\right]$$
So, Rouché tells you that the system only has solution if
\begin{equation}
c-a-(\zeta_3+1)(b-a)=0,\tag{1}
\label{eq:abc}
\end{equation}
and in that case there are infinite solutions. Then, call $z=\lambda$, and the solutions are:
$$x=a-\frac{b-a}{\zeta_3-1},\quad y=\frac{b-a}{\zeta_3-1}-\lambda,\quad z=\lambda,\quad\forall\lambda\in\mathbb{C}.$$
If you use the relation \eqref{eq:abc} I think that it can be shown that one of the solutions is exactly the answer you have, but obviously there are (infinitely) many more.

I just read in a comment from Jyrki Lahtonen that the equations may have an error. That is likely to be the case.

Title: Solve a Linear System


问:
For 52 cards in a line, what is the probability that the card immediately to the right of the first queen is a queen or a seven?
For a standard deck of 52 cards, lay down all 52 cards in a line. What is the probability that the card immediately to the right of the first
queen is a queen or a seven?

I tried 51!*28/52! but I dont think that I'm accounting for finding the first queen.

51! ways to order the items so that queen is next to another queen or seven (treat as 1 item).

28 ways to rearrange first queen with another queen or seven

52! possible outcomes.

答:
Consider only the queens and $7s$, every permutation of them will necessarily have the first queen followed by a queen or $7$

If we glue the desired combo together, we shall have $7$ objects which can serve as bars creating $8$ "compartments" where the remaining $44$ cards can be put, whereas unglued, they create $9$ compartments.

Using stars and bars,

$Pr = \dfrac{\dbinom{51}{7}}{\dbinom{52}{8}}= \dfrac 2 {13}$

It later dawned on me that "stars and bars" (of which I am perhaps inordinately fond) need not be invoked, and the issue can be looked at more simply as below:

If we glue the desired combo together, we shall have a string of $7$ to be put among the $7+44= 51$ "cards", whereas unglued, we need a string of $8$ to be put among $8+44=52$ cards. [Since only $Pr$ has been asked for, we need not permute.]

And we get exactly the same formula as before


答:
Rename $0=a,1=b,x=c,1+x=d$.

Adding $0$ induces the identity on $\{a,b,c,d\}$.

Adding $1$ induces the permutation $a\mapsto b,b\mapsto0,c\mapsto d,d\mapsto c$.

Adding $x$ induces the permutation $a\mapsto c,b\mapsto d,c\mapsto a,d\mapsto b$.

Adding $1+x$ induces the permutation $a\mapsto d,b\mapsto c,c\mapsto b,d\mapsto a$.

So, as a permutation group, it is $\{id,(ab)(cd),(ac)(bd),(ad)(bc)\}$.


问:
Let $(x_n)_n$ be a sequence of positives st $\exists n_0$ st $0<\frac{x_{n+1}}{x_n}\leq c<1,\ \forall n\geq n_0$. Show that $\lim_{n\to\infty}x_n=0$
I can show that this sequence converges because it is monotonic and bounded, but I'm having trouble showing that the limit is 0.

My attempt:

Let $\lim_{n\to\infty}x_n = L$, thus $$\forall\epsilon>0\exists n_0\in\mathbb{N}\forall n\geq n_0 \ (|x_n - L|<\epsilon)$$
For each natural $k$, let $\epsilon=\frac{c}{k}$, hence there exist $n_{0,k}$ s.t. $\forall n>n_{0,k}$

$$L-\frac{c}{k} < x_n < L+\frac{c}{k}$$
and
$$L-\frac{c}{k} < x_{n+1} < L+\frac{c}{k}$$

but $$0<x_{n+1} < cx_n<x_n$$
so
$$L-\frac{c}{k} < cx_n < L+\frac{c}{k}$$
$$\frac{L}{c}-\frac{1}{k} < x_n < \frac{L}{c}+\frac{1}{k}\implies |x_n-\frac{L}{c}|<\frac{1}{k} ,\ \forall n\geq n_{0,k}$$

I thought of using this to somehow show that $L$ cannot be greater than 0 by making $k$ large enough, but don't know exactly how.

答:
Hint

For $n>n_0$ we have,

$$x_{n}\le cx_{n-1} \to x_n\le c^2x_{n-2}\to ...\to x_n\le c^{n-n_0}x_{n_0}$$
so,
$$0<x_n\le c^{n-n_0}x_{n_0}$$

What about using Squeeze theorem?

Or, if you want to use the definition of limit, choose $n$ s.t., $c^{n-n_0}x_{n_0}<\epsilon.$


问:
Interior of Self-Intersecting Curve
Is there a natural analogue of the notion of the interior (and thus exterior) for a closed curve that is self-intersecting? For some curves I’ve come up with I can visualise it, as shown in the diagram below, and might have a definition. Red depicts the interior and blue the exterior, arrows indicate the nature of the curve as I think multiple curves with the same image could have different interiors.

I think for a closed curve $f:S^1\to\mathbb{R}^2$ you could define it as $\bigcap\left\{\textrm{range}\left(f’\right)\ \middle|\ f’:D^2\to\mathbb{R}^2\land f’|_{S^1}=f\right\}\setminus\textrm{range}\left(f\right)$. However, I don’t know if this set corresponds to the interior for non-intersecting curve, or if it is even non-empty for non-degenerate curves.

答:
There is.

The matter resolves to what a surface means, and how this moves into the next dimension.

Suppose you take a figure of 8.  This is self-intersecting at the middle, yet it holds two areas of positive area (the loops).  If you suppose that $S = \nabla D$, that is, surface is the gradient of density, the surface becomes an 'outwards-pointing' vector or out-vector, the intensity of which is the differences in density.

But if you suppose that the fabric ("manifold") of the surface is unaware of its aroundings, (ie the space not in the surface), then it has a density of one throughout, and is thus completely closed (ie there are no vertices since $S = \nabla D = 0$ throughout).

One possibility is that at the crossing of the surface there is a 'skew marginoid', or partition of the surface where the out-vector reverses but retains the same intensity.  This allows certain polytopes, such as the seven-sided 'Thah', to keep a stable volume pattern.

Non-orientability becomes 'there exists a closed path that crosses an odd number of skew marginoids'.

This was the proposition I advanced on a mail-list to people like John Conway and Norman Johnson.

Title: Interior of Self-Intersecting Curve


问:
Taylor series for tetration
I would like to know what is known about Taylor series for tetration (and other hyper-exponentiations).

Surprisingly, such information is rare on internet. Numerical values for expansion in hyperexponent for tetration with basis $e$ can be found here:

http://en.citizendium.org/wiki/Tetration#Taylor_expansion_at_zero

However I am interested in expansion in the basis of tetration. I look for

$$ x \uparrow \uparrow m = \sum_{n=0}^{\infty} c_n(m)x^n $$

Wolfram mathematica gives me wired result (obviously a bug) at $x=0$ for $x \uparrow 3$:

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)

where in the output $log(x)$ appears (not a polynomial).

It is better at $x=1$

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)+at+x%3D1

I imagine the point $x=0$ may be "peculiar", but at least expansion at $x=1$ should be possible.

List of my questions:

1) Is tetration analytic at $x=0$? At $x=1$?

2) If yes, is an explicit closed-form formula for $c_n(m)$ known? (at any of these points)

3) If not, is an explicit closed-form formula for $c_n(m)$ know for some specific values of $m$?

4) Same questions for extension to higher hyper-exponentiations...

Thank you.

Title: Taylor series for tetration

Body:
Just for your curiosity.

Expanding as Taylor series aound $x=1$ as
$$f_n=\sum_{k=0}^\infty c_k\,(x-1)^k$$ see how fast converge the coefficients
$$f_1=\left\{1,1,1,\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{3}{40},-\frac{1}{120},\frac{59}
   {2520},-\frac{71}{5040},\frac{131}{10080}\right\}$$
$$f_2=\left\{1,1,1,\frac{3}{2},\frac{4}{3},\frac{3}{2},\frac{53}{40},\frac{233}{180},\frac{56
   27}{5040},\frac{2501}{2520},\frac{8399}{10080}\right\}$$
$$f_3=\left\{1,1,1,\frac{3}{2},\frac{7}{3},3,\frac{163}{40},\frac{1861}{360},\frac{33641}{504
   0},\frac{8363}{1008},\frac{22391}{2160}\right\}$$
$$f_4=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{243}{40},\frac{3421}{360},\frac{71861}{504
   0},\frac{54371}{2520},\frac{69281}{2160}\right\}$$
$$f_5=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4321}{360},\frac{102941}{50
   40},\frac{85871}{2520},\frac{61333}{1080}\right\}$$
$$f_6=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{118061}{50
   40},\frac{106661}{2520},\frac{81583}{1080}\right\}$$
$$f_7=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50
   40},\frac{115481}{2520},\frac{93013}{1080}\right\}$$
$$f_8=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50
   40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$
$$f_9=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50
   40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$