问:
Question about Rings and Fields.
We have proved before that $\mathbb Q$ is a prime field. Can we say the same for $\mathbb R$ (prime field)? Using the same proof we used for $\mathbb Q$?

Thank you.

答: Prime fields are $\mathbb{Q}$ and $\mathbb{F}_p$ for some prime $p$, depending on whether $1$ has finite additive order or not...None of them are isomorphic to $\mathbb{R}$.

Title: Question about Rings and Fields.


问:
Question about field extension.
Let $K$ be a field with $\operatorname{char}K =p$, where $p$ is a prime number and  $a \ne 0,1 \in K $. Assume that $P(x) = x^p-a$ is irreducible in $K[x]$ and let $\alpha$ a root of $P(x)$, show that $x^p - \alpha$ is irreducible over $K(\alpha)$.

I have tried to solve that, but I have some questions:

IF $\operatorname{char}K =p$, then $\operatorname{char}K(\alpha) = p??$

Suppose that  $\operatorname{char}K(\alpha) = p$,  we have $x^p- \alpha = x^p - (\sqrt[p]{\alpha})^p = (x - (\sqrt[p]{\alpha}))^p$. So the only root is $\sqrt[p]{\alpha}$, but why $\sqrt[p]{\alpha} \notin K(\alpha)$??

答:
Lemma: Let $q=p^k$ and $c\in K$ such that there is no $d\in K$ with $d^p=c$. Then $f(x)=x^q-c$ is irreducible over $K$.

Proof: Let $b$ be a root of $f$ in some extension field. Assume that $f$ factors over $K$, let $g\in K[x]$ be a non-trivial factor of $f$. Since $f(x)=(x-b)^q$ by unique factorization $g$ is of the form $(x-b)^l$ with some $1\leq l<q$. In particular $b^l\in K$. If $l$ was coprime to $p$, then $b=b^{mp+nl}\in K$ (for suitable $n,m\in\Bbb Z$), so $(b^{p^{k-1}})^p=c$ which contradicts your assumption. Otherwise $l=p^n$ for some $n<k$. Then we also have $c=((b^l)^{p^{k-n-1}})^p$, again a contradiction. Hence $f$ is irreducible.

Applying the lemma with $k=1$ we see that it is sufficient to prove $\beta:=\sqrt[p]{\alpha}\notin K(\alpha)$ (as OP already noted). Assume that $\beta\in K(\alpha)$. By assumption the polynomial $x^p-a$ is irreducible over $K$, in particular $a$ has no $p$-th root in $K$ (i.e. $\alpha\notin K$). By the lemma with $k=2$ this implies that $x^{p^2}-a$ is irreducible over $K$. In particular if $z$ is any root of that polynomial in some extension field then $[K(z):K]=p^2$. But $\beta$ is a root of that polynomial and $[K(\beta):K]\leq [K(\alpha):K]=p$ which is a contradiction. Therefore $\beta\notin K(\alpha)$ and hence $x^p-\alpha$ is irreducible over $K(\alpha)$.

ContentLicense: CC BY-SA 4.0

Title: Question about field extension.


问:
Given a probability density function $f(x) = \begin{cases} 0 & x < 100 \\ \frac{a}{x^2} & x \geq 100 \end{cases}$. Probability distribution function?
Given a probability density function:

$$ f(x) = \begin{cases} 0 & x < 100 \\ \\ \frac{a}{x^2} & x \geq 100 \end{cases} $$

What is the probability distribution function F(x)?

We know that: $F(y) = \int_{-\infty}^{y}f(x)dx$

When $x < 100$:
$$
\begin{equation}
\begin{split}
F(y) &= \int_{-\infty}^{y}0dx \\\\
     &= 0 \Biggr|_{-\infty}^{y} \\\\
     &= 0
\end{split}
\end{equation}
$$

When $x \geq 100$:
$$
\begin{equation}
\begin{split}
F(y) &= \int_{-\infty}^{100}0dx + \int_{100}^{y}\frac{a}{x^2}dx \\\\
     &= 0 \Biggr|_{-\infty}^{100} + \left( -\frac{a}{x} \right) \Biggr|_{100}^{y} \\\\
     &= -\frac{a}{y} + \frac{a}{100} \\\\
\end{split}
\end{equation}
$$

So:
$$ F(y) = \begin{cases} 0 & y < 100 \\ \\ -\frac{a}{y} + \frac{a}{100} & y \geq 100 \end{cases}$$

When $x >= 100$, $F(x)$ should be equal to 1 but it contains $a$.

I think that $F(x) = 1$ and $a = 100$ but do not know how to prove it.

答:
You have to first find $a$ such that $\int_{\mathbb R} f(x)dx=1$. This gives $\frac a {100}=1$ or $a=100$.

[$1= \int_{\mathbb R} f(x)dx=0+\int_{100}^{\infty} \frac a {x^{2}}dx=\frac a {100}$. Hence, $a=100$].

Once you have  this value of $a$ your computation of $F(y)$ is correct.


问:
Mean of inverse of sum of squared standard normal variate
I am working on standard normal distribution and stuck with one derivation.

The problem is I have $10$ independent SNV (standard normal variate) $X_1, X_2,\dots, X_{10}$

What I need to find is Expectation of inverse of sum of squared value these SNV.

$$E\left(\frac{1}{X_1^2+X_2^2+\dots}\right)$$

My views:
I know sum of squared SNV is chi-square, and inverse of which is inv chi square.

Mean of inverse chi square is $1/(\lambda-2)$, where $\lambda
$ is degrees of freedom of chi square.

So my answer to current question is $1/(10-2)=1/8$ is it correct?
Also, what exact theory is applied here?

答:
$\dfrac18$ is correct.

You can get from the density of a chi-squared distribution $\frac{2^{-\nu/2}}{\Gamma(\nu/2)}\; x^{\nu/2-1} e^{-x/2}$ to the density of a so-called inverse chi-squared distribution (reciprocal chi-squared distribution might be a better name) of $\frac{2^{-\nu/2}}{\Gamma(\nu/2)}\,x^{-\nu/2-1}  e^{-1/(2 x)}$ by standard change of variables: essentially you replace $x$ by $\frac1x$, and multiply by the absolute value of the derivative of the inverse (in its wider meaning), so by $\frac1{x^2}$.

Knowing the density and that it integrates to $1$ makes finding the expectation easy since $x\cdot x^{-\nu/2-1}=x^{-(\nu-2)/2-1}$ and thus  the expectation is $\dfrac{\frac{2^{-\nu/2}}{\Gamma(\nu/2)}}{\frac{2^{-(\nu-2)/2}}{\Gamma((\nu-2)/2)}}=\dfrac{1}{\nu-2}$ when $\nu > 2$, and here $\nu=10$.


问:
Why is the volume of a parallelepiped equal to the area of its base times its height?
Is there a formal proof? or is it by definition?</p>

答:
There are proofs, but what approach best suites your setup depends on how you define volume in the first place.

One possible approach is to define volume as an integral over areas. Conceptually, you split your shape into slices of the same thickness, compute the area of each slice, multiply each by its thickness and add them up. In the limit, the slices will become infinitesimally thin, but you'll get an finfinite number of them, and end up with an integral instead of a sum (don't let the formula distract you if you don't feel at home with this kind of notation):

$$V = \int_0^h A(z)\,\mathrm dz$$

If you consider $z=0$ to be the plane containing the base, then you can think of other $z$ values as different distances from that base plane, resulting in parallel planes intersecting your body. One thing is readily apparent from this definition is the following: all that matters is the area of the intersection of your body with such a plane. It does not matter where that area is located, or how it is shaped. This is also known as Cavalieri's principle, and is sometimes used to derive a formula for the volume of the sphere.

Applied to your case, every intersection with a plane parallel to the base results in a parallelogram shape congurent to the base itself. So all slices have the same area $A$, and it doesn't matter whether they are offset against one another, or placed directly one above the other like in a prism. You can either use this to get back to a cuboid of same volume (via a prism as intermediate shape), or plug the constant area into the formula above:

$$ V = \int_0^hA\,\mathrm dz = A\,\int_0^h\mathrm dz = A\,h $$

ContentLicense: CC BY-SA 3.0


问:
sub-algebra of inductive limit(II)
This question is similar to my question here, but not the same question.

Let $(A_{i},\alpha_{i})$ be directed system of C* algebras and *-homomorphisms. Let the $\beta_{i}:A_{i}\rightarrow A$ are the canonical *-homomorphisms. Consider some subalgebra $B$ belongs to direct limit $A$. Does it always exists a $j\in I$ such that $B\subseteq \beta_{j}(A_{j})$?

My guess is that is true. Since $A$ is quotient algebra of disjoint union of algebras, we can write every elements $x\in A$ in the form $x=(x_{1},x_{2},...)+N$, with $x_{i}\in A_{i}$, $N$ is the ideal form by equivalence relation. If $B$ is a sub-algebra, then every component of $B$ form a sub-algebra. Is my thought correct?

答:
The inductive limit of C$^*$-algebras is often not the union of the ranges of the $\beta _j$.

For instance, consider the
C$^*$-algebra $K$, formed by all compact operators on $\ell ^2$.  Also,  for each $n$,  consider the subset $K_n\subseteq K$, formed by
all  operators whose matrix $(a_{i, j})_{i, j}$, relative to the canonical basis of $\ell ^2$,  have nonzero entries  only in
the top left $n\times n$ block.

Then

each $K_n$ is a closed $^*$-subalgebra of $K$ (isomorphic to $M_n(\mathbb C)$),

$K_n\subseteq K_{n+1}$, and

the union
$\bigcup_nK_n$ is dense in $K$.

With this
much information you are  able to deduce that $K$ is the inductive limit of the $K_n$,  with the connecting maps being
the inclusions $K_n\hookrightarrow K_{n+1}$.

Observing that every operator in $\bigcup_nK_n$ has finite rank, we see that $\bigcup_nK_n$ is not equal to $K$.   It
is only a dense $^*$-subalgebra (and hence can't be the inductive limit in the category of C$^*$-algebras since we want the
inductive limit to be a C$^*$-algebra).

Any
infinite rank compact operator  will therefore  generate a subalgebra that is not contained in the union of the ranges
of the $\beta _j$.

Title: sub-algebra of inductive limit(II)


答: A field is prime if and only if it has no proper subfields. $\mathbb R$ has $\mathbb Q$ as a proper subfield, so it is not prime. Looking at it another way, nontrivial extensions of prime fields (like $\mathbb R$ is of $\mathbb Q$) cannot be prime, since they always have their prime field as a proper subfield.

Title: Question about Rings and Fields.

Body:
We have proved before that $\mathbb Q$ is a prime field. Can we say the same for $\mathbb R$ (prime field)? Using the same proof we used for $\mathbb Q$?

Thank you.


问:
How can I show that (A−B)−(B−C)=(A−B)?
I'm stuck in this step, can someone show me how to complete the proof with each step name?

答: You have $(B \setminus C) \subset B$, so $(A \setminus B) \cap (B \setminus C) =\emptyset$, so $(A \setminus B) \setminus (B \setminus C) = A \setminus B$.

Title: How can I show that (A−B)−(B−C)=(A−B)?


问:
Evaluating B(3/4,5/4) without the Legendre duplication formula?
Suppose that one wanted to evaluate $B(3/4,5/4)$ where $B$ is the beta function.  One approach is as follows.

$$B(3/4,5/4)=\frac{\Gamma(3/4)\Gamma(5/4)}{\Gamma(3/4 + 5/4)}=\Gamma(3/4)\Gamma(5/4).$$

There is an identity, the Legendre duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$, and applying this with $z=3/4$ we get that our function equals $2^{-1/2}\sqrt{\pi}\Gamma(3/2)$.  Since $\Gamma(3/2)=(1/2)\Gamma(1/2)=\sqrt{\pi}/2$, we get a final answer of $\frac{\pi}{2\sqrt{2}}$.

Is there a more direct way to get this result? The integral formulas for the beta function all see to give horrible messes, e.g., $2\int_0^{\pi/2} \sin(\theta)^{1/2}\cos(\theta)^{3/2}\,\mathrm{d}\theta$, and I don't see a reasonable substitution to make for that.

答: As @Svyatoslav noted, you can use the reflection formula $\Gamma(s)\Gamma(1-s)=\pi\csc(\pi s)$:$$\operatorname{B}(\tfrac34,\,\tfrac54)=\tfrac{\Gamma\left(\tfrac34\right)\Gamma\left(\tfrac54\right)}{\Gamma(2)}=\tfrac14\Gamma\left(\tfrac14\right)\Gamma\left(\tfrac34\right)=\tfrac{\pi}{4}\csc\tfrac{\pi}{4}=\tfrac{\pi}{\sqrt{8}}.$$


问:
Definition of Complex Convolution?
The convolution of two real value functions f and g is defined to be

$$(f \ast g)(t) = \int_{-\infty}^\infty f(\tau)g(t-\tau)\,d\tau$$
If I extend this idea into higher dimension (Let's say in a unit disc or maybe $
\mathbb{C}$, a set of complex number), can we defined a convolution of two complex value function?

My guess is going to be a double integral in some way. Could you please precisely state it?

答:
The convolution of functions has a natural and very important generalization to functions defined on any locally
compact topological group (such as $\mathbb C$, $\mathbb R^n$, the circle, $\text{GL}_n$, $\text{SL}_n$, $\text{O}_n$,
$\text{U}_n$, $\mathbb Z$, $\mathbb Z_n$, any discrete group,  and many many others).

If the group is called $G$, then the convolution  is defined by
$$
  (f * g)(x) = \int_Gf(y)g(y^{-1} x) \, d\mu(y),
  $$
whrere $\mu $ is the Haar measure on $G$, namely the unique (up to a scalar multiple) left-invariant measure on $G$.

To be more precise, in the case of

$G=\mathbb C$
$$
  (f * g)(x+iy) = \int_{-\infty}^\infty \int_{-\infty}^\infty  f(z+iw)g(x+iy - z-iw) \, dzdw,
  $$

$G=\mathbb T = \{z\in \mathbb C: |z|=1\}$
$$
  (f * g)(e^{i\theta }) = \frac 1{2\pi}\int_{0}^{2\pi } f(e^{i\tau })g(e^{i(\theta -\tau )}) \, d\tau ,
  $$

$G=\mathbb Z$
$$
  (f * g)(n) = \sum_{k=-\infty }^{\infty } f(k)g(n-k).
  $$

Finally,  since the unit disk in the complex plane is not a group,  there is no natural convolution for functions
defined there,  unless you extend these functions  to the whole plane by setting them to be zero outside the disk.

Title: Definition of Complex Convolution?


答:
Distribute and conclude like this $=(A\cap\underbrace{\bar B\cap \bar B}_{=\bar B})\cup(\underbrace{A\cap\bar B\cap C}_{\subset \ A\cap\bar B})=A\cap\bar B=A-B$

By using $(F\subset E)\implies E\cup F=E$.

Title: How can I show that (A−B)−(B−C)=(A−B)?

Body: I'm stuck in this step, can someone show me how to complete the proof with each step name?


问:
Area of $ABCD$ where $A,B,C$ lie on circle with center $D$
Let $\mathrm{A}\left(z_{1}\right), \mathrm{B}\left(z_{2}\right)$ and $\mathrm{C}\left(z_{3}\right)$ lie on the circle $|z-\mathrm{i}|=1$ and satisfy the equation $3 \mathrm{z}_{1}+\mathrm{i}=2 \mathrm{z}_{2}+2 \mathrm{z}_{3}$. If $\mathrm{D}$ is the centre of the circle $|z-\mathrm{i}|=1$, then area of quadrilateral $\mathrm{ABCD}$ equals---

I tried to analyze by using geometry but am not able to reach to final answer. Please help,

答:
If we shift the circle down by $1$ unit, we get $D=0$,
$$|z|=1$$
and $A(z_1),B(z_2),C(z_3)$ satisfy
$$3(z_1+i)+i=2(z_2+i)+2(z_3+i)$$
$$\Rightarrow 3z_1=2z_2+2z_3 \tag{1}$$

Since the expression is symmetric in $z_2,z_3$, $B,C$ lie symmetrically wrt to $DA$ (see remark below). The quadrilateral has diagonals $AD$ and $BC$ where radius $AD$ is perpendicular to chord $BC$. Its area is then half of product of diagonals.

Let arguments of $A,B,C$ be $\theta_1,\theta_2,\theta_3$. Then $BC$ subtends $2\theta=|\theta_2-\theta_3|$ at the center. So $BC=2\sin \theta$

From $(1)$, we have
$$3\cos \theta_1=2\cos \theta_2 + 2\cos \theta_3 \tag{2}$$
$$3\sin \theta_1=2\sin \theta_2 + 2\sin \theta_3 \tag{3}$$

Squaring and adding we get
$$\cos (\theta_2 - \theta_3)=\frac{1}{8}=\cos 2\theta$$

As pointed by @user, the above can also be obtained by multiplying $(1)$ by its complex conjugate equation.

Now $$\text{Area} = \frac{1}{2} \cdot DA \cdot BC = \sin \theta$$
$$=\sqrt{\frac{1-\cos 2\theta}{2}}$$
$$=\boxed{\frac{\sqrt{7}}{4}}$$

Remark :

If this is not convincing, note that $(1)$ can be rewritten as $$5z_1=2(z_1+z_2+z_3)$$
where term in brackets is coordinates of orthocenter of $\triangle ABC$. It says $A$ is collinear with orthocenter and $D$, the circumcenter $\Rightarrow \triangle ABC$ is isosceles with $AB=AC$ i.e., $B,C$ lie symmetrically on either side of $DA$.


问:
Finding the distance between a unit circle and a line analytically
Let $A = \{(x, y) \in \mathbb{R}^2\mid x^2 + y^2 = 1\}$ and $B = \{(x, y) \in \mathbb{R}^2 \mid y = 3 - x\}$. I am currently taking my analysis/metric spaces class and would like to find the distance between these two sets in a more "analytical" way. That is, using only the definition $\mathrm{dist}(A, B) = \inf \{d_{\mathbb{R}^2}(a, b) \mid a \in A, b \in B\}$. I know this might sound like heresy, but I would prefer for the proof to not contain any argumentation by pictures etc. Just by the tools of analysis. The issue is that I honestly do not even know where to start. Sure, $d_{\mathbb{R}^2}(a, b) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$ and points in $A$ and $B$ satisfy the properties of a circle and a line. But by using only the tools of analysis, how can you derive the distance between the sets?

I do know that this problem is trivial with the use of the formula $\frac{\left|ax_0 + by_0 + c\right|}{\sqrt{a^2 + b^2}} - r$, where $(x - x_0)^2 + (y - y_0)^2 = r^2$ is a circle and $ax + by + c  = 0$ is a line. But so far (through HS) I have only ever used these ready made tools and argumentation by e.g. geometry to find the answers. Therefore I do not know how to approach this problem in the "analysis" way.

答:
If you can accept the result in post as a lemma, then there we is an easy proof for how to get the shortest distance between a curve and a circle. I plan on using some 'uncommon' results simply to make this post 'fun'

I will center my circle at point $(h,k)$ and it's equation will be denoted as $C(x,y)=0$, now my second curve I will denote at $S(x,y)$, now from the result on page-92 of Joseph Edward's calculus book see here(*), we get the tangent line equation of S at $(x,y)$ as:
$$\frac{X-x}{S_x} = \frac{Y-y}{S_y} ---- (**)$$

We can think of $(X,Y)$ as the free variable of tangent equation after we plug in point on curve as $(x,y)$ , now for shortest length, it must be that the normal line passes through the centre. So, plug in $(X,Y)$ as $(h,k)$:

$$ \frac{h-x}{S_x} = \frac{k-y}{S_y} \tag{2}$$

Now from (2) and $S(x,y)=0$(***), we can solve for the point $(x,y)$ such that distance is minimized, after that, simply take the distance from circle's center for this point and substract the radius Q.E.D.

*:You can find how he got result by reading few pages back , requires only some knowledge on partials and eulers theorem (also simple)

**: $ S_x = \frac{\partial S}{\partial x}$

***: Ideally you want to parameterize the curve $S(x,y)$ from the algebraic equation that will make solving much much easier since that'll directly reduce equation (2) into one variable.


问:
Antiderivatives such as $\int \frac{dx}{\sqrt{e^{2x} +c}}$
Is there a good strategy for calculating antiderivatives such as$$\int \frac{1}{\sqrt{e^{2x} +c}}dx\enspace?$$

Right now I'm substituting the entire radical expression, which sort of simplifies the calculation. But nevertheless, I need to make another substitution after, since I have the integral $$\int \frac{1}{u^2-c}du$$

I can solve it, but are there other ( perhaps shorter ) methods?

答:
You can write this integral as
$$ I = \int \frac{e^{-x}dx}{\sqrt{1+ce^{-2x}}}$$
and make the substitution (for $c>0$):
$$ \sqrt{c}e^{-x} = \sinh t$$
$$ \sqrt{1+ce^{-2x}} = \cosh t$$
$$ -\sqrt{c}e^{-x}dx = \cosh t\,dt $$
then you immediately get
$$ I = -\frac{1}{\sqrt{c}}\int dt = -\frac{1}{\sqrt{c}}{\rm arsinh}(\sqrt{c}e^{-x}) + const.$$
For $c<0$ there's a similar substitution, $\sqrt{-c}e^{-x} = \sin t$.


问:
Symmetry in the quadratic form matrix
The following is an excerpt from Greene's Econometric Analysis (7th Edition). Therein, the author states the matrix in quadratic forms must be symmetric. I would like to know why. What if, for instance, $a_{12} \neq a_{21}$?

答:
Therein, the author states the matrix in quadratic forms must be symmetric.

No, he didn't. And he didn't say the $ij$-th component in $A$ is $a_{ij}$. Here the quadratic form can be written as $x^{\top}Ax$ such that $A$ is a symmetric matrix. The representation is not unique without posing any restriction on $A$. The point is for any set of real $a_{ij}$, there always exists a symmetric $A$ such that $x^{\top}Ax=\sum_{i=1}^{n}\sum_{j=1}^{n}x_{i}x_{j}a_{ij}$, hence we can choose the matrix to be symmetric if we want.

Title: Symmetry in the quadratic form matrix


问:
How to create an inverted $S$ cruve
I'm looking for a way to create an inverted $S$ curve function. Below is an image of what I want. Left is a regular S curve, on the right is the kind of curve I want. Here are the allowed set of operations that I can use:

答:
Possible with Inverse function where  function inversion is possible. The given graph indicates an odd function, straight forward function inversion solution. In polar coordinates $ r(\theta) = r(-\theta).$

In other cases absolute values to be taken and later on modified suitably.

$$ y= f(x)\to x= f(y) $$

$$ or$$

$$ y= f(x)\to y_{inv}= f^{-1}(x_{inv}) $$

In the above we have inverse tanh brown  function solution graphed along with starting green tanh function. Similarly the sine function.

Title: How to create an inverted $S$ cruve


问:
Find bound for $\sum_{n=0}^\infty\frac{(-1)^n sin \pi s}{n!(n+1-s)\pi}$
We have the sum
$$\sum_{n=0}^\infty\frac{(-1)^n \sin \pi s}{n!(n+1-s)\pi}$$ for all $s$ with $|im(s)|\le1$
This is a step from the proof of growth order of Gamma function by E.Stein in his Complex analysis.

Here for each $|im(s)\le1|$ exists integer $k$ such that $k-1/2\le Re(s)<k+1/2$ (here I only need to consider $k>0$, otherwise trivial). Now truncate this sum into $$\sum_{n\neq k-1}^\infty\frac{(-1)^n\sin \pi s}{n!(n+1-s)\pi}+\frac{(-1)^{k-1} \sin \pi s}{（k-1）!(k-s)\pi}.$$
For the first term, it is clear that $1/2 \le n+1-s$ therefore the whole term is bounded by $C|sin \pi s|$ for some constant $C$ and therefore bounded by $Ce^{\pi|s|}$. But for the second term, I know $sin\pi s$ also has a simple zero at $k$， so $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}=\frac{(-1)^{k-1} h(s)}{（k-1)!\pi}$ for some analytic $h$, with $h$ bounded by $C_s$ on $\{(x,y): k-1/2\le x<k+1/2, |y|\le1 \}$ but $C_s$ depends on $k$ and $k$ depends on $s$, so $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$ is bounded by $C_s$ is saying nothing but $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$ well defined on $\{(x,y):|y|\le1\}$. Is there a way to find the bound independent of $s$ for $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$?

答:
I believe you wanted to say $k - 1/2 \le \Re s \le k + 1/2$. Note also that $\sin(\pi s)$ has a simple zero at $k$, not a simple pole. Furthermore,
$$
( - 1)^{k - 1} \frac{{\sin (\pi s)}}{{(k - s)\pi }} = \frac{{\sin ((k - s)\pi )}}{{(k - s)\pi }}
$$
and
$$
\left| {\frac{{\sin ((k - s)\pi )}}{{(k - s)\pi }}} \right| \le \sum\limits_{n = 0}^\infty  {\frac{{\left| {(k - s)\pi } \right|^{2n} }}{{(2n + 1)!}}}  \le e^{\left| {(k - s)\pi } \right|^2 }  \le K
$$
in a compact neighbourhood of $k$. If, for example, $k - 1/2 \le \Re s \le k + 1/2$ and $|\Im(s)|<1$, then
$$
\left| {(k - s)\pi } \right|^2  = \pi ^2 \left| {\Re (k - s)} \right|^2  + \pi ^2 \left| {\Im (k - s)} \right|^2  \le \frac{{\pi ^2 }}{4} + \pi ^2 ,
$$
i.e., $K = e^{\frac{{5\pi ^2 }}{4}}$ works.


问:
Modular exponentiation with exponent that has repeating pattern
If i have to count a large exponent modulo N, that has some repeating pattern. Is there some way how to simplify the exponent before actually counting the remainder?

$131^{123456789876543212345678987654321 } \;(mod \;12)$

答: $131=120+11 \equiv 11 \equiv-1\;(\bmod\; 12)$. As the exponent is odd then $131^{123456789876543212345678987654321} \equiv-1\;(\bmod\; 12)$, this is a way of simplifying the basis before counting the remainder.


问:
Direct Proof: For all n(n is a natural number) 5 divides n^5 - n
In order for $n^5 - n$ divisible by $5$, $n^5 - n = 5 x + 0$ (for some $x$, $x$ is a natural number)
I simplified the $(n^5 - n) = n(n^2+1)(n+1)(n-1)$ and I do not know what to do next.

And I tried something else like $(n^5 - n) = n (n^4 - 1)$, but now I need to show $(n^4 - 1)$ is divisible by $5$, how do I do that? Thanks in advance.

答:
If we look at the factorization modulo 5, we can factor that quadratic term as:

$$x^2+1 \equiv (x-2)(x-3) \mod 5$$

That means your factorization can be written as:

$$n^5 - n \equiv n(n-1)(n-2)(n-3)(n-4) \mod 5$$

Since every residue class is a root, no matter what $n$ we pick, we will always get a number that is $0 \mod 5$.


答:
For simplicity let us assume that the common difference of the arithmetic progression is equal to one (the same calculation works in the general case with small modifications).

You correctly noticed that the middle expression can be written as
$$
\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+\cdots+\frac{1}{a_{2n-1}a_{2n}} \\
= \frac{1}{a_1}-\frac{1}{a_2} + \frac{1}{a_3}-\frac{1}{a_4} + \cdots + \frac{1}{a_{2n-1}} - \frac{1}{a_{2n}} \, .
$$
Now the “trick“ is to group this sum as
$$
= \frac{1}{2a_0} - \left( \frac{1}{2a_0}-\frac{1}{a_1}+\frac{1}{2a_2}\right) - \cdots - \left( \frac{1}{2a_{2n-2}}-\frac{1}{a_{2n-1}}+\frac{1}{2a_{2n}}\right) - \frac{1}{2a_{2n}} \, .
$$
and show that all terms in the parentheses are positive. It follows that the whole expression is
$$
 < \frac{1}{2a_0}- \frac{1}{2a_{2n}} = \frac{n}{a_0 a_{2n}}
$$
which proves the right inequality.

The proof of the left inequality is similar.

Title: Inequality proof $\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$

Body:
Can someone give me an idea about how to prove the inequality?
Mention that $a_0, a_1, a_2, ..., a_{2n}, a_{2n+1}$ is an AP and $0<a_0<a_1<...<a_{2n}<a_{2n+1}$
$$\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$$
Mention that I tried ideas like: $$\frac{1}{a_{2n+1}}<\frac{1}{a_1}<\frac{1}{a_{0}};$$
and then add, but with no result.
A possible solution is:
$$\frac{n}{a_1a_{2n+1}}<\frac{1}{r}(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_3}-\frac{1}{a_4}+...+\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})<\frac{n}{a_0a_{2n}}, r=a_n-a_{n-1}
$$

$$\frac{rn}{a_1a_{2n+1}}<\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}}-(\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}})<\frac{rn}{a_0a_{2n}}
$$
Let be $S_1>I_1$:
$$\frac{1}{a_1}+\frac{1}{a_1}+...+\frac{1}{a_1}>\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}}
$$
Let be $S_2>I_2$:
$$\frac{1}{a_2}+\frac{1}{a_2}+...+\frac{1}{a_2}>\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}}
$$
The following difference can be taken into account in this case?
$$S_1-S_2>I_1-I_2 \rightarrow S_1-S_2=\frac{nr}{a_1a_2}
$$
Then
$$
\frac{nr}{a_1a_{2n+1}}<\frac{nr}{a_1a_2}<\frac{nr}{a_0a_{2n}}
$$
Inequality comes out immediately if this difference occurs. Can give me some advice to solve it or if this approach is wrong, why is it wrong?
Thank you!


问:
Is the quadratic character, unique multiplicative character over $\mathbb Z_{p^n}$, for odd $p$?
Let $p$ be odd and $\mathbb Z_{p^n}$ denote the ring of integers modulo $p^n$. Let  the quadratic character, $\eta$, be the function defined on $\mathbb Z_{p^n}^*$ (multiplicative group of $\mathbb Z_{p^n}$) with $\eta(c)=1$ if $c$ is the square of an element of $\mathbb Z_{p^n}^*$  and $\eta(c)=-1$ otherwise.

Let $G$ be a finite abelian group. A character of $G$ is simply a homomorphism $\chi$ from $G$ into the multiplicative group $\mathbb{C}^*$ of complex numbers. That is, $\chi(g_1g_2)=\chi(g_1)\chi(g_2)$ for all $g_1,g_2 \in G$.

We call a character of $\mathbb Z_{p^n}^*$ as multiplicative character of $\mathbb Z_{p^n}$.

Can we say that, $\eta$ is a multiplicative character for $\mathbb Z_{p^n}$ and all multiplictive characters are obtained in this way?

Thank you.

答:
As $p$ is odd, the group $G=\mathbb{Z}^*_{p^n}$ is cyclic of order $p^{n-1}(p-1)$. So if $g$ is a generator of $G$, and $\omega$ is any complex number that satisfies the equation $\omega^{p^{n-1}(p-1)}=1,$
then we get a homomorphism $\chi=\chi_\omega$ from $G$ to $\mathbb{C}^*$ from the recipe
$$
\chi_\omega(g^k)=\omega^k
$$
for all $k=0,1,2,\ldots,p^{n-1}(p-1)-1$.

There are $p^{n-1}(p-1)$ choices for $\omega$, and each choice gives you a different character. It is not too hard to see that there are no others.

If you are only interested in characters that take values $\pm1$ only, then you must pick $\omega=\pm1$. The choice $\omega=+1$ gives you the trivial character (everything is mapped to one), and the choice $\omega=-1$ gives the character $\eta=\chi_{-1}$ that you described:

$$
\eta(g^k)=(-1)^k=\begin{cases}+1,\ \text{if $k$ is even,}\\-1,\ \text{if $k$ is odd}.\end{cases}
$$
So your $\eta$ is a multiplicative character all right (the square are exactly the even powers of $g$, as the order of $g$, $p^{n-1}(p-1)$, is an even number.

For example, when $p^n=3^2=9$, then $|G|=3^{2-1}(3-1)=6$, and it is easy to see that we can use $g=2$. If we select $\omega$ to be a primitive sixth root of unity, say $\omega=e^{\pi i/3}=(\sqrt3+i)/2$, then we get the character $\chi=\chi_\omega$:
$$
\begin{eqnarray*}
\chi(1)=\chi(2^0)&=\omega^0=1,\\
\chi(2)=\chi(2^1)&=\omega^1=\omega,\\
\chi(4)=\chi(2^2)&=\omega^2,\\
\chi(8)=\chi(2^3)&=\omega^3=-1,\\
\chi(7)=\chi(2^4)&=\omega^4=-\omega,\\
\chi(5)=\chi(2^5)&=\omega^5=\overline{\omega}.
\end{eqnarray*}
$$
This time $\chi(0),\chi(3)$ and $\chi(6)$ are left undefined as those numbers are divisible by three. Sometimes a convention is used, when those are defined to be zero. Those three residue classes do not belong to the group $G$, but number theorists sometimes want to define these functions outside of $G$ as well.
This time you get the quadratic character $\eta$ as the third power $\chi^3$.

Hope this helps. Ask for clarifications if/where needed.


问:
Is the closure of $S^1-\{(0,1)\}$ compact?
My question is very simple : Is the closure of $S^1-\{(0,1)\}$ compact ? (where $S^1\subset \mathbb R^2$ is the unit circle). In somehow, I think it is, since it should be the circle, but in on other way, $S^1-\{(0,1)\}$ is homeomorphic to $\mathbb R$, so in somehow, if the closure of $S^1-\{(0,1)\}$ is compact, then so is the closure of $\mathbb R$ (which I guess it's not). So, at the end, I'm not so sure if the closure of $S^1-\{(0,1)\}$ is really compact. Any idea ?

答:
The closure of $S^1 - \{(0,1)\}$ as a subspace of $\mathbb{R}^2$ is the circle. Since it is closed and bounded we know it is compact. As other users pointed out, you need to consider the topological space involved, because the notion of closure depends on the space, but also on the topology of the space.

Regarding homeomorphisms, the circle is homeomorphic to the one point compactification of the real line $\mathbb{R}$. As you said, $S^1 - \{(0,1)\}$ is homeomorphic with the real line and both spaces are Hausdorff and locally compact. For this reason you can extend this homeomorphism to another homeomorphism between the one point compactifications, that is between $S^1$ and $\mathbb{R} \cup \{\infty\}$. Hence, we conclude again that $S^1 = Cl(S^1 - \{(0,1)\})$ is compact.


问:
Compute the following improper integral by complex integration
Here I tried to compute such integral by Complex Integration.  $\int \frac{z^2}{1+z^4} $ It follows that the solution to $1+z^4$ on such domain is $e^{1/4\pi i}, e^{3/4\pi i}$
.

I tried to find a decomposition, it follows that $P(x) = (z-e^{1/4\pi i})(z- e^{3/4\pi i}）Q(x)$, and I let $ Q(x) = z^2 + iBz + 1$, but then the equation yields a strange B = $\sqrt(2)$ which does not satisfy the equation(why?) and I can proceed no further.

The original post is the entire function, then I edited it into an integral on a semi-circle so that it would be simpler. But now I switch back to the OP to be consistent with the accepted answer. Sorry if causing any confusion!

答:
If I'm not mistaken you want to calculate
$$\oint_\mathcal{C}\frac{z^2}{1+z^4}\,\mathrm{d}z,$$
where every singularity of the integrand $f(z)$ is contained in $\mathcal{C}$. First of all - what are the poles of $f(z)$? Solving $z^4+1=0$ gives us $z_1=e^{(i\pi)/4}, z_2=-e^{(i\pi)/3}, z_3=e^{(3i\pi)/4}, z_4 = -e^{(i\pi)/4}$. Now the the only thing left is to calculate the residues. Note that all poles are of fourth order. If $f$ has a pole of fourth order at $z_j$ we have
$$ \operatorname{Res}_{z_j} f = \cfrac{1}{6}\lim\limits_{z\rightarrow z_j}\cfrac{\partial^{3}}{\partial z^{3}}[(z-z_j)^4f(z)].$$

If we want to calculate $f(z)$ along a semi-circle in the upper half plane we will parametrize $z=Re^{i\varphi},\varphi\in(0,\pi)$. We get
$$\int\limits_{0}^{\pi}\frac{R^2e^{2i\varphi}}{1+R^4e^{4i\varphi}}iRe^{i\varphi}\,\mathrm{d}z.$$
For $R\to\infty$, this integral vanishes. This is because
$$\left\lvert\int\limits_{0}^{\pi}\frac{R^2e^{2i\varphi}}{1+R^4e^{4i\varphi}}iRe^{i\varphi}\,\mathrm{d}z\right\rvert \leq \int\limits_{0}^{\pi}\frac{R^3}{R^4-1}\,\mathrm{d}z\to0.$$


问:
How to prove that two convex functions have the same minimizer?
I am looking for a general method to show that two convex functions have the same minimizer. That is, for a pair of arbitrary convex functions $f$ and $g$, I want to know when it is possible to show that

$x^* = \text{argmin}_x f(x) = \text{argmin}_x g(x)$

Is this possible? Others have asked this kind of question, but they always asked about a specific pair of functions, and the answers they got were fairly specific: see here and here. What I am looking for is a general answer to this question.

答:
I am not sure I understood the generality level you are looking for with your question. I suppose that it is pretty difficult to say something in general, but if you are looking for a simple criterion, then I think immediately at the following:

Let $f$, $g$ two convex real-valued functions. If you can show that $f$ can be obtained from $g$ composing $g$ with a strictly increasing function $h$, then the minimizer of $f = h \circ g$ and $g$ is going to be the same.


答:
$$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+(b+c)xy+dy^2$$ can be symmetrized as

$$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&\frac{b+c}2\\\frac{b+c}2&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+2\frac{b+c}2xy+dy^2.$$

So a quadratic form has a symmetric matrix WLOG.

Title: Symmetry in the quadratic form matrix

Body: The following is an excerpt from Greene's Econometric Analysis (7th Edition). Therein, the author states the matrix in quadratic forms must be symmetric. I would like to know why. What if, for instance, $a_{12} \neq a_{21}$?


问:
Additive group of all polynomials in x with integral coefficients is isomorphic to $\mathbb{Q}^*$

Let G be the additive group of all polynomials in $x$ with integer coefficients.
Show that G is isomorphic to the group $\mathbb{Q}$* of all positive rationals (under
multiplication).

This question is from my abstract algebra assignment and I am unable to  prove it. I am not able to deduce what should I map   $a_0 +a_1 x + a_2 x^2 +...+ a_n x^n $ to so not able to proceed.

Can you please give a hint?

答:
Hint: The fundamental theorem of arithmetic implies that
$$
\mathbb{Q}^\times_+ \cong \bigoplus_p \mathbb Z
$$


答:
Hint

Consider the map

$$\begin{array}{l|rcl}
\phi : & \mathbb Z[x] & \longrightarrow & \mathbb Q^* \\
    & q(x)=q_0 + q_1 x + \dots + q_n x^n& \longmapsto & p_0^{q_0}p_1^{q_0} \cdots p_n^{q_n}\end{array}$$

where $p_i$ stands for the $i$-th prime number.

Title: Additive group of all polynomials in x with integral coefficients is isomorphic to $\mathbb{Q}^*$

Body:
Let G be the additive group of all polynomials in $x$ with integer coefficients.
Show that G is isomorphic to the group $\mathbb{Q}$* of all positive rationals (under
multiplication).

This question is from my abstract algebra assignment and I am unable to  prove it. I am not able to deduce what should I map   $a_0 +a_1 x + a_2 x^2 +...+ a_n x^n $ to so not able to proceed.

Can you please give a hint?


问:
Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled

Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).

I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?

答:
Here is a second approach: Let $x_n$ $(n\geq0)$ denote the number of admissible words having $n$ letters. Then
$$x_0=1,\quad x_1=2,\qquad x_n=2x_{n-1}+x_{n-2}\quad (n\geq2)\ .$$
The characteristic equation of the recursion is $\lambda^2-2\lambda-1=0$ with roots $\lambda=1\pm\sqrt{2}$. It follows that
$$x_n=c(1+\sqrt{2})^n+c'(1-\sqrt{2})^n\qquad(n\geq0)\ ,$$
where the constants $c$, $c'$ have to be determined from the initial conditions. The computation gives
$$c={2+\sqrt{2}\over4},\qquad c'={2-\sqrt{2}\over4}\ .$$
Now $\bigl|c'(1-\sqrt{2})^n\bigr|<{1\over2}$ for all $n\geq0$. We therefore can write
$$x_n={\tt round}\left({2+\sqrt{2}\over4}(1+\sqrt{2})^n\right)\qquad(n\geq0)\ .$$
This gives
$$x_{10}={\tt round}\bigl(5740.9999782267896753\bigr)=5741\ .$$
This coincides with the value obtained by ${\tt user10354138}$.


问:
Showing that the given sequence is bounded as follows
Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.

Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows:
To show $x_{n+1}<x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that is s, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$

Back to the problem. It's clear that $A$ is bounded above by $\frac{1}{3^6}$ (Should I show this first?).
Then, to show that $A$ is bounded, it's suffices to show that $A$ is bounded below by $0$.
I show this one again by induction. Indeed, it's true for $n=1$.
Assume that it's true for $n=k$; that is $0 < x_k$. Then, $x_{k+1} = \frac{1}{3^{(k+1)+5}}
= \frac{1}{3^{k+5}} \cdot \frac{1}{3} > 0$. Hence, $0< x_n$ for all $n \in \Bbb Z^+$. Therefore, $A$ is bounded below by $0$. Thus, $A$ is bounded, as desired.

Now, I claim that $\sup A = \frac{1}{3^6}$ and $\inf(A) = 0$.
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.

For the supremum, let $M$ be an another upper bound of $A$. To show: $\frac{1}{3^6} \le M$. Suppose $M < \frac{1}{3^6}$. Then, by the Density Theorem, there exists $s \in \Bbb Q$ such that
$M < s < \frac{1}{3^6}$. Hence, $s \in A$, contradiction with the fact that $M$ is an upper bound of $A$.
Therefore, $\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.

Does those approach true?

答:
Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that $0
< r < m$.

This is true, but irrelevant.

Hence, $r \in A$. A contradiction, since $m$ is a lower
bound of $A$.

No, you haven't show that $r \in A,$ so no, you haven't shown a contradiction. You correctly showed that there is a rational number between $0$ and $m$, but you haven't (and you won't be able to) get from "$r$ is a rational number" to "$r$ is a member of $A$".

In summary, you haven't shown that $\ r \in A.$ Think about what it means to show that $\ r \in A $ [Hint: what is $A$?]

You could, however, amend your argument as follows:

For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, since [$\color{blue}{\text{*insert reasoning here*}}$], there exists $\color{blue}{r \in A}$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.


问:
Let $\lim_{n\to\infty}a_n=a$. Given $a>0$, prove that there exists some $n_0\in\mathbb{N}$ such that for all $n\geq n_0$, $a_n>0$.
Note. $\{a_n\}$ is a sequence in R.

I need to prove the claim in the title. The following is what I have got so far:

Suppose for contradiction that for all $n\in\mathbb{N}$, $a_n\leq 0.$ Let $a$ be greater than $0$. Since $\lim_{n\to\infty}a_n=a$, it follows that there exists $n_0\in\mathbb{N}$ such that given $\epsilon>0$, $|a_n-a|<\epsilon$ for all $n\geq n_0.$ Since $a>0,$ $a+\epsilon>0$. Therefore, we have $-\epsilon<a-\epsilon<a_n\leq 0<\epsilon<a+\epsilon.$
In particular, $-\epsilon<a_n<\epsilon$.

Note. I am aware of the following theorems:

All convergent sequences in R is bounded.

The limit of a sequence in R is unique.

答: Hint : Apply the definition of the convergence with $$\varepsilon = \frac{a}{2} > 0$$


问:
Lewis Carroll's logic problem: Proving that opium eaters do not wear white kid gloves.

If he goes to a party, he does not fail to brush his hair. To look fascinating it is necessary to be tidy. If he is an opium eater, then he has no self-command. If he brushes his hair, he looks fascinating. He wears white kid gloves only if he goes to a party. Having no self-command is sufficient to make one look untidy. Therefore, opium eaters do not wear white kid gloves.

P : He goes to a party.

B: He does not brush his hair.

F: He looks fascinating.

T: He looks tidy.

O: He is an opium eater.

S: He has self-command.

G: He wears white kid gloves.

I have been trying to solve this logic problem for days, but every time I get stuck. I have also tried to solve it with 'Proof by contradiction' but it was worthless.

答:
With (please, note change in B):

P : He goes to a party.

B: He brushes his hair.

F: He looks fascinating.

T: He looks tidy.

O: He is an opium eater.

S: He has self-command.

G: He wears white kid gloves.

IMO we must have:

$P \to B$

$F \to T$

$O \to \lnot S$

$B \to F$

$G \to P$

$\lnot S \to \lnot T$

Conclusion: $O \to \lnot G$.

The solution is straightforward, with some contrapositions.

Added

"B is a necessary condition for A" is $A \to B$.

Why so? Because it means that we cannot have $A$ without $B$, and thus the case $A$ True and $B$ False is not possible.

Thus, "(in order) To look Fascinating, it is necessary to be Tidy" reads (thanks to ACheca): "To be Tidy is a necessary condition to look Fascinating", that sounds quite natural.


问:
What is wrong with this theorem about closed and bounded non empty subsets of $\mathbb{R}$?
I am self-studying metric spaces from the book by Satish Shirali. One of the theorems from the chapter on "Topology of Metric Spaces" states:

Let $F$ be a nonempty bounded closed subset of $\mathbb{R}$. Then $F$ is either
a closed interval or is obtained from some closed interval by removing a countable
family of pairwise disjoint open intervals whose endpoints belong to $F$.

Now, let us consider the set $F = \{0\} \cup \left\{\frac{1}{n} ; n \in \mathbb{N}\right\} = \left\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \cdots\right\}$. Clearly, F is both bounded as well closed.

But, it is neither a closed interval nor is it obtained from some closed interval by removing a countable family of pairwise disjoint open intervals whose endpoints belong to $F$.

If we consider
$F = \left\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \cdots\right\} = [0,1] / S $, where  $S = (\frac{1}{2},1) \cup (\frac{1}{3},\frac{1}{2}) \cup (\frac{1}{4},\frac{1}{3})\cup \cdots$, even then we shall have an uncountable family of pairwise disjoint open intervals.

This seems to contradict the given theorem. Where am I mistaken ? Any insight shall be highly appreciated.

答:
The family $$\mathcal H:=\left\{\left(\frac12,1\right); \left(\frac13,\frac12\right); \left(\frac13,\frac14\right);\cdots\right\}=\left\{\left(\frac1{n+2},\frac1{n+1}\right)\,:\, n\in\Bbb N\right\}$$ seems quite countable to me and $S=[0,1]\setminus\bigcup\mathcal H$, where $$\bigcup\mathcal H:=\{x\in\Bbb R\,:\, \exists I\in \mathcal H,\, x\in I\}=\left\{x\in\Bbb R\,:\, \exists n\in\Bbb N,\, x\in\left(\frac1{n+2},\frac1{n+1}\right)\right\}.$$

Now, the set $\bigcup \mathcal H$ itself may be uncountable (and in fact it kind of has to be, since it's an union of open intervals of $\Bbb R$), but the important fact is that there is a countable family of intervals the union of which is $\bigcup\mathcal H$, because here countable refers to the cardinality of one such $\mathcal H$ (if there is one).


问:
If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$.
Question from Math Olympiad:

Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.

My approach:

$a\sqrt{a} + b\sqrt {b} = 183$

$a\sqrt{b} + b\sqrt {a} = 182$

Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$

I only see two integral possibilities:

$1 \times 365 $  and $5 \times 73 $

And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.

Please help me with the general method to go about the same.

This is the picture from the source.

答:
Hint : If you make the difference of the two equations, you get
$$(\sqrt{a}-\sqrt{b})(a-b)=0$$

Probably you can conclude.


问:
Why there is no closed form solution to $\tan(x) + x = \frac{\pi}{2}$?
I want to solve for x...

$$\tan(x) + x = \frac{\pi}{2}$$
where $0 < x < 1$

...but i can't.

Someone wrote to me saying that this have no closed form solution (very sad).

It's really true?
No closed form solution?
Maybe sine or cosine of a rational or a rational part of  $\pi$

EDIT: Approximation
$$x \approx 0.710462737775516856747428268$$

EDIT: I've tried to solve this equation:

$$\tan(\tan(x)) = \frac{1}{\tan(x)}$$

I suppose they have the same solution.

答:
The general rule is: equations have no closed-form roots.

When the equations are polynomials of degree no exceeding $4$, the roots have always a closed-form. But for higher degrees, this is exceptional.

For transcendental equations that cannot be reduced to algebraic ones, AFAIK we have no general resolution method (not even to just tell the number of roots).


答:
Using the "Archimedean property" , we get for every $\epsilon \gt 0 $ , there exists $m \in \mathbb{N}$ such that $\frac{1}{n} \lt \epsilon$ for every $n\ge m $ .

As, $3^{n+5}\gt n \implies \frac{1}{3^{n+5}} \lt \frac{1}{n} \lt \epsilon $ for all $n\ge m $

By definition of convergence of a sequence, sequence

$\{\frac{1}{3^{n+5}}\}$ converge to $0$.

So, the given sequence is bounded and $limsup=liminf=0$

Title: Showing that the given sequence is bounded as follows

Body:
Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.

Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows:
To show $x_{n+1}<x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that is s, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$

Back to the problem. It's clear that $A$ is bounded above by $\frac{1}{3^6}$ (Should I show this first?).
Then, to show that $A$ is bounded, it's suffices to show that $A$ is bounded below by $0$.
I show this one again by induction. Indeed, it's true for $n=1$.
Assume that it's true for $n=k$; that is $0 < x_k$. Then, $x_{k+1} = \frac{1}{3^{(k+1)+5}}
= \frac{1}{3^{k+5}} \cdot \frac{1}{3} > 0$. Hence, $0< x_n$ for all $n \in \Bbb Z^+$. Therefore, $A$ is bounded below by $0$. Thus, $A$ is bounded, as desired.

Now, I claim that $\sup A = \frac{1}{3^6}$ and $\inf(A) = 0$.
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.

For the supremum, let $M$ be an another upper bound of $A$. To show: $\frac{1}{3^6} \le M$. Suppose $M < \frac{1}{3^6}$. Then, by the Density Theorem, there exists $s \in \Bbb Q$ such that
$M < s < \frac{1}{3^6}$. Hence, $s \in A$, contradiction with the fact that $M$ is an upper bound of $A$.
Therefore, $\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.

Does those approach true?


问:
Area under the curve visually
We know the area function $A(x)$ is an antiderivative of $f(x)$.Meaning there could be numerous area functions whose derivative will be $f(x)$.Could you please show 2 area functions in graph which only differ by a constant?

答:
By integrating $ y= \cos x $ we obtain $ y= \sin x + C$ where $C$ is an arbitrary constant in $y$

The extra area added is the arbitrary constant multiplied by their domain interval.

$$ \int y(x) dx + C (x_2-x_1) $$

When we say area under a curve, it is implied that the constant of integration is zero.

Graph shows difference between two integrands yellow patch which arise from constants $ C_1=0, C_2=2 $ for $ y,y_2$

$$ y_1= \cos x  + 1, y_2= \cos x +2$$

The wavy region can be made to descend as shown to an equal area rectangular patch as shown.. where each ordinate goes down to touch the x- axis.

Title: Area under the curve visually


问:
Discretization of nonlinear curvilinear dynamics equations
I am considering a tracking model of the form
$$
\dot{x} = v\cos\psi\\
\dot{y} = v\sin\psi\\
\dot{v} = a\\
\dot{a} = 0\\
\dot{\psi} = \omega\\
\dot{\omega} = 0
$$
which is considering 2D position in $x$ and $y$, speed $v$ in direction $\psi$ from pos $x$ with turn rate $\omega$.

I am trying to discretize this. A similar model, assuming $a=0$, has been discretized in doi.org/10.1109/cdc.1996.573612. In the paper it is proposed that
$$
x_{k+1} = x_{k} + \int_{t}^{t+T} v\cos\psi d\tau
$$
where if $\dot{v}=0$ then
$$
x_{k+1} = x_{k} + \frac{v}{\omega}\big(\sin(\omega(t+T)) - \sin(\omega t)\big)\\
\quad= x_{k} + \frac{2v}{\omega}\sin(\frac{\omega T}{2}) \cos(\psi + \frac{\omega T}{2})
$$
leveraging $\psi = \int \dot{\psi} dt = \int \omega dt = \omega t$.

I am trying to expand this in order to allow for a constant, but non-zero, $a$. So I set up the integral
$$
\begin{aligned}
x_{k+1} &= x_{k} + \int_{t}^{t+T} v\cos\psi d\tau\\
&= x_{k} + \int_{y}^{t+T} v\cos(\omega\tau)d\tau
\end{aligned}
$$
I figured this can be solved using integration by parts, where
$$
u=v\\
du = a\\
d\nu=\cos(\omega\tau)d\tau\\
\nu = \frac{1}{\omega} \sin(\omega\tau)
$$ such that
$$
\begin{aligned}
\int_{y}^{t+T} v\cos(\omega\tau)d\tau &= \frac{v}{\omega}\sin(\omega\tau)\bigg|_{t}^{t+T} - \int_{t}^{t+T} \frac{a}{\omega}\sin(\omega\tau)d\tau\\
&= \frac{v|_{t+T}}{\omega}\sin(\omega(t+T)) - \frac{v|_{t}}{\omega}\sin(\omega t) + \frac{a}{\omega^{2}} \bigg(\cos(\omega(t+T)) - \cos(\omega t) \bigg)\\
&= 2\frac{v|_{t+T}-v|_{t}}{\omega} \sin(\frac{\omega T}{2}) \cos(\psi+\frac{\omega T}{2}) - \frac{2a}{\omega^{2}} \sin(\frac{\omega T}{2})\sin(\psi + \frac{\omega T}{2})
\end{aligned}
$$
I figure here that $v|_{t}$ is just $v$ and $v_{t+T}$ is $v+aT$? But I am in doubt if this is even a correct way to integrate this? As well I imagine there will be some numerical problems in evaluating $\frac{a}{\omega^{2}}$ for small turn rates. Is this a correct discretization of the original dynamics assuming that $a\neq 0$?

答:
Indeed you get
$$
x\rvert_{t+T}=x\rvert_t+\frac{(v\rvert_t+aT)\sin(\psi\rvert_t+\omega T)- v\rvert_t\sin(\psi\rvert_t)}{\omega}+\frac{a}{\omega^2}[\cos(\psi\rvert_t+\omega T)-\cos(\psi\rvert_t)]
$$
but you shouldn't compute it this way when $\lvert\omega\rvert$ is small.  Instead, we have
\begin{align*}\require{color}
x\rvert_{t+T}&=x\rvert_t
+\frac{v\rvert_t[\sin(\psi\rvert_t+\omega T)-\sin(\psi\rvert_t)]}{\omega}\\
&\hphantom{{}=x\rvert_t}
+\frac{a}\omega \frac{\omega T\sin(\psi\rvert_t+\omega T)+\cos(\psi\rvert_t+\omega T)-\cos(\psi\rvert_t)}{\omega}\\
&=x\rvert_t
+v\rvert_t T\cos(\psi\vert_t+\tfrac12\omega T)\operatorname{sinc}(\tfrac12\omega T)\\
&\hphantom{{}=x\rvert_t}
+\tfrac12 aT^2 \cos(\psi\rvert_t+\tfrac12\omega T)\operatorname{sinc}(\tfrac12\omega T)\\
&\hphantom{{}=x\rvert_t}
+\frac{aT}{\omega}
\sin(\psi\rvert_t+\tfrac12\omega T)
\underbrace{\color{blue}[\cos(\tfrac12\omega T)-\operatorname{sinc}(\tfrac12\omega T)]}_{= -\frac1{12}\omega^2T^2+\frac1{480}\omega^4T^4+\dots }\\
&=x\rvert_t
+(v\rvert_t T+\tfrac12aT^2)\cos(\psi\vert_t+\tfrac12\omega T)\operatorname{sinc}(\tfrac12\omega T)\\
&\hphantom{{}=x\rvert_t}
-\tfrac1{12}{\color{red}(1-\tfrac1{40}\omega^2T^2+\dots)}a\omega T^3\sin(\psi\rvert_t+\tfrac12\omega T).
\end{align*}
The relative error of approximating the term in red by $\cos(\omega T/\sqrt{20})$ is approximately $(\omega T)^4/8400$ (or with $\operatorname{sinc}(\omega T\sqrt{3/20})$ is $(\omega T)^4/28000$), so should be small enough practically when $\omega T$ is small.


问:
On the centres of the dihedral groups
In an proof that I recently read, the following 'fact' is used, where $D_{2n}$ denotes the dihedral group of order $2n$:

If $n$ is even, then $D_{2n} \cong C_2 \times D_n$.

The (short) given justification is that the centre $Z(D_{2n}) \cong C_2$, whenever $n$ is even, and is trivial provided that $n$ is odd.

However, here is a result that a friend of mine found in the litterature which contradicts the previous one.

Suppose that 4 divides $n$. Then $D_{2n}$ is not isomorphic to $C_2 \times D_n$.

Proof: Suppose otherwise. We know that $Z(D_{2n}) \cong C_2$. By assumption, $D_{2n} \cong C_2 \times D_n$, therefore $Z(D_{2n}) \cong Z(C_2 \times D_{n})$. Since in a direct product $A \times B$, the groups $A$ and $B$ commute, we obtain that $Z(C_2 \times D_{n}) = Z(C_2) \times Z(D_n) \cong C_2 \times C_2$, a contradiction. QED.

Now comes my first question : Is the above proof correct?

Second question : Given a finite group $G$ whose centre is not trivial, does it always exist a group $H$ such that $G \cong Z(G) \times H$?

If so, then given $n$ a multiple of 4, the dihedral group $D_{2n}$ could be written as a direct product $C_2 \times H$. Here comes my third question : What should be $H$ since we know that it cannot have nontrivial centre (and, in particular, since $H$ cannot be $D_n$)?

答:
I like your questions.

Yes, and it is very silly.

No, for example the quaternion group of order 8 is not like this.

(So, there is no such H).

To be clear, the dihedral group of order 8k+4 is isomorphic to the direct product of a cyclic group of order 2 with a dihedral group of order 4k+2, but all other dihedral groups are "directly indecomposable", that is, they do not have a decomposition as a direct product of two non-identity groups.

In case you want to prove the stronger indecomposability:  Basically, you look at the normal subgroups.  Assuming the order of the dihedral group is at least 6, they are all contained in the subgroup of rotations, and so the only way you can have a direct product is if that subgroup of rotations has order 4k+2, so that its Sylow 2-subgroup is centralized by a flip.

Title: On the centres of the dihedral groups


问:
Basic ring isomorphism question
Let $R=\mathbb{Z}[x], I=(x^2+1,x+1).$ Prove that $R/I \cong \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}/2\mathbb{Z}$.

I am confused with the rather messy looking of $R/I$. My first step is to define the homorphism $f: R/I \rightarrow \mathbb{Z}[i]$ but then I don't know what to do, please help!

答:
Remember the third ismorphism theorem for rings: $\,R\,$ is a ring with ideals $\,I\le J\le R\, $ ,then

$$\left(R/I\right)/\left(J/I\right)\cong R/J$$

In your problem  we have

$$\langle x^2+1\,\rangle\le\langle\,x^2+1\,,\,x+1\,\rangle\le\Bbb Z[x]\implies$$

$$\left(\Bbb Z[x]/\langle\,x^2+1\,\rangle\right)/\left(\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\right)\cong\Bbb Z[x]/\langle\,x^2+1\,,\,x+1\,\rangle$$

Now just apply that under the same isomorphism , we have

$$\Bbb Z[x]/\langle\,x^2+1\,\rangle\cong\Bbb Z[i]\;,\;\;\langle\,x^2+1\,,\,x+1\,\rangle/\langle\,x^2+1\,\rangle\cong\langle\,x+1\,\rangle$$

the isomorphism being the one we get from the first isomorphism theorem and the homomorphism

$$\phi:\Bbb Z[x]\to\Bbb Z[i]\;,\;\;\phi(f(x)):=f(i)\;\;\;(\text{i.e., determined by}\;\;x\mapsto i\,)$$

soo that under the above, $\,\Bbb Z[x]\ge\langle\,x+1\,\rangle\mapsto\langle i+1\,\rangle\le\Bbb Z[i]\;$

Title: Basic ring isomorphism question


问:
Why do we ignore the negative sign when finding basic angle when solving trigonometric equations?
Suppose I'm asked to solve for $\cos\theta=−0.5$, for $\theta$ between $0^\circ$ and $360^\circ$ inclusive.

I am told that the first step would be to buy the basic angle, $\alpha$. The way I am told to do this is by omitting the negative sign from -0.5 and hence finding sine inverse of 0.5 to get $60^\circ$.

I understand the next steps of identifying the quadrant $\theta$ lies in etc: what I don't understand is why I need to omit the negative sign when finding the basic angle.

答:
What you are searching for is the reference angle $\theta$ which is in the first quadrant.  Once we have that, then the candidate angles are  $\theta, 180-\theta, 180+\theta,$ and $360-\theta$.  But since the trig functions for angles in the first quadrant are always positive, we want to consider the absolute value of the ratio.

There is also probably a historical artifact.  Before scientific calculators, one would have to use trig tables to calculate inverse trig values, and those would only contain positive ratios and angles from $0$ to $90$.  Nowadays, it is much more natural to find at least one solution for $\cos\theta=-0.5$ using a scientific calculator, but it probably still makes sense to find the reference angle in the first quadrant so that you can find the other solution without too much drama.


问:
Integrating a rational function with square roots
Solve :$$\int _ 0 ^ {\infty}{x^2\over (2+x^2)^{3/2}}dx  = \int _ 0 ^ {\pi/2} \tan^2 {\theta} \cos {\theta} \,d{\theta}=[\sec{\theta}+\cos {\theta}]_0 ^{\pi/2}$$

I've tried to do it and got the answer as upper limit is becoming divergent, then what to do? Please help to solve this problem. Thanks in advance.

答:
Hint:

Let's try integration by parts

$$\int x\cdot\dfrac x{(2+x^2)^{3/2}}\ dx=x\int\dfrac x{(2+x^2)^{3/2}}\ dx-\int\left(\dfrac{dx}{dx}\cdot\int\dfrac x{(2+x^2)^{3/2}}\ dx\right)dx$$

Set $\sqrt{2+x^2}=y\implies2+x^2=y^2,x\ dx=y\ dy$

$$\int\dfrac x{(2+x^2)^{3/2}}\ dx=\int\dfrac{y\ dy}{y^3}=-\dfrac1y=?$$

Use this for $$\int\dfrac{dx}{\sqrt{2+x^2}}=\ln|x+\sqrt{x^2+2}|+K$$

The definite integral $[0,\infty]$ is divergent as $$\lim_{x\to\infty}\ln|x+\sqrt{x^2+2}|=?$$


答:
Hint: Let $x=\sqrt a$ and $y=\sqrt b$. You know that
$$x^3+y^3=183$$
and
$$xy(x+y)=182.$$
Since
$$x^3+3xy(x+y)+y^3=(x+y)^3,$$
you can use this to find $x+y$; see if you can use this to find $\frac 95(x^2+y^2).$

Title: If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$.

Body:
Question from Math Olympiad:

Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.

My approach:

$a\sqrt{a} + b\sqrt {b} = 183$

$a\sqrt{b} + b\sqrt {a} = 182$

Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$

I only see two integral possibilities:

$1 \times 365 $  and $5 \times 73 $

And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.

Please help me with the general method to go about the same.

This is the picture from the source.


问:
Solving $\frac{\cos5x \;\cos3x-\sin3x\;\sin x}{\cos2x}=1$
What is general solution of $$\cfrac{\cos5x \;\cos3x-\sin3x\;\sin x}{\cos2x}=1$$
$$1)\frac{k\pi}3\quad\quad\quad\quad\quad\quad2)\frac{k\pi}2\quad\quad\quad\quad\quad\quad3)\frac{2k\pi}5\quad\quad\quad\quad\quad\quad4)\frac{2k\pi}3$$

The numerator of the fraction is $\cos(5x+3x)$. so I should find general solution of $\cos8x=\cos2x$. I'm not sure how to do it, I can write $\cos8x $ in term of $\cos 2x$:

$$\cos8x=2\cos^24x-1=2(2\cos^22x-1)^2-1$$
After substituting it in the equation and using $\cos2x=t$ we have degree four equation . Is there easier way to do it?

答:
$$0=\cos5x \;\cos3x-\sin3x\;\sin x-\cos 2x$$ $$=\cos5x \;\cos3x-\sin3x\;\sin x-\cos (5x-3x)$$ $$=\sin 3x\sin 5x-\sin 3x\sin x$$ $$\to -2\sin ^2 3x\cos 2x=0$$ Can you finish it?

Note that $\cos 2x\neq 0$ has to be included(why?)  $$\to \sin 3x=0$$


问:
When is the diagonal a closed immersion with open image?
Let $A$ be a commutative unital $R$-algebra and consider the  multiplication map $m:A\otimes_R A\rightarrow A$ given by $m(a\otimes b) = ab$ with kernel $I$. The map $m$ is surjective, hence the image of $f: \operatorname{Spec}A\rightarrow \operatorname{Spec}A\times\operatorname{Spec}A$ is $V(I)$.

If we assume that $V(I)$ is also open, then is $f$ an open immersion?

As far as I know, an immersion with closed image is a closed immersion, but an immersion with open image maybe not an open immersion?

Thanks.

答:
I believe the following holds (by an exercise in Atiyah-Macdonald (AM)): Let $B:=A \otimes A$ and let $I:=ker(m)$ be the kernel of the multiplication map.
It follows $V(I) \cup V(I)^c=S:=Spec(B)$. If $V(I)^c=V(J)$ is closed it follows by AM Exercise I.15 that

$$V(I) \cup V(J)= V(I\cap J)=S$$

hence $I \cap J \subseteq \mathfrak{p}$ for all prime ideals $\mathfrak{p}\subseteq B$,
hence $I\cap J \subseteq nil(B)$. Since $V(I) \cap V(J)=V(I+J)=\emptyset$ it follows
$I+J=(1)$ is the unit ideal. Hence if $B$ is a domain it follows $I\cap J =(0)$ and $I+J =(1)$ hence there is by the Chinese remainder theorem an isomorphism

$$ B:=A\otimes A \cong A\otimes A/I \oplus A\otimes A/J \cong A \oplus A\otimes A/J.$$

Hence

$$S:=Spec(A\otimes A) \cong Spec(A) \cup Spec(A\otimes A/J):=S_1 \cup S_2$$

is a disjoint union. It seems the diagonal map $\Delta$ is an open immersion in this case, since $Spec(A)\cong S_1 \subseteq Spec(B)$ is an open subscheme. The two schemes $S_1:=Spec(A),S_2:=Spec(B/J)$ are open and closed in $S$.

Question: "If we assume that $V(I)$ is an open subscheme, then is $f$ an open immersion?"

Answer: If the product $S$ is a reduced scheme it seems the following holds:

Lemma. If $V(I)$ is an open subscheme  it follows the diagonal map $\Delta$ is an open immersion.

Proof.  This holds since $nil(B)=0$ and hence by the above argument it follows the diagonal map $\Delta$  induces an isomorphism between $Spec(A)$ and the image $\Delta(Spec(A))\cong S_1 \subseteq S$.

Example. In general if $S:=Spec(A)$ is a reduced affine scheme and $Z:=V(I)$ is open and closed it follows $V(I)^c=V(J)$ and $I \cap J =(0), I+J=(1)$ hence by the Chinese remainder theorem we get a direct sum decomposition

$$A \cong A/I \oplus A/J$$

and a disjoint union of schemes

$$Spec(A) \cong Spec(A/I) \cup Spec(A/J):=V(I) \cup V(I)^c.$$

Conversely if $A \cong A_1 \oplus A_2$ there are  ideals $I_i\subseteq A$
with $A/I_i \cong A_i$ and $I_1 \cap I_2=(0), I_1+I_2=(1).$ The subschemes $Spec(A_i)\subseteq Spec(A)$ are open and closed.


答:
Hint:

Using Werner Formulas,

$$2\cos2x=2\cos5x\cos3x-2\sin3x\sin x=\cos2x+\cos8x-(\cos2x-\cos4x)$$

Now use Prosthaphaeresis Formulas $$\cos8x+\cos4x=2\cos6x\cos2x$$

Hope you can take it from here?

Title: Solving $\frac{\cos5x \;\cos3x-\sin3x\;\sin x}{\cos2x}=1$

Body:
What is general solution of $$\cfrac{\cos5x \;\cos3x-\sin3x\;\sin x}{\cos2x}=1$$
$$1)\frac{k\pi}3\quad\quad\quad\quad\quad\quad2)\frac{k\pi}2\quad\quad\quad\quad\quad\quad3)\frac{2k\pi}5\quad\quad\quad\quad\quad\quad4)\frac{2k\pi}3$$

The numerator of the fraction is $\cos(5x+3x)$. so I should find general solution of $\cos8x=\cos2x$. I'm not sure how to do it, I can write $\cos8x $ in term of $\cos 2x$:

$$\cos8x=2\cos^24x-1=2(2\cos^22x-1)^2-1$$
After substituting it in the equation and using $\cos2x=t$ we have degree four equation . Is there easier way to do it?


问:
complex analysis -bounded but not holomorphic function
What are some examples of functions that are bounded but not holomorphic?

I know that some the not holomorphic functions are

Absolute value function, Piecewise functions (not analytic where the pieces meet)

答:
You can start with the absolute value function and make it bounded, like for example $$
  f(z) = \frac{|z|}{1+|z|} \leq 1 \text{.}
$$

You get (assuming that $z=u+iv$ and using that $|z|=\sqrt{u^2+v^2}$) $$\begin{aligned}
  \frac{\partial \Re f}{\partial u} &= \frac{-u|z|^{-1}(1+|z|) + u}{(1+|z|)^2} = -\frac{u}{\sqrt{u^2+v^2}\left(1+\sqrt{u^2+v^2}\right)^2} \\
  \neq\frac{\partial \Im f}{\partial v} &= 0 \text{.}
\end{aligned}
$$
i.e. $f$ is not holomorphic.


答:
For large $x$, the integrand behaves as $1/x$ and thus your integral to infinity diverges logarithmically. The indefinite integral is

$$
I(x)=-\frac{x}{\sqrt{x^2+2}} + \sinh^{-1}(x/\sqrt{2})
$$

By trig substitution. $I(0)=0$. For large $x$, the first term on the right is finite (and equal to $-1$), but the second term is singular. Using the series for large argument

$$
\sinh^{-1}(z)=\ln(2z) + \mathcal{O}(1/z^2) \ \ , \ z \to \infty
$$

So we have for the principle part of your integral

$$
\int\limits_0^a dx \frac{x^2}{(2+x^2)^{3/2}} = \ln(a\sqrt{2}) -1 +\mathcal{O}(1/a^2) \ \ , \ \ a \to \infty
$$

Here is a plot:

Title: Integrating a rational function with square roots

Body:
Solve :$$\int _ 0 ^ {\infty}{x^2\over (2+x^2)^{3/2}}dx  = \int _ 0 ^ {\pi/2} \tan^2 {\theta} \cos {\theta} \,d{\theta}=[\sec{\theta}+\cos {\theta}]_0 ^{\pi/2}$$

I've tried to do it and got the answer as upper limit is becoming divergent, then what to do? Please help to solve this problem. Thanks in advance.


问:
For what value of a and b will the given sequence and series converge, given that a, b > 1?

$S_n$ = $\frac{n^a+3}{n^b-1}$

I know that if we take $lim_{n \to \infty} S_n$, and get a value L then it converges. If we get something that is either undefined or tends to $\infty$, then the sequence does not converge.

The question asks us for the value of a and b (which has to be greater than 1) which can be applied to make the sequence convergent, but upon applying the limits, I am ending up with an indeterminate form. How should I proceed with this?

Body:
Case 1: $a=b$. Then

$$S_n=\frac{1+ \frac{3}{n^a}}{1-\frac{1}{n^a}} \to 1.$$

Case 2: $a<b$. Then

$$S_n=\frac{n^{a-b}+ \frac{3}{n^b}}{1-\frac{1}{n^b}} .$$

Since $n^{a-b}= \frac{1}{n^{b-a}}$ and $b-a >0$, we get

$$S_n \to 0.$$

Case 3: $a>b$. This is your turn.


答:
$$\cos (5 x) \cos (3 x)-\sin (3 x) \sin (x)=\cos (2 x)$$
$$\frac{1}{2} (\cos (2 x)+\cos (8 x))-\frac{1}{2} (\cos (2 x)-\cos (4 x))=\cos(2x)$$
$$\frac{1}{2} (\cos (4 x)+\cos (8 x))-\cos(2x)=0$$
$$\cos (8 x)+\cos (4 x)-2 \cos (2 x)=0$$
$$2 \cos (2 x) \cos (6 x)-2 \cos (2 x)=0$$
$$\cos(2x)\left[\cos(6x)-1\right]=0$$
$$\cos(2x)=0\to x=\pm\frac{\pi}{4}+k\pi$$
discarded because this solution will make zero the denominator of the original equation
$$\cos(6x)=1\to x=\frac{k\pi}{3}$$


问:
How to go back from the stabilization of the pendulum around $\theta=\frac{\pi}{2}$ to stabilization around $\theta=0$
Let the pendulum equation: $\ddot \theta+\sin{\theta}+b\dot\theta=cu$, that in state space ($x_1=\theta$, $x_2=\dot \theta$) representation will be:

$\begin{cases}
\dot x_1=x_2\\
\dot x_2=-\sin{x_1}-bx_2+cu
\end{cases}$

$\textbf{The aim is to stabilize the system around $\theta=\frac{\pi}{2}$}$.

My idea so is the following:

in order to go back from the stabilization technique to the stabilization around the origin, I consider the design $u=u_{SS}+u_{\delta}$, with $u_{SS}$ for which the translated system , has the origin for $x_1=\frac{\pi}{2}$, and so at the origin an equilibrium point, so:

$\begin{cases}
 z_1=x_1-\frac{\pi}{2}\\
z_2=x_2
\end{cases}$$\iff$ $\begin{cases}
\dot z_1=z_2\\
\dot z_2=-\sin{z_1+\frac{\pi}{2}}-bz_2+cu_{SS}
\end{cases}$

So the $u_{SS}$ that realizes the requirement to have at the origin an equilibrium point is given by $u_{SS}=\frac{\sin{\frac{\pi}{2}}}{c}$.

$\textbf{My problem is:}$ in Khalil NonLinear Control Theory, it is not considered this $u_{SS}$, but only considering the system traslated in the coordinates $z$ it is designed then the control $u$, as it is sufficient to translate the system without considering this $u_{SS}$ to go back from the stabilization around $\theta\neq 0$ to the stabilization around $\theta=0$.

Please can you tell me if my idea is correct or I am doing some mistakes?

$\textbf{EDIT:}$ the parameter $c$ is uncertain...can this information give the explaination for my problem?

答:
Since I do not own this book, it is hard for me to directly assess what Khalil has to say about it. Either way, linearizing a system around any point requires one to define this point as an equilibrium (whether that is stable is not important). This is (as you show) done by offsetting the state and the input accordingly:
$$z_1 = x_1 - \theta_0, ~~ z_2 = x_2, ~~ u = u_{SS} + u_\delta$$
$$\dot{z_2} = -\text{sin}(z_1 +\theta_0) - bz_2 + c(u_{SS} + u_\delta)$$
Where $\theta_0$ is the chosen equilibrium point, $u_\delta$ is the parameter you control using $z_1$ as input (which you offset manually). The actual control input is indeed $u_{SS} + u_\delta$ in which $u_{SS}$ is calculated manually using your equation: $$u_{SS} = \frac{\text{sin}(\theta_0)}{c}$$
The fact that $c$ is an uncertain parameter troubles the control goal massively, as you are uncertain about the offset input which ensures the equilibrium (i.e. the actual equilibrium can be somewhere else now). This can however be solved by implementing an integral action in the controller and discarding the input offset (this should be computed iteratively with the integral action). Now what I think answers your question is the following: The input $u_{SS}$ equals zero if the offset angle equals 0. Hence it is not required to mention it.


问:
Dirac delta from polar coordinates to cartesian coordinates
I have:

$$k_x = k \cos\theta\\k_y=k\sin\theta$$

I would like to rewrite in terms of $k_x$ and $k_y$:

$$\exp(in\theta)\,\frac{\delta(k-\alpha)}{k}$$

I start from:

$$\exp(in\theta)=\left(\cos\theta+i\sin\theta\right)^n=\left(k_x+ik_y\right)^n\,k^{-n}$$

but I don't if it helps and if yes how to go on.

答:
There are two other formulae you need, which express $k$ and $\theta$ in terms of $k_x$ and $k_y$:
$$k = \sqrt{k_x^2 + k_y^2}$$
and
$$ \theta = \tan^{-1}\left(\frac{k_y}{k_x}\right) $$
Note that the second one always gives a value for $\theta$ between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$. To get the actual value you should eg add $\pi$ whenever $k_y$ is negative (and renormalize if necessary). Programming languages and computer algebra systems usually have a two-argument atan2 that does this for you.


问:
There is a point on the road from where the car with initially empty fuel tank can complete a loop.
I need help proving the following statement:

There are $n$ gas stations along a circular road. The total amount of fuel at these stations is exactly the amount needed for a car to complete one loop. Show that there is a point on the road from where the car with initially empty fuel tank can complete a loop.

Here is my progress:

Note that a successful starting point must always be at one of the gas stations since; otherwise, the car will not be able to move due to the empty fuel tank. Before tackling the proof, consider the following example with four gas stations. Label the gas stations

$g_1, g_2, g_3, g_4$

so that the amount of fuel at gas station $g_i$ is $x_i$ for $1 \leq i \leq 4$. Define the distance between consecutive gas stations to be $d(g_i, g_{i+1})$. Then $x_1 + x_2 + x_3 + x_4 = 1$ and $d(g_1, g_{2}) + d(g_2, g_{3}) + d(g_3, g_4) + d(g_4, g_1) = 1$. We want to make $d(g_i, g_{i+1})$ and $x_1, ..., x_4$ arbitrary so; for example, consider $x_1 =0.2,x_2 = 0.3 ,x_3 = 0.4, x_4= 0.1$ and $d(g_1, g_{2}) = 0.3 , d(g_2, g_{3}) = 0.1 , d(g_3, g_4)= 0.2, d(g_4, g_1) = 0.4$. We want to find a starting gas station so that we can get back to it. If the starting point is at $g_2$ and fill the car with fuel, call it $X$, we have $X=0.3$. We can then travel to the next gas station $g_3$ but will be left with $X = 0.3 - d(g_2, g_{3}) = 0.2$. After fueling at $g_3$, $X = 0.7$ and traveling to $g_4$ we have $X= 0.7 - d(g_3, g_4) = 0.5$. After fueling at $g_4$, $X = 0.6$ and traveling to $g_1$ we have $X= 0.6 - d(g_4, g_1) = 0.2$. Therefore, we have completed a loop so at least one starting point exists and that is at $g_2$.

claim:

If $\sum_{i =1}^k x_i = 1$ and $\sum_{i=1}^k d(g_i, g_{i+1}) = 1$

then there exists an $x_i$ where $i = 1, ..., k$ such that $x_i \geq d(g_i, g_{i+1})$.

proof of claim:
By way of contradiction suppose $x_i < d(g_i, g_{i+1})$

for some $1 \leq i \leq k$. Then, $\sum_{i=1}^k d(g_i, g_{i+1}) = 1$ implies $\sum_{i = 1}^k x_i < 1$ which is a contradiction.

proof of problem:
We induct on the number of gas stations. Suppose there is only one gas station along the circular road such that the total amount of fuel at this station is exactly the amount needed for a car to complete one loop. Therefore, there is only one possible starting point for the car which is at the single gas station. Now, suppose there is always a successful starting point when there are $k$ gas stations. That is, for gas stations

$g_1, ..., g_k$ with fuel amounts of $x_1, ..., x_k$, respectively, such that $d(g_1, g_2) + ...+ d(g_k, g_1) = 1$ and $x_1 + ... + x_k = 1$.
The total amount of fuel at the
$k$ stations is exactly the amount needed for a car to complete one loop. We want to show that there is always a successful starting point when there are $k+1$ gas stations so consider a gas station with $k+1$ gas stations. By the claim above, there is at least one gas station that contains enough gas to reach the next gas station, call this gas station $h_1$ and consider the gas station that comes
after
$h_1$; call it, $h_2$.

I am not sure where to go from here.

An idea:

Combining the fuel from $h_1$, $h_2$ and proceeding up to $h_k$, we have enough fuel for a car with an initial empty fuel tank to make a loop so that a starting point exists, namely at $h_1$.

But how do we know that the fuel from $h_2$ will get us to the next gas station?

答:
You don't know that the fuel from $h_2$ will get you to the next station.

The inductive hypothesis is that there is always a successful loop for any $k$ stations. You do know that given $k+1$ stations with total fuel exactly equal to the amount of fuel required to complete one loop, there must be at least one station that has enough fuel to get to the next station.

I find it more convenient if you number the stations so that station $g_k$ is a station that has enough fuel to reach the next station, $g_{k+1}$.
Now if the fuel at $g_{k+1}$ were added to the fuel at $g_k$ instead of being placed at $g_{k+1}$, we would have a road with $k$ stations with exactly enough fuel to complete one loop, and we know we can start the loop somewhere and complete it successfully.

Suppose the solution for $g_1,\ldots,g_k$ is to start the loop at $g_i$ for some $1 \leq i \leq k.$
Then we know the gas at $g_i, \ldots, g_{k-1}$ is distributed so that we can get to $g_k$ starting at $g_i$; the original fuel stores at $g_k$ and $g_{k+1},$ now combined at $g_k$, together with the leftover gas from $g_i, \ldots, g_{k-1}$ are sufficient to get to $g_1$; and the leftover gas from $g_i, \ldots, g_k$ at $g_1$ together with the gas stored at $g_1$ is a sufficient amount for the car to continue traveling all the way to $g_i$ without running out of gas between stations.

Now restore the amounts of fuel at $g_{k+1}$
Then for the original configuration $g_1,\ldots,g_k,g_{k+1}$,
show that you can get from $g_i$ to $g_k$ without running out of fuel,
from $g_k$ to $g_1$ without running out of fuel, and from $g_1$ to $g_i$ without running out of fuel.

Note that this still works if $i = 1$ or $i = k$; it is trivial to get from $g_1$ to $g_1$ (for example) without running out of fuel regardless of the fuel already in the car's tank.


问:
Generalized rule behind equating real and imaginary components
When two complex numbers $z$ and $w$ are said to be equal, that implies that

Re$(z)$=Re$(w)$

and that

Im$(z)$=Im$(w)$.

I believe this, and always have, but now I wonder what the generalized rule is behind it. I know that if $z$ and $w$ were two-dimensional vectors, one could appeal to the linear independence of their two components. But I know that the set of complex numbers forms a field, not a vector space (correct?) so what would be the appropriate statement in this case?

答: As well as being a field in its own right, $\Bbb{C}$ is a two-dimensional vector space over $\Bbb{R}$ with scalar $\times$ vector multiplication given by restricting complex multiplication to $\Bbb{R} \times \Bbb{C}$.  $1$ and $i$ are a basis, leading to the rule about equating real and imaginary parts.


答:
Case-$1$,

Now,, for $b\gt a $, $$\frac{1+\frac{3}{n^a}}{n^{b-a}-\frac{1}{n^a}} \to 0 $$ as $n \to \infty$

Case-$2$

for $b=a$ , $$\frac{1+\frac{3}{n^a}}{n^{b-a}-\frac{1}{n^a}} \to 1 $$ as $n \to \infty$

Case-$3$

For $a\gt b$ , $$\frac{n^{a-b}}{1-\frac{1}{n^b}} + \frac{\frac{3}{n^b}}{1-\frac{1}{n^b}}$$

We show divergence using the first part $$\frac{n^{a-b}}{1-\frac{1}{n^b}}$$,  ( since, $$\frac{\frac{3}{n^b}}{1-\frac{1}{n^b}} \to 0 $$ as $n \to \infty $)

As, $$\frac{n^{a-b}}{1-\frac{1}{n^b}} \gt n^{a-b}$$

Now, however large $G$ , we take , we always find $n=\lceil (G+1)^{\frac{1}{a-b}} \rceil $ such that $n^{a-b} \gt G $

Proving the sequence is divergent for $a \gt b $

Hence , the result.

Title: For what value of a and b will the given sequence and series converge, given that a, b > 1?

Body:
$S_n$ = $\frac{n^a+3}{n^b-1}$

I know that if we take $lim_{n \to \infty} S_n$, and get a value L then it converges. If we get something that is either undefined or tends to $\infty$, then the sequence does not converge.

The question asks us for the value of a and b (which has to be greater than 1) which can be applied to make the sequence convergent, but upon applying the limits, I am ending up with an indeterminate form. How should I proceed with this?


问:
strictly positive iid random variables: a property
Let $X_n$ be i.i.d and $>0\forall n$. Could anyone tell me how to show that $\mathbb P\left(\frac{X_{2n}}{X_{2n+1}}\ge 1 \text{ i.o}\right)=1$ Thanks.

答:
Apply the second Borel-Cantelli Lemma as follows:

Note that since the $X_n$ are i.i.d. (so in particular independent), the sets $\{X_{2n} \geq X_{2n + 1}\} = \{\frac{X_{2n}}{X_{2n+1}} \geq 1\}$ are independent for any $n \in \mathbb{n}$.

Now we also have that for any $n \in \mathbb{n}$

$$ \begin{align}
\mathbb{P}(\{X_{2n} \geq X_{2n + 1}\}) &= \mathbb{P}(\{X_{2n} > X_{2n + 1}\}) + \mathbb{P}(\{X_{2n} = X_{2n + 1}\}) \\
&= \frac{1}{2}(1 - \mathbb{P}(\{X_{2n} = X_{2n + 1}\})) + \mathbb{P}(\{X_{2n} = X_{2n + 1}\}) \geq \frac{1}{2}
\end{align}
$$

where in the second line we used that the $X_n$ are i.i.d. and thus $\{X_{2n} > X_{2n + 1}\} = \{X_{2n+1} > X_{2n}\}$.

Now we have

$$ \sum_{n = 1}^{\infty} \mathbb{P}(\{X_{2n} \geq X_{2n + 1}\}) \geq \sum_{n = 1}^{\infty} \frac{1}{2} = \infty$$

Hence since the sets are independent, by Borel-Cantelli:

$$\mathbb{P}(\limsup_{n \rightarrow \infty} \{X_{2n} \geq X_{2n + 1}\}) = \mathbb{P}(X_{2n} \geq X_{2n + 1} ~~ \text{i.o.} \}) = 1$$


问:
Convergence of the sequence $a_n=\frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(2n)^2}$
What I did is,
$$n\frac{1}{(2n)^2} \le a_n \le n\frac{1}{n^2}$$
Can we conclude that $a_n \rightarrow 0$ using squeeze theorem. Is there any mistake?

答:
$$\frac{n+1}{(2n)^2}\le a_n \le \frac{n+1}{n^2}$$

And, yes your conclusion is right!


问:
Evaluate $I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1}$

$$I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1},$$

I used
$$u=\frac{c}{xt}-\frac{mt}{x}$$

and

$$du=-\left(\frac{c}{xt^2}+\frac{m}{x}\right)dt$$
but I'm not able to find the limits at $t=0$ and $t=\infty$. Neither are straight forward to use as
$$\int_0^\infty\frac{1}{1+t^2}dt=\pi/2$$

Any hints on how to proceed?

答:
Rearrange the integral as follows

\begin{align}
& \int_0^\infty\frac{1}{x^2t^2}\frac{1}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt \\
=& \frac1{2cx} \int_0^\infty \frac{(\frac{c}{xt^2}+\frac mx)+ (\frac{c}{xt^2}-\frac mx )}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt\\
=& \frac1{2cx} \int_0^\infty \frac{d(\frac {mt }x-\frac{c}{xt})}{(\frac{c}{xt}-\frac{mt}{x})^2+1}
 -\frac1{2cx} \int_0^\infty \frac{d(\frac{c}{xt}+\frac {mt }x )}{(\frac{c}{xt}+\frac{mt}{x})^2+1-\frac{4cm}{x^2}}\\
\end{align}
where the first integral is of the form $\int_0^\infty\frac{du}{1+u^2}$ and the second integral vanishes.


问:
Write an exponential model - A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.
I am to write an exponential model of the scenario "A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour".

With that, I am to find the amount remaining after 3 hours.

My textbook solution says  $A=125e^{−0.3567t}$; $A≈43$ mg

I instead got $y=125e^{0.3*3}$; $A=307.45$ mg

Why is the correct solution $A=125e^{−0.3567t}$? And specifically where did $−0.3567$ come from?

答:
Let $A\,\mathrm{mg}$ be the amount remaining after $t\,\mathrm{h}.$ Your expression $A=125e^{0.3t}$ seems to have arisen from two modelling errors. Do note that:

The decay rate being $30\%$ (of itself) per hour means that by $t=1,$ $70\%$—not $30\%$—of it remains.

Since the drug is decreasing by a constant proportion, it is decreasing by a decreasing amount (as $A$ decreases over time). In other words, $A$ is decreasing exponentially—not linearly—with time. An analogy is compound interest versus simple interest.

The correct model is $$A=125(0.7^t)\\=125\exp\left(\ln0.7^t\right)\\=125\exp\left(t\ln0.7\right).$$

P.S. While we're on the topic of exponential decay: it follows from the above formula that this drug's half-life (time taken for half of any given amount of it to decay away) is $\displaystyle\frac{\ln(2)}{\ln(\frac{10}7)}$ hours.


问:
Phase portrait for a 2D system of ODEs
I have been asked to draw the phase portrait for the following system of ODEs $A =\begin{bmatrix}-1 & -4\\1 & -1\end{bmatrix}$, $\frac{d\vec{y}}{dt} = A\vec{y}$

I am able to characterise the phase portrait but I cannot seem to find a way to draw the exact behaviour of the system.

By calculating the trace to be $-2$ and the determinant to be $5$, I know that the phase portrait is an unstable spiral, but how would I know which direction the spiral goes in, which vectors are parallel to it etc.

The only method that seems valid to me is to just plug and chug certain vector values and see what happens to the gradient.

Is there any other way of easily identifying the behaviour from the eigenvectors or eigenvalues?

答: Evaluate $x^\prime, y^\prime$ above the critical point and determine the slope of the vector field.

Title: Phase portrait for a 2D system of ODEs


问:
Prove $a ≤ x ≤ b ⇒ x ≤ a +b ∀a,b ∊ R^+$
It seems too obvious to be proved algebraically.
What i thought was:

We know that the sum of two positive numbers is a positive number so:

$a + b > b$ and $a + b > a$

Using the hypotesis of $a > x$ and $b > x$ then

$a + b > b > x$ and $a + b > a > x$

Is it correct? If you know, please tell me another way to prove it.

答:
It seems you are on the right track, but you made one error in the final line with the $a + b > a > x$. I would solve this as:

$\forall a,b \in \mathbb{R}^+$, $b<a+b$ and $a<a+b$ (definition of addition between two positive numbers)

Given $x, a, b$ such that $a \leq x \leq b$ and $b<a+b$ then we can conclude through the transitive property $a \leq x < b$ or

$$x<a+b \ \ \  \text{and} \ \ \ x \geq a$$

The only item I can't seem to prove is that $x \leq a+b$. I would say by the above proof, $x$ is strictly less than $a+b$.

Title: Prove $a ≤ x ≤ b ⇒ x ≤ a +b ∀a,b ∊ R^+$


答:
Under the interchange $t\leftrightarrow \frac{1}{t}$ we have that

$$I = \int_{0}^\infty \frac{dt}{c^2\left(t-\frac{m}{ct}\right)^2+ x^2}$$

The integral is even and converges so we can say

$$I = \frac{1}{2}\int_{-\infty}^\infty \frac{dt}{c^2\left(t-\frac{m}{ct}\right)^2 + x^2}$$

Lastly we'll use the theorem

$$p.v.\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)dx = p.v.\int_{-\infty}^\infty f(x)dx$$

for $a>0$ to get

$$I = \frac{1}{2}\int_{-\infty}^\infty \frac{dt}{c^2t^2+x^2} = \frac{\pi}{2|xc|}$$

assuming $mc>0$

Title: Evaluate $I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1}$

Body:
$$I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1},$$

I used
$$u=\frac{c}{xt}-\frac{mt}{x}$$

and

$$du=-\left(\frac{c}{xt^2}+\frac{m}{x}\right)dt$$
but I'm not able to find the limits at $t=0$ and $t=\infty$. Neither are straight forward to use as
$$\int_0^\infty\frac{1}{1+t^2}dt=\pi/2$$

Any hints on how to proceed?


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
If $n^2$ is even, then it can be factorized as

$$ n^2 = 2^{2k}C^2, $$

where $k$ and $C$ are a positive integers, and $C$ is odd. $k$ cannot be non-positive, because if it is zero or negative, n cannot be integer.

Given $n^2$, $n$ can be written as

$$ n = \mp 2^k C. $$

Which implies that $n$ must also be even.


问:
Area under the curve visually
We know the area function $A(x)$ is an antiderivative of $f(x)$.Meaning there could be numerous area functions whose derivative will be $f(x)$.Could you please show 2 area functions in graph which only differ by a constant?

答:
Consider the function $f(x)=2x$. This has an area function describing the signed area under the curve from $0$ to $x$ given by: $A_0(x)=\int_0^x2t\,\mathrm dt=x^2$. It also has another area function describing the signed area under the curve from $1$ to $x$ given by $A_1(x)=\int_1^x2t\,\mathrm dt=x^2-1$. The constant difference of $1$ corresponds to the area of the triangle with vertices $(0,0)$, $(1,0)$ and $(1,2*1)=(1,2)$. Both of these area functions are two different antiderivatives of $f(x)=2x$, and all (real-valued) antiderivatives arise in this way.

The following graph may help show what's going on here:

Title: Area under the curve visually


问:
How to express a formula that iterates over itself n number of times?
I'm not very good at math so I might use the wrong words when searching for an answer (i even don't know what tags to apply).

I have an expression in a loop and the only way I can explain it is something like this.

```<code>x = h
Loop n times
  x = k*x
end loop
Result = x
</code>```

h is the initial value of x. For each loop x = k*x
n is the number of loops
After n loops x now hold the result.

How do I express this mathematically?

答:
First of all: there are no problems in "not being good at math". If you want a math solution, them you're free to come at SE and ask.

For your question, we can start by $x_0 = h$.

Then, we define $f(x)$ (or $f\circ x$) as whatever is in the loop: $f\circ x = k\cdot x$.

As the result is the end of the iteration of the loop (with start at $x_0$), we can use the recomposition¹ notation. Letting $y$ be your result:

$$
y = \underbrace{f(f(\cdots f(x_0)\cdots ))}_{n \textrm{ times } f(\textrm{ something })}
= \underbrace{f\circ f\circ\cdots f\circ x_0}_{n \textrm{ times } f\,\circ\textrm{ something}}
= f\overset n\circ (x_0)
$$

But for your case, in specific, as $f\circ x = kx$, we have

$$
y = f\overset n\circ (x_0)
= (kx)\overset n\circ (x_0)
= (k^n x)\circ(x_0)
= k^n x_0
$$

Note:

Recomposing is composing again

Some prefer notating $f^{\circ n} x$

f \overset{n}\circ x evaluates to $f \overset{n}\circ x$


答:
We know that,

$n^2=n\times n $, We also know that

even $\times$ even = even

odd  $\times$ odd  = odd

odd  $\times$ even = even

Observation $1$: As $n^2$ is even, we also get an even result in the 2nd and 4th case .

Observation $2$: In the expression "$n \times n$" both operands are same i.e. '$n$', hence we get the result even in the $2$nd and $3$rd case

Since the $2$nd case is common in both the operations we take the first case.Hence comparing even $\times$ even = even and $n^2=n\times n $,
Hence proved if $n^2$ is even, then $n$ is even.

Title: Prove if $n^2$ is even, then $n$ is even.

Body:
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?


问:
Show that $|E(x)|\le 0.65 (x_1-x_0)^3M$ for the error of interpolation of a 2nd degree polynomial

Show that on the polynomial interpolation of a function $f \in C^3$ in
3 points $x_0 < x_1 < x_2$ with $x_2-x_1 = x_1 - x_0$, the error of
interpolation verifies $|E(x)| \le 0.65(x_1-x_0)^3M, \forall x$
between $x_0$ and $x_2$, where $M$ is the absolute value of the
maximum of the 3rd derivative of $f$ on the interval $[x_0,x_2]$.

I did

$$\begin{align}
|E(x)| &\le 0.65(h)^3M \\
|E(p_2(x))| &\le \max_{x_2,x_0}|(x-x_0)(x-x_1)(x-x_2)|M/3! = \\
\end{align}$$
$$\begin{align}
 |(x_0+h/2-x_2)(x_0+h/2-x_1)(x_0+h/2-x_2)|M/6&\le 0.65(h)^3M  = \\
|(h/2)(-h/2)(3h/2)|M/6 &\le 0.65h^3M \Leftrightarrow \\
3h^3/8* M/6 &\le 0.65h^3M
\end{align}$$

...q.e.d?

How I got to x being equal to $x_0+h/2$ where h = $x_1-x_0$: I drew some 3rd degree polynomials and "kind of" noticed they were symmetrical and that $x_0$, $x_1$ and $x_2$ were always the zeroes. I also noticed that $f(x_1)$ was always equal to zero and that it stood halfway between the other two, and between each, there was a maximum or minimum. So I chose $x = x_0+h/2$.

Is this correct?

答:
So, you wrote the error formula
$$
|E_2(x)| \leq |(x-x_0)(x-x_2)(x-x_2)| \frac{M}{3!}.
$$

Now you can simply maximize the 3rd degree polynomial... The maximum will be attained at a stationary point, that you can compute explicitly. The stationary points are
$x_0 + (1 \pm \frac{\sqrt{3}}{3})h$, and the corresponding maximum value will be $\frac{2h^3}{3 \sqrt{3}}$. In general, your choice of $x = x_0 +\frac h2$ is wrong and does not yield the maximum.

Using this information,

$$
|E_2(x)|\leq \frac{M h^3}{9 \sqrt{3}}\approx 0.06415 (x_1-x_0)^3 M.
$$

Finally, if this estimate holds, so does the (much worse) proposed estimate.

Regarding the maximization of $|(x-x_0)(x-x_1)(x-x_2)|$ in the interval $[x_0, x_2]$, as noted by Lutz Lehmann, it is equivalent to the maximization of $g(s)=|(s^3-s)h^3|$ for $s \in [-1,1]$. Since $g$ is zero on the boundary and at the branching point for the absolute value, the maximum will occur at a point where $g'(s)=0$, i.e. $s= \pm \frac{\sqrt{3}}{3}$. Substituting back in the function, we conclude that the maximum is $g(\pm \frac{\sqrt{3}}{3}) = \frac{2h^3}{3\sqrt{3}}$.


答:
Initially, $x=h$. After one iteration, $x'=kx=kh$. After two iterations, $x''=kx'=kkh=k^2h$. More generally, after $n$ iterations,

$$x^{(n)}=k^nh.$$

In a more mathematical language, you have the recurrent sequence defined by

$$x_0=h,\\x_n=kx_{n-1}.$$

Title: How to express a formula that iterates over itself n number of times?

Body:
I'm not very good at math so I might use the wrong words when searching for an answer (i even don't know what tags to apply).

I have an expression in a loop and the only way I can explain it is something like this.

```<code>x = h
Loop n times
  x = k*x
end loop
Result = x
</code>```

h is the initial value of x. For each loop x = k*x
n is the number of loops
After n loops x now hold the result.

How do I express this mathematically?


问:
Making $\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$ stationary using Euler's equation
9.2.6 - Mary L Boas:

This problem asks us to make the following integral stationary using Euler's equation.

$$\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$$

If we have $$F=y'^2\,+\,\sqrt{y}$$ then $$\frac{d}{dx}\,\frac {\partial F}{\partial y'}-\frac {\partial F}{\partial y}=0\, \implies \frac{d}{dx}2y'\,-\,\frac{1}{2}y^{-1/2}=0 \, \implies 2y''\,-\,\frac{1}{2}y^{-1/2}=0$$also multiplying by $y'$ and integrating each term we get $$\require{cancel}y'^2\,-\sqrt{y}\,=\,k\implies \int \frac{dy}{(k+\sqrt {y})^\cancelto{1/2}{2}}=x+\mathrm{constant} $$ Here i stumble, because my answer comes nowhere close to the form of given solution: $x+a=\frac{4}{3}(y^{1/2}-2b)(b+y^{1/2})^{1/2}$. Please help.

Edited according to user10354138, now it's ok.

Body:
We have $$y'^{2}=k+\sqrt{y}\implies\int\frac{1}{({k+\sqrt{y}})^{\frac{1}{2}}}dy=x+\text{constant}$$

Then substitute $u={k+\sqrt{y}}$ on the LHS so $du=\frac{1}{2\sqrt{y}}dy=\frac{1}{2(u-k)}dy$ and we have

$$x+\text{constant}=\int \frac{2(u-k)}{u^{\frac{1}{2}}}du=2\int u^{\frac{1}{2}}du-2k\int u^{-\frac{1}{2}}du$$
$$=\frac{4}{3}u^{\frac{3}{2}}-4ku^{\frac{1}{2}}=\frac{4}{3}u^{\frac{1}{2}}(u-3k)=\frac{4}{3}(k+y^{\frac{1}{2}})^{\frac{1}{2}}(y^{\frac{1}{2}}-2k)$$

Or $$x+a=\frac{4}{3}(y^{\frac{1}{2}}-2b)(b+y^{\frac{1}{2}})^{\frac{1}{2}}$$

since $a$ and $b$ are constants.


问:
Antiderivatives such as $\int \frac{dx}{\sqrt{e^{2x} +c}}$
Is there a good strategy for calculating antiderivatives such as$$\int \frac{1}{\sqrt{e^{2x} +c}}dx\enspace?$$

Right now I'm substituting the entire radical expression, which sort of simplifies the calculation. But nevertheless, I need to make another substitution after, since I have the integral $$\int \frac{1}{u^2-c}du$$

I can solve it, but are there other ( perhaps shorter ) methods?

答:
Firstly, the integral
$$\int\frac{1}{u^2-c^2}du=-\frac{1}{c}\operatorname{artanh}\frac{u}{c}+C$$
is quite standard, so really if you'd have known it your method would be pretty efficient! :)

Secondly, you could first make the substitution $u=e^x/{\sqrt{c}}$, so that $e^x=u\sqrt{c}$ and $du/dx=e^x/\sqrt{c}=u\implies dx=du/u$. This means that if $I$ is the integral then
$$\begin{align}
I&=\int \frac{1}{\sqrt{e^{2x} +c}}dx\\
&=\int\frac{1}{\sqrt{u^2c+c}}\times\frac{1}{u}du\\
&=\frac{1}{\sqrt c}\int\frac{1}{u\sqrt{u^2+1}}du
\end{align}$$
Here we have a not-very-standard integral:
$$\int\frac{1}{u\sqrt{u^2+1}}du=-\operatorname{arcosech}u+C=-\operatorname{arsinh}\frac{1}{u}+C$$Can you finish it from there?

I hope you found that helpful. If you have any questions please don't hesitate to ask. :)


答:
Set $y=u^2$ so that $\sqrt{y}=u$, then
$$
4u^2u'^2-u=k,\\~\\
x+c=\int\frac{2udu}{\sqrt{u+k}}
=2\int\sqrt{u+k}\,du-2k\int\frac{du}{\sqrt{k+u}}
\\
=\frac43(u+k)^{\frac32}-4k\sqrt{u+k}
$$

Title: Making $\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$ stationary using Euler's equation

Body:
9.2.6 - Mary L Boas:

This problem asks us to make the following integral stationary using Euler's equation.

$$\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$$

If we have $$F=y'^2\,+\,\sqrt{y}$$ then $$\frac{d}{dx}\,\frac {\partial F}{\partial y'}-\frac {\partial F}{\partial y}=0\, \implies \frac{d}{dx}2y'\,-\,\frac{1}{2}y^{-1/2}=0 \, \implies 2y''\,-\,\frac{1}{2}y^{-1/2}=0$$also multiplying by $y'$ and integrating each term we get $$\require{cancel}y'^2\,-\sqrt{y}\,=\,k\implies \int \frac{dy}{(k+\sqrt {y})^\cancelto{1/2}{2}}=x+\mathrm{constant} $$ Here i stumble, because my answer comes nowhere close to the form of given solution: $x+a=\frac{4}{3}(y^{1/2}-2b)(b+y^{1/2})^{1/2}$. Please help.

Edited according to user10354138, now it's ok.