问:
prove an inequality containing digamma function
I wonder how to prove this or if it is correct since that from numerical result, it is correct
$$
(2 \log 2-1) x^{2}+(2 \log 2+2) x+2+\left(2 x^{2}+2 x\right) \Psi(x)>0
$$
where $x\in(0,+\infty)$

答:
We split into two cases:

$x > 1$:

By using $\Psi(x) \ge \ln x - \frac{1}{x}, \ \forall x > 1$, it is easy to prove the desired inequality.

$0 < x \le 1$:

Let
$$f(x) = \frac{(2 \log 2-1) x^{2}+(2 \log 2+2) x+2}{ 2 x^{2}+2 x} + \Psi(x).$$
We have
\begin{align}
f'(x) &= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \Psi'(x) \\
&= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \int_0^\infty \frac{t\mathrm{e}^{-xt}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\
&= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \int_0^\infty t\mathrm{e}^{-xt} \mathrm{d} t
+ \int_0^\infty \frac{t\mathrm{e}^{-xt}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\
&= - \frac{1}{2(x+1)^2} + \int_0^\infty \frac{t\mathrm{e}^{-xt}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\
&> - \frac{1}{2} + \int_0^\infty \frac{t\mathrm{e}^{-t}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\
&= - \frac{1}{2} - 1 + \frac{1}{6}\pi^2 \\
&> 0
\end{align}
where we have used $\Psi'(x) = \int_0^\infty \frac{t\mathrm{e}^{-xt}}{1 - \mathrm{e}^{-t}} \mathrm{d} t$.

Also, $\lim_{x\to 0^{+}} f(x) > 0$. Thus, $f(x) > 0$ for all $x \in (0, 1]$.

We are done.


问:
Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$
I hit a wall on this question. Below are my steps

$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$

Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$

$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$

Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$

Solve for $A$ and $B$:

$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$

Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$

Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.

My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$

Feedback says my answer is wrong. Where did I mess up?

ContentLicense: CC BY-SA 4.0

Body:
You are right up to

$$\frac{7}{2a^{2}}(\int \frac{du}{2u+a^{2}}-\int \frac{du}{2u-a^{2}})
= \frac7{4a^2}\ln\frac{2u+a^2}{|2u-a^2|}= \frac7{4a^2}\ln\frac{x^2+a^2}{|x^2-a^2|}
$$


问:
are non-degenerate critical points always isolated?
I have a question regarding the isolation of critical points of a function:

Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is a $C^\infty$ function such that $f$ has a non - degenerate critical point at $0 \in \mathbb{R}^n$. That is, we have $\triangledown f (0) = 0$, and the Hessian $\left((\partial^2 f/ \partial x_i \partial x_j ) (0) \right)$ is invertible.

Can I deduce from this that the critical point at $0$ is an isolated critical point ? My guess is to say yes, because the fact that the Hessian is non-degenerate forces $f$ to change its value in the vicinity, and in all directions. But I am unsure, in particular with regards to the last statements ("in all directions") - though that should be encoded by the fact all eigenvalues of the Hessian are non - zero.

Is this the right way to think about the question of whether the critical point is isolated ?
Thanks for your thoughts !

答:
In Neal answer ,there is a flaw in the proof. Hessian(x0) being non-degenerate does not imply f is P.D or N.D hence x(0) not necessarily a local extremum. e.g saddle point.
Here is my version: for x in N(0), the Hessian(x) is non-degenerate since eigenvalues of matrix continously depends on entry hence x.
in N(0) any x,y we have
grad(x)-grad(y)=H(x-y)  +o(|x-y|^2)
H is non-degenrate ,then the map from x to grad(x) is one-to one linear map in N(0).therefore grad(x) implies x=0</p>

ContentLicense: CC BY-SA 3.0


问:
Finding Variance from given table using formula
Im having trouble calculating the variance for $X_m$ and $X_k$ for the following table:

"We now consider a random household and let the stochastic variables $X_k$ og $X_m$ be defined in the following way: $X_k$: The woman's working hours measured in hours per. week $X_m$: The man's working hours measured in hours per. week"

I calculated the marginal distribution for Xm:
$$ P(X_m = 0) = 0.08$$
$$P(X_m = 20) = 0.12$$
$$P(X_m = 40) = 0.8$$
I've already calculated the mean for $X_m$, which i did by using the marginal distribution values $$\text{mean} = 0.08 \cdot 0 +0.12 \cdot 20 +0.8 \cdot 40 =34.4$$

Now i am to calculate the Variance, however it seems as though there are multiple variance formulas available however im not sure which to use.

答:
There are two possible formulas for the variance.
I will illustrate both.

Assuming $\mu = E(X_m) - 34.4,$ as you say,
$V(X_m) = (.08)(0-34.4)^2 + (.12)(20-34.4)^2 +(.80)(40-34.4)^2 = 144.64.$

```<code>p = c(.08, .12, .8);  mu = 34.4
x = c(0,20,40)
sum(p*(x-mu)^2)
[1] 144.64
</code>```

An alternate formula for the variance:
$E(X_m^2) = (.08)0^2+(.12)20^2+(.8)40^2 = 1328$
$V(X_m) = E(X_m^2) - \mu^2 = 144.64.$

```<code>sum(p*x^2)
[1] 1328
1328 - 34.4^2
[1] 144.64
</code>```

Simulation: As a check. With 10 million iterations, we
can expect about four significant digits of accuracy. Thw mean of a very large sample should be approximately the same as the population mean. The variance of a very large sample should be approximately the same as the population variance. Agreement is satisfactory.

```<code>set.seed(226)
p = c(.08, .12, .8)
x.m = sample(c(0,20,40), 10^7, rep=T, p=p)
mean(x.m);  var(x.m)
[1] 34.39397     # aprx E(X.m) = 34.4
[1] 144.7988     # apro V(X.m) = 144.64
</code>```

Body:
There are two possible formulas for the variance.
I will illustrate both.

Assuming $\mu = E(X_m) - 34.4,$ as you say,
$V(X_m) = (.08)(0-34.4)^2 + (.12)(20-34.4)^2 +(.80)(40-34.4)^2 = 144.64.$

p = c(.08, .12, .8);  mu = 34.4
x = c(0,20,40)
sum(p*(x-mu)^2)
[1] 144.64

An alternate formula for the variance:
$E(X_m^2) = (.08)0^2+(.12)20^2+(.8)40^2 = 1328$
$V(X_m) = E(X_m^2) - \mu^2 = 144.64.$

sum(p*x^2)
[1] 1328
1328 - 34.4^2
[1] 144.64

Simulation: As a check. With 10 million iterations, we
can expect about four significant digits of accuracy. Thw mean of a very large sample should be approximately the same as the population mean. The variance of a very large sample should be approximately the same as the population variance. Agreement is satisfactory.

set.seed(226)
p = c(.08, .12, .8)
x.m = sample(c(0,20,40), 10^7, rep=T, p=p)
mean(x.m);  var(x.m)
[1] 34.39397     # aprx E(X.m) = 34.4
[1] 144.7988     # apro V(X.m) = 144.64


问:
Can the minimal polynomial have multiple roots?
Let $F$ be an extension field of $K$. I'm trying to think of an example of an element $\alpha$ which is algebraic over $K$ and whose minimal polynomial over $K$ has a root of multiplicity $> 1$ in some splitting field. Is this possible? Nothing comes to mind immediately, but I might have to looked hard enough.

答:
I will explain a little bit the comment of @Lubin.

In your notations, let's say the minimal polynomial of an element $\alpha \in F$ is a polynomial $f \in K[x]$. Then $f$ is an irreducible polynomial over $K$: if $f = gh$ with $g, h \in K[x]$ of smaller degree than $f$, then we would have $g(\alpha)h(\alpha) = 0$, which implies that at least one of $g(\alpha)$ and $h(\alpha)$ is zero, contradicting the minimality of $f$.

On the other hand, if $f$ has multiple roots, then the $\gcd$ of $f$ and its derivative $f'$ is not equal to $1$ (as any multiple root is a root of both). Since $f$ is irreducible, we must have $\gcd(f, f') = f$, i.e. $f$ divides $f'$.

However, the degree of $f'$ is strictly smaller than that of $f$, thus the only possibility is $f' = 0$.

If $K$ has characteristic zero, then $f' = 0$ implies that $f$ is a constant, hence cannot be a minimal polynomial. This is the reason that you never find an example in characteristic zero.

In positive characteristic, we may have $f' = 0$ while $f$ is non-constant. The example provided by @Lubin in the comment is a typical one.


问:
Passivity of a dynamical system
Let a dynamical system , in particular a RLC circuit. I want to study the passivity of this.

In general the notion of $\textbf{passivity}$ comes out form the statics physics for which by a mathematical point of view it can be said that a static system $y=h(t,u)$ is said to be passive if $u^Ty\geq 0$ (for instance let a resistor and let $u=V$, $y=I$, it is passive since $VI\geq 0$).

Now it is possible to extend the notion of passivity to the dynamical system in the sense that follows with an example:

let a RLC circuit for which the energy absorbed by
the network over any period of time $[0, t]$ such that is greater than or equal to the change in
the energy stored in the network over the same period; that is,
$$\int_{0}^{t}u(s)y(s)ds\geq V(x(t)-V(x(0))$$ where $V(x)=\frac{1}{2}Lx_1^2+\frac{1}{2}Cx_2^2=\text{Total energy stored in the network}$, with $x_1=I_L$ and $x_2=V_C$.

My book indicates that the inequality above it is a matter of passitivity in the sense that this is due by the fact that the total energy stored by the network is greater than the variation of this in the period $[0,t]$.

$\textbf{My problems:}$

$1)$ It is true to think about the term $\int_{0}^{t}u(s)y(s)ds$ (that is the integral of the power flows the network over the period $[0,t]$) $\textit{equal to the energy stored in this period}$?

$2)$I can't understand the fact that the total energy stored by the network is greater than the variation of this in the period $[0,t]$ indicates dissipation of energy.

Can you help me please?

答:
I'll answer the questions first and proceed to my explanations after.

The integral term tells you how much work you perform on the system. In principle, if there are no losses of any sort, then this is precisely equal to the change in the system's energy. Of course, energy is relative. You can always view $E(0)$ has being no energy stored and then you could view that integral as the amount of energy the system stores over that period.

Briefly, the integral tells you exactly how much energy you've put into the system. If the constraint $$V(t) - V(0) \leq \int_{0}^{t} u(s) y(s)\,ds$$ is satisfied, it is telling you that the system is not going to magically increase in energy greater than the amount you are applying. That is, you can strictly bound the growth of the system's energy using your input and, in particular, force it to decrease if you like!

It helps to consider a physical example. Consider the position of a particle $x(t) \in \mathbb{R}^3$ with mass $M$ and velocity $v(t) \in \mathbb{R}^3.$ We will take the output of the system to be $y(t):= v(t).$ The energy of the cart (there is no potential energy here) is $$E(t) = \frac{1}{2} M v(t)\cdot v(t) = \frac{1}{2} M y(t)\cdot y(t).$$ Suppose we could apply an input force $u(t) \in \mathbb{R}^3.$

In this case, with no potential energy, we know that the work is the change in kinetic energy. The work induced by the applied force $u(t)$ is the integral of the applied force inner producted with the velocity. That is, $$W(t) = \int_{0}^{t} u(t) \cdot v(t)\,dt = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$ As discussed, we have that the change in kinetic energy is the work (because of a lack of potential energy) so we get the relation $$E(t) - E(0) = W(t) = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$

The kinetic energy is energy the object we wish to control has (the particle) that is "stored" in the sense that the system has energy it can lose. What happens if there is friction or some other force that "steals" energy from the system? We must have $$E(t) - E(0) \leq W(t) = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$

Why? Suppose we did positive work $W(t) > 0$ over this period. This would be equivalent to adding energy into the system, such as adding force in the direction of motion making the particle move faster. If it was the case that $E(t) - E(0) < W(t)$ then somehow, although we added some $W(t)$ amount of energy, we don't see it in $E(t),$ implying some was stolen through losses. You can see a similar argument in the case $W(t) < 0.$

Of course, if you do satisfy the equality (as in the above example), you of course don't have any internal losses but it still is a useful property for a system to have. It implies that you can essentially determine how much the energy the system gains or loses by your input $u(t).$

Back to your example, the question you must ask yourself --- if you seek the intuition --- is what is the work in a circuit? We have that power $P(t)$ is the derivative of the work, $$P(t) = \frac{dW}{dt}(t).$$ We know the instantaneous power of an ideal linear circuit is $$P(t) = I(t) V(t)$$ where $I(t)$ and $V(t)$ are the instantaneous current and voltage respectively measured from the single input to the single output. Solve for $W(t),$ $$W(t) = \int_{0}^{t} I(t) V(t)\,dt = \int_{0}^{t} u(t) y(t)\,dt$$ and we have that the change in energy, the work, is the integral in question!

Hope this helps.

Title: Passivity of a dynamical system


问:
Find an angle in an isosceles triangle.
Question

Given $\triangle ABB'$, where $\angle BAB'=108^{\circ}$.
$\overline{BC}$ is the bisector of $\angle ABB'$, and $\overline{AB}=\overline{AB'}=\overline{B'C}$. Find $\angle{AB'C}$.

I prefer solutions without trigonometric functions, but answers with them are also fine.

I have tried constructing a point $D$ on $\overline{AB}$ such that $\overline{AB}=\overline{BD}$. Therefore, both $\triangle{AB'D}$ and $\triangle{ACD}$ are isosceles. However, the problem remains unsolved after such construction. I can't see the correct auxiliary lines to make.

Any suggestions, hints, or even full solutions (you don't have to) are appreciated.
Thanks for reading my post.

答:
A circle can be imagined with center at B' and radius AB'... etc. But sincere apologies,  as...

after constructing a perpendicular to one side of regular pentagon I thought we obtain point C and  $x^0 =24^0$ by direct angle  chase alone.. although one consideration is still missing.

Title: Find an angle in an isosceles triangle.


问:
Find the matrix of the orthogonal projection in R2 onto the line x=−2y.
Find the matrix of the orthogonal projection in $\mathbb{R}^2$ onto the line $x=−2y$.

If I had to decompose this matrix into a rotation projection rotation matrix, do I first start with the negative rotation of the angle made by the line $y=-x/2$ (call it -theta), then project down the $x$ axis, and then rotate by $\theta$ again? So the order would be T(theta)T(projection on $x$ axis)$T(-\theta)$

答:
It's not exactly clear what mean by "rotating negatively", or even which angle you're measuring as $\theta$. Let's see if I can make this clear.

Note that the $x$-axis and the line $y = -x/2$ intersect at the origin, and form an acute angle in the fourth quadrant. Let's call this angle $\theta \in [0, \pi]$.

You start the process by rotating the picture counter-clockwise by $\theta$. This will rotate the line $y = -x/2$ onto the $x$ axis. If you were projecting a point $p$ onto this line, you have now rotated it to a point $R_\theta p$, where
$$R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix}.$$

Next, you project this point $R_\theta p$ onto the $x$-axis. The projection matrix is
$$P_x = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix},$$
giving us the point $P_x R_\theta p$.

Finally, you rotate the picture clockwise by $\theta$. This is the inverse process to rotating counter-clockwise, and the corresponding matrix is $R_\theta^{-1} = R_\theta^\top = R_{-\theta}$. So, all in all, we get
$$R_{-\theta} P_x R_\theta p = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix}p.$$


问:
Prove that $\text{Hom}(\Bbb Z/m \Bbb Z, \Bbb Z/n\Bbb Z) \cong \Bbb Z/(m, n)\Bbb Z$

$\text{Hom}(\Bbb Z/m \Bbb Z, \Bbb Z/n\Bbb Z) \cong \Bbb Z/(m, n)\Bbb Z$

$\text{Hom}(\Bbb Z/m \Bbb Z, \Bbb Z/n\Bbb Z)$ is the group consisting of all homomorphisms from $\Bbb Z/m\Bbb Z$ to $\Bbb Z/n\Bbb Z$. I want to show that this is isomorphic to $\Bbb Z/(m,n)\Bbb Z$, where $(m,n)$ is notation for $\gcd(m,n)$.

How should I go about constructing an isomorphism $\phi: \text{Hom}(\Bbb Z/m \Bbb Z, \Bbb Z/n\Bbb Z) \to \Bbb Z/(m, n)\Bbb Z$?

I started with thinking about the objects in $\text{Hom}(\Bbb Z/m \Bbb Z, \Bbb Z/n\Bbb Z)$. Suppose $f: \Bbb Z/m\Bbb Z \to \Bbb Z/n\Bbb Z$ is a homomorphism. It is completely determined by $f(1)$ alone, which has $n$ options for itself in $\Bbb Z/n\Bbb Z$. This is because: if $f(1) = a \in \Bbb Z/n\Bbb Z$, we have $f(x) = f(1) + f(1) + ... + f(1)$ ($x$ times) $= ax$.

What do I do next? Perhaps I am overcounting in some sense, or maybe some of these choices of $f(1)$ do not result in homomorphisms, because $\Bbb Z/(m,n)\Bbb Z$ has only $(m,n)$ elements, which in general is less than $n$. What am I missing? How do I work through this problem?

Thanks a lot for your help!

P.S. I'm familiar with elementary group theory, i.e. topics including group actions, Cauchy and Sylow's theorems, etc. - nothing too fancy - so simpler answers would be nicer!

答: You're on the right track, but when you try to define $ f: \mathbb{Z} / m\mathbb{Z} \rightarrow \mathbb{Z} / n\mathbb{Z} $ by $ f(1) = a $, then you need to check that this is well-defined. For this, you need $ f(m) = ma $ to be equal to $0 \in \mathbb{Z} / n\mathbb{Z}$. (Convince yourself why this is necessary and also why this is sufficient for $f$ to be well-defined.) This will constrain what values of $a$ actually produce a well-defined $f$.


问:
How many epsilon numbers $<\omega_1$ are there?
An epsilon number is an ordinal $\epsilon$ such that $\epsilon=\omega^\epsilon.$ What is the cardinality of the set of all epsilon numbers less than $\omega_1$?

I'm asking this because of a proof I've just read that seems to presuppose that there are countably many such ordinals, and it seems to me intuitively that there should be uncountably many (although I don't know how to prove it).

Added. OK, I've just understood that the proof I mentioned is OK even if there are uncountably many such ordinals, but I still don't see how I can find their number.

答:
Another answer that uses choice:

Since $\varepsilon_\alpha$ is countable if $\alpha$ is countable by this other SE question or by the arguments in the other answers (and obviously $\varepsilon_\alpha\ge\alpha$), the map $f:\beta\to\omega_1$ defined by $f(\alpha)=\varepsilon_\alpha$ (where $\beta$ is the number of countable $\varepsilon$ numbers) is a cofinal map. But by the axiom of choice, $\omega_1$ is regular. Thus there are $\beta=\omega_1$ countable $\varepsilon$ numbers.


问:
Number of conjugacy classes of a group of order $5^4$ whose center is $25$ order
How many conjugacy classes does a group of order $625$ have, if its center is of order $25$?

I know that the orders of conjugacy classes should divide the order of the group, so this leaves with each centralizer class order being either $5$ or $25$. But, now how do we proceed further? Will all conjugacy classes have a uniform order? How can we use the Burnside lemma here? Any hints? Thanks beforehand.

答:
If $x \notin Z(G)$, then $\vert C_G(x) \vert =125$.

That's because $x \notin Z(G) \Rightarrow C_G(x) \neq G$, but $(Z(G) \subseteq C_G(x) \land x \in C_G(x) \setminus Z(G)) \Rightarrow \vert C_G(x) \vert \gt 25$.

The size of the conjugacy class containing $x$ is therefore $5$.  There are $600$ non-central elements, so $G$ has $145$ conjugacy classes.

Note that this proof works for any group of size $p^4$ with center of order $p^2$, where $p$ is prime.


问:
Python coding with numpy sympy
New to python, and I am working on a problem with the ultimate goal of being able to evaluate

$$\frac{\sqrt{x}\bigg[\prod_{k=0}^{n-1}(\sqrt{x}+2^k)- \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)\bigg]}{\prod_{k=0}^{n-1}(\sqrt{x}+2^k)+ \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)}$$

at whichever value of $x, n$ I choose.

First, I probably need to expand the function: how should I go about doing so? Perhaps sympy?
Then, how would I evaluate the sympy function?

答:
```<code>import numpy as np
def prod(n,a):
    p=1
    for k in range(n):
       p*=a+2**k
    return p
def f(n,x):
    return np.sqrt(x)*(prod(np.sqrt(x))+prod(-np.sqrt(x)))/(prod(np.sqrt(x))-prod(-np.sqrt(x)))
</code>```

The first function prod computes $\Pi_{k=0}^{n-1}(a+2^k)$ for given $n$ and $a$. The function f is the one you specified, which uses prod 4 times, each time with $\pm\sqrt{x}$ as $a$ in the argument.

Title: Python coding with numpy sympy

Body:
import numpy as np
def prod(n,a):
    p=1
    for k in range(n):
       p*=a+2**k
    return p
def f(n,x):
    return np.sqrt(x)*(prod(np.sqrt(x))+prod(-np.sqrt(x)))/(prod(np.sqrt(x))-prod(-np.sqrt(x)))

The first function prod computes $\Pi_{k=0}^{n-1}(a+2^k)$ for given $n$ and $a$. The function f is the one you specified, which uses prod 4 times, each time with $\pm\sqrt{x}$ as $a$ in the argument.


答: The problem is a standard application of the fact that the center of the group is a factor of the centralizer of every element of the group, as the center is the normal subgroup of the centralizers. Thus, the only possible order for centralizer of each element is $125$, as otherwise the group should have been abelian. Hence, the indices of centralizers are all $5$, whence the class equation is $1+1\ldots (25\ \text{times})+1+5+5+\ldots (120\ \text{times})+5$, thus giving us a total of $145$ conjugacy classes.

Title: Number of conjugacy classes of a group of order $5^4$ whose center is $25$ order

Body:
How many conjugacy classes does a group of order $625$ have, if its center is of order $25$?

I know that the orders of conjugacy classes should divide the order of the group, so this leaves with each centralizer class order being either $5$ or $25$. But, now how do we proceed further? Will all conjugacy classes have a uniform order? How can we use the Burnside lemma here? Any hints? Thanks beforehand.


问:
Prove or disprove: If $A$ and $A ∩ B$ are regular, then $B$ is regular.
I suspect the Myhill–Nerode theorem will come into play, but not certain. If this was a union as opposed to an intersect, I'd be pretty confident that this would be true, but with it being an intersect, I'm less sure (albeit, I may be wrong about that too). I know that the empty language is a regular subset of all non-regular languages (or any language), so I was thinking that maybe if $A$ and $B$ shared only the empty language, then $B$ could be non-regular and this statement would be false.

答:
Disprove: let $A = (ab)^*$, and $B = \{a^n b^n : n \in \Bbb{N}\}$. Recall that $B$ is famously not regular. Then $A \cap B = \{\varepsilon, ab\}$, a finite language, which is hence regular.

EDIT: Come to think of it, your thoughts lead to a solution that is a bit more elegant. If we just take $A = \emptyset$ and $B$ to be any non-regular language, then $A \cap B = \emptyset$ is regular.


问:
Find if $W=\lbrace z\in \mathbb{C}\mid 2 \leq |z| \leq 3 \rbrace$ set is connected and compact

Let $W=\lbrace z\in \mathbb{C}\mid 2 \leq |z| \leq 3 \rbrace$. Prove that $W$ is connected and compact.

Let $z=x+iy,\: x,y\in\mathbb{R}$. Notice that $$W=\lbrace z\in \mathbb{C} \mid 4 \leq x^2+y^2 \leq 9\rbrace.$$
Now consider $$U=\lbrace z\in \mathbb{C}\mid x^2+y^2 \leq 9\rbrace ,\, \, V=\lbrace  z\in \mathbb{C}\mid
 x^2+y^2 \geq 4 \rbrace. $$

Note that $U$ is closed and $\mathbb{R}^2\setminus V$ is open, hence $V$  too closed and
$W=U \cap V$ hence $W$ is closed since is finite intersection of closed sets.

Now is pretty easy see that for any $z\in W$ we have the following inequality holds $|z| \leq 9$, it show that $W$ is bonded and hence $W$ is compact.

Now for see that the set is path connected I think that I should prove that $W$ is path connected. I dont´t have a formal proof the unique idea that comes to my mind is:

Let $w_1,w_2\in W$ there are two cases.

Case I

If $|w_1|=|w_2|$ then we only need consider the circunference  with center in $(0,0)$ and radious $|w_1|$ let us denote by $C(0,|w_1|)$ is easy see that $C(0,|w_1|)\subset W$ hence the arc of circunference that connect $w_{1}$ with $w_2$ is our path.

Case II

If $|w_1|\neq |w_2|$.Then consider the straight line that join $w_1$ with $w_2$ let us denote by $\ell$
Then consider $m=Min\lbrace|w_1|,|w_2| \rbrace $. WLOG assume that $m=|z_1|$  and consider $C(0,m)$ notice that $\ell \cap C(0,m)\neq \emptyset $ since it contains at most the complex number with norm $m$.Frome here since $\ell$ is a straight line must intersect $C(0,m)$ in a point $P\neq w_1$.

WLOG let us take the arc of circunference from $w_1$ to $P$ and the path from $P$ to $w_2$ which is contained in $W$(dangerous argue),hence the union of both path is the path requiered.

Is my answer right? Is possible find  a more formal proof of the set $W$ is connected? In my proof I have a due in a argue: Is always true that the path from $w_1$ to $P$ and the path from $P$ to $w_2$ is always contained in $W$?

答:
Your argument is correct for Case I but not for Case II. Your point $P \neq w_1$ many not exist.

Consider $C(0,|w_1|)$ and show that there  is a point  $P$ on this which is a positive multiple of $w_2$. To be precise, the point is $w_1e^{it}$ where $t$ is the argument of $\frac {w_2} {w_1}$.  Now go along the circle $C(0,|w_1|)$ from $w_1$ to $P$ and then go to $w_2$ using the line segment from $P$ to $w_2$.


问:
Pigeonhole principle application sums and differences
Let $A \subset \{1,2,...,99\}$, prove or disprove the following:

a. For $|A| = 27$

b. For $|A| = 26$

There are $2$ different numbers in $A$ that their sum or their difference can be divided with $50$.

I am sure the question points me to use Pigeonhole Principle, but I'm not quite sure which are the pigeons and which are the holes because the question deals with or so proving that a sum (or a difference) can be divided with $50$ closes the proof without handling the other case.

I've started with saying that there are $(99+98)-(1+2)+1=195$ different sums,
and $(99-1)-(1-99)+1=197$ differences (including negatives) in $\{1,2,...,99\}$.

Also, when dividing with $50$ you can get maximum 50 remainders (let $r \in \{0,1,...,49\}$ mark the reminders)

Can anyone hint me with how to proceed from here?

答:
For part b) consider the set $A = \{n \mid 25 \leq n \leq 50\}$. It has $26$ elements and no sum or difference of two different elements is divisible by $50$.

For part a) consider the remainders of $\pm a$ ($\bmod$ $50$) for all $a \in A$,  so $54$ remainders in total.  Therefore there must be at least four pairs $$(a_1,b_1), (a_2,b_2), (a_3,b_3), (a_4,b_4)$$ with $a_k \in A$ all distinct and either $b_k \in A$ or $-b_k \in A$ such that $a_k-b_k$ is divisible by $50$ for $k \in \{1,\dotsc, 4\}$.  If $b_k \in A$ then $b_k \neq a_k$ since each number in $A$ is used only once.  The only possible pairs that do not lead to a solution are $$(25,-25), (50,-50), (75,-75)$$ since for these pairs the same number from $A$ is used twice. So there must be at least one pair left that leads to a proper solution.


答:
This answer uses a somewhat miraculous construction, so it may be less direct than the "direct angle chase" suggested by the other answer.

Construct $D$ such that $D$ lies on $AB$ and $AB' = B'D$. Then $\angle ADB' = \angle B'AD = 72^\circ$.

As $\angle ABB' = 36^\circ$, $\triangle BDB'$ is isosceles as well. Hence $BE$ is the perpendicular bisector of $B'D$.

$C$ lies on $BE$, hence $B'C = CD$. As we are given $DB' = AB' = CB'$, $\triangle B'CD$ is equilateral, and thus $\angle AB'C = 24^\circ$ after some calculations.

Title: Find an angle in an isosceles triangle.

Body:
Question

Given $\triangle ABB'$, where $\angle BAB'=108^{\circ}$.
$\overline{BC}$ is the bisector of $\angle ABB'$, and $\overline{AB}=\overline{AB'}=\overline{B'C}$. Find $\angle{AB'C}$.

I prefer solutions without trigonometric functions, but answers with them are also fine.

I have tried constructing a point $D$ on $\overline{AB}$ such that $\overline{AB}=\overline{BD}$. Therefore, both $\triangle{AB'D}$ and $\triangle{ACD}$ are isosceles. However, the problem remains unsolved after such construction. I can't see the correct auxiliary lines to make.

Any suggestions, hints, or even full solutions (you don't have to) are appreciated.
Thanks for reading my post.


问:
How to compare means from a CLT exercise and a Normal distribution if i have $n=4896$ random samples
I have the central limit theorem exercise solved but the normal distribution not. i want to compare the means and now I am stucked.

the exercise is a discrete random variable modelling in a table with his $E(x)$, population mean and standart deviation. after that I have to run with $4896$ samples. all good the sum and average.

but i dont know how to compare this mean with the normal distribution. anyone knows?

答: If you take the sample mean of each of your $4896$ samples, they should form a sampling distribution that is normally distributed with mean $\mu$ with $\mu$ being the population mean and standard deviation $\sigma/\sqrt n$ with $\sigma$ being the population standard deviation. That is, each of your samples has lots of observations in it. Once you take the average of the observations in each sample, and plot the resulting 4896 sample means, they should form an approximately normal distribution with aforementioned mean and standard deviation.


问:
What is the deal with the three isomorphism theorems?
I've been having lectures in group theory with Hungerford's book. We were presented with the following theorem:

And then with:

Previous to the lectures about them, I was understanding most of the stuff, that is: I kinda could figure out the motivation for things. But when I came into these theorems, things got pretty weird for me. I don't understand the motivation for them. I have the following guesses:

Theorem 5.6 is used to prove Corollary 5.7 but I think it has a utility on its own: If given an homomorphism $f: G \to H$ with $N\lhd G$, we have a unique homomorphism $\overline{f} : G/N\to H$, then we could use this to know what homomorphisms could exist betwen $G$ and $H$ and this depends only on the normal subgroups of $G$. Does that make sense?

For the isomorphism theorems, I noticed they are corollaries so they must be very important. One of the reasons I conjectured for them is that we can construct very interesting isomorphisms. But why are those isomorphisms interesting? It's not clear to me what interesting stuff one could do with them.

I know my guesses are maybe too obvious or wrong, but I can't figure out on my own the answer to that question. Can you help me?

答:
Not sure, how much below information can help you. But may help others and I too can get more intuitive insight of  this very important theorem, by some experts.

The way I understand the group is a kind of Symmetries of an object. On which one can perform some actions.
Now when it comes to comparing two groups or symmetries, what mathematical approach we can take, to  find it if they are 'same' or 'similar' to some extent.
we understand these by a function which maps elements(vertices,edges etc..) of symmetries to other objects symmetries, and then apply action on elements and compare
how do they appear on other symmetries.

As you know isomorphism and homomorphism are the tool to compare two symmetries.  Most of the time we would like to know when two symmetries doesn't result
same kind of o/p , then  what extent they behave same? i.e., if we give up bijection constraint and try to find what we loose and still behave somewhat similar
we use a tool called homomorphism.

Now, one can ask , that what are the properties that is lost when we move from isomorphism to homomorphism, and what is still preserved?
and two of the important properties are ${Ker(\phi)}$ and ${Im(\phi)}$ , So image of ${Im(\phi)}$, which help us to find out what part
of the other symmetry H is going to hit by this function.    ${Ker(\phi)}$ : If we sacrifice one-to-oneness, then in our structure preserving function
more than one elements will hit to identity of the other symmetry. i.e, some non identity element from the domain symmetry is going to be mapped on the identity of the other symmetry.
And it is the most important property which help us to find out how many homomorphism is possible between two group.

so mathematically any function which follow ${\phi (a \cdot b)= \phi(a) \cdot \phi(b)}$, has enough power to tell us something about these two groups even if it is not bijection.
for ex. ${D_4 \rightarrow Z_4}$ are two different order group, but still we can find homomorphism between them.

Continuing same example of ${D_4 \rightarrow Z_4}$ if we partition D4 into rotation and reflection and send rotation to 0 ${(e_H)}$ in ${Z_4}$
and reflection to 2 in ${Z_4}$, These partition acts almost similar to sub group {0,2} of ${Z_4}$ (is it normal subgroup?)

Having these two piece of information ${Im(\phi)}$ and ${Ker(\phi)}$ we can compare narrower part of symmetries and gain some information about the structure of other group.

First Isomorphism Theorem:  Here we use homomorphism to tell us something about of the structure of that symmetry via this theorem.

which tells that image of that homomorphism is isomorphic to the factor group of a domain by the kernel of it.(Kernel is a property of homomorphism and not of a group)
This also, tell us, that every homomorphism is hiding isomomorphism inside them.

for ex. ${Z_{12} \rightarrow Z_3}$ by mapping ${\phi(x)=x mod3}$. ${Ker(\phi)= \{0,3,6,9\} \cong Z_4}$, using this we can get more information
i.e, factoring out ${Z_{12}}$ with kernel, there are still 3 cosets are hiding under the domain group can be achieved via ${Im(\phi)}$ which is isomorphic to ${Z_3}$
these 3 coset form a factor group which is isomorphic to cyclic group ${Z_3}$

Hence, if you give me the homomorphism between two group, I look at the Kernel of that homomorphism, that is the normal subgroup of domain group and they are exactly
the building blocks of symmetries(groups).

Use homomorphism to discover normal subgroups and  to determine factor group. 2. building a cosets out of this and which will be in one to one correspondence with other group that you get via ${Im(\phi)}$.


问:
$n^5-n$ is divisible by $10$?
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>

答:
Your proof is good enough. There's a slight improvement, if you want to avoid modular arithmetic / considering cases.

$n^5 - n$ is a multiple of 5
$\Leftrightarrow$  $ n^5 + 10 n^4 + 35n^3 + 50 n^2 + 24 n = n^5 -n + 5(2n^4 + 7n^3 + 10n^2 + 5n) $ is a multiple of 5. The latter is just $n(n+1)(n+2)(n+3)(n+4)$, which is the product of 5 consecutive integers, hence is a multiple of 5.

Note: You should generally be able to do the above transformation, and can take the product of any 5 (or k) consecutive integers, if you are looking at a polynomial of degree 5 (or k).

Title: $n^5-n$ is divisible by $10$?


问:
Closed form for limit of a sum $\lim\limits_{m\to\infty}(\ln m-4\sum_{n=1}^{m}\frac{(-1)^{n}}{2n+1}\sum_{k=1}^{2n}(-1)^{k}k\ln k)\approx-0.092$
I'm looking for a closed form of

$$\lim\limits_{m\to\infty}\left(\ln m-4\sum_{n=1}^{m}\frac{\left(-1\right)^{n}}{2n+1}\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k\right)\approx-0.092$$

for even $m$.

(For context this is a continuation of my ongoing work from this answer.)

Here's what I've tried so far. Wolfram alpha gives gives me

$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\zeta^{(1,0)}\left(-1,n+\frac{1}{2}\right)+2\zeta^{(1,0)}\left(-1,n+1\right)+n\ln2+3\ln A-\frac{1}{12}\ln2-\frac{1}{4}$$

From Wolfram's function site I get

$$\zeta^{(1,0)}(-1,x)=-\frac{1}{12}\left(-6x^{2}+18x-12\left(x-1\right)\ln\left(x-1\right)+12\ln A+6\left(x-1\right)\ln2\pi-13\right)+\psi^{(-2)}\left(x-1\right)$$

Combining these gets me

$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\psi^{(-2)}\left(n-\frac{1}{2}\right)+2\psi^{(-2)}\left(n\right)-\left(2n-1\right)\ln\left(n-\frac{1}{2}\right)+2n\ln n+n+n\ln2-\frac{1}{2}\ln2\pi+3\ln A-\frac{1}{12}\ln2-1$$

I could keep trying to find more expressions, but it feels like this is making things more complicated and possibly going to bring me back where I started.

Body:
I do not know if this could be of any help.

For large values of $x$, we have
$$\zeta ^{(1,0)}(-1,x)=x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)-\frac{1}{2} x \log
   (x)+\frac{1}{12} (\log (x)+1)+$$ $$\frac{1}{720 x^2}-\frac{1}{5040
   x^4}+\frac{1}{10080 x^6}-\frac{1}{9504
   x^8}+\frac{691}{3603600 x^{10}}+O\left(\frac{1}{x^{12}}\right)$$ Using your fiirst formula, this would give
$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=n \log (2 n)+\frac{1}{4}\log (n)+\left(3 \log (A)-\frac{\log (2)}{12}\right)+\frac{1}{192 n^2}-$$ $$\frac{1}{1280 n^4}+\frac{17}{43008 n^6}-\frac{31}{73728
   n^8}+\frac{691}{901120 n^{10}}+O\left(\frac{1}{n^{12}}\right)$$ which is almost exact even for small values of $n$
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{exact} \\
 5 & 12.603993216766530412 & 12.603993216758928837 \\
 10 & 31.221522180442592703 & 31.221522180442590647 \\
 15 & 52.383497574013162289 & 52.383497574013162273 \\
 20 & 75.215036332678595144 & 75.215036332678595144
\end{array}
\right)$$


问:
Flip a fair coin 5 times. $A$ is the event that the first flip is heads, and $B$ is the event that you get at least three heads. What is $P(A|B)$?
I'm trying to use the conditional probability formula, $P(A|B)= \frac{P(A \cap B)}{P(B)}$. I've determined that $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2^5}(\binom{5}{3} + \binom{5}{4} + \binom{5}{5}) = \frac{1}{2}$, but I can't figure out how to solve for $P(A \cap B)$.

答:
Given the event $B$, that you get at least three heads in five flips, there are $\binom{5}{3} + \binom{5}{4} + \binom{5}{5}$ equiprobable such outcomes.  How many of these have heads in the first flip?  If there were exactly $3$ heads flipped and the first flip is heads, then there are $\binom{4}{2}$ ways to distribute the remaining $2$ heads and $2$ tails; if there are exactly $4$ heads flipped, there are $\binom{4}{1}$ ways to distribute the remaining $3$ heads and $1$ tail; and if $5$ heads are flipped there is only $1$ outcome.  So the desired probability is $$\frac{\binom{4}{2} + \binom{4}{1} + \binom{4}{0}}{\binom{5}{3} + \binom{5}{4} + \binom{5}{5}} = \frac{6+4+1}{10+5+1} = \frac{11}{16}.$$

The above is a pure enumeration approach.  We can consider a more general approach in which the probability of heads is some $p \in (0,1)$ rather than the fair case $p = 1/2$.  This is done by Bayes' theorem:
$$\Pr[A \mid B] = \frac{\Pr[B \mid A]\Pr[A]}{\Pr[B]}.$$  The unconditional number of heads is a binomial random variable, $$X \sim \operatorname{Binomial}(n = 5, p), \\ \Pr[X = x] = \binom{5}{x} p^x (1-p)^{5-x},$$ hence event $B$ is equivalent to $X \ge 3$, and $$\Pr[B] = \Pr[X \ge 3] = \sum_{x=3}^5 \binom{5}{x} p^x (1-p)^{5-x}.$$  The conditional random variable $B \mid A$ corresponds to the outcome $Y \ge 2$ where $Y$ is binomial but with $n = 4$, because flips are independent; i.e. $$Y \sim \operatorname{Binomial}(n = 4, p).$$  And of course, $\Pr[A] = p$.  So it follows that $$\Pr[A \mid B] = \frac{\Pr[Y \ge 2]p}{\Pr[X \ge 3]} = \frac{p \sum_{y=2}^4 \binom{4}{y} p^y (1-p)^{4-y}}{\sum_{x=3}^5 \binom{5}{x} p^x (1-p)^{5-x}} = \frac{6-8p+3p^2}{10-15p+6p^2}.$$  Note when $p = 1/2$ we recover the original answer $11/16$.


问:
How to explain these features of the squares from 01-99?
The last two digits of the square of numbers from 1-99 have some interesting  features as observed below. Do we have simple explanations for them? Or are there any other patterns we find?

Let  the square be $(10X+Y)^2$  (X from 0-9 and Y from 1-9)

[Case X=0 and 5]The last two digits of the squares of $({Y})^2$ and  $({50+Y})^2$ are the same (example: $01^2= 01,51^2=2601$):     List 1
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 01 & 04&09&16&25&36&49&64&81
 \\\hline\hline
\end{array}

[Case X=4 and 9]The last two digits of the squares of $({40+Y})^2$ and  $({90+Y})^2$  are the same((example: $41^2= 1681,91^2=8281$): List 1'
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 81 & 64&49&36&25&16&09&04&01
 \\\hline\hline
\end{array}
Note List 1' is List1 in reverse order.

[Case X=1 and 6]The last two digits of the squares of $({10+Y})^2$ and  $({60+Y})^2$ are the same(example: $11^2= 121,61^2=3721$): List 2
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 21& 44&69&96&25&56&89&24&61
 \\\hline\hline
\end{array}

[Case X=3 and X=8]The last two digits of the squares of $({30+Y})^2$ and  $({80+Y})^2$ are the same(example: $31^2= 961,81^2=6561$):
List 2'
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 61& 24&89&56&25&96&69&44&21
 \\\hline\hline
\end{array}
Note List 2' is List 2 in reverse order.

[Case X=2 and 7]The last two digits of the squares of $({20+Y})^2$ and  $({70+Y})^2$ are the same(example: $21^2= 441,71^2=5041$): List 3
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 41 & 84&29&76&25&76&29&84&41
 \\\hline\hline
\end{array}

Note List 3 is symmetric about the center value of 25 (example: $21^2= 441,29^2=841$).

Summary:

When the tenth digits differ  by 5, last two digits of squares of  two digit numbers are the same.

When the tenth digits sum to 4,9, or 14, the lists of last two digits of squares of  two digit numbers are in reverse order.

The list of last two digits of squares for $({20+Y})^2$ and  $({70+Y})^2$  is symmetric about its enter of 25.

Body: $(10a +i)^2 = 100a^2 + 20ai + i^2$ while $(10(a+ 5) + i)^2 = 100(a^2 + 10a + 25) + 20(a+5)i + i^2 = 100(a^2 + 10a +25 + i) + 20ai + i^2$.


问:
If $2+\log_2a=3+\log_3b=\log_6(a+b)$, find $\frac1a+\frac1b$ and $\log_{ab}\left(\frac1a+\frac1b\right)$
I have a kind of system of equations with logarithms.
How do I get to the expression that the problem asks for?
(Taken from the net.)

If $a>0$ and $b>0$ and it is given that
$$2+\log_{2}{a} =3+\log_{3}{b}=\log_{6}{(a+b)}$$

find the value of

a) $\quad\dfrac{1}{a}+\dfrac{1}{b}$

b) $\quad\log_{ab}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}$

Edit 1
I have related the first with the third and the second with the third to form a system of equations, but the values that come out are not very comfortable to deal with, I do not know if I am wrong or if the problem is like that.

$$2+\log_{2}{a}=\log_{6}{(a+b)}$$
$$3+\log_{3}{b=}\log_{6}{(a+b)}$$

Edit 2 :I have reformulated the problem by splitting it in two

答:
This answer attempts to find part b).

Let us prove that $a^{\log_b c} = c^{\log_b a}$. From the LHS:

$$\exp(\ln (a^{\log_b c}) ) = \exp(\log_b (c \ln a)) = \exp(\ln c /\ln b \cdot\ln a) = \exp(\ln a / \ln b \cdot \ln c) = \exp(\log_b( a \ln c))$$
$$ = \exp(\ln (c^{\log_b a})) = c^{\log_b a} = \text{RHS}$$

Raising $6$ to the power of both sides:

$$36 \cdot 6^{\log_2 a} = 216 \cdot 6^{\log_3 b} = a+b$$
$$\Rightarrow 36 \cdot a^{\log_2 6} = 216 \cdot b^{\log_3 6} = a+b \tag{using the above property}$$
$$\Rightarrow 36a \cdot a^{\log_2 3} = 216b \cdot b^{\log_3 2} = a+b$$

and taking the logarithm base-$ab$ of both sides:

$$\frac{\log_a (36a \cdot a^{\log_2 3})}{\log_a ab} = \frac{\log_b(216b \cdot b^{\log_3 2})}{\log_b ab} = \log_{ab} (a+b)$$
$$\Rightarrow \frac{1 + \log_a 36 + \log_2 3}{1 + \log_a b} = \frac{1 + \log_b 216 + \log_2 3}{1 + \log_b a} = \log_{ab} (a+b)$$
$$\Rightarrow \frac{\log_a 36/b + \log_2 3}{1 + \log_a b} = \frac{\log_b 216/a + \log_2 3}{1 + \log_b a} = \log_{ab}(a+b) - \log_{ab}(ab) = \log_{ab}(1/a+1/b)$$

where in the last step, we have subtracted $1$ from each term.


答:
Let’s say you have a group you are interested in. One way to try to understand the group is to stare at it intently until you see interesting things about it.

But the big insight of abstract algebra in general (and of Category Theory) is that you can also learn a lot about the group by considering morphisms, both into and out of the group. In fact, the Yoneda embedding tells you that understanding all the ways that all groups can map into $G$ will completely determine the isomorphism type of $G$, and that understanding all the ways in which you can map from $G$ to all other groups will completely determine the isomorphism type of $G$ as well.

In addition, one can view these maps as casting “shadows” of the object $G$; with some luck, the different shadows will each be simpler than the group, and perhaps we can piece together enough “shadows” to get a full picture of what the group looks like without having to look at the full group (in turns out that is not always possible, as witnessed by simple groups, but it works for many other types of groups).

The First Isomorphism Theorem tell you that if you want to understand the way that $G$ maps into other groups, you don’t need to go scour the universe to find those groups: you can “realize” any map $f\colon G\to H$ by finding a suitable normal subgroup $K$ of $G$, and then the image of $G$ under $f$ will behave exactly the same way as the image of $G$ under the canonical (and usually well-understood) quotient map $\pi\colon G\to G/K$ behaves. So in a sense you can understand all those images of $G$ by staring at suitable shadows of $G$: just the quotient ones.

The Third Isomorphism Theorem tells you that you that iterating the process won’t get you any new information: if you take a quotient of $G/K$, you are still just taking a quotient of $G$.

The Second Isomorphism Theorem tells you how the subgroups of $G$ behave under the map $G\to G/N$: namely the image of $K$ in the map will be $K/(N\cap K)$, but this will be isomorphic to $KN/N$. When you connect it with the Fourth (or Lattice) isomorphism theorem, which you did not quote, it provides complete information about what happens to the subgroups of $G$ under the map $G\to G/N$: the subgroups that contain $N$ are mapped to the subgroups of $G/N$ in a one-to-one, inclusion preserving, normality preserving manner, while the subgroups of $G$ that do not contain $N$ correspond, under the Second Isomorphism Theorem, to certain subgroups that do contain $N$.

These theorems are valid in a much more general setting: they apply, suitably modified to account for some technical issues, to any kind of algebraic structure: semigroups, monoids, rings, lattices, and more. They form a backbone of investigations, tools used over and over again, because one is always looking at homomorphisms to better understand the structures, and because one is always looking at quotients to simplify matters and try to get a better sense of the object one is studying.

Title: What is the deal with the three isomorphism theorems?

Body:
I've been having lectures in group theory with Hungerford's book. We were presented with the following theorem:

And then with:

Previous to the lectures about them, I was understanding most of the stuff, that is: I kinda could figure out the motivation for things. But when I came into these theorems, things got pretty weird for me. I don't understand the motivation for them. I have the following guesses:

Theorem 5.6 is used to prove Corollary 5.7 but I think it has a utility on its own: If given an homomorphism $f: G \to H$ with $N\lhd G$, we have a unique homomorphism $\overline{f} : G/N\to H$, then we could use this to know what homomorphisms could exist betwen $G$ and $H$ and this depends only on the normal subgroups of $G$. Does that make sense?

For the isomorphism theorems, I noticed they are corollaries so they must be very important. One of the reasons I conjectured for them is that we can construct very interesting isomorphisms. But why are those isomorphisms interesting? It's not clear to me what interesting stuff one could do with them.

I know my guesses are maybe too obvious or wrong, but I can't figure out on my own the answer to that question. Can you help me?


问:
If a complex number has infinitely many solutions why is it a line?
Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z = x+iy \text{ },\:x,y \in \mathbb{R}$ of the equation $sz+t\bar{z}+r=0$ where $\bar{z} = x-iy$.

Prove the number of elements in $L \cap \{z: |z-1+i|=5\}$ is at most $2$.

What they did in the solution is say $z(|s|^2-|t|^2) = \bar{r}t-r\bar{s}$. They did this by taking conjugate of the original equation and eliminating $\bar{z}.$

Now how is $L \cap \{z: |z-1+i|=5\}$ is at most $2$?

If $|s| \neq |t|$, $z = \dfrac{\bar{r}t-r\bar{s}}{(|s|^2-|t|^2)}$ which is one solution and it can intersect the circle once. If $|s| = |t|$ and  $\bar{r}t-r\bar{s}\neq0$ then $L$ is empty. But if $|s| = |t|$ and  $\bar{r}t-r\bar{s}=0$, then $z\cdot0=0 \implies$ $z$ has infinitely many solutions. Why is $L$ a line in this case? Isn't it the entire plane? Why does it intersect the circle at most twice?

Here Z Ahmed told its a line but I can't seem to wrap my head around why this is the case.

答:
The equation $sz + t\bar{z} + r = 0$ can be rewritten in terms of $x,y$ as
$$s(x+iy) + t(x-iy) + r = 0 \\ (s+t)x + i(s-t)y + r = 0$$

If $s-t \neq 0$, we may solve for $y$ as $$y = \frac{i(s+t)}{s-t}x + \frac{ir}{s-t}$$
which describes either a line, a point, or an empty set.

If $s-t = 0$, then the equation becomes $$2sx + r = 0 \\ x = -\frac{r}{2s}$$
which describes either a line or an empty set.

Alternatively, if you're having trouble seeing how these are the solution sets, it may be easier to just go back to $$(s+t)x + i(s-t)y + r = 0 \\ \iff \\ \begin{cases}\Re(s+t) x + \Im(t-s)y + \Re(r) = 0 \\ \Im(s+t)x + \Re(s-t)y + \Im(r) = 0\end{cases}$$
so the solution set is the intersection of two real linear equations, each of which is either empty, a line, or the whole plane.  The intersection of two such sets is either empty, a point, a line, or the whole plane, and we may ignore the whole plane as a possibility since, e.g., $r\neq 0$.


问:
Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3.
This is the answer given:

I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start  with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.

答:
You are right. The given answer is not correct.

But what is more:
The way how to find the reflection matrix is far too cumbersome.

The standard way is as follows:

Let $d=  \begin{pmatrix} 3\\ -2 \end{pmatrix}$ be the direction vector of the line. Then, the projection $P_d$ of any point with positional vector $v$ onto the line is given by:

$$P_d(v) = \frac{d\cdot v}{|d|^2}v$$

So, you get the reflection $T$ of $v$  in your line by flipping the sign of the part of $v$ which is orthogonal to $d$:

$$v = P_dv \color{blue}{+} (v-P_dv) {\longrightarrow} Tv = P_dv \color{blue}{-} (v-P_dv) = 2P_dv - v = (2P_d-I)v$$

So, you only need the matrix representation of $P_d$ which is quickly done by plugging in the unit vectors $e_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $ e_2 =\begin{pmatrix} 0\\ 1 \end{pmatrix}$:

$$P_de_1 = \frac 1{13}\begin{pmatrix} 9\\ -6 \end{pmatrix},P_de_2 = \frac 1{13}\begin{pmatrix} -6\\ 4\end{pmatrix} \Rightarrow P_d =\frac 1{13}\begin{pmatrix} 9 & -6\\ -6 & 4\end{pmatrix} $$

Plug this into

$$T = 2P_d - I \Rightarrow T = \frac 1{13}\begin{pmatrix} 5 & -12\\ -12 & -5\end{pmatrix}$$

I recommend to check - for example - with Geogebra that this matrix nicely reflects points in the given line.


问:
Integrate function $\int_{0}^{\infty}\lambda^{x}e^{-2\lambda}d\lambda$
How can I integrate this function? It's originated by an exponential prior and a poisson likelihood.

$\int_{0}^{\infty}\lambda^{x}e^{-2\lambda}d\lambda$

答:
This is related to the gamma function:

$$\Gamma(z) = \int_0^{\infty} dt \, t^{z-1} e^{-t}$$

In your case, sub $t=2 \lambda$, $d\lambda = t/2$ and get

$$2^{-(x+1)} \int_0^{\infty} dt \, t^x \, e^{-t} = 2^{-(x+1)} \Gamma(x+1) = \frac{x}{2^{x+1}} \Gamma(x)$$


问:
Prove or disprove: If $A$ is regular and $A ∩ B$ is not regular, then $B$ is not regular.
I suspect the Myhill–Nerode theorem may come into play, but not certain. If this was a union instead of an intersect, I'd be almost 100% sure it was true. I'm relatively confident that this statement is true and thus, is to be proven, but not certain how to formally prove it. Of course, I could be wrong and it's false. Anything to help me better understand this would be greatly appreciated.

答: It simply follows from the known fact that the intersection of two regular languages is regular. Thus if $B$ were regular, then $A \cap B$ would be regular, a contradiction.


问:
Find the Norm of the an operator
Let $\{e_n\}$ be the basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$ on $l^2$.

Show that $\|A\|=\sup|\alpha_n|$.

Attempt:

We have $\| A\| = \sup \{ \| Ax \|: \| x\|=1 \} \geq \sup\|Ae_n \| = \sup \|\alpha_n e_n \|_{\infty} = \sup_n | \alpha|$.

Also, let $x\in \ell^2$ with $\| x\| =1$.

Then we have $\|Ax \| = \| A (\sum_{n=1}^{\infty} x_n e_n)\| \leq \|A(\|x_n \|_{\infty} \sum_{n=1}^{\infty}e_n) \| \leq  \|A (\sum_{n=1}^{\infty}e_n) \| = \| \sum_{n=1}^{\infty} \alpha_n e_n\|_{\infty} \leq \sup_n |\alpha_n|$.

Could someone please let me know if my work is correct? If not, how should I fix it?

Thank you!

答:
No. Your inequalities are not correct.

$\|A(\sum x_ne_n)\|^{2}=\|\sum Ax_ne_n\|^{2}=\|\sum \alpha_nx_ne_n\|^{2}=\sum |\alpha_n|^{2} |x_n|^{2}\leq (\sup_n |\alpha_n|)^{2} \sum |x_n|^{2} $ so $\|A\| \leq \sup_n |\alpha_n|$.

Title: Find the Norm of the an operator


问:
How to show this limit approaches $0$?
For the sequence $ a_n=\dfrac{2^n}{n!} $, is there a way to show that the $\lim_{n \to \infty }\dfrac{2^n}{n!}=0$?

I know that it is bounded above by $2$ because it is a decreasing sequence. I also know that the closest number it will every get to is $0$ because it won't be negative.

However, is there a way of showing that the limit goes to $0$?

答:
Intuition

$\frac {2^n}{n!} = \frac 21 \cdot \frac 22 \cdot \frac 23 \cdot \frac 24..... \cdot \frac 2n$. Surely as the terms of the product go to $0$ the product will go to $0$.

Well, intuition and \$2.75 will buy you a cup of coffee.

But if $a_n = \frac {2^n}{n!}$ then $a_1 = a_2 = 2$ and $a_{n+1}=a_n\cdot 2n$ and $\frac 2n < 1$ if $n \ge 3$ so for significantly large $n$ we have $a_n = (a_{n-1}) \cdot \frac 2n$ and $a_{n-1} < 2$ so $a_n < 2\cdot \frac 2n =\frac 4n$ and as $\frac 4n \to 0$ then be squeeze theorem $a_n \to 0$.

Title: How to show this limit approaches $0$?


答:
No, you don't take $||x_n||_{\infty}$ !

It should be,

$$\Big|\Big|Ax\Big|\Big|_2=\Big|\Big|A\Big(\sum x_ne_n\Big)\Big|\Big|_2$$ $$=\Big|\Big|\sum x_nAe_n\Big|\Big|_2$$ $$=\Big|\Big|\sum \alpha_nx_ne_n\Big|\Big|_2$$ $$ \le sup_n |\alpha_n|\Big|\Big|\sum x_ne_n\Big|\Big|_2$$ $$=\sup_n|\alpha_n|\Big|\Big|x\Big|\Big|_2$$

Title: Find the Norm of the an operator

Body:
Let $\{e_n\}$ be the basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$ on $l^2$.

Show that $\|A\|=\sup|\alpha_n|$.

Attempt:

We have $\| A\| = \sup \{ \| Ax \|: \| x\|=1 \} \geq \sup\|Ae_n \| = \sup \|\alpha_n e_n \|_{\infty} = \sup_n | \alpha|$.

Also, let $x\in \ell^2$ with $\| x\| =1$.

Then we have $\|Ax \| = \| A (\sum_{n=1}^{\infty} x_n e_n)\| \leq \|A(\|x_n \|_{\infty} \sum_{n=1}^{\infty}e_n) \| \leq  \|A (\sum_{n=1}^{\infty}e_n) \| = \| \sum_{n=1}^{\infty} \alpha_n e_n\|_{\infty} \leq \sup_n |\alpha_n|$.

Could someone please let me know if my work is correct? If not, how should I fix it?

Thank you!


问:
What is the domain and co-domain of this set?
If $R = \{(1,2)(1,1),(1,3),(3,2),(2,3)\}$ what is the domain and codomain of $R$?

In class, the professor did not give a clear definition and straight-forward example of what domain and co-domain mean. This is a discrete math class I'm asking for.

EDIT: R is a relation on G, where G is {1,2,3}

答:
So you have this relation with you, now the set of first elements of all the ordered pairs is called the domain, whereas the set of the second elements of the ordered pairs is called the range.
In a relation from A to B, the entire set B is called the co-domain. It's basically the entire values of what can come out as an answer,whereas R is what came out as an answer.
One important thing here is that R is a subset of the Co-domain.
Refer to the example below
A={1,2,3,4} and B={5,6,7,8}
R={(1,5),(2,6),(3,7)}
Domain={1,2,3}
Range={5,6,7}
Co-domain={5,6,7,8}=Set B</p>


问:
Using integral to find the area under a portion of a semicircle
I've done quite a bit of research, but I always fail right before the end.
I know the integral of a circle, I know the substitution $$\theta = \arcsin\left(\frac x  r\right)$$ However, I never get the right results.

My question is:

How do I find the area under a semicircle from $-r$ to $x_1; x_1 \in[-r;+r]$?

Also if I want to find the are under a semicircle offset in $$y= f(x) = \sqrt{r^2 - x^2} - d$$

I would simply have to use the above formular and subtract $(x + r) * d$?

Body:
Integrate the area by parts directly

\begin{align}
Area &=\int_{-r}^{x_1}\sqrt{r^2-x^2}dx
 =\int_{-r}^{x_1}\frac{\sqrt{r^2-x^2}}{2x}d(x^2)\\
& = \frac12 x \sqrt{r^2-x^2}\bigg|_{-r}^{x_1} + \frac12\int_{-r}^{x_1}\frac{r^2}{\sqrt{r^2-x^2}}dx\\
&= \frac12 x_1\sqrt{r^2-x_1^2}+ \frac12r^2 \sin^{-1}\frac xr \bigg|_{-r}^{x_1}\\
&= \frac12 x_1\sqrt{r^2-x_1^2}+ \frac12r^2 (\sin^{-1}\frac {x_1}r +\frac\pi2)
\end{align}


问:
Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$
given two positive numbers $a, b$ so that $a+ b= 1$

Sui Zhen Lin ; @szl6208 gave a very beautiful proof for the following inequality
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}+ \sqrt{\frac{4b}{a+ 4}}\leq 1$$
Source: AoPS/@szl6208_ on.AoPS

Proof. We have
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}\leq\frac{\sqrt{a\left ( 9b^{2}- 8b+ 4 \right )\cdot a\left ( a+ 1 \right )^{2}}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}\leq\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}$$
$$\sqrt{\frac{4b}{a+ 4}}\leq\frac{\sqrt{4b\left ( a+ 4 \right )\cdot\left ( 2b+ 2 \right )^{2}}}{\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq\frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}$$
Hence, we need to prove that
$$\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}+ \frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq 1$$
$$\Leftrightarrow RHS- LHS= \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}\geq 0$$
wait a minute, actually
$$RHS- LHS- \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}= \left ( a+ b- 1 \right )f\left ( a, b \right )= 0$$
but how.. why is that so perfect ? I think there maybe are solutions constructed by substitutions like $g\left ( a \right )- g\left ( 1- b \right )\!, g\left ( 2a \right )- g\left ( a+ 1- b \right )\!= \left ( a+ b- 1 \right )f\left ( x \right ).$ What's the best perfect substitution here ? Someone teaching me, huh ?

答:
Using algebra.

From the constraint $b=1-a$. So, we face
$$f(a)=\sqrt{\frac{a^2}{9 a^2-10 a+5}}+2 \sqrt{\frac{1-a}{a+4}}$$
$$f'(a)=\frac{5-5 a}{(9 a^2-10 a+5)^{3/2}}-\frac{5}{\sqrt{1-a} (a+4)^{3/2}}$$

Square and factor; the numerator is just
$$(2 a-1) (5 a-1) \left(a^2-a+1\right) \left(73 a^2-118 a+61\right)$$ So, the derivative only cancels for $a=\frac 15$ and $a=\frac 12$. The second derivative test shows the the first is a minimum and the second a maximum. So, we have
$$f\left(\frac{1}{5}\right)=\frac 32\sqrt{\frac{3}{7}} \qquad \text{and} \qquad f\left(\frac{1}{2}\right)=1$$

Around $a=\frac 12$, we have
$$f(a)=1-\frac{20}{27} \left(a-\frac{1}{2}\right)^2-\frac{1480}{729}
   \left(a-\frac{1}{2}\right)^3+O\left(\left(a-\frac{1}{2}\right)^4\right)$$


问:
Function to produce a vector field with two focal areas: one in the top right and one in the bottom left quadrant.
I'm trying to make a conceptual visualization of a system with two stable areas in a $2$-dimensional possibility space.  To do this I want to produce a vector field showing the direction the system will tend to evolve towards from any given point in possibility space. What I want to end up with is something like this:

All the vectors point towards either the center of the bottom left hand quadrant or the top right hand quadrant. I can make a nice visualization of a vector function using $R$, but I need a mathematical function that describes the vector field that I want. Is there a simple function that will produce such a vector field?

答:
Well, in case anyone is wondering, an answer is:

enter image description here

and the working R script is here.


问:
A polynomial function has degree 3 and a positive lead coefficient. Which graph could represent the function? *

A polynomial function has degree 3 and a positive lead coefficient. Which graph could represent the function? *

答:
How does it behave for large x?

How many roots can it have?


问:
Solve determinants with $|AB| = |A||B|$
How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.

(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$

(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$

(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$

(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$

My Attempt:

(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i  \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.

But it's not helpful at all. :(

(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.

(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$

(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.

Plz give me some hints. Thx in advance.

Title: Solve determinants with $|AB| = |A||B|$

Body:
for D_3 case;I think LU decomposition can help to rewrite as $AB$ and use $|AB|=|A||B|$

$$\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
0 & -2*a & 0 & -2*a \\
0 & 0 & -2*a & -2*a \\
0 & 0 & 0 & 4*a
\end{matrix}\right)$$ so det=$$(1.1.1.1).(a.(-2a).(-2a).(4a))=16a^4$$so det$D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}=-16a^4$


问:
$n^5-n$ is divisible by $10$?
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>

答:
$$n^5-n=n(n^2+1)(n+1)(n-1)= n(n^2-4)(n+1)(n-1)+5n(n-1)(n+1)=(n-2)(n-1)n(n+1)(n+2)+5n(n-1)(n+1)$$

$(n-2)(n-1)n(n+1)(n+2)$ is even and divisible by 5, since it is the product of 5 consecutive integers.

$5(n-1)n(n+1)$ is also even and divisible by $5$.

Note: Both expressions are also divisible by $3$, so $n^5-n$ is actually divisible by $30$!

Title: $n^5-n$ is divisible by $10$?


问:
How to graph $y=4x^2-4x^4$
In physics when learning simple harmonic motion they gave us the x coordinates dependence on time and similarly for the y coordinates

I got the resulting equation of path taken as $y=4x^2-4x^4$ which I have no idea how to graph

Is this a significant graph like the ellipse and hyperbola whose equation would be helpful to memorize .

Although to solve the question this graph is not required ,my query is just that if given this graph how would I plot it

答: You could just study it as a function: see where it equals zero, take the derivatives and study their signs finding local maxima/minima an then study it's behaviour towards infinity. This should give you all the informations about the graph.</p>

Title: How to graph $y=4x^2-4x^4$


问:
Prime ideals in ring of real analytic maps correspond to closed submanifolds?
In affine algebraic geometry, prime ideals in a ring $R$ correspond to irreducible subschemes of $Spec(R)$. For a real, compact, smooth manifold $M$, each prime ideal in $\mathcal{C}^\infty(M)$ is contained in a unique maximal ideal (corresponding to a single point), so we can’t use prime ideals to understand submanifolds.

But what if we restrict to the ring $\mathcal{C}^{an}(M)$ of real, analytic functions on $M$? I understand that the zero sets of analytic functions are unions of submanifolds of lower dimension, so it seems like we’re on the right track. But can it be shown (or proven false) that for all prime ideals $P$, the zero set $Z(P)$ is a closed, irreducible submanifold of $M$, and that for any closed, irreducible submanifold $N$ in $M$, $I(N)$ (the set of analytic functions vanishing on $N$) is prime in $\mathcal{C}^{an}(M)$?

Note that I use “irreducible” to mean “unable to written as a finite union of closed submanifolds”, not just closed subsets.

答:
Whitney's umbrella provides a counterexample to your first statement. Let $M=\Bbb R^3$ and $P=(x^2-y^2z)$; then the zero set is not even pure-dimensional. $P$ is prime for the following reason: $\Bbb R^2$ maps to the Whitney umbrella by $\varphi:(u,v)\mapsto (uv,u,v^2)$ with real-analytically dense image. So if $fg\in P$, then pulling back we have that $\varphi^*(fg)=(\varphi^*f)(\varphi^*(g))$ is a function on $\Bbb R^2$ which is zero and real analytic functions on $\Bbb R^2$ are a domain. Thus one of $f$ and $g$ is already zero on the Whitney umbrella, and $P$ is prime.

The second statement is true. Suppose $f,g\in C^{an}(M)$ with $fg\in I(N)$. Then $N=V(f)\cup V(g)$ is a decomposition in to closed analytic subsets, and by irreducibility we have that one of $V(f)$ and $V(g)$ is all of $N$, so one of $f$ and $g$ is in $I(N)$.


问:
Why does the inverse Fourier transform differs from the Laplace inverse Bromwich integral?
This might be a repeated question, but I am looking for a more in depth explanation for the relation between inverse Fourier and Laplace transforms. We all know that the inverse Laplace transform is

\begin{equation}
f(t) = \frac{1}{2\pi j} \int_{\gamma-j\infty}^{\gamma+j\infty} F(s) e^{st} ds
\end{equation}

On the other hand, the inverse Fourier transform is

\begin{equation}
f(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(\omega) e^{j\omega t} d\omega
\end{equation}

It is easy to substitute $s = j\omega $ into the inverse Laplace transform and obtain a similar inverse Fourier transform like this

\begin{equation}
f(t) = \frac{1}{2\pi} \int_{-\infty-j\gamma}^{+\infty-j\gamma} F(\omega) e^{j\omega t} d\omega
\end{equation}

However, I have always seen $\gamma=0$, just like the second expression from top. Am I missing something here? Can someone explain why it is not possible to move between the two transforms directly?

答:
In the domain of convergence of $$F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt$$
(for simplicity assuming the integrals converge absolutely, although this is not necessary: convergence in $L^2$ sense, in the sense of distributions..)

then $F(\gamma+i.)$ is the Fourier transform of $e^{-\gamma t}f(t)$ and
$$\frac{1}{2i\pi } \int_{\gamma-i\infty}^{\gamma+i\infty} F(s) e^{st} ds=\frac{1}{2i\pi } \int_{-\infty}^{\infty} F(\gamma+i\omega) e^{(\gamma+i\omega)t} di\omega
$$
is $e^{\gamma t}$ times the inverse Fourier transform of the Fourier transform of $e^{-\gamma t}f(t)$.

Does it answer your question?


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
This one goes hand in hand with the enumerability of $\mathbb{Q}$

The fact that though most of the real numbers are transcendental, it is extremely difficult to find one. (excluding some slight modification of the already known ones)

LastEditorDisplayName: anon


答:
Clearly, $n$ and $n^5$ are of the same parity. Hence, $2 \vert (n^5-n)$.

To check for divisibility by $5$, note that
\begin{align}
n^5 - n & = (n^2+1)n(n+1)(n-1)\\
& = (n^2-4+5)n(n+1)(n-1)\\
& = (n^2-4)n(n+1)(n-1) + 5n(n+1)(n-1)\\
& = (n-2)(n-1)n (n+1)(n+2) + 5n(n+1)(n-1)\\
\end{align}

Clearly, $5 \vert 5n(n+1)(n-1)$. Also, $(n-2)(n-1)n (n+1)(n+2)$ is a product of $5$ consecutive numbers and hence $5$ divides it.

Title: $n^5-n$ is divisible by $10$?

Body: I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>


问:
Let $f$, $g$ be multiplicative functions, not identically $0$ such that $f(p^k)=g(p^k)$ for each prime $p$ and $k \ge 1$. Prove that $f = g$.
Let $f$ and $g$ be multiplicative functions that are not identically $0$ and such that $f(p^k)=g(p^k)$ for each prime $p$ and $k\ge1$. Prove that $f=g$.

Source: Elementary Number Theory by David M. Burton

I just know that any function is multiplicative if $f(mn)=f(m)f(n)$ where $gcd(m,n) = 1$ ,I just need some ideas or hints to solve problems like these.

答:
This is a sketch.  Many details are missing.  You should be able to fill those in by following the patterns you are seeing in the various proofs in your source material.

$f = g$ means: for all $n$, $f(n) = g(n)$.

So let $n$ be arbitrary.  By unique factorization,
$$  n = p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m}  \text{,}  $$
for some nonnegative integer $m$, a collection of distinct primes, $p_1, \dots, p_m$, and a collection of positive integers $n_1, \dots, n_m$.

You want to show the proposition
$$  P_m: f(p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m}) = g(p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m})  \text{.}  $$
You are given that it is true when $m = 1$.  Can you show that if $P_M$ is true (for some $M \geq 1$), then $P_{M+1}m$ is true using multiplicativity?  If you put these two facts together, you have a complete proof.


问:
On powers of Trigonometric functions using Complex numbers
For a Calculus course i am taking, we are going through successive differentiation($n^{th}$-derivatives if you will). I came across this technique, given below, for successive derivatives of functions of the form $cos^px,sin^px ,sin^pxcos^qx$:

First,| we let z=cosx+isinx $\implies z^{-1}$=cosx-isinx | so we have that $$z+z^{-1}=2cosx,and z-z^{-1}=2i*sinx....(1) $$
Further, we have: $ z^r+z^{-r}=2cos(rx),z^r-z^{-r}=2i*sin(rx)$......(2)

this first step was okay to understand. But then they did something i didnt quite understand:

let $y=cos^px \implies 2^p*y=(2cosx)^p$ $$=(z+z^{-1})^p=(z^p+z^{-p})+C^p_1(z^{p-2}+z^{-(p-2)})+C^p_2(z^{p-4}+z^{-(p-4)})+\cdots$$ then by using (1)&(2) we will go on to evaluate the successive derivatives.

My question is, how did they expand out the $y=cos^px \implies 2^p*y=(2cosx)^p$ term into a seemingly binomial theorem-esque expansion?. What theorem(s) did they employ to write it out like that? I haven't taken a complex analysis course yet and what i do know about complex numbers is from Intro. ODE courses for my major requirement. so any further resources from which i could learn more about this would be appreciated.

答: I think it's just the binomial theorem. $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$.  When $b = a^{-1}$ then this is $\sum_{k=0}^n \binom{n}{k} a^{n-k} a^{-k} = \sum_{k=0}^n \binom{n}{k} a^{n-2k}$.  Because $\binom{x}{y} = \binom{x}{x-y}$, when $n$ is odd, you can perfectly pair each positive power of $a$ (e.g. $a^3$) in this sum with the corresponding same negative power of $a$ (e.g. $a^{-3}$) having the same coefficient.  When $n$ is even you can also do this but there is also a constant term.


答:
Assuming $f $ and $g$ are  non-zero functions, for any $m\in \mathbb{N^{>1}}$,
let $m=p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k}$ be its prime factorization. Then clearly,

$f(m)=f(p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k})=f(p_1^{\alpha_1})f(p_2^{\alpha_2})\cdot \cdot f(p_k^{\alpha_k})=g(p_1^{\alpha_1})g(p_2^{\alpha_2})\cdot \cdot g(p_k^{\alpha_k})=g(p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k})=g(m).$

Title: Let $f$, $g$ be multiplicative functions, not identically $0$ such that $f(p^k)=g(p^k)$ for each prime $p$ and $k \ge 1$. Prove that $f = g$.

Body:
Let $f$ and $g$ be multiplicative functions that are not identically $0$ and such that $f(p^k)=g(p^k)$ for each prime $p$ and $k\ge1$. Prove that $f=g$.

Source: Elementary Number Theory by David M. Burton

I just know that any function is multiplicative if $f(mn)=f(m)f(n)$ where $gcd(m,n) = 1$ ,I just need some ideas or hints to solve problems like these.


Title: Solve determinants with $|AB| = |A||B|$

Body:
How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.

(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$

(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$

(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$

(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$

My Attempt:

(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i  \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.

But it's not helpful at all. :(

(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.

(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$

(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.

Plz give me some hints. Thx in advance.

Body:
for the case $\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)$
also LU decopmostion
$$\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 \\
\frac{x_2*y_1+1}{x_1*y_1+1} & 1 & 0 \\
\frac{x_3*y_1+1}{x_1*y_1+1} & \frac{x_1-x_3}{x_1-x_2} & 1
\end{matrix}\right)\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
0 & \frac{x_1*y_1-x_2*y_1-x_1*y_2+x_2*y_2}{x_1*y_1+1} & \frac{x_1*y_1-x_2*y_1-x_1*y_3+x_2*y_3}{x_1*y_1+1} \\
0 & 0 & 0
\end{matrix}\right)$$so $$\left|\begin{matrix}
1+x_1*y_1 & 1+x_1*y_2 & 1+x_1*y_3 \\
1+x_2*y_1 & 1+x_2*y_2 & 1+x_2*y_3 \\
1+x_3*y_1 & 1+x_3*y_2 & 1+x_3*y_3
\end{matrix}\right|=0$$


问:
Parameterizing both branches of a hyperbola
Recently I have been studying parametric equations of surfaces and curves, specifically hyperbolic functions. Given by the equations $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad\frac{(x-\alpha)^2}{a^2}-\frac{(x-\beta)^2}{b^2}=1$$

denoting hyperbolic functions centred at $(0,0)$ and $(\alpha,\beta)$ respectively. I am however, unable to derive a parametrisation that represents the entire function. Help to further understand this topic would be greatly appreciated.

答:
Here's a geometric way to derive a parametrization. Consider the family of all lines passing through the point $(a,0)$. Such a line is of the form $y = t(x-a)$ for some slope $t$. Since the hyperbola has degree $2$, each line will intersect it in exactly two points: $(a,0)$ and one other point. (This is an example of Bézout's Theorem.) As we vary the slope $t$, this other point will trace out the curve.

Let's use this observation to compute equations for the parametrization. Substituting $y = t(x-a)$ into the equation for the hyperbola, we have
\begin{align*}
0 &= \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1 = \frac{x^2}{a^2} - \frac{(t(x-a))^2}{b^2} - 1 \, .
\end{align*}
Multiplying through by $a^2 b^2$ to clear denominators yields
\begin{align*}
0 &= b^2 x^2 - a^2 (t(x-a))^2 - a^2 b^2 = b^2 x^2 - a^2 t^2(x^2 - 2ax + a^2) - a^2 b^2\\
&= (b^2 - a^2 t^2)x^2 + 2 a^3 t^2 x - a^4t^2 - a^2 b^2
\end{align*}
and dividing through by the leading coefficient, we find
$$
0 = x^2 + \frac{2 a^3 t^2}{b^2 - a^2 t^2} x - \frac{a^2 (a^2 t^2 + b^2)}{b^2 - a^2 t^2} \, .
$$

One of the roots is $x = a$ and since the constant term is the product of the roots, then the other is
\begin{align*}
x = -\frac{a (a^2 t^2 + b^2)}{b^2 - a^2 t^2}
\end{align*}
which, together with
$$
y = t(x-a) = t\left(-\frac{a (a^2 t^2 + b^2)}{b^2 - a^2 t^2} - a\right)
$$
gives a parametrization of the whole hyperbola, except for the point $(a,0)$. Note that since $b^2 - a^2 t^2 = (b - at)(b+at)$, then $x$ and $y$ have poles at $t = b/a$ and $t=-b/a$. Here's a SageMath cell plotting this parametrization, and the resulting plot with the ranges $t < -b/a$, $-b/a < t < b/a$, and $b/a < t$ colored differently.

$\hspace{4cm}$

For more on parametrizing curves, see Ch. $1$, $\S3$ of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms. It's a wonderful book, and they use fun examples like this to motivate later topics in the book.


问:
Why can't you solve $0 = -5\sqrt{x+2}-2$
So i did this,
\begin{aligned}
2&=-5\sqrt{x+2},\\
\tfrac{2}{-5}&=\sqrt{x+2},\\
\tfrac{4}{25}&=x+2,\\
x&=-\tfrac{46}{25}.
\end{aligned}
Is there something wrong with my algebra?

答:
Remember that, $\sqrt{x+2}≥0$ and multiplying each side by $(-1)$  then you get,

$$-5\sqrt{x+2}-2=0 \Longleftrightarrow 5\sqrt{x+2}+2=0$$ which is impossible.

The sum of  non-negative and positive  numbers can not equal to zero.

EDIT:

Your argument is not generally correct. Because, $$a=b \not \Longleftrightarrow a^2=b^2.$$ For example,

$$-1=1 \\  (-1)^2=1^2 \\  1=1 $$ which is invalid.


问:
How did they obtain $15-15i$
This might be a silly question, but how did they get $15-15i$ for example 1? The operation they did didn't follow any of the definitions listed (a-d). Did they use operation a? If so, where did they get $z$ from because the question only gave $x$ & $y$ within $C^2$, and why did they use the conjugate of $y$?

Lastly, why does $\langle x,y\rangle=\langle y.x\rangle$ in definition c.

Title: How did they obtain $15-15i$

Body:
Expanding the product and using $i^2=-1$,

$$(1+i)(2+3i)+4(4-5i)=2+3i+2i+3i^2+16-20i$$

$$=2+5i-3+16-20i=15-15i.$$


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
The problem is solving $x^2\equiv x\pmod{10}$, or $x(x-1)\equiv 0\pmod{10}$, which means finding integers $x$ such that $10$ is a factor of $x(x-1)$.   For that to hold, either $x$ or $x-1$ must be a multiple of $5$, which means the last digit of $x$ is $0,1,5,$ or $6$.  Then it is a simple verification that the equation holds in each of these cases.

Rephrased without "$\pmod{10}$" notation, this could be expressed as follows.  We are looking for integers $x$ such that $x$ and $x^2$ have the same last digit.  This is the same as saying that the last digit of $x^2-x=x(x-1)$ is $0$.  That means that $x(x-1)$ is a multiple of $10$.  See above for the rest.


问:
Maximum and minimum of $x^2+2y^2+3z^2$ subject to $x^2+y^2+z^2=100$

Find the maximum and minimum values of the function $f(x,y,z)=x^2+2y^2+3z^2$ subject to the constraint $x^2+y^2+z^2=100$.

I know to find the critical points I need to solve the system of equations

$$\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$$

I ended up with

$$\begin{aligned} 2x &= \lambda 2x\\ 4y &= \lambda 2y\\ 6z &= \lambda 2z\end{aligned}$$

I don't know to go from here since $\lambda$ ends up as $1$, $2$, and $3$.

答:
This is a sort of "dual" of the extremization problem that is often asked here [for example:  Lagrange Multipliers to find the maximum and minimum values ] , with the function being $ \ x^2  +  y^2  +  z^2 \ $ under the constraint $ \ ax^2 + by^2 + cz^2 \ = \ d \ $ with specified coefficients.  The geometrical interpretation is similar, though. The constraint surface $ \ x^2  +  y^2  +  z^2 \ = \ 100 \  $  represents a spherical surface of radius $ \ 10 \ $ centered on the origin.  Level surfaces of $ \ x^2 + 2y^2 + 3z^2 \ $ are triaxial ellipsoids of varying sizes, with all of them being similar.

For a particular ellipsoid   $ \ x^2 + 2y^2 + 3z^2 \ = \ C \ \ \rightarrow \ \ \frac{x^2}{C} \ + \ \frac{y^2}{C/2} \ + \ \frac{z^2}{C/3} \ = \ 1 \ \ , $ the semi-axis lengths are $ \ \sqrt{C} \ , \ \sqrt{C/2} \ $ and $ \ \sqrt{C/3} \ $ .  The smallest of this "family" of ellipsoids which will be tangent to the spherical surface is the one for which $ \ \sqrt{C} \ = \ 10 \ \Rightarrow \ C \ = \ 100 \ \ ; $ this gives us an ellipsoid inscribed within the sphere.  The largest ellipsoid that can be tangent to the spherical surface is the one for which its shortest axis makes contact (the ellipsoid circumscribes the sphere); for this, $ \ \sqrt{C/3} \ = \ 10 \ \Rightarrow \ C \ = \ 300 \ \ . $  So the constrained minimum and maximum values of $ \ x^2 + 2y^2 + 3z^2 \ $ are $ \ 100 \ $ and $ \ 300 \ $ , respectively, as also shown in the other answers.  [The ellipsoid which has the endpoints of its "middle-length" axis tangent to the spherical surface has $ \ \sqrt{C/2} \ = \ 10 \ \Rightarrow \ C \ = \ 200 \ \ , $ which has no special importance here.]


答:
$\sqrt{M} \ge 0$ always.  So you can't have the statement true.  If you assume the statement is true (which it can't be) then your math does nothing wrong but the result won't mean anything.

It's bit like saying solve $\frac x{0} = 7$.  That can't be true as nothing can be divided by $0$, but if you assume it is true then doing valid math $\require{cancel}$

$\frac x0 = 7$

$0\cdot \frac x0 = \cancel{0}\frac x{\cancel0} = 0\cdot 7$

$x = 0$ is a meaninless result.

This error gets hidden.

When every you have $a = M$ and you square both sides to get $a^2 = M^2$ you add in the possibility that $a = -M$ which was not an initial case.

In this case $-\frac 25 = \sqrt{x+2}$ is impossible as $\sqrt{x+2} \ge 0$.

But when you square both sides you add in the possibility $\sqrt{x+2} = \frac 25$ and because that is possible that is what you end up solving.

Once you get $x = -\frac {46}{25}$ you can check and see that it doesn't sovlve

$-5\sqrt{x+2} -2=$

$-5\sqrt{-\frac {46}{25}+2}-2=$

$-5\sqrt{-\frac {46}{25}+\frac {50}{25}} -2=$

$-5\sqrt{\frac 4{25}} - 2=$

$-5\frac 25 -2=$

$-2-2 = -4 \ne 0$.

But it is a solution to $0 = 5\sqrt{x +2} -2$.

Title: Why can't you solve $0 = -5\sqrt{x+2}-2$

Body:
So i did this,
\begin{aligned}
2&=-5\sqrt{x+2},\\
\tfrac{2}{-5}&=\sqrt{x+2},\\
\tfrac{4}{25}&=x+2,\\
x&=-\tfrac{46}{25}.
\end{aligned}
Is there something wrong with my algebra?


Title: Solve determinants with $|AB| = |A||B|$

Body:
For the fourth determinant start with
$$
V=\pmatrix{1 & 1 & 1 & 1\\a_1 & a_2 & a_3 & a_4\\a_1^2 & a_2^2 & a_3^2 & a_4^2\\a_1^3 & a_2^3 & a_3^3 & a_4^3\\}
$$
and notice that
$$
\begin{pmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{pmatrix}=VV^{T}.
$$

The determinant of $V$ is well-known to be $\prod_{i<j}(a_i-a_j)$, so $D_4=(\prod_{i<j}(a_i-a_j))^2$.


问:
How to taylor expand $1/(1- \frac{k}{r})$ where k is some constant
Hi i am doing some integrations and there came across one step that i couldn't understand. The step is as follows:
$$\frac{dr}{1 - \displaystyle\frac{k}{r}}=dr\left(1 + \frac{k}{r-k}\right)$$. I don't know how we can come up with right hand side from left hand side. I know that using taylor series expansion, we have $\frac{1}{1-x}=1 +x+x^2 + x^3+...$ but i am not able to apply this here. How did we get the right hand side $dr(1 + \frac{k}{r-k})$ from left hand side of the same equation?

PS: k is a constant here and r is the variable term.

Body:
$$\frac{dr}{1 - \displaystyle\frac{k}{r}}=\frac{dr}{\frac{r-k}{r}}=\\dr(\frac{r}{r-k})=\\dr(\frac{r-k+k}{r-k})=\\
\\dr(\frac{r-k}{r-k}+\frac{k}{r-k})=\\
\\dr(1+\frac{k}{r-k})$$


答: You've incorrectly assumed $f(x)=f(y)\implies x = y$ (line 2 to line 3) which is not true when $f$ is non injective.


答:
Modular arithmetic is the most convenient and powerful language to express the explanation in, but there is a low-tech version. The desired condition is equivalent to just $x$ and $x^2$ having the same units digit, which is equivalent to $10\mid (x^2-x)=x(x-1)$, which is equivalent to

$$10\mid x(x-1)\implies \begin{cases}10\mid x \\ 2\mid x, ~ 5\mid (x-1) \\ 5\mid x,~2\mid(x-1) \\ 10\mid (x-1)\end{cases}\implies \rm units~digit=\begin{cases}0 \\ 6 \\ 5 \\ 1\end{cases} $$

In base $b$, an integer $x$ has the same units digit no matter what power you raise it to if and only if $x$ has the same units digit as $x^2$, or equivalently $x^2\equiv x\bmod b$. This occurs when $b\mid x(x-1)$. The set of $x$ for which this occurs can be precisely characterized.

For each divisor $d\mid b$ such that $\gcd(d,b/d)=1$, the set of solutions $x$ to the system

$$\begin{cases} x\equiv 0 \bmod d \\ x\equiv 1\bmod b/d\end{cases}$$

satisfy the desired condition, and every desired $x$ is attained in this way. Using SZ aka CRT,

$$\{d\cdot\big(d^{-1}\bmod b/d\big)+bn:~d\mid b,\,\gcd(d,b/d)=1,n\in{\bf Z}\}$$

is the set of all $x$ s.t. $x,x^2,x^3,\cdots$ all have the same units digit in base $b$.

Title: What makes $5$ and $6$ so special that taking powers doesn't change the last digit?

Body: Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>


问:
When does a prime $p=m^2+n^2$ divide the term $ m^3+n^3-4$?
I came across a math olympiad type question that goes like this:

For what primes $p$ will $p=m^2+n^2$ and $p$ divide $m^3+n^3-4$?

I tried a few examples and think that $p=5$ is the only solution but am unable to prove it. The property that $p \equiv 1 \mod 4$ seems to not be helpful so far. Any clues or solutions are welcome.

答:
Sorry for waking up old post...
Anyway, I think the following works.

Since $p\mid m(m^2+n^2)$ and $p\mid n(m^2+n^2)$ we have $p|3(m^3+m^2n+n^2m+n^3)$ Since \begin{align}3(m^3+m^2n+n^2m+n^3)&=(m+n)^3+2(m^3+n^3)\\&=(m+n)^3+8+ 2(m^3+n^3-4)\end{align}
we obtain $p|(m+n)^3+8$，while $$(m+n)^3+8=(m+n+2)(m^2+n^2+2mn-2m-2n+4)$$
Since $p$ is prime, it divides one of the factors.

The case $p|(m+n+2)$ is easily dealt and gives $p=2,5$.

The latter case reduces to $p|2(mn-m-n+2)$, and to $p|(mn-m-n+2)$ if $p\neq 2$.
Now $m^2+n^2\leq mn-m-n+2$, but $m^2+n^2\geq 2|mn|$, so some simple comparison gives the only possibility $m=-2,n=-3,p=13$.

Hence the answer $p=2,5,13$.


问:
If a complex number has infinitely many solutions why is it a line?
Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z = x+iy \text{ },\:x,y \in \mathbb{R}$ of the equation $sz+t\bar{z}+r=0$ where $\bar{z} = x-iy$.

Prove the number of elements in $L \cap \{z: |z-1+i|=5\}$ is at most $2$.

What they did in the solution is say $z(|s|^2-|t|^2) = \bar{r}t-r\bar{s}$. They did this by taking conjugate of the original equation and eliminating $\bar{z}.$

Now how is $L \cap \{z: |z-1+i|=5\}$ is at most $2$?

If $|s| \neq |t|$, $z = \dfrac{\bar{r}t-r\bar{s}}{(|s|^2-|t|^2)}$ which is one solution and it can intersect the circle once. If $|s| = |t|$ and  $\bar{r}t-r\bar{s}\neq0$ then $L$ is empty. But if $|s| = |t|$ and  $\bar{r}t-r\bar{s}=0$, then $z\cdot0=0 \implies$ $z$ has infinitely many solutions. Why is $L$ a line in this case? Isn't it the entire plane? Why does it intersect the circle at most twice?

Here Z Ahmed told its a line but I can't seem to wrap my head around why this is the case.

答:
Brian seems to have given a complete solution, but to address your specific confusion: You're right that the equation $$z(|s|^2-|t|^2) = \bar{r}t - r\bar{s}$$ will have infinitely many solutions if $|s|^2 - |t|^2 = 0$ and $\bar{r}t - r\bar{s}= 0.$  But, in that case, the elimination method step (where we solve for $z$ by eliminating the conjugate) is producing the trivial equation $0 = 0,$ which loses the original solution set.  The arbitrary $z$ values you choose won't satisfy the original equation - only this secondary equation.

For example, take $s = t = \frac{1}{2}$ and $r = -1$ so that the original equation becomes $$\frac{z}{2} + \frac{\bar{z}}{2} - 1 = 0,$$ which is equivalent to $\Re(z) = 1$.  Clearly that solution set is a line.

If you take the conjugate equation and use it to eliminate $\bar{z}$, then $z$ will be eliminated as well, and you'll get the equation $0z + 0\bar{z} = 0$, which does have infinitely many solutions but isn't equivalent to the original equation.  We've lost the original equation's information.

The overall problem is arising because the conjugate equation is a multiple of the original equation in this degenerate case, so it's not useful for elimination.  For example, if you start with the system of equations \begin{array}{11}x + y = 0\\2x + 2y = 0,\end{array} then there's a line (not plane) of solutions: $\{(x, -x): x \in \mathbb{R}\}$, but the elimination method will kill both variables and lose this information, because this is a dependent system of equations.  You've identified the conditions under which the conjugate trick/elimination method fails.


问:
Flip a fair coin 5 times. $A$ is the event that the first flip is heads, and $B$ is the event that you get at least three heads. What is $P(A|B)$?
I'm trying to use the conditional probability formula, $P(A|B)= \frac{P(A \cap B)}{P(B)}$. I've determined that $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2^5}(\binom{5}{3} + \binom{5}{4} + \binom{5}{5}) = \frac{1}{2}$, but I can't figure out how to solve for $P(A \cap B)$.

答:
Flip a fair coin 5 times.  A  is the event that the first flip is heads, and  B  is the event that you get at least three heads. What is  P(A|B) ?

For the event $(A\cap B)$ you need at least $2$ heads in the next $4$ tosses, thus

$P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac {n(A\cap B)}{n(B)}$

$=\dfrac{\binom 4 2 + \binom 4 3 + \binom 4 4}{\binom 5 3 + \binom 5 4 + \binom 5 5} = \dfrac {11}{16}$

PS

If you want to use probabilities, $P(A\cap B) = P(A)\cdot P(B|A)$, but you can further simplify the computation by dealing with number of ways rather than probabilities as shown above


问:
Stretching a point cloud at a certain angle; is this right, and is there a form not using trigonometric functions?
I have a group of points $x, y$ on an elastic sheet and need the new coordinates after "stretching" the sheet in one dimension by a fraction $a$ in a direction defined by $\theta$.

It should be easy for me to calculate the position of the points after stretching $x_s, y_s$ but I'm stuck.

I am guessing that the points move a distance away from the origin by a fraction $f$ where

$$f = \frac{a}{2} (1 + \cos ( \tan^{-1}(y/x) - \theta))$$

(which equals $a$ when $x, y$ fall on the line $y = x \tan(\theta)$ and zero when perpendicular to it) and then I can use

$$x_s = (1+f)x$$

$$y_s = (1+f)y$$

to get the stretched coordinates.

Question(s):

Have I got it right?

Can this be written in a simple form without the use of trigonometric functions? (no approximations)

答:
With stretch direction $\mathbf{\hat{s}} = \cos \theta \ \hat{x} + \sin \theta \ \hat{y}$, then $\mathbf{r} \cdot \mathbf{\hat{s}}$ is the distance of each point from the line of zero stretch in the $\mathbf{\hat{s}}$ direction. Extend it by a fraction $a$ further in the  $\mathbf{\hat{s}}$ direction:

$$\mathbf{r_s} = \mathbf{r} + a \ (\mathbf{r} \cdot \mathbf{\hat{s}}) \ \mathbf{\hat{s}}$$

Here's a hexagonal lattice stretched with $a=0.2$ at an unsettling 80 degrees:

```<code>import numpy as np
import matplotlib.pyplot as plt

def stretch(points, a, theta):
    to_degs, to_rads = 180/np.pi, np.pi/180
    r_hat = points / np.sqrt((points**2).sum(axis=1))[:, None]
    s_hat = np.array([f(to_rads * theta) for f in (np.cos, np.sin)])
    print('s_hat: ', s_hat)
    dot = (points * s_hat).sum(axis=1)
    points_s = points + a * dot[:, None] * s_hat
    return points_s

def get_points(vecs, nmax=3):
    nmax = int(nmax)
    i = np.arange(-nmax, nmax+1)
    I, J = [thing.flatten() for thing in np.meshgrid(i, i)]
    keep = np.abs(I + J) <= nmax
    I, J = [thing[keep] for thing in (I, J)]
    v1, v2 = vecs
    points = I[:, None] * v1 + J[:, None] * v2
    return points

r3o2 = np.sqrt(3) / 2
vecs = np.array([[1, 0], [0.5, r3o2]])
points = get_points(vecs, nmax=5)
points_s = stretch(points, a=0.2, theta=80)

plt.figure()
x, y = points.T
plt.plot(x, y, 'ok')
x, y = points_s.T
plt.plot(x, y, 'or')
for p, ps in zip(points, points_s):
    plt.plot([p[0], ps[0]], [p[1], ps[1]], '-b')
plt.gca().set_aspect('equal')
plt.show()
</code>```

Body:
With stretch direction $\mathbf{\hat{s}} = \cos \theta \ \hat{x} + \sin \theta \ \hat{y}$, then $\mathbf{r} \cdot \mathbf{\hat{s}}$ is the distance of each point from the line of zero stretch in the $\mathbf{\hat{s}}$ direction. Extend it by a fraction $a$ further in the  $\mathbf{\hat{s}}$ direction:

$$\mathbf{r_s} = \mathbf{r} + a \ (\mathbf{r} \cdot \mathbf{\hat{s}}) \ \mathbf{\hat{s}}$$

Here's a hexagonal lattice stretched with $a=0.2$ at an unsettling 80 degrees:

import numpy as np
import matplotlib.pyplot as plt

def stretch(points, a, theta):
    to_degs, to_rads = 180/np.pi, np.pi/180
    r_hat = points / np.sqrt((points**2).sum(axis=1))[:, None]
    s_hat = np.array([f(to_rads * theta) for f in (np.cos, np.sin)])
    print('s_hat: ', s_hat)
    dot = (points * s_hat).sum(axis=1)
    points_s = points + a * dot[:, None] * s_hat
    return points_s

def get_points(vecs, nmax=3):
    nmax = int(nmax)
    i = np.arange(-nmax, nmax+1)
    I, J = [thing.flatten() for thing in np.meshgrid(i, i)]
    keep = np.abs(I + J) <= nmax
    I, J = [thing[keep] for thing in (I, J)]
    v1, v2 = vecs
    points = I[:, None] * v1 + J[:, None] * v2
    return points

r3o2 = np.sqrt(3) / 2
vecs = np.array([[1, 0], [0.5, r3o2]])
points = get_points(vecs, nmax=5)
points_s = stretch(points, a=0.2, theta=80)

plt.figure()
x, y = points.T
plt.plot(x, y, 'ok')
x, y = points_s.T
plt.plot(x, y, 'or')
for p, ps in zip(points, points_s):
    plt.plot([p[0], ps[0]], [p[1], ps[1]], '-b')
plt.gca().set_aspect('equal')
plt.show()