问:
Two sets A, B of size n. Each element selects one element of the other set. Probability of no mutual selection?
Came up in a challenge list of probability interview Q's.

You have two sets A, B of size n each. Each member in set A designates a member in B at random. Each member in set B designates a member in A at random. What is the probability of no mutual designation?

Where I got to:
Use an n-tuple (k_1, k_2, ...k_n) to represent how many members of A are designated to the i-th member of B.

I get that each tuple is created by (n!)/[(k_1!)(k_2!)...(k_n!)] designations of A. Or more simply ${n \choose k_1 ... k_n}$

For each n-tuple, I also get there are $\Pi_k (n-k_i)^{k_i}$ to satisfy the no mutual requirement from B's designations.

Putting that altogether I get:

$\Sigma_{k_1 + k_2 .... k_n = n}  {n \choose k_1 ... k_n} * \Pi_k (n-k_i)^{k_i}$

Which seems to be really close to multinomial expansion, but not quite.

Tried a few other methods, too. But they keep running into the issue that each tuple allocation (barring permuting the order) results in different number of combinations (ways to reach via A times ways to preserve no mutual from B).

答:
When I encounter problems like these, my instinct is always to:

Use the method of moments to prove that the number of mutual selections (or whatever) converges to a Poisson random variable in distribution;

Conclude that as $n \to \infty$, the probability of having no mutual selections goes to $e^{-\mu}$, where $\mu$ is the expected number of mutual selections (here, $1$).

If $\mathbf X$ is the number of mutual selections, we can show that for a constant $k$ and $n \ge k$, $\mathbb E[\binom{\mathbf X}{k}] = k! \binom nk^2 n^{-2k} \sim \frac1{k!}$, which is what we expect from a Poisson random variable with mean $1$. So that ticks off the first box.

(Why this expected value? Well, $\binom{\mathbf X}{k}$ counts the number of $k$-sets of mutual selections. There are $k! \binom nk^2$ ways to describe a potential such $k$-set in terms of which $A$-elements are matched with which $B$-elements, and each one is actually a $k$-set of mutual selections with probability $n^{-2k}$.)

In particular, $\mathbb E[X] = 1$, so $\Pr[\mathbf X = 0] \to e^{-1}$ as $n \to \infty$.


问:
Why isn't 1/2 a limit point of sequence $X_n = 1/n$
There's one confusion I have been having regarding Limit points of a sequence.
The definition dictates that $p$ is called a limit point of a sequence $(X_n)$ if every neighborhood of $p$ contains infinitely many terms of the given sequence.

I took one example.
$$X_n = \frac{1}{n}$$

Can someone give me an example of a neighborhood of 1/2 that doesn't contain infinitely many terms of the sequence? I know this sequence only has one limit point which is 0 but I want examples as to why the other members of the sequence can't be called limit points.

Please help guys

答:
$(\frac 1 2 -\frac 1 6, \frac 1 2 -\frac 1 6)$ does not contain $X_n$ for any $n \neq 2$.

More generally, $(\frac 1 n-\frac 1 {n(n+1)},\frac 1 n-\frac 1 {n(n+1)})$ is an interval around $X_n$ which contains no other $X_m$. So $X_n$ is not a limit point for any $n$.

ContentLicense: CC BY-SA 4.0


问:
Lax Equation formulation of the Focusing Non-Linear Schrodinger Equation.
For a project I am writing at university, I have a section where I am writing about rational solutions to the  NLSE, which use Lax pairs as the starting point for the derivation. I therefore wish to derive the Lax pair for the NLSE from first principles. In a paper I have been reading, the $\bf{KdV}$ equation was expressed in terms of the linear operators $\mathbf{L}$ and and $\mathbf{B}$ as follows: $$\frac{d\mathbf{L}}{dt} + [\mathbf{L},\mathbf{B}] = 0$$
where $$\mathbf{L} = -6\frac{d^{2}}{dx^{2}}$$
and $$\mathbf{B} = -4\frac{d^{3}}{dx^{3}} - u\frac{d}{dx} - \frac{1}{2}u_{x}.$$
I have been seraching around for a simmilar formulation for the focusing NLSE and have not been able to come up with anything. I am therefore wondering whether anyone could show me what the Lax equation formulation of the NLSE is and how I might go about deriving it.

答:
The Lax pair of the Nonlinear Schrödinger equation was first given in 1972 by Zakharov and Shabat – see Equations (4) and (5) in the PDF of their paper https://www.semanticscholar.org/paper/Exact-Theory-of-Two-dimensional-Self-focusing-and-Zakharov-Shabat/73ef172d46291e1b2235d4fea6d80078f5e705ef?p2df

The Lax pair in this case is a $2\times 2$ matrix linear system, rather than a pair of scalar differential equations as in the case of the KdV.


问:
Show that $X$ is compact iff $X_n$ compact for all $n\in \mathbb N$.
Let $(X_n,d_n) $  a family of metric spaces and $(X=\prod_{n \in \mathbb N} X_n,\mathcal d ) $ a metric space with $\mathcal T_p$ the product topology is equal to the topology defined by the distance $d$, such that for all $x=(x_n)_n, y=(y_n)_n\in X$ we have  $d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}d_n(x_n,y_n) $ and $d_n(x,y) \leq M$ for all $x, y\in X_n$.

Show that $X$ is compact iff $X_n$  compact for all $n\in \mathbb N$.

For the first implication, I have used the the fact that $X$ is compact and the projection maps are continuous. For the other one, I get stuck with it. any help is appreciated.

答: One direction is the fact that the projections are continuous, and continuous maps preserve compactness.  The other (nontrivial) direction is the theorem of Tychonoff, which incidentally is equivalent to the axiom of choice.</p>


问:
All closed balls are compact each isometry is bijective
Let $(X,d)$ be a metric space in which all closed balls are compact and such that for any two points $x,y \in X$ there exists a function $u \in Iso(X,d)$ such that $u(x)=y$.

Prove that then each isometry $u: X \rightarrow X$ is bijective.

How can I prove this?

I know that if $X$ is compact, then each isometry is bijective.

Could you help me wih that?

答:
Let $u$ be an isometry of $X$, and $x$ some point of $X$.
It is enough to show that $u$ is surjective.

Taking $u'$ an isometry such that $u'(u(x)) = x$ given by your hypothesis, then $u' \circ u$ is an isometry that fixes $x$. Therefore, it stabilizes the balls centered at $x$, and you can show that $u' \circ u$ is surjective because the balls are compact.
The surjectivity of $u$ follows directly :

If $y \in X$, then $\exists x \in X$ such that $(u' \circ u) (x) = u'(y)$ by surjectivity of $u' \circ u$. Then $u(x) = y$ by injectivity of $u'$.

EDIT : If you remove one of the hypothesis, the assumption fails :

Take $\mathbb{N}$ with the classical metric. Then $n \mapsto n+1$ is an isometry which isn't surjective. Its closed balls are finite and therefore compact ; but the isometries group of $\mathbb{N}$ is trivial, so its action can't be transitive.

And in $l^2$, take the shift operator $e_i \mapsto e_{i+1}$. It is not surjective either. The action of its isometries group is transitive because it contains the translations ; but it is known that its closed balls aren't compact.

ContentLicense: CC BY-SA 3.0


问:
Is dealing with mathematics without understanding philosophy behind it good aproach?
I was good at mathematics in high school , I can make and understand proofs , and if there is something wrong in proof I can recognise it but all of this without knowing what is the real reason , I just see that "this makes sense" .. For example no one explained to me What "assumption" or "definition" means I have seen it several times so I understood it  .. the thing is I'm confused is the way I'm dealing with mathematics correct? Does this have something to do with philosophy? And if so should I start learning philosophy of mathematics before learning basic mathematic topic like calculus ?</p>

答:
Good question.

In my opinion, it greatly enhances the whole mathematical experience to find out the history behind (and hence the reason for) the development of a mathematical idea (so as to find out, for example, what the motivation behind it was).

If nothing else, it enhances the understanding.

Many good text books include some of the historical notes behind the development. As an example of this, the textbooks of George Finlay Simmons are among the best.


问:
Analytic Proof of Wilson's Theorem
I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:

We start by considering the Maclaurin series of the function $f(x)=\ln \left( \frac{1}{1-x} \right)$, that is,
$\displaystyle \ln \left( \frac{1}{1-x} \right)= \sum_{n=1}^{\infty} \frac{x^n}{n} \tag*{}$
for $x \in [-1, 1).$
By simply using the definition of $\ln$ on the equation above, we get
$\displaystyle e^{x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots}=\frac{1}{1-x}=1+x+x^{2}+ \cdots + x^{p} + \cdots \tag{1}$
We can then rewrite the leftmost side of $(1)$ as
$\displaystyle \begin{align} e^{x+\frac{x^2}{2}+\frac{x^3}{3}+…} & = e^{x}e^{\frac{x^2}{2}}e^{\frac{x^3}{3}} \cdots \\ & = \prod_{n=1}^{\infty} e^{\frac{x^n}{n}} \\ & = \prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\left( \frac{x^n}{n} \right)^k}{k!} \\ & = \left( 1+\frac{x}{1!}+\frac{x^2}{2!}+ \cdots \right) \cdot \left( 1+\frac{x^{2}/2}{1!}+\frac{(x^{2}/2)^2}{2!} + \cdots \right) \cdots \left(1+\frac{x^{p}/p}{1!}+\frac{(x^{p}/p)^2}{2!}+ \cdots \right) \cdots \\ & = 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots \end{align} \tag{2}$
Combining the rightmost sides of $(1)$ and $(2)$ together gives us
$\displaystyle 1+x+x^{2}+ x^{3}+ \cdots + x^{p} + \cdots= 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots , \tag{3}$
from which follows that the coefficient of $x^{p}$ on the RHS of $(3)$ is $1$. That is,
$\displaystyle \frac{1}{p!}+\frac{1}{(p-2)!} \frac{1/2}{1!}+ \cdots + \frac{1/p}{1!} = 1. \tag{4}$
Notice that such a coefficient is of the form $\frac{1}{p!}+\frac{r}{s}+\frac{1}{p}$, where $\frac{r}{s}$ is the sum of a finite number of rationals, none of which has the factor $p$ in its denominator. Therefore, if $\frac{r}{s}$ is irreducible, then $p$ doesn’t divide $s$. We then proceed by rewriting $(4)$ as
$\displaystyle \frac{1}{p!}+\frac{r}{s}+\frac{1}{p}=1. \tag*{}$
Equivalently,
$\displaystyle 1-\frac{r}{s}=\frac{1}{p!}+\frac{1}{p}=\frac{1+(p-1)!}{p!}, \tag*{}$
and
$\displaystyle (s-r) \cdot (p-1)! = \frac{s(1+(p-1)!)}{p}. \tag*{}$
Hence, $\frac{s(1+(p-1)!)}{p}$ is an integer. Moreover, since $p$ doesn’t divide $s$, $\frac{1+(p-1)!}{p}$ is also an integer. Equivalently, $(p-1)! \equiv -1 \text{ (mod } p)$, as desired.

答:
An alternative viewpoint.

Analytic part

In $\Bbb{Q}[[x]]$ we have
$$\frac1{1-x}=\prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{x^{nk}}{n^k k!}$$

Algebraic part

Let $R=\Bbb{Z}_{(p)}[x,x^p/p]+x^{p+1}\Bbb{Q}[[x]]$ the subring of formal series such that $p$ doesn't divide the denominator of the coefficients of $x^m,m<p$ and $p^2$ doesn't divide the denominator of the coefficient of $x^p$.

In the quotient ring $$R/(p,x,x^{p+1}\Bbb{Q}[[x]])\cong \Bbb{F}_p+\frac{x^p}p\Bbb{F}_p$$ we get

$$1=\frac1{1-x}=\prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{x^{nk}}{n^k k!}=(1+\frac{x^p}p)(1+\frac{x^p}{p!})=1+(1+\frac{1}{(p-1)!}) \frac{x^p}p$$
And hence $$1+\frac{1}{(p-1)!}=0\bmod p$$

Title: Analytic Proof of Wilson's Theorem


问:
Realized Value, What is it? Does it have a formula?
Your class is divided into groups with three members. If X is the random variable representing  the number of males in the group, find its realized values.</p>

答:
The realized value is the values that your random variable can take:

$$\{0;1;2;3\}$$

each value with a certain probability to be calculated.


问:
Is it possible for some results to be "more" statistically significant than others provided they both have p values less than 0.05?
Say you have two sets, and the mean of one is higher than the other. A two-sample t test yields a result of p = 0.01, indicating a significant difference in means. Another pair of sets also have different means, but a two-sample t test yields a result of p = 0.04. Would it be correct to call the results from the first pair of sets "more" statistically significant since the p value is lower?</p>

答:
Would it be correct to call the results from the first pair of sets "more" statistically significant since the p value is lower?

Yes, absolutely right. The p-value represents a measure of significance's level of the test

As a Statistical practice

if $p<5\%$ the test is significant

if $p<1\%$ the test is Highly significant

if $p>5\%$ the test is not significant


答:
Definition of limit points: Let $A \subseteq X$ Then, $x \in X$ is a limit point of $A$ iff
$$
\forall \epsilon > 0: \exists y \in A: |x - y| < \epsilon \land y \neq x
$$

That is, any open ball centered at $x$ has at least one element of $A$ different from $x$.

If we choose $\epsilon = \frac{1}{2} - \frac{1}{3}$, there is no element different from $\frac{1}{2}$ whose distance from $\frac{1}{2}$ is less than $\epsilon$. So, there cannot be a subsequence that converges to $\frac{1}{2}$.

Title: Why isn't 1/2 a limit point of sequence $X_n = 1/n$

Body:
There's one confusion I have been having regarding Limit points of a sequence.
The definition dictates that $p$ is called a limit point of a sequence $(X_n)$ if every neighborhood of $p$ contains infinitely many terms of the given sequence.

I took one example.
$$X_n = \frac{1}{n}$$

Can someone give me an example of a neighborhood of 1/2 that doesn't contain infinitely many terms of the sequence? I know this sequence only has one limit point which is 0 but I want examples as to why the other members of the sequence can't be called limit points.

Please help guys


问:
Is it a Cauchy sequence?
Let $(X,d)$ be a metric space and $(x_n)$ is a sequence in $X$. Then
 $\sup_{p\ge 1} d(x_n,x_{n+p}) \rightarrow 0$ as $n\rightarrow \infty$ implies  $(x_n)$ is a Cauchy sequence.
Any hints or counterexample to prove or disprove this implication? Thanks in advance.

答: Given $\epsilon>0$ there is $N$ such that $\sup_{p\ge 1}d(x_nx_{n+p})<\epsilon$ for all $n>N$. Then for all $m>n>N$, we have $d(x_n,x_m)\le \sup_{p\ge 1}d(x_n,x_{n+p})<\epsilon$.</p>

Title: Is it a Cauchy sequence?


问:
Linear factors of non homogeneous PDE
Is there a general rule for finding linear factor of auxiliary equation of non homogeneous partial differential equation with constant coefficients like algebraic equation.

Find Linear factors of each of the following questions.

Question 1

$(D^2 - DD' + D' - 1)z =0$

Question 2

$(D^2 - D'^2 + D + 3D' - 2)z = 0$

Question 3

$(D^2 + DD' - 6D'^2)z = 0$

Question 4

$(D^2 - D'^2 + D - D')z = 0$

My Attempt:

Auxiliary equation of question 4 is

$D^2 - D'^2 + D - D' = 0$ or $(D - D')(D + D') + (D - D') = 0$ or
$(D - D')(D + D' + 1) = 0$
So linear factors of
$D^2 - D'^2 + D - D' $
are given by

$(D - D')(D + D' + 1) $

答:
Generally the factorization can be done by inspection.

If you cannot find it that simple way, write down the expected form. For example in case (2) :
$$D^2-D'^2+D+3D'-2=(D+aD'+b)(D+cD'+d)$$
$a,b,c,d$ are constants to be determined.

$$(ac+1)D'^2+(a+c)DD'+(b+d-1)D+(ad+bc-3)D'+(bd+2)=0$$
$$\begin{cases}
ac+1=0\\
a+c=0\\
b+d-1=0\\
ad+bc-3=0\\
bd+2=0
\end{cases}$$
Solving the system for the unknown $a,b,c,d$ leads to two solutions:
$$\begin{cases}
a=1 \\
b=-1 \\
c=-1 \\
d=2
\end{cases} \quad\text{and}\quad
\begin{cases}
a=-1 \\
b=2 \\
c=1 \\
d=-1
\end{cases}$$
The both solutions are equivalent :
$$(D+D'-1)(D-D'+2)z=0\quad\text{and}\quad (D-D'+2)(D+D'-1)z=0$$

Title: Linear factors of non homogeneous PDE


问:
Is it surjective because of $K \subset H$?
Let $G$ be a group and $H,K$ normal subgroups of $G$ s.t. $K \subset H$. If I define the following map: $\phi:G /K \to G/H, gK \mapsto gH$ then the claim is that it is surjective. Is it because we have that $K \subset H$ and hence $G /H \subset G/K$?

答:
Since we aim to get a map between coset sets, but we can't avoid using coset representatives, we need to ensure that $g'\in gK\Rightarrow \phi(g'K)=\phi(gK)$ (good definition). But this is indeed the case, because by $K<H$ (see $(*)$ step):

\begin{alignat}{1}
g'\in gK &\Rightarrow g^{-1}g'\in K \\
&\stackrel{(*)}{\Rightarrow} g^{-1}g'\in H \\
&\Rightarrow g'\in gH \\
&\Rightarrow g'H=gH \\
&\Rightarrow \phi(g'K)=\phi(gK) \\
\end{alignat}

About surjectivity, notice that, by $K<H$, it is $\#G/K=[G:K]=[G:H][H:K]>[G:H]=\#G/H$ and use the pigeonhole principle.

Finally, as long as you interpret $G/H$ and $G/K$ as quotient sets (not necessarily groups), the normality of $H$ and/or $K$ is not necessary for the map to be (well defined and) surjective.


问:
$n$-th derivatives and Taylor series of $\sqrt{1-x^2}$ around $0$
I am trying to write down Taylor's series of $\sqrt{1-x^2}$ around $0$.

After few computations, what I arrived at is: if $f(x)=\sqrt{1-x^2}$, then
$$
f(0)=1, \,\,\, f'(0)=0, \,\,\, f^{(2)}(0)=-1, \,\,\, f^{(3)}(0)=1, \,\, f^{(4)}(0)=0, \ldots,
$$
So, for the derivatives at $0$, pattern looking is $(1,0,-1,1,0,-1,\ldots)$; but, I want to determine by some tools the explicit formula for $n$-th derivative at $0$. Any trick here is applicable?

答:
I suggest the following way which is probably easier than finding a closed expression for $f^{(n)}(x)$. Take $g(x) = \sqrt{1+x}$. Inspecting a couple of derivatives, you can come up with the expression
$$
g^{(n)}(x) = (1+x)^{1/2-n}\prod_{i=0}^{n-1} \left(\frac{1}{2}-i\right)
$$
which you might check by induction. The Taylor series of $g$ around $x_0=0$ is then given by
$$
\sum_{k=0}^\infty \frac{\prod\limits_{i=0}^{k-1} \left(\frac{1}{2}-i\right)}{k!}x^k.
$$
From here you get your series by replacing $x$ by $-x^2$. (Note that the empty product takes the value 1 by convention).

EDIT: From this expansion you can of course find an explicit expression for $f^{(n)}(0)$.


答:
"[S]hould I start learning philosophy of mathematics before learning basic mathematic topic like calculus?" No -- and I say this as a one-time philosophy don! If anything it is the other way around: to understand serious work on the philosophy of mathematics you already need to understand something of the mathematics which is being philosophised about.

However ...

Maybe by "philosophy" you don't so much mean philosophy as conceptual understanding. And yes, it is hugely important in getting into an area of mathematics to understand why various basics concepts of the area are defined as they are. Why, for just one example, is a topology defined as it is? As  @PrimeMover remarks, knowing the history behind a mathematical idea can be conceptually illuminating in this way. An explicitly  historical approach isn't the only one: but still, decent textbooks should indeed aim to engender conceptual understanding, to make it clear why the definitions of key notions aren't arbitrarily pulled out the air, etc.

Title: Is dealing with mathematics without understanding philosophy behind it good aproach?

Body: I was good at mathematics in high school , I can make and understand proofs , and if there is something wrong in proof I can recognise it but all of this without knowing what is the real reason , I just see that "this makes sense" .. For example no one explained to me What "assumption" or "definition" means I have seen it several times so I understood it  .. the thing is I'm confused is the way I'm dealing with mathematics correct? Does this have something to do with philosophy? And if so should I start learning philosophy of mathematics before learning basic mathematic topic like calculus ?</p>


问:
Number of right triangles formed by the diagonals of an $n$-sided regular polygon
I am trying to derive a general equation to get the number of right triangles formed by the diagonals of an $n$-sided regular polygon.

For $n\in(\text{odd}),$
this number is $0$, as we can join any two vertices of the regular polygon but they will not form the diameter of the circumscribed circle and hence according to Thales' theorem, the triangle formed by connecting that segment with any other vertex will not be a right triangle.

But, for $n\in(\text{even})$, we join the opposite vertices of the polygon. Hence we get a diameter of the circumscribable circle. Now, the polygon is divided into two parts and hence the segment can be joined to    $\frac{n}{2}-1$ vertices on either side. So the number of ways of forming the right triangle here would be ${\frac{n-1}{2}}\choose{1}$$=(\frac{n-1}{2})$ ways.

$\therefore$ The total number of right-triangles in the regular polygon of $n$-sides would be equal to $(\frac{n}{2})(\frac{n-1}{2})=\frac{n(n-1)}{4}$ ways.

Am I right in this logic and calculation?

Thanks very much!

答:
Your logic is correct but the calculation is wrong; everything is fine until "so the number of ways of forming the right triangle here would be $\dfrac {n-1}2$ ways."

There are $\dfrac n2$ ways to choose a diameter, and as you have observed, there are $\dfrac n 2 - 1$ vertices to connect to on either side. This gives:

$$\frac n2\left(\frac n2 - 1\right) \times 2 = \frac {n(n-2)}{2}$$

distinct right triangles.


问:
Would this algebra be still closed?
I have a problem understanding when an algebra is closed.

For example the angular momentum algebra closes:

$$[L_i,L_j]=\epsilon_{ijk}L_k$$

but would the algebra still be closed if, for example, there would be a change like this:

$$[L_1,L_2]=L_3 + a$$

all the other brackets unchanged?

答:
So two answers here.

Yes it would still be closed if you do one of two things: redefine the other Lie brackets to contain $L_3 +a$ instead of $L_3$ OR if you add $1$ to your Lie algebra so that you can break the sum apart. Technically $[L_3+a,L_1] = [L_3, L_1] + [a, L_1]$ requires that both $L_3$ and $1$ be in your Lie algebra. $1$ is not a standard element of the angular momentum Lie algebra ($\mathfrak{su}(2)$). Of course in a quantum mechanical world where these are operators, the Lie algebra actually exists inside of a larger structure, so maybe by borrowing from that, you can get away with not doing the above and the closure exists a fortiori. (Note that adding $L_3+a$ to your Lie algebra is the same as adding $1$ since $L_3+a-L_3 = a$.)

The other answer is: provided you have an actual Lie bracket, you can always close your Lie algebra when given a generating set. You just might not like what you end up with.. For instance, a Lie algebra closely related to yours which appears in mathematical physics is what I call a Smith type (based on the person that originally heavily studied them). It's generated by $[H, L] = -L$, $[H, R] = +R$, and $[L, R] = H^2$. At first glance this doesn't seem terrible, but using a certain trick (which I independently rediscovered almost 30 years later (no I'm not bummed out about it..)) that Smith employs, you can show that this Lie algebra, despite being finitely generated, is not finite dimensional. In other words, you have three base relations but from those you can create an infinite number of linearly independent elements in the algebra.

Title: Would this algebra be still closed?


问:
Proper definition of a function
I'm sorry if this is a pedantic question, but I want to be sure I'm using terminology correctly.

Without a second thought, I would make statements of the following form: "Consider the function $f: A \to B$ defined by $f(a) = b$."

This isn't fully correct, however, because the full definition of the function $f$ should be the domain, codomain, and the rule. Without one, the definition is ambiguous.

Am I incorrect on this, or is this a commonly accepted shorthand? Would it make more sense to say "consider the function $f: A \to B$ given by" or "governed by"?

答: You should say "with $f(a)=b$" or  "satisfying $f(a)=b$". Some people might say  "where $f(a)=b$", but I recommend "where" Immediately before definitions.

Title: Proper definition of a function


问:
Solving for matrix in multiplication of matrix by vector
I have a specific problem I have been trying to solve for some time, and can't figure out how to even start on it. (this is not for homework)
The original problem states:

Let a and b be 3D vectors. Find a 3x3 matrix R such that Ra is equal to the rejection of a onto b.

I know that the projection of a onto b is equal to

$a-\frac{(a\cdot b)}{b^2} b$

and thats basically as far as I get. I would try to divide both sides by a, but as far as I know, vector division is not possible, therefore wouldn't yield any results. I've also tried just writing out the multiplication for the first row, and trying to solve for $R_{11}$, $R_{12}$, and $R_{13}$, but that turns into a big mess really quickly that I don't think is meant to be solved.

I think there is supposed to be a relatively elegant way to solve this, but I can't see what that could be.

答:
The strategy here is to manipulate the expression for the projection of $a$ onto $b$ into the form $(\text{some matrix})a$. Note that $a \cdot b$ can be written as $b^Ta$. Also, any scalar "commutes" with a vector, so $(b^Ta)b = bb^Ta$. Hence,

$$a-\dfrac{a \cdot b}{\|b\|^2}b = a - b\dfrac{b^Ta}{\|b\|^2} = Ia - \dfrac{bb^T}{\|b\|^2}a = \left(I - \dfrac{bb^T}{\|b\|^2}\right)a.$$


问:
Linear Algebra Question
My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system

$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$

is consistent for any values of $a,b$, and $c$ that satisfy that equation.

Thanks,

答:
Perform Gaussian elimination on the system. For each row with 3 zeros on the left, the corresponding equation on the right is a restriction on $a$, $b$, $c$.

Here are the details:
$$
\begin{array}{rrrl}
2 & 1 & -1 & a \\
1 & -2 & -3 & b \\
-3 & -1 & -2 & c
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
2 & 1 & -1 & a \\
-3 & -1 & -2 & c
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 5 & 5 & a-2b \\
0 & -7 & -7 & c+3b
\end{array}
\quad\to\quad
$$
$$
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 35 & 35 & 7a-14b \\
0 & -35 & -35 & 5c+15b
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 35 & 35 & 7a-14b \\
0 & 0& 0 &7a+b+5c
\end{array}
$$
Hence we must have $7a+b+5c=0$ if the original system has a solution.

Title: Linear Algebra Question


问:
If the limit as $x$ approaches $3$ of $x^2$ is $9$, how do I use the epsilon delta definition to prove this?
If the limit as $x$ approaches $3$ of $x^2$ is $9$, how do I use the epsilon delta definition to prove this? I was trying to work it out this way but could do figure out how.</p>

答: Hint: $$|x^2-9|=|x-3||x+3|=|x-3||x-3+6|\le|x-3|^2+6|x-3|,$$ so given $\epsilon>0,$ we need only find $\delta>0$ small enough so that $|x-3|^2<\frac\epsilon2$ and $6|x-3|<\frac\epsilon2$ whenever $|x-3|<\delta$.


答:
Saying "given by", "verifying", "satisfying", etc. is more suitable for defining functions using words. Otherwise, remember that you can also use the notation:
\begin{align*}
f: & A  \longrightarrow  B \\
& a  \longmapsto f(a)=b
\end{align*}
And we'll say it's well defined if any point $a\in A$ has an image through $f$ that satisfies $f(a)\in B$, and also that it gives no contradiction, to be said: $a=a'$ implies $f(a)=f(a')$. If that's not verified, we say it's not well defined, and you may better change it's definition to make clear where does $f$ have sense. For example, the function
\begin{align*}
f: & \mathbb{R}  \longrightarrow  \mathbb{R}_+ \\
& x  \longmapsto f(x)=1/x
\end{align*}
isn't really well defined, since $x=0$ has no image, and not al images from points in $\mathbb{R}$ are positive, so a right definition of it will be
\begin{align*}
f:  \mathbb{R}\setminus\{0\} & \longrightarrow  \mathbb{R} \\
 x \ \ \ & \longmapsto  \ f(x)=1/x
\end{align*}
though sometimes domain and codomain specifications are omitted because they can be deduced by the rule or the context

Title: Proper definition of a function

Body:
I'm sorry if this is a pedantic question, but I want to be sure I'm using terminology correctly.

Without a second thought, I would make statements of the following form: "Consider the function $f: A \to B$ defined by $f(a) = b$."

This isn't fully correct, however, because the full definition of the function $f$ should be the domain, codomain, and the rule. Without one, the definition is ambiguous.

Am I incorrect on this, or is this a commonly accepted shorthand? Would it make more sense to say "consider the function $f: A \to B$ given by" or "governed by"?


问:
How many N-digit numbers have the sum of odd places digits equal the sum of even places digits?
We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.

My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.

Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.

We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.

After you find a nice way, please provide some example calculations for N=4 and N=6.

答:
This answer is intended to be simple and self-contained but please ask if it has not been expressed sufficiently clearly.

The expansion of $(1+x+x^2+...+x^9)^n$ as $\sum a_ix^i$

Consider multiplying out  $(1+x+x^2+...+x^9)(1+x+x^2+...+x^9)...(1+x+x^2+...+x^9).$

For a given integer $M$, a term $a_Mx^M$ in the product is given by the sum of terms $$x^ux^v...$$
where $x^u$ is from the first bracket, $x^v$ is from the second bracket and so on. Thus $a_M$ is precisely the number of ways of expressing the number $M$ as the sum of $n$ digits.

The sum of squares

We require $\sum a_M^2$. This is the constant coefficient in the product $$\sum a_Mx^M\sum a_Mx^{-M}$$ or, equivalently, the coefficient of $x^{9n}$ in $(1+x+x^2+...+x^9)^{2n}$

The calculation for $N=4$ i.e. n=2

We require the coefficient of $x^{18}$ in $(1+x+x^2+...+x^9)^{4}=(1-x^{10})^4(1-x)^{-4}$.

Expanding, we obtain

$$(1-4x^{10}+...)(1+4x+...+\begin{pmatrix}11\\8\\\end{pmatrix}x^{8}+...+\begin{pmatrix}21\\18\\\end{pmatrix}x^{18}+...)$$
The required number is $\begin{pmatrix}21\\18\\\end{pmatrix}-4\begin{pmatrix}11\\8\\\end{pmatrix}=670.$

The calculation for $N=6$ i.e. n=3

We now require the coefficient of $x^{27}$ in $(1-x^{10})^6(1-x)^{-6}$. This is
$$(1-6x^{10}+15x^{20}-...)(1...+\begin{pmatrix}12\\7\\\end{pmatrix}x^{7}+...+\begin{pmatrix}22\\17\\\end{pmatrix}x^{17}+...\begin{pmatrix}32\\27\\\end{pmatrix}x^{27}+...)$$
The required number is $\begin{pmatrix}32\\27\\\end{pmatrix}-6\begin{pmatrix}22\\17\\\end{pmatrix}+15\begin{pmatrix}12\\7\\\end{pmatrix}=55252.$

A general formula

Let $K=9N/2$. Then the calculation we require is just $$\begin{pmatrix}K+N-1\\K\\\end{pmatrix}-\begin{pmatrix}N\\1\\\end{pmatrix}\begin{pmatrix}K+N-11\\K-10\\\end{pmatrix}+\begin{pmatrix}N\\2\\\end{pmatrix}\begin{pmatrix}K+N-21\\K-20\\\end{pmatrix}-....$$


问:
Finding dimension of the subspace of $\mathbb{P_{3}}$
The question is that(from Bernard Kolman, Linear Algebra). Find the dimension of the subspace of $\mathbb{P_{3}}$ consisting of all vectors of the form, $at^3+bt^2+ct+d$, where $b=3a-5d$ and $c=d+4a$.

First I substitute the values that I have given in the question. That is:
$at^3+(3a-5d)t^2+(d+4a)t+d$=$a(t^3+3t^2+4t)+d(-5t^2+4t+1)$ and because we can write this as a linear combination I concluded that it spans, say $\mathbb{W}$ subspace of $\mathbb{P_{3}}$ and I wrote it in the form $\{t^3+3t^2+4t, -5t^2+4t+1\}$. After that I thought that it must be linearly independent(properties of being a basis). So,
$a_{1}(t^3+3t^2+4t)+a_{2}(-5t^2+4t+1)=0$ and then, I made a matrix:
$\left[
  \begin{matrix}
    1 & 2 \\
    3 &-5 \\
    4 & 4 \\
    0 & 2 \\
  \end{matrix}
  \left|
    \,
    \begin{matrix}
      0  \\
      0  \\
      0  \\
      0  \\
    \end{matrix}
  \right.
\right] \equiv \left[
  \begin{matrix}
    1 & 0 \\
    0 & 0 \\
    0 & 1 \\
    0 & 0 \\
  \end{matrix}
  \left|
    \,
    \begin{matrix}
      0  \\
      0  \\
      0  \\
      0  \\
    \end{matrix}
  \right.
\right]$. So one can clearly see that $a_{1}=a_{2}=0$. And finally I conclude that $\dim(\mathbb{W})=1$, but then I thought that it may $2$. So I confused. Thanks for helping me, any contribution will make me happy.

答: A basis for the space is $\{t^{3}+3t^{2}+4t, -5t^{2}+t+1\}$ and the correct dimension is $2$.


问:
Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$

I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.

答:
Here is a simple and direct way using just some rough estimates:

For $x>2$ you have surely
$$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} \stackrel{x>2}{<} \frac{5x^2 + 3x^2 + x^2}{x^3 - (x^2 +1)} = \frac{9x^2}{x^3 - (x^2 +1)}$$

Since you want to have a $10$ in the numerator, note that

$$\frac 1{10}x^3 > x^2+1 \Leftrightarrow x^2(\frac x{10}-1)>1$$
which is surely satisfied for $x>20$. Hence,

$$\frac{9x^2}{x^3 - (x^2 +1)} \stackrel{x>20}{<} \frac{9x^2}{x^3 - \frac 1{10}x^3} = \frac{10}x$$

Done.


答:
Introduction

Here is the formal definition of a function.

You need two sets $A,B$ (domain, codomain) and a subset $f$ of $A\times B$, that is, a collection of pairs of the form $(a,b)$. A pair $(a,b)$ in this set means that "$a$ is in relation to $b$". To obtain a function you need two additional constraints:

every element of $A$ is in relation to at least one element of $B$ (i.e. the function is defined on all of the domain)

an element of $A$ can be in relation to only one element of $B$ (i.e. you can compute $f(a)$ unambiguously)

If the subset $f$ of the set $A\times B$ satisfies all of these rules it's called a function and we write $f: A\rightarrow B$ and define the notation "$f(a)=b$" $\iff (a,b) \in A\times B$.

This definition makes sense if we view functions as being arrows that put in relation elements of the domain to elements of the codomain.

Answer

TLDR: the "rule of computation" is not sufficient to define a function.

TL: if you consider "$f$ is defined by the rule $f(a)=b$" to also encode the information about domain and codomain than the definition is complete.

Sometimes domain and codomain are know from the context or declared in advance and omitted.

But if you want to define a function correctly, the only way is to specify domain, codomain and rule.

Examples

Consider the rule $x\mapsto \exp(x)$. If viewed from $\mathbb{R}\rightarrow \mathbb{R}$ it defines a function that is not surjective. If viewed from $\mathbb{R}\rightarrow (0,+\infty)$ it's a function that is surjective. So you can already see that giving only the computation rule is not sufficient.

And lasly, if the same computation rule is seen from $M_{\mathbb{R}}(n,n) \rightarrow M_{\mathbb{R}}(n,n)$ then we have a totally different object ($M_{\mathbb{R}}(n,n)$ is the set of real $n\times n$ matrices. See the matrix exponential map for more details about this last example: https://en.wikipedia.org/wiki/Matrix_exponential).

Title: Proper definition of a function


问:
Estimating $p$ from $A1$ and $A1$, system of equations, an estimation problem
Does anyone here know how $\rho$ is estimated in terms of $A_1, A_2 $ in the below equations, any mathematical procedure for this ..

$$ A_1 =  \left |  \alpha\; \left(\frac{ 1- e^{-j\cdot 2\cdot\pi \rho}}{j2\pi \rho}\right)  \right|^2   $$

$$ A_2 = \left |\alpha\; \left(\frac{ 1- e^{-j\cdot 2\cdot\pi \rho}}{j2\pi+j2\pi \rho}\right) \right|^2   $$
$$ \rho=\frac{A_2+\sqrt(A_1A_2)}{A_1-A_2}  $$

I am working on an estimation problem where from $A_1 , A_2 $ they estimated $p$ that is inside $A1$ and $A2$.

Anyone who could tell me about any mathematical procedure that how they did so?

答:
According to what you wrote in comments
$$\frac{A_1}{A_2}=\frac{(\rho +1)^2}{\rho ^2}$$ which is a quadratic in $\rho$.

Solve it and select the root you need.


问:
How to fit a 2D ellipse to given points and a given focus
This question is similar to the question, How to fit a 2D ellipse (in x-y plane) to given points? (See the link below)
(https://stackoverflow.com/questions/47873759/how-to-fit-a-2d-ellipse-to-given-points)

Now, we know how to use least square method to fit a 2D ellipse with given points by the code provided by Casey. (The code is also provided below.) Based on this code, if I want to fit not only the given points but also a given focus at (0,0), how could I do it? Or is there any better way to do it?

I was thinking about if we can derive the focus (as a function of x and y) based on the ellipse equation,$  Ax^2 + Bxy + Cy^2 + Dx + Ey=1 $, where A, B, C, D, E are coefficients, and use it as a constrain to the least square method. Buts sadly, I don't know how to derive the focus either.

CODE

```<code>import numpy as np
import matplotlib.pyplot as plt
alpha = 5
beta = 3
N = 500
DIM = 2

np.random.seed(2)

# Generate random points on the unit circle by sampling uniform angles
theta = np.random.uniform(0, 2*np.pi, (N,1))
eps_noise = 0.2 * np.random.normal(size=[N,1])
circle = np.hstack([np.cos(theta), np.sin(theta)])

# Stretch and rotate circle to an ellipse with random linear tranformation
B = np.random.randint(-3, 3, (DIM, DIM))
noisy_ellipse = circle.dot(B) + eps_noise

# Extract x coords and y coords of the ellipse as column vectors
X = noisy_ellipse[:,0:1]
Y = noisy_ellipse[:,1:]

# Formulate and solve the least squares problem ||Ax - b ||^2
A = np.hstack([X**2, X * Y, Y**2, X, Y])
b = np.ones_like(X)
x = np.linalg.lstsq(A, b)[0].squeeze()

# Print the equation of the ellipse in standard form
print('The ellipse is given by {0:.3}x^2 + {1:.3}xy+{2:.3}y^2+{3:.3}x+{4:.3}y = 1'.format(x[0], x[1],x[2],x[3],x[4]))

# Plot the noisy data
plt.scatter(X, Y, label='Data Points')

# Plot the original ellipse from which the data was generated
phi = np.linspace(0, 2*np.pi, 1000).reshape((1000,1))
c = np.hstack([np.cos(phi), np.sin(phi)])
ground_truth_ellipse = c.dot(B)
plt.plot(ground_truth_ellipse[:,0], ground_truth_ellipse[:,1], 'k--', label='Generating Ellipse')

# Plot the least squares ellipse
x_coord = np.linspace(-5,5,300)
y_coord = np.linspace(-5,5,300)
X_coord, Y_coord = np.meshgrid(x_coord, y_coord)
Z_coord = x[0] * X_coord ** 2 + x[1] * X_coord * Y_coord + x[2] * Y_coord**2 + x[3] * X_coord + x[4] * Y_coord
plt.contour(X_coord, Y_coord, Z_coord, levels=[1], colors=('r'), linewidths=2)

plt.legend()
plt.xlabel('X')
plt.ylabel('Y')
plt.show()
</code>```

答:
If the ellipse has its focus at $(0,0)$ and its major axis on $x$-axis, then its equation is simply:
$$
{(x+c)^2\over a^2}+{y^2\over b^2}=1,
\quad\text{with}\quad
c=\sqrt{a^2-b^2},
$$
and you can proceed as in the usual case.

If the ellipse has a different orientation, however, then its equation gets more complicated.


问:
Why $ \int_{-\infty}^{\infty}|x| e^{-|x|} d x=-\int_{-\infty}^{0} x e^{-x} d x+\int_{0}^{\infty} x e^{x} d x $
I dont understand why the following is true?
$$
=\int_{-\infty}^{\infty}|x| e^{-|x|}  d x=-\int_{-\infty}^{0} x e^{-x}  d x+\int_{0}^{\infty} x e^{x}  d x
$$

I thought it should be $$
=\int_{-\infty}^{\infty}|x| e^{-|x|}  d x=-\int_{-\infty}^{0} x e^{- (-x)}  d x+\int_{0}^{\infty} x e^{-x}  d x
$$
I am not sure I understand why the variable in the exponent is positive in the negative interval and vice versa

答: You are right. It's easy to check this since obviously $\displaystyle \int_{-\infty }^0 e^{-x} x \, dx$ and $\displaystyle  \int_{0}^{\infty} x e^{x}  d x$ diverges but the original integral $\displaystyle  \int_{-\infty}^{\infty}|x| e^{-|x|}  d x$ converges.


问:
Let X be any set and let (Y, T*) be a topological space. Describe the smallest topology for X which makes f continuous.
Can I take coarser topology here to describe this? I need this solution.</p>

答: In the topological sense a function $f:X\mapsto Y$ is continuous if the preimage of every open set (i.e. every set $A\in T^*$) is open, which implies that the coarsest topology to make f a continuous function is $f^{-1}(T^*)$.


答: Note that $x^2-9=(x-3)(x+3)$. Given $\epsilon>0$ try $\delta=\min\{1,\frac\epsilon4\}$. Note that then $|x-3|<\delta$ implies $0<x+3<4$.</p>

Body: If the limit as $x$ approaches $3$ of $x^2$ is $9$, how do I use the epsilon delta definition to prove this? I was trying to work it out this way but could do figure out how.</p>


问:
Finding nth derivative of $\frac{1}{x^4+4}$
I am supposed to find the nth order derivative of:
$$\frac{1}{x^4+4}$$

I tried to resolve into partial fractions. But it didn't work out for me.

Edit- where I am stuck
$$\frac{1}{x^4+4}=\frac{1}{(x-1+i)(x-1-i)(x+1+i)(x+1-i)}= \frac{A}{(x-1+i)} + \frac{B}{(x-1-i)} + \frac{C}{(x+1+i)} + \frac{D}{(x+1-i)} $$
where A,B,C,D are to to be found.
I am unable to proceed further.
If I am able to find these values, the rest is easy to handle.

I am a beginner in this subject. Please help.

答:
Hint

As the roots of $p(x)=x^4+4$ are $\xi_k = \sqrt 2 e^{i(\frac{\pi}{4} + k \frac{\pi}{2})}$ where $k \in \{0, 1, 2, 3\}$ you can write

$$\frac{1}{x^4+4} = \sum_{k=0}^3 \frac{a_k}{x-\xi_k}$$

where you have to find the $a_k$ using partial fraction decomposition. Based on the formula $a_k = \frac{1}{p^\prime(\xi_k)} = \frac{1}{4\xi_k^3}$, you get

$$\frac{1}{x^4+4} = \frac{1}{16}\sum_{k=0}^3 \frac{\xi_k}{x-\xi_k}$$
Then the $n$th-derivative of $\frac{a_k}{x-\xi_k}$ is easy to find. And you're done.


问:
Checking if the complement of an undirected graph is bipartite
Is there any linear time algorithm that says if the complement of a given undirected graph is bipartite?

I've tried using König's theorem, with no success.

答:
Testing if a given graph is bipartite is easy. Informally, one just has to iterate the following for each connected component: start with any vertex in the component, color it red, then color blue the neighbors of red vertices, then color red the neighbors of blue vertices, and so on until all the edges have been used. The graph is bipartite if and only if each vertex ends up with one color. This can be done in linear time and space (it suffices to avoid coloring any vertex twice the same color).

Computing the complement of a given graph is easy. It suffices to start with the empty graph on the same set of vertices, and then add each edge $(u,v)$ that does not exist in the original graph.

Therefore, it is easy to test if the complement of a graph is bipartite.

However, making it in linear time and space is not that trivial. Indeed, if the original graph has $n$ vertices and $m$ edges, then its size is $n+m$. The complement has of the order of $n^2 - m$ edges, and so if $m$ is small, typically if $m$ is in $O(n)$, then building the complement has a $n^2$ time and space cost.

It turns out that a closely related question was posted here 6 years ago and received no satisfactory answer.


问:
How to show the equivalence of a compound and a binomial distribution in a proper manner?
I'm a student working on the master's in biology. My thesis is based on a simulation I wrote. Reading the literature I found a discrepancy between my implementation and some published implementations. To resolve this discrepancy I attempted to show that the compound random variable $X \sim Hypergeometric(N, K, n \sim Binomial(N, p))$ is equivalent to the binomial random variable $X \sim Binomial(K,p)$. Numerically this seems to hold.

This is what I managed to come up with:

On the one hand
\begin{align}
P(X=k) &= \sum_{n=k}^{N}\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\binom{N}{n}p^{n}(1-p)^{N-n} \\
&= \binom{K}{k}\sum_{n=k}^{N}\binom{N-K}{n-k}p^{n}(1-p)^{N-n}
\end{align}
on the other hand
$$P(X=k) = \binom{K}{k}p^{k}(1-p)^{K-k}.$$
Note that it's sufficient to start the summation with $n = k$ because $n \geq k$.

\begin{align}
\binom{K}{k}p^{k}(1-p)^{K-k} &= \binom{K}{k}\sum_{n=k}^{N}\binom{N-K}{n-k}p^{n}(1-p)^{N-n} \\
p^{k}(1-p)^{K-k} &= \hphantom{\binom{K}{k}}\sum_{n=k}^{N}\binom{N-K}{n-k}p^{n}(1-p)^{N-n}
\end{align}

With the help of Wolfram Mathematica I found the following:
\begin{align}
p^{k}(1-p)^{K-k} &= p^{k} (1-p)^{N-k} \left( \frac{1}{1-p}\right)^{N-K} - \frac{p^{N+1}}{1-p} \binom{N-K}{N-k+1} {}_{2}F_{1}(1,K-k+1;N-k+2;-\frac{p}{1-p}) \\
p^{k}(1-p)^{K-k} &= p^{k}(1-p)^{K-k} - \frac{p^{N+1}}{1-p} \binom{N-K}{N-k+1} {}_{2}F_{1}(1,K-k+1;N-k+2;-\frac{p}{1-p})
\end{align}

$\binom{N-K}{N-k+1}=0$ because $k \leq K$, so
$$p^{k}(1-p)^{K-k} = p^{k}(1-p)^{K-k}.$$

Is this correct?

Is this a proper presentation of a proof? Does this count as a proof at all? Should I include more assumptions? For example: $N,K,n,k$ are nonnegative integers, $p$ is a real, $N \geq K, N \geq n, N \geq k, K \geq k, n \geq k, 0 \leq p \leq 1$. Are there any conventions of notation/terminology that would make it more professional? What should be done when a step was done by Mathematica? Should I look for the appropriate identities to make the workings of that step explicit?

Are there any shortcuts/tricks that would make it more elegant?

答:
I do not know much about the Wolfram Mathematica solution. It becomes laborious.

You could rewrite

$$\sum_{n=k}^{N}\binom{N-K}{n-k}p^{n}(1-p)^{N-n}$$

as

$$\begin{array}{}\sum_{i=0}^{N-k}\binom{N-K}{i}p^{i+k}(1-p)^{N-(i+k)}
&=& \hphantom{p^k(1-p)^{K-k} }\sum_{i=0}^{N-K}\binom{N-K}{i}p^{i+k}(1-p)^{N-(i+k)+K-K} \\
 &=& p^k(1-p)^{K-k} \sum_{i=0}^{N-K}\binom{N-K}{i}p^{i}(1-p)^{N-K-i} \\
&=& p^k(1-p)^{K-k}\end{array}$$

The first equation changes the upper limit because you can not have $n-k>N-K$. The second takes a factor $p^k(1-p)^{K-k}$ outside the sum (this is actually similar to how you took the binomial coefficient outside the sum). The third equation sets the summation equal to one (you may recognize the term in the summation as a binomial distribution and the sum is over the entire range).

Intuitive

Here is an intuitive viewpoint of why the compounded distribution should equal a binomial distribution.

Imagine the following process:

You have $K$ red balls and $N-K$ blue balls. You place them randomly in a row and for each spot in the row you decide with probability $p$ to select the ball or not.

Note that the randomly distributing of the red and blue balls can be done just as well before as after deciding the spots for which the balls will be selected.

If you randomly distribute the red and blue balls afterwards then this process relates to your compound distribution:

The number of balls that you pick is similar to a Binomial distribution.

If you do it afterwards then it is just like a Hypergemetric distribution (selecting the number of picked balls out of the $K$ red and $N-K$ blue balls without replacement)

If you randomly distribute the red and blue balls before then this process relates to a binomial distribution:

For each colored ball you independently decide to pick it or not based on a Bernoulli variable. This is similar to a Binomial distribution.

Alternative relationship

When you have two Binomial distributed variables (with equal $p$) then the distribution conditional on their sum is a Hypergeometric distribution.

$$P(X = x, Y = n-x | X+Y = n) =  \frac{P(X=x)P(Y=y)}{P(X+Y = n)}$$

with $P(X=x)$, $P(Y=y)$ and $P(X+Y = n)$ each being binomial distributed variables. You will get that the power terms of $p$ and $(1-p)$ cancel and what is left are the three binomial coefficients that you have in the formula for the hypergeometric distribution.

From this, you can imagine how you could reverse this construction of the hypergeometric distribution by compounding with the conditional term $P(X+Y = n)$.

$$P(X=x) = \sum P(X = x, Y = n-x | X+Y = n)P(X+Y = n)  $$


问:
Calculating the circle between two quadratic curves
I have two quadratic curves sharing one control point $A$. The control points for the first curve are $A, B, C$ and for the second curve: $A, D, E$.

The two quadratic curves are defined as:

$P(t) = (t - 1)^2A - 2t(t - 1)B + t^2C$

$P(u) = (u - 1)^2A - 2u(u - 1)D + u^2E$

I need to find the $t$ value(s) on the curve $P(t)$ for the input value $r$ which is the radius of the circle that touches the tangent lines of both curves (note that there can be more than one circle that matches the criteria). In other words, I need to find the $t$ value where the offset curves (with offset $r$) of the two curves intersect.

I've got far enough to get the equations for the offset curves and a relation between $u$ and $t$ for the intersection points but I'm having trouble getting some sort of relation between $r$ and $t$, which will help me calculate or approximate the intersection points and the maximum radius of the circle that fits between the curves.

答:
Just a suggestion:

You can construct the offset curve of one of the Beziers (with offset $2r$) and intersect with the other. To implement this, flatten the two curves and intersect the segments pairwise (do not flatten then offset).

A brute-force solution will take effort $O(nm)$ for $n$ and $m$ segments in the respective polylines, but you can reduce this by plane-sweep techniques (Bentley-Ottmann).


问:
Question on the base of a topology
So I am new to topology and I've been trying to understand the concept of a basis and am a bit stuck on this statement:
The topology $\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$ on the set $\{a,b\}$ has the following bases:

1.$\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$

2. $\big\{\{a\},\{b\},\{a,b\}\big\}$

My confusion arises in the 2nd base how can $\big\{\{a\},\{b\},\{a,b\}\big\}$  be a base for the topology since no unions of subsets give $\varnothing$?.

Thanks in advance.

答: The union of an empty collection is defined to be empty. Hence, the empty set is a union of members of any family of sets.</p>

Title: Question on the base of a topology


问:
Need help with proof: $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$
As the title says I am trying to prove $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$.

My steps so far:

$$a \mathbb Z + b \mathbb Z = \{ a z_1 + bz_2 : z_1, z_2 \in \mathbb Z \} = \{ kp z_1 + lpz_2 : z_1, z_2 \in \mathbb Z \}$$

where $p=gcd(a,b)$, $kp=a$ and $lp=b$.

$$\ldots = \{ p \cdot (k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \}$$

So far, so good. What remains to show is that

$$\{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = \mathbb Z $$

since then it would follow that

$$p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \mathbb Z $$

but here I am stuck. I suspect it has something to do with the fact that $k$ and $l$ are coprime? Maybe there is a theorem about linear combinations of coprime numbers that somehow generates all integers? In any case, help is much appreciated!

答:
I think the most direct way to show this is by dual inclusion, as this is set equality after all. Let $d$ be the gcd of $a,b$. We want to show that
$$
d\mathbb{Z} = a\mathbb{Z} + b\mathbb{Z}
$$
Fix some $ax+by \in a\mathbb{Z} + b\mathbb{Z}$. Since $d|a$ and $d|b$ we have that:
$$
ax+by = da'x+db'y  = d(a'x+b'y) \in d \mathbb{Z}
$$
for some $a',b' \in \mathbb{Z}$. Thus we have that $a\mathbb{Z} + b\mathbb{Z} \subseteq d \mathbb{Z}$.

Now fix some $dx \in d\mathbb{Z}$. By Bezout's identity, we know that there exists some $u,v \in \mathbb{Z}$ such that $ua+vb = d$, so we have that:
$$
dx = (ua+vb)x = (ux)a + (uv)b \in a\mathbb{Z}+b\mathbb{Z}
$$
so we have that $d\mathbb{Z} \subseteq a\mathbb{Z}+b\mathbb{Z}$ and thus the equality of sets.


问:
intermediate value theorem convex combination
Let $f \colon [a,b] \to \mathbb{R}$ be continuous, $p,q>0$. We want to show that there is a $c\in [a,b]$ s.t. $pf(a)+qf(b)=(p+q)f(c)$.
$\frac{p}{p+q}, \frac{q}{p+q}$ are both in $(0,1)$. Then
$\frac{p}{p+q} f(a) + \frac{q}{p+q} f(b) $ is in $(0, f(a)+f(b))$.
If we want to apply the intermediate value theorem we better would need it to be in $[f(u), f(w)]$, where we know that $f(u)$ is the Minimum and $f(w)$ is the Maximum.
But we have not any info about what the Min/Max would be?

答: By the intermediate value theorem, $f([a, b]) \subset \mathbb R$ is an interval, which is always a convex subset. Thus $\frac{p}{p + q} f(a) + \frac{q}{p + q} f(b) \in f([a, b])$ by convexity as $f(a), f(b) \in f([a, b])$, so there exists a $c \in [a, b]$ such that $f(c) = \frac{p}{p + q} f(a) + \frac{q}{p + q} f(b)$.


问:
Number of conditional types
Let $A = \{1, 2, ..., J \}$ and $B = \{1,2, ..., K \}$.

A sequence $x^n \in A^n$ induces a probability distribution $t$ called the type where
\begin{align}
   t(j) = \frac{1}{n} |\{x_i:x_i = j \}| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ for all } j \in A
\end{align}

Two sequences $x^n \in A^n$ and $y^n \in B^n$ induce a joint probability distribution on $A \times B$ called the joint type call it $s$ given by
\begin{align}
   s(j, k) &= \frac{1}{n} |\{(x_i, y_i): (x_i, y_i) = (j, k) \}|
\end{align}

We can also define a conditional type of $y^n$ given $x^n$ call it $W$ by
\begin{align}
 W(k|j) &= \frac{s(j, k)}{t(j)}
\end{align}
Let's ignore the case when $t(j) = 0$. It might be irrelevant to the question I want to ask.

The number of types on $A$ is exactly equal to $$\binom{n + J - 1}{J-1}$$

Let's fix $x^n$ of type $t$. What is the number of conditional types $W$ given $x^n$ of type $t$ ? Any lower or upper bounds would help me also .

答:
The question is when $j$ is fixed, how many types $W(\cdot|j)$ are there ? given that there are exactly $nt(j)$ element of $x$ that are equal to $j$ by a dot and bar counting argument similar to what you did, there are exactly $nt(i)+B-1\choose B-1$ such types. So now to compute the total amount of types you just need to multiply all those :
\begin{align*}
\prod_{j=1}^J{nt(j)+B-1\choose B-1}
\end{align*}

Title: Number of conditional types


问:
How to define this pattern as $f(n)$
Given a binary table with n bits as follows:

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &0&0&0&1 \\ &0&0&1&2 \\ &0&1&0&3 \\ &0&1&1&4 \\ &1&0&0&5 \\ &1&0&1&6 \\ &1&1&0&7 \\ &1&1&1&8
\end{array}
$$

If I replace each $0$ with a $1$ and each $1$ with $g(n)$ as follows

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &1&1&1&1 \\ &1&1&g(0)&2 \\ &1&g(1)&1&3 \\ &1&g(1)&g(0)&4 \\ &g(2)&1&1&5 \\ &g(2)&1&g(0)&6 \\ &g(2)&g(1)&1&7 \\ &g(2)&g(1)&g(0)&8
\end{array}
$$

I want to define a $$f(n) = \sum_{m=1}^{2^n}\prod_{i=1}^n row_mcol_i$$

I'm not sure if I wrote the above equation correctly but in words it is the following, for each row multiply the columns values together and add it to the result of the previous row.

e.g from the table above we would have the following:

$$f(3) = 1 + g(0) + g(1) + g(0)g(1) + g(2) + g(2)g(0) + g(2)g(1) + g(2)g(1)g(0)$$

Any help in generalising the equation for $f(n)$ would be much appreciated. The tables above are merely to help illustrate the pattern, if possible I would prefer not to reference the rows and columns of some matrix.

Thanks in advance.

答:
This is just $$(1+g(0))(1+g(1))(1+g(2))\dots(1+g(n-1))$$

This could be written as:

$$\prod_{k=0}^{n-1} (1+g(k))$$

Your expression could be written by defining $G(m,k)$ as $1$ if $m$ has zero in the $k$th bit and $g(k)$ otherwise.  Then you want:

$$\sum_{m=0}^{2^n-1}\prod_{k=0}^{n-1} G(m,k)$$

But this simplifies to my above expression.

An alternative formulation is to define $[n]=\{0,1,\dots,n-1\}$ and then you can write your formula as:

$$\sum_{S\subseteq [n]} \prod_{k\in S} g(k)$$

This uses the fact that the numbers from $0$ to $2^n-1$ encode the subsets of $[n]$.

Title: How to define this pattern as $f(n)$


问:
Composite of two purely inseparable extensions is purely inseparable.
Let $L,F$ be extensions of the field $K$ and are contained in a common field. Prove that, if $L$ and $F$ are purely inseparable extensions over $K$ then $LF$ is also a purely inseparable extension over $K$. Is the converse true?

How do I prove it?

Thanks in advanced.

答:
Proposition:

The class of purely inseparable extensions is distinguished.

Proof. Let $k\subset F\subset E$. Suppose that $E/k$ is purely inseparable. Every element of $F$ lies in $E$ and is therefore purely inseparable, so $F/k$ is purely inseparable. The minimal polynomial of any $\alpha\in E$ over $F$ divides its minimal polynomial over $k$, which is of the form $x^{p^n}-a$, which factors in $k^a$ into $(x-\alpha)^{p^n}$. A polynomial of form $(x-\alpha)^{p^kr}$ where $r$ is prime to $p$ may be written as $(x^{p^k}-\alpha^{p^k})^r$, and upon expanding the binomial, one sees the coefficient of the $(x^{p^k})^{r-1}$ term is $r\alpha^{p^k}=0$. Since $r\ne 0$, we must have $\alpha^{p^k}=0$, so any irreducible factor of $x^{p^n}-a$ must be of the form $x^{p^k}-a$. Hence the minimal polynomial of $\alpha$ over $F$ is of this form, and $E$ is then purely inseparable over $F$.

Conversely, suppose that $F/k$ is purely inseparable and $E/F$ is purely inseparable. Then for any $\alpha\in E, a=\alpha^{p^n}\in F$ for some $n$, and then $a^{p^m}=(\alpha^{p^n})^{p^m}=\alpha^{p^{n+m}}\in E$ for some $m$. Thus $E/k$ is purely inseparable.

If $K/k$ and $L/k$ are extensions of $k$ and $K$ is purely inseparable over $k$, then $KL$ is generated over $L$ for a set of elements in $K$, and thus they are purely inseparable over $L$. Then $KL$ is purely inseparable over $L$.

Using the towers $K\subset L\subset LF, K\subset F\subset LF$ and the above proposition, it follows that $LF/K$ is purely inseparable iff $F/K$ and $L/K$ are both purely inseparable.


答: $\frac{p}{p+q} f(a) + \frac{q}{p+q} f(b) $ is also in $[m,M]$ where $m$ is the minimum and $M$ is the maximum.  For example, $f(a) \leq M$ and $f(b) \leq M$ implies $\frac{p}{p+q} f(a) + \frac{q}{p+q} f(b)\leq  \frac{p}{p+q} M+ \frac{q}{p+q} M=M$ and similarly for the minimum.

Title: intermediate value theorem convex combination

Body:
Let $f \colon [a,b] \to \mathbb{R}$ be continuous, $p,q>0$. We want to show that there is a $c\in [a,b]$ s.t. $pf(a)+qf(b)=(p+q)f(c)$.
$\frac{p}{p+q}, \frac{q}{p+q}$ are both in $(0,1)$. Then
$\frac{p}{p+q} f(a) + \frac{q}{p+q} f(b) $ is in $(0, f(a)+f(b))$.
If we want to apply the intermediate value theorem we better would need it to be in $[f(u), f(w)]$, where we know that $f(u)$ is the Minimum and $f(w)$ is the Maximum.
But we have not any info about what the Min/Max would be?


答:
Sometimes Bezout's identity is derived from the equality $a\Bbb Z+b\Bbb Z =\gcd(a,b)\Bbb Z$, so here here is a proof that avoids that. (Of course it is very similar arguing and if you already proved Bezout's identity, e.g. using the euclidean algorithm, then the other solution is probably better).

By the euclidean algorithm $\Bbb Z$ is a PID, hence there is some $c\in\Bbb Z$ such that $$a\Bbb Z+b\Bbb Z=c\Bbb Z$$ We want to show that $c=\varepsilon\gcd(a,b)$ for some unit $\varepsilon\in\{-1,1\}$. Since $a,b\in c\Bbb Z$ we see $c\mid a,b$ so $c$ is a common divisor of $a,b$. We need to verify the universal property in the sense that for any other common divisor $d\mid a,b$ we have $d\mid c$. So let $d\mid a,b$. Then $a\Bbb Z,b\Bbb Z\subseteq d\Bbb Z$. Therefore $$c\Bbb Z=a\Bbb Z+b\Bbb Z\subseteq d\Bbb Z$$ This implies $d\mid c$. Hence $c$ is a greatest common divisor of $a,b$ and the claim follows.

Title: Need help with proof: $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$

Body:
As the title says I am trying to prove $a \mathbb Z + b \mathbb Z = gcd(a,b) \mathbb Z$.

My steps so far:

$$a \mathbb Z + b \mathbb Z = \{ a z_1 + bz_2 : z_1, z_2 \in \mathbb Z \} = \{ kp z_1 + lpz_2 : z_1, z_2 \in \mathbb Z \}$$

where $p=gcd(a,b)$, $kp=a$ and $lp=b$.

$$\ldots = \{ p \cdot (k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \}$$

So far, so good. What remains to show is that

$$\{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = \mathbb Z $$

since then it would follow that

$$p \{(k z_1 + l z_2) : z_1, z_2 \in \mathbb Z \} = p \mathbb Z $$

but here I am stuck. I suspect it has something to do with the fact that $k$ and $l$ are coprime? Maybe there is a theorem about linear combinations of coprime numbers that somehow generates all integers? In any case, help is much appreciated!


答:
Define a function $x \mapsto h(x) = f(x)-\frac{p}{p+q}f(a)-\frac{q}{p+q} f(b)$. Observe that this is continuous in $[a,b]$. Then:

$$h(a) = \frac{q}{p+q} \left(f(a)-f(b) \right)$$

$$h(b) = \frac{p}{p+q} \left(f(b)-f(a) \right)$$

If $f(a) = f(b)$, then we can let $c = a$ and we are done. Suppose that $f(a) \neq f(b)$. Then, $h(a) \cdot h(b) < 0$ so by the Intermediate Value Theorem:

$$\exists c \in [a,b]: h(c) = 0$$

$$\implies f(c) = \frac{p}{p+q} f(a)+\frac{q}{p+q} f(b)$$

$$\implies p f(a)+q f(b) = (p+q) f(c)$$

as was desired. $\Box$


答:
Take the table to be a matrix $A\in \mathbb{R}^{m\times n}$.

$$f(n)=\sum_{k=1}^m\prod_{l=1}^ng(l)^{A_{kl}}$$

Title: How to define this pattern as $f(n)$

Body:
Given a binary table with n bits as follows:

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &0&0&0&1 \\ &0&0&1&2 \\ &0&1&0&3 \\ &0&1&1&4 \\ &1&0&0&5 \\ &1&0&1&6 \\ &1&1&0&7 \\ &1&1&1&8
\end{array}
$$

If I replace each $0$ with a $1$ and each $1$ with $g(n)$ as follows

$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &1&1&1&1 \\ &1&1&g(0)&2 \\ &1&g(1)&1&3 \\ &1&g(1)&g(0)&4 \\ &g(2)&1&1&5 \\ &g(2)&1&g(0)&6 \\ &g(2)&g(1)&1&7 \\ &g(2)&g(1)&g(0)&8
\end{array}
$$

I want to define a $$f(n) = \sum_{m=1}^{2^n}\prod_{i=1}^n row_mcol_i$$

I'm not sure if I wrote the above equation correctly but in words it is the following, for each row multiply the columns values together and add it to the result of the previous row.

e.g from the table above we would have the following:

$$f(3) = 1 + g(0) + g(1) + g(0)g(1) + g(2) + g(2)g(0) + g(2)g(1) + g(2)g(1)g(0)$$

Any help in generalising the equation for $f(n)$ would be much appreciated. The tables above are merely to help illustrate the pattern, if possible I would prefer not to reference the rows and columns of some matrix.

Thanks in advance.


问:
If F is normal over E and E is normal over K then F need not be normal over K
The following question was given as homework by the prof. and I am not able to deduce it. So, I am asking for help here.

If F is normal over E and E is normal over K then  prove that F  need not be normal over K.

Prof. also gave hint to consider $\mathbb{Q} ((2)^{1/4})$ , $\mathbb{Q}(\sqrt(2))$ and $\mathbb{Q}$.

I proved that $\mathbb{Q} ((2)^{1/4})$ is normal over  $\mathbb{Q}(\sqrt(2))$ and $\mathbb{Q}(\sqrt(2))$ is normal over $\mathbb{Q}$ but I am not able to prove that $\mathbb{Q} ((2)^{1/4})$ is nt normal over $\mathbb{Q}$.

My attempt for what I am not able to prove, If I assume $x^4-2 \in \mathbb{Q}$ then $(2)^{1/4}$ lies in $\mathbb{Q} ((2)^{1/4})$ and all the other roots of the equation $x^4-2=0$  also lie in $\mathbb{Q} ((2)^{1/4})$. So, It is normal in $\mathbb{Q}$.

which is wrong . SO, Can you please tellme what is wrong in my argument and what would be the correct argument.

答:
$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2})$.
Thus there are two real roots  $\pm\sqrt[4]{2}$ and two complex roots $\pm\sqrt[4]{2}i$.

Note that $\Bbb{Q}(\sqrt[4]{2})$ is the smallest field containing $\sqrt[4]{2}$ and $\Bbb{Q}$.

Therefore $\Bbb{Q}(\sqrt[4]{2})$ is not the splitting field of the polynomial $x^4-2$.

The extensions$\Bbb{ Q}(\sqrt[4]{2})/\Bbb{Q}(
\sqrt{2})$ and $\Bbb{Q}(
\sqrt{2})/\Bbb{Q }$are normal but the extension $\Bbb{Q}(
\sqrt[4]{
2})/\Bbb{Q}$is not a normal extension since the complex roots of
the polynomial $x^4-2$ are not in $\Bbb{Q}(\sqrt{2}).$

It is not hard to see that $ \Bbb{Q}(\sqrt[4]{2},\sqrt[4]{2}i)$ is the splitting field of the said polynomial.


问:
How to find perimeter of right angle triangle given area and three angles

Hello everyone,

I’ve come across a question where I need to find the perimeter of a right angle triangle given its area and three sides (the only angle written in the picture is 40 degrees, but the other must be 50 degrees given that it is a right triangle). I honestly am at a complete loss at how to solve this, I’d appreciate any help.

Thanks!

答:
Since its a right-angled triangle, follow these 4 steps to get the solution:

Step-1: Angle is given use the Trigonometric relation of Tangent:

```<code>tan(40) = Height/Base ------eq(1)
</code>```

Step-2: Triangle area formula i.e.

```<code>1/2 * Height * Base = 20 ------eq(2)
</code>```

Now you can substitute one unknown variable from eq(1) to the eq(2) or vice-versa to obtain both the Base and Height of the triangle.

Step-3: Use the Pythagoras theorem

```<code>Hypotenus^2 = Height^2 + Base^2
</code>```

Step-4: Perimeter = Hypotenus + Height + Base

Body:
Since its a right-angled triangle, follow these 4 steps to get the solution:

Step-1: Angle is given use the Trigonometric relation of Tangent:

tan(40) = Height/Base ------eq(1)

Step-2: Triangle area formula i.e.

1/2 * Height * Base = 20 ------eq(2)

Now you can substitute one unknown variable from eq(1) to the eq(2) or vice-versa to obtain both the Base and Height of the triangle.

Step-3: Use the Pythagoras theorem

Hypotenus^2 = Height^2 + Base^2

Step-4: Perimeter = Hypotenus + Height + Base


问:
REVISTED$^1$: Circumstantial Proof: $P\implies Q \overset{?}{\implies} Q\implies P$
To prove that if a matrix $A\in M_{n\times n} ( F )$ has $n$ distinct eigenvalues, then $A$ is diagonalizable is enough to show that the opposite holds? That is, if $A$ is diagonalizable, then $A$ has distinct eigenvalues?

Please don't answer the actual question. I want to do it myself. I just want to know if it suffice to show the opposite in this particular case.

EDIT$^1$:

If $A$ is similar to a diagonlizable matrix, then $A$ is diagonalizable, right?

答:
No. It would only prove the converse.

$$\underbrace{P \implies Q}_{\text {implication}} \quad\not\equiv \underbrace{Q \implies P}_{\text{converse of implication}}$$

If you need to prove $P \implies Q$, you can prove its equivalent:

$$\underbrace{\lnot Q \implies \lnot P}_{\text{contrapositive of implication}}$$


问:
In how many ways can $6$ different candies be divided between three children?
I am the freshman in Computer Science at university, and currently struggling with some tasks from the Combinatorics part of the Discrete Math class. Any help would be appreciated.

In how many ways can $6$ different candies be divided between three children?

My vision of solving this problem:

For every candy there is $3$ children to give it to $\implies 3^6$ possible ways of candy distribution

OR

Each child can get from $0$ to $6$ candies:
$\implies 7$ possible distributions $\implies 7^3$ possible ways of candy distribution between the $3$ children?

But I feel like I am missing something which leads to the duality of my solutions. (Probably that the candies are different?) And I don't understand how to count that in.

答:
Your first solution is correct. There are three possible recipients for each of the six candies, so there are $3^6$ possible distributions of the candies.

If you wish to consider how many candies each child receives, you need to consider cases based on partitions of $6$ into at most three parts.
\begin{align*}
6 & = 6\\
  & = 5 + 1\\
  & = 4 + 2\\
  & = 4 + 1 + 1\\
  & = 3 + 3\\
  & = 3 + 2 + 1\\
  & = 2 + 2 + 2
\end{align*}

One child receives six candies:  There are three ways to select the recipient of all six candies.

One child receives five candies and another child receives one candy:  There are three ways to select the child who will receive five candies, $\binom{6}{5}$ ways to select which five of the six candies that child receives, and two ways to choose the child who receives the remaining candy.  Hence, there are
$$\binom{3}{1}\binom{6}{5}\binom{2}{1}$$
such distributions.

One child receives four candies and another child receives two candies:  There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose the child who receives the remaining two candies.  Hence, there are
$$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$
such distributions.

One child receives four candies and each of the other children receive one candy each:  There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose which of the remaining two candies the younger of the two remaining children receives.  The other child must receive the remaining candy.  Hence, there are
$$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$
such distributions.

Two children each receive three candies: There are $\binom{3}{2}$ ways to select which two children receive three candies each and $\binom{6}{3}$ ways to select which three of the six candies the younger of the two selected children will receive.  The other selected child must receive the three remaining candies.  Hence, there are
$$\binom{3}{2}\binom{6}{3}$$
such distributions.

One child receives three candies, a second child receives two candies, and the third child receives one candy: There are three ways to select the child who will receive three candies, $\binom{6}{3}$ ways to select which three candies that child will receive, two ways to decide which of the remaining children will receive two candies, and $\binom{3}{2}$ ways to select which two of the remaining three candies that child will receive.  The remaining child must receive the remaining candy.  Hence, there are
$$\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}$$
such distributions.

Each of the three children receives two candies:  There are $\binom{6}{2}$ ways to select which two candies are received by the youngest child and $\binom{4}{2}$ ways to select which two of the remaining four candies are received by the next youngest child.  The oldest child must receive the two remaining candies.  Hence, there are
$$\binom{6}{2}\binom{4}{2}$$
such distributions.

Total:  Since the above cases are mutually exclusive and exhaustive, the number of ways of distributing six different candies to three children is
$$\binom{3}{1} + \binom{3}{1}\binom{6}{5}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{2}\binom{6}{3} + \binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2} + \binom{6}{2}\binom{4}{2} = 3^6$$


问:
Does the subspace topology of rationals have the cofinite topology?
Just to be clear, we are talking about the subspace topology of all rational numbers as a subset of $\bf{R}$. The answer is clearly no right? Because every open set in Q is the intersection of open intervals in $\bf{R}$ with Q and their complement are the the remaining rationals which are infinite.

答: Yes. With the usual (standard) topology on $\Bbb R,$ the interval $(0,1)$ is open in $\Bbb R.$ So $\Bbb Q\cap (0,1)$ is not empty and is not co-finite in $\Bbb Q$ but it is open in $\Bbb Q.$


问:
Newton's Method on $f(x)=\arctan(x)$ with oscillation
Consider the function $f(x)=\arctan(x)$ defined on $\mathbb{R}$. Let $x_0\neq0$ be an initial guess for the root of $f$ and apply Newton's Method with $x_0$. Further assume that the iterations produce an oscillating sequence $(x_n)_n=(x_0,x_1,x_0,x_1,...)$ with period $2$ and $x_0,x_1$ as repeating terms.

Problem: How to prove that $x_0$ and $x_1$ are roots of the function $g(x):=\frac{2x}{x^2+1}-f(x)$ ?

答:
As $g(x)$ is an odd function, if $x_0$ is a root, then $x_1=-x_0$ is a root as well.

The periodic sequence is

$(x_n)=x_0,-x_0,x_0,\ldots$

Apply Newton method to find the root of $f(x)=\arctan x$

$x_1=x_{0}-\frac{f(x_{0})}{f'(x_{0}}= -x_0$

$x_1=-x_0$ so we get

$-x_0=x_0-\frac{\arctan x_0}{\frac{1}{x_0^2+1}}$

expand

$2 x_0=\left(x_0^2+1\right) \arctan x_0$

Which proves that $\pm x_0$ are roots of $g(x)=\frac{2 x}{x^2+1}-\arctan x$

$x_0\approx 1.39175$

Hope this helps

Edit

They say

$x_0,x_1$ form a periodic sequence when used in Newton method to find roots of $\arctan x$. Show that exactly those two values are roots of $g(x)$.

I say: if they are roots of $g(x)$ then $x_1=-x_0$ because $g(x)$ is odd (as shown in the image below). Then I prove that those values are exactly the values found in the first part.

$$...$$


问:
Derivation of Hamilton’s equations
The general derivation of Hamilton’s equations involve the change in the hamiltonian and consequently the change in the Lagranian that is a function of q and q_dot, L(q,q_dot). This is shown in this simple video https://www.youtube.com/watch?v=jXu6zIItnLM

When deriving Hamilton’s equations for q_dot and p_dot, the lagranian is only a function of q and q_dot but not time. Therefore the Lagranian used to derive Hamilton’s equations is time-independent as it appears. Does that means that Hamilton’s equations for q_dot and p_dot are only right for systems for which the lagranian is time-independent (and therefore the hamiltonian- or energy is conserved) ?

答:
Lagrange movement equations $(n)$ extracted from $L(q,\dot q,t)$ are equivalent to the $2n$ first order equations

$$
\cases{
\dot p = -H_q\\
\dot q = H_p
}
$$

with $H(p,q, t) = p\dot q-L(q,\dot q,t)$. This last equation is known as  the Legendre transform. Here $L(q,\dot q,t)$ and $H(p,q, t)$ are respectively the Lagrangian and the Hamiltonian. Both can be non autonomous.

Title: Derivation of Hamilton’s equations


问:
Understanding homotopy equivalence in a special case
I'm beginning to learn algebraic topology from Hatcher, and I'm trying to get an intuitive grasp on homotopy equivalence. Hatcher says these two shapes are homotopy equivalent:

Let's call the left one A and the right one B. I can easily see visually how to "stretch" A to get B, but I'm having trouble understanding how to apply the actual, formal definition of homotopy equivalence.

Specifically, we need continuous functions $f : A\to B$ and $g : B\to A$ such that $fg$ is homotopic to $id_B$ and $gf$ is homotopic to $id_A$. But I can't really see how to do this. For $g$, we can map the whole vertical bar of $B$ (including the endpoints) to the point at the center of $A$, and the other parts of $B$ besides the vertical bar to the round parts of $A$ in the obvious way. This is continuous. But what can we do for a continuous $f$, such that the compositions $fg$ and $gf$ are homotopic to the identity maps?

答:
Let $S(p)$ denote the plane circle with radius $1$  and center $p \in \mathbb R^2$ and $D = \{0\} \times [-1,1]$. Let $S_l(p)$ and $S_r(p)$ denoet the left and right closed halfcircle of $S(p)$. Let $p^- = (-1,0), p^+ = (1,0)$ and $p^0 = (0,0)$.

Then
$$A = S(p^-) \cup S(p^+) , B = S(p^0) \cup D .$$
Define $f : A \to B$ by projecting $S_r(p^-)$ and $S_l(p^+)$ horizontally to $D$ and by translating $S_l(p^-)$ horizontally to $S_l(p^0)$ and translating $S_r(p^+)$ horizontally to $S_r(p^0)$.

Define $g : B \to A$ by contracting $D$ to $p^0$ and mapping $S_l(p^0)$ in the obvious way to $S(p^-)$ and $S_r(p^0)$ to $S(p^+)$.

The explicit construction of homotopies $gf \simeq id_A$ and $fg \simeq id_B$ is somewhat tedious, but I am sure you can see how that works. The essence is to understand that the identity $id_S$ on a circle $S$ is homotopic to map $\phi : S \to S$ contracting a closed half circle $H  \subset S$ to a point $h \in H$. This homotopy keeps $h$ fixed and "stretches" the open halfcircle $S \setminus H$ to $S \setminus \{h\}$.


问:
Combinatorics - prove white can always force a win or draw in double chess
The game of double chess is played like regular chess, except
each player makes two moves in their turn (white plays twice,
then black plays twice, and so on). Show that white can always
win or draw.

Credit: Olympiad Combinatorics - Pranav Sriram

EDIT:
I tried a few thing's like finding the specific strategy, but nothing seems to work. I dont know how to approach this question.

答:
This is specific to chess. Hint: At the start, there is a chess piece that you can move twice in the beginning to return to the starting board position. Effectively passing your turn.

We show white (the first player) has a non-losing strategy. Suppose not, then there is a winning strategy for black (the second player). But as white, at the very beginning you can move a certain chess piece twice, and still have the same starting chess configuration. Thus white effectively becomes the second player and having a winning strategy, a contradiction.

(I'll let you figure out which chess piece this is.)


问:
Exercise: The measure of the boundary of a level set of a Holder continuous function is positive

Let $M$ be a compact manifold in $R^3$ and consider a non negative
function $f\in C^{1,\alpha}(M)$ for some $0<\alpha<1$. I am interested
in showing that $|\partial \{f>0\} |>0$, where $|\cdot|$ stands for
the Hausdorff measure and $\partial A$ denotes
the boundary of $A$.

My Attempt:

By definition, $x^* \in M$ is a boundary point of $\{f>0\}$, if for every $r>0$ it holds that $B_r(x^*) \cap \{f>0\} \neq \emptyset$ and $B_r(x^*) \cap \{f=0\} \neq \emptyset$. In other words, if for every ball with center at $x^*$, I can find not only points of $\{f>0\}$ but also of its complement. Now, it is also true that $M=\{f>0\}\cup \{f=0\} $. Hence, if I could find/construct a dense subset of $\{f>0\}\cup \{f=0\}$, let's say $D$, I could obtain that $B_r(x^*) \cap D \neq \emptyset$ by density. This in particular would yield that $B_r(x^*) \cap (\{f>0\}\cup \{f=0\})=(B_r(x^*) \cap \{f>0\})\cup (B_r(x^*) \cap\{f=0\}) \neq \emptyset$. At this point, two issues arise:

The latter does not imply that both intersections must be non empty, but only at least one between them.

I am not sure if such a dense set exists.

DISCLAIMER: I have seen some similar question here in which the answer is given by a Cantor set. However, since I am not that familiar with topology, I would prefer to prove this (if possible) using the "slightest" tools.

I feel that density is a key part here but I am not able to tackle this problem properly. Any help or hint will be much appreciated!

Thanks in advance.

答:
I do not understand why the dense set would help you in this situation as nonempty sets can still have zero Hausdorff measure. In this answer I will denote the $2$-dimensional Hausdorff measure by $\mathcal{H}^2$ (using the absolute value seems very misleading to me). In the comments I gave some heuristics. Let me back this up by something a bit more rigorous.

As I pointed out in the comments it depends on the desired notion of boundary (either withing the manifold with the subspace topology or in $\mathbb{R}^3$). As you are using $\{ f> 0 \}^c = \{ f=0 \}$, you seem to take the boundary within the manifold. In this case, your claim is not true. Consider
$$\mathbb{T}^2 = \{ (x,y,z) \in \mathbb{R}^3 \ : \ (x^2+y^2-2)^2+z^2 =1 \}$$
(some 2-dimensional torus) embedded in $\mathbb{R}^3$. This is a compact, smooth manifold. Now consider the map $f: \mathbb{T}^2 \rightarrow \mathbb{R}, f(x,y,z)=z^2$. Then $f$ is a smooth, nonnegative map. However, the boundary of the zero locus is equal to two circles (1-dimensional manifolds) and thus, the $2$-dimensional Hausdorff measure will vanish.

If you were to pick the boundary in the topology of $\mathbb{R}^3$, (meaning, the balls should intersect $\{ f>0 \}$ and $\{ f>0 \}^c$), then you get $\{f>0 \} \subseteq \partial \{ f>0 \}$ (that is because any ball in $\mathbb{R}^3$ intersects $M^c$). However, $\{ f>0 \}$ is an open subset of a $2$-dimensional, $C^{1,\alpha}$ manifold and thus is a $2$-dimensional $C^{1,\alpha}$ manifold itself. However, nonempty $2$-dimensional $C^{1,\alpha}$ manifolds have positive $2$-dimensional Hausdorff measure. As $M$ is embedded, for every point $p\in M$ there exists a diffeomorphism $\varphi: B_1(0) \rightarrow \mathbb{R}^3$ such that $$\varphi((-1,1)^3) \cap \{ f>0\} = \{ \varphi(x,y,0) \ : \ (x,y,0)\in (-1,1)^3 \}$$
and such that $\varphi(0)=p$. By the monotonicity of measures, we get
$$ \mathcal{H}^2( \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})) \vert \leq \mathcal{H}^2( \{ f>0 \} ) \leq \mathcal{H}^2( \partial \{ f>0 \} ). $$
However, as $\varphi$ is $C^1$, we have that its restriction to $[-0.5,0.5]^3$ is bilipschitz (lets say with the inverse having lipschitz constant equal to $L$). By Measure of image of Lipschitz function is bounded? (the $C^{1,\alpha}$ regularity actually would allow you for a simpler proof of this fact, in fact we do not really need $M$ to be more than a topological manifold, see the next paragraph) we have
$$ 0<1 = \mathcal{H}^2([-0.5,0.5] \times [-0.5,0.5]\times \{0\})=\mathcal{H}^2(\varphi^{-1} \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\}))
\leq L^2 \mathcal{H}^2(\varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})). $$
Hence, we get
$$ 0< 1/L^2 \leq \mathcal{H}^2(\partial \{ f>0\}). $$

Added: The $C^{1, \alpha}$ assumption is not really need. Everything works in the topological setting if your manifold is locally given as a graph. Then we can use the projection map. I.e. we have
$$ \mathcal{H}^2((-1,1)^2)=\mathcal{H}^2(P(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})) \leq \mathcal{H}^2(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})  $$
where $g$ is any map (we need no regularity, only the fact that we can write it as a graph) and $P: \mathbb{R}^3\rightarrow \mathbb{R}^2, (x,y,z)=(x,y)$ is the projection. So you only need the estimate for the projection which is much easier to show (also because it has all the regularity one can hope for).


问:
Use of Chain rule in PDE problem
We study the one dimensional heat equation

$$u_{t}(x, t) = u_{xx}(x, t), \quad x \in \mathbb{R}, t> 0.$$

I our textbook, in order to prove that the solution is a function of one variable, they define a function

$$v(x, t) = u(cx, c^2t)$$

for a $c > 0$ and claims

$$v_t = c^2 u_t \quad \text{and} \quad v_{xx} = c^2 u_{xx}.$$

This is probably completely trivial, but I fail to see how they have applied the chain rule here. Here is my reasoning. If we let $\phi_1 = cx$ and $\phi_2 = c^2 t$, we can apply the chain rule to find $v_t$ as follows:

$$v_t = \frac{\partial u}{\partial \phi_1}\frac{\partial \phi_1}{\partial t} + \frac{\partial u}{\partial \phi_2}\frac{\partial \phi_2}{\partial t} = c^2\frac{\partial u}{\partial \phi_2}.$$

What I don't quite understand is how $\frac{\partial u}{\partial \phi_2}$, the partial with respect to $\phi_2 = c^2 t$, is equal to $u_t = \frac{\partial u}{\partial t}$. What am I missing?

答:
Let us use the following notation. Suppose $u(X, T)$ solves the heat equation, i.e.
\begin{align}
u_{T}(a, b)= u_{XX}(a, b).
\end{align}
for any $(a, b) \in \mathbb{R}\times \mathbb{R}$.

Now, consider
\begin{align}
v(x, t) = u(cx, c^2t)
\end{align}
i.e. set $X = cx$ and $T = c^2t$. Then we see that
\begin{align}
\frac{\partial v}{\partial t} = \frac{\partial u}{\partial T}\frac{\partial T}{\partial t} = c^2u_T(cx, c^2t)
\end{align}
and
\begin{align}
\frac{\partial v}{\partial x} =  \frac{\partial u}{\partial X}\frac{\partial X}{\partial x} = cu_X(cx, c^2t)
\end{align}
which means
\begin{align}
\frac{\partial}{\partial x}(cu_X(cx, c^2t)) = c^2u_{XX}(cx, c^2t).
\end{align}
Hence we have that
\begin{align}
v_t-v_{xx} = c^2\left(u_T(cx, c^2t)-u_{XX}(cx, c^2t)\right) = 0.
\end{align}

Title: Use of Chain rule in PDE problem


问:
Do there exists $p$ and $q$ primes for which $p^3+1=2q^2$?
Do there exists $p$ and $q$ primes for which
$p^3+1=2q^2$?

答:
Answer:

No

Proof:

First we shall prove that $p=2$ can't be a solution.

If $p=2$ then we need to find a prime $q$ such that $9=2^3+1=2q^2$, obviously there is no such prime.

Now let us assume that there are prime solutions, we will call them $(p,q)$, therefore:
$$(p+1)(p^2-p+1)=p^3+1=2q^2$$
Therefore
$$\frac{p+1}{2}(p^2-p+1)\mid q^2$$
where $\frac{p+1}{2}\in \mathbb{N}\setminus\{1\}$ since $p$ is an odd prime.

Now let's assume that $p^2-p+1 = 1$.

In this case we get no prime solutions because the solutions to $p^2-p= 0$ are $1$ and $0$ both are not prime.

Therefore because both of the factors are in $\mathbb{N} \setminus\{1\}$ and their product divide $q^2$ where $q$ is a prime number we get:
$$\frac{p+1}{2}=q \  \land \ p^2-p+1 = q \ \Rightarrow \ p^2-p+1 = \frac{p+1}{2} \ \Rightarrow \ 2p^2-3p+1 = 0 \ \Rightarrow \ p=1 \lor p=\frac{1}{2}$$
And again we get that $p$ be is not a prime resulting in a contradiction to the assumption that there are solutions.


问:
Grad(f) in index notation?
If $f(\mathbf{r})=\vert\mathbf{r}\vert^4$

How would you calculate $\operatorname{grad} f$ in index notation?

I get that $\vert\mathbf{r}\vert^2=x_ix_i$ but how do I represent $f$ in index notation?

答:
The gradient of a scalar field is contravariant (known as one-form). In index notation
$$[grad(f)]^n=\nabla^nf=g^{rn} \nabla_{r}f=g^{rn}\partial_{n}f \tag{1}$$
where $\nabla_n $ is the covariant derivative and $g^{ij}$ is the contravriant metric tensor, and we have used the fact that $\nabla_nf=\partial_n f$ for a scalar field.

Your function $f(x^j)=|\bf{r}|^4$ can be written as
$$f=(g_{ij}x^{i}x^j)^2$$
Now $$\partial_nf=\partial_n(g_{ij}x^ix^j)^2=2(g_{ij}x^ix^j)((\partial_ng_{ij})x^ix^j+g_{ij}(\delta^i_nx^j+x^i\delta^j_n))$$
$$=2(g_{ij}x^ix^j)((\partial_ng_{ij})x^ix^j+2g_{nj}x^j)$$

$$g^{rn}\partial_nf=2(g_{ij}x^ix^j)(g^{rn}(\partial_ng_{ij})x^ix^j+2x^r )$$
So the answer is dependent on your metric.

Assuming that we're working under the Cartesian metric, $g_{ij}=\delta_{ij}=constant$, the $\partial_ng_{ij} $ term vanishes, then
$$[grad(f)]^r=4g_{ij}x^ix^jx^r=4x_jx^jx^r$$
$$$$

Title: Grad(f) in index notation?


问:
Calculating $\int_0^\infty Mn x^{n-1} \cdot \exp(-Mx^n-Ax^b)dx$
I am trying to solve $$\int_0^\infty Mn x^{n-1} \cdot \exp(-Mx^n-Ax^b)dx=C$$ wrt $A$ and $b$.
Here $M>0$, $n>0$ and $0<C<1$ are known constants and $A>0$ and $b>0$ are unknown.

I was not able to calculate this integral using WolframAlpha.

I thought about using numerical methods, but I am not sure how to use them considering I have these unknown constants in the expression.

How should I approach this problem?

答: This is not an answer, it just does not fit in comments. Numerically, it seems that you get an infinite number of solutions. In the picture below, you can see (for $M=1, n=2, C=0.5$) the $(A,b)$ solutions as the zero contour in the center.


问:
Maximal ideal in noetherian local ring
Given a noetherian domain $A$ of dimension 1, I know that $A/p$ is a noetherian local domain of dimension 1 for any non-zero prime ideal $p$. But what is the maximal ideal of $A/p$?

答:
The statement is wrong. Consider for example the integers, then $\dim(\mathbb{Z})=1$ and it is noetherian as it is a PID but for any prime number we have that $\mathbb{Z}/p\mathbb{Z}$ is a field, thus has dimension zero with the only prime ideal $(0)$.
Your problem here is that when factoring out a nonzero prime ideal your dimension reduces, in this case by one as any nonzero prime ideal in a domain of dimension one has height one. So you always end up with a field where the maximal ideal is just $(0)$.

Title: Maximal ideal in noetherian local ring


问:
We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?
We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?

My attempt:
For the first draw of 2 dice together, the sample space has 36 possible outcomes and the ones with the same number in both dice are $(1,1), (2,2)...(6,6)$ so a total of 6. The probability for one throw is $P_1 = \frac {6}{36} = \frac {1}{6}$

Therefore for 2 such throws, the probability to get doubles is $(\frac {1}{6})^2$.

The probability to get doubles in 3 out of 6 throws is

$(\frac {1}{6})^3*(\frac {5}{6})^3$ ?? Is this correct?

or maybe $\frac {6^3}{36^6}$?

Thank you!

答:
As you observed, the probability of obtaining doubles on any one throw of a pair of dice is $1/6$ since $6$ of the $36$ possible outcomes result in doubles.  Since the probability of obtaining doubles is the same for each of the six trials, this is a binomial distribution problem.

The probability of obtaining exactly $k$ successes in $n$ trials is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ counts the number of ways in which exactly $k$ of the $n$ trials could result in a success.

Here, we will define a "success" as a double, so $p = 1/6$.  Since there are six trials, $n = 6$.  Hence, the probability to get exactly three doubles in six trials is
$$\Pr(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$
In your attempt, you did not take into account the number of ways exactly three of the rolls could result in a success.

If you were instead interested in the probability of at least three doubles in six trials, you would add the probabilities of three or more successes.
$$\Pr(X \geq 3) = \sum_{k = 3}^{6} \binom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6 - k}$$


问:
Is it true that $\Gamma(x+1) \leq x^x$ for every $x>0$?
I've seen the inequality $\Gamma(x+1) \leq x^x$ for $x > 0$ used without proof or reference and I'm not sure it's true.

Thanks for your help.

答:
Here is a proof relying on Euler´s integral representation of the Gamma function

$$\Gamma(z+1)=\int_{0}^{\infty}x^{z}e^{-x}dx$$

The function $x^{z}e^{\frac{-x}{2}}$ drops off to $0$ as $x\rightarrow \infty$. It´s not hard to find that its maximum is attained when $x=2z$. From this fact we can find an upper bound for $\Gamma(z+1)$ as folowing.

$$\Gamma(z+1)=\int_{0}^{\infty}x^{z}e^{-x}dx$$
$$=\int_{0}^{\infty}\left(x^{z}e^{\frac{-x}{2}}\right)e^{\frac{-x}{2}}dx$$
$$\leq\int_{0}^{\infty}(2z)^{z}e^{-z}e^{\frac{-x}{2}}dx$$
$$=(2z)^{z}e^{-z}\int_{0}^{\infty}e^{\frac{-x}{2}}dx$$
$$=2(2z)^{z}e^{-z}$$
dividing both sides of the inequality by $z^{z}$
$$\frac{\Gamma(z+1)}{z^z}\leq 2\left( \frac{2}{e}\right)^z$$

Since $2<e$, the right hand side tends to $0$ as $z\rightarrow \infty$, and we can conclude that $\frac{\Gamma(z+1)}{z^z}$.

Another more intuitive way to see it is by considering $x$ a positive integer number $n$, than

$$\Gamma(n+1)=n!= 1\cdot2\cdot3\cdots(n-1)\cdot n$$

$$n^n = \underbrace{n \cdot n \cdots n}_{n times}$$

$$\frac{n!}{n^n}=\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}$$

From the last equation it is plausible that $n^n$ grows a lot faster than $n!$!


问:
Determinant of associated real matrix from a complex one
Given $X\in \operatorname{GL}(n, \mathbb{C}) $, let $X_r\in \operatorname{GL}(2n, \mathbb{R}) $ be the real matrix obtained by substituting to each complex entry $a+ib$ the matrix $\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$.
Does someone can give me the details of the proof that $\det(X_r) =|\det(X)|^2$?
The only thing I have tried is to use the fact that the determinant is a wedge product but it seems not work.

答:
Another option is to think about the matrix $X_r$ as a $2n \times 2n$ complex matrix. Motivated by the $2 \times 2$ case, over the complex numbers, we have the identity

$$ \begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix} = P^{-1} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} P$$

where

$$ P = \frac{1}{\sqrt{2}} \begin{pmatrix} -iI_n & iI_n \\ I_n & I_n \end{pmatrix}, \,\,\,
 P^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix} iI_n & I_n \\ -iI_n & I_n \end{pmatrix}. $$

Since similar matrices have the same determinant, we have

$$ \det(X_r) = \det \begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix} = \det(a - ib) \det(a + ib) = \\
\det(X) \cdot \det(\overline{X}) = \det(X) \cdot \overline{\det(X)} = \left| \det(X) \right|^2. $$


问:
Simple Derivative $\frac{x}{\ln x}$
Okay,
I have find the derivative of $$\frac{x}{\ln x}$$ so I used the quotient rule and got: $$\frac{\ln x(1)-x(\frac{1}{x})}{(\ln x)^2}$$

I then simplified this to: $$\frac{-x}{\ln x}.$$

Now here come the problem, when I put $e$ into $f'(x)$ I am supposed to get $-1.5271$ and I do not and I cannot figure out why.  Could someone point out what I am doing wrong?

Thanks

答: You incorrectly cancelled during your last step.  It is not true that $$\frac{ab+c}{a^2}=\frac{b+c}{a}$$

Title: Simple Derivative $\frac{x}{\ln x}$


问:
A sequence that grows faster than the ackermann function?
Ackermann's function and all the up-arrow notation is based on exponentiating. We know for a fact that the factorial function grows faster than any power law so why not build an iterative sequence based on factorials? I am thinking of the simple function defined by
$$
a(n)=a(n-1)!
$$
and to start things off we need $a(1)=3$. The numbers generated are $3,6,24,720$ and the next one ($720!$) is roughly $10^{1746}$ and I can't even start to imagine what happens next!

Is it primitive recursive? Can it be coded up using for loops avoiding calls to itself? I think not but this is just a hunch. Computer scientists might have better chances of answering that than I.

Thanks in advance, I appreciate it.

答:
So you mean the function $a(n)$ defined by $a(1)=3$ and $a(n) = a(n-1)!$.

This function is primitive recursive.
Multiplication can be defined by a primitive recursive scheme. Based on this, the factorial function can be defined by a primitive recursive scheme.
From here, you can use induction on $n$ to show that $a(n)$ is primitive recursive.


问:
Assume $f:[a,b] \to \mathbb R$ is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$
Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :

$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$

Taking the limit of both sides:

$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$

I have some questions:

How do we know that $M-\epsilon$ is not negative?

We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$

Why that's true??

答:
There are two cases. If $f \equiv 0$ then the claim follows immediately and there's no need to consider the inequality. In all other cases, there exists some point in the interval in which $f$ is positive and so $M > 0$. Hence, for small enough $\varepsilon$, the expression $M - \varepsilon$ will be positive.

You are using the monotonicity of the integral. If $f(x) \leq g(x)$ for all $x \in [a,b]$ and they are integrable then $\int_{a}^b f(x) \, dx \leq \int_a^b g(x) \, dx$. This follows from the fact that $g(x) - f(x) \geq 0$ for all $x \in [a,b]$ and that the integral of a non-negative function is non-negative.


问:
Proof explanation of a simple sum
Can someone explain to me (step by step), how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n(1-(1-p)^r)/p$.

I'm having trouble with maths..

答:
Hint

You can proof by induction that $$\sum_{i=0}^np^i=\frac{1-p^{n+1}}{1-p},$$ or just observe by simple calculation that $$(1-p)\sum_{i=0}^np^i=(1-p)(1+p+p^2+\cdots +p^n)=1-p^{n+1}.$$

Title: Proof explanation of a simple sum


答:
First question: if $f(x)=0$ for all $x \in [a,b]$, then you are done.

Hence we assume that $f(x)>0$ for some $x \in [a,b]$. Then we have that $M>0.$

Then consider $ \epsilon >0 $ such that $\epsilon <M.$

Second question: if $f_1$ and $f_2$ are integrable and $f_1 \le f_2$ on $[a,b]$ , then we have

$$\int_a^b f_1(x) dx \le \int_a^b f_2(x) dx.$$

Body:
Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :

$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$

Taking the limit of both sides:

$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$

I have some questions:

How do we know that $M-\epsilon$ is not negative?

We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$

Why that's true??