问:
Finding a 3-regular graph "with least no. of vertiecs" containing P6 as an induced-subgraph
Can you tell me a 3-regular graph with the least no of vertices, that contains P-6 as an induced subgraph?

A 3-regular graph is one in which the degree of every vertex is 3.

P-6 looks like: o--o--o--o--o--o

H is an induced subgraph of G if the vertex set of H is a subset of the vertex set of G and uv is an edge connecting vertices 'u' and 'v' in H only if uv is an edge in G.

The main problem is the 'least no. of vertices' clause.

I have already tried for all the following 3-regular graphs and have found that none of these graphs contain P6 as an induced subgraph. So the order of the graph has to more than 8 for sure.

答:
The vertices of the induced $P_6$ have degrees $1, 2, 2, 2, 2, 1$ in the path, and need to have total degree $3,3,3,3,3,3$, so they need $2+1+1+1+1+2=8$ edges more coming out of them. None of these can go to other vertices in the $P_6$, because it's induced, so they need to go to other vertices in the graph.

These $8$ edges need at least $\lceil \frac83 \rceil = 3$ vertices to go to. But we can't have just $3$ other vertices, because there is no $3$-regular graph on $9$ vertices. (The handshake lemma says that such a graph would have $\frac{3\cdot 9}{2} = 13.5$ edges.) So we must have at least $4$ other vertices.

With $4$ other vertices, their total degree must be $4 \cdot 3 = 12$, so they need to take the $8$ edges coming out of the $P_6$, and have $2$ more internal edges. (Each internal edge contributes $2$ to the sum of their degrees.) There are plenty of ways to do this; one of them is shown below.

Here's a more symmetric picture of this graph, with an induced $P_6$ highlighted:


问:
How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$
Given $27w^2+20w+35 \equiv  0 \pmod{23}$

How to solve this  quadratic congruence?
All hints are welcome.
I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv  0 \pmod{23} $
I need to get 2 solutions in the least residue system which are the correct answers:  $w\equiv  6 \pmod{23}$ and  $w \equiv  12 \pmod{23}$

答:
For ease of manual computation we rewrite the equation as
$$4w^2-3w+12\equiv0\bmod23$$
Divide by the leading coefficient, i.e. multiply by $4^{-1}=6$:
$$w^2+5w+3\equiv0\bmod23$$
Now apply the quadratic formula:
$$w\equiv\frac{-5\pm\sqrt{13}}2\bmod23$$
We need to work out the square roots of $13$ in $\mathbb Z_{23}$. $6$ is easily verified as one root, so $-6$ is the other:
$$w\equiv\frac{-5\pm6}2\equiv9\pm3\bmod23$$

ContentLicense: CC BY-SA 4.0


问:
About the infinite solutions of a Diophantine equation
Consider the following problem:
$$\sum_{k=1}^N k^2=q^2$$
where q is an integer number. This can be written as:
$$\frac{1}{3}N^3+\frac{1}{2}N^2+\frac{1}{6}N=q^2$$
In the same way we can write:
$$\sum_{k=1}^N k^3=q^3$$
that means:
$$\frac{1}{4}N^4+\frac{1}{2}N^3+\frac{1}{4}N^2=q^3$$
In general, we can consider the following equation:
$$\sum_{k=1}^N k^r=q^r$$
with $r$ integer number.
The question is: does this equation has infinite solutions for every $r$ integer?
Thanks.</p>

答:
Generally, Schaffer proved in 1956 that the equation
$$
\sum_{k=1}^K k^r = q^n
$$
has at most finitely many solutions unless
$$
(r,n) \in \{ (1,2), (3,2), (3,4), (5,2) \}.
$$
The proof uses Siegel's theorem (though, as Erick notes, Faltings would lead to the same conclusion).</p>

ContentLicense: CC BY-SA 3.0


答:
Hint:

$$\pmod{23}: 4w^2-3w+12\equiv 0 \implies 8w^2-6w+1\equiv 0  \implies (2w-1)(4w-1)\equiv 0. $$

Update To justify why I multiply 2 to $4w^2-3w+12$, it's easier to work with whole numbers than fractions, so to complete square while keeping every coefficients integer we multiply by 16:

$$16(4w^2-3w+12)=64w^2-48w+192=(8w-3)^2+183\equiv (8w-3)^2-1 = (8w-2)(8w-4)=8(4w-1)(2w-1) \pmod{23}$$

and now you see why.

Update 2: I like Parcly Taxel's way of making the quadratic monic first:

$$w^2+5w+3\equiv0\pmod{23}$$

After that it can be done a little faster:

$$w^2-18w+3\equiv 0 \implies (w-9)^2 = 78\equiv 9 =3^2 \implies (w-6)(w-12) \equiv 0 \pmod{23}$$

Title: How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$

Body:
Given $27w^2+20w+35 \equiv  0 \pmod{23}$

How to solve this  quadratic congruence?
All hints are welcome.
I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv  0 \pmod{23} $
I need to get 2 solutions in the least residue system which are the correct answers:  $w\equiv  6 \pmod{23}$ and  $w \equiv  12 \pmod{23}$


问:
Is $\mathbb{R}^{n} - B\left[ 0;r\right]$ connected, with $n>1$ and $r > 0$?
I need to prove that the complement in $\mathbb{R}^{n}$, $n>1$ of a limited subset $X \subset \mathbb{R}^{n}$ has only one ilimited connected component. This lies in prove that $\mathbb{R}^{n} - B\left[ 0;r\right]$ (where $B\left[0;r\right]$ is the closed ball with center in the origin and radius $r$) is connected, since it will be the only one ilimited connected component (any another component different from this will be in the closed ball).

How to prove that? There is another way to prove the statement whitout the usage of this? Thanks in advance!

答:
$ \mathbb R \setminus B[0,r]$ is not connected, because it is union of two (cl)open sets, $(-\infty, - r) \cup (r, +\infty) $.
For $n\ge 2 $, $\mathbb R^n \setminus B[0,r]$ is connected because it is path-connected.


问:
In what sense is the tensor algebra the most general (unital, associative) algebra
Terminology:

Since I want to consider only unital, associative algebras, algebra will be unital and associative unless otherwise stated.

An ideal will be a two-sided ideal unless otherwise stated.

Vector spaces will be over an infinite field $\mathbb F$ with characteristic 0.

Background:

Let $V$ be a finite dimensional vector space and let $T(V)$ be its tensor algebra. It is often said that $T(V)$ is the "most general" algebra that contains $V$ as a subspace. This can also be seen by noting that intuitively $T(V)$ is the algebra generated by $1,e_1,...,e_n$ ($e_1,...,e_n$ is a basis of $V$) in a way that $1$ is the unit, the multiplication is associative but from that point on, no additional relations are assumed to hold about the generators.

The tensor algebra also observes the universal property, namely that if $A$ is an algebra, $\phi :V\rightarrow A$ is a linear map, then there exists a unique algebra homomorphism $\phi_\otimes:T(V)\rightarrow A$ that satisfies $\phi=\phi_\otimes\circ i$ where $i:V\rightarrow T(V)$ is the inclusion.

I would like to understand that in what sense the tensor algebra is the most general algebra containing $V$. In particular, this can mean one of several things:

Hypothesis 1: There is no such algebra $A$ that contains $V$ as a subspace and $T(V)$ is a sub algebra of $A$.

Hypothesis 2: There is no such algebra $A$ that contains $V$ as a subspace and it has an ideal $I$ such that $T(V)=A/I$.

Hypothesis 3: If $A$ is an algebra that contains $V$ as a subspace, then there is an ideal $I\vartriangleleft T(V)$ of $T(V)$ such that $A=T(V)/I$.

Hypothesis 3 seems at least believable, since pretty much all famous algebras containing $V$ arise as quotients of the tensor algebra (the exterior algebra, the symmetric algebra, the Clifford and Weyl algebras, the universal enveloping algebra if $V$ is a Lie algebra etc.).

On the other hand, I am not convinced that the universal property stated above actually implies Hypothesis 3.

In particular, assume that $A=T(V)/I$. If $A$ is to contain the entirety of $V$ then we must have $I\cap V=0$. I don't know if $T(V)$ has a nontrivial ideal that intersects $V$ nontrivially or not, but nonetheless let's make this assumption. Then the quotient map $\pi:T(V)\rightarrow T(V)/I$ is surjective. Moreover if $\pi:T(V)\rightarrow A$ is a surjective algebra morphism then $\ker\pi$ is an ideal of $T(V)$ and $A=T(V)/\ker\pi$.

If the universal property is to imply Hypothesis 3, then we must be able to show that if $\phi:V\rightarrow A$ is injective (i.e. the image of $V$ in $A$ is faithful), then $\phi_\otimes :T(V)\rightarrow A$ must be surjective. The universal property does guarantee that the image of $V$ via $\phi_\otimes$ is faithful but to have $A=T(V)/\ker\phi_\otimes$, we must have surjectivity. I don't see it being implied.

Question:

Are any of the hypotheses presented above true? If not, then in what sense is the tensor algebra the most general algebra containing $V$? The "raw" statement of the universal property does not convince me that $T(V)$ is "the most general", what are some additional implications of the universal property that makes $T(V)$ "the most general"?

答:
All three of your hypotheses are false.

For hypothesis 1, let $W$ be another nonzero vector space and consider $T(V \oplus W) \cong T(V) \otimes T(W)$, which contains $V$ as a subspace and $T(V)$ as a proper subalgebra.

For hypothesis 2, consider $T(V) \times k$, which contains $V$ as a subspace and $T(V)$ as a proper quotient.

For hypothesis 3, consider $T(V \oplus W)$ again.

The correct statement is that any algebra $A$ equipped with a map $i : V \to A$ (which need not be injective) and which is generated by the image of $V$ must be a quotient of $T(V)$. This follows directly from the universal property.

Loosely speaking, the tensor algebra is the "most general" algebra containing $V$ and nothing "extra." There are ways in which "most general" is a bit misleading, as you've discovered; the way I prefer to describe it is that the tensor algebra is the "laziest" way to build an algebra from $V$, in the sense that you don't impose any conditions on how elements multiply and add except the ones forced on you by the axioms. "Free-est" would also be appropriate, which is why it would also be appropriate to call $T(V)$ the free (associative) $k$-algebra on $V$, and why we call constructions like this free objects in general.

Nothing I've just said requires any hypotheses on the base field $k$, and in fact everything applies without modification to a module $V$ over an arbitrary commutative ring $k$.


问:
Finding $P(y|x)$ given $P(y), P(x|y)$, and $P(\lnot x|\lnot y)$.
I have a probability question that asks me to find $P(y|x)$ given that $P(y) = 0.6, P(x|y) = 0.8$, and $P(\lnot x|\lnot y) = 0.3$.

I've tried applying Bayes's theorem and using it I can find $P(x\wedge y) = 0.48$ and $P(\lnot x\wedge \lnot y) = 0.12$. The roadblock I run into is that I seem to need to know $P(x)$ to find it. Any help would be greatly appreciated.

答: From $P(x\mid \lnot y)+P(\lnot x\mid\lnot y)=1$ find $P(x\mid \lnot y)$. Then from $P(x\mid \lnot y)=\frac{P(x, \lnot y)}{P(\lnot y)}$ find $P(x, \lnot y)$. Finally, $P(x, \lnot y) + P(x, y) = P(x)$.


问:
A company hires $11$ new employees, and they will be assigned to four different departments, A, B, C, D

A company hires $11$ new employees, and they will be assigned to four different departments, A, B, C, D. Each department has at least one new employee. In how many ways can these assignments be done?

I know that for each section (A,B,C,D) I should add a () and as long as every section must get a new employee
we should start like this:
$$(x+x^2/2!+x^3/3!+...)^4$$
then if we look  it's $(e^x-1)^4$.
After this step I don't know what to do.

Body:
Using $$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 +4ab^3 + b^4$$

$$(e^x-1)^4 = e^{4x} - 4e^{3x} + 6e^{2x} -4e^x + 1$$

and
$$e^{kx} = 1+kx+\dfrac{(kx)^2}{2!}+\ldots+\dfrac{(kx)^{11}}{11!}+\ldots$$

$$\Rightarrow [\frac{x^{11}}{11!}] \, (e^{kx}) = k^{11}$$
Therefore
$$[\frac{x^{11}}{11!}] \, (e^x-1)^4 =$$ $$ 4^{11} - 4\cdot 3^{11} + 6\cdot 2^{11} -4\cdot 1^{11}$$


问:
show $ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!} $
While doing an exercise and playing around in wolfram alpha I stumbled upon this identity, but I dont see how one could show it. Any hint? $(a)_{k}$ is the Pochhammer symbol so rising factorial.

$$ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!}  $$

答:
If we expand the LHS we get
$$\frac{(-a)(-a-1)\cdots(-a-(k-1))}{k!}$$
Take out all the minus signs from the numerator – there are $k$:
$$=(-1)^k\frac{a(a+1)\cdots(a+(k-1))}{k!}=\frac{(-1)^k(a)_k}{k!}$$


答:
\begin{align*}
\mathbb{P}(Y|X) = \frac{\mathbb{P}(Y\cap X)}{\mathbb{P}(X)} & = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(X\cap Y) + \mathbb{P}(X\cap Y^{c})}\\\\
& = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c}) - \mathbb{P}(X^{c}\cap Y^{c})}\\\\
& = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c}) - \mathbb{P}(Y^{c})\mathbb{P}(X^{c}|Y^{c})}\\\\
& = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c})(1 - \mathbb{P}(X^{c}|Y^{c}))}\\\\
& = \frac{0.6\times 0.8}{0.6\times 0.8 + (1 - 0.6)\times(1 - 0.3)}
\end{align*}

Title: Finding $P(y|x)$ given $P(y), P(x|y)$, and $P(\lnot x|\lnot y)$.

Body:
I have a probability question that asks me to find $P(y|x)$ given that $P(y) = 0.6, P(x|y) = 0.8$, and $P(\lnot x|\lnot y) = 0.3$.

I've tried applying Bayes's theorem and using it I can find $P(x\wedge y) = 0.48$ and $P(\lnot x\wedge \lnot y) = 0.12$. The roadblock I run into is that I seem to need to know $P(x)$ to find it. Any help would be greatly appreciated.


问:
length of sum of two submodule
Let $M$ be a $R$-module with finite length and $K$ and $N$ be a submodule of $M$. Prove that
$l(K+N)+l(K\cap N)=l(K)+l(N)$.

My proof: First, by assuming that $K\cap N=\{0\}$, we can conclude that $l(K+N)=l(K)+l(N)$.

The detail: Let $n=l(K)$ and $m=l(N)$.
Let $\{0\}\subseteq K_0\subsetneq K_1 \subsetneq \cdots \subsetneq K_n=K$ be a composition series of $K$. Let $\{0\}\subseteq N_0\subsetneq N_1 \subsetneq \cdots \subsetneq N_m=N$ be a composition series of $N$.
Define $\phi: K \rightarrow K+N $ by $\phi(k)=k$ and $\psi:K+N \rightarrow N$ by $\psi(k+n)=n$. For $0 \leq i \leq n$, let $M_i=\phi(K_i)$ and for $n+1\leq i\leq n+m$, let $M_i=\psi^{-1}(N_i)$, then $\{M_i\}_{i=0}^{n+m}$ be a composition series for $K+N$. Hence, $l(K+N)=n+m=l(K)+l(N)$.

Could you help me proving the general cases, that is $K\cap N \neq \{0\}.$

答:
The second isomorphism theorem says we have an exact sequence

$$0 \to K \cap N \to  K  \to (K+ N)/N \to 0$$

and so $l(K) = l(K\cap N) + l\left( K+N/N\right) = l(K \cap N)+ l(K+N) - l(N)$ . Thus

$$l(K) + l(N) = l(K \cap N) + l(K+N).$$

Title: length of sum of two submodule


问:
How to solve this trigonomectric equation $2^{\lvert\sin x\rvert} + 2^{\lvert\cos x\rvert} = 3$?
I thought its solutions are $(\sin x;\cos x) = \{(\pm1;0),(0;\pm 1)\}$

But how can I solve it?

Please help :(

答:
Give the periodicity and symmetries of the trig functions, it suffices to show that $f(u)=2^u+2^{\sqrt{1-u^2}}$ is increasing for $0\lt u\le1/\sqrt2$.  That will imply that the only solutions to $2^{|\sin x|}+2^{|\cos x|}=3$ occur at multiples of $\pi/2$.

We have

$$f'(u)=\ln2\left(2^u-{u\over\sqrt{1-u^2}}2^{\sqrt{1-u^2}}\right)$$

so it's enough to show that

$${2^u\over u}\gt{2^{\sqrt{1-u^2}}\over\sqrt{1-u^2}}$$

for $0\lt u\lt1/\sqrt2$. But notice that in this range we actually have $0\lt u\lt1/\sqrt2\lt\sqrt{1-u^2}\lt1$ (i.e., $0\lt\sin x\lt1/\sqrt2\lt\cos x\lt1$ for $0\lt x\lt\pi/4$). So we need only prove that the function $g(u)=2^u/u$ is decreasing for $0\lt u\lt1$. This is equivalent to showing that $h(u)=(\ln2)u-\ln u$ is decreasing on the same range, which is clear since

$$h'(u)=\ln2-{1\over u}\lt0\iff0\lt u\lt{1\over\ln 2}$$

and $\ln2\lt1$.


问:
Regarding finite simple groups of type $^2D_n$
I would like to known if is there any list (even partial) of (maximal) subgroups of a finite simple group of type $^2D_n$ ($n>3$). I tried to search on Google and MathSciNet, but it seems either there is nothing or I used the wrong keywords. Is there such a list?

答:
These groups are orthogonal of type $O^-$. A recent, quite explicit, description of the maximal subgroups can be found in:
Holt, Roney-Dougal: Constructing maximal subgroups of orthogonal groups.
LMS J. Comput. Math. 13, 164-191 (2010)

Link:
https://homepages.warwick.ac.uk/~mareg/download/papers/maxorthoggps/maxorth.pdf


问:
Extension of morphism of quasiprojective varieties
I have $\phi:U⊂\mathbb{P_{\mathbb{C}}^{1}}→X$ a morphism, where U is a open dense subset and X is a projective variety in $\mathbb{P_{\mathbb{C}}^{n}}$. I want to prove that $\phi$ extends in a unique form to the morphism $\overline{\phi}:$ $\mathbb{P_{\mathbb{C}}^{1}} \rightarrow X$ such that $\overline{\phi}\restriction_U=\phi$.
My question is the same as in the post https://mathoverflow.net/questions/296794/extension-of-morphism-of-quasiprojective-varieties. My problem is that I don't understand the answers there and I am looking forward to a easier one.

答:
Near some $p\in U$ your morphism is given by some rational functions $f_j$ $$\phi([x:y]) = [f_0(x/y):\ldots :f_n(x/y)]$$ which is a morphism at every $[a:b]$ such that $a/b$ is not a pole of any $f_j$ nor a zero of all the $f_j$.

If $b\ne 0$ then for some $k$, replacing $f_j(x/y)$ by $f_j(x/y)(x/y-a/b)^k$ we can repair those two cases.

Can you deal with the case $b=0$ ?


问:
What does "translation invariant" mean for linear differential functionals?
I am not super experienced in solving PDEs. Today I was reading about Green's functions, from Wolfram Mathworld and Wikipedia, since the notes in my course focus a bit to much on a special case and I would like to have something a bit more general. I understand most of the theory, but there is a particular part that I can't seem to find any good resources for. So first, I'll present the problem and then discuss my issues with what Wikipedia has.

Given a linear differential functional $\mathscr{L}:F\mapsto \mathscr{L}[F]$ and a corresponding PDE
$$\mathscr{L}[u]=f\tag{1}$$
What is the most general solution for $u:\mathbb{R}^n\to\mathbb{R}$ ?

I'm sure many of you are familiar with Green's functions and whatnot so I won't go through the theory. It's the following part I am having trouble understanding -

If the operator is translation invariant, that is when $\mathscr{L}$ has constant coefficients with respect to $x$, then the Green's function can be written as a convolution kernel,
$$G(x,s)=G(x-s)$$

First question - what is the exact meaning of "translation invariant" with regard to functionals? To me based on intuition of linearity from linear algebra it seems pretty obvious that $\text{constant coefficients}\implies \text{translation invariant}$ but I am doubtful of the converse, namely
$$\text{translation invariant}\overset{?}{\implies}\text{constant coefficients}$$

Second question - What is the meaning of $G(x,s)=G(x-s)$ ? $G$ has gone from being a function $G:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}$ to instead being a function $G:\mathbb{R}^n\to\mathbb{R}$. What's going on here?

Final question - Once we have found the Green's function, what is the most general solution $u_{\text{G}}$ to the above? Letting $u_{\text{H}}(x)$ be the solution of $\mathscr{L}[u]=0$ and $u_{\text{P}}$ be the solution to (1), is it simply
$$u_{\text{G}}=u_{\text{H}}+u_{\text{P}}$$
As in the case of linear ordinary differential equations? Or is it not that simple?

Thanks everyone.

答:
I have been working with Green's function for the heat equation lately, so maybe I can help with some points of your question. I'm not sure about the translation part, thus, I will jump straight to your second question.

The heat equation $$u_t -\Delta u = 0$$ is in my opinion translation invariant. The fundamental solution to the heat equation is sometimes stated as $$\Gamma(x,y,t) = \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x-y|^2}{4t}}$$
for $x,y\in \mathbb{R}^n$, $t>0$. We can now build a solution for the heat equation with initial data $u(x,0)=u_0$ as described in the Wikipedia article $$u(x,t) = \int_{\mathbb{R}^n} \Gamma(x,y,t) u_0(y) \, \mathrm{d}y.$$ But often the fundamental solution of is stated as $$\Phi(x,t) = \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}}.$$ In this case we have $\Gamma(x,y,t)=\Phi(x-y,t)$ and the solution becomes $$u(x,t) = \int_{\mathbb{R}^n} \Phi(x-y,t) u_0(y) \, \mathrm{d}y,$$ thus, the solution has the form of a convolution which wasn't the case for $\Gamma$. But a general parabolic differential operator might not have a fundamental solution $\Gamma$ that can be written that way.

For your last question, I'm not quite sure what you mean by most general solution, but solutions using Green's function can often be represented as you describe. For exmaple for evolution equations this is basically Duhamel's principle. Duhamel's principle states that a solution of the problem \begin{align} u_t(x,t)-\Delta u(x,t) &= f(x,t) \\ u(x,0)&=u_0(x)\end{align} is given by $$u(x,t) = \int_{\mathbb{R}^n} \Phi(x-y,t) u_0(y) \, \mathrm{d}y + \int_0^t\int_{\mathbb{R}^n} \Phi(x-y,t-s) f(y,s) \, \mathrm{d}y\mathrm{d}s,$$ where the first term solves the homogeneous heat equation with initial data $u_0$ and the second term solves the non-homogeneous problem with zero initial data.

Edit: In general if $L$ is linear and you have a solution $L u_H=0$ and a solution $L u_P = f$, then $u=u_H+u_P$ is another solution for $Lu=f$.

In the setting of the Wikipedia article, where $L G(x,s)=\delta(x-s)$ we can build a solution for $L u(x)= f(x)$ by setting $$u(x) = \int G(x,s) f(s) \, \mathrm{d}s.$$ This means if you want to build a solution for $L u=0$ then $$u(x) = \int G(x,s) \cdot 0 \, \mathrm{d}s =0,$$ thus, using a green's function this results in $u_H=0$ in your setting.


问:
Find $[n, n+1]$ such that $f(n)<0<f(n+1)$ where $f(x)=x^3-x+3<0$
$x^3-x+3=0$ doesn't factor, though $x^3-x=x(x-1)(x+1)$ and $x^3-x<0$ when $x<-1$. The constant term $3$ translates the graph up, and I check the value of $f(x)$ at $-1$ and $-2$ assuming that after the translation it can be positive at $-1$ and negative at $-2$, which is indeed the case. The answer is $n=-2$.

Is there a better way to solve this, or I just need to check all $x$ in some interval to find where $f(x)$ changes the sign?

答:
For me it's a little easier to work with positive numbers. Obviously $f(x)>0$ if $x \ge 0, x \in \mathbb{Z}$, so let's assume $x<0, y:=-x, g(y) = f(x)=-y^3+y+3=3-(y^3-y), y \in \mathbb N.$

Now it's easy to show $y^3-y$ is monotone increasing with $y\in \mathbb N$ (via factoring or taking first derivative), so $g(y)$ is monotone decreasing and $g(1)=3>0>g(2)=-3$.


问:
Geometric sequence, finding the first term using only the sum, the number of terms and value of one term.
In Geometric series: S = 56, a(2) = 16 and n = 3

S - sum, a(2) - second term, n - number of terms

Is it possible to get a(2) and a(3) from here? (If yes, hints would be awesome)

Thank You!

答:
So we have $S = a_1 + a_2 + a_3$, as we are considering a geometric sequence, we have $a_2 = ra_1$, $a_3 = r^2a_1$. The first one gives $a_1 = r^{-1}a_2$, plugin this into the second gives us $a_3 = ra_2$. So
$$ S = r^{-1}a_2 + a_2 + ra_2 = \left(\frac 1r + 1 + r\right)a_2 $$
From this we can compute
$$ \frac 1r + 1 + r = \frac S{a_2} $$
hence
$$ 1 + r + r^2 = \frac {Sr}{a_2} $$
which is a quadratic equation for $r$. I'm sure you can do it from here.</p>


问:
How do I find sample size from large sample with known proportion?
I came across this question in my stats book in the Estimation chapter and I can quite figure it out.

The question in book is as follows:

The proportion of European men who are red-green colour-blind is 8%. How large a sample would need to be selected to be 95% certain that it contains at least this proportion of red-green colour-blind men?

I have tried equating the width of the confidence interval with the proportion that is 8%, but it doesn't seem to work. Does anybody know how to go about this question?

答:
I think this is a something of a trick question.  Since only $8\%$ of men in the population are color-blind, and we expect the sampled proportion to be normally distributed about the mean, there's really no way to be $95\%$ confident that $8\%$ of the sample will be colored blind, except by sampling the whole population, or perhaps very close to the whole population.

To take a concrete example, suppose the population size is $100$ and there are $8$ color-blind individuals.  With a sample size of $90$, in order for at least $8\%$ of the sample to be color-blind, we need to sample all the color-blind men.  The probability that this happens is $$\frac{\binom{92}{82}}{\binom{100}{90}}\approx.41655$$

That said, I'm not sure exactly what answer is expected.


Title: show $ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!} $

Body:
While doing an exercise and playing around in wolfram alpha I stumbled upon this identity, but I dont see how one could show it. Any hint? $(a)_{k}$ is the Pochhammer symbol so rising factorial.

$$ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!}  $$


问:
How to interpret $|x|^{-1}$ as a distribution
Currently I'm reading a little about regularization of distributions which usually have annoying singularities like $x^{-1}$ or $|x|^{-1}$. To the best of my understanding,
$$
\frac{1}{x}
$$
is interpreted as a distribution by considering it's Cauchy principal value instead of doing the bruteforce integration. This way, it's possible to recover a meaningful result by carefully cancelling contributions to the singularity. So if $g$ is, say, a function from Schwartz space, then
$$
\left\langle \text{p.v.}\, \frac{1}{x}, \, g(x) \right\rangle = \int _{0}^{\infty} \frac{g(x) - g(-x)}{x} \, dx.
$$
Intuitively, this makes total sense to me. However, when reading about $|x|^{-1}$ I get slightly confused. The definition states that $|x|^{-1}$ (or, rather, it's regularized cousin $\mathcal{P}|x|^{-1}$) acts on test functions as per
$$
\left\langle \mathcal{P}\frac{1}{|x|}, \, g(x) \right\rangle = \int _{|x|<1} \frac{g(x)-g(0)}{|x|} \, dx + \int_{|x|>1} \frac{g(x)}{|x|} \, dx
$$
The second integral is pretty straightforward; no singularities to sneak around. But the first one...
Does it solve the singularity problem? Yes, I can see that, since
$$
\frac{g(x)-g(0)}{|x|}
$$
remains bounded as $x \rightarrow0$. But how do we get to it? What's the reasoning behind it? I've been struggling to see it for a little while. Any and all help will be appreciated :).

答:
Since $u(x):=\operatorname{sign}(x)\ln|x| \in L^{1}_{\text{loc}}(\mathbb{R})$ we can take
$$
\frac{1}{|x|}:=u'(x),
$$
i.e. for $\varphi\in C_c^\infty(\mathbb{R})$ we set
$$\begin{align}
\left< \frac{1}{|x|}, \varphi(x) \right>
&:= - \int_{-\infty}^{\infty} \operatorname{sign}(x)\ln|x| \, \varphi'(x) \, dx
\\&= \lim_{\epsilon\to 0} \left( \int_{-\infty}^{-\epsilon} \ln|x|\,\varphi'(x)\,dx - \int_{\epsilon}^{\infty} \ln|x|\,\varphi'(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \left[\ln|x|\,\varphi(x)\right]_{-\infty}^{-\epsilon} - \int_{-\infty}^{-\epsilon} \frac{1}{x} \varphi(x)\,dx - \left[\ln|x|\,\varphi(x)\right]_{\epsilon}^{\infty} + \int_{\epsilon}^{\infty} \frac{1}{x}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \ln\epsilon\,\varphi(-\epsilon) - \int_{-\infty}^{-\epsilon} \frac{1}{x} \varphi(x)\,dx + \ln\epsilon\,\varphi(\epsilon) + \int_{\epsilon}^{\infty} \frac{1}{x}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \ln\epsilon\,(\varphi(-\epsilon)+\varphi(\epsilon)) + \int_{|x|>\epsilon} \frac{1}{|x|}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( 2\ln\epsilon\,\varphi(0) + \int_{|x|>\epsilon} \frac{1}{|x|}\,\varphi(x)\,dx \right)
.\end{align}$$


问:
Compact operators in $\bigoplus_{i \in I} H_i$
Let $\{H_i: i \in I\}$ be a collection of Hilbert spaces. We can form the Hilbert space direct sum
$$H:= \bigoplus_{i \in I} H_i$$

Question: Is there a "nice" dense subset of $B_0(H)$ (compact operators on the direct sum $H$)? For example, something like the operators
$$\left\{\bigoplus_{i \in I}  T_i \ \Bigg| \ T_i \in F(H_i), \text{all but  finitely many $T_i$ are zero}\right\}$$

where $F(H_i)$ are the finite-rank operators on $H_i$.

答:
Let me begin with a discussion of "operators vs. matrices", striving to make precise statements based on adjoints and
compositions   of operators, while avoiding the need to make  identifications between  sets, such as between a factor of a direct sum and its
canonical image.

Let $\{H_i\}_{ i \in I}$ be a collection of Hilbert spaces and put
$$
  H:= \bigoplus_{i \in I} H_i.
  $$
For each $i$, denote by $\iota _i:H_i\to H$ the  inclusion map and   observe that $\iota _i^*$ is precisely the projection from
$H$ onto $H_i$.
Incidentally, since $\iota _i$ is an isometry, then $\iota _i^*\iota _i$ is the identity on the domain of $\iota _i$ , while $\iota _i\iota _i^*$ is the orthogonal
projection onto $\iota_i (H_i)$, a.k.a. the canonical image of $H_i$ in $H$.  See $(\dagger)$ below regarding
a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$.

Given any bounded linear operator $T\in B(H)$, put
$$
  T_{i,j}=\iota _i^*T\iota _j, \quad \forall i,j\in  I,
  $$
so that each $T_{i,j}\in  B(H_j,H_i)$.  The "matrix" $(T_{i,j})_{i,j}$ may then be seen as the matrix of $T$ relative to the
above decomposition of $H$, and it can be used to recover $T$ in the following way:
given any vector $\xi \in  H$, write $\xi =(\xi _i)_{i\in I}$, so that each $\xi _i=\iota _i^*\xi $.
Noticing that $\xi =\sum_{j\in I} \iota _j\xi _j$, one then has for every $i$   that
$$
  \iota _i^*T\xi  = \sum_{j\in I} \iota _i^*T\iota _j\xi _j,
  $$
so, writing $T(\xi )=(\eta _i)_{i\in I}$, we see that the $i^{th}$ component $\eta _i$ of $T(\xi )$  may be computed in the familiar way as
$$
  \eta _i=\sum_{j\in I} T_{i, j}\xi _j.
  $$

Conversely, if $(T_{i,j})_{i,j}$ is any given matrix, with $T_{i,j}\in  B(H_j,H_i)$,  assume that $(T_{i,j})_{i,j}$  has finitely many
nonzero entries.   One can then define a bounded operator $T$ on $H$ by
$$
  T = \sum_{i, j} \iota _iT_{i,j}\iota^* _j,
  \tag 1
  $$
and it is easy to see that the matrix of $T$ coincides with the originally given matrix.
In case one moreover has that each $T_{i,j}$ lies in the set $K(H_j,H_i)$ of compact operators,  then $T$ is clearly
compact.  If instead each $T_{i,j}$ lies in the set $F(H_j,H_i)$ of finite rank operators, then $T$ has finite rank.

Directly addressing the question posed by the OP, the set formed by all operators built as in (1), from finitely
supported matrices $(T_{i,j})_{i,j}$, with each $T_{i,j}\in F(H_j,H_i)$, is a dense subspace of $K(H)$.  The proof may be
easily done by observing that the identity operator $1_H$ may be writen as
$$
  1_H = \sum_i \iota _i\iota _i^*,
  $$
with convergence in the strong operator topology,   and moreover noticing that left- or right-multiplication by compact operators turns
strong convergence into norm convergence.

$(\dagger)$ There is a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$ in the sense that both project
onto $H_i$, but the former  has $H$ for co-domain while the latter's  co-domain is $H_i$.  Co-domains are often
irrelevant in Math, but since the co-domain of an operator  becomes the domain of its adjoint, it is sometimes important to pay
attention to co-domains too!  In this note  all compositions $TS$ of operators are such that the domain of $T$ is equal
to the co-domain of $S$, even though for $TS$ to make sense, it is enough for the domain of $T$ to contain the range of $S$.


问:
How to solve this differential equation.
I need to solve this eqn $\partial_x g=1 $.
If instead of partial derivatives I had total derivative, I would do separation of variables and get the answer.

Is doing something like $\int \partial g =\int \partial x$ correct...
Can we do this separation, when the derivatives are partial rather than total.

Is there a thing as partial differential. Can we integrate with respect to a partial differential.....

答:
$$\partial_r (r e^{a(r) })=1$$
Simply gives after integration:
$$r e^{a(r)}=r+C$$
Now it depends on what the second variable is. Then $C$ is just a function of that second variable.
$$r e^{a(r)}=r+C(\theta)$$
For the other equation:
$$\partial_x g(x,y)=1$$
$$\implies g(x,y)=x+C(y)$$
Where $C(y)$ is any finction of $y$.

Consider the equation as an ordinary differential equation:
$$\partial_x g=1$$
$$\dfrac {dg}{dx}=1$$
Integrate:
$$\int dg=\int dx$$
$$g=x+C$$
Now consider the constant as a function of the second variable:
$$g(x,y)=x+C(y)$$

$$\partial_r (r e^{a(r) })=1$$
Is like an ordinary DE:
$$\dfrac {d (r e^{a(r) })}{dr}=1$$
$${d (r e^{a(r) })}=dr$$
Integrate:
$$\int {d (r e^{a(r) })}=\int dr$$
$$r e^{a(r) }=r+C$$
Consider the constant $C$ as a function of a second variable $\theta $:
$$r (e^{a(r) }-1)=C(\theta)$$


问:
Intuition for multivariable functions
Normal equations are easy enough to perceive as geometrical shapes or volumes when defined explicitly in the form $y=f(x)$.

But how do I understand equations like $g(x+y)=e^yg(x)+e^xg(y)$? I can make some sense of it algebraically, with it being a definition of a function and $x$ and $y$ being two variables. But how is it intuitively represented?

答:
A hint:

The functional equation in your question reminded me of the following: The exponential function satisfies the functional equation $e^{x+y}=e^x\cdot e^y\>$.  Therefore I would bring this into the picture by trying the Ansatz
$$g(x):= e^x\cdot h(x)$$
with a new unknown function $h$. Maybe a simpler problem for $h$ results.

Title: Intuition for multivariable functions


问:
Cardinality of preimages of finite mappings
Let $1\leq n,r$ be integers and $f:[1,n]\rightarrow[1,r]$ a mapping. For $1\leq j\leq r$, let $\epsilon_j:=|f^{-1}(j)|$. I want to show that
$$|f^{-1}(\epsilon_j)|=\epsilon_j$$
for all $1\leq j\leq r$. It seems like it might hold but I'm having difficulty proving/disproving it. Any suggestions?

答:
Any non surjective function has a $j$ such that $\epsilon _j=0,$ the inverse is not even defined there. Another example would be just $f=1\mapsto 2, 2\mapsto 2,3\mapsto 3$

$$|f^{-1}(3)|=1,|f^{-1}(1)|=0.$$
It seems really strong your condition. Why you think is true?


问:
Difference between Rician distribution and Gaussian distribution
could any one please tell me the difference between Rician and Gaussian Distribution and the advantages of using one over other please.With some mathematical proof would be truly appreciated
Thank you.</p>

答:
This Wikipedia article begins as follows:

In probability theory, the Rice distribution or Rician distribution is the probability distribution of the magnitude of a circular bivariate normal random variable with potentially non-zero mean. It was named after Stephen O. Rice.

That should make clear that they're two different things.  The normal (or "Guassian") distribution has as its support the whole real line, whereas the distribution of magnitudes of any random vector has to be supported within the half-line $[0,\infty)$.  And if you look at these graphs of densities, you see that (as the above statement would lead you to think) they are asymmetric, unlike normal (i.e. "Gaussian") distributions, which are symmetric:

http://upload.wikimedia.org/wikipedia/commons/a/a0/Rice_distributiona_PDF.png


问:
What is the probability that the Red ﬁsh are the ﬁrst species to become extinct from the pond? A Combinatorial Approach
A pond contains 3 distinct species of ﬁsh, which we will call the Red, Blue, and Green ﬁsh. There are r Red, b Blue, and g Green ﬁsh. Suppose that the ﬁsh are removed from the pond in a random order. (That is, each selection is equally likely to be any of the remaining ﬁsh.) What is the probability that the Red ﬁsh are the ﬁrst species to become extinct from the pond?

I came across this answer here

Answer: Consider an fish tank with n black fishes and m white fishes,  the total number of outcomes = (n+m)!.  Number of ways the last fish is white is m.(m+n-1)!.  Thus the probability is $\frac{m.(m+n-1)!}{(m+n)!} = \frac{m}{m+n}$.
In your problem, if there are r red fishes, b blue fishes, and g green fishes,
Based on the same reasoning as (a), the probability that the Blue fish and the Green fish are the last species to become extinct is $\frac{b}{r+b+g}\text{ and }\frac{g}{r+b+g}$ respectively. Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is$ \frac{b}{r+b}$. Similarly, the conditional probability that the Green fish are the second species to become extinct, given that the Blue fish are the last remaining fish in the pond, is $\frac{g}{r+g}$
Let R be the event that the Red fish are the first species to become extinct in the pond,RBG denote the event that the Red fish first become extinct, followed by the Blue fish, then by the Green fish, and RGB denote the event that the Red fish first become extinct, followed  by the Green fish, then by the Blue fish. Then we can obtain

$P(RBG) = P(G-last)P(RBG/G-last) =\frac{g}{r+b+g}\frac{b}{r+b}$

$P(RGB) = P(B-last)P(RGB/B-last) =\frac{b}{r+b+g}\frac{g}{r+g}$
Finally, we can compute the probability that the Red fsh are the first species to become
extinct in the pond by adding these two.

In the answer, the author used a combinatorial approach when there are only $2$ types of fish. I understood much of the argument except this :

Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is$ \frac{b}{r+b}$.

I am not quite able to visualize why this is true, especially the part in which the green fish are plain ignored. I think ignoring can be a bad idea because the left over green fish can contribute to the number of permutations. Can someone please give a combinatorial argument to prove this?

Thanks a lot!

答:
The probability that the blue fish are the second-to-last species, given that the green fish are the last species is simply the probability that the blue fish outlast the red fish.  If you want a combinatorial argument, think of permutations of the red and blue fish where the last fish is blue.

It obviously doesn't matter where we put the green fish.

EDIT

Suppose we have $g$ green fish and $p=b+r$ purple fish.  Take any permutation of the green and purple fish.  Now consider the number of ways to color $b$ of the purple fish blue and $r$ of them red.  This can be done in $\binom{b+r}{b}$ ways. Now consider the number of ways to color $b$ of them blue and $r$ red if the last one must be blue.  There are $\binom{b+r-1}{b-1}$ to do this.  If all permutations are equally likely, we get that the probability that the last fish is blue is $\frac b{b+r}$.  Since this is true for every permutation of the green and purple fish, and every permutation of the green, red, and blue fish corresponds to a unique permutation of the green and purple fish, we see that the probability that the blue outlast the red fishes is the same, no matter what we assume or don't assume, about the green fishes.


问:
How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$
Given $27w^2+20w+35 \equiv  0 \pmod{23}$

How to solve this  quadratic congruence?
All hints are welcome.
I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv  0 \pmod{23} $
I need to get 2 solutions in the least residue system which are the correct answers:  $w\equiv  6 \pmod{23}$ and  $w \equiv  12 \pmod{23}$

答:
Since $27 \equiv 4$ we can write the equation as $4w^2 + 20w + 35 \equiv 0.$ Completing the square gives $(2w+5)^2 + 10 \equiv 0,$ i.e. $(2w+5)^2 \equiv -10.$ But $-10 \equiv -10+2\cdot 23=36=6^2,$ so $2w+5\equiv\pm 6,$ i.e. $2w=-5\pm 6.$

Case $+$: $2w=-5+6=1\equiv 1+23=24=2\cdot12$ so $w\equiv12.$

Case $-$: $2w=-5-6=-11\equiv -11+23=12=2\cdot6$ so $w\equiv6.$

Thus the solutions are $w=12$ and $w=6$.


答:
First question - what is the exact meaning of "translation invariant" with regard to functionals?

Many, many people write "invariant" when they mean "equivariant," and these are very different conditions. Write $L_t$ for the operator given by translation by $t$, which means

$$(L_t u)(x) = u(x - t).$$

Invariance would mean that the differential operator $\mathscr{L}$ satisfies

$$\mathscr{L}(L_t u) = \mathscr{L}(u).$$

This is not the intended condition; for example, it is already not satisfied by $\mathcal{L} = \frac{d}{dx}$ when $n = 1$. Actually this condition is not satisfied by any nontrivial differential operator. The intended condition is equivariance, which means the differential operator satisfies

$$\mathscr{L}(L_t u) = L_t \mathscr{L}(u).$$

It's a nice exercise to check that (under mild hypotheses, e.g. $\mathscr{L}$ has smooth coefficients or something like that) this condition is equivalent to the condition that $\mathscr{L}$ commutes with every partial derivative operator $\frac{\partial}{\partial x_i}$. This in turn is equivalent to the condition that every coefficient of $\mathscr{L}$ (writing $\mathscr{L}$ as a linear combination of monomials in the partial derivatives, with coefficients given by smooth functions) is annihilated by every partial derivative operator $\frac{\partial}{\partial x_i}$, which gives that they must be constant.

Second question - What is the meaning of $G(x,s)=G(x-s)$ ? $G$ has gone from being a function $G:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}$ to instead being a function $G:\mathbb{R}^n\to\mathbb{R}$. What's going on here?

This is imprecise. What is meant here is that there is some function (really a distribution in general) $F : \mathbb{R}^n \to \mathbb{R}$ such that $G(x, s) = F(x - s)$.

Final question - Once we have found the Green's function, what is the most general solution $u_{\text{G}}$ to the above? Letting $u_{\text{H}}(x)$ be the solution of $\mathscr{L}[u]=0$ and $u_{\text{P}}$ be the solution to (1), is it simply
$$u_{\text{G}}=u_{\text{H}}+u_{\text{P}}$$
As in the case of linear ordinary differential equations? Or is it not that simple?

There is something to be careful about with boundary conditions here. I know very little about PDEs so I'll wait for someone else to comment, though.

Title: What does "translation invariant" mean for linear differential functionals?

Body:
I am not super experienced in solving PDEs. Today I was reading about Green's functions, from Wolfram Mathworld and Wikipedia, since the notes in my course focus a bit to much on a special case and I would like to have something a bit more general. I understand most of the theory, but there is a particular part that I can't seem to find any good resources for. So first, I'll present the problem and then discuss my issues with what Wikipedia has.

Given a linear differential functional $\mathscr{L}:F\mapsto \mathscr{L}[F]$ and a corresponding PDE
$$\mathscr{L}[u]=f\tag{1}$$
What is the most general solution for $u:\mathbb{R}^n\to\mathbb{R}$ ?

I'm sure many of you are familiar with Green's functions and whatnot so I won't go through the theory. It's the following part I am having trouble understanding -

If the operator is translation invariant, that is when $\mathscr{L}$ has constant coefficients with respect to $x$, then the Green's function can be written as a convolution kernel,
$$G(x,s)=G(x-s)$$

First question - what is the exact meaning of "translation invariant" with regard to functionals? To me based on intuition of linearity from linear algebra it seems pretty obvious that $\text{constant coefficients}\implies \text{translation invariant}$ but I am doubtful of the converse, namely
$$\text{translation invariant}\overset{?}{\implies}\text{constant coefficients}$$

Second question - What is the meaning of $G(x,s)=G(x-s)$ ? $G$ has gone from being a function $G:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}$ to instead being a function $G:\mathbb{R}^n\to\mathbb{R}$. What's going on here?

Final question - Once we have found the Green's function, what is the most general solution $u_{\text{G}}$ to the above? Letting $u_{\text{H}}(x)$ be the solution of $\mathscr{L}[u]=0$ and $u_{\text{P}}$ be the solution to (1), is it simply
$$u_{\text{G}}=u_{\text{H}}+u_{\text{P}}$$
As in the case of linear ordinary differential equations? Or is it not that simple?

Thanks everyone.


问:
Determine conic section $x^2-4xy+4y^2-6x-8y+5=0$ and its center
So I got a task to determine the conic section of the following:

$$x^2-4xy+4y^2-6x-8y+5=0$$

I started using matrices and got to the equation :

$$5x`^2 + \frac {10x`}{\sqrt5}-\frac {20y`}{\sqrt5}+5=0$$

I completed the square and got to the equation :

$$\frac {\sqrt5(x`+\frac {1}{\sqrt5})^2}{4} + \frac {1}{\sqrt5}=y`$$

So I can see it's most likely a parabola.

But I'm confused - How can I surely know ? How can I find the "center" (tasked asked for center but probably meant vertex).

I know that somehow I need to find $x`$</span> and <span class="math-container">$y`$ and then multiply it by my orthogonal $P$ matrix and this will be the center.

答:
For large $x$ and $y$, the equation becomes

$$x^2-4xy+4y^2=(x-2y)^2=0$$

indicating a parabola, with the symmetry line $y=\frac12x$. Let the tangential line to the vertex  be $y=-2x+b$ and substitute it into the conic $x^2-4xy+4y^2-6x-8y+5=0$ to get
$$25x^2+10(1-2b)x+4b^2-8b+5=0$$
Its discriminant has to be zero, yielding $b=1$, and the equation reduces to

$$25x^2-10x+1=(5x-1)^2=0$$

Then, solve for the vertex $(\frac15,\frac35)$.


问:
some number theory method can be used to solve this example quickly?
I have extract some notes from my notes:

one way to found which of four example is uniform function is that try by hand and take some examples. I will search for a method that we can easily infer just the second one has uniform capability. is there any hint or idea to quickly choose the second one instead of try by hand and examples one by one?

for the last one:
$h(1)=1, (h3)=4, h(2)=4$ so means not uniform because $x=3$ and $x=2$ has the same slot.

答:
You want to distribute stuff to $k$ slots. Computing over modulo $k-1$ would not make you attain $k$ slots, hence immediately, we reject the first and third choice.

You can then try some simple number, say when $k=4$ to reject the last case.

Note that $k-1$ and $k$ are coprime. if $a$ and $k$ are coprime, we know that $ax \pmod{k}$ would achieve a bijection.


问:
Can a direct product $R\times X$ of two integral domains $R$ and $X$ ever be an integral domain?
Assuming we are using the standard definition of a direct product, I think this is never the case.

An integral domain is a commutative ring with unity that has no zero divisors. Just the fact that both $X$ and $R$ have a multiplicative identity element, allows us to pick two elements of this direct product set, specifically $(r, 0_X)$ and $(0_R, x)$ for any $r\in R, x\in X$ and we will always be able to get a zero as a product of two non-zero elements.

Am I correct here, or is there perhaps something that I'm missing?

Note that I did consider the case where both $R$ and $X$ are trivial, but by definition of integral domain, trivial rings cannot be considered as integral domains.

Any and all help is appreciated! Thank you in advance.

答:
A product of rings is an integral domain if and only if exactly one of them is an integral domain and all the others are the zero ring. Otherwise there's indeed a phenomenon like $(1,0)\cdot (0,1)=(0,0)$.

Added: I should point out that my assertion works under the convention that the $0$ is not an integral domain: for, if it were, then the product of the empty family would be the "$0$ domain", but it wouldn't be true that exactly one of the rings in the family is a domain (since exactly zero of them are). So the correct formulation in the (non-standard) case where we allow $0$ to be a a domain would be that a product of rings is "domain-or-zero" if and only if all of them are domains-or-zero and at most one of them is non-zero.


问:
$E[P(X|Y)]=\sum_iP(X|Y=i)P(Y=i)$?
I found an equality, but I don't see why it is true. I hope you can help me see it.

Why is it true that $E[P(X|Y)]=\sum_iP(X|Y=i)P(Y=i)$?

Thank you!

答:
$P(X|Y)$ is a function of $Y$, and we could write $f(Y) = P(X|Y)$ for some function $f$. Then
$$
\mathbb{E}[P(X|Y)] = \mathbb{E}(g(Y)) = \sum_{y} g(y)\mathbb{P}(Y=y) = \sum_{y} P(X|Y=y)\mathbb{P}(Y=y)
$$

Title: $E[P(X|Y)]=\sum_iP(X|Y=i)P(Y=i)$?


问:
Backward and forward Kolmogorov equation for pure birth
I am dealing with a pure birth process and trying to solve the forward and backward kolmogorov equations. I am stuck solving the backward equation.

$\begin{bmatrix}P'_{0,0}(t)&P'_{0,1}(t)\\P'_{1,0}(t)&P'_{1,1}(t)\end{bmatrix}=
\begin{bmatrix}- \lambda& \lambda\\0&0\end{bmatrix} \begin{bmatrix}P_{0,0}(t)&P_{0,1}(t)\\P_{1,0}(t)&P_{1,1}(t)\end{bmatrix} $

Where, $P'_{i,j}$ is the derivative.

I did try to attempt this and got the following

$P'_{00}(t) = \lambda - 2\lambda P_{00}(t)$

I am not sure how to proceed further, and would love if you could show 1) that the back ward and forward Kolmogorov equations are equal 2) How to solve the backward kolmogorov equations

答:
Note that $P_{1,0}(t)$ is actually a constant since the derivative is $0$.

$$P'_{0,0}(t) = -\lambda P_{0,0}(t) + \lambda P_{1,0}$$

$$P'_{0,0}(t) +\lambda P_{0,0}(t) = \lambda P_{1,0}$$

$$e^{\lambda t}P'_{0,0}(t) +\lambda e^{\lambda t}P_{0,0}(t) = \lambda e^{\lambda t}P_{1,0}$$

$$\frac{d}{dt}\left( e^{\lambda t}P_{0,0}(t) \right)= \lambda e^{\lambda t}P_{1,0}$$

$$e^{\lambda t}P_{0,0}(t) = e^{\lambda t} P_{1,0} + A $$

$$P_{0,0}(t) = P_{1,0}+Ae^{-\lambda t}$$

Also by symmetry,

$$P_{0,1}(t) = P_{1,1}+Be^{-\lambda t}$$

Try to think of what relationship can you say about $P_{0,0}$ and $P_{0,1}$ to relate $A$ and $B$ together.


问:
In what sense is the tensor algebra the most general (unital, associative) algebra
Terminology:

Since I want to consider only unital, associative algebras, algebra will be unital and associative unless otherwise stated.

An ideal will be a two-sided ideal unless otherwise stated.

Vector spaces will be over an infinite field $\mathbb F$ with characteristic 0.

Background:

Let $V$ be a finite dimensional vector space and let $T(V)$ be its tensor algebra. It is often said that $T(V)$ is the "most general" algebra that contains $V$ as a subspace. This can also be seen by noting that intuitively $T(V)$ is the algebra generated by $1,e_1,...,e_n$ ($e_1,...,e_n$ is a basis of $V$) in a way that $1$ is the unit, the multiplication is associative but from that point on, no additional relations are assumed to hold about the generators.

The tensor algebra also observes the universal property, namely that if $A$ is an algebra, $\phi :V\rightarrow A$ is a linear map, then there exists a unique algebra homomorphism $\phi_\otimes:T(V)\rightarrow A$ that satisfies $\phi=\phi_\otimes\circ i$ where $i:V\rightarrow T(V)$ is the inclusion.

I would like to understand that in what sense the tensor algebra is the most general algebra containing $V$. In particular, this can mean one of several things:

Hypothesis 1: There is no such algebra $A$ that contains $V$ as a subspace and $T(V)$ is a sub algebra of $A$.

Hypothesis 2: There is no such algebra $A$ that contains $V$ as a subspace and it has an ideal $I$ such that $T(V)=A/I$.

Hypothesis 3: If $A$ is an algebra that contains $V$ as a subspace, then there is an ideal $I\vartriangleleft T(V)$ of $T(V)$ such that $A=T(V)/I$.

Hypothesis 3 seems at least believable, since pretty much all famous algebras containing $V$ arise as quotients of the tensor algebra (the exterior algebra, the symmetric algebra, the Clifford and Weyl algebras, the universal enveloping algebra if $V$ is a Lie algebra etc.).

On the other hand, I am not convinced that the universal property stated above actually implies Hypothesis 3.

In particular, assume that $A=T(V)/I$. If $A$ is to contain the entirety of $V$ then we must have $I\cap V=0$. I don't know if $T(V)$ has a nontrivial ideal that intersects $V$ nontrivially or not, but nonetheless let's make this assumption. Then the quotient map $\pi:T(V)\rightarrow T(V)/I$ is surjective. Moreover if $\pi:T(V)\rightarrow A$ is a surjective algebra morphism then $\ker\pi$ is an ideal of $T(V)$ and $A=T(V)/\ker\pi$.

If the universal property is to imply Hypothesis 3, then we must be able to show that if $\phi:V\rightarrow A$ is injective (i.e. the image of $V$ in $A$ is faithful), then $\phi_\otimes :T(V)\rightarrow A$ must be surjective. The universal property does guarantee that the image of $V$ via $\phi_\otimes$ is faithful but to have $A=T(V)/\ker\phi_\otimes$, we must have surjectivity. I don't see it being implied.

Question:

Are any of the hypotheses presented above true? If not, then in what sense is the tensor algebra the most general algebra containing $V$? The "raw" statement of the universal property does not convince me that $T(V)$ is "the most general", what are some additional implications of the universal property that makes $T(V)$ "the most general"?

答: Further to Qiaochu Yuan's answer, another way of understanding the claim that $T(V)$ is the most general $k$-algebra containing $V$ as a subspace is this: any equation (possibly involving parameters in $V$) that holds in $T(V)$ holds in any $k$-algebra containing $V$.


问:
How to solve a binomial expression up to and including $x^2$
I'm doing a binomial expansion that uses $(1+ax)^2$ form and I was wondering how to solve the bottom expression up to and including the term $x^2$

$$\frac{19x -3}{(1+2x)(3-4x) }$$

would I rearrange $(19x-3)(1+2x)^{-1} 3^{-1}\left(1-\frac{4x}{3}\right)^{-1}$ and just expand it all out and multiply them together?

Thank you!

答:
Hint: It is often convenient to start with a partial fraction decomposition
\begin{align*}
\frac{19x-3}{(1+2x)(3-4x)}=-\frac{5}{2}\cdot\frac{1}{1+2x}+\frac{9}{2}\cdot\frac{1}{3-4x}
\end{align*}
and continue with a geometric series expansion.


问:
How to solve $5x+9y=n$?
Show that $n_0 = 32$ is the smallest value of $n$ for which the equation $5x+9y=n$ has a solution in $(\mathbb N \cup \{0\})^2$ for all $n \geq n_0$.

Wouldn't $(x,y)=(1,1)$ be a valid solution for $n=14$?

The example that appears before this question consider $(x,y)=(1,1)$ as a solution for $3x+5y=n$ in $(\mathbb N \cup \{0\})^2$ for $n \geq 8$.

答:
$(x,y)=(1,1)$ is a valid solution for $n=14$, but I think the question asks for the smallest $n_0$ such that $5x+9y=n$ has a soution for all $n\ge n_0$, and there's no solution for $5x+9y=16>14$ in non-negative integers.

Addendum to answer OP's additional question:

$31$ cannot be expressed as $5x+9y$ with non-negative integers $x,y$, but

$32=5(1)+9(3)$,
$33=5(3)+9(2)$,
$34=5(5)+9(1)$,
$35=5(7)+9(0)$,

and
$36=5(0)+9(4)$,

and  if $n-5$ can be expressed as $5x+9y$ with non-negative integers then $n$ is $5(x+1)+9y$.

Title: How to solve $5x+9y=n$?


问:
Hilbert space homomorphisms and operators
When studying Hilbert spaces, the functions I've mostly seen being dealt with is linear functions. Of course, a big part of that is that differential operators are linear and many other interesting examples are too.

However, thinking about mathematical spaces as being "dual" to their set of "structure preserving functions" (I guess the Yoneda lemma makes this rigorous? I'm not strong in category theory), this doesn't sit right with me. Actually, what should the morphisms on  Hilbert spaces be? What I mean is that a bijective morphism from a Hilbert space to itself, with an inverse function that is also a morphism , should be a Hilbert space isomorphism.

Come to think about it, in metric spaces I don't know what the morphisms inducing the "correct" isomorphisms (isometries) should be.

I might have to learn some more category theory besides the basic stuff, and if that's the cause, feel free to point that out :).

答:
It is actually surprisingly complicated to give an answer to this that is even halfway satisfactory; Hilbert spaces are very unusual. First, your question about metric spaces is easier to answer: you can take the morphisms to be short maps, which are maps $f : X \to Y$ satisfying

$$d_Y(f(x_1), f(x_2)) \le d_X(x_1, x_2).$$

This turns out to be a nice and natural choice for several reasons; for more about this you can read about Lawvere metric spaces. But for starters it's clear that an isomorphism in the category of metric spaces and short maps is an isometry (not a homeomorphism as in the usual category of metric spaces and continuous maps, which is better understood as the category of metrizable spaces).

It is also a nice and natural choice for morphisms in the category of Banach spaces; here short linear maps can equivalently be characterized as linear maps with operator norm $\le 1$. The resulting category of Banach spaces has many pleasant features, such as being complete, cocomplete, and closed symmetric monoidal. As two tantalizing hints about how this category works, in this category the coproduct of $S$ many copies of $\mathbb{C}$ is $\ell^1(S)$ (so this is also the free Banach space on $S$), and the product is $\ell^{\infty}(S)$. You can read more about this in my blog post Banach spaces (and Lawvere metrics, and closed categories) which also includes some relevant material about Lawvere metric spaces.

Hilbert spaces are in particular Banach spaces, so we can talk about the category of Hilbert spaces and short maps. But this isn't quite right: among other things, there's no natural way to talk about adjoints in this category (as far as I know). The fix to this is to axiomatize adjoints as a fundamental feature: this leads to the notion of a dagger category, which is a category equipped with a notion of adjoints

$$(-)^{\dagger} : \text{Hom}(a, b) \to \text{Hom}(b, a)$$

satisfying some axioms.

The dagger category of Hilbert spaces and continuous linear maps, together with the dagger structure given by adjoints, recovers everything: the dagger can be used to recover the inner product (this is a nice exercise), and in any dagger category we can talk about unitary maps and hence unitary isomorphisms (as well as self-adjoint maps, skew-adjoint maps, isometries, and other fun stuff like that). We also get the $C^{\ast}$-algebra structure on every endomorphism algebra. You can read a bit more about this in my blog post Hilbert spaces (and dagger categories), although I don't go into a ton of detail on the dagger structure.

Unlike the case of Banach spaces, there is no categorical formalism I'm aware of which has the property that $\ell^2(S)$ is the free Hilbert space on a set $S$, which I'm a little unhappy about.


问:
Writing Fourier transform integral in terms of real-valued functions
I am interested in the following integral.

$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$

One way (for example, by using Mathematica) to solve it is in the following way.

$$ f(x,t) = \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}} $$

I believe that $f(x,t)$ should be real valued, but it is not clear to me how to simplify given expression so that it consists only of real valued functions.

答:
The two parts of your answer are obviously a complex conjugate pair.

Let's just take one of the terms and double its real part:

$$\begin{align*}f(x,t) &= \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}}\\
\\
&= \sqrt{\pi}\;\Re\left[\frac{e^{-ipt}e^{-\frac{x^2}{4}\frac{1}{1+iqt}}}{\sqrt{1+iqt}}\right]\\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{1-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{re^{i(\theta +2n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[{\sqrt[4]{1+q^2t^2}e^{i(\theta/2 +n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[e^{i\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\mathrm{arctan2}(-qt,1)}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
\end{align*}$$

You can take $n=0$ for the principal value of the square root.  Or you can take $n=1$ and effectively just change the sign of the answer.


问:
Calculate two vectors given their norms and angle

For two vectors $\mathbf{u,v}$ in $\mathbb{R}^n$ euclidean space, given:

$\|\mathbf{u}\| = 3$

$\|\mathbf{v}\| = 5$

$\angle (\mathbf{u,v})=\frac{2\pi}{3}$

Calculate the length of the vectors

$4\mathbf{u}-\mathbf{v}$

$2\mathbf{u}-7\mathbf{v}$

I'm not sure how to approach this with the given information

With the formula for the angle between the two vectors being

$$\cos \theta=\frac{\mathbf{u\cdot v}}{\|\mathbf{u}\|\cdot\|\mathbf{v}\|}$$

I already have the denominator, but how do I get the point product of u and v in this case?

A point to start would be most appreciated

答: Hint:  You should be able to use the formula you quote to calculate $\mathbf{u\cdot v}$.  Then use the linearity of the dot product to expand out $\mathbf{(4u-v)\cdot (2u-7v)}$</p>


问:
Comparing two box plots
Say, these are two boxplots for ratings for a movie (1 to 10) say Star Wars The Phantom Menace.
The first boxplot are ratings received by various experts for the same movie
The second boxplot are ratings received by the audience for the same movie again.

I see that the
First Range: 7-1=6
Second Range: 10-0=10

First max: 7
Second max: 10

Comparing the two box plots above, Is it fair to say that that there's more variation in the Second boxplot data?

Also, Is it fair to say that on an average the rating giving by experts is higher than that of the audience?

答:
I think you can say that the variation is higher in the second boxplot data. However, I don't think you can say that about the average rating since the second boxplot data has so much more variation, and the first boxplot's data has its entire range fall within the second box plots.

Is this data that you have on hand? If so, you may want to adjust your whiskers for the box plots so that the upper/lower fences fall +/- 1.5*IQR, and mark the remaining data points as outliers. Then we will have a better grasp about the distribution of the data. Otherwise, hard to say.

Title: Comparing two box plots


答:
Adding to @QiaochuYuan's good answer:

As he notes, really "the tensor algebra" (on a module $M$ over a commutive ring $R$ with identity) is "the universal associative $R$-algebra...

That is, to my mind, things are clearest when we first characterize the universal associative algebra of a (unital) $R$-module $M$ as an associative $R$-algebra $A(M)$ with $R$-module map $j:M\to A(M)$ (not assumed injective...) such that every $R$-module map $\varphi:M\to B$ to an $R$-algebra $B$ gives rise to a unique $R$-algebra map $\Phi:A(M)\to B$ through which $\varphi$ factors.

Then, separately, we can prove existence, granting the existence/properties of tensor products, by showing that the so-called tensor algebra $\bigoplus_n M^{\otimes n}$ (with suitable multiplication) fits the requirements of the universal associative algebra.

Sure, it's convenient to be able to write elements of the universal associative algebra as tensors (as opposed to having to make up some other notation), but this does create potential confusions about tangential issues.

EDIT: also, as a significantly simpler analogue of the Poincare'-Birkhoff-Witt theorem for universal enveloping algebras of Lie algebras, the construction of universal associative algebras as direct sums of tensor powers shows that it is "as large as possible"...

Title: In what sense is the tensor algebra the most general (unital, associative) algebra

Body:
Terminology:

Since I want to consider only unital, associative algebras, algebra will be unital and associative unless otherwise stated.

An ideal will be a two-sided ideal unless otherwise stated.

Vector spaces will be over an infinite field $\mathbb F$ with characteristic 0.

Background:

Let $V$ be a finite dimensional vector space and let $T(V)$ be its tensor algebra. It is often said that $T(V)$ is the "most general" algebra that contains $V$ as a subspace. This can also be seen by noting that intuitively $T(V)$ is the algebra generated by $1,e_1,...,e_n$ ($e_1,...,e_n$ is a basis of $V$) in a way that $1$ is the unit, the multiplication is associative but from that point on, no additional relations are assumed to hold about the generators.

The tensor algebra also observes the universal property, namely that if $A$ is an algebra, $\phi :V\rightarrow A$ is a linear map, then there exists a unique algebra homomorphism $\phi_\otimes:T(V)\rightarrow A$ that satisfies $\phi=\phi_\otimes\circ i$ where $i:V\rightarrow T(V)$ is the inclusion.

I would like to understand that in what sense the tensor algebra is the most general algebra containing $V$. In particular, this can mean one of several things:

Hypothesis 1: There is no such algebra $A$ that contains $V$ as a subspace and $T(V)$ is a sub algebra of $A$.

Hypothesis 2: There is no such algebra $A$ that contains $V$ as a subspace and it has an ideal $I$ such that $T(V)=A/I$.

Hypothesis 3: If $A$ is an algebra that contains $V$ as a subspace, then there is an ideal $I\vartriangleleft T(V)$ of $T(V)$ such that $A=T(V)/I$.

Hypothesis 3 seems at least believable, since pretty much all famous algebras containing $V$ arise as quotients of the tensor algebra (the exterior algebra, the symmetric algebra, the Clifford and Weyl algebras, the universal enveloping algebra if $V$ is a Lie algebra etc.).

On the other hand, I am not convinced that the universal property stated above actually implies Hypothesis 3.

In particular, assume that $A=T(V)/I$. If $A$ is to contain the entirety of $V$ then we must have $I\cap V=0$. I don't know if $T(V)$ has a nontrivial ideal that intersects $V$ nontrivially or not, but nonetheless let's make this assumption. Then the quotient map $\pi:T(V)\rightarrow T(V)/I$ is surjective. Moreover if $\pi:T(V)\rightarrow A$ is a surjective algebra morphism then $\ker\pi$ is an ideal of $T(V)$ and $A=T(V)/\ker\pi$.

If the universal property is to imply Hypothesis 3, then we must be able to show that if $\phi:V\rightarrow A$ is injective (i.e. the image of $V$ in $A$ is faithful), then $\phi_\otimes :T(V)\rightarrow A$ must be surjective. The universal property does guarantee that the image of $V$ via $\phi_\otimes$ is faithful but to have $A=T(V)/\ker\phi_\otimes$, we must have surjectivity. I don't see it being implied.

Question:

Are any of the hypotheses presented above true? If not, then in what sense is the tensor algebra the most general algebra containing $V$? The "raw" statement of the universal property does not convince me that $T(V)$ is "the most general", what are some additional implications of the universal property that makes $T(V)$ "the most general"?


问:
Bivariate function that produces "ridge"-like surface
I'm trying to find the right vocabulary to describe a "ridge" like distribution that depends on two variables and how similar they are to each other. I'm imagining it looks something like the following. The utility of such a "ridge" like function is motivated by physical modeling. If $x$ is some expected distance and $y$ is the observed distance, then the probability $z$ surface should peak when $x - y = 0$ .

I admit this sounds really silly to ask, but I'm not finding relevant answers by searching for "ridge" functions. Would someone be able to point me in the right direction?

答:
Would a sufficiently rotated/scaled Two-dimensional Gaussian function work?  Below, I've plotted $$f(x,y) = \exp\left(- \left(a(x - x_o)^2 + 2b(x-x_o)(y-y_o) + c(y-y_o)^2 \right)\right)$$ where

\begin{align}
a & = \frac{\cos^2\theta}{2\sigma_X^2} + \frac{\sin^2\theta}{2\sigma_Y^2} \\[4pt]
b & = -\frac{\sin2\theta}{4\sigma_X^2} + \frac{\sin2\theta}{4\sigma_Y^2} \\[4pt]
c & = \frac{\sin^2\theta}{2\sigma_X^2} + \frac{\cos^2\theta}{2\sigma_Y^2}.
\end{align}
Here $\theta$ is the angle of clockwise rotation.

See CalcPlot3D for the full example, here I've used the variables $a = \sigma_{x} = 2$ and $b = \sigma_{y} = 1$ centered at the origin and rotated an angle of $t = \theta = \frac{\pi}{4}$.

Body:
Would a sufficiently rotated/scaled Two-dimensional Gaussian function work?  Below, I've plotted $$f(x,y) = \exp\left(- \left(a(x - x_o)^2 + 2b(x-x_o)(y-y_o) + c(y-y_o)^2 \right)\right)$$ where

\begin{align}
a & = \frac{\cos^2\theta}{2\sigma_X^2} + \frac{\sin^2\theta}{2\sigma_Y^2} \\[4pt]
b & = -\frac{\sin2\theta}{4\sigma_X^2} + \frac{\sin2\theta}{4\sigma_Y^2} \\[4pt]
c & = \frac{\sin^2\theta}{2\sigma_X^2} + \frac{\cos^2\theta}{2\sigma_Y^2}.
\end{align}
Here $\theta$ is the angle of clockwise rotation.

See <a href="https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/?type=z;z=exp(-(1/2)(((cos(t)/a)%5E2+(sin(t)/b)%5E2)x%5E2+(-sin(2*t)/a%5E2+sin(2*t)/b%5E2)xy+((cos(t)/b)%5E2+(sin(t)/a)%5E2)y%5E2));visible=true;umin=-5;umax=5;vmin=-5;vmax=5;grid=30;format=normal;alpha=-1;constcol=rgb(255,0,0);view=0;contourcolor=red;fixdomain=false&type=slider;slider=t;value=0.7853975;steps=100;pmin=-2;pmax=2;repeat=true;bounce=true;waittime=1;careful=false;noanimate=false;name=-1&type=slider;slider=a;value=2;steps=100;pmin=-2;pmax=2;repeat=true;bounce=true;waittime=1;careful=false;noanimate=false;name=-1&type=slider;slider=b;value=1;steps=100;pmin=-2;pmax=2;repeat=true;bounce=true;waittime=1;careful=false;noanimate=false;name=-1&type=window;hsrmode=3;nomidpts=true;anaglyph=-1;center=-6.021743100045566,12.938427845716461,10.414267324804355,1;focus=0,0,0,1;up=0.31492374305611076,-0.501852743862459,0.805584793511676,1;transparent=false;alpha=140;twoviews=false;unlinkviews=false;axisextension=0.7;xaxislabel=x;yaxislabel=y;zaxislabel=z;edgeson=true;faceson=true;showbox=true;showaxes=true;showticks=true;perspective=true;centerxpercent=0.5;centerypercent=0.5;rotationsteps=30;autospin=true;xygrid=false;yzgrid=false;xzgrid=false;gridsonbox=true;gridplanes=false;gridcolor=rgb(128,128,128);xmin=-5;xmax=5;ymin=-5;ymax=5;zmin=0;zmax=2;xscale=1;yscale=1;zscale=1;zcmin=-4;zcmax=4;zoom=0.333333;xscalefactor=1;yscalefactor=1;zscalefactor=3" rel="nofollow noreferrer">CalcPlot3D for the full example, here I've used the variables $a = \sigma_{x} = 2$ and $b = \sigma_{y} = 1$ centered at the origin and rotated an angle of $t = \theta = \frac{\pi}{4}$.


问:
Solution to $y′=x+y,\, y(0)=1$
I am trying to learn numerical method for differentials. It is key for me to understand this stuff right now, but I've come across an obstacle.

The books I read about these all seem to often say 'exact solution is easily found to be' in more or less the same way. I clearly have an issue with this, because I do not know any practical method, or even the name of the method I am suppose to learn, to find this sort of solution 'easily' or sometimes at all.

We shall illustrate our methods by applying them to the simple problem
$y′=x+y, y(0)=1, (3)$ which we call our benchmark problem. This
differential equation in $(3)$ is clearly linear, and the exact solution
is easily found to be $y=2e^x − x − 1$.

I look at this, and I don't understand. I guess I can say I covered one of these ideas which is that $y'=y, y(0) = 1$ because it is glearingly obvious that the only thing that differentiates to itself is $e^x$.

Now, $y' = x+y$, how would I approach that? What is the method behind the solution to get there? I tried to look at the solution instead, but $y'$ of the solution for me is $2e^x - 1$.

This must be something really trivial, but I just don't get it.

答:
I solve

$y' = y + x, \; y(0) = 1 \tag 1$

in another "real way", as follows: write the equation as

$y' - y  = x; \tag 2$

make the clever observation that

$(e^{-x}y)' = -e^{-x}y + e^{-x}y'$
$= e^{-x}y' - e^{-x}y = e^{-x}(y' - y); \tag 3$

in light of (2) this becomes

$(e^{-x}y)' = xe^{-x}; \tag 4$

integrate 'twixt $0$ and any value of $x$:

$e^{-x}y(x) - y(0) = \displaystyle \int_0^x (e^{-u}y(u))'  \; du = \int_0^x ue^{-u} \; du; \tag 5$

the right-hand integral may be integrated by parts to yield

$\displaystyle \int_0^x ue^{-u} \; du = -(x + 1)e^{-x} + 1; \tag 6$

substituting this and the initial condition $y(0) = 1$ into (5):

$e^{-x}y(x) - 1 = -(x + 1)e^{-x} + 1, \tag 7$

or

$y(x) = 2e^x - x - 1. \tag 8$

Check:

from (8),

$y' = 2e^x - 1, \tag 9$

$y'- y = x, \tag{10}$

$y' = y + x, \tag{11}$

$y(0) = 1; \tag{12}$

(11) and (12) together are (1). $ \checkmark$

Title: Solution to $y′=x+y,\, y(0)=1$


问:
Is $\mathbb{R}^{n} - B\left[ 0;r\right]$ connected, with $n>1$ and $r > 0$?
I need to prove that the complement in $\mathbb{R}^{n}$, $n>1$ of a limited subset $X \subset \mathbb{R}^{n}$ has only one ilimited connected component. This lies in prove that $\mathbb{R}^{n} - B\left[ 0;r\right]$ (where $B\left[0;r\right]$ is the closed ball with center in the origin and radius $r$) is connected, since it will be the only one ilimited connected component (any another component different from this will be in the closed ball).

How to prove that? There is another way to prove the statement whitout the usage of this? Thanks in advance!

答:
It is not hard to check that $\mathbb{R}^2 \backslash B[0,r]$ is connected.

Let $p$, $q$ be points in $\mathbb{R}^n\backslash B[0,r]$.  Let $V$ a two dimensional subspace of $\mathbb{R}^n$ containing $p$, $q$. By the above, $V\backslash B[0,r]$ is connected. We conclude that $p$, $q$ are in the same component of $\mathbb{R}^n \backslash B[0,r]$.

Let $X$ be a bounded subset of $\mathbb{R}^n$.  Consider $r>0$ such that $X \subset B[0,r]$.  Now  $\mathbb{R}^n \backslash B[0,r]$ is contained in one component of $\mathbb{R}^n\backslash X$. The other components (if they exist) are contained in $B[0,r]$, so bounded.


问:
Supremum of a bounded sequence is accumulation point.
Let $(a_n)_{n \in \mathbb{N}}$ be a bounded sequence in $\mathbb{R}$. Denote by $L := \sup\{a_n : n \in \mathbb{N}\}$ the Supremum. Suppose that $L \notin \{a_n : n \in \mathbb{N}\}$. How can I show that $L$ is an accumulation point,i.e., I have to show hat there is a subsequence $(a_{n_{k}})_{k \in \mathbb{N}}$ of $(a_n)_{n \in \mathbb{N}}$ such that
$$
\lim\limits_{k \to \infty} a_{n_{k}} = L.
$$
In other words
$$
\limsup\limits_{n \to \infty} a_n = \sup\{a_n : n \in \mathbb{N}\}.
$$
My idea was to give a proof by contradiction. But this seems hopeless. First I have to show that for $\varepsilon > 0$ there exist a $m \in \mathbb{N}$ such that for all $k > m$ we have
$$
|a_{n_{k}} - L| < \varepsilon.
$$
This would show me that there exists an infinite set of terms of $(a_n)_{n \in \mathbb{N}}$ which satisfy $|a_n - L| < \varepsilon$. How can I prove this?
Then I think it is not difficult to construct a subsequence which converges to $L$. I only need help for the first part.

答:
Suppose $$
a_0 = \limsup_{n \to \infty} a_n = \lim_{k \to \infty} \sup_{n \geq k} a_n \neq L.$$ When viewing $\tilde{a}_k = \sup_{n \geq k} a_n$, we find that $\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \tilde{a_n}$. Then, $a_0 < L$ since $\{a_i\}$ is bounded above by $L$ so if we take $\varepsilon = |a_0  - L|$, then there is a natural number $N$ so that we have that $|\tilde{a}_N - a_0| < \varepsilon.$ This then gives $$|\tilde{a}_N - L| = |(L - a_0) - (\tilde{a}_N - a_0)| \geq |a_0 - L| - |\tilde{a}_N - a_0|  > 0.$$ That means that $\sup_{k \geq N} a_k  < L$. Since $$L = \sup_{n \geq 1} a_n = \max\left(\sup_{k \geq N} a_k, \max_{k \leq N} a_k\right)$$ but $\sup_{k \geq N} a_k  \neq L$, it must be the case that $L = \max\{a_1, \dots, a_N\}$ which implies that $L = a_j$ for some $j \leq N$. This proves the claim by contrapositive.


问:
MGF Expect Random
If I have a two norm distributed RV, i.e. $L \sim N(m,s^2)$ and $M \sim N(0,1)$, with coo coefficient $ \operatorname{Corr}(L,M) =\rho$, what is then:

$$ E[e^L M^2] = \mathop{?} $$

答:
Assuming that $A$ and $B$ are jointly normal, then $X=(A,B)$ has a normal distribution with the mean $\mu$ and the covariance matrix $\Sigma$ given by

$$ \mu = (m, 0) \qquad\text{and}\qquad \Sigma = \begin{pmatrix} s^2 & \rho s \\ \rho s & 1 \end{pmatrix}. $$

Then we know that the moment generating function of $X$ is given by

$$ \mathbb{E}[e^{\xi \cdot X}] = e^{\xi\cdot\mu + \frac{1}{2}\xi\cdot\Sigma \xi}, \qquad \xi \in \mathbb{R}^2. $$

Now plugging $\xi = (1, t)$, we have

$$ \mathbb{E}[e^{A}e^{tB}] = e^{m + \frac{1}{2}(s^2 + 2\rho s t + t^2)}. $$

Differentiating both sides with respect to $t$ twice and plugging $t = 0$, we get

$$ \mathbb{E}[e^{A}B^2] = e^{m + \frac{1}{2}s^2}(1 + s^2\rho^2). $$

Title: MGF Expect Random


问:
How to prove that the subsequential limits are the same?
Given two sequences: {$a_n$}$_{n=0}^{\infty}$  and {$b_n$}$_{n=0}^{\infty}$

Prove that: if $ \lim_{n\to\infty}$ $ (a_n-b_n ) = 0$  then all the subsequential limits of both sequences are the same

I tried to prove it by proving that each option of subsequence limit of these both sequences are equal, tried to take two subseqenece limits k and l for both of them and proved that they are the same, but I have a problem with that if the subsequence limit is infinity then i didn't know what to do...
Any ideas of how to solve it?

答:
It’s very similar to the way you deal with finite limits.

Let $\langle a_{n_k}:k\in\Bbb N\rangle$ be a subsequence of $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_{n_k}:k\in\Bbb N\rangle$ the corresponding subsequence of $\langle b_n:n\in\Bbb N\rangle$, and suppose that $\lim_{k\to\infty}a_{n_k}=\infty$. For each $x\in\Bbb R$ there is a $k_x\in\Bbb N$ such that $a_{n_k}>x$ whenever $k\ge k_x$, and since $\lim_{n\to\infty}(a_n-b_n)=0$, there is an $\ell\in\Bbb N$ such that $|a_{n_k}-b_{n_k}|<1$ whenever $k\ge\ell$. Now let $x\in\Bbb R$, and suppose that $k\ge\max\{\ell,k_{x+1}\}$; then $a_{n_k}>x+1$, and $|a_{n_k}-b_{n_k}|<1$, so $b_{n_k}>x$. Thus, $\lim_{k\to\infty}b_{n_k}=\infty$, as desired.


问:
Infinite examples of non isomorphic Lie groups with Lie algebra $\mathfrak{so}(3)$
I'm introducing myself to Lie Groups theory, and I read Ado's Theorem says that every finite real Lie algebra is isomorphic to a Lie subalgebra of $\mathfrak{gl}(n,\mathbb{R})$.

I found the following problem as a consequence of Ado's Theorem:
"Give infinite examples of non isomorphic Lie groups with Lie algebra  $\mathfrak{so}(3)$."

Because of the following statement:

"If a Lie group has Lie algebra $\mathfrak{g}$, then it is isomorphic to $G/\Gamma$, where $G$ is the simply connected Lie group with Lie algebra $\mathfrak{g}$, and $\Gamma$ is a discrete subgroup of the center of $G$, $Z(G)$."

I think I must work with a group $G$ non simpy connected, but I don't achieve the examples. Could you help me?

答:
One cannot find an infinite number of connected Lie groups with Lie algebra $\mathfrak{so}(3)$ because such a Lie group is either $\mathrm{SO}(3)$, either $\mathbb{S}^3$, the universal cover of $\mathrm{SO}(3)$. This is because $\mathbb{S}^3$ is a simply connected Lie group with Lie algebra $\mathfrak{so}(3)$ and has center $\{\pm 1\}$, thus the only discrete subgroups of its center are $\{1\}$ and $\{1,-1\}$, and the results follows from what you stated.

On the other hand, there are infinitely many non-connected Lie groups with Lie algebra $\mathfrak{so}(3)$. For example, $\mathbb{S}^3 \times \mathbb{Z}/n\mathbb{Z}$ for all $n \geqslant 2$.


问:
A Problem on Time Complexity of Algorithms
I want no know if the following problem is solved or not, or how can I solve it?

Problem: For every integer $t$, Is there any problem that can be verified in $O(n^{s})$ but its solution can be found in $T(n)=\omega(n^{st})$, i.e., its solution can not be found in $O(n^{st})$.

By verifying, I mean that given a candidate solution $y$, we can judge whether $y$ is correct or not in time $O(n^s)$.

答: You have an array of $n^{st}$ integers and want to check if there is a zero. You check it in $O(1)$, if the index is given. You find it just by the brute-force search in $O(n^{st})$.</p>


答:
You can apply standard divisibility tests for certain prime factors, and thereby divide out any factors that "pass" to get smaller numbers.  With base 10 representation we have reasonably easy tests for $2,3,5,7,11,13,37$.  Letting $a=2021019$ and $b=1431471$ we apply these tests:

$2$ -> both odd final digits, both fail.

$3$ -> both give a digit sum divisible by $3$, divide both inputs by $3$ and note that the hlgcd willthen also be divided by $3$.  Thus $a\to673673, b\to477157$.

$3$ again -> we see if we can extract another factor of $3$, but the sum of digits no longer passes for either input.

$5$ -> neither input ends in $0$ or $5$.

$11$ -> alternating sum of digits is divisible by $11$ for $a$ but not for $b$.  Divide $a$ by $11$.  $a\to61243,b=477157$.

$7,13$ -> alternating sum of three-digit groups is divisible by both of these factors for $a$ ($182=13×14$) but not for $b$.  Divide $a$ by both factors or in one step by their product $91$.  $a\to673,b=477157$.

$37$ -> sum of groups of three digits.  This gives just $673$ back for $a$ but since that is $111×6+7$ it will not be divisible by $37$.  For $b$ we get $634=111×6-32$, so this fails.  Retain $a=673,b=477157$.

So we reduce to finding the gcd of 673 and 477157, then we have to multiply by $3$ because we commonly divided $a$ and $b$ by $3$ during the above reductions.  It happens that $477157$ gives a zero remainder when divided by $673$, therefore render $673×3=2019$.

Title: Are there any other ways to find GCD faster?

Body:
Find $GCD$ of $2021019$ and $1431471$

My solution:

$2021019 = 1\cdot(1431471) + 589548$

$1431471 = 2\cdot(589548) + 252375$

$589548 = 2\cdot(252375) + 84798$

$252375 = 2\cdot(84798) + 82779$

$84798 = 1\cdot(82779) + 2019$

$82779 = 41\cdot(2019)$

So $GCD$ of $2021019$ and $1431471$ is $2019$

Are there any other ways to solve this faster?


问:
How this condition work for this set of number?
$\left| \{ A : (A\subseteq  \{1,2,3,..n\})\ \wedge\ (\{3\} \subseteq A)\ \wedge\ (\{5,8\}\cap A = \emptyset) \}\right |$

I need to find set A size for all $n$ cases.

I don't know if i got the question right but for example

$n=0$, $|A|=1$ because of the 2nd conditionn?

$n=3$, does that mean $|A|=3$?

and if $n <5$ would that be $|A|=4$ and etc..?

and what if $ n = 5$, would that be $|A| = 4$ because by the 3rd condition 5 and 8 dont count?

What i've came to so far:

$n<8, |A|=2^{n-2}$

$0 \leq n \leq 8, 1 \leq|A| \leq 6$

答: If $n \ge 8$ you are told that $3$ is in $A$ and $5,8$ are not.  How many binary choices does that leave you?  If $n \lt 3$ you cannot satisfy the requirement that $3$ be in $A$, so there are none.  How about the range from $3$ to $7$?


问:
What is the probability that the Red ﬁsh are the ﬁrst species to become extinct from the pond? A Combinatorial Approach
A pond contains 3 distinct species of ﬁsh, which we will call the Red, Blue, and Green ﬁsh. There are r Red, b Blue, and g Green ﬁsh. Suppose that the ﬁsh are removed from the pond in a random order. (That is, each selection is equally likely to be any of the remaining ﬁsh.) What is the probability that the Red ﬁsh are the ﬁrst species to become extinct from the pond?

I came across this answer here

Answer: Consider an fish tank with n black fishes and m white fishes,  the total number of outcomes = (n+m)!.  Number of ways the last fish is white is m.(m+n-1)!.  Thus the probability is $\frac{m.(m+n-1)!}{(m+n)!} = \frac{m}{m+n}$.
In your problem, if there are r red fishes, b blue fishes, and g green fishes,
Based on the same reasoning as (a), the probability that the Blue fish and the Green fish are the last species to become extinct is $\frac{b}{r+b+g}\text{ and }\frac{g}{r+b+g}$ respectively. Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is$ \frac{b}{r+b}$. Similarly, the conditional probability that the Green fish are the second species to become extinct, given that the Blue fish are the last remaining fish in the pond, is $\frac{g}{r+g}$
Let R be the event that the Red fish are the first species to become extinct in the pond,RBG denote the event that the Red fish first become extinct, followed by the Blue fish, then by the Green fish, and RGB denote the event that the Red fish first become extinct, followed  by the Green fish, then by the Blue fish. Then we can obtain

$P(RBG) = P(G-last)P(RBG/G-last) =\frac{g}{r+b+g}\frac{b}{r+b}$

$P(RGB) = P(B-last)P(RGB/B-last) =\frac{b}{r+b+g}\frac{g}{r+g}$
Finally, we can compute the probability that the Red fsh are the first species to become
extinct in the pond by adding these two.

In the answer, the author used a combinatorial approach when there are only $2$ types of fish. I understood much of the argument except this :

Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is$ \frac{b}{r+b}$.

I am not quite able to visualize why this is true, especially the part in which the green fish are plain ignored. I think ignoring can be a bad idea because the left over green fish can contribute to the number of permutations. Can someone please give a combinatorial argument to prove this?

Thanks a lot!

答:
The probability that red will be the first to have all removed is the same as the probability that red will be the last to have any removed. This is because reversing the order of removal is a bijection. For any permutation $\sigma$ that results in red being the first to have all removed, we can take $\sigma'$ to be that permutation in reverse order. Since all the fish in the tail of $\sigma$ are non-red, all the fish in the head of $\sigma'$ are non-red.  Under this bijection, the claim

Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is $\frac b{r+b}$.

becomes

Furthermore, the conditional probability that the Blue fish are the second species to have any removed, given that Green fish is the first species to have any removed, is $\frac b{r+b}$.

There are a total of $b+r$ non-green fish, and all of them are equally likely to be the first non-green fish removed, so the probability that the first non-green fish is blue is $\frac b{r+b}$.


问:
Show that a limit set is connected?
How do i prove this question:

Show that a limit set is connected (that is, not the union of two disjoint non- empty closed sets)

This question is in the book (Morris W. Hirsch, Stephen Smale - Differential Equations, Dynamical Systems,

and Linear Algebra)page241

答:
As stated in my comment this is not generally true. For example define the vector field
$$
f(x,y)=(y+x(1-x^2-y^2),-x+y(1-x^2-y^2))
$$
on the open set $A=\mathbb{R}^2 \setminus \{q,r\}$ where $q,r$ are two distinct points on the unit circle $C$. Then if $p \notin \{0\} \cup C \setminus \{q,r\}$ one can show that $\omega(p)=C \setminus \{q,r\}$ which is not connected.

If you suppose $\omega(p)$ is compact than we can prove this as follows:

Suppose $\omega(p)$ is not connected. Then there are $A,B$ both closed and non empty such that $\omega(p)=A\cup B$ and $A \cap B= \emptyset$. Take $a \in A$ and since $a \in \omega(p)$ there is a sequence $(s_n)$ such that $s_n \to \infty$ and $\varphi(s_n,p) \to a$ as $n \to \infty$. By the same argument take $b \in B$ and there is a sequence $(r_n)$ such that $r_n \to \infty$ and $\varphi(r_n,p) \to b$ as $n \to \infty$. Denote $d=d(A,B)$. Notice $d>0$ (why?). Using the sequences above we can define (do you see how?) a sequence $(t_n)$ such that $d(\varphi(t_n,p),A)<\frac{d}{2}$ and $d(\varphi(t_{n+1},p),A)>\frac{d}{2}$ for every $n$ odd. Hence, for all odd $n$, the function defined by $h(t)=d(\varphi(t,p),A)$ is continuous on $t_n \leq t \leq t_{n+1}$ and $g(t_n)<\frac{d}{2}$, $g(t_{n+1})>\frac{d}{2}$. By intermediate value theorem there is some $t_n<\overline{t}_n < t_{n+1}$ such that  $d(\varphi(\overline{t}_n,p),A)=g(\overline{t}_n)=\frac{d}{2}$. Notice the sequence $(\varphi(\overline{t}_n,p))$ is contained in the set S=$\{x \mid d(x,A)=\frac{d}{2}\}$. Since $S$ is compact (why?) we can find a subsequence of $(\varphi(\overline{t}_n,p))$ that converges to a point $\overline{p} \in S$ (without lost of generality assume that $\varphi(\overline{t}_n,p) \to \overline{p}$). Hence by definition $\overline{p} \in \omega(p)$. But $\overline{p} \in S$ implies that $d(\overline{p},A)=\frac{d}{2}>0$. Then $\overline{p} \notin A$. On the other hand
$$
d(A,B) \leq d(\overline{p},A)+d(\overline{p},B)
$$
$$
d(\overline{p},B)\geq d - d(\overline{p},A)=d-\frac{d}{2}=\frac{d}{2} > 0.
$$
And $\overline{p} \notin B$. This is a contradiction because $\omega(p)=A \cup B$.

Title: Show that a limit set is connected?


问:
Existence of a graph G?
The Q i am asked is as follows.

Q:Let G=(V,E) be a graph with nine vertices  such that each vertex has a degree 5 or 6. Show that G has at least 5 vertices of degree 6, or at least 6 vertices of degree 5.

first lets bust the middle 5 of6 = 9-5 thus 4of 5 and  6 of 5 = 9-6 hence 3 of 6.

so they are claiming that 4 vertex of degree 6 and 5 of degree 5 cannot be a graph. and anything else is fine.
clearly $4*6 + 5*5 = 24+25 = 49$  thus $E= 24.5$ which i would like to see you show me half an edge?

Now i could draw pictures of all the other graphs ( me being lazy i want to find a better way)  or i could ask is the existence of an even sum here sufficient to prove that there exists a simple connected graph? i know that having half an edge is sufficient to disprove the existence  but for a set number of vertices if there exists an even number of edges does a graph need to exist?

EDIT: ( to downvote and comment on duplicate)

My question is for a set number of vertices if there exists an even number of edges does a graph need to exist?

答:
I almost overlooked the additoinal question near the end of yourp post:
Given integers $r_1,\ldots ,r_n$ with $r_i\ge 0$ and $\sum r_i=2e$ even, does there exist a graph with $n$ vertices of degrees $r_1,\ldots, r_n$?

Unless you allow multiple edges and loops, the answer is no:
Firstly, there are certainly a few more necessariy conditions, for example $r_i\le n-1$ as no vertex can be connected to more than all other vertices, and $r_i<e$ as all edges incident with a vertex are distinct. Many more conditions arise, for example $r_1=1$ implies that at most one $r_i$ can equal $n-1$.

Title: Existence of a graph G?


问:
Proving a matrix is a vector subspace and determining dimensions
Let
$A=\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}$ be a matrix in $M_2(\mathbb{R})$. Then $U$ and $V$ are defined by:

$U=\{M\in M_2(\mathbb{R})|MA=AM\}$

$V=\{M\in M_2(\mathbb{R})|MA^T=A^TM\}$

I want to show that U and V are subspaces of $M_2(\mathbb{R})$ but I am unsure how to do it. I know that in order for a subset to be a subspace, it has to be closed under addition, scalar multiplication and must contain the zero vector. I know how to prove scalar multiplication and the zero vector, but I am unsure about addition.

$\begin{bmatrix}
\mu & \lambda \\
0 & \mu
\end{bmatrix}$ some $\mu,\lambda\in\mathbb{R}$.  How does this work under addition?

Also, I am not sure that I understand the concept of dimension correctly, because I thought it was just the product of the number of columns and rows in a matrix, but then I stumbled upon this formula:

$dim(U+V)=dim(U)+dim(V)-dim(U\cap{V})$ where $U, V$ are some sets of matrices. So when calculating dimension of a set of matrices, do we perform operations with the dimensions of the individual matrices? (This is probably a confusing question, I probably don't understand the whole concept of dimension properly).

答:
For adittion, if $M_1, M_2\in U$ you get $(M_1+M_2)A=M_1A+M_2A=AM_1+AM_2$, because $M_1$ and $M_2$ are in $U$. Then $U$ is a subspace of $M_2(\mathbb{R})$. The process is similar for $V$.

For the dimensions, you need to find a base for every subspace. If $M\in U$ and $M$ is like

$$M=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$$

Then you get the following relation

$$\left[\begin{matrix}a&a+b\\c&c+d\end{matrix}\right]=\left[\begin{matrix}a+c&b+d\\c&d\end{matrix}\right]$$

And with this $c=0$, $a=d$ and $b\in\mathbb{R}$. So your matrix $M$ is like

$$M=\left[\begin{matrix}a&b\\0&a\end{matrix}\right]=a\cdot\left[\begin{matrix}1&0\\0&1\end{matrix}\right]+b\cdot\left[\begin{matrix}0&1\\0&0\end{matrix}\right]$$

You can see that the base of $U$ has two elements, and with this $dim(U)=2$. Sorry for my english.


问:
Are all product topologies/spaces over real numbers Euclidean spaces?
Until earlier today, when I thought of "real numbers" I thought of only the pure set of real numbers.
It seems I was mistaken to do so, as this seems to be the set-theoretical Baire space.
Instead, real numbers seem to be the set and what you can do with it, namely their field operations (addition and multiplication, their algebraic inverses, namely subtraction and division, and all the rules regarding associativity, commutativity and so on) and ordering.
So the real numbers can be written as $(R, +, \cdot, <)$. Usually we write them as $\mathbb R$ though and know that we can do these things with the real numbers.

Next, we can build a product space (not inner product space!), which

is the Cartesian product of a family of topological spaces equipped with a natural topology called the product space,

from a finite number ($n\in\mathbb N$) of real numbers, which would be $\mathbb R^n$.
A couple of people in How do I formally write down a Euclidean space with symbols? state that $\mathbb R^n$ is already the Euclidean space.
In other words, every finite dimensional product space of the real numbers is a Euclidean space.

But I do not see what that should be the case.
For a Euclidean space I would also need the

completeness

linear combination

Euclidean distance as metric,

Euclidean norm

the "normal" dot product

Now it seems to me that

completeness is inherited by every product space over the real numbers from the completeness of the real numbers

linear combination can be derived using the direct product. Even though I am not sure if the direct product is always implied for product topologies!

the Euclidean distance is induced by the Euclidean norm

the Euclidean norm is induced by the "normal" dot product, or according to J.W.Tanner in How do I formally write down a Euclidean space with symbols?, the dot product is induced by the Euclidean norm. However, one of them must be derived from somewhere else.

So, it seems to me that many of the properties of the Euclidean space are properties every product space over the real values has.
However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.
For example I could choose to use a different inner product, which is not the Euclidean dot product, without having the dot product induced

So, are all product spaces over real numbers Euclidean spaces? If so, how is the dot product necessarily induced?

答:
I don't understand your comment about the Baire space.

For example I could choose to use a different inner product, which is not the Euclidean dot product

Yes, that's true. What you are learning is that "$\mathbb{R}^n$" is actually extremely ambiguous notation: depending on context it could refer to any of

a set

a topological space (conventionally with the Euclidean topology)

a smooth manifold (conventionally with the "usual" smooth structure)

a metric space (conventionally the Euclidean metric)

a vector space over $\mathbb{R}$

a normed vector space (conventionally the Euclidean norm)

an inner product space (conventionally the diagonal inner product $\sum_{i=1}^n x_i y_i$)

a commutative $\mathbb{R}$-algebra (with pointwise multiplication); this is probably the least common option.

This is part of a standard convention in mathematics, which is to refer to a structured set using the underlying set (also known as the carrier set) only, without naming explicitly the rest of the structure. This is for convenience; mostly it would be too annoying to do this and mostly people understand what you mean from context anyway.

However, I do not see why every product space over the real values should necessarily have the dot product defined as the Euclidean space has.

It's not necessary; it's a convention that if someone says "the inner product space $\mathbb{R}^n$" with no further elaboration they're referring specifically to the diagonal inner product defined above. This convention is relatively harmless because, among other things, all inner products on $\mathbb{R}^n$ (here I mean $\mathbb{R}^n$ the real vector space!) are related by a linear change of coordinates, so it doesn't really matter which one you pick and the diagonal one is maximally easy to compute with.