问:
Let V, W, and U be vector spaces over a field F, and let f,g: V → W and h,j: W → U be linear maps.
Let V, W, and U be vector spaces over a field F, and let f,g: V → W and h,j: W → U be linear maps. Prove that

(a) h ◦ (f + g) = (h ◦ f) + (h ◦ g)

(b) (h + j) ◦ f= (h ◦ f) + (j ◦ f)

I think I get the idea of this problem, but I really don't know where to start. The question basically asks to deduce for every vector space V, the set of all linear transformations from V to V (i.e. Hom(V,V)) forms a ring with point-wise addition and composition.

答:
Hint:

Start with the definition of compositions and the addition of maps.

For instance,
$$
\begin{align}
(h\circ(f+g))(x)
&=h((f+g)(x))
&(\textrm{$F\circ G(x):=F(G(x))$})
\\
&=h(f(x)+g(x))
&(\textrm{$(F+G)(x):=F(x)+G(x)$})\\
&=\cdots
\end{align}
$$

Then use the linearity.


问:
What does "well defined" mean?
On page 153 of Linear Algebra Done Right the second edition, it says:

Define a linear map $S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$ by:

7.43:   $S_1 (\sqrt{T^* T}v)=Tv$

First we must check that $S_1$ is well defined. To do this, suppose $v_1, v_2 \in V$ are such that $\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$. For the definition given by 7.43 to make sense, we must show that $Tv_1=T v_2$.

It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?

Thanks

答: A function $f: A \to B$ is well defined if for every $x \in A$, $f(x)$ is equal to a single value in $B$. So if $f(x)$ could equal two different values in $B$, then it is not well defined. Or if $f(x)$ is not equal to a value in the set $B$, then it is not well defined.

ContentLicense: CC BY-SA 4.0

Title: What does "well defined" mean?


问:
Let $f(x)$ be continuous in $[0,1]$, differentiable in $(0,1)$, and $f(0)=0, f(1)=1$, and $f'(x)\le 2x$ for every $0<x<1$, prove that $f(x)=x^2$.

Let $f(x)$ be continuous in $[0,1]$, differentiable in $(0,1)$, and $f(0)=0, f(1)=1$, and $f'(x)\le 2x$ for every $0<x<1$, prove that $f(x)=x^2$.

Basically I have no idea how to start my proof, I can understand it intuitively but I don't know what kind of proof this is meant to be.

I would really appreciate a push in the right direction.

答:
Let $g:[0,1]\rightarrow\mathbb{R}$ be defined by $g(x)=f(x)-x^{2}$,
then $g$ is continuous on $[0,1]$, differentiable on $(0,1)$, $g(0)=g(1)=0$,
and $g'(x)\leq0$ for all $x\in(0,1)$. We go to prove that $g(x)=0$
for all $x\in[0,1]$.

Let $x\in(0,1)$. By Mean-Value Theorem, there exists $\xi\in(0,x)$
such that
\begin{eqnarray*}
g(x) & = & g(x)-g(0)\\
 & = & g'(\xi)(x-0)\\
 & \leq & 0.
\end{eqnarray*}
By Mean-Value Theorem again, there exists $\eta\in(x,1)$ such that
\begin{eqnarray*}
-g(x) & = & g(1)-g(x)\\
 & = & g'(\eta)(1-x)\\
 & \leq & 0.
\end{eqnarray*}
Therefore, $g(x)=0$ and hence $f(x)=x^{2}$.


问:
Bound on complex triples satisfying some simple equations.
Let $a,b,c \in \mathbb{C}$ be three complex numbers satifying the following equations :
$$\left\{ \begin{array}{rl} a+b+c &= 3 \\ a^2+b^2+c^2 &=9 \end{array} \right.$$

I want to prove that $\max(|a|, |b|, |c|)\geq 2$.

I can formulate this as an optimisation problem for a multiple real variable function and use Lagrange multipliers, but it would probably be extremely long. The problem seems simple enough that there should be a direct and elegant proof.

So, any help would be very appreciated. Thanks.

答:
Plotting the Regions

Since $ab+bc+ca=\frac12\left((a+b+c)^2-\left(a^2+b^2+c^2\right)\right)=0$, Newton's identities say that $a$, $b$, and $c$ are roots of
$$\newcommand{\Re}{\operatorname{Re}}
z^3-3z^2-abc=0\tag1
$$
Suppose that $|a|\le|b|\le|c|\lt2$
$$
\begin{align}
|c^3-3c^2|
&=|c|^2|c-3|\tag{2a}\\
&=|abc|\tag{2b}\\
&\le|c|^3\tag{2c}
\end{align}
$$
That is, $|c-3|\le|c|$, which means that $c$ is closer to $3$ than it is to $0$. The dividing line between "closer to $0$" and "closer to $3$" is $\Re(z)=\frac32$. Therefore, in addition to $|c|\lt2$, we know that $\Re(c)\ge\frac32$.

Given $c$, we can use $(1)$ and Newton's identities to show that $a+b=3-c$ and $ab=c^2-3c$. Then we can then compute
$$
\begin{align}
a&=\frac{3-c-\sqrt{9+6c-3c^2}}2\tag{3a}\\
b&=\frac{3-c+\sqrt{9+6c-3c^2}}2\tag{3b}
\end{align}
$$
These could be switched, but it will be seen that, given $(3)$ and the possible locus of $c$, $\Re(b)\ge2$.

Here is a diagram of the regions where $c$, $a+b$, $ab$, $b$, and $a$ lie.

The yellow moon-shaped region, where $b$ lies, belies the assumption that $|a|\le|b|\le|c|\lt2$. Let us now prove that $\Re(b)\ge2$.

Getting Real

Since
$$
\begin{align}
\cos(\theta)
&=2\cos^2(\theta/2)-1\tag{4a}\\
&\le\cos^2(\theta/2)\tag{4b}
\end{align}
$$
if we use the square root with the branch cut along the negative real axis, where $\cos(\theta/2)\gt0$, we have
$$
\overbrace{\Re\left(\sqrt{z}\right)}^{\sqrt{r}\cos(\theta/2)}\ge\overbrace{\sqrt{\Re(z)_+}}^{\sqrt{r}\sqrt{\cos(\theta)_+}}\tag5
$$
where $x_+=\max(x,0)$ for $x\in\mathbb{R}$.

With $c=x+iy$,
$$
\Re(9+6c-3c^2)=9+6x-3x^2+3y^2\tag6
$$
Therefore, for $\frac32\le x\lt2$
$$
\begin{align}
\Re(b)
&=\Re\left(\frac{3-c+\sqrt{9+6c-3c^2}}2\right)\tag{7a}\\
&\ge\frac{3-x+\sqrt{9+6x-3x^2}}2\tag{7b}\\[12pt]
&\ge2\tag{7c}
\end{align}
$$
Explanation:

$\text{(7a)}$: derived in $(3b)$

$\text{(7b)}$: apply $(5)$ and $(6)$

$\text{(7c)}$: the function is decreasing on $[1,3]$ and equals $2$ at $x=2$

$\text{(7c)}$ is a contradiction since we assumed $|a|\le|b|\le|c|\lt2$.


问:
Solving Exact Differential Equations Short Cut/Second method
I remember reading something about another way to solve an exact differential equation by ignoring particular terms in the equation and then determining if what you have left is the derivative of another function.

Let me illustrate:
Suppose we have the first order differential equation $2xy-9x^2+2yy'+x^2y'+y'=0$. If we ignore the following types of terms:

Terms only involving $x$

Terms involving $y'$ and $x$

Scalar terms

Which in this case bring us down to the equation $2xy+x^2y'=0$, then we can check for exactness if what we are left with is the derivative of a function. In this case, $(x^2y)'=2xy+x^2y'=0$, so our equation is exact.

The solution method is also different:

Once we have figured our the function, we can very simply solve the DE:

Re-write the DE like so: $(x^2y)'-9x^2+2yy'+y'=0$

Integrate both sides with respect to $x$: $\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx=0$

$\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx = x^2y-3x^3+y^2+y+c$

And you're done! (This is indeed the correct solution for this differential equation).

I cannot seem to find an example where this method doesn't work, but I am also having trouble understanding WHY it works. Can someone please help? I would also love to hear from whoever I read this from in the first place!

答:
If the equation is exact, then that means that there is some function $\phi(x,y)$ such that

$$ \phi(x,y)=c\tag{1} $$

and such that

$$ \frac{\partial\phi}{\partial x}dx+\frac{\partial\phi}{\partial y}dy=0 $$

Alternately,

$$ \frac{\partial\phi}{\partial x}+\frac{\partial\phi}{\partial y}y^\prime=0\tag{2} $$

So we can parse an equation such as

$$ 2xy-9x^2+2yy'+x^2y'+y'=0 $$

into terms containing $y^\prime$ and those not containing $y^\prime$.

$$ 2xy-9x^2 +(2y+x^2+1)y^\prime=0$$

So we know that we are looking for a function $\phi(x,y)$ which simultaneously satisfies

$$ \frac{\partial\phi}{\partial x}=2xy-9x^2 $$
and
$$ \frac{\partial\phi}{\partial y}=2y+x^2+1 $$

For the first equation we see that it must be the case that

$$ \phi(x,y)=x^2y-3x^3+c_y $$

where $c_y$ is a 'constant' function of $y$, as far is partial differentiation with respect to $x$ is concerned. Likewise

$$ \phi(x,y)=x^2y+y^2+y+c_x $$

Note that any term of $\phi$ which contains both $x$ and $y$ will occur in both places, even though it occurs only once in the function.

So, using these two clues, we conclude that the solution to equation (1) is

$$ x^2y-3x^3+y^2+y=c$$

and equation (2) is merely a means of parsing the expression to separate out the contributions of the two partials.

Notice that your method is doing the same thing but in a way that makes the process seem somewhat mysterious.


问:
Material biconditional and substitution
Does the material biconditional have the substitution property?

As an example, if I was able to derive $A \leftrightarrow B$ during an argument, may I substitute $A$ for $B$ (or vice-versa) in any previous formula?

答:
Yes. If $\phi(A)$ is some statement with zero or more instances of $A$ as part of it, and $\phi(B)$ the result of substituting any number of those (you don't have to substitute all of them), then if $A \leftrightarrow B$ is true, and $\phi(A)$ is true, then $\phi(B)$ will be true as well.

So yes, if in some proof you have both $A \leftrightarrow B$ and $\phi(A)$, then you can infer $\phi(B)$. Of course, to be able to do this in a purely formal proof requires an explicit inference rule to this effect.

Title: Material biconditional and substitution


Title: Solving Exact Differential Equations Short Cut/Second method

Body:
I remember reading something about another way to solve an exact differential equation by ignoring particular terms in the equation and then determining if what you have left is the derivative of another function.

Let me illustrate:
Suppose we have the first order differential equation $2xy-9x^2+2yy'+x^2y'+y'=0$. If we ignore the following types of terms:

Terms only involving $x$

Terms involving $y'$ and $x$

Scalar terms

Which in this case bring us down to the equation $2xy+x^2y'=0$, then we can check for exactness if what we are left with is the derivative of a function. In this case, $(x^2y)'=2xy+x^2y'=0$, so our equation is exact.

The solution method is also different:

Once we have figured our the function, we can very simply solve the DE:

Re-write the DE like so: $(x^2y)'-9x^2+2yy'+y'=0$

Integrate both sides with respect to $x$: $\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx=0$

$\int(x^2y)' dx - \int9x^2 dx+\int2yy'dx+\int y'dx = x^2y-3x^3+y^2+y+c$

And you're done! (This is indeed the correct solution for this differential equation).

I cannot seem to find an example where this method doesn't work, but I am also having trouble understanding WHY it works. Can someone please help? I would also love to hear from whoever I read this from in the first place!

Body:
$$(x^2y)'-9x^2+2yy'+y'=0$$
It works because you have only derivatives:
$$(x^2y)'-3(x^3)'+(y^2)'+y'=0$$
$$(x^2y-3x^3+y^2+y)'=0$$
$$x^2y-3x^3+y^2+y=C$$


问:
Herstein Topics in Algebra, $\pi \mid a$ or $(\pi,a) = 1$
In Herstein’s Topics in Algebra, p.147, there is a sentence

If $\pi$ in $R$ is a prime element of $R$ and $a \in R$, then either $\pi \mid a$ or $(\pi,a) = 1$, for, in particular, $(\pi,a)$ is a divisor of $\pi$ so it must be $\pi$ or $1$ (or any unit).

I’m having a bit of difficulty seeing why the last part of this sentence is true (“for, in particular…”).

Note:

Herstein’s definition of a prime element in a Euclidean ring $R$ is a nonunit $\pi$ such that whenever $\pi = ab$, where $a, b$ are in $R$, then one of $a$ or $b$ is a unit in $R$.

This result comes before the Unique Factorization Theorem, so that can’t be used. However, we have proved that every element of a Euclidean ring $R$ is either a unit or can be written as a product of finitely many primes. (I don’t know if this is useful, but it’s what I tried to use, to no avail.)

My attempt:

I tried supposing to the contrary, so that $(\pi,a)$ is neither $\pi$ nor $1$ nor a unit. Since $(\pi,a)$ divides $\pi$ we can write $\pi = (\pi,a)k$ for some $k \in R$. Then by definition of $\pi$ being prime, $k$ must be a unit, since $(\pi,a)$ isn’t. So $\pi$ and $(\pi,a)$ are associates. This is where I tried a few other things and got stuck each time.

答: Hint: a prime $\pi$ is irreducible so its divisor $\,(\pi,a)\,$ is either a unit, or else associate to $\pi$ (then $\,\pi\mid (\pi,a)\Rightarrow\, \pi\mid a)$


答:
Ultimately, well-defined means simply: defined, however in a special kind of defining functions.
Namely, we want to define a map $f\colon A\to B$ in terms of given maps $ g\colon C\to A$ and $h\colon C\to B$. We attempt to do so by saying that for given $x\in A$, we pick $z\in C$ with $g(z)=x$ and then set $f(x):=h(z)$. But is that really a definition of a function? We need two properties:

For every $x\in A$, there exists $z\in C$ with $g(z)=x$.

If there are multiple choices for $z$, say $g(z_1)=g(z_2)=x$, then $h(z_1)=h(z_2)$

By showing these two properties, we prove that our attempted definition is indeed a definition. In this case we say that $f$ is well-defined.

Title: What does "well defined" mean?

Body:
On page 153 of Linear Algebra Done Right the second edition, it says:

Define a linear map $S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$ by:

7.43:   $S_1 (\sqrt{T^* T}v)=Tv$

First we must check that $S_1$ is well defined. To do this, suppose $v_1, v_2 \in V$ are such that $\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$. For the definition given by 7.43 to make sense, we must show that $Tv_1=T v_2$.

It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?

Thanks


问:
Trying to graph $x^{\frac12} + y^{\frac12} =1$
I am a bit rusty with math and I am not sure what the steps would be to graph the function  $x^{\frac12} + y^{\frac12} =1$. I tried doing implicit differentiation but then I wasn't sure where to go from there. Maybe I'm using the wrong approach.

Thanks, everyone.

答:
It's always nice to restrict the domain and range for implicit equations. If you're graphing this in the real numbers, than you know that neither x nor y can be negative, but they can be zero.

$0 \leq x,y $

Additionally, since the square root of x or y can be both positive and negative, we can restrict the domain and range based on each possibility:

sign$(\sqrt{x}) = $ sign$(\sqrt{y}) = 1\;|\;0 < x,y < 1$

sign$(\sqrt{x}) = $ sign$(\sqrt{y}) = -1\;|\;$Impossible

sign$(\sqrt{x}) = 1$ , sign$(\sqrt{y}) = -1\;|\;1<x$

sign$(\sqrt{x}) = -1$ , sign$(\sqrt{y}) = 1\;|\;1<y$

$x = 0\;|\;y = 1$

$y = 0\;|\;x = 1$

After this exercise, we should algebraically manipulate the original equation into an explicit equation.

$\pm\sqrt{x}\pm\sqrt{y}=1$

$\pm\sqrt{y}=1\pm\sqrt{x}$

$y=(1\pm\sqrt{x})^2$

This can be split into four functions:

$y=(1+\sqrt{x})^2\;|\;0\leq\sqrt{y}$

$y=(1+\sqrt{x})^2\;|\;0\geq\sqrt{y}$

$y=(1-\sqrt{x})^2\;|\;0\leq\sqrt{y}$

$y=(1-\sqrt{x})^2\;|\;0\geq\sqrt{y}$

Since the actual equations don't depend on sign$(\sqrt{y})$, then we can further condense this into two equations:

$y=(1+\sqrt{x})^2$

$y=(1-\sqrt{x})^2$

TADA! We have a system of two explicit functions that graph the original implicit equation.


问:
let $F(X)=\int_0^x \sin\frac{1}{1+t^2}\,dt$, then which of these is correct?
a) $0 \le F(x)$ for all $x$.

b) $\lim_{x \to 0} \frac {F(x)}{x} = 1$

c) $\lim_{x \to \infty} F(x)= \infty $

d) $F$ has an inverse and $(F^{-1})'(0)=\frac{1}{\sin 1}$

What I did: for (a)  I went for the derivative $F'(x)=\sin\frac{1}{1+x^2}$, I saw that it's bigger than zero and it's decreasing, could just infer that the function $F(x)$ is increasing, and not $F(x)\ge0$ for all x, so I left it aside.

(b) I took the limit and it was $0$, so it's wrong.

(c) I think I thought a little too broad here and just said that $\sin\frac{1}{1+x^2}$, is like $\frac{1}{x^2}$, which converges if the integral was from $1 \to \infty$, not from zero.

(d) I tried to do this $F((F^{-1})'(0))$, which is supposed to give me $0$, but I didn't get zero. so I assumed its wrong.

 In the end the answer was (d), I would really appreciate an explanation of why it is true, hence $F((F^{-1})'(0))\ne0$, also I would appreciate any tips about my thinking and the way I tried to solve the question.
 Thanks in advance.

答:
That comes from the inverse function theorem that
$$
(F^{-1})'(0)=\frac{1}{F'(F^{-1}(0))}
$$

Notes.

By the fundamental theorem of calculus, $\displaystyle F'(x)=\sin \frac{1}{1+x^2}$. On the other hand, $F(0)=0$, which tells you what $F^{-1}(0)$ is.


问:
Find a maximum number for failed students (Probability)
i've benn stuck on this question for a while. Any help would be appreciated:

The number of students examined is $N = 134$. It's known that the average grade is $61$ and the standard deviation is $4$.
Let $k$ be the number of students that failed the test (got a grade lower than $56$). What is the largest possible value for k?

答:
Some starting places:
Note that the standard deviation grows the further students are away from the mean.   So if you have students failing by more than the minimum failure (55),  they would increase the standard deviation by more.   Since 4 is effectively a constraint in an optimization problem, this means the optimal case will be when all your failure students got 55s.   Each one contributes 36 to the variation,    so you have part of your variation being 36k.   The rest comes from the students who scored above average.  Note that for an average of 61 among 134 students, the sum of their scores is 13461.  So the total scores of passing students is 13461-55k,  divided up amongst 134-k students.  I"m pretty sure the variation would be minimized when each of those students had an equal score,  so you should be able to take that average passing score minus 61,  square them,  add the variations to the variations of the failing scores,  take the square root, and have an equation in k.


问:
The unit $\eta$ of an adjunction satisfies $\sigma \circ \eta = 1$ if and only if $\Gamma_{C, C'} \circ \mathbb F_{C, C'} = 1$.
Let $F: \mathcal C \to \mathcal D$ a functor. This induces
$$\mathbb F_{C, C'} :\text{Hom}_\mathcal C(C, C') \to \text{Hom}_\mathcal D(FC, FC'): f \mapsto Ff,$$
which is natural in $C, C'$. Suppose that $F$ has a right adjoint $R: \mathcal D \to C$ and consider the associate unit and counit $\eta :1 \to RF, \varepsilon:FR \to 1$.

Now I want to show that there is a natural transformation $\sigma: RF \to 1$ such that $\sigma \circ \eta = 1$ if and only if there is $\Gamma_{C, C'}:\text{Hom}_\mathcal D(FC, FC')\to \text{Hom}_\mathcal C(C, C')$ such that $\Gamma_{C, C'} \circ \mathbb F_{C, C'} = 1$. The "only if" part is not too hard by defining
$$\Gamma_{C, C'}  = (- \circ \eta_C) \circ (\sigma_{C'} \circ -) \circ R,$$
we do indeed get $(\Gamma_{C, C'} \circ \mathbb F_{C, C'})(f) = f$ using the naturality of $\eta$ and $\sigma$. But I am really stuck at the "if" part: If we have $\Gamma_{C, C'}$ satisfying the above condition, the more natural way to define $\sigma$ is (I think)
$$\sigma_C := \Gamma_{RFC, C} (\varepsilon_{FC}).$$
I've tried to show that $\sigma_C \circ \eta_C = 1_C$ by using some property of the unit and the counit but I haven't been able to conclude. Does anyone know how to do that ? Am I on the right path ?

答:
Define $$ \sigma_C = \Gamma_{RFC,C}(\varepsilon_{FC}).$$ Then we have $$\sigma_C\circ\eta_C = \Gamma_{RFC,C}(\varepsilon_{FC})\circ \eta_C=\Gamma_{C,C}(\varepsilon_{FC}\circ F(\eta_C))=\Gamma_{C,C}(\operatorname{id}_{FC}) = \operatorname{id}_C.$$

where we used the naturality of $\Gamma$ and then the triangle identity.


答:
Hint

Let $$f : t\mapsto \sin(\frac{1}{1+t^2})$$

$ f $ is defined and is continuous at $ \Bbb R $, thus $ F $ is differentiable at $ \Bbb R $ and
$$(\forall x\in \Bbb R)\;\; F'(x)=f(x)>0$$
because
$$(\forall x\in \Bbb R)\;\;0<\frac{1}{1+x^2}\le 1<\frac{\pi}{2}$$

So,  $ F $ is continuous and strictly increasing at $ \Bbb R$ , then it has an inverse satisfying

$$(F^{-1})'(0)=\frac{1}{F'(F^{-1}(0))}$$

$$=\frac{1}{F'(0)}=\frac{1}{f(0)}=\frac{1}{\sin(1)}$$

Remark
$$F(0)=0\implies F^{-1}(0)=0$$

Title: let $F(X)=\int_0^x \sin\frac{1}{1+t^2}\,dt$, then which of these is correct?

Body:
a) $0 \le F(x)$ for all $x$.

b) $\lim_{x \to 0} \frac {F(x)}{x} = 1$

c) $\lim_{x \to \infty} F(x)= \infty $

d) $F$ has an inverse and $(F^{-1})'(0)=\frac{1}{\sin 1}$

What I did: for (a)  I went for the derivative $F'(x)=\sin\frac{1}{1+x^2}$, I saw that it's bigger than zero and it's decreasing, could just infer that the function $F(x)$ is increasing, and not $F(x)\ge0$ for all x, so I left it aside.

(b) I took the limit and it was $0$, so it's wrong.

(c) I think I thought a little too broad here and just said that $\sin\frac{1}{1+x^2}$, is like $\frac{1}{x^2}$, which converges if the integral was from $1 \to \infty$, not from zero.

(d) I tried to do this $F((F^{-1})'(0))$, which is supposed to give me $0$, but I didn't get zero. so I assumed its wrong.

 In the end the answer was (d), I would really appreciate an explanation of why it is true, hence $F((F^{-1})'(0))\ne0$, also I would appreciate any tips about my thinking and the way I tried to solve the question.
 Thanks in advance.


问:
Why not any linearly independent set of vector is basis
I think this is a question asked several times but I am still confusing about some detail. Many of answer for this question give an example such that give an basis $V = \{v_1,v_2,...,v_n\}$, then removing one element from $V$, the rest of set $\{v_1,v_2,...,v_{n-1}\}$ is still a linearly independent set but no longer a basis. However this is where I am confused, for simplicity, restrict V to be a basis of $\mathbb{R^3}$. If we remove $v_3$ from $V$, the set $\{v_1,v_2\}$ can form a plane, which should be a subspace of $\mathbb{R^3}$. Then according to our question, the $\{v_1,v_2\}$ may not be the basis of this plane subspace, why this is true?

答: A basis is a maximal linearly independent set, which means it must span the entire space to be a basis. Since $\mathbb{R}^3$ is three dimensional it requires three basis vectors. While two linearly independent sets do form a basis for a plane since they are not a maximal linearly independent set since I can find a third independent vector in $\mathbb{R}^3$. This means it cannot be a basis.


答: Adding onto David's comment with an example, $\{\ (1,0,0),\ (0,1,0) \}$ is the basis of a plane, but no linear combination of $(1,0,0)$ and $(0,1,0)$ equals $(0,0,1)$, which means that $\{\ (1,0,0),\ (0,1,0) \}$ is not a basis of $\mathbb{R}^3$.

Title: Why not any linearly independent set of vector is basis

Body: I think this is a question asked several times but I am still confusing about some detail. Many of answer for this question give an example such that give an basis $V = \{v_1,v_2,...,v_n\}$, then removing one element from $V$, the rest of set $\{v_1,v_2,...,v_{n-1}\}$ is still a linearly independent set but no longer a basis. However this is where I am confused, for simplicity, restrict V to be a basis of $\mathbb{R^3}$. If we remove $v_3$ from $V$, the set $\{v_1,v_2\}$ can form a plane, which should be a subspace of $\mathbb{R^3}$. Then according to our question, the $\{v_1,v_2\}$ may not be the basis of this plane subspace, why this is true?


问:
power series quotient of polynomial functions
I have given $g(x)=\sum_{k=1}^\infty k^2x^k$. Why can you now write $g:(-1,1)\rightarrow\mathbb R$ as a quotient of two polynomial functions?

I just know the radius of convergence is $\lim\limits_{n\rightarrow\infty}\sup\sqrt[k]{k^2}=1$ and by using the fact for $|x|=1$ is $k^2x^k$ not a null sequence I get convergence for $|x|<1$.

ContentLicense: CC BY-SA 3.0

Body:
Hints: for $\,|x|<1\,$ :

$$\frac1{1-x}=\sum_{k=0}^\infty x^k\implies\frac1{(1-x)^2}=\sum_{k=1}^\infty kx^{k-1}\implies$$

$$\implies\frac2{(1-x)^3}=\sum_{k=2}^\infty k(k-1)x^{k-2}=\frac1{x^2}\sum_{k=1}^\infty k^2x^k-\frac1x\sum_{k=1}^\infty kx^{k-1}\;\ldots\ldots$$


问:
Understanding the base case for the following formula: $\varphi(n):= n \geq 3 \rightarrow (n!)^2 \gt n^n$
In my book The Foundations of Mathematics by Kenneth Kunen, the following passage is stated:

To prove $\forall n \in \omega \ \varphi(n)$, it is sufficient to prove $\varphi(0)$ (the basis) and $\forall n \in \omega [\varphi(n) \rightarrow \varphi(S(n))]$ (the induction step); then, apply the Principle of Ordinary Induction to $X:=\{n \in \omega: \varphi(n)\}$

where the Principle of Ordinary Induction is the theorem:

For any set $X$: If $\emptyset \in X$ and $\forall y \in X (S(y) \in X)$, then $X$ contains all natural numbers.

I wondered how the above descriptions come into play for a $\varphi(n)$ of the following form:

$\varphi(n):= n \geq 3 \rightarrow (n!)^2 \gt n^n$...i.e. when proving the statement: $\forall n \in \omega [n \geq 3 \rightarrow (n!)^2 \gt n^n]$

In more informal settings, I often read analogies of the "domino effect", which tries to illustrate "what happens" after proving of the base case and inductive case $\forall n \in \omega [\varphi(n) \rightarrow \varphi(S(n))]$.

However in the statement $\forall n \in \omega [ n \geq 3 \rightarrow (n!)^2 \gt n^n]$, the first three natural numbers $0,1,$ and $2$ are are vacuously true.

Said differently, the first three numbers are, in essence, true for a different reason (as compared to numbers like $3, 4, 5, ...$ etc).

So my question is: What should be my base case for a statement like $\forall n \in \omega [n \geq 3 \rightarrow (n!)^2 \gt n^n]$?

Should it be $\varphi(0)$? Or should it be $\varphi(3)$?

I am tempted to think that the base case should be $\varphi(3)$. And then after my inductive step, I essentially backfill $\varphi(0), \varphi(1),$ and $\varphi(2)$ as being true because of their vacuous qualification. In this fashion, my set $\{n \in \omega : \varphi(n)\}$ would still qualify for the Principle of Ordinary Induction and thus conclude $\omega = \{n \in \omega : \varphi(n)\}$.

Is that the right idea?

答: To show $\forall n\ in \omega$, the base case has to be $n=0$, at least formally. However, the proof of $\phi(n)\to \phi(n+1)$ can be made in form of a case distinction, one trivial for small $n$, one explicit check for when $n+1$ is the wannabe base case, and one normal induction step for larger $n$. But in effect, this works exactly as if we consider $n=3$ the base case right away. It would therefore be overly formal to use the full method. But it is perhaps instructive to try that once, but not more than once.


答:
For a function to be well-defined, its output needs to be unambiguous. So if $a$ = $b$, then we need $f(a) = f(b)$. Let's look at something that's not well-defined. Let's try to define addition on the rational numbers as
$$ \frac{a}{b} + \frac{c}{d} = \frac{a + c}{b + d}. $$

Let's look at $\frac{1}{2} + \frac{1}{3}$. Under this proposed definition of addition we have $\frac{1}{2} + \frac{1}{3} = \frac{2}{5}$. We know that $\frac{1}{2} = \frac{2}{4}$. So we have $\frac{2}{4} + \frac{1}{3} = \frac{3}{7} \neq \frac{2}{5}$. This definition of addition isn't well-defined because our output is ambiguous.

Edit: I hope this makes sense.

Title: What does "well defined" mean?


问:
closed point with stalk being a field is open?
Let $X$ be a scheme with $p$ a closed point of $X$. If the stalk of $p$ is a field, then is $p$ an isolated point?(this means $\{p\}$ is open in $X$)

The question is local, so we assume $X=\operatorname{Spec}A$, then $p$ is closed means $\mathfrak{p}$ is maximal in $A$, $\mathcal{O}_{X,p}$ is a field means that for any $t\in\mathfrak{p}$, there exists an element $s\in A-\mathfrak{p}$ such that $ts=0$.

But how to find an ideal $I$ of $A$ such that $V(I)=\operatorname{Spec}A-\{p\}$?

Edit: If the above condition is not sufficient to prove $\{p\}$ is open, I can add the condition that $X$ is locally of finite presentation over a field $k$.

答:
You have a topological problem so you can assume that $A$ is reduced. The fact that $O_{X,p}$ is a field means that $P$ is a minimal prime ideal. If you assume that $A$ is noetherian then it has finitely many minimal prime ideals. Take an element from $P_i-P$ for each minimal prime ideal and take their product $a$ now $a$ is contained in every minimal prime ideal but $P$. The fact that $P$ is also maximal means that $a$ is contained in every prime ideal but $P$ which is what you want. (I think I didn't use the reduced condition!)

I think this is not true for non-noetherian ring for example any ring whose underlying topological space of its spec is $\mathbb{Z}_p$.


答: Note that $x,y\in [0,1]$, restricting curve in the first quadrant with the two endpoints $(0,1)$ and $(1,0)$. Another reference point would be $x=y= \frac1{4}$. These three special points would help sketch the curve, symmetric around $y=x$,

Title: Trying to graph $x^{\frac12} + y^{\frac12} =1$

Body:
I am a bit rusty with math and I am not sure what the steps would be to graph the function  $x^{\frac12} + y^{\frac12} =1$. I tried doing implicit differentiation but then I wasn't sure where to go from there. Maybe I'm using the wrong approach.

Thanks, everyone.


问:
By making the substitution $x=\pi-t$, show that: $ \int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx $
I have a question regarding the following problem:

By making the substitution $x=\pi-t$, show that:

$$
\int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx
$$

This is my work so far:

$$
x=\pi-t
\\dx=-dt
$$

$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}(t-\pi)f(\sin(\pi-t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}tf(\sin(t))dt-\pi\int_{0}^{\pi}f(\sin(t))dt
$$

I noticed that
$$\int_{0}^{\pi}xf(\sin{x})dx$$
and
$$\int_{0}^{\pi}tf(\sin(t))dt$$
are very similar and
$$\pi\int_{0}^{\pi}f(\sin(t))dt$$
is also close to what I want to find.

However, I am unsure how to make further progress.

Any help would be much appreciated

EDIT:
As the comment mentioned, I forgot to change the limits of integration. After changing the limits, I got:
$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}tf(\sin(t))dt-\pi\int_{\pi}^{0}f(\sin(t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt
$$

I can see that this is very close to the final answer. However, I still don't know how I can eliminate the variable $t$

答:
$$I=\int_0^\pi x\,f(\sin x)dx$$
$x=\pi-t\Rightarrow dx=-dt$ so:
$$I=\int_\pi^0(\pi-t)f\left[\sin(\pi-t)\right](-dt)=\int_0^\pi(\pi-t)f(\sin t)dt$$
$$=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi t\,f(\sin t)dt$$
which is the same thing as saying:
$$I=\pi\int_0^\pi f(\sin t)dt-I$$
and so rearrange this equation and we get:
$$I=\frac\pi2\int_0^\pi f(\sin x)dx$$


问:
How to integrate $\int e^{-(x-a)^2} \cos(bx) \, dx$?
How can I integrate the following? By the way a and b are just contants.

$$\int e^{-(x-a)^2} \cos(bx) \, dx$$

答: Hint: assuming $a,\,b\in\Bbb R$, the integrand is$$\Re\exp[-x^2+(2a+ib)x-a^2]=\exp(b^2/4)\Re\left(\exp(iab)\exp[-(x-a-ib/2)^2]\right).$$Now substitute $y=x-a-ib/2$, then write the integral in terms of the error function.


问:
In Euclidean Neighborhood Retracts, the neighborhood have the $\mathbb{R}^n$ topology?
I have $X$ retract by retraction $r$ of its open neighboorhood $Y\in\mathbb{R}^n$.

I'd like to write that, if $O\subset X$ is an open set, so $r^{-1}(O)$ is an open set in $\mathbb{R}^n$. But $r$ continuous only means that $r^{-1}(O)$ is an open set in $Y$. However, I feel that in this case the topology of $Y$ is the same that the one usual in $\mathbb{R}^n$.

Could someone give me a light on this?

Thank you so much.

答: The setup is that there is an embedding $j:X\hookrightarrow\mathbb{R}^n$, an open set $Y\subset\mathbb{R}^n$ with $X\cong j(X)\subseteq Y$, and a continuous map $r:Y\rightarrow X$. The question regards the topology on $Y$. Suppose that $V\subset Y$ is any subset which is open in $Y$. Because $Y$ has the subspace topology inherited from $\mathbb{R}^n$, this means that there is an open set $\widetilde V\subset\mathbb{R}^n$ such that $V=\widetilde V\cap Y$. But $\widetilde V$ and $Y$ are both open in $\mathbb{R}^n$, and hence so is their intersection $V$. In particular we can apply this when $V=r^{-1}(U)$ with $U\subseteq X$ open.


问:
colouring a graph
In a given graph with $n$ vertices ,the degree of each vertex is less than or equal to $5$. Prove that the vertices can  be coloured using only three colors such that the number of edges which have endpoints of same colour is atmost $n/2$

Attempts:
If the graph is a tree than it can be coloured using only 2 colours such that there is no SCE(edge with endpoints of same colour)

If there exists a cycle of 4 vertices , no. of sce is at most 1(after coloring)
If 5 vertices, then at most 2
If 6 vertices than at most 3.
I wasn't able to proof for more and also for combinations of cycles

答:
Let $G$ be a graph with $n$ vertices and no vertex of degree greater than $5$. Then $3$-colour $G$ so as to minimise the number of SCE.

First suppose a vertex $v$ of colour $c$ is incident with two SCE. Then there is a colour $d$ such that $v$ is adjacent to at most one vertex of colour $d$. Changing $v$s colour to $d$ then reduces the number of SSE, a contradiction.

Each vertex is now joined to at most $1$ SCE. The total number of SCE is therefore less than or equal to $n/2$, as required.

Title: colouring a graph


问:
How can I solve this limit with a definite integral inside?
This is the problem:

$$ \lim_{m \to 0}\left(\frac1m \int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx\right) $$

I don't have an idea how to solve it. I was thinking maybe it's an indeterminate form $ \infty \cdot 0$ and we can do something with l'hopital's rule. But I don't know what to do with the integral. What is the trick here?

答:
The result follows from the fundamental theorem of calculus. The limit is equal to
$$
\frac{d}{dx}\int_a^{x}\sin(y^2)\,dy\bigg|_{x=a}=\sin(a^2)\tag{1}
$$

If you use l'Hopital,
then the limit is equal to
$$
\lim_{m\to 0}\frac{\dfrac{d}{dm}\int_a^{a+m}\sin(x^2)dx}{1}
$$
and you end up with the same result in (1).


问:
The distance from a point on a line to a plane
Assume you have a line e and a plane alpha that are not parallel, How do you calculate the coördinates where point Q will be on that line e at a distance d from the plane.

Assuming that both are given in cartesian equations, I can't seem to figure out how to approach this.

答:
Write the equation of the planes parallel to $\alpha$ having a distance $d$ from it. Then find the intersection between the line $e$ and the planes.

For instance
$$\alpha:x+2y+2z=1;\;d=3;\;e=t(1,1,1)$$

distance from $\alpha$ and the two parallel planes must be $3$ therefore
$$\alpha_1:x+2y+2z=h$$
$$d(\alpha,\alpha_1)=\frac{|1+h|}{\sqrt{1^2+2^2+2^2}}=\frac{|1+h|}{3}$$
$$\frac{|1+h|}{3}=3\to |1+h|=9\to h_1=8;\;h_2=-10$$
the parallel planes are
$$\alpha_1:x+2y+2z=8;\;\alpha_2:x+2y+2z=-10$$
intersection with the line $e=(t,t,t)$

with $\alpha_1$ plugging the coordinates of the points of the line

$ t+2t+2t=8\to t=\frac{8}{5}$ point is $\left(\frac{8}{5},\frac{8}{5},\frac{8}{5}\right)$

with $\alpha_2$ we get $t+2t+2t=-10\to t=-2$ point is $(-2,-2,-2)$.

Hope this helps


问:
Measure preserving transformation exercise
I'm trying to solve this exercise but I need some hints because my teacher didn't give me the theory necessary to solve it.

Let $(X, M, \mu)$ be a measurement space such that $\mu (X) = 1$. Suppose
that $T: X → X$ is measurable and $\mu (T^{−1} (E)) = \mu (E)$ for all $E\in M$. Prove that

For all $E ∈ M$ such that $\mu (E)> 0$ there exists a natural $n$ such that $\mu (E \cap T^n (E))> 0$. Here $T^0$ is the identity map in $X$ and $T^n = T \circ T^{n − 1}$ for $n \geq 1$.

For every $E$ the set of points for which there is a
natural $n_0$ such that $T^n(x)$ $\notin$ $E$ for all $n \geq n_0$ has measure $0$.

答:
Let $\mu(E) = \alpha > 0$.  Consider the sets $E, T^{-1}E, T^{-2}E, \dots$.  This is a countable list of sets, each with measure $\alpha$.  Therefore, since $\mu(X) = 1$, some pair of them must overlap in a set of positive measure, i.e. there exists $n > m$ such that $\mu(T^{-n}E \cap T^{-m}E) > 0$.  But this is also equal to $\mu(T^{-n}(E \cap T^{n-m}E)) = \mu(E \cap T^{n-m}E)$.

We need to show that $\mu$-a.e. $x \in E$ satisfies the statement: "for all $k$ there exists $n>k$ such that $T^n x \in E$."  Define $E' = \bigcup_{n=0}^{\infty} T^{-n} E$.  Fix an arbitrary $k$ and note that $T^{-k}E' = \bigcup_{n=k}^{\infty} T^{-n}E$, so we have
$$
E' = (E \cup T^{-1}E \cup \dots \cup T^{-k+1}E) \cup T^{-k}E'.
$$
Because $T$ is measure preserving we must have $\mu(E') = \mu(T^{-k}E')$, so the above implies that $(E \cup T^{-1}E \cup \dots \cup T^{-k+1}E) \setminus T^{-k}E'$ has measure zero.  In particular, almost every $x \in E$ is also in $T^{-k}E'$.  This shows that almost every $x \in E$ satisfies $T^n x \in E$ for some $n\geq k$.  Thus we've proved that for each $k$ there is a set of full measure in $E$ with the desired property.  To finish, take a countable intersection over all $k$ to obtain a fixed set of full measure in $E$ that has the desired property for all $k$, as required.


问:
Proving an identity using formal power series

4.

(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.

(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?

Body:
Hints:

$$1-x^2=(1-x)(1+x)$$

$$1+x^3=(1+x)(1-x+x^2)\;\;\ldots$$


答:
There are a few ways of doing this:
$$I=\int e^{-(x-a)^2}\cos(bx)dx\tag{1}$$
we can introduce the substitution $u=x-a\Rightarrow dx=du$ to get:
$$I=\int e^{-u^2}\cos(bu+ab)du\tag{2}$$
now we can try and split this cosine up:
$$\cos(bu+ab)=\cos(bu)\cos(ab)-\sin(bu)\sin(ab)$$
now let us define the following:
$$I=\cos(ab)I_1-\sin(ab)I_2\tag{3}$$
$$I_1=\int e^{-u^2}\cos(bu)du\tag{4}$$
$$I_2=\int e^{-u^2}\sin(bu)\tag{5}$$
now we will define a new integral:
$$J=I_1+jI_2=\int e^{-u^2}e^{bju}du$$
then combine it and express it that way.

Alternatively you could say:
$$I=\int e^{-(x-a)^2}\cos(bx)dx=\Re\int e^{-(x-a)^2+ibx}dx$$
$$=\Re\int e^{-\left[x^2-(2a+ib)x+a^2\right]}dx=\Re\int e^{-\left[(x+(a+ib/2))^2+(iab-b^2/4)\right]}dx$$
Or you could try integration by parts, I feel like whichever way you do it could be quite messy

Title: How to integrate $\int e^{-(x-a)^2} \cos(bx) \, dx$?

Body:
How can I integrate the following? By the way a and b are just contants.

$$\int e^{-(x-a)^2} \cos(bx) \, dx$$


问:
$m \subset R$ is maximal $\iff$ $0 \subset R / m$ is maximal
Let $R$ be a ring and $m$ be a maximal ideal. The goal is to show $m \subset R$ is maximal $\iff$ $0 \subset R / m$ is maximal.

The proof is given as: Let $\pi: R \to R/m$, $r \mapsto r +m$.
Then is

$$\Rightarrow \; \{I \subset R \, \vert\;m \subset I \;\text{ideal}\, \} \to \{J \subset R/m \,\vert \, J \;\text{ideal}\;\}, \; I \mapsto \pi(I)$$
a bijective map. Hence we have that $m \subset R$ maximal $\iff$ $0=m/m \subset R/m$.

The last part is quite fast. I don't understand why this map should be bijective, and how can we conclude the proof with this?

Many thanks for some help!

答:
If $I\subseteq R$ is an ideal which contains $m$ then $\pi(I)$ is an ideal of $R/m$. We'll show this correspondence is bijective.

Surjective: Suppose $J\subseteq R/m$ is an ideal. It is easy to check that $\pi^{-1}(J)$ is an ideal of $R$ which contains $m$. Since $\pi$ is a surjective map, the equality $J=\pi(\pi^{-1}(J))$ holds.

Injective: Suppose $I\subseteq R$ is an ideal which contains $m$. We'll show that $\pi^{-1}(\pi(I))=I$. Obviously, $I\subseteq\pi^{-1}(\pi(I))$, this is immediate from the definitions. Conversely, if $x\in\pi^{-1}(\pi(I))$ then $\pi(x)\in\pi(I)$, and so there is some $i\in I$ such that $x+m=\pi(x)=\pi(i)=i+m$. Then $x-i\in m\subseteq I$, and so $x=i+(x-i)\in I$, as a sum of elements from $I$.

It follows that if $I_1, I_2\subseteq R$ are ideals which contains $m$ such that $\pi(I_1)=\pi(I_2)$ then:

$I_1=\pi^{-1}(\pi(I_1))=\pi^{-1}(\pi(I_2))=I_2$

And so the correspondence is also injective.

So we have shown there is a bijection between the ideals of $R/m$ and the ideals of $R$ which contains $m$. In particular, if $m$ is a maximal ideal then there are only $2$ ideals of $R$ which contain it, namely $m$ and $R$. In that case, the only ideals of $R/m$ are the trivial ideals, and so the zero ideal of this ring is maximal. The converse is similar.


问:
How to reduce a quadric into canonical form without rigorous calculations?
Consider the quadric in $(x,y,z)$ given by $x^2+y^2+z^2-2xy-2yz+2zx+x-4y+z+1=0$.I am asked to reduce it into canonical form and describe the nature.Here it is a parabolic cylinder.But the method taught to us by the college prof. is based on calculation of eigenvalues and unit eigenvectors,then applying a rotation.But this becomes very long,I am looking for a smarter approach to this problems which is based less on calculation and more on linear algebra concepts.I thought first that I can use congruence operation for this but I found that normal form under congruence is not the same as the canonical form of a quadric.

答:
Even if this is (now) an old question, for completeness I would like to give an answer.

As pointed out in the comments, rewriting the quadric as
$$\Big(x-y+z+\frac{1}{2}\Big)^2=3\Big(y-\frac{1}{4}\Big)$$
makes clear that we are dealing with a parabolic cylinder, but we want a (metric) canonical form, we have to do something more.

First, let's try to perform all the calculations.
Let $B$ be the "complete" matrix of the quadric, i.e.
$$
B=\left( \begin{matrix}1 & -1 & 1 & \frac{1}{2}\\ -1 & 1 & -1 & -2\\ 1 & -1 & 1 & \frac{1}{2}\\ \frac{1}{2} & -2 & \frac{1}{2} & 1\end{matrix} \right)
$$
and $A$ the of the "quadratic part", i.e.
$$
A=\left( \begin{matrix}1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{matrix} \right)
$$
Since the eigenvalues of $A$ are $3$, $0$ and $0$ (I mean, $0$ has algebraic multiplicity $2$) and an orthonormal basis of $\mathbb{R}^3$ made by corresponding eigenvectors is $\{\frac{1}{\sqrt{3}}(1,-1,1),\frac{1}{\sqrt{2}}(1,1,0),\frac{1}{\sqrt{6}}(-1,1,2)\}$, then the matrix
$$
R=\left( \begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \end{matrix} \right)
$$
represents a rotation that trasform that avoid the "rectangular" term from the quadric. Indeed, if we perform the substitution
$$
\left( \begin{matrix} x \\ y \\ z \end{matrix} \right)=R\left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right)
$$
we get the equation (of course, it is easy if one first notices that $x-y+z=\sqrt{3}X$)
$$
3X^2+2\sqrt{3}X-\sqrt{6}Z+1=0.
$$
Now, completing the square, we get
$$
3(X+\frac{\sqrt{3}}{3})^2-\sqrt{6}(Z-\frac{2}{\sqrt{6}})=0.
$$
and then, with the obvious translation, finally
$$
3X^2-\sqrt{6}Z=0.
$$

Now, let's see if we can avoid some annoying calculations.
First of all, it's is well known that the nature of a quadric depends on the inertia of $B$ and $A$, i.e., from the number of positive, negative and zero eigenvalues of $B$ and of $A$. Inertia can be easily calculeted from Descartes' rule of signs: in this case we have for $B$ the inertia $(2,1)$ ($2$ positive eigenvalues, $1$ negative and $4-2-1=0$ equal to $0$) and for $A$, $(1,0)$, so a parabolic cylinder. Such a quadric has a canonical form like
$$
\Gamma:\ \lambda X^2+2pY=0
$$
with $\lambda$ and $p$ different from $0$ and $\lambda$ positive (the $2$ in $2p$ just serves to avoid fractions).
Indeed, if $B'$ and $A'$ are the matrix of $F$, i.e.
$$
B'=\left( \begin{matrix}\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & p\\ 0 & 0 & 0 & 0\\ 0 & p & 0 & 1\end{matrix} \right)
$$
and
$$
A'=\left( \begin{matrix}\lambda & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0  \end{matrix} \right)
$$
then the inertia of $B'$ is $(2,1)$ and that of $A$ is $(1,0)$, as one expects.

We also know that there are four invariants under translation and rotation of the coordinate system, namely, if a quadric has complete matrix
$$
\mathcal{B}=\left( \begin{matrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{11} & a_{22} & a_{23} & a_{24} \\ a_{13} & a_{23} & a_{33} & a_{34} \\ a_{14} & a_{24} & a_{34} & a_{44} \end{matrix} \right)
$$
then the quantities $\det(B)$, $\det(A)$, $\mathrm{tr}(A)$ and
$$
I_2=\det\left( \begin{matrix} a_{11} & a_{12}\\ a_{11} & a_{22}  \end{matrix} \right)+\det\left( \begin{matrix} a_{22} & a_{23}\\ a_{23} & a_{33}  \end{matrix} \right)+\det\left( \begin{matrix} a_{33} & a_{13}\\ a_{13} & a_{11}  \end{matrix} \right)
$$
do not change under translations and rotations.
From this, and in particular from $\mathrm{tr}(A)=\mathrm{tr}(A')$, one can easily obtain $\lambda=3$, but there is no way to get a value for $p$.

Anyway, there are also two other quantities, which, unluckily, are invariant only under rotations, but not under traslations, namely the semi-invariants
$$
\Delta_B=B_{11}+B_{22}+B_{33}
$$
(where $B_{ij}$ is the algebraic complement of the element $a_{ij}$ in $B$) and
$$
\Delta'_B=\det\left( \begin{matrix} a_{11} & a_{14}\\ a_{14} & a_{44}  \end{matrix} \right)+\det\left( \begin{matrix} a_{22} & a_{14}\\ a_{14} & a_{44}  \end{matrix} \right)+\det\left( \begin{matrix} a_{33} & a_{34}\\ a_{34} & a_{44}  \end{matrix} \right).
$$

Now, if we set $\Delta_B=\Delta_{B'}$ and $\Delta'_B=\Delta'_{B'}$, we get
$$
\left\{
\begin{array}{l}
3-p^2=-\frac{3}{2}\\
3p^2=\frac{9}{2}
\end{array}
\right.
$$
which is impossible, and so clearly wrong. Why?

As said, $\Delta_B$ and $\Delta'_B$ are invariants only under rotation, but to bring the quadric in canonical form, we also need a traslation. But we know that after the rotation, the quadric becomes something like $3X^2+2aX+2pY=0$, as in general we cannot avoid a linear term for $X$ (again, the $2a$ is just for convenience). Let $B''$ be the matrix for this quadric, so
$$
B''=\left( \begin{matrix}3 & 0 & 0 & a\\ 0 & 0 & 0 & p\\ 0 & 0 & 0 & 0\\ a & p & 0 & 1\end{matrix} \right)
$$
then from $\Delta_B=\Delta_{B''}$ and $\Delta'_B=\Delta'_{B''}$, we get the system
$$
\left\{
\begin{array}{l}
(3-a^2)-p^2=-\frac{3}{2}\\
3p^2=\frac{9}{2}
\end{array}
\right.
$$
from which we have $p=\pm\frac{\sqrt{3}}{2}$: since we do not need the value of $a$, the only useful equation is $\Delta'_B=\Delta'_{B''}$.

Lastly, there is also a third possibility.
Let $\lambda^4-D_1\lambda^3+D_2\lambda^2-D_3\lambda+D_4$ the characteristic polynomial of $B$ and $\lambda^3-d_1\lambda^2+d_2\lambda-d_3$ the characteristic polynomial of $A$, important: the two polynomial bust be written in order to be monic.

We have the following:  if our quadric is a elliptic (possibly imaginary) or hyperbolic cylinder, then its canonical form is
$$
\lambda_1X^2+\lambda_2Y^2+\frac{D_3}{d_2}=0
$$

(where $\lambda_1$ and $\lambda_2$ are the non zero eigenvalues of $A$) while if it is a parabolic cylinder, then a canonical form is
$$
\lambda_1X^2-2Y\sqrt{\frac{-D_3}{d_1}}=0
$$
With the same rationale, in the case of a pair of parallel straight lines, a canonical form is
$$
\lambda_1X^2+\frac{D_2}{d_1}=0
$$
(again, here and above, $\lambda_1$ is the non zero eigenvalue of $A$).


问:
If $X$ has mean $\mu_x$, variance $\sigma_x^2$, and $Y$ has mean $\mu_y$, variance $\sigma_y^2$, what are the mean and cov of $(X, Y)$?
If r.v. $X$ has mean $\mu_x$, variance $\sigma_x^2$, and r.v $Y$ has mean $\mu_y$, variance $\sigma_y^2$. I'm interested in the expectation and covariance of the random vector $\begin{bmatrix} X \\ Y \end{bmatrix}$. Without any additional information about the joint distribution, is it always true that

$$\mathbb{E}\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix}$$
$$Cov\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \sigma_x^2 & \sigma_{xy} \\ \sigma_{xy}& \sigma_y^2 \end{bmatrix}$$

where $\sigma_{xy}$ is the covariance $Cov(X, Y)$?

答:
Yes, the formula is true by the definitions of expectation of random vectors and covariance matrices.

But when you calculate $E(XY)$, which is needed when you calculate the covariance $\textrm{Cov}(X,Y)$, you need information for the joint distribution.


问:
If a structure is generated by a Set $S$, then every homomorphism is determined by its value on $S$
Let $\scr A$ be a structure generated by $S$, then every homomorphism $\pi: \scr A \to \scr{B}$ is determined by its value on $S$.

Proof:

Let $\pi '$ be another homomorphism, then $C=\{b: \pi (b)= \pi '(b) \}$ is either empty or a substructure. If $\pi$ and $\pi'$ coincide on $S$, then $S\subseteq C$ and hence $C=A$.

I am not sure I see that $C$ is a substructure, if it's not empty?

The rest I understand, if $\pi=\pi '$ for each $s$ in $S$, then $S$ is a subset of $C$ since the two functions are homomorphisms.

[It also follows that $C\subseteq S$. So, $C=A$.] This is wrong, as Mr. Alex Kruckman pointed out

Since $S\subseteq C$ and $\scr A$ is generated by $S$, then $A$ must be $C$;

答:
Let $C = \{b : \pi(b) = \pi'(b)\}$ where $\pi, \pi' : \mathcal{A} \to \mathcal{B}$ are two homomorphisms betweeen structures $\mathcal{A}$ and $\mathcal{B}$ (over some signature $\Sigma$). Then $C$ is either empty or a substructure of $\mathcal{A}$, because if $c_1, \ldots, c_n \in C$ then so is $f(c_1, \ldots, c_n)$ for any $n$-ary function symbol $f$ in $\Sigma$. (It follows as you state in the question that if $\pi$ and $\pi'$ agree on a set $S$ of generators for $\mathcal{A}$, then $\pi = \pi'$.)

A few comments are in order:

Treating the case when $C$ is empty as a special case is presumably because structures are required to have non-empty universes in your context. Some authors (notably Hodges in his book Model Theory) allow structures to have empty universes.

If the signature $\Sigma$ includes any constant symbols, then their interpretations in $\mathcal{A}$ necessarily belong to $C$ by the definition of homomorphism (so $C$ is non-empty in that case).

If the signature $\Sigma$ has no constant symbols, then the notion of a substructure of a structure generated by $S$ is undefined (in general) if $S = \emptyset$ (unless you allow empty universes).


问:
Prove that $T^{*}$ is injective

Let be $V$ and $W$ vector spaces of finit dimension over $\mathbb{C}$
with inner product $\left \langle    ,  \right \rangle_{v}$ and $\left
 \langle , \right \rangle_{w}$ respectively. Consider $T:V \rightarrow
 W$ a surjective linear transformation such that exists $T^{*}:W
 \rightarrow V$ and $\left \langle Tv,w \right \rangle_{w}=\left
 \langle v,T^{*}w \right \rangle_{v}$,  $\forall v\in V$ and $w \in W$

Prove that $T^{*}$ is injective

My attempt to prove if $T^{*}$ is injective:

Let be $w_1, w_2 \in W$ such that $T^{*}w_1=T^{*}w_2$, that implies that

\begin{align}
\left \langle v, T^{*}w_1\right \rangle_{v}&=\left \langle v, T^{*}w_2\right \rangle_{v} \ \ \ \ \ \ \ \ \  \forall v \in V\\ \iff \left \langle Tv, w_1\right \rangle_{w}&=\left \langle Tv, w_2\right \rangle_{w}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ w_1&=w_2\\ \therefore T^{*}& \text{ is injective}
\end{align}

Am I correct? I would really appreciate your help!

Update:

Let be $w_1, w_2 \in W$ such that $T^{*}w_1=T^{*}w_2$.

$\forall v \in V$, we have that:
\begin{align}
0&=\left \langle v, 0\right \rangle_{v}\\&=\left \langle v, T^{*}w_1-T^{*}w_2\right \rangle_{v}\\&=\left \langle v, T^{*}(w_1-w_2)\right \rangle_{v}\\&=\left \langle Tv, w_1-w_2\right \rangle_{w}
\end{align}

We have, $\left \langle Tv, w_1-w_2\right \rangle_{v}=0$, but $T$ is surjective, and that means that $\forall w\in W, \exists v \in V$, such that $w=Tv$.

That implies that there exists $v^{'} \in V$ such that $Tv^{'}=(w_1-w_2)\in W$. And

\begin{align}
\left \langle Tv^{'}, w_1-w_2\right \rangle_{w}&=\left \langle w_1-w_2, w_1-w_2\right \rangle_{w}=0\\ \iff w_1-w_2&=0 \iff w_1 = w_2
\end{align}
Would it be okay like this?

答:
Why do folks insist on not taking advantage of the fact that injectivity of linear transformations can be checked at $\mathbf{0}$?

Say $T^*\mathbf{w}=\mathbf{0}$. We just need to show that this implies that $\mathbf{w}=\mathbf{0}$.

Let $\mathbf{v}$ be such that $T\mathbf{v}=\mathbf{w}$ (using surjectivity of $T$). Then
$$0 = \langle \mathbf{v},\mathbf{0}\rangle = \langle \mathbf{v},T^*\mathbf{w}\rangle = \langle T\mathbf{v},\mathbf{w}\rangle = \langle\mathbf{w},\mathbf{w}\rangle.$$
Therefore, $\mathbf{w}=\mathbf{0}$, so $T^*$ is injective.

Title: Prove that $T^{*}$ is injective


问:
Show that if $\lambda A + \mu B$ is nilpotent therefore $tr(A^kB)=0$.

Let $A,B \in \Bbb M_n(\Bbb C)$. Show that if $\forall \lambda, \mu \in \Bbb C, \lambda A + \mu B$ is nilpotent therefore $\forall k \ge 0, tr(A^kB)=0$.

I looked at some particular cases ($k=0$ ; $B=I_n$ ; $k=1$ ; when $AB=BA$) but I don't see how to generalize my ideas. Does someone have a hint or insights on how to tackle this kind of problems? I'm really more interested in how someone could get to the answer, than the answer itself. Thanks for you advice :D

EDIT: @user299843 showed that we can get the result using a certain polynomial. Does someone know another proof?

答: Hint: Let $k\ge 0$, look at $f(t)=\text{Tr}((A+tB)^{k+1})$. What kind of function is this, what are its properties? If you need another hint, let me know!


问:
Non vanishing smooth vector field on the solid $g-$torus that points inwards
If I think to the surface genus $g$-torus as the boundary of a solid genus $g$-torus in $\mathbb{R}^{3}$, is it possible to have a nonvanishing smooth vector field such that in every point is pointing inward ?

I think the answer is no if $g \ne 1$ : this should follows from the fact that for the Hopf-Lemma given such field than the sum of the index, i.e. $0$ for the sake of contradiction should be equal the gauss map on the boundary, which means the genus $g$-torus, which is $1-g$ for a known fact.

So $g=1$ is the only problem to deal with, and I think that even for $g = 1$ the statement can't hold but I couldn't find any contradiction in that,

Any help would be appreciated.

答:
Such a vector field exists: The solid torus is the surface of revolution of the closed disc, so it can be described with three coordinates $(r,\theta,\varphi)\in[0,1]\times[0,2\pi)^2$, e.g. with the embedding
$$
\begin{pmatrix}r\\\theta\\\varphi\end{pmatrix}\mapsto\begin{pmatrix}(2+r\cos\varphi)\cos\theta\\(2+r\cos\varphi)\sin\theta\\r\sin\varphi\end{pmatrix}\in\mathbb{R}^3
$$
Using $\partial_r,\partial_\theta,\partial_\varphi$ as the corresponding tangent frame, I claim the vector field $V=\partial_\theta-r\partial_r$ is a smooth nonvanishing vector field that is inward pointing on the boundary. One can check smoothness easily enough by writing $V$ in $\mathbb{R}^3$.


问:
Limit of functions and improper integral
As a criterion for the existence of the limit of a function we have:

$\lim\limits_{x\to a} f(x)$ exists iff for each sequence $(a_n)$ with $\lim\limits_{n\to \infty}a_n=a$ the limit $\lim\limits_{n\to \infty}f(a_n)$ exists.

Let's consider an improper integral $\int\limits_a^{b}f(t)dt$ where $a<b$ and $b\in\mathbb{R}\cup\{\infty\}$.

The improper integral exists iff $f(t)$ is Riemann integrable on $[a,\beta]$ for each $\beta$ with $a<\beta<b$ and the limit $\lim\limits_{\beta\to b} \int\limits_a^{\beta}f(t)dt$ exists. In fact, $\lim\limits_{\beta\to b} \int\limits_a^{\beta}f(t)dt$ can be interpreted as a limit of a function, let's call it $F(\beta):=\int\limits_a^{\beta}f(t)$. Hence, it must be possible to prove the existence of $\lim\limits_{\beta\to b}F(\beta)$ by the above criterion. However, in our lecture it says that in the context of improper integrals it is sufficient to find only one sequence such that $\lim\limits_{n\to \infty} F(\beta_n)=\lim\limits_{n\to \infty} \int\limits_a^{\beta_n}f(t)dt$ exists.

Why is it sufficient to show it only for one sequence?

答:
For a counterexample, take $f(x) = \cos x$, $a = 0$, $b = \infty$, and $\beta_n = n\pi$.

We have $\beta_n \to \infty$, and

$$\lim_{n \to \infty}\int_a^{\beta_n}f(x) \, dx = \lim_{n \to \infty}\int_0^{n\pi} \cos x\,dx = \lim_{n \to \infty}(\sin n\pi - \sin 0) = 0$$

However, the improper integral $\displaystyle\int_0^\infty \cos x \, dx$ does not converge, since $\underset{\beta \to \infty}\lim \sin \beta$ does not exist.

If $f \geqslant 0$, then $F(\beta)$ is nonnegative and nondecreasing. Suppose there exists a sequence $\beta_n$ where $\beta_n \to \infty$ and

$$\lim_{n \to \infty} F(\beta_n) = \lim_{n \to \infty}\int_a^{\beta_n} f(x) \, dx = I$$

For any $\epsilon > 0$, there exists $N$ such that $|F(\beta_n) - I| < \epsilon$ for all $n \geqslant N$. If $\beta > \beta_N$, then since $\beta_n \to \infty$, there exists $m > N$ such that  $F(\beta_N) \leqslant F(\beta) \leqslant F(\beta_m),$ and

$$-\epsilon < F(\beta_N) - I < F(\beta)-I < F(\beta_m)- I < \epsilon$$

Thus,

$$\lim_{\beta \to \infty} F(\beta) = \lim_{\beta \to \infty}\int_a^{\beta} f(x) \, dx = I$$


问:
Exponential distribution in queue model M/M/1
Let's say I have $60$ time intervals and I want to generate appearance of people where time between their appearances is selected using exponential distribution with parameter $\lambda$. How do I do it?

答:
If the intervals between successive arrivals are independent and follow an $\text{Exponential}(\lambda)$ distribution, we call this a Poisson process with rate $\lambda$.

The number of arrivals in any time interval of length $t$ has the $\text{Poisson}(\lambda t)$ distribution.


问:
Missing Quadrilateral Angle
Please find the missing green angle for the quadrilateral using simple geometry and explain your steps. I am not able to use trigonometry for this problem. Thank you.

答:
$ABCD$ is cyclic so the opposite angles are supplementary

$\angle BCD=130°\to \angle ACD=130°-90°=40°$

Or since it's cyclic we can use the inscribed angle. The central angle is already given as $80°$.

So $\angle ACD=\frac12\,80°=40°$

Title: Missing Quadrilateral Angle


问:
Applying Jensen's inequality to a convex tarsnformation of a non-convex simple random variable
My objective is to (numerically) maximize an expectation of the form of
$$
E[\exp(g(\mathbf{X}))]
$$ where $\mathbf{X}$ is random variable and $g$ is a non-convex function on $\mathbf{X}$. If I consider the random variable $\mathbf{Y} := g(\mathbf{X})$, it seems to me that due to the convexity of the $\exp$ function, I can apply Jensen's inequality to write:
$$
\exp({E[\mathbf{Y}]}) \leq E[\exp({\mathbf{Y}})] \\
\exp(E[g(\mathbf{X})]) \leq E[\exp(g(\mathbf{X}))]
$$ and then simply choose to maximize the logarithm of the LHS, i.e. $E[g(\mathbf{X})]$, despite the whole transformation not being convex.

Is there a fault in my argument?

Also, if the above is correct, supposing that $g(\mathbf{X}) = g_1(g_2(...g_n(\mathbf{X})))$ where all of the $g_i$ apart from $g_n$ are convex, can I repeatedly apply the same argument to ultimately maximize $E[g_n(\mathbf{X})]$?

Thanks.

答:
I'm assuming X is a parameterized random variable and you are trying to maximize with respect to the parameters. Otherwise, I don't see what we are trying to optimize. E(X) is simply E(X), there's nothing to optimize over...

That being said, although it is true (by Jensen's Inequality) that:

$exp(E[Y])≤E[exp(Y)]$

Maximizing LHS is not necessarily equivalent to maximizing the RHS. So no, we cannot do this unless you prove that maximizing the LHS is equivalent to maximizing the RHS in your particular case.


问:
Finishing the solution to this functional equation: $ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y $
I am trying to solve the below functional equation for $f$ (where $f: \mathbb{R} \to \mathbb{R}$):
$$ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y \text. $$
(Source: https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf: "Iran TST 1996")

Here's what I've got so far:

Substituting $y=0$ we have $f\left(x^2\right) = f\bigl(f(x)\bigr)$.

Substituting $y=f(x)$ we have
$$f\left(x^2+f(x)\right) = f(0) + 4f(x)^2 \text.$$
$$f(0) = 0 \text. $$

Letting $x=0$ we can show the function is even.

Finally letting $y = -x^2$ we can simplify to yield
$$f(x) \left(x^2 - f(x)\right) = 0 \text. $$

And this is where I'm stuck. Given that it's an Olympiad I'm fairly sure that actually the only solutions are $f(x) = x^2$ and $f$ is identically zero and that there will not be any pathological solutions that have discontinuities - though I have not been able to show this.

My ideas that I've tried:

Showing $f$ is injective when not identically zero (on $\mathbb{R}^+$ that is), this doesn't seem to go anywhere useful.

Showing that $f$ is surjective (again onto $\mathbb{R}^+$) when not identically zero, as if that is the case we always must choose $x^2$ instead of zero other we miss out a positive number. However I haven't been able to find any way to show this either.

Use inequalities to show that if $f$ is not identically zero then the inequality $f(x) \ge 0$ that we have from our current information (either $f$ is zero or the square of a real) is strict when $x \ne 0$, but again I couldn't do anything with this either.

答:
Hint Take
\begin{equation}
y = \frac{f(x)-x^2}{2}
\end{equation}
This leads to $\forall x, f(x)\in \{0, x^2\}$, hence we can write $f(x) = x^21_x$ where $\forall x\in{\mathbb R}, 1_x\in\{0,1\}$. Let us rewrite the functional equation as
\begin{equation}
y^2(1_{x^2+y}-1_{x^2 1_x - y})+2x^2 y(1_{x^2+y}+1_{x^2 1_x-y}-2. 1_x)
+ x^4(1_{x^2+y}-1_{x^2 1_x-y}) = 0
\end{equation}
This equation must be satisfied for all $x, y$ but the coefficient of $y^2$ here can be only $-1,0 ,1$, the coefficient of $2x^2 y$  can be only $-2, -1, 0, 1, 2$ and the coefficient of $x^4$ can be only $-1,0,1$.

Let $x$ be given such that $x\not=0$ and $1_x=1$ (or equivalently $f(x) = x^2\not=0$). Clearly the coefficients of $y^2$ and $y$ can be non zero for at most a finite number of $y$'s because in this case, $y$ must be a solution of a polynomial equation of degree 1 or 2 chosen among 44 such non trivial equations. It follows that for all $y$ but a finite number of them one has
\begin{equation}
1_{x^2+y} = 1_{x^2 1_x-y} = 1
\end{equation}
Hence $1_z = 1$ for all $z\in{\mathbb R}$ except on a finite set and $f(z) = z^2$ except perhaps on this finite set.


问:
Show that the single matrix that commutes with all invertible matrices is the scalar matrix
Show that the scalar matrix $\lambda E$, with $E$ being the identity matrix and $\lambda \in K$ a constant, is the only matrix $A\in K^{n\times n}$ that commutes with all invertible matrices.

I am not sure how to tackle this question. I would appreciate if you could help me. Thanks!

答:
Assume the square matrix $A=(a_{ij})$ of size $n$ commutes with all invertible square matrices of the same size $n$.

Take this particular matrix $B = {(b_{ij})}$ where $b_{ij} = i$ when $i=j$ and $b_{ij} = 0$ if $i \ne j$.

$B$ is obviously invertible.

Use the fact that it commutes with your matrix A. So we have $AB=BA$ When we write this down in details, this easily gives us that $a_{ij} = 0$ when $i\ne j$. So all elements of $A$ which are not on the main diagonal are zeros. Now we just need to prove that all elements along the main diagonal of $A$ are equal...

I am thinking on this part myself right now. Should be some simple trick too, I just haven't figured it out yet...

EDIT:

Oh, here it is, the solution for the final part... Multiply now the matrix $A$ to the Vandermonde matrix $V$ (you can pick the alphas in $V$ pretty much as you wish, just take them all distinct). Vandermonde matrix is invertible (as we know from theory, from its determinant) if all alphas in $V$ are pairwise distinct.

OK... Now use the fact that $A$ commutes with $V$ and compare the first columns of the two result matrices $VA$ and $AV$. From comparing the two first columns, you easily get now that the elements along the main diagonal of $A$ are all equal (and this completes the proof). You just need to write down $VA$ and $AV$ (e.g. for $n=3$) and you will notice the pattern and why the desired statement follows.

https://en.wikipedia.org/wiki/Vandermonde_matrix


问:
How can I solve this limit with a definite integral inside?
This is the problem:

$$ \lim_{m \to 0}\left(\frac1m \int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx\right) $$

I don't have an idea how to solve it. I was thinking maybe it's an indeterminate form $ \infty \cdot 0$ and we can do something with l'hopital's rule. But I don't know what to do with the integral. What is the trick here?

答:
Observe that we have both

$$\lim_{m\to0} m=0\;\;\;\text{ and}\;\;\;\lim_{m\to0}\int_{\sqrt\frac \pi2}^{\sqrt\frac \pi2+m}\sin x^2\,dx=0$$

so by L'Hospital's rule, with $\;F(x)\;$ being a primitive function of $\;\sin x^2\;$ :

$$\lim_{m\to0}\frac{\int_{\sqrt\frac \pi2}^{\sqrt\frac \pi2+m}\sin x^2\,dx}m=\lim_{m\to0}\left[F\left(\sqrt{\frac \pi2}+m\right)-F\left(\sqrt{\frac \pi2}\right)\right]'=$$

$$=\lim_{m\to0}F'\left(\sqrt{\frac \pi2}+m\right)=\lim_{m\to0}\sin\left(\sqrt{\frac \pi2}+m\right)^2=\sin\frac\pi2=1$$

In the above, the derivative is with respect to $\;m\;$ , of course.


问:
Finding a function that satisfies this property.
Suppose that $f\in L^2([0,1],\mathbb{C})$, that $0 < \alpha < 1$ and that
$$(T_{\alpha}f)(x) = \int_0^{x^{\alpha}}f(y)dy$$
is continuous (which is actually something you can prove)
I need to find an $f\not = 0$ a.e. such that
\begin{equation}(1) \mbox{     ....... } \int_0^{x^{\alpha}}f(y)dy = (1-\alpha) f(x) .\end{equation}
In tecnical terms I want to show $(1-\alpha)$ is an eigenvalue for the operator $T_{\alpha}$ by showing there is an associated eigenvector.

I could only observe that since the left hand side in (1) is continuous, then $f$ (on the right) is too, so that going back on the left we have $f$ is $C^1$,... and so on; so I think we can deduce that $f\in C^\infty$, but also deriving (1) I could not get anywhere.

Would you be able to give me some hints or a way to find such a function? Thanks in advance

答:
As mentioned in the question, the usual method of differentiating the integral equation does not really help here (it leads to a functional differential equation). I am not aware of a general strategy for solving problems of this kind, but in this particular case simply guessing that the solution has the form $f(x) = x^\beta$ works out. Plug this ansatz into the integral equation and you should get

 $$  \beta = \frac{\alpha}{1-\alpha} \, . $$


问:
Distributing 10 books to 5 schools
The problem asks:

```<code>10 copies of a book are distributed in 5 schools.
1. In how many ways can these books be distributed among schools?
2. What if at least one book is assigned to each school?
</code>```

What I'm thinking is "I have 10 books to distribute in 5 schools, no matter how, so perhaps I can even give all the books to a single school"
So my answer to the first question is $C'_{10,5}=C_{14,5}$.
For the second question I was thinking "I have to give at least one book at school (so $C_{10,5}$), the rest (5) to any school, no matter how, so $C'_{5,5}$ my final answer is $C_{10,5} C'_{5,5}$".
Actually,  I was wrong on both answers, the answers are: $$1)\ C_{14,10}\\2)\ C_{9,5}$$
Why?

答:
Denote the five schools by $1,2,3,4,5$ and let $x_1,\ldots,x_5$ be the number of books distributed to each school, respectively. Then we always have an equation: $$x_1+x_2+x_3+x_4+x_5=10.$$ Our goal is to find the number of non-negative integer solutions under some specific conditions.

There is no restriction, so by the star-and-bars formula, the number of non-negative integer solutions is precisely $$\binom{10+5-1}{10}=\binom{14}{10}.$$

Now we have a restriction that $x_i\geq 1$ for $i=1,2,3,4,5$, but the star-and-bars formula only works for the restriction $x_i\geq 0$. Therefore, we define $y_i:=x_i-1$ for each $i$. Then the system becomes $$y_1+y_2+y_3+y_4+y_5=5,$$ where $y_i\geq 0$ for all $i$. Clearly, there is a one-to-one correspondence between the solution to $y_i$'s and the solution to $x_i$'s. By the star-and-bars formula, the number of non-negative integer solutions is $$\binom{5+5-1}{5}=\binom{9}{5}.$$

For the star-and-bars formula, you can check the Wikipedia.

Title: Distributing 10 books to 5 schools


问:
Prove $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$
Show that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$ for $x>0$. My idea is:

$(1+\frac{1}{x})^x(1+\frac{1}{x})>(1+\frac{1}{x})^x\rightarrow e$, as $x\rightarrow \infty $. So from the permanence theorem of the sign I conclude that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$.

to do: check my idea

答:
It doesn't work, because
$$
\Bigl(1+\dfrac{1}{x}\Bigr)^{\!x}<e
$$
You want to prove that, for $x>0$,
$$
\Bigl(1+\dfrac{1}{x}\Bigr)^{\!x+1}>e
$$
which is equivalent to
$$
f(x)=(x+1)\log\Bigl(1+\dfrac{1}{x}\Bigr)>1 \tag{*}
$$
Consider
$$
f'(x)=\log\Bigl(1+\dfrac{1}{x}\Bigr)+(x+1)\Bigl(\frac{1}{x+1}-\frac{1}{x}\Bigr)
=1-\Bigl(1+\frac{1}{x}\Bigr)+\log\Bigl(1+\dfrac{1}{x}\Bigr)
$$
OK, let's simplify the thing: set $g(t)=1-t+\log t$, for $t>1$; then $g'(t)=(1-t)/t$<0. Hence $g$ is decreasing and its limit at $1$ is $0$. Thus $g(t)<0$ for $t>1$ and we conclude that $f'(x)<0$.

Therefore $f$ is decreasing; since its limit at $\infty$ is $1$, we have verified the inequality (*).

We can also verify the first statement in a similar way. Consider
$$
F(x)=x\log\Bigl(1+\frac{1}{x}\Bigr)
$$
Then
$$
F'(x)=\log\Bigl(1+\frac{1}{x}\Bigr)+x\Bigl(\frac{1}{x+1}-\frac{1}{x}\Bigr)=-\log\Bigl(\frac{x}{x+1}\Bigr)+\frac{x}{x+1}-1
$$
Consider $G(t)=t-1-\log t$, so that $G'(t)=(t-1)/t$, which is negative for $0<t<1$ and $G(1)=0$. Therefore $F'(x)>0$ and so $F$ is increasing. Since its limit at $\infty$ is $1$, we have the desired inequality.


问:
"Evenness" of weak k-compositions of n
For every weak $k$-composition of $n$, $x_0 + x_1 + ... + x_{k-1} = n$ with $x_i \geq 0$, we define the "evenness" of that composition to be $\left\lfloor{\frac{x_0}{2}}\right\rfloor + \left\lfloor{\frac{x_1}{2}}\right\rfloor + ... + \left\lfloor{\frac{x_{k-1}}{2}}\right\rfloor$, i.e. how many times 2 fits into its individual terms. I'm interested in how many weak $k$-compositions of $n$ there are with evenness $p$.

It is fairly easy to define a function $E_p(k,n)$ to compute this. To construct a $k$-composition of $n$ with evenness $p$, we can either set $x_0 = 0$ and follow up with a $k-1$-composition of $n$ with evenness $p$, or set $x_0 = 1$ and follow up with a $k-1$-composition of $n-1$ with evenness $p$, or set $x_0 = 2$ and follow up with a $k-1$-composition of $n-2$ with evenness $p-1$, etc.:

$$\begin{align*}
E_p(k,n) \;&=\; \sum_{i\,=\,0}^{p} {E_{p-i}(k-1,\, n-2i) + E_{p-i}(k-1,\, n-2i-1)} \\
           &=\; E_p(k-1,\, n) + E_p(k-1,\, n-1) \\ &\;\;\;\;+ \sum_{i\,=\,1}^{p} {E_{p-i}(k-1,\, n-2i) + E_{p-i}(k-1,\, n-2i-1)} \\
           &=\; E_p(k-1,\, n) + E_p(k-1,\, n-1) \\ &\;\;\;\;\;+ \sum_{i\,=\,0}^{p} {E_{p-1-i}(k-1,\, n-2-2i) + E_{p-1-i}(k-1,\, n-2-2i-1)} \\
           &=\; E_p(k-1,\, n) + E_p(k-1,\, n-1) + E_{p-1}(k,\, n-2)
\end{align*}$$

with the initial conditions

$$\begin{align*}
E_{-1}(k,n) &= 0, \\
E_0(k,0) &= 1, \\
E_p(k,0) &= 0, \\
E_p(0,n) &= 0.
\end{align*}$$

This gives $E_0(k,n) = {k\choose n}$, since the recursive formula for $E_0(k,n)$ reduces to that of the binomial coefficient. (This result has been described before in Number of compositions of $n$ into $k$ parts with each part at most $1$.) It also seems like $E_1(k,n) = k{k\choose n-2}$ for $n \geq 3$, and $E_p(k, 2p) = {k+p-1\choose p}$ and $E_p(k, 2p + 1) = k{k+p-1\choose p}$ for $p \geq 2$. The latter must surely be connected to multisets somehow (the number of ways you can choose $k$ elements from $p$ total with repetitions), but I don't have proofs for any of those results and arrived at them only after looking up the series $E_p(0,n), E_p(1,n), E_p(2,n), ...$ for different values of $p$ and $n$ in OEIS and noticing the similarities.

Is there a closed-form expression for $E_p(k,n)$? Looking up some of the series for larger $n$ on OEIS, the expressions get rather ugly, but they are indeed listed and described by another author. Here, $E_p(k,n)$ seems to have been identified as "the number of $n$-subsequences of $[1,k]$ with exactly $p$ contiguous pairs" (A027777, A027765, A027789, A027766). This sounds similar enough to my question to obviously be equivalent, but I don't know enough about the theory here (what is a $n$-subsequence in this context?) to make much further progress with that information, and I cannot find any relevant resources linked either.

答:
This can be done with standard generating function techniques. The generating function for weak $k$-compositions is $(1 + z + z^2 + \dots)^k = \frac{1}{(1 - z)^k}$; here each term $z^i$ in each factor corresponds to one of the terms in the composition taking value $i$. Keeping track of evenness can be done by introducing a second variable $t$ and replacing each $z^i$ with $t^{\lfloor \frac{i}{2} \rfloor} z^i$, which gives

$$\sum_{n, p \ge 0} E_p(k, n) t^p z^n = (1 + z + tz^2 + tz^3 + \dots)^k = \left( \frac{1 + z}{1 - tz^2} \right)^k.$$

We have $(1 + z)^k = \sum_{i=0}^k {k \choose i} z^i$ and

$$\frac{1}{(1 - tz^2)^k} = \sum_{i \ge 0} {i+k-1 \choose k-1} t^i z^{2i}$$

and isolating the coefficient of $t^p z^n$ in their product gives

$$\boxed{ E_p(k, n) = {p + k - 1 \choose k - 1} {k \choose n - 2p} }.$$

This specializes to all of the special cases you listed; probably there is a nice combinatorial proof to be extracted from this argument also.

Edit: Okay, the combinatorial argument is quite straightforward. If the total evenness is $p$ then we can first "distribute the evenness" among the $x_i$; that is, the number of ways to write $\sum \lfloor \frac{x_i}{2} \rfloor = p$ is the number of weak $k$-compositions of $p$ which is counted by the first factor. Having fixed the evennesses, what remains is a choice of whether each $x_i$ is even or odd, subject to the constraint that $\sum \left( x_i - \lfloor \frac{x_i}{2} \rfloor \right) = n - 2p$, which is counted by the second factor. (This is actually exactly what the generating function argument is saying although it takes a little familiarity with these techniques to make the translation.)

Title: "Evenness" of weak k-compositions of n


问:
Laplacian matrix and the eigenvalue 0 of bipartite graphs with no perfect matching
Let $G$ be a bipartite graph with no perfect matching. Show that $\lambda = 0$ is an eigenvalue of the adjacency matrix of $G$.

Hint: I do know that I should use the Laplacian matrix of $G$ and also $\lambda =0$ is the smallest eigenvalues of the Laplacian matrix.

答:
Well, this answer was obtained independently of anything, and the proof is actually fairly simple--no need to consider permanents.

So let us write the adjacency matrix of $G=(V,E)$ as $A_G$ where the entries are $a_{uv}$; $u,v \in V(G)$ and also:

For every pair of distinct vertices $u,v$, the equation $a_{uv}=a_{vu}=1$ holds iff $uv$ is in $E$; otherwise $a_{uv}=a_{vu}=0$.

The diagonal entries of $A(G)$ are 0. Equivalently, for every vertex $u$, the equation $a_{uu}=0$ holds.

We establish the following:

Prop: Let $G$ be a bipartite graph that does not have a perfect matching. Then $A_G$ has an eigenvalue of 0.

Convention: For each vector $w \in \mathbb{R}^V$ and each vertex $v \in V$ we write the $v$-th component of $w$ as $w(v)$. For each $v \in V$ let us write $\chi_v$ to be the vector in $L_V = \mathbb{R}^V$ where $\chi_v(v)=1$ and $\chi_v(u)=0$ for each $u \in V(G)\setminus \{v\}$. We prove the following:

If $G$ does not have a perfect matching, then there is a subset $S$ of one side of $G$ that satisfies $|N_G(S)| < |S|$ by Hall's Matching Theorem. So let $S$ be a subset of one side of $G$ satisfying $|N_G(S)| < |S|$

Claim: Then as $|S| > |N_G(S)|$ the following holds: $\{A_G\chi_v; v \in S \}$ is a linearly dependent set of vectors. Precisely, there exists scalars $c_v; v \in S$; at least one of the $c_v$s nonzero, such that $\sum_{v \in S} A_G (c_v\chi_v) = 0$ is satisfied.

To see this, for each subset $U \subseteq V$ let us define $L_{U}$ as follow: $L_{U} \doteq \{w \in \mathbb{R}^V; w(v) = 0$ for all $v \not \in U\}$. Then for each subset $U$ of $V$, the set $L_{U}$ is a linear subspace $\mathbb{R}^V$ of dimension $|U|$.

Now for each $v \in S$, the vector $A_G \chi_v$ is in $L_{N(S)}$. But $L_{N(S)}$ has dimension $|N(S)| < |S|$. So the $|S|$ vectors $A_G \chi_v$s; $v \in S$, are not linearly independent, and thus indeed there exists scalars $c_v; v \in S$ at least one of the $c_v$s nonzero, such that $\sum_{v \in S} A_G (c_v\chi_v) = A_G(\sum_{v \in S} c_v\chi_v) = 0$ is satisfied.

Note however, that the set $\{\chi_v; v \in S\}$, is a linearly independent set of vertices in $\mathbb{R}^V$, and so $\sum_{v \in S} c_v\chi_v$ is a nonzero vector as at least one of the $c_v$s is nonzero. Thus $\sum_{v \in S} c_v \chi_v$ is an eigenvector of $A_G$, with eigenvalue 0.


问:
What are the steps in breaking down the exponent in this limit analysis?
I'm trying to understand the reasoning in the following step of a limit analysis:

$$\lim_{n \to \infty} n\left(\left[1- \frac{1+c}{\frac{n}{\ln(n)}} \right]^{\frac{n}{\ln (n)}}\right)^{(n-1)\ln n/n} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}$$

I understand the "inner" part; $\lim_{n \to \infty} [1-\frac{1+c}{\frac{n}{\ln (n)}}]^{\frac{n}{\ln n}} = e^{-(1+c)}.$ And I sort of see that outer exponent $((n-1)\ln n) /n = (\ln n) - (\ln n / n)$ and the second part goes to 0, but it's not clear to me what rules actually justify "bringing the limit to the exponent". What are the actual steps involved in deducing this limit?

More generally, these types of asymptotic analysis show up in comp sci all the time and I feel there is a bag of tricks that I am missing.

答:
For this limit, you can bring up the $\ln(n)$ from the denominator, and then make the exponent match by using $(a^b)^c = a^{bc}$:

$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{\frac{n}{\ln (n)}}\right)^{\frac{(n-1)\ln n}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)^{\frac{1}{\ln(n)} \cdot \frac{(n-1)\ln n}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)^{\frac{n-1}{n}} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}
$$

This particular problem can be simiplied further into a form that uses the common limit property $\lim_{x \to a}\left[\frac{f(x)}{g(x)}\right] = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}$, so that we don't have to worry about whether a corresponding rule holds for exponents.

$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n-1}\right) = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\frac{\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right)}{\lim_{n \to \infty} \left[1- \frac{(1+c)\ln(n)}{n} \right]} = \lim_{n \to \infty} ne^{-[(1+c)\ln(n)]} \\
\frac{\lim_{n \to \infty} ne^{-[(1+c)\ln(n)]}}{1-0}
$$

Breaking down that last step for the numerator requires the chain rule for limits, which states that if $\lim_{u \to b} f(u) \to L$ and $\lim_{x \to a} g(x) = b$, and $f(x)$ is continuous at $x=b$, then $\lim_{x \to a} f(g(x)) \to L$.  Colloquially, as the input to a function approaches some input limit, its output approaches some output limit, and we can link up the output limit of one function to be the input limit of another function.

$$
\lim_{n \to \infty} n\left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right) \\
\lim_{n \to \infty} n \cdot \lim_{n \to \infty} \left(\left[1- \frac{(1+c)\ln(n)}{n} \right]^{n}\right) \\
$$

Extracting out the inner function, define:

$$
g(n) = (1+c)\ln n \\
\lim_{n \to \infty} g(n) \to G
$$

And now we can write our expression as:

$$
\lim_{n \to \infty} n \cdot \lim_{x \to G}\left[\lim_{n \to \infty} \left(\left[1- \frac{G}{n} \right]^{n}\right)\right] \\
$$

But whatever G may end up being, this portion of behavior as $n \to \infty$ is well defined as $e^x$, whatever the argument $x$ approaches.  G need not vary with n for this next step:

$$
\lim_{n \to \infty} n \cdot \lim_{x \to G} e^G
$$

Running the chain rule back the other direction, substituting back in for G:

$$
\lim_{n \to \infty} n \cdot \lim_{n \to \infty} e^{(1+c)\ln n}
$$

And running the product rule in reverse, we have the desired

$$
\lim_{n \to \infty} ne^{(1+c)\ln n}
$$


问:
Composition of Reflection Is a Rotation?
Prove that if $T_{1},T_{2}$ are reflections in lines through O then $T_{1}\circ T_{2}$ is a rotation about O.

Once again, a hint would be preferable to an answer. I'm not familiar with these types of linear transformations, as I am not accustomed to thinking of them as functions. Thanks in advance for everyone who answers.

答: Since you requested a hint, go ahead and think of the transformations as matrices, and treat composition as multiplication. Otherwise, think about how reflection changes chirality, but two reflections would leave chirality unchanged.</p>


问:
Quadratic Variation of $X$ defined by a differential equation
Let $X = (X_t)_{t\geq 0}$ be a stochastic process defined by
$$
d X_t = 3dt + 2 \sqrt X_t d B_t,
$$
where $B$ is a standard Brownian motion.
What is its quadratic variation? I have a subtle intuition that is defined by $d\langle X,X\rangle_t = 2\sqrt X_t d t$, but I do not know how to justify it correctly.

答:
The quadratic variation of the process is given by
$$\mathrm d \langle X,X\rangle_t=4 |X_t|\mathrm dt.
$$
In general, for processes $Z_t=\int_0^tH_s\mathrm dX_t$ and $Q_t=\int_0^tK_s\mathrm dY_t$ holds
$$\mathrm d \langle Z,Q\rangle_t=H_tK_t\mathrm d\langle X,Y\rangle_t,
$$
for $X,\ Y$ semimartingales and $H,\ K$ adapted càglàd processes.


问:
Construction of functions in $L^{1}$ with certain properties
I'm trying to find two functions with the following properties:

I'm trying to find a function $f \in L^{1}([0, \infty))$ such that $f$ is continuous and not bounded. I know that if such an $f$ exists, then it must be essentially bounded, but of course this doesn't say anything about whether $f$ exists or not.

On the other hand, I'm trying to find a function $h \in L^{1}([0,1])$ such that $h$ is not bounded in any neighbourhood of $x \in [0,1]$, for every x.

If such functions exist, I would like to compute the value of the corresponding integrals. I suspect that both functions exist, and are constructed as a limit of a certain sequence of functions, but I can't come up with examples. Any help will be appreciated.

答:
As suggested in the comments, for (1), let $f(x)$ have taller and taller spikes that get thinner and thinner as $x \to \infty$. Explicitly, for $n \ge 2$, suppose that $$f_n(x) = \left\{\begin{matrix}n^4(x-n), & n \le x < n + \tfrac 1 {n^3}, \\ n^4(n+\tfrac 2 {n^3} - x), & n + \tfrac 1 {n^3} \le x < n + \frac  2 {n^3}, \\ 0, & \text{ otherwise}. \end{matrix} \right. $$ Then $f_n$ is zero except on the interval $[n,n+2/n^3]$, where the graph of $f_n$ is a triangle of height $n$ and base $2/n^3$ and thus $\int_{0}^\infty f_n(x) dx = \frac 1 {n^2}$. Now let $f = \sum_{n=2}^\infty f_n$. Then $f$ is continuous and unbounded but $\int^\infty_{0} f(x) dx = \sum^\infty_{n=2} \frac 1 {n^2} < \infty$.

For (2), you should try to exploit the facts that the rationals are dense and yet still have measure zero. Take $$h(x) = \left \{\begin{matrix} q, & \text{ if } x \in \mathbb Q \text{ and } x = \tfrac p q \text{ where } \text{gcd}(p,q)=1, \\ 0, & \text{ if } x \not \in \mathbb Q. \end{matrix} \right.$$ The function integrates to zero (in the Lebesgue sense) since it equals $0$ almost everywhere. However, it is arbitrarily large in any neighborhood since near any irrational number $y$, you can find a rational $x$ which is arbitrarily close to $y$ and has an arbitrarily large denominator when reduced.

EDIT: as an aside, this second example is closely related to Thomae's Function. There the goal is to make a function which is continuous only at the irrationals, and you do so in the same way, but by using $1/q$ rather than $q$.


问:
Showing $\int_{-\infty}^{\infty}\frac{x^3\sin{x}\;dx}{x^4+5x^2+4} = \frac{(4-e)\pi}{3e^2}$ via contour integration
I have a complex integral $I$ to solve to show that

$$I=\int_{-\infty}^{\infty}\frac{x^3\sin{x}\;dx}{x^4+5x^2+4} = \frac{(4-e)\pi}{3e^2}$$

I know that you can integrate over the contour of a semicircle where and represent sin as an infinite sum but not sure what to do next. Should the polynomial be converted to a partial fraction too?

答:
On a canonical semicircle of radius $\;R>0\;,\;\;C: \{z\in\Bbb C\;|;|z|=R\;,\;\;\text{Im z}>0\}\;$ , we get

$$\left|\int_C\frac{z^3\,e^{iz}}{z^4+5z^2+4}\right|\le \max_{z\in C}\frac{|z|^3|e^{iz}}{|z^4+5z^2+4}\,\cdot \ell(C)\le\frac{R^3\,e^{-\text{Im}\,z}}{R^4-5R^2-4}\cdot\pi R\xrightarrow[R\to\infty]{}0$$

since exponential kills polynomial and $\;e^{-\text{Im}\,z}\xrightarrow[R\to\infty]{}0\;$ because $\;\text{Im}\,z>0\;$ and $\;\text{Im}\,z\to\infty\;$ when $\;R\to\infty\;$ ...You could use Jordan's Lemma

Thus, if $\;D\;$ is the contour $\;C\cup [-R,R]\;$ (the semicircle and the diameter  from $\;-R\;$ to $\;O\;$ in the real axis), you get

$$2\pi\left[\sum \text{Res}(F,z_k)\right]=\oint_D\frac{z^3\,e^{iz}}{z^4+5z^2+4}dz=\int_{-R}^R\frac{x^3\,e^{ix}}{x^4+5x^2+4}dx+\int_C\frac{z^3\,e^{iz}}{z^4+5z^2+4}dz$$

and on passing to the limit $\;R\to\infty\;$ you get your real improper integral on the right, and its value on the left.

Also, for the singularities (in fact, simple poles in this case so the residue is easily calculated):

$$z^4+5z^2+4=(z^2+1)(z^2+4)=(z-i)(z+i)(z+2i)(z-2i)\ldots\text{ and etc.}$$

Last thing: by Euler, $\;e^{ix}=\cos x+i\sin x\;$ , so you get two improper integrals, one of the real part , and the other a purely imaginary one (which is your integral). Figure out what happens here (further hit: odd and even functions)


问:
since a symmetric tridiagonal matrix contains only two distinct vectors
I don't understand meaning of "since a symmetric tridiagonal matrix contains only two distinct vectors" I write example for symmetric tridiagonal matrix. The 1st row vector and 1st column vector are the same, the 2nd row vector and 2nd column vector are the same. This way they are all eliminated. What does "contains only two distinct vectors" mean?

答:
The subdiagonal elements and the superdiagonal elements can be represented as the vector $(\beta_1, \beta_2, \dots \beta_{n-1})$. It has  length $n-1$. The diagonal elements can be represented as the vector $(\alpha_1, \alpha_2, \dotsc, \alpha_n)$. It has length $n$. Tridiagonal solvers such dpttrs (based on the LDLT-factorization) from LAPACK capitalize on this fact to reduce the storage requirement from $n^2$ words of memory needed for a dense matrix to a mere $2n-1$ words of memory.

Personally, I would prefer to write: "since a symmetric tridiagonal matrix can be represented using only two distinct vectors, it customary to replace the generic notation ...", but this is a matter of taste.


答: Using @EeveeTrainer's tips, $I=\Im w$ with$$\begin{align}w&:=\int_{\Bbb R}\frac{z^3e^{iz}dz}{(z^2+1)(z^2+4)}\\&\stackrel{\star}{=}2\pi i\left(\left.\tfrac{z^3e^{iz}}{(z+i)(z^2+4)}\right|_{z=i}+\left.\tfrac{z^3e^{iz}}{(z+2i)(z^2+1)}\right|_{z=2i}\right)\\&=2\pi i\left(\tfrac{-1}{6e}+\tfrac{2}{3e^2}\right)\\\implies I&=\frac{(4-e)\pi}{3e^2},\end{align}$$where $\stackrel{\star}{=}$ uses a semicircle in the half-plane $\Im z>0$ (@DanAntonio's answer verifies that the contour integral along the arc vanishes in the infinte-radius limit).

Title: Showing $\int_{-\infty}^{\infty}\frac{x^3\sin{x}\;dx}{x^4+5x^2+4} = \frac{(4-e)\pi}{3e^2}$ via contour integration

Body:
I have a complex integral $I$ to solve to show that

$$I=\int_{-\infty}^{\infty}\frac{x^3\sin{x}\;dx}{x^4+5x^2+4} = \frac{(4-e)\pi}{3e^2}$$

I know that you can integrate over the contour of a semicircle where and represent sin as an infinite sum but not sure what to do next. Should the polynomial be converted to a partial fraction too?


问:
Show that $\int f(x)dx=\int g(u)du$
Dose anyone know how to do this?

If a function $f(x)$ can be written $f(x)=g(u(x))du(x)/dx$ for a suitable function $u(x)$, show that $\int f(x)dx=\int g(u)du$.

答: Use the substitution $z = u(x)$ on the right integral.</p>

Title: Show that $\int f(x)dx=\int g(u)du$


Body:
$$\frac{x^3}{x^4+5x^2+4}=\frac{x^3}{(x^2+4)(x^2+1)}=\frac43\frac{x}{x^2+4}-\frac13\frac{x}{x^2+1}$$
then you could maybe split this up further:
$$x^2+4=(x+2i)(x-2i)$$
$$x^2+1=(x+i)(x-i)$$

$$\frac43\frac x{x^2+4}=\frac23\frac1{x+2i}+\frac23\frac1{x-2i}$$
$$\frac13\frac x{x^2+1}=\frac16\frac1{x+i}+\frac16\frac1{x-i}$$
and so:
$$\frac{x^3}{x^4+5x^4+4}=\frac23\frac1{x+2i}+\frac23\frac1{x-2i}+\frac16\frac1{x+i}+\frac16\frac1{x-i}$$


问:
Probability with conditional replacements
I am having difficulty understanding the answer to this problem due to the conditional replacement.  My initial thought is that I could draw a tree diagram and use the multiplication rule for independent events to find probability of white being drawn but I'm confused by not knowing the first selection and the conditional nature of the problem?

An urn contains three blue and seven white chips.  A chip is selected at random.  If the color of the chip selected is white, it is replaced and two more white chips are added to the urn.  However, if the chip drawn is blue, it is not replaced and no additional chips are put in the urn.  A second chip is drawn after the first (problem does not specify which color chip was drawn first).  What is the probability that it is white?  Supposing that we are given that the chip selected for the second time is white, what is the probability that the first chip selected is blue?

答:
As @JMoravitz says you should forget about 'independent events' ideas. You can still multiply along branches of a probability tree and that is a good method to use for this problem.

From the initial scenario of $3$B,$7$W you have a branch leading to $2$B,$7$W and one leading to $3$B,$9$W.

Is that sufficient info. for you to be able to draw the tree and insert probabilities?

For the first part,  can you now see that the probability Chip 2 is white is $$3/10\times 7/9+7/10\times 9/12?$$


答:
Let $A$ be the event that a white was drawn on the first pull.  Let $B$ be the event that a white was pulled on the second pull.

You are asked to find $\Pr(B)$ and you are asked to find $\Pr(A^c\mid B)$

To do this, note that it is clear how to calculate $\Pr(A)$ as this is just going to be the ratio of white chips compared to total number of chips in the initial configuration of the urn.  It is also apparent how to calculate $\Pr(B\mid A)$ and $\Pr(B\mid A^c)$ as being the ratio of white chips to all chips in the configurations of the urn had we picked a white on the first pull or not respectively.

 $\Pr(A)=\frac{7}{10},\Pr(B\mid A)=\frac{9}{12},\Pr(B\mid A^c)=\frac{7}{9}$

Similarly, we can calculate the opposites of these events by subtracting away from $1$

 $\Pr(A^c)=\frac{3}{10},\Pr(B^c\mid A)=\frac{3}{12},\dots$

Next piece of information we want is to calculate $\Pr(B)$.  For this, recall the Law of Total Probability and multiplication principle

 $\Pr(B)=\Pr(A\cap B)+\Pr(A^c\cap B) = \Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)$

That should get you far enough to have $\Pr(B)$ calculated.

Now, to calculate $\Pr(A^c\mid B)$, apply Bayes' Theorem which is really just repeated application of the definition of conditional probability to get:

 $\Pr(A^c\mid B)=\dfrac{\Pr(A^c\cap B)}{\Pr(B)}=\dfrac{\Pr(B\mid A^c)\Pr(A^c)}{\Pr(B)}$

Each value of which we have figured out earlier in the problem.

Title: Probability with conditional replacements

Body:
I am having difficulty understanding the answer to this problem due to the conditional replacement.  My initial thought is that I could draw a tree diagram and use the multiplication rule for independent events to find probability of white being drawn but I'm confused by not knowing the first selection and the conditional nature of the problem?

An urn contains three blue and seven white chips.  A chip is selected at random.  If the color of the chip selected is white, it is replaced and two more white chips are added to the urn.  However, if the chip drawn is blue, it is not replaced and no additional chips are put in the urn.  A second chip is drawn after the first (problem does not specify which color chip was drawn first).  What is the probability that it is white?  Supposing that we are given that the chip selected for the second time is white, what is the probability that the first chip selected is blue?


答:
We know that for each $x\in\mathbb R$, either $f(x)=x^2$ or $f(x)=0$.

Let $A=\{x\neq0\mid f(x)=0\}$ and $B=\{x\neq0\mid f(x)=x^2\}$. Then $A$, $B$ partition $\mathbb R\setminus\{0\}$.

Suppose both are non-empty. Take $a\in A$ and $b\in B$. Sub $x=a,y=-b$ to get $f(a^2-b)=b^2$. Since $b\neq0$, we must have $b^2=(a^2-b)^2\implies a^2=2b$ (as $a\neq0$).

This implies $\lvert A\rvert\leq2$ and $\lvert B\rvert=1$, a clear contradiction. So one of $A$, $B$ is empty, and we recover the two solutions.

Title: Finishing the solution to this functional equation: $ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y $

Body:
I am trying to solve the below functional equation for $f$ (where $f: \mathbb{R} \to \mathbb{R}$):
$$ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y \text. $$
(Source: https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf: "Iran TST 1996")

Here's what I've got so far:

Substituting $y=0$ we have $f\left(x^2\right) = f\bigl(f(x)\bigr)$.

Substituting $y=f(x)$ we have
$$f\left(x^2+f(x)\right) = f(0) + 4f(x)^2 \text.$$
$$f(0) = 0 \text. $$

Letting $x=0$ we can show the function is even.

Finally letting $y = -x^2$ we can simplify to yield
$$f(x) \left(x^2 - f(x)\right) = 0 \text. $$

And this is where I'm stuck. Given that it's an Olympiad I'm fairly sure that actually the only solutions are $f(x) = x^2$ and $f$ is identically zero and that there will not be any pathological solutions that have discontinuities - though I have not been able to show this.

My ideas that I've tried:

Showing $f$ is injective when not identically zero (on $\mathbb{R}^+$ that is), this doesn't seem to go anywhere useful.

Showing that $f$ is surjective (again onto $\mathbb{R}^+$) when not identically zero, as if that is the case we always must choose $x^2$ instead of zero other we miss out a positive number. However I haven't been able to find any way to show this either.

Use inequalities to show that if $f$ is not identically zero then the inequality $f(x) \ge 0$ that we have from our current information (either $f$ is zero or the square of a real) is strict when $x \ne 0$, but again I couldn't do anything with this either.