问:
$\lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} \sin(\alpha x)\cos(\beta x) \,dx$
I have to compute this limit:

Suppose that $|\alpha| \neq |\beta |$, compute the following limit:
\begin{align} \lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T}
 \sin(\alpha x)\cos(\beta x) \, dx \end{align}

What I've done is this:
\begin{align}
& \int_0^T \sin(\alpha x)\cos(\beta x) \, dx \\[8pt]
= {} & \frac{1}{2(\alpha-\beta)} + \frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)}
\end{align}

Then,

\begin{align}
\lim_{T \rightarrow \infty}\frac{1}{T}\int_0^T \sin(\alpha x) \cos(\beta x) \,dx=\lim_{T \rightarrow \infty}\frac{1}{T}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]
\end{align}

But I'm not sure how can I conclude. What I think that I can do is this:

\begin{align}
\underbrace{\lim_{T \rightarrow \infty}\frac{1}{T}}_{=0}\cdot \underbrace{\lim_{T \rightarrow \infty}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]}_{\text{I don't know how to compute it}}
\end{align}

So intuitively I think this must be zero. How can I conclude? Is there any other way to solve it? I really appreciate your help.

答:
Your last step doesn't work because $\lim_{T \rightarrow \infty}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]$ does not exist. It doesn't exist becuase the cosines oscillate rather than approaching a limit.

But you can compute
$$
\lim_{T \rightarrow \infty}\frac{1}{T}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]
$$
by squeezing. You have
$$
\lim_{T\to\infty} \frac 1 T \Big[\cdots\cdots\cdots \Big]
$$
where the expression in the $\displaystyle\Big[\cdots\text{ large square brackets }\cdots\Big]$ remains bounded, i.e. remains between two identifiable bounds, as $T$ changes. Thus you seek the limit of an expression that is squeezed between two things that approach $0.$


问:
How to show that $z=0$ is an essential singularity of $\sin(\frac{1}{z})$
My attempt: $0$ is clearly a singularity since $f$ is unbounded as $z\rightarrow 0$.

Using Taylor expansion we see that $\sin(\frac{1}{z})= -\frac{1}{z} + \frac{1}{3!z^3} - \frac{1}{5!z^5}+\dots=\frac{1}{z}[-1 + \frac{1}{3!z^2} - \frac{1}{5!z^4}+\dots] $ so that as $z\rightarrow 0$ we have $f(z)$ is unbounded and therefore $0$ is not a removable singularity.

My question: How do we know that this is not a pole and hence it is an essential singularity. If I check out $1/f$ nothing really simplifies and it's just $1$ over a messy expansion. I can see that the polynomial on the bottom will diverge faster than the linear $z$ will converge to $0$ on the top so that $1/f$ diverges so it can't be a pole, but this doesn't seem rigorous enough.

Are there any better ways to see this, or to think about this problem?

答:
If it was a pole, then this would be valid: $\lim_{z\to 0}\sin\left(\frac{1}{z}\right)=\infty$.

Then, for every sequence $z_n\to0$ ($n\to\infty$) we would have $\lim_{n\to\infty}\sin\left(\frac{1}{z_n}\right)=\infty$.

However - compare that with $z_n=\frac{1}{\frac{\pi}{2}+n\pi}\to 0$ and $\sin\left(\frac{1}{z_n}\right)=(-1)^n$.

ContentLicense: CC BY-SA 4.0


问:
The convergence of $\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}$
I need to evaluate the following integral

$$\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}dx$$
where $\alpha\in (0,1)$ and $[x]$ is the floor function.
Without $x-[x]$, we can evaluate the integral easily using analytical continuation.
$$\int_0^{\infty} \sin(x)/x^{\alpha}dx=\Im \int_0^{\infty} e^{ix}/x^{\alpha}dx=\Im [\Gamma(1+\alpha)/i^{\alpha-1}]$$

In the presence of $x-[x]$, my idea is that one could use inequality to upper bound the integral if it converges or lower bound the integral if it diverges. But I'm not sure what kind of inequality I should use.

I tried Hölder's inequality, but it will square the sin function in the upper bound, which makes it diverge.

答:
To answer to @ProfessorVector

For $a\in (1,2)$, $$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx= \lim_{b\to 0} \int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$ is analytic for $a < 2$.

For $a \in (0,1/2)$

$e^{-bx/2}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})$  converges in $L^2(0,\infty)$

and $x^{-a}e^{-bx/2}$ is $L^2(0,\infty)$

Thus no problem to switch $\int,\sum$
$$\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $$= \int_0^{\infty} e^{-bx}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})x^{-a}dx$$
$$ = \Gamma(1-a)\Re(\frac{(b+i)^{a-1}}{2i}-\sum_{n\ge 1} \frac{(b+i(2\pi n-1))^{a-1}-( b+i(2\pi n+1))^{a-1}}{2\pi n})$$

The latter series extends analytically to $\Re(a)\in (0,2)$  so it must stay equal to $\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$.

Thus for $a\in (1,2)$ we can let $b\to 0$ to obtain

$$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx = \Gamma(1-a)\Re(\frac{i^{a-1}}{2i}-\sum_{n\ge 1} \frac{(i(2\pi n-1))^{a-1}-(i(2\pi n+1))^{a-1}}{2\pi n})$$


问:
Consequences of Axiom of Choice
In the proof of Zorn's lemma by Jonathan Lewin published in AMM in 1991, the following reasoning appears:

Suppose every chain of a set $\mathcal{P}$ has a strict upper bound in $\mathcal{P}$(an element strictly greater than all the elements ).Then by Axiom of choice, there exists a function $f$ mapping every chain in $\mathcal{P}$ to its strict upper bound.

I am not sure how is this derived directly from Axiom of Choice, which only states there exists some function mapping every chain to its element. Nothing is said about choosing a particular element, and let alone the fact the upper bound is not even an element of the set.
It might be a simple consequence of Axiom of Choice, but this is a new field for me so I would appreciate a detailed explanation.

答:
Assume that every chain has a strict upper bound.

Therefore given a chain $C$, the set $U_C=\{x\mid x\text{ is a strict upper bound of }C\}$ is non-empty.

Therefore the family $\{U_C\mid C\text{ is a chain}\}$ is a set of non-empty sets.

Therefore it admits a choice function, $f$, that is $f(U_C)\in U_C$.

Therefore the function $F(C)=f(U_C)$ is a function mapping a chain to a strict upper bound.

Title: Consequences of Axiom of Choice


答:
If $h$ is an analytic function on a punctured disc $D - \{c\}$, then $h$ has an absolutely convergent Laurent expansion

$$h(z) = \sum\limits_{n \in \mathbb Z} a_n (z-c)^n$$

for all $z \neq c$ sufficiently near $c$.

If $a_n = 0$ for all $n < 0$, the singularity at $h$ is removable.

If $a_n \neq 0$ for finitely many $n < 0$, the singularity is a pole.

If $a_n \neq 0$ for infinitely many $n < 0$, the singularity is essential.

The proofs of these statements are easy, but depend on what definitions you are using of pole and essential singularity.

Title: How to show that $z=0$ is an essential singularity of $\sin(\frac{1}{z})$

Body:
My attempt: $0$ is clearly a singularity since $f$ is unbounded as $z\rightarrow 0$.

Using Taylor expansion we see that $\sin(\frac{1}{z})= -\frac{1}{z} + \frac{1}{3!z^3} - \frac{1}{5!z^5}+\dots=\frac{1}{z}[-1 + \frac{1}{3!z^2} - \frac{1}{5!z^4}+\dots] $ so that as $z\rightarrow 0$ we have $f(z)$ is unbounded and therefore $0$ is not a removable singularity.

My question: How do we know that this is not a pole and hence it is an essential singularity. If I check out $1/f$ nothing really simplifies and it's just $1$ over a messy expansion. I can see that the polynomial on the bottom will diverge faster than the linear $z$ will converge to $0$ on the top so that $1/f$ diverges so it can't be a pole, but this doesn't seem rigorous enough.

Are there any better ways to see this, or to think about this problem?


问:
Invalid Operations on Trig Equations
Reviewing Trig I come across this problem :
$\text{Solve for all real $x$ such that } 2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$

The first thing I did was use the cosine half-angle identity get this look...

$$2\sqrt{2}\sqrt{\frac{\cos(x)+1}{2}}=\cos(x)+2$$

I then squared both sides of the equation (I believe this is an error but can't pinpoint why).

The rest looks like this :

$$8\frac{\cos(x)+1}{2}=\cos^2(x)+4\cos(x)+4$$

$$4\cos(x)+4=\cos^2(x)+4\cos(x)+4$$

$$0=\cos^2(x)$$

Then $x = \left\{\pm \frac{\pi}{2}+2\pi n \ \middle|\ n \in\mathbb Z \right\}$

Now, I believe this set does contain all $x$ which satisfy the orignial equation, but it definitely contains invalid solutions as well.

What specifically did I do wrong here and, if possible, are there any hard and fast rules about when algebraic operations on trig equations will change the solution set?

答:
If you are trying to solve an equation of the type $f(x)=g(x)$, it is perfectly fine to do $f^2(x)=g^2(x)$. The solutions of the first equation will also be solutions of the second one. But there is a real possibility of creating new ones. An extreme case of this is the equation $x=-x$, whose only solution is $x=0$. But every real number is a solution of the equation $x^2=(-x)^2$.

In your specific case, take all solutions  that you got and check which ones are solutions of the original equation.

Title: Invalid Operations on Trig Equations


问:
Can't solve this integral $ \int_{- \pi}^{\pi} \sin(x^{13})dx$
Hi I don't know how to solve this integral. Wolfram alpha says its equal to $0$ and it's the answer to my college exam but I don't know how to solve it. Anyone mind taking a guess?

$ \int_{- \pi}^{\pi}\sin(x^{13})dx $

Thanks

Body:
Hint:

$$\int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx=\int_0^a f(-x)\,dx+\int_0^a f(x)\,dx.$$


问:
In any finite group $G$, the number of elements not equal to their own inverse is an even number.
In any finite group $G$, the number of elements not equal to their own inverse is an even number.

I'm not sure how to go about this. I was considering assuming the number of elements not equal to their own inverse is an odd number (contradiction), but I'm not sure how to get a contradiction. I was considering thinking about when $|G|$ has even order and odd order, and then finding a way to do it directly, but I'm not sure. I think that in a finite group the number of elements equal to their own inverse should be even since they come in pairs. Any hints or solutions are greatly appreciated.

Body: HINT: $a\ne a^{-1}$ if and only if $a^{-1}\ne(a^{-1})^{-1}$


问:
How to solve the matrix equations: $\frac{1}{2}(\Omega Q - Q\Omega)=F$ for $\Omega$
How to solve the following matrix equation: $$\frac{1}{2}(\Omega Q + Q\Omega^T)=F$$

$Q$ is of rank $1$, $\text{tr}(Q)=1$, $Q\succeq0$ (positive semidefinite).  So we can werite $Q = qq^T$ for some $q$ with $\|q\|=1$.

$\Omega$ is skew-symmetric, i.e., $\Omega^{T} = -\Omega$. So diagonal entries are all $0$.

$\Omega$, $Q$, $F$ are $4\times 4$ square matrices. $\Omega$ is the variable I want to solve.

I cannot find any method to solve it. Can anyone suggest me how to solve $\Omega$ in terms of $Q$ and $F$?

Even though we cannot obtain the closed form, any approximation or trick can we use?

Thanks!

Note:  $\Omega$ should look like
$$\begin{bmatrix}0 & \omega_3 & -\omega_2 & \omega_1 \\
  -\omega_3 & 0 & \omega_1 & \omega_2 \\
   \omega_2 & -\omega_1 & 0 & \omega_3 \\
   -\omega_1 & -\omega_2 & -\omega_3 & 0 \end{bmatrix}$$

答:
Presumably the matrices are real. Since $Q$ is a rank-one positive semidefinite matrix with unit trace, we can write $Q=uu^T$ for some unit vector $u$. Let $S=2F$. The equation can then be rewritten as
$$
\Omega uu^T+uu^T\Omega^T=S.\tag{1}
$$

When $S=0$, the equation is clearly solvable and its solutions are given by those skew-symmetric matrices $\Omega$ such that $\Omega u=0$. So, if we pick an orthogonal matrix $U$ whose first column is $u$, the general solution in this case is given by $\Omega=U(0\oplus K)U^T$ where $K$ is an arbitrary skew-symmetric matrix.

When $S\ne0$, we will prove that the equation is solvable if and only if $S$ is a traceless rank-$2$ symmetric matrix such that $\{u,Su\}$ is an orthogonal basis of $\operatorname{range}(S)$. When this condition is satisfied, the solutions to $(1)$ are given by those skew-symmetric matrices $\Omega$ such that $\Omega u=Su$.

For necessity, note that when $S\ne0$, the LHS of $(1)$ is a traceless symmetric matrix whose rank is at most two. Since skew-symmetric matrices have no nonzero real eigenvalues, the LHS of $(1)$ cannot have rank one. Therefore the equation is solvable only when $S$ is a traceless symmetric matrix of rank $2$. In this case, if we right-multiply both sides of $(1)$ by $u$, we get $\Omega u=Su$. Since $S$ has rank two, $(1)$ shows that $\Omega u$ and $u$ must be linearly independent and they span the range of $S$. As we also have $u^TSu=0$ (this can be obtained directly from $(1)$ or from $u^TSu=u^T\Omega u=0$), $\{u,Su\}$ is also an orthogonal basis of $\operatorname{range}(S)$.

Conversely, suppose $S$ is a traceless rank-$2$ symmetric matrix such that $\{u,Su\}$ is an orthogonal basis of $\operatorname{range}(S)$. By a change of orthonormal basis, we may assume that $u=e_1$ and $Su$ is parallel to $e_2$. Hence $S=\pmatrix{0&a\\ a&0}\oplus0$ and it is straightforward to verify that $Suu^T+uu^TS=S$. It follows that $(1)$ is satisfied when $\Omega u=Su$. In other words, if we put $a=\|Su\|$ and pick an orthogonal matrix $U$ whose first two columns are $u$ and $\frac{Su}{\|Su\|}$ respectively, then
$$
S=U\left[\begin{array}{cc|c}0&a&\mathbf0\\ a&0&\mathbf0\\ \hline \mathbf0&\mathbf0&\mathbf0\end{array}\right]U^T
$$
and the solutions to $(1)$ are given by skew-symmetric matrices in the form of
$$
\Omega=U\left[\begin{array}{cc|c}0&-a&\mathbf0\\ a&0&-y^T\\ \hline \mathbf0&y&K\end{array}\right]U^T.
$$


问:
Convolution of r.v.'s
Suppose a sequence $\{X_{n} \}$ of pairwise independent r.v.'s. If $F_{X_n}$ is the distribution function of $X_n$ then $ F_{X_1 + \ldots + X_{N}} = F_{X_1} \ast F_{X_2 + \ldots + X_{N}}$ (convolution) for some $N>1$ fixed? </p>

答:
Let $X_1$ and $X_2$ be independent Bernoulli$(0.5)$ random variables ($1$ with probability $\frac 12$, $0$ with probability $\frac 12$) and set $X_3 = |X_1-X_2|$.

Then $X_1$, $X_2$ and $X_3$ are pairwise independent Bernoulli$(0.5)$ random variables, but $X_1 + X_2 + X_3$ is always even. This is not the same as the iid case.

ContentLicense: CC BY-SA 3.0

Title: Convolution of r.v.'s


答:
$$2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$$

By the double-angle formula for the cosine, which says $\cos(2\theta)= 2\cos^2\theta-1,$ applied in the case where $\theta = x/2,$ we get:

$$ 2\sqrt{2} \cos\left(\frac{x}{2}\right)= \left[ 2\cos^2\left(\frac x 2\right) - 1\right] + 2. $$

$$
2\sqrt 2 \, u = 2u^2 + 1.
$$

This is a quadratic equation. Solve it for $u$ and then write $\cos\dfrac x 2 = \text{the solution for $u$}$ and solve that for $x.$

Title: Invalid Operations on Trig Equations

Body:
Reviewing Trig I come across this problem :
$\text{Solve for all real $x$ such that } 2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$

The first thing I did was use the cosine half-angle identity get this look...

$$2\sqrt{2}\sqrt{\frac{\cos(x)+1}{2}}=\cos(x)+2$$

I then squared both sides of the equation (I believe this is an error but can't pinpoint why).

The rest looks like this :

$$8\frac{\cos(x)+1}{2}=\cos^2(x)+4\cos(x)+4$$

$$4\cos(x)+4=\cos^2(x)+4\cos(x)+4$$

$$0=\cos^2(x)$$

Then $x = \left\{\pm \frac{\pi}{2}+2\pi n \ \middle|\ n \in\mathbb Z \right\}$

Now, I believe this set does contain all $x$ which satisfy the orignial equation, but it definitely contains invalid solutions as well.

What specifically did I do wrong here and, if possible, are there any hard and fast rules about when algebraic operations on trig equations will change the solution set?


问:
Ellipse equation equal to zero, how to define foci?
I will appreciate if you could help me with Equation $10x^2+y^2+200x=0$ . The answer key is giving foci $(-10, 30)$ and $(-10,-30).$ I am confused: ellipse equation must be equal to $1.$ Or it is in a form of a line? I tried to make it standard form:
$(10(x+10)^2)/(1/\sqrt {10})^2)+y^2=0$. But this zero...And where did $30$ and $-30$ come from?
P.S. Sorry for the formating, this is my first time using this web-site, I couldnt figure how to put square root sign.

Body:
Completing squares we get
$$10x^2+y^2+200x=0\iff (\sqrt{10}x+10\sqrt{10})^2+y^2=1000.$$

We write $1$ on the rigth hand side
$$(\sqrt{10}x+10\sqrt{10})^2+y^2=1000\iff \dfrac{(\sqrt{10}x+10\sqrt{10})^2}{1000}+\dfrac{y^2}{1000}=1.$$

Simplifying we obtain $$\dfrac{(x+10)^2}{10^2}+\dfrac{y^2}{(10\sqrt{10})^2}=1.$$


问:
How many surjective functions satisfies the given property?
I've been struggling on this problem for quite a bit and was wondering if someone could help me.
The question goes as follows:
$X$ is a finite set with $|X|=n$ and $n$ is a odd number. How many surjective functions $f:X\rightarrow\{0,1\}$ are they with the property: $$|f^{-1}(0)|<|f^{-1}(1)|$$
If someone knows the answer please tell me how you got to your answer so I can understand it.

My thougts: I thougth of $arcsin(x)$ because $arcsin(0)$ < $arcsin(1)$, and $arcsin(x)$ is a inverse function, but how could I possibly find all functions?

Thanks in advance :)

PS: could someone also explain to me what they exactly mean with $|X|=n$, do they just mean absolute value?

答:
$|X|=n$ just means that there are $n$ elements in $X$.

Note that for any surjective function $f:X\to2$ ($2$ is the set $\{0,1\}$), either $$|f^{-1}(0)|<|f^{-1}(1)|$$or vice versa, since $n$ is odd, and $$|f^{-1}(0)|+|f^{-1}(1)|=n$$And, for any surjective function $f$, $g(x)=1-f(x)$ is also a surjection $X\to2$, and it satisfies the property iff $f$ doesn't

In other words, this implies that exactly half of all surjective functions $X\to2$ have the property. There are $2^n-2$ surjective functions from $X\to 2$ (since there are only $2$ non surjective functions), and hence, $2^{n-1}-1$ surjective functions with the property you asked for.


问:
If $\omega$ is a complex cube root of unity, show that the following equals null matrix.

If $\omega$ is a complex cube root of unity, show that

  $$ \left(  \begin{bmatrix}
        1 & \omega & \omega^2 \\
       \omega & \omega^2 & 1 \\
        \omega^2 & 1 & \omega \\
        \end{bmatrix} + \begin{bmatrix}
        \omega & \omega^2 & 1 \\
        \omega^2 & 1 & \omega \\
        \omega & \omega^2 & 1 \\
        \end{bmatrix} \right)
     \begin{bmatrix}
        1  \\
        \omega  \\
        \omega^2  \\
        \end{bmatrix}=\begin{bmatrix}
        0  \\
        0  \\
        0  \\
        \end{bmatrix}$$

I tried to solve this and I reduced the L.H.S. to
$$\begin{bmatrix}
        2 +2\omega+2\omega^2 \\
        2 +2\omega+2\omega^2 \\
        2 +2\omega+2\omega^2 \\
        \end{bmatrix}
$$ (since $\omega^3=1$)

but couldn't equate it to R.H.S.

Please provide your assistance.

Thank you

答:
$$\omega^3=1 \Rightarrow ω^3−1=0 \Rightarrow (ω−1)(ω^2+ω+1)=0 \Rightarrow ω−1=0 \lor  ω^2+ω+1= 0$$

Since, $\omega \neq 1 $ as it is complex , $ω^2+ω+1=0$

Hence, LHS = RHS

- Answer originally posted as comment by Git Gud


问:
Multivariate generating function for coprimes
I have learnt $f(x,y)=\sum_{m,n \in \mathbb{N}}x^ny^m$ has a closed form $\frac{1}{1-x}.\frac{1}{1-y}$, but if I modified a little bit by defining a charateristic function for coprime indexes: $a(m,n)=1$ if $(m,n)=1$ and $a(m,n)=0$, otherwise, then the situation is completely different. I am completely newbie in this area and it seems there are not much research on multivariate generating functions. So my question is: Is there a closed form for the generating function as follows:

$g(x,y)=\sum_{\gcd(m,n)=1}x^ny^m$

If it exists, could you please give me the general methods to get there?

After searching for a while, I found a related open question in Does there exist a generating function for the rational numbers?

答:
I don't believe this generating function has a closed form in any usual sense; already setting $x = y$ produces

$$f(x) = \sum_{\gcd(m, n) = 1} x^{m+n} = \sum_{n \ge 1} \varphi(n) x^n$$

(since $\gcd(m, n) = 1$ is equivalent to $\gcd(m, m+n) = 1$) and I don't believe this generating function has a closed form in any usual sense; in particular it is not rational because $\varphi(n)$ has the wrong growth rate. It's known that

$$\liminf_{n \to \infty} \frac{\varphi(n) \log \log n}{n} = e^{- \gamma}$$

but this limit is either $0$ or $\infty$ for the coefficients of a rational generating function, because the coefficients of a rational generating function necessarily grow like $Cn^r \lambda^n$ where $r$ is a non-negative integer. Actually this asymptotic behavior even implies that $f(x)$ is neither meromorphic nor algebraic.


问:
How to show that there exists a way to color the edges of a non-complete graph such that there is no monochromatic triangle?
If you were to color the edges of a non complete Graph in two colors, how could one show that there is at least one way that there is no monochromatic triangle. Or in the reverse how could one show that there always must be at least one monochromatic triangle?

These are the questions I have.

Somewhat recently I started learning about graph theory and I have of course come across Ramsey's theorem but as far as I understand it, it only works for complete graphs. Could anyone please give me hints and such on how to do the same for non-complete graphs? It is just for general learning and interest.
Here is an example I am refering to:

答:
The ability to color a given graph is (as we discussed in comments) likely to depend on the existence on complete subgraphs within any particular incomplete graph, and the interaction between any such distinct graphs

In your example, it's fairly easy to see two disjoint $K_4$ graphs - I've picked out one of them with green nodes below, the other left black - which need coloring so as to eliminate triangles, and with a careful choice taking account of the links between those two $K_4$ subgraphs we can show a triangle-free 2-coloring. Obviously the symmetry helps a lot here.


问:
Why is addition and multiplication associative but not exponentiation and beyond?
The most similar question I found on math.se was concerning commutativity (Why are addition and multiplication commutative, but not exponentiation?). If this ostensibly simple (but not obvious to me) question has already been discussed, please post the link and I apologize for duplicate. I am wondering too how other rules of arithmetic in basic and other algebras (distributivity, transitivity, additive and multiplicative inverses, etc) apply or don't apply to operators and any rhyme to their properties' applicability or hierarchy.

associativity of addition: (α+β)+γ ≡ (α+β+γ) := α+β+γ ≡ α+(β+γ)

associativity of multiplication: (a×b)×c := (ab)⋅c = abc ≡ a(bc)

non-associativity of exponentiation: A^B^C := (A^B)^C := $(A^B)^C$ = ${A^B}^C$ ≢ $A^{B^C}$ = $A^{(B^C)}$ := A^(B^C)

答:
Hmmm.... Well, lets think....

in the natural numbers $a+b$ is defined to be $a + \underbrace{1+1+....+1}_b$ and so $a+(b+c) = a+\underbrace{\underbrace{1+1+.....+1}_b + \underbrace{1+1+....+1}_c}_{b+c}=\underbrace{a+\underbrace{1+1+.....+1}_b}_{a+b} + \underbrace{1+1+....+1}_c=(a+b) + c$ which by the construction of natural numbers is consistent and meaningful

(googles Peano Postulates: https://en.wikipedia.org/wiki/Peano_axioms  mumbles to self  Hmm.... induction...seems to work)

And multiplication defined as $a\times b = \underbrace{b+b+ .....+b}_a$ so

$a\times (b\times c) = \underbrace{b\times c + b\times c+ .... + b\times c}_a=$

$\underbrace{\underbrace{c+c+....c}_b + \underbrace{c+c+....c}_b+..+\underbrace{c+c+....c}_b}_a=$

$\underbrace {c+c+c+c+ .....+ c+c}_{\underbrace{b+b+...+b}_a} =\underbrace{c+c+c+c+...+c+c}_{a\times b}=$

$(a\times b)\times c$.

(mumble,mumble  so far so good.)

Now $a^b$ is defined as $\underbrace{a\times a\times ...\times a}_b$ and things just wont work.

We can see that $b^a\times b^c = \underbrace{b\times b....\times b}_a\times \underbrace{b\times b....\times b}_c=\underbrace{b\times b\times ....\times b}_{a+c}=b^{a+c}$,

And that $(b^a)^c = \underbrace{b^a\times b^a\times ..... \times b^a}_c =$

$\underbrace{\underbrace{b\times b\times... \times b}_a\times\underbrace{b\times b\times... \times b}_a\times....\times \underbrace{b\times b\times... \times b}_a}_c=$

$\underbrace{b\times b\times .... \times b}_{\underbrace{a + a+....+a}_c = c\times a}=b^{c\times a}$

but from there it just breaks down.

$(a^b)^c = a^{b\times c}=\underbrace{a\times a\times ....\times a}_{b\times c}$

whereas $a^{(b^c)}=\underbrace{a\times a\times .....\times a}_{b^c}$.

And multiplying $a$ by itself by two different number of times result in different numbers (we haven't proven that it always results in different numbers (although if $a \ne 0$ that will be the case) but we have no reason to think it would result in the same an we can find many examples where they don't) and that $b^c$ need not equal $b\times c$ (ditto about proving it... for all but a finite number of cases and plenty of counter examples) they just don't work.

Actually though.... As addition was defined as simply listing $1$ out first $a$ times, then $b$ times; and multiplication by listing $a$ out $b$ times, they are naturally associative.  If we defined an operation by doing the same thing repetitively that would be associative.  For example if we defined $a Eb$ as $a^{(a^{(a^{.....a)})})_{b\text{ times}}}$ that would be associative.


问:
inner product (real or imaginary?)
I have a simple question. Is the inner product $\langle x | z\rangle$ always real or does this hold only for an inner product with itself $\langle x | x \rangle$?</p>

答:
In real vector spaces, both are real, and the latter is also positive.

In complex vector spaces, the first can be anything (e.g. $\left< x|(i | x\right>)$, but the latter is still real and positive.

Title: inner product (real or imaginary?)


问:
Why does the column space equal $\mathbb{R}^m$ (where $m$ is the number of rows) when the rank $= m$?
Let $m$ be the number of rows and $n$ be the number of columns. I read that if the rank of some matrix $A$ is equal to $m$, then we know that $Ax=b$ is solvable for every $b$. I believe this means that the column space spans all of $\mathbb{R}^m$.

But I have not been able to come up with a proof to show that rank$(A) =m\Rightarrow C(A)=\mathbb{R}^m$, where $C(A)$ denotes the column space of $A$. Can someone give me the intuition behind why this is be true? The rank is defined based off the rows, so how can it can say anything about the columns?

答:
We can interpret the matrix equation $$\begin{pmatrix}a_{11}&\cdots&a_{1n}\\\vdots&&\vdots\\a_{m1}&\cdots&a_{mn}\end{pmatrix}\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}0\\\vdots\\0\end{pmatrix}$$ as meaning that $\mathbf{x}$ is in the nullspace of $A$ or that it is perpendicular to the row space (it is 'killed off' by all the row vectors, viewed as column vectors). The two -- the nullspace of $A$ and the perpendicular of the row space -- are the same. Hence the nullity of $A$ is the same as the complement of the row rank of $A$, i.e., $$\textrm{nullity}(A)=n-\textrm{row-rank}(A).$$

By the rank-nullity formula, the nullity is the complement of the dimension of the image of $A$, which is the number of linearly independent columns, i.e., $$\textrm{nullity}(A)=n-\textrm{col-rank}(A).$$ Hence it follows that the column rank and row rank are the same.


答:
HINT:

Divide by 10 and add 100 on both sides

$$\dfrac{(x+10)^2}{1^2}+\left( \dfrac{y}{\sqrt {10}} \right) ^2
=100$$
Divide by $100 =10^2$
$$\dfrac{(x+10)^2}{10^2}+\left( \dfrac{y}{10\sqrt {10}} \right) ^2 =1.$$

$$ a=10; b=10\sqrt{10};c=\sqrt{a^2-b^2}=30;$$

Push the ellipse to the left by 10 units along $x-$axis.

Title: Ellipse equation equal to zero, how to define foci?

Body:
I will appreciate if you could help me with Equation $10x^2+y^2+200x=0$ . The answer key is giving foci $(-10, 30)$ and $(-10,-30).$ I am confused: ellipse equation must be equal to $1.$ Or it is in a form of a line? I tried to make it standard form:
$(10(x+10)^2)/(1/\sqrt {10})^2)+y^2=0$. But this zero...And where did $30$ and $-30$ come from?
P.S. Sorry for the formating, this is my first time using this web-site, I couldnt figure how to put square root sign.


Title: Calculate the LU-decomposition $PA=LU$ : How do we calulate $L$?

Body:
Calculate the LU-decomposition $PA=LU$ for the matrix $$A=\begin{pmatrix}3 & 1 & -3 & 2 \\ -2 & 1 & 0 & 0 \\ 2 & -2 & 4 & 1 \\ 0 & -1 & -1 & 3\end{pmatrix}$$ with column pivoting.

First we apply Gauss-elimination :
\begin{align*}\begin{pmatrix}3 & 1 & -3 & 2 \\ -2 & 1 & 0 & 0 \\ 2 & -2 & 4 & 1 \\ 0 & -1 & -1 & 3\end{pmatrix} & \ \overset{R_2:R_2+\frac{2}{3}\cdot R_1}{\longrightarrow} \ \begin{pmatrix}3 & 1 & -3 & 2 \\ 0 & \frac{5}{3} & -2 & \frac{4}{3} \\ 2 & -2 & 4 & 1 \\ 0 & -1 & -1 & 3\end{pmatrix}  \overset{R_3:R_3-\frac{2}{3}\cdot R_1}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2 \\ 0 & \frac{5}{3} & -2 & \frac{4}{3} \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & -1 & -1 & 3\end{pmatrix}  \\ & \overset{R_2\leftrightarrow R_3}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & \frac{5}{3} & -2 & \frac{4}{3} \\ 0 & -1 & -1 & 3\end{pmatrix} \overset{R_3:R_3+\frac{5}{8}\cdot R_2}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & 0 & \frac{7}{4} & \frac{9}{8}\\ 0 & -1 & -1 & 3\end{pmatrix} \\ & \overset{R_4:R_4-\frac{3}{8}\cdot R_2}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & 0 & \frac{7}{4} & \frac{9}{8}\\ 0 & 0 & -\frac{13}{4} & \frac{25}{8}\end{pmatrix}  \overset{R_3 \leftrightarrow R_4}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & 0 & -\frac{13}{4} & \frac{25}{8} \\ 0 & 0 & \frac{7}{4} & \frac{9}{8} \end{pmatrix} \\ & \overset{R_4 : R_4+\frac{7}{13}\cdot R_3}{\longrightarrow} \begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & 0 & -\frac{13}{4} & \frac{25}{8} \\ 0 & 0 & 0 & \frac{73}{26} \end{pmatrix} \end{align*}
After the Gauss elimination we get \begin{equation*}U=\begin{pmatrix}3 & 1 & -3 & 2  \\ 0 & -\frac{8}{3} & 6 & -\frac{1}{3} \\ 0 & 0 & -\frac{13}{4} & \frac{25}{8} \\ 0 & 0 & 0 & \frac{73}{26} \end{pmatrix}\end{equation*}
From each step at the  Gauss eliminaation we get the matrices \begin{equation*}G_1=\begin{pmatrix}1 & 0 & 0 &0 \\ -\frac{2}{3} & 1 & 0 & 0 \\ \frac{2}{3} & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \ , \ G_2=\begin{pmatrix}1 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 & -\frac{5}{8} & 1 & 0 \\ 0 &  \frac{3}{8} & 0 & 1\end{pmatrix} \ \text{ and } \ G_3=\begin{pmatrix}1 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 &  0 & -\frac{7}{13} & 1\end{pmatrix}\end{equation*}
We have the permutation matrices \begin{equation*}P_0=\begin{pmatrix}1 & 0 & 0 & 0   \\ 0 & 0 & 1 & 0  \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \ \text{ und } \ P_1=\begin{pmatrix}1 & 0 & 0 & 0   \\ 0 & 1 & 0 & 0  \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\end{equation*} We have that\begin{equation*}P=P_1\cdot P_0=\begin{pmatrix}1 & 0 & 0 & 0   \\ 0 & 0 & 1 & 0  \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{pmatrix}\end{equation*}
Now I am confused about how to calculate the matrix $L$.

Is it $L= P\cdot (G_3\cdot P_1\cdot G_2\cdot P_0\cdot G_1)^{-1}$ or $L=  (G_3\cdot P_1\cdot G_2\cdot P_0\cdot G_1)^{-1}$ or somehow else?

Body:
With an algebraic calculator, we have

A1:rowop(A,2,1,-2/3)

A2:rowop(A1,3,1,2/3)

A3:rowop(A2,3,2,-8/5)

A4:rowop(A3,4,2,-3/5)

A5:rowop(A4,4,3,-11/14)

That gives the matrices:

$G_1=\begin{pmatrix}1 & 0 & 0 & 0\\2/3 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{pmatrix}$

$G_2=\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\-2/3 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{pmatrix}$

$G_3=\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 8/5 & 1 & 0\\0 & 0 & 0 & 1\end{pmatrix}$

$G_4=\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 3/5 & 0 & 1\end{pmatrix}$

$G_5=\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 11/14 & 1\end{pmatrix}$

We have $G_5 G_4 G_3 G_2 G_1 A = U$

So $L=(G_5 G_4 G_3 G_2 G_1)^{-1}$


答:
Hint: Any inner product $\langle {-} | {-} \rangle$ on a complex vector space satisfies $\langle \lambda x | y \rangle = \lambda^* \langle x | y \rangle$ for all $\lambda \in \mathbb{C}$.

You're right in saying that $\langle x|x \rangle$ is always real when the field is defined over the real numbers: in general, $\langle x|y \rangle = \overline{\langle y|x \rangle}$, so $\langle x|x \rangle = \overline{\langle x|x \rangle}$, so $\langle x|x \rangle$ is real. (It's also always positive.)

Title: inner product (real or imaginary?)

Body: I have a simple question. Is the inner product $\langle x | z\rangle$ always real or does this hold only for an inner product with itself $\langle x | x \rangle$?</p>


问:
Exponential Distribution - memoryless
A post office has 2 clerks. Alice enters the post office while 2 other customers,
Bob and Claire, are being served by the 2 clerks. She is next in line. Assume
that the time a clerk spends serving a customer has the Exponential(x)
distribution.

(a) What is the probability that Alice is the last of the 3 customers to be done
being served Hint: no integrals are needed.

(b) What is the expected total time that Alice needs to spend at the post
office?

I understand that no matter who leaves the office first, the remainer has the same expo(x) like Alice, but don't have intuitive thinking why the answer for the 1st is 1/2.

答:
I have not seen an answer for (b) so I will provide one here. So Alice will get served by the clerk that was serving the first person that leaves, i.e the person that takes the least amount of time between Bob and Claire. Denote $T_i=$ Take that a clerk spends seving costumer $i$. Notice $$L=\min(T_B,Tc) \implies L \sim Expo(\lambda+\lambda)$$

Now, since $$W=\text{Total time spent by Alice}=\min(T_B,Tc)+ T_A$$
we obtain:

$$E(W)=\frac{1}{2\lambda}+\frac{1}{\lambda}=\frac{3}{2\lambda}$$
Only needed to change the final answer since it was added up wrong first

Title: Exponential Distribution - memoryless


问:
Bounded subsets (supremum and infimum)
The problem is as follows:

Let $S$ and $T$ be nonempty, bounded subsets of the real numbers. Prove that if $T$ contains $S$, then $\inf T\leq\inf S\leq \sup S\leq\sup T.$

My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound. Help. Please and thank you.

答:
The first thing you need to know is that since the sets are nonempty and bounded, the Completeness of the Reals says that they have suprema and infima.

The main insight you need to have is that if $S\subseteq T$, then every upper (resp. lower) bound of $T$ is also an upper (resp. lower) bound of $S$. Then explain how this gives you the inequality involving the sups and infs.

Title: Bounded subsets (supremum and infimum)


问:
Inverse matrices in an equation
I know that $AX = b \iff X = A^{-1}b$, but how can I solve for X in this equation: $ACX = b$?

Is the answer $X = C^{-1}A^{-1}b$?

答:
You are correct!

If $A$ and $C$ are both invertible matrices then there is a unique solution to the equation $AC\hat{x} = \hat{b}$ for which we can solve.

\begin{align*}
AC\hat{x} &= \hat{b}\\
\implies C\hat{x} &= A^{-1}\hat{b}\\
\implies \hat{x} &= C^{-1}A^{-1}\hat{b}.
\end{align*}

Or you could think of the matrix $AC$ as a single matrix (the result of computing the matrix multiplication). Since $A$ and $C$ are invertible, then $AC$ is invertible.

\begin{align*}
(AC)\hat{x} &= \hat{b}\\
\implies \hat{x} &= (AC)^{-1}\hat{b}\\
\implies \hat{x} &= C^{-1}A^{-1}\hat{b}.
\end{align*}

The credit really goes to Don Thousand for his comment that preceded my answer, I only chose to elaborate as an answer so the question can close.

Title: Inverse matrices in an equation


问:
Distribution in $L^p$ space with continuity
Given $f \in L^p(\mathbb{R})$, $T_f(\phi) = \int f(x)\phi(x)dx, \phi \in  L^p(\mathbb{R})$, I want to show that $T_f$ is continuous.

Since both $f,\phi \in L^p(\mathbb{R})$, they are continuous and satisfying the delta-epsilon definition. By definition, I need to show
$$
\forall \epsilon>0, \exists \delta>0,s.t. |x-y|<\delta, \Rightarrow |\int f(x)\phi(x)dx - \int f(y)\phi(y)dy|<\epsilon.
$$
I am confused on how to separate the integral and functions to use the continuity of $f, \phi$.

Any hint/help is appreciated.

答:
I am not too sure about your knowledge of functional analysis. So i will try to use the epsilon delta method. Notice, you want to show that $T_f:L^p(\mathbb{R})\to\mathbb{R}$ is continous. Namely the domain you are working with is $L^p(\mathbb{R})$ not $\mathbb{R}$ (Note that your two integrals are the same with variable change $y=x$).

So to show continuity of $T_f$, you need to show
$$\forall \epsilon>0, \exists \delta>0,s.t. ||\phi_1-\phi_2||_{L^p}<\delta, \Rightarrow \left|\int f(x)\phi_1(x)dx - \int f(x)\phi_2(x)dy\right |<\epsilon.$$

Now see that $|\int f(x)\phi_1(x)dx - \int f(x)\phi_2(x)dx| \leq \int f(x)|\phi_1(x)-\phi_2(x)|dx$, by triangle inequality.

I will leave the rest for you to discover. Else, I can help out a bit more.


问:
To show that orthogonal complement of a set A is closed.
To show that orthogonal complement of a set A is closed.

My try: I first show that the inner product is a continuous map. Let $X$ be an inner product space. For all $x_1,x_2,y_1,y_2 \in X$, by Cauchy-Schwarz inequality we get,
$$|\langle x_1,y_1\rangle - \langle x_2,y_2\rangle| = |\langle x_1- x_2,y_1\rangle + \langle x_2, y_1-y_2\rangle| $$
$$\leq \|x_1- x_2\|\cdot\|y_1\| +\|x_2\|\cdot\| y_1-y_2\|$$

This implies continuity of inner products.

Let $A \subset X$ and $y \in A^\perp$. To show that $ A^\perp$ is closed, we have to show that if $(y_n)$ is convergent sequence in $ A^\perp$, then the limit $y$ also belong to $ A^\perp$.

Let $x \in A$, then using that inner product is a continuous map,
$$\langle x,y\rangle = \langle x, \lim_{n\to \infty} (y_n)\rangle = \lim_{n\to \infty} \langle x, y_n\rangle = 0.$$

Since $\langle x, y_n\rangle = 0$ for all $x \in A$ and $y_n \in A^\perp$. Hence $y \in A^\perp$.

Is the approach\the proof correct??

Thank You!!

答:
Let's denote the orthogonal complement of $A$ by $A^{\perp}$. Also, we denote the scalar product $\langle \cdot, y \rangle : V \rightarrow \mathbb{F}$ as the function $\varphi_y$. So, all the orthogonal elements corresponding to $y$ is nothing but $Ker \; \varphi_y$.
Hence,

$$
A^{\perp} = \cap_{y \in A} \varphi_y
$$

Since $Ker \; \varphi_y$ is closed subset of $V$ (why?) and arbitrary intersection of closed set is again a closed set, we have the result.


问:
Bounds on the elements of the Leontief inverse
Let $X$ be an $n\times n$ real matrix with coefficients $0\leq X_{ij}\leq 1$ such that $\sum_{j=1}^n X_{ij} \leq \bar{X}_i$, with $\bar{X}_i<1$ for all $i$. Consider the matrix
$$
Y=(I-X)^{-1},
$$
where $I$ is the identity matrix. I would like to find bounds $\underline{Y}$ and $\overline{Y}$ such that $\underline{Y}\leq Y_{ij} \leq \overline{Y}$ for all $i,j$.

What I could find so far: $I-X$ is diagonally dominant so its eigenvalues are positive. In addition, its off-diagonal terms are nonpositive. It is therefore an $M$-matrix and the elements of its inverse are nonnegative. So one lower bound for $Y_{ij}$ is $\underline{Y}=0$, although there might be a tighter one.

答:
As can be seen by considering the induced $\infty$-norm, the bound on the row-sums is enough to guarantee that the Neumann series
$$
(I - X)^{-1} = \sum_{k=0}^\infty X^k
$$
is convergent. Because $X^k$ is non-negative for all $k$, we have $Y_{ij} \geq X^{k}_{ij}$ for all $k$. In particular, if $X$ is irreducible, then we can guarantee that
$$
Y_{ij} \geq \underline{Y} := r^{n},
$$
where $r = \min \{X_{ij} : X_{ij} \neq 0\}$ (the smallest non-zero entry of $X$). On the other hand, because the induced $\infty$-norm is multiplicative, we have
$$
\left\| \sum_{k=0}^\infty X^k \right\|_\infty \leq \sum_{k=0}^\infty \|X\|_\infty^k = \frac{1}{1 - \|X\|_\infty}.
$$
In other words, we have
$$
Y_{ij} \leq \overline{Y} := (1 - \|X\|_\infty)^{-1} \leq \max_{i=1,\dots,n} (1 - \bar X_i)^{-1}.
$$


问:
How is this 20square root3 here?

The question that we are given a polygon on left on diagram with each side 20m. Then in the right side I made a triangle of ABC. I drew head and tail of the vector on my own. Angle I got between vector AB and BC is 60. Therefore I drew components of BC . Then the triangle (green colour) I considered as my triangle whose length AC I have to find.

My teacher got the answer 20$\sqrt{3}$. So what I did is that $20^2 + 10^2 + 10$$\sqrt{3}$ As sum of Base and perpendicular = $ H^2$

My sir answer is 20$\sqrt{3}$ but I am not getting the same answer.Where am I wrong ?

I have tagged vectors since I want to confirm whether the way I drew head and tail of vector and components is right or not.

答:
$$\angle A=30^\circ\implies BD=10\implies
AD= \sqrt{20^2-10^2}=10\sqrt{3}\implies AC=20\sqrt{3}$$

Title: How is this 20square root3 here?


问:
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

I tried many method, such as AM-GM, but all of them failed.

Thank you.

答:
You can rewrite the equation as $$(x+\frac12)^2+(y+1)^2+(z+\frac32)^2=\frac{27}4$$

As others have mentioned, this is the equation of a sphere. When $x=y=z=0$ the left hand side of this equation is $\dfrac72$, and this is too low. Increasing $x$ or $y$ or $z$ increases the value, and to keep $x+y+z$ as small as possible you need to increase the value as fast as possible. Thinking about the square function, the gradient gets steeper as you move away from $0$, so you want to be as far out as possible. $z+\dfrac32$ starts larger than the other two squares, so you increase fastest by setting $x=y=0$ and $z$ as large as possible.

This comes out as $$\frac14+1+(z+\frac32)^2=\frac{27}4$$

$$z+\frac32=\sqrt{\frac{11}2}$$


问:
The composition of a measurable function with a continuous function
I know that the composition of a continuous function with a measurable function is measurable, however the composition of a measurable function with a continuous function is not necessarily measurable (I suppose). What example can you give? Thank you!

Detail. I mean function f is measurable if set A = $\{x: f(x) < C \}$ for any $C \in \mathbb{R}$ is measurable

答: If you mean that measurable function $f:\mathbb{R}\to\mathbb{R}$ is a function such that $f^{-1}(B)$ is a Borel set for any Borel set $B\subseteq \mathbb{R}$, then if $g$ is continuous, $g$ is measurable, and so both $f\circ g$ and $g\circ f$ are measurable.


答:
We know that $S \subseteq T$, let's see what happens if $\inf S < \inf T$. If this were true, then $\inf T - \inf S > 0$, so there exists an $s \in S$ such that $\inf S \leq s < \inf T$ (to see this: define $\varepsilon = \inf T - \inf S > 0$ which allows us to find some $s \in S$ such that $\inf S \leq s < \inf S + \varepsilon$ and finally conclude $\inf S \leq s < \inf S + \inf T - \inf S = \inf T$).

As we know that $\inf S \leq s < \inf T$, we can state that $s$ is an strict lower bound for $T$, i.e., for all $t \in T$, $s < t$, which is a contradiction since $S \subseteq T$ and thus $s \in T$ ($\implies s<s$). We conclude that $\inf S \geq \inf T$.

By definition we know $\inf S \leq \sup S$ and using the above argument for the supremum we finally have $\inf T \leq \inf S \leq \sup S \leq \sup T$.

I hope it helps :)

Title: Bounded subsets (supremum and infimum)

Body:
The problem is as follows:

Let $S$ and $T$ be nonempty, bounded subsets of the real numbers. Prove that if $T$ contains $S$, then $\inf T\leq\inf S\leq \sup S\leq\sup T.$

My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound. Help. Please and thank you.


答:
The composition of a Lebesgue measurable function with a continuous function need not be Lebesgue measurable.

Take our measurable set to be the Cantor set $C$, and let $h:[0,1]\rightarrow [0,1]$ be the Cantor-Lebesgue function. This is continuous and $h(C)=[0,1].$ Let $g:[0,1]\rightarrow [0,2]$ be defined via $g(x)=x+h(x),$ which is both continuous and injective; hence, its inverse is continuous.  Since $g(C)$ has positive measure there is a non-measurable subset $D\subset g(C)$.  Call $E=g^{-1}(D).$ This must have measure zero, as it's a subset of $C$. If you define $f(x)=\chi_E(x)$, then this is a measurable function, as $E$ is measurable. However, $f\circ g^{-1}:[0,2]\rightarrow [0,1]$ is non-measurable. Indeed, it equals $\chi_D$, and you can check that $(1/2,1]$ is open in $[0,1]$, but $(f\circ g^{-1})^{-1} ((1/2,1])=D,$ which is not measurable.

Title: The composition of a measurable function with a continuous function

Body:
I know that the composition of a continuous function with a measurable function is measurable, however the composition of a measurable function with a continuous function is not necessarily measurable (I suppose). What example can you give? Thank you!

Detail. I mean function f is measurable if set A = $\{x: f(x) < C \}$ for any $C \in \mathbb{R}$ is measurable


问:
Prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$
I am trying to prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$. So far I know that $\text{dist}\left({x,y}\right) \leq d_1\left({x,y}\right)$ where $\text{dist}\left({x,y}\right)$ is the standard distance function in k-dimensional Euclidian space. I also know that $d_1\left({x,y}\right) \leq \text{dist}\left({x,y}\right) \leq \sqrt{k}\cdot d_1\left({x,y}\right)$. I have already proven that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = 0 \iff x=y$, and that $d_1\left({x,y}\right) = d_1\left({x,y}\right)$, but I am stuck on the triangle inequality property of this particular metric. Any hints or suggestions would be greatly appreciated. </p>

答:
Three properties need to be checked:

*(1) $d(x,x)=0$

*(2) $d(x,y)=d(y,x)$

*(3) $d(x,z)\le d(x,y)+d(y,z)$

Usually the first two are easy to see.

So we check (3) suppose that when choose n,then the left side gets max,
$$ ∣∣x_n−z_n∣∣ =||(x_n-y_n) + (y_n-z_n)|| \le ||x_n-y_n|| +||y_n-z_n||$$
$$\le max\{∣∣x_j−y_j∣∣:j=1,2,...,k\} + max\{∣∣y_j−z_j∣∣:j=1,2,...,k\} $$

so $d$ is metric on $R^k$


问:
Intuition about coreflective subcategories.
When we have an inclusion of a full subcategory $I:B\hookrightarrow C$ which has a left adjoint we call it a reflector and say we have a reflective subcategory. An example is the inclusion of the category of complete metric spaces into metric spaces, where the reflector is given by the completion, another one is the inclusion of compact Hausdorff spaces in topological spaces, where the reflector is the Stone-Cech compactification. These examples build an intuition of a reflective subcategory as one that contains 'nice' objects , which can be built canonically for each object in the ambient category. Moreover such construction is idempotent.

Now, the dual notion of coreflective subcategory is that of an inclusion of a full subcategory $I:B\hookrightarrow C$ which has a right adjoint we call it a coreflector. And honestly, I have not found much more explanation.

How can we intuitively deal with this concept? Does it somehow express a concept dual to completion?

答:
Your intuitive description of reflective subcategories doesn't distinguish left adjoints from right adjoints so you could apply it equally well to either. Anyway, I think the best way to get a handle on these sorts of things is through a bunch of examples.

Examples of reflective subcategories:

The subcategory of abelian groups in the category of groups, with reflector the abelianization $G \mapsto G/[G, G]$. Note that it doesn't really make sense to think of this as a "completion." This is one of my preferred "prototypical" examples.

The subcategory of torsion-free abelian groups in the category of abelian groups, with reflector the torsion-free quotient $A \mapsto A/A_{tors}$.

The subcategory of fields in the category of integral domains and injections, with reflector the fraction field $D \mapsto \text{Frac}(D)$.

Let $f : R \to S$ be an epimorphism of commutative rings (for example, a quotient or a localization). Then the category of $S$-modules is a reflective subcategory of the category of $R$-modules, with reflector the extension of scalars $M \mapsto M \otimes_R S$ (for example, a localization). Taking $R = \mathbb{Z}, S = \mathbb{Q}$ produces an example similar to the second one: the subcategory of torsion-free divisible abelian groups in the category of abelian groups, with reflector $A \mapsto A \otimes \mathbb{Q}$ (which has kernel $A_{tors}$).

Examples of coreflective subcategories:

The subcategory of torsion abelian groups in the category of abelian groups, with coreflector the torsion subgroup $A \mapsto A_{tors}$. This example has the nice feature of looking quite dual to the $A \mapsto A/A_{tors}$ example; I think it's worth chewing on in detail.

More generally, let $R$ be a commutative ring and let $S$ be a multiplicative subset of it. Call an element $m \in M$ of an $R$-module $S$-torsion if it is annihilated by some element of $S$; equivalently, if it lies in the kernel of the localization map $M \mapsto S^{-1} M \cong M \otimes_R S^{-1} R$. Then the subcategory of $S$-torsion modules is coreflective in the category of $R$-modules, with coreflector the $S$-torsion submodule.

The subcategory of divisible abelian groups in the category of abelian groups, with coreflector the divisible subgroup $A \mapsto \{ a \in A : \forall n \in \mathbb{N}, \exists b \in A : a = nb \}$.

The subcategory of groups in the category of monoids, with coreflector the group of units $M \mapsto M^{\times}$.

The subcategory of connected topological groups in the category of topological groups, with coreflector taking the connected component of the identity $G \mapsto G_0$. If we restrict our attention to compact Hausdorff abelian groups then this example is related to the example of torsion-free groups by Pontryagin duality: the opposite of the category of abelian groups is equivalent to the category of compact Hausdorff abelian groups and the subcategory of torsion-free abelian groups (which are equivalently the filtered colimits of copies of $\mathbb{Z}^n$) is sent to the subcategory of connected compact Hausdorff abelian groups (which are equivalently the cofiltered limits of copies of $T^n$).

These algebraic examples work loosely as follows. A reflective subcategory consists of "nice" objects such that every object has a maximal "nice quotient" (with reflector given by this quotient), and a coreflective subcategory consists of "nice" objects such that every object has a maximal "nice subobject" (with coreflector given by this subobject). Probably there are examples a bit stranger than this (for example one could take the opposite category of metric spaces or topological spaces) but I think this is a decent place to start.


答:
You have the row operations right, but you store in $G_1$ the opposite of the row coefficients, that's why you end up with everything inverted ($G_1^{-1}$, etc.). But as you will see, it's the correct approach.

Let's start from $A$ and let's forget about row swaps for a moment.

$$A=\begin{pmatrix}3 & 1 & -3 & 2 \\ -2 & 1 & 0 & 0 \\ 2 & -2 & 4 & 1 \\ 0 & -1 & -1 & 3\end{pmatrix}$$

The row ops will be $R_2\leftarrow R_2+\frac23R_1$, $R_3\leftarrow R_3-\frac23R_1$, $R_4\leftarrow R_4+0R_1$ (the last is useless but let's keep it for completeness).

These row operations are exactly equivalent to a left matrix product $A\leftarrow M_1A$ with

$$M_1=\begin{pmatrix}1 & 0 & 0 & 0 \\ \frac23 & 1 & 0 & 0 \\ -\frac23 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$

See how the signs in the first column are reversed compared to $G_1$. And notice that the inverse of $M_1$ is obtained by reversing the signs of the coefficients of the first column, below the diagonal. That is, $M_1^{-1}=G_1$.

So, when computing $M_1A$, the first column is annihilated below the pivot:

$$M_1A=\begin{pmatrix}3 & 1 & -3 & 2 \\ 0 & \frac53 & -2 & \frac43 \\ 0 & -\frac83 & 6 & -\frac13 \\ 0 & -1 & -1 & 3\end{pmatrix}$$

Now, let's do the second column: $R_3\leftarrow R_3+\frac85 R_2$, $R_4\leftarrow R_4+\frac35R_1$, and the matrix $M_2$ that describes these row operations is

$$M_2=\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & \frac85 & 1 & 0 \\ 0 & \frac35 & 0 & 1\end{pmatrix}$$

And

$$M_2M_1A=\begin{pmatrix}3 & 1 & -3 & 2 \\ 0 & \frac53 & -2 & \frac43 \\ 0 & 0 & \frac{14}5 & \frac95 \\ 0 & 0 & -\frac{11}5 & \frac{19}5\end{pmatrix}$$

Same remark as above about the inverse of $M_2$: change the signs of the coefficients. However, since we didn't swap rows, we don't have the same coefficients up to sign as in your $G_2$, now.

Finally, we need the row operation $R_4\leftarrow R_4+\frac{11}{14}R_3$, and the matrix:

$$M_3=\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & \frac{11}{14} & 1\end{pmatrix}$$

$$M_3M_2M_1A=\begin{pmatrix}3 & 1 & -3 & 2 \\ 0 & \frac53 & -2 & \frac43 \\ 0 & 0 & \frac{14}5 & \frac95 \\ 0 & 0 & 0 & \frac{73}{14}\end{pmatrix}$$

So far, we have the decomposition:

$$M_3M_2M_1A=U$$

Now, a product of lower triangular matrices with ones on the diagonal has the same form, so $M_3M_2M_1$ is lower triangular, and its inverse is also lower triangular, so we have the decomposition:

$$A=(M_3M_2M_1)^{-1}U=M_1^{-1}M_2^{-1}M_3^{-1}U=G_1G_2G_3U$$

(here $G_2$ and $G_3$, while similar to what you did, are different since we didn't swap rows, but I keep your notation, as I think it makes things clearer)

That is, $A=LU$ with $L=G_1G_2G_3$. Notice how it's similar to your case, except the permutations matrices $P_0,P_1$ do not appear. Theoretically these permutation matrices are only required when you get a zero pivot. For numerical computations it's better to always pick the largest pivot to reduce numerical errors.

But we are not ready to permute rows yet.

This product $G_1G_2G_3$ has an interesting property, that I'll show on an example. Let:

$$T_1=\begin{pmatrix}1 & 0 & 0 & 0 \\ u_2 & 1 & 0 & 0 \\ u_3 & 0 & 1 & 0 \\ u_4 & 0 & 0 & 1\end{pmatrix}$$

$$T_2=\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & v_3 & 1 & 0 \\ 0 & v_4 & 0 & 1\end{pmatrix}$$

Then

$$T_2T_1=\begin{pmatrix}1 & 0 & 0 & 0 \\ u_2 & 1 & 0 & 0 \\ u_3+v_3u_2 & v_3 & 1 & 0 \\ u_4+v_4u_2 & v_4 & 0 & 1\end{pmatrix}$$

But

$$T_1T_2=\begin{pmatrix}1 & 0 & 0 & 0 \\ u_2 & 1 & 0 & 0 \\ u_3 & v_3 & 1 & 0 \\ u_4 & v_4 & 0 & 1\end{pmatrix}$$

See? The product $T_2T_1$ is relatively difficult to compute, whereas the product $T_1T_2$ is trivial: just put the coefficients together! It's a general result when you multiply these transformation matrices: if the product starts with the leftmost columns and goes to the right, you just put the coefficient together. Otherwise you have to compute the matrix product.

Now remember that $L=G_1G_2G_3$: that is, we just need to put the coefficients together to write $L$. That is,

$$L=\begin{pmatrix}1 & 0 & 0 & 0 \\
-\frac23 & 1 & 0 & 0 \\
\frac23 & -\frac85 & 1 & 0 \\
0 & -\frac35 & -\frac{11}{14} & 1\end{pmatrix}$$

So, if no row swap is involved, we have an easy way to compute the $LU$ decomposition: write the modified $A$ on one side, the $L$ coeficients on the other, and you can build $L$ and $U$ during the computation. We don't even need to write them apart: when $A$ is transformed to $U$, only the upper triangular part is useful, so the lower triangular array can be used to store $L$. This way, you don't need extra space to compute $L$ and $U$: just overwrite $A$. It's how $LU$ is implemented in practice (with more subtleties if you want to be cache-friendly, but it's another story).

Hence, we compute the $LU$ decomposition as:

$$\begin{pmatrix}3 & 1 & -3 & 2 \\
\color{red}{-\frac23} & \frac53 & -2 & \frac43 \\
\color{red}{\frac23} & \color{red}{-\frac85} & \frac{14}5 & \frac95 \\
\color{red}{0} & \color{red}{-\frac35} & \color{red}{-\frac{11}{14}} & \frac{73}{14}\end{pmatrix}$$

In black: the $U$ part. In red: the $L$ part. The diagonal of $L$ is missing, but we know it's all made of ones.

Now, what happens it swaps intervene? Surely, it must be complicated, with the storage in $A$ not working anymore.

Let's see.

We are doing the same process, but we will choose each time the largest pivot. For the first column, it's already in place, so the first permutation matrix is $P_1=I$, and we compute $M_1P_1A$ as previously. However, we will now store $L$ inside the same matrix. Since this completed matrix is not really equal to the product, I write the formula in brackets to remember we are doing something more.

$$[M_1P_1A]=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{-\frac23} & \frac53 & -2 & \frac43 \\
\color{red}{\frac23} & -\frac83 & 6 & -\frac13 \\
\color{red}{0} & -1 & -1 & 3
\end{pmatrix}$$

In red, the $L$ part, in black, the $A$ part that it transformed to $U$. Remember that the $L$ part has the coefficients of $M_1^{-1}$, that is the coefficients of $G_1$ (signs swapped).

Now the second column: we would like to take $-\frac83$ as pivot, so we need to swap the rows in the black part. How do we keep track of this in the $L$ part? Simple. Swap the $L$ part too!

So, first the swap:

$$P_2=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}$$

$$[P_2M_1P_1A]=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{\frac23} & -\frac83 & 6 & -\frac13 \\
\color{red}{-\frac23} & \frac53 & -2 & \frac43 \\
\color{red}{0} & -1 & -1 & 3
\end{pmatrix}$$

Note how the black bart is identical to what you actually did. Now the pivot, with

$$M_2=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & \frac58 & 1 & 0\\
0 & -\frac38 & 0 & 1
\end{pmatrix}$$

Now apply this transformation. Wait, to the $A$ part, or also to the $L$ part? Now it's only the $A$ part: we want to swap the rows in $L$ to keep track of the permutation, we don't want to change the values.

$$[M_2P_2M_1P_1A]=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{\frac23} & -\frac83 & 6 & -\frac13 \\
\color{red}{-\frac23} & \color{red}{-\frac58}  & \frac74 & \frac98 \\
\color{red}{0} & \color{red}{\frac38}  & -\frac{13}4 & \frac{25}8
\end{pmatrix}$$

Let's do it again with the fourth column. This time I don't write $P_3$ and $M_3$, but I write the two steps separately to make it clearer.

First swap rows $3$ and $4$:

$$[P_3M_2P_2M_1P_1A]=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{\frac23} & -\frac83 & 6 & -\frac13 \\
\color{red}{0} & \color{red}{\frac38}  & -\frac{13}4 & \frac{25}8 \\
\color{red}{-\frac23} & \color{red}{-\frac58}  & \frac74 & \frac98
\end{pmatrix}$$

Now $R_4\leftarrow R_4+\frac7{13}R_3$:

$$[P_3M_2P_2M_1P_1A]=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{\frac23} & -\frac83 & 6 & -\frac13 \\
\color{red}{0} & \color{red}{\frac38}  & -\frac{13}4 & \frac{25}8 \\
\color{red}{-\frac23} & \color{red}{-\frac58}  & \color{red}{-\frac{7}{13}} & \frac{73}{26}
\end{pmatrix}$$

And we are done. We can write down the matrices $L$ and $U$:

$$L=\begin{pmatrix}
1 & 0 & 0 & 0 \\
\color{red}{\frac23} & 1 & 0 & 0 \\
\color{red}{0} & \color{red}{\frac38}  & 1 & 0 \\
\color{red}{-\frac23} & \color{red}{-\frac58}  & \color{red}{-\frac{7}{13}} & 1
\end{pmatrix}$$

$$U=\begin{pmatrix}
3 & 1 & -3 & 2 \\
\color{red}{0} & -\frac83 & 6 & -\frac13 \\
\color{red}{0} & \color{red}{0}  & -\frac{13}4 & \frac{25}8 \\
\color{red}{0} & \color{red}{0}  & \color{red}{0} & \frac{73}{26}
\end{pmatrix}$$

And now, $PA=LU$, with $P=P_3P_2P_1$.

In practice, $L$ and $U$ are computed in place, in $A$, and the matrix $P$ is not stored, but you store enough information to know the permutation matrix $P$. Since there is a bijection between permutation matrices and permutations of integers, you can simply store the integer vector $P=[1,2,\dots,n]$, and swap elements in $P$ when you swap rows. It's not the only possible storage: if I remember correctly, the storage in LAPACK is a bit different to allow more efficient computations with $L$ and $U$.

For instance, to compute the determinant of $A$, you compute the product of diagonal elements of $U$, and multiply by $\sigma(\pi)$, the parity of the permutation $\pi$. Even though it's not difficult to compute, LAPACK's storage allows for a faster computation of this parity. See https://stackoverflow.com/questions/47315471/compute-determinant-from-lu-decomposition-in-lapack


问:
For a commutative exact sequence show that $V_1$ and $V_3$ are finite dimensional iff $V_2$ and $V_4$ also are.
For the following commutative exact sequence:
\begin{array}\\
 &V_1 & \stackrel{A_1}{\longrightarrow}  & V_2\\
 & \uparrow_{A_4} &&\downarrow _{A_2}\\
 &V_4 & \stackrel{A_3}{\longleftarrow} & V_3
\end{array}
Let $V_1$, $V_2$, $V_3$, $V_4$ be vector spaces and $A_1$, $A_2$, $A_3$, $A_4$ be homomorphism. Show that $V_1$ and $V_3$ are finite dimensional if and only if $V_2$ and $V_4$ also are.

As the diagram is exact and commutative, it can be said hat:
\begin{array}\\
\ker(A_1) = \operatorname{Im}(A_4) & \ker(A_2) = \operatorname{Im}(A_1) \\
\ker(A_3) = \operatorname{Im}(A_2) & \ker(A_4) = \operatorname{Im}(A_3)
\end{array}
Also, if $V_1$ and $V_3$ are finite-dimensional, then:
\begin{array}\\
\dim(V_1)&=\dim(\ker(A_1))+ \dim(\operatorname{Im}(A_1))\\
\dim(V_3)&=\dim(\ker(A_3))+ \dim(\operatorname{Im}(A_3))
\end{array}

I don't know what should I do next. I would really appreciate your answers.

答:
Hint

The short exact sequence $\;0\longrightarrow \ker A_2\longrightarrow V_2\longrightarrow \operatorname{Im}A_2\longrightarrow 0$ is split since we have vector spaces. Now $\ker A_2\simeq \operatorname{Im}A_1$, which is finite dimensional, and similarly $ \operatorname{Im}A_2\simeq\ker A_3$, which is also finite dimensional.

Can you take it from there?


问:
How can we show this convolution identity for the exponential of measures on a group?
Let $(G,\mathcal G)$ be a measurable group (think of a metric vector space, where the group operation is the addition, if you like) and $\mu,\nu$ be finite signed measures on $\mathcal M(G,\mathcal G)$.

Let $$\exp(\mu):=\sum_{k=0}^\infty\frac{\mu^{\ast k}}{k!},$$ where $\mu^{\ast k}$ denotes the $k$fold convolution.

I would like to show the identity $$\exp(\mu+\nu)=\exp(\mu)\ast\exp(\nu)\tag1.$$

I'm only able to show this under the assumption that $G$ is a normed $\mathbb R$-vector space, $\mathcal G=\mathcal B(G)$ and that $\mu$ and $\nu$ are tight$^1$, since it's easy to see that the characteristic functions of the left- and right-hand side of $(1)$ coincide and under the stated assumptions we know that the family $\{e^{{\rm i}\varphi}:\varphi\in G'\}$ is separating for the class of tight finite signed measures.

$^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called (uniformly) tight if for all $\varepsilon>0$, there is a compact $K\subseteq E$ with $\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon$. A finite signed measure $\mu$ is called tight if $\{\mu\}$ is tight.

答:
I assume that your group is commutative (otherwise I am not sure that the result is true). Recall that by definition of convolution $\mu * \nu$ is the unique measure such that for all mesurable function f :
$$ \int_{g \in G} f(g) ~ d(\mu * \nu)(x) = \iint_{g_1,g_2 \in G} f(g_1.g_2) ~ d\mu(g_1) \otimes d\nu(g_2).$$
Hence one has :
$$ \int_{g \in G} f(g) ~ dexp(\mu)(x) = \sum_{k=0}^{\infty} \frac{1}{k!} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ d\mu(g_1) \otimes \dots \otimes d\mu(g_k).$$

Let $f$ be a measurable function. We have to prove that :
$$ \int_{g \in G} f(g) ~ d(exp(\mu + \nu)(g) = \int_{g \in G} f(g) ~ d(exp(\mu)*exp(\nu)(g).$$

The RHS is equal to :
$$ \begin{eqnarray*} \int_{g \in G} f(g) ~ d(exp(\mu)*exp(\nu)(g)
&=& \iint_{x,y \in G} f(x.y) ~ dexp(\mu)(x) \otimes dexp(\nu)(y) \\
&=& \sum_{p=0}^{\infty} \sum_{q=0}^{\infty} \frac{1}{p!} \frac{1}{q!} \iiint_{x_1,\dots,x_p \in G} \iiint_{y_1,\dots,y_q \in G} f(x_1 \dots x_p.y_1 \dots y_q) ~  \\
& & \quad \quad \quad \quad  d\mu(x_1) \otimes \dots \otimes d\mu(x_p) \otimes d\mu(y_1) \otimes \dots \otimes d\mu(y_q) \\
\end{eqnarray*}$$

And the LHS is equal to :
$$ \begin{eqnarray*}
\int_{g \in G} f(g) ~ dexp(\mu + \nu)(g)
&=& \sum_{k=0}^{\infty} \frac{1}{k!} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k)
\end{eqnarray*}$$
Note that for every $k$, the product $d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k)$ expands to :
$$ d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k) = \sum_{I,J} \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j),$$
where $(I,J)$ runs over all partitions of $\{1,2,3,...,k\}$.
Hence the LHS is equal to :
$$ \begin{eqnarray*}
\int_{g \in G} f(g) ~ dexp(\mu + \nu)(g)
&=& \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{I,J} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j)
\end{eqnarray*}.$$
Now for every $k$ and every partition $(I,J)$, since we assumed that the group is commutative, the integral $\iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j)$ depends only on the cardinal of $I$ and $J$ and is equal to
$$\iiint_{x_1,\dots,x_p,y_1,\dots,y_q \in G} f(x_1 \dots x_p.x_1 \dots x_p) ~ d\mu(x_i) \otimes \dots d\mu(x_p) \otimes d\nu(x_j) \dots \otimes d\nu(x_q),$$
where $p = \mathrm{Card}(I)$ and $p = \mathrm{Card}(J)$. And since for every pairs of integers $(p,q)$ such that $p+q=k$, there are $\frac{k!}{p!q!}$ partitions $(I,J)$ such that $p = \mathrm{Card}(I)$ and $p = \mathrm{Card}(J)$, you can see that LHS is equal to RHS.


问:
What are some interpretations of Von Neumann's quote?
John Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. </p>

<p>Did he mean that with experience and practice, one obtains understanding? </p>

答:
Poincare once wrote that all of mathematics is one plus one equals two, and a lot of definitions.  Speaking as a physicist, I believe the most important thing to understand about mathematics is whether those definitions are useful in describing reality (i.e., observations). I don't believe mathematicians are similarly concerned, though certainly von Neumann himself is a most prominent counter-example.

Just to be provocative, I have found Newton's calculus to be sufficient for my needs and have never understood \delta -- \epsilon proofs.  There seems to be no place for "infinity" in physics. On the other hand, nonstandard analysis has proven quite useful to me.

ContentLicense: CC BY-SA 2.5


问:
How to prove that this map is not a quotient map
This question was aksed in my topology quiz and I was unable to solve it.

Let $X=\bigcup_{n\in \mathbb{N}} (\mathbb{R} \times \{n\})$ and let $Y = \bigcup_{n\in \mathbb{N}} \{(x,y)\in \mathbb{R}^2 : y=nx\}$. Suppose both $X$ and $Y$ have the subspace topology induced by usual topology on $\mathbb{R}^2$. Define $p:X \to Y$ by $p((x,n))=(x,nx)$ for each $(x,n)\in X$.
Show that $p$ is not a quotient map.

I have proved the map to be surjective but I am not able to get an idea on how to obtain acontradiction if the map is assumed to be quotient map.

Kindly help me.

答:
Notice that $p$ is almost a bijection: it sends all of the points $\langle 0,n\rangle$ to $\langle 0,0\rangle$, but everywhere else it’s bijective. Thus, the place to look for problems is around $\langle 0,0\rangle$. What if we can find a sequence $\langle y_n:n\in\Bbb Z^+\rangle$ in $Y$ that converges to the origin and has just one point on each line $y=nx$? Then $\{y_n:n\in\Bbb Z^+\}$ won’t be closed in $Y$, because it’s missing the limit point $\langle 0,0\rangle$, but its preimage under $p$ will have just one point on each line $\Bbb R\times\{n\}$ and will be closed. If $p$ were a quotient map, this would be impossible: $p^{-1}[C]$ would be closed in $X$ if and only if $C$ were closed in $Y$.

Since we want the sequence to have just one point on each line $\Bbb R\times\{n\}$, we must have $y_n=p(\langle x_n,n\rangle)$ for some $x_n\in\Bbb R$, so that $y_n=\langle x_n,nx_n\rangle$. If these points are to converge to $\langle 0,0\rangle$ in $Y$, clearly the points $x_n$ must converge to $0$ in $\Bbb R$, and they must do so fast enough that the points $nx_n$ also converge to $0$ in $\Bbb R$. An easy way to ensure this is to let $x_n=\frac1{n^2}$, so that $nx_n=\frac1n$. Then

the set $A=\left\{\left\langle\frac1{n^2},\frac1n\right\rangle:n\in\Bbb Z^+\right\}$ is not closed in $Y$, since it has $\langle 0,0\rangle$ as a limit point, but

the set $p^{-1}[A]=\left\{\left\langle\frac1{n^2},n\right\rangle:n\in\Bbb Z^+\right\}$ is closed in $X$.


问:
PGF of a Poisson process using differential equations - telephone answering service
Could really do with help on this particular problem. I am comfortable with finding PGFs for most process but have no idea where to start for this question which begins with difference equations. Thanks.

A telephone answering service receives calls whose frequency varies with time but independently of other calls perhaps with a daily pattern – more during the day than the night. The rate $\lambda(t)\geq 0$ becomes a function of the time $t.$ The probability that a call arrives in the small time interval $(t,t+\delta t)$ when $n$ calls have been received at time $t$ satisfies

$$p_{n+1}(t+\delta t) = p_{n-1}(t)(\lambda(t)\delta t+o(t\delta))+p_{n}(t)(1-\lambda(t)\delta t+o(\delta t))$$

with

$$p_o(t+\delta t) = (1-\lambda(t)\delta t+o(\delta t))p_o(t)$$

It is assumed that the probability of two or mote calls arriving in the time interval $(t,t+\delta t)$ is negligible. Find the set of differential-difference equations for $p_n(t)$ Hence show that the probability generating function for the process is given by $$G(s,t)= \exp\left[(s - 1)\int_0^t\lambda(u) \, du\right].$$

答:
To start with since we have been asked to generate differential difference equations, we should be looking at how derivate might come up here. We know by definition that
$$ \frac{dp_n}{dt} = lim_{\delta \rightarrow 0} \frac{p_n(t+\delta t)-p_n(t)}{\delta t}$$
Then substituting in what you know $p_n(t)$ and $p_n(t+\delta t)$ to be, we see that
$$ \frac{dp_n}{dt} = lim_{\delta \rightarrow 0} (p_{n-1}\lambda - p_n\lambda + O(\delta t))=p_{n-1}\lambda - p_n\lambda$$
And using the definition  for $p_0$,
$$ \frac{dp_0}{dt} = - p_0\lambda$$
This is our set of differential difference equations.
Then we have:
$$ G(s,t)=\sum\limits_{n=0}^{\infty} p_n(t)s^n$$
So that, assuming we can pass the derivative through the sum (which is true for $|S|<1$?):
$$
\frac{dG(s,t)}{dt} = \frac{d}{dt} \sum\limits_{n=0}^{\infty} p_n(t)s^n = \sum\limits_{n=0}^{\infty} \frac{dp_n(t)}{dt}s^n \\
= \sum\limits_{n=0}^{\infty} (p_{n-1}\lambda - p_n\lambda)s^n \\
$$
We then recognise the summand as being similar to that in the definition of G, so that we can write:
$$
\frac{dG(s,t)}{dt} = \lambda s G(s,t) - \lambda G(s,t) = \lambda (s-1) G(s,t)
$$
Then you can solve that (partial) differential equation in G using an integrating factor to give the desired result.


问:
Fixed Point Iteration Convergence
We wish to solve the equation $f_a(x) = 0$, where $a ∈ R$ is a parameter and $f_a(x) = (1−a)x+ax^3$. We will study the fixed point iteration $x^{k+1} = φ(x^k)$, where the function $φ(x) = ax(1 − x^2)$.

(i) Find the positive values of $a$ such that the fixed point iteration converges to the root $α_1 = 0$ for any initial guess $x^0 ∈ [0, 1]$.

(ii) Find a condition on $a$ under which a second zero $α_2 > 0$ exists in the interval [0, 1] and determine for which $a$ the fixed point iteration can approximate $α_2$.

(iii) For what value of $a$ can the fixed point iteration approximate $α_2$ with second order convergence?

I tried taking the limit for part (i) using the proof of Lemma: Let $x^k$ be a fixed point iteration. Suppose that $x^k \to \alpha$ as $k \to \infty$, and that $\phi$ is continuous in the neighbourhood of $\alpha$. Then $\alpha=\phi(\alpha)$.

Where the proof stated $\alpha = \lim_{k \to \infty} x^{k+1} = \lim_{k \to \infty} \phi(x^{k}) = \phi(\lim_{k \to \infty} x^{k}) = \phi(\alpha)$. I keep getting $a=0$ which is evidently wrong. Not sure where to go from part (ii) and (iii) either.

答:
I believe the fixpoint iteration is with respect to $x$, not $a$. In order to find the fixpoints, you can do as you mentioned, which leads to the following equation:
$$\alpha = a\alpha(1-\alpha^2).$$
Here, either $\alpha = 0$ (which would be one possible fixpoint) or $\alpha \neq 0$, in which case we can divide the $\alpha$ away to obtain
$$1 = a (1-\alpha^2).$$
Solving this for $\alpha$ yields
$$\alpha = \pm \sqrt{1-\frac{1}{a}}$$
which should already give you a criterion to see when other fixpoints can appear.

Concerning the convergence of the iteration, I would use the Banach fixpoint theorem, i.e. show that the function has Lipschitz constant (that's just the maximum absolute value of the derivative) strictly smaller than 1 over some neatly-chosen domain, and that it sends this domain to itself; I tried this directly on $x \in [0,1]$ but this is not good enough, so you might need to look at the maximum value of $x_1 \in \phi([0,1]) = [0, m]$, and work with $[0,m]$ instead, since $m$ should be smaller than 1. It may also be that there is a better way to do this, depending on what you have seen in your lecture.

Title: Fixed Point Iteration Convergence


问:
Serre's vanishing theorem and global generation.
Given a short exact sequence of coherent sheaves on a projective variety if we tensor the sequence with $\mathcal{O}(n)$ with a high enough $n$, everything becomes globally generated and by Serre's vanishing theorem taking the global sections becomes exact. I wonder whether it is possible to prove this without Serre's vanishing theorem. Serre's theorem applies for the proper varieties, I wonder whether the fact mentioned above remains true for let's say if we remove a high codimension subvariety from a projective variety? Like $\mathbb{P}^2\setminus \text{pt}$.

答: Let $L$ be a line through $p\in\mathbb{P}^2$ and let $X=\mathbb{P}^2-\{p\}$ and $M=L-\{p\}$. You have an exact sequence of coherent sheaves, $0\to O_X(-1)\to O_X\to O_M\to 0$. Easy to see that $H^0(O_X(n))\to H^0(O_M(n))$ is not surjective for any $n$, since $H^0(O_M(n))$ is not finite dimensional over the base field.


问:
Prove $(S,\cdot)$ is a group.
Let $G$ be a group and $H \leq \text{Aut}(G)$. On Cartesian product $S=G \times H$ define multiplication: $$ \text{For} \,\, a,b \in G \,\, \text{and} \,\, \alpha,\beta\in \text{Aut}(G) \,\, \text{let} \,\, (a,\alpha)\cdot(b,\beta)=(a\alpha^{-1}(b),\alpha \beta).$$

Prove $(S,\cdot)$ is a group.

In my attempt I wasn't sure if I was taking the proper approach so I left out the identity.

$\textbf{Proof:}$ Since $G$ is a group and $H \leq \text{Aut}(G)$ is a group, then $\forall a,b \in G, \,\,\, a \cdot b \in G$ by closure. So, $\alpha, \beta \in H \implies \alpha, \beta \in \text{Aut}(G)$. Thus, $\alpha \cdot \beta \in H,\text{Aut}(G)$ by closure. So, $(a,\alpha) \cdot (b,\beta)=(a\alpha^{-1}(b),\alpha \beta) \in S \,\, \forall (a, \alpha),(b,\beta) \in S$.

$\therefore S$ is closed under multiplication.

For associativity, we want to show $((a,\alpha) \cdot (b,\beta)) \cdot (d, \delta)=(a,\alpha) \cdot ((b,\beta)\cdot(d,\delta)) \,\,\, \forall (a,\alpha),(b,\beta),(d,\delta) \in S$.

$((a,\alpha) \cdot (b,\beta)) \cdot (d,\delta) = (a\alpha^{-1}(b),\alpha \beta) \cdot (d,\delta)=(a\alpha^{-1}(b)(\alpha \beta)^{-1}(d),\alpha \beta \delta)$

$(a,\alpha)\cdot ((b,\beta)\cdot (d,\delta))=(a\alpha^{-1}(b\beta^{-1}(d),\alpha \beta \delta)$ At this point I am not sure if I am on the right track and need to show these last two steps are equivalent.

Need to show there exists inverse and identity for $S$ which I said is true since it exists for groups $G$ and $H$.

答:
Well-definedness (closure): Given $a,b \in G$ and $\alpha,\beta \in H$, obviously $a\alpha^{-1}(b) \in G$ and $\alpha \circ \beta \in H$ since $H$ is a subgroup of $\operatorname{Aut}(G)$. Thus $(a,\alpha) \cdot (b,\beta) = (a\alpha^{-1}(b),\alpha \circ \beta) \in G \times H$.

Associativity: Given $a,b,c \in G$ and $\alpha,\beta,\gamma \in H$, $$(a\alpha^{-1}(b),\alpha \circ \beta) \cdot (c,\gamma) = (a\alpha^{-1}(b)(\alpha \circ \beta)^{-1}(c),(\alpha \circ \beta) \circ \gamma)$$ and, on the other hand $$(a,\alpha) \cdot (b\beta^{-1}(c),\beta \circ \gamma) = (a\alpha^{-1}(b\beta^{-1}(c)),\alpha \circ (\beta \circ \gamma)) = (a\alpha^{-1}(b)\alpha^{-1}(\beta^{-1}(c)),\alpha \circ (\beta \circ \gamma)),$$
so, they are the same.

Identity: If $1_G$ is the identity of $G$, just consider $(1_G,\operatorname{id}_G) \in G \times H$, and check that $$(a,\alpha) \cdot (1_G,\operatorname{id}_G) = (a,\alpha) = (1_G,\operatorname{id}_G) \cdot (a,\alpha).$$

Inverses: Given $a \in G$ and $\alpha \in H$, check that $$(a,\alpha) \cdot (\alpha(a^{-1}),\alpha^{-1}) = (\alpha(a^{-1}),\alpha^{-1}) \cdot (a,\alpha) = (1_G,\operatorname{id}_G).$$

Title: Prove $(S,\cdot)$ is a group.


问:
Why does $\operatorname{Proj} k[x, y]/(xy)$ consist of only two points?
In the book Introduction to Schemes by Ellingsrud and Ottem it's claimed that if $R = k[x, y]/(xy)$, then $\operatorname{Proj} R$ consists of just two points.

I assume that the authors here are working with an algebraically closed field $k$. The argument that the authors give is the following -

"$\operatorname{Proj} R$ is obtained by gluing $\operatorname{Spec} (R_{(x)})$ and $\operatorname{Spec} (R_{(y)})$ together. Now, $$R_{(x)} = k[x, y]_{(x)}/(xy) =  k[x, y]_{(x)}/(y) = k[x]_{(x)} =k$$ and the corresponding chart of $\operatorname{Proj} R$ is just $\operatorname{Spec} k$. Similarly, the other chart $\operatorname{Spec}(R_{(y)})$ also equals $\operatorname{Spec} k$. We have $R_{(xy)} = 0$, so the overlap is empty, and it ensures that $\operatorname{Proj} R$ consists of the two points."

So here (partly) what the authors are saying is that $D_+((x)) = \operatorname{Spec} (R_{(x)}) = \operatorname{Spec} k$ which is a singleton and similarly $D_+((y)) = \operatorname{Spec} (R_{(y)}) = \operatorname{Spec} k$. Also $\operatorname{Spec}(R_{(xy)}) = D_+((x\cdot y)) = D_+((x))\cap D_+((y))$ (I think this equality holds as it holds in the case of the usual Zariski topology for $\operatorname{Spec}$ but I am not fully sure) and this is what the authors are talking about when they say "overlap".

Now I have essentially two main questions.

How do we know that $\operatorname{Proj} R$ is obtained by gluing $\operatorname{Spec} (R_{(x)})$ and $\operatorname{Spec} (R_{(y)})$ together?

Why do the two equalities $R_{(x)} = k[x, y]_{(x)}/(xy) =  k[x, y]_{(x)}/(y)$ hold?

答:
In addition to the formality described in the previous answers, say we take any cone  $C \subset \mathbb{A}^n$, where cone means heuristically that if a point is in $C$ then so is the whole line through the origin and that point. More algebraically, we could say C is a closed subscheme defined by a homogeneous ideal. Then Proj C is just the space of lines in C.

So here k[x, y]/(xy) corresponds to the union of the x and y axes in the plane. What are the lines of this union? Just these two axes, the two points of the corresponding Proj.


问:
Let $g(x)=\sin(x)$ and $f(x)=x$,show that the function $fg$ defined by $(fg)(x)=f(x)g(x)$ is not uniformly continuous on $\mathbb R$.
Let $g(x)=\sin(x)$ and $f(x)=x$,show that the function $fg$ defined by $(fg)(x)=f(x)g(x)$ is not uniformly continuous on $\mathbb R$.

A function $f:D \to \mathbb R$ is not uniformly continuous on $D$ if and only if $$\exists (x_n),(y_n) \subset D: \lim_{n\to \infty} (x_n-y_n)=0 \;\;\text{But} \;\;\lim_{n\to \infty} (f(x_n)-f(y_n))\ne 0$$

Take $\left\{x_n\right\}=\left\{2 \pi n\right\},\left\{y_n\right\}=\left\{2\pi n +1/n\right\} \subset \mathbb R$,now $$\lim_{n\to \infty} (x_n-y_n)=\lim_{n\to \infty}(2 \pi n-2 \pi n-1/n)=0$$

But $$\lim_{n\to \infty} (f(x_n)g(x_n)-f(y_n)g(y_n))$$$$=\lim_{n\to \infty} (2\pi n \sin(2\pi n)-(2\pi n +1/n)\sin(2\pi n +1/n))$$
$$=-\lim_{n\to \infty}(2\pi n +1/n)\sin(1/n)=-2\pi \ne 0$$

And so the function is not uniformly continuous on $\mathbb R$.

I would kike to know if my proof is correct.

答:
The if and only if statement in your question is false. Uniform continuity implies Cauchy continuity but Cauchy continuity does not imply uniform continuity. Both of the definitions I quote below.

Uniform continuity according to wikipedia:

Given metric spaces $\left(X, d_{1}\right)$ and $\left(Y, d_{2}\right),$ a function $f: X \rightarrow Y$ is called uniformly continuous if for every real number $\varepsilon>0$ there exists real $\delta>0$ such that for every $x, y \in X$ with $d_{1}(x, y)<\delta,$ we have that $d_{2}(f(x), f(y))<\varepsilon$
If $\mathrm{X}$ and $\mathrm{Y}$ are subsets of the real line, $d_{1}$ and $d_{2}$ can be the standard one-dimensional Euclidean distance, yielding the definition: for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x, y \in X,|x-y|<\delta \Longrightarrow|f(x)-f(y)|<\varepsilon$

and Cauchy continuity:

For a function between metric spaces, uniform continuity implies Cauchy continuity (Fitzpatrick 2006). More specifically, let A be a subset of $\mathbf{R}^{n}$. If a function $f: A \rightarrow \mathbf{R}^{m}$ is uniformly continuous then for every pair of sequences $x_{n}$ and $y_{n}$ such that
$$
\lim _{n \rightarrow \infty}\left|x_{n}-y_{n}\right|=0
$$
we have
$$
\lim _{n \rightarrow \infty}\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|=0
$$

The negation of Cauchy continuity in your comment is false but correct in question statement with the exception of absolute values.

All in all, your proof would be correct if you said: suppose $fg$ is uniform continuous which implies Cauchy continuous which implies contradiction (essentially, not Cauchy cont. implies not uniform cont.) while justifying each step including your limits.


问:
Finding the range of a vector valued function
For a single valued function, I can infer if the function is monotone from its derivative.

For a vector valued function, is it possible to infer monotonicity from the directional derivative?

For example, define
$$
D=[1,2]\times[1,2],
$$
and
$$
f(x,y)=\left( \frac{2}{1/x+1/y},\sqrt{xy} \right).
$$
Is it possible to show that $f$ maps $D$ to $D$ from its gradient $\nabla f$?

The gradient is
$$
\nabla f = \begin{pmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{pmatrix},
$$
whence, for $(x,y)$ in $D$, the directional derivative
$$
\left(\nabla f(x,y)\right)\begin{pmatrix} x \\ y \end{pmatrix},
$$
is always positive and I would like to conclude that, on $D$,
$$
\text{$f$ is minimal at $(1,1)$},\\
\text{$f$ is maximal at $(2,2)$}.
$$

Is it the right way to proceed?

The graph of each component of $f$ looks like this

答:
I'm not quite sure what you mean by monotonicity for a function of several variables, let alone a vector-valued function. My approach would be to start by seeing where $f$ maps the boundary of $D$. Note that, by the symmetry, it covers the curve
$$\left\{\left(\frac{2x}{x+1},\sqrt x\right): 1\le x\le 2\right\} \cup \left\{\left(\frac{4x}{x+2},\sqrt{2x}\right): 1\le x\le 2\right\}$$
twice. The rest of the boundary of the image comes from the diagonal:
$$\left\{(x,x): 1\le x\le 2\right\} \,.$$

A Mathematica image of the boundary of $f(D)$ is this:

Some thought about the Inverse Function Theorem will tell you that the mapping is open away from critical points (points where $Df$ is singular).


问:
Intersection of open sets
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott.

In the introduction to the topology of $\mathbf{R}$, the author proves that countably finite unions and intersections of open sets results in an open set.

Exercise 3.2.1 asks for the following rather interesting questions.

After giving these answers a thought, I would like to ask for some help; verify if my intuition is correct. I could be entirely wrong, in which case, please share some hints. (Please feel free to add any asides/interesting remarks.)

(a) Where in the proof of the theorem part (ii) does the assumption that the collection of open sets be finite get used?

(b) Give an example of a countable collection of open sets $\{O_1,O_2,O_3,\ldots\}$ whose intersection $\cap_{n=1}^{\infty}O_n$ is closed, not empty and not all of $\mathbf{R}$.

(Story) Proof.

(a) In the proof of the theorem, "The intersection of an arbitrary (finite) collection of open sets is open", we argue that if $x \in \cap_{k=1}^{N}O_k$, then for all $1 \le k \le N$, because $O_k$ is open, there is an $\epsilon_k$ neighbourhood of $x$, $V_{\epsilon_k}(x) \subseteq O_k$. If we choose $\epsilon = \min \{\epsilon_1,\epsilon_2,\ldots,\epsilon_N\}$, then $V_\epsilon(x) \subseteq V_{\epsilon_k}$.

But, instead, if we had a countably infinite sequence of sets, such as $A_n = (0,\frac{1}{n})$, then $\cap_{n=1}^{\infty}A_n = \{0\}$. Every $\epsilon$-neighborhood around this point will not by contained in  $\cap_{n=1}^{\infty}A_n$, so the countably infinite intersection is not open.

I know that, $\lim_{n \to\infty}\frac{1}{n} = 0$. We can make this set as small as we like. I am not very confident, about $\cap_{n=1}^{\infty}A_n$, if its a isolated point or $\emptyset$.

(b) I am tempted to think that the Cantor set $C$ is an open set. Basically, its Cantor dust. So, I wonder if $\left[\left(\frac{1}{3},\frac{2}{3}\right) \cup  \left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right) \ldots \right] $ formed by the intersection of the open middle one-thirds of the sets $C_1,C_2,\ldots$, that is $[0,1] - C$ is a closed set.

答: If there were say countably infinite number of open sets, so that the sequence $(\epsilon_1,\epsilon_2,\cdots)$ went on forever with no end, then the minimum of the $\epsilon_k$ might be $0$ [for example if $\epsilon_k=1/k$ for each $k$].

Title: Intersection of open sets


问:
Is the converse of $ R \cong R^{op} $ true?
If R is commutative, then trivially $R \cong R^{op} $. Is the converse true ?

I understand that for every ring R we can define the opposite ring  $R^{op}$ by reversing the multiplication in R. Given the multiplication * in R, the multiplication ° in $R^{op}$ is defined as $a°b = b*a$. The "identity map" from R to $R^{op}$ , mapping each element to itself, is an isomorphism if and only if R is commutative.

答:
No. A nice example is the quaternions $\mathbb{H}$, which are noncommutative but have an involution given by

$$(a + bi + cj + dk)^{\dagger} = a - bi - cj - dk$$

which reverses the order of multiplication and gives an isomorphism $\mathbb{H} \cong \mathbb{H}^{op}$. Similarly we can consider matrix rings $M_n(k)$ over commutative rings, with involution the transpose $X \mapsto X^T$.


问:
Bounding a Complex Polynomial
Given the complex polynomial $P(z) = z^2 + a_1z + a_0$ and the constraint that $|z| > 1$, I'm trying to show that $|P(z)| \geq |z|^2 - |a_1||z| - |a_0|$. The obvious thing to do here of course is to apply the triangle inequality which yields
$$
|P(z)| = |z^2 + a_1z + a_0| \leq |z|^2 + |a_1||z| + |a_0|
$$
Just as obvious though is the fact that this is not remotely close to what I am trying to show. I'm sure I'm missing a relevant detail. I have attempted, and failed, to productively apply the constraint on $|z|$. This is effectively the start of a proof of the Fundamental Theorem of Algebra.

Would appreciate any tips.

答:
Hint. The triangle inequality also shows that
$$|a+b|\geq |a|-|b|.$$

To see this, note that
$$|a| = |a+b-b| \leq |a+b|+|b|.$$

(In point of fact, I don't see that the constraint is needed at all to establish what you want, though it would be useful to obtain further lower bounds on $|P(z)|$...)

Title: Bounding a Complex Polynomial


问:
Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$
I have this integral to evaluate:

$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$

I have tried substitution, trig identity and integration by parts but I'm just going round in circles.

Can anyone explain the method I need to work this out?

答:
HINT:

As $x^4-1=(x^2)^2-1=(x^2+1)(x^2-1),$

$$\frac{x^4 + 1}{x^2 +1}=\frac{x^4-1+2}{x^2 +1}=x^2-1+\frac2{x^2+1}$$

Generalization :

If the integrand is $$\frac{a_0+a_1x+a_2x^2+a_3x^3+\cdots}{ax^2+bx+c},$$

just divide to get the quotient of the form $b_0+b_1x+b_2x^2+\cdots$ which is easily integrable and the remaining part will be of the form  $\frac{px+q}{ax^2+bx+c}$

If $p=0$ check the $I_2$ below.

else we set $px+q=r\frac{d(ax^2+bx+c)}{dx}+ s=r(2ax)+rb+s$

Comparing the coefficients of $x,2ar=p\implies r=\frac p{2a}$

and Comparing the constants, $rb+s=q\implies s=q-rb=q-\frac{pb}{2a}$

$$\text{So, }\int \frac{px+q}{ax^2+bx+c} dx$$

$$=r \int \frac{d(ax^2+bx+c)}{dx}\frac1{ax^2+bx+c}dx+s\int\frac1{ax^2+bx+c}dx $$

$$=\frac p{2a}\int \frac{d(ax^2+bx+c)} {ax^2+bx+c}+\left(q-\frac{pb}{2a}\right)\int\frac1{ax^2+bx+c}dx $$

$$\text{Now, } \int \frac{d(ax^2+bx+c)} {ax^2+bx+c}=\ln|ax^2+bx+c|+C$$

$$\text{and } I_2=\int\frac1{ax^2+bx+c}dx =\int\frac{4a}{(2ax+b)^2+4ac-b^2}dx $$

If $4ac-b^2=0,$ put $2ax+b=u$ in $I_2$

If $4ac-b^2>0, 4ac-b^2=t^2$(say, ) $I_2=4a\int \frac{dx}{(2ax+b)^2+t^2}$ and put $2ax+u=t\tan\theta$

If $4ac-b^2<0, 4ac-b^2=-t^2$(say, ) $I_2=4a\int \frac{dx}{(2ax+b)^2-t^2}$ and put $2ax+u=t\sec\theta$


问:
Bounded subsets (supremum and infimum)
The problem is as follows:

Let $S$ and $T$ be nonempty, bounded subsets of the real numbers. Prove that if $T$ contains $S$, then $\inf T\leq\inf S\leq \sup S\leq\sup T.$

My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound. Help. Please and thank you.

答:
"My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound."

Well, both sets are bounded.

Both sets are bounded so upper and lower bounds exist for both sets.

And as they are both subsets of the Real Numbers (and neither is empty) it is the nature (actually the definition almost) of real numbers that all bounded non-empty sets do have least upper bounds and greatest lower bounds, so both the sets have a $\sup$ and an $\inf$.

The main thing to realize is that if $S\subset T$ and $u$ is greater than all elements of $T$ then $u$ is greater than all elements of $S$ because all elements of $S$ are elements of $T$.

From there if you understand the definitions, everything falls into place.

$\sup S$ must exist because $S$ is bounded above.  $\sup T$ must exist because $T$ is bounded above.  $\sup T \ge x$ for all elements of $T$ because $\sup T$ is an upper bound of $T$.  So $\sup T \ge y$ for all elements of $S$ because if $y \in S$ then $y \in T$ because $S\subset T$.  So $\sup T$ is an upper bound of $S$ (as well as for $T$).  But $\sup S$ is the least upper bound so $\sup S \le u$ for all possible upper bounds of $S$.  And that include $\sup T$ which is also an upper bound of $S$.  So $\sup S \le \sup T$.

The same arguments can be made about $S$ and $T$ being bounded below so $\inf S$ and $\inf T$ exist and $\int T$ being a lower bound of $T$ means it is at least as small as any element of $T$ and that includes the elements of $S$ so $\int T$ is a lower bound of $S$ and as $\inf S$ it the greatest lower bound of $S$ than $\inf T \le \inf S$.  So $\inf T \le \inf S$.

Just have to show $\inf S \le \sup S$.

which should be obvious.  Every lower bound of $S$ is at least as small as any element of $S$, and every upper bound is at least as large as that element of $S$. so every lower bound of $S$ is smaller or equal to every upper bound of $S$.  So $\inf S \le \sup S$.

======

Alternatively:

$S$ is bounded above.  $S$ is not empty.  And $S\subset \mathbb R$. So because the reals have the least upper bound property $\sup S$ exists.

But the same reasoning $T$ is bounded above and non-empty and a subset of the realls so $\sup T$ exists.

Now either $\sup S\le \sup T$ or $\sup S > \sup T$.

Suppose $\sup s > \sup T$.  Now the definition of $\sup S$ says that if anything is less than $\sup S$ then it can not be an upper bound of $S$.  So if $\sup T < \sup S$ then $\sup T$ is not an upper bound of $S$.  That means there is an $s \in S$ so that $\sup T < s$.

But $\sup T \ge t$ for all $t \in T$ so if $s > \sup T$ then $s \not \in T$.

And if $s \in S$ but $s \not \in T$ then $S\not \subset T$.

But we were told $S\subset T$ so we have reached a contradiction.  $\sup S > \sup T$ is impossible so $\sup S \le \sup T$.

.....

$S$ is non-empty and bounded below so $\inf S$ exist.  $S$ is not empty so there exists a $k\in S$.  By the definition of $\sup S$ and $\inf S$ we know $\inf S \le s$ and $s \le \sup S$.  And above we proved $\sup S \le \sup T$.

so $\inf S \le s \le \sup S \le \sup T$.

....

Now we repeat the argument two sections above to prove $\inf T$ exists and is $\le \inf T$.

So $\inf T\le \inf S \le s \le \sup S\le \sup T$.

Title: Bounded subsets (supremum and infimum)


问:
Find the number of integer non-negative solutions to $3 x_1+5 x_2+x_3+x_4=10.$

Find the number of integer non-negative solutions to
$$3 x_1+5 x_2+x_3+x_4=10.$$

My proposed solution: First find the possible values for the pairs $(x_1,x_2)$ which are allowed: $(0,0),(0,1),(0,2),(1,0),(1,1),(2,0),(3,0)$, and count solutions for $x_3, x_4$ in each case. Counting for each case is, respectively: 11,6,1,8,3,5,2. They add to a total of 36 solutions.

My problems: Given solution in a book is 30 and I would like to know whether my solution is indeed wrong. Is there a more elegant way to solve this problem and similar problems? I believe my approach is kind of brute force.

Body:
your result is true the number of solutions are $36$ for sure.
all solutions are listed here:

$ 1 :    x_1=0, x_2=0, x_3=0, x_4=10$

$ 2 :    x_1=0, x_2=0, x_3=1, x_4=9$

$ 3 :    x_1=0, x_2=0, x_3=2, x_4=8$

$ 4 :    x_1=0, x_2=0, x_3=3, x_4=7$

$ 5 :    x_1=0, x_2=0, x_3=4, x_4=6$

$ 6 :    x_1=0, x_2=0, x_3=5, x_4=5$

$ 7 :    x_1=0, x_2=0, x_3=6, x_4=4$

$ 8 :    x_1=0, x_2=0, x_3=7, x_4=3$

$ 9 :    x_1=0, x_2=0, x_3=8, x_4=2$

$ 10 :    x_1=0, x_2=0, x_3=9, x_4=1$

$ 11 :    x_1=0, x_2=0, x_3=10, x_4=0$

$ 12 :    x_1=0, x_2=1, x_3=0, x_4=5$

$ 13 :    x_1=0, x_2=1, x_3=1, x_4=4$

$ 14 :    x_1=0, x_2=1, x_3=2, x_4=3$

$ 15 :    x_1=0, x_2=1, x_3=3, x_4=2$

$ 16 :    x_1=0, x_2=1, x_3=4, x_4=1$

$ 17 :    x_1=0, x_2=1, x_3=5, x_4=0$

$ 18 :    x_1=0, x_2=2, x_3=0, x_4=0$

$ 19 :    x_1=1, x_2=0, x_3=0, x_4=7$

$ 20 :    x_1=1, x_2=0, x_3=1, x_4=6$

$ 21 :    x_1=1, x_2=0, x_3=2, x_4=5$

$ 22 :    x_1=1, x_2=0, x_3=3, x_4=4$

$ 23 :    x_1=1, x_2=0, x_3=4, x_4=3$

$ 24 :    x_1=1, x_2=0, x_3=5, x_4=2$

$ 25 :    x_1=1, x_2=0, x_3=6, x_4=1$

$ 26 :    x_1=1, x_2=0, x_3=7, x_4=0$

$ 27 :    x_1=1, x_2=1, x_3=0, x_4=2$

$ 28 :    x_1=1, x_2=1, x_3=1, x_4=1$

$ 29 :    x_1=1, x_2=1, x_3=2, x_4=0$

$ 30 :    x_1=2, x_2=0, x_3=0, x_4=4$

$ 31 :    x_1=2, x_2=0, x_3=1, x_4=3$

$ 32 :    x_1=2, x_2=0, x_3=2, x_4=2$

$ 33 :    x_1=2, x_2=0, x_3=3, x_4=1$

$ 34 :    x_1=2, x_2=0, x_3=4, x_4=0$

$ 35 :    x_1=3, x_2=0, x_3=0, x_4=1$

$ 36 :    x_1=3, x_2=0, x_3=1, x_4=0$


问:
Variance of Brownian Bridge
With $W(t)$ denoting the standard Wiener process, Brownian Bridge $B(t)$ can be defined on $t \in [0,T]$ as:

$$B(t):=W(t)-\frac{t}{T}W(T)$$

Computing the expectation gives $\mathbb{E}[B(t)]=\mathbb{E}[W(t)]-\frac{t}{T}\mathbb{E}[W(T)]=0-\frac{t}{T}0=0.$ From this we can deduce that $Var(B(t))=\mathbb{E}[B(t)^2]$. Here is my attempt:

$$\mathbb{E}[B(t)^2]=\mathbb{E}\left[\left(\sqrt{t}W(1)-\frac{t}{\sqrt{T}}W(1)\right)^2\right]=\mathbb{E}[W(1)^2]\left(\sqrt{t}\left(1-\sqrt{\frac{t}{T}}\right)\right)^2=\\=t\left(1-2\frac{\sqrt{t}}{\sqrt{T}}+\frac{t}{T}\right)=t\left(\frac{T-2\sqrt{t}\sqrt{T}+t}{T}\right)=\frac{t}{T}\left(\sqrt{T}-\sqrt{t}\right)^2$$

Now wikipedia gives the variance as $\frac{t}{T}(T-t)$, so I must be making a mistake somewhere, but I cannot figure out where? Any hints would be greatly appreciated,

答:
Although $W(t)$ and $\sqrt{t}W(1)$ have the same distribution, they are not equal. The correct computation is
$$\mathbb{E}[B(t)^2] = \mathbb{E}[(W(t) - (t/T)W(T))^2]$$
$$= \mathbb{E}[W(t)^2] - 2 \frac{t}{T} \mathbb{E}[W(t) W(T)] + \left(\frac{t}{T}\right)^2 \mathbb{E}[W(T)^2]$$

We know $\text{Var}(W(t)) = \mathbb{E}[W(t)]^2 = t$ and
$$\text{Cov}(W(t), W(T)) = \mathbb{E}[W(t)W(T)] = \min(t,T) = t$$
Thus, we conclude
$$\mathbb{E}[B(t)^2] = t - 2 \frac{t}{T} t + \left(\frac{t}{T}\right)^2 T = \frac{t}{T} (T-t)$$
as on Wikipedia.

Title: Variance of Brownian Bridge


问:
How to calculate the growth when old number is negative?
Sorry, I feel like an idiot for not knowing this, but assume my old number is -10. How would I calculate a 50% growth of that -10? Multiplying -10 by 1.50 results in -15 which is going in the wrong direction.

But multiplying 1.50 with the absolute value of -10 is not correct either. I think I am missing something fundamental here..

答:
If you start with minus ten and increase it by 50%, you will end up with -15.  Imagine you have £10,000 in the bank.  Increase it by 50% and you'll get £15,000.  But if you owe the bank £10,000 and increase that by 50% the logic works just the same the other way around.

If you start with -10 and want to increase it to get a less negative number, you need to be looking at addition rather than multiplication.  I am guessing there's some context here.  What is the -10 you are trying to increase/reduce?


问:
Second-Order Tikhonov Regularization
In the second-order Tikhonov regularization approach

$\min\left\|Gm - d \right\|_2^2 + \alpha\ ^2 \left\|\Gamma x\right\|_2^2$         (1)

given that $\Gamma\ $ contains second order derivatives, $f_i $ are the filter factors, $G=U \Lambda\ X^T $ and $\Gamma\ =V M X^T $ and $Y=X^{-T} $ the solution for model parameter $m$ is

$m_{\alpha\ \Gamma\ }=\sum_{i=1}^n f_i \dfrac{U_{.,i+k}^T d}{\lambda\ _i} Y_{.,i}\ $     (2)

where $k=n-m$ when $m \leq n$, and $m$ and $n$ are dimensions of $G$.

In my case I have a $m\times n=16\times84$ matrix of $G$, so $U$ should be a $16\times 16$ matrix.

Now, the question is how to calculate $m_{\alpha\ \Gamma\ }$ from equation (2) while the index $i+k$ is always greater than the dimension of $U^T$?!

答:
Are you sure that the solution for model parameter is given with that form? I'd suppose it is only if $\Gamma = \mathbb{1}$, so it applies to the zeroth-order Tikhonov regularisation.

In fact, you should rather solve the normal equation, so rather $m_{\alpha} = (G^T G + \alpha^2 \Gamma^T \Gamma)^{-1} G^T d$. If your approach is correct, it is new to me.

Title: Second-Order Tikhonov Regularization