问:
System of congruences with moduli not coprime and chinese remainder theorem application
Find all solutions, if any, to the system of congruences
$$\Bigg\{
 \begin{array}{c}
     x \equiv 1 (\mod 6)\\
     x \equiv 7 (\mod 9)\\
     x \equiv 4 (\mod 15)\\
    \end{array}$$
we can see that

$ x \equiv 1 \pmod 6 \implies
    \left\{
    \begin{array}{l}
        x\equiv 1 \pmod 2 \\
        x \equiv 1 \pmod 3
    \end{array}
    \right.$

as well as
$ x \equiv 7 \pmod 9\\
    \qquad \Rightarrow \quad x=7+9k\\
   \qquad \Rightarrow \quad x=1+3(2+3k)\\
   \qquad \Rightarrow \quad x\equiv 1 \pmod 3\\
$

and
$  x \equiv 4 \pmod{15} \implies
    \left\{
    \begin{array}{l}
        x\equiv 4 \pmod 5 \\
        x \equiv 1 \pmod 3
    \end{array}
    \right.
$

After removal of redundant congruences and picking the one with the highest power of 3 we are left with:

$
\Bigg\{
 \begin{array}{c}
     x \equiv 1 \pmod 2\\
     x \equiv 7 \pmod {3^2}\\
     x \equiv 4 \pmod 5\\
    \end{array}
$

Now we reduced it to standard CRT problem.

$ x=2k+1 \\
2k+1 \equiv 7\pmod{9}\\
2k \equiv 6\pmod{9}\quad \gcd{(2,9)}=1\\
k\equiv 3\pmod{9}\\
k=9\cdot l+3$

$x=2\cdot(9\cdot l+3)+1=18l+7$

$
18l+7\equiv 4\pmod{5}\\
18l\equiv -3\pmod{5}\\
18l\equiv 2\pmod{5}\\
3l\equiv 2\pmod{5}\\
\quad \Rightarrow \quad 7\pmod{5}\\
\quad \Rightarrow \quad 12\pmod{5} \quad \gcd{(3,5)}=1\\
l\equiv 4\pmod{5}\\
l=5\cdot m +4\\
$

The solution to the original system:

$
x=2\cdot(9\cdot (5\cdot m +4)+3)+1=90\cdot m+79\\
x\equiv 79\pmod{90}
$

with the smallest x = 79

Is it a correct approach?

答:
It is correct, but you can make the solution a bit shorter using the explicit inverse isomorphism in the C.R.T. for two coprime  moduli $a$ and $b$, given a Bézout's relation $ua+vb=1$:

$$x\equiv \begin{cases}\alpha\mod a,\\\beta\mod b,\end{cases}\iff x\equiv \beta ua+\alpha vb\mod ab$$

As $4\cdot 3^2-7\cdot 5=1$, the last two final congruences yield as a first step $$x\equiv 4\cdot 4\cdot 9-7\cdot 7\cdot 5=144-245=-101\mod 45.$$
Next, from $45-22\cdot 2=1$, you deduce from this congruence and the first above that
$$x\equiv 45+101\cdot 22\cdot 2=4489\equiv 79\mod 90.$$


问:
About Weak law for triangular arrays

Definition: to truncate a random varialbe X at level M means to consider $\hat{X}=X 1_{(|X| \le M)}=\begin{cases} X, & \textbf{if} |X| \le M \\ 0,& \textbf{if} |X|>M \end{cases}$

Theorem(*): Let $X_1,...,X_n$ have $E({X_i}^2)<\infty$ and be uncorrelated. Then $\textbf{var}(X_1+...+X_n)=\textbf{var}(X_1)+...\textbf{var}(X_n)$ where $\textbf{var}(Y)$ is the variance of Y.

Theorem (Weak law for triangular arrays): For each n, let $X_{n,k}, 1 \le k \le n$ be independent. Let $b_n>0$ with $b_n \rightarrow \infty$, and let $\hat{X}_{n,k}=X_{n,k} 1_{(|X_{n,k} \le b_n )}$. Suppose that as $n \rightarrow \infty$. $$(i). \sum_{k=1}^n P(|X_{n,k}|>b_n) \rightarrow 0$$, and $$(ii). b^{-2}_n \sum_{k=1}^n E\hat{X}^2_{n,k} \rightarrow 0.$$ If we let $S_N=X_{n,1}+...+X_{n,n}$ and put $a_n=\sum_{k=1}^n E\hat{X}_{n,k}$, then $(S_n-a_n)/b_n \rightarrow 0$ in probability.

Proof: Let $\hat{S}_n=\hat{X}_{n,1}+...+\hat{X}_{n,n}$. Clearly, $P(|\frac{S_n-a_n}{b_n}|>\epsilon) \le P(S_n \neq \hat{S}_n)+P(\frac{\hat{S_n}-a_n}{b_n}|>\epsilon)$. To estimate the first term, we note that $$P(S_n \neq \hat{S}_n \le P(\cup_{k=1}^n \{\hat{X}_{n,k} \neq X_{n,k} \}) \le \sum_{k=1}^n P(|X_{n,k}>b_n) \rightarrow b_n) \rightarrow 0$$ by property (i). For the second term, we note that Chebyshev's inequality, $a_n=E\hat{S}_n$, by theorem (*), and $\textbf{var}(X) \le EX^2$ imply $$P(|\frac{\hat{S}_n-a_n}{b_n}|>\epsilon) \le \epsilon^{-2} E{|\frac{\hat{S}_n-a_n}{b_n}|}^2=\epsilon^{-2} b^{-2}_n \textbf{var}(\hat{S}_n)=(b_n \epsilon)^{-2} \sum_{k=1}^n \textbf{var}(\hat{X}_{n,k}) \le (b_n \epsilon)^{-2} \sum_{k=1}^n E(\hat{X}_{n,k})^2 \rightarrow 0$$ by property (ii).

Question: I don't see how "$(S_n-a_n)/b_n \rightarrow 0$ in probability" comes true. And all those inequalities in the proof, like $$P(S_n \neq \hat{S}_n \le P(\cup_{k=1}^n \{\hat{X}_{n,k} \neq X_{n,k} \}) \le \sum_{k=1}^n P(|X_{n,k}>b_n) \rightarrow b_n) \rightarrow 0$$ and $$P(|\frac{\hat{S}_n-a_n}{b_n}|>\epsilon) \le \epsilon^{-2} E{|\frac{\hat{S}_n-a_n}{b_n}|}^2=\epsilon^{-2} b^{-2}_n \textbf{var}(\hat{S}_n)=(b_n \epsilon)^{-2} \sum_{k=1}^n \textbf{var}(\hat{X}_{n,k}) \le (b_n \epsilon)^{-2} \sum_{k=1}^n E(\hat{X}_{n,k})^2 \rightarrow 0$$ don't make sense to me, especially $\epsilon^{-2}$. Thanks a lot in advance. (The proof is a bit technical to me, so I would like it to be simpler to understand).

答:
For a random variable $X$ with $E[X]=\mu$, Chebyshev's inequality is:

$$P(|X-\mu|>\epsilon)\leq \frac{\mbox{Var}(X)}{\epsilon^2}=\frac{E[|X-\mu|^2]}{\epsilon^2}.$$

If you're unfamiliar with this, then hopefully you've seen Markov's inequality:

$$P(|Y|\geq \epsilon)\leq \frac{E[Y]}{\epsilon},$$

from which you can derive Chebyshev's Inequality by substituting $Y:=X-\mu$, and using the fact that since $\epsilon>0$, $P(|Y|\geq \epsilon)=P(|Y|^2\geq \epsilon^2)$.

That should hopefully clear up the convergence in probability issue.

For $P(S_n\neq \hat{S}_n)$, notice that both $S_n,\hat{S}_n$ are sums of $X_{k,n},\hat{X}_{k,n}$ respectively. Thus, the event $\{S_n\neq \hat{S}_n\}$ implies that there is at least one $j$ such that $X_{j,n}\neq \hat{X}_{j,n}$. So:

$$\{S_n\neq \hat{S}_n\}\subseteq \cup_{j=1}^n\{X_{j,n}\neq \hat{X}_{j,n}\}.$$

Then applying $P$ to both sides, and using countable additivity, you get:

$$P(\{S_n\neq \hat{S}_n\})\leq \sum_{j=1}^nP(\{X_{j,n}\neq \hat{X}_{j,n}\}).$$

Now stare at the definition of $\hat{X}_{j,n}$ until you see that you need $X_{j,n}>b_n$ to enforce $X_{j,n}\neq \hat{X}_{j,n}$.

ContentLicense: CC BY-SA 4.0

Title: About Weak law for triangular arrays


问:
Estimating the $L^2$ norm of convolution with $\dfrac{1}{|x|^2}$ of a function $f$ which is in $L^2\cap L^1$
Let $f\in L^1(\mathbb{R}^3)\cap L^2(\mathbb{R}^3)$ and let $g=\displaystyle\int_{\mathbb{R}^3} \dfrac{f(x+y)}{|y|^2}dy$. Show that
$$\left \|g \right \|_{L^2}\le C\left \| f \right \|_{L^1}^{2/3} \cdot \left \| f \right \|_{L^2}^{1/3} $$
for some constant $C$.

Note that directly applying Young's convolution inequality to $g=f*\dfrac{1}{|x|^2}$ only gives a bound $\left \| g \right \|_{L^2}\le C \left \| f \right \|_{L^1}$.

答:
By Hardy-Littlewood-Sobolev's inequality (see e.g. E. Lieb "Functional Analysis"):
$$
\|g\|_{L^2} ≤ C\,\|f\|_{L^{5/6}}
$$
and your result then follows from Hölder's inequality.


问:
Quadratic form positive definite if limits are positive infinity?
Let
$$q(x_1,\ldots,x_n) = \sum_{i,j=1}^{n}a_{ij}{x_i}{x_j}, \quad a_{ij}\in \mathbb{R}.$$
be a quadratic form with real coefficients. Suppose that
$$ \lim_{x_i \rightarrow \pm \infty} q(x_1,\ldots,x_n) = + \infty$$
for every $i$ while the remaining variables are fixed. Is it true that $q$ is necessarily positive definite?</p>

答:
Consider the form $$q(x,y)=x^2+3y^2+4xy$$

In fact this is the form $$q(\bar x,\bar y)=\bar x^2-\bar y^2$$ in the basis $(1,0), (2,1)$.

ContentLicense: CC BY-SA 3.0


问:
moment generating function uniquely determines distribution
Suppose that X and Y are random variables both taking possible values in {$0, 1, 2 ...$}.
Further, suppose that X and Y have the same mgf for all t around $0$:
Then it holds:
$$\sum_{j=0}^{\infty}e^{tj}f_X(j)-\sum_{j=0}^{\infty}e^{tj}f_Y(j)=0$$

$$\Rightarrow\sum_{j=0}^{\infty}e^{tj}[f_X(j)-f_Y(j)]=0$$
$$\Rightarrow \sum_{j=0}^{\infty}e^{tj}c_j=0 $$
with $c_j:=f_X(j)-f_Y(j)$.

Is there a way to justify that  $c_j=0$  for all $j \in \{0,1,...\}$

答:
First, notice that from the definition if $c_j$ we have $-1 \le c_j \le 1$ for all $j$.  Now suppose $c_0 \ne 0$, and WLOG $c_0 > 0$.  Then we have

\begin{align*}
0 &= \sum_{j=0}^\infty e^{tj}c_j = c_0 + \sum_{j=1}^\infty e^{tj}c_j
\end{align*}

so $c_0 = -\sum_{j=1}^\infty e^{tj}c_j$ for all $t \in \mathbb{R}$.  Since each $c_j$ is at least $-1$ this implies $$ 0 < c_0 \le \sum_{j=1}^\infty e^{tj} = \sum_{j=1}^\infty (e^t)^j,$$

but this is a geometric series and can be made arbitrarily small by choosing $t \ll 0$.  This implies $c_0 = 0$, and we can use the same reasoning to show that $c_j = 0$ for all $j$ with induction.


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答: Imagine the tea in the wine in the form of a tiny ball within the wine. Then that ball must be exactly the amount of wine that is missing from the wine. Ergo, it is the amount of wine that is in the tea. So, the two amounts are the same, with exactly as much tea in the wine as wine in the tea.</p>


问:
Solving $u=2x+y$ and $v=\frac{5x}{x+y}$ for $x$ and $y$
This feels like a stupid question but I'm struggling with it nevertheless. I have the following variables
$$u=2x+y \qquad v=\frac{5x}{x+y}$$ I have been trying in vain to put $x$ and $y$ in terms of $u$ and $v$, but with the variables I have the method for eliminating variables is not clear to me.

Sorry if this is a little too elementary for this exchange, but is there a useful method for combining variables in situations such as this?

Thanks.

Body:
First, note that $y = u - 2x$. Then, substituting:

$$v = \frac{5x}{u - x}$$

$$v(u - x) = 5x$$

Here, we assume $x\neq u$.

$$uv - vx = 5x$$

$$(5 + v)x = uv$$

$$x = \frac{uv}{5 + v}$$

$$y = u - 2\frac{uv}{5 + v}$$

$$y = \frac{uv + 5u}{5 + v} - \frac{2uv}{5 + v}$$

$$y = \frac{5u - uv}{5 + v}$$

Thus:

$$\boxed{(x,y) = \bigg(\frac{uv}{5 + v}, \frac{5u - uv}{5 + v}\bigg)}$$


答:
Upon inspection, it is best to solve for $(x + y)$ as a first step (instead of solving for $x$ or $y$).
$$\begin{align}
u &= 2x + y\\
u-x&=x+y
\end{align}
$$
Substituting in to the second equation...
$$\begin{align}
v&=\frac{5x}{u-x} \tag{ where $u\neq x$ }\\
v(u-x)&=5x \\
uv-vx&=5x \\
uv&=5x+vx\\
uv&=x(v+5)\\
x &= \frac{uv}{v+5}
\end{align}$$
We then substitute this back into the first equation to get $y$.
$$\begin{align}
u&=2\bigg(\frac{uv}{v+5}\bigg)+y\\
y&=u-\frac{2uv}{v+5}\end{align}
$$

Title: Solving $u=2x+y$ and $v=\frac{5x}{x+y}$ for $x$ and $y$

Body:
This feels like a stupid question but I'm struggling with it nevertheless. I have the following variables
$$u=2x+y \qquad v=\frac{5x}{x+y}$$ I have been trying in vain to put $x$ and $y$ in terms of $u$ and $v$, but with the variables I have the method for eliminating variables is not clear to me.

Sorry if this is a little too elementary for this exchange, but is there a useful method for combining variables in situations such as this?

Thanks.


问:
Showing that $D^2\setminus S^1$ is diffeomorphic to $\mathbb{R}^2$
I'm studying some manifolds and I'm stuck at the exercise 1.5c of Tu's Introduction To Manifolds.

There he asks us to verify that the function $h \colon D^2 \setminus S^1 \to \mathbb{R}^2$ given by $$h(a,b) = \left( \frac{a}{\sqrt{1-a^2-b^2}}, \frac{b}{\sqrt{1-a^2-b^2}}\right)$$ is a diffeomorphism.

How one should go about that? I know that we need to check that both $h$ and $h^{-1}$ are bijective smooth maps, but I can't think on how I should go about proving that all derivatives exist and are continuous, I can't find a general formula.

答:
Finding the partial derivatives of $h_1(a,b)=\frac{a}{\sqrt{1-a^2-b^2}}$ and $h_2(a,b)=\frac{b}{\sqrt{1-a^2-b^2}}$ is just an application of the quotient rule. Computing each:
$$\frac{\partial h_1}{\partial a}=\frac{-b^2+1}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_1}{\partial b}=\frac{ab}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_2}{\partial a}=\frac{ab}{(1-a^2-b^2)^{3/2}}$$
$$\frac{\partial h_2}{\partial b}=\frac{-a^2+1}{(1-a^2-b^2)^{3/2}}$$
The numerators are all everywhere continuous and the denominators are only zero on $S^1$. Thus $h$ is differentiable on all of $D^2\setminus S^1$. Moreover, when we compute the determinate of the Jacobian we get:
$$\frac{\partial h_1}{\partial a}\frac{\partial h_2}{\partial b}-\frac{\partial h_1}{\partial b}\frac{\partial h_2}{\partial a}=\frac{a^2b^2-a^2-b^2+1-a^2b^2}{(1-a^2-b^2)^{3}}=\frac{1}{(1-a^2-b^2)^{2}}$$
From this we see that the determinate is non-singular on all of $D^2\setminus S^1$, and so by the inverse function theorem $h$ has an everywhere differentiable inverse $h^{-1}:h(D^2\setminus S^1)\to D^2\setminus S^1$. Thus, all that is left is to show that $h(D^2\setminus S^1)=\mathbb{R}^2$. Notice that the function $h(a,b)=\frac{1}{\sqrt{1-a^2-b^2}}(a,b)$ just scales the vector $(a,b)$ by $\frac{1}{\sqrt{1-r^2}}$ where $r$ is the magnitude of $(a,b)$. Picking any $\hat{s}\in S^1$ we can thus view $h$ as a map from $[0,1)\times S^1$ to $\mathbb{R}^2$ by $h(r,\hat{s})=\frac{r}{\sqrt{1-r^2}}\hat{s}$ or just as a map $h_\hat{s}$ from $[0,1)$ to $[0,\infty)$ via $h_\hat{s}(r)=\frac{r}{\sqrt{1-r^2}}$. Notice $h_\hat{s}(0)=0$ for all $\hat{s}$ so this way of viewing $h$ is well defined. $h_\hat{s}$ is a surjection and so we see that $h$ is a surjection from $D^2\setminus S^1\to\mathbb{R}^2$.


问:
Find an $a(z)>0$ such that $an^{z+1} \geq (an+a-1)(n+1)^z$ holds for all $z\in \mathbb{R}_{\geq 0}$ and $n \in \mathbb{N}$.
Find a real number $a(z)>0$ such that $an^{z+1} \geq (an+a-1)(n+1)^z$ holds for all $z\in \mathbb{R}_{\geq 0}$ and $n \in \mathbb{N}$. The value of $a$ is only allowed to be dependent of $z$.
This seems rather hard since $z$ only appears as an exponent.

答:
From $a \leqslant \frac{(n+1)^z}{(n+1)^{z+1}-n^{z+1}} \tag 1$ what if $n \to \infty$?

Try a few $z \in \mathbb N$ if you want to get an idea.

Also note that if $a_0>0$ is such a number, then any positive $a$ less than $a_0$ is also a solution.


问:
Linear algebra: proof of sufficient conditions for $x,Ax,A^{2}x,\dots,$ being linearly independent
This questsion is an extension of a previously asked question that I think needs a more rigorous proof (please see the link and the answer by user Aaron).

The question is, given a little bit of different notation (in accordance with control theory), for $b\in\mathbb{R}^{n}$ and $A\in\mathbb{R}^{n\times n}$, what is a sufficient condition for

\begin{align}
b,Ab,A^{2}b,\dots,A^{n-1}b
\end{align}

all being linearly independent? Furthermore, if we assume that $A$ has $n$ distinct eigenvalues $\lambda_{i}$, $i=1,\dots,n$, is this condition sufficient for the above to hold and if so how do we prove it?

In the above linked question Aaron's answer was:

If the characteristic polynomial of $A$ has no repeated roots, then we have a basis $\{v_{i}\;\vert\;1\leq i\leq n\}$ of eigenvectors for $\mathbb{C}^{n}$, and we can express $x=\sum \alpha_{i}v_{i}$ in terms of the basis. If all of the $\alpha_{i}$ are non-zero, then all of your vectors will be linearly independent. In fact, with this condition on $A$, this is a necessary and sufficient condition on $x$.

If $A$ has an eigenvalue with geometric multiplicity greater than 1, then no $x$ will work.

If $A$ has repeated eigenvalues but all eigenvalues have geometric multiplicity of 1, then it is still possible to find such an $x$, but things are a bit more complicated.

I would be very happy if someone could fill in the blanks for his statements, it is not immediately evident that a $b$ with nonzero entries leads to all vectors $b,Ab,\dots,A^{n-1}b$ being linearly independent.

答:
No assumption is made on the field $k$ and $A\in M_n(k)$.

For $b\in k^n$, the $A^0 b,\ldots, A^{n-1}b$ are linearly dependent iff there is some non-zero polynomial $h\in k[x]$ of degree $<n$ such that $h(A)b=0$.

With $p\in k[x]$ the minimal polynomial of $A$ and $g=\gcd(p,h)$ then $h(A)b=0,p(A)b=0$ implies $g(A)b=0$.

Thus the necessary and sufficient condition for the $A^0 b,\ldots, A^{n-1}b$ to be linearly independent is that $\deg(p)=n$ and $f(A)b\ne 0$ for each $f\in k[x]$ strictly dividing $p$.

When the $\deg(p)=n$ condition is satisfied, factorize in irreducibles $p=\prod_j q_j^{e_j}\in k[x]$.
Taking $b_j \in \ker(q_j(A)^{e_j}) ,\not \in \ker(q_j(A)^{e_j-1})$ then $b = \sum_j b_j$ works.

(take $f$ such that $p = q_j f$, then $0=f(A)b = f(A)b_j=\gcd(f,q_j^{e_j})(A)b_j=q_j(A)^{e_j-1}b_j$ is a contradiction)


问:
If $||x||_2 = 3$ and $||y||_2 = 7$ find the minimum of $\langle x, y\rangle$
If $||x||_2 = 3$ and $||y||_2 = 7$, find the minimum of $\langle x, y\rangle$

My thought process is that it would be 3 since if $||x||_2 = 3$, we can have $|x_1| = 1$, $|x_2| = 1$, $|x_3| = 1$ and every other term can equal $0$. Similarly, if $||y||_2 = 7$ then we can have $||y_1|| = 1$, $||y_2|| = 1$, ... , $||y_7|| = 1$ and every other term can equal $0$. So, we would have that $$\langle x, y\rangle =  |x_1| |y|_1 + |x_2| |y|_2 + |x_3| |y|_3 + |x_4| |y|_4 + |x_5| |y|_5 + |x_6| |y|_6 + |x_7| |y|_7 + ... + |x_k| |y|_k\\=(1\times1) + (1\times1) + (1\times1) + (0\times1) + (0\times1) + (0\times1) + (0\times1) + (0\times0) + ... + (0\times0) = 3$$. So our minimum is 3.

Is is train of logic correct?

答:
Take $x$ and $y$ to be in opposite directions.  For example $x=(3,0)$ and $y=(-7,0)$.

The minimum is $$-3\cdot 7 = -21.$$

Or are $x$ and $y$ $n-$dimensional?:

$x=(3,0,0,\cdots,0)$ and $y=(-7,0,0,\cdots,0)$ and again the minimum is $-21.$

UPDATE:  To derive this result, we can use Lagrange multipliers.

$$ L = \langle x,y \rangle - \lambda (  \langle x,x \rangle-9) - \mu( \langle y,y \rangle-49)$$

Computing $0=\frac{\partial L}{\partial x}=y-2\lambda x$ and $0=\frac{\partial L}{\partial y}=x-2\mu y$ leads to the condition that $y= \alpha x$, for some $\alpha$, so $x$ and $y$ are in the same direction (maximizing the inner product) or in opposite directions (minimizing the inner product).


问:
$f\colon X\to \mathbb{R}$ is continuous at $a$ iff we have $\lim{f(x_n)}=f(a)$ for every $\{x_n\}_{n\in\mathbb{N}}$ in $X$ with $\lim{x_n}=a$
A function $f\colon X\to \mathbb{R}$ is continuous at $a\in X$ if and only if we have $\lim{f(x_n)}=f(a)$ for every sequence $\{x_n\}_{n\in\mathbb{N}}$ in $X$ with $\lim{x_n}=a$.

Scratch.

($\Rightarrow$). As $f$ is continuous at $a\in X$, then for all $\varepsilon>0$ we're going to have an interval $(f(a)-\varepsilon,f(a)+\varepsilon)$ such that $(f(a)-\varepsilon,f(a)+\varepsilon)\cap f[X]\neq \emptyset$, so $f(a)$ is accumulation point for some sequence $\{y_n\}$ in $\mathbb{R}$.

But how do I proceed from here to show that there's also a sequence $\{x_n\}$ in $X$ such that $\lim{x_n}=a$? In order to be a sequence there, we're going to allow say $\delta>0$ to be any value, but $\delta$ is dependent on $\varepsilon>0$, as the definition of continuity tells us.

And after I do this, how to show that $\lim{f(x_n)}=f(a)$? Because although $f(a)$ is an accumulation point, it needs to be an accumulation point for the image of $f$ with $\{x_n\}\subset X$

($\Leftarrow$). Again, I'm having a hard time connecting the fact that a sequence $\{x_n\}$ in $X$ would need an $\delta>0$ to be any $\delta$ such that there'll be $N\in\mathbb{N}$ with $n>N\implies \left|x_n-a\right|<\delta$. And this $\delta$ will define our $\varepsilon>0$...

答:
This is a corollary of a more general result: $f(x)\to b$ when $x\to a$ iff $f(x_n)\to b$ for each $(x_n)$ such that $x_n\to a$.

($\Rightarrow$). Consider $(x_n)$ such that $x_n\to a$ and $\epsilon>0$ arbitrary.

Since $f$ is continuous, there exist $\delta>0$ such that if $|x_n - a|<\delta$ then $|f(x_n) - b|<\epsilon$.

Since $x_n\to a$, there exists $n_0$ such that if $n\ge n_0$, then $|x_n - a|<\delta$.

Therefore, there exists $n_0$ such that $n\ge n_0$ implies $|f(x_n)-b|<\epsilon$.

($\Leftarrow$). Suppose the $f(x)\not\to b$. This means there exists $\epsilon>0$ such that for any $\delta>0$ there exists $x_\delta$ such that $|x_\delta - a|<\delta$ and $|f(x_\delta) - b|\ge \epsilon$.

In particular, for each $n\in\mathbb N$ we may take $\delta_n = \frac 1n$ and consider the sequence $(x_n)$ given by $x_n = x_{\delta_n}$. In this case, $|x_n - a|<\frac 1n\to0$ but $|f(x_n) - b|\ge\epsilon$ and therefore $f(x_n)\not\to b$, which is a contradiction.


答:
Another trick to eliminate one of the variables is to take linear combinations of the two equations to eliminate a variable. I rearrange the 2nd equation a bit

$u = 2x+y$ (1)

$\frac{5x}{v} = x+y$ (2)

I take eqn (1) and subtract (2) to eliminate y. So I take the left-side of (1) and subtract the left-side of (2). and the same with the right side. so I subtract out the y's.

$u-\frac{5x}{v} = x$ (notice there's no y)

$u = x(1+\frac{5}{v})$

multiply both sides by v to get rid of that v in the denominator

$uv = x(v+5)$

$$x=\frac{uv}{v+5}$$

Now you can substitute this x back into either of your original two equations to solve for y.


问:
Minimizing deviations from threshold value from a given group of numbers
Given a set of numbers $a_n$, a threshold level $t$, how do I find the combination of numbers that will sum to at least the threshold with minimum deviation? Added: That is, they must always exceed the threshold, but we want to minimize this excess.

For example, if I have $5$, $15$, $4$, $8$, $16$, $12$, and $9$; and my threshold is $17$. How do I find the combination of numbers with minimum deviation from this threshold?

One way I'm thinking about expressing the question is to minimize the sum of the squares of the differences between the threshold and the sum of that group. Like, if my groups were $5$ and $16$; 8 and 15; 9, 4, and 12; then I want to minimize this way: $$[(16 + 5)^2 - 17^2] + [(15 + 8)^2 - 17^2] + [(12 + 4 + 9)^2 - 17^2] = 728$$

But the groups could be grouped like this, so this is a better combination:

$$[(16 + 4)^2 - 17^2] + [(15 + 5)^2 - 17^2] + [(12 + 4 + 6)^2 - 17^2] = 417$$

How do I find the best combination in the general case? Preferably, if someone could show me the linear program formulation, that would be great. I think this is most likely an integer program though, with binary variables. Am I in the right direction when I say:

$$Min Z = \sum_n {(a_nx_n)^2 - t^2}$$
$$st$$
$$\sum_n{a_nx_n} > t$$
$$x_n = 1,0$$

The above seems to be wrong because some of the x's will be set to 0, but I'm trying to use all the numbers I have, but grouped in different ways. But I'm out of ideas on how to address that.

Added: To be clearer, I want to use all the numbers in $a_n$. I don't want to just pick a few numbers from it. The situation I have is I'm trying to assign jobs to people, and all jobs have to be worked. The $a_n$ represent money they make for doing 1 unit of the job (piecework rate), and the threshold is the minimum wage.

Also, in extension, how do I ban/disallow specific combinations from appearing? For example, if I wanted to ban $a_5 = 16$ and $a_7 = 9$ because the processes they represent can't happen simultaneously, so that due to that constraint, if that combination was part of the optimal solution, I'd have to change my solution; how do I do that? Would it be:

$$x_5 + x_7 = 1$$

Again, it doesn't seem to be that above because one off the x's will be set to 0, but I need all the numbers to be used. The program would be good if I was just finding the combination of numbers that would reach the threshold, but I'm trying to use all the numbers and collectively trying to minimize all their deviations.

答:
I am (still) not sure if I understand the problem correctly. I'm considering the following problem: given a set of numbers $a_n$, find the group of numbers whose sum is minimal, given that it exceeds a certain threshold.

$$\min_{x_n}\sum_n x_na_n\\
\text{s.t. } \sum_n x_na_n \ge t,\quad x_n\in\{0,1\}$$

where $t$ is the threshold and $x_n$ are the unknown weights.

[Need for clarification: I do not understand how you want to use all the numbers. What are the allowed weights then in the sum? Is the threshold supposed to be a lower limit on the (weighted) sum of $a_n$ or on $a_n$ themselves?]


问:
How to find estimator for $\lambda$ for $X\sim \operatorname{Poisson}(\lambda)$ using the 2nd method of moment?
So an estimator for for $\lambda$ using the first method of moment is simply the sample mean:

$$\mu_1 = E(X) = \lambda = \bar{X} = \hat{\mu_2}$$

For the second moment,

$$\operatorname{Var}(X) = E(X^2) - E(X)^2$$
$$\lambda = E(X^2) - \lambda^2$$
$$\mu_2 = E(X^2) = \lambda + \lambda^2$$
$$s^2 = \frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2$$
$$s^2 = \frac{1}{n}\sum_{i=1}^n X_i^2 - \bar{X}^2$$
$$\frac{1}{n}\sum_{i=1}^n X_i^2 = s^2 + \bar{X}^2$$
$$\hat{\mu_2} = s^2 + \bar{X}^2$$
Making $\mu_2 = \hat{\mu_2}$, we get:

$$\mu_2 = \lambda + \lambda^2 = s^2 + \bar{X}^2 = \hat{\mu_2}$$
$$ \lambda + \lambda^2 = s^2 + \bar{X}^2$$

From here I've no idea how to factorise it so I have $\lambda$ all on one side.

答:
So you have three potential estimators:

The first, $\bar X$ is unbiased and has the smallest
variance among unbiased estimators.

The second, $S^2$
is also unbiased, but has much larger variance than $\bar X.$

The third is given in @JimB's comment. It is unbiased
and its variance is not much larger than the variance of $\bar X.$

in the R simulation below, SDs of estimators are given instead of their variances. With $m = 100\,000$ iterations we can expect about two significant digits
of accuracy.

```<code>set.seed(2020)
m = 10^5;  n = 10;  lam = 3
x = rpois(m*n, lam)
MAT = matrix(x, nrow=m)
a = rowMeans(MAT)
v = apply(MAT, 1, var)
est.1 = a
mean(est.1);  sd(est.1)
[1] 3.0008      # aprx 3 (unbiased)
[1] 0.5484283   # smallest variability
est.2 = v
mean(est.2);  sd(est.2)
[1] 2.99409     # aprx 3 (unbiased)
[1] 1.518167    # largest variability of the 3
est.3 = .5*(sqrt(4*v+4*a^2+1)-1)
mean(est.3);  sd(est.3)
[1] 2.994443    # aprx 3 (unbiased)
[1] 0.581836    # slightly more variable than mean
</code>```

```<code>par(mfrow=c(3,1))
mn = min(est.1, est.2, est.3)
mx = max(est.1, est.2, est.3)
 hist(est.1, prob=T, xlim=c(mn,mx), col="skyblue2")
 hist(est.2, prob=T, br=60, xlim=c(mn,mx), col="skyblue2")
 hist(est.3, prob=T, xlim=c(mn,mx), col="skyblue2")
par(mfrow=c(1,1))
</code>```

Body:
So you have three potential estimators:

The first, $\bar X$ is unbiased and has the smallest
variance among unbiased estimators.

The second, $S^2$
is also unbiased, but has much larger variance than $\bar X.$

The third is given in @JimB's comment. It is unbiased
and its variance is not much larger than the variance of $\bar X.$

in the R simulation below, SDs of estimators are given instead of their variances. With $m = 100\,000$ iterations we can expect about two significant digits
of accuracy.

set.seed(2020)
m = 10^5;  n = 10;  lam = 3
x = rpois(m*n, lam)
MAT = matrix(x, nrow=m)
a = rowMeans(MAT)
v = apply(MAT, 1, var)
est.1 = a
mean(est.1);  sd(est.1)
[1] 3.0008      # aprx 3 (unbiased)
[1] 0.5484283   # smallest variability
est.2 = v
mean(est.2);  sd(est.2)
[1] 2.99409     # aprx 3 (unbiased)
[1] 1.518167    # largest variability of the 3
est.3 = .5*(sqrt(4*v+4*a^2+1)-1)
mean(est.3);  sd(est.3)
[1] 2.994443    # aprx 3 (unbiased)
[1] 0.581836    # slightly more variable than mean

par(mfrow=c(3,1))
mn = min(est.1, est.2, est.3)
mx = max(est.1, est.2, est.3)
 hist(est.1, prob=T, xlim=c(mn,mx), col="skyblue2")
 hist(est.2, prob=T, br=60, xlim=c(mn,mx), col="skyblue2")
 hist(est.3, prob=T, xlim=c(mn,mx), col="skyblue2")
par(mfrow=c(1,1))


问:
Is it true that the integral of $\delta(x)/x$ between symmetrical limits is zero?
My professor is claiming that the following is true:

$$\int_{-\infty}^{\infty}\frac{\delta(x)}{x}dx=0,$$

where $\delta(x)$ is the Dirac delta "function", as he calls it. I think the integral diverges from the definition of the delta "function", but his rational is that the solution must be zero because the integrand is "an odd function of $x$".

I think that if he is correct then the definition of the delta distribution is basically meaningless.I know it is true that $\delta(x)=\delta(-x)$, but I think that the reason his explanation that the integrand is an odd function fails because because the delta distribution isn't a function at all, and (presumably) distributions don't have this usual integration property.

Would somebody please confirm or deny, and if possible explain why a distribution doesn't have to integrate to zero in the same way as a function when it is odd-valued?

答:
As written in the other answers, "$\delta_0(x)/x$" does not mean a priori anything (since there is no defintion for the product of two singular distributions in general). In a general procedure, one can only define the multiplication of $\delta_0$ with functions $g$ continuous at $0$ as the distribution acting on smooth functions $\varphi$ as
$$
\langle g\,\delta_0,\varphi\rangle = g(0)\,\delta_0.
$$

However, one can look in the particular case of "$\delta_0(x)/x$" what is the distribution $T$ that has the closest properties to what we would expect. A way is to take $T$ as a solution of the equation
$$
x\,T(x) = \delta_0(x)
$$
However there are several solutions to this equation (since if $T$ is solution, then $T + c\, \delta_0$ is also solution). An additional constraint can be to require the solution to be homogeneous. In this case, we are left with only one solution (see e.g. here for more details):
$$
T = -\delta_0',
$$
which is a well defined distribution defined by $\langle T,\varphi\rangle = \varphi'(0)$. Remarking that this defines a linear functional not only on smooth  compactly supported functions, but also on $C^1$ functions, we can take $\varphi  = 1$, which yields
$$\boxed{
\langle T,1\rangle = 0}
$$

And this is the rigorous version of the result claimed by your professor. Indeed, with less rigorous notations, we could define the (possibly misleading) notation $"\frac{\delta_0(x)}{x}" := T = - \delta_0'$ and then the (possibly misleading) notation $"∫_{-\infty}^\infty T(x)φ(x)\,\mathrm d x" := \langle T,φ\rangle$. With these notations, $\langle T,1\rangle = 0$ becomes
$$
"∫_{-\infty}^\infty \frac{\delta_0(x)}{x}\,\mathrm d x" = 0.
$$


问:
How many selections contain exactly $4$ green balls and $2$ yellow balls if you pick a subset of $6$ balls?
Question: Given a set of $13$ balls, $9$ of which are green and the other $4$ are yellow, if you pick a subset $6$ balls, how many selections contain exactly $4$ green balls and $2$ yellow balls?

I know that there are $\binom{13}{6}$ ways to pick a subset of 6 balls and that the number of ways to pick exactly $4$ green balls and $2$ yellow balls is equal to the number of ways to pick exactly 4 yellow balls out of 6 picks (since the other two will have to be yellow) and vice versa. Besides that, I have no idea how to solve this problem (though I'm pretty sure the answer is $756$).

答:
As Don Thousand said, all I needed to do was select $4$ green balls from the green pile and $2$ yellow balls from the yellow pile.

So $\binom{9}{4}$$\binom{4}{2}$ = $(126)(6) = 756$


问:
Perfect squares using 20 1's, 20 2's and 20 3's.
How many perfect squares can be formed using 20  1's, 20  2's and 20  3's. This is a recent exam question, which I had no clue how to solve?

There is some kind of trick here, since time allotted to solve it was just 4 minutes.

Body:
\begin{array}{ll}
    20 *1 &= 20 \\
    20 *2 &= 40 \\
    20 *3 &= 60 \\
    ---------------
 \text{total} &= 120
 \end{array}

Now, Perfect square(s)...
$1+4+9+16+25+36+49 = 140$ ( I can remove $2$ number with total of $20$ ($4+16$))

that will lead to

$1+9+25+36+49$ (i.e. $5$ numbers)

Answer: $5$ perfect squares can be formed..


问:
What are the steps to parametrize a curve and solve its integral?

Let C be the piece of the curve of intersection of the cylinder $y = x^2$
and the plane
$x + y + z = 0$ between the points $(0, 0, 0)$ and $(2, 4, −6)$.

Parametrize C and compute $$\int _C\sqrt{2+4x+8y}\:ds$$

How would I do this?

Here's what I tried:

Let $x=t$

So then: $y=t^2$

Then I use a random vector function $r(t) = (t,t^2)$

So:

$r(0) = (0,0)$ and $r(2) = (2,4)$

And those equal my points. But how do I get the Z value? Do I just throw in another value like:

$r(t) = (t,t^2,-t^2-2)$

How does it work? I have no clue how to solve this. I'm guessing my bounds are $0->2$ when I differentiate in respect to dt.

So then:

$$ds = ||r'(t)|| = ||(1,2t,-2t)|| = \sqrt{1+8t^2}$$

Therefore my integral is:

$$\int _0^2\:\left(t^2\right)\left(\sqrt{1+8t^2}\right)dt$$

Is this right?

But like. Even if it IS right, how do I get the parametrization without guessing and checking? I feel like I just used random logic to get to my integral instead of actual steps. So essentially I'm wondering, how to parametrize the curve of intersection.

Like what if it wasn't just $x=t$? Do you just eyeball the parametrization?

答:
Let us find a parametrisation $r(t)=(x(t),y(t),z(t))$ such that $r(0)=(0,0,0)$ and $r(1)=(2,4,-6)$. We also require that $$x(t)+y(t)+z(t)=0\quad\text{and}\quad y(t)=x(t)^2\quad\forall\ t\in[0,1].$$ If we set $x(t)=2t$, then the two equations above demand that $$y(t)=x(t)^2=4t^2\quad\text{and}\quad z(t)=-(x(t)+y(t))=-2t-4t^2.$$ So we get the parametrisation
$$r(t)=(2t,4t^2,-2t-4t^2).$$
As for "how do I get the parmetrisation without guessing and checking" - well, there is a level of choice. I chose the limits of $t$, then I chose an $x(t)$ that works. From there $y(t)$ and $z(t)$ are determined by those choices.

I should add that you should check that $r'(t)\ne 0$ for all $t\in[0,1]$. Otherwise this would correspond to the curve having a 'cusp'.


问:
How many ways can you add the numbers 1, 2, and 3 to create a number n?
I would like to find an equation in which, given a number n, you can find the amount of ways to add 1, 2, and 3 to create the number. The commutative rule doesn't apply; for example, 1 + 1 + 3 is a different arrangement than 1 + 3 + 1.
Let's take the number 4 for an example. There are 7 ways to combine the numbers 1, 2, and 3 to create 4:
1 1 1 1
1 1 2
1 2 1
2 1 1
1 3
3 1
2 2

Any ideas?

答:
Let $a_n$ be the number of such partitions.  Then $a_1 = 1, a_2 = 2, a_3 = 4$, and by conditioning on the size of the first part, we find that
$$a_n = a_{n-1} + a_{n-2} + a_{n-3}.$$
Do you know how to solve this recurrence relation?


问:
Polynomial and complex numbers.
Let $ f (x) $ be a polynomial with real coefficients.  Show that the remainder of dividing $ f (x) $ by $ x² + 2x + 2 $ is given by $$ r (x) = \alpha x + \beta $$ where $ \alpha = \Im(f (1 + i  )) $ y $ \beta = \Re (f (1 + i)) - \Im (f (i + 1)) $.

I assumed that the remainder of $ f (x) $ when dividing by $ x ^ 2 + 2x + 2 $ is $ r (x) = \alpha x + \beta $.  And using the roots of $ x ^ 2 + 2x + 2 $ I am trying to find linear equations in which the unknowns would have to be $ \alpha $ and $ \beta $.  I don't know how to continue to find those Equations.

Title: Polynomial and complex numbers.

Body:
I think that the core of your problem is to relate $f(1+i)$, $f(-1-i)$ and $f(-1+i)$.

My attempt:

$$f(1+i)$$
$$f(-1-i) = f(-(1+i))$$
$$f(-1+i) = f(-(1+i)^*)$$

It's important to note that

$$
\begin{array}{lll}
(1+i)^1 = (1+i)\\
(1+i)^2 = (2i)\\
(1+i)^3 = -2+2i=-2(1-i)=-2(1+i)^*\\
(1+i)^4 = (-4)\\
(1+i)^5 = -4-4i=-4(1+i)\\
(1+i)^6 = -8i=(2i)^3\\
(1+i)^7 = 8-8i=8(1-i)=8(1+i)^*\\
(1+i)^8 = 16=(-4)^2\\
(1+i)^9 = 16+16i=16(1+i)\\
(1+i)^{10} = 32i=(2i)^5\\
(1+i)^{11} = -32+32i=-32(1-i)=-32(1+i)^*\\
(1+i)^{12} = -64=(-4)^3\\
.\\
.\\
.\\
\end{array}
$$

$$
\begin{array}{lll}
(1+i)^n = (-1)^{\frac{n+7}{4}}(+2)^{\frac{n-1}{2}}(1+i) \text{ for } n = 1,5,9,...\\
(1+i)^n = (-1)^{\frac{n+6}{4}}(+2)^{\frac{n}{2}}\,i \text{ for } n = 2,6,10,...\\
(1+i)^n = (-1)^{\frac{n+1}{4}}(+2)^{\frac{n-1}{2}}(1+i)^* \text{ for } n = 3,7,11,...\\
(1+i)^n = (-2)^{\frac{n}{2}} \text{ for } n = 4,8,12,...\\
\end{array}
$$

We know that $f(x)$ is a polynomial with real coefficients

$$
\begin{align}
f(x) & = a_0 + \sum_{n=1,5,9,...} a_n \, x^n + \sum_{n=2,6,10,...} a_n \, x^n + \sum_{n=3,7,11,...} a_n \, x^n + \sum_{n=4,8,12,...} a_n \, x^n
\end{align}
$$

$$
\begin{align}
f(-x) & = a_0 - \sum_{n=1,5,9,...} a_n \, x^n + \sum_{n=2,6,10,...} a_n \, x^n - \sum_{n=3,7,11,...} a_n \, x^n + \sum_{n=4,8,12,...} a_n \, x^n
\end{align}
$$

$$
\begin{align}
f(-x^*) = [f(-x)]^* & = a_0 - \sum_{n=1,5,9,...} a_n \, (x^n)^* + \sum_{n=2,6,10,...} a_n \, (x^n)^* - \sum_{n=3,7,11,...} a_n \, (x^n)^* + \sum_{n=4,8,12,...} a_n \, (x^n)^*
\end{align}
$$

Maybe there is sufficient information now.


问:
Proving $f(f^{-1}(x))=f^{-1}(f(x))$
I have to prove $f^{-1}(f(x))=f(f^{-1}(x))$ for the following function :
$$f(x)=-\sqrt{x-1}$$
But I got $-x$ on the one side and $x$ on the other.
What am I missing here?

答:
Actually,
$$f(f^{-1}(x))=f^{-1}(f(x))$$
doesn't make sense. The domain and the range of $f(x)$ are $[1,\infty)$ and $(-\infty,0]$, respectively. So its inverse is
$$ f^{-1}(x)=x^2+1 $$
whose domain and range are $(-\infty,0]$ and $[1,\infty)$. For $x\in(-\infty,0]$, $\sqrt{x^2}=-x$ and hence
$$ f(f^{-1}(x))=f(x^2+1)=-\sqrt{x^2}=-(-x)=x. $$
For $x\in[1,\infty)$,
$$ f^{-1}(f(x))=f^{-1}(-\sqrt{x-1})=(-\sqrt{x-1})^2=x. $$

Title: Proving $f(f^{-1}(x))=f^{-1}(f(x))$


答:
$f(x)=-\sqrt{x-1}\;\;,\qquad f:\left[1,+\infty\right[\to\left]-\infty,0\right]\;.$

In order to obtain the inverse function $\;f^{-1}(x)\;,\;$ we have to solve the following equation for $\;x\in\left[1,+\infty\right[\;$ and then substitute $\;x\;$ for $\;f^{-1}(x)\;$ and $\;y\;$ for $\;x\;.$

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

$y^2=x-1$

$x=y^2+1$

Hence

$f^{-1}(x)=x^2+1\;\;,\qquad f^{-1}:\left]-\infty,0\right]\to\left[1,+\infty\right[\;.$

Consequently we get that

\begin{align}
f^{-1}\left(f(x)\right)&=f^2(x)+1=\left(-\sqrt{x-1}\right)^2+1=\left|x-1\right|+1=\\
&=x-1+1=x
\end{align}

for all $\;x\in\left[1,+\infty\right[\;.$

Moreover

\begin{align}
f\left(f^{-1}(x)\right)&=-\sqrt{f^{-1}(x)-1}=-\sqrt{x^2+1-1}=-\sqrt{x^2}=\\
&=-|x|=x
\end{align}

for all $\;x\in\left]-\infty,0\right]\;.$

Therefore

$f^{-1}\left(f(x)\right)=x\;\;$ for all $\;x\in\left[1,+\infty\right[\;,$

$f\left(f^{-1}(x)\right)=x\;\;$ for all $\;x\in\left]-\infty,0\right]\;.$

Nevertheless there does not exist any $\;x\in\left]-\infty,+\infty\right[\;$ such that $\;f^{-1}\left(f(x)\right)=f\left(f^{-1}(x)\right)\;$ indeed $\;f^{-1}\left(f(x)\right)\;$ is defined for $\;x\ge1\;,\;$ where as $\;f\left(f^{-1}(x)\right)\;$ is defined for $\;x\le0\;.$

Title: Proving $f(f^{-1}(x))=f^{-1}(f(x))$

Body:
I have to prove $f^{-1}(f(x))=f(f^{-1}(x))$ for the following function :
$$f(x)=-\sqrt{x-1}$$
But I got $-x$ on the one side and $x$ on the other.
What am I missing here?


问:
How to verify the results of the Present Value of a Growing Annuity Formula?
I am working on building a financial planning tool and need to verify the results of the Present Value of A Growing Annuity formula. I'm using the following formula:

$PV = Pmt x (1 - (1 + g)n x (1 + i)-n ) / (i - g)$

Where:

PMT = Periodic Payment

g = periodic growth rate of the payment

i = periodic growth rate of the principal

n = number of periods

While this formula itself works as expected, when I construct a table to show the change for each period, the numbers never balance. I've tried several approaches:

(beginning of period principal x (1+i)) - (periodic payment x (1+g))

(beginning of period principal - (periodic payment x (1+g))) x (1+i)

(beginning of period principal - periodic payment) x (1+(i-g))

and other less memorable approaches

Here's a link to a table I've built with some of these approaches. I've used sample data of:

PMT = 48,000

g = 3.22%

i = 7.8%

n = 50

This produces a value of $928,462.67.

When I try to reproduce the periodic steps of this formula in Excel, the principal is exhausted in period 33. What I want is a table that can show annual or monthly payments that align to this result, OR I would appreciate guidance on how to break down the formula to recreate the steps myself so that I can show the periodic payment and principal by period.

答: The problem with my approach above was an issue of sequencing the steps. Since the payment is received at the end of the first year, the principal should grow for the first period while the payment should not. Once this is correctly structured, the PVGA formula matches the deconstructed results. You can see this laid out in the file I shared above.</p>


问:
How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$?
I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert}$ and show it is equal to $\frac{1}{\sqrt{x^2+r^2}}$, but I ended up with $$ \frac{d}{dx} \ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert} = \dots = \frac{1}{x + \sqrt{x^2+r^2}}\left(1+\frac{1}{\sqrt{x^2+r^2}}\right)$$ and I do not see how it can be simplified to $\frac{1}{\sqrt{x^2+r^2}}$.
Any help would be appreciated (even involving complex calculus, if needed).

答:
HINT

What about the substitution $x = r\sinh(u)$? In such case, one has that
\begin{align*}
\int\frac{1}{\sqrt{x^{2}+r^{2}}}\mathrm{d}x & = \int\frac{r\cosh(u)}{\sqrt{r^{2}\sinh^{2}(u) + r^{2}}}\mathrm{d}u = \int1\mathrm{d}u = u + c
\end{align*}

Can you take it from here?

EDIT

Since the $\sinh$ function is bijective, there is always a solution to
\begin{align*}
z = \sinh(w) = \frac{e^{w} - e^{-w}}{2} & \Longleftrightarrow (e^{w})^{2}  - 2ze^{w} - 1 = 0\\\\
& \Longleftrightarrow e^{w} = \frac{2z + 2\sqrt{z^{2}+1}}{2}\\\\
& \Longleftrightarrow w = \ln(z + \sqrt{z^{2}+1})
\end{align*}
At your case, we have that
\begin{align*}
\int\frac{\mathrm{d}x}{\sqrt{x^{2}+r^{2}}} = \sinh^{-1}\left(\frac{x}{r}\right) + c = \ln\left(\frac{x + \sqrt{x^{2} + r^{2}}}{r}\right) + c = \ln\left(x + \sqrt{x^{2} + r^{2}}\right) + C
\end{align*}
and we are done.

Hopefully this helps!


问:
basic claim about composite function limit.
is it always true that if $$\lim\limits_{x \to a} f(x)=b$$
then $$\lim\limits_{x \to a} g(f(x))= \lim\limits_{f(x) \to b}g(f(x))$$
given there is no variables outside of f(x) inside g(x)?

is it acceptable saying "the condition ${x \to a}$ is equivalent to ${f(x) \to b}$"?

whether its true or false please explain, simple words if possible,

thanks a lot, hope you understand my english and math language /=

答:
Firstly let me state definition of limit: If for $\forall \epsilon >0, \exists \delta > 0$, such that for $\forall x, 0<|x-a|<\delta $ holds $|f(x)-b|<\epsilon $, then we say, that $b$ is limit of $f$ in point $a$ and wrote $\lim\limits_{x \to a} f(x)=b$. Sometimes as shorthand for this definition is used sentence "$f$ tends to $b$, when $x$ tends to $a$" or "$f(x) \to b$ for $x \to a$", but it is good to remember, that such type of sentences are only designation for definition.

Set $U_\delta(a)=\{x: 0<|x-a|<\delta  \}$ is called punctured neighbourhood or deleted neighbourhood of $a$.

There is well known theorem for limit of composition:

Suppose exists $\lim\limits_{x \to a} f(x)=b$ and $\lim\limits_{y \to b} F(y)$ and suppose that for $f$ in some punctured neighbourhood of $a$ holds $f(x)\ne b$. Then exists limit of composition $F \circ f$ and holds $\lim\limits_{y \to b} F(y)=\lim\limits_{x \to a} F(f(x))$.

With some extra conditions there is possible to state, also, equivalence of limits: suppose $f$ is invertible or one-to-one function on some punctured neighbourhood of $a$ and mapped it on some punctured neighbourhood of $b$, so, that $f(x) \ne b$ and $f^{-1}(y) \ne a$ and $\lim\limits_{x \to a} f(x)=b$ and $\lim\limits_{y \to b} f^{-1}(y)=a$. Then for $F$ defined on punctured neighbourhood of $b$ existence of limit $\lim\limits_{y \to b} F(y)$ is equivalent to  existence of limit $\lim\limits_{x \to a} F(f(x))$ and they are equal in case of existence.


问:
Correlated noise in Kalman filter
Assume a standard state-space setting:
\begin{align}
x_{t+1} &= Ax_t + w_t\\
y_t &= Cx_t+v_t,
\end{align}
where the noise sequences are i.i.d. Gaussian with $w_i\sim N(0,W)$ and $v_i\sim N(0,V)$.

I am interested in the pre- and post- error covariance matrices recursions when there is a correlation $\mathbb{E}[w_tv_t^\top]= S$.

Specifically, using the notation
\begin{align}
\Sigma_{t|t} &= \mathbf{cov}(x_{t}|y^t)\\
\Sigma_{t+1|t} &= \mathbf{cov}(x_{t+1}|y^t),
\end{align}
how does the standard propagations:
\begin{align}
\Sigma_{t|t}&= \Sigma_{t|t-1} - \Sigma_{t|t-1} C^\top (C\Sigma_{t|t-1}C^\top+V)^{-1}C \Sigma_{t|t-1}\\
\Sigma_{t+1|t} &= A\Sigma_{t|t}A^\top+W
\end{align}
change. Thanks.

答:
I will use your notation, and also define the conditional mean $\hat x_{t|s} = \mathbf{E}(x_t \mid y^{s})$.

The way you wrote the KF updates in your question have them split in two steps, the "measurement update" $(t|t-1) \to (t|t)$ and the "time update" $(t|t) \to (t+1|t)$.

In general, you can just combine the two updates into a single update $(t|t-1) \to (t+1|t)$. Here is what you get for the case $S\neq 0$ when you do that; these are the standard KF update equations.

$$
\begin{aligned}
\Sigma_{t+1|t} &= \left(A \Sigma_{t|t-1} A^\top + W\right) - \left( A \Sigma_{t|t-1} C^\top + S \right) \left( C \Sigma_{t|t-1} C^\top + V \right)^{-1} \left( C \Sigma_{t|t-1}A^\top + S^\top\right) \\
L_t &= \left( A \Sigma_{t|t-1} C^\top + S \right) \left( C \Sigma_{t|t-1} C^\top + V \right)^{-1} \\
\hat x_{t+1|t} &= A \hat x_{t|t-1} + L_t (y_t - C\hat x_{t|t-1})
\end{aligned}
$$

If you want to look at the measurement and time updates separately, that's also possible. The measurement update when $S\neq 0$ is the same as in the case $S=0$. For completeness sake, here is the measurement update:

\begin{align}
\hat x_{t|t} &= \hat x_{t|t-1} + \Sigma_{t|t-1} C^\top (C\Sigma_{t|t-1}C^\top +V)^{-1}(y_t - C \hat x_{t|t-1})\\
\Sigma_{t|t} &= \Sigma_{t|t-1} - \Sigma_{t|t-1} C^\top (C\Sigma_{t|t-1}C^\top+V)^{-1}C \Sigma_{t|t-1}
\end{align}

The time update is different when $S\neq 0$. Actually, you can't write $(t+1|t)$ neatly in terms of just $(t|t)$. I'll try to give some insight. When $S=0$, you can take the state update equation $x_{t+1} = Ax_t + w_t$ and take conditional expectations given $(y_0,\dots,y_t)$ and directly obtain $\hat x_{t+1|t} = A \hat x_{t|t}$ and $\Sigma_{t+1|t} = A\Sigma_{t|t}A^\top + W$. This works because $x_t$ and $w_t$ are independent. But when $S\neq 0$, although $x_t$ and $w_t$ are still independent by assumption, they become correlated when you condition given $(y_0,\dots,y_t)$, since $y_t$ depends on $x_t$ and $v_t$ (which is correlated with $w_t$), i.e. $y_t$ is correlated with both $x_t$ and $w_t$. Therefore, your terms involving $(w_t|y^t)$ will need to be expressed in terms of $(w_t|y^{t-1})$, which is simply $w_t$, but this introduces dependency on $(x_t|y^{t-1})$, which we were trying to avoid in the first place. Working through this, you get:

\begin{align}
\hat x_{t+1|t} &= A\hat x_{t|t} + S (C\Sigma_{t|t-1}C^\top +V)^{-1}(y_t - C \hat x_{t|t-1})\\
\Sigma_{t+1|t} &= A\Sigma_{t|t}A^\top + W - \left( A \Sigma_{t|t-1} C^\top + S \right) \left( C \Sigma_{t|t-1} C^\top + V \right)^{-1} \left( C \Sigma_{t|t-1}A^\top + S^\top\right)\\
&\hspace{1cm}+A\Sigma_{t|t-1} C^\top (C\Sigma_{t|t-1}C^\top+V)^{-1}C \Sigma_{t|t-1}A^\top
\end{align}

Note that combining the measurement and time updates recovers the standard KF update equations at the top of the post.

Title: Correlated noise in Kalman filter


问:
Prove that if a sequence is a Cauchy sequence, then all of its subsequences converge.
Suppose that we do not know that every real Cauchy sequence converges. Since we know that if every subsequential limit of a sequence of real numbers converges to the same number, then the sequence itself converges to that number, let us try to get from a Cauchy sequence to this result.

Question: Suppose $(x_n)$ is a Cauchy sequence of real numbers. Show that every subsequential limit converges to the same real number.

I have been struggling to come up with a solid proof for this question. I keep getting stuck after applying the fact that a Cauchy sequence is bounded and therefore has a convergent subsequence. How in the world do I show that all the subsequences converge to the same number?

Thank you in advance.

答:
Part $1$: Convergent subsequence

From the definition of Cauchy Sequence, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_N-x_n|<1$. This then implies that for all $n\geq N$ we have

$$x_N-1<x_n<x_N+1$$

and therefore for all $n\in\mathbb{N}$

$$\min\{x_1,x_2,...,x_{N-1},x_N-1\}\leq x_n \leq \max\{x_1,x_2,...,x_{N-1},x_N+1\}$$

This implies that $x_n$ is a bounded sequence. Then by the Bolzano–Weierstrass theorem we may conclude that $x_n$ has at least one convergent subsequence. Let us denote this subsequence by $x_{n_k}$ and let its limit be

$$\lim_{k\to\infty} x_{n_k}=L$$

Part $2$: Entire sequence converges

Now, let $\epsilon>0$ be given. By definition, there exists $N_1\in\mathbb{N}$ such that $k\geq N$ implies $|x_{n_k}-L|<\frac{\epsilon}{2}$. Additionally, there exists $N_2\in\mathbb{N}$ such that $N_2\leq n\leq m$ implies $|x_n-x_m|<\frac{\epsilon}{2}$. Now, choose $N=\max\{n_{N_1},N_2\}$ and let $l$ be the smallest natural such that $n_l\geq N$ and $x_{n_l}$ is a member of the convergent subsequence. Then for all $n\geq N$ we have

$$\bigg||x_n-L|-|x_{n_l}-L|\bigg|\leq |(x_n-L)-(x_{n_l}-L)|=|x_n-x_{n_l}|<\frac{\epsilon}{2}$$

(the first step in this equation is the reverse triangle inequality). This then implies

$$0\leq  |x_n-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

(since $|x_{n_l}-L|<\frac{\epsilon}{2}$). We conclude that

$$\lim_{n\to\infty}x_n=L$$

Part 3: Convergent subsequences

Let $\epsilon>0$ be given. By defintion there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_n-L|<\epsilon$. Now, let $x_{n_k}$ be any subsequence of $x_n$ and note that $n_k\geq k$. Therefore, for all $k\geq N$ we have $n_k\geq k\geq N$. This implies

$$|x_{n_k}-L|<\epsilon$$

We conclude

$$\lim_{k\to\infty}x_{n_k}=L$$


问:
Prove that if every family $\mathcal{C}$ of ideals in $R$ has a maximal element then every ideal is finitely generated.
This is what I want to prove:

Every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ implies that every ideal in $R$ is finitely generated.

I found this answer here Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.

Proof. We take an ideal $I$ of $R$ and we define the set
$$F:=\{J: J\text{ finitely generated ideal of } R  \text{ and } J\subseteq   I\}.$$
We can see that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). So, from hypothesis, there exists a maximal element $M\in F$.

We want to show that $I=M$.

Obviously, $M\subseteq I$, because $M\in F$.
We suppose now that $M\subsetneq I$. Then, there exists an element $a\in I\backslash M\iff a\in I$, with $a\notin M$.

We consider the ideal
$$J:=M+\langle a\rangle.$$
Then, if $M=\langle m_1,...,m_n \rangle$, we have $J=M+\langle a\rangle=\langle m_1,..., m_n,a \rangle \implies J\in F$

But I am not fully understanding the idea of the solution. specifically,

1- Why we took $J\subseteq I$ in the set we defined? i.e. why we need all the f.g. ideals to be inside our arbitrary ideal $I$? Are we going to use Zorn's Lemma in our proof?

2- why we need to show that $M= I$?

3- why we are considering the case $M \subsetneq I$ in the proof?

4- Why $M+\langle a\rangle=\langle m_1,..., m_n,a \rangle$?

could anyone explain this to me please?

Zorn's Lemma which states that:

A partially ordered set in which every chain (i.e.totally ordered subset) is upper bounded has a maximal element.

答:
You are not using Zorn's Lemma. In fact, no choice is necessary for this proof, since we already start by assuming that a maximal element exists, and choice is necessary exactly when trying to prove the existence of maximal elements.

The idea behind the proof is simple, but it might be clearer if we prove the following theorem of Tarski:

$A$ is finite if and only if every non empty collection of subsets of $A$ has a maximal element.

Proof. One direction is trivial, if $A$ is finite. In the other direction, let $J$ be the set of finite subsets of $A$, and let $M$ be a maximal element in $J$. If $M\neq A$, then there is some $a\in A\setminus M$, and then $M\cup\{a\}$ is a finite subset of $A$ which is a proper superset of $M$. But $M$ was maximal, so no such $a$ can exist, so $A=M$. $\square$

The idea is to look at the set of all finitely generated subobjects, and are that if the maximal, which exists by assumption, is not equal to the whole object, then we derive a contradiction.


答:
Your calculation of the derivative is incorrect.

Note $$\begin{align}
\frac{d}{dx}\left[ x + \sqrt{x^2 + r^2} \right]
&= 1 + \frac{1}{2}\left(x^2 + r^2 \right)^{-1/2} \cdot \frac{d}{dx}\left[x^2 + r^2\right] \\
&= 1 + \frac{1}{2 \sqrt{x^2 + r^2}} \cdot 2x \\
&= 1 + \frac{x}{\sqrt{x^2 + r^2}}. \end{align}$$  You are missing the $x$ in the numerator of the second term.

Then we obtain $$\begin{align}
\frac{d}{dx}\left[\log \left| x + \sqrt{x^2 + r^2} \right| \right]
&= \frac{1}{x + \sqrt{x^2 + r^2}} \left(1 + \frac{x}{\sqrt{x^2 + r^2}}\right) \\
&= \frac{1}{x + \sqrt{x^2 + r^2}} \left( \frac{\sqrt{x^2 + r^2} + x}{\sqrt{x^2 + r^2}} \right) \\
&= \frac{1}{\sqrt{x^2 + r^2}},
\end{align}$$ as claimed.

In order to perform the integration in the first place, there are a variety of substitutions that can be performed.  One is via the tangent identity $$\tan^2 \theta + 1 = \sec^2 \theta,$$ hence the choice $$x = r \tan \theta, \quad dx = r \sec^2 \theta \, d\theta$$ yields

$$\int \frac{1}{\sqrt{x^2 + r^2}} \, dx = \int \frac{1}{\sqrt{r^2 (\tan^2 \theta + 1)}} \cdot r \sec^2 \theta \, d\theta = \int \frac{\sec^2 \theta}{\sec \theta} \, d\theta = \int \sec \theta \, d\theta.$$

Now we perform the algebraic manipulation

$$\sec \theta = \frac{\sec \theta (\sec \theta + \tan \theta)}{\sec \theta + \tan \theta} = \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta}.$$  Then the derivative of the denominator equals the numerator; i.e. $$\frac{d}{d\theta}[ \sec \theta + \tan \theta] = \sec \theta \tan \theta + \sec^2 \theta,$$ from which we may choose the substitution $u = \sec \theta + \tan \theta$ and obtain $$\begin{align}
\int \sec \theta \, d\theta &= \int \frac{1}{u} \, du \\ &= \log |u| + C \\
&= \log |\sec \theta + \tan \theta| + C \\
&= \log \left| \sqrt{(x/r)^2 + 1} + x/r \right| + C \\
&= \log \left| \frac{x + \sqrt{x^2 + r^2}}{r} \right| + C \\
&= \log \left| x + \sqrt{x^2 + r^2} \right| - \log |r| + C \\
&= \log \left| x + \sqrt{x^2 + r^2} \right| + C,
\end{align}$$
since $\log |r|$ is a constant with respect to $x$ and may be absorbed into $C$.

Title: How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$?

Body:
I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert}$ and show it is equal to $\frac{1}{\sqrt{x^2+r^2}}$, but I ended up with $$ \frac{d}{dx} \ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert} = \dots = \frac{1}{x + \sqrt{x^2+r^2}}\left(1+\frac{1}{\sqrt{x^2+r^2}}\right)$$ and I do not see how it can be simplified to $\frac{1}{\sqrt{x^2+r^2}}$.
Any help would be appreciated (even involving complex calculus, if needed).


问:
Representations of a C*-algebra of bounded Borel functions
Let $X$ be a compact Hausdorff space. Let $B(X)$ be the C*-algebra of bounded Borel measureable functions on $X$ (under the supremum norm). I am curious whether the (say unital) $*$-representations of $B(X)$ are completely classified. One way to get these is from spectral measures on $X$. See IX.1.12 in A Course in Functional Analysis by John B. Conway. I'm not sure whether this is all of them, though. What this all boils down to is a question about the following continuity condition.

Let $\pi : B(X) \to B(H)$ be a unital $*$-representation of $B(X)$ on a Hilbert space $H$. Suppose that $E_1,E_2,\ldots$ is a countable pairwise disjoint collection of Borel measurable subsets of $X$. Let $E = \bigcup_{n=1}^\infty E_n$. Is it necessarily true that $$\pi(\chi_E) = \sum_{n=1}^\infty \pi(\chi_{E_n})$$ with the sum converging in the strong operator topology?

答:
Not all representations of $B(X)$ satisfy the continuity condition mentioned by the OP.

To exhibit an example, let us
first notice that $B(X)$ is a unital  commutative C*-algebra, and hence it is isomorphic to $C(K)$, where $K$ is its
spectrum.

Given any Borel set $E\subseteq X$, we have that the characteristic function $\chi_E$ is an idempotent element
in $B(X)$.   So, once $\chi_E$ is  viewed as an element of $C(K)$ via Gelfand's transform,  it is necessarily given by   the characteristic function of some
clopen subset of $K$, which we will denote by $\bar E$.

To make things more concrete, let us suppose that $X=[0,1]$, although similar counter-examples can be given in any
infinite compact space.

Letting $E_n=(0,1/n)$, we have that
$$
  \chi_{E_n}\chi_{E_{n+1}}=\chi_{E_{n+1}},
  $$
so
$$
  \chi_{\bar E_n}\chi_{\bar E_{n+1}}=\chi_{\bar E_{n+1}},
  $$
which means that $\bar E_{n+1}\subseteq \bar E_n$, for every $n$.  Since the $\bar E_n$ are nonempty and compact, their
intersection is nonempty, so we may choose a point $\omega $ lying in every $\bar E_n$.

We then consider the character $\pi $ of $C(K)$ given by point evaluation at $\omega $, namely
$$
  \pi :g\in C(K)\mapsto g(\omega )\in \mathbb C,
  $$
which we view as a *-representation by identifying $\mathbb C$ with the algebra of all operators on a one-dimensional
Hilbert space.

Viewing $\pi $  as a representation of $B(X)$ by composing it with the Gelfand transform, we then have that
$\pi (\chi_{E_n})=1$, for every $n$.

We next claim that the alleged continuity condition fails for $\pi $,  regarding the  following pairwise disjoint union of
measurable sets:
$$
  E_1 = \bigcup_{n=1}^\infty  E_n\setminus E_{n+1}.
  $$
Indeed,
$$
  \pi \left(\chi_{E_n\setminus E_{n+1}}\right) =
  \pi \left(\chi_{E_n}(1-\chi_{E_{n+1}})\right) =
  \pi \left(\chi_{E_n}\right) -   \pi \left(\chi_{E_{n+1}}\right) = 1 - 1 = 0,
  $$
while clearly   $\pi \left(\chi_{E_1}\right) = 1$.

Combining IX.1.12 and IX.1.13 in Conway's book,  one has that every *-representation $\pi $ of $C(X)$ on $H$ extends to a
*-representation $\tilde\pi $ of $B(X)$ on $H$, which moreover satisfies the following continuity property:  if
$\{f_n\}_n$ is a uniformly bounded sequence in $B(X)$, converging pointwise to some $f$,  then $\tilde\pi  (f_n)$ converges
strongly to $\tilde\pi  (f)$ (see Lemma 3.5.5  in Sunder's "Functional Analysis - Spectral Theory").
In particular $\tilde\pi $ satisfies the desired continuity condition.  The counter-example described therefore shows that
not all
representations of $B(X)$  arise as the extension of a
representation of $C(X)$, as above.


问:
SVD Singular Value Issue
I am having an issue with a SVD problem and would like advice. The matrices given are

$$
  A= \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]\ \ \
  A^TA= \left[ \begin{matrix}
    10 & 20\\
    20 & 40 \\
    \end{matrix}\right]\ \ \
  AA^T= \left[ \begin{matrix}
    5 & 15\\
    15 & 45 \\
    \end{matrix}\right]
$$

I initially determined my $λ$ values to be $λ_1=50$ and $λ_2=0$ by computing the following:

$$
det(A^TA-λI)=0\\
\begin{vmatrix}
    10-λ & 20\\
    20 & 40-λ \\
    \end{vmatrix}\\
(10-λ)(40-λ)-20^2=0\\
400-50λ+λ^2-400=0\\
λ^2-50λ=0\\
λ(λ-50)=0
$$

This in turn means that my singular values will be  $\sigma_1=\sqrt {50}$ and $\sigma_2=0$, and this is where I run into my issue. When I am trying to solve for the orthonormal set with $u_1$ and $u_2$, I was told to use the equation $u_n=\frac{Av_n}{\sigma_n}$. I cannot compute $u_2$ because that would make the denominator $0$, and the fraction would become undefined.

I discussed with my professor and they said I should try to calculate $u_2$ using the eigenvectors from $AA^T$, but $AA^T$ and $A^TA$ have the same eigenvalues. I thought maybe they meant to compute the $u$ for the $\sigma=\sqrt {50}$, but this is not the same answer as the one the book gives. The answers for this and Wolfram Alpha show that the SVD should be

$$
  \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]=
\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 & -3\\
    3 & 1 \\
    \end{matrix}\right]
  \left[ \begin{matrix}
    \sqrt{50} & 0\\
    0 & 0 \\
    \end{matrix}\right]
\frac{1}{\sqrt{5}}
  \left[ \begin{matrix}
    1 & -2\\
    2 & -1 \\
    \end{matrix}\right]
$$

I have gotten that $u_1=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 \\
    3 \\
    \end{matrix}\right]$, but I do not know how to get $u_2=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    -3 \\
    1 \\
    \end{matrix}\right]$. Any help would be appreciated.

答:
Your professor was right.

Have you tried the definition of an eigenvector?

$$(AA^T)v=\lambda v$$

$$
\left[ \begin{matrix}
    5 & 15\\
    15 & 45 \\
\end{matrix}\right]
\left[ \begin{matrix}
    v_1\\
    v_2\\
\end{matrix}\right]
=\left[ \begin{matrix}
    0\\
    0\\
\end{matrix}\right]
$$

$$
\begin{align}
& 5v_1+15v_2=0\\
& 15v_1+45v_2=3(5v_1+15v_2)=0
\end{align}
$$

So you have one degree of freedom to choose $v_2 = 1/\sqrt(10)$ and obtain $v_1 = -15/5 \, v_2 = -3/\sqrt(10)$.

Title: SVD Singular Value Issue


问:
Fourier Transform of $\log{|x|} \text{sign}(x)$
Is there a way to define the Fourier transform of $\log{|x|} \text{sign}(x)$? Naively I thought that one could define it through a convolution, namely
\begin{equation}
\mathcal{F}(\log{|t|}\text{sign}(t))=\mathcal{F}(\log{|t|})*\mathcal{F}(\text{sign}(t))\,.
\end{equation}
However, we have that the Fourier transform of each distribution is given by
\begin{equation}
\mathcal{F}\left(\text{sgn}(t) \right)=\mathcal{P}\frac{2}{i \omega}\,,
\end{equation}
\begin{equation}
\mathcal{F}\left(\log{|t|}\right)=-\mathcal{P} \frac{\pi}{|\omega|}-\sqrt{2\pi}\gamma \delta(\omega)\,,
\end{equation}
where $\mathcal{P}$ denotes the Cauchy principal value. That means that we need to compute the convolution of two Cauchy principal value distributions, namely
\begin{equation}
\mathcal{P}\frac{2}{i \omega}*\mathcal{P} \frac{\pi}{|\omega|}\,.
\end{equation}
However, as far as I could understand, one can only define the convolution of distributions if one of them has compact support. That's the the case for $\delta(\omega)$ but neither of this seems to have compact support so I guess their convolution cannot be defined. Is it because  $\log{|x|} \text{sign}(x)$ is not a tempered distribution. How does one tell if some distribution is tempered or not in general? I do not know much about distribution theory... Just wanted to define this Fourier transform.

答:
9 months late to the party, but I've been deriving Fourier transforms all day and I stumbled upon this and it's begging for closure.

To obtain the distributional form of the Fourier transform one can to consider a more general integral that converges absolutely ($\epsilon>0$):

$$F_\epsilon(\omega)=\int_{-\infty}^{\infty}\log|x|\text{sgn}|x|e^{-\epsilon|x|-i\omega x}dx=\int_{0}^{\infty}\log x~e^{-(\epsilon+i\omega) x}dx-\int_{0}^{\infty}\log x~e^{-(\epsilon-i\omega) x}dx$$

It is known and easy to derive that

$$\int_0^{\infty}\log x ~e^{-sx}dx=\frac{-\gamma-\log s}{s}~~,~ ~\text{Re}(s)>0$$

With $\Theta=\arg(\epsilon+i\omega)\in(-\pi,\pi)$ we can write

$$F_\epsilon(\omega)=2i\gamma\frac{\omega}{\epsilon^2+\omega^2}+2i\frac{\log(\sqrt{\epsilon^2+\omega^2})}{\sqrt{\epsilon^2+\omega^2}}\sin\Theta+2i\frac{\Theta\cos\Theta}{\sqrt{\epsilon^2+\omega^2}}$$

Now in the limit $\epsilon\to 0^+$, $\Theta\to \frac{\pi}{2}\text{sgn}(\omega)$ and wherever we get $\epsilon$ as a cutoff to a pole we rename the function as the "principal value" equivalent of the function we would get with $\epsilon=0$. With all these understood, we obtain the final result
$$F[\log|x|\text{sgn}(x)]=2i\gamma ~\mathcal{P}\frac{1}{\omega}+2i~\mathcal{P}\frac{\log|\omega|}{\omega}$$

which is odd, as prescribed above. In fact I think that with careful treatment of the convolution integrals (everything is well defined since there are principal values regulating the poles and furthermore the integrals are elementary) you would be able to compute the Fourier transform in that way too. However, I will leave this as an exercise to you. Cheers!


问:
Do you know of a reference for Aitken's graphical construction of points on a Parabola?
I have been using Aitken's construction in my work. I have seen several references to the construction but never a direct reference to an article in a Journal.
I am preparing a comprehensive report to show the direct equivalence of the construction to the second-order Lagrange polynomial.
I will be including an extension of the construction that generates the local slopes at points on the parabola.</p>

答:
I was able to find my answer.
https://www.cambridge.org/core/journals/proceedings-of-the-edinburgh-mathematical-society/article/on-interpolation-by-iteration-of-proportional-parts-without-the-use-of-differences/482BBF2A12045CB996628BE8C5B42FFC

Aitken, A. (1932). On Interpolation by Iteration of Proportional Parts, without the Use of Differences. Proceedings of the Edinburgh Mathematical Society, 3(1), 56-76. doi:10.1017/S0013091500013808

Regards
Alfred
Here are more details about Aitken's Construction.
https://www.dropbox.com/s/hhtos2l85wa0qfr/Aitken%E2%80%99s%20Graphical%20Construction%20for%20a%20Parabola%20V4.pdf?dl=0
https://www.dropbox.com/s/iv6d52uymu3xslv/Step%20by%20Step%20%20Aitken%20Construction.pdf?dl=0


问:
Average of dice roll with chance of rerolling
My question regards a simple dice game. You roll one six sided die and win, say, a dollar for each dot on the roll. e.g. five dollars for rolling a five, two dollars for rolling a two, etc.
You are given the option to reroll once.

Letting $X$ be the outcome of the first roll, and $Y$ be the outcome of the (potential) second roll, consider the strategy $W$
\begin{equation}
  W=\begin{cases}
  X&\text{if $X\geq 4$}\\
  Y&\text{if $X< 4$}
  \end{cases}
 \end{equation}
I.e. if you roll 4 or higher on first try you keep it, if you roll lower you try again.
What is the average winnings using this strategy?

Solution

My issue here is, really, that I don't have more mathematically stringent way to prove this, and I got the answer rather intuitively.

I consider two "macroscopic" outcomes;

a) We roll a 4, 5, or 6 in the first attempt and stop.

b) We take a second attempt, and roll either 1, 2, 3, 4, 5, or 6.

The average winnings is the average of these two options, which in itself is the average of a roll of 4/5/6 and the average of a roll of 1/2/3/4/5/6. That is

\begin{equation}
E[W] = (4+5+6)\frac{1}{6}+(1+2+3+4+5+6)\frac{1}{2} = \frac{15}{6} + \frac{3.5}{2} = 4.25
\end{equation}

Which indeed is a higher expectation than winning than rolling the die just once ($\mu=3.5$)

I feel like my argumentation is mostly just intuition, and could benefit from a more mathematically sound description.

Body:
First note that $$E[Y]=\sum_{k=1}^6 k P[Y=k] = \frac{1}{6}\sum_{k=1}^6 k = \frac{7}{2}.$$
Now apply the law of total expectation:
\begin{align}
E[W] &= E[E[W|X]] \\
&= \sum_{k=1}^6 E[W|X=k] P[X=k] \\
&= \frac{1}{6} \sum_{k=1}^6 E[W|X=k] \\
&= \frac{1}{6} \sum_{k=4}^6 E[W|X=k] + \frac{1}{6} \sum_{k=1}^3 E[W|X=k] \\
&= \frac{1}{6} \sum_{k=4}^6 E[X|X=k] + \frac{1}{6} \sum_{k=1}^3 E[Y|X=k] \\
&= \frac{1}{6} \sum_{k=4}^6 k + \frac{1}{6} \sum_{k=1}^3 E[Y] \\
&= \frac{15}{6} + \frac{3}{6}\cdot \frac{7}{2} \\
&= \frac{17}{4}
\end{align}


问:
Level set that is both closed and open.
Let $M$ be an abstract manifold, and let $f \in C^{\infty}(M)$. Assume that $df = 0$.

Prove that each level set $\{p \in M | f(p) = c\}$ is both closed and open. Conclude that $f$ is constant on each component of $M$, and show that the dimension of the kernel of $d$ in $C^{\infty}(M)$ is the number of connected components (which is either a finite number or infinite).

I consider $d: C^{\infty}(M) \to \Omega^1(M)$

$dim(N (d)) =$ number of connected components of $M$

if $df =0$, then $f$ is constant in each connected component of $M$.
If $f \in N (d)$, $f = (c_i)$ such that

$i \in$ {connected components of $M$ }

If $f \in N (d)$; $f = \sum c_i f_i$  where $f_i: M \to \Bbb R.$

I think the road is good, but I don't know how to continue

答:
$f$ is continuous therefore the preset of a closed set ($\{c\}$ in our case) is closed.

If $(U_p,\phi_p)$ is a chart containing $p$ then clearly $df=0$ implies $f_{U_p}$ is constant, hence our set $f^{-1}(c) = \cup_{p \in f^{-1}(c)} U_p $ is an union of open sets.

Title: Level set that is both closed and open.


答:
Note that another way to write the singular value decomposition is to write it as the sum of rank one matrices:
$$A = U D V^T = \sum_{k = 1}^n \sigma_i \vec{u}_i \vec{v}_i^T$$
where $\vec{u}_i$ is the $i$th column of $U$ and $\vec{v}_i$ is the $i$th column of $V$. What does this tell us? It tells us that whenever $\sigma_i = 0$, the resulting left and right singular vectors don't really matter. So you have two choices:

If $A$ is $m \times n$ (here $m = n = 2$) You can complete $\vec{u}_1$ to an orthonormal basis $\{\vec{u}_1, \vec{u}_2, \cdots, \vec{u}_m\}$ of $\mathbb{R}^m$ and complete $\vec{v}_1$ to an orthonormal basis $\{\vec{v}_1, \vec{v}_2, \cdots, \vec{v}_m\}$ of $\mathbb{R}^m$ of $\mathbb{R}^n$ (this can be done with Gram-Schmidt). As a result, $D$ will be an $m \times n$ diagonal matrix. That's what the solutions have done. Then we may write
$$A = \begin{bmatrix} \vec{u}_1 & \cdots & \vec{u}_m \end{bmatrix} \cdot \text{diag}_{m \times n}(\sigma_1, \sigma_2, \cdots, \sigma_{\min(m,n)}) \cdot \begin{bmatrix} \vec{v}_1 & \cdots & \vec{v}_n \end{bmatrix}^T$$

Calculate the reduced singular value decomposition. This involves only worrying about the nonzero singular values, so we won't need to complete any set of singular vectors to a basis. Here $D$ will become an $r \times r$ diagonal matrix, where $r$ is the number of nonzero singular values ($r$ is also the rank of $A$). $U$ will be $n \times r$, and $V$ will be $m \times r$. In this case, we know $r = 1$ and you've calculated the corresponding left and right singular vectors of $\sigma_1 = \sqrt {50}$. So you can write
$$A = \frac{1}{\sqrt{10}}\begin{bmatrix} 1 \\ 3\end{bmatrix} \begin{bmatrix} \sqrt{50} \end{bmatrix} \left(\frac{1}{\sqrt 5} \begin{bmatrix} 1 & -2\end{bmatrix}\right)$$
Here $U$ and $V$ are partially orthogonal, in the sense that $U^T U = I_r = V^T V$ but $UU^T \neq I_n$ and $V V^T \neq I_m$.

Title: SVD Singular Value Issue

Body:
I am having an issue with a SVD problem and would like advice. The matrices given are

$$
  A= \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]\ \ \
  A^TA= \left[ \begin{matrix}
    10 & 20\\
    20 & 40 \\
    \end{matrix}\right]\ \ \
  AA^T= \left[ \begin{matrix}
    5 & 15\\
    15 & 45 \\
    \end{matrix}\right]
$$

I initially determined my $λ$ values to be $λ_1=50$ and $λ_2=0$ by computing the following:

$$
det(A^TA-λI)=0\\
\begin{vmatrix}
    10-λ & 20\\
    20 & 40-λ \\
    \end{vmatrix}\\
(10-λ)(40-λ)-20^2=0\\
400-50λ+λ^2-400=0\\
λ^2-50λ=0\\
λ(λ-50)=0
$$

This in turn means that my singular values will be  $\sigma_1=\sqrt {50}$ and $\sigma_2=0$, and this is where I run into my issue. When I am trying to solve for the orthonormal set with $u_1$ and $u_2$, I was told to use the equation $u_n=\frac{Av_n}{\sigma_n}$. I cannot compute $u_2$ because that would make the denominator $0$, and the fraction would become undefined.

I discussed with my professor and they said I should try to calculate $u_2$ using the eigenvectors from $AA^T$, but $AA^T$ and $A^TA$ have the same eigenvalues. I thought maybe they meant to compute the $u$ for the $\sigma=\sqrt {50}$, but this is not the same answer as the one the book gives. The answers for this and Wolfram Alpha show that the SVD should be

$$
  \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]=
\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 & -3\\
    3 & 1 \\
    \end{matrix}\right]
  \left[ \begin{matrix}
    \sqrt{50} & 0\\
    0 & 0 \\
    \end{matrix}\right]
\frac{1}{\sqrt{5}}
  \left[ \begin{matrix}
    1 & -2\\
    2 & -1 \\
    \end{matrix}\right]
$$

I have gotten that $u_1=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 \\
    3 \\
    \end{matrix}\right]$, but I do not know how to get $u_2=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    -3 \\
    1 \\
    \end{matrix}\right]$. Any help would be appreciated.


答: The first term of your $E(W)$ represents the contribution to the expectation of rolling at least a $4$ and is correct.  The second term should be $\frac 12 \cdot \frac 16(1+2+3+4+5+6)$ where the factor $\frac 12$ represents the chance that you roll less than $4$ and then accept the second roll.  This correction would give you $\frac {15}6+\frac {3.5}2$, which does equal $4.25$.  Your second term comes out to $\frac {15}6$, then your addition is not correct as it should come out $\frac {18.5}6 \approx 3.0833$

Title: Average of dice roll with chance of rerolling

Body:
My question regards a simple dice game. You roll one six sided die and win, say, a dollar for each dot on the roll. e.g. five dollars for rolling a five, two dollars for rolling a two, etc.
You are given the option to reroll once.

Letting $X$ be the outcome of the first roll, and $Y$ be the outcome of the (potential) second roll, consider the strategy $W$
\begin{equation}
  W=\begin{cases}
  X&\text{if $X\geq 4$}\\
  Y&\text{if $X< 4$}
  \end{cases}
 \end{equation}
I.e. if you roll 4 or higher on first try you keep it, if you roll lower you try again.
What is the average winnings using this strategy?

Solution

My issue here is, really, that I don't have more mathematically stringent way to prove this, and I got the answer rather intuitively.

I consider two "macroscopic" outcomes;

a) We roll a 4, 5, or 6 in the first attempt and stop.

b) We take a second attempt, and roll either 1, 2, 3, 4, 5, or 6.

The average winnings is the average of these two options, which in itself is the average of a roll of 4/5/6 and the average of a roll of 1/2/3/4/5/6. That is

\begin{equation}
E[W] = (4+5+6)\frac{1}{6}+(1+2+3+4+5+6)\frac{1}{2} = \frac{15}{6} + \frac{3.5}{2} = 4.25
\end{equation}

Which indeed is a higher expectation than winning than rolling the die just once ($\mu=3.5$)

I feel like my argumentation is mostly just intuition, and could benefit from a more mathematically sound description.


问:
Sequence of compactly supported functions converging to rectangle
I have the following problem and would like to know if the following solution is right:

I have to find for every $\varepsilon\in\mathbb{R}_{>0}$ a function $f_{\varepsilon}\in C_{c}((0,1))$, such that $f_{\varepsilon}\to \chi_{(a,b)}$ in $L_{p}((0,1))$ for $\varepsilon\to 0$, where $\chi_{(a,b)}$ denotes the characteristic function, e.g. $\chi_{(a,b)}(x)=1$, if $x\in (a,b)$ and 0 otherwise.

After some tries I have constructed the following sequence of functions:

$$f_{\varepsilon}(x):=\begin{cases}\exp\bigg\{-\frac{\varepsilon}{1-\big(\frac{2}{b-a}\big(x-\frac{b+a}{2}\big )\big)^{2}}+\varepsilon\bigg\}&\text{if $x\in (a,b)$}\\0, &\text{otherwise}\end{cases}$$

These functions are continous, since we have that
$$\lim_{x\to a}\exp\bigg\{-\frac{\varepsilon}{1-\big(\frac{2}{b-a}\big(x-\frac{b+a}{2}\big )\big)^{2}}+\varepsilon\bigg\}=0=\lim_{x\to b}\exp\bigg\{-\frac{\varepsilon}{1-\big(\frac{2}{b-a}\big(x-\frac{b+a}{2}\big )\big)^{2}}+\varepsilon\bigg\}$$

Furhtermore, they are compactly supported in $(0,1)$, since we have that

$$\mathrm{supp}(f_{\varepsilon})=\overline{(a,b)}=[a,b].$$

If I draw this functions with smaller and smaller $\varepsilon$ they also indeed seem to approach the rectangle $[a,b]\times [0,1]$.

So we have now to show that $\lim_{\varepsilon\to 0}\Vert f_{\varepsilon} - \chi_{(a,b)}\Vert_{p}=0$

This can be done with help of the dominant convergence theorem:

$$\lim_{\varepsilon\to 0}\Vert f_{\varepsilon} - \chi_{(a,b)}\Vert_{p}^{p}=\lim_{\varepsilon\to 0}\int_{(0,1)}\vert f_{\varepsilon}(x) - \chi_{(a,b)}(x)\vert_{p}^{p}dx=\int_{(0,1)}\lim_{\varepsilon\to 0}\vert f_{\varepsilon}(x) - \chi_{(a,b)}(x)\vert_{p}^{p}dx=\int_{(a,b)}(1-1)dx=0$$

答:
You have proclaimed, but have not effectively demonstrated, that
$$\lim_{\varepsilon\to 0}\vert f_{\varepsilon}(x) - \chi_{(a,b)}(x)\vert_{p}^{p} = 1 - 1$$ for $x \in (a,b)$.

It might be true, but "I've graphed it for several $\varepsilon$ and it appears to head that direction" isn't enough of an argument.

You could probably come up with one, but let me point out something easier: $f_\epsilon$ doesn't have to be pieced together only at $a$ and $b$. For example, if $$g(x) = \begin{cases}0,& x \in (0,1]\\x-1,& x\in [1,2]\\1,& x\in [2,3]\\4-x,& x\in [3,4]\\0,& x \in [4,5)\end{cases}$$
then $g \in C_c((0,5))$.

If you were wanting smooth functions, then playing around with $e^{-1/|x|}$ or similar functions would be the way to go (but again, piecing them only at $a, b$ is not necessary, and will only cause you extra woes). But there is nothing here that calls for smoothness.


问:
Find integral of $1/z^2$ from $-4i$ to $4i$
Find integral of $1/z^2$ from $-4i$ to $4i$ and the path given is "any contour that doesn't pass through the origin". Now I understand that the book wants me to use the fundamental theorem of complex integral because I am on the section about path independence. But the theorem itself states that a function is only path independent if it's smoothly connected, but $0$ is not defined so the function is multiply connected no? So can we still assume path independence here?

答:
Well, if you integrated along a closed circular path that enclosed the origin, the result would be zero, so as long as you avoid the origin, the integral is path independent:

$$\oint_C \frac{dz}{z^2}=0,$$ where $C$ is any closed curve whose interior may or may not contain the origin, but does not pass through the origin.

Contrast this with $$\oint_C \frac{dz}{z}=\left\{
\begin{aligned}
&2\pi i, & C \text{ surrounds the origin} \\
&0, & C \text{ does not surround the origin}
\end{aligned}
\right.$$

The function $f(z) = \frac{1}{z^2}$ has a primitive $F(z)=-\frac{1}{z}$,

so $$\int_{z_1}^{z_2} \frac{dz}{z^2}  = \left. -\frac{1}{z}\right|_{z_1}^{z^2} = \frac{1}{z_1} - \frac{1}{z_2}. $$

$$\int_{-4i}^{4i} \frac{dz}{z^2} = \frac{i}{4} - \left(-\frac{i}{4}\right)=\frac{i}{2}.$$


问:
Is my logic correct in this proof
Prove
Let {$B_i$}i∈ I and {$C_i$}i∈ I each be a family of mutually disjoint sets. If $B_i≈C_i$ for each i ∈ I, prove that
$\bigcup$$B_i$= $\bigcup$$C_i$

First I know this theorem is a general
version of

7.8 Theorem Let A, B, C, and D be sets where A ∩ C = Ø and B ∩ D = Ø. If f : A → B and g : C → D are bijective functions, then f ∪ g is a bijective function A ∪ C → B ∪ D.

I found that out in Lin’s book, Set Theory:An intuitive approach

Referenced Theorems

7.6 Lemma There exists an injective function f : A → B if and only if there exists a surjective function g : B → A.

5.4 Theorem

Let A be a set and let f:A$\mapsto$ B be a function;f:A$\mapsto$ B is surjective iff there exists a function g:B→A∋f $\circ$g=$1_B$

AC (Axiom of Choice). Every set has a choice function

Proof attempt

Let $B_i \cap C_i =\emptyset$ for $i\in I$

There exists a function $f_i$:$B_i \mapsto C_i$ which is
bijective and function $g_i$:$C_i \mapsto B_i$

Now $B_i \subseteq P(B)$ ; $C_i \subseteq P(C)$ for i $\in$ I ,so that by AC,
B=$\bigcup B_i$ and C=$\bigcup C_i$
There exists a choice function$g_i$:B$\mapsto$ C where g(b)=$g_i(b)$ for an unique i for
b$\in$ B

To show the above is bijection use 5.4
and show it is an injection.
After this I am lost.I don’t know how to
create f$\cup$g assuming what I stated is
correct

答:
Corrected.

If $I$ is infinite you do need the axiom of choice to select a bijection $f_i$ for each $i\in I$, but after that it’s straightforward: just let $f=\bigcup_{i\in I}f_i$ and check that $f$ is a bijection from $\bigcup_{i\in I}B_i$ to $\bigcup_{i\in I}C_i$. For instance, suppose that $f(b)=c$ for some $b\in\bigcup_{i\in I}B_i$. There is a unique $i_0\in I$ such that $b\in B_{i_0}$, and clearly $c=f(b)=f_{i_0}(b)\in C_{i_0}$. Now suppose that $f(b')=c$ for some $b'\in\bigcup_{i\in I}B_i$; there is an $i_1\in I$ such that $b'\in B_{i_1}$. Then $c=f(b')=f_{i_1}(b')\in C_{i_1}$, and $C_{i_1}\cap C_{i_0}\ne\varnothing$ iff $i_0=i_1$, so $i_1=i_0$, and $b'\in B_{i_0}$. Finally, $f_{i_0}$ is a bijection, so $b'=b$, and $f$ is therefore injective. It’s even easier to verify that $f$ is surjective.

Title: Is my logic correct in this proof


问:
What is the Hausdorff dimension of this disjoint object?
I am looking to find the Hausdorff dimension of the following object. It is composed of self-similar disjoint annuli. This is my current work.

If anybody can help describe the steps to find the Hausdorff dimension of this object I would really appreciate it. Thank you.

答:
Dilating $\Omega$ by a factor of $\frac14$ yields a copy, say $\Omega'$, of $\Omega$. Because of that, The area of $\Omega'$, $A(\Omega') = \frac{1}{16} A(\Omega)$.

Thus, the scaling factor is $\frac{1}{16}$. The Hausdorff dimension $D$ of $\Omega$ is:

$$\frac{1}{16} = \Big(\frac{1}{4}\Big)^D \Leftrightarrow D=2.$$


问:
Is $μ_\bar{X} = μ$ the same thing as saying the mean of the population is the same as the mean of the sample?
My e-textbook is telling me that, when finding a sample from a population using a normal probability distribution, $\mu_\bar{X}$ is roughly equal to $μ$. It then gives me a problem asking me to find the probability that the sample mean is $110$ given that the parent population's mean = $100$ and the standard deviation = $15$. I thought, "surely the probability is $0$, since the population and sample mean should roughly equal each other," but this is clearly not the correct thinking, as evidenced by the video in the textbook giving a completely reasonable explanation and solution of $2%$. I'm clearly missing some crucial information, or the textbook just worded the explanation and question strangely.

答:
The expected value of a sample is equal to the expected value of the population (assuming the latter exists). Using linearity of expectation:

$$\mu_{\bar{X}}=\mathbb E\left[\bar X\right]=\mathbb E\left[\frac1n \sum_1^n X_i\right] = \frac1n\mathbb E\left[ \sum_1^n X_i\right] = \frac1n \sum_1^n \mathbb E\left[X_i\right] = \frac1n \sum_1^n \mu = \frac1n n \mu = \mu$$

This is exact for the expectation of the sample mean.  What may be approximate is that the actually observed sample mean can be close to the population mean, as a consequence of the law of large numbers, at least when the standard error of the sample mean is small, often as the result of a large enough sample.


问:
The proof of Leads for tied-down random walk in Grimmett's Probability and Random Processes
The Theorem and the proof are as follows, where $L_{2n}$ is the number of steps of the walk which were not within the negative half-line, and $T_0$ is the first time of the walk revisit the position $0$.
And the reference to $(1b)$ of the proof is
$$
P_0(s)=\sum_{n=0}^{\infty}P(S_n=0)s^n=\sum_{n=0}^{\infty}p_0(n)s^n=(1-4pqs^2)^{-\frac{1}{2}}.
$$
I understand the proof except fail to verify the last equation.
Why
$$
\frac{2\big[\sqrt{1-s^2t^2}-\sqrt{1-t^2}\big]}{t^2(1-s^2)}=\sum_{n=0}^{\infty}t^{2n}\mathbb{P}(S_{2n}=0)\bigg(\frac{1-s^{2n+2}}{(n+1)(1-s^2)}\bigg)?
$$
I replaced $\sqrt{1-s^2t^2}\;$ by $\;(1-s^2t^2)\cdot P_0(st)\;$ and $\sqrt{1-t^2}\;$ likewise, but couldn't get the same result.

The following content is in Page 169 of Grimmett's Probability and Random Processes

答:
I don't know much about what you learn but I think you have overthought this problem and chosen a not true approach.

So you see if we have to deal with an analytic function $f(x)$ with the convergence radius $1$ :
$$ f(x) := \sum_{n \ge 0} a_nx^n$$
in order to obtain a compact form of the analytic expansion as you wrote at the beginning of your post, we can integrate $f$ and do some substitution and subtraction :

Integrating from $0$ to $t$
$$ F(t) := \int_{0}^t f(x)dx = \sum_{n \ge 0} \dfrac{a_n}{n+1}t^{n+1} $$

Substitution and subtraction:
$$ \dfrac{F(t^2)-F(t^2s^2)}{t^2(1-s^2)} = \sum_{n \ge 0} \dfrac{a_n}{n+1} t^{2n} \left( \dfrac{ 1-s^{2n+2}}{1-s^2} \right)$$

By letting $a_n= C_n$; $f(x)= (1-x^2)^{-1/2} $, you get what you want.


问:
Homomorphism from $\mathbb{Z}/n\mathbb{Z}$
Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly? </p>

答: Let $\varphi : \mathbb{Z}_n \to  \mathbb{Z}$ be a homomorphism. Using isomorphism theorem, $\mathbb{Z}_n / \text{ker}(\varphi)$ is isomorphic to $\text{Im}(\varphi)$. Because $\text{Im}(\varphi) \leq\mathbb{Z}$, either $\text{Im}(\varphi)= \{0\}$ or $\text{Im}(\varphi) \simeq \mathbb{Z}$; on the other hand, $\mathbb{Z}_n/ \text{ker}(\varphi)$ is necessarily a finite group. Therefore, $\varphi =0$.</p>


问:
How to use the $\;\epsilon-\delta\;$ definition to show that $\lim\limits_{x \to 1} \frac{1}{3x+1}=\frac14$
I have tried evaluating the $\;\delta\;$ part of the definition. So far I have reached the following point:

$0<|x-1|<\delta\;$ implies $\;\left|\frac{1}{3x+1}-\frac14\right|<\epsilon$

which I simplified to be $\;3|1-x|\left|\frac{1}{12x+4}\right|<\epsilon\;$,

but I am unsure where to go from here to obtain $\;\delta$.

答:
If $0<|x-1|<\delta$ and $\delta<1$, then $x\in(0,2)$ so $12x+4\in(4,28)$, so $\left|\frac{1}{12x+4}\right|<\frac14$.

So take $\delta=\min\left(1,\frac{4\epsilon}3\right)$.


答:
Let $\delta >0$, being small.

Let $0<\lvert x-1\rvert<\delta.$

$$\left\lvert \frac{1}{3x+1} -\frac{1}{4}\right\rvert =\left\lvert \frac{3-3x}{12x+4}\right\rvert = \frac{3}{4}\left\lvert \frac{1-x}{3x+1}\right\rvert \le\\ \le\frac{3}{4}  \frac{\delta}{2-3\delta}\le\frac{3}{4}\delta \to0$$ as $\delta \to 0,$ where we have used the fact that $\lvert x \rvert>1-\delta$ and $\lvert 3x+1\rvert \ge \lvert 3x\rvert -1 \gt 2-3\delta.$

Title: How to use the $\;\epsilon-\delta\;$ definition to show that $\lim\limits_{x \to 1} \frac{1}{3x+1}=\frac14$

Body:
I have tried evaluating the $\;\delta\;$ part of the definition. So far I have reached the following point:

$0<|x-1|<\delta\;$ implies $\;\left|\frac{1}{3x+1}-\frac14\right|<\epsilon$

which I simplified to be $\;3|1-x|\left|\frac{1}{12x+4}\right|<\epsilon\;$,

but I am unsure where to go from here to obtain $\;\delta$.


问:
Construct an elementary chain of order type $\mathbb Q$
Let $T$ be a complete first-order theory over a countable language $L$, such that $T$ has an infinite model.

It is well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type the reverse ordering of any countable ordinal, see e.g. here. Simply, a possible proof uses Gödel's compactness theorem as well as augmenting the language with suitable Skolem functions.

It is even more well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type any countable ordinal. Moreover, the ordinary or reversed ordering of any countable ordinal continuously embeds into the ordering of $\mathbb Q$.

Question 1: Is it always possible to construct an elementary chain $\{M_q:q\in\mathbb Q\}$ of models of $T$, such that every $\|M_q\|=\aleph_0$, every $M_p\precneqq M_q$ whenever $p<q$, and moreover for all $q\in\mathbb Q$, it holds that
\begin{equation}\tag{$\dagger$}\bigcup_{i<q}M_i=M_q=\bigcap_{j>q}M_j
\end{equation}

Question 2: If the answer to Question 1 is NO, then under what conditions for $T$ can the answer become YES?

Thanks in advance!

Simple observations:

(1) Every discontinuity of the chain must happen at Dedekind cuts of irrarional numbers. For example, if $T$ is the theory of infinite sets, then let $S$ be the set of all irrational algebraics, and the models $M_q=\{s\in S:s<q\}$ work as desired.

(2) By repeating the argument mentioned above, it is clear that the answer is YES when $(\dagger)$ is weakened to
$$\bigcup_{i<q}M_i\subseteq M_q\subseteq \bigcap_{j>q} M_j
$$
and it's easy to find examples such that both subset relations modified as above are cofinitely strict. Thus by replacing each $M_q$ with their upper/lower bounds, we see that the answer is also YES when either of the two equalities in $(\dagger)$ is omitted.

(3) It suffices for $(\dagger)$ to hold only for a dense subset of an interval in $\mathbb Q$, since $\mathbb Q$ would continuously embed into it.

答:
Here's a quick sketch (assuming WLOG that $T$ has definable Skolem functions):

Let $\mathbb{A}$ be the set of rational multiples of $\pi$. Suppose $\mathcal{N}$ is a "big" model of $T$. Pick some $\mathbb{A}$-indexed "very indiscernible" sequence $X=\{x_i: i\in\mathbb{A}\}$ of elements of $\mathcal{N}$. For each $q\in\mathbb{Q}$ we can then set $M_q$ to be the definable closure of $\{x_i: i\in\mathbb{A}, i<q\}$ in $\mathcal{N}$.

Clearly we have $p<q\implies M_p\preccurlyeq M_q$ and $M_q=\bigcup_{p<q}M_p=\bigcap_{p>q}M_p$ (this last bit is where $\mathbb{A}\cap\mathbb{Q}=\emptyset$ comes in), so we just have to ensure that $M_p\subsetneq M_q$. This will come down to the appropriate choice of $\mathcal{N}$ and $X$. Choosing $\mathcal{N}$ is easy since all we want is sufficient "richness:" we'll just pick any sufficiently saturated model of $T$.

Now we just have to pick an appropriate $X$. Here we run into the issue Alex Kruckman highlighted in the linked question: it's not the case that any old $\mathbb{A}$-indexed indiscernible sequence in $\mathcal{N}$ will do. Instead, we need something like a "coheir sequence of ordertype $\mathbb{Q}$." Here we can use a cute trick: if I have a structure $\mathcal{S}$ and a sequence $\sigma=(s_i)_{i\in\omega}$ of elements of $\mathcal{S}$, I can get a richer sequence similar to $\sigma$ by taking an ultrapower of $\mathcal{S}$ with $\sigma$. Precisely, let $\mathcal{S}_\sigma$ be the expansion of $\mathcal{S}$ by adding a unary predicate $U$ naming the elements in $\sigma$ and a binary relation $\triangleleft$ describing the ordering of those elements according to $\sigma$, and consider an appropriate ultrapower $\mathcal{U}$ of $\mathcal{S}_\sigma$ (or really just any sufficiently saturated model of $Th(\mathcal{S}_\sigma)$). The "expanded sequence" $(U^\mathcal{U},\triangleleft^\mathcal{U})$ now has very rich ordertype, and in particular has a suborder of type $\mathbb{A}$.

Here we can use this trick as follows: "blow up" a coheir sequence in $\mathcal{N}$ to get a $\mathbb{A}$-indexed sequence $X$ in some larger $\mathcal{N}'$ which we can check works for the same reason that a coheir sequence helped solve the problem in the linked question. OK fine, we're now using $\mathcal{N}'$ instead of $\mathcal{N}$ after all, but that's okay.


问:
A proportion question regarding video game players
I am trying to find a proportion of console players that play a certain number of call of duty games. 40% of players use xbox, 50% use playstation, and 10% use wii u. Of those who play on xbox, 60% play more than 3 games. Of those who play on playstation, 15% play more than 3 games. Of those who play on wii u, 0% play more than 3 games. What proportion of the players playing on wii u play less than 3 games?

So far, I Let total players be 100.

Then no. of players that use XBOX, Playstation and Wii U are 40 (40% of 100), 50 (50% of 100), and 10 respectively.

The no. of players that play more than 3 games on XBOX, Playstation and Wii U are 24 (60% of 40) , 25 (50% of 50) and 0 respectively.

The no. of players that play less than 3 games on XBOX, Playstation and Wii U are 16 (40-24), 25 (50-25) and 10 (10-0) respectively.

The proportion of the players who plays less than 3 games on Wii U is given as: 10/100 = 0.1 (my answer)

答:
Are you sure you read the question right? It's impossible to answer if taken literally.

"Of those who play on wii u, 0% play more than 3 games. What proportion of the players playing on wii u play less than 3 games?"

You cannot find out how many players play exactly 3 games, so there's no way to answer the question.

If you play exactly 3 games, you don't play more than 3 games, but you also don't play less than 3 games. You play exactly 3 games.


问:
The number of subspaces of a vector space forming direct sum with a given subpace
Let $V$ be a vector space of dimension $n$ over $\mathbb{F}_q$, and let $U$ be a subspace of dimension $k$. I want to compute the number of subspaces $W$ of $V$ of dimension $m$ such that $W\cap U=0$.

I know why the number of subspaces of $V$ that contain $U$ and have dimension $m$ is $\binom{n-k}{m-k}_q$, but I don't understand why $q^{km}\binom{n-k}{m}_q$ is number of these subspaces?

答:
Hint

Consider the standard projection
$$\pi : V \to V/U,\quad v \mapsto v + U.$$
For a subspace $W$ of $V$, $W\cap U = \{0\}$ is equivalent to $\dim(\pi(W)) = \dim(W)$.

Using this characterization, you have to count the possible $\pi$-images of $W$ in $V/U$, and for a fixed such $\pi$-image $M$ the suitable $\pi$-preimages of $M$ in $V$.

Let me remark that in the spacial case $m = n - k$, you get the number of complements of $U$ in $V$ as $q^{k(n-k)}$.