问:
Intersection of family of compact set is compact
Let{Cj:j∈J} be a family of closed compact subsets of a topological space(X,τ). Prove that {⋂Cj:j∈J} is compact.

I realized this is not a metric space, so compactness in general topology does not imply closed or boundedness. But if we use the subcover definition of compactness, it should always by possible find a finite open subcover right since each C is compact and each has a finite subcover, the intersection of finite open subcover is still open and is still a subcover for the intersection of all the compact sets.

One concern I have: even though each C has a finite subcover, if there are an uncountable number of C, then we are taking an infinite intersection of open subcovers which is not open.

答:
HINT: Fix $j_0\in J$, and let $\mathscr{C}=\{C_j\cap C_{j_0}:j\in J\}$. Then

$$\bigcap\{C_j:j\in J\}=\bigcap\mathscr{C}\,,$$

and $\bigcap\mathscr{C}$ is a closed subset of $C_{j_0}$.


问:
Commutator property
Can you please show me how to prove this? If H,K,L are normal subgroups of G then $$[[H,K],L]\subseteq [[K,L],H][[L,H],K]$$ Thanks in advance!</p>

答: i took an element from the first and show that it is contained in $$H\cap K\cap L$$ but the second is contained in $$K\cap L$$.is it ok?</p>

ContentLicense: CC BY-SA 3.0

Title: Commutator property


问:
Sigmoid function with a longer, straighter middle
Can I modify the following function $y = 1 / (1 + e^{-x})$ such that I can make the curve straighter in the middle?

I would like the function to resemble a straight line until it approaches $0$ or $1$.

Here is an example diagram, the red line is desired:

答:
An Answer: There is no Answer

We could try to generalize the expression a bit to get a "straighter" curve. For example, the expression can be generalized to
$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\brl}[1]{{\left.#1\right|}}$
$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
$\newcommand{\R}{\Bbb{R}}$
$$y \eqd \frac{a_1}{a_2 + a_3e^{-bx}} + a_4$$
And the minimum constraints are
$$\begin{align*}
    \lim_{x\to-\infty}y &= 0
  \\\lim_{x\to+\infty}y &= 1
  \\\brl{y}_{x=0} &= \frac{1}{2}
\end{align*}$$
But under the constraints given/implied, there is no solution; that is, there is no way, under the above constraints, to "straighten" it while holding its "steepness" unchanged. Here is the reason$\ldots$

$$\begin{align*}
  0 &= \lim_{x\to-\infty}y
  \\&\eqd \lim_{x\to-\infty} \brs{\frac{a_1}{a_2 + a_3e^{-bx}} + a_4 }
    && \text{by definition of $y$}
  \\&= \brs{0 + a_4 }
  \\&\implies \boxed{a_4=0} \implies y = \frac{a_1}{a_2 + a_3e^{-bx}}
  \\\\
  1 &= \lim_{x\to\infty}y
  \\&\eqd \lim_{x\to\infty} \brs{\frac{a_1}{a_2 + a_3e^{-bx}} + 0 }
  \\&= \lim_{x\to\infty} \brs{\frac{a_1}{a_2 + 0}}
  \\&\implies \boxed{a_1=a_2}\implies y = \frac{a_1}{a_1 + a_3e^{-bx}}
  \\\\
  \frac{1}{2} &= \brl{y}_{x=0}
  \\&= \frac{a_1}{a_1 + a_3e^{-b\cdot0}} + 0
  \\&= \frac{a_1}{a_1 + a_3}
  \\&\implies 2a_1=a_1+a_3
  \\&\implies\boxed{a_1=a_3}
  \\&\implies y = \frac{a_1}{a_1 + a_1e^{-bx}}
  \\&\implies y = \frac{1}{1 + e^{-bx}}
\end{align*}$$

You still have one parameter $b$ that you can change$\ldots$ but changing that will change the ``steepness" of the curve, where here steepness is defined as the derivative $\deriv{y}{x}$ at $x=0$:
$$\brl{\deriv{}{x}y}_{x=0} = \brl{\frac{-(-be^{-bx})}{(1+e^{-bx})^2}}_{x=0}=\frac{b}{4}$$
But if you don't want to change that (as you implied in your previous comment), then you have no degrees of freedom left and there is no way to straighten the curve without resorting to changing to what may be considered more "drastic" measures like changing the function altogether.

ContentLicense: CC BY-SA 4.0


问:
Proof for $\boldsymbol{2}^{\mathbb{N}} \preceq \mathbb{R}$
Define a function $\theta: \boldsymbol{2}^{\mathbb{N}} \to \mathbb{R}$ such that
\begin{equation*}
\theta\left(f\right) = \left\{q \in \mathbb{Q}: q < \sum_{i=0}^{n}\frac{f\left(i\right)}{10^{i}} \textrm{ for some } n \in \mathbb{N}\right\},
\end{equation*}
where $\mathbb{Q}$ is the set of rational numbers. I cannot come up with the proof that this $\theta$ is one-one. Can anyone provide thoughts?

P.S. It should be noted that $\boldsymbol{r} \in \mathbb{R}$ is defined as a Dedekind cut. The first problem was to show that the set $\left\{q \in \mathbb{Q}: q < \sum_{i=0}^{n}\frac{f\left(i\right)}{10^{i}} \textrm{ for some } n \in \mathbb{N}\right\}$ is a Dedekind cut, which I proved myself.  The function $\theta$ is used in proving $\boldsymbol{2}^{\mathbb{N}} \preceq \mathbb{R}$.

I have organized a proof based on comments from this post. It goes as follows.

First we shall prove the following statement:

For all $n,m\in\mathbb{N}$, if $m > n$, then
\begin{equation*}
\frac{1}{10^{n}} > \sum_{i=n+1}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+1}}.
\end{equation*}

I am using mathematical induction to prove this statement. First define the following set with a fixed $m \in \mathbb{N}$:
\begin{equation*}
A = \left\{ n \in \mathbb{N}: m > n \textrm{ implies } \frac{1}{10^{n}} > \sum_{i=n+1}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+1}} \right\}.
\end{equation*}
If $m = 0$, then $m > n$ has to be false, thus $m = 0$ trivially implies $A = \mathbb{N}$. Now assume that $m > 0$. For $n = 0$,
\begin{equation*}
\sum_{i = 1}^{m}\frac{1}{10^{i}} + 0.5 = \frac{\frac{1}{10}\left(1-\frac{1}{10^{m}}\right)}{\frac{9}{10}} + 0.5 = \frac{1-\frac{1}{10^{m}}}{9} + 0.5 < 1.5 = \frac{1}{9} + \frac{1}{2} = \frac{11}{18} < 1 = \frac{1}{10^{0}}.
\end{equation*}
Thus, $0 \in A$. Further, assume that for an arbitrary $n \in \mathbb{N}$,
\begin{equation*}
\frac{1}{10^{n}} > \sum_{i=n+1}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+1}}.
\end{equation*}
Now consider the case of $n + 1$.  If $m \leq n$, then trivially the induction relation holds at $n + 1$. If $m = n + 1$, then
\begin{equation*}
\sum_{i=n+1+1}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+1+1}} = \sum_{i=n+2}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+2}} = \frac{5}{10^{n+2}} < \frac{1}{10^{n+1}}.
\end{equation*}
The relation also holds at $n + 1$. If $m > n+1$, then multiplying $1/10$ on both sides of the induction hypothesis, we have
\begin{equation*}
\frac{1}{10^{n+1}} > \sum_{i=n+2}^{m}\frac{1}{10^{i}} + \frac{5}{10^{n+2}}.
\end{equation*}
We may conclude that for an arbitrary $m \in \mathbb{N}$, $0 \in A$ and an arbitrary $n \in A$ implies $n + 1 \in A$. Consequently, $A = \mathbb{N}$ for an arbitrary $m \in \mathbb{N}$.

Next we shall use the proved statement to prove that $\theta$ is one-one. Assume that $f_{1},f_{2} \in \boldsymbol{2}^{\mathbb{N}}$ and $f_{1} \neq f_{2}$. Define the following set:
\begin{equation*}
S = \left\{n \in \mathbb{N}: f_{1}\left(n\right) \neq f_{2}\left(n\right)\right\}.
\end{equation*}
We have $S \neq \emptyset$ as $f_{1} \neq f_{2}$. Then assume that $n_{0} = \min{\left(S\right)}$. Then $n_{0}$ is the first index where $f_{1}$ and $f_{2}$ diverges. As $f_{1}, f_{2} \in \boldsymbol{2}^{\mathbb{N}}$, without loss of generality, assume $f_{1}\left(n_{0}\right) = 1$, $f_{2}\left(n_{0}\right) = 0$ and $f_{1}\left(n\right) = f_{2}\left(n\right)$ for all $n < n_{0}$. Then we can write
\begin{equation*}
\sum_{i=0}^{n_{0}}\frac{f_{1}\left(i\right)}{10^{i}} = s + \frac{f_{1}\left(n_{0}\right)}{10^{n_{0}}} = s + \frac{1}{10^{n_{0}}},
\end{equation*}
where
\begin{equation*}
s =
\begin{cases}
0,\ n_{0} = 0\\
\sum_{i=0}^{n_{0}-1}\frac{f_{1}\left(i\right)}{10^{i}},\ n_{0} > 0
\end{cases}.
\end{equation*}
Further, for an arbitrary $m > n_{0}$ we have
\begin{equation*}
\begin{aligned}
\sum_{i=0}^{m}\frac{f_{2}\left(i\right)}{10^{i}} &= s + \frac{f_{2}\left(n_{0}\right)}{10^{n_{0}}} + \sum_{i=n_{0}+1}^{m}\frac{f_{2}\left(i\right)}{10^{i}} = s + \sum_{i=n_{0}+1}^{m}\frac{f_{2}\left(i\right)}{10^{i}}\\
&\leq s + \sum_{i=n_{0}+1}^{m}\frac{1}{10^{i}} \leq s + \frac{1}{10^{n_{0}}} - \frac{5}{10^{n_{0}+1}} = \sum_{i=0}^{n_{0}}\frac{f_{1}\left(i\right)}{10^{i}}-\frac{5}{10^{n_{0}+1}}.
\end{aligned}
\end{equation*}
It can be seen that
\begin{equation*}
\left(\sum_{i=0}^{n_{0}}\frac{f_{1}\left(i\right)}{10^{i}}-\frac{5}{10^{n_{0}+1}}\right) \in {f_{1}}^{*}
\end{equation*}
and
\begin{equation*}
\left(\sum_{i=0}^{n_{0}}\frac{f_{1}\left(i\right)}{10^{i}}-\frac{5}{10^{n_{0}+1}}\right) \not\in {f_{2}}^{*}.
\end{equation*}
As a result, $\theta\left(f_{1}\right) \neq \theta\left(f_{2}\right)$ for $f_{1} \neq f_{2}$ and $\theta$ is injective.

答:
Observe that

$$\theta(f)=\left\{q\in\Bbb Q:q<\sum_{i\ge 0}\frac{f(i)}{10^i}\right\}\,,$$

the Dedekind cut corresponding to the real number $\sum_{i\ge 0}\frac{f(i)}{10^i}$. No two real numbers in $[0,1]$ have have distinct decimal expansions whose digits are all $0$ or $1$, so $\theta$ must be injective.

To prove it, suppose that $f,g:\Bbb N\to\{0,1\}$, and $f\ne g$. Let $k\in\Bbb N$ be minimal such that $f(k)\ne g(k)$. Without loss of generality we may assume that $f(k)=0$ and $g(k)=1$. Then

$$\sum_{i=0}^kf(i)+\frac2{10^{k+1}}\in\theta(g)\setminus\theta(f)\,,$$

so $\theta(f)\ne\theta(g)$, and $\theta$ is an injection.


问:
Using the Chebyshev Inequality
This is the Q:

A 20 fair coins tosses, (f means the "H" of the coin).
I have to block the probability that I will get n/2+n/100 "H"-s by Chebyshev Inequality. [n=20 in this case...], so:

n/2+n/100 = 20/2+20/100 = 10.2

How I'm doing it?
the result it's much higher then one.

What I get is:

Click Here

I'm right?

答:
OK, if I got the question right, it should be
$$
P(|S_{20} - 10|>10.2)<\frac{20}{4 \cdot 10.2^2} \approx 0.048
$$</p>

Title: Using the Chebyshev Inequality


问:
Three quotient-ring isomorphism questions
I need some help with the following isomorphisms.

Let $R$ be a commutative ring with ideals $I,J$ such that $I \cap J = \{ 0\}$. Then

$I+J \cong I \times J$

$(I+J)/J \cong I$

$(R/I)/\bar{J} \cong R/(I+J) \quad \text{where} \quad  \bar{J}=\{x+I \in R/I: x \in J \}$

For the first item $\theta: I \times J \ \rightarrow I+J: \ (x,y) \mapsto x+y$ is clearly a surjective homomorphism. It's injective because $I \cap J = \{ 0\}$.

For the second item, the mapping $\eta \ : \ I+J \rightarrow I \ : \ x+y \mapsto x$ is well-defined by item 1, surjective, and the kernel equals $0+J$. Now we can use the first isomorphism theorem.

For the third item, I tried to find a define a mapping:
$$\phi: \quad R/I \ \rightarrow \ R/(I+J) \quad : \quad x + I \ \mapsto \ x+I+J $$
And I tried to show that $\bar{J}$ is the kernel, but it didn't totally feel okay because I got confused. Is the following correct?
$$ x \in \ker(\phi) \ \iff \ x+ I \in I+J \iff x+I \in \bar{J} $$

I would appreciate it if you could tell me if I made mistakes. Could you provide me a little information about the third item? It seems like a blur to me.

答:
The justification for why $ker(\phi) = \bar{J}$ is faulty, as Dune correctly points out.
So let's drill down on what the definition of each of these things really is.

Let $x + I$ be an element in $ker(\phi)$ this means that $\phi(x + I) = x + I + J = I + J$ in $R/(I + J)$.

From $x + I + J = I + J$ we get that $x\in I + J$. This means we can write $x$ as $x' + i$ for some $i\in I$ and $x' \in J$.

Note that the coset $x + I = x' + i + I = x' + I$. This satisfies the definition of $\bar{J}$. Therefore $x + I \in \bar{J}$.


问:
Let X ∼ exp (λ) and let a> 0. Calculate the expectation and variance of Y = min {X, a}.
Let X ∼ exp (λ) and let a> 0.
Calculate the expectation and variance of   Y = min {X, a}.

where if X ∼ exp (λ), that is, if P (X ≤ x) = 1 - e − λx

答:
I believe this is one way to set it up:

As Y takes on two values, use indicator functions to choose its values.
$\displaystyle Y=X\cdot\textbf 1_{X<a}+a\cdot \textbf 1_{X\ge a}$

The indicator functions determine the limits of integration.$\displaystyle E(Y)=\int_{-\infty}^\infty x\cdot\textbf 1_{x<a}f_X(x)dx+\int_{-\infty}^\infty a\cdot \textbf 1_{x\ge a} f_X(x)dx=\int_0^a xf_X(x)dx+a\int_a^\infty f_X(x)dx$

Use the standard formula for variance. $\displaystyle var(Y)=E(Y^2)-(E(Y))^2$

The indicator variables cancel each other out in the cross term. $\displaystyle Y^2=X^2\cdot\textbf 1_{X<a}+a^2\cdot \textbf 1_{X\ge a}$

The only issue is that you have to do a lot of integration by parts


问:
Example to show that $\mathbb Z^3$ is not a quotient of $\mathbb Z[X]$
I am asked to prove that $\mathbb{Z}^n$ is not a quotient of $\mathbb{Z}[X]$ for any integer $n \geq 3$ by showing that there is no surjective ring homomorphism $\mathbb{Z}[X] \to \mathbb{Z}^n.$

We can just focus on the $n = 3$ case since we can always look at the first 3 entries of the image. If $\phi: \mathbb{Z}[X] \to \mathbb{Z}^3$ is a ring homomorphism, then restricted to each entry, it is still a ring homomorphism $\phi_i: \mathbb{Z}[X] \to \mathbb{Z}, i = 1, 2, 3$. Now, $\phi_i$ must be the identity on $\mathbb{Z}$, and it is completely determined by $\lambda_i := \phi_i(X)$, so it is essentially the evaluation homomorphism $p \mapsto p(\lambda_i)$.

But $p(\lambda_i)$ is the remainder of $p(X)$ modulo $X - \lambda_i,$ so it reminds me of how in $\mathbb{Z},$ there is no integer $n$ satisfying $n \equiv 0 \bmod 3$ and $n \equiv 1 \bmod 6$. However, if the $\lambda_i$s are distinct, then the polynomials $X - \lambda_i$ look "coprime" to me, and I was wondering about whether there is a version of Chinese Remainder Theorem for ideals. Moreover, when looking at the examples, I noticed that (for instance) there is a surjective homomorphism $p \mapsto (p(0), p(1))$ since $(m, n)$ has pre-image $(m-n)X + n$, so why can we not extend it to a third remainder? More generally, how would the homomorphism fail to be surjective for any $\lambda_1, \lambda_2, \lambda_3$?

答:
Unfortunately, in order for the ideals $(X - a),$ $(X - b),$ and $(X - c)$ to be pairwise comaximal (i.e., in order for $(X - a) + (X - b) = \mathbb Z[x],$ etc.), we would need \begin{align*} b - a &= (X - a) - (X - b) = \pm 1, \\ c - a &= (X - a) - (X - c) = \pm 1, \text{ and} \\ c - b &= (X - b) - (X - c) = \pm 1. \end{align*} Considering that $c - b = (c - a) - (b - a)$ (for example), it cannot be the case that all three of these equations hold simultaneously.

On the other hand, observe that $X - (X - 1) = 1$ so that $(X) + (X - 1) = \mathbb Z[x].$ Consequently, by the Chinese Remainder Theorem, we have that $$\frac{\mathbb Z[x]}{(X) \cap (X - 1)} \cong \frac{\mathbb Z[X]}{(X)} \times \frac{\mathbb Z[X]}{(X - 1)} \cong \mathbb Z \times \mathbb Z.$$


答:
A function is not the same thing as a mathematical expression.  Every curve that you can draw/think of is already a function (as long as it does not cross the same vertical line twice), so the picture in your question already defines a function.

Once a function is defined, such as in your case, there are lots of mathematical methods to help you describe it via algebraic expressions.  Examples are Stone's Theorem (polynomial approximations), Fourier series (sines and cossines) and all sorts of other powerful methods!

Title: Sigmoid function with a longer, straighter middle

Body:
Can I modify the following function $y = 1 / (1 + e^{-x})$ such that I can make the curve straighter in the middle?

I would like the function to resemble a straight line until it approaches $0$ or $1$.

Here is an example diagram, the red line is desired:


问:
Laplace transform and Integral $f(t)$
I calculate Laplace transform.
$$f\left( t\right) =\dfrac{e^{λ_1t}-e^{λ_{2}t}}{t}$$ $$λ_1,λ_2>0$$

I think

$$\int _{0}^{\infty }f\left( t\right) e^{-st}dt=\int _{0}^{\infty }\dfrac{e^{\lambda _{1}t}-e^{\lambda _{2}t}}{t}e^{-st}dt$$
$$=\int _{0}^{\infty }\dfrac{e^{\left( λ_{1}-s\right) t}}{t}dt-\int _{0}^{\infty }\dfrac{e^{\left( λ_{2}-s\right) t}}{t}dt$$

I don't know what to do next. Please tell me hint.

答: There is a property of the Laplace transform where $\mathscr{L}\Big\{\frac{f(t)}{t}\Big\}= \int_{s}^{\infty}F(u)\,du$ where $F(s)=\mathscr{L}\{f(t)\}$

Title: Laplace transform and Integral $f(t)$


问:
A basic question about prenex normal form
I meet the following formula:

$$(\exists xAx\rightarrow\exists yBy)\rightarrow(\exists x\Box Ax\rightarrow\exists y\Box By)$$

I wonder what amounts to a prenex form of this formula, that is, putting all quantifiers in front of a quantifier-free formula.

Following the prenex rules for implication, I obtain the following:

$$\exists y\forall x(Ax\rightarrow By)\rightarrow\exists y\forall x(\Box Ax\rightarrow\Box By)$$

But now, $x$ and $y$ occur in both formulas. I don't know how to proceed. Can anyone help?

答:
Here's one way to do it, step by step.

$$ (\exists x A x \to \exists y B y) \to (\exists x \square A x \to \exists y \square B y) $$

replace $x \to y $ with $(\lnot x) \lor y$.

$$ (\exists x A x \not\to \exists y B y) \lor (\forall x \lozenge \lnot A x \lor \exists y \square B y ) $$

Replace $x \not\to y$ with $x \land \lnot y$.

$$ (\exists x A x \land \lnot \exists y B y) \lor \forall x \lozenge \lnot A x \lor \exists y \square B y  $$

Move $\lnot$ inside the quantifier.

$$ (\exists x A x \land \forall y \lnot B y) \lor \forall x \lozenge \lnot A x \lor \exists y \square B y $$

Rename the variables so that no bound variables have the same name.

$$ (\exists z A z \land \forall w \lnot B w) \lor \forall x \lozenge \lnot A x \lor \exists y \square B y $$

None of the variables are shared between multiple subexpressions, so we can extract them in whatever order we want.

$$ \exists z \mathop. \exists y \mathop. \forall w \mathop. \forall x \mathop. (Az \land \lnot B w) \lor \lozenge \lnot A x \lor \square B y $$


Body:
Let X ∼ exp (λ) and let a> 0.
Calculate the expectation and variance of   Y = min {X, a}.

where if X ∼ exp (λ), that is, if P (X ≤ x) = 1 - e − λx

Body:
$Z=\min(x,a)=\dfrac{(x+a)-|x-a|}{2}$

$Z = \begin{cases}\dfrac{(x+a)-(x-a)}{2}=a; & x>a\\\dfrac{(x+a)+(x-a)}{2}=x;& x<a \end{cases}$

$E(Z)$ is?


问:
Orthogonal on an ellipse
Consider two points $a \in \mathbb{R}^2$ and $b \in \mathbb{R}^2$ on the boundary of an ellipse such that $a^\top \Sigma^{-1} a = b^\top \Sigma^{-1} b$. I am trying to understand which points on the boundary of this ellipse satisfy $a^\top \Sigma^{-1} b = 0$. One possibility for this to happen is when $a \perp v_1$ and $b \perp v_2$, where $v_1$ and $v_2$ are eigenvectors of $\Sigma$, then $a^\top \Sigma^{-1} b = 0$. It is also clear that if $a$ and $b$ are valid, then $(-a, b), (a, -b) \text{ and } (-a, -b)$ are also valid solutions. Is it possible to find other points on the boundary of the ellipse such $a^\top \Sigma^{-1} b = 0$ or show that no other points besides these four exist, where this condition is satisfied ?

答:
Start with the simplest case:  ${\bf \Sigma} = {\bf I}$ (the identity matrix):

Here's what you get with ${\bf \Sigma} \neq {\bf I}$:

Clear?

Title: Orthogonal on an ellipse


问:
Functional Notation Confusion
This is a fairly basic question but I still can't seem to wrap my head around it, so any help would be appreciated.

In my homework, I have seen this functional notation being used:

$$\ln(1-n)$$

I am confused on what this means, because usually I am more familiar with functional notation like this:

$$f(n) = \ln n$$

So would that mean the $1-n$ in the first expression used is an input into the $\ln n$ function? Like so:

$$f(1-n) = \ln(1-n) = \text{constant}$$

Please let me know as I believe I'm interpreting it wrong.

答:
$\ln(1-n)$ is shorthand for the following calculation:

Suppose you have a number called $n$ take for example $0.1$.

You compute $1-n$, this would be $1-0.1 = 0.9$.

Then you compute the logarithm of that number. So in this case $\ln(1-n) = \ln(1-0.1) = \ln 0.9 $ $= -0.10536051565$.

If your book defines a function using $f(n) = \ln(1-n)$ they mean it maps a number $n$ to $\ln(1-n)$.

Title: Functional Notation Confusion


答:
Recall Euler's theorem: $a^{\phi(n)} \equiv 1 \pmod n$, where $\gcd(a,n)=1$. We hence have
$$48^{20} \equiv 1 \pmod{25}$$
Hence,
$$48^{322} \equiv \left(48^{20}\right)^{16} \cdot 48^2 \pmod{25} \equiv 48^2 \pmod{25} \equiv (-2)^2 \pmod{25} \equiv 4 \pmod{25}$$</p>

Title: Evaluating $48^{322} \pmod{25}$

Body: How do I find $48^{322} \pmod{25}$?</p>


问:
How can I prove that if $A$ is a singular square matrix, then $\det A = 0$ without using Binet's Theorem?
Well, I need to prove that if $A$ is a $n \times n$ matrix and it is singular, then $\det A = 0$, in order to show Binet's Theorem $\det(AB) = \det(A)\cdot\det(B)$ in the case when both $A$ and $B$ are singular square matrices.

I'll thank any help with this issue!

答:
I would use two proof

Permutation change the sign of the determinant

Multiplying a row by a constant multiply the determinant by this constant

Then, you can just say if you have a singular matrix S, 2 row $r_i$ and $r_j$ are linearly dependent. You can multiply $r_i$ by a constant C to get a new matrix S' where $r_i$ = $r_j$.

We know
\begin{equation}
det(S') = C det(S)
\end{equation}
Then, you can swap $r_i$ and $r_j$ but you have the same matrix has before so
\begin{equation}
det(S') = -det(S') = -C det(S) = C det(S)
\end{equation}
The equation det(S) = -det(S) means det(S) is 0.


问:
Can anyone explain this to me? (Epsilon and Delta)
If I have the statement $\lim_{x\to 0}f(x)=0$ and
$$
f(x)=\begin{cases}
x,&x>0\\
x-0.3,&x<0
\end{cases}
$$
how do I prove that the statement is false?

Is this correct?
$[f(x)-0]$ < e for every $[x-0]<d$

So if we choose e= 0.3 then it is $[f(x)]<0.3$ For every $[x]<d$

$-0.3<f(x)<0.3$ for x<d
But where do i go next?

答:
Hint.

The definition $\lim_{x\to 0}f(x)=0$ implies that for every $\epsilon>0$, there exists $\delta>0$ such that $-\delta<x<0$ implies $|f(x)-0|<\epsilon$.

But when $x<0$ and close to $0$, $f(x)$ is close to $-0.3$.

So your goal is showing the following:

there exists $\epsilon>0$ such that for every $\delta>0$, there exists $x\in(-\delta,0)$ such that $|f(x)-0|\ge \epsilon$.

Let $\epsilon=0.1$. For every $\delta>0$, $x=-\frac{\delta}{2}\in(-\delta,0)$ and
$$
|f(x)-0|=|x-0.3|=\frac\delta2+0.3>\epsilon.
$$
So the statement must be false.


问:
What is the distribution of residual in simple linear regression?
Suppose that
$$Y_i=\beta_0+\beta_1x_i+\epsilon_i,$$
where $\epsilon_1,\ldots,\epsilon_n$ are independent random variables with identical normal distribution $N(0,\sigma^2)$.
Let
$$\hat{Y}_i=B_0+B_1x_i$$
be the least squares fit of the data $\{(x_1,Y_1),\ldots,(x_n,Y_n)\}$. My question is, what is the distribution of the residual
$$e_i=Y_i-\hat{Y}_i \ ?$$
If it is not $N(0,\sigma^2)$, then in what sense are we justified to say that the residual $e_i$ "mimics" the error $\epsilon_i$ ? Are we justified to infer from the residual plot of $e_i$ anything about $\epsilon_i$ ?
I know that
$$\epsilon_i=Y_i-E[\hat{Y}_i],$$
but this is hardly a convincing reason to treat $e_i$ as if it "behaves like $\epsilon_i$."

答:
The $i$th residual has normal distribution, with mean zero and variance
$$
\operatorname{Var}(e_i) = \sigma^2\left(1-\frac1n-\frac{(x_i-\bar x)^2}{\operatorname{SSX}}\right)
$$
where SSX is shorthand for $\sum_k(x_k-\bar x)^2$. It has normal distribution because of the formulas
$$
e_i=y_i-\hat y_i=(\epsilon_i-\bar\epsilon)-(B_1-\beta_1)(x_i-\bar x)
$$
and
$$B_1-\beta_1=\frac{\sum_k(x_k-\bar x)(\epsilon_k-\bar \epsilon)}{\operatorname{SSX}}
$$ which express $e_i$ as a linear combination of the independent variables $\epsilon_1,\ldots,\epsilon_n$. The mean is computed as zero from the same formulas. An elementary derivation of the variance is found in this answer. A slicker but more advanced derivation is found in this answer.

More can be said: It turns out that the correlation between the residual $e_i$ and the error $\epsilon_i$ is
$$\sqrt{1-\frac1n-\frac{(x_i-\bar x)^2}{\operatorname{SSX}}}.
$$ This tells us that as $n$ gets large, the correlation tends to $1$, so that the residual is virtually identical to the error. This makes sense, because as the sample gets bigger, the estimators $B_0$ and $B_1$ converge to the true $\beta_0$ and $\beta_1$, hence the regression line $y=B_0+B_1 x$ converges to the theoretical line $y=\beta_0+\beta_1 x$.


答:
The variable $n$ in the expression $f(n)=\ln(n)$ is a "dummy variable"; this $n$ is different from the $n$ in $\ln(1-n)$.

You can write your function as $f(m)=\ln(m)$, and interprets $\ln(1-n)$ as the value of the function $f$ evaluated at $m=1-n$.

Alternatively, $\ln(1-n)$ is the composition $f\circ g$ of the function $f(n)=\ln(n)$ and the function $g(n)=1-n$.

Title: Functional Notation Confusion

Body:
This is a fairly basic question but I still can't seem to wrap my head around it, so any help would be appreciated.

In my homework, I have seen this functional notation being used:

$$\ln(1-n)$$

I am confused on what this means, because usually I am more familiar with functional notation like this:

$$f(n) = \ln n$$

So would that mean the $1-n$ in the first expression used is an input into the $\ln n$ function? Like so:

$$f(1-n) = \ln(1-n) = \text{constant}$$

Please let me know as I believe I'm interpreting it wrong.


问:
How do we calculate the proposed limit when $\theta\to 0^{+}$ and when $\theta\to 1^{-}$?
So here it is the function which I am interested in:
\begin{align*}
f(\theta) = \frac{6}{\theta^{2}}\left[-5\theta + (1+3\theta)\ln(1-\theta) + (3+\theta)\sqrt{\theta}\ln\left(\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right)\right]
\end{align*}

wher $\theta\in(0,1)$.

I would like to know how to calculate its limit when $\theta\to 0^{+}$ and when $\theta\to 1^{-}$.

One possible approach consists in applying the L'Hôpital rule when $\theta$ approaches $0$.

But isn't there any other "clever" way to calculate it?

Still, it remains to solve the problem when $\theta\to 1^{-}$.

This is part of my research project. Any hint is appreciated.

答:
I think that composition of Taylor series is convenient (working one piece at the time). If you have numerator and denominator present, work each of them separately.

Your case
\begin{align*}
f(\theta) = \frac{6}{\theta^{2}}\left[-5\theta + (1+3\theta)\log(1-\theta) + (3+\theta)\sqrt{\theta}\log\left(\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right)\right]
\end{align*} is rather simple since around $\theta=0$ since you need to use
$$\log(1-\theta)=-\theta -\frac{\theta ^2}{2}-\frac{\theta ^3}{3}-\frac{\theta ^4}{4}+O\left(\theta^5\right)$$ For the most delicate piece, I should first use $\theta=x^2$
$$A=\sqrt{\theta}\log\left(\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right)=x \log \left(\frac{x+1}{1-x}\right)=x \left(2 x+\frac{2 x^3}{3}+\frac{2 x^5}{5}+\frac{2 x^7}{7}+O\left(x^9\right) \right)$$
$$A=2 x^2+\frac{2 x^4}{3}+\frac{2 x^6}{5}+\frac{2 x^8}{7}+O\left(x^{10}\right)$$ Back to $\theta$
$$A=2 \theta +\frac{2 \theta ^2}{3}+\frac{2 \theta ^3}{5}+\frac{2 \theta
   ^4}{7}+O\left(\theta ^5\right)$$ It just remains to multiply, axpand and simplify before deviding by $ \theta ^2$to finally obtain
$$f(\theta) =3+\frac{\theta }{5}+\frac{3 \theta ^2}{70}+O\left(\theta ^3\right)$$ which shows the limit and how it is approached. Moreover, it gives an approximation which could be very useful for other purposes.

For example, use $\theta =0.1$ (far way from $0$). The exact value is, in decimals, $f(0.1)=3.02044$ while the truncated series gives $3.02043$.

For the limit when $\theta\to 1^-$, I should let $\theta=1-x$ and work around $x=0$ (this is always easier). I let you doing the work to arrive at
$$f(\theta) =(24 \log (4)-30)+x (30 \log (4)-42)+x^2 \left(36 \log
   (4)-\frac{99}{2}\right)+O\left(x^3\right)$$ Jus replace $x$ by $1-\theta$.

Here again, try with $\theta=0.9$.  The exact value is, in decimals, $f(0.9)=3.23294$ while the truncated series gives $3.23401$.


问:
A Viéte-like infinite product for $\sin (\pi/7)$
In the article "The unruly $\sin (\pi/7)$ of Samuel Moreno, the following infinite product is given:

Also, the same article shows that $\sin (\pi/7)$ is equal to the infinite nested radical

$ \sin (\pi/7) = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2+...}}}}}}}}}}$

The question is: are these two expressions fundamentally different for $\sin(\pi/7)$, or is it possible to rewritte the infinite nested radical into that infinite product by simple algebraic manipulations?

答:
With the help of Half angle cosine formula it is possible to express many sine and cosine angles as finite or infinite nested square roots of 2 here.

In this perspective $2\sin\frac{\pi}{7}$ can be solved as follows

$2\sin\frac{\pi}{7}$ = $\sqrt{2-2\cos\frac{2\pi}{7}}$ = $\sqrt{2-\sqrt{2+2\cos\frac{4\pi}{7}}}$ = $\sqrt{2-\sqrt{2-2\cos\frac{3\pi}{7}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{6\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-2\cos\frac{\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{2\pi}{7}}}}}$

One can observe repetition of $2\cos\frac{2\pi}{7}$ subsequently and $\therefore$ $2\sin\frac{\pi}{7}$ can be represented simply as

$$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+...2\cos\frac{2\pi}{7}}}}}}}}$$ with repeating pattern $--+$ infinitely and first $-$ as single nonrepeating sign

But getting infinite product has to be worked out


问:
Scalar dot product with directional derivative
I'm dealing with a problem of the form

$\phi \cdot \left[(A \cdot \nabla)B\right]$

in which $\phi$ is a scalar function and both $A$ and $B$ are vector functions.

Can I simply "distribute" the dot product with scalar function inside the expression to give

$ (C \cdot \nabla)B$

in which $C = \phi \cdot A$

Is this correct? I often get confused working with vector calculus operators.

On the side...

It seems that the parenthesis are important when reading the wikipedia identities. For example, would it be true that

$A \times (\nabla \times B) \neq (A \times \nabla) \times B$

Is there a "trick" to understanding how to manage the parenthesis?

答:
Actually, A⋅(∇B) is NOT the same as (A⋅∇)B [where A and B are vectors].

A⋅(∇B) is $ A_i \partial B_i/ \partial x_j $

whereas

(A⋅∇)B is $ A_i \partial B_j/ \partial x_i $

[also stated here: https://en.wikipedia.org/wiki/Vector_calculus_identities

(look at the portion just above the 'Second derivative identities' sub-topic in the page)]


问:
Is the set of surjective functions from $\mathbb{N}$ to $\mathbb{N}$ uncountable?
I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form $\Bbb{N} \to \Bbb{N}$ is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function $p$.The problem that I'm having is that I don't know in which way to change the diagonal elements so that $p$ should, but cannot be in the matrix.

Can someone help me, or at least give me some tips?

答: The following is essentially Asaf's construction except that I think it's a little closer to the standard "diagonal" method.  I assume that, as in the usual picture of the diagonal method, we're given an infinite matrix and that we want a new sequence that differs from every row of that matrix, but that we also want our new sequence to contain every natural number at least once.  Let me start by producing every second term of my new sequence; I'll fill in the omitted terms later.  The key idea is to use a slightly different "diagonal".  Instead of the "main diagonal", consisting of the $n$-th entry of the $n$-th row for all $n$, I'll use a "diagonal" of half the "slope".  That is, in the $n$-th row, I'll concentrate on the $2n$-th entry, and I'll make sure my new sequence differs from all these.  To be specific, I'll define the $2n$-th entry of my new sequence to be $1$ plus the $2n$-th entry in the $n$-th row.  (So I'm defining the even-numbered terms of my new sequence here; the odd-numbered ones will be defined later.)  Notice that, by doing this for every $n$, I've already ensured that my new sequence will differ from every row in the given matrix; specifically, it differs from row $n$ in position $2n$, and this difference is guaranteed no matter what I do with the odd-numbered positions in my new sequence.  But now it's easy to finish the job and take care of the surjectivity requirement; just put $k$ into the $k$-th odd-numbered position (i.e., into position $2k+1$) for every $k$.</p>


问:
Splitting exterior derivatives on product manifolds
Let $M_1 \times M_2$ be the product of two smooth manifolds $M_1, M_2$, and let
$\mu$ be a $k$-form on the product. Set $n_1 := \dim M_1$ and $n_2 := \dim M_2$.

The question is whether the product manifold structure induces a canonical splitting of the exterior derivative in the sense that
$$ \operatorname{d}\negthinspace \mu = \operatorname{d}_1 \negthinspace \mu +
\operatorname{d}_2 \negthinspace \mu \, .$$

Indeed, if one looks at the coordinate expression
$$\mu = \frac{1}{k!} \, \mu_{i_1 \dots i_k} \, \operatorname{d}\negthinspace x^{i_1}\wedge
\dots \wedge \operatorname{d}\negthinspace x^{i_k} \, ,$$
in product coordinates
$$\left(x^1, \dots, x^{n_1},x^{n_1+1}, \dots, x^{n_1 +n_2} \right)
= \left(x^1_1, \dots, x^{n_1}_1,x^{1}_2, \dots, x^{n_2}_2 \right) \, , $$
then
\begin{align}
\operatorname{d}\negthinspace \mu
&= \frac{1}{k!}  \, \left( \sum_{j_1=1}^{n_1}  \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_1}_1} \,
\operatorname{d} \negthinspace x_1^{j_1}
 + \sum_{j_2=1}^{n_2} \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_2}_2} \,
\operatorname{d} \negthinspace x_2^{j_2}
\right)\wedge \operatorname{d}\negthinspace x^{i_1}\wedge
\dots \wedge \operatorname{d}\negthinspace x^{i_k}
\\
&=  \frac{1}{k!}  \, \left( \sum_{j_1=1}^{n_1}  \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_1}_1} \,
\operatorname{d} \negthinspace x_1^{j_1}
\right) \wedge \operatorname{d}\negthinspace x^{i_1}\wedge
\dots \wedge \operatorname{d}\negthinspace x^{i_k}
 +  \frac{1}{k!}  \,
\left( \sum_{j_2=1}^{n_2} \frac{\partial \mu_{i_1 \dots i_k}}{\partial x^{j_2}_2} \,
\operatorname{d} \negthinspace x_2^{j_2}
\right)\wedge \operatorname{d}\negthinspace x^{i_1}\wedge
\dots \wedge \operatorname{d}\negthinspace x^{i_k}  \, .
\end{align}
(Of course, one needs to anti-symmetrize to get the components of the $(k+1)$-forms.)

Is there a way to define this invariantly e.g. by using projections onto some subbundles of
the tangent bundle/tensor bundles of the product? A quick search in the literature did not really yield anything. One would expect this to be more common, if this were a thing.

EDIT: To maybe give some motivation/context for why one would want to do this, consider the projection mappings onto the respective factor
$$\operatorname{pr}_i \colon M_1 \times M_2 \to M_i \colon (m_1,m_2) \mapsto m_i \, ,$$
a differential form $\alpha$ on say $M_1$, as well as a vector field $X$ on the product $M_1 \times M_2$. One shows easily that formally
$$X \lrcorner \left( \operatorname{pr}_1^* \alpha \right) = \operatorname{pr}_1^* \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr)  \, .$$
($\lrcorner$ is contraction, the upper and lower stars denote pullback and pushforward, respectively)

However, the expression in parentheses on the right hand side is not a differential form on $M_1$, so
$$\operatorname{d} \left( \negthinspace \operatorname{pr}_1^* \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr) \right)
\neq
\operatorname{pr}_1^* \left( \operatorname{d}   \Bigl( \bigl((\operatorname{pr}_1)_* X \bigr) \lrcorner \alpha \Bigr) \right)$$
It is the splitting I was asking for that would allow this equality to hold, if $\operatorname{d}$ were replaced by $\operatorname{d} \negthinspace _1$ on the left hand side.

答:
Let $\pi_1,\pi_2$ denote the canonical projections and $A_1=\pi_1^*TM_1$, $A_2=\pi_2^*TM_2$. We have the canonical decomposition of vector bundles which splits $T(M_1\times M_2)$ into "horizontal" and "vertical" parts.
$$
T(M_1\times M_2)\cong A_1\oplus A_2
$$
This in turn gives rise to a decomposition of exterior powers
$$
\Lambda^kT^*(M_1\times M_2)\cong\bigoplus_{p+q=k}\Lambda^pA_1^*\otimes\Lambda^qA_2^*
$$
which gives us projections $P^{p,q}$ to each factor of this direct sum. We can use these to write the decomposition as
$$
d_1=\sum_{p+q=k}P^{p+1,q}\circ d\circ P^{p,q},\ \ \ \ \ d_2=\sum_{p+q=k}P^{p,q+1}\circ d\circ P^{p,q}
$$
That said, this decomposition may not do what you want it to. The equality you give:
$$
X\lrcorner\pi_1^*\alpha=\pi_1^*(\pi_{1*}X\lrcorner\alpha)
$$
Doesn't make sense, since the pushforward $\pi_{1*}X$ isn't generally well defined. In fact, $X\lrcorner\pi_1^*\alpha$ cannot generally be expressed as a pullback. One can write a similar expression using Cartan's magic formula, but an extra Lie derivative appears:
$$
d(X\lrcorner\pi_1^*\alpha)=\mathcal{L}_X(\pi_1^*\alpha)-X\lrcorner\pi_1^*(d\alpha)
$$


问:
Does a bounded sequence with diminishing increment converge?
In my research, I have the following discrete-time dynamical system

$M_k=\frac{1}{k+1} I_k+\frac{k}{k+1}M_{k-1},$

where $|I_k|<c$ is absolutely bounded, and $I_k$ depends on $M_{k-1}$.

Now, I want to prove that $M_k$ has a limit. I can rewrite the dynamic as follows

$M_k=\frac{1}{k+1} M_0+\frac{1}{k+1}\sum_{t=1}^k I_t.$

Since I know the initial condition $|M_0|<c$ is absolutely bounded, it means $|M_k|<c$ is also absolutely bounded. I can also rewrite the original dynamic as follows

$M_k-M_{k-1}=\frac{1}{k+1}(I_k-M_{k-1}),$

thus the increment $|M_k-M_{k-1}|<\frac{2c}{k+1}$ is diminishing.

How can I proceed from here to prove/disprove that $M_k$ has a limit?

Thank you in advance for any help.

Edit 1: After seeing the counter-examples some people constructed, I think it is better to illustrate a little bit more on the dynamic of $I_k$. In my modeling, $I_k$ is governed by a random process. Let $\omega_k$ denotes the realization of the state at time $k$ which is sampled $i.i.d.$ from a known distribution $\mu$. In addition to that, $I_k$ also depends on $M_{k-1}$. The simplified version of the dynamic of $I_k$ can be written as follows

$I_k=\alpha^{\omega_k} f(M_{k-1})+\beta^{\omega_k},$

where $\alpha^{\omega_k}, \beta^{\omega_k}$ are state-dependent parameters, $f(\cdot)$ is a real function and is Lipschitz continuous.

Will the additional information of $I_k$'s dynamic help to prove the existence of the limit of $M_k$? Maybe some variations of the Law of Large Numbers might work here?

答:
Not in general. Consider the sequence whose first terms are
$$
1,1+1/2,1+1/2+1/3, \dots 1+1/2+1/3+\dots 1/n
$$
where $n$ is chosen in such a way that $a_n\le 2$, $a_n+1/(n+1)>2$. Let the next group of terms be
$$
a_n-1/(n+1), a_n-1/(n+1)-1/(n+2),..a_n-1/(n+1)-1/(n+2)-\dots -1/m
$$
where $m$ is chosen in such a way that $a_{n+m}\ge -2$, $a_{m+n}-1/(m+1)<-2$
and so on. Then clearly
$$
|a_k-a_{k-1}|=1/k\to 0
$$
but your sequence oscillates between $-2$ and $+2$ without limit. Observe that this construction is possible because the harmonic series is divergent.


问:
Function "evaluation" just means "composition"?
I am self-studying and have a basic or naive question that follows from a simple observation. I have also included tags for type theory, etc because "evaluation" probably has a different meaning in these constructive approaches, which I'm also curious about.

Any set $X$ with elements $x_i$ can be characterized by a set denoted $(X)$ of pairs $(\star, x_i)$, alternatively denoted $(\star \mapsto x_i)$ (for all $x_i \in X$). Each one of these latter pairs encodes a distinct function from common domain $\{ \star \}$ to (each different element of) $X$. If we name such functions after their image, i.e. $(x_i):\star \mapsto x_i$, then the symbol "$f(x_i)$", traditionally interpreted as function evaluation, is reinterpreted as function composition, i.e. $f\circ (x_i)$ and the $\circ$ is often notationally suppressed. And roughly, $f = \cup_xf(x)$, every function decomposes to a unique set of singleton functions (better word?).

Question: Is there anything more to evaluation, or can I exactly equate evaluation with composition and forever de-clutter my mind of the term "evaluation"? Is the above perspective unusual or fruitful beyond reducing terminological clutter?

The notion of evaluation above - perhaps "extensional" is the right term? - means essentially subsetting the function at the chosen argument $x$. Perhaps I will see that a full answer to my question requires considering an intentional (?) or constructive perspective on "evaluation" (as in lambda calculus, etc). Perhaps only there will composition and evaluation legitimately have distinct meanings.

Edit: The notation $(x)$ seemed the most natural way to denote any function with singleton domain because it coheres with the conventional way to write evaluation as $f(x)$. The notation seems unambiguous because there is no context I can think of, other than "evaluation", where a single symbol, say $x$, is enclosed in brackets "()"? It does seem useful to notationally distinguish functions on singleton domain in some way: I would like the notation to be a mnemonic for whether the resulting function composite is a singleton function (i.e. a "value" like $f(x_i) = f \circ (x_i) = (\star, y_j)$) versus a non-singleton function. Perhaps there is a conventional notation for functions from singleton domain, other than $(x)$ and a conventional name for such functions, other than "singleton functions"?

答:
Is there anything more to evaluation, or can I exactly equate evaluation with composition and forever de-clutter my mind of the term "evaluation"?

Frankly, there is no way to non-circularly 'get rid' of the notion of "evaluation". A rose by any other name is still a rose. Even Lawvere's axiomatization of ETCS explicitly axiomatizes the evaluation mapping in Axiom 2, just phrased in category theory terms.

I think what you must realize is that certain mathematical notions are core notions and whatever implementation you use for them does not make them disappear at all. For example, it makes no sense to try to argue that ZFC set theory has eliminated the notion of function application because instead of "$f(x)$" we can use "$\bigcup \{ y : ⟨x,y⟩∈f \}$" given by the Union axiom and the Replacement schema (or a slightly more complicated expression if you use Specification and Powerset instead of Replacement).

The whole point is that we design foundational systems to be able to do the mathematics that we want to do. Since we want to have the usual natural numbers with the usual induction, we also want our foundational system to be able to construct a model of PA (for suitable encoding of "model"). And since we want to be able to reason about functions on $ℕ$, functions on $ℝ$ and so on, with the usual notion of "function", including evaluation, composition, function types and equality, we make sure that our foundational system is able to construct representations of the usual functions and prove those usual properties! See also this related post explaining why we care about abstractions rather than how things are concretely implemented in the foundational system.

There is nothing wrong with thinking about weaker systems that cannot do some of these things, but most logicians would not consider these systems viable as foundations for all mathematics, precisely for that reason!


问:
When does $H^{n-k}(M,\Bbb R)\simeq \Bbb R$ imply $H_{k}(M,\Bbb Z)\simeq \Bbb Z$?
I am not sure that my question is a trivial fact or not or even make sense or not. Anyway I want to know

When does cohomology group $H^{n-k}(M,\color{blue}{\Bbb R})\simeq \Bbb R$ imply homology group $H_{k}(M,\color{red}{\Bbb Z})\simeq \Bbb Z$ for a closed smooth manifold $M^n$?

I also want to see a general version of this statement but I don't know what it should be! Here are two related posts:

Does trivial cohomology imply trivial homology? Does $\operatorname{Hom}(A,\mathbb Z) = \operatorname{Ext}^1(A, \mathbb Z) = 0$ imply $A = 0$?

Is homology determined by cohomology?

答:
The Universal Coefficient Theorem for cohomology states that there is an exact sequence

$$0 \to \operatorname{Ext}^1_{\mathbb{Z}}(H_{n-k-1}(M; \mathbb{Z}), \mathbb{R}) \to H^{n-k}(M; \mathbb{R}) \to \operatorname{Hom}_{\mathbb{Z}}(H_{n-k}(M; \mathbb{Z}), \mathbb{R}) \to 0.$$

As $\mathbb{R}$ is a divisible abelian group (and hence, an injective $\mathbb{Z}$-module), the first group vanishes and $H^{n-k}(M; \mathbb{R}) \cong \operatorname{Hom}_{\mathbb{Z}}(H_{n-k}(M; \mathbb{Z}), \mathbb{R}) \cong \mathbb{R}^{b_{n-k}(M)}$. So if $H^{n-k}(M; \mathbb{R}) \cong \mathbb{R}$, we see that $b_{n-k}(M) = 1$.

If $M$ is orientable, then by Poincaré duality $b_k(M) = b_{n-k}(M) = 1$, so $\operatorname{rank}H_k(M; \mathbb{Z}) = 1$, i.e. $H_k(M; \mathbb{Z})$ is of the form $\mathbb{Z}\oplus T$ for some finitely generated abelian torsion group $T$. Note that $T$ need not be zero. For example, let $M = (S^1\times S^2)\#\mathbb{RP}^3$. Then one can show that $H^2(M; \mathbb{R}) \cong \mathbb{R}$ and $H_1(M; \mathbb{Z}) = \mathbb{Z}\oplus\mathbb{Z}_2$; see this answer for the computation of (co)homology groups of a connected sum.

If $M$ is non-orientable, then we need not even have $b_k(M) = 1$. For example, let $M = \mathbb{RP}^4\#\mathbb{RP}^4$. Then $H^3(M; \mathbb{R}) \cong \mathbb{R}$ but $H_1(M; \mathbb{Z}) \cong \mathbb{Z}_2\oplus\mathbb{Z}_2$.

What follows addresses the discussion in the comments.

Let $M$ be a closed manifold. Then there are finitely generated abelian torsion groups $T_i$ and $T^i$ such that $H_i(M; \mathbb{Z}) \cong \mathbb{Z}^{b_i(M)}\oplus T_i$ and $H^i(M; \mathbb{Z}) \cong \mathbb{Z}^{b_i(M)}\oplus T^i$. It follows from the Universal Coefficient Theorem that $T^i \cong T_{i-1}$. If $M$ is orientable, then we also have $T^i \cong T_{n-i}$ by Poincaré duality. If $\dim M = 4$, this leaves only one possible torsion group: $T^3 \cong T_2 \cong T^2 \cong T_1$. In particular, a closed orientable four-manifold has torsion in its homology/cohomology if and only if it has torsion in $H_1$. It follows that a simply connected closed four-manifold has no torsion in its homology/cohomology because such a manifold is orientable and has $H_1 = 0$. This conclusion is no longer true in higher dimensions. For example, the Wu manifold $W = SU(3)/SO(3)$ is simply connected, but $H_2(W; \mathbb{Z}) \cong \mathbb{Z}_2$; see this answer.


问:
If $R$ is a commutative ring, can we show that the set of nilpotent elements of $R$ form an ideal without using the Binomial Theorem?
All answers I have seen regarding this question (showing that the set of nilpotent elements of $R$ form an ideal) make use of the Binomial theorem in order to prove that $(N,+)$ is an additive subgroup of $(R,+)$. My book (Abstract Algebra A First Course by Dan Saracino) makes no reference to the Binomial theorem, and so I am unsure as to whether I can use it (perhaps that is a silly question).

答:
Yes, you could show that the set of nilpotent elements is equal to the intersection of all the prime ideals!  The intersection of a family of ideals is an ideal.

But typically showing that everything in that intersection is nilpotent uses the concept of localization, which is more advanced than just the binomial theorem.

If I were you and squeamish about assuming the binomial Theorem, I’d just prove it.


答:
I'm afraid it's not true that $M_k$ will always have a limit. The sequence $I_k$ can be chosen in such a way that it won't. I'll take $c>1$ in this example, and $M_k=0$. Define $I_k$ such that

$I_0=1$, if $M_k \ge \frac{1}{2}$ then $I_k=-1$, if $M_k \le -\frac{1}{2}$ then $I_k = 1$ and if $-\frac{1}{2} < M_k < \frac{1}{2}$ then $I_k = I_{k-1}$.

This sequence will "attempt to converge" to $1$ in the sections where $M_k < \frac{1}{2}$ and to $-1$ in the sections where $M_k > -\frac{1}{2}$, therefore oscillating between $\frac{1}{2}$ and $-\frac{1}{2}$.

Title: Does a bounded sequence with diminishing increment converge?

Body:
In my research, I have the following discrete-time dynamical system

$M_k=\frac{1}{k+1} I_k+\frac{k}{k+1}M_{k-1},$

where $|I_k|<c$ is absolutely bounded, and $I_k$ depends on $M_{k-1}$.

Now, I want to prove that $M_k$ has a limit. I can rewrite the dynamic as follows

$M_k=\frac{1}{k+1} M_0+\frac{1}{k+1}\sum_{t=1}^k I_t.$

Since I know the initial condition $|M_0|<c$ is absolutely bounded, it means $|M_k|<c$ is also absolutely bounded. I can also rewrite the original dynamic as follows

$M_k-M_{k-1}=\frac{1}{k+1}(I_k-M_{k-1}),$

thus the increment $|M_k-M_{k-1}|<\frac{2c}{k+1}$ is diminishing.

How can I proceed from here to prove/disprove that $M_k$ has a limit?

Thank you in advance for any help.

Edit 1: After seeing the counter-examples some people constructed, I think it is better to illustrate a little bit more on the dynamic of $I_k$. In my modeling, $I_k$ is governed by a random process. Let $\omega_k$ denotes the realization of the state at time $k$ which is sampled $i.i.d.$ from a known distribution $\mu$. In addition to that, $I_k$ also depends on $M_{k-1}$. The simplified version of the dynamic of $I_k$ can be written as follows

$I_k=\alpha^{\omega_k} f(M_{k-1})+\beta^{\omega_k},$

where $\alpha^{\omega_k}, \beta^{\omega_k}$ are state-dependent parameters, $f(\cdot)$ is a real function and is Lipschitz continuous.

Will the additional information of $I_k$'s dynamic help to prove the existence of the limit of $M_k$? Maybe some variations of the Law of Large Numbers might work here?


问:
extending a norm on an $\mathbb{R}$-vector space

Let $(V,\lVert\cdot \rVert_0)$ be a normed vector space and let $(\overline{V}, \overline{d_0}) $ be a completion of $(V,d_0)$ where $d_0(x,y) = \lVert x - y \rVert_0.$ Show that $\overline{V}$ is an $\mathbb{R}$-vector space and the norm $\lVert\cdot \rVert_0$ extends to a norm $\lVert\cdot \rVert$ on $\overline{V}.$

To show that $\overline{V}$ is an $\mathbb{R}$-vector space, one should show that it satisfies the properties of a vector space (e.g. associativity, distributivity, commutativity of addition, etc.) and I believe that is relatively routine, though shouldn't this use the property that $\overline{V}$ is complete and contains a dense vector space that's isomorphic to $V$ ?

Also, how should I extend the norm? Should I use the completion $\overline{d_0}$ and define $\lVert x\rVert := \overline{d_0}(x,0)$ (I'm quite sure all normed vector spaces contain zero)?

Or is it necessary to consider the functions $A: V\times V\to V, A(v,w) = v+w$ and $S_\alpha : V\to V, S_\alpha(v) = \alpha v, \alpha\in \mathbb{R}$?

答: The metric completion is done in such a way that $V$ is dense in $\bar V$. You extend the norm to elements $x\in \bar V\setminus V$ by setting $\|x\|=\bar d_0(x,0)$ as you suggested. The fact that this is a norm follows from the continuity of the extended metric $d_0$ and the fact that this metric, restricted to $V$, is associated to a norm (you can take limits in all the properties: triangle inequality, homogeneity, etc.).


答:
Suppose $a,b$ are nilpotent elements of $R$, say $a^n=0=b^m$. Consider the ring $\overline{R}=R/Ra$ and use $\overline{\phantom{b}}$ to denote cosets in $\overline{R}$. Then $\overline{b}\,^m=0$. Since $\overline{a+b}=\overline{b}$, we have $\overline{a+b}\,^m=\overline{0}$. Thus $(a+b)^m=ra$ for some $r\in R$, and so $(a+b)^{mn}=r^na^n=0$.

This proof can be modified slightly to prove that any finite sum of nilpotent ideals in any ring (including noncommutative rings) is nilpotent and any (possibly infinite) sum of nil ideals is nil.

Title: If $R$ is a commutative ring, can we show that the set of nilpotent elements of $R$ form an ideal without using the Binomial Theorem?

Body: All answers I have seen regarding this question (showing that the set of nilpotent elements of $R$ form an ideal) make use of the Binomial theorem in order to prove that $(N,+)$ is an additive subgroup of $(R,+)$. My book (Abstract Algebra A First Course by Dan Saracino) makes no reference to the Binomial theorem, and so I am unsure as to whether I can use it (perhaps that is a silly question).


问:
Writing a Recursive Sequence and finding out limit by a "weird" property
I apologize for lacking of a better title and for potencial math language mistakes as I'm not an english native so there's some terms I might miss. I found out a note of my college that had the following written on:

Calculate $\lim(x_n)$ if $$x_n=\frac{n}{2^{n+1}}\sum_{i=1}^{n}\frac{2^i}{i}$$

Then their resolution follows as:

This sequence can be defined by recursion as
$$x_{n+1}= \frac{n+1}{2n}.\underbrace{\frac{n}{2^{n+1}}\sum_{i=1}^{n}\frac{2^i}{i}}_{x_n} + \frac{n+1}{2^{n+2}}\frac{2^{n+1}}{n+1} = $$
$$\frac{n+1}{2n}x_n + \frac{1}{2} = x_{n+1}$$

I really don't understand how they defined the recursive sequence, I tried to calculate $x_{n+1}$ and had a different result than the solution everytime I tried (if I even did it correctly). The next step is perhaps the most confuse to me, as they said:

$\frac{n+1}{2n} \to \frac{1}{2}\hspace{1.5mm}\text{(obvious to me)}$ and as stated in a theorem we studied before (I really don't recognize this theorem and can't search for it) $x_n$ has a limit and it is given by $$\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$

Besides not getting how they defined the recursive sequence I believe I'm not aware of the theorem they stated as well (even though the outcome looks like the sum of a geometric series of ratio $\frac{1}{2}$ starting on the first term) so I would really appreciate it if someone could help me figuring out both things.

Thank you so much for your attention.

答:
Perhaps it will be clearer if I include a few more steps:

$$\begin{align*}
x_{n+1}&\overset{(1)}=\frac{n+1}{2^{n+2}}\sum_{i=1}^{n+1}\frac{2^i}i\\
&\overset{(2)}=\left(\frac{n+1}{2n}\cdot\frac{n}{2^{n+1}}\right)\left(\sum_{i=1}^n\frac{2^i}i+\frac{2^{n+1}}{n+1}\right)\\
&\overset{(3)}=\frac{n+1}{2n}\color{red}{\left(\frac{n}{2^{n+1}}\sum_{i=1}^n\frac{2^i}i\right)}+\left(\frac{n+1}{2n}\cdot\frac{n}{2^{n+1}}\cdot\frac{2^{n+1}}{n+1}\right)\\
&\overset{(4)}=\frac{n+1}{2n}x_n+\frac12
\end{align*}$$

Step $(1)$ is just the definition of $x_{n+1}$. Step $(2)$ is rewriting the coefficient $\frac{n+1}{2^{n+2}}$ in what will prove to be a more useful form and splitting off the $i=n+1$ term of the summation. Step $(3)$ is just multiplying out and emphasizing the red part as a single unit. And step $(4)$ is identifying that unit as $x_n$ and simplifying the last term.

It follows that if $\langle x_n:n\in\Bbb Z^+\rangle$ converges to some limit $x$, we must have

$$\begin{align*}
x&=\lim_{n\to\infty}x_{n+1}\\
&=\frac12+\lim_{n\to\infty}\frac{n+1}{2n}x_n\\
&=\frac12+\left(\lim_{n\to\infty}\frac{n+1}{2n}\right)\lim_{n\to\infty}x_n\\
&=\frac12+\frac{x}2
\end{align*}$$

and hence $x=1$. Thus, we have only to make sure that the sequence actually does converge. I don’t know what theorem was intended, but here is a (somewhat brute force) argument to show that the sequence converges.

By direct calculation $x_3=\frac54$. An easy induction on $n$ applied to the recurrence

$$x_{n+1}=\frac{n+1}{2n}x_n+\frac12$$

shows that $x_n>1$ for all $n\ge 3$: if $x_n>1$, then $\frac{n+1}{2n}x_n>\frac12$, and $x_{n+1}>1$.

Note that $x_n-x_{n+1}=\frac{n-1}{2n}x_n-\frac12$ is non-negative if and only $x_n\ge\frac{n}{n-1}$. By direct calculation $x_4=\frac43=\frac4{4-1}$. Suppose that $x_n\ge\frac{n}{n-1}$ for some $n\ge 4$. Then

$$\begin{align*}
x_{n+1}&=\frac{n+1}{2n}x_n+\frac12\\
&\ge\frac{n+1}{2n}\cdot\frac{n}{n-1}+\frac12\\
&=\frac{n+1}{2(n-1)}+\frac{n-1}{2(n-1)}\\
&=\frac{2n}{2(n-1)}\\
&=\frac{n}{n-1}\\
&>\frac{n+1}n\,,
\end{align*}$$

so by induction $x_n\ge\frac{n}{n-1}$ for all $n\ge 4$ (and in fact the strict inequality at the last step means that $x_n>\frac{n}{n-1}$ for all $n\ge 5$). It follows that $x_n-x_{n+1}\ge 0$ for all $n\ge 4$ (and in fact that $x_n-x_{n+1}>0$ for $n\ge 5$), so the sequence $\langle x_n:n\ge 4\rangle$ is decreasing and bounded below and therefore must converge.


问:
Boundedness of the inverse (Evans) why the chosen subsequence converge in $L^2(U)$

THEOREM 6 (Boundedness of the inverse). If $\lambda \notin \Sigma$, there exists a constant $C$ such that

$$
\|u\|_{L^{2}(U)} \leq C\|f\|_{L^{2}(U)} \tag{29}
$$
whenever $f \in L^{2}(U)$ and $u \in H_{0}^{1}(U)$ is the unique weak solution of
$$
\left\{\begin{array}{c1}
L u=\lambda u+f & \text { in } U \\
u=0 & \text { on } \partial U
\end{array}\right.
$$
The constant $C$ depends only on $\lambda, U$ and the coefficients of $L .$
This constant will blow up if $\lambda$ approaches an eigenvalue.

Proof. If not, there would exist sequences $\left\{f_{k}\right\}_{k=1}^{\infty} \subset L^{2}(U)$ and $\left\{u_{k}\right\}_{k=1}^{\infty} \subset$ $H_{0}^{1}(U)$ such that
$$
\left\{\begin{aligned}
L u_{k} &=\lambda u_{k}+f_{k} & & \text { in } U \\
u_{k} &=0 & & \text { on } \partial U
\end{aligned}\right.
$$
in the weak sense, but
$$
\left\|u_{k}\right\|_{L^{2}(U)}>k\left\|f_{k}\right\|_{L^{2}(U)} \quad(k=1, \ldots)
$$
As we may with no loss suppose $\left\|u_{k}\right\|_{L^{2}(U)}=1,$ we see $f_{k} \rightarrow 0$ in $L^{2}(U)$. According to the usual energy estimates the sequence $\left\{u_{k}\right\}_{k=1}^{\infty}$ is bounded in $H_{0}^{1}(U) .$ Thus there exists a subsequence $\left\{u_{k_j}\right\}_{j=1}^{\infty}$ $\subset \left\{u_{k}\right\}_{k=1}^{\infty}$ such that
$$
\left\{\begin{array}{c1}
u_{k_j}&→u  & & \text {weakly in } H_{0}^{1}(U) \\
u_{k_j} &→u & & \text {in } L^2(U)\\ \tag{30} \label{eq30}
\end{array}\right.
$$
(See §D.4 for weak convergence.)Then $u$ is a weak solution of
$$
\left\{\begin{array}{c1}
L u=\lambda u & \text { in } U \\
u=0 & \text { on } \partial U
\end{array}\right.
$$
since $λ \notin \Sigma, u \equiv 0 \dots$

Why could the subsequence $ u_{k_j} $ converge in $L^2(U)$ and why it converges to $u$? (in \eqref{eq30})

Why could we take the limit in \eqref{eq30}?

答:
The reason is stated in the appendix D4. Let me give slightly more detail. By Banach-Alaoglu, a bounded sequence in a Banach space has a subsequence that converges in the weak* topology. Since $H^1_0$ is a (separable) Hilbert space (in particular reflexive), this subsequence, considered as elements in the dual of $H^1_0$, converge in the weak topology. Regarding why convergence is equal to $u$, see 3 below.

This is guaranteed by the Rellich-Kondrachov compactness theorem(see previous chapter): $L^2(\Omega)$ is compactly embedded into $H^1_0(\Omega)$. So norm-boundedness in $H^1_0$ implies strong convergence of a subsequence in $L^2$.

You first take a subsequence $u_{k_i}$ of $u_k$ such that $u_{k_i}$ is weakly convergent in $H^1_0$ to something, call the limit $u$. Now you take a subsequence $v_j=u_{k_{i_j}}$ of $u_{k_i}$ such that $u_{k_{i_j}}$ converges to something in $L^2$, say it is $v$. In particular it converges weakly to $v$ in $L^2$. But $L^2$ is a subset of the dual of $H^1_0$. So by uniqueness of weak limits, $v=u$.


问:
Example to show that $\mathbb Z^3$ is not a quotient of $\mathbb Z[X]$
I am asked to prove that $\mathbb{Z}^n$ is not a quotient of $\mathbb{Z}[X]$ for any integer $n \geq 3$ by showing that there is no surjective ring homomorphism $\mathbb{Z}[X] \to \mathbb{Z}^n.$

We can just focus on the $n = 3$ case since we can always look at the first 3 entries of the image. If $\phi: \mathbb{Z}[X] \to \mathbb{Z}^3$ is a ring homomorphism, then restricted to each entry, it is still a ring homomorphism $\phi_i: \mathbb{Z}[X] \to \mathbb{Z}, i = 1, 2, 3$. Now, $\phi_i$ must be the identity on $\mathbb{Z}$, and it is completely determined by $\lambda_i := \phi_i(X)$, so it is essentially the evaluation homomorphism $p \mapsto p(\lambda_i)$.

But $p(\lambda_i)$ is the remainder of $p(X)$ modulo $X - \lambda_i,$ so it reminds me of how in $\mathbb{Z},$ there is no integer $n$ satisfying $n \equiv 0 \bmod 3$ and $n \equiv 1 \bmod 6$. However, if the $\lambda_i$s are distinct, then the polynomials $X - \lambda_i$ look "coprime" to me, and I was wondering about whether there is a version of Chinese Remainder Theorem for ideals. Moreover, when looking at the examples, I noticed that (for instance) there is a surjective homomorphism $p \mapsto (p(0), p(1))$ since $(m, n)$ has pre-image $(m-n)X + n$, so why can we not extend it to a third remainder? More generally, how would the homomorphism fail to be surjective for any $\lambda_1, \lambda_2, \lambda_3$?

答:
Suppose $p(x) \in \mathbb{Z}[x]$ a polynomial and $deg(p) > 1$.

Let $\lambda_{1} \neq \lambda_{2}$ be integers such that $p(\lambda_{1}) \neq p(\lambda_{2})$, for a certain $p(x) \in \mathbb{Z}[x]$.

Then there are $f_{1}(x), f_{2}(x) \in \mathbb{Z}[x]$ such that, for maximal values of the exponents $m_{1}, m_{2}$ we can write:

$$p(x) = p(\lambda_{1}) + (x-\lambda_{1})^{m_{1}}f_{1}(x),$$

$$p(x) = p(\lambda_{1}) + (x-\lambda_{1})^{m_{1}}[(x-\lambda_{2})^{m_{2}}f_{2}(x) + f_{1}(\lambda_{2})].$$

Using this last equation, we have that:

$$p(\lambda_{2}) - p(\lambda_{1}) = (\lambda_{2} - \lambda_{1})^{m_{1}}f_{1}(\lambda_{2}).$$

Because $deg(p) > 1$ and $p(\lambda_{2}) \neq p(\lambda_{1})$, we have that $m_{1} \geq 1$
and $f_{1}(\lambda_2) \neq 0$ so, we have that $(\lambda_{2} - \lambda_{1}) | (p(\lambda_{2}) - p(\lambda_{1}))$.

Now let us get back to your problem: if there is such a surjective homomorphism $p \mapsto (p(\lambda_{1}), p(\lambda_{2}), p(\lambda_{3}))$ (all $\lambda_{i}$'s are distinct), then there would exist infinite values $(a,b,c) \in \mathbb{Z}^{3}$ that are the image of polynomials of degree at least 2 (since the polynomials of degree $\leq 1$ form a $\mathbb{Z}$-module of rank 2, and $\mathbb{Z}^{3}$ has rank 3). Now, for such polynomials, if $p(\lambda_{i}) \neq p(\lambda_{j})$ we know that:

$$(\lambda_{i} - \lambda_{j})|(p(\lambda_{i}) - p(\lambda_{j})).$$

We can assume that $\lambda_{2} - \lambda_{1} = k > 1$. Therefore, for every $(a, a+k+1, c)$, there must be a linear polynomial $p$ such that $p(\lambda_{1}) = a, p(\lambda_{2}) = a + k + 1$ and $p(\lambda_{3}) = c$, and this cannot be true for all values of $c$. Hence, we reached our contradiction.


问:
Minimal Red-Black tree with depth 3
I'd like to ask what is minimal RBT with black depth 3. Is this following RBT ? Values are not important. And that tree can't have depth 2 or 1.</p>

答:
According to Wikipedia, a red-black tree should be "roughly balanced".  The rule is "that the path from the root to the furthest leaf is no more than twice as long as the path from the root to the nearest leaf".  In your example, every node has a leaf.  The root has a leaf, so the path in between has length $1$.  Also, the furthest node has a leaf, with the path from the root having length $3$.  Thus, that tree does not meet the requirements of a red-black tree.

You can obtain a red-black tree by adding another node to the root.  Since your tree was minimal as a binary tree of depth $3$, the tree you obtain must be minimal as a red-black tree of depth $3$.

Title: Minimal Red-Black tree with depth 3


问:
$ B=\left\{(\beta_1,\beta_2,\dots,\beta_n)\in \mathbb{R^n}\colon |\beta_1|+|\beta_2|+\dots+|\beta_n|=1\right\} $ is closed & bounded
How can I show that
$$
B=\left\{(\beta_1,\beta_2,\dots,\beta_n)\in \mathbb{R^n}\colon |\beta_1|+|\beta_2|+\dots+|\beta_n|=1\right\}
$$
is closed and bounded?

答:
Note that the function $| \beta_1 | + |\beta_2| + ... + |\beta_n|$ is a continuous function, so the pre-image of a closed set, in this case the singleton $1$ is a closed set. Therefore the set $B$ is closed.

It is bounded, because $\|(\beta_1, \beta_2, ..., \beta_n)\| > 1$ implies that $|\beta_1| + |\beta_2| + ... + |\beta_n|$ > 1.


答:
Hint: consider the continuous function

$$
f \colon \Bbb R^n \to \Bbb R, f(x_1,\ldots,x_n) = |x_1|+\cdots +|x_n|
$$

Title: $ B=\left\{(\beta_1,\beta_2,\dots,\beta_n)\in \mathbb{R^n}\colon |\beta_1|+|\beta_2|+\dots+|\beta_n|=1\right\} $ is closed & bounded

Body:
How can I show that
$$
B=\left\{(\beta_1,\beta_2,\dots,\beta_n)\in \mathbb{R^n}\colon |\beta_1|+|\beta_2|+\dots+|\beta_n|=1\right\}
$$
is closed and bounded?


问:
Explanation on Google’s PageRank is Webpages as Eigenvectors
Help understand what is the matrix A and the vector x discussed below.

Mathematics for Machine Learning Example 4.9

Google uses the eigenvector corresponding to the maximal eigenvalue of
a matrix A to determine the rank of a page for search. The idea for
the PageRank algorithm, developed at Stanford University by Larry Page
and Sergey Brin in 1996, was that the importance of any web page can
be ap- proximated by the importance of pages that link to it. For
this, they write down all web sites as a huge directed graph that
shows which page links to which. PageRank computes the weight
(importance) x i > 0 of a web site a i by counting the number of pages
pointing to a i . Moreover, PageR- ank takes into account the
importance of the web sites that link to a i . The navigation behavior
of a user is then modeled by a transition matrix A of this graph that
tells us with what (click) probability somebody will end up on a
different web site. The matrix A has the property that for any ini-
tial rank/importance vector x of a web site the sequence $x, Ax, A^2x$,
. . . converges to a vector $x*$ . This vector is called the PageRank
and satisfies Ax∗ = x∗ , i.e., it is an eigenvector (with
corresponding eigenvalue 1 ) of A . After normalizing $x*$ , such that
||$x*$|| = 1 , we can interpret the entries as probabilities. More
details and different perspectives on PageRank can be found in the
original technical report (Page et al., 1999).

I suppose A is a linking from a source web page FROM which has a link(s) to another page TO. Is x like a one-hot-encoding of a web page to identify a specific page, then $Ax$ will result in multiple pages? Then the next step $A(Ax)$ will be multiple concurrent steps?

A web page is an eigenvector means that the matrix A has as many eigenvectors as the web pages existing in the world?

答:
The element $x_i$ of the vector $x$ is the probability that a random web surfer is visiting web page $i$ on a given time step. The elements of $x$ should sum to $1$. The $i$th element of $Ax$ is the probability that the random web surfer is on page $i$ on the next time step. The columns of $A$ are also probability vectors, and each should sum to $1$: the $j$th element of column $i$ is a transition probability, the probability that a random surfer visiting page $i$ on one time step will find herself on page $j$ on the next time step. In the simplest model, this probability is $1/L_i$ if there is a link from page $i$ to page $j$ and $0$ otherwise; $L_i$ is the number of links on page $i$. This simple model has some defects, so some tweaks are made to give the matrix better properties.

A matrix $A$, as described, will have at least one eigenvalue $1$ and all eigenvalues less than or equal to $1$. Corresponding to eigenvalue $1$ is an eigenvector $s$ satisfying $As=s$. Assuming $A$ satisfies suitable conditions (and one of the purposes of tweaking the model above is to ensure that it has those conditions), the $i$th element of $s$ represents the probability that the random surfer ends up on page $i$ after many clicks. The idea is that this probability will be higher for more important web pages. That is, the link structure of the web is being used to determine which pages are important; if random clicking leads you to a certain page with high probability, that page is important.


问:
Difficult Laplace inverse transform
Let $$F(s)=-\frac{1}{s}+\arctan(s+2),\ \ \mathrm{with\ } \ \Re  (s)>0$$
I want to compute the inverse Laplace transform.

attempt: I need to find $f$ such that $L(f)(s)=F(s)$, i.e.,
$$\int_0^\infty e^{-ts}f(t)\ dt = -\frac{1}{s}+\arctan(s+2) $$

but how can I clear $f(t)$ from here? I know for example that if $F(s)=\frac{1}{s}$ for $s\geq 0$ then we should have $f(t)=1$, for $t\geq 0$.

答:
I have not studied inverse Laplace transforms much, so my knowledge on the matter is very limited. But giving Mathematica the code

```<code>InverseLaplaceTransform[-1/s + ArcTan[s + 2], s, t]
</code>```

It outputs (in regular typesetting)
$$-1+e^{-2t}\left(\frac{\pi}{2}\delta(t)-\operatorname{sinc}(t)\right)$$

Title: Difficult Laplace inverse transform

Body:
I have not studied inverse Laplace transforms much, so my knowledge on the matter is very limited. But giving Mathematica the code

InverseLaplaceTransform[-1/s + ArcTan[s + 2], s, t]

It outputs (in regular typesetting)
$$-1+e^{-2t}\left(\frac{\pi}{2}\delta(t)-\operatorname{sinc}(t)\right)$$


问:
Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$?
Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$? I suspect it does from plots, but I want to use the ratio test to prove that
\begin{align}
\sum_{n=0}^\infty \exp(cn-vn^x)
\end{align}
converges for $c,v>0$ and $x>1$. Thanks.

答:
If you want to use the ratio test for $$a_n=(n+1)^x-n^x$$
$$R=\frac{a_{n+1}}{a_n}=\frac{(n+2)^x-(n+1)^x } {(n+1)^x-n^x }=\frac{\left(\left(1+\frac{2}{n}\right)^x-\left(1+\frac{1}{n}\right)^x\right) n^x } {\left(\left(1+\frac{1}{n}\right)^x-1\right) n^x  }$$
Let $\epsilon=\frac 1n$ to make
$$R=\frac{(1+2 \epsilon )^x-(1+\epsilon )^x }{(1+\epsilon )^x-1 }=1+(x-1)\epsilon+O(\epsilon^2)$$
$$R\sim1+\frac {x-1} n > 1 \qquad \text{if} \qquad x>1 \implies \text{divergence}$$


问:
The $q$ Multilinear Theorem
Let $R$ be the skew polynomial ring $k_\mathfrak{q}[x_1,\ldots,x_m]$ where $x_ix_j=qx_jx_i$ with $q\in k^*$ and for all $i<j$.

The $q$ Multinomial Theorem states that $$(x_1+\ldots+x_m)^r=\Sigma_{r_1+\cdots+r_m=r;r_i\geq0}\binom{r}{r_1,\cdots,r_m}_qx_1^{r_1}\cdots x_m^{r_m}$$

Where $\binom{r}{r_1,\ldots,r_m}_q=\frac{[r]_q!}{[r_1]_q!\cdots[r_m]_q!}$, and $[l]_q!=\begin{cases}[l]_q[l-1]_q\cdots[1]_q\quad\mathrm{if}\;l\geq1\\1\quad\mathrm{if}\;l=0\end{cases}$, and $[l]_q=\frac{1-q^l}{1-q}$.

I could not find a proof of this fact online. Could you show me a proof or give me a reference?

答:
Suppose $x_jx_i=q(x_ix_j)$ when $i<j$. The observation

$$ x_m(x_1+\cdots+x_{m-1})=q(x_1+\cdots+x_{m-1})x_m $$

means the $m=2$ case implies all of the $m>2$ cases by induction:

$$ \begin{array}{ll} (x_1+\cdots+x_m)^n  & \displaystyle =\sum_{k=0}^n \left[\begin{matrix} n \\ k \end{matrix}\right](x_1+\cdots+x_{m-1})^{n-k}x_m^k \\ & = \displaystyle \sum_k \left[\begin{matrix} n \\ k \end{matrix}\right] \left(\sum \left[ \begin{matrix} n-k \\ k_1, \cdots,k_{m-1} \end{matrix}\right] x_1^{k_1}\cdots x_{m-1}^{k_{m-1}} \right)x_m^{k} \\ & = \displaystyle \sum \left[ \begin{matrix} n \\ k_1 , \cdots, k_m \end{matrix} \right]x_1^{k_1}\cdots x_m^{k_m}.  \end{array} $$

There are numerous proofs of the $q$-binomial theorem online.

Title: The $q$ Multilinear Theorem


问:
From the system of equations prove $a+b+c=0$

$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:

$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
 {c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$

I added all the equations together and get:

$$a^3+b^3+c^3+(a+b+c)x+3y=0$$

It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:

$$3abc+3y=0$$
But it seems that doesn't work.

答:
Hint: $a,b,c$ are the roots of cubic $$p(t)=t^3+tx+y$$

so by vieta $a+b+c=0$


问:
Calculus - Calculate Work done to lift water out of tank

I need help setting up the integral so that I can calculate the work done. I've tried it many times and have referred to Youtube, slader, the textbook, and also this site, but I still don't get how to solve it. I came up with a few solutions but they were all wrong. Please help, I've been at this for 2 hours.

答: Your $P(x)$ is wrong because you took the base to be the top face, presumably because it is larger than the bottom face.  The base is the bottom face.  Instead of $5-(2/3)x$ it should be $3+(2/3)x$.  Good on you for using parentheses.  I didn't do the second part.


答:
Here's an alternative solution to Albus Dumbledore's. Note that the three equations imply that the nonzero vector $(1,x,y)$ is in the kernel of
$$\begin{pmatrix}a^3&a&1\\b^3&b&1\\c^3&c&1\end{pmatrix}$$
of determinant (after factoring)
$$(a-b)(a-c)(b-c)(a+b+c).$$
But also, the determinant must be 0 since it has nontrivial kernel. So $(a-b)(a-c)(b-c)(a+b+c)=0$ and by the assumption that $a,b,c$ are distinct, the result follows.

Title: From the system of equations prove $a+b+c=0$

Body:
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:

$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
 {c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$

I added all the equations together and get:

$$a^3+b^3+c^3+(a+b+c)x+3y=0$$

It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:

$$3abc+3y=0$$
But it seems that doesn't work.


答:
Subtract consecutive equations to get$$a^3-b^3+(a-b)x=0\iff a^2+b^2+ab+x=0~~\because a\ne b\\b^3-c^3+(b-c)x=0\iff b^2+c^2+bc+x=0~~\because b\ne c$$
Subtract these equations to get$$a^2-c^2+b(a-c)=0\iff a+b+c=0~~\because a\ne c$$


问:
To cut a line in two so that the squares of ... (geometric construction)

Cut a given straight line so that the sum of the square of one part and twice the square of the other part equals a given size.

Given a line of length $L$ which is cut in two pieces, lengths $x$ and $L-x$, I first thought of constructing the squares with sides $x$ and $\sqrt{2}(L-x)$ but I cannot relate the total area to a given length.

If $M$ is the given length the length to construct is
$$\frac{2L\pm\sqrt{3M-2L^2}}{3}$$
but I assume there is an easier way to solve this problem using only straight-edge and compass.

Any idea how to solve this problem? TIA.

答:
Given the square of one part plus twice the square of the other part equals the whole.

$$L=(x)^2+2(L-x)^2=2 L^2 - 4 L x + 3 x^2\\
\implies 3 x^2 - 4 L x + (2 L^2 - L) = 0$$

$$x=\frac{4L\pm\sqrt{16L^2-4(3)(2 L^2 - L)}}{2(3)}
=\frac{2L\pm \sqrt{L(3 - 2 L)}}{3}\quad x\in\mathbb{R}\iff 0\le  L\le \frac{3}{2}$$

Any value of $L$ in this range will work but the only (L,x) solutions that "contain" integers are
$$
(0,\space0)\qquad
(1,\space1)\qquad (3/2,\space1)$$

So $$0^2+2(0-0)^2=0\qquad
1^2+2(1-1)^2=1\qquad
 1^2+2(1.5-1)^2=1+2(.25)=1.5$$

Outside of this range, solutions are complex (containing imaginary parts),
i.e. $x\in\mathbb{C}$

$\textbf{Update}$ WolframAlpha shows the solution to be different from mine where I stayed in the limits.

$$(x)^2+2(L-x)^2-L=0,\quad L=414\\
\implies x = 276 + 5 i \sqrt{1518}\qquad L-x=138 + 5 i \sqrt{1518}$$

Multiplied back, it yields an enormous number for $L$, even with plus-minus the imaginary part here, and here.


问:
How to find Mean and Standard Deviation from 2 data points
The question reads "Suppose that the BMIs of American women of ages 50 to 59 also follow a normal distribution approximately but with different mean and standard deviation. It has been found that 5.05% American women of ages 50 to 59 have BMIs less than or equal to 20.4 kg/m^2 and 10% have BMIs greater than or equal to 38.9% kg/m^2. What are the mean and standard deviation of the BMI distribution for American women of ages 50 to 59?" I found that
x-mu/sigma = z
So I plugged my numbers in and got two equations 20.4 - mu/sigma = -1.64 and 38.9 - mu/sigma = 1.28. My question is how do I go from here to solving mu and sigma?

答:
Using your data you have that

$$\begin{cases}
\mathbb{P}[X<20.4]=5.05\%\\
\mathbb{P}[X>38.9]=10\%
\end{cases}$$

$$\begin{cases}
\mathbb{P}[X<20.4]=5.05\%\\
\mathbb{P}[X<38.9]=90\%
\end{cases}$$

Now using Z-tables you can find the two quantiles

$$\begin{cases}
\frac{20.4-\mu}{\sigma}=-1.640\\
\frac{38.9-\mu}{\sigma}=1.282
\end{cases}$$

Solving the system in $\mu;\sigma$ you get the solution


答:
There is a pretty straightforward way to see why. As you correctly pointed out, a ring homomorphism $\varphi :\mathbb{Z}[X]\rightarrow\mathbb{Z}^3$ is nothing more than an evaluation homomorphism in each component.
So there are $\alpha$, $\beta$, $\gamma\in\mathbb{Z}$ such that $\varphi(p)=(p(\alpha), p(\beta), p(\gamma))$ for every $p\in\mathbb{Z}[X]$.

Now, observe that $\varphi$ is surjective if and only if its image contains the elements $(1,0,0), (0,1,0), (0,0,1)$. This means that there are three polynomials $p_1, p_2, p_3 \in\mathbb{Z}[X]$ such that:

-$p_1(\alpha)=p_2(\beta)=p_3(\gamma)=1$;

-$p_1(\beta)=p_1(\gamma)=p_2(\alpha)=p_2(\gamma)=p_3(\alpha)=p_3(\beta)=0$.

What does it tell you about $\alpha, \beta, \gamma$?

Title: Example to show that $\mathbb Z^3$ is not a quotient of $\mathbb Z[X]$

Body:
I am asked to prove that $\mathbb{Z}^n$ is not a quotient of $\mathbb{Z}[X]$ for any integer $n \geq 3$ by showing that there is no surjective ring homomorphism $\mathbb{Z}[X] \to \mathbb{Z}^n.$

We can just focus on the $n = 3$ case since we can always look at the first 3 entries of the image. If $\phi: \mathbb{Z}[X] \to \mathbb{Z}^3$ is a ring homomorphism, then restricted to each entry, it is still a ring homomorphism $\phi_i: \mathbb{Z}[X] \to \mathbb{Z}, i = 1, 2, 3$. Now, $\phi_i$ must be the identity on $\mathbb{Z}$, and it is completely determined by $\lambda_i := \phi_i(X)$, so it is essentially the evaluation homomorphism $p \mapsto p(\lambda_i)$.

But $p(\lambda_i)$ is the remainder of $p(X)$ modulo $X - \lambda_i,$ so it reminds me of how in $\mathbb{Z},$ there is no integer $n$ satisfying $n \equiv 0 \bmod 3$ and $n \equiv 1 \bmod 6$. However, if the $\lambda_i$s are distinct, then the polynomials $X - \lambda_i$ look "coprime" to me, and I was wondering about whether there is a version of Chinese Remainder Theorem for ideals. Moreover, when looking at the examples, I noticed that (for instance) there is a surjective homomorphism $p \mapsto (p(0), p(1))$ since $(m, n)$ has pre-image $(m-n)X + n$, so why can we not extend it to a third remainder? More generally, how would the homomorphism fail to be surjective for any $\lambda_1, \lambda_2, \lambda_3$?


问:
Using residues to evaluate a sum involving the square of $\text{csch}$
I have been trying to evaluate the following sum using residues:

$$ \sum_{n=1}^{\infty}\frac{1}{\sinh^{2}(\pi
 n)}=\frac{1}{6}-\frac{1}{2\pi}$$

I am mainly interested in using residues to do this.  I can do it using real methods.

I tried using $\displaystyle \oint\frac{\pi\cot(\pi z)}{\sinh^{2}(\pi z)}dz$

The residue at $z=0$ is $\frac{-2}{3}$

The residue at $\displaystyle z=n, \;\ (n=\pm 1, \pm 2, \pm 3, ....)$ is $\displaystyle \lim_{z\to n}\frac{(z-n)\cos(\pi z)}{\sin(\pi z)\sinh^{2}(\pi z)}=\frac{1}{\sinh^{2}(\pi n)}$

The residue at $\displaystyle z=ni, \;\ (\pm 1, \pm 2, \pm 3, .....)$ is

$\displaystyle \lim_{z\to ni}\frac{(z-ni)\pi\cot(\pi z)}{\sinh^{2}(\pi z)}=\frac{1}{\sinh^{2}(\pi n)}$

So, by the residue theorem:

$\displaystyle \oint\frac{\pi \cot(\pi z)}{\sinh^{2}(\pi z)}dz=\frac{-2}{3}+4\sum_{n=1}^{N}\frac{1}{\sinh^{2}(\pi n)}$

As $N\to \infty$, the left side goes to 0, then solve for the sum at hand.

It would appear the 1/6 is in there, but I have failed to arrive at the correct solution.

Where does the $\frac{1}{2\pi}$ come into play?.

No doubt, I am doing it incorrectly. Can someone point me in the right direction?

Thanks a lot.

答:
The inverse Mellin transform provides another approach that uses residues.

I had previously asked about evaluating a similar series using this method.

For $\operatorname{Re} (s)>2$, we have

$$ \begin{align} \left\{\mathcal{M} \ \frac{1}{\sinh^{2} (\pi x)} \right\}(s) &= \int_{0}^{\infty} \frac{x^{s-1}}{\sinh^{2}(\pi x)} \, dx \\ &= \pi^{-s}\int_{0}^{\infty} \frac{u^{s-1}}{\sinh^{2} u} \, du \\ &=  4 \pi^{-s} \int_{0}^{\infty} u^{s-1} \, \frac{e^{-2u}}{(1-e^{-2u})^2} \, du \\  &= 4 \pi^{-s} \int_{0}^{\infty} u^{s-1} \sum_{n=1}^{\infty} n e^{-2nu} \,du \\ &=  4 \pi^{-s} \sum_{n=1}^{\infty} n \int_{0}^{\infty} u^{s-1} e^{-2nu} \, du \\ &= 4 \pi^{-s} \sum_{n=1}^{\infty}n \,  \frac{\Gamma(s)}{(2n)^{s}}\\  &= \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1). \end{align}$$

So according to the Mellin inversion theorem,

$$\frac{1}{\sinh^{2} (\pi x)} = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i\infty} \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1) x^{-s} \, ds \, , $$ where $c= \frac{5}{2}$.

(Since the Mellin transform of $\frac{1}{\sinh^{2}(\pi z)}$ converges absolutely in the right half-plane $\operatorname{Re}(s) >2$, we could have chosen any value of $c$ greater than $2$.)

Replacing $x$ with $n$ and summing both sides, we get

$$\begin{align} \sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} &= \frac{1}{2 \pi i} \int_{5/2 - i \infty}^{5/2 + i\infty} \pi^{-s} \ 2^{2-s} \ \Gamma(s) \zeta(s-1) \zeta(s) \, ds \\ &= \frac{1}{2 \pi i} \int_{5/2 - i\infty}^{5/2+ i\infty} f(s) \, ds. \end{align}$$

Since the magnitude of $\Gamma(z)$ decays exponentially fast to $0$ as $\Im(z) \to \pm \infty$, we can shift the contour to the line $\operatorname{Re}(s) =1$ and conclude that $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} =  \frac{1}{2 \pi i} \left( \operatorname{PV} \int_{-\infty}^{\infty} f(1+it) i \, dt +  \pi i \, \text{Res}[f,1] + 2 \pi i \, \operatorname{Res}[f,2]  \right),$$ where

$$\begin{align} \operatorname{Res} [f(s),1] &= \lim_{z \to 1} (s-1) \zeta(s) \pi^{-s} 2^{2-s} \Gamma(s) \zeta(s-1) \\ &= (1) \left( \frac{1}{\pi} \Big)(2)(1)\Big(-\frac{1}{2} \right) \\ &= - \frac{1}{\pi } \end{align}$$

and

$$ \begin{align} \operatorname{Res}[f(s),2] &= \lim_{z \to 2} (s-2) \zeta(s-1) \pi^{-s} 2^{2-s} \Gamma(s) \zeta(s) \\ &= (1) \left(\frac{1}{\pi^{2}} \Big) (1)(1) \Big(\frac{\pi^{2}}{6} \right) \\ &= \frac{1}{6}. \end{align}$$

But it can be shown using the functional equation of the Riemann zeta function that $f(s)$ is odd (and purely imaginary) along the line $\operatorname{Re}(s)=1$.

Specifically, $$f(1+it) = 4i \cosh \left(\frac{\pi t}{2} \right)\operatorname{csch}(\pi t) \left|\zeta(it) \right|^{2} \, , \quad t \in \mathbb{R}.  $$

Therefore, $$\operatorname{PV} \int_{-\infty}^{\infty} f(1+it) i \, dt =0 $$ and the result follows.

EDIT:

Alternatively, it can be shown that $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)} = 4 \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}. $$

See this answer. This is the series I had asked about in my question.

Also see this paper for the evaluation of similar sums using the inverse Mellin transform.


问:
How is the averaging of this stochastic process determined?
I am reading a physics paper dealing with a random fluctuation $R(t)$ satisfying:

$\langle R(t)\rangle = 0$.

$\langle R(t) R(t')\rangle = \delta(t - t')$ where $\delta$ is the Dirac-delta function

The paper says:

$R(t)$ is not a well-behaved function. Instead of its instantaneous value, which is not well-defined, one needs to take its average over the interval $\delta t$, which depends on $\delta t$ and is $r/\sqrt{\delta t}$, where $r$ is a random number with $\langle r \rangle = 0$ and $\langle r^2\rangle = 1$.

My Question is: How was this average determined?

Thank you for your response.

答: In this paper, $R(t)$ is a gaussian white noise which is formally the derivative of a Wiener process. When integrated over time $\delta t$, it becomes a normally distributed random variable with mean zero and standard deviation $\sqrt{\delta t}$. So its average, (its integral over $\delta t$) would be $1/\sqrt{\delta t}$ times a normalized random variable that is called $r$ here.


问:
Algebra II online courses
Can anyone give me recommendations for Algebra II online courses. I do know about Khan academy but looking for other courses that are also really good for learning Algebra II.

Thank you

答:
Try these courses:

https://brilliant.org/courses/graphing-and-modeling/landing/?utm_medium=cpc&utm_source=search&utm_term=%2Balgebra%20%2B2%20%2Bcourse_&utm_content=486381395220_&utm_campaign=11315342810_consolidated_sem&gclid=CjwKCAiAgc-ABhA7EiwAjev-jy3fNTwe-Bw0QmyYiSr8R0j5_7_KxtKh5AyamzQMS87Z2Bl6FoVasRoCkpYQAvD_BwE

https://www.thegreatcoursesplus.com/algebra-ii

https://www.udemy.com/course/algebra2-course/?utm_source=adwords&utm_medium=udemyads&utm_campaign=LongTail_la.EN_cc.US&utm_content=deal4584&utm_term=_._ag_81829991947_._ad_436603485109_._kw__._de_c_._dm__._pl__._ti_dsa-1007766171592_._li_9053025_._pd__._&matchtype=b&gclid=CjwKCAiAgc-ABhA7EiwAjev-j91v0deqTa2fiPjOlRhYfyhuDWwWuluPjzHQore8SH2VbSbA5DwL1BoChMYQAvD_BwE

Brilliant.org and Udemy are highly reputable. While the great course plus has a free trial.

I hope this helps.

Title: Algebra II online courses


问:
Given $z^{z^x}=x$, estimate $\frac{dx}{dz}$
I'm reading Knoebel's Exponentials Reiterated, page 242 and have very hard time understanding the following:

If $z^{z^x}=x$, then $\frac{dx}{dz}\gt 0$ on the open line
$$L=\{<z,x>|0\lt z\lt e^{-e}\,\text{and}\, x=e^{-1}\}.$$

This doesn't make sense to me. Because if $z^{z^x}=x$, then $x=e^{-1}$ implies $z=e^{-e}$, which is not in the open interval $\left(0,e^{-e}\right)$.

I know that, if $j(z,x)=z^{z^x}-x$, then
$$dj(z,x)=z^{z^x}z^{x-1}(x\ln z+1)\,dz+\left(z^{z^x}z^x\ln ^2z-1\right)\,dx,$$
and there is a unique trajectory satisfying $j(z,x)=0$ when it is not true that $z=e^{-e}$ and $x=e^{-1}$ simultaneously by the Implicit function theorem,
but how could that be used to prove $\frac{dx}{dz}\gt 0$ on the line?

答:
Consider the implicit function
$$F(x,z)=z^{z^x}-x=0$$
$$\frac{\partial F(x,z)}{\partial x}=z^{z^x+x} \log ^2(z)-1$$
$$\frac{\partial F(x,z)}{\partial z}=z^{z^x} \left(z^{x-1}+x z^{x-1} \log (z)\right)$$
$$\frac {dx}{dz}=-\frac{\frac{\partial F(x,z)}{\partial z} } {\frac{\partial F(x,z)}{\partial x} }=-\frac{z^{z^x+x-1} (x \log (z)+1)}{z^{z^x+x} \log ^2(z)-1}$$

More pleasant would be the reverse way using
$$z=\left(\frac{x \log (x)}{W(x \log (x))}\right)^{\frac{1}{x}}$$ then logarithmic differentiation and chain rule with $t=x \log(x)$. We should end with
$$\frac {dx}{dz}=f(x)$$ which is positive for $0 \leq x \lt e$ with a vertical asymptote at $x=\frac 1e$. For $x>e$, the derivative is always negative.

Edit

For the calculation of the derivative, start with
$$z=\left(\frac{t}{W(t)}\right)^{\frac{1}{x}} \qquad t=x\log(x)$$
$$\log(z)={\frac{1}{x}} \log\left(\frac{t}{W(t)}\right)$$
$$\frac {z'} z=-\frac 1{x^2}\log\left(\frac{t}{W(t)}\right)+\frac 1x \frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]\frac {dt} {dx}$$
$$\frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]=\frac{W(t)}{t (1+W(t))}$$ All of the above gives
$$z'=\frac z {x^2 t (1+W(t)) }\Bigg[x W(t) t'-t (1+W(t)) \log \left(\frac{t}{W(t)}\right) \Bigg]$$
$$\frac {dx}{dz}=\frac 1 {z'}$$


问:
Difference Between Feynman-Kac PDE and Kolmogorov Backward Equation
Suppose $X_u$ satisfies the SDE
$$
dX_u = a(u, X_u) du + b(u,X_u) dW_u\;.
$$
Fix $T > 0$ and let $X_t = x$ for $0 \leq t \leq T$ be given.  For any measurable function $h$, define
$$
g(t,x) = E^{t,x}(h(X_T))\;.
$$
Then $g$ satisfies the Feynman-Kac equation
$$
g_t(t,x) + a(t, x) g_x(t,x) + \frac{1}{2}b^2(t,x)g_{xx}(t,x) = 0
$$
with terminal condition
$$
g(T,x) = h(x) \;.
$$
Fair enough, but now consider the Kolmogorov backward equation.  Let $p(t,x; T, y)$ be the transition density of $X_u$; i.e., if $X_t = x$, then $X_T$ has density $p(t,x; T,y)$ in the $y$ variable.  Then for each $T$ and $y$, $p$ satisfies the Kolmogorov backward equation
$$
p_t(t,x; T,y) + a(t,x)p_x(t,x;T,y) + \frac{1}{2}b^2(t,x)p_{xx}(t,x;, T,y) = 0
$$
with terminal condition
$$
p(T,x; T,y) = u_T(x)
$$
for some function $u_T(x)$

As far as I can tell, these are the same equations; the only difference between the solutions being the terminal conditions.  Is this a correct understanding?  Is there some greater insight as to why these aren't both called the same equation?  I see that $g$ and $p$ are related by
$$
g(t,x) = \int_{-\infty}^\infty h(y) p(x,t;T,y)\,dy
$$
but I'm not seeing anything insightful from this relationship, and am afraid I'm missing something.  I've read the answer here but am still a bit unsure of the difference.

答:
I hate to resurrect a four year old question with an only-partially-helpful response, but this paper shows that Feynman-Kac is the same as a Kolmogorov backward equation. I haven't read it in detail, so I can't summarise and do it any justice, but hopefully it helps you. In case the link doesn't work, paper is: VERSIONS OF THE FEYNMAN-KAC FORMULA by A N Borodin.

Great question!


问:
Atiyah-Mcdonald, Exercise 1.6

A ring $A$ is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element $e$ such that $e^2=e\ne0$). Prove that the nilradical and Jacobson radical of $A$ are equal.

Solutions for this book exist but I would if you help me with where I specifically going wrong instead of providing a solution. There are similar questions on this site, but I specific help with understanding what is going wrong in the quotient. Here is my attempt:

It is enough to show that every prime ideal is maximal. We do this by showing for arbitrary prime ideal $p$, $A/p$ is a field. Let $x\in A-p$, its image $\overline{x}$ is nonzero. Then the ideal generated by $x$ is not in the nilradical. By our assumption there is a nonzero $e=e^2$ in $(x)$, write $e=ax$. Then we see that $e(e-1)=0$. Then $\overline{ax}\cdot\overline{ax-1}=0\in A/p$. Since the quotient is an integral domain, we either have $\overline{ax}=0$ or $\overline{ax-1}=0$. Here starts my confusion. We would like to conclude that since $e$ was non zero $\overline{ax}$ is non zero, but it could be possible if $e\in p$ that $e$ be non zero and $\overline{ax}$ be zero.

Similar questions:
Check my proof that the nilradical and the Jacobson radical are equal (A&M 1.6) and  Atiyah-Macdonald, Problem 6 of Chapter 1

答:
Edit: There are mistakes in the first paragraph (see below). But looking at a finite product of fields is still sufficient to understand where your proof goes wrong.

If we look at $A/\operatorname{nil}(A)$ then that is a ring in which every ideal contains an idempotent. So that is a semisimple ring and hence it is a finite product of fields (there's a proof of this somewhere on this site, just make sure you're specifically looking for commutative rings). When you have a finite product of fields, every prime ideal is necessarily maximal.

Where your proof comes short is that you take just any idempotent but there might be multiple idempotents and you need to somehow identify one that will work.

For instance, consider $A = \mathbb{R}^2$ with $e_1 = (1,0), e_2 = (0,1)$. Let $x = (1,1)$ (so the identity) and let $P = \ker((u,v) \mapsto u)$.

We have three idempotents in $(x)$ but let us focus just on $e_1$ and $e_2$. And now, let us look at what happens in your proof depending on which of the idempotents we take. Since $x$ is the identity, in every case we are going to decompose $e = ax$ with $a = e$. So now:

$e = e_1$ then $ax \equiv 1 \pmod P$

$e = e_2$ then $ax \equiv 0 \pmod P$.

So you see it is possible that $ax \equiv 0 \pmod P$ if we choose the wrong idempotent.

Title: Atiyah-Mcdonald, Exercise 1.6


问:
Show the collection of subsets is a basis.
Here is my attempt at showing this collection forms a basis. Have I done all the necessary cases? Is this correct?

Problem:We construct a topological space. Let $\mathbb{R}_{+00}$ be the set consisting of $\mathbb{R}_+$ together with two points called $0'$ and $0''$. Put a topology on it generated by a basis consisting of all intervals in $\mathbb{R}_+$ of the form $(a,b)$ or else of the form $(0,b) \cup \{0'\}$ or $(0,b) \cup \{0''\}$, $a,b \in \mathbb{R}_+$. Show this collection of sets forms a basis for a topology.

Attempt:
Each point in $\mathbb{R}_{+00}$ is in at least one basis element. If $x \in \mathbb{R_+}$,the basis element $(0,x+1) \cup \{0'\}$ certainly contains $x$. $0' \in (0,b) \cup \{0'\}$ for any $b \in \mathbb{R_+}$ and $0'' \in (0,b) \cup \{0''\}$ for any $b \in \mathbb{R_+}$. The intersection of any two open intervals $(a,b),(c,d),a,b,c,d \in \mathbb{R_+},a<b,c<d$  is either $\varnothing$, or a basis element. If  $a<b$ and $x \in ((0,a) \cup \{0'\}) \cap ((0,b) \cup \{0''\})$, choose $\epsilon$ such that $0<\epsilon<x$. Then $x\in (x-\epsilon,a)  \subset ((0,a) \cup \{0'\}) \cap ((0,b) \cup \{0''\})$. The case where $b>a$ is similar. $(c,d) \cap ((0,a) \cup \{0'\})$ is either $\varnothing$ or an open interval of the form  $(e,f)$, a basis element. Similarly for $(c,d) \cap ((0,a) \cup \{0''\}).$

答:
Without knowing the sign of $0',0''$, your assumption that $(c,d)\cap[(0,a)\cup\{0'\}]$ being either empty or an open interval is incorrect. Note that it is equal to$$A=[(c,d)\cap(0,a)]\color{pink}\cup[(c,d)\cap\{0'\}]=(\max\{0,c\},\min\{a,d\})\color{pink}\cup[(c,d)\cap\{0'\}]$$where the former is an open interval (including the empty set) and the latter may give you an isolated point $0'$ in case $0'$ is positive. In this case there is no basis element containing $0'$ which is a subset of $A$ unless $c=0$. For example, $(c,d)=(5,15),(0,a)\cup\{0'\}=(0,6)\cup\{12\}$.

So you must remark that the given set forms a basis iff $0',0''\le0$.

For $a<b$,$$[(0,a) \cup \{0'\}] \cap[(0,b) \cup \{0''\}]=(0,a)\color{pink}\cup[\{0'\}\cap\{0''\}]$$There is a possibility that this intersection contains the point $0'$ iff $0'=0''$, in which case $(0'-\epsilon,a)$ is not be a suitable choice. For $0'$, select the basis element  $(0,a)\cup\{0'\}$ and you are done.


问:
Finding the MLE of $\theta$ where $\theta \leq x$
consider the following PDF</p>

<p>$$
\begin{eqnarray}
f(x;\theta) &=&
\left\{\begin{array}{ll}
2\frac{\theta^2}{x^3} & \theta \leqslant x\\
0 & x< \theta; 0 < \theta
\end{array}\right.\\
\end{eqnarray}
$$
Now the answer stats $X_{1:n}$ so the minimum of $X$, but this cannot be deduced from the answer via my normal method (log likelihood), how am I to approach this, and other questions like it? </p>

答: Hint: At what value of $\theta$ is the likelihood maximized keeping in mind that $\theta \le x_i$ for all $i$?</p>


问:
Are these sets compact in $\ell^2$?
Hey everyone I have another analysis question hoping you guys could give me some insight on it. Basically, I am trying to determine whether or not these sets are compact in $\ell^2$.

$$A = \{(a_1,a_2,a_3,\dots):|a_n| \leq \frac{1}{\sqrt{n}}\} \\ B=\{(a_1,a_2,a_3,\dots):|a_n| \leq \frac{1}{n}\}$$

(We are assuming the elements in these sets actually belong to $\ell^2$). Anyways I think the first set is not compact since it seems we can find a sequence in here that does not have a converging subsequence. The reason I say this is because of the bound the terms have.

For the second set, I do not even know where to start honestly. I tried to do a diagonalization argument and argue that there is a converging subsequence but failed. I tried showing that every Cauchy sequence in here convergences and we can cover the set by some $\delta$-net, but failed too.

Any help would be appreciated.
Thanks.

答:
$(1,\frac 1 {\sqrt 2},,...,\frac 1 {\sqrt n},0,0,0,...), n=1,2,3...$ is a  sequence in $A$ . If there is  a subsequence which converges then the limit sequence has to be $(1,\frac 1 {\sqrt 2},,...,\frac 1 {\sqrt n},....)$ but this sequence does not belong to $\ell^{2}$. Hence $A$ is not compact.

$B$ is compact. Given any sequence in $B$ you can use Cantor's diagonal procedure to  extract a  subsequence which conveges coordinatewise. The limiting sequence is in  $\ell^{2}$ because $\sum  \frac 1 {n^{2}} <\infty$. Using the convergence of this series you can show that the subsequence we have obtained converges in the norm of $\ell^{2}$. [I will add more details if necessary].

Title: Are these sets compact in $\ell^2$?


问:
When rolling two dice, which of the following events are independent of the event that the first die is 4:
When rolling two dice, which of the following events are independent of the event that the first die is 4:

(a) The second is 2

(b) The sum is 6

(c) The sum is 7

(d) The sum is even

I thought only option (a) is independent but that wasn't the case, so want to find out which ones are independent and which ones are dependent along with their reasoning.

There are multiple correct answers.

答:
As an example, also (d) is independent of the event "first die is 4"

In fact

$$\mathbb{P}(\text{Sum is even})=\frac{1}{2}=\mathbb{P}(\text{Sum is even}|\text{First die is 4})$$

Check the rest in the same way finding a "tricky" result...


问:
Is there a solution to $\int \frac{x^7-2}{\ln x}dx$ without the $\text{Ei}(x)$ and $\text{Li}(x)$ integral?
I found this question in writing so perhaps I have made an error in the handwriting interpretation. I think it should involve a simple $u$-substitution but Wolfram Alpha gives an answer involving the $\text{Ei}(x)$ and $\text{Li}(x)$ integral. Is there a substitution if the exponent was to power $2$ instead of $7$?

答:
As a hint you can take $x=e^u$ so
$dx=e^udu$ and $$x^y=(e^u)^7=e^{7u}\\ \ln(x)=u$$ Can you take over ?


问:
If the scalar product of two vectors is equal to the magnitude of their vector product, find the angle between them.
I found some similar questions but none could satisfy me. I am not given any other conditions except those mentioned above. Please help me with this question. I am mainly confused on the formula for the magnitude of vector product between 2 vectors.</p>

答:
With two vectors $u,v\in\Bbb R^3$ and angle $\alpha$ between them, We know
$$ u\cdot v = |u||v|\cos\alpha$$
and
$$ |u\times v|=|u||v|\sin \alpha$$
Hence the given conditions imply
$$ |u||v|\cos\alpha=|u||v|\sin \alpha.$$
This holds trivially when at least one of $u,v$ is the zero vector (in which case we say they are orthogonal, though actually speaking of an angle between them makes no sense). If we only consider the case that $u,v$ are both non-zero, we can divide by the lengths and find
$$ \cos\alpha=\sin\alpha.$$
As additionally $0\le\alpha\le\pi$, the only solution to this is $$\alpha=\frac\pi4,$$ where $\sin\alpha=\cos\alpha =\frac{\sqrt 2}2$.


答: It would seem the events b and c are not independent. So let's check. For example P(sum is 7 given first roll is 4) is 1/6. If by chance it were independent P(sum is 7 given first roll is 4) would have been the same as P(sum is 7) which is 1/6 thus indeed independent</p>

Body:
When rolling two dice, which of the following events are independent of the event that the first die is 4:

(a) The second is 2

(b) The sum is 6

(c) The sum is 7

(d) The sum is even

I thought only option (a) is independent but that wasn't the case, so want to find out which ones are independent and which ones are dependent along with their reasoning.

There are multiple correct answers.


问:
How many different ways are there for a person to take the trip in one full week if the person wants to bike at least once?
From Monday to Friday, there are three ways to take a trip: driving, biking, or taking the bus. How many different ways are there for a person to take the trip in one full week if the person wants to bike at least once?

It seems that there are $3^5$ ways for the whole week and if there is no bike appears in a whole week, there are $2^5$ options. So the result seems $3^5-2^5$, right?

答:
Your options are $S:=\{1,2,3\}$. You're looking for
$$R:=\{(t_1,...,t_5)\in S^5\mid \sum_{i=1}^5 \mathbb{1}_{\{1\}}(t_i)\ge 1\}$$

Therefore we have
$$
R= S^5-R\setminus S^5
$$
or, written more familiarly
$$
\mathbb P(R) = 1- \mathbb P(R^C)
$$
, using the Laplace-distribution over the Sample space $S^5$.

Now we have
$$
R\setminus S^5= \{(t_1,...,t_5)\in S^5\mid \sum_{i=1}^5 \mathbb{1}_{\{1\}}(i)<1\} = \{(t_1,...,t_5)\in S^5\mid \sum_{i=1}^5 \mathbb{1}_{\{1\}}(t_i)= 0\}
 = \{(t_1,...,t_5)\in S^5\mid t_i\neq 1\} = \{(t_1,...,t_5)\in \{2,3\}^5\}= 2^5$$

And therefore we arrive at your answer $3^5-2^5$.