问:
Estimating Binomial random variable with Normal distribution
Given $n = 10^4$ trials of independent Bernoulli experiments with $p = 0.6$, I'm trying to estimate the probability that the number of successes fall between $7901$ and $8100$.

I defined $S_n = \sum_{i=1}^{n}X_i$ where $X_i$ is the i-th trial. We then have $X_i \sim Be(p)$ and therefore $S_n \sim B(n, p)$. Since we know $E[X_i] = p = 0.6$ and $Var[X_i] = p(1-p) = 0.24$, I tried using the Central Limit Theorem:

$$
P(7901 \leq S_n \leq 8100) = P \left( \frac{7901 - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \leq \frac{S_n - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \leq \frac{8100 - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \right) = P \left( \frac{1901}{\sqrt{2400}} \leq \frac{S_n - 6000}{\sqrt{2400}}\leq \frac{2100}{\sqrt{2400}} \right)
$$

Since $\frac{S_n - 6000}{\sqrt{2400}} \xrightarrow[]{d} N(0,1)$,
$$
P \left( \frac{1901}{\sqrt{2400}} \leq \frac{S_n - 6000}{\sqrt{2400}}\leq \frac{2100}{\sqrt{2400}} \right) = P \left( \frac{S_n - 6000}{\sqrt{2400}} \leq \frac{2100}{\sqrt{2400}} \right) - P \left( \frac{S_n - 6000}{\sqrt{2400}} \leq \frac{1900}{\sqrt{2400}}\right) \approx \phi \left( \frac{2100}{\sqrt{2400}} \right) - \phi \left( \frac{1900}{\sqrt{2400}} \right)
$$

Where $\phi$ is the cumulative distribution function for the standard normal distribution. But this of course yields $$ \approx 1 - 1 = 0$$

I can't seem to pin down where I made a mistake, any hints/tips will be greatly appreciated. Thanks in advance.

答:
There's nothing wrong; you just haven't used enough precision. The correct answer is, rounded to four significant figures,
$$2.434×10^{-329}$$
which tells you that it is negligible that between $7901$ and $8100$ heads will occur.


问:
Find the Probability??
A deck of playing cards contains 52 cards. What is the probability of pulling out a King of Diamonds and without replacing it, then an ace of spades?

1/52 x 1/51 = 1/ 2652

My thought process is that there is one king diamond card so you take it out (1/52). then you have to pull one ace of spades being 1/51.

答:
Prob. of pulling out a King of Diamonds = n(King of Diamonds)/n(Cards in the deck) = 1/52

Now, due to no replacement, the prob. of pulling out an Ace of Spades = n(Ace of Spades)/n(Cards in the deck - 1) = 1/51

So final probability = (1/52)(1/51) = 1/2652.
Note: We're multiplying because of the use of "and". Hope this helps

ContentLicense: CC BY-SA 4.0

Title: Find the Probability??


问:
Linear Algebra - Vector Spaces of Vector Spaces
If we have a Vector Space such as $\Bbb R$, we can make Vector Spaces out of it. For example, let $v_1,v_2,v_3\in \Bbb R$. We know that $(v_1,v_2,v_3)$ is a Vector Space - it is $\Bbb R_3$ - and its dimension is $3\times \dim {(\Bbb R)} = 3$. Thus we can see that $\Bbb R_3$ is composed of three elements of $\Bbb R$.

Any "Container Space" ($V_1$) is made of a Vector Space of elements whose elements are vectors in another Vector Space ($V_2$). Also, $(v_1,v_2,...,v_n) \oplus_1 (v'_1,v'_2,...,v'_n) = (v_1\oplus_2v'_1,v_2\oplus_2v'_2,...,v_n\oplus_2v'_n)$ and $c\odot_1(v_1,v_2,...) = (c\odot_2 v_1, c\odot_2 v_2...)$. In other words, Operators are passed to the component vectors.

In general, if we have a Vector Space $V$ and a Set $C$ made from vectors in $V$, is $C$ always a Vector Space? If so, will the dimension of $C$ always be $\dim{(P)}\times\dim{(V)}$ where $P$ is the space that forms elements in the way that the Set $C$ does?

I have found a few more Vector Spaces of the same type; if you want more examples, I can post them.

答:
"Piano tiger", all of these things have normal names, which would be much more preferable to use than these new ones you've invented. (Actually, I'm not even sure two of those names are necessary to invent, unless your intention is to boggle the reader.)

Given a vector space $V$, you can always form the product $V\times V\times V$ (which you could choose to denote as $V^3$). This is the product space of three copies of the space $V$.

The dimension of the product is always the sum of the dimensions of the pieces of the product... so if $\dim(v)=n$, then $\dim(V^3)=3n$.

In general, $\dim(V^k)=k\cdot \dim(V)$.

Rather than saying "$C$ is a container space of $S$", it would be more conventional to say that $S$ is a subspace of $V$.

Your latest revision of your third paragraph can now be translated as: "If $P$ is a subspace of $V$, and $C$ is a set made from vectors of $V$, is $C$ a vector space?"

If $C$ is just some random subset of $V$, then no, it is unlikely that it is a vector space. For one thing, it might not have the zero vector.

Latest addition for latest edition.

The set you described forming from elements of $P$ is just another product space $P^n$. That $P$ is a subspace of $V$ does not change anything about forming the product... it certainly does not need a newly invented name. A $n\times n$ matrix with entries from $V$ is just a fancy way of writing a vector of length $n^2$. It is the same as $V^{n^2}$ except for the way you write it.

ContentLicense: CC BY-SA 3.0


问:
Commutative Ring Epimorphism on quotients
Let $R$ be a commutative ring and $I$,$J$ be ideals of $R$.

Define $f: R \rightarrow R/I \times R/J$ as $f(a) = (a+I,a+J)$.

Prove that if $f$ homomorphism is surjective, then $I+J = R$.

..

I think $\gcd([R:I],[R:J])$ must be 1.

I've found Ker$f$ = $I \cap J$ and tried first isomorphism theorem but could not go any further.

答:
Remember that epimorphisms need not be surjective (consider $\mathbb{Z} \to \mathbb{Q}$, for example).

In general, $R \to S \times T$ is an epimorphism if and only if $R \to S$ and $R \to T$ are epimorphisms and $S \otimes_{R} T=0$.

In particular, $R \to R/I \times R/J$ is (a priori possibly non-surjective) epimorphism iff $(R/I) \otimes_{R} (R/J)=0$ iff $R/(I+J)=0$ iff $I+J=R$.

The Chinese Remainder Theorem then implies that $R \to R/I \times R/J$ will in fact be surjective if $I+J=R$.


问:
Morphism between rings of Laurent polynomials.
Let $I_n = \text{Set}([1,n],\mathbb{Z})$. We define a ring homomorphism $f_{nn-1}:\mathbb{Z}[x_1,\dots,x_n] \to \mathbb{Z}[x_1,\dots,x_{n-1}]$ by
\begin{equation}
x_i \mapsto \begin{cases} x_i, & i \leq n-1, \\
                          0, & i=n
\end{cases}
\end{equation}
extended algebraically. Suppose we now want to define "the same" ring homomorphism  between rings
$\mathbb{Q}[x_1^{\pm1},\dots,x_n^{\pm1}]$ and $\mathbb{Q}[x_1^{\pm1},\dots,x_{n-1}^{\pm1}]$ of Laurent polynomials. Clearly, we can no longer send $x_n$ to $0$. Instead, we define $g_{nn-1}:\mathbb{Q}[x_1^{\pm1},\dots,x_n^{\pm1}] \to \mathbb{Q}[x_1^{\pm1},\dots,x_{n-1}^{\pm1}]$ as:
\begin{equation}
g_{nn-1}\bigg(\displaystyle\sum_{\alpha \in I_n}c_{\alpha}x^{\alpha}\bigg)=\displaystyle\sum_{\beta \in I_{n-1}}d_{\beta}x^{\beta},
\end{equation}
with $d_{\beta}=c_{\alpha(\beta)},$ where \begin{equation}
I_n  \ni \alpha(\beta)_i=\begin{cases} \beta_i, &i \leq n-1, \\
                                       0, i=n.
\end{cases}
\end{equation}
Esentially, the coefficients of monomials involving non-zero powers of $x_n$ are ignored. Is this ok or am I doing something which is not allowed?

答:
Consider the polynomial $x_n\in\mathbb{Q}[x_1^{\pm 1},\dots,x_n^{\pm 1}]$. This has $c_\alpha=0$ for all $\alpha\neq(0,\dots,0,1)$, and $c_{(0,\dots,0,1)}=1$. Since $(0,\dots,0,1)\neq\alpha(\beta)$ for any $\beta\in I_{n-1}$, this means that $g_{nn-1}(x_n)=0$. This means that $g_{nn-1}$ cannot be a ring homomorphism, since $x_n$ is a unit in $\mathbb{Q}[x_1^{\pm 1},\dots,x_n^{\pm 1}]$ and so its image under any ring homomorphism must be a unit in $\mathbb{Q}[x_1^{\pm 1},\dots,x_{n-1}^{\pm 1}]$.

Here is any easy way to define maps out of $\mathbb{Q}[x_1^{\pm 1},\dots,x_n^{\pm 1}]$. In my view, it is easier to think about this ring as the quotient ring $$R:=\frac{\mathbb{Q}[x_1,\dots,x_n,y_1,\dots,y_n]}{(x_1y_1-1,\dots,x_ny_n-1)}.$$ By the universal property of quotient rings, defining a ring homomorphism from $R$ to another ring $S$ is the same as defining a ring homomorphism $f:\mathbb{Q}[\bar{x},\bar{y}]\rightarrow S$ such that $(x_i y_i-1)_{i\leqslant n}\subseteq\ker{f}$.

Now, sending the generators of an ideal to $0$ will be enough to send the whole ideal to $0$, so the condition $(x_i y_i-1)_{i\leqslant n}\subseteq\ker{f}$ is really the same as requiring $$0=f(x_iy_i-1)=f(x_i)f(y_i)-1$$ for each $i$. In other, $f$ sends the ideal $(x_i y_i-1)_{i\leqslant n}$ to $0$ if and only if $f(x_i)$ is the multiplicative inverse of $f(y_i)$ for each $i$. Thus, any ring morphism $f:\mathbb{Q}[\bar{x},\bar{y}]\rightarrow S$ such that each $f(x_i)$ is a unit with $f(y_i)$ as its inverse can be turned into a ring morphism $R\rightarrow S$.

If you're only interested in defining maps to commutative rings $S$, then we can simplify even further. Indeed, by the universal property of polynomial rings, defining a ring homomorphism from $\mathbb{Q}[\bar{x},\bar{y}]$ to a commutative ring $S$ is the same as defining a ring homomorphism from $\mathbb{Q}$ to $S$ and then choosing any elements $s_1,\dots,s_n\in S$ and $t_1,\dots,t_n\in S$ to send the $x_i$ and the $y_i$ to. (We take $x_i\mapsto s_i$ and $t_i\mapsto t_i$.)

So, putting these two facts together, here is a recipe for defining maps $f:R\rightarrow S$, where $S$ is commutative. First, define a map $g:\mathbb{Q}\rightarrow S$. Then choose any units $s_1,\dots,s_n\in S^\times$. Then we can glue these two ingredients together to get a map $f:R\rightarrow S$ such that $f(q)=g(q)$ for any $q\in\mathbb{Q}$, $f(x_i)=s_i$ for each $i$, and $f(y_i)=s_i^{-1}$ for each $i$.

In your notation for the Laurent ring, defining a map $f:\mathbb{Q}[x_1^{\pm 1},\dots,x_n^{\pm 1}]\rightarrow S$ where $S$ is commutative is then the same as defining a map $g:\mathbb{Q}\rightarrow S$ and choosing any units $s_1,\dots,s_n\in S^\times$ to map each $x_i$ to. Once you have chosen the image of $x_i$ to be $s_i$, that forces the image of $x_i^{-1}$ to be $s_i^{-1}$.

So, for example, when $S$ is another Laurent ring $\mathbb{Q}[x_1^{\pm 1},\dots,x_{n-1}^{\pm 1}]$, the natural thing to do is to define $g:\mathbb{Q}\rightarrow S$ by $g(q)=q$, and let $s_i=x_i$ for each $i<n$. Since each $x_i$, is invertible in $S$ for $i<n$, we have each of these $s_i\in S^\times$, as needed. Let's assume you're only interested in maps that act this way, ie that act like the identity on the subring $\mathbb{Q}[x_1^{\pm 1},\dots,x_{n-1}^{\pm 1}]$ of $\mathbb{Q}[x_1^{\pm 1},\dots,x_{n}^{\pm 1}]$

We still need to choose $s_n$, which is to be the image of $x_n$. By our discussion above, we can choose $s_n$ to be any unit of $S$; for instance, we can let $x_n$ map to $x_i$, for any $i<n$, or let $x_n$ map to any non-zero element of $\mathbb{Q}$. More generally, note that any monomial $x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}\in S$, with $\alpha_i\in\mathbb{Z}$, is a unit, and has inverse given by $x_1^{-\alpha_1}\dots x_{n-1}^{-\alpha_{n-1}}\in S$. Thus we can let $x_n$ map to $x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}\in S$ if we wish.

In fact, every unit of $S$ is of the form $\lambda x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}$ where $\lambda\in\mathbb{Q}\setminus\{0\}$ and each $\alpha_i\in\mathbb{Z}$, so for every such element of $S$ there is a ring homomorphism from $R$ to $S$ that sends $x_n$ to $\lambda x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}$. Conversely, no ring homomorphism from $R$ to $S$ can send $x_n$ to a non-unit, so these are all of the possible images for $x_n$, and thus this gives a complete characterization of the maps you are interested in. Concretely, once we've chosen $s_n\in S^\times$, the induced map will act by replacing every instance of $x_n$ in a polynomial of $\mathbb{Q}[x_1^{\pm 1},\dots,x_{n}^{\pm 1}]$ by $s_n$.

If you enjoy these constructions, I recommend looking into localization, which is a big generalization of some of the ideas above and serves as a very powerful/important tool in commutative algebra. In this language, the Laurent polynomial ring $\mathbb{Q}[x_1^{\pm 1},\dots,x_n^{\pm 1}]$ is the localization of the polynomial ring $\mathbb{Q}[x_1,\dots,x_n]$ at the multiplicative set generated by $\{x_1,\dots,x_n\}$.


问:
Question about Fourier cosine series
I'm trying to find Fourier cosine series for function
$$f(x) = \begin{cases}
        \frac{\pi}{2}-x,\quad &\text{if } x\in\left[0,\frac{\pi}{2}\right) \\
        \pi,\quad &\text{if } x\in\left[\frac{\pi}{2},\pi\right]
        \end{cases}$$
in $[0,\pi]$.

I found coefficients
$$b_k=0,\quad a_0=\frac{5\pi}{4}\quad \text{and}\quad a_k=\frac{2}{\pi}\cdot\frac{1-\cos\frac{k\pi}{2} + k\pi\sin k\pi - k\pi\sin\frac{k\pi}{2}}{k^2}$$
where $k\in\mathbb{N}$, but I don't know how to simplify $a_k$ to write cosine series as
$$\frac{5\pi}{4}+\sum_{k=1}^{\infty}\ ...\ \cos kx$$
and investigate pointwise convergence in $[0,\pi]$.

Can anyone help me with this?

答:
$a_k=0$ if $k$ is a multiple of $4$. $a_k=\frac 1 {\pi j^{2}}$ if $k=2j$ with $j$ odd.
$a_k=\frac 1 {\pi} \frac { 1-(2j-1)\pi(-1)^{j+1}} {(2j-1)^{2}}$ if $k=2j-1$.

In all cases $|a_k| \leq \frac C {k^{2}}$ for some $C$ so $\sum |a_k| <\infty$. This imlies that the Fourier series converges to $f$ at very point.

I have just used basic properties sine and cosine functions.

Title: Question about Fourier cosine series


问:
Show whether the sequence converges (sequence given as a sum)
Here is the problem:

Show whether $a_n$ converges:

$$ \frac1{2^2}+\frac2{3^2}+\dots+\frac n{(n+1)^2}$$

So what I tried to do is I tried to show that $a_{n+1}-a_n>0$ which can be simply proven given that $n>0$. What I am having trouble is proving if $a_n$ has an upper boundary. I tried fractional it to partial fractions and the best I can get is the sum of two separate sequences which I don't think does anything useful.

Can anyone help?

答:
This is immediate with (elementary) asymptotic analysis:
$$\frac n{(n+1)^2}\sim_\infty\frac n{n^2}=\frac1n,$$
and the harmonic series diverges.


问:
Find random variable of double exponential

Let $X$ be a random variable with the double exponential distribution with the pdf
$$f(x,\theta)=(2\theta)^{-2}e^{(-|x|/\theta)}.$$
Consider the random variable $Y=c|X|$. Find the constant $c$ and integer number $r$ such that $Y \sim \chi^2(r)$.

I am not sure how to do this.

答:
The Laplace distribution (centred at $\mu=0$) has PDF $$f(x,\theta) = \frac{1}{2\theta}e^{-\frac{|x|}{\theta}}$$

and MGF (for $\theta>0$ and $\theta |t| < 1$)

\begin{align*}
\mathbb{E}[e^{tX}] &= \frac{1}{2\theta} \left (\int_{-\infty}^{0} e^{tx+\frac{x}{\theta}}\, dx + \int_{0}^{\infty} e^{tx-\frac{x}{\theta}}\, dx \right ) \\
&= \frac{1}{2} \left (\int_{-\infty}^{0} e^{u(\theta t +1)}\, du + \int_{0}^{\infty} e^{u(\theta t-1)}\, du \right ) \\
&= \frac{1}{2} \left (\frac{1}{1 + \theta t} + \frac{1}{1 - \theta t} \right )\\
&= \frac{1}{1-\theta^{2}t^{2}}.
\end{align*}

Next, as homework, follow the example and (i) compute $\mathbb{E}[e^{ct|X|)}] = \frac{1}{1-c\theta t}$.

Then (ii) note that, if $Y \sim \chi^{2}(r)$, then $\mathbb{E}[e^{tY}] = \frac{1}{(1-2t)^{k/2}}$.

Finally (iii) match coefficients so that $c\theta = 2$ and $\frac{k}{2}=1$.


问:
How to construct an increasing sequence in an ordinal $\alpha$ from a bijection $f: |\alpha| \rightarrow \alpha$
I have a question related to cofinality:

Let $\alpha$ is a limit ordinal and is not a cardinal. How can I construct an increasing sequence $\alpha_{\xi}$ in $\alpha$ with the length $|\alpha|$ from a bijection $f: |\alpha| \rightarrow \alpha$ such that $\lim_{\xi \rightarrow |\alpha|} \alpha_{\xi} = \alpha$.

I have read one construction from here where the sequence was built based on a set $S$:

$S = \{ \beta \in |\alpha| \mid \forall \gamma \prec \beta: f(\gamma) \prec f(\beta)  \}$

However, let take an example for $\alpha = \omega +1$ where the bijection $f$ is defined with $f(0) = \omega$ and $f(n+1) = n$ for all $n \in \omega$. We can easily get $S = \emptyset$.

I cannot comment to that question so my post (as an answer) has been deleted by moderators. I hope your helps in this case and so sorry if I made a mistake.

Edited: $\alpha$ is a limit ordinal.

答: In general this is not possible: the best that can be hoped for is a cofinal increasing sequence of length $\operatorname{cf}\alpha$. It certainly isn’t possible if $\alpha$ is not a limit ordinal, but it isn’t possible for all limit ordinals, either. For instance, let $\alpha=\omega_1+\omega$; then $|\alpha|=\omega_1$, but $\operatorname{cf}\alpha=\omega$. There are certainly increasing $\omega$-sequences cofinal in $\alpha$, the simplest being $\langle \omega_1+n:n\in\omega\rangle$, but no increasing $\omega_1$-sequence can be cofinal in $\alpha$.


问:
Can any number of squares sum to a square?
Suppose

$$a^2 = \sum_{i=1}^k b_i^2$$

where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).

Can any positive integer be the value of $k$?

The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we have $a^2A = \sum_{i=1}^k b_i^2A$, where we have $k$ pieces scaled by $b_i$ to tile the big figure, which is scaled by $a$. I am wondering what constraints there are on the number of pieces.

Here is an example tiling that realizes $4^2 = 3^2 + 7 \cdot 1^2$, so $k = 8$.

答:
Yes, any number of squares can be a perfect square. Take the example of Pythagorean triples generated by the formula with a table of triples shown below it where $n$ is a set and where $k$ is a set member.

\begin{equation}
  A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}
\begin{array}{c|c|c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4 & k=5   &  k=6  \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 &  75,100,125 & 85,132,157 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline
\end{array}
Since $3^2+4^2=5^2$, we can replace the $5$ in $5,12,13$ with $3,4$ so $3^3+4^2+12^2=13^2$. The value of $A$ can be any odd number greater than $1$ so there is always a triple where side-A of one equals side-C of another. Extending to a couple more squares, $3^2+4^2+12^2+84^2+132^2+12324=12325$. The process can be extended to infinite numbers of squares summed.

Let me know if you would like to know how to find "extensions".


问:
Approximation of characteristic function of compact sets via $C_0$ functions
In Kallenberg Proposition 19.14 this purely topological statement seems to be used and I don't understand it.

Let $\Omega$ be a completely regular locally compact space. Denote by $\hat{\Omega}$ the one-point compactification by the element ${*}$. Assume that we are given a Radon probability measure $\mu$ on $\hat{\Omega}$ with the property that
$$
\int_\mu \hat{u} dx=0
$$
for all $u\in C_0(\Omega)$ extended by $\hat{u}(*)=0$.

Show: All the mass lies with $*$, i.e. $\mu(*)=1$.

My approach: Let $K\subseteq\Omega$ be an arbitrary compact set. By "some" approximation theorem, be it Tietzsche or Urysohn, we find a $C_0(\Omega)$ function $u$ with $\chi_K\leq u$. Hence $\mu(K)=0$, but since the measure $\mu$ is Radon, we know that
$$
\mu(\Omega)=\sup_{K\subseteq\Omega \text{ compact}}\mu(K)=0.
$$
We conclude $\mu(*)=1-\mu(\Omega)=1$.

Question: Can somebody provide me the "some" approximation theorem with precise conclusions to this particular case? I somehow fail to find the correct argumentation here. Perhaps because the statement is false?

答: As mentioned in comments, the correct statement is the lemma of Urysohn, as in Rudin's "Real and complex analysis", §2.7, §2.12.</p>


问:
License plate combinations 4 letters 3 numbers
How many license plates can be created if they have $4$ letters and $3$ digits in any order?

My thinking is $26^4 \times 10^3 \times {7 \choose 4} \times {7 \choose 3}$:

$26$ possibilities for each of 4 letters, $10$ possibilities for each of 3 digits, $4$ places out of $7$ to place the letters, and $3$ places out of $7$ to place the numbers.

Is this correct?

答: There are 28 letters and 10 digts from 0 to 9. You want to chose 7 of them when the order in which you select is important. So there are $P_{38}^7$ possibilites. ($P$ stands for permutation)


问:
Plot of $f'(x)$ against $f(x)$
What does the plot of $f'(x)$ against $f(x)$ represent?

答:
In conservative and non-conservative dynamical system analysis such as ( Limit cycles, Van der Pol equation of oscillation, Lissajou's figures involving vibration in two directions/frequencies etc.) DEs of two, three variables are together considered coupled or decoupled. They  occur in mechanical or electrical dynamic systems.

They can be represented as phase portraits which are displacement/velocity plots. We  see how much deviation is there from simplest harmonic model situation (ellipse) due to oscillations in-phase, out of phase or partial offset at start points of prescribed boundary values.The plot can visually display effects of stiffness or damping.

In Lyapunev Stability $( \dot x, \ddot x) $ of state variables  we see elliptic points, saddle points etc., characterizing the dynamics coming out of the governing ODE or transfer function.

Phase portraits are a characteristic of the entire dynamical process.

Van der Pol Oscillator

OwnerDisplayName: GitGoodCodes

Title: Plot of $f'(x)$ against $f(x)$


问:
How to solve trigonometric inequality $\cos\left(\frac{2\pi t}{12}\right) \ge \frac12$ without using calculators
I have a trig question. How do you I solve this, my professor put this kind of inequality in the exam where we can't watch table or use calculator so I suppose there is a fast way to do it.
$$\cos\left(\frac{2\pi t}{12}\right) \ge \frac12$$
Thank you so much

答:
Here is an outline: First solve $\cos\left(\frac{2\pi t}{12}\right) = \frac{1}{2}$ which comes down to $\cos\left(\frac{2\pi t}{12}\right) = \cos\left(\frac{\pi}{3}\right)$.

This gives two branches to solve: $\frac{2\pi t}{12}=\frac{\pi}{3}+2k\pi$ and $\frac{2\pi t}{12}=-\frac{\pi}{3}+2k\pi$.

You should be able to find $t=12k+2$ and $t=12k-2$. Since for $t=0$ the cosine is above $\frac{1}{2}$, should now be able to put all the pieces together, which is what I want you to try now.


问:
Notation for joins/meets in lattice theory
Most often I see $\lor$ and $\land$ used for the join and meet operations respectively in a lattice, however what I am writing makes concurrent use of both symbolic logic and a particular lattice, thus I am already using these operations to express logical conjunction/disjunction and am unsure what to use for the joins/meets. For example I don't want to use $\cup$ and $\cap$ since I think this might cause confusion with sets or enable some unconscious error where I treat them like their set counter parts i.e. I might accidentally distribute them over each other despite not dealing with a distributive lattice, likewise for similar reasons I'd rather not use $\sqcup$ and $\sqcap$ as I often use the latter for 'disjoint' unions. So with all of that in mind what are some standard alternatives to these that I could use for joins/meets in a lattice?

答:
Supremum and infimum, or greatest lower bound and least upper bound.

$$a\vee b=\sup(a,b)=\text{lub}(a,b)$$

$$a\wedge b=\inf(a,b)=\text{glb}(a,b)$$


答:
A simple sketch with the trigonometric circle will show this inequality is equivalent to
\begin{align}
\cos\Bigl(\frac{\pi t}6\Bigr)\ge \cos\frac\pi 3&\iff -\frac\pi3\le\frac{\pi t}6\le \frac\pi3 \pmod{2\pi\mathbf Z} \\
&\iff-2\le t\le 2\pmod{12\mathbf Z}.
\end{align}

Title: How to solve trigonometric inequality $\cos\left(\frac{2\pi t}{12}\right) \ge \frac12$ without using calculators

Body:
I have a trig question. How do you I solve this, my professor put this kind of inequality in the exam where we can't watch table or use calculator so I suppose there is a fast way to do it.
$$\cos\left(\frac{2\pi t}{12}\right) \ge \frac12$$
Thank you so much


问:
Stolz-Cesàro Theorem
Recently I've been trying to find a satisfactory proof of the Stolz-Cesàro Theorem but I havent found any. As I remember the claim is as follows:

Let $${\left\{ {{b_n}} \right\}_{n \in {\Bbb N}}}$$ be a sequence such that

$${b_{k + 1}} - {b_k} > 0  $$ and $$ \mathop {\lim }\limits_{k \to \infty }\sum_{n=0}^{k} {b_n} = \infty   $$

Then if $${\left\{ {{a_n}} \right\}_{n \in {\Bbb N}}}$$ is another sequence and the limit

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} = \ell_1 $$

exists, then

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = {\ell _2}$$

exists too and

$${\ell _1} = {\ell _2}$$

答:
Here's a more general situation:

THM Let $\langle a_n\rangle$ be any sequence of real numbers and suppose that $\langle b_n\rangle $ is a sequence of positive numbers such that $b_n$ is strictly monotone increasing to $\infty$. Then $$\liminf_{n\to\infty}\frac{a_n}{b_n}\geq \liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ $$\limsup_{n\to\infty}\frac{a_n}{b_n}\leq \limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

PROOF We prove the case for $\liminf$; the $\limsup$ case is analogous. Take $$\alpha <\liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

Then there exists $N$ such that for each $k\geq 0$ we have $$\alpha <\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$$ Since $b_{n+1}>b_n$, we have for $k\geq 0$ that
$$\alpha \left( {{b_{N + k}} - {b_{N + k - 1}}} \right) < {a_{N + k}} - {a_{N + k - 1}}$$

Thus, for any $m\geq 0$, $$\eqalign{
   \alpha \sum\limits_{k = 0}^m {\left( {{b_{N + k}} - {b_{N + k - 1}}} \right)} & < \sum\limits_{k = 0}^m {\left( {{a_{N + k}} - {a_{N + k - 1}}} \right)}   \cr
   \alpha \left( {{b_{N + m}} - {b_{N - 1}}} \right) &< {a_{N + m}} - {a_{N - 1}} \cr} $$

It follows that $$\alpha \left( {1 - \frac{{{b_{N - 1}}}}{{{b_{N + m}}}}} \right) < \frac{{{a_{N + m}}}}{{{b_{N + m}}}} - \frac{{{a_{N - 1}}}}{{{b_{N + m}}}}$$ and taking $m\to\infty$ $$\alpha  \leq \mathop {\lim \inf }\limits_{m \to \infty } \frac{{{a_m}}}{{{b_m}}}$$

It follows that, for each $\alpha <\liminf\limits_{n\to\infty}\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ we have $\alpha  \leq \liminf\limits_{m \to \infty } \dfrac{{{a_m}}}{{{b_m}}}$, which means $$\mathop {\liminf }\limits_{n \to \infty } \dfrac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} \leq  \liminf\limits_{m\to\infty} \frac{{{a_m}}}{{{b_m}}}$$

COR Let $\langle a_n\rangle$ and $\langle b_n\rangle$ be as before. Then if $$\ell=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists, so does  $$\ell'=\lim_{n\to\infty}\frac{a_n}{b_n}$$ and $\ell=\ell'$

COR Let $x_n$ be any sequence. If $$\lim_{n\to\infty} x_n=\ell$$ then $$\lim_{n\to\infty}\frac 1 n \sum_{k=1}^n x_k=\ell$$

P By the first corollary with $b_n=n$ and $a_n=\sum_{k=1}^n x_k$, we have $$\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}}  - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {x_{n + 1}} = \mathop {\lim }\limits_{n \to \infty } {x_n}$$

which means $$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{x_k}}  = \mathop {\lim }\limits_{n \to \infty } {x_n}$$

Title: Stolz-Cesàro Theorem


问:
Bijection example given set of integers to natural numbers with condition.
State an example of a bijection $f:\mathbb{Z}\to\{ 2n \space|\space n\in\mathbb{N} \}$.

I know for a bijection the two sets must have the same cardinality. An example I came up with for this condition is:
$$
f(n)=2n^2, f:\{ ...,-3,-2,-1,0,1,2,3,... \}\to\{ 2,4,6,8,... \}
$$

But I'm not sure if this is valid or if I am missing something?

答: To answer you question about whether your function $f$ is bijective, the anwer is no. In fact, $f(n)=2n^2$ is very well surjective, since $f(\mathbb{Z})\subset \{2,4,6,...\}$. How every, it is not injective, since, for example, $f(2)=f(-2)$ while $2\not=-2$.


问:
How to find a number when given the $n$-th root
If I am given something like $\sqrt[n]{100} = 10$ then it is obvious that $n=2$.

But say I get something like $\sqrt[n]{2} = 36$; how do I find what $n$ is then?

答:
Fleshing out Jet Chung's comment, writing $\sqrt[n]2$ as $2^{1/n}$, we have

$$2^{1/n}=36\iff{1\over n}\log2=\log36\iff n={\log2\over\log36}$$

Note, it doesn't matter what base you use for the logarithm, the quotient is the same. Note also, the $n$th root notation $\sqrt[n]{x}$ is usually reserved for positive integer values of $n$, which is not the case here. (Since $36\gt32=2^5$, it's easy to see that $n\lt1/5$.)


问:
An eely function $\mu (n):\;\;\prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)} = 1$
Time ago, dealing with a generalization of the Stirling numbers, I stumbled on the following implicit recurrence
$$
\mu (n):\;\;\prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)}  = 1\quad \left| \matrix{
  \,0 \le n \in Z \hfill \cr
  \,0 \le \mu (n) \in R \hfill \cr
  \,\mu (0) = 0 \hfill \cr}  \right.
$$

Using a good CAS it is not difficult to compute the first few values and plot them

Clearly, the sequence is monotonically increasing, and its finite difference is monotonically decreasing (1).

Such a regular behaviour leads me to expect that it might be extended over the reals and that
$\mu (x)$ might be expressible through a combination of conventional functions.

From time to time I am returning to this challenge with some inspiration for a new approach, but
the combination difference & product has frustrated all my attempts.

I couldn't even succeed to establish its asymptotic behaviour (now, thanks to @AMarino answer I know it's logarithmic) .

So I am asking for hints, suggestions.

-- addendum  --

Putting $\rho _{\,n,\,m}  = \mu _{\,n + 1}  - \mu _{\,m} $ , an alternative way to express the problem is
$$
\left\{ \matrix{
  \prod\limits_{0\,\, \le \,k\, \le \,n} {\rho _{\,n,\,k} }  = 1 \hfill \cr
  \rho _{\,n,\,k}  - \rho _{\,n - 1,\,k}  = \mu _{\,n + 1}  - \mu _{\,n}  \hfill \cr}  \right.
$$
which means to find a family of functions whose product wrt $k$ is $1$
and whose difference wrt $n$ is constant, as shown

My last tentative has been to take two discrete pmf's on the support $[0,n]$ and put
$$
\rho \left( {n,m} \right) = e^{\,h(n)\left( {p(m\,|n) - q(m\,|n)} \right)}
$$
which by definition gives the unitary product, but cannot go  yet through keeping the difference constant.

-- note 1 --

That the difference is monotonically decreasing comes from being
$$
\eqalign{
  & \prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)}
  = \prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (n - 1) + \mu (n - 1) - \mu (k)} \right)}  =   \cr
  &  = \prod\limits_{k = 0}^{n - 1} {\left( {x + \left( {\mu (n - 1) - \mu (k)} \right)} \right)}
  = p_n (x)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ \matrix{
  p_n (0) = 0 \hfill \cr
  p_n (x) < p_{n + 1} (x)\quad \left| {\,0 < x} \right. \hfill \cr
  1 = p_n (\mu (n) - \mu (n - 1)) < p_{n + 1} (\mu (n + 1) - \mu (n)) \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \mu (n + 1) - \mu (n) < \mu (n) - \mu (n - 1) \cr}
$$
which tells the interesting fact that $\Delta \mu (n-1)$ comes as the root of $p_n(x)=1$, which in turn is added
to its zeros, shifting them to the left as to start from $0$ and making them the zeros of $p_{n+1}(x)$.

答:
This is only a partial answer.

If your claim is correct, i.e. differences are decreasing, then the sequence is sublogarithmic. Indeed, let $s(n) := \mu(n+1) -\mu(n) $. The recurrence relation, using the claim $s(j) > s(n)$ for $j<n$, rewrites as

$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) > \prod_{k=0}^n (n-k+1) s(n) = (n+1)! s(n) ^{n+1} $$

Which implies that

$$ s(n) < \left ( \frac{1}{(n+1)! } \right ) ^{1/(n+1) }$$

The Stirling approximation can be written as $m! \approx (m/e) ^{m}$. Substituting we get that $s(n) < \frac{e}{n+1}$ . Thus we have

$$ \mu(n) = s(n-1) +\ldots+s(0) < $$
$$ e \sum_{k=0}^{n-1} \frac{1}{k+1} \approx e \log(n) $$

We also have a sort of converse to this. Suppose that for all n we have
$$(*) \ \ s(n) > \left ( \frac{1}{(n+2)! } \right ) ^{1/(n+2) }$$

That is, the sequence is also superlogarithmic. We want to show that differences are decreasing. Suppose that at some point $n$ we have for the first time that the difference increase, aka $s(n) \ge s(n-1) $. Then we get

$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) \ge \prod_{k=0}^n (n-k+1) s(n-1) = (n+1)! s(n-1) ^{n+1} $$

From which we get, using again Stirling approximation, the opposite (*) inequality, yielding a contradiction.

This argument can be made precise with inequalities instead of plain Stirling approximation, but since it is not a strong result with respect to your requests, I think it's not worth. Hope it can inspire someone else for a more sophisticated approach :)

Edit. I have an approach for the lower bound. By substituting each factor with the biggest $s$ appearing we get a reversed inequality

$$ 1\le n s(0) \cdot  (n-1)s(1) \ldots 1 \cdot s(n-1) $$

Rearranging terms and taking the logarithms we get

$$ 0 \le \log ns(n-1) + \ldots +\log 1\cdot s(0) $$

Let us define
$$\sigma(n) := \frac{1}{n} \sum_{j=0} ^{n-1} \log (j+1) s(j) $$
Equation above ensures that $\limsup \sigma(n) \ge 0$. Now the plot twist: by Stolz Cesaro the limsup of the general term is greater or equal than the average one, so that

$$ \limsup \log s(n) (n+1) \ge \limsup \sigma(n) \ge 0$$

Which implies that

$$ \limsup s(n) (n+1) \ge 1$$

Which provides a lower bound like estimate. If this can be refined to be a limit, than by summing over n we would get that $\mu(n) \ge \log(n) $ asymptilotically.


答:
Reiterating what was said earlier in a comment:

Your answer is almost correct however you made the mistake of picking the positions twice rather than only once.  By choosing the four positions to be occupied by letters and then again choosing which three positions were to be occupied by digits and having chosen said positions out of the original seven, you will have run the risk of picking the same position multiple times designating it to be simultaneously used by letter and digit.

Instead, the digits' positions should have been selected out of the remaining unused positions giving a total of $\binom{3}{3}=1$ way in which this step could be accomplished, this being merely $1$ way it can be left out of the final product.  This leaves the final answer as being:

$$26^4\cdot 10^3\cdot \binom{7}{4}$$

Compare this to a slightly larger problem where you have $26$ letters to choose from, $10$ digits, and $15$ special symbols (e.g. !@#\$%^ and so on...) and we want to make a ten-character password consisting of $6$ letters, $3$ digits, and $1$ special symbol with repetition allowed.

We first pick the positions used by letters, then from those remaining positions pick the positions used by digits, letting the final remaining position be occupied by the symbol, and then picking which letters, digits, and symbol they happened to be for a total count of:

$$26^6\cdot 10^3\cdot 15\cdot \binom{10}{6}\cdot \binom{4}{3}$$

Title: License plate combinations 4 letters 3 numbers

Body:
How many license plates can be created if they have $4$ letters and $3$ digits in any order?

My thinking is $26^4 \times 10^3 \times {7 \choose 4} \times {7 \choose 3}$:

$26$ possibilities for each of 4 letters, $10$ possibilities for each of 3 digits, $4$ places out of $7$ to place the letters, and $3$ places out of $7$ to place the numbers.

Is this correct?


问:
Finding the exact value of $P(X \leq np)$ for $k = 10$
Let $X_1,......,X_k$ be iid random variables from a Binomial distribution with n=4 and p=0.1.

Let $\bar{X} = \frac{1}{k}\sum_{i=1}^k X_i$.
Find the exact value of P($\bar{X}$ $\leq np)$ for $k = 10$.

I have found out the value of this problem using the central limit theorem (CLT) using randomly generated values but this problem specifically states to not use CLT. At first, I thought to write it like this:
$$\frac{1}{k}\sum_{i=1}^k P(X_i \leq np),$$
but I am not sure this is allowed.

Is this problem only possible if we randomly generated $X_1$ to $X_k$ random variables with $\operatorname{Bin}(4,0.1)$?

答:
Let $S$ be the sum of independent random variables $$S = \sum_{i = 1}^k X_i = k\overline X.$$

Then $S$ also follows binomial distribution: $S \sim Bin(kn, p) = Bin(40, 0.1)$. The required probability is

$$\begin{align*}
P(\overline X \le np) &= P(k\overline X \le knp)\\
&= P(S \le 4)\\
&= \sum_{i=0}^4 \binom{40}{i}0.1^i(1-0.1)^{40-i}\\
&= \cdots
\end{align*}$$


问:
Divergence of a sequence proof
Let $(x_n)$ be a sequence with $x_1=1 $ and $x_{n+1}=x_n+\frac{1}{x_n^2}$ for all integers n. I want to formally prove that this sequence is not limited.

I started with showing that $x_n$ is strictly monotonically increasing because:

$x_{n+1}-x_n=\frac{1}{x_n^2} > 0$

Now I want to show that for all real numbers $a$ I will find an integer $n_0$ thus $x_n>a$ for all integers $n_0$. But I am stuck at this point. Can someone give me a hint how to go on? :)

答: Since $(x_n)$ is a non decreasing sequence; either $(x_n)$ converges, either its limit is $+\infty$. If $(x_n)$ converges towards $\ell\geqslant x_1=1$ we would have $\ell=\ell+\frac{1}{\ell^2}$ which is not, thus $\lim\limits_{n\rightarrow +\infty}x_n=+\infty$.

Title: Divergence of a sequence proof


问:
Is there a way to deal with this singularity in numerical integration?
I would like to compute numerically, e.g., using the standard method of trapezes the following definite integral over the interface $[0,1]$:
$$
I = \int_0^1 \frac{f(x)}{\sqrt{1-x^2}} \, \mathrm{d} x \, ,
$$

where $f(x)$ is continuous in the interval [0,1].
It can readily be shown analytically that the integral is well convergent.
However, when proceeding numerically, difficulties arise since the integrand diverges at the upper limit of integration for $x=1$.

I was wondering whether there exists a procedure that can help to remove the singularity in this integral.
Any help is highly appreciated

Thank you,

hartmut

Body:
The standard way is $x = 1 - t^{2}$:
$$
\int_{0}^{1}{\operatorname{f}\left(x\right) \over
\,\sqrt{\,{1 - x^{2}}\,}\,}\,\mathrm{d}x =
2\int_{0}^{1}{\operatorname{f}\left(1 - t^{2}\right) \over
\,\sqrt{\,{2 - t^{2}}\,}\,}\,\mathrm{d}t
$$


问:
What exactly does this physically mean?
Let X(w) be a real random variable on ($\Omega$ , P). The image X($\Omega$) the set of all the values X(w) can take ,written $\Omega^{X}$. For any set $ B \subset \Omega^{X}$ the probability of the event that the value of X lies in B is equal to $P\{w|X(w) \in B\} = P(X^{-1}(B))$.

What I understood from that is that the probability measure P is the probability of the set of all ws such that the event X(w) is realized and which from that is an element of B, namely is equal to $P\{w|X(w) \in B\}$ however I do not understand why $P(X^{-1}(B))$ is used? Does it actually mean the inverse of event X or inverse composite? How physically can I vindicate the use of $X^{-1}$? Any ideas?

答: The sentence beginning "What I understood" is difficult to parse, but from the last part of it I get the impression that you are unaware that $\{w\mid X(w)\in B\}$ is the same thing as $X^{-1}(B)$.  If that's the problem, then you should review the definition of $X^{-1}(B)$, perhaps not in a specifically probabilistic context but in the general context of any function $X$ that maps into a set that includes another set $B$ as a subset.</p>

Title: What exactly does this physically mean?


问:
Probability of intersection of dependent events, having probability approaching 1 in random graph
For $\mathcal{G}(n,p)$ (Erdős–Rényi graph) let $A$ be a property of pair of vertices. And let $A_{uv}$ be the event that property $A$ holds for vertices $u$ and $v$. If the following holds
$$\forall u,v \quad \lim_{n \to \infty} \Pr[A_{uv}] = 1$$

then would it be correct to say that
$$\lim_{n \to \infty} \Pr[\bigcap\limits_{u,v}A_{uv}]=1$$

The events $A_{uv}$ are not independent. Intuitively it seems correct, but how I can formally prove/disprove the above.

答:
No it's not correct : If $\Pr[A_{uv}]$ tends to 1 too slowly (e.g. in $1/n$) then this is an issue as the number of pairs grows too fast (in $n^2$) :

For instance assume that for any $u,v$, $\Pr[A_{uv}]=1-\frac{1}{n}$ and assume that the events are independent for simplicity. Then

$$ \Pr[\bigcap_{u,v}A_{uv}] = \left(1-\frac{1}{n}\right)^{\binom{n}{2}} \leq e^{-\binom{n}{2}\frac{1}{n}} = e^{-\frac{n-1}{2}} \to 0$$

This might be understood intuitively by saying that if you want $G\in\mathcal{G}(n,p)$ to be with high probability a complete graph $K_n$, then you need $p$ to tend to 1 very quickly, otherwise you have too many edges, and one of them will not be present in $G$.


答:
Note that for $m > 0$, we have $$x_{n+m} - x_n = \sum\limits_{k=1}^m\frac{1}{x_{n+k-1}^2} \geq \frac{m}{x_n^2}$$

But if $x_n$ converges, then $|x_{n+m} - x_n| \to 0$. In particular, if $(x_n)$ is bounded above, say by $C$ (WLOG, $C > 0$), then we have $$x_{n+m} - x_n \geq \frac{m}{C^2} > 0,$$ so in particular $$x_{m} \geq 1 + \frac{m-1}{C^2} \to \infty$$ for all $m$,
and hence $(x_m) \to \infty$, contradicting our assumption that $(x_n)$ is bounded above. Thus, it is not so bounded.

Title: Divergence of a sequence proof

Body:
Let $(x_n)$ be a sequence with $x_1=1 $ and $x_{n+1}=x_n+\frac{1}{x_n^2}$ for all integers n. I want to formally prove that this sequence is not limited.

I started with showing that $x_n$ is strictly monotonically increasing because:

$x_{n+1}-x_n=\frac{1}{x_n^2} > 0$

Now I want to show that for all real numbers $a$ I will find an integer $n_0$ thus $x_n>a$ for all integers $n_0$. But I am stuck at this point. Can someone give me a hint how to go on? :)


Title: Divergence of a sequence proof

Body:
Well, first we can prove that the sequence is positive, then, let $ n\in\mathbb{N}^{*} $, for any $ k\in\mathbb{N} $, we have : \begin{aligned} x_{k+1}&=x_{k}+\frac{1}{x_{k}^{2}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{k+1}^{3}&=x_{k}^{3}+3+\frac{3}{x_{k}^{3}}+\frac{1}{x_{k}^{6}}\\&\geq x_{k}^{3}+3 \\ \Longrightarrow\  \sum_{k=1}^{n-1}{\left(x_{k+1}^{3}-x_{k}^{3}\right)}&\geq 3\left(n-1\right)\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{n}^{3}&\geq 3n-2\end{aligned}

Thus : $$ \lim_{n\to +\infty}{x_{n}}=+\infty $$


问:
Show whether the sequence converges (sequence given as a sum)
Here is the problem:

Show whether $a_n$ converges:

$$ \frac1{2^2}+\frac2{3^2}+\dots+\frac n{(n+1)^2}$$

So what I tried to do is I tried to show that $a_{n+1}-a_n>0$ which can be simply proven given that $n>0$. What I am having trouble is proving if $a_n$ has an upper boundary. I tried fractional it to partial fractions and the best I can get is the sum of two separate sequences which I don't think does anything useful.

Can anyone help?

答:
Using that $k\geq \frac{k+1}{2}$ for all positive integers, one gets
$$
a_n = \sum_{k=1}^{n} \frac{k}{(k+1)^2} \geq \frac{1}{2}\sum_{k=1}^{n} \frac{1}{k+1}.
$$
The lower bound sequence obviously diverges, and so the initial sequence diverges as well!


问:
Determine the commutator of the non-abelian group of order $pq$
By the Sylow theorem, I know that there exists normal subgroup of order $p$. But the commutator is the smallest normal subgroup $G'$ such that $G/G'$ is abelian, so it suffices to show that Sylow $q$-subgroup is not normal. But how can I see this?

答:
Although it is the case that the Sylow $p$-subgroup cannot be normal (I’m assuming that $p\lt q$, though you never say so), you don’t need that.

Recall that if $G$ is any group, and $N$ is any normal subgroup of $G$ such that $G/N$ is abelian, then $G’\leq N$. Thus, if the $q$-Sylow subgroup $Q$ of $G$ is normal, then $G/Q$ is of order $p$ and hence abelian, so we know that $G’\leq Q$. As $Q$ is of prime order, this means either $G’$ is trivial of $G’=Q$. But $G$ itself is nonabelian by assumption, so you know that $G’$ cannot be trivial. Thus, $G’=Q$.


答:
No, this isn't guaranteed.

Consider $\mathcal G(n, \frac1n)$, and let $A_{uv}$ be the event "there is no edge between $u$ and $v$". Then $\Pr[A_{uv}] = 1 - \frac1n \to 1$ as $n \to \infty$, but $\bigcap_{u,v} A_{uv}$ is the event that the graph has no edges, and the probability of that goes to $0$.

The most typical way to solve the problem is to let $\mathbf X_{uv} = 1$ if $A_{uv}$ does not hold, and $X_{uv}=0$ if $A_{uv}$ does hold. Then $\mathbb E[\mathbf X_{uv}] \to 0$ as $n \to \infty$. Let $\mathbf X = \sum_{u,v} \mathbf X_{u,v}$. If we can show the stronger statement that $\mathbb E[\mathbf X] = \sum_{u,v} \mathbb E[\mathbf X_{u,v}] \to 0$ as $n \to \infty$, then we can conclude that $\mathbf X = 0$ with probability tending to $1$, and so $\bigcap_{u,v} A_{uv}$ holds with probability tending to $1$.

Title: Probability of intersection of dependent events, having probability approaching 1 in random graph

Body:
For $\mathcal{G}(n,p)$ (Erdős–Rényi graph) let $A$ be a property of pair of vertices. And let $A_{uv}$ be the event that property $A$ holds for vertices $u$ and $v$. If the following holds
$$\forall u,v \quad \lim_{n \to \infty} \Pr[A_{uv}] = 1$$

then would it be correct to say that
$$\lim_{n \to \infty} \Pr[\bigcap\limits_{u,v}A_{uv}]=1$$

The events $A_{uv}$ are not independent. Intuitively it seems correct, but how I can formally prove/disprove the above.


问:
The Fundamental Theorem of Arithmetic for $\mathbb Z[i]$
I understand how to prove the Fundamental Theory of Arithmetic, but I do not understand how to further articulate it to the point where it applies for $\mathbb Z[I]$ (the Gaussian integers).

I started by first defining norm/modulus and proving basic properties.

Next, I am going to show that the division algorithm works in $\mathbb Z[I]$.

After this, I will then define gcd and show that its two definitions agree. This here will complete my proof.

My question is, where do I begin to show that the division algorithm works in $\mathbb Z[I]$?

答: Say you want to divide $b$ into $a$. Let $q$ be the true quotient $a/b$ rounded off to the nearest Gaussian integer. Then $a/b-q$ has real and imaginary components with absolute value less than or equal to $1/2.$ From here, you should show that $|r|^2 <|b|^2,$ where $r=a-bq.$


问:
Basic likelihood quest: finding the likelihood function for a simple experiment.
The question:

In order to determine the probability of success of a certain experiment, each of one
hundred persons perform a series of these experiments until one results in success.
A model to describe this: the number of attempts $N$ required until the first success
follows a $Geo(p)$-distribution. We wish to estimate $p$ using the maximum likelihood
method.

The data:

My reasoning:

Let $X_i$ with $i=1, ..., 100$ denote the number of attempts required until the first succes for person $i$. We now consider the follwing probability mass function

$$ P(X_1=x_1, ..., X_{100}=x_{100})=\prod_{i=1}^{100}P(X_i=x_i)=p^{100}\prod_{i=1}^{100}(1-p)^{x_i-1} \\
=p^{100}(1-p)^{30}[(1-p)^{2}]^{15}[(1-p)^{\sum_{i'}x_{i'}-1}]$$

where $x_{i'}\geq4$ and $i'$ the index for the corresponding group of people. The possible answers are

So the only possible answer would be d, is my reasoning correct?

答:
No, your reasoning is not correct.

Recall that the PMF for a geometric distribution with probability $p$ of success that counts the total trials needed to observe the first success, is given by $$\Pr[X = x] = (1-p)^{x-1} p, \quad x \in \{1, 2, \ldots \}.$$
From this we conclude that $$\Pr[X \ge x] = \sum_{k=x}^\infty \Pr[X = x] = p (1-p)^{x-1} \sum_{k=0}^\infty (1-p)^k = p (1-p)^{x-1} \frac{1}{1-(1-p)} = (1-p)^{x-1}.$$  We can also see this intuitively since it takes at least $x$ tries to obtain the first success if and only if the first $x-1$ tries are all failures.

Therefore, the likelihood for $p$ given the sample is $$\begin{align}
\mathcal L(p \mid \boldsymbol x) &= \Pr[X = 1]^{40} \Pr[X = 2]^{30} \Pr[X = 3]^{15} \Pr[X \ge 4]^{15} \\
&= ((1-p)^0 p)^{40}((1-p)^1 p)^{30} ((1-p)^2 p)^{15} ((1-p)^3)^{15} \\
&= (1-p)^{30 + 2(15) + 3(15)} p^{40 + 30 + 15} \\
&= p^{85} (1-p)^{105}.
\end{align}$$
Your error lies in counting every person's experiment as contributing a factor of $p$ to the likelihood, when in fact only those people who reported an exact number of trials contribute $p$ to the likelihood.  Those $15$ people who reported at least four trials were needed because their first three failed, do not contribute a factor of $p$.  If instead they all reported exactly four trials were needed, then you would get $p^{100}(1-p)^{105}$ as you claim.


问:
Derivative of a smooth function as an equivalence class
In my differential manifolds class, the derivative of a function $f \in C^\infty (M)$, where $M$ is a manifold, at $p \in M$ was defined as the image of the linear map
$$(df)_p := C^\infty (M) \mapsto C^\infty (M)/Z_p$$

where $Z_p$ is the set of all smooth functions that have zero derivative at $p$. In other words, the derivative is an equivalence class of $C^\infty (M)$ defined by the relation that two functions are related if their derivative vanishes at $p$, the derivative of a function at a point is an equivalence class.

I'm struggling to wrap my head around this. How can a derivative become an equivalence class? Does this hold in elementary calculus too?

答:
The practice of defining mathematical objects as quotients, i.e. sets of equivalence classes of some other, more elementary objects, is widespread in mathematics. Here are a few examples: homology (where two chains are considered equivalent if they share a boundary), projective space (where two vectors are considered equivalent if one is a scalar multiple of the other) and rational numbers (where two pairs of suitable integers $(a, b)$ and $(c, d)$ are considered equivalent if $ad = bc$).

One way to help wrap one's mind around this is to distinguish between and adopt two perspectives. In one view, you consider an object's "internal structure", i.e. how it is built from other, more elementary objects. In the other view, you consider the object's "external properties", i.e. what relationships it enters with other abstract objects.

You generally use the first perspective to derive basic properties and then you begin treating the object as a "black box", forgetting how it has been built and just using its known properties to derive more properties. (An extreme form of the latter perspective occurs when you postulate a set of basic properties as axioms. This allows you to build a theory that applies to any construction that satisfies them.)

Responding more directly to your question: I believe the definition you have in mind is not in fact a definition of a derivative, but one of the definitions of the contangent space. In any case, in this setting, the first perspective tells you how the contangent space and the differential 1-forms that make it up are constructed and allows you to establish that they have some basic properties (e.g. that they satisfy the Leibniz product rule). Soon after that you generally stop thinking much about their internal structure and adopt the second perspective.

Regarding the relationship to elementary calculus: differential geometry intentionally raises the level of abstraction to increase the power of the theory beyond what you can achieve in elementary calculus. Many concepts familiar from calculus become special cases for the concepts in differential geometry. In particular, in calculus you can exploit the linear structure of $\mathbb{R}^n$ to pretend that tangent vectors live within the manifold under study. This allows you to maintain certain comfortable illusions, e.g. that a derivative of a smooth function $\mathbb{R} \to \mathbb{R}$ is also a function $\mathbb{R} \to \mathbb{R}$.


问:
Determining end coordinates of line with the specified length and angle?
I have the point $(x_1, y_1)$, the angle $0 \leq a < 360^o$ and the length $l > 0$.

How do I determine the end point $(x_2, y_2)$ if there is a line between $(x_1, y_1)$ and $(x_2, y_2)$ of length $l$ and with angle $a$?

答:
$$(x_2,y_2)=(x_1+l\cdot\cos(a),y_1+l\cdot\sin(a)) \; .$$

There is an ambiguity in your question though: what is the reference w.r.t. which the angle is measured? I assumed it's the X-axis, since that seemed the most natural without any further specifications.


问:
Rationalizing Denominators with Radicals
How do you rationalize this fraction:

$$\frac{1}{\sqrt{n+1}}$$

答:
To rationalize the denominator given:
$$\frac{1}{\sqrt{n+1}}$$

1) Multiply both the numerator and the denominator by $\sqrt{n+1}$
$$=\frac{1}{\sqrt{n+1}}\times\frac{\sqrt{n+1}}{\sqrt{n+1}}$$

2) This leaves us with:
$$\frac{\sqrt{n+1}}{\sqrt{n+1}{\sqrt{n+1}}}$$

3) Simplifying we are left with:

$$\frac{\sqrt{n+1}}{{n+1}}$$

Because:

$$\sqrt{n+1}\times\sqrt{n+1} = n+1$$

OwnerDisplayName: Joseph


问:
A question about two definitions of covering numbers
I am reading the book "Greedy approximation"(Page 208) written by V. Temlyakov and thinking about the following question of covering numbers:

Let $X$ be a Banach space and let $B_X(y,r)$ denote the ball of $X$ whose center is at $y$ with radius $r$.

Denote $N_{\epsilon}(A,X):=\min\{n: \exists  y^1, y^2, ... ,y^n, where\ \ each \ \ y^j\in X: A\subset \cup_{j=1}^n B_X(y^j,\epsilon)\}$.

Denote $N(A,\epsilon,X):=\min\{n: \exists  y^1, y^2, ... ,y^n, where\ \  each \ \   y^j\in A: A\subset \cup_{j=1}^n B_X(y^j,\epsilon)\}$.

Prove that:
$$N_{\epsilon}(A,X)\leq N(A,\epsilon,X)\leq N_{{\epsilon}/2}(A,X)$$

I can understand that the first inequality but I am a bit of confusion about the second inequality.

My idea for the first inequality is because for each $j$, $y^j\in A\subset X$. This means the minimum cover number would be smaller if we consider a bigger set. I have no idea why the second inequality is correct.

Any suggestions would be welcome and thank you !!

答:
Suppose $y^1, \ldots, y^n \in X$ such that the $B_X(y^j, \epsilon/2)$ cover $A$.

Each ball contains some point of $A$ (else you can throw away that ball to get a smaller cover). Let these points be $z^1, \ldots, z^n$.

Use the triangle inequality to argue that the balls $B_X(z^j, \epsilon)$ cover $A$.


问:
Probability of Choosing a Unique Sensor of Specific Type During Sampling
Given total number of Sensors in a region = S=100. Each sensor has a unique ID due to its unique location in the region of interest and also has a specific category T due to kind of measurement it takes. We have 5 categories as, 10% sensors belong to T1, 20% to T2,  50% to T3,  10% to T4 and 10% to T5. We have 10 sampling intervals, we randomly choose a sample of 4 sensors at each interval. In the sample the sensors must have unique ID but could be of any category. (For Example : All could be of same category or all could be of different categories or mixed). In the next sampling interval, the same holds and we can have a repetition of the sensor having a unique ID. We represent each sensor as ID(#)T(#). For example we could have the samples as:

Sampling Interval 1: SAMPLE= {ID(45)T(5), ID(33)T(4), ID(10)T(4), ID(1)T(4)}

Sampling Interval 2: SAMPLE= {ID(88)T(1), ID(45)T(5), ID(99)T(3), ID(1)T(4)}

Question No 1:

1). Given the aforesaid scenario, what is the probability at each sampling interval that a sensor with a specific ID is likely to be part of the sample of 4 picked from total of 100 ?

Question No 2:

2). Given the aforesaid scenario, what is the probability at each sampling interval that a sensor with a specific type is likely to be part of the sample of 4 picked from total of 100?

Question No 3:

Given the aforesaid scenario, what is the probability at each sampling interval that a sensor with a specific type and ID is likely to be part of the sample of 4 picked from total of 100

I need help in my scenario against the combinatorics and probability assessment expressions.

答:
Probability that the one ID you chose is represented is the same as the compliment of the one ID you chose is not represented:

$$1-\dfrac{\dbinom{99}{4}}{\dbinom{100}{4}} = \dfrac{4}{100} = \dfrac{1}{25}$$

Probability that the one type you chose is represented is the same as the compliment of the one type you chose is not represented. Since there are $10, 20, 50, 10, 10$ of each of types 1,2,3,4,5 respectively, you have:

$$\begin{array}{rr}\text{Type 1, Type 4, Type 5}: & 1-\dfrac{\dbinom{90}{4}}{\dbinom{100}{4}} = \dfrac{8279}{23765} \approx 34.84\% \\ \text{Type 2}: & 1 - \dfrac{\dbinom{80}{4}}{\dbinom{100}{4}} = \dfrac{6077}{10185} \approx 59.67\% \\ \text{Type 3}: & 1-\dfrac{\dbinom{50}{4}}{\dbinom{100}{4}} \approx 94.13\% \end{array}$$

For (3), it depends. Is it possible that whomever is guessing will guess an ID and Type that are mismatched? Like, in your example, they will guess ID(45)T(1), which does not exist since ID(45)T(5) exists? If guesses with type mismatches are not possible, then guessing the ID is the same as guessing the ID + Type. If type mismatches are possible, then what is the probability the guesser guesses a valid ID/type combo?


答:
Here's a direct proof of divergence
showing the rate of divergence.

$x_{n+1}=x_n+\frac{1}{x_n^2},
x_1>0
$

$\begin{array}\\
x_{n+1}^3
&=(x_n+\frac{1}{x_n^2})^3\\
&=x_n^3+3+\frac{3}{x_n^3}+\frac1{x_n^6}\\
&\gt x_n^3+3\\
\end{array}
$

so

$x_n^3
\gt 3(n-1)
$
or
$x_n
\gt \sqrt[3]{3(n-1)}
$.

Title: Divergence of a sequence proof


问:
Analytic continuation of power series of holomorphic with real nonnegative coefficients
Consider a holomorphic function $f(z)$ defined as power series $$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n},$$ with radius of convergence $R=1$.

I want to show that

If $a_{n}\geq 0$ for all $n$, then $f(z)$ cannot be analytically continued to $1\in \partial\mathbb{D}$.

I don't quite have an idea about where to start. As the exercise supposes $a_{n}\geq 0$ for all $n$, it means that the exercise also implicitly assumes that $a_{n}\in\mathbb{R}$ for all $n$.

From the radius of convergence, I know that $$\limsup_{n\rightarrow\infty}a_{n}^{\frac{1}{n}}=1\ \ \text{and}\ \ \lim_{n\rightarrow\infty}\dfrac{a_{n}}{a_{n+1}}=1.$$ (I get rid of the norm because $a_{n}$ is real and non-negative.)

Suppose $f(z)$ can be analytically continued to $1\in\partial\mathbb{D}$, then it means that there exists a $r=r(1)>0$ such that there exists a holomorphic function $g(z)$ on $B(1,r)$  such that $g(z)=f(z)$ on $B(1,r)\cap\mathbb{D}$.

But then I don't know how to proceed..

I tried to argue that if I define $h(z):=f(z)$ for $z\in\mathbb{D}$ and $h(z):=g(z)$ for $z\in B(1,r)$, then $h(z)$ is holomorphic in the whole $\mathbb{D}\cup B(1,r)$, so it has a power expansion $$h(z)=\sum_{n=0}^{\infty}b_{n}z^{n},$$ and in particular, at $z=1$, we must have $$\sum_{n=0}^{\infty}b_{n}<\infty.$$

But it seems no way to get back to $f(z)$. What should I do? Thank you!

答:
Assuming $f$ is analytic at $1$, since everything is increasing $$f^{(k)}(1)=\lim_{z\to 1^-}f^{(k)}(z)= \lim_{z\to 1^-} \sum_{n=k}^\infty a_n \frac{n!}{(n-k)!} z^{n-k}=\sum_{n=k}^\infty a_n \frac{n!}{(n-k)!}$$
Next we are told that for any $s>0$ $$\sum_{n=0}^\infty a_n (1+s)^n= \infty$$
Everything is non-negative so we can change the order of summation as we want obtaining
$$\infty=\sum_{n=0}^\infty a_n \sum_{k=0}^n s^k\frac{ n!}{k!(n-k)!}  = \sum_{k=0}^\infty \frac{s^k}{k!} \sum_{n=k}^\infty a_n\frac{n! }{(n-k)!}=\sum_{k=0}^\infty \frac{s^k}{k!} f^{(k)}(1)$$
contradicting that $f$ is analytic at $1$.


问:
Inequality for the square of difference

is it true the following inequality?
$$(a-b)^2 \leq C |a^2-b^2|$$
for some $C>0$, for every $a,b \in \mathbb{R}$ ?

Thanks in advance.

答: This is false in general, otherwise you would have $(a-(-a))^2\leqslant C|a^2-(-a)^2|$ for all $a\in\mathbb{R}$, that is $4a^2\leqslant 0$ and thus $\mathbb{R}=\{0\}$ (and they are not quite the same sets).

Title: Inequality for the square of difference


问:
Use Baire’s theorem to show that the set of transitive configurations is dense in $X$.
Let $X :=$ {$0, 1$}$^\mathbb N$. Let $x = x_1x_2 · · · ∈ X$ be a configuration and $w =
w_1w_2 · · · w_n ∈$ {$0, 1$}$^∗$ be a word. We say that $w$ occurs in $x$ at position $k$ if $x_kx_{k+1}· · · x_{k+n−1} = w_1w_2 · · · w_n$. For instance, $w := 1011$ occurs in the configuration
$x := 01101101100001010001···$ at positions $3$ and $6$ but not at position $1$. Let $C_k(w)$ denote the set of all configurations in which $w$ occurs at position $k$, and note that $C_1(w) = C(w)$ is simply the (standard) cylinder set with base $w$.
A configuration $x ∈ X$ is said to be transitive if every word $w ∈$ {$0, 1$}$^∗$ occurs in $x$ at some position.

$(i)$ Provide an example of a transitive configuration in $X$.

$(ii)$ Use Baire’s theorem to show that the set of transitive configurations is dense in $X$.

$(iii)$ Provide a direct construction based on part $(i)$ of a dense set of transitive configurations.

$(iv)$ Show that in every transitive configuration $x$, every word $w$ occurs at infinitely many different positions.

My attempt:

$(i)$ example: For $n = 4$, let $w = 0110$ and $x = 01101101110101101... $ is it correct?

$(ii)$ Baire's Theorem: if $X$ is a compact Hausdorff space, then the intersection of any countable collection of dense open sets is dense; i.e., if $A_1,A_2,...A_n \subset X$ open and dense, then $\cap_{n=1}^∞ A_n$ is dense and in particular non-empty.

Here, $X=${$0,1$}$^\mathbb N$ is compact and Hausdorff, and $C(w)$ is the cylinder set which is a basis, and hence is open by definition of basis. How do I proceed from here?

答:
You cannot produce a transitive configuration by looking only at words of length $4$, let alone at just one of those words. A transitive configuration must contain at least one instance of every word in $\{0,1\}^*$. Probably the easiest way to produce such a configuration is to concatenate the non-empty words in shortlex order:

$$0\color{red}100\color{red}{01}10\color{red}{11}000\color{red}{001}010\color{red}{011}100\color{red}{101}110\color{red}{111}000\ldots$$

Here I’ve alternated between black and red in order to make it easier to identify the individual words being concatenated.

As for (ii), for each $w\in\{0,1\}^*$ let $C(w)=\bigcup_{n\in\Bbb N}C_n(w)$; this is a union of open sets, so it is open, and it contains every configuration that has $w$ as a substring.

Show that each of the sets $C(w)$ is dense in $X$.

Explain why there are only countably many of these sets $C(w)$.

The Baire category theorem will then tell you that $\bigcap_{n\in\Bbb N}C(w)$ is dense in $X$, and it should be clear that $\bigcap_{n\in\Bbb N}C(w)$ is precisely the set of transitive configurations.

Added: For (iii) let $c_0$ be the transitive composition shown above. For each $n\in\Bbb Z^+$ be the composition

$$0^nc_0=\underbrace{00\ldots00}_nc_0\,,$$

and let $c_{-n}$ be the result of left-shifting $c_0$ $n$ places. For instance,

$$c_{-5}=\color{red}{1}10\color{red}{11}000\color{red}{001}010\color{red}{011}100\color{red}{101}110\color{red}{111}000\ldots\,,$$

where I’ve kept the colors to make it easier to match this up with $c_0$ as displayed above.

Show that for each $w\in\{0,1\}^*$ and $n\in\Bbb N$ there is a $k\in\Bbb Z$ such that $c_k\in C_n(w)$. Conclude that $\{c_k:k\in\Bbb Z\}$ is dense in $X$.

Finally, for (iv) let $c$ be a composition in which some word $w$ occurs at only $n$ different positions. Does the word $w^{n+1}$ occur in $c$?


问:
Strategically $\kappa$-closed poset that is not $\kappa$-closed.
I'm trying to figure out the difference between being $\kappa$-closed and strategically $\kappa$-closed (in the context of forcing).

A poset is $\kappa$-closed if every $\alpha$-chain with $\alpha<\kappa$ has a lower bound. A poset is strategically $\kappa$-closed if player I has a winning strategy in the game where both players successively choose an element below all previous moves, where player I moves first at limit steps and wins if the game lasts $\kappa$ rounds.

My initial question is to have an example of a strategically $\kappa$-closed poset that is not $\kappa$-closed.

My follow-up question would be if there is any significant difference in terms of forcing with a $\kappa$-closed poset instead of merely a strategically $\kappa$-closed poset.

答:
Here's a silly example:

Let $\mathbb{P}$ be the poset consisting of all countable-ordinal-length sequences of countable ordinals $p$ satisfying $$p(\alpha)\le\alpha\mbox{ for every $\alpha\in\operatorname{dom}(p)$}$$ and $$\sup\{p(\alpha): \alpha<\lambda\}=\lambda \mbox{ for every limit }\lambda\in\operatorname{dom}(p),$$ ordered by reverse inclusion as usual. This is clearly not $\omega_1$-closed, but it is $\omega_1$-strategically closed.


问:
Volumes in calculus [Difference in Washers Techniques]
The volume of solid of revolution that results when $ f(x)=x^2 + 1$, $g(x)=3-x^2$ revolved around $y=1$

What's the difference in the two solutions below?

Solution 1:

$$\pi \int_{-1}^{1} [1-g(x)]^2 - [1-f(x)]^2 \,dx $$
Solution 2:

$$\pi \int_{-1}^1 [g(x) - 1]^2 - [f(x) -1]^2 \,dx $$

Also, are the succeeding solutions the same as above ones?

Solution 1:
$$\pi \int_{-1}^1 [1-f(x)]^2 - [1-g(x)]^2 \,dx $$
Solution 2:
$$\pi \int_{-1}^1 [f(x) - 1]^2 - [g(x) -1]^2 \,dx $$

答:
Within each pair of solutions 1 and 2, they are the same because $$[1-f(x)]^2 = [f(x)-1]^2.$$

Conceptually though, since $f(x)\ge 1$ and $g(x)\ge1$ in that range $-1\le x\le 1$, the $[f(x)-1]^2$ form may be easier to understand.

The first pair of solutions is different from the second pair of solution, and the first pair is the correct one.

By comparing $1 \le f(x)\le g(x)$ in the range $-1\le x \le 1$, $f(x)$ is the "inner edge of the washers" and $g(x)$ is the "outer edge of the washers", hence the integrand should be

$$[g(x)-1]^2 - [f(x)-1]^2 = [1-g(x)]^2 - [1-f(x)]^2.$$


问:
Let $f: [0,6] \rightarrow \mathbb R$ be a three times differentiable function such that $f(0)=f(1.4)=f(3.9)=f(5.2)$.
Let $f: [0,6] \rightarrow \mathbb R$ be a three times differentiable function such that $f(0)=f(1.4)=f(3.9)=f(5.2)$. Prove that there is c $\epsilon$ (0,6) such that $f^{'''}(c)=0$.

Now using Rolle Theorem I know that from (0,5.2) I have that $f^{'}(x)=0$. But I am struggling with what to do from (5.2,6). I figure I can use the fact that we are looking for $f^{'''}(c)=0$, but I don't know what to do.

答: From Rolle's theorem, you know that there is a $c_1 \in (0, 1.4)$ such that $f'(c_1) = 0$ and that there is a $c_2 \in (1.4, 3.9)$ such that $f'(c_2) = 0$ and that there is a $c_3 \in (3.9, 5.2)$ such that $f'(c_3) = 0$. Then since you know that $f'(c_1) = f'(c_2) = f'(c_3) = 0$, you can apply Rolle's theorem again to show that there is $c_4, c_5$ such that $c_1<c_4<c_2<c_5<c_3$ and $f''(c_4) = f''(c_5) = 0$. Finally, apply it again to show that there is a $c$ such that $f'''(c) = 0$.


问:
Integral involving gaussian and triangle function
I would like to calculate the following integral:

$$\int_{-\infty}^{+\infty} \quad tri(x)\exp\left(-\frac{(x-x_0)^2}{a}\right)dx$$

Thanks!

答:
Define

$$\text{tri}(x) = \begin{cases}1-\left| \frac{x}{2} \right | & |x| \le 2\\0 & |x| \gt 2 \end{cases}$$

Then, after making the substitution $u=x-x_0$ and doing some algebra, you can show that the stated integral is equal to

$$\left ( 1+\frac{x_0}{2}\right) \int_{-2-x_0}^{-x_0} du \, e^{-u^2/a} + \left ( 1-\frac{x_0}{2}\right) \int_{-2-x_0}^{-x_0} du \, e^{-u^2/a} + \\ \frac12 \int_{-2-x_0}^{-x_0} du \, u\,e^{-u^2/a} - \frac12 \int_{-x_0}^{2-x_0} du \, u\,e^{-u^2/a}$$

which in turn is equal to

$$\left ( 1+\frac{x_0}{2}\right)  \frac{\sqrt{\pi a}}{2} \left [\text{erf}\left(-\frac{x_0}{\sqrt{a}}\right) - \text{erf}\left(\frac{-2-x_0}{\sqrt{a}}\right) \right ] + \left ( 1-\frac{x_0}{2}\right)  \frac{\sqrt{\pi a}}{2} \left [\text{erf}\left(\frac{2-x_0}{\sqrt{a}}\right) - \text{erf}\left(-\frac{x_0}{\sqrt{a}}\right) \right ]\\-\frac{a}{4} \left [e^{-(2+x_0)^2/a} + e^{-(2-x_0)^2/a}-2 e^{-x_0^2/a} \right ] $$


问:
$Var (X_n)= Var(X_0) +\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$, Martingale Variance
i'm already diving through the martingales properties and try to prove the following problem:

Let $X_{n}, Y_{n}$ two martingales squared integrable respect to the filtred probability  space $(\Omega, \mathcal{F},(\mathcal{F}_n)_{0\leq n}, \mathbb{P}),$ prove:

$Var (X_n)= Var(X_0) +\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$

$X_0$ is independent from $(X_k-X_{k})$

I tried to begin asumming that $X_0=0$ and for $0\leq t, \Delta_T=X_t-X_{t-1}$ then we can write $X_t$ as $\sum_{i=1}^{t}\Delta$, where $\mathbb{E}(\Delta|\Delta_1 \cdots \Delta_{j-1})=\mathbb{E}(X_j-X_{j-1}|\mathcal{F})=0$. In other words, $\Delta_i$ is uncorrelated with $\Delta_1 \cdots \Delta_{j-1}$ This is ok to prove the independence or i have to go to the definition on independent increments?

My attempt:
  If we suppose we want $Var(S_n)=\mathbb{E}[(X_t-\mathbb{E}(X_t))^2]=\mathbb{E}(X_t)^2$ (Recall where assuming $X_t=0$, so $\mathbb{E}(X_t)=0$ by the martingale property). Then we have $Var[X_t]=Var[\sum_{i\leq t}\Delta_j]$=$\sum_{i,j}Cov(\Delta_i,\Delta_{j})=\sum_{i,j}Var(\Delta_j)+2\sum_{i<j\leq t}Cov(\Delta_i,\Delta_j)$.Now then $Var(\Delta_{i})=\mathbb{E}[\Delta_{i}^2]$ but we have $\mathbb{E}(\Delta_{i})=0$, then $Cov(\Delta_i,\Delta_j)=\mathbb{E}(\Delta_i,\Delta_j)$ by the tower property of conditional expectation over $\Delta_i$, $\mathbb{E}[\mathbb{E}[\Delta_j|\mathcal{F_{j-1}}]|\Delta_j]=0$, it follows in this case $Var[X_t]=\sum_{i\leq t}\mathbb{E}(\Delta_i^2)$

As you see i didn't get $\sum_{k=0}^{n} Var (X_{k}-X_{k-1})$, i'm thinking that the correct aproach of this is using the fact that $Var(X)= E(Var(X|Y))+Var(E(X|Y))$

Anyone could help me out? And even more to specify what really means the variance of a martingale, or is just analogue to random variables, thanks.

答:
The key is to telescope the right sequence:

\begin{align*} X_k^2&= X_0^2 +  \sum_{j=1}^k (X_j^2 - X_{j-1}^2) \\
 & = X_0^2 + \sum_{j=1}^k (X_j - X_{j-1})(\underset{=X_{j}}{\underbrace{X_{j-1} + (X_j - X_{j-1})}} +X_{j-1}).
\end{align*}

Taking expectations and using the martingale property  $E[X_j - X_{j-1} | {\cal F}_{j-1} ] =0$ to obtain

$$E[X_k^2] = E[X_0^2] + \sum_{j=1}^k E[ (X_j-X_{j-1})^2 ].$$

Hope this help.


问:
Formula for the number of naturals whose prime factors lie in a finite set $A \subset \Bbb{P}$ and that have exactly $r$ prime factors.
Let $A$ be a finite set of prime numbers.  Then the set $X_r = \{x \in \Bbb{N}: \Omega(x) = r,$ and all the primes comprising $x$ are in $A\}$.

Has size formula ?

For example:

$$
|A| = \{2,3,5,7\} \\
r = 2\\
$$

We have:

$$
 2 \cdot 2 \\
 2 \cdot 3 \\
 2 \cdot 5 \\
 2 \cdot 7 \\
 3 \cdot 3 \\
  3\cdot 5\\
 3 \cdot 7\\
  5\cdot 5 \\
  5 \cdot 7 \\
 7 \cdot 7 \\
$$

or $10$ possibilities.  Does the formula for $|X_r|$ have a closed form expression in terms of $|A|$ and $|r|$?

I currently think it's ${|A|r\choose r}$ since there are up to $r$ copies of each of $|A|$ things, and you are choosing a combination (set) of the items of size $r$.  However computing it on $|A|=4, r=2$ produces $4\cdot 3 = 12$ as an answer not $10$.

答:
Let $x$ be a number that you form doing this. Then
$$x=p_1^{\alpha _1}\cdots p_k^{\alpha _k},$$
where $\sum _{i=1}^k\alpha _i=r$
notice that you can treat this as a tuple $(b_1,\cdots ,b_{|A|})$ such that $b_{i}\geq 0$ and $\sum _{i=0}^{|A|}b_i=r.$ By the stars and bars method, you get
$$\binom{|A|+r-1}{|A|-1}.$$ In the case you have you are getting
$$\binom{4+2-1}{3}=\binom{5}{2}=10.$$


问:
What exactly does this physically mean?
Let X(w) be a real random variable on ($\Omega$ , P). The image X($\Omega$) the set of all the values X(w) can take ,written $\Omega^{X}$. For any set $ B \subset \Omega^{X}$ the probability of the event that the value of X lies in B is equal to $P\{w|X(w) \in B\} = P(X^{-1}(B))$.

What I understood from that is that the probability measure P is the probability of the set of all ws such that the event X(w) is realized and which from that is an element of B, namely is equal to $P\{w|X(w) \in B\}$ however I do not understand why $P(X^{-1}(B))$ is used? Does it actually mean the inverse of event X or inverse composite? How physically can I vindicate the use of $X^{-1}$? Any ideas?

答:
Several inconsistencies, which might hinder your understanding:

Let X(w) be a real random variable on ($\Omega$ , P).

No, a real random variable is a mapping $X:\Omega\to\mathbb R$, not the image $X(\omega)$ of a specific element $\omega$ of $\Omega$ by the mapping $X$.

The image X($\Omega$) the set of all the values X(w) can take ,written $\Omega^{X}$.

This is a very bad idea to use $\Omega^X$ to denote $X(\Omega)=\{X(\omega)\mid\omega\in\Omega\}$. Let us stick to $X(\Omega)$.

For any set $ B \subset \Omega^{X}$ the probability of the event that the value of X lies in B is equal to $P\{w|X(w) \in B\} = P(X^{-1}(B))$.

Yes, replacing $ B \subset \Omega^{X}$ by $B\subseteq\mathbb R$ (and in fact, by $B\in\mathcal B(\mathbb R)$ the Borel sigma-algebra on $\mathbb R$).

What I understood from that is that the probability measure P is the probability of the set of all ws such that the event X(w) is realized and which from that is an element of B, namely is equal to $P\{w|X(w) \in B\}$ however I do not understand why $P(X^{-1}(B))$ is used? Does it actually mean the inverse of event X or inverse composite? How physically can I vindicate the use of $X^{-1}$? Any ideas?

This is a quite general notation, not specific to probability theory: if $X:\Omega\to\mathbb R$ and $B\subseteq\mathbb R$, then $X^{-1}(B)=\{\omega\in\Omega\mid X(\omega)\in B\}$. Thus, such that the event X(w) is realized and inverse of event X mean nothing: $X$ is not an event (events are (some) subsets of $\Omega$) and $X^{-1}$ does not mean an inverse of $X$. Rather, for every $B\subseteq\mathbb R$, $X^{-1}(B)\subseteq\Omega$ is defined as above.

Title: What exactly does this physically mean?


问:
Are all normality-preserving homomorphisms/endomorphisms normal?
This is Exercise 4.9 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.

The Details:

From page 140 of Roman's book,

Definition: Let $G$ be a group. A homomorphism $\sigma:G\to H$ is normality-preserving if $$N\unlhd G\implies \sigma N\unlhd H.$$

On the same page,

Definition: An endomorphism $\sigma:G\to G$ of a group $G$ is normal if $\sigma$ commutes with all inner automorphisms $\gamma_g$ of $G$, that is, for any $a\in G$, $$\sigma(a^g)=(\sigma a)^g$$ for all $g\in G$.

NB: The latter is the only definition of a homomorphism being normal that I could find in the book so far.

The Question:

Are all normality-preserving homomorphisms normal? Hint: Use the fact that $S_3$ is simple.${}^{\dagger}$

NB: I think the question should be, "are all normality-preserving endomorphisms normal".

Thoughts:

I suspect that the answer is "no" but I can't prove it yet. The group $S_3$ might be a counterexample.

I think I need to use the fact that $S_3$ is the smallest of the nonabelian groups.

The identity map ${\rm id}_{S_3}$ is normality-preserving trivially and normal.

The map

$$\begin{align}
\varphi :S_3 & \to S_3,\\
\tau &\mapsto e
\end{align}$$

is an endomorphism, and, since $\{e\}\unlhd \{e\}$, $\varphi$ is normality-preserving. It is also clearly normal.

Cheating a bit, the endomorphism monoid of $S_3$ has ten elements${}^{\dagger\dagger}$. They're given, I suppose, by their actions on the generators, so consider the presentation

$$S_3\cong\langle a,b\mid a^3, b^2, bab=a^{-1}\rangle\tag{1}$$

of $S_3$. I'm not sure what to do next (besides list all ways $a$ and $b$ can be mapped to elements in $S_3$; but surely there's a more sophisticated way than that, right?). Presentations aren't covered in the book so far too.

Please help :)

$\dagger$: But $S_3$ is not simple . . .

$\dagger\dagger$: See the Groupprops page "Endomorphism structure of symmetric group:S3". (I couldn't link to it because the address has a close-bracket.)

答:
The question as stated has an easy answer: “no”, because “normal” is a property of endomorphisms, and normality preserving morphisms need not be endomorphisms.

So presumably, the actual question should be: “Is every normality preserving endomorphism a normal morphism?”

As suggested by Moishe Kohan, let $\sigma$ be a normal endomorphism. Given $g\in G$, we know that $\sigma(a^g) = \sigma(a)^g$; i.e.,
$$\sigma(g)^{-1}\sigma(a)\sigma(g) = g^{-1}\sigma(a)g,$$
and therefore, that
$$(\sigma(g)g^{-1})^{-1}\sigma(a)(\sigma(g)g^{-1}) = \sigma(a).$$
Therefore, for every $g\in G$, $\sigma(g)g^{-1}\in C_G(\sigma(G))$; it centralizes the image of $\sigma$.

You can construct an example with an automorphism of $S_3$ (the nontrivial endomorphism that are not automorphisms have image of order $2$, so they are never normality-preserving). For instance, consider conjugation by $(12)$. This is normality preserving, but not normal: because $\sigma(13) = (12)(13)(12) = (123)$, and $\sigma(13)(13)^{-1} = (123)(13) = (23)\notin Z(S_3)$, so it is not normal.

P.S. My first counterexample was with $Q_8$, using the map $i\mapsto j\mapsto k\mapsto i$; I picked $Q_8$ because every subgroup is normal, so every endomorphism is normality preserving.


问:
Equivalence classes of adjunction space
Let $X,Y$ be topological spaces, $A\subset X$ and $f:A\rightarrow Y$ a continuous mapping. Define
$$R:=\{z\in(X\sqcup Y)\times(X\sqcup Y)\ |\ (\exists x\in A)(\text{pr}_1(z)=(0,x)\land\text{pr}_2(z)=(1,f(x)))\}.$$
Clearly $R\subset(X\sqcup Y)\times(X\sqcup Y)$. Let $\sim$ be the equivalence relation on $X\sqcup Y$ generated by $R$. Write $X\cup_f Y:=(X\sqcup Y)/R$ and let $\pi:X\sqcup Y\rightarrow X\cup_f Y$ be the canonical surjection.

How can I show that
$$\pi(0,x)=f^{-1}(\{f(x)\})\sqcup\{f(x)\}$$
for all $x\in A$?

Attempt: Let $z\in X\sqcup Y$ and $x\in X$ such that $\pi(z)=\pi(0,x)$. There exists $n\in\mathbb{N}$ and $\alpha\in(X\sqcup Y)^{[0,n]}$ such that $\alpha_0=(0,x)$, $\alpha_n=z$ and
$$(\alpha_i,\alpha_{i+1})\in R\text{ or }(\alpha_{i+1},\alpha_i)\in R$$
for all $i\in[0,n-1]$. On the other hand, we have $\pi(1,f(x))=\pi(0,x)$. How to proceed from here? Any suggestions?

Edit: $X\sqcup Y:=\{0\}\times X\cup\{1\}\times Y$.

答:
Your equivalence relation $\sim$ is generated by the basic relation
$$z \equiv z' \text{ iff there exists } a \in A \text{ such that } z = (0,a), z '= (1,f(a)) .$$
$z \equiv z$ is only possible if $z  \in \{0\} \times A$ ("$z$ has type $0$") and $z ' \in \{1\} \times f(A)$ ("$z'$ has type $1$"). Note that the concept of type is only defined for points of $A^* = \{0\} \times A \cup \{1\} \times  f(A)$.

By a chain we mean a finite sequence of points $(z_1, \ldots, z_n)$ such that

$z_i \equiv z_{i+1}$ or $ z_{i+1} \equiv z_i$ for $i = 1,\ldots, n-1$.

Thus $z \sim z'$ iff there exists a chain $(z_1, \ldots, z_n)$  such that $z = z_1, z' = z_n$. Note that for $n = 1$ we get $z \sim z$. Also observe that if $(z_1, \ldots, z_n)$ is a chain from $z$ to $z'$, then the reverse sequence $(z_n, \ldots, z_1)$ is a chain from $z'$ to $z$.

By definition of $\equiv$ each chain of length $n > 1$ must alternate between points of type $0$ and points of type $1$. Thus the equivalence class of each $z \in \{0\} \times (X \setminus A) \cup \{1\} \times (Y \setminus f(A)) = X \sqcup Y \setminus A^*$ is a singleton. Only equivalences classes of points $z \in A^*$ may contain more than one element. Note that each chain between two points of the same type must have odd length, each chain between two points of different type must have even length.

By a simple chain of type $1$ we mean a chain of length $3$ such that $z_1, z_3$ have type $1$. Write $z_1 = (1,b), z_3 = (1,b')$ with $b, b' \in f(A)$. The intermediate point $z_2$ must have the form $z_2 = (0,a)$ with $a \in A$. We have $z_2 \equiv z_1$, i.e. $b = f(a)$ and  $z_2 \equiv z_3$, i.e. $b' = f(a)$. Thus $z_1 = z_3$.

Now let us denote by a chain of type $1$ any chain of length $n \ge 3$ such that $z_1, z_n$ have type $1$. Thus $n = 2m+1$. In each subchain $z_{2i-1}, z_{2i}, z_{2i+1}$, $i = 1,\ldots, m$, the points $z_{2i-1},z_{2i+1}$ have type $1$, thus $z_{2i-1} =z_{2i+1}$. We conclude that $z_1 = z_n$.

Now let us consider an arbitrary chain $(z_1,\ldots, z_n)$ of length $n > 3$. There are the follwing cases:

$z_1, z_n$ have type $1$. Then $z_1 = z_n$.

$z_1$ has type $1$ and $z_n$ has type $0$. Then $(z_1,\ldots, z_{n-1})$ has type $1$, i.e. $z_1 = z_{n-1}$, and $z_n \equiv z_{n-1}$. Thus $z_n \equiv z_1$.

$z_1$ has type $0$ and $z_n$ has type $1$. Then $(z_2,\ldots, z_n)$ has type $1$, i.e. $z_2 = z_n$, and $z_1 \equiv z_2$. Thus $z_1 \equiv z_n$.

$z_1, z_n$ have type $0$. Then $(z_2,\ldots, z_{n-1})$ has type $1$, i.e. $z_2 = z_{n-1}$, and $z_1 \equiv z_2$, $z_n \equiv z_{n-1}$. Thus we get a length $3$ chain $(z_1,z_2,z_n)$. We have $z_1 =(0,a), z_n = (0,a')$ with $a, a' \in A$. The intermediate point $z_2 = (1,b)$ must satisfy $z_1 \equiv z_2$, i.e. $b = f(a)$ and  $z_3 \equiv z_2$, i.e. $b  = f(a')$. Thus we get $f(a) = f(a')$.

This gives us a very transparent description of $\sim$. In fact, $z \sim z'$ means that either

$z = z'$

$z, z'$ have different type, where $z \equiv z'$ or $z' \equiv z$

$z, z'$ have type $0$, i.e. $z =(0,a), z' = (0,a')$ with $a, a' \in A$, where $f(a) = f(a')$.

Therefore the equivalence class of $(0,a)$ with $a \in A$ is
$$\pi(0,a)  = \{(0,a') \mid a' \in A, f(a) = f(a') \} \cup \{(1,f(a))\} = \{0\} \times f^{-1}(f(a)) \cup \{1\} \times \{f(a)\} \\ = f^{-1}(f(a)) \sqcup \{f(a)\}.$$
The equivalence class of $(1,b)$ with $b \in f(A)$ is
$$\pi(1,b)  = f^{-1}(b)) \sqcup \{b\}.$$
All other equivalence classes are singletons.

Title: Equivalence classes of adjunction space


问:
Demonstrating equivalency between Runge-Kutta and Simpson's Rule
I am trying to show that a fourth order Runge-Kutta method is equivalent to Simpson's method for approximating an integral over an interval [a,b]. I've read that it is, but I'm unsure how to demonstrate this. I would appreciate any help in figuring out how to show this equivalency.</p>

答:
The Runge-Kutta method I think you are talking about is
\begin{align}
&k_1 = f(t_n, y_n) \\
&k_2 = f(t_n + h/2, y_n + hk_1/2) \\
&k_3 = f(t_n + h/2, y_n + hk_2/2) \\
&k_4 = f(t_n + h, y_n + hk_3) \\
&y_{n + 1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4).
\end{align}
If the ode is just integration, i.e. $y'(t) = f(t)$, this reduces to
\begin{align}
&k_1 = f(t_n) \\
&k_2 = f(t_n + h/2) \\
&k_3 = f(t_n + h/2) \\
&k_4 = f(t_n + h) \\
&y_{n + 1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) = y_n + \frac{h}{6}(f(t_n) + 4f(t_n + h/2) + f(t_n + h))
\end{align}
which is identical to the 3 point Simpsons rule.


问:
Linear Algebra - Vector Spaces of Vector Spaces
If we have a Vector Space such as $\Bbb R$, we can make Vector Spaces out of it. For example, let $v_1,v_2,v_3\in \Bbb R$. We know that $(v_1,v_2,v_3)$ is a Vector Space - it is $\Bbb R_3$ - and its dimension is $3\times \dim {(\Bbb R)} = 3$. Thus we can see that $\Bbb R_3$ is composed of three elements of $\Bbb R$.

Any "Container Space" ($V_1$) is made of a Vector Space of elements whose elements are vectors in another Vector Space ($V_2$). Also, $(v_1,v_2,...,v_n) \oplus_1 (v'_1,v'_2,...,v'_n) = (v_1\oplus_2v'_1,v_2\oplus_2v'_2,...,v_n\oplus_2v'_n)$ and $c\odot_1(v_1,v_2,...) = (c\odot_2 v_1, c\odot_2 v_2...)$. In other words, Operators are passed to the component vectors.

In general, if we have a Vector Space $V$ and a Set $C$ made from vectors in $V$, is $C$ always a Vector Space? If so, will the dimension of $C$ always be $\dim{(P)}\times\dim{(V)}$ where $P$ is the space that forms elements in the way that the Set $C$ does?

I have found a few more Vector Spaces of the same type; if you want more examples, I can post them.

答:
Wait a moment. I think there are some things confused here. Given $(v_1,v_2,v_3)$ for each $v_i \in \mathbb{R}$ this tuple isn't a vector space, rather it's the element of another vector space, namely $\mathbb{R}^3$. So, how do we construct new vector spaces from old ones? Well, there are two methods that I think are very interesting, and that probably will interest you.

The first method which is the simpler (and which is what you are looking for, looking at your question) is the so called Direct Sum. Given a family of vector spaces $\{V_i : i \in I\}$ we define the direct sum as the result of taking the cartesian product of the $V_i$ and adding the operations componentwise as defined in each $V_i$. In other words, elements of the direct sum are of the form $(v_1, \dots, v_n)$ with each $v_i \in V_i$ and we define the operations componentwise in the sense that we have:

$$(v_1,\dots, v_n)+(w_1,\dots,w_n)=(v_1+w_1,\dots,v_n+w_n)$$

$$a(v_1,\dots, v_n)=(av_1,\dots,av_n)$$

Note that the operation in the $i$-th component is the operation defined in the space $V_i$. This vector space is denoted by the symbol $V_1 \oplus \cdots \oplus V_n$ or simpler with the notation:

$$\bigoplus_{i\in I}V_i$$

It's fairly easy to prove that $\dim(\oplus_{i\in I}V_i) = \sum_{i \in I}\dim V_i$ (and I invite you to do that). This is the way $\mathbb{R}^3$ is construct in truth. You can see that with this definition we have:

$$\mathbb{R}^3 = \bigoplus_{i=1}^3\mathbb{R}$$

The other method is called the tensor product. It's construction is kind of sofisticated (and in truth it's motivation is only clear when we come to study multilinear constructions), but in the end it's simple when you understand it: it's meant to mimic usual products. This space is denoted by $V_1 \otimes \cdots \otimes V_n$ or simpler:

$$\bigotimes_{i \in I}V_i$$

And it satisfies $\dim(\otimes_{i\in I}V_i) = \prod_{i\in I}\dim V_i$ so in some intuitive sense the tensor product is bigger than the direct sum. To read more about the construction of the tensor product consult Kostrikin's Linear Algebra and Geometry chapter 4 about Multilinear Algebra.


问:
Is this proof of a ring being Noetherian correct or we should prove that it is left Noetherian and then right Noetherian and why?
Here is the question I want to answer:

Let $\varphi : R \rightarrow S$ be a surjective homomorphism of commutative rings. Show that, if $R$ is Noetherian, then $S$ is Noetherian.

Here is a trial to the solution:

Let $\varphi : R \rightarrow S$ be a surjective homomorphism of commutative rings and assume that $R$ is Noetherian. We want to show that $S$ is Noetherian.

Now, since we know the following definition of a Noetherian ring : a commutative ring $S$ is Noetherian iff it satisfies the $ACC$ condition. And since we know that a commutative ring $S$ has the ascending chain condition $(ACC),$ if, for every ascending chain of ideals $$I_1 \subset I_2 \subset I_3 \subset \dots $$ in $S,$ there is an $N > 0,$ so that $I_n = I_{n+1}$ for $n \geq N.$

So, let $I_1 \subset I_2 \subset I_3 \subset \dots $ be an ascending chain of ideals in $S,$ Now since we have a surjective ring homomorphism, then the preimage of the ideal $I_k$ in $S$ (i.e. $f^{-1}(I_k)$) is an ideal of $R$(to be proved at the end). And similarly for all other ideals. So we get the following ascending (we will prove later why it is ascending) chain of ideals in $R,$ $$f^{-1}(I_1) \subset f^{-1}(I_2) \subset f^{-1}(I_3) \subset \dots \subset f^{-1}(I_k) \subset \dots $$

Now, since by assumption, $R$ is Noetherian, so its ascending chain of ideals terminate. That is $\exists N \in \mathbb Z$ such that  $f^{-1}(I_n) = f^{-1}(I_{n+1})$ for all $n \geq N.$

Now since $f$ is surjective, we have $$f (f^{-1}(I_k)) = I_k, \quad  \forall k > 0.$$

Therefore, it follows that we have $$ I_n = I_{n+1}, \quad \forall n \geq N.$$
So, each ascending chain of ideals of $S$ terminates, as we have taken an arbitrary ascending chain of $S$ and so $S$ is Noetherian as required.

My questions are:

1-Is this proof correct? or we should prove that $S$ is left Noetherian and then right Noetherian? If so, why?

2- why we are sure that we get an ascending chain of ideals in $R$?

答:
Considering that $R$ and $S$ are commutative, there is no need to talk about left- or right-Noetherian because all ideals of $R$ and $S$ are two-sided. Consequently, $S$ is left-Noetherian if and only if it is right-Noetherian if and only if it is Noetherian.

For the rest of the proof, we need a few key ingredients. Consider a homomorphism $\varphi : R \to S$ of commutative rings.

1.) For any ideal $I$ of $S,$ we have that $\varphi^{-1}(I)$ is an ideal of $R$ (called the contraction of $I$ in $R$).

2.) If $\varphi$ is surjective, then for any ideal $J$ of $R,$ we have that $\varphi(J)$ is an ideal of $S$ (called the extension of $J$ in $S$).

3.) If $I \subseteq J$ are ideal of $S,$ then $\varphi^{-1}(I) \subseteq \varphi^{-1}(J).$

4.) If $I \subseteq J$ are ideal of $R,$ then $\varphi(I) \subseteq \varphi(J).$

5.) If $\varphi$ is surjective, then for any ideal $I$ of $S,$ we have that $\varphi(\varphi^{-1}(I)) = I.$

Putting these all together (as you have done) gives the proof: for any ascending chain $I_1 \subseteq I_2 \subseteq \cdots$ of ideals of $S,$ we have that $\varphi^{-1}(I_1) \subseteq \varphi^{-1}(I_2) \subseteq \cdots$ is an ascending chain of ideals of $R.$ By hypothesis that $R$ is Noetherian, there exists an integer $N \gg 0$ such that $\varphi^{-1}(I_n) = \varphi^{-1}(I_{n + 1})$ for each integer $n \geq N.$ Considering that $\varphi$ is surjective, it follows that $\varphi(\varphi^{-1}(I_k)) = I_k$ for all integers $k \geq 1.$ By viewing the original chain $I_1 \subseteq I_2 \subseteq \cdots$ of ideals in $S$ as the ascending chain of ideals $\varphi(\varphi^{-1}(I_k))$ and using the fact that $\varphi^{-1}(I_n) = \varphi^{-1}(I_{n + 1})$ for all integers $n \geq N,$ we conclude that $I_1 \subseteq I_2 \subseteq \cdots$ stabilizes. QED.

By the way, if you know the equivalent condition that a commutative ring $R$ is Noetherian if and only if all of its (two-sided) ideals are finitely generated, then there is an alternate proof.

Proof. Consider an ideal $I$ of $S.$ By the first point above, we have that $\varphi^{-1}(I)$ is an ideal of $R.$ By hypothesis that $R$ is Noetherian, there exist elements $i_1, \dots, i_n$ of $\varphi^{-1}(I)$ such that every element of $\varphi^{-1}(I)$ can be written as $r_1 i_1 + \cdots + r_n i_n$ for some elements $r_1, \dots, r_n$ of $R.$ Considering that $\varphi$ is surjective, by the fifth point above, we have that $I = \varphi(\varphi^{-1}(I)),$ hence every element of $I$ can be written as $\varphi(r_1 i_1 + \cdots + r_n i_n) = \varphi(r_1) \varphi(i_1) + \cdots + \varphi(r_n) \varphi(i_n)$ for some elements $\varphi(r_1), \dots, \varphi(r_n)$ of $S$ and some elements $\varphi(i_1), \dots, \varphi(i_n)$ of $I.$ Consequently, the ideal $I$ of $S$ is finitely generated by the elements $\varphi(i_1), \dots, \varphi(i_n)$ of $I.$ QED.


问:
Quotient map from Sorgenfrey plane to Real Line
I have been told that $\alpha: \mathbb{R}_\ell^2 \longrightarrow \mathbb{R} $, defined by $\alpha(x,y) = x+y$ is a quotient map. From the definition in Munkres, it needs to be the case that $U \subset \mathbb{R}$ is open if and only if $\alpha^{-1}(U) \subset \mathbb{R}_\ell^2$ is open. I am trying to understand what $U$ and $\alpha^{-1}(U)$ "look like".

From another book, I've found that a basis for $\mathbb{R}_\ell^2$ is the set of squares with side-length $\epsilon$ that is closed on the bottom and left-hand side. But I'm not sure how I can use that to obtain the set of possible $U$'s.

答:
You don’t really have to look at all open $U\subseteq\Bbb R$: it’s enough to consider open intervals, since they form a base for the topology of $\Bbb R$. So let $U=(a,b)$ for some $a,b\in\Bbb R$ with $a<b$. Then

$$\alpha^{-1}[U]=\left\{\langle x,y\rangle\in\Bbb R_\ell^2:x+y\in U\right\}\,.$$

This is an easy set to visualize: it’s an open strip with a slope of $-1$ that intersects the $x$-axis in the interval $(a,b)\times\{0\}$ and the $y$-axis in the interval $\{0\}\times(a,b)$. Such a strip is open in the Euclidean topology on $\Bbb R^2$, which is a subset of the Sorgenfrey topology, so it’s open in $\Bbb R_\ell^2$.

Thus, if $V$ now is any open set in $\Bbb R$, $\alpha^{-1}[V]$ is a union of such strips and therefore open in $\Bbb R_\ell^2$. To complete the argument, you need only verify that every Euclidean open set in the plane is also open in $\Bbb R_\ell^2$, which follows from the fact that every Euclidean open set in $\Bbb R$ is open in $\Bbb R_\ell$.


问:
Finding the limit in an $∞ - ∞$ case
$$\lim_{x \to ∞}6x^{2/3} - 2x + 1$$
How can I find the right side and left side limit to this function? It is an ∞ - ∞ case. I know I need to convert this function into a $∞/∞$ or $0/0$ case so that I can use the l'hopital's rule, but I am not sure how.

答:
You can make a change of variables $u=x^{1/3}$. Then $u$ goes to infinity as $x$ does, and your limit becomes
$$\lim_{u\to\infty}-2u^3+6u^2+1.$$
This is a cubic with negative leading term, so it goes to $-\infty$.

In other words, $u^3$ grows more quickly than $u^2$ as $u\to\infty$, so the cubic term dominates both the quadratic and constant terms.

Title: Finding the limit in an $∞ - ∞$ case


问:
Show that the trace of T*T is the sum of squares
Let $a_{jk} \in \mathbb{C}$ such that
$$\sum_{(j, k) \in \mathbb{N}^{2}}^{\infty}\left|a_{j k}\right|^{2}<\infty.$$
For $f\in l^2(\mathbb N)$ define $Tf$ by
$$Tf(j)=\sum_{k=1}^{\infty} a_{j k} f(k), \quad j \in \mathbb{N}.$$
By spectral theorem, we know that there is an orthonormal basis $\{v_n\}$ which consists of eigenvectors of $T^*T$. Let $\{\mu_n\}$ be the corresponding eigenvalues. We want to show that

$$\sum_{(j, k) \in \mathbb{N}^{2}}\left|a_{j k}\right|^{2}=\sum_{n=1}^{\infty}\left\|T v_{n}\right\|^{2}=\sum_{n=1}^{\infty} \mu_{n}.$$

The right part is trivial from the definition of eigenvalues. However, I don't know how to do the left part. It looks like the trace of $T^*T$. In the finite dimensional case, we can use the fact that the trace is invariant under orthogonal transformation. Any hints?

答:
If $\{v_n\}$ and $\{w_n\}$ are two orthonormal bases then
$$
\sum_{n=1}^\infty \|Tv_n\|^2 =
\sum_{n=1}^\infty\sum_{m=1}^\infty |\langle Tv_n, w_m\rangle|^2 = $$$$ =
\sum_{m=1}^\infty\sum_{n=1}^\infty |\langle v_n, T^*w_m\rangle|^2 =
\sum_{m=1}^\infty \|T^*w_m\|^2.
$$

This shows that $\sum_{n=1}^\infty \|Tv_n\|^2$ does not depend on the choice of basis.

On the other hand, if we take the canonical basis $\{e_n\}$, then we have
$$
  \sum_{m=1}^\infty \|Te_n\|^2 =
  \sum_{(j,k)\in\mathbb N^2}|a_{jk}|^2,
  $$
thus proving the "left part", as required by the OP.


答:
Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):

$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C  \tag{1}$$

Good, up to this point... $\uparrow$.

So the error was in your last equality at the very end:

You made an error by ignoring the fact that the first term with $\ln(2x+1)$ as a factor also has $x$ as a factor, so we cannot multiply the arguments of $\ln$ to get $\ln(2x+1)^{3/2}$. What you could have done was first express $x\ln(2x+1) = \ln(2x+1)^x$ and then proceed as you did in your answer, but your result will then agree with your text's solution.

Alternatively, we can factor out like terms.

$$ = x\ln(2x + 1) + \frac 12 \ln(2x + 1) - x + C \tag{1}$$
$$= \color{blue}{\bf \frac 12 }{\cdot \bf 2x} \color{blue}{\bf \ln(2x+1)} + \color{blue}{\bf \frac 12 \ln(2x+1)}\cdot {\bf 1} - x + C$$

Factoring out $\color{blue}{\bf \frac 12 \ln(2x + 1)}$ gives us

$$= \left(\dfrac 12\ln(2x + 1)\right)\cdot \left(2x +1\right) - x + C $$ $$= \frac 12(2x + 1)\ln(2x+1) - x + C$$

Title: Integrate by parts: $\int \ln (2x + 1) \, dx$

Body:
$$\eqalign{
  & \int \ln (2x + 1) \, dx  \cr
  & u = \ln (2x + 1)  \cr
  & v = x  \cr
  & {du \over dx} = {2 \over 2x + 1}  \cr
  & {dv \over dx} = 1  \cr
  & \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
  &  = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}}   \cr
  &  = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|)  \cr
  &  = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C  \cr
  &  = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$

The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$

Where did I go wrong?

Thanks!


问:
$\lim_{r\to\infty} \int_0^\infty rx/(exp(x) + r^2x)dx = 0$
I've got a feeling I need to use the Dominated Convergence Theorem, but I'm honestly not sure how to even begin tackling this.

I realize the limit of the integrand will ultimately be $0$ as the denominator will be dominated by n, but I'm not sure how to continue.

答:
You can find a dominating function  by differentiating the integrand with respect to $n$. If not clear, here are more details.

Fix $x>0$. For $t\in [1,\infty)$, define $f_t (x) = \frac{t x} {e^x + t^2 x}$.

$$\frac{d}{dt} \frac{1}{f_t (x)} = \frac{d}{dt} (\frac{e^x + t^2x}{tx}) = -\frac{1}{t^2} \frac{e^x}{x}+1.$$

This derivative is increasing in $t$, is negative at $t=1$ ($e^x>x$ for $x>0$) and tends to $1$ as $t\to\infty$.

Therefore the unique minimum  over $t \ge 1$ is attained when $t^2 = e^{x}/x$. As a result, for all $t\ge 1$ we have

$$f_t (x) \le f_{e^{x/2}/\sqrt{x}}(x) = \frac {\frac{e^{x/2}} { \sqrt{x} }x} {e^x + e^x} =\frac{1}{2} \sqrt{x} e^{-x/2}.$$

This upper bound clearly holds for all $t\in{\mathbb N}$. Furthermore the function on the righthand side is integrable. Therefore we can apply the dominated convergence theorem with it as the dominating function.


问:
Show that basis of $\mathbb{C}^2$ is (1, j)
I'm trying to understand this section: https://en.wikipedia.org/wiki/Quaternion#:~:text=Quaternions%20can%20be%20represented%20as,as%20pairs%20of%20real%20numbers.

One of the things they claim is that the basis of $$\mathbb{C}^2$$ is (1, j).

For real numbers, to prove that the standard basis vectors are a basis for $R^4$, we would say: for a vector [a,b,c,d], we can always decompose the vector into a*[1, 0, 0, 0] + b*[0, 1, 0, 0] ...

To show that (1, j) is a basis for $\mathbb{C}^2$, we would first have to define some arbitrary vector, analogous to  how we defined [a,b,c,d] above for the real case, in $\mathbb{C}^2$ using variables, and then analogously show that we can decompose that vector into $(a+bi)*1 + (c+di)*j$.

But what is that arbitrary vector? How do we show that (1,j) makes sense as a basis (compared to, for example, [1,0] and [0, 1]) without making circular arguments?

答:
There is no circular argument; the natural argument is that given an arbitrary quaternion vector $a+bi+cj +dk $, where $1,i,j,k$ are the standard quaternion basis vectors, then from the identity $ij=k$ we simply obtain
$$a+bi+cj+dk=(a+bi)\cdot 1+(c+di)\cdot j\,,$$
which gives a canonical identification of quaternions with $\mathbb{C}^2$—— that is, $(a,b,c,d)\leftrightarrow(a+bi,c+di)$. You can then identify $(1,0)\in \mathbb{C}^2$ with the quaternion $1$ and $(0,1)\in \mathbb{C}^2$ with the quaternion $j$.


答:
What you want to have is the following:

$$f(x)-g(x)= \frac{1}{\frac{1}{f(x)}}-\frac{1}{\frac{1}{g(x)}}=\frac{\frac{1}{g(x)}-\frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}$$ so you have the form $\dfrac{0}{0}$ from the form $\infty-\infty$.

Title: Finding the limit in an $∞ - ∞$ case

Body:
$$\lim_{x \to ∞}6x^{2/3} - 2x + 1$$
How can I find the right side and left side limit to this function? It is an ∞ - ∞ case. I know I need to convert this function into a $∞/∞$ or $0/0$ case so that I can use the l'hopital's rule, but I am not sure how.