问:
Why is $12$ the smallest weight for which a cusp form exists?
On wikipedia (here) I have read the following:

Twelve is the smallest weight for which a cusp form exists. [...] This fact is related to a constellation of interesting appearances of the number twelve in mathematics ranging from the value of the Riemann zeta function at $−1$ i.e. $\zeta(-1) = -1/12$, the fact that the abelianization of $SL(2,\mathbb{Z})$ has twelve elements, and even the properties of lattice polygons.

I know the dimension formula and the weight formula for modular forms as well as the proof of the weight formula via the residue theorem, but I do not understand how this relates to any of the three concepts mentioned in the quote above. Can anybody tell me more?

答:
Regarding the question in the title I must admit that, for me personally, "why" questions are always very hard to answer so I'll simply give some evidence that the (non-) existence of cusp forms can be viewed as a geometric phenomenon. What I am about to relate is very well known and details can be found in many places but I must say that I found the treatment offered in R. Hain's, LECTURES ON MODULI SPACES OF ELLIPTIC
CURVES to be particularly instructive.

Let $\mathcal{M}_{1,1}=\operatorname{SL}_2(\mathbb Z)\backslash\!\!\backslash \mathfrak{H}$ be the moduli space of complex elliptic curves, where $\operatorname{SL}_2(\mathbb Z)$ acts on the upper half-plane $\mathfrak{H}$ in the usual way through Möbius transformations. Warning: this is not a complex manifold, rather a more complicated object called a complex orbifold. Nevertheless, all the algebraic invariants that can be associated to complex manifolds (such as fundamental group, homology groups, et cetera) can also be associated to $\mathcal{M}_{1,1}$. In particular, there is a good notion of (holomorphic) line bundle over $\mathcal{M}_{1,1}$ and, as per usual, we denote by $\operatorname{Pic}(\mathcal{M}_{1,1})$ the abelian group of isomorphism classes of such.

Now for our purposes the key statement is that:

$$
\operatorname{Pic}(\mathcal{M}_{1,1}) \cong \mathbb Z/12\mathbb Z
$$

How does this relate to the (non-) existence of cusp forms? The point is that there exists a line bundle $\mathcal{L}$ on $\mathcal{M}_{1,1}$ such that global sections of its $k$-th tensor power $\mathcal{L}^{\otimes k}$, for $k \in \mathbb Z$, are so-called ``weakly holomorphic'' (:=holomorphic on $\mathfrak{H}$ and meromorphic at the cusp $\infty$) modular forms of weight $k$. On the other hand, a line bundle is trivial if and only if there exists a nowhere vanishing global section and since $\mathcal{L}^{\otimes 12}$ is necessarily trivial by the above result, there exists a weakly holomorphic modular form of weight $12$ that vanishes nowhere on $\mathfrak{H}$. The $k/12$ formula then forces it to vanish at $\infty$ as well and that is your cusp form.

In fact the class of $\mathcal{L}$ even generates $\operatorname{Pic}(\mathcal{M}_{1,1})$ so that $12$ is indeed the smallest weight in which a non-zero cusp form can exist.

A final comment, since you mention $\zeta(-1)$ in this context: That is actually equal to the Euler characteristic $\chi(\mathcal{M}_{1,1})$ of $\mathcal{M}_{1,1}$. More generally Harer and Zagier proved that
$$
\chi(\mathcal{M}_g)=\zeta(1-2g),
$$
for all $g\geq 2$, where $\mathcal{M}_g$ denotes the orbifold of compact Riemann surfaces of genus $g$.

PS: I would have liked to vote up this question but since at the time of my answer it had exactly $12$ upvotes already I light-heartedly decided against that.


问:
Prove $\lim_{n\to \infty} (\sqrt{n^2+2021}-\sqrt{n^2+1})=0$
Got stuck in the middle of proving a limit and I don't know what to do.

Prove $\lim_{n\to\infty} (\sqrt{n^2+2021}-\sqrt{n^2+1})=0$

I wrote the equation as $\Rightarrow(\sqrt{n^2+2021}+\sqrt{n^2+1})\,\frac{({\sqrt{n^2+2021}-\sqrt{n^2+1}})}{\sqrt{n^2+2021}-\sqrt{n^2+1}}$

=$\frac{(n^2+2021)-(n^2+1)}{\sqrt{n^2+2021}-\sqrt{n^2+1}} = \frac{2020}{\sqrt{n^2+2021}-\sqrt{n^2+1}}$

and now I'm stuck , I don't know exactly how to proceed.

Any hints?
Thank you and have a nice day!

ContentLicense: CC BY-SA 4.0

Body:
$$L=\lim_{n \to \infty} [\sqrt{n^2+2021}-\sqrt{n^2+1}] =\lim_{n\to \infty} n[(1+2021/n^2)^{1/2}-(1+1/n^2)^{1/2}$$
Use $(1+z)^k =1+k$ if $|z|<<1$.
Then
$$\lim_{n\to \infty} n[1+2021/(2n^2)-1-1/(2n^2)]= \lim_{n \to \infty} 1010/n=0$$


问:
Find the limit of the series $6^n/n!$ as $n$ tends to infinity.
I need to find the limit of the following serie:

$\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$

I was thinking of the following solution, but i'm not sure it's correct, please let me know your thoughts :)

$0$ $<$ $\frac{6^n}{n!}$ $<$ $\frac{6}{1}\cdot$$\frac{6}{2}\cdot$ $\frac{6}{3}\cdot$$\frac{6}{4}\cdot$$\frac{6}{5}\cdot$$\frac{6}{6}\cdot$ $1\cdot$ $1\cdot$ $1\cdot$ .... $\cdot1\cdot$ $\frac{6}{n}$ $=$ $\frac{1944}{5n}$

$\lim_\limits{n\to \infty}$$0$ $= 0$

$\lim_\limits{n\to \infty}$$\frac{1944}{5n}$ $= 0$

Hence, by using the sandwich theorem, $\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$ $= 0$.

答:
Perhaps overkill, but it works:
$$\lim_{n\to\infty}\frac{6^n}{n}=\lim_{x\to\infty}\frac{6^{x-1}}{\Gamma(x)}=L$$
$$\log(L)=\lim_{x\to\infty}\left(\log(6^{x-1})-\log\Gamma(x)\right)$$
Series expansion of log Gamma:
$$=\lim_{x\to\infty}\left[(x-1)\log6-(-\gamma x-\ln x+...)\right]=-\infty$$
$$\log(L)=-\infty\implies L=0.$$


答:
If you know about the exponential function, you can argue as follows:

The series $\displaystyle\sum_{n=0}^{\infty}\frac{6^n}{n!}$ converges to $e^6$ and so $\displaystyle\lim_{n\to\infty}\frac{6^n}{n!}=0$.

Title: Find the limit of the series $6^n/n!$ as $n$ tends to infinity.

Body:
I need to find the limit of the following serie:

$\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$

I was thinking of the following solution, but i'm not sure it's correct, please let me know your thoughts :)

$0$ $<$ $\frac{6^n}{n!}$ $<$ $\frac{6}{1}\cdot$$\frac{6}{2}\cdot$ $\frac{6}{3}\cdot$$\frac{6}{4}\cdot$$\frac{6}{5}\cdot$$\frac{6}{6}\cdot$ $1\cdot$ $1\cdot$ $1\cdot$ .... $\cdot1\cdot$ $\frac{6}{n}$ $=$ $\frac{1944}{5n}$

$\lim_\limits{n\to \infty}$$0$ $= 0$

$\lim_\limits{n\to \infty}$$\frac{1944}{5n}$ $= 0$

Hence, by using the sandwich theorem, $\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$ $= 0$.


问:
For any two sets how to prove: (1) $A ⊆ B$ (2) $A ∪ B = B$ (3) $A − B = B − A$
I'm trying to prove these statements are equivalent and I'm stuck at proving the second one. I started by proving that  $A \cup B \subseteq B$.

$x \in (A \cup B)  $ then $ x \in B$

$x \in A$ or $x \in B$ then $ x \in B$

$ x \in A $ then $x \in B$ is true from the first assumption

$x \in B$ then $x \in B$ is true because B is a subset of itself

This was in the first direction. ($A \cup B \subseteq B$) But when I try to do it in the other direction I get this.

$B \subseteq A \cup B$

$x \in B$ then $x \in (A \cup B)$

$x \in B$ then $x \in A$ or $x \in B$

I can prove that $x \in B$ then $x \in B$ again but how do I prove $x \in B$ then $x \in A$? ($B \subseteq A$)

答:
Let $A, B$ be any sets.

$(1) \implies (2).\quad$ Suppose that $A \subseteq B.$ Let $x \in A \cup B.$ Then

$$\begin{align}
x \in A \cup B & \implies x \in A \vee x \in B\\
& \implies x \in B \vee x \in B & \text{($A \subseteq B$)}\\
& \implies x \in B.
\end{align}$$

Hence, $A \cup B \subseteq B.$ Now, let $y \in B.$ Then

$$\begin{align}
y \in B & \implies y \in A \vee y \in B\\
& \implies y \in A \cup B.
\end{align}$$

Hence $B \subseteq A \cup B.$ Therefore $A \cup B = B.$

$(2) \implies (3)? \quad$ No. Let $A = \{1\}$ and $B = \{1,2\}.$ Note that $A \cup B = \{1,2\} = B$ But $A - B = \emptyset \neq \{2\} = B -A.$

$(3) \implies (1). \quad$ Suppose that $A - B = B - A.$ Then $A = B.$ Therefore $A \subseteq B.$

$(2) \implies (1). \quad $ Suppose that $A \cup B = B.$ Let $x \in A.$ Then

$$\begin{align}
x \in A & \implies x \in A \vee x \in B\\
& \implies x \in A \cup B\\
& \implies x \in B & \text{($A \cup B = B$)}.
\end{align}$$

Therefore $A \subseteq B.$

$(3) \implies (2). \quad$ Suppose that $A \cup B = B.$ Hence $A = B.$ Then

$$A \cup B = B \cup B = B.$$

$(1) \implies (3)? \quad$ No. Use the same example used in proving that $(2) \implies (3)$ is false.

So the implications $(1) \implies (2), (3) \implies (1), (2) \implies (1)$ and $(3) \implies (2)$ are true, while $(2) \implies (3)$ and $(1) \implies (3)$ are false.

More, you can see that $(1) \iff (2)$ so $(1)$ and $(2)$ are the only equivalent statements. $\quad \square$


问:
Understanding the RSI Formula - $RSI = {\color {red}{100} - { \color {red}{100} \over {\color {red} 1} + {n_{up} \over n_{down}} }}$
RSI (Relative Strength Index) is an Oscillator Formula used in trading to detect overbought/oversold conditions. High values suggests overbought and low values suggest oversold conditions in the market.

While I understand what the denominator shown in the formula below does, I am having trouble understanding what the values highlighted in red does to the overall picture from a mathematical standpoint.

$$RSI = {\color {red}{100} -  { \color {red}{100} \over {\color {red} 1} + {n_{up} \over n_{down}}  }}$$

Any help is appreciated. Thanks.

答:
Extending on @Henry 's answer. The idea behind using this notation for $\textbf{Relative Strength Index}$ rather than simply $$100*\frac{n_{up}}{n_{up}+n_{down}}$$
is because

The RSI is classified as a momentum oscillator, measuring the velocity and magnitude of price movements. Momentum is the rate of the rise or fall in price. The RSI computes momentum as the ratio of higher closes to lower closes

In finance, momentum is generally termed as $\textbf{Relative Strength}$.

The ratio of these averages is the relative strength or relative strength factor:

$$RS={\frac{{\text{SMMA}}(U,n)}{{\text{SMMA}(D,n)}}}$$

then,

The relative strength factor is then converted to a relative strength index between 0 and 100:

$$RSI=100-{100 \over {1+RS}}$$

The rationale is basically to have RSI in terms of RS.

Please refer https://en.wikipedia.org/wiki/Relative_strength_index.


问:
How to determine the maximum of the following function?
Assume $B,C \in \mathbb{C}^{5 \times  5}$ are Hermitian matrices, and $B,C \ge 0$, then we want to calculate
$$
\max_{x \in \mathbb{C}^5} \frac{\left\langle Cx,x \right\rangle}{\left\langle (I+B)x,x \right\rangle},
$$
and the answer tells me that it equals to the largest eigenvalue of the Hermitian matrix
$$
(I+B)^{-1/2} C (I+B)^{-1/2},
$$
but how can I check it?

答:
Note that
$$
\langle (I + B)x,x\rangle = \langle (I + B)^{1/2}x, (I + B)^{1/2}x \rangle.
$$
With the substitution $y = (I + B)^{1/2}x$, we have
$$
\max_{x \in \mathbb{C}^5} \frac{\left\langle Cx,x \right\rangle}{\left\langle (I+B)x,x \right\rangle} =
\max_{x \in \mathbb{C}^5}\frac{\langle C(I + B)^{-1/2}y,(I + B)^{-1/2}y\rangle}{\langle y,y \rangle}
\\= \max_{x \in \Bbb C^5} \frac{\langle (I+B)^{-1/2}C(I + B)^{-1/2}y,y\rangle}{\langle y,y \rangle}.
$$
By the Rayleigh-Ritz theorem, this maximum is the largest eigenvalue of $(I+B)^{-1/2}C(I + B)^{-1/2}$.


问:
$\log((n^2!))>n^2$
I need to prove that $\log\bigl((n^2)!\bigr)$ is bigger than $c\mkern1mu n^2$ for some $c$ and starting from a specific location.

How can I do that?

I know the following:

Note: My logarithm is to the base of 2.

答:
Actually you are almost done. Look at each term of
\begin{align*}
\log (n^2) + \log (n^2 - 1) + \dotsm + \log (2) + \log (1) \\
= 2 \log (n) + \underbrace{\log (n^2 - 1) + \dotsm + \log (2)}_{n^2 - 2 \text{ terms}} + \log (1)
\end{align*}
Of course, we can assume that $n > 2$, so that $2 \log (n) > 2$. Now, the first term is $2 \log (n) > 2$. The last term is $\log(1) = 0$. The remaining terms $(n^2-2)$ terms are all at least $1$.
Thus, this whole thing is greater than
$$ 2 + \underbrace{1 + \dotsm + 1}_{n^2 - 2 \text{ terms}} + 0 = n^2$$
Can you do the rest?

Title: $\log((n^2!))>n^2$


问:
Continuity of function with parameter
I have this function. I need to find all the parameters a (real number) so that the function is continuous.

$${f(x)=\begin{cases}
sin(ax-a)/(x-1) & \text{, } x< 1 \\
(a/(x+2))+ax & \text{, } x\geq1
\end{cases}} \\
$$

When I substitute first I get a limit of type 0/0. If I extend the expression by a/a and make adjustments I get as a result a. For the second limit I get a/3 + a. So I get a = a / 3 + a which is probably not good. How should I work with the first limit?

答:
So if we have a limit from right and left, I finally get the equation
a + a / 3 = a.

and the result is a = 0 ?

Title: Continuity of function with parameter


问:
Working with functions using modulo operator
I have a function defined in a computer program that I would like to study mathematically:

$y = \frac{x}{2} + (\frac{5}{2}x + 1)*(x\%2)$

This function is used on strictly positive integers and also returns integers only.

The main issue is that it's using the modulo operator ($\%$) as part of the function calculation, not as in a modular equation, and I have no clue how to translate into a formula that is suitable for mathematical analysis...

I have found various topic related to modulo/floor and how to translate them into mathematical form:

Solving equations involving modulo operator

Solving modular equations with a variable in the modulus divisor

How to represent the floor function using mathematical notation?

But none of them seems to match this particular situation.

Also, to clarify what I mean by "studiy mathematically", it would be for example to determine the range of output values if input values are taken in a given range, or to determine limits/periodicity if this function was called recursively.

答:
x % 2 is equivalent to $\frac{1-(-1)^x}{2}$.  On positive integer $x$, x % 2${}\in \{0,1\}$.  (Some programming languages do stupid things for negative $x$.  Example.)  Consequently, you have
$$  y = \begin{cases}
x/2 &, \text{$x$ even}  \\
3x + 1 &, \text{$x$ odd}
\end{cases}  \text{.}  $$
The Collatz conjecture covers the study of this function under recursion.


问:
Finding variance of random variable in trial with sampling without replacement
We have 20 cards, 5 green, 5 yellow, 5 red and 5 blue.
We take out cards until we first encounter the same color of card as the first card that was sampled.

Let $X$ be a random variable, defined by the number of cards that was taken out in the trial. (From the beggining to the end).

If we sample the cards without replacement what is the variance of $X$.

What i tried:

I tried to look for a distribution that will fit.

Negative binom doesnt fit -> the success chance is not constant.

Hypergeometric doesnt fit -> the size of the sample is not constant.

So i tried to calculate $E[X]$ in order to solve it using $V[x] = E[X^2]-(E[X])^2$ and it is getting realy complicated:

I probably miss something, can anyone help?

Thanks.

答:
So the hard part of this question is determining $\operatorname{Pr}(X=x)$, as is often the case. It is often first useful to identify $\mathcal{X}$, the possible values $X$ can take. So let's ask ourself - Can $X=0$ ? No, because we definitely draw at least one card. can $X=1$ ? No, because we definitely draw another card after the first one. Can $X=20$? No, because then we would have drawn more than two of the starting card, which doesnt make sense as the experiment would have ended by then. It's pretty obvious to see that the possible values of $X$ are $\mathcal{X}=\{2,...,17\}$. The fact that there are four colors here is kind of confusing - let's just assume that $5$ of the cards are red and the rest are black, and we assume that the first card we draw is red. What is the probability the next card drawn is red? Well, this is clearly $4/19$. So,
$$\operatorname{Pr}(X=2)=4/19$$
Assume the second card was black. What is the probability the third card is red? The number of red cards is the same, but the number of total cards is now smaller since we are sampling without replacement. Thus,
$$\operatorname{Pr}(\text{third card is red | all other cards black})=4/18$$
But in order to reach this point, the second card must have been black, which means
$$\operatorname{Pr}(X=3)=(1-4/19)\cdot4/18$$
Now assume the third card was black. What is the probability the fourth card is red? Once again, there is one fewer black card, so this is
$$\operatorname{Pr}(\text{fourth card is red | all other cards black})=4/17$$
Once again, in order to reach this point, all previous cards needed to be black. Thus
$$\operatorname{Pr}(X=4)=(1-4/19)\cdot(1-4/18)\cdot4/17$$
We can carry on with this pattern and see that $\forall x\in\mathcal{X}$,
$$\operatorname{Pr}(X=x)=\frac{4}{20-x+1}\prod_{k=2}^{x-1}\left(1-\frac{4}{20-k+1}\right)$$
Hence,
$$\mathrm{E}[X]=\sum_{x=2}^{17}x\cdot\Pr(X=x)=\sum_{x=2}^{17}\left[x\cdot\frac{4}{20-x+1}\prod_{k=2}^{x-1}\left(1-\frac{4}{20-k+1}\right)\right]=5$$
And,
$$\mathrm{E}[X^2]=\sum_{x=2}^{17}x^2\cdot\Pr(X=x)=\sum_{x=2}^{17}\left[x\cdot\frac{4}{20-x+1}\prod_{k=2}^{x-1}\left(1-\frac{4}{20-k+1}\right)\right]=33$$
Thus
$$\operatorname{Var}[X]=\mathrm{E}[X^2]-\mathrm{E}[X]^2=33-5^2=8.$$


问:
Find $\int_0^bg(x,u)\,du$ given $\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$
I encounter the following equation, which holds for any $x\in[0,b]$:
$$\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$$
I would like to solve for $G(x,b)=\int_0^b g(x,u)du$, subject to the condition that $G(x,0)=0$. Intuitively, $G(x,b)$ is the column sum of an (infinite dimensional) symmetric matrix $g$ with entries given by the $g(x,y)$ function for $x,y\in[0,b]$, and the integral on the l.h.s. says that pre-multiplying the $x$-th column of $g$ by the vector $u^2+(b^2-1)u+(b-1)^2$ yields a known function of column sum in terms of $x$.

It seems that integration by parts won't really simplify the problem as the integral of $G$ then enters into the problem. I am not sure whether this question is solvable at all, and frankly, I do not even know where to start with. Maybe I have missed something very simple. Any hints or suggestions are highly appreciated.

Body: We have \begin{align}\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)\,du&=\frac{x^2+b^2x+b^2-2b}{x-1}\tag1\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_x(x,u)\,du&=1-\frac{2b^2-2b+1}{(x-1)^2}\tag2\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_{xx}(x,u)\,du&=\frac{2(2b^2-2b+1)}{(x-1)^3}\end{align} so assume that $(x-1)^3g_{xx}(x,u)$ is constant. Then $g_{xx}(x,u)=t(b)/(x-1)^3$ where \begin{align}t(b)=\frac{2(2b^2-2b+1)}{\int_0^bu^2+(b^2-1)u+(b-1)^2\,du}=\frac{12(2b^2-2b+1)}{b(3b^3+8b^2-15b+6)}\end{align} so $$g_x(x,u)=h_1(u)-\frac{t(b)}{2(x-1)^2}\implies g(x,u)=h_2(u)+xh_1(u)+\frac{t(b)}{x-1}$$ where $h_1,h_2$ can be determined (with considerable freedom) using $(1)$ and $(2)$.


问:
The set of real numbers is a subset of the set of complex numbers?
So, I was taught that $\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}$.

But since the set of complex numbers is by definition $$\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},$$ doesn't this mean $\mathbb{R}\subseteq\mathbb{C}$, since for each $x \in \mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$?

答: Yes, $\mathbb R \subset \mathbb C$, since any real number can be expressed as a complex number with $b=0$ (as you state).</p>

ContentLicense: CC BY-SA 3.0


问:
Show that, without determining them, that $\ker (h)$ and $\text{im} (h)$ are orthogonal in $\mathbb {R}^3$ and $E=\ker (h)\oplus \text{im} (h)$
The space $ \mathbb {R}^3 $ is endowed with its usual scalar product, noted $ \langle. ,. \rangle $ for which the canonical basis $ \mathcal {B} = (e_1, e_2, e_3) $ is orthonormal.

We denote by $ h $ the endomorphism of $ \mathbb {R}^3 $ whose canonical matrix is

$$H=\left(\begin{matrix}1& -2& 3\\-2 &3 &-4\\3& -4 &5\end{matrix}\right)$$

Problem

Show that, without determining them, that $\ker (h)$ and $\operatorname{im} (h)$ are orthogonal in $\mathbb {R}^3$ and $E=\ker (h)\oplus \operatorname{im} (h)$.

An idea please.

答:
The matrix $H$ is symmetric that is $H^T=H$. Now if $x\in\text{ker}(h)$ and $h(y)\in\text{im}(h)$ we have
$$h(y)\cdot x=(Hy)^Tx=y^TH^Tx=y^THx=y\cdot h(x)=0$$
since $h(x)=0$. We make use of the formula
$$a\cdot b=a^Tb$$
The direct sum should be obvious since if $F$ is any subspace we have
$$E=F\oplus F^{\perp}$$


问:
Does the Axiom of Substitution Have Limits?
According to Terence Tao's book Analysis I, the axiom of substitution is one of the four axioms of equality for objects (which are basically eligible set elements). In particular, it says that for any two objects $x$ and $y$ of the same type, if $x$ = $y$, then $f(x)=f(y)$ for all functions and operations $f$, and that furthermore, $P(x)$ and $P(y)$ are logically equivalent propositions (i.e. they imply each other) for all  propositions $P$ whose truth-value depends on $x$ and $y$.

Sometimes, the axiom of substitution makes sense (i.e. is correct), such as when considering sets to be the object in question (they are indeed valid objects according to the first set-theory axiom presented in the book), and considering the operation $f(X) = X \cup C$, where $C$ is some arbitrary set. Then indeed, one may verify that if two sets $A$ and $B$ are equal to each other (i.e. $A=B$), then $f(A)=f(B)$ for the operation $f$ defined above.

But there are times when it starts acting up. Still considering sets to be the object in question, let $P(X)$ be the proposition that the elements of $X$ are ordered from least to greatest (this proposition's truth value clearly depends on what the set $X$ is that you "feed" it so to speak). Evidently however, as Tao himself acknowledges, even if two sets $A$ and $B$ were equal to each other, because sets are unordered collections of elements, there are plenty of counterexamples to give that would thoroughly disprove the statement: $P(A)$ iff $P(B)$. So my question then is: does the axiom of substitution have limits to its applicability, and if so, can someone very explicitly clarify what they are? Thanks.

答:
In short, it is not a problem of the axiom of equality but a problem that your "property" $P$ is not a property.

First of all, since you are "considering the object of sets", you cannot in general talk about "least" or "greatest". For instance, it is meaningless to talk about what element is larger for the set $\{\text{cat, banana}\}$.

Given a finite subset $S$ of integers, "whether the elements of $S$ are ordered from least to greatest" is not a property of $S$. This is similar to the scenario that
$$
f(\frac{a}{b})=a+b\qquad a\in\mathbb{Z},\ b\in\mathbb{Z}\setminus\{1\}.
$$
is not a well-defined function on the set of rational numbers. The value of the "function" depends on the concrete representations of the object of rational numbers. For instance, $\frac{1}{2}$ and $\frac{2}{4}$ are the same element in the set of rational numbers, but $1+2\ne 2+4$.

Returning to your case, $\{1,2\}$ and $\{2,1\}$ are two different "representations" (whatever that means) of the same object of sets. But your "property" depends on different "representations" and is thus not a property on a finite subset of integers.


问:
Limit set of a function is a closed set
Let $f: M\rightarrow N$ be a continous function between topological spaces, in the case that I am interested we are working with manifolds so we have nice properties. We define the limit set of $L(f)$ as the set of $y\in N$ such that $y=\lim_{n\rightarrow \infty}f(x_n)$ for some sequence $\{x_n\}$ in $M$ that has no convergent subsequence. Is this set going to be closed ? I have tried taking $y\in \bar L(f)$ and take a sequence $y_n\rightarrow y$ such that $y_n=\lim_{k\rightarrow \infty} f(x_n^k)$. Now I have tried constructing a sequence for $y$ of $f(x_i)$ such that $x_i$ has no convergent subsequence, I have tried a diagonalization argument and so on, but I got nothing.  Does anyone know if this is true or not and if so why ?

New edit : As was mentioned in the comments there will be a counterexample if $M$ is not a manifold , so I guess I can change the question to if anyone knows if this is true if we assume that $M$ and $N$ are manifolds?

And if this is not true I guess a thing that would help me is that if $f(M)\cap L(f)= \emptyset$ will we have that $f(M)\cap \bar L(f) =\emptyset$
?

This is because I am trying to prove that if we have that $f(M)\cap L(f)=\emptyset $ then we will have that there exists an open set $V$ such that $f(M)\subset V$ and $f:M\rightarrow V$ is proper, and $L(f)$ is exactly the points where properness fails, and this are the obvious first choices to create the open set I think.

Thanks in advance.

答:
In general spaces, this property fails. To see this, consider the set $\mathbb{N}\times\mathbb{N}\cup\{\infty\}$ with the topology $\tau=\{A\subseteq\mathbb{N}\times\mathbb{N}\cup\{\infty\}\vert A\subseteq\mathbb{N}\times\mathbb{N}\lor(\infty\in A\land\exists k\in\mathbb{N}\colon\{(n,m)\colon n,m\ge k\}\subseteq A)\}$. It is easily checked that this is a first-countable topology such that $(n_k,m_k)_k\rightarrow(n,m)$ iff $n_k=n$ and $m_k=m$ for large enough $k$ and $(n_k,m_k)_k\rightarrow\infty$ iff $n_k\rightarrow\infty$ and $m_k\rightarrow\infty$. It is then easy to see that the function $f\colon\mathbb{N}\times\mathbb{N}\cup\{\infty\}\rightarrow\mathbb{R},(n,m)\mapsto\frac{1}{n}+\frac{1}{m},\infty\mapsto0$ is continuous.

Consider $y\in L(f)$ and a sequence $(x_k)_k$ in $\mathbb{N}\times\mathbb{N}\cup\{\infty\}$ containing no convergent subsequence and such that $f(x_k)\rightarrow y$.
If $x_k=\infty$ for infinitely many $k$, $(x_k)_k$ possesses a constant, hence convergent, subsequence, so $x_k=\infty$ for only finitely many $k$ and, by passing to a subsequence, we may WLOG assume that $x_k=(n_k,m_k)$ with $n_k,m_k\in\mathbb{N}$ for all $k$. Then, $(n_k)_k$ and $(m_k)_k$ cannot be both be bounded, since that would mean they only take on finitely many values, whence $(n_k,m_k)_k$ would possess a constant, hence convergent, subsequence by pingeonholing. Since $f(n,m)=f(m,n)$, we may WLOG assume that $(m_k)_k$ is unbounded, so that there is a subsequence $(m_{k_l})_l$ s.t. $m_{k_l}\rightarrow\infty$. Then $(n_{k_l})_l$ is bounded, since otherwise $(n_{k_l},m_{k_l})_l$ possesses a subsequence converging to $\infty$, hence $(n_{k_l})_l$ possesses a constant subsequence by pigeonholing, so that $(n_k,m_k)_k$ ultimately possesses a subsequence of the form $(n,m_{k_{l_i}})_i$ for some fixed $n$. It follows that $y=1/n$. Conversely, the sequence $(n,k)_k$ for a fixed $n$ possesses no convergent subsequence and $f(n,k)\rightarrow1/n$, so that we obtain $L(f)=\{1/n\colon n\in\mathbb{N}\}$, which is not closed in $\mathbb{R}$. (The idea behind this construction is that adding the point $\infty$ precisely makes those sequences convergent that would otherwise testify $0$ to be in $L(f)$.)

On the other hand, I believe the diagonalization argument can be made to work in a large number of cases:

Proposition: If $M$ is a topological space admitting an exhaustion by sequentially compact sets, $N$ is a first-countable topological space and $f\colon M\rightarrow N$ is any function, then $L(f)$ is closed in $N$.

To say that $M$ admits an exhaustion by sequentially compact sets means that there is an ascending chain $K_1\subseteq K_2\subseteq ...$ of sequentially compact subspaces of $M$ such that $\bigcup_{i=1}^{\infty}K_i=M$ and $K_i\subseteq\operatorname{int}K_{i+1}$ for all $i$. This is relevant due to the following

Lemma: Let $M$ be a topological space admitting an exhaustion $K_1\subseteq K_2\subseteq...$ by sequentially compact sets. Then a sequence $(x_n)_n$ in $M$ contains no convergent subsequence iff $(x_n)_n$ is eventually in $K_i^c$ for all $i$.

Proof: To see this condition is necessary, assume there is an $i$ s.t. $(x_n)_n$ is not eventually in $K_i^c$. Then there is a subsequence $(x_{n_k})_k$ contained in $K_i$. Since $K_i$ is sequentially compact, this admits a convergent subsequence $(x_{n_{k_l}})_l$, but this is also a convergent subsequence of the original sequence, contrary to hypothesis. To see sufficiency, assume $(x_n)_n$ contains a convergent subsequence $(x_{n_k})_k$ with limit $x$. By exhaustion, there is an $i$ such that $x\in K_i$. Since $K_i\subseteq\operatorname{int}K_{i+1}$, there is an open neighborhood $U\subseteq K_{i+1}$ of $x$. By convergence, there is a $K$ s.t. $x_{n_k}\in U$ for $k>K$. For each $j=1,...,K$, there is an $i_j$ such that $x_{n_j}\in K_{i_j}$. Thus, $(x_{n_k})_k$ lies in $K_{\max\{i_1,...,i_K,i+1\}}$, contrary to hypothesis.

Proof of the Proposition: Since $N$ is first-countable, it suffices to check that if $(y_n)_n$ is a sequence in $L(f)$ converging to some $y\in N$, then $y\in L(f)$. By definition, for each $n$, we can find a sequence $(x_k^n)_k$, containing no convergent subsequence, such that $f(x_k^n)\rightarrow y_n$ as $k\rightarrow\infty$. Both of these conditions are preserved by passing to a subsequence, so, by the Lemma, we can WLOG assume that $x_k^n\in K_i^c$ for all $n$ and $k\ge i$. Since $N$ is first-countable, there is a local neighborhood base $(U_n)_n$ of $y$. We assume $U_1\supseteq U_2\supseteq...$ WLOG. For each $n$, by definition of convergence, there is a $y_{k_n}$ such that $y_{k_n}\in U_n$. By definition of convergence, there is a $x_{l_n}^{k_n}$ such that $f(x^{k_n}_{l_n})\in U_n$. We also take $l_{n+1}>l_n$ WLOG. Then, $f(x^{k_n}_{l_n})\rightarrow y$ as $n\rightarrow\infty$. Also $x^{k_n}_{l_n}\in K_i^c$ for $l_n\ge i$ by construction, so, by the Lemma, $(x^{k_n}_{l_n})_n$ contains no convergent subsequence, hence $y\in L(f)$, as desired.

Lastly, of course, any manifold admits an exhaustion by sequentially compact sets and is first-countable, so the theorem in particular applies when $M,N$ are manifolds.

Title: Limit set of a function is a closed set


问:
$\frac{n(n+1)+2(n+1)}{2}$ to $\frac{(n+1)((n+1)+1)}{2}$
So somehow my professor converts $\frac{n(n+1)+2(n+1)}{2}$ to  $\frac{(n+1)((n+1)+1)}{2}$ and I don't understand how. Can somebody explain the steps to do this please?

答:
The denominators are the same (namely, $2$).  For the numerators,

$$\color{blue}n(n+1)+\color{blue}2(n+1)=(\color{blue}n+\color{blue}2)(n+1)=(n+1+1)(n+1)=(n+1)((n+1)+1).$$


问:
Find a general solution
$$(y+e^x)\frac{dy}{dx}=-\frac{y^2}{2}-2ye^x$$

I attempted to turn this inexact eq. to exact:

I multiply by IF = $e^t$ and got:

$$(ye^x+e^{2x})dy+(\frac{y^2}{2}e^x+2ye^{2x})dx = 0$$

which is exact.

Then I integrate, and I get $y^2e^x+2ye^{2x} + c = 0$.

My question is if this is considerate a general solution? or does this expression need to be expressed in terms of y?

If so, how can I do this?

Thanks

答:
Re-write The ODE as: $M dx +N dy=0$, then
$$[(y^2e^{-x}/2+2y]dx+(ye^{-x}+1) dy~~~~(1)$$
$$\frac{\partial M}{\partial y}=2ye^{-x}+2, ~~\frac{\partial N}{\partial x}=-ye^{-x}$$
So the integrating factor $I$ to make it exact is given by
$$I=\exp \left[\int \frac{1}{N}\left(\frac{ \partial M}{\partial y}-\frac{\partial N}
{\partial x} \right)dx\right]=e^{2x}.$$
Multiplying (1) by I we get an exact ODE as
$$[(y^2 e^{x}/2)+2ye^{2x}]dx+(ye^x+e^{2x}) dy=0 \implies \bar M dx+ \bar Ndy=0$$
Then the solution is given by
$$\int [(y^2 e^{x}/2)+2ye^{2x}] dx~\text{(treat $y$ as constant)}+ \int \text{terms of $\bar N$ not containing $x$]} dy=C$$
$$(y^2e^x/2)+ye^{2x}=C. $$
The general solution of (1) can be written as:
$$F(y^2e^x/2+ye^{2x})=A,$$
where $F$ is an arbitrary function and $A$ is and arbitrary constant.
These two could be fixed by specifying consitions,

Title: Find a general solution


问:
Prove that a subset is a subspace (classic one)
Given a subset S, $$ S =  \left\{ f \in  \mathbb{R}^{ \mathbb{R} } : f(1) = 1 + f(0)  \right\}  \subset \mathbb{R}^ \mathbb{R}  $$
the goal is to check whether S is a subspace or not.

I already know that it's not a subspace, but my questions are:

Why does the element $0$ doesn't belong to S? Isn't the function $f(0) = 0$ $\subset$ S? (e.g. f(x)=x) [I.e. what is the element $0$ in concrete and how I can I tell which element is the element $0$? (is it when x + (element zero) = x? My guess in this case would be f(x) = 0; nevertheless I would like to hear some clear explanation];

Can I prove S isn't closed under addiction as it follows?:$f(1)+g(1)=(f+g)(1)=(f+g)(0)+2  \not \subset S$ (because it's adding two and not one it doesn't belong to S)

Why is not $\alpha$f(x)=($\alpha$f)(x) true? Isn't it trivial?

Thanks in advance, I'm really trying to get into Linear Algebra...

答:
If $f$ is the null function, then $f(1)=0\ne1=1+f(0)$. So, $f\notin S$.

No, you cannot. Asserting that $(f+g)(0)+2\not\subset S$ makes no sense, since $(f+g)(0)+2$ is a real number, not a set of functions from $\Bbb R$ to $\Bbb R$.

Of course that $\alpha f(x)=(\alpha f)(x)$; that's how $\alpha f$ is defined. What is not true in general is that if $f\in S$ and $\alpha\in\Bbb R$, then $\alpha f\in S$.


问:
Show that Wiener process with drift is a Levy process
Hey I have to check if process $X_t=\mu t+\sigma W_t$ is a Levy process where $W_t$ is a Wiener process and $\sigma>0,\mu\in\mathbb{R}$. First, $X_0=0$ so it is ok. Then, for $s<t$ we have $X_t-X_s=\mu(t-s)+\sigma(W_t-W_s)\sim N(\mu(t-s),\sigma(t-s))$ so $X_t-X_s$ has the same distribution as $X_{t-s}$. Now I take $0<u<v<s<t$ and we have that $X_v-X_u=\mu(v-u)+\sigma(W_v-W_u)$ and $X_t-X_s=\mu(t-s)+\sigma(W_t-W_s)$ so they are independent since $W_v-W_u$ and $W_t-W_s$ are independent. Now I have to check that $X_t$ is stochastic continuity. This condition states that
$$\lim_{s\to t}P(|X_s-X_t|>\epsilon)=0$$
Can anyone show me how to chech this condition?

EDIT: I try sth like this:
$$P(|X_s-X_t|>\epsilon)=1-P(|\mu(s-t)+\sigma(W_s-W_t)|\le \epsilon)$$
$$=1-P(|\mu(t-s)+\sigma W_{t-s}|\le \epsilon)=1-P(-\epsilon\le\mu(t-s)+\sigma\sqrt{t-s}W_1\le \epsilon)$$
$$=1-P(\frac{-\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}}\le W_1\le \frac{\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}})=$$
$$=1-\Phi(\frac{\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}})+\Phi(\frac{-\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}})$$
And now I have to take limit as $s\to t$ but because there is square root, the limit can be only one sided.

答: Note that $$1-\Phi\left(\frac{\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}}\right)+\Phi\left(\frac{-\epsilon-\mu(t-s)}{\sigma\sqrt{t-s}}\right)$$ is equivalent to $$1-\Phi\left(\frac{\epsilon}{\sigma\sqrt{t-s}}-\frac\mu\sigma\sqrt{t-s}\right)+\Phi\left(\frac{-\epsilon}{\sigma\sqrt{t-s}}-\frac\mu\sigma\sqrt{t-s}\right)$$ so we have $$\lim_{s\to t^-}P(|X_s-X_t|>\epsilon)=1-\Phi(+\infty)+\Phi(-\infty)=1-1+0=0.$$ It does not matter that the limit is one-sided as WLOG you set $s<t$, similar calculations hold when you set $s>t$.


问:
How many ways are there to perform the 83 serves when 3 types of serves are being used?

The player hits 83 serves in his first round match. For each
one he randomly chooses between the variants slice, kick and flat. How
many ways are there to perform the 83 serves?

We have an algorithm to answer such questions:

Decide whether to look at all elements or just a sample.

Decide if the order / arrangement is important.

Decide if the elements can be repeated.

Select formula.

Choose n and k .

I'm getting a bit lost here. We're viewing all 83 serves, correct? Or are we viewing the subset of 3? Does the order matter or not? I dont think it does. The player can choose any type of server he wants to.

答:
Yes. You are viewing all the $83$ serves.

Notice that the wording is: "How many ways are there to perform the 83 serves?" implying that you want to know the sequence of serves. Because it is a sequence, the order does matter.

Yes, but the player has a choice out of $3$ in every one of the $83$ serves. If there were $2$ serves, then how many choices you have? Extend to $83.$


问:
If for all $x \in B$, either $a \leq x$ or $a \leq x^c$ then $a$ is an atom
I saw a related answer here, but was unable to figure out what's going on.

If for all $x \in B$, either $a \leq x$ or $a \leq x^c$ then $a$ is an atom The converse holds as well. $B$ is a boolean algebra.

I was able to prove the converse. Here's how:

Consider $a \land x$, for $x \in B$ where $a$ is an atom. We know that $a \land x \leq a$ by definition of $\land$. Since $a$ is an atom, it is the  minimal element in $B\backslash\{0\}$, so $a \land x \leq a$ means either $a \land x = 0$ or $a \land x = a$. From $a \land x = a$, we get (i) $a \leq x$ (since the g.l.b of $a$ and $x$ is $a)$ and from $a \land x = 0$ we get $a \leq x^c$. Is this correct?

To prove that: if for all $x \in B$, either $a \leq x$ or $a \leq x^c$ then $a$ is an atom, I'm trying to retrace my steps back in the above proof.

If (i) $a \leq x^c$, then $a \land x = 0$ & if (ii) $a \leq x$ then $a \land x = a$. I'm kinda stuck here, it seems.

答:
Depending on your axioms for Boolean algebras, your proof of the converse may be complete or not.

Perhaps it is convenient to show that $a \wedge x = 0$ implies that $a \leq x'$:
if $a \wedge x = 0$, then
\begin{align}
a
&= a \wedge 1 = a \wedge 0' = a \wedge (a \wedge x)'\tag{by hypothesis}\\
&= a \wedge (a' \vee x') = (a \wedge a') \vee (a \wedge x')\tag{by distributivity}\\
&= 0 \vee (a \wedge x') = a \wedge x',
\end{align}
whence $a \leq x'$.

For the direct implication, suppose that $a \in B$ is such that for every $x \in B$ we have that $a \leq x$ or $a \leq x'$.

In order to prove that $a$ is an atom, take $b \in B$ such that $0 \leq b < a$.

It is enough to show that $b=0$.

As $b \in B$, it must be $a \leq b$ or $a \leq b'$ (because that's our hypothesis about $a$);
but $b < a$ so $a \nleq b$, and hence $a \leq b'$.
Now it follows that
$$b = b \wedge a \leq b \wedge b' = 0,$$
and therefore $b=0$.

We conclude that $a$ is an atom.


问:
Vector Space and Linear Map
Let $m,n \in \mathbb{N}$ and let $V = \mathbb{Q^{n}}$ and $W = \mathbb{Q^n}$ be vector spaces over $\mathbb{Q}$. Suppose that $f: V \to W$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in V$ . Prove that $f$ is a linear map.

答:
Hint

Then for all $l \in \mathbb N$ and $x \in V$ $f(l x) = l f(x)$, $f(-lx)= -l f(x)$. And if $l \neq 0$ $f(\frac{1}{l} x) = \frac{1}{l} f(x)$.

Conclude as any rational $\lambda$ can be written $\lambda = \frac{p}{q}$ where $p \in \mathbb Z$ and $q \in \mathbb N$.

Title: Vector Space and Linear Map


问:
Convergence of $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ depending on $a$
Calculate the convergence of $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ depending on the values of $a \in\mathbb R$.

If a = 1, then $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^1 > \sum_{n=2}^{\infty} (\frac{1}{n})$, so it is divergent.

What should I do in case $a < 1$ or $a > 1$?

Edit: Can I say for a < 0 $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a $ = $\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{(-a)}  $  = $\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{b} $ (where b = -a and b > 0 )

$\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{b}  $ = $\sum_{n=2}^{\infty} (\frac{n^b}{b\ln n})  $ > $\sum_{n=2}^{\infty} (\frac{1}{b\ln n})  >  \sum_{n=2}^{\infty} (\frac{1}{bn})  $, so it is divergent.

and for a > 0 $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ = $\sum_{n=2}^{\infty} (\frac{a\ln n}{n^a})$ < $\sum_{n=2}^{\infty} (\frac{a n}{n^a})$ = $\sum_{n=2}^{\infty} (\frac{a}{n^{a-1}})$, so for a -1 > 1 => a>2 it is convergent. And for 0<a<2 is it divergent

Is this correct?

答:
If $a\leq 0$ then $\left(\dfrac{\ln(n)}{n}\right)^a$ doesn't tend to $0$ as $n\rightarrow\infty$ and the series is obviously divergent. Your argument is correct but the series is so far from being convergent that is is unnecessarily complicated.

If $0<a<1$ we have since $\dfrac{\ln(n)}{n}<1$ (is that fact obvious enough?)
$$\left(\dfrac{\ln(n)}{n}\right)^a>\dfrac{\ln(n)}{n}$$
The RHS is divergent by your analysis so the LHS is divergent as well by a comparison test.

If $a>1$, let $b$ be any number such that $1<b<a$. Then:
$$\dfrac{\left(\dfrac{\ln(n)}{n}\right)^a}{\dfrac{1}{n^b}}=\dfrac{\ln(n)^a}{n^{a-b}}$$
Since $a-b>0$ we obtain by standard results on limit comparisons that
$$\dfrac{\left(\dfrac{\ln(n)}{n}\right)^a}{\dfrac{1}{n^b}}\rightarrow 0$$
as $n\rightarrow\infty$. Moreover the series
$$\sum_{n=2}^{\infty}\dfrac{1}{n^b}$$
is convergent since $b>1$. It follows from comparison test on series that
$$\sum_{n=2}^{\infty}\left(\dfrac{\ln(n)}{n}\right)^a$$
is convergent as well - its summand is dominated by a convergent series.


答:
To show linearity, it only remains to show that $f(\lambda x)=\lambda f(x)$ for any \lambda\in\Bbb Q$.

1.- For $n\in\Bbb N$, by induction it is easy to show that $f(nx)=nf(x)$:

$n=1$ is trivial.

$n\to n+1$: $f((n+1)x)=f(nx+x)=f(nx)+f(x)\overset{HI}{=}nf(x)+f(x)=(n+1)f(x)$.

2.- For $n\in\Bbb N$ it holds that $f(x/n)=f(x)/n$. Indeed, $f(x)=f(n\cdot (x/n))=nf(x/n)$.

3.- Finally, $f(-x)=-f(x)$: $0=f(0)=f(x+(-x))=f(x)+f(-x)$, so $f(-x)=-f(x)$. (Observe that this, togeteher with 1, we get that $f(nx)=nf(x)$ for any $n\in\Bbb Z\setminus\{0\}$, but the case $n=0$ still holds).

Putting altogether, for any $\lambda\in\Bbb Q$, $\lambda=\frac{m}{n}$ for some $n\in\Bbb N$ and $m\in\Bbb Z$. So $f(\lambda x)=f(mx/n)=mf(x/n)=(n/m)f(x)=\lambda f(x)$.

Title: Vector Space and Linear Map

Body: Let $m,n \in \mathbb{N}$ and let $V = \mathbb{Q^{n}}$ and $W = \mathbb{Q^n}$ be vector spaces over $\mathbb{Q}$. Suppose that $f: V \to W$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in V$ . Prove that $f$ is a linear map.


答:
$f(0+0)=f(0)+f(0)\Rightarrow f(0)=0\Rightarrow f(0x)=f(0)=0=0f(x)$.

$f(-x)=f(-x+(-x)+x)=f(-x)+f(-x+x)=f(-x)+f(-x)+f(x)\Rightarrow f(-x)=-f(x)$

Let $n\in\Bbb N$. Then (using induction) $f(nx)=f(x+x+\ldots+x)=f(x)+f(x+\ldots+x)=\ldots=f(x)+f(x)+\ldots+f(x)=nf(x)$.

Also $f(-nx)=f(-x+(-x)+\ldots+(-x))=f(-x)+f(-x)+\ldots+f(-x)=-f(x)+-f(x)+\ldots+-f(x)=-nf(x)$

Finally, for $\frac{p}{q}\in \Bbb Q\setminus\{0\}$, we have $pf(x)=f(px)=f(q\frac{p}{q}x)=qf(\frac{p}{q}x)\Rightarrow f(\frac{p}{q}x)=\frac{p}{q}f(x)$

Title: Vector Space and Linear Map


问:
Using area=$\int_{y = a}^{y = b} x \, dy $ find the following shaded area:

So I did:

$$\eqalign{
  & y = {1 \over x^2}  \cr
  & {x^2} = {1 \over y}  \cr
  & x = y^{ - {1 \over 2}}  \cr
  & \int_a^b x \, dy  = \left[ 2{y^{1 \over 2}} \right]_{1 \over 4}^1  \cr
  &  = 2\sqrt 1 - 2\sqrt {1 \over 4} \cr
  &  = 2 - 1  \cr
  &  = 1 \cr} $$

The answer however is 2, where have I gone wrong and misunderstood the question?

Thank you!

答:
The cross-section is twice as large as you think it is. You have only calculated the first quadrant area. To put it another way, the distance between $x=-y^{-1/2}$ and $x=y^{-1/2}$ is $2y^{-1/2}$.

Remark: Sometimes, because we are so accustomed to integrating with respect to $x$, it can be useful to interchange the roles of $x$ and $y$. If you do that, and draw the appropriate picture, you are more likely to get things right.


问:
Linear algebra exercise about linear maps
Determine for each of the following sets of vectors $\{v_1, v_2, v_3\}\in V$, $\{w_1, w_2, w_3\}\in W$, how many linear maps $F:V\longrightarrow W$ there exist such that $\forall i$, $F(v_i)=w_i$:

$\{t^2+1, t+1, t^2+t\}\in V=\mathbb R[t]_{2}$, $\{1, t-1, t^2-t\}\in W=\mathbb R[t]_{2}$;

$\{t^2+1, t+1, t^2+t\}\in V=\mathbb R[t]_{2}$, $\{1, t-1, t^2-t\}\in W=\mathbb R[t]_{3}$;

$\{t^2+1, t+1, t^2+t\}\in V=\mathbb R[t]_{3}$, $\{1, t-1, t^2-t\}\in W=\mathbb R[t]_{2}$.

I cannot start the exercise. Some helps?

Thank you

答:
The set of vectors  $\{v_1,v_2,v_3\}$ is a basis of $V$in the first two cases. Therefore a linear map  from $V$ to any vector space is uniquely determined by the  values of  the $F(v_i)$s.

In the third case, this set can be completed  by a fourth vector, say $v_4$, to obtain a basis of $V$, and the linear map is uniquely determined by the image of $v_4$ (and of course the prescribed images of $v_1, v_2,v_3$). Therefore the set of linear maps we seek for corresponds bijectively to $W$.


问:
Prove that the function has a smallest positive period...
Let $ f \colon \mathbb{R} \rightarrow \mathbb{R} $ be continuous, periodic and non-constant. Prove that $ f $ then has a smallest positive period $ p $ and all other periods are integer-valued multiples of $ p $.

no ideas... please give me a hint

答:
Let $\;P\;$ be the set of periods of the function $\;f:\mathbb{R}\to\mathbb{R}\;.$

$P\;$ is non-empty and bounded below by $0$.

Let us consider $\;p^*=\inf P\;.$

There exist a sequence $\;\left\{p_n\right\}_{n\in\mathbb{N}}\subseteq P\;$ such that $\;p_n\to p^*$.

Let $\;x\;$ be any real number. Then $\;x+p_n \to x+p^*$ and $f(x+p^*)=\lim_\limits{n\to\infty}\left(x+p_n\right)=f(x)$.

If $\;p^*$ were equal to zero, there would be the positive numbers $p_1,p_2,\dots$, tending to zero such that $f(x+p_n) = f(x)$ for all $x\in\mathbb{R}$. Moreover, if $x\neq y$ were any two distinct points, we could make $x$ and $y + m\cdot p_n$ as close as we like by choosing appropriate values of $m$ and $n$. This would imply, by the continuity of $f$, that $f$ should take the same value at $x$ and $y$, i.e., that $f$ would be constant.

So $\;p^*>0\;$, then $\;p^*\in P\;$ and $\;p^*= \min P\;.$

It means that the function $\;f\;$ has a smallest positive period that is $\;p^*.$

Now we are going to prove that $\;\forall\;p\in P\;$ there exists $\;\lambda\in\mathbb{N}\;$ such that $\;p=\lambda\cdot p^*$.

Let $\;p\;$ be any element of $\;P\;$ and let $\;\lambda\in\mathbb{N}\;$ the greatest positive integer number such that $\;p-\lambda\cdot p^*\ge0\;.$

If $\;q=p-\lambda\cdot p^*$ were positive, then it would result that $\;0<q<p^*$ and $\;f(x+q)=f(x)\;$ for all $\;x\in\mathbb{R}\;.$

It means that $\;q\in P\;,$ but it is impossible because $\;q<p^*=\min P\;.$

Hence $\;q=0\;,$ so $\;p=\lambda\cdot p^*.$


问:
Finding $\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$$

Okay this seems weird to me-- couldn't i just write down the answer 0 because $\int_{0}^{0} f(x)dx=0$? (The answer is $-1$ btw)
There was even a hint for the question: $\displaystyle \lim_{x\to0}\frac{e^{-x^2}-1}{x^2}$ may be useful.

Am I supposed to go  $$\displaystyle \int_{\varepsilon}^{0}\frac{e^{x^2}}{x^2}dx+\int_{0}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{x^2}}{x^2}dx\,\,\,?$$ And I couldn't even go L'Hospital with that.....
Any help would be appreciated ! Thanks!

Of course, it would be nice if the answer was simply 0!

答:
I don't know if you like this method or not. By the MVT for definite integrals, there is $\xi\in(\varepsilon,\frac{\varepsilon}{1+\varepsilon})$ such that
$$\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx=e^{-\xi^2}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{1}{x^2}dx=e^{-\xi^2}(-\frac{1}{x})\bigg|_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}=-e^{-\xi^2}$$
and hence
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx=\lim_{\varepsilon\to 0}(-e^{-\xi^2})=-1.$$


问:
Every finite ring $R$ has a nilpotent ideal $I$ such that the only nilpotent of $R/I$ is the zero ideal
Just like the title.

How to show that every finite ring $R$ has a nilpotent ideal $I$ such that the only nilpotent ideal of $R/I$ is the zero ideal?

I just know that every finite ring has a nilpotent ideal, but how to construct a ideal $I$ satisfiying the additional condition?

I guess that the $I$ need to be large enough, but what's more?

答: Consider a nilpotent ideal $I$ whose cardinal is maximal. If $R/I$ has a non trivial nilpotent ideal $J$, $p^{-1}(J)$ is nilpotent, where $p:R\rightarrow R/I$ is the quotient map.


问:
How to calculate it without l'Hopital?
How to calculate this limit without using l'Hopital rule?
$$\lim_{x\to 1 }\frac{3^{5x}-3^{2x^2+3}}{\sin(\pi x)}$$
I know how to make it using L'Hopital and that the result is $-\frac{243\log(3)}{\pi}$, but I'm getting nowhere when I try without it. Any advice? I also tried to change ${\sin(\pi x)}$ to ${\tan x}*{\cos x}$, but it didn't help.

Title: How to calculate it without l'Hopital?

Body:
Hint:

Set $x-1=h$

$$\lim_{h\to0}\dfrac{3^{5+5h}-3^{2(h+1)^2+3}}{\sin\pi(1+h)}$$

$$=-\dfrac{5\cdot3^5\lim_{h\to0}\dfrac{3^{5h}-1}{5h}-3^5\lim_{h\to0}\dfrac{3^{2h^2+4h}-1}{2h^2+4h}\cdot\lim_{h\to0}\dfrac{2h^2+4h}h}{\pi\cdot\lim_{h\to0}\dfrac{\sin\pi h}{\pi h}}$$

Now use for  $a>0,$ $$\lim_{h\to0}\dfrac{a^h-1}h=\ln a$$


问:
How do I see that $F$ is a vector field defined on all of $\mathbb{R}^3$?
$$\vec{F}(x,y,z)= y^3z^3\mathbf{i} + 2xyz^3\mathbf{j} + 3xy^2z^2\mathbf{k}$$

How do I see that $F$ is a vector field defined on all of $\mathbb{R}^3$? And then is there an easy way to check if it has continuous partial derivatives?

I am looking at a theorem and it states, If F is a vector field defined on all of R^3 whose component functions have continuous partial derivatives and curl F = 0, then F is a conservative vector field.

I don't know how to check those conditions. Could someone show with that problem given?

答: A vector field on $\mathbb{R}^3$ is the same as a function $\mathbb{R}^3\to \mathbb{R}^3$. Your $F$ as $\mathbb{R}^3\to \mathbb{R}^3$ is clearly continuous.</p>


问:
Show that projections of element on orthogonal subspaces are summable.
Let $H$ be a Hilbert space and $\{H_i: i \in I\}$ be a collection of closed orthogonal subspaces such that

$$H = \bigoplus_{i \in I} H_i:= \overline{\sum_{i \in I} H_i}$$
where the closure is the norm-closure. I.e. $H$ is the inner orthogonal direct sum of these orthogonal subspaces.

Given $i \in I$, consider the orthogonal projection $p_i: H \to H_i$. If $x \in H$, is it true that
$$x= \sum_{i \in I} p_i(x)$$

Here the sum is considered to be in the norm-topology.

I don't have really much progress on this problem (I'm not even sure it's true!). I tried to show that $\sum_i p_i(x)$ converges by showing that $\sum_i \| p_i(x) \|^2 < \infty$ but could not even succeed at this!

Thanks in advance.

答:
This easily follows from the following theorem:

Let $(q_j)_{j \in J}$ be an increasing net of projections on the
Hilbert space $H$. Then this net is strongly convergent to the
projection of $H$ onto $$\overline{\bigcup_{j \in J} q_j(H)}$$

For a reference, see [1] theorem 4.1.2.

Let us see how this implies the theorem:

Given a finite subset $F \subseteq I$, put $p_F:= \sum_{i \in F} p_i$. Then note that
$$\overline{\bigcup_{F \subseteq I} p_F(H)}= H$$

So the net $p_F$ converges strongly to the identity function on $H$, that is
$$x= \sum_{i\in I} p_i(x)$$
for all $x \in H$.

[1] Murphy, G. J. (2014). C*-algebras and operator theory. Academic press.


问:
If initial condition gives $\dot{x_i}=0$ then $x_i$ is constant for all $t$
This is probably a silly question but I am struggling to get my head around it.

Suppose I have the $2D$ system $\dot{x}=1-y(1+x)$, $\dot{y}=y(y-1)(1+2x^2+y^2)$. Here $x,y$ are both functions of $s\in(-\infty,\infty)$.

Clearly if $y=0$ at $s=0$ then $\dot{y}=0$ at $s=0$. But how can I show that, given this condition $y(0)=0$, that $y(s)=0$ for all $s<0$ and $s>0$?

All I can think to do is Taylor expand $y(0±s_0)$. If I can show that $\frac{d^ny}{ds^n}$ vanishes at $s=0$ for $n\geq2$ then this approach would work, but I am struggling to come up with a suitable induction hypothesis for this. I am guessing there is a much simpler approach?

答:
It should be clear from the equations that
$$x(s)=s+a,\,\, y(s)=0$$
is a solution to the system with initial values $x(0)=a$, $y(0)=0$. Are there some uniqueness theorems you could apply to this system? If you the know the solution is unique for all $s\in\mathbb{R}$ then any solution for which $y(0)=0$ is of the above form and $y(s)=0$ for all $s$.


问:
How I can solve this limit only with algebra
I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.

$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$

first all I use the logarithms properties and rewrite like this:

$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$

and tried to make to this expression goes to  $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2

I can't reach to this because the limit goes to 0 instead of ∞

答:
\begin{align*}
L &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 +(-x+x)+ x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 -x +2x}{1 - x}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt{1 + \frac{2x}{1 - x}}}  \\
    &= \lim_{x\to 0} \frac{1}{x} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2}}  \\
    &= \lim_{x\to 0} \log{\left(1 + \frac{2x}{1 - x} \right)^{1/2x}}  \\
    &= \log \lim_{x\to 0} \left(1 + \frac{2x}{1 - x} \right)^{1/2x}  \text{.}
\end{align*}
Let $x = \frac{1}{2u+1}$, so that $u = \frac{1-x}{2x}$.  Then as $x \rightarrow 0^+$, $u \rightarrow \infty$ and as $x \rightarrow 0^-$, $u \rightarrow -\infty$.  We investigate both.  First,
\begin{align*}
M^+ &= \log \lim_{u \rightarrow \infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u}  \\
    &= \log \lim_{u \rightarrow \infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right)  \\
    &= \log \left( 1 \cdot \mathrm{e} \right)  \\
    &= 1  \text{.}
\end{align*}
Then,
\begin{align*}
M^- &= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{\frac{1}{2} + u}  \\
    &= \log \lim_{u \rightarrow -\infty} \left( \left(1 + \frac{1}{u}\right)^{1/2} \left(1 + \frac{1}{u}\right)^{u} \right)  \\
    &= \log \lim_{u \rightarrow -\infty} \left( 1 + \frac{1}{u}\right)^{u}  \\
    &= \log \lim_{v \rightarrow \infty} \left( 1 + \frac{1}{-v}\right)^{-v}  &  &  \text{u = -v}  \\
    &= \log \lim_{v \rightarrow \infty} \left(\left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1}  \\
    &= \log \left(\lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v}\right)^{-1}  \\
    &= -\log \lim_{v \rightarrow \infty} \left( 1 + \frac{-1}{v}\right)^{v}  \\
    &= -\log \mathrm{e}^{-1}  \\
    &= --1  \\
    &= 1  \text{.}
\end{align*}
Since $M^- = M^+ = 1$, $L$ exists and $L = 1$.


问:
Pattern with Square Numbers
I have noticed two patterns with square numbers.

$1^2\equiv 1\pmod{10}$

$2^2\equiv 4\pmod{10}$

$3^2\equiv 9\pmod{10}$

$4^2\equiv 6\pmod{10}$

$5^2\equiv 5\pmod{10}$

$6^2\equiv 6\pmod{10}$

$7^2\equiv 9\pmod{10}$

$8^2\equiv 4\pmod{10}$

$9^2\equiv 1\pmod{10}$

This set of numbers ${1, 4, 9, 6, 5, 6, 4, 1}$ is palindromic (meaning it is the same backwards and forwards) and it rotates around $5$. In fact, it rotates around all odd multiples of $5$. And the set of ones digits of squares $6, 9, 4, 1, 0, 1, 4, 9, 6$ rotates around all even multiples of $5$.

That is the first pattern. If $|x-5|=|y-5|,$ then $x^2\equiv{y^2}\pmod{10}$.

The second pattern is similar, except it is bigger. If $|x-25|=|y-25|,$ then $x^2\equiv{y^2}\pmod{100}$.

My questions are: Why do these patterns work? Are the reasons related? And does this pattern appear with any other numbers?

答:
The single digit symmetry can be understood from the fact that $(-1)^2=1$.  $8 \equiv -2 \pmod {10},$ so $8^2 \equiv (-2)^2 \equiv 2^2 \pmod {10}$ and similarly for the others.  The symmetry around $0$ will be true in any base.  The symmetry around the midpoint will be true in any even base because $(5-a)^2=25-10a+a^2, (5+a)^2=25+10a+a^2$ and $+10a \equiv -10a \equiv 0 \pmod {10}$.

The previous paragraph will prove the two digit case is symmetric around $50$, but is not sufficient to prove it is symmetric around $25$.  For that, write $(25+a)^2=625+50a+a^2, (25-a)^2=625-50a+a^2$ and $50a \equiv -50a \pmod {100}$.  This pattern will work for any base that is a multiple of $4$.

Title: Pattern with Square Numbers


问:
The set of real numbers is a subset of the set of complex numbers?
So, I was taught that $\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}$.

But since the set of complex numbers is by definition $$\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},$$ doesn't this mean $\mathbb{R}\subseteq\mathbb{C}$, since for each $x \in \mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$?

答:
Strictly speaking (from a set-theoretic view point), $\mathbb{R} \not \subset \mathbb{C}$. However, $\mathbb{C}$ comes with a canonical embedding of $\mathbb{R}$ and in this sense, you can treat $\mathbb{R}$ as a subset of $\mathbb{C}$.

On the same footing, $\mathbb{N} \not \subset \mathbb{Z} \not \subset \mathbb{Q} \not \subset \mathbb{R}$. However, there is an embedding of $\mathbb{N}$ in $\mathbb{Z}$, and similarly an embedding of $\mathbb{Z}$ in $\mathbb{Q}$ and an embedding of $\mathbb{Q}$ in $\mathbb{R}$.

You may want to look at this post for more details.


问:
Sublevel set is an absorbing convex set

Let $|| \cdot ||$ be a sublinear functional on a linear vector space $X$. Then the sublevel set $$K=\{ x \in K: ||x|| \leq 1\}$$ is an absorbing convex set and $0 \in K$.

What I've got so far:

It is easy to see that a set $K$ is convex since it contains balls $||x||\leq 1$ that are convex as well.

Definition: A subset $K$ of a linear vector space $X$ is called absorbing if $$X= \bigcup_{t\geq 0} tK$$ where $tK=\{tx:x\in K\}$.

Any help would be greatly appreciated.

答:
$K$ is absorbing if for every $x\in X$ you have a $t>0$ and an $a\in K$ with $t\cdot a= x$. So look at some $x\in X$ and try to find the relevant $t$ and $a$.

First take care of the case $\|x\|=0$. If this is the case then clearly $\|x\|≤1$ and then $x\in K$, so take $a= x$ and $t=1$.

Whats left is $\|x\|>0$, in this case note by sub-linearity that $\left\|\frac x{\|x\|}\right\|=1$ and hence $\frac{x}{\|x\|}\in K$, let this be your $"a"$, now choose $t=\|x\|$, then:
$$t\cdot a= \|x\|\cdot \frac{x}{\|x\|}=x$$
and the condition is verified also in the case that $\|x\|\neq 0$, hence $K$ is absorbing.

Title: Sublevel set is an absorbing convex set


问:
Proving covering relation for a product of ordered sets
Prove that if L and M are two ordered sets, then $(a_2, b_2)$ covers $(a_1, b_1)$ in $L × M$ iff $(a_1 = a_2$ and $b_2$ cover $b_1$ ) or ($a_2$ cover $a_1$ and $b_1 = b_2$).

答:
For the direct implication, take $(a_1,b_1) \prec (a_2,b_2)$.

It follows that $a_1 \leq a_2$ and $b_1 \leq b_2$.

If $a_1 < a_2$ and $b_1 < b_2$, then $(a_1,b_1) < (a_2,b_1) < (a_2,b_2)$, a contradiction.

Hence $a_1 = a_2$ or $b_1 = b_2$.

If $a_1 = a_2$ and $b_1 < c < b_2$, then $(a_1,b_1) < (a_1,c) < (a_2,b_2)$, a contradiction.
Hence $b_1 \prec b_2$.

Analogously if $b_1 = b_2$ then $a_1 \prec a_2$.

Now for the converse, if $b_1 \prec b_2$, and $(a,b_1) \leq (c,d) \leq (a,b_2)$, then
$$a \leq c \leq a,$$
whence $a=c$, and
$$b_1 \leq d \leq b_2,$$
whence $d=b_1$ or $d=b_2$.
It follows that $(a,b_1)\prec(a,b_2)$.

Analogously, if $a_1 \prec a_2$, then $(a_1,b_1) \prec (a_2,b_2)$.


问:
Probability - distribute people in a bus
(a) In how many ways can 6 people be lined up to get on a bus?

       Answer = 6!

(b) If 3 speciﬁc persons, among 6, insist on following each other, how many ways are possible?
 Answer = 4! * 3!

(c) If 2 speciﬁc persons, among 6, refuse to follow each other, how many ways are possible ?

Answer = 6! - 5! * 2

I can't figure out how we got the answer of part (c). May you explain how to think in order to solve such question

答:
(a) In how many ways can 6 people be lined up to get on a bus?
Answer = 6P6 = 6! = 720 ways

(b) If 3 speciﬁc persons, among 6, insist on following each other, how many ways are possible?
Answer = 4P4* 3P3 = 4! * 3!= 144 ways

(c) If 2 speciﬁc persons, among 6, refuse to follow each other, how many ways are possible ?
Answer = 6P6 - 5P5 * 2P2 = 6! - 5! * 2 = 480 ways


问:
Is $H$ a subgroup of $\Gamma$?
The question is in the picture, I understand the requirements for a subset to be a subgroup (closure, associativity under the same operation etc.), however I am struggling with this question. I would be grateful if someone could walk me through this. Thanks in advance.

1.3. Let $\Gamma$ denote the set of all mappings $f:\mathbb R\to\mathbb R$ such that $f(x)\ne 0$ for all $x\in\mathbb R$. For $f,g\in\Gamma$, define $fg$ by $$fg(x)=f(x)g(x)$$ for all $x\in\mathbb R$. You may assume that the set $\Gamma$ forms a group with respect to the above composition.

Let $H$ be the subset of $\Gamma$ defined to be the set of all mappings $f:\mathbb R\to\mathbb Z\setminus \{0\}$. Is $H$ a subgroup of $\Gamma$? Remember to justify your answer fully.

答: The constant mapping $h_2$ equal to $2$ belongs to $H$. Its inverse (namely the constant mapping equal to $1/2$) doesn't belong to $H$ though. So $H$ isn't a subgroup of $\Gamma$.

Title: Is $H$ a subgroup of $\Gamma$?


答:
You need to account for symmetry, because, recall, the square root function returns only the positive root. So you actually found the area under the curve strictly within the first quadrant. That is you found the area under only $x = +\frac{1}{\sqrt y}$

But actually, $x = \pm \frac{1}{\sqrt y}$, so we need $2 \times$ the area you computed. Using just $x = +\frac{1}{\sqrt y}$ is fine, but then double the area computed.

Title: Using area=$\int_{y = a}^{y = b} x \, dy $ find the following shaded area:

Body:
So I did:

$$\eqalign{
  & y = {1 \over x^2}  \cr
  & {x^2} = {1 \over y}  \cr
  & x = y^{ - {1 \over 2}}  \cr
  & \int_a^b x \, dy  = \left[ 2{y^{1 \over 2}} \right]_{1 \over 4}^1  \cr
  &  = 2\sqrt 1 - 2\sqrt {1 \over 4} \cr
  &  = 2 - 1  \cr
  &  = 1 \cr} $$

The answer however is 2, where have I gone wrong and misunderstood the question?

Thank you!


问:
Bounded, well-defined linear operators on complex matrices in $l^2$ - necessary conditions for the matrix.

Suppose $A = [a_{ij}]_{i,j=1,2,\dots}$ is a matrix of complex numbers and lets define an operator $(T_{A}x)_i = \sum_{j=1}^{\infty}a_{ij}x_j$ for $i = 1,2,\dots$ and $x = (x_j)_{j\geq1} \in l^2$. What rules must A obey for T to be a well defined bounded operator on $l^2$?

Well, this was asked in $l^1$, here but with assumption of well-defined operator and here in a slight another context but none of them gave me a straight answer.

I see that the operator basically takes the sequence $x \in l^2$ and creates a new sequence where i-th element is made from a "dot product" of i-th row and the sequence.

For a $x \in l^2$
$$
\|Tx\|_2 = \sqrt{\sum_{i=1}^{\infty}\left(\sum_{j=1}^{\infty}a_{ij}x{j}\right)^2} \leq \sqrt{\sum_{i=1}^{\infty}\left(\sum_{j=1}^{\infty}a_{ij}^2\right)\left(\sum_{j=1}^{\infty}x_{j}^2\right)} \leq
$$
by Cauchy Schwarz. Then:
$$
\leq \sum_{i=1}^{\infty}\left(\sqrt{\sum_{j=1}^{\infty}a_{ij}^2}\sqrt{\sum_{j=1}^{\infty}x_{j}^2}\right) = \sum_{i=1}^{\infty}\sqrt{\sum_{j=1}^{\infty}a_{ij}^2}\|x\|_2 \leq \\
\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}|a_{ij}|\|x\|_2
$$
since $\|x\|_2 < \infty$ the condition is that modules of entire  matrix must be summable (i.e. $\forall_i \sum_{i,j=1}^\infty|a_{ij}| < \infty$).

Is that correct? How do we prove that the operator is well defined or what are the necessary matrix conditions?

答:
For well-definedness of $T_A$, it is necessary that $(a_{ij})_{j\ge1}\in\ell^\infty$ for all $i$ because $\lim_{j\to\infty}a_{ij}x_j$ must vanish for each $i$ for the series to converge. Now, let $\alpha_i:=(a_{ij})_{j\ge1}\in\ell^\infty$ and let $\langle\cdot,\cdot\rangle\colon\ell^2\times\ell^\infty\to\mathbb{C}$ be a partial pairing defined as $\langle x,y\rangle=\sum_jx_jy_j :=\lim_{n\to\infty}\sum_{j=1}^nx_jy_j $, whenever the limit exists. Note that $T_A(x)=(\langle x\,,\alpha_i\rangle)_{i\ge1}$. Now, we have the following:

$\alpha_i\in\ell^1$ for all $i$ is sufficient for $T_A$ to be well-defined (and for continuity, $(\|\alpha_i\|_2)_{i\ge 1}\in\ell^2$ suffices.

Indeed, if $\alpha_i\in\ell^1$, then by the Triangle Inequality, $$|(T_Ax)_i|=|\langle x\,,\alpha_i\rangle|\le\|x\|_\infty\|\alpha_i\|_1<\infty$$ and hence $T_A$ is well-defined once $\alpha_i\in\ell^1$ for each $i$. For the continuity, this follows from Cauchy-Schwarz Inequality, that is
$$\|T_Ax-T_Ay\|_2^2=\sum_i|\langle x-y\,,\alpha_i\rangle|^2\le\|x-y\|_2^2\sum_i\|\alpha_i\|_2^2\,.$$

$\alpha_i\in\ell^2$ for all $i$ is necessary for $T_A$ to be well-defined and continuous.

To see this, observe that functional $$f_i\colon\ell^2\to\mathbb{C}\,,\,~\,~\,~x\mapsto(T_A(x))_i$$ is linear and continuous; indeed, if $x_n\to x$ then $$\lim_{n\to\infty}f_i(x_n)= \lim_{n\to\infty}(T_A(x_n))_i=(T_A(x))_i=f_i(x)\,,$$ thereby establishing continuity.
Linearity follows equally, that is for scalars $\beta,\gamma$, we have
$$f_i(\beta x+\gamma y)=(T_A(\beta x+\gamma y))_i= \beta (T_A(x))_i+\gamma(T_A(y))_i= \beta f_i(x)+\gamma f_i(y)\,.$$
Since $(\ell^2)^*=\ell^2$, it follows that $ f_i(\cdot)=\sum_ja_{ij}(\cdot)\in\ell^2$; that is, $\alpha_i\in\ell^2$.


问:
In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a point on $AB$ and $D$ is the midpoint of $OB$ . Find $\angle BCD$ .

In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a  point on $AB$ and $D$ is the midpoint of $OB$ . The circle centered at $O$ with radius $OD$ intersects $AC$ at $T$. Find $\angle BCD$ .

What I Tried: Here is a picture :-

You can see what I tried. The first part is to do angle-chasing and find the respective angles in the picture. I also find that $OT \parallel BC$, since $OBCT$ is a trapezium . Now I think that to find $\angle BCD$ , somehow I have to use the fact that $OD = DB$ (which I have not used) and maybe the parallel lines, but I seem to be stuck here.

Can anyone help me figure this out? Thank You.

答:
Construct $TD$ and a line parallel to $OT$ passing through $D$.

By Intercept theorem, $TO// GD//CB$ and $OD = DB$ implies $TG = GC$.

Since $GD$ is common side and $TC \perp GD$, $\triangle TGD \cong \triangle CGD$ via SAS. This gives $\angle TDG = \angle CDG$.

Now $\angle TDO = \frac12 \angle TOA$ by circle properties and $\angle GDA = \angle TOA$ by parallel lines, giving $$\angle TDO = \angle TDG = \angle CDG = \frac12 \angle TOA$$ This gives $$\angle CDO = \frac32 \angle TOA$$

Finally by considering the exterior angle of $\triangle BCD$, $$\angle BCD + \angle CBD = \angle CDO = \frac32 \angle TOA$$
but $\angle CBA = \angle TOA$, giving $\angle BCD = \frac12 \angle TOA$.

In this case, it is equal to $\frac12(70^\circ) = 35^\circ$.


问:
Help Implicit funtions theorem question
Let

\begin{cases}
u_t(x,t) + u(x,t)u_x(x,t) = 0,  & \text{for } x\in\mathbb{R},\; t > 0 \\
u(x,0) = h(x), & \text{for } x\in\mathbb{R}.
\end{cases}

Suppose that $h\in \mathcal C^2 (\mathbb{R})$ and that $h' (x) > 0$ for all $x\in \mathbb{R}$. Check, with the help of implicit function theorem, for $(x, t)$ sufficiently close to $(x_0, 0)$  (where
$x_0\in\mathbb{R}$ is a fixed number), that the equation
$u = h(x - tu)$
defines a solution to the Cauchy problem (for $x$ sufficiently close to $x_0$ and $t$ close to zero).

答:
Let
$$
F=F(x,t,u)=u-h(x-tu)
$$
Then $F: \mathbb R^2\times\mathbb R^1\to\mathbb R^1$ is $C^2$, and $F_u\in C^1$ and
$$
F_u=1+th'(x+tu)>0\quad
$$
is $x$ $t$ near $0$ and $x$ near $x_0$. Hence, IFT provides the existence of $u(x,t)\in C^1$,
for an open set of $(x_0,0)$.

Note. If fact, this is Burger's equation, and if $h'>0$, then this IVP possesses a unique global solution ($t\ge 0$). But if $h'<0$, then it develops shocks in finite time.

Title: Help Implicit funtions theorem question


问:
Show that $T^{-1}(M)+T^{-1}(N) \leq T^{-1}(M+N)$ for $T:V \to W$ where $M,N \leq W$

Let $V$ and $W$ be vector spaces over the same field $\mathbb{F}$. Let $T:V \to W$ be a linear transformation. Denote the preimage of $Y \leq W$ under $T$ by $T^{-1}(Y)$. Suppose that $M,N \leq W$. Show that $T^{-1}(M)+T^{-1}(N) \leq T^{-1}(M+N)$.

 What I have so far is the following:

 Let $x \in T^{-1}(M)$ and $y \in T^{-1}(M)$. Then $x,y \in V$ and $T(x) \in M$ and $T(y) \in N$. Then $T(x) +T(y) \in M+N$, so that $x,y \in T^{-1}(M+N)$. So $T^{-1}(M)+T^{-1}(N) \subseteq T^{-1}(M+N)$.

 Where I'm struggling is showing that $T^{-1}(M)+T^{-1}(N)$ is a subspace of $T^{-1}(M+N)$.

 Do I fist need to show that $T^{-1}(M+N)$ is a subspace of $V$, then since $T^{-1}(M)+T^{-1}(N) \subseteq T^{-1}(M+N)$, then $T^{-1}(M)+T^{-1}(N) \leq T^{-1}(M+N)$ ?

答:
$T^{-1}(M)+T^{-1}(N)$ is the sum of two linear subspaces of $V$ and is therefore a linear subspace of $V$.

Now if $a + b \in T^{-1}(M)+T^{-1}(N)$ where $T(a) \in M$ and $T(b) \in N$, you have $T(a)+T(b) = T(a+b) \in T(M+N)$. Therefore $a+b \in T^{-1}(M+N)$ proving the desired inclusion.


问:
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
The question is:

When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?

Edit

I should add that $f:\mathbb R\to\mathbb R$ to the question.

答:
Partial answer. If $f(x)$ is the a polinomial function say $f(x)=\sum_{k=0}^na_kx^k$ then
$$
\sum_{k=0}^na_k(x+1)^k-\sum_{k=0}^na_kx^k=\sum_{k=0}^nka_kx^{k-1}
$$
 For $x=1$ we have
$$
\sum_{k=0}^na_k2^k-\sum_{k=0}^na_k=\sum_{k=0}^nka_k
$$
If $n=2$  then
$
a_0+2a_1+4a_2-a_0-a_1+a_2=a_1+2a_2\implies a_2=0.
$
In general case for all $n>1$ we have for any no root $x_0$ of $f(x)$ or $p(x)= (x+1)^n-x^n$ that
$$
\sum_{k=0}^na_k(x_0+1)^k-\sum_{k=0}^na_kx_0^k-\sum_{k=0}^nka_kx^{k-1}=0 \implies [(x_0+1)^n-x_0^n]a_n=0\implies a_n=0
$$

Soon the solution can not be a polynomial of degree greater than 1.


问:
Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)?
Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is
$$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$
I am wondering why this is non-negative?

答:
(Not a complete answer but hopefully close enough)

You want to show that
$$ x \log(x)+ (1-x)\log(1-x) \le - \log(x) \log(1-x)$$
Divide on both sides by $\log(x) \log(1-x)$ and this is
$$ \frac{x}{\log(1-x)} + \frac{1-x}{\log(x)} \le -1 $$

This is of the form
$$ f(x) + f(1-x) \le c$$
where $f(x) = \frac{x}{\log(1-x)}$. I don't know if there is a nice way to show this.

But the not-so-nice way is to Taylor expand
$$ \frac{x}{\log(1-x)} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dotsm $$
$$ \frac{1-x}{\log(x)} = -1 + \frac{1-x}{2} + \frac{(1-x)^2}{12} + \frac{(1-x)^3}{24} + \dotsm $$

Now you add them up, you see that the RHS becomes
$$ -2 + \frac{1}{2} + \frac{x^2 + (1-x)^2}{12} + \frac{x^3 + (1-x)^3}{24} + \dotsm $$
$$ \le -2 + 1 + \frac{1}{12} + \frac{1}{24} + \dotsm $$

But then showing that the last sum is $\le -1$, seems clear to me but maybe you might want more concrete justification...


问:
Is there a uniform way to define an upper bound on $|X|$ union of $|X|$-sized sets in ZF?
Working in ZF (without choice); Let $A \times B$ denote, as usual, the cartesian product of sets $A$ and $B$. Let $|X|$ denote the cardinality of $X$ defined after Scott's. Lets define: $$|X|^2= | X \times X |$$

Does ZF prove that any set union of $|X|$ many sets each of size $|X|$, is subnumerous to $|X|^2$?

If that fails then is there a definable function on $X$ such that every union of $|X|$ many |X|-sized sets is subnumerous to? Could that be for example $|\mathcal P(X)|^2$?

答:
No, not even remotely, not even ever.

$\sf ZF$ does not prove that a countable union of countable sets is countable, so for $X=\omega$ your question is already proved false. Moreover, the results of Morris and many others show that $\sf ZF$ cannot even place a bound on cardinalities of countable unions of countable sets.

In other words, it is consistent with $\sf ZF$ that for every set $X$, there is a set $Y$ which is a countable union of countable sets and $|Y|\nleq|X|$. Specifically, Morris proved that it is consistent:

For every ordinal $\alpha$, there is a set $A_\alpha$ which is a countable union of countable sets, but $\mathcal P(A_\alpha)$ can be mapped onto $\omega_\alpha$.

Recently, I've extended said result to replace $\omega_\alpha$ by $V_\alpha$.

This style of failure of choice can be extended in many different ways, all of which are very terrible.


答: You can use the subgroup criterion. We just need to check whether $e\in H$, and whether $x,y\in H\implies xy^{-1}\in H$.

Title: Is $H$ a subgroup of $\Gamma$?

Body:
The question is in the picture, I understand the requirements for a subset to be a subgroup (closure, associativity under the same operation etc.), however I am struggling with this question. I would be grateful if someone could walk me through this. Thanks in advance.

1.3. Let $\Gamma$ denote the set of all mappings $f:\mathbb R\to\mathbb R$ such that $f(x)\ne 0$ for all $x\in\mathbb R$. For $f,g\in\Gamma$, define $fg$ by $$fg(x)=f(x)g(x)$$ for all $x\in\mathbb R$. You may assume that the set $\Gamma$ forms a group with respect to the above composition.

Let $H$ be the subset of $\Gamma$ defined to be the set of all mappings $f:\mathbb R\to\mathbb Z\setminus \{0\}$. Is $H$ a subgroup of $\Gamma$? Remember to justify your answer fully.


问:
Existence of rainbow paths in graphs?
Let $G$ be a graph, and let $c: V(G) \to \{1,\dots,k\}$ be a proper $k$-colouring of $G$. We say that a path $v_1 \dots v_k$ in $G$ is a rainbow path of this colouring if $c(v_i) = i$ for every $i \in \{1,\dots,k\} $. Now suppose that $k = \chi(G)$, the chromatic number of $G$; I am trying to determine whether or not a rainbow path exists for this colouring.

I don't really know where to start, though. If $k = 2$, then a rainbow path certainly exists; for the case $k = 3$, I've tried looking at cycles of odd length and wheels with an odd number of spokes and it seems that such graphs must always have rainbow paths as well, but I'm not sure how to generalize to general graphs with $\chi(G) = 3$ or to higher values of $k$.

Any help is appreciated!

答:
The answer is yes.

We may without loss of generality assume that $G$ is connected; otherwise we throw away each component with lower chromatic number, and consider every other component individually.

For $k=3$: Let $E_1$ be the set of all vertices with the color $1$. Let $E_2$ be the set of all vertices adjacent to a vertex in $E_1$ with the color $2$. If there is no such vertex, then all vertices in $E_1$ could be recolored to the color $2$, and we have a $(k-1)$-coloring of $G$; contradiction. Let $E_3$ be the set of all vertices adjacent to a vertex in $E_2$ with the color $3$. If there is no such vertex, then we could recolor $E_1$ to the color $2$ and $E_2$ to the color 3, and obtain a 2-colorable graph; contradiction. Therefore $E_1,E_2,E_3$ are nonempty, so we can find a rainbow path $e_1,e_2,e_3$ where $e_i\in E_i$.

This argument can be carried on for general $k$: Let us continue. Let $E_4$ be the set of vertices adjacent to $E_3$ which have the color $4$. If it were empty, then we could recolor $E_1,E_2,E_3$ to the colors $2,3,4$, respectively, and obtain a $(k-1)$-coloring of $G$; contradiction. We can define $E_5,\ldots,E_k$ analogously. When we are done, we can choose a rainbow path $e_1,\ldots,e_k$ where $e_i\in E_i$ for each $i$.

Title: Existence of rainbow paths in graphs?


问:
is $\omega$ in every model of ZFC always isomorphic to the standard natural numbers
ZFC can have some very odd models one of them being countable. Is it true that in any model of ZFC  $\omega$ with the empty set as zero the successor function $S(x):= x\cup\{x\}$ and the addition and moltiplication functions defined by recursion is isomorphic to the "real" natural numbers. I know that Second order arithmetic with quantifiers over all the subsets of it's domain is categorical but from my understanding in some ZFC models not all of the subsets of $\omega$ are sets

答: No (assuming ZFC is consistent).  This is essentially immediate from the compactness theorem.  For instance, add uncountably many constant symbols $c_i$ to the language of ZFC, and add axioms saying that $c_i\neq c_j$ for each $i\neq j$ and $c_i\in\omega$ for each $i$.  Every finite subset of the resulting theory is consistent, since it will only involve finitely many of the $c_i$ and so we can send those $c_i$ to distinct natural numbers in some model of ZFC and the rest to $0$.  So by compactness, this theory has a model $M$.  Then $M$ is a model of ZFC such that $\omega^M$ has (externally) uncountably many elements.


问:
$(I-B)^{-1} (I+B)$ always orthogonal if $B$ anti-symmetric
I saw this claim in Strang Intro to linear algebra 5th edition, page 336, and am finding it surprisingly difficult to prove.

The claim is that $A = (I-B)^{-1} (I+B)$ is orthogonal if $B^T = -B$.

We have $A^T = (I + B^T) (I - B^T)^{-1} = (I - B)(I + B)^{-1}$. Multiplying, we have

$$ A^T A = (I - B)(I + B)^{-1} (I-B)^{-1} (I+B).$$

Since the claim is that $A$ is orthogonal, this is supposed to be $I$, but it is unclear how to proceed from here. Do $(I + B)^{-1}$ and $(I-B)^{-1}$ commute? If yes, then the result follows trivially. However, I don't see any reason why they would commute.

It's certainly easy to show that $\det A = 1$, since $\det (I - B) = \det (I + B^T) = \det (I + B)$ and thus $\det A = \det (I + B) / \det (I - B) = 1$. However, I think that's necessary but not sufficient for $A$ to be orthogonal.

答:
$(I+B)^{-1}$ and $(I-B)^{-1}$ indeed commute (and thus the result follows): we have

$$(I+B)^{-1} (I-B)^{-1} = [(I-B)(I+B)]^{-1} = (I - B^2)^{-1}.$$

The result is clearly invariant under sign flip.


问:
why $\sigma$ finite is necessary to show that $f=g$ a.e.?
Let $(X, \mathcal{B}, m)$ be the measure space,show that if the measure is $\sigma$ finite,we have for any $f, g: X \to [0,+\infty]$

We have $$\int_X 1_E f\ dm = \int_X 1_E g\ dm $$

for any $E\in \mathcal{B}$ if and only if  $f = g$ a.e.

We can prove by this if $f,g\in L^1$ then the result above holds for $f$ and $g$ since we can subtract $\int_X 1_E g\ dm$ to the otherside.For $\sigma$ finite case just restrict the measure on each pieces of the partition and we know one each pices the result holds so globally the result holds.

The question is why $\sigma$ finite is necessary here?

If taking $f = \chi_{x}$ and $g = 2\chi_{x}$ then we can construct the measure that $\delta({x}) = \infty$ and $X=\{x\}$ then $\int_X 1_E f\ d\delta = \int_X 1_E gd\delta = \infty$ correct?

答:
I assume that $f$ and $g$ shall both be measurable, so that the integrals are well-defined.

$\sigma$-finiteness is a stronger property than necessary to have the conclusion. As your example shows, some assumption has to be made about $m$. The necessary and sufficient assumption is that $m$ be semi-finite, i.e. that for every $A \in \mathcal{B}$ with $m(A) = \infty$ there is an $F \in \mathcal{B}$ with $F \subset A$ and $0 < m(F) < \infty$. Informally, $m$ shall have no atoms of infinite measure.

The necessity is essentially shown by your example, of course we need to adapt it to the general situation. If $m$ is not semi-finite, then there is $A \in \mathcal{B}$ with $m(A) = \infty$ such that for every measurable $B \subset A$ either $m(B) = 0$ or $m(B) = \infty$. Then taking $f = \chi_A$ and $g = 2\cdot \chi_A$ works. We have
$$\int f\chi_E\,dm = \int g\chi_E\,dm = 0$$
if $m(E\cap A) = 0$, and both integrals are $\infty$ if $m(E\cap A) = \infty$.

To see sufficiency, suppose $f$ and $g$ differ on a set of nonzero measure. Without loss of generality suppose $A = \{x : f(x) > g(x)\}$ has positive measure. If $A' = \{ x \in A : f(x) < \infty\}$ has positive measure, then there is an $n \in \mathbb{N}$ such that $A'' = \{x \in A' : g(x) < n,\, 0 < f(x) - g(x) < n\}$ has positive measure, and by semi-finiteness there is a measurable $B \subset A''$ with $0 < m(B) < \infty$. Then
$$\int_B f\chi_B\,dm > \int g\chi_B\,dm\,.$$
If $m(A') = 0$, then for some $n$ the set $A''' = \{x \in A : g(x) < n\}$ has positive measure and there is $B \in \mathcal{B}$, $B \subset A'''$ with $0 < m(B) < \infty$. Again $\int g\chi_B\,dm < \int f\chi_B\,dm$.

The converse direction, $f = g$ a.e. implies $\int f\chi_E\,dm = \int g\chi_E\,dm$ for all $E \in \mathcal{B}$, holds without any assumptions on the measure.

The reason that $\sigma$-finiteness was given as the (sufficient) assumption is probably that $\sigma$-finiteness is a better-known property, and one mostly cares about $\sigma$-finite measures anyway, thus in practice not much is lost by choosing the stronger assumption.


问:
Normal bundle and blow-up
Let us work over the complex projective space: consider a smooth variety $X$ and a subvariety $Y$. I learnt that, if we do the blow-up of $X$ with center $Y$, we obtain a new variety $\tilde{X}$, along with a map $\pi: \tilde{X}\to X$, which is an isomorphism outside the exceptional locus, that is $Y$.

Altough I don't have a precise reference, it's been told me that the exceptional divisor of $Y$, that is the inverse image $\pi^{-1}(Y)$, coincides with the projective bundle of the normal bundle, that is,

$$\tilde{Y}=\pi^{-1}(Y)\simeq \mathbb{P}(\mathcal{N}_{Y\mid X}^\vee)=(\mathcal{N}_{Y\mid X}\setminus Y)/\sim,$$

where $\sim$ is the standard action of $\mathbb{C}$.

Questions:

What is a good reference of this construction? I know it is the content of Theorem II.8.24 of Hartshorne's Algebraic geometry, but without a knowledge of scheme theory (and proj construction, and coherent sheaves) it is a bit difficult, so maybe there's a more accessible text;

In pp. 86-87 of these notes ( https://www.math.ens.fr/~debarre/M2.pdf ), we start with a rational curve $\Gamma^+$ in $X^+$ with normal bundle $\mathcal{O}(-1)\oplus\mathcal{O}(-2)$: then the authors does the blow-up along $\Gamma^+$, and it claims that the exceptional divisor is $$S^+_1=\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(1))$$
but using the above formulas should be $\mathbb{P}(\mathcal{O}(1)\oplus \mathcal{O}(2))$: what am I missing?

答:
Your statement is missing a hypothesis: we also need $Y$ to be nonsingular. Anyway, an alternate reference is section 22.3 of Vakil's "The Rising Sea: Foundations of Algebraic Geometry", which is generally considered a more accessible reference than Hartshorne. Vakil doesn't give a proof of the statement you want—it's Exercise 22.3.D—but the leadup to the exercise may give some useful guidance on how to prove it.

Unfortunately, I'm not aware of a reference for this statement that doesn't use the language of schemes, the Proj construction, and quasicoherent sheaves at least to some extent.

To answer your second question, the two projective bundles are isomorphic, because projective bundles don't change if you tensor by a line bundle. This is exercise 17.2.G in Vakil's notes, and Lemma II.7.9 in Hartshorne. Another reference for the general statement is Stacks Project, tag 02NB. In particular, for all $c \in \mathbb{Z}$, we have
$$\mathbb{P}(\mathcal{O}(a) \oplus \mathcal{O}(b)) \cong \mathbb{P}(\mathcal{O}(a + c) \oplus \mathcal{O}(b + c)).$$

Title: Normal bundle and blow-up


问:
Homeomorphism between $\mathbb{R}^2$ and the open unit disc
Given the function $f(x,y)=(\frac{x}{1+\sqrt{x^2+y^2}},\frac{y}{1+\sqrt{x^2+y^2}})$, I have to prove $f$ is a homemorphism between $\mathbb{R}^2$ and the open unit disc. Proving this function is conitunuous is trivial, proving it's a bijection is a bit harder (I wasn't able to find it's inverse). I did find the same function in a different notation - $f(z)=\frac{z}{1+\|z\|}$ where $z\in\mathbb{R}^2$, but since it's my first time dealing with two variable functions and homemorphisms, I can't see why $f(z)=f(x,y)$, other than $\|z\|=\|(x,y)\|$.
What is the best way to approach this with given notation ($f(x,y)$) - more generally how can I prove this function is a bijection and find its inverse.

(I was able to show that $lim_{(x,y)\rightarrow(\infty,\infty)}f(x,y)$ is (1,1) which I guess shows its a surjection but otherwise had no idea).

答:
It is easy to find the inverse in complex notation---
\begin{align}
&f(z) = \dfrac{z}{1+||z||}\\
\implies &||f(z)|| = \dfrac{||z||}{1+||z||}\\
\implies &||z|| = \dfrac{||f(z)||}{1-||f(z)||}\\
\implies &z = {f(z)}\times\left({1+\dfrac{||f(z)||}{1-||f(z)||}}\right)=\dfrac{f(z)}{1-||f(z)||}\\
\end{align}
So the inverse function is
\begin{equation}
f^{-1}(z) = \dfrac{z}{1-||z||}
\end{equation}
Now you can do whatever you wish with this. Note that proving this function  continuous is also trivial when $(x,y)$ takes values in open unit disk. Now just find the domain and range of the two functions and you're done.


问:
Convergence of the improper integral $\int_0^2\frac x{(16-x^4)^{1/3}}\,\mathrm dx$
Need your help in the checking convergence of this improper integral :$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}$$I tried my luck and somehow was able to establish a relationship between  $(1-t^4)$  and $t^4$ i.e $$(1-t^4)\geq t^4$$
$$\frac{1}{1-t^4} \leq \frac{1}{t^4}$$
$$\Rightarrow \frac{1}{(1-t^4)\frac{1}{3}} \leq \frac{1}{t^\frac{4}{3}}$$ $$\Rightarrow \frac{t}{(1-t^4)\frac{1}{3}} \leq \frac{t}{t^\frac{4}{3}}$$
for some $$t\in\left[0,\frac{1}{2^\frac{1}{4}}\right]$$ It can be shown that$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}=2^\frac{2}{3}\int_0^1\frac{tdt}{(1-t^4)^\frac{1}3} $$
Therefore ,
$$F(x) =\int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{(1-t^4)^\frac{1}3} \leq \int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{t^\frac{4}{3}} = g(x)$$ For this particular interval I was able to prove its convergence as $g(x)$ converges but I am unable to comment on the remaining interval. Please help and also provide some insight which might help me in future. Thanks in advance...

答:
The problem with your approach, as you observed, is that the inequality $1-t^4\ge t^4$ doesn't hold near $t=1$, which is where the impropriety in the integral occurs. Instead it's better to argue along the following lines:

For $0\le x\le2$, we have

$$16-x^4=(2-x)(8+4x+2x^2+x^3)\ge8(2-x)$$

and thus

$${x\over(16-x^4)^{1/3}}\le{2\over2(2-x)^{1/3}}=(2-x)^{-1/3}$$

so that

$$\int_0^2{x\over(16-x^4)^{1/3}}\,dx\le\int_0^2(2-x)^{-1/3}\,dx=-{3\over2}(2-x)^{2/3}\big|_0^2={3\over2^{1/3}}$$


Title: $(I-B)^{-1} (I+B)$ always orthogonal if $B$ anti-symmetric

Body:
I saw this claim in Strang Intro to linear algebra 5th edition, page 336, and am finding it surprisingly difficult to prove.

The claim is that $A = (I-B)^{-1} (I+B)$ is orthogonal if $B^T = -B$.

We have $A^T = (I + B^T) (I - B^T)^{-1} = (I - B)(I + B)^{-1}$. Multiplying, we have

$$ A^T A = (I - B)(I + B)^{-1} (I-B)^{-1} (I+B).$$

Since the claim is that $A$ is orthogonal, this is supposed to be $I$, but it is unclear how to proceed from here. Do $(I + B)^{-1}$ and $(I-B)^{-1}$ commute? If yes, then the result follows trivially. However, I don't see any reason why they would commute.

It's certainly easy to show that $\det A = 1$, since $\det (I - B) = \det (I + B^T) = \det (I + B)$ and thus $\det A = \det (I + B) / \det (I - B) = 1$. However, I think that's necessary but not sufficient for $A$ to be orthogonal.

Body:
Observe that

$$(I-B)(I+B)=(I-B^2)=(I+B)(I-B)$$
and
$$(I-B)^{-1} (I+B)(I - B)(I + B)^{-1}= (I-B)^{-1} (I-B)(I+B)(I + B)^{-1}.$$


问:
How to simplify addition of quotients with variables
Recently I was given a problem where I had a quotient with variables and a regular quotient and had to simplify it, but had no clue how to do it. For example,
$$  \left|\frac{x-1}{x+3}-\frac{1}{5}\right|=\frac{4}{5}\left|\frac{x-2}{x+3}\right|$$

Can you please explain how this works?
I am having trouble understanding this, any help will be appreciated.
Original Image

答:
As Nick indicates in the comment, you do this with variables the exact same way you do it without variables. Remember that variables are just place-holders for numbers (or whatever else they are varying over). They behave just like those numbers.

There is one rule for mixing multiplication/division with addition/subtraction. That is the distributive law:

$$A(B + C) = AB + AC$$

To apply this to adding fractions, you need to rewrite them into a form where it applies. If you have $$\frac ab + \frac cd$$
you don't have a common multiplier. But, you can rewrite the fractions in a form where they do. Note that $\frac AA = 1$ and multiplying by $1$ does not change values. So you can multiply the first term by $\frac dd$ and the second term by $\frac bb$:
$$\begin{align}\frac ab + \frac cd &= \frac dd\frac ab + \frac bb\frac cd \\&= \frac{ad}{bd} + \frac{bc}{bd} \\&= \frac 1{bd}ad +\frac 1{bd}bc \\&= \frac{1}{bd}(ad + bc) \\&= \frac{ad+bc}{bd}\end{align}$$

In your example, $a = x-1, b = x+3, c = -1, d = 5$
$$\begin{align}\frac{x-1}{x+3}-\frac{1}{5} &= \frac 55 \frac{x-1}{x+3} + \frac{x+3}{x+3}\frac{-1}5\\
&=\frac{5x-5}{5x+15} + \frac{-x-3}{5x+15}\\
&=\frac 1{5x+15}\left((5x-5) + (-x - 3)\right)\\
&=\frac {4x-8}{5x+15}\end{align}$$

That combines the fractions, but then your calculation goes a step further, factoring the combined fraction:
$$\frac {4x-8}{5x+15} = \frac{4(x-2)}{5(x+3)} = \frac 45\frac{x-2}{x+3}$$

And finally also handles the absolute values:
$$\left|\frac{x-1}{x+3}-\frac15\right|=\left|\frac45\frac{x-2}{x+3}\right| = \frac45\left|\frac{x-2}{x+3}\right|$$

Of course, one normally does not write out the steps as I've done here - particularly in pulling out the denominators as a multiplication by the inverse. I choose to show it this way to tie it directly to the distributive law, so that you can see how this is justified.


问:
How to prove something is a subset using only the subspace test and abstract properties of dot products?
My question is if $v_0\in\Bbb R^n$ is a fixed vector and we define $W\subset\Bbb R^n$ to be the set of all vectors in $\Bbb R^n$ that are orthogonal to the particular vector $v_0$. In set builder notation it would be $W=\{w\in\Bbb R^n:w\cdot v_0=0\}$. (the dot represents the dot product). Then using the subspace test, and abstract properties of dot products how does one prove that $W$ is a subspace of $\Bbb R^n$?

Note: To use abstract properties means you should not use the formula for a dot product. Instead, use properties, such as the distributive property, in abstract form: a.(b+c) = a.b + a.c (there are probably more properties that you would need to use). Also, make sure you use the particular fixed vector v0 as you work through your proof.

答:
Well, check whether $w\in W$ and $c\in\Bbb R\implies cw\in W$.  Since $cw\cdot v_0=c(w\cdot v_0)=c\cdot0=0$, it's yes.

Secondly, if $w,v\in W$ do we have $w+v\in W$?  Well, $(w+v)\cdot v_0=w\cdot v_0+v\cdot v_0=0+0=0$.  So yes.

Thus it's a subspace.


答:
First, your definition of $W$ is not correct. You have

$$W = \{w \in \mathbb R^n \mid w \cdot v_0=0 \}$$ (you missed the $=0$ part).

Then the dot product is bilinear. So if $\lambda \in \mathbb R$ and $w, w_1,w_2 \in W$ with $ w_1 \cdot v_0= w_2 \cdot v_0=0$ you get

$$(w_1+w_2) \cdot v_0 = w_1 \cdot v_0 + w_2 \cdot v_0=0$$ and

$$(\lambda w) \cdot v_0 = \lambda (w \cdot v_0) = 0.$$

Proving that $W$ is a linear subspace.

Title: How to prove something is a subset using only the subspace test and abstract properties of dot products?

Body:
My question is if $v_0\in\Bbb R^n$ is a fixed vector and we define $W\subset\Bbb R^n$ to be the set of all vectors in $\Bbb R^n$ that are orthogonal to the particular vector $v_0$. In set builder notation it would be $W=\{w\in\Bbb R^n:w\cdot v_0=0\}$. (the dot represents the dot product). Then using the subspace test, and abstract properties of dot products how does one prove that $W$ is a subspace of $\Bbb R^n$?

Note: To use abstract properties means you should not use the formula for a dot product. Instead, use properties, such as the distributive property, in abstract form: a.(b+c) = a.b + a.c (there are probably more properties that you would need to use). Also, make sure you use the particular fixed vector v0 as you work through your proof.


问:
Orthogonal linear transformations and the dot product
I am trying to solve a problem on the dot product, and I do some manipulations and come to the conclusion that

$\langle x, A^{T} Ax \rangle = \langle x,x \rangle$. $x$ is a column vector with $n$ rows, and $A$ an $n \times n$ matrix.

I know in general that if $a \cdot b = a \cdot c$, then this does not mean that $b = c$. However, from the above concerning the matrix $A$, can I conclude that the only way the left and right hand sides are equal is when $A^{T}A = I$, the identity matrix? Otherwise how can the two quantities be equal?

If the statement were not true, I would be glad if someone could provide a counter example.

Thanks

答:
$\langle x, A^{T} Ax \rangle = \langle x,x \rangle$ implies $\langle x, (A^{T} A-I)x \rangle = 0$. Now it is true that $\langle x, Tx \rangle = 0$ for all $x$ implies $T=0$ because you can  write $4\langle x, Ty \rangle = \langle x+y, T(x+y) \rangle - \langle x-y, T(x-y) \rangle = 0$. (This is a polarization identity.) Taking $x=Ty$ implies $Ty=0$ for all $y$, and so $T=0$.

There's a related question, which you can use once you realize that $\langle x, A^{T} Ay \rangle$ defines an inner product.


问:
When are groups of order 12 non-abelian?
I am currently reading http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/group12.pdf, and have a quick question about the group being non-abelian. Let me explain:

Let $|G|=12=2^2\cdot 3$ and let $n_2,$ $n_3$ denote the number of Sylow $2,3$-subgroups respectively. By using Sylow's Theorems and counting elements we see that either $n_3=1$, or $n_3=4$ (and $n_2=1$). So we always have a normal Sylow $2$-subgroup or a normal Sylow $3$-subgroup.

On page 2 the author turns to the case $n_2\ne 1$ (so $n_2=3$, $n_3=1$) and says "Since $n_2\ne 1$, the group is non-abelian". Why is this true?

答: By the second Sylow theorem, any Sylow 2-subgroups are conjugates of each other.  But in an abelian group, every conjugate is the same subgroup again.</p>


问:
Show that the polynomial $x^{2}+1$ has no fixed prime divisor
An integer-valued polynomial $Q(x)$ has a fixed divisor $m$ if there is an integer $m>1$ such that
$\frac {Q(x)}{m}$ is also an integer-valued polynomial. See https://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H#Fixed_divisors. If $m$ is a prime then we use the term fixed prime divisor.

I am interested in this problem:

Show that the polynomial $x^{2}+1$ has no fixed prime divisor.

答:
If $p$ is the desired prime factor, just set $x=p$. $p$ is surely not a divisor of $p^2+1$, hence $p$ cannot be a prime factor of $x^2+1$ in general.

Or formulated in terms of the definition of "fixed divisor" : $\frac{x^2+1}{p}$ is not integer-valued as the case $x=p$ shows.

We do not even need that $p$ is prime. This holds for every integer $p>1$.


问:
Parametric equation of a curve
What is the curve that has a parametric equation as, $(a\cos^{4}t, a\sin^{4}t)$? Does it have any special name?

答:
This parametric equation is a special case of a family of curves called superellipses, which have equations:

$$x^{n} + y^{n} = a^{n}$$

Or, parametrically:

$$x = a\cos^{\frac{2}{n}}(t)$$

$$y = a\sin^{\frac{2}{n}}(t)$$

Thus, the parametric equation you are inspecting is a superellipse with $n = \frac{1}{2}$. It has equation:

$$\sqrt{x} + \sqrt{y} = \sqrt{a}$$

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

$$y = a + x - 2a\sqrt{x}$$

With $0\leq x,y\leq a$. It happens that this equation produces a segment of a parabola with its axis of symmetry as the line $y=x$.

Many curves are superellipses (for example, a circle is $n=2$, a astroid is $n=\frac{2}{3}$, and a line segment is $n=1$.)

Title: Parametric equation of a curve


答: If $x^2+1$ has a fixed prime divisor $p$, then $p \mid 1^2+1 = 2$ and $p \mid 2^2+1 = 5$, which is a contradiction.

Body:
An integer-valued polynomial $Q(x)$ has a fixed divisor $m$ if there is an integer $m>1$ such that
$\frac {Q(x)}{m}$ is also an integer-valued polynomial. See https://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H#Fixed_divisors. If $m$ is a prime then we use the term fixed prime divisor.

I am interested in this problem:

Show that the polynomial $x^{2}+1$ has no fixed prime divisor.


答: Consider $Q(0)$. Can $Q(0)$ have any prime divisors, or in general any divisor $>1$?

Title: Show that the polynomial $x^{2}+1$ has no fixed prime divisor


问:
Choosing a linear map $(\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ which is nonzero on half of a sequence of vectors
Let $v_1,\ldots,v_m \in (\mathbb{Z}/2\mathbb{Z})^n$ be nonzero vectors.  Is it always possible to choose a linear map $f : (\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that

$$|\{\text{$i$ $|$ $1 \leq i \leq m$ and $f(v_i) \neq 0$}\}| \geq \frac{1}{2}m?$$

My guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.

答:
My coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known.  I'm amazed that it was asked here so recently!

I'm answering now to record a short alternative proof that we found.  I'll prove the same more general result that Landscape proved, namely:

Claim : Let $\vec{v}_1,\ldots,\vec{v}_m \in \mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct).  Then there
exists a linear map $f : \mathbb{F}_q^n \rightarrow \mathbb{F}_q$ such that $f(\vec{v}_i) = 1$ for at least $\frac{q-1}{q}$
of the $\vec{v}_i$, i.e. such that $\{\text{$i$ $|$ $1 \leq i \leq m$, $f(\vec{v}_i) \neq 0$}\}$ has cardinality at least $\frac{q-1}{q}m$.

Proof : Let $\Omega$ be the probability space consisting of all linear maps from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$, each given
equal probability.  Let $\mathcal{X} : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to the
cardinality of the set $\{\text{$i$ $|$ $1 \leq i \leq m$, $f(\vec{v}_i) \neq 0$}\}$.  We will prove that the expected value $E(\mathcal{X})$ of $\mathcal{X}$
is $\frac{q-1}{q} m$, which clearly implies that there exists some element $f \in \Omega$ such that $\mathcal{X}(f) \geq \frac{q-1}{q} m$.

To prove the desired claim, for $1 \leq i \leq m$ let $\mathcal{X}_i : \Omega \rightarrow \mathbb{R}$ be the random variable that
takes $f \in \Omega$ to $1$ if $f(\vec{v}_i) \neq 0$ and to $0$ if $f(\vec{v}_i) = 0$.  Viewing $\vec{v}_i$ as an element of the double dual
$(\mathbb{F}_q^n)^{\ast \ast}$, the kernel of $\vec{v}_i$ consists of exactly $\frac{1}{q}$ of the elements of $(\mathbb{F}_q^n)^{\ast}$.  This
implies $E(\mathcal{X}_i) = \frac{q-1}{q}$.  Using linearity of expectation (which recall does not require that the random
variables be independent), we get that
$$E(\mathcal{X}) = E(\sum_{i=1}^m \mathcal{X}_i) = \sum_{i=1}^m E(\mathcal{X}_i) = \sum_{i=1}^m \frac{q-1}{q} = \frac{q-1}{q} m,$$
as desired.


问:
$58^{5745} \pmod{59}$
Trying to solve the modular equation $$x \equiv 58^{5745} \pmod{59}$$

I've seen a solution to this that involves using that $58 \equiv (-1) \pmod{59}$, and this in turn means that $58^{5745} \equiv (-1)\pmod{59}$, but I'm missing the rule that explains this step.

Can someone clarify?

Title: $58^{5745} \pmod{59}$

Body: Because $58 \equiv 59-1\equiv -1\pmod{59}$, $58^{5745}\equiv (-1)^{5745}\equiv (-1)\cdot ((-1)^{2})^{2871}\equiv \boxed{-1\pmod{59}}$