问:
How do I prove dim($A^t A$)=dim($A$) on proving rank($A^TA$)=rank($A$)?
While I'm studying on Linear Algbera, I was stucked in:

Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$

I understood what above means, and I am now curious about

Yes, I understood : N($A^TA$)=N($A$)

Then, how can I convince that  rank($A^TA$)=rank($A$)?

In the above question, they say it is by rank-nullity theorem

But in my book, it seems like there is a no rank-nullity theroem, but dimension theorem:

Let $V,W$ be vector spaces, and let $T:V->W$ be linear. If $V$ is finite-dimensional, then

nullity($T$) +rank($T$) = dim($V$)

I thought like:

nullity($A^TA$) + rank($A^TA$) =dim($A^TA$) and

nullity($A$) +rank($A$) = dim($A$) So both of the rank would be the same.

But I think the dimensions of $A^TA$ and $A$ are different.

$A^TA$ becomes ${n \times n}$ matrix and $A$ is ${m \times n}$ matrix so the dimension is different.

Where I was wrong? Or I am misunderstood?

答:
Rank is the maximum number of linearly independent columns of $A$, and dim($A$) is the number of columns of $A$. Same goes for $A^TA$. Since they both have $n$ columns, the proof you linked is correct.

If you look at $A$ and $A^TA$ as linear transformations, then we have $A: \mathbb R^n \rightarrow \mathbb R^m$, and $A^TA: \mathbb R^n \rightarrow \mathbb R^n$. So the right hand side of the dimension theorem is equal to $n$ for both $A$ and $A^TA$.


问:
Is a single element a chain or an anti-chain?
Is a single element in a Hasse Diagram a chion or an anti-chain or both?

As a Hasse Diagram contains reflexive elements, it is compararble to itself. So it is a chain.

But I read somwhere that a single element is also an anti-chain

How is it an anti-chain if the Hasse diagram does not contain any other element to which it might be incomparable?

答:
An anti-chain is a subset $A$ of a poset $P$ such that, if $p, q \in A$ and $p \neq q$, then $p \not\le q$ and $q \not \le p$. If $P$ is a singleton, then this is vacuously true.

Vacuous truth can occur when we have a "for all" statement or an "if-then" statement. An "if-then" statement is vacuously true if the premise of the if-then statement is never true. For example,

If $x \in \Bbb{R}$ such that $x^2 + x + 1 = 0$, then $x = 6$ and $x = -3$.

is a vacuously true statement. Even though the conclusion can never be true, the premise is always false. It is a logical convention to call this implication true (but we consider its truth "vacuous", because it's only true due to an absence of a counterexample).

You can sort of think of it as being harmless to consider it true. A true if-then statement, whenever the premise is true, allows us to safely conclude the conclusion. If the premise is never true, we never are allowed to conclude the conclusion (at least, using this statement).

Similarly, a "for-all" statement is vacuously true if we are considering all things in a set that turns out to be empty. For example,

For all unbounded continuous functions on $[0, 1]$, there exists some $c \in [0, 1]$ such that $f(c) = 12$.

is another vacuously true statement, simply because every continuous function on $[0, 1]$ is bounded. Again, it's harmless to consider this true, simply because we will never get an unbounded continuous function on $[0, 1]$, so the conclusion is ultimately irrelevant.

So, back to anti-chains, if we have $A \subseteq P$ is a singleton (or even empty!), we can never find $p, q \in A$ such that $p \neq q$. This is vacuous truth, and hence $A$ is indeed an anti-chain.

ContentLicense: CC BY-SA 4.0


问:
Integral in n-dimensional spherical coordinates
I have to calculate the following integral:

$\int_{B_1(0)} \frac{1}{|x|^m}  dx $

where $x \in \mathbb{R}^d$ and  $B_1(0)$ is a $d$ dimensional ball around origin with radius equal to $1$.

I know I can use $\int_{R^d}f(x)dx=\int_{S^{d-1}}\left(\int_{0}^{\infty} f(r \gamma) r^{d-1} dr \right) d \sigma(\gamma)$, where $r=|x|$ and $\gamma=\frac{x}{|x|}$. But, I don't know how to apply it on my problem and obtain the results.

答:
Another way to solve this to use the alternate polar coordinates formula: $$\int_{B_r(x_0)} f(x) dx = \int_0^r \int_{\partial B_t(x_0)} f d \mathcal{H}^{n-1} dt.$$ (See Appendix C.3 in Partial Differential Equation by Lawrence C. Evans). Then \begin{align*} \int_{B_1 } \frac 1 {\vert x \vert ^m} d x &= \int_0^1 \int_{\partial B_t} \frac 1 {\vert x \vert^m } d \mathcal{H}^{n-1} d t.   \end{align*} Since $\vert x \vert = t$ on $\partial B_t$ it follows this equals \begin{align*} \int_0^1  \frac 1 {t^m }  \mathcal{H}^{n-1}(\partial B_t) d t.   \end{align*} Using that $\mathcal{H}^{n-1}(\partial B_t) = n \omega_n t^{n-1}$ where $\omega_n $ is the volume of the unit ball in $\mathbb{R}^n$, we conclude \begin{align*}
\int_{B_1 } \frac 1 {\vert x \vert ^m} d x &= n \omega_n \int_0^1  t^{n-m-1}   d t\\
&= \begin{cases}
\frac{n \omega_n}{n-m}, & \text{if } n >m \\
\text{undefined}, & \text{if } n \leqslant m.
\end{cases}
\end{align*}

ContentLicense: CC BY-SA 3.0


问:
Calculus Made Easy: Case of a Negative Exponent
In Calculus Made Easy, on pg. 56, the author performs the following algebra and I can't seem to figure out how they did it.

$$
\begin{align}
y + dy & = (x + dx)^{-2} \\[5pt]
       & = x^{-2} \Bigg(1 + \frac{dx}{x}\Bigg)^{-2}
\end{align}
$$

It's been a minute since I've been in a calculus class, let alone an algebra class. I'm able to do the follwing, but can't seem to figure out how the 2 was eliminated as well as how the second factor is raised by -2 instead of -1.

$$
\begin{align}
y + dy & = (x + dx)^{-2} \tag 1 \\[5pt]
       & = \frac{1}{x^2 + 2 x \cdot dx + dx^2} \tag 2 \\[5pt]
       & = \frac{1}{x^2 + 2 x \cdot dx} \tag 3 \\[5pt]
       & = \frac{1}{x^2 \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)} \tag 4 \\[5pt]
       & = x^{-2} \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)^{-1} \tag 5
\end{align}
$$

答: The first computation is exact. Your computation is correct up to infinitesimals of order, higher than $dx$. Up to such infinitesimals, both computations are correct.


答:
There's nothing wrong.

$(1 + \frac {dx}{x})^2 =(1 + \frac {2dx}x + \frac {dx^2}{x^2}) =  (1 + \frac {2dx}{x})^{1}$

So either expression is correct.

Title: Calculus Made Easy: Case of a Negative Exponent

Body:
In Calculus Made Easy, on pg. 56, the author performs the following algebra and I can't seem to figure out how they did it.

$$
\begin{align}
y + dy & = (x + dx)^{-2} \\[5pt]
       & = x^{-2} \Bigg(1 + \frac{dx}{x}\Bigg)^{-2}
\end{align}
$$

It's been a minute since I've been in a calculus class, let alone an algebra class. I'm able to do the follwing, but can't seem to figure out how the 2 was eliminated as well as how the second factor is raised by -2 instead of -1.

$$
\begin{align}
y + dy & = (x + dx)^{-2} \tag 1 \\[5pt]
       & = \frac{1}{x^2 + 2 x \cdot dx + dx^2} \tag 2 \\[5pt]
       & = \frac{1}{x^2 + 2 x \cdot dx} \tag 3 \\[5pt]
       & = \frac{1}{x^2 \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)} \tag 4 \\[5pt]
       & = x^{-2} \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)^{-1} \tag 5
\end{align}
$$


答:
We have $$(x+dx) = \left[x\left( 1+\frac{dx}{x}\right)\right]$$

Hence
\begin{align}
(x+dx)^{-2} &= \left[x\left( 1+\frac{dx}{x}\right)\right]^{-2} \\
&=x^{-2}\left( 1+\frac{dx}{x}\right)^{-2}
\end{align}

Also notice that

$$\left( 1+\epsilon\right)^2 \approx 1+2\epsilon$$

if $\epsilon$ is small enough, it is the first order approximation.


问:
Proving an inverse function is continuous

Let $f: (a,b) \to \mathbb{R}$ be a continuous function with $f'(x) \neq 0$ for all $x$. Prove that $g: f(a,b) \to (a,b)$ is continuous.

Here is my attempt.

Let $\epsilon > 0$, $x \in (a,b)$. and consider $[x - \epsilon, x + \epsilon]$. Since $f'(x) \neq 0$, $f$ is either increasing or decreasing. Suppose, WLOG, that $f$ is increasing. Then the maximum of $f$ on $[x - \epsilon, x + \epsilon]$ is $f(x + \epsilon)$ and the minimum of $f$ is $f(x-\epsilon)$. Set $\delta = \min{|f(x+ \epsilon) - f(x)|, |f(x - \epsilon) - f(x)|}$. For $x'$ such that $|f(x') - f(x)| < \delta$, we have $x' \in (f(x - \epsilon), f(x + \epsilon)$, and since $f$ is increasing, $x - \epsilon < x' < x + \epsilon$, so $|x' - x| < \epsilon$, so $g$ is continuous at $x$.

Is this proof correct? My two doubts are, first, going backwards from $f(x - \epsilon) < x' < f(x + \epsilon)$ using that $f$ is increasing. I know that the converse is true by definition. My second doubt is that $\epsilon$ might be so large that $[x - \epsilon, x + \epsilon] \not \subset (a,b)$. Do I need to impose an additional restriction on $\delta$?

答: For your doubts: 1. a function is increasing iff its inverse is increasing (within the domain where it is defined). 2. You can always shrink $\epsilon$ a little bit to make that interval included in $(a,b)$. Get the $\delta$ for that smaller $\epsilon$, and it works for the situation where the difference required need not be so narrow.


问:
Intuition behind the flow of a vector field
Lemma : Let $X$ be a smooth vector field on a a smooth manifold $M$.

For all $m\in M,$ there exists $a_m$ and $b_m$ in $\mathbb R \cup\{\pm \infty\}$ and a smooth curve $c_m :]a_m,b_m[ \rightarrow M$ such that :

$0 \in ]a_m,b_m[$ and $c_m(0)=m$;

$c_m$ is an integral curve of $X$;

If $\mu :]c,d[\rightarrow M$ is a smooth curve satisfying $1.$ and $2.$, then $]c,d[ \subset ]a_m,b_m[$ and $\mu=c_{m{|]c,d[}}.$

Now we define fot all $m\in M$ and $t\in \mathbb R$ the set $$\mathcal D_t=\{m\in M/ t\in ]a_m,b_m[\}$$
and the application $$\theta_t :\mathcal D_t \rightarrow M : \theta_t(m)=c_m(t),$$
and also $$\mathcal D(X)=\{(t,m)\in \mathbb R\times M / t\in]a_m,b_m[\}$$
and finally the application $$\theta : \mathcal D(x) \rightarrow M : \theta(t,x)=\theta_t(x)$$.
$\theta$ is called the flow of $X$.

I can't visualize the sets $\mathcal D_t, \mathcal D(X)$ and the flow of $X$ and what's the intuition behind all this ? Any help is highly appreciated.

答:
The expression $\theta_t(x)$ simply denotes the point where you end up if you start at $x$ and “follow the flow” for $t$ units of time. That is, at each instant you go with the velocity that the vector field (at the point where you happen to be at that instant) tells you to go. That is, you follow the integral curve through $x$.

For some vector fields and some starting points, it may happen (if the manifold is not compact) that you “go off to infinity” in finite time, so that $\theta_t(x)$ is not defined for all $t$ but only in some interval around $t=0$. The set $\mathcal{D}$ is just the set of pairs $(t,x)$ such that $\theta_t(x)$ is defined.


问:
$\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$

Evaluate
$$S=\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$

Its quite simple to conclude that this limit does not exist.

Now here comes the interesting part. Consider the limit

$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{1-\frac{\sin x}x}{1+\frac{\cos x}x}=1$$

But this limit is of the form $\frac \infty\infty$. So from L'Hopital's

$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{(x-\sin x)'}{(x+\cos x)'}=S$$

$$\implies S=1$$

Is this reverse reasoning correct? Is there any faulty assumption that is made here?

答: L'Hopital's rule says, in this case, that if the limit$$\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$exists, then it is equal to$$\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}.$$You tried to apply this implication in the wrong direction, which is not valid.


问:
Find integrating factor for $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$
There's differential equation $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$. Then what is integrating factor. First I convert it to standard form $\frac{dy}{dx}+P(x)y=Q(x)$. And integrating factor is $e^{\int P(x)dx}$. But first to find $P(x)$ I should rearrange it.

Body:
The3 Integrating factor is $I=\exp(J)$, use $t=\tan^2 x$
$$J=\int \frac{\sin 2x dx}{2(1+2\sin^2 x)} =\int \frac{\tan x \sec^2x}{(1+2\tan^2 x)(1+\tan^2 x)} =\frac{1}{2}\int \frac{dt}{(1+t)(1+2t)}.$$
$$J=\frac{1}{2}\int \left(\frac{2}{1+2t}- \frac{1}{1+t}\right)dt=  \frac{1}{2}\ln \frac{1+2t}{1+t}$$
So $$I =\sqrt{\frac{1+2\tan^2 x}{1+\tan^2x}}= \sqrt{1+\sin ^2x}$$


问:
Find the Var[X] for the continuous random variables with given joint probability density function.

In this question I was supposed to find the $E[X]$ and $\text{Var}[X]$.

I found $E[X] = 0.6$ and $\text{Var}[X] = 0.3696$ but when I checked the answer it shows that my $E[X]$ is correct but my $\text{Var}[X]$ is wrong!!!

What is the correct answer for $\text{Var}[X]$?

ThankYou in advance.

答:
Integrating

$$\int_x^{\infty}f(x,y)dy=f_X(x)=\frac{125}{2}x^2e^{-5x}=\frac{5^3}{\Gamma(3)}x^{3-1}e^{-5x}$$

we immediately realize that

$$X\sim \text{Gamma}[3;5]$$

Thus

$$\mathbb{E}[X]=\frac{3}{5}$$

and

$$\mathbb{E}[X]=\frac{3}{5^2}$$

wihtout any other calculations


问:
If a subgroup of an algebraic group is solvable, is its closure necessarily solvable?
$G$ is an algebraic group, and $H$ is a subgroup which is solvable. $\overline{H}$ is its closure in $G$.

Then $\overline{H}$ is also a subgroup of $G$. Is it also solvable?

For any algebraic group $G$, denote $[G,G]$ the derived subgroup of $G$. Then is it true that $\overline{[H,H]} = [\overline{H}, \overline{H}]$? If this is true, I think the solvability of $\overline{H}$ might be proved by dimension comparision.

Many thanks~ Special thanks to @awllower for the enlightening comments.

答:
Following the proof of @ShinyaSakai, try to prove this result:

For $H_{1}\subset H_{2}$ two subgroups (not necessarily closed) of $G$, we have $[\overline{H_{1}},\overline{H_{2}}]=\overline{[H_{1},H_{2}]}$.
The reason for the assumption $H_{1}\subset H_{2}$ is that the commutator of two closed subgroups is not closed generally.

There is a tricky point in the proof of this result. As is pointed out by @evgeny under the answer of @ShinyaSakai, generally speaking, $[H_{1},H_{2}]\neq \phi(H_{1}×H_{2}) $. A correct proof is the following,

$H_{1}×H_{2}$ is dense in $\overline{H_{1}}×\overline{H_{2}}$.

$\phi(H_{1}×H_{2})$ is dense in $\phi(\overline{H_{1}}×\overline{H_{2}})$.

$[\overline{H_{1}},\overline{H_{2}}]$ is a closed subgroup of $G$, generated by $\phi(\overline{H_{1}}×\overline{H_{2}})$, hence containing $\overline{[H_{1},H_{2}]}$.

$\overline{[H_{1},H_{2}]}$ is the closure of the geoup generated by $\phi(H_{1}×H_{2})$, hence containing $\phi(\overline{H_{1}}×\overline{H_{2}})$. By 3, it contains $[\overline{H_{1}},\overline{H_{2}}]$.

You can use this result to prove that the closure of a solvable (nilpotent) algebraic group is still solvable (nilpotent).


问:
The probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 is rolled?
A 6-sided die is weighted so that the probabilities of rolling a 1, 2, or 3 are equally likely; the probabilities of rolling a 4, 5, or 6 are equally likely; and the probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 is rolled?

How I solved it:

$P(\text{rolling a } 1, 2, 3) = x$

$P(\text{rolling a } 4, 5, 6) = 4x$ (because rolling a 4, 5, 6 is 4x the probability of rolling a 1, 2, 3)

$3(x)+3(4x)=15x$ (because 1,2,3 each has $x$ probability of being rolled and same with 4,5,6 except each of them has a $4x$ chance)

$15x = 1 ,  x=\frac{1}{15}$

Therefore, there's a $\frac{1}{15}$ chance of rolling a 1, rolling a 2, rolling a 3 and a $\frac{4}{15}$ chance of rolling a 4, rolling a 5, rolling a 6.

So to roll a 2 or 5, that would be $\frac{1}{15} + \frac{4}{15} = \boxed{\frac{1}{3}}$.

My teacher told me I got the answer correct, but my solution wasn't right and she said the part where it mentions that rolling a 5 is four times the probability of rolling a 3 is irrelevant. Can someone tell me what I'm doing wrong? Thanks!

答:
Your solution was correct but there was an easier way to see it.

The probability of a $1$ or $4$ is the same as the probability of a $2$ or $5$, which is the same as the probability of a $3$ or $6$.  Thus, the answer has to be $\frac 13$ no matter what the relative probability of the two groups are.


问:
How to find all roots of nonlinear function - an example.
How would I find all the roots of the function $f(x) = \sin (10x) - 2x$?

I know all sin functions have multiple roots and so this function can also have multiple roots but how would I find all these roots? Using matlab fzero which uses the bracket criteria for the existence of solution only gives one root to the function which is 0.25957.

Can I report all the roots from the graph? Or How would I do to find all the multiple roots from the graph? Could there a better way of finding all these multiple roots?

Graph is due to MATLAB

```<code>fplot( @(x) sin(10*x)-2*x, [-.5,.5])
%ploting the function with interval [-.5, .5]
grid on
</code>```

答:
Since the function is odd, you can limit the problem to the positive solution(s) for $0 \leq x \leq \frac 12$.

Since $x=0$ is a trivial solution, just focus on the next one. You have
$$f(x) = \sin (10x) - 2x$$
$$f'(x)=10 \cos(10x)-2$$

The first derivative cancels at
$$x_*=\frac{1}{10} \cos ^{-1}\left(\frac{1}{5}\right)$$ and the second derivative test confirms that this is a maximum.

Expand $f(x)$ around $x_*$
$$f(x)=\left(\frac{2 \sqrt{6}}{5}-\frac{1}{5} \cos
   ^{-1}\left(\frac{1}{5}\right)\right)-20 \sqrt{6} \left(x-\frac{1}{10} \cos
   ^{-1}\left(\frac{1}{5}\right)\right)^2+O\left(\left(x-\frac{1}{10} \cos
   ^{-1}\left(\frac{1}{5}\right)\right)^3\right)$$ then, as an estimate,
$$x=\frac{1}{10} \cos ^{-1}\left(\frac{1}{5}\right)+\frac{1}{10}\sqrt{2 -\frac 1{\sqrt 6} \cos ^{-1}\left(\frac{1}{5}\right)} \sim 0.2570$$ while the "exact" solution, obtained using Newton method, is $0.2596$.

Edit

For the fun of it, let $x=\frac t{10}$ and use the $1{,}400$ years old approximation
$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$ and you just need to solve a quadartic equation in $t$ and the approximate solution is
$$x=-1+\frac{\pi }{20}+\frac{1}{10} \sqrt{100+(10-\pi) \pi }\approx 0.259560$$ instead of $0.259574$.


问:
Signs in the tensor product and internal hom of chain complexes
Let $R$ be a commutative ring and $\text{Ch}(R)$ the category of chain complexes of $R$-modules. $\text{Ch}(R)$ is first of all an abelian category, but it can also be equipped with the structure of a (symmetric?) closed monoidal category: it has

a tensor product $A \otimes B$ whose component in degree $n$ is $\bigoplus_{i+j=n} A_i \otimes_R B_j$ and where the differential is defined by $d(a \otimes b) = da \otimes b + (-1)^{|a|} a \otimes db$, and

an internal hom $\text{hom}(A, B)$ whose component in degree $n$ is $\prod_i \text{Hom}_R(A_i, B_{i+n})$ and where the differential is defined by $(df)(a) = d(fa) - (-1)^{|f|} f(da)$

which are related via the adjunction $\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{hom}(B, C))$.

How can I motivate the sign convention in either of these definitions?

I can sort of answer that question: both sign conventions are just the graded Leibniz rule. As Theo Johnson-Freyd explains on MO, this is because $\text{Ch}(R)$ is precisely the category of modules over a one-dimensional graded-commutative Lie algebra over $R$ concentrated in degree $1$. But this is unsatisfying because I don't know the answer to this question:

What do graded Lie algebras have to do with the topological motivation for studying chain complexes?

By "the topological motivation" I mean the idea, made precise by the Dold-Kan correspondence, that a chain complex is, roughly speaking, a linearization of a combinatorialization of a topological space. I think that the sign convention for the tensor product can be motivated by thinking about simplicial chains, but this doesn't have quite the air of inevitability that I would like out of so fundamental a definition.

I have the feeling that everything goes back to Euler characteristic. A bounded complex can be sent to an alternating sum of elements of $K_0(R)$, and this sum is invariant under chain homotopy and additive in short exact sequences and possibly satisfies a universal property of some kind, and it clearly shows there is a real distinction to be made between the even and odd elements of a chain complex. But I don't know the conceptual route from this idea to the graded Leibniz rule.

答:
One algebraic way to motivate this is to observe that the signs in the differential for the Hom are precisely what is needed for 0-cycles in the $\hom(A,B)$ complex to be the set of morphisms of complexes $A\to B$ (and also, that the 0th homology group $H_0(\hom(A,B))$ is the set of homotopy classes of morphisms $A\to B$). This is quite great.

Once you decide you want this, all the other signs you mention follow because you need various things to hold. For example, you want the adjuntion between $\hom$ and $\otimes$ to hold for the internal versions, so this forces you to add signs to the $\otimes$, and so on.

A topological, indirect explanation for the appearence of most signs is that in the long exact sequence of bases maps corresponding to a map $f:X\to Y$, which looks like
$$X\to Y\to C(f)\to S(X) \to S(Y) \to S(C(f)) \to \cdots$$ there is a "sign" which you cannot get rid of. This sign reproduces itself in every algebraic version.


问:
I wanted to know in the statement $5$ added to a number divided by $3$ whether we divide after adding $5$ or the number is already divided?
If we were to convert the the statement

$5$ added to a number divided by $3$

to equation.

Would it be $5+\frac x3$ or $\frac{5+x}3$?

答:
"five added to a number barely perceptible pause divided by three" would be $\frac {5+x}3$.

"five added to barely perceptible pronouncing of the word 'to' for just a nanosecond longer then normal a number divided by three with the words 'a number divided by three' just barely noticeably rushed" would be $5 + \frac x3$.

English words are not necessarily well-formed mathematical statements and the meaning of those words is ambiguous.  We must ask the speaker to elaborate.


答:
No, your proof is not correct, unfortunately. There is one subtle, yet vitally important issue that needs to be addressed. The definition of continuity of $g$ goes like this:

For all $\varepsilon > 0$ and $x_0 \in f(a, b)$, there exists some $\delta > 0$ such that
$$|x - x_0| < \delta \implies |g(x)- g(x_0)| < \varepsilon.$$

In particular, if anyone picks their favourite positive number $\varepsilon$, you should be able to respond with an appropriate $\delta > 0$ so that the above holds true.

Your proof doesn't do this. Your $\varepsilon$ value for $g$ isn't assumed to be arbitrary, but instead assumed to be a $\delta$ value for $f$. At no point in your proof am I allowed to pick my favourite positive number $\varepsilon$, and your proof gives me no instruction as to what $\delta$ would be appropriate.

To fix this, I would instead go for an indirect proof. Suppose that $g$ is not continuous at a point $x_0$. Because $g$ is monotone, such a point must be a jump discontinuity: both $\alpha := \lim_{x \to x_0^-} g(x)$ and $\beta := \lim_{x \to x_0^+} g(x)$ exist, but they're not equal to each other. You can then show that $f(\alpha) = f(\beta)$, which would contradict $f$ being strictly increasing.

I'd also be careful about your conclusion that $f$ is increasing or decreasing. While this is absolutely true, it's not quite so obvious. You know that if $f'$ maintains the same sign over an interval, then $f$ is increasing or decreasing, but you don't know (without citing a theorem) that $f'$ maintains the same sign over the interval. For example, could $f'(x) > 0$ for rational $x$, and $f'(x) < 0$ for irrational $x$?

There is a theorem to help you here: Darboux's Theorem. Either reference it, or follow its proof, in order to plug that hole.

Title: Proving an inverse function is continuous

Body:
Let $f: (a,b) \to \mathbb{R}$ be a continuous function with $f'(x) \neq 0$ for all $x$. Prove that $g: f(a,b) \to (a,b)$ is continuous.

Here is my attempt.

Let $\epsilon > 0$, $x \in (a,b)$. and consider $[x - \epsilon, x + \epsilon]$. Since $f'(x) \neq 0$, $f$ is either increasing or decreasing. Suppose, WLOG, that $f$ is increasing. Then the maximum of $f$ on $[x - \epsilon, x + \epsilon]$ is $f(x + \epsilon)$ and the minimum of $f$ is $f(x-\epsilon)$. Set $\delta = \min{|f(x+ \epsilon) - f(x)|, |f(x - \epsilon) - f(x)|}$. For $x'$ such that $|f(x') - f(x)| < \delta$, we have $x' \in (f(x - \epsilon), f(x + \epsilon)$, and since $f$ is increasing, $x - \epsilon < x' < x + \epsilon$, so $|x' - x| < \epsilon$, so $g$ is continuous at $x$.

Is this proof correct? My two doubts are, first, going backwards from $f(x - \epsilon) < x' < f(x + \epsilon)$ using that $f$ is increasing. I know that the converse is true by definition. My second doubt is that $\epsilon$ might be so large that $[x - \epsilon, x + \epsilon] \not \subset (a,b)$. Do I need to impose an additional restriction on $\delta$?


问:
How to subtract two rectangles?
I am wondering how to get the Set Subtraction between two rectangles from the programming point of view. I would really appreciate if anyone could help!

We know that in $\mathbb R^2$ each rectangle can be encoded by two points at the left-bottom and right-top corner, respectively. Now, assume that we have 2 rectangles encoded by 4 points. We observe that the set subtraction of the two rectangles is a union of some other rectangles. How can we encoded these resulting rectangles?

Of course, I can do it by considering all the possible positions of the two rectangles (there are 10 cases), but I have to do it manually and without any formula.
Are there any more efficient ways to do it?

答:
As a programming point of view is mentioned, I assume that we are only interested in lattice points (e.g., "pixels") in $\Bbb Z^2$ and need not worry about the distinction between open and closed rectangles, not to mention those containing only part of their boundary.

Suppose you have rectangles $R_1=[x_1,x_2)\times[y_1,y_2)$ and $R_2=[x_3,x_4)\times[y_3,y_4)$, where $x_1<x_2$ etc.
To simplify, replace
$$x_3\leftarrow\max\{x_1,x_3\},\quad y_3\leftarrow\max\{y_1,y_3\},\quad
x_4\leftarrow\min\{x_2,x_4\},\quad y_4\leftarrow\min\{y_2,y_4\}.$$
This way, we make sure that $R_2$ does not extend outside $R_1$ and the following formulations become simpler insofar as we now have
$$ x_1\le x_3<x_4\le x_2,\quad y_1\le y_3<y_4\le y_2,$$
except that the above may make  $x_3\ge x_4$ or $y_3\ge y_4$, but then we are already done because $R_1\setminus R_2=R_1$.
Now $R_1\setminus R_2$ can be covered by four pairwise disjoint rectangle:
$$\tag1[x_1,x_3)\times[y_3,y_2)\;\cup  \;
[x_3,x_2)\times[y_4,y_2) \;\cup  \;
[x_4,x_2)\times[y_1,y_4) \;\cup  \;
[x_1,x_4)\times[y_1,y_3).$$
The only unusual thing that may happen is that one or more (in fact, perhaps all!) of these degenerate to a line (e.g., the first when $x_1=x_3$) and can thus be removed from the list.

If after removing degenerate rects, we are left with two rectangles
$$\tag2[u_1,u_2)\times [v_1,v_2)\;\cup \;[u_3,u_4)\times [v_3,v_4),$$ it may happen that we can simplify further if we want (but the two-rectangle partition is valid), namely

If the horizontal intervals are the same (i.e., $u_1=u_3$ and $u_2=u_4$) and the vertical ones are juxtaposed (i.e., $v_2=v_3$ or $v_1=v_4$), thenh $(2)$ can be simplified to
$$ [u_1,u_3)\times [\min\{v_1,v_3\},\max\{v_2,v_4\}).$$

If the vertical intervals are the same (i.e., $v_1=v_3$ and $v_2=v_4$) and the horizontal ones are juxtaposed (i.e., $u_2=u_3$ or $u_1=u_4$), thenh $(2)$ can be simplified to
$$ [\min\{u_1,u_3\},\max\{u_2,u_4\})\times [v_1,v_2).$$

Title: How to subtract two rectangles?


问:
Standard approach for critical values when degree of freedom is missing in a T-table?
I am a bit confused about what is generally done (without the help of any kind of software) to obtain critical values when exact degrees of freedom values are missing in a T-table. Most T-tables tend to have degree of freedom/df values up to 100 and then jump straight to 200 and so on. But what is the best approach when suppose I have to consider df=199 where alpha/2= 0.025 in a two-tail test?

Should I take an average of t0.025,100 and t0.025,200?

Should I only consider t0.025,100 to get the critical values?

答:
When you don't see the exact degrees of freedom in the table, then the most appropriate thing to do is apply a linear interpolation to adjacent table entries.  For instance, the critical value for a two-sided test with $\alpha = 0.0001$ corresponds to the last column in the table.  If the degrees of freedom is $\nu = 32$, we only have the critical values for the following:

$$\begin{array}{c|c}
\nu & t_{\nu}^* \\
\hline
30 & 4.48  \\
35 & 4.39  \\
\end{array}$$

So linear interpolation for $\nu = 32$ would look like this:  $$t_{32}^* \approx \frac{35 - 32}{35 - 30} t_{30}^* + \frac{32 - 30}{35 - 30} t_{35}^*  = \frac{3}{5} (4.48) + \frac{2}{5} (4.39) \approx 4.44.$$

In fact, the exact value is $4.44105393811587\ldots$.  So it's not too bad.  However, interpolation is not advisable if the table values are too far apart.


问:
How do integrals add up all the areas within their bounds?
Before I proceed,
I am a high school student and have just started learning calculus.

OK

Now, I understand that an integral is basically the opposite of a derivative. I also understand that it finds the area under the graph of a function.

Based on what I understood, while integrating you take a very small change in the variable $x$ (or maybe $\mathrm{d}x$) such that the change in $f\left(x\right)$ is constant and therefore calculate the area of the rectangle and add the areas...

However, my question here is how does the integration tool include all the areas
bound under its limit, it makes less sense to me.
For example:
$$\int_2^4 x\,\mathrm{d}x$$

Solving the integral:

You get
$$\frac{x^2}{2}$$

with limits $4$ and $2$.
So, now you substitute the limits and subtract them .....
but this is what feels weird, it looks like I'm only subtracting the area at $4$ and $2$,
How does this tool work

(Any response would be helpful)
EDIT:
(i want to make my question a bit more clear)

[My question is basically, how does integration include all the areas between the interval, i have seen a couple of proofs on the fundamental theorem of calculus
however, that just seemed to explain why we subtract while integrating (to remove the constant term (i think).The problem i guess i am facing is that there is no hint of a summation of areas happening,and this is exactly what my question is -how do we know the areas are included??]

答: First, you have to understand what a definite integral really is. First, you have to do is learn Riemann integral, you'll see $\int_{a}^{b} f(x)dx$ just denotes the area under the curve between the points a and b. As for your main question, why $\int_{a}^{b} f(x)dx=F(b)-F(a)$ where $F'(x)=f(x)$. You will need to learn Lagrange's theorem. I believe after that, you might be able to understand how the Fundamental Theorem of Calculus works


问:
How can one the cardinality of function space $C(X, Y)$ equipped with uniform topology, where $X$ is a Tychonoff space and $(Y, d)$ a metric space.
How can one the cardinality of function space $C(X, Y)$ equipped with uniform topology, where $X$ is a Tychonoff space and $(Y, d)$ a metric space.

I mean how does the cardinality of $X$ and $(Y, d)$ help in finding the cardinality of function space $C_u(X, Y)$.

答:
$|C(X,Y)|$ will depend on the combination of the topologies on $X$ and $Y$ and $|X|$ and $|Y|$ as well.

Some examples: if $X$ is $\Bbb R$ in the usual topology and $Y = \Bbb Q$, in the usual metric, then $|C(X,Y)| = \aleph_0$, because only the constant functions from $X$ to $Y$ are continuous. We also have the trivial bound $|C(X,Y)| \ge |Y|$ because we always have those constant functions in the set of continuous functions.

If $X=\Bbb R$ in the discrete topology (indeed Tychonoff) and $Y$ is any metric space, then all $f: X \to Y$ are continuous, so $|C(X,Y)| = |Y|^{|X|}$ so when $Y=\Bbb Q$  this equals $\aleph_0^\mathfrak{c} = 2^\mathfrak{c}$, e.g.

So the size can differ quite dramatically, and has to be considered on a case by case basis.


问:
Finding $a \pmod c$ if $a \pmod b$ is known
Suppose that:

$Y \pmod B = 0$

$Y \pmod C = X$

I know $B$ and $C$. $Y$ is unknown, it might be an extremely large number, and it does not interest me.

The question is: Is it possible to find $X$, and if so, how?

答:
$\bigg\lbrace\begin{eqnarray}y\equiv 0\pmod b\\ y\equiv x\pmod c\end{eqnarray}\bigg\rbrace $ $\!\iff\! \bigg\lbrace\begin{eqnarray} y &=& j b,\ \ &{\rm  some}\ \ j\in \Bbb Z\\ y &=& x\! +\! k c,\ \ &{\rm some}\ \ k\in \Bbb Z\end{eqnarray}\bigg\rbrace\!\iff\! x = jb\!-\!kc,\ \ {\rm some}\ \ j,k\in\Bbb Z$

Thus, by Bezout, the possible values of $\,x\,$ are precisely the multiples of $\,\gcd(b,c).$


问:
How to solve these equations with $x^2$, $xy$, and $y^2$?
Is there any good method to solve equations like this?
$$
\begin{cases} 2y-2xy-y^2=0\\2x-x^2-2xy=0\end{cases}
$$
This is what I did:
$$
\begin{cases} y(2-2x-y)=0\\x(2-x-2y)=0\end{cases}
$$
now I see, that:
$$
x=0
$$
$$
y=0
$$
and it's the first solution, now:
$$
\begin{cases} 2-2x-y=0\\2-x-2y=0\end{cases}
$$
and:
$$
x=\frac{2}{3}, y=\frac{2}{3}
$$
I have 2 solutions, but wolfram found more...
Is there any good method for equations like this?</p>

答:
HINT:

From the first equation either $y=0$ or $2-2x-y=0$

If  $y=0,$ from the second equation $2x-x^2=0\implies x(x-2)=0\implies x=0,2$

If $2-2x-y=0,y=2-2x$ from the second equation  $2x-x^2-2x(2-2x)=0\implies x(3x-2)=0\implies x=0$ (which has been found already) and $x=\frac23$

Similarly start from the second equation


问:
Are cardinal functions topological properties?
As we know that cardinality itself is a topological property. But now i am a bit confused about the other cardinal functions like character, extent, weight, cellularity pseudoextent. I want to know if these all are preserved during homeomorphism. Can anybody explain?</p>

答:
Of course all cardinal functions that are usually considered for topological spaces are topological invariants: they are purely defined in set theory terms and notions that depend on the open or closed sets of $X$. If $X \simeq Y$, then anything in $X$ can be "translated" to $Y$ using a given homeomorphism $h: X \to Y$, e.g. if $X$ has character $\kappa$ and $y \in Y$, then there is a unique $x \in X$ with $h(x)=y$ and if $\mathcal{B}$ is a local base at $x$, $\{h[B]\mid B \in \mathcal{B}\}$ is a local base at $y$ etc. so we can choose one of size $\le \kappa$ in $X$ and this translated local base will be of the same size, so also $\le \kappa$ etc. So the character of $Y$ is also $\kappa$ (not smaller, because we can also go back and we'd contradict the minimality of $\kappa$ for $X$).
As to cellularity: the images of a disjoint open family in $X$ are also a disjoint open family in $Y$ and vice versa, so maximal sizes of them are also preserved. For all cardinal functions we can hold similar arguments, but in all papers it's usually omitted as being to obvious to mention: if the definition only uses set theory and (properties derived from) open sets, it's obviously a topological invariant.

The chapter by Hodel on cardinal functions in the Handbook of Set-theoretic Topology has a nice first overview of the most common ones and their interrelations (a diagram at the end of the chapter), Engelking's General Topology has a series of exercises on them (if you're self-studying) and the two books by Juhász on Cardinal Functions in Topology are seminal classics on this topic.


问:
Show that if $D$ is a dense subset in $[0, π]$, then $\sin(D)$ is dense in $[0, 1]$.
Working my way through my first topology course. Would anyone be able to break this example down for me? We were never shown anything concrete like this to work with, just abstract intuition, so I don't really know how to start...

This is the definition of dense we were given: Let $X$ be a metric space, $A \subseteq X$ and $D \subseteq A$. $D$ is dense in $A$ if for every $a \in A$ and $r > 0$, $B(a, r) \cap D$ is non-empty.

答:
Let $y \in [0,1]$. Then  there exists $x \in [0,\pi]$ such that $\sin x =y$. There is  sequence $(d_n)$ in $D$ converging to $x$. Now $\sin d_n -y\to \sin x-y=0$. This proves that $\{\sin d: d\in D\}$ is dense in $[0,1]$.

In general continuous surjective functions map dense sets to dense sets.


问:
Coordinates reachable from $(x, y) = (2,5)$
If we are on coordinate $(m,n)$ then we can either go to coordinate $(m + n, n)$ or $(m, m + n)$. If we start at $(2,5)$ then what all coordinates we can reach in finitely many moves?

In our case we have: From coordinate $(2,5)$ we can go to $(7,5)$ or $(2,7)$, now if you choose to go to $(7,5)$ then from $(7, 5)$ you can go to $(12,5)$ or $(7,12)$ and so on we keep on iterating this process and we need to find all such reachable coordinates. Any help would be highly appreciated. Thanks.

答:
Given $(x,y)$, can it be reached?
Clearly (or formally: by induction), $x,y$ must be integers and $x\ge 2$, $y\ge 5$.
Also, we see (again formally: by induction) that $x\ne y$ because the allowed steps can only produce different coordinates $m+1\ne n$ or $m\ne m+n$ from positive $n,m$. Thus $(x,y)\ne(2,5)$ can be reached only via $(x-y,y)$ if $x>y$ and only via $(x,y-x)$ if $y>x$ and not at all if $x=y$.

Thus to check if $(x,y)$ can be reached, we only have to repeatedly go backwards and subtract the smaller from the larger coordinate and have won iff the last position before making $x<2$ or $y<5$ is  $(2,5)$.
Note that these steps are really just Euclid's algorithm in disguise, except that we stop a few steps before one coordinate becomes $0$ and the other the gcd. In particular, necessarily $\gcd(x,y)=\gcd(2,5)=1$. But this is not sufficient as many such pairs will rather go via points like $(3,7)$ for example.


答:
The only thing we need is that the function is continuous: if $f: X \to Y$ is continuous and onto and $D$ is dense in $X$, $f[D] = \{f(d)\mid d \in D\}$ is dense in $Y$.

Let $y \in Y$ and $r>0$, for some $x \in X$, $f(x)=y$ and by continuity at $x$ there is some $\delta >0$ so that $$x' \in B(x,\delta) \to f(x') \in B(f(x),r)$$ Now we can pick $d \in D$ with $d \in B(x,\delta)$ ($D$ is dense) and then the continuity condition gives that $f(d) \in B(y,r) \cap f[D]$. So $f[D]$ is dense.

Now use that $x \to \sin(x)$ is continuous and onto from $[0,\pi]$ to $[0,1]$.

Title: Show that if $D$ is a dense subset in $[0, π]$, then $\sin(D)$ is dense in $[0, 1]$.

Body:
Working my way through my first topology course. Would anyone be able to break this example down for me? We were never shown anything concrete like this to work with, just abstract intuition, so I don't really know how to start...

This is the definition of dense we were given: Let $X$ be a metric space, $A \subseteq X$ and $D \subseteq A$. $D$ is dense in $A$ if for every $a \in A$ and $r > 0$, $B(a, r) \cap D$ is non-empty.


问:
How can I find all incongruous system solutions mod n?
$\begin{array}{r}
x+y+z \equiv 1 \bmod 5 \\
2 x+4 y+3 z \equiv 1 \bmod 5
\end{array}$

I have this system and I want to find all incongruous system solutions.

Well, I know that

$x+y+z-1=5k_1$ and $2x+4y+3z-1=5k_2$.

Can I find the solutions before without taking $\bmod 5$?
I mean $x+5q,y+5q_1,z+5q_2$.

答:
Your system is equivalent to this one:
$$
x+y+z\equiv 1\pmod5 \\
2y+z\equiv -1\pmod5
$$
for $-2$ times the first congruence added to the original second one yields the second congruence in this system; and $2$ times the first congruence added to the second one is the original congruence.

Solve the second one for $y, z$ by discussion or listing out all possibilities. Then let $x=1-y-z$ for these solutions.


答:
You are on the right track, but need to consider your cases carefully.

Case 1. $y= 0$, $x=0 $.

Done.

Case 2. $y=0 $, $2-x-2y = 0$.

Do this.

Case 3. $2-2x-y = 0$, $x = 0$.

Do this.

Case 4. $2-2x-y = 0$, $2-x-2y = 0$.

Done.

Title: How to solve these equations with $x^2$, $xy$, and $y^2$?

Body:
Is there any good method to solve equations like this?
$$
\begin{cases} 2y-2xy-y^2=0\\2x-x^2-2xy=0\end{cases}
$$
This is what I did:
$$
\begin{cases} y(2-2x-y)=0\\x(2-x-2y)=0\end{cases}
$$
now I see, that:
$$
x=0
$$
$$
y=0
$$
and it's the first solution, now:
$$
\begin{cases} 2-2x-y=0\\2-x-2y=0\end{cases}
$$
and:
$$
x=\frac{2}{3}, y=\frac{2}{3}
$$
I have 2 solutions, but wolfram found more...
Is there any good method for equations like this?</p>


问:
Understanding open set in product topology
Let $X,Y$ are topological spaces and $U\times V\subset X\times Y$ be a subset where $X\times Y$ with product topology. If $U\times V$ is open then $U$ and $V$ are open sets in $X, Y$ respectively?

I know that if $U\times V$ is open then $U\times V =\bigcup_{i\in I}U_i\times V_i$ where $U_i\subset X$: basic open set and $V_i\subset Y$: basic open set. But I'm not sure this implies the above statement. Could you help?

答:
Using that images of functions commute with unions:

$$U = \pi_1[U \times V]= \pi_1[\bigcup_i U_i \times V_i] = \bigcup_i \pi_1[U_i \times V_i] = \bigcup_i U_i$$

and $U$ is open as the union of the $U_i$. W can do the same for $V$ with $\pi_2$, the other projection.


问:
Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$
I'm trying to compute $$\int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx$$ i.e. the Fourier transform of $x\mapsto \frac{\sinh(kx)}{\sinh(x)}$, where $0<k<1$ is fixed.

But I'm having trouble with it.

Motivation: I'm trying to derive an expression for the solution of the Dirichlet problem $\Delta u = 0$ on the strip $[0,1]\times \mathbb R$ with values on the boundary $f_0, f_1$ (assuming necessary niceness conditions for all functions involved).

For this I took the Fourier transform of the solution $u$ and got a formula for $\hat u$ in terms of $\hat{ f_0}, \hat{ f_1}$ and now I want to transform it back. Doing this led (more or less) to the integral expression: $u(x,y) = \int \frac{\sinh(kx)}{\sinh(k)}e^{-iky} \hat f(k) \ dk$.

Now I'm trying to apply the product formula $\int f \hat g = \int \hat f g$ to get everything in terms of $f$. This is why I'm interested in computing the above integral.

My attempt: (which led nowhere, so you may actually ignore everything below)

I think it should be possible using residues. For this I thought of the path having the following components:

$$[-R,R], \ [R,R+i\pi], \ [R+i\pi, \delta + i\pi],$$
$$ \ \text{semicircle from $\delta + i\pi$ to $-\delta + i\pi$ below $i\pi$},$$
$$[-\delta + i\pi, -R + i\pi], \ [-R+i\pi, -R]$$

with the intention of letting $\delta \to 0$ eventually.

The integrals over the vertical components will vanish for $R\to\infty$, so the integral over the path then becomes

\begin{align}
0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx + \left(\int_{\infty}^{\delta} + \int_{-\delta}^{-\infty}\right) \frac{\sinh(k(x+i\pi))}{\sinh(x+i\pi)}e^{-i\omega (x+i\pi)} \ dx \\ & \qquad + \int_{\text{semicircle}}  \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}

Using $\sinh(a+ib) = \sinh(a)\cos(b) + i \cosh(a)\sin(b)$ for real $a,b$, we get

\begin{align}
0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ &\qquad + \left(\int_{\delta}^{\infty} + \int_{-\infty}^{-\delta}\right) \frac{\sinh(kx)\cos(k\pi) + i \cosh(kx)\sin(k\pi)}{\sinh(x)}e^{-i\omega x}e^{\omega \pi} \ dx \\ & \qquad + \int_{\text{semicircle}}  \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}

The integral over the semicircle should go to $$(-\pi i) \ \mathrm{Res}_{x = \pi i}\left(\frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x}\right) = -\pi \sin(k\pi)e^{\omega \pi}$$
as $\delta \to 0$. Therefore

\begin{align}
\pi \sin(k\pi) e^{\omega \pi} &= \int_{-\infty}^\infty \frac{\sinh(kx)(1+\cos(k\pi)e^{\omega \pi}) + i \cosh(kx) \sin(x\pi)e^{\omega \pi}} {\sinh(x)} e^{-i\omega x} \ dx \\
&= (1+\cos(k\pi)e^{\omega \pi}) \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\
& \qquad + i \sin(k\pi)e^{\omega \pi}\int_{-\infty}^\infty \frac{\cosh(kx)}{\sinh(x)}e^{-i\omega x} \ dx
\end{align}

I don't see whether this has brought me any closer to my goal?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}
\expo{-\ic\omega x}
\,\dd x\,\right\vert_{0\ <\ k\ <\ 1}}
\\[5mm] = &\
\int_{-\infty}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\cos\pars{\omega x}\,\dd x
\\[5mm] = &\
2\int_{0}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\cos\pars{\omega x}\,\dd x
\\[5mm] = &\
2\,\Re\int_{0}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\expo{-\ic\omega x}\,\dd x
\\[5mm] = &\
2\,\Re\int_{0}^{\infty}
{\expo{-\pars{-k + 1 + \ic\omega}x}\,\,\, -
\expo{-\pars{k + 1 + \ic\omega}x}\,\,\over 1 - \expo{-2x}}\,\dd x
\\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,&\
\Re\int_{0}^{\infty}
{\expo{-\pars{-k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\,\, -
\expo{-\pars{k/2 + 1/2 + \ic\omega/2}x}\,\,\, \over
1 - \expo{-x}}\,\dd x
\\[5mm] = &\
\Re\left[\int_{0}^{\infty}
{\expo{-x} - \expo{-\pars{k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\, \over
1 - \expo{-x}}\,\dd x -\right.
\\[2mm] &\ \left.\phantom{\Re\left[\right.}\int_{0}^{\infty}
{\expo{-x} - \expo{-\pars{-k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\, \over
1 - \expo{-x}}\,\dd x\right]
\\[5mm] = &\
\Re\bracks{\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic + {1 \over 2}\,k} -
\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k}}
\\[5mm] = &\
{1 \over 2}\,\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic + {1 \over 2}\,k} -
{1 \over 2}\,\Psi\pars{{1 \over 2} - {1 \over 2}\omega\ic - {1 \over 2}\,k}
\\[2mm] + &\
{1 \over 2}\,\Psi\pars{{1 \over 2} - {1 \over 2}\omega\ic + {1 \over 2}\,k} -
{1 \over 2}\,\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k}
\\[5mm] = &\
{1 \over 2}\,\pi\cot\pars{\pi\bracks{{1 \over 2} - {1 \over 2}\omega\ic - {1 \over 2}\,k}}
\\[2mm] + &\
{1 \over 2}\,\pi\cot\pars{\pi\bracks{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k}}
\\[5mm] = &\
{1 \over 2}\,\pi\tan\pars{{1 \over 2}\pi\,k + {1 \over 2}\pi\omega\ic} +
{1 \over 2}\,\pi\tan\pars{{1 \over 2}\pi\,k - {1 \over 2}\pi\omega\ic}
\\[5mm] = &\
\pi\,\Re\tan\pars{{1 \over 2}\pi\,k + {1 \over 2}\pi\omega\ic}
\\[5mm] = &\
\bbx{\pi\sin\pars{\pi k} \over \cos\pars{\pi k} + \cosh\pars{\pi\omega}} \\ &
\end{align}


问:
How to find the orthogonal vector in Lagrange Method of solving PDE (Quasilinear equations)

Example 2.7 Solve the equation $u_x + 3y^{2/3}u_y = 2$ subject to the initial condition $u(x,1) = 1+x$.

It's an example in "An introduction to partial differential equations" book by Yehuda Pinchover, to be solved by the method of characteristics. And I'm trying to solve it by Lagrange's method. I know that I need to find 2 orthogonal vectors so I can find $Ψ$ and $Φ$. But my struggles is finding the 2 orthogonal vectors.

答:
$$u_x + 3y^{2/3}u_y = 2$$
The Charpit-Lagrange equations are :
$$\frac{dx}{1}=\frac{dy}{ 3y^{2/3}}=\frac{du}{2}$$
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{ 3y^{2/3}}$ :
$$x-y^{1/3}=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{2}$ :
$$u-2x=c_2$$
The general solution of the PDE is : $\quad c_2=F(c_1)=u-2x=F(x-y^{1/3})$ :
$$\boxed{u=2x+F(x-y^{1/3})}$$
$F$ is an arbitrary function in the general case.

In the particular case with condition $u(x,1)=1+x$ the function F is non longer arbitrary.
$$1+x=2x+F(x-1^{1/3})=2x+F(x-1)$$
Let $X=x-1\quad\implies\quad x=X+1$
$$1+(X+1)=2(X+1)+F(X)$$
$$F(X)=-X$$
Now the function $F(X)$ is known. We put it into the above general solution where $X=x-y^{1/3}$ :
$$u=2x-(x-y^{1/3})$$
$$\boxed{u(x,y)=x+y^{1/3}}$$


答:
Just do it $\mod 5$.

$x \equiv 1 - y - z \pmod 5$ and

Substitute

$2x + 4y + 3z \equiv 1$

$2(1-y-z) + 4y + 3z \equiv 1$

$2 + 2y + z \equiv 1$

$z \equiv -1 -2y$.  And substitute back

And $x \equiv 1-y -(-1-2y) \equiv 2+y$.

So there are five classes of solutions.

$y \equiv 0; x \equiv 2; z\equiv 4$

$y\equiv 1; x\equiv 3; z \equiv  2$

$y \equiv 2; x\equiv 4; z \equiv 0$

$y \equiv 3; x\equiv 0; z\equiv 3$

$y\equiv 4; x\equiv 1; z\equiv 1$

Title: How can I find all incongruous system solutions mod n?

Body:
$\begin{array}{r}
x+y+z \equiv 1 \bmod 5 \\
2 x+4 y+3 z \equiv 1 \bmod 5
\end{array}$

I have this system and I want to find all incongruous system solutions.

Well, I know that

$x+y+z-1=5k_1$ and $2x+4y+3z-1=5k_2$.

Can I find the solutions before without taking $\bmod 5$?
I mean $x+5q,y+5q_1,z+5q_2$.


问:
Rudin Theorem 2.41 - Heine-Borel Theorem
When proving Theorem 2.41 in Principles of Mathematical Analysis:

Let $E \subset \mathbb{R}^k$. If every infinite subset of $E$ has a limit point in $E$, then $E$ is closed and bounded.

Rudin says,

"If $E$ is not bounded then $E$ contains points $x_n$ with

$$\vert x_n \vert > n \hspace{2cm} (n = 1,2,3,...).$$

The set $S$ consisting of these points $x_n$ is infinite and clearly has no limit point in $\mathbb{R}^k$, hence has none in $E$."

My first reaction to this was that it is indeed obvious as $S$ is composed of discrete points so I could find a ball around any given point not containing a point of $S$. So, if $p \in \mathbb{R}^k$ is the point I'm constructing a ball around, taking the radius of the ball as $r = \min_{x_n \neq p \in S} \{ \vert x_n - p \vert \}$ should work and as $p$ was arbitrary $S$ doesn't have a limit point. However, I'm not sure if I can take the minimum over an infinite set? I honestly don't know why I shouldn't be able to, but nonetheless I decided to try another proof:

Suppose that $S$ has a limit point $p \in \mathbb{R}^k$. Then for any $r > 0$ $B_r(p)$ has infinitely many points of $S$. So in particular, $B_{\frac{1}{2}}(p)$ contains infinitely many points of $S$. Therefore, there exists a point $y_m \in B_{\frac{1}{2}}(p)$ with $\vert y_m \vert > m > \vert p \vert + 1$ (as otherwise, there would be only finitely many points of $S$ in $B_{\frac{1}{2}}(p)$). Therefore, we have

$$\vert y_m - p \vert \geq \big\vert \vert y_m \vert - \vert p \vert \big\vert > 1 $$

However, this is a contradition as we also have $\vert y_m - p \vert < \frac{1}{2}$. Therefore, as $p$ was arbitrary, $S$ has no limit point in $\mathbb{R}^k$.

My questions:
Can I take the min over an infinite set as in my first argument? Is my second argument okay?

I appreciate any answers. Thanks.

答:
Fix an arbitrary point $p\in{\mathbb R}^k$. Put $n_p:=\bigl\lceil|p|\bigr\rceil+1$. All $x_n\in S$  with $n\geq n_p$ satisfy
$$|x_n|>n\geq n_p\geq|p|+1\ .$$ Therefore all these $x_n$ are outside  the ball of radius $1$ around $p$. It follows that the chosen $p$ cannot be a limit point of $S$, and since $p$ was arbitrary the set $S$ cannot have a limit point in ${\mathbb R}^k$ whatsoever.</p>


问:
Questions about Rudin's proof of Lebesgue's Monotone Convergence Theorem
I just finished working through Papa Rudin's proof of Lebesgue's Monotone Convergence Theorem, and I have some questions:

Why are each $E_n$ measurable?

How is $(6)$ concluded, i.e. $\displaystyle \alpha \ge \int_X s\ d\mu$?
 $(5)$ holds for all $0<c<1$, and not $c=1$ itself. Did we do something like taking $\displaystyle \lim_{c\to 1}$ on both sides of the inequality $(5)$? Is that allowed (and why)?

How do we get from $(6)$ to $(7)$? $(6)$ holds for every simple measurable $s$ satisfying $0\le s\le f$, but $f$ itself may not be simple! Are we supposed to use Theorem 1.17 (approximating measurable functions on $[0,\infty]$ by simple measurable functions) here - and construct a sequence of simple measurable functions $\{s_n\}_{n\in\Bbb N}$ such that $0\le s_1 \le s_2 \le \ldots \le f$, and $s_n \stackrel{n\to\infty}{\longrightarrow} f$ pointwise? Then $(6)$ would hold for every $s_n$ ($n\in \Bbb N$), and taking $\displaystyle \lim_{n\to\infty}$ should give $(7)$. Am I correct about this?

Finally, why are the $E_n$ constructed in such a way? $$E_n = \{x:f_n(x) \ge cs(x)\}$$ seems out of the blue! What is the motivation behind this (it definitely does the job, no doubt - but why this particular choice of $E_n$)?

Proof attached for reference:

Thanks a lot for your time and help!

答:
The set $E_n = \left(f_n-c\cdot s \right)^{-1}\left([0,+\infty)\right)$ is measurable as the preimage of a measurable set by a measurable function.

Suppose $x$ and $y$ are two real numbers such that $x \geqslant c \cdot y$ for all $c <1$. Then $x \geqslant y$ (just make $c \to 1$, as the inequality is a closed condition, it remains true for the limit).

By definition, $\int f = \sup \int s$ where $s$ is a simple measurable function with $s \leqslant f$. At a fixed $s \leqslant f$, $\int s \leqslant \alpha$. Taking the supremum, the inequality remains true for $f$ (again, $\leqslant$ is a closed condition, it stays true while taking limits).

As $f_n$ converges pointwisely to $f$, for almost all $x$, there exists $n$ such that $f_n(x) > c s(x)$ (recall that $c<1$ and $s(x) \leqslant f(x)$). This suggests that $f_n$ will be bigger that $cs$ on $E_n$, and what as been done above is justified.


问:
V has a linearly independent subset S = {s1, ... , sk} & basis β = {b1, ... ,bn}. Why is {s1, ... , sk, b1, ... , bn} a spanning set of V?
I know that for the linearly independent subset S, for λ1s1 + λ2s2 + .. + λksk = 0, the coefficients λ1 = λ2 = .. = λk = 0. I also know that for the basis β, the vectors b1, .. , bn are linearly independent and also span(β) = V. However, I do not know how to link both ideas that I know to conclude that the set {s1, .. ,sk, b1, .. ,bn} is a spanning set for V. Thanks!</p>

答: Since $\beta$ is a basis, $\operatorname{span}(\beta)=V$. So, since $S\cup\beta\supset\beta$, $\operatorname{span}(S\cup\beta)=V$.


答:
You have two equations, each of which offers you two possibilities.
$$y(2−2x−y)=0 \implies  y=0 \text{ or }2-2x-y=0$$
$$x(2−x−2y)=0\implies x=0 \text{ or }2-x-2y=0$$

You have to try each possibility from the first equation with each option from the second to see which of the pairs give a solution.


问:
On the topology of disjoint union of topological spaces
I wonder if two definitions of disjoint union of topological spaces are equivalent:

If $X,Y$ are topological spaces, then the disjoint union of topological space $X\sqcup Y$ is the set $X\times\{0\}\cup Y\times\{1\}$ with the topology making $X\times\{0\}$ and $Y\times\{1\}$ clopen and the inclusions $i_X:X\to X\sqcup Y$ and $i_Y:Y\to X\sqcup Y$ homoeomorphisms to their images.

Same setting as above, we give topology on $X\sqcup Y$ by the largest topology making the inclusions $i_X:X\to X\sqcup Y$ and $i_Y:Y\to X\sqcup Y$ continuous.

The latter definition is the usual definition of disjoint union of topological spaces but in Bredon's Topology and Geometry textbook, he defined in a former way.

Could you explain this?

答:
They’re equivalent. Let $\tau_1$ be the topology on $X\sqcup Y$ defined in the first definition, and note that because $X\times\{0\}$ is open in $\tau_1$, a subset of $X\times\{0\}$ is open in the relative topology on $X\times\{0\}$ if and only if it is actually open in $X\sqcup Y$, i.e., if and only if it is in $\tau_1$. Similarly, a subset of $Y\times\{1\}$ is open in the relative topology on $Y\times\{1\}$ if and only if it is actually in $\tau_1$.

It’s clear that $i_X$ and $i_Y$ are continuous with respect to $\tau_1$, so it only remains to show that $\tau_1$ is maximal with respect to this property. Suppose that $\tau$ is a topology on $X\sqcup Y$ making $i_X$ and $i_Y$ continuous, and let $U\in\tau$. Let $V=i_X^{-1}[U]$ and $W=i_Y^{-1}[U]$. Then $V$ is open in $X$, and $i_X$ is a homeomorphism from $X$ to $X\times\{0\}$ with respect to $\tau_1$, so $i_X[V]\in\tau_1$. And $i_X[V]=U\cap(X\times\{0\})$, so $U\cap(X\times\{0\})\in\tau_1$. Similarly, $i_Y[W]=U\cap(Y\times\{1\})\in\tau_1$, so $U=i_X[V]\cup i_Y[W]\in\tau_1$. Thus, $\tau\subseteq\tau_1$, and $\tau_1$ is indeed the largest topology on $X\sqcup Y$ making $i_X$ and $i_Y$ continuous.

The second definition is more elegant and expresses an important fact about the topology, but the first definition makes it a bit easier to understand exactly what the open sets are.


问:
Basic confusion in real analysis
I have just started real analysis (first time) and self studying from book Mathematical analysis by Apostol.

In the first chapter of real and complex number system, it is written that

We assume there exist a non empty set R of objects ,called real numbers which satisfy the 10 axioms listed below .

1.But what is objects in this context ,is it number or could be a function or it could be anything ?

2.And if let's say it is given mathematical operation (addition) can be perform on the elements of sets,then objects must  be a number or function only?

答:
"Objects" is another word for thingamabob.

You have these thingamabobs that obey the axioms.  That's the only thing  you know about them.  They are thingamabobs that obey the axioms.

ut what is objects in this context ,is it number or could be a function or it could be anything ?

"nummm-burrs"?  "Funk-shun"? What are those words we have never hear of those words and don't know what they mean. (Actually maybe I have heard of "functions". If so these aren't "functions".  They are just things that exist in their own rights.)   This things are thingamabobs we just made up that obey those axioms.

But wait  "We assume there exist a non empty set $\mathbb R$ of thingamabobs ,called real numbers which satisfy the 10 axioms listed below ."

So these thingwhazzits are "numbers".  Every time we use the word "number" we will be talking these doohickies.

Title: Basic confusion in real analysis


问:
finding codomain and range of a function
I am solving function exercises and i encounter a question. I tried to solve it but i am not sure if it is true or not. Can you help me ?

Question: Find the codomain and the range of the following function , $f:(-4,4) \rightarrow R $  and $f(x)=x^2-6x+4$

My work:

I said that if $-4<x<4$ then $0 \leq x^2 <16$ , $-24<-6x<24$ , so $-20<x^2 -6x +4 <44$. Hence , the codomain of f is equal to $(-20,44)$

To find the range of f , i should find the greatest and lowest value which f has taken , so i placed the end points and critical value of f into the function, for $x=-4 , f(-4)=44$ ,$x=4 , f(4)=-4$  $x=3 , f(3)=-5$ so range of f is equal to $(-5,44)$

Is my solutions correct ? If not , can you help ?

答:
This is a parabolic function, it is opens upward, need to find its valley at $x=-\frac{b}{2a}$ $= -\frac{-6}{2}$ = 3 , check f(-4) = 44, f(3)= -5, f(4)=-4, so the range is (-5,44).

In your writing, you can not multiply negative values w/o changing the direction of the inequity sign. That is why you are wrong.


问:
Prove that $\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$
$$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$$
It tells me to prove this. I inputed both and they both equale to -0.0866. Is this enough to prove it or do I have to make the left side equaled to the right- and if so how do you do this? Any help is much appricated. Thank you!

答:
So, just calculate this integral:

notice that $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$

in your integral put $\cos(2x) = t$ then $-2\sin(2x)dx = dt$
What will we have?


答:
Question 1.

For any measurable functions $f,g:X\to[0,\infty]$, it is a standard exercise (which you should definitely attempt by yourself; otherwise there are certainly questions about this on this site too) that $[f\leq g]:=\{x\in X\,|\,f(x)\leq g(x)\}$ is measurable (from this it follows that $[f\geq g], [f<g], [f>g], [f=g]$ are all measurable). Do you see how this applies to your case?

Question 2.

Yes, you can take limits in $(5)$. For every $c<1$, we have $\alpha\geq c\int_X s\,d\mu$. So, take the limit as $c\to 1^-$ on both sides. You should have definitely seen in a basic analysis course that inequalities are preserved under limits.

Question 3.

From $(6)$ to $(7)$ it's simply applying the definition of the Lebesgue integral as a supremum. We have defined $\alpha$ in the beginning, and what $(6)$ is showing is that for every simple $0\leq s\leq f$, we have $\int_Xs\,d\mu\leq \alpha$. In other words, the set of numbers $\left\{\int_Xs\,d\mu\,:\, \, \text{$0\leq s\leq f$ is simple}\right\}$ has $\alpha$ as an upper bound. Thus by definition of supremum, we have that
\begin{align}
\sup\left\{\int_Xs\,d\mu\,:\, \, \text{$0\leq s\leq f$ is simple}\right\} \leq \alpha.
\end{align}
But the LHS is none other than $\int_Xf\,d\mu$, by definition.

Question 4.

The inequality $\lim\limits_{n\to \infty}\int_Xf_n\,d\mu =\alpha \leq \int_Xf\,d\mu$ is a very trivial consequence of monotonicity of the integral. The other direction is not so trivial, because the definition of $\int_Xf\,d\mu$ involves a huge supremum over all simple functions. Also, up to this point in the treatment, the "only" thing we really know about integrals are basic facts (theorem 1.24), and the only integrals we can explicitly calculate are those of simple functions. The idea is therefore to somehow reduce the more complicated problem of proving the reverse inequality to something more familiar.

By definition of the Lebesgue integral using supremum, showing that $\int_Xf\,d\mu\leq \alpha$ is thus equivalent to showing that for every simple $0\leq s\leq f = \lim f_n$, we have $\int_Xs\,d\mu \leq \alpha$. It would be nice if we could say something like "because $s$ is smaller than $f$ and since $f_n$ increases to $f$, thus for large $n$, we have $s\leq f_n$". If we could say this, then we have $\int_Xs\,d\mu \leq \int_Xf_n\,d\mu$ for all large $n$, and hence by taking limits, $\int_Xs\,d\mu \leq \alpha$, thereby completing the proof.

Unfortunately, this isn't quite right. We CANNOT deduce that for large $n$, that $s\leq f_n$. Thus, what can we do? We give ourselves some room for error, by introducing the scaling factor $0<c<1$. Then, $cs$ will be an overall scaled down function, and for this, we can obtain a nice control: $cs\leq f_n$ on $E_n$, and $\bigcup E_n = X$. So by Rudin's argument, we thus deduce $c\int_Xs\,d\mu \leq \alpha$. Then, finally, we let $c\to 1^-$.

This final idea of "giving yourself some room for error" is a VERY VERY VERY common idea in analysis. If you've studied baby Rudin (or Spivak or any other introductory book), you must have surely seen such ideas. The simplest example of this I can think of is that given $z\in \Bbb{C}$, we have $z=0$ if and only if for every $\epsilon>0$, we have $|z|\leq \epsilon$. These are equivalent statements, but sometimes, the second statement is easier to prove, because you have an $\epsilon$ amount of wiggle room to establish the inequality $|z|\leq \epsilon$.

Title: Questions about Rudin's proof of Lebesgue's Monotone Convergence Theorem

Body:
I just finished working through Papa Rudin's proof of Lebesgue's Monotone Convergence Theorem, and I have some questions:

Why are each $E_n$ measurable?

How is $(6)$ concluded, i.e. $\displaystyle \alpha \ge \int_X s\ d\mu$?
 $(5)$ holds for all $0<c<1$, and not $c=1$ itself. Did we do something like taking $\displaystyle \lim_{c\to 1}$ on both sides of the inequality $(5)$? Is that allowed (and why)?

How do we get from $(6)$ to $(7)$? $(6)$ holds for every simple measurable $s$ satisfying $0\le s\le f$, but $f$ itself may not be simple! Are we supposed to use Theorem 1.17 (approximating measurable functions on $[0,\infty]$ by simple measurable functions) here - and construct a sequence of simple measurable functions $\{s_n\}_{n\in\Bbb N}$ such that $0\le s_1 \le s_2 \le \ldots \le f$, and $s_n \stackrel{n\to\infty}{\longrightarrow} f$ pointwise? Then $(6)$ would hold for every $s_n$ ($n\in \Bbb N$), and taking $\displaystyle \lim_{n\to\infty}$ should give $(7)$. Am I correct about this?

Finally, why are the $E_n$ constructed in such a way? $$E_n = \{x:f_n(x) \ge cs(x)\}$$ seems out of the blue! What is the motivation behind this (it definitely does the job, no doubt - but why this particular choice of $E_n$)?

Proof attached for reference:

Thanks a lot for your time and help!


答:
Let's consider that:  $\sin{x}\cos{x}=\frac{1}{2}\sin{2x}$ and $(-\cos{2x})'=2\sin{2x}$. So:
$$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= \int_{π/3}^{π/4}\frac{{\frac{1}{2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=\frac{1}{8}\log{(1-2\cos{2x}})\lvert_{\pi/3}^{\pi/4}=0-\frac{\log{2}}{8}$$

$\textbf{EDIT:}$ to be more clear let's observe that,

You should have at the numerator the derivative of $(1-2\cos{2x})$ to be able to solve the integral with the logarithm as I have shown...but the numerator it is not exactly the derivative of the denominator, why? $$(1-2\cos{2x})'=(-2\cos{2x})'=-2\cdot(\cos{2x})'=4\sin{2x}$$ But at the numerator we have $\frac{1}{2}\sin{2x}$ so in effect we can multiply and divide by 4 to obtain the desidered form, in this following way:
$$\int_{π/3}^{π/4}\frac{{\frac{1}{2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=\int_{π/3}^{π/4}\frac{{\frac{\color{red}4}{\color{red}4 \cdot 2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=$$
$$=\frac{1}{8}\int_{π/3}^{π/4}\frac{{4\sin{2x}}}{{1 - 2\cos (2x)}}dx=\Big[\frac{1}{8}\log{(1-2\cos{2x})}\Big]\Big\lvert_{\pi/3}^{\pi/4}$$

Title: Prove that $\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$

Body:
$$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$$
It tells me to prove this. I inputed both and they both equale to -0.0866. Is this enough to prove it or do I have to make the left side equaled to the right- and if so how do you do this? Any help is much appricated. Thank you!


问:
$f$ Open, Closed and continuous but not a local homeomorphism
Give an example of topological spaces $X$ and $Y$ and a map $f: X \to Y$ that
$f$ is an open, closed, and continuous but not a local homeomorphism.
(The map and the topological spaces have to satisfy these 4 characteristics simultaneously)

答:
Let $f: [0,1] \times [0,1] \to [0,1]$, all in the usual topology, be given by $f(x,y)=x$ (the projection).

$f$ is open and continuous, as all projections are.

$f$ is closed as the domain is compact and the codomain is Hausdorff.

$f$ is not a local homeomorphism for dimension reasons, basically. Or: all open sets in the domain cannot be disconnected by removing a point, while (basic) open sets in the codomain can be, so they can never be homeomorphic.


答:
Part of the point is that you want numbers which have certain particular properties - but when you start it is not necessarily clear that the properties are all compatible with each other.

Taking an axiomatic approach (ie beginning with the properties), there are two questions to answer: "are there any objects at all which have this combination of properties?" (the existence question) and "do the objects which have this combination of properties behave like the objects I want to call 'real numbers'?" (the question of whether I have defined the right thing, and specified enough properties)

Once we answer those questions we can start calling the objects which have the properties 'real numbers'. It is a question of taking a step back and noticing the assumptions which are made when we call things 'real numbers'.

Title: Basic confusion in real analysis

Body:
I have just started real analysis (first time) and self studying from book Mathematical analysis by Apostol.

In the first chapter of real and complex number system, it is written that

We assume there exist a non empty set R of objects ,called real numbers which satisfy the 10 axioms listed below .

1.But what is objects in this context ,is it number or could be a function or it could be anything ?

2.And if let's say it is given mathematical operation (addition) can be perform on the elements of sets,then objects must  be a number or function only?


问:
Marginal Distribution Calculation of Discrete Random Vector

In the question I know how to calculate marginal distribution for $X_1$ and $X_2$
Like for $X_1$ the marginal distribution for each column is the sum of each joint probability mass function in that column.
For example the marginal distribution for $0$ column of $X_1 = 0.343$
But I'm not able to understand how to use marginal Probability of $X_1$ in the binomial distribution as asked in the question in the image!

答:
For $X_1$~$Bi(3,p)$
the probability for $X_1=3$ is $\binom{3}{0}p^3(1-p)^0 = p^3$ therefore
$p^3=0.027$ which will give $p=0.3$

$X_1 \sim Bi(3,0.3)$ is the answer & similarly we can find
$X_2 \sim Bi(3,0.1)$.


问:
Product of two first countable sets is first countable
Prove that the product of two first countable spaces $X$ and $Y$ is first countable.

Attempt:Let $X$ and $Y$ be two first countable spaces. Let $(x,y) \in X \times Y$. Then $x \in X$ and $y \in Y$. Then $x$ has a countable neighborhood basis, $\beta_x$.Also $y$ has a countable neighborhood basis $\beta_y$. Then the set $\beta_x \times \beta_y=\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$ is a countable neighborhood basis at $(x,y)$. The set is countable since the cardinality is $|\beta_x \times \beta_y|$ and the cartesian product of two countable sets is countable. Each basic open set $U \times V$ containing $(x,y)$ contains some $B_x \times B_y$, since if $x \in U$ and $y \in V$, there is some $B_x$ with $x \in B_x\subset U$ and some $B_y$ with $y \in B_y \subset V$.So $(x,y) \in B_x \times B_y \subset U \times V$.

Comment: I looked at the proof of this for countable products and noticed the proof to be much more demanding, which makes me unsure if this attempt suffices.However I noticed this is generally the case for proofs involving infinite/finite products. Is this ok, if not how could I change/improve it?Also I am unsure if it appropriate to use the notation $\beta_x \times \beta_y$ for the set $\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$.

答:
Yes, it's correct. You rightly realised that you only need to show that a basic open neighbourhood of $(x,y)$ contains a set from $\beta_x \times \beta_y$ (this is quite a standard notation BTW) to show that the latter forms a local base at $(x,y)$.

Finite products now follow by induction of course, and we can get the countable index case as well (we cannot go higher), but we need a bit more set theory to count the base where we use basic open sets but with all non-trivial components from the correct countable component local bases. It's a bit more notation but not really a lot harder.


问:
Evaluating $\int_{E_{ij}} (s+t)\ d\mu$ in Proposition $1.25$ (Rudin's RCA)
I am unsure how $$\int_{E_{ij}} (s+t)\ d\mu$$ has been evaluated in Proposition 1.25, and I've attached a picture of the same for context:

(Please scroll down to below the image to see my question and thoughts!)

$s,t$ are simple measurable functions, i.e. they take the form $$s = \sum_{i=1}^n \alpha_i \chi_{A_i}, t = \sum_{i=1}^m \beta_i \chi_{B_i}$$
where $A_i = \{x: s(x) = \alpha_i\}$ and $B_i = \{x:t(x) = \beta_i\}$. $\chi_S$ denotes the indicator function of set $S$, for every $S$. On the previous page, Rudin defined
$$\int_E s\ d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$$

Firstly, let's think about $s+t$. Since $s,t$ are simple measurable, so is $s+t$. In fact, since $s$ takes $n$ distinct values, and $t$ takes $m$ distinct values, $s+t$ can take at most $mn$ distinct values, right? Let's say these are $\gamma_i$ for $i\in\{1,2,\ldots,p\}$ where $p\le mn$. Now they define $E_{ij} = A_i \cap B_j$.

So, by definition, one must have $$\int_{E_{ij}} (s+t)\ d\mu = \sum_{k=1}^p \gamma_k \mu(C_k\cap E_{ij})$$ where $C_k = \{x: (s+t)(x) = \gamma_i\}$. So we must show
$$\sum_{k=1}^p \gamma_k \mu(C_k\cap E_{ij}) \stackrel{??}{=} (\alpha_i + \beta_j) \mu(E_{ij})$$

Is it right to say that $C_k \cap E_{ij} = \varnothing$ for all $k$ for which $\gamma_k \neq \alpha_i + \beta_j$? If yes, then I think that solves it and we obtain $(\alpha_i + \beta_j) \mu(E_{ij})$ on the RHS as required.

Is similar reasoning employed when claiming in the next line that $\int_{E_{ij}} s\ d\mu = \alpha_i\mu(E_{ij})$ and $\int_{E_{ij}} t\ d\mu = \beta_j\mu(E_{ij})$?

Thank you!

答:
Yes, $C_k \cap E_{ij} = \varnothing$ for all $k$ for which $\gamma_k \neq \alpha_i + \beta_j$. This is because $x\in C_k \cap E_{ij}$ implies $s(x)=\alpha_i$, $t(x)=\beta_j$ and $(s+t)(x)=\gamma_k$ but $\gamma_k \neq \alpha_i + \beta_j$, which is not possible.

Yes, a similar reasoning can be applied to these situations. In fact, one may observe that the simple functions are constant on these sets. Take a look at the general definition:
$$\int_E s\ d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$$
except for $A_i$ where $\alpha_i$ is equal to that constant (in which case $A_i\cap E=E$), $A_j\cap E=\varnothing$.


问:
What will be $\mu \left (f (\partial \left ([0,1]^k \right ) \right )\ $?

Let $f : [0,1]^k \longrightarrow [0,1]^k$ be an one to one and onto $C^1$-map. We know that the Lebesgue measure of $\partial \left ( [0,1]^k \right )$ is zero. Can we conclude that the Lebesgue measure of $f \left ( \partial \left ([0,1]^k \right ) \right )$ is also zero?

Any help in this regard will be highly appreciated. Thanks in advance.

答: Yes, any Lispchitz function maps sets of measure $0$ to sets of measure $0$. And any $C^{1}$ map on $[0,1]^{k}$ is Lipschitz.


问:
Maximize quadratic function over unit sphere
In lecture, we proved that, given a symmetric matrix $A$, the

$$\max_{\|x\|_2 = 1} x^T A x$$

is the largest eigenvalue $\lambda_{\max}$ of matrix $A$: we diagonalize the matrix $A$ and show that for every unit vector x the inequality $x^T A x \leq \lambda_{\max}$ holds. I understand every part of the computation.

What I do not understand is why it suffices to consider unit vectors for $x$? If the inequality is true for every $x$ in the unit sphere and we find one for which the inequality is an equality, then of course we have found a maximum. But in the unit sphere do not only lie the unit vectors.

答:
Let us consider a Lagrangian function of the following form
\begin{equation}
\mathcal{L}(\vec{x},\lambda) = -\vec{x}^{\intercal}A\vec{x} + \lambda(\vec{x}^2 - 1).
\end{equation}
The first derivatives of the Lagrange function are
\begin{equation}
\frac{\partial\mathcal{L}}{\partial\vec{x}} = -2A\vec{x} + 2\lambda\vec{x} = 0,\\
\frac{\partial\mathcal{L}}{\partial\lambda} = \vec{x}^2 - 1 = 0.
\end{equation}
So,
\begin{equation}
A\vec{x} = \lambda\vec{x},\\
\vec{x}^2 = 1.
\end{equation}
It is seen from the last equation that the Lagrangian multiplier is an eigenvalue of the matrix $A$. At a stationary point, the Lagrange function can be rewritten as follows
\begin{equation}
\mathcal{L} = -\lambda\vec{x}^2,\\
\vec{x}^2 = 1.
\end{equation}
The last equation shows that the Lagrange function has the minimum value at $\lambda = \lambda_{\rm max}$. The min value of this function is equal to $-\lambda_{\rm max}$ only if $\vec{x}^2 = 1$. So the max value of the $\vec{x}^{\intercal}A\vec{x}$ is equal to $\lambda_{\rm max}$ only if $\vec{x}^2=1$. For $\vec{x}^2\neq1$, the value of the $\vec{x}^{\intercal}A\vec{x}$ at the maximum will be different.


问:
$dx/dt=rx-\ln(1+x)$
I have the equation $\frac{dx}{dt}=rx-\ln(1+x)$
I know the bifurcation point is at $r=1$ but I saw this through the geometric approach. How would I get this from a purely algebraic method?

答: There is always a root of $f\colon x\in(-1,\infty)\mapsto rx-\log(1+x)$ at $x=0$.  Since $f$ is convex, and $\lim_{x\to -1}f(x)=\lim_{x\to+\infty}f(x)=+\infty$, there is one other root unless the horizontal line is the support line of $f$, i.e., $f'(0)=0$, which is the same as saying $r=1$.

Title: $dx/dt=rx-\ln(1+x)$


答: A spanning set remains a spanning set when further vectors are added.</p>

Title: V has a linearly independent subset S = {s1, ... , sk} & basis β = {b1, ... ,bn}. Why is {s1, ... , sk, b1, ... , bn} a spanning set of V?

Body: I know that for the linearly independent subset S, for λ1s1 + λ2s2 + .. + λksk = 0, the coefficients λ1 = λ2 = .. = λk = 0. I also know that for the basis β, the vectors b1, .. , bn are linearly independent and also span(β) = V. However, I do not know how to link both ideas that I know to conclude that the set {s1, .. ,sk, b1, .. ,bn} is a spanning set for V. Thanks!</p>


问:
Irreducible triangulations of manifolds
Does there exist a closed Riemann manifold $M$, two distinct irreducible triangulations $S_1$ and $S_2$ of $M$, and a triangulation $T$ of $M$ such that there exists a sequence of edge contractions on $T$ such that $T$ reduces into $S_1$ and another seuquence of edge contractions on $T$ such that $T$ reduces into $S_2$?

Perhaps being a closed manifold is an unnecesary condition, but I am partial to closed Riemann manifolds.  My assumption is that the answer to my question is "Yes", but if the answer is "No", then that would mean that the irreducible triangulations of a manifold form an equivalence relation between triangulations of that manifold.

答: Here is a triangulation of the real projective plane which reduces into both irreducible triangulations of the real projective plane.

Title: Irreducible triangulations of manifolds


问:
The number of distinct real roots of the folllowing determinant

The number of distinct real roots of determinant $$
        \begin{vmatrix}
        \csc x & \sec x & \sec x \\
        \sec x & \csc x & \sec x \\
        \sec x & \sec x & \csc x \\
        \end{vmatrix}
=0$$ lies in the interval $\frac{-\pi}{4} \le x \le \frac{\pi}{4}$ is
(a) $1$
(b) $2$
(c) $3$
(d) $0$

I tried to solve the determinant and solved it till I got:

 $$(\csc x + 2\sec x)(\csc x - \sec x)^2=0$$

How do I get the final answer?
Please offer your assistance,
 Thank you

答:
For the root of given determinant, you need either $\csc x+2\sec x=0$ or $\csc x-\sec x=0$.

i.e. $\tan x=-1/2$ or $tan x=1$

As $\tan x$ is bijective in $(-\pi/4,\pi/4)$ so both conditions give you a unique value of $x$ in prescribed interval. Hence the correct answer is $2$.


问:
If $X$ is second countable and $A \subset X$ is uncountable, $A$ has a limit point
If $X$ is second countable and $A \subset X$ is uncountable, $A$ has a limit point.

Attempt:Suppose $A$ is uncountable and each point $a \in A$ is isolated. Let $\beta$ be a countable basis for $X$. Then for each $a \in A$ there is a $B_a \in \beta$ with $B_a \cap A=\{a\}$. Then $a \neq b \implies B_a \neq B_b$, and for each $x \in A$ there is a unique $B_x \in B$ containing $x$, with $B_x \cap A=\{x\}$. Thus $|\beta| \geq |A|$. But $A$ was assumed to be uncountable and $\beta$ countable, a contradiction, thus the assumption that $A$ has no limit points was false.

Comment: Could I get a proof verification if this attempt was completely correct? Thanks.

答:
The "streamlined" version, without duplication etc. :

Let $\beta$ be a countable base for $X$ and suppose $A$ is uncountable and such that no point of $A$ is a limit point of $A$.

Then by definition for each $a \in A$ we have an open set, that we can take from $\beta$ (!), say $B_a \in \beta$, such that $B_a \cap A = \{a\}$. (We always have such an open set $O$ containing $a$, but then we pick $B_a \in \beta$ so that $a \in B_a \subseteq O$ of course, using the property of a base.)

The assignment $a \to B_a$ is injective from $A$ to $\beta$:

If $U_a =U_b$ then $\{a\} = U_a \cap A= U_b \cap A= \{b\} \to a = b$. So $\aleph_0 < |A| \le |\beta| = \aleph_0$, which is a contradiction.

So for an uncountable $A$ there must be some $a \in A$ that is a limit point of $A$.


问:
Independence of $T$ and $B_T$
Let $\{B_t:t\ge0\}$ be a real brownian motion such that $B_0=0$. Let $T=\inf \{t:B_t \notin (-a,a)\}$ with $a>0$. Are $T$ and $B_T$ independent?

I tried the following and I would like your opinion. Let $t>0$,
$P(T<t)=P(T<t, B_T=a)+P(T<t, B_T=-a)=2 P(T<t, B_T=a)$.

答:
Let $\mathcal G$ the $\sigma$-algebra generated by $\|B_t\|$.
Then $T$ is $\mathcal G$-measurable.

Now, using what you've already tried, see if you can show that $\mathbb E(B_T\cdot 1_A) = 0$ for every $A\in\mathcal G$.

Therefore you can show that $\mathbb E(B_T|\mathcal G) \equiv 0$ hence $B_T$ is independent from $\mathcal G$.

Title: Independence of $T$ and $B_T$


问:
Determine if the set $A=\{(\frac{1}{n},1-\frac{1}{n}) \mid n\in\mathbb{N}, n\geqslant 2 \}$ is closed in the open unit disk $B$.

Determine if the set $A=\{(\frac{1}{n},1-\frac{1}{n}) \mid n\in\mathbb{N}, n\geqslant 2 \}$ is closed in the open unit disk $B$.

A set is closed if it contains all it's limit points right? So if $x_n=(\frac{1}{n}, 1-\frac1n) \to (0,1)$, and $(0,1) \notin B$ thus the set must be open?

答: $A$ has no limit points inside $B$.  Hence it contains all its limit points and it is a closed subset of $B$.


问:
Existence of non-constant continuous function
Let both $(X , d)$ and $(Y , d)$ are non-trivial connected metric spaces. That is both $X$ and $Y$ contains at least two elements each.

Is it true that there always exist a non-constant continuous function from $(X ,d)$ to $(Y ,d)$. I know that if $(X ,d)$ is connected and $(Y ,d)$ is totally disconnected (i.e connected components of $(Y ,d)$ are the one-point sets) then there cannot exist a non-constant continuous function from $(X , d)$ to $(Y , d)$. Also if $(X ,d)$ is not connected then no matter what is the structure of $(Y ,d)$ there always exist a non-constant continuous function. Furthermore, with the mentioned assumptions if the cardinality of $X$ is less than or equal to $Y$ then is it true that there always exist an injective continuous map from $(X ,d)$ to $(Y ,d)$.

Any help please. Thanks in advance.

答:
Let me address all the points here:

If $X$ is connected and $Y$ totally disconnected, then only constant functions $X\to Y$ are continuous.

If $X$ is disconnected and $Y$ has at least two points then there exists a non-constant continuous function $X\to Y$.

If $X$ and $Y$ are connected then it doesn't mean that there is a non-constant continuous function $X\to Y$. Consider $X=[0,1]$ and let $Y$ be the pseudo-arc. The pseudo-arc is well known to be connected but totally path-disconnected (any such space will work) and so every continuous map $X\to Y$ is constant. This example also answers the question about cardinalities.

If $X$ is any metric space and $Y$ is path connected, both with at least two points, then there is always a non-constant map $X\to Y$. Let $x,y\in X$, $x\neq y$. By Tietze extension theorem the function $\{x,y\}\to[0,1]$ that maps $x$ to $0$ and $y$ to $1$ extends to $A:X\to [0,1]$. We then compose that $A$ with any path $[0,1]\to Y$ connecting two distinct points.


问:
Evaluate the limiting value using mean-value theorem

Prove that
$$\lim_{n\rightarrow\infty}\int_{n}^{n+ 1}\sin e^{x}{\rm d}x= 0$$

I already have a solution but I want to see a proof(s) using mean-value theorem, it will be travelled ! Now let me show
$$\left | \int_{n}^{n+ 1}\sin e^{x}{\rm d}x \right |= \left | \int_{e^{n}}^{e^{n+ 1}}\frac{\sin x}{x}{\rm d}x \right |\leq\left | \left ( \frac{\cos e^{n}}{e^{n}}- \frac{\cos e^{n+ 1}}{e^{n+ 1}} \right )- \int_{e^{n}}^{e^{n+ 1}}\frac{\cos x}{x^{2}}{\rm d}x \right |\leq$$
$$\leq\left | \frac{\cos e^{n}}{e^{n}}- \frac{\cos e^{n+ 1}}{e^{n+ 1}} \right |+ \int_{e^{n}}^{e^{n+ 1}}\frac{{\rm d}x}{x^{2}}\leq\frac{1}{e^{n+ 1}}+ \frac{1}{e^{n}}+ \frac{1}{e^{n+ 1}}+ \frac{1}{e^{n}}\rightarrow 0\,{\rm as}\,n\rightarrow\infty$$

答:
MVT for integrals:

$|\displaystyle{\int_{e^n}^{e^{n+1}}}\dfrac{\sin x}{x}dx|=$

$|\dfrac{\sin t}{t}\displaystyle{ \int_{e^n}^{e^{n+1}}}1dx|=|\dfrac{\sin t}{t}(e^{n+1}-e^n)|$

where $t \in [e^n,e^{n+1}].$

Take the limit.

There is a problem here : Taking the limit gives you an upper bound $(e-1)$, not the desired result.


问:
how can we prove that this function is homomorphism?
If $G_1 , ... , G_n$ are groups
let $G=G_1\times G_2\times\dotsb\times G_n$.

$E_p\colon G\to G_{p^{-1}(1)}\times G_{p^{-1}(2)}\times\dotsb\times G_{p^{-1}(n)}$
where $p \in S_n$ is defined as
$(g_1 , g_2 , \dotsc , g_n )\mapsto(g_{p^{-1}(1)} , g_{p^{-1}(2)} , \dotsc ,g_{p^{-1}(n)})$.

Now in the case $G_1 = G_2 = \dotsb = G_n$, show that the function $A\colon S_n\to\operatorname{Aut}(G)$ defined by $p\mapsto E_p$ is homomorphism.

I tried to solve it "the idea is not difficult" but every time, I found it's not a homomorphism!

So I need help

thanx

This is an exercise #8 in Dummit and Foote's abstract algebra book
page 156 but i have changed the symbols because I can't use the greek symbols in this site.

答:
Just compute. Let $\sigma, \tau\in S_n$ and $g = (g_1, \ldots, g_n) \in G$, then
we have denoting by $()_i\colon G \to G_i$ the projection onto the $i$-th coordinate:
\begin{align*} (E_\sigma E_\tau g)_i &= (E_\tau g)_{\sigma^{-1}(i)}\\
  &= g_{\tau^{-1}\sigma^{-1}(i)}\\
  &= g_{(\sigma\tau)^{-1}(i)}\\
  &= (E_{\sigma\tau}g)_i
\end{align*}
Hence $E_\sigma E_\tau = E_{\sigma\tau}$, as wished.</p>


问:
How can I force the auxiliary angle method to work on this question?
If $\tan{a}=\frac{x}{y}$, when what is the value of $x\cos{2a}+y\sin{2a}$? I know you can simply draw a trialge and expand the trig identities but the auxiliary method should also work in this case, but Im not sure how to continue. I have so far $R\cos{(2a-\beta)}$ where $R = \sqrt{a^2+b^2}$ and $\beta = \arctan{\frac{y}{x}}$. How can I simplify further to get the answer of $x$? I am stuck on $$\sqrt{a^2+b^2}\big{(}(2\arctan{\frac{x}{y}})-\arctan{\frac{y}{x}}\big{)}$$

Body:
Remember the relations
$$\cos 2a=\frac{1-\tan^2a}{1+\tan^2 a};\;\sin 2a=\frac{2\tan a}{1+\tan^2 a}$$
set $\tan a=t$
$$x\cos{2a}+y\sin{2a}=\frac{1-t^2 }{t^2+1}x+\frac{2 t }{t^2+1}y$$
$t=\frac{x}{y}$
$$x\cos{2a}+y\sin{2a}=\frac{2 x}{\frac{x^2}{y^2}+1}+\frac{x \left(1-\frac{x^2}{y^2}\right)}{\frac{x^2}{y^2}+1}=\frac{3 x y^2-x^3}{x^2+y^2}$$


问:
Prove equality of cauchy schwarz if one vector is a multiple of other from this proof

I need to show that the inequality is an equality iff x is a multiple of y or the other way around and I have to make use of this proof presented.

So for the forward direction assume x is a multiple of y. Then there is some c where x-cy=0 so |x-cy|=0. Let t=c. Then 0 = |x-cy| = |x|^2 - 2c(x,y) + c^2|y|^2 = 0  We note the right side is a quadratic that is greater than or equal to 0, thus 0 must be the minimum and the quadratic achieves its minimum at t=c. And we note that t=(x,y)/|y|^2 is the minimum so plugging that in for t, we can carry out the same proof of the inequality but this time with the equal sign.

Backward direction
Assume |(x,y)|=|x||y|. How do I show that either x or y is a scalar multiple of the other?

答:
Hint: Show that there is a $t$ for which $\|x\|^{2}-2t(x,y)+t^{2}\|y\|^{2}=0$ (by solving the quadratic) . This makes $\|x-ty\|=0$ so $x=ty$.

[In fact, $t=\frac {\|x\|} {\|y\|}$ will do].


问:
Find tangent plane for mean squared error loss function
I would really appreciate your help on a rather simple issue that I just can't solve on my own. I'd like to visualize gradient descent for a simple linear regression problem with two parameters $\theta_0$ (intercept) and $\theta_1$ (slope) using mean squared error as the cost function. What I'd like to visualize is a 3d plot of the cost function and the corresponding tangent plane for different $\theta_0$, $\theta_1$.

Here's what I have so far:

Cost function:
$$J(\theta_0,\theta_1)=\frac{1}{2m}\sum_{i=1}^{m}{(\theta_0 + \theta_1 x_{1}^{(i)}-y^{(i)})^2}$$

Partial derivative wrt $\theta_0$: $$ J_{\theta_0}(\theta_0, \theta_1) = \frac{1}{m} \sum_{i=1}^{m}{(\theta_0 + \theta_1 x_{1}^{(i)} - y^{(i)})} $$

Partial derivative wrt $\theta_1$: $$ J_{\theta_1}(\theta_0, \theta_1) = \frac{1}{m} \sum_{i=1}^{m}{(\theta_0 + \theta_1 x_{1}^{(i)} - y^{(i)})(x_{1}^{(i)})} $$

Let's say my tangent plane kisses the loss function surface plot at $(\theta_0,\theta_1,z)=(1,2,3)$. So if I want to know the z-value at $(\theta_0,\theta_1)=(4,5)$ I would to the following:
$$ z = J_{\theta_0}(1,2)(\theta_0-1)+J_{\theta_1}(1,2)(\theta_1-2)+3 $$
$$ z = J_{\theta_0}(1,2)(4-1)+J_{\theta_1}(1,2)(5-2)+3 $$

However, if I implement all of this in python my tangent plane won't end up where I would expect it to be. Here's my code:

```<code>import sympy as sp
import numpy as np
import plotly.graph_objects as go
import dash_html_components as html
import dash_core_components as dcc

# Generate a linear dataset
m = 100
X = 2 * np.random.rand(m, 1)
y = 4 + 3 * X + np.random.randn(m, 1)

# Append ones for bias term
X_b = np.c_[np.ones(len(X)),X]

# Loss function (j), its derivatives (djt0, djt1) and equation for tangent plane (z)
j = lambda theta0,theta1: 1/2*1/m*((X_b.dot(np.array([[theta0],[theta1]]))-y)**2).sum()
djt0 = lambda theta0,theta1: 1/m*(X_b.dot(np.array([[theta0],[theta1]]))-y).sum()
djt1 = lambda theta0,theta1: 1/m*((X_b.dot(np.array([[theta0],[theta1]]))-y)*X_b[:,1]).sum()
z = lambda theta0,theta1: djt0(10,10)*(theta0-10)+djt1(10,10)*(theta1-10)+j(10,10)

# setup sympy
theta0,theta1 = sp.symbols('theta0,theta1')
loss_surface = sp.lambdify((theta0,theta1), j(theta0,theta1))
tangent_plane = sp.lambdify((theta0,theta1), z(theta0,theta1))

# grid and plot
points = np.linspace(-20,20,90)
xgrid,ygrid=np.meshgrid(points,points)

fig = go.Figure(
    data=[
        go.Surface(z=loss_surface(xgrid,ygrid),x=xgrid,y=ygrid),
        go.Surface(z=tangent_plane(xgrid,ygrid),x=xgrid,y=ygrid)
    ]
)

fig.update_layout(
    scene = dict(
        zaxis = dict(range=[0,1600]))
)

fig.show()
</code>```

Note: the z-lambda function is set for $(\theta_0,\theta_1)=(10,10)$. So I would expect to see to see a tangent plane touching the loss surface at $(10,10)$.

Here's my current result:
loss function looks okay, tangent plane not really

Is there anything obviously wrong in the equations? If not, then I probably messed up the matrix computation in python.

Thank you!

答:
Solved it! One of the numpy arrays had the wrong shape, which messed up the $J_{\theta_1}(\theta_0,\theta_1)$ computation.

This fixed the problem:

```<code>djt1 = lambda theta0,theta1: 1/m*((X_b.dot(np.array([[theta0],[theta1]]))-y)*X_b[:,1].reshape(-1,1).sum()
</code>```

Now the results look perfect and I can plot tangent planes for different $\theta_0$, $\theta_1$ values.

Hope that helps.

Body:
Solved it! One of the numpy arrays had the wrong shape, which messed up the $J_{\theta_1}(\theta_0,\theta_1)$ computation.

This fixed the problem:

djt1 = lambda theta0,theta1: 1/m*((X_b.dot(np.array([[theta0],[theta1]]))-y)*X_b[:,1].reshape(-1,1).sum()

Now the results look perfect and I can plot tangent planes for different $\theta_0$, $\theta_1$ values.

Hope that helps.


问:
Why can't this submanifold be a Leaf of a Foliation?
I am studying geometric theory of foliations from Camacho and Lins and I have stumbled upon an afirmation I can't understand.

There is a theorem which states:

Let F be a leaf of a foliation $\mathcal{F}$ on M  and $\Sigma$ a transverse section
of $\mathcal{F}$  such that $\Sigma \cap F \neq \varnothing$. We have three possibilities:

$\Sigma \cap F$ is discrete,  and in this case F is an imbedded leaf.

The closure of  $\Sigma \cap F$ in $\Sigma$ contains an open set.
This occurs if and only if the closure $\bar{F}$ has non-empty interior and
$int (\bar{F}) = \bar{F} - \partial \bar{F}$ is an open set which contains F. In this case we say
F is locally dense.

The closure of  $\Sigma \cap F$ is a perfect set  (i.e., without isolated points) with empty interior.
In this case we say F is an exceptional leaf.)

After this theorem comes the part I don't get. The authours state that the following submanifold $F$ in the picture bellow can't be a leave of any foliation deffined at an open set of $\mathbb{R}^2$

The reason being the point $x$ is an accumulation point of $\Sigma \cap F$ where $\Sigma$ is a transverse segment passing through x, while the same
does not happen with the point y.

I really don't get this. Suppose that upper horizontal part of $F$ containing $y$ was not there. In this case I guess I would have the case (2) of the theorem, right? To me, that upper part simply means I would have another transverse segment $\Sigma'$ passing through $y$, (vertical to the figure), giving me case (1) of the theorem.

Since we are talking about 2 distinct transverse segments, I don't see what is the problem.

I would appreciate any help since I just recently started studying the topic (please be kind).

答:
Let $F$ be the space indicated in Figure 2. Note that $F$ is closed, i.e. , $\bar{F}=F$. The boundary $\partial \bar{F}$ is given by
\begin{align*}
\partial \bar{F} = \bar{F}\cap (\overline{\mathbb{R}^2\setminus F}) = \bar{F}\cap \mathbb{R}^2=\bar{F}
\end{align*}
Hence we obtain for the interior of $F$ that $\mathrm{int}(\bar{F})=\bar{F}\setminus\bar{F}=\emptyset$ and $2.$ cannot hold.

Let us assume that $\Sigma$ is the vertical line through $x$. Then $F\cap \Sigma$ has isolated points but it is not discrete since $x$ is an accumulation point. Hence neither $1.$ nor $3.$ can hold which implies that $F$ cann't be a leaf of a foliation.


问:
grothendieck group over flat family
Say $f:X\to Y$ is a flat family of finite length objects. If $K_0(f^{-1}(y))$ stays constant for all $y$ in an open dense set of $Y$, I believe it is then true that $K_0(f^{-1}(y))$ stays constant for all $y\in Y$. But I don't know how to prove this. Any hint on this will be appreciated.

答: Let $f\colon X=\operatorname{Spec} (\Bbb{C}[x,y]/(x^2-y)) \to Y=\operatorname{Spec} (\Bbb{C}[y])$ induced by the embedding of rings, which is flat. If $y\neq 0$ then $K_0(f^{-1}(y))=K_0(\Bbb{C}\oplus \Bbb{C})\cong \Bbb{Z} \oplus \Bbb{Z}$, and $K_0(f^{-1}(0))=K_0(\Bbb{C}[x]/(x^2))\cong \Bbb{Z}$, where the last isomorphism follows from $K_0(R)=K_0(R_{red})$ for any commutative ring $R$.

Title: grothendieck group over flat family


问:
Commutativity in rings of matrices
Let n be arbitrary positive integer and R arbitrary ring (perhaps, non-associative). Let's denote $M_n(R)$ the set of all $n х n$ matrices with entries from R.

As i know if R is non-trivial commutative ring without zero divisors then scalar matrices are the only matrices which commutate with all other matrices over R.

By scalar matrix i mean diagonal matrix such that all its diagonal entries are equal.

I have the following question: what if R is non-trivial ring WITH zero divisors? Is it possible that there is matrix $M \in M_n(R)$ such that $\forall A\in M_n(R) ~~ AM = MA$ but M is non-diagonal or at least non-scalar.

Thanks in advance.

答: $M$ is in the center of $M_n(R)$ iff $M$ commutes with all matrices $E_{ij}$ (the $(i,j)$th entry equals $1$ and others $0$). From here you can easy obtain the answer.</p>

Title: Commutativity in rings of matrices


问:
Criterion of map $M_f\to Z$ to be continuous
I was about to prove the following proposition

A map $M_f\to Z$ is continuous if and only if the induced maps $X\times I\to Z$ and $Y\to Z$ are both continuous.

Here, $M_f$ is a mapping cylinder of $f:X\to Y$ i.e. $M_f = X\times I\sqcup Y/f(x)\sim (x,0)$ for $x\in X$. I first let $p:X\to X\times I$ by $x\mapsto (x,0)$. Letting $f_1:X\times I\to M_f$ and $f_2:Y\to M_f$ be induced map by inclusion, I considered the following diagram:

If $g$ is continuous, then clearly $i_1,i_2$ are continuous. To see the converse, first see the following diagram:

It's a bit complicated but readable. The upper half is a universal property of disjoint union and $q$ is a quotient map. Now, I want to show $g\circ q \equiv \tilde{f}$. Let $x\in M_f$. If $x\in X\times I$, then $g\circ q(x) =g\circ q\circ j_1(x) = g\circ f_1(x) = i_1(x)= \tilde{f}\circ j_1(x) = \tilde{f}(x)$. Similarly to $x\in Y$. Now the following fact:

The surjection $f:X\to Y$ is an identification map if and only if for any function $g:Y\to Z$, $g\circ f$ is continuous if and only if $g$ is continuous.

Hence, $g$ is continuous.

Is this a valid proof?

Note. I don't know anything about Category theory.

答:
Your proof is correct. Trust yourself!

Note that the mapping cylinder is constructed as the pushout of the pair of maps $(f,p)$ which is a general concept of category theory. It would be worth to learn same basic category theory to understand that point.

Anyway, you do need that. As you said, we have a quotient map $q : (X \times I) \sqcup Y \to M_f$. The universal property of the quotient shows then that $g : M_f \to Z$ is continuous iff $g \circ q$ is continuous. But $g \circ q$ lives on the disjoint union of $X \times I$ and $Y$, thus it is continuous iff $g \circ q \circ j_1 = q \circ f_1$ and $g \circ q \circ j_2 = q \circ f_2$ are continuous.