问:
IMO long list spatial Combinatorial problem
Can anyone send a solution or hint to these problem? I tried a greedy algorithm but couldn’t make it work

答:
Of the finitely many possible pairings, choose one for which the sum of the lengths of the line segments is least.

Claim:$\;$For such a pairing, no two line segments intersect.

Suppose for such a pairing the line segments $AC$ and $BD$ intersect at $P$.

Then $ABCD$ is a convex quadrilateral with diagonals intersecting at $P$.

But then if we replace the pairs $A,C$ and $B,D$ by $A,B$ and $C,D$ we would have
$$
\left\lbrace
\begin{align*}
|AB| &< |AP|+|PB|\\[4pt]
|CD| &< |CP|+|PD|\\[4pt]
\end{align*}
\qquad\qquad\;\;\;\;\;
\right.
$$
hence
\begin{align*}
|AB| + |CD| &< (|AP|+|PB|)+(|CP|+|PD|)\\[4pt]
&= (|AP|+|CP|)+(|PB|+|PD|)\\[4pt]
&= (|AP|+|PC|)+(|BP|+|PD|)\\[4pt]
&= |AC|+|BD|\\[4pt]
\end{align*}
so the new pairing would have a smaller sum of segment lengths, contradiction.


问:
Probability depending on two uniformly distributed variables
Apologies if this is a duplicate, the problem is very hard to word in a search

The bivariate distribution of random variables X and Y is uniform over the triangle with vertices (1,0), (1,1), (0,1). A pair of values $x,y$ is chosen at random from this distribution and a (perhaps degenerate) triangle ABC is constructed with BC=$x$ and CA=$y$ and AB = $2 -x -y$. Show that the construction is always possible and that $ABC$ is obtuse if and only if
$$y>\frac{x^2-2x+2}{2-x}$$
Deduce that the probability ABC is obtuse is $3-4 \ln 2$

This question regards the "Deduce that" part, and I will quote the obtuse result

I'm going to (WLOG?) roll X first and say x=t. With this paramaterisation, I can write
$$P(y>\frac{x^2-2x+2}{2-x})=\int_0 ^1{P(X=t)P(Y>\frac{t^2-2t+2}{2-t}|X=t)}$$
Since $X\sim U[0,1]$, then $P(X=t)=t dt$ and if you draw a sketch of the triangle, it is clear $Y\sim U[1-X,1]$

therefore $P(Y\le u|X=t)=\frac{u-(1-t)}{t}$ for $1-t\le u \le 1$, so $P(Y>\frac{t^2-2t+2}{2-t}|X=t)=1-\frac{1}{2-t}$ after taking the compliment and letting u equal the rational function.

Putting these into the integral I get
$P(A)=\int^1_0{t(1-\frac{1}{2-t})}dt=\frac{1}{2}(3-4\ln2)$

I can't see at all where this factor of one half came in. It feels hard to believe that my method was wrong since my answer is very similar to the correct answer. I think I'm missing one last bit and I can't see what it is.

答:
Sketch for the solution: an uniform distribution over some set means that the probability of some point to be chosen from some subset is proportional to the area or volume of this subset, hence in this case if $A$ is a subset of the triangle $T$ then
$$
\Pr [(X,Y)\in A]=\frac{\text{ area of }A}{\text{ area of }T}=2 \cdot \text{area of }A\tag1
$$
as the area of the triangle is one-half. Now, we have that
$$
\begin{align*}
A:=&\left\{(x,y)\in T: y>\frac{x^2-2x+2}{2-x}\right\}\\
=&T\cap \left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\}\\
=&\left\{(x,y)\in \mathbb{R}^2: 0\leqslant x\leqslant 1 \,\land\, 1-x\leqslant y\leqslant 1\right\}\cap \left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\}
\end{align*}\tag2
$$
Therefore setting $B:=\left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\}$ we find that
$$
\Pr [(X,Y)\in A]=2\int_{0}^1\int_{1-x}^1 \mathbf{1}_{B}(x,y)dydx\tag3
$$

The value of the inner integral is just the length of the interval
$$
I_x:=\left[\max\left\{1-x,\frac{x^2-2x+2}{2-x}\right\},1\right]\tag4
$$
that is
$$
\Pr [(X,Y)\in A]=2\int_{0}^1 \ell (I_x)dx\tag5
$$
where $\ell (I_x)$ is the length of $I_x$. So, depending of the value of $x\in[0,1]$ it length will be $1-(1-x)=x$ or $1-\frac{x^2-2x+2}{2-x}=\frac{x -x^2}{2-x}$, then the integral on (5) can be separated in (maybe just two) integrals where it integrand will be $x$ or $\frac{x-x^2}{2-x}$.

ContentLicense: CC BY-SA 4.0


问:
a second order ordinary differential equation to be solved by method of variation of parameters
how to solve $$x^2y"-2x(1-x)y'+2(x+1)y=x^3$$ by method of variation of parameters.
how do you find solution of the above equation?
iam familiar with method of variation of parameters but dont know how to find $y_1$ and $y_2.$ please help.

答: You might first need to set an interval for $x$, as the equation degenerates as $x\rightarrow 0$.


问:
Find the limit $\lim_{x \to 1}(\frac{3x}{2+x})^\frac{x}{1-x}$
Find the limit
$$
\lim_{x \to 1}\left(\frac{3x}{2+x}\right)^\frac{x}{1-x}
$$

I've transformed the function by changing limit of 1 to zero and become the following:
$$
\lim_{x \to 0}\left(1 + \frac{3x-3}{x-3}\right)^\frac{1 - x}{x}
$$

I spent many time and tried to transform the power part $\frac{1 - x}{x}$ to $\frac{x-3}{3x-3}$ but still can't remove the $x$'s on final. How can I find this limit? Thanks a lot!

答:
Hint: try to match
$$
\frac{3x}{2+x}=1+\frac1t\iff t=\frac{x+2}{2(x-1)}.
$$
Now
$$
\frac{x}{1-x}=t\cdot\frac{-2x}{x+2}.
$$
What is the limit of
$$
\left[\left(1+\frac1t\right)^t\right]^{-\frac{2x}{x+2}}?
$$


问:
How to prove that $\log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq 2 k\pi)$, $k$ is an integer
This question number 11.15(a) Tom M Apostol's MAthematical Analysis From Chapter Fourier analysis.

Prove that  $ \log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq  2 k\pi)$.

So, I tried by using $a_0 = 2/\pi \int_{0}^{\pi} \log|\sin(x/2)| dx$ and I used $\sin(2x) =2\sin x\cos x$.

So Using it I get $a_0 /2 =\log 2 + 2/\pi [\int_{0}^{\pi}\log|\sin(x/4)|dx +\int_{0}^{\pi}\log|\cos (x/4) |dx]$.

But the problem is how to compute $\int_{0}^{\pi}\log|\cos (x/4) |dx$. The limit of integral can be taken from $0$ to $2 \pi $

Similar problem will also arise when I will compute $a_n = \int_{0}^{\pi} \log|\sin(x/2)| \times\cos(nx) dx$.

Can you please tell how to compute this integral OR If there is alternate way of computing the fourier series.

Thank you.

Body:
$$S=\Re \sum_{n=1}^{\infty} \frac{e^{inx}}{n}=-\Re\ln(1-e^{ix})=-\Re \ln[1-\cos x+i\sin x]=-\frac{1}{2}\ln (2-2\cos x)$$$$ \implies S=-\frac{1}{2}\ln 4\sin^2(x/2)=- \ln 2- \ln{|\sin(x/2)|}=-\ln 2-\ln |\sin (x/2)|$$
Here we have used $-\ln(1-z)=\sum_{n=0}^{\infty} \frac{z^n}{n}$ and $\ln(x+iy)=\frac{1}{2}\ln(x^2+y^2)+i\operatorname{atan2}(y,\,x)$


问:
Sum of the powers of roots of unity
I was reading this proof of the roots of unity filter and I point can't understand is why:

$$ S = \sum_{j=0}^{n-1} w^{kj} =0$$

If $ \omega $ is the nth root of unity and $k$ is some positive integer not equal to zero and n doesn't divide k

I understand that the sum of nth roots of unity are zero as in:

$$ S = \sum_{j=0}^{n-1} w^j = 0$$

But I can't understand the powers of them should be as well. The reason I find it strange is that the nth roots of unity solves the following polynomial equation

$$ P(x) =(x-1)( \sum_{i=0}^{n-1} x^i)$$

So why should it be that any of their powers solve the above equation as well?

答:
You can use the typical trick for geometric series. You have that if

$$S = \sum_{j=0}^{n-1} w^{kj}$$

then

$$w^k S = \sum_{j=1}^n w^{kj} \implies S = \frac{1-w^{nk}}{1-w^k} = 0 $$

Title: Sum of the powers of roots of unity


问:
Sum of Legendre symbols for divisors of a quadratic non-residue modulo an odd prime.
Let $\big(\frac{n}{p}\big)=-1$ (Legendre's symbol) for an odd prime p. Then show that
$$\sum_{d,\ \ d\mid n}\Big(\frac{d}{p}\Big)=0.$$
I know that if $d$ is a divisor of $n$, then $-d$ is also its divisor. Therefore, by properties of Legendre's symbol, we have
$$\Big(\frac{-d}{p}\Big)=\Big(\frac{-1}{p}\Big)\times \Big(\frac{d}{p}\Big).$$Since $p$ is an odd prime, we have that $$ \Big(\frac{-1}{p}\Big)=(-1)^{\frac{p-1}{2}}=\pm 1.$$  If $\frac{p-1}{2}$ is odd, then we can prove. How to deduce the result when $\frac{p-1}{2}$ is even.

Is my way of solving correct? Please give hint or solution.

答:
Legendre symbol is a completely multiplicative function whose values are in $\{-1,1,0\}$. Therefore
$$\Big(\frac{d}{p}\Big)\Big(\frac{n/d}{p}\Big)=\Big(\frac{n}{p}\Big)=-1\implies\Big(\frac{d}{p}\Big)=-\Big(\frac{n/d}{p}\Big).$$
It follows that
$$\sum_{d,\ \ d\mid n}\Big(\frac{d}{p}\Big)=\frac{1}{2}\left(\sum_{d,\ \ d\mid n}\Big(\frac{d}{p}\Big)+\sum_{d,\ \ d\mid n}\Big(\frac{n/d}{p}\Big)\right)
=\frac{1}{2}\sum_{d,\ \ d\mid n}\left(\Big(\frac{d}{p}\Big)+\Big(\frac{n/d}{p}\Big)\right)=0$$


问:
Bounds on $S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+\dots+\frac{1}{3001}$

$S =  \frac{1}{1001} +   \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.

Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.

My Attempt:

$S<\dfrac{500}{1000} +   \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$

(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)

Using AM-HM inequality gave me $S > 1$, but the bounds are weak.

Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$

Inequality with sum of inverses of consecutive numbers

The answers to these questions are nice, but the bounds are weak.

Any help without calculus and without calculations involving calculators would be appreciated.

(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))

答:
Dividing the sum into groups as you have done will obtain the limits if you take enough buckets (around 10).

A similar approach is to add $1/1000$ and then pair the terms, with two terms left over added to the end
\begin{align}
S&=\frac{1}{1000} + \frac{1}{3000} + \frac{1}{1001} + \frac{1}{2999} + \cdots +
\frac{1}{1999} + \frac{1}{2001} + \frac{1}{3001} - \frac{1}{1000} \\
&= \frac{4000}{1000\cdot 3000} + \frac{4000}{1001\cdot 2999} + \cdots +\frac{4000}{1999\cdot 2001} + \frac{1}{3001}-\frac{1}{1000}
\end{align}

Then try grouping these into $m$ buckets, each with $n$ terms.  The lower bound is
\begin{align}
\frac{4000n}{(1000+n-1)(3000-n+1)}+\cdots+\frac{4000n}{(1000+mn-1)(3000-mn+1)} \\ \quad+\frac{1}{3001}-\frac{1}{1000}
\end{align}
and the upper bound,
\begin{align}
\frac{4000n}{(1000)(3000)}+\cdots+\frac{4000n}{(1000+(m-1)n)(3000-(m-1)n)}\\ \quad+\frac{1}{3001}-\frac{1}{1000}
\end{align}
Using a calculator, and (still) choosing $m=4, n=250$ can create bounds comparable to the ones you have, but it is still cumbersome.

With $n=250$, the upper bound is
\begin{align}
U &= \frac{1}{3}+\frac{16}{55}+\frac{4}{15}+\frac{16}{63}+\frac{1}{3001}-\frac{1}{1000} \\
&\leqslant \frac{1}{3}+\frac{16}{55}+\frac{4}{15}+\frac{16}{63} \\
&\leqslant \frac{20 + 18+16+16}{60} \\
&= \frac{7}{6}
\end{align}


问:
Euler's method for second order ODE
I'm new to numerical analysis and got stuck on an ODE:
$$y'' + y^3 - 0.5 = 0$$
with $h = 0.1$, $y(0) =1$ and $y'(0)=1$.

How should I begin?

答:
Start by defining a new variable satsfiying $z=y'$. Then rewrite your second order equation as a system of first order ODEs:
$$
\left\{
\begin{array}[rl]
zz' &=& -y^3 + 0.5 \\
y' &=& z
\end{array}
\right.
$$
You can then solve these two equations simultaneously with the Euler method.

Title: Euler's method for second order ODE


问:
Recurrence relations of the form $T(n) = T(n-1) + (n-1)! n^k$
In the context of recursive permutation-related algorithms, the following form of recurrence relations often comes up:
$$
T(n) = 1 \\
T(n) = T(n-1) + (n-1)! \cdot n^k
$$
where $k$ is usually 0, 1, or 2.

What can be said about the asymptotic behaviour of $T(n)$ (in terms of $\Theta$ or $O$ notation, and in general or for specific $k$)?

Body:
We have

$\sum_{p=2}^{n} (T(p)-T(p-1))=\sum_{p=2}^{n} (p-1)!.p^k$

By telescoping

$T(n)-T(1)=\sum_{p=2}^{n} (p-1)!.p^k$

As $T(1)=1$,

$T(n)=1+(n-1)!.n^k+(n-2)!.(n-1)^k+(n-3)!.(n-2)^k+ \sum_{p=2}^{n-3} (p-1)!.p^k$

Considering the last sum

$\sum_{p=2}^{n-3} (p-1)!.p^k=(n-3)!.(n-1)^k.\sum_{p=2}^{n-3} \frac{(p-1)!}{(n-3)!}.\frac{p^k}{(n-1)^k}$

$\leq(n-3)!.(n-1)^k.\sum_{p=2}^{n-3} \frac{(n-4)!}{(n-3)!}.\frac{(n-3)^k}{(n-1)^k}$

However $\frac{(n-4)!}{(n-3)!}.\frac{(n-3)^k}{(n-1)^k} \leq 1$ ($k\geq 1$)

So $\sum_{p=2}^{n-3} (p-1)!.p^k \leq (n-3)!(n-1)^k. (n-3-2+1)$

$\leq (n-2)!(n-1)^k$

We deduce that

$\sum_{p=2}^{n-3} (p-1)!.p^k = \mathcal{O}\left((n-2)!.(n-1)^k\right)$

Hence $T(n)\sim (n-1)!.n^k$ as n tends to $+\infty$


问:
Number of conjugated to (12)(34) in $S_6$ of rank 6
I need to find the number of conjugated to the permutation (12)(34) in the symmetric group $S_6$ of rank 6

My answer is 6! = 720

Is this correct?

I concluded that (12)(34)=(12)(34)(5)(6) and the number of combinations for $S_6$ is 6! as they need to be the same partition type

Edit:

It seems to be $6! / (2*2*1*1) = 180$

答: Hint: An element of $S_6$ is conjugated with $(12)(34)$ if it has the same disjoint cycle type, that is you have to count the amount of elements that look like $(a b)(cd)$ such that $(ab)$ and $(cd)$ are disjoint transpositions.


答:
Here's the polynomial which solves it:

$$  1 + x^k + x^{2k}..+x^{(n-1)k} = \frac{x^{nk} -1}{x-1}$$

Or,

$$ (x-1)\left[ 1 + x^k + x^{2k}..+x^{(n-1)k}  \right]= x^{nk}-1 = P(x)$$

Clear to see power of nth roots of unity solves this equation

Title: Sum of the powers of roots of unity

Body:
I was reading this proof of the roots of unity filter and I point can't understand is why:

$$ S = \sum_{j=0}^{n-1} w^{kj} =0$$

If $ \omega $ is the nth root of unity and $k$ is some positive integer not equal to zero and n doesn't divide k

I understand that the sum of nth roots of unity are zero as in:

$$ S = \sum_{j=0}^{n-1} w^j = 0$$

But I can't understand the powers of them should be as well. The reason I find it strange is that the nth roots of unity solves the following polynomial equation

$$ P(x) =(x-1)( \sum_{i=0}^{n-1} x^i)$$

So why should it be that any of their powers solve the above equation as well?


答:
$w=e^{2i j\pi/n}, j=0,1,2,....n-1,$
$$S_n=\sum_{j=0}^{n-1} w^{k}= \sum_{j=0}^{n-1} e^{2\pi  k j/n}~~~~(1) $$
Use sun of a finite GP, to get
$$\implies S_n=\frac{e^{2i\pi k }-1}{e^{2i\pi k/n}-1}$$
If $k\ne pn$, $S_n=0$,

If $k=p n$ then $S_n=0/0$. So go (1) and put $k=pn$, where $p$ is an integer.
We get $S_n=n$ when $k$ is an integer multiple of $n$.

Title: Sum of the powers of roots of unity


问:
How to do this discrete math problem?
Construct a sequence of $16$ positive integers that has no increasing or decreasing subsequence of $5$ terms.

I know the answer but I fail to see how it is arrived at.

答:
Clearly due to Erdős–Szekeres theorem this sequence must have an monotonic subsequence of length $4$. So the first idea shall be dividing the numbers into groups of size $4$.
WLOG assume the 16 numbers to be $1,2,...,16$.
Then group them like the following:
$A=\{1,2,3,4\}$,
$B=\{5,6,7,8\}$,
$C=\{9,10,11,12\}$,
$D=\{13,14,15,16\}$
Now if they don't come in the order $D,C,B,A$ it will surely have an increasing subsequence of length $5$.
So we set them like $D,C,B,A$ and it's easy to check that it's true, too.

Title: How to do this discrete math problem?


问:
Why does $\hat{\boldsymbol e_i} = \frac{\frac{\partial\mathbf r}{\partial q_i}}{\left\lvert\frac{\partial\mathbf r}{\partial q_i}\right\rvert}$?
So I'm trying to understand the derivation for the line element in spherical coordinates as well as the derivation for the basis vectors. Both hinge on the question posed in the title. So here's what I have so far in figuring this out:

In general, $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr+\frac{\partial\mathbf r}{\partial\theta}\mathrm d\theta+\frac{\partial\mathbf r}{\partial\varphi}\mathrm d\varphi$, but if we make $\mathrm d\theta=\mathrm d\varphi=0$ then the equation becomes $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr$, and also $\mathrm d\mathbf r$ points in the $\hat{\boldsymbol r}$ direction, since there is no change in the $\theta$ or $\varphi$ directions of the infinitessimal displacement. The next step would be to make the infinitesimal displacement a unit vector, and we accomplish this by diving by the magnitude. And here's where I'm getting stuck. I feel like the magnitude of the infintessimal displacement should be $\left\lvert\frac{\partial\mathbf r}{\partial r}\mathrm dr\right\rvert$ since $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr$, but of course the book is writing the magnitude simply as $\left\lvert\frac{\partial\mathbf r}{\partial r}\right\rvert$. What am I missing here?

答: Okay, that was a stupid question, and the answer was staring me in the face the whole time. If $\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr$ points in the same direction as $\hat{\boldsymbol r}$ then so does $\frac{\partial\mathbf r}{\partial\mathrm r}$ and therefore $\hat{\boldsymbol r}=\frac{\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr}{\left\lvert\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr\right\rvert}=\frac{\frac{\partial\mathbf r}{\partial\mathrm r}}{\left\lvert\frac{\partial\mathbf r}{\partial\mathrm r}\right\rvert}$.


问:
Is the intermediate value theorem saying that if f is continuous, then it is onto?
How can I prove

if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $\overline{f(\mathbb{N})} = \mathbb{R}$, then $f$ is onto (surjective)

by the intermediate value theorem?

答: Yes, if $y$ in $\mathbb{R}$, there exist $n_1,n_2\in\mathbb{N}$ such that $f(n_1)<y, f(n_2)>y$. IVT implies the existence of $x\in [n_1,n_2]$ or $[n_2,n_1]$ such that $f(x)=y$.


答: The number of permutations with cycle type $(2,2,1,1)$ is ${6\choose2}\cdot{4\choose2}/2=45$.  (I divided by two because there is double counting.)

Title: Number of conjugated to (12)(34) in $S_6$ of rank 6

Body:
I need to find the number of conjugated to the permutation (12)(34) in the symmetric group $S_6$ of rank 6

My answer is 6! = 720

Is this correct?

I concluded that (12)(34)=(12)(34)(5)(6) and the number of combinations for $S_6$ is 6! as they need to be the same partition type

Edit:

It seems to be $6! / (2*2*1*1) = 180$


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
There are a lot of "inclusions" of a similar character, for example $\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R$. It depends on your interpretation of what these sets are whether you regard $\subset$ as a genuine inclusion or if you regard $\subset$ as a notation indicating that there is a canonical embedding of the object on  left hand side as a subobject of the object on the right hand side.

Frequently one starts on the level of axiomatic set theory and then successively constructs $\mathbb  N$, $\mathbb Z$, $\mathbb Q $, $\mathbb R$ etc. The usual constructions produce an "ascending  sequence" of algebraic objects which are not related by genuine inclusions, but the respective construction yields a new object plus an embedding of the original object into the new one. Recall the construction of $\mathbb Q$ as a set of equivalence classes of pairs in $\mathbb Z \times (\mathbb Z \setminus \{0\})$.

However, usually there are various alternative constructions supplying specific models of the sets $\mathbb N, \mathbb Z, \mathbb Q, \mathbb R$ etc. But I think  most people have a more abstract (or, if you want,  more intuitive) understanding of these objects. We know what their purpose is, which operations can be performed and which rules are satisfied. The specific construction is then fairly irrelevant. I would call that an axiomatic point of view. For example, the reals can be constructed based on $\mathbb Q$ via Dedekind cuts,  via nested intervals or via Cauchy sequences in  $\mathbb Q$, and certainly there are even more approaches. But I doubt the anybody imagines a real number as a Dedekind cut or as an equivalence class of Cauchy sequences.

Therefore the concrete model of an object as $\mathbb R$ is fairly uninteresting, it can be replaced without problems by any other model having the same properties.

In your case we start with some model of $\mathbb R$ and construct $\mathbb C$ as the set of real pairs. Then $\mathbb R' = \mathbb R \times \{0\}$ is a subfield of $\mathbb C$ which is canonically isomorphic to the original $\mathbb R$ and we may now work with the new model $\mathbb R'$ which is a genuine subset of $\mathbb C$.

If you do not like this replacement, be aware that the standard construction of $\mathbb C$ is just one possible approach (but a very transparent one). It can be easily modified as follows. The simple idea is to add new (non-real) numbers to $\mathbb R$ instead of embedding $\mathbb R$ into $\mathbb R^2$ (which produces the isomorphic subfield $\mathbb R' = \mathbb R \times \{0\} \subset \mathbb C$ and causes the "philosophical discussion" whether, or in what sense, $\mathbb R$ is a genuine subset of $\mathbb C$).

Given $\mathbb R$, the sets $\mathbb R$ and  $\mathbb R \times \mathbb R^*$ (where $\mathbb R^* =  \mathbb R \setminus \{0\}$) are disjoint. Define
$$\mathbb C  = \mathbb R \cup (\mathbb R \times \mathbb R^*)$$
and
$$\phi : \mathbb C \to \mathbb R \times \mathbb R, \phi(\zeta) = \begin{cases} (x,0) & \zeta = x \in \mathbb R \\ \zeta & \zeta \in \mathbb R \times \mathbb R^*\end{cases} $$

This is a bijection. The usual complex field structure on $\mathbb R \times \mathbb R$ is then transferred via $\phi$ to our modified $\mathbb C$. Doing so, we get a genuine inclusion $\mathbb R \subset \mathbb C$. But, to be honest, I can't see that this has an added value. Anyway, our model of $\mathbb C$ is a real vector space with basis $\mathcal B  = \{1 \in \mathbb R, i = (0,1) \in \mathbb R \times \mathbb R^* \}$. Noting that $x\cdot 1 = x$ for $x \in \mathbb R$ and writing $y \cdot i = i y$ for $y \in \mathbb R$ we get
$$\mathbb C = \{x\cdot 1 + y\cdot i = x + iy  \mid x,y \in \mathbb R\} .$$
Here the complex number $x + i0$ is literally the same as the real number $x$.

Using similar methods we can of course construct a chain of genuine set inclusions
$$\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C .$$


问:
Proving that $\lim_{x \to 0} \frac{f(x)}{|x|}=1$ implies $f$ is not differentiable at $0$.
$$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$.
I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient?

Assume toward contradiction that $f$ is differentiable at $0$.

Then by definition we know that $f$ is continuous at $0$. Therefore

$\lim _{x\to 0}f\left(x\right)=f\left(0\right)$

We know that

$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$

Therefore we can conclude

$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$

But we can also rewrite this as

$\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$

Therefore $f(0) = 0$.

Since $f$ is differentiable at $0$ we have

$\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$

Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have

$\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$

$\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$

Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$.

Therefore $f$ is not differentiable at 0.

答: You are right. Assuming $f$ is differentiable at $x=0$ and since $f\sim \lvert x\rvert$ in a neigborhood of $0$, necessarily $f(0)=0$. Then taking the derivatives from both sides, you get the contradiction.


问:
A typo in Proposition 2.2, Eisenbud commutative algebra?
I am trying to understand the formulation of Proposition 2.2 in Eisenbud's "Commutative algebra with the view...". Here is the place that is not clear to me

Proposition 2.2. Let $\varphi: R\to R[U^{-1}]$ be the natural map $r\to r/1$.

a. For any ideal $I\subset R[U^{-1} ]$ we have $I=\varphi^{-1}(I)R[U^{-1}]$. ...

Do I understand correctly that instead of $I=\varphi^{-1}(I)R[U^{-1}]$ we should have $I=\varphi^{-1}(I)[U^{-1}]$? Namely that we localize $\varphi^{-1}(I)$ at $U$? If my interpretation is wrong, how am I supposed to understand the notation $\varphi^{-1}(I)R[U^{-1}]$?

答:
Disclaimer: I am unfamiliar with Eisenbud's book.

The ring $R[U^{-1}]$ can be naturally viewed as an $R$ module: you can define $s\frac{a}{b}$ as $\frac{s}{1}\frac{a}{b}$, which is the same as $\varphi(s)\frac{a}{b}$. Now remember that for any morphism $f: R\to S$ and any ideals $I\subset R$, $J\subset S$ we have ideals $f^{-1}(J)\subset R$ (contraction) and $f(I)S\subset S$ (extension).

What Eisenbud seems to be saying is that $I$ coincides with the extension of its contraction, which shows that any ideal of the localized ring is an extension of some ideal of the original ring.

But you are also not wrong: this extension coincides with the localization of the ideal. If you have access to Atiyah's Commutative Algebra, you can look up Proposition 3.11 and the text just preceding it.


问:
How to find inverse of the function $f(x)=\sin(x)\ln(x)$
My friend asked me to solve it, but I can't.

If $f(x)=\sin(x)\ln(x)$, what is $f^{-1}(x)$?

I have no idea how to find the solution. I try to find

$$\frac{dx}{dy}=\frac{1}{\frac{\sin(x)}{x}+\ln(x)\cos(x)}$$

and try to solve it for $x$ by some replacing and other things, but I failed.

Can anyone help? Thanks to all.

答: This function isn't one-to-one, so you can't inverse it.</p>

ContentLicense: CC BY-SA 3.0


问:
Showing that if $\lim_{n \to \infty} x_n=a$, then for any $k\in \mathbb{N}$, $\lim_{n \to \infty} x_n^{2k}=a^{2k}$
$x_n$ is a sequence. The only thing I can do here is just write the definition of $\lim_{n \to \infty} x_n=a$, but that doesn' t seem helpful to me. I tried looking for some inequalities that would help, and the only thing  that I found that I thought could be a little bit helpful, was the reverse triangle inequality, but I don' t know how to use that here.Could you please give some advice or a hint?

Thanks in advance

答:
This follows immediately, by induction on $k$, from the product rule for limits; if the limits $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ exist, then
$$\lim_{n\to\infty}(a_n\cdot b_n)=\left(\lim_{n\to\infty}a_n\right)\cdot\left(\lim_{n\to\infty}b_n\right).$$


问:
Solve the partial differential equation in two variables.
Find the general solution of the partial differential equation
                                         $$u_{yy}-2u_y+u=e^x.$$
My effort: Auxiliary equation corresponding to this equation is
$$D'^2-2D'+1=0\implies D'=1, 1$$ Therefore, the complementary function is given by
$$Ae^y+Bye^y,$$ where A and B are constants.
Particular integral (P.I) is
$$P.I.=\frac{1}{D'^2-2D'+1}(e^x)=(1-D')^2e^x$$ $$=(1+2D'+3D'^2+\cdots)(e^x)$$ $$=e^x.$$
The solution is $$Ae^y+Bye^y+e^x.$$
Is my solution correct?

答:
Your answer is almost correct, if you replace $A$ and $B$ by two arbitrary functions of $x$.

Using a transformation $u\to v e^{x}$, you will get $v_{yy}-2v_y+v=1$;

The general solution of $v_{yy}-2v_y+v=1$ is $e^y c_1(x)+e^y y c_2(x)+1$;

The full solution is then $u=e^{x}[ c_1(x) e^y+c_2(x) e^y y +1]$.


问:
If $x=\max(0,y+\alpha)$, how can we write $y$?
If we have $x=\max(0,y+\alpha)$
so we have $x=0$ or $x=y+\alpha$.

How can we write $y$? Only as $y=x-\alpha$, or do we add $y=-\alpha$, too?

答:
If $y+\alpha\ge 0$, then $x=y+\alpha$ and $y=x-\alpha$.

If, however, $y+\alpha<0$, then $x=0$ and it is impossible from that information to deduce $y$. All you can say is that $y<-\alpha$, but there are many such $y$'s and all of them will produce the same $x$.


问:
To show the equivalence of two statements related to $B^n_2 \subset K \subset dB^n_2$

Consider the polytope $$K = \{x\in\mathbb{R}^n: \langle x,v_i\rangle \leq 1, 1\le i\leq m\}$$ with $m$ facets, for which $$B^n_2 \subset K \subset dB^n_2$$ where $1<d\in\mathbb{R}^{+}$ and $B^n_2 := \{ x\in\mathbb{R}^n: \| x\| \leq 1\}$, i.e. the Euclidean ball of unit radius. $dB^n_2$ is the scaled version of this.

What I have so far:

The first inclusion implies that each $\|v_i\| \leq 1$ for all $1\leq i\leq m$ (since $\frac{v_i}{\|v_i\|} \in B^n_2$).

The second inclusion gives us that if $\|x\| > d$ $(x\notin dB^n_2)$, then there exists $i$ such that $\langle x,v_i\rangle > 1$ (so that $x\notin K$).

What I want to show:

I want to show that the second conclusion above is equivalent to:
For every unit vector $\theta$, there exists an $i$ such that $$\langle \theta,v_i\rangle \geq \frac{1}{d}$$

How should I go about this? I'm unable to relate the statements I have so far to unit vectors in any sense - and hence I haven't had much progress proving the above equivalence. I tried starting off with $\langle d\theta,v_i\rangle \geq 1$ instead, so that $\|d\theta\| = d$ and $d\theta\in dB^n_2$.

I'd appreciate any help, thanks!

Clarification:

$d\theta$ = $d \times \theta$, i.e. $d$ times the unit vector $\theta$. It is not the infinitesimal $d\theta$ seen often in analysis.

答:
Which part are you struggling with?

The backward direction follows immediately, since for any $x$ with $\|x\|>d$, you know that there exists some $i$ such that $\left\langle \frac{x}{\|x\|},v_i\right\rangle \geq \frac{1}{d}$. Multiplying this by $\|x\|$ on either side gives the required.

For the forward direction, for any $d'>d$, you know that there exists $i$ such that $\langle \theta,v_i\rangle>\frac{1}{d'}$. Are you able to conclude from here that there exists $i$ such that $\langle \theta,v_i\rangle\geq\frac{1}{d}$?


问:
Double integration using a change of variables for $\iint_D\sqrt{x}\ dx\ dy $
I need to calculate this double integral using change of variables

$$\iint_D\sqrt{x}\ dx\ dy $$

with $D=\{(x,y) \mid x^2+y^2 < x\}$.

I thought of putting $x=r\cos\theta$ and $y=r\sin\theta$. So $x^2+y^2<x$ will become $r<\cos\theta$ (should I suppose here that $r \ne 0$?). I'll just rewrite the integral as

$$\int_0^{2\pi}\int_0^{\cos\theta}r\sqrt{r\cos\theta}$$

which is double with a bit of calculation.

I would be very thankful if someone can tell me if my method is right. If not, how can I solve it?

Thank you very much.

答:
Your method is correct but your limits for $\theta$ is incorrect.

The correct integral is:

$$\int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta} r \, \sqrt{r\cos\theta} \,  dr \, d\theta$$

Circle $r = \cos \theta$ with center at $(\frac{1}{2}, 0)$ and with radius of $\frac{1}{2}$ is between $ - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.


问:
inscribed vs. circumscribed randomly generated similar triangles: which one has more ‘excess’ area on average? by how much
Obtain two similar triangles with uniformly-randomly chosen angle measures, assigning the larger one an area of 1 and the smaller one an area such that its circumcircle has an area of 1. Which area on average would you expect to be greater: the area inside the larger triangle but outside its incircle or the area inside the larger circle but outside its inscribed triangle? (And what are these average areas? Minimum and maximum?)</p>

答:
Not an answer, just a walk-through of the formula reduction from a comment ...

OP is comparing the ratio of excess triangle area ($\Delta$) to incircle area ($I$) to the ratio of excess circumcircle area ($K$) to triangle area, so we can write

$$\begin{align}
\require{cancel}
\frac{\Delta-I}{\Delta}&\;\lesseqgtr\;\frac{K-\Delta}{K} \tag1\\[4pt]
\text{K}\cdot(\Delta-I) &\;\lesseqgtr\; \Delta\cdot(K-\Delta) \tag2 \\[4pt]
\cancel{K\Delta}-KI &\;\lesseqgtr\; \cancel{\Delta K}-\Delta^2 \tag3\\[4pt]
\Delta^2 &\;\lesseqgtr\; KI =  \pi R^2\cdot\pi r^2=\left(\pi rR\right)^2 \tag4\\[4pt]
\Delta &\;\lesseqgtr\; \pi r R \tag5
\end{align}$$
where $r$ and $R$ are inradius and circumradius. Note that the order of the inequalities has been (and will be) preserved; the left-hand side is smaller than (or equal to, or larger than) the right-hand side in $(1)$ whenever this is true in $(2)$, $(3)$, $(4)$, etc.

From here, we can invoke some relations between the quantities, and a little trig. (See, for instance, MathWorld's "Inradius", "Law of Sines", and "Double-Angle Formulas" entries.)

$$\begin{align}
r &= 4R\sin\tfrac12A\sin\tfrac12B\sin\tfrac12C \tag6 \\[4pt]
\Delta &= \tfrac12ab\sin C =\tfrac12\cdot 2R\sin A\cdot 2R\sin B\cdot \sin C  \tag7 \\[4pt]
&= 2 R^2 \cdot 2\sin\tfrac12A\cos\tfrac12A\cdot 2\sin\tfrac12B\cos\tfrac12B\cdot 2\sin\tfrac12 C\cos\tfrac12C \tag8 \\[4pt]
&= 4 R \sin\tfrac12A\sin\tfrac12B\sin\tfrac12C \cdot 4R\cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \tag9 \\[4pt]
&= 4 r R \cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \tag{10}
\end{align}$$

Substituting $(10)$ into $(5)$, and dividing-through by $4rR$, yields

$$\cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \;\lesseqgtr\; \frac\pi4\tag{$\star$}$$

as claimed. $\square$


问:
Prove $ \left| \int_{a}^{\infty} \{x\} \frac{1 - \log x}{x^2} dx \right| \le 2 \int_{a}^{\infty} \frac{\log x}{x^2} dx $
It this inequality true for $ a > 1 $?

$$
\left| \int_{a}^{\infty} \{x\} \frac{1 - \log x}{x^2} dx \right|
\le 2 \int_{a}^{\infty} \frac{\log x}{x^2} dx.
$$

The expression $ \{ x \} $ denotes the fractional part of a real number $ x $. So $ \{ x \} = x - \lfloor x \rfloor. $

Is the above inequality true? If true, how can I prove it?

I realize that $ 0 \le \{ x \} \lt 1 $, so it will play a role in proving the inequality but I have been unable to figure out how to do the algebraic manipulation to use this fact to prove the inequality. Any help will be appreciated.

答:
Here's a proof for $a\geq e$:

$$
\begin{align*}
\left|\int_a^\infty \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x \right| &\leq \int_a^\infty \left|\{x\}\right| \frac{\left|1-\log x\right|}{x^2} \, \mathrm{d}x \\
&\leq \int_a^\infty \frac{1+\left|\log x\right|}{x^2} \, \mathrm{d}x \\
&=\int_a^\infty \frac{1+\log x}{x^2}\, \mathrm{d}x \\
&\leq \int_a^\infty \frac{2\log x}{x^2}\, \mathrm{d}x.
\end{align*}
$$

First inequality is from the triangle inequality for integrals, second from $\{x\}<1$ and the ordinary triangle inequality, third inequality holds for $x\geq 1$ and the final inequality comes from $\log x\geq 1$ for $x\geq e$.

For $a\in \left]1,e\right[$,

$$
\left|\int_a^e \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x\right|\leq \int_a^e \frac{\left|1-\log x\right|}{x^2} \, \mathrm{d}x=\int_a^e \frac{1-\log x}{x^2} \, \mathrm{d}x=\frac{1}{e}-\frac{\log a}{a}
$$

and

$$
\int_a^e \frac{2\log x}{x^2} \, \mathrm{d}x=\frac{2}{a}-\frac{4}{e}+\frac{2\log a}{a}.
$$

Define a function $f:\left[1,e\right]\to \mathbb{R}$, $f(a)=\frac{2}{a}-\frac{5}{e}+\frac{3\log a}{a}$. Then $f'(a)=\frac{1-3\log a}{a^2}$. Since $f'(a)=0$ iff $x=e^{1/3}$ and $f(e^{1/3}), f(1), f(e)\geq 0$, we have that $f(a)\geq 0$ for all $a\in \left[1,e\right]$ and therefore

$$\left|\int_a^e \{x\} \frac{1-\log x}{x^2} \, \mathrm{d}x\right|\leq\int_a^e \frac{2\log x}{x^2} \, \mathrm{d}x.$$

A similar approach could've been used when proving the inequality for $a\geq e$, but I find it unnecessarily complicated. The reason I brute-forced the case $a<e$ is that once the fractional part is estimated away, one is left with the integrand $\frac{1-\log x}{x^2}$. This, however, is not always smaller than $2\log x$, and so one must somehow account for the entire integration interval.


答:
Use the equality

$$x^n-a^n =(x-a)(x^{n-1} +x^{n-2}a + \dots + a^{n-1})$$

To prove the inequality

$$\vert x^n -a^n \vert \le n \vert x -a\vert \vert \vert 2a \vert^{n-1}$$

valid for $\vert x \vert \le \vert 2a\vert$.

Then apply it for $n=2k$.

Title: Showing that if $\lim_{n \to \infty} x_n=a$, then for any $k\in \mathbb{N}$, $\lim_{n \to \infty} x_n^{2k}=a^{2k}$

Body:
$x_n$ is a sequence. The only thing I can do here is just write the definition of $\lim_{n \to \infty} x_n=a$, but that doesn' t seem helpful to me. I tried looking for some inequalities that would help, and the only thing  that I found that I thought could be a little bit helpful, was the reverse triangle inequality, but I don' t know how to use that here.Could you please give some advice or a hint?

Thanks in advance


问:
Trying to figure out a fitting function for a curve where the fitting function always passes through the origin (0,0)
I have a curve that looks like this.

Acceleration v Velocity

I am trying to figure out a function I can use to fit to this curve. I have tried using a curve of the form $Ax^2 + Bx + C$ and while that does a good job of fitting the points it does not go through the origin. I am trying to construct a function of some sort that fits the points in the image and also goes through or is as close to the origin as possible.

I'd really appreciate some guidance!

答:
We observe that the function is on the power kind for small $x$ :

This draw us to look for a function of this kind :
$$y(x)=x^p f(x)$$
With $p\simeq 0.4$ :

The curve $f(x)$ appears like a piecewise linear function :

$$f(x)\simeq c+(ax+b)\text{H}(x-r)$$
H is the Heaviside step function.

In order to have a smooth transition between the two linear functions one can replace the Heaviside function by an approximate function of the logistic kind.
$$H(X)\simeq \frac{1}{1+e^{-q\,X}}$$
So, the function considered is :
$$y(x)=x^p\left(\frac{ax+b}{1+e^{-q\,(x-r)}}+c \right)$$

Then the parameters are computed thanks to non-linear regression :


问:
Latent Dirichlet Allocation: Newton-Raphson method
I am trying to understand one step in the Newton-Raphson method, used in the paper on LDA by Blei, Ng and Jordan. Namely, how does taking the derivative of $L$, w.r.t. $\alpha_j$ result in this equation. Also, what exactly is $\delta(i,j)$?

The terms which contain $\alpha$ are:

$$ L_{[\alpha]} =  \sum_{d=1}^{M} \bigg(  \log\Gamma(\sum_{j=1}^k \alpha_j ) - \sum_{i=1}^{k} \log \Gamma(\alpha_i)  + \sum_{i=1}^{k}((\alpha_i-1)( \Psi(\gamma_{d_i}) - \Psi\sum_{j=1}^{k} \gamma_{d_j}))) \bigg)$$
Taking the derivative with respect to $\alpha_i$ gives:
$$\frac{\partial L}{\partial \alpha_i} = M ( \Psi( \sum_{j=1}^{k} \alpha_j) - \Psi (\alpha_i)) + \sum_{d=1}^{M}( \Psi(\gamma_{d_i}) - \Psi(\sum_{j=1}^{k} \gamma_{d_j}))$$
The derivative depends on $\alpha_j$ where $j \neq i$, and we therefore must use and iterative method to find the maximal $\alpha$. In particular, the Hessian is in the form found in Eq. (10):
$$  \frac{\partial L}{\partial \alpha_i \alpha_j} = \delta(ij)M \Psi'(\alpha_i) - \Psi'(\sum_{j=1}^{k} \alpha_j)$$

答:
It seems like there is some typo in the paper. The actual Hessian should be of the form:

$$  \frac{\partial L}{\partial \alpha_i \alpha_j} = M(\Psi'(\sum_{j=1}^{k} \alpha_j)-\delta(ij) \Psi'(\alpha_i))$$

Note the difference.

Here, "M" is multiplied by both the terms instead of just one term.

The signs are reversed.


问:
Find the antiderivative of $\sqrt{3x-1} dx$

Find the antiderivative of $\sqrt{3x-1} dx$.

I got $\frac{2}{3}(3x-1)^{3/2}+c$ but my book is saying $\frac{2}{9}(3x-1)^{3/2}+c$

Can some one please tell me where the $2/9$ comes from?

答: I think all you missed was to divide by the inside derivative, that is the derivative of $3x - 1$</p>


问:
How to solve ODE with sine involved (comparison of coefficients)
I thought I grasped it thanks to an answer to a similar question I posed, but it seems a little trickier:

Given $y' = y\,\sin(x)+\sin(2\,x)$, trying to solve with the same approach:

$y_p = a_0\,\sin(2\,x)+b_0\,\cos(2\,x)$

so that:

$\begin{align}
{y_p}' &= 2\,a_0\cos(2\,x) -2\,b_0\,\sin(2\,x)\\
 &= y\,\sin(x)+\sin(2\,x) \\
&= a_0\,\sin(2\,x)\,\sin(x)+b_0\,\cos(2\,x)\,\sin(x) + \sin(2\,x)
\end{align}$

Unlike the other inquiry no comparison of coefficients seems possible due to the multiplication.

Is there something to be done still?

答:
The method of undetermined coefficients works for linear ODE with constant coefficients. The given ODE is a linear ODE with variable coefficients.

After multiplying both sides by the integral factor $e^{\cos(x)}$, the given linear ODE is equivalent to
$$D(e^{\cos(x)}y(x))=e^{\cos(x)}\sin(2x).$$
Now integrate the RHS:
$$\begin{align}
\int e^{\cos(x)}\sin(2x)\,dx&=2\int e^{\cos(x)}\cos(x)\sin(x)\,dx\\
&=-2\int e^{t}t\,dt=-2e^t(t-1)+c\\
&=-2e^{\cos(x)}(\cos(x)-1)+c.
\end{align}$$
Hence the general solution is
$$y(x)=ce^{-\cos(x)}+\underbrace{2(1-\cos(x))}_{\text{particular solution}}.$$
Note that the particular solution $y_p(x)=2(1-\cos(x))$ is not of the form:
$a_0\sin(2x)+b_0\cos(2x)$. Indeed $y_p(0)=0=b_0$ and $y_p(\pi)=4=b_0$ which is a contradiction.


问:
How to show the von Neumann trace inequality?
Let $A,B$ have the appropriate size. How can we show the von Neumann trace inequality?

$$ \mbox{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $$

Also, what is the intuition behind this inequality?

答:
A different (non-standard) proof:

Consider the SVD of $A$, with singular values $\sigma_i(A)$ and associated orthonormal bases $e_i$, $\tilde{e}_i$; similarly for $B$, with bases $f_j$, $\tilde{f}_j$. Then \begin{align}
\mathrm{tr}(A^*B)&=\mathrm{tr}\langle Ae_i,Be_i\rangle\\
&=\sum_i\sigma_i(A)\langle\tilde{e}_i,Be_i\rangle\\
&=\sum_{ij}\sigma_i(A)\langle\tilde{e}_i,\tilde{f}_j\rangle\langle \tilde{f}_j,Be_i\rangle \\
&=\sum_{ij}\sigma_i(A)\sigma_j(B)\langle\tilde{e}_i\tilde{f}_j\rangle\langle f_j,e_i\rangle\\
|\mathrm{tr}(A^*B)|&\le\sum_{ij}\sigma_i(A)\sigma_j(B)(|\langle\tilde{e}_i,\tilde{f}_j\rangle|^2+|\langle f_j,e_i\rangle|^2)/2\\
&=\vec{\sigma}(A)^*\,T\,\vec{\sigma}(B)
\end{align}
$T$ is a doubly stochastic matrix since $$\sum_iT_{ij}=\tfrac{1}{2}\sum_i|\langle\tilde{e}_i,\tilde{f}_j\rangle|^2+\tfrac{1}{2}\sum_i|\langle f_j,e_i\rangle|^2|=\tfrac{1}{2}\|\tilde{f}_j\|^2+\tfrac{1}{2}\|f_j\|^2=1$$ (and similarly for columns).

But by Birkhoff's theorem, every doubly stochastic matrix is the convex sum of permutation matrices, $$T=\sum_k\alpha_kP_k,\qquad\qquad \sum_k\alpha_k=1,$$ hence $$|\mathrm{tr}(A^*B)|\le\sum_k\alpha_k(\vec{\sigma}(A)^*P_k\vec{\sigma}(B))\le\max_k\vec{\sigma}(A)^*P_k\vec{\sigma}(B)$$ By the rearrangement inequality, the largest sum of $(\sigma_i(A))$ dotted with the permuted values of $(\sigma_i(B))$ is the one in which the values align from biggest to smallest. This is precisely the von Neumann trace inequality.


问:
How to find function for a parabola that goes though 3 points and is rotated 90° clockwise?
Short background: I am building a laser engraving system that can engrave on wood. The power Is always at its maximum, what changes is the speed of the laser on the surface. slow movement produces dark burns, fast movements light burns. Currently the speed is proportional and linear to the brightness. But once a spot gets lightly burned, it absorbs the light faster, and burns very quickly. Therfore I need my speed to vary according to the function shown below:

i know the values of 2 points and the vertex.

P1 = (0,300)

P2 = (255,1000)

Pv = P1 - 1 = (-1,299)

The values 300 and 1000 can change based on the material, Pv is alway P1 - 1. The other values stay the same. How can I calculate the speed for any given brightness?

答:
The general equation for such a parabola is $$(y-a)^2 = bx+c $$

Plug in the three points in this equation to get three equations in three variables, which you can solve.

For example in your specific case, the three equations are $$(300-a)^2=c \\ (1000-a)^2 =255b+c \\ (299-a)^2 =-b+c$$ which give $$a= \frac{151451}{78}, b =-\frac{98140}{89}, c=\frac{9613998601}{31684} $$


问:
Noah and his ark - inequalities
By spending an emerald, Noah turns a swan into a frog or a duck into a lizard, and by spending a ruby, he turns a swan into a duck or a frog into a lizard. Today he spent 20 emeralds and 23 rubies, and the number of frogs increased by 5. How and how much can the number of ducks change? (Give all possible results and explain why).

I found this in a Facebook group.

Let's name $S_0, F_0, D_0, L_0$ the initial number of each kind and $S_e, F_e, D_e, L_e$ their numbers after the exchange with the use of emeralds.
Clearly we have, for 1 emerald, $S_0-S_n = F_n-F_0$, therefore for $x$ emeralds, $x(S_0-S_n) = x(F_n-F_0)$.

Also for $y$ emeralds,
$y(D_0-D_n) = y(L_n-L_0)$.

We have $x+y=20$.

We get similar equations for the rubies:
$z(S_0-S_m) = z(D_m-D_0)$.

$w(F_0-F_m) = w(L_m-L_0)$.

$z+w=23$.

The difference in number of frogs after all transformations is 5 but I don't know how to take into account the consecutive transformations; it is getting very complicated.

答:
We define $r_1$ as the number of rubies used to transform swans to ducks and $r_2,e_1,e_2$ in a similar way according to the following diagram.

```<code>      e2
   D ----> L
   ^       ^
   |       |
r1 |       |r2
   |       |
   S ----> F
      e1
</code>```

then the following holds, where $\Delta X$ is the change in the expression $X$ after all transformations are applied.

$$\begin{eqnarray}
e=e_1+e_2\\
r=r_1+r_2
\end{eqnarray}$$

$$\begin{eqnarray}
\Delta(D+L)=e\\
\Delta(F+L)=r
\end{eqnarray}$$

$$\begin{eqnarray}
   \Delta(D+L)=\Delta D + \Delta L\\
   \Delta(F+L)=\Delta F + \Delta L
\end{eqnarray}$$

$$\begin{eqnarray}
e=20\\
r=23
\end{eqnarray}$$

Additionally we know that
$$ \Delta F= 5$$

From the above we get
$$\begin{eqnarray}
\Delta F + \Delta L&=&20\\
\Delta D + \Delta L&=&23\\
\Delta F&=&5
\end{eqnarray}$$
and therefore
$$\begin{eqnarray}
\Delta L&=& 15\\
\Delta D&=& 8 \\
\Delta F&=&5
\end{eqnarray}$$
So the number of Ducks changes by $8$.

We also see that the number of lizard increased by $15$ The number of animals did not change. So the number of swans decreases by $\Delta L+\Delta D+\Delta F=28$.

Title: Noah and his ark - inequalities

Body:
We define $r_1$ as the number of rubies used to transform swans to ducks and $r_2,e_1,e_2$ in a similar way according to the following diagram.

      e2
   D ----> L
   ^       ^
   |       |
r1 |       |r2
   |       |
   S ----> F
      e1

then the following holds, where $\Delta X$ is the change in the expression $X$ after all transformations are applied.

$$\begin{eqnarray}
e=e_1+e_2\\
r=r_1+r_2
\end{eqnarray}$$

$$\begin{eqnarray}
\Delta(D+L)=e\\
\Delta(F+L)=r
\end{eqnarray}$$

$$\begin{eqnarray}
   \Delta(D+L)=\Delta D + \Delta L\\
   \Delta(F+L)=\Delta F + \Delta L
\end{eqnarray}$$

$$\begin{eqnarray}
e=20\\
r=23
\end{eqnarray}$$

Additionally we know that
$$ \Delta F= 5$$

From the above we get
$$\begin{eqnarray}
\Delta F + \Delta L&=&20\\
\Delta D + \Delta L&=&23\\
\Delta F&=&5
\end{eqnarray}$$
and therefore
$$\begin{eqnarray}
\Delta L&=& 15\\
\Delta D&=& 8 \\
\Delta F&=&5
\end{eqnarray}$$
So the number of Ducks changes by $8$.

We also see that the number of lizard increased by $15$ The number of animals did not change. So the number of swans decreases by $\Delta L+\Delta D+\Delta F=28$.


问:
Probability that 3 heads will occur on 6 flips of fair coin given that head occurs on fifth as well as sixth toss.
Suppose an experiment consists of tossing a fair coin until three heads occur. What is the probability that experiment ends after exactly six flips of the coin with a head on the fifth toss as well as sixth?

Now I know how to solve for the case; when I have to find probability of three heads to occur in 6 independent tosses of fair coin. I use the binomial distribution with $n$=6 , $x$=3 and get result $P(x=3)$ = 5/16

What confuses me is the last condition of getting head on fifth and sixth trial. How do I proceed?

答:
Hint:

To get the experiment to end with three heads after exactly six flips of the coin with a head on the fifth toss as well as sixth, you need

exactly one of the first four to be heads

the fifth to be heads

the sixth to be heads


答: It comes from $2/3$ from the power and $1/3$ from the coefficient at $x$.</p>

Body:
Find the antiderivative of $\sqrt{3x-1} dx$.

I got $\frac{2}{3}(3x-1)^{3/2}+c$ but my book is saying $\frac{2}{9}(3x-1)^{3/2}+c$

Can some one please tell me where the $2/9$ comes from?


答:
I would suggest the following:

Flip the diagram along the diagonal. In other words declare the vertical axis to be $x$ and the horizontal axis to be $y$.

Given any three points without common $x$-value you can find the quadratic polynomial through these points by solving $y=ax^2+bx+c$ for $a,b,c$

Solve for $x$. Note that as $x$ appears as a square this will include solving for a root, hence making a decision of whether taking $+\sqrt{...}$ or $-\sqrt{...}$. Computing both solutions $x_0$ and $x_1$ for $x$, take their maximum ie. $x=\max (x_0,x_1)$

Title: How to find function for a parabola that goes though 3 points and is rotated 90° clockwise?

Body:
Short background: I am building a laser engraving system that can engrave on wood. The power Is always at its maximum, what changes is the speed of the laser on the surface. slow movement produces dark burns, fast movements light burns. Currently the speed is proportional and linear to the brightness. But once a spot gets lightly burned, it absorbs the light faster, and burns very quickly. Therfore I need my speed to vary according to the function shown below:

i know the values of 2 points and the vertex.

P1 = (0,300)

P2 = (255,1000)

Pv = P1 - 1 = (-1,299)

The values 300 and 1000 can change based on the material, Pv is alway P1 - 1. The other values stay the same. How can I calculate the speed for any given brightness?


问:
Let $x$ be a $p$-letter word, $f^{n}(x)=x \implies x$ must be a 'boring' word
Let $D$ be the set of permutation of $p$-letter word, where $p$ is prime number. And let $f$ be the function $f:D \rightarrow D$ that moves the last letter to the front. For example, if $p=7$,
$$ f(abcdefgh) = habcdefg  $$

We may call a word $x$ is a 'boring' word, only if $f(x) = x$. Notice that there exist only 26 boring words: all $a$s $\underbrace{aa...a}_{p}$, all $b$s, ..., and all $z$s $\underbrace{zz...z}_{p}$.

Proof that if $f^{n}(x) = x$, where $1 \le n \le p-1$, then $x$ is a 'boring' word.

My attempt, if $n=1$ is clear. Now let $p>2$, $n=2$ and $f^{2}(x)=x$. Define $\alpha_{i}$ to be the letter at position $i$th in $x$.

$$ x = \alpha_{1}, \alpha_{2}, ..., \alpha_{p} $$
$$ \implies f(x) = \alpha_{p}, \alpha_{1} , ..., \alpha_{p-1} $$
$$ \implies f^{2}(x) = \alpha_{p-1} , \alpha_{p}, ..., \alpha_{p-2} $$

This means that
$$ \alpha_{p} = \alpha_{p-2} = \alpha_{p-4} = ... = \alpha_{1} $$
but
$$ \alpha_{1} = \alpha_{p-1} = \alpha_{p-3} = ... = \alpha_{2} $$

so all the letters must be the same, which means a 'boring' word.

Now if $n > 2$ and $f^{n}(x)=x$. Notice that we always have $p = qn + r$, where $0<r<n$ is the remainder, then
$$ \alpha_{p} = \alpha_{p-n} =  ... = \alpha_{(p -qn) = r} $$

how to continue..?

答: Let $x=a_0\cdots a_{p-1}$ so $f^n(x)=a_{p-n}\cdots a_{p-1}a_0\cdots a_{p-n-1}$. By Lagrange we have $\langle p-n\rangle=\Bbb Z_p^+$ whenever $0<n<p$ and since $f^n(x)=x$ the result follows.


问:
Equation of sphere
Find equation of the sphere if $$\frac{3}{2}x-\frac{1}{2}y+\frac{7}{2}z=14 $$ is a tangent plane at the point $(2,-1,3)$.

What I have done so far.

from the tangent plane, the normal is $(\frac{3}{2},-\frac{1}{2},\frac{7}{2})$ and the radius is $\sqrt{14}$

I think what I need now is the center point but how can I find that

答: Take any real number $t\ne0$ and consider the point $(x_0,y_0,z_0)=(2,1,-3)+t\left(\frac32,-\frac12,\frac72\right)$. Then the plane $\frac32x-\frac12y+\frac72z=14$ is tangent to the sphere$$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=t^2\left(\left(\frac32\right)^2+\left(-\frac12\right)^2+\left(\frac72\right)^2\right)$$at $(2,1,-3)$. The radius of such a sphere is $|t|\sqrt{\left(\frac32\right)^2+\left(-\frac12\right)^2+\left(\frac72\right)^2}$.

Title: Equation of sphere


答:
Tangent plane to $f(x,y,z) = k$ at the point $P = (a,b,c)$ is given by
$$
f_x (x-a) + f_y(y-b) + f_z(z-c) = 0
$$
where each partial derivatives of $f$ is evaluated at the point $P$.

Let $f(x,y,z) = (x-p)^2 + (y-q)^2+ (z-s)^2 = r^2$ then we have $f_x = 2(x-p)$, $f_y = 2(y-q)$ and $f_z = 2(z-s)$, at the point $P$ we have $f_x = 4 - 2p$ , $f_y = -2 - 2q$ and $f_z = 6 -2s$. Then the tangent plane equation becomes
$$
(4-2p)(x-2) + (-2-2q)(y+1) + (6-2s)(z-3) = 0
$$
Compare this with the given tangen equation and determine the value of $p$,$q$ and $s$. Note that for any real number $t \ne 0$ , the equation $3tx-yt+7zt = 28t$ is still tangent equation (This explains why there are infinite such spheres!).
Then it is straightforward to find the value of $p$,$q$ and $s$. To evaluate $r$, observe that the point $(2,-1,3)$ is part of the sphere.

I hope you can finish rest of the calculations.

Title: Equation of sphere

Body:
Find equation of the sphere if $$\frac{3}{2}x-\frac{1}{2}y+\frac{7}{2}z=14 $$ is a tangent plane at the point $(2,-1,3)$.

What I have done so far.

from the tangent plane, the normal is $(\frac{3}{2},-\frac{1}{2},\frac{7}{2})$ and the radius is $\sqrt{14}$

I think what I need now is the center point but how can I find that


问:
Generalized Recursion vs. Turing Completeness
$\newcommand{\NN}{\mathbb N}\newcommand{\UU}{\mathcal U}$I'm currently reading through Homotopy Type Theory: Univalent Foundations of Mathematics. In Exercise 1.10, we construct the Ackermann function using (their version of) primitive recursion, which surprised me because I'm used to having the Ackermann function being a function between primitive recursion and computability.

Of course, what makes the exercise possible is that we can recurse over not just $\NN$ but over functions $\NN\to\NN.$ Roughly speaking, the fact we can define $\textsf{ack}(m+1)$ in terms of $\textsf{ack}(m)$ is enough to recursively define the Ackermann function $\textsf{ack}:\NN\to\NN\to\NN.$

Anyways, being able to recurse over arbitrary types seems like a lot of power, and I was wondering exactly how much power.

Is there a total computable function $\NN\to\NN$ which cannot be formed from the above "generalized" primitive recursion?

I suspect one to exist. If there's a finer line between what generalized primitive recursion gives and computable enumerability, I'd be interested in that too.

答:
The range of total computable $\mathbb{N} \to \mathbb{N}$ functions definable in a type theory corresponds to the proof-theoretic strength of the theory.

There is no total language for all total computable $\mathbb{N} \to \mathbb{N}$ functions. This is a constructive analogue to Gödel's second incompleteness theorem. If such a language existed, it would also contain its self-interpreter function for a suitable $\mathbb{N}$-encoding of programs, but that would make it possible to write general recursive programs, contradicting totality.

For some type theories, the range of definable functions can be concisely characterized by an ordinal number. In this case, we have an ordinal number $\alpha$ such that the fast-growing function $f_{\alpha} : \mathbb{N} \to \mathbb{N}$ grows faster than any definable function in the theory. Here's a listing of ordinals for some type theories.

For any type theory with an impredicative universe, no proof-theoretic ordinal is known. Here the strength of the theory can be explained by comparison to other systems. As Dan Doel mentioned, System F (a non-dependent stratified theory with an impredicative universe) has the same strength as second-order arithmetic.

Simple type theory with $\mathbb{N}$ is also known as Gödel's System T, and has strength $\epsilon_0$. This is stronger than primitive recursive arithmetic; the Ackermann function is definable in System T but not in PRA. The System-T-definable functions are also precisely the functions which are provably total in Peano Arithmetic. Calling the recursion principle of System T (or HoTT) "primitive recursion" is a bit confusing, because it's not the same thing as the recursion principle in PRA.

Infinitary inductive types (such as W-types) increase strength in predicative theories significantly. Intuitively, that's because their induction principles immediately provide a form of transfinite induction.

Is there a total computable function ℕ→ℕ which cannot be formed from
the above "generalized" primitive recursion?

It depends on what other features are supported in the type theory. If we only have $\mathbb{N}$ and functions, $f_{\epsilon_0}$, Goodstein sequences and hydra games are examples for undefinable functions. If we have infinitary inductive types (or W-types), we get a power boost and we can easily define these functions. For example, in Agda we can define $f_{\epsilon_0}$ in the following way:

```<code>data ℕ : Set where
  zero : ℕ
  suc  : ℕ → ℕ

data O : Set where
  zero : O
  suc  : O → O
  lim  : (ℕ → O) → O     -- ℕ-branching node (infinitary)

iter : {A : Set} → ℕ → (A → A) → A → A
iter zero    f a = a
iter (suc n) f a = f (iter n f a)

f : O → ℕ → ℕ
f zero    n = n
f (suc α) n = iter n (f α) n
f (lim g) n = f (g n) n

_+_ : O → O → O
α + zero  = α
α + suc β = suc (α + β)
α + lim f = lim (λ n → α + f n)

_*_ : O → O → O
α * zero  = zero
α * suc β = (α * β) + α
α * lim f = lim (λ n → α * f n)

_^_ : O → O → O
α ^ zero  = suc zero
α ^ suc β = (α ^ β) * α
α ^ lim f = lim (λ n → α ^ f n)

ω : O
ω = lim (λ n → iter n suc zero)

fε₀ : ℕ → ℕ
fε₀ = f (lim (λ n → iter n (ω ^_) ω))
</code>```

Body:
The range of total computable $\mathbb{N} \to \mathbb{N}$ functions definable in a type theory corresponds to the proof-theoretic strength of the theory.

There is no total language for all total computable $\mathbb{N} \to \mathbb{N}$ functions. This is a constructive analogue to Gödel's second incompleteness theorem. If such a language existed, it would also contain its self-interpreter function for a suitable $\mathbb{N}$-encoding of programs, but that would make it possible to write general recursive programs, contradicting totality.

For some type theories, the range of definable functions can be concisely characterized by an ordinal number. In this case, we have an ordinal number $\alpha$ such that the fast-growing function $f_{\alpha} : \mathbb{N} \to \mathbb{N}$ grows faster than any definable function in the theory. Here's a listing of ordinals for some type theories.

For any type theory with an impredicative universe, no proof-theoretic ordinal is known. Here the strength of the theory can be explained by comparison to other systems. As Dan Doel mentioned, System F (a non-dependent stratified theory with an impredicative universe) has the same strength as second-order arithmetic.

Simple type theory with $\mathbb{N}$ is also known as Gödel's System T, and has strength $\epsilon_0$. This is stronger than primitive recursive arithmetic; the Ackermann function is definable in System T but not in PRA. The System-T-definable functions are also precisely the functions which are provably total in Peano Arithmetic. Calling the recursion principle of System T (or HoTT) "primitive recursion" is a bit confusing, because it's not the same thing as the recursion principle in PRA.

Infinitary inductive types (such as W-types) increase strength in predicative theories significantly. Intuitively, that's because their induction principles immediately provide a form of transfinite induction.

Is there a total computable function ℕ→ℕ which cannot be formed from
the above "generalized" primitive recursion?

It depends on what other features are supported in the type theory. If we only have $\mathbb{N}$ and functions, $f_{\epsilon_0}$, Goodstein sequences and hydra games are examples for undefinable functions. If we have infinitary inductive types (or W-types), we get a power boost and we can easily define these functions. For example, in Agda we can define $f_{\epsilon_0}$ in the following way:

data ℕ : Set where
  zero : ℕ
  suc  : ℕ → ℕ

data O : Set where
  zero : O
  suc  : O → O
  lim  : (ℕ → O) → O     -- ℕ-branching node (infinitary)

iter : {A : Set} → ℕ → (A → A) → A → A
iter zero    f a = a
iter (suc n) f a = f (iter n f a)

f : O → ℕ → ℕ
f zero    n = n
f (suc α) n = iter n (f α) n
f (lim g) n = f (g n) n

_+_ : O → O → O
α + zero  = α
α + suc β = suc (α + β)
α + lim f = lim (λ n → α + f n)

_*_ : O → O → O
α * zero  = zero
α * suc β = (α * β) + α
α * lim f = lim (λ n → α * f n)

_^_ : O → O → O
α ^ zero  = suc zero
α ^ suc β = (α ^ β) * α
α ^ lim f = lim (λ n → α ^ f n)

ω : O
ω = lim (λ n → iter n suc zero)

fε₀ : ℕ → ℕ
fε₀ = f (lim (λ n → iter n (ω ^_) ω))


问:
How to prove $\lvert z \rvert \geq \Re(z)$?
I try to prove
$$
\lvert z \rvert \geq \Re(z) \tag 1
$$
For a complex number $z= x+iy$ we have
\begin{align}
\lvert z \rvert &=\sqrt{x^2+y^2} \tag 2 \\
\lvert z \rvert^2 &=x^2+y^2 \tag 3
\end{align}
And for att real number $x$ we have
$$
\lvert x \rvert = \sqrt{x^2} \tag 4
$$
The proof:
A number squared is always non-negative, so we start with $y^2\geq 0$, i.e.
\begin{align}
y^2 &\geq 0
\tag 5
\\
y^2 + x^2 &\geq x^2
\tag 6
\\
\lvert z\rvert ^2 &\geq x^2
\qquad (\text{from } (3))
\tag 7
\\
\sqrt{\lvert z\rvert ^2} &\geq \sqrt{x^2}
\tag 8
\\
\lvert z \rvert &\geq x
\qquad
\tag 9
\\
\lvert z \rvert &\geq \Re(z)
\tag{10}
\end{align}

I don't know if $(9)$ is correct, should it instead be $\lvert z \rvert > \lvert x\rvert$, due to the absolute value in $(4)$? But, if so, I instead end up with $\lvert z \rvert \geq \lvert \Re(z) \rvert$?

答: That is correct, that is, you proved correctly that $|z|\geqslant|\operatorname{Re}(z)|$. And, since $|\operatorname{Re}(z)|\geqslant\operatorname{Re}(z)$, you get that $|z|\geqslant\operatorname{Re}(z)$.


问:
Proving that $\lim_{x \to 0} \frac{f(x)}{|x|}=1$ implies $f$ is not differentiable at $0$.
$$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$.
I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient?

Assume toward contradiction that $f$ is differentiable at $0$.

Then by definition we know that $f$ is continuous at $0$. Therefore

$\lim _{x\to 0}f\left(x\right)=f\left(0\right)$

We know that

$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$

Therefore we can conclude

$\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$

But we can also rewrite this as

$\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$

Therefore $f(0) = 0$.

Since $f$ is differentiable at $0$ we have

$\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$

Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have

$\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$

$\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$

Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$.

Therefore $f$ is not differentiable at 0.

答:
Your proof looks almost perfect to me. There's a couple of minor things I would change in the last few lines. You wrote:

Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$
exist, and are both equal to 1, we have

$\lim \:_{x\to
 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot
 \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to
 \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot
 \:\:\:\frac{x}{x}=1\cdot 1=1$

$\lim \:_{x\to
 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot
 \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to
 \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot
 \:\:\:\frac{-x}{x}=1\cdot -1=-1$

Since the limit doesn't exist, we have a contradiction to the fact $f$
is differential at $0$.

Therefore $f$ is not differential at 0.

I don't see the need to remove the absolute value signs in the following places in red:

$f'(0)= \lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} = \lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{\color{red}{|}x\color{red}{|}}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$, but

$f'(0) =\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =  \lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{ \color{red}{|}x\color{red}{|}} \cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$

Since $\lim \:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$ only exists if both
$\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$ and $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}$
exist and are equal - which they aren't - we have a contradiction to the fact $f$ is differentiable at $0$.

Therefore $f$ is not differentiable at $0$.


问:
Prove that $||f'|| \geq ||f||$

Assume that $f$ has a continuous derivative on $[0,2\pi]$, that $f(0)=f(2\pi)$ and that $\int_{0}^{2\pi} f(t) dt =0$. Prove that $||f'||\geq ||f||$ with equality iff $f(x)= a \cos x + b \sin x$.

If  $f(x) = a\cos x+ b\sin x$ then $||f'||=||f||$ holds but I have no idea on how to prove the converse and also I got no intution on which result should be used.

This is a second course of real analysis quiz question.

Can you please help me with this?
I am not adding much information regarding attempt but that's due to the reason that I can't think of any known result on how to approach this question and couldn't even start this.
Thanks a lot.

答:
I think the norm here is the  $L^{2}$ norm. We have $\hat {f'} (n)=-in \hat {f} (n)$. Hence, using Parseval's identity,  $\|f'\|^{2}=\sum_n | {f'} (n)|^{2}= \sum_n n^{2}| {f} (n)|^{2}\geq \sum_n |\hat {f} (n)|^{2} =\|f||^{2}$ where I have used the  fact that the constant term is missing (since $\int f(t)dt=0$) and $n^{2} \geq 1$ for the remaining  terms.

Obviuosly, equality can hold here only when  all the coefficients correspoinding to $n$ with $|n|>1$ are zero. Hence, $f (x)=a\cos x+b\sin x$ for some $a$ and $b$.

Title: Prove that $||f'|| \geq ||f||$


问:
How would one approximate the surface area of a curved shape as on these ETFE pillows?
I need to get the surface area of these ETFE pillows:

答:
There are surely different types of modelization, some of them based on physics. Here is one based on a visually satisfying similarity. Indeed (white) cross sections appear to be well modelized by parabolas ($z=k(W^2-y^2)$); the way these coefficients $k$ vary as a function of $x$ is rather arbitrary; here we have taken:

$$z=f(x,y)=\underbrace{\frac{h}{W^2}  \sqrt{\cos(\pi \frac{x}{2L})}}_{k}(W^2-y^2) \ \ \text{for}  \ \ -L<x<L, \ -W<y<W $$

where $2L$ is the length and $2W$ the width of the mattress and $h$ the measured maximal height in the middle of the mattress.

It remains to use the following formula for the surface area (see here):

$$A=\int_{x=-L}^L \int_{y=-W}^W \sqrt{1+\left(\partial f/\partial x\right)^2 + \left(\partial f/\partial y\right)^2}dx dy$$

For given values of $L$ and $W$, using a numerical software, one gets a numerical value for $A$.

Remarks: One could instead have considered NURBS surfaces.

Here is the family of parabolas with common roots $\pm a$ (here with $a=1$) with common equation $y=k(a^2-x^2)=k(a-x)(a+x)$. Here $k$ is linearly increasing. In the case of the mattress, $k$ must stay a long time almost constant then decrease rather sharply.


问:
Olympiad question: How many permutations with no fixed letters does BHARAT have?
We will say that a rearrangement of the letters of a word has no fixed letters if, when the rearrangement is placed directly below the word, no column has the same letter repeated.  For instance, HBRATA is a rearrangement with no fixed letters of BHARAT.  How many distinguishable rearrangements with no fixed letters does BHARAT have?  (The two As are considered identical.)

A p&c question is given with its solution which was asked in mathematics olympiad , but the solution is very tricky to understand so how will we understand the solution ?

答:
There are $\binom4 2$  ways = $6$ ways in which the two $A's$ are out of position.

The remaining $4$ distinct letters with $2$ of them (call them $X$ and $Y$) "in position"

We now apply inclusion-exclusion, to find total permutations minus those with at least $X$ still in position minus at least $Y$ still in position plus both $X$ and $Y$ still in position (because we subtracted such cases twice), thus

$4!-3!-3!+2! = 14$ and multiply by $6$


问:
Proposition about regular functions on an affine variety
I have the following basic issue:

Let $X$ be an affine variety in $\mathbb{A}_{k}^ {n}$. Co-ordinate ring of $X$ be  $A(X):= k[x_1,\ldots ,x_n]/I(X)$ whereas $K(X)$ be its quotient field.
$$ \mathcal{O}_{X,p}:=\{f/g :g(p)\neq0; f,g\in A(X)\} \\
\mathcal{O}_{X}(U):= \bigcap_{p \in U} \mathcal{O}_{X,p}$$
Where $\mathcal{O}_{X}(U)$ be known as the regular functions on open subset $U$.

Now consider the proof of following proposition: A set map $\phi:U \rightarrow k$ is regular, then for each $p \in U$ it has a neighborhood $V$ such that there are $f,g \in k[x_1, \ldots ,x_n]$ with $g(q)\neq 0$ and $\phi (q)= f(q)/g(q) \; \; \forall q \in V$

By definition $\phi \in \mathcal{O}_{X}(U) \implies \phi \in \mathcal{O}_{X,p}\;\; \forall p\in U$. Hence $\phi(p)=f_{p}(p)/g_{p}(q)$[Here the subscripted $p$ is for indexing the representation at a point $p$]. I can also find a neighborhood $V_{p}$ where $g_{p}(q)\neq 0$. But how do I relate representation of $\phi$ at $p$ with representation at another point $q\in V_{p}$? It might happen that $\phi (q)= f_{q}(q)/g_{q}(q)\neq f_{p}(q)/g_{p}(q).$

答:
Your comment was: "Let $X$ be an affine variety in $\mathbb{A}^n_k$. Let the co-ordinate ring of $X$ be $A(X):=k[x_1,…,x_n]/I(X)$ where $K(X)$ is its quotient field.

$\mathcal{O}_{X,p}:=\{f/g:g(p)≠0;f,g∈A(X)\}$ and let $ \mathcal{O}_{X}(U):=\cap_{p \in U}\mathcal{O}_{X,p}$

Where $\mathcal{O}_X(U)$ by definition is the ring of regular functions on the open subset $U$."

A possible answer is: If $A:=k[x_1,..,x_n]/I$ where $I$ is a prime ideal and $X:=Spec(A)$ we define the stalk

I0. $\mathcal{O}_{X,p}:= lim_{p \in U} \mathcal{O}_X(U)$ as a limit.

Since $X$ is an integral scheme ($A$ is an integral domain) it follows there is an isomorphism

I1. $\mathcal{O}_{X,p} \cong \cap_{p \in U}\mathcal{O}_X(U)$

for any point $p\in X$.

Your question: "I completely understand what you are trying to say. But you see, I am trying to solve the above mentioned issue directly from the definition of $\mathcal{O}_{X,p}$ and the definition of stalk I've presented here doesn't have any mention of neighborhood. Thank for your replies btw!"

Answer:

We may take intersection in I1 since there is an inclusion of rings $\mathcal{O}_X(U) \subseteq K(A)$ for any open set $U$. Hence the intersection in I1 takes place inside the quotient field $K(A)$ of the integral domain $A$. If the rings $\mathcal{O}_X(U)$ for varying open sets $U$ are not embedded into one fixed ring, you cannot "take intersections". In general when $X:=Spec(A)$ is any affine scheme you cannot in general embed $\mathcal{O}_X(U)$ into one fixed ring $K$ for all open sets $U$. This is why the stalk in I0 is defined using "limits".

When you define the stalk $\mathcal{O}_{X,p}$ using a limit as in I0, it follows any element $u\in \mathcal{O}_{X,p}$  has a representative on the form $(U,s)$ where $p\in U \subseteq X$ is an open set and $s\in \mathcal{O}_X(U)$ is a section on $U$. You may find this on page 62 in Hartshorne. Hartshorne comments that any element in $\mathcal{O}_{X,p}$ is an equivalence class $\overline{(U,s)}$ where $(U,s) \cong (V,t)$ iff there is an open set $p\in W \subseteq U \cap V$ with $t_W=s_W$.

In general if $X$ is any topological space and $F$ any sheaf of abelian groups on $X$ with $x \in X$ any point, let

$i_x: \{x\} \rightarrow X$

be the inclusion map. It follows by definition

$i_x^{-1}(F):= lim_{x\in U} F(U)\cong F_x$,

where $F_x$ is the stalk of $F$ at $x$.
Hence we may use the canonical map $i_x$ to define the stalk $F_x$. The inverse image $i_x^{-1}F$ is defined on page 65 in Hartshorne, and there you will find the precise definition of this "limit".


问:
The notation of the second fundamental theorem of Calculus
I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\Large\frac{d}{dx}(\int f(x)\,dt) =  f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function.
this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

is how a very small change in $x$ affects that area under $f(t)$ between $a$ (a constant) and $x$. how do I get from that to the right interpretation?

答:
It should be
$$\frac{\text{d}}{\text{d}x} \int_a^x f(t)\:\text{d}t=f(x).$$
The variable $t$ has no meaning outside the context of the integral.

Concerning your question about indefinite integrals, they may be a convenient abuse of notation if you're trying to solve an ODE, but they are confusing at best and misleading at worst and aren't/shoudn't be a thing. The integral is designed to produce a number given a function (and maybe a set to integrate over, but that one can be done away with), and conditioning students to mechanically add a $+C$ to integrals does way more harm than good in my experience.


答:
$$
\int \sqrt{3x-1}~dx=\int(3x-1)^{1/2}~dx
$$
Let $u=3x-1$, $du=3~dx$, so
$$
\int(3x-1)^{1/2}~dx=\frac{1}{3}\int u^{1/2}~du
$$
Add one to the power of $u$, and divide by the new power
$$
\int \sqrt{3x-1}~dx=\frac{1}{3}\int u^{1/2}~du=\frac{1/3}{3/2}u^{3/2}+c=\frac{2}{9}(3x-1)^{3/2}+c
$$

Your problem is that you need to take into consideration the derivative of $3x-1$, which means you need to divide by $3$, giving you the factor of $\frac{1}{3}$ missing from your answer.

As was suggested in the comments of your previous question, if you differentiate the answer that you got with the chain rule, then you'll see why you've made a mistake
$$
\frac{d}{dx}\left(\frac{2}{3}(3x-1)^{3/2}\right)=\frac{3}{2}\cdot\frac{2}{3}(3x-1)^{1/2}\underbrace{\frac{d}{dx}(3x-1)}_{=3}=3\sqrt{3x-1}
$$

Title: Find the antiderivative of $\sqrt{3x-1} dx$


问:
Does Itô's formula define a semimartingale?
We may define a semimartingale as:
$$X_{t}=X_{0}+M_{t}+A_{t}$$
Where $X_0$ is $\mathcal{F}_0$ measurable, $M$ is a continuous
local martingale $M_0 = 0$ and $A$ is an adapted continuous finite variation process
with $A_0 = 0$. The Itô's formula reads:
$$df(t,X_t) =\frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial X_s}\,dX_t+\frac{1}{2}\frac{\partial^2f}{\partial X_s^2}d[X]_t$$

$[X]_t$ is a quadratic variation and hence a finite variation process, is $\frac{\partial^2f}{\partial x^2}d[X]_t$ also a finite variation process?

答: Since $[X]_t$ is of finite variation, $\int f_{xx}(s, X_s)d[X]_s$ is interpreted in the Lebesgue-Stieltjes sense. This integral itself produces finite variation functions. You can see that by decomposing $f_{xx}(s, X_s)$ into positive and negative parts and $d[X]_s$ as the difference of two nondecreasing functions.


问:
$\mathbb{C}$-point of a variety and automorphism of $\mathbb{C}$
Let $X$ be a scheme of finite type over $\mathbb{C}$.

A closed point is the topologically closed point. A $\mathbb C$-rational point is given by $\mathrm{Spec}(\mathbb C)\to X$ a morphism of scheme over $\mathbb{C}$.

Now, let's just take $X=\mathrm{Spec}(\mathbb C)$ as well for simplicity. It seems that there will be more than one rational points because we have a lot of automorphisms $\mathbb{C}\to\mathbb{C}$.

I also see this (cf. Algebraic and Analytic Geometry by Amnon Neeman)
I think it is wrong because of the automorphisms on $\mathbb C$. The closed point in this case corresponds to the maximal ideal of $R$. The correspondence given by taking $\ker\varphi$ to obtain the maximal ideal. But when we concatenate an automorphism $\psi$ of $\mathbb C$ or embed $\mathbb C$ into $\mathbb C$ (there are some subfield of $\mathbb C$ isomorphic to $\mathbb C$). Say the structure homomorphism $j:\mathbb C\to R$. Then we still get a $\mathbb C$-algebra homomorphism $\psi\varphi:R\to\mathbb C$ by setting the structure homomorphism on the second $\mathbb C$ to be $\psi\varphi j:\mathbb C\to\mathbb C$. And the kernel of $\psi\varphi$ is just the maximal ideal so we get more homomophisms than closed points.

答:
You should not allow all morphisms $\mathrm{Spec}(\mathbb C)\to X$, only the morphisms over $\mathrm{Spec}(\mathbb C)$.

Let me clarify: to say that $X$ is a scheme over $\mathrm{Spec}(\mathbb C)$ means there is a structural morphism $X\to \mathrm{Spec}(\mathbb C)$. Then for any commutative $\mathbb{C}$-algebra $A$, an $A$-point of $X$ should be a morphism $\mathrm{Spec}(A)\to X$ which gives a commutative triangle with the structural morphisms $\mathrm{Spec}(A)\to \mathrm{Spec}(\mathbb C)$ and $X\to \mathrm{Spec}(\mathbb C)$.

In particular, when $A=\mathbb C$, a $\mathbb C$-point is a morphism $\mathrm{Spec}(\mathbb C)\to X$ such that the composition $\mathrm{Spec}(\mathbb C)\to X\to \mathrm{Spec}(\mathbb C)$ is the identity morphism of $\mathrm{Spec}(\mathbb C)$.

If you take that definition, $X=\mathrm{Spec}(\mathbb C)$ has only one $\mathbb C$-point over $\mathbb C$.

To answer your edit, you cannot compose with automorphisms of $\mathbb C$, because then you don't have $\mathbb C$-algebra automorphisms anymore. The only $R$-algebra automorphism of a ring $R$ is the identity. All the other automorphisms are only ring automorphisms, not $R$-algebra automorphisms.


问:
No determinant of a homomorphism even when $\dim V = \dim W$?

Why is there no determinant of a homomorphism $f: V \to W$, even when $\dim V = \dim W$?

I don't know how the concepts in linear algebra are related to each other and I am new to it.

Thanks

答:
Generally you can only take the determinant of matrices. The reason why this definition carries over to endomorphisms is because this way it turns out that it doesn't depend on the basis: If you choose a different basis of your vector space, that affects both domain and codomain of your endomorphism, or put differently, as matrices you'd have to basis-transform before and after applying your endomorphism, which cancels after taking the determinant. If you have two different vector spaces, the representing matrix depends on both bases, and changing one of them would change the determinant.

We can see this using the following construction: if you have an isomorphism $\phi: V\to W$, then you could choose a basis $(v_1,\dots,v_n)$ on $V$ and $(\phi(v_1),\dots,\phi(v_n))$ on $W$. Then, written in these bases, $\phi$ would just correspond to the identity matrix, so it would have determinant 1. But the same would be true for $2\phi$, but then the determinant wouldn't be multilinear.

One way in which you could achieve what you want to do is to fix an isomorphism $\psi$ between $V,W$. If you now have a homomorphism $f: V\to W$, then $\psi^{-1}\circ f: V\to V$ is an endomorphism and has a determinant.


答: Because the determinant should be independent of basis. Intuitively, the determinant measures how volumes are scaled under a linear transformation. The important thing here is that it's only about scaling, not about absolute volumes. The determinant only compares volumes before and after transforming. It says: Whatever volume the set $X\subset V$ has, the set $f(X)$ has $\vert\det f\vert$ times that volume. We don't actually have a natural way to give $X$ a volume. We could choose some basis and say that the parallelepiped spanned by it has volume $1$. But we could also choose a completely different basis and say that the parallelepiped spanned by that basis has volume 1. In reality, we do so by switching between cubic feet, cubic meters, liters, etc.: We choose orthogonal vectors of length 1 ft., 1m or 10cm which span 3d space, and then say that the parallelepiped spanned by them has volume 1 (cubic foot, cubic meter or liter). If we stay within the same vector space, this is not a problem: The determinant doesn't care which basis we choose, as long as we stick with it, since it only measures how volumes are scaled, not their absolute values. But this breaks if we have two different vector spaces. Now we have to specify two bases to compare volumes, and how volume changes depends on our choices. For instance, if, say, the unit cube wrt. some basis in $V$ is mapped to the parallelepiped spanned by the basis $B_W=\{v_1,\dots,v_n\}$ of $W$ and we measure how volume changes here, we I will get something different when measuring wrt. this same basis than when we choose the basis $B'_W=\{2v_1,\dots,2v_n\}$, by a factor of $2^{-n}$. So a determinant could no longer be independent of basis, which is bad. So we just don't define it in this case.

Title: No determinant of a homomorphism even when $\dim V = \dim W$?

Body:
Why is there no determinant of a homomorphism $f: V \to W$, even when $\dim V = \dim W$?

I don't know how the concepts in linear algebra are related to each other and I am new to it.

Thanks


问:
How to find 3 x 3 matrix inverses
Is there a way of finding the inverse of a $3 \times 3$ matrix without forming an augmented matrix with the identity matrix? Also, is there a quick way of checking that a  $3 \times 3$ matrix's inverse exists, without trying to compute it? Thanks.</p>

答:
Steps to find Inverse

1. Find determinant of $3\times 3$ Matrix

2. Find minor

3. Find Cofactor

4. Find Adjoint

5. Replace results in below formula

$A^{-1} = \frac{1}{\det(A)} adj(A)$

As you are unaware of these terms so let me first define it for you.

If A is a square matrix, $(3\times 3)$ for example, then the minor of entry $a_{ij}$ , denoted by $A_{ij}$  is defined to be the determinant of the submatrix that remains after the $i$th row and $j$th column are deleted from A. The number $(-1)^{i+j} A_{ij}$, denotd by $C_{ij}$ and is called the cofactor of entry $a_{ij}$.

You need to form cofactor matrix given as

$C   = \left(
  \begin{array}{ccc}
    C_{11} & C_{12} & C_{13} \\
    C_{21} & C_{22} &  C_{23} \\
   C_{31}  & C_{32} & C_{33} \\
  \end{array}
\right)$

To find the adjoint of a matrix denoted by $adj(A)$, just transpose the cofactor
matrix.

This link will help you to understand this

Title: How to find 3 x 3 matrix inverses


问:
Euler's method for second order ODE
I'm new to numerical analysis and got stuck on an ODE:
$$y'' + y^3 - 0.5 = 0$$
with $h = 0.1$, $y(0) =1$ and $y'(0)=1$.

How should I begin?

答:
After reducing to first order, you have to apply the Euler method in each component, i.e.

$$\vec{x_{n+1}} = \vec{x_n} + h \vec{f(x_n)}$$

where $x_n=[y_n,z_n]$ and $\vec{f(x_n)}=[-y_n^3+0.5,z_n]$. Of course $x_0=[y_0,z_0]=[1,1]$

The following C++ snippet illustrates how to write it (using valarray)

```<code>#include <iostream>
#include <cmath>
#include <valarray>

std::valarray<double> rhs(std::valarray<double>& x){
    return std::valarray<double>{-x[1]*x[1]*x[1]  + 0.5, x[0]};
}

std::valarray<double> forw_euler(const double& h, const double& tf,const std::valarray<double>& x0){

    const int n = ceil(tf/h);
    std::valarray<double> y = x0;

    for (auto i=0; i<=n; ++i) {
        y+=h*rhs(y);
    }
    return y;
}

int main(){
    std::valarray<double> x0 {1.0, 1.0};
    const double h = 0.1;
    const double tf = 1.0;
    std::valarray<double> sol = forw_euler(h,tf,x0);
    std::cout << "Solution at time T = " << tf << "\n"
    << "first component: " << sol[0] << "\t" << "second component " << sol[1] << std::endl;

    return 0;
}
</code>```

Title: Euler's method for second order ODE

Body:
After reducing to first order, you have to apply the Euler method in each component, i.e.

$$\vec{x_{n+1}} = \vec{x_n} + h \vec{f(x_n)}$$

where $x_n=[y_n,z_n]$ and $\vec{f(x_n)}=[-y_n^3+0.5,z_n]$. Of course $x_0=[y_0,z_0]=[1,1]$

The following C++ snippet illustrates how to write it (using valarray)

#include <iostream>
#include <cmath>
#include <valarray>

std::valarray<double> rhs(std::valarray<double>& x){
    return std::valarray<double>{-x[1]*x[1]*x[1]  + 0.5, x[0]};
}

std::valarray<double> forw_euler(const double& h, const double& tf,const std::valarray<double>& x0){

    const int n = ceil(tf/h);
    std::valarray<double> y = x0;

    for (auto i=0; i<=n; ++i) {
        y+=h*rhs(y);
    }
    return y;
}

int main(){
    std::valarray<double> x0 {1.0, 1.0};
    const double h = 0.1;
    const double tf = 1.0;
    std::valarray<double> sol = forw_euler(h,tf,x0);
    std::cout << "Solution at time T = " << tf << "\n"
    << "first component: " << sol[0] << "\t" << "second component " << sol[1] << std::endl;

    return 0;
}


问:
Prove that there is an $n \times n$ matrix $B$ such that $A=B^{2}+B .$
Given a positive integer $n,$ prove that there is $\varepsilon>0$ such that for every $n \times n$ matrix $A$ with $|A|<\varepsilon$ (Hilbert-Schmidt norm), there is an $n \times n$ matrix $B$ such that $A=B^{2}+B .$ Hint:
Differentiate $f(B)=B^{2}+B$

I am not getting any clue here. As given in the hint I can take $f(x)=x^2+x.$ I also think that I have to use Implicit function theorem by taking $g(A, B)=A-B^2-B$. But I am confused on how to use implicit function theorem or inverse function theorem on matrices.

答:
The inverse and implicit function theorems are equivalent, so you can use either one of them to solve your question; it's a matter of taste (though sometimes one or the other may be simpler for the particular problem at hand)... in this case though, they're about the same amount of work.

Before we can even talk about how to use these theorems, let's recall what exactly they say. As you will see, these theorems are valid in a very general context, and I think seeing them in this generality (and working through some examples carefully) is the clearest approach, so here goes:

Inverse Function Theorem on Banach Spaces

Let $r\geq 1$ be an integer, $E,F$ be Banach spaces, $U\subset E$ open, and $f:U\to F$ a $C^r$ mapping. If $p\in U$ is some point such that $Df_p:E\to F$ is a linear isomorphism then there exist open neighbourhoods $\Omega,\Omega'$
with $p\in \Omega\subset U$ and $f(p)\in \Omega'\subset F$ such that the restriction $f:\Omega\to \Omega'$ is a $C^r$ diffeomorphism (i.e has a $C^r$ inverse).

Next, we state the implicit function theorem

Implicit Function Theorem on Banach Spaces

Let $r\geq 1$ be an integer, $E,F, G$ be Banach spaces, $U\subset E\times F$ open, and $\phi:U\to G$ a $C^r$ mapping. If $(a,b)\in U$ is some point such that the partial derivative with respect to second variable, $\partial_2\phi_{(a,b)}:F\to G$, is a linear isomorphism then there is an open neighbourhood $\Omega_1\times \Omega_2$ of $(a,\phi(a,b))$ in $E\times G$ and a $C^r$ function $g:\Omega_1\times \Omega_2\to F$ such that for all $(x,w)\in \Omega_1\times \Omega_2$, we have $(x,g(x,w))\in U=\text{domain}(\phi)$ and $\phi(x,g(x,w))=w$.

These may sound abstract but the point is this: the inverse function theorem tells us that if the derivative at a point is invertible (is an isomorphism) then locally the function is also invertible. The implicit function theorem (in essence) tells us that if part of a function has invertible derivative then we can actually for some of the variables as functions of the remaining ones; explicitly we can locally solve the equation $\phi(x,y)=w$ for $y$ as a smoothly behaving function of $x$ and $w$ (in most applications we take $w=0$).

The best example to keep in mind for the inverse function theorem is saying that you can solve the equation $y=Ax$ as $x=A^{-1}y$ (if $x,y\in\Bbb{R}^n$ and $A\in M_{n\times n}(\Bbb{R})$ is invertible). For the implicit function theorem, the canonical example to remember is again a linear one: the equation $w=Ax+By$ can be solved for $y$ as $y= B^{-1}(w-Ax)$ (if $w,y\in\Bbb{R}^n, x\in \Bbb{R}^m$, $A\in M_{n\times m}(\Bbb{R})$ and $B\in M_{n\times n}(\Bbb{R})$ is invertible). These two examples are the linear algebraic versions of these calculus theorems, and I found it greatly beneficial to understand this analogy.

You may have mostly encountered these theorems in the context of functions between say $\Bbb{R}^n\to \Bbb{R}^n$ or $\Bbb{R}^n\times \Bbb{R}^m\to \Bbb{R}^m$, but clearly this formulation is not the best way of doing things for your question. Also, before getting to the solution, one final comment I have is that it doesn't really matter what norm you use on the space of matrices; it could be Hilbert Schmidt/ max norm/ operator norm whatever... it doesn't matter because it is a nice theorem that all norms on a finite-dimensional vector space are equivalent (i.e they all induce the same topology).

Solution Using Inverse Function Theorem.

Here, we take $U=E = F = M_{n\times n}(\Bbb{R})$. Consider $f(B):= B^2+B$. Then, $f(0)=0$, and it is even a polynomial in $B$ hence it is $C^{\infty}$ (indeed, the third derivative will vanish identically). The first derivative is calculated as for all $C\in E=M_{n\times n}(\Bbb{R})$,
\begin{align}
Df_B(C)&= C B + BC + C
\end{align}
This is a consequence of product (or chain rule)... the "quick way" of doing this calculation is to write $f(B)=B^2+B=B\cdot B +B$, then $df=dB\cdot B + B \cdot dB + dB$; now we interpret this as a pointwise equality of linear transformations (i.e we regard $dB$ as the linear mapping $C\mapsto C$), in which case, this says exactly what I wrote above in the more explicit notation. Since we have $f(0)=0$, it is a good idea to consider $B=0$, doing so tells us $Df_0(C)=C$, i.e $Df_0 = \text{id}_{M_{n\times n}(\Bbb{R})}$, which is clearly an isomorphism.

So, the inverse function theorem now tells us that there exist open neighbourhoods $\Omega$ of $0\in U= E = M_{n\times n}(\Bbb{R})$, and $\Omega'$ of $f(0)=0$ in $F=M_{n\times n}(\Bbb{R})$ such that the restriction $f:\Omega\to \Omega'$ is a $C^{\infty}$ diffeomorphism. So, now, for any $A\in \Omega'$, define $B:= f^{-1}(A)$. Then, $A=f(f^{-1}(A))=f(B)=B^2+B$.  i.e we have shown that for every $A\in \Omega'$, there exists (a unique) $B\in \Omega$ such that $A=B^2+B$, which completes the proof.

(and if you really want to connect things back to the $\epsilon$, then just say that since $\Omega'$ is an open neighbourhood of $0$ in $M_{n\times n}(\Bbb{R})$, and all norms are equivalent, there exists $\epsilon>0$ such that the $\epsilon$ ball $|A|<\epsilon$ is contained in $\Omega'$).

Solution Using Implicit Function Theorem.

Here, $E=F=G=M_{n\times n}(\Bbb{R})$ and $U=E\times F$. We consider $\phi(A,B):= B^2+B-A$. Again, this is a polynomial function of its arguments so it is $C^{\infty}$ (the third derivative will vanish identically). A similar calculation shows that the partial derivative with respect to the second variable (i.e wrt $B$) is the linear mapping $\partial_2\phi_{(A,B)}:F\to G$ (i.e $M_{n\times n}(\Bbb{R})\to M_{n\times n}(\Bbb{R})$)
such that for all $C\in F=M_{n\times n}(\Bbb{R})$
\begin{align}
\partial_2\phi_{(A,B)}(C) &= (CB + BC) + C - 0
\end{align}
Since $\phi(0,0)=0$, it is a good idea to take $(a,b)=(0,0)$, in which case we note that $\partial_2\phi_{(0,0)} = \text{id}_{M_{n\times n}(\Bbb{R})}$ is an isomorphism. Now we can use the implicit function theorem to tell us that there is an open neighbourhood $\Omega_1 \times \Omega_2$ of $(a,\phi(a,b)) = (0,0)$ in $E\times G = M_{n\times n}(\Bbb{R})\times M_{n\times n}(\Bbb{R})$ and a $C^{\infty}$ function $g:\Omega_1\times \Omega_2\to F=M_{n\times n}(\Bbb{R})$ such that for all $(A,w)\in \Omega_1\times \Omega_2$,
\begin{align}
\phi(A,g(A,w)) &= w
\end{align}
Now, plug in $w=0$, and see what the equation says: $[g(A,0)]^2 + g(A,0) - A = 0$, i.e if you take $B=g(A,0)$, then $A=B^2+B$, so for any $A\in \Omega_1$, we have found a corresponding $B$ (namely $g(A,0)$) which works.


答:
Simply knowing that the dimensions are equal merely tells you that there is an isomorphism between the two vector spaces. However, there are, typically, infinitely many such isomorphisms and there is no canonical one. Now, what is the determinant? Algebraically, it is given by a formula (whatever is your favourite one). Geometrically, it is measuring volume. In particular, the determinant of $n$ vectors in $\mathbb R^n$ measures the (signed) area of the parallelepiped determined by these vectors (in the given order). However, volume is not something that exists in a vector space without having fixed a frame of reference. So, to speak of volume we must first choose a basis (usually in $\mathbb R^n$ this would be the standard basis) and decide that the parallelepiped they form has volume one (thus effectively normalising the determinant) and measuring volumes of other parallelepipeds relative to it.

So, what is going on here is that once you speak of the determinant in $V$ you really implicitly chose a basis for $V$. Now, if you have another vector space $W$ and you wish to speak of determinants there, then you also chose a basis for $W$. A linear transformation $T\colon V\to W$ need not care at all about these chosen bases. So, if you measure the volume deformation borne by $T$ (this is what the determinant of $T$ is --- the factor by which it deforms volumes of parallelepipeds) and you don't make sure $T$ respects the chosen bases, then you will really be comparing apples and oranges. If, however, $T$ does preserve the bases, then the whole situation is isomorphic to $T$ being an endomorphism on a single vector space $V$ with just a single chosen basis in it to give meaning to determinants. All is well, and the usual notion of the determinant of an endomorphism is restored.