问:
Prove if $f'$ is a Riemann-integrable function then $f(x)= a_{0} +cx + \sum_{n=1}^{\infty } a_{n} \cos(nx)+ \sum_{n=1}^{\infty } b_{n} \sin(nx)$
I am trying to prove the following, but I ran into problems:

Suppose that $f:[-\pi, \pi] \rightarrow \mathbb{R}$ is differentiable and $f':[-\pi, \pi] \rightarrow \mathbb{R} $ is Riemann-integrable. Show that there exists $c, a_{0}, a_{1}, a_{2},.... b_{1}, b_{2} \in \mathbb{R}$ such that

$\sum_{n=1}^{\infty } |a_{n}| $ and $ \sum_{n=1}^{\infty } |b_{n}|$

converge, and

$f(x)= a_{0} +cx + \sum_{n=1}^{\infty } a_{n} \cos(nx)+ \sum_{n=1}^{\infty } b_{n} \sin(nx)$

for all $ x \in [-\pi, \pi] $

where should I start?

Can I start by stating if is  if $f'$ is Riemann integrable then
$f'(x)= d_{0} + \sum_{n=1}^{\infty } d_{n} \cos(nx)+ \sum_{n=1}^{\infty } e_{n} \sin(nx)$?

答:
If $f^{'}$ is Riemann-integrable it is bounded so it is in $L^{2}$.
It means that its Fourier coefficients are in $l_{2}$.
At the same time if $a_{n}(f)$ is a Fourier coefficient $f$ and $a_{n}(f^{'})$ is a Fourier coefficient $f^{'}$ then $$|a_{n}(f^{'})| = n*|a_{n}(f)|$$. The rest is just Cauchy-Schwartz inequality, that is:
$$\sum |a_n| = \sum n* |a_{n}(f)| *(\frac{1}{n}) = \sum |a_{n}(f^{'})| *(\frac{1}{n}) <=
\sqrt(\sum |a_{n}(f^{'})|^{2}) * \sqrt(\sum (\frac{1}{n})^{2})$$


问:
How can I prove that if $a < b$ then $a^2$ < $b^2$ for $a,b \in \Bbb N$
How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.

答:
Use the elementary rules for inequalities, as seen in middle school!

From $a<b$, you deduce instantly that $a^2<ab$ (multiplication by $a>0$).

And also $ab<b^2$ (multiplication by $b>0$).

Last, by transitivity,
$\:a^2<ab$ and $ab<b^2$ imply $a^2<b^2$.

Another way: $a<b\:$ means $\:b-a>0$. Multiply by $b+a$ (which is positive) to obtain
$$(b-a)(b+a)=b^2-a^2>0\iff a^2<b^2.$$

ContentLicense: CC BY-SA 4.0


问:
Prove that a vector space $V = \mathbb{F}^3_2$ over $\mathbb{F}_2$ has 16 subspaces.
$V$ also has 256 subsets and I do not know how.

答:
$V$ has one subspace of dimension 0 - the zero space.

The subspaces of $V$ of dimension 1 correspond to non-zero elements of $V$. There are $2^3$ elements of $V$, one of which is 0. So there are $7$ dimension-1 subspaces.

Note that every subspace $U \subseteq V$ has an orthogonal complement $U^\perp = \{v \in V : \forall u \in U (u \cdot v = 0)\}$. By the rank-nullity theorem, $dim(U) + dim(U^\perp) = dim(V)$, so $dim(U^\perp) = dim(V) - dim(U)$. Therefore, $dim((U^\perp)^\perp) = dim(U)$. And $U$ is a subspace of $(U^\perp)^\perp$. Then $U = (U^\perp)^\perp$.

Therefore, we see that the subspaces of $V$ of dimension 2 correspond to the subspaces of $V$ of dimension 1 in a bijective way (the bijeciton $U \mapsto U^\perp$). There are thus 7 of them.

Similarly, the subspaces of $V$ of dimension 3 correspond exactly to subspaces of $V$ of dimension 0. There is thus 1 of them.

This is a total of 16.


问:
Vector problem - intersecting lines
Q: Does the line through $(3,4,7)$ in the direction $(2,3,-6)$ intersect the line passing through $(5,3,-2)$ and $(1,3,-9)$?

I said that the equation of the first line is $r_1=\begin{bmatrix}3\\4\\7\end{bmatrix}+s\begin{bmatrix}2\\3\\-6\end{bmatrix}=\begin{bmatrix}3+2s\\4+3s\\7-6s\end{bmatrix}$, and the equation of the second is $r_2=\begin{bmatrix}5-4t\\3\\-2-7t\end{bmatrix}$. In order for these to intersect, each row must agree with each other, so from row 2 we have $4+3s=3$ and so $s=-\frac{1}{3}$. Subbing into row 3 gives $t=-\frac{11}{7}$. Now substituting into the first row, we have $3+2s=\frac{7}{3}$ whereas $5-4t=\frac{79}{7}$ so the lines do not intersect. Is this correct?

答:
Yes, you're right.

You can set these out using Left Hand Side and Right Hand side,  solve the simultaneous equations using the two easiest rows and then check in the remaining row, e.g

$y:L.H.S = 32  , R.H.S. = 20$, therefore lines do not intersect, or

$y:L.H.S = 32  , R.H.S. = 32$, therefore lines intersect at...

and find the point by substituting $s$ or $t$ back into either line. If you hold two pencils up and look from the right angle, they'll seem to cross each other, but then if you move around, you might see a gap between them, that's the third row not matching, the lines are skew...or they can properly intersect in all 3 dimensions, or they can be parallel.

Title: Vector problem - intersecting lines


问:
Fredholm Equations
I have the following problem to solve

$$\phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+ \sin x$$

following the instructions from the following link early to conclude that:

$$\phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+\sin x$$ $$=\lambda \int_{o}^{\pi }\cos^2(x)\cos(y)-\sin^2(x)\cos(y)-2\sin(x)\cos(x)\sin(y)\phi (y) dy+ \sin x$$

then start to calculate $a_{11}, a_{12},a_{13},a_{21},a_{22},a_{23},a_{31},a_{32},a_{33}$, such that (these will be the elements of the matrix A):

$$a_{1}=\cos^2(x);a_{2}=-\sin^2(x) ;a _{3}=-2\sin(x)\cos(x)$$

$$b_{1}=\cos(y);b_{2}=\cos(y) ;b_{3}=\sin(y)$$

thus calculated:

$$a_{11}=\int_{o}^{\pi }b_{1}(y)a_{1}(y)dy$$

$$a_{12}=\int_{o}^{\pi }b_{1}(y)a_{2}(y)dy$$

$$a_{13}=\int_{o}^{\pi }b_{1}(y)a_{3}(y)dy$$

calculating and so on until all $a_{nm}$ with $n =1,2,3$ and $m=1,2,3$.

After calculations we obtain the matrix

$$A=\begin{bmatrix}
a_{11} &a_{12}  & a_{13}\\
a _{21}&a_{22}  & a_{23}\\
a_{31} &a_{32}  & a_{33}
\end{bmatrix}$$

making: $\;(I-\lambda A)$,

and calculating the determinant originated by $\;(I-\lambda A)$, get all values of $\lambda$.

The solutions they have there is only one $\lambda=3/2\sqrt{2}$ and $\lambda=-3/2\sqrt{2}$

But never get to that result. I appreciate any and all help. I do not even know if I was wrong in whole or in calculating the determinant of the matrix.

答:
Related problem. First note this, the solution is

$$  \phi \left( x \right) =\sin \left( x \right) +12\,{\frac {{\lambda}^{2
}\pi \,  \cos^2\left( x \right) }{8\,{\lambda}^{2}-9
}}-6\,{\frac {{\lambda}^{2}\pi }{8\,{\lambda}^{2}-9}}+9\,{\frac {\pi
\,\lambda\,\sin \left( x \right) \cos \left( x \right) }{8\,{\lambda}^
{2}-9}}.$$

Now, you can see the singular values or the wanted $\lambda$ is given by

$$ 8\,\lambda^{2}-9 = 0  \implies \lambda = -\frac{3}{2\sqrt{2}},\,\frac{3}{2\sqrt{2}}.$$

Added: Here is a start

$$ \phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+\sin x  $$

$$ \phi(x)  =  \sin(x) +\lambda \cos(2x)\int_{0}^{\pi }\cos(y)\phi (y) dy - \sin(2x) \int_{0}^{\pi }\sin(y)\phi (y) dy   $$

$$  \phi(x)  =  \sin(x) +\lambda c_1\cos(2x) - \lambda c_2 \sin(2x). $$

Now, just follow the technique in the link and you should not have a problem.

ContentLicense: CC BY-SA 3.0

Title: Fredholm Equations


问:
How do you show a function is constant within a disc?
If f is holomorphic at every point on the open disc $$D=\{z:|z|\lt1\}$$
I want to show that f is constant

答: Consider $Re(f(z) \cdot (1 - i))$. This quantity always equals $10$. But if a holomorphic function has a constant real part in a connected open set, it must be constant.


答:
Firstly, let us prove that it has $256$ subsets. As $V=\mathbb{F}_2^{3}$ its cardinality is equal to $8$. Hence it has $2^8=256$ subsets (as $2^n$ is the number of subsets of a $n$-set). Now lets prove that it has $16$ subspaces.
The elements of $V$ are : $(0,0,0);(0,0,1);(0,1,0);(1,0,0);(1,1,0);(1,0,1);(0,1,1);(1,1,1)$.
In addition we know that any subspaces of $V$ must contain $(0,0,0)$ and $(1,1,1)$. As $dim(V)=3$ the dimension of any subspace $W$ of $V$ is $\le 3$.
So you just have to consider the different cases $dim(W)=1,2$ or $3$.
Hope this will help you. If you have any more questions dont hesitate to comment.
Hubert

Title: Prove that a vector space $V = \mathbb{F}^3_2$ over $\mathbb{F}_2$ has 16 subspaces.

Body: $V$ also has 256 subsets and I do not know how.


问:
Is $\Vert \max(0, M-x)\Vert^2$ differentiable?
I think $\max(0, M-x)^2$ is differentiable everywhere, but what if $x\in \mathbb{R}^n$, can we prove that $f(x) = \Vert \max(0, M-x)\Vert^2, x\in \mathbb{R}^n$ is differentiable? $M \in \mathbb{R}^n$ is a constant, when

$$x \in \mathbb{R}^n,  \max(0, M-x) = \begin{bmatrix}\max(0, M_1 - x_1) \\ \vdots \\ \max(0, M_n - x_n)\end{bmatrix} \in \mathbb{R}^n.$$

答:
The function $g(x) = \lVert x \rVert^2 = \sum_{i=1}^n x_i^2$ is differentiable. If $h(x):=max(0,M-x)$ is also differentiable, then by the chain rule:
$$\frac{\partial}{\partial x_i}f\bigl(h(x)\bigr) = \frac{\partial f}{\partial x_i}\bigl(g(x)\bigr)\frac{\partial g}{\partial x_i}(x).$$
Hence, since both $f$ and $h$ are differentiable over $\mathbb R^n$, $f\circ h$ is also differentiable. But that is assuming $max(0, M-x)$ is differentiable.


问:
How to get the maximum integer value for $1+\sin^2\alpha$?
The problem is as follows:

Assume: $\alpha \in \left[ \frac{\pi}{2},\, 4 \right\rangle$
(Read as alpha belonging to the interval closed in $\frac{\pi}{2}$ and open in $4$).

$B=1+\sin^2\alpha$

Find the maximum integer value that it can take $B$.

The choices given are:

$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{3}\\
3.&\textrm{0}\\
4.&\textrm{2}\\
\end{array}$

I'm stuck here from the very beginning. How exactly should I find the maximum integer value for that expression here?.

Does it exist a criteria for this?. I'm requesting for help here because I feel lost.

I think I could use:

$-1 < \sin \alpha < 1$

But I doubt that's the approach right for here. Can someone help me here?. This topic isn't well covered in my precalculus book. But I'm certain that such simple approach would be used. As I mentioned, it is not much that I can offer. Some other ideas which had come to my mind are perhaps the use of.

$AM\geq GM$

But I don't know if such would be used or have to be used?.

Other approach could be derivatives.

Trying this would be:

$B(\alpha)=1+\sin^2\alpha$

$B'(\alpha)=2\sin\alpha\cdot \cos\alpha$

Equating this to zero yields:

$2\sin\alpha\cdot \cos\alpha = 0$

$\sin 2\alpha = 0$

Hence this can take values in the given range:

$2\alpha \in \left[ \pi,\, 8 \right\rangle$

Gee, $8$ is somewhat close to $2.54 \pi$

Then this means I have to look for values equal to zero in that range:

Those would be:

$\pi$, $2\pi$

But since it is divided by two for \alpha would be:

$\frac{\pi}{2}, \pi$.

This means:

Inserting any of these values in the function would give me the maximum?.

But again this would make any of those values $B=1$ and none of those values appears in the choices. Thus can someone help me here?.

答:
Since $-1\leq \sin(x)\leq 1$ for any $x$, then $0\leq \sin^2(x)\leq 1$ and $1\leq 1+\sin^2(x)\leq 2$ for any $x$. That immediately gets rid of all but one option.

I guess the only remaining thing is to make sure the question is not posed under a false premise and that you actually do achieve the value $2$ at some point between $\frac{\pi}{2}$ and $4$. That would require you to find some $x$ in that interval where $\sin^2(x)=1$, or equivalently where $\sin(x)=1$ or $\sin(x)=-1$. We know exactly at which points $\sin(x)$ takes those values, so then all you need to do is check whether any of them are in the interval you are interested in.


答: HINT: $\sin^2 \frac{\pi}{2} = 1$. And of course $\sin^2 x \le 1$ for all other $x$.

Title: How to get the maximum integer value for $1+\sin^2\alpha$?

Body:
The problem is as follows:

Assume: $\alpha \in \left[ \frac{\pi}{2},\, 4 \right\rangle$
(Read as alpha belonging to the interval closed in $\frac{\pi}{2}$ and open in $4$).

$B=1+\sin^2\alpha$

Find the maximum integer value that it can take $B$.

The choices given are:

$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{3}\\
3.&\textrm{0}\\
4.&\textrm{2}\\
\end{array}$

I'm stuck here from the very beginning. How exactly should I find the maximum integer value for that expression here?.

Does it exist a criteria for this?. I'm requesting for help here because I feel lost.

I think I could use:

$-1 < \sin \alpha < 1$

But I doubt that's the approach right for here. Can someone help me here?. This topic isn't well covered in my precalculus book. But I'm certain that such simple approach would be used. As I mentioned, it is not much that I can offer. Some other ideas which had come to my mind are perhaps the use of.

$AM\geq GM$

But I don't know if such would be used or have to be used?.

Other approach could be derivatives.

Trying this would be:

$B(\alpha)=1+\sin^2\alpha$

$B'(\alpha)=2\sin\alpha\cdot \cos\alpha$

Equating this to zero yields:

$2\sin\alpha\cdot \cos\alpha = 0$

$\sin 2\alpha = 0$

Hence this can take values in the given range:

$2\alpha \in \left[ \pi,\, 8 \right\rangle$

Gee, $8$ is somewhat close to $2.54 \pi$

Then this means I have to look for values equal to zero in that range:

Those would be:

$\pi$, $2\pi$

But since it is divided by two for \alpha would be:

$\frac{\pi}{2}, \pi$.

This means:

Inserting any of these values in the function would give me the maximum?.

But again this would make any of those values $B=1$ and none of those values appears in the choices. Thus can someone help me here?.


问:
Fastest path for $A$ to catch $B$
I was wondering if is there a way to attack with Euler-Lagrange equation the following problem.

Suppose $B$ is moving in straight line with costant velocity $\mathbf{u}=u\,\hat{x}$. What is the fastest path for an $A$, which can move with costant speed $v>u$ (and variable velocity $\mathbf{v}=v\hat{\mathbf{e}}$), to catch $B$?
Suppose $A(0)=(0,0)$ and $B(0)=(b_1,b_2)$.

I didn't do much progress. What I think is that, since the speed of $A$ is costant, the best path must be the one with minimal lenght. However, the fact that $B$ is moving is quite disorienting and I don't know how to set up the problem.

Do you have any idea?

答:
You can do it without calculus.  The coordinates of $B$ are $(b_1+ut,b_2)$.  For $A$ to catch $B$, it should move in a straight line to the intercept point.  The distance from the origin to $B$'s location is $\sqrt{(b_1+ut)^2+b_2^2}$.  We can equate this to $A$'s range to get $$vt=\sqrt{(b_1+ut)^2+b_2^2}\\v^2t^2=b_1u^2t^2+2b_1ut+b_1^2+b_2^2\\(v^2-b_1^2u^2)t^2-2b_1ut-(b_1^2+b_2^2)=0$$
which can be solved for $t$ using the quadratic equation.  Then plug that $t$ into $B$'s trajectory and you have the intercept point.  $A$ should move straight toward that point.</p>

Title: Fastest path for $A$ to catch $B$


问:
Definite integral of a function with finite $x$ such that $f(x)>\gamma$
Hi I'm having a little bit of difficulty in proving the following:

$f:[a,b]\to\mathbb{R}$ is non-negative and for every $\gamma>0$, the set of points $\mathcal{X}=\{x\in[a,b]|f(x)>\gamma\}$ is finite. In other words, $|\mathcal{X}|=n$ for some $n\in\mathbb{Z}_{\geq0}$. Show that $\int_a^b f =0$. Is there any relationship between $n$ and $\gamma$?

How could this be done using the upper and lower Riemann sums? I've tried drawing out a picture of this and I can't seem to figure out how to construct such a function, unless this is just the zero function, or with certain values of $x\in[a,b]$ such that there are only $n$ of them such that $f(x)>0$ (so removing a finite amount of elements from the image of $f$).

For the second part, the only relationship I can observe between $n$ and $\gamma$ is that obviously as $\gamma$ increases $n$ decreases, and we must have that $\max\{\mathcal{X}\}$ exists since it is finite and clearly for some $\gamma>\gamma'$, we will have that $n=0$, since $\mathcal{X}$ has to eventually be $\emptyset$.

I believe my intuition behind this is correct? But I'm at a loss as to where to even begin beyond what I described above (which is definitely not rigorous at all).

Any help is appreciated, thank you!

Best,
Adam

答:
Let $\varepsilon > 0$ be arbitrary. Let $\chi_{\varepsilon} = \{x_1, \dots, x_n\}$ be the finite set of points where $f > \varepsilon$. Let $P$ be a partition of $[a, b]$ where each $x_i$ is covered with an interval of length $\leq \frac{\varepsilon}{n}$. Let $M > 0$ be a constant that bounds $f$ on $[a, b]$, e.g. $M = \max_{1 \leq i \leq n}f(x_i)$. Then
\begin{align}
U(f, P) &\leq nM\frac{\varepsilon}{n} + \varepsilon(b - a) \\
&= (M + (b - a))\varepsilon.
\end{align}
Hence $U(f) \leq U(f, P) \leq (M + (b - a))\varepsilon$. Letting $\varepsilon \to 0$ gives $U(f) \leq 0$. Since $f \geq 0$, $U(f) \geq 0$. Thus $U(f) = 0$.


答:
Since $f$ is holomorphic, it satisfies the Cauchy–Riemann equations. Decompose $f$ in the usual way, into real and imaginary parts:
$$f(x+iy)=u(x,y)+v(x,y)i$$
We have $u+v=10$, so the starred equalities below hold:
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=^*\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=^*-\frac{\partial u}{\partial y}$$
Hence all partial derivatives involved are zero and $f$ is constant.

Title: How do you show a function is constant within a disc?

Body:
If f is holomorphic at every point on the open disc $$D=\{z:|z|\lt1\}$$
I want to show that f is constant


问:
About Riemann rearrangement theorem!
How to show that $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt n}$ converges to $+\infty$, by changing the order of its terms.

I think that the idea is changing the order of terms such as: taking enough positive terms that there sum is bigger than 1/2 ,then the first negative term in the series, and so on:
$(1+1/\sqrt 3 -1/\sqrt2)+(1/\sqrt 5+1/\sqrt7+\cdots+1/\sqrt 21-1/\sqrt 4)+...$.
Now, if our series is $\sum_{k=1}^{\infty} b_k$.
Then, for every $k\in N$, $b_k \geq 1/2- 1/\sqrt(2k)$.

For $k\geq 8 , b_k \geq 1/2-1/4=1/4$.
So, $S_K=\sum_{k=1}^{K} b_k$ does not converge to 0.
Which lead to that the series converges to $\infty$.

答: Well, if we always put two positive values in front of one negative, that would be groups of three, $\displaystyle\frac1{\sqrt{4n-3}}+\frac1{\sqrt{4n-1}}-\frac1{\sqrt{2n}}$. That would be $\sim\frac{1-\sqrt{1/2}}{\sqrt{n}}$, clearly a divergent series.

Title: About Riemann rearrangement theorem!


问:
Is it possible to transform a Riemann sphere into the complex plane by removing a single point?
It's well know that the Riemann sphere is constructed by adding a single point at infinity to the complex plane.  The distinguishing feature of the point at infinity is that it's the inverse of the point $z=0$

$$0^{-1}=\infty$$

It's impossible to define what a point at infinity means unless you have previously defined what doing an inverse operation means.  In other words, the point at infinity comes with an implicit definition of an inverse mapping, which other points don't possess.

If I want to remove any other point besides zero or infinity from the Riemann sphere, it's required to remove 2 points, since otherwise the inverse would not be well defined for the image of that point under the inverse operation, and the inverse operation is required for the definition of the point at infinity.

It's not possible to delete the definition of inverse while our space contains a point at infinity, and it's not possible to remove a point at infinity without without also removing the point 0 from the Riemann sphere, and then deleting the definition of inverse, and then adding the point $0$ back to the complex plane.

Is this wrong?

答:
Your text is far away from the way we think about these things.

Adding a single point $\infty$ to ${\mathbb C}$ means that we define at free will new set $\bar{\mathbb C}:={\mathbb C}\cup\{\infty\}$. A priori this is a "purely geometric" operation, and there are no algebraic rules associated with it. In particular, $\bar{\mathbb C}$ is not a field, and we cannot expect that the special "number" $\infty$ has a multiplicative inverse in $\bar{\mathbb C}$, or that expressions like $0\cdot\infty$ give a nice result in $\bar{\mathbb C}$.

Nevertheless there are certain "exception handling rules". This point $\infty$ also has obtained its "neighborhood filter". In this way formulas like
$$\lim_{z\to0} {1\over z}=\infty, \qquad\lim_{z\to\infty}{1\over z}=0\ ,$$
and similar, make sense. When we deal with Moebius transformations $T$ we, e.g.,  find that
$$\lim_{z\to\infty} T(z)=\lim_{z\to\infty}{3z-4\over2z+25}={3\over2}\ ,$$
and therefore put $T(\infty):={3\over2}$.

Similarly, the function $$f(z):={z-3\over (z+1)(z-2)}$$
is taking the value $\infty$ at $z:=-1$ and $z:=2$.


问:
Let $\{y_n\}$ be an unbounded sequence of positive terms. Does it necessarily follow that its limit equals infinity?
I need to either prove or disprove the following: Let $\{y_n\}$ be an unbounded sequence of positive terms. Does it necessarily follow that its limit is equal to infinity.

I think this is false, and requires a counterexample to disprove it.

However, I am unsure how to go about creating a counterexample that would help disprove it.

答:
Take the sequence defined by

$$(\forall n\ge 0)\;\; u_{2n}=n\text{ and } u_{2n+1}=1$$

then

$(u_n) $ is unbounded  and  has no limit.


问:
Multiplying out by a minus sign under an absolute value function
$\require{\cancel}$

My original problem was:

$$I=\int_0^\infty \frac{1}{x^2+b^2}dx$$

where $b>0$

which i solved the following way

$$I=\int_0^\infty\left(\frac{1}{2ib(x-ib)}-\frac{1}{2ib(x+ib)}\right)dx=$$

$$=\frac{1}{2ib}\lim_{s\to\infty}\int_0^s\left(\frac{1}{(x-ib)}-\frac{1}{(x+ib)}\right)dx=$$

$$=\frac{1}{2ib}\lim_{s\to\infty}\left\{\left[\ln|(x-ib)|\right]_0^s-\left[\ln|(x+ib)|\right]_0^s\right\}=$$

$$=\frac{1}{2ib}\lim_{s\to\infty}\left\{\ln|\frac{s-ib}{-ib}|-\ln |\frac{s+ib}{ib}|\right\}=$$

$$=\frac{1}{2ib}\ln\lim_{s\to\infty}\left\{\dfrac{|\frac{s-ib}{-ib}|}{|\frac{s+ib}{ib}|}\right\}=$$

$$=\frac{1}{2ib}\ln\lim_{s\to\infty}\left\{|\frac{s-ib}{\color{red}{\boxed{-}}\cancel{ib}}|\cdot|\frac{\cancel{ib}}{s+ib}|\right\}=\frac{1}{2ib}\ln\underbrace{\left(\color{green}{\boxed{-}}\lim_{s\to\infty} \left|\dfrac{s-ib}{s+ib}\right|\right)}_{-1}$$

Why am i allowed to multiply out by the minus sign, but why only until inside the logarithm?
If i don't multiply out with it the logarithm's argument will become $1$ so then i get a wrong answer. If I multiply out to the front my final result will be negative.

$\frac{\pi}{2b}$ is the correct solution by the way.

$$\ln(-1)=x\Rightarrow e^{\ln(-1)}=e^x\Rightarrow-1=e^x\Rightarrow e^x+1=0\Rightarrow x=i\pi$$

$$I=\frac{1}{2\cancel{i}b}\cancel{i}\pi=\frac{\pi}{2b}$$

答:
I use that $\log z=\log|z|+i\arg(z).$

$$\int_0^{\infty}\left(\frac{1}{x-ib}-\frac{1}{x+ib}\right)dx=$$
$$\log(x-ib)-\log(x+ib){\large|}_{x=0}^{\to \infty}=$$

$$\log{{x-ib\over x+ib}}{\large |}_{x=0}^{\to \infty}=$$

$$\log\left|{{x-ib\over x+ib}}\right| +i \arg {x-ib\over x+ib}
{\large|}_{x=0}^{\to \infty}=$$
$$0-\{\log\left|{-ib\over ib}\right|+i(-\pi/2-\pi/2)\}=i\pi.$$
Thus, $I={i\pi\over 2ib}={\pi\over 2b}.$


答:
It is correct that
$$
 \Bbb R \ni t \mapsto \max(0, M-t)^2
$$
is differentiable. It follows that
$$
 \Bbb R^n \ni x \mapsto f(x) = \Vert \max(0, M-x)\Vert^2
 = \sum_{j=1}^n \max(0, M_i - \pi_i(x))^2
$$
is differentiable as well, because:

Each projection: $\pi_i : \Bbb R^n \to \Bbb R, \pi_i(x) = x_i$ is differentiable (because it is a linear mapping), and

the sum of differentiable functions is differentiable.

Title: Is $\Vert \max(0, M-x)\Vert^2$ differentiable?

Body:
I think $\max(0, M-x)^2$ is differentiable everywhere, but what if $x\in \mathbb{R}^n$, can we prove that $f(x) = \Vert \max(0, M-x)\Vert^2, x\in \mathbb{R}^n$ is differentiable? $M \in \mathbb{R}^n$ is a constant, when

$$x \in \mathbb{R}^n,  \max(0, M-x) = \begin{bmatrix}\max(0, M_1 - x_1) \\ \vdots \\ \max(0, M_n - x_n)\end{bmatrix} \in \mathbb{R}^n.$$


问:
Calculating $\nabla\left(\nabla \cdot \boldsymbol{r}_{0} e^{i\boldsymbol{k} \cdot \boldsymbol{r}}\right)$

In a longer derivation I ran into the following quantity:
$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right]
$$
( i.e., the gradient of the divergence ) where ${\bf k}$ is a vector of constants and ${\bf r}$ is a position vector.

Can someone help explaining how to calculate this?
I am hoping it gives:
$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right] =
-{\bf k}\left({\bf k}\cdot{\bf r}_{0}\right)
$$
( because then the rest of my equations add up ).

答: Summing over repeated indices, the divergence is $r_{0i}\partial_ie^{\text{i}k_jr_j}=r_{0i}\text{i}k_ie^{\text{i}k_jr_j}=\text{i}(k\cdot r_0)e^{\text{i}k_jr_j}$. Applying $\partial_l$ pulls down another $\text{i}k_l$ factor, so the gradient is $-k(k\cdot r_0)e^{\text{i}k\cdot r}$. Your desired result drops the exponential, which I suspect is a typo.


问:
Can we count past $ \infty $?
Let's say, $ \infty $ is not the biggest number, in fact, there is something bigger, somewhere. We want to try and count past $ \infty $, because it is not really an ordinal one, but a definition of something uncountable or too big to even count.

What numbers are bigger than $ \infty $?

答: $\infty$ can't be the biggest number, because it isn't a number. It's just a symbol used in limits. Ok, you can consider compactifications of real or complex numbers, but that's certainly not what you had in  mind, because both are very far from "counting".

Title: Can we count past $ \infty $?


答: Hint $(-1)^n+1$ is alternately $0$ and $2$. Multiplying this by any sequence which goes to infinity creates an unbounded sequence that is half of the times $0$. Add a positive constant to make it positive.

Title: Let $\{y_n\}$ be an unbounded sequence of positive terms. Does it necessarily follow that its limit equals infinity?

Body:
I need to either prove or disprove the following: Let $\{y_n\}$ be an unbounded sequence of positive terms. Does it necessarily follow that its limit is equal to infinity.

I think this is false, and requires a counterexample to disprove it.

However, I am unsure how to go about creating a counterexample that would help disprove it.


答:
More straightforward :

If $a>b$ and $a,b>0$ (there is no need to take $a,b\in\mathbb Z^{+}$), then you have

$$\ln a> \ln b$$

$$\implies 2 \ln a> 2\ln b$$

$$\implies  \ln a^2> \ln b^2$$

$$\implies a^2 >b^2.$$

This is an additional answer that I believe will be helpful to you.

Title: How can I prove that if $a < b$ then $a^2$ < $b^2$ for $a,b \in \Bbb N$

Body:
How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.


问:
What is the Idea of a Relative Open Set?
My real-analysis text gave the following defintion:

Let U be a subset of E.  U is open relative to E if

for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.

Although the idea that U is open in $\mathbb R$ follows the definition,

I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.

U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.

Intuitively, each point, t, is contained in a "bubble".

So, what is the significance of specifying the intersection?

Apparently, I can't think of a relative open set in terms like interior points and "bubbles".

答:
The idea is that you are giving the subset $E$ of $\mathbb{R}$ a topology. A topology on a set is just a collection of subsets $\mathcal{T}$ that satisfy the following three conditions:

$\emptyset$ and $E$ are in $\mathcal{T}$.

$\mathcal{T}$ is closed under finite intersections (i.e. if $U_1,\ldots U_n$ are in $\mathcal{T}$, then so is $U_1\cap\cdots\cap U_n$) .

$\mathcal{T}$ is closed under arbitrary unions.

call the elements of $\mathcal{T}$ open sets of $E$. In $\mathbb{R}$ you know what sets are open. When you take a set together with a topology (called a topological space), you can determine what it means to be open in a space that is not something comfortable like $\mathbb{R}$.

In your example, you are declaring which subsets of $E$ are open. In fact, the topology you are giving $E$ in this context is what is called the subspace topology.  You defined what it means for a set to be relatively open in $E$. A nice exercise is to show that the collection of relatively open subsets of $E$ satisfy the properties above.


问:
For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution?
Here's my system of linear equations:

$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$

Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$

By row reduction, I obtain:

$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$

Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.

Update: I have managed to solve this with the use of the determinant.

Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$

Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$

Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$

Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$

$(a - 5)(a + 1) = 0$

Thank you all for your help!

答: Hint:  Set the determinant equal to zero and solve for $a$.


问:
Is there a general representation of $u[n-a]-u[n-b]$ in terms of unit impulses $\delta$?
Consider in the discrete-time domain, the unit-step function :
$$
u[n]=\begin{cases}
1&\text{if $n\geq 0$}\\
0&\text{if $n<0$}
\end{cases}
$$
We know that the first-order difference equation describes a relation between the unit-impulse function and the unit-step function as follow :
$$
\delta[n]=u[n]-u[n-1]
$$
I was wondering that in general, how can we express a difference equation of this kind :
$$
u[n-a]-u[n-b]
$$
In terms of unit impulses $\delta$? With $a,b\in\mathbb{Z}^{+}$

答:
$u(x)=\int\limits\limits_{-\infty}^x\delta(x)\,dx$ where $u(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac delta function.

So $u(x-a)-u(x-b)=\int\limits\limits_{-\infty}^x(\delta(x-a)-\delta(x-b))\,dx$


答:
The easiest thing is to use determinants: the system has a unique solution if and only if the matrix is invertible, i.e., if and only if the matrix has determinant not equal to $0$. So you can compute the determinant (either in your original or row-reduced matrix) in terms of $a$, set it equal to $0$, and solve for $a$. But if you do not know this yet, you can proceed as follows:

If $a=2$, then the system clearly has a unique solution, so you may assume $a\neq 2$. The reason to do this is so that we can continue the row reduction by multiplying by $a-2$. Multiply the second row by $9$ and the last row by $a-2$ to get:
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 9(a-2) & (a-2)^2 & 2-a
\end{array}\right).$$
Subtracting the second row from the third row, we get
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 0 & (a-2)^2-9 & -16-a
\end{array}\right).$$
So: you will have a unique solution if $(a-2)^2-9\neq 0$. If $(a-2)^2-9=0$, you either get infinitely many solutions (if $16+a=0$ as well), or no solutions (if $-16-a\neq 0$). So now we just need to figure out when that bottom right entry of the coefficient matrix is $0$.

Title: For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution?

Body:
Here's my system of linear equations:

$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$

Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$

By row reduction, I obtain:

$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$

Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.

Update: I have managed to solve this with the use of the determinant.

Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$

Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$

Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$

Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$

$(a - 5)(a + 1) = 0$

Thank you all for your help!


问:
Solve $x^\prime=x+2y, y^\prime=4x+3y, x(0)=−1, y(0)=1$
Solve $x^\prime=x+2y,  y^\prime=4x+3y,  x(0)=−1,  y(0)=1$

$x(t)= $

$y(t)= $

Find the limits

$\lim_{t \to \infty} x(t)=$

$\lim_{t \to \infty} y(t)= $

Select the appropriate (A, B, or C) that best represents the long time behavior of the solution.

(A) Equilibrium solution

(B) Arc solution

(C) Periodic Solution

This example comes with answers as follows:

$x(t)=-e^{-t}$

$y(t)=e^{-t}$

$\lim_{t \to \infty} x(t)=0$

$\lim_{t \to \infty} y(t)=0$

B

However, the example has no work behind it and I can't find a similar example online or in my textbooks. Would somebody be willing to show me the procedure of how to solve something like this?
Thank you!

答:
You can write the system of ODEs under the following form:

$$
\left[\begin{array}{c}
x^{\prime}\\
y^{\prime}
\end{array}
\right]=\left[\begin{array}{cc}
1&2\\
4&3\end{array}\right]\,\left[\begin{array}{c}
x\\
y
\end{array}\right]
$$

and the solution is given as follows

$$
\left[\begin{array}{c}
x(t)\\
y(t)
\end{array}\right]=\exp(A\,t)\,\left[\begin{array}{c}
x(0)\\
y(0)
\end{array}\right]
$$$$
\exp(A\,t)=L^{-1}\left[(s\,I_{2\times2}-A)^{-1}\right]
$$

where $s$ is the Laplace variable and $L^{-1}$ is the Laplace inverse transform. An example of the calculation of $\exp(A\,t)$ is given in this video : https://www.youtube.com/watch?v=RqK_m7qJ_R0


问:
How will I prove that determinant $det: GL (R) → R$ is a homomorphism? where R is a ring and $G=GL(R)_{n}$ a group
My Algebra teacher made that statement and asked why it was true. I am not able to formulate a correct demonstration. I thought of something using Det (AB) = Det (A). Det (B) but I think I'm wrong.</p>

答:
In fact you’re absolutely right!

The formula $$\det AB = \det A \det B$$ is exactly saying that $\det$ is an homomorphism between the groups $(GL_n(\mathbb R), \times)$ and $(\mathbb R^*, \cdot)$.


答:
0)If $y_n$ is not bounded, say above, there is a subsequnce $y_{n_k} \rightarrow \infty$.

That's all you can say.

1)Example:

$y_{2k}=2k$, and $y_{2k+1}=(1/2)^{2k+1},$ where $k$ is a non-negative integer


问:
Formula for length of diagonal in a Lambert quadrilateral
Given a Lambert quadrilateral $AOBF$ where the angles $ \angle FAO ,  \angle AOB , \angle OBF $ are right, and $F$ is opposite $O , \angle AFB$ is the acute angle , and the Gaussian curvature = -1 (so all is standard)

Then a whole lot of equations hold see https://en.wikipedia.org/wiki/Lambert_quadrilateral#Lambert_quadrilateral_in_hyperbolic_geometry

(these are the relations I found in different sources)

There is also another relation (follows from the Klein Disk model)

$$ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $$

But I was not able to deduce this formula from the others.

How can $ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $ be deduced from the other formula's?

答:
Maybe take a look at the paper "The Right Right Triangle on the Sphere" by Dickinson and Salmassi.  In spherical (rather than hyperbolic) geometry, they derive and champion the similar-looking $\sin^2 OF = \sin^2 FA + \sin^2 FB$.  I'm sure that in the hyperbolic case one would have $\sinh^2 OF = \sinh^2 FA + \sinh^2 FB$, although this is still using the 'wrong' (from your perspective) legs FA and FB.  But perhaps with further application of the 'usual' Pythagorean Theorem you can deduce what you want.

As a remark, I worked out the relation $\sin^2 OF = \sin^2 FA + \sin^2 FB$ myself from the same Lambert quadrilateral Wikipedia formulas you cited (assuming they held also on the sphere), and since it's so simple and pleasant-looking I assumed it was a "well known" fact.  But it doesn't seem to be; it took me a long time to find the Dickinson--Salmassi paper, and it's the only source I managed to find.


问:
How to estimate the number of decimal places required for a division?
Given two decimal numbers, is it possible to estimate the number of decimal places required to fit the result of their division? Provided that the division yields a finite number of decimals, of course.

For example:

1234.5678 / 2 = 617.2839, 4 decimal places required

1234.5678 / 4 = 308.64195, 5 decimal places  required

1234.5678 / 8 = 154.320975, 6 decimal places  required

1234.5678 / 6.4 = 192.90121875, 8 decimal places  required

By estimate, I don't necessarily need the exact number of decimals in the result, but a number of decimals at least equal to the required amount, so that it is guaranteed that the result fits.

What I've tried

I was able to roughly solve my problem using rational numbers & prime factorization, but this is very compute expensive. Here are the steps:

Take the original division: 1234.5678 / 6.4

Convert it to a rational number: 12345678 / 64000

Simplify this fraction using the GCD of the two numbers: 6172839 / 32000

Take the denominator: 32000

Compute the factors of 2 and 5 by dividing successively by these two numbers:

32000= 28 * 53

(if it is found at this step that the number has other factors than 2 and 5, then stop here: the division yields an infinite number of digits)

Take the maximum of the two exponents: max(8,3) = 8

⇒ 8 decimal places is enough to fit the result of the division.

How I came to the conclusion above

Out of all the prime numbers, only dividing by 2 and 5 yields a finite number of digits.

Each division by 10 extends the scale of the decimal number by 1 digit.
Each combination of 2 and 5 yields a 10, so an extra digit.

In 2x * 5y, there are min(x,y) times 10.

Now each division by 2 or 5 can potentially (although not always) require an extra digit. So I will carefully add an extra digit for each remaining 2 or 5 factor:

Maximum required digits = min(x,y) + (x - min(x,y)) + (y - min(x,y))

Which simplifies to: x + y - min(x,y)

Which further simplifies to max(x,y).

I feel like my approach, although it works, is overly complex. The direct consequence on my software is the slowness of the algorithm.

Is there a more straightforward approach to estimating the number of decimal places required for the result of the division to fit?

Note that I've read this question: Number of decimal places to be considered in division but it didn't help.

答:
Have you tried calculating the maximum precision needed to reserve a place for the result?

Look how Java's BigDecimal does that:

https://github.com/frohoff/jdk8u-dev-jdk/blob/da0da73ab82ed714dc5be94acd2f0d00fbdfe2e9/src/share/classes/java/math/BigDecimal.java#L1674

Then the non-terminating numbers will cause an overflow of this area, so just check for this corner case during calculation and if that will happen, you can be sure it cannot be precisely expressed as decimal number.


问:
l'Hôpital's Rule and Multivariable Limits
I decided to post another message regarding this problem because I still didn't understand it at all:
Can someone give me an example of function $f(x,y),g(x,y)$ for which:
$\lim\limits_{r\to 0^+} \dfrac {f(r\cos \theta , r\sin \theta ) }{ g(r\cos\theta , r\sin \theta)} =\dfrac{0}{0}$, and $\lim\limits_{r\to 0^+} \dfrac {\frac{\mathrm df(r\cos \theta , r\sin \theta )}{\mathrm dr} }{ \frac{\mathrm dg(r\cos\theta , r\sin \theta )}{\mathrm dr} } = C $ for some constant $C$ , but the actual limit $\lim\limits_{(x,y)\to (0,0)} \dfrac {f(x,y)}{g(x,y)}$ does not exist at all?

All I need is an example of a case where l'Hôpital's rule for multivariable limits when appearing in polar coordinates is not helpful (the reason such an example must exist is because $\theta$ can also depend on $r$ , but when using l'Hôpital's rule wrt $r$, we consider $\theta$ to be a constant... ).

Hope someone will be able to help me this time.

Thanks in advance.

Just to clarify things-
This is not a homework question... Only something I thought about...

答:
The classic example of a function that is continuous along lines but discontinuous will do it, if you allow repeated L'Hôpital:
$$f(x,y)=x^2y \quad\text{and}\quad g(x,y)=x^4+y^2\,.$$
So, in polar coordinates, it looks like $\dfrac{r^3(...)}{r^2(...)}$ and approaches $0$ as $r\to 0$ (with $\theta\ne 0,\pi$).

Recall that if we let $y=cx$, then $$\frac{f(x,cx)}{g(x,cx)} = \frac{cx^3}{x^4+c^2x^2} =\frac{cx}{c^2+x^2}\to 0 \quad\text{as }x\to 0\,,$$
unless $c=0$, but that case is easily handled directly. On the other hand, if we let $y=x^2$, then
$$\frac{f(x,x^2)}{g(x,x^2)} = \frac{x^4}{x^4+x^4} = \frac12 \quad\text{for all }x\ne 0\,,$$

If you want a single application of L'Hôpital to do it, let's try
$$f(x,y) = x|y|^{1/2} \quad\text{and}\quad g(x,y)=x^2+|y|\,.$$

The same sort of analysis works here: Along $y=cx$, $\dfrac{f(x,cx)}{g(x,cx)} \to 0$, and along $y=x^2$, $\dfrac{f(x,x^2)}{g(x,x^2)}=\dfrac{|x|}{x} = \pm 1$.

I believe that if your limit $C\ne 0$ and $f(r,\theta)$ and $g(r,\theta)$ are actually differentiable functions, then your result works, because we'll have
$$f(r,\theta) = arh(\theta) + o(r) \quad\text{and}\quad g(r,\theta)=brk(\theta) + o(r)\,,$$
from which we'll have
$$\frac{f(r,\theta)}{g(r,\theta)} = \frac{a r h(\theta) + o(r)}{b r k(\theta) + o(r)} \to \frac{ah(\theta)}{bk(\theta)}\,;$$
if this is a constant $C$, we must have $h(\theta) = k(\theta)$, and it works.


答:
Well if $a = 2$ you have $z=2; y=-\frac 19$ and $x = whatever$.

But if $a-2\ne 0$ keep at it.

Multiply row 2 by $\frac 9{a-2}$ and subtract getting rows.

$\begin{array}\\
1&2&2&|\,\,1\\
0&a-2&1&|\,\,2\\
0&0&(a-2)-\frac 9{a-2}&|-\frac 9{a-2}\\
\end{array}$

Now if $(a-2)-\frac 9{a-2} = 0$ we have a case with not solutions.

That is if $(a-2)^2 = 9$ of $a = 5,-1$.

If we assume $a\ne 2,5,-1$ then we can reduce

$\begin{array}\\
1&2&2&|\,\,1\\
0&1&\frac 1{a-2}&|\,\,\frac 2{a-2}\\
0&0&1&|-\frac 9{a-2}\cdot\frac 1{(a-2)-\frac 9{a-2}}=-\frac{9}{(a-2)^2-9}\\
\end{array}$

..... and that's in the whatchamacallit form "upper triangle"?) so we know this results in a unique solution.

So $a = 5$ or $1$ are the cases without solutions.

(because we'd get $((a-2)-\frac 9{a-2})z=-\frac 9{a-2}$ which if $a = 5$ is

$(3-3)z = -3$ or $0 = 3$ and if $a=-1$ we get $(-3 + 3)z = 3$ or $0 = 3$.


问:
Minimum Edge Cover of the Hypercube
I want to find the minimum edge cover of the n-dimensional hypercube $Q_n$. I think the answer is $2^{n-1}$ after doing some casework, but I am not entirely convinced: for $n=2,3,4$, the size of min edge cover is $2,4,8$ respectively. I also have an idea for the proof: because there is $n2^{n-1}$  edges in $Q_n$, and the degree of each vertex is $n$, as long as each of the vertices cover $n$ unique edges of $Q_n$( that is, the edges that each vertex cover is entirely different from the edges that other vertex covers) then the minimum edge cover is $2^{n-1}$. However, I can't figure out how to show it is possible to arrange the vertices such that they cover $n$ unique edges of $Q_n$. Some hint will be greatly appreciated!

答:
Construction:
You can construct a $Q_{n+1}$ by placing two $Q_n$ s side-by side, each with $2^n$ vertices and $n.2^{n-1}$ edges (since $\sum\limits_i degree(v_i)=2^n.n=2.e)$, and joining vertex $i$ from a $Q_n$ to the corresponding vertex $i^{\prime}$ for the other $Q_n$, with another $2^{n}$ edges, so that total number of edges in the resulting volume formed  $=2.|E(Q_n)|+2^n=2.n.2^{n-1}+2^n=(n+1).2^n=|E(Q_{n+1})|$.

Now, completing the induction proof on the dimension $n$ of the hypercube:

Basis: for $n=2$, trivially select $2=2^{n-1}$ $||$ edges to cover 4 vertices and it's the minimum edge cover size.

Induction hypothesis: Let's assume till $n \leq m$, the minimum edge cover size be $2^{n-1}$

Induction Step: for $n=m+1$, construct $Q_{m+1}$ as described above by placing two $Q_m$s side by side and adding $2^{m}$ edges to connect the corresponding vertices.

Now, notice that in $Q_{m+1}$ there is exactly $2^{m+1}=2.2^m$ vertices obtained by combining two $Q_m$s and no extra vertex.

Also, by induction hypothesis, each of the $2^m$ vertices from the individual $Q_m$ s can be covered by selecting $2^{m-1}$ edges from the respective $Q_m$s.

Hence, if we combine those edges used to cover the individual $Q_m$s, we get $2.2^{m-1}=2^m$ edges, that cover all the $2^{m+1}$ vertices of $Q_{m+1}$, since there is no additional vertex.

Hence using $2^m=2^{(m+1)-1}$ edges we can cover the vertices in $Q_{m+1}$.

Also, notice that the edge cover size can't be less, otherwise we shall not be able to cover all the vertices of at least one of the component $Q_m$s, hence this is the minimum possible edge cover size.

It completes the proof that $2^{n-1}$ is the minimum edge cover size for a hypercube of dimension $n$, $\forall{n} \in N$.

Title: Minimum Edge Cover of the Hypercube


答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\nabla\bracks{%
\nabla\cdot\pars{{\bf r}_{0}\expo{\ic{\bf k}\cdot{\bf r}}}}} =
\nabla\bracks{%
\overbrace{\pars{\nabla\cdot{\bf r}_{0}}}^{\ds{=\ 0}}\
\expo{\ic{\bf k}\cdot{\bf r}} +
{\bf r}_{0}\cdot\nabla\pars{\expo{\ic{\bf k}\cdot{\bf r}}}} =
\ic\nabla\pars{{\bf r}_{0}\cdot{\bf k}\,\expo{\ic{\bf k}\cdot{\bf r}}}
\\[5mm] = &\
\ic\,{\bf r}_{0}\cdot{\bf k}\,\nabla\expo{\ic{\bf k}\cdot{\bf r}} =
\ic\,{\bf r}_{0}\cdot{\bf k}\,\pars{\expo{\ic{\bf k}\cdot{\bf r}}
\,\ic{\bf k}} =
\bbx{-{\bf k}\pars{{\bf k}\cdot{\bf r}_{0}}\expo{\ic{\bf k}\cdot{\bf r}}}
\\ &
\end{align}
There are a lot of vectorial identities over here.

Title: Calculating $\nabla\left(\nabla \cdot \boldsymbol{r}_{0} e^{i\boldsymbol{k} \cdot \boldsymbol{r}}\right)$

Body:
In a longer derivation I ran into the following quantity:
$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right]
$$
( i.e., the gradient of the divergence ) where ${\bf k}$ is a vector of constants and ${\bf r}$ is a position vector.

Can someone help explaining how to calculate this?
I am hoping it gives:
$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right] =
-{\bf k}\left({\bf k}\cdot{\bf r}_{0}\right)
$$
( because then the rest of my equations add up ).


问:
if a set S is convex then $\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $ for any z in the set
let S be a convex set, and let $x\notin S;y,z\in S$.
such y is the closest to x in S.
prove:
$\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $.
The inner product here is of course the euclidean one. and the distance is with the regular Euclidean distance.
I tried using Cauchy Schwartz, but it has not been successful.

答:
You want to show that
$$
 \langle x-y,\frac{x}{2}+\frac{y}{2}\rangle - \langle x-y,z\rangle
= \langle x-y,\frac{x}{2}-\frac{y}{2}+y-z\rangle \\
= \frac 12\Vert x-y \Vert^2 + \langle x-y, y-z \rangle
$$
is strictly positive. It suffices to show that
$$ \tag {*}
\langle x-y, y-z \rangle \ge 0
$$
and that is true because $y$ is the closest point to $x$ in $S$: The function
$$
 [0, 1] \ni t \mapsto f(t) = \Vert x- (y+t(z-y))\Vert^2 \\= \Vert x-y \Vert^2 -2 t \langle x-y, z-y \rangle + t^2 \Vert z-y \Vert^2
$$
is minimal at $t=0$, therefore
$$
 f'(0) = -2  \langle x-y, z-y \rangle \ge 0
$$
and that proves $(*)$.


答:
Set ordinalities:

$0,1,2,...,\omega,\omega+1,\omega+2,...,\omega^n,...,\omega^{\omega},...,\omega^{\omega^{\omega^{...}}}=\epsilon_0,...$

One might say it's hugely, vastly, mind-bogglingly big.

Title: Can we count past $ \infty $?

Body:
Let's say, $ \infty $ is not the biggest number, in fact, there is something bigger, somewhere. We want to try and count past $ \infty $, because it is not really an ordinal one, but a definition of something uncountable or too big to even count.

What numbers are bigger than $ \infty $?


问:
Generalized Card Flipping Problem
The problem goes as follow. Suppose that I have a total of $n$ cards, which $p$ are upwards, while $n-p$ are downwards. Let's say that I randomly choose $k$ of these cards to flip (upwards turn into downwards and downwards into upwards), so what is the probability that I am going to have $q$ of them upwards after this? The case for $k=1$ is fairly trivial, as we can only have $p+1$ or $p-1$ cards upward and it is fairly easy to see that the probabilities are going to be $(n-p)/n$ and $p/n$ respectively and all others are zero. For $k=n$ is also trivial, but for other values I am having some trouble.

答:
First of all note that $q-p$ should have the same parity as $k$. Let $k_1$ and $k_2$ be the numbers of flipped upward- and downward cards, respectively.

From the equations:
$$\begin{align}
k_1+k_2&=k\\
k_1-k_2&=p-q
\end{align}$$
we find $k_1=\frac{k+p-q}2$, $k_2=\frac{k+q-p}2$, so that the probability in question is:
$$
\mathbb P_k(p\to q)=\frac{\binom{p}{\frac{k+p-q}2}\binom{n-p}{\frac{k+q-p}2}}{\binom nk}.
$$

Title: Generalized Card Flipping Problem


问:
Isomorphism between two finite fields
We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic.

Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then

$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$

where

$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$

Thus

$$(x+y\beta)^2 = \phi(\alpha)^2  = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$

So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that

$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$

Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?

答:
An idea: since

$$k_1=\Bbb F_7[x]/\langle\;x^2-3\;\rangle\;,\;\;k_2=\Bbb F_7[x]/\langle\;x^2+1\;\rangle$$

Now, in both cases there exists a primitive element, i.e.: an element that generates the multiplicative group $\,k_i^*\;$ , so if you find both elements in both fields you shall be done...

For example: let $\,\alpha\in\overline{\Bbb F_7}\;$ be a root of $\,x^2-3\in\Bbb F_7[x]\,$ , then in $\,k_1\,$ we have
$$\alpha^2=3\implies k_1:=\{p(\alpha)\;;\;p(x)\in\Bbb F_7[x]\;,\;\;\deg p\le 1\}$$

For example, we have that $\,\alpha +5\;,\;3\alpha\;,\;2\alpha -5\,$ , etc. are element in $\,k_1\,$ .

Since $\,|k_1^*|=48\;$ ,every non-zero element in the field has multiplicative order a divisor of $\,48\,$ , so we can try the following: pick (nd remember: we work modulo $\,7\,$ all the time!)

$$\begin{align*}r:&=\alpha+1\\
 r^2&=\alpha^2+2\alpha+1=3+2\alpha+1=2\alpha+4\\
r^4&=4\alpha^2+2\alpha+2=2\alpha\\
r^6&=(2\alpha)^3=\alpha(3)=3\alpha\\
r^{12}&=(3\alpha)^2=2\cdot 3=-1\end{align*}$$

and we've found that $\;r\;$ is not a generator of $\,k_1^*\,$ (why? check what happens with $\,r^{24}\,$ ...) . Try other elements until you find an generator and then try to do the same with $\,k_2\,$ and then you'll have an easy choice for your isomorphism...

Title: Isomorphism between two finite fields


问:
If $G=G_1\times\dots\times G_n$ for abelian $G_i$, then ${\rm Aut}(G)$ is isomorphic to a specific general linear group.
This is Exercise 1.5.10 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

The Details:

On page 11, ibid., the set product of subsets $X,Y$ of a group is defined as

$$XY=\{xy\mid x\in X, y\in Y\}.$$

On page 17, ibid., for groups $G,H$, the definition of ${\rm Hom}(G, H)$ is given as the set of all homomorphisms from $G$ to $H$.

The definition of a direct product on pages 20 to 21, ibid., is a little more involved.

Let $\{G_\lambda\mid \lambda\in\Lambda\}$ be a given set of groups. The cartesian (or unrestricted direct) product,

$$C=\underset{\lambda\in\Lambda}{{\rm Cr}}\, G_\lambda,$$

is the group whose underlying set is the set product of the $G_\lambda$s [. . .] and whose group operation is multiplication of components: thus

$$(g_\lambda)(h_\lambda)=(g_\lambda h_\lambda),$$

$g_\lambda, h_\lambda\in G_\lambda$. [. . .]

The subset of all $(g_\lambda)$ such that $g_\lambda=1_\lambda$ for almost all $\lambda$ [. . .] is called the external direct product,

$$D=\underset{\lambda\in\Lambda}{{\rm Dr}}\, G_\lambda,$$

[. . .] In case $\Lambda=\{\lambda_1,\dots,\lambda_n\}$, a finite set, we write

$$D=G_{\lambda_1}\times\dots\times G_{\lambda_n}.$$

Of course $C=D$ in this case.

On page 26, ibid., we have that

The set of all automorphisms of [a group] $G$ is denoted by

$${\rm Aut}(G).$$

It is a group under function composition.

The Question:

This is on page 30, ibid.

Let $G=G_1\times\dots\times G_n$ where the $G_i$ are abelian groups. Prove that ${\rm Aut}(G)$ is isomorphic with the group of all invertible $n\times n$ matrices whose $(i,j)$ entries belong to ${\rm Hom}(G_i, G_j)$, the usual matrix product being the group operation.

Thoughts:

I'm used to thinking of the direct product of groups as the group defined by a certain presentation, or else simply as the obvious generalisation of $$H\times K=\{(h,k)\mid h\in H, k\in K\}$$ under the operation $(h,k)\cdot(h',k')=(hh', kk')$.

My linear algebra is a little rusty, which is probably why I'm struggling here.

Examples:

When $n=1$:

We have just the abelian group $G$. The automorphism group is

$${\rm Aut}(G)=\{\varphi\in {\rm Hom}(G,G)\mid \varphi \text{ is an isomorphism}\}$$

and the $1\times 1$ matrices are simply $(\varphi)$ with $(\varphi)^{-1}=(\varphi^{-1})$.

The isomorphism is trivial.

When $n=2$:

We have $G=G_1\times G_2$. Now the $(i,j)$ entries of the $2\times 2$ matrices are such that . . .

Wait a Minute . . .

What operation is the addition of the homomorphisms?

I guess this is my main stumbling block.

Let's try working it out & see . . .

So, we have

$${\rm Aut}(G)\stackrel{?}{\cong}\left\{
\begin{pmatrix}
\varphi_{(1,1)} & \varphi_{(1,2)}\\
\varphi_{(2,1)} & \varphi_{(2,2)}
\end{pmatrix} : \varphi_{(1,1)}\stackrel{?}{\times}\varphi_{(2,2)}\stackrel{?}{-} \varphi_{(1,2)}\stackrel{?}{\times}\varphi_{(2,1)}\neq \stackrel{?}{0} \right\}.$$

Okay, so I'm not sure on multiplication either, nor am I sure of what the zero element is.

This is a mess. I'm sorry.

Please help :)

答:
Hints:

When $G$ is any group and $B$ is an abelian group, then $\mathrm{Hom}(G,B)$ is more than just a set, it is actually an abelian group. The operation is "pointwise addition", so that $(f+g)(a) = f(a)+g(a)$ for all $a\in G$. This is a homomorphism as well, since
$$\begin{align*}
(f+g)(a+a') &= f(a+a') + g(a+a')\\
&= f(a)+f(a') + g(a)+g(a')\\
&= f(a)+g(a)+f(a')+g(a')\\
&= (f+g)(a)+(f+g)(a').
\end{align*}$$
The identity element of this abelian group is the zero map, sending everything to $0_B$, and the additive inverse of $f$ is the function $g(a)=-f(a)$ for all $a\in A$. (When $G=B$, this is actually a ring, with the multiplication being composition of functions).

For any family of groups $\{G_i\}_{i\in I}$, a morphism $f$ from a group $H$ into the direct product $\prod_{i\in I}G_i$ is equivalent to a family of morphisms $\{f_i\}_{i\in I}$, with $f_i\colon H\to G_i$. This is the universal property of the product: given the family $\{f_i\}_{i\in I}$, there exists a unique morphism $F\colon H\to \prod G_i$ such that $f_i =\pi_i\circ F$, where $\pi_i$ is the projection onto the $i$th coordinate. Namely, $F(h) = (f_1(h),\ldots,f_n(h))$.

For a finite family of abelian groups, we also have maps going the other way. The embeddings $\iota_i\colon A_i\to \prod_{j\in J}A_j$ that sends $a\in A_i$ to the element of the product that has $a$ in the $i$th coordinate and $0$s in all the other coordinates have the following universal property: given any abelian group $K$, and given any morphisms $g_i\colon A_i\to K$ for each $i$, there exists a unique morphism $G\colon \prod A_i\to K$ such that $g_i = G\circ \iota_i$ for each $i$. Namely, $G(a_1,\ldots,a_n) = g_1(a_1)+\cdots + g_n(a_n)$. Thus, maps from a finite direct product of abelian groups are equivalent to a family of maps from the factors.

Putting those together, it should be clear that any homomorphism $G_1\times\cdots\times G_n\to G_1\times\cdots\times G_n$ corresponds to a family of morphisms $\varphi_{ij}\colon G_j\to G_i$. You should then verify that if you arrange them in a matrix in the "obvious" way, and what the corresponding morphism is, then the image of an element $(g_1,\ldots,g_n)$ under the corresponding morphism $\varphi\colon G_1\times\cdots\times G_n\to G_1\times\cdots\times G_n$ is given by computing
$$\left(\begin{array}{ccc}
\varphi_{11}& \cdots & \varphi_{1n}\\
\vdots & \ddots & \vdots \\
\varphi_{n1} & \cdots & \varphi_{nn}
\end{array}\right)
\left(\begin{array}{c}
g_1\\ \vdots\\ g_n\end{array}\right).$$

That will give you a description of the homomorphisms. It also tells you that the composition of homomorphisms "should" correspond to matrix multiplication. And that will tell you exactly when a homomorphism is actually an automorphism.


Body:
Solve $x^\prime=x+2y,  y^\prime=4x+3y,  x(0)=−1,  y(0)=1$

$x(t)= $

$y(t)= $

Find the limits

$\lim_{t \to \infty} x(t)=$

$\lim_{t \to \infty} y(t)= $

Select the appropriate (A, B, or C) that best represents the long time behavior of the solution.

(A) Equilibrium solution

(B) Arc solution

(C) Periodic Solution

This example comes with answers as follows:

$x(t)=-e^{-t}$

$y(t)=e^{-t}$

$\lim_{t \to \infty} x(t)=0$

$\lim_{t \to \infty} y(t)=0$

B

However, the example has no work behind it and I can't find a similar example online or in my textbooks. Would somebody be willing to show me the procedure of how to solve something like this?
Thank you!

Body:
$$x^\prime=x+2y\\ y^\prime=4x+3y
$$ with $x(0)=−1$ and $ y(0)=1$.

Sum both DE:
$$(x+y)'=5(x+y)$$
$$\ln |x+y|=5t+C$$
$$x+y=C_1e^{5t}$$
Now solve the second DE:
$$y'=4x+3y$$
$$y'=4C_1e^{5t}-y$$


问:
choice set for equivalence classes
I'm trying to understand the proof of the Paradox of Banach-Tarski. I was reading Wagon's work about the paradox and tried to understand the proof of Theorem 1.5, only I haven't warmed up to group theory yet. Maybe someone can explain to me what exactly Wagon means by:

Let M be a choice set for equivalence classes of the relation on $S^1$ given by
calling two points equivalent if one is obtainable from the other by a
rotation about the origin through a (positive or negative) rational
multiple $2\pi$ radians.

I would appreciate it if someone could explain the statements so that I can continue to work on the proof and to understand it better. Thanks for your help!

答:
So you have a point $A=(x,y) = (\cos \theta, \sin \theta)$ and a point $B=(x',y') = (\cos \theta', \sin\theta')$.  We say the two points are equivalent if $\theta' = \theta + 2r\pi$ for some $r \in \mathbb Q$.

(That is to say, if you can get from point $A$ to point $B$ by rotating the circle by rational proportion of a full turn they are equivalent.  If, however, you can only get to $B$ by rotating an irrational proportion they are not equivalent.)

(Or if it's any easier you can get from any point $A$ to any point $B$ by rotating the circle.  If the amount you rotate be is rational multiple of a full turn the points are equivalent and consider to be essentially the same.  If the amount you rotate by is an irrational proportion of a full turn they are not equivalent.)

As there are uncountably many points and only countably many rationals there will be an uncountable number of equivalence classes.

Are you familiar with the concept of equivalence relations and equivalence classes?

Title: choice set for equivalence classes


问:
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$
I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1}  \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.

I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding  indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.

答:
Too long for a comment

Playing a little bit with expression one can see that

$$\begin{align}
\sin x+\cos x+\tan x&+\cot x+\csc x+\sec x \\
\end{align}
=\sin x+\cos x +\frac{2}{\sin x+\cos x -1}=\sqrt{2}\sin\Big(x+\frac{\pi}{4}\Big)+\frac{2}{\sqrt{2}\sin\Big(x+\frac{\pi}
{4}\Big)-1}$$

Plugging this in integral we get,

$$I=\int_{0}^{\frac{\pi}{2}}
\frac{\sqrt{2}\sin\Big(x+\frac{\pi}
{4}\Big)-1}{2\sin^{2}\Big(x+\frac{\pi}
{4}\Big)-\sqrt{2}\sin\Big(x+\frac{\pi}
{4}\Big)+2}dx$$

Let $x=t-\frac{\pi}{4}$ $\implies dx=dt$

$$ I= \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}
\frac{\sqrt{2}\sin t-1}{2\sin^{2}t-\sqrt{2}\sin t+2}dt$$

We can split this in the following two integrals

$$I= \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}
\frac{\sqrt{2}\sin t}{2\sin^{2}t-\sqrt{2}\sin t+2}dt-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}
\frac{1}{2\sin^{2}t-\sqrt{2}\sin t+2}dt$$

I'm not sure how to calculate above two integrals separately.
Initially I thought that we have two integrals and two nested square root terms In given result so they should be equal but none of the individual integral is equal to the terms in result.Numerically these two integrals evaluate $0.8448$ and $0.6688$ and terms in result are $0.6054$ and $-0.4294$. Notice their sum are equal to each other so we are in right direction.


问:
Full Binary Trees - Maximizing the arithmetic mean of powers of leaf node levels.
I have a full binary tree with $n$ leaf nodes. Therefore we  get the following constraint $\sum_{i=1}^{n}2^{-l_{i}} = 1$ where $l_i$ is the level occupied by each leaf node.

I have the metric $\frac{1}{n}\sum_{i=1}^{n}2^{l_{i}}$.

What I would like to show is, that the more unbalanced the tree is, the higher the metric. Since there is no universal metric for tree unbalance-ness, I dont know how to go about it.

Possible Directions -

I tried to show that an act of unbalancing a tree increases the
metric.

I can show that the metric is lower bounded by $n$ using the
AM-HM inequality.

答:
Let $\ell_1$ be the minimum depth of a leaf and $\ell_n$ the maximum depth; since the tree is full, there must be a sibling pair of leaves at depth $\ell_n$. If $\ell_n\ge\ell_1+2$, remove a sibling pair at depth $\ell_n$, and give a leaf at depth $\ell_1$ a pair of children; this loses two leaves at depth $\ell_n$, adds one at depth $\ell_n-1$, adds two at depth $\ell_1+1$, and loses one at depth $\ell_1$, for a net change in $\sum_{k=1}^n2^{\ell_k}$ of

$$2^{\ell_n-1}-2\cdot 2^{\ell_n}+2\cdot 2^{\ell_1+1}-2^{\ell_1}=3\left(2^{\ell_1}-2^{\ell_n-1}\right)<0\,.$$

(I ignore the factor of $\frac1n$, since $n$ is not changed by this procedure.) Thus, as long as the tree is not balanced, this procedure will reduce your metric.


问:
Can you solve for the identity matrix where A is non-invertible?
If you have an $n\times m$ matrix $A$, and we assume that those dimensions are something like $3\times6$, is there a matrix $D$ such that $AD = I$?

答:
If $AA^T$ is invertible, then $D=A^T(AA^T)^{-1}$. This is the Moore-Penrose inverse; see https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse.

Below is a series of Python commands to find the pseudo-inverse of a 2 x 4 matrix.

```<code>>>> import numpy as np
>>> A = np.array([[2, 4, -6, -10], [4, 8, -8, -8]])
>>> A
array([[  2,   4,  -6, -10],
       [  4,   8,  -8,  -8]])
>>> np.linalg.pinv(A)
array([[-0.06060606,  0.06818182],
       [-0.12121212,  0.13636364],
       [ 0.02272727, -0.05681818],
       [-0.17424242,  0.10227273]])
C = np.linalg.pinv(A)
>>> C.dot(A)
array([[ 1.00000000e+00, -2.22044605e-16],
       [ 8.88178420e-16,  1.00000000e+00]])
</code>```

Note that since $n < m$, we can only find a left-inverse. Rank-nullity quickly implies that no right-inverse exists.

Body:
If $AA^T$ is invertible, then $D=A^T(AA^T)^{-1}$. This is the Moore-Penrose inverse; see https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse.

Below is a series of Python commands to find the pseudo-inverse of a 2 x 4 matrix.

>>> import numpy as np
>>> A = np.array([[2, 4, -6, -10], [4, 8, -8, -8]])
>>> A
array([[  2,   4,  -6, -10],
       [  4,   8,  -8,  -8]])
>>> np.linalg.pinv(A)
array([[-0.06060606,  0.06818182],
       [-0.12121212,  0.13636364],
       [ 0.02272727, -0.05681818],
       [-0.17424242,  0.10227273]])
C = np.linalg.pinv(A)
>>> C.dot(A)
array([[ 1.00000000e+00, -2.22044605e-16],
       [ 8.88178420e-16,  1.00000000e+00]])

Note that since $n < m$, we can only find a left-inverse. Rank-nullity quickly implies that no right-inverse exists.


答:
There is no general answers. It is clearly impossible if $A=0$ for example. However that is possible for the example

$$A=\begin{pmatrix}
1 &0&0&0\\
0& 1&0&0
\end{pmatrix}$$

Title: Can you solve for the identity matrix where A is non-invertible?

Body: If you have an $n\times m$ matrix $A$, and we assume that those dimensions are something like $3\times6$, is there a matrix $D$ such that $AD = I$?


问:
How does one derive $O(n \log{n}) =O(n^2)$?
I was studying time complexity where I found that time complexity for sorting is $O(n\log n)=O(n^2)$. Now, I am confused how they found out the right-hand value. According to this $\log n=n$. So, can anyone tell me how they got that value?

Here is the link where I found out the result.

答: What this equation means is that the class $O(n\log n)$ is included in the class $O(n^2)$. That is, if a sequence is eventually bounded above by a constant times $n \log n$, it will eventually be bounded above by a (possibly different) constant times $n^2$. Can you prove this? The notation is somewhat surprising at first, yes, but you get used to it.</p>


问:
Meaning of the geometric mean in the power mean inequality
The power-mean inequality states that for positive real numbers $a_i$ and all real numbers $k_1,k_2$: $$(\frac{1}{n}\sum_{i=1}^{n}a_i^{k_1})^{\frac{1}{k_1}} \ge (\frac{1}{n}\sum_{i=1}^{n}a_i^{k_2})^{\frac{1}{k_2}}$$
if $k_1>k_2$.
https://artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality

When $k_1=2,k_2=1$, one can prove the RMS-AM inequality (RMS is the root mean square). And when $k_1=1,k_2=-1$, one can prove the AM-HM inequality (HM is the harmonic mean).

This site(https://brilliant.org/wiki/power-mean-qagh/) states that by convention when $k_1$ or $k_2$ is $0$, then that side of the inequality is equal to the geometric mean.

So, in some sense could one say that the statement below is true?

$$\sqrt[0]{(\frac{1}{n}\sum_{i=1}^{n}a_i^{0})} = \sqrt[n]{\prod_{i=1}^na_i}$$

Though this may simply be a complete coincidence, it seems to fit too perfectly for me not to wonder about it. I am asking, besides simply being a nice notational trick, is it useful for mathematicians to define the zeroeth root as a way to invert operations or something similar? For example, in the example above, the zeroeth root seems to turn $\frac{1}{n}$ into the nth root and it turns the summation into a product.

Also, are there any other situations or theorems in math where a zeroeth root could mean something?

答:
There is a reason for that. The limit of the $k$-th power mean as $k\to 0$ is precisely the geometric mean. here is a proof with $n=2$:

Call
$$
f(k)=\ln \left(\frac{a_1^k+a_2^k}{2}\right)^{1/k}
$$.
Then
$$
\lim_{k\to 0}f(k)=\lim_{k\to 0}\frac{\ln\left(1+\frac{a_1^k+a_2^k}{2}-1\right)}{k}=\lim_{k\to 0}\frac{a_1^k+a_2^k-2}{2k}.
$$
By L'Hopitals rule then

$$
\lim_{k\to 0}f(k)=\lim_{k\to 0}\frac{a_1^k\ln a_1+a_2^k\ln a_2}{2}=\frac{\ln a_1a_2}{2}
$$
Therefore, the limit of the power means is
$$
e^{\frac{\ln a_1a_2}{2}}=\sqrt{a_1a_2}.
$$
A similar proof works for any $n$.


问:
Proof of if $|a-b| < \frac{\beta}{2}$ and $|b-c| < \frac{\beta}{2}$, then $|a-c| < \beta$
So I'm reading Real Analysis by Jay Cummings and he says that

If $|a-b| < \frac{\beta}{2}$ and $|b-c| < \frac{\beta}{2}$, then $|a-c| < \beta$

because
$|a-c| = |a-b+b-c| \leq |a-b|+ |b-c|= \frac{\beta}{2}+ \frac{\beta}{2}=\beta$

But if you follow the above statement, doesn't this imply that $|a-c| \leq \beta$, not $|a-c| < \beta$?

because the triangle inequality gives us the $\leq$ sign, not the $<$ sign. Did Cummings make a mistake here?

答:
I'm assuming the argument is
$$
|a-c| = |a-b+b-c| \leq |a-b|+ |b-c|\color{red}< \frac{\beta}{2}+ \frac{\beta}{2}=\beta
$$
with $\color{red}<$ in place of $=$. If so, then you can conclude that $|a-c|<\beta$. The final statement connecting $|a-c|$ and $\beta$ is governed by the "weakest link" (or is that the strongest link?), which is the strict inequality $\color{red}<$.

The statement $|a-c|<\beta$ does imply that $|a-c|\le \beta$, though, since $x<y$ implies $x\le y$. Given that $x<y$ implies $x\le y$, the above argument can also be written
$$
|a-c| = |a-b+b-c| \leq |a-b|+ |b-c|\color{red}\le \frac{\beta}{2}+ \frac{\beta}{2}=\beta
$$
since every link in the chain is a true statement. From this reasoning you would conclude $|a-c|\le\beta$.


答:
If $A$ is $n\times m$, with $n\lt m$, then you can find an $m\times n$ matrix $B$ such that $AB=I_n$, the $n\times n$ identity, if and only if $A$ has rank $n$. You cannot find a matrix $C$ such that $CA=I_m$, however.

If $A$ is $n\times m$ with $m\lt n$, then you can find an $m\times n$ matrix $B$ such that $BA=I_m$ if and only if $A$ has rank $m$; you cannot find a matrix $C$ such that $AC=I_n$, however.

If $A$ is square, $n\times n$, then you can find an $n\times n$ matrix $B$ such that $AB=I_n$ if and only if you can find an $n\times n$ matrix $C$ such that $CA=I_n$, if and only if $A$ has rank $n$; and in this case, $C=B$.

In summary, you can find an inverse on one side if and only if $A$ has full rank; and in that case, the side will be the one that gives you the "smaller" identity matrix.


答:
You can prove the following result:

Claim Let $A$ be a $2 \times n$ matrix. Then, there exists a $n \times 2$ matrix $B$ so that $AD=I_2$ if and only if $\mbox{Rank}(A)=2$.

More generally, for an $m \times n$ matrix $A$, there exists a $n \times m$ matrix $B$ so that $AD=I_m$ if and only if $\mbox{Rank}(A)=m$.

This can be proven by using the fact that the corresponding linear transformation $T :\mathbb R^n \to \mathbb R^2$ is onto and hence has a right inverse.

A more elementary solution, which also tells you how to find $D$, is the following:
The systems
$$Ax= \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\
Ay= \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\
$$
are consistent (why?). Let $D$ be the matrix with columns $x$ and $y$ respectively.

Then
$$
AD = \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}
$$


问:
imaginary numbers - how can I understand them - especially as they occur in 'roots' of polynomials?
In another question here, about roots of equations being imaginary, the accepted answer
said something interesting about "imaginary" (as a technical word in math) not meaning "not real".  I understand that imaginary numbers are "manipulable and usable to do all kinds of things".   And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots.  In maxima, $\sqrt{-1}$ is represented as %i.

Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results:
[$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)?  (At first I thought it had reported the same root twice as I overlooked the negative sign.)

One can easily plot a real root and see how it intercepts the x axis.  How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

What uses do imaginary roots of polynomials have in math?

答:
You might want to take a look at this as well: Do complex numbers really exist?.

Additionally, you can feel proud of yourself for inquiring along the same lines (with the aid of a computer) that Cardano & other mathematicians did 500 years ago, which led to the acceptance and elaboration of %i. Your observation about $\pm i$ is perspicacious and important in the theory of differential equations (dynamical systems), as well as other places.

We think about $\sqrt{-1}$ as part of a larger entity called the complex plane (google is your friend). $\sqrt{-1}$ is 90 degrees around the complex plane, whereas $-1$ is 180 degrees around the complex plane -- a fully turned-around $+1$. The 180-degree turn is the number line you learned in school; going only 90 degrees enters the complex plane.

The unifying idea is that a pair of imaginary and "real" numbers $x + i \cdot y$ are equivalent to a radius and an angle. Simple enough if you're familiar with rectangular and polar coordinates. But the key observation by de Moivre which unlocked a deep and mysterious world is that this can all be done with the natural base $e$.

$\exp (i \cdot x) = \cos x + i \sin x$.

Therefore $x+\sqrt{-1} \cdot y = \sqrt{x^2 + y^2} \cdot \exp (\sqrt{-1} \cdot \arctan {y \over x})$, which we think of as

$\mathrm{radius} \cdot e^{i \cdot \mathrm{angle}}$.

Here are some concrete examples of writing %i numbers in polar ($\exp$) form:

$3 + 4i = 5 \cdot e^{\pi \over 4}$

$-1 = e^{ \sqrt{-1} \cdot \pi }$

$\sqrt{-1} = e^{\sqrt{-1} \cdot {\pi \over 2}}$

$\sqrt{-25} = 5 \cdot e^{\sqrt{-1} \cdot {\pi \over 2}}$

As far as uses within mathematics, %i:

makes any polynomial equation soluble

matrices with %i model certain groups (google hermitian matrix)

complex eigenvalues tell you about the kind of solution to a (system of) differential equation(s)

helps model fluid flow

%i reveals more about the derivative (google amplitwist or Cauchy-Riemann equations)

%i reveals a fundamental to unity to the Moebius transformations (watch this)

you need %i for Fourier decomposition

the "angle" allows us to model the phase of a signal or quantum state as well as its magnitude

some previously non-integrable functions become integrable when you make use of the complex plane (google Laurent series or calculus of residues)

twistors and spinors in mathematical physics use %i

you can factor any polynomial all the way down using %i

the characteristic function of a probability distribution uses %i

If you've got the time, look at Tristan Needham's Visual Complex Analysis or Roger Penrose's The Road to Reality. Two of the best maths books that have been written, period. And they both show really cool things you can do with %i.

You're asking all the right questions to begin a magnificent journey of discovery.


答:
No, $O(n\log n)=O(n^2)$ doesn't mean $\log n=n$.

Recall that $f(n)=O(g(n))$ means there exists $C$ and $N$ such that $|f(n)|\leq C g(n)$ for $n\geq N$. For example, $\log n$ is $O(n)$. Thus, $n\log n$ is $O(n^2)$.

What I guess they are trying to say is that if something is $O(n\log n)$ then it is $O(n^2)$, that is $n\log n$ is $O(n^2)$. Note that the reverse usage $$O(n^2)=O(n\log n)$$ would be incorrect. A better notation might be $$O(n\log n)\subseteq  O(n^2)$$

Title: How does one derive $O(n \log{n}) =O(n^2)$?

Body:
I was studying time complexity where I found that time complexity for sorting is $O(n\log n)=O(n^2)$. Now, I am confused how they found out the right-hand value. According to this $\log n=n$. So, can anyone tell me how they got that value?

Here is the link where I found out the result.


问:
Boundness of an integral of a function
Let $f(z,y)$ a positive continuous function with respect to its arguments $z\in[0,1]$ and $y\in[0,\infty)$ and $\alpha >0$. If we have

$$
\lim_{y\rightarrow+\infty}\int\limits_0^1f(z,y)\,dz\leq \alpha
$$

Is it sufficient to conclude that $f(z,y)$ is bounded?

答:
Hint

The answer is negative.

The idea is to build a map with unbounded narrow bumps (using bump maps) as $y \to \infty$. Narrow enough for the integral to be always bounded by $\alpha$.

Title: Boundness of an integral of a function


答:
Given an abelian group $G$ noted additively (or more generally a module over any ring), its endomorphisms form a ring $\text{End}(G)$ under pointwise addition ($(\phi+\psi)(x)=\phi(x)+\psi(x)$ and composition (you can check the axioms yourself; distributivity follow from the functions being morphisms). An automorphism is an invertible endorphism, so $\text{Aut}(G)$ is the Group of units of $\text{End}(G)$.

Now, you just want to represent an endomorphism of the direct sum $G_1\oplus \cdots \oplus G_n$ with a matrix (direct sum is the same as direct product in this case but its more common to talk about sums when working with abelian groups/modules and it makes clearer the analogy with linear algebra).
You can see elements of $G_1\oplus \cdots \oplus G_n$ as column vectors with an element of $G_i$ in the $i$'th coordinate.
Then a matrix of endomorphisms acts on such column vectors just by usual matrix multiplication. For example when $n=2$:
$$\left(\begin{array}{ccc}
\varphi_{11} & \varphi_{12}\\
\varphi_{21} & \varphi_{22}
\end{array}\right)
\left(\begin{array}{c}
x_1\\ x_2\end{array}\right)=\left(\begin{array}{c}
\varphi_{11}(x_1) + \varphi_{12}(x_2)\\ \varphi_{21}(x_1) + \varphi_{22}(x_2)\end{array}\right)$$
It's not too hard to show that all endormorphims can be written in matrices. The proof is very similar to the analogous result in linear algebra that every linear map is given by a matrix.

Title: If $G=G_1\times\dots\times G_n$ for abelian $G_i$, then ${\rm Aut}(G)$ is isomorphic to a specific general linear group.

Body:
This is Exercise 1.5.10 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

The Details:

On page 11, ibid., the set product of subsets $X,Y$ of a group is defined as

$$XY=\{xy\mid x\in X, y\in Y\}.$$

On page 17, ibid., for groups $G,H$, the definition of ${\rm Hom}(G, H)$ is given as the set of all homomorphisms from $G$ to $H$.

The definition of a direct product on pages 20 to 21, ibid., is a little more involved.

Let $\{G_\lambda\mid \lambda\in\Lambda\}$ be a given set of groups. The cartesian (or unrestricted direct) product,

$$C=\underset{\lambda\in\Lambda}{{\rm Cr}}\, G_\lambda,$$

is the group whose underlying set is the set product of the $G_\lambda$s [. . .] and whose group operation is multiplication of components: thus

$$(g_\lambda)(h_\lambda)=(g_\lambda h_\lambda),$$

$g_\lambda, h_\lambda\in G_\lambda$. [. . .]

The subset of all $(g_\lambda)$ such that $g_\lambda=1_\lambda$ for almost all $\lambda$ [. . .] is called the external direct product,

$$D=\underset{\lambda\in\Lambda}{{\rm Dr}}\, G_\lambda,$$

[. . .] In case $\Lambda=\{\lambda_1,\dots,\lambda_n\}$, a finite set, we write

$$D=G_{\lambda_1}\times\dots\times G_{\lambda_n}.$$

Of course $C=D$ in this case.

On page 26, ibid., we have that

The set of all automorphisms of [a group] $G$ is denoted by

$${\rm Aut}(G).$$

It is a group under function composition.

The Question:

This is on page 30, ibid.

Let $G=G_1\times\dots\times G_n$ where the $G_i$ are abelian groups. Prove that ${\rm Aut}(G)$ is isomorphic with the group of all invertible $n\times n$ matrices whose $(i,j)$ entries belong to ${\rm Hom}(G_i, G_j)$, the usual matrix product being the group operation.

Thoughts:

I'm used to thinking of the direct product of groups as the group defined by a certain presentation, or else simply as the obvious generalisation of $$H\times K=\{(h,k)\mid h\in H, k\in K\}$$ under the operation $(h,k)\cdot(h',k')=(hh', kk')$.

My linear algebra is a little rusty, which is probably why I'm struggling here.

Examples:

When $n=1$:

We have just the abelian group $G$. The automorphism group is

$${\rm Aut}(G)=\{\varphi\in {\rm Hom}(G,G)\mid \varphi \text{ is an isomorphism}\}$$

and the $1\times 1$ matrices are simply $(\varphi)$ with $(\varphi)^{-1}=(\varphi^{-1})$.

The isomorphism is trivial.

When $n=2$:

We have $G=G_1\times G_2$. Now the $(i,j)$ entries of the $2\times 2$ matrices are such that . . .

Wait a Minute . . .

What operation is the addition of the homomorphisms?

I guess this is my main stumbling block.

Let's try working it out & see . . .

So, we have

$${\rm Aut}(G)\stackrel{?}{\cong}\left\{
\begin{pmatrix}
\varphi_{(1,1)} & \varphi_{(1,2)}\\
\varphi_{(2,1)} & \varphi_{(2,2)}
\end{pmatrix} : \varphi_{(1,1)}\stackrel{?}{\times}\varphi_{(2,2)}\stackrel{?}{-} \varphi_{(1,2)}\stackrel{?}{\times}\varphi_{(2,1)}\neq \stackrel{?}{0} \right\}.$$

Okay, so I'm not sure on multiplication either, nor am I sure of what the zero element is.

This is a mess. I'm sorry.

Please help :)


问:
Non-alternating knot diagrams
Start with a closed, self-intersecting curve, where every crossing is
transverse. Now form something like the opposite of an alternating knot
diagram as follows. Starting anywhere, traverse the curve, and
at each previously unvisited crossing, go over/above. If the crossing has been previously visited, leave the assigned crossing designation.

Two examples are shown below. (a) is clearly the unknot.
(b) is also the unknot, perhaps not as obviously.

Red circle indicates starting point, arrow the traversal direction.

I expected these diagrams to obviously represent the unknot, but
I am not seeing a clear proof. So:

Q. Prove (or disprove) that such a knot diagram always represents
the unknot.

答: A diagram such as you describe is called a descending diagram, and indeed always results in the trivial knot. For a proof, see Lemma 3.2.10 of http://www.math.ucsd.edu/~justin/Roberts-Knotes-Jan2015.pdf. The previous answer has the right idea.

Title: Non-alternating knot diagrams


问:
How can I prove that if $a < b$ then $a^2$ < $b^2$ for $a,b \in \Bbb N$
How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.

答:
Assume $a^2 \ge b^2;$

Then

$a^2-b^2\ge0;$

$(a-b) (a+b) \ge 0$; since $a+b \in \mathbb{Z^+}$

we get by multiplying the two sides of the inequality by $(a+b)^{-1}>0$:

$a-b \ge 0,$ or $a\ge b,$ a contradiction


答:
That's not what $O$ means!

When people write $f(x) = O(g(x))$ (for large $x$) they're not saying something about the specific, exact form of $f$. What this notation means is that there's some (possibly unknown number) $M$ such that
$$|f(x)| \le M |g(x)|$$
for all big enough $x$. Intuitively,

$f(x)$ is no bigger than $g(x)$ for large $x$ - at least up to a constant factor.

In this case, they're saying that

the number of steps sorting takes is no more than $\text{something}\times n\log n$.

the function $n\log n$ isn't any bigger than $An^2$ for large $n$, for some $A$.


问:
Prove that $\int_0^a |f(x)|^2|f'(x)|dx \le \frac{a^2}{3}\int_0^a |f'(x)|^3dx$

If $f \in C^1([0,a])$ and $f(0) = 0$, prove that
$$\int_0^a |f(x)|^2|f'(x)|dx \le \frac{a^2}{3}\int_0^a |f'(x)|^3dx$$
when does the equality hold?

My attempt: Since $f(0) = 0$, we have $$f(x) = \int_0^x f'(t)dt$$
Plugging it into the LHS of the inequality the LHS becomes
$$\int_0^a |f(x)|^2|f'(x)|dx  = \int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx$$
Due to the Cauchy-Bunyakovskiy inequality for $1$ and $f'(t)$, we have
$$\begin{align}
\int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx
&\le \int_0^a\left(\int_0^x 1^2 dt\right)\left(\int_0^x [f'(t)]^2dt \right)f'(x)dx \\
&= \int_0^a \int_0^x xf'(x)[f'(t)]^2dt dx \\
&= \int_0^x [f'(t)]^2dt\int_0^axf'(x)dx \\
&= \left(xf(x)|_0^a - \int_0^a f(x)dx \right)\int_0^x [f'(t)]^2dt \\
&= af(a)\int_0^x [f'(t)]^2dt - \int_0^a f(x)dx \int_0^x [f'(t)]^2dt
\end{align}$$
I got stuck here. I cannot figure out the best way to observe the result.

Any help is greatly appreciated.

答:
One can proceed similarly as in Let $f \in AC[0,1],f(0)=0 $. Show that $\int_0^1 \lvert f(x)f'(x)\rvert\,dx \leq \int_0^1\lvert f'(x)\rvert^2 \, dx$:

We have
$$
|f(x)| = \left| \int_0^1 f'(t) \, dt \right| \le \int_0^x |f'(t)| \, dt =: F(x) \, ,
$$
with equality for all $x$ if and only if $f'$ has constant sign on $[0, a]$.

Then $F'(x) = |f'(x)|$ and
$$
\int_0^a |f(x)|^2|f'(x)|dx  \le \int_0^a F(x)^2 F'(x) \, dx = \frac 13 F(a)^3 \, .
$$

Then $F(a)$ can be estimated Hölder's inequality (with $p=3/2$ and $q=3$), this gives
$$
 F(a) = \int_0^a 1 \cdot |f'(x)|\, dx \le a^{2/3} \cdot
\left( \int_0^a |f'(x)|^3 \, dt \right)^{1/3} \, ,
$$
with equality if and only if $|f'|$ is constant on $[0, a]$.

Combining these estimates gives the desired result.

Remark: In the same way one can prove that
$$
\int_0^a |f(x)|^p|f'(x)| \, dx \le \frac{a^p}{p+1}\int_0^a |f'(x)|^{p+1} \, dx
$$
for any $p \ge 1$.


问:
Variance of non-biased estimator $S^2$ for $\theta$ in $X \sim \text{Poisson}(\theta)$
I am having some trouble finding the $\text{Var}(S^2)$.

I am using this formula: $$\text{Var}(S^2)=\frac{1}{n}\left[\mu_4-\frac{n-3}{n-1}\sigma^4\right]$$

With, $$\mu_4=E\left[(x-\mu)^4\right]$$

To calculate $\mu4$, I have used the central moment generating function: $$M_{x}^{c}(t)=E\left[e^{t(x-\mu)}\right]=\exp{\{-\theta t\}}\exp{\{\theta(e^t-1)\}}=\exp{\{\theta(e^t-1)-\theta t\}}$$

So, according to Wolfram Alpha, $$\mu_4=\frac{\partial^4 M_{x}^{c}(t)}{\partial t^4}\Bigr|_{t=0}=\theta(3\theta+1)$$

Therefore, $$\text{Var}(S^2)=\frac{1}{n}\left[3\theta^2+\theta-\frac{n-3}{n-1}\theta^2\right]=\frac{1}{n}\left[\frac{2n\theta^2+n\theta-\theta}{n-1}\right]$$

But I know that the variance should be: $$\text{Var}(S^2)=\frac{\theta}{n}+\frac{2\theta^2}{n-1}$$

Am I missing something in the process?

Thank you very much!

答:
\begin{align*}
\text{var}(S^2) &= \frac{1}{n}\left( \frac{2n\theta^2 + n\theta - \theta}{n -1} \right)\\
&= \frac{2n}{n(n - 1)}\theta^2 + \frac{n - 1}{n(n - 1)}\theta \\
&= \frac{2}{n - 1}\theta^2 + \frac{1}{n}\theta.
\end{align*}
I have no doubt you can take it from there, comparing this with your desired solution.


问:
Can we count past $ \infty $?
Let's say, $ \infty $ is not the biggest number, in fact, there is something bigger, somewhere. We want to try and count past $ \infty $, because it is not really an ordinal one, but a definition of something uncountable or too big to even count.

What numbers are bigger than $ \infty $?

答:
The symbol $\infty$, as is currently used, simply means that the value is beyond the range supported by the real numbers.  It is not itself a number.  For instance, let's say that we only knew single-digit numbers.  We might have values, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and $\infty$, to represent a value that is outside the numbers we were familiar with.  That's what $\infty$ does.  You can't do any sort of arithmetic with it, because it isn't actually specific enough to be a value.  It's like when your calculator says "error".  You can't actually add two "error" values together, because they aren't really numbers.

However, there are number systems that are beyond the real numbers, and include larger values than are included in the real numbers.  Some of these systems include Cantor's transfinite numbers, the surreal numbers, and various hyperreal systems.

Hyperreal numbers are my favorite, as you can basically treat infinities as ordinary algebraic objects.  You simply have a symbol to represent some particular infinite value (since we can't actually count to infinity, I can't actually tell you which infinite value it represents), and then treat it algebraically.  I usually use $\omega$ for this.

So, in this system, $\omega + 1$ is a distinct number from $\omega$, as is $2\omega$ or $\omega^2$.  You can get infinitesimal numbers by dividing by $\omega$.  $\frac{1}{\omega}$ is closer to zero than any real number, but is still non-zero.  It's really easy to work with, and also simplifies many cases where real numbers have to have a lot of theorems to decide if the value can be contained by the real numbers.  In hyperreal numbers, all numbers essentially act as "finite" numbers even if they are infinite, so all of theorems around deciding if a value is finite or not are largely moot.

Title: Can we count past $ \infty $?


问:
$ \Bbb Q_p^×/(\Bbb Q_p^×)^p$ is isomorphic to $\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$
I would like to prove
$ \Bbb Q_p^×/（\Bbb Q_p^×）^p$ is isomorphic to  $\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$ as a group.

I know $\Bbb Q_p^×$ is isomorphic to $\Bbb Z×\Bbb Z_p×\Bbb Z/（p-1）\Bbb Z$ using formal logarithm.

I tried to find group homomorphism $\Bbb Q_p^×$→$\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$ of which kenerl is $（\Bbb Q_p^×）^p$, but I couldn't.

Thank you in advance.

答:
Is it clear to you that (Hensel lemma) for $p$ odd: $$\Bbb{Q}_p^\times = p^\Bbb{Z}\times (1+p\Bbb{Z}_p)\times \langle \zeta_{p-1}\rangle$$

Then $(1+p\Bbb{Z}_p)^p=1+p^2\Bbb{Z}_p$ because it contains $(1+cp^k)^p\equiv 1+ c p^{k+1}\bmod p^{k+2}$ so any $1+p^2 a$ is of the form $\prod_{k\ge 1} (1+c_k p^k)^p$ with $c_k\in 0 \ldots p-1$.

Whence $$\Bbb{Q}_p^\times/(\Bbb{Q}_p^\times)^p\cong p^\Bbb{Z}/p^\Bbb{pZ}\times (1+p\Bbb{Z}_p)/(1+p^2\Bbb{Z}_p)\cong \Bbb{F}_p\times \Bbb{F}_p$$

For $p=2$ it is the same except that $(1+2\Bbb{Z}_2)^2=1+8\Bbb{Z}_2$ so $\Bbb{Q}_2^\times/(\Bbb{Q}_2^\times)^2\cong \Bbb{F}_2\times \Bbb{F}_2\times \Bbb{F}_2$.


问:
Prove that $N(Z;Y)$ is trivial if and only if there exists a set of $k$ independent functions $g_1,…,g_k$ for $Z$ on some set $U$ in $Y$.
Prove that $N(Z;Y)$ is trivial if and only if there exists a set of $k$ independent global defining functions $g_1,…,g_k$ for $Z$ on some set $U$ in $Y$. That is $Z=\{y∈U:g_1 (y)=0,…,g_k (y)=0\}$

Local submersion theorem: Suppose that $f:X\to Y$ is submersion at $x$ and $y=f(x)$. Then there exist local coordinate around $x$ and $y$ such that $f(x_1,...x_k)=(x_1,...,x_k)$.

$\epsilon$ neighborhood theorem: For a compact boundaryless manifold $Y$ in $R^M$ and $\epsilon >0$, let $Y^\epsilon$ be the open set of points in $R^k$ with distance less than $\epsilon$ from $Y$. If $\epsilon$ is sufficiently small then each point $w\in Y^\epsilon$ possesses a unique closest point in $Y$, denoted $\pi(w)$. Moreover the map $\pi : Y^\epsilon \to Y$ is a submersion.

Tubular neighborhood theorem there exist a diffeomorphism from an open neighborhood $Z$ in $N(Z;Y)$ onto an open neighborhood of $Z$ in $Y$

Trivial normal bundle proposition: Let $Z$ be a submanifold of codimension $k$ in $Y$, then $N(Z;Y)$ is trivial if there exists a diffeomorphism $\phi :N(Z;Y) \to Z\times R^k$ that restrict to a linear isomorphism $N_z (Z;Y)\to {z}\times R^k$ for each point $z\in Z$.

Here is what I got so far.

=>

Assume that $N(Z;Y)$ , the normal bundle of $Z$ in $Y$, is trivial, then by Tubular neighborhood theorem, there exist a diffeomorphism from an open neighborhood $Z$ in $N(Z;Y)$ onto an open neighborhood of $Z$ in $Y$. Let $U$ be an open neighborhood in $Y$, then $g:U \to R^k $ is submersion, thus $U$ can be cut out by many independent functions $g_1, \dots, g_k$

I feel like I'm saying something really stupid here.

<=

Assume that there exists a submersion $g:U \to R^k $ with $g^{-1} (0)=Z$. For this part, the book give me a hint that for each $z\in Z$ the transpose map $dg_z ^t:R^k \to T_z(Y)$ carries $R^k$ isomorphically onto the orthogonal complement of $T_z(Z) $ in $T_z(Y)$. But I don't know how they get this.

答:
Let $S$ be an embedded submanifold of $\mathbb{R}^n$ of codimension $k$. Let $NS$ denote the normal bundle and $\pi_{NS}:NS \rightarrow S$ the projection.

Suppose first that there is a smooth defining function $\Phi:U \rightarrow \mathbb{R}^k$ on a neighborhood $U$ of $S$ in $\mathbb{R}^n$, say $S = \Phi^{-1}(c)$. Consider the map $\Psi: NS \rightarrow S \times \mathbb{R}^k$ (identifying $T_c\mathbb{R}^k$ with $\mathbb{R}^k$) given by $\Psi(v) = (p, d\Phi_p v)$ for $v \in N_pS$ and $p \in S$. If $\delta: NS \rightarrow NS \times NS$ is the diagonal map, then $\Psi = (\pi_S \times d\Phi|_{NS}) \circ \delta$, so $\Psi$ is smooth. Note that $\pi_S \circ \Psi = \pi_{NS}$, where $\pi_S:S \times \mathbb{R}^k \rightarrow S$ is the projection, so $\Psi$ is a smooth bundle homomorphism over $S$. For $p \in S$, $T_pS = \text{Ker}(d\Phi_p)$, so $d\Phi_p|_{N_pS} = (d\Phi|_{NS})_p:N_pS \rightarrow \mathbb{R}^k$ is a linear isomorphism. The map $\Gamma: S \times \mathbb{R}^k \rightarrow NS$ defined by $\Gamma(p,w) = (d\Phi|_{NS})_p^{-1} w$ is thus an inverse for $\Psi$.
We now show that $\Gamma$ is smooth. Let $(E_1, \ldots, E_k)$ be a smooth local frame for $NS$ on an open set $V$ in $S$. Then $d\Phi|_{NS} \circ E_i = A_i^j \frac{\partial}{\partial x^i}\big|_c$ for some smooth functions $A^j_i$ on $V$, namely
\begin{equation*}
 A^j_i(p) = \big((d \Phi|_{NS})_p \circ E_i|_p \big)(x^j) = E_i|_p(\Phi^j) = (E_i \Phi^j)(p)
\end{equation*}
for $p \in V$. Since $(d\Phi|_{NS})_p$ is an isomorphism for $p \in V$, the matrix $A^j_i$ is (smoothly) invertible. Let $B$ denote the inverse matrix. Then, for $p \in V$,
\begin{equation*}
 \Gamma(p,w) = w^i B_i^j(p) E_j|_p,
\end{equation*}
so $\Gamma$ is smooth. Since $\Gamma$ is also a bundle homomorphism over $S$, $NS$ is trivial.

Suppose conversely that $NS$ is trivial. By the tubular neighborhood theorem, there is a neighborhood $U$ of $S$ in $\mathbb{R}^n$ and a diffeomorphism $E:V \rightarrow U$, where $V \subset NS$ is an open set of the form
\begin{equation*}
 V = \{(p,v) \in NS : |v|< \delta(p)\}
\end{equation*}
for some continuous function $\delta:M \rightarrow \mathbb{R}$. By construction, $E(p,0) = p$ for $p \in S$, so $E^{-1}(q) \in N_{\pi_{NS}(E^{-1}(q))}S$ is nonzero for any $q \in U \smallsetminus S$. Let $\Theta: NS \rightarrow S \times \mathbb{R}^k$ be a smooth bundle isomorphism, and let $\pi_k:S \times \mathbb{R}^k \rightarrow \mathbb{R}^k$ denote the projection onto $\mathbb{R}^k$. Then $\Phi = \pi_k \circ \Theta \circ E^{-1}:U \rightarrow \mathbb{R}^k$, as a composition of smooth submersions, is a smooth submersion. For $p \in S$, since $\Theta|_{N_pS}: N_pS \rightarrow \{p\} \times \mathbb{R}^k$ is a linear isomorphism and $E^{-1}(q) \in N_{\pi_{NS}(E^{-1}(q))}S$ is zero if and only if $q \in S$, $S = \Phi^{-1}(0)$.


问:
Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$

$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$

Find $x$.

I am not sure how to proceed.

Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?

Any help would be sincerely appreciated.

答:
$$x = \sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4-\cdots }}}} $$
If I put write the recurring equation of this statement
$f[n] = \sqrt{n-(-1)^{n}\cdot f[n+1] }$
But the value of $x$ is a complex number so recurring equation can't reveal much, But instead if we mirror $x$, by a change of sign from $+ $ to $ - $

We get it's conjugate
$$ y = -\sqrt{1-\sqrt{2+\sqrt{3-\sqrt{4+\cdots }}}} $$
$y$ is also a complex number, but conjugate to $x$
$x$ and $y$ can therefore appear as the roots of some quadratic, so that $x + y $ and $ x \cdot y $ are real

What remains is to reduce the identities


答:
$f(x)=\mathcal{O}(g(x))$ means that $f$ is asymptotically smaller or equal to $g$. This means that $|f(x)|\le c|g(x)|\quad \forall x$ for some constant c.

Now in your example, $$n\log n \le n^2$$

This does not say that $n=\log n$, but instead, that $\log n\le n$. In other words, that $\log n$ is asymptotically smaller or equal to $n$, or, in Big-O notation, $\log n = \mathcal{O}(n)$.