问:
Complex conjugate doubts
$$
\frac{\bar{z}+i}{1-i\bar{z}}
$$

I wanted to know what is the conjugate to simplify the equation from above, i mean:

Would it be like this:
$$
\overline{(1-i\bar{z})}=1+i\bar{z}
$$
Or:
$$
\overline{(1-i\bar{z})}=1+iz
$$
since $\overline{\bar{z}}=z$

答:
$$\overline{(1-i\bar{z})}=1+iz
$$
is the correct way

So the fraction will now become

$$
\frac{z-i}{1+iz}
$$

Title: Complex conjugate doubts


问:
Proving that this right triangle $ABC$ is isosceles
Having sides of lengths $a, b$ and hypotenuse $c$, show that if $\sin(A)=\sqrt{a/(2b)}$, then the triangle $ABC$ is isosceles.
Here's a picture:

The only things I can think of is that $\sin A=a/c$ and the  pythagorean theorem but I don't know where to start. I would appreciate any hints.

答:
Hint

$$\sin^2 A=\frac{a^2}{c^2} =\frac{a}{2b}\to c^2=2ab$$

by Pithagoras,

$$c^2=a^2+b^2$$

can you finish?

ContentLicense: CC BY-SA 4.0


问:
A more fair way to evaluate performance than average and mode
Hi,

I want to evaluate the overall performance of a soccer team.
The team has participated in 5 tournaments so far and this is their rank in each of them

```<code>#4
#5
#6
#7
#200
</code>```

you cant get the mode because there is none as they got a different rank each time, whereas using average to evaluate their overall performance would be very unfair since that single time they got a 200th would completely overshadow their other performances which were quite good. I think a fair number to represent their performance would be something between 5 and 7.

What is used in this cases?

Thank you.

答:
Median sounds like what you are looking for. Using individual tournament ranks though would be reasonable only if each tournament was almost identical, otherwise the ranks would have to be weighted somehow.

You may be better off looking at individual performance game by game in each of the tournaments.


问:
$y^{(4)}+3y''-4y=\sinh(x)-\sin^2(x)$
Hello to all whom might to assist!
$$
y^{(4)}+3y''-4y=\sinh(x)-\sin^2(x)\\
$$
for the homogenous part i come with the following solutions based on the roots:
$$
s_{1,2}= \pm 1\\
s_{1,2}= \pm i2 \\
thus:\\
y_c = C_1e^x+C_2e^{-x}+C_3cos(2x)+C_4sin(2x)
$$
continue to solve for the particular solution i substituted
$$
\sinh(x)-\sin^2(x) = \frac{e^x-e^{-x}}{2}+\frac{1}{2}-\frac{1}{2}\cos(2x)\\
$$
so the "guess" for the particular solution is in the form of:
$$
y_p= A+Bxe^x+Cxe^{-x}+Dx\sin(2x)+Ex\cos(2x)
$$
so, my inquiry is basically if i am on the right way, i.e. all what is left is derivative 4 times(!) and to plug in..
or is there simpler way?
Thanks  advance!

答:
You are on the right path.
$$y^{(4)}+3y''-4y=\sinh(x)-\sin^2(x)$$
$$(y^{(4)}+4y'')-(y''+4y)=\sinh(x)-\sin^2(x)$$
$$z''-z=\sinh(x)-\sin^2(x)$$
Where $z=y''+4y$.

Solve the first DE:
$$z''-z=\sinh(x)$$
The guess for the particular solution should be:
$$\implies z_p=Ax \cosh x$$

Title: $y^{(4)}+3y''-4y=\sinh(x)-\sin^2(x)$


问:
How can I evaluate this sum?
How can I evaluate this sum ?

$$\sum_{n=1}^{m}n^p$$

I know when $p\in \Bbb N$ but when $p\in\Bbb R$ what do I do ?

please help and thanks for all

答:
If $p \in \mathbb{N}$, you have the Faulhaber's formula, i.e.,
$$\sum_{n=1}^m n^p = \dfrac1{p+1} \sum_{j=0}^p (-1)^j \dbinom{p+1}j B_j m^{p+1-j}$$
where $B_j$ are the Bernoulli numbers, with $B_1 = - \dfrac12$. If $p \not\in \mathbb{N}$, we can still obtain the asymptotic behavior as
$$\sum_{n=1}^m n^p \sim \begin{cases}\dfrac{m^{p+1}}{p+1} & \text{ if } p \neq -1\\ \log(m) + \gamma & \text{ if } p =-1\end{cases}$$</p>

ContentLicense: CC BY-SA 3.0

Title: How can I evaluate this sum?


问:
$\lim_{n\to\infty} x^{1+\frac{1}{2n+1}} = |x|$ on the set $[-1,1]$
How does $\lim_{n\to\infty} x^{1+\frac{1}{2n+1}} = |x|$ on the set $[-1,1]$?

I have $$\lim_{n\to\infty} x^{1+\frac{1}{2n+1}} = x(\lim_{n\to\infty}x^{\frac{1}{2n+1}}) = x\cdot x^0 = x$$

What am I doing wrong?

答:
Check this similar question.
Basically we have $x^{\frac{2n+2}{2n+1}}=(x^2)^ {\frac{n+1}{2n+1}}$. Then take the limit and you will have your answer.


问:
defining conjugate diameters
I was asked the following question:

"Prove analytically Proposition VII-12 that in any ellipse the sum of the squares on any two of its conjugate diamters is equal tot the sum of the squares on it's two axes."

What are conjugate diameters? Are they diameters that make right angles with each other?

答: In a comment, Blue already mentioned the definition from Wikipedia. To illustrate things, I created a GeoGebra file. You can download it here. Save the file with a .ggb extension and open it with GeoGebra.

Title: defining conjugate diameters


问:
Minimizing quadratic objective over convex set
Given the vectors $x_0, \theta \in \mathbb{R}^n$, $\varepsilon\in \mathbb{R}^+$, and symmetric positive definite matrices $A, \Sigma\in \mathbb{R}^{n\times n}$, how the following optimization problem can be solved (possibly in closed form)?

$$\begin{array}{ll} \underset{x \in \mathbb{R}^n}{\text{minimize}} & (x - x_0)^\top A (x - x_0)\\ \text{subject to} & \sqrt{x^\top \Sigma x} + x^\top \theta \leq \varepsilon\end{array}$$

What I tried:
First, note that the problem is convex since both the objective is convex in $x$, and the constraint defines a convex set in $x$.

Write the Lagrangian of the problem as

$$L(x,\lambda) = (x-x_0)^\top A (x-x_0) + \lambda (\sqrt{x^\top \Sigma x} + x^\top \theta  - \varepsilon). $$ The KKT conditions for this problem leads to the system:
$$
\begin{cases}
A (x-x_0) + \lambda \left(\frac{\Sigma x}{\sqrt{x^\top\Sigma x}}+ \theta \right) = 0
\\ \lambda\left(\sqrt{x^\top \Sigma x} + x^\top \theta  - \varepsilon \right) = 0 \\ \sqrt{x^\top \Sigma x} + x^\top \theta  - \varepsilon \leq 0\\\lambda\geq 0
\end{cases}
$$
However, the solution of this system gets complicated in my calculation and I am not able to find a closed-form solution. Any suggestion?

答:
If you change coordinates to the eigenbasis of $A$ with lengths of eigenvectors equal to square roots of the corresponding eigenvalues, the objective becomes distance from a point. The constraint set is cut out by a quadratic equation (and remains such after this linear coordinate change), which by further orthonormal coordinate change (diagonalizing $\Sigma-\theta^T\theta$) and a shift you can \textbf{usually} make have equation $d_1 x_1^2+\ldots d_n x_n^2\leq1$. So your problem is equivalent, up to affine coordinate changes, to finding the closest point of the "standard quadratic" with the equation above, to some fixed point in space. If "the point in space" is inside the constraint set (i.e. in the original coordinates $x_0$ satisfies the constraint) then the distance is zero; otherwise the optimum is on the boundary, but while you can reduce the resulting constrained optimization to a single-variable problem of finding the Lagrange multiplier, see this question, for example, it seems unlikely that you will get a closed-form solution (not a guarantee, perhaps the optimal value can be found, even if finding the optimal point is harder).

Here are some details:

The initial coordinate change is from $A=U^T \Delta U=(U \Delta^{1/2})^T (U\Delta^{1/2})=M^TM$  so $(x-x_0)^TA(x-x_0)=(U \Delta^{1/2}(x-x_0))^T (U\Delta^{1/2} (x-x_0)$ and we set $y=U \Delta^{1/2}(x-x_0)$ so that the objective is $y^Ty$.
The original $x$ is recoverable as $x=(\Delta^{-1/2}U^Ty)+x_0=M^{-1}y+x_0$.

The constraint set is

$\sqrt{x^T \Sigma x} + x^T \theta \leq \varepsilon$

$\sqrt{x^T \Sigma x} \leq   (\varepsilon - x^T \theta)$

Since $\Sigma$ is positive definite, the expression under the root is non-negative and this is equivalent to

$x^T \Sigma x  \leq   (\varepsilon - x^T \theta)^2=\varepsilon^2-2 \varepsilon x^T\theta+ x^T \theta \theta^T x$

$x^T(\Sigma - \theta \theta^T )x +2 \varepsilon x^T\theta -\varepsilon^2\leq 0$

$(M^{-1}y+x_0)^T(\Sigma- \theta \theta^T)(M^{-1}y+x_0)^T +2 \varepsilon (M^{-1}y+x_0)^T\theta -\varepsilon^2\leq 0$

Collecting terms

$$y^T(M^{-1})^T(\Sigma- \theta \theta^T) (M^{-1})y$$

$$+  y^T (M^{-1})^T (\Sigma- \theta \theta^T)x_0+ x_0^T (\Sigma- \theta \theta^T) (M^{-1}) y+ 2 \varepsilon (M^{-1}y)^T\theta$$

$$x_0^T (\Sigma- \theta \theta^T) x_0+ 2 \varepsilon x_0^T\theta -\varepsilon^2<0$$

Or $$y^T Q y + y^T\beta +c\leq 0,$$ where $Q=(M^{-1})^T(\Sigma- \theta \theta^T) (M^{-1})$. Now, $Q$ is symmetric, so $Q=V^TDV$ with orthogonal $V$, and we set $z=Vy$. The objective is still $y^Ty=z^Tz$. The constraint is now in the form

$z^TDz+z^T\gamma+k \leq 0$.

Now we pass to coordinates and play "complete the square":

$\sum D_i z_i^2 + \gamma_i z_i \leq -k$

For those $D_i$ that are non-zero we get

$D_i z_i^2 + \gamma_i z_i=D_i (z_i-\frac{\gamma_i}{2D_i})^2-(\frac{\gamma_i}{2D_i})^2=D_i w_i^2+ n_i$

So if all $D_i$ are non-zero (i.e. if $\Sigma-\theta^T\theta$ is non-singular), we have  the constraint rewritten as $\sum D_i w_i^2< N$  and if $N\neq 0$ the constraint is indeed
$\sum d_i w_i^2\leq  1$  (where $d_i=D_i/N$).

The objective is $z^Tz=(w-w_0)^T(w-w_0)$ where $(w_0)_i=-\frac{\gamma_i}{2D_i}$.

So, overall we are optimizing distance from a point to a region defined by a quadratic - which can be somewhat standartized, and, depending on various parameter values, "often" completely diagonalized(in practice one would need to pay attention to various conditioning issues).


问:
Why do these two volume integrals express different values?
This is a question from a textbook I am reading:

Find the volume when the region R is revolved about the y-axis
  when R is defined as the region bounded by $x = \sqrt{16-y^2}$ and
  $x = 0$.

Because the function is symmetrical about the y-axis and being revolved about the y-axis, why don't these two integrals evaluate to the same value?

Integral one: $$\int_0^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{128\pi}3$$

Integral two: $$\int_{-4}^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{256\pi}3$$

It seems to me that they represent the same volume--like creating a semicircular volume by revolving a fourth of a circle about its base, or by revolving a semicircle along its base.

I think that the second equation overestimates the volume because it counts the same volume twice. (The textbook says otherwise!)

答:
The function is not symmetrical about the $y$-axis, indeed the curve "lives" in the first and fourth quadrants.

There is symmetry about the $x$-axis, which is why one integral is twice the other.

Remark: Since you are much more familiar with integration with respect to $x$, you might interchange the roles of $x$ and $y$ (geometrically: reflect in the line $y=x$). Then you will see more clearly what is going on. The first integral gives the volume of a half-sphere. The second gives the volume of a full sphere.


问:
Alternative solution for hypergeometric equation
I need to get another solution for

$$ x(1-x) \frac{\partial^{2}}{\partial x^2} y(x) + (c-(a+b+1)x)\frac{\partial}{\partial x} y(x) - aby(x) = 0$$

by substituting with $y=x^{1-c}z$. When I do the substitution I get really long derivatives there, but in the end the answer should look like

$$ x(1-x) \frac{\partial^{2}}{\partial x^2} z(x) + ((2-c)-((a-c+1)+(b-c+1)+1)x)\frac{\partial}{\partial x} z(x) - (a-c+1)(b-c+1)z(x) = 0$$

So there is no $x^{1-c}$ in this equation which leads me to think that I am doing something wrong.

Thanks for any hints

答:
You did not differentiate properly $y=z(x)x^{1-c}$

You should get $y'=x^{1-c} z'(x)+(1-c) x^{-c} z(x)$

and $y''=x^{1-c} z''(x)+2 (1-c) x^{-c} z'(x)-(1-c) c x^{-c-1} z(x)$


问:
Proof critique of least number principle, please!
I am independently working through Elliot Mendelson's "Number Systems and the Foundations of Analysis," which I find very well-written and rigorous, along with providing a healthy set of exercises. I have recently finished reading the order relation in Peano Systems section. The section finishes with the Least Number Principle theorem; the proof is (to me) somewhat jumbled and, at times, not precisely rigorous. As such, I wanted to rewrite it myself, filling in the gaps. EDIT- I can now see why many explicit steps were perhaps omitted or left to the reader =), this got quite tedious, upon completion.

(Let $P$ denote the underlying set of a Peano system with distinguished element $1$, unary operation S (successor) on $P$, and the familiar binary operation and relation $+$ and $<$, respectively, on $P$.)

Theorem 4.5: (Least Number Principle) Any non-empty subset of $P$ has a least element.

Proof: Let $A$ be a non-empty subset of $P$. Suppose, to the contrary, $A$ has no least element. Let $B := \{x \in P\  |\ (\forall u \in P)(u \leqslant x \implies u \notin A\}$. First, we show by mathematical induction that $B=P$. Note that $1\in B$, for if not: $1\notin P \lor (\exists u \in P)(u\leqslant 1\ \land u\in A)$. $1\in P$ by the definition of Peano system, so it must be the case that $(\exists u \in P)(u\leqslant 1\ \land u\in A)$. By Lemma 1, it follows that $u=1$, so $1\in A$. But by Lemma 2, then, $1$ would be a least element of $A$, a contradiction. Thus, $1\in B$. Next, we must show that $(y\in B \implies S(y) \in B)$, so assume $y\in B$.
This implies $(\forall u \in P)(u \leqslant y \implies u \notin A\}$. Suppose $S(y) \in A$. But if so, we can prove $S(y)$ to be a least element of $A$. To do so, let $v\in P$ and assume too that $v \in A$. But by the contrapositive of $(\forall u \in P)(u \leqslant y \implies u \notin A\}$ applied to $u=v$, this implies $y < v$. But by lemma 3, $y<v \iff S(y) \leqslant v$. But this means $S(y)$ is a least element of $A$, a contradiction. Hence, $S(y) \notin A$. Now, to prove that $S(y) \in B$, we must show that $(\forall u \in P)(u \leqslant S(y) \implies u \notin A\}$, so let $u\in P$ be arbitrary and suppose $u \leqslant S(y)$. If $u = S(y)$, then $S(y) \notin A$, as we have shown above. If $u < S(y)$, then by Lemma 4, $u \leqslant y$, in which case $u \notin A$ by the inductive step hypothesis, i.e., $(\forall u \in P)(u \leqslant y \implies u \notin A\}$. Hence, $S(y)\in B$. We have thus shown by induction that $B=P$. Note by Lemma 5 that $A\ \cap B = \emptyset$. But $B=P$, so $A\ \cap P = \emptyset$. By lemma 6, this implies $A = \emptyset$, since $P \ne \emptyset$, as $1\in P$. But $A$ is non-empty by hypothesis above, thus yielding a contradiction. As a result, $A$ must possess a least element.

$\\$

$\\$
Theorem 1.1 Every element different from $1$ is a successor.

Proof: (provided in text)

Lemma 1. $(\forall x\in P)(x\leqslant 1 \implies x=1)$.

Proof: Let $x\in P$ and suppose $x\leqslant 1$. If $x=1$, then we have proven our desired result. If $x < 1$, then by the definition of $<$, $x + u = 1$ for some $u\in P$. If $u=1$, then $x+1 = 1$, or $S(x) = 1$. But this contradicts an axiom of Peano systems; namely, that $(\forall x \in P)(1 \ne S(x))$ If $u\ne 1$, then by theorem 1.1 above, $u = S(w)$ for some $w\in P$. So $x  + u = x + S(w) = x + (w+1) = (x+w) + 1$, where the second equality holds by the definition of $+$ in $P$ (defined and proven in text, i.e., $(\forall x\in P)(x+1 = S(x))$), and the third equality holds via the associativity of $+$ on $P$ (also proven in text). Hence, $1 = S(x+w)$, which is again a contradiction to the same Peano axiom referenced in the previous sentence. As a result, $\lnot(x < 1)$, so it must be that $x=1$.

Lemma 2. $(\forall N\subseteq P)(1\in N \implies 1$ is a least element of $N)$.

Proof: Let $N\subseteq P$ be arbitrary and suppose $1\in N$. To prove that $1$ is a least element of $N$, we must prove that $(\forall u\in P)(u\in N \implies 1 \leqslant u)$. Thus, let $u\in P$ and suppose also $u\in N$. By theorem 1.1 above, either $u=1$ or $u=S(x)$ for some $x\in P$. If $u=1$, then clearly $1\leqslant u$, since $1 \leqslant 1$. If $u=S(x)$ for some $x\in P$, then $u = x+ 1 = 1 + x$, so by the definition of $<$, $1<u$; so $1\leqslant u$. In either case, $1\leqslant u$, so $1$ is a least element of $N$.

Lemma 3. $(\forall x,y \in P)(x < y \iff S(x) \leqslant y)$.

Proof: (provided in text)

Lemma 4. $(\forall x,y \in P)(x < S(y) \implies x \leqslant y)$.

Proof: (provided in text)

Lemma 5. $A\ \cap B = \emptyset$.

Proof: We must show that $(\lnot\exists z\in P)(z \in A\ \land z \in B)$. In other words, $(\forall z \in P)(z\in A \implies z\notin B)$. To prove this statement, let $z\in P$ be arbitrary. Suppose, furthermore, $z\in A$. If $z\in B$, then by the definition of $B$, $(\forall u \in P)(u \leqslant z \implies u \notin A\}$. Applying this statement for $u=z$, and noting that $z=z$, so clearly $z\leqslant z$, we may conclude that $z\notin A$. But this is a contraction, as $z\in A$ by hypothesis above. Hence, it must be the case that $z\notin B$. Since $z$ was arbitrary, $(\forall z \in P)(z\in A \implies z\notin B)$. So $A\ \cap B = \emptyset$.

Lemma 6. $(\forall X \subseteq P)(X \cap P = \emptyset \implies X = \emptyset)$.

Proof: Suppose $X\subseteq P$ and let $X \cap P = \emptyset$. Then $(\lnot \exists x\in P)(x\in P \wedge x\in X)$. Suppose $X  \ne \emptyset$. Then $\exists y\in X$. But $X\subseteq P$, so $y\in P$. But this means $y\in X$ and $y\in P$, or $y\in (X \cap P)$. But this is a contradiction, as $X \cap P = \emptyset$ by hypothesis. Thus, $X = \emptyset$.

答:
Your proof seems correct; I can't find any errors other than a few typos. However, it's not totally clear to read, nor is it the nicest thing to look at. Here are some tips on how to make it better:

First, excessive use of symbols can obscure your message. Your definition of $B$ made me do a double-take. I would re-write a lot of these in plain English, maybe followed by the symbolic definition if you think it's necessary. Something along the lines of "Let $B$ be the set of all $x\in P$ such that no $u\leq x$ is an element of $A$; in other words, $B=\{x\in P\,|\, (u\leq x)\implies (u\not\in A)\}$." (Incidentally, I'm not sure that your use of $\forall u\in P$ is strictly necessary; the domain of discourse is unambiguous to me.)

Further, paragraphs are critically important with any proof that's more than a few sentences. Make a point, prove it, new paragraph. This is especially important when you're working with something I like to call "nested contradiction," where you have a proof by contradiction with cases that also use contradiction. That statement "Suppose $S(y)\in A$" is where I got confused and needed to re-read a few times. To that end, I might structure the proof like this:

Let $A$ be a nonempty subset of $P$ and suppose to the contrary that $P$ has no least element. Define the set $B$ as (insert definition here). We will proceed by induction to show that $B=P$.

First, we show that $1\in B$. (Reasons, etc). So $1\in B$.

Now, we show that $(y\in B)\implies (S(y)\in B)$. To do this, let $y\in B$. We will consider two cases: either $S(y)\in A$ or $S(y)\not\in A$. (Telling us what you will do before you do it adds organization and makes it easier to follow along.)

In the case $S(y)\in A$, (insert reasons here) which means that $S(y)$ is a least element of $A$. But this is a contradiction because $A$ has no least element, so clearly $S(y)\not \in A$.

So, $S(y)\not\in A$, and (reasons reasons reasons). Therefore, $S(y)\in B$.

(You just completed a big step in the proof. Recap it.) We showed above that $1\in B$ and $(y\in B)\implies (S(y)\in B)$. Therefore, $B=P$.

(Finish the proof here) and so every subset of $P$ has a least element. QED


问:
Statistics task
I have very simple math task to solve but as am not a math person, I would kindly ask for your help.
The task: 30% of students failed test.the test average score was 7.5 and standart deviation was 1.3. What was the needed score to pass the test?
That's all info I have. Hope it makes sense. Thank you</p>

答:
Assume a normal distribution of test scores. Approximately 68% of the marks will lie inside 1 standard deviation. So the approximately 16% of the students will have marks less than 7.5-1.3=6.2

Since 50% of the students have marks less than 7.5, it's reasonable to set a passing score of 7.5-(1.3/2)=6.85

Title: Statistics task


问:
Finding the number of non-neg integer solutions?
How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.</p>

答:
Hint: Consider the following $3$ cases.

$x_1=0$, $x_1=1$ and $x_1=2$.

Now you have reduced the problem to a standard distribution question where $$x_2+x_3+x_4=12,11,10 \; \; \forall x_2,x_3,x_4 \in \mathbb{N_0^+}$$


问:
Killing singularities of differential forms via multiplying with a function which is 1 on the boundary of the domain of integration.
The most famous differential form is probably

$$\textrm{d}\theta =  \frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y$$

Warning:  $\textrm{d}\theta$, despite appearances, is not the exterior derivative of any function $\theta : \mathbb{R}^2 \to \mathbb{R}$.  It is locally of such a form however.  On any simply connected open set $U$ of $\mathbb{R}^2-\{(0,0)\}$ one can define a function $\theta_U : U \to \mathbb{R}$ whose differential agrees with $\textrm{d}\theta$ on $U$.

There are several ways to integrate this around the unit circle:

Parameterize the unit circle and integrate in the parameter space.

Cover the unit circle with simply connected open sets where the form is exact (for instance $\textrm{d}\theta = \textrm{d} \arctan(\frac{y}{x}) $ when $x>0$), chop the circle into pieces whose endpoints are in the overlaps of the open sets, and use the fundamental theorem of calculus for line integrals on each segment.

There is a third way which doesn't seem to be discussed as often:  multiply this form by a smooth function which kills the singularity at the origin, but does not alter the form on the unit circle, and then apply Stoke's theorem.  Here is that calculation in detail:

$$
\begin{align}
\int_{S^1}  \frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y & = \int_{S^1}  (x^2+y^2)\left(\frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y\right)\\
&=\int_{S^1} -y \textrm{d}x + x\textrm{dy}\\
&= \int_{D(0,1)} \textrm{d}\left( -y \textrm{d}x + x\textrm{dy} \right)\\
&= \int_{D(0,1)} 2 \textrm{d}x \wedge \textrm{d}y\\
&= 2\pi
\end{align}
$$

Here the first equality is justified by the fact that $x^2+y^2 = 1$ on $S^1$, the third equality is Stoke's theorem, and the last equality is true because the area of the unit disk is $\pi$.

It is interesting that this argument relates the circumference of the circle to its area via Stoke's theorem.  A slight variant of this calculation can prove that $C = 2A/r$ for any circle of radius $r$.

This trick seems widely applicable:  if you have a form which is defined off of a singular set, you can integrate along a boundary by first multiplying the form by a function which is 1 on the boundary and kills the singularity, and then applying Stoke's theorem to the new "singularity free" form.  It has the feeling of being an important trick to me, one which can be used over and over again in many diverse situations.

My question:  where has this trick been used?  I would like a "big list" of applications.

答:
Consider the differential equation $Mdx+Ndy=0$ where $M,N$ have domain of definition $U \subseteq \mathbb{R}^2$. If $\partial_y M = \partial_x N$ on $U$ and $U$ is simply connected then there exists $F: U \rightarrow \mathbb{R}$ such that $dF = Mdx+Ndy$ and hence the solution set of $F(x,y)=c$ constitutes a solution of the exact differential equation $Mdx+Ndy=0$. We say $I$ is an integrating factor of the inexact differential equation $Mdx+Ndy=0$ if the corresponding differential equation $IMdx+INdy=0$ is exact. That is, there exists $\phi$ for which $d \phi = IMdx+INdy$. Upon writing this it is clear my initial hunch seems completely off-base.

In fact, this example is almost exactly backwards of the integrating factor method! The initial form
$$ \omega = \frac{-ydx+xdy}{x^2+y^2} $$
is closed on its domain of definition as follows from the calculus below:
$$ \frac{\partial}{\partial y} \left[ \frac{-y}{x^2+y^2} \right]  = \frac{y^2-x^2}{(x^2+y^2)^2} \qquad \& \qquad \frac{\partial}{\partial x} \left[ \frac{x}{x^2+y^2} \right]  = \frac{y^2-x^2}{(x^2+y^2)^2}. $$
In fact, multiplying by $x^2+y^2$ spoils the closed condition. Note, multiplication by $x^2+y^2$ in the OP's example yielded $-ydx+xdy$ and $d(-ydx+xdy) = 2dx \wedge dy$ hence $-ydx+xdy$ is not closed which indicates $-ydx+xdy$ is not exact.

Instead, I think the way to understand the calculation is more rooted in the theory of residues.

In complex notation $z = x+iy$ and $dz = dx+idy$ so formally $x = \frac{1}{2}(z+\bar{z})$ and $y = \frac{1}{2i}(z- \bar{z})$ and $dx = \frac{1}{2}(dz+d\bar{z})$ and $dy = \frac{1}{2i}(dz- d\bar{z})$
\begin{align}
\frac{-ydx+xdy}{x^2+y^2} &= \frac{-\frac{1}{2i}(z- \bar{z})\frac{1}{2}(dz+d\bar{z})+\frac{1}{2}(z+\bar{z})\frac{1}{2i}(dz- d\bar{z})}{z\bar{z}} \\
&= \frac{-1}{4iz\bar{z}}\left(zdz-\bar{z}dz+zd\bar{z}-\bar{z}d\bar{z}
-(zdz+\bar{z}dz-zd\bar{z}-\bar{z}d\bar{z}) \right) \\
&= \frac{-1}{4iz\bar{z}}\left(-2\bar{z}dz+2zd\bar{z}\right) \\
&= \frac{\bar{z}dz}{2iz\bar{z}} -  \frac{zd\bar{z}}{2iz\bar{z}} \\
&= \frac{1}{2i}\left[\frac{dz}{z} -  \frac{d\bar{z}}{\bar{z}}\right] \\
\end{align}
Then integration around a CCW picks up $2\pi i$ from the $\frac{dz}{z}$ term and $-2\pi i$ from the $\frac{d\bar{z}}{\bar{z}}$ term. Hence $\frac{1}{2i}(2\pi i-(-2\pi i)) = 2\pi$.

Can we come up with an example which moves beyond just a single singularity ? It's a bit silly, but, I'd try:
\begin{align} \gamma &= \omega+ \omega_1  \\
&= \frac{-ydx+xdy}{x^2+y^2} + \frac{-(y-y_1)dx+(x-x_1)dy}{(x-x_1)^2+(y-y_1)^2} \\
&= \frac{1}{2i} \left( \frac{1}{z} + \frac{1}{z-z_1}\right)dz -
\frac{1}{2i} \left( \frac{1}{\bar{z}} + \frac{1}{\bar{z}-\bar{z}_1}\right)d\bar{z}.
\end{align}
This form is closed except at its singularities $z=0$ and $z = z_1$. This gives us the freedom to adjust line integrals by deforming curves until they encircle the appropriate singularities. Suppose $C$ is a CCW-curve which encloses both $z=0$ and $z=z_1$. Then, I propose we can deform $C$ into CCW-oriented circles $C_0$ and $C_1$ of radii $R_0$ and $R_1$ respective. Then on each circle I can use the OP's trick: multiply by $\frac{1}{R_0^2}z\bar{z} = 1$ for $C_0$ and multiply by $\frac{1}{R_1^2}(z-z_1)(\bar{z}-\bar{z}_1) = 1$ for $C_1$. Then,
$$ \frac{1}{R_0^2}z\bar{z}\gamma = \frac{1}{2iR_0^2} \left( \bar{z}dz-zd\bar{z}+\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1} \right)$$
likewise,
$$ \small\frac{1}{R_1^2}(z-z_1)(\bar{z}-\bar{z}_1)\gamma = \frac{1}{2iR_1^2} \left( (\bar{z}-\bar{z}_1)dz-(z-z_1)d\bar{z}+\frac{(z-z_1)(\bar{z}-\bar{z}_1)dz}{z}-\frac{(z-z_1)(\bar{z}-\bar{z}_1)d\bar{z}}{\bar{z}} \right)$$
I suspect the non-singular (on $C_0$ and $C_1$ respective) terms $\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1}$ and $\frac{(z-z_1)(\bar{z}-\bar{z}_1)dz}{z}-\frac{(z-z_1)(\bar{z}-\bar{z}_1)d\bar{z}}{\bar{z}}$ are closed and hence vanish upon application of Stokes' Theorem. In short, we recover the calculation given by the OP, but now for a pair of singularities.

I hope someone will improve on my non-answer here.

EDIT: (12-16-20), indeed, as per Steven Gubkin's comment, my "suspect" closed forms do not seem to be closed. I also calculate:
$$ d\beta = d \left[\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1} \right] = \left( \frac{\bar{z}}{\bar{z}-\bar{z}_1} + \frac{z}{z-z_1} \right) d\bar{z} \wedge dz$$
Then, I thought, maybe this integrates to zero for the area $|z| \leq R$. Well,
after some algebra,
$$ d\beta = \frac{4i(x^2+y^2+xx_1)}{(x-x_1)^2+(y-y_1)^2} \, dx \wedge dy. $$
If $z \neq z_1$ (as I have always implicitly assumed) then the denominator is positive over the disk $|z| \leq R$ and only the $xx_1$-term vanishes in the integral. It looks to me like that integral does not vanish. Apparently, I cannot have my cake and eat my antiholomorphic cake too.


问:
Seeing if this Function is Bijective.
$f: \mathbb{R^+} \rightarrow \mathbb{R}: f(x) = \log_{10}(x)$

How is this function onto? I know it is one-to-one because every image has a pre-image. Can someone explain to me why this is also onto? I thought that everything in the codomain must be mapped to to be onto?

答: $f$ is onto $\mathbb R$, because, for any $y\in \mathbb R$, $f$ maps $10^y$ to $y$.

Title: Seeing if this Function is Bijective.


问:
5 points determine a conic uniquely
This is exercise 3.5.5 in Gathman's notes. We proved so far that zero set of degree $2$ homogeneous polynomials in variables $x_0,x_1,x_2$ can be identified as $\mathbb{P}^5$, and conics can be identified as an open subset $U\subset\mathbb{P}^5$. Also, given a point $P\in\mathbb{P}^2$, and let $F: \mathbb{P}^2\mapsto\mathbb{P}^5$ be the Veronese embedding. Then a conic $C\in U$ passing through $P$ if and only if $F(P)\cdot C = 0$. The last part requires to prove that given $P_1,\ldots, P_5\in\mathbb{P}^2$ such that no three of them is on the same line, then there is a unique conic passing through the $5$ points.

What I prove so far is that the matrix $$\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$$ is a $5\times 6$ matrix, and hence the null set have dimension at least $1$ which corresponds to vanishing of degree $2$ homogeneous polynomials passing though $P_1,\ldots, P_5$. I also showed that if no three of $P_1,\ldots, P_5$ lies on the same line, then that degree $2$ homogeneous polynomial must be a conic (irreducible).

However, to prove the uniqueness, I need to show that $F(P_i)$ are linearly independent, and I got stuck on this. I read answers from another post, and it suggests that linear relation on $F(P_1),\ldots, F(P_5)$ can be pulled back to linear relation on $P_1,\ldots, P_5$. I don't see why this is true.

答:
You will need the condition that $P_i$'s are in general position, which means an open subset (in the Zariski sense) of all the possible configurations $\{(P_1,\ldots,P_5)\}$.

Precisely, you need to show the condition that $\{F(P_i)\}$ linearly independent is an open condition. Indeed, by basic linear algebra, the dependency of $\{F(P_i)\}$ is equivalent to that the determinants $D_i=D_i(P_1, \ldots, P_5)$ $(i=1,\ldots,6)$ of all the $5\times 5$ submatrices of
$\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$ are zero, which is a closed condition.

Title: 5 points determine a conic uniquely


问:
Show the sum of a sequence of trace is convergent. $\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S)$
I encountered this problem when doing research.

Matrices are all real-valued. $\Sigma_S$ is a covariance matrix. One can assume $\Sigma_S$ is positive definite. $A$ is a square matrix with dimension $n\times n$. $C$ is a $n\times m$ matrix  where $m\leq n$. $C$ is a bit redundant here. One can assume that $C'\Sigma_S C$ is positive definite.

Now I want to find the conditions on $A$ and $C'\Sigma_S C$ such that
$$
\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S) < \infty
$$
where $S$ is positive definite. Here, prime $'$ means transpose and $A^t$ refers to the $t$th power of matrix $A.

Ideas: We can write $C\Sigma_SC = LL'$ and assume that $(A,L)$ is controllable. We can see $\sum_{t=0}^\infty (A')^t LL'A^t$ is actually the infinite-horizon controllability Gramian. If all eigenvalues of $A$ have magnitudes less than $1$, then there is a unique positive definite solution of
$$
W_\infty  -A'W_\infty A = LL'.
$$
And the controllability Gramian can be expressed as $W_\infty = \sum_{t=0}^\infty (A')^t LL'A^t.$ Then, we have $trace( \sum_{t=0}^{T} (A')^t C' \Sigma_S C A^t S)\rightarrow \operatorname{trace}(W_\infty S)$ as $T\rightarrow \infty$.

But this convergence requires that $A$ is table and $(A,L)$ is controllable. Can we conclude $$
\operatorname{trace}( \sum_{t=0}^{\infty} (A')^t C' \Sigma_S C A^t S) < \infty
$$ using only the assumption that $(A,L)$ is controllable?

More Background:

Consider a linear-quadratic-gaussian optimal control problem with a discrete-time linear system
$$
x_{t+1} = Ax_t + Bu_t + Cw_t,
$$
where $x_t$ is the state, $u_t$ is the control and $w_t$ is Gaussian noise with zero mean and variance $\mathbf{E}[w_t' w_\tau ]=\Sigma_S\delta(t-\tau)$. Here, $\delta(\cdot)$ is the Kronecker delta function. Given $x_0 =x$, let $\mathcal{F}_t = \{x_0,u_0,\dotsc,u_{t-1}\}$. And the problem is associated with quadratic stage cost $x_t'Qx_t + u_t' Ru_t$ with $Q\geq 0$ and $R\geq 0$.

We have $\hat{x}_t = \mathbf{E}\left[ x_t \middle \vert   \mathcal{F}_t \right] = A^{t-1}x +\sum_{\tau=0}^{t-1}A^{t-1-\tau}B u_\tau$. And the variance of estimation error $P_t(\mathcal{F}_t) := \mathbf{E}\left[(x_t -\hat{x}_t)'(x_t - \hat{x}_t)\middle\vert \mathcal{F}_t\right]$ is propagated as $P_{t+1} = A' P_t A + C'\Sigma_SC$ with $P_0 =0$. Thus, $P_t(\mathcal{F}_t)$ can be expressed as
$$
P_t(\mathcal{F}_t) = \sum_{\tau=0}^{t-1} (A')^\tau C' \Sigma_S C A^\tau.
$$

Let $S = A' PB (R+ B' P B)^{-1}B' PA$ where $P$ is a positive definite solution of an algebraic Riccati equation. Then the control performance degradation at time $t$ by not having observations from times $0$ to $t$ is
$$
\mathbf{E}\left[ (x_t - \hat{x})'S(x_t - \hat{x}) \middle\vert \mathcal{F}_t\right] = \operatorname{trace}(P_{t}(\mathcal{F}_t) S) = \operatorname{trace}(\sum_{\tau=0}^{t-1} (A')^\tau C' \Sigma_S C A^\tau S).
$$

Wants to find the loosest conditions such that the degradation is bounded as $t\rightarrow \infty$.

Added Questions: Under what conditions will the accumulated degradation with a discounted factor $\beta<1$
$$
\sum_{t=0}^\infty \beta^t \operatorname{trace}(P_{t}(\mathcal{F}_t) S)
$$
be bounded?

答:
Let $x_{t+1} = Ax_t + Bu_t$ be a DT control system. Usually, $BB'$ is not invertible and thus not positive definite (for which case Ben presented his answer). Here, I'd like to show the divergence in the general case, where $(A,B)$ is controllable.

The $T$-horizon controllability gramian is defined by $W_T := \sum_{t=0}^TA^tBB'(A')^t$. Its eigenvalues tell us something about how far we can reach at time $T$ from zero. However, what you have here is the expression $Tr(W_TS)$ and you would like to know whether is has a limit as $T\to\infty$ if $A$ is not stable. As Ben already showed for the case that $BB'$ is positive definite (in which case $(A,B)$ is trivially controllable), the answer is no. Here is my reasoning:

If $C$ is positive semi-definite, then $Tr(C)\ge \langle Cx,x\rangle$ for any unit vector $x$.

Indeed, if $C = U'DU$ is the eigenvalue decomposition of $C$, then, setting $y := Ux$, we have
$$
\langle Cx,x\rangle = \langle DUx,Ux\rangle = \sum_k\lambda_ky_k^2\le\sum_k\lambda_k = Tr(C)
$$
since $y_k^2\le\|y\|^2 = \|Ux\|^2 = \|x\|^2\le 1$ for each $k$.

Assume that $A$ (and thus also $A'$) has an eigenvalue $\lambda$ with $|\lambda|\ge 1$. Choose a corresponding eigenvector $x\neq 0$ for $A'$. Then
$$
\langle W_T x,x\rangle = \sum_{t=0}^T\|B'(A')^tx\|^2 = \sum_{t=0}^T|\lambda|^{2t}\|B'x\|^2\,\ge\,(T+1)\|B'x\|^2.
$$
As $(A,B)$ is controllable, the Kalman matrix $K = (B,AB,\ldots,A^{n-1}B)$ has full rank $n$ and so its transpose $K'$ has trivial kernel. If $B'x = 0$ and $A'x = \lambda x$, then $K'x=0$, which is impossible. So, $\|B'x\|\neq 0$.

We now come to the conclusion. Let $y$ be a vector such that $S^{1/2}y = x$, where $x$ is as in 2. above. Then, by 1. and 2.,
\begin{align*}
Tr(W_TS)
&= Tr(S^{1/2}W_TS^{1/2})\,\ge\,\|y\|^{-2}\langle S^{1/2}W_TS^{1/2}y,y\rangle\\
&= \|y\|^{-2}\langle W_Tx,x\rangle\ge C\cdot T
\end{align*}
with $C = \|y\|^{-2}\|B'x\|^2$. This shows that $Tr(W_TS)\to\infty$ as $T\to\infty$.


问:
Tetration limit
For each $n$, define $f_n:\mathbb R^+\rightarrow \mathbb R^+$ by
$f_n(x) = \underbrace{x^{x^{x^{...^{x^x}}}}}_n$

Is it true that $\lim\limits_{n \to \infty} f_n(\frac{n+1}{n}) = 1$ ?

A few computations using Wolfram Alpha seem to suggest so, but I am unsure of how to prove it.

One way to think about why it might be true intuitively is to compare it to the limit definition of $e$, but even this intuition is handwavy (although it seems like it is true based on computations).

Furthermore, I'm unsure if the Binomial expansion actually helps us here.

I'm interested in hearing your thoughts.

答:
Let $g(x)$ be the inverse function of $x\to x^{1/x}$, which is strictly increasing on $[1,e]$, hence $g$ exists and is increasing on $[1,e^{1/e}]$. This definition implies that $x^{g(x)} = g(x)$ because $g(x)^{1/g(x)} = x$.

For any $x\in[1,e^{1/e}]$, We can use induction to show $$
f_n(x) \le g(x)
$$
for all $n$. For the base case, obviously $f_0(x) = x\le g(x)$. If the inequality holds for $n$, then we have that $$
f_{n+1}(x) = x^{f_n(x)} \le x^{g(x)} = g(x)
$$
Therefore, we can conclude by observing $$
1 \le f_n(\frac{1+n}{n}) \le g(\frac{1+n}{n})
$$
$g$ is continuous, so taking the limit on the RHS, we can see that the squeeze theorem implies the limit of $f_n(\frac{1+n}{n})$ is $1$.

Title: Tetration limit


问:
Average curvature of a curve?
Given an a.l.p parametric curve $r(t)$ we know that $r''(t)$ is the curvature scaled normal to the curve at $r(t)$.

Thus $\|r''(t)\|$ is the actual curvature. Consider then a discretization of the domain of $r$ such that the domain is $U = \bigcup\{u_i \mid u_i \cap u_j = \emptyset \text{ if } i \neq j\}$ and let $S = \{t_i \in u+i\}$. Or simply put we grab a lot of points through the domain.

Then $\frac{\sum \|r''(t_i)\|}{N}$ where $N$ is the total of all intervals is a discrete approximation of the global average of the curvature of the curve.

What would the analytic formulation be? i.e. what is the global average of the curvature of a curve?

答:
What does "a.l.p" mean in the question? I would note that in general, $||r''(t)||$ is not the curvature; this is only true when $r$ is an arclength parametrization, or equivalently $||r'(t)||\equiv1$. In general, we have

$$\kappa=\left|\left|\frac{dT}{ds}\right|\right|=\frac{||T'(t)||}{||r'(t)||}$$

where $T$ is the unit tangent vector. Of course, $\kappa$ is a scalar, and so the best way to capture a meaningful "average" (keeping in mind that different parametrizations can give different results) is to use an arclength line integral, and then divide by the total arclength $s$:

$$\kappa_{avg}=\frac1s\int_C\frac{||T'||}{||r'||}ds$$

But remember that $ds=||r'(t)||dt$, so:

$$\kappa_{avg}=\frac1s\int_0^T||T'(t)||dt=\frac{\int_0^T||T'(t)||dt}{\int_0^T||r'(t)||dt}$$

(EDIT: I've been informed in the comments that "a.l.p" means arclength parametrized. In this case, many of the points I make about parametrizations are moot. Nonetheless, I'll leave the original here for posterity. Note that the result I give is true regardless of parametrization, which may be useful in general.)

Title: Average curvature of a curve?


问:
Birkhoff spectrum is not compact
Let $f:X\to X$ be a homemmorphim on a compact metric space $X$. Assume that $T:X \to \mathbb{R}$ is continuous. We consider
$$
\sigma_{f}(x):=\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}T(f^{k}(x).$$

$\textbf{Question}$: Why is
$$W(\alpha):=\{x \in X, \sigma_{f}(x)=\alpha\}$$
not compact?

$W(\alpha)$ is the set of point where the limit of $\sigma_{f}(x)$ exists.
I know the set might be an empty set(consider the identity functions), so consider the question in case the set is not empty.

答:
Perhaps you need to realize that $\sigma_f$ need not be continuous under your assumptions.

Actually, in many situations it turns out that $W(\alpha)$ is dense but not compact for $\alpha$ in some interval. This is the case for example for the dynamics on a hyperbolic set and Hölder continuous functions.

The main reference for this type of results is L. Barreira and J. Schmeling, Sets of “non-typical” points have full topological entropy and full Hausdorff dimension, Israel J. Math. 116 (2000), 29–70.

Incidentally, you need of course to clarify what you mean by "the limit exists for almost every point".

Title: Birkhoff spectrum is not compact


问:
How to deal with polynomial quotient rings
The question is quite general and looks to explore the properties of quotient rings of the form $$\mathbb{Z}_{m}[X] / (f(x)) \quad \text{and} \quad \mathbb{R}[X]/(f(x))$$
where $m \in \mathbb{N}$

Classic examples of how one can treat such rings is to  find relationships like $$\mathbb{Z}[x]/(1-x,p) \cong \mathbb{Z_{p}}$$ for prime $p$.

However, I often struggle to intuitively understand what the elements of such rings are, and they to compute using them.

For example, what do the elements of the set $\mathbb{Z}_{2}[X] / (x^4+1)$ actually look like. Can this set be described in a clearer way to help understand the way the rings work.

Is there some general way of describing elements of such rings so that isomorphisms and computations become easier to handle?

Apologies if this question is not sufficiently clear.

答:
To treat your special case
$$\mathbb{Z}_{2}[X] / (x^4+1),$$
its elements are the (classes) of the remainders of Euclidean divisions by $x^4+1$, so
$$\mathbb{Z}_{2}[X] / (x^4+1) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{2} \}.$$
(Here I am using brackets to denote residue classes.)

You sum the classes normally, and as to the product, you first take the product, and then take the remainder of Euclidean division by $x^4+1$.
Alternatively, you take the product, and then use the relation $x^4 \equiv -1$ (which is the same as $x^4 \equiv 1$ in this case, as $1 = -1$ in $\mathbb{Z}_{2}$) to reduce the result. So it's not really different from computing in $\mathbb{Z}_{m}$, where you first take an ordinary sum or product, and then take the remainder modulo $m$.

So for instance
$$
[1 + x^2] \cdot [1 + x^3] = [1 + x^2 + x^3 + x^5] = [1 + x + x^2 + x^3],
$$
using one of the two methods above.

The general case of $F[x] / (f(x))$, with $F$ a field, is similar, you get the classes of the remainders of Euclidean division by $f(x)$.


问:
Surjective mapping from $\mathbb{R}^m$ to the set of symmetric positive (semi-)definite matrices of size $n\times n$
I want to map from $\mathbb{R}^m$ (for $m=n\cdot (n+1) / 2$, however, if another $m$ is necessary, that would also be ok) to symmetric positive (semi-)definite matrices.

I know that I can obtain a random positive semi-definite matrix by multiplying a random matrix by its transpose.
However, I assume that there are positive semi-definite matrices that cannot be produced by this procedure.
That is why I am looking for a subjective mapping to the space of symmetric positive (semi-)definite matrices of a given size.

I am aware that one could use a recursive function that checks whether the input is a symmetric positive (semi-)definite matrix and, if it is not tries some pseudo-random number in order to just check again until it finds one that is a symmetric positive (semi-)definite matrix.
However, I want to find a "nice" mapping that can produce all symmetric positive (semi-)definite matrices and is also a.e. differentiable.

答: Define $l:\mathbb{R}^{{1 \over 2}n (n+1)} \to \mathbb{R}^{n \times n}$ to be a lower triangular matrix in the obvious way. Then $f(x)=l(x) l(x)^T$ is a surjective map onto the positive semidefinite matrices.


问:
Finding the multiplicity of eigenvalues

Let $T: V \to V$ for a finite-dimensional vector space $V$ be a linear operator whose matrix relative to the standard basis consists of all $1's$. find all eigenvalues and eigenvectors of $T$ and their geometric and algebraic multiplicities.

I didn't have much trouble finding the eigenvectors and eigenvalues, but I am not totally certain on the multiplicities. Here is what I have.

Let $A$ denote the matrix of $T$ relative to the standard basis, so $A_{ij} = 1$ for all $i,j$. If $v = (v_1, \ldots, v_n)$ is an eigenvector of $A$ with eigenvalue $\lambda$, then
$$Av = \begin{pmatrix} \sum\limits_{i} v_i \\ \vdots \\ \sum\limits_{i} v_i \end{pmatrix} = \lambda v = \begin{pmatrix} \lambda v_1 \\ \vdots \\ \lambda v_n \end{pmatrix}.$$
So
$$\sum\limits_{i} v_i = \lambda v_k $$
for all $k$, so
$$\lambda v_1 = \lambda v_2 = \ldots = \ldots,$$
which is true if and only if $\lambda = 0$ or $v_1 = v_2 = \ldots = v_n$. If $\lambda = 0$, then  eigenvectors must satisfy $\sum\limits_{i} v_i = 0$. If $\lambda \neq 0$, then eigenvectors must satisfy $v_1 = \ldots = v_n$. Let $t$ equal this common term, then we have $\sum_i t = tn = \lambda t$, so $\lambda = n$.

So the eigenvalues are $\lambda_1 = 0$, in which case the sum of the eigenvectors must be $0$, and $\lambda_2 = n$, in which case all components of the eigenvector must sum to $1$.

By definition, the algebraic multiplicity is the dimension of the nullspace of $(T - \lambda I)^{\dim V}$ and the geometric multiplicity is the dimension of the nullspace of $T - \lambda I$. I cannot figure out how to compute these from my above work, though.

答: Hint: the matrix has rank $1$.

Title: Finding the multiplicity of eigenvalues

Body: Hint: the matrix has rank $1$.


问:
Proving Levy's theorem using 0-1 law
Suppose $X_1, X_2, \dots$ are independent. Let $S_n = \sum_{i=1}^nX_i$. Levy's theorem states that $S_n$ converges in probabilty implies $S_n$ converges a.s. I am aware of Ottaviani's proof using maximal inequality. But I want to prove this using Kolmogrov's 0-1 law. I am wondering if my proof idea is correct.

My idea:
Suppose $S_n \to S_\infty$ in probability. Now since the event $\lim_{n\to\infty}S_n = S_\infty$ is a tail event, therefore, it only takes probability $0$ or $1$. Suppose that
$$
P(\lim_{n\to\infty}S_n = S_\infty) = 0,
$$
i.e.
$$
\lim_{n\to\infty}S_n \neq S_\infty \ a.s.
$$
Then there exists a subsequence $S_{nk}$ that has no further subsequence that converges to $S_\infty$ a.s. Hence, it contradicts the claim that $S_n \to S_\infty$ in probability. Therefore,
$$
P(\lim_{n\to\infty}S_n = S_\infty) = 1.
$$

Edit:
After some thought, I don't think my proof is correct since the same argument can be used upon convergence of $X_n$ since $\lim_{n\to\infty}X_n$ is also a tail event. But we don't have $X_n$ converges in probability implies a.s. convergence.

I think the issue is that the subsequence $S_{nk}$ we picked is not fixed as it can change when we pick a different $\omega$ from the set
$$
A = \{\omega: \lim_{n\to\infty}S_n(\omega) \neq S_\infty(\omega)\}.
$$
So we can't really conclude there is a subsequence with no further subsequence that converges a.s.

Does anyone have an idea how to fix this? Or using zero-one law just doesn't work.

答:
I don't think that $\{\lim_{n \rightarrow \infty} S_n = S_\infty \}$ is a tail event.  I agree that $\{\lim_{n \rightarrow \infty} S_n \text{ exists}\}$ is a tail event, but since it's a sum the specific value that it converges to depends on the early terms like $X_0$ as well.  For example, if $X_n = 0$ for $n \ge 2$ then $S_\infty = X_1$ and $X_1$ is not measurable with respect to the tail $\sigma$-algebra.

I think the mistake in your proof is around the line "there exists a subsequence $S_{n_k}$ that has no further subsequence that converges to $S_\infty$ a.s."  That subsequence depends on $\omega$, so we won't be able to use it to reach the contradiction you're looking for.

Title: Proving Levy's theorem using 0-1 law


问:
Is $(x-t)^p$ irreducible?
$f(x) = x^p-s = x^p - t^p = (x-t)^P$

I don't know how to show that it is irreducible... How can I use Eisenstein to show this? Is it suitable to say by Gauss' lemma, since in the question $p$ is prime, $p=(t)$ and $t \notin p^2$ so it is irreducible?

答:
If $t$ is transcendental over $\mathbb{F}_p$, then so is $t^p$. Thus, $\mathbb{F}_p(t^p)\cong\mathbb{F}_p(y)$ for a transcendental variable $y$ (I switch so we are not distracted by the exponent in "$t^p$"; we want to treat it as a transcendental variable, not the $p$th power of something.)

So... is $x^p-y$ irreducible over $\mathbb{F}_p(y)$? Yes. To see this, we can apply Gauss's Lemma. The polynomial $x^p-y$ is irreducible in $\mathbb{F}_p(y)[x]$ if and only if it is irreducible in $\mathbb{F}_p[y][x]=\mathbb{F}_p[x,y]=\mathbb{F}_p[x][y]$. But as a polynomial in $(\mathbb{F}_p[x])[y]$ it is clearly irreducible, since it is of degree $1$. Therefore, it is also irreducible over $\mathbb{F}_p(y)$.

Title: Is $(x-t)^p$ irreducible?


问:
Determinant and inverse matrix calculation of a special matrix
Is there any smart way of calculating the determinant of this kind matrix?

\begin{pmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}

I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.

I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.

I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.

答:
On a related note, the inverse of the closely related matrix $(a_{ij}) = (\frac{n}{4}- \frac{1}{2} |i-j|)$ is still Toeplitz and symmetric with first row $(2,-1, 0, \ldots, 0, 1)$, determinant $4$, and eigenvalues coming in pairs ( and a $4$, when $n$ odd). Moreover, one can explicitely find the eigenvectors of this inverse, using trigonometric functions. Now, there is a formula for the inverse of a peturbation of a matrix by a matrix of rank $1$. So finding the inverse is feasible.

As an example, check out this inverse ( case $n=5$): WA link .


问:
Show that the set $\{(x,y): y > 0\} \subset \mathbb{R}^{2}$ is open - verification and criticism
Show that the set $C = \{(x,y): y > 0\} \subset \mathbb{R}^{2}$ is open

I know a very similar question has already been asked, but I wanted criticism on what I did because I felt very frustrated in articulating my solution formally for an idea that I visually was able to put together quickly.

Attempt
In order to show that the set $C$ is open requires us to show that any neighbourhood of a point will be fully contained in the set. That is we must show that a ball, $B(\textbf{a}, \delta) \subset C$, where $\textbf{a} = (a_{1},a_{2})$.

My issue was in choosing a $\delta$ for this scenario and showing that it worked. I went off the tracks here thinking about open rectangles and balls inscribed in open rectangles with their radius being $\frac{\delta}{\sqrt{n}}$ and I frazzled myself in the process.

So using the "notion" of the open rectangle I settled on letting $\delta = \frac{|y|}{2}$, so in my particular case this would mean $\delta = \frac{|a_{2}|}{2}$.

With this as my $\delta$, let's take an arbitrary point $\textbf{x} = (x,y) \in B(\textbf{a},\delta)$. Now I have to show that $\textbf{x} \in C$.

I know that:

$$||\textbf{x} - \textbf{a}|| = ||(x-a_{1})^{2} + (y-a_{2})^{2}|| < \delta = \frac{|a_{2}|}{2}$$.

Now to simplify things we can assume that $x = a_{1}$, which then leaves me with

$$||(y-a_{2})^{2}|| < \delta = \frac{|a_{2}|}{2} \\
|(y-a_{2})| < \frac{|a_{2}|}{2}$$

which we can do by rewriting the form of the norm being used w.r.t absolute values.

This can then be rewritten as:

$$\frac{-a_{2}}{2} < y - a_{2} < \frac{a_{2}}{2} \\
 \frac{a_{2}}{2} < y < \frac{3a_{2}}{2}$$

Because we are only concerned in the positive value of $a_{2}$ we can remove the absolute value bars.

Since it was assumed $0 < a_{2}$, this shows that $\textbf{x} \in C$.

So my issues:

I feel my choice of $\delta$ is not right. At least it doesn't "feel" right in terms of how I ended up phrasing it. It took too much mental strain to articulate it properly, but when drawing the picture I could see the idea clearly

I think I missed a step because am I not supposed to also have shown that $\textbf{x} \in B(\textbf{a},\delta)$? I kind of just assumed it was there without establishing it.

My manipulation of the series of inequalities also feels off and not as fluid as I feel they should be.

Comments on my concerns would be appreciated.

答:
This might help: Suppose $(x,y)\in C.$ Then $y>0.$ Draw a picture! Doesn't it look like the entire $B((x,y),y)\subset C?$ (You chose the radius to be $y/2$ but it seemed like you were taking a stab in the dark.)

Let's verify. Let $(x',y')\in B((x,y),y).$ We want to show $(x',y')\in C.$ All we need to verify is that $y'>0.$ But we know $|x'-x|^2 + |y'-y|^2<y^2.$ This implies $|y'-y|<y$, which implies $-y < y'-y.$ Add $y$ to both sides to get $0<y'.$ We're done


问:
First order logic equivalence
Given some initial formula, a Universe $U$ and a model $A$ for this formula I arrived at the statement $A(\neg(\forall x F)) = 1$. How could I transform this without using any theorems, just by the semantics of $\neg$ and $\forall$ into $A_{[x\rightarrow u]}(F) = 0$ for some $u \in U$ ?

答:
It really is nothing more than that -- unpacking the definitions of semantics of the two operators:

$\def\iff{\Leftrightarrow\ }
\phantom{\iff} A(\neg \forall x F)) = 1\\
\iff A(\forall x F)) = 0 \quad \text{(def. $\neg$)}\\
\iff \text{not:} \text{ for all } u \in U: A_{|x \to u|}(F) = 1 \quad \text{(def. $\forall$)}\\
\iff \text{for some } u \in U: A_{|x \to u|}(F) = 0 \quad \text{(paraphrasing)}
$

The last two steps implicitly include complementarity of $1$ and $0$, i.e. swapping between $0$ and $1$ by prepending resp. removing "$\text{not}$" on the conditions. This can be taken for granted, given that the definitions are of the form "$\ldots = 1$ if some condition, else $ = 0$"; the "$\text{else}$" part makes for the "$\text{not}$".

Title: First order logic equivalence


问:
Find the average temperature between $t=0$ and $t=24$ when $T(t) = 49+8t-(1/2)t^2$ degrees.
What was the average temperature during that period?

My initial thought was to take the derivative of the problem, plug in 24 for $t$ and solve.  I was wrong.  This is what I have

$T'=8-t=8-24=-16$ deg.

Should I have taken the integral instead of the derivative?

答: The derivative gives the rate of change of temperature. Instead, you should compute the integral from 0 to 24 hours and then divide by 24 to get the average temperature.</p>


问:
Combinatorics - How many ways to arrange 4 different dumbbells in 7 different drawers?
How many ways to arrange 4 different dumbbells in 7 different drawers?

My attempt:

There are 5 cases to arrange 4 different dumbbells call them {A,B,C,D}

1 2 3 4 5 6 7 : Drawers

2  2  0  0  0  0   0 (1

3 0 1 0 0 0 0 (2

4 0 0 0 0 0 0 (3

2 1 1 0 0 0 0 (4

1 1 1 1 0 0 0 (5

To distribute 2 sets each over 7 different drawers so we have
P(7,2)*C(4,2) when P(7,2) number of ways to choose a place for each 2 sets of size 2 and (4,2) to arrange the dumbbells

-Same idea for the followings

P(7,2) * C(4,3)

P(7,1)*C(4,4)

P(7,3)*C(4,2)

P(7,4) * C(4,3)*C(3,2)C(2,1)

And at the end we sum all the cases 1+2+3+4+5

Would that be a correct solution to the question?

*Subside question (for whoever wants to answer) : same question but how to arrange them when the heaviest dumbbell =A and lightest =D can't be in the same drawer?

I thought about cases (1+2+3+4+5) - (All the possibilities when A and D are together)

To count A and D as one body so we will only have 3 cases when

1 2 3 4 5 6 7 : Drawers

1  2  0  0  0  0   0 (1

1 1 1 0 0 0 0 (2

3 0 0 0 0 0 0 (3

So at end the cases from first question (1+2+3+4+5) - (1+2+3) = possibilities of A and D not together?

答:
Each (distinguishable) dumbell can be placed in one of seven (distinguishable) drawers, and hence the number is $7^4$.

The first dumbell can be put in drawer 1, or drawer 2, or ... 7.

The second dumbell can be put in drawer 1, or drawer 2, or ... 7, independent of the first.

Thus there are $7^2 = 7 \times 7 = 49$ possible distributions of the two distinguishable dumbells.

The third dumbell...


问:
Find force of gravity
Material point describes an ellipse (with axes $a$ and $b$) under the action of force of gravity. At the end of bigger axis the velocity of point is equal to $v_0$. I need to find the force of gravity as a function of radius vector. So the force is conservative therefore the law of conservation of energy is right. Also we can describe the velocity (in cartesian coordinates, from the equation of ellipse) $v^2=\dot{x}^2+\dot{y}^2$ as a $v^2=\dot{y}^2+\frac{a^4}{4x^2} (1-\frac{2 y}{b^2}\dot{y})^2$ From which we can derive that $v_0^2=\dot{y}^2+\frac{a^2}{4}$ (simply input $x=a$ and $y=0$ and use the condition that the velocity at the end of bigger axis is equal to $v_0$. But I'm stuck, I don't know wthat to do next.

答: An ellipse is often written as $r=\frac{l}{1+e\cos\theta}$, with $e$ its eccentricity and $l$ its semi-latus rectum, but the Cartesian coordinates $X,\,Y$ associated with these polar coordinates aren't the ones you're using. We have$$r=\tfrac{l}{1+eX/r}\implies r=l-eX\implies\tfrac{\left(X-\tfrac{el}{1-e^2}\right)^2}{(l/(1-e^2))^2}+Y^2=1.$$Comparing this to your coordinates gives $x=X-ea,\,y=Y,\,a=\frac{l}{1-e^2}$. With $u=\frac1r$, the Binet equation $\frac1l=u_{\theta\theta}+u=-\frac{F}{mh^2u^2}$ implies the force on a mass $m$ of specific orbital angular momentum $h$ is$$F=-\frac{mh^2}{l}u^2=-\frac{mh^2a^{-1}/(1-e^2)}{(x+ea)^2+y^2}$$(the $-$ sign denotes an attractive force). At $x=a,\,y=0$ we have $\dot{x}_0=0,\,v_0=\dot{y}_0$, so$$mh\vec{k}=ma\vec{i}\times\dot{y_0}j\implies h=av_0.$$If $a$ is the semi-major axis,$$1-e^2=\frac{b^2}{a^2}\implies F=-\frac{ma^3b^{-2}v_0^2}{(x+ea)^2+y^2}.$$

Title: Find force of gravity


问:
Fourier series for $x\cos(x)$
I am attempting to find the fourier series for $f(x) = x\cos x$ for $-\pi < x < \pi$. The method is fine however I find that for $n = 1$ I am unable to evaluate the coefficient $b_1$.

$f(x) = \sum_{n=1}^{\infty}{b_n\sin(nx)}$ since $f(x)$ is an odd function.

Calculating my coefficients yields $$b_n = -\left(\frac{\cos((n+1)\pi)}{n+1}+\frac{\cos((n-1)\pi)}{n-1}\right)$$

This is clearly not defined at $n=1$, how should I proceed?

Note: this is an intermediary step in order to find the fourier series of $x(1+\cos x)$ if you wanted context.

答:
Your assertion comes from produt-to-sum formula $$\sin s\cos t=\frac12\sin(s+t)+\frac12\sin(s-t)$$

So the integrand of interest in the various passages is still $\frac12\sin((n+1)x)+\frac12\sin((n-1)x)$. The only difference is that for $n=1$ its antiderivative is no longer $-\frac{\cos((n-1)x)}{2(n-1)}-\frac{\cos((n+1)x)}{2(n+1)}$.

Title: Fourier series for $x\cos(x)$


问:
Finding the number of non-neg integer solutions?
How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.</p>

答:
Without the restriction on $x_1$ this is a standard stars-and-bars question and can be answered using the formula (or the reasoning) at the link. However, the resulting count will include the unwanted solutions that have $x_1>2$. You can finish the problem by counting these and subtracting that number from the number of unrestricted solutions.

To count the solutions with $x_1>2$, let $y_1=x_1-3$, $y_2=x_2$, $y_3=x_3$, and $y_4=x_4$. Then each solution to $$x_1+x_2+x_3+x_4=12\tag{1}$$ with $x_1>2$ corresponds to a solution to $$y_1+y_2+y_3+y_4=9\tag{2}$$ with no upper limit on any of the variables, and vice versa. (In all cases I’m assuming that the solutions must be in non-negative integers.) Thus, you need only get the number of unrestricted solutions to $(1)$ and subtract from it the number of unrestricted solutions to $(2)$, and you’ll have the number of solutions to $(1)$ with $x_1\le 2$.


问:
Does this subgroup of $\mathrm{Core}_G(H)$ have a name?
Let $G$ be a group and $H\le G$. The normal core of $H$ in $G$ is the kernel of the (left transitive) action of $G$ by left multiplication on the set $X$ of the left cosets of $H$, namely $\mathrm{Core}_G(H)=\bigcap_{a\in G}aHa^{-1}$. Similarly, if we take into account also the set of the right cosets of $H$, say $Y$, we can build up the (left) action $G\times (X\times Y)\to X\times Y$ defined by $g\cdot (aH,Hb):=(gaH,Hbg^{-1})$. Unless I've messed up anything, the kernel of this action ought to be $K:=\bigcap_{a,b\in G}(aHa^{-1}\cap b^{-1}Hb)=\mathrm{Core}_G(H)\cap\biggl(\bigcap_{a,b\in G, b\ne a^{-1}}(aHa^{-1}\cap b^{-1}Hb)\biggr)$, which would then be a subgroup of $\mathrm{Core}_G(H)$ normal in $G$.

So as to googling for it, in case, does the subgroup $K$ have a name?

答:
It is just the core of $H$ (which is the largest normal  subgroup of $G$ contained in $H$). You did not make it any smaller.

Note that each $b^{-1}Hb$ is of the form $aHa^{-1}$ (by taking $a=b^{-1}$), and vice-versa. So your intersection is just the intersection of all conjugates of $H$. Any element of $\mathrm{Core}_G(H)$ lies in both $aHa^{-1}$ and $b^{-1}Hb$ for all  $a,b\in G$.


问:
Complex Numbers Binomial Expansion
Let $a, b, n$ be any complex numbers.

For $(a\pm b)^n$, is there a binomial formula with a set number of finite terms that define this?

答:
A working example, as the OP asked.

Let's consider the simplest possible case that is still relevant to the question:

$$x = (3+2i) ^i$$

The exact value is something awful, just check with Wolfram Alpha, if you want.

Can we use binomial series here?

Let's try.

$$x = ^? 3^i \sum_{k=0}^\infty \binom{i}{k} \left( \frac{2i}{3} \right)^k$$

First, let's deal with the prefix.

$$3^i = e^{i \log 3}=\cos(\log 3)+i \sin (\log 3)$$

The binomial coefficients are defined as follows:

$$  \binom{i}{k}={\frac {i(i-1)(i-2)\cdots (i-(k-1))}{k(k-1)(k-2)\cdots 1}} $$

$$  \binom{i}{0}=1$$

$$  \binom{i}{1}=i$$

I suggest you try to calculate several terms of these series, add them, and then compare the results with the numerical value Wolfram Alpha gives.

For faster convergence we can use $13$ instead of $3$.

Title: Complex Numbers Binomial Expansion


问:
What is the largest finite number you can make using no more than 6 characters?
(I am new to the Math Stack Exchange community so tell me if this question is not allowed, however, I checked on Meta first.)

(Also, I don't really know which tags I should assign this question, so please offer tag suggestions if you don't agree with the tags I tagged this question with)

Sorry for the many edits to this question, but I needed to clarify what I was asking.

I have done some research and several google searches but I have not found the answer to this question:

What is the largest finite number you can make using no more than 6 characters? Characters include numbers and operations, etc. Rules: Parenthesis do not count as characters, and letters are not allowed. You can use the following operations: Basic Arithmetic, Exponents, Factorials, and Tetration (although you may also use Knuth's up-arrow notation).

Clarification Edit:
If you are using exponents - keep in mind this rule:
9^9 (9 to the power of 9) would be 3 characters. However, 9^99 (9 to the power of 99) would be 4 characters.

The following are some guesses that I have come up with, but I'm not confident in any of them

((((9!)!)!)!)!

((((9⁹)!)!)!)

$9^{9^{9{9{{}}}}}$

If I am not clear enough with the rules regarding what counts as a character please ask.

Any help is appreciated!

Thank you in advance!

Edit

Gerry Myerson has (quiet repeatedly) asked me to include the following links for reference: This link and this link.

答:
I submit:$$
9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9
$$
Using iterated Knuth's up-arrow notation, where we take $a \uparrow^n b$ to be $$a\,\underbrace{\uparrow\cdots\uparrow}_n\,b$$

It depends a lot on what notation we're allowed to use. I have proven all the following are optimal for the given notation sets (in addition to digits of course):

$$
\{!,+,-,\div,\times, a^b\} \hspace{26pt} \left(\left(\left(\left(9^9\right)!\right)!\right)!\right)!
$$

$$\{!,+,-,\div,\times, a^b, {^b} a\} \hspace{26pt} {^{^{^{^{^9 9}9}9}9}}9$$

$$\{!,+,-,\div,\times, a^b, \uparrow\}\hspace{26pt}9\uparrow\uparrow\uparrow\uparrow 9
$$

$$\{!,+,-,\div,\times, a^b, \uparrow,\uparrow^{k}\} \hspace{26pt} 9\uparrow^{((9^9)!)}9
$$

$$
\{!,+,-,\div,\times, a^b, {^b} a, \uparrow,\uparrow^{k}\} \hspace{26pt}9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9
$$

Proofs:
1: Let $s_n$ be the largest number achievable in $n$ symbols. Clearly $s_1=9$, and it's not hard to check $s_2 = 9^9$ (which is quite a bit larger than $9!$). Note that $s_{n+1}$ is always in the following set:$$
\left\{s_n!, 9^{s_n}, s_n^9\right\} \cup \{s_m + s_{n+1-m}\mid 1 \le m\le n\} \cup \{s_m \times s_{n+1-m}\mid 1 \le m< \le n\}
$$
(Note: if $n$ becomes too large, this might not work anymore, as it might be possible to get an extremely small number by some complicated process and then divide by it, proving that this doesn't ever happen is probably impossible. $n=6$ should be small enough to not worry about that).

Observe that the elements of the two sets on the right side there are smaller than $2s_n$ and $s_n^2$ respectively. However, if $s_n\ge 1$, then $2s_n$ and $s_n^9$ are smaller than all three of $\{s_n!, 9^{s_n}, s_n^9\}$. Hence $s_{n+1}$ is among $\{s_n!, 9^{s_n}, s_n^9\}$. Obviously $x! > 9^x > x^9$ for $x$ large enough, and one can check $x=22$ is large enough (see Wolfram Alpha https://www.wolframalpha.com/input/?i=9%5Ex+%3C+x%21). Since $9^9 > 22$, we have that for $n>2$, $$
s_{n+1} = s_n!
$$

2:
Same as above, but note that $^x 9 > x!$ for $x$ larger than about $2$.

3: Same idea, but we would also have the possibility $$
s_{n+1} \in \left\{s_i\uparrow^j s_k \mid i+j+k = n+1\right\}
$$
I claim that for $n\ge 5$, with this symbol set, $$
s_n = 9\uparrow^{n-2} 9
$$
is optimal. For $n=6$ this gives dbx's solution of $$
9\uparrow\uparrow\uparrow\uparrow 9
$$ We can prove inductively. Observe that for $n=5$ $$
9 \uparrow\uparrow\uparrow 9 = \underbrace{9\uparrow\uparrow\left(9\uparrow\uparrow\left(9\cdots9\right)\right)}_{9 \text{ times}}
$$
however, the biggest number achievable with 5 non-arrow symbols is $$
^{^{^{^9 9}9}9}9 = \underbrace{9\uparrow\uparrow\left(9\uparrow\uparrow\left(9\cdots9\right)\right)}_{5 \text{ times}} = 9\uparrow\uparrow\uparrow 5 < 9\uparrow\uparrow\uparrow 9
$$
Now, for each $n$ we'll have $s_{n+1}$ is either $s_n!$ or something involving arrows. If $s_n = 9 \uparrow^{n-2} 9$, the $(n+1)$symbol arrow-forms other than $9\uparrow^{n-1} 9$ are all less than $$
s_{n-1} \uparrow^{n-2} s_{n-1}
$$
Now for some magic:$$
s_{n-1} \uparrow^{n-2} s_{n-1} < s_n \uparrow^{n-2} s_n = \left(9 \uparrow^{n-2} 9\right) \uparrow^{n-2} \left(9 \uparrow^{n-2} 9\right) \le 9 \uparrow^{n-2}\left(9 \uparrow^{n-2} \left(9 \uparrow^{n-2} 9\right)\right) = 9 \uparrow^{n-1} 4 < 9\uparrow^{n-1}9
$$
Hence if we allow arrows but not powers of arrows, optimal is $$
9\uparrow\uparrow\uparrow\uparrow 9
$$

4 & 5: We can use similar reasoning if we're allowed to do powers of arrows instead of writing all arrows consecutively: My claim is, with either symbol set (4 or 5), $s_{n+3} = 9 \uparrow^{s_n} 9$ for all $n\ge 1$. The proof is almost identical to before using induction. It's easy to check that this is true for $n=1$; none of the other 4-symbol expressions are even close to $9\uparrow^9 9$. For $n\ge 2$, if $s_{n+3}$ isn't $9\uparrow^{s_n} 9$, then it must be less than \begin{eqnarray}
s_{n+2} \uparrow^{s_{n-1}} s_{n+2} &=& \left(9\uparrow^{s_{n-1}} 9\right) \uparrow^{s_{n-1}} \left(9\uparrow^{s_{n-1}} 9\right)\\ &<&9 \uparrow^{s_{n-1}} \left(9 \uparrow^{s_{n-1}} \left(9 \uparrow^{s_{n-1}} 9\right)\right) = 9\uparrow^{s_{n-1}+1} 4 < 9\uparrow^{s_n}9
\end{eqnarray}

This proves that my solutions at the top
$$
9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9 \hspace{20pt}\text{and}\hspace{20pt}9\uparrow^{((9^9)!)}9
$$
are optimal if we allow arrow-powers and tetration or arrow powers but not tetration, respectively.

I used a couple of times the inequality for $x,y,z \ge 2$, $k\ge 1$ $$
(x \uparrow^k y) \uparrow^k z \le x \uparrow^k \left(y\uparrow^k z\right)
$$
which is easy to show.


问:
Conditional Gaussian Probability

The random variable $x$ has normal distribution $N(10,1)$. Find $p(x\mid (x-10)^2 < 4)$.

I have been trying it for hours. I tried to solve $\mathbb{E}[(x-10)^2]$ using $\mathbb{E}[ay + b]$ but I'm not sure if it works for $x^2$.

Maybe with the $\mathbb{E}$ and $\mathbb{V}ar$ I can plug into another formula for conditional Gaussians.

答: This plot may help you see the appropriate integral:

Title: Conditional Gaussian Probability


问:
How to bound the norm of this matrix?
Let $(A_n)$, $(B_n)$, $(C_n)$ be sequences of $n\times k$ matrices, where $k<n$ is fixed and

$$A'_nA_n=B'_nB_n=C'_nC_n=I_k \hspace{1cm} $$

$$A_n'C_n=0$$

$$\|A_n\|_F=\|B_n\|_F=\|C_n\|_F=\sqrt{k}$$

for each $n$.

How can I bound the Frobenius norm of the matrix $A_nA'_nB_nB'_nC_nC'_n$ as $n\to\infty$?

I feel it should be quite small when $n$ is large. But I don't know how to show it. Any help is greatly appreciated.

答:
We have
$$
\|AA'BB'CC'\|_F^2 =  \operatorname{tr}[AA'BB'(CC')^2BB'AA'] \\ =  \operatorname{tr}[(AA')^2BB'(CC')^2BB'] =  \operatorname{tr}[AA'BB'CC'BB']\\ \leq \|AA'BB'\|_F \|CC'BB'\|_F,
$$
and $\|AA'BB'\|_F^2 = \operatorname{tr}[AA'BB'] = \|A'B\|_F^2$.

With that established, note that the matrix $W = [A \ \ C]$ has orthogonal columns. It follows that
$$
\|B\|_F^2 \geq \|W'B\|_F^2 = \left\|\pmatrix{A'B\\C'B} \right\|_F^2 = \|A'B\|_F^2 + \|C'B\|_F^2.
$$
Now, we note that the maximum
$$
\max_{a,b\geq 0} ab \quad \text{s.t.} \quad a^2 + b^2 = \|B\|_F^2
$$
is attained with $a = b = \|B\|_F/\sqrt{2}$ (as can be seen with the AM-GM inequality for instance). Putting all this together, we have
$$
\|AA'BB'CC'\|_F^2 \leq \|A'B\|_F \cdot \|C'B\|_F \leq \frac 12 \|B\|_F^2 = k/2.
$$

Title: How to bound the norm of this matrix?


问:
Proof that the plots of the fitted values vs. residuals yields parabola when needed quadratic term omitted?
There is a step I do not understand in my professor's class. In his slide, he is trying to formally show us that when the true, unknown model contains a squared predictor, and that this predictor is omitted when estimating the model parameters, then the plots of a) the fitted values vs. the residuals and b) the predictor vs. the residuals will yield a parabolic outcome. It is the last step [from (1) to (2)] of his reasoning that I simply don't understand.

He proceeds as follows:

"Assume for instance that the true model is quadradic in $X_1$:
$$y_i=\beta_0+\beta_1x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i$$
but we fit only a linear model
$$\hat y_i=\hat \beta_0+ \hat \beta_1x_{i1}$$
Then,
\begin{align}
e_i & =y_i-\hat y_i \\
& = (\beta_0-\hat \beta_0)+(\beta_1-\hat \beta_1)x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i \qquad (1)\\
& = \gamma_0+\gamma_1(\hat \beta_0 +\hat \beta_1x_{i1})+\gamma_2(\hat \beta_0 +\hat \beta_1x_{i1})^2+\epsilon_i \qquad (2)\\
& = \gamma_0+\gamma_1\hat y_i+\gamma_2\hat y_i^2+\epsilon_i
\end{align}

for certain values of $\gamma_0$, $\gamma_1$ and $\gamma_2$. This shows that both a $(x_{i1},e_i)$ and a $(\hat y_i,e_i)$ plot will show a quadratic curve."

As mentioned above, does anyone see the manipulation he performed to deduce (2) from (1)? How did he rid himself of the "true", unknown $\beta$'s?

I tried to consult a textbook (Sheather's "Modern Approach") to find this out, but his way of thinking about this is rather simplistic. They simply assume that the least squares estimates are close to the unknown population parameters $\beta_0$ and $\beta_1$, and that therefore we find that
$e_1=y_i-\hat y_i=(\beta_0-\hat \beta_0)+(\beta_1-\hat \beta_1)x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i\approx\beta_{11}x_{i1}^2+\epsilon_i$. I'm quite unsatisfied with this simplistic assumption and it doesn't answer my question above.

Can anyone help me out? If it is simple algebra, sincere apologies!

Many thanks in advance for your help

答:
The assertion is that (1) can be written in form (2) provided $\gamma_0$, $\gamma_1$, and $\gamma_2$ are chosen properly.

To show this, expand out (2) and collect terms so that the result is a constant plus something times $x_{i1}$ plus something times $x_{i1}^2$. The goal is obtain a revised version of (2) that
looks closer to (1). By matching the coefficients between (1) and the revised (2), you'll obtain three equations involving $\gamma_0$, $\gamma_1$, and $\gamma_2$ as well as the true $\beta_0$, $\beta_1$ and $\beta_{11}$ and the estimators $\hat\beta_0$ and $\hat\beta_1$. For example, in (1) the coefficient on $x_{i1}^2$ is $\beta_{11}$ while in the revised (2) it's $\gamma_2\hat\beta_1^2$; these two quantities must be equal.

Assuming these three equations are consistent (and almost surely they will be, except in degenerate cases), you'll be able to solve for the desired $\gamma$'s in terms of the $\beta$'s and the $\hat\beta$'s, thereby verifying that (1) can be written in form (2).


问:
Integral of the sequence of functions diverge imply supreme of the function is not integrable
Let $(X,\mathscr{M},\mu)$ be a measure space and consider a sequence of non-negative measurable
functions $f_n$ convergent to zero in measure and such that the sequence $a_n := \int f_n$ does not
converge to zero. Prove that the function $g(x) = sup_n f_n(x)$ is not Lebesgue integrable

答:
If $g$ is integrable, then, because $|f_n| \leq |g|$, the dominated convergence theorem implies that
$$\int f_n d\mu \to 0,$$
which is a contradiction.


答:
The average value of $f(t)$ over the interval $[a,b]$ is
$$\frac{\int_a^b f(t)\,dt}{b-a}.$$
To see that derivative has not much to do with average value, suppose that $f(t)$ is the constant $K$ over our interval. Then the average value must be $K$. But $f'(t)=0$ for all $t$.</p>

Title: Find the average temperature between $t=0$ and $t=24$ when $T(t) = 49+8t-(1/2)t^2$ degrees.

Body:
What was the average temperature during that period?

My initial thought was to take the derivative of the problem, plug in 24 for $t$ and solve.  I was wrong.  This is what I have

$T'=8-t=8-24=-16$ deg.

Should I have taken the integral instead of the derivative?


问:
Probability of coin streak at start of sequence
I am trying to work out how to express the probability of getting a streak of either heads or tails at the start of a series of coin tosses, considering those tosses may not be fair. For example, $Y= 3$ if a sequence starts with $HHHTHT....$ or $TTTHHHHT....$.

So far I think that:

$$
P(Y)=P(H)+P(T)
$$

and thus

$$
P(Y=k)=\sum_{i=1}^kP(H=i)P(T=k-i)
$$

but I have got myself confused and don't know where to go now.

Any thoughts would be greatly appreciated.

答:
Suppose $P(H) = p$ and $P(T) = 1-p$.

You are looking for a streak of length $k$. We can classify all streaks of length $k$ into two categories of outcomes. You have the first $k$ flips are heads followed by a tails, or the first $k$ flips are tails followed by a heads. These outcomes have respective disjoint probabilities: $$p^k (1-p)\qquad \text{and} \qquad p(1-p)^k$$

So, the total probability that you start with a streak of length $k$ is given by:

$$p(1-p)\big( p^{k-1}+(1-p)^{k-1} \big)$$

Sorry, don't know why I divided. Probability is multiplicative.


问:
Bertrand's postulate Strong proof
Bertrand's postulate

claim :

for all $ 5< n \in \mathbb{N}$ there is at least two prime numbers different in the opening interval $ (n,2n)$.

Use reinforcement of Bertrand's postulate in order to prove that :

for all $10<n\in \mathbb{N} $ Exists that has at least two Prime factors in the factorization of $n!$ which appear with a power of 1.

Example : $n = 11$ the primes  $7,11$ are such prime's meet the conditions.

But , for $n=10$ it dosen't meet the conditions because only $7$ appears with power 1 in the factorization of $10!$.

Hint : Consider two cases, $n$ even or $n$ odd

Attempt:

we need to use the claim in order to Implement the solution

Case (1): $n$ even

if $n$ is even we can rewrite $n = 2t$

if we use the claim we can get $(2t,4t)$

Case (2): $n$ odd

if $n$ is odd we can rewrite $n = 2t + 1$

if we use the claim we can get $(2t+1,2(2t+1))$

答:
You want to find primes that divide $n!$ exactly once, i.e., primes such that $p\le n$ but $2p>n$. On the other hand, Bertrand will give us primes with $t<p<2t$ for suitable $t>5$.
So in our situation, we want $2t\ge n$ (to make $2p>n$) and $2t\le n+1$ (because $p<n+1$ means the same as $p\le n$). This suggests taking $t=\lceil \frac n2\rceil$. This will make $t>5$ as soon as $n>10$.

Title: Bertrand's postulate Strong proof


答:
This can be boiled down to the stars and bars problem: http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

Let us consider just the case where $x_1 = 0$
Then $x_2 + x_3 + x_4 = 12$
Imagine there are 12 "stars" on the table. We need all the possible ways of putting 2 bars within these stars. From the wiki article, this is equal to: $\binom{2+12-1}{12}$

Similarly, do this work for $x_1 = 1$ and $x_1 = 2$

Title: Finding the number of non-neg integer solutions?

Body:
How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.</p>


问:
Linearly dependence of a vector set in a real vector space.
Is a vector set that belongs to a real vector space defined linearly dependent, iff the span of the set is equal to the zero vector?</p>

答:
$
\newcommand{\span}{\operatorname{span}}
\newcommand{\set}[1]{\{ #1 \}}
$
Note that if $S$ contains any nonzero vector, then $\span(S) \supsetneq \set{\vec 0}$. Namely, because if $s \ne \vec 0$ is in $S$, then $\set{\vec 0} \ne \set s \subseteq \span(S)$.

So, we have to ask, what of the cases $S = \set{\vec 0}$ and $S = \varnothing$? (At least in my experience, we often define $\span(\varnothing) = \set{\vec 0}$, so it should be considered too.)

You can show that any singleton set is linearly independent. You can also show that $\varnothing$ is linearly independent, through vacuous reasoning and the fact that $\varnothing$ cannot be linearly dependent.

I'll leave the details of proving these facts, and seeing how they apply to your situation, to you.


问:
Variance of the empirical mean
Let $X_1,...,X_n$ be i.i.d. R.V. with variance $\sigma^2$
Is the $$\text{Var}{\frac{X_1+...+X_n}{n}}=\frac{\sigma^2}{n}$$ or rather
$$\text{Var}{\frac{X_1+...+X_n}{n}}=\frac{\sigma^2}{n-1}$$ ?

I just remember that sometimes $n$ is decreased to $n-1$ by loosing one degree of freedom when computing moment's, is this that case ?

答:
It's the first. When you compute the variance you get with $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$ that
$$
var[\bar{X}_n] = \frac{1}{n^2} \sum_{i=1}^n var[X_i] = \frac{n}{n^2} var[X_1] = \frac{\sigma^2}{n},
$$
where $\sigma^2$ is the variance of the $X_i$.
The reduced $n-1$ in the denominator usually comes into play when estimating the variance of the $X_i$. In this case
$$
\widehat{\sigma}^2 = \frac{1}{n-1} \sum_{i=1}^n(X_i - \bar{X}_n)^2
$$
is an unbiased estimator for $\sigma^2$, i.e., $\mathbb{E}[\widehat{\sigma}^2] = \sigma^2$.
Another place where $n-1$ can be found is in the construction of confidence intervals for the mean when using a t-statistic.

Title: Variance of the empirical mean


问:
The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$
The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.

By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$

This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?

答:
Cubing the given equation $x^4=x^3+1$
$$x^{12} -3x^6-1=x^9+3x^3$$
and then squaring above
$$x^{24}-7x^{18}+x^{12}-3x^6+1=0$$
which is quartic in $x^6$. Thus,
$$\alpha^6+\beta^6+\gamma^6+\delta^6=7$$


问:
Seeing if this Function is Bijective.
$f: \mathbb{R^+} \rightarrow \mathbb{R}: f(x) = \log_{10}(x)$

How is this function onto? I know it is one-to-one because every image has a pre-image. Can someone explain to me why this is also onto? I thought that everything in the codomain must be mapped to to be onto?

答:
Actually, $g\colon \Bbb R\to \Bbb R^+$, $x\mapsto 10^x$ turns out ot be a two-sided inverse of $f$, i.e.,

$g\circ f\colon \Bbb R^+\to \Bbb R^+$, $x\mapsto 10^{\log_{10}x}$ is the identity map of $\Bbb R^+$

$f\circ g\colon \Bbb R\to \Bbb R$, $x\mapsto \log_{10}(10^x)$ is the identity map of $\Bbb R$

Having a two-sided inverse is equivalent to being injective.

Title: Seeing if this Function is Bijective.


问:
Difference between dot product, inner product, cross product, outer product. And what are their symbols?
When I was learning machine learning, I often encounter these terms, yet I don't the difference and relationship between them. And what are the symbols for them? Can someone help me?</p>

答:
The inner product is denoted $\langle x,\,y\rangle$ (or occasionally $(x,\,y)$) or $\langle x|y\rangle$. It mustn't be cofused with an outer product, which is basically a square matrix. You may see it denoted as $\sum_{ij}|i\rangle\langle j|$.

For your purposes, all these products will be on a finite-dimensional vector space, and with respect to some orthonormal basis the inner product will just be a dot product $x\cdot y=\sum_ix_i^\ast y_i$ (you only need the ${}^\ast$ if the space is complex). In a space of uncountable dimension, the $\sum$ in the dot (outer) product becomes a single (double) integral.

The cross product $x\times y$ almost certainly refers to something exclusive to $3$ dimensions, but sometimes it's something related but different.


问:
Intervals in measure theory
I am going through Rosenthal's A First Look at Rigorous Probability Theory but I am stuck on one particular part. On page 9, when constructing the uniform distribution, he says

It seems clear that we should choose $\Omega = [0,1]$. But what about $\mathcal{F}$? We know from proposition 1.2.6 that $\mathcal{F}$ cannot contain all intervals of $\Omega$, but it should certainly contain all the intervals $[a,b],[a,b)$, etc. That is, we must have $\mathcal{J} \subseteq \mathcal{F}$, where $\mathcal{J} = \{\textrm{all intervals contained in [0,1]}\}$

I am confused when he says that $\mathcal{F}$ cannot contain all intervals of $\Omega$, yet must contain all intervals contained in $\Omega$. What exactly are the intervals that are not contained in the set $\mathcal{J}$, but are in the set of all intervals of $\Omega$?

Thank you

答: Assuming you are working with the first printing, there is an errata correcting that mistake that can be found here. It should say that due to proposition $1.2.6$, that $\mathcal{F}$ can not contain all subset of $\Omega$, but it should contain all intervals.

Title: Intervals in measure theory


问:
How can I show that Mathematics is not all about numbers?
My goal is to be able to show that many concepts in Mathematics do not necessarily need numbers, and they can be formalized for sets that do not involve numbers. To explain what I mean, suppose I define the following set of objects:

$$
G = \{phone,volleyball,shoe,water \ bottle\}
$$

I want to know what questions I need to answer so that I can be able to treat set $G$ the same as the set of real numbers $\mathbb{R}$. More precisely, I want to be able to perform the basic operations (addition, subtraction, etc), I want to be able to define functions from the set $G$ to another arbitrary set of objects, and I want to be able to differentiate and integrate these functions.

However, as I have no formal background in Mathematics, I'm interested in resources that can help me do this. Honestly, I am not even sure where to start, so any help would be greatly appreciated.

EDIT

For future reference, based on the comments, I found a free book called Abstract Algebra: Theory and Applications by Thomas W. Judson that I think is helpful.

答:
Let do a little Frame challenging here:

I want to know what questions I need to answer so that I can be able to treat set $G$ the same as the set of real numbers $\Bbb R$.

You can't. One characteristic of $\Bbb R$ is that it has infinitely many elements. Many of the things you do with $\Bbb R$ require infinitely many elements to do. But $G$ has only four elements. You can set up a lot of things on $G$ that look like similar operations on $\Bbb R$, but only if you do not look closely. No matter what you define on $G$, it will not be long before you discover differences in behavior from $\Bbb R$.

Okay, so we abandon the $G$ you defined, and instead pick one with as many elements in it as $\Bbb R$. (This is infinite - indeed a bigger infinity than even the rationals have, so it cannot be physical items. But it doesn't have to be a subset of $\Bbb R$ or $\Bbb C$ or some other acknowledged set of numbers.) Can we define operations on $G$ so that it acts just like $\Bbb R$?

Yes, in theory. Because it is infinite, how one goes about that is going to depend exactly how the set is picked. But we'll just assume that part can be worked out. What you need to define is:

An element "$0$" of $G$ and an operation "$+$" on $G\times G$ with values in $G$ which has the properties

(Additive Identity) For all $x \in G, 0 + x = x$

(Additive Inverse)  For all $x \in G$ there exists a $y \in G$ with $x + y = 0$

(Commutivity of Addition) For all $x, y \in G, x + y = y + x$

(Associativity of Addition) For all $, y, z \in G, (x + y) + z = x + (y + z)$

An element "$1$" of $G$, $1 \ne 0$, and an operation "$\cdot$" on $G \times G$ with values in $G$ which has the properties:

(Multiplicative Identity) For all $x \in G, 1 \cdot x = x$

(Multiplicative Inverse)  For all $x \in G, x \ne 0$, there exists a $y \in G$ with $x \cdot y = 1$

(Commutivity of Multiplication) For all $x, y \in G, x \cdot y = y \cdot x$

(Associativity of Multiplication) For all $, y, z \in G, (x \cdot y) \cdot z = x \cdot (y \cdot z)$

(Distributivity of Multiplication over Addition) For all $x, y, z \in G, x\cdot(y + z) = (x\cdot y) + (x\cdot z)$

A binary relation $\le$ between elements of $G$ satisfying

(Reflexivity) For all $x\in G, x \le x$

(Antisymmetry) For all $x,y \in G$, if $x \le y$ and $y \le x$, then $x = y$.

(Transitivity) For all $x, y ,z \in G$, if $x \le y$ and $y \le z$, then $x \le z$.

(Linearity) For all $x, y \in G$, either $x\le y$ or $y \le x$.

If $x, y, z \in G$ and $x \le y$, $x + z \le y + z$.

If $x, y, z \in G, z > 0$ and $x \le y$, then $x \cdot z \le y \cdot z$.

(Completeness) If $A, B \subseteq G, A \ne \varnothing, B \ne \varnothing, A\cup B = G, A \cap B = \varnothing$ and for all $a \in A, b \in B, a \le b$, then there is some $x \in G$ such that if $y < x$, then $y \in A$ and if $y > x$, then $y \in B$.

(That is, if $G$ is divided into two non-empty sets $A, B$ such that all elements of $A$ are $<$ all elements of $B$, there is a point $x \in G$ where $A$ and $B$ meet. Either $A = \{r\mid r < x\}, B = \{r\mid r \ge x\}$ or $A = \{r\mid r \le x\}, B = \{r\mid r > x\}$. This property is what differentiates $\Bbb R$ from $\Bbb Q$. The sets $\{r\in \Bbb Q\mid r < \sqrt 2\}$ and $\{r\in \Bbb Q\mid r > \sqrt 2\}$ divide $\Bbb Q$ with no point in the middle.)

You get all of that done, and you have a set $G$ that can be used in every way exactly like $\Bbb R$, no matter what the elements of $G$ actually are. So you've shown that Math doesn't require numbers, right? Wrong.

This is where the frame challenge comes in. No matter what the elements of $G$ were. Now they are numbers. Defining those operations turned them into numbers. What makes $\Bbb R$ to be $\Bbb R$ is not the elements of the set. Instead, it is exactly those properties listed. It can be shown that given any two sets that meet all of those properties, you can match up the elements of one to the other in such a way that every property is perfectly preserved by the match.

This is how $\Bbb R$ is unique. You can define the same operations and properties on other sets, but all you get is something that is effectively just $\Bbb R$ again. The underlying objects do not matter.


问:
Lebesgue-measure in $\mathbb{R}^2$
For $f : [0,1] \to \mathbb {R}_{\ge 0}$ let $$A = \{ (x,y) \in [0,1] \times \mathbb {R}_{\ge 0}\,|\,0 \le y \le f(x) \}$$

and $$B = \{ (x,y) \in [0,1] \times \mathbb {R}_{\ge 0}\,|\,f(x) \le y \le f(x)+1 \} .$$

Let $\lambda$ be the Lebesgue-measure in $\mathbb{R}^2$. Which of these following claims is true for all continuous $f\colon [0,1] \to \mathbb {R}_{\ge 0}$?

a. $\lambda (B) = 1$

b. $\lambda (A) = \int _{[0,1]} f(x)\,\mbox {d}x$

c. $\lambda (A) = 1$

d. $\lambda (B) = \int _{[0,1]} f(x)\,\mbox {d}x$

I think that a and b are true, and that c and d are false. Is that correct?

答:
It is correct. We just use Fubini:
$$
\lambda(B) = \int_0^1 \int_{f(x)}^{f(x)+1} 1~\mathrm{d}y ~\mathrm{d}x = \int^1_0 f(x)+1-f(x)~\mathrm{d}x = \int^1_0 1~\mathrm{d}x = 1
$$
$$
\lambda(A) = \int^1_0 \int^{f(x)}_0 1~\mathrm{d}y \mathrm{d}x = \int^1_0 f(x)~\mathrm{d}x
$$
By the way: To see that c and d are false consider $f(x) = 2$ and the (correct) formulas we calculated above.

Title: Lebesgue-measure in $\mathbb{R}^2$


问:
Differential equation system solution: do I get the right solution?
I am very stuck with differential equation systems.

For example:

$
Y'(x) =
    \begin{pmatrix}
    2 & 0 & 1 \\
    0 & 2 & 0 \\
    0 & 1 & 3 \\
    \end{pmatrix}
Y(x)
$

I get the eigenvalues and eigenvectors:

$ \lambda = 2  (double)  \rightarrow   \vec v_{\lambda2} =  \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
$

and

$ \lambda = 3   \rightarrow   \vec v_{\lambda3} =  \begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix}
$

I obtain just one vector asociated with $ \lambda = 2 $ value, so that I suppose I've to get another vector.

For this eigenvalue:

$ (A-\lambda I) = \begin{pmatrix}
    0 & 0 & 1 \\
    0 & 0 & 0 \\
    0 & 1 & 1 \\
    \end{pmatrix}
$

So that the vector I'm looking must meet:

$
\begin{pmatrix}
    0 & 0 & 1 \\
    0 & 0 & 0 \\
    0 & 1 & 1 \\
    \end{pmatrix}
\begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
$

From doing this I get:
$
\begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
\sim
\begin{pmatrix}
    \alpha \\
    -1 \\
    1 \\
    \end{pmatrix}
$, and I choose the vector $
\begin{pmatrix}
    0 \\
    -1 \\
    1 \\
    \end{pmatrix}
$

The problems come from this point. I think the general solution should be:

$
y(x) = C_{1}
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
C_{1}x
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
    0 \\
    -1 \\
    1 \\
    \end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix}
e^{3x}
$

But this result is wrong. I've tried to proove it and it doesn't work.

I've read in some books about it but I think I'm following the mathematic method fine...
Does anyone know what I'm doing wrong?

Thanks in advance.

答:
The eignevalues and eigenvectors looks correect to me.
$$y(x) = C_{1}
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
C_{1}x
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
    0 \\
    -1 \\
    1 \\
    \end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix}
e^{3x}$$
It should be:
$$y(x) = C_{1}
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
\color {red}{C_{2}x}
\begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
    0 \\
    -1 \\
    1 \\
    \end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix}
e^{3x}$$


问:
Tips for solving equations for a variable
I am trying to solve the following equation for $V_{12}$:

$$\sqrt{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}\sqrt{\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}}=\sqrt{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}$$

(this is not really relevant, but it is the derivation of the formula for the relativistic addition of velocities using the Doppler k-factor as presented in this paper).

For me as a highschool student, this isn't an easy equation to solve. Of course, I would square everything to remove the square roots. Then I thought that it might be good to have $V_{12}$ (the variable I'm trying to solve for) on one side and the two fractions with $V_1$ and $V_2$ on the oder side. This way, I got

$$\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}=\frac{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}$$

However, I am a bit clueless what to do next. I thought of multiplying by $\left(1-\frac{V_{12}}{c}\right)$ to get rid of the fraction on the right side, but then I would again have $V_{12}$ on the right side. Whatever I tried seemed not to be helpful (for example, I had the idea to multiply out everything after squaring and then multiply by the denominators to remove the fractions, but this would only lead to overly complicated expressions with even more fractions like $\frac{V_1V_2V_{12}}{c^3}$.

The correct result should be

$$V_{12}=\frac{V_2-V_1}{1-\frac{V_1V_2}{c^2}}$$

but I cannot figure out how to get there.

So: What are gerneral reccomendations and tips to approach solving an equation like this for one variable? (Note that I'm not specifically asking for a derivation of the last mentioned formula from the first since this would probably be outside the scope of this site)

(I also tried using Wolfram Alpha to solve for $V_{12}$ and then analyse the step-by-step solution, but it didn't work)

答:
Ultimately, it looks like your goal is to solve for $x$ in the equation

$$y = \frac{1+x}{1-x}$$

While an oversimplification, I think it should be clear how it applies to your situation. Take $y$ to be your complex fraction, and $x$ to be $V_{12}/c$.

(So I suppose this would be my main tip: boil down your equation into its "essence", in a sense, like the above equation does. Solving for $x$ in the above equation makes it trivial to find $V_{12}$ in your case. A mass of symbols can be daunting and mislead the mind, but by attributing  names and definitions to larger structures, you can simplify something messy into something familiar. For instance, this is a quotient of linear functions, whose inverses are well known.)

To find $x$, simply work as follows:

\begin{align*}
y = \frac{1+x}{1-x}
&\iff y(1-x) = 1+x \\
&\iff y - yx = 1 +x \\
&\iff -x - yx = 1 - y \\
&\iff x(-1-y) = 1 - y \\
&\iff x = \frac{1-y}{-1-y} \\
&\iff x = \frac{y-1}{y+1}
\end{align*}

Of course, bear in mind that $x \ne 1$ and $y \ne -1$ throughout this process. (These represent asymptotes of the function and are never attained, and result in divisions by zero when plugged in.)


问:
If $X_1,X_2,\dots$ are independent, $S_n=X_1+\cdots+X_n$ and $S_n\to S_\infty$ a.s., $\varphi_{S_\infty}(t)=\prod_{j=1}^\infty\varphi_{X_j}(t)$
Problem: Let $X_1,X_2,\dots$ be independent and let $S_n=X_1+\cdots+X_n$. Suppose that $S_n\to S_\infty$ almost surely. Then $S_\infty$ has characteristic function $\prod_{j=1}^\infty\varphi_{X_j}(t).$

My Thoughts: Since the random variables are independent, we have that
$$\varphi_{S_n}(t)=\prod_{j=1}^n\varphi_{X_j}(t).$$
That $S_n\to S_\infty$ almost surely implies that $S_n\to S_\infty$ in distribution as well, so that $\varphi_{S_n}(t)\to\varphi_{S_\infty}(t)$ for all $t\in\mathbb R$. But
$$\varphi_{S_\infty}(t)=\lim_{n\to\infty}\varphi_{S_n}(t)=\lim_{n\to\infty}\prod_{j=1}^n\varphi_{X_j}(t)=\prod_{j=1}^\infty\varphi_{X_j}(t),$$
so we are done.

I think this solution I have cooked up is too easy to be a problem in Rick Durrett's Probability: Theory and Examples. Therefore, I would like to ask if anyone can spot any mistakes in the above argument.

Thank you for your time and feedback.

答: As pointed out by  Kavi Rama Murthy, your thoughts are correct; not all the problems in Durrett's book are hard, although a large part of them are tricky.</p>


问:
Finding the Jordan Form of matrices with mostly 0
I have been studying linear algebra and came across this question which I'm not quite sure how to solve.
The problem asks whether the matrices are similar. I wanted to find the Jordan form of 3 matrices and compare to see if they are similar:
$A=\begin{bmatrix} 0 & 1 & 1 & 0\cr 0 & 0 & 0& 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , }
B=\begin{bmatrix} 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , }
C=\begin{bmatrix} 0 & 0 & 1 & 1\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\end{bmatrix}$

I know that you have to find the eigenvalues from the characteristic polynomial but I'm not sure what the next step is from there. Also, since the matrices mostly contain $0$s is there a shortcut?

Any help would be greatly appreciated!

答:
The magic words are "sparse matrices" (matrices with "most" entries equal to zero, with some specific measure for "most), but here the really important fact is that they are all strictly upper triangular, not that they are sparse.

All three matrices are upper triangular, so the eigenvalues are just the constants in the diagonal. In this case, each of them has a single eigenvalue $\lambda=0$, with algebraic multiplicity $4$.

The dimension of the eigenspace corresponding to $\lambda$ is the nullity. From the Rank-Nullity Theorem/Dimension Theorem, it is easier to get the ranks. It is easy to see that in all three cases the rank is $2$, so the nullity is $2$. That means that the eigenspace has dimension $2$.

The dimension of the eigenspace is equal to the number of blocks in the Jordan canonical form corresponding to $\lambda=0$. Thus, you have only two possible Jordan Canonical forms for these matrices: either two $2\times 2$ blocks, or one $3\times 3$ block and one $1\times 1$ block.

You have two $2\times 2$ blocks if and only if the square of the matrix is $0$. By thinking about what happens to the standard basis upon two applications of the given matrix, it is easy to verify that $A^2\neq 0$ (see what happens to $\mathbf{e}_4$), and that $B^2=C^2=0$.


问:
Identity with Bernoulli numbers: $\sum\limits_{k=1}^{n}k^p=\frac{1}{p+1}\sum\limits_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j}$
How I can prove that
$$\sum_{k=1}^{n}k^p=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j},$$
where $B_j$ is the $j$th Bernoulli number?

I hope to find the answer. Thanks for help.

答:
This should be a quick application of Euler MacLaurin summation formula.

See my previous answer here: https://math.stackexchange.com/a/18989/1102

(Note: I haven't really verified if it is true).