问:
Zig Zags from origin to any point of triangle
There's triangle below. The line is zig zags, and $P_1,P_2,P_3,...P_n$ is where the zig zags is touching the hypotenuse.

I ask, is the distance from the origin, $O$, to any $P_1, P_2...P_n$ irrational? Can the distance from $O$ to $P_n$ be integers, ect.? So I can calculate it for the first triangle, $\Delta OP_1$. The base is 0.5, and height can be found too. The answer is irrational. But what about from $O$ to $P_2$? $O$ to $P_3$? ect.

答:
Let $x_i$ be the distance from $P_i$ to the right edge, with $x_0 = h$.  Each triangle where $P_i$ is a right angle vertex is an isosceles right triangle, its both sides fraction $\sqrt{\frac{1}{2}}$ of the horizontal hypotenuse.  Therefore
$$x_i = \frac{x_{i-1}}{\sqrt{2}} = \frac{h}{2^{i/2}}, \quad 1 \le i \in \mathbb{N}$$
and the distance $d_i$ between $P_{i-1}$ and $P_i$ is
$$d_i = \sqrt{2 (x_{i-1} - x_i)^2} = h \frac{2 - \sqrt{2}}{2^{i/2}}, \quad 1 \le i \in \mathbb{N}$$
Expanding it for odd and even $i$, via $0 \le k \in \mathbb{Z}$,
$$d_i = \begin{cases}
\displaystyle \frac{h}{2^k} \frac{2 - \sqrt{2}}{\sqrt{2}}, & i = 2 k + 1 \\
\displaystyle \frac{h}{2^k} \frac{2 - \sqrt{2}}{2}, & i = 2 k + 2 \\
\end{cases} \tag{1}\label{G1}$$
If $h = 1$, $1 \gt d_1 \gt d_n$, so there are no integer $d_i$ for $i \ge 1$, as OP noted.

Let $\delta_i$ be the distance between $P_0$ and $P_i$, $1 \le i \in \mathbb{N}$:
$$\delta_i = \sqrt{2 (h - x_i)^2} = h 2^{(1-i)/2} \left( 2^{i/2} - 1 \right)$$
and again expanding separately for odd and even $i$ via $0 \le k \in \mathbb{Z}$,
$$\delta_i = \begin{cases}
\displaystyle h \left( \sqrt{2} - \frac{1}{2^k} \right), & i = 2 k + 1 \\
\displaystyle \frac{h}{\sqrt{2}} \frac{2^{k+1} - 1}{2^k}, & i = 2 k + 2 \\
\end{cases} \tag{2}\label{G2}$$
If $h = 1$, then $\delta_3 \lt 1$ and $\delta_4 \gt 1$, and $\delta_i \lt 2$ for all $1 \le i \in \mathbb{Z}$, so there are no integer $\delta_i$ for $1 \le i \in \mathbb{N}$.

If you want to examine whether there are rational $d_i$ or $\delta_i$ depending on $h$, you only need to examine $\eqref{G1}$ and $\eqref{G2}$ for all $0 \le k \in \mathbb{Z}$: for any rational $h$, $d_i$ and $\delta_i$ are irrational.


问:
On "the Hessian is the Jacobian of the gradient"
According to Wikipedia,

The Hessian matrix of a function $f$ is the Jacobian matrix of the gradient of the function $f$; that is: $H(f(x)) = J(\nabla f(x))$.

Suppose $f : \Bbb R^m \to \Bbb  R^n,x \mapsto f(x)$ and $f \in C^2 (\Bbb  R^m)$. Here, I regard points in $\Bbb R^m, \Bbb R^n$ as column vectors, therefore $f$ sends column vectors to column vectors. When $n=1$, we can define $\nabla f: \Bbb R^m \to (\Bbb R^m)^t,x\mapsto\nabla f(x)$, which sends column vectors to row vectors. I  use $(\Bbb R^m)^t$ to denote row vector space, which is just a random notation.

We do have a good definition for functions that sends column vectors to column vectors, but what can we say about functions that sends column vectors to row vectors?

I discovered that if I manipulate $\nabla f(x)$ as a column vector, then I know how to calculate, and my calculation agree with Wiki. But I don't think we can "manipulate $\nabla f(x)$ as a column vector".

答:
The relationship between gradients and Jacobians is the transpose.  Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is some $C^1$ function.

$J_f$ takes a point of $\mathbb{R}^n$ and produces a $1\times n$ matrix that is able to calculate directional derivatives.  That is to say, if $\vec{v}\in\mathbb{R}^n$ is a vector and $p\in\mathbb{R}^n$ is a point, then $J_f(p)\vec{v}$ is a $1\times 1$ matrix whose entry is the directional derivative at $p$ in the $\vec{v}$ direction.

$\nabla f$ takes a point of $\mathbb{R}^n$ and produces a vector in $\mathbb{R}^n$ that can be used to calculate directional derivatives using the dot product.  That is, with the same $p$ and $\vec{v}$ above, $\vec{v}\cdot\nabla f(p)$ is the directional derivative at $p$ in the $\vec{v}$ direction.

Recall that if $\vec{v},\vec{w}\in\mathbb{R}^n$ are vectors, then $\vec{v}\cdot\vec{w}=\vec{v}^T\vec{w}$, where we pretend that $1\times 1$ matrices are the same as scalars for this equation to make sense.  Then, we have that $\vec{v}\cdot\nabla f(p) = (\nabla f(p))^T\vec{v}$, where the latter is commonly written as $(\nabla^Tf(p))\vec{v}$.  Since this represents the directional derivative, too, and it holds for all $\vec{v}$ and $p$, then $\nabla^Tf=J_f$.

Let's take a look at the equation $H_f=J_{\nabla f}$ for the Hessian, where $f:\mathbb{R}^n\to\mathbb{R}$ is a $C^2$ function.  We have that $\nabla f$ is a function $\mathbb{R}^n\to\mathbb{R}^n$, taking points to column vectors, and so $J_{\nabla f}$ is going to be an $n\times n$ matrix.  Using that $(\nabla f(p))_i = \tfrac{\partial f}{\partial x_i}(p)$, then
$$(H_f(p))_{ij} = (J_{\nabla f}(p))_{ij} = \frac{\partial}{\partial x_j}(\nabla f(p))_i = \tfrac{\partial^2f}{\partial x_j\partial x_i}(p),$$
as expected.

ContentLicense: CC BY-SA 4.0


问:
Is this a circular proof of Pythagorean Theorem? If not, are there better ones?
I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":

In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?

If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.

答:
Harry Helson showed us this proof my first semester at Berkeley, using the dot product: $(x+y)\cdot(x+y)=x\cdot x+y\cdot y+2x\cdot y$, which gives the result when $x$ and $y$ are perpendicular.

I like even better the one where a square of side length $c$, the hypotenuse, is offset in a square of side length $a+b$, the sum of the lengths of the legs.


问:
How do you solve the following integral?
How do you solve the following integral
$$\int_0^{2\pi}\sqrt{\frac{5}{4} + \cos(t)} \, dt.$$

It has been a while since I have done integration and I think somehow I could multpily by the conjugate but I am unsure.  Any assistance would be appreciated.

This problem is part of a larger problem to find the length of the cardiod.

$$r(\theta)=\frac{1}{2}+\cos(\theta).$$
This integral is the length integral and I originally just found a numerical approximation but after reviewing my problem my teacher said there is a nice way to antidifferentiate but I was unsure how to go about that.

答:
Let $a\ge1$ and note that, because of symmetry of the integrand, we can write
$$J(a)=\int_0^{2\pi}\sqrt{a+\cos x}\,dx=2\int_0^\pi\sqrt{a+\cos x}\,dx.$$
Write $x=2u$ to get
$$J(a)=2\int_0^{\pi/2}\sqrt{a+\cos2u}\,du=2\int_0^{\pi/2}\sqrt{a+1-2\sin^2u}\,du,$$
which is $$J(a)=2\sqrt{a+1}\int_0^{\pi/2}\sqrt{1-\frac{2}{a+1}\sin^2u}\,du=2\sqrt{a+1}\,E\left(\tfrac{\pi}{2},\sqrt{\tfrac{2}{a+1}}\right),$$
with $$E(\phi,k)=\int_0^\phi\sqrt{1-k^2\sin^2\theta}\,d\theta$$ being the elliptic integral of the second kind. Your integral is $$J(5/4)=3E\left(\tfrac\pi2,\tfrac{2\sqrt2}{3}\right)\approx 6.68244661028.$$


问:
Crossings in $K_n$
I was wondering if we draw a $n$-regular $K_n$, is there a way to determine the crossings in this regular drawing? I know that the crossing number of $K_n$ has a crossing number of

$\frac{1}{4}\lfloor\frac{n}{2} \rfloor \lfloor \frac{n-1}{2} \rfloor \lfloor \frac{n-2}{2} \rfloor\lfloor \frac{n-3}{2} \rfloor$.

However this is not what I am looking for, since I am simply looking for the formula for number of crossings in a $n$-regular $K_n$.

答:
I assume a "regular $K_n$" is the drawing of the complete graph with the vertices equally spaced on a circle and the edges being straight lines connecting the vertices. For clarity, I define the crossing number as the number of pairs of edges which intersect at a point which is not an endpoint of both edges.

There is a nice trick to count the number of crossings in this nice circular drawing of $K_n$: every four vertices yields exactly one crossing, and vice versa. The proof is a simple picture. Four points on a circle span six line segments, with only one interior crossing. Thus the number of crossings for this drawing is $\binom{n}{4}$.

The formula you give is only a conjecture. The above drawing shows that the crossing number is at most $\binom{n}{4} \approx n^4/24$. The conjecture is that the true crossing number should be about $3/8$ times this.

Title: Crossings in $K_n$


问:
Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

This exercise I received on a test, and I didn't resolve it. I would be curious in any demonstration with explanations. Thanks!

答: Hint: Make a Newton polygon for the prime $p=3$. Use the corollary at the top of page 2 in these notes by Paul Garrett (alternatively, here are screenshots: page 1, page 2).

ContentLicense: CC BY-SA 3.0


问:
Finding a cdf of a random variable X given it's pdf
A random variable $X$ is given with pdf $f(x) =\Bigg\{cxe^\frac{-x}{10}$ if $x\geq0$ and $0$ otherwise

(a) find c to make this pdf correct

(b) find the cdf F(x)

For (a) I used integration by parts for $\int_0^\infty cxe^\frac{-x}{10}dx$ to get the answer $-10cxe^\frac{-x}{10} - 100ce^\frac{-x}{10}$. Since the total integral of the pdf must equal 1, I found that $c = \frac{1}{100}$

For (b) integrated $\int_0^\infty \frac{1}{100}xe^\frac{-x}{10}dx$ and came out with

$$F(x) = \begin{cases} -\frac{1}{10}xe^\frac{-x}{10} - e^\frac{-x}{10} & \text{if } x\geq0, \\0 & \text{otherwise} \end{cases} $$

Obviously this must be wrong because the values of the cdf would be negative for all $x\geq0$. Can anyone help me out?

答:
Your value for $c$ is correct.

For the cdf, you don't integrate to infinity, you integrate up to $x$. And you have the correct formula except for a $+ 1$ from evaluating the second integration by parts at the limit $x=0$.


问:
Non-standard model of naturals and Löwenheim–Skolem
I'm taking a beginner course in mathematical logic. In the proof of some properties of non-standard natural numbers, the lecturer has used the downward Löwenheim–Skolem theorem, which I didn't understand. The version of Löwenheim–Skolem theorem that I know is:

Suppose $L$ is countable first order language, and $\mathcal{A}$ is an $L$-structure. Let $T$ be the theory of $\mathcal{A}$, i.e. $T = Th(\mathcal{A}) =\{ \theta: \mathcal{A} \models \theta \text{ where } \theta \text{ is an L-sentence}\}$. Then, there is a countable model $\mathcal{B}$ such that $\mathcal{B}\models T$.

In the lecture, the Peano Arithmetic Axioms are expressed in logical formulas. These axioms are denoted $P_1, \dots, P_n$. We define a model $\mathcal{N} =\langle \mathbb{N}, <, \operatorname{succ}, 0\rangle$. We define the set $\Sigma:= \{c>t_m (0): m \in \mathbb{N}\}$, where $c,0$ are $L$-constants, and $t_m:= \operatorname{succ}(\operatorname{succ}(\dots \operatorname{succ}(0)\dots)$, where $\operatorname{succ}$ has been applied $m $ times.

In the lecture, we prove that $\Sigma \cup Th(\mathcal{N})$ has a model, which is not the model $\mathcal{N}$. Then, we use the "downward Löwenheim–Skolem theorem to conclude that there is a countable model $\mathcal{M} \models \Sigma \cup Th(\mathcal{N})$". This is the part that I don't understand. No additional explanation was given, so this is probably something simple that I'm missing here. Can anyone help me understand this bit? (How can I know that $\Sigma \cup Th(\mathcal{N})$ is some theory of some $L$-structure, which is a condition of being able to use the Löwenheim–Skolem theorem?)

答: Let $\mathcal{A}$ be a model of $\Sigma \cup Th(\mathcal{N})$.  Then $Th(\mathcal{A})$ contains $\Sigma \cup Th(\mathcal{N})$.  So, if $\mathcal{M}$ is a countable model of $Th(\mathcal{A})$, it is also a model of $\Sigma \cup Th(\mathcal{N})$.


Title: Finding a cdf of a random variable X given it's pdf

Body:
A random variable $X$ is given with pdf $f(x) =\Bigg\{cxe^\frac{-x}{10}$ if $x\geq0$ and $0$ otherwise

(a) find c to make this pdf correct

(b) find the cdf F(x)

For (a) I used integration by parts for $\int_0^\infty cxe^\frac{-x}{10}dx$ to get the answer $-10cxe^\frac{-x}{10} - 100ce^\frac{-x}{10}$. Since the total integral of the pdf must equal 1, I found that $c = \frac{1}{100}$

For (b) integrated $\int_0^\infty \frac{1}{100}xe^\frac{-x}{10}dx$ and came out with

$$F(x) = \begin{cases} -\frac{1}{10}xe^\frac{-x}{10} - e^\frac{-x}{10} & \text{if } x\geq0, \\0 & \text{otherwise} \end{cases} $$

Obviously this must be wrong because the values of the cdf would be negative for all $x\geq0$. Can anyone help me out?

Body:
\begin{align}
\text{For } x\ge0, \\[8pt]
F(x) = {} & \int_0^x \left( \frac u {10} \right) e^{-u/10} \, \frac{du}{10} = \int_0^{x/10} v \Big( e^{-v} \, dv\Big) \\[8pt]
= {} & \int v\, dw = vw - \int w\,dv \\[8pt]
= {} & \Big[{-v}e^{-v} \Big]_0^{x/10} - \int_0^{x/10}-e^{-v} \, dv \\[8pt]
= {} & - \frac x {10} e^{-x/10} + \Big[ {-e}^{-v} \Big]_0^{x/10} \\[8pt]
= {} & -\frac x {10} e^{-x/10} -e^{-x/10} + 1. \\[8pt]
& \begin{cases} =0 & \text{when } x=0, \\
>0 & \text{when } x>0, \\
\to1 & \text{as } x\to+\infty. \end{cases}
\end{align}


问:
Leap year probability
A year in the 2020s (i.e., from 2020 to 2029 inclusive) is selected uniformly at random, a month
is selected uniformly at random from that year, and a day of the month is selected uniformly at
random from that month.

(a) What is the (exact) probability that the day is the 29th of February?
->I use 3/10 * 1/12 * 1/29 [not very sure]

(b) Given that the day is the 29th, what is the probability that the month is February?
For this one, I know I should use Baye's rule, but cannot proceed further. Is anyone willing to help?

答:
Your answer (a) is correct.

You can calculate the probability $P_{31}$ that the day you draw is the 29th of a month with exactly 31 days:
$$P_{31}= \frac{7}{12}×\frac{1}{31}.$$ [Duly note that every year has exactly 7 months with 31 days.]

And then you can calculate the probability $P_{30}$ that the day you draw is the 29th of a month with exactly 30 days:
$$P_{30}=\frac{4}{12}×\frac{1}{30}.$$

You already calculated above in (a) correctly that the probability $P_{29}$ that the day you draw is the 29th of February, 2020, 2024, or 2028.
$$P_{29}=\frac{3}{10}×\frac{1}{12} ×\frac{1}{29}.$$

Then the probability $P$ that you draw is the 29th of a month, satisfies
$$P = P_{31}+P_{30}+P_{29}.$$ And so, the probability that you the day you draw is the 29th of some February given that the day you drew is the 29th of some month, is $\frac{P_{29}}{P} = \frac{P_{29}}{P_{31}+P_{30}+P_{29}}.$ Check to make sure you see why this is.

Can you use this to work out the numbers and finish (b).

Title: Leap year probability


问:
Finite group with a special property
Let $(G,\cdot)$ be a finite group  with the identity element $e$ such that for each proper subgroup $H$ of $G$ there exists a unique element $g_H\in H$, $g_H\ne e$, such that the following properties hold:

(a) $g_H^2=e$;

(b) for any two distinct proper subgroups $H,K$ of $G$ we have $g_H\ne g_K$.

Then the group $G$ is abelian.

Thus far, I have considered a prime number $p$ dividing the order of $G$. Then, by Cauchy's theorem, there exists an element of $G$ of order $p$ and consider the subgroup $H:=\langle x\rangle=\{e,x,\dots,x^{p-1}\}$. By hypothesis, there exists $g_H=x^r$ (with $r\in \{1,\dots,p-1\}$) and $g_H^2=x^{2r}=e$. Then, we have $p\mid 2r$ and it follows $p=2$. Therefore, the order of $G$ is a power of 2.

答:
There is a simpler way than the accepted answer.

Again as you say in the question, the group has order a power of $2.$ I claim that every element has order $2.$ Indeed, if not, there is an element $g$ of order $4.$  The subgroups generated by $g$ and $g^2$ have the same element of order $2$ ($g^2$) contradicting the hypothesis. So every element has order $2.$ But such a group is abelian ($ab(ba)^{-1} = a b a^{-1} b^{-1} = a b a b = (ab)^2 = e.$)

Title: Finite group with a special property


答:
Your answer to a is correct. You must choose a leap year, and then you must choose February, and then you must choose the 29th.

For part (b), $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ Setting $A = $ "February" and $B = $ "29th", we find $P(A) = 1/12$, regardless of the year chosen.

Finding $P(B)$ is the sum of all the ways it can happen.  $$\frac{1}{10}\cdot \frac{1}{12} \cdot \frac{1}{31}$$ is the probability of getting January 29th, 2020. Likewise there are another $112$ possibilities. We just need to group them.  $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{1}{31} + \frac{1}{31} + \frac{1}{30} + \frac{1}{31} + ...]$$ has eleven terms, $7$ of which are $\frac{1}{31}$ and $4$ of which are $\frac{1}{30}$, giving us $$\frac{7}{10}\cdot \frac{1}{12}\cdot [\frac{7}{31} + \frac{4}{30}]$$ gives the probability for non-leap years. Add to that $$\frac{3}{10} \cdot \frac{1}{12} \cdot [\frac{7}{31} + \frac{4}{30} + \frac{1}{29}]$$ to get the total for $P(B)$.

Title: Leap year probability

Body:
A year in the 2020s (i.e., from 2020 to 2029 inclusive) is selected uniformly at random, a month
is selected uniformly at random from that year, and a day of the month is selected uniformly at
random from that month.

(a) What is the (exact) probability that the day is the 29th of February?
->I use 3/10 * 1/12 * 1/29 [not very sure]

(b) Given that the day is the 29th, what is the probability that the month is February?
For this one, I know I should use Baye's rule, but cannot proceed further. Is anyone willing to help?


答:
You have the indefinite integral correct: $$\int c x \mathrm e^{-x/10}\mathrm d x = -10c\,(x+10)\,\mathrm e^{-x/10}$$

Therefore you evaluate $c=1/100$ correctly because the positive infinte limit of the cummulative distribution function shall be 1. $~$ $\lim\limits_{x\to\infty} F(x)=1$ $$\begin{align}1&=\int_0^\infty c x\mathrm e^{-x/10}\mathrm d x \\&= \left[-10c\,(x+10)\,\mathrm e^{-x/10}\right]_{x=0}^\infty\\&=(0)-(-100c)\\&=100 c\end{align}$$

To find the Cummulative Distribution Function you simply do not integrate all the way to the infinite:

$$\begin{align}F(x)&=\mathbf 1_{0\leqslant x}~\int_0^x 10^{-2} s\mathrm e^{-s/10}\mathrm d s \\&= \mathbf 1_{0\leqslant x}~\left[-10^{-1}\,(s+10)\,\mathrm e^{-s/10}\right]_{s=0}^x\\&=((-10^{-1}(x+10)\mathrm e^{-x/10})-(-1))~\mathbf 1_{0\leqslant x}\\[2ex]&=(1-10^{-1}(x+10)\mathrm e^{-x/10})~\mathbf 1_{0\leqslant x}\end{align}$$


问:
Probability that the product of two numbers in $\{0,1,...,10^{n - 1}\}$ contains $2n - k$ digits, $k = 0,1,...$
How to find the probability that the product of two numbers in ${0,1,...,10^n - 1}$ contains $2n-k$ digits, $k = 0,1,...$? As I understand, the sum of all the combinations of two numbers is equal to $10^{2n}$. But I can't understand how to find the sum of all the positive combinations. I understand that the product of these numbers is in $10^{2n-k-1} \le x*y < 10^{2n-k}$. But it doesn't help me. Explain me please, how I can find the sum of all the positive combinations.

答:
I would express the probability in terms of logarithms, base $(10)$.

The positive integer $n$ will have exactly $k$ digits if and only if

$10^{k-1} \leq n < 10^k \implies (k-1) \leq \log_{10} n < k.$

Further, $\log_{10}(ab) = \log_{10}a + \log_{10}b.$

For a given fixed $n$ and fixed $k$, let $r$ denote $(2n-k).$

Let $S$ denote the set $\{1,2,\cdots, 10^{n} -1\}.$

Notice that I have necessarily exluded the element $0$ from the set $S$,
because (among other things) $\log_{10} (0)$ is undefined.

It shouldn't be too onerous to modify the discussion below to include consideration
of what happens if the set $S$ is appended with the element $(0)$, and
one or both of the numbers chosen is $(0)$.

There are exactly $D = (10^n -1)^2$ ways of selecting two numbers at random from
the set $S$ such that the selection is done with replacement.

Thus, when attempting to compute a probability, $D$ will represent the denominator.

Therefore, the problem is reduced to computing the numerator $N$, for a specific value of
$r$.

Edit

Thanks to prostak for indicating that my original computation of $N$ (below) was flawed.

Answer has been edited.

Given any element $s \in S$:

Let $f(s)$ to denote the smallest positive integer $a$ such that

$\log_{10}s + \log_{10}a \geq (r-1).$

Let $g(s)$ to denote the largest positive integer $b$ such that

$\log_{10}s + \log_{10}b < (r).$

Note that $g(s)$ is undefined, for $s \geq 10^r.$

Note:

To facilitate the summation formula at the end of this answer, I am going to employ
the following contrivance:

For $s < (10)^r,$ let $h(s)$ denote $g(s) - f(s) + 1.$

For $s \geq (10)^r,$ let $h(s)$ denote $0$.

Then $$N = \sum_{i=1}^{10^n - 1} h(i) = \sum_{i=1}^{10^r - 1} h(i).$$

Edit

While preparing my Addendum-2 response to prostak's comment/question, I realized that the expression above is wrong.  It overlooks that

$r = (2n - k)$ might be larger than $n$.

In such a circumstance, given the fact that $s$ is required to be $\leq (10^n - 1),$ it would not make any sense to have the summation exceed $(10^n - 1).$
Therefore, the corrected expression is:

$$N = \sum_{i=1}^{10^n - 1} h(i).$$

Since the artificial contrivance that
$i \geq 10^r \implies h(i) = 0$ is being used,

the situations where $r = (2n - k) < n$ continue to be handled okay.

Addendum

Responding to the comment/question of prostak

As I understand

$$N ~=~ \sum_{i=1}^{10^n - 1} \frac{\left(10^{2n-k} - 10^{2n-k-1}\right)}{i} ~+~ 1.$$

But how can we get this sum if this row is divergent?

First of all, my expression for $N$ is

$$N ~=~ \sum_{i=1}^{10^n - 1} h(i) = \sum_{i=1}^{10^r - 1} h(i)\tag1$$

My expression is intended to represent the exact number of ordered pairs

$(a,b)$ such that $a,b \in \mathbb{Z^+}$, $a,b$ are each less than $(10^n)$,

and the product $(a \times b)$ has exactly $r = (2n - k)$ digits.

The idea is that there are exactly $D = (10^n - 1)^2$ ordered pairs $(a,b)$

that one might choose at random, where each component of the ordered pair is in

$\{1,2,\cdots, (10^n - 1)\}$.

Therefore, the chance of a
a satisfying ordered pair occurring, at random, is $\frac{N}{D}.$

If I understand your comment/question correctly, you are comparing the
expression in equation (1) above to the following expression:

$$N ~=~ \sum_{i=1}^{10^n - 1}
\frac{\left(10^{2n-k} - 10^{2n-k-1}\right)}{i} ~+~ 1.\tag2$$

You are then indicating that the expression in equation (2) above is
"divergent".

I will be happy to respond by adding a second (Addendum-2) section to
my answer.  But first, I need you to edit your original query, adding your own
Addendum, that does the following:

Demonstrate, in very clear detail, how you derived equation (2) above, from
equation (1) above.

My description of $N$ is that it is the number of ordered pairs $(a,b)$ such that $(a \times b)$ has exactly $r = (2n - k)$ digits.
I need you to explain in English, whether you are also using the variable $N$
to signify the same thing.  If not, I need you to explain what you are
intending that the variable $N$ should signify.

You have also indicated that the expression in equation (2) above is
"divergent".  That expression refers to a finite number of terms,
each of which is well defined (and therefore finite).  Therefore, I am
confused by what you intend by the description "divergent".  Please
explain further, again providing very clear mathematical detail.

Given the questions that I am posing, I feel that it is not appropriate for you to attempt to provide the requested details in
one or more comments.  Instead, please edit your query to provide the details.

Addendum-2

Responding to the comment/question of prostak:

Nice detail provided as you were following up your question.  One minor point:

The comments don't really lend themselves to complicated math discussions, although
you've obviously mastered using MathJax nicely.  It is better to edit your query,
as I edit mine, for this type of back and forth.

Chat room is another alternative, but it is somewhat difficult because we have
conflicting schedules and because I (for one) haven't found a way to render
math nicely in the chat room.  So, if we need to continue going back and forth,
let's do it

using query <--> answer (i.e. you edit your query and I will edit my answer), and simply leave short flagging comments, to
notify each other.

$\underline{\text{Your comment/question}}$

You indicated that:

$$\text{if} ~s = 2, f(s) \geq \frac{10^{2n-k-1}}{2}~ \text{and} ~g(s) < \frac{10^{2n-k}}{2}.\tag3$$

$$\text{if} ~s = 3, f(s) \geq \frac{10^{2n-k-1}}{3}~ \text{and} ~g(s) < \frac{10^{2n-k}}{3}.\tag4$$

You then concluded that

$$\text{if} ~s = 10^n - 1, f(s) \geq
\frac{10^{2n - k - 1}}{10^n - 1}~ \text{and}
~g(s) < \frac{10^{2n-k}}{10^n - 1}.\tag5$$

You then inferred that

$$h(i) = \frac{10^{2n-k} - 10^{2n-k-1}}{i} + 1\tag6$$

and that we need to sum $h(i)$ in $[1,10^n - 1]$ which is a problem, because
this sum hasn't a well defined evaluation.

$\underline{\text{My Responses}}$

Best to take these points one at a time.

First, notice the correction that I made in the specification of $N$, just before the (first) Addendum.

For any real number $r$, let $\lfloor r\rfloor$ denote the largest
integer $\leq r$ (i.e. the floor of $r$).

For any real number $r$, let $\lceil r\rceil$ denote the smallest
integer $\geq r$ (i.e. the ceiling of $r$).

Assuming that $n,k$ fixed, with $r = (2n-k)$, then, for any positive integer $s$,

my definition of $f(s)$ is that it is the smallest positive integer $a$ such that
$\log_{10} s + \log_{10} a \geq (2n - k - 1)$.

From this definition, you know that $s \times a \geq 10^{2n - k - 1}.$

This implies that $a \geq \frac{10^{2n - k - 1}}{s}$.

Since $a$ is required to be a positive integer, my definition of $f(s)$ may be
equivalently expressed as

$$f(s) = \left\lceil \frac{10^{2n - k - 1}}{s}\right\rceil$$

Similarly, my definition of $g(s)$ is that

if $~s \geq 10^{2n - k},~$ then $~g(s)~$ is undefined.

Otherwise $g(s)$ is the largest positive integer $b$ such that
$\log_{10} s + \log_{10} b < (2n - k)$.

From this definition, you know that when $s < 10^{2n - k},$
and $b$ is as specified above

then $s \times b < 10^{2n - k}$, which implies that
$b < \frac{10^{2n-k}}{s}.$

Since $b$ is required to be a positive integer, my definition of $g(s)$ may be
equivalently expressed as

$$ s \geq 10^{2n-k} \implies g(s) ~~\text{is undefined.  Otherwise}~~
g(s) = \left\lfloor \frac{10^{2n - k} - 1}{s}\right\rfloor.$$

The inequalities expressed in equations (3) and (4) are accurate, though (perhaps) misleading, since they don't focus on the fact that both $f(s)$ and $g(s)$ are required to be positive integers.

With respect to equation (5) above, I would also point out that $g(s)$ is undefined when $s \geq 10^{2n - k}.$  Therefore, since equation (5) is specifically focusing on $(s = 10^n - 1)$, the specification of $g(s)$ in equation (5) is incorrect, in those cases where $(2n - k) < (n).$

Finally, your specification for $h(i)$ in equation (6) is wrong and should be:

$$\text{When} ~~i \geq 10^{2n -k}, ~h(i) = 0.$$

$$\text{Otherwise} ~~h(i) ~=~
\left\lfloor \frac{10^{2n - k} - 1}{i}\right\rfloor ~-~
\left\lceil \frac{10^{2n - k - 1}}{i}\right\rceil ~+1.$$

This should resolve your inference that for some values of $i, ~h(i)$ can't be evaluated.  If you have further questions, please edit your query and then leave me another flagging comment.


问:
Is the space of polynomial isometric to $\mathbb R^n$?
Consider the space of polynomials of degree $n-1$ defined on $[0,1]$ with $\| p(x)\| = \max_{x\in[0,1]}\vert p(x) \vert$. Is it isometric to $\mathbb R^n$? Is it possible to find a bijective isometry?

答:
I think you meant polynomial of degree at most $n-1.$ And, I also feel, you meant absolute maximum, to be your norm.

Can of find extreme points of the unit ball of $\mathbb{R}^n$ and the space you mentioned. Note that number of extreme points is "something" preserved under isometric isomorphism.


问:
In this sample, are the two events independent?
This equation comes from Edgenuity's course of Statistics, and I am taking the course as a high school senior. I understand that for two events to be independent, $P(A) = P(A|B)$. I also know how to find $P(B|A)$. I attempt to do this in my solving of the problem below:

According to a recent survey of adults, approximately 62% carry cash on a regular basis. The adults were also asked if they have children. Of the 46% who have children, 85% carry cash on a regular basis. Is carrying cash independent from having children in this sample?

First, I assume $P(A)$ is alike to $P$(carrying cash), which would equal $0.62$. $P(A|B)$ then is alike to $P$(carrying cash|having children). I am unsure of how to find this second probability. I divided $P$(carrying cash and having children), which was assumed to be $0.2852$ when multiplying $0.62$ and $0.46$, by $P$(carrying cash), which is $0.62$. If my method was correct, then $P$(carrying cash|having children) $= 0.46$, and my conclusion would then be that carrying cash is not independent from having children in this sample because $0.46 ≠ 0.62$. Are my methods and answer correct? If not, please explain why.

答: If the events were independent, then $62$% of those having children would carry cash.  Since $85$% of them carry cash, the events are not independent.


问:
Struggling with integration in a Differential Equations problem
This is part of a hw problem for school. (the original problem is the DE $x' = x^2 -4$)
It is a differential equations course but I'm stuck on a calculus part.
I plugged the integral to wolfram alpha and can't follow why the
    integral of $\displaystyle \int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \left(\log(u+1)-\log(1-u)\right)+ C$</p>

答: Hint: $\dfrac{1}{1-u^2}=\dfrac{A}{1-u}+\dfrac{B}{1+u}$, for some $A,B\in \Bbb R$. Find $A$ and $B$.


问:
What is so special about the generator chosen for Reed solomon Codes?
I am a bit confused about the Reed Solomon codes which we are dealing with in our course right now.
We first had linear codes than cyclic codes and at the end we had the Reed Solomon codes.

First I would like to tell you what I have understanded until now.

Cyclic Codes

Let $C$ be a $(n,k)$-cyclic code with $C \subset \mathbb{F}_q^n$ for $q$ prime.

$(c_0,c_1,\ldots,c_{n-1}) \leftrightarrow c_0+c_1x+\ldots+c_{n-1}x^{n-1} \in \mathbb{F}_q[x]/(x^n-1)$

Under these circumstances the set of all these polynomials constructed by the codewords of $C$, we call it $C[x]$, forms a principal ideal, since $\mathbb{F}_q[x]$ is a principal ideal ring. So there is an element in $C[x]$ which generates this ideal and this generator is, as far as I understood, crucial for the decoding process. I know that this generator $g(x)$ is a divisor of $x^n-1$ in $\mathbb{F}_q[x]$. My first question here is:

Is this generator unique and what is the best way to calculate it?

If a have messages of length $k$ than the message polynomials have degree $k-1$ and so the generator polynomial needs to have degree $n-k$. Would it be possible to factor $x^n-1$ and than just use a product of these factors so that I get a polynomial of degree $n-k$ ? Would this polynomial be a generator which fulfills my needs ?

After finding the generator we can construct a decoding technique. Let $r(x)$ be my received polynomial. We divide $r(x)$ by $g(x)$ and if the remainder is zero, than there was no error. But if it is not, than we just look at our syndrom table of remainders which completely depends on the error polynomial. So when we know the remainder we know exactly where the error happened.

Reed Solomon Codes

So now we are discussing the Reed Solomon Code. Here my professor first specifies the minimum distance as $d = 2t + 1$ so that we can correct $t$ errors. From what I know Reed Solomon Codes also fulfill the singleton bound("in the best way"). It is $d = n-k+1$. So if we know $k$, what we usually do and we want to correct $t$ errors than $n = k+2t$. So now here comes the thing that I don't really understand. Reed Solomon Codes are cyclic codes and so we also need some generator polynomial and here my professor just states:

$g(x) = (x-\alpha)(x-\alpha^2)\cdot\ldots\cdot(x-\alpha^{2t})$, where $\alpha$ is a primitve Element of $\mathbb{F}_q$.

It would be good to know what the point of choosing the generator like this is. I know that for Reed Solomon Codes we use another method for decoding. From what I understanded we use the fact, that we exatly know the zeros of our generator. If I have a received message $r(x) = c(x) + e(x)$, than I know that:

$r(\alpha^i) = c(\alpha^i)+e(\alpha^i) = e(\alpha^i)$, $i\in\{1,\ldots,2t\}$

As far as I understand, we now try to find the error polynomial through this system of equations, which seems to be the hardest part compared to the other steps. I am asking myself why we are considering this special generator. We maybe could have used some other generator where we also know the zeros and proceed in the same way.

答:
Is this generator unique?

The generator polynomial is unique if $q=2$ but not if $q>2$. This because if $g(x)$ is a generator polynomial, then so is $\alpha g(x)$ a generator polynomial for any nonzero $\alpha \in \mathbb F_q$.  Usually, but not always, the generator polynomial is defined as the monic polynomial of least degree in $\mathbb{F}_q[x]/(x^n-1)$ which makes it unique.

With regard to Reed-Solomon codes, there are numerous questions, many with detailed answers, about Reed-Solomon codes on math.SE. Do a search, and you will likely get a lot more helpful information.  Reed-Solomon codes are important because they are decodable using straightforward though somewhat time-consuming computation, a notion that has not entered your consciousness as yet since you blithely say about decoding a general linear code

..... if the remainder is zero, than there was no error. But if it is not, than we just look at our syndrom table of remainders which completely depends on the error polynomial. So when we know the remainder we know exactly where the error happened.

while completely ignoring the enormous storage requirements that real-life syndrome tables require. It is one thing to use a toy example as is often done in coding theory courses and textbooks to illustrate a syndrome decoding table, and quite another to use one in a practical system.  For example, the "NASA standard" $[255,223]$ Reed-Solomon code over $\mathbb F_{2^8}$ has a syndrome of $32$ $\mathbb F_{2^8}$ symbols (that is, $32\times 8 = 256$ bits, so the syndrome table would have $2^{256} >0.64\times 10^{77}$ entries of $255$ (or $223$) bytes each (say $2000$ bits each) for a total exceeding $10^{88}$ bits of storage.  In comparison, consider that the number of atoms in this planet has been estimated at around $10^{50}$ and the number of atoms in the known universe to be about $10^{80}$ (see here on physics.SE).   Thus, the syndrome table idea is not feasible in real life; don't go about touting it as the solution and denigrating Reed-Solomon codes as being very hard to decode in comparison to the simplicity of syndrome table lookup. Reed-Solomon decoders are very practical and appear in the innards of many devices including some that might be on your worktable or stereo cabinet or even your shirtpocket.


问:
How to find bounds of integration for this question?
I am currently studying joint pdfs and am struggling with finding the bounds of integrations for this problem.
Consider a pair of random variables X, Y with constant joint
density on the triangle with vertices at $(0, 0)$, $(3, 0)$, and $(0, 3)$. Find $P(X +Y > 2)$.

The fact that it says constant density tells me that this is continuous uniform with area $0.5 \cdot 3 \cdot 3 = 4.5$ so $f(x) = 2/9$

Since $P(X +Y > 2)$ I thought that this would mean that the bounds of integration would be $\displaystyle\int_2^3\int_{2-x}^3\cfrac{2}{9}dydx = \cfrac{7}{9}$

But my book says the answer is $\cfrac{5}{9}$

What am I misunderstanding here? What are the proper bounds of integration and why?

答:
The endpoints of the integral are wrong in a few places: it should be
$$
\int_0^3\int_{\max\{2-x,0\}}^{3-x}\cfrac{2}{9} \, dy \, dx = \frac59.
$$

When you put the lower limit of the first integral equal to $2$, that prohibited $X$ from being less than $2$;

Putting $2-x$ as the lower limit of the second integral is an error because $2-x$ can be negative, and $f(x,y)=0$ for $y<0$;

Having $3$ as the upper limit of the second integral allowed pairs $(x,y)$ with $x+y>3$, contrary to the initial domain of the random variables.


问:
For two odd primes $p<q$, can we deduce positive integers $a,b$ solving $a^2+b^4=pq$ without trial & error (brute force)?
Let us fix two primes $p,q$ with $2<p<q$. How can we find positive integers $a,b$ which solve the equation $a^2+b^4=pq$ without brute force?

Interestingly there exist sometimes two solutions:

$5\cdot13=7^2+2^4=8^2+1^4$

$5\cdot821=3^2+8^4=53^2+6^4$

$17\cdot113=25^2+6^4=36^2+5^4$

In more rare cases I even obtain three solutions, e.g.:

$73\cdot89=49^2+8^4=64^2+7^4=79^2+4^4$

Is there a systematic way to obtain the solutions $a,b$ directly? Or at least, can we establish a relationship between the two primes and these natural solutions? I generated a CSV file containing some more cases, which maybe help in finding patterns or connections between $p,q$ and $a,b$.

Can we anticipate (state in advance) how many positive integer solutions we will get depending on $p,q$?

答:
Given $\quad a^2+b^4=pq\quad$
we can use a formula for generating Pythagorean triples where $C-A=2\qquad$ Here, $\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\qquad$ If we let $\quad b^4=4n\quad$ we find integer $n-values$ when $B\in\big\{2,4,6,8,\cdots\big\}$ These correspond to the triples

$$(63,16,65)\qquad (16383,256,16385)\qquad (419903,1296,419905)\\ (4194303,4096,4194305)\qquad (24999999,10000,25000001)\qquad  \cdots$$

In these cases, we can see the $C$=values are composite and easily represented as $pq$, but whether or not $pq$ are prime must be determined for each case.

We can find more "candidates" but still have to check $pq$ for primality by using Euclid's formulas shown here as $ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2.\qquad$ If we solve the $B$-function for $k$, we need only test a defined range of $m$=values to see which yield integers for $k$.

\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$
$$B=16\implies\qquad \bigg\lfloor \frac{1+\sqrt{32+1}}{2}\bigg\rfloor =3 \le m \le \frac{16}{2}=8\\ \land \quad m\in\{8\}\implies k\in\{1\}$$
$$F(8,1)=(63,16,65)\qquad $$

This generates many non-primitives and a few more primitives where each $C$-value must be screened as above. Those primitives corresponding to the above $B$-values are.

$$f(8,1)=(63,16,65)\qquad  f(128,1)=(16383,256,16385)\\ f(81,8)=(6497,1296,6625)\qquad f(648,1)=(419903,1296,419905)\\ f(2048,1)=(4194303,4096,4194305)\quad f(625,8)=(390561,10000,390689)\\
 f(5000,1)=(24999999,10000,25000001)\qquad $$


问:
Question about the relation between integration and differentiation (From Calculus by Apostol)
I got stuck while doing exercise of the Apostol's Calculus, the exercise 28 of Section 5.5.

Here's the question

Given a function $f$ such that the integral $A(x) = \int_a^xf(t)dt$ exists for each $x$ in an interval $[a, b]$. Let $c$ be a point in the open interval $(a, b)$. Consider the following ten statements about this $f$ and this A:

And there are five (a) ~ (e) statements on the left, and five ($\alpha$) ~ ($\epsilon$) statements on the right. The author asks the reader to decide the implicative relation from statements on the left to statements on the right. I thought I answered correctly but the solution at the end tells different. I don't know why this is wrong.

(d) $f'(c)$ exists.  $\implies$ ($\epsilon$) $A'$ is continuous at c.

This is my argument:
By the Example 7 of Section 4.4, the differentiability of $f$ at c implies the continuity of $f$ at c. Since $f$ is differentiable at c, $f$ is continuous at c, so that $A'$, which equals to $f$, should continuous at c.

But the solution at the end says (d) does not implies ($\epsilon$).

Sorry for the partializing the problem, it maybe tough to point out what is wrong.

答:
Paramanand gives an explicit counterexample, which proves that your claim is false. But I want to add this further answer to expose the flaw in your reasoning.

Your argument appears to rely on the following missing assumption: if $g$ and $h$ are two functions defined on a set containing $c$, $g(c)=h(c)$, and $g$ is continuous at $c$, then $h$ is continuous at $c$. (If this were true, your conclusion would follow. In your case, $g=f$ and $h=F’$.)

This assumption is false. Here’s a very simple counterexample. Take $g(x)=0$ on the real line and let $h(x)=1$ for all $x$ except zero, where $h$ jumps to zero. Then $g(0)=0=h(0)$ and $g$ is continuous at zero, but $h$ is not continuous at zero.


问:
If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod a$ and $z\equiv y\pmod b$

If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod{a}$ and $z\equiv y\pmod{b}$

What I have so far:

Let $z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$ and $
z \equiv y\pmod b
 $. Then by the chinese remainder theorem there is a unique $z\pmod{\text{lcm}(a,b)}$ which satisfies this...

Is this the right approach here? I can't figure out how to get from $$z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$$ what I need.

答:
Existence

Bezout's Identity says that we have $g_a,g_b$ so that
$$
g_aa+g_bb=\gcd(a,b)\tag{1}
$$
Thus,
$$
\begin{align}
g_bb&\equiv\gcd(a,b)&\pmod{a}\\
g_bb&\equiv0&\pmod{b}
\end{align}\tag{2}
$$
Since $x\equiv y\pmod{\gcd(a,b)}$, we have $\gcd(a,b)\mid x{-}y$. Multiplying $(2)$ by $\frac{x-y}{\gcd(a,b)}$ gives
$$
\begin{align}
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}\gcd(a,b)=x-y&\pmod{a}\\[6pt]
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}0=0&\pmod{b}
\end{align}\tag{3}
$$
Adding $y$ to $(3)$ and setting $z=\frac{x-y}{\gcd(a,b)}g_bb+y$, we have
$$
\begin{align}
z&\equiv x&\pmod{a}\\
z&\equiv y&\pmod{b}
\end{align}
$$

Uniquenes

Suppose that $z_1\equiv x\pmod{a}$ and $z_2\equiv x\pmod{a}$. Then $z_1\equiv z_2\pmod{a}$.

Suppose that $z_1\equiv y\pmod{b}$ and $z_2\equiv y\pmod{b}$. Then $z_1\equiv z_2\pmod{b}$.

Therefore, we have $z_a,z_b$ so that
$$
z_1-z_2=z_aa=z_bb\tag{4}
$$
Multiplying $(1)$ by $(4)$ yields
$$
\begin{align}
g_aa\color{#C00000}{z_bb}+g_bb\color{#C00000}{z_aa}&=\gcd(a,b)\color{#C00000}{(z_1-z_2)}\\
(g_az_b+g_bz_a)\frac{ab}{\gcd(a,b)}&=z_1-z_2
\end{align}
$$
Therefore, since $\gcd(a,b)\,\mathrm{lcm}(a,b)=ab$, we have
$$
z_1\equiv z_2\pmod{\mathrm{lcm}(a,b)}
$$


问:
write a finite rank operator as sum of a nilpotent and diagonal operators

In the highlighted part, I don't see why we can write a finite rank operator as sum of a nilpotent and diagonal operators.

I know that if the space is of finite dimension, then we can write any operator as a upper triangular matrix. But here, the space could be infinite dimensional...

Thank you for your help.

答: The operator $B_n$ is finite-rank. So, as an operator $(\ker B_n)^\perp\to\operatorname{Ran}B_n$, it is an invertible operator between finite-dimensional Hilbert spaces of the same dimension. Which allows you to see $B_n$ as an $m\times m$ matrix, where $m=\dim\operatorname{Ran} B_n$.


答: A direct calculation can be made by observing that the entire domain is an isosceles right triangle with side length $3$ so the area is $\frac{9}{2}$.  The condition is exclusion of a similar right triangle with side length $2$ and area$\frac{4}{2}$ leaving the area for $P(X+Y\gt 2)$ as $\frac{5}{2}$ and a probability of $\frac{5}{9}$.

Title: How to find bounds of integration for this question?

Body:
I am currently studying joint pdfs and am struggling with finding the bounds of integrations for this problem.
Consider a pair of random variables X, Y with constant joint
density on the triangle with vertices at $(0, 0)$, $(3, 0)$, and $(0, 3)$. Find $P(X +Y > 2)$.

The fact that it says constant density tells me that this is continuous uniform with area $0.5 \cdot 3 \cdot 3 = 4.5$ so $f(x) = 2/9$

Since $P(X +Y > 2)$ I thought that this would mean that the bounds of integration would be $\displaystyle\int_2^3\int_{2-x}^3\cfrac{2}{9}dydx = \cfrac{7}{9}$

But my book says the answer is $\cfrac{5}{9}$

What am I misunderstanding here? What are the proper bounds of integration and why?


问:
Probability of choosing an element in an array
I'm working on a randomized algorithm question that essentially applies rejection sampling, but I'm mostly struggling with the proof of a specific probability part in the algorithm.

Consider a vector $v \in \mathbb{R}^n$. Let $v_i$ denote a weight. Let $v_i \in [a, b]$, where $a$ and $b$ are positive lower and upper bounds, respectively.
We would like to choose some $v_i$ such that $i \in \{1,2,\ldots, n\}$ with probability $\frac{v_i}{\sum_{i=1}^n v_i}$. That is, we're performing weighted sampling.

An algorithm for doing so is to randomly sample an index with uniform probability. e.g., let $j$ denote the sampled index, then we sample $j \sim U(1, n)$. After we sample this index, we then sample a value $t \sim U(0, b)$. If $v_j > t$, we have found our element. If not, then we repeat the sampling of $j$ and $t$. Until we find a pair such that $v_j > t$.

So for a given iteration, the probability of deleting the $i$-th element is $\frac{1}{n}\frac{v_i}{b}$. It appears that this is independent of $a$.

I believe that the probability of succeeding in an iteration in picking an element in a single iteration is $\frac{1}{n} \sum_{i=1}^n \frac{v_i}{b}$, because if we define $E$ to be the probability of success (picking a $j$ and $t$ that satisfies the criteria) and $c_i$ be the event of choosing the $i$-th index, then we have

$$
P(E) = \sum_{i=1}^n P(E|c_i)p(c_i) = \frac{1}{n} \sum_{i=1}^n P(E|c_i) = \frac{1}{n} \sum_{i=1}^n \frac{v_i}{b}
$$

This is where I'm getting stuck. Apparently, the probability should be $\frac{a}{b}$. This is totally different from my expression as it is independent of $n$ and dependent on $a$. What is wrong with my thought process?

答:
$P(E)$ must be strictly greater-than $\frac{a}{b}$ if any $v_i>a$; it can't be exactly $\frac{a}{b}$.

$t<a$ is an automatic success because $v_j\geq a$, which means there is an $\frac{a}{b}$ chance of success even before looking at $v_j$, and still a non-zero chance otherwise if $v_j>a$.

There is a $\frac{b-a}{b}\frac{v_j-a}{b-a}=\frac{v_j-a}{b}$ chance of success with $t\geq a$. So, your per-iteration probability for the $i$th element is $\frac{1}{n}\left(\frac{a}{b}+\frac{v_i-a}{b}\right)=\frac{1}{n}\frac{v_i}{b}$. This result is not independent of $a$, because you can't set $v_i<a$. (Framed differently, $\frac{v_i-a}{b-a}$ is only a valid probability if $v_i\in[a,b]$.)

If you substitute $v_i\geq a$ into your final result, you get $\frac{1}{n}\sum_{i=1}^n\frac{a}{b}=\frac{a}{b}$ as a lower bound for $P(E)$, given what is known a priori about $v_i$.

Further, given that all iterations are independent and that you'll iterate until you succeed, you get $P(i)=\frac{v_i}{\sum_{j=1}^n v_j}$ as expected, but again, this is not independent of $b$ because $v_i\leq b$.


问:
Find all polynomials $P(x)$ satisfying this functional equation

Find all polynomials $P(x)$ which have the property
  $$P[F(x)]=F[P(x)], \quad P(0) = 0$$
  where $F(x)$ is  a given function with the property $F(x)>x$ for all $x\geq 0$.

This is an exercise from my homework. I would appreciate any kind explanations. Thanks!

答:
Denote $x_0=0$ and $x_{n+1}=F(x_n)$ for every $n\ge 0$. Since $F(x)>x$ for every $x\ge 0$, $x_{n+1}>x_n$ for every $n\ge 0$. Note that $P(x_0)=x_0$ and
$$P(x_{n+1})=P(F(x_n))=F(P(x_n)),\quad\forall n\ge 0.$$
Then by induction, it is easy to see that
$$P(x_n)=x_n,\quad\forall n\ge 0.\tag{1}$$
$(1)$ implies that the polynomial $P(x)-x$ has infinitely many roots, i.e. $P(x)\equiv x$.</p>


问:
How to know convexity from matrix?
Given $a \in \mathbb R^3$, let $f(X) := a^T X a$. How to know whether the following minimization problem is convex or not?

$$ \min_{X \text{ is PSD}} \, f(X) $$

I know that if $f''(X) > 0$, it is convex. However, for my given function calculating $f''(X)$ gets too complicated and not straightforward. Please help.

答:
Essentially, you have the semidefinite program (SDP)

$$\begin{array}{ll} \text{minimize} & \langle \mathrm a \mathrm a^\top, \mathrm X \rangle\\ \text{subject to} & \mathrm X \succeq \mathrm O_3\end{array}$$

Does this answer your question?

Title: How to know convexity from matrix?


答:
If you are using integral, then notice it is the shaded area you need to integrate over. So easier is to integrate over the unshaded part and subtract from $1$.

So the integral should be

$P(X+Y \lt 2) = \displaystyle\int_0^2\int_{0}^{2-y} \cfrac{2}{9} \ dx \ dy = \cfrac{4}{9}$

So $P(X+Y \gt 2) = \displaystyle 1 - \frac{4}{9} = \frac{5}{9}$.


问:
Why must a free throw shooter starting out cold, have an 80% make rate on their way to 90+%
Our fantastic free throw shooter, Stephanie Spicy, ends her season with her free throw percentage above 90%, as she expects. However, the season didn't start out so great, she went 0 for her first 8. She got nervous, because an oddity in her contract stated she had to be exactly 80% accurate at some point in her season in order to get her bonus. However, she breathed a sigh of relief because when she simulated out the season in Excel (being a fellow math geek), she noticed she always crossed 80% exactly at some point, so long as she played her typical season with over 300 attempts.

Why, in mathematical terms, did she always cross 80% exactly?

One possible example scenario below, but adding additional misses only moves the 80%, never eliminates it. On the other hand, 70% is sometimes present, and sometimes not.

答:
Define $a_n$ to be the number of shots made in the first $n$ shots attempted. Consider the quantity $b_n = 5 a_n - 4n$. Note that $b_n \geq 0$ iff $\frac{a_n}{n} \geq \frac{4}{5}$.

Now note that $a_{n + 1} - a_n \in \{0, 1\}$. Therefore, $b_{n + 1} - b_n \in \{1, -4\}$.

In particular, this means that $b$ can only increase one at a time. And $b_n$ is always an integer. So if $b_n$ ever goes from negative to nonnegative, it must have been zero at some point. At this point, we will have an exact 80% record.


问:
Selecting at least three non consecutive integers from $\{x:x\le n; x,n\in\mathbb{N}\}$
The question requires me to find a general formula for selecting at least three non-consecutive integers within a range of natural numbers for any $n \in \{1,2,3,...\}$

Added Restrictions:
Any selection with both 1 and n is not a valid selection.

MY ATTEMPT:

I went about this question by trying to calculate the number of ways of selecting consecutive pairs and subtracting that from total possible selections, where total possible selections
$$T(n)= {n\choose 3}\cdot 2^{n-3}$$
The difficulty arose in calculating consecutive selections. I soon realised that accounting for all possible consecutive selections would be pretty difficult, and appears to be quite exhaustive.

Am I missing a crucial aspect in my approach, or is there a better method to approach this question?

答:
I will help you get started. For starters, ignore the "at least three" and "any set with both $1$ and $n$ is not valid" conditions.

The number of subsets of $\{1,\dots,n\}$ with no consecutive elements is $F_{n+2}$, where $F_n$ are the Fibonacci numbers, defined by $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for all $n\in\mathbb Z$. This can be proved by induction; letting $a_n$ be the number of nonconsecutive subsets of $\{1,\dots,n\}$, then $a_n=a_{n-1}+a_{n-2}$ as well, which can be seen by conditioning on whether $n$ is in the subset or not.

You then have to subtract out the nonconsecutive subsets with only zero, one or two elements, and subtract out the subsets containing both $1$ and $n$, being careful to add back in any doubly subtracted subsets. Once you have a formula, you can check it works by calculating the number of valid subsets by brute force for several small values of $n$.


答:
Consider a line of $n$ balls from which $3$ have been removed, so the line now consists of $(n-3)$ balls with $(n-3+1) = (n-2)$ uparrows where they can be inserted

$\uparrow \bullet\,\uparrow \bullet\,\uparrow \bullet\,\uparrow\bullet\, \uparrow \bullet\,\uparrow$

If you now re-number the balls after reinsertion, the count of the triplets  will be $\dbinom{n-2}3$, but you need to exclude combos where $1, n,$ and one number from $3\;\;thru\;\;(n-2)$  are disallowed

You should be able to complete the formula from here, selecting $4,5,...$ are non- consecutive.

Title: Selecting at least three non consecutive integers from $\{x:x\le n; x,n\in\mathbb{N}\}$

Body:
The question requires me to find a general formula for selecting at least three non-consecutive integers within a range of natural numbers for any $n \in \{1,2,3,...\}$

Added Restrictions:
Any selection with both 1 and n is not a valid selection.

MY ATTEMPT:

I went about this question by trying to calculate the number of ways of selecting consecutive pairs and subtracting that from total possible selections, where total possible selections
$$T(n)= {n\choose 3}\cdot 2^{n-3}$$
The difficulty arose in calculating consecutive selections. I soon realised that accounting for all possible consecutive selections would be pretty difficult, and appears to be quite exhaustive.

Am I missing a crucial aspect in my approach, or is there a better method to approach this question?


问:
How can I find the standard deviation of the sum of a series of rolls of a weighted die?
Let's say you possess an $n$-sided weighted die with probabilities $p_1, p_2, ...,p_n$ of landing on sides $1,2,...,n$ such that $\sum\limits_k p_k = 1 $. What is the standard deviation of the sum of $k$ rolls?

Example: $n=3, p_1=0.25, p_2=0.25, p_3=0.5, k=2$.

The sum is $2$ with probability $\frac{1}{16}$, $3$ with probability $\frac{1}{8}$, $4$ with $\frac{5}{16}$, $5$ with $\frac{1}{4}$, $6$ with $\frac{1}{4}$.

$\mu= \frac18+\frac38+\frac54+\frac54+\frac32=4.5$

$\sigma = \sqrt{\frac{2.5^2}{16}+\frac{1.5^2}{8}+5\frac{0.5^2}{16}+\frac{0.5^2}{4}+\frac{1.5^2}{4}}=\sqrt{1.375}\approx 1.172$.

My question is how to generalize this so that I can find it quickly for very large n and k ($n=100, k=1000000$ or so). I don't necessarily need an exact answer (although one is preferred), a good approximation will suffice.

Thanks!

答:
Whenever we have independent events $X_1, X_2, ..., X_k$, we have the following formula:

$Var(X_1 + X_2 + ... + X_k) = Var(X_1) + Var(X_2) + ... + Var(X_k)$

Where $Var(X)$ is the variance of $X$. In the case where the $X$s are IID (independent and identically distributed) with distribution $X$, we have

$Var(X_1 + X_2 + ... + X_k) = k Var(X)$

Using the fact that $Var(X) = E[X^2] - (E[X])^2$, we end up with a variance for rolling a single die of

$\sum\limits_{i = 1}^n p_i i^2 - (\sum\limits_{i = 1}^n p_i i)^2$

And thus the variance of rolling $k$ dice is

$k (\sum\limits_{i = 1}^n p_i i^2 - (\sum\limits_{i = 1}^n p_i i)^2)$

The standard deviation will be the square root of this quantity.


问:
Finding equilibrium point from a differential equation
I have the differential equation of a system as given below, $$ m\frac{d^2H}{dt^2} = mg-k\frac{I^2}{H^2} $$

How do I find the equilibrium points of this?

答:
HINT

You can solve such ODE by multiplying both sides by $\dot{H}$.


问:
If $f$ is integrable on $[a,b]$ with $F=\int_a^xf$, and $f$ is differentiable at $c\in(a,b)$, must $F'$ be continuous at $c$?
I'm doing the exercise 3b, chapter 14 in Spivak's Calculus. It asks to prove or disprove the following statement:

If $f$ is differentiable at $c$, then $F'$ is continuous at $c$.

where $F(x)=\int_a^x f(y)dy$

$f$ is continuous at $c$ since it's differentiable at $c$ (by Theorem 9-1 "If $f$ is differentiable at $a$ then $f$ is continuous at $a$, page 156).

By the Fundamental Theorem of Calculus, $F$ is differentiable at $c$ and $F'(c)=f(c)$. And since $F$ is differentiable at $c$, $F$ is continuous at $c$.

But in the solutions the author says that we must assume that $f$ is continuous in an interval around $c$, otherwise $F'$ might not be continuous. How can $F$ be differentiable and not continuous at $c$? It contradicts the Theorem 9-1.

答:
Partial answer

How can $F$ be differentiable and not continuous at $c$?

It can’t. But this isn’t what Spivak claims; you’re mixing up $F$ and $F’$.

Spivak claims that $F’$, not $F$, need not be continuous at $c$. And in the solution manual he gives an explicit counterexample: $f$ can be Riemann integrable on $[a,b]$ and differentiable at an interior point $c$ without $F’$ being continuous at $c$. But as I discuss here, this counterexample only works because of Spivak’s very stringent definition of a limit at a point, which requires the function to be defined on an interval around the point (except possibly at the point itself). In his counterexample, the reason $F’$ isn’t continuous at $c$ is that $F’$ isn’t even defined at a series of points that converges to $c$.

Because of this, I find Spivak’s example “morally” disappointing: it works only because of a technicality. It’s more a consequence of a bad definition than a statement about the regularity of the function $F’$.

Indeed there is a more general definition of the limit of a function at a point that doesn’t require the function to be defined on a deleted interval around the point, and according to that definition, Spivak’s counterexample no longer works: his $F’$ is indeed continuous at $c$ using that other definition. In other words, near $c$ his $F’$ is pretty well-behaved wherever it’s actually defined.

I don’t know yet whether the claim “$f$ differentiable at $c$ implies $F’$ continuous at $c$” is true using this more general definition of a limit.

EDIT

User zhw proves here that the claim becomes true if we use the more general definition of continuity.


问:
Left Compatible with Ordinal Addition (Multiplication) for alephs
I have a question about strictly increasing properties of addition and multiplication on Ord. Do they work for alephs?

My attempt:

Subset is Left Compatible with Ordinal Addition:

$\forall \alpha, \beta, \gamma \in Ord: \beta \prec \gamma \Rightarrow \alpha + \beta \prec \alpha + \gamma$

Subset is Left Compatible with Ordinal Multiplication:

$\forall \alpha, \beta, \gamma \in Ord: [(\beta \prec \gamma \wedge 0 \prec \alpha) \Rightarrow \alpha \cdot \beta \prec \alpha \cdot \gamma]$
;

For alephs, the sum and product operations are trivial:

$\forall \alpha, \beta \in Ord: \aleph_{\alpha} + \aleph_{\beta} = \aleph_{\alpha} \cdot \aleph_{\beta} = \max \{\aleph_{\alpha}, \aleph_{\beta} \}$

My questions:

$0 \prec \aleph_{\alpha} \Rightarrow \aleph_{\alpha} + 0 \prec
    \aleph_{\alpha} + \aleph_{\alpha} = \aleph_{\alpha} \Rightarrow
    \aleph_{\alpha} \prec \aleph_{\alpha}$  ?

$1 \prec \aleph_{\alpha}    \Rightarrow \aleph_{\alpha} \cdot 1
    \prec \aleph_{\alpha} \cdot    \aleph_{\alpha} = \aleph_{\alpha}
    \Rightarrow \aleph_{\alpha} \prec    \aleph_{\alpha}$ ?

Are there something wrong? I hope your helps.

答:
Cardinal addition and multiplication are not cancellative: $\kappa<\lambda$ does not imply $\mu+\kappa<\mu+\lambda$ or $\mu\cdot\kappa<\mu\cdot\lambda$ (note that cardinal addition and multiplication is commutative so we don't have to specify that we're adding on the left). Your examples demonstrate this.

Note that we see this already in the realm of the natural numbers, once we replace $+$ and $\cdot$ each with $\max$ (which after all is how they behave at infinite cardinals).


问:
Cosine integral question
In the books i am asked to show that $\int \limits_{0}^{\infty} \frac{\cos(x) - \cos(y x)}{x} dx = \ln (y) $ for all $y>0$ ?.

I did solve the question $\int \limits_{0}^{\infty} \frac{\cos(z x) - \cos(y x)}{x^2} dx = \frac{\pi}{2} (y-z)$, i tried to find a connection between them but with no success

答:
This is one of the formulas of Froullani

Theorem If $f：[0,+\infty)\to [0,+\infty)$ is continuous at $x\in [0,+\infty)$,and $\int_1^{+\infty} \frac{f(x)}{x} \mathrm{d}x$ is convergent.Then
$$\int_0^{+\infty}\frac{f(ax)-f(bx)}{x}\mathrm{d}x＝f(0)\log{\frac{b}{a}}$$
for $b>a>0$.

Title: Cosine integral question


问:
Permutation of Elements of a Countable Infinite Set Such That No Element Repeats a Position
Let $ A = \{a_1,a_2, \ldots\}$, $B = \{b_1, b_2, \ldots\}$ and $C = \{c_1, c_2, \dots\}$ be three countable infinite sets. Let's say for each element $a_i \in A$ we want to pair an element $b_j \in B$ with an element $c_k \in C$ where $i$, $j$, and $k$ can be equal. Let's say we want to impose a restriction on this and that is if the pair $(b_j, c_k)$ has been paired before, we can't pair them again. Also every element of $B$ and $C$ has to be paired.

My understanding so far has been that we need $|\mathbb{N}|$ number of bijections from $\mathbb{N} \rightarrow \mathbb{N}$. Essentially what I thought about doing is to have a fixed permutation of $(b_1, b_2, b_3, \ldots )$ and just permute $(c_1, c_2, c_3, \ldots )$ countably infinite times such that no $c_i$ ever appears in the same spot. I was thinking of somehow using the unique prime factorization theorem to ensure each element $c_i$ appears in a unique spot each time, but I can't figure out the bijection.

答: Your question amounts to finding an injection $g : \mathbb{N} \to \{f : \mathbb{N} \to \mathbb{N} | f$ is a bijection$\}$. It suffices to instead find an injection $g : \mathbb{N} \to \{f : \mathbb{Z} \to \mathbb{Z} | f$ is a bijection$\}$ since $\mathbb{Z}$ and $\mathbb{N}$ are in a canonical bijection. This is quite simple: for each $n \in \mathbb{N}$, define $g_n(k) = n + k$.


问:
Find the closed formula for the following sequence $3, 3, 4, 8, 26\cdots$
Find the closed formula for the following sequence$:\quad 3, 3, 4, 8, 26\cdots$

I have been struggling with this one for a while, and am failing to see what I'm missing. It looks like it might be related to Fibonacci's number, but I'm not sure how.

答:
Following up on my comments: If $a_0=3, a_1=3, a_2=4, a_3=8, a_4=26$, then this is matched by

$$a_n= \frac{1}{24}(9n^4-46n^3+87n^2-50n+72).$$

(If you want to start at index $a_1$ instead of $a_0$, just replace each $n$ in this formula by $n-1$.)

Now this formula gives out $a_5= 78$ as the next term, which is well and nice, but actually, I would like to have a formula $a'_n$ where the next term is $a'_5=18$. Well, fortunately I can add any constant multiple of

$$n(n-1)(n-2)(n-3)(n-4)= n^5-10n^4+35n^3-50n^2+24n$$

to the above formula without changing the terms $a_0, .., a_4$. So with $c:= -\frac{1}{2}$ I get

$$a'_n= \frac{1}{24}(-12n^5+129n^4-466n^3+687n^2-338n+72)$$

which gives me my new favourite sequence $3,3,4,8,26, 18, ...$.

Do you see now why this kind of question makes no sense?


问:
Let {$r_n$} be an enumeration of all rationals in [0, 1]. Show that {$r_n$} is not convergent.
Let {$r_n$} be an enumeration of all rationals in [0, 1]. Show that {$r_n$} is not convergent.

My approach
Let $r_n$ be the sequence.It can't be monotone . Thus there exists monotonic subsequnces (both increasing and decreasing) in it.Also it is bounded.
Thus it has two limit points.

答: Suppose on the contrary that the given sequence is convergent (i.e., $\limsup r_n =\liminf r_n$). Consider subsequences defined by $a_n=\frac 1{n}, b_n=\frac n{n+1}$ and note that $a_n\to 0, b_n\to 1$, which is a contradiction as $\limsup r_n \ne \liminf r_n $.


问:
Quick questions of infinite series
Are these true:

$$1.\sum_{n=1}^{\infty}\frac{1}{n^2+88n}=\frac{1}{88}\sum_{n=1}^{88}\frac{1}{n}$$

$$2.\sum_{n=1}^{\infty}\frac{1}{n^2+4n}=\frac{25}{48}$$

For first one, it's factored. ¿It's not factored correctly? Because the bound is from $\infty$ to $88$ and if you multiply $\frac{1}{8}$ with the other fraction, it's not the same.

For second one, it's basic finding if it's converge. I find pattern $\frac{1}{5}+\frac{1}{12}+\frac{1}{21}$, etc.

Title: Quick questions of infinite series

Body:
$$\frac{1}{n^2 + 88n} = \frac{1}{n(n+88)} = \frac{1}{88}\left( \frac{1}{n} - \frac{1}{n+88} \right).$$  Therefore, $$\sum_{n=1}^\infty \frac{1}{n^2 + 88n} = \frac{1}{88} \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+88} = \frac{1}{88} \sum_{n=1}^{88} \frac{1}{n}.$$

The other sum is handled in a similar fashion.


问:
How can I show that an element does not belong to a certain ideal?
Let
$$
R = \frac{\mathbb{Z}[x_1, x_2, x_3, \cdots]}{(x_1 - x_2^2, x_2 - x_3^2, \cdots)}.
$$
I'm trying to show that $R$ does not satisfy the ascending chain condition for principal ideals, though I'm not entirely sure that it doesn't. So far, I've come up with letting $ f_n(x) = x_n - x_{n+1}$. Then we have
\begin{align}
f_{n}(x) &= x_n - x_{n+1} \\
         &= x_{n+1}^2 - x_{n+2}^2 \\
         &= (x_{n+1} + x_{n+2}) (x_{n+1} - x_{n+2}).
\end{align}
That is, $ (f_n) \subseteq (f_{n+1})$. Then we have the chain $ (f_1) \subseteq (f_2) \subseteq (f_3) \subseteq \cdots$. If I can show that $ (f_n) \neq (f_{n+1}) $ then it becomes clear that the ascending chain does not stabilize. I think the most logical way to approach this is to show that $ f_{n+1} \notin (f_n) $. Here's where I'm having trouble, I have no clue how to show that an element doesn't belong to an ideal.

答:
From the comments, we've shifted to considering $f_n=x_n$ and identified that if $x_{n+1}\in (x_n)$, then $x_{n+1}(p\cdot x_{n+1}-1)=0$, so it's enough to show that $R$ is a domain.

Let $a,b\in R$ be nonzero, and our goal is to show that $ab\neq 0$. Pick some $n\gg0$ so that $a=p(x_n)$ and $b=q(x_n)$ for polynomials $p,q$, where we know by the assumption $a,b\neq 0$ that $p,q\neq 0$. Then $ab=p(x_n)q(x_n)\neq 0$ since $\Bbb Z[x_n]\cap (x_1-x_2^2,\cdots)=0$ inside $\Bbb Z[x_1,\cdots]$.


答:
Suppose that the sequence does converge to some limit $\ell$.  Choose $\epsilon>0$ so that at least one of $\ell-\epsilon>0$ and $\ell+\epsilon<1$ holds.

Then there exists $N$ such that for $n>N$, we must have $r_n\in(\ell-\epsilon,\ell+\epsilon)$.  However, this in turn implies that all the rational numbers in $[0,\ell-\epsilon]$ (or $[\ell+\epsilon,1]$, depending on the above) must be enumerated by $r_1,\ldots,r_N$.  However, there are clearly infinitely many such rationals, so this is impossible.

Body:
Let {$r_n$} be an enumeration of all rationals in [0, 1]. Show that {$r_n$} is not convergent.

My approach
Let $r_n$ be the sequence.It can't be monotone . Thus there exists monotonic subsequnces (both increasing and decreasing) in it.Also it is bounded.
Thus it has two limit points.


问:
Is it possible to find out $x^2$ parabola and function from 3 given points?
I am programming a ball falling down from a cliff and bouncing back. The physics can be ignored and I want to use a simple $y = ax^2$ parabola to draw the falling ball.

I have given two points, the edge of the cliff at $C(-0.9; 0.8)$ and the point where the ball hits the bottom $B(0.1; -1.05)$. Due to its symmetry we know there is another point at $A(-1.9; -1.05)$. So that are 3 points I could work with. $C$ is the vertex.

I've tried this approach but my parabola is not as exact as I need it to estimate for example the intersects with the x- or y-axis.

As the legs of the parabola are down, all I can tell is that $a$ is negative:

$$
 y = ax^2 + bx +c; a < 0
$$

I tried drawing it in Excel which allows to add an polynomial trendline. This is the best approximation I could do so far (dotted line).

But I need the function. Is it somehow possible to calculate the equation somehow?

答:
Yes, we can find the equation for the parabola with the given information.

In this problem there are three unknowns, the coefficients $a,b$ and $c$ so to find them we will need three pieces of information. In this case we know that the parabola must pass through the three given points.

Substitute the given points into the equation for the parabola. Then:

$$ 0.8 = a(-0.9)^2 + b(-0.9) + c, $$
$$ -1.05 = a(0.1)^2 + b(0.1) + c, $$
$$ -1.05 = a(-1.9)^2 + b(-1.9) + c.$$

This is a system of three equations for three unknowns which we can solve in the following way.

From the first equation we can write $c = 0.8 - a(-0.9)^2 - b(-0.9).$ This expression for $c$ can be substituted into the second and third equations. Then:

$$ -1.85 = a\left[(0.1)^2 - (-0.9)^2\right] + b(-0.8), $$
$$ -1.85 = a\left[(-1.9)^2-(-0.9)^2\right] + b(-2.8) + 0.8.$$

From the first of these expressions $b = -\left[-1.85 - a\left[(0.1)^2 - (-0.9)^2\right]\right/0.8$

Substitute this into the final expression and solve for $a$, then $b$, then $c$. In this case:

$$a = -1.850,$$
$$b = -3.330,$$
$$c = -0.6985.$$


问:
Brownian motion property on interval
Say $B$ is a Brownian motion. Knowing that for some constant $a\gt0$
$$P(\sup_{[0,n]} |B|\leq 1) \leq e^{-{an}}$$
how can I deduce that

$$P\left(\sup_{[0,1]} |B| \leq \frac{1}{n}\right)\leq e^{-{an^2}}$$

答:
First, rewriting the inequality for $n' \equiv n^2$, we have that $$ \mathbb P(\sup_{[0,n^2]} |B|\leq 1) \leq e^{-{an^2}} \tag1$$
Now, if $(B_t)_{t\ge0}$ is a Brownian Motion, then $W \equiv \left(nB_{t/n^2}\right)_{t\ge0}$ is also a Brownian Motion for all $n >0$. Therefore, applying inequality $(1)$ to the process $W$, we get :
$$\begin{align}\mathbb P(\sup_{[0,n^2]} |W|\leq 1) &= \mathbb P(\sup_{t\in[0,n^2]} n|B_{t/n^2}|\leq 1)\\
&=\mathbb P\left(\sup_{t\in[0,n^2]} |B_{t/n^2}|\leq \frac1 n\right)\\
&=\mathbb P\left(\sup_{s\in[0,1]} |B_{s}|\leq \frac1 n\right) \\
&\le e^{-{an^2}} \end{align} $$

Title: Brownian motion property on interval


问:
Elliptic curves and primes
I have a question related to elliptic curves and primes:

For which primes $p \geq 5$ does the equation

$$y^2 = x^3 + 7x + 3 $$

define an elliptic curve over $F_p$?

How should I approach this problem?

答:
Let $K$ be a field. An elliptic curve $E$ over $K$ is a nonsingular projective curve of genus one, equipped with a fixed $K$-rational point $O$. With this definition, your curve is technically never an elliptic curve, as you have written down an affine curve and not given a chosen point!

But let's ignore that issue. It is common to present elliptic curves in the form you've given, and your curve in particular is an affine patch of the projective curve
$$
Y^2Z = X^3 + 7XZ^2 + 3Z^3.
$$
Here, we typically choose the point $O$ to be the "point at infinity," which has projective coordinates $[0:1:0].$

So now we must address when the curve you've given is nonsingular. It is a well-known fact that the curve
$$
y^2 = x^3 + ax + b
$$
is nonsingular if and only if the discriminant $\Delta = −16(4a^3 + 27b^2)$ is nonzero. See, for example [1, Proposition III.1.4], but be aware that there is a typo in the statement: (i) in part (a) should read "It is nonsingular if and only if $\Delta\neq0$."

Now we can address the question: for your curve, we have
$$
\Delta = -16(4\cdot 7^3 + 27\cdot 3^2) = -25840 = -2^4\cdot 5\cdot 17\cdot 19.
$$
This will be nonzero in $\Bbb{F}_p$ if and only if $p\neq 2,5,17,19.$ So, the curve you've given is not an elliptic curve if $p = 2,5,17,$ or $19.$

Finally, you might ask what is the genus of this curve? Well, we may calculate this in the non-singular case using the genus-degree formula, which states that for an irreducible nonsingular plane curve of degree $d,$ the genus $g$ is
$$
g = \frac{(d - 1)(d - 2)}{2}.
$$

The curve you've written down has degree $d = 3,$ so $g = 1.$

Thus, we may conclude that $y^2 = x^2 + 7x + 3$ defines an Elliptic curve over $\Bbb{F}_p$ when $p\neq 2,5,17,19.$

[1] Silverman, Joseph. (2009). The Arithmetic of Elliptic Curves.

Title: Elliptic curves and primes


问:
Probability and source emission.
I am not sure how does probability enter in this question, but anyway:

"A source emits particles at an average K particles per second. Each
emission is stochastically independent of the previous emission.
Calculate the probability that exactly n particles will be emitted
within a time interval t"

I just thought that it could be: $P(n) = \frac{n}{\int_{0}^{t} K dt}$. But i am not sure how does the stochastically independence enter in my answer.

Stochasticaly independence means independent events.

答:
The common way here is to assume that the particles emitted follow a poisson law $Po(k)$ per second.
This assumption is reasonable observing that is also reasonable that increasing the number of particles per second the probability tends to decrease ( i.e., the probability to observe k particles is assumed greater than observing 4k particles).

Thus the requested probability is

$$\mathbb{P}[X=n]=\frac{e^{-kt}(kt)^n}{n!}$$

Where

$n=0,1,2\dots$

$t$ is expressed in seconds

Stochastic independence implies that $\sum_{i=1}^t X_i\sim Po(kt)$

This if the particles are few (law of rare events); if $kt$ becomes large the distribution can be approximated with a gaussian $N(kt;kt)$

Title: Probability and source emission.


问:
$\mathbb{R}/{\sim}$ is Frechet Urysohn, where $x \sim y \Leftrightarrow x = y \lor x,y \in \mathbb{Z}$
With this equivalence relation on R:

$x \sim y \Leftrightarrow x = y \lor x,y \in \mathbb{Z}$

I have to proof that the topologic quotient $X/\sim$ is Frechet Urysohn, which means that for evert $A \subset X/\sim$ and $p \in Cl(A)$, there is a sequence of elements of A such that $(a_n) \Rightarrow p$.

First, I studied $X/\sim$ and I can show that it contains an equivalence class for $\mathbb{Z}$ and the others look like $[x] = \{x\}$ for $x \in \mathbb{R} - \mathbb{Z}$. Now, about the proof, I have seen other answers in this forum that differentiate when $p \in A$ or p is a limit point of A, but I have been asked to do it using the definition $p \in Cl(A) \Leftrightarrow \forall U$ neighborhood of p $U \cap A \not = \emptyset$. Up until now, I chose to differentiate two situations: $p \in \mathbb{Z}$ or $p \not \in \mathbb{Z}$. For the second case, I can see easily that $p \in (a, a+1)$ for some $a \in \mathbb{Z}$ and that $(a, a+1) \cap A \not = \emptyset$. Is it enough to take the elements in $(a, a+1) \cap A$ and construct a sequence with them. How can I define it? For the first case, I am actually a little confused because I am not sure how would be A in the case that $p \not \in A$ due to the fact that it could be the limit point of other subset and I guess it is not possible to characterize $A$ to a general case. If you could help to see what I'm missing in this proof, I would really appreciate it.

答:
HINT: Let $X=\Bbb R/\!\sim$, let $q:\Bbb R\to X$ be the quotient map, and let $z=q(0)$ (so that $q[\Bbb Z]=\{z\}$. Let $E$ be the set of functions $\epsilon:\Bbb Z\to\left(0,\frac12\right)$, and for each $\epsilon\in E$ let $U_\epsilon=\bigcup_{n\in\Bbb Z}\big(n-\epsilon(n),n+\epsilon(n)\big)$; it’s not too hard to show that $\{q[U_\epsilon]:\epsilon\in E\}$ is a local base of open nbhds of $z$ in $X$. Use this to show that if $z\in\operatorname{cl}_XA$, there must be an $n\in\Bbb Z$ such that

$$n\in\operatorname{cl}_{\Bbb R}\big(q^{-1}[A]\cap(n-1,n+1)\big)\,.$$

This is of course trivial if $z\in A$, so you can concentrate on the case $z\in\operatorname{cl}_XA\setminus A$.


问:
How many numbers greater than $4000$ can be formed using $0,2,4,5,7$?

How many four digit or five digit numbers greater than $4000$ can be formed using $0,2,4,5,7$, if none of the digits are repeated and the numbers formed are even?

I've tried all the ways I can think of to do this question but I always get a different answer to the answer sheet (the correct answer is 108 according to the book). I'm not really sure how to guarantee that it ends in an even number. Any help would be greatly appreciated!

答:
The strategy should be taking care of the first and last digits, then permutate the rest of the numbers for the middle digits. To end with an even number, you can only use 0,2,4 for the last digits. You can not use 0 as the first digit.

Choosing from 0,2,4,5,7

For 5 digits even number

a. First digit even (2 choices), last digit even (2 choices), the rest 3 permutate for the middle 3 digits (3x2 choices)
$\to$ 2x2x3x2=24

b. First digit odd (2 choices) , last digit even  (3 choices ), the rest 3 permutate for the middle 3 digits(3x2 choices)
$\to$ 2x3x3x2= 36

For 4 digits even number greater than 4000

a. First digit odd (2 choices ), last digit even (3 choices ), the rest 3 permutate for the middle 2 digits (3x2 choices)
$\to$ 2x3x3x2=36

b. First digit even (1 choice), last digit even (2 choices), the rest 3 permutate to for the middle 2 digits (3x2 choices)
$\to$ 1x2x3x2=12

Total:  24+36+36+12=108


问:
How can I prove that $\frac{\sin\theta+ \tan\theta}{2}> \theta$ when $0 < \theta < \frac{\pi}{2}$
I don’t know how to do it without calculus, could someone help ? How do I prove it geometrically and/or algebraically ?

EDIT: $0 < \theta < \frac{\pi}{2}$

My teacher never explained it. Someone tried explaining it to me with calculus, but I would appreciate a different explanation.

答:
Since $\sin \theta, \tan \theta > 0$ in the given domain, AM > HM so we have:

$$\frac{\sin\theta+ \tan\theta}{2} > \frac{2}{1/\sin \theta + 1/\tan \theta}$$

and $\csc \theta + \cot \theta = \cot(\theta/2)$, thus the RHS is $2 \tan(\theta/2) > 2 \cdot \theta/2 = \theta$.

In the last line, we have used the fact that $\tan x > x$ when $0 < x < \pi/2$.


问:
Show that $f(x)$ have even degree

$f(x)$, $g(x)$ are nonzero polynomials and $f(x^2+x+1)=f(x)g(x)$. Show that $f(x)$ has even degree.

答:
I'll show that $f$ cannot have any real roots and hence cannot be of odd degree (every odd degree polynomial has at least one real root). Hence it is of even degree.

Consider the roots of $x^2+x+1=x$. They are $\pm i$.

Now let $f$ have a real root, say $x_0$. Then $f(x_0) = 0$. Substituting this in the given relation, we get:
$$f(x_0^2+x_0+1) = f(x_0)g(x_0) = 0$$
Which tells us that if $x_0$ is a root of $f$, then so is $x_0^2+x_0+1$. Continuing in this way, if $x_0$ is a root, so is $x_0^2+x_0+1, [(x_0^2+x_0+1)^2+(x_0^2+x_0+1)+1],$ etc. (which are all distinct if $x_0 \neq \pm i$) leading to an infinite number of distinct roots. As $f$ is a polynomial, this is a contradiction.

Thus $f$ cannot have a real root. In fact, the only roots it can have are those that satisfy $x_0^2+x_0+1=x_0$, as then $x_0$ and $x_0^2+x_0+1$ will be the same, i.e. if $x_0 = \pm i$.

Notice that the argument does not depend on what $g$ is at all: $f$ can always be shown to have no real roots.

Title: Show that $f(x)$ have even degree


问:
Probability of at least one person in a car
There are 7 people, and 4 cars (the cars have no limit to how many people they can fit). Each person randomly chooses a car, and their decision is independent of what the others choose.
I need to find the probability that there is at least one person in each car.

I'm thinking to do 1 - probability there is at least 1 car with 0 people in, but I'm not sure how to do this?

答:
Here is my attempt at simulation. With a million
embarkations of the cars, one can expect about
two or three places of accuracy. The simulated
answer is in good agreement with $525/1024.$
(+1) for @User's Answer.

```<code>set.seed(2021)
m = 10^6;  event = logical(m)
for(i in 1:m){
 c = sample(1:4,7,rep=T)
 t[i] = c(tabulate(c),0,0,0)[1:4]
 event[i] = (sum(t >= c(1,1,1,1))==4)
}
mean(event)
[1] 0.513738
525/1024
[1] 0.5126953
</code>```

Illustration of R code in terms of the last iteration.
Cars are numbered 1 through 4. Modification to 'tabulate' is needed in case
higher numbered cars happen to be empty.
Vector event consists of 1 million TRUEs and
FALSEs; its mean is the proportion of its
TRUEs.

```<code>c
[1] 2 1 3 2 2 2 2  # cars chosen by 11 passengers
t
[1] 1 5 1 0        # numbers in each car, 1 - 4
event[10^6]
[1] FALSE          # event FALSE (last car empty)

</code>```

Body:
Here is my attempt at simulation. With a million
embarkations of the cars, one can expect about
two or three places of accuracy. The simulated
answer is in good agreement with $525/1024.$
(+1) for @User's Answer.

set.seed(2021)
m = 10^6;  event = logical(m)
for(i in 1:m){
 c = sample(1:4,7,rep=T)
 t[i] = c(tabulate(c),0,0,0)[1:4]
 event[i] = (sum(t >= c(1,1,1,1))==4)
}
mean(event)
[1] 0.513738
525/1024
[1] 0.5126953

Illustration of R code in terms of the last iteration.
Cars are numbered 1 through 4. Modification to 'tabulate' is needed in case
higher numbered cars happen to be empty.
Vector event consists of 1 million TRUEs and
FALSEs; its mean is the proportion of its
TRUEs.

c
[1] 2 1 3 2 2 2 2  # cars chosen by 11 passengers
t
[1] 1 5 1 0        # numbers in each car, 1 - 4
event[10^6]
[1] FALSE          # event FALSE (last car empty)


问:
Evolution of angle between two triangles as they tilted towards each other
My problem is as follows. I have two flat isosceles triangles (marked in yellow below) -- with sides of $L$ and base of width $w$ -- touching at a common point (marked in red), such that the angle between them is $\theta_i$. Note that $\theta_i$ is also the apex angle of another isosceles triangle, which I have depicted in lavender.

Now, I tilt these triangles (shown in deep blue) towards each other, by an angle $\alpha$. Then what is the equation relating the final angle $\theta_f$ to $\theta_i$ and $\alpha$? I know from measurements that $\theta_f \leq \theta_i$.

I realise that I only need to figure out how the lateral sides of the yellow triangles behave under the tilt $\alpha$, but I am unable to do so. Can someone please help?

Note: In the picture, the entire deep blue configuration is tilted off the horizontal plane, but this is just for viewing convenience.

答:
Pick coordinates centred at the red point with the $z$-axis being the internal angle bisector of $\theta_i$, $x$-axis being the orthogonal on the plane and $y$-axis the vertical out-of-plane direction.  So the two initial apex are $(\pm L\sin\frac12\theta_i,0,L\cos\frac12\theta_i)$.  If the bases of the two triangles are on the same line, then we have $\sin\frac12\theta_i=w/(2L)$.  It isn't clear whether this is the case from your description.

Then rotate the triangles by angle $\pm\alpha$ about $z$-axis so $y$ becomes positive, we have the two points $(\pm L\sin\frac12\theta_i\cos\alpha,L\sin\frac12\theta_i\sin\alpha,L\cos\frac12\theta_i)$.  The angle they made is $\cos^{-1}(1-2\sin^2\frac12\theta_i\cos^2\alpha)$.

If you rotate about other axes parallel to this, you will get parallel line, so the angle is unchanged (except you get skewed lines instead of intersecting lines in general).


答:
Solve the system of linear equations
$$\begin{cases}
ax_1^2 + bx_1 +c=y_1; \\
ax_2^2 + bx_2 +c=y_2; \\
ax_3^2 + bx_3 +c=y_3;
\end{cases}$$ w.r.t. unknown variables $a,\ b,\ c.$</p>

Title: Is it possible to find out $x^2$ parabola and function from 3 given points?

Body:
I am programming a ball falling down from a cliff and bouncing back. The physics can be ignored and I want to use a simple $y = ax^2$ parabola to draw the falling ball.

I have given two points, the edge of the cliff at $C(-0.9; 0.8)$ and the point where the ball hits the bottom $B(0.1; -1.05)$. Due to its symmetry we know there is another point at $A(-1.9; -1.05)$. So that are 3 points I could work with. $C$ is the vertex.

I've tried this approach but my parabola is not as exact as I need it to estimate for example the intersects with the x- or y-axis.

As the legs of the parabola are down, all I can tell is that $a$ is negative:

$$
 y = ax^2 + bx +c; a < 0
$$

I tried drawing it in Excel which allows to add an polynomial trendline. This is the best approximation I could do so far (dotted line).

But I need the function. Is it somehow possible to calculate the equation somehow?


问:
Ant walking cube
This is a modification of the ant crossing to the opposite diagonal problem:

An ant traverse from a cube vertex and does so on the edge. It has equal probability to choose all the edge including the original one. What is the expected number of edge it will traverse before returning to the original vertex?

My intuition is: there are 8 vertices, so if the ant just walking non-stop randomly to infinity, it would spend 1/8 of the time hitting each vertex. So intuitively, my answer is the average edge it would walk between each visit to the original vertex is 8 edges. But I don't know how to set it up in mathematically rigorous way.

答:
This is the approach Ross Millikan gave in the comments, but with more detail.

Let the vertices of the cube be $\{0,1\}^3$, and let the ant start at $(0,0,0)$.

For each vertex $v \in \{0,1\}^3$ (aside from $(0,0,0)$), let $E(v)$ denote the expected number of edges the ant will traverse before hitting $(0,0,0)$ if it starts from $v$.

By symmetry, $E(1,0,0) = E(0,1,0) = E(0,0,1) = A$ and $E(1,1,0) = E(1,0,1) = E(0,1,1) = B$. Also, let $E(1,1,1) = C$.

If the ant is currently at $(1,0,0)$, then it will take one step to get to its next vertex, which will be $(0,0,0)$ or $(1,1,0)$ or $(1,0,1)$ with equal probability. If the ant moves to $(0,0,0)$, the ant is done. If the ant moves to $(1,1,0)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,1,0)$ (by definition). Similarly, if the ant moves to $(1,0,1)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,0,1)$. Hence $$E(1,0,0) = \dfrac{1}{3} \cdot 1 + \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3} \cdot (E(1,0,1)+1)$$ $$A = \dfrac{2}{3}B+1$$

Using the same logic, you can work out that $$E(1,1,0) = \dfrac{1}{3} \cdot (E(1,0,0)+1)+\dfrac{1}{3} \cdot (E(0,1,0)+1) + \dfrac{1}{3}\cdot (E(1,1,1)+1)$$ $$B = \dfrac{2}{3}A+\dfrac{1}{3}C+1$$ and that $$E(1,1,1) = \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3}\cdot(E(1,0,1)+1) + \dfrac{1}{3}\cdot(E(0,1,1)+1)$$ $$C = B+1$$

You can easily solve the system of equations for $A,B,C$. Finally, since the ant starts at $(0,0,0)$, its first step takes it to either $(1,0,0)$ or $(0,1,0)$ or $(0,0,1)$ with equal probability. So the answer to the question is $$\dfrac{1}{3}\cdot(E(1,0,0)+1)+\dfrac{1}{3}\cdot(E(0,1,0)+1)+\dfrac{1}{3}\cdot(E(0,0,1)+1) = A+1.$$

Title: Ant walking cube


问:
Again quick questions of infinite series
These are series:

1.$\sum_{n=2}^{1000}\frac{2}{n^3-n}=\sum_{n=2}^{1000}\frac{1}{n-1}-\sum_{n=2}^{1000}\frac{2}{n}+\sum_{n=2}^{1000}\frac{1}{n+1}$.

$\sum_{n=2}^{\infty}\frac{2}{n^3-n}=\sum_{n=2}^{\infty}\frac{1}{n-1}-\sum_{n=2}^{\infty}\frac{2}{n}+\sum_{n=2}^{\infty}\frac{1}{n+1}$

Both is to express the sum in different ways possibly. ¿But it's not the same? Which is correct, or is both is correct. ¿Should I factor to solve it? I don't have access to know it. Can I have help? Thanks.

答:
$$\sum_{n=2}^{k}\frac{2}{n^3-n}=\sum_{n=2}^{k}\frac{1}{n-1}-\sum_{n=2}^{k}\frac{2}{n}+\sum_{n=2}^{k}\frac{1}{n+1}$$

is true for all $k\ge2,k\in I$.

The result immediately follows after a quick partial fraction decomposition


问:
How many integral solutions does $2x + 3y + 5z = 900$ have when $ x, y, z \ge 0$?
Solution: Let $2x + 3y = u.$ Then we must solve $\begin{align} u + 5z = 900 \tag 1 \\ 2x + 3y = u \tag 2 \end{align}$

For $(1),$ a particular solution is $(u_0, z_0) = (0, 180).$ Hence, all the integral solutions of $(1)$ are $\begin{cases} u = 5t \\ z = 180 - t \end{cases} (t \in \mathbb Z)$

Substituting $u = 5t$ into $(2)$ gives $2x + 3y = 5t$ whose particular solution is $(x_0, y_0) = (t, t).$ Hence all the integral solutions of $(2)$ are $\begin{cases} x = t - 3s \\ y = t + 2s \end{cases} (t \in \mathbb Z)$

Thus all the integral solutions of $2x + 3y + 5z = 900$ are given by $$\begin{cases} x = t - 3s \\ y = t + 2s \\ z = 180 - t \end{cases} (s,t \in \mathbb Z)$$

Now suppose $x, y, z \ge 0.$

Note, $180 - t \ge 0 \implies t \le 180$ and so $t + 2s \ge 0 \implies s \ge -90$ and $t - 3s \ge 0 \implies s \le 60.$ Thus we have $-90 \le s \le 60.$

Consider $0 \le s \le 60.$ Now $t \le 180, \ t \ge 3s \implies 3s \le t \le 180$. Thus in this range of $s$, there are $180 - 3s + 1 = 181 - 3s$ of $t$'s.

Consider $-90 \le s < 0.$ Now $t \le 180, \ t \ge -2s \implies -2s \le t \le 180$. Thus in this range of $s$, there are $180 + 2s + 1 = 181 + 2s$ of $t$'s.

Range $0 \le s \le 60$ has the following points:

$(0, 181 - 3(0)), \ (1, 181 - 3(1)), (2, 181 - 3(2), \ldots (60, 181 - 3(60))$ of which there are $61.$

The range $-90 \le s < 0$ must have $91$ points. In sum, we have $61 + 91 = 152$ points for $x, y, z \ge 0.$

My question:

According to the book the answer is $\displaystyle{\sum_{s = 0}^{60}(181 - 3s) + \sum_{s = -90}^{-1}(181 + 2s) = 13651.}$ I don't understand why they took the sum of all $t$'s in the range of $s$. That means some of my denotations and labels above must be incorrect. Where's the mistake? Thanks.

edit:

I think I see my mistake. The number $181 - 3s$ is the number of $t$'s, not necessarily the form of $t$. Given that, the number of ordered pairs (in the given range) must be $(181 - 3s)*61$ by the product rule.

答:
You can solve this problem by creating equals Ax = b. Where A - transition matrix by x and b.
Let's consider this example.
Introduce F(m, i) where F(m, 1) - count of solutions 2x = m, F(m, 2) - count of solutions 2x + 3y = m, F(m, 3) - count of solution 2x + 3y + 5z = m ...

The following equations are valid:

F(m, 1) = F(m - 2, 1)

F(m, 2) = F(m, 1) + F(m - 3, 2) = F(m - 2, 1) + F(m - 3, 2)

F(m, 3) = F(m, 2) + F(m - 5, 3) = F(m - 2, 1) + F(m - 3, 2) + F(m - 5, 3)

Vector b we can construct as [F(m, 1), F(m - 1, 1), F(m, 2), F(m - 1, 2), F(m - 2, 2), F(m, 3), F(m - 1, 3), F(m - 2, 3), F(m - 3, 3), F(m - 4, 3)].

Vector x we can construct as [F(m - 1, 1), F(m - 2, 1), F(m - 1, 2), F(m - 2, 2), F(m - 3, 2), F(m - 1, 3), F(m - 2, 3), F(m - 3, 3), F(m - 4, 3), F(m - 5, 3)].

Matrix A is easily constructed by x and b.

Start vector x is [F(0, 1), F(-1, 1), F(0, 2), F(-1, 2), F(-2, 2), F(0, 3), F(-1, 3), F(-2, 3), F(-3, 3), F(-4, 3)] and have value [1, 0, 1, 0, 0, 1, 0, 0, 0, 0]

And it's finish.
We interested in F(900, 3). We need calculate A^900 * x and take appropriate b[5].
We can raise the matrix to a power by O(2 * lb(900) * 10^3)


问:
Reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$?
So I am trying to find the reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$.

My attempt: Taking $x^m$ as the second function and integrating by parts, I got,

$I_n=\frac{(n-1)}{m+1}I_{n-2,m+1}$. But the given answer is $\frac{(n-1)!m!}{(m+n)}$. How do I go from my answer to the one given in the textbook? I tried manipulating my answer, but it isn't the same as the one in the textbook. Thanks in advance.

答:
I will assume that the integral in question is from $0$ to $1$. The reduction formula you obtained is correct, except for the fact that you should write $I_{n-1,m+1}$ instead of $I_{n-2,m+1}$. The answer in the textbook intends obviously to show the value of the integral (after the reduction is repeatedly done till $I_{0,m+n} $).

However there is an error in the textbook formula (if you correctly cited it). The correct one is
$$\frac{(n-1)!m!}{(m+n)\color{red}!}.
$$


问:
Fundamental Theorem of Calculus

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable at every point, and that $f'(x)$ is continuous.

1) Prove that $\displaystyle\int_0^1 \left( \int_0^1 \left(f'(x)f(y)-f(x)f'(y)\right)^2 \ dx \right) \ dy \geq 0$

2) Hence prove that $2\displaystyle\int_0^1 f'(t)^2 dt \int_0^1 f(t)^2 \ dt \geq \dfrac{1}{2} \left(f(1)^2 - f(0)^2 \right)^2$

Ok, so I've done 1) quite easily using the fact that since the inside integral is a square, this implies that the integral is $\geq 0$ then using the same theorem this implies that the outside integral also gives something $\geq 0$.

Now, in regards to 2) I've tried doing this a number of ways but it doesn't seem to fall out properly. Clearly the Fundamental Theorem of Calculus needs to be used at some point but every way I've tried doing it doesn't actually help. Thanks for any help!

Title: Fundamental Theorem of Calculus

Body:
Using the Cauchy-Schwarz inequality, $2\int_0^1 f'(t)^2\,dt\int_0^1 f(t)^2\,dt\ge 2(\int_0^1 f(t)f'(t)\,dt)^2=\frac{1}{2}(\int_0^1 (f(t)^2)'\,dt)^2=\frac{1}{2}(f(1)^2-f(0)^2)^2$ where I have used that $(f(t)^2)'=2f(t)f'(t)$.

Note that if you don't want to use the Cauchy-Schwarz inequality, you can use 1) and write $0\le \int_0^1(\int_0^1(f'(x)f(y)-f(x)f'(y))^2\,dx)\,dy=\int_0^1(\int_0^1 f'(x)^2f(y)^2+f(x)^2f'(y)^2-2f'(x)f(x)f'(y)f(y)\,dx)\,dy\implies 2\int_0^1 f'(x)f(x)\,dx\int_0^1 f'(y)f(y)\,dy\le2\int_0^1\int_0^1 f'(x)^2f(y)^2\,dx\,dy=2\int_0^1 f'(x)^2\,dx\int_0^1 f(y)^2\,dy$

this gives $\frac{1}{2}(\int_0^1 (f(x)^2)'\,dx)^2\le \int_0^1 f'(t)^2\,dt\int_0^1 f(t)^2\,dt$, or using the fundamental theorem of calculus, $\frac{1}{2}(f(1)^2-f(0)^2)^2\le \int_0^1 f'(t)^2\,dt\int_0^1 f(t)^2\,dt$.


问:
homeomorphism in $(-\infty,-1) ∪ [0,\infty)$ of $\mathbb{R}$
Let X be the subspace $(-∞, -1) ∪ [0, ∞)$  of $\mathbb{R}$. We define $f∶ X → \mathbb{R}$ as

$f(x)=\begin{cases}
x+1 & \text{ if } x< -1 \\
x & \text{ if } x\geq 0
\end{cases}$

Prove that f is bijective, preserves order, and is continuous. Is it f a
homeomorphism?

Now, I'm a little confused by the concept of "piecewise-continuous" I mean my function f is continuous in $(-∞, -1) ∪ [0, ∞)$, or not? :c
I appreciate any help

答:
Yes, $f$ is continuous: $f^{-1}[(a,b)] = (a,b)$ when $0 < a < b$ and $f^{-1}[[(a,b)] = (a-1,b-1)$ when $a < b <0$ while $f^{-1}[(-a,a)] = (-a-1,-1) \cup [0,a)$ which is open in $X$ (as it equals $(-a-1,a) \cap X$). These $3$ types of open intervals form a base for $\Bbb R$ (to take care of the special role of $0$ here).

$g:=f^{-1}$ is not continuous: in $\Bbb R$, $-\frac1n \to 0$ while $g(-\frac1n) = -1 - \frac1n$ does not converge to $g(0) = 0$ in $X$ (in the subspace topology:  $[0,\frac12)$ is open in $X$ but contains no points of $(g(-\frac1n))_n$).

$g$ would be continuous if $X$ had the order topology wrt the inherited order, though.. Here is an example where the subspace topology on $X$ differs from it.


问:
Jensen's Formula application $f(z)$ has no zeros on $|z|=r, r<1$
I am reading the Balazard Saias and Yor paper where he defines,
$s=s(z)=\frac{1}{1-z}$ and $$f(z)=(s-1)\zeta(s)$$ where $\zeta$ is the Riemann zeta function.Then by Jensen's formula we have for $f(0)\neq 0$ and $r<1$,
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}log\ |f(re^{i\theta})|d\theta=log\ |f(0)|+\sum_{|\alpha|<r,f(\alpha)=0} log \ \frac{r}{|\alpha|}$$

For applying Jensen's formula we need that $f(z)$ has no zeros on $|z|=r$.

Question Show that $f(z)$ has no zeros on $|z|=r$, $r<1$.

Attempt
$$f(z)=(s-1)\zeta(s)$$
$$f(z)= \frac{z}{1-z}\zeta(\frac{1}{1-z})$$
Since $|z|=r$ so write $z=re^{i\theta}$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-re^{i\theta}})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-rcos\theta-irsin\theta})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{(1-rcos\theta)^2+r^2sin^2\theta})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{r^2-2rcos\theta+1})    $$
How to show that $f(re^{i\theta})\neq 0$ on $|z|=r, r<1$?.

答:
The question is equivalent to showing that $\zeta(s)\ne0$ for $|s|<1$.

When $\Re s\le 0$, we know that $\zeta(s)$ has only trivial zeros at $s=-2,-4,-6,\dots$, by the functional equation for $\zeta(s)$.

When $\Re s>0$, we can use the fact that
$$
\sum_{\substack{\rho \\ \zeta(\rho)=0 \\ 0<\Re\rho<1}} \Re\frac1\rho = \frac{C_0}2+1-\frac12\log4\pi\approx0.0231
$$
to show that $\zeta(s)$ has no nontrivial zeros $\rho$ with $|\Im\rho|\le6.5$. (See for example Exercise 2 in Section 10.2 of Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory.)

Together these two results prove the desired statement and then some.


问:
How to determine the envelope of some curves, that is a curve tangent to many ellipses
In the attached figures, we have two kinds of curves; 1- a big closed curve which is somehow a limaçon. 2- many ellipses which are the same by the formula but different by the center. By putting some rotate and translated elliple, that is $$4\big(\frac{u}{\tau}\big)^2+\frac{13}{16}\big(\frac{v+\frac13}{\tau}\big)^2+\frac{7}{16}\big(\frac{1}{\tau}\big)^2-\frac{\sqrt{3}}{48}\big(\frac{3v+1}{\tau^2}\big)=1,$$ I found the figure 3. Now to continue I have to consider another rotated ellipse. Thats is I have to put the new rotated ellipse on different points belonging to the last closed curve in figure 3, and then find the envelope. By envelope, it meas the curve which is tangent to all of these ellipses. Supposing that the rotated ellipse is $ax^2+bxy+cy^2=1$ whose centers vary on the points on the last closed curve in figure 3, which are shown in next figures, what would be the equation or location of the curve tangent to all these ellipses?   I have no idea how to determine such a curve.

Any help is appreciated in advance.

I should say that

1- I can add as many as ellipses I want.

2- I do not have the equation of the big closed curve, the last curve in fugure 1. In fact, I plotted it with the points which I had from it. I think I can have as many as points on it I want.

My effort

1- First of all, I thought I can find such a curve tangent to all the ellipses by considering the outer vertices of them, but it does not work for the top and bottom of the curve.

答:
Let's assume (as an example) the equation of the ellipse is $x^2+2\,x\,y+3\,y^2=1$. We can write that as $(x+y)^2+(\sqrt{2}\,y)^2=1$. So if we make an affine transformation
$$u=x+y,\quad v=\sqrt{2}\,y\tag1,$$ the curve, given by some parametric equations $$x=x(t),\quad y=y(t)\tag2$$ in the $x,y$-plane, becomes $$u=u(t),\quad v=v(t)\tag3$$ in the $u,v$-plane. But in the latter, your ellipse became a circle of radius $1$ ($u^2+v^2=1$), and you can find the envelope simply by computing the normal vector (in the usual, Euklidean geometry of the $u,v$-plane, i.e. with components $\displaystyle\left(\frac{-v'(t)}{\sqrt{u'^2(t)+v'^2(t)}},\frac{u'(t)}{\sqrt{u'^2(t)+v'^2(t)}}\right)$) and going a distance $1$ in the outer or inner direction. Of course, you'll get an outer and an inner envelope. Even if you have your curve only as a sufficiently dense list of points $(x_i,y_i)$, they'll become a list of points $(u_i,v_i)$, and you can approximate the normal vector, there, just using $u_{i+1}-u_i$ instead of $u'(t)$, and $v_{i+1}-v_i$ instead of $v'(t)$. And after finding the envelope(s) in the $u,v$-plane, you transform back via $$x=u-\frac1{\sqrt{2}}\,v,\quad y=\frac1{\sqrt{2}}\,v\tag4.$$