问:
Nested sums - reciprocal of a sum : Find exact value
$$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2}$$

(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)

At first I thought it has something to do with Riemann's zeta function $\zeta$ and thus the solution is:

$$ \zeta \left( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right ) $$

However this totally seems incorrect because we cannot find this sum as we have an unknown, $k$.

Which also is being summed to $\infty$.

I thought (however was unsure) of the fact that:

$$ \left ( \sum f \right )^2 \ge  \sum f^2  $$

Thus (maybe) it is trivial the sum is converging to some value, which according to a little program I wrote is about: $$ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2} \approx 1.596$$

But this is not a rigorous proof, is there a way to solve this so we get a solution by hand - calculating the exact value of this sum if it is finite (if not, why? )

Any mathematical tools (not programs) are acceptable, because this is not a question from a specific course ( I don't have context for this).

Body:
Let
$$a_k=\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}=H_{k^2}^{\left(\frac{1}{2}\right)}-H_{k-1}^{\left(\frac{1}{2}\right)}  $$ and

$$S_p=\sum_{k=1}^{p} \frac{1}{ a_k^2}$$

Using asymptotics of the harmonic numbers and continuing with Taylor series
$$a_k=2 k-2 \sqrt{k}+\frac 1{2\sqrt k}+\frac 1{2k}+\frac 1{24 k\sqrt k}-\frac 1{24k^3}+O\left(\frac{1}{k^{7/2}}\right)$$
$$\frac{1}{ a_k^2}=\frac 1{4k^2}+\frac 1{2k^{5/2}}+\frac 3{4k^{3}}+O\left(\frac{1}{k^{7/2}}\right)$$
$$\sum_{k=1}^{\infty} \frac{1}{ a_k^2}=\sum_{k=1}^{p-1} \frac{1}{ a_k^2}+\sum_{k=p}^{\infty} \frac{1}{ a_k^2}$$


问:
Apps for practicing math (all levels)
I am looking for an app that I can use to PROVIDE me with math problems for practice and to stay fresh on various subjects in mathematics. This includes all levels of math (from low grade school to differential equations, linear algebra, ect). These practice problems would be random and new each time so that they continue to challenge me and provide an endless supply of practice.

That is the ideal, if this is not realistic I would settle for a handful of apps that provide most (if not all) of the subjects for practice examples. I am finding many options for subjects up to basic grade school math and one that goes up to precalculus but I would like to practice the upper level problems as well.

Note: What I mean by "provide me with problems" is if I were to "generate" a problem for a particular subset of mathematics I would be presented with an appropriate problem that I would be able to solve and put in an answer to be given feedback for. This app does not need to provide a calculator or any math solving tools/software.

Thank you.

答:
My top picks:

Mathcounts trainer: It gives you lots and lots of problems from MATHCOUNTS middle-school competition, and it lets you draw in the app, so you can write your work down their without going through lots of pencils and papers.

Alcumus: This isn't really an app, but its a website you can access on mobile so what's the difference? It gives you problems more difficult than the ones in MATHCOUNTS. I believe that it covers problems in Prealgebra, Algebra, Number Theory, Counting/Probability, Geometry, Intermediate Algebra, and Precalculus. It also lets you pick a difficulty as well.
You can access it here https://artofproblemsolving.com/alcumus.

You can also find many other resources here https://artofproblemsolving.com/resources.

ContentLicense: CC BY-SA 3.0

Title: Apps for practicing math (all levels)

ContentLicense: CC BY-SA 4.0


问:
Trouble solving $\lim _{x\to -\infty }\frac{3x^7-25x^5+e^{-x^3/(1-x)^3}}{-5x^7+\left(111!\right)\cdot \:x}$
$$\lim _{x\to -\infty }\frac{3x^7-25x^5+e^{-\frac{x^3}{\left(1-x\right)^3}}}{-5x^7+\left(111!\right)\cdot \:x}$$

not sure how to approach this limit. I have tried simplifying the function but I can't get it into a usable form. am I missing something when it comes to limit rules on how to approach this?

答:
Use L's hopital rule. Also note that (111)! Is as good as any other constant.

 Neglect all terms except the highest Power. ( This is the same if you apply L's hopital too. Answer will be -3/5, as they the powers will get cut. This works because e^ term converges to 1


问:
How to find the limit of the following question when x approaches 0 from positive side?
\begin{equation}
\lim _{x \rightarrow 0^{+}}(\sin 2 x)^{\tan 2 x}
\end{equation}

答:
You can also just use the standard limit $\lim_{t\to 0^+}t^t=1$ which can be proved quickly by taking the logarithm and using L'Hospital.

Now, just substituting $t=\sin 2x$ you get

$$(\sin 2 x)^{\tan 2 x}\stackrel{t=\sin 2x}{=} \left(t^t\right)^{\frac 1{\sqrt{1-t^2}}}\stackrel{t\to 0^+}{\longrightarrow}1^1 = 1$$


问:
Prove that $\max_x \frac{(x^Tv)^2}{x^TBx}=v^TB^{-1}v$

Let $B>0$ be positive definite $d \times d$ matrix, and $v \in \mathbb R^d$.
Prove that $$\max_x \frac{(x^Tv)^2}{x^TBx}=v^TB^{-1}v$$ Hint: Change of variable $z=B^{1/2}x$

$$\begin{split}\max_z\frac{(v^TB^{-1/2}z)^2}{z^Tz}\end{split}$$

Take the derivative

$$\begin{split}\frac{(z^Tz)2(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-(v^TB^{-\frac 1 2}z)^2(2z)}{(z^Tz)^2}
&=\frac{2(z^Tz)(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}z)(v^TB^{-\frac 1 2}z)z}{z^Tzz^Tz}\\
&=\frac{2(z^T\not{z)(z^T}B^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}\not{z)(z^T}B^{-\frac 1 2}v)z}{z^T\not{zz^T}z}\\
&=\frac{2(z^TB^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-1}v)z}{z^Tz}\end{split}$$

Probably the last cancellation is invalid, wasn't sure what to do, or how to proceed.

答:
Following Semiclassical's comment, it suffices to show $\max_z \frac{(u^\top z)^2}{z^\top z} = u^\top u$.
The Cauchy-Schwarz inequality implies $\max_z \frac{(u^\top z)^2}{z^\top z} \le u^\top u$. To show equality is attained, take $z=u$ (or $z=cu$ for any scalar $c$).


问:
Proving that the intersection of $(a-4)^{2}+(b-1)^{2}<4$ and $(a-1)^{2}+(b-5)^{2}<4$ is empty
I´m trying to prove that these two inequalities intersection is empty, but i don´t know how to proceed i know they are circles, but i´m trying to do it with only algebra

Let $a,b\in\mathbb{R}$ and the inequalities are

$$(a-4)^{2}+(b-1)^{2}<4 \quad\land\quad (a-1)^{2}+(b-5)^{2}<4$$

Any insight would be appreciated.

答:
The distance between the centers of the two circles $= \sqrt{(4-1)^2 + (1-5)^2} = 5$

is less than the sum of the radii of the two circles $= 2 + 2 = 4$


问:
Factorisation of a given polynomial into another with integer roots
Source: Challenge and Thrill of Pre-College Mathematics

"Find all integers $a$ such that

$$(x-a)(x-12)+2$$

can be factored into $(x-b)(x-c)$, such that $b$ and $c$ are integers."

My attempt: Simplifying the given polynomial and by the given condition, we get

$$x^2+(-12-a)x+(12a+2)=x^2-(b+c)x+ bc,$$ from where we obtain the system

$$b+c= (12+a)$$

$$bc= 12a+2$$

I can't proceed with the same process. A hint would be appreciated.

答:
Hint: Using comment by MEEL, we have:

$(b-a)(b-12)=-2$

Following cases can be considered:

1): $b-12=-1\Rightarrow b=11\Rightarrow 11-a=2\Rightarrow a=9$

Similarly:

2): $b-12=-2 \Rightarrow b=10\Rightarrow 10-a=1\Rightarrow a=9$.

3): $b-12=2$, $b-a=1$

4): $b-12=1$, $b-a=-2$

These two cases give $a=15$


答: Put A = (4,1), B = (1,5), C = (a,b). Then by property of triangle ABC : 5 = AB < CA + CB < 2+ 2 = 4. Thus : 5 < 4 : contradiction. Thus you can’t have both of inequalities above to hold .</p>

Title: Proving that the intersection of $(a-4)^{2}+(b-1)^{2}<4$ and $(a-1)^{2}+(b-5)^{2}<4$ is empty

Body:
I´m trying to prove that these two inequalities intersection is empty, but i don´t know how to proceed i know they are circles, but i´m trying to do it with only algebra

Let $a,b\in\mathbb{R}$ and the inequalities are

$$(a-4)^{2}+(b-1)^{2}<4 \quad\land\quad (a-1)^{2}+(b-5)^{2}<4$$

Any insight would be appreciated.


问:
How to find the number of trousers sold by a merchant on a specific day of the month?
The problem is as follows:

The owner of a bazaar in Daegu obtained for the sale of pants of the
same price, $6804$ usd in September, $10800$ usd in October and $7938$
usd in December. If the price of each pants is greater than $14$ usd
and less than $21$ usd, how much did he earned on the first day of
January, assuming that he sold a number of pants equal to the product
of the digits of its price?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{360 usd}\\
2.&\textrm{252 usd}\\
3.&\textrm{486 usd}\\
4.&\textrm{144 usd}\\
\end{array}$

The official solution according to my precalculus workbook is as follows:

First, find the $\operatorname{gcd}$ of $6804$, $10800$ and $7938$.

Therefore:

$\operatorname{gcd}(6804, 10800, 7938)=54$

This can be obtained from the prime factorization of each number:

$6804=2^2\cdot 3^5\cdot 7$

$10800=2^4 \cdot 5^2 \cdot 3^3$

$938=3^4\cdot 7^2 \cdot 2$

Then $\operatorname{gcd}(6804, 10800, 7938)=2\cdot 3^3 = 54$

Therefore the pants price is $18$ usd.

Because the quantity sold on January 1st coincides with the product of the digits of the price then:

$\textrm{number of pants}=1\times 8=8$

Since the price per pant is $18$ usd. Then the quantity he earned is:

$8\times 18 =144\,\textrm{usd}$

Therefore the answer is option 5.

This is the part where it ends the official solution.

Now my problem with this answer is I don't know why did the author used the greatest common divisor? I mean at some point he does obtain $54$, but what is it used for? Can someone help me here?.

There is a very imporant aspect and missing step here and that is how did he got $18$ usd the price per pant?

I'm not very savvy with the correct usage of the greatest common divisor, but I believe the intented approach is that to get the price I must find a common divisor for the three. But why should this be the maximum?

The numbers which $54$ multiplies, with those being 126, 200 and 147 I assume that these must be the number of pants sold on the months of september, october and december. But from this logic I'm getting that the price of the pants is 54 usd.

This itself causes a contradiction as the boundaries set by the problem constrain the value of the price of the pants between:

$14<p<21$

If this were the case, then I believe that the price of the pants might be a sub multiple of 54?. But which?, 2?, 3?, 6?, 9?.

Because if any of those were the case the price could be, 15,16,18 and 20. This part confuses me the most.

So in brief the sort of answer which I'm aiming to get is a one which first and foremost explain why was it used the greatest common divisor then how was it obtained the price per pant and on which cases does the greatest common denominator is used? and how should it be avoided its confusion with the least common multiple?. In other words why using the least common multiple isn't used here. By the way I've read the wikipedia entry on the subject but it did not helped me much as it only covers two applications none of which seem to properly explain its usage in word problems.

Since what I'm lacking is the conceptual explanation it would really help me a step by step explanation for this problem. So can someone help me here?.

答:
The price per pants must be a divisor of each of the amounts of money made in each of the months, as presumably he sold an integral number of pants. Thus, the price per pant must divide the greatest common common divisor of 6804 (money made in September), of 10800 (money made in October), and of 7938 (money made in December). Since the greatest common divisor is 54, the price per pant must be a divisor of 54.

(The defining property of the greatest common divisor of a set of numbers is that (i) it divides each of the numbers in the set; and (ii) any number that divides all numbers in the set divides the greatest common divisor. So the price per pant, which must divide each of the three amounts, must also divide $54$)

Now, what are the divisors of $54$? As $54=2\times 3^3$, the divisors are: $1$, $2$, $3$, $6$, $9$, $18$, $27$, and $54$. The price must be one of them.

We are told the price is “more than $14$ and less than $21$”. Well, the only possibility then is that the price per pant is $18$.


答:
Lets suppose that $(a,b)$ belongs to the set of points where
$$ \sqrt{(a-1)^2+(b-5)^2} <2. $$
By the triangle inequality for vectors in $\mathbb R^2$, in the form $|x|\ge |x-y|-|y| $, where $|x|:=\sqrt{x_1^2+x_2^2}$,

$$\sqrt{(a-4)^2+(b-1)^2} \ge \sqrt{(4-1)^2+(1-5)^2}-\sqrt{(a-1)^2+(b-5)^2}   > 5 - 2 =3 $$

which implies that $(a,b)$ cannot belong to the set of points where
$$ \sqrt{(a-4)^2+(b-1)^2} < 2.$$ So there are no points in the intersection.


问:
Solving System of Equations using transformation rotation
I've never had to post the same question twice, but my last post is getting filled out with work and I'm going about it a different way so I figured i'd try a whole different question

So This is the problem I've been working with: $$ \text{Let} \ a,b \ \text {be a positive real number with} \ a\neq b \ \text{Solve the system:}$$ $$a^2x^2-2abxy+b^2y^2-2a^2bx-2ab^2y+a^2b^2=0$$$$\text {and}$$$$abx^2+(a^2-b^2)xy-aby^2+ab^2x-a^2by=0$$

I have solved the top for $x$ plugged into the bottom and was working on solving the quartic for $y$. When someone gave me a new Idea to go about this.

They said to try and substitue $$x=\frac {u-v}{\sqrt 2} \ \text {and} \ y=\frac {u+v}{\sqrt 2}$$

How did they pick those two equations for $x$ and $y$ and how does that even help solving the two systems for x and y? Because If I use those substitutions to rotate the system, won't I have to somehow rotate them back?

Thanks for any and every idea!

FIRST EDIT:

So now I have used the substitutions $$x = \hat{x}\cos\theta - \hat{y}\sin\theta$$ and $$y = \hat{x}\sin\theta + \hat{y}\cos\theta$$
I also used the exact value of $$\cos(\frac{\pi}8)=\frac{\sqrt{2+\sqrt2}}2 \qquad \text {and also} \ \qquad \cos^2(\frac{\pi}8)=\frac{2+\sqrt2}4 $$ and $$\sin(\frac{\pi}8)=\frac{\sqrt{2-\sqrt2}}2 \qquad \text {and also} \ \qquad \sin^2(\frac{\pi}8)=\frac{2-\sqrt2}4$$

Which I was able to simplify my equation down some, but how does that help the overall process of solving the systems?

SECOND EDIT:

So I substituted and rearranged the top equations and got: $$a^2\hat{x}^2\frac {2+\sqrt{2}}4-a^2\hat{x}\hat{y}a^2\hat{y}^2\frac{2-\sqrt2}4-ab\hat{x}^2-2ab\hat{x}\hat{y}+ab\hat{y}^2+b^2\hat{x}^2\frac{2-\sqrt2}4+b^2\hat{x}\hat{y}+b^2\hat{y}2\frac{2+\sqrt2}4-2ab\hat{x}\frac{\sqrt{2+\sqrt2}}2+2ab\hat{y}\frac{\sqrt{2-\sqrt2}}2-2ab^2\hat{x}\frac{\sqrt{2-\sqrt2}}2+2ab^2\hat{y}\frac{\sqrt{2+\sqrt2}}2+a^2b^2=0$$
Then I did the same thing with the bottom equation and got: $$\frac{a^2\hat{x}^2}2+a^2\hat{x}\hat{y}\frac{\sqrt2}2-\frac{a^2y^2}2+ab\hat{x}^2\frac{2+\sqrt2}4-ab\hat{x}\hat{y}+ab\hat{y}^2\frac{2-\sqrt2}4-\frac{b^2\hat{x}^2}2+\frac{b^2\hat{y}^2}2b^2\hat{x}\hat{y}\frac{\sqrt2}2-ab\hat{x}\frac{\sqrt{2-\sqrt2}}2-ab\hat{y}\frac{\sqrt{2+\sqrt2}}2+ab^2\hat{x}\frac{\sqrt{2+\sqrt2}}2-ab^2\hat{y}\frac{\sqrt{2-\sqrt2}}2-a^2b\hat{x}\frac{\sqrt{2-\sqrt2}}2+a^2b\hat{y}\frac{\sqrt{2+\sqrt2}}2=0$$

Now once I have it simplified what can I do now?

答:
My approach is to solve both equations for $x$ and then set them equal. The I go through stages of expansion and collection of terms. As I write, I'm stopping for the night. The next stage will be separation of radicals to one side and of non-radicals to the other side. Then both sides will be squared, the remaining radical will be separated, and both sides will be squared again. At that point a solution for $y$ should be possible even if it takes WolframAlpha to find it.

$$a^2x^2-2abxy+b^2y^2-2a^2bx-2ab^2y+a^2b^2=0\\ \implies
x = \frac{(2 a^2 b + 2 a b y) ± \sqrt{(-2 a^2 b - 2 a b y)^2
- 4 a^2 (a^2 b^2 - 2 a b^2 y + b^2 y^2)}}{2 a^2} \\$$

$$abx^2+(a^2-b^2)xy-aby^2+ab^2x-a^2by=0\\ \implies
x = \frac{(a^2 (-y) - a b^2 + b^2 y) ± \sqrt{(a^2 y + a b^2 - b^2 y)^2
- 4 a b (a^2 (-b) y - a b y^2)}}{2 a b}\\ $$

$$b\bigg((2 a^2 b + 2 a b y) ± \sqrt{(-2 a^2 b - 2 a b y)^2
- 4 a^2 (a^2 b^2 - 2 a b^2 y + b^2 y^2)}\bigg) \\
=a\bigg( (a^2 (-y) - a b^2 + b^2 y) ± \sqrt{(a^2 y + a b^2 - b^2 y)^2
- 4 a b (a^2 (-b) y - a b y^2)}\bigg)\\
\implies\\
b (\bigg(2 a^2 b + 2 a b y) ± 4 \sqrt{a^3 b^2 y}\bigg)=\\
a \bigg((a^2 (-y) - a b^2 + b^2 y) ±
\sqrt{4 a^2 b^2 y (a + y) + (a^2 y + a b^2 - b^2 y)^2}\bigg) \implies \\
4 b \sqrt{a^3 b^2 y} + a^3 y + 3 a^2 b^2 - a \sqrt{a^4 y^2 + 6 a^3 b^2 y + a^2 b^4 + 2 a^2 b^2 y^2 - 2 a b^4 y + b^4 y^2} + a b^2 y = 0$$


答: Expand both equations to get: $$a^2-8a+16+b^2-2b+1 < 4$$ $$a^2-2a+1+b^2-10b+25 < 4$$ Add these two together and divide by $2$ to get: $$a^2-5a+b^2-6b+\dfrac{43}{2} < 4.$$ Now, we use the "complete the square" trick on the left side twice: $$a^2-5a+\dfrac{25}{4}+b^2-6b+9+\dfrac{43}{2} < 4+\dfrac{25}{4}+9$$ $$\left(a-\dfrac{5}{2}\right)^2+(b-3)^2+\dfrac{43}{2} < \dfrac{77}{4}$$ $$\left(a-\dfrac{5}{2}\right)^2+(b-3)^2 < -\dfrac{9}{4}$$ This is clearly a contradiction, since the LHS must be positive. Therefore, there are no reals $(a,b)$ which satisfy both inequalities.


问:
Find the general solution to the ODE:
I was asked to find the general solution to these ODEs.

$$\frac{dy}{dx}=\frac{(-8x+3y-31)}{(-3x+y-11)}$$
I tried by rearranging to the form $M(x,y)+N(x,y)\frac{dy}{dx}=0$ and let $y=vx$, but what I got was $$(v^2x-6vx-11v+8x+31)dx+x(-3x+vx-11)dv=0$$
and I could not separate $x$ and $v$ for integration.

$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xexp(-\frac{2y}{x})+\frac{9}{2}xexp(\frac{2y}{x})$$
I have totally no idea which method to be used in Question 2.

May I know if I'm on the wrong track, and how should I solve them? Thanks.

答:
We have
$$
\frac{dy}{dx} = \frac{-8x+3y-31}{-3x+y-11}.
$$
One way to solve it is to introduce time variable $t$, obtaining a system
$$
\left\{
\begin{align*}
&\dot{y} = -8x+3y-31 \\
&\dot{x} = -3x+y-11
\end{align*}
\right. ,
$$
which is linear. A matrix of this system
$$
A =
\begin{pmatrix}
-8 & 3\\
-3 & 1
\end{pmatrix}
$$
is nondegenerate: $\det A = 1$.
Thus we can find a particular solution almost immediately substituting
$$
\left\{
\begin{align*}
&y = Y \\
&x = X
\end{align*}
\right. .
$$
That gives us a system of linear equations:
$$
\left\{
\begin{align*}
&-8X+3Y = 31\\
&-3X+Y = 11
\end{align*}
\right. ,
$$
which has a unique solution $Y = 5$, $X = -2$. Now everything what we we have to do is to find a general solution of the homogeneous system
$$
\left\{
\begin{align*}
&\dot{y} = -8x+3y\\
&\dot{x} = -3x+y
\end{align*}
\right. .
$$
So we consider a characteristic equation $\det (A - \lambda I) = \lambda^2 + 7\lambda + 1 = 0$. We have two distinct roots:
$$
\left\{
\begin{align*}
&\lambda_1 = \frac{3\sqrt{5}-7}{2}\\
&\lambda_2 = \frac{-3\sqrt{5}-7}{2}
\end{align*}
\right. .
$$
Corresponding eigenvectors are
$$
e_1 = \begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix}, \quad e_2 = \begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix}.
$$
Hence the general solution is
$$
\begin{pmatrix} y \\ x \end{pmatrix} = C_1\begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)},
$$
$C_1, C_2 = \mathrm{const}$. Finally, we combine two parts together:
$$
\begin{pmatrix} y \\ x \end{pmatrix} = C_1\begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)} + \begin{pmatrix} 5 \\ -2 \end{pmatrix}.
$$
To proceed to the answer we have to say a few words about singularities where the right hand side of our original equation doesn't exist. A condition is $-3x + y - 11 = 0$ (we look for singularities). It gives us
$$
C_1\left(\frac{-\sqrt{5}+3}{2}  - 3 \right)\exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\left(\frac{\sqrt{5}+3}{2} - 3 \right)\exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)} = 0,
$$
$$
C_1\frac{\sqrt{5}+3}{2}\exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} =  C_2\frac{\sqrt{5}-3}{2}\exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)},
$$
$$
C_1\frac{\sqrt{5}+3}{\sqrt{5}-3}\exp{\left(3\sqrt{5}t\right)} =  C_2.
$$
But those points are not a real problem: in those points a tangent line to our integral curves is vertical. Not a big deal. So the answer is above in the form of a family of curves parametrised by time.

Title: Find the general solution to the ODE:


问:
The value of a limit of a power series: $\lim\limits_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$
What is the answer to the following limit of a power series?

$$\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$$

答:
It is straightforward to reproduce the integral representations of the sum found by @deoxygerbe and @user17762 in their interesting paper and by @sos440,
$$\begin{eqnarray*}
\sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k
&=& -x\int_0^1 t^{x t} dt \\
&=& -x \int_0^\infty e^{-z(xe^{-z}+1)} dz.
\end{eqnarray*}$$
Here we verify @sos440's result, that the sum is $-1$, by examining the last integral.

We have
$$\begin{eqnarray*}
x \int_0^\infty e^{-z(xe^{-z}+1)} dz
&=& \underbrace{\int_0^1 f(z)dz}_{I_1} + \underbrace{\int_1^\infty f(z)dz}_{I_2},
\end{eqnarray*}$$
where $f(z) = x e^{-z(xe^{-z}+1)}$.
On $(1,\infty)$, $f(z)$ has a global max at $z_0$, the solution to
$x(z_0-1)=e^{z_0}$.
Note that $z_0 \sim \log x + \log\log x$
so that $\lim_{x\to\infty} z_0 = \infty$.
Applying Laplace's method we find
$$\begin{eqnarray*}
I_2
&\sim& \sqrt{\frac{2\pi}{(z_0-2)(z_0-1)}} e^{-z_0/(z_0-1)} \\
&\sim& \frac{\sqrt{2\pi}}{e z_0}.
\end{eqnarray*}$$
Thus, in the limit, $I_2 = 0$.

But
$$\begin{array}{ccccc}
\displaystyle x\int_0^1 e^{-z(x+1)} dz
&\le&
\displaystyle x \int_0^1 e^{-z(xe^{-z}+1)} dz
&\le&
\displaystyle x\int_0^1 e^{-z[x(1-z)+1]} dz.
\end{array}$$
(Here we use that on $[0,1]$, $1-z\le e^{-z}\le 1$.)
The bounding integrals can be calculated explicitly and in the limit they are unity.
Thus, in the limit, $I_1 = 1$.

Therefore,
$$\begin{eqnarray*}
\lim_{x\to\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k
&=& -\lim_{x\to\infty} x \int_0^\infty e^{-z(xe^{-z}+1)} dz \\
&=& -\lim_{x\to\infty} (I_1+I_2) \\
&=& -1.
\end{eqnarray*}$$


问:
Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6...

Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6, find distance of focus from tangent at P.

Let P and Q be $(at^2,2at)$ and $(\frac{a}{t^2}, \frac{-2a}{t})$ respectively, for parabola $y^2=4ax$

So $$0= 2(-\frac{2a}{t})+2at$$
$$\implies t^2=2$$

Now tangent at Q is

$$\frac{y}{\sqrt 2} =x+\frac a2$$

So point R is $(-a, -\frac{a}{\sqrt 2})$

Then $$QR^2 = 36$$
$$\implies a=\frac{12}{\sqrt {11}}$$

So now, tangent at P is $$\sqrt 2y=x+2a$$ whose distance from $(a,0)$ is
$$d=\left | \frac{a+2a}{\sqrt 3}\right|$$
$$d=\sqrt 3 a$$
$$d=\frac{\sqrt 3 \times 12}{\sqrt {11}}$$

But given answer is 4

Where am I going wrong?

答:
The mistake lies in taking incorrect sign of $t$ for point $Q$. It lies on opposite side of $P$. So for $Q$, parameter is $-1/t=-1/\sqrt{2}$ but in the tangent equation at $Q$ you took $1/t=1/\sqrt{2}$ which gave wrong coordinates for $R$.

Since $(x,y)=(at^2,2at)$, $t$ has same sign as $y$. So the entire upper branch of a parabola has $t$ positive and entire lower branch has $t$ negative - something to keep in mind.


答:
The way via the greatest common divisor is absolutely not necessary. Actually the revenues for the given three months aren't necessary at all to solve the problem. So, I add this short answer, even if it does not address the OP's question:

We know for the price $p=10a+b$ and $a\cdot b \in \{5,6,7,8,9\}$ since a price of $20$ would mean no pants sold.

Given the options, the revenue $p\cdot a\cdot b$ must be even. So only $16$ or $18$ are to be considered. Checking these, $p= 18$, hence, a revenue of $144$ is the only possible answer.

Title: How to find the number of trousers sold by a merchant on a specific day of the month?

Body:
The problem is as follows:

The owner of a bazaar in Daegu obtained for the sale of pants of the
same price, $6804$ usd in September, $10800$ usd in October and $7938$
usd in December. If the price of each pants is greater than $14$ usd
and less than $21$ usd, how much did he earned on the first day of
January, assuming that he sold a number of pants equal to the product
of the digits of its price?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{360 usd}\\
2.&\textrm{252 usd}\\
3.&\textrm{486 usd}\\
4.&\textrm{144 usd}\\
\end{array}$

The official solution according to my precalculus workbook is as follows:

First, find the $\operatorname{gcd}$ of $6804$, $10800$ and $7938$.

Therefore:

$\operatorname{gcd}(6804, 10800, 7938)=54$

This can be obtained from the prime factorization of each number:

$6804=2^2\cdot 3^5\cdot 7$

$10800=2^4 \cdot 5^2 \cdot 3^3$

$938=3^4\cdot 7^2 \cdot 2$

Then $\operatorname{gcd}(6804, 10800, 7938)=2\cdot 3^3 = 54$

Therefore the pants price is $18$ usd.

Because the quantity sold on January 1st coincides with the product of the digits of the price then:

$\textrm{number of pants}=1\times 8=8$

Since the price per pant is $18$ usd. Then the quantity he earned is:

$8\times 18 =144\,\textrm{usd}$

Therefore the answer is option 5.

This is the part where it ends the official solution.

Now my problem with this answer is I don't know why did the author used the greatest common divisor? I mean at some point he does obtain $54$, but what is it used for? Can someone help me here?.

There is a very imporant aspect and missing step here and that is how did he got $18$ usd the price per pant?

I'm not very savvy with the correct usage of the greatest common divisor, but I believe the intented approach is that to get the price I must find a common divisor for the three. But why should this be the maximum?

The numbers which $54$ multiplies, with those being 126, 200 and 147 I assume that these must be the number of pants sold on the months of september, october and december. But from this logic I'm getting that the price of the pants is 54 usd.

This itself causes a contradiction as the boundaries set by the problem constrain the value of the price of the pants between:

$14<p<21$

If this were the case, then I believe that the price of the pants might be a sub multiple of 54?. But which?, 2?, 3?, 6?, 9?.

Because if any of those were the case the price could be, 15,16,18 and 20. This part confuses me the most.

So in brief the sort of answer which I'm aiming to get is a one which first and foremost explain why was it used the greatest common divisor then how was it obtained the price per pant and on which cases does the greatest common denominator is used? and how should it be avoided its confusion with the least common multiple?. In other words why using the least common multiple isn't used here. By the way I've read the wikipedia entry on the subject but it did not helped me much as it only covers two applications none of which seem to properly explain its usage in word problems.

Since what I'm lacking is the conceptual explanation it would really help me a step by step explanation for this problem. So can someone help me here?.


问:
Find the $\mathcal{o}\left(n\right)$ so that $a_n\sim\frac{\mathcal{o}\left(n\right)}{n}+\mathcal{O}\left(\frac{1}{n}\right){\rm as}\lim a_n=\infty$

Given a recursion $a_{n+ 1}= \dfrac{1}{2}\left ( a_{n}+ \dfrac{1}{a_{n}} \right )$ with $a_{1}> 0.$ Find the $\mathcal{o}\left ( n \right )$ so that
$$a_{n}\sim\frac{\mathcal{o}\left ( n \right )}{n}+ \mathcal{O}\left ( \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$

Remark. The original one is proving $\lim a_{n}= 1.$ I see two approaches but which are kind of luckily nonsense and unuseful here.

We have that
$$a_{n+ 1}- 1= \frac{1}{2}\left ( a_{n}+ \frac{1}{a_{n}} \right )- 1= \frac{1}{2}\left ( a_{n}- 1 \right )^{2}+ \mathcal{O}\left ( \left ( a_{n}- 1 \right )^{2} \right )\,{\rm as}\,a_{n}\rightarrow 1$$
We easily find the $\mathcal{o}\left ( n \right )$ but actually I find $\mathcal{o}\left ( n \right )$ to prove that $\lim a_{n}= 1,$ that's a bad idea !

From the example * Estimations of some new recurrence sequences *, for the recursion $a_{1}= 1$ and $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}:$
$$a_{n}\sim\frac{1}{2}\left ( n+ \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$
I need to the help to change the problem of the disconvergences to the convergence problem by using $n$ cleverly, I know it's too inconvenient, but I can't think such a better way at all. Thanks for all your nice and valuable comments a real lot !

Body:
We have:
$$a_{n+1}-1 = \frac{1}{2}\frac{(a_n-1)^2}{a_n}$$
$$a_{n+1}+1 = \frac{1}{2}\frac{(a_n+1)^2}{a_n}$$
Hence
$$\frac{a_{n+1}-1}{a_{n+1}+1}=\left( \frac{a_{n}-1}{a_{n}+1} \right)^2$$
We deduce that

$$\frac{a_{n}-1}{a_{n}+1} = \left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}$$
or the closed form expression of $a_n$
$$a_n =\frac{1-\left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}}{1+\left( \frac{a_{1}-1}{a_{1}+1} \right)^{2^{n-1}}}$$


答:
i know they are circles,

And that's key.

The are the interior of circles (not actually circles themselves) with radius $2$.  And centered and $(4,1)$ and $(1,5)$. If the distance between the centers is more than sum of the radii the circles must be disjoint.

And indeed $||(4, 1),(1,5)|| = \sqrt{(4-1)^2 + (1-5)^2}=\sqrt{3^2 + (-4)^2}= \sqrt{9+16}=\sqrt{25} = 5 > 2+ 2$

And that proves the circles are disjoint.

You can prove this directly.  If $(a-4)^2 +(b-1)^2 < 4$ Then as squares are non negative we must have $(b-1)^2 < 4$ so $|b-1| < 2$ and $-1 < b < 3$.

This means that $-6 < b-5 <-2$ and therefore $36 > (b-5)^2 > 4$.

But that means $(a-1)^2 + (b-5)^2 > 4$.

So you can't have both $(a-4)^2 +(b-1)^2 < 4$ and  $(a-1)^2 + (b-5)^2< 4$


问:
Question regarding the proof of a topological claim
The lecturer in the Topology course I'm taking defined the following:
Given a topological space $X$
  we say that:

$X$ is weakly locally compact if for all $x\in X$
there exists a compact nbhd.

$X$ is strongly locally compact if every nbhd of $x$
contains a compact nbhd of $x$
.

We then made the following claim: A weakly locally compact (wlc) Hausdorff space is strongly locally compact.

Briefly the proof went as follows:

Given $U$ a nbhd of $x\in X$  since $X$ is wlc there is a compact nbhd of $x$
, $C\subseteq X$.

Since $U,C$
are both nbhds of $x$ then $U\cap C$
is also a nbhd of $x$
and thus there is an open set $V\subseteq X$
such that $x\in V\subseteq U\cap C$
.

Since $C$
is a compact Hausdorff space (Hausdorff being hereditary) we know $C$
is regular.

Since regularity is hereditary and $V\subseteq C$
we know $V$
is also regular and thus there is an open set $W\subseteq V$
such that $x\in W\subseteq\overline{W}\subseteq V\subseteq C$ .

Since $C$
is compact and $\overline{W}$
is closed in $C$
we know that $\overline{W}$
is also compact.

Finally $x\in W\subseteq\overline{W}\subseteq V\subseteq U$
and thus $\overline{W}$
is a compact nbhd of $x$
contained in $U$ .

The lecturer then noted that it's important to notice the proof hangs on the fact that the closure of $W$ in $V$ and $C$ is the same. That is since we can only deduce compactness of $\overline{W}$ since it is closed in $C$. However, we used the regularity of $V$ in order to find $W$ and thus the closure of $W$ is relative to the topology in $V$ and not in $C$. He also noted that in fact from the way we carried out the construction the closure of $W$ is the same in all the groups in which it is contained, that is $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$
  (the substring marking closure relative to which space).

My question is why is it in fact true that $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$ ?

答:
There is a useful formula: If $A$ is a subset of $X$ and $W$ is another subset, then $\text{cl}_A(W\cap A)=\text{cl}_X(W\cap A)\cap A$. So let $A$ be the $V$ in your example.

If we take the closure of $W$ with respect to $C$, then $\text{cl}_C(W)=\text{cl}_X(W)$ since $C$ is closed in $X$.

On the other hand $\text{cl}_V(W)=\text{cl}_X(W)\cap V$ which is just $\text{cl}_X(W)$ since $W$ was chosen to have its closure contained in $V$.

Finally, $\text{cl}_U(W)=\text{cl}_X(W)\cap U=\text{cl}_X(W)$ since $\text{cl}(W)\subset U$.

I should also mention that the above formula simplifies one more time if $A$ is open. In this case we have $\text{cl}_A(W\cap A)=\text{cl}_X(W\cap A)\cap A=\text{cl}_X(W)\cap A$. Using this the last two equalities follow automatically since $U$ and $V$ were open.


问:
Proving the product topology and standard topology on $\mathbb{R}^2$ are the same
I was trying to prove that the product topology on $\mathbb{R}^2$ is the same as the topology induced by the euclidean metric. My attempt was to find a square that fits inside an open ball in the first case, and an open ball that fits inside a rectangle in the second case to do this. Let me know if my attempt is on the right track.

Attempt:Yes the product topology on $\mathbb{R} \times \mathbb{R}$ is the same as the standard topology on $\mathbb{R}^2$. Let $U$ be an open set in $\mathbb{R}^2$. Let $p=(x_1,y_2) \in U$. Then there is an $\epsilon>0$ such that $B_{\epsilon}(p)=\{(x_2,y_2) \in \mathbb{R}^2|\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}<\epsilon\} \subset U$. Then $(x_1-\frac{\epsilon}{\sqrt{2}},x_1+\frac{\epsilon}{\sqrt{2}}) \times (y_1-\frac{\epsilon}{\sqrt{2}},y_1+\frac{\epsilon}{\sqrt{2}}) \subset B_{\epsilon}(p)=\{(x_2,y_2) \in \mathbb{R}^2|\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}<\epsilon\} \subset U$

So $U$ is open in $\mathbb{R} \times \mathbb{R}$.

Conversely suppose that $W$ is open in $\mathbb{R} \times \mathbb{R}$. Then for each point $(x,y) \in U$ there is a basis element $(a,b) \times (c,d)$ centered about $(x,y)$ such that $(a,b) \times (c,d) \subset W$. Then $(x,y)=(\frac{a+b}{2},\frac{c+d}{2})$. Choose $\epsilon=\text{min}\{\frac{|b-a|}{2},\frac{|c-d|}{2}\}$. Then $B_{\epsilon}((x,y))\subset (a,b) \times (c,d) \subset W$. So $W$ is open in $\mathbb{R}^2$.

答:
Yes, this is indeed a valid way to show that these topologies are the same. Missing details: You should show by computation that indeed

$$(x_1-\frac{\epsilon}{\sqrt{2}},x_1+\frac{\epsilon}{\sqrt{2}}) \times (y_1-\frac{\epsilon}{\sqrt{2}},y_1+\frac{\epsilon}{\sqrt{2}}) \subset B_{\epsilon}(p)$$

and also that

$$B_{\epsilon}((x,y))\subset (a,b) \times (c,d)$$ for your choice of $\epsilon$.

This might be clear (and inspired from) a picture, but the computations here aren't that hard.

Add that and I'd give full marks.


问:
help solving limit with integral inside
$\lim_{x\to 3} \dfrac{\int_9^{x^2} \frac{\sqrt{t}(\sqrt{t}-3)}{t-5}\ dt}{(x-3)^2}$

not sure how to approach this limit. I have tried simplifying the function by solving the integral but I can't get it into a usable form. The integral is not coming out nicely at all so I'm not sure what to do with it. can I get this in a form for L'Hopital?

答: You can apply L'Hopital's rule by differentiating the definite integral.</p>

Title: help solving limit with integral inside


问:
Having trouble with plugging in a function to a limit question
ORIGINAL QUESTION:

find
$$\lim _{Δx\to 0}\frac{f\left(x+Δx\right)-f\left(x\right)}{Δx}$$

for the function
$$f(x)=3x-2$$

I tried plugging it in and got
$$\lim _{Δx\to 0}\frac{3\left(x+Δx\right)-2-\left(3x-2\right)}{Δx}$$

I'm not sure if this is the correct way to plug it in, I would assume that I need to cancel out the ΔX but am unsure of the steps that I need to take in order to do so.

答:
Treat $\Delta x$ as a variable separate from $x$. I usually use $h$ instead so:
$$\frac{3(x+h)-2-3x+2}{h}=\frac{3h}{h}=3$$
Then clearly $$\lim_{h \to 0} 3 =3$$


问:
How to prove that for any two natural numbers a, b, there exist natural numbers c,d with a + d = b +c
I understand the intuition for how to do this, but I don't know how to formally prove it. For example, I know that if you take (a,b) to be (1,2), then a sample (c,d) could be (1,2) since 1 + 2 = 2 + 1. But how do I formally prove that such numbers c,d exist for an arbitrary c,d such that a,b,c,d are all in the naturals.

Thanks in advance!

答:
Well, you can always let $c=a$ and $d=b$. Then $a+d = c+d$.

If you mean a different pair of numbers.

Well, that isn't true.   $0 +0$ will have not $\{c,d\} \ne \{0\}$ so that $c + d = 0+ 0 =0$.

Nor is it true for $\{a,b\}=\{0,1\}$ and $a+b = 1$ will only have $\{c,d\} =\{0,1\}$ if $c+d =a+b= 1$.

However if we rule out that $\{a,b\} = \{0,1\}$

Then we can say let $\alpha = \min(a,b)$ and $\beta = \max (a,b)$.

Case 1:  $\alpha = 0$.  Then $\beta \ne 0, 1$ so $\beta \ge 2$.

Let $c = 1$ and $d = \beta -1$.

Case 2: $\alpha > 0$.  Then let $c = \alpha -1$ and $d = \beta + 1$.


问:
Showing the Regularity of any Positive Borel Measure $\lambda$ in Theorem 2.20 of Rudin Real and Complex Analysis
$\lambda$ is given as any positive Borel measure on $R^{k}$ and $\lambda\left(E\right)=m\left(E\right)$ for all boxes $E$.

A box is defined in 2.19 of RCA (Please let me know if I should provide it here). The measure $m$ satisfies the properties of the Riesz representation theorem (Theorem 2.14).

Rudin concludes that $\lambda$ is regular using Theorem 2.18. This theorem (2.18) gives conditions for $\lambda$ to be regular which is that every open set is $\sigma$-compact (countable union of compact sets) and $\lambda\left(K\right)<\infty$, where $K$ is any compact set belong to the corresponding $\sigma$-algebra.

But I am not sure how Theorem 2.18 can be used as above (as it is done in Theorem 2.20) without knowing that $\lambda\left(K\right)<\infty$, where $K$ is any compact set belong to the corresponding $\sigma$-algebra.

Any pointers would be highly appreciated. I could be missing something simple, hence am happy to delete the question if this is too trivial.

Related Posts:

Questions about construction of Lebesgue measure in Rudin's RCA

答:
Any compact set $K$ is  subset of some box $B$. So $\lambda (K) \leq \lambda (B)=m(B) <\infty$.

Any compact set is bounded. Hence it is contained in a box $\left(0,\delta\right)$ for large enough $\delta$. Hence, $\lambda\left(Q\left(0,\delta\right)\right) =m\left(Q\left(0,\delta\right)\right) = \delta^{k}<\infty$. Here, we are using property (a) of Theorem 2.20, which is that $m\left(W\right)=\text{vol}\left(W\right)$ for every $k$-cell $W$.


问:
How to find the greatest amount of savings from two people when it is related with a division?
The problem is as follows:

A bank manager calculates the greatest common divisor of the numbers
that represent the amounts of money in usd that Jenny and Gwen have,
using Euclid's algorithm, the successive quotients are $3$; $1$ and
$2$ respectively. If the first division was made by excess and also
the least common multiple of the quantities they have is 315, how many
dollars represents the greatest savings among the two girls?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{105 usd}\\
2.&\textrm{98 usd}\\
3.&\textrm{112 usd}\\
4.&\textrm{77 usd}\\
\end{array}$

The official solution according to my precalculus workbook is as follows:

(Comment: The author arranged the information in this sort of weird table which I don't understand)

$\begin{array}{|c|c|c|c|}
\hline
& \textrm{3(excess)} & 1 & 2 \\ \hline
a=7d & b=3d & 2d & d\\ \hline
 & 2d & 2d& 0\\ \hline
\end{array}$

Assume the numbers representing the savings for Jenny and Gwen are $a$ and $b$.

From the given information in the problem:

Use the least common multiple:

$\operatorname{lcm}(a,b)=315$

$(7d)(3d)=315$

Then $d=15$

Therefore $a=7\cdot 15 =105$

Therefore the greatest saving it is $105$ usd.

The answer is option 1.

This is the part where it ends the official solution.

Now my problem is I'm stuck here at the very beginning. Does it exist a way to reverse engineering what it was intended to be done here by this method?

From what I do recall the Euclid's algorithm of the division it goes as this:

$D=dq+r$

In an inexact division the $d$ represents the divisor and the $q$ the quotient and the $r$ represents the remainder.

I'm assuming that the second row in the table seems to point the first term of the algorithm. But from then on I don't get it.

A division by excess implies that the remainder is like as follows:

$D=dq_{\textrm{excess}}-r_{\textrm{excess}}$

But I don't know if this was used here?. The part where it mentions the least common multiple seems to imply primes because it places as $(7d)(3d)=315$

But the prime factorization of $315$ is equal to $3^2 \cdot 5 \cdot 7$. But from these there can be different combinations such as:

$3\cdot 7$ or $3 \cdot 5$

Since it implies that both must be the same I mean $d$ there isn't sufficient $5$'s to get $15$ on both sides of the multiplication from the prime factorization of $315$.

All and all, can someone help me here?. I'm just lost. Perhaps does it exist a better argument or method of solution for this problem?. I don't know how is it established that the greatest of both savings is $105$. As it is mentioned the sort of answer which it would help me the most is a step-by-step explanation of the solution.

答:
Note that the calculations which agree with the answer to this question are
$$105=3\times45-30$$
$$45=30+15$$
$$30=2\times 15$$

We can work this out as follows, where $a$ is the larger of the two amounts $a,b$.
$$a=3\times b-R$$
$$b=R+S$$
$$R=2\times S$$

Solving in terms of $S$ gives $$R=2S,b=3S,a=7S.$$
The LCM of $7S$ and $3S$ is $315$ and so $$21S=315.$$

Therefore $S=15$ and $a=105$.


问:
Collatz Conjecture, can the following skip a prime number?
Following my previous questions at: Collatz Conjecture, why an increment of $+6$ in the following? and Collatz Conjecture, why a rate of change of $*4$ in the following?

Following the rules of the Collatz Conjecture, in this experiment I have created a list of all odd numbers until $33333$. The list includes 3 columns, such as in the following sample:

A) Starting Odd $(X)$
B) $(X * 3) +1$
C) $X/2$ repeat until odd

1
4
2, (1)

3
10
(5)

5
16
8, 4, 2, (1)

7
22
(11)

9
28
14, (7)

11
34
(17)

13
40
20, 10, (5)

15
46
(23).

17
52
26 (13)

19
58
(29)

21
64
32, 16, 8, 2. (1)

23
70
(35)

25
76
38, (19)

...

You will notice that all the final odd results in column C) represent a list of all the prime numbers. as denoted in the () in column C): 1, 5, 7, 11, 13, 17, 19, 23, 29.

Is it possible to skip a prime number in that list (with the exception of $3$)?

答:
No, you should actually get every odd number which isn't a multiple of $3$ (and, in particular, every odd prime save $3$).

Note that any prime $p\ne2,3$ is either $1$ or $5$ mod $6$. Suppose $p=6k+5$. Then set $X=4k+3$, so that $3X+1=12k+10$ and dividing by $2$ gives $6k+5=p$. Now suppose $p=6k+1$. Then let $X=8k+1$, so that $3X+1=24k+4$. Dividing by $2$ twice gives $6k+1=p$, as desired.

In fact, since nothing here depends on $p$ being prime, this shows that any odd number which isn't a multiple of $3$ is a final number in your column (c).

Conversely, note that no multiple of $3$ can ever be a final number in (c). (In fact, no multiple of $3$ can ever be an element in that column!) After all, if $X=2k+1$, then the final number is of the form $2^{-n}(6k+4)$. Note that $6k+4$ isn't divisible by $3$, and so no number in the last column can be of the form $3k$.


答:
That substitution is correct, yes, since

$$\require{cancel}f(x+\Delta x) = 3(x+\Delta x) - 2$$

So now we just need to do some algebra to simplify:

\begin{align*}
&\lim _{Δx\to 0}\frac{3\left(x+Δx\right)-2-\left(3x-2\right)}{Δx}\\
&=\lim _{Δx\to 0}\frac{3x+3Δx-2-3x+2}{Δx} \tag{distribute}\\
&=\lim _{Δx\to 0}\frac{\cancel{3x}+3Δx\cancel{-2}\cancel{-3x}\cancel{+2}}{Δx} \tag{like terms cancel}\\
&=\lim _{Δx\to 0}\frac{3 \Delta x}{Δx} \tag{simplify the previous}\\
&=\lim _{Δx\to 0}\frac{3 \cancel{\Delta x}}{\cancel{Δx}} \tag{cancel common factor}\\
&=\lim _{Δx\to 0}3 \tag{simplify}\\
&=3 \tag{trivial limit}
\end{align*}

Title: Having trouble with plugging in a function to a limit question

Body:
ORIGINAL QUESTION:

find
$$\lim _{Δx\to 0}\frac{f\left(x+Δx\right)-f\left(x\right)}{Δx}$$

for the function
$$f(x)=3x-2$$

I tried plugging it in and got
$$\lim _{Δx\to 0}\frac{3\left(x+Δx\right)-2-\left(3x-2\right)}{Δx}$$

I'm not sure if this is the correct way to plug it in, I would assume that I need to cancel out the ΔX but am unsure of the steps that I need to take in order to do so.


问:
Interesting sequence question (modified)
Let $N_0\in \mathbb{N}.$ If a sequence of complex numbers $\{F_N\}_{N \in \mathbb{N}}$ has the following properties:
$$\lim_{N \rightarrow \infty} |F_N|^{1/N}=0$$ and for all $N \geq N_0$,
$$|F_N|^2\leq \sum_{k=N+1}^{\infty} |F_k|,\quad \quad \quad  $$
then there exists $N_1\in \mathbb{N}$ such that $F_N=0$ for all $N \geq N_1.$

I would like to know if this is true or false. Could you provide me an example if it is false?.

答: False. Take, say, $F_N=2^{-2^N}$.

Body: False. Take, say, $F_N=2^{-2^N}$.


问:
Right-continuous and Progressively measurability of Lebesgue-Stieltjes integrals
In Karatzas&Shreve Brownian Motion and Stochastic Calculus P.23 Remarks 4.6: If  $A$ is an increasing process and  $X$ is progressively measurable, define the Lebesgue-Stieltjes integrals $I_{t}^{\pm}\left(\omega\right)=\int_{\left(0,t\right]}X_{s}^{\pm}\left(\omega\right)dA_{s}\left(\omega\right)$, then  $I_{t}=I_{t}^{+}-I_{t}^{-}$ is well defined and finite for all $t\geq0$, then $I$ is right-continuous and progressively measurable.

Here is the definition of an increasing process： An adapted process A is called increasing if for $P$-a.e. $\omega\in\Omega$ we have

• (a) $A_{0}\left(\omega\right)=0 $

• (b) $t\mapsto A_{t}\left(\omega\right)$ is a nondecreasing, right-continuous function,

• (c) $E\left(A_{t}\right)<\infty$ holds for every $t\in\left[0,\infty\right)$.

Now I want to show that $I$ is right-continuous and progressively measurable. It is well known that every right-continuous adapted process is progressively measurable. So I just need to show that $I$ is right-continuous and adapted. But how can I show that $I$ is right-continuous and adapted?

答:
I assume you also need the property that $X$ has bounded variation paths. Then right-continuity follows from
$$
\bigg| \int_{(0;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega)
- \int_{(0;t]} X^+_s(\omega) \mathrm dA_s(\omega) \bigg|
\\=
\bigg| \int_{(t;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega) \bigg|
\le
\int_{(t;t+\varepsilon]} X^+_s(\omega) \mathrm dA_s(\omega)
\le
\int_{(t;t+\varepsilon]} X^+_{t+\varepsilon}(\omega) \mathrm dA_s(\omega)
=
X^+_{t+\varepsilon}(\omega) (A_{t+\varepsilon}(\omega)  - A_t(\omega) )
$$
This converges to $0$ because $A$ is right-continuous and $X^+_{t+\varepsilon}(\omega) < \infty$.
Adaptedness is a little bit trickier from scratch.


问:
Confusion with limits with respect to a certain metric
I was looking at some proofs that certain metric spaces are not complete, and I have been confused.

Here is a link to an example of a main source of my confusion:Show that this metric is not complete

Here is the line that drove my confusion:

No, there is a error in the proof. In fact the functions $f_n(x) = x^n$ converge to the zero function in the metric $\rho$ on $C([0,1]):$

$$ \sqrt {\int_0^1 |x^n-0|^2\,dx} = 1/\sqrt {2n+1} \to 0.$$

If I understand this correctly, if we have a metric space equipped with a certain metric $d$, all limits will be computed with respect to this metric. We can pretend no other metrics exist within the space, as we did with limits in $\mathbb{R}$ in the beginning. I do not understand the last line because it seems to me that the user is taking the limit with respect to the metric $d(x,y)=|x-y|$ in $\mathbb{R}$ when he writes $\frac{1}{\sqrt{2n+1}} \rightarrow 0$. I know I am really confused here, and would appreciate if someone could help clear this confusion.

答:
As Brian showed, $f_n \to \mathbf{0}$ in $C([a,b])$ so it's not a counterexample to completeness of $C([a,b])$. You need a Cauchy sequence that is not convergent in $C([a,b])$ instead.

Note that this neglected answer gives a better example of $f_n$ that form a Cauchy sequence (and in fact a convergent sequence in the larger space $L^2[0,1]$) but without a continuous limit $f$ (the limit in $L^2[0,1]$ is the function that is $0$ on $[0,\frac12)$ and $1$ on $[\frac12,1]$, which has no continuous representative in its class, put technically), and this sequence does illustrate the non-completeness, in a similar way as a sequence of rationals converging to $\sqrt{2}$ does the incompleteness of $\Bbb Q$..


问:
Proving that the function $\frac{1}{q^2}$ is essential to Dirichlet's Approximation Theorem
First post on math.stackexchange so I apologize in advance.

In my number theory class, we've been handed 3 things in order to do the following:

Prove using Dirichlet’s theorem and Liouville’s lemma that one cannot replace the function $\frac{1}{q2}$ in the statement of Dirichlet’s theorem by a faster decaying function by considering quadratic algebraic numbers (degree 2).

In this same assignment, we've been given three things - I believe as hints:

Let $\phi, \psi : \mathbb{N} \rightarrow \mathbb{R}$. We say that:

a) $\phi$ is equivalent to $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=C$, where $C \neq 0$ is some real number.

b) $\phi$ decays faster than $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=0$

c) $\phi$ decays slower than $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=\infty$

Recall Dirichlet's theorem: For every irrational $\alpha$, there exists infinitely distinct rationals $\frac{p}{q}$ such that $$|\alpha - \frac{p}{q}|<\frac{1}{q^2}$$

Recall Liouville's Lemma: Let $\alpha$ be an irrational algebraic number of degree $n$. Then there exists $M\in \mathbb{N}$ such that for every $\frac{p}{q} \in \mathbb{Q}$, $$|\alpha-\frac{p}{q}|>\frac{1}{Mg^n}$$

Now I've already submitted my best attempt at this but I know for a fact it's wrong and I'd prefer not to embarrass myself on my first post on this website. I'm really just looking for an explanation that I can study so I don't miss it again.

答:
The main idea is just to note that Liousville's lemma tells you that you can only get so close to an irrational algebraic number. So if you could improve Dirichlet's theorem, you would eventually end up violating the lemma.

More rigorously, suppose $f(q)$ is a faster decaying function than $\frac1{q^2}$, so that $\lim_{q\to\infty}q^2f(q)=0$, and suppose for the sake of contradiction that we can replace $\frac1{q^2}$ with $f(q)$ in Dirichlet's theorem. Now let $\alpha$ be an irrational number of degree two, as hinted.

Liouville's lemma tells us that there exists some $M$ so that $\left\lvert\alpha-\frac pq\right\rvert>\frac1{Mq^2}$ for every $\frac pq$. By hypothesis, we know, however, that for infinitely many $\frac pq$, we must have $\left\lvert\alpha-\frac pq\right\rvert<f(q)$. In particular, it follows that there are infinitely many fractions $\frac pq$ such that
$$1=\frac{\left|\alpha-\frac pq\right|}{\left\lvert\alpha-\frac pq\right\rvert}<\frac{f(q)}{\frac1{Mq^2}}=Mq^2f(q).$$
But because $q^2f(q)\to0$, we know that there exists some $Q\in\mathbb N$ so that for all $q>Q$ we have $|q^2f(q)|<\frac1M$.

However, this would mean that $q\le Q$ for each of the infinitely many fractions $\frac pq$ with $\left|\alpha-\frac pq\right|<f(q)$. But since $f(q)$ is finite, there are only finitely many $p$ for any given $q$ which satisfy this. And since we just said that $q\in\{1,2,\dots,Q\}$, it follows that there are only finitely many pairs $(p,q)$ which satisfy that inequality. Obviously, this is impossible, so the $\frac1{q^2}$ in Dirichlet's theorem is the best possible.


问:
What is the meaning of the 'or' in a axiom or a definition?
'A or B' means one or the other or both.

So I think $\{x | x = u$ or $ x = v\}$ can be equal to $\{u\}$ or $\{v\}$ or $\{u, v\}$.
Similary, I think $\{x | x\in a $ or $ x\in b\}$ can be equal to any subset of $a \cup b$ except empty set.

But actually $\{x | x = u$ or $ x = v\} = \{u, v\}$ and $\{x | x\in a $ or $ x\in b\} = a \cup b$. What am I missing?

答:
$x$ is variable and is a hypothetical.  It is a place value of everything that could be.   It is not an actual thing and doesn't actually equal anything.

$\{x|x= u$ or $x=v\} = \{$ all possible numbers so long as the number we are considering is either $u$ or $v\}$.

As $u$ is a possible value $u$ must bein the set. And as $v$ is a possible value it is in the set.  Any other value is neither $u,v$ so no other value is in the set.

So the set $\{x|x=u$ or $x=v\}$ is precisely $\{u,v\}$.


问:
Exact Definition of Non-Isomorphic Posets
This screenshot is from Invitation to Discrete Mathematics by Matoušek. :

I understand the definition when two posets (or ordered sets) are isomorphic or non-isomorphic. However, I am confused about question (a), the selected line in the screenshot.

More generally, when we talk about Non-Isomorphic n-element posets, do we mean that they are pair-wise non-isomorphic, or do we mean that for any of these posets, we cannot find another poset that is isomorphic to it (which would imply the previous I guess)?

Thanks!

答:
You have to draw $n$ many Hasse diagrams for posets $P_1, P_2, \ldots P_n$, all having $3$ elements, such that if $i \neq j$ then $P_i$ is not isomorphic to $P_j$ and moreover if $P$ is any three-element poset there is an $i\in \{1,2,\ldots,n\}$ so that $P$ is isomorphic to $P_i$.

So you have then a "catalogue" of all "types" of posets of that size with a unique representative for all. Your job is to figure out $n$ and the $P_i$.


问:
Does box topology satisfy $f:Y \to \prod X_i$ is continuous iff the composition $p_if$ is continuous
In my text, it is stated that the following definition below is another way of defining the product topology:

i) for each factor, $p_i$ is continuous,
ii) A function $f:Y \to \prod X_i$ is continuous iff the composition $p_if$ is continuous for all $i$

They two conditions don't look that restrictive so I was wondering why this is not true for the box topology. Certainly, i) is satisfied so the problem must be with ii). The forward direction is certainly satisfied and there is no difference with the product topology. So I guess the problem must be with the backward direction right?

答:
The box topology does not satisfy that criterion.

Let $X= \Bbb R^{\Bbb N}$ in the box topology (so countably many copies of $\Bbb R$ in the standard topology).

Then $f: \Bbb R \to X$ defined by $f(x)=(x,x,x,\ldots)$ so that $p_n \circ f = \textrm{id}$ fulfills the right hand side of ii, but $f$ is not continuous as

$f^{-1}[\prod_n (-\frac1n, \frac1n)] = \{0\}$ is not open in $\Bbb R$.

The product topology has the handy property that you can conclude the continuity of $f$ going into a product, just from its compositions with all projections, and it's the unique topology on the product with that property. This is a common thing in so-called initial topologies.


问:
What is the probability that the circuit doesn't work?
First, apologies for the image, I couldn't find a good way to explain without including it.

I have the following circuit:

Number on each component represents the probability this component works

Assuming all its components are independent, what is the probability that the circuit doesn't work?

Note: for the circuit to work we need at least one line from left to right with all components working

I am posting this for two reasons:

I think my solution is long and I might be potentially missing something to make things easier

I got stuck somewhere while solving

My attempt:

I tried doing $1-P(W)$

Let $W$ be the event that the circuit works. For this, we always need $A$ and work, and then either $B$ OR $C$ AND $D$ OR $E$ AND $F$, mathematically this can be written $P(A \cap (B \cup (C\cap D) \cup (E \cap F)))$, and I am not sure how I could proceed with it, the $\cap$ is easy to do since these are independent events I can just multiply them but I don't know how to go ahead with all this $\cup$, I thought about the general formula of $P(A\cap B \cap C=P(A)+P(B)+P(C)-P(A\cap B) \dots $  but I am not quite sure of it

I also tried doing $P(W')$ directly but it seemed longer.

Any help would be appreciated.

答:
Let's write some general rule for this kind of probability circuits. Assume for simplicity that the events are all of the kind "work".

$(G1) \qquad $  If you have two or more independent events $C_1, C_2, \dots, C_n$ that are "in series" then you can collapse them all togheter in an event, say $C'$, that has probability of success $$P(C') = \prod_{i=1,\dots, n} P(C_i)$$

$(G2)  \qquad$ If you have two or more independent events $C_1, C_2, \dots, C_n$ "in parallel" then you can collapse them all togheter in an event, say $C'$, that has probability of success
$$P(C') = 1- \prod_{i=1,\dots, n} (1-P(C_i))$$

A basic proof of $(G2)$ can be seen like this:
The probability that a parallel circuit will work is exactly the probability that at least one component will work. Using the "at least one" principle we can conclude that the probability that the whole circuit will work is $1$ minus the probability that all the component will not work simultaneously.

In your probem you have the couples $C,D$ and $E,F$ in series, so they can be combined in two events $C'$ and $E'$ that have probability of success $$P(C') = P(C)P(D) = 0.525 \hspace{1.5cm} P(E') = P(E)P(F) = 0.68$$

Now you have $B$, $C'$, $E'$ in parallel. Theese can be sees as a unique event $B'$ that has probability of success
$$P(B') = 1 - (1-P(B))(1-P(C'))(1-P(E')) \sim  0.94$$

Finally we arrived at two events $A$ and $A'$ in series and we can conclude that the probability that the event $W$ "the whole circuit works" is:

$$P(W) = P(A)P(B') \sim 0.56$$


答: It sounds like what you're missing is that your example uses the set builder notation $\{x|P(x)\}$, which is defined to mean "the set of all objects $x$ for which the sentence $P(x)$ is true". The "or" in your example is part of the $P(x)$ formula - it's incorrect to interpret the expression in a different way (such as the way you might interpret an English sentence with "or" in it).

Title: What is the meaning of the 'or' in a axiom or a definition?

Body:
'A or B' means one or the other or both.

So I think $\{x | x = u$ or $ x = v\}$ can be equal to $\{u\}$ or $\{v\}$ or $\{u, v\}$.
Similary, I think $\{x | x\in a $ or $ x\in b\}$ can be equal to any subset of $a \cup b$ except empty set.

But actually $\{x | x = u$ or $ x = v\} = \{u, v\}$ and $\{x | x\in a $ or $ x\in b\} = a \cup b$. What am I missing?


问:
How to derive Inverse Exponential Fourier Transform for Groundwater Flow Equation?
I'm working through equations in "Nonsteady Flow To a Well in an Infinite Anisotropic Aquifer" (Papadopulos, 1965). I've have gotten stuck trying to figure out how the inverse transform given below is derived. I've looked through the transform tables in Churchill (1958) and Erdelyi (Vol1, 1954) but haven't made much headway. Is there anyone who can help me figure it out?

Eqn(10), the transform solution for the presented groundwater flow equation, is given as:

$$z = {\frac{Q}{2\pi p}} {\frac{1}{T_{xx}\alpha^2+2T_{xy}\alpha\beta+T_{yy}\beta^2+Sp}}$$

where $\beta$ is the parameter of transformation and the remaining parameters can be considered as constants for transform purposes.

Eqn(11) is stated to be the inverse transform of the above with respect to y:

$$w={\frac{Q}{2 \sqrt{2\pi}}}{\frac{exp\{(iy\alpha T_{xy}-\mid y\mid [\ (T_{xx}T_{yy}-T_{xy}^2) \alpha^2+ST_{yy}p]\ ^{1/2})/T_{yy}\}}{p[\ (T_{xx}T_{yy}-T_{xy}^2) \alpha^2+ST_{yy}p ]\ ^{1/2}}}$$

答: The answer is that the inverse was derived using the Residue Method. The author of the paper was kind enough to pass on his notes to me; I'll include an update here once I've had a chance to sit down and work through them.</p>


问:
What is the probability of being selected for execution?
Thirty Arkton hostages in a Brumton prison in occupied Arkland need to select three of them to be executed by their captives in retribution to the killing of three Brums by the Ark army.
They tear down an old letter into thirty pieces, draw a cross on three of them, and alphabetically one after the other draw them out of a shoe.
I need the answer of the following,

What is the probability you are chosen to be executed if you draw 1st from the shoe? Second? Third?

答:
Consider the thirty pieces laid out in a line in order of being drawn. The probability of the $i$th piece having a cross is $\frac{3}{30}.$

Nothing else needs to be said.


问:
Discrete Mathematics Enumeration problem
It is assumed that Valley street has 4 people living alone, 8 couples, 4 families of 3 people, 5 families of 4 people, 6 families of 5 people and 3 families of 6 people. How many ways are there to distribute these inhabitants in the 30 houses of the street?

I got this question that i've been stuck on for a while, any idea how to solve it?

答:
You appear to have $30$ houses and $30$ 'families'.

So you can choose a house for the first family in $30$ ways.

There are then $29$ houses for the next family and so on.

In total there are $30\times 29\times 28 \times ... \times 1=30! $ ways.


问:
probability theorem for continuous distribution

Let $C$ be the unit circle $x^2 + y^2 = 1.$ A point $p$ is chosen randomly on the circumference of $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly at random over their domains). Let $R$ be the rectangle with sides parallel to the $x-$ and $y-$axes with diagonal $pq.$ What's the probability that no point of $R$ lies outside of $C$?

I know the answer's $\frac{4}{\pi^2}.$ Here's the reasoning. Given a point $p$ with an angle $\theta$ from the positive $x$-axis, one can show that $q = (x,y)$ where $|x|\leq |\cos \theta|$ and $|y| \leq |\sin\theta|$ so the probability that $q$ lies in the right region is $\frac{|2\sin 2\theta|}{\pi}.$ Integrating over all possible values of $\theta$ gives that the desired probability is $\frac{1}{2\pi}\int_0^{2\pi} \dfrac{|2\sin 2\theta|}{\pi}d\theta = \frac{4}{\pi^2}.$

Alternatively, I could compute the volumes of the regions $A=\{(\theta, x,y) : \theta \in [0,2\pi], |x|\leq |\cos\theta|, |y|\leq |\sin \theta|\}$ and $\Omega = \{(\theta, x,y) : \theta \in [0,2\pi], x^2 + y^2 = 1\}.$ Then dividing the volume of $A$ by the volume of $\Omega$ should give the result.

Edit: It seems it's not entirely clear what I'm asking. I'm asking for a formal justification of the above solution. I know how to formally show that a point $q=(x,y)$ satisfies the conditions if $p$ makes an angle $\theta$ with the positive $x$-axis iff $-|\cos \theta|\leq x\leq |\cos\theta|$ and $-|\sin\theta| \leq y\leq |\sin\theta|.$ I also know that the probability $q$ lies in the above region is given by the above result. The thing I'm struggling to justify well is why we have to divide by $2\pi$ at the end. Could I get an answer that at least defines random variables and uses probability density functions to justify the final step or some other formal thing?

I would also like some response as to whether this problem can be further generalized. For example, does the following problem make sense and if so what would be a solution? If it doesn't I presume there are multiple such rectangular prisms?

Now consider the unit sphere $C'$. A point $p$ chosen on the surface and a point $q$ chosen in the interior (these points are chosen uniformly and independently over their domains). What is the probability that no point of the rectangular prism with diagonal $pq$ and whose sides are parallel to the $x,y,z$ axes lies outside of $C'$?

答:
The probability density function (pdf) is:
$$
\rho(\omega)=\lim_{\substack{d\Omega\to0\\ \omega\in d\Omega}}\frac{dP}{d\Omega}
$$
where $dP$ is the probability of the random variable $\omega$ to fall within a particular volume of the sample space $d\Omega$.

The pdf of the event which certainly happened in some point of the sample space satisfies the relation:
$$
\int_{\Omega}\rho(\omega)d\omega=1.
$$
If we know only the general "relative" behavior (the ratio of the probability of the event $\omega$ to that of the event $\omega_0$) $$\rho_r(\omega)=\dfrac{\rho(\omega)}{\rho(\omega_0)}$$ we may restore the pdf simply by the relation:
$$
\rho(\omega)=\frac{\rho_r(\omega)}{\int_{\Omega}\rho_r(\omega)d\omega}.
$$
The important point is the following. The expected value of a function of a random variable is:
$$
\bar f=\int_{\Omega}f(\omega)\rho(\omega)d\omega.
$$

Particularly simple is the pdf of uniform distribution (every point in the sample space is equiprobable):
$$
\rho(\omega)=\frac1{\Omega},
$$
where $\Omega$ is the volume of the whole sample space (it cannot be infinite for uniform distribution).

Let now consider your example. Without loss of generality we assume that the radius of the sphere is 1. Let the coordinates of the first point be:
$$
(x,y,z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)
$$
where $(\theta,\phi)$ are the spherical coordinates of the same point. The coordinates of the second point $(X,Y,Z)$ have to satisfy:
$$
|X|\le|x|,\quad |Y|\le|y|,\quad |Y|\le|y|
$$
which results in the volume
$$
V(\theta,\phi)=8|\sin^2\theta\cos\theta\cos\phi\sin\phi|.
$$
To obtain the probability in question we integrate over the coordinate of the first point:
$$
P=\frac1{V_2}\int_S\frac{V(\theta,\phi)}{V_3}\,dS,
$$
where $V_2=4\pi$, $V_3=\frac{4\pi}3$ are the area and the volume of the sphere, respectively, and $dS=\sin\theta\,d\theta\,d\phi$ is the infinitesimal element of the surface.

Combining everything together we have:
$$
P=\frac3{2\pi^2}\int_0^\pi\sin^3\theta|\cos\theta|\,d\theta\int_0^{2\pi}|\sin\phi\cos\phi|\,d\phi=\frac3{2\pi^2}.
$$

In this example $\frac1{V_2}$, $\frac1{V_3}$ are the pdf's of the event that the corresponding point is at a certain point of the surface or volume of the sphere, respectively.


问:
If an ODE has only periodic solutions and one equilibrium point, is that equilibrium point Lyapunov stable?
Consider the non-linear IVP
$$\dot x=f(x),$$
$$x(0)=x_0.$$
where $f$ is locally Lipschitz and
$$f(0)=0$$
and $$f(x)\ne 0\ \forall x\ne 0.$$
If all solutions to this IVP for various initial conditions are periodic does that mean that the origin is Lyapunov stable, i.e.,
$$\forall \varepsilon>0, \ \exists\delta>0, \ |x(0)|<\delta\implies|x(t)|<\varepsilon \ \forall t\ge 0.$$

My professor said yes in class and he "justified" this by drawing periodic orbits that are circular. I know that periodic solutions are bounded, but what if a solution close starts very close to the origin, moves far away from the origin, then returns to the same point? Is there an example of such a system or is my professor right?

I was thinking of a solution like
$$x(t)=\begin{cases}\frac{x(0)}{|x(0)|}\sin(t)+x(0)(1-\sin(t))&x\ne 0\\0&x=0\end{cases}.$$
$x(t)$ oscillates between $x(0)$ and a point with norm 1, so no matter how small $\delta$ is, the solution will never satisfy the definition of Lyapunov stability with $\varepsilon=\frac{1}{2}$. The origin is an equilibrium point and the only equilibrium point and $x(t)$ satisfies the initial condition. $x(t)$ is differentiable, $2\pi$ periodic and satisfies the time-invariance property. I know that $x(t)$ does not work because it is not continuous with respect to the initial condition and $f$ is locally Lipschitz but I'm wondering if there is a way to alter $x(t)$ to make it a solution to an autonomous ODE and still have the same properties.

答:
For 2-d only:  Suppose not.  Then there is an $\epsilon >0$ such that there is no such $\delta >0$ with $|x(0)| < \delta \implies |x(t)|<\epsilon$.

Take such an $\epsilon > 0$ and consider a sequence $x_1, x_2, \cdots$ approaching the origin.  By periodicity with equilibrium point $0$, the orbit with initial $x_1$ must exit the $\epsilon$ ball and reenter at least once.  Label the first exit and first reentrance as $E_{1,A}, E_{1,B}$.  Since this orbit encompasses the origin, we can choose the sequence such that each successive point is contained in the interior of the orbit of the previous.  By existence-uniqueness, the orbits origininating from these point cannot cross.  In turn, this then creates sequences of points on the $\epsilon$ ball $E_{i,A}, E_{i,B}$.  These must converge to limit points $E_{\infty,A},E_{\infty,B}$ on the ball.

If $E_{\infty,A} = E_{\infty,B}$, this point is an equilibrium, which contradicts the condition $f(x) \not= 0$.  If not, then consider a point $S$ on the arc of the ball defined by $E_{\infty,A}, E_{\infty,B}$. By index theory, the region enclosed periodic orbit in the plane must contain an equilibrium point. The orbit of $S$ must contain $0$, which contradicts existence-uniqueness from those orbits defined by our sequence.


问:
Ring Homomorpism
Consider the polynomial ring $R[X]$ and the ring $M_2(R)$ of $2\times 2$ matrices with entries in R. We define a map $t: R[X] \to M_2(R)$ by

$$t(f) = \begin{pmatrix}f(0)&  f'(0)\\0 &  f(0)\end{pmatrix}$$

where $f'$ denotes usual derivative of $f$.

Show that t is a homomorphism. Determine the kernel and the image of t. Is ker(t) a principal ideal.

I have shown that t is a homomorphism, however am stuck on the rest of the question. Thank you for any help in advance.

答:
Hints:

$$\ker T=\{f(x)\in\Bbb R[x]\;;\;f(0)=f'(0)=0\iff x^n\mid f(x)\;,\;\;2\le n\in\Bbb N\}\;\ldots$$

Check now that

$$\forall\;\begin{pmatrix}a&b\\0&a\end{pmatrix}\in M_2(\Bbb R)\;\;\exists\,f(x)\in\Bbb R[x]\;\;s.t.\,\,f(0)=a\;,\;f'(0)=b$$

Title: Ring Homomorpism


问:
Studying the nature of an improper integral
How can I determine the nature of this integral and also, how can I calculate it?

I've been trying to do it for like an hour, but I can't figure it out.

$$
\int_{1}^{\infty} \frac{\cos x \ln x}{x} dx
$$

答:
Hint

Use integration by parts integrating $\cos x$ and differentiating $\frac{\ln x}{x}$. Then the remaining integral is absolutely convergent.


问:
Solving the system with $y=x$ and $ y=(1-40x)^{-1/51.33}-1$

Solve for $(x,y)$:
\begin{align}
y &= x\\
y &= (1-40x)^{-1/51.33}-1\
\end{align}

I don't know if finding a closed-form solution is possible, have been trying for almost half an hour but still get nothing.

Please tell me some idea about how to solving it, or if it is possible to do.

Edit: I did manage to get the answer via Geogebra, but I would love to see how people solve it analytically.

Body:
Using whole numbers, consider that you look for the zero of function
$$f(x)=(1-40x)^{-\frac {100}{5133}}-x-1$$ It is clear that $x=0$ it a trivial solution. Discarding this one, we can approximate the other building the $[1,n]$ Padé approximant of $\frac{f(x)}x$ around $x=0$. This will write
$$\frac{f(x)}x \sim \frac{ a^{(n)} x -\frac{1133}{5133}} {1+\sum _{i=1}^n b_i^{(n)} x^i }$$ where the coefficients $ a^{(n)}$ and $b_i^{(n)}$ are compued from the function and derivative values at $x=0$.

Then the approximate solution is just given by
$$x_{(n)}=\frac{1133}{5133\,a^{(n)}}$$

This will generate a sequence of rational numbers, the very first ones being
$$\left\{\frac{5815689}{418640000},\frac{17447067}{1725707120},\frac{24440327087}{243
   9163381610},\frac{21844121555788638898671}{2184051507726409040779072}\right\}$$

Let us look at their decimal representation
$$\left(
\begin{array}{cc}
 n & x_{(n)} \\
 1 &  0.0101100974 \\
 2 &  0.0100199631 \\
 3 &  0.0100016513 \\
 4 &  0.0099970864 \\
 5 &  0.0099958134 \\
 6 &  0.0099954326 \\
 7 &  0.0099953131 \\
 8 &  0.0099952742 \\
 9 &  0.0099952613 \\
 10 &  0.0099952569 \\
 11 &  0.0099952553 \\
 12 &  0.0099952548 \\
 13 &  0.0099952546 \\
 14 &  0.0099952546 \\
 15 &  0.0099952545
\end{array}
\right)$$

Just for the fun
$$x_{(8)}=\frac{2076678199429529816581224448972052669538078027653922391}{20776600532441929227
   8132724022546069652413472361607062796}$$


问:
Find the line parallel to $x + y = 1$ and tangent to $x^2 + xy + y^3 = 3$
Can someone please help me?

How can I find a line that is parallel to $x + y = 1$ and tangent to $x^2 + xy + y^3 = 3$?
 (at the same time)

Thank you.

答:
Hints: implicit differentiation

$$2x\,dx+y\,dx+x\,dy+3y^2\,dy=0\implies \frac{dy}{dx}=-\frac{2x+y}{x+3y^2}$$

Now you want points $\,(x_0,y_0)\;$ on the graph of the function s.t.

$$-\frac{2x_0+y_0}{x_0+3y_0^2}=-1\;\ldots\ldots$$


问:
Accumulation points with respect to Zariski Topology on $\mathbb{R}$.
Consider the topology on $\mathbb{R}$  : the closed sets are given by the finite sets and $\mathbb{R}$ (basically the Zariski Topology on $\mathbb{R}$).

Let $(x_n)_{n \in \mathbb{N} }$ be a sequence in $\mathbb{R}$.

How do we show : if $l$ is a limit of the sequence $(x_n)_{n \in \mathbb{N} }$  if and only if for all $x \neq l$, the number of indices $n$ such that $x_n = x$ is finite.

Determine the accumulation points of the sequence $(x_n)_{n \in \mathbb{N} }$ w.r.t. Zariski topology

答:
You can translate the definition of “being a limit” using closed sets:

$ $          for every closed set $C$ not containing $l$, there exists $m$ such that, for $n>m$, $x_n\notin C$

The closed sets non containing $l$ are necessarily finite.

Suppose $l$ is a limit of the sequence. Then, taking $C=\{x\}$, with $x\ne l$, we see that the set $\{n\in\mathbb{N}:x_n=x\}$ is upper bounded, hence finite.

Suppose $l$ satisfies the condition about finiteness. Take a closed set $C$ not containing $l$. Then $C=\{r_1,r_2,\dots,r_k\}$ is finite and you should be able to finish up.

What about accumulation points of the sequence? Using Kelley's terminology, a sequence is convergent to $l$ if (and only if) it is eventually in every (open) neighborhood of $l$ (it belongs there from a certain point on). The point $l$ is an accumulation point if (and only if) it is frequently in every (open) neighborhood of $l$ (for every $m$, there is $n$ such that $n>m$ and $x_n$ is in the given neighborhood).

If $l$ is an accumulation point of the sequence and $x\ne l$, for every $m$ there is $n>m$ such that $x_n\ne x$. In particular, $\{n\in\mathbb{N}:x_n\ne x\}$ is infinite.

Can you finish with the converse?


问:
In a course there's 10 boys and 10 girls and we need to divide them into 3 different groups..
We have $3$ groups $A,B,C$.

We have to divide all the boys and the girls into these three groups such that in group $C$, there's exactly $6$ or $7$ members, there's atleast one girl, and the number of the boys must be greater than the number of girls.

in how many ways we can do that?

My attempt: I'm trying to split this problem into cases:

Case 1: in group $C$ theres $5$ boys and $1$ girl, or $6$ boys and $1$ girl. for the rest of the boys and girls I would have to choose a group for each one from $A,B$ so we get $2^{14}+2^{13}$.

Case 2: in group $C$ theres $4$ boys and $2$ girls or $5$ boys and $2$ girls. so $2^{14}+2^{13}$.

Case 3: in group $C$ theres $4$ boys and $3$ girls: $2^{13}$.

Then I summed up all the results and got $57344$ ways and the answer was $50610$.

What am I missing?
Appreciate any help, thanks in advance.

答:
It seems that the question is only about the possible combinations for group $C$.

This is where the answer 50610 comes from:

$$\binom{10}{5}\cdot\binom{10}{1} + \binom{10}{6}\cdot\binom{10}{1} + \binom{10}{4}\cdot\binom{10}{2} + \binom{10}{5}\cdot\binom{10}{2} + \binom{10}{4}\cdot\binom{10}{3} = 50610$$

For example: to choose 5 boys (out of 10) and 1 girl (out of 10) to put into group $C$ we have $\binom{10}{5}\cdot\binom{10}{1}$ possibilities.


问:
How was this really handy bijection thought up of?
Here's a proof from my combinatorics textbook.

Theorem 1.6.5. For any positive integer $k$,

$$(1-x)^{-k}=\sum_{n\geq 0}\binom{n+k-1}{k-1}x^n$$

Proof: Let $S_n$ be the set of all $k$-tuples $(a_1,\dots ,a_k)\in N^k_{\geq 0}$ with $a_1+\cdots +a_k=n$. We previously showed that

$$|S_n|=[x^n](1-x)^{-k}$$Let $T_n$ denote the set of all $(k-1)$-element subsets of ${1,2,\dots ,n+k-1}$, thus

$$|T_n|={n+k-1 \choose k-1}$$
We will establish a bijection between $T_n$ and $S_n$.

For $A=\{a_1,a_2,\dots ,a_{k-1}\}\in T_n$, with $a_1<a_2<\cdots<a_{k-1}$, we define $f(A)=\sigma$ where

$$\sigma=(a_1-1,a_2-a_1-1,\dots ,a_k-a_{k-1}-1,(n+k)-a_k-1)$$

(proof that it is bijection follows)

My question is, how in earth would somebody derive such a useful bijection. Just by staring at $\sigma$, where this came from is very opaque - leading to this theorem being really hard to proof without giving you the bijection to use.

答:
It’s easier to think about it the other way around. Assume that you’ve gathered enough numerical data (by brute force, if necessary) to suspect that $|S_n|$ is a binomial coefficient with lower number $k-1$. Then you want somehow to relate solutions to $x_1+\ldots+x_k=n$ in non-negative integers to $(k-1)$-element subsets of something. The key insight is that a solution to $x_1+\ldots+x_k=n$ in non-negative integers determines a non-decreasing sequence of partial sums, one of which is automatically known and can therefore be discarded without loss of information, leaving only $k-1$ integers.

Suppose that $a_1+\ldots+a_k=n$, where each $a_i\in\Bbb N$. For $i=1,\dots,k$ let $s_i=a_1+\ldots+a_i$. Then $0\le s_1\le s_2\le\ldots\le s_k=n$. We know that $s_k=n$, so in order to reconstruct $a_1,\dots,a_k$, we need only the first $k-1$ partial sums, $s_1,\dots,s_{k-1}$. This sets up a nice correspondence between solutions to $x_1+\ldots+x_k=n$ in non-negative integers and $(k-1)$-term sequences non-decreasing sequences $\langle s_1,\dots,s_{k-1}\rangle$ of non-negative integers such that $s_{k-1}\le n$.

This doesn’t quite give me $(k-1)$-element sets, since it’s entirely possible that $s_i=s_{i+1}$ for some $i$, but I can always simply add $1$ to the first term, $2$, to the second, and so on: let $t_i=s_i+i$ for $i=1,\dots,k-1$. Then $1\le t_1<t_2<\ldots<t_{k-1}\le n+k-1$, and $\{t_1,\dots,t_{k-1}\}$ is a genuine $(k-1)$-element subset of $\{1,\dots,n+k-1\}$.

It’s not hard to see that if I start with any $(k-1)$-element subset of $\{1,\dots,n+k-1\}$ and list it in increasing order as $\{t_1,\dots,t_{k-1}\}$, I can reverse the process by setting $s_i=t_i-i$ for $i=1,\dots,k-1$ and $s_k=n$, and then letting $a_1=s_1$ and $a_i=s_i-s_{i-1}$ for $i=1,\dots,k$.


问:
Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6...

Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6, find distance of focus from tangent at P.

Let P and Q be $(at^2,2at)$ and $(\frac{a}{t^2}, \frac{-2a}{t})$ respectively, for parabola $y^2=4ax$

So $$0= 2(-\frac{2a}{t})+2at$$
$$\implies t^2=2$$

Now tangent at Q is

$$\frac{y}{\sqrt 2} =x+\frac a2$$

So point R is $(-a, -\frac{a}{\sqrt 2})$

Then $$QR^2 = 36$$
$$\implies a=\frac{12}{\sqrt {11}}$$

So now, tangent at P is $$\sqrt 2y=x+2a$$ whose distance from $(a,0)$ is
$$d=\left | \frac{a+2a}{\sqrt 3}\right|$$
$$d=\sqrt 3 a$$
$$d=\frac{\sqrt 3 \times 12}{\sqrt {11}}$$

But given answer is 4

Where am I going wrong?

答:
As you rightly obtained, $t^2 = 2 \implies t = \pm \sqrt2$

If we take $t = \sqrt2$ and rewrite,

$P(2a, 2a \sqrt2), Q(\frac{a}{2}, - a \sqrt2)$

Slope of tangent at point $Q = \frac{2a}{y} = - \sqrt2$

So equation of tangent line $y + a \sqrt2 = -\sqrt2(x-\frac{a}{2})$

or $y = - \sqrt2 x - \frac{a}{\sqrt2} $

Point $R$ where the line intersects the directrix, $R(-a, \frac{a}{\sqrt2})$.

So $QR^2 = \frac{9a^2}{4} + \frac{9a^2}{2} = 36 \implies a = \frac{4}{\sqrt3}$.

Now slope of tangent line at $P = \frac{2a}{y} = \frac{1}{\sqrt2}$.

So equation of tangent line at $P$, $y - 2a\sqrt2 = \frac{1}{\sqrt2}(x - 2a)$

or $y = \frac{x}{\sqrt2} + \sqrt2 a$

Its distance from focus $(a,0)$ is

$d = \frac{3a/ \sqrt2}{\sqrt{3/2}} = a \sqrt3 = 4$


问:
Stereographic Projection: Does Conformality Imply Circle Preservation?
This may be a silly question. I have learned to prove that circles are preserved at the infinitesimal scale, however, does this ALONE imply that circles are mapped as circles for the stereographic projection? Why/Why not.</p>

答:
Consider the map $z \mapsto z^2$, which is conformal on the punctured plane $\mathbb{C} \setminus \{0\}$. A circle in this domain is $C : |z-2| = 1$. Suppose $z \in C$, so $z = 2+e^{i \theta}$ for $\theta \in [0, 2\pi)$. We have

$$z^2 = (2+e^{i \theta})^2= 4 + 4e^{i \theta} + e^{2i \theta},$$

which is clearly not a circle. So the answer is no, conformality alone does not imply that circles are preserved. Conformality means that the angle between contours are preserved. So for example, if you have two orthogonal circles, they will be mapped to two orthogonal contours under a conformal map, albeit not necessarily to two new circles.

However, circles are preserved under stereographic projection, or more generally circles and lines are mapped to circles under the steoreographic projection, which can be shown by simply applying the formula for the steoreographic projection to a generalized circle.


问:
$UL$ factorization from $LU$ factorization with permutation matrix
The answer at A = UL factorization shows that, if a matrix $A$ has an $LU$ factorization, then one can also get a $UL$ factorization. But it does so by starting with an $LU$ factorization not of $A$ but of the matrix $B = PAP$, where $P$ is the permutation matrix \begin{equation*}\begin{pmatrix}0 & \dots & 0 & 1 \\ 0 & \dots & 1 & 0 \\ \vdots & \ddots & \vdots & \vdots \\ 1 & \dots & 0 & 0\end{pmatrix}\end{equation*} My question is how do we know that an $LU$ factorization of $B$ actually exists? It does not seem to easily follow just from the fact that $A$ has an $LU$ factorization. Thanks in advance for assistance.

答:
UL factorization from LU factorization with permutation matrix

According to the preprint Necessary And Sufficient Conditions For Existence of the LU Factorization of an Arbitrary Matrix, a both necessary and sufficient condition for the existence of an LU factorization of a square $n\times n$ matrix $A$ is given by:

$$(1)\,\text{$A$ has LU factorization} \iff \forall k\colon\, \operatorname{rk}(A[\colon k,\, \colon k]) \ge \operatorname{rk}(A[\colon k,\, \colon n]) + \operatorname{rk}(A[\colon n,\, \colon k]) -k$$

Where $A[\colon i, \colon j]$ is the matrix consisting of the first $i$ rows and $j$ columns of $A$. We shall see that:

$$(2)\,\text{$A$ has UL factorization} \iff \forall k\colon\, \operatorname{rk}(A[k\colon,\, k\colon]) \ge \operatorname{rk}(A[k\colon,\, n\colon]) + \operatorname{rk}(A[n\colon,\, k\colon]) -k$$

Where $A[i\colon, j\colon]$ is the matrix consisting of the last $i$ rows and $j$ columns of $A$.

Now, without further notice, given any $k\times l$ matrix  $X$ let $\overline X = P_k^T X P_l$, be the flip of $X$, where $(P_n)_{ij}=\delta_{i, n-i}$ is the exchange matrix of size $n$. We will show:

$$(3)\qquad \text{$A$ has UL factorization} \iff \text{$\overline A$ has LU factorization}$$

So the existence of an LU factorization of $B=\overline A$ is in fact equivalent to $A$ having a UL factorization.

Proof of (2): Note that the way the preprint proves (1) is by splitting $A$ into a block matrix at index $k$ for every $k=1 \ldots n$

$$
A =\left[\begin{array}{cc}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]
=\left[\begin{array}{ll}
L_{11}U_{11} & L_{11}U_{12} \\
L_{21}U_{12} & L_{21}U_{12} + L_{22} U_{22}
\end{array}\right]
=\left[\begin{array}{cc}
L_{11} & 0 \\
L_{21} & L_{22}
\end{array}\right]\left[\begin{array}{cc}
U_{11} & U_{12} \\
0 & U_{22}
\end{array}\right]
=LU
$$

and then apply rank inequalities of the type $\operatorname{rk}(XY) \ge \operatorname{rk}(X) + \operatorname{rk}(Y) -k$ to the blocks.

Note that $A=LU$ is equivalent to $P^TAP = (P^T L P)(P^T U P)$, i.e. $\overline A = \overline L \cdot \overline U$, hence

$$
\overline A =\left[\begin{array}{cc}
\overline A_{22} & \overline A_{21} \\
\overline A_{12} & \overline A_{11}
\end{array}\right]
=\left[\begin{array}{ll}
\overline L_{21}\overline U_{12} + \overline L_{22} \overline U_{22} & \overline L_{21}\overline U_{12} \\
\overline L_{11}\overline U_{12} & \overline L_{11}\overline U_{11}
\end{array}\right]
=\left[\begin{array}{cc}
\overline L_{22} & \overline L_{21} \\
0 & \overline L_{11}
\end{array}\right]\left[\begin{array}{cc}
\overline U_{22} & 0 \\
\overline U_{12} & \overline U_{11}
\end{array}\right]
=\overline L \overline U
$$

which is a block UL factorization! So we can in principle apply the exact same argumentation of the proof of (1) given in the preprint "module flipping" to conclude (2).

Proof of (3): This now follows immediately because conditions (1) and (2) are equivalent modulo flipping, since for any matrix $X$ holds $\operatorname{rk}(X) = \operatorname{rk}(\overline X)$, and the submatrices in the block decomposition of $\overline A$ are precisely the flips of the submatrices of the block decomposition of $A$ since

$$\begin{aligned}
\forall ij:\; (P_n^T AP_n)[\colon i,\, \colon j] = P_{i}^TA[n-i\colon,\, n-j\colon] P_{j}
\end{aligned}$$

I.e. $\overline{A}[\colon i,\, \colon j] = \overline{A[n-i\colon,\, n-j\colon]}$, which allows us to conclude

$$\begin{aligned}
   &\text{$A$ has LU factorization}
\\ &\iff \forall k\colon\, \operatorname{rk}(A[\colon k,\, \colon k]) \ge \operatorname{rk}(A[\colon k,\, \colon n]) + \operatorname{rk}(A[\colon n,\, \colon k]) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A[\colon k,\, \colon k]}) \ge \operatorname{rk}(\overline{A[\colon k,\, \colon n]}) + \operatorname{rk}(\overline{A[\colon n,\, \colon k]}) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A}[n-k\colon,\, n-k\colon ]) \ge \operatorname{rk}(\overline{A}[n-k\colon,\, n\colon]) + \operatorname{rk}(\overline{A}[n\colon,\, n-k\colon]) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A}[k\colon,\, k\colon]) \ge \operatorname{rk}(\overline{A}[k\colon,\, n\colon]) + \operatorname{rk}(\overline{A}[n\colon,\, k\colon]) -k
\\ &\text{$\overline A$ has UL factorization}
\end{aligned}$$


问:
show that $\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n}f(x_{k})$ is exists
let $f:[0,1]\to R^{+}$ real-valued continuous functions,and such
$$\int_{0}^{1}f(x)dx=2019,~~\int_{0}^{1}f^2(x)dx=20181027$$
(1):show that:there exists unique sequence $x_{0},x_{1},\cdots,x_{n}\in [0,1]$,such $x_{0}<x_{1}<\cdots<x_{n}$ and for any postive integer $k=1,2,\cdots,n$.such
$$\int_{x_{k-1}}^{x_{k}}f(t)dt=\dfrac{1}{n}\int_{0}^{1}f(t)dt$$

(2):and show that $$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})$$ is exists.and find this value.

I can do it $(1)$ exists.I think it is First mean value theorem for integration。but How to this sequence is unique. How to solve this (2),if this limt has exists.it seem use Stolz-Cesaro's Lemma:$$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})=\lim_{n\to\infty}f(x_{n})$$

this problem is my teacher gave me the exercise, these two problems I can not solve all , so I want to ask the teacher here,Thanks

答:
First note that if the stated conditions hold, then
\begin{align}
     \int_{x_0}^{x_k}f(x)dx&=\sum_{i=1}^k\int_{x_{i-1}}^{x_i}f(x)dx\\
    &=\frac{k}{n}\int_0^1f(x)dx\ ,
   \end{align}
and, in particular,
$$
     \int_{x_0}^{x_n}f(x)dx=\int_0^1f(x)dx\ .
   $$

Subtracting the left side of the final equation above from both sides gives
\begin{align}
    0&=\int_0^1f(x)dx-\int_{x_0}^{x_n}f(x)dx\\
  &=\int_0^{x_0}f(x)dx+\int_{x_n}^1f(x)dx\ ,
\end{align}
and since $\ 0\le x_0\le x_n\le1\ $, and $\ f\ $ is positive and continuous, it follows from this that $\ x_0=0\ $ and $\ x_n=1\ $ if the stated conditions hold

Now, following the suggestion of Kelenner in the first of the comments, put
$$
    F(x)=\int_0^xf(t)dt\ .
 $$
Then $\ F\ $ is strictly increasing and continuous on $\ [0,1]\ $, and if the stated conditions hold, then the first equation above tells us that $\ x_k\ $ must satisfy the equation
$$
 \ F\big(x_k\big)=\frac{k}{n}\int_0^1f(x)dx\ .
$$
But since
$$
 F(0)=0<\frac{k}{n}\int_0^1f(x)dx<\int_0^1f(x)dx=F(1)\ ,
   $$
for $\ k=1,2,\dots,n-1\ $ , then by the intermediate value theorem, therefore, there must exist $\ x_{nk}\in(0,1)\ $ with
$$ F\big(x_{nk}\big)=\frac{k}{n}\int_\limits{0}^1f(x)dx\ ,$$
and since $\ F\ $ is strictly increasing, $\ x_{nk}\ $ is uniquely determined.  Moreover,
\begin{align}
  F\big(x_{nk}\big)-F\big(x_{n\,(k-1)}\big)&=\int_{x_{n\,(k-1)}}^{x_{nk}}f(x)dx\\
&=\frac{1}{n}\int_0^1f(x)dx\\
&\ge m\big(x_{nk}-x_{n\,(k-1)}\big)\ ,
   \end{align}
where $\ m=\min_\limits{x\in[0,1]}f(x)>0\ $.  Therefore, $\ 0<x_{nk}-x_{n\,(k-1)}\le$$\frac{1}{mn}\int_\limits{0}^1f(x)dx\ $, and so $\ x_{nk}-x_{n\,(k-1)}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, as Paramanand Singh has observed in the comments.  For notational convenience, put $\ x_{n0}=x_0=0\ $ and $\ x_{nn}=x_n=1\ $.

We are now in a position to prove that
$$ \lim_{n\rightarrow\infty}\sum_{k=0}^nf\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}=\frac{20181027}{2019}\ .
 $$
Let $\ M=\max_{x\in[0,1]}f(x)\ $ and $\ \epsilon\ $ be any positive real number.  Since $\ f\ $ must be uniformly continuous on $\ [0,1]\ $, there exists a positive real number $\ \delta\ $ such that $\ \left|f(x)-f(y)\right|\le\frac{\epsilon}{M}\ $ for all $\ x,y\in[0,1]\ $ with $\ |x-y|\le\delta\ $.  Therefore, if $\ n>\frac{1}{m\delta}\int_\limits{x=0}^1f(x)dx\ $, then $\ \left|x-x_{nk}\right|\le\delta\ $, and hence $\ \left|f(x)-f\big(x_{nk}\big)\right|\le\frac{\epsilon}{M}\ $, for all $\ x\in\left[x_{nk}, x_{n\,(k+1)}\right]\ $. Hence, for all such $\ n\ $ we have
\begin{align}
\left|\,\int_\limits{0}^1f(x)^2dx-\right.&\left.\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)\right|\\
&=\left|\,\int_\limits{0}^1f(x)^2dx- \sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}f\big(x_{nk}\big)f(x)dx\right|\\
&=\left|\,\sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}\left(f(x)-f\big(x_{nk}\big)\right)f(x)dx\right|\\
&\le\sum_{k=0}^{n-1}\epsilon\big(x_{n\,(k+1)}-x_{nk}\big)\\
  &=\epsilon\ .
  \end{align}
Thus,
$$
\lim_{n\rightarrow\infty}\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)=\int_\limits{0}^1f(x)^2dx\ ,
$$
from which, since $\ \int_\limits{0}^1f(x)dx\ne0\ $, and $\ \frac{f\left(x_{nn}\right)}{n}=\frac{f(1)}{n}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, it follows that
$$
\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{k=0}^nf\big(x_{nk}\big)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}\ .
$$


答:
Parts (2) and (3) are not math questions.

For (1), you're right, you have a $\frac3{30}$ chance of drawing a marked piece regardless of your position in line.

Here's an attempt to elaborate on the "obvious" symmetry argument: The shoe ensures that all $30!$ permutations of the 30 pieces are equally likely, so by symmetry, each of the $\binom{30}3$ sets of 3 positions is equally likely to be the set that gets the marked pieces. For a given position $i$, there are $\binom{29}2$ such sets that include $i$, so the probability that $i$ ends up in the marked set is $\binom{29}2/\binom{30}3=\frac3{30}$.

Alternatively, you could compute it using conditional probabilities: Let $E_i$ be the event that the $i$th piece drawn is marked. Clearly $P(E_1)=\frac3{30}$. Then $$P(E_2)=P(E_2|E_1)P(E_1)+P(E_2|\lnot E_1)P(\lnot E_1)=\frac2{29}\frac3{30}+\frac3{29}\frac{27}{30}=\frac3{30}.$$ And so on.

Title: What is the probability of being selected for execution?

Body:
Thirty Arkton hostages in a Brumton prison in occupied Arkland need to select three of them to be executed by their captives in retribution to the killing of three Brums by the Ark army.
They tear down an old letter into thirty pieces, draw a cross on three of them, and alphabetically one after the other draw them out of a shoe.
I need the answer of the following,

What is the probability you are chosen to be executed if you draw 1st from the shoe? Second? Third?


问:
If $ a=e^{i\alpha}$ and $b=e^{i\beta} $ Show the following
If $ a=e^{i\alpha}$ and $b=e^{i\beta} $, Show that
$$
    \frac{(a+b)(ab-1)}{(a-b)(ab+1)} = \frac{\sin(\alpha)+\sin(\beta)}{\sin(\alpha)-\sin(\beta)}
$$

答:
Hint

$$\dfrac{a^2b-b}{ab^2-a}=\cdots=\dfrac{ab\left(a-\dfrac1a\right)}{ab\left(b-\dfrac1b\right)}=\cdots=\dfrac{2i\sin\alpha}{?}$$

Now apply componendo dividendo


问:
Poset vs Set and Some Notations
I apologize if I am splitting hairs here.

First of all, is a poset a set?

The official definition on my textbook says,

If $X$ is a set and $R$ is an ordering on $X$, the pair $(X, R)$ is called an ordered set.

I mean a poset is a set with a partial order, whereas a set itself, by definition, does not have order. So I guess a poset is not a set? But if this is true, then I have another confusion:

There's a line from the book that says,

A finite linearly ordered set has a (unique) largest element.

So if $x \in X$ and this $x$ is this largest element, is it ok to also write $x \in (X, R)$?

答:
A poset is the combination (ordered pair) of a set and a partial order $R$ on that same set.

So you could say that a poset $(X,R)$ has underlying set $X$ and an order $R$. Often the order is understood from context and we just say the poset $X$, which is not entirely correct formally. A poset is a set with extra structure, like a group, or a field, or a topological space, etc.

If a poset has a maximum, just say $x \in X$, not $x \in (X,R)$. It is an element of the underlying set, with special properties wrt the order $R$.

Title: Poset vs Set and Some Notations


答:
$(X, R)$ is an ordered pair. Which in set theory is a set.

You have $x \in X$. But you can’t say $x \in (X,R)$ as $(X,R)$ is indeed a set, but not a set that contains $x$.

Title: Poset vs Set and Some Notations

Body:
I apologize if I am splitting hairs here.

First of all, is a poset a set?

The official definition on my textbook says,

If $X$ is a set and $R$ is an ordering on $X$, the pair $(X, R)$ is called an ordered set.

I mean a poset is a set with a partial order, whereas a set itself, by definition, does not have order. So I guess a poset is not a set? But if this is true, then I have another confusion:

There's a line from the book that says,

A finite linearly ordered set has a (unique) largest element.

So if $x \in X$ and this $x$ is this largest element, is it ok to also write $x \in (X, R)$?


问:
Functions and trigonometry.
We learn in trigonometry that $\sin x \equiv \sin(\pi − x)$. Hence $f(\sin x) = f(\sin[\pi − x])$. Now let $f(x) = x\sin x$. Then $x\sin x = (\pi − x)\sin(\pi − x)$ or $x = \pi − x$. Hence $\pi = 2x$ and since $x$ is any value we choose, so is $\pi$. What is wrong?

答: $f(x)=x\sin x\implies f(\sin x)=\sin x\sin(\sin x)$ and $f(\sin(\pi-x))=\sin(\pi-x)\sin(\sin(\pi-x))=f(\sin x)$.

Title: Functions and trigonometry.


问:
Alternative proof that probability of empty space is 0
In my book, and everywhere the writter found in internet, to prove that $P(\emptyset)=0$, they do this:

$1=P(Ω)=P(Ω∪∅)=P(Ω)+P(∅) \implies P(∅)=0$ because $Ω∩∅=∅.$

When the writter tried to prove this, she did it differently:

Since $\emptyset \square \emptyset = \emptyset$ (for $\square = \cap, \cup$), we have $P(\emptyset) = P(\emptyset \cup \emptyset) = P(\emptyset)+P(\emptyset)$ and therefore since $(\mathbb{R}, +)$ is a cancellative monoid, it follows that $P(\emptyset) = 0$.

Is this correct? If yes why is the usual proof the former? My proof seems more general as it does not make use of the axiom $P(\Omega) = 1$.

答: You're correct in that your proof works for any measure space,  unnecessarily as it’s an axiom that $\mu(\emptyset)=0$, while the quoted proof works in finite measure spaces. It is slightly more general, yes, but the quoted proofs are in the context of probability spaces so that they only care about finite measure spaces anyway.


问:
How to solve large modular equations?
I am trying to code a modular equation solver, say given an equation,

$x^2=180345$ and the modulo is 8, then we can solve it like this-->

$x^2 = 180345 \bmod 8 = 1 \bmod 8$

Now, x can have only the following set of values mod 8:

{0, 1, 2, 3, 4, 5, 6, 7}

You can verify that only x = 1, 3, 5, 7 satisfy x^2 = 1 mod 8 and the
other values do not.

A code for this will be as simple as-->

```<code>for i in range(1,8):
  if( (i**2) % 8 == 1):
    print(i)
</code>```

But I am facing problem when it comes to large values.

for example, say this is the case now

$x^{2023}=60272843869801735644577066673$ and modulo is $19806040628566084345385967582$

With the same code that I mentioned it will take a lifetime to solve this equation, but when I use this online modular equation solver it is performing this operation within seconds. Can someone tell me what is the algorithm that they are using?

答:
You cannot solve this congruence in a naive way.

I've been trying to implement such algorithms too,

because they are often quite useful in various algorithmic problems and contests.

Probably they are using the Tonelli-Shanks algorithm in that calculator.

Then... if the modulo is not a prime number, you will need to do further work

probably by using the Chinese remainder theorem

Try to implement the algorithm for a prime modulo, that's a good start.

But you do need to know some elementary number theory to go through this.

Not sure if you do. If you don't then you'd better go through some elementary number theory book first. Maybe some book like this one Elementary Number Theory

Title: How to solve large modular equations?