问:
Evaluate $ \int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\:\mathrm dx$
Evaluate the following integral
$$
\tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\mathrm dx
$$

My Attempt:

Letting $x=\frac{\pi}{2}-x$ and using the property that

$$
\int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx
$$

we obtain

$$
\tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\mathrm dx
$$

Now, add equation $(1)$ and $(2)$. After that I do not understand how I can proceed further.

答:
Let us try to understand the computer-based answer of Oliver Oloa.

Rational decomposition.

Is known the rational decomposition of the trigonometric functions in the forms of
$$\tan z = \sum\limits_{n=0}^\infty\dfrac{8z}{\pi^2\left(2n+1\right)^2 - 4z^2},\tag{R1}$$
$$\sec z = \sum\limits_{n=0}^\infty (-1)^n\dfrac{4\pi(2n+1)}{\pi^2\left(2n+1\right)^2 - 4z^2},\tag{R2}$$
with the corollaries
$$\tanh z =-i\tan(iz) = \sum\limits_{n=0}^\infty\dfrac{8z}{\pi^2\left(2n+1\right)^2 + 4z^2},\tag{R3}$$
$$\operatorname{sech}z = \sec(iz) = \sum\limits_{n=0}^\infty (-1)^n\dfrac{4\pi(2n+1)}{\pi^2\left(2n+1\right)^2 + 4z^2}.\tag{R4}$$

Preparation.

$$I=\int\limits_0^{\Large^\pi/_2}\dfrac{\text dx}{(1+x^2)(1+\tan x)}
=\dfrac12\int\limits_0^{\Large^\pi/_2}\left(\dfrac{1+\tan x}{1+\tan x}
-\dfrac{\tan x-1}{1+\tan x}\right)\,\dfrac{\text dx}{1+x^2}$$
$$=\dfrac12\int\limits_0^{\Large^\pi/_2}\dfrac{\text dx}{1+x^2}
-\dfrac12\int\limits_0^{\Large^\pi/_2}\tan\left(x-\dfrac\pi4\right)\,\dfrac{\text dx}{1+x^2},$$
$$I=\dfrac12\int\limits_0^{\Large^\pi/_2}\dfrac{\text dx}{1+x^2}
-\dfrac12\int\limits_0^{\Large^\pi/_2}\sum\limits_{n=0}^\infty g_n(x)\,\text dx,\tag1$$
where, taking in account $(R1),$
$$g_n(x)=\dfrac{2x-\large\frac\pi2}{\pi^2\left(n+\large\frac12\right)^2 - \left(x-\large\frac\pi4\right)^2}\cdot\dfrac1{1+x^2}.\tag2$$

Partial fractions.

The functions $\;g_n(x)\;$ allow decomposition in the form of
$$g_n(x)=\dfrac{A_nx+B_n}{1+x^2} + \dfrac {C_n}{\pi n+\frac34\pi-x} + \dfrac {D_n}{\pi n+\frac14\pi+x}.\tag3$$
Then
$$C_n=\lim_{\large x\to \pi n+\frac34\pi}\, \left(\pi n+\frac34\pi-x\right)g_n(x)
= \dfrac{16}{(4 n\pi+3\pi)^2 +16},$$
$$D_n=\lim_{\large x\to -\pi n-\frac14\pi}\, \left(\pi n+\frac14\pi+x\right)g_n(x)
= -\dfrac{16}{(4n\pi+\pi)^2 +16},$$
$$A_n-C_n+D_n=\lim_{\large x\to \infty}\, x g_n(x) = 0,\quad A_n=C_n-D_n,$$
$$B_n+ \dfrac {C_n}{\pi n+\frac34\pi} + \dfrac {D_n}{\pi n+\frac14\pi} =g_n(0) = -\dfrac8{\pi(16n^2+16n+3)},$$
$$B_n = \dfrac4{4\pi n+3\pi} -\dfrac4{4\pi n+\pi} -  \dfrac {4C_n}{4\pi n+3\pi} - \dfrac {4D_n}{4\pi n+\pi}.$$
$$B_n = 4(4\pi n+3\pi)C_n + 4(4\pi n+\pi)D_n.$$

Let
$$U_k = \dfrac{16}{(2k\pi+\pi)^2+16},$$
then, taking in account $(R3)-(R4),$
$$D_n = -U_{2n},\quad C_n=U_{2n+1},\;$$
$$\sum\limits_{n=0}^\infty A_n = \sum\limits_{n=0}^\infty U_n = \tanh2,\tag4$$
$$\sum\limits_{n=0}^\infty B_n = \sum\limits_{n=0}^\infty (-1)^{n+1}4\pi(2n+1)U_n
= -\operatorname{sech}2.\tag5$$

Integration.

$$I=\dfrac12\int\limits_0^{\Large^\pi/_2}\dfrac{\text dx}{1+x^2}
-\dfrac12\int\limits_0^{\Large^\pi/_2}\sum\limits_{n=0}^\infty g_n(x)\,\text dx$$
$$ = \dfrac12 \arctan \,\dfrac\pi2-\dfrac12\sum\limits_{n=0}^\infty \left(\dfrac12 A_n \ln\left(\dfrac{\pi^2}4+1\right) +B_n \arctan \,\dfrac\pi2+(C_n+D_n) \ln\dfrac{4n + 3}{4n+1}\right),$$
$$I=-\dfrac14\tanh 2\ln\left(\dfrac{\pi^2}4+1\right) + \dfrac{1+\operatorname{sech}2}2 \arctan \,\dfrac\pi2-\dfrac12 I_1,\tag6$$
where
$$I_1=\sum\limits_{n=0}^\infty (C_n+D_n) \ln\dfrac{4n + 3}{4n+1},\tag7$$
with the representations
$$I_1=\sum\limits_{n=0}^\infty \dfrac{128\pi^2(2n+1)}{\pi^4((4n+2)^2-1)^2 +32\pi^2((4n+2)^2+1)+256}\,\ln\dfrac{4n+3}{4n+1}\tag{7a}$$

and
$$I_1=\sum\limits_{n=0}^\infty \Im\left(-\dfrac4{4\pi n+3\pi+4i}+\dfrac4{4\pi n+\pi+4i}\right) \ln\dfrac{4n + 3}{4n+1}.\tag{7b}$$

The poly-Stieltjes constants using.

The poly-Stieltjjes constants do not belong to the standard special functions, but they are defined in the Oliver Oloa answer as the known specific functions.

In particular, from $(7b)$ and the given definition in the form of
$$\gamma_1(a,b) = \lim\limits_{N\to\infty}\left(\sum\limits_{n=1}^N \dfrac{\ln(n+a)}{n+b} - \dfrac{\ln N^2}2\right)\tag{$\large\diamond$}$$
should
$$\begin{align}
&I_1= \dfrac{16}{16+9\pi^2}\,\ln3 - \dfrac1\pi \Im\left(\gamma_1\left(\small\dfrac34,\frac34+\frac i\pi\right)
- \gamma_1\left(\small\dfrac14,\frac34+\frac i\pi\right) \right)\\[4pt]
&-\dfrac{16}{16+\pi^2}\,\ln3 + \dfrac1\pi \Im\left(\gamma_1\left(\small\dfrac34,\frac14+\frac i\pi\right)
- \gamma_1\left(\small\dfrac14,\frac14+\frac i\pi\right) \right),
\end{align}\tag8$$

$$\begin{align}
&I=-\dfrac14\tanh 2\ln\left(\dfrac{\pi^2}4+1\right) + \dfrac{1+\operatorname{sech}2}2 \arctan \,\dfrac\pi2
+\dfrac{64\pi^2}{(16+\pi^2)(16+9\pi^2)}\\[4pt]
&+\dfrac1{2\pi}\Im\left(\gamma_1\left(\small\dfrac34,\frac34+\frac i\pi\right)
- \gamma_1\left(\small\dfrac14,\frac34+\frac i\pi\right) \right)\\[4pt]
&- \dfrac1{2\pi} \Im\left(\gamma_1\left(\small\dfrac34,\frac14+\frac i\pi\right)
- \gamma_1\left(\small\dfrac14,\frac14+\frac i\pi\right) \right).
\end{align}\tag9$$

Formula $(9)$ gives the closed-form representation, which cannot be used in the popular package Wolfram Alpha.

Results.

Obtained the alternatiive notation of the poly-Stieltjes closed form $(9)$ for the given integral.

Obtained the calculating formulas $(6),(7a)$ via the elementary functions series. Evaluated value

corresponds to the numerical calculations of the given integral

.


问:
Pulling out a function from an integral. When is this allowed?
If I have a function which does not depend on the variable which is the integrating variable, can I pull that function out? That is, suppose I have $\int g(x,t)f(u_i - u(x,t))du_i$. Can I pull out the $g(x,t)$?

What about for when I have this: $\int \frac{\partial g}{\partial t}(x,t) f(u_i - u(x,t))du_i$? I know that by Leibniz's rule I can pull out the $\frac{\partial }{\partial t}$, but I do not know if I am allowed to pull out the entire term $\frac{\partial g}{\partial t}(x,t)\int f(u_i - u(x,t))du_i$.

答:
Yes this is a valid step. You can think of
$$\frac{\partial}{\partial t}g(x,t)$$
as a constant in terms of $u_i$.

ContentLicense: CC BY-SA 4.0


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
Computational instability of the Quadratic Formula. Who would have thought?

Due to this computational stability an alternative formula is also employed. Here is the relevant quote from the Wikipedia article:

“The alternative formula can reduce loss of precision in the numerical evaluation of the roots, which may be a problem if one of the roots is much smaller than the other in absolute magnitude.”

And here is the link to the full article:

http://en.wikipedia.org/wiki/Quadratic_equation

ContentLicense: CC BY-SA 3.0

LastEditorDisplayName: anon


问:
Evaluating a definite integral with e
Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

Title: Evaluating a definite integral with e

Body:
Use $u=x^3$, then $dx = \frac{du}{3x^2}$. This turns the integral into:
$$\frac{1}{3}\int\limits_{0}^{1}e^{-2u}du = \frac{-1}{6}e^{-2u}+C=\frac{-1}{6}e^{-2x^3} +C$$


问:
Slowly decaying sequences are dense in $l^1$

Consider $S:=\{x \in l^1 : \sup_n|nx_n|=\infty\}$. Show this set is dense in $l^1$.

I am a bit stuck on how to proceed, but I think I managed to simplify this problem slightly.

By Baire category theorem we could show that $S$ is co-meager, and thus it would suffice to show that $C_N:=\{x \in l^1 : \sup_n|nx_n|\leq N\}$ are nowhere dense ($N$ here is a natural number).

It's easy to show these sets are closed. How could I show that their complement is dense, or that they have empty interior?

Addendum: do you also have in mind an explicit element of $S$?

答:
Consider the sequence

$$a_n=\begin{cases}
0 &\text{ if } n \neq p^3\\
\frac{1}{p^2} &\text{ if } n =p^3
\end{cases}$$

$\{a_n\}$ belongs to $S$. For $x=\{x_1, x_2, \dots\} \in \ell^1$, consider the $S$ sequence $\{X_n\}$ defined as

$$X_n(m)=\begin{cases}
x_m & m \le n\\
a_m & m \gt n
\end{cases}$$

We have $X_n \to x$ in $\ell^1$ which allows to conclude to the desired result.


问:
Independence of 2 discrete random variables
If I have 2 discrete random variables $X$ and $Y$, and I know their probability mass functions, then how do I show that the $2$ random variables are independent? I know that $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)$, but I don't quite understand what I can do with this to show that they are independent. I think I am just confused by the LHS. What does that even mean? If I have the $2$ pmfs, and I multiply them together to get the RHS of the equation, how would I show that it equals the LHS and are therefore independent?

答:
First you want to calculate $P(X=x,Y=y) = P(X=x \,\&\, Y=y) = P(\{X=x\} \cap \{Y= y\})  $

Next calculate $P(X=x)\cdot P(Y=y)$

If you get the same result then they are independent, denoted $X \perp \!\!\! \perp Y$.

A few ways to possibly compute $P(X=x, Y =y)$:

Bayes Rule: $P(X=x, Y=y) = P(X=x\mid Y=y)P(Y=y)$

Addition/Union Rule: $P(\{X = x \} \cup \{Y=y\}) = P(X=x) + P(Y=y) - P(X=x,Y=y)$


答:
Using the intuitive definition of joint probability in terms of conditional probability, we have

$P(X=x, Y=y)=P(X=x|Y=y).P(Y=y)$

$=P(X=x)P(Y=y)$ (given)

Assuming $P(Y=y) \neq 0$ (the trivial case for an impossible event) it implies that $P(X=x|Y=y)=P(X=x)$, which tells us that probability of $X$ does not depend upon the probability of occurence of $Y$, hence they are independent.

Title: Independence of 2 discrete random variables

Body: If I have 2 discrete random variables $X$ and $Y$, and I know their probability mass functions, then how do I show that the $2$ random variables are independent? I know that $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)$, but I don't quite understand what I can do with this to show that they are independent. I think I am just confused by the LHS. What does that even mean? If I have the $2$ pmfs, and I multiply them together to get the RHS of the equation, how would I show that it equals the LHS and are therefore independent?


问:
Is there a definition of Riemann integration in $\mathbb{R^n}$?
I know that Riemann integration is well-defined for a function $f: \mathbb{R} \to \mathbb{R}$.

I would like to ask whether there is a definition of Riemann integration for a function $f: \mathbb{R^n} \to \mathbb{R}$ ? If there is not, are there technical reasons to prevent such a definition ?

The reason for the question is to understand more about the motivation to construct the Lebesgue integration.

Thanks for your explanation!

答:
There is such a notion, which is used often in multivariable calculus before one introduces the Lebesgue integral. The motivation is the same - we want the volume under the graph of $f: \mathbb R^n \longrightarrow \mathbb R$. The approach of summing the area of rectangles with Riemann sums generalizes to summing the volume of higher dimensional rectangles.

Formally, a single variable Riemann sum of a bounded function $f: [a, b] \longrightarrow \mathbb R$ is a sum of the form $\sum_{i=0}^n f(t_i^*) (t_{i + 1} - t_i)$ where $a = t_0 < \dots < \ t_n = b$ is a partition of $[a, b]$ and $t_i^* \in [t_i, t_{i+1}]$. That is, we sample $f$ on these subintervals and sum the corresponding rectangle's areas. We say that $f$ is Riemann integrable if the limit of these expressions as the mesh size $\delta = \sup (t_{i+1} - t_i)$ goes to $0$. This limit is called the Riemann integral of $f$ and is denoted $\int_a^b f(x) dx$.

With this review aside, we can see a clear path to generalizing this to multivariable calculus. Take a rectanglesm $R = [a_1, b_1] \times \dots \times [a_n, b_n]$ and let $f: R \longrightarrow \mathbb R$ be bounded. Previously, we took smaller and smaller partitions of our interval $[a, b]$ into subintervals. Here, we'll partition our rectangle $R$ into subrectangles. Specifically, we consider partitions $R = \bigcup R_i$ where $R_i$ is a product of subintervals of $[a_i, b_i]$. We consider only "almost disjoint" unions $R_i$, meaning that if $i \neq j$ then $R_i$ and $R_j$ can intersect only on their boundaries. You can picture a partition of, say, a rectangle in $\mathbb R^2$ by drawing finitely many lines parallel to the axes which start from one edge and go to the parallel edge. The resulting "brick wall" is a partition. For higher dimensions, you do this with hyperplanes instead of lines.

Now, given such a partition, a Riemann sum for this partition is a sum of the form $\sum_i f(R_i^*) m(R_i)$ where $R_i^* \in R_i$ and $m(R_i)$ is the volume of the rectangle $R_i$, which is defined as the product of its side lengths. This is just like the single variable case - we sample $f$ along a partition and approximate the volume under its graph by these Riemann sums.

To finally define an integral, we need a limit of these sums. We define the diamter of a rectangle to be the largest distance between any two points in the rectangle (which is realized as the long diagonal). The mesh size of a partition $\bigcup R_i$ is the largest diameter of the $R_i$. Then we say $f$ is Riemann integrable if the limit of these Riemann sums as the mesh size goes to $0$ exists, and this limit is called the Riemann integral of $f$ and is denoted $\int_R f(x_1, \dots, x_n) dx_1 \dots dx_n$.


问:
Solivng differential equation by subtracting Y.
I have the equation $$ \frac{dy}{dx} = y + 2$$ And normally u would isolate the y's and the x's but what if I subtract y and then solve the equation. That would like the following. $$ \frac{dy}{dx} - y =2 $$ $$
 \int(\frac{dy}{dx} - y)dx = \int2dx$$ $$ \int\frac{dy}{dx}dx - \int ydx = \int 2dx $$ $$ \int1dy - \int ydx = \int2dx $$ $$ y - yx = 2x + C$$ $$ y = \frac{2x+C}{1-x}$$ But when you check this u get an equation where your only solution that works is when x=0. What am is wrong with the math.

Body:
$$\frac{dy}{dx} - y =2$$
Multiply by $e^{-x}$ then you can integrate both sides:
$$\frac{dy}{dx}e^{-x} - ye^{-x} =2e^{-x}$$
$$\int \left(\frac{dy}{dx}e^{-x} - ye^{-x}\right)dx =2\int e^{-x} dx$$
$$\int (ye^{-x})'dx =2\int e^{-x} dx$$
$$\int d(ye^{-x}) =2\int e^{-x} dx$$
$$ye^{-x}=-2 e^{-x} +C$$
$$y(x)=-2+ Ce^{-x} $$


Title: Evaluating a definite integral with e

Body:
Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

Body:
$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

Let $u=2x^3$. Then $du = 6x^2 dx $, and we have:

$$\frac16 \int_0^1 \dfrac{6x^2}{e^{2x^3}} dx$$
$$\frac16 \int_0^2 \dfrac{1}{e^u} du \tag{Limits change}$$
$$\frac16 \int_0^2 \dfrac{1}{e^u} du$$


问:
Show that the series $\sum_{n=0}^{\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$ is divergent
In this thread, I use the "C+iS" summation method for trigonometric series to sum it:

lab bhattacharjee pointed out that:

$\left|\dfrac{e^{ix}}{\cos(x)}\right|=|\sec(x)|\geq1$

What I don't understand is why $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, I thought it is $e^{ix}\sec(x)$. By looking at the graph of $\sec(x)$, it is clear that $\sec(x)\geq1 $ on the interval $(0,\pi/2)$ and $(0,5\pi/2)$ and so on. $\sec(x)\leq1$ on the interval $(\pi, 3\pi/2)$ and $(3\pi/2, \pi)$. So I get the part if $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, then the series diverge.

Also, how can I show that the series $\left(\dfrac{e^{-ix}}{\cos(x)}\right)^n$ is a divergent series?

答:
$$\lim_{n\to\infty}\frac{\cos (n x)}{\cos ^n(x)}$$
fails to exist, therefore the general term does not tend to zero, necessary condition for the convergence. Thus the series diverges.


问:
prove that $\frac{|z-a|}{|\overline{a}z-1|}=1$ if |z|=1 and a $\in \mathbb{C}$
I started with that: $\frac{|z-a|}{|\overline{a}z-1|}=\frac{|(z-a)(\overline{\overline{a}z} -1)|}{|\overline{a}z-1|^2}=\frac{|a|z|^2 - z -a^2\overline{z} +a|}{|\overline{a}z-1|^2}= \frac{|a^2\overline{z}-2a+z|}{|\overline{a}z-1|^2} = ..?$ How to prove this equality with 1?

答: Since $|z|=1$, $|\bar{a}z-1|=|\bar{a}-1/z|=|\bar{a}-\bar{z}|=|z-a|$. The ratio is undefined if $a=z$; otherwise, it's $1$.


问:
Show that $f(x)=x$ if $x\in\mathbf{Q}$ and $f(x)=1-x$ if $x \in\mathbf{R}/\mathbf{Q}$ is Riemann integrable on $[0,1]$
Let $f(x)=x$ if $x\in\mathbf{Q}$ and $f(x)=1-x$ if $x \in\mathbf{R}/\mathbf{Q}$. Show that $f$ is Riemann integrable on $[0,1]$.

I know that there are a few posts on this function, but I didn't see the same approach as mine. So, I would like to know if my prove holds, please and have a feedback.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.

First, consider the interval $[0,\frac{1}{2}]$. We know that $\int_{0}^{1}f(x)=\int_{0}^{1/2}f(x)+\int_{1/2}^{1}f(x)$. So, if one of the integrals is undefined, we could conclude that it is not integrable on the whole interval (*I'm not sure that is true. Should I show that the integral doesn't exist for both intervals, or one is sufficient?)

Let $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$ and $m_i=\inf\{f(x):x\in[x_i,x_{i+1}]\}$ (We work on $[0,\frac{1}{2}]$ now). We remark by density of $\mathbf{Q}$ and $\mathbf{R}/\mathbf{Q}$, that $M_i=(1-x_i)$ and $m_i=x_i$. Thus,

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)=\sum_{i=0}^{n}(1-x_i)(x_{i+1}-x_i)-\sum_{i=0}^{n}x_i(x_{i+1}-x_i)$.

But, $\overline{S}_{\sigma}(f)\ge\int_{0}^{1/2}(1-x)\ dx$ and $\underline{S}_{\sigma}(f)\le \int_{0}^{1/2}x\ dx$ whatever the partition. So, $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)\ge \int_{0}^{1/2}(1-x)\ dx - \int_{0}^{1/2}x\ dx=\frac{1}{4}$.

Thus, there is no partition on the interval $[0,\frac{1}{2}]$ such that $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)<\epsilon$ for $0<\epsilon<\frac{1}{4}$ and we conclude that $f$ is not integrable on $[0,1/2]$ and so on $[0,1]$ as well.

答:
So, if one of the integrals is undefined, we could conclude that it is
not integrable on the whole interval (*I'm not sure that is true.
Should I show that the integral doesn't exist for both intervals, or
one is sufficient?)

If $f$ is Riemann integrable on $[a,b]$, then for any $\epsilon> 0$ there is a partition $P$ of $[a,b]$ such that $U(P,f) - L(P,f) < \epsilon$.

Let $P'$ be the refinement of $P$ obtained by adding the point $c \in (a,b)$ if not already there.  Otherwise let $P' = P$. Let $P'' $ be the partition of $[a,c]$ induced by $P'$.  It follows that

$$U(P'',f) - L(P'',f) \leqslant U(P',f) - L(P',f) \leqslant U(P,f) - L(P,f) < \epsilon,$$

and by the Riemann criterion $f$ must be Riemann integrable on $[a,c]$.


答:
This happens because:

$$11^{n} = (10 + 1)^{n} = 10^{n} + \binom{n}{1}10^{n-1}+\binom{n}{2}10^{n-2} + \dots + \binom{n}{n-1}10^{1} + 10^{0}$$

By binomial expansion. The pattern first breaks down at $n=5$ because it is the first time one of the binomial coefficients is at least $10$. If you used a higher base (say, hexadecimal), the pattern would continue for longer.

You have associated it with Pascal's Triangle because each row of Pascal's Triangle contains the respective binomial coefficients.

Title: Pascal's triangle curiosity

Body:
I noticed the following pattern in the rows of Pascal's triangle:

$$
1 = 11^0\\
11 = 11^1\\
121= 11^2\\
1331= 11^3\\
14641=11^4
$$

at this point I thought maybe this pattern would follow indefinitely, but the next row is
$15101051=7\times2157293$. The next two rows aren't powers of 11 either, they are $1615201561=43\times 37562827$ and $172135352171=29\times5935701799$. Are the first 5 rows the only rows in the Pascal triangle that are powers of 11?
The following row 18285670562881 is a prime (at least according to Linux factor function).
How often primes appear in these rows?
This might be fun to show or investigate. I am just curious.

I just wrote a Python script to generate a few more rows of the Pascal triangle. Each row was also factored up to row n=19:

```<code>from scipy.special import binom
from sympy import factorint

for n in range(5, 20):
  row = ''
  print('n=%d' % n)
  for k in range(n+1):
    binom_nk = int(binom(n, k))
    row = row+str(binom_nk)
  print(row)
  row_number = int(row)
  factors=factorint(row_number)
  print(factors)
</code>```

n=5
15101051
{7: 1, 2157293: 1}

n=6
1615201561
{43: 1, 37562827: 1}

n=7
172135352171
{29: 1, 5935701799: 1}

n=8
18285670562881
{18285670562881: 1}

n=9
193684126126843691
{5647: 1, 34298587945253: 1}

n=10
1104512021025221012045101
{13: 1, 197: 1, 4649: 1, 92768668286052709: 1}

n=11
1115516533046246233016555111
{11: 1, 101410593913295112092414101: 1}

n=12
1126622049579292479249522066121
{523: 1, 4637: 1, 464557485113006356820471: 1}

n=13
11378286715128717161716128771528678131
{369268429: 1, 180642383: 1, 170574866715037030033: 1}

n=14
11491364100120023003343230032002100136491141
{41: 1, 86399681: 1, 3910184543: 1, 829618322366629399154147: 1}

n=15
11510545513653003500564356435500530031365455105151
{17: 1, 113: 1, 149: 1, 337: 1, 5362678549: 1, 3533441: 1, 2118359: 1, 2972851397279413777: 1}

n=16
116120560182043688008114401287011440800843681820560120161
{7: 1, 13: 1, 10567: 1, 835035230611: 1, 144614294599341081090704192088907914583: 1}

n=17
1171366802380618812376194482431024310194481237661882380680136171
{71: 1, 89: 1, 1583: 1, 11969: 1, 82361: 1, 8401361: 1, 6505783: 1, 2173376383488959250244793817754469: 1}

n=18
118153816306085681856431824437584862043758318241856485683060816153181
{31: 1, 59: 1, 4421: 1, 28979: 1, 248303053: 1, 2030710802068204317879252665132300005649014383107: 1}

n=19
1191719693876116282713250388755829237892378755825038827132116283876969171191
{4506133: 1, 37569563: 1, 70854794496642568411: 1, 99349280033387580503673432191635187608339: 1}


答: $11^5 = 161051$ but you can get this by using the 1 from the 10s as a carry figure on your 15101051 and the pattern will still work

Title: Pascal's triangle curiosity


答:
you can simplify this by assuming WLOG that $\mathbf v$ has length one.

simple case:

suppose $\mathbf v := \mathbf e_1$, the first standard basis vector. Then by direct computation

$\|A(I - \mathbf{e_1e_1}^T)\|_F^2  =\sum_{j=2}^n\sum_{i=1}^n a_{i,j}^2$  and

$\|A\|_F^2 - \|A\mathbf e_1\|_2^2 =\big(\sum_{j=1}^n\sum_{i=1}^n a_{i,j}^2\big)  -\big(\sum_{j=1}^1\sum_{i=1}^n a_{i,j}^2\big)= \sum_{j=2}^n\sum_{i=1}^n a_{i,j}^2$

hence $\|A(I - \mathbf{e_1e_1}^T)\|_F^2=\|A\|_F^2 - \|A\mathbf e_1\|_2^2$

general case:

Real symmetric matrices may be orthogonally diagonalizaed, so $\big(I-\mathbf {vv}^T\big) =Q\big(I -\mathbf e_1\mathbf e_1^T\big)Q^T$, where $Q$ is real orthogonal.  Define $B:=Q^TAQ$

$\|A(I - \mathbf{vv}^T)\|_F^2$

$=\|AQ(I - \mathbf e_1\mathbf e_1^T)Q^T\|_F^2$

$=\|Q^TAQ(I - \mathbf e_1\mathbf e_1^T)\|_F^2$

$=\|B(I - \mathbf e_1\mathbf e_1^T)\|_F^2$

$=\|B\|_F^2 - \|B\mathbf e_1\|_2^2$

$=\|Q^TAQ\|_F^2 - \|Q^TAQ\mathbf e_1\|_2^2$

$=\|A\|_F^2 - \|A(Q\mathbf e_1)\|_2^2$

$=\|A\|_F^2 - \|A\mathbf v\|_2^2$

Title: Show that $\|A(I - vv^T/v^Tv)\|_F^2 = \|A\|_F^2 - \|Av\|^2_2/(v^tv)$

Body:
Show that $\|A(I - vv^T/v^Tv)\|_F^2 = \|A\|_F^2 - \|Av\|^2_2/(v^tv), A \in R^{n \times n}, v \in R^n$

$$
\begin{aligned}
\|A(I - vv^T/v^Tv)\|_F^2
&= \sum_{i = 1}^{n}\sum_{j = 1}^{n}(A - Avv^T/c)^2 \\
&= \sum_{i = 1}^{n}\sum_{j = 1}^{n}(A^2 + (Avv^T)^2/c^2 - 2A^2vv^T/c) \\
&= \sum_{i = 1}^{n}\sum_{j = 1}^{n}A^2 - \sum_{i = 1}^{n}\sum_{j = 1}^{n}(vv^TA^TAvv^T/c^2 - 2A^TAvv^T/c)\\
&= \|A\|_F^2 - \sum_{i = 1}^{n}\sum_{j = 1}^{n}(vv^TA^TAvv^T/c^2 - 2A^TAvv^T/c)
\end{aligned}
$$


答:
I found this Magic 11's

This even works until the 7th row

Title: Pascal's triangle curiosity


问:
Find a basis for the subspace $W= \left \{ f \in V | f(x)=0 \text{ } \forall x \in X, \text{ except for a finite number of elements of } X\right \}$
I need your help to check if my proof is correct please. This is what I have to prove:

Let be $X$ a not empty set and let be $F$ a field. Consider the vector
space given by the set $V= \left \{ f : X \rightarrow F \right \}$
with the usual operations. Find a basis for the subspace $W= \left \{
 f \in V |  f(x)=0 \text{ } \forall x \in X, \text{ except for a finite
 number of elements of } X\right \}$

My attempt:

With $i,j \in \left \{ 1,2,...,n \right \}$, we can define this functions:

\begin{equation*}
f_{x_{j}}(x_i)=
   \left\{
\begin{array}{rcl}
     1 & ~~~~~~~~~if & x_i=x_j, \text{ with } x_i,x_j\in X\\
     0 & ~~~~~~~~~&\text{another case}
\end{array}
\right.
\end{equation*}

Let's see if the set $S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$ is a basis of $W$. First,
we note that if:

\begin{align}
\lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i)=0 & &\text{ for any $x_i \in X$}\\ & &\lambda_i \in F \\& &i \in \left \{ 1,...,n \right \}
\end{align}
then,
\begin{align}
\lambda_1 +\lambda_2 +\cdots+\lambda_n=0 & &\text{ for any $x_i \in X$}
\end{align}
So, that implies that $S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$ is linearly independent (1)

By the other hand, let be $f \in W$, then $f(x_i)=0$, except for a finite number of elements of $X$. Then, we can propose to write $f$ as:

\begin{align}
f(x_i)=\lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i)
\end{align}

since the quantity of $\lambda_i$ that satisfies that $\lambda_i \neq 0$ is the number of finite elements of $X$ where $f(x_i)\neq 0$, that implies that $f \in span(S)$. So, $S$ is a generating set of $W$. (2)

By (1) and (2), $S$ is a basis of $W$.

答:
You seems to have the main ideas. However your « proof » seems to mention a finite number of maps to be a basis of $W$ which isn’t the case if $X$ is infinite.

$$\mathcal B=\{f_x \mid x \in X\}$$

where

$$f_x(v)=\begin{cases}
1 & v =x\\
0 & v \ne x
\end{cases}$$ is indeed a basis of $W$. To prove that $\mathcal B$ is linearly independent, suppose that a finite linear combination of elements of $\mathcal B$
$$\lambda_1 f_{a_1} + \dots +\lambda_n f_{a_n}=0$$ always vanishes. Then by plugging in $v=a_i$ you get $\lambda_i=0$ as desired.

You were almost there to prove that $\mathcal B$ spans $W$.

Note: the cardinality of $\mathcal B$ is the one of $X$.


问:
Best linear approximation in a neighborhood property for Frechet derivative
Let $D$ denote a Frechet derivative at a point $a$. Then for $f:\mathbb{R}\to\mathbb{R}$ it holds for all $x$ from some neighborhood of $a$:
$$|f(x)-(f(a)+D(x-a))|\leq|f(x)-(f(a)+M(x-a))|,\ M\neq D$$
Circular definition of tangent line and derivative

Prove that $T(x)$ is the most accurate linear approximation to $f(x)$ near $x_0$...

Do we have this for Frechet derivative in other normed spaces $f:\mathbb{S}\to\mathbb{P}$ also?

I can only get that for some $\delta$ and all $x$ s.t. $||x-a||_S\leq \delta$
$$||f(x)-(f(a)+D(x-a))||_{P}\leq\frac{1}{3}||D-M||||x-a||_S$$
$$\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+D(x-a))||_{P}\leq\frac{1}{3}\delta||D-M||$$
$$\delta||D-M||=\operatorname{sup}_{||x-a||_S\leq\delta}||(D-M)(x-a)||_P\leq\frac{1}{3}\delta||D-M||+$$
$$+\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+M(x-a))||_{P}$$
So it follows
$$\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+D(x-a))||_{P}<\operatorname{sup}_{||x-a||_S\leq\delta}||f(x)-(f(a)+M(x-a))||_{P}$$
But this is a weaker result of course.

Upd: changed the result to strict inequality.

答:
Ok, it seems that in general the best local linear approximation doesn't need to exist.
I found the counterexample here If there is a linear function $g$ which is at least as good of an approximation as any other linear $h$, then $f$ is differentiable at $x_0$. in Milo Brandt's answer and comments.

Let $f(x,y)=x^2+y^2$. Then the linear approximation with the derivative at point $(0, 0)$ is $l(x,y)=0$. But the function $g(x,y)=x$ coincides with $f(x,y)$ at points along the curve $x=\frac{1-\sqrt{1-4y^2}}{2}$ in any neighborhood of $(0,0)$ because of continuity. So the linear approximation with the derivative is not the best linear approximation in any neigborhood of $(0,0)$.

Another note.
Existence of the local best linear approximation in the sense of suprema defined in the question doesn't imply differentiability. Taking $f(x)=|x|$ it can be seen that for $g(x)=0$ for any $\delta>0$ and for all $|x|<\delta$:
$$\operatorname{sup}_{|x|<\delta}|f(x)-g(x)|=\operatorname{sup}_{|x|<\delta}|x|=\delta$$
And for any $l(x)=mx$ with $m\neq0$ it holds
$$\operatorname{sup}_{|x|<\delta}|f(x)-l(x)|=\operatorname{max}\{\delta|1+m|;\delta|1-m|\}>\delta=\operatorname{sup}_{|x|<\delta}|f(x)-g(x)|$$
But $|x|$ is not differentible at $0$.


问:
How do continue/end this infinite limit expression? Shall I use the Reciprocal Rule? - Almost finished
The picture of my work is here

I am almost done with this, but I do not know how to end it? Shall I use the Reciprocal rule?

答:
The final step is to write it as
$$
\frac{\lim_{x \to \infty}2+\frac{4}{x}+\frac{1}{x^3}}{\lim_{x \to \infty}5+\frac{2}{x^2}}=\frac{2}{5} \, .
$$
If you were doing things completely rigorously, then you would also have to justify why
$$
\lim_{x \to \infty}2+\frac{4}{x}+\frac{1}{x^2}=2
$$
and likewise for the denominator. One approach would be to use the sum rule:
$$
\lim_{x \to \infty}2+\frac{4}{x}+\frac{1}{x^2}=\lim_{x \to \infty}2+\lim_{x \to \infty}\frac{4}{x}+\lim_{x \to \infty}\frac{1}{x^2}
$$
and then use the formal definition of a limit. For instance, $\lim_{x \to\infty}\frac{1}{x^2}=0$ if and only if
$$
\forall \varepsilon > 0 \exists N>0 : x>N \implies \left|\frac{1}{x^2}\right|<\varepsilon
$$
If we take $N=1/\sqrt{\varepsilon}$, then
\begin{align}
&x > 1/\sqrt{\varepsilon} \\
\implies & 1/x < \sqrt{\varepsilon} \\
\implies & 1/x^2 < \varepsilon
\end{align}
as required.


问:
mutidimensional local homeomorphism and inverse map
Let $f: X^N \to Y$ be a continuous function such that $f(x_1, \cdots, x_{N-1}, \cdot): X\to Y$ is a local homeomorphism. Is it true that there exists a continuous function $h: X^{N-1}\times Y \to X$ such that
$$
f(x_1,\cdots, x_{N-1}, h(x_1,\cdots, x_{N-1}, y))=y ?
$$
Thanks in advance.

答: As freakish said in the comments, there are counterexamples already when $N=1$. Let $f:\Bbb R\to S^1:x\mapsto e^{ix}$; this is easily seen to be a local homeomorphism that just wraps $\Bbb R$ repeatedly around the unit circle. Let $h:S^1\to\Bbb R$ be any continuous map. Then $h[S^1]$ is a compact, connected subset of $\Bbb R$, so there are $a,b\in\Bbb R$ such that $a\le b$, and $h[S^1]=[a,b]$. Clearly $f\circ h\ne\mathrm{id}_{S^1}$ if $a=b$, so assume that $a<b$, and let $p,q\in S^1$ be such that $h(p)=a$ and $h(q)=b$. Let $C_0$ and $C_1$ be the two closed arcs of $S^1$ with endpoints $p$ and $q$; then $h[C_0]=[a,b]=h[C_1]$. Let $x\in(a,b)$; there are $u\in C_0$ and $v\in C_1$ such that $h(u)=h(v)=x$. Then $(f\circ h)(u)=f(x)=(f\circ h)(v)$, and $u\ne v$, so either $(f\circ h)(u)\ne u$, or $(f\circ h)(v)\ne v$, and again $f\circ h\ne\mathrm{id}_{S^1}$.


答:
You have the right idea but you're not quite there. I'll define $S = \{f_x | x \in X\}$ where $f_x: X \longrightarrow F$ is defined by $f_x(y) = 1$ if $x = y$ and $0$ otherwise. This is like your definition, but you only allowed $S$ to be finite. But that won't work in case $X$ is infinite, so you need to consider all of these elements.

Now we show linear independence. Your proof is wrong, in part because showing $\sum \lambda_i = 0$ is not what you need for linear independence. You want all of the $\lambda_i = 0$. Let's suppose we have $\sum_{i = 1}^n \lambda_i f_{x_i} = 0$ for distinct $x_i \in X$. Now, let $1 \leq j \leq n$ and evaluate this at $x_j$. This yields $\sum_i \lambda_i f_{x_i}(x_j) = 0$. For every $i \neq j$ we have $f_{x_i}(x_j) = 0$. Furthermore, $f_{x_j}(x_j) = 1$. Hence, we have $0 = \sum_i \lambda_i f_{x_i}(x_j) = \lambda_j$. Thus, each coefficient $\lambda_j = 0$ so $S$ is linearly independent.

Now let's show $S$ spans $W$. Take some $f: X \longrightarrow F$ in $W$. We claim that $f = \sum_{x \in X} f(x) f_x$. First of all, why is this sum even well defined? Well the set of $x$ such that $f(x) \neq 0$ is finite by definition of $W$, so this sum only has finitely many nonzero elements. Now, let's prove that these are equal. Take a $y \in X$ and compute $\left( \sum_x f(x) f_x \right)(y) = \sum_x f(x) f_x(y)$. Again, $f_x(y) = 0$ if $x \neq y$ and $1$ if $x = y$. Thus, this sum equals $f(y)$. Hence, $f = \sum_x f(x) f_x$. The right hand side is a linear combination of elements of $S$ (recall that we showed that this is a finite sum), so $f \in span(S)$.

Hence, $S \subseteq W$ is linearly independent and spans, so it is a basis.

Body:
I need your help to check if my proof is correct please. This is what I have to prove:

Let be $X$ a not empty set and let be $F$ a field. Consider the vector
space given by the set $V= \left \{ f : X \rightarrow F \right \}$
with the usual operations. Find a basis for the subspace $W= \left \{
 f \in V |  f(x)=0 \text{ } \forall x \in X, \text{ except for a finite
 number of elements of } X\right \}$

My attempt:

With $i,j \in \left \{ 1,2,...,n \right \}$, we can define this functions:

\begin{equation*}
f_{x_{j}}(x_i)=
   \left\{
\begin{array}{rcl}
     1 & ~~~~~~~~~if & x_i=x_j, \text{ with } x_i,x_j\in X\\
     0 & ~~~~~~~~~&\text{another case}
\end{array}
\right.
\end{equation*}

Let's see if the set $S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$ is a basis of $W$. First,
we note that if:

\begin{align}
\lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i)=0 & &\text{ for any $x_i \in X$}\\ & &\lambda_i \in F \\& &i \in \left \{ 1,...,n \right \}
\end{align}
then,
\begin{align}
\lambda_1 +\lambda_2 +\cdots+\lambda_n=0 & &\text{ for any $x_i \in X$}
\end{align}
So, that implies that $S=\left \{ f_{x_1},f_{x_2},\cdots,f_{x_{n}} \right \}$ is linearly independent (1)

By the other hand, let be $f \in W$, then $f(x_i)=0$, except for a finite number of elements of $X$. Then, we can propose to write $f$ as:

\begin{align}
f(x_i)=\lambda_1 f_{x_1}(x_i)+\lambda_2 f_{x_2}(x_i)+\cdots+\lambda_n f_{x_n}(x_i)
\end{align}

since the quantity of $\lambda_i$ that satisfies that $\lambda_i \neq 0$ is the number of finite elements of $X$ where $f(x_i)\neq 0$, that implies that $f \in span(S)$. So, $S$ is a generating set of $W$. (2)

By (1) and (2), $S$ is a basis of $W$.


问:
Doubt about a proof of "Every absolutely convergent series is commutative convergent"
In the prove of my book for the theorem which says that every absolutely convergent series is commutative convergent, where $\Sigma a_n$ is absolutely convergent, $s_n = a_1 + ... + a_n$, $\Sigma b_n$ is a series which elements are the $\Sigma a_n$ elements switched and $t_n = b_1 + ... + b_n$, it is written:

"Being $\sigma:\Bbb{N}\rightarrow\Bbb{N}$ a bijection and defining $b_n = a_{\sigma(n)}$. For each $n\in\Bbb{N}$, we will call by $m$ the biggest number of $\sigma(1),\sigma(2),...,\sigma(n)$. So ${\sigma(1),...,\sigma(n)}\subset[1,m]$. Then we have that $t_n=\sum_{i = 1}^na_{\sigma(i)}\le\sum_{j = 1}^ma_j=s_m$. Thus
for each $n\in \Bbb{N}$ there is a $m\in \Bbb{N}$ such that $t_n \ge s_m$ and, by an analogous way, for each $m\in \Bbb{N}$ there is a $n\in \Bbb{N}$, such that $s_m \ge t_n$. So $\lim s_n=\lim t_n$"

The part that I didn't understand:

"for each $n\in \Bbb{N}$ there is a $m\in \Bbb{N}$ such that $t_n \ge s_m$ and, by an analogous way, for each $m\in \Bbb{N}$ there is a $n\in \Bbb{N}$, such that $s_m \ge t_n$. So $\lim s_n=\lim t_n$"

(Sorry if something is grammatically wrong, the book isn't in English and I am trying to translate it.)

答:
I assume that the book that you were studying was "Curso de Análise Vol.1" by Elon Lages Lima, since I had the same problem of not understanding the middle steps of the proof.

It is important to note that the proof given is only for $a_n \geq 0$, and that the general case for $\sum a_n$ absolutely convergent follows by the separation of $a_n$ by positive terms and negative terms such that $\sum a_n = \sum p_n - \sum q_n$.

Let $s_n = a_1 + ... + a_n$ and $t_n = b_1 + ... + b_n$. Since $a_n \geq 0$ for all $n$, the sequences $\{s_n\}$ and $\{t_n\}$ are non-decreasing. Furthermore, since $\{s_n\}$ is convergent, we have that:

$$
S = \sum^\infty a_n = \lim s_n = \sup\{s_n: n\in \mathbb{N}\}
$$

then, it suffices to prove that:

i) $\{t_n\}$ is bounded above, which implies that $T = \sum^\infty b_n = \lim t_n = \sup\{t_n: n\in \mathbb{N}\}$

ii) $\sup \{s_n: n \in \mathbb{N}\} = \sup \{t_n: n\in \mathbb{N}\}$

Now for every $n\in \mathbb{N}$, let $m = \max\{\sigma(1), ...., \sigma(n)\}$. It follows that:

$$
t_n = \sum^n_{i = 1} a_{\sigma(i)} \leq \sum_{i=1}^m a_k = s_m \leq S
$$

thus, $\forall n \in \mathbb{N}\exists{m}\in \mathbb{N}: t_n \leq s_m \leq S$, so $\{t_n\}$ is bounded above, and then $\sum b_n$ is convergent with $\sum b_n = \sup \{t_n: n\in \mathbb{N}\}$.

The last inequality also shows that $\sup \{t_n\} \leq S$. By an analogous way (using $\sigma^{-1}$) we can show that $\forall m \in \mathbb{N}\exists n \in \mathbb{N}: s_m \leq t_n \leq T$, which implies that $S \leq T$. Therefore, $S = T$, that is $\lim t_n = \sum b_n = \sum a_n = \lim s_n$.


问:
How to remove roots from an equation?
The question here is how (if it is even possible) to remove the square root terms and transform the following equation to a polynomial with one unknown $x$.
The coefficients $a$, $b$, $c$, and $d$ are known and also $r$.

$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$$

答:
Yes it is possible. Take $\color{red}{x=r \cos t, r>0}$. Because $r^2=(-r)^2$. So, we can accept $r>0$ . Then we have,

$$a \sqrt {r}\sqrt {\cos t}+br\sqrt {\cos t}|\sin t|+cr|\sin t|+d=0$$

$$\sqrt {\cos t}\left(a\sqrt r+br |\sin t|\right)=-d-cr|\sin t|$$

$$\cos^2 t\left (a\sqrt r+br |\sin t|\right)^4=\left(-d-cr|\sin t|\right)^4$$

$$(1-|\sin t|^2)\left(a\sqrt r+br |\sin t|\right)^4=\left(d+cr|\sin t|\right)^4$$

Then $\color{red}{|\sin t|=y}$,  you get

$$(1-y^2) (a\sqrt r+bry)^4-(d+cr y)^4=0.$$

Finally, you get $6$ degree polynomial respect to $y.$

I believe you can take from here.

Title: How to remove roots from an equation?


问:
Do Cauchy-Riemann Equations Hold?
Consider $f(z) = \frac{(\overline{z})^2}{z}$ if $z \neq 0$ and $f(z) = 0$ if $z = 0$

Do Cauchy-Riemann Equations only hold at (0,0)?

I'm confused as to how to start this problem. I have tried converting the function $\frac{(\overline{z})^2}{z}$ into the form U(x,y) + iV(x,y) by letting $z = x+yi$ and $\overline{z} = x-yi$ but that just makes things over complicated and I end up with division by zero at the point (0,0).

Can anyone point me in the right direction?

答:
Hint:

In the polar coordinates $z=re^{i\theta},\;\;(r>0),$
$$f(re^{i\theta})=re^{-3i\theta}=r\cos{3\theta}-3i\sin{3\theta}.$$
Then substitute $f$ into the
Cauchy-Riemann equations in polar coordinates
$$\begin{cases}{ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta}, \\ { \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}. \end{cases}$$

Title: Do Cauchy-Riemann Equations Hold?


问:
Find $\int \frac 1 {(x^2 +a^2)^2} dx $.
I am trying to integrate $\int \frac 1 {(x^2+a^2)^2} \ dx$.  The only thing that I can think to try is substitution, $u=x^2+a^2$ so that $\frac{du}{dx}=2x \Rightarrow du = 2x\ dx = 2\sqrt{u-a^2}\ du$ and then the integral becomes

$$ \int \frac{1}{u^2} (2\sqrt{u-a^2} \ du) $$

which does not seem productive.

I could try perhaps to separate with integration by parts, setting $u = \frac{1}{x^2 +a^2}$ and $dv = \frac{dx}{x^2+a^2}$.  Then we obtain $\frac{du}{dx} = -\frac{1}{(x^2+a^2)^2}(2x)$ and $v = \frac 1 a \tan^{-1}(x/a)$.  Then the integral becomes

$$ \begin{align*}
uv - \int v \ du &= \left(\frac{1}{x^2 +a^2}\right)\left(\frac 1 a \tan^{-1}(x/a)\right) - \int \left(\frac 1 a \tan^{-1}(x/a)\right) \left( -\frac{1}{(x^2+a^2)^2}(2x) \right) \ du
\end{align*}$$

but this also looks like it's headed nowhere good.  Advice?

答:
hint

Write the integral as

$$\frac{1}{a^2}\int \frac{a^2+x^2-x^2}{(a^2+x^2)^2}dx$$

$$\frac{1}{a^2}\Bigl(\int\frac{dx}{a^2+x^2}-\frac 12\int x\frac{2xdx}{(a^2+x^2)^2}
\Bigr)$$

Put $ x=at $ for the first integral and use by parts for the second.

Title: Find $\int \frac 1 {(x^2 +a^2)^2} dx $.


问:
Mathematical modelling of flu on an island

Given a flu on an island of 10000 people. Every day 15% of healthy people become infected and have a less illness (not require hospitalization), 12% of healthy people become infected and have a more illness (need to get hospitalized). At the same time, 12% of people with a less illness recover completely, and 15% go into the category of more ill patients. In the category of more ill patients, the situation is as follows: 20% go into the category of patients with a less form of the illness and 10% completely recover. Recovered patients may become infected again. At the initial time on the island, 500 patients were identified in a less illness of the disease and 100 patients in a more illness. Find mathematical of the number of patients in less and more illness with respect to time.

I formulated the problem by making four states: Infected, Less Illness, More Illness and Recovered. The transitions between them were modelled as part of the above problem statement. For getting the state of the model with respect to time, i am thinking of using the page rank algorithm to calculate the same. For less and more illness, state probabilities can be calculated at the initial time by the given data but for recovered and infected state how to calculate the same? Also for the transition probabilities would it be $\frac{1}{d_{i}}$ * percentage given in the problem statement where $d_{i}$ is the outdegree of node $i$ for the above model?

答: I have done it by using Markov chain but have a question should i consider damping factor in the above formulation while calculating the transition matrix?</p>


问:
Need help with proving $x_n = (−1)^n (n^2 + 3) / (n^2 + 5n − 1)$ is NOT a Cauchy sequence using only definition
The Cauchy Sequence is that: $\forall \epsilon > 0 \,\,\exists n_0 \in \mathbb{N}$ such that $\forall m>n \ge n_0 \,\,|x_m - x_n| \le \epsilon$. Let's negate it, we will get ∃>0 ∀0∈ℕ  such that ∃>≥0 |−|≥.

I have tried to estimate it from above, but could not take it to the end.

Body:
Well, we can feel that $x_{2k} \to 1$ while $x_{2k + 1} \to -1$ so for a large enough $N$ if we have $n > N; n$ even and $m > N; m$ odd we should have

$|x_n - x_m| \approx 2>  \epsilon$ for a small epsilon.

Let's get that in detail.

Let $\epsilon > 0$.  Suppose $2k, 2k+ 1 > N$. with

$|x_{2k} - x_{2k+1}| < \epsilon$.

That means $|\frac {(2k)^2+3}{(2k)^2+ 5(2k) + 1} -(-\frac {(2k+1)^2+3}{(2k+1)^2+ 5(2k+1) + 1})| =$

$|\frac {4k^2 + 3}{4k^2 + 10k + 1} + \frac {4k^2 + 4k +4}{4k^2 + 14k + 7}|< \epsilon$.

But $\frac {4k^2 + 4k + 4}{4k^2 + 14k +7} > \frac {4k^2 +3}{4k^2 + 14k +7}$ and $\frac {4k^2 + 3}{4k^2 + 10k + 1} > \frac {4k^2 +3}{4k^2 + 14k +7}$ so (assuming $k > \frac {N}2 > 0$) we have

$2\frac {4k^2 +3}{4k^2 + 14k +7} < |\frac {4k^2 + 3}{4k^2 + 10k + 1} + \frac {4k^2 + 4k +4}{4k^2 + 14k + 7}|< \epsilon$

So $\epsilon$ must be larger than $2\frac {4k^2 +3}{4k^2 + 14k +7}=2(1 - \frac {14k+4}{4k^2+14k + 7})> 2(1-\frac {14k}{4k^2}) = 2- \frac {14}{k}$.

So if $k > 14$ then $\epsilon > 1$ .

so for any $\epsilon < 1$  we can't have an $N$ where $n, m > N$ imply $|x_n - x_m| < \epsilon$ as taking any $2k, 2k+1 > \max(N, 28)$ we have $|x_n - x_m| > \epsilon$.

SO not cauchy.


答:
As Oiler has mentioned in the comments, setting $x=a\tan\theta$ is the best approach because we can use the identity $\tan^2\theta + 1 = \sec^2\theta$ to our advantage:
$$
(x^2+a^2)^2=(a^2\tan^2\theta+a^2)^2=(a^2\sec^2\theta)^2=a^4\sec^4\theta \, .
$$
Since $dx=a\sec^2\theta \, d\theta$, the integral becomes
\begin{align}
\int \frac{a\sec^2\theta \, d\theta}{a^4\sec^4\theta}&=\frac{1}{a^3}\int\frac{d\theta}{\sec^2\theta} \\
&= \frac{1}{a^3} \int \cos^2\theta \, d\theta \, .
\end{align}
From here you can use integration by parts or, more simply, use the identity
$$
\cos(2\theta)=2\cos^2\theta - 1 \, .
$$

Title: Find $\int \frac 1 {(x^2 +a^2)^2} dx $.

Body:
I am trying to integrate $\int \frac 1 {(x^2+a^2)^2} \ dx$.  The only thing that I can think to try is substitution, $u=x^2+a^2$ so that $\frac{du}{dx}=2x \Rightarrow du = 2x\ dx = 2\sqrt{u-a^2}\ du$ and then the integral becomes

$$ \int \frac{1}{u^2} (2\sqrt{u-a^2} \ du) $$

which does not seem productive.

I could try perhaps to separate with integration by parts, setting $u = \frac{1}{x^2 +a^2}$ and $dv = \frac{dx}{x^2+a^2}$.  Then we obtain $\frac{du}{dx} = -\frac{1}{(x^2+a^2)^2}(2x)$ and $v = \frac 1 a \tan^{-1}(x/a)$.  Then the integral becomes

$$ \begin{align*}
uv - \int v \ du &= \left(\frac{1}{x^2 +a^2}\right)\left(\frac 1 a \tan^{-1}(x/a)\right) - \int \left(\frac 1 a \tan^{-1}(x/a)\right) \left( -\frac{1}{(x^2+a^2)^2}(2x) \right) \ du
\end{align*}$$

but this also looks like it's headed nowhere good.  Advice?


问:
What would be the smallest/lowest Big-O set possible of these $2$ functions
Here i have $2$ functions and i was just wanting to know what the smallest big-oh set possible is for these functions, my guess would be the lowest possible of 1 would be the big-ph set $O(n^3)$ but was unsure if it may be $O(1)$, and for the 2nd function i was unsure if it was $O(n \log n)$ or $O(n)$

$\frac {n^3} 4 + 6n^2 – 10000$

$5n\log 2n + 6n$

Any help and explanation is appreciated.

答:
You have:

$\begin{align*}
    n^3 / 3 + 6 n^2 - 10000
      &= O(n^3) \\
    5 n \log 2 n + 6 n
      &= O(n \log n)
\end{align*}$

You prove the above and take it from here.


答:
Another thing you can do is to compute first
$$
\int\frac{\text{d}x}{a^2+x^2}
$$
and i'm sure you can compute it.
Then you can use integration by parts on this previous integral, that will make $\displaystyle \frac{1}{\left(a^2+x^2\right)^2}$ appear.

Title: Find $\int \frac 1 {(x^2 +a^2)^2} dx $.


问:
What would be the worst case for this algorithm
I was unsure what would be the worst case running time for this algorithm considering comparisons of values and assignments to variables would it be having max at n-1 element in the val array and min at n-2 element and having the array sorted in ascending order from element 0 to n-3?

This is the algoritm:

```<code>int max= -1; int min = 10000;
int [] val = {SOME_VALUES_FROM_0_TO_9999};
for (int i=0; i<val.length; i++)
 if (val[i]>max) max = val[i];
for (int i=0; i<val.length; i++)
 if (val[i]<min) min = val[i];
</code>```

答:
You are right with your idea that the order of the elements in $val$ effects the total running time costs of your program. However, it has no effect on its asymptotic running time complexity in terms of Big-O analysis.

For the sake of simplicity let us introduce something what we will call time unit.

Accessing an element in an array via index $i$ takes one time unit. Doing a single comparision (like $x < y$ or $x > y$) takes one time unit. An assignment statement (like $a=b$) takes one time unit. Adding two values ($a+b$) takes one time unit.

Now, let us proceed with analyzing the provided program step by step. The first two statements take two time units as they are simple assignments. Now, in the first for-loop we do a single assignment at the very beginning (int $i=0$) which takes one time unit. At each iteration we do one comparision and accessing one value (the length of the array) $i < val.length$ which together takes two time units (of course modern compilers will optimize this, but we do not care at the moment). Then we increment $i$, which is actually the same as $i = i + 1$, thus we add two values and assign them back to $i$. Together this takes two time units. So all in all for each iteration we need four time units.

The body of our loop does a comparision by accessing the value in $val$ at index $i$ and comparing it to the value stored in $max$. Thus, for each iteration we need two time units. Now, here comes the "tricky" part. How often do we enter the if-statement? Assuming the worst case, then $val$ is ordered in ascending order, hence we enter the statement in the first loop at each iteration. Hence, we need four time units for each iteration (two for the comparision, one for the assignment of $max$ and one for accessing the $i$th element of $val$ again). And this is set regardless what values are in $val$ (sorted integers in ascending order of course).

Now, consider that our array contains $n$ values. Then, we perform our loop $n$-times. Combining the time units, which we have examined above, yields to $2+1+n(2+2+2+2)=8n+3$.

Now, the second loop. Since $val$ only contains values in the range $[0 \cdots 9999]$ and $min=10000$ and our list $val$ is ordered in ascending order, we do exactly one assignment (and one array accesss) inside the second if-statement. So, instead of performing this $n$ times (as in the first loop), we only do this a single time. Hence our second loops takes $2 + n (2+2+2)=6n+2$.

We have a total cost of $8n+3+6n+2=14n+5$. However, in terms of Big-O analysis we of course have $14n+5 \in O(n)$.

Edit: Notice, that having alternating values (as you have suggested) does not effect this running time. Assume that $val$ is not ascending (meaning that it does not hold that $val[i] < val[i+1]$, then the $i+1$ time the first if-statement of the first loop will not be entered (because of the first assignment). Then, the second loop could be entered at $i+1$ instead. However, this would just sum up to the same running time.


答: If you do $x=ay$ and $\mathrm dx=a\,\mathrm dy$, your integral becomes$$\int\frac a{(a^2y+a^2)^2}\,\mathrm dy=\frac1{a^3}\int\frac1{(y^2+1)^2}\,\mathrm dy.$$On the other hand\begin{align}\int\frac1{(y^2+1)^2}\,\mathrm dy&=\int\frac1{y^2+1}\,\mathrm dy-\int\frac{y^2}{(y^2+1)^2}\,\mathrm dy\\&=\arctan(y)-\frac12\int\frac{2y}{(y^2+1)^2}y\,\mathrm dy\\&=\arctan(y)-\frac12\left(-\frac y{y^2+1}+\int\frac1{y^2+1}\,\mathrm dy\right)\\&=\frac12\arctan(y)+\frac12\frac y{y^2+1}\\&=\frac12\arctan\left(\frac xa\right)+\frac12\frac{ax}{x^2+a^2},\end{align}and therefore$$\int\frac1{(x^2+a^2)^2}\,\mathrm dx=\frac1{2a^3}\arctan\left(\frac xa\right)+\frac1{2a^2}\frac x{x^2+a^2}.$$

Title: Find $\int \frac 1 {(x^2 +a^2)^2} dx $.


问:
calculating this implicit function $\ln(x)(\log_{10}y)=1$
I'm working through this implicit function and would warmly appreciate your feedback.

$$\ln(x)(\log_{10}y)=1$$

my working out:

I take the derivative with respect to $x$ then $y$

$$\frac{1}{x}\log_{10}e\cdot\frac{1}{y}\frac{dy}{dx}=0$$

Then calculate for $\frac{dy}{dx}$

$$\frac{dy}{dx}=ln(e^x)\cdot y = xy?$$

答:
You need to apply the product rule.  For the sake of convenience, I will use $\ln$ to denote the natural logarithm, and $\log$ to denote the common logarithm.

$$\frac{d}{dx}[\ln x \log y] = \frac{d}{dx}[\ln x] \log y + (\ln x) \frac{d}{dx}[\log y] = \frac{1}{x} \log y + (\ln x)\frac{1}{y \ln 10} \frac{dy}{dx}.$$  Equating this to $0$ and solving for $dy/dx$ yields
$$\frac{dy}{dx} = -\frac{\log y}{x} \frac{y \ln 10}{\ln x} = - \frac{y \log y}{x \log x}. \tag{1}$$

You may also compute the explicit derivative: $1 = \ln x \log y$ implies $$y = 10^{1/\ln x} = \exp\left(\frac{\ln 10}{\ln x} \right),$$ hence $$\frac{dy}{dx} = \frac{d}{dx}\left[\frac{\ln 10}{\ln x}\right] \exp \left(\frac{\ln 10}{\ln x}\right) = -\frac{\ln 10}{x (\ln x)^2} \exp\left(\frac{\ln 10}{\ln x}\right) = -\frac{10^{1/\ln x} \ln 10}{x (\ln x)^2}. \tag{2}$$  I leave it as an exercise to demonstrate that Equations $(1)$ an $(2)$ are equivalent.


问:
Using Egorov's Theorem when the sequence of functions converges pointwise to infinity
I need help with the following question:

Suppose $(X, \mathcal{S}, \mu)$ is a measure space with $\mu(X) < \infty$. Suppose $f_1,f_2,\dots$ is sequence of $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$ such that $\lim\limits_{k \rightarrow \infty} f_k(x) = \infty$ for each $x \in X$. Prove that for every $\epsilon > 0$, there exists a set $E \in \mathcal{S}$ such that $\mu(X \setminus E) < \epsilon$ and $f_1,f_2,\dots$ converges uniformly to $\infty$ on $E$ (meaning that for every $t>0$, there exists $n \in \mathbb{Z}^+$ such that $f_k(x) > t$ for all integers $k \geq n$ and all $x \in E$).

I also have access to Egorov's Theorem which states the following:

Suppose $(X,\mathcal{S},\mu)$ is a measure space with $\mu(X)<\infty$. Suppose $f_1,f_2,\dots$ is a sequence of $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$ that converges pointwise on $X$ to a function $f: X \rightarrow \mathbb{R}$. Then for every $\epsilon>0$, there exists a set $E \in \mathcal{S}$ such that $\mu(X \setminus E) < \epsilon$ and $f_1,f_2,\dots$ converges uniformly to $f$ on $E$.

I think that the above question is just a direct application of the theorem, but that convergence to infinity messes me up a little bit. Would someone be able to help, please?

答:
HINT

Assume that $f_n \to +\infty$ pointwise.

Try mimicking Egorov's theorem proof for the sets $$A_{m,n}=\bigcup_{k=n}^{\infty}\{x:f_k(x) \leq m\}$$

https://en.wikipedia.org/wiki/Egorov%27s_theorem Here is a link to the proof,for the case of uniform convergence to a function.Mimic this proof.

Do the same ,when $f_n \to -\infty$ pointwise,using the sets $$A_{m,n}=\bigcup_{k=n}^{\infty}\{x:f_n(x)\geq-m\}$$


问:
Prove that $\left(\prod\limits_{2<p}^{p_i} \frac{p-1}{p}\right) \cdot \left( p_{i + 1}^2 - p_i^2 \right) \geq \pi(p_{i + 1}^2) - \pi(p_i ^2)$
Conjecture:

$$\left(\prod_{2
 \pi(p_{i + 1}^2) - \pi(p_i ^2) \tag{1}$$

The LHS expression is the product of Mertens's third theorem truncated at $p_i$ (subsequently just $M_x$ for truncation at $x$) times the length of the interval between the squares of $p_i$ and $p_i+1$, and the RHS is the actual number of primes between those squares. For instance, for $p_i = 7$, the expression is:

$$\frac{8}{35} \cdot \left(11^2 - 7^2\right) > \pi(121) - \pi(49)$$

And in fact $16.46 > 15$. Inequality $(1)$ is (perhaps surprisingly) empirically true for all $p_i$. The inequality seems intuitively likely to be true, as we would expect the result of $M_xx^2$ to be larger than $\pi(x^2)$ simply because $e^{-\gamma} > \frac12$ and therefore

$$\pi(x^2) \sim \frac{x^2}{\log x^2} = \frac12\frac{x^2}{\log x} < e^{-\gamma}\frac{x^2}{\log x} \tag{2}$$

(Inequality $(2)$ is empirically true only above $x \approx 100$, because the difference between $M_x \log x$ and $e^{-\gamma}$ is large for small $x$)

We can show (much more easily than I expected, in this question where I briefly failed basic algebra) that the density of primes between $p_{i-1}^2$ and $p_i^2$ is lower than the density of primes between $0$ and $p_i^2$, that is:

$$\frac{\pi\left(p_{i+1}^2) - \pi(p_i^2\right)}{p_{i+1}^2 - p_i^2} < \frac{\pi(p_i^2)}{p_i^2} \tag{3}$$

With this, we expect difference between the LHS and RHS of $(1)$ to be greater than $(2)$ alone would imply.

But... I'm roughly 110% certain that $(2)$ and $(3)$ together don't qualify as proof. They might be proof of asymptotic behavior as $p_i \to \infty$, but I'm wondering if this can be (or already has been!) proven for smaller values. Any links or thoughts from the community?

答:
$(2)$ and $(3)$ can be combined to form a proof of the "for large enough $i$'s" style and everything else can be checked with a computer program.
$$\frac{\pi\left(p_{i+1}^2) - \pi(p_i^2\right)}{p_{i+1}^2 - p_i^2} <
\frac{\pi(p_i^2)}{p_i^2}=
\frac{\pi(p_i^2)\cdot \log{p_i^2}}{p_i^2}\cdot \frac{1}{\log{p_i^2}}\leq ...$$
from PNT, for large enough $i$'s we have
$\frac{\pi(p_i^2)\cdot \log{p_i^2}}{p_i^2}\leq 1+\varepsilon$, thus
$$...\leq \frac{1+\varepsilon}{2\log{p_i}}=
\frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}\cdot \frac{e^{-\gamma}\cdot (1-\varepsilon)}{\log{p_i}}\leq ...$$
from Mertens' 3rd theorem, for large enough $i$'s we have
$\prod\limits_{p_i\leq p}\frac{p-1}{p}\geq\frac{e^{-\gamma}\cdot(1-\varepsilon)}{\log{p_i}}$, thus
$$...\leq \frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}\cdot\left(\prod\limits_{p_i\leq p}\frac{p-1}{p}\right) \tag{4}$$
What is left is to find a suitable $\varepsilon>0$ such that
$$\frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}<1 \iff
1+\varepsilon < \frac{2}{e^{\gamma}}\cdot(1-\varepsilon)$$
which works for $e\leq0.01$ (it's easy to see that $f(x)=\frac{1+x}{1-x}$ is ascending).


问:
What is the point of countable vs. uncountable sets?
I understand how to use these concepts and how to prove certain sets are countable or uncountable. However I don't get the point of it. What difference does it make whether a set is countable? People say that Cantor's proof that the real numbers are uncountable is a milestone of mathematics. Having read through the proof, I'm still struggling to understand what these ideas are important.</p>

答:
Sample a real number from $\mathbb{R}[0,1]$ as an infinite series of fair coin tosses.  Then, each sample is not only a refinement, also the start of a new sample.  The result being a rational may seem much less likely in the ergodic, but, sampling a rational sees sampling infinitely many copies of the terminus.  Sampling an irrational (in this manner) sees infinitely many different, thus that the irrational is represented less, of the balance that the rationals and irrationals are each dense in the reals.  It takes infinitely many fair coin tosses to sample a real number, that samples infinitely many, the sample is always plural.

Consider the function $\frac{n}{d}$ for naturals n,d with $0 \le n \le d$.  As $d \to \infty$, $lim_{n\to d} f = 1$.  Here $f$ ranges between zero and one and always includes zero and one.  Also, the differences $f(n+1)-f(n)$ are constant (uniform).  Then it is a CDF, uniform, with domain $\mathbb{N}$.

As that's in the countable it's a way to build notions of uniform probability from the natural integers, satisying various intuitions, and reasonably for application.  The point of the uncountable here is to see there are various ways to see what would be facts about these systems.

Matters of the infinite are relevant to mathematics, in the pure for what it is and the applied for what it is and does.

In terms of the applied and for practical purposes for real analysis:  additivity of measure is countable.  And, much work is done in formalizing analysis in the constructible and countable.

It's good to know because other people believe it, to understand.  But, except for its own sake, there aren't results solely due the uncountable.  Results not of the uncountable are available via asymptotics, methods of exhaustion, and concrete mathematics.  Here for example, point discontinuities can be discarded in solutions due their vanishing without appeal to their zero measure.  And, the computable can be described in the countable constructible, the concrete mathematics is quite heavy.  For example, given a wager whether a random integer is even or odd, the asymptotic density of even integers in integers is $\frac{1}{2}$, as an example of that the cardinalities of those two set being the same does not reflect their densities being different.

So, there is much intuition about the real numbers, not all satisifed with the uncountable.  However, it is quite standard and in the general curriculum not knowing the general features as instructed and thus shared would be ignorant, regardless their eventual disposition or lack thereof.

The ideas are important as an example of a way to consider the mathematically infinite.  It is important that their relevance is qualified, in their consideration.


答:
Well the $r-x^2$ screams that they want a trig substitution as lone student's answer suggest.

But you can always remove roots but bringing terms over and squaring.

$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$

$\sqrt{x}(a + b\sqrt{r^2-x^2}) = - d - c\sqrt{r^2-x^2}$

$x(a^2 + b(r^2-x^2) + 2ab\sqrt{r^2-x^2})) = d^2 + c^2(r^2 - x^2) - 2cd\sqrt{r^2-x^2}$

$x(a^2 + b(r^2-x^2))- d^2 - c^2(r^2-x^2) = -(2cd+2abx)\sqrt{r^2 -x^2}$

$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 = (2cd+2abx)^2(r^2-x^2)$ so

$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 -(2cd+2abx)^2(r^2-x^2) =0$

Which is a $6$th degree polynomial

which 1) Answers exactly what you asked and 2) makes things much, much, much worse.

Title: How to remove roots from an equation?

Body:
The question here is how (if it is even possible) to remove the square root terms and transform the following equation to a polynomial with one unknown $x$.
The coefficients $a$, $b$, $c$, and $d$ are known and also $r$.

$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$$


问:
Prove $\left\{u \# v | u,v \in \{0,1\}^{*} \text{ and } u \text{ is a substring of } v \right\}$ is not context free
I'm practising for my CS exam and got stuck on this problem

$\left\{u \# v | u,v \in \{0,1\}^{*} \text{ and } u \text{ is a substring of } v \right\}$

So I've tried using following pumping lemma:

Pumping Lemma would be: $0^n1^n\#0^n1^n$. v and x cannot contain #, otherwise when pumping more or less # appear and the new string is not in the language.

Now suppose $vwx$ are in the left $0^n1^n$, then $v^2$ and $x^2$ will make it longer than the R.H.S.;

However I'm not really sure how to proceed here, since u is a substring I believe the N on RHS must be lower than on LHS, am I going about it the right way or not?

答:
I’m assuming that your $n$ is the pumping length, and that you’re suppose that $0^n1^n\#0^n1^n$ can be decomposed as $uvwxy$ so that $|vwx|\le n$, $|vx|\ge 1$, and $uv^kwx^ky\in L$ for all $k\ge 0$, where $L$ is the language in question. This will work, but there are several cases to be considered.

If $vwx$ is contained in the first $0^n1^n$, taking $k=2$ gives you a word $z_1\#z_2$ with $|z_1|>|z_2|$, and such a word cannot be in $L$.

If $vwx$ is contained in the second $0^n1^n$, taking $k=0$ gives you a word that is not in $L$ for the same reason.

If the $\#$ is in $vx$, any $k\ne 1$ gives you a word that is not in $L$, since it does not have exactly one $\#$.

Those cover what you’ve already done.

Now show that the only other possibility is that $v=1^p$, $w=1^q\#0^r$, and $x=0^s$ for some $p,q,r,s\ge 0$ such that $p+q+r+s\le p$, and $p+s\ge 1$. Then split this into two case, $p\ne 0$ and $p=0$. In each case there is a way to choose $k$ so that $uv^kwx^ky\notin L$ for the same kind of reason that we’ve seen before.


问:
Evaluating a definite integral with e
Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

Title: Evaluating a definite integral with e

Body:
Do this:

$ u= 2x^3$ , then $du = 6x^2 dx$ . when $x = 0 $ we have $u = 0$. When $x=1$ we have $u=2$.
 Then

$$ \int_{0}^{1} \frac{x^2}{e^{2x^3}} dx =   \frac{1}{6}\int_{0}^{1} 6 \frac{x^2}{e^{2x^3}} dx$$

$$ \int_{0}^{2} \frac{1}{e^u} du = \int_{0}^{2}e^{-u} du =  - e^{-2} + e^{-0}$$


问:
I'm new to Calc 3 and I don't know the difference between ijk and <> for vectors
Title

I am quite sure that each have their uses but I am confused exactly when and how to use either form

i+j+k
or
<1,1,1>

Thanks

答:
Both ideas, if I read you right, are expressing vectors in three dimensions in terms of three basis vectors. In one case these are called $i, j, k$ and in the other $\langle 1,0,0\rangle; \langle 0,1,0\rangle; \langle 0,0,1\rangle$. The obvious mapping between the two systems is an isomorphism, the two systems are completely equivalent, and the notation to use is that which is most convenient.

Alternatively the notation your tutor uses and which might appear in your exam should be preferred.

You will find that there are other notations too (!) because vectors are versatile things and crop up in different contexts and the notations developed from the contexts before a general notation could be agreed. The second notation $\langle 1,0,0\rangle$ etc specifies components in a convenient way which is easier to generalise than the other, and is more readily evolved into the expression of linear mappings (with respect to a basis) as matrices.


问:
Sheldon Axler Measure Integration Real Analysis Section 2E Exercise 14
As you can tell from the title, this is an exercise from Axler's measure theory book, and I am struggling with the following problem:

Suppose $b_1,b_2,\dots$ is a sequence of real numbers. Define $f: \mathbb{R} \rightarrow [0,\infty]$ by
$$f(x) = \left\{\begin{array}{cc}
 \sum\limits_{k=1}^\infty \frac{1}{4^k|x-b_k|} & \text{ if } x \notin \{b_1,b_2,\dots\},\\
 \infty & \text{ if } x \in \{b_1,b_2,\dots\}
\end{array}\right.$$
Prove that $|\{x \in \mathbb{R} : f(x) < 1\}| = \infty$.

I am not sure what to do. Can someone help, please?

答:
Consider the sets
$$
E_k = \{x : |x-b_k| > 2^{-k}\},\qquad k=1,2,3\dots,\\
E = \bigcap_{k=1}^\infty E_k .
$$
If $x \in E_k$ then
$$
|x-b_k| > 2^{-k} ,\\
4^k|x-b_k| > 2^{k} ,\\
\frac{1}{4^k|x-b_k|} < 2^{-k} .
$$
If $x \in E$ then
$$
f(x) = \sum_{k=1}^\infty \frac{1}{4^k|x-b_k|} < \sum_{k=1}^\infty 2^{-k} = 1 .
$$
What is the measure of $E$?  To compute that, consider complements:
$$
E_k^c = \{|x-b_k| \le 2^{-k}\},\qquad |E_k^c| = 2\cdot 2^{-k} ,\\
|E^c| = \left|\bigcup_{k=1}^\infty E_k^c\right| \le \sum_{k=1}^\infty |E_k^c|
= \sum_{k=1}^\infty 2\cdot 2^{-k} < \infty ,
$$
and therefore $|E| = \infty$.


答:
${\bf i},{\bf j},{\bf k}$ is the standard basis of the 3d space, and the coordinates of the vector $a{\bf i}+b{\bf j}+c{\bf k}$ are $(a,b,c)$ with respect to the standard basis.

That ${\bf i},{\bf j},{\bf k}$ is a basis means that every vector in the space can be uniquely written in the above form for some real numbers $a,b,c$.

Title: I'm new to Calc 3 and I don't know the difference between ijk and <> for vectors

Body:
Title

I am quite sure that each have their uses but I am confused exactly when and how to use either form

i+j+k
or
<1,1,1>

Thanks


问:
How to linearize a constraint that containing square of sum of product?
I am working on an optimization problem. In order to solve it using CPLEX, we need to linearize the non-linear constraint stated in the following.

$$\left(\sum_{i,j \in A}\mu _{i,j}X_{i,j,t}\right) ^2 \ge \sum_{i,j \in A}\sigma_{i,j}^2X_{i,j,t}$$

Where $X_{i,j,t}$ is binary variable and all remain ones are parameters.

答:
One approach is to expand the square and linearize each resulting product of binary variables in the usual way.

Alternatively, just use the MIQCP solver in CPLEX.


问:
If lim inf $a_n$ = $\infty$ and lim sup $a_n$ = $-\infty$, $a_n$ diverges?
I know that a sequence is a convergent sequence if and only if it is a Cauchy sequence so lim inf $a_n$ = $\lim sup a_n$.

Then, if $\liminf a_n = \infty$, then can we conclude $\lim a_n=\infty$ ?

What about, if $\limsup a_n= -\infty$, then what can we conclude?

My understanding so far is, if $\liminf a_n=\infty$, then $\limsup a_n\neq \infty$, so $\lim a_n \neq \infty$.

I am not sure about the second if statement.

答:
If $\liminf_na_n=\infty$, then, since$$\liminf_na_n=\lim_n\inf_{k\geqslant n}a_k,$$you have that, for any $M>0$, $\inf_{k\geqslant n}a_k>M$ if $n$ is large enough; in particular, $a_n>M$ if $n$ is large enough. In other words, $\lim_na_n=\infty$.

A similar argument shows that if $\limsup_na_n=-\infty$, then $\lim_na_n=-\infty$.


问:
Proof that $1/\sqrt{x}$ is itself its sine and cosine transform
As far as I understand, I have to calculate integrals

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x \operatorname{d}\!x$$

and

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\sin \omega x \operatorname{d}\!x$$

Am I right? If yes, could you please help me to integrate those? And if no, could you please explain me.

EDIT: Knowledge of basic principles and definitions only is supposed to be used.

答:
First, use the change of variables $ y=\omega \,x $
$$I=\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x\, dx = \frac{1}{\sqrt{\omega} }\int_{0}^{\infty} \frac{1}{\sqrt{y}}\cos y\,dy\,.$$

Then, use the Mellin transform method  (using the tables)

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x) dx .$$

Now, the Mellin transform of $\cos(y)$ is $$ \Gamma  \left( s \right) \cos \left( \frac{\pi}{2} \,s \right) .$$

Then subs $s=\frac{1}{2}$, since $s-1=-\frac{1}{2}$, gives

$$ I =  \frac{1}{\sqrt{\omega}}\Gamma  \left( \frac{1}{2} \right) \cos \left( \frac{\pi}{4}  \right) . $$

You can do the same for the other one.


问:
Universal Algebra & Free Algebra
Q: Let $f$ be the embedding (injective homomorphism) of $A$ into $Str(A): a \in A$ is sent to the corresponding string of length $1$.

Prove that $(f; \langle Str(A), \hat{}, ()\rangle )$ is the free monoid where $\hat{} $ is an associative binary operation with identity $()$ on $A$.

Notes:

i) Question is in the setting of Universal Algebra.

ii) $\hat{}$ is concatenation of string, i.e. attach two strings together.

iii) I have already shown $\langle Str(A), \hat{}, () \rangle$ is a monoid, only need to show $(f; \langle Str(A), \hat{}, ()\rangle )$ is a free algebra.

答:
You state you have shown it is a monoid,

So you have shown $\hat{}$ is associative and that the structure has identity which you denote as $()$, seemingly the empty string

This is related to Kleene star

Appears all that is left is to show that the elements of $A$ under $f$ are free generators for $(A \text{ under } f)^*$ (the Kleene star of $A$ under $f$ and since $f$ embeds $A$ into $Str(A)$, we have $f(a) = a$ which is the string of length 1 as you say,
we can take $A = \{a_1,...a_n\} =\{f(a_1), \ldots, f(a_n)\} $ as your free-generators.

Perhaps $A$ is not finite by assumption which is fine, I just use a finite set to be illustrative in my explanation.

Title: Universal Algebra & Free Algebra


问:
Asymptotics of $\sum_{i=0}^n e^{-i^2/n + i^a}$ as $n \to \infty$
To what function is $s(n) = \sum_{i=0}^n e^{-i^2/n + i^a}$ with $a \in ]0,1]$ asymptotic (as $n \to \infty$)? or what are relatively tight lower and upper bounds?

Here is what I tried so far: go over to a "Riemann sum"
$$
n \sum_{i=0}^n \tfrac{1}{n} e^{ -n (i/n)^2 + n^a (i/n)^a}
\approx n \int_0^1 e^{ -n x^2 + n^a x^a}
$$
when $a = 1$ this integral is actually easy to evaluate (it's asymptotic to $\sqrt{\pi} \cdot e^{n/4}/\sqrt{n}$) and gives a very tight bound for the asymptotics of the sum.

When $a<1$, I used the following estimates: $-n x^2 + n^a x^a >0 \iff 0< x < n^{(a-1)/(2-a)}$. The maximum of the exponent happens when $-2nx + an^ax^{a-1} =0 \iff x = (\tfrac{a}{2})^{1/(2-a)}n^{(a-1)/(2-a)}$. This gives a maximal value of the exponent of the integrand
$$
-n (\tfrac{a}{2})^{2/(2-a)}n^{(2a-2)/(2-a)}  + n^a (\tfrac{a}{2})^{a/(2-a)}n^{(a^2-a)/(2-a)}
=
\bigg( -(\tfrac{a}{2})^{2/(2-a)} + (\tfrac{a}{2})^{a/(2-a)} \bigg)n^{a/(2-a)}
=
( 1-\tfrac{a}{2}) (\tfrac{a}{2})^{a/(2-a)} n^{a/(2-a)}
$$
Up to a mistake of $1$, one can ignore the part of the area of the integral which is the region of the unit square.
There remains at most a rectangle with dimensions $n^{(a-1)/(2-a)} \times \mathrm{exp}\bigg( ( 1-\tfrac{a}{2}) (\tfrac{a}{2})^{a/(2-a)} n^{a/(2-a)} \bigg)$.

Hence $s(n) \leq n^{1/(2-a)} \mathrm{exp}\bigg( ( 1-\tfrac{a}{2}) (\tfrac{a}{2})^{a/(2-a)} n^{a/(2-a)} \bigg)$ (for large $n$).

The lower bound [for $n$ large] $s(n) \geq \mathrm{exp}\bigg( ( 1-\tfrac{a}{2}) (\tfrac{a}{2})^{a/(2-a)} n^{a/(2-a)} \bigg)$ is also not too hard to get (just restrict the sum to the a few values of $i$ so that $i/n \approx (\tfrac{a}{2})^{1/(2-a)}n^{(a-1)/(2-a)}$).

Numerics indicate that $s(n) \approx K n^{1/2} \mathrm{exp}\bigg( ( 1-\tfrac{a}{2}) (\tfrac{a}{2})^{a/(2-a)} n^{a/(2-a)} \bigg)$ for some $K \in [1,2]$

答:
I suppose that you can also evaluate $K \in [1,2]$.
Consider the power of exponent near the maximum: $$e^{f(k)}=e^{f(k_{max})+\frac{1}{2}f''(k_{max})(k-k_{max})^2+...}$$
It is supposed that other terms of the series in the power of the exponent can be omitted (do not contribute into the main asymptotics term). In many cases such approximation works well.

$$f(k)=-\frac{k^2}{n}+k^a$$
$$k_{max}=(\frac{a}{2}n)^\frac{1}{2-a}$$
$$ f(k_{max})=\Bigl((\frac{a}{2})^\frac{a}{2-a}-(\frac{a}{2})^\frac{2}{2-a}\Bigr)*n^\frac{a}{2-a}$$
$$f''(k)=-\frac{2}{n}+a(a-1)k^{a-2}$$
$$f''(k_{max})=-\frac{2}{n}+a(a-1)(\frac{an}{2})^\frac{a-2}{2-a}=-\frac{2}{n}(2-a)$$

$$S(n)=\sum_{k=0}^n e^{-k^2/n + k^a}\sim{e}^{\Bigl((\frac{a}{2})^\frac{a}{2-a}-(\frac{a}{2})^\frac{2}{2-a}\Bigr)n^\frac{a}{2-a}}\sum_{k=0}^n e^{\frac{1}{2}f''(k_{max})(k-k_{max})^2}$$

Switching from summation to integration and expanding integration to $\pm\infty$ we get the main asymptotics term:
$$S(n)\sim{e}^{\Bigl((\frac{a}{2})^\frac{a}{2-a}-(\frac{a}{2})^\frac{2}{2-a}\Bigr)n^\frac{a}{2-a}}\int_{0}^n{e}^{-\frac{2-a}{n}(k-k_{max})^2}dk$$
$$S(n)\sim{e}^{\Bigl((\frac{a}{2})^\frac{a}{2-a}-(\frac{a}{2})^\frac{2}{2-a}\Bigr)n^\frac{a}{2-a}}\sqrt\frac{\pi{n}}{2-a}$$

You can check the expression at $a=1$.

$\mathsf{Supplement}$

We can also dig a bit deeper - to get an idea about the limitations of the method.
$$F(k)=\exp(-\frac{k^2}{n}+k^a)$$
$$F'(k)=(-\frac{2k}{n}+\frac{a}{k^{1-a}})\exp(-\frac{k^2}{n}+k^a)$$
$k_{max}=(\frac{a}{2}n)^\frac{1}{2-a}=\frac{n}{2}$ at $a=1$, so at $a=1$ the function is exactly the Gauss curve with a maximum at $k=\frac{n}{2}$.
At $a\to0$ $k_{max}=(\frac{a}{2}n)^\frac{1}{2-a}\to\sqrt{\frac{a}{2}n}$ and is shifting to the left from the center of the interval $(0,n)$. The function becomes asymmetrical, and at $a=0$ we end up with a half of Gauss curve with $k_{max}=0$.
So we may suppose that the method works as soon as $a$ is not too close to zero.

In order to omit higher powers in  the expansion $f(k)=f(k_{max})+\frac{1}{2}f''(k_{max})(k-k_{max})^2+\frac{1}{6}f'''(k_{max})(k-k_{max})^3+...$

there should be $|f'''(k_{max})|<<|f''(k_{max})|$, or $\frac{2}{n}>>a(1-a)(\frac{an}{2})^{\frac{a-3}{2-a}}$

We consider the case $a<<1$, so the limitation is $\frac{2}{n}>>a(\frac{an}{2})^{-\frac{3}{2}}$, or $\sqrt{na}>>1$. We can check that when this requirement is met $f^{(l+1)}(k_{max})<<f^{(l)}(k_{max})$ for all derivatives.

Therefore the main asymptotics term
$$S(n)\sim{e}^{\Bigl((\frac{a}{2})^\frac{a}{2-a}-(\frac{a}{2})^\frac{2}{2-a}\Bigr)n^\frac{a}{2-a}}\sqrt\frac{\pi{n}}{2-a} (n>>1, \sqrt{na}>>1)$$


问:
Proving certain properties of limits of sequences using the definition of limits
I have been trying to prove the following properties for limits of sequences:

Let $b \neq 0$. If $\lim x_n = a$ and $\lim \frac{x_n}{y_n} = b$, then $\lim y_n = \frac{a}{b}$.

Let $a \neq 0$. If $\lim x_n = a$ and $\lim x_ny_n = b$, then $\lim y_n = \frac{b}{a}$.

I know that these are straightforward to prove using other elemental properties of limits of sequences, but I want to prove them using the definition and I have been not able to do it.

So far I have, for (1):

Let $\epsilon > 0$. We know that:

i)$\exists n_1 \in \mathbb{N}: n >n_1 \implies |x_n - a| < \epsilon$

ii)$\exists n_2 \in \mathbb{N}: n >n_2 \implies |x_n/y_n - b| < \epsilon \implies |x_n - by_n| < |y_n|\epsilon$

Now let $n_0 = \max\{n_1, n_2\}$. Then, for $n > n_0$:

\begin{aligned}
|y_n - a/b| &= |1/b||by_n-a| \\
&= |1/b||by_n + x_n - x_n - a| \\
&\leq |1/b|(|x_n - by_n| + |x_n - a|) \\
&<|1/b|(|y_n| + 1)\epsilon
\end{aligned}

which doesn't work, since it depends on the value $|y_n|$, so what I need to prove is that $|y_n|$ is bounded above by an $M > 0$. I know that both $x_n$ and $x_n/y_n$ are bounded since they are convergent, but I can't prove the last step that $y_n$ is bounded. Thanks in advance!

答:
You're definitely on the right track.

Because $x_n/y_n \to b$ and $b$ is nonzero, there is a positive integer $W$ so that whenever $n \geq W$ one has $|x_n/y_n| \geq |b|/2$.

(Please note, this is a step away from the "epsilon-N" definition of the assumed limit, but you get there by putting $|b|/2$ in for the "epsilon" in that definition and manipulating a few inequalities to see that $|x_n/y_n - b| < |b|/2$ implies $|x_n/y_n - 0| \geq |b|/2$.  This is also clear from a picture of the number line.  Anyway, you'll later want to toss $W$ into that $\max$ that you've got going.)

For all $n \geq W$ it follows that $|y_n| \leq 2 |x_n|/|b|$.

And if you know that $x_n$ that is bounded...


问:
What are some examples of induction where the base case is difficult but the inductive step is trivial?
According to Wikipedia:

...proofs by mathematical induction
  have two parts: the "base case" that
  shows that the theorem is true for a
  particular initial value such as n = 0
  or n = 1 and then an inductive step
  that shows that if the theorem is true
  for a certain value of n, it is also
  true for the value n + 1. The base
  case is often trivial and is
  identified as such, although there are
  cases where the base case is difficult
  but the inductive step is trivial.

What are some examples of proofs by induction where the base case is difficult but the inductive step is trivial?

答: Define the function $E(x)=\displaystyle\sum_{i=0}^\infty\frac{x^n}{n!}$ (you can forget about the convergence issue for this problem). To prove the fact that $E(mx) = E(x)^m$ for all integers $m$, you need a relatively straightforward induction step. The more laborious part is proving that $E(x+y)=E(x)E(y)$, which allows for the induction. I got this example from "How We Got from There to Here: A Story of Real Analysis" by Eugene Bowman and Robert Rogers.


问:
Question about dense subsets
How can we show that for the topological space $X$ that $X$ is the only dense subset of $X$.

I feel as if this is true, and the definition I have of being dense is as follows: Let $X$ be a space, and a subspace $D \subseteq X$ is dense if and only if $\overline{D} =X$.

If we want to show that $X$ is the only dense subset of $X$ here is what I have thought up:

Since we have that $X = \mathbb{R}$, we can pick $D=\mathbb{R}$. Then clearly the closure of $D$ is still $\mathbb{R}$ which is $X$. So we would have that $X = \overline{X}$, but this doesn't specify that this is the $only$ dense subset of $X$. What am I missing. I don't want any solutions here, I just want more of an understanding of what's going on.

答:
If $X$ has a non-isolated point $x$, so that $\{x\}$ is not open, then $X\setminus \{x\}$ is dense and a proper subset of $X$.

So the only spaces where all dense subsets must be $X$ are the discrete ones.

Title: Question about dense subsets


问:
A non-empty subset of integers bounded above has a maximum
Suppose the set $\mathrm A$ $\neq$ $\emptyset$ , $\mathrm A$ $\subseteq$ $\Bbb Z$ is bounded above. Then since $\Bbb Z$ $\subseteq$ $\Bbb R$, I know that by the completeness axiom there exists a $supremum$ for the set $\mathrm A$, say $s$ $=$ $sup$($\mathrm A $). But, I need to show this is in fact the maximum of this set $\mathrm A$. For that, I know it has to be an element of the set $\mathrm A$. How can we show that $s$ $\in$ $\mathrm A$ ?

答:
Let non-empty subset of integers be B.

$\mathtt \enspace Assume \enspace that \enspace B \enspace has \enspace no \enspace maximum.$

B is non-empty and B is bounded above → B has a supremum. (By the completeness axiom).
Let sup(B)=s.

So $$∀x ∈ B , s>x$$(s can't be equal to x.Because if s=x then x will be the maximum).

Also $$∀ϵ>0, ∃x_1 \enspace s.t\enspace x_1 + ϵ > s → x_1 > s - ϵ$$
So $$∃x_2 ∈ B \enspace s.t\enspace x_2>x_1 $$ (Because B has no maximum. So there can be element higher than $x_1$).

So $$s> x_2 > x_1 > s- ϵ $$

$\mathtt Let  \enspace ϵ = 1.$ Then
$$s> x_2 > x_1 > s- 1$$

So observe this inequality. You can see that there are two integer numbers s.t there's difference is lower than 1(Two integers difference must be equal to 1 or higher). So that can't happen. So it is a contradiction. So our assume is false. Because of that B has a maximum element.


问:
Linear Integer Programing: fill the package with products $A, B, C, D$
A linear integer programming problem ask to consider the next restriction: we want to fill as much as possible a package that has a capacity of $1m^3$ and we have to choose between a variety of products $A, B, C, D$ with the following dimensions:

Product
Dimensions (cm)

A
$8\times 2.5 \times 0.5$

B
$7\times 2 \times 0.5$

C
$3\times 2 \times 0.5$

D
$3\times 3 \times 0.5$

My try:
$$8x_1+7x_2+3x_3+3x_4\leq100$$
$$2.5x_1+2x_2+2x_3+3x_4\leq100$$
$$0.5(x_1+x_2+x_3+x_4)\leq100$$
With $x_1,...,x_4\in\mathbb{Z}^+$.

But I think I'm not considering combinations between different dimensions. Any suggestions would be great!

答: The idea you are attempting is correct for one dimension but not for three dimensions.  I suggest considering a much smaller problem in two dimensions to gain some insight: two products $A$ (dimensions $2 \times 1$) and $B$ (dimensions $1 \times 1$) in a $2 \times 2$ package.


问:
What is the decomposition of $[G/H]$ in terms of irreducible representations?
Let $G$ be a finite group and $$ be a field. Any $G$-set can be linearized to give a $$-representation of $G$. Each $G$-set is decomposed into a coproduct of indecomposable (transitive) $G$-sets $G/H$ for $H ⊆ G$ a subgroup. For $H=1$, the decomposition of the linearization of $G/H = G$ (the regular representation) contains exactly $\dim(V)$ copies of each irreducible representation $V$. What can be said in the more general case where $H$ is any subgroup?

PS: I'm sorry if it's a standard question but I don't really know what to call this representation $[G/H]$.

答:
This is simply a hard question in general. As an example, let $G=S_n$ and $H = S_{\lambda_1} \times S_{\lambda_2} \times \cdots \times S_{\lambda_k}$ where $\lambda_1 + \cdots + \lambda_k = n$. The multiplicities of the irreducible decomposition of $\mathbb{C}[G/H]$ are the Kostka numbers. When $H = C_a \wr S_b$ viewed as a subgroup of $S_{ab}$, the multiplicities are a generalization of Thrall's problem, which has been open for 80 years.

There's the character-theoretic calculation

$\langle V, \mathbb{K}[G/H]\rangle_G = \frac{1}{|G|} \sum_{g \in G} \chi_V(g) \chi_{\mathbb{K}[G/H]}(g^{-1}) = \frac{1}{|H|} \sum_{h \in H} \chi_V(h).$

When $H=1$ this gives the regular representation answer you mentioned. When $H=G$, applying the orthogonality relations gives the expected result.

Continuing from the final equation gives an instance of Frobenius reciprocity,

$$\langle V, \mathbb{K}[G/H]\rangle_G = \langle V|_H, 1_H\rangle_H.$$

That is, your question is equivalent to determining the number of copies of the trivial representation in the restrictions of the irreducible representations of $G$ to $H$. Questions along these lines are known as "branching rules".


问:
Isometries of a hyperbolic quadratic form
I am reading an article that says "The group of isometries (of a hyperbolic space) of a hyperbolic quadratic form in two variables is isomorphic to the semi-direct product $\mathbb{R} \rtimes \mathbb{Z}/2\mathbb{Z}$".

Could someone help me to understand that fact? First, it seems that the semi-direct product $\mathbb{R} \rtimes \mathbb{Z}/2\mathbb{Z}$ is equal to the direct product $\mathbb{R} \times \mathbb{Z}/2\mathbb{Z}$. And, what means a hyperbolic quadratic form? Is it simply a quadratic form defined in a hyperbolic space?

Thanks a lot.

答:
A hyperbolic quadratic form is a quadratic form such that the matrix, in a suitable basis, is

$$
A =
 \left(  \begin{array}{rr}
  0  &  1       \\
  1  &  0
\end{array}
  \right).
  $$
A two dimensional vector space over $\mathbb R$ with this bilinear form is often called a hyperbolic plane. Given two column vectors $x,y$ the value of the bilinear form is just
$B(x,y) = y^t A x.$ The associated quadratic form is $q(x) = x^t A x = B(x,x).$ We are using this way of writing the quadratic form as in page 13 of GERSTEIN and page 31 of GROVE. The result, for fields not of characteristic 2, is $$ B(x,y) = \frac{q(x+y) - q(x) - q(y)}{2} $$

Let's see, you did not type the result correctly. The orthogonal group, or group of isometries, or automorphism group, or group of automorphs of $q$ are those matrices $T$ of deteminant $\pm 1$ such that
$$ T^t A T = A,   $$ page 26 of Gerstein.

You can work this part out yourself: the isometries with positive determinant, also called the special orthogonal group or the rotation group, are
 $$
T =
 \left(  \begin{array}{rr}
  a  &  0       \\
  0  &  \frac{1}{a}
\end{array}
  \right)
  $$
for nonzero $a.$

The isometries with negative determinant are
 $$
T =
 \left(  \begin{array}{rr}
  0  &  b       \\
   \frac{1}{b} & 0
\end{array}
  \right)
  $$
for nonzero $b.$

The thing you missed in the statement is that it is not $\mathbb R$ being used, it is the multiplicative group $\mathbb R^\ast$ of nonzero reals. Otherwise you have it. This is Theorem 6.1 on page 45 of Grove, or example 2.31 on page 27 of Gerstein.

The two other books I would recommend here are LAM and my favorite (because of number theoretic information) CASSELS

==================================

page 45 from Grove. If anyone knows how to crop the image, please feel free.

===================================


问:
Formal proof of ($A$ and ($A$ implies $B$)) implies $B$?

$$(A ∧ (A → B)) ⇒ B$$

I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks

答:
A semantic reasoning ( not sure it counts as a " formal" proof)

Suppose the antecedent does not imply logically the consequent.

It means there is ( at least) one interpretation inwhich the antecedent is true and the consequent false.

Hence, an interpretation in which $B$ is false and $( (A\land (A\rightarrow B))$ is true ( meaning is particular that $A$ is true).

The only way to have $(A\rightarrow B)$ true is not to have both  $A$ true and $B$ false.

But the only way to have this when you alredy have $B$ false is not to have $A$ true, hence to have  $A$ false.

By what precedes,however,  you have $A$ true.

This shows that there is no possible  interpretation that falsifies the original material conditional, and, consequently, that the antecedent logically implies the consequent.


问:
Find $\int \frac{e^{-x}\sin x}{x} \ dx$.
I'm trying to compute $\int \frac{e^{-x}\sin x}{x} \ dx$.  This resembles integrating $\int e^{-x}\sin x \ dx$ so my main thought about how to approach this is by finding some connection with that.  I know that $\int e^{-x}\sin x \ dx = -\frac 1 2 e^{-x}(\sin x + \cos x)$. If I try to use integration by parts setting

$$u=x^{-1}$$

and

$$dv=e^{-x}\sin x \ dx$$

then we get

$$\frac{du}{dx}=-\frac{1}{x^2}$$

and also

$$v=-\frac 1 2 e^{-x}(\sin x + \cos x)$$

Then the integral becomes

$$ \int u \ dv = uv-\int v\ du = x^{-1}\left(-\frac 1 2 e^{-x}(\sin x + \cos x) \right) - \int \left( -\frac 1 2 e^{-x}(\sin x + \cos x) \right) \left( -\frac{1}{x^2} \right) \ dx $$

I seem to have made things worse.  I am sorry.

If I try to make any other assignment of $u$ and $dv$ in integration by parts, I don't foresee them going any better.  No substitution seems helpful, either $u$ or trig.  I don't think I can integrate twice and get back to where I started.  That almost seems possible if I take what I have done above and now choose $\frac 1 {x^2}$ for integration, except that the derivative part will now have much more in it than I started with.  I think my bag of tricks may be empty.

答:
Since no one is going to post an answer I will prove that
$$I = \int \frac{e^{-x}\sin x}{x}\,\mathrm{d}x = -\frac{\mathrm{i}\left(\operatorname{Ei}\left(\left(\mathrm{i}-1\right)x\right)-\operatorname{Ei}\left(-\left(\mathrm{i}+1\right)x\right)\right)}{2}+C,$$
where $\operatorname{Ei}$ is the expontial integral and $C$ is an arbitrary real constant.

Lemma. Let $z$ be a complex number. Then
$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}.$$
If we rewrite the sine in the integral, we may add exponents and use linearity of the integral to get to
$$I = \frac{1}{2i}\int \frac{e^{ix-x}}{x}\,\mathrm{d}x-\frac{1}{2i}\int \frac{e^{-ix-x}}{x}\,\mathrm{d}x.$$
Substitute in the first integral $u_1=(i-1)x$ and in the second one $u_2=-(i+1)x$.
$$\frac{1}{2i}\operatorname{Ei}(u_1)-\frac{1}{2i}\operatorname{Ei}(u_2).$$
Substituting $u_{1,2}$ back yields
$$\frac{\operatorname{Ei}((i-1)x)-\operatorname{Ei}(-(i-1)x)}{2i}+C.$$
The last step is a matter of taste. You can write $\tfrac{1}{i}$ as $-i$, so
$$\frac{-i(\operatorname{Ei}((i-1)x)-\operatorname{Ei}(-(i-1)x))}{2}+C.$$


问:
Evaluating $a_j$ sum given the $n$-th partial sum. Not working.

Evaluate the sum $$\sum_{j=1}^\infty a_j$$ (or determine that it diverges) given the formula for its $n$-th partial sum $$S_n = \frac{n^2-4n+5}{n^3+7n-9}.$$

There are $2$ other $S_n$ given and both I am able to get the $a_j$ term by using $a_j = S_n - S_n-1$,
however when I attempted to get the $a_j$ for this $S_n$ I got a nasty looking poly that I can't seem to figure out what to do with. I tried to take the integral and the limit.

$$a_j=\left(\frac{n^2-4n+5}{n^3+7n-9}\right)-\left(\frac{\left(n-1\right)^2-4\left(n-1\right)+5}{\left(n-1\right)^3+7\left(n-1\right)-9}\right)=\frac{-n^4+10n^3-21n^2-6n+5}{\left(n^3+7n-9\right)\left(n^3-3n^2+10n-17\right)}.$$

答: $\sum a_j$ is by definition just the limit of $S_n$ as $n \to \infty$. Dividing numerator and denominator by $n^{2}$ we can write $S_n$ as $\frac {1-4/n+5/n^{2}} {n+7/n-9/n^{2}}$ The numerator tends to $1$ and the denominator tends to $\infty$ and the limit is $0$.


问:
Lower bound on $P(X \geq \theta E [X] )$
Let $X$ be a non-negative random variable and $\theta \in [0,1]$. Show that:

$$\mathbb P\left(X \geq \theta \, \mathbb E X\right ) \geq (1-\theta)^2\frac{\mathbb E [X]^2}{\mathbb E[ X^2] - \theta(2-\theta)\mathbb E[X]^2}$$

What I've tried

This is very similar to the Paley-Zygmund inequality (in which case the denominator is missing the second term). I don't think the proof for PZ is adaptable here (unless we can apply something stronger than Cauchy-Schwarz):

$$\mathbb E X = \mathbb E X \mathbb 1(X < \theta \mathbb E X) + \mathbb E X \mathbb 1(X\geq \theta \mathbb E X) \leq \theta \mathbb E X + \sqrt {\mathbb E X \mathbb P(X \geq\theta \, \mathbb EX)}$$

Where the inequality is from cauchy-schwarz. Simplifying gives the PZ inequality. If I can get a tighter bound on the second term then I should be able to show the inequality of interest.

Any tips would be appreciated.

答:
Note that your denominator is exactly $\mathbb{E}[(X - \theta \mathbb{E}[X])^2]$. So we have the following:

$$
\begin{aligned}
(1 - \theta) \mathbb{E}[X] &= \mathbb{E}[X - \theta E[X]] \leq \mathbb{E}\big[(X - \theta \mathbb{E}[X]) 1\{X - \theta \mathbb{E}[X] \geq 0\}\big] \\
&\overset{\text{(C.S.)}}{\leq} \sqrt{\mathbb{P}(X \geq \theta \mathbb{E}[X])} \sqrt{\mathbb{E}[(X - \theta \mathbb{E}[X])^2]}
\end{aligned}
$$

Rearranging will give you the result.


问:
The Lebesgue Monotone Convergence theorem on function domains
I am trying to prove an analogous statement of the Lebesgue Monotone Convergence theorem on function domains.

Let $A \subset \mathbb{R}^n$ be a bounded set that has volume. Let $f : A \mapsto \mathbb{R}, f \ge 0$ be an unbounded function. Suppose $C_i$ is a sequence of compact sets with volume, $C_i \subset A$ with $C_i$ increasing to $A$. Then $f$ is integrable iff $f$ is integrable on each $C_i, \lim_{i \to \infty} \int_{C_i} f$ exists, and $\int_A f = \lim_{i \to \infty} \int_{C_i} f$.

This is intuitively clear. If $f$ is integrable on each $C_i$ and $\lim_{i \to \infty} \int_{C_i} f = I$, then I should be able to find $N_1, N_2$ such that $vol(A)-vol(C_i) < \epsilon$ if $i > N_1$ and $|\int_{C_i} f - I| < \epsilon$ if $i > N_2$ for some $\epsilon > 0$. Then take $N = max\{N_1,N_2\}$, I should be able to show $f$ is integrable and $|\int_A f - I| < \epsilon$. But I got stuck in trying to work out the details of partitions.

Conversely, if $f$ is integrable, I should be able to apply Darboux's theorem and show that $|\int_{C_i} f - \int_A f| < \epsilon$. But I think Darboux's theorem only applies to bounded functions, and I don't know that how to use the condition that each $C_i$ is compact. It would be nice if $f$ is continuous, then $f$ would be bounded on each $C_i$, but the proposition does not mention this either. Can someone help with these details or give me a hint? Thanks a lot!

答: Let $f_i=f\chi_{C_i}$. Then $(f_i)$ is a sequence of non-negative measurable functions increasing to $f$. By Lebesgue's Monotone convergence Theorem $\int_A f=\lim \int_A f_i=\lim \int_{C_I} f$ (and this does not require any further hypothesis like integrability, boundedness etc). The conclusion follows immediately from this.


问:
Are Ramsey numbers defined for $n=0$?
I'm familiar with the trivial case $R(1,a)=1$, as there are no edges to colour.

Is $R(0,a)$ defined at all? Or would it similarly be trivially 0 since there are no edges and no vertices?

答: It would be trivially $0$. The goal of Ramsey Theory is to study how large an ambient space (e.g., a graph) needs to be in order for prescribed substructures (e.g., subgraphs) to appear. So in that sense, it doesn't make much sense to talk about $R(0, a)$.

Title: Are Ramsey numbers defined for $n=0$?


问:
Matrix change of basis, versus vector
I am under the impression, that changing the basis of a vector involves multiplication by a matrix:

$$
\mathbf{x}'=\mathbf{A}\mathbf{x}
$$

where $\mathbf{x}',\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}\in\mathbb{R}^{n\times n}$.

But the same concept of a change of basis over a matrix involves the multiplication be a similarity transformation;

$$
\mathbf{A}'=P\mathbf{A}P^{-1}
$$

Is this correct? Why not $\mathbf{A}'=\mathbf{B}\mathbf{A}$?

The reason I am asking about the difference, is when it comes to integrating the determinant of a matrix, say

$$
1=\int \det \mathbf{M}(x,y)dxdy
$$

To perform a change of coordinates, such that the integral remains invariant, must one write it as such:

$$
1=\int \det \mathbf{M}(x,y)\sqrt{|g|}dxdy
$$

How then is the transformation done with both a similarity transformation and a matrix-multiplication vector transformation? Must I do both $M'=PMP^{-1}$ and a coordinate transformation $(x,y)\to (f(x,y),g(x,y))$ for instance:

$$
M'=PM(f(x,y),g(x,y)) P^{-1}
$$

Is the function

$$
1=\int \det \mathbf{M}(x,y)\sqrt{|g|}dxdy
$$

invariant with respect to arbitrary coordinate changes?

答:
To answer your first question:

This 'similarity' transformation for a matrix arises because (!) of the way a vector is transformed. Consider a vector $x \in \mathbb{R}^n$ and a matrix $A \in \mathbb{R}^{n\times n}$.

Now, consider a basis transformation, such that $x' = Bx$ with $B \in \mathbb{R}^{n \times n}$. The matrix $A$ is defined with respect to the old basis, meaning that the operation $A x'$ will not(!) yield the desired result. To obtain your desired result, you transform your vector into its original form, i.e. $(B^{-1}x')$ and THEN do the matrix multiplication $A (B^{-1} x')$. However, the resulting vector will again be with respect to the OLD basis, therefore you transform it such that it is described with respect to the NEW basis. The whole operation thus looks like

$$ B \cdot A \cdot (B^{-1} x') = (BAB^{-1})x' $$

Consequently, your matrix with respect to the NEW basis is of the form $A' = (BAB^{-1})$.

$${}$$

A remark to the second question:

Under a change of basis $A' = BAB^{-1}$, the determinant of the matrix $A'$ transforms as follows:

$$ \det A' \enspace = \enspace \det \big( B A B^{-1} \big) \enspace = \enspace \det B \cdot \det A \cdot \det B^{-1} \enspace = \enspace \det B \cdot \det A \cdot \frac{1}{\det B} \enspace = \enspace \det A $$

The determinante is therefore invariant under a change of basis.

Title: Matrix change of basis, versus vector


问:
Counting ways when first position is covered with a tie

Six horses are entered in a race. If two horses are tied for first
place and there are no ties among the other four horses, in how many
ways can the six horses cross the finish line?

My approach:

We choose $2$ out of $6$ for the first place and $4$ out of the remaining $4$:

$$_6P_2\cdot _4P_4 = \frac{6!}{(6 -2)!}\cdot\frac{4!}{(4 - 4)!}=720$$

But the solution states $360$.

What am I doing wrong here?

答:
There are 6 choices for one of the two horses to finish in a tie and 5 choices for the other- 6*5= 30.  But since that puts an order on the two horse that tied, that we don't want, divide by 2: 30/2= 15.

That leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth.  The number of ways this can happen is
$15(4)(3)(2)(1)= \frac{6!}{2}= 720/2= 360$,


问:
Help solving basic modular arithmetic problem
I'm stuck on this basic problem,

Show that $39$ divides $17^{48}-5^{24}$

I am familiar with the basic arithmetic properties of modular. But I have absolutely no idea how I'm supposed to apply them to solve this type of question. Some advice on the pattern I'm supposed to be looking for would be very helpful.

答:
Hint:

You only have to prove that $3$ and $13$ both divide $17^{48}-5^{24}$, i.e. that $2^{48}-2^{24}\equiv 0\mod 3$, and that $\:4^{48}-5^{24}\equiv 0\mod 13$.

Remember that Fermat's theorem can be written as
$$ a^k\equiv a^{k\bmod p-1}\mod p\quad\text{ for any $a$ coprime to }p. $$