问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
An alternate approach for $\det(B_n)$: The Schur Complement:

At the recommendation  Rodrigo de Azevedo, I used his approach of the Schur complement -- which they used for $\det(A_n)$ -- in order to determine $\det(B_n)$ and was recommended to post it as a separate answer. We will follow a similar approach here, though it's marginally messier algebraically than in the case of $A_n$.

So, recall: let us write a matrix $M$ as a block matrix by

$$M = \m{A & B \\ C & D }$$

Then provided $A$ is invertible, we have

$$\det(M) = \det(A) \det(D - CA^{-1} B)$$
In the case of $B_n$, we focus on $B_{n+1}$. Recall that, if $e_n$ is the $n$th unit column vector in the usual basis of $\Bbb R^n$, then $Me_n$ gives the $n$th column of $M$ and $e_n^T M$ gives $M$'s $n$th row. With this in mind, we may write

$$B_{n+1} =
\left(
\begin{matrix}
B_{n} & \frac{n+1}{n} B_ne_n\\
\frac{n+1}{n} e_n^T B_n & n+1
\end{matrix}
\right)$$

Then the determinant can be found recursively by simplifying the resulting determinant relation. With some minor abuses of notation for simplicity's sake, e.g. treating scalars as analogous to $1 \times 1$ matrices,

\begin{align*}
\det(B_{n+1})
&= \det(B_{n}) \det\left( \underbrace{n+1 - \frac{n+1}{n} e_n^T B_n B_{n}^{-1} \frac{n+1}{n} B_n e_n}_{\text{this is a $1\times 1$ matrix, may drop $\det(\cdot)$}} \right)\\
&= \det(B_{n}) \left( n+1 - \left( \frac{n+1}{n} \right)^2 e_n^T \underbrace{B_n B_{n}^{-1} B_n}_{=B_n} e_n \right)\\
&= \det(B_{n}) \left( n+1 - \left( \frac{n+1}{n} \right)^2 e_n^T B_n e_n \right)\\
&= \det(B_{n}) \left( n+1 - \left( \frac{n+1}{n} \right)^2 e_n^T \m{ n \\ \vdots \\ n }\right)\\
&= \det(B_{n}) \left( n+1 - \left( \frac{n+1}{n} \right)^2 n \right)\\
&= \det(B_{n}) \left( - \frac{n+1}{n} \right)\\
\end{align*}

The final equality follows from some basic algebra, but, for completeness:

\begin{align*}
n+1 - \left( \frac{n+1}{n} \right)^2 n
&= n+1 -  \frac{(n+1)^2}{n}\\
&= \frac{n (n+1) - (n+1)^2}{n} \\
&= \frac{n^2 + n -n^2 - 2n - 1}{n^2} \\
&= -\frac{n+1}{n}
\end{align*}

A backward-iteration argument in the vein of my original answer's alternate approach, alongside $\det(B_1) = 1$, then allows us to express this by

$$\det(B_n) = \prod_{k=1}^{n-1} \left( - \frac{n+1}{n} \right) = (-1)^{n-1}n$$

as desired.


问:
How to deal with negative Z score problems
Given mean life time of 40 days and standard deviation of $1.25$ days. If $12\%$ of the cars produced are having a maximum lifespan of $n$ days, find the value of n.

My work::

$Z = \dfrac{X- 40}{1.25}$

$P (X<n) = 0.12$

Using normal table conversion,

$\dfrac{X- 40}{1.25} = -1.175$

$X = -1.175(1.25)+40$

$X = 38.5325 $ days

Is the $Z$ value correct? As per the statement, the maximum of $n$  days $(X<n)$ but it gives negative $Z$ score.

答:
Negative Z score is perfectly fine.

Z-distribution (=Standard Normal Distribution) follows
Normal Distribution with mean=0 and standard deviation=1.

That is, Z ~ N(0, 1)

The support of Z-distribution is in between -infinity and +infinity.

Thus, your Z-score can be any negative numbers or positive numbers.

This Wikipedia will show you the Z-table of negative and positive Z-values as well as the density function of Z-distribution on the right side if your scroll down a bit. https://en.wikipedia.org/wiki/Standard_normal_table

I hope you find this helpful. Thanks.

ContentLicense: CC BY-SA 4.0


问:
Compact subset of a compact topological space
Let $X$ be a compact topological space and $K$ be a compact subset of $X$. I want to prove that $K$ is a closed subset of $X$, or not. I tried to show that the complement of $K$ in $X$, is open. But, I am not able prove that. I also tried to show that $K$ contains all its limit points. If $K$ is not closed in $X$, please give an example.

答:
For a counterexample, let $K$ be the Sierpinski space, whose underlying point set is $\{0,1\}$

and whose open sets are
$ \{\varnothing ,\{1\},\{0,1\}\}$.  Since $K$ is finite, $K$ is compact.

The subset $\{1\}$ is compact but not closed.


问:
Last non zero digit of $n!$
What is the last non zero digit of $100!$?

Is there a method to do the same for $n!$?

All I know is that we can find the number of zeroes at the end using a certain formula.However I guess that's of no use over here.

答: This question gets asked fairly frequently. There was originally a problem on the AMC asking for the last two digits. Here is my post on the single digit case, which is not all that much simpler than the two digit case. </p>

ContentLicense: CC BY-SA 3.0

Title: Last non zero digit of $n!$


问:
Find the minimum value of $7x-24y$
LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :

$x^2+y^2+Ax+By+C=0$

with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$

can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$

Stuck,:>

答:
Since
$$
x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\
\left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C
$$
Let $\displaystyle r=\sqrt{\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C}$, we can set $x=r\cos \theta -\dfrac{A}{2},y=r\sin \theta -\dfrac{B}{2}$, then
$$
\begin{aligned}
7x-24y&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} (7 \cos \theta-24 \sin \theta)-7 A+24 B\right)\\
&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} \sqrt{7^2+24^2}\sin(\theta+\phi)-7 A+24 B\right)\\
&\geq \frac{1}{2} \left(25\sqrt{A^2+B^2-4 C} -7 A+24 B\right)
\end{aligned}
$$
The second step uses The auxiliary Angle formula.

Title: Find the minimum value of $7x-24y$


问:
Why is inter/extrapolation technique on finite difference table different than taylor's expansion of FDM?
I am studying finite difference methods on my free time.

Finite Difference table

So based on the table, if I want to extrapolate the next value of my function, I have to use ...
$$f(x+dx)=f(x)+f'(x)+f^2(x)+f^3(x)+...+f^n(x)$$
which will be ...
$$f(6)=125+61+24+6+0=216=6^3$$

Taylor's expansion on Forward Difference Method

$$f(x+dx)=f(x)+dxf'(x)+dx^2\frac{f^2(x)}{2!}+dx^3\frac{f^3(x)}{3!}+...+dx^n\frac{f^n(x)}{n!}$$

Say my $dx=1$, so if I want to find $f(6)$, I have to divide the derivatives I found with their respective factorial before summing all of the results up, no? If so, ... $$f(6)=\frac{125}{0!}+\frac{61}{1!}+\frac{24}{2!}+\frac{6}{3!}+\frac{0}{4!}=199\ne216$$

Why are their inter/extrapolation between these two different? Which one should I use to perform inter/extrapolation? Also, is the benefit of finding higher derivatives is just to increase the accuracy of inter/extrapolation? Is there any other benefits of finding higher derivatives? or, did I misunderstood?

答: Apparently you want to use $x=5$. For $f(x)=x^3$, we have $f(5)=125$, $f‘(5)=75$, $f‘‘(5)=30$, $f‘‘‘(5)=6$, which does lead to $216$.


问:
Self-dual Lie Algebra representation
If $V$ is a Lie algebra $L$ module, then I want to know whether if $V$ and $V^*$ are isomorphic as $L$ modules then there is a basis for $V$ in which the matrices representing the action of $L$ are all skew-symmetric. If there is a basis $(e_i)$ for which the isomorphism takes $e_i$ to $e_i^*$ then I can do the problem. But is there such a basis?

答:
While Tsemo Aristide's answer shows via striking examples that the answer is "no", I would like to quickly explain what I think is the mistake in the question. Full disclosure: This is partly due to me making the same mistake at first when thinking about the related question Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices? today.

An isomorphism of $\mathfrak g$-modules $V \simeq V^*$ indeed is equivalent to the existence of a non-degenerate $\mathfrak g$-invariant bilinear form $\langle \cdot , \cdot \rangle$ on $V$ (NB, $\mathfrak g$-invariant for Lie algebra representations i.e. $\mathfrak g$-modules means $\langle x v ,w\rangle = -\langle v, x w\rangle$). (Compare also https://math.stackexchange.com/a/3781591/96384 .)

Further, in the case of a simple $\mathfrak g$-module, and $\mathfrak g$ being absolutely simple, this form is even unique up to a scalar (as follows from Schur's Lemma.)

Now in general this form does not need to be symmetric, it can be alternating. E.g. for the fundamental representation of $\mathfrak {sl}_2(\mathbb C)$ on $V:=\mathbb C^2$, check that up to a scalar, that non-degenerate bilinear form is the alternating $\langle \pmatrix{x\\y}, \pmatrix{w\\z} \rangle := \det(\pmatrix{x &w \\y &z})$. However e.g. for the adjoint representation, one can show that it is always symmetric. Indeed, up to a scalar it is the Killing form.

So why is the assertion still not (generally) true in that case, as Tsemo Aristide's answer and this here show? Because to get to skew-symmetric matrices from a symmetric bilinear form with the method shown by Malkoun in their answer there, you need an orthonormal basis, and for that you need the form to be much more than non-degenerate, you need it to have an orthonormal basis i.e. being equivalent up to scaling to the form given by the symmetric matrix $S= Id$ (where in general, you represent a symmetric BLF $b(\cdot, \cdot)$ by any symmetric matrix $S$ via $b(x,y) := x^{tr}Sy$). Now over an algebraically closed field like $\mathbb C$, every symmetric BLF is equivalent to that standard one. But e.g. over $\mathbb R$, where due to Sylvester's Law of Inertia symmetric BLFs are classified by their index, this is equivalent to the form being anisotropic i.e. (positive or negative) definite. Which over the real numbers it is only in the case of semisimple Lie algebras which come from compact Lie groups. Indeed, since the Killing form is indefinite as soon as our Lie algebra contains nonzero nilpotent elements, this shows that over any subfield of $\mathbb R$ there is no hope to represent all $ad(x)$ for $x$ in a given semisimple Lie algbera with skew-symmetric matrices, unless the Lie algebra contains only semisimple elements i.e. is anisotropic ("compact").

Title: Self-dual Lie Algebra representation


问:
If $\lceil\frac{x}{3}\rceil < \lceil\frac{y}{3}\rceil$ what can you conclude about $x$ and $y$?
Certainly if there were no ceiling functions around I would know what that inequality implies. I'm just not sure what else I have to be worried about.

My intuition tells me that I should just be able to conclude that $x<y$, however I'm concerned there's more to the ceiling function than meets the eye. I know that if the input is an integer the function will return itself, if it's not an integer it will round up to the nearest integer.

Edit: Certainly if $x<y$ then it's not necessarily true that $\lceil\frac{x}{3}\rceil < \lceil\frac{y}{3}\rceil$. For example take $x=2$ and $y=3$, obviously $x<y$ but $\lceil\frac{x}{3}\rceil = \lceil\frac{y}{3}\rceil$.

答: The most general conclusion you can draw is that there exists an integer $n$ such that $x\le 3n< y$.


问:
Proof by induction, $2^{n} > n^{2} - 2$
So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.

I start of with the "basecase" for n = 3:

RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$

Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.

$2^{k} > k^{2} - 2$

We multiply both sides with 2:

$2^{k+1} > 2(k^{2} - 2)$

$2^{k+1} > 2k^{2} - 4$

$2^{k+1} > k^{2} + k^{2} - 2 - 2$

According to our assumption,  $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $

Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:

$2^{k+1} > k^{2} + 2k + 1 - 2$

Thus:

$2^{k+1} > (k+1)^{2} - 2$

QED.

How does my proof look, does it have any deficits, and how would you prove it?

Thank you!

Title: Proof by induction, $2^{n} > n^{2} - 2$

Body:
I think this is a little clearer.

$2^{k+1}\ge2k^2-4$

$=k^2+2k+1-2+k^2-2k-3$

$=(k+1)^2-2+(k-3)(k+1)$

$\ge(k+1)^2-2$ for $k\ge3$


问:
A little doubt on the usage of the $(=\; elim)$ rule?
I'm reading Language, Proof and Logic. I'm trying to understand this proof:

"= Intro" and "= Elim" are:

I understand the use of "= Intro" in the proof, but how exactly are we using "= Elim"? What is $p(n)$? I suspect $p(n) := [n=b]$. So we have in the fourth line:

$p(b)$ which means $b=b$

$b=a$

$p(a)$ which means $a=b$

Is that it?

答:
Equality elimination is a rule of substitution:  When $n=m$ is given then we may substitute any (some or all) occurrences of $n$, in a statement, for $m$ .

When $P(n)$ is a statement containing some occurrences of $n$, then $P(m)$ is that statement with those occurrences replaced with $m$, and so when we have the equality $n=m$ we may 'eliminate' this to say: $$\begin{array}{l}P(n)\\n=m\\\hline P(m)\end{array}$$

So your proof is using equality elimination to make the inferences that:

$$\begin{array}{l}3.~b=b\\2.~b=a\\\hline4.~a=b\hspace{4ex}{=}\textsf{E 3,2}\end{array}\qquad\begin{array}{l}1.~\operatorname{SameRow}(a,a)\\4.~a=b\\\hline5.~\operatorname{SameRow}(b,a)\hspace{2ex}{=}\textsf{E 1,4}\end{array}$$

Where the equality being 'eliminated' is the second line referenced in the rule, and has the form [original] = [substitute].

The first inference is a common subproof using equality introduction to 'flip' an equality around so we have the required order for the second inference.

So we put this together with the equality introduction to claim:

$$\begin{array}{|l}1.~\operatorname{SameRow}(a,a)\\2.~b=a\\\hline3.~b=b\hspace{12ex}{=}\textsf{I}~\\4.~a=b\hspace{12ex}{=}\textsf{E 3,2 }\\5.~\operatorname{SameRow}(b,a)\hspace{2ex}{=}\textsf{E 1,4}\end{array}$$


问:
Find all integers n such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.
I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.

Using the theorem, I get $\frac{p}{q} = \pm 1, \pm 13, \pm \frac{1}{5}$, and $\pm \frac{13}{5}$ as possible roots. Then I use synthetic division and Horner's method to get a remainder of $-(n+8)$. For this to be a solution, $-(n+8)=0$, so $n = -8$. Then I could do this for $+1, +13, -13,$ etc.

Is this the correct approach to answering the original question? Original question:

Find all integers $n$ such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.

Why would I need to have a rational root to answer the problem? Couldn't I have complex solutions where I can express $5x^2 + nx - 13$ (where n is an integer) as a product of two linear factors with integer coefficients?

I greatly appreciate any insight you could provide on this. It's been about 2 years since I've done any mathematics (a brief foray into Neuroscience turned into a longer expedition than intended) and I am longing to return to the beautiful realm of mathematics. Thanks for your time in reading through this jumbled mathematical thought!

答:
As mentioned in the comments, the discriminant $\sqrt{n^2-4 \cdot 5 \cdot -13} = \sqrt{n^2+260}$ needs to be an integer (such as $k$) for both factors to have integer coefficients.

Let $n^2+260 = k^2$ where $k$ is an integer. Then $260 = k^2-n^2 = (k+n)(k-n)$, and factorising $260 = 2^2 \times 5 \times 13$, the factor pairs of $260$ are $(1, 260); (2, 130); (4, 65); (5, 52); (10, 26); (13, 20)$.

Discard all pairs where the numbers have opposite parity: (odd, even) or (even, odd), which leaves only $(2,130)$ and $(10,26)$. As to why this is, what happens to $k$ and $n$, which must be integers?

Therefore:

$$k + n = 130, k - n = 2 \Rightarrow n = 64 \tag{1}$$
$$k + n = 2, k - n = 130 \Rightarrow n = -64 \tag{2}$$
$$k+n=26, k-n=10 \Rightarrow n=8 \tag{3}$$
$$k+n=10, k-n=26 \Rightarrow n=-8 \tag{4}$$

which is no surprise, as substituting $n \to -n$ into $(1)$ gives $(2)$, and the same into $(3)$ gives $(4)$, and vice versa.


问:
Finding sequence in a set $A$ that tends to $\sup A$
I have been reading the book at http://www.neunhaeuserer.de/short.pdf, and have noticed that in the proof of the intermediate value theorem (Theorem 5.8 in the book), it seems to be quietly assumed that you can always find a sequence of points in a set that tend to the supremum of the set (in $\mathbb{R}$).

I presume this is true, but how would one go about proving it? I can't seem to find the result easily, but maybe I'm searching the wrong. What does one call this property (of there being a sequence converging to you?). It seems "limit point" would be the natural term, but I've discovered that the definition doesn't quite mean that--e.g. an isolated point $a$ cannot be limit point, but the sequence $a,a,a,a,a,a \dots$ converges to it (or is that a matter of how we define convergence?).

答: Actually I don't think you really need an axiom of countable choice: suppose there's no sequence that tends to $\sup A$, then $\exists \epsilon >0$ such that $A \cap (\sup A-\epsilon,\sup A + \epsilon)=\emptyset$. This is an absurd since it will give you that $\sup A-\epsilon/2$ is an upper bound for A, violating the minimality of $\sup A$.</p>


问:
Finite Unions of Dendrites
The question is a bit specific, but seems to be the most general question to ask after handling some obvious counterexamples.

Initially, I was wondering the following.  Let $X$ be a one-dimensional Peano continuum.  If $X$, aside from its points whose neighborhoods are arcs, has only finitely many local cut points, can $X$ be written as a finite union of dendrites $D_1, \dots, D_n$?  If so, does $n$ depend on $X$ specifically, or just the number/orders of the local cut points?

A Peano Continuum is a compact, connected, locally connected metric space.  A Dendrite is a Peano continuum which contains no copies of the circle.  By a local cut point I mean a point $x \in X$ such that $x$ has an open, connected neighborhood $U$ such that $U \setminus x$ is not connected.  By the 'order' of a local cut-point $x$ I mean the maximal number of components it cuts a neighborhood $U$ into as $U \rightarrow x$.

Answer 1: No.  The Sierpinski Carpet is a one-dimensional Peano continuum which contains all one-dimensional, planar continua: It happens to be locally connected with no local cut-points.  In particular, it contains the Cantor Fan, which is just $C \times [0,1] / [\sim]$ where $C$ is the Cantor Set and $\sim$ collapses $C \times \lbrace 0 \rbrace$ to a point.  Its vertex has order the continuum, clearly impossible for a finite union of dendrites (or any subset thereof).

A useful concept is hereditarily locally connected (hlc), i.e. every subcontinuum is locally connected.  Such a continuum is also regular (a stronger condition): Every point has a neighborhood base whose elements have finite boundary.  However, it is known that the union of two dendrites need not be hlc; there is a counterexample constructed in Nadler's Continuum Theory.

It is also known that the union of rational continua is rational: A rational continuum is one whose points have neighborhood bases with countable boundary.  Dendrites are rational.  Thus "connected, finite unions of dendrites" forms a class of curves nicer than rational continua, but either independent of, or weaker than, hlc.  To my knowledge it has not been studied, but such a class sits in a very useful place in the hierarchy of curves.

Main question: What continua can be written as the finite union of dendrites?  What hlc (resp. regular) continua cannot be written as the finite union of dendrites?

In particular, can anyone give an example of an hlc continuum which can't be written as the union of a finite collection of dendrites?  I tried looking at examples of hlc-but-not regular continua; however, the standard examples I found can be written as the finite union of dendrites.  In each case I was able to construct them as the union of only two dendrites.

Examples of continua which can be written as the union of three dendrites, but not two, would be useful.  A trivial example would be three copies of the Cantor Tree identified at the endpoints - visually trivial, I should say, but if someone wants to write the details then go for it.  In a sentence, too many arcs are produced at the Cantor Set for one dendrite to 'seep' into one of the others and pick up some extra space.

Q2: Suppose that $X$ is locally cyclically connected (every point is contained in arbitrarily small open neighborhoods $U$ each of whose pair of points is connected by a simple closed curve in $U$).  Can $X$ be written as the union of two dendrites?  Three?

Q2b: Suppose that $X$ is locally cyclically connected and known to be the finite union of dendrites.  Can it be two?  Three?

Q3: Can the Sierpinski Triangle be written as the finite union of dendrites?  How many are required, if so?

I made this one a separate question, since it can potentially be solved without any topological sophistication.  Sierpinski Triangle as Finite Union of Dendrites

It's been solved, now.  It's not possible.

Q4: (Stronger than Q2) If $X$ is locally $n$-connected (every point is contained in arbitrarily small open neighborhoods $U$ each of whose pair of points is connected by $n$ arcs in $U$ intersecting only in their end points), can we bound the necessary number of dendrites?  Is it $n$ (or $n+1$)?

Q5: What continua can be written as the union of two (or finitely many) dendrites each of which intersect at most in a subset of their end points?  This problem has some tangential relevance to conformal welding theory and SLE, so has probably been studied in the geometric function theory literature if it hasn't already been studied back in the early 1900's when curve theory was a hot area.

Regarding Q5, I don't think there's anything interesting to ask about dendrites which only intersect in endpoints or vertices.  It seems to be an equivalent question to Q1, as any 'vertex' intersection can be obtained as a regular intersection of two dendrites augmented by small triods attached along arcs where we'd want a vertex.  But this might be different:

Q5b: Classify finite Dendrite unions $D = D_1 \cup \cdots \cup D_n$ such that if $x \in D_i \cap D_j$ for $i \neq j$ then $x$ is an endpoint of some $D_k$.  Here we don't require that $x$ be an endpoint of any other dendrite it belongs to.

Q6: What if we restrict how many dendrites can intersect in a single point to 2?  Or $k < n$?  What if we require any intersection point to belong to at least three (or $k$) dendrites?  This second part has some problems with (local) duplicates to iron out to get an interesting question, probably.

Q7: Anything nice happen if you restrict to continua in the plane?

Have fun!

答:
Ok, here's an hlc continuum which isn't the finite union of dendrites.  Not going to give the explicit proof, but it's similar to one of the spaces mentioned in the question.

Let $C_i$ be copies of the Cantor Tree, with end point sets $E_i$.  To $C_1$ attach $C_2, C_3$ to the 'left' and 'right' halves of $E_1$ along $E_2, E_3$ respectively.  Now iterate: Attach two more copies to the left and right halves of $E_2$, then to $E_3$ etc. so that $C_i \rightarrow E_1$.  Continue and take the limit.  This limit space $X$ is clearly a continuum, and is hlc, which can be observed due to the following theorem:

A continuum is hlc if and only if it contains no convergence continuum.

A convergence continuum is a non-trivial (i.e. not just a point) subcontinuum $Y$ such that there is a sequence of pairwise-disjoint subcontinua converging to $Y$ (in the Hausdorff topology) no one of which intersects $Y$.

Each point $p$ of $X$ is either in one of the dendrites away from $E_1$ or in $E_1$ itself.  Dendrites are hlc, so there are no convergence continua containing any $p$ outside of $E_1$.  Thus a convergence continuum would have to be contained in $E_1$.  But $E_1$ is the Cantor Set.

This shows that the class "finite union of dendrites" and the class "hlc continua" are independent of each other.  So the question is now more interesting.  Incidentally, it also shows that "hlc continua which are the countable union of dendrites" is independent from "finite union of dendrites" as well.  Actually, the continuum $X$ is regular and even completely regular, i.e. every non-degenerate subcontinuum has non-empty interior in $X$.

The tricky part is showing that it's not the finite union of dendrites.  It's 'geometrically obvious' but I'm still trying to think of a slick way to prove it in just a line or two.

Title: Finite Unions of Dendrites


问:
How to understand the object in a category
Am a real beginner in category theory... I read that a category can be defined as a $4$-tuple:

$$
\textbf{C}=(\text{Ob}(\textbf{C}),\text{Ar}(\textbf{C}),\mu(\textbf{C}),\text{id}(\textbf{C}))
$$

where $\text{Ob}(\textbf{C})$ is any collection of elements, $\text{Ar}(\textbf{C})$ is a family of sets $\textbf{C}(X,Y)$ indexed by pairs of elements of $\text{Ob}(\textbf{C})$, and $\text{id}(\textbf{C})$ are in $\textbf{C}(X,X)$ that satisfy the identity laws ...

But am still trying to understand $\text{Ob}(\textbf{C})$ ... I read that given a concrete group $G$, with the category of $G$ as $G_{cat}$, we have that $\text{Ob}(G_{cat})$ has only one member (which can be indexed as $\{0\}$) ... However, this member does not consist of $G$, but rather is a fictitious mathematical object on which $G$ acts...

The description above sounds very nebulous ... Does this mean that the single object ($\{0\}$) can simply be any set such that the $G$ can act on it ? In this case, can an infinite set $S$ isomorphic to $\mathbb{N}$ be considered also as "a specific case of" $(\{0\})$?

答:
The key idea to view a group as a category with one object is to treat the elements in the set of morphisms as elements in the group. I think some examples will help you understand it better.

If you consider the symmetry group $S_n$, then you can take the object as the set $\{1,2,\dots,n\}$ and you can define the morphisms are the bijection from $\{1,2,\dots,n\}$ to itself.

If you consider the dihedral group, then the object can be a regular n-gon and the morphisms are rotations and reflections of the regular n-gon. I hope this helps you.


问:
Stability of equilibrium points in Gradient Systems , Lyapunov functions and Hartman-Grobman Theorem
So I have learned about Lyapunov theory to study the stability of equilibrium points are now we want to apply it to the study of gradient systems. So suppose we have $x'=-\nabla V(x)$ and we have that $a$ is an equilibrium point for this equation that is $\nabla V(a)=0$.  Now if we have that $a$ is an isolated local minimum we can use the Lyapunov function $H(x):=V(x)-V(a)$ to see that this is an asymptotically stable point. If $a$ is an isolated local maximum we can use $-H(x)$ to see that it is unstable, but what happens if $a$ is an isolated saddle point ? How can we study the stability in this case? One way I thought about it would be to use the Hartman-Grobman theorem and we know that the linearization of this dynamical system will be unstable and so since they have homeomorphic flows I guess this would also be unstable, but I am not completely sure this works, or if there is another way to see this.
I guess my biggest doubt is that if we can use the Hartman-Grobman theorem to study the stability of the sistem from the linearized equation.

Any help is appreciated, thanks in advance.

答: You could use Hartman–Grobman (if $a$ is hyperbolic), but it's perhaps overkill. There's the simpler Lyapunov instability theorem which says that if there is a differentiable function $H$ which is defined in a neighbourhood of $a$ and does not have a local minimum at $a$, and $\dot H<0$ on a punctured neighbourhood of $a$, then $a$ is unstable.


问:
What is the geometric significance of the total differential being zero?
If $U$ is a function of $x,y,z$ then the total increment of $U$ due to a transition of a point from $(x,y,z)$ to $(x+\delta x,y+\delta y,z+\delta z)$ is given by the total differential given by

$$dU = \frac {\partial U}{\partial x} dx + \frac {\partial U}{\partial y} dy +\frac {\partial U}{\partial z} dz$$

which implies

$$dU = ( \frac {\partial U}{\partial x} \hat i + \frac {\partial U}{\partial y} \hat j +\frac {\partial U}{\partial z} \hat k) \cdot (dx \hat i+dy \hat j+dz \hat k)$$

$$dU = \nabla U \cdot dl$$

where $$\nabla U = \frac {\partial U}{\partial x} \hat i + \frac {\partial U}{\partial y} \hat j +\frac {\partial U}{\partial z} \hat k$$ and  $$dl= (dx \hat i+dy \hat j+dz \hat k)$$

but if the total differential is zero what does it say about the gradient other than the fact that this implies that there is no change in U.

Edit:

$dU=0$ does not necesaarily mean change in U is zero but is or order higher than 2

答:
It does NOT imply that there is no change in $U$.

What it implies is that the change in $U$ is of order 2 or higher; or to put it another way, the best "polynomial" approximation is a polynomial of degree 2 or higher. Think for example of $f(x,y,z)=x^2+y^2+z^2$ at $(x,y,z)=(0,0,0)$.


问:
alternating series test for $\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n}}{n+4}$
I must determine if this series converges (using specifically the alternating series test) $$\sum_{n=4}^{\infty}(-1)^n\frac{\sqrt{n}}{n+4}.$$

I know the necessary and sufficient conditions are:
The series $\displaystyle\sum_{n=1}^{\infty}(-1)^na_n$ converges if $a_{(n+1)} \leq a_n$ (or if $f'(x) < 0$) and $\displaystyle\lim_{n\to\infty}a_n = 0.$

So first I tried to show the derivative, $\frac{d}{dx}(\frac{\sqrt{x}}{x+4}) < 0$ and after a bunch of algebra got $\displaystyle\frac{2-x}{\sqrt{x^5} + \sqrt{x^3} + 16 \sqrt{x}} < 0$, but I don't know how to demonstrate if that's really true for all $x \geq 1$.

Next I tried to show $a_{n+1} \leq a_n\  \forall \ n \geq 1$, or
$\frac{\sqrt{n+1}}{4(n+1)^2 + 1} \leq \frac{\sqrt{n}}{4n^2+1}. \ $ After cross multiplying and some rearranging I get $(4n^2 + 1)\sqrt{n+1} \leq 4n^{\frac{5}{2}} + 8n^\frac{3}{2} + 2\sqrt{n}$, and I don't know how to tell if that's true for all $n \geq 1$.

So I'm stumped.  I know there are other tests.  But this is in the alternating series section so either the alternating series test can show it is convergent, or if not then I know it's divergent.
But the answer in the back of the book says it's convergent.  I just don't know how to prove it.

Any suggestions?

n.b.  The original series goes from $n=4\to\infty$ but I don't know if the 4 is a typo or is intended.

答:
There are two conditions for an alternating series to be convergent.
1) For some particular n, the corresponding graph must be declining and by using the derivative, this is easy to verify.
2) The infinite term must go towards zero, which it does.

If you remove the (-1) part, the series will diverge, hence your series is Conditionally Convergent. (Rearranging the terms affect the sum)
Source: Stewart Calculus Chapter 11, section 5


答:
Hint:

Let $7x-24y=z\iff x=?$

Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$

As $y$ is real, the discriminant must be $\ge0$

Title: Find the minimum value of $7x-24y$

Body:
LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :

$x^2+y^2+Ax+By+C=0$

with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$

can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$

Stuck,:>


问:
Find the value of $\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx$
I'm trying to figure out how to evaluate the following:
$$
J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx
$$
I'm tried considering $I(s) = \int_{0}^{\infty}\frac{x^3}{(e^x-1)^s}\,dx\implies J=-I'(1)$, but I couldn't figure out what $I(s)$ was. My other idea was contour integration, but I'm not sure how to deal with the logarithm. Mathematica says that $J\approx24.307$. </p>

<p>I've asked a similar question and the answer involved $\zeta(s)$ so I suspect that this one will as well. </p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
J & \equiv \bbox[5px,#ffd]{\int_{0}^{\infty}{x^{3} \over
\expo{x} - 1}\ln\pars{\expo{x} - 1}\,\dd x}
\\[5mm] & =
\left.\partiald{}{\nu}\int_{0}^{\infty}x^{3}
\pars{\expo{x} - 1}^{\nu}\,\dd x
\,\right\vert_{\,\nu\ =\ -1}
\\[5mm] & =
\left.\partiald{}{\nu}\int_{0}^{\infty}x^{3}\expo{\nu x}
\pars{1 - \expo{-x}}^{\nu}\,\dd x
\,\right\vert_{\,\nu\ =\ -1}
\\[5mm] & =
\left.\partiald{}{\nu}\sum_{k = 0}^{\infty}{\nu \choose k}
\pars{-1}^{k}\int_{0}^{\infty}x^{3}\expo{-\pars{k - \nu}x}
\,\,\dd x\,\right\vert_{\,\nu\ =\ -1}
\\[5mm] & =
\left.\partiald{}{\nu}\sum_{k = 0}^{\infty}
{-\nu + k - 1\choose k}\,{6 \over \pars{k - \nu}^{4}}\,\right\vert_{\,\nu\ =\ -1}
\\[5mm] & =
24\sum_{k = 0}^{\infty}{1 \over \pars{k + 1}^{5}} -
6\sum_{ k = 0}^{\infty}{H_{k} \over \pars{k + 1}^{4}}
\\[5mm] & =
30\,\underbrace{\sum_{k = 1}^{\infty}{1 \over k^{5}}}
_{\ds{\zeta\pars{5}}}\ -\
6\
\underbrace{\sum_{ k = 1}^{\infty}{H_{k} \over k^{4}}}
_{\ds{3\zeta\pars{5} - \pi^{2}\zeta\pars{3}/6}}
\\[5mm] & =
\bbx{\pi^{2}\,\zeta\pars{3} + 12\,\zeta\pars{5}}
\approx 24.3070 \\ &
\end{align}
$\ds{\sum_{ k = 1}^{\infty}{H_{k} \over k^{4}}}$: See $\ds{\pars{20}}$ in MW.


问:
Introduction to p-adic numbers
I am a freshman and for a final project of a subject I have to give an introduction to p-adic numbers, I look for some sources (books, videos,  articles) to be able to do my work, the only bases I have is a very basic "number theory" (construction of $\mathbb{N}$,  $\mathbb{Z}$, $\mathbb{Q}$, through equivalence relations and the construction of $\mathbb{R}$ by Dedekind cuts. And dense sets in $\mathbb{R}$ and $\mathbb{R}^n$), also basic things of modular arithmetic and diophantic equations and more things that are seen in the first year (calculus I, basic set theory etc.)

Could you give me some recommendations?

Thanks in advance.

答:
I first learned about $p$-adic numbers from (the first chapter of) W. Schikhof's Ultrametric Calculus, I was in my second semester with only very basic undergraduate knowledge in calculus and linear algebra back then, and I think I was lucky there. If you like R. Borcherds' videos in Noah Solomon's answer, this is doubly recommended because just like that, it starts with infinite digit representations. (Later one kinds of needs to forget that, but I still think it is a very good way to first learn about these beauties.)

On a personal note, I think the moment $p$-adics got me hooked was when I realised that something I had encounterd in high school suddenly made rigorous sense. Namely, little Torsten had realised that if you look at factorials of higher and higher numbers ($10!, 15!, 17!, 27!, ...$) they always end in longer and longer strings of zeroes. More precisely, he convinced himself that once some factorial $n!$ ends in a certain number of zeroes, then the factorials $m!$ of all $m \ge n$ must end in at least as many zeroes, and eventually more. The smartass that he was, he thought of that as

$$\infty ! = 0$$

and was very pleased with himself. Then he went to university, learned rigorous $\epsilon$-$\delta$ calculus and laughed about his younger self. Then he learned about $p$-adics, and after a while ... well, long story short, here he is on a math forum, writing

$$\text{$p$-adic} \lim_{n\to \infty} n! = 0 \quad \text{ (for any prime $p$)}$$

and again is kind of pleased with himself.

Title: Introduction to p-adic numbers


问:
The divergence of a series implies the divergence of limit superior of absolute value of sums of random variables
I'm here for another problem. I need to show the following:

If $\{X_n\}$ is a sequence of independent random variables, and $S_n=\sum_{k=1}^{n}X_k$ then
$$\limsup_{n}|S_n|=\infty \quad a.s. \text{ if } \quad \forall r>0: \sum_{n=1}^{\infty}\mathbb{P}\left[|X_n|>r\right]=\infty$$

And I had two ideas:

IDEA 1: Using Borel-Cantelli I can conclude that the event $$A_r=\{|X_n|>r \ \text{infinitely often}\}$$
has probability one. But this implies that $\bigcap_{r=1}^{\infty} A_r$ has probability one. So, I can choose a subsequence $\{X_{n_j}\}$ such that $|X_{n_j}|\rightarrow \infty$ as $j\rightarrow \infty$. How I can use this subsequence to show that $\limsup |S_n|=\infty$?

The trouble that I have with this is that I don't know the sign of $X_{n_j}$: If for all $j$ the $X_{n_j}$ have the same sign, then I'm done. The problem is when there exists a $j$ such that $X_{n_j}$ has different sign.

IDEA 2: The event $\{\limsup|S_n|=\infty\}$ is a tail event and since the $X_n$ are independent, it follows that it has probability zero or one. I have the idea to proceed by contradiction; but when I start, I don't see how to work with it.

答: If $\lim \sup |S_n| \leq r$ for some $r <\infty$ then $|X_n|=|S_n-S_{n-1}| \leq 2r$ for all $n$. But this has probability $0$ by Borel-Cantelli Lemma applied to the condition $\sum P(|X_n| >2r)=\infty$. Hence $(\lim \sup |S_n| <\infty)=\bigcup_{r=1}^{\infty} (\lim \sup |S_n| \leq r)$ has probability $0$.


问:
$φ_t : \mathbb{R}^3[x] → \mathbb{R}^3[x]$ is a linear transformation: $φ_t(ax^3 + bx^2 + cx + d) = (a - bt)x^3 + (dt^2 - c)x$. Find $im φ_5$.
$φ_t : \mathbb{R}^3[x] → \mathbb{R}^3[x]$ is a linear transformation of the form:
$$φ_t(ax^3 + bx^2 + cx + d) = (a - bt)x^3 + (dt^2 - c)x$$
where $t \in \mathbb{R}$ is the parameter. Determine the bases of space $ker φ_3$ and $im φ_5$.

Now, I did simillar exercises, but never in the field of polynominals.

So I thought to do that in a similar manner. For $im φ_5$ I get:

$$φ_5(ax^3 + bx^2 + cx + d) = (a - 5b)x^3 + (25d - c)x $$
$$φ_5(a, b, c, d) = (a - 5b, 0, 25d - c, 0) = (a, 0, 25d, 0) + (-5b, 0, -c, 0)$$

\begin{bmatrix}
    1       & 0 & 25 & 0 \\
    -5       & 0 & -1 & 0 \\
\end{bmatrix}

after some transformations I get:

\begin{bmatrix}
    1       & 0 & 0 & 0 \\
    0       & 0 & 1 & 0 \\
\end{bmatrix}

So (1, 0, 1, 0), (0, 1, 0, 1) is a basis of $im φ_5$? If I got $im φ_5$ right I think I can do $ker φ_3$ as well. So, is that right?

答:
You can just work with $\varphi_t(a,b,c,d)=(a-bt,0,dt^2-c,0)$ as a map $\mathbb R^4\to\mathbb R^4$; but you should convert your final answers back to polynomials.

The matrix of $\varphi_t$ with respect to the standard basis has $$\varphi(1,0,0,0),~ \varphi(0,1,0,0),~ \varphi(0,0,1,0),~ \varphi(0,0,0,1)$$ as the columns, i.e.
$$\begin{bmatrix}1 &-t & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & t^2 \\ 0 & 0 & 0 & 0\end{bmatrix}.$$

You can row reduce this (with $t=5$ if you want), and from there you can determine bases for the kernel and image.


问:
calculate kernels of matrices with angles
So my professor gave me this question:

I have to find the basis of the eigenvalues of this matrix

\begin{pmatrix}
  \cos(q) & \sin(q)\\
  \sin(q) & -\cos(q)\\
  \end{pmatrix}

so I calculate the eigenvalues and I found it is 1 and -1.
so now I need to find the basis of the kernel of those matrices

\begin{pmatrix}
  1-\cos(q) & -\sin(q)\\
  -\sin(q) & 1+\cos(q)\\
  \end{pmatrix}

\begin{pmatrix}
  -1-\cos(q) & -\sin(q)\\
  -\sin(q) & -1+\cos(q)\\
  \end{pmatrix}

so actually I need to find

\begin{pmatrix}
  a \\ b\\
  \end{pmatrix}

that will be function as basis for the kernel of each one of those matrices.

but how do I do it ?

I know how to do it when there is no angles meaning I would just compare it to zero.

for example in the first matrix I get these two equations :

$x-xcos(q)-ysin(q)=0$

$-xsin(q)+y+ycos(q)=0$

so I get this equation:

$x(-sin(y)+((1-cos(y))/(sin(y)))+((cos(y)-cos(y)cos(y))/(sin(y))))=0$

which is true for every $x$ and then I can assign some value to $x$ for example zero and then $y$ equal zero also but after I am checking it, it is not true.
could u please help me ?
so how can I do it ?

All he told us is that $q\neq{0}$ and $q\neq\pi$ and $q\neq{2}\pi $

and nothing more.

答:
Observe that $\cos(q-\pi/2)=\sin q$, and $\sin(q-\pi/2)=-\cos q$. So, the linear transformation corresponding to the matrix sends ${\bf i}\,(=\pmatrix{1\\0})$ to $\bf u$ in the unit circle at angle $q$ (the first column) and sends ${\bf j}\,(=\pmatrix{0\\1})$ to $\bf v$ in the unit circle at angle $q-\pi/2$.

If you draw it, it could be clear that the reflection through the line $e$ at angle $q/2$ also does the same to the basis vectors $\bf i$ and $\bf j$. As both are linear and coincide on a basis, we get that this is the matrix of reflection through $e$. From this, we easily get the eigenvectors: vectors parallel to $e$ will be fixed (i.e. they correspond to eigenvalue $1$) and vectors orthogonal to $e$ will be reversed (i.e. they correspond to eigenvalue $-1$).


问:
Finding the nth integral of $\ln(1+z/k)$
I can find the nth integral of $\ln(z)$ as follows:
\begin{aligned}
\left(\frac d{dz}\right)^{-n}\ln(z)&=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln(t)dt\\
&=\frac1{n!}\left[\int\limits_0^z\frac1t(z-t)^ndt-z^n\ln(0)\right]\\
&=\frac1{n!}\left[\int\limits_0^1\frac{z^nt^n-z^n+z^n}{1-t}dt-z^n\ln(0)\right]\\
&=\frac{\ln(z)-H_n}{n!}z^n,
\end{aligned}
but I can't get very far with $\ln(1+z/k)$. I was able to come up with this but it is only a conjecture:
\begin{aligned}
\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln\left(1+\frac tk\right)dt&=\frac{\ln\left(1+\frac{z}{k}\right)-H_n}{n!}(k+z)^n+\sum ^{n}_{i=0}\frac{H_{i}}{i!(n-i)!}z^{n-i}k^i\\
&=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=0}\frac{H_{n} -H_{i}}{i!( n-i) !} z^{n-i} k^{i}.
\end{aligned}
I'm only using $k,n\in\mathbb{N}$ and $z\in\mathbb{R}$ for now. Any help proving this would be appreciated.

答:
This is not a proof

Consider
$$f_1=\int_0^z \log \left(1+\frac{z}{k}\right)\,dz\qquad \text{and} \qquad f_{n}=\int_0^z f_{n-1}\,dz$$
Define
$$a_n=n\, a_{n-1} \qquad \text{with} \qquad a_1=1$$ and
$$g_n=\frac{a_n} z  \left(f_n-\frac{(k+z)^n }{n!}\log
   \left(1+\frac{z}{k}\right)\right)+k^{n-1}$$

The terms $g_n$ appears to be
$$g_n=-\frac { z\,P_{n-2}} {\text{lcm}(1,2,\cdots,n)} $$ where the polynomials $P$ are homogeneous is $k$,$z$, all coefficients being positive. None of these polynomials can be factored.

Tested up to extremely large values of $n$, you conjecture is verified.

In any manner, if
$$I_n=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\log\left(1+\frac tk\right)dt$$
$$I_n=\frac{z^{n+1} }{n (n+1) (z+k) \Gamma(n)}\,\, _2F_1\left(1,n+1;n+2;\frac{z}{z+k}\right)$$


问:
How to show that $\sum_{i=1}^{n} (x_i - a)^2 = \sum_{i=1}^{n}(x - \overline{x})^2 + n(\overline{x} - a)^2$ for all real $a$?
Suppose $x_1, \cdots, x_n$ are real numbers, and let $\mu = \frac{1}{n}\sum_{i}x_i$ be the arithmetic mean of the $x_i$. I am trying to show that for each real $a$, the following identity holds:
$$\sum_{i=1}^{n} (x_i - a)^2 = \sum_{i=1}^{n}(x - \mu)^2 + n(\mu - a)^2.$$

If I think of $\vec{x} = (x_1, \cdots, x_n)$ as a vector in $\mathbb{R}^n$ and similarly let $\vec{\mu} = (\mu, \cdots, \mu) \in \mathbb{R}^n$ and $\vec{a} = (a, \cdots, a) \in \mathbb{R}^n$, then this identity is equivalent to:
$$||\vec{x} - \vec{a}||^2 = ||\vec{x} - \vec{\mu}||^2 + ||\vec{\mu} - \vec{a}||^2.$$
The triangle inequality implies that $||\vec{x} - \vec{a}|| \leq ||\vec{x} - \vec{\mu}|| + ||\vec{\mu} - \vec{a}||$.
Is there some geometric relationship between $\vec{x}$ and $\vec{\mu}$ I can make use of?

答:
Just expand the sum: $\mu$ is a very specific value, and you need to use that. For every $a\in\mathbb{R}$
$$\begin{align*}
\sum_{i=1}^n (x_i-a)^2
&= \sum_{i=1}^n x_i^2-2a\sum_{i=1}^n x_i + na^2\\
&= \sum_{i=1}^n x_i^2-2na\mu + na^2 \tag{1}
\end{align*}$$
where we used for the last equality that $\mu = \frac{1}{n}\sum_{i=1}^n x_i$.
Similarly, we can compute (re-using the first derivation to obtain the first equality)
$$\begin{align*}
\sum_{i=1}^n (x_i-\mu)^2 + n(\mu-a)^2
&= \sum_{i=1}^n x_i^2-2n\mu^2 + n\mu^2 + n(\mu-a)^2\\
&= \sum_{i=1}^n x_i^2-2n\mu^2 + n\mu^2 + n\mu^2 -2na\mu + na^2 \\
&= \sum_{i=1}^n x_i^2-2na\mu + na^2 \tag{2}
\end{align*}$$
You can see that (1) and (2) are identical.


问:
Do people agree on what dx is in Riemann integration?
I've been rather curious about what the official rigorous understanding of the dx term in a Riemann integral is. Of course, we could use the formal definition of a Riemann integral but I have seen on Wikipedia other proposed ways of defining the differential of some function (we'll call x), dx. The classic high school teacher approach is to say it's a "very small change in x", or rather a small change in a linear approximation of x. However, as someone who studies differential geometry, I am wondering how this can be interpreted as the exterior derivative of a function x, in which dx is a 1-form. Do we then reinterpret $\int{f(x)}\,dx$ as being the integral of a 1-form? And how then can we consider a 1-form to be a "small change in x"?

答:
The title doesn't really match the question, but perhaps it suffices to say that $d x$ is a 1-form which is the exterior derivative of the function $x$, i.e. the identity function on $\mathbb{R}^1$.

This function takes $x$ to $x$, hence the name $x$; compare e.g. when we write $\sin(x)$ for the function that takes $x$ to $\sin(x)$.  (These are both abuses of notation; the latter should just be called $\sin$ when we're talking about the function, while the former should be called ${\rm id}$ or something.)


问:
Is the world-wide web a category? Question posed in "Category Theory for Programmers" by Bartosz Milewski.
Challenge 1.4.4 of Bartosz Milewski's "Category Theory for Programmers" poses the following question:

Is the world-wide web a category in any sense? Are the links morphisms?

I believe it is a category, where the objects are pages and the morphisms are links between pages (specifically URIs of pages, since the content itself might change upon page refresh). To prove this then it must be established that 1) there is an identity morphism for every object (page URIs), and 2) morphims (links between those pages/URIs) compose.

The identity morphism is page refresh, or even more simply, doing nothing at all. My preference would be page refresh, since the pages identity is its current content if that differs from the current page load.

Links compose because page A links to B and B to C, then one can navigate from A to C by first following the link from A to B and then immediately following the link from B to C. However, another way of looking at it is that for links to compose, then anytime page A links to B and B to C, then A must have a link directly to page C. That stronger condition generally isn't true. My inclination is to say that the awkward composition of navigating through intermediate pages should count as a composition function in its own right, but taken to an extreme in which two pages require thousands (or millions, etc) of intervening redundant link navigation, perhaps it shouldn't qualify. This raises another question, though it's off topic: what is the expected number of links required for that awkward protocol on average, and in the worst case? If it isn't too awful, then perhaps it would be worth tolerating. In any case, just wondering what others think about this interesting question.

答:
You got it right in your question. It's a category if you define the morphisms as reachability through 0 or more links (or reachability through 1 or more links OR the refresh button). As Qiaochu Yuan mentioned any graph has a corresponding category where A -> B is a morphism iff B is reachable from A in 0 or more steps.

Alternatively if you were to say A -> B is a morphism iff A links to B, the internet would not be a category, since the identity morphism won't exist for all objects (a page may not link to itself), and morphisms don't compose (A may link to B and B to C but A does not necessarily link to C).


答:
What the single object is need not matter much. It is just some mathematical object... literally anything.

Where the group enters the picture is that it is encoded as the morphisms, and the group multiplication is encoded as the composition operation on the morphisms. So, for instance if we're trying to represent the group $(\mathbb Z, 0,+)$ as a category, the category will have a single object $*$ and will have morphisms $$\{f_n:*\to *\mid n\in \mathbb Z\}.$$ And then $f_0$ is the identity morphism, and for the composition operation, we have $f_n\circ f_m = f_{n+m}.$ There is nothing special about $\mathbb Z$ here. For a general group $(G,e,\cdot),$ we have morphisms $\{f_g: g\in G\},$ and $f_e=id_*$ and $f_{g^{-1}}\circ f_g=f_g \circ f_{g^{-1}}=id_*,$ and $f_{g_1}\circ f_{g_2}=f_{g_1\cdot g_2}.$

So as with many things in category theory, it's really all about the morphisms. The object $*$ is only there as a formality so there is a source and a target for these morphisms. It's not doing much of anything.

That said, there's nothing wrong about imagining that $*$ is some set that the group $G$ acts on (e.g. the group itself acted on by left-multiplication, or some set the group represents permuations of), but it's not strictly necessary to envision it that way. This is really just the (immaterial) distinction between groups in the abstract and groups as symmetry transformations, only in the perhaps unfamiliar territory where the group operation is being understood as composition of morphisms in a category with one object.

Title: How to understand the object in a category

Body:
Am a real beginner in category theory... I read that a category can be defined as a $4$-tuple:

$$
\textbf{C}=(\text{Ob}(\textbf{C}),\text{Ar}(\textbf{C}),\mu(\textbf{C}),\text{id}(\textbf{C}))
$$

where $\text{Ob}(\textbf{C})$ is any collection of elements, $\text{Ar}(\textbf{C})$ is a family of sets $\textbf{C}(X,Y)$ indexed by pairs of elements of $\text{Ob}(\textbf{C})$, and $\text{id}(\textbf{C})$ are in $\textbf{C}(X,X)$ that satisfy the identity laws ...

But am still trying to understand $\text{Ob}(\textbf{C})$ ... I read that given a concrete group $G$, with the category of $G$ as $G_{cat}$, we have that $\text{Ob}(G_{cat})$ has only one member (which can be indexed as $\{0\}$) ... However, this member does not consist of $G$, but rather is a fictitious mathematical object on which $G$ acts...

The description above sounds very nebulous ... Does this mean that the single object ($\{0\}$) can simply be any set such that the $G$ can act on it ? In this case, can an infinite set $S$ isomorphic to $\mathbb{N}$ be considered also as "a specific case of" $(\{0\})$?


问:
Principal value of $\int_0^\infty \frac{x^{-p}}{x-1}dx$ for $|p|<1$
If $|p|<1$, how to find the Cauchy Principal Value of $$\int_0^\infty \frac{x^{-p}}{x-1}dx$$

I tried spliting the integration from $0\to 1$ and $1 \to \infty$ and switching $x = 1/u$, but no luck getting desired result which the book says is $ \pi \cot p \pi$. Please help.

ADDED:: using above method I got $\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right) $, it turned out to be $p \cot p\pi$ from Mathematica, anyone any ideas?

ADDED:: It seems that $$\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right)  = \sum_{k=-\infty}^\infty \frac{1}{k +p} = - \text{Res}[ \pi \cot(z \pi)/(p+z), z=-p] = \pi \cot p\pi$$

Also I was having trouble doing this via contour. Is this the shape of contour?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{\Re\pars{p} \in \pars{0,1}}$:
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}{x^{-p} \over
x - 1}\,\dd x} =
\Re\int_{0}^{\infty}{x^{-p} \over x - 1 + \ic 0^{+}}\,\dd x
\\[5mm] = &\
-\,\Re\int_{\infty}^{0}{\ic^{-p}\,y^{-p} \over
\ic y - 1 + \ic 0^{+}}\,\ic\,\dd y =
\Im\bracks{\ic^{-p}\int_{0}^{\infty}
{y^{\pars{\color{red}{1 - p}} - 1} \over 1 - \ic y}\,\dd y}
\end{align}
I'll evaluate the last integral with Ramanujan's Master Theorem: Note that
$\ds{{1 \over 1 - \ic y} =
\sum_{k = 0}^{\infty}\pars{\ic y}^{k} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}\expo{-k\pi\ic/2}}
\,\,\,{\pars{-y}^{k} \over k!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}{x^{-p} \over
x - 1}\,\dd x}
\\[5mm] = &\
\Im\bracks{\expo{-p\pi\ic/2}\,\,\Gamma\pars{\color{red}{1 - p}}
\Gamma\pars{1 - \bracks{\color{red}{1 - p}}}
\expo{\pars{\color{red}{1 - p}}\pi\ic/2}}
\\[5mm] = &\
\Im\bracks{\ic\expo{-p\pi\ic}\,{\pi \over \sin\pars{\pi p}}} =
\cos\pars{p\pi}\,{\pi \over \sin\pars{\pi p}}
\\[5mm] = &\
\bbx{\pi\cot\pars{\pi p}}\,,\qquad
\Re\pars{p} \in \pars{0,1} \\ &
\end{align}


问:
Prove $\sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p$
For $p> 1$ and $a_1,a_2,...,a_n$ positive, show that
\begin{equation}
    \sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p
\end{equation}

I was hoping to use the convexity of $f(x) = x^p$, or some induction argument. But I can't seem to figure it out. Some hints would be greatly appreciated!

答:
This is exactly the Hardy's inequality. Here's one possible proof:

Let $A_k=\dfrac{a_{1}+a_{2}+\cdots+a_{k}}{k}$ and define $A_0=0$. From Hölder's inequality, we have
$$
\left(\sum_{k=1}^{n}A_k^p\right)^{\frac{p-1}{p}}\left(\sum_{k=1}^{n}a_k^p\right)^{\frac{1}{p}}\ge\sum_{k=1}^{n}A_k^{p-1}a_k
$$
Hence we need to prove
$$
\sum_{k=1}^{n}A_k^{p-1}a_k\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p
$$
Since
$$
\begin{align} LHS&=\sum_{k=1}^{n}A_k^{p-1}(kA_k-(k-1)A_{k-1})\\ &=\sum_{k=1}^{n}kA_k^{p}-\sum_{k=1}^{n}(k-1)A_k^{p-1}A_{k-1}\\ &\ge\sum_{k=1}^{n}kA_k^{p}-\sum_{k=1}^{n}(k-1)\left(\frac{(p-1)A_k^p}{p}+\frac{A_{k-1}^p}{p}\right)(\text{Young's inequality})\\ &=\frac{p-1}{p}\sum_{k=1}^{n-1}A_k^p+(n-\frac{(p-1)(n-1)}{p})A_n^p\\ &\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p \end{align}
$$
where Young's inequality is:

(Young's inequality) $$\text{For }x,y,p,q>0,\frac{1}{p}+\frac{1}{q}=1, \text{ we have }\frac{x}{p}+\frac{y}{q}\ge x^\frac{1}{p}y^\frac{1}{q}$$

Thus
$$
\left(\sum_{k=1}^{n}A_k^p\right)^{\frac{p-1}{p}}\left(\sum_{k=1}^{n}a_k^p\right)^{\frac{1}{p}}\ge\sum_{k=1}^{n}A_k^{p-1}a_k\ge\frac{p-1}{p}\sum_{k=1}^{n}A_k^p
$$
Reorganize, we get
$$
\sum_{k=1}^{n}\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{p} \leq\left(\frac{p}{p-1}\right)^{p} \sum_{k=1}^{n} a_{k}^{p}
$$


答:
By induction, $n=1\implies a_1^p\lt (\dfrac{p}{p-1})^p a_1^p$, since $p\gt 1$.

For $n\gt 1$, may assume $a_{n+1}\ge a_k, \forall k=1,\cdots,n$. Then $\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\le a_{n+1}^p$

$\sum_{k=1}^{n+1} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p=\sum_{k=1}^{n} \left(\dfrac{a_1+a_2+\cdots a_k}{k}\right)^p+\left(\dfrac{a_1+a_2+\cdots a_{n+1}}{n+1}\right)^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p+a_{n+1}^p\lt \left(\dfrac{p}{p-1}\right)^p \sum_{k =1}^{n+1} a_k^p$

Title: Prove $\sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p$

Body:
For $p> 1$ and $a_1,a_2,...,a_n$ positive, show that
\begin{equation}
    \sum_{k = 1}^n \left(\frac{a_1+a_2+\cdots a_k}{k}\right)^p < \left(\frac{p}{p-1}\right)^p \sum_{k =1}^n a_k^p
\end{equation}

I was hoping to use the convexity of $f(x) = x^p$, or some induction argument. But I can't seem to figure it out. Some hints would be greatly appreciated!


问:
Parametrization of curve with linearly increasing curvature
I'm looking for a curve that has a steadily increasing curvature, that is for $\gamma: I \subset \mathbb{R}^+ \rightarrow \mathbb{R}^2$ and $C>0$ constant.

\begin{equation}\label{eq:1} \frac{d}{dt} || \gamma^{\prime \prime}(t) || = C, \end{equation} with $|| \gamma^{\prime}(t) || = 1 $ and some initial conditions (say $\gamma(0) = (0,0), \gamma^{\prime}(0) = (1,0), \gamma^{\prime \prime} = (0,0)$).

I know that such a problem most likely has not a closed form, but numerically I could try to search for something. This is what I got to:

Lets name $\gamma(t) = (\gamma_x(t), \gamma_y(t)) = (\gamma_x, \gamma_y)$. By integrating the curvature equation up to time $t$ and taking the square of the norm we get $$ (\gamma_x^{\prime\prime})^2 + (\gamma_y^{\prime\prime})^2 = C^2t^2.$$

Consider the equation for the derivative of $\gamma$, by taking the square and differentiating we obtain:
\begin{align} (\gamma_x^\prime)^2 + (\gamma_y^\prime)^2 = 1 \\
\Rightarrow \gamma_x^\prime \gamma_x^{\prime \prime} + \gamma_y^{\prime} \gamma_y^{\prime \prime} = 0 \\
\Rightarrow \gamma_x^{\prime \prime} = - \frac{\gamma_y^{\prime} \gamma_y^{\prime \prime}}{\gamma_x^\prime} = - \frac{\gamma_y^{\prime} \gamma_y^{\prime \prime}}{\sqrt{1 - (\gamma_y^{\prime})^2}}  \end{align}

So returning to the equation derived from the curvature, we get a nonlinear differential equation for $\gamma_y$:

\begin{align}
\frac{(\gamma_y^{\prime}\gamma_y^{\prime \prime})^2}{1-(\gamma_y^{\prime})^2} + (\gamma_y^{\prime \prime})^2 = C^2 t^2 \\  \Rightarrow
(\gamma_y^{\prime \prime})^2 = C^2t^2(1 - (\gamma_y^{\prime})^2) \\
\Rightarrow \gamma_y^{\prime \prime} = - Ct \sqrt{1 - (\gamma_y^{\prime})^2
}
\end{align}

Where in the last step I just took one of the solutions. Now I don't know how to proceed as I'm not familiar with solving nonlinear differential equations. How would you suggest to proceed? or is there any alternative way to obtain such a curve?

答:
Perhaps an Euler spiral (also known as a clothoid)?  Quoting the Wikipedia entry for "Euler spiral": An Euler spiral is a curve whose curvature changes linearly with its curve length.

In particular, in terms of your notation, reframe $\gamma$ as being the angle of turn after traversing length $t$.  Then compare with this formulation (at the Wikipedia entry) which uses $\theta$ where you have used $\gamma$ and $s$ where you have used $t$.  What you have called $(\gamma_x, \gamma_y)$ turn out to be obtainable by Fresnel integrals.  Also see this question and answers from it.


问:
Find the length of the radius of a inscribed semicircles
This is a problem that came from AoPS Online Facebook page......can anyone find the solution to it?

答:
$R$ is the center of the bigger semicircle. $S$ is the center of the smaller semicircle.

Radius $RC = 6, $ radius $CS = r$.

Then $RS^2 = 6^2 - r^2 = 36 - r^2$

$PR = 3, PT = \frac{7}{2}, RT = \frac{1}{2}$.

$ST^2 = RS^2 - RT^2 = \frac{143}{4} - r^2$

But also, $ST^2 = PS^2 - PT^2 = r^2 - \frac{49}{4}$

So, $2r^2 = \frac{192}{4} \implies r = \sqrt {24}$

$CD = 2r = 4 \sqrt6$


问:
Greatest value of $n^{1/n}$ without calculus where n is a natural number.
I want to find the greatest value of $n^{1/n}$ where n is a natural number without using calculus

As n tends to infinity $n^{1/n}$ will get very small.

$3^{1/3}$ looks like the largest because $1^1<2^{1/2}<3^{1/3}>4^{1/4}$

The only thing I could think of was using binomial theorem to prove that

${(n)}^{1/n}>(n+1)^{1/{(n+1)}}$ for $n \ge 3$

However, after applying the binomial theorem, I am not able to prove the inequality.

Is using binomial theorem, not the right approach?
What could be the other approach?

Can anybody help?

答:
Notice that $n^{1/n} > 1$ for all $n \in \mathbb N$.

Raise both sides of $n^{1/n} \stackrel?> (n+1)^{1/(n+1)}$ to the $n(n+1)$-th power, now we just need to show that

$$n^{n+1} > (n+1)^n \text{ for } n \ge 3$$

which can be shown by induction quite easily.

See Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$.


问:
Inclusion–exclusion principle, password combinations
Co-workers are required to create 6-character long passwords. The letters must be from lowercase letters or digits. Each password must start with a lowercase letter and end with a digit and contain at most 2 digits. How many valid passwords are there?

Is my answer correct? What I'm doing wrong if not?
$$26 \cdot 36^3 \cdot 10 - 36^2 + 26^2$$

答:
Just consider two cases

Case 1, only one digit at end : $\;26\cdot36^4\cdot10$

Case 2, an extra digit somewhere in the middle:$\;26\cdot36^3\cdot\binom41\cdot10\cdot10$

Add up


问:
Finding other linearly indpendent solution other that $J_0(x)$
I'm trying to find other linear independent solution given $J_0(x)$ is the solution of the Bessel's equation.

Bessel's equation looks like
$$x^2y''+xy'+(x^2-n^2)y=0$$
Now it's given $n=0$
$$x^2y''+xy'+x^2y=xy''+y'+xy=0$$
Now I'm thinking to solve it with frobenius method. But that would leave to again $J_0(x)$.

I might write it directly by relation
$$Y_n(x)=\frac{J_n(x)\cos(n\pi)-J_{-n}(x)}{\sin(nx)}$$
But How do I find not knowing this?

答:
The solution of
$$x^2y''+xy'+x^2y=xy''+y'+xy=0$$ is given by
$$y=c_1 J_0(x)+c_2 Y_0(x)$$ You cannot find the $Y_0(x)$ using Frobenius method since they cannot be expanded as polynomial.

Take care that
$$Y_\alpha(x) = \frac{J_\alpha(x) \cos (\alpha\pi) - J_{-\alpha}(x)}{\sin (\alpha\pi)}$$ only applies for non-integer values of $\alpha$.

For integer values of $n$, the expansion of $Y_0(x)$ is given as
$$Y_0(x)=\frac 1 \pi\sum_{n=0}^\infty (a_n+b_n \log(x))\, x^{2n}$$


答:
First note that

$$\sum_{i=1}^{n}(x_i - \overline{x})= \sum_{i=1}^{n}x_i -n\bar x = 0$$

Now, you can prove the rearranged equation using differences of squares:

\begin{eqnarray*} \sum_{i=1}^{n} (x_i - a)^2 - n(\overline{x} - a)^2
& = & \sum_{i=1}^{n} \left[(x_i - a)^2 - (\overline{x} - a)^2\right]\\
& = & \sum_{i=1}^{n} \left((x_i - a) -(\overline{x} - a)\right)\left((x_i - a) +(\overline{x} - a)\right)\\
& = & \sum_{i=1}^{n} (x_i -\overline{x})(x_i + \overline{x}- 2a)\\
& = & \sum_{i=1}^{n} (x_i -\overline{x})(x_i-\bar x - (\overline{x}+ 2a))\\
& = & \sum_{i=1}^{n} (x_i -\overline{x})^2 -(\overline{x}+ 2a)\underbrace{\sum_{i=1}^{n} (x_i -\overline{x})}_{=0}\\
& = & \sum_{i=1}^{n} (x_i -\overline{x})^2
\end{eqnarray*}

Title: How to show that $\sum_{i=1}^{n} (x_i - a)^2 = \sum_{i=1}^{n}(x - \overline{x})^2 + n(\overline{x} - a)^2$ for all real $a$?

Body:
Suppose $x_1, \cdots, x_n$ are real numbers, and let $\mu = \frac{1}{n}\sum_{i}x_i$ be the arithmetic mean of the $x_i$. I am trying to show that for each real $a$, the following identity holds:
$$\sum_{i=1}^{n} (x_i - a)^2 = \sum_{i=1}^{n}(x - \mu)^2 + n(\mu - a)^2.$$

If I think of $\vec{x} = (x_1, \cdots, x_n)$ as a vector in $\mathbb{R}^n$ and similarly let $\vec{\mu} = (\mu, \cdots, \mu) \in \mathbb{R}^n$ and $\vec{a} = (a, \cdots, a) \in \mathbb{R}^n$, then this identity is equivalent to:
$$||\vec{x} - \vec{a}||^2 = ||\vec{x} - \vec{\mu}||^2 + ||\vec{\mu} - \vec{a}||^2.$$
The triangle inequality implies that $||\vec{x} - \vec{a}|| \leq ||\vec{x} - \vec{\mu}|| + ||\vec{\mu} - \vec{a}||$.
Is there some geometric relationship between $\vec{x}$ and $\vec{\mu}$ I can make use of?


问:
Least upper bounds and Dedekind cuts
I am working an exercise in Pugh's real analysis book.

Let $b = l.u.b. S$, where $S$ is a bounded nonempty subset of $\mathbb{R}$.

(a) Given $\epsilon > 0$ show that there exists an $s \in S$ with $b - \epsilon \leq s \leq b$.

(b) Can $s \in S$ always be found so that $b - \epsilon < s < b$?

(c) If $x = A|B$ is a cut in $\mathbb{Q}$, show that $x = l.u.b. A$.

My first point of confusion is understanding the difference between (a) and (b). I can find a strict inequality immediately, which should solve both of these problems simultaneously. Here is my attempt.

Given any $\epsilon > 0$, $b - \epsilon < b$, so $b - \epsilon$ is not an upper bound of $S$. That means that there exists some element of $S$ strictly larger than it; call it $s$. Since $s \in S$ and $b$ is an upper bound of $s$, $s \leq b$. Therefore, $b - \epsilon < s \leq b$. Of course, that implies that $b - \epsilon \leq s \leq b$.

Something tells me that Pugh did not want me to solve (a) by first solving (b), but I do not know another way to solve (a) first. Contradiction, perhaps?

Suppose there dd not exist such an $s$. So for all $s \in S$. $s < b - \epsilon$ or $s > b$. The latter is impossible since $b$ is the supremum of $s$. If the former held, then $b - \epsilon$ would be an upper bound of $S$, but $b - \epsilon < b$, which contradictions the definition of least upper bound.

Is that a better proof for the purposes of this problem?

As for part (c), Pugh gives a direct proof in the textbook that I had some difficulty following, so I thought it was easier to do by contradiction. Even with that said, I am having difficult understanding exactly what a cut is or how it makes sense to talk about a cut as a supremum of a set.

Since any element $b \in B$ is an upper bound of $A$, $A$ is bounded above. By the first property of cuts, $A$ is nonempty. So $A$ has a supremum by the completeness axiom. For a contradiction, suppose that $x$ is not the least upper bound of $A$. So there exists $y = l.u.b. A$ such that $y < x$ and $y \geq a$ for all $a \in A$. By density, there exists $z$ such that $y < z < x$. Then $z > y \geq a$. Since $z > y$, $z \not \in A$, so $z \in B$. This is a contradiction.

答:
You correctly proved that there is an $s\in S$ such that $b-\epsilon<s\le b$. This does indeed imply the result in (a) and is the simplest way to prove that result, but it does not imply the result in (b): what if, for instance, $S=\{1\}$? Then $\operatorname{lub}S=1$, and no matter what $\epsilon>0$ you consider, the only possible $s\in(1-\epsilon,1]$ is $1$ itself. This sort of thing will happen whenever $S$ has an isolated maximum element. For instance, if $S=[0,1]\cup\{2\}$, then $\operatorname{lub}S=2$, and if $0<\epsilon\le 1$, the only member of $S$ in $(2-\epsilon,2]$ is $2$ itself.

I can’t judge your argument for (c) without access to Pugh’s book: it’s not clear how much of the work of constructing $\Bbb R$ you’ve already done and what properties you have available. If you’re constructing the reals from the rationals by means of Dedekind cuts, you shouldn’t really be talking about the completeness axiom: completeness of $\Bbb R$ should be a theorem. I assume that for each $q\in\Bbb Q$ you’ve identified $q$ with the cut $A_q\mid(\Bbb Q\setminus A_q)$ , where $A_q=\{r\in\Bbb Q:r<q\}$. If so, in order to show that $A\mid B=\operatorname{lub}A$ in $\Bbb R$ I would expect you to have to show that $A_q\subseteq A$ for each $q\in A$, and that if $C\mid D$ is any cut such that $A_q\subseteq C$ for each $q\in A$, then $A\subseteq C$; i.e., that $q\le x$ for each $q\in A$, and $x\le y$ for any cut $y=C\mid D$ such that $q\le y$ for all $q\in A$. (In fact you can show that $A=\bigcup_{q\in A}A_q$.)

If you’ve proved enough of the properties of $\Bbb R$ at this point, your argument can be made to work, but you have to say a bit more. Specifically, you have to use the fact that $\Bbb Q$ is dense in $\Bbb R$ (if you’ve proved it) to say that there is a rational number $z$ such that $y<z<x$; it’s not enough to know that there is a real number between $y$ and $x$. Then you know that $a\le y<z$ for all $a\in A$, so $z\notin A$ and therefore $z\in B$. But then $x<z$, which is a contradiction.

Title: Least upper bounds and Dedekind cuts


问:
What does it mean to evaluate $i^{786}$?
In my book ,

To evaluate $i^{786}$.

They did. $i^{4*196 + 2}$

Why do we have to do it this way? I have to write in the exam in a subjective format. So want to know is this the only correct way to write it. Since if the exponent is sometimes bigger, why would anyone even think of doing it like this?

I would have simply said that

$-1^{786/2}$ since the power is 393. Therefore the answer is $-1$.

I understand it is more of a homework question but I just couldn’t find the tag for it.

答:
It's probably just a way to emphasize that the result will be $-1$, as
$$i^{4+2}=i^4\cdot i^2 = 1\cdot-1$$That's easier to comprehend.


答:
$i^n$ is periodic with period of $4$

$$i^1=i$$
$$i^2=-1$$
$$i^3=-i$$
$$i^4=1$$

So we only need to know the remainder when $786$ is divided by $4$:

So $i^{786}=i^{4*196+2}=i^2=-1$

Title: What does it mean to evaluate $i^{786}$?

Body:
In my book ,

To evaluate $i^{786}$.

They did. $i^{4*196 + 2}$

Why do we have to do it this way? I have to write in the exam in a subjective format. So want to know is this the only correct way to write it. Since if the exponent is sometimes bigger, why would anyone even think of doing it like this?

I would have simply said that

$-1^{786/2}$ since the power is 393. Therefore the answer is $-1$.

I understand it is more of a homework question but I just couldn’t find the tag for it.


答:
You can use the binomial theorem but in a slightly adopted manner.

Consider

$$\left(\frac{(n+1)^{\frac 1{n+1}}}{n^{\frac 1n}}\right)^{n(n+1)}=\frac 1n\left(1+\frac 1n\right)^n$$

Now use the binomial theorem on $\left(1+\frac 1n\right)^n$ and note that for $n\geq 3$ and $2\leq k\leq n$ you have

$$\binom{n}{k}\frac 1{n^k}<\frac 1{k!}\leq\frac 1{2^{k-1}}$$
Summing up the binomial expansion and using these estimates, you get for $n\geq 3$

$$\left(1+\frac 1n\right)^n<3 \stackrel{n\geq 3}{\Rightarrow} \frac 1n \left(1+\frac 1n\right)^n<1$$

Checking the cases $n=1$ and $n=2$ you get $\boxed{n=3}$ as maximum.

Title: Greatest value of $n^{1/n}$ without calculus where n is a natural number.

Body:
I want to find the greatest value of $n^{1/n}$ where n is a natural number without using calculus

As n tends to infinity $n^{1/n}$ will get very small.

$3^{1/3}$ looks like the largest because $1^1<2^{1/2}<3^{1/3}>4^{1/4}$

The only thing I could think of was using binomial theorem to prove that

${(n)}^{1/n}>(n+1)^{1/{(n+1)}}$ for $n \ge 3$

However, after applying the binomial theorem, I am not able to prove the inequality.

Is using binomial theorem, not the right approach?
What could be the other approach?

Can anybody help?


问:
How do I find the transformation between the following random variables?
Let $U$ $\sim$ $U(0,1)$ .

The Probability Integral Transform theorem states that for any continuous random variable $Y$ with cumulative distribution function $F(y)$ and inverse cumulative distribution function $F^{-1}$:

$F(Y)$ is a $U(0,1)$ random variable.

$F^{-1}(U)$ is a random variable with distribution function $F$.

I am being asked to apply this theorem to find the following transformation between the random variable $U$ $\sim$ $U(0,1)$ and $V$ which has the following cumulative distribution function:

$$F_{V}(v) = e^{({\frac{v-\alpha}{\beta})}^{-\alpha}}$$ for $\alpha$ $\lt$ $v$ $\lt$ $\infty$,

and $$F_{V}(v) = 0$$ otherwise.

So far, I have begun by:

Let $U$ $\sim$ $U(0,1)$.

$$u = F(v)$$
$$u = e^{({\frac{v-\alpha}{\beta})}^{-\alpha}}$$

but I am finding it hard to rearrange for $v$ to find the transformation between $u$ and $v$

答:
Hint:

Taking natural logarithms gives you
$$
\ln u=\left(\frac{\beta}{v-\alpha}\right)^\alpha\ .
$$
Can you see how to proceed from here?


问:
A simple relation on constants
Suppose $\lambda > 0 $ and $\epsilon >0$ are two constant such that $ \lambda + \epsilon <1$, I want to show that $\lambda^{-1} - \epsilon >1$.

Could anyone help?

答: Since $\lambda + \epsilon <1$ and $\lambda>0,\epsilon>0$, we have $0<\lambda<1$, which indicates $\epsilon/\lambda>\epsilon$. Divide both sides by $\lambda$ we get $1+\epsilon/\lambda<\lambda^{-1}$ i.e. $\lambda^{-1}-\epsilon/\lambda>1$. And hence $\lambda^{-1}-\epsilon>\lambda^{-1}-\epsilon/\lambda>1$.

Title: A simple relation on constants


Body:
Notice that
$$ i = i$$
$$i^2 = i * i= -1$$
$$i^3 = i^2*i = -1*i = -i$$
$$i^4 = i^3*i = -i*i = -(-1) = 1$$
$$i^5 = i^4 * i = 1 * i = i$$

You can notice that for any integer k you'll have that:
$$i^{4k} = 1$$
$$i^{4k+1} = i$$
$$i^{4k+2} = -1$$
$$i^{4k+3} = -i$$

This is commonly written as $i^k = i^{k \mod 4}$


问:
quadratic equation precalculus
from Stewart, Precalculus, 5th, p56, Q. 79

Find all real solutions of the equation

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$

my solution

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$

$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$

official answer at the back of the book has only one real solution of $-4$

where did I go wrong?

答: You multiplied both sides by $(x-2)(x+2)$.  If this is zero, you may introduce extra solutions, hence you need to check your final answer to see if you have any extraneous solutions.  In this case, you do!  For $x=2$, two of the three fractions in the original equation are undefined.</p>

Title: quadratic equation precalculus


答:
$\lambda+\epsilon <1 \implies {\lambda}^{-1}(\lambda+\epsilon)<{\lambda}^{-1} \implies 1+{\lambda}^{-1} \epsilon < {\lambda}^{-1} \implies {\lambda}^{-1}-{\lambda}^{-1}\epsilon>1$

Now as $0<\lambda <1$, we have $\lambda^{-1}>1$ and thus

$\lambda^{-1} \epsilon > \epsilon \implies -\epsilon > -\lambda^{-1} \epsilon \implies \lambda^{-1} -\epsilon > \lambda^{-1} -\lambda^{-1} \epsilon  >1$

Title: A simple relation on constants

Body:
Suppose $\lambda > 0 $ and $\epsilon >0$ are two constant such that $ \lambda + \epsilon <1$, I want to show that $\lambda^{-1} - \epsilon >1$.

Could anyone help?


答:
Your answer raises a new question: How do you know that $(-1)^{393}=-1$? Well, you could argue:
$$
(-1)^{393} = (-1)^{2\times 196+1} = 1^{196}\times(-1)^1=-1
$$
This is basically doing with $-1$ what the book did with $i$. Since you're making a similar argument either way, the book's author probably decided that it's simpler to skip past $(-1)^{393}$ and operate directly on the expression $i^{786}$.


答:
$0<\lambda + \epsilon <1$ so $0<\lambda<1$. Thus

\begin{eqnarray}
\lambda + \lambda\epsilon&<&\lambda+\epsilon<1\\
1+\epsilon&<&\frac{1}{\lambda}\\
1&<&\frac{1}{\lambda}-\epsilon
\end{eqnarray}

Title: A simple relation on constants


问:
Is Borel algebra generated by all $G_\delta$ sets?
I am wondering this since the Borel algebra is generated by all open sets, or by all closed sets, or by all half-open intervals. My questions:

However, is it also generated by all $G_\delta$ sets? Similar question for by $F_\sigma$ sets, or by the finitely many combinations, e.g., $G_{\delta\sigma\delta\sigma}$, which means countable intersections of $G_{\delta\sigma\delta}$ sets and so on, etc.

Trying to answer my Q1 for $G_\delta$ set: Firstly, an open set is also (generated by) a $G_\delta$ set. Now, if I have a $G_\delta$ set, then it is also formed by (intersections of) open sets. So the answer to my question is yes for $G_\delta$ set?

If Q2 is correct, can this be continued to $G_{\delta\sigma\delta\sigma...\delta}$ and still result in the answer "yes"?

Any suggestion is appreciated. Thanks!

答:
As noted, any open set is by definition a $G_{\delta}$ set and a closed set is a $F_{\sigma}$ set. Hence, the Borel $\sigma-$algebra, $\mathscr{B}$ is of course contained in the $\sigma$-algebra generated by the $G_{\delta}$ sets, say $\mathscr{C}$ i.e. $\mathscr{B} \subset \mathscr{C}$.  But note that any $G_{\delta}$ set is contained in $\mathscr{B}$ and thus $\mathscr{C} \subset \mathscr{B}$ hence $\mathscr{C} = \mathscr{B}$.

Same argument works for your second question


答:
Here is a way to show it using the geometric series for $0<\epsilon <1$:
$$\frac 1{1-\epsilon} =\sum_{n=0}^{\infty}\epsilon^n>1+\epsilon$$

$$\lambda+\epsilon<1 \Rightarrow \frac 1{\lambda} >\frac 1{1-\epsilon}>1+\epsilon$$

Title: A simple relation on constants


答:
Why do we have to reject the solution $x=2$?

Hint: What happens when we put $x=2$ in the original equation?

Review your equations. Make sure that you didn't multiply by 0.

Title: quadratic equation precalculus

Body:
from Stewart, Precalculus, 5th, p56, Q. 79

Find all real solutions of the equation

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$

my solution

$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$

$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$

official answer at the back of the book has only one real solution of $-4$

where did I go wrong?


问:
How to prove this inequality $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$
How to prove this inequality in interval: $(0,\frac{\pi}{2})$
$
2\cos{x} + \sec^{2}{x} - 3 \ge{0}
$

I have to use it to show that the differentiation of a function is greater than or equal to 0 so that I can prove that the function is an increasing function.

Edit: Added the interval

答:
By Fermat's theorem, we examine zeroes of derivative of the function $f(x)=2\cos{x} + \sec^{2}{x} - 3 $ to find the extrema, that is, to solve the following:
$$
f'(x)=2 \tan x \sec ^2 x-2 \sin x = 0
$$
The solution is $x=\{k\pi | k\in\mathbb Z\}$. When $x=\pi$, we get
$$
2 \cos (x)+\sec ^2(x)-3=-4
$$
Thus $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$ doesn't hold.

Since the interval is added, we can prove the claim is true. For $x=0$, $f(x)=0$. And when $x\in(0,\pi/2)$, it is trivial that $f'(x)\geq 0$, thus $f(x)$ monotonously increases on $(0,\pi/2)$. Hence $f(x)\geq f(0)=0$ for all $x\in(0,\pi/2)$ and we finished the proof.


问:
The probability that the drawn card are face card of of same suits.
Three cards are drawn successively,without replacement from the pack of 52 cards.Then the probability that the drawn card are face card of same suits , is ?

My approach :
Probability of select one suit among 4 is 1/4.

Among each suits, there are 3 face card so the probability of face cards among each suits is (3/13)(2/12)(1/11).

So Required probability is 1/4 * (3/13)(2/11)(1/11).

Is the above approach correct?

Please help me.

答:
First, one thing to note is that the question indicates no replacement this is usually a good hint for you to think about the hypergeometric distribution as that is the only probability distribution with no replacement.

In general, the distribution takes the following form:

$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $

Where:

$N$ is the population size,

$K$ is the number of success states in the population,

$n$ is the number of draws (i.e. quantity drawn in each trial),

$k$ is the number of observed successes

The probability of drawing $3$ face cards of the same suit is then the following:

$\frac{{4 \choose 3} {48 \choose 0}}{52 \choose 3} =\frac{{4 \choose 3} }{52 \choose 3}$

Note that this counts the ace as a face card, if you don't want to count it that would be

$\frac{{3 \choose 3} {48 \choose 0}}{52 \choose 3} =\frac{{3 \choose 3} }{52 \choose 3}$

Then this should be multiplied by $4$ since that is the probability of drawing all $4$ face cards from one specific suit (so for example that is the probability of drawing all heart face cards, we multiply by $4$ to take into account ace, spades, and I forgot what the last suit is called.

so our final answer would be $4 \cdot \frac{{4 \choose 3} }{52 \choose 3}$ if we count the ace and $4 \cdot \frac{{3 \choose 3} }{52 \choose 3}$

Note that this isn't the only way to do it, the other way would be using multiplication rule, we can do the following (here I am not counting ace as a face card:

$\frac{3}{52}\cdot \frac{2}{51}\cdot\frac{1}{50}$ that is, probability of getting the first face card from a specific deck multiplied by the probability of getting the second face card from the same deck multiplied by the probability of getting the third face card from the same deck and then multiply the final answer by $4$ again (due to the same explanation) which would give the same result, note that since you are drawing these cards with no replacement the denominator is decreasing.


问:
Please help to find out the general solution of the following equation : ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$
For this equation :

${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

The integer solutions of $x$ found by WolframAlpha using inverse (logarithmic) function are $ 2 , 5 , 6 , 7 .$ Why it cannot find the other solutions 3 and 4 ? Is there any full proof method to find all the solutions at once ?

答:
I bet the reason why Wolfram can't find solutions 3 and 4 for the equation

$$
(x^2-7x+11)^{x^2-13x+42}=1
$$

is because it thinks that the domain of the function in LHS is $x^2-7x+11 \gt 0$. And this does make sense. In my school textbook the domain of exponential function $a^x$ is given by $a>0, a \ne 1, x \in \mathbb{R}$. There is no definition to calculate raise of some arbitrary negative base to the power of some arbitrary real number. Your function clearly is not the exponential, since the base is not a constant.

But you can do it manually. Firstly, let's assume that $x^2-7x+11>0$. Then let's take the logarithm from both sides(since it is monotonic, $\ln a= \ln b \iff a=b$):
$$
(x^2-13x+42)\ln(x^2-7x+11)=0.
$$

The equality holds when $x^2-13x+42=0 \space \text{and in the same time} \space x^2-7x+11>0$ or if simply $x^2-7x+11=1$.

$$
x^2-13x+42=0 \iff x=6 \space \text{or} \space x=7. \\
6^2-7 \cdot 6+11=36-42+11>0 \\
7^2-7 \cdot 7+11>0
$$

Then both $x=6$ and $x=7$ satisfy our conditions.

Now consider another case:
$$
x^2-7x+11=1 \iff x=2, x=5.
$$

We solved our equation assuming the domain is $x^2-7x+11>0$. We can not solve it when $x^2-7x+11 \le 0$ since the function is not well-defined for arbitrary case. But we do know how to raise negative integer to the positive integer power.It means that if we assume that $x^2-7x+11$ is a negative integer, then obviously it should be equal to $(-1)$ and the power the should be positive even integer:

$$
x^2-7x+11=-1 \iff x=3, x=4 \\
\text{then} \space x^2-13x+42 \space \text{should be positive even integer. Let's check it: }\\
3^2-13 \cdot 3 + 42=9-39+42>0 \\
4^2-13 \cdot 4 + 42=16-52+42> 0
$$

So, your favourite $x=3$ and $x=4$ also satisfy our equation. We are done now! Any questions are greatly appreciated!


答: As pointed out by @Andrew Chin that it does not hold for all real numbers. Thus it is reasonable to restrict the domain to some meaningful interval such as $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that $\cos x > 0$. In that case, using the AM-GM inequality you have: $LHS = \cos x+ \cos x + \sec^2x - 3 \ge 3\sqrt[3]{\cos x\cdot \cos x\cdot \sec^2 x} - 3 = 3 -3 = 0 = RHS $.

Title: How to prove this inequality $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$

Body:
How to prove this inequality in interval: $(0,\frac{\pi}{2})$
$
2\cos{x} + \sec^{2}{x} - 3 \ge{0}
$

I have to use it to show that the differentiation of a function is greater than or equal to 0 so that I can prove that the function is an increasing function.

Edit: Added the interval


答:
We want to solve in $\mathbb{Z}$ the equation ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1.$

Since $a^b=1$ for some $a,b\in \mathbb{Z}$ only if

$a=1$ or

$a\neq 0$ and $b=0$ or

$a=-1$ and $b$ is even,

we solve for each of listed cases:

${x^2 - 7x + 11}=1 \iff x=2 \quad \text{or}\quad x=5,$

${x^2 - 13x + 42}=0 \iff x=6 \quad \text{or} \quad x=7,$

${x^2 - 7x + 11}=-1 \iff x=3 \quad \text{or} \quad x=4.$

It is easy to check that each of the above numbers satisfies the equation.

REMARK

The "difficulty" of WolframAlpha is related to the use of a function inverse to exponential function. It is not defined if the basis of the exponential is not positive. In the above solution, this corresponds to the case 3.

Title: Please help to find out the general solution of the following equation : ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

Body:
For this equation :

${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

The integer solutions of $x$ found by WolframAlpha using inverse (logarithmic) function are $ 2 , 5 , 6 , 7 .$ Why it cannot find the other solutions 3 and 4 ? Is there any full proof method to find all the solutions at once ?


问:
how can calculate discrete value of $sinc^2x$
how to calculate discrete values of $sinc^2x = sin^2x/x^2$ in general form? what if x is odd (certainly function parameter will not be an integer)? can it be calculated as some coefficients of $\pi$ ?
$$
sinc^2(x/2)=\begin{cases}
1 & \text{ if } x= 0\\
0 & \text{ if } x= \pm 2,\pm 4,\pm 6, ..., 2k+2\\
? & \text{ if } x= \pm 1,\pm 3,\pm 5, ..., 2k+1
\end{cases}
$$

答:
In the following links there are many information about the function $sinc(x)$ and it's values:

Definition of Sinc function
(another question with complete answers)

https://en.m.wikipedia.org/wiki/Sinc_function
(the wikipedia article)

Hope that's helping you.


问:
Finding the Fourier Series of $\sin(x)^2\cos(x)^3$
I'm currently struggling at calculation the Fourier series of the given function

$$\sin(x)^2 \cos(x)^3$$

Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.

What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$

Transforming it back, the result is:

$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$

(I've checked my calculations multiple times, I'm pretty sure it's correct.)

Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.

Could someone help me find my mistake?

答:
I used trig idendities and came to
\[ \sin(x)^2 \cdot \cos(x)^3 = \frac{1}{16} \cdot \left( 2\cos(x) - \cos(3x)-\cos(5x)\right)\]
This surely works for $x=0$, and Mathematica gives the same.</p>


问:
Dot product of two vectors with different degrees
I have vector $\mathbf r$ which is defined as
$$
\mathbf r=(\mathbf r_1-\mathbf r_2)
$$
I need to calculate $\mathbf r^3 $ namely $(\mathbf r_1-\mathbf r_2)^3$ somewhere, so I obtain
$$
\mathbf r_1^3-3\mathbf r_1^2 \mathbf r_2-\mathbf r_2^3+3\mathbf r_2^2 \mathbf r_1
$$
Now I don't know "Should I write the angle between $\mathbf r_1$ and $\mathbf r_2$ in second and fourth terms?" namely:$-3\mathbf r_1^2 \mathbf r_2 \cos[\theta]$ or not?

答:
Your result $\mathbf r_1^3-3\mathbf r_1^2 \mathbf r_2-\mathbf r_2^3+3\mathbf r_2^2 \mathbf r_1$ is wrong. You must make it clear which is a dot product and which is a scalar product in your calculations. You cannot expand the expression $(\mathbf r_1-\mathbf r_2)^3$ as expanding $(a-b)^3$, where $a,b$ are scalars.

Here is the correct way:
$$
\mathbf{r}^3=\left( \mathbf{r}_1-\mathbf{r}_2 \right) \left( \mathbf{r}_1-\mathbf{r}_2 \right) \left( \mathbf{r}_1-\mathbf{r}_2 \right) =\left( \mathbf{r}_1^2-2\mathbf{r}_1\cdot \mathbf{r}_2+\mathbf{r}_2^2 \right) \left( \mathbf{r}_1-\mathbf{r}_2 \right)
\\
=\mathbf{r}_1^2\mathbf{r}_1-2\left( \mathbf{r}_1\cdot \mathbf{r}_2 \right) \mathbf{r}_1+\mathbf{r}_2^2\mathbf{r}_1-\mathbf{r}_1^2\mathbf{r}_2+2\left( \mathbf{r}_1\cdot \mathbf{r}_2 \right) \mathbf{r}_2-\mathbf{r}_2^2\mathbf{r}_2\\
=\mathbf{r}_1^2\left( \mathbf{r}_1-\mathbf{r}_2 \right) +2\left( \mathbf{r}_1\cdot \mathbf{r}_2 \right) \left( \mathbf{r}_2-\mathbf{r}_1 \right) +\mathbf{r}_2^2\left( \mathbf{r}_1-\mathbf{r}_2 \right)
$$
Write the above in the form of scalar coefficients, we obtain
$$
\mathbf{r}^3=\color{blue}{r_1^2}\left( \mathbf{r}_1-\mathbf{r}_2 \right) +\color{blue}{2r_1r_2\cos \theta} \left( \mathbf{r}_2-\mathbf{r}_1 \right) +\color{blue}{r_2^2}\left( \mathbf{r}_1-\mathbf{r}_2 \right)
$$


问:
Do people agree on what dx is in Riemann integration?
I've been rather curious about what the official rigorous understanding of the dx term in a Riemann integral is. Of course, we could use the formal definition of a Riemann integral but I have seen on Wikipedia other proposed ways of defining the differential of some function (we'll call x), dx. The classic high school teacher approach is to say it's a "very small change in x", or rather a small change in a linear approximation of x. However, as someone who studies differential geometry, I am wondering how this can be interpreted as the exterior derivative of a function x, in which dx is a 1-form. Do we then reinterpret $\int{f(x)}\,dx$ as being the integral of a 1-form? And how then can we consider a 1-form to be a "small change in x"?

答:
This question has been asked a million times on this site in one form or another. But I suppose it can’t hurt to answer it again since it is so natural a question but also so typically badly answered.

The short answer to your title question is yes: there is agreement about what the $dx$ means — but only once you get precise about exactly what kind of integral you’re talking about. The one-dimensional baby’s-first-integral in $\mathbb{R}$ can be interpreted in many different ways that only come apart once you move to integration in more general spaces.

If you’re thinking about $\int_{[a,b]}f dx$ as giving the integral of a density $f$ over an unoriented interval $[a,b]$, then $dx$ generally denotes a measure. Example: if $f$ measures mass per unit length over a rod, then the integral gives the mass of the rod. This generalizes easily to higher-dimensional Euclidean spaces: the “area element” $dA$ and the “volume element” $dV$ are really measures. In differential geometry you want to integrate such things not just over subsets of Euclidean space but over abstract manifolds. In that case you don’t have a measure, but you do have the differential geometric analog of a measure called a density.

If, however, you’re thinking about $\int_a^bfdx$ as giving the integral of $f$ over the oriented interval $[a,b]$, then $dx$ generally denotes a differential form. Example: if $f$ measures the rate of change of some quantity as a function of $x$, then the integral gives the total change in the quantity as $x$ ranges from $a$ to $b$ — that’s the fundamental theorem, which is just a special case of Stokes theorem, which is a theorem that depends crucially on orientations and thus on differential forms. There is a connection between differential forms and densities, however. Densities are in some sense the natural objects to integrate: they’re defined whether you have an orientation or not. You can only integrate a differential form if you have an orientation, by contrast — and indeed the whole point of the orientation is that it provides enough information to turn the differential form into a density, which can then be integrated.

If you reject all this seeming abstract nonsense and insist on an answer in the case of the most familiar, calculus 101 one-dimensional Riemann integral of a bounded real function $f$ on $[a,b]$, the answer is that the $dx$ is literally meaningless, a mere piece of suggestive notation. The value of the integral, if it exists, is solely a function of $f$ and of $[a,b]$; the best notation to emphasize this is $\int_{[a,b]}f$, with no $dx$ at all.


问:
Euler lagrange equation is a constant
I'm working through exercises which require me to find the Euler-Lagrange equation for different functionals.
I've just come across a case where the Euler Lagrange equation simplifies to
$$1=0.$$
Please could someone explain what can be concluded about the set of extremals to problems where this is the case.</p>

答:
Notation $$L[u]=\int(f(u)u_{x}+u)dx=\int F(u,u_{x})dx $$

$$\frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0 $$

As $F(u,u_{x})=f(u)u_{x}+u$  we have

$$\frac{\partial F}{\partial u_{x}}=f(u) $$

$$ \frac{\partial F}{\partial u}=\frac{df}{du}\frac{du}{dx}+1$$

so $$ $$

Hence $$\frac{d}{dx}(f(u)+1)-\frac{df(u)}{du}\frac{du}{dx}=0 $$

$$\frac{df}{du}\frac{du}{dx}-\frac{df}{du}\frac{du}{dx}=0 \rightarrow 0=0$$

Title: Euler lagrange equation is a constant


问:
What is the difference between countable infinity and uncountable infinity?
What is the difference between countable infinity and uncountable infinity? Are there any examples? How can I imagine it?
Can you offer some assistance? please.</p>

答:
A denumerable set is in one-to-one correspondence with the set of natural numbers. Such sets may also be called countably infinite, see https://mathworld.wolfram.com/DenumerableSet.html

Such a set has the form $A = \{a_0,a_1,a_2,\ldots\}$ and so can be tabulated by the natural numbers (indices). Examples are the natural numbers, the even or odd natural numbers, the integers and the rational numbers.

Basic non-denumerable sets are the real numbers or the power set of the set of natural numbers which are equipotent.


答:
Intuitively speaking, if $A$ is countable, you have a way to list out the items in $A$. If $A$ is uncountable, you will not have such ways, and if you try to make a list of $A$, the list must have some elements missing(between the elements in the list).

You can take $\mathbb Q$ and $\mathbb R$ for corresponding examples.

Body:
What is the difference between countable infinity and uncountable infinity? Are there any examples? How can I imagine it?
Can you offer some assistance? please.</p>


问:
Portmanteau theorem for signed measures
Let $(E,d)$ be a metric space, $(\mu_t)_{t\in I}$ be a net of finite nonnegative measure on $\mathcal B(E)$ and $\mu$ be a finite nonnegative measure on $\mathcal B(E)$ such that $(\mu_t)_{t\in I}\to\mu$ weakly, i.e. $$\left(\int f\:{\rm d}\mu_t\right)_{t\in I}\to\int f\:{\rm d}\mu\tag1.$$

By the Portmanteau theorem, $(1)$ implies (it is even equivalent to) $$\lim_{t\in I}\mu_t(B)=\mu(B)\;\;\;\text{for all }B\in\mathcal B(E)\text{ with }\mu(\partial B)=0\tag2.$$

Are we able to show an analogous result for finite signed measures: If $(\nu_t)_{t\in I}$ is a net of finite signed measure on $\mathcal B(E)$ and $\nu$ be a finite signed measure on $\mathcal B(E)$ such that $(\nu_t)_{t\in I}\to\nu$ weakly, does it follow that $$\lim_{t\in I}|\nu_t-\nu|(B)=0\;\;\;\text{for all }B\in\mathcal B(E)\text{ with }\mu(\partial B)=0\tag3$$ holds?

答: The example I quoted for your previous question works here too. On $[0,1]$ let $\nu_n=\delta_{\frac 1 2+\frac1  n}-\delta_{\frac 1 2-\frac1  n}$ and $\nu=0$. Take $B=[0,1]$.

Title: Portmanteau theorem for signed measures